问:
Direction of the spiral trajectories for coupled differential equation system
I am looking at the coupled system $x'=ax+by,$ $y'=cx+dy$ with non-real eigenvalues.

Do I know it correctly that the trajectories are clockwise if $b-c$ is positive and anticlockwise if $b-c$ is negative? I am quoting this claim from http://wwwf.imperial.ac.uk/metric/metric_public/differential_equations/second_order/qualitative_methods_1.html

Can you also please point me to an online resource where I can find the proof of the condition?

答:
Wew can diagonalize the matrix $$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \to \begin{bmatrix} m_1 & 0 \\ 0 & m_2 \end{bmatrix}$$
Then we work with $(X,Y)$ instead of $(x,y)$. Yhe eigenvalues are gicven by $$m_1,m_2=\frac{1}{2}[(a+d)\pm \sqrt{(a-d)^2+4bc}]$$ Suppose

Case (1): $m_1=1, m_2=2 \implies X=e^t, Y=e^{2t}\implies Y=X^2,$ we get parabolic trajectory.
Next, suppose

Case(2):$m=-1, m_2=-2 \implies X=e^{-t}, Y=e^{-2t}$, we again get get parabolic trajectory $Y=X^2$.

But the difference is that when time increases to $\infty$ in case (1) both $X,Y$ diverge away from origin (0,0). So the arrow goes out and the orbit is unstable. In case (2) the arrow will direct towards origin, orbit will be stable.

When eigenvalues are non-real (Complex) we get spiral trajectories if $\Re (m_1,m_2)>0$, $X,Y$ diverge, we get unstable spiral. Otherwise, we get stable spiral.

One may call unstable spiral as clockwise and stable one as anti-clockwise.

Any way the sign of real part of eigenvalues decides this.

When eigenvalues are real but of opposite sign the origin becomes a saddle point.

You find a good discussion on this in Problems in ODE by M.L. Krasnov et al (MIR Publication)


问:
exponential distribution - get some meaning
I read that a continuous random variable having an exponential distribution can be used to model inter-event arrival times in a Poisson process. Examples included the times when asteroids hit the earth, when earthquakes happen, when calls are received at a call center, etc. In all these examples, the expected value of the number of events per unit of time, lambda, is known and is constant over time. Moreover, each event's occurrence is independent of previous events' occurrences. And the exponential variable that models inter-event arrivals has the same lambda parameter as the Poisson variable that models the number of events.

And now, my problem. It don't get the connection between the intuition behind the exp distrib and its pdf. It seems obvious that the more time it passes by without an earthquake happening, the more likely it is than an earthquake will happen. Assume my understanding of lambda is correct, i.e., lambda = the rate at which the event happens, e.g., 5 earthquakes per minute on some remote, angry planet. On the pdf graphic, the prob of generating a value between 0 and 1 is greater than the prob of choosing a value between 4 and 5 for instance. How is this graphic related to the fact that on average we need to have 5 earthquakes per minute?

答:
The key property of the exponential distribution is that it is memoryless, so that translation of the PDF in time $x\to x-c$ is just scaling of the PDF $f\to e^{\lambda c} f$ plus cutting it off at zero. Probabilistically this says $P(X>s+t|X>s)=P(X>t)$. That is, it doesn't matter how long it was since the last event, the probability that it takes a certain time from now is the same as it would be in any other situation. For somebody who has been waiting the whole time it might seem very unlikely that it will take 20 minutes, but given that it doesn't happen in the first quarter an hour, their new estimate of the likelihood of the  total wait (not the wait stating at quarter an hour) being at least 20 minutes has gone up - it is now the same they thought the probability of it's taking over 5 minutes was at the start. Thus you have to be careful what you are asking.

Your feeling that the likelihood of an earthquake goes up over time is indeed contrary to the principle of the exponential dist. as Did said. However, one thing is true: e.g. let $p$ be the prob. of the next earthquake being between 15-20 minutes at the start, say at midnight; let $X$ be the time past midnight. Let $q$ be the the probability it happened in that same time given that you know it doesn't happen in the first 10 minutes. Then $q>p$. But let $Y$ be the time taken for the first earthquake after $00:10$. The probability that $Y$ is between 5 and 10 is unchanged by the knowledge that there was nothing between midnight and 10 past midnight. It was always $q$.

As to the mean, well, there's no reason why earlier times being more likely can prevent the mean being anything in particular - if there are large times with some probability then it can all balance out.

ContentLicense: CC BY-SA 3.0


问:
Prove that a Truncated Gaussian is log-concave
Is the truncated Gaussian a log-concave distribution?

I read here at Section 2.2.5 that whatever log-concave truncated distribution (and hence also Gaussian) is log-concave and the proof is trivial. How to prove this statement?

答:
the density of a truncated rv is the following

$$f_{X|X>c}=\frac{1}{1-F_X(c)}f_X(x)$$

This prove your statement

Proof of the truncated density

Let's$X$ a rv (assuming it is continuous) with CDF $F_X(x)$ and let $B=\{X>c\}$

Thus, by definition

$$F(x|B)=\frac{\mathbb{P}[c<X<x]}{\mathbb{P}[X>c]}=\frac{F_X(x)-F_X(c)}{1-F_X(c)}$$

if $x>c$ and zero  if $x\leq c$

derivating F you get the density I showed before

ContentLicense: CC BY-SA 4.0


问:
Probability space and random variable
Let us assume that we have a coin that is deformed and weighted. When you toss the coin, it lands on its side in
20% of the tosses. The coin lands on heads 30% of the time, and otherwise on tails.
1.We are now tossing our deformed, weighted coin several times until it lands on its side. Let us denote Y as the
number of times the coin was actually tossed. Is Y a random variable? Explain your reasoning.
2. Assuming now that Y = 9 (where Y is from the previous assignment), what is the expected number of heads
tossed before the coin landed on its side? Use the conditional expectation and explain your result.

1.I can say that y is a random variable because it is F-B(R) measurable but how to prove that it is F-B(R) measurable?
2. Let Y = no. Of toss
X= no. Of heads tossed before the coin landed on its side.
If now we compute E(X|Y=9) that means on 9th toss the coin landed on its side
So we have X~B(n,p) with n=8 and p=0.3
E(X|Y=9) = np = 8x0.3 = 2.4
Is this approach right?

答:
Yes, you are almost there.

$Y$ is a random variable and it is also a known rv: it's a geometric with density

$$f_Y(y)=0.8\cdot0.2^{y-1}$$

and CDF

$$F_Y(y)=1-0.2^y$$

$y=1,2,3,\dots$

Having defined a density (and a CDF too) it is enough as a proof.

Given $Y=9$ the number of heads have a Binomial $Bin(8;0.3)$ as you stated

Title: Probability space and random variable


问:
Norm of multiplication operator by a Fourier transform of an $L^1$ function
I'm trying to compute the operator norm of the multiplication operator $\mathcal{M}_{\hat{g}}:L^2(\mathbb{R}^d)\to L^2(\mathbb{R}^d)$ given by $\mathcal{M}_{\hat{g}}(f)=f\cdot \hat{g}$, where $\hat{g}$ is the Fourier transform of $g\in L^1(\mathbb{R}^d)$. I think that the operator norm is equal to $\Vert g\Vert_{L^1}$, but I was only able to show it is a bound for $\Vert \mathcal{M}_{\hat{g}}\Vert_{op}$.

I know that $\Vert \mathcal{M}_{\hat{g}}\Vert_{op}=\Vert \hat{g}\Vert_{L^\infty}$ and that $\Vert \hat{g}\Vert_{L^\infty}\leq \Vert \mathcal{F}\Vert_{op}\cdot \Vert g\Vert_{L^1}=\Vert g\Vert_{L^1}$, where $\mathcal{F}$ is the Fourier transform. I want to find a sequence of $L^2$ functions such that

$$ \Vert f_n\cdot \hat{g}\Vert_{L^2} \overset{n\to \infty}{\to}\Vert g\Vert_{L^1} \quad \text{and} \quad \Vert f_n\Vert_{L^2}=1 \quad \text{for all} \quad n, $$

but I'm not sure how to go about this.
This might be a trivial question but I'm not sure how to solve this point. I'm also not sure whether this is indeed true.

答:
As you remarked, for $g∈ L^1$, we have $\|\mathcal{M}_{\widehat g}\|_{L^2\to L^2} = \|\widehat g\|_{L^\infty} ≤ \|g\|_{L^1}$.

$\bullet$ If $g≥ 0$, then $\|g\|_{L^1} = ∫ g = \widehat{g}(0) ≤ \|\widehat g\|_{L^\infty}$.

$\bullet$ However, in the general case, this reverse inequality is false. See for example the answer of Giuseppe Negro here: Estimate the $L^1$-norm of the Fourier transform, or the answer of David C Ullrich here: Is the inverse of the Fourier transform $L^1(\mathbb R)\to (C_0(\mathbb R),\Vert \cdot \Vert_\infty)$ bounded?.


问:
Calculate $ \lim_{n\to\infty}\left(\frac{1}{\sqrt{n^{2}+1}}+\frac{1}{\sqrt{n^{2}+2}}+...+\frac{1}{\sqrt{n^{2}+n}}\right) $
I have to calculate the limit :

$ \lim_{n\to\infty}\left(\frac{1}{\sqrt{n^{2}+1}}+\frac{1}{\sqrt{n^{2}+2}}+...+\frac{1}{\sqrt{n^{2}+n}}\right) $

Now here's what I've tried:

$ \lim_{n\to\infty}\left(\frac{1}{\sqrt{n^{2}+1}}+\frac{1}{\sqrt{n^{2}+2}}+...+\frac{1}{\sqrt{n^{2}+n}}\right)=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{\sqrt{n^{2}+k}}=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{n}\frac{1}{\sqrt{1+\frac{k}{n^{2}}}}$

Now to me it looks like a Riemman sum of the function  $ f\left(x\right)=\frac{1}{\sqrt{1+x^{2}}} $, the partition $ P_{n}=\left\{ 0,\frac{1}{n},\frac{2}{n},...,\frac{n}{n}\right\}  $ and for the interval $ [\frac{j-1}{n},\frac{j}{n}] $ we choose that point $ \frac{\sqrt{j}}{n} $ so the sequence of points would be $ \theta_{n}=\left\{ \frac{\sqrt{1}}{n},\frac{\sqrt{2}}{n},...,\frac{\sqrt{n}}{n}\right\}  $.

Thus

$ \lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{n}\frac{1}{\sqrt{1+\frac{k}{n^{2}}}}=\intop_{0}^{1}\frac{1}{\sqrt{1+x^{2}}}dx=\ln\left(|\sqrt{x^{2}+1}+x|\right)|_{0}^{1}=\ln\left(\sqrt{2}+1\right) $

But I doubt myself because the calculation of the integral $ \intop_{0}^{1}\frac{1}{\sqrt{1+x^{2}}}dx $ is not very easy and Im not sure if this is the right way.

So I'd appreciate if someone can tell if this solution holds.

If my solution is not correct I'd happy to see a correct solution.

答:
The solution is incorrect because in general $$\frac{\sqrt{j}}{n} \notin [\frac{j-1}{n},\frac{j}{n}] $$

To find this limit notice that for $1\le k\le n$
$$ \frac{1}{\sqrt{n^2+n}} \le \frac{1}{\sqrt{n^2+k}} \le \frac{1}{n} $$
so
$$ \frac{1}{\sqrt{1+\frac{1}{n}}}=\frac{n}{\sqrt{n^2+n}} \le \sum_{k=1}^n\frac{1}{\sqrt{n^2+k}} \le n \cdot \frac{1}{n} =1$$
And use the squeeze theorem.


问:
Show that operator is diagonalizable

Let $T$ be the self-adjoint operator defined by

$$(T x)_{n} = x_{n + 1} + x_{n - 1} \text{ for } n \in \mathbb{Z}, (x_{n})_{n \in \mathbb{Z}}$$
on $\ell^{2}(\mathbb{Z})$.

Show that $T$ is diagonalizable by giving an explicit unitary operator $B : L^{2}(\mathbb{T}, \frac{d \theta}{2 \pi}) \rightarrow \ell^{2}(\mathbb{Z})$ such that $B M_{f} = TB$, where $f : \mathbb{T} \rightarrow \mathbb{R}$ is $f(e^{i \theta}) = 2 \cos(\theta)$. Here $M_f$ is multiplication by $f$ such that $(M_f g)(\xi) = f(\xi)g(\xi)$.

Note also that $\mathbb{T}$ here means the unit circle.

I am not sure what $B$ should be. Usually, to show diagonalizability I use the Spectral Theorem which states that every normal operator (acting on a separable
Hilbert space) is diagonalizable. But here one asks to explicitly give a unitary operator.

答:
Let $\{\epsilon_n:n\in\mathbb{Z}\}$ denote the standard ONB of $\ell^2(\mathbb{Z})$, so $\epsilon_n=(\dots,0,0,\dots,0,1,0,\dots,0,0,\dots)$ where $1$ lies in the $n$-th position. Also, let $e_n(z)=z^n$ for $n\in\mathbb{Z}$, so $\{e_n: n\in\mathbb{Z}\}$ is the standard orthonormal basis for $L^2(\mathbb{T})$.

Note that $T\epsilon_n=\epsilon_{n-1}+\epsilon_{n+1}$ for all $n\in\mathbb{Z}$, since

$$T(\epsilon_n)=(\dots,0,0,\dots,0,1,0,1,0,\dots,0,0,\dots)$$
where the aces lie on the $n-1$ and the $n+1$ positions.

Also, observe that if $z=e^{i\theta}$ then $$f(z)=2\cos(\theta)=e^{i\theta}+e^{-i\theta}=z^1+z^{-1}=e_1(z)+e_{-1}(z).$$

Now observe that $M_f(e_n)(z)=((e_1+e_{-1})\cdot e_n)(z)=(z+z^{-1})\cdot z^n=z^{n+1}+z^{n-1}=e_{n+1}(z)+e_{n-1}(z)$, so $M_f(e_n)=e_{n+1}+e_{n-1}$.

Define $B: L^2(\mathbb{T})\to \ell^2(\mathbb{Z})$ on the orthonormal basis and extend to the entire space: simply set $Be_n=\epsilon_n$. Note that $B$ maps the ONB onto an ONB, so $B$ is a unitary operator.

The relation $BM_f=TB$ is easily verified: it suffices to do so on the vectors of the ONB, and then continuity and linearity will yield the result on the entire space. Let $n\in\mathbb{Z}$. We have
$$BM_f(e_n)=B(e_{n+1}+e_{n-1})=\epsilon_{n+1}+\epsilon_{n-1}=T\epsilon_n=TB(e_n).$$

Title: Show that operator is diagonalizable


问:
A variation on the induced topology (subspace topology)
One of the easiest ways to find a topology on a set is to used the Induced Topology.

Induced Topology: Suppose you have a set $X$ and a topology on $X$ called $\mathcal{T}_X$. Suppose $Y \subseteq X.$ Then we can find a topology for $Y$ (call it $\mathcal{T}_Y$) as given by:

$\mathcal{T}_Y=\{O \cap Y  \mid O \in \mathcal{T}_X\}$.

My question is: does the induced topology still hold for sets $Y \not \subseteq X.$ Suppose $Y \cap X = \emptyset.$ Can we still inherit a topology on $Y$ if $X$ and $Y$ are disjoint? I'm going to say no.

Here's my test:

Let $X =\{1,2\}$ and $\mathcal{T}_X = \{\emptyset,\{1\},\{1,2\}\}$

Let $Y = \{5\}$

Then, following from our induced topology method,

$\mathcal{T}_Y=\{\{5\} \cap \emptyset, \{5\} \cap \{1\}, \{5\} \cap \{1,2\}\} = \{\emptyset,\emptyset,\emptyset\}$

But this fails to satisfy definition of topology since $Y=\{5\}$ and $Y \not \in \mathcal{T}_Y.$

So it would seem that we need $Y \subseteq X$ so that $Y \in \mathcal{T}_Y$, otherwise we end up with a bunch of empty sets.

Is my reasoning here correct?

答:
No, this concept makes no sense unless $Y \subseteq X$. It will only be a topology if this is the case: We must have $Y \in \mathcal{T}_Y$ so $\exists O \in \mathcal{T}_X: O \cap Y = Y$ from which it follows that $Y \subseteq O \subseteq X$.

Note that the definition of $\mathcal{T}_Y$ is not an arbitrary one, it's the natural choice in that it is the smallest topology on $Y$ that makes the inclusion map $i: Y \to X$ continuous. Extending it to any $Y$, unrelated to $X$ does not solve such a natural problem, or is not especially useful in constructing new spaces. So don't consider it, it won't get accepted.


问:
Lemma 1 - Banach-Mazur Game - Dan Ma's topology Blog
I have been given a presentation for a course based on this particular post in Dan Ma's topology blog. The post itself is very clear and educational, but there is one proof that I'm not sure why it is complete.

Lemma 1 has a rather standard application of Zorn's lemma, but I feel that I'm missing something in the second paragraph of the proof.

Lemma 1: Let $X$ be a space, Let $O \subset X$ be a nonempty open set. Let $\tau$ be the set of all nonempty open subsets of $O$. Let $f: \tau \longrightarrow \tau$ be a function such that for each $V \in \tau$, $f(V) \subset V$. Then, there exists a disjoint collection $\mathcal{U}$ consisting of elements of $f(\tau)$ such that $\bigcup \mathcal{U}$ is dense in $O$.

The proof begins by explaining the trivial case of $O$ having just one point, then if I understand correctly everything holds in a rather obvious way, namely: the property of $f$ gives me the same set as an image and clearly this is dense in $O$.

Then, we assume $O$ has two points and the author shows that $\mathcal{P}$ (the set of all collections $\mathcal{F}$ such that $\mathcal{F}$ is a disjoint collection of elements of $f(\tau)$) is non-empty. For this, two disjoint open subsets of $O$ are taken based on the assumption that $O$ has at least two points.

Why must these subsets be disjoint? Isn't the author assuming anything in addition?

I underestand that by applying $f$ later those sets become exactly of the form of an element of $\mathcal{P}$ which is the goal of that particular part of the proof. I thought we could possibly assume they are disjoint and otherwise apply $f$ again and again until they are but no assumption really guarantees this process would stop.

答:
In the introduction, before the heading Banach-Mazur Game, he says that all spaces are assumed to be Hausdorff.

And then if $O$ has at least two points, say $p \neq q$, we can find disjoint open  sets $U_p$ and $U_q$ around them and then $\{U_p \cap O, U_q \cap O\}$ is a starting point, one of the members of your poset of disjoint open subfamilies of $O$ etc.


问:
Density of continuous knots in the plane transversal to some circles
This is an exercise from the book "Knots and Links" by Rolfsen (exercise 6 in section 2C)

Let $\kappa : S^1 \rightarrow \mathbb{R}^2-(0,0)$ be a continuous imbedding.  Let $M := \{ x \in \mathbb{R}^2-(0,0) : |x| \in \mathbb{Z} \}$.   Let $\epsilon > 0$.  Show that there exists an imbedding $\lambda : S^1 \rightarrow \mathbb{R}^2-(0,0) $ satisfying

$|\kappa - \lambda | < \epsilon$

$\lambda$ and $\kappa$ are homotopic in $\mathbb{R}^2-(0,0)$

$\lambda$ anx $\kappa$ have disjoint images

$\lambda$ is transversal to $M$

The image of $\lambda$ meets $M$ in only finitely many points.

Here transversality means that each point of intersection has a neighborhood homeomorphic to the plane such that one set (really its intersection with the neighborhood) is mapped to the x axis and the other to the y axis.

Of course in the smooth setting one has tubular neighborhoods and the density of transversal maps to work with.  In the purely topological setting I'm completely lost.   Thank you for reading my question.

答:
A quick way to get $\lambda$ is to use complex analysis. Let $f$ be a conformal map of the unit disk $D=\{z:|z|<1\}$ onto the domain $\Omega$ bounded by $\kappa(S^1)$. Let   $K=\{w\in\Omega: \operatorname{dist}(w,\partial\Omega)\ge \epsilon /2 \}$. The preimage $f^{-1}(K)$ is a compact subset of $D$. Therefore, it is contained in $|z|<r$ for some $r<1$. The embedding $\lambda$ obtained by restricting  $f$ to $|z|=r$ satisfies 1 and 3 (for 1, one has to use Caratheodory's theorem which makes $|f(re^{it})-f(e^{it})|$ uniformly small). It automatically satisfies 2 (by straight-line homotopy) if you make $\epsilon$ smaller than the distance from $\kappa$ to the origin $(0,0)$. Now you are in the smooth setting where you can work toward 4 and 5.

An alternative approach is to use a smooth distance-type function, that is, a positive smooth function $\rho$ defined in $\Omega$  such that $\rho(x)\to 0$ as $x\to\partial \Omega$. Such a function can be constructed without Riemann mapping theorem, by using the Whitney decomposition of $\Omega$. See, e.g., Singular integrals by Stein. Almost every level set of $\rho $ is a disjoint union of smooth $1$-manifolds (closed curves). For sufficiently small $\delta$ the set $\rho=\delta$ separates $\partial\Omega$  from the origin; therefore one of its components does the same. This component provides an approximating curve $\lambda$, but unfortunately the task of finding its parametrization that satisfies 1 remains somewhat nontrivial.


问:
Uniformization property of $\mathbf{\Sigma}^0_n(\mathscr X\times\omega)$ for $n>1$
This question is about exercise 1C.6 from Moschovakis.

The exercise is to prove if $P\subset\mathscr X\times \omega$ is in $\mathbf{\Sigma}^0_n$ for $n>1$, then there is some $P^*\subset P$ that is also in $\mathbf{\Sigma}^0_n$ which uniformizes $P$.

Since $P$ is $\mathbf\Sigma^0_n$, there exists $Q\subset \mathscr X\times\omega\times\omega$ in $\mathbf \Pi^0_{n-1}$ such that $P=\exists^\omega Q$. The hint of the exercise defines $R\subset Q$ such that $(x,s_0,s_1)\in R$ iff $(x,s_0,s_1)\in Q$ and $(x,t_0,t_1)\notin Q$ for all $(t_0,t_1)=t<s=(s_0,s_1)$, where I assume $<$ is a well-order on $\omega\times\omega$. I understand why $P^*=\exists^\omega R$ uniformizes $P$, but I don't understand why $R$ is $\mathbf{\Pi}^0_{n-1}$.

答:
I am not sure the order-type of $<$ over $\omega\times\omega$, but if its order-type is $\omega$, then $R$ is $\mathbf{\Delta}^0_n$, since it is a conjunction of

a $\mathbf{\Pi}^0_{n-1}$ formula ($(x,s_0,s_1)\in Q$) and

a $\mathbf{\Sigma}^0_{n-1}$ formula ($(x,t_0,t_1)\notin Q$ for all
$(t_0,t_1)<(s_0,s_1)$. Adding the quantifier bounded by $(s_0,s_1)$
does not increase the complexity.)

Hence $P^*=\exists^\omega R$ is
$\mathbf{\Sigma}^0_n$.


问:
Probability that Brownian Motion hits a point
I have some question regarding the notation from the book on Brownian Motion by Peres and Mörters found online here. They define $\mathbb{P}_x$ to mean a probability measure which makes $(B(t))$ a Brownian motion started at $x$ (p. 36).

In Corollary 2.26 on p. 47 in the book on Brownian Motion they want to show that $\mathbb{P}_x\{y\in B(0,1]\} = 0$ for any $x,y\in \mathbb{R}^2$. If $\Omega$ is the underlying probability space on which the Brownian motion is defined this means that for fixed $x,y\in \mathbb{R}^2$
$$\mathbb{P}\{\omega\in \Omega:y\in B((0,1],\omega)\} = 0$$
where $B$ is a Brownian motion starting at $x$.
They do this by first showing, using their notation, that $\mathbb{P}_x\{y\in B[0,1]\} = 0$ for Lebesgue- almost every $x$ where $y$ is arbitrary. They then say that $\mathbb{P}_{B(\varepsilon)}\{y\in B[0,1]\} = 0$ almost surely where $\varepsilon>0$ is arbitrary.

Now I have some trouble understanding what $\mathbb{P}_{B(\varepsilon)}\{y\in B[0,1]\}$ means.

I read this notation as follows: the event consists of all $\omega$ such that $y\in B([0,1],\omega)$ where $B$ is a Brownian motion starting at $B(\varepsilon,\omega)$. However I don't see how to make sense of this based on previous definitions. Previously Brownian motions start at fixed non-random points in $\mathbb{R}^n$. Furthermore saying that a Brownian motion should start from the point where the Brownian motion hits at $\varepsilon$ is also strange, since in defining the Brownian motion we need to simultaneously know where it hits at $\varepsilon$.

Do they mean that if $\tilde{B}$ is another Brownian motion, then almost surely for $\hat{\omega}$ it holds that
$$\mathbb{P}_{\tilde{B}(\varepsilon,\hat{\omega})}\{\omega: y\in B([0,1],\omega)\} = 0$$
for this I see follows from the previous result. Could someone help me with this?

答:
Now I have some trouble understanding what $\mathbb{P}_{B(\varepsilon)}\{y\in B[0,1]\}$ means.

It is nothing but $\mathbb{P}_{x}\{y\in B[0,1]\}|_{x=B(\varepsilon)}$.

I read this notation as follows: the event consists of all $\omega$ such that $y\in B([0,1],\omega)$ where $B$ is a Brownian motion starting at $B(\varepsilon,\omega)$.

No.  The event precisely consists of all $\omega$ such that $y \in B([0,1],\omega)$. For instance, if $B$ is defined canonically (i.e. with $\Omega = C([0,1];\mathbb R^2)$ and $B(\cdot,\omega) =\omega(\cdot)$), then this event is nothing else but the set of functions attaining value $y$ at some point. This has nothing to do with the probability $\mathbb P_{x}$ (actually, probabilities, since there are many of them, and they are mutually singular). Naturally, since the value $\mathbb P_x(A)$ is zero for almost all $x$ and since the law of $B(\varepsilon)$ is absolutely continuous, then $\mathbb P_{B(\varepsilon)}(A) = 0$ almost surely.

Actually, the authors are indeed a bit sloppy here: as I've already mentioned, there are many probability measures, so "almost surely" may a priori sound pretty ambiguous. A diligent way of writing this is "for any $x$, $\mathbb P_{B(\varepsilon)}(A) = 0 \ $   $\ \mathbb P_x$-almost surely". And they are using precisely this when they write further (thanks to the Markov property and homogeneity) that
$$
\mathbb P_x (y \in B ([\varepsilon,1])) = \mathbb E_x \mathbb P_{B(\varepsilon)} (y \in B ([0,1-\varepsilon]))=0.
$$

Some elaboration of the last formula:
$$
\mathbb P_x (y \in B ([\varepsilon,1])) = \mathbb P_x (\{\exists t\in [\varepsilon,1]: B(t) = y\}) = \mathbb P_x (\{\exists t\in [\varepsilon,1]: B(t) - B(\varepsilon) + B(\varepsilon) = y\})
$$
Thanks to the ($\mathbb P_x$-)independence of $B(t) - B(\varepsilon)$ and $B(\varepsilon)$, this is equal to
$$
\mathbb E_x \mathbb P_x (\{\exists t\in [\varepsilon,1]: B(t) - B(\varepsilon) + z = y\})\big|_{z = B(\varepsilon)}.
$$
Now using the fact that (under $\mathbb P_x$) $B(t) - B(\varepsilon)$ is, up to a time shift by $\varepsilon$, a Brownian motion starting from $0$, we can further rewrite this as
$$\mathbb E_x \mathbb P_0 (\{\exists s\in [0,1-\varepsilon]: B(s) + z = y\})\big|_{z = B(\varepsilon)} = \mathbb E_x \mathbb P_z (\{\exists s\in [0,1-\varepsilon]: B(s) = y\})\big|_{z = B(\varepsilon)}.
$$


问:
Is the following derivation for the Normal Equation correct?
Let $\displaystyle J(\theta):=\frac{1}{2}(X \theta-\vec{y})^{T}(X \theta-\vec{y})$.

I wish to solve the normal equations that is to say finding partial derivative of $J$ with respect to $\theta$ and setting it equal to $0$. I proceed as follow:

\begin{align*}
\nabla_{\theta}J(\theta)&=\nabla_{\theta} \frac{1}{2}(X \theta-\vec{y})^{T}(X \theta-\vec{y}) \\
&=\frac{1}{2} \nabla_{\theta}\left((X \theta)^{T} X \theta-(X \theta)^{T} \vec{y}-\vec{y}^{T}(X \theta)+\vec{y}^{T} \vec{y}\right) \\
&=\frac{1}{2} \nabla_{\theta}\left(\theta^{T}\left(X^{T} X\right) \theta-\vec{y}^{T}(X \theta)-\vec{y}^{T}(X \theta)\right) \\
&=\frac{1}{2} \nabla_{\theta}\left(\theta^{T}\left(X^{T} X\right) \theta-2\left(X^{T} \vec{y}\right)^{T} \theta\right) \\
&=\frac{1}{2}\left(2 X^{T} X \theta-2 X^{T} \vec{y}\right) \\
&=X^{T} X \theta-X^{T} \vec{y}
\end{align*}
So this means $\theta=\left(X^{T} X\right)^{-1} X^{T} y$

Is this derivation correct or is there any missing step?

Note that:

$\theta$ is an $(n+1)\times1$ column vector.

$y$ is an $m\times1$ column vector.

$X$ is an $\mathbb{R^{m\times(n+1)}}$ matrix.

答:
The derivation is correct. You could also use the definition
\begin{equation}
h^T\nabla_\theta F(\theta) = \lim_{t\to 0} \frac{1}{t}(F(\theta + t h) - F(\theta))
\end{equation}
for all $h\in {\mathbb R}^{n+1}$
Applying this with $F(\theta) = J(\theta)$ gives
\begin{equation}
\begin{array}{rcl}
\frac{1}{t}(J(\theta + t h) - J(\theta)) &=& \frac{1}{2}(X h)^T (X\theta - y) + \frac{1}{2}(X\theta - y)^T (X h) + O(t)\\
&=&h^T X^T (X\theta -y) + O(t)\\
\end{array}
\end{equation}


问:
In the process of calculating the limit, why is this substitution equal
On the limit of quotient, I know that infinity is comparable.
Functions and infinity are too abstract for me,I don't know how to understand it.

Excuse me for taking a screenshot of an answer in the forum.
enter image description here

enter image description here(There is my red mark in the picture)

$\lim_{y\to\infty}{f^{-1}(8y)\over y^{1/3}}=2\lim_{y\to\infty}{f^{-1}(8y)\over(8y)^{1/3}}=2\lim_{z\to\infty}{f^{-1}(z)\over z^{1/3}}\ $ $Q=\lim_{z\to\infty}{f^{-1}(z)\over z^{1/3}}=\lim_{x\to\infty}{f^{-1}\bigl(f(x)\bigr)\over \bigl(f(x)\bigr)^{1/3}}=\lim_{x\to\infty}{x\over 2x\bigl(1+{3\over 8x^2}\bigr)^{1/3}}={1\over2}\ .$

This is a copy command, and I don't know how to handle it

Question :Why is the above replacement equivalent? What can we learn from？

A better description of the problem:

$\lim_{ab\rightarrow \infty} \frac{a}{b}=1$

$\lim_{ab\rightarrow \infty} \frac{f(a)}{f(b)}=1/2$

I don't know if there is any necessary connection above

If so,what will be the result?

ab is another number and $\lim_{ab\rightarrow \infty} \frac{a}{b}=2$

答:
If you multiply the numerator and the denominator by 2, you get $$\lim_{y\rightarrow \infty} \frac{f^{-1}(8y)}{y^{1/3}} = \lim_{y\rightarrow \infty} \frac{2 \cdot f^{-1}(8y)}{2\cdot y^{1/3}}$$ And because $2 = 8^{1/3}$, it simplifies to $$\lim_{y\rightarrow \infty} \frac{f^{-1}(8y)}{y^{1/3}} = 2 \lim_{y\rightarrow \infty} \frac{f^{-1}(8y)}{8^{1/3}\cdot y^{1/3}} = 2\lim_{y\rightarrow \infty} \frac{f^{-1}(8y)}{(8y)^{1/3}}$$
Then you can just substitute $z = 8y$, which clearly approaches infinity when $y \rightarrow \infty$.


问:
Number of equivalence relation on $A=\{1,2,...,11\}$ such that $|A\setminus R| = 2$.
I am still trying to understand the question, I know that if a relation is reflexive (and equivalence relation is surely reflexive), then for every $a\in A$ , $(a,a)\in R$. But that means every equivalence relation on $A$ has atleast $11$ elements, which makes $|A \setminus R=0|$, because the cardinality of $R$ is surely $11$ or more.

The final answer was $2^{10}-1$, so seems like I have some big misunderstanding that I don't understand where it comes from, I would really appreciate any explanation.

Thanks in advance.

答: The notation $A/R$ represents the set of equivalence classes (with respect to $R$). So, the question is: how many equivalence relations can you define on $A$ such that there are exactly two equivalence classes? And the answer is: it's half the number of subsets $B$ of $A$ such that neither $B$ nor $A\setminus B$ is empty, because, for any such set $B$, you can define the equivalence relation$$a_1\mathrel Ra_2\iff a_1,a_2\in B\text{ or }a_1,a_2\notin B$$and the equivalence classes of $R$ and $B$ and $A\setminus B$; besides, $B$ and $A\setminus B$ induce the same equivalence relation. There are $2^{11}-2$ such sets, and therefore the answer is $2^{10}-1$.


问:
The concept of the creation of a trapdoor in NP-complete or NP-hard problems
I am reading the book An Introduction to Mathematical Cryptography. In its chapter 7, there is the following statement:

In real world scenarios, cryptosystems based on NP-hard or NP-complete problems tend to rely on particular subclass of problems either to achieve efficiency or to allow the creation of a trapdoor.

Questions:

Is is possible to explain what does it mean the phrase "to allow the creation of a trapdoor"?

Thanks for any help

答: The knapsack problem is a case where you can see it: a genral knapsack problem is NP-hard and worst-case NP-complete, but in developing the cryptosystem we must be able to mask an easily solvable instance to look like a hard one, so the problems that were used were easy ones (super-increasing) then modified by a modular multiplication and modular reduction to make them look like "random" instances (this is the Merkle-Hellman variant). But if you knew the original problem plus the modifications (this is the trapdoor) solving them would be easy. So choosing special problems allowed for a trapdoor (extra info that allows the decryption algorithm) to exist. This kind of example is what the writer of that chapter had in mind, probably.


答:
You can define

$\tau_Y=\{O\cap Y: O\in\tau_X\}\cup\{Y\}.$

This is a topology on $Y$. It is the topology generated by $\{O\cap Y: O\in\tau_X\}$, i.e. the smallest topolgy containing that family of sets.
(And in fact that topology carries informations only about $Y\cap X$.)

Basically you have a topology on the subset $Y\cap X$, and you promote it to a topology on $Y$ by adding $\{Y\}$.

If $Y\cap X=\emptyset$, then $\tau_Y=\{\emptyset, Y\}$ is just the trivial topology.

Title: A variation on the induced topology (subspace topology)

Body:
One of the easiest ways to find a topology on a set is to used the Induced Topology.

Induced Topology: Suppose you have a set $X$ and a topology on $X$ called $\mathcal{T}_X$. Suppose $Y \subseteq X.$ Then we can find a topology for $Y$ (call it $\mathcal{T}_Y$) as given by:

$\mathcal{T}_Y=\{O \cap Y  \mid O \in \mathcal{T}_X\}$.

My question is: does the induced topology still hold for sets $Y \not \subseteq X.$ Suppose $Y \cap X = \emptyset.$ Can we still inherit a topology on $Y$ if $X$ and $Y$ are disjoint? I'm going to say no.

Here's my test:

Let $X =\{1,2\}$ and $\mathcal{T}_X = \{\emptyset,\{1\},\{1,2\}\}$

Let $Y = \{5\}$

Then, following from our induced topology method,

$\mathcal{T}_Y=\{\{5\} \cap \emptyset, \{5\} \cap \{1\}, \{5\} \cap \{1,2\}\} = \{\emptyset,\emptyset,\emptyset\}$

But this fails to satisfy definition of topology since $Y=\{5\}$ and $Y \not \in \mathcal{T}_Y.$

So it would seem that we need $Y \subseteq X$ so that $Y \in \mathcal{T}_Y$, otherwise we end up with a bunch of empty sets.

Is my reasoning here correct?


问:
A question about 3D Trigonometry

The questions I am stumped on are parts c) and d).
I am bewildered as to which triangles I am meant to use. For context, I have got an answer of 54.7 for part b and this is correct.

Solutions to c and d are below:

Solutions

c) 35 degrees
d) 24 degrees

I would really appreciate any help about where I am going wrong here. Thank you.

答:
In cross section through points A, F and M by a plane containing these points, is triangle AMF which gives first angle $\alpha$:

$\tan (\alpha)=\frac{15}{10}=1.4\Rightarrow \alpha=56.3$

In cross section through points A, B and M by a plane containing these points, is triangle AMB which gives first angle $\beta$:

$\tan (\beta)=\frac{15}{\sqrt{10^2+10^2}}=1.06 \Rightarrow \beta= 46.7$

Second question:

a):

$AG=\sqrt{AB^2+BC^2+CG^2}=\sqrt{4^2+4^2+8^2}=9.7$

b):Consider a plane contanig vertices A,C,E and G. we have right angled triangles AGC and AGE.

$tan(\gamma)=\frac8{\sqrt{4^2+4^2}}=1.41\Rightarrow \gamma=54.7$

c):Angle between AG and EFGH is equale to angle between AG and ABCD:

$\theta=\gamma=54.7$

d):

$AH^2=AE^2+EH^2=4^2+8^2=80 \Rightarrow AH=4\sqrt 5$

$\tan (\alpha)=\frac 4{4\sqrt 5}=0.447\Rightarrow \alpha= 24^o$

Title: A question about 3D Trigonometry


问:
Converse of Heine-Cantor Theorem
Heine-Cantor theorem asserts that every continuous function on a compact metric space is uniformly continuous.

The converse is , if every continous function is uniformly continuous on a metric space X then X is compact. I have got the answer that the converse is false but I am unable to find a counter example.

The other question is if the converse is false then its negation must be true.

Negation- if every continous function is uniformly continuous on a metric space X and X is not compact.

How do we prove this?

答:
Theorem 1 of this old paper gives conditions on a metric space so that all continuous functions on it are uniformly continuous.

It seems to me that $\Bbb Z$ (as a subspace of $\Bbb R$) is an easy example of such a non-compact space. And also $\Bbb Z \cup \{\frac1n: n =1,2,3,\ldots\}$ is one that is non-discrete too. As the theorem 1  quoted above shows, we do need lots of "far away" isolated points for this to happen, and in "nice" metric spaces this will imply compactness.

BTW the negation of the converse is: there exists a non-compact $X$ such that every continuous function on $X$ is uniformly continuous.

Title: Converse of Heine-Cantor Theorem


问:
Convergence in distribution for joint vector and maximum
I know that $X_n\xrightarrow{d} X$ and  $Y_n\xrightarrow{d} Y$ does not necessarily imply $\max(X_n,Y_n)\xrightarrow{d} \max(X,Y)$, where $\xrightarrow{d}$ is convergence in distribution.

However, if the joint vector $(X_n,Y_n)$ is such that $(X_n,Y_n)\xrightarrow{d} (X,Y)$, is it then true that $\max(X_n,Y_n)\xrightarrow{d} \max(X,Y)$?

Edit: I think this is true as:

$$(X_n,Y_n)\xrightarrow{d} (X,Y)$$
$$\implies \forall x, y, \lim_{n \rightarrow \infty} F_{(X_n, Y_n) }(x, y) = F_{(X, Y)} (x, y) $$
$$\implies \forall x \lim_{n \rightarrow \infty} F_{(X_n, Y_n) }(x, x) = F_{(X, Y)} (x,x) $$
$$\implies \forall x \lim_{n \rightarrow \infty} F_{\max(X_n, Y_n) }(x) = F_{\max(X, Y)} (x) $$
$$\implies\max(X_n,Y_n)\xrightarrow{d} \max(X,Y)$$

where $F_A(a)$ represents the cumulative distribution function of A evaluated at $a$.  Please could someone confirm if my reasoning is correct?

答:
Yes, the result is true. I assume the random variables are real-valued.

The function $\mathbb{R}^2\to\mathbb{R}, (x,y) \mapsto \max(x, y)$ is continuous.
Hence, applying the Continuous mapping theorem to $(X_n, Y_n) \overset{d}{\to} (X,Y)$ yields
$$\max(X_n,Y_n) \overset{d}{\to} \max(X,Y) .$$

Regarding your proof, you probably have to be a bit more careful; remember that convergence in distribution is not equivalent to pointwise convergence of the distribution functions!

For example, consider deterministic random variables $X_n = \frac{1}{n}$, $X = 0$.
Then we have
$X_n \overset{d}{\to} X$,
but the distribution functions do not converge at the point $0$:
$$\lim_{n\to\infty} F_{X_n}(0) = \lim_{n\to\infty} 0 \neq 1 = F_{X}(0) $$


问:
Surface integrals of scalars
just having trouble with this question:

Find $\iint_s z\,dS$ where $s$ is the triangular plate with vertices $(1,0,0)$, $(0,2,0)$, and $(0,1,1)$.

I started by getting the two vectors between the points and getting the normal.

I ended up getting $\sqrt{6}$ as the magnitude and $2x + y + z = 2$ as the plane.

Now this is the point where I'm not fully sure. I projected onto the $x$-$y$ plane and attempted to get the boundaries for integration. But as the points go up to $z=1$ I'm not sure if I should have done that:

$$0\le y\le 2-2x$$
$$0\le x\le 1$$

I took the integral and I got $\frac{2}{\sqrt{6}}$, but the answer is sadly $\frac{1}{\sqrt{6}}$.

I'm not sure where to go from there. I find it really difficult to visualize these boundaries.

答:
Ok so after extensive thought I have come up with a more refined answer, the existing plane is fine, but because of its position, to integrate it as a function of x and y, or a function of z and x would be very messy seeing as the plane must then be cut into two different planes instead. Noting this i found that it is actually best to orient the usual Cartesian plane so that x points upwards, z points out of the paper and y points to the right (so as to preserve the normal right handed sense of the axes) in this regard the plan takes on the definition
$$ x = 1 - \frac{1}{2}y - \frac{1}{2}z$$
Thus the new definition of the surface integral becomes
$$ \int \int f(x(y,z), y ,z) \sqrt{ \left(\frac{\partial x}{ \partial y} \right)^2 + \left(\frac{\partial x}{ \partial z} \right)^2 + 1 } d A $$
Using the above definition of the plane we have the partial derivatives are:
\begin{eqnarray*}
\left(\frac{\partial x}{ \partial y} \right)^2 &=& \frac{1}{4} \\
\left(\frac{\partial x}{ \partial z} \right)^2 &=& \frac{1}{4} \\
\end{eqnarray*}
so that the square root as you almost correctly calculated is $ \frac{\sqrt{6}}{2}$
Now for the bounds involving y and z, draw the y-z plane with y on the vertical axis ( in a similar way you might draw x horizontal and y vertical, like you are treating z as the x axis) The resulting shape is a triangular region made of the two lines
\begin{eqnarray*}
 y &=& z \; \text { (lower bound)} \\
 y &=& 2 - z \; \text{ (upper bound)} \\
\end{eqnarray*}
Thus we have that for the double integral over the area is given by
$$ \int_{0}^{1} \int_{z}^{2-z} z \frac{\sqrt{6}}{2}dydz $$
If you compute this integral you should find the correct answer of $\frac{1}{\sqrt{6}}$</p>

Title: Surface integrals of scalars


问:
How to integrate $\frac{1}{x\ln(1+x)}$?
I tried to evaluate the indefinite integral

$$\int{\frac{dx}{x\ln(1+x)}}.$$

Even Wolfram Alpha can't give any result in terms of standard mathematical functions. Here are some related integrals:

$$\int{\frac{\ln(x)}{\ln(1+x)}}\,dx$$

$$\int{\frac{Li(x)}{1+x}}\,dx,$$

Unfortunately, I'm not able to solve any of them. Is there any special function related with this kind of integrals?

Title: How to integrate $\frac{1}{x\ln(1+x)}$?

Body:
Probably dangerous

For an approximation, expand
$\frac{1}{\log (a+x)}$ as a Taylor series around $a=0$ and later make $a=1$.

This will give
$$\frac{1}{x\log (1+x)}=\sum_{n=1}^p (-1)^{n+1}\frac {P_n(t)}{b_n\,x^n} \qquad \text {where} \qquad t=\frac 1{\log(x)}$$

The first $b_n$ are
$$\{1,1,3,12,60,360,210,1680,15120,151200,\cdots\}$$ and the very first polynomials are
$$\left(
\begin{array}{cc}
 n & P_n(t) \\
 1 & t \\
 2 & t^2 \\
 3 &  t^2 (2 t+1) \\
 4 & t^2 \left(3 t^2+3 t+1\right) \\
 5 &  t^2 \left(12 t^3+18 t^2+11 t+3\right) \\
 6 &  t^2 \left(60 t^4+120 t^3+105 t^2+50 t+12\right) \\
 7 &  t^2 \left(360 t^5+900 t^4+1020 t^3+675 t^2+274 t+60\right) \\
 8 &  t^2 \left(210 t^6+630 t^5+875 t^4+735 t^3+406 t^2+147
   t+30\right)
\end{array}
\right)$$
Now
$$\int \frac {dx} {x^p \log^q(x)}=\int e^{(1-p) y} y^{-q}\,dy=-(p-1)^{q-1} \Gamma (1-q,(p-1) y) $$

For a quick check, using $p=10$, I computed
$$I_n=\int_e^{e^n} \frac{dx}{x \log (x+1)}$$
$$\left(
\begin{array}{ccc}
 n & \text{approximation} & \text{"exact"} \\
 2 & 0.59692 & 0.59703 \\
 3 & 0.98844 & 0.98854 \\
 4 & 1.27345 & 1.27355 \\
 5 & 1.49600 & 1.49610 \\
 6 & 1.67817 & 1.67827 \\
 7 & 1.83228 & 1.83239 \\
 8 & 1.96581 & 1.96591 \\
 9 & 2.08359 & 2.08369 \\
 10 & 2.18895 & 2.18905\\
 20 & 2.88209 & 2.88219 \\
 30 & 3.28756 & 3.28766 \\
 40 & 3.57524 & 3.57534 \\
 50 & 3.79485 & 3.79494
\end{array}
\right)$$


问:
Solving a 2D partial differential equation from an old science paper
I am reading an atmospheric science paper (B. J. Hoskins, 1985) and stumble on a second-order partial differential equation that after some simplification roughly reads
\begin{align*}
\frac{\partial}{\partial r} (\frac{1}{r} \frac{\partial (rv)}{\partial r}) + \frac{\partial^2 v}{\partial t^2} &= \frac{\partial P}{\partial r} \\
\frac{\partial^2 v}{\partial r^2} + \frac{1}{r}\frac{\partial v}{\partial r} - \frac{v}{r^2} + \frac{\partial^2 v}{\partial t^2} &= \frac{\partial P}{\partial r}
\end{align*}
where $v(r, t)$ is some function of $r$ and $t$. $P$ is a quantity that has a value of $1+\epsilon$ inside the circular region $r^2 + t^2 = C^2$, and and $1$ outside the circle, where $\epsilon$ is a very small number. The boundary conditions should be $v = 0$, when $r \to 0$ and $r \to \infty$. The solution given in the paper is
\begin{align*}
v &= \frac{1}{3} \epsilon r & r^2 + t^2 < C^2 \\
v &= \frac{1}{3} \epsilon r (\frac{C^2}{r^2 + t^2})^{3/2} & r^2 + t^2 > C^2
\end{align*}
The paper claims that this is a straight forward theoretical exercise, but I have no idea how to arrive at the given solution. I have tried Separation of Variables but I can at most deduce that $v \propto r$ inside the circular region. Laplace Transform and Fourier Transform seems futile. I have tried to use Duhamel's Principle but my answer is wrong. I am thinking if some transformations can be applied.  Can anyone give me some suggestions? Any help would be appreciated.

答: I have found a way (although not perfect) to derive the solution in a reasonable way. First make a transformation $Q = v/r$ to make the equation simpler, leaving three terms in the Equation. Then, by observation, one solution for $Q$ is a constant. Another solution for $Q$ is motivated by the circular boundary. Let $z = r^2 + t^2$ and assume that the solution is solely a function of $z$. By Chain Rule, the equation can be transformed into an ordinary equation in $z$ and it is possible to get another solution $1/z^{3/2}$. Due to the physics, we know that $Q = \alpha$ inside the circle, and $Q = \beta/z^{3/2}$ outside the circle. Finally, we consider the difference in P across the interface. By integrating over a small neighborhood at the interface, $\alpha$ and $\beta$ can be determined.


问:
Why is $\int_{|z|=1} \dfrac{\log(1-az)}{z}dz = 0$ for $|a|<1$?
As the title suggests, I am a bit unsure why that is so? Log is not holomorphic so Cauchy seems a bit off.

If anyone could provide a hint that would be very appreciated.

Edit: I have tagged the question with "cauchy-integral formula" as I suspect it has something to do with it. If it is clearly wrong however, I will remove it if so.

答: The principal branch of logarithm, which I denote by $\log z$, is analytic in the disk $D(1,1)$. So $\log (1-az)$ is analytic in $D(0,\frac 1 {|a|})$ which includes the closed unit disk. Hence, Cauchy's integral formula is applicable and the value of the integral is $2\pi i$ times the value of $\log (1-az)$ at  $0$ which is $0$.


问:
Show that transposition is an odd permutation
This is in symmetric groups $S_n$. I don't know how to approach this problem. I guess I have to use the definition of
$$\mathrm{sgn}(\sigma)=\prod_{1 \leq i < j \leq n} \frac{\sigma(j)-\sigma(i)}{j-i}$$

I know I have to use this on the permutation $(i \ j)$ which represents a transposition of the $i$th and $j$th elements of the permutation, but I don't know how to approach this.

答:
Take wlog the sequence $1, 2, \ldots, n$. An inversion is defined as a pair of elements in a sequence which are not in natural order. Just as a permutation can be decomposed into transpositions, let us decompose a transposition into something even simpler: adjacent transpositions, where two adjacent elements swap places. This is helpful, since it can easily be seen that an adjacent transposition always increments or decrements the inversion number by precisely 1.

A transposition makes two elements swap places, so how can we go about this using adjacent transpositions? One way is to first move the smaller element to the larger element's position then move the larger element to the smaller element's position.

As an example, take the sequence $1, 2, 3, 4, 5, 6, 7$ where we want to swap $2$ and $5$.

First step: $$1, 2, 3, 4, 5, 6, 7 \rightarrow 1,3,2,4,5,6,7 \rightarrow 1,3,4,2,5,6,7 \rightarrow 1,3,4,5,2,6,7$$

$2$ is in the right position and we have used three adjacent transpositions. Notice however that $5$ has been moved to the left one position, so one less adjacent transposition than were necessary for $2$ will be needed.

Second step: $$1,3,4,5,2,6,7 \rightarrow 1,3,5,4,2,6,7 \rightarrow 1,5,3,4,2,6,7$$

As expected, two adjacent transpositions were needed.

In fact, for any arbitrary transposition, we need an odd number of adjacent transpositions, since the last adjacent transposition in the first step "benefits" both steps. Note that although we started from a sequence in natural order, we can relabel the elements any way we like, we still have to use the same number of adjacent transpositions.

Now armed with the fact that an adjacent transposition increments or decrements (but does not leave constant!) the inversion number and the fact that a transposition can be decomposed into an odd number of adjacent transpositions, we can state that a transposition is an odd permutation.


问:
Proof of the Lebesgue Density Theorem in $\Bbb{R}^n$
I am looking for a simple proof of the Lebesgue density theorem for $\Bbb{R}^n$. The Wikipedia page on the Lebesgue differentation theorem leads me to a proof of that more general theorem, but mentions that there are simpler proofs of the density theorem giving a reference to the book Measure and Category by Oxtoby. Because of the pandemic, I don't have access to a library just now and I can't find a simpler proof online except for the case $n = 1$. Can anyone point me to a source that is accessible online.

Note that where to find a proof of the Lebesgue Density Theorem asks about the general case, but the answer leads to a broken link to a paper by Faure. If that was Faure's paper in the American Mathematical Monthly from February 2002, then it only covers the case $n = 1$.

答:
As commented in your other post, the proof of the Lebesgue density theorem from the book Measure and Category by Oxtoby, refers to case $n = 1$, despite being in the chapter regarding $r$-Spaces. The author also suggests the article An Elementary Proof of the One-Dimensional Density Theorem, which as the title suggests, also refers to the $1$ dimensional case.

About the Faure's paper it actually refers to the $1$ dimensional case. I will be leaving two links to this article. Link 1 -- Link 2.

As Bruno Stonek commented in the post you left referenced, the $n$ dimensional case is left as an exercise in the book Real Analysis: Modern Techniques and Their Applications by Folland, more precisely, the exercise 3.4.25.

A reference to the $n$-dimensional case would be the book Geometry of Sets and Measures in Euclidean Spaces by Mattila, the Lebesgue density theorem is the Corollary 2.14 in page 38, where the author uses the Lebesgue differentiation theorem to prove it. Another interesting reference is The Lebesgue Density Theorem in Abstract Measure Space by Miriam Yevick which can be found at this link.

On the Lebesgue differentiation theorem, I suggest, chapter 11 of the book Measure Theory and Integration by Taylor. I will reference the article An Elementary Proof of Lebesgue's Differentiation Theorem by Botsko and this pdf.

I will present the Folland exercise solution.

If $E$ is a Borel set in $\mathbb{R}^n$, the density of $E$ at $x$ is defined as,

$$D_E (x) = \lim_{r\to 0} \frac{m(E\cap B(r,x))}{m(B(r,x))}$$
whenever the limit exists.

Show that $D_E (x)=1$ for a.e. $x\in E$ and $D_E (x) = 0$ for a.e. $x \in E^{c}$

Define $\mu:\mathcal{B}_{\mathbb{R}^n}\to [0,\infty]$ by $\mu(A)=m(E\cap A$), this is a measure and the Lebesgue-Radon-Nikodym representation is $d\mu=\chi_E\, dm$.

Let $A\in\mathcal{B}_{\mathbb{R}^n}$ and suppose $m(A)<\infty$. Given $\varepsilon\in (0,\infty)$, there exist a open set $U\subseteq\mathbb{R}^n$ containing $A$ such that $m(U)<m(A)+\varepsilon$ and hence $m(U\setminus A)<\varepsilon$. It follows that

$$\mu(U)=m(E\cap U)\le m(E\cap A)+m(U\setminus A)<\mu(A)+\varepsilon$$

Given $A\in \mathcal{B}_{\mathbb{R}^n}$ and $\varepsilon\in (0,\infty)$, chose a sequence $\{A_k\}_k$ of Borel sets such that $m(A_k)<\infty$ fo all $k\in\mathbb{N}$ and

$$\bigcup_{k=1}^{\infty} A_k = A$$

For each $k\in\mathbb{N}$ there exist a open set $U_k\subseteq\mathbb{R}^n$ containing $A_k$ such that $\mu(U_k)<\mu(A_k) + 2^{-k}\varepsilon$. We have,

$$A\subseteq\bigcup_{k=1}^{\infty} U_k\quad\text{and}\quad \mu \left(\bigcup_{k=1}^{\infty} U_k\,\setminus A\right)<\varepsilon$$

This implies that $\mu(A)=\inf\{\mu(U) : U\subseteq\mathbb{R}^n\text{ is open and } A\subseteq U\}$. Therefore $\mu$ is regular, so for almost every $x\in\mathbb{R}^n$

$$D_E (x) = \lim_{r\downarrow 0} \frac{m(E \cap B(r,x))}{m(B(r,x))}=\lim_{r\downarrow 0} \frac{m(B(r,x))}{m(B(r,x))}=\chi_E (x)$$

In particular, $D_E (x)= 1$ for almost all $x\in E$ and $D_E(x)=0$ for almost all $x\in E^{c}$.


答:
If $z=x+iy$, then
\begin{align}
  log(1-az)&=log(1-a(x+iy))\\
  &=log(1-ax-i(ay))\\
&\Rightarrow \mathfrak{R}(f(z))=1-ax\geq 0 \Rightarrow x\leq\frac{1}{a} \\
&\Rightarrow \mathfrak{I}m(f(z))=0\Rightarrow y=0\\
\end{align}

then with conditions it'll verify that the domain analyticity of $f(z)$ like this in $z=0$ is  inside the curve $\gamma:=|z|=1$, so yo can apply the Cauchy integral formule.

Title: Why is $\int_{|z|=1} \dfrac{\log(1-az)}{z}dz = 0$ for $|a|<1$?

Body:
As the title suggests, I am a bit unsure why that is so? Log is not holomorphic so Cauchy seems a bit off.

If anyone could provide a hint that would be very appreciated.

Edit: I have tagged the question with "cauchy-integral formula" as I suspect it has something to do with it. If it is clearly wrong however, I will remove it if so.


问:
Splitting ideals generated by prime numbers in $\mathbb{Z}[\sqrt{-5}]$
Let $p\in \mathbb{Z}$ a prime such that $\left(\frac{-5}{p}\right)=1$. Then as
$$\mathbb Z[\sqrt{-5}]/(p)\cong \mathbb{Z}[X]/(X^2+5)/(p)\cong \mathbb{F}_p[X]/(X^2+5),$$
we have some $a$ such that $(X-a)(X+a)=X^2+5$ in $\mathbb{F}_p[X]$. Then by Chinese remainder theorem,
$$\mathbb Z[\sqrt{-5}]/(p)\cong \mathbb{F}_p[X]/(X+a)\times\mathbb{F}_p[X]/(X-a)$$
How does this imply $(p)=(p, \sqrt{-5}+a)(p, \sqrt{-5}-a)$? This might be a trivial question, but I am really confused. I tried to find similar questions, but I failed to understand the rationale behind this immediate inference. I couldn't see why the isomorphic relation suggests that $(p)$ is a product of two ideals.

答:
Disclaimer: This argument might be somewhat overcomplicated (and/or wrong), but I've written it now so I might as well post. I think sometimes the details of this sort of thing are glossed over, so it's possible this is what the author intended. It is also possible that I'm just an idiot.

Answer: If $R$ is a ring and $P$ is an ideal, then $P$ is prime if and only if $R/P$ is an integral domain. Therefore $(p)$ is not prime, so it must factor as a product of other prime ideals. The norm of $(p)$ is $p^2$, so it has two prime factors $\mathfrak{a}$ and $\mathfrak{b}$ of norm $p$.

Let $A, B$ be the images of $\mathfrak{a}$ and $\mathfrak{b}$ under the natural isomorphism

$$\mathbb{Z}[\sqrt{-5}] \to \mathbb{Z}[x]/(x^2+5).$$

Note that $p\in \mathfrak{a},\mathfrak{b}$ so $p \in A,B$. Therefore we may reduce mod $p$ to obtain ideals $\bar{A}, \bar{B}$ of $\mathbb{F}_p[x]/(x^2+5)$ with $\bar{A}\bar{B} = 0$. These ideals lift to ideals $\tilde{A},\tilde{B}$ of $\mathbb{F}_p[x]$ with $\tilde{A}\tilde{B} = (x^2+5)$, so by unique factorisation in $\mathbb{F}_p[x]$, we have (without loss of generality) $\tilde{A} = (x+\sqrt{-5}),\tilde{B} = (x-\sqrt{-5})$, where we take the square root of $-5$ in $\mathbb{F}_p$.

Therefore $\bar{A} = (x+\sqrt{-5})$ and $\bar{B} = (x-\sqrt{-5})$ modulo the ideal $(x^2+5)$. Lifting to $\mathbb{Z}[x]/(x^2+5)$, we see that $A = (p, x+\sqrt{-5})$ and $B=(p, x-\sqrt{-5})$, and therefore we are done, because your $a$ is the same thing as my $\sqrt{-5}$.


问:
Use the definition of continuous function to show that $f(x) = x^2 + 3$ is continuous at $x=3$.
anyone can help me with this question? I know that to prove continuity the left side limit must equal the right side limit, in this case is 3. But there is only one equation for me to do so, so any idea how to solve this question?

Is it possible to choose a value less than 3 and a value more than 3 and sub it into the equation? If that’s the case the limit will not be the same.

Really confused with this question, any help is appreciated. Thank you.

答:
We are interested in the behavior of $f$ "near" the point $3$, hence we can assume that $2 <x <4.$

Then we have

$$|f(x)-f(3)|=|x^2-9|=|x-3| \cdot |x+3|=|x-3|(x+3) \le 7|x-3|.$$

Can you proceed with $ \varepsilon , \delta$ ?


问:
exponential distribution - get some meaning
I read that a continuous random variable having an exponential distribution can be used to model inter-event arrival times in a Poisson process. Examples included the times when asteroids hit the earth, when earthquakes happen, when calls are received at a call center, etc. In all these examples, the expected value of the number of events per unit of time, lambda, is known and is constant over time. Moreover, each event's occurrence is independent of previous events' occurrences. And the exponential variable that models inter-event arrivals has the same lambda parameter as the Poisson variable that models the number of events.

And now, my problem. It don't get the connection between the intuition behind the exp distrib and its pdf. It seems obvious that the more time it passes by without an earthquake happening, the more likely it is than an earthquake will happen. Assume my understanding of lambda is correct, i.e., lambda = the rate at which the event happens, e.g., 5 earthquakes per minute on some remote, angry planet. On the pdf graphic, the prob of generating a value between 0 and 1 is greater than the prob of choosing a value between 4 and 5 for instance. How is this graphic related to the fact that on average we need to have 5 earthquakes per minute?

答:
It seems obvious that the more time it passes by without an earthquake happening, the more likely it is than an earthquake will happen.

As explained in the comments, to use an exponential distribution is to assume the opposite: that whatever time passed by without an earthquake happening, still as likely as ever it is than an earthquake will happen.

So, if one thinks this rate should not be constant, how to model the time $X$ until the first earthquake happens? The hypothesis is that, for every $t\geqslant0$, when $s\to0^+$,
$$
P[X\leqslant t+s\mid X\geqslant t]=r(t)s+o(s),
$$
for some function $r$. Thus, the function $\bar F:t\mapsto P[X\geqslant t]$ is such that
$$
\bar F(t+s)=\bar F(t)\cdot(1-r(t)s+o(s)),
$$
for every $t\geqslant0$, that is,
$$
\bar F'(t)=-r(t)\bar F(t).
$$
Since $\bar F(0)=1$ by definition, one gets
$$
\bar F(t)=\exp\left(-\int_0^tr(u)\,\mathrm du\right),
$$
and finally, $X$ has PDF
$$
f(t)=\exp\left(-\int_0^tr(u)\,\mathrm du\right)r(t)\,\mathbf 1_{t\geqslant0}.
$$
One sees that every density $f$ can be written like that, since one recovers the instantaneous rate $r$ through the formula
$$
r(t)=\frac{f(t)}{\bar F(t)}=\frac{f(t)}{1-F(t)}.
$$


问:
Norm of multiplication operator by a Fourier transform of an $L^1$ function
I'm trying to compute the operator norm of the multiplication operator $\mathcal{M}_{\hat{g}}:L^2(\mathbb{R}^d)\to L^2(\mathbb{R}^d)$ given by $\mathcal{M}_{\hat{g}}(f)=f\cdot \hat{g}$, where $\hat{g}$ is the Fourier transform of $g\in L^1(\mathbb{R}^d)$. I think that the operator norm is equal to $\Vert g\Vert_{L^1}$, but I was only able to show it is a bound for $\Vert \mathcal{M}_{\hat{g}}\Vert_{op}$.

I know that $\Vert \mathcal{M}_{\hat{g}}\Vert_{op}=\Vert \hat{g}\Vert_{L^\infty}$ and that $\Vert \hat{g}\Vert_{L^\infty}\leq \Vert \mathcal{F}\Vert_{op}\cdot \Vert g\Vert_{L^1}=\Vert g\Vert_{L^1}$, where $\mathcal{F}$ is the Fourier transform. I want to find a sequence of $L^2$ functions such that

$$ \Vert f_n\cdot \hat{g}\Vert_{L^2} \overset{n\to \infty}{\to}\Vert g\Vert_{L^1} \quad \text{and} \quad \Vert f_n\Vert_{L^2}=1 \quad \text{for all} \quad n, $$

but I'm not sure how to go about this.
This might be a trivial question but I'm not sure how to solve this point. I'm also not sure whether this is indeed true.

答:
Just wanted to add a relatively easy counter-example.

Let $M>0$ and suppose $g(x)=1_{[-M,M]}\cos(x)$.

Then, using that $\sin$ and $\cos$ is odd, you get that

$$
\hat{g}(\xi)=2\int_0^M \cos(x)\cos(\xi x)\textrm{d}x
$$
And applying the age-old double integration by parts trick, we get that
$$
\int_0^M \cos(x)\cos(\xi x)\textrm{d}x=[\sin(x)\cos(\xi x)]_0^M-\xi\left([-\cos(x)\sin(\xi x)]_0^M +\xi\int_0^M \cos(x)\cos(\xi x)\textrm{d}x\right)
$$
Gathering everything, we get that
$$
\hat{g}(\xi)=\frac{2}{1+\xi^2}\left(\sin(M)\cos(\xi M)+\xi\cos(M) \sin(\xi M\right)),
$$
which is uniformly bounded by 4, independently of $M$.

However, $\|g\|_{L^1}\geq \lfloor\frac{M}{2\pi}\rfloor \frac{\pi}{\sqrt{2}}$, so indeed, it is not true that $\|g\|_{L^1}=\|\hat{g}\|_{L^{\infty}}$ for $M$ sufficiently large.


问:
Finding $\lim_{x\rightarrow \infty}x\ln\left(\frac{x+1}{x-1}\right)$
I'm trying to find out :

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

First I tried by inspection. I suspect that
$$\lim_{x\rightarrow \infty} \left( \frac{x+1}{x-1}\right)=1$$
So I expect $\ln (\cdots)\rightarrow 0$ as $x\rightarrow \infty$. While If I multiply this by $x$, One might expect this to go infinite.

Next what I have done is to use a substitution, I don't know if it's right to do or not.
$$x'=\frac{x+1}{x-1}\Rightarrow x=\frac{x'+1}{x'-1}$$

So that
$$\lim_{x\rightarrow \infty}x\ln\left(\frac{x+1}{x-1}\right)\Rightarrow \lim_{x'\rightarrow 1}\left(\frac{x'+1}{x'-1}\right)\ln x'$$
Now I can use L'Hospital rule
$$\lim_{x'\rightarrow \infty}\frac{\ln x'+\frac{x'+1}{x'}}{1}=2$$

Can I use such a substitution always to solve problems? What's wrong with the reasoning I did earlier?

答:
My attempt

$\lim\limits_{x\to \infty} xln(\frac{x+1}{x-1})=\lim\limits_{x\to \infty} xln(x+1)- xln(x-1) $

Suppose $t=x+1$

So $x=t-1$

$\lim_{t\to \infty}tln(t) - ln(t) -tln(t-2)+ln(t-2)$

$=lim_{t\to \infty} tln(\frac{t}{t-2})+ln(\frac{t-2}{t})$

$\lim\limits_{t\to \infty} (t(\frac{t} {t-2}-1)\frac{ln(\frac{t} {t-2})}{(\frac{t} {t-2}-1)}+ln(\frac{t-2}{t})=\lim\limits_{t\to \infty} ((\frac{2t} {t-2})\frac{ln(\frac{t} {t-2})}{(\frac{t} {t-2}-1)}+ln(\frac{t-2}{t})$

Suppose $ w=\frac{t} {t-2}$

$\lim_{w\to1}2w(\frac{ln(w)}{w-1})+ln(\frac{1}{w})=2$


问:
Sum of stopping times
Consider a probability space $(\Omega, \mathcal F, \mathbb P)$, endowed with a filtration $(\mathcal F_t)_{t\geq 0}$. A stopping time $T$ is a random variable taking values in $[0,\infty]$ such that for all $t\geq 0$ we have $\{T\leq t\}\in\mathcal F_t$.

I want to show that the sum of two stopping times $T$ and $S$ is a stopping time. I know that I can rewrite

$$\{S+T<t\} = \bigcup_{r,s \in \mathbb{Q}^+;\; r+s<t} \{S < r\} \cap \{T < s\}\,.$$

But to make it a proof I would need to show that an equivalent definition for stopping time is that $\{T<t\}\in\mathcal F_t$ for all $t\geq 0$.

Using $\{T<t\} = \bigcup_{n\in\mathbb N^+}\{T\leq t- 1/n\}$ I see that if $T$ is a stopping time than $\{T<t\}\in\mathcal F_t$. But is the reverse implication true? It seems to me that it requires right continuity of the filtration... But as far as I know the sum of two stopping times is a stopping time with no further assumption on the the filtration.

So how to prove that $S+T$ is a stopping time? Should I use a different kind of argument or am I wrong about the necessity of the right continuity?

答:
Hint

$$\{\sigma +\tau> t\}=\{\sigma =0,\tau>t\}\cup\{\sigma >t,\tau=0\}\cup\{\sigma >0, \tau\geq t\}\cup\{0<\tau<t, \tau+\sigma >t\}.
$$

Title: Sum of stopping times


问:
can you determine the distinct variable bit packets from within this bitstream?
11001101111011000011000111001011010011010111011011100011100111101011
the above bit stream should break up into these variable bit packets
11 0011 0111 1011 000011 000111 001011 010011 010111 011011 100011 100111
101011
basically it is a base 3 number represented in base 2 as 2 bits base 2 number
of variable length, with a 11 as a stop pattern, as long as you are looking at this 2 bits at a time, can you determine the second list of values from the bit stream I gave at the top part?</p>

答:
If you're allowed to remember some state between the processing steps, then yes.

Just do the following, assuming that the state bit is $s$ and the current input bits are $b_0$, $b_1$

Append $b_0$ to the output

If $s \land b_0$

Append space to the output

Append $b_1$ to the output

If $\lnot s \land b_0 \land b_1$

Append space to the output

Set $s \leftarrow 0$

Otherwise

Set $s \leftarrow b_1$

Thus, $s = 1$ if the last bist of the previous 2-bit chunk might be the first (but not the second!) bit of a stop pattern. By comparing $s$ and $b_0$, stop pattern which are split between two such chunks are detected and handled appropriately. By comparing $b_0$ and $b_1$, stop patterns within a chunk are detected, and the additional check for $s=0$ ensures that $0111$ isn't decoded to "011 1 " but rather to "011 1". Setting $s$ to $0$ in 4.2 prevents "1110" from being decoded to "11 1 0".

Here's an example. Say the input is $$
  011110001110100110 \text{.}
$$
The processing then does this $$
\begin{array}{cc|ll|ll|ll|ll|ll|ll|ll|ll|ll}
b_0 &b_1                          & 0&1 & 1&1 & 1&0 & 0&0 & 1&1 & 1&0 & 1&0 & 0&1 & 1&0 \\
s_{\textrm{in}} &s_{\textrm{out}} & 0&1 & 1&1 & 1&0 & 0&0 & 0&0 & 0&0 & 0&0 & 0&1 & 1&0 \\
\hline
                                 && 0&1 & 1.&1& 1.&0& 0&0 & 1&1.& 1&0 & 1&0 & 0&1 & 1.&0
\end{array}
$$
where the last line is the output, and dots represent spaces.

Notation-wise, $\lnot$ means logical not, i.e. $\lnot x$ is true if $x=0$, and $\land$ means logical and, i.e. $x \land y$ is true if $x=1$ and $y=1$. Logical or (which isn't used here) would be written as $x \lor y$ and would be true if $x=1$ or $y=1$.


答: Your observations are correct. The trick is to look at $\{S+T >u\}=\cup_r \{S>r\} \cap \{T>u-r\}$ for $u>0$ and the take complements on both sides. Also $(S+T=0) =(S=0)\cap (T=0) \in \mathcal F_0$.

Title: Sum of stopping times

Body:
Consider a probability space $(\Omega, \mathcal F, \mathbb P)$, endowed with a filtration $(\mathcal F_t)_{t\geq 0}$. A stopping time $T$ is a random variable taking values in $[0,\infty]$ such that for all $t\geq 0$ we have $\{T\leq t\}\in\mathcal F_t$.

I want to show that the sum of two stopping times $T$ and $S$ is a stopping time. I know that I can rewrite

$$\{S+T<t\} = \bigcup_{r,s \in \mathbb{Q}^+;\; r+s<t} \{S < r\} \cap \{T < s\}\,.$$

But to make it a proof I would need to show that an equivalent definition for stopping time is that $\{T<t\}\in\mathcal F_t$ for all $t\geq 0$.

Using $\{T<t\} = \bigcup_{n\in\mathbb N^+}\{T\leq t- 1/n\}$ I see that if $T$ is a stopping time than $\{T<t\}\in\mathcal F_t$. But is the reverse implication true? It seems to me that it requires right continuity of the filtration... But as far as I know the sum of two stopping times is a stopping time with no further assumption on the the filtration.

So how to prove that $S+T$ is a stopping time? Should I use a different kind of argument or am I wrong about the necessity of the right continuity?


问:
number of interesting divisors of $9!$

We call natural number $a$, "interesting" if natural numbers $x,y$
(not necessary distinct) exist so that $a=x^2y^3$. how many
interesting divisors the number $9!$ has?

To avoid calculations errors question brings this hint: "sum of the digits of the answer is divisible by $10$."

Here is my work:

for $x=y=1$ we have $a=1$ so it is one of the number we are looking for. also we know that  $9!=2^7\times3^4\times5\times7$, from this, we can conclude that $x$ and $y$ can't be $5$ or $7$ (because their exponent is $1$) so we should working on $2^7\times3^4$ and (also should considering $1$).

I don't know how to continue from here. I tried counting all the answers but my result is $16$ which is wrong. (maybe I missed some cases and answer should be $19$ according to the hint)

EDIT:

I got these answers:

$$x=2 , y=1,3$$
$$x=4 , y=1,3$$
$$x=8, y=1,3$$
$$x=3, y=1,2,4$$
$$x=9,y=1,2,4$$
$$x=1, y=1,2,3,4$$

答:
Suppose that $p^r$ is an exact prime power divisor of $a^2b^3$ then if the exact prime power divisors of $a$ and $b$ are $p^s$ and $p^t$ respectively, then $$r=2s+3t$$

Also note that if $r=2s+3t$ then $p^r=(p^s)^2(p^t)^3$

So the exponent for every prime has to be of the form $2s+3t$ and any non-negative exponent of this form will do.

That should help your counting.

Title: number of interesting divisors of $9!$


问:
Help with inverse trigonometric substitutions $ \int x^2\sqrt{a^2+x^2}\,dx $.
how would I go about integrating this? It is a lecture exercise.

$$
\int x^2\sqrt{a^2+x^2}\,dx
$$
I used a substitution of $x=a\tan\theta$ and ended up with $$\int a^4\tan^2\theta \sec^3\theta \,d\theta.$$ I was thinking by parts but then I would have to integrate a $\ln|\sec\theta+\tan\theta|$. Can't seem to think of the appropriate identity to collapse it. I am given the answer to verify but I don't know any mental algorithms to even begin doing inverse trigonometric substitution. The answer given is:
$$
\frac{x}{8}(a^2+2x^2)\sqrt{a^2+x^2}-\frac{a^2}{8}\ln\left(x+\sqrt{a^2+x^2}\right)
$$
Also, what are some things I should be thinking about first when looking at such questions when I need to substitute trigonometric functions into the x-variable? Should I attempt to make odd powers into even? How do I do so?
Thank you very much for your help.

答:
Let us consider the integral $$\int\tan^2x\sec^3xdx=I.$$ Then using the identity $1+\tan^2x=\sec^2x,$ this becomes $$\int\sec^5xdx-\int\sec^3xdx.$$

Then integrating by parts, we find that $$\int\sec^5xdx=\sec^3x\tan x-3I.$$ Also we find that $$\int\sec^3xdx=\sec x\tan x-\int\sec^3xdx+\int\sec xdx.$$ Since $$\int\sec xdx=\log|\sec x+\tan x|+c,$$ we can easily find $$\int\sec^3xdx.$$

Finally, we get that $$4I=\sec^3x\tan x-\frac 12\left(\sec x\tan x+\log|\sec x+\tan x|\right)+C.$$


问:
Show that $\triangle ABC$ satisfying $B^2=AC$ and $2b=a+c$ is equilateral

A $\triangle ABC$ satisfies the conditions below:
$$B^2=AC \qquad 2b=a+c$$
Show that it's an equilateral one, where $a=|BC|$, $b=|AC|$, $c=|AB|$ and $B=\angle ABC$, $A=\angle BAC$, $C=\angle ACB$.

What I have done:

When trying to solve these question, I want to find a solution in all triangles which satisfy $2b=a+c$. It means that point $B$ can be regarded as a point in an ellipse.

Then I try to prove that for all points $B$ in that ellipse, we have $B^2\leq AC$. When proving it I think I can get the equal conditions and then show that $AC=B^2$ if and only if $A=B=C$ or $AC=0$.

Then I meet a terrible problem, meaning that I need to show ($x=A$):

$$
\frac{4-5\cos x}{5-4\cos x}\leq\cos\frac{(2\pi-x)-\sqrt{x(4\pi-3x)}}{2}=-\cos \frac{x+\sqrt{x(4\pi -3x)}}{2}，x\in[0,\frac{\pi}{3}]\\
$$

I have never seen such a hard problem before. I try to prove the inequality by segment amplification and minification, derivative, and many other ways but all failed.

Can you help me or give me some hints on this problems?

答:
I am assuming $A\ge C$ without loss of generality. Therefore $A=Br$ and $C=\frac Br$ for some $r\ge1$, by the first equation. From the second equation, note that
\begin{align}
&2b=a+c\\
\implies &\dfrac{2b}{2R} = \dfrac{a}{2R}+\dfrac{c}{2R}\\
\implies &2\sin B = \sin A+\sin C&&(\text{By sine rule})\\
\implies &4\cos\dfrac B2\sin\dfrac B2 = 2\sin\left(\dfrac{A+C}{2}\right)\cos\left(\dfrac{A-C}{2}\right)\\
\implies &2\sin\dfrac B2=\cos\left(\dfrac{A-C}{2}\right)&&\left(\because\cos\frac B2 = \sin\left(\dfrac{A+C}{2}\right)\right)\\
\implies &2=\dfrac{\cos\left(\dfrac{A-C}{2}\right)}{\cos\left(\dfrac{A+C}{2}\right)}\\
\implies &3=\cot\left(\dfrac{A}{2}\right)\cot\left(\dfrac{C}{2}\right)&&(\text{By componendo and dividendo})\\
\implies &\tan\left(\frac A2\right)\tan\left(\frac C2\right)=\frac13\\
\implies &\tan\left(\dfrac{Br}{2}\right)\tan\left(\dfrac{B}{2r}\right)=\frac13\tag1\label{1}
\end{align}
Also note that
\begin{align}
&A+B+C=\pi\\
\implies &Br+B+\dfrac Br=\pi\\
\implies &B = \dfrac{\pi r}{r^2+r+1}\tag2\label{2}
\end{align}
Using \eqref{2} in \eqref{1}, we get,
\begin{equation}
\tan\left(\dfrac{\frac\pi2\times r^2}{r^2+r+1}\right)\tan\left(\dfrac{\frac\pi2}{r^2+r+1}\right)=\frac13
\end{equation}
Now, since LHS is a decreasing function of $r$ for all $r\ge 1$ and equality holds at $r=1$, hence $r=1$ and we're done.


问:
$d$ forms a metric space on The set of bounded sequences.
Let $X =\{(x_n) \subset \Bbb R \}$ such that the sequences are bounded.
Define $d$: $X×X \to \Bbb R $ where $d((x_n),(y_n))=sup\{|x_n-y_n|\}$ $n \in \Bbb N$.

Prove $(X,d)$ is a metric space.
The solution isn't difficult as I only need to check three trivial condition and the triangular inequality. That's not my problem. I actually wonder if $d$ is a function or not.
So suprimum does exist because the sequences are bounded, right? Is that enough to say the relation $d$ is a function? Is that even necessary?
Thank you very much.

答: Since both sequence $(x_n)_{n\in\Bbb N}$ and $(y_n)_{n\in\Bbb N}$, the sequence $(|x_n-y_n|)_{n\in\Bbb N}$ is bounded too. What this means is that the set $\{|x_n-y_n|\mid n\in\Bbb N\}$ is bounded. And, clearly, it's not empty. Since every bounded non-empty subset of $\Bbb R$ has a supremum, the definition of $d$ makes sense. In other words, yes, $d$ is a function.


问:
Show existence and find the value of the limit
Show that $$\frac{\tanh(x)-1}{e^{-2x}}$$ has a limit for $x \rightarrow \infty$ and the method to finding it.

I am guessing I have to use epsilon-delta-definitions for limits and then find a inequality that narrows the limit down. I am having such a hard time using the definitions to find the limit as it does not intuitively make sense to me and the ammount of notation throws me off as I can not use a generel method to solving such a problem.

Body: $$L=\lim_{x \to \infty} \frac{\tanh x-1}{e^{-2x}}=\lim_{x\to \infty} \frac{-2e^{-x}}{e^{-2x}(e^{x}+e^{-x})}=\lim_{x\to \infty} \frac{-2}{(1+e^{-2x})}=-2.$$


问:
$y'=\frac{x-y+2}{x+y-1}$
First, I substitute $x=\xi + h$, $y=\eta+k$. The equation becomes $$y'=\frac{\xi+h-(\eta+k)+2}{\xi+h+\eta+k-1}$$

and this becomes homogenous if we let $h=-\frac{1}{2}$ and $k=\frac{3}{2}$ ( We solve the constant terms such that they become zero ). I think it's correct this far, but what next? At this point we'd have $$\eta'=\frac{\xi-\eta}{\xi+\eta}$$

which is homogenous, and I tried substituting $\eta = u\xi$. This would lead to the equation $$u'\xi+u=\frac{1-u}{1+u}$$

I reduce $u$ from both sides. This will give $$u'\xi=\frac{1-u}{1+u}-u=\frac{1-2u-u^2}{1+u}$$

or $$\frac{du}{d\xi}\xi=\frac{1-2u-u^2}{1+u}$$

inverting this gives the relation $$\frac{1}{\xi}d\xi= \frac{1+u}{1-2u-u^2}du$$

So there has been some mistake made. The correct answer does not involve logarithms, and $\ln(\xi)$ will inevitably be a part of the solution. Can somebody spot my mistake?

答:
The integral on the RHS also involves logarithm. When you take antilogs on both sides after integration, you will be free of any logarithms

Another method:

$$y'=\frac{x-y+2}{x+y-1}$$

Cross multiply:

$$xdy+ydy-dy=xdx-ydx+2dx$$

$$d(xy)+ydy-dy=xdx+2dx$$

Title: $y'=\frac{x-y+2}{x+y-1}$


问:
Uniqueness part of the structure theorem for finitely generated p-primary modules over a PID
I'm having trouble understanding the proof of the uniqueness part of the following theorem:

Let $R$ be a PID and $M$ be a finitely generated $p$-primary module over $R$ (i.e. $p$ is a prime in $R$ and for every $m\in M$ there exists an $n\in\mathbb{N}$ such that $p^nm=0$). Then there exists an isomorphism $M\cong\bigoplus_{i=1}^kR/p^{l_i}R$ with $l_1\ge l_2\ge\ldots\ge l_k$ and such that $k\in\mathbb{N}$ as well as $(l_i)_{i=1}^k$ are uniquely determined.

The proof of the uniqueness part, as given in my lecture notes, begins as follows: For every $n\in\mathbb{N}$, the $R$-module $M_n:=p^nM/p^{n+1}M$ is annihilated by $p$ and thus has a unique structure of an $R/pR$ vector space with $r\cdot \mu=\pi(r)\cdot \mu$ for all $\mu\in M_n$ where $\pi$ denotes the projection from $R$ to $R/pR$. Since the isomorphism classes of the $M_n$ are uniquely determined by those of $M$, the same holds for $\dim_{R/pR}(M_n)$.

I don't understand the last sentence above: Why are the isomorphism classes of the $M_n$ uniquely determined by those of $M$? How can I show that there exists no isomorphism from $M_n$ to an $R/pR$ vector space $N$ which doesn't have the form $N=p^n\tilde{N}/p^{n+1}\tilde{N}$ and how can I determine the dimension of $M_n$ from that?

答:
A construction is uniquely determined when there is no flexibility in what is done.  For example, if $M$ is an $R$-module then $p^nM$ is a completely intrinsic construction: all multiples of $p^n$ in $M$. The quotient modules $p^nM/p^{n+1}M$ for each $n \geq 0$ are each intrinsic o the module structure of $M$, depending on nothing artificial like a particular way of viewing $M$.

It's  similar to the torsion submodule of $M$ being intrinsic: an element is a torsion element or it isn't, and you have no choice in the matter.  When you write $M = F \oplus T$ for "some" free submodule and "some" torsion submodule, the $T$ part is actually completely ("uniquely") determined because it must be the torsion submodule of $M$. The free part, in contrast, is not uniquely determined. Consider, when $R = \mathbf Z$, the decompositions $\pm 2^\mathbf Z = \pm(-2)^\mathbf Z$, which have different free parts.


问:
Verifying the interpretation of stopping times and stopping time $\sigma$-algebras
I have been thinking about the intuition of stopping times and stopping time $\sigma$-algebras. While I feel more or less comfortable with the former notion, I would like to get more insight in the latter. Having read different intuitive explanations, I tried to come up with the following interpretations which I would like to be verified.

Let $\mathbb{F} = {(\mathcal{F}_n)}_{n \in \mathbb{N}_0}$ be a filtration in $(\Omega, \mathcal{F})$. A random variable $\tau : \Omega \rightarrow \mathbb{N}_0 \cup \{ \infty \}$ is called a stopping time if
$\{ \tau \leq n \} \in \mathcal{F}_n$ for all $n \in \mathbb{N}_0$.

It can easily be shown that $\{ \tau \leq n \} \in \mathcal{F}_n \quad \forall n \in \mathbb{N}_0 \iff \{ \tau = n \} \in \mathcal{F}_n \quad \forall n \in \mathbb{N}_0$.

Interpretation I have come up with:

The relation $\{ \tau \leq n \}\in \mathcal{F}_n$ means that all the elementary events $\omega$ in the case of which I stop before time $n$ or at $n$ comprise an event in $\mathcal{F}_n$. This means, in particular, that at time $n$, having observed the current event I am at, I know precisely whether I have or whether I have not stopped before time $n$ or at $n$.

For example, suppose that at time $n$ I am on some event $A \in \mathcal{F}_n$. Two mutually exclusive cases are possible:

$A \cap \{ \tau \leq n \} = \emptyset\in \mathcal{F}_t$, so the decision to stop before time $t$ or at $t$ has not been made.

$A \cap \{ \tau \leq n \} \neq \emptyset$.  Additionally, $A \cap \{ \tau \leq n \} \in \mathcal{F}_n$ so  $A \cap \{ \tau \leq n \}$ is also an event which I can distinguish at time $n$. So depending on whether I am on $A \cap \{ \tau \leq n \}$ or on $(A \setminus A \cap \{ \tau \leq n \})$, I can tell whether I have or whether I have not stopped before time $n$ or at $n$.

A similar interpretation and example can be given for the relation $\{ \tau = n \} \in \mathcal{F}_n$.

Now consider the stopping time $\sigma$-algebra:
$$
\mathcal{F}_{\tau} := \{ A \in \mathcal{F}: A \cap \{ \tau \leq n\} \in \mathcal{F}_n \quad \forall n \in \mathbb{N}_0\}
$$
It can indeed be verified that the above family is a $\sigma$-algebra and that $$A \in \mathcal{F}_{\tau} \iff A \cap \{ \tau = n\} \in \mathcal{F}_n \quad \forall n \in \mathbb{N}_0.$$

In literature, usually, $\mathcal{F}_{\tau}$ is described as the $\sigma$-algebra of the events observed up to the stopping the $\tau$, in analogy to $\mathcal{F}_n$, which represents the events observable up to time n.

Interpretation I have come up with:

Suppose that at some arbitrary but fixed time $n$ the event  $\{ \tau \leq n \} \in \mathcal{F}_{n}$ has occurred which means that there has been a decision to stop before time $n$ or at $n$. Then for every $A \in \mathcal{F}_{\tau}$ I can tell whether $A$ has occurred or not depending on whether I am at the event $A \cap \{ \tau \leq n\} \in \mathcal{F}_{n}$ or not. So the events in $\mathcal{F}_{\tau}$ are those for which I can tell whether they have occurred or not provided that the event $\{\tau \leq n \}$ for some $n \in \mathbb{N}_0$ has occurred (i.e. there has been a decision to stop).

Conversely, take any $A \in \mathcal{F}_{\tau}$ and arbitrary but fixed time $n$. Further assume that event $A$ has occured at time t, in the sense that some $B \in \mathcal{F}_n$ has occured with $B \subset A$. So I can tell whether I have or whether I have not stopped before time $n$ or at $n$ depending on whether the event $B \cap (A \cap \{ \tau \leq n \}) \in \mathcal{F}_n$ has occurred or not.
(However I think this second point is irrelevant since at every time $n$ I know whether the event $\{\tau \leq n\} \in \mathcal{F}_n$ has occurred or not.)

I feel more or less assured regarding the first interpretation for the stopping time however I am unsure of the interpretation for the $\sigma$-algebra, namely, whether it actually corresponds to the description stated earlier:

$\mathcal{F}_{\tau}$ is the $\sigma$-algebra of the
  events observed up to the stopping time $\tau$.

So here are my questions:

Do you agree with the two interpretations?

Can you add something to make them better (especially to that of the $\sigma$-algebra)?

Can you come up with different interpretations?

答:
I don't think we actually know whether a specific event occurred just because it is in the $\sigma$ algebra. Its complement is in $\mathfrak{F}_n$ too.

As was detailed more thoroughly in the answer of Nate Eldredge on the
Intuition behind Conditional Expectation, I'd say that $\{\tau \leq n\} \in \mathfrak{F}_n$ gives us the ability to ask whether we stopped. If we had the realization of our stochastic process nothing would be random. That is just a deterministic function depending on n.

So I think the sentence

"This means, in particular, that at time $n$, having observed the current event I am at, I know precisely whether I have or whether I have not stopped before time $n$ or at $n$."

is wrong since the $\sigma$-Algebra does not contain what you observe but rather what is observable.

Suppose you are blind and pick coloured balls with a unique number from an urn and you have a machine that cannot tell you the number on the ball but only the colour. You cannot see the concrete realization of the process. You can't have full information but the machine could tell you the colour for each pick you make. So the $\sigma$-Algebra would contain something like

$\mathfrak{F}_n = \{ \emptyset, \Omega, R, B \}$

In case we only have two colours Red and Blue. Now suppose your machine got an update and now it can tell you whether the number on the ball is even or not as well. Now we have

$\mathfrak{F}_{n+1} = \sigma (\{R, B, E, \neg E \})$

So I think the $\sigma$-Algebra represents more the capacities of the observer than our knowledge at a certain time. For the stopping time $\tau$ with $\{\tau \leq n\} \in \mathfrak{F}_n$ that would then mean that the observer actually has the capacity to say whether the process is stopped. So if we want to calculate probabilities or conditional expectations we can use this capacity in our analysis.


问:
Easy to read partial differential equations book?
I'm looking for an easy to read undergraduate book on partial differential equations, ideally something that is not much harder than a multivariable calculus/ordinary differential equations book.

I am preparing for a course which is using the text by Walter Strauss, but I found this text a bit difficult to read. Specifically, I found that many of the derivations were missing steps that were not obvious to me or provided little justification for the manipulations. When I searched for introductory books however the Strauss book seems to be recommended.

Google seems to recommend, among others:

Partial Differential Equations for Scientists and Engineers by Farlow

Introduction to Partial Differential Equations by Peter Olver

I have ruled out:

Partial Differential Equations: An Introduction by Walter Strauss

An Introduction to Partial Differential Equations by Michael Renardy

Partial Differential Equations by Fritz John

Partial Differential Equations by Lawrence C Evans

My background is having read A First Course in Differential Equations with Modelling Applications by Dennis Zill. Would I be better off reading the extended version of this book (Differential Equations with Boundary Value Problems)? I was a little bit hesitant because I was not sure how relevant the book is to PDEs. The chapters I have not read are Fourier Series, Boundary Value Problems in Rectangular Coordinates, Boundary Value Problems in Other Coordinate Systems, Integral Transforms and Numerical Solutions of Partial Differential Equations.

EDIT: I ended up finishing about half the book, which is everything covered in my course, Chapters 1 through 7.

答:
Personally I am quite fond of Evan's Partial differential equations as an introductory textbook.

I would not expect to find something really "easy to read", though, simply because the subject matter tends not be all that easy.


问:
Prove that $ \mathbb{P}\left(Y_{n}\geq0\right)=\mathbb{P}\left(Z\geq0\right) $
Let $ \left(X_{n}\right)_{n} $ be a sequence of random variables, $X$ a random variable and $ Z $ a random variable, such that for any $ n $ we have $ |X_{n}|\leq Z  $, and $ X_{n}\overset{a.s}{\longrightarrow}X $. Now define $ Y_{n}=2Z-|X-X_{n}| $.

Is it true that all the given details that I wrote here is enough to conclude that $ \mathbb{P}\left(Y_{n}\geq0\right)=\mathbb{P}\left(Z\geq0\right)=1 $ ?

If so, I'll be glad if someone can show how to prove it. Thanks in advance

答:
It's evident that $Z \ge 0$ almost sure because $Z \ge |X_1| \ge 0$. So,   $P(Z \ge 0) = 1$

Besides, we have
\begin{align}
Z \ge |X_1| &\iff  -Z<X_n<Z
\end{align}

Hence, $-Z<\lim X_n<Z $ or $-Z<X<Z$ or $$|X| \le Z$$
Now we have
$$Y_n = 2Z-|X-X_n| \ge 2Z-(|X|+|X_n|) \ge 0$$

We can conclude that  $P(Y_n \ge 0) = 1$.


答:
Just restating my comment as an answer:

You are confusing "long interarrival times are rare" with "the more time passes by, the more likely it is an earthquake will happen". The first is true for an exponential distribution if your mean interarrival time is short. The second one is never true for an exponential distribution, as Did and the others already have noted.

I find it an interesting question though because this is something you will come across relatively often in newspapers. Statements like "a big eruption of Yellowstone park is overdue" or "the last big asteroid crash dated from ..., considering the mean rate is ..., we are overdue for a crash.".

They create a false sense of impending doom. What they are really stating is that we are experiencing an unusually long interarrival time, but insofar as the process is Poisson, the probability of the event happening in the next days, months or years is still independent of how long it has been since the last event.

Add to that the fact that these processes are only approximately Poisson or even that the rates are variable and you realize that one should really relativize all this sensationalist press. For instance, asteroid crashes become ever rarer as planetary orbits get gradually cleared of asteroids through the solar system's history. There are just less and less asteroids to crash. So the rate of the process is not constant.

Title: exponential distribution - get some meaning

Body:
I read that a continuous random variable having an exponential distribution can be used to model inter-event arrival times in a Poisson process. Examples included the times when asteroids hit the earth, when earthquakes happen, when calls are received at a call center, etc. In all these examples, the expected value of the number of events per unit of time, lambda, is known and is constant over time. Moreover, each event's occurrence is independent of previous events' occurrences. And the exponential variable that models inter-event arrivals has the same lambda parameter as the Poisson variable that models the number of events.

And now, my problem. It don't get the connection between the intuition behind the exp distrib and its pdf. It seems obvious that the more time it passes by without an earthquake happening, the more likely it is than an earthquake will happen. Assume my understanding of lambda is correct, i.e., lambda = the rate at which the event happens, e.g., 5 earthquakes per minute on some remote, angry planet. On the pdf graphic, the prob of generating a value between 0 and 1 is greater than the prob of choosing a value between 4 and 5 for instance. How is this graphic related to the fact that on average we need to have 5 earthquakes per minute?


问:
Why is $(\vec{b} \times \vec{a}) \times \vec{b}$ non zero?
If I use the rule of vector triple product, it becomes:

$$ \vec{b} \times \vec{a} \times \vec{b} = \vec{a} ( |\vec{b}|^2) - \vec{b}(\vec{b} \cdot \vec{a})$$

which is generally non-zero, but suppose I use properties of cross product:

$$ \vec{a} \times \vec{b} = - \vec{b} \times \vec{a}$$

Hence,

$$ \vec{b} \times \vec{a} \times \vec{b} = - \vec{a} \times \vec{b} \times \vec{b} =  \vec{a} \times (\vec{b} \times \vec{b})=0$$

What did I do wrong?

答:
$\vec S= \vec P \times (\vec Q  \times \vec R)= (\vec P. \vec R) \vec Q-(\vec P. \vec Q) \vec R$

So $(\vec b \times a)\times \vec b=-\vec b\times(\vec b \times \vec a)=-[(\vec b. \vec a) \vec b- (\vec b . \vec b) \vec a] \ne 0$


问:
Find a formula for $\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$
I need to find a clear formula (without summation) for the following sum:

$$\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$$

Well, the first few elements look like this:

$1,1,1,2,2,2,2,2,3,3,3,...$

In general, we have $(2^2-1^2)$ $1$'s, $(3^2-2^2)$ $2$'s etc.

Still I have absolutely no idea how to generalize it for $n$ first terms...

答:
The way I approached this problem is dividing the sum into two halves. Where the first half contains all the numbers starting from $1$ and reaching upto $m$ such that $\sqrt{m}$ is greatest integer $<\lfloor\sqrt{n}\rfloor$, and second half contains the leftover numbers. Intuitively $m$ can be calculated by using simple formula $\lfloor\sqrt{n}\rfloor - 1$Mathematically:

$\displaystyle\sum_{k=1}^n\lfloor\sqrt{k}\rfloor = \underbrace{\displaystyle\sum_{p=1}^{\lfloor\sqrt{n}\rfloor-1}p\times(2p + 1)}^{\text{first part}} +  \underbrace{(n - {\lfloor\sqrt{n}\rfloor}^2 + 1)\times{\lfloor\sqrt{n}\rfloor}}^{\text{second part}}$

The same can be resolved using the following identities.

$\displaystyle\sum_{x = 1}^n x = \frac{x\times(x+1)}{2}$

and

$\displaystyle\sum_{x = 1}^n x^2 = \frac{x\times(x+1)\times(2 \times x+1)}{6}$

I didn't tested this solution with every value, but the solution seems convincing.

I am taking an example to demonstrate it, taking $n = 5$

The answer should contain $\underbrace{1+1+1+2+2}+2+2+2+3+... = 7$

The $\lfloor\sqrt{5}\rfloor-1 = 1$(upper limit for the first part)
So, our first part is calculated as $1\times (2+1) = 3$

Similarly second part is calculated as
$(5-{\lfloor\sqrt{5}\rfloor}^2)\times \lfloor\sqrt{5}\rfloor = (5 - 2^2+1)\times 2 = (2)\times 2 = 4$

So we get $4 + 3 = 7$ as our answer, which is correct.

The same intuition can be used to derive formula for $\displaystyle\sum_{k=1}^n\lfloor\sqrt[x]{k}\rfloor$.

i.e.

$\displaystyle\sum_{k=1}^n\lfloor\sqrt[x]{k}\rfloor = \displaystyle\sum_{p=1}^{\lfloor\sqrt[x]{p}\rfloor}p\times ((p+1)^x-p^x)+(n - (\rfloor\sqrt[x]{n}\rfloor)^x + 1)\times\lfloor\sqrt[x]{n}\rfloor$

P.S. This is just an intuition, if anything wrong, feel free to comment.


问:
Prove that $F[x,y] / (y^2-x)$ is an integral domain
I want to show that this quotient $F[x,y] / (y^2-x)$ is an integral domain.

The first idea I had is to show that $y^2 - x$ is prime in the polynomials with variable $x$ and $y$,however, this seems to be rather hard to verify.

Then I was thinking that would it be possible to construct some isomorphisms between this quotient ring and some integral domain, but I got stuck.

答: $$f(x,y)=y^2-x$$ is irreducible in $\Bbb F[x][y]$ because it is a monic quadratic polynomial in $y$ with no roots in $\Bbb F(x)$. Equivalently, $f(x,y)$ is irreducible over $\Bbb F[x,y]$ because $$g(x,y)\cdot h(x,y)=f(x,y)$$ implies one of $g(x,y),h(x,y)$, say $g(x,y)$, is a $\Bbb F[x][y]$ unit and thus a $\Bbb F[x,y]$ unit as well. Therefore, the ideal $\langle f(x,y)\rangle$ is a prime $\Bbb F[x,y]$ ideal and thus $\frac{\Bbb F[x,y]}{\langle f(x,y)\rangle}$ is an integral domain.


答:
Removing the parenthesis makes you perform invalid transformations, because the cross product is not associative.

In fact $$\vec a\times\vec b\times\vec c$$ has no defined meaning.

Unless $\vec a,\vec b$ are parallel, $\vec b\times\vec a$ is orthogonal to $\vec b$, hence not parallel, and $(\vec b\times\vec a)\times\vec b$ is nonzero.

Title: Why is $(\vec{b} \times \vec{a}) \times \vec{b}$ non zero?

Body:
If I use the rule of vector triple product, it becomes:

$$ \vec{b} \times \vec{a} \times \vec{b} = \vec{a} ( |\vec{b}|^2) - \vec{b}(\vec{b} \cdot \vec{a})$$

which is generally non-zero, but suppose I use properties of cross product:

$$ \vec{a} \times \vec{b} = - \vec{b} \times \vec{a}$$

Hence,

$$ \vec{b} \times \vec{a} \times \vec{b} = - \vec{a} \times \vec{b} \times \vec{b} =  \vec{a} \times (\vec{b} \times \vec{b})=0$$

What did I do wrong?


问:
What is the difference between surface area and scalar surface integrals?
What is the difference between the surface area of a paremetrized surface and the scalar surface integral of a function in $\mathbb{R}^3$? Are they not the same thing? </p>

答:
The problem is when usually calculus teachers don't expand into multidimensional calculus in they way that might make sense. See a triple integral of a function might also seem strange until you realize that it is more like the volume integral of a 4 dimensional function. If you recall how a triple integral is turned into a double integral by integrating the function $f(x,y)$ over the region in question, in a same sense a 4 dimensional integral is turned into a triple integral over a three dimensional region (the function acts like the shadow of a 4 dimensional function in the same way as f acted like the shadow of a three dimensional function). This can be extended to surface integrals in the same sense, its simply the surface area of a higher dimensional object.
A parameterized surface represents the normal three dimensional object whereas, a scalar function represents a higher dimensional object.
And yes the above post is correct, when the function in question is constant at 1 they are the same.</p>


问:
Dimension of a matrix subspace given a rank
Let A be a $m*n$ real valued matrix, $r$ the rank of A and let $S$ be the set of all real $n*k$ matrices such that AB=0 (the $m*k$ zero matrix).
Show that S is a vector subspace of $\mathbb{R^{n,k}}$ and find the dimension of $S$

I am stuck with the second question.
My reasoning has been that I can see both B and A as represenations of linear functions, $B$ being a function f from $\mathbb{R}^{k}$ to $\mathbb{R}^{n}$ and $A$ being a function g from  $\mathbb{R}^{n}$ to  $\mathbb{R}^{m}$, $AB$ being the fucntion composition, therefore if $AB$=0
Then g(f(v))=0 for every vector v, thus $im(f) \subseteq ker(g)$.
By the Rank–nullity Theorem, $dim(ker(g))$=n-r, and $dim(im(f))$=$rank(B)$.

Therefore S is just the set of all matrices such that $rank(B) \leq n-r$ but I can't continue.

Am i on the right track?

答:
$AB=0$ holds if and only if each column of $B$ is in the kernel of $A$.

This observation gives a vector space isomorphism $S\cong(\ker A)^k$.


答: Another way to proceed with the original integral is to use the substitution $x=a\sinh y,$ so that the integral now is $$a^4\int \sinh^2y\cosh^2y dy=\frac{a^4}{4}\int\sinh^22y dy=\frac{a^4}{8}\int(1-\cosh 4y)dy=\frac{a^4}{8}\left(y-\frac14\sinh 4y\right)+k,$$ for some constant $k.$ Reverting the substitution now gives $$\frac{a^4}{8}\left[\log\left(\frac x a+\sqrt{\frac{x^2}{a^2}+1}\right)-\frac x a\left(1+\frac{x^2}{a^2}\right)-\frac{x^3}{a^3}\sqrt{1+\frac{x^2}{a^2}}\right]+C,$$ for some constant $C.$

Title: Help with inverse trigonometric substitutions $ \int x^2\sqrt{a^2+x^2}\,dx $.

Body:
how would I go about integrating this? It is a lecture exercise.

$$
\int x^2\sqrt{a^2+x^2}\,dx
$$
I used a substitution of $x=a\tan\theta$ and ended up with $$\int a^4\tan^2\theta \sec^3\theta \,d\theta.$$ I was thinking by parts but then I would have to integrate a $\ln|\sec\theta+\tan\theta|$. Can't seem to think of the appropriate identity to collapse it. I am given the answer to verify but I don't know any mental algorithms to even begin doing inverse trigonometric substitution. The answer given is:
$$
\frac{x}{8}(a^2+2x^2)\sqrt{a^2+x^2}-\frac{a^2}{8}\ln\left(x+\sqrt{a^2+x^2}\right)
$$
Also, what are some things I should be thinking about first when looking at such questions when I need to substitute trigonometric functions into the x-variable? Should I attempt to make odd powers into even? How do I do so?
Thank you very much for your help.


答:
If $\mathbf{x}$ and $\mathbf{y}$ are perpendicular, then $\|\mathbf{x} \times \mathbf{y}\| = \|\mathbf{x}\|\;\|\mathbf{y}\|$.  So in that case, if the two factors are nonzero, so is the cross product.  But $\mathbf{b} \times \mathbf{a}$ is perpendicular to $\mathbf{b}$, so we can conclude: if $\mathbf{b} \times \mathbf{a}$  is nonzero, then so is $(\mathbf{b}\times\mathbf{a})\times \mathbf{b}$.

Note, however, that if $\mathbf{a}$ and $\mathbf{b}$ are parallel, then
$\mathbf{b} \times \mathbf{a} = \mathbf{0}$ and your conclusion fails.


问:
Alternative proof for $1+2w+3w^2+\ldots+nw^{n-1}= \frac{n}{w-1}$
I want to prove that if $w\neq 1$ is a root of unity, then
$$F(w)=1+2w+3w^2+\ldots+nw^{n-1}= \frac{n}{w-1}$$
I have already proved it in two different ways but I am looking for a shorter and more elegant one, using derivatives. I already know that $1+w+w^2+\ldots +w^{n-1}=w+w^2+\ldots+w^n=0$. It turns out that if
$$f(w) = w+w^2+\ldots+w^n \quad \mbox{then} \quad f'(w)=F(w)$$
But $f(w)=0$ and therefore $f'(w)=0 \neq F(w)$. Am I missing something? Can't I use derivatives for this argument?

答:
Your development can be understood in two ways:

$\omega$ is a constant, namely a root of unity. Then taking the derivative on $\omega$ is pointless.

$\omega$ is a variable, which you will later identify to a root of unity. In this case, $f(\omega)=0$ does not hold yet.

For an alternative proof, observe that

$$F(\omega)-\omega F(\omega)=1+\omega+\omega^2+\cdots\omega^{n-1}-n\omega^n=-n$$ and

$$F(\omega)=-\frac n{1-\omega}.$$