问:
Notions of continuity for stochastic processes
I would like to receive some clarification regarding the difference between continuous in probability and continuous almost surely. Using the definition of the wikipedia page (that match the one I have seen in other references), we have

Continuous in probability
for all $\varepsilon>0$
\begin{equation}
\lim_{s\rightarrow t} \mathbb{P}\left(\left\{\omega \in \Omega : |X_s(\omega) - X_t(\omega) | \geq \varepsilon\right\} \right)= 0
\end{equation}

and

Continuous with probability 1 (almost surely)
\begin{equation}
\mathbb{P}\left(\left\{\omega \in \Omega : \lim_{s\rightarrow t} |X_s(\omega) - X_t(\omega) | = 0 \right\} \right)= 1
\end{equation}

Now, I can see that the second condition is stronger than the first one, but to me they seem analogous, since the first should be valid for arbitrarily close $\varepsilon$. Obviously, my intuition is wrong, but I cannot understand how the difference can be significant. Could you please explain how the two definitions give raise to different processes? Thank you!

答:
[Note: I am answering this question because I see no replies and I was also interested, like the OP, in understanding better this difference and the intuitive aspects. Take this as just an attempt and please, feel free to correct, add, make precise, as appropriate, so that I can also learn from my errors.]

The almost sure continuity is stating that each of the functions $X_\omega(s)$ that you have, for each one of the ω in the space of outcomes $\Omega$, is continuous (in the usual math sense). And that the possible exceptions $\omega's$ have an overall probability equal to $0$.

The continuity in probability does not guarantee the continuity of "almost all" (i.e., with probability $1$) the sample paths $X_\omega(s)$, as we have seen before.
On the contrary, each sample-path, for each $\epsilon>0$ you use, and no matter how close you get $s$ to $t$, is allowed to have still one index $s < s' < t$ (and, therefore, infinite of them) where it keeps violating the inequality $| X_\omega(t) - X_\omega(s) | < \epsilon$ (so that the continuity limit does not hold).
However, if we consider, at each index $s$, all the $\omega$ where the corresponding paths, on that index, are having the "violation" $| X_\omega(t) - X_\omega(s) | < \epsilon$, their probability must tend to zero, as $s$ tends to $t$.

Example (please, correct me if I got this one wrong):

```<code>    X(s)= Be( min(1, t-s) ) , for s < t
    X(t) = 0
    X(s)= U(0, s-t)         , for s > t
</code>```

where $Be(p)$ denotes a Bernoulli with success probability $p$ and $U(a,b)$ denotes a Uniform in $(a,b)$.

From the right, this indexed family $X(s)$ has a limit of $0$, surely.

From the left, the limit does not hold, because infinite $1's$ will keep preventing the function from getting definitely close to $0$. However, the probability of getting those sporadic "violations" is tending to $0$, as $s$ tends to $t$.

[UPDATE: more details]

Thank you for the upvotes. I am encouraged to put in the answer some more details, which I previously glossed over because I was mostly concerned with focusing on the "intuitive" aspects for the OP and myself. Again, feel free to jump in and correct/make precise where I am exaggerating with "intuition" and hurting rigor.

The reason for the infinite $1's$ which keep popping out in the left limit to $t$ in my example above (Bernoulli with decreasing success probability), is essentially the fact that:

1.) for each $s$, the probability of having a "violation" $| X_\omega(s) - X_\omega(t) | < \epsilon$ is non zero

and ($1.$ alone would not suffice, because the probability might decrease, as it does in this case)

2.) as $s$ tends to $t$, the probability of having a violation decreases, but not fast enough" to quit "fuelling" the infinite appearances of the value $1$. More precisely, the limit of the sum of the violation probabilities, over all $s < t$, is exploding to infinity, instead of converging (a continuous analog of the harmonic series).

If, for instance, instead of the $1/(t-s)$ probability, we consider its square, the overall "violation" probability from $s$ to $t$ would instead converge (a continuous analog of Basel problem) to a constant and there would be no probability left to "fuel" an infinite sequence of $1's$ to cause the limit inequality violations.

And this is, essentially, at least as I see it, the intuition with the Borel-Cantelli Lemma (which originally addressed the discrete case, though).

This is also the reason why the fact that the "individual" (for each $s$) violation probability goes to $0$ is not enough to prevent the phenomenon of occurrence of "infinite violations" (which will prevent the continuity limit to happen). In other words, the reason why convergence in probability does not cause convergence almost sure.

What you will need to make the convergence in probability imply the convergence almost sure, is that the integral, over all $s < t$, of the violation density (the overall probability to get a violation on $(s,t)$) to be finite, in order to stop "fuelling" the infinite appearance of those violations and thus allow the continuity limit to take place.

Body:
[Note: I am answering this question because I see no replies and I was also interested, like the OP, in understanding better this difference and the intuitive aspects. Take this as just an attempt and please, feel free to correct, add, make precise, as appropriate, so that I can also learn from my errors.]

The almost sure continuity is stating that each of the functions $X_\omega(s)$ that you have, for each one of the ω in the space of outcomes $\Omega$, is continuous (in the usual math sense). And that the possible exceptions $\omega's$ have an overall probability equal to $0$.

The continuity in probability does not guarantee the continuity of "almost all" (i.e., with probability $1$) the sample paths $X_\omega(s)$, as we have seen before.
On the contrary, each sample-path, for each $\epsilon>0$ you use, and no matter how close you get $s$ to $t$, is allowed to have still one index $s < s' < t$ (and, therefore, infinite of them) where it keeps violating the inequality $| X_\omega(t) - X_\omega(s) | < \epsilon$ (so that the continuity limit does not hold).
However, if we consider, at each index $s$, all the $\omega$ where the corresponding paths, on that index, are having the "violation" $| X_\omega(t) - X_\omega(s) | < \epsilon$, their probability must tend to zero, as $s$ tends to $t$.

Example (please, correct me if I got this one wrong):

    X(s)= Be( min(1, t-s) ) , for s < t
    X(t) = 0
    X(s)= U(0, s-t)         , for s > t

where $Be(p)$ denotes a Bernoulli with success probability $p$ and $U(a,b)$ denotes a Uniform in $(a,b)$.

From the right, this indexed family $X(s)$ has a limit of $0$, surely.

From the left, the limit does not hold, because infinite $1's$ will keep preventing the function from getting definitely close to $0$. However, the probability of getting those sporadic "violations" is tending to $0$, as $s$ tends to $t$.

[UPDATE: more details]

Thank you for the upvotes. I am encouraged to put in the answer some more details, which I previously glossed over because I was mostly concerned with focusing on the "intuitive" aspects for the OP and myself. Again, feel free to jump in and correct/make precise where I am exaggerating with "intuition" and hurting rigor.

The reason for the infinite $1's$ which keep popping out in the left limit to $t$ in my example above (Bernoulli with decreasing success probability), is essentially the fact that:

1.) for each $s$, the probability of having a "violation" $| X_\omega(s) - X_\omega(t) | < \epsilon$ is non zero

and ($1.$ alone would not suffice, because the probability might decrease, as it does in this case)

2.) as $s$ tends to $t$, the probability of having a violation decreases, but not fast enough" to quit "fuelling" the infinite appearances of the value $1$. More precisely, the limit of the sum of the violation probabilities, over all $s < t$, is exploding to infinity, instead of converging (a continuous analog of the harmonic series).

If, for instance, instead of the $1/(t-s)$ probability, we consider its square, the overall "violation" probability from $s$ to $t$ would instead converge (a continuous analog of Basel problem) to a constant and there would be no probability left to "fuel" an infinite sequence of $1's$ to cause the limit inequality violations.

And this is, essentially, at least as I see it, the intuition with the Borel-Cantelli Lemma (which originally addressed the discrete case, though).

This is also the reason why the fact that the "individual" (for each $s$) violation probability goes to $0$ is not enough to prevent the phenomenon of occurrence of "infinite violations" (which will prevent the continuity limit to happen). In other words, the reason why convergence in probability does not cause convergence almost sure.

What you will need to make the convergence in probability imply the convergence almost sure, is that the integral, over all $s < t$, of the violation density (the overall probability to get a violation on $(s,t)$) to be finite, in order to stop "fuelling" the infinite appearance of those violations and thus allow the continuity limit to take place.


问:
Continuous function defined on compact set is uniformly continuous
Here's what I'm trying to prove:

Let $K \subseteq \mathbb{R}^n$ be a compact set. If $f: K \to \mathbb{R}^p$ is a continuous map, then $f$ is uniformly continuous.

Proof Attempt:

Let $v(x)$ denote the norm of $x$. $v_1$ will denote the norm on $\mathbb{R}^n$ and $v_2$ will denote the norm on $\mathbb{R}^p$.

Suppose that $f$ is not uniformly continuous. Then:

$$\exists \epsilon > 0: \forall \delta > 0: \exists x,y \in K: v_1(x-y) < \delta \land v_2(f(x)-f(y)) \geq \epsilon$$

Construct two sequences $(x^{(k)})_{k \in \mathbb{N}}$ and $(y^{(k)})_{k \in \mathbb{N}}$ such that:

$$\forall k \in \mathbb{N}: v_1(x^{(k)}-y^{(k)}) < \frac{1}{k} \land v_2(f(x^{(k)})-f(y^{(k)})) \geq \epsilon$$

The compactness of $K$ guarantees the existence of two subsequences $(x^{(k_m)})_{m \in \mathbb{N}}$ and $(y^{(k_m)})_{m \in \mathbb{N}}$ with limits $x$ and $y$. By the definition of the two sequences formed earlier, we can see that $x = y$ (using the Squeeze Theorem, for example).

By the continuity of $f$, it follows that:

$$\lim_{m \to \infty} [f(x^{(k_m)})-f(y^{(k_m)})] = f(x)-f(y) = 0$$

which is impossible because the above would guarantee that the norm of $f(x^{(k_m)})-f(y^{(k_m)})$ would be less than the $\epsilon$ given above for large enough $m$. $\Box$

Does the proof above work? If it doesn't, then why? How can I fix it?

Edit:

For that statement in the middle, I can justify it properly. Let $(X,d)$ be a metric space and suppose that $(x_n)$ and $(y_n)$ are sequences with limits $x$ and $y$ respectively. Then, $d(x_n,y_n) \to d(x,y)$.

We have, by the triangle inequality:

$$d(x_n,y_n) \leq d(x_n,x)+d(x,y)+d(y,y_n)$$

$$d(x,y) \leq d(x,x_n)+d(x_n,y_n)+d(y_n,y)$$

So, both inequalities above give us:

$$d(x,y)-d(x,x_n)-d(y,y_n) \leq d(x_n,y_n) \leq d(x_n,x)+d(x,y)+d(y,y_n)$$

Since $d(x_n,x) \to 0$ and $d(y,y_n) \to 0$, it follows that $d(x_n,y_n) \to d(x,y)$ and this is exactly what I used in the proof above.

答:
I'm posting as an answer because I need more reputation to comment.. Anyway, I did'nt see nothing wrong except for the part: "Construct two sequences $(^{()})$ and $(^{()})$ such that:

$$_1(^{()}−^{()})<\frac{1}{k}$$

and:

$$v_2(f(x^{(k)})-f(y^{(k)})) \geq \epsilon$$

Thats dont break the hypothesis of continuity?

ContentLicense: CC BY-SA 4.0


问:
Solving for a differential equation Gompertz growth equation
What is the general solution of this differential equation?
$$ \frac{dy}{dt} = k \enspace  y \enspace \ln(\frac{a}{y})$$ where $a$ and $k$ are positive constants.

So far, my solution is:
$$ \frac{dy}{y \enspace \ln(\frac{a}{y})} = k \enspace dt$$
When I let $u=lny$,
$$ \int \frac{1}{y(\ln a-\ln y)} \,dy =\int k dt $$
$$ \int \frac{1}{\ln a-u} \,du= kt + C $$

How to continue this?

答:
$$y'=k y (\log a-\log y)$$
$$kdt=\frac{dy}{y(\log a-\log y)}$$
Let $\log y=u\to\frac{dy}{y}=du$
$$kt=\int \frac{du}{\log a-u}$$
$$kt=-\log (\log a-u)+C$$
$$C-kt=\log (\log a-\log y)$$
$$C-kt=\log\log\frac{a}{y}$$
$$e^{C-kt}=\log\frac{a}{y}$$

The general solution is $$y=a \exp(-e^{C-k t})$$


问:
$a/p+b/q>1$ with certain conditions
Let $a$ and $b$ be positive integers and $p$, $q$ distinct primes such that $aq\equiv 1\pmod p$ and $bp\equiv 1\pmod q$ prove that $a/p+b/q>1$. My approach was to notice that $pn+aq=qm+bp$ for some integers $n$ and $m$ (from the congruences), which is equivalent to $q(a-m)=p(b-n)$, so $a-m = px$ and $b-n = qy$, for some integers $x$ and $y$. Using the last two equalities we have $a/p+b/q = [(a-m)+m]/p+[(b-n)+n]/q>px/p+qy/q>2>1$.
My biggest concern is if I can use $x,y,n,m$ as $natural$ $numbers$ and im also concerned about the fact that I got that the expression is $>2$ which seems a very huge bound. If there is a flaw in my solution pls. point it. If the solution is completly incorrect pls show me yours. Any help appreciated.

答:
Your concerns are not unfounded.

First of all, if $0 < a < p$ and $0 < b < q$, which certainly is possible, then
$$\frac{a}{p} + \frac{b}{q} < \frac{p}{p} + \frac{q}{q} = 2\,.$$
It follows that in your argument, if everything works up to that point, you cannot have both of $x$ and $y$ be positive integers. More precisely it follows that $x + y \leqslant 1$.

We can explicitly parametrise the pairs $(m,n)$ such that $pn + aq = qm + bp$: One immediately verifies that $m_0 = a$, $n_0 = b$ works. Since changing $m$ changes the right hand side by a multiple of $q$ and changing $n$ changes the left hand side by a multiple of $p$, any two possible choices of $m$ differ by some multiple of $p$ and any two possible choices of $n$ differ by some multiple of $q$. Hence we have $m = m_0 + sp$ and $n = n_0 + tq$ for some integers $s,t$. Plugging this in, we see that we have a solution if and only if $s = t$, thus the solutions to $pn + aq = qm + bp$ are the pairs $(m_t,n_t)$, $t \in \mathbb{Z}$, where $m_t = a + tp$ and $n_t = b + tq$.

Now writing
$$\frac{a}{p} + \frac{b}{q} = \frac{(a - m_s) + m_s}{p} + \frac{(b - n_t) + n_t}{q} \tag{1}$$
is certainly possible, we don't even need to have $s = t$ in that. But to have
$$\frac{(a - m_s) + m_s}{p} + \frac{(b - n_t) + n_t}{q} \geqslant \frac{a - m_s}{p} + \frac{b - n_t}{q} \tag{2}$$
we need
$$\frac{m_s}{p} + \frac{n_t}{q} \geqslant 0 \tag{3}$$
which only obviously holds if $s \geqslant 0$ and $t \geqslant 0$. However, in that case the right hand side of $(2)$, namely $(-s) + (-t)$, is nonpositive, thus certainly not $> 1$. A less obvious but still easy to see condition for $(3)$ is $s + t \geqslant 0$, but that again renders the right hand side of $(2)$ nonpositive. We are left with the option $s + t = -1$ (see above, $s = -x$, $t = -y$), but then $(3)$ is equivalent to
$$\frac{a}{p} + \frac{b}{q} \geqslant 1\,.$$
Since $p$ and $q$ are distinct primes, this inequality is necessarily strict (if it holds), thus we are back at square one — to make the argument work, we must first prove what the argument is supposed to prove.

So let's take a different route. Get rid of the denominators, i.e. multiply the inequality with $pq$. Thus we want to show
$$aq + bp > pq\,. \tag{$\ast$}$$
Now we know everything in $(\ast)$ is a positive integer, which can be useful. And we know some congruences:
\begin{align}
aq + bp &\equiv aq \equiv 1 \pmod{p}\,,\\
aq + bp &\equiv bp \equiv 1 \pmod{q}\,.
\end{align}
Since $p$ and $q$ are coprime, it follows that
$$aq + bp \equiv 1 \pmod{pq}\,. \tag{$\ast\ast$}$$
And also $aq + bp \geqslant 1\cdot q + 1\cdot p > 1$. Together with $(\ast\ast)$ it follows that $aq + bp \geqslant pq + 1$, which is $(\ast)$.

(And in case $a < p$, $b < q$ we have $aq + bp \leqslant (p-1)q + (q-1)p < 2pq$, hence $aq + bp = pq+1$.)

Title: $a/p+b/q>1$ with certain conditions


问:
Are my examples of False $\implies$ False being true correct?
I know that $p \implies q$ is true if $p$ is false. I am trying to wrap my brain around this by thinking about examples. I have two examples that I would like to understand.

Example 1: Consider the set

$$
X = \{x\in \mathbb{R}: x\geq 0 \implies x^2\leq 0\}.
$$

This set is equal to the interval $(-\infty, 0]$ because, for example, $-1 \geq 0 \implies -1\leq 0$ is a true statement. Is that correct?

<s>Example 2: I know that the definition of limit is
$$
\forall \epsilon>0 \exists \delta > 0: 0<\lvert x-a\rvert < \delta \implies \lvert f(x) - L\rvert < \epsilon.
$$

Would this make this a true statement for all functions $f$? Say $\epsilon$ is given, then pick $\delta = -1$. Then $0<\lvert x- a\rvert < \delta$ is false, so it is true that it does imply that $\lvert f(x) - L\rvert < \epsilon$. Have I just misunderstood the definition?</s>

EDIT: I just realized that in my second example the requirement is that $\delta$ be positive. I guess I am down to the first example.

答:
Be careful about mixing a general mathematical statement with a specific instance.  There are typically two ways to interpret $\Rightarrow$.

As an operator: This is thinking of $\Rightarrow$ as a function on a pair of Boolean values.  In Example 1, you're thinking of $\{x\in\mathbb{R}:x\geq 0\Rightarrow x^2\leq0\}$ as the set of all real $x$ such that for fixed $x$, the statement $x\geq 0\Rightarrow x^2\leq 0$ is true.  This is where $(-\infty,0]$ comes from.  In David K.'s answer, this would correspond to $x\geq 0\rightarrow x^2\leq 0$.  I think that using this notation would make this example much clearer.

As a mathematical statement: Typically, when we write $p\Rightarrow q$ in math, we're thinking about a universally quantified statement.  In other words, we say that a statement is true if for all possible instances, the corresponding implication is true.  In this case, the statement $x\geq 0\Rightarrow x^2\leq 0$ is false since there exist counterexamples, i.e., when $x=1$, the hypothesis is true, but the conclusion is false.  Therefore, as a universally quantified statement, this is false.  In this case, the set would be empty.  This may not be what is intended in example since the same $x$ is used in both places in the set-builder notation.

Overall, I think that your Example 1 is correct, but the use of $\Rightarrow$ can be quite confusing when used in this way.  I would write this as $\{x\in\mathbb{R}:(x<0)\vee(x^2\leq 0)\}$ to be absolutely clear about what you're intending.


问:
for which $x$, is $f'(x)=lim_{n \to \infty}f_n'(x)$?
let $f_n(x)=\dfrac{x}{1+nx^4} \quad n=1,2... \quad$ and $f(x)=\lim_{n \to \infty}f_n(x)$

I have to determine for  for which $x$, is $f'(x)=\lim_{n \to \infty}f_n'(x)$?

Here is what I have done:

$f_n'(x)=\dfrac{1-3nx^4}{(1+nx^4)^2}$ and I have found that $f(x)=\lim_{n \to \infty}\dfrac{x}{1+nx^4}=0.$

Now I don't really understand the question

"determine for which $x$, is $f'(x)=\lim_{n \to \infty}f_n'(x)$?"

Any suggestions?

答:
Assuming you calculated $f_n'$ correctly.

As you pointed out, $f(x)=0$, so $f'(x)=0$.
Now $\lim\limits_{n \to \infty} f_n'(x) = \lim\limits_{n \to \infty} \frac{1-3nx^4}{(1+nx^4)^2}$; this is $0$ if $x\neq0$ and $1$ if $x=0$.

Therefore the answer is for every $x\neq0$.


答:
Some people make a distinction between $\implies$ implication and $\rightarrow$ implication. But if you're treating $\implies$ as the simple kind of material conditional that might appear in a truth table, as I think you are,
then your first example is correct, because the sentence

$$ x\geq 0 \implies x^2\leq 0 $$

(with free variable $x$) is equivalent to the sentence

$$ x \leq 0. $$

In this example, because of the use of the set-builder notation
$\{x\in \mathbb{R}:\cdots\}$, there is implicitly a quantification that says $x$ must be in $\mathbb R$ but does not force $x$ to be in any particular proper subset of $\mathbb R$. Only the implication removes the positive values of $x.$

The set-builder notation is also what tells us that
$x\geq 0 \implies x^2\leq 0$ is not universally quantified,
that is, we are not being asked whether there are any counterexamples.
We're merely being told that the counterexamples (if they exist) don't belong to the set that is being constructed.

Edit: I see you figured out the following on your own just before I posted, but here is what I had to say about the original version of the question:

In example 2, however, the implication is quantified by $\exists \delta > 0$.
This is what prevents you from supposing the implication could be true by virtue of setting $\delta = -1.$

Title: Are my examples of False $\implies$ False being true correct?

Body:
I know that $p \implies q$ is true if $p$ is false. I am trying to wrap my brain around this by thinking about examples. I have two examples that I would like to understand.

Example 1: Consider the set

$$
X = \{x\in \mathbb{R}: x\geq 0 \implies x^2\leq 0\}.
$$

This set is equal to the interval $(-\infty, 0]$ because, for example, $-1 \geq 0 \implies -1\leq 0$ is a true statement. Is that correct?

<s>Example 2: I know that the definition of limit is
$$
\forall \epsilon>0 \exists \delta > 0: 0<\lvert x-a\rvert < \delta \implies \lvert f(x) - L\rvert < \epsilon.
$$

Would this make this a true statement for all functions $f$? Say $\epsilon$ is given, then pick $\delta = -1$. Then $0<\lvert x- a\rvert < \delta$ is false, so it is true that it does imply that $\lvert f(x) - L\rvert < \epsilon$. Have I just misunderstood the definition?</s>

EDIT: I just realized that in my second example the requirement is that $\delta$ be positive. I guess I am down to the first example.


问:
Understanding totally ordered abelian groups
Let $(\Lambda, \leq ,+)$ be a totally ordered abelian group. Say such an abelian group is simple if it has no nontrivial quotients (the only quotients are $0$ and itself).

One might wish to understand totally ordered abelian groups in terms of the simple ones. We may wish to embed a totally ordered abelian group $\Lambda$ as $\Lambda \rightarrow \prod_{i \in I} \Lambda_i$, where the $\Lambda_i$ are simple and the product is given some kind of dictionary order.

My question is, is there a terminal simple totally ordered abelian group? I would suspect it is $(\mathbb{R}, \leq, +)$, but I cannot seem to show this.

Note: by simple, I mean there are no subgroups $S \subset \Lambda$ such that $[a, b] \subset S$ for each $a, b \in S$.

Lemma 1: A simple totally ordered abelian group $\Lambda$ is Archimedian.

By contrapositive, suppose that $\Lambda$ is not archimedian, and take $a, b \in \Lambda$ such that $an < b$ for each $n$. Then consider the order closure of the subgroup $\langle a \rangle$, i.e.
$$ \{ x \in \Lambda : x \in [-na, na] \text{ for some } n \in \mathbb{N}_{\geq 0} \} $$
This is a subgroup of $\Lambda$ which does not include $b$. Hence it is a proper subgroup, so that $\Lambda$ is not simple.

Lemma 2: The terminal simple totally ordered group $\Lambda$ has all suprema and infima.

This is clear enough, as we can embedd any simple totally ordered abelian group $\Lambda$ into its completion $\tilde{\Lambda}$ (where we addded in suprema and infima). Note: we add in the supremum of any set bounded above.

Edit: it was pointed out in the comments that we need to show that the completion $\tilde{\Lambda}$ is simple. Arguing by contrapositive, suppose that the completion is not simple, and take the preimage of a proper nontrivial order closed subgroup $T \subset \tilde{\Lambda}$, and call this preimage $S$. $T$ has a nonzero element $t > 0$ in it, and therefore contains $[-t, t]$. The preimage $S$ of $T$ under $\Lambda \rightarrow \tilde{\Lambda}$ therefore contains at least one nonzero element, as there must be some subset $X \subset \Lambda$ such that $\text{sup}(X) = t$, and such a set must contain an element of $[-t, t]$, which would then be contained in $S$. Hence $S \neq \{ 0 \}$, so that $S = \Lambda$ since $\Lambda$ is simple. Since $T$ is order closed, it must contain all of the elements of the order closure $f(S)$, which is all of $\tilde{\Lambda}$.

Hence we can assume that are are looking for a complete archimedian simple totally ordered abelian group. I would really imagine it is $\mathbb{R}$ at this point, but I cannot manage to show it.

Let's try to construct a map from $\Lambda$ to $\mathbb{R}$. Take $a \in \Lambda$ not equal to $0$ and define $f(a) = 1$. Define $f(na) = n$. Now, for each $m$ and each $b$ such that $mb = na$, define $f(b) = n/m$. Define $f$ on suprema by setting $f( \text{sup} S  ) = \text{sup} f(S)$. Now we need to show that this is well defined and defined on all of $\Lambda$.

For positive $b \in \Lambda$, let $\rho(n)$ be the minimal integer such that $\rho_b(n) b \geq n a$. Suppose $b$ and $b'$ are such that $\text{lim}_{n \rightarrow \infty} n/ \rho_b(n) = \text{lim}_{n \rightarrow \infty} n/ \rho_{b'}(n)$. Without loss of generality, $b < b'$. Then we can show that $N(b - b') < a$ for each $N$, violating the archimedian property. This shows that the map constructed is injective.

Well definedness is probably a bit tedious, but I suspect it is true.

答:
Let $A$ be a simple linearly ordered complete divisible archimedean abelian group.

It is in particular a $\mathbb Q$-vector space. For any $a\in A$, we have an embedding $\mathbb Qa\to A$ which automatically extends to $\mathbb R\cong \mathbb Ra\to A$, as $A$ is complete.

I claim that this map must be an isomorphism. Clearly it must be injective as $\mathbb R$ is simple. Moreover, if $b\in A$, $b\leq na$ for some $n$, and so we can actually consider $\{r\in \mathbb R, ra < b\}$ and  this has an upper bound, so it has a least upper bound, $r_b$.

Now for any $r< r_b$ we have $(r_b-r)a > r_ba - b$, so for any $n\in \mathbb N$, $a> n(r_ba-b)$. By the archimedean character of $A$, it follows that $r_ba - b \leq 0$, i.e. $r_ba \leq b$ as well.

But now one can also similarly prove $r_ba \geq b$, and so $r_ba = b$

(a simpler way to phrase surjectivity would be to use Dedekind reals, but let me stick to this formulation in case you don't know these).

The claim follows, and so by embedding your linearly ordered group into its rationalization and then the completion of that, you get that $\mathbb R$ is weakly terminal among simple linearly ordered abelian groups.


问:
Elliptic integrals, general form
According to https://mathworld.wolfram.com/EllipticIntegral.html, Elliptic integrals are integrals of the form of equation (1).

I am trying to convert an integral of the form

$$\int_{|x_+|}^\infty \frac{dx}{\sqrt{(x^2-x_-^2)(x^2-x_+^2)}}$$

To express as a function of elliptic integrals, where $|x_+|>|x_-|$, we can take $x_+>0$ for simplicity. We can easily see that this integral converges, and that it is real.

I have trying following the method outline in the webpage I linked, with no success, because of the arising of imaginary roots in the process. However, plugging this expression into mathematica with $x_-=1$, $x_+=2$, yields me the expression :

$$\int_{2}^\infty \frac{dx}{\sqrt{(x^2-1)(x^2-4)}}=\frac{i}{2}K(\frac{3}{4})+K(4)$$

Where $K(m) = \int_0^{\pi/2}\frac{1}{\sqrt{1-m\sin^2(\phi)}}d\phi$. Now clearly the integral I'm looking to re-express is well-defined, and real, but it would appear I get an complex number as an answer. However, looking closer, $K(4)$ will be complex because the parameter $m>1$. Thus, if we numerically compute the expression the imaginary parts cancel and we are left with a real answer, which is correct.

So my question is as follows : when we say that Elliptic integrals (as defined in my link) are expressible as a function of the three "kinds" of elliptic integrals (+ elementary functions), do we allow "complex" elliptic integrals ? In other words, is there any way to express my example as a function of "well defined" elliptic integrals, in the sense that we only integrate real numbers ?

I am unsure if I expressed myself clearly, I hope it makes sense.

答:
Byrd and Friedman 215.00 gives the answer as
$$\frac1{x_+}K\left(m=\frac{x_-^2}{x_+^2}\right)$$
By the constraints on the variables, the argument always lies in $[0,1]$. This expression is related to the one in the OP's answer by a Gauss transformation.

Nevertheless, elliptic integrals may be evaluated at complex numbers with no hassle, and indeed they are required in a few applications like parametrising minimal surfaces.

Title: Elliptic integrals, general form


问:
Ideal correspondence
I'm confusing the ideal correspondence theorem. Is the following  right?

Ideal correspondence: Let $f:A \to B$ be a ring homomorphism. Then there is a one-to-one order-preserving correspondence between ideals of $f(A)$ and ideals of $A$ which contain $\ker(f)$ by $\alpha \mapsto f(\alpha)$ (conversely $\beta \mapsto f^{-1}(\beta)$), and prime ideals correspond to prime ideals, and maximal ideals correspond to maximal ideals.

答:
Yes. For me, everything is much less confusing if you think of ideals $I$ in terms of the corresponding quotient maps $A \to A/I$. Then the ideals of $f(A)$ are precisely the quotient maps $f(A) \to f(A)/I$, which composed with the map $A \to f(A)$ are precisely those maps $A \to f(A) \to f(A)/I$ which factor through $f(A)$ - in other words, which have kernel containing $\ker(f)$. </p>

<p>Moreover, an ideal $I$ is prime if and only if $A/I$ is an integral domain and maximal if and only if $A/I$ is a field. Hence if $f(A) \to f(A)/I$ has image an integral domain then the composition $A \to f(A) \to f(A)/I$ also has image an integral domain, and the same is true if you replace "integral domain" by "field." Other properties of ideals can also be stated in this way, hence also pull back: for example, an ideal $I$ is radical if and only if $A/I$ has no nilpotents. </p>

ContentLicense: CC BY-SA 3.0

Title: Ideal correspondence

Body:
Yes. For me, everything is much less confusing if you think of ideals $I$ in terms of the corresponding quotient maps $A \to A/I$. Then the ideals of $f(A)$ are precisely the quotient maps $f(A) \to f(A)/I$, which composed with the map $A \to f(A)$ are precisely those maps $A \to f(A) \to f(A)/I$ which factor through $f(A)$ - in other words, which have kernel containing $\ker(f)$.

Moreover, an ideal $I$ is prime if and only if $A/I$ is an integral domain and maximal if and only if $A/I$ is a field. Hence if $f(A) \to f(A)/I$ has image an integral domain then the composition $A \to f(A) \to f(A)/I$ also has image an integral domain, and the same is true if you replace "integral domain" by "field." Other properties of ideals can also be stated in this way, hence also pull back: for example, an ideal $I$ is radical if and only if $A/I$ has no nilpotents.


答:
Suppose $a_0$ and $b_0$ are the smallest positive integers, respectively, such that $a_0q \equiv 1 \pmod p$ and $b_0p \equiv 1 \pmod q$.

Note that $a_0 q \equiv 1 \equiv aq \pmod p \iff p|a-a_0$. This means $a_0<p$ and all other $a$'s are greater than $p$. Similarly $b_0 < q$ and all other $b$'s are greater than $q$.

Assume $a_0 q=pm+1, m\in \mathbb N$. Then $$pm < pm+1=a_0q  < pq \implies m < q$$ Now, $$ (q-m)p\equiv -mp=1-a_0q \equiv 1 \pmod q \\ \text{ and } 0< q -m < q$$

Therefore $b_0=q-m$. And it follows that

$$\frac ap + \frac bq \ge \frac{a_0}{p} + \frac{b_0}{q} = \frac{pm+1}{pq}+\frac{q-m}{q}=\frac{pm+1+pq-pm}{pq} = 1+\frac{1}{pq}>1.\blacksquare
$$

(We note that $p, q$ need not be prime numbers. As long as $\gcd(p,q)=1$, the inequality holds.)

Title: $a/p+b/q>1$ with certain conditions

Body:
Let $a$ and $b$ be positive integers and $p$, $q$ distinct primes such that $aq\equiv 1\pmod p$ and $bp\equiv 1\pmod q$ prove that $a/p+b/q>1$. My approach was to notice that $pn+aq=qm+bp$ for some integers $n$ and $m$ (from the congruences), which is equivalent to $q(a-m)=p(b-n)$, so $a-m = px$ and $b-n = qy$, for some integers $x$ and $y$. Using the last two equalities we have $a/p+b/q = [(a-m)+m]/p+[(b-n)+n]/q>px/p+qy/q>2>1$.
My biggest concern is if I can use $x,y,n,m$ as $natural$ $numbers$ and im also concerned about the fact that I got that the expression is $>2$ which seems a very huge bound. If there is a flaw in my solution pls. point it. If the solution is completly incorrect pls show me yours. Any help appreciated.


问:
Linear functional on $M_n(\mathbb{C})$
Any linear functional on $M_n(\mathbb{C})$ is of the form $X⟼Tr(AX)$ for some $A∈M_{n}(\mathbb{C})$.
Can this statement be proved with the help of the Riesz representation theorem
and also by any other method?

答:
The bilinear form $\varphi : \mathcal{M}_n(\mathbb{C}) \times \mathcal{M}_n(\mathbb{C}) \rightarrow \mathbb{C}$ defined by
$$\varphi(A,B)=\mathrm{Tr}(AB)$$

is non-degenerated, hence it induces a canonical isomorphism $ \mathcal{M}_n(\mathbb{C}) \rightarrow \left( \mathcal{M}_n(\mathbb{C})\right)^*$.

Title: Linear functional on $M_n(\mathbb{C})$


问:
How to multiply time in different denominations?
I am writing some software that must do generalized time multiplication of various denominations. I can't wrap my head around how to generalize it.

[edit] The use case is this. You have a clock measuring time and you want to provide a mapping (ratio) to accelerate or decelerate the clock. For example, every seconds passed is equivalent to 1 days passed. Thus, $clocktime = xseconds * ndays$, with n days probably represented in seconds.

For example, lets say I have :

$1min * 2min = 2min\\
60s * 120s = 7200s^2\\
7200s^2 / 60s = 120s = 2min$

Now, if I multiply time in days, for example :

$1days * 2days = 2days$

And in seconds :

$1 day = 86'400s\\
86'400s * 172'800s = 14'929'920'000s^2\\
14'929'920'000s^2 / 60s = 248'832'000s$

But, that is incorrect.

$248'832'000s / 60 = 4'147'200min\\
4'147'200min / 60 = 69'120h\\
69'120h / 24 = 2'880days$

So my question is, how do you generalize time multiplication from/to various denominations? Ultimately, what am I missing here? Ty

答:
OK so here is what I believe to be the answer and the problem with the question assumptions.

When representing days as seconds, the units were wrong.

$1day \neq 86400s\\
1day = 24h * 60m *60s = 86400hms$

And so, when multiplying $1day * 2days$ represented as "seconds", the correct equation is :

$86400hms * 172800hms = 14929920000h^2m^2s^2$

To convert back to days and get the correct result :

$14929920000h^2m^2s^2 / 24h / 60m / 60s = 172800hms\\
172800hms / 24h / 60m / 60s = 2days$


问:
Verification for maximum principle
Given optimal control problem
$$
\dot x = f(t,x(t),u(t)), \quad x(0) = x_0,\\
J(u) = \int_0^T f^0(t,x(t),u(t))dt \to \min,
$$
we can apply Pontryagin's maximum principle to get a necessary condition for an optimum.

My question is when maximum principle is sufficient and how do we verify that an obtained control function is optimal?

答:
It depends on restrictions of control $u$. If $u$ belongs to open set (for example $u\in\mathbb{R}$) then you can apply Legendre–Clebsch condition and Jacobi equation to prove weak optimality. Also you can combine them with Weierstrass condition to prove strong optimality. For your case you can find these conditions in this book

If $u$ belongs to not open set (for example $u\in [-1;1]$) then above methods do not work. If the problem is convex or concave then you can use Karush–Kuhn–Tucker theorem for Banach spaces -- it is equvilent to Pontryagen maximum principle but gives you sufficient conditions for optimality. If there is no convexity you may try to apply some second order conditions (see Ioffe Tikhomirov) but it is usually hard to verify them. Also there is Bellman PDE which gives you sufficint condition but it is necessary to construct optimal synthesis to verify it.

Title: Verification for maximum principle


问:
Build Taylor series of functions
I would like to ask what is the general procedure for calculating such examples.

Build Taylor series of functions $f(x)=\frac{1}{x^2}$
at the point $a = −1$ and determine all $x\in\mathbb{R}$ for which it converges.

答:
If $f$ is a function that is infinitely differentiable at some point $a$ in its domain, then the Taylor series of $f$ at $a$ is the series

$$\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$$

This means that you can find the Taylor series of any function  by simply computing the numbers $f^{(n)}(a)$. In your case, you'll want to calculate $f^{(n)}(-1)$ for all integers $n\geq 0$.

The case $n=0$ is understood to be the value of $f$ at $-1$, so

\begin{align*}
f^{(0)}(-1) &= f(-1)\\
&= \frac{1}{(-1)^2}\\
&= 1
\end{align*}

For all other values of $n$, you'll need to start differentiating.

\begin{align*}
f^{(1)}(-1) &= \left[\frac{d}{dx}\frac{1}{x^2}\right]_{x=-1}\\
&= \left[-\frac{2}{x^3}\right]_{x=-1}\\
&= -\frac{2}{(-1)^3}\\
&= 2 = 2!
\end{align*}

\begin{align*}
f^{(2)}(-1) &= \left[\frac{d^2}{dx^2}\frac{1}{x^2}\right]_{x=-1}\\
&= \left[\frac{6}{x^4}\right]_{x=-1}\\
&= \frac{6}{(-1)^4}\\
&= 6 = 3!
\end{align*}

\begin{align*}
f^{(3)}(-1) &= \left[\frac{d^3}{dx^3}\frac{1}{x^2}\right]_{x=-1}\\
&= \left[-\frac{24}{x^5}\right]_{x=-1}\\
&= -\frac{24}{(-1)^5}\\
&= 24 = 4!
\end{align*}

Using induction, it's easy to prove that $f^{(n)}(-1)=(n+1)!$ in general, so the Taylor series is

\begin{align*}
& \sum_{n=0}^{\infty}\frac{(n+1)!}{n!}\left(x-(-1)\right)^n\\
&= \sum_{n=0}^{\infty}(n+1)(x+1)^n
\end{align*}

To find the radius of convergence of the series (and hence the set of all $x\in\mathbb{R}$ where the series converges), you can use the ratio test or the root test. Here, it's better to use the ratio test.

\begin{align*}
\lim_{n\to\infty}\left|\frac{(n+2)(x+1)^{n+1}}{(n+1)(x+1)^n}\right| &= \lim_{n\to\infty}|x+1|\cdot\frac{n+2}{n+1}\\
&= |x+1|\cdot 1\\
&= |x+1|
\end{align*}

It follows that the series converges if $|x+1|<1$ ($-2<x<0$), and diverges if $|x+1|>1$ ($x<-2$ or $x>0$), so the radius of convergence is $1$. To determine if the series converges for $x=-2$ or $x=0$, you'll need to test

$$\sum_{n=0}^{\infty}(n+1)(-2+1)^n=\sum_{n=0}^{\infty}(-1)^n (n+1)$$

and

$$\sum_{n=0}^{\infty}(n+1)(0+1)^n=\sum_{n=0}^{\infty}(n+1)$$

for convergence. Since they both clearly diverge, the series converges for $x\in(-2,0)$.

Whether or not the series actually converges to $f$ is another question. The answer happens to be yes in this case, so for $x\in(-2,0)$, you can finally write

$$\frac{1}{x^2}=\sum_{n=0}^{\infty}(n+1)(x+1)^n$$

Title: Build Taylor series of functions


答:
If you want to use Riesz theorem, indeed
$$\langle A, B\rangle=\mathrm{Tr}(A^*B)$$

is an inner product on $M_n(\mathbb C)$, the Frobenius inner product. So any linear form $h(X)$ on $M_n(\mathbb C)$ can be written as

$$h(X)=\mathrm{Tr}(A^*X).$$

Riesz representation theorem is however as mentioned in a comment an over killer as the result is quite obvious if you just write what $\mathrm{Tr}(A^*B)$ is for any matrix $A$.

Title: Linear functional on $M_n(\mathbb{C})$

Body:
Any linear functional on $M_n(\mathbb{C})$ is of the form $X⟼Tr(AX)$ for some $A∈M_{n}(\mathbb{C})$.
Can this statement be proved with the help of the Riesz representation theorem
and also by any other method?


答:
For your Example 1, for $x\in \mathbb{R}$ let $S(x)$ be the statement $[x\ge 0 \implies x^2\le 0]$. The set can be written as $X=\left\{x\in \mathbb{R}: S(x)\,\text{is true}\right\}$. Since $-1\ge 0$ is false, the statement $S(-1)=[-1\ge 0\implies (-1)^2\le 0]$ is true. Therefore, I do believe that you are right.

The truth behind logical implication seems to be subtle. I have always accepted truth behind the statement $p\implies q$ as a matter of definition, the definition coming from the truth table for $p\implies q$. My guess is that a logician will give a better answer to this.

This all said, I do think you can develop intuition as to why false $\implies$ false should be true. For example, consider the statement "If you score a touchdown, I will buy you a gift." You can write this as:
$$\text{You score touchdown}\implies\text{I buy you a gift}.$$
If you end up not scoring a touchdown, and, as a result I do not buy you a gift, then both statements above are false, and therefore the implication is true. The fact of the matter is that I did not break my promise of buying you a gift, because you did not score a touchdown for me! So nothing "false" occurred. On the other hand, if you had scored the touchdown, and I didn't buy you a gift, then I broke my promise. In this case it was a "false promise." Although advanced mathematics is more abstract than real world settings like these, I like to think about such examples to understand $p\implies q$ more generally.


问:
Can an irregular tetrahedron be inscribed in a cuboid or prism?
A regular tetrahedron can be inscribed in a cube in a way that the tetrahedron edges are diagonals of the cube faces.

Is it possible to similarly inscribe any irregular tetrahedron in a cuboid or prism?

Image source: http://mathworld.wolfram.com/RegularTetrahedron.html

答:
Affine transformations of the cube allow inscribing absolutely any tetrahedron in a parallelepiped, the unit cell of the triclinic crystal system. Now, for the three opposite pairs of edges of a tetrahedron, consider the vectors formed as cross products of each pair. Extending the previous line of reasoning, we see that

a tetrahedron may be inscribed in a parallelogram prism, unit cell of the monoclinic crystal system, iff one vector is orthogonal to the other two

it may be inscribed in a cuboid, unit cell of the orthorhombic crystal system, iff the vectors are mutually orthogonal.


问:
Two cowboys A and B decide to solve a dispute with a duel.
Two cowboys $A$ and $B$ decide to solve a dispute with a duel.
Cowboy $A$ hits his target $\frac13$ of the time.
Cowboy $B$ hits the target $\frac23$ of the time.

It is decided that $A$ will take the first shot, cowboy $B$ will take the second shot (if still alive).
This will continue until there is only one left alive. Also, a cowboy can not shoot two times in a row. What are cowboy $A$ chances of winning the duel?

My logic is to simulate the cases where $A$ wins:

I Case:  $A$ wins: $\frac13$

II Case: $A$ loses, $B$ loses, $A$ wins: $\frac23\cdot\frac13\cdot\frac13$

...

But I don't know what to do from here. Any help?

答:
Let $p$ be the probability that A wins. With probability ${1\over3}$ he wins on his first shot. With probability ${2\over3}$ he misses. When B hits in his first shot, A has lost. When B misses, A again has a chance of $p$.  It follows that
$$p={1\over3}+{2\over3}\cdot{1\over3}\cdot p\ .$$
This implies $p={3\over7}$.


问:
Maximal number of generators of first homology in Vietoris-Rips complex
For a point cloud $P$ with $n$ vertices is there a nice formula for the maximum number of points in a persistence diagram of the Vietoris-Rips complex on this point cloud?

Since in a VR complex a $1$-cycle must consist of $4$ distinct vertices at any filtration value of $\epsilon$ there should be $\binom{n}{4}$ cycles maximum. But of course this isn't actually true because if all sets of $4$ points make cycles then we will end up having to fill in many triangles as well.

Any thoughts on this would be much appreciated.

答:
You will likely be interested in the paper "Extremal Betti Numbers of Vietoris–Rips Complexes" by Michael Goff, from DCG in 2011: https://link.springer.com/article/10.1007/s00454-010-9274-z. See in particular Section 3 for results on 1-dimensional homology, although higher-dimensional homology is also considered.

A clarifying comment is that I suspect you are asking about Vietoris-Rips complexes of a point cloud $P$ embedded in some Euclidean space $\mathbb{R}^d$. This is also the context considered in the Goff paper, and the dependence on $d$ is considered. One could also ask about Vietoris-Rips complexes of a point cloud $P$ embedded in an arbitrary metric space, which might be equivalent to asking what is the largest Betti number possible from an arbitrary clique simplicial complex with $n$ vertices.

While searching again for the above paper by Goff, I also ran across the following slides: https://plv.colorado.edu/dmoon/assets/docs/nzp.pdf

Best, Henry


问:
$\int_{0}^{2}\int_{0}^{2}\sqrt{2-4z^2}\,dz\,dy$
How do you evaluate this integral? I have tried to start this, but I have not seen an integral like this in forever. Please help me out!

$$\int_{0}^{2}\int_{0}^{2}\sqrt{2-4z^2}\,dz\,dy$$

答:
$$\int_0^2 \sqrt{2-4 z^2} \, dy=2 \sqrt{2-4 z^2}$$
$$\int 2 \sqrt{2-4 z^2}\,dz=z \sqrt{2-4 z^2}+\arcsin\left(\sqrt{2} z\right)+C$$
when plugging $z=2$ we get $\arcsin(2\sqrt 2)$ which is not a real number

Anyway the result is
$$\int_0^2 2 \sqrt{2-4 z^2} \, dz=\arcsin\left(2 \sqrt{2}\right)+2 i \sqrt{14}$$


问:
The Product of Power Series
I understand the Cauchy product for power series, but we have slightly different notation here. Suppose we have the following power series:

$$f(x) = \sum_{n=0}^\infty a_nx^n$$
$$g(x) = \sum_{m=0}^\infty b_mx^m$$

Their product is written as follows:
$$g(x) \times f(x) = \sum_{n=0}^\infty \sum_{m=0}^\infty  a_n b_{m}x^{n+m}$$
$$= \sum_{k=0}^\infty x^k (\sum_{n=0}^k a_n b_{k-n})$$

where, $k=n+m, m=k-n, 0\le n \le k$.

Could you please tell me how we got $$g(x) \times f(x) = \sum_{k=0}^\infty x^k (\sum_{n=0}^k a_n b_{k-n})$$

答:
Here is an algebraic derivation.

We obtain
\begin{align*}
g(x)\times f(x)&=\sum_{n=0}^\infty\sum_{m=0}^\infty a_nb_mx^{n+m}\\
&=\sum_{k=0}^\infty\left(\sum_{{n+m=k}\atop{n,m\geq 0}} a_nb_m\right)x^k\tag{1}\\
&=\sum_{k=0}^\infty\left(\sum_{n=0}^ka_nb_{k-n}\right)x^k\tag{2}
\end{align*}

Comment:

In (1) we reorder the sums  by increasing powers of $x$ (provided the reordering is feasible, e.g. due to absolute convergence of the series).

In (2) we eliminate the variable $m$ by using $m=k-n$.

Title: The Product of Power Series


答:
In example 1, rather than reading $X = \{x\in \mathbb{R}: x\geq 0 \implies x^2\leq 0\}$ as if it says that $X$ is the set of non-negative reals whose squares are less than or equal to $0$, you are reading it as the set of reals which if non-negative have squares less than or equal to $0$.  So instead of $X=\{0\}$ you end up with the non-positive reals and this is easy to demonstrate:

It is a property of an ordered field that if $a \le 0$ and $b \ge 0$ then $ab \le 0$.

So, for any $a \le 0$, you have $a \ge 0 \implies a^2 \le 0$

and this might be justified as a literal if unintended reading

For the limit of $f(x)$ being $L$ as $x \to a$, I think a better definition might be $$\forall \epsilon>0 \,\exists \delta > 0\, \forall x\in \mathbb R: 0<\lvert x-a\rvert < \delta \implies \lvert f(x) - L\rvert < \epsilon$$ which would not allow your reading


问:
Reference for Deligne-Mumford
What is a good reference for someone new to the theory of Deligne-Mumford stacks, other than the original Deligne-Mumford paper itself? The paper itself seems readable with some effort; but the fear is that the reader will miss out on whatever happened in the later years. Is there a good introduction at the graduate level?</p>

答:
You can take a look at the notes by Daniel Halpern-Leistner which can be found at</p>
```<code>http://pi.math.cornell.edu/~danielhl/moduli_theory_notes.pdf
</code>```

Title: Reference for Deligne-Mumford

Body:
You can take a look at the notes by Daniel Halpern-Leistner which can be found at

http://pi.math.cornell.edu/~danielhl/moduli_theory_notes.pdf


答:
Your first example is technically correct. I say technically because there seems to be a subtle difference between the actual definition of $\implies$ and the way most mathematicians use it. Of course, $x\geq 0 \implies x^2 \leq 0$ is a well-formed formula, and by the axiom of specification it follows that $X$ exists. What is in it? Elements of $\mathbb{R}$ that satisfy the above implication. And these are, as you correctly pointed out, $(-\infty, 0]$. Call this the axiomatic interpretation if you like. For most mathematicians, however, material implication isn't a definitive truth function, but rather a common tool, a shorthand for saying that if the antecedent is correct, then the consequent is, too. As for the other truth configurations, they just don't bother, intuitively brushing them off as "undefined" or simply irrelevant (note that this usually doesn't prevent them from reaping the mathematical power of contradiction proofs, contrapositives etc.) Whenever used axiomatically or in logical contexts, you'd often find that the skinny arrow $\to$ notation is preferred, whereas the common usage seems to rely on $\implies$.

For your second example, things are not quite so straightforward. Here, we have to prove the implication holds for all $x$ in the domain of $f$. The last part is often suppressed in the $\varepsilon$-$\delta$ limit definition, but it is there. Observe that the only thing we actually need to prove is that whenever the antecedent is true, the consequent is true as well (if the antecedent is false, we don't care either way, as the implication is satisfied). Note, however, that now we have quantifiers bounding some of the variables, and for us to claim our proof as successful, the valuation of the overall expression has to be $\mathbb{T}$, not just the implication bit. For instance, this is precisely why your example of $\delta = -1$ is false: $\delta$ is bounded by $\forall \delta > 0$. But suppose you pick another $\delta$, positive this time. And this is where the hidden bit $\forall x\in \text{Domain}(f)$ comes in. No matter how small $\delta$ is, we'll always be able to find $x$ close enough to $a$ to make $0< \lvert x-a\rvert <\delta$ true, and unless you had a very lucky choice of $\delta$, you'd end up with $\mathbb{T}\implies \mathbb{F}$, i.e. a false implication.


问:
How can the center of mass of a region with density $d[x, y]$ be understood as as a kind of (average x-value, average y-value)?
For a region of uniform density, the center of mass is equal to centroid, which is equal to:

$$(\text{average x-value}, \text{average y-value}) = \left(\frac{\int\int_R{x}\,dxdy}{ \int\int_R{1}\,dxdy}, \frac{\int\int_R{y}\,dxdy}{ \int\int_R{1}\,dxdy}\right)$$

For density $d[x, y]$, the center of mass is instead:

$$\left(\frac{\int\int_R{x d[x, y]}\,dxdy}{ \int\int_R{d[x, y]}\,dxdy}, \frac{\int\int_R{y d[x, y]}\,dxdy}{ \int\int_R{d[x, y]}\,dxdy}\right)$$

Is there a way to understand the center of mass with variable density using this same idea of an average value? The uniform density center of mass is analogous to finding the mean value in 1D ($\frac{\int_a^b f[x]\,dx}{b - a} = \frac{F[b] - F[a]}{b -a}$); is there a way a similar way to understand center of mass for variable density?

When I look online (such as this and this), they use moment and mass, which I don't really grasp. So, preferably, I would like an explanation that doesn't use these concepts; or if they are necessary, an explanation that explains what they are and why they are necessary. What is the connection between $\left(\frac{M_y}{m}, \frac{M_x}{m}\right)$, and $\left(\frac{\int\int_R{x}\,dxdy}{ \int\int_R{1}\,dxdy}, \frac{\int\int_R{y}\,dxdy}{ \int\int_R{1}\,dxdy}\right)$ and $\left(\frac{\int\int_R{x d[x, y]}\,dxdy}{ \int\int_R{d[x, y]}\,dxdy}, \frac{\int\int_R{y d[x, y]}\,dxdy}{ \int\int_R{d[x, y]}\,dxdy}\right)$?

(One thought I had was: the reason the uniform density center of mass works is because all surrounding areas around the point balance out. (If you draw a straight line through the point, two mass of the two sides are the same. Or maybe another way of thinking of it is, if you take the integral of both sides created by the line, those two masses are equal). Perhaps the center of mass with variable density can be derived similarly?)

Sorry if this is very rambling.

答:
The idea of center of mass is that you have a point in space that behaves like if all the mass of the given object is concentrated in that point. So how would you describe this? Let's suppose that you divide the object into infinitesimal masses with density $d(x,y,z)$ and volume $dV=dxdydz$. Then, if the acceleration of the object is $\vec a$, we can write $$\iiint_V d(x,y,z) \vec a dxdydz=\vec a\iiint_V d(x,y,z) dxdydz=M\vec a$$
Here $M$ is the mass of the object.

But there is one more condition. That is that the sum of torques with respect to the center of mass. This is equivalent of saying that all points in the object move with the same acceleration. So the infinitesimal torque is $$d\vec F\times(\vec r-\vec r_{CM})=dxdydz\ d(x,y,z)\vec a\times(\vec r-\vec r_{CM})$$
You can get the integral equal to zero for any $\vec a$ only if $$\iiint_Vdxdydz\ d(x,y,z)(\vec r-\vec r_{CM})=0$$
We can write the last equation by components. I will just do it for $x$:
$$\iiint_Vdxdydz\ d(x,y,z)(x-x_{CM})=0\\\iiint_Vdxdydz\ d(x,y,z)x-\iiint_Vdxdydz\ d(x,y,z)x_{CM}=0$$
So from here $$x_{CM}=\frac{\iiint_Vdxdydz\ d(x,y,z)x}{\iiint_Vdxdydz\ d(x,y,z)}=\frac{M_x}M$$
What this expression tells you is that the center of mass position is the weighted average of the positions of all the points in the object. If the density is constant, then it reduces straight to the simple average.

Note for your problem the density function is two dimensional, so you can ignore the $z$ direction for integration. Also, use for example the gravity along the $-\hat z$ as $\vec a$, if you want to reproduce the calculations in 2D.


问:
Write $\left\Vert T\right\Vert _{OP}$ using eigenvalues of orthogonal basis of $\mathbb{R}^k$.
I need help with the following question:

If  $T\in Hom(\mathbb{R}^k,\mathbb{R}^k)$ and $v_1,\dots,v_k$ orthogonal basis of  $\mathbb{R}^k$ and $\lambda_1,\dots,\lambda_k \in \mathbb{R}^k$   such as $T(v_i)=\lambda_i\cdot v_i$ for every $i\in {1,\dots,k}$.
Write $\left\Vert T\right\Vert _{OP}$ using $\lambda_1,\dots,\lambda_k$.

My solution: I tried many things

$$
\left\Vert T\right\Vert _{OP}=\sup_{\left\Vert x\right\Vert _{\mathbb{R}^{k}}=1}\left\Vert T\left(x\right)\right\Vert _{\mathbb{R}^{k}}=\sup_{x\neq0}\frac{\left\Vert T\left(x\right)\right\Vert _{\mathbb{R}^{k}}}{\left\Vert x\right\Vert _{\mathbb{R}^{k}}}=\sup_{x\neq0}\frac{\left\Vert \sum_{i=1}^{k}\alpha_{i}\lambda_{i}v_{i}\right\Vert _{\mathbb{R}^{k}}}{\left\Vert \sum_{i=1}^{k}\alpha_{i}v_{i}\right\Vert _{\mathbb{R}^{k}}}=\sup_{x\neq0}\frac{\left\Vert \sum_{i=1}^{k}\alpha_{i}\lambda_{i}v_{i}\right\Vert _{\mathbb{R}^{k}}}{\left\Vert \sum_{i=1}^{k}\alpha_{i}v_{i}\right\Vert _{\mathbb{R}^{k}}}
$$

and then I tried to use the norm and I got

$$
\sup_{x\neq0}\frac{\sqrt{\sum_{i=1}^{k}\left(T\left(x\right)\right)_{i}^{2}}}{\sqrt{\sum_{i=1}^{k}\left(x_{i}\right)^{2}}}=\sup_{x\neq0}\sqrt{\frac{\sum_{i=1}^{n}\left(T\left(x\right)\right)_{i}^{2}}{\sum_{i=1}^{n}\left(x_{i}\right)^{2}}}
$$

I also tried this
$$
\sup_{\left\Vert x\right\Vert _{\mathbb{R}^{k}}=1}\left\Vert T\left(x\right)\right\Vert _{\mathbb{R}^{k}}=\sup\left\{ \left\Vert T\left(x\right)\right\Vert _{\mathbb{R}^{k}}:\left\Vert x\right\Vert _{\mathbb{R}^{k}}=1\right\} =\sup\left\{ \sqrt{\sum_{i=1}^{n}\left(\lambda_{i}\alpha_{i}\right)^{2}}:\sqrt{\sum_{i=1}^{n}\alpha_{i}^{2}}=1\right\}
$$

but I didn't know how to continue. I would like to get some help :).

Thank you very much in advance.

答:
Hint: Continue with your second approach. If we order the eigenvalues so that $|\lambda_1| \geq \cdots \geq |\lambda_k|$, then we have
$$
\|T(x)\|_{\Bbb R^k}^2 = \sum_{i=1}^n \alpha_i^2\lambda_i^2 \leq \sum_{i=1}^n \alpha_i^2 \color{red}{\lambda_1^2}.
$$


问:
Residues odd real function issue
I'm trying to solve this integral

$$\int_{0}^{\infty}\frac{x^3}{x^6+1}dx$$

using the residues theorem, but I failed noisily...

Here, I show the process I've followed, getting 0 as a result.

Thank you in advance,

A.

答:
Let $C$ be the closed contour in the upper-half plane that is comprised of $(i)$ the straight line segment from $0$ to $R$, $(ii)$ the circular arc from $R$ to $Re^{i\pi/3}$, and the straight line segment from $Re^{i\pi/3}$ to $0$.  Then, we can write

$$\oint_C \frac{z^3}{z^6+1}\,dz=\int_0^R \frac{x^3}{1+x^6}\,dx+\int_0^{\pi/3}\frac{(Re^{i\theta})^3}{(Re^{i\theta})^6+1}\,iRe^{i\theta}\,d\theta+\int_R^0 \frac{-t^3}{t^6+1}\,e^{i\pi/3}\,dt\tag1$$

From the residue theorem, we have for $R>1$

$$\begin{align}
\oint_C \frac{z^3}{z^6+1}\,dz&=2\pi i \text{Res}\left(\frac{z^3}{z^6+1}, z=e^{i\pi/6}\right)\\\\
&=2\pi i \frac{1}{6e^{i\pi/3}}\tag2
\end{align}$$

Letting $R\to\infty$, we find that

$$\begin{align}
\int_0^\infty \frac{x^3}{1+x^6}\,dx&=\frac{2\pi i}{6e^{i\pi/3}(1+e^{i\pi/3})}\\\\
&=\frac{\pi}{3\sqrt{3}}
\end{align}$$

Title: Residues odd real function issue


问:
finding nth term
Let
  3,8,17,32,57 . . . . .
be a pattern.How do we find the nth number?My brains are completely jammed,I am tired.I do not even recognize the pattern.I calculated a few ways,but all I want is a little hint,not the whole solution.</p>

答: Hint: One solution can be obtained by taking differences-of-differences.

Title: finding nth term


问:
Solving for a differential equation Gompertz growth equation
What is the general solution of this differential equation?
$$ \frac{dy}{dt} = k \enspace  y \enspace \ln(\frac{a}{y})$$ where $a$ and $k$ are positive constants.

So far, my solution is:
$$ \frac{dy}{y \enspace \ln(\frac{a}{y})} = k \enspace dt$$
When I let $u=lny$,
$$ \int \frac{1}{y(\ln a-\ln y)} \,dy =\int k dt $$
$$ \int \frac{1}{\ln a-u} \,du= kt + C $$

How to continue this?

答:
$$
-\frac{(\ln a-\ln y)'}{(\ln a-\ln y)} = k dt\ \ \Rightarrow \ \ -\ln(\ln a-\ln y) = k t + C_0
$$

so

$$
\ln a-\ln y = C_1 e^{-k t}\ \ \Rightarrow y = a e^{-C_1 e^{-k t}}
$$


问:
Moment generating function of Laplace distribution step by step
There are two similar posts but none of them helped me to get through the full derivation of the very simple MGF that should be according to Wikipedia:

$$
\frac {\exp(\mu t)}{1-b^{2}t^{2}}
$$

Here's my attempt:

Definition of MGF
$$
M_X(t) = \mathbb{E}\left[\exp(t X)\right]
$$

by LOTUS

$$
= \int \exp(t x) \cdot p(x) dx
$$

Plugging-in PDF

$$
= \int \exp(t x) \cdot \frac{1}{2b}\exp \left(- \frac{\mid x - \mu \mid }{b} \right) dx
$$

$$
= \frac{1}{2b} \int \exp(t x) \cdot \exp \left(- \frac{\mid x - \mu \mid }{b} \right) dx
$$

$$
= \frac{1}{2b} \int \exp \left(t x - \frac{\mid x - \mu \mid }{b} \right) dx
$$

getting rid of abs. value

$$
= \frac{1}{2b} \int_{-\infty}^{\mu} \exp \left(t x - \frac{ x - \mu  }{b} \right) dx
+ \frac{1}{2b} \int_{\mu}^{\infty} \exp \left(t x - \frac{- x + \mu  }{b} \right) dx
$$

the left integral first

$$
= \frac{1}{2b} \int_{-\infty}^{\mu} \exp \left(\frac{bt x - x - \mu  }{b} \right) dx
$$

$$
= \frac{1}{2b} \int_{-\infty}^{\mu} \exp \left(\frac{bt x - x  }{b} \right) \exp \left(\frac{- \mu  }{b} \right) dx
$$

$$
= \frac{1}{2b} \cdot \exp \left(\frac{- \mu  }{b} \right) \int_{-\infty}^{\mu} \exp \left(\frac{bt x - x  }{b} \right)  dx
$$

$$
= \frac{1}{2b} \cdot \exp \left(\frac{- \mu  }{b} \right) \int_{-\infty}^{\mu} \exp \left(\frac{ (bt  - 1)  }{b}  x \right)  dx
$$

Now since $\int e^{ax}\,dx={\frac {1}{a}}e^{ax}+C$

$$
= \frac{1}{2b} \cdot \exp \left(\frac{- \mu  }{b} \right) \cdot \left|  \frac{b}{ (bt  - 1)  } \cdot \exp \left(\frac{ (bt  - 1)  }{b}  x \right)  \right|_{-\infty}^{\mu}
$$

Let's evaluate the integral first

$$
\left|  \frac{b}{ (bt  - 1)  } \cdot \exp \left(\frac{ (bt  - 1)  }{b}  x \right) \right|_{-\infty}^{\mu}
$$

$$
= \frac{b}{ (bt  - 1)  } \cdot \exp \left(\frac{ (bt  - 1)  }{b}  (\mu) \right)
-
\frac{b}{ (bt  - 1)  } \cdot \exp \left(\frac{ (bt  - 1)  }{b}  (-\infty)\right)
$$

Since $\lim_{x \rightarrow 0} e^{-x} = 0$, the second term is 0

$$
= \frac{b}{ (bt  - 1)  } \cdot \exp \left(\frac{ (bt  - 1)  }{b}  (\mu) \right)
- 0
$$

Double check from WorlframAlpha:

$$
= \frac{b}{bt-1} \exp \left( \mu \left( t - \frac{1}{b} \right) \right)
$$

for $\frac{1}{b} < t$

Now the right-hand side integral:

$$
\frac{1}{2b} \int_{\mu}^{\infty} \exp \left(t x - \frac{- x + \mu  }{b} \right) dx
$$

$$
\frac{1}{2b} \int_{\mu}^{\infty} \exp \left(\frac{b t x + x - \mu  }{b} \right) dx
$$

$$
= \frac{1}{2b} \int_{\mu}^{\infty} \exp \left(\frac{b t x + x   }{b} \right) \exp \left(\frac{-\mu  }{b} \right)dx
$$

$$
= \frac{1}{2b} \cdot \exp \left(\frac{ -\mu  }{b} \right) \int_{\mu}^{\infty} \exp \left(\frac{b t x + x   }{b} \right) dx
$$

$$
= \frac{1}{2b} \cdot \exp \left(\frac{ -\mu  }{b} \right) \int_{\mu}^{\infty} \exp \left( \frac{b t + 1 }{b} x\right) dx
$$

Now evaluate the integral

$$
\int_{\mu}^{\infty} \exp \left( \frac{b t + 1 }{b} x\right) dx
$$

According to wolfram alpha

$$
= - \frac{b}{bt + 1}\exp\left(\mu \frac{1}{b + t}\right)
$$

for $\frac{1}{b} + t < 0$

Now sum-up these two integrals:

$$
= \frac{1}{2b} \cdot \exp \left(\frac{- \mu  }{b} \right) \cdot \frac{b}{ (bt  - 1)  } \cdot \exp \left(\frac{ (bt  - 1)  }{b}  (\mu) \right)  + \\
\frac{1}{2b} \cdot \exp \left(\frac{ -\mu  }{b} \right) \cdot \left( - \frac{b}{bt + 1}\exp\left(\mu \frac{1}{b + t}\right) \right)
$$

And simplify

$$
= \frac{1}{2b} \cdot \exp \left(\frac{- \mu  }{b} \right) \cdot \frac{b}{ (bt  - 1)  } \cdot \exp \left(\frac{ (bt  - 1)  }{b}  (\mu) \right)  - \\
\frac{1}{2b} \cdot \exp \left(\frac{ -\mu  }{b} \right) \cdot  \frac{b}{bt + 1} \cdot \exp\left(\mu \frac{1}{b + t}\right)
$$

$$
= \frac{1}{2b} \cdot  \frac{b}{ (bt  - 1)  } \cdot \exp \left(\frac{- \mu  }{b} \right) \cdot \exp \left(\frac{ (bt  - 1)  }{b}  (\mu) \right)  - \\
\frac{1}{2b} \cdot   \frac{b}{bt + 1} \cdot \exp \left(\frac{ -\mu  }{b} \right) \cdot \exp\left(\mu \frac{1}{b + t}\right)
$$

$$
= \frac{1}{b} \cdot  \frac{1}{ (bt  - 1)  } \cdot \exp \left(\frac{- \mu  }{b} \right) \cdot \exp \left(\frac{ (bt  - 1)  }{b}  (\mu) \right)  - \\
\frac{1}{b} \cdot   \frac{1}{bt + 1} \cdot \exp \left(\frac{ -\mu  }{b} \right) \cdot \exp\left(\mu \frac{1}{b + t}\right)
$$

$$
= \frac{1}{b} \cdot  \frac{1}{ (bt  - 1)  } \cdot \exp \left(\frac{- \mu  }{b} +\frac{ (bt  - 1)  }{b}  (\mu) \right)  - \\
\frac{1}{b} \cdot   \frac{1}{bt + 1} \cdot \exp \left(\frac{ -\mu  }{b} + \mu \frac{1}{b + t}\right)
$$

$$
= \frac{1}{b} \cdot  \frac{1}{ (bt  - 1)  } \cdot \exp \left( \mu \left(t - \frac{2}{b} \right) \right)  - \\
\frac{1}{b} \cdot   \frac{1}{bt + 1} \cdot \exp \left(\frac{ -\mu  }{b} + \mu \frac{1}{b + t}\right)
$$

But I cannot move away from this ugly formula to the simple beauty. Is there just a stupid mistake above or am I doing something substantially wrong?

答:
Alright, here's the full correct solution if anyone is interested.

Definition of MGF
$$
M_X(t) = \mathbb{E}\left[\exp(t X)\right]
$$

by LOTUS

$$
= \int \exp(t x) \cdot p(x) dx
$$

Plugging-in PDF

$$
= \int \exp(t x) \cdot \frac{1}{2b}\exp \left(- \frac{\mid x - \mu \mid }{b} \right) dx
$$

$$
= \frac{1}{2b} \int \exp(t x) \cdot \exp \left(- \frac{\mid x - \mu \mid }{b} \right) dx
$$

$$
= \frac{1}{2b} \int \exp \left(t x - \frac{\mid x - \mu \mid }{b} \right) dx
$$

For integrating absolute value, we split the integral at $\mu$, such that it change the absolute values to

for $x < \mu$: $\mid x - \mu \mid = -(x - \mu)$

for $x \geq \mu$: $\mid x - \mu \mid = x - \mu$

$$
= \frac{1}{2b} \int_{-\infty}^{\mu} \exp \left(t x - \frac{-(x - \mu) }{b} \right) dx +
\frac{1}{2b} \int_{\mu}^{\infty} \exp \left(t x - \frac{x - \mu}{b} \right) dx
$$

$$
= \frac{1}{2b} \left( \int_{-\infty}^{\mu} \exp \left(\frac{b t x + x - \mu }{b} \right) dx +
\int_{\mu}^{\infty} \exp \left(\frac{btx - x + \mu}{b} \right) dx \right)
$$

$$
= \frac{1}{2b} \left(  \int_{-\infty}^{\mu} \exp \left(\frac{b t x + x - \mu }{b} \right) dx +
\int_{\mu}^{\infty} \exp \left(\frac{btx - x + \mu}{b} \right) dx \right)
$$

Let's solve the left-hand side integral first

$$
\int_{-\infty}^{\mu} \exp \left(\frac{b t x + x - \mu }{b} \right) dx
$$

$$
= \exp \left( \frac{-\mu}{b} \right) \int_{-\infty}^{\mu} \exp \left(\frac{b t x + x}{b} \right) dx
$$

$$
= \exp \left( \frac{-\mu}{b} \right) \int_{-\infty}^{\mu} \exp \left(\frac{x (b t + 1)}{b} \right) dx
$$

Now since $\int e^{ax}\,dx={\frac {1}{a}}e^{ax}+C$

$$
= \exp \left( \frac{-\mu}{b} \right) \frac{b}{bt + 1} \left| \exp \left(\frac{x (b t + 1)}{b} \right) \right|_{-\infty}^{\mu}
$$

When evaluating for $-\infty$, we need to have this limit converge:

$$
\lim_{x \rightarrow - \infty} \exp \left(x \frac{(b t + 1)}{b} \right)
$$

The constants have to be positive, so

$$
\frac{(b t + 1)}{b} > 0 \\
t > -\frac{1}{b}
$$

So let's evaluate and move on

$$
= \exp \left( \frac{-\mu}{b} \right) \frac{b}{bt + 1} \left( \exp \left(\frac{\mu (b t + 1)}{b} \right) - 0 \right)
$$

$$
= \exp \left( \frac{-\mu + \mu (b t + 1)}{b} \right) \frac{b}{bt + 1}
$$

$$
= \exp \left( \frac{-\mu + \mu b t + \mu}{b} \right) \frac{b}{bt + 1}
$$

$$
= \exp \left( \frac{\mu b t}{b} \right) \frac{b}{bt + 1}
$$

$$
= \exp \left( \mu  t \right) \frac{b}{bt + 1}
$$

Now the right-hand side integral:

$$
\int_{\mu}^{\infty} \exp \left(\frac{btx - x + \mu}{b} \right) dx
$$

$$
= \exp \left( \frac{\mu}{b} \right) \int_{\mu}^{\infty} \exp \left(\frac{btx - x}{b} \right) dx
$$

$$
= \exp \left( \frac{\mu}{b} \right) \int_{\mu}^{\infty} \exp \left(\frac{x(bt - 1)}{b} \right) dx
$$

And again since $\int e^{ax}\,dx={\frac {1}{a}}e^{ax}+C$

$$
= \exp \left( \frac{\mu}{b} \right) \frac{b}{bt - 1} \left| \exp \left(\frac{x(bt - 1)}{b} \right) \right|_{\mu}^{\infty}
$$

Again, for the integral to converge, we need to make sure that this goes to zero

$$
\lim_{x \rightarrow \infty} \exp \left(x \frac{(b t - 1)}{b} \right)
$$

So the constant have to be negative such that

$$
\frac{(b t - 1)}{b} < 0 \\
t < \frac{1}{b}
$$

Let's evalate and move on

$$
= \exp \left( \frac{\mu}{b} \right) \frac{b}{bt - 1} \left( 0 -\exp \left(\frac{\mu(bt - 1)}{b} \right)  \right)
$$

$$
= - \exp \left( \frac{\mu}{b} \right) \frac{b}{bt - 1} \exp \left(\frac{\mu(bt - 1)}{b} \right)
$$

$$
= - \exp \left( \frac{\mu}{b} \right) \frac{b}{bt - 1} \exp \left(\frac{\mu bt - \mu)}{b} \right)
$$

$$
= - \exp \left( \frac{\mu + \mu bt - \mu}{b} \right) \frac{b}{bt - 1}
$$

$$
= - \exp \left( \mu t  \right) \frac{b}{bt - 1}
$$

Now plug these two results in the original equation

$$
\frac{1}{2b} \left(  \int_{-\infty}^{\mu} \exp \left(\frac{b t x + x - \mu }{b} \right) dx +
\int_{\mu}^{\infty} \exp \left(\frac{btx - x + \mu}{b} \right) dx \right)
$$

$$
= \frac{1}{2b} \left(  \exp \left( \mu  t \right) \frac{b}{bt + 1}
- \exp \left( \mu t  \right) \frac{b}{bt - 1} \right)
$$

$$
= \frac{1}{2b} \exp \left( \mu t  \right) \left(   \frac{b}{bt + 1}
-  \frac{b}{bt - 1} \right)
$$

$$
= \exp \left( \mu t  \right) \left(   \frac{b}{2b(bt + 1)}
-  \frac{b}{2b(bt - 1)} \right)
$$

$$
= \exp \left( \mu t  \right) \left(   \frac{1}{2(bt + 1)}
-  \frac{1}{2(bt - 1)} \right)
$$

$$
= \frac{1}{2 }\exp \left( \mu t  \right) \left(   \frac{1}{bt + 1}
-  \frac{1}{bt - 1} \right)
$$

$$
= \frac{1}{2 }\exp \left( \mu t  \right) \left(   \frac{(bt - 1) - (bt + 1)}{b^2 t^2 - 1}  \right)
$$

$$
= \frac{1}{2 }\exp \left( \mu t  \right) \left(   \frac{bt - 1 - bt - 1)}{b^2 t^2 - 1}  \right)
$$

$$
= \frac{1}{2 }\exp \left( \mu t  \right) \left(   \frac{- 2}{b^2 t^2 - 1}  \right)
$$

$$
= \exp \left( \mu t  \right) \left(   \frac{- 1}{b^2 t^2 - 1}  \right)
$$

$$
= \exp \left( \mu t  \right) \left(   \frac{1}{- b^2 t^2 + 1}  \right)
$$

$$
=     \frac{\exp ( \mu t  )}{1 - b^2 t^2} \qquad \square
$$

for $t < \frac{1}{b}$ and  $t > -\frac{1}{b}$, which we can rewrite as $t < \frac{1}{b}$ and  $- t < \frac{1}{b}$ and thus simplify to $|t| < \frac{1}{b}$.


问:
Look for a special example: $a(n)  \lim b(n)$ as $n \rightarrow \infty$
I understand that there are already proofs that show if $a(n) < b(n)$, then $\lim a(n) < \lim b(n)$ as $n$ approaches infinity

But the question asks to find a special case in which even if $a(n) < b(n)$, yet $\lim a(n) > \lim b(n)$ as $n$ approaches infinity. I wonder if that is even possible?

Thank you very much beforehand!

答:
Error: If $a(n)<b(n)$ for all $n$ (or for all but finitely many $n$) then $\lim a(n)\le \lim b(n).$ The limits may be equal. E.g. if $a(n)=0$ for every $n,$ and $b(n)=1/n.$

But this means there can be no "special case", i.e. no exceptions.

If $\lim a(n)=a>b,$ and $a(n)\le b(n)$ for all but finitely many $n$, let $(a-b)=2\epsilon.$ Then $\epsilon>0. $

Let $S=\{n: a(n)>b(n) \}.$ Let $T=\{n: a(n)\le a-\epsilon \}.$ Then $S$ and $T$ are finite, so $S\cup T$ is finite.

If $n\not\in S\cup T$ then $$b(n)-b=(b(n)-a(n))+(a(n)-a)+(a-b)\ge$$ $$\ge (a(n)-a)+(a-b)>$$ $$>(-\epsilon)+(a-b)=\epsilon.$$ So for all but finitely many $n$ (i.e. for all $n\not\in S\cup T)$ we have $$|b(n)-b|=b(n)-b>\epsilon$$ with $\epsilon >0$, so $b(n)$ $cannot$ converge to $b.$

Title: Look for a special example: $a(n) < b(n)$, but $\lim a(n) > \lim b(n)$ as $n \rightarrow \infty$


问:
How to show that a equation is a vector space?
Let $W$ be the set $W := \{(p,q,r,s) \in \mathbb{R}^{4} \mid  x+3y+4q=0 \}$. How can I show that $W$ is a vector space? Is $(6,8,6,4)$ in $W$? How and why?
Please do all the works I am really confused about it.</p>

答:
The two essential properties for a vector space W are that the "zero vector" should be a member and that if vectors $ \ \overrightarrow{v_1} \ $ and $ \ \overrightarrow{v_2} \ $ are in W , then any linear combination  $ a \  \overrightarrow{v_1} \ + \ b \ \overrightarrow{v_2} \ $ is also in W.

Since $ \ 0 \ + \ 3 \cdot 0 \ + \ 4 \cdot 0 \ = \ 0 \ , $ the vector $ \ \langle 0 , 0 , 0 , 0 \rangle \   $ is indeed in W.

For vectors $ \ \langle -(3y_1 + 4q_1) , y_1 , z_1 , q_1 \rangle $ and $ \ \langle -(3y_2 + 4q_2) , y_2 , z_2 , q_2 \rangle \ ,  $ is it the case that

$$  \langle  \ a \ [ -(3y_1 + 4q_1) ] \ + \ b \ [ -(3y_2 + 4q_2) ]  \  , ay_1 \ + \ by_2 \ , \  az_1 \ + \ bz_2 \ , \ aq_1  \ + \ bq_2  \ \rangle $$

is also in W?  Does it still satisfy the relationship between the first, second, and fourth components?  If so, then W is closed under linear combinations and is a vector space.


问:
Notation to make a negative answer equal 0
So basically, I'm making a big equation, and one component of it is this

$$(n-5)+(n-7)$$

The n-value can be anything from 1 to far away. So, that obviously leads to some negative numbers. However, this part of the equation is only applicable whenever the n-value is greater than the subtracting numbers (respectively). So, is there some kind of notation I can use to signal that if the answer is a negative number, then it is to be dismissed, or just turned to 0? A kind of notation similar to the ceiling and floor functions, where a non-integer is turned to the closest integer (either up or down).

答:
We can use Iverson brackets in this case.

\begin{align*}
\left((n-5)+(n-7)\right)\color{blue}{[[n\geq 6]]}:=
\begin{cases}2n-12&\qquad n\geq 6\\0&\qquad n< 6\end{cases}
\end{align*}


问:
Why the direct sum of an ideal and its orthogonal is equal to the whole Lie algebra if it is semisimple?
There is a theorem that any finite dimensional semisimple Lie algebra $L$ is in fact a direct sum of simple ideals $I_1, I_2, ..., I_n$, and I am now studying its proof in Humphreys' "Introduction to Lie algebras" (section 5.2).
It starts with any ideal $I$ in $L$. We introduce $$I^{\bot}=\{x \in L: \kappa(x,y) = \mathrm{Trace}(ad_x ad_y) = 0, \forall y\in L\},$$
which is an ideal, and by the Cartan's Criterion and semi-simplicity of L, we have $I \cap I^{\bot} = 0$, which I understand.$\\$

The fact I need is that $I \oplus I^{\bot} = L$, or, equivalently $\mathrm{dim} \  I + \mathrm{dim} \  I^{\bot} = \mathrm{dim} \ L$. $\\$

In other source I found that it is implied by the non-degenarativity of a Killing form $\kappa(x,y) = \mathrm{Trace}(ad_x ad_y)$ for a semisimple $L$
$\left( \text{ non-degenerativity means that the set } S = \{x \in L: \kappa (x,y) = 0, \  \forall \ y \in L \} \text{ is equal to 0} \right)$, but I don't see how.

答:
The fact that the bilinear form $b$ defined on $L$ is non degenerated is equivalent to the fact that $d(b):L\rightarrow L^*$ defined by $d(b)(x)(y)=b(x,y)$ is an isomorphism.
$I^{\perp}$ is identified with the anihilator $I^0$ of $I$ with this isomorphism  and $dim(I)+dim(I^0)=dim(L)$.


问:
What does the syntax '$f(a)$' mean in the context of FOL set theory
Now that I have a little bit of set theory and first order logic under my belt, I have revisited some older concepts. One of these is the statement: $f(a)=b$.

Previously, I always treated this statement as purely a sort of aliasing statement, where I am claiming that the syntax $f(a)$ is an alternative way of referring to the variable $b$. However, I no longer think that is the only thing going on here.

Firstly, the syntax $'f(a)'$ is not intrinsically part of the language of set theory. Therefore, I assume it is a naming convention. That is to say:

$$f(a): = \text{something}$$

I played around with this a little bit and here is what I came up with:

Let $A$ and $B$ be sets, and let $a \in A$ and $b \in B$. Further, let $f \subseteq A \times B$ and $f$ be a function (i.e. $\forall x,y,z (\langle x,y \rangle \in f \land \langle x,z \rangle \in f \rightarrow y=z)$

In the canonical sense, we want $f(a)$ to be a reference to the second component of the ordered pair in $f$ whose first component is $a$.

Consider the set $G_{a,f}:=\{x \in B | \langle a,x \rangle \in f \}$ where $a$ and $f$ are fixed.

Because $f$ is a function and $a \in A$, $G_{a,f}$ only contains one element. Call that component $b$. So we have $G_{a,f}=\{b\}$...and we want $f(a)$ to effectively pluck out that element $b$.

Therefore, let $f(a):=\bigcup G_{a,f}$. Note, however, that:

$$\bigcup G_{a,f} = \bigcup \{b\}$$

Universally, you can prove the following statement: $$\forall S ( \bigcup \{S\}=S)$$

Knowing this, $\bigcup \{b\}=b$...but this is just the same thing as saying $f(a)=b$.

A corollary to the above comments is the following:

$\langle a,b \rangle \in f \rightarrow f(a)=b \land f(a)=b \rightarrow \langle a,b \rangle \in f$, which is a straightforward proof. Therefore:

$\langle a,b \rangle \in f \iff f(a)=b$

Is all of this in the right ballpark?

Thanks!

答:
You're making this more complicated than it needs to be: "$f(a)=b$" is simply shorthand for $$\langle a,b\rangle\in f$$ (which of course is itself a shorthand since we need to further unpack the pairing operation).

Note that this does not fold in the hypothesis that $f$ is a function - that's tacitly assumed when we use this notation. You could interpret "$f(a)=b$" as "$f$ is a function and $\langle a,b\rangle\in f$" if you prefer, though.


问:
Real-valued function solving $f(x+y)=\frac{f(x)+f(y)}{1-f(x)f(y)}$
Let $f$ be a real valued differentiable function of a single variable s.t $$f(x+y)=\frac{f(x)+f(y)}{1-f(x)f(y)},$$ $\forall x,y$ with $f(x)f(y)$ is not equal to $-1$ & $f(0)=0.$ Then prove that $f(t)=\tan(gt)$, $g$ being a constant.

答:
Let

$$f(x+y)=\frac{f(x)+f(y)}{1-f(x)f(y)} \tag{1}$$

We note that $f$ is odd; since

\begin{align*}
f(x-x) = \frac{f(x) + f(-x)}{1 - f(x) f(-x)} &\Rightarrow f(0) = \frac{f(x) + f(-x)}{1 - f(x) f(-x)}\\
 &\Rightarrow f(x) + f(-x) =0 \\
 &\Rightarrow f(-x) = - f(x)
\end{align*}

Since $f$ is differentiable the limit $\displaystyle \lim_{x \rightarrow x_0} \frac{f(x) - f(x_0)}{x-x_0} \in \mathbb{R}$. Hence,

\begin{align*}
\lim_{x \rightarrow x_0} \frac{f(x) - f(x_0)}{x-x_0} &= \lim_{h \rightarrow 0} \frac{f \left ( x_0 + h \right ) - f(x_0)}{h} \\
 &=\lim_{h \rightarrow 0} \frac{\frac{f(x_0) + f(h)}{1-f(x_0)f(h)} - f(x_0)}{h} \\
 &=\lim_{h \rightarrow 0} \frac{\frac{f(x_0) + f(h) - f(x_0) \left ( 1 - f(x_0) f(h) \right )}{1-f(x_0) f(h)}}{h} \\
 &= \lim_{h \rightarrow 0} \frac{f(x_0) + f(h) - f(x_0) + f^2(x_0) f(h)}{h\left ( 1 - f(x_0) f(h) \right )} \\
 &= \lim_{h \rightarrow 0} \frac{f(h) + f^2(x_0) f(h)}{h \left ( 1 - f(x_0) f(h) \right )} \\
 &= \lim_{h \rightarrow 0} \frac{f(h) \left ( 1 + f^2(x_0) \right )}{h \left ( 1 - f(x_0) f(h) \right )} \\
 &= \lim_{h \rightarrow 0} \frac{f(h)}{h} \cdot \frac{1 + f^2(x_0)}{1 - f(x_0) f(h)} \\
 &= f'(0) \left ( 1 + f^2(x_0) \right )
\end{align*}

Therefore ,

$$f'(x) = f'(0) + f’(0) f^2(x) \quad \forall x $$

The last differential equation is classical. Can you take it from here?


问:
Sequence version of L'Hospital's Rule
Consider two sequences  $A_n$ and $B_n$ such that $B_n$ is monotonically decreasing and both $A_n$ and $B_n$ tend to zero.

Now let us consider the limits $\lim_{n\to\infty}\left(\frac{A_n}{B_n}\right)$ and $\lim_{n\to\infty}\left(\frac{A_n-A_{n-1}}{B_n-B_{n-1}}\right)$. I guess that these two limits are equal when n tends to infinity if the second limit exists .

I tried to prove this using the following logic :

For any convergent sequence $C_n$, tending to limit $K$, $\lim_{n\to\infty}(C_n-k)= \lim_{n\to\infty}(C_n-C_{n-1})$. So as $A_n$ and $B_n$ tend to zero so $\lim_{n\to\infty}(A_n-0)= \lim_{n\to\infty} (A_n-A_{n-1})$ and the same for $B_n$. Hence if the limit second limit exists the first limit exists and is equal to the second limit.

But my stance is pretty skeptic and I feel an absence of rigor and strong reasoning.

Hope the members here can help me in strengthening the logic of this proof.

答:
This is much like the Stolz-Cesaro theorem:

Consider sequences $\{a_n\}_{n=1}^\infty,\{b_n\}_{n=1}^\infty$ such that $b_n$ is strictly increasing and unbounded.

If $\lim\limits_{n\to\infty}\left(\frac{a_n-a_{n-1}}{b_n-b_{n-1}}\right)=l
$ exists, then $\lim\limits_{n\to\infty}\dfrac{a_n}{b_n}$ exists and is equal to $l$.

A proof can be found here http://www.imomath.com/index.php?options=686

Title: Sequence version of L'Hospital's Rule


问:
$\int_{\partial R} \, f(z)\,dz=0$ if $z_0 \notin R$ for a rectangle
Let $z_0 \in \Bbb C$, $A\subset \Bbb C$ and $f:A \setminus \{z_0\} \to \Bbb C$ be holomorphic and continuous in $G$. Suppose furthermore that $R \subset A $ is a rectangle and $z_0 \notin R$. Then $$\int_{\partial R} \, f(z)\,dz=0$$

I guess this should follow from the cauchy theorem but I don't see the relation. Many thanks for your help!

答:
The intuitive thinking is, the point is outside the rectangle so if you take an open set U that covers the rectangle but not the point. Then f is holomorphic in U, and since $\partial R$ is a closed loop, because of Cauchy it integrates 0.

Thus, you need to prove there is such set: since $z_0\not\in R$ and R is closed and $\lbrace z_0\rbrace$ is compact $\implies d(z_0,R)>0$ it follows naturally that you can get an open set U.


问:
Olympiad-style question about functions satisfying condition $f(f(f(n))) = f(n+1) + 1$
QN: What functions (from non-negative integers to non-negative integers) satisfy the condition

$$f(f(f(n))) = f(n+1) + 1$$

Comment: Evidently $f(n) = n+ 1$ is one solution. Equally evidently no other polynomial will work. But there, having played around for a while, I get stuck.

There's a story behind the question, which makes me as interested in the question whether there is a systematic way of tackling questions like this, as in the question of what the answer is.

答:
Let $m=f(n+1)$, apply $f(\cdot)+1$ to both sides:
$$\begin{align} f\bigl(\,f(f(f(n)\,\bigr)+1&=f\bigl(\,f(n+1)+1\,\bigr)+1\\
&= f(m+1)+1\\
& = f(f(f(m)))\\
&=f(f(f(f(n+1)))),\end{align}$$
so with $g:=f^{\circ 4}\colon \Bbb N_0\to\Bbb N_0$, we have
$$ g(n+1)=g(n)+1,$$
and by induction
$$ g(n)=\underbrace{g(0)}_{=:c}+n.$$
Then $$f(n+c)=f(g(n))=g(f(n))=f(n)+c. $$
In particular, $g$ (and with it, $f$) is injective.
As $f^{\circ 3}(n)\ne 0$ for all $n$, $f$ and $g$ cannot be onto, hence $c>0$.
We note that this implies
$$\tag1 f^{\circ k}(n)\ne n\quad \text{if }k>0$$
as otherwise $n=f^{\circ 4k}(n)=g^{\circ k}(n)=n+kc\ne n$, contradiction.

Let $S_0=\Bbb N_0\setminus f[\Bbb N_0]$ be the points missing from the image of $f$ and $S_n=f ^{\circ n}[S_0]$. Then the $S_n$ are pairwise disjoint and $f$ induces bijections $S_n\to S_{n+1}$. As
$$\{0,\ldots, c-1\}=\Bbb N_0\setminus g[\Bbb N_0]=S_0\cup S_1\cup S_2\cup S_3,$$
we see that $c=4|S_0|$.
If $x\in (S_0\cup S_1\cup S_2)\setminus\{0,f^{-1}(c-1)\}$, then $f(x)\in (S_1\cup S_2\cup S_3)\setminus\{c-1\}$, so $f^{\circ 3}(x-1)=f(x)+1<c$ and finally $x-1\in S_0$.
We conclude that $$\tag23|S_0|-2\le \left|(S_0\cup S_1\cup S_2)\setminus\{0,f^{-1}(c-1)\}\right|\le |S_0|,$$ so $|S_0|\le 1$.
As $c>0$, we conclude $|S_0|=1$ and $c=4$.
We can write $S_n=\{a_n\}$ and thus $(a_0,a_1,a_2,a_3)$ is a permutation of $(0,1,2,3)$, and $f$ maps
$$ a_0\mapsto a_1\mapsto a_2\mapsto a_3\mapsto a_0+4\mapsto a_1+4\mapsto a_2+4\mapsto a_3+4\mapsto a_0+8\mapsto \cdots$$
By sharpness of inequality $(2)$, $0$ and $f^{-1}(3)$ are distinct elements of $S_0\cup S_1\cup S_2$ (and $f^{-1}(3)$ exists in the first place!), so one of $a_0,a_1,a_2$ is $0$, one of $a_1,a_2,a_3$ is $3$, and $f(0)\ne 3$.
So $f(0)=1$ or $f(0)=2$. If $f(0)=2$, then $4=g(0)=f(f(f(2)))=f(3)+1$, contradicting $(1)$. Therefore $f(0)=1$.
This leaves us with $(a_0,a_1a_2,a_3)$ one of
$$ (0,1,2,3), (0,1,3,2), (2,0,1,3), (2,3,0,1).$$
The first case leads to the solution
$$\tag A f(n)=n+1. $$
The second case leads to $2=f(f(f(0)))=f(1)+1=4$, contradiction.
The third case leads to $3=f(f(f(2)))=f(3)+1=7$, contradiction.
The fourth case leads to the valid solution
$$\tag B f(n)=\begin{cases}n+2&n\equiv 0,1\pmod 4\\n-2&n\equiv 2,3\pmod 4\end{cases}. $$


问:
Prove radial vector field is not tangent to $S^n$ rigorously
Well, this seems to be a very trivial fact that the radial vector field $x^i\frac{\partial}{\partial x^i} $is an outward pointing vector field on $S^n$, but after some thinking, I’m not sure how can I rigorously prove this. The first thought comes into my mind is to check this is true for the north pole, and then using symmetric argument to conclude the result, but how can I claim this symmetry argument rigorously? The second thought is to prove this by using coordinate chart, but I’m not sure what should I calculate here. Any hints?

答:
It's just the usual calculus argument. If you accept the fact that $S^n$ is defined by the equation $f(x)=\|x\|^2-1=0$, then $df_p(v)=0$ for any $v$ in the tangent space at $p\in S^n$. But $df_p(v) = 2\langle \sum x^i\frac{\partial}{\partial x^i},v\rangle$, so the given vector is in fact orthogonal to the tangent space. (This is the same thing as applying the vector field to $f$ and getting $0$.)

Now, what does outward-pointing mean? Has the assignment given an orientation on the sphere? Of course, we can all look and see it points radially outward, but you need to provide an orientation for the sphere.


问:
Compact connected surfaces such that non-null homotopic map exists from itself to $\mathbb S^1$.
First, I know that any compact connected surface is either $\mathbb S^2$, $\mathbb P^2 \# \cdots \# \mathbb P^2$ ($n$ copies), or $\mathbb T^2 \# \cdots \# \mathbb T^2$ ($n$ copies).

Also, suppose I know the following result:

-Given $X$ is a connected space with a contractible universal covering
space, $f : Y \to X$ is null homotopic if and only if the induced homomorphism $f_* : \pi_1(Y, y) \to \pi_1(X, f(y))$ is trivial for each $y \in Y$.

For the first case, I know that since the fundamental group of $\mathbb S^2$ is trivial, it must be that any continuous $f$ from $\mathbb S^2$ to $\mathbb S^1$ should be trivial.

For the second case, when $n$ is just one, since the fundamental group of the projective plane is just a group of order $2$.

However, I am not sure how to proceed with other cases, and show that there exists/does not exist non-null homotopic maps.

Also, what should I do to prove the fact I am assuming?

答:
(This should be a comment but I dont have enough reputation to make a comment)

The non-orientable surface $\Sigma$ has a contractible covering space hence it is $K(\pi_{1}(\Sigma),1)$ in order to give a map to $S^1$ which is $K(\mathbb{Z},1)$ you can give a map at the level of groups. As you pointed out the fundemental group of $\mathbb{RP}^2$#$\mathbb{RP}^2$ is the infinite dihedral group, hence you get a map to $\mathbb{Z}$.

Even in the case of orientable surfaces you can produce a lot of map by repeating the arguement above, as the universial cover of orientable surfuce of genus $\geq$ 2 is disc (or equivalently the upper half plane). Hence the question boils down to a purely group theoretical question. In particular if you have elements of infinite order in the fundemental group you have a map. Which might have several interesting geometric intepretations.


答:
If you want the smallest degree polynomial that gives these numbers (which is only one of the infinitely many possible explanations for them), we can use the calculus of finite differences:
$$\fbox{3}\quad 8\quad 17\quad 32\quad 57 \\
\fbox{5}\quad 9\quad 15\quad 25\\
\fbox{4}\quad 6\quad 10\\
\fbox{2} \quad 4\\
\fbox{2}$$
Define
$$\begin{align*}
f(n)&=3\binom{n}{0}
+5\binom{n}{1}+4\binom{n}{2}+2\binom{n}{3}+2\binom{n}{4}\\\\
&=3 + \frac{19 n}{6} + \frac{23 n^2}{12} - \frac{n^3}{6} + \frac{n^4}{12}
\end{align*}$$
Then
$$f(0)=3,\quad f(1)=8,\quad f(2)=17,\quad f(3)=32,\quad f(4)=57$$</p>

Title: finding nth term

Body:
Let
  3,8,17,32,57 . . . . .
be a pattern.How do we find the nth number?My brains are completely jammed,I am tired.I do not even recognize the pattern.I calculated a few ways,but all I want is a little hint,not the whole solution.</p>


问:
Solving a Secret List (Inspired by 'Are you the One?')
Let's say there's a list of the numbers $1$ through $N$ in some sequence, unknown to us. This list is the "correct" order. You are able to arrange those numbers in any way, and after each step you are told how many of the items on your list are in the correct position, but not which ones. The goal is to guess the order of the secret list in as few steps as possible.
The 2 questions I've been thinking of are these:

Is there a strategy that guarantees you find the correct order in $M<N!$ steps?

Given a maximum number of guesses $G<M$, is there a strategy that maximizes your probability of finding the right order in $\leq G$ steps?

So far I haven't been able to make much headway. I though I could do something by keeping a fixed order and rotating it $N-2$ times, which would put every item in all but one spot once without double counting, and the $N-1$st could be deduced given the number of correct spaces would have to add up to $N$.

I don't think it gives useful information though. I imagine in order to find out which positions are correct you'd have to have that spot repeated at least once. Thoughts?

答:
Great question! I have no idea.

What I came up with in the past 10 minutes is the following strategy, that (if my math is correct) guarantees finding the correct move in $O(N^3)$ moves. This answers your first question, but your second still remains.

To start, take any single sequence, and check how many numbers are in the correct spot. Now we keep the first number fixed, and rotate all other numbers in the sequence. We again note how many numbers are in the correct spot. We keep doing this for all the $N-1$ different rotations of all but the first number.

Now there are two possibilities: Either the sum of the correct counts equals $N-2$, or it equals $2N-2$.

In the first case, the first number in our sequence was not correct. That means that $N-2$ of the numbers after the first number reached their correct spot exactly once. The last of the $N-1$ numbers that was rotated was the number that should have been the first number of the hidden sequence, and therefore it didn't reach its correct position in this rotation.

In the second case, the first number was in the correct spot. It was therefore counted $N-1$ times, once after each rotation we did. The other $N-1$ elements of the sequence were all rotated once into the correct spot, so in total we should have $2N-2$ correct counts.

If we are in the second case, we have found the first number, and we can repeat this algorithm for the remaining $N-1$ numbers in the sequence. If we are in the first case, we choose another number to be at the front and repeat this algorithm again, hoping we have now chosen the correct number. Since we have $N$ different possible values for the beginning, so determining this first number should take at most $(N)(N-1)$ guesses: $N$ different starting numbers, and $N-1$ guesses per starting number. Since we have to do this for every length of the sequence we get that the maximum number of guesses necessary for this algorithm equals:

$$\sum_{n=2}^N (n)(n-1) = \sum_{n=2}^N n^2 - n = \frac{N(N+1)(N-1)}{3} \in O(N^3)$$

Example:

If our sequence is four digits long and the correct order would be 2314, we might try putting the 1 on front first. This would lead to the following counts:

$1234 \rightarrow 1$

$1423 \rightarrow 0$

$1342 \rightarrow 1$

Since $1 + 0 + 1 = 2 = N - 2$, we have found that $1$ is not the start of the hidden sequence. Let's now try with 2:

$2134 \rightarrow 2$

$2413 \rightarrow 2$

$2341 \rightarrow 2$

Since $2 + 2 + 2 = 6 = 2N - 2$, we have found that $2$ is indeed the first number of the hidden sequence. We now keep the 2 fixed, and check again for the part of the sequence consisting of the 1, 3 and 4.


问:
Proving that a function doesn't have a limit given $|a-b| = |f(a)-f(b)|$ for all $a,b\in \mathbb R$
For all $a,b\in \mathbb R$,

$|a-b| = |f(a)-f(b)|$.

Prove that $\;\lim\limits_{a\to \infty}|f(a)|=\infty$

What I was trying to do is show that if we assume that there is a limit, $\;\lim\limits_{a\to \infty}|f(a)|=L\;$ and therefore $\big||f(a)|-L\big|=\big||f(a)|-|L|\big| \leq |f(a)-L| < \epsilon$

Then we set $\epsilon$ to some arbitrary value and try to get a contradiction.

I also tried proving directly but couldn't get much out of it.

Is there a better approach for proving it?

Body:
Hint By the triangle inequality
$$|f(x)| \geq |f(x)-f(0)|-|f(0)| = |x-0|-|f(0)|=|x|-|f(0)|$$


问:
$(f(X_n))_{n \in \mathbb{N}_0}$ martingale implies $f'' = 0$?
I want to understand why the following statement holds:  $f \in \mathcal{C}^2(\mathbb{R})$. If for every martingale $(X_n)_{n \in \mathbb{N}_0}$, $(f(X_n))_{n \in \mathbb{N}_0}$ is also a martingale, then the second derivative of $f$ is zero.

答:
Alternative solution, which only shows that $f'' \equiv 0$ and not that $f(x) = mx+b$.

We only consider the case where $X=B$ is a Brownian Motion. By Ito's Lemma, we obtain $$df(B_t) = f'(B_t)dB_t + \frac{1}{2}f''(B_t)dt$$
since it is a martingale, we obtain that almost surely $\int_0^tf''(B_s)ds = 0$ for every $t \in \mathbb{R}_{\geq 0}$. Since $f''$ is continuous, this implies $f''(B_t) = 0$ for every $t$, and thus $f''\equiv0$.


问:
How can we prove that a set $A$ of subsets of $X$ is an algebra of $X$?
If $X$ is a finite set, and $A$ is a set of subsets of $X$, is $A$ an algebra set of $X$?

For example: If $X$ contains $5$ elements, can an algebra $A$ of $X$ have only $31$ subsets of $X$?

I tried to prove that it's wrong by contradiction, that is, not every subset of $X$ belongs to $A$ have a complement in $A$.

答:
The power set $P(X)$ contains exactly $2^{|X|}$ elements: subsets of $X$ correspond to functions $X\to\{0,1\}$ (assign $1$ to exactly those elements that are in the given subset, and for the other direction, take the preimage of $\{1\}$).

So, here if $A=P(X)$, then it is certainly a set algebra, and indeed $|A|=32$ is even.

You are right that any finite nontrivial set algebra must have an even number of elements, since no subset can be its own complement.

Moreover, any finite set algebra has cardinality $2^k$, because they naturally carry a vector space structure over the two element field $\Bbb Z/2\Bbb Z$, with the symmetric difference as addition.


问:
Are the elements of this ideal $I = (x, xy, xy^2, xy^3, ...)$ polynomials with polynomial coefficients?
I am trying to answer this question:

Let $k$ be a field and $k[x,y].$ Define the subring $A \subset k[x,y]$ by $A = k[x, xy, xy^2, xy^3, ...].$ Show that $A$ is not Noetherian.

And I got the following hint:

Hint: Consider the ideal $I = (x, xy, xy^2, xy^3, ...).$ Assume $xy^{n+1}= f_{0}x + f_1xy + \dots + f_n xy^n$ for $f_i \in A.$ Divide by $x$ and evaluate at $x=0.$}

But I have the following question:

Is this form of the elements in $I$ correct (If $I$ is assumed to be a finitely generated ideal):

Do we have that each element is a polynomial(in the picture $f_i$) in those variables $(x, xy, xy^2, xy^3, ...)$ with coefficients (in the picture $g_{i_{s_i}}$) also polynomials in $(x, xy, xy^2, xy^3, ...)$?

Could anyone clarify this to me please?

答: There is an important piece I think you are missing here: you begin by supposing $I$ has some finite generating set $\{f_1,\ldots, f_t\}$. However, in a Noetherian ring, any generating set of any ideal contains a finite generating set. The hint suggests you should begin with $\{x,xy,xy^2,\ldots\}$ as a generating set, and suppose that $I$ is actually generated by $\{x,xy,\ldots, xy^n\}$. Then $xy^{n+1}$ must be in the span of these generators, so for some $f_0,\ldots, f_n$, $xy^{n+1}=f_0x+f_1xy+\cdots+f_nxy^n$. These ``coefficients'' $f_i$ are simply elements of $A$, i.e., certain polynomials in $x$ and $y$.