答:
Let's start with the notion of proof in a formal system. This is a sequence of formulas $A_1,\ldots,A_n$ such that each formula on the list is either an axiom, or follows from earlier items on the list via a rule of inference.

You mentioned modus ponens, which is the standard example of a rule of inference. It can almost serve as the only rule of inference, depending on how you set up your formal apparatus.

Now let's contrast this with the formula you gave as a translation, $(a\wedge(a\rightarrow b))\rightarrow b$. This is a single formula. It could be included in a proof as one of the $A_i$'s but it doesn't give you the power to add $b$ once you already have $a$ and $a\rightarrow b$.

Lewis Carroll gave a droll but profound explanation of why you need rules of inference, why axioms alone won't suffice: [What the Tortoise Said to Achilles]. 1. Here's an excerpt.

Achilles had overtaken the Tortoise, and had seated himself comfortably on its
back.

“So you’ve got to the end of our race-course?” said the Tortoise. “Even though it
does consist of an infinite series of distances? I thought some wiseacre or other had proved that the thing couldn’t be done?”

“It can be done,” said Achilles. “It has been done! ... You see
the distances were constantly diminishing; and so—”

“But if they had been constantly increasing?” the Tortoise interrupted “How then?” ... [W]ould you like to hear of a race-course, that most people
fancy they can get to the end of in two or three steps, while it really consists of an infinite number of distances, each one longer than the previous one?”

“Very much indeed!” said the Grecian warrior...

“That beautiful First Proposition of Euclid!” the Tortoise murmured dreamily. ...

(A) Things that are equal to the same are equal to each other.

(B) The two sides of this Triangle are things that are equal to the same.

(Z) The two sides of this Triangle are equal to each other.

"Readers of Euclid will grant, I suppose, that Z follows logically from A and B, so that any one who accepts A and B as true, must accept Z as true?”
...“Well, now, I want you to ... force me, logically, to accept Z as true.”

“I’m to force you to accept Z, am I?” Achilles said musingly. “And your present
position is that you accept A and B, but you don’t accept the Hypothetical—”

“Let’s call it C,” said the Tortoise.

“—but you don’t accept

(C) If A and B are true, Z must be true.”

“That is my present position,” said the Tortoise.

“Then I must ask you to accept C.”

“I’ll do so,” said the Tortoise, “as soon as you’ve entered it in that note-book of yours. ...  Now write as I dictate:—

(A) Things that are equal to the same are equal to each other.

(B) The two sides of this Triangle are things that are equal to the same.

(C) If A and B are true, Z must be true.

(Z) The two sides of this Triangle are equal to each other.”

“You should call it D, not Z,” said Achilles. “It comes next to the other three. If you accept A and B and C, you must accept Z.”

“And why must I?”

“Because it follows logically from them. If A and B and C are true, Z must be true. You don’t dispute that, I imagine?”

“If A and B and C are true, Z must be true,” the Tortoise thoughtfully repeated. “That’s another Hypothetical, isn’t it? And, if I failed to see its truth, I might accept A and B and C, and still not accept Z, mightn’t I?”

"I must ask you to grant one more Hypothetical.” ...

(D) If A and B and C are true, Z must be true.

You can see where this is going. We want to infer (C) from (A) and (B), not add a new formula $A\wedge (A\rightarrow B)\rightarrow B$ to the list.

Title: Formal relationship between rules of inference and the material conditional

Body: I am not $100\%$ clear as to what constitutes the difference between a rule of inference and the material conditional, at least in classical logic. I am using the truth-functional definition of the material conditional, commonly visualised through its truth table, but I'm not entirely sure what the formal definition of a rule of inference is. The wikipedia article defines it to be a particular kind of logical form, which seems to be a term from philosophical logic that I'm not familiar with, but reading that article didn't really answer my question. It pertains more to the mathematical side of things, and I am specifically interested in the interplay between the concepts on the syntactic and semantic level. As far as I can tell, any rule of inference can be 'captured' by a corresponding material conditional: if we take modus ponens as a well-known example, what is the difference between $$(a\land (a\to b))\to b$$ and $${a\to b,\text{ } a \over b}?$$ On a functional level, both statements seem to be expressing the same thing. What determines the need to use two separate terms and notations, and what, if anything, separates them?


问:
Defining $dy=f'(x)dx$
In integration,we often use $u$ substitution method which says if $u=f(x)$ then $du=f'(x)dx$. Here $dx, du$ are infinitesimals. But $dx,du$ aren't real numbers, then how can we multiply them? Is this independently derived from non standard analysis?

Also is a constant multiplied by an infinitesimal also an infinitesimal?

答:
Actually setting $u = f(x)$ and $du = f'(x) dx$ is just a mnemonic, a notation trick. The actual theorem is

$$\int_a^b g(f(x))f'(x)\,dx = \int_{f(a)}^{f(b)}g(x)\,dx$$

ContentLicense: CC BY-SA 4.0

Title: Defining $dy=f'(x)dx$


问:
Jump distribution of a continuous time Markov chain
Given the following jump chain:

\begin{pmatrix}
0 & 1 & 0 & 0 & 0 & 0\\
1 & 0 & 0 & 0 & 0 & 0\\
\frac{1}{4} & 0 & 0 & \frac{1}{4} & \frac{1}{2} & 0\\
0 & 0 & 0 & 0 & 0 & 1\\
0 & 0 & \frac{1}{2} & 0 & 0 & \frac{1}{2}\\
0 & 0 & 0 & 1 & 0 & 0\\
\end{pmatrix}

And that $q_1=q_3=q_5=12$ and $q_2=q_4=q_6=6$.
Given as well that $X_0=3$ what are:

$\lim_{t\to\infty} P(X_t=2|X_0=3)$

the distribution of the number of jumps to state 3

I have found the Q matrix and computed part 1 as follows but I am not sure if it is correct and I do not understand how to do part 2.

Q-matrix:

\begin{pmatrix}
-12 & 12 & 0 & 0 & 0 & 0\\
6 & -6 & 0 & 0 & 0 & 0\\
3 & 0 & -12 & 3 & 6 & 0\\
0 & 0 & 0 & -6 & 0 & 6\\
0 & 0 & 6 & 0 & -12 & 6\\
0 & 0 & 0 & 6 & 0 & -6\\
\end{pmatrix}

$\lim_{t\to\infty} P(X_t=2|X_0=3)= h_3^{\{1,2\}}\lim_{t\to\infty} P(X_t=2|X_0\in \{1,2\}) = 1 h_3^{\{1,2\}} = \frac{1}{3}$ \

where $h_3^{\{1,2\}}$ is the probability to $\{1,2\}$ starting from 3.

答:
I highly recommend that you draw a state transition diagram to solve this problem. If you do this you'll clearly see that $\{1,2\}$ and $\{4,6\}$ are absorbing and $P(X_{2n-1}=2|X_0=3)=0$ for any $n\in \mathbb{N}$ i.e. the Markov chain can only reach state $2$ on an even$-$numbered step originating at state 3. In fact, we have for large $n$ that $$P(X_{2n}=2|X_0=3)\approx P\Big(\text{Absorbed in \{1,2\}}\Big|X_0=3\Big)$$ To calculate the probability on the right hand side, set $$a_i=P\Big(\text{Absorbed in \{1,2\}}\Big|X_0=i\Big)$$ for $i\in\{1,\ldots,6\}$. Clearly $a_1=a_2=1$ while $a_4=a_6=0$. Using the total law of probability with recursion we have $$a_3=p_{31}a_1+p_{34}a_4+p_{35}a_5=\frac{1}{4}+\frac{1}{2}a_5=\frac{1}{2} \\ a_5=p_{56}a_6+p_{53}a_3=\frac{1}{2}a_3$$ Solving this system yields $a_3=\frac{1}{3}$ and $a_5=\frac{1}{6}$. To summarize, $$\lim_{n\rightarrow \infty}P(X_{2n-1}=2|X_0=3)=0 \\ \lim_{n\rightarrow \infty}P(X_{2n}=2|X_0=3)=\frac{1}{3}$$
For the second part, let $X$ represent the number of jumps to state #$3$. Clearly $X$ is supported on the set $\{0,1,2,\ldots\}$. Notice $$p_X(0)=p_{31}+p_{34}+p_{35}p_{56}=\frac{3}{4}$$ and for $k \geq 1$ we have $$p_{X}(k)=\frac{3}{4}\times\Big(\frac{1}{2}\Big)^{2k}$$


问:
Calculating $\sum_{k=0}^{n} {n\choose k} \sqrt{k}$
I am trying to calculate the following expression but not sure how this can be solved yet.
$$\sum_{k=0}^{n} {n\choose k} \sqrt{k}$$
The wiki page (https://en.wikipedia.org/wiki/Binomial_coefficient) contains a similar expression with $\sqrt{k}$ replaced by $k$ or $k^2$. If the above expression can be solved, I also wonder if we can solve it for any $k^j$ with any real number $j>0$?

答:
Let $X$ be a binomial random variable, with $n$ trials and probability $1/2$.  Your expression is exactly $2^nE[X^j]$. Using the central limit theorem, for large $n$, $X$ is well approximated by $\frac{\sqrt{n}}2Z+\frac{n}2$, where $Z$ is a standard normal random variable. I will then use the approximation
\begin{align}
E[X^j]
&\approx E\left[\left(\frac{\sqrt{n}}2Z+\frac{n}2\right)^j\right]\tag 1
\end{align}
to get an approximation for $E[X^j]$. I am not at this time able to quantify the error of the approximation in $(1)$, which is necessary to determine how good of an approximation the final answer is.
\begin{align}
2^nE[X^j]
  &\approx 2^nE\left[\left(\frac{\sqrt{n}}2Z+\frac{n}2\right)^j\right]
\\&            =2^{n}(n/2)^jE\left[(1+Z/\sqrt{n})^j\right]
\\&\stackrel{1}=2^{n}(n/2)^j\sum_{i=0}^\infty \binom{j}{i}E[Z^i]n^{-i/2}
\\&\stackrel{2}=\boxed{2^{n}(n/2)^j\sum_{i=0}^\infty \binom{j}{2i}(2i-1)!!\cdot n^{-i}}
\end{align}

In $1$, we use the Taylor series for $f(x)=(1+x)^j=\sum_{i\ge 0}\binom{j}ix^i$, where the binomial coefficient is defined for non-integer $j$ as $$\binom{j}i=\frac{j(j-1)\cdots(j-i+1)}{i!}.$$

In $2$, we reindex $i\gets 2i$, and use the known moments of the standard normal random variable.

We can then further approximate by only taking the first several terms of the infinite summation. For example, when $j=1/2$, you get
$$
\sum_{k=0}^n \binom{n}kk^{1/2}=2^nE[X^{1/2}]\approx 2^n(n/2)^{1/2}\left(1-\frac18n^{-1}-\frac{15}{128}n^{-2}-\dots\right),
$$
so that
$$
\sum_{k=0}^n \binom{n}kk^{1/2}\approx 2^n(n/2)^{1/2}(1+O(1/n))
$$
Again, use this with caution, because the goodness of this approximation depends on the the quality of the approximation $(1)$, which I am not able to quantify.


问:
Definite integration with natural logarithm
$$\int_0^a \ln(x)\ln(a-x)\,dx$$

How to do this? I couldn't proceed at all.

($\ln$ is natural logarithm.)

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{0}^{a}\ln\pars{x}\ln\pars{a - x}\,\dd x\,\right\vert_{\,a\ >\ 0}}
\\[5mm] \stackrel{x/a\ \mapsto\ x}{=}\,\,\,\,\,&
a\int_{0}^{1}\bracks{\ln\pars{x} + \ln\pars{a}}
\bracks{\ln\pars{a} + \ln\pars{1 - x}}\,\dd x
\\[5mm] = &\
2a\ln\pars{a}\int_{0}^{1}\ln\pars{x}\,\dd x +
a\int_{0}^{1}\ln^{2}\pars{a}\dd x
\\[2mm] + &\
\left.a\,{\partial^{2} \over \partial\mu\partial\nu}
\int_{0}^{1}x^{\mu}\,\pars{1 - x}^{\nu}\,\dd x
\,\right\vert_{\substack{\mu\ =\ 0\\[0.75mm] \nu\ =\ 0}}
\\[5mm] = &\
-2a\ln\pars{a} + a\ln^{2}\pars{a} +
a\
\underbrace{\left.{\partial^{2} \over \partial\mu\partial\nu}
{\Gamma\pars{\mu + 1}\Gamma\pars{\nu + 1} \over \Gamma\pars{\mu + \nu + 2}}
\,\right\vert_{\substack{\mu\ =\ 0\\[0.75mm] \nu\ =\ 0}}}
_{\ds{2 - {\pi^{2} \over 6}}}
\\[5mm] = &\
\bbx{a\ln^{2}\pars{a} - 2a\ln\pars{a} + 2a
- {\pi^{2} \over 6}\,a} \\ &
\end{align}

ContentLicense: CC BY-SA 3.0


问:
Inverse of a exponential function
I try to calculate the inverse of the following exponential function (it's a CDF):  $$ f(x) = { 1 \over 3} e^{-x/2} + { 2 \over 3} e^{-x/4}  $$

I used a website to calculate it, they provided me the following: $$f^{-1}(x) = 4\cdot \ln\left({\sqrt{3x+1} +1 \over x}\right) + 4 \cdot \ln(3), $$ which seems to be correct. I now try to understand the way it was calculated but I can't figure it out.

I started with:

\begin{align}
y &= { 1 \over 3} e^{-x/2} + { 2 \over 3} e^{-x/4}\\
y &= 3 \cdot ( e^{-x/2} + 2 \cdot e^{-x/4} )\\
\ln(y) &= \ln(3 \cdot ( e^{-x/2} + 2 \cdot e^{-x/4} ))\\
\ln(y) - \ln(3) &= \ln( e^{-x/2} + 2 \cdot e^{-x/4} )
\end{align}

At this point I tried $\ln(a + c) = \ln(a) + \ln(1 + { c \over a})  $ which gave me:

\begin{align}
\ln(y) - \ln(3) &= -{x \over 2} + \ln\left(1 + 2 \cdot {e^{-x/4} \over e^{-x/2}  }\right)\\
&= -{x \over 2} + \ln(1 + 2 \cdot {e^{x/4}})
\end{align}
At this point, I'm stuck. Any help is appreciated.

Regards

答:
Hint. Let $z=e^{-x/4}$ and solve the quadratic equation with respect to $z$.
$$y = { 1 \over 3} e^{-x/2} + { 2 \over 3} e^{-x/4}={ 1 \over 3} z^2 + { 2 \over 3} z\Leftrightarrow (z+1)^2=1+3y.$$

P.S. As pointed out by Yves Daoust, starting from the hint we easily find
$$x=f^{-1}(y)=-4\log\left(\sqrt{3y+1}-1\right)$$
which can be written also in the following way
\begin{align}
x=f^{-1}(y)&=-4\log\left(\sqrt{3y+1}-1\right)=4\log\left(\frac{1}{\sqrt{3y+1}-1}\right)\\
&=4\log\left(\frac{\sqrt{3y+1}+1}{(3y+1)-1}\right)
=4\log\left(\frac{\sqrt{3y+1}+1}{y}\right)-4\log(3).
\end{align}
So there is a missing minus sign in your formula. Note that $f(0)=1$ and therefore $f^{-1}(1)=0$.

Title: Inverse of a exponential function


答:
Let $z:=e^{-x/4}$. We can write the equation as

$$3y=z^2+2z.$$

The solution of this quadratic equation is

$$z=\pm\sqrt{3y+1}-1$$

or

$$x=-4\log\left(\sqrt{3y+1}-1\right)$$ because the negative solution must be rejected.

Title: Inverse of a exponential function

Body:
I try to calculate the inverse of the following exponential function (it's a CDF):  $$ f(x) = { 1 \over 3} e^{-x/2} + { 2 \over 3} e^{-x/4}  $$

I used a website to calculate it, they provided me the following: $$f^{-1}(x) = 4\cdot \ln\left({\sqrt{3x+1} +1 \over x}\right) + 4 \cdot \ln(3), $$ which seems to be correct. I now try to understand the way it was calculated but I can't figure it out.

I started with:

\begin{align}
y &= { 1 \over 3} e^{-x/2} + { 2 \over 3} e^{-x/4}\\
y &= 3 \cdot ( e^{-x/2} + 2 \cdot e^{-x/4} )\\
\ln(y) &= \ln(3 \cdot ( e^{-x/2} + 2 \cdot e^{-x/4} ))\\
\ln(y) - \ln(3) &= \ln( e^{-x/2} + 2 \cdot e^{-x/4} )
\end{align}

At this point I tried $\ln(a + c) = \ln(a) + \ln(1 + { c \over a})  $ which gave me:

\begin{align}
\ln(y) - \ln(3) &= -{x \over 2} + \ln\left(1 + 2 \cdot {e^{-x/4} \over e^{-x/2}  }\right)\\
&= -{x \over 2} + \ln(1 + 2 \cdot {e^{x/4}})
\end{align}
At this point, I'm stuck. Any help is appreciated.

Regards


问:
Use of definition of limit to prove $\lim_{x→2} \dfrac{x}{x^2-2}=1$
I know that by definition I have to prove that $$\lim_{x→2}\dfrac{x}{x^2−2}=1⟺∀ϵ>0,∃δ>0,\, 0<|x−2|<δ⟹\left|\frac{x}{x^2−2}−1\right|<ϵ.$$

I have:
$\left|\dfrac{-x^2+x+2}{x^2-2}\right|= \left|\dfrac{-(x^2-x-2)}{x^2-2}\right|= \left|\dfrac{-(x^2-x+1-3)}{x^2-2}\right|=$
$\left|\dfrac{-(x-1)^2+3}{x^2-2}\right|$.

So, I don't know how to continue.

答:
Two points: 1) It's often much easier to deal with $x\to 0$ than $x\to c\neq 0$; and  2) you may want to avoid fractions whenever you can.

So let's denote $d=x-2$ and prove $\lim_{d\to 0} \dfrac{d+2}{(d+2)^2-2} = 1$.

$$
\left| \dfrac{d+2}{(d+2)^2-2} - 1 \right|=\left| \dfrac{d+2-d^2-4d-2}{d^2+4d+2} \right| = \left| d \right| \left| \dfrac{d+3}{d^2+d+2} \right|
$$

If $|d|<1, |d+3| \le |d| + 3<4$. Also note $|d^2+d+2|=\left(d+\dfrac{1}{2}\right)^2+\dfrac{7}{4} > 1$.

Therefore, if $|d| < \delta = \min(1, \dfrac{1}{4} \varepsilon )$
$$\left| d \right| \left| \dfrac{d+3}{d^2+d+2} \right| < \dfrac{1}{4} \varepsilon \cdot \dfrac{4}{1} = \varepsilon.$$


问:
How to Prove or disprove : Let I be an ideal of R and R/I be a integral domain then R is an integral domain?
Prove or disprove : Let I be an ideal of R and R/I be a integral domain then R is an integral domain.

I am confused this is true or not. My approach is if I assume that R is not a integral domain then there exist some a,b$\in$ R-I such that ab=0 then (a+I)(b+I)=ab+I=I,which contradicts the fact that R/I is a integral domain. But I am not sure if my proof is right or wrong. It will be enough if you give me some hint and if there is a better solution I will appreciate that. Thank you.

答:
Your mistake is in assuming that if $R$ is not an integral domain then you must be able to find a pair of zero divisors, neither element living in $I$.  You wrote $a, b \in R-I$, but there is no guarantee of that.  Instead, all you can say is that $a, b \in R$ are both non-zero yet $ab =0$.  This does not yield a contradiction once you carry out your coset multiplications.

Instead, the claim is false.  You can quotient out $\{0,2\}$ from the ring $\mathbb{Z}_4$ (not an integral domain) and get (something isomorphic to) $\mathbb{Z}_2$, which is a field.


问:
why $I_n \subset G_\alpha$ contradicts $(b)?$
I have some confusion in RUDIN book Theorem $2.40$ :Every $k-$cell is compact

My confusion written in red colour given below

Here is an outline of Rudin's proof(Theorem $2.40$):

Every k-cell is compact.

Proof. Let $I$ be a k-cell consisting of all points $x = (x_1, \dots, x_k)$ such that $a_j \leq x_j \leq b_j$ for $1 \leq j \leq k$. Put
$\delta = \{ \sum\limits_1^k(b_j - a_j)^2\}^{1/2}$
Then $|x-y| \leq \delta$ if $x, y \in I$.

Suppose there is an open cover $\{G_\alpha \}$ which contains no finite subcover. Put $c_j = \frac{a_j + b_j}{2}$. The intervals $[a_j, c_j]$ and $[c_j, b_j]$ determine $2^k$ k-cells whose union is $I$. At least one of these subsets of $I$, say $I_1$, cannot be covered by any finite subcollection of $\{ G_\alpha \}$. So we begin again with the k-cell $I_1$ and subdivide further to achieve a sequence of k-cells such that

$(a)$ $I \supset I_1 \supset I_2 \supset I_3 \supset \dots$

$(b)$ $\color{red}{ I_n \text{is not covered by any finite subcollection of}\ G_\alpha}$

$(c)$ If $x \in I_n$ and $y \in I_n$ then $|y-x| \leq 2^{-n}\delta$

by $( a)$ and theorem $2.39$ , there is  a point $x^*$ which lie in every $I_n$. For some  $\alpha , x^* \in G_\alpha.$ Since $G_{\alpha}$ is open , there exist $r >0$ such that $|y-x^*|<r $ implies  that $y \in G_{\alpha}$.If $n$ is so large that $2^{-n} \delta < r$(there is such  an $n$ , for otherwise  $2^n \le \frac{\delta}{r}$ for all positive integr $n$ , which is  absurd  since $R$ is archimedean , then $(c)$ implies that  $\color{red}{I_n \subset G_{\alpha}}$ , which  $\color{red}{\text{contradict} (b)}$

This completes the  proof

My doubt : in b) it already said that $I_n$ is not covered by any finite subcollection of $G_{\alpha}$

This implies that  $I_n \subset \bigcup_{\alpha}G_{\alpha}$ doesn't have a finite subcover

Then why $I_n \subset G_\alpha$ contradicts $(b)?$

答:
Take  neighborhood $N_r(x^*) \subset G_\alpha$. If $n$ is large enough that $2^{-n}\delta < r$ then given $p \in I_n$ $|x^* - p | < 2^{-n}\delta < r \implies p \in N_r(x^*) \implies I_n \subset N_r(x^*)\subset G_\alpha$

Here $I_n \subset N_r(x^*)$ where  $N_r(x^*)$ is bounded open ball this implies that  $I_n $is covered  by a finite subcover   so it contradicts $(b)$


问:
Convention for locally compact groups?
$\bf{\text{Suppose I find the phrase:}}$

Let $G$ is a locally compact group, and $\mathcal{U}$ a basis of neighborhoods of $1$.

$\bf{\text{Question:}}$

Is it a convention to automatically take each $U\in\mathcal{U}$ to be compact?

Clearly this can be done if the need arises by the locally compact assumption on $G$, but when reading material, should this be my first interpretation such a statement?

答: I wouldn't think so, unless it is explicitly stated as a convention in what you are reading.  (Usually, an author would explicitly signal that $U$ is a compact neighbourhood if this is what was meant.)</p>

Title: Convention for locally compact groups?


问:
Invariance of generalized eigenspace
I have a lemma saying that each of the generalized eigenspaces of a linear operator $T$ is invariant under $T$. This means that if $E_j$ is a generalized eigenspace then $T:E_j \rightarrow E_j.$

The proof of this goes like this.

Take a $v\in E_j$ so that $(T-\lambda_j I)^{n_{j}}v=0$.

Now we are going to show that the same holds for $T(v)\in E_j$.

$(T-\lambda_j I)^{n_{j}}T(v)=(T-\lambda_j I)^{n_{j}-1} T(T-\lambda_j I)v=T(T-\lambda_j I)^{{n_j}}v= T(0)=0$.

Can someone help me with the proof?

答: Consider the polynomials $p(x)=(x-\lambda_j)^{n_j}$ and $q(x)=x$. Note that $p(T) = (T-\lambda_jI)^{n_j}$ and that $q(T)=T$. Then, as $p(x)q(x) = q(x)p(x)$, we have $p(T)q(T) = q(T)p(T)$, that is, $$(T-\lambda_jI)^{n_j}T = T(T-\lambda_jI)^{n_j}.$$ Then, for $v \in E_j$ we have $(T-\lambda_jI)^{n_j}v = 0$, and then $Tv \in E_j$ since $$(T-\lambda_jI)^{n_j}Tv = T(T-\lambda_jI)^{n_j}v = T0 = 0.$$

Title: Invariance of generalized eigenspace


答: Notice that $T$ commutes with $T-\lambda_jI$. So pull one factor out of $(T-\lambda_jI)^{n_j}$ to get the first equality.

Title: Invariance of generalized eigenspace

Body:
I have a lemma saying that each of the generalized eigenspaces of a linear operator $T$ is invariant under $T$. This means that if $E_j$ is a generalized eigenspace then $T:E_j \rightarrow E_j.$

The proof of this goes like this.

Take a $v\in E_j$ so that $(T-\lambda_j I)^{n_{j}}v=0$.

Now we are going to show that the same holds for $T(v)\in E_j$.

$(T-\lambda_j I)^{n_{j}}T(v)=(T-\lambda_j I)^{n_{j}-1} T(T-\lambda_j I)v=T(T-\lambda_j I)^{{n_j}}v= T(0)=0$.

Can someone help me with the proof?


问:
Specific element in a set
Given a set $A=\{a,b,c,d\}$ how can I take, for an example, the first element?
Like this:
$A(0)=a;
A(1)=b;
A(0).A(1)=a.b$.
I think there's no way to do that with sets, because it isn't an ordered set.</p>

答:
For a set $A$, there is no way to refer to its first element, second elements, or any such enumeration of elements, simply because a set is not ordered. You can ask whether an element is in a given set, and the answer is either yes or no. You can't ask where an element is placed in a given set, since no ordering exists in a set. </p>

<p>It is possible to consider some extra structure on a set, turning it into a poset (short for partially ordered set). Then you might be able to speak of first element, second element, and so on, but that is also not guaranteed. Posets that most closely resemble an ordering like you are alluding to are called well-orders. A finite well-order really looks just like a sequence of elements in order. But infinite well-orders can be very complicated. </p>

Title: Specific element in a set

Body:
For a set $A$, there is no way to refer to its first element, second elements, or any such enumeration of elements, simply because a set is not ordered. You can ask whether an element is in a given set, and the answer is either yes or no. You can't ask where an element is placed in a given set, since no ordering exists in a set.

It is possible to consider some extra structure on a set, turning it into a poset (short for partially ordered set). Then you might be able to speak of first element, second element, and so on, but that is also not guaranteed. Posets that most closely resemble an ordering like you are alluding to are called well-orders. A finite well-order really looks just like a sequence of elements in order. But infinite well-orders can be very complicated.


问:
Is showing $\text{Tr}\Omega U = \text{Tr} \Omega$ equivalent to showing $\text{Tr}\Omega = \text{Tr}(U^{\dagger} \Omega U)$?
On pg. 30 of Principles of Quantum Mechanics (by Shankar), we are asked:

Show that the trace of an operator is unaffected by a unitary change of basis $|i \rangle \rightarrow U |i\rangle$. [Equivalently, show $\text{Tr}\Omega = \text{Tr}(U^{\dagger} \Omega U)$].

I have already convinced myself that

$$
\text{Tr} \Omega = \text{Tr}(U^{\dagger} \Omega U)
$$

by using the fact that

$$
\text{Tr} ( \Omega \Lambda ) = \text{Tr} ( \Lambda \Omega )
$$

implies that

$$
\text{Tr} \Omega = \text{Tr}(\Omega (U U^{\dagger})) = \text{Tr}((\Omega U) U^{\dagger}) = \text{Tr}(U^{\dagger} (\Omega U)) = \text{Tr}(U^{\dagger} \Omega U)
$$

as desired.

Question: But why is showing this equivalent to showing that the trace of an operator is unaffected by a unitary change of basis $|i \rangle \rightarrow U | i \rangle$?  It seems to me this means that

$$
\text{Tr} \Omega = \text{Tr}(\Omega U)
$$

which is a distinct condition from showing that

$$
\text{Tr} \Omega = \text{Tr}(U^{\dagger} \Omega U)
$$

答: In an orthonormal basis, an operator $\Omega=\sum_{ij}\Omega_{ij}|i\rangle\langle j|$ has trace $\sum_i\Omega_{ii}=\sum_i\langle i|\Omega|i\rangle$, which becomes transformed to $\sum_i\langle i|U^\dagger\Omega U|i\rangle$.


答:
Just use the definition: you need to verify that
$$\Bigg|\frac{x}{x^2-2}-1\Bigg| < \epsilon$$
gives you a neighbourhood of $2$.

This means to solve

$$\begin{cases}
\dfrac{x}{x^2-2}-1 < \epsilon \\\\
\dfrac{x}{x^2-2}-1 > -\epsilon
\end{cases}
$$

Solve this and you will get

$$\sqrt{\dfrac{2\epsilon-4}{\epsilon-1}} < x < \sqrt{\dfrac{-2\epsilon-4}{-\epsilon-1}}$$

Which, being $\epsilon$ small ad libitum, gives you $2 < x < 2$.

Title: Use of definition of limit to prove $\lim_{x→2} \dfrac{x}{x^2-2}=1$

Body:
I know that by definition I have to prove that $$\lim_{x→2}\dfrac{x}{x^2−2}=1⟺∀ϵ>0,∃δ>0,\, 0<|x−2|<δ⟹\left|\frac{x}{x^2−2}−1\right|<ϵ.$$

I have:
$\left|\dfrac{-x^2+x+2}{x^2-2}\right|= \left|\dfrac{-(x^2-x-2)}{x^2-2}\right|= \left|\dfrac{-(x^2-x+1-3)}{x^2-2}\right|=$
$\left|\dfrac{-(x-1)^2+3}{x^2-2}\right|$.

So, I don't know how to continue.


问:
Unicity of solution to a one parameter system of ODE
Assume that  $m,p,$ and $\rho$ are known constants. The functions $\theta(s)$ and $y(s)$ are smooth and defined on the interval $[0,1].$ They also satisfy the system:
\begin{equation}\label{ODE}\left\{ \begin{array}{lll}\theta'(s)(c-my(s))=p+\rho y(s)-m\cos\theta(s) \\ y'(s)=\sin \theta (s) \\ y(0)=0 \\ y(1)=0\\ \theta(0)=0 \\ \theta(1)=2\pi \end{array}\right.\end{equation}

Is the constant $c$ uniquely determined by $m,p,$ and $\rho?$

答:
As long as $y(s) \ne c/m$ you have a first-order system with a smooth right side, and
the solution with initial conditions $y(0)=0$, $\theta(0)=0$ is unique.  This unique solution may or may not satisfy the boundary conditions at $s = 1$.  It's not clear what happens if $y(s)$ hits $c/m$.


问:
Error in convservation laws integral equality argument
I'm attempting to solve the following problem from Evans on conservation laws,

using the following definition of weak solution

I came up with the following test function construction.
Let $0 < \hat{t} < T$, $K$ be compact such that $\text{supp}(u) \subset \subset K$
$\newcommand{\ep}{\varepsilon}$
$$ v_t = \frac{\chi_{B(\hat{t}, \ep)}}{|B(\hat{t}, \ep)|}\chi_K \qquad v(x,t) = \int_0^t v_t(s) ds $$

then $v_x = 0$ on the support of $u$ and thus $F(u)$ ($v$ is constant there), so by the definition of weak solution
$$ \int_0^T \int_{-\infty}^\infty u(x,t) v_t(t) dxdt = \int_{-\infty}^\infty g(x) v(x, 0) dx = 0$$
and
$$ \int_0^T \int_{-\infty}^\infty u(x,t) v_t(t) dxdt = \int_{-\infty}^\infty \int_0^T u(x,t) v_t(t)dtdx = \int_{-\infty}^\infty \frac{1}{|B(\hat{t},\ep)|} \int_{B(\hat{t},\ep)} u(x,t) dtdx$$
so letting $\ep \rightarrow 0$ we have
$$ \lim_{\ep \rightarrow 0} \int_{-\infty}^\infty \frac{1}{|B(\hat{t},\ep)|} \int_{B(\hat{t},\ep)} u(x,t) dtdx = \int_{-\infty}^{\infty} u(x,\hat{t}) dx $$
(by the Lebesgue differentiation theorem and bounded convergence)
however, this implies that the integral of g has to be zero, which apparently is not required. While this construction isn't itself smooth, it is fairly trivial to regularlize it by mollification. Where is the error here?

答:
The error comes from assuming that v has compact support. While the space dimension doesn't really matter (a smooth cutoff function can be used to smooth the characteristic of the support of u), in time, the test function $v$ needs to have have compact support (w/r/t time) in $[0, \infty)$ (as opposed to just $[0,T]$). Alternatively, the notion of a weak solution can be reformulated to include the boundary terms, and so would become
$\newcommand{\R}{\mathbb{R}}$
$$\int_0^T \int_{\R}  u(x,t) v_t dt + \int_0^T \int_{\R}  F(u(x,t))v_x + \int_{\R} u(x,0) v(x,0) - \int_{\R} u(x,T) v(x,T)  $$


问:
What are some examples of "visually beautiful" math texts?
Are there good examples of math textbooks that seamlessly combines art, cartoon drawing or computer graphics with the texts? I know that there are a few books on the market, which are not text-book per se, that are geared towards general audience, where there are plenty of artistic figures naturally generated by the phenomena under study, usually things of geometric nature, Escher tiling, or things like chaos.

Wolfram's a New Kind of Science comes to mind. What are some other examples of "visually beautifu" or "artistic" math textbooks or texts.

答: Homotopical Topology by A. Fomenko and D. Fuchs.</p>


问:
Finding a linear recurrence regarding strings
The question is

Let $T(n)$ be the number of length-$n$ strings of letters $a$, $b$ and $c$, that do
not contain three consecutive $a$'s. Give a recurrence relation for $T(n)$ and
justify it. (You do not have to solve it.)

How would I go about solving this problem. I found the initial conditions of

$n = 0, T(n) = 1$

$n = 1, T(n) = 3$

$n = 2, T(n) = 9$

$n = 3, T(n) = 26$

But I do not know what to do after that point.

答:
Hint: Call such strings good. Let $p(n)$ be the number of good strings of length $n$ that end in a letter other than $a$. Let $q(n)$ be the number of good strings that end in a single $a$. Let $r(n)$ the number of good strings that end in two $a$'s. Note that $p(n)+q(n)+r(n)=T(n)$.

We have
$$p(n+1)=2(p(n)+q(n)+r(n))=2T(n).\tag{$1$}$$
This is because we get a good word of length $n+1$ that doesn't end in $a$ by appending either $b$ or $c$ to any good word of length $n$. By similar reasoning, we have
$$q(n+1)=p(n),\tag{$2$}$$
and
$$r(n+1)=q(n).\tag{$3$}$$
Bump up the indices in Equation $(3)$ by $1$. We get $r(n+2)=q(n+1)=p(n)$.

Add the the right-hand sides and left-hand sides of the three displayed equations. We get
$$T(n+1)=2T(n)+p(n)+q(n)=3T(n)-r(n).$$
But $r(n)=p(n-2)=2T(n-3)$. Thus
$$T(n+1)=3T(n) -2T(n-3).$$


问:
Metric on semi-Riemannian space
Here we have some (blue) curves $\ln(x)\ln(y)=s$ for several values of $s$ and some (green) curves $\ln(1-x)\ln(y)=s$ for several values of $s.$ The (red) is the multiplication of the green and blue.

The blue and green curves are the invariant "hyperbola."

Let $(M_1,g_1)$ be the blue semi-Riemannian manifold and $(M_2,g_2)$ the green semi-Riemannian manifold with the following metrics:

$g_1=\frac{dxdy}{xy}$ and $g_2=\frac{dudv}{v-uv}.$

Upon taking the Cartesian product of blue curves and green curves, what metric can be put on this new space (of red curves)? Will it be some sort of combination of $g_1$ and $g_2$ or something completely different?

答:
For two semi-Riemannian manifolds $(M_{1}, g_{1})$ and $(M_{2}, g_{2})$, we have the product metric tensor $g_{} = g_{1} \bigoplus g_{2}$. Before doing that, however, we need to obtain $g_{1}$ and $g_{2}$ in terms of $x$, $y$, $dx$, and $dy$. We now have that
\begin{equation}
ds_{1}^{2} = \frac{dx dy}{xy} \implies xy = \frac{dx dy}{ds_{1}^{2}}
\end{equation}
\begin{equation}
ds_{2}^{2} = \frac{dx dy}{x - xy} \implies x - xy = \frac{dx dy}{ds_{2}^{2}}
\end{equation}
We can now add them and find a suitable metric
\begin{equation}
xy + x - xy = x = \frac{dx dy}{ds_{1}^{2}} + \frac{dx dy}{ds_{2}^{2}} = \frac{ds_{2}^{2} dx dy}{ds_{1}^{2}ds_{2}^{2}} + \frac{ds_{1}^{2} dx dy}{ds_{1}^{2}ds_{2}^{2}} = \frac{ds_{2}^{2} dxdy + ds_{2}^{1} dxdy}{ds_{1}^{2}ds_{2}^{2}}
\end{equation}
\begin{equation}
\implies ds_{1}^{2}ds_{2}^{2} = \frac{ds_{2}^{2} dxdy + ds_{2}^{1} dx dy}{x} = dxdy  \frac{ds_{2}^{2} + ds_{2}^{1}}{x} = dxdy \left( \frac{dxdy}{x^{2}y} +  \frac{dx dy}{x^{2} - x^{2}y}\right)
\end{equation}
by substituting in the original metrics $ds_{1}^{2}$ and $ds_{2}^{2}$.

Title: Metric on semi-Riemannian space


问:
Prove that in a triangle $ABC$ $\forall P, P\neq M_{a},M_{b} ,M_{c}:min\{ \frac{|PA|}{|PM_{a}|},\frac{|PB|}{|PM_{b}|},\frac{|PC|}{|PM_{c}} \}\le 2.$
Suppose a triangle $ABC$. Define $M_{a},M_{b} ,M_{c}$ as the midpoints of the sides $a,b,c$ respectively. Then prove that for any point $P$ different from $M_{a},M_{b} ,M_{c}$, the smallest ratio of magnitudes of lines from $P$ to $M_{x}$ and from $P$ to $x$ is $\le 2$.

I have had certain success in prooving it for special cases, such as triangle $ABP$ being isosceles, but I cannot see a way to generalize this properly. I would think this is a pretty important (or at least very interesting result) but I have not seen it anywhere before (or its proof) so I'm intrigued to hear your ideas.

答:
Hint: "3-way Equality"
occurs at the centroid $G$.

Hint: Construct Apollonius circles about $ A, M_a$ with the desired $2:1$ ratio.

Show that these 3 circles will cover triangle $ABC$.

 From the first hint, these 3 circles go through a common point.

 Find the other point of intersection of 2 of these circles.

 This other point is the reflection of the centroid in the corresponding side.

 Hence show that if $ P \in \Delta AGD_b$ (where $D_b$ needs to be suitably defined, then $ \frac{ PA}{PM_a} \leq 2$.


问:
Nonsensical question on recurrence relations. Can it be fixed?
Problem. A biologist takes $3$ cells and observes them in his lab as they duplicate themselves.  Each step of time he observes that the new cells that are "born" is the double of the number of cells in the previous step, but a third of the number of cells of the previous step actually die.  Write a recurrence relation for the number of cells that are alive at time step $n$.

Solution. Straight from the problem, we know

$$c_n = \begin{cases}3 & \text{if } n = 0\\
                   c_{n - 1} + 2c_{n -1} - c_{n - 1}/3 & \text{if } n > 0.
\end{cases}$$

So far, so good.  But the problem is that $c_1 = 8$ and $c_2 = 8 + 16 - 8/3 = 24 - 8/3$.  In other words, at step two, we need to kill six cells and then kill $2/3$ of another.  How am I going to kill $2/3$ of a cell?

Is it possible to ignore this imperfection?  Can I replace "cell" with something that's "always" (if you will) divisible, like ounces of water?

答:
Allow fractions of cells. After very few time steps the number will be large enough so that the fractions will be insignificant.

This is a toy problem that's poorly written. The initial condition makes little biological sense. And "time instant" should be "time step".


问:
Maximum Likelihood Problem of a variation of a Poisson random variable
I currently have to solve exercise 3.2.

Can someone give me a hint on how to come up with a solution for 3.2? It should be similar to 3.1, which I have solved, not?

3.1. Given a realization (observation) $x_1,\dots,x_n$ of these (Poisson $(\lambda)$ i.i.d. random variables $X_1,\dots,X_n$) find the maximum likelihood estimate of $\lambda.$

$$\begin{align}
L(\lambda) &= P(x_1,\dots,x_n)\\
&=\prod_{i=1}^n P(x_i)\\
&=\prod_{i=1}^n \frac{e^{-\lambda}\lambda^{x_i}}{x_i!}\\
&=e^{-n\lambda} \lambda^{\sum x_i}\frac 1 {\prod x_i!}
\end{align}$$

Hence,

$$\begin{align}
\log(L(\lambda))&=\log\left(e^{-n\lambda}\right)\log\left( \lambda^{\sum x_i}\right)\log\left(\frac 1 {\prod x_i!}\right)\\
&=-n\lambda +\sum x_i \log \lambda + \log \frac{1}{\prod x_i!}
\end{align}$$

Getting the maximum,

$$\begin{align}
\frac{d\log(L(\lambda))}{d\lambda}&=-n+\frac{\sum x_i} \lambda + 0\\
&\implies \lambda= \frac{\sum x_i} n
\end{align}$$

3.2. Assume that we only observe events $X_i=0$ or $X_i>0$ for each random variable. Given a realization $y_i,\dots, y_n$ of these events ($y_i=0$ if $x_i=0$ and $y_i=1$ if $x_i>0$), find the MLE of $\lambda.$

答:
So the probability of getting $y_i=1$ can be considered a success of a Bernoulli experiment. And this $\Pr(Y_i=1)=1- \Pr(X_i=0) = 1-\exp(-\lambda).$

The MLE of a Bernoulli rv is the same as for a Poisson - the mean of the sample, which corresponds to the estimated probability of success.

Therefore,

$$\begin{align}
\bar y&=1-\exp(-\lambda)\\[2ex]
\lambda&=-\log \left(  1 -{\bar y}\right).
\end{align}$$

This seems to hold on a quick simulation:

```<code>set.seed(561)

m = matrix(0,12,2)

for(i in 1:12){
l = i
n = 1e6
sam = rpois(n,l)
ybar = mean(sam!=0)
m[i,1] <- i
m[i,2] <- round(-log(1-ybar), 3)
}
plot(1:12, m[,2], xlab="λ", ylab= "Estimated λ", main="MLE", pch=19, col='tan4')
</code>```

Body:
So the probability of getting $y_i=1$ can be considered a success of a Bernoulli experiment. And this $\Pr(Y_i=1)=1- \Pr(X_i=0) = 1-\exp(-\lambda).$

The MLE of a Bernoulli rv is the same as for a Poisson - the mean of the sample, which corresponds to the estimated probability of success.

Therefore,

$$\begin{align}
\bar y&=1-\exp(-\lambda)\\[2ex]
\lambda&=-\log \left(  1 -{\bar y}\right).
\end{align}$$

This seems to hold on a quick simulation:

set.seed(561)

m = matrix(0,12,2)

for(i in 1:12){
l = i
n = 1e6
sam = rpois(n,l)
ybar = mean(sam!=0)
m[i,1] <- i
m[i,2] <- round(-log(1-ybar), 3)
}
plot(1:12, m[,2], xlab="λ", ylab= "Estimated λ", main="MLE", pch=19, col='tan4')


问:
Proving that $\nu^+(E) = \sup\{\nu(C) :C \in \mathcal{A}, C\subset E\}$
If $\nu$ is a signed measure on $(X, \mathcal{A})$, how to prove that $\nu^+(E) = \sup\{\nu(C) :C \in \mathcal{A}, C\subset E\}$? I am thinking of using the Hahn decomposition, so $\nu^+(E) = \nu(E \cap P)$. But I am not sure how to proceed.

答:
Since $E\cap P\subset E$,
$$
\nu^+(E)=\nu(E\cap P)\le \sup\{\nu(C): C\in \mathcal{A},C\subset E\}.
$$
Also for $C\subset E$, $\nu(C)= \nu^+(C)-\nu^-(C)\le \nu^+(E)$. Thus,
$$
\sup\{\nu(C): C\in \mathcal{A},C\subset E\}\le \nu^+(E).
$$


问:
Reference request: about consistency of adding any amount Quine atoms to ZF-Reg.?
It is known that we can add to ZF-Reg. any amount of Quine atoms (singletons that are elements of themselves), we can even add a proper class of them.

When that result was first proved?

Does it depend on forcing?

答:
I don't know the history of this result, but it is much easier than forcing and I imagine it probably was known by the 1940s if not earlier.  Basically, you can just formally construct a cumulative hierarchy starting with some collection of Quine atoms.

Here is one way to formalize the construction.  Let $(M,\epsilon)$ be a structure in the language of set theory.  We define an extension $M'$ of $M$ as follows.  Let $S$ be the set of all subsets of $M$ that are not the extension of any element of $M$.  We then define $M'=M\sqcup S$, with $\epsilon$ extended to say that additionally $m\mathbin{\epsilon}s$ when $m\in M, s\in S,$ and $m\in s$.

Now start with a structure $M_0$ consisting of some set of Quine atoms.  Recursively define $M_{\alpha+1}=M_\alpha'$ and let $M_\alpha$ be the direct limit of $M_\beta$ for $\beta<\alpha$ if $\alpha$ is a limit ordinal.  The direct limit of these structures $M_\alpha$ over all ordinals $\alpha$ is then easily verified to be a class model of ZF without regularity, with $M_0$ as its set of Quine atoms.  With a little care you can carry out a similar construction where the initial $M_0$ is a proper class to get a model with a proper class of Quine atoms (for instance, let $N_\alpha$ be the model constructed above starting from $M_0=\alpha$ and take the direct limit of the $N_\alpha$).


问:
Closed form for continued fraction
I am working with a recursive algorithm and I have realized that in each step it computes something equivalent to:

$A_{0}=\frac{a}{b}$

$A_{1}=a/(b-\frac{a^{2}}{b})$

$A_{2}=a/(b-\frac{a^{2}}{b-\frac{a^{2}}{b}})$

$\vdots$

$A_{n+1}=a/(b-aA_{n})$

And so on. I was wondering if there could exist some closed form for $A_{n}$ to avoid unnecessary iterations

答:
The closed form does exist, I found it with Mathematica and it's amazing, to use in a program I mean.

I found it with this function

```<code>RSolve[{A[n + 1] == a/(b - a A[n]), A[0] == a/b}, A[n], n]
</code>```

Set:
$$
\begin{cases}
D=\sqrt{b^2-4 a^2}\\
x=\frac{b-D}{a}\\
y=\frac{b+D}{a}\\
\end{cases}
$$
Then
$$A_n=\frac{a (b+D ) y^n-a (b-D ) x^n}{x^n \left(2 a^2-b (b-D )\right)+y^n \left(b (b+D )-2 a^2\right)}$$

You will surely find a way to optimize better this formula in order to use it in a program.

Edit

This is the formula as output by Mathematica
$$A_n=\frac{a \left(b \left(\frac{\sqrt{b^2-4 a^2}+b}{a}\right)^n+\sqrt{b^2-4 a^2} \left(\frac{\sqrt{b^2-4 a^2}+b}{a}\right)^n+(-b) \left(-\frac{\sqrt{b^2-4 a^2}-b}{a}\right)^n+\sqrt{b^2-4 a^2} \left(-\frac{\sqrt{b^2-4 a^2}-b}{a}\right)^n\right)}{-2 a^2 \left(\frac{\sqrt{b^2-4 a^2}+b}{a}\right)^n+b^2 \left(\frac{\sqrt{b^2-4 a^2}+b}{a}\right)^n+b \sqrt{b^2-4 a^2} \left(\frac{\sqrt{b^2-4 a^2}+b}{a}\right)^n-b^2 \left(-\frac{\sqrt{b^2-4 a^2}-b}{a}\right)^n+2 a^2 \left(-\frac{\sqrt{b^2-4 a^2}-b}{a}\right)^n+b \sqrt{b^2-4 a^2} \left(-\frac{\sqrt{b^2-4 a^2}-b}{a}\right)^n}$$

Title: Closed form for continued fraction

Body:
The closed form does exist, I found it with Mathematica and it's amazing, to use in a program I mean.

I found it with this function

RSolve[{A[n + 1] == a/(b - a A[n]), A[0] == a/b}, A[n], n]

Set:
$$
\begin{cases}
D=\sqrt{b^2-4 a^2}\\
x=\frac{b-D}{a}\\
y=\frac{b+D}{a}\\
\end{cases}
$$
Then
$$A_n=\frac{a (b+D ) y^n-a (b-D ) x^n}{x^n \left(2 a^2-b (b-D )\right)+y^n \left(b (b+D )-2 a^2\right)}$$

You will surely find a way to optimize better this formula in order to use it in a program.

Edit

This is the formula as output by Mathematica
$$A_n=\frac{a \left(b \left(\frac{\sqrt{b^2-4 a^2}+b}{a}\right)^n+\sqrt{b^2-4 a^2} \left(\frac{\sqrt{b^2-4 a^2}+b}{a}\right)^n+(-b) \left(-\frac{\sqrt{b^2-4 a^2}-b}{a}\right)^n+\sqrt{b^2-4 a^2} \left(-\frac{\sqrt{b^2-4 a^2}-b}{a}\right)^n\right)}{-2 a^2 \left(\frac{\sqrt{b^2-4 a^2}+b}{a}\right)^n+b^2 \left(\frac{\sqrt{b^2-4 a^2}+b}{a}\right)^n+b \sqrt{b^2-4 a^2} \left(\frac{\sqrt{b^2-4 a^2}+b}{a}\right)^n-b^2 \left(-\frac{\sqrt{b^2-4 a^2}-b}{a}\right)^n+2 a^2 \left(-\frac{\sqrt{b^2-4 a^2}-b}{a}\right)^n+b \sqrt{b^2-4 a^2} \left(-\frac{\sqrt{b^2-4 a^2}-b}{a}\right)^n}$$


答:
It seems as though you have a fundamental misunderstanding of what a "change of basis" means in this context. In short: as copper.hat says, you need to change the basis in the domain and in the range.

The operator $\Omega$ of interest is represented by a matrix whose entries are $\omega_{ij}$. This means that
$$
\Omega |j \rangle = \sum_{i=1}^n \omega_{ij} |i \rangle.
$$
In other words, the $j$th column of the matrix encodes the destination of the $j$th basis vector.  Correspondingly, if $M$ is the matrix of $\Omega$ relative to the new basis $U |1 \rangle,\dots U|n\rangle$ and has entries $m_{ij}$, then we should have
$$
\Omega (U |j \rangle) = \sum_{i=1}^n m_{ij} (U |i \rangle).
$$
We can relate this expression to the original as follows: if we multiply both sides by $U^\dagger$, then we have
$$
U^\dagger \Omega (U |j \rangle) = \sum_{i=1}^n m_{ij} U^\dagger U|i \rangle \implies\\
(U^\dagger \Omega U) |j \rangle = \sum_{i=1}^n m_{ij} |i \rangle.
$$
In other words, saying that $M$ is the matrix of the operator $\Omega$ relative to the new basis $\{U |i \rangle\}$ is the same as saying that $M$ is the is the matrix of the operator $U^\dagger \Omega U$ relative to the original basis $\{|i \rangle\}$.

Title: Is showing $\text{Tr}\Omega U = \text{Tr} \Omega$ equivalent to showing $\text{Tr}\Omega = \text{Tr}(U^{\dagger} \Omega U)$?

Body:
On pg. 30 of Principles of Quantum Mechanics (by Shankar), we are asked:

Show that the trace of an operator is unaffected by a unitary change of basis $|i \rangle \rightarrow U |i\rangle$. [Equivalently, show $\text{Tr}\Omega = \text{Tr}(U^{\dagger} \Omega U)$].

I have already convinced myself that

$$
\text{Tr} \Omega = \text{Tr}(U^{\dagger} \Omega U)
$$

by using the fact that

$$
\text{Tr} ( \Omega \Lambda ) = \text{Tr} ( \Lambda \Omega )
$$

implies that

$$
\text{Tr} \Omega = \text{Tr}(\Omega (U U^{\dagger})) = \text{Tr}((\Omega U) U^{\dagger}) = \text{Tr}(U^{\dagger} (\Omega U)) = \text{Tr}(U^{\dagger} \Omega U)
$$

as desired.

Question: But why is showing this equivalent to showing that the trace of an operator is unaffected by a unitary change of basis $|i \rangle \rightarrow U | i \rangle$?  It seems to me this means that

$$
\text{Tr} \Omega = \text{Tr}(\Omega U)
$$

which is a distinct condition from showing that

$$
\text{Tr} \Omega = \text{Tr}(U^{\dagger} \Omega U)
$$


问:
Evaluation of integral with residue theorem
picture

I tried to solve this integral using the residue theorem but wolframalpha says it's incorrect, and I can't find any mistakes in my algebra, so I'm wondering where I messed up. Wolframalpha says the residual should be what I have but with -27 instead of -36

答:
You seem to be trying to evaluate the residue at $\;z=3i\;$ ...but the limit that you wrote above is not the way to do it as that is a pole of order two, so the correct way to do it is with

$$\lim_{z\to3i}\left[(z-3i)^2\frac{e^{iz}}{(z-3i)^2(z+3i)^2}\right]'$$

Give it a try now...


问:
Is a finite metric space complete?
If $(X,d)$ is a metric space with a finite number of elements, is $X$ complete?

I know that every finite metric space is compact, and also compact spaces are complete. So, is this true?

答:
Here is a quick way to see this is true:

Since $X$ is finite, you can find a distance $\rho > 0$ such that no two points are within $\rho$ of each other. For instance $\rho = \frac{1}{2} \min_{x \neq y} d(x,y)$ works. Notice this $\rho$ is strictly greater than $0$ since we're taking the minimum of finitely many nonzero things.

But now fix a cauchy sequence in $X$. Eventually the terms get closer together than any $\epsilon$. In particular, they get closer together than $\rho$. After this point, there is no choice, the sequence must be constant. Then it clearly converges.

The argument for compact spaces is a variation on this theme, so your higher level argument more or less agrees with this one once you unpack the content of the theorem.

I hope this helps ^_^

Title: Is a finite metric space complete?


问:
Why does the beta distribution become U shaped when $\alpha$ and $\beta$ <1?
In the Beta distribution (used to model Bernoulli probabilities), the $\alpha$ and $\beta$ parameters can be interpreted as the number of heads$+1$ and the number of tails$+1$ seen. So, if they were both $2$, it would lean towards the coin being fair and have a maximum at $0.5$. If they are both $20$, the distribution would become even surer we're dealing with a fair coin and peak even more at $p=0.5$.

What I don't get is its behavior when $\alpha$ and $\beta$ both become $<1$.

In that case, it becomes U-shaped and the density peaks at $p=0$ and $p=1$. Meaning the coin is likely to be two-sided. I know there is an intuition for this since I think I had an idea about it a long time ago. However, I've been trying to recollect all day and can't piece it together. Does anyone have an intuition?

答:
As Sample Size Decreases, Variance Increases, which Requires a Bimodal Distribution

Thinking about variance provides one explanation for the U-shaped Beta.  As always, a larger sample size (${\displaystyle \alpha + \beta }$) decreases a distribution's variance, and a smaller sample size increases it.  If Betas were limited to unimodal distributions, their variance could never reach its full potential.  In order to maximize a Beta distribution's variance for a particular mean, the distribution must become bimodal, with its density concentrated at the two extremes.  At the limit, as the variance approaches its maximum (for any given mean), the Beta distribution approaches a Bernoulli distribution and its variance likewise approaches the variance of a Bernoulli with its same mean.

Different Interpretations of Alpha and Beta: Mean-Based vs. Mode-Based

The Wikipedia entry for Conjugate Prior (note 3) offers crucial advice about the interpretation of Beta parameters:

"The exact interpretation of the parameters of a beta distribution in terms of number of successes and failures depends on what function is used to extract a point estimate from the distribution. The mean of a beta distribution is ${\displaystyle {\frac {\alpha }{\alpha +\beta }},}$ which corresponds to $\alpha$  successes and $\beta$  failures, while the mode is ${\displaystyle {\frac {\alpha -1}{\alpha +\beta -2}},}$ which corresponds to ${\displaystyle \alpha -1}$ successes and ${\displaystyle \beta -1}$ failures. Bayesians generally prefer to use the posterior mean rather than the posterior mode as a point estimate, justified by a quadratic loss function, and the use of ${\displaystyle \alpha }$  and ${\displaystyle \beta }$  is more convenient mathematically, while the use of ${\displaystyle \alpha -1}$ and ${\displaystyle \beta -1}$ has the advantage that a uniform ${\displaystyle {\rm {Beta}}(1,1)}$ prior corresponds to $0$ successes and $0$ failures."

A similar point is made by Tom Minka in his answer to a related question.

The contrast between these two interpretations becomes especially stark in the case of bimodal Betas, since they have two modes yet only a single mean.  Focusing on the example of a fair coin, as this question does, hides the issue because that's the unusual case where the difference between the mean and mode disappears.

In Doing Bayesian Data Analysis, John Kruschke notes that a bimodal Beta would mean we "believe that the coin is a trick coin that nearly always comes up heads or nearly always comes up tails, but we don't know which." (p. 83, 1st ed.)  And since that's a rather contrived scenario, it confirms the limitations of the coin tossing example.

Note that if we interpret ${\displaystyle \alpha }$ as successes + 1 and ${\displaystyle \beta }$ as failures + 1, then the success count and failure count must both turn negative when ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ are less than 1.  By contrast, if we interpret ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ as successes and failures, respectively, without subtracting 1, then we sidestep the seemingly nonsensical idea of negative counts.  Even when ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ are both less than 1, the mean-based interpretation of them poses no issues since the mean remains a single value even when the mode splits off into two.

Just as there are some contexts when it makes sense to focus on a distribution's mean and other contexts when it makes sense to focus on its mode(s), so too will our interpretation of ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ depend on what central tendency is of interest in a particular context.  As Wikipedia's Conjugate Prior entry entry puts it:

"It is often useful to think of the hyperparameters of a conjugate prior distribution as corresponding to having observed a certain number of pseudo-observations with properties specified by the parameters. For example, the values ${\displaystyle \alpha}$ and ${\displaystyle \beta}$ of a beta distribution can be thought of as corresponding to ${\displaystyle \alpha -1}$ successes and ${\displaystyle \beta -1}$ failures if the posterior mode is used to choose an optimal parameter setting, or ${\displaystyle \alpha}$ successes and ${\displaystyle \beta}$  failures if the posterior mean is used to choose an optimal parameter setting."

You might find this discussion of bimodal Betas helpful too.

The Polya Urn Interpretation Yields Nice Intuitions about U-Shaped Betas

A lesser-known, but surprisingly accessible interpretation of the Beta distribution views it as the result of draws from a Polya urn model.  Rather than attempting a full proof here, I will simply explain how this alternative interpretation yields an attractively intuitive explanation of U-shaped Betas.

The basic idea is that an urn initially contains S number of success balls and F number of failure balls, which correspond to the ${\displaystyle \alpha}$ and ${\displaystyle \beta}$ parameters (we are interested in the distribution's mean).  After drawing a single ball from the urn, you not only replace it, but add an additional ball of its same type.  In the limit, drawing and then adding an infinite number of balls in this way yields a single proportion drawn from a Beta(success, failure).

One can see that every successive draw will have slightly less impact over the resulting, limit ratio than did the draw before it.  Starting with a Beta(1,1) means that the urn's ratio will shift from 1/2 to either 1/3 or 2/3 once a third ball gets introduced.  With each successive introduction of a new ball, that new ball's influence over successive draws shrinks.

Now, this same Polya urn procedure can be applied with fractional balls if one stipulates (1) that the likelihood of drawing a fractional ball remains proportional to its size--a half ball remains half as likely to be drawn as any whole ball--and (2) that when drawn and replaced, fractional balls are nevertheless accompanied by whole balls of their same type.  Fractional balls thereby acquire an influence disproportionate to their actual size.

Consider an example where ${\displaystyle \alpha}$ and ${\displaystyle \beta}$ both begin at .1, so that the initial draw provides even odds of drawing either the one-tenth success ball or the one-tenth failure ball.  Because fractional balls are accompanied by whole balls when replaced, the first whole ball introduced will dominate all subsequent draws.  What started off as a ratio of .1 success balls to .2 balls immediately veers toward a lopsided ratio of either ${\displaystyle {\frac {.1}{1.2}}}$ or ${\displaystyle {\frac {1.1}{1.2}}}$.  Indeed, that initial draw so dominates all subsequent draws that the ratio is likely to grow increasingly lopsided over time.  Once the ratio tilts decisively away from ${\displaystyle {\frac {.1}{.2}}}$ it is exceedingly unlikely to ever return to anything comparably balanced.  And, of course, that effect becomes even more pronounced if one starts with a Beta(.001, .001): the U-shape becomes thinner and thinner in the middle and thicker and thicker at the extremes as the sum of ${\displaystyle \alpha}$ and ${\displaystyle \beta}$ gets smaller because the initial draw more completely dominates the subsequent draws.

While the equivalence between Betas and Polya urns is hardly obvious, the Polya urn offers elegant insights into the Beta distribution.


答:
Direct proof.

Let $r=\min\{d(x,y):x,y\in X, x\ne y\}$. Then $r>0$. Suppose $(x_n)$ is a Cauchy sequence. Then there exists $k$ such that, for $m,n\ge k$, $d(x_m,x_n)<r$.

This implies that the sequence is constant from $k$ onwards.

Title: Is a finite metric space complete?

Body:
If $(X,d)$ is a metric space with a finite number of elements, is $X$ complete?

I know that every finite metric space is compact, and also compact spaces are complete. So, is this true?


问:
Is $f_n$ uniformly convergent on $(-\infty,\infty)$
Let $f_n(x)=\frac{2n^2x}{1+n^4x^4}$ where $n\in\mathbb{N}$ and $x\in(-\infty,\infty)$.

I've already shown that this function is pointwise convergent to the limit function $f(x)=0$.  My question is whether it is uniformly convergent anywhere on $(-\infty,\infty)$.

答:
Note that $f_n(\frac{1}{n})=n$, which is very far from the limit function. So for every $n$ we have:

$\sup \{x\in\mathbb{R}: |f_n(x)-f(x)|\}\geq f_n(\frac{1}{n})-f(\frac{1}{n})=n$

So in particular the sequence of supremums doesn't converge to zero. There is no uniform convergence.


问:
Which approach is correct?
Find an augmented matrix with non-zero entries whose system
$$
\begin{align*}
ax + by &= c   \\
dx + ey &= f
\end{align*}
$$
has

no solutions,

exactly one solution,

infinitely many solutions.

I'm considering firstly change the matrix into reduced echelon form using $a,b,c,d,e,f$ to represent. So I get $ea=bd$ and $fa\neq cd$ for no solutions. I'm sure whether I'm on the right track.

答:
$$\begin{cases}x+y=1,\\x+y=2\end{cases}$$
(impossible because $1\ne2$)

$$\begin{cases}x+y=1,\\x+2y=1\end{cases}$$
(equivalent to $\begin{cases}x+y=1,\\\ \ \ \ \ \ \ y=0\end{cases}$)

$$\begin{cases}x+y=1,\\x+y=1\end{cases}$$
(indeterminate because the equations are the same)

Title: Which approach is correct?


问:
Short exact sequence of groups acting on a space
Let $X$ be a metrizable topological space, and suppose we have three groups $\Gamma$, $\Gamma'$, and $\Gamma''$, acting properly on, respectively $X$, $X$, and $X/\Gamma'$. What can we say about them if we know that $X/\Gamma\cong (X/\Gamma')/\Gamma''$?

My intuition is that the latter two contains together all the information of the first, so I would be tempted to say that we have a short exact sequence $0\to\Gamma'\to\Gamma\to\Gamma''\to 0$. In fact this would be the analogue of the third isomorphism theorem.

This setting appears when studying orbifold coverings, so we might assume all groups being finite subgroups of $O(n)$, and $X$ an open subset of $\mathbb{R}^n\times\mathbb{R}^{\geq 0}$, but I suspect the answer could hold much more generally.

答:
Take $\Gamma'' = 1$ for simplicity. Then you are asking whether $X/\Gamma \cong X/\Gamma'$ implies $\Gamma \cong \Gamma'$. Even if we assume that the actions of $\Gamma$ and $\Gamma'$ are free as I suggested in the comments this need not be true in general; for example, any cyclic group $C_n$ acts freely by rotations on $X = S^1$ with quotient $S^1$ again.

However, it is true if $X$ is simply connected and we assume the actions of $\Gamma$ and $\Gamma'$ are properly discontinuous, since then $\Gamma \cong \Gamma'$ is the fundamental group. This gives us a salvage of the general case $\Gamma'' \neq 1$; if we assume that $X$ is simply connected and all three actions are properly discontinuous then the fundamental group of $(X/\Gamma')/\Gamma''$ sits in a short exact sequence $1 \to \Gamma' \to \pi_1 \to \Gamma'' \to 1$ (e.g. by the long exact sequence in homotopy) so we have a short exact sequence $1 \to \Gamma' \to \Gamma \to \Gamma'' \to 1$ as desired.


问:
Iterated integral $\int_D \sqrt{x^2 +y^2} \ dx \ dy \ dz$

Compute $$\int_D \sqrt{x^2 +y^2} \ dx \ dy \ dz$$ when $D = \{(x,y,z) \in \mathbb{R}^3 \mid x^2+y^2 \leqslant 1, 0\leqslant z \leqslant 7 \}.$

I managed to get the triple integral

$$\int_{0}^{7}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \sqrt{x^2+y^2} \ dx \ dy \ dz$$

but this isn't trivial to compute at all. $\sqrt{x^2+y^2}$ seems to define a surface in $\mathbb{R^3}$ that looks like a cone, should I consider polar or cylindrical coordinates here or is there something else I'm missing?

答: Your cylindrical-coordinates instinct is right. It gives $7\int_0^{2\pi}d\theta\int_0^1r^2dr$, which I'll leave to you. The shape is a length-$7$ cylinder of radius $1$; we're computing its volume, times the mean of $r$ throughout that volume.


答:
You need to use the Gaussian method and do row reduction, then at the end of the process you do an analysis on the equation of the form $ax = b$ and analyze the different cases.

Indeed, note that $$(S.E.L): \left\{\begin{aligned} ax+by=c\\
 dx+ey=f  \end{aligned} \right. \iff  \begin{pmatrix} a & b & | & c \\ d & e & | & f \end{pmatrix}\sim \begin{pmatrix} a & b & | & c \\ 0 & \frac{-bd+ae}{a} & | & \frac{-cd+af}{a}\end{pmatrix} $$
Now,

The S.E.L doesn't have solution if $$ \frac{-bd+ae}{a}=0 \quad \text{and} \quad \frac{-cd+af}{a}\not=0$$

The S.E.L exactly one solution if $$\frac{-bd+ae}{a}\not=0 \quad \text{and} \quad \frac{-cd+af}{a}\not=0$$

The S.E.L have infinitely many solution if $$\frac{-bd+ae}{a}=0 \quad \text{and} \quad \frac{-cd+af}{a}=0$$

Title: Which approach is correct?

Body:
Find an augmented matrix with non-zero entries whose system
$$
\begin{align*}
ax + by &= c   \\
dx + ey &= f
\end{align*}
$$
has

no solutions,

exactly one solution,

infinitely many solutions.

I'm considering firstly change the matrix into reduced echelon form using $a,b,c,d,e,f$ to represent. So I get $ea=bd$ and $fa\neq cd$ for no solutions. I'm sure whether I'm on the right track.


问:
How to find the limit for the quotient of the least number $K_n$ such that the partial sum of the harmonic series $\geq n$
Let $$S_n=1+1/2+\cdots+1/n.$$ Denote by $K_n$ the least subscript $k$ such that $S_k\geq n$. Find the limit $$\lim_{n\to\infty}\frac{K_{n+1}}{K_n}\quad ?$$</p>

答:
Consider that, for $p>k>1$,
$$
S_k+\int_{k+1}^{p}\frac1xdx\leq S_{p} \leq S_k+\int_{k}^{p-1}\frac1xdx
$$
Now, consider $S_{K_n}=n+\epsilon_n$, where $0<\epsilon_n<\frac1{K_n}$ for $n>1$. If we let $k=K_n$ and $p=K_{n+1}$, then we have
$$
S_{K_n}+\log(K_{n+1})-\log(K_n+1)\leq S_{K_{n+1}}\leq S_{K_n}+\log(K_{n+1}-1)-\log(K_n)
$$
or
$$
n+\epsilon_n+\log\left(\frac{K_{n+1}}{K_n+1}\right)\leq n+1+\epsilon_{n+1}\leq n+\epsilon_n+\log\left(\frac{K_{n+1}-1}{K_n}\right)
$$
As such, we can write that
$$
\log\left(\frac{K_{n+1}}{K_n+1}\right)\leq 1+\epsilon_{n+1}-\epsilon_n\leq \log\left(\frac{K_{n+1}-1}{K_n}\right)
$$
or
$$
\frac{K_{n+1}}{K_n+1}\leq e^{1+\epsilon_{n+1}-\epsilon_n}\leq \frac{K_{n+1}-1}{K_n}
$$
Now, we may write the left inequality as
$$
K_{n+1} \leq e^{1+\epsilon_{n+1}-\epsilon_n}(K_n+1)
$$
and the right inequality as
$$
e^{1+\epsilon_{n+1}-\epsilon_n}K_n+1\leq K_{n+1}
$$
and so
$$
e^{1+\epsilon_{n+1}-\epsilon_n}+\frac1{K_n}\leq \frac{K_{n+1}}{K_n} \leq e^{1+\epsilon_{n+1}-\epsilon_n}(1+\frac1{K_n})
$$
Taking the limit as $n\to\infty$, we note that $\lim_{n\to\infty} \epsilon_n=0$ and $\lim_{n\to\infty} \frac1{K_n}=0$, and so
$$
e\leq \lim_{n\to\infty} \frac{K_{n+1}}{K_n} \leq e
$$
Therefore,
$$
\lim_{n\to\infty} \frac{K_{n+1}}{K_n} = e
$$</p>


问:
Normed Vector Space, open ball contained in subspace
I was reading my measure theory book and came across a statement that I was a bit unsure of. I really do not have too much training in linear algebra, but was hoping to get some guidance on this:

If $U$ is a subspace of a normed vector space $V$ such that some open ball of $V$ is contained in $U$. Then is it true that $U=V$?

答:
Yes, it is. Intuitively, since a subspace is closed under scalar multiplication you can extend this ball as much as you want. For example, assume $U$ contains a ball with radius $\epsilon$ around $0$. Then for every $0\ne v\in V$ we have $\frac{\epsilon}{2}\frac{v}{||v||}\in U$, and so:

$v=\frac{2}{\epsilon}||v||(\frac{\epsilon}{2}\frac{v}{||v||})\in U$

The proof for a ball around any other point is similar.


问:
Boundedness in probability of Stochastic Process
This question is related to some other similar ones I did in the recent past.

Let $X_t$ be a Stochastic Process defined through the equation $$\text{d}X_t=f(X_t,t)\text{d}t+\text{d}W_t,$$
where $f$ is a twice differentiable function such that $f(x)<a<0$ for all $x>k$, and $X_0<k$.

What I am wishing to show is:

For all $\epsilon>0$ there is $M>0$ (which we can WLOG assume to be bigger than $k$) such that for all $t\ge0$ it holds that $$\mathbb{P}[X_t>M]\le\epsilon.$$

What I tried to do is define $\tau_t$ as $\sup\{0<s<t: X_s=k\}$, i.e. the last hitting time of the level $k$ before time $t$, where the sup is set equal to $0$ if the set $\{0<s<t: X_s=k\}$ is empty. This should be a stopping time, even if last hitting times in general are not.

Then, we can manipulate the given probability in this way:
\begin{align*}\mathbb{P}[X_t>M]&=\underbrace{\mathbb{P}[X_t>M,\tau_t=0]}_{=0}+\underbrace{\mathbb{P}[X_t>M,\tau_t\ne0, X_t<k]}_{=0}+\mathbb{P}[X_t>M,\tau_t\ne0, X_t>k]\\
&\le\mathbb{P}[X_t-X_{\tau_t}>M-k,\tau_t\ne0]\\
&\le \mathbb{P}[a(t-\tau_t)+W_t-W_{\tau_t}>M-k, \tau_t\ne0]\\
&\le \mathbb{P}[a(t-\tau_t)+W_t-W_{\tau_t}>M-k].\end{align*}

Since $a<0$ we can bound this probability uniformly over $t$, and the claim should be proven. However, I think I've been too slick with some steps, and I want to make sure they are correct (in particular, the $\tau_t$ seems to be not a stopping time). Can someone help me find any potential errors? Also, is there a more straightforward way to prove this? Maybe I just didn't see an easier solution.

EDIT: the $f$ in my specific problem satisfies the hypothesis $|f(x,t)-k_0|<\theta\cdot |x|+ \mu,$ for some constants $k_0, \theta$ and $\mu$: is there some kind of comparison principle with the absolute value bound? I can't find it online, but if it exists, then I would be done, as the Ornstein-Uhlenbeck process is bounded in the sense I am looking for.

答:
Summarization

Consider a stochastic process $(X)$ driven by a Brownian motion $(W)$ which sastifies the following SDE:
$$dX_t = f(t,X_t)dXt +dW_t$$

then

Theorem 1:

If $f(x,t) \le 0 \quad \forall x \ge m$, then for all t, we have :
$$\mathbb{E}((X_t)^2_+) \le 2m^2+2t$$
$\square$

Theorem 2

If:

$f(x,t) \le 0$  for all $ x \ge m$ and

There are two number $m<m_1<m_2$ such that $X$ upcrosses the invertal $[m_1,m_2]$ infinitely often.
then,
$$\mathbb{E}(X_t-m_2)_+^2 ) \le \mathbb{E}( t\vee \tau_{N(t)-1} -\tau_{N(t)-1})$$
where $\tau_{N(t)-1}$ is the first time $X$ upcrosses $m_2$ such that the next time $X$ downcrosses $m_1$ is after $t$

$\square$

Theorem 3

If:

$f(x,t) \le a <0 $  for all $ x \ge m$ and

There are two number $m<m_1<m_2$ such that $X$ upcrosses the invertal $[m_1,m_2]$ infinitely often.

$f(t,x)=f(x)$

then there is a constant $C$ not depending on $t$ such that:
$$\mathbb{E}(X_t-m_2)_+^2  \le C$$

$\square$

Discussion

Perhaps, there is a simpler way to obtain the same result but I would like to choose the one whose rigour I can be sure about. It would be nice if anyone can help me expand my humble knowledge.

The conditions on the upcrossings and the homogeneity seem relaxable, yet the relaxing work can quickly become complicated with my approach.

(Exclamation) This work is much longer than I expected ( it is not really long in my draft) even though the solution is here is also a shortened version.


答:
We have uniform convergence on $[a,\infty)$ and $(-\infty, -a]$ for any $a > 0$.

Note that

$$\left|\frac{2n^2x}{1+n^4x^4}-0 \right| = \frac{2n^2|x|}{1 + n^4||x|^4} \leqslant \frac{2n^2|x|}{n^4||x|a^3} = \frac{2}{n^2a^3} \underset{n \to \infty}\longrightarrow 0$$

Convergence is not uniform on any set $S$  with $0$ as an interior or boundary   point -- and, thus, $(-\infty,\infty)$ -- since

$$\sup_{x \in S}\left|\frac{2n^2x}{1+n^4x^4}-0 \right| \geqslant \frac{2n^2\cdot \frac{1}{n}}{1 + n^4 \cdot \left(\frac{1}{n} \right)^4} = n \underset{n \to \infty}\longrightarrow \infty $$

Title: Is $f_n$ uniformly convergent on $(-\infty,\infty)$

Body:
Let $f_n(x)=\frac{2n^2x}{1+n^4x^4}$ where $n\in\mathbb{N}$ and $x\in(-\infty,\infty)$.

I've already shown that this function is pointwise convergent to the limit function $f(x)=0$.  My question is whether it is uniformly convergent anywhere on $(-\infty,\infty)$.


答:
Yes.

Let's say that $B(a,r)$ denotes the ball of center $a$ and radius $r$, included in your subspace $U$. Then for every $x \in V \setminus \lbrace 0 \rbrace$, you have $$x= \frac{2||x||}{r}\left[\left( \frac{rx}{2||x||}+a\right)-a \right]$$

which belongs to $U$, since both $\left( \frac{rx}{2||x||}+a\right)$ and $a$ belong to $B(a,r) \subset U$, and $U$ is a vector space. So $V \setminus \lbrace 0 \rbrace \subset U$.

And of course $0 \in U$, so you get $V \subset U$ and so $V=U$.

Title: Normed Vector Space, open ball contained in subspace

Body:
I was reading my measure theory book and came across a statement that I was a bit unsure of. I really do not have too much training in linear algebra, but was hoping to get some guidance on this:

If $U$ is a subspace of a normed vector space $V$ such that some open ball of $V$ is contained in $U$. Then is it true that $U=V$?


问:
Can the Fibonacci sum $\sum_{i=0}^N x^i F_{a+ib}$, for integers $x$, $a$, $b$, be determined by a direct formula or generating function?
Suppose we have a variable $x$ such that $x\in\mathbb{N}$. Let

$$S=\sum_{i=0}^N  F_{a+i.b}. x^i$$

where $F_i$ is the $i$-th Fibonacci number, $a$ and $b$ are constants having range between $0\leq a,b\leq10^{9}$, and $N$ can be a big value up to $10^{18}$

Here's an example let's suppose $a=0,b=2,x=1,N=4$ then $$S=F_0+F_2+F_4+F_6+F_8=33$$

Can $S$ be solved using a direct formula or maybe a generating function? If yes, how?

答:
The $n$-th Fibonacci number is
$$F_n=\frac{1}{\sqrt5}(\phi^n-\psi^n),$$
where $\phi=\frac{1}{2}(1+\sqrt5)$ and $\psi=\frac12(1-\sqrt5)$. Thus
$$\begin{align}
S&=\sum_{k=0}^NF_{a+bk}x^k\\
&=\sum_{k=0}^N\frac{1}{\sqrt5}(\phi^{a+bk}-\psi^{a+bk})x^k\\
&=\frac{\phi^a}{\sqrt5}\sum_{k=0}^N(\phi^bx)^k-\frac{\psi^a}{\sqrt5}\sum_{k=0}^N(\psi^bx)^k\\
&=\frac{\phi^a}{\sqrt5}\cdot\frac{(\phi^bx)^N-1}{\phi^bx-1}-\frac{\psi^a}{\sqrt5}\cdot\frac{(\psi^bx)^N-1}{\psi^bx-1}.
\end{align}$$
This is of course making the assumption that $x\ne \phi^{-b},\psi^{-b}$.


问:
To find maximum value of a function
I am new to mathjax so couldn't format so I attached images regarding the problem this is the question.

Find the maximum value of
$$\left| \sqrt{x^4-3 x^2-6 x+13}-\sqrt{x^4+5 x^2+4}\right|$$

And this is my approach.In my approach I found that the equation is of the form $|PA-PB|$ which will be maximum when $P,A,B$ are collinear and $P$ must divide $A,B$ externally
But in this particular problem the points can't be collinear.I tried using derivative but that is big mess .so how to proceed further or any other methods?

答:
Since we're dealing with the absolute value of a function, we can interpret the absolute value as giving us the distance from the function to the $x$ axis independently from the direction. In other words, we just want to find the point that's furthest away from the $x$ axis, independently if it's above or below the axis since the distance (absolute value) is all we care about maximizing. With this in mind, it becomes clear that the maximum of $\lvert f(x) \rvert$ has to be either a maximum or a minimum of $f(x)$.

From here it's a fairly straightforward (although lengthy) calculation to find the result. We see that given $f(x) = \sqrt{x^4-3 x^2-6 x+13}-\sqrt{x^4+5 x^2+4}$, the extrema points are found when the derivative equals $0$, hence
\begin{align*}
\frac{d}{dx} \left(\sqrt{x^4-3 x^2-6 x+13}-\sqrt{x^4+5 x^2+4}\right)& = \frac{4x^3-6x-6}{2\sqrt{x^4-3 x^2-6 x+13}} - \frac{4x^3 +10x}{2\sqrt{x^4+5 x^2+4}} \tag{1}
\end{align*}
And by equating equation $(1)$ to $0$ we get
$$
\frac{4x^3-6x-6}{2\sqrt{x^4-3 x^2-6 x+13}} = \frac{4x^3 +10x}{2\sqrt{x^4+5x^2+4}}
$$
$$
\implies  \frac{\left(2x^3-3x-3\right)^2}{x^4-3 x^2-6 x+13} = \frac{\left(2x^3 +5x\right)^2}{x^4+5 x^2+4}
$$
$$
\implies \left(x^4+5 x^2+4\right)\left(4 x^6 - 12 x^4 - 12 x^3 + 9 x^2 + 18 x + 9\right) = \left(x^4-3 x^2-6 x+13 \right)\left(4 x^6 + 20 x^4 + 25 x^2 \right)
$$
$$
\implies \color{blue}{4 x^{10}} \mathbin{\color{brown}{-}} \color{brown}{12 x^{8}} - 12 x^{7} + 9 x^{6} + 18 x^5 + 9x^4 +  \color{green}{20 x^8} \mathbin{\color{purple}{-}} \color{purple}{60 x^6} - 60 x^5 + 45 x^4 + 90 x^3 + 45 x^2 + 16 x^6 - 48 x^4 - 48 x^3 + 36 x^2 + 72 x + 36 = \color{blue}{4 x^{10}} +  \color{green}{20 x^8} + 25 x^6 \mathbin{\color{brown}{-}} \color{brown}{12 x^{8}}  \mathbin{\color{purple}{-}} \color{purple}{60 x^6} - 75 x^4 -24 x^7 - 120 x^5 - 150 x^3 + 52 x^6 + 260 x^4 + 325 x^2
$$
$$
\implies  12 x^7 - 52 x^6 + 78 x^5 - 179 x^4 + 192 x^3 - 244 x^2 + 72 x + 36 =0
$$
$$
\implies (3 x^2 - 4 x + 6) (4 x^5 - 12 x^4 + 2 x^3 - 33 x^2 + 16 x + 6)=0
$$

This last implication can be verified using long division. Since we can easily show (using the quadratic formula, for example) that the roots of $3 x^2 - 4 x + 6$ are complex, we can rule them out since we're only interested in real solutions. This simplifies our problem to finding the roots of
$$
4 x^5 - 12 x^4 + 2 x^3 - 33 x^2 + 16 x + 6
$$
Unfortunately, this last expression can't be factored any further, and since there isn't a quintic formula to solve for the roots the best we can do is find the roots using numerical methods. By doing so, you find that the real roots are given by
\begin{align}
x_1 &\approx -0.24569\\
x_2 & \approx 0.687122\\
x_3 & \approx 3.44225
\end{align}
However, if we actually plug in $x_2$ into equation $(1)$ we see that we don't get $0$ (in fact it evaluates to approximately $-3.18$). This in fact is a "fake" root, since it is indeed a root of an equation that we obtained later in our procedure to find the roots, but it's not a cero to our derivative, which is what we're interested in. This leaves us with two extrema points of our function $f(x)$: $-0.24569$ and $3.44225$. By evaluating the function $f(x)$ at these points we see that
$$
 \sqrt{(-0.24569)^4-3 (-0.24569)^2-6 (-0.24569)+13}-\sqrt{(-0.24569)^4+5 (-0.24569)^2+4} \approx 1.706
$$
$$
\sqrt{(3.44225)^4-3 (3.44225)^2-6 (3.44225)+13}-\sqrt{(3.44225)^4+5 (3.44225)^2+4} \approx -4.1146
$$
And recalling that, because of the absolute value, we're just interested in the absolute value of these quantities, we see that
$$
\lvert f(3.44225) \rvert = \lvert -4.1146 \rvert = 4.1146 \mathbin{\color{red}{>}} 1.706 = \lvert f(-0.24569) \rvert
$$
And so, we can conclude that $x \approx 3.44225$ is the maximum value.

We can visually check our result by plotting the function $\left| \sqrt{x^4-3 x^2-6 x+13}-\sqrt{x^4+5 x^2+4}\right|$ as follows

In conclusion, the final result for the maximum can be stated as

The maximum value of $\left| \sqrt{x^4-3 x^2-6 x+13}-\sqrt{x^4+5 x^2+4}\right| $ is given by the largest positive root of $4 x^5 - 12 x^4 + 2 x^3 - 33 x^2 + 16 x + 6$ and has an approximate value of $3.44225$.

Title: To find maximum value of a function


问:
Compression of a von Neumann algebra
Let $M$ be a von Neumann algebra on $B(H)$. and let $p \in M$ be a projection How can we prove that the compression $pMp$ is a von Neumann algebra on $B(pH)$?

I saw a result that if $p \in M$ is a projection, then $(pMp)'=pM'p$. Using this can we conclude that

$(pMp)''=pM''p=pMp$ since $M$ is a von neumann algebra? I know if $M$ is a von Neumann algebra so is $M'$. But to apply the result  $(pMp)'=pM'p$ once more we need $p\in M'$ right?

Please help me.

答:
One cannot expect $p\in M'$ to hold in general.  (Take for example $M=M_2$ and $p=\bigl(\begin{smallmatrix}1&0 \\ 0&0\end{smallmatrix}\bigr)$.)

But why not instead prove that $pMp$ is a von Neumann algebra directly by showing that $pMp$ is a weakly closed $C^*$-subalgebra of $B(pH)$?

Interestingly, showing that $pMp$ is a von Neumann algebra by proving that it is the commutant in $B(pH)$ of the von Neumann algebra $pM'$ is more difficult but possible, see Corollary 5.5.7 of Kadison and Ringrose, and has the advantage of giving an expression of the commutant of $pMp$ in $B(pH)$.

Title: Compression of a von Neumann algebra


问:
Permutation matrix for $(13)(12)(23)(34)$?
for some reason I'm having a lot of trouble getting the correct Permutation matrix to the amalgamation of transpositions: $(13)(12)(23)(34)$

I'm pretty sure it's supposed to be:

\begin{bmatrix}
 0&0  & 0 & 1\\
0 & 1 & 0 &0 \\
 1& 0 & 0 & 0\\
0 & 0 & 1 &0
\end{bmatrix}

Is this true? I don't know why I'm having so much trouble getting this! If somebody could walk me through this that would be great. Thank you.

https://en.wikipedia.org/wiki/Permutation_matrix#Examples

答:
We assume left-to-right multiplication and simplify the product of transpositions in three ways. This might help to better see what's going on.

Cycle multiplication:

\begin{align*}
\left(\left((13)(12)\right)(23)\right)(34)&=\left((132)(23)\right)(34)\\
&\,\,\color{blue}{=(12)(34)}
\end{align*}

Two line notation:

\begin{align*}
&\left(\left(
\begin{pmatrix}1&2&3&4\\3&2&1&4\end{pmatrix}\circ
\begin{pmatrix}1&2&3&4\\2&1&3&4\end{pmatrix}
\right)\circ
\begin{pmatrix}1&2&3&4\\1&3&2&4\end{pmatrix}\right)\circ
\begin{pmatrix}1&2&3&4\\1&2&4&3\end{pmatrix}\\
&\qquad=
\left(
\begin{pmatrix}1&2&3&4\\3&1&2&4\end{pmatrix}
\circ
\begin{pmatrix}1&2&3&4\\1&3&2&4\end{pmatrix}\right)\circ
\begin{pmatrix}1&2&3&4\\1&2&4&3\end{pmatrix}\\
&\qquad=
\begin{pmatrix}1&2&3&4\\2&1&3&4\end{pmatrix}\circ
\begin{pmatrix}1&2&3&4\\1&2&4&3\end{pmatrix}\\
&\qquad\,\,\color{blue}{=
\begin{pmatrix}1&2&3&4\\2&1&4&3\end{pmatrix}}\\
\end{align*}

Matrix notation:

\begin{align*}
&\left(\left(\begin{pmatrix}0&0&1&0\\0&1&0&0\\1&0&0&0\\0&0&0&1\end{pmatrix}\cdot
\begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\right)\cdot
\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}\right)\cdot
\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}\\
&\qquad=\left(\begin{pmatrix}0&0&1&0\\1&0&0&0\\0&1&0&0\\0&0&0&1\end{pmatrix}\cdot
\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}\right)\cdot
\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}\\
&\qquad=\begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\cdot
\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}\\
&\qquad\,\,\color{blue}{=\begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}}
\end{align*}


问:
Prove that in a triangle $ABC$ $\forall P, P\neq M_{a},M_{b} ,M_{c}:min\{ \frac{|PA|}{|PM_{a}|},\frac{|PB|}{|PM_{b}|},\frac{|PC|}{|PM_{c}} \}\le 2.$
Suppose a triangle $ABC$. Define $M_{a},M_{b} ,M_{c}$ as the midpoints of the sides $a,b,c$ respectively. Then prove that for any point $P$ different from $M_{a},M_{b} ,M_{c}$, the smallest ratio of magnitudes of lines from $P$ to $M_{x}$ and from $P$ to $x$ is $\le 2$.

I have had certain success in prooving it for special cases, such as triangle $ABP$ being isosceles, but I cannot see a way to generalize this properly. I would think this is a pretty important (or at least very interesting result) but I have not seen it anywhere before (or its proof) so I'm intrigued to hear your ideas.

答:
This follows from the fact that centroid divides median into $2:1$.

WLOG, point $P \in \triangle ACG$ (or in $\triangle BCG$ or in $\triangle ABG$) where $G$ is the centroid of the $\triangle ABC$.

So $AG + CG \geq AP + CP \implies AG \geq AP  \, $ or $\, CG \geq CP$ or both.

If $AG \geq AP,$ (you can show the same for $CG \geq CP$)

Using triangular inequality,

$AP + PD \geq AG + GD \implies PD \geq GD \,$ (as $AP \leq AG$)

So, $\frac{AP}{PD} \leq \frac{AG}{GD} = 2$

Equality occurs when point $P$ is the centroid.


问:
$u$-substitution seems arbitrary

Evaluate: $$ \int \frac{x^2}{\sqrt{1-x^2}}\,dx$$

The solution I came across does a $u$-substitution by letting $x = \sin(t)$. But why $\sin(t)$? It seems a lot like guessing to me - should I just guess the right $x$ for substitution? Why not $\cos(t)$? Is there any way that I could identify that "this" expression is solved by "that" trig substitution?

答:
I think that the OP's question refers to how to distinguish why to make this change of variable. Now, I will write the answer to that question.

Suppose you want to solve an integral of the form $$\color{blue}{\boxed{\int R(x,\sqrt{a^{2}-x^{2}})dx}}$$
As in your problem, that you have $$\color{red}{\text{Example:} \quad  \int \frac{x^{2}}{\sqrt{1^{2}-x^{2}}}dx}$$So, this integral becomes a trigonometric integral with the change of variable $$\color{green}{\boxed{x=a\cos(t)}} \quad \color{green}{\boxed{x=a\sin(t)}}$$
As the MathS users indicate in the comments. Also note that you can also perform the variable change $$\color{green}{\boxed{x=a\tanh(t)}}$$ and this variable change transforms the integral into a hyperbolic integer.

Title: $u$-substitution seems arbitrary


答: Here's your figure.  Just let $r = \sqrt{x^2 + y^2}$ and proceed.

Title: Iterated integral $\int_D \sqrt{x^2 +y^2} \ dx \ dy \ dz$

Body:
Compute $$\int_D \sqrt{x^2 +y^2} \ dx \ dy \ dz$$ when $D = \{(x,y,z) \in \mathbb{R}^3 \mid x^2+y^2 \leqslant 1, 0\leqslant z \leqslant 7 \}.$

I managed to get the triple integral

$$\int_{0}^{7}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \sqrt{x^2+y^2} \ dx \ dy \ dz$$

but this isn't trivial to compute at all. $\sqrt{x^2+y^2}$ seems to define a surface in $\mathbb{R^3}$ that looks like a cone, should I consider polar or cylindrical coordinates here or is there something else I'm missing?


答: Draw a triangle!  When looking at the term $\sqrt{1 - x^{2}}$ in the integrand, you should immediately be reminded of the Pythagorean Theorem.  If you draw a right triangle with hypotenuse length $1$, and side lengths $x$ and $\sqrt{1 - x^{2}}$, you end up with the following triangle.  I've labelled the remaining two angles $s$ and $t$.  So, if you want to isolate $x$ for a potential change of variables, based on the triangle, you can either use $\sin(s) = x$ or $\cos(t) = x$.  Either would give you a valid change of variables.

Title: $u$-substitution seems arbitrary

Body:
Evaluate: $$ \int \frac{x^2}{\sqrt{1-x^2}}\,dx$$

The solution I came across does a $u$-substitution by letting $x = \sin(t)$. But why $\sin(t)$? It seems a lot like guessing to me - should I just guess the right $x$ for substitution? Why not $\cos(t)$? Is there any way that I could identify that "this" expression is solved by "that" trig substitution?


问:
Klein bottle is a manifold
How do we prove Klein bottle is a 2-manifold?

Define $f(x,y)=(x,y+1)$ and $g(x,y)=(x+1,1-y)$, and $G$ is the group generated by $f$ and $g$.

I know Klein bottle can be defined by the quotient $q:\mathbb{R}^2\rightarrow\mathbb{R}^2/G$.

How do we find an open set $U$ for each point in the Klein bottle s.t. $U\approx\mathbb{R}^2$?

I know $q$ is a covering space map. Will that help?

答:
$q$ being a covering space map is indeed very useful. It means that each point in the quotient has a neighborhood $U$ that is mapped by $q^{-1}$ to a discrete collection of open sets, each of them homeomorphic to $U$, and each of them an open subset of $\Bbb R^2$.

By possibly taking an even smaller neighborhood inside $U$, you can easily make this homeomorphic to $\Bbb R^2$.

Title: Klein bottle is a manifold


问:
How should I approach this integral?
I am trying to work out the following integral:
$$
\int_{-\infty}^{\infty}
\left\vert\frac{1}{1 + \alpha\,\iota}\,
\exp\left(-\,\frac{\pi x^{2}}
{\,\sqrt{\,{1 + \alpha\,\iota}\,}\,}\right)
\,\right\vert\mathrm{d}x
$$
Edit: $\alpha$ is just some constant and $\iota$ is supposed to be the imaginary unit. I was thinking that I can take the absolute of $\frac{1}{1+\alpha\iota}$ and throw that outside the integral. Then for ${\,\sqrt{\,{1 + \alpha\,\iota}\,}\,}$ I could find the square root, find the reciprocal and exploit $\left|e^{c\iota}=1\right|$ for any constant c and the fact that what remains looks like the pdf of a standard normal? Does that sound reasonable?

Thanks.

答:
As a hint:

$$\left|\frac{1}{1+ a I}\right| = \frac{1}{\sqrt{1+a^2}}$$

and
$$\left|\exp\left(-\pi \frac{x^2}{\sqrt{1+ a I}}\right)\right|  = \exp\left(-\pi x^2 Re\left(\frac{1}{\sqrt{1 + a I}}\right)\right),$$
where
$$Re\left(\frac{1}{\sqrt{1 + a I}}\right) = \frac{\cos(\arctan(a)/2)}{(1 + a^2)^{1/4}}.$$

Now the integral can be solved as a usual squared exponential integral.

Title: How should I approach this integral?


问:
Find the area of the surface defined by $x-y-z=1$ and $2x^2+y^2 \leq 1$

Find the area of the surface defined by $x-y-z=1$ and $2x^2+y^2 \leq 1$

I've tried to solve this using Green's theorem, but I'm not getting the correct answer.

The curve I get when I intersect those two surfaces is $2x^2+y^2=1$, so I get the parametric equation $\alpha(t)=(\frac{1}{\sqrt2}\cos(t), \sin(t))$ with $t \in [0, 2\pi]$.

And I choose the field $X(x,y)=(0,x)$ such that $Q_x-P_y=0$, where $Q=x$ and $P=0$

So, by Green's theorem: $A(S)=\int\int _{S} dA = \int _{0} ^{2\pi} X(\alpha(t)) \cdot \alpha'(t) dt=\frac{1}{\sqrt2}\int _{0} ^{2\pi} \cos^2(t)dt=\frac{\pi}{\sqrt2}$, which is incorrect

The correct answer is $\frac{\sqrt3\pi}{\sqrt2}$

答: This is an oval with primary axes of length $2$ and $\sqrt{6}$, so by the formula for the area of an elliptical region the answer is $\sqrt{3/2} \pi$.


问:
Is a finite metric space complete?
If $(X,d)$ is a metric space with a finite number of elements, is $X$ complete?

I know that every finite metric space is compact, and also compact spaces are complete. So, is this true?

答: It's easy to see that what you actually have here is a discrete space.  Hence every Cauchy sequence is eventually constant.  Hence convergent.</p>

Title: Is a finite metric space complete?


问:
How to show that $\det(A+I)\ne 0$
How to show that for any skew symmetric real matrix $A$, we have $\det(A+I)\ne 0?$

Where to begin? I'm looking for some clue only.

答: Hint: What are the eigenvalues of a real skew symmetric matrix like? If $v$ is an eigenvector of $A$, what is $(A+I)v$? </p>

Title: How to show that $\det(A+I)\ne 0$


问:
What happens to a mapping when ad - bc = 0?
So, I know that when you have a mapping of the form $\frac{az + b}{cz + d}$, with $ad − bc \neq 0$, this gives you a Möbius transformation. But if $ad − bc = 0$, this gives you a constant mapping. I'm trying to figure out how to prove that latter statement - for a mapping $m(z) = \frac{az + b}{cz + d}$, if $ad − bc = 0$, then $m(z)$ is a constant map. I've been trying to think of the best way to prove this, but I keep getting stuck because I find myself making assumptions that potentially result in loss of generality (i.e. assuming $c$ and/or $a$ is $0$).

What is the best way to prove this? Should I maybe try a proof by contradiction? That is, assume that the mapping is not constant, and arriving at a contradiction?

答:
$ad-bc$ is the determinant of the following matrix:

$A = \begin{bmatrix}
    a       & b\\
    c       & d\\ \end{bmatrix}$

Since the determinant is zero, the rows of this matrix are linearly dependent. So there is some $\lambda\in\mathbb{C}$ such that $\lambda c=a$ and $\lambda d=b$. And then:

$m(z)=\frac{az+b}{cz+d}=\lambda\frac{cz+d}{cz+d}=\lambda$


答:
Let's take the square root which preserves the quadrant; i.e.
$$
\sqrt{1+\alpha i}=\sqrt{\frac{1+\sqrt{1+\alpha^2}}2}+i\operatorname{sgn}(\alpha)\sqrt{\frac{-1+\sqrt{1+\alpha^2}}2}
$$
Then, define $\beta$ as the real part of the reciprocal:
$$
\frac1{\sqrt{1+\alpha i}}=\underbrace{\sqrt{\tfrac{1+\sqrt{1+\alpha^2}}{2\left(1+\alpha^2\right)}}}_\beta-i\operatorname{sgn}(\alpha)\sqrt{\tfrac{-1+\sqrt{1+\alpha^2}}{2\left(1+\alpha^2\right)}}
$$
Furthermore, note that $\left|e^{x+iy}\right|=e^x$
$$
\begin{align}
\int_{-\infty}^\infty\left|\frac1{1+\alpha i}\exp\left(-\frac{\pi x^2}{\sqrt{1+\alpha i}}\right)\right|\,\mathrm{d}x
&=\frac1{\sqrt{1+\alpha^2}}\int_{-\infty}^\infty\exp\left(-\pi\beta x^2\right)\,\mathrm{d}x\\
&=\frac1{\sqrt{\beta\left(1+\alpha^2\right)}}\\
&=\sqrt[\large 4]{\frac2{\left(1+\alpha^2\right)\left(1+\sqrt{1+\alpha^2}\right)}}
\end{align}
$$

Title: How should I approach this integral?

Body:
I am trying to work out the following integral:
$$
\int_{-\infty}^{\infty}
\left\vert\frac{1}{1 + \alpha\,\iota}\,
\exp\left(-\,\frac{\pi x^{2}}
{\,\sqrt{\,{1 + \alpha\,\iota}\,}\,}\right)
\,\right\vert\mathrm{d}x
$$
Edit: $\alpha$ is just some constant and $\iota$ is supposed to be the imaginary unit. I was thinking that I can take the absolute of $\frac{1}{1+\alpha\iota}$ and throw that outside the integral. Then for ${\,\sqrt{\,{1 + \alpha\,\iota}\,}\,}$ I could find the square root, find the reciprocal and exploit $\left|e^{c\iota}=1\right|$ for any constant c and the fact that what remains looks like the pdf of a standard normal? Does that sound reasonable?

Thanks.


问:
Solution to non-linear PDE
I think I have found a solution for a PDE of the form

$u_t + g(u)u_x = 0$

where $u(x, 0) = g^{-1}(x)$

The solution is  $u(x,y) = g^{-1}\left(\frac x{t+1}\right)$

This solution satisfies 1 and 2 under the assumption that $\forall z, g\left(g(z)^{-1}\right) = z$

However I am worried about the effects of discontinuities in $g$ or its inverse, and issues where the function is not 1-1.

What sort of problems should I watch out for and how can I get around them?

答:
The classical solution $$
u(x,t) = g^{-1}\left(\frac{x}{t+1}\right)
$$
follows from the method of characteristics, which requires that the solutions remains smooth along the characteristic curve as time increases. Smoothness is lost at the breaking time $$
t_b = \inf \frac{-1}{(g\circ g^{-1})'(x)} = -1 \, .
$$
This can be seen in the solution's denominator which vanishes at $t=-1$. Therefore, if only positive times are considered, no shock wave will form!
Now, if there are some discontinuities in $g$ or $g^{-1}$, the problem statement makes little sense as $g$ is no longer invertible over its whole domain of definition.

Title: Solution to non-linear PDE


问:
Why we have for any choice of descent direction $s_n$, the (n+1)th iterate obtained with exact line search satisfies $\nabla f(x_{n+1}).s_n=0$?
Why we have for any choice of descent direction $s_n$, the (n+1)th iterate obtained with exact line search satisfies $\nabla f(x_{n+1}).s_n=0$ ?

I don't see why they should be orthogonal at all.

Thanks in advance for any help!

答: If $t^*$ minimizes $g(t)=f(x_n + t s_n)$, then $g'(t^*)=0$. Use the chain rule to show that $g'(t^*) = \nabla f(x_n + t^* s_n) \cdot s_n$.


问:
Number of Zonotope Edges Parallel to Generator
Suppose we have a zonotope $Z$ that is the Minkowski sum of line segments $U_1+\dots +U_n$. All the edges of $Z$ are parallel to some $U_i$. Is it also true that the number of edges parallel to $U_i$ equals the number of edges parallel to $U_j$ for every $j\neq i$? Surely this question has been asked and probably answered before. I looked through a page of zonohedra and verified that they all answer the question true.

https://www.ics.uci.edu/~eppstein/junkyard/ukraine/ukraine.html

It is known that $Z$ is a zonotope iff $Z$ is the projection of a cube from $\mathbb{R}^n$. Of course, a cube in $\mathbb{R}^n$ is the sum of $n$ orthonormal vectors and has the same number of edges parallel to each of its generators. I suspect there may be a good induction proof down this way by adding one generator, $v$, at a time and  but haven't found one. The new generator divides the old zonotope into two pieces. The edges parallel to $v$ will stretch between the two halves.

Any help is appreciated!
Thanks! Aaron

答:
The thing is more or less obvious as long as the set of your spanning vectors $\{U_i\}_{i\in I\ }$ is such that none of those is being contained within the spanned cone of the others. For then all edges will be exterior to the produced zonotope.

Now consider the simple case of 3 coplanar vectors. Any 2 of them would span a rhomb and the effect by the respective third is to expand that into an hexagon. So again your proposition is true.

--- rk


问:
Does proof of FOL undecidability require tacit appeal to the Church-Turing Thesis?
We can prove that FOL is undecidable using a strategy based on the undecidability of Q. But does this latter proof require tacit appeal to the Church-Turing Thesis?</p>

答:
No, it does not. However, if you're not careful about interpreting the statement

$(*)\quad$ $\mathsf{Q}$ is undecidable,

it may seem necessary.

See also my answer to essentially the same question on philosophy.stackexchange.

"Undecidability" is a technical term referring to the specific formal notion of computation given by Turing machines, and $(*)$ above is a genuine formal theorem to which the Church-Turing thesis is totally irrelevant. This will become clear if you read the proof in detail.

Note btw that Godel's incompleteness theorem predated the introduction of Turing machines, and hence Godel's acceptance of the Church-Turing thesis, by five years. Of course Godel wasn't talking about $\mathsf{Q}$, but Robinson introduced $\mathsf{Q}$ as a finitely axiomatizable theory for which Godel(/Rosser)'s argument goes through without serious change. So we can see right off the bat that Church-Turing can't possibly be relevant here.

On the other hand, we also have the statement

$(**)\quad$ There is no algorithm for determining whether a sentence is a theorem of $\mathsf{Q}$. Or put another way, the problem of telling whether a sentence is a theorem of $\mathsf{Q}$ is not effectively solvable.

Now there's a crucial issue here, in that $(**)$ is an informal statement: "algorithm" and "effective solvability" are informal terms, referring to some imagined ideal notion of human computation. One consequence of the Church-Turing thesis is that $(*)\iff (**)$, and so the formal (and thesis-free) proof of $(*)$ tells us that $(**)$ is true. But this is the only place where the thesis comes in: it's not relevant to the proof of $(*)$, only to the claim that $(*)$ and $(**)$ "mean the same thing."


答:
You can do it by cases. Suppose $c=0$. Then $ad=bc$ implies $a=0$ or $d=0$. But for $d=0$, the map is well-defined, so $a=0$ and you're done. Similarly, if $a=0$, then $b$ or $c$ is $0$, and you're done.

If $a,c \neq 0$, then
$$\frac{az+b}{cz+d} = \frac{a(az+b)}{acz+ad} = \frac{a(az+b)}{acz+bc}= \frac ac \frac{az+b}{az+b} = \frac ac.$$

Title: What happens to a mapping when ad - bc = 0?

Body:
So, I know that when you have a mapping of the form $\frac{az + b}{cz + d}$, with $ad − bc \neq 0$, this gives you a Möbius transformation. But if $ad − bc = 0$, this gives you a constant mapping. I'm trying to figure out how to prove that latter statement - for a mapping $m(z) = \frac{az + b}{cz + d}$, if $ad − bc = 0$, then $m(z)$ is a constant map. I've been trying to think of the best way to prove this, but I keep getting stuck because I find myself making assumptions that potentially result in loss of generality (i.e. assuming $c$ and/or $a$ is $0$).

What is the best way to prove this? Should I maybe try a proof by contradiction? That is, assume that the mapping is not constant, and arriving at a contradiction?


答:
You do have to worry about picky corner cases where you're dividing by zero. To be thorough you must rule out the case where both $c$ and $d$ are zero. With that case out of the way, assume $c\ne0$. This means you can divide by $c$, obtaining $b=\frac{ad}c$, which you then plug in:
$$
\frac{az+b}{cz+d}=\frac{az+\frac{ad}c}{c(z+\frac dc)}=\frac ac.$$
The remaining case $d\ne0$ is handled similarly.


问:
why the increment doesnt affect the randomness?
I'm doing some homework and I need to answer why the increment (b) doesn't affect randomness in the mixed congruential method.

The formula is

$$X_{n+1} \equiv (a X_n + b) \mod m$$

答:
Answer given by Robert Israel in the comments:

Assuming $\mathrm{gcd}\,(a−1,m)=1$, there is some $d$ such that $Y_n=X_n+d \mod m$ satisfies $Y_{n+1}=aY_n \mod m$, so all the increment accomplishes is a constant shift in the sequence.


问:
$ \lim\limits_{n \to \infty} \| x_{n}\| = \| x \|$ if $\left(x_{n}\right)$ weakly convergent
I want to show that for a banach space $X, U \subset X$ non-empty and compact and $\left(x_{n}\right) \subset U$ a weakly convergent sequence with limit $x$, it converges in norm to $\mathrm{x}$

We know that  $\quad f\left(x_{n}\right) \rightarrow f(x) \: \:  \forall f \in X’$.

So

$$ \| x \| = sup_{\| f \| = 1} \| f(x) \| \\
 = sup_{\| f \| = 1} \lim\limits_{n \to \infty}  \| f(x_{n}) \| \\
= \lim\limits_{n \to \infty}sup_{\| f \| = 1} \| f(x_{n}) \| \\
= \lim\limits_{n \to \infty}  \| x_{n}\|$$

However, is it valid that I can interchange the limit and supremum?

答: Indeed it is true. However, I would argue differently. Pick some subsequence $(x_{n_k})_{k}$. By compactness it admits a convergent subsequence. However, as the strong and the weak limit coincide, it comverges strongly to $x$. This implies that every subsequence of $(x_n)_n$ admits a subsequence converging strongly to $x$ and thus the whole sequence converges strongly to $x$.


问:
What's other possible ways to justify that?
Let A be an n×m matrix and let D be a matrix such that AD=In (where In is the n×n identity matrix). Show that the equation Ax = b has a solution for every b ∈ Rn.

I tried to assume the matrix A is invertible, but it's wrong because it may not even be square.

答: Let $C(A)$ denote column space of $A$ and $R(A)$ denote row space of $A$. We know that $C(AB) \subset C(A)$ and $R(AB) \subset R(B)$. By using this fact we know $C(A) \supset C(I_n)=\mathbb{R}^n$, so thus $C(A)=\mathbb{R}^n$. This means that the set of all linear combination of the columns of $A$ is $\mathbb{R}^n$. Hence for any $b \in \mathbb{R}^n$, we have $x_1,...,x_m$ such that $x_1v_1+...+x_mv_m=b$ where $v_i$ is $i^{th}$ column vector of $A$. If we choose $x=(x_1,...,x_m)^T$, we have $Ax=b$