问:
why $\overline {A} \cup \overline{B}=\mathbb{R}^n\setminus\{0\}?$
My main  motivation came from my previous post

After that i read this solution
After reading i have some confusion in my mind given below

Here is an outline of the Aloizio Macedo solution

If you want to avoid usage of path-connectedness, do the following. For $n>1$, take $A,B \subset \mathbb{R}^n \backslash \{0\}$ given by $A:=\{x \in \mathbb{R}^n \mid x_n > 0\}$ and $B:=\{x \in \mathbb{R}^n \mid x_n<0\}$. $A$ and $B$ are clearly connected (both in fact homeomorphic to $\mathbb{R}^n$). Since the closure of connected sets is connected, $\overline{A}$ and $\overline{B}$ are connected. But $\overline{A}$ and $\overline{B}$ have points in common. Therefore, their union is connected. But their union is the entire $\mathbb{R}^n\backslash \{0\}$ .

Fun exercise: Where does this fail for $\mathbb{R}$?

"But $\overline{A}$ and $\overline{B}$ have points in common" - They don't, for $n=1$.

My thinking : If $A:=\{x \in \mathbb{R}^n \mid x_n > 0\}$ the  $\overline{A}=\{x \in \mathbb{R}^n \mid x_n \ge 0\}$ simliarly  if $B:=\{x \in \mathbb{R}^n \mid x_n<0\}$ then $\overline B:=\{x \in \mathbb{R}^n \mid x_n\le0\}$

Now $\overline {A} \cap \overline{B}= \{0\}\neq \emptyset$

$\overline A \cup \overline{ B}= \mathbb{R}^n$

for $n=1$ we have $A=\{x \in \mathbb{R}^1 \mid x_1 > 0\} $and $B:=\{x \in \mathbb{R}^1 \mid x_1<0\}$

$\overline{A}=\{x \in \mathbb{R}^1 \mid x_1 \ge 0\} $and $\overline{B}:=\{x \in \mathbb{R}^1 \mid x_1\le 0\}$  then it give $\overline {A}\cap \overline {B} =\{0\} \neq \emptyset$

Here I have two confusion given below

$1.$ why $\overline {A} \cup \overline{B}=\mathbb{R}^n\setminus\{0\}?$

$2.$ why it fail in $\mathbb{R} ?$

答:
He meant the closure in $\Bbb R^n\setminus\{0\}$ (with its relative topology).

In $\Bbb R$ we would have $A=(0,\infty), B=(-\infty,0)$, and taking closures in $\Bbb R\setminus\{0\}$ we would get $\overline A=A$ and $\overline B=B$, since we are in the relative topology of $\Bbb R\setminus\{0\}$. This sets don't have common points, so the argument doesn't work.


问:
Is there extension of function from a curve on the whole space preserving smoothness?
Assume that $\alpha: (a,b) \rightarrow \mathbb R^3$ and $f: (a,b) \rightarrow \mathbb R$ are given smooth functions. Let $t_0 \in (a,b)$.

Do there exist a $\delta>0$ and a smooth function $V: \mathbb R^3 \rightarrow \mathbb R$ such that
$V(\alpha(t))=f(t)$  for $t\in (t_0-\delta, t_0+\delta)$ ?

Edit. Assume additionally that $\alpha'(t)\neq 0$ for all $t\in (a,b)$.

答:
Yes. It is enough to have $v:=\alpha'(t_0)\ne 0$.
Then in a neighbourhood of $t_0$, the map $t\mapsto \langle \alpha(t)-\alpha(t_0),v\rangle$ is injective and smooth and has a smooth inverse around $0$ with $0\mapsto t_0$. Let $g\colon \mathbb R\to\mathbb (a,b)$ be a smooth function that coincides with this inverse near $0$.
Define $$V(x)=f(g(\langle x-\alpha(t_0),v\rangle)).$$
Then for $t\approx t_0$, we have
$$ V(\alpha(t))=f(g(\langle \alpha(t)-\alpha(t_0),v\rangle))=f(t).$$</p>

ContentLicense: CC BY-SA 3.0


问:
Is the linear approximation of the product of two functions the same as the product of the linear approximations of the two functions?
I have just seen a lecture about linear approximation, in which it was established that when $x$ is near $0$:

The linear approximation of $e^x$ is $1+x$

The linear approximation of $(1+x)^r$ is $1+rx$

Afterwards, the lecturer explained how to find—using 1 and 2—the linear approximation of:

$e^{-3x} \times (1+x)^{-1/2}$.

His next line is:

$(1-3x)(1-0.5x)$

So it seems that he was acting under the assumption that the linear approximation of the product of two functions is the product of the linear approximations of the functions.

Is that true in general for any order of approximations? Maybe just for linear ones? Maybe just when x is near zero? Maybe only when the approximation is linear and x near 0?

Here is a link to the lecture. The comments were made at 26:30.

答:
This is commonly used, and that is because the difference between the two is negligible.

In general, if we have a function $f(x)$ approximated by $f(x_0)+f'(x_0)(x-x_0)$ and another function $g(x)$ approximated the same then you can do some algebra to multiply the two approximations in a part and multiply the two functions then approximate in the other part and compare the results, it won't be the same but the difference would be pretty negligible as long as $x$ is close to $x_0$ so you can use the two methods interchangeably, to be exact the difference would be $f'(x_0)g'(x_0)(x-x_0)^2$.

ContentLicense: CC BY-SA 4.0


问:
Question about the FG algorithm (convex optimization) (Flury 1984)
I am currently trying to understand the FG algorithm which was presented by Flury in 1984.

You can find here a description of the algorithm : https://www.researchgate.net/publication/248861527_An_Algorithm_for_Simultaneous_Orthogonal_Transformation_of_Several_Positive_Definite_Symmetric_Matrices_to_Nearly_Diagonal_Form

And here the background for the algorithm (on which I have an issue): https://www.jstor.org/stable/2288721?seq=2#metadata_info_tab_contents

Its objective is to find a common othorgonal matrix to make as diagonal as possible several positive definite symmetric (p.d.s) $p \times p$ matrices $A_1,A_2,...,A_g$.

The goal is to minimize $\phi$ defined as:

\begin{equation}
\phi(F_1,...,F_g)=\prod_{i=1}^{g}{\left(\frac{|\text{diag}(F_i)|}{|F_i|}\right)^{n_i}}
\end{equation}
where $F_i=B'A_iB$ for $B$ an orthogonal matrix, diag($F_i$) is the p$\times$p diagonal matrix with the same diagonal components as $F_i$, and $n_i$ are fixed weights.

It can be shown (Flury 1983) that the above problem is equivalent to minimizing the following function:
\begin{equation}
g(\Sigma_1,...,\Sigma_g)=g(B_{1},...,B_p,\lambda_{1,1},...,\lambda_{1,p},...,\lambda_{g,1},...,\lambda_{g,p})=\sum_{i=1}^{g}{n_{i} \left[ \sum_{j=1}^{p}{log\lambda_{i,j}+B_{j}'A_{i} B_{j}/\lambda_{i,j}} \right]}
\end{equation}
where $B_i$ is the i-th column of an orthogonal matrix, $\lambda_{i,j}$ are stricly positive and $\Sigma_i$ are p.d.s matrices for which we want to find $B$ such that $B'\Sigma_iB=\text{diag}(\lambda_{i,1},...,\lambda_{i,p})$.

Until here I understand the reasoning, my issue comes in the following.
To solve the above problem, the author use the Lagrange multipliers method, with the following constraints:

$B_i'B_j=0$ if i$\neq$j

$B_i'B_j=1$ else

I have two problems:

First, for $\Sigma_i$ to be p.d.s the $\lambda_{i,j}$ should be strictly positive, so why is not in the constraints of the Lagrange multiplier methods?

Second, I do not understand why the Lagrange methods leads to a necessary condition, because the constraint $B_i'B_i=1$ is not convex I believe. ( $f: u,v \rightarrow u'v-1$ is not convex). The theorem that I know for Lagragian multipliers on convex functions uses the fact that the constraints are convex.

I know that because we want to minimize a convex function over a convex set (the set of p.s.d functions), then global minimums are found where its gradient is null. But in the case of Lagrange multipliers methods, if we add non convex functions I don't see why this result would hold.

Thank you very much in advance, and of course feel free to ask me if anything is not clear.

答:
It seems that you confused the Lagrange multiplier method for equality constrained optimization problems, with the Lagrange multiplier method for convex optimization problems.

The former deals with problems in the following form (e.g., see http://en.wikipedia.org/wiki/Lagrange_multiplier):
$$\begin{align}\min\;\;\;& f(x)\\
\mathrm{s.t.}\;\;\;& g(x)=0,
\end{align}$$
where $f,g$ are continuous functions.
This does not require either $f$ or $g$ to be convex, so it is quite general. On the other hand, it only gives you a necessary condition. What you get are just candidates for optimal points.

The latter is used for convex programs with inequality constraints. And under some constraint qualification, the KKT conditions are necessary and sufficient for optimality.


问:
Graph minus a point has at most 2 connected components
Let $G(V,E)$ be a connected undirected graph (undirected=easier), such that, $S_x=V\setminus\{x\}, \forall x \in V$ is a subset of vertices of G. Then the induced subgraph $G[S_x]$ is the graph whose vertex set is $S_x$ and whose edges set consists of all of the edges in $E$ that have both endpoints in $S_x$.

Prove that a graph $G(V,E)$ has a Hamiltonian path (not necesarily a cycle) if the induced subgraphs $\forall x \in V, G[S_x]$ have at most 2 connected components.

答: Below is probably the simplest counterexample.


答:
Let's consider the function in question, $f(x)=e^{-3x} \cdot (1+x)^{-1/2}$. To find the linear approximation we need to compute $f(0)$ and $f'(0)$:

$f(0)=1$

Using the product rule, we find that $$f'(x)=-\frac{e^{-3x}(6x+7)}{2(1+x)^{3/2}} \, .$$Hence, $f'(0)=-7/2$.

The linear approximation formula is $$f(a+h) \approx f(a)+f'(a)h \, .$$ At $a=0$, it is
\begin{align}
f(h)&\approx f(0)+f'(0)h \\
f(h)&\approx 1-\frac{7}{2}h \, .
\end{align}
Since you used the variable $x$ in your question, we'll rewrite this as $f(x) \approx 1-\frac{7}{2}x$. This is not the same as the approximation given in the lecture. As HallaSurvivor notes in the comments, the approximation given is a quadratic approximation. If we expand the brackets, it is
$$
f(x) \approx 1.5x^2 - 3.5x + 1 \, .
$$
But this approximation is only very rough and ready. If we use the formula
$$
f(a+h) \approx f(a) + f'(a)h + \frac{f''(a)}{2}h^2
$$
we obtain a different result, which is better than the one obtained by the lecturer. Still, the quadratic approximation obtained by the lecturer is better than the linear approximation we computed earlier.

Title: Is the linear approximation of the product of two functions the same as the product of the linear approximations of the two functions?

Body:
I have just seen a lecture about linear approximation, in which it was established that when $x$ is near $0$:

The linear approximation of $e^x$ is $1+x$

The linear approximation of $(1+x)^r$ is $1+rx$

Afterwards, the lecturer explained how to find—using 1 and 2—the linear approximation of:

$e^{-3x} \times (1+x)^{-1/2}$.

His next line is:

$(1-3x)(1-0.5x)$

So it seems that he was acting under the assumption that the linear approximation of the product of two functions is the product of the linear approximations of the functions.

Is that true in general for any order of approximations? Maybe just for linear ones? Maybe just when x is near zero? Maybe only when the approximation is linear and x near 0?

Here is a link to the lecture. The comments were made at 26:30.


问:
Some questions on random step processes
Given $(f(t)) _{t \ge 0}$, a stochastic process with almost sure continuous paths adapted to the filtration $\mathcal{F}_t = \sigma(W_s : s\le t)$, where $W_s$ is a wiener process.

If we define
$$
f_n(t) = \begin{cases}
n\int_{\frac{k-1}{n}}^{\frac{k}{n}}f(s)ds, & \frac{k}{n} \le t < \frac{k+1}{n} ~\text{for}~ k = 1, 2, \cdots, n^2-1 \\
0, & \text{otherwise}
\end{cases}
$$
why is $(f_n(t))_{t \ge 0}$ a sequence of random step processes? Furthermore, why do we have
$$
\lim_{n\to\infty} \int_{0}^N |f(t)-f_n(t)|^2dt = 0,
$$
for any fixed $N$ by the continuity path property of $f$?

答:
1) From Jensen, $$\left(\frac{1}{b-a}\int_a^b f\right)^2\leq \frac{1}{b-a}\int_a^b f^2.$$
Take $a=\frac{k-1}{n}$ and $b=\frac{k}{n}$ yields $$\left(\int_{(k-1)/n}^{k/n} f\right)^2=\frac{1}{n^2}\left(n\int_{(k-1)/n}^{k/n}f\right)^2\leq \frac{1}{n}\int_{(k-1)/n}^{k/n}f^2,$$
and thus your result.

2) If $\mathbb E[\int_0^\infty f^2]<\infty $, then $\int_0^\infty f^2<\infty $ a.s. Now, you can prove that if $\int_0^\infty f^2$ converges, then $\lim_{N\to \infty }\int_N^\infty f^2=0$ (which is rather straightforward...)

3) \begin{align*}
\int_0^N|f(t)-f_n(t)|^2\,\mathrm d t&=\int_0^N\left|n\int_{(\lfloor tn\rfloor-1)/n }^{\lfloor nt\rfloor/n }f(t)-f(u)\,\mathrm d u\right|^2\,\mathrm d t\\
&\leq \int_0^N\underbrace{\int_{(\lfloor tn\rfloor-1)/n }^{\lfloor nt\rfloor/n }|f(t)-f(u)|^2\,\mathrm d u}_{:=I_n(t)}\,\mathrm d t
\end{align*}

where the last inequality follow from 1). Finally, you have that $I_n(t)\to 0$ when $n\to \infty $ and $|I_n(t)|\leq 4\|f\|_{L^\infty (0,N)} $, and thus $(I_n)$ is uniformly bounded. Therefore, Dominated convergence theorem allow you to conclude.

Title: Some questions on random step processes


问:
How to find the minutes and the seconds of an angle in sexagesimal units when a condition is given?
The problem is as follows:

A certain angle $\alpha$ is measured by a goniometer in sexagesimal units. This results in $a^{\circ}b'c''$ assuming $\textrm{a, b, c} \in \mathbb{Z}$ and $0\leq b < 60\,0\leq c < 60$ and the number of sexagesimal minutes plus the double of the number of its sexagesimal degrees equals to $155$, find $a+b+c$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{30}\\
2.&\textrm{32}\\
3.&\textrm{31}\\
4.&\textrm{60}\\
\end{array}$

How exactly should this question should be approached?. What I've attempted is to account from the condition this:

Since it mentions:

```<code>the number of sexagesimal minutes plus
the double of the number of its
sexagesimal degrees equals to 155
</code>```

Then this is translated as follows:

$b+2a=155$

But from here its where I'm stuck. I do believe that you can put this as:

$155=2k+1$

Hence it would meant that $b$ might be $1^{\circ}$?

Upon further inspecting this I thought it could meant:

$2\left(a+\frac{b}{60}+\frac{c}{3600}\right)+b+\frac{c}{60}=155$

Which would become into:

$a+\frac{b}{30}+\frac{c}{1800}+b+\frac{c}{60}=155$

$\frac{1800a+60b+c+30b+30c}{1800}=155$

Then this becomes into:

$\frac{1800a+90b+31c}{1800}=155$

But as it can be seen there isn't a way to simplify this expression.

Can someone help me here?. What exactly should be the right approach?.

答:
I've interpreted the phrase you pulled out as

$$2(a + \frac{b}{60}+\frac{c}{3600})+(60a+b+\frac{c}{3600})=155 $$
Which simplifies to
$$3600a+60b+c=9000$$
With 3 unknowns normally this couldn't be taken much further. But, there are restrictions on $a, b, c$, namely they must all be integers. $b, c$ have further restrictions. I am also assuming $a≥0$.

A quick calculation shows that $a$ can only be 0, 1, or 2.

If $a=0$ then $60b+c=9000$ which cannot be since $b$ and $c <60$

Similarly for $a=1$. $\therefore a=2$, $\therefore 60b+c=1800$

By following a similar process for $b$ it quickly becomes apparent that $b=30$, leaving $c=0$.

Hence $a+b+c=32$

Not a direct approach but without further information I'm not sure what else could be done.


问:
martingales, almost sure convergence
I am given a sequence of independent random variables $(X_n)$ with respective laws given by $P(X_n=-n^2)=\dfrac{1}{n^2}$ and $P(X_n=\dfrac{n^2}{n^2-1})=1-\dfrac{1}{n^2}$, and letting $S_n=X_1+...+X_n$ I am asked to show $\dfrac{S_n}{n}$ tends to 1 as $n$ tends to $\infty$. Furthermore I am asked to deduce $(S_n)$ is a martingale which converges to infinity.

For the first part, I tried using the strong law of large numbers but I keep getting $0=1$ which is definitely wrong. Could somebody give me a hint?
Thanks.
Harry

答:
You can not use SLLN because $(X_n)_{n \geq 2}$ in your case is a sequence of independent but not identically distributed random variables. You can review the conditions of SLLN here. According to the hint of Morris, I will complete the proof:

$$
\because \sum_{n=2}^\infty\mathbb{P}(X_n=-n^2)<\infty\,,
$$

By first Borel-Cantelli lemma, we arrive that

$$
\mathbb{P}(X_n=-n^2 \quad \text{infinitely often}) = 0\,,
$$

thus

$$
X_n = \frac{n^2}{n^2-1} \text{ eventually, a.s} \,,
$$

that is

$$
X_n \to 1 \text{ a.s}\,,
$$

which implies

$$
\frac{S_n}{n} \to 1 \text{ a.s}.
$$

We also arrive that the martingale $(S_n)_{n \geq 2}$ converges to (positive) infinity since $X_n \to 1$ almost surely.

Title: martingales, almost sure convergence


问:
What's the fundamental group of $E^2\setminus Q^2$
Here $E^2$ is the two-dimensional Euclid space and $Q$ is the set of all rational numbers. Regard $E^2\setminus Q^2$ as a subspace of $E^2$. So what's its fundamental group and how to represent it? I suppose that it's not a free group. Thanks a lot for your help.</p>

答: This has been discussed quite a bit here and here.</p>


问:
Calculate expected value in a gambling game
In a game the players can earn or lose $1$ dollar every round, the probability of winning is $\frac{1}{4}$. All the games are independent from each other. A player enter the game and play until he win for the first time (after that he leave the game). Let $X$ be the profit of the player. calculate $\mathbb{E}[X]$

My attempt:

I understand that $X \leq 1$ so I assume that $\mathbb{E}[X]<0$
however I dont understand if $X$ is discrete or continuity.
The way that the player is playing is seems like $Geo(\frac{1}{4})$. I dont understand what I should do now

答:
The expectation of your geometric is

$$\mathbb{E}[Y]=\frac{1}{p}=4$$

Thus, the average loss is

$$-1\times 3+1=-2$$

This because the expected trials are 4: 3 losses before the first winning game


问:
Inequality with mean inequality

If $a,b,c > 0$ and $a+b+c = 18 abc$, prove that
$$\sqrt{3 +\frac{1}{a^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}+\sqrt{3 +\frac{1}{c^{2}}}\geq 9$$

I started writing the left member as $\frac{\sqrt{3a^{2}+ 1}}{a}+ \frac{\sqrt{3b^{2}+ 1}}{b} + \frac{\sqrt{3c^{2}+ 1}}{c}$ and I applied AM-QM inequality, but I obtain something with $\sqrt[4]{3}$.

答:
From Minkowski inequality: $\sqrt{x^2 + a^2} + \sqrt{y^2 + b^2} \ge \sqrt{(x + y)^2 + (a + b)^2}$

applied to the first two terms, with the notation of @Albus, we get:
$\sqrt{\sqrt{3}^2 +x^2} + \sqrt{\sqrt{3}^2 +y^2} \ge \sqrt{(2\sqrt{3})^2 + (x+ y)^2}$. Let's denote RHS of the inequality by $X$.

Then, apply Minkowski inequality again to $ X + \sqrt{3 + z^2}$, to obtain:
$\sqrt{27 + (x + y + z)^2} \ge 9$, which is equivalent to (1) from @Albus Dumbledore's answer.

Title: Inequality with mean inequality


答:
Let $W$ denote the event on winning at first game and let $L$ denote the event on losing at first game.

Then:$$\mathbb EX=P(W)\mathbb E(X\mid W)+P(L)\mathbb E(X\mid L)=\frac14\cdot1+\frac34\cdot\mathbb E(X\mid L)$$

Now realize that: $$\mathbb E(X\mid L)=-1+\mathbb EX$$

This because we are back in start position with one dollar less in our pocket.

So:$$\mathbb EX=\frac14+\frac34(-1+\mathbb EX)$$

From this equality we deduce easily that: $$\mathbb EX=-2$$

Another route:

Let $Y$ denote the number of games that in total will be played.

Then $Y$ has Geometric distribution with parameter $p=\frac14$.

Further we observe that: $$\mathbb E[X\mid Y=n]=2-n$$
From this we are allowed to conclude that: $$\mathbb E[X\mid Y]=2-Y$$

Then applying the general rule:$$\mathbb EX=\mathbb E[\mathbb E[X\mid Y]]$$we find:$$\mathbb EX=\mathbb 2-Y=2-\frac1{\frac14}=2-4=-2$$

Title: Calculate expected value in a gambling game

Body:
In a game the players can earn or lose $1$ dollar every round, the probability of winning is $\frac{1}{4}$. All the games are independent from each other. A player enter the game and play until he win for the first time (after that he leave the game). Let $X$ be the profit of the player. calculate $\mathbb{E}[X]$

My attempt:

I understand that $X \leq 1$ so I assume that $\mathbb{E}[X]<0$
however I dont understand if $X$ is discrete or continuity.
The way that the player is playing is seems like $Geo(\frac{1}{4})$. I dont understand what I should do now


问:
Show by induction $n^2 - 3n + 4$ is even
I have this question I am working on. I am not sure If I am doing the correct thing.

Show by induction $n^2-3n+4$ is even.

Basis step
$n = 1$ $\implies$$1^2 - 3(1) + 4 = 2$

$n^2 - 3n + 4$ is even means

$n^2 - 3n + 4 = 2m$ for some $m$

Induction step $n=k$

$k^2 - 3k + 4=2m$

$(k+1)^2 - 3(k+1) + 4 = 2m$

From step 5, $k^2 - 3k = 2m - 4$

$\frac{k^2 - 3k} {2} = m - 2$

Which comes to $0=0$ when $n=1$

Now I want to prove the same for $k+1$

From step 6, $(k+1)^2 - 3(k+1) = 2m - 4$

$\frac{(k+1)^2 - 3(k+1)} {2} = m - 2$

Which is $0=0$ when $k=2$

What I have done above, is the best I could come up with but I know it's wrong somehow. Can you correct me if there are any mistakes and if there is somewhere I am going wrong.

答:
The $\implies$ sign makes no sense here.

Right.

Right.

Right.

Right.

Wrong: $(k+1)^2-3(k+1)+4=2n$ or $2j$ or $2l$. Any letter but $m$ (or $k$), since $m$ is the natural number such that $k^2-3k+4=2m$.

Right.

Right.

This makes no sense, since there is no $n$ here.

And neither the rest makes sense.

Note that$$\bigl((k+1)^2-3(k+1)+4\bigr)-(k^2-3k+4)=2k-2,$$which is an even number. So, since $(k+1)^2-3(k+1)+4$ is equal to $k^2-3k+4$ plus an even number, if you assume that $k^2-3k+4$ is even, then so will be $(k+1)^2-3(k+1)+4$.


问:
Notation $2^T$ for the operator T of the Hilbert space
I came across an unknown to me notation $2^T, 2^{-T}$ for a positive operator $T$ on a Hilbert space (more precisely, in $\mathbb{C}^n$).

What does this mean? I can't find it anywhere. Is it the image and the kernel?

答:
Since $T$ is a positive operator, it is a normal operator and we can consider the continuous functional calculus
$$C(\sigma(T)) \to  B(\mathbb{C}^n) \cong M_n(\mathbb{C})$$

Then $2^T$ is the operator obtained as the image of $\sigma(T)\ni x \mapsto 2^x$ under the above map.

In some sense, it is unnatural to consider the operator $2^T$ since $\sigma(T)$ is a finite space, but I think it suffices for your purposes.


答:
To prove something by Induction we must:

Show that it is true for the smallest value on the set we are proving it

Suppose it is true for $k$

Show it is true for $k+1$

Then say it is therefore true for all $n$

I assume $n \in \mathbb{Z}^+$ meaning that our smallest value is $0$. So,
$$P_0 = 0^2 -3\cdot0 + 4 = 2\cdot 2 $$
Meaning that this is true for $P_0$

We now assume that $P_k$ is true meaning that:
$$ P_k = k^2 -3k + 4 = 2a\, , \quad a \in \mathbb{Z}^+$$
We now want to show that $P_{k+1}$ is true:
$$ P_{k+1} = (k+1)^2 -3(k+1) + 4 = 2b \, , \quad b \in \mathbb{Z}^+$$
Expanding:
$$ P_{k+1} = k^2 + 2k + 1 -3k -3 + 4 = k^2 -k + 2$$
We can now make a substitution. Remember that
$$ P_k = k^2 -3k + 4 = 2a$$
so
$$ k^2 = 2a +3k -4$$
Plugging this in $P_{k+1}$
$$2a + 3k -4 -k + 2 = 2a + 2k -2 = 2(a+ k -1) $$
$P_{k+1}$ was set to be equal to $2b$. We have now shown that $P_{k+1}$ is true meaning that by induction $P_n$ is true.

Title: Show by induction $n^2 - 3n + 4$ is even

Body:
I have this question I am working on. I am not sure If I am doing the correct thing.

Show by induction $n^2-3n+4$ is even.

Basis step
$n = 1$ $\implies$$1^2 - 3(1) + 4 = 2$

$n^2 - 3n + 4$ is even means

$n^2 - 3n + 4 = 2m$ for some $m$

Induction step $n=k$

$k^2 - 3k + 4=2m$

$(k+1)^2 - 3(k+1) + 4 = 2m$

From step 5, $k^2 - 3k = 2m - 4$

$\frac{k^2 - 3k} {2} = m - 2$

Which comes to $0=0$ when $n=1$

Now I want to prove the same for $k+1$

From step 6, $(k+1)^2 - 3(k+1) = 2m - 4$

$\frac{(k+1)^2 - 3(k+1)} {2} = m - 2$

Which is $0=0$ when $k=2$

What I have done above, is the best I could come up with but I know it's wrong somehow. Can you correct me if there are any mistakes and if there is somewhere I am going wrong.


问:
If $X_n \to X$ in probability and $E[X_n] < \infty$, is $E[X] < \infty$?
I am trying to prove that

If $X_n \sim N(\mu_n, \sigma_n^2)$ converges to $X$ in probability, then $X$ is also normal, and the convergence is indeed in $L^2$.

by following the argument in https://stats.stackexchange.com/questions/24362/convergence-in-probability-and-l-2-for-normal-random-variables and without using the fact that $X_n \to X$ in distribution and tightness of $X_n$.

I believe it boils down to say $E[X] := \mu < \infty$. If that's the case, I could consider the characteristic function of $X_n - \mu_n\sim N(0,\sigma_n^2)$, and show the limit $X-\mu$ has normal distribution $N(0,\sigma^2)$. Could anyone give me some hint on this?

答:
There were 2 question.

Question 1:

If $X_n \to X$ in probability and $EX_n < \infty$, is $EX< \infty$?

No. Consider $\xi \sim Cauchy(1)$, $X = |\xi|$,  $X_n = X I_{|X|<n}$. Hence $EX_n < \infty$, $X_n \to X$ in probability and $EX = \infty$.

Question 2:

How to prove ..., and without  using the fact that $X_n \to X$ in distribution and tightness of $X_n$?

If  we don't have condition $X_n \to X$ then we can't derive any properties of $X$.


问:
Algebraic transformations to continuously extend functions
Lately I was browsing through my analysis lecture notes (since right know I'm somewhat rusty in analysis) and the proof that $x \mapsto \frac{1}{x}$ is differentiable at every $x'\neq 0$ captured my attention. The easy proof is based on the following algebraic manipulation  $$\frac{\frac{1}{x+h}-\frac{1}{x}}{h}=\frac{1}{h}\frac{-h}{x^{2}+xh}=\frac{-1}{x^{2}+xh}.$$ Letting $h$ tend to zero, we get the limit that we sought.

What aroused my interest was the idea that we can (continuously) extend a (continuous) function to a larger domain by simple algebraic manipulations (since in essence this is what we do, when calculating a derivative -- continuously extending a function): The second equality from above is key: By dividing by $h$ the domain changes from $\mathbb{R}\setminus \{0\}$ to $\mathbb{R}$ (if we consider the functions $h\mapsto \frac{1}{h}\frac{-h}{x^{2}+xh}$ and $h\mapsto \frac{-1}{x^{2}+xh}$, for $x\in \mathbb{R}\setminus \{0\}$).

Questions:

1. Is there some general theory of "algebraic transformations" that allow the extension  of functions (even if it extends the functions only for one single point, as for in the case of calculating derivatives) ?

(I don't know what algebraic geometry deals with, but to the uneducated ear it sounds like this would be it)

2. What other kinds of purely algebraic tricks, like above, do you know, that allow you (in the context of finding derivatives) to continuously extend functions $h\mapsto \frac{f(x+h)-f(x)}{h}$ ?

[From the examples of my notes (and 2 books I browsed through) the only tricks seems to be:

$  \quad$- Manipulating the numerator long enough, until you can factor an $h$ out, so that  you can write $\frac{f(x+h)-f(x)}{h}=\frac{h}{h}\cdot s_x(h)$ for some function $s_x$, since the functions $h\mapsto \frac{h}{h}$ is preventing you from continuously  extending $h\mapsto \frac{f(x+h)-f(x)}{h}$ from $\mathbb{R}\setminus \{0\}$ to $\mathbb{R}$ (and the function $h\mapsto \frac{h}{h}$ is trival to continuously extend to $\mathbb{R}$).

$ \quad$- Writing $f$ as the product of $f=p_1 \cdot p_2$ and then adding and substracting again $p_1(x)\cdot p_2 (x+h)$ as in the proof that $(p_1 \cdot p_2)'=p_1' p_2 + p_1 p_2'$.

(Estimating $\frac{f(x+h)-f(x)}{h}$ etc. - anything that does not involve pure algebraic transformations to extend it - I'm not interested in. Also, very simple algebraic manipulations, like rearranging terms as in the proof that $(\frac{1}{f})'=\frac{-f}{f^2}$, don't count.]

答:
In the context of finding derivatives (Question 2), the only problem with $$f(x+h)-f(x)\over h$$ is that at $h=0$ it gives you $0/0$. And the only way to fix that is to find an $h$ in the numerator to cancel that $h$ in the denominator; or, to finesse the problem by doing manipulations to bring it to a form already handled, as in the proof of the product rule.

As an example of what you may have to do to get cancelation of $h$, if you have $$\sqrt{x+h}-\sqrt x\over h$$ you have to multiply top and bottom by $\sqrt{x+h}+\sqrt x$.

I think the 1st question is too vague to admit of any useful answer.


问:
What's the use of having both a proper subset and subset definition when we have the of notion equality?
By definition a set $A$ is a proper subset of another set $B$ if we have $A$ is a subset of $B$ and $A\neq B$, I understand the need for such a definition however what I don't understand what's the use of the subset definition when we can say the A is a proper subset of $B$ in case they are not equal and $A=B$ in case they care, what does the subset definition really add to the table here?

答:
It is useful to have the $\leq$ relation defined on numbers; it conveys the idea that one number is at most as big as another one.

But , arguably, numbers are sets . For example $0=_{df} \emptyset$ and $1=_{df}\{\emptyset\}= \{0\}$ , and $2=_{df}\{0,1\}$

So how can we define the $\leq$ relation on sets? For example suppose that sets N and M are natural numbers. How are we going to express the fact that $N\leq M$ ?

We will say that $N\leq M \iff N\subseteq M$.


答:
You can prove that the approximation at each order, is exactly equal:
Let $H(x)=f(x)g(x)$, by approximating $f$ and $g$ by their taylor series:
$$H(x)\approx \left(f(x_0)+f^{\prime}(x_0)(x-x_0)+\frac{f^{\prime \prime}(x_0)}{2}(x-x_0)^2+\dots\right)\left(g(x_0)+g^{\prime}(x_0)(x-x_0)+\frac{g^{\prime \prime}(x_0)}{2}(x-x_0)^2+\dots\right)$$
Multiplying and grouping by $(x-x_0)^k$:
$$H(x)\approx f(x_0)g(x_0)+(x-x_0)\left(f(x_0)g^{\prime}(x_0)+f^{\prime}(x_0)g(x_0)\right)+(x-x_0)^2\left(\frac{f(x_0)g^{\prime\prime}(x_0)}{2}+\frac{g(x_0)f^{\prime\prime}(x_0)}{2}+f^{\prime}(x_0)g^{\prime}(x_0)\right)$$ which, using the product rule, you can check that equals the taylor expansion of $H(x)$.
So, yes, approximating each function separately up to order $k$ and then multiplying them will give you the same answer as approximating the function as a whole up to the same order $k$. In your example, you are approximating the functions up to linear terms, therefore, your approximation of $e^{-3x}(1+x)^{-1/2}$ will be correct up to linear terms, and the quadratic one will be incorrect and should be discarded.


答:
We can make the induction step much shorter:

Set $P(k)=k^2-3k+4$. Then $\:\Delta P(k)=P(k+1)-P(k)=\dots= 2(n-1)$, so $P(k+1)$ and $P(k)$ have the same parity, and the induction step follows.


问:
Irreducible element in $D$ is irreducible in $D[x]$
This is question from abstract algebra.

Question : If $p$ is irreducible in UFD $D$, show that $p$ is also irreducible in $D[x]$.

But, I think this question is quite strange. I think every single element in $D$ is irreducible in $D[x]$.

Every element in $D$ cannot be factored into the product of two non-constant polynomials. Because it could be only factored by constants, not polynomials...

Something I misunderstood?

答: Recall that an element $a$ of an integral domain $A$ is reducible if it can be written as $a=bc$ with $b,c$ non-units in $A$. Note that e.g $6$ is not irreducible in $\mathbb{Z}[X]$ as $6=3\cdot2$ so your conjecture about all elements being irreducible is false. To answer the question, suppose that $p=fg$ in $D[X]$, then we can reduce this to a question about irreducibility in $D$ by thinking about possible degrees of $f,g$.


问:
Simplifying the second moment of PDF
Suppose I have a continuous PDF $f(x)$ such that $x>0$ and $\int_0^\infty f(x)\,dx = 1$ .

Given the second moment about $0$ is $$E[X^2]=\int_0^\infty x^2f(x)\,dx$$

I try to simplify this using the product rule:

$g(x)=x^2$

$g'(x)=2x$

$h(x)=\int_0^\infty f(x)dx$

$h'(x)=f(x)$

Hence:

$$E[X^2]=x^2\int_0^\infty f(x)\,dx-\int_0^\infty 2x\left(\int_0^\infty f(x)\,dx\right) \,dx$$

But because $\int_0^\infty f(x)\,dx = 1$ this just simplifies to $x^2-\int_0^\infty 2x\,dx = 0$

Which isn't right. Am I doing something completely wrong? Thank you.

答:
We have the indefinite integral$$\int hg=h\int g-\int\left(h'\int g\right)+C$$If we take $\int g=G$, we have$$\int hg=hG-\int h'G+C$$

So when you apply the limits, you get$$\int_a^b hg=h(b)G(b)-h(a)G(a)-\int_a^bh'G$$In your case, $h(x)=x^2,g(x)=f(x)$ so$$\int_0^\infty x^2f(x)=\left[x^2\int f(x)\right]\Big|_{\text{at }x=\infty}-\left[x^2\int f(x)\right]\Big|_{\text{at }x=0}-\int_0^\infty\left(2x\int f(x)\right)$$

Title: Simplifying the second moment of PDF


问:
Find a confidence interval using the particular statistics function
This problem was a part of my assignment, which is over now, but I would like to know the right answer to it. Assume that $X_1,...,X_n$ are non-negative i.i.d. r.v.s with p.d.f.
$$f(x | θ) = 2 θ e^{−2θx}$$

Find the distribution of the statistics $T := X1 + · · · + Xn$.

Find the distribution of the statistics $4θT$.

Using the above, find a $100(1 − α)\%$ confidence interval for $θ$.

I know how to solve all of the above, but how is point 2 supposed to help find the confidence interval?

答:
It' s a chi square distribution

$4\theta T\sim\chi_{(2n)}^2$

This is useful to calculate the Confidence interval using chi square-table.

As an example, fix $n=10$ and $\alpha\%=10\%$ and try to calculate the proposed CI (with equiprobable tails)

Solution

Using the chi-square table we immediately get

$$10.85<4\theta\Sigma_iX_i<31.41$$

That is

$$\frac{10.82}{4\Sigma_iX_i}<\theta<\frac{31.41}{4\Sigma_iX_i}$$


问:
Area under the curve $y=2x^4-x^2$

Find the area between the curve $y=2x^4-x^2$, the x-axis and the ordinates of the two minima of the curve.

$y=2x^4-x^2=x^2(2x^2-1)\implies$ function is zero at $0,\pm\frac1{\sqrt2}$.

$y'=8x^3-2x\implies$ critical points are $0,\pm\frac12$. We also note that local minima are at $\pm\frac12$ and local maxima at $0$. At, $x=\pm\frac12$, $y=-\frac18$ i.e. ordinate of the minima is $-\frac18$.

So, the graph is similar (not same) to the following:

So, in my opinion, answer should be $2\int_{\frac1{\sqrt2}}^0(2x^4-x^2)dx$. But the answer is given as $2\int_{\frac1{2}}^0(2x^4-x^2)dx$. Looks like they are considering ordinates as $x=\pm\frac12$ but isn't the ordinate $y=-\frac18$?

答:
First of all, the question is a bit vague

Well, if I understand the question correctly I think that they want to find:

This is also shown in a bigger picture:

And that blue area is given by:

$$\left|\int_{-\frac{1}{2}}^\frac{1}{2}-\frac{1}{8}\space\text{d}x\right|-\left|\int_{-\frac{1}{2}}^\frac{1}{2}\left(2x^4-x^2\right)\space\text{d}x\right|=\frac{1}{15}\tag1$$

Title: Area under the curve $y=2x^4-x^2$


问:
Finding the equation of the sphere that passes through a circle and a point
The problem is the following:
I am given a circle by the equations $$x^2+y^2+z^2+2ax+2by+2cz+d=0, Ax+By+Cz+D=0$$ (The first equation is a sphere, the second one a plane, so as I understand the plane cuts the sphere so we get a circle). I have to find the equation of a sphere that passes through the given circle and contains the point $(x_0;y_0;z_0)$, which does not belong to the given plane.

The answer is $$(Ax_0+By_0+Cz_0+D)(x^2+y^2+z^2+2ax+2by+2cz+d)-(x_0^2+y_0^2+z_0^2+2ax_0+2by_0+2cz_0+d)(Ax+By+Cz+D)=0$$
I don't understand how this can be the answer. Why do we have to pug in that point then subtract? I tried rewriting the equation of the given sphere so that I can find the center and the radius, but nothing happened. Could you please give me some hints or methods to approach this problem?
Thank you very much!

答:
HINT:

The idea is: any solution of the system
$$f=0\\
g=0$$
is also a solution of $f+ \lambda g =0$ for any $\lambda$. So now consider the sphere
of equation
$$x^2 + y^2 +z^2 + 2 a x + 2 b y + 2 c z + d + \lambda(A x + B y + C z + D)=0$$
that passes through the point $(x_0, y_0, z_0)$. $\lambda$ can be easily obtained now.


答:
Consider the circumference $\mathcal C$ of equations

\begin{cases}
x^2+y^2+z^2+2ax+2by+2cz+d=0 &\\
Ax+By+Cz+D=0
\end{cases}

A sphere containing $\mathcal C$ is therefore $\mathcal S_1$ of equation $x^2+y^2+z^2+2ax+2by+2cz+d=0$.

Any other sphere $\mathcal S$ containing $\mathcal C$ must have an equation of the form:

$$x^2+y^2+z^2+2a'x+2b'y+2c'z+d'=0 \tag 1$$

such that
$$\scriptstyle x^2+y^2+z^2+2a'x+2b'y+2c'z+d'-(x^2+y^2+z^2+2ax+2by+2cz+d)
=\lambda (Ax+By+Cz+D) \tag 2\\
$$ where $\lambda \in \Bbb R$.

If, for example, we want to determine the sphere containing $\mathcal C$ and passing through $P_0 = (x_0, y_0, z_0)$ we must choose $λ$ such that the assigned point satisfies the
$$x^2+y^2+z^2+2ax+2by+2cz+d-\lambda (Ax+By+Cz+D)=0$$

Title: Finding the equation of the sphere that passes through a circle and a point

Body:
The problem is the following:
I am given a circle by the equations $$x^2+y^2+z^2+2ax+2by+2cz+d=0, Ax+By+Cz+D=0$$ (The first equation is a sphere, the second one a plane, so as I understand the plane cuts the sphere so we get a circle). I have to find the equation of a sphere that passes through the given circle and contains the point $(x_0;y_0;z_0)$, which does not belong to the given plane.

The answer is $$(Ax_0+By_0+Cz_0+D)(x^2+y^2+z^2+2ax+2by+2cz+d)-(x_0^2+y_0^2+z_0^2+2ax_0+2by_0+2cz_0+d)(Ax+By+Cz+D)=0$$
I don't understand how this can be the answer. Why do we have to pug in that point then subtract? I tried rewriting the equation of the given sphere so that I can find the center and the radius, but nothing happened. Could you please give me some hints or methods to approach this problem?
Thank you very much!


问:
Flux of the field F(x,y,z)=(x,y,z) throught the surface of cylinder
So I recently learned about flux and work of the field, but I cannot really solve this problem:

What is the flux of the field $F(x,y,z)=(x,y,z)$ through the surface of the cylinder $ \{ (x,y,z), x^2+y^2 \leq 0, |z| \leq 1 \}$

I am confused because usually in the examples i saw it was $x^2+y^2 = z^2 $ or a number. Could you please tell me how to solve this? I was thinking i can say it is less than $z$  or just say $x^2+y^2=0$.But i am not sure what to do after this.

答:
Take a look at: $x^{2} + y^{2}\leq 0 \Leftrightarrow (x,y)=(0,0)$. Then, the surface is: $$ S= \left \{ (0,0,z)\in \mathbb{R}^{3} : -1\leq z\leq 1 \right \} $$ which is really a curve... You can parametrize it as: $$ \alpha (t) = (0,0, t), \quad -1\leq t\leq 1 $$

I'm sure you can take it from here. :)


Title: Find $\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}$

Body:
As stated in the title.

My attempt. Dividing through $(x-2)^{\frac{2}{3}}$.
$$L=\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}=\lim_{x \to \infty}\frac{(\frac{x+1}{x-2})^{\frac{2}{3}}-(\frac{x-1}{x-2})^{\frac{2}{3}}}{(\frac{x+2}{x-2})^{\frac{2}{3}}-1}$$
L'Hopital
$$L=\lim_{x \to \infty}\frac{\frac{2}{3}(\frac{x+1}{x_2})^{-\frac{1}{3}}(\frac{x-2-(x+1)}{(x-2)^2})-\frac{2}{3}(\frac{x-1}{x-2})^{-\frac{1}{3}}(\frac{x-2-(x-1)}{(x-2)^2})}{\frac{2}{3}(\frac{x+2}{x-2})^{-\frac{1}{3}}(\frac{x-2-(x+2)}{(x-2)^2})}=\lim_{x\ \to \infty}{\frac{3(x+1)^{-\frac{1}{3}}-(-1)(x-1)^{-\frac{1}{3}}}{(x+2)^{-\frac{1}{3}}(-4)}}$$
$$=\lim_{x \to \infty}{\frac{3(1+\frac{1}{x})^{-\frac{1}{3}}-(1-\frac{1}{x})^{-\frac{1}{3}}}{4(1+\frac{2}{x})^{-\frac{1}{3}}}}=\frac{3-1}{(4)(1)}=\frac{1}{2}.$$

Is this correct and is there a more elegant way of doing it?

EDIT: strictly speaking L'Hopital is not applicable with $x \to \infty$ so just got lucky here...

Body:
Here is another way to do it. Dividing by $x^{2/3}$ yields $$ \frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}} =  \frac{(1+\frac 1x)^{\frac{2}{3}}-(1-\frac 1x)^{\frac{2}{3}}}{(1+\frac 2 x)^{\frac{2}{3}}-(1-\frac 2x)^{\frac{2}{3}}}
$$

Substitute $t = 1/x$. Then $$\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}} = \lim_{t \to 0+}\frac{(1+ t)^{\frac{2}{3}}-(1-t)^{\frac{2}{3}}}{(1+ 2t)^{\frac{2}{3}}-(1- 2t)^{\frac{2}{3}}}  $$

Putting $f(t) = (1+t)^{2/3}$ yields \begin{align} \lim_{t \to 0+}\frac{(1+ t)^{\frac{2}{3}}-(1-t)^{\frac{2}{3}}}{(1+ 2t)^{\frac{2}{3}}-(1- 2t)^{\frac{2}{3}}} &= \lim_{t \to 0
+} \frac{f(t)-f(-t)}{f(2t)-f(-2t)} \\ &= \frac 12 \cdot\lim_{t \to 0+} \frac{f(t)-f(-t)}{t} \cdot \lim_{t \to 0+} \frac{2t}{f(2t)-f(-2t)}\end{align}

Note that $$ \lim_{t \to 0+} \frac{f(t)-f(-t)}{t}= \lim_{t \to 0+} \frac{f(t)-f(0)}{t} + \lim_{t \to 0+} \frac{f(-t)-f(0)}{-t} = 2f'(0)=\frac 23$$

Thus $$\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}} = \frac 12 \cdot\lim_{t \to 0+} \frac{f(t)-f(-t)}{t} \cdot \lim_{t \to 0+} \frac{2t}{f(2t)-f(-2t)} = \frac 12 \cdot \frac 23 \cdot \frac 32 = \frac 12$$


问:
Probability of uniform random subset

Let $X$ be a uniform random subset of $\left \{ 1, 2, ..., n \right \}$ (i.e, $\forall A \subseteq X, \mathbb{P}(X=A)=2^{-n})$

And similarly $Y$ a uniform random subset of $X$.

a. For $n=9$, what is the probability of $|X|$ being even?

b. For any $n$, what is the probability that $|X|=|Y|$

For a. we can notice there is combinatorial symmetry between $0 \sim 9, 1 \sim 8, ... ,4 \sim 5$ and therefore the answer is $\frac{1}{2}$. But if someone could show how to solve it in a more technical way it would help me.

For b. we can notice that given some $|X|$, $\mathbb{P}(|Y|=|X|)=\mathbb{P}(Y=X) = 2^{-|X|}$ but I'm not sure how to continue from here, help would be appreciated.

答:
Hint: Formulate the questions in terms of binomial coefficients.

Especially for a., there is a one-line proof using the binomial theorem.

 Edit: Forgot to write, $\mathbb{P}[|Y| = |X|]$ is not the same as $\mathbb{P}[Y = X]$ (maybe you are considering the case when $X = \{1,\ldots,n\}$). (This is wrong as pointed out in the comment by OP below. However, the earlier part of the answer stands. Use the binomial thoerem.)

Title: Probability of uniform random subset


答:
With difference of squares and difference of cubes, you get that the limit simplifies to

$$\lim_{x\to\infty} \frac{(x+1)-(x-1)}{(x+2)-(x-2)}\cdot \frac{f(x)}{g(x)} = \frac{1}{2}$$

because $f$ and $g$ both grow at the rate $\sim 6 x$ for large $x$.


问:
Integrate $\int_0^\infty \frac{\log x}{x^2+2x+4}\ dx$
What is $$\int_0^\infty  \frac{\log x}{x^2+2x+4}\ dx$$

Here's a hint that came along with the problem,

Substitute $x$ as $2t$ and write it as a sum of two integrals. Then try to simplify.

答:
Well, I will present another method:

Compute:

$$\mathscr{S}_\text{n}:=\frac{\partial}{\partial\text{n}}\left\{\int_0^\infty\frac{\ln\left(\text{n}x\right)}{x^2+2x+4}\space\text{d}x\right\}=\int_0^\infty\frac{1}{x^2+2x+4}\cdot\frac{\partial}{\partial\text{n}}\left(\ln\left(\text{n}x\right)\right)\space\text{d}x=$$
$$\int_0^\infty\frac{1}{x^2+2x+4}\cdot\frac{\partial}{\partial\text{n}}\left(\ln\left(\text{n}\right)+\ln\left(x\right)\right)\space\text{d}x=$$
$$\int_0^\infty\frac{1}{x^2+2x+4}\cdot\left\{\frac{\text{d}}{\text{dn}}\left(\ln\left(\text{n}\right)\right)+\ln\left(x\right)\cdot\frac{\partial}{\partial\text{n}}\left(1\right)\right\}\space\text{d}x\tag1$$

Now, find:

$$\mathcal{I}:=\int_1^2\mathscr{S}_\text{n}\space\text{dn}\tag2$$

Prove that the integral you want to find is equal to $\mathcal{I}$;

You will/must find:

$$\mathcal{I}=\frac{\pi\ln\left(2\right)}{3\sqrt{3}}\tag3$$


答: From the binomial theorem, we know $(1+y)^n\approx1+ny$ for $y\approx0$. Divide both numerator and denominator by $x^{2/3}$. The limit is$$\lim_{x \to \infty}{\frac{\left(1+\frac1x\right)^{\frac{2}{3}}-\left(1-\frac1x\right)^{\frac{2}{3}}}{\left(1+\frac2x\right)^{\frac{2}{3}}-\left(1-\frac2x\right)^{\frac{2}{3}}}}=\lim_{x \to \infty}\frac{1+\frac{2/3}x-1+\frac{2/3}x}{1+\frac{4/3}x-1+\frac{4/3}x}=1/2.$$


问:
Computing the order of elements in Dihedral Groups

I am working through "Abstract Algebra" by Dummit & Foote. Exercise 1.2.1 states:

Compute the order of each of the elements in ... $D_6$, $D_8$, and $D_{12}$.

I have found that:
$$\begin{align*}
D_6 &= \{ 1, r, r^2, s, sr, sr^2 \},\\
D_8 &= \{ 1, r, r^2, r^3, s, sr, sr^2, sr^3 \}, \qquad\text{and}\\
D_{12} &= \{ 1, r, r^2, r^3, r^4, r^5, s, sr, sr^2, sr^3, sr^4, sr^5 \}.
\end{align*}$$

In these descriptions, $r$ is rotation of $\frac{2\pi}{n}$ radians, and $s$ is reflection through vertex "1" and the origin.

Also, this book uses $D_{2n}$ for the dihedral group of rigid motions of the regular $n$-gon (that is, the index is the order of the group).

Determining the orders of powers of $r$ is easy (and $|s| = 2$), but I get confused when trying to compute, say, $|sr^5|$.

I think the key is the statement: $r^is = sr^{-i}$ for $0 \leq i \leq n$.

Then we would have
$$ (sr^5)(sr^5) = s(r^5s)r^5 = s(sr^{-5})r^5 = (ss)(1) = s^2  = 1 $$

So $|sr^5| = 2$ ?

Am I doing this correctly? Is there a better way?

Thanks.

答:
Yes, you are perfectly right.
If you think at the problem geometrically, probably you will get a sharper picture of the situation.

Every element of the form $sr^k$ is in fact a reflection around some axes, thus it's very natural to expect it has order 2.
Your proof is correct, indeed; just replace the $5$ with a $k$ and you've done!

bye


问:
Complex root argument justification
Show that the roots of ${(1+z)^n}$ = ${(1 -z)^n}$ are the values of $$i \tan (rπ/n)$$ where $$r = 0, 1, 2, ... n - 1,$$ but omitting n/2 if n is even.

I have got the expression of root I just want a concrete explanation of why $\frac{n}{2}$ can be omitted when n is even. The thing i deduced was that when n is even the expression has $(n-1)$ roots with 0 being one so the rest $(n-2)$ roots must be complex conjugate pairs,so exclude n/2. But I don't think it's concrete or the right reason. Also if i include $\frac{n}{2}$ in $n$ being odd case
then there's no complex conjugate of it $i(\infty)$ so how the integer coefficient is restored. Please explain the case of $\frac{n}{2}$ for both the cases. I don't want any worked out process of answer but just to enlighten me on my confusion and doubt . Please help.

答:
Even case

You need to omit $n/2$ if $n$ is even otherwise the tangent diverges (at $ \pi/2$) indeed you will have :

$$r=n/2$$

so

$$ x_r=i\tan(r\pi/n)=i\tan(\pi/2)=\pm\infty$$

and you work to find finite complex number.

When you use a reasonement using analysis-synthesis

You first suppose that you have solution, you find a certain form here the form you write

You have to verify that they all work or otherwise exclude the case that don't work (for example here $r=n/2$ when $n$ is even)

Odd case

You want to avoid infinite values so if $n$ is odd, we have to look at :

$$ i\tan(r\pi/n) $$ when $r$ goes from $0$ to $n-1$

Suppose that there exists $r \in [|0,n-1|], \ r\pi/n=(2k+1)\pi/2 , \ k \in \mathbb{Z}$ (it is the case of divergence i.e. $\tan(r\pi/n)=\infty$)

You would have $r=n(2k+1)/2$ which is not whole number because $n$ and $(2k+1)$ are odd so their product divided by 2 is not whole.

So it is not possible to have $\tan(r\pi/n)=\infty$ is the case of $n$ odd.

Hope it helps you to see clearer.

Title: Complex root argument justification


问:
Approximating pi using the $\int_0^1\sqrt{1-x^2}dx$
The first step would be to find the series expansion for $\sqrt{1-x^2}$ using the binomial theorem
$$\sqrt{1-x^2} = \sum_{n=0}^\infty {1/2\choose n}(-x^2)^n $$
Expanding and simplifying the first terms
$$\sqrt{1-x^2} = 1 -\left(\frac{1\cdot x^2}{2 \cdot1!} + \frac{1 \cdot 3 \cdot x^4}{2^2 \cdot 2! } + \frac{1 \cdot 3 \cdot 5 \cdot x^6}{2^3 \cdot 3!} +... \right)$$
Here I run into a problem. How come all the terms are negative while $1$ isn't? This would mean I can't add it to the infinite sum. So continuing:
$$\sqrt{1-x^2} = 1- \sum_{n=0}^\infty \frac{(2n+1)!!}{2^n n!}x^{2n} = 1-\sum_{n=0}^\infty \frac{(2n+1)!!}{(2n)!!}x^{2n} $$
When I plot this on desmos my approximation is a) shifted by $1$ unit down b) inaccurate. On the image below I have used 50 terms so I doubt the error is due to a lack of terms.

Where have I made a mistake. Thank you

Body: For $n\ge2$,$$[x^{2n}]\sqrt{1-x^2}=(-1)^n\binom{1/2}{n}=(-1)^n\frac{1}{n!}\frac12\prod_{k=1}^{n-1}(\tfrac12-k)=-\frac{(2n-3)!!}{n!2^n},$$so$$\sqrt{1-x^2}=1-\tfrac12x^2-\sum_{n\ge2}\frac{(2n-3)!!}{n!2^n}x^{2n}.$$See how that looks in Desmos.


问:
Primes represented by $x^3-21xy^2+35y^3$.
What do we know about the primes represented by this special binary cubic form $x^3-21xy^2+35y^3$?

I know that my question is very short, but I don't have any idea about it, and I don't know where can I find the answer in the literature.

I searched through the net to find a program to check whether a binary cubic equation $f(x, y)=n$ has a solution or not, but I didn't find anything. In the case of the absence of an answer, or a reference to my question, introducing any program/ engine would be welcome.

Let $\alpha$ be a root of the polynomial $x^3-21x-35=0$, and let $K:=\mathbb{Q}(\alpha)$. Then it is easy to show that $$Norm(x+y\alpha+z\alpha^2)=x^3+35y^3+1225z^3-105xyz-21xy^2+441xz^2+42x^2z-735yz^2.$$ That Binary cubic form is just $Norm(x+y\alpha)$.

Note that the discriminant of $P(x)=x^3-21x-35$ is
$-(4\times(-21)^3+27\times(-35)^2)=3969=3^4\times7^2$, so discriminant of $K$ is a square, so it is a cyclic cubic Galois extension, so we can conclude that $r_1=3$ and $r_2=0$. By Dirichlet's unit theorem, we can conclude that $\mathcal{O}_K^{\times}=\{\pm1\}\times\mathbb{Z}^2$. Also, note that $P(x)=x^3-21x-35$ is $7$-Eisenstein, and $P(x-1)=x^3-3x^2+3x-1-21x+21-35=x^3-3x^2-18x-15$ is $3$-Eisenstein; so we can conclude that $\mathcal{O}_K=\mathbb{Z}[\alpha]=\mathbb{Z}\oplus\mathbb{Z}\alpha\oplus\mathbb{Z}\alpha^2$.

Is the answer to the following question positive?

Assume that $Norm(a+b\alpha+c\alpha^2)=p$. Does there exist a unit $u \in \mathcal{O}_K^{\times}$ such that $(a+b\alpha+c\alpha^2)\times u = A+B\alpha$ for some integers $A, B$? Assume that $a+b\alpha+c\alpha^2$ is given. Can we find a suitable unit, such that after the multiplication, we can write the product as linear combination of $1$, and $\alpha$? without need of $\alpha^2$?

答:
Let $\alpha$ be a root of $x^3-21x+35=0$. Then it is easy to characterize primes of form $$N(x + y\alpha + z\alpha^2) = x^3+35y^3+1225z^3-105xyz-21xy^2+441xz^2+42x^2z-735yz^2$$
which was already eluded in Will Jagy's answer.

(Theorem) A prime $p\neq 3,7$ can be represented by above cubic form iff $p\equiv \pm 1, \pm 8 \pmod{63}$.

Proof of Theorem: Let $K$ be number field of $x^3-21x+35$. I assume the following facts: $K$ has class number $3$, contained in $\mathbb{Q}(\zeta_{63})$.

Let $H$ be the Hilbert class field of $K$, then $H/\mathbb{Q}$ is abelian of degree $9$ ($H/\mathbb{Q}$ is Galois and any group of order $9$ is abelian).

Claim: $H\subset \mathbb{Q}(\zeta_{63})$. This follows from a general (but not well-known) fact of cyclotomic fields. We have the following proposition, proved in the answer here: If $F/\mathbb{Q}(\zeta_m)$ is unramified (at finite primes) and $F/\mathbb{Q}$ abelian, then $F=\mathbb{Q}(\zeta_m)$. Because $H/\mathbb{Q}$ is abelian, applying this proposition to $F=H\mathbb{Q}(\zeta_{63})$ shows that $H\mathbb{Q}(\zeta_{63}) = \mathbb{Q}(\zeta_{63})$, so $H\subset \mathbb{Q}(\zeta_{63})$.

Claim: $H$ corresponds to $\{\pm 1,\pm 8\} \subset (\mathbb{Z}/63\mathbb{Z})^\times$. $H$ corresponds to a subgroup of order $4$ of $(\mathbb{Z}/63\mathbb{Z})^\times = C_6 \times C_6$, such a subgroup is unique, and this is the only one.

Finally $p\neq 3,7$ can be represented as $N(x + y\alpha + z\alpha^2)$ iff $p$ splits into principle ideals in $K$, iff $p$ splits completely in $H$, completing the proof.

Restricting to $z=0$ of the cubic form is more tricky, and likely has no simple answer. If $\pi(n)$ denotes the prime counting function, then

$p$
No. of $p \equiv 1, 8, 55, 62 \pmod{63}$
No. of $p=x^3-21xy^2+35y^3$

$\pi(p)\leq 3000$
326
61

$3001\leq \pi(p)\leq 6000$
344
42

$6001\leq \pi(p)\leq 9000$
326
32

The equation of form $N(x+y\alpha)$ is a Thue equation. For each individual $p$, there is an algorithm to check whether $N(x+y\alpha) = p$ has integral solution. The following Magma code checks above table for small $p$:

```<code>R<x> := PolynomialRing(Integers());
f := x^3 -21*x+35;
T := Thue(f);
list := {71, 127, 181, 197, 251, 307, 379, 433, 449, 503, 631, 701, 757, 811};
t := { n : n in list | Solutions(T, n) ne [] };
t
</code>```

which outputs { 71, 127, 197, 307, 379, 449, 757 }. The complete list of primes $p$ with $\pi(p)\leq 9000$ which can be written as $p=x^3-21xy^2+35y^3$ is

```<code>{71,127,197,307,379,449,757,827,1259,1511,1637,1693,1889,2017,2339,2393,3221,3851,4283,4591,4789,5417,5419,5923,6047,6229,6553,6679,6733,7127,7253,7309,7687,7993,8387,8819,9883,10151,11593,11717,11719,12781,13033,14057,14923,15121,15749,16057,16829,17891,19081,19853,20593,21617,21673,22877,23633,24373,24697,24877,26641,28351,28547,28909,29287,30241,30493,31193,32381,32507,34469,35279,35281,35603,37799,37997,38611,38737,39439,40123,41887,42013,42407,44281,44729,45863,46187,47431,47881,49391,51659,51913,52289,53171,53857,54181,54559,55061,55763,55817,57457,57709,58897,60103,61487,62047,62189,62819,66403,67481,68041,70309,72269,72577,72883,77813,78569,79813,81017,81019,81703,82727,83719,84239,84869,86491,87443,87697,89767,90019,90271,92177,92357,92413,92861}
</code>```

Body:
Let $\alpha$ be a root of $x^3-21x+35=0$. Then it is easy to characterize primes of form $$N(x + y\alpha + z\alpha^2) = x^3+35y^3+1225z^3-105xyz-21xy^2+441xz^2+42x^2z-735yz^2$$
which was already eluded in Will Jagy's answer.

(Theorem) A prime $p\neq 3,7$ can be represented by above cubic form iff $p\equiv \pm 1, \pm 8 \pmod{63}$.

Proof of Theorem: Let $K$ be number field of $x^3-21x+35$. I assume the following facts: $K$ has class number $3$, contained in $\mathbb{Q}(\zeta_{63})$.

Let $H$ be the Hilbert class field of $K$, then $H/\mathbb{Q}$ is abelian of degree $9$ ($H/\mathbb{Q}$ is Galois and any group of order $9$ is abelian).

Claim: $H\subset \mathbb{Q}(\zeta_{63})$. This follows from a general (but not well-known) fact of cyclotomic fields. We have the following proposition, proved in the answer here: If $F/\mathbb{Q}(\zeta_m)$ is unramified (at finite primes) and $F/\mathbb{Q}$ abelian, then $F=\mathbb{Q}(\zeta_m)$. Because $H/\mathbb{Q}$ is abelian, applying this proposition to $F=H\mathbb{Q}(\zeta_{63})$ shows that $H\mathbb{Q}(\zeta_{63}) = \mathbb{Q}(\zeta_{63})$, so $H\subset \mathbb{Q}(\zeta_{63})$.

Claim: $H$ corresponds to $\{\pm 1,\pm 8\} \subset (\mathbb{Z}/63\mathbb{Z})^\times$. $H$ corresponds to a subgroup of order $4$ of $(\mathbb{Z}/63\mathbb{Z})^\times = C_6 \times C_6$, such a subgroup is unique, and this is the only one.

Finally $p\neq 3,7$ can be represented as $N(x + y\alpha + z\alpha^2)$ iff $p$ splits into principle ideals in $K$, iff $p$ splits completely in $H$, completing the proof.

Restricting to $z=0$ of the cubic form is more tricky, and likely has no simple answer. If $\pi(n)$ denotes the prime counting function, then

$p$
No. of $p \equiv 1, 8, 55, 62 \pmod{63}$
No. of $p=x^3-21xy^2+35y^3$

$\pi(p)\leq 3000$
326
61

$3001\leq \pi(p)\leq 6000$
344
42

$6001\leq \pi(p)\leq 9000$
326
32

The equation of form $N(x+y\alpha)$ is a Thue equation. For each individual $p$, there is an algorithm to check whether $N(x+y\alpha) = p$ has integral solution. The following Magma code checks above table for small $p$:

R<x> := PolynomialRing(Integers());
f := x^3 -21*x+35;
T := Thue(f);
list := {71, 127, 181, 197, 251, 307, 379, 433, 449, 503, 631, 701, 757, 811};
t := { n : n in list | Solutions(T, n) ne [] };
t

which outputs { 71, 127, 197, 307, 379, 449, 757 }. The complete list of primes $p$ with $\pi(p)\leq 9000$ which can be written as $p=x^3-21xy^2+35y^3$ is

{71,127,197,307,379,449,757,827,1259,1511,1637,1693,1889,2017,2339,2393,3221,3851,4283,4591,4789,5417,5419,5923,6047,6229,6553,6679,6733,7127,7253,7309,7687,7993,8387,8819,9883,10151,11593,11717,11719,12781,13033,14057,14923,15121,15749,16057,16829,17891,19081,19853,20593,21617,21673,22877,23633,24373,24697,24877,26641,28351,28547,28909,29287,30241,30493,31193,32381,32507,34469,35279,35281,35603,37799,37997,38611,38737,39439,40123,41887,42013,42407,44281,44729,45863,46187,47431,47881,49391,51659,51913,52289,53171,53857,54181,54559,55061,55763,55817,57457,57709,58897,60103,61487,62047,62189,62819,66403,67481,68041,70309,72269,72577,72883,77813,78569,79813,81017,81019,81703,82727,83719,84239,84869,86491,87443,87697,89767,90019,90271,92177,92357,92413,92861}


答:
For (a), another approach (which works for any $n$) is to realise that if we go through each element in order to decide whether we include it in $X$ or not, the parity of $|X|$ is determined by whether $n\in X$, so the probability is $\frac{1}{2}$.

More explicitly: let $p$ be the probability that a random subset of $\{1,\dots,n-1\}$ has even size. If $n\in X$, we need $|X\cap[n-1]|$ to be odd; else $n\not\in X$, so $|X\cap[n-1]|$ must be even. Then the probability we seek is $\frac{1}{2}\cdot(1-p)+\frac{1}{2}\cdot p=\frac{1}{2}$.

For (b), we can say using the law of total probability
$$\mathbb P(|X|=|Y|)=\sum_{k=0}^n\mathbb P(X=Y\mid |X|=k)\cdot\mathbb P(|X|=k)=\sum_{k=0}^n2^{-k}\cdot\mathbb P(|X|=k).$$
Can you finish from here?

Title: Probability of uniform random subset

Body:
Let $X$ be a uniform random subset of $\left \{ 1, 2, ..., n \right \}$ (i.e, $\forall A \subseteq X, \mathbb{P}(X=A)=2^{-n})$

And similarly $Y$ a uniform random subset of $X$.

a. For $n=9$, what is the probability of $|X|$ being even?

b. For any $n$, what is the probability that $|X|=|Y|$

For a. we can notice there is combinatorial symmetry between $0 \sim 9, 1 \sim 8, ... ,4 \sim 5$ and therefore the answer is $\frac{1}{2}$. But if someone could show how to solve it in a more technical way it would help me.

For b. we can notice that given some $|X|$, $\mathbb{P}(|Y|=|X|)=\mathbb{P}(Y=X) = 2^{-|X|}$ but I'm not sure how to continue from here, help would be appreciated.


问:
Why is Strömberg Wavelet section in 2018 book an amost exact copy of 2016 Wikipedia article?
Does anyone know why Section 2.10 Strömberg Wavelet of the 2018 book Wavelet: Analysis and Methods seems to be an almost exact copy of the 2016 Wikipedia article of the same name?

There are some small differences:

The book section is missing citations while the Wikipedia article cites Strömberg (1983) and Wojtaszczyk (1997).

The book at one point appears to mis-typeset $P^{(m)}(V)$ as $P(m)(V)$

The Wikipedia article does not cite the book (which would be difficult anyway given the 2016 versus 2018 timestamps). The book meanwhile claims "All rights reserved.", and a Google Book search of the book for "Wikipedia" returns "No results found in this book for Wikipedia".

答: I have never heard of this book's publisher and, searching for the authors, couldn't find a university page or anything (as a matter of fact, this and one other book seem to be their only meaningful trace on the internet - and I believe their names are sufficiently rare to where one should be able to stumble upon at least some mention of their work). Unless I'm terribly mistaken, all this suggests it may indeed be a predatory publication, trying to earn a quick buck on a poorly produced textbook. This is only a conjecture, however, and should be treated as such. Whether there is any point in pursuing this matter further depends entirely on your own disposition and availability of resources. In either case, emailing the publisher about it should be a good first step.


答:
We can use

$$\frac{1}{\sqrt{1-4x}} = \sum_{n=0}^\infty {2n \choose n} x^n$$

Since both

$$\int_{-1}^1\frac{dx}{\sqrt{1-x^2}}$$

and

$$\int_{-1}^1 2\sqrt{1-x^2}dx$$

converge to $\pi$, you can either use

$$\frac{1}{\sqrt{1-x^2}} = \sum_{n=0}^\infty {2n \choose n} \left(\frac{x}{2}\right)^{2n}$$

or

$$2\sqrt{1-x^2} = \frac{2(1-x^2)}{\sqrt{1-x^2}} = 2+2\sum_{n=1}^\infty \left[{2n \choose n} - {2n-2 \choose n-1}\right]\left(\frac{x}{2}\right)^{2n}$$

$$= 2+2\sum_{n=1}^\infty {2n-2 \choose n-1}\left[3-\frac{2}{n}\right]\left(\frac{x}{2}\right)^{2n}$$

Title: Approximating pi using the $\int_0^1\sqrt{1-x^2}dx$

Body:
The first step would be to find the series expansion for $\sqrt{1-x^2}$ using the binomial theorem
$$\sqrt{1-x^2} = \sum_{n=0}^\infty {1/2\choose n}(-x^2)^n $$
Expanding and simplifying the first terms
$$\sqrt{1-x^2} = 1 -\left(\frac{1\cdot x^2}{2 \cdot1!} + \frac{1 \cdot 3 \cdot x^4}{2^2 \cdot 2! } + \frac{1 \cdot 3 \cdot 5 \cdot x^6}{2^3 \cdot 3!} +... \right)$$
Here I run into a problem. How come all the terms are negative while $1$ isn't? This would mean I can't add it to the infinite sum. So continuing:
$$\sqrt{1-x^2} = 1- \sum_{n=0}^\infty \frac{(2n+1)!!}{2^n n!}x^{2n} = 1-\sum_{n=0}^\infty \frac{(2n+1)!!}{(2n)!!}x^{2n} $$
When I plot this on desmos my approximation is a) shifted by $1$ unit down b) inaccurate. On the image below I have used 50 terms so I doubt the error is due to a lack of terms.

Where have I made a mistake. Thank you


问:
Prove that every abelian group is a subgroup of divisible group.
I need a proof that every abelian group is a subgroup of divisible group (to make sure that every object of the category of $\mathbb Z$-modules has injective resolution).

I found a proof on group props but it uses wreath product which is out of my range at that point in my life.

Can you show me a simpler proof?

答:
Let $G$ be a group. It can be written as the quotient of a free group, say $G=F/N$. If $G$ is abelian, the commutator subgroup $F'$ is included in $N$, so $G$ is a quotient of the abelianization $F^{ab} \simeq \bigoplus\limits_{i \in I} \mathbb{Z}$, say $F^{ab}/M$. Clearly, $F^{ab}$ can be embedded into $Q=\bigoplus\limits_{i \in I} \mathbb{Q}$, so $G=F^{ab}/M$ is embedded into $Q/M$. But $Q/M$ is divisible as the quotient of a divisible group.

Edit: The source is this post.


问:
Determinant of block matrix with commuting blocks + ring homomorphism
Let $A, B$ be commutative rings with unity and let $p$ be a positive integer. Suppose $\varphi:A\to M_p(B)$ is a ring homomorphism. Is it true that for any positive integer $q$ and any $M\in M_q(A)$
$$\det(\varphi(M)) = \det(\varphi(\det(M)))~?$$
where the leftmost $\det$ is the determinant on $M_{pq}(B)$, the middle one that on $M_{q}(B)$ and the rightmost one that on $M_{p}(A)$? (And we denote $\varphi$ the natural extension of $\varphi$ to a ring homomorphism $M_q(A)\to M_{pq}(B)$.)

This should be true (with only a few possible exceptions) when $A$ is a field. Indeed every matrix over a field is equivalent to a projection; it follows that $\varphi(M)$ invertible iff $M$ invertible iff $\det(\varphi(M))\neq 0$. Every invertible matrix $M$ can be written as $UD$ where $U\in\mathrm{SL}_p(A)$ and $D=\mathrm{Diag}(\theta_1,\dots,\theta_p)$. Thus when ever $\mathrm{SL}_p(A)=\mathrm{GL}_p(A)'$ holds (and there are only a few exceptions to this) we will get $\det(\varphi(M))=\det(\varphi(\det(M)))$.

I initially thought this was true for $A$ a PID since we can use elementary operations on lines to put $M$ into echelon form. But upon inspection I don't know how to deal with the images under $\varphi$ of the matrices used when improving the pivot in the Smith Normal Form algorithm.

To wit, we essentially would need to compare, for $\sigma,\tau, \gamma,\delta\in A$ with $\sigma\delta+\tau\gamma = 1_A$,
$$
1=
\det\Bigg(
\underbrace{
\varphi\left(
\det\left(
\begin{matrix}
\sigma&\tau\\
-\gamma&\delta
\end{matrix}
\right)
\right)
}_{=I_n\in M_n(B)}
\Bigg)
\overset?=
\det
\underbrace{
\left(
\begin{matrix}
\varphi(\sigma)&\varphi(\tau)\\
-\varphi(\gamma)&\varphi(\delta)
\end{matrix}
\right)}_{\in M_{2n}(B)}.$$

EDIT. Equivalently (and justifying the title) the question asks whether when $M_{ij}$, for $1\leq i,j\leq q$ are pairwise commuting matrices in $M_p(B)$ if $$\det(M)=\det\Big(\sum_{\sigma\in\mathfrak{S}_q}(-1)^\sigma\prod_{1}^qM_{i,\sigma(i)}\Big)$$
where $M$ is the $pq$-square matrix given in block form as $M=(M_{ij})_{i,j}\in M_{pq}(B)$.

Second EDIT. A Google search revealed a paper proving something almost identical to my question (and identical if we consider the second form from the previous EDIT): Determinants of Commuting-Block Matrices by Istvan Kovacs, Daniel S. Silver, and Susan G. Williams. Their proof works over any commutative ring but asks for $A=B$.

答:
I reproduce here the proof from the paper Determinants of Commuting-Block Matrices by Istvan Kovacs, Daniel S. Silver, and Susan G. Williams.

Theorem. Let $M_{ij}$, $1\leq i,j\leq q$, be pairwise commuting matrices in $M_p(A)$ for some commutative ring $A$ and $B\subset M_p(A)$ the (commutative) subring generated by the $M_{ij}$, and $M=(M_{ij})_{ij}$ the associated block matrix. Then
$$\det_{M_{pq}(A)}(M) = \det_{M_{p}(A)}(\det_{M_{q}(B)}(M))$$

The proof is by induction on $q$ and uses operations on lines. Viewing $M$ as a matrix in $M_q(B)$ with lines $L_1,\dots,L_q$ we perform the following operations:
$$\begin{cases}
L_2&\leftarrow\quad M_{11}L_2\\
L_3&\leftarrow\quad M_{11}L_3\\
&\ddots\\
L_q&\leftarrow\quad M_{11}L_q
\end{cases}
\qquad\textrm{followed by}\qquad
\begin{cases}
L_2&\leftarrow\quad L_2-M_{21}L_1\\
L_3&\leftarrow\quad L_3-M_{31}L_1\\
&\ddots\\
L_q&\leftarrow\quad L_q-M_{q1}L_1
\end{cases}$$
In matrix terms this is
$$
\begin{pmatrix}
1\\ -M_{21}&1\\ -M_{31}&&1\\ &&&\ddots \\ -M_{q1}&&&&1
\end{pmatrix}
\times
\begin{pmatrix}
1\\&M_{11}\\&&M_{11}\\ &&&\ddots \\&&&&M_{11}
\end{pmatrix}
\times
M
=
M'
=
\begin{pmatrix}
M_{11}&*&\cdots&*\\
0\\
\vdots&&N\\
0
\end{pmatrix}
$$
Next compute both $\displaystyle\det_{M_{pq}(A)}(\bullet)$ and $\displaystyle\det_{M_{p}(A)}(\det_{M_{q}(B)}(\bullet))$ on this product:

Computing $\displaystyle\det_{M_{pq}(A)}(\bullet)$: We find
$$
\begin{array}{rcl}
\displaystyle\det_{M_p(A)}\Big(M_{11}\Big)^{q-1}\cdot\det_{M_{pq}(A)}\Big(M\Big)
& = &
\displaystyle\det_{M_{pq}(A)}\Big(M'\Big)\\
& = &
\displaystyle\det_{M_p(A)}\Big(M_{11}\Big)\cdot
\color{orange}{\det_{M_{p(q-1)}(A)}\Big(N\Big)}
\end{array}$$

Computing $\displaystyle\det_{M_{p}(A)}(\det_{M_{q}(B)}(\bullet))$: first compute in $M_q(B)$:
$$
\begin{array}{rcl}
\displaystyle\Big(M_{11}\Big)^{q-1}\cdot\det_{M_{q}(B)}\Big(M\Big)
& = &
\displaystyle\det_{M_{q}(B)}\Big(M'\Big)\\
& = &
\displaystyle M_{11}\cdot\det_{M_{q-1}(B)}\Big(N\Big)
\end{array}$$
Then apply $\displaystyle\det_{M_p(A)}$:
$$
\begin{array}{rcl}
\displaystyle\Big(\det_{M_p(A)} M_{11}\Big)^{q-1}
\cdot \det_{M_p(A)}\Big(\det_{M_{q}(B)}\Big(M\Big)\Big)
& = &
\displaystyle\det_{M_p(A)}\Big(\det_{M_{q}(B)}\Big(M'\Big)\Big)\\
& = &
\displaystyle\det_{M_p(A)}\Big(M_{11}\Big)\cdot
\color{orange}{\det_{M_p(A)}\Big(\det_{M_{q-1}(B)}\Big(N\Big)\Big)}
\end{array}$$

We now apply our induction hypothesis: the two orange terms are identical. Therefore
$$\Big(\det_{M_p(A)} M_{11}\Big)^{q-1}
\cdot\bigg[
\det_{M_{pq}(A)}\Big(M\Big)
-
\det_{M_p(A)}\Big(\det_{M_{q}(B)}\Big(M\Big)\Big)
\bigg]
=
0
$$
To deal with $\displaystyle\Big(\det_{M_p(A)} M_{11}\Big)$ which may very well be a zero divisor in $A$ the authors shift to working with matrices with coefficients in $A[T]$ and replace $M_{11}$ with $\widetilde{M_{11}}=T I_n + M_{11}$: then $\displaystyle\Big(\det_{M_p(A[T])} \widetilde{M_{11}}\Big)$ is a monic polynomial of degree $p$ and thus not a zero divisor in $A[T]$. Thus, defining $\widetilde{M}$ to have the same coefficients as $M$ except $\widetilde{M}_{11}=\widetilde{M_{11}}$ yields
$$\Big(\det_{M_p(A[T])} \widetilde{M_{11}}\Big)^{q-1}
\cdot\bigg[
\det_{M_{pq}(A[T])}\Big(\widetilde{M}\Big)
-
\det_{M_p(A[T])}\Big(\det_{M_{q}(B[T])}\Big(\widetilde{M}\Big)\Big)
\bigg]
=
0
$$
i.e.
$$\det_{M_{pq}(A[T])}\Big(\widetilde{M}\Big)
=
\det_{M_p(A[T])}\Big(\det_{M_{q}(B[T])}\Big(\widetilde{M}\Big)\Big)$$
and evaluating at $T=0$:
$$\det_{M_{pq}(A)}\Big(M\Big)
=
\det_{M_p(A)}\Big(\det_{M_{q}(B)}\Big(M\Big)\Big).$$


问:
Notation $2^T$ for the operator T of the Hilbert space
I came across an unknown to me notation $2^T, 2^{-T}$ for a positive operator $T$ on a Hilbert space (more precisely, in $\mathbb{C}^n$).

What does this mean? I can't find it anywhere. Is it the image and the kernel?

答:
For a positive (more generally, normal) operator $T$ on a Hilbert space $H$, there is a unique unital *-homomorphism (a linear map that preserves products and involutions) from $C(\sigma(T))$ to $B(H)$ that maps $\operatorname{id}$ to $T$, where $\operatorname{id}$ is the map $C(\sigma(T))\to \mathbb{C}, z\mapsto z$.

For a continuous function $f$ on the spectrum $\sigma(T)$ of $T$, $f(T)$ means the image of $f$ under the unique *-homomorphism.

For example, let $p(x)=x^2+2x+1$,  you can check $p(T)=T^2+2T+e$ where $e$ is the  identity of $B(H)$.

For $f(x)=2^x$, it can be approximated by polynomials $p_N(x)=\sum_{n=0}^N\frac{(\ln 2)^n}{n!}x^n$ with respect to the uniform norm. $f(T)$ can be proved to be the limit of $p_N(T)$.


问:
P-trivial $\sigma-$algebra
I want to proof the following:

Consider an probability space with a P-trivial $\sigma-$Algebra $G$. Show that for any real-valued $G$-measurable random variable $X$, there exists $c \in \mathbb{R}$ such that$X=c$ almost surely.

My idea:

Considering cdf implies: $F_X(t)=P(X\leq t) \in \{0,1\}$

Hence $G$ is trivial, there is $x_0 := \inf\{x: F_X(x)=1\}$. Thus $P(X=x_0)= F_X(x_0)- \lim_{x<x_0} F_X(x)=1$.

Therefore $X$ is constant. But why does this hold almost surely?

答:
Your argument is essentially correct. It needs just a small adjustments. Here is a detailed proof, using your argument.

Consider $F_X(t)$  for $t$ in the extended real line, $\overline{\Bbb R}= \Bbb R \cup \{-\infty , +\infty\}$.

Then $F_X(+\infty)=P(X\leq +\infty) =1$, and so let $c := \inf \{x\in \overline{\Bbb R} : F_X(x)=1\}$.

Claim: $F_X(c) = 1$.

In fact, since $c := \inf \{x\in \overline{\Bbb R} : F_X(x)=1\}$, there is a decreasing sequence $\{x_n\}_n$ in  $\{x\in \overline{\Bbb R} : F_X(x)=1\}$, such that $x_n \to c$. So
$$F_X(c) = P(X\leq c) = \lim_{n \to \infty} P(X\leq x_n) = \lim_{n \to \infty} F_X(x_n) = \lim_{n \to \infty} 1=1$$
(You can also see the claim is true, directly from the fact that $F_X$ is continuous from the right.)

Now, $P(X=c)= F_X(c)- \lim_{x \to a^{-}} F_X(x)=1$.

So, $P(X \neq c)=0$, which means that $X=c$ almost surely.

Remark: Note that we can not conclude that $\{x: X(x) \neq c\} =\emptyset$. So we can not conclude that $X=c$. We can only conclude that $P(X \neq c)=0$, that is why we have $X=c$ almost surely.

Title: P-trivial $\sigma-$algebra


答:
Here is how I would do it without using L'hopital, or series expansion :

Using the identity $ a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right) $, we have : $$ \left(x+1\right)^{2}-\left(x-1\right)^{2}=\left(\left(x+1\right)^{\frac{2}{3}}-\left(x-1\right)^{\frac{2}{3}}\right)\left(\left(x+1\right)^{\frac{4}{3}}+\left(x^{2}-1\right)^{\frac{2}{3}}+\left(x-1\right)^{\frac{4}{3}}\right) $$

And : $$ \left(x+2\right)^{2}-\left(x-2\right)^{2}=\left(\left(x+2\right)^{\frac{2}{3}}-\left(x-2\right)^{\frac{2}{3}}\right)\left(\left(x+2\right)^{\frac{4}{3}}+\left(x^{2}-4\right)^{\frac{2}{3}}+\left(x-2\right)^{\frac{4}{3}}\right) $$

Thus : \begin{aligned}\lim_{x\to +\infty}{\frac{\left(x+1\right)^{\frac{2}{3}}-\left(x-1\right)^{\frac{2}{3}}}{\left(x+2\right)^{\frac{2}{3}}-\left(x-2\right)^{\frac{2}{3}}}}&=\lim_{x\to +\infty}{\frac{\left(\left(x+1\right)^{2}-\left(x-1\right)^{2}\right)\left(\left(x+2\right)^{\frac{4}{3}}+\left(x^{2}-4\right)^{\frac{2}{3}}+\left(x-2\right)^{\frac{4}{3}}\right)}{\left(\left(x+2\right)^{2}-\left(x-2\right)^{2}\right)\left(\left(x+1\right)^{\frac{4}{3}}+\left(x^{2}-1\right)^{\frac{2}{3}}+\left(x-1\right)^{\frac{4}{3}}\right)}}\\ &=\lim_{x\to +\infty}{\frac{\left(1+\frac{2}{x}\right)^{\frac{4}{3}}+\left(1-\frac{4}{x^{2}}\right)^{\frac{2}{3}}+\left(1-\frac{2}{x}\right)^{\frac{4}{3}}}{2\left(\left(1+\frac{1}{x}\right)^{\frac{4}{3}}+\left(1-\frac{1}{x^{2}}\right)^{\frac{2}{3}}+\left(1-\frac{1}{x}\right)^{\frac{4}{3}}\right)}}\\ &=\frac{1}{2}\end{aligned}


问:
Find a closed form for the following integral:
I have to evalute a closed form for the following integral:

$$\int_0^1\left\{\frac{(-1)^{\lfloor\frac{1}{x}\rfloor}}{x}\right\}dx$$

where $\{x\}$ is the fractional part.

I thought of using a substitution $y=1/x$ this will lead me to:

$$\int_\infty^1\frac{\left\{y(-1)^{\lfloor y\rfloor}\right\}}{-y^2}dy$$

And now I have no idea how to proceed.

答:
Assignment:

Find a closed form for the following integral:

$$\mathscr{I}:=\int_0^1\left\{\left(-1\right)^{\left\lfloor\frac{1}{x}\right\rfloor}\cdot\frac{1}{x}\right\}\space\text{d}x\tag1$$

Where $\left\lfloor x\right\rfloor$ is the Floor function and $\left\{x\right\}$ is the Fractional part.

Well, let's first consider the substitution $\text{u}:=\frac{1}{x}$, this let's us rewrite $(1)$ as follows:

$$\mathscr{I}=\int_\infty^1-\frac{1}{\text{u}^2}\cdot\left\{\left(-1\right)^{\left\lfloor\text{u}\right\rfloor}\cdot\text{u}\right\}\space\text{du}=\int_1^\infty\frac{1}{\text{u}^2}\cdot\left\{\left(-1\right)^{\left\lfloor\text{u}\right\rfloor}\cdot\text{u}\right\}\space\text{du}\tag2$$

Now, we can break it into the sum of sub-integrals (because if $\text{p}\le\text{s}<\text{p}+1$ we get $\lfloor\text{s}\rfloor=\text{p}$). So:

$$\mathscr{I}=\sum_{\text{k}\ge1}\int_\text{k}^{\text{k}+1}\frac{1}{\text{u}^2}\cdot\left\{\left(-1\right)^{\left\lfloor\text{u}\right\rfloor}\cdot\text{u}\right\}\space\text{du}=\sum_{\text{k}\ge1}\int_\text{k}^{\text{k}+1}\frac{1}{\text{u}^2}\cdot\left\{\left(-1\right)^\text{k}\cdot\text{u}\right\}\space\text{du}\tag3$$

In order to deal with the term $(-1)^\text{k}$ it is not hard to see that we can split the sum further into the following two sums:

$$\mathscr{I}=\sum_{\text{n}\ge1}\int_{2\text{n}-1}^{2\text{n}}\frac{\left\{-\text{u}\right\}}{\text{u}^2}\space\text{du}+\sum_{\text{n}\ge1}\int_{2\text{n}}^{2\text{n}+1}\frac{\left\{\text{u}\right\}}{\text{u}^2}\space\text{du}\tag4$$

We can use the fact that $\left\{-\text{u}\right\}=1-\left\{\text{u}\right\}=1-\left(\text{u}-\lfloor\text{u}\rfloor\right)$:

$$\mathscr{I}=\sum_{\text{n}\ge1}\int_{2\text{n}-1}^{2\text{n}}\frac{1-\left(\text{u}-\lfloor\text{u}\rfloor\right)}{\text{u}^2}\space\text{du}+\sum_{\text{n}\ge1}\int_{2\text{n}}^{2\text{n}+1}\frac{\text{u}-\lfloor\text{u}\rfloor}{\text{u}^2}\space\text{du}\tag5$$

Now, we can use the fact that we are summing over the positive integers. So we can rewrite the floor function terms in the following way:

$$\mathscr{I}=\sum_{\text{n}\ge1}\int_{2\text{n}-1}^{2\text{n}}\frac{1-\left(\text{u}-\left(2\text{n}-1\right)\right)}{\text{u}^2}\space\text{du}+\sum_{\text{n}\ge1}\int_{2\text{n}}^{2\text{n}+1}\frac{\text{u}-2\text{n}}{\text{u}^2}\space\text{du}\tag6$$

You can evaluate the integrals and combining the sums (because they sum over the same interval), in order to get:

$$\mathscr{I}=\sum_{\text{n}\ge1}\left(\frac{1}{2\text{n}-1}-\frac{1}{2\text{n}+1}+\ln\left(\frac{\left(2\text{n}+1\right)\left(2\text{n}-1\right)}{\left(2\text{n}\right)^2}\right)\right)\tag7$$

Split the sum (because they are both absolutely convergent) and use the fact that a sum of $\ln$ functions can be written as a product of the argument:

$$\mathscr{I}=\underbrace{\sum_{\text{n}\ge1}\left(\frac{1}{2\text{n}-1}-\frac{1}{2\text{n}+1}\right)}_{=\space 1}+\ln\left(\underbrace{\prod_{\text{n}\ge1}\frac{\left(2\text{n}+1\right)\left(2\text{n}-1\right)}{\left(2\text{n}\right)^2}}_{=\space\frac{2}{\pi}}\right)=1+\ln\left(\frac{2}{\pi}\right)\tag8$$

Where $\frac{2}{\pi}$ is found using the Wallis product.


问:
What does $f(x,y) = y' = \frac{\Delta y}{\Delta x}$ mean? $\partial_x f(x,y)$ or $\partial_y f(x,y)$
I'm a little unsure about this case but curious to know.</p>

答:
Neither $\partial_x f(x,y)$ nor $\partial_y f(x,y)$. I imagine that the context is the one of ODE.

And you're looking to find $y(x)$ as a map of $x$. You just want to have at any point $x$ the derivative $y^\prime(x) = \frac{dy}{dx}(x)$ to be equal to $f(x,y(x))$.


问:
Green's formula for area
Using Green theorem I need to calculate the area bounded with $(x+y)^4=x^2y$. First(after converting to polar coordinates) i get $x=\cos^6\phi\sin^2\phi$ and $y=\cos^4\phi\sin^4\phi$. And after i plug that into Green's formula i get $\int\limits_{0}^{2\pi} \cos^9\phi\sin^5\phi d\phi$ and that is $0$. But when i set bounds for $\phi$ to be $0, \frac{\pi}{2}$ i get correct solution. Why we have $\frac{\pi}{2}$ for $\phi$ when we don't have any restrictions?

答: The closed curve is drawn once as $\phi$ goes from  $0$ to $\pi/2$, because both the functions for $x$ and $y$ have period $\pi/2$.  Then as $\phi$ goes from $\pi/2$ to $\pi$ the same curve is traced backward.  So integrating from $0$ to $\pi$ should give $0$.   This act is repeated from $\pi$ to $2\pi$.

Title: Green's formula for area


问:
Why does $\frac{V_n}{A_n}$ represent the concentration of volume around the boundary?
I read this answer about the volume of a ball in $\mathbb{R}^n$ being concentrated around the boundary.

I don't understand why:

$$\frac{V_n}{A_n}$$ is a measure of concentration around the boundary?

$$\frac{V_n}{V_{n-1}}$$ is a measure of concentration around the equator?

$V_n$ and $A_n$ denotes the volume and surface area of a (unit) ball in $\mathbb{R}^n$. We know that $A_n = nV_n$.

Also what exactly is the difference between the equator and the boundary in this context? Is there a diagram that can help me?

Thanks a lot!

答:
Intuitively, A$_n$ * r  should be equal to volume of the shell V$_n$(1) - V$_n$(1-r).
Here V$_n$(x) stands for volume of n-dimensional ball with radius x; also, V$_n$ is just V$_n$(1).
In words: area of base * height = volume.
The strange thing here is that one can take small r (= 1/n) to get almost whole unit ball volume, hence the claim that volume is concentrated around surface.
Similarly, one can cut V$_n$ around equator (to get "base" with area V$_{n-1}$) and ask about height of a cylinder with base V$_{n-1}$ so that its volume is equal to volume of V$_n$.
The answer is 1/sqrt(n), i.e. quantity much smaller than radius of the ball (=1).
These things entered modern (graduate level) analysis and probability under the name
"concentration of measure" in the last few decades.


问:
finding the directional derivative at the origin
A certain scalar quantity is governed by the equation $\Phi(x, y, z) = x+y+z^2$. Evaluate the gradient of $\Phi$ along the direction of the
vector, $x{\bf\hat  i} + y{\bf\hat  j} + z{\bf\hat  k}$. Now, evaluate the value of the directional derivative you obtained at the origin of the coordinate system.

I know how to solve the question, but I am not sure of the answer as I am getting it in terms of $0/0$.

edit 1:
the method/the way i did it

答:
I think your confusion is that you are mixing up $(x,y,z)$ the coordinates of an arbitrary point (which you later set to the origin) and $(x,y,z)$ the vector with respect to which you are taking the directional derivative. Not entirely your fault, as the notation in the question is ripe to such misunderstandings. I will set the direction we are taking the derivative in as $(a,b,c)$, to make things clearer.

$\nabla\Phi=(1,1,2z)$, so the directional derivative for the vector $\mathbf n=(a,b,c)$ is $$\mathbf n\cdot\nabla\Phi=\frac{a+b+2cz}{\sqrt{a^2+b^2+c^2}}.$$ Now evaluate this at $(x,y,z)=(0,0,0)$.


问:
Fredholm counterexample for two operators
Let $A$ be bounded a Fredholm operator on a separable Hilbert space, and let $B$ be another bounded operator on a separable Hilbert space.

Can anyone provide an example where $AB^\ast$ is Fredholm such that $\mathrm{index}(AB^\ast)=0$ yet $B$ is not Fredholm?

If $AB^\ast$ is Fredholm then it has a parametrix, say $C$, such that $CAB^\ast - 1,AB^\ast C-1$ are compact, so $CA$ could be a parametrix of $B^\ast$, but we don't know that $B^\ast CA-1$ is compact...

答:
The answer is no, that is, $B$ is necessarily Fredholm.

The reason is that the image of $A$ in the Calkin algebra $B(H)/K(H)$ is invertible, so any right inverse of it must be itself invertible.

To follow through with the OP's attempt, let $L=AB^*C-1$, hence a compact operator, and let $D$ be a parametrix of $A$, so that $DA=1+K$, where $K$ is compact.  Then
$$
  B^*CA = (DA-K)B^*CA = DAB^*CA - KB^*CA = $$$$ =
  D(L+1)A - KB^*CA = DLA + DA - KB^*CA = DLA + 1+K - KB^*CA\in 1+\mathscr K(H).
  $$


Body:
Hint:

$$\lim_{x\to\infty}\dfrac{(x+1)^{2/3}-(x-1)^{2/3}}{(x+2)^{2/3}-(x-2)^{2/3}}=\dfrac12\cdot\lim_{t\to0}\dfrac{\dfrac{(1+t)^{2/3}-(1-t)^{2/3}}t}{\dfrac{(1+2t)^{2/3}-(1-2t)^{2/3}}{2t}}=\dfrac12\cdot\dfrac{\lim_{t\to0}f(t)}{\lim_{t\to0}f(2t)}$$

Now for $\lim_{t\to0}f(t)=\lim_{t\to0}\dfrac{(1+t)^{2/3}-(1-t)^{2/3}}t,$  set $(1+t)^{1/3}=p,(1-t)^{1/3}=q$

$\implies p^3-q^3=2t$ and $\implies p\to q, q\to1$

$$\lim_{t\to0}f(t)=\lim_{p\to q, q\to1}\dfrac{2(p-q)}{p^3-q^3}=\lim_{q\to1}\dfrac2{3q^2}=?$$


问:
Is zero odd or even?
Some books say that even numbers start from $2$ but if you consider the number line concept, I think zero($0$) should be even because it is in between $-1$ and $+1$ (i.e in between two odd numbers). What is the real answer?

答:
The AP sequence
$$ (-6,-4,-2,0,2,4,6)$$
shows that it is even.

If $k$ is an even integer, then
$ (-1)^ k = 1 $

Division of $0$ by $2$ leaves no remainder.

&c. are further confirmations.

Title: Is zero odd or even?


问:
Counterexample for why the $||\cdot||_1$ norm doesn't work for the product rule i.e. $f_n \to f$, $g_n \to g$ $\implies f_ng_n \to fg $ in C[0,1].
Is the product rule  $f_n \to f$, $g_n \to g$ $\implies f_ng_n \to fg $ true in C[0,1]?
It depends on the norm.

I have to examine the 1-norm $||\cdot||_1$.

I know this isn't true but I am struggling to come up with a simple counterexample on [0,1].

So far I have seen similar threads on here giving counterexamples however I don't understand them. See this thread.

Can anyone come up with a good counterexample and show detailed working as to why it works? Thank you.

答:
Take $f_n=g_n$ piece-wise linear defined by $f_n(0)=f_n(1/n^2)=f_n(1)=0$ and $f_n(1/2n^2)=n$.

$\{f_n\}$ converges pointwise and for $\Vert \cdot \Vert_1$to the always vanishing map.

However

$$\Vert f_n g_n \Vert_1 = \int_0^1 \vert f_n g_n \vert = \int_0^1f_n^2 = \frac{1}{3} \neq 0 = \Vert fg \Vert_1$$


问:
Linearization of $ m \dfrac{dy^2}{dt^2} = u(t) - C_d \left( \dfrac{dy}{dt} \right)^2-mg $
$$ m \frac{dy^2}{dt^2} = u(t) - C_d \left( \frac{dy}{dt} \right)^2-mg $$

where
$$\begin{align*}
y(t)&=\text{missile altitude}\\
u(t)&= \text{force}\\
m&= \text{mass}\\
C_d&= \text{aerodynamic drag coefficient}
\end{align*}$$
How do I linearize this beast? I want to obtain a transfer function so that I can create a PID controller for it..

I'm really stumped and could use some help.

答:
If you want to design a PID controller for a nonlinear system based on linearization, the first step is to have to set an operation point for the altitude $x_1^*$, and another operation point for the velocity $x_2^*$.

Rewrite your system in state variable form

$$
\dot{x}_1 = x_2 \\
\dot{x}_2 = -\frac{C_d}{m}x_2^2 -g + \frac{u}{m}
$$
, where $x_1$ is the altitude and $x_2$ is the velocity. In matrix form

$$
\begin{bmatrix}\dot{x}_1 \\ \dot{x}_2 \end{bmatrix}=
\begin{bmatrix}0 & 1 \\ 0 &  -\frac{C_d}{m}x_2\end{bmatrix}
\begin{bmatrix}x_1 \\x_2\end{bmatrix}
+ \begin{bmatrix}0 \\  \frac{u}{m} - g\end{bmatrix} \\
y = Ix
$$

I have assumed here that you can observe both, altitude and velocity.

Then, you have to compute the Jacobian of the system evaluated at your operation points (you will realize that the altitude does not matter), and you will have your constant matrices A, B, C and D.

To compute the desired transfer function, you can check the equivalences here State Space Representation

Note that your PID is only valid for a neighborhood about your desired velocity. I.E. you want to keep your desired velocity starting from this desired velocity, and not from another velocity.


问:
Combinatorics question - find how many different ways
Entrapta has $20$ clones of her best friend, Hurdak. Each of the clones can be dressed either in a hat, or in a shirt, or in pants (exactly one clothing detail per clone) and each of the clothing can be painted in one of six colors (blue, white, black, red, yellow or pink).

$1)$ In how many different ways can Entrapta dress up all the clones so that exactly two clones get the same item of clothing that is painted the same color?

$2)$ In how many different ways can Entrapta dress all the clones so that at least two of them will be dressed in the exact same item of clothing?

I'm stuck here, don't know even how to start.

答:
$1-)$ Let say that two clones exactly dressed with hats whose colors same, there are $6$ choice for it.Then,the rest cannot be dressed with hat ,so $2$ choice remains for them such as shirt and pants with six colors.

$\therefore C(6,1).(C(2,1).C(6,1))^{18}$

However , we should multiply it by $3$ because the first two clones may have been dressed with pants or shirt.

$\therefore 3. C(6,1).(C(2,1).C(6,1))^{18}$

$2-)$ HINT: All situations - all clones dressed with different


问:
Show that $\left(a_n\right)_{n=1}^{\infty }$ is a Cauchy sequence of rational numbers.
let $\left(d_k\right)_{k=1}^{\infty }\:$ be a sequence of numbers in the set $\left(1,2,3,4,5,6,7,8,9\right)\:$  Assume $a_n=\sum _{k=1}^n\left(\frac{d_k}{10^k}\right)$ for $n\in \mathbb{Z}^+$. Show that  $\left(a_n\right)_{n=1}^{\infty }$ is a Cauchy sequence of rational numbers.

答:
For $n \in \mathbb N$ and $m \gt 0$ you have

$$\vert a_{n+m} - a_n \vert= \sum_{k=n+1}^{n+m} \frac{d_k}{10^k} \le \sum_{k=n+1}^\infty \frac{9}{10^k} = \frac{1}{10^n}$$

Proving that $\{a_n\}$ is Cauchy. It is also obviously a sequence of rational numbers.


问:
When is $X$ $P$-embedded in $\beta X$?
Let $X$ be a space and $A\subseteq X$ a subspace. Recall that $A\subseteq X$ has the extension property with respect to a space $Z$ if every continuous map $A\rightarrow Z$ extends over all of $X$. We say that:

$A$ is $C^*$-embedded in $X$ if it has the extension property with respect to $[0,1]$.

$A$ is $C$-embedded in $X$ if it has the extension property with respect to $\mathbb{R}$.

$A$ is $P^\kappa$-embedded in $X$ for some infinite cardinal $\kappa$ if it has the extension property with respect to every Banach space $E$ of weight $\leq\kappa$.

$A$ is $P$-embedded in $X$ if it has the extension property with respect to every Banach space $E$.

Then $P$-embeddded $\Rightarrow$ $P^\omega$-embeddded $\Leftrightarrow$ $C$-embedded $\Rightarrow$ $C^*$-embedded. In general only the one indicated arrow is reversible.

Here are some examples. A space $X$ is normal if and only if each closed $A\subseteq X$ is $C$-embedded if and only if each closed $A\subseteq X$ is $C^*$-embedded (this is the Tietze extension theorem). Every closed subspace of a paracompact space is $P$-embedded, while in general a space $X$ is collectionwise normal if and only if each closed $A\subseteq X$ is $P$-embedded.

More to the point, every Tychonoff space $X$ is densely $C^*$-embedded in its Stone-Cech compactification $\beta X$, and in turn this property characterises the Stone-Cech compactification. On the other hand it is possible for a Tychonoff space $X$ to be $C$-embedded in its Stone-Cech compactification. In fact this will occur if and only if $X$ is pseudocompact.

This brings me to my main question.

Which Tychonoff spaces are $P$-embedded in their Stone-Cech compactifications? Are these already the compact spaces? As remarked above, a pseudocompact Tychonoff space $X$ is $P^\omega$-embedded in $\beta X$. What can be said in general about the Tychonoff spaces which are $P^\kappa$-embedded in their Stone-Cech compactifications for any given $\kappa>\omega$?

According to Hewitt a space $X$ is pseudocompact if and only if it is $G_\delta$-dense in $\beta X$. My guess would be that if $X$ is $P^\kappa$-embedded in $\beta X$, then it should be $G_\kappa$-dense in $\beta X$, and hence if $X$ is $P$-embedded in $\beta X$, then it should already equal $\beta X$ (i.e. be compact).

答:
Theorem. A Tychonoff space $X$ is $P$-embedded in its Stone-Čech compactification iff  $X$ is pseudocompact.

Proof. $(\Rightarrow)$ If $X$ is $P$-embedded in $\beta X$ then $X$ is $C$-embedded in $\beta X$, so $X$  is pseudocompact.

$(\Leftarrow)$ Let $X$ be any pseudocompact space and $f$ be any continuous map from $X$ to a metrizable (in particular, Banach) space $E$. Since $f(X)$ is a pseudocompact subspace of a metrizable space, $f(X)$ is compact, see an implication ii$\Rightarrow$ i here.
Since $f(X)$ is Tychonoff space, there exists a homeomorphic embedding $g$ of $f(X)$ into a Tychonoff cube $I^A$, where $I$ is the segment $[0,1]$ endowed with the usual topology. For each $\alpha\in A$ let $\pi_\alpha:I^A\to I$ be the projection onto the $\alpha$-th factor. A map $\pi_\alpha gf:X\to I$ has a continuous extension $h_\alpha$ to $\beta X$. It is easy to show that a diagonal $h:\Delta_{\alpha\in A} h_\alpha:\beta X\to I^A$, $x\mapsto (h_ \alpha(x))_{\alpha\in A}$  is a continuous map (see, for instance,  Proposition 2.3.6 form [Eng]).
It is easy to check that $h(x)=fg(x)$ for each point $x\in X$. Since the space $f(X)$ is compact, a $gf(X)$ is compact too, so $gf(X)$ is a closed subset of $I^A$. Since $gf(X)$ is a dense subset of a compact set $h(\beta X)$, $gf(X)=h(\beta X)$. Then $g^{-1}h:\beta X\to f(X)\subset E$ is a required continuous extension of the map $f$ from $X$ to $\beta X$. $\square$

References

[Eng]  Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.

Title: When is $X$ $P$-embedded in $\beta X$?


问:
Continuity of a Single Point
My problem is :Find the points at which the the mentioned function is continuous
$$f(x) = \begin{cases} x & \text{if $x$ is a Rational Number}
                   \\ -x & \text{if $x$ is not a Rational Number}
\end{cases}$$
I was asked to learn that this function is continuous at $x = 0$ and the LHL and RHL were equated as follows

LHL
$$ \lim_{h \to 0} f(0 - h) = \lim_{h \to 0}-(0 - h) = 0$$
,

RHL
$$ \lim_{h \to 0} f(0 + h) = \lim_{h \to 0}-(0 + h) = 0$$
and

$$f(0) = 0$$
Now since
$$ LHL = RHL = f(0)$$
Therefore the function is continuous.

My question is why are we taking a point just before Zero to be Irrational.
In my opinion it could be Rational as well as Irrational making the function oscillatory and hence making it discontinuous.Please help.

I possible please state your educational qualifications(It will help me when I discuss the solution with my teacher).

答:
Limit value should be same irrespective of the way x approaches a (=0 in this case). Here it is totally incorrect to apply RHL and LHL.
Correct way would be as follows: To ensure continuity at 0 we should have lim x tends to 0, such that x belongs to rational f(x)= lim x tends to 0, such that x belongs to irrational f(x)= f(0). Note that we can do this since we can always find a rational as well as irrational number AS CLOSE to any number (0 here) AS WE PLEASE.
So, lim x tends to 0, such that x belongs to rational x= lim x tends to 0, such that x belongs to irrational -x= f(0) Or 0= -0= 0 which is clearly true. Hence     proved.</p>

Title: Continuity of a Single Point


问:
What does it mean to equate the coefficients of like terms when solving for A and B in partial fractions?
I'm trying to step myself through solving partial fractions in a year 10 book by Cambridge. This is a concept they're introducing early for students who want to challenge themselves and it's pretty light on the explanation.

For example: 7/( x+2 ) ( 2x-3 ) = A/2x-3 + B/x+2. I understand how to work this to the point where I reach 7=x(A+2B)+2A−3B. From there I've read that I need to do something called "equating coefficients. The coefficients near the like terms should be equal, so the following system is obtained: A+2B=0 2A−3B=7.

But I don't understand WHY or how it is valid that we set these parts of the equation to these values. Why not A+2B=7 2A−3B=0 for instance. I've tried looking at YouTube and asking friends, but I can't seem to get my head around it.

I can do it and I can solve for A and B using this method. But I'm really struggling to understand what it is I'm doing at that point in the process.  The phrase that keeps popping up when I look into this is "we can equate the coefficients of like terms".  For example on the wikipedia page on Fraction Decomposition it says "Equating the coefficients of x and the constant (with respect to x) coefficients of both sides of this equation...".  Second example: it says "The coefficients near the like terms should be equal, so the following system is obtained: " on the emathhelp page when I enter the equation 7/( x+2 ) ( 2x-3 ).

答:
Assume you were working with

$$ax+b=3x+2.$$

We mean that this holds for any $x$. So in particular, we could write

$$x=0\to b=2,\\x=1\to a+b=5,\\ x=-1\to -a+b=-1,\\ x=2\to 2a+b=8,\ \ \ \ \ \ \ \ \ \ \ \ \\ x=5000\to 5000a+b=1502,\\\cdots$$

This is a system of two unknowns and infinitely many equations. But it turns out that if you solve for a minimum number of equations (with the first two, $a=3, b=2$), the solution is valid for all equations, because the symbolic expressions are fully equivalent.

The same holds for rational fractions or any kind of identification.


问:
Is there an efficient formula for computing point halving on elliptic curves in $\operatorname{char}F=p\ne2$?
Consider the elliptic curve in $F_p$ and let $n$ be the order of the group.

Suppose the curve is the same as bitcoin: $y^2 = x^3 + 7$, I have a point $P=(x,y)$，how to compute $\frac{1}{2}P$.

Suppose $P=aG$ for some base point $G$(we know the points $P,G$, but the coefficient $a$)，then is there efficient method to compute $\lfloor\frac{a}{2}\rfloor$G

答:
Case $P = \lfloor \frac a 2 \rfloor G$;

As said in the comments if possible this can help to break the DLog.

Case $P = \frac a 2 G$;

We may expect that it has two solutions $$P =\frac{x}{2} \cdot G =\frac{x + q}{2} \cdot G$$, however, the base point $G$ of the bitcoin curve has odd order. Therefore

```<code>int("FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141", 16)
115792089237316195423570985008687907852837564279074904382605163141518161494337
</code>```

Therefore if there exists half of a point then it is unique.

The derivative of the equation $y^2 = x^3 + 7$

$\lambda = \dfrac{3x^{2}}
    {2(x^{3}  + 7)^{1/2}} \tag{1} \label{1}$

Write down the equation for the tangent line to $E$ at G:

$$y-p_y = \lambda(x-p_1) => y = \lambda(x - p_x) + p_y \tag{2}$$

Write the equation for the intersection of the tangent line with E:

$$y^{2} = (\lambda(x - p_x) + p_y)^{2} = x^{3} + 7 \tag{3}\label{3}$$

Write down $\lambda$ in terms of $p_x$:

If we extend the equation \ref{3} we will get a monic cubic polynomial, so the coefficient of $x^2$ will be the minus the sum of the roots. Therefore, $p_x$ will be double and $g_x$ will be a single root.  $$(\lambda (x - p_x) + p_y)^{2} = x^{3} + 7$$  $$\lambda^2 (x - p_x)^2 + 2 \lambda (x - p_x) p_y + p_y = x^{3} + 7$$ as we can see the term for $x^2$ is $\lambda$. Since the roots are the intersection point then $p_x$ will be a double point and $g_x$ will be a single point, so $2 p_x + q_x = \lambda^2$ $$\lambda^2 = 2 p_x + q_x  \label{4}\tag{4}$$

Form equation using equation \ref{4} and and replacing $x$ with $p_x$ in the equation \ref{1};

$$\lambda^{2} = 2p_x + q_x  =  \dfrac{3p_x^{4}}{4(p_x^{3}  + 7)} \tag{5} \label{5}$$ Simplify this equation and move everything over to the left-hand side will yield $$ (2p_x + q_x)(4(p_x^{3}  + 7))  =  3p_x^{4} $$  $$ 5 p_x^4 + 4 q_x p_x^3 + 28 q_x + 56 p_x = 0$$ a quartic polynomial in $p_x$. The roots of this polynomial are the x-coordinates of the half points of $G$. We are in the odd case, the factors of the quartic polynomial will be a linear equation and an irreducible resolvent cubic.

Body:
Case $P = \lfloor \frac a 2 \rfloor G$;

As said in the comments if possible this can help to break the DLog.

Case $P = \frac a 2 G$;

We may expect that it has two solutions $$P =\frac{x}{2} \cdot G =\frac{x + q}{2} \cdot G$$, however, the base point $G$ of the bitcoin curve has odd order. Therefore

int("FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141", 16)
115792089237316195423570985008687907852837564279074904382605163141518161494337

Therefore if there exists half of a point then it is unique.

The derivative of the equation $y^2 = x^3 + 7$

$\lambda = \dfrac{3x^{2}}
    {2(x^{3}  + 7)^{1/2}} \tag{1} \label{1}$

Write down the equation for the tangent line to $E$ at G:

$$y-p_y = \lambda(x-p_1) => y = \lambda(x - p_x) + p_y \tag{2}$$

Write the equation for the intersection of the tangent line with E:

$$y^{2} = (\lambda(x - p_x) + p_y)^{2} = x^{3} + 7 \tag{3}\label{3}$$

Write down $\lambda$ in terms of $p_x$:

If we extend the equation \ref{3} we will get a monic cubic polynomial, so the coefficient of $x^2$ will be the minus the sum of the roots. Therefore, $p_x$ will be double and $g_x$ will be a single root.  $$(\lambda (x - p_x) + p_y)^{2} = x^{3} + 7$$  $$\lambda^2 (x - p_x)^2 + 2 \lambda (x - p_x) p_y + p_y = x^{3} + 7$$ as we can see the term for $x^2$ is $\lambda$. Since the roots are the intersection point then $p_x$ will be a double point and $g_x$ will be a single point, so $2 p_x + q_x = \lambda^2$ $$\lambda^2 = 2 p_x + q_x  \label{4}\tag{4}$$

Form equation using equation \ref{4} and and replacing $x$ with $p_x$ in the equation \ref{1};

$$\lambda^{2} = 2p_x + q_x  =  \dfrac{3p_x^{4}}{4(p_x^{3}  + 7)} \tag{5} \label{5}$$ Simplify this equation and move everything over to the left-hand side will yield $$ (2p_x + q_x)(4(p_x^{3}  + 7))  =  3p_x^{4} $$  $$ 5 p_x^4 + 4 q_x p_x^3 + 28 q_x + 56 p_x = 0$$ a quartic polynomial in $p_x$. The roots of this polynomial are the x-coordinates of the half points of $G$. We are in the odd case, the factors of the quartic polynomial will be a linear equation and an irreducible resolvent cubic.


问:
Summation of integration terms
If $I = \sum\limits_{k = 1}^{98} {\int\limits_k^{k + 1} {\frac{{k + 1}}{{x\left( {x + 1} \right)}}} dx} $ then

(A) $I>\ln99$

(B) $I<\ln99$

(C) $I<\frac{49}{50}$

(D) $I>\frac{49}{50}$

The official answer is B and D

My approach is as follow

$$I = \sum\limits_{k = 1}^{98} {\int\limits_k^{k + 1} {\frac{{k + 1}}{{x\left( {x + 1} \right)}}} dx} $$

$$\frac{A}{x} + \frac{B}{{x + 1}} = \frac{{A\left( {x + 1} \right) + Bx}}{{x\left( {x + 1} \right)}} = \frac{{x\left( {A + B} \right) + A}}{{x\left( {x + 1} \right)}}$$

$$A = k + 1\text{ and } B =  - \left( {k + 1} \right)$$

$$I = \sum\limits_{k = 1}^{98} {\int\limits_k^{k + 1} {\left( {\frac{{k + 1}}{x} - \frac{{k + 1}}{{x + 1}}} \right)} dx}  \Rightarrow I = \sum\limits_{k = 1}^{98} {\left( {k + 1} \right)\int\limits_k^{k + 1} {\left( {\frac{1}{x} - \frac{1}{{x + 1}}} \right)} dx} $$

$$I = \sum\limits_{k = 1}^{98} {\left( {k + 1} \right)\left. {\ln\frac{x}{{x + 1}}} \right|_k^{k + 1}}  \Rightarrow I = \sum\limits_{k = 1}^{98} {\left( {k + 1} \right)\left( {\ln\frac{{k + 1}}{{k + 2}} - \ln\frac{k}{{k + 1}}} \right)}$$  $$\Rightarrow I = \sum\limits_{k = 1}^{98} {\left( {k + 1} \right)\left( {\ln\frac{{{{\left( {k + 1} \right)}^2}}}{{k\left( {k + 2} \right)}}} \right)} $$

Not able to approach from here

Title: Summation of integration terms

Body:
You have
$$\sum_{k = 1}^{98} \int_k^{k + 1} \frac{{k + 1}}{{x\left( {x + 1} \right)}} dx  \le \sum_{k = 1}^{98} \int_k^{k + 1} \frac{dx}{x} = \int_1^{99} \frac{dx}{x} = \ln 99$$

hence B is true. Also

$$\begin{aligned}\sum_{k = 1}^{98} \int_k^{k + 1} \frac{{k + 1}}{{x\left( {x + 1} \right)}} dx  &\ge \sum_{k = 1}^{98} \int_k^{k + 1} \frac{dx}{x+1} \\
&\ge \sum_{k = 1}^{98} \int_k^{k + 1} \frac{dx}{100}\\
&=\frac{98}{100}= \frac{49}{50}
\end{aligned}$$

hence D is also true.