问:
A question about SDE and geometric Brownian motion.
In Bernt Oksendal's Stochastic Differential Equations, Chapter 4, one has the following stochastic differential equation (whose solution is geometric Brownian motion):
$$dN_t=rN_tdt+\alpha N_tdB_t\;\;\;\text{ ie } \;\;\; N_t-N_0=r\int_0^t N_sds+\alpha\int_0^tN_sdB_s,$$
where $\alpha,r\in\mathbb{R}$ and $B_t$ is standard Brownian motion (ie $B_0=0$). After assuming that $N_t$ solves the above equation, the author abruptly deduces that
$$\frac{dN_t}{N_t}=rdt+\alpha dB_t \;\;\;\text{ie}\;\;\; \int_0^t\frac{1}{N_s}dN_s= rt+\alpha B_t. \;\;\;(*)$$

I don't understand how he obtained this directly.

What I understand for sure is that if we seek to compute
$$\int_0^t\frac{1}{N_s}dN_s $$
we apply Itô's formula for $Y_t=\ln(N_t)$ (assuming $N_t$ satisfies all the needed conditions). After some computation this yields

$$\frac{1}{N_t}dN_t=d\ln N_t+\frac{1}{2}\alpha^2dt \;\;\text{ i.e } \;\;\int_0^t\frac{1}{N_s}dN_s=\ln(N_t)-\ln(N_0)+\frac{1}{2}\alpha^2t.\;\;\;(**)$$
But at first glance, it does not seem that $(**)$ implies $(*)$. How did he obtain $(*)$? Did he use a method other than the Ito formula or am I missing something?

Thank you for the helpful answers! I've upvoted both and will accept whichever has more votes (in case of tie I'll just leave them be).

It turns out that what I was missing is the definition of the Itô integral with respect to an Itô process, which I could not find in the book. So actually by definition one has that for any Itô process of the form
$$dX_t=\alpha dt+\sigma dB_t,$$
and $Y_t$ an appropriate integrand that
$$\boxed{\int_0^tY_sdX_s:=\int_0^t\alpha Y_s ds + \int_0^t\sigma Y_sdB_s}$$
This justifies the formal notation
$$\frac{1}{N_t}dN_t=\frac{1}{N_t}(rN_tdt+\sigma N_tdB_t)= \alpha dt+\sigma dB_t,$$
and automatically gives $(*)$ when $Y_t=1/N_t$, of course, while assuming $Y_t$ meets all the necessary requirements.

答:
I find notation such as $\frac{dN_t}{N_t}=rdt+\alpha dB_t$ incredibly unfortunate, it leads to the type of confusion you have encountered (and that I used to encounter myself).

That is why I advise against using short-hand notation (at least until one becomes very confident in stochastic calculus techniques).

In long-hand notation:

$$N_t=N_0+\int_{h=0}^{h=t}rN_hdh+\int_{h=0}^{h=t}\alpha N_hdB_h$$

From the above, it should be obvious that $N_h$ cannot be taken out of the integral and brought to the LHS (which is not obvious in the short-hand notation). Ito's Lemma needs to be used, applied to $ln(N_t)$ as you point out. Let $F(N_t,t):=ln(N_t)$, then taking the derivatives:

$$\frac{\partial F}{\partial N_t}=\frac{1}{N_t}, \frac{\partial^2 F}{\partial N_t^2}=\frac{-1}{N_t^2}, \frac{\partial F}{\partial t}=0$$

Then:

$$F_t=F(N_0)+\int_{h=0}^{h=t}\left(\frac{\partial F}{\partial h}+\frac{\partial F}{\partial N_h}rN_h+0.5\frac{\partial^2 F}{\partial N_h^2}\alpha^2 N_h^2\right)dh+\int_{h=0}^{h=t}\frac{\partial F}{\partial N_t}\alpha N_hdB_h=\\=ln(N_0)+\int_{h=0}^{h=t}\left(r-0.5\alpha^2 \right)dh+\int_{h=0}^{h=t}\alpha dB_h=\\=ln(N_0)+(r-0.5\alpha^2)t+\alpha B_t$$

Because $F_t=ln(N_t)$, the final answer is (by exponentiating both sides):

$$N_t=N_0e^{(r-0.5\alpha^2)t+\alpha B_t}$$


问:
Discete Math propositional law solution
Simplify ¬(w∨(m∧¬w)) to ¬w∧¬m

So I'm just starting to get into Discrete Math and have overall had a good grasp on everything thus far, but this problem is wracking my brain. I've tried using different laws in different ways to try to work around this. I'm sure it's something simple, but I'm sure you guys have heard of zybooks that a lot of online classes use and it's not the most explanatory book in the world. So in a nutshell I'm looking for a bit of further explanation into finding this solution.

Much appreciated!

答:
''wracking my brain'' interesting.

Simplify ¬(w∨(m∧¬w)) to ¬w∧¬m

$\neg(w\vee (m\wedge \neg w))$ gives by DeMorgan

$\neg w \wedge \neg (m\wedge \neg w)$ and by DeMorgan again

$\neg w \wedge (\neg m\vee w)$ and by distributivity

$(\neg w \wedge \neg m)\vee (\neg w \wedge w)$ and since $\neg w \wedge w$ is false,

$\neg w \wedge \neg m$ as claimed.

ContentLicense: CC BY-SA 4.0

Title: Discete Math propositional law solution


问:
Why is $\frac{Z_n}{\mu^n}$ a martingale? (Galton-Watson Process.)
The following images taken from Durrett Pg 200 explain what a Galton Watson Process is and its corresponding martingale $\frac{Z_n}{\mu^n}$.

However, I don't see why $\frac{Z_n}{\mu^n}$ is a martingale based on the proof given. I know we need to show that $E[\frac{Z_{n+1}}{\mu^{n+1}}|F_n]=\frac{Z_n}{\mu^n}$ but why does it suffice to only consider the event $\{Z_n=k\}$. Don't we have to consider all events in $F_n$ to show equality? Can someone explain why the proof is valid?

Body:
We note that

$\begin{eqnarray}
\mathbb{E}(\frac{Z_{n+1}}{\mu^{n+1}}|\mathcal{F}_n) &=& \frac{1}{\mu^{n+1}}\mathbb{E}(\xi_1^{n+1}+...+\xi_{Z_n}^{n+1}|\mathcal{F}_n)\\
&=& \frac{1}{\mu^{n+1}}\sum_{k=1}^{\infty}\mathbb{E}((\xi_1^{n+1}+...+\xi_{k}^{n+1})\mathbf{1}_{\{Z_n = k\}}|\mathcal{F}_n)\\
&=& \frac{1}{\mu^{n+1}}\sum_{k=1}^{\infty}\mathbf{1}_{\{Z_n = k\}}\mathbb{E}((\xi_1^{n+1}+...+\xi_{k}^{n+1})|\mathcal{F}_n); \ \text{because } \{Z_n = k\} \text{ is }  \mathcal{F}_n\text{-measurable}\\
&=& \frac{1}{\mu^{n+1}}\sum_{k=1}^{\infty}\mathbf{1}_{\{Z_n = k\}}(\mathbb{E}(\xi_1^{n+1})+...+\mathbb{E}(\xi_{k}^{n+1})); \ \text{because } \xi_j^{n+1} \text{ is independent of } \mathcal{F}_n\\
&=& \frac{1}{\mu^{n+1}}\sum_{k=1}^{\infty}\mathbf{1}_{\{Z_n = k\}}k\mu = \frac{1}{\mu^n}\sum_{k=1}^{\infty}k\mathbf{1}_{\{Z_n = k\}} = \frac{1}{\mu^n}Z_n
\end{eqnarray}$
And we have proved that $\frac{Z_n}{\mu^n}$ is a martingale.


问:
Differential equation whose solution are circles
I want to find a differential equation, which has as solutions circles of radius $r$. And the equation needs to be of the form $F(y'',y',y)=0$.

The equation $$(x-x_0)^2 +(y-y_0)^2 = r$$

describes a circle of radius $r$ centered at the point $(x_0,y_0)$. If we differentiate this relation twice with respect to $x$, we have the equation $$2+2y''(y-y_0)+2(y')^2=0$$

which is in the correct form. I think this is too simple to be correct. Any suggestions?

答:
Your differential equation is satisfied by any circle of any radius whose centre has $y$ coordinate $y_0$.  I would interpret the question as saying that you want $x_0$ and $y_0$ to be arbitrary but $r$ to be fixed.  So you want to take your original equation and its first and second derivatives and eliminate $x_0$ and $y_0$. I get

$$ (y')^6 + 3 (y')^4 - r^2 (y'')^2 + 3 (y')^2 + 1 = 0$$


问:
Calculating the probability of drawing two cards of the same suit from a deck of dual-suit cards
I'm making a tabletop roleplaying game which uses a handmade deck of cards as a chance generator. All we need to know is that there are 16 cards and four suits, but:

Each card is of two suits;

Six cards in the deck are of two different suits;

And one card in the deck is of the same suit twice.

Thus, our cards look like this (with Hearts highlighted as an example):

Card
Suits

1
Hearts and Clubs

2
Hearts and Spades

3
Hearts and Diamonds

4
Hearts and Hearts

5
Clubs and Hearts

6
Clubs and Spades

7
Clubs and Diamonds

8
Clubs and Clubs

9
Spades and Hearts

10
Spades and Spades

11
Spades and Diamonds

12
Spades and Clubs

13
Diamonds and Spades

14
Diamonds and Diamonds

15
Diamonds and Hearts

16
Diamonds and Clubs

If I choose Hearts as my suit, and turn over a Hearts card, that's called a success. If I turn over the card with both suits as Hearts, that's two successes.

I need to calculate the probability of getting at least n successes when I draw x number of cards without replacement. For instance, it is quite common to need two successes and to draw three cards. The following sets of cards would all qualify:

One Hearts card, one Hearts card, and one other card (2 successes)

One double Hearts card and two other cards (2 successes)

One double Hearts card, one Hearts card, and one other card (3 successes)

Three Hearts cards (3 or 4 successes)

My knowledge of probability maths is basic at best, and I've set up what feels like quite a challenging system, so I'm getting nowhere calculating it. I'd be grateful for the specific probability of getting 2 successes on 3 cards, but a general formula for me to calculate successes would be even better!

答:
There are $6$ single-Heart cards and $1$ double-Heart card, so to get at least $n$ successes in $x$ draws, we must either

Draw at least $n$ single-Heart cards and no double-Heart card, or

Draw the double-Heart card and get at least $n-2$ single-Heart cards in the other $x-1$ draws.

In the first case, if we are to draw $k$ single-Heart cards, there are $\binom 6k$ ways to choose them, and there are $\binom 9{x-k}$ ways to choose the remaining $x-k$ cards from among the $9$ non-Heart cards,  so there are $$\sum_{k=n}^x{\binom 6k\binom 9{x-k}}$$ ways in all.

We can do the second case in much the same way.  The double-Heart cards must be chosen, and then we have $\binom 6k\binom 9{x-1-k}$ ways to choose $k$ single-Hearts and $x-1-k$ non-Hearts, giving $$\sum_{k=n-2}^{x-1}{\binom 6k\binom 9{x-k-1}}$$

There are $\binom{16}{x}$ ways to draw $x$ cards, so the probability is
$$\frac1{\binom{16}{x}}\left(\sum_{k=n}^x{\binom 6k\binom 9{x-k}}+\sum_{k=n-2}^{x-1}{\binom 6k\binom 9{x-k-1}}\right)$$


问:
Pointwise limit,$f$, of the sequence is not bounded
Question: Let $f_n(x) := \frac{nx}{1+nx^2}$ for $x \in A := [0, \infty)$. Show that each $f_n $is bounded on $A$, but the point-wise limit of $f$ of the sequence is not bounded on $A$.

Does $(f_n)$ converge uniformly to $f$ on $A$?

I'm having trouble showing the first part of the question, let alone completing it.

I know that $f_n(x)\leq \frac{1}{x},$ but that doesn't seem to getting me anywhere.

答:
1.) If $x=0$ then for all $n\in\mathbb{N}$ we have $f_n(0)=0$. If $x\in(0,\infty)$ we have that $1\leq\frac{1}{x}+nx\implies \frac{1}{\frac{1}{nx}+x}=\frac{nx}{1+nx^2}\leq n$. So each $f_n(x)$ is bounded by $n$.

2.) Now, consider $\lim\limits_{n\to\infty}f_n(x)=\lim\limits_{n\to\infty}\frac{nx}{1+nx^2}=?$. If we plug in $x$ values in a "certain manner" we see that the limit function goes to $\infty$. Let's plug in $x$-values according to the sequence $x_n:=\frac{1}{\sqrt{n}}$, clearly $x_n\in(0,\infty)$ for all $n\in\mathbb{N}$. Then we see that  $f_n(x_n)=\frac{nx_n}{1+nx_n^2}=\frac{\sqrt{n}}{2}$ and hence $\lim\limits_{n\to\infty}f_n(x_n)=\infty$. So the limit function $\lim\limits_{n\to\infty}f_n(x)$ is unbounded and therefore can't be uniformous convergent.

ContentLicense: CC BY-SA 3.0


问:
Convergence of $\sum \frac{\sqrt{a_n}}{n^r}$, given $\sum a_n$.
I'm preparing for the preliminary exam, and some practice problems are all of the same form. One problem is to prove that convergence of $\sum a_n$ where $a_n \geq 0$ implies convergence of $\sum \frac{\sqrt{a_n}}{n}$. I know how to prove that case using Cauchy-Schwarz (or AM-GM) as is discussed here. Now I have another problem:

Suppose $a_n \geq 0$ and the series $\sum a_n$ converges. Prove the series $\sum \frac{\sqrt{a_n}}{n^{2/3}}$ also converges.

Give an example of a divergent series $\sum a_n$ such that the series $\sum \frac{\sqrt{a_n}}{n^{2/3}}$ converges.

By Abel's test it's easy to see that, in general, convergence at $r$ implies convergence at any $r' > r$.
I'm not sure how to go about it for $r = 2/3$. Maybe it follows from the case $r = 1$ but I don't see how.

答:
You can use G.M $\leq$ A.M here too on the numbers $\frac{1}{n^{4/3}}$ and $a_n$.      \

For the second part you can take $a_n=\frac{1}{n}$. \

Use the convergence of $\sum\frac{1}{n^p}$ for $p>1$.


问:
A Probability Question of Infinite situation
Consider this situation:

The computer will randomly output 1 or 0, the probability of each one is 1/2.

The computer will continuously output until the number of 1 reaches 100.

Now,the number of 0 will be 1 or 2 or 3 or ...

My question is the expected number of 0s.

(

There is another explanation:

Let N be the number of zeros before there are a hundred ones

I just want to know the expected value of N
)

Maybe it does not meet the definition, I just want to say----which number of 0 is the most possible?

I do this question with the binormial distribution,I think it's may be

$$\displaystyle\sum_{n=0}^{\infty} C_{100+n}^{n}\displaystyle\frac{n}{2^{100+n}}$$

However,I don't know how to deal with it.

On the other hand, I'm not sure , I think my method is strange...

答:
Let $X_i \in \{0,1\}$, $i=1, 2 \cdots M$, with $\sum_{i=1}^{M-1} X_i=99$, $X_{M}=1$

Then $N=M-100$ and $E[N \mid M] = M - 100$ and

$$E[N]= E [E[N \mid M]] = E[M] - 100$$

To compute $M$, partition the sequence in subsequences consisting on $k$ (perhaps zero) $0$'s followed by one $1$. Let $Y_j$ be the length of these subsquences, hence $Y_j = 1,2 \cdots$ with
$M=\sum_{j=1}^{100} Y_j$

Now each $Y_j$ follows a geometric distribution  (number of tries till first success) with $E[Y_j] = 2$

Hence $$E[N]= 100 E[Y_j] - 100 = 100$$

Put in another way, for each $1$ we expect before a run of $0$'s which have average length one. Hence, in average, we expect the same number of zeroes and ones.


问:
Uniform Distributions Ratio
Let $\xi$ is $U(0, 1)$, $\nu$ is $U(0, 1)$. What type of distribution is $g = \frac{\xi}{\nu}$? I have build logarithm of $g$, here is the plot but it seems like it is not normal distribution:
Distribution Hist

答:
How did you receive the last statement? –

after calculating the jacobian that is $|J|=\frac{u}{g^2}$ you conclude that the joint density is

$$f(u,g)=\frac{u}{g^2}$$

but in the system you see that

$$0<\nu=\frac{u}{g}<1$$

Now, when $0<g<1$ this means that $0<u<g$ thus in this interval to get $f(g)$ you have to integrate the joint density in

$$f(g)=\int_0^g \frac{u}{g^2}du=\frac{1}{2}$$

When $g>1$ there are no problem for $\frac{u}{g}$ to be less than one thus you integrate

$$f(g)=\int_0^1 \frac{u}{g^2}du=\frac{1}{2g^2}$$

... put all together and get the result

this is the drawing of your density

Title: Uniform Distributions Ratio


问:
Pairwise Attracting Sequences Bounded Proof
We say that a pair of sequences {$x_n$}, {$y_n$} are pairwise attractive if there is a positive number $\rho<1$ such that $|x_{n+1}-y_{n+1}|<\rho |x_n-y_n|$

Prove that a pairwise attracting sequences $\{x_n\}$ and $\{y_n\}$ with $\{x_n\}$ bounded implies that $\{y_n\}$ is bounded.

I'm struggling with this proof. I know that if $x_n$ is a bounded sequence then it's true that there exists some numbers $a,b$ such that $a\leq x_n \leq b$. But I'm having difficulty showing how this relates to the pairwise formula. Is there some other relation that I am missing regarding bounded sequences and relations between terms?

答:
By induction, you find that $|x_{n}-y_{n}| \leq \rho^n |x_0-y_0| \to 0$ since $\rho < 1$.

Then $|y_n| \leq |y_n-x_n| + |x_n|$. For $n$ large the first term is at most $1$ and the second term is at most $B$ where $B$ is the bound on $(x_n)$.


答:
Applying the Cauchy-Schwarz inequality

$$\left|\sum_{n=1}^N x_ny_n\right|\le \sqrt{\sum_{n=1}^N x_n^2\sum_{n=1}^N y_n^2}$$

to the case for which $x_n=\sqrt{a_n}$ and $y_n=\frac1{n^r}$, we find that

$$\sum_{n=1}^N \frac{\sqrt a_n}{n^r}\le \sqrt{\sum_{n=1}^N a_n \sum_{n=1}^N \frac1{n^{2r}}}\tag1$$

As $N\to \infty$, the first sum on the right-hand side of $(1)$ converges by assumption while the second converges whenever $r>1/2$.  Inasmuch as $2/3>1/2$, the the series $\sum_{n=1}^\infty \frac{\sqrt a_n}{n^{2/3}}$ converges.

The condition that $\sum_{n=1}^\infty a_n$ converges is not a necessary one for the series $\sum_{n=1}^\infty \frac{\sqrt a_n}{n^r}$ to converge.   For example, if $a_n=\frac1{n^p}$, $p<1$, then the series $\sum_{n=1}^\infty \frac{\sqrt a_n}{n^r}$ converges for $r>1-p/2$.

Title: Convergence of $\sum \frac{\sqrt{a_n}}{n^r}$, given $\sum a_n$.

Body:
I'm preparing for the preliminary exam, and some practice problems are all of the same form. One problem is to prove that convergence of $\sum a_n$ where $a_n \geq 0$ implies convergence of $\sum \frac{\sqrt{a_n}}{n}$. I know how to prove that case using Cauchy-Schwarz (or AM-GM) as is discussed here. Now I have another problem:

Suppose $a_n \geq 0$ and the series $\sum a_n$ converges. Prove the series $\sum \frac{\sqrt{a_n}}{n^{2/3}}$ also converges.

Give an example of a divergent series $\sum a_n$ such that the series $\sum \frac{\sqrt{a_n}}{n^{2/3}}$ converges.

By Abel's test it's easy to see that, in general, convergence at $r$ implies convergence at any $r' > r$.
I'm not sure how to go about it for $r = 2/3$. Maybe it follows from the case $r = 1$ but I don't see how.


问:
Product of two compact topological spaces is compact.
Proof:

I am following this. But, I feel, I am missing something.

Consider two compact spaces $X_1$ and $X_2$ and some cover $U$ of their product space. Consider an element $x\in X_1$. The sets $A_{x,y}$ within $U$ contain $(x,y)$ for each $y\in X_2$.
Now, define $\pi_2:A_{(x,y)}$ which forms an open cover for $X_2$ and has finite subcover $\{A_{(x,y_i)}\}$.

```<code>It's clear up to this point.
</code>```

Let $\{B_x\}$ be the intersection of $\{\pi_1:(A_y)\}$ within $A_{y_i}$, which is open. Thus, $\{B_x\}$ forms an open cover, which has a finite subcover, $\{B_{x_i}\}$. The corresponding sets $\{A_{x_i,y_i}\}$ is finite, and forms an open subcover of the set.

First, we know the covers cover $X_1$ and $X_2$ respectively. But, how do we know that they cover $X_1*X_2$? Second, I believe $\pi_1$ is not necessarily non-empty. So how can we be sure that it is open?

I would appreciate any comments or other opinions.

P.S. The notations may be, like in original source, somewhat ambigious too.

答: Each $x\in X$ is element of some $B_{x_j}=\bigcap_{i=1}^{n_{x_j}}\pi_X(A_{x_j,y^{x_j}_i})$. It suffices to show that $\{x\}\times Y\subseteq\bigcup_{i=1}^{n_{x_j}}A_{x_j,y^{x_j}_i}$. (I put some super-index on the $y_i$'s because they depend on $x_j$.) At this point it is crucial that the $A\in S$ are products of open sets in $X$ and $Y$, and not just open sets in $X\times Y$. Then, since $x\in\bigcap_{i=1}^{n_{x_j}}\pi_X(A_{x_j,y^{x_j}_i})$ and for each $y\in Y$ there is a $y^{x_j}_k$ ($1\le k\le n_{x_j}$) such that $y\in\pi_Y(A_{x_j,y^{x_j}_k})$, you can derive that $(x,y)\in A_{x_j,y^{x_j}_k}$.</p>


问:
A recursive sequence $c_n =\sqrt{2+\sqrt{2+...+\sqrt{2}}}$
We have a recursive sequence $$c_n =\sqrt{2+\sqrt{2+...+\sqrt{2}}}$$ with $n$ square roots.

We can obtain a recursive formula: $$c_n =\sqrt{2+c_{n-1}}\\
c_{n+1}=\sqrt{2+c_n}$$

Now we show that the sequence is increasing:
$$c_{n+1} \geq c_n \rightarrow\text{  This is what we need to prove}$$
$$
\sqrt{2+c_n} \geq \sqrt{2+c_{n-1}}\\
2+c_n \geq 2+c_{n-1}\\
c_n \geq c_{n-1} \implies c_{n+1} \geq c_n
$$
Let's assume that a limit $c$ exists:
$$c = \sqrt{2+c}$$
$$c^2-c-2 = 0 \iff (c-2)(c-1)=0$$
So if the limit exists it is either 2 or 1. We know that it is not 1 since $c_1 > 1$ and the sequence is increasing.
$$c_n \leq 2\\
c_1 \text{ holds}\\$$
Now let's see for $n\rightarrow n+1$:
$$
c_{n+1} \leq 2\\
\sqrt{2+c_{n}} \leq \sqrt{2+2}\leq 2$$
Is the proof that the sequence is increasing  and is bounded sufficient?

答:
We you define the recursive formula you need to define $c_0 = \sqrt 2$

you don't need to define both $c_n$ and $c_{n+1}$.  Defining $c_n = \sqrt{2 + c_{n-1}}$ when $n>1$ is enough.  Or defining $c_{n+1}=\sqrt{2 + c_n}$ is enough.

We can streamline our prove that $c_n$ is increasing and bounded by $2$ in one fell induction step.

$0 \le \sqrt{2} \le c_0 < 2$.

And if $\sqrt {2} \le c_n < 2$ then $c_{n+1} =\sqrt {2 +c_n} > \sqrt {c_n^2 + c_n} > \sqrt{c_n^2} = c_n \ge \sqrt 2$.

And $c_{n+1}= \sqrt{2+c_n} < \sqrt{2 + 2} = 2$.

(Your inductions had no base cases!)

Then as we know it is bounded above and increasing $\lim_{n\to \infty} c_n=c$ exists.

And as $\lim_{n\to \infty} c_n = \lim_{n\to \infty} c_{n+1} = \lim_{n\to\infty} \sqrt {2 + c_n} = \sqrt{2 + \lim_{n\to\infty}{c_n}}$ so

$c = \sqrt {2+c} \ge 0$ so

$c^2 - c -2 =0$

$(c-2)(c+1) = 0$ so $c = 2$ or $c = -1$ but $c> 0$ so $c = 2$.


问:
Numerical method to solve PDE system $\mathbf{A}\dot{\mathbf{U}} + \mathbf{B}\mathbf{U}^{'} + F(\mathbf{U}) = 0$
I would like to numerically solve a system of PDEs of the form

$$\mathbf{A}\dot{\mathbf{U}} + \mathbf{B}\mathbf{U}^{'} + F(\mathbf{U}) = 0$$

Where, $\dot{\mathbf{U}}(s, t)$ represents the time derivative, $\mathbf{U}^{'}(s, t)$ represents the space derivative, $\mathbf{U}$ is the vector of unknowns, and $F(\mathbf{U})$ is a nonlinear function of $\mathbf{U}$.

Both $\mathbf{A}$, and $\mathbf{B}$ are constant singular matrices.

So, the given system is a coupled system of PDEs and ODEs.

How to numerically solve such system of PDEs given initial and boundary conditions?. Which numerical method will be suitable?

答:
There are several related posts on this site where a possible and convenient method is presented. A possible approach known as operator splitting consists in writing two PDE systems
\begin{aligned}
&(S_1):\qquad {\bf A}\dot{\bf U} + {\bf B} {\bf U}' = 0 \\
&(S_2):\qquad {\bf A}\dot{\bf U} + F({\bf U}) = 0
\end{aligned}
which are then integrated successively. The splitting error resulting from this process is often small enough compared to the numerical errors corresponding to the integration of $S_1$ and $S_2$, see this post. For the  numerical resolution of $S_1$ a finite-difference method could be used, or a finite-volume scheme for instance. For the numerical resolution of $S_2$, a well-suited ODE solver will do. However, for most schemes the matrix $\bf A$ is required non-singular. One may need to introduce a change of variable or a perturbation of the initial system to rewrite it in a non-singular form. The most appropriate approach is very likely to depend on the form of the matrices and functions involved.


问:
Square Brackets With Superscript & Subscript on Vector Space
Does anybody know what the following notation means?
(I'm referring to the subscript & superscript to the right of the square brackets)

$\psi(T)=[T]_β^γ$

In this case, $T: V \rightarrow W$ is a linear map on finite-dimensional vector spaces $V$ & $W$ (over the same field $F$) with dimensions $n$ and $m$ resp. And, $\beta$ & $\gamma$ are ordered bases for resp. $V$ & $W$. Lastly, $\psi$ is the isomorphism $\psi:L(V,W)→M_{m\times n}(F)$.

(Full context here Proving isomorphism between linear maps and matrices)

答:
The expression on the right represents the matrix of the operator $T:V\to W$ with respect to the bases $\beta$ of $V$ and $\gamma$ of $W$.

More specifically, one evaluates T at the $j$-th basis vector in $\beta$ and expresses the output as a linear combination of basis vectors from $\gamma$. The respective coefficients in the linear combination form the $j$-th column of the matrix $[T]_{\beta}^{\gamma}$.


问:
Intersection of two vectors with tails?
I'm working on a 2 degree of freedom simulation of gas dynamics.

Suppose you have two vectors in the x-y plane. Each vector has two tails, or coordinates where they start. Because I'm looking for a point in the future where the vectors will intersect, I can treat the vectors as rays, as long as they maintain the same direction and tail. How can you find where, or if, they intersect?

The Vectors in the program are stores as having an x-Magnitude, y-Magnitude, and a coordinate position for their tails.

答:
Each vector can be represented as $V_k=a_k+s_kb_k$ where $a_k$ are the start points, $b_k$ the direction vectors and $s_k$ distances (scalar) along the vectors.  Use $V_1=V_2$ to find values for $s_k$, where the vectors meet.

Note that $a_k$ and $b_k$ are known.  Answer is given by solving for $s_k$.

Title: Intersection of two vectors with tails?


问:
Prove that every compact subspace of a locally compact space $X$ has a compact neighbourhood

Let $X$ be locally compact space.Prove that every compact subspace has a compact neighbourhood.

We should prove that for any $K$ compact there exists $H\subset X$ compact such that $K \subset H^{\circ}$

Proof

Let $K\subset X$, consider $x\in K$ since $X$ is locally compact then exits an open set  $U_{x}$ that contains $x$ and compact $K_{x}$ such that $x\in U_{x}\subset K_{x}$.

Let $H=\bigcup_{x\in K}K_{x}$ which is compact.

Now we want prove that  $H^{\circ}=\bigcup_{x\in K}U_{x}$

let $m\in H^{\circ}$ then exists a open neightbourhood $U_{m}\subset H$ that contains $m$
but then $m\in\bigcup_{x\in K}U_{x} $

The conversely is always true.

Now the results follows as well.
I never use that $K$ is compact, I don´t know where I made mistake.

答:
Use that $K \subseteq \bigcup_{x \in K} U_x$ which is an open so there is a finite subcover, i.e. a finite $F \subseteq K$ such that

$$K \subseteq \bigcup_{x \in F} U_x$$

and then we can use that this union of open sets is an open neighbourhood of $K$ and is contained in

$$K=\bigcup_{x \in F} K_x$$

which is compact as a finite union of compact sets. You make the mistake of saying that an infinite union of compact sets (your $H$) is compact while this need not be the case at all: recall that any set is a union of (compact) singletons...

So we use compactness to reduce to finitely many open neighbourhoods and corresponding compact supersets. $K^\circ$ is then as required.


问:
Simple graph with odd circuit can't be 2-colored.
Show that a simple graph that has a circuit with an od number of vertices in it cannot be coloured using two colours.</p>

答:
Use proof by contradiction; suppose that you could color the graph with 2 color and then consider a circuit with an odd number of vertices, what will be happened then?

P.S : because of the homework tag I rather not to clarify more.


问:
Let $a>R>0$. Calculate the volume of the body that meets the differences $x^2+y^2+z^2 \leq a^2$ and $x^2+y^2 \leq R^2$
Let $a > R > 0$. Calculate the volume of the body constricted by $x^2+y^2+z^2 \leq a^2$ and $x^2+y^2 \leq R^2$

I don't know where to start with this problem. The constraints form a cylinder which body I need to integrate the volume of. Only hint that I've gotten is that it might be helpful to use polar-coordinates. Would really appreciate help!

答:
I'm assuming you mean $x^2+y^2+z^2\leq a$. In this case we are looking for the volume between the cylinder and the sphere. I would use cylindrical coordinates (https://en.wikipedia.org/wiki/Cylindrical_coordinate_system) which are the same thing as polar except there is also a z coordinate. Then $x^2+y^2\leq R$ becomes $r^2\leq R$. Similarly, $x^2+y^2+z^2 \leq a$ becomes $r^2+z^2 \leq a$. Therefore, $z^2 \leq a-r^2$. This gives you an upper and lower bound on $z$. Finally, if you visualize the shape you can figure out bounds for $\theta$. Your integral should be in the form
$$\int_{\text{lower theta bound}}^{\text{upper theta bound}} \int_{\text{lower r bound}}^{\text{upper r bound}} \int_{-\sqrt{a-r^2}}^{\sqrt{a-r^2}} 1 dz dr d\theta$$
I'll let you figure out the remaining bounds.

We integrate over $1$ because that gives us volume in the same way that in $\mathbb{R}^2$ integrating over 1 gives the area of the region we are integrating over.

Note that the square roots are real because $r^2\leq R^2 \leq a^2$.


问:
How to calculate module $a\cdot7 \equiv1\pmod8$?
How should i calculate following module ?

$a\cdot7 \equiv1\pmod8$

is value of $a = 1$ ?

Thanks.

答: Hint: $ $ mod $\,8\!:\ 8\equiv 0\,\Rightarrow\,\color{#c00}{7\equiv -1},\ $ so $\ \color{#c00}7a\equiv 1\,\Rightarrow (\color{#c00}{-1})a\equiv 1\,\Rightarrow\,a\equiv\, \ldots$</p>


问:
Sufficiency and completeness in two distribution
I have two variables $X_1$ and $X_2$ which follows exponential and double exponential with rate $\theta$

$X_1\sim\frac1\theta \exp((-x/\theta)),\quad X>0$

$X_2\sim\frac2\theta \exp(-2x/\theta)),\quad X>0$

$X_1$ and $X_2$ are independent. Then

How can we prove $X_1 + 2X_2$ is sufficient?

And how to check that $X_1 + X_2$ is sufficient?

$X_2$ is sufficient for $f_2$ but I am confused

答: $X_1+2X_2$ is sufficient by the factorization theorem. It is also a minimal sufficient statistic.  Therefore, $X_1+X-2$ is not sufficient because it is not a function of the minimal sufficient statistic. You can also prove it is not sufficient directly using the same idea described here.


问:
A question about SDE and geometric Brownian motion.
In Bernt Oksendal's Stochastic Differential Equations, Chapter 4, one has the following stochastic differential equation (whose solution is geometric Brownian motion):
$$dN_t=rN_tdt+\alpha N_tdB_t\;\;\;\text{ ie } \;\;\; N_t-N_0=r\int_0^t N_sds+\alpha\int_0^tN_sdB_s,$$
where $\alpha,r\in\mathbb{R}$ and $B_t$ is standard Brownian motion (ie $B_0=0$). After assuming that $N_t$ solves the above equation, the author abruptly deduces that
$$\frac{dN_t}{N_t}=rdt+\alpha dB_t \;\;\;\text{ie}\;\;\; \int_0^t\frac{1}{N_s}dN_s= rt+\alpha B_t. \;\;\;(*)$$

I don't understand how he obtained this directly.

What I understand for sure is that if we seek to compute
$$\int_0^t\frac{1}{N_s}dN_s $$
we apply Itô's formula for $Y_t=\ln(N_t)$ (assuming $N_t$ satisfies all the needed conditions). After some computation this yields

$$\frac{1}{N_t}dN_t=d\ln N_t+\frac{1}{2}\alpha^2dt \;\;\text{ i.e } \;\;\int_0^t\frac{1}{N_s}dN_s=\ln(N_t)-\ln(N_0)+\frac{1}{2}\alpha^2t.\;\;\;(**)$$
But at first glance, it does not seem that $(**)$ implies $(*)$. How did he obtain $(*)$? Did he use a method other than the Ito formula or am I missing something?

Thank you for the helpful answers! I've upvoted both and will accept whichever has more votes (in case of tie I'll just leave them be).

It turns out that what I was missing is the definition of the Itô integral with respect to an Itô process, which I could not find in the book. So actually by definition one has that for any Itô process of the form
$$dX_t=\alpha dt+\sigma dB_t,$$
and $Y_t$ an appropriate integrand that
$$\boxed{\int_0^tY_sdX_s:=\int_0^t\alpha Y_s ds + \int_0^t\sigma Y_sdB_s}$$
This justifies the formal notation
$$\frac{1}{N_t}dN_t=\frac{1}{N_t}(rN_tdt+\sigma N_tdB_t)= \alpha dt+\sigma dB_t,$$
and automatically gives $(*)$ when $Y_t=1/N_t$, of course, while assuming $Y_t$ meets all the necessary requirements.

答:
Typically associativity of the integral is proved early on. If $X$ is a semimartingale and the integral $K \cdot X = \int K dX$ makes sense then the integral $(HK) \cdot X = \int HK dX$ makes sense if and only if the integral $H \cdot (K\cdot X) = \int H d(\int K dX)$ makes sense, in which case they are equal. In "differential form", this is $H\, d(K dX) = (HK) dX$. I am assuming you are already comfortable with linearity of stochastic integrals.

Thus taking $dN_t = N_t r dt + N_t\alpha dB_t$ and integrating $1/N_t$ with respect to the semimartingales defined by either side gives
$$
(1/N_t) dN_t = (1/N_t) d(N_t r dt + N_t\alpha dB_t) = (1/N_t) d(N_t r dt)+ (1/N_t)d( N_t\alpha dB_t) = r dt + \alpha dB_t.
$$

There is no need to use Ito here, it is simply associativity of the integral. The key idea is that we are not "dividing by $N_t$", instead we are integrating $1/N_t$ with respect to two (equal) semimartingales. In differential form, it just looks like dividing.


问:
Convex set. Proof.
Prove that if A is convex set and $\alpha, \beta ≥ 0$ then $(\alpha + \beta)A = \alpha A + \beta A$

What came first on my mind is that I have to show that $(\alpha + \beta)A\subset \alpha A + \beta A$ and $\alpha A + \beta A\subset (\alpha + \beta)A$. But I don't know how to use convexity of A to show this.

答:
One direction doesn't require convexity.  Let $(\alpha+\beta)a\in (\alpha+\beta)A$.  Then $\alpha a\in \alpha A$ and $\beta a\in \beta A$, and $(\alpha+\beta)a=\alpha a+\beta a$, so $(\alpha+\beta)a\in \alpha A + \beta A$.

For the other direction, let $\alpha a_1 + \beta a_2\in \alpha A + \beta A$.  We need to find some $a_3\in A$ such that $\alpha a_1+\beta a_2=(\alpha+\beta)a_3$.  To find such an $a_3$, divide both sides by $(\alpha+\beta)$. (footnote*).  We set $a_3=\frac{\alpha}{\alpha+\beta}a_1+\frac{\beta}{\alpha+\beta}a_2$.  Because $A$ is convex and these fractions are in $(0,1)$ and add up to 1, $a_3\in A$ and we're done.

(footnote*) This proof frays at the edges if $\alpha=\beta=0$, so this special case must be handle separately; it's easy though since $(0+0)A=0A+0A=\{0\}$ for any $A$.

Title: Convex set. Proof.


问:
Probability of a family name surviving based on number of offspring
If a family patriarch (e.g name = "Adams") has x number of children, and the probability of having a son or daughter is 50%, then what is the probability of his family name surviving for 3 or 4 or more generations?

(Assume all children marry, and only sons carry on the Adams name.)

E.g. If he has 1 child , probability  to survive 1 generation is 50%.

If he has 2 children, who each have 2 children,
75% (?) chance to survive 1 generation, then ? % chance to survive 2 generations, 3 generations, and so on?

Are there any models based on probability that can present this, e.g. in a simple program with 2 inputs, # of children per generation, and # of generations?

If this could be turned into a formula that can be charted, would the line approach an asymptote of 100% and resemble a mirrored exponential line?

答:
Each male child has (passes the family name to) $n$ children and each child has probability $p=1/2$ of being male. That is, each male child is modelled by a binomial random variable $X\sim B(n,p=1/2)$.

To investigate extinction (survival), we model a branching stochastic process. That is, we have a Galton–Watson process - one of the most common formulations of a branching process. If $Z_t$ is the state in generation $t$, we have the recurrence

$$
Z_{t+1}=\sum_{i=1}^{Z_{t}} X_{t, i}.
$$

Let $P_t$ be the extinction probability by the generation $t$. This extinction process can be analyzed using the method of probability generating function  $G$. We have that

$$
P_{t}=G\left(P_{t-1}\right), P(0)=0.
$$

In our case, we have random variables $X_{n,i}$ with binomial distribution $X_{n,i}\sim B(n,p=1/2)$ and we know that the PGF of a binomial random variable is $G(z)=[(1-p)+p z]^{n}$. It follows that

$$
P_{t+1}=\left(\frac{P_t+1}{2}\right)^n, P(0)=0.
$$

Finally, probability of the family name surviving for at least first $t$ generations is the complementary probability of the extinction probability $P_t$ by the generation $t$, which is $1-P_t$.

When $n\le 2$, the survival probability goes to zero $(1-P_t)\to0$. This is not hard to find, as seen in "Calculating the extinction probability for $X_{n,i}\sim B(2,p)$".

When $n=3$, the survival probability goes to  $(1-P_t)\to [1-(\sqrt{5}-2)]\approx 0.763932\dots$ as already calculated in "Branching process and calculating the probability of extinction  for $X_{n,i}\sim B(3,1/2)$".

For larger $n$, the survival probability $(1-P_t)$ indeed approaches larger values.

For example, for $n=3,4,5,6$ you can plot $1-P_t$ for $t=1,2,\dots,10$,


问:
Determining compactness of the set $S=\{\frac{1}{m}+\frac{1}{n}:m\in \mathbb N, n\in \mathbb N\}\cup \{0\}$ in metric space $\mathbb R$
Given a set $S=\{\frac{1}{m}+\frac{1}{n}:m\in \mathbb N, n\in \mathbb N\}\cup \{0\}$.

My attempt:

Claim $1:0$ is a limit point of set $S$.

Proof: Given any $\epsilon \gt 0, \exists N_\epsilon \in \mathbb N: \frac1{N_\epsilon}\lt \epsilon/2$. Choosing $m,n \ge N_\epsilon$,we have $\frac1{n}\le \frac1{N_\epsilon}\lt \epsilon/2$ and $\frac1{m}\le \frac1{N_\epsilon}\lt \epsilon/2 $. Thus $0\lt\frac{1}{m}+\frac{1}{n}\lt \epsilon$. Since $\epsilon \gt 0$ is arbitrary, we conclude that $0$ is a limit point of $S$.

Claim $2$: All elements of set $T=\{1/n: n\in \mathbb N\}$ are also limit points of $S$.

Proof: Given any $\epsilon \gt 0, \exists M_\epsilon \in \mathbb N: \frac1{M_\epsilon}\lt \epsilon\implies \frac1{m}\lt \epsilon \;\;\forall m\ge M_\epsilon\implies -\epsilon\lt \frac1{m}\lt \epsilon$. Hence for any $\frac1{n}\in T$, we have $\frac 1{n}-\epsilon\lt \frac 1{n}+ \frac1{m}\lt \frac1{n}+\epsilon$. Just as previosuly, it follows that all elements of $T$ are limit points of $S$.

Heine-Borel theorem states that a set $S$ in $\mathbb R$ is compact $\iff S$ is closed and bounded. Indeed, in the present case, $S$ is bounded as $2$ is one upper bound.

Since, $1/n \notin S$ for any $n\in \mathbb N$, $S$ is not closed and therefore, $S$ can't be compact.

Is my conclusion correct? Thanks.

Edit: How do I show that $S$ does not have limit points except $0$ and $\{1/n: n\in \mathbb N\}$?

答:
As noted in the comments, $T\subseteq S$: for each $\frac1n\in T$ we have $\frac1n=\frac1{2n}+\frac1{2n}$. Even if the definition of $S$ were changed to require $m\ne n$, we could write

$$\frac1n=\frac1{n+1}+\frac1{n(n+1)}$$

for any $n\in\Bbb Z^+$. Thus, if you can show that $0$ and the numbers $\frac1n$ for $n\in\Bbb Z^+$ are the only limit points of $S$, you can conclude that $S$ is compact. I’ll sketch an argument and let you fill in the details.

Let $x\in\Bbb R\setminus S$. If $x<0$, then $(x-1,0)$ is an open interval around $x$ disjoint from $S$, so $x$ is not a limit point of $S$. Similarly, if $x>2$, then $(2,x+1)$ is an open interval around $x$ disjoint from $S$.

Now suppose that $1<x<2$. Since $\frac1n+\frac1m\le 1$ when $n,k\ge 2$,

$$S\cap(1,2)\subseteq\left\{1+\frac1n:n\in\Bbb Z^+\right\}\,,$$

and therefore

$$(1,2)\setminus S=\bigcup_{n\ge 1}\left(1+\frac1{n+1},1+\frac1n\right)\,,$$

so there is a unique $n\in\Bbb Z^+$ such that $\left(1+\frac1{n+1},1+\frac1n\right)$ is an open interval around $x$ disjoint from $S$.

Finally, suppose that $0<x<1$. There is a unique $n\in\Bbb Z^+$ such that

$$\frac1{n+1}<x<\frac1n\,.$$

Suppose that $\frac1k+\frac1\ell\in S\cap\left(\frac1{n+1},\frac1n\right)$, with $k\le\ell$.

Show that $k\le 2n+2$.

Show that for each positive integer $k\le 2n+2$ there are only finitely many $\ell\in\Bbb Z^+$ such that $\frac1k+\frac1\ell<\frac1n$.

Conclude that there are $a,b\in S$ such that $a=\max\{s\in S:s<x\}$ and $b=\min\{s\in S:x<s\}$, so that $(a,b)$ is an open interval around $x$ disjoint from $S$.


问:
When is $\beta(G)=2\alpha'(G)$ for a graph $G$?
From a question (see link) here at math stackexchange I saw that $\beta(G)\leq 2\alpha '(G)$ and I think I understand the proof. What I am not sure is when this inequality turns into an equality, i.e. for what graph $G$ is $\beta(G)=2\alpha'(G)$? Is there any such graph?

$\alpha'(G)$  is the number of edges in a maximum matching of G.

$\beta'(G)$ is the minimum cardinality of a vertex cover.

For every graph $G$, prove that $\beta(G)\leq2\alpha'(G)$

答:
The complete graph $K_5$ on $5$ vertices is such an example. It has $\beta'(K_5)=4$ and $\alpha'(G)=2$.

In fact all disjoint unions of odd complete graphs are examples in which equality holds. See this discussion on MathOverflow.


问:
Product formula for sign of permutation
I would like help (a complete proof) that $$ (-1)^{Inv(\sigma)} = \prod_{1\leq i<j\leq n} \frac{\sigma(j)-\sigma(i)}{j-i} ,$$ where $Inv(\cdot)$ is the number of inversions of the permutation received as argument and $\sigma$ is a permutation in $S_n$.

The formula seems intuitive somehow,  but I cannot manage to fully write a proof for it. The factors of the numerator product seem to coincide with the ones in the denominator, albeit with a negative sign where inversions are happening, but I can't fully convince myself each element at the numerator has exactly one correspondent in the denominator and vice versa.

EDIT: My book states as an obvious fact that the product in the numerator is equal to the one in the denominator, where each factor is multiplied by $\epsilon(i,j)$ ($-1$ iff (i,j) is an inversion for $\sigma$).. I can finish the proof from here, but it is not obvious to me.

答:
Let $P = \{(i,j) \mid 1 \le i < j \le n \}$. Instead of $\prod_{1\leq i<j\leq n}$ we shall now write $\prod_{(i,j) \in P}$

Let $Q$ denote the set of all sets $\{i, j\}$ with $1 \le i, j \le n$ and $i \ne j$, i.e. the set of all subsets of $I = \{ 1, \ldots, n\}$ that have exactly two elements. Define
$$s : P \to Q, s((i,j)) = \{i,j\} .$$
This is a bijection. Therefore for each function $f : Q \to \mathbb R$
$$\prod_{(i,j) \in P} f(s((i,j))) =  \prod_{M \in Q} f(M) \tag{1}.$$

For a subset $M \subset I$ we define its diameter as $d(M) = \max( \lvert j - i \rvert \mid i, j \in M\}$. Clearly, if $M = \{i, j\} \in Q$, then $d(M) = \lvert j - i \rvert$. Hence for $\mathfrak p = (i,j) \in P$ we  have $d(s(\mathfrak p)) = j  - i$.

$\sigma$ induces a bijection $\sigma' : Q \to Q, \sigma'(M) =$ image of $M \in Q$ under $\sigma$, i.e. $\sigma'(M) = \{ \sigma(m) \mid m \in M \}$.

We now recall the following general commutativity law for real multiplication:

If $\varphi : F \to F$ is a bijection on a finite set $F$ and $x_\alpha \in \mathbb R$ for $\alpha \in F$, then $\prod_{\alpha \in F} x_\alpha = \prod_{\alpha \in F} x_{\varphi(\alpha)}$.

We apply it for $F  = Q$, $\varphi = \sigma'$ and $x_M = d(M)$ and use $(1)$ to get

$$\prod_{(i,j) \in P}(j - i) = \prod_{\mathfrak p \in P} d(s(\mathfrak p)) = \prod_{M \in Q } d(M) = \prod_{M \in Q } d(\sigma'(M)) = \prod_{\mathfrak p \in P} d(\sigma'(s(\mathfrak p))) \\= \prod_{(i,j) \in P} d(\{\sigma(i), \sigma(j)\}) .$$

An inversion of $\sigma$ is a pair $(i,j) \in P$ such that $\sigma(i) > \sigma(j)$. Let $\text{inv}(\sigma) \subset P$ denote the set of all inversions of $\sigma$ and $\text{Inv}(\sigma)$ the number of elements of $\text{inv}(\sigma)$. Morever let $\epsilon(i,j) = 1$ for $(i,j) \in \text{inv}(\sigma)$ and $\epsilon(i,j) = 0$ for $(i,j) \in P \setminus \text{inv}(\sigma)$. We have
$$\sum_{(i,j) \in P} \epsilon(i,j) = \text{Inv}(\sigma) ,$$
$$d(\{\sigma(i), \sigma(j)\}) = \lvert \sigma(j) - \sigma(i)\rvert = (-1)^{\epsilon(i,j)}(\sigma(j) - \sigma(i)) .$$
Therefore

$$\prod_{(i,j) \in P} d(\{\sigma(i), \sigma(j)\}) = \prod_{(i,j) \in P } (-1)^{\epsilon(i,j)}(\sigma(j) - \sigma(i)) \\ = \prod_{(i,j) \in P } (-1)^{\epsilon(i,j)} \prod_{(i,j) \in P } (\sigma(j) - \sigma(i)) = (-1)^{\sum_{(i,j) \in P } \epsilon(i,j)}  \prod_{(i,j)\in P } (\sigma(j) - \sigma(i)) \\ = (-1)^{\text{Inv}(\sigma)} \prod_{(i,j) \in P} (\sigma(j) - \sigma(i)) .$$
This proves your claim.

Title: Product formula for sign of permutation


问:
Convergent integral implies convergent sum
$f(x)$ is a continuous positive function s.t. the sequence $xf(x)$ is nonincreasing. Then show that $\int_1^\infty f(x) < \infty$ implies $\sum_1^\infty f(n)< \infty$?

I try to bound $f(x)-f(n)$ for $x\in [n,n+1]$ by something summable?

答:
For every $n \ge 2$ is
$$
 \int_{n-1}^{n} f(x) \, dx =  \int_{n-1}^{n} \frac{ x f(x)}x \, dx
\ge  \int_{n-1}^{n} \frac{ n f(n)}x \, dx
= n  \ln \left( \frac{n}{n-1}\right) f(n) \ge f(n) \, .
$$
The last inequality holds because, using $\ln (1+u) \le u$ for $u > -1$,
$$
\ln \left( \frac{n}{n-1}\right) = -\ln \left( 1 - \frac 1n\right) \ge - \left( -\frac{1}{n}\right) = \frac 1n \, .
$$
It follows that
$$
\sum_{k=1}^n f(k) \le f(1) + \int_1^n f(x) \, dx
$$
for all $n$, and the conclusion follows.


问:
Mergers and ultrafilters
F - filter over I.
Please, prove, F ultrafilter <=> when $\forall$
X, Y $\subseteq$ I, if X $\notin$ F, Y $\notin$ F, $\Rightarrow$ X $\cup$ Y $\notin$ F.</p>

答: Hint: If $\cal F$ is an ultrafilter, pick $X,Y\notin\cal F$ and use the fact that their complements are in $\cal F$ and it is closed under intersections; in the other direction, if $\cal F$ is not an ultrafilter there is some $X$ such that $X,I\setminus X\notin\cal F$. Use these two sets to contradict the statement you have on the RHS.

Title: Mergers and ultrafilters


问:
Comparison test proof of convergence
I want to prove that $\sum_{n=1}^\infty\frac{3^n+4^n}{3^n+5^n}$ converges. To do this, I said that $\sum_{n=1}^\infty\frac{3^n+4^n}{3^n+5^n}<3\sum_{n=1}^\infty (\frac{4}{5})^n$, which indeed converges. Hence by the comparison test, the series we are given converges. Is this valid?

答:
Hint:
$$\frac{3^n+4^n}{3^n+5^n} \le \frac{3^n+4^n}{5^n} = \left(\frac 35 \right)^n + \left(\frac 45 \right)^n \le 2\left(\frac 45 \right)^n $$

Title: Comparison test proof of convergence


答: This is a good idea. You can use that $\sum_{n=1}^\infty \int_n^{n+1}f(x)dx<\infty$ and therefore $\int_n^{n+1}f(x)dx = \int_n^{n+1}\frac{xf(x)}{x}dx \geq (n+1)f(n+1)\int_n^{n+1}\frac{1}{x}dx \geq (n+1)f(n+1)\ln(1+\frac{1}{n})$ but $(n+1)\ln(1+\frac{1}{n})\to 1$ when $n\to\infty$ so there exists $M>0$ such as $f(n+1)\leq M\int_n^{n+1}f(x)dx, \ \forall n\in\mathbb{N}$ and you finish by comparison criterion.

Title: Convergent integral implies convergent sum

Body:
$f(x)$ is a continuous positive function s.t. the sequence $xf(x)$ is nonincreasing. Then show that $\int_1^\infty f(x) < \infty$ implies $\sum_1^\infty f(n)< \infty$?

I try to bound $f(x)-f(n)$ for $x\in [n,n+1]$ by something summable?


问:
How do you find an analytical solution from a graph?
Given is a graph with three functions, g(x), G(x) and g'(x).

I don't need simply the answer. I want to know how to do it, or at least how to begin.

答:
Starting with the red graph:

This is obviously a sine wave. A sine wave is
$$A\sin(\omega x+\varphi),$$
where $A$ is the amplitude, $f$ the angular frequency and $\varphi$ the phase. We can easily see that $A=1$ and $\varphi=0$. The angular frequency is defined as $\frac{2\pi}{T}$, where $T$ is the lenght of a period, which in this case is $\pi$.
So the red wave is the graph of
$$\sin(2x).$$

Continuing with the blue graph

First calculate the antiderivative of $\sin(2x)$. It is $-\cos(2x)/2+C_1$. At $x=0$ the function,with $C_1=0$, is $-1/2$. To bring it up to $y=0$ set $C=1/2$.

Finishing with the green graph

Calculate the antiderivative again. This yealds $x/2-\sin(2x)/4+C_2$. At $x=0$ this function is equal to the green graph if $C_2=0$.

Conclusion

We have
$$g'(x)=\sin(2x), \quad g(x) = -\frac{\cos(2x)}{2}+\frac{1}{2},\quad G(x) = \frac{x}{2}-\frac{\sin(2x)}{4}.$$


问:
Continuous functions $f$, $g$ from $X$ to $\mathbb{R}^n\setminus \left\{0\right\}$ are homotopic
Suppose we have $f$, $g$ continuous mappings from a space $X$ to $\mathbb{R}^n\setminus \left\{0\right\}$ with $\|f(x)-g(x)\|\leq \|f(x)\|$, prove that $f$, $g$ are homotopic.

Is anyone willing to help me solve this? I thought about using the straight line homotopy, but I'm not sure it would work.

答:
Yes, it will work. Define $F:X\times I\to\mathbb{R^n}\setminus\{0\}$ by $F(x,t)=(1-t)f(x)+tg(x)$. The only thing which is not immediately clear is why $F(x,t)$ is never equal to $0$. Well, if $t=1$ then $F(x,t)=g(x)$ and so it isn't $0$. Now let $t\in [0,1)$. Assume there is some $x\in X$ such that $F(x,t)=0$. Then:

$0=(1-t)f(x)+tg(x)=f(x)-t(f(x)-g(x))$

$f(x)=t(f(x)-g(x))$

By taking norm on both sides we obtain $||f(x)||=t||f(x)-g(x)||\leq t||f(x)||$. Since $t\in [0,1)$ this implies $||f(x)||=0$, which is a contradiction.


问:
Find a locus of a centroid point of a triangle.
When doing another geometry proof, I have this problem:
Let $A$ is the point outside the circle $(O;R)$ such that $OA=2R$. The line $d$ goes through $A$ and cut the circle at $2$ distinct points $E$ and $F$. What is the locus of the centroid point of the triangle $OEF$?

I used Geogebra to draw the trace of this centroid, it's an arc of a circle but i can't find out which point is the center.

答:
Assume $O$ is origin. Let $M$ be the midpoint of all chords $EF$. $\angle OMA = 90$. $M$ lies on arc of circle of diameter $OA$.

The centroid $G$, of $\triangle OEF$ lies two-thirds away from $O$ along $OM$.

Find $B$ on $OA$ so that $GB\parallel MA$. Clearly $\triangle OGB \sim \triangle OMA$. So $B$ lies two-thirds away from $O$ along $OA$.
$$B=\frac{2A}{3}$$

Hence $G$ lies on the circle whose diameter is $OB$. Its center $C$ is midpoint of $OB$.

$$C=\frac{O+B}{2}=\frac{A}{3}$$

Thus $C$ is $2R/3$ away from $O$ along $OA$.


问:
Dense subalgebra intersection with essential ideal is dense?
Consider $A$ is an unital C* algebra, $D$ is a dense *-subalgebra, $B$ is a essential,maximal ideal of $A$. Is $D\cap B$ dense in $B$?

I have seen this question, and this question, but my assumption is little bit different, so I think I am asking a different question here.

答:
Certanily not.  Consider $A=C([0, 1])$,  $$B=\{f\in  A: f(0)=0\},$$ and $D=\mathbb C\cdot 1$.

EDIT.
OK, so here is a counter-example satisfying  the new requirement that $D$ be a dense subalgebra.

$\newcommand{\F}{\mathbb F_2}$
Let $\F$ be the free group on two generators, and let:

$A$ be the full group C$^*$-algebra of $\F$, often also denoted by $C^*(\F)$,

$B$ be the kernel of the left-regular representation $\lambda :C^*(\F)\to C_r^*(\F)$, and

$D$ be the canonical copy of the complex group algebra $\mathbb C(G)$ within $C^*(\F)$.

Then

$B\cap D$ is not dense in $B$,  and in fact  $B\cap D=\{0\}$.

Reason.  This is because the restriction of $\lambda $ to $\mathbb C(G)$ is injective.

$B$ is a maximal ideal of $C^*(\F)$.

Reason.  This is because $C_r^*(\F)$ is simple,  thanks to a well known result by Powers [1].

$B$ is an  essential ideal.

Reason.  This is the hardest point to justify:  arguing by contradiction,  suppose that $J$ is a nontrivial ideal in $C^*(\F)$,
trivially intersecting $B$.  Then $\lambda (J)$ is a nontrivial ideal in  $C_r^*(\F)$,  so necessarily $\lambda (J)=C_r^*(\F)$, by
simplicity.

It follows that $\lambda $ restricts to an isomprphism from $J$ to $C_r^*(\F)$, and hence  the inverse of that restriction is  an
injective *-homomorphism
$$
  \mu :C_r^*(\F)\to C^*(\F).
  $$

By Choi's Theorem [2], $C^*(\F)$ admits a separating family of finite dimensional representations (this is to say that  $C^*(\F)$ is
residually finite dimensional), so the same applies to $C_r^*(\F)$.  However, since the latter is simple and infinite
dimensional, we see that it has no finite dimensional representation whatsoever.  This is a contradiction.

[1] Powers, Robert T., Simplicity of the C$^*$-algebra associated with the free
group on two generators, Duke Math. J. 42, 151-156
(1975). ZBL0342.46046.

[2] Choi, Man-Duen, The full C$^*$-algebra of the free group on two
generators, Pac. J. Math. 87, 41-48
(1980). ZBL0463.46047.


问:
difference between the integral and the sum over a function
I have a function of $\sigma> 0$,

$$ f(\sigma) = \int d^3 \mathbb{u} \frac{e^{-u^2 \sigma^2}}{u^2} - \sum_{\mathbb{n}\in \mathbb{Z}^{3*}} \frac{e^{-n^2 \sigma^2 }}{n^2} . $$

Here $n = (n_1, n_2, n_3) $, $n^2 = n_1^2 + n_2^2 + n_3^2 $. Similarly,  $u= (u_1, u_2, u_3)$, $u^2 = u_1^2 + u_2^2 + u_3^2 $.

The integral can be done easily, it is $2\pi\sqrt{\pi}/\sigma $. This should also be the lead term of the sum as $\sigma \rightarrow 0^+ $.

I want to find out the value $f(\sigma \rightarrow 0^+ )$. Numerically, it is close to 8.9148. For this value, I take $\sigma = 0.02$ and sum in a cube of width $400$. It takes huge memory. I am wonder whether there is a smart way to get the value.

答:
$\xi:=\lim_{\sigma\to0^+}f(\sigma)$ has the following integral representations: $$\xi=2\pi\int_0^\infty\big(1+t^{-3}-\vartheta(t)^3\big)\,dt=2\pi\int_0^\infty t\big(1+t^{-3}-\vartheta(t)^3\big)\,dt,$$ where $\vartheta(t)=\sum_{n\in\mathbb{Z}}e^{-\pi(nt)^2}$ for $t>0$ (a kind of theta functions). The second of these integrals is obtained from the trivial $\xi=-\int_0^\infty g'(t)\,dt$, where $g(t)=f(t\sqrt\pi)$; the substitution $t=1/x$, together with the known identity $\vartheta(1/x)=x\vartheta(x)$, gives the first integral.

While the above doesn't give a closed form for $\xi$, we can compute it using numerical integration. A simple but efficient recipe here is to combine the above with the substitution $t=e^x$: $$\xi=\pi\int_{-\infty}^\infty F(x)\,dx,\qquad F(x)=e^x(1+e^x)\big(1+e^{-3x}-\vartheta(e^x)^3\big).$$ Here we have $F(-x)=F(x)$, and $F(x)$ decays exponentially as $x\to\pm\infty$. The last property implies that the simple trapezoidal rule $\xi\approx\pi h\sum_{n\in\mathbb{Z}}F(nh)$ converges very quickly as $h\to 0$: $$\xi=\underbrace{8.91363291}_{h=0.2}\underbrace{7585151272}_{h=0.1}\underbrace{687120136008681131187}_{h=0.05}\underbrace{6260368122105976533005323519\cdots}_{h=0.02}$$


问:
Minimum number $s$ such that all the unions of $s$ of the $p$ subsets are equal
I'm asking for references in literature on the following problem.

Given a positive integer $n$, $S=\{1,\dots,n\}$ and $p$ distinct subsets of $S$, $A_1,\dots,A_p$, what is the minimum number $s>0$ such that
$$
\bigcup_{i\in L}A_i=\bigcup_{j\in M}A_j
$$
for all subsets $L,M\subseteq S$ of size $s$.

Clearly, if the sets $A_i$ are pairwise disjoint the number $s$ is $p$. Is it possible to determine the integer $s$ a priori, or to find upper and/or lower bounds for $s$.

Your help is appreciated.

答:
The minimal $s$ is $\min(p,2^{n-1}+1)$.

You know that $s=p$ will always work. Similarly, $s=2^{n-1}+1$ will work, because $|L|=2^{n-1}+1$ implies $\bigcup_{i\in L} A_i=\{1,\dots,n\}$. Indeed, $\bigcup_{i\in L} A_i$ cannot be missing any element $i$ of $\{1,\dots,n\}$, because there are only $2^{n-1}$ subsets of $\{1,\dots,n\}$ which are missing $i$.

To see this is optimal, suppose $s<\min(p,2^{n-1}+1)$. Choose the sets $A_1,\dots,A_p$ such that

$A_1=\{1\}$,

$1\notin A_i$ for $i\in \{2,3,\dots,\min(p,2^{n-1}+1)\}$.

With this setup, if $L$ contains $1$, then $\bigcup_{i\in L}A_i$ will contain $1$, but if $L\subseteq \{2,3,\dots,\min(p,2^{n-1}+1)\}$, then $\bigcup_{i\in L}A_i$ will not contain $1$. Therefore, $\bigcup_{i\in L}A_i$ is not always the same.


问:
How to create an inverted $S$ cruve
I'm looking for a way to create an inverted $S$ curve function. Below is an image of what I want. Left is a regular S curve, on the right is the kind of curve I want. Here are the allowed set of operations that I can use:

答:
Using complex variables, we can transform the exact $S$-curve you started with very easily. Let us assume the inflection point passes through the origin and let the curve be expressed parametrically as $z(t)$. The first thing we want to do is to flip the curve about the $x$-axis, then rotate it $90^{\circ}$ counterclockwise, the the desired curve becomes

$$w(t)=z^*(t)e^{i\pi/2}=iz^*(t)$$

In Cartesian coordinates then

$$w(t)=i\big(x(t)-iy(t)\big)=y(t)+ix(t)$$

Which means that all you have to do is to switch $x$ and $y$. You can see as much by just flipping your original image about.

Title: How to create an inverted $S$ cruve


问:
How should I find the four vertices of a rectangle if I have its center of gravity and it length and width?
basically all I want to know is in the question.

I know that in a square if I have the length of it and its center of gravity I can find the vertices by this formula:

If the coordinates of the center of gravity is G(1, 2) and its length is 3, and suppose the four vertices are A, B, C and D:

XA = 1 - 3/2

YA = 2 - 3/2

Etc...

答:
If you have a rectangle $ABCD$ where

$AB$ is parallel to $x-$axis

$|AB| = |CD| = a, |BC| = |AD| = b; a, b \gt 0$

Coordinates of $G (h, k)$

Then the coordinates of vertices are given by $(h \pm \frac{a}{2}, k \pm \frac{b}{2})$. Going anticlockwise starting from $A$, the signs are $(+, +), (-, +), (-, -) $ and $(+, -)$.

Note that the vertices of the rectangle are on a circle with radius $\frac{\sqrt{a^2+b^2}}{2}$ with center at $G(h, k)$. If we rotate the rectangle by angle $x$ in anticlockwise direction by point $G$, $A'$ is our new $A$ and coordinates of $A'$ is given by,

$\displaystyle \big(h + \frac{a \cos x - b \sin x}{2}, k + \frac{a \sin x + b \cos x}{2}\big)$

Other vertices follow the same pattern.


问:
Prove that $\left(\prod\limits_{2<p}^{p_i} \frac{p-1}{p}\right) \cdot \left( p_{i + 1}^2 - p_i^2 \right) \geq \pi(p_{i + 1}^2) - \pi(p_i ^2)$
Conjecture:

$$\left(\prod_{2
 \pi(p_{i + 1}^2) - \pi(p_i ^2) \tag{1}$$

The LHS expression is the product of Mertens's third theorem truncated at $p_i$ (subsequently just $M_x$ for truncation at $x$) times the length of the interval between the squares of $p_i$ and $p_i+1$, and the RHS is the actual number of primes between those squares. For instance, for $p_i = 7$, the expression is:

$$\frac{8}{35} \cdot \left(11^2 - 7^2\right) > \pi(121) - \pi(49)$$

And in fact $16.46 > 15$. Inequality $(1)$ is (perhaps surprisingly) empirically true for all $p_i$. The inequality seems intuitively likely to be true, as we would expect the result of $M_xx^2$ to be larger than $\pi(x^2)$ simply because $e^{-\gamma} > \frac12$ and therefore

$$\pi(x^2) \sim \frac{x^2}{\log x^2} = \frac12\frac{x^2}{\log x} < e^{-\gamma}\frac{x^2}{\log x} \tag{2}$$

(Inequality $(2)$ is empirically true only above $x \approx 100$, because the difference between $M_x \log x$ and $e^{-\gamma}$ is large for small $x$)

We can show (much more easily than I expected, in this question where I briefly failed basic algebra) that the density of primes between $p_{i-1}^2$ and $p_i^2$ is lower than the density of primes between $0$ and $p_i^2$, that is:

$$\frac{\pi\left(p_{i+1}^2) - \pi(p_i^2\right)}{p_{i+1}^2 - p_i^2} < \frac{\pi(p_i^2)}{p_i^2} \tag{3}$$

With this, we expect difference between the LHS and RHS of $(1)$ to be greater than $(2)$ alone would imply.

But... I'm roughly 110% certain that $(2)$ and $(3)$ together don't qualify as proof. They might be proof of asymptotic behavior as $p_i \to \infty$, but I'm wondering if this can be (or already has been!) proven for smaller values. Any links or thoughts from the community?

答:
$\prod \limits_{2\leq p \leq p_i } (1-\frac{1}{p})  (p_{i+1}^2-p_{i}^2) \geq li(p_{i+1}^2)-li(p_{i}^2)$ for all $i \geq 4$

By noticing that $ \prod \limits_{2\leq p \leq p_i } (1-\frac{1}{p})  (p_{i+1}^2-p_{i}^2)  \approx \frac{e^{-\gamma}}{\ln p_i} (p_{i+1}^2-p_{i}^2)\geq  \frac{p_{i+1}^2-p_{i}^2}{2\ln p_i}\geq \int \limits_{p_{i}^2}^{p_{i+1}^2} \frac{dt}{\ln t}=li(p_{i+1}^2)-li(p_{i}^2)$

If you assume R.H. and let $ p_{i+i^{\epsilon}}^2-p_{i}^2$ with $i^{\epsilon} = O(\ln^k i)$ for $ k \geq 2$ then since $ |\pi(x)- li(x)| \leq \sqrt{x} \ln x$ this would imply the correctness of your conjecture for all $ i\geq 3000$ give or take according to the choice of $k$.

But without R.H. i don't see a way.


问:
Groups $A \leq B \trianglelefteq G$. Prove that $A \trianglelefteq G$
Groups $A \leq B \leq G$ are given, and for all automorphisms $\kappa \in \text{Aut}(B)$, $\kappa(A) = A$.

We are also given $B \trianglelefteq G$.

I proved that $A \trianglelefteq B$ by using the above property of an isomorphism.

$$\phi(a) = bab^{-1}, b \in B, \forall a \in A$$ And got stuck.

Thus, how do I show $A \trianglelefteq G$.

答: Let $g\in G, a\in A$. We have to show that $gag^{-1}\in A$. First define the map $\varphi:B\to B$ by $\varphi(b)=gbg^{-1}$. It is well defined because $B\trianglelefteq G$, and it is an automorphism of $B$. By assumption $\varphi(A)=A$, so in particular $\varphi(a)\in A$.


问:
Path difference on a sphere due to geodesic deviation
Between two points of same latitude but different longitudes [spherical coordinates $(\theta,\phi)$] we measure two lengths of arcs  $(s_1,s_2)$ along (1) the geodesic or red great circle (a graphic of numerical integration) and (2) along the constant $ \phi$ non-geodesic thicker purple arc  or latitude.

There is no difference of these distances at the equator or poles but is a maximum at some intermediate latitude value.

How to find the $\phi$ at which the difference of these arc distances $ (s_2-s_1) $  is a maximum?

Thanking you in advance.

答:
(Not about my answer, regarding part of discussion with user's answer).

```<code>F[x_, t_] := t Cos[x] - ArcCos[1 - (1 - Cos[t]) Cos[x]^2]
Plot3D[F[XX, TT], {XX, 0, Pi / 2}, {TT, 0, Pi },
 AxesLabel -> Automatic, PlotLabel -> " Geodetic_ Devn"]
</code>```

Still I am missing something in the unmodified $\phi$ latitude domain of user's function mentioned above .. ( geodesic deviation is more than half earth radius ! ). Please comment.

Body:
(Not about my answer, regarding part of discussion with user's answer).

F[x_, t_] := t Cos[x] - ArcCos[1 - (1 - Cos[t]) Cos[x]^2]
Plot3D[F[XX, TT], {XX, 0, Pi / 2}, {TT, 0, Pi },
 AxesLabel -> Automatic, PlotLabel -> " Geodetic_ Devn"]

Still I am missing something in the unmodified $\phi$ latitude domain of user's function mentioned above .. ( geodesic deviation is more than half earth radius ! ). Please comment.


问:
Convex set. Proof.
Prove that if A is convex set and $\alpha, \beta ≥ 0$ then $(\alpha + \beta)A = \alpha A + \beta A$

What came first on my mind is that I have to show that $(\alpha + \beta)A\subset \alpha A + \beta A$ and $\alpha A + \beta A\subset (\alpha + \beta)A$. But I don't know how to use convexity of A to show this.

答: The "$\subset$" part is easy: $(\alpha + \beta)x=\alpha x + \beta x \in \alpha A +\beta A \ \forall x \in A$;  on the other hand $\alpha x + \beta y=(\alpha+ \beta)(\frac{\alpha x}{\alpha + \beta} + \frac {\beta y}{\alpha + \beta})\in(\alpha+\beta)A$  where the membership relation is due to the convexity of A.</p>

Title: Convex set. Proof.


答:
You can rewrite your formula as $\prod_{i=1}^{n}\prod_{j=i+1}^{n}{1 \over {j-i}} \times \prod_{i=1}^{n}\prod_{j=i+1}^{n} {\sigma(j)-\sigma(i)}$.

Let $i< j$ and $\varepsilon(i,j) = -1$ if $\sigma(i) > \sigma(j)$, $1$ otherwise.

For each pair $i < j$, there exists a (unique) pair $i'< j'$ such that $|\sigma(i')-\sigma(j)| = |i'-j'|$ (this is basically because $\varphi : (i,j) \mapsto \varepsilon(i,j)(\min({\sigma(i),\sigma(j)}),\max({\sigma(i),\sigma(j)}))$ is a bijection (the reciprocal is $\varphi^{-1}(1,i,j) = (\sigma^{-1}(i), \sigma^{-1}(j)) $ and $\varphi^{-1}(-1,i,j) = (\sigma^{-1}(j), \sigma^{-1}(i)) $.

Hence, we have $\sigma(j)-\sigma(i) = \varepsilon(i,j)(j'-i')$ for some $i'<j'$.

We get

\begin{align*}
 \prod_{1\leq i < j \leq n} {\sigma(j)-\sigma(i)\over j-i}&=\prod_{i=1}^{n}\prod_{j=i+1}^{n}{1 \over {j-i}} \times \prod_{i=1}^{n}\prod_{j=i+1}^{n} {\sigma(j)-\sigma(i)} \\
&=\prod_{i=1}^{n}\prod_{j=i+1}^{n}{1 \over {j-i}} \times \prod_{i=1}^{n}\prod_{j=i+1}^{n} {\sigma(j)-\sigma(i)} \\
&=\prod_{i=1}^{n}\prod_{j=i+1}^{n}{1 \over {j-i}} \times \prod_{i=1}^{n}\prod_{j=i+1}^{n} \varepsilon(i,j)({j'-i'}) \\
&=\prod_{i=1}^{n}\prod_{j=i+1}^{n}{1 \over {j-i}} \times \prod_{i=1}^{n}\prod_{j=i+1}^{n} \varepsilon(i,j)\times \prod_{i=1}^{n} \prod_{j'=i+1}^{n}({j'-i}) \\
&= \prod_{i=1}^{n}\prod_{j=i+1}^{n} \varepsilon(i,j)\\
&= (-1)^{Inv(\sigma)}
\end{align*}

Title: Product formula for sign of permutation

Body:
I would like help (a complete proof) that $$ (-1)^{Inv(\sigma)} = \prod_{1\leq i<j\leq n} \frac{\sigma(j)-\sigma(i)}{j-i} ,$$ where $Inv(\cdot)$ is the number of inversions of the permutation received as argument and $\sigma$ is a permutation in $S_n$.

The formula seems intuitive somehow,  but I cannot manage to fully write a proof for it. The factors of the numerator product seem to coincide with the ones in the denominator, albeit with a negative sign where inversions are happening, but I can't fully convince myself each element at the numerator has exactly one correspondent in the denominator and vice versa.

EDIT: My book states as an obvious fact that the product in the numerator is equal to the one in the denominator, where each factor is multiplied by $\epsilon(i,j)$ ($-1$ iff (i,j) is an inversion for $\sigma$).. I can finish the proof from here, but it is not obvious to me.


问:
Solving a simple quadratic in modular arithmetic

I've been trying to solve this congruence> $$x^2 \equiv 9\pmod{13}$$

I keep coming back to $x =3$ but it seems to be wrong, any help on what's the correct technique to find the answer?

答: One way to see this is solving $x^2=9$ over the integers yields $x = \pm 3$. Clearly, $3$ is one solution, and similarly, $-3 \equiv 10 \pmod{13}$ so $10$ is another solution. But if $3$ is a solution, so must be $3+13=16$, etc. Can you find a general form?


问:
Prove the limit equation: $\lim_{x\rightarrow\infty}e^{-x}\sum_{n=0}^{\infty}f_{n}\left(x\right)=-\gamma.$ FYI$,\quad\gamma$ is Euler's constant
Define a function sequence $\left \{ f_{n} \right \}_{n= 0}^{\infty}$ on $\left ( 0, \infty \right )$ as
$$f_{0}\left ( x \right )= \ln x,\quad f_{n+ 1}\left ( x \right )= \int_{0}^{x}f_{n}\left ( t \right ){\rm d}t\quad\left ( n= 0, 1, 2\cdots \right )$$
Prove the limiting equation:
$$\lim_{x\rightarrow\infty}e^{-x}\sum_{n= 0}^{\infty}f_{n}\left ( x \right )= -\gamma$$
For your information$,\quad\gamma$ is Euler's constant defined
$$\gamma:=\lim\left ( \sum_{k= 1}^{n}\frac{1}{k}- \ln n \right )$$
This problem belongs to my friend@twelve_sakuya, I haven't touched Euler's constant of course we have no approaches from me. But I had done a similar work. Maybe, there is a relationship between both problems, I need to the help. Here we go.

Given a function $f\left ( x \right )$ continuous on the real interval. Prove that
$$\lim_{x\rightarrow\infty}\left ( e^{-x}\int_{-x}^{x}e^{t}f\left ( t \right ){\rm d}t \right )= 0$$
Source: Hungarian Math Contest Revision

Integral's limit, which I think that required us to divide and evaluate the integral function. By using the integral form of Cauchy-Schwarz inequality, we have
$$\left | e^{-x}\int_{-x}^{x}e^{t}f\left ( t \right ){\rm d}t \right |\leq e^{-x}\left | \int_{-x}^{-\frac{x}{2}}e^{t}f\left ( t \right ){\rm d}t \right |+ e^{-x}\left | \int_{-\frac{x}{2}}^{\frac{x}{2}}e^{t}f\left ( t \right ){\rm d}t \right |+ e^{-x}\left | \int_{\frac{x}{2}}^{x}e^{t}f\left ( t \right ){\rm d}t \right |\leq$$
$$\leq e^{-x}\left ( \int_{-x}^{-\frac{x}{2}}e^{2t}{\rm d}t \right )^{\frac{1}{2}}\left ( \int_{-x}^{-\frac{x}{2}}f^{2}\left ( t \right ){\rm d}t \right )^{\frac{1}{2}}+ e^{-x}\left ( \int_{-\frac{x}{2}}^{\frac{x}{2}}e^{2t}{\rm d}t \right )^{\frac{1}{2}}\left ( \int_{-\frac{x}{2}}^{\frac{x}{2}}f^{2}\left ( t \right ){\rm d}t \right )^{\frac{1}{2}}+$$
$$+ e^{-x}\left ( \int_{\frac{x}{2}}^{x}e^{2t}{\rm d}t \right )^{\frac{1}{2}}\left ( \int_{\frac{x}{2}}^{x}f^{2}\left ( t \right ){\rm d}t \right )^{\frac{1}{2}}\leq e^{-\frac{3}{2}x}\left ( \int_{-\infty}^{-\frac{x}{2}}f^{2}\left ( t \right ) \right )^{\frac{1}{2}}+ e^{-\frac{x}{2}}\left ( \int_{-\infty}^{\infty}f^{2}\left ( t \right ) \right )^{\frac{1}{2}}+$$
$$+ \left ( \int_{\frac{x}{2}}^{\infty}f^{2}\left ( t \right ) \right )^{\frac{1}{2}}\underset{x\rightarrow\infty}{\rightarrow}0\therefore\left | \int_{-\infty}^{\infty}f^{2}\left ( x \right ){\rm d}x \right |< \infty\Rightarrow\lim_{x\rightarrow\infty}\left ( e^{-x}\int_{-x}^{x}e^{t}f\left ( t \right ){\rm d}t \right )= 0$$

答:
Induction shows that
$$
f_n(x)=\frac{x^n}{n!}(\log(x)-H_n)\tag1
$$
where $H_n$ are the Harmonic Numbers.

Define
$$
s(x)=\sum_{n=0}^\infty f_n(x)\tag2
$$
$(1)$ shows that the sum in $(2)$ converges for all $x\ne0$.

Compute the derivative:
$$
\begin{align}
s'(x)
&=\sum_{n=0}^\infty f_n'(x)\tag{3a}\\
&=\frac1x+\sum_{n=1}^\infty f_n'(x)\tag{3b}\\
&=\frac1x+\sum_{n=0}^\infty f_n(x)\tag{3c}\\
&=\frac1x+s(x)\tag{3d}
\end{align}
$$
Rearranging $(3)$, multiplying by the integrating factor $e^{-x}$, and subtracting $\frac1x$ yields
$$
\left(e^{-x}s(x)-\log(x)\right)'=\frac{e^{-x}-1}x\tag4
$$
Integration by parts gives
$$
\begin{align}
e^{-x}s(x)
&=\log(x)+\int_0^x\frac{e^{-t}-1}t\mathrm{d}t\tag{5a}\\
&=\log(x)\,e^{-x}+\int_0^x\log(t)\,e^{-t}\,\mathrm{d}t\tag{5b}
\end{align}
$$
As shown in this answer,
$$
\int_0^\infty\log(t)\,e^{-t}\,\mathrm{d}t=-\gamma\tag6
$$
Applying $(6)$ to the limit of $(5)$ gives
$$
\lim_{x\to\infty}e^{-x}s(x)=-\gamma\tag7
$$


问:
If $a^x \equiv a^y \pmod{m}$, does this imply $x \equiv y \pmod{\varphi(m)}$?
I know that $x \equiv y \pmod{\varphi(m)} \implies a^x \equiv a^y \pmod{m}$, so I was wondering if the converse also holds. If it does not hold in general, then are there any conditions where it does (such as assuming $(a, m)=1$)?

*Here $\varphi$ denotes the Euler totient function.

Edit: I see that the answer is no. However, the reason I was wondering about this is that I was watching this (https://youtu.be/f1oO9dEkqso?t=1386) lecture and don't know why else would Prof. Borcherds claim that $a_i \equiv 1 \pmod{p_{i}^{k_i}}$ is solvable if $\varphi(p_{i}^{k_i}) \mid n$ if the property in question is not true. In other words, how else is what's written justified?

答:
No it does not. A trivial example is $a = 1$. Slightly non-trivial examples can be found whenever $\operatorname{ord}_m(a) \ne \varphi(m)$. (But it does hold when $\operatorname{ord}_m(a) = \varphi(m)$).

For example, $\operatorname{ord}_7 (2) = 3 \ne \varphi(7)$. Hence $2^1 \equiv 2^4 \pmod 7$ (and $\gcd(2,7) = 1$) but $1 \ne 4 \pmod 6$.

EDIT: In the context of the YouTube lecture, we are trying to find a smaller $n \le \varphi(m)$ such that $a^n \equiv 1 \pmod m$ holds whenever $\gcd(a,m)=1$ (hinting at the Carmichael Function perhaps?). To do this the prime factorization of $m$ was considered, and the relation $a^n \equiv 1 \pmod {p^k}$ holds whenever $\varphi(p^k) \mid n$ holds. The converse was never asserted. (Although if one considers all $a$, there would be elements of multiplicative order $\varphi (p^k)$, called the primitive roots, but it is not needed here.)


问:
How to prove that $\lim\limits_{x \to 0 }\;x^{-a}e^{\frac{-1}{x^{2}}} =0$ for all a?
I'd like your help with this:

I tried using L'Hôpital's Rule and all kinds of arithmetic to prove that
$$\lim_{x \to 0 }\left(x^{-a}e^{\left(\frac{-1}{x^{2}}\right)}\right) = 0$$ for every $a$, and it didn't work.

($a=0$ is trivial)

Any hints?

Thank you.

答: HINT $\ $ Take the $\rm\:log\:,\:$ put $\rm\: z = 1/x\:$ to reduce it to $\rm\: z^2/\:log(z)\to \infty\:$ as $\rm\: z\to\infty\:,\:$ by L'Hôpital.


问:
Prove the monotonicity of the expectation of a binary random variable function
Consider $R$ independent binary random variables $y^1, \ldots, y^R$ over the space $\{-1, +1\}$ such that $\Pr(y^j = 1) = p^j \geq 0.5$ and $\Pr(y^j = -1) = 1 - p^j$, $\forall j = 1,\ldots,R$. Moreover, let $\mathcal{J}$ denote a subset of $\mathcal{Y} = \{1, \ldots, R\}$, i.e., $\mathcal{J} \subseteq \mathcal{Y}$. I would like to prove that:

$f(\mathcal{J} \cup \{j'\}) \geq f(\mathcal{J}) \quad \forall \mathcal{J} \in 2^{\mathcal{Y}}, \forall j' \in \mathcal{Y} \setminus \mathcal{J}$

where the function $f( \cdot )$ is defined as follows:

$f(\mathcal{J}) = E\left(| \sum_{j \in \mathcal{J}} y^j  |\right)$

In other words, I want to verify that the afore-mentioned expectation is monotonically non decreasing in the set $\mathcal{J}$. Thanks.

EDIT: I posed a new question about the same proof in case $f(\mathcal{J})$ is altered so that $f(\mathcal{J}) = E\left(| \sum_{j \in \mathcal{J}} logit(p^j) y^j  |\right)$. The new question can be found here: Where is the fallacy in this coupling argument of two Bernoulli variables?

答:
wlog $\mathcal T = 1,...,n,.X = \sum^n Y_j, j^{\prime} = n+1$, and that is wlog because you can always go from a smaller set to a larger adding one index at a time, and just number them in the order in which you want to add.
First use Jensen's inequalty/ submartingales to show that if you add a symmetric $Y_{n+1} $ the expected value of the absolute value increases.  Then use the following argument to show that if $\mathbb P (X > 0) > \mathbb P ( X < 0), \mathbb E(|X + Y_{n+1}|) $ is an increasing function of $p_{n+1}$

let $X$ be integer valued, $Y$ be $\pm 1$ with $\mathbb P(Y=1) = p$.  $\mathbb E (|X + Y|) = \mathbb P (X=0) + \sum_{k  > 0} \mathbb P(X = k)((k+1) p + (k-1) (1-p)+ \mathbb P(X = -k)((k+1) (1-p) +( k-1)p)$.

$\frac  d {dp}$ of that is something like  $ 2 \sum_{k > 0} (\mathbb P (X=k) - \mathbb P(X=-k))$ which  is  positive in your case.  Also, if $p_j = \frac 12$ the expectation is larger by the fact that $|X|, |X+Y_j|$ is a submartingale.


问:
Is the Radon Nikodym derivative conditioned on a filtration monotonically increasing?
Let $(\Omega, \mathcal F, (\mathcal F_t)_{t\geq 0},\mathbb P)$ a filtered probability space satisfying the usual conditions (especially $\mathcal F_0$ contains all null sets) and $\mathbb Q$ a further probability measure with $\mathbb Q\ll\mathbb P$. If we now restict the measures for each $t$ to the sub-$\sigma$-algebras $\mathcal F_t$ of the filtration, the property $$\mathbb Q|_{\mathcal F_t}\ll\mathbb P|_{\mathcal F_t}$$ is conserved. Thus the Radon Nikodym derivatives
$$S_t:=\frac{\mathrm d\mathbb Q|_{\mathcal F_t}}{\mathrm d\mathbb P|_{\mathcal F_t}}=\mathbb E_{\mathbb P}\left(\left.\frac{\mathrm d\mathbb Q}{\mathrm d\mathbb P}\right|\mathcal F_t\right)
$$
exist and form a martingale. As the Radon Nikodym derivative is only unique up to a $\mathbb P$-null set, I'm asking myself:

1. Question: Can the stochastic process $S:=(S_t)_{t\geq0}$ be choosen monotonically increasing?

The background of this question is: As $S$ is a martingale converging a.s. and in $L^1$ to $S_\infty:=\frac{\mathrm d\mathbb Q}{\mathrm d\mathbb P}$, I want to show that for a monotonacally decreasing convex function $\phi$ the process $\phi(S)$ is a submartingale converging in $L^1$ to $\phi(S_\infty)$. But to deduce that $\phi(S)$ is a submartingale I need $\phi(S)$ to be integrable (as a requirement for the Jensen inequality).

2. Question: Can be shown without any further assumptions that $\phi(S)$ is integrable?

In the final step I want to use something like the monotone convergence theorem to deduce the $L^1$-convergence of $S$.

答:
1. No, outside the trivial case $\Bbb Q=\Bbb P$.  Because $S$ is a martingale, $\Bbb E_{\Bbb P}[S_t]=\Bbb E_{\Bbb P}[S_u]$ for all $0\le t<u$. If it were true that $S_t\le S_u$, then you would have $\Bbb P[S_t=S_u]=1$.

2. If $\phi$ is understood to be a decreasing convex function mapping $[0,\infty)$ to $\Bbb R$, then the integrability of each $\phi(S_t)$ is assured. On the one hand, $\phi(x)\le\phi(0)$ for each $x\ge 0$, so $\phi(S_t)$ is bounded above. In the other direction, by convexity there are reals $a$ and $b$ such that $\phi(x)\ge a+bx$ for all $x\ge 0$. Thus, $\phi(S_t)\ge a+bS_t$. Because $S_t$ is integrable, these two bounds ensure the integrability of $\phi(S_t)$, which is therefore a submartingale. In addition, you have the bounds
$$
a+bS_t-\phi(0)\le\phi(S_t)-\phi(S_\infty)\le \phi(0)-a-bS_\infty,
$$
which together with the uniform integrability of $(S_t)_{0\le t\le\infty}$ is enough to guarantee that $\phi(S_t)$ converges to $\phi(S_\infty)$ in $L^1$.


答: take $m=8$ and $a=2$, so $\phi(m)=4$ and $2^{3}\equiv2^{4}\equiv0(mod8)$ yet $3\not \equiv4(mod4)$

Title: If $a^x \equiv a^y \pmod{m}$, does this imply $x \equiv y \pmod{\varphi(m)}$?

Body:
I know that $x \equiv y \pmod{\varphi(m)} \implies a^x \equiv a^y \pmod{m}$, so I was wondering if the converse also holds. If it does not hold in general, then are there any conditions where it does (such as assuming $(a, m)=1$)?

*Here $\varphi$ denotes the Euler totient function.

Edit: I see that the answer is no. However, the reason I was wondering about this is that I was watching this (https://youtu.be/f1oO9dEkqso?t=1386) lecture and don't know why else would Prof. Borcherds claim that $a_i \equiv 1 \pmod{p_{i}^{k_i}}$ is solvable if $\varphi(p_{i}^{k_i}) \mid n$ if the property in question is not true. In other words, how else is what's written justified?


问:
Countable linear combinations of indicator functions has a canonical form and form a vector space
Generalizing simple functions, let us say we have countable linear combinations of indicator functions over $\mathbb{R}^n$, that is $f(x)=\sum_{i=1}^{\infty}a_i\chi_{E_i}(x)$, where $\{E_i\}$ is a set of countable measurable sets and $a_i$ are scalars.

Then, will such functions have a canonical form, like the simple functions, that is, can they be represented as $f(x)=\sum_ib_i\chi_{G_i}(x)$ for constants $b_i$ and pairwise disjoint measurable sets $G_i$. Also, do these functions form a vector space over $\mathbb{R}^n$. The problem seems quite immediate, but I think the real problem here is convergence, as the summands are infinite. Any hints? How to proceed? Thanks beforehand.

答: The set isn't well defined. Take $\bigcap E_i\ne\emptyset$, $a_i = 1$ for all $i$.


问:
If $\tan x\tan y=\frac{b}{a},\,a,\,b\ne 0$, prove that $\frac{\sec^2x}{a\tan^2x+b}+\frac{\sec^2y}{a\tan^2y+b}=\frac{a+b}{ab}$
If $\tan x\tan y=\frac{b}{a},\,a,\,b\ne 0$, prove that $\frac{\sec^2x}{a\tan^2x+b}+\frac{\sec^2y}{a\tan^2y+b}=\frac{a+b}{ab}$

I solved it in the following way:

$\frac{1+\tan^2x}{a\tan^2x+b}+\frac{1+\tan^2y}{a\tan^2y+b}$

$=\frac{a\tan^2y+b+a\tan^2x\tan^2y+b\tan^2x+a\tan^2x+a\tan^2x\tan^2y+b+b\tan^2y}{2b^2+ab(\tan^2x+\tan^2y)}$

$=\frac{a(\tan^2y+\tan^2x)+b(\tan^2x+\tan^2y)+tb+2a\tan^2x\tan^2y}{2b^2+ab(\tan^2x+\tan^2y)}$

$=\frac{(\tan^2x+\tan^2y)(a+b)+2a\tan^2x\tan^2y+2b}{2b^2+ab(\tan^2x+\tan^2y)}$

$2b^2=2ab\tan x\tan y$

$\therefore \frac{1+\tan^2x}{a\tan^2x+b}+\frac{1+\tan^2y}{a\tan^2y+b}=\frac{(\tan^2x+\tan^2y)(a+b)+2a\tan^2x+\tan^2y+2b}{ab(\tan^2x+\tan^2y+2\tan x\tan y)}$

It remains to prove that:

$2a\tan^2x\tan^2y+2b=2(a+b)\tan x\tan y$

$2a\tan^2x\tan^2y+2b=ta\tan x\tan y+2b\tan x\tan y$

$2a\tan^2x\tan^2y+2b=2b+2b\tan x\tan y$

$2a\tan^2x\tan^2y=2b\tan x\tan y$

$\tan^2x\tan^2y=\frac{b}{a}\tan x\tan y$

Which is true.

$\therefore \frac{1+\tan^2x}{a\tan^2x+b}+\frac{1+\tan^2y}{a\tan^2y+b}=\frac{(a+b)(\tan x+\tan y)^2}{ab(\tan x+\tan y)^2}=\frac{a+b}{ab}$

Which completes the proof.

This proof is overly complex and took me over an hour to think of and write. Could you please explain to me a simpler, less messy and more intuitive proof?

Body: Let $t:=\tan x$ so $\tan y=\frac{b}{at}$ and we want$$\frac{1+t^2}{at^2+b}+\frac{1+b^2/(a^2t^2)}{b^2/(at^2)+b}=\frac{b(1+t^2)+at^2+b^2/a}{b(at^2+b)}=\frac{a+b}{ab},$$taking out a factor of $t^2+b/a$.


问:
Finding distributions
For the following, determine the distribution that the random variable $X$ follows (be sure to include appropriate parameters):

A particle starts at $(0, 0)$ and moves in one-unit steps that are independent and with probabilities in two directions: a probability of $0.5$ going North and $0.5$ going West. $X$ is the $x$ coordinate after $3$ steps.

A country club has $4$ halls: halls $A, B$, $C$ and $D$. Each hall has $25$ tables for a total of $50$ guests. Let $X$ be the number of tables in hall $A$ that are occupied after the first $18$ guests arrive. Guests arrive at the club at random, each order equally likely.

There are two random variables $X$ and $Y$ such that they have a joint distribution given by

$$f(x, y)= \left\{
  \begin{array}{lr}
      \frac{1}{e^y} & \mbox{if } x \in [0, \infty), y \in [x, \infty) \\
      0 & \mbox{otherwise}
      \end{array}
\right.$$

My attempt:

$X$ has a binomial distribution with $n = 3$ and $p = 0.5$.

I think is either exponential with $\theta = 25$ or gamma with $\theta = 25$ and $\alpha = 18$.

$X$ has a marginal distribution?

Are my solutions correct? If not, what is/are the correct distributions? Any assistance is appreciated.

答:
I think

$X\sim B(3,0.5)$

$X\sim B(18,1/4)$  (we have 18 trials / persons, each person with 1/4 prob. goes to room A, $X$ is total number of persons in room A - you can verify with simulation)

$X$ has marginal distribution (integrate $y$ out): $f(x)=\int_{x}^{\infty}e^{-y}dy=e^{-x}$

Title: Finding distributions


问:
What is the expected value of the estimator?
I would like to clear some doubts about estimators in statistics. By definition, an estimator $\hat{\theta}$ is a function that 'estimates' the value of the parameter $\theta$, itself being a random variable. An estimator is unbiased if it's expected value is $\theta$.

What I'm confused about is, what exactly does it mean, the expected value of the estimator? Is it the usual

$$\sum_i \hat{\theta_i} \cdot P\left(\hat{\theta}(X_i) = \hat{\theta_i}\right), \int_{-\infty}^{\infty} \hat{\theta} \cdot P\left(\hat{\theta}(X_i) \leq \hat{\theta}\right)$$
?

Or, is it simply taking the expected value of the parent distribution? As in,

$\mathbb{E}(\hat{\theta}) = \int \hat{\theta}\cdot f(x) dx $?

答:
Any estimator is a function of (only of) the data. thus tipically

$$\hat{\theta}=t(\mathbf{x})$$

and thus, by definition,

$$\mathbb{E}[\hat{\theta}]=\int_T tf(t)dt$$

Example...

given a simple random sample $X_1,\dots,X_n$ from a Uniform $U(0;\theta)$ the optimal estimator of $\theta$ is $t(\mathbf{x})=max(\mathbf{x})$

Its distribution is (easy to prove)

$$f_T(t)=\frac{n t^{n-1}}{\theta^n}$$

thus

$$\mathbb{E}[T]=\int_0^{\theta}\frac{nt^n}{\theta^n}dt=\frac{n}{n+1}\theta$$


问:
Adjoint Functors Operating On Natural Transformations
Suppose we have a natural transformation $\alpha$ from functor $L: \mathbb{C} \rightarrow \mathbb{D}$ to functor $L': \mathbb{C} \rightarrow \mathbb{D}$. Suppose $R: \mathbb{D} \rightarrow \mathbb{C}$ is the right adjoint of $L$. Suppose $R': \mathbb{D} \rightarrow \mathbb{C}$ is the right adjoint of $L'.$ Then is it possible to use $\alpha$ to somehow construct a natural transformation from $R'$ to $R$ ? If so how ?

答:
You have, for any $x\in \mathbb D$, the map $\bar\alpha_x: R'x\to RLR'x\to RL'R'x\to Rx $
given by composing the unit of the $(L,R)$ adjunction at $R'x$ with $R(\alpha_{R'X})$ and $R$ of the counit of the $(L',R')$ adjunction at $x$. It is straightforward to check that the $\bar\alpha_x$ assemble into a natural transformation $R'\to R$, called the mate of $\alpha$ in the literature.

In fact, the mate is the key ingredient needed to construct an identity-on-objects equivalence, contravariant on 1-morphisms and 2-morphisms, between the 2-category of categories, left adjoint functors, and natural transformations between them, and the 2-category of categories, right adjoint functors, and natural transformations between them.


问:
Showing that a sequence is Cauchy w.r.t. $d(x, y) = \left|e^{x} - e^{y}\right|$
I am trying to demonstrate that a metric space $X$ is not complete with the metric $d(x, y) = \left|e^x - e^y\right|$. To this end I am first trying to show that a sequence $x_n = \frac{1}{n}$ is Cauchy w.r.t. the said metric. My current mental block is with the "final touch", where one shows that the distance between any two elements of the sequence is at most $\epsilon$. Specifically, let $n_\epsilon > \left|\frac{1}{\ln(\epsilon)}\right|$. Then, for $l, k \geq n_\epsilon$ it holds that $\left|e^{x_l} - e^{x_k}\right| = \left|e^{\frac{1}{l}} - e^{\frac{1}{k}}\right| < \left|e^{\frac{1}{\min\{l, k\}}}\right| \leq \left|e^{\frac{1}{n_\epsilon}}\right| < \left|e^{\left|\ln(\epsilon)\right|}\right|$.

This might seem as an elementary step to most, but how to you go about with the absolute value around the logarithm in the exponent?

答: Actually, the sequence $(1/n)_{n \geq 1}$ is Cauchy in $\mathbb{R}$, and $\exp$ is uniformly continuous over $[0,1]$, so the sequence $(\exp(1/n))_{n \geq 1}$ is also Cauchy in $\mathbb{R}$, which means that $(1/n)_{n \geq 1}$ is Cauchy for your metric $d$.


问:
On the requirements of the Jensen inequality for conditional expectation
I'm often reading this version of the Jensen inequality for conditional expectation:

Let $(\Omega,\mathcal A, P)$ a probability space and $X$ a integrable random variable. Then for any convex function $\phi:\mathbb R\to\mathbb R$ holds $$\phi\big(E(X|\mathcal A)\big)\leq E\big(\phi(X)|\mathcal A\big).$$

I'm wondering if there isn't the requirement "$\phi(X)$ is integrable" missing? Otherwise I can't imagine why the RHS should even exist.

For the ordinary case (not conditioned) I understand why we don't need $\phi(X)$ to be integrable, as the RHS would be $\infty$ and the inequality would hold trivially. But here we might have something on the RHS that doesn't even exist.

答: Because $\phi$ is convex, it is bounded below by a function of the form $x\mapsto mx+b$, so the integrability of $X$ implies that of $\phi(X)^-$ (the negative part of $\phi(X)$). Even though $\phi(X)^+$ may not be integrable, the conditional expectation $E[\phi(X)\mid\mathcal A]$, as a random variable with values in $(-\infty,+\infty]$, exists in a generalized well-defined sense (for example, as the increasing limit $\lim_nE[\phi(X)\wedge n\mid\mathcal A]$). As such Jensen's inequality holds, even on the event $\{E[\phi(X)\mid\mathcal A]=+\infty\}$ because $\phi(E[X\mid\mathcal A])$ is a.s. finite.


答:
I've had a look at the video. In it, Prof. Borcherds never uses the converse you are mentioning, and is instead using just the statement in the "ordinary" direction. The essence is: if you break $m$ into primes, $m=p_1^{k_1}p_2^{k_2}\cdots$, and then, if you set:

$$n=\text{lcm}(\varphi(p_1^{k_1}),\varphi(p_2^{k_2}),\ldots)$$

then $n\le \varphi(p_1^{k_1})\cdot\varphi(p_2^{k_2})\cdots=\varphi(m)$ (lcm is at most as big as the product!), and most often this inequality is strict, yet each $\phi(p_i^{k_i})\mid n$, so, as per Euler's theorem, for each $a$ coprime with $m$ we have:

$$a^n\equiv 1\pmod {p_i^{k_1}}$$

and thus, for each $a$, we have $a^n\equiv 1\pmod m$.


问:
Rank of a permutation given k things are arranged out of n things
There are some standard dictionary problems like given a word bottle, one can find the rank of this word alphabetic order by making sperate cases, but my question is that if you can choose any three letters from this word bottle and arranged it, then how could one can find the rank of this word for example find the rank of the word btt or bol assuming any letter in  b,o,t,l,e can come in this new word and t can come two times out of three letters in this new word or permutation formed.

So in general if you have a word with n letters which can repeat and you form a another word or a permutation with k letters out of given n word letter how to find the rank of that word.

答:
As far as I know, there is no nice answer if the word can have repeated letters.

For any multiset $M$ of letters, let $P(M,k)$ be the set of $k$ letter words you can make using letters of $M$. For example,

$$P\big(\{A,A,B,C\} , 2\big)=\{AA,AB,AC,BA,BC,CA,CB\}$$

If $M$ consists of $m$ distinct letters, then $|P(M,k)|=m!/(m-k)!$.

Assuming you can compute $|P(M,k)|$, then you can compute the rank of a word $w$ in $P(M,k)$ has follows. Let $\ell$ be the first letter of $w$, and let $w'$ be the word obtained by removing $\ell$ from the beginning of $w$. Then
$$
\text{rank of $w$ in $P(M,k)$}
= \left(\sum_{a\in M,a<\ell}|P(M-\{a\},k-1)|\right)
+\text{rank of $w'$ in $P(M-{\ell},k-1)$}
$$
The first summation ranges over all letters $a$ in $M$ which are alphabetically before $\ell$, and $M-\{a\}$ means you remove one copy of $a$ from $M$. This is a recursive formula, so you need to apply the same formula to compute the rank of $w'$, and then apply this to $w''$, etc.

Edit: To explain why this is true; for each $a<\ell$, $|P(M-\{a\},k-1)|$ counts all of the words in $P(M,k)$ which start with $a$. These all come before $w$, so they increase its rank. The recursive part is comparing $w$ with the words in $P(M,k)$ that start with the same first letter.

Unfortunately, there is no simple "formula" to compute $|P(M,k)|$. This problem gets asked a lot: see Permutation Problem: Select m objects from n ones of r types by sampling without replacement and the linked questions in my comment for some discussion.

Example: Let $M=\{A,A,A,B,C,C\}$, and say you want to find the rank of $CBA$ among all three-letter words made of the letters of $M$. Then
\begin{align}
\text{rank of $CBA$ in $P(AAABCC,3)$}
&=
\overbrace{|P(AABCC,3)|}^{\text{remove A}} +\overbrace{|P(AAACC,3)|}^{\text{remove B}}
\\&\quad+ \text{rank of $CBA$ in $P(AAABC,2)$}
\\\\
\text{rank of $BA$ in $P(AAABC,2)$}
&= |P(AABC,3)|
\\&\quad+ \text{rank of $A$ in $P(AAAC,1)$}
\\\\
\text{rank of $A$ in $P(AAABC,1)$}
&= 0^{\quad\text{no letters before $a$, so this is empty sum}}
\\&\quad+ \text{rank of $\varnothing$ in $P(AAC,0)$}
\\\\
\text{rank of $\varnothing$ in $P(AAC,0)$} &= 1^{\quad\text{since length is $0$, only word is empty word, which has rank $1$}}
\end{align}
Finally, you get the rank of $CAB$ is $|P(AABCC,3)|+|P(AAACC,3)|+|P(AABC)|+1$. Again, to compute these sizes, look at the links I mentioned.


问:
Intersection of two vectors with tails?
I'm working on a 2 degree of freedom simulation of gas dynamics.

Suppose you have two vectors in the x-y plane. Each vector has two tails, or coordinates where they start. Because I'm looking for a point in the future where the vectors will intersect, I can treat the vectors as rays, as long as they maintain the same direction and tail. How can you find where, or if, they intersect?

The Vectors in the program are stores as having an x-Magnitude, y-Magnitude, and a coordinate position for their tails.

答:
Let's say particle $i$ is at $(\mathcal{X}_{i} ,\, \mathcal{Y}_{i})$ at time $t = 0$, with velocity $(\dot{\mathcal{X}}_i, \dot{\mathcal{Y}}_i)$, so that unless it collides with another particle, it's trajectory at time $t \ge 0$ is
$$\begin{aligned}
x_i(t) & = \mathcal{X}_i + t \dot{\mathcal{X}}_i \\
y_i(t) & = \mathcal{Y}_i + t \dot{\mathcal{Y}}_i \\
\end{aligned}$$

There are two possible questions here: whether particles $i$ and $j$ intersect, i.e.
$$\left\lbrace \begin{aligned}
x_i(t) & = x_j(t) \\
y_i(t) & = y_j(t) \\
\end{aligned} \right. \tag{1}\label{G1}$$
or whether the lines extending the particle trajectories intersect,
$$\left\lbrace \begin{aligned}
x_i(t_1) & = x_j(t_2) \\
y_i(t_1) & = y_j(t_2) \\
\end{aligned} \right. \tag{2}\label{G2}$$

Solving $\eqref{G1}$ we find that if at least one of particles $i$, $j$ is moving,
$$\lvert \dot{\mathcal{X}}_i \rvert + \lvert \dot{\mathcal{Y}}_i \rvert + \lvert \dot{\mathcal{X}}_j \rvert + \lvert \dot{\mathcal{Y}}_j \rvert \gt 0 \tag{1a}\label{G1a}$$
and an intersection is possible,
$$(\mathcal{X}_i - \mathcal{X}_j)(\dot{\mathcal{Y}}_j - \dot{\mathcal{Y}}_i) - (\mathcal{Y}_i - \mathcal{Y}_j)(\dot{\mathcal{X}}_j - \dot{\mathcal{X}}_i) = 0 \tag{1b}\label{G1b}$$
then the particles intersect at time $t$,
$$t = \frac{\mathcal{X}_i - \mathcal{X}_j}{\dot{\mathcal{Y}}_j - \dot{\mathcal{Y}}_i} = \frac{\mathcal{Y}_i - \mathcal{Y}_j}{\dot{\mathcal{X}}_j - \dot{\mathcal{X}}_i} \tag{1c}\label{G1c}$$
noting that if the particles' trajectories diverge, above yields $t \lt 0$ (indicating a collision would occur if time passed backwards); and also that the first right side is undefined if $\dot{\mathcal{Y}}_i = \dot{\mathcal{Y}}_j$, and the second right side is undefined if $\dot{\mathcal{X}}_i = \dot{\mathcal{X}}_j$.

(Numerically, you'll need to use $\lvert \dots \rvert \le \epsilon_2$ for $\eqref{G1b}$, where $\epsilon_2$ is the largest value squared that you still consider zero.  For example, if your coordinates and velocity components have three decimals, then $\epsilon_2 = 0.0005^2 = 0.00000025$. For $\eqref{G1c}$, use the side whose denominator has the largest magnitude: the first right side if $\lvert \dot{\mathcal{Y}}_j - \dot{\mathcal{Y}}_i \rvert \ge \lvert \dot{\mathcal{X}}_j - \dot{\mathcal{X}}_i \rvert$, the second right side otherwise.)

Solving $\eqref{G2}$ for $t_1$ and $t_2$ yields
$$\begin{aligned}
d &= \dot{\mathcal{X}}_i \dot{\mathcal{Y}}_j - \dot{\mathcal{Y}}_i \dot{\mathcal{X}}_j \\
t_1 &= \displaystyle \frac{ ( \mathcal{X}_j - \mathcal{X}_i ) \dot{\mathcal{Y}}_j - ( \mathcal{Y}_j - \mathcal{Y}_i ) \dot{\mathcal{X}}_j }{d} \\
t_2 &= \displaystyle \frac{ ( \mathcal{X}_j - \mathcal{X}_i ) \dot{\mathcal{Y}}_i - ( \mathcal{Y}_j - \mathcal{Y}_i ) \dot{\mathcal{X}}_i }{d} \\
\end{aligned}$$
and if $d = 0$, the trajectories are parallel and do not intersect.  One or both of the times $t_1$ or $t_2$ may be negative, as well.

Title: Intersection of two vectors with tails?


问:
Parametrizing straight lines intersecting a circle
I've got a point $ P (-1,1,2) $ and the circle $ C: ~ (x-1)^2+(y-1)^2 = \dfrac {25} {4} $ (which is contained in the plane: $ z=0 $). I need to write the parametric equations of all straight lines that go through $P$ and intersect $C$.

I know it must be something like: $ \dfrac {x+1} {\lambda_1} =  \dfrac {y-1} {\lambda_2} =  \dfrac {z-2} {\lambda_3} $ and then transform it to parametric. But my knowledge here weakens and I don't know how to indicate the straight lines that intersect $C$.

Furthermore, after that I need one straight line from that bundle that cuts $s: ~ \{x=a, y=3.5, z=-2; ~ a \in \mathbb{R}_0^+ \} $

Any help would be great. Thanks in advance!

答: So you have the flat circle ${(x-1)}^2 + {(y-1)}^2 = \frac{25}{4}$, not a cylinder, as z=0.  You need to choose a parameter.  Angle is often a good choice for circles.  Let $0 ≤ \theta ≤ 2\pi$.  Then $x = 1+\frac{5}{2}cos \theta $ and $y =1+\frac{5}{2}sin \theta $. Find the equation of the line through such a point and your given point.


问:
Let $a>R>0$. Calculate the volume of the body that meets the differences $x^2+y^2+z^2 \leq a^2$ and $x^2+y^2 \leq R^2$
Let $a > R > 0$. Calculate the volume of the body constricted by $x^2+y^2+z^2 \leq a^2$ and $x^2+y^2 \leq R^2$

I don't know where to start with this problem. The constraints form a cylinder which body I need to integrate the volume of. Only hint that I've gotten is that it might be helpful to use polar-coordinates. Would really appreciate help!

答:
Here is an easy way to calculate this volume, IMO.

Let's denote $I$ be this volume , and $J$ be the complement of $I$ (in the sphere). We have then $I+J = \frac{4}{3}\pi a^3$.

The volume $J$ is defined by the 2 inequalities: $x^2+y^2+z^2 \le a^2$ and $x^2+y^2 >R^2$. We deduce then $a^2-z^2 > R^2$ or $ -\sqrt{a^2-R^2}< z<\sqrt{a^2-R^2} $

We have
\begin{align}
J &= \int_{-\sqrt{a^2-R^2}< z<\sqrt{a^2-R^2}} \left( \int_{R^2<x^2+y^2 \le a^2-z^2}dxdy  \right) dz\\
\end{align}
It's evident that the inner integral is the surface are of a circle ring and is equal to $\pi \left((a^2-z^2) -R^2\right)$. Then
\begin{align}
J &= \int_{-\sqrt{a^2-R^2}< z<\sqrt{a^2-R^2}} \pi \left((a^2-R^2) -z^2\right) dz \\
 &=  2\pi (a^2-R^2)\sqrt{a^2-R^2}- \frac{2}{3} \pi (a^2-R^2)^{\frac{3}{2}} \\
 &=  \frac{4}{3} \pi (a^2-R^2)^{\frac{3}{2}} \\
\end{align}

Finally, we have
$$I =  \frac{4}{3}\pi a^3 - J =  \frac{4}{3}\pi a^3 -\frac{4}{3} \pi (a^2-R^2)^{\frac{3}{2}} $$


问:
Quadratic forms into sums of squares of linear forms
I would like to know whether there exists software or online calculators that turn quadratic forms directly into sums/differences of squares of linear forms? For instance:

For $q(x,y,z) = 2x^2 - 2y^2 - 6z^2 + 3xy - 4xz + 7yz$

Returns $q(x,y,z) = 2(x + \frac{3}{4}y - z)^2 - 8(z - \frac{5}{8}y)^2$

I know how to compute it using the Gauß reduction method, but I wish I could get it swifter...

Thanks — I'm a French student, sorry if I made any language mistakes.

答:
With MATHEMATICA you can proceed with

```<code>Clear[SquareIt]
SquareIt[expr_, vars_List] := Module[{n = Length[vars], rels, coefs, sol, varsint, obj},
  obj = Sum[a[i] Sum[b[i, j] vars[[j]], {j, 1, n}]^2, {i, 1, n}];
  rels = CoefficientRules[expr - obj, vars];
  coefs = Map[Last, rels];
  varsint = Variables[coefs];
  sol = FindInstance[coefs == 0, varsint][[1]];
  Return[obj /. sol]
]

SquareIt[2 x^2 - 2 y^2 - 6 z^2 + 3 x y - 4 x z + 7 y z, {x, y, z}]
</code>```

$$
\frac{25}{8} \left(x+\frac{z}{5}\right)^2-\frac{9}{8} \left(x-\frac{4 y}{3}+\frac{7 z}{3}\right)^2
$$

NOTE

Regarding

$$
2 a x^2-6 a z^2-2 b y^2+3 c x y-4 x z+7 y z
$$

Manually first we separate the squares as

$$
2 a x^2-6 a z^2-2 b y^2
$$

and then we use the identity

$$
x_i x_j = \frac 14\left((x_i+x_j)^2-(x_i-x_j)^2)\right)
$$

obtaining

$$
2 a x^2-6 a z^2-2 b y^2+\frac{3}{4} c \left((x+y)^2-(y-x)^2\right)+(z-x)^2-(x+z)^2+\frac{7}{4} \left((y+z)^2-(z-y)^2\right)
$$

Gauss method.

Body:
With MATHEMATICA you can proceed with

Clear[SquareIt]
SquareIt[expr_, vars_List] := Module[{n = Length[vars], rels, coefs, sol, varsint, obj},
  obj = Sum[a[i] Sum[b[i, j] vars[[j]], {j, 1, n}]^2, {i, 1, n}];
  rels = CoefficientRules[expr - obj, vars];
  coefs = Map[Last, rels];
  varsint = Variables[coefs];
  sol = FindInstance[coefs == 0, varsint][[1]];
  Return[obj /. sol]
]

SquareIt[2 x^2 - 2 y^2 - 6 z^2 + 3 x y - 4 x z + 7 y z, {x, y, z}]

$$
\frac{25}{8} \left(x+\frac{z}{5}\right)^2-\frac{9}{8} \left(x-\frac{4 y}{3}+\frac{7 z}{3}\right)^2
$$

NOTE

Regarding

$$
2 a x^2-6 a z^2-2 b y^2+3 c x y-4 x z+7 y z
$$

Manually first we separate the squares as

$$
2 a x^2-6 a z^2-2 b y^2
$$

and then we use the identity

$$
x_i x_j = \frac 14\left((x_i+x_j)^2-(x_i-x_j)^2)\right)
$$

obtaining

$$
2 a x^2-6 a z^2-2 b y^2+\frac{3}{4} c \left((x+y)^2-(y-x)^2\right)+(z-x)^2-(x+z)^2+\frac{7}{4} \left((y+z)^2-(z-y)^2\right)
$$

Gauss method.


答:
Your approach won't work since $e^{1/n} > e^0 = 1$ for any $n \in \mathbb{N}$, and  we can find no bound $e^{1/n_\epsilon} < \epsilon <1$.

Instead, for $l > k > n$, we have

$$\left|e^{\frac{1}{l}} - e^{\frac{1}{k}} \right|= e^{\frac{1}{k}}(1 - e^{\frac{1}{l}- \frac{1}{k}}) \leqslant e \cdot \left(1 - e^{-\left(\frac{1}{k}- \frac{1}{l}\right)} \right)$$

Since $e^{-x} \geqslant 1 - x $ for $0 \leqslant x < 1$, it follows that with $n > 1/(e\epsilon)$,

$$\left|e^{\frac{1}{l}} - e^{\frac{1}{k}} \right|\leqslant e \left(\frac{1}{k} - \frac{1}{l}\right) < \frac{e}{k}< \frac{e}{n}< \epsilon$$

Title: Showing that a sequence is Cauchy w.r.t. $d(x, y) = \left|e^{x} - e^{y}\right|$

Body:
I am trying to demonstrate that a metric space $X$ is not complete with the metric $d(x, y) = \left|e^x - e^y\right|$. To this end I am first trying to show that a sequence $x_n = \frac{1}{n}$ is Cauchy w.r.t. the said metric. My current mental block is with the "final touch", where one shows that the distance between any two elements of the sequence is at most $\epsilon$. Specifically, let $n_\epsilon > \left|\frac{1}{\ln(\epsilon)}\right|$. Then, for $l, k \geq n_\epsilon$ it holds that $\left|e^{x_l} - e^{x_k}\right| = \left|e^{\frac{1}{l}} - e^{\frac{1}{k}}\right| < \left|e^{\frac{1}{\min\{l, k\}}}\right| \leq \left|e^{\frac{1}{n_\epsilon}}\right| < \left|e^{\left|\ln(\epsilon)\right|}\right|$.

This might seem as an elementary step to most, but how to you go about with the absolute value around the logarithm in the exponent?


问:
Possible Maximum or Minimum??
I was solving some curve sketching questions and wanted to ask something that I am a little confused about.
Why, when we find the first derivative of an equation and set it equal to zero to find the critical points, we refer to the points as "possible" max and min. Are there times when the critical point is not a max or min? Thank you in advance!</p>

答:
The graph of $f(x)=x^2$ has a minimum at $x=0$.

The graph of $f(x)=-x^2$ has a maximum at $x=0$.

The graph of $f(x)=x\sin x$ does not have a minimum or maximum at $x=0$. Rather, it has a local minimum.

The graph of $f(x)=x^3$ does not have a minimum, maximum, local minimum, or local maximum at $x=0$. It has a point of inflection.

Just knowing that $f'(x)=0$ at some point is not enough. We need more information. Moreover, not all critical points occur when $f'(x)=0$. The graph of $f(x)=|x|$ is not even differentiable at $x=0$, but it certainly has a minimum there.

Title: Possible Maximum or Minimum??