问:
Galois Theory Basic Algebra I Jacobson
I'm having troubles with this problem.  I saw it in Jacobson's "Basic Algebra I".
I hope you can help me.

-Show that $E=\mathbb{Q}(\sqrt{2}, \sqrt{3}, u)$ where $u^2=(9-5\sqrt{3})(2-\sqrt{2})$ is normal. Determine $\mathrm{Gal}(E/\mathbb{Q})$.

答:
To show that $E/\mathbb Q$ is normal (in fact Galois), it suffices to show that $|\mathrm{Aut}(E/\mathbb Q)|=[E:\mathbb Q]$, so we shall first determine $\mathrm{Aut}(E/\mathbb Q)$ and $[E:\mathbb Q]$. To do so consider the tower of fields $E/\mathbb Q(\sqrt2,\sqrt3)/\mathbb Q(\sqrt2)/\mathbb Q$. We see that each extension in this tower has degree $2$, since they are the splitting fields of $x^2-(9-5\sqrt3)(2-\sqrt2),x^2-3$ and $x^2-2$ respectively (show this), so $[E:\mathbb Q]=2\cdot 2\cdot 2=8$. Each extension has two automorphisms, the identity and the maps
$$\begin{align}
a+bu&\mapsto a-bu &a,b\in\mathbb Q(\sqrt2,\sqrt3)\\
a+b\sqrt3&\mapsto a-b\sqrt3 &a,b\in \mathbb Q(\sqrt2)\\
a+b\sqrt2&\mapsto a-b\sqrt2 &a,b\in\mathbb Q
\end{align}$$
Clearly these automorphisms fix $\mathbb Q$. It remains to check that the last two extend to all of $E$. I leave it to you to check that the maps
$$\begin{align}
a+b\sqrt3&\mapsto a-b\sqrt3 &a,b\in \mathbb Q(\sqrt2,u)\\
a+b\sqrt2&\mapsto a-b\sqrt2 &a,b\in\mathbb Q(\sqrt3,u)
\end{align}$$
are well-defined field automorphisms. This gives you three non-identity automorphisms, which I'll call $\alpha,\beta,\gamma$. It is easy to see that $\mathrm{id}, \alpha,\beta,\gamma,\alpha\beta,\alpha\gamma,\beta\gamma,\alpha\beta\gamma$ are all distinct. Since $|\mathrm{Aut}(E/\mathbb Q)|\leq [E:\mathbb Q]=8$ it follows that these must be all automorphisms of $E/\mathbb Q$, and so $|\mathrm{Aut}(E/\mathbb Q)|=[E:\mathbb Q]$ hence the extension is Galois and
$$\mathrm{Gal}(E/\mathbb Q)=\{\mathrm{id}, \alpha,\beta,\gamma,\alpha\beta,\alpha\gamma,\beta\gamma,\alpha\beta\gamma\}.$$</p>

Title: Galois Theory Basic Algebra I Jacobson


问:
Assume $T$ is a linear operator over a vector space $V$ with a finite dimension and assume a linear operator $U$ on $V$ does exist such that $TU=I$
Assume $T$ is a linear operator over a vector space $V$ with a finite dimension and assume a linear operator $U$ on $V$ does exist such that $TU=I$,where $I$ is the identity linear transformation.

Show that $T$ is  invertible and that $U=T^{-1}$.

From the assumption we have that $TU=I$ and it follows that $U$ is surjection and $T$ is injection,moreover we have the following lemma:

If $V$ and $W$ are finite vectors spaces over the field $\mathbb F$ such that $\text{dim}(V)=\text{dim}(W)$ and $T:V \to W$ is a linear transformation then the following statements are equivalent:

$T$ is invertible

$T$ is nonsingular,that is $T( \alpha)=0_w \implies  \alpha=0_v$

$T$ is surjection

So now from the lemma we know that $T$ has an inverse $T^{-1}$ such that $TT^{-1}=I=TU$,the linear transformation $T$ is well-defined and so $T^{-1}TT^{-1}=T^{-1}TU \implies IT^{-1}=IU \implies T^{-1}=U$.

I used the fact that $T$ is well-defied,but is it really well-defined? since the mapping has not been given explicitly then I think it's kinda meaningless to talk about the well-definedness,however without that I'm not able to finish the proof.

答:
Let $m(x)=x^k+a_{k-1}x^{k-1}+\cdots+a_1 x + a_0$ be the minimal polynomial for $T$. Then
$$
       (T^{k-1}+a_{k-1}T^{k-2}+\cdots+a_1 I)T=-a_0 I.
$$
So, $V=T^{k-1}+a_{k-1}T^{k-2}+\cdots+a_1 I$ either satisfies $VT=TV=0$, which would imply that $T$ has a non-trivial null space, or else $a_0\ne 0$, and $T$ is invertible with inverse
$$
           T^{-1}=\frac{1}{a_0}(T^{k-1}+a_{k-1}T^{k-2}+\cdots+a_1 I).
$$
Given that $TU=I$, it must be that $T$ is invertible, and $U=T^{-1}$.

ContentLicense: CC BY-SA 4.0


问:
$U \cap \operatorname{cl}(A) \neq \emptyset$ implies $U \cap A \neq \emptyset$, is $U$ open?
In a metric space $(X,d)$, let $U\subseteq X$. If $$U \cap \operatorname{cl}(A)\neq \emptyset \implies U\cap A\neq \emptyset$$ for all set $A\subseteq X$ , is $U$ open? (Notation: $\operatorname{cl}(A)$ is the closure of $A$)

I am currently stuck at this problem for a long time, the converse is easy and there are many similar ones out there.

Any proofs or hints? Or any counter examples if not true? Thanks in advance.

答: You could also prove the contrapositive. Suppose that $U$ is not open then we can find a set $A$ such that $cl(A)\cap U \neq \emptyset$ but $A \cap U = \emptyset$. To see this let $x \in U$ be such that $B_{\frac{1}{n}}(x)$ is not a subset for all $n \in \mathbb{N}$ then we can construct a sequence $(x_n)_n$ by letting $x_n \in B_{\frac{1}{n}}(x) \setminus U \neq \emptyset$. Then let $A=\{x_n : n \in \mathbb{N}\}$ and notice that $A \cap U = \emptyset$ but $x \in cl(A)$ so $cl(A) \cap U \neq \emptyset$.


问:
Variant of balls bin problem
I am stuck at the following variant of the balls bin problem.

The task is to count the number of ways of placing $n$ identical balls into $k$ ordered bins under the constraint: the bin number $i$ must have strictly less than $i$ balls.

Any help/suggestions would be appreciated.

答:
$HINT:$ As far as  I understood, It is said that balls are identical but bins are different.Then it reminds us Stirling numbers with second types look at:https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind#:~:text=Stirling%20numbers%20of%20the%20second%20kind%20are%20one%20of%20two,to%20the%20parameters%20n%2C%20k.

Moreover , it is said that $i$ th bin have at most $i$ balls so apply it to stirling numbers from $1$ to $i$

Title: Variant of balls bin problem


问:
Maximum of $\prod_{k=1}^{n} { f(c_k) }$ where $f(m) = \sum_{k=1}^{m} k^2, \sum_{i=1}^{n} c_i = 2020$.
What's the maximum of  $\prod_{k=1}^{n} { f(c_k) }$ where $f(m) = \sum_{k=1}^{m} k^2, \sum_{i=1}^{n} c_i = 2020$. $n$ can be any integers?

Since $f(1) = 1, f(2) = 5, f(3) = 14, f(4) = 30$.

It seems like we are better off using $f(3)$ as much as possible, followed by $f(4)$ and followed by $f(2)$. But is there a rigorous proof?

答:
The idea is, indeed, to find the optimal way to split $2020$ into smaller parts.
Capitalizing on the fact that the number of parts is unconstrained. Below I will frequently compare various ways of splitting a total $N$ into smaller parts, looking for the maximal product.

A starting point is the well known formula
$$
f(m)=\frac16m(m+1)(2m+1).
$$
Using this we easily see that the inequality
$$
f(2)f(m)>f(m+2)
$$
holds whenever $m\ge3$. This is hardly surprising. After all, the left hand side is a cubic polynomial of $m$ with leading coefficient $2f(2)/6=5/3$ but the right hand side is a cubic polynomial of $m$ with leading coefficient $2/6=1/3$. So the inequality automatically holds for large values of $m$, and we only need to find when.

This has the immediate consequence that at the maximum we always have $c_i<5$ for all $i$. This is because if $c_i$ were $\ge5$, the product $\prod_i f(c_i)$ increases, when we split $c_i$ to $(c_i-2)+2$. All because $f(2)f(c_i-2)>f(c_i)$.

As you already observed, we don't want to have two $2$s among the $c_i$s given that $f(4)>f(2)^2$; $4$ gives a larger product than $2+2$.

At this point we know that the optimal collection of $c_i$s has $3$s and $4$s and at most a single $2$.

Let's look at $4$s next. We have $4+4=8$ and
$$f(4)^2=900<980=5\cdot14^2=f(2)f(3)^2,$$
where also $8=2+3+3$. Therefore we can conclude that we never want to have more than a single $4$; it pays to use $2+3+3$ instead of $4+4$.

Do we want to have both $2$ and $4$? As above,
$$
f(2)f(4)=150<196=f(3)^2.$$
It follows that instead of using $2+4$ we get a higher product with $3+3$ instead.

So, as you predicted, it is all $3$s and then, if necessary, we can use a single $2$ or a single $4$.

Notice that so far we have not used the number $2020$ at all. This holds for any number $N\ge8$ in place of $2020$, and the arguments above all go through.

All that remains is to observe that
$$
2020=673\cdot3+1.
$$
So to make $\sum_ic_i$ equal to $2020$ we need an extra four, and the answer is
$$
f(3)^{672}f(4).
$$

If, next week, somebody asks the same question about $2021$ that is congruent to $2$ modulo $3$, and the answer will be $f(3)^{673}f(2)$ instead.


问:
Is there a formula for $\int_0^\pi\frac{\sin(nx)}{\cos(\theta)-cos(x)}dx$?
I'm working on a school project about the aerodynamics of an helicopter blade. I'm trying to adapt my aerodynamics class ( centered around planes wings mostly ) to this case.

In our class, we used what we called the integrals of Glauert and more specifically, this integral :
$$
\int_{0}^{\pi}\frac{\cos\left(nx\right)}
{\cos\left(\theta\right) - \cos\left(x\right)}\,{\rm d}x = -\pi\,\frac{\sin\left(n\theta\right)}{\sin\left(\theta\right)}
$$
In the helicopter case, I need to evaluate a similar integral but with $\sin\left(nx\right)$ instead of $\cos\left(nx\right)$.

I found a beautiful proof of the first result in the book Inside Interesting Integrals by Paul Nahin ( page $60$ ) which uses recursivity. I tried to adapt it to the other integral using the formula
$$\sin\left(\left(n + 1\right)x\right) +
\sin\left(\left(n - 1\right)x\right)
= 2\sin\left(nx\right)\cos\left(x\right)$$
$$
\mbox{Calling}\quad
I_{n} =
\int_0^\pi\frac{\sin\left(nx\right)}{\cos\left(\theta\right) - \cos\left(x\right)}\,{\rm d}x,
$$
I find that
$$
I_{n+1} - 2\cos\left(\theta\right)I_{n} +
I_{n - 1} = \int_{0}^{\pi}2\sin\left(nx\right)\,{\rm d}x
$$ but I then have a problem to solve the recursive equation be the result is $0$ or $\frac{4}{n}$ depending on the parity of n. I decoupled the system to get two recursive equations for the different parities :

For $a_n=I_{2n}$, I get $a_{n+1}-2\cos(2\theta)a_n+a_{n-1}=8\cos\theta(\frac{1}{2n+1}+\frac{1}{2n-1}$)

And for $b_n=I_{2n+1}$, I get $b_n-2\cos(2\theta)b_{n-1}+b_{n-2}=\frac{8}{2n-1}\cos(\theta)$

I entered these formula on Wolfram Alpha and got some really ugly results. I was thus wondering if a more simple formula could exist for my integral.

I also quickly looked at the wikipedia proof which uses residual theorems but I got lost quite fast...

Do you know some other way to evaluate this integral or maybe if it's possible to solve the recursive equations to get to a simpler result than the one from Wolfram ?

Thank you very much !

答:
Probably, there is no a manageable general expression but we can get an exact answer for  moderate values of n.

The idea is to get rid of singularity in the integrand.

So we replace $\theta$ with $i\theta$ ($i$ is the imaginary unit).

Since $\cos(i\theta)=\cosh(\theta)$ we get

$$J_n(\theta)=\int_0^\pi\frac{\sin(nx)}{\cosh(\theta)-\cos(x)}dx$$
Now, $J_1(\theta)=-2\ln(\tanh\frac{\theta}{2})$

To return to the original integral we replace $\theta$ with $i\theta$ again and equate the real parts on both sides of the equal sign:

$$I_1(\theta)=P\;V\int_0^\pi\frac{\sin(x)}{\cos(\theta)-\cos(x)}dx=-2\ln(\tan\frac{\theta}{2})$$

Completely analogously:

$$J_2(\theta)=\int_0^\pi\frac{\sin(2x)}{\cosh(\theta)-\cos(x)}dx$$

$$=2\cosh(\theta)\ln\frac{\cosh(\theta)+1}{\cosh(\theta)-1}-4$$

and $$I_2(\theta)=P\;V\int_0^\pi\frac{\sin(2x)}{\cos(\theta)-\cos(x)}dx$$

$$=4\cos(\theta)\ln(\cot\frac{\theta}{2})-4$$

For $n=3,4,...$ the procedure is the same.

REMARK:

In similar manner we can establish the initial integral.

Namely, consider

$$\int_0^\pi\frac{\cos(nx)}{\cosh(\theta)-\cos(x)}dx=\pi\frac{e^{-n\theta}}{\sinh(\theta)}$$

Replacing here $\theta$ with $i\theta$ ($i$ is the imaginary unit) and equating real parts gives

$$P\;V\int_0^\pi\frac{\cos(nx)}{\cos(\theta)-\cos(x)}dx=-\pi\frac{\sin(n\theta)}{\sin(\theta)}$$


问:
The $2$nd, the $1$st and the $3$rd term of an arithmetic progression form a geometric progression
An arithmetic progression is given with a common difference $\ne0.$ The $2nd$, the $1st$ and the $3rd$ term of the ap form a geometric progression. Find the common ratio.

So we have the ap: $a_1,a_1+d,a_1+2d$ and the gp: $a_1+d,a_1,a_1+2d.$ How can we find $q?$ Thank you in advance!

I have tried to solve the problem using a system, but I am not sure which equations we can use. For example, $a_1^2=(a_1+d)(a_1+2d)$ and $\dfrac{a_1+d}{2}=a_1+(a_1+2d)$ hold, but I don't see how to use them.

Why is it important for the common difference of the AP to be $\ne0?$ Then we will have a constant sequence in which all terms are equal, right?

答:
For the geometric progression, we have $$a_1^2=(a_1+d)(a_1+2d)=a_1^2+3a_1d+2d^2.$$
Cancelling $a_1^2$ from both sides, we can see that $$d(3a_1+2d)=0.$$ If $d=0$, then $q=1$.

If $d=-\dfrac{3a_1}{2}$, then $a_1+d=-\dfrac{a_1}{2}$, $a_1+2d=-2a_1$. The common ratio $q=-2$.


答: This implies that $$a_1+d=\frac1qa_1$$$$a_1+2d=qa_1$$So, we have $$d=\left(q-\frac1q\right)a_1$$$$a_1\cdot\left(1+q-\frac1q\right)=\frac1qa_1$$Assuming $a_1\neq0$, which is pretty obvious, we have $$1+q=\frac2q$$You can solve from here.

Title: The $2$nd, the $1$st and the $3$rd term of an arithmetic progression form a geometric progression

Body:
An arithmetic progression is given with a common difference $\ne0.$ The $2nd$, the $1st$ and the $3rd$ term of the ap form a geometric progression. Find the common ratio.

So we have the ap: $a_1,a_1+d,a_1+2d$ and the gp: $a_1+d,a_1,a_1+2d.$ How can we find $q?$ Thank you in advance!

I have tried to solve the problem using a system, but I am not sure which equations we can use. For example, $a_1^2=(a_1+d)(a_1+2d)$ and $\dfrac{a_1+d}{2}=a_1+(a_1+2d)$ hold, but I don't see how to use them.

Why is it important for the common difference of the AP to be $\ne0?$ Then we will have a constant sequence in which all terms are equal, right?


问:
How can I prove that $\sin (10^\circ), \sin(1^\circ), \sin(2^\circ), \sin(3^\circ), \tan(10^\circ)$ are irrational

Prove that $\sin (10^\circ)$, $\sin(1^\circ)$, $\sin(2^\circ)$, $\sin(3^\circ)$, and $\tan(10^\circ)$ are irrational.

My Attempt:

Let $x = 10^\circ$. Then

$$
\begin{align}
x &= 10^\circ\\
3x &= 30^\circ\\
\sin (3x) &= \sin (30^\circ)\\
3\sin (10^\circ)-4\sin^3(0^\circ) &= \frac{1}{2}
\end{align}
$$

Now let $y = \sin (10^\circ)$. Then

$$
\begin{align}
3y-4y^3 &= \frac{1}{2}\\
6y-8y^3 &= 1\\
\tag18y^3-6y+1 &= 0
\end{align}
$$

How can I calculate the roots of $(1)$?

答:
Note: All that you want to show is that it's not rational.

Hint: Apply the rational root theorem. What possible rational roots are there? Now check if those are indeed roots.

ContentLicense: CC BY-SA 3.0


Body:
$a_1 = q(a_1 + d)$ and $(a_1 + 2d) = qa_1 = q^2 (a_1 + d)$

(So $q = \pm \sqrt{\frac{a_1 + d}{a_1 + 2d}}$.)

We can remark that $a_1 (1-q) = qd$ and $a_1 (1-q) = -2d$ so $-2d = qd$, so $q=-2$ if $d\neq 0$.


问:
Linear Regression: Expectation Proof
I found the following proof in my notes:

$E(Y_i) = E[\beta_0 + \beta X_i + \varepsilon_i] =\cdots= \beta_0 + \beta X_i$.  This does not seem right to me, however.  Why would $E(\beta_1 X_i) = \beta_1 X_i$?  I wonder if i might have written it down incorrectly, with the actual proof meaning to be for the estimated value Yi hat (I don't know how to code this unfortunately).  Does anyone recall this property of linear regression?

答: If $Y_i=\beta_0+\beta X_i+\epsilon_i$, where $\beta_0$ and $\beta$ are constants and $\epsilon_i$ is an "error" random variable with mean $0$, then $E(Y_i)=\beta_0+\beta E(X_i)$. </p>

Title: Linear Regression: Expectation Proof


答:
Hint/Comment: Is easier to control the opposite problem, meaning you place at least $i$ balls in the $i-$th bin. Call $$A = \{x=(x_1,\cdots , x_k):\sum _{i=1}^kx_i=n\},$$ and $A_i = \{x\in A:x_i\geq i\}$ then you want exactly $$\left |A\setminus \bigcup _{i=1}^kA_i\right |=\sum _{i=0}^k(-1)^i\sum _{X\in \binom{[k]}{i}}\left |\bigcap _{x\in X}A_x\right |.$$
Use Stars and Bars to show that $|A|=\binom{n+k-1}{k-1}$ and $|A_i|=\binom{n-i+k-1}{k-1}$ so, in general, $|\bigcap _{x\in X}A_x|=\binom{n-\sum _{x\in X}x+k-1}{k-1}.$

If you are interested in going further, notice that the number of times $\binom{n-\ell+k-1}{k-1}$ appears in that sum is the number of ways that you can write $\ell$ with distinct parts.

Title: Variant of balls bin problem

Body:
I am stuck at the following variant of the balls bin problem.

The task is to count the number of ways of placing $n$ identical balls into $k$ ordered bins under the constraint: the bin number $i$ must have strictly less than $i$ balls.

Any help/suggestions would be appreciated.


问:
How to solve equation for a variable that is inside and outside an exponent?
I have this equation

$$y=x\exp\big(a(1−xk)\big).$$

It is possible to solve for $x$?
I can see from this post that sometimes there is no solution. Is this also the case? Is there a numerical approach that can be used?
Thank you

答:
$\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$

The solution in terms of the Lambert $\W$ function is found as follows:

\begin{align}
x\exp(a(1-xk))&=y
,\\
x\exp(a-akx)&=y
,\\
x\exp(a)\exp(-akx)&=y
,\\
x\exp(-akx)&=y\exp(-a)
,\\
-akx\exp(-akx)&=-aky\exp(-a)
,\\
\W\left(-akx\exp(-akx)\right)&=W(-aky\exp(-a))
,\\
-akx&=W(-aky\exp(-a))
,\\
x&=-\frac1{ak}\,W(-aky\exp(-a))
.
\end{align}

The number of real solutions is defined by the value of the argument
$-aky\exp(-a)$ of $\W$.

$\endgroup$


问:
normal states on a type $III_{\lambda}$ factor
Let $M$ be of a type $III_{\lambda}$ factor with separable predual.

If $\lambda=1$, given two normal states $\tau_1$ and $\tau_2$ on $M$.For any $\epsilon>0$, there exists a unitary $u\in M$ such that $\|\tau_1–u\tau_2 u^*\|<\epsilon$.

If $\lambda\neq 1$, does there exist similiar conclusions?

答:
The paper you are reading is devoted to prove that in a III$_\lambda$-factor $M$
$$
\operatorname{diam}(S_0(M)/\operatorname{Int}(M))=2\,\frac{1-\lambda^{1/2}}{1+\lambda^{1/2}}.
$$
This shows that what you want can only happen in a type III$_1$-factor.

Edit: why faithful states are enough. Let $\phi,\psi$ be states, with supports $p$ and $q$. Because we are in a type III factors, $p$ and $q$ are equivalent; in fact, unitarily equivalent. So $q=vpv^*$ for some unitary $v$. Then the state $v\psi v^*$ has support $p$ and is equivalent to $\psi$. This reduces the problem to states with the same support $p$. And these are faithful in $pMp$.


问:
A function that is both continuous and discontinuous
I was asked to find a function that is continuous on Z but discontinuous on R\Z.
as I'm new to continuity I want a feedback on the function I've created, and also tips on what to look for in these types of question!

$f: \mathbb{R}\to \mathbb{R}$ such that
\begin{align}
f(x) =
\begin{cases}
1 & \text{if $x\notin \mathbb{Z}$} \\
\lfloor x \rfloor & \text{if $x \in \mathbb{Z}$}
\end{cases}
\end{align}

答:
Actually, the function is discontinuous at every integer other than $1$.

You can take, for instance,$$f(x)=\begin{cases}\sin(\pi x)&\text{ if }x\in\Bbb Q\\0&\text{ otherwise.}\end{cases}$$Can you check that it works?


问:
Why is it important to have a discrepancy between image and codomain?
A function $f:\mathbb{R}\rightarrow \mathbb{R}$ given by $f(x)=x^2$ has $\mathbb{R}_{\geq0}$ as its image and $\mathbb{R}$ as its codomain.

What's the need for this discrepancy? Why don't we just write $f:\mathbb{R}\rightarrow \mathbb{R}_{\geq0}$?

I believe the reason for this is that there may be some functions in which stating the image this way may be impractical - if this hypothesis is true, what is this class of functions? If yes or no, is there any other reason for this?

答:
I think the point of a co-domain is so we can say things like

$f : X \to Y$ is a surjection, as in Samuel's answer

$j : V \to M$ is an elementary embedding (of structures)

$f : X \to Y$ is an embedding (of topological spaces).

Considering a function as a statement about ordered pairs, these statements are not just statements about $f$, but also about the given codomains. If we wanted to say the same things without using the notion of co-domain, we could say things like

$f$ is a function on $X$ whose range is $Y$

$j$ is an embedding from $V$ to an elementary substructure of $M$

$f$ is a homeomorphism from $X$ to a subspace of $Y$.

However, such statements are often more cumbersome than the natural formulations using co-domains.


问:
Ito's lemma 2nd order term notation.
I have a notation question here.

In the simplest form of Ito's lemma, we have this

$ df(Y_t) = f'(Y_t) dY_t + \frac{1}{2} f''(Y_t) d\langle Y \rangle_t$

I know how to calculate the $ d\langle Y \rangle_t $ term, but I always want to ask

what is the name of the term, and what exactly it means?

why is it written in such a special way but not using $ Cov() $, or $ Var() $?

Conceptually to me that's the variance of the process but I just don't understand the notation. Why the subscript $ t $ is being put outside the $ \langle \cdot \rangle $.

Can I write it like any of these below?

$ \langle dY_t \rangle $

$ d \langle Y_t \rangle $

If there are two processes involved, following the pattern I guess it should be written like this $ d\langle X, Y \rangle_t $, but can I write it like these below?

$ \langle dX_t, dY_t \rangle $

$ d\langle X_t, Y_t \rangle $

Also can I write it in integral form? Where should I put the $ t $ if I am writing it in integral form?

Thanks a lot

答:
Long-hand / Short-hand notation:

I personally have always found the short-hand notation confusing and to this day try to avoid it whenever possible. Below, I will try to demonstrate why it is confusing and leads to commonly made mistakes.

In the "long-hand" notation, an Ito process $X_t$ is defined as follows:

$$X_t:=X_0+\int_{h=0}^{h=t}a(X_h,h) dh + \int_{h=0}^{h=t}b(X_h,h) dW_h $$

Above, $a(X_t,t)$ and $b(X_t,t)$ are some square-integrable processes.

It is worth noting that the Quadratic variation of $X_t$ would then be:

$$\left<X\right>_t=\int_{h=0}^{h=t}b(X_h,h)^2dh $$

(this follows from the definition of Quadratic variation for Stochastic Processes, see edit at the end of this post)

Now, in short-hand notation, we can write the equation for $X_t$ above as:

$$dX_t=a(X_t,t) dt + b(X_t,t) dW_t$$

Firstly, what does the short-hand notation really mean? We could define $\delta X_t$ as follows:

$$\delta X_t:=X_t-X_0=\int_{h=0}^{h=\delta t}a(X_h,h) dh + \int_{h=0}^{h=\delta t}b(X_h,h) dW_h$$

And then $dX_t$ could be (intuitively, not rigorously) understood along the lines of:

$$\lim_{\delta t \to 0} \delta X_t = dX_t$$

But I think it's best to just understand the short-hand notation for what it really is: i.e. a short-hand for the stochastic integrals.

Ito's Lemma:

Now Ito's Lemma states that for any such Ito process $X_t$, any twice-differentiable function $F()$ of $X_t$ and $t$ would obey the following equation:

$$F(X_t,t)=F(X_0,t_0)+\int_{h=0}^{h=t} \left( \frac{\partial F}{\partial t}+\frac{\partial F}{\partial X}*a(X_h,h) + 0.5\frac{\partial^2 F}{\partial X^2}*b(X_h,h)^2 \right)dh+\int_{h=0}^{h=t}\left(\frac{\partial F}{\partial X}b(X_h,h)\right)dW_h$$

Above, you can spot the "quadratic variation" term:

$$\int_{h=0}^{h=t}0.5\frac{\partial^2 F}{\partial X^2}b(X_h,h)^2 dh$$

(which, in "short-hand" notation could be written as $0.5F''(X_t)d\left<X\right>_t$, i.e. exactly the same as yours $0.5f''(Y_t) d\langle Y \rangle_t$, I just use $F$ instead of $f$ and $X_t$ instead of $Y_t$: again, I find the short-hand much less intuitive than the long-hand notation, even after years of playing around with Ito processes).

Why not to use Short-hand notation

Now I would like to show an example of why I think the short-hand notation can be super-confusing: Let's go with the Ornstein-Uhlenbeck process (below, $\mu$, $\theta$ and $\sigma$ are constant parameters):

$$X_t:=X_0+\int_{h=0}^{h=t}\theta(\mu- X_h)dh + \int_{h=0}^{h=t}\sigma dW_h $$

We have $a(X_t,t)=\theta(\mu- X_h)$ and $b(X_t,t) = \sigma$.

The trick to solving the above is to apply Ito's lemma to $F(X_t,t):=X_t e^{\theta t}$, which gives:

$$X_te^{\theta t}=F(X_0,t_0)_{=X_0}+\int_{h=0}^{h=t} \left( \frac{\partial F}{\partial t}_{=\theta X_h e^{\theta h}}+\frac{\partial F}{\partial X}_{=e^{\theta h}}*a(X_h,h) + 0.5\frac{\partial^2 F}{\partial X^2}_{=0}*b(X_h,h)^2 \right)dh+\int_{h=0}^{h=t}\left(\frac{\partial F}{\partial X}_{=e^{\theta h}}b(X_h,h)\right)dW_h=\\=X_0+\int_{h=0}^{h=t}\left(\theta X_h e^{\theta h}+e^{\theta h}\theta(\mu- X_h)\right)dh+\int_{h=0}^{h=t}\left(e^{\theta h} \sigma\right)dW_h=\\=X_0+\int_{h=0}^{h=t}\left(e^{\theta h}\theta\mu\right)dh+\int_{h=0}^{h=t}\left(e^{\theta h} \sigma\right)dW_h$$

Now, to get the solution for $X_t$, the final step is simply to divide both sides by $e^{\theta t}$, to isolate the $X_t$ term on the LHS, which gives:

$$X_t=X_0e^{-\theta t}+\int_{h=0}^{h=t}\left(e^{\theta(h-t)}\theta\mu\right)dh+\int_{h=0}^{h=t}\sigma e^{\theta(h-t)} dW_h$$

I have seen many people trying to solve the Ornstein-Uhlenbeck writing everything out using the "short-hand" notation, and in the last step, when we divide through by $e^{\theta t}$, I have seen people "cancelling out" the terms that would normally be written as $e^{\theta h}$ inside the integrals: because the short hand notation fails to distinguish between what is an integration dummy variable (i.e. "$h$") and what had already been integrated to "$t$".

In conclusion, I wouldn't recommend using the short hand notation for SDEs, and if you come across it, I would encourage "translating it" into what it really means (i.e. the "long-hand" notation): at least for me, it has made things a lot easier to comprehend.

Edit on Quadratic Variation: Quadratic variation for Stochastic Processes is defined as a limit in Probability as the mesh-size gets finer and finer, specifically for a Brownian motion, we could write $\forall \epsilon > 0$:

$$\left<W\right>_t:=\lim_{n \to \infty} \mathbb{P}\left(\left|\sum_{i=1}^{i=n}\left(W_{t_i}-W_{t_{i-1}}\right)^2-t\right|>\epsilon\right)=0$$

I.e. the probability that the Quadratic variation converges to $t$ goes to 1 as the mesh size gets infinitely fine (the proof is rather technical, see for example here, where they actually seem to prove convergence almost surely (which implies convergence in probability)).

Notice that we can then simply write:

$$t=\int_{h=0}^{h=t}dh$$ and thereby obtain the well-known formula:

$$ \left< W \right>_t=\int_{h=0}^{h=t}dh=t$$

Title: Ito's lemma 2nd order term notation.


答:
You see, the basic idea for an AP is the common difference .
Similarly, for the GP(geometric progression) the basic idea is the common ratio .

Now,
If you assume an AP (a, a+d, a+2d) then you get the GP as-
(a+d, a, a+2d) .
Here the common ratio will always be same.... So,
Your first approach may be to equate the terms, find a relation and get the values and this completely works!

Now coming to the method you used............
(a+d)²=a(a+2d) is the correct and the same equation which you get if you apply the method I suggested above.
This is because of the fact that the equation you used is the formula for the geometric mean which is derived from the way I suggested......
But you see,
The second equation you wrote is not correct.
The equation should have been this-
2(a+d)=a+(a+2d) and let me clear you that this formula won't help you do this question as it is always true for any a and d .

Now if you ask how to solve this question, then you should understand that we have one equation and two variables, so there are infinite solutions of this equation(infinite values of a and d).
So you just need to find a relation between a and d(which you will get from the equation) and then form the GP which involves only one variable (either a or d).......
Then find the common ratio of that GP......
As easy as that........        :)


问:
Do there exist planar curves that can be constructed from geometric figures beyond conic sections?
Conic sections are algebraic curves that can be constructed from different cross-sections of a cone in $\mathbb{R}^3$. Do there exist other geometric shapes in $\mathbb{R}^3$ besides the cone whos cross-sections can yield algebraic curves analogous to conic sections? Are there higher dimensional equivalents for 3D curves or sheets in 3D that can be defined in by some geometric figure like a cone (or potential alternative geometric figures) in $\mathbb{R}^4$?

Maybe I can be a bit more concrete. Conic sections are an example of a two-variable second degree polynomial of the form
$$
Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0.
$$
I can construct a similar two-variable higher-degree polynomials as such
$$
\text{Curve}_n(x,y) =
A_0x^n + A_2x^{n-1}y + \ldots + A_{n-1}xy^{n-1} + A_ny^n + \text{Constant} = 0
$$
Does there exist something analogous to a "cone" for such a polynomial? What if I add another variable $z$?

答:
Let us consider a torus (2nd figure) with a vertical axis of revolution with $c$ = main radius, $d$ = radius of revolving circle ($c>d$ for a non-self intersecting torus).

The sections by vertical planes at distance $f$ from the axis of revolution are called "spiric sections".

For certain combinations of $c,d,f$ (see Appendix below) we obtain Cassinian ovals (first figure). Recall: A Cassinian oval $(C_b)$ with foci $F_1, F_2$ is the set of points $M$ such that $MF_1.MF_2=b^2$.

Please note 1) that this locus has 2 connected components if $b<\frac{F_1F_2}{2}$, 2) That the famous Bernoulli's lemniscate is the particular oval where $b=\frac{F_1F_2}{2}$ 3) the similarity with the definition of an ellipse (with product of lengths instead of their sum).

There are other particular curves of intersection with planes: circles... (meridians, parallels and Villarceau circles) but as well other curves like the hippopede(s).

Reference: https://www.lucamoroni.it/toric-sections/

Appendix: When vertical sections of a torus are Cassinian ovals ?

The torus equation is

$$(\sqrt{x^2 + y^2} - d)^2 + z^2 = c^2$$

Expanding the square, isolating the square root and squaring again, we can give it the fourth degree implicit polynomial equation:

$$(x^2 + y^2 + z^2-c^2)^2=4d^2(x^2+y^2)\tag{1}$$

Besides, by definition, the Cassinian oval implicit equation in a vertical plane parallel to any plane with equation $y=k$ is

$$\sqrt{(x-a)^2+z^2}\sqrt{(x+a)^2+z^2}=b^2$$

Squaring this relationship and expanding it, we get:

$$(x^2+z^2+a^2-2ax)(x^2+z^2+a^2+2ax)=b^4$$

$$(x^2+z^2+a^2)^2-4a^2x^2=b^4\tag{2}$$

It remain to express that the section of (1) by vertical plane with equation $y=f$ has equation:

$$(x^2 + f^2 + z^2-c^2)^2=4d^2(x^2+f^2)\tag{3}$$

Therefore (2) and (3) can be identified under the following conditions:

$$f^2-c^2=a^2, \ \ a=d, \ \ b^2=2df.$$

meaning that if $c,d$ are given,

$$f=\sqrt{c^2+d^2} \ \ \text{and} \ \   b=\sqrt{2df}$$

Therefore, for a given torus, there is exactly one vertical section that can be a Cassinian oval ; moreover, due to the fact that $f>c$, this section has a single component.

Edit: Another example, this time with a cubic curve called the folium of Descartes, with equation:

$$x^3+y^3-3xy=0$$

that can be considered as the intersection of surface defined by $$z=-\operatorname{atan}(x^3+y^3-3xy)$$ with plane $z=0$ (see Fig.) (with the interest that the surface is bounded ($-\dfrac{\pi}{2}<z<\dfrac{\pi}{2}$).

We could as well have taken $z=x^3+y^3-3xy$ or $z=|x^3+y^3-3xy|$, etc.


问:
Proof Verification : Continuous mapping is determined by the values on a dense subset.
Just wanted to check this proof real quick, I haven't found this solution.

$E$ is dense in $X$. Let $f,g : X \rightarrow Y$ be continuous maps. ( X and Y are metric spaces). Suppose $f(p)=g(p) \forall p \in E$. Then $f(x)=g(x) \forall x \in X$

My solution: Fix $x \in X/E$ and $f(p)$ in $f(X)$. Let $\varepsilon$ be arbitrary. Then there is a $\delta_f$ such that $d_Y(f(p),f(x)) < \varepsilon$ whenever $d_X(p,x)< \delta_f$. Similarly there is a $\delta_g$ such that $d_Y(g(p),g(x))<\varepsilon$ whenever $d_X(p,x)<\delta_g$.

Set $\delta = \mathrm{min}(\delta_g, \delta_f)$. Then for every $x$ such that $d_X(x,p)< \delta$, we have $$d_Y(g(x),f(x))\leq d_Y(g(x),g(p)+d_Y(g(p),f(x))$$
$$=d_Y(g(x),g(p)+d_Y(f(p),f(x))$$ $$< 2\varepsilon$$.

Arbitrariness of $\varepsilon$ implies the result.

Thanks for your help!!

P.S : All the solutions I have seen till now involve sequences so I wanted to see if this works.

答:
Your idea is basically correct except some typos in your proof. Let me reorganize this for you:

Fix an arbitrary $x\in X\setminus E$ and $\varepsilon>0$. Since $f$ is continuous, there exists $\delta_f>0$ such that $$d_X(x,x')<\delta_f\implies d_Y(f(x),f(x'))<\varepsilon.$$
Similarly, we can find a $\delta_g>0$ such that $$d_X(x,x')<\delta_g\implies d_Y(g(x),g(x'))<\varepsilon.$$
Note we set $\delta:=\min\{\delta_f,\delta_g\}$. Since $E$ is dense in $X$, there exists $p\in E$ such that $d_X(x,p)<\delta$. As a result, by triangle inequality, (You also made a mistake here.)$$d_Y(f(x),g(x))\leq d_Y(f(x),f(p))+d_Y(f(p),g(p))+d_Y(g(x),g(p))<2\varepsilon.$$
By setting $\varepsilon\to 0$, we indeed have $f(x)=g(x)$. Q.E.D.

P.S. An alternative way is to show that the fixed point set $\{x\in X\mid f(x)=g(x)\}$ is closed, or equivalently the set $\{x\in X\mid f(x)\neq g(x)\}$ is open. Recall that the closure of any dense subset is the whole space.


问:
off diagonal Ramsey number (4,k) lower bound probabilistic method asymptotic reasoning
I want to show that $R(4,k)\ge\Omega((k/\ln k)^2)$, where $R(4,k)$ is the Ramsey number.

This question is quite close to what I'm after, the asymptotic part is only missing (and they talk about $3$ instead of $4$).

Similar to that question we define $Y$ and $Z$ as the number of $4-$cliques and the number of empty (no edges) sets of size (number of vertices) $k$ in a random Erdos-Renyi graph (a graph over $n$ vertices with edge probability $p$).<-- All of that is written in the answer to the quoted question.

So here is what I did to show that $n-{n\choose 4}p^6-{n\choose k}(1-p)^{k\choose 2}=\Omega((k/\ln k)^2)$.

Note: $p^6$ comes from an analogous argument as in the quoted question, $6$ is the number of edges in the complete graph on $4$ vertices. And I also suppose writing $\ge\Omega (...)$ is redundant, equality is OK.

First I restrict $n$ to be of the form $\frac{k^2}{(\ln k)^2}$ and I set $p:=1/n$. We get $${n\choose 4}{p^6}\le n^4p^6=n^{-2}\ll n$$ For the second term we have $${n\choose k}(1-p)^{k\choose 2}\le \frac{n^k}{k!}(1-\frac{\frac{1}{2}(\ln k)^2}{\frac{1}{2}k^2})^{k(k-1)/2}\\ \sim \frac{n^k}{k!} (\frac{1}{\sqrt k})^{\ln k}$$

This divided by $n$ is $\frac{n^{k-1}}{k!}(\frac{1}{\sqrt k})^{\ln k}$ which we want to go to zero. This would imply it is $o(n)$.

This value is equal to $e^{2(k-1)\ln k-2(k-1)\ln\ln k -\ln k!-\frac{1}{2}\ln k\ln k}$ where the exponent goes to $-\infty$, which conlcludes the proof.

But I'm afraid I only showed $n-{n\choose 4}p^6-{n\choose k}(1-p)^{k\choose 2}= n-o(n)$ which is not the same as showing it is $\Omega(n)$.

Although I (think I) did that $n-{n\choose 4}p^6-{n\choose k}(1-p)^{k\choose 2}=\Theta (n)$ which is a subclass of $\Omega(n)$.

答:
If the asymptotic claims you've made really were true, then your proof would be fine. In particular:

Yes, $n - o(n)$ is $\Omega(n)$. Even showing that $\binom nk (1-p)^{\binom k2} < 0.99n$ for large enough $n$ would be fine, and showing that it's $o(n)$ is stronger than that.

Yes, assuming $n = (\frac{k}{\log k})^2$ without a constant is fine - if it works. We set $n = c (\frac{k}{\log k})^2$ for flexibility later on, so that we can choose a $c$ that will make our argument work. If $c=1$ happens to be fine, that's okay.

However, your asymptotic claims are false. In particular, $\frac{n^{k-1}}{k!}(\frac{1}{\sqrt k})^{\log k} \to \infty$ as $k \to \infty$. That's because

$\log (n^{k-1}) = 2(k-1) \log k - 2(k-1) \log\log k \sim 2k \log k$ as $k \to \infty$;

$\log(k!) \sim k \log k$ as $k \to \infty$ (this follows from Stirling's formula but here, it's enough that $\log(k!) < \log (k^k) = k \log k$);

$\log\left(\sqrt{k}^{\log k}\right) \sim \frac12(\log k)^2 \ll k \log k$ as $k \to \infty$.

So you have $2 k \log k$ to start with in the exponent, and cancel out roughly $k \log k$ of it.

Your primary mistake is setting $p = \frac1n$. This is way too small. You need to set $p$ to be just small enough that you don't have to worry about the $\binom n4 p^6$ term; in particular, $p = \frac1{\sqrt n}$ is good enough. The larger $p$ is, the easier it will be to deal with the last term, so we don't want to make $p$ any smaller.

However, even then, setting $n = (\frac{k}{\log k})^2$ will not work - the constant matters here! I can tell you that $n = \frac14  (\frac{k}{\log k})^2$ and $p = \frac1{\sqrt n}$ will work; you should do the asymptotic analysis of that case yourself. You can get better results by playing around with the constants on $n$ and $p$.


问:
Is the class of planar graphs axiomatizable?
In first order logic, consider the language $L$ consisting of one binary relation symbol. Consider the class $P$ of binary relations satisfying the following properties:

Antireflexive

Antisymmetric

Planar (the relations can be drawn on a plane as digraphs without the edge-intersection)

Is $P$ (finitely/decidably) axiomatizable?

Can we use here the fact that finite graphs are planar iff they do not contain some subdivision (obtained by adding vertices to the edges) of $K_5$ and $K_{3,3}$?

答:
For any finite graph $G$, there is a sentence $\varphi_G$ which holds exactly in the graphs containing no copy of $G$ as a (possibly non-induced) subgraph. Basically, fix an enumeration $\{a_i: 1\le i\le n\}$ of the vertices of $G$, let $X\subseteq\{1,...,n\}^2$ be the pairs of indices corresponding to edges in $G$, and let $$\varphi_G\equiv \neg\exists x_1,...,x_n([\bigwedge_{1\le i<j\le n}x_i\not=x_j]\wedge [\bigwedge_{\langle i,j\rangle\in X}x_iEx_j]).$$ This is very similar to the construction of a sentence characterizing a given finite structure (in a finite language) up to isomorphism.

The construction of $\varphi_G$ from $G$ is computable, and so we get that for any computable set of finite graphs $\mathcal{G}$ there is a computable theory $T_\mathcal{G}$ whose models are exactly the graphs not containing any element of $\mathcal{G}$ as a subgraph. The characterization of planarity by forbidden subgraphs corresponds to such a $\mathcal{G}$ (namely the set of all subdivisions of $K_5$ and $K_{3,3}$), and so we get:

The class of planar graphs is computably axiomatizable.

Of course this leaves open the question of finite axiomatizability. In fact the class of planar graphs is not finitely axiomatizable, and this can be shown by proving that the class of non-planar graphs is not axiomatizable at all. This in turn can be proved via compactness - think about "stretching" a copy of (say) $K_5$ by subdividing the edges more and more. (Or if you prefer, run an ultraproduct construction: let $G_0=K_5$, let $G_{i+1}$ be gotten from $G_i$ by adding a vertex in the middle of each edge, and consider $\prod_{i\in\mathbb{N}}G_i/\mathcal{U}$ for some nonprincipal ultrafilter $\mathcal{U}$ on $\mathbb{N}$.)


答:
The three terms of a geometric series $a_0, a_1= a_0 + d, a_2 = a_0+2d$ or if it is easier $a_0= a_1-d, a_1, a_2= a_1 + d$.

The three terms of a geometric series is $b_0, b_1=b_0\cdot r, b_2=b_0\cdot r^2$ or if it is easier $b_0=\frac {b_1}r, b_1, b_2= b_1\cdot r$.

We are told that $a_1, a_1 -d, a_1 +d$ form a geometric series. so

$\frac {b_1}r = a_1$.  $b_1 = a_1-d$, $b_1r = a_1 + d$.

So Solve for $r$.

========

substitute $b_1$ with $a_1 -d$ and

$\frac {a_1 -d}r = a_1$ and $(a_1 -d)r = a_1 + d$.

So $r = \frac {a_1 -d}{a_1} = 1 - \frac d{a_1}$ and $r = \frac {a_1 + d}{a_1-d}=1 + \frac {2d}{a_1-d}$

So $-\frac d{a_1} = \frac {2d}{a_1-d}$ so $-\frac 1{a_1} = \frac 2{a_1-d}$ so

$-(a_1 -d) = 2a_1$ so $d = 3a_1$.

Substituting $3a_1$ for $d$ we get

$r = \frac {a_1 -d}{a_1}  = \frac {-2a_1}{a_1} = -2$ and  $r = \frac {a_1 + d}{a_1-d}= \frac {4a_1}{-2a_1} = -2$.

So $r = -2$.

......

Furthermore the three terms as an arithmetic series $a_0 = a_1-d =-2a_1; a_1 = a_1; a_2= a_1 + d = 4a_1$.

And as a geometric series they are $b_0 = a_1; b_1 =-2a_1; b_2 = 4a_1$.


问:
Integral of $\sqrt{1 + \sqrt{x}}$
My professor wants us to do this problem to refresh ourselves with substitution. We have to solve:

$$\int\sqrt{1 + \sqrt{x}}\,\mathrm dx$$

  $$\int\sqrt{1 + \sqrt{1 + \sqrt{x}}}\,\mathrm dx$$

  ...etc...

If someone could point me in the right direction with the first one, I think I can handle the others.

Thanks for the help guys!

答: Let $f(y):=\sqrt{1+y}$, with inverse $g(y):=y^2-1$. You want to evaluate$$\int f^{\circ n}(\sqrt{x})dx$$(the exponent with $\circ$ denoting repeated application of a function), where your first integral has $n=0$. With$$u=f^{\circ n}(\sqrt{x})\iff x=(g^{\circ n}(u))^2$$(where the ${}^2$ denotes squaring), integration by parts proves the result is$$xu-\int2xg^{\circ n}(u)\prod_{k=0}^{n-1}g^\prime(g^{\circ k}(u))du.$$The last integral is just a polynomial in $u$, albeit one we'd have to work out for a specific (hopefully small, or it'll take ages) $n$.

Title: Integral of $\sqrt{1 + \sqrt{x}}$


问:
$U \cap \operatorname{cl}(A) \neq \emptyset$ implies $U \cap A \neq \emptyset$, is $U$ open?
In a metric space $(X,d)$, let $U\subseteq X$. If $$U \cap \operatorname{cl}(A)\neq \emptyset \implies U\cap A\neq \emptyset$$ for all set $A\subseteq X$ , is $U$ open? (Notation: $\operatorname{cl}(A)$ is the closure of $A$)

I am currently stuck at this problem for a long time, the converse is easy and there are many similar ones out there.

Any proofs or hints? Or any counter examples if not true? Thanks in advance.

答: Let $x \in U$. If $x$ is not an interior point of $U$, this means that every neighbourhood of $x$ intersects $X\setminus U$. So $x \in A \cap \operatorname{cl}(X\setminus U)$. By the assumption on $U$, applied to $A=X\setminus U$, we would conclude that $U \cap (X\setminus U) \neq \emptyset$, a contradiction. It follows that $x$ is an interior point of $U$ and as $x$ was arbitrary, $U$ is open. QED


问:
Base for Y topology
Be $X$ a topological space, $B$ a base of open for $X$ and $Y \subseteq X$. Prove that $B_Y = \{U \cap Y : U \in B\}$ is an open base for $Y$ (considering, about $Y$ , the subspace topology).

Be $A\in \tau_{Y}$, we need to show that there is a $U_i\cap Y$ such as $A=\bigcup_{i\in I} (U_i\cap Y)$. In fact if $\mathcal{B}$ is an open base for $X$ then for all open $U\in \tau_{X}$ there is a collection of open $B_i\subset \mathcal{B}$ such as $U=\cup_{i\in I} B_i$. Thus, $U_i \in B$. How can I show $Y\cap B_i\in B_Y$?

答: Because the $B_i$ are from $\mathcal{B}$, $B_i \cap Y \in \mathcal{B}_Y$ by definition of that collection.

Title: Base for Y topology


问:
problem on matrix derivatives
If I have two $3\times3$ rotation matrices $R_1$ and $R_2$, what is the matrix derivatives of $\frac{\partial(R_1 R_2^T)}{\partial R_1}$?

Also, I have found a website which can answer my question

http://www.matrixcalculus.org/
The following is the answer:

I am not sure whether this is the correct answer? Also,I don't understand the meaning of the notation ∏. Can someone explain it for me？Thanks

答:
$\def\E{{\cal E}}\def\pcolor#1#2#3{\frac{\partial #1}{\color{#3}{\partial #2}}}$
Consider the fourth-order tensor $\E$ whose components
(in terms of Kronecker deltas) are
$$\E_{ijk\ell} = \delta_{ik} \delta_{j\ell}$$
This tensor is useful for rearranging matrix products, e.g.
$$ABC = A\E C^T:B$$
where a colon denotes the double-contraction product
$$\eqalign{
{\cal C} &= {\cal A}:{\cal B} \\
{\cal C}_{ijpq}
  &= \sum_{k=1}^m\sum_{\ell=1}^n {\cal A}_{ijk\ell}\,{\cal B}_{k\ell pq} \\
}$$
Then consider the following matrix-valued function and its gradient
$$\eqalign{
F &= F(X) &\doteq XA \\
dF &= dX\,A &= \E A^T\color{red}{:dX} \\
\pcolor{F}{X}{red} &=\E A^T&\qquad\big({\rm tensor\,gradient}\big) \\
}$$
Another way to tackle the problem is to use vectorization, which is what the website has implicitly done
$$\eqalign{
{\rm vec}(dF) &= (A^T\otimes I)\,{\rm vec}(dX) \\
\pcolor{\big({\rm vec}(F)\big)}{\big({\rm vec}(X)\big)}{black}
  &= A^T\otimes I \;\qquad\big({\rm matrix\,gradient}\big) \\
}$$
where $\otimes$ denotes the Kronecker product and $I$ is the $3\times 3$ identity matrix.

In either case, setting $$X=R_1 \qquad A=R_2^T$$ yields the desired gradient.

However, neither solution takes into account the orthogonality constraint
$$\eqalign{
X^TX = I \quad\implies\quad dX &= -X\;dX^TX \\
}$$

Title: problem on matrix derivatives


问:
Geometric Sums - Equivalency
Where $ar^0+ar^1+ar^2+ar^3+ar^4...=\sum_{n=0}^\infty ar^{n-1}=\frac{a(1-r^n)}{(1-r)}$

My question is: by this same definition, is it also true that $\sum_{n=0}^\infty ar^{n}=\frac{a(1-r^{n+1})}{(1-r)}$?

答:
I assume $a \neq 0$, and $|r|<1$ then:

$$\sum_{n=1}^{+\infty} ar^{n-1} = \dfrac{a}{1-r}$$

This is the same thing as:

$$ \sum_{n=0}^{+\infty}ar^n = \dfrac{a}{1-r}$$

More generally if $N \in \mathbb{N}$ and $q_0 \in \mathbb{N}$ such that $q_0\leq N$ then:

$$ \sum_{n=q_0}^N ar^{n} = ar^{q_0}\dfrac{(1-r^{N-q_0+1})}{1-r}$$ and by changing the indexes:
$$ \sum_{n=0}^{N-q_0} ar^{n+q_0} = ar^{q_0}\dfrac{(1-r^{N-q_0+1})}{1-r}$$ so : $$\forall q_0 \in \mathbb{N}, \quad \sum_{n=q_0}^{+\infty}ar^n = \sum_{n=0}^{+\infty} a r^{n+q_0} =\dfrac{ar^{q_0}}{1-r}$$

Title: Geometric Sums - Equivalency


问:
I have a question about the dimension of the Kernel of a transposed matrix
I've been studying linear algebra and I came across this reasoning that is certainly not correct, but I can't see why. If A is a matrix, I know that $$Row(A)^{\perp} = Kernel(A)$$. Consequently, $$Col(A)^{\perp} = Kernel(A^T)$$ I also know that $$dim(Row(A)) = dim(Col(A))$$. It would be natural for me to assume that if the above is true, it should be true that $$dim(Row(A))^{\perp} = dim(Col(A))^{\perp}$$ This would imply that the dimension of the kernel of A (nullity of A) is equal to the dimension of the kernel of its transpose, which is not the case unless A is a square matrix. Where is the flaw in this reasoning? Thank you all in advance!

答: For subspaces $U \subset \Bbb R^m$ and $V \subset \Bbb R^n$, it is not the case that $\dim(U) = \dim(V)$ implies that $\dim(U^\perp) = \dim(V^\perp)$. Note that $\dim(U^\perp) = m - \dim(U)$ and $\dim(V^\perp) = n - \dim(V)$, so we will only have $\dim(U^\perp) = \dim(V^\perp)$ if $m=n$.


答:
As I said in the comment, you have all of the pieces, but you need to put them together a bit more clearly.

If $A\in\tau_Y$, there is a $U\in\tau$ such that $A=U\cap Y$. $\mathcal{B}$ is a base for $X$, so there is a family $\{B_i:i\in I\}\subseteq\mathcal{B}$ such that $U=\bigcup_{i\in I}B_i$, and therefore

$$A=U\cap Y=\left(\bigcup_{i\in I}B_i\right)\cap Y=\bigcup_{i\in I}(B_i\cap Y)\,.$$

By definition $B_i\cap Y\in\mathcal{B}_Y$ for each $i\in I$, so we’ve shown that $A$ is the union of members of $\mathcal{B}_Y$. $U$ was an arbitrary element of $\tau_Y$, so

every open set in $Y$ is the union of members of $\mathcal{B}_Y$.

Finally, $\mathcal{B}_Y\subseteq\tau_Y$, so

every union of members of $\mathcal{B}_Y$ is open in $Y$.

These two points together show that $\mathcal{B}_Y$ is a base for $\tau_Y$.

Title: Base for Y topology

Body:
Be $X$ a topological space, $B$ a base of open for $X$ and $Y \subseteq X$. Prove that $B_Y = \{U \cap Y : U \in B\}$ is an open base for $Y$ (considering, about $Y$ , the subspace topology).

Be $A\in \tau_{Y}$, we need to show that there is a $U_i\cap Y$ such as $A=\bigcup_{i\in I} (U_i\cap Y)$. In fact if $\mathcal{B}$ is an open base for $X$ then for all open $U\in \tau_{X}$ there is a collection of open $B_i\subset \mathcal{B}$ such as $U=\cup_{i\in I} B_i$. Thus, $U_i \in B$. How can I show $Y\cap B_i\in B_Y$?


问:
Big o notation on both sides
Can someone explain what exactly means $o(o(f(x))) = o(f(x))$?
I do understand what little oh means by definition(fraction limit of two functions converges to zero) , but somehow i don't get the idea of Big o or little o on both sides.

答:
I understand this expression in the following way: "any function $h(x)$ which is infinitely small compared to some other function $g(x)$, which, in turn, is infinitely small compared to function $f(x)$ (interpretation of left part of expression ends here) is itself infinitely small compared to $f(x)$".

Note, that "=" is not usual equality sign here, if you swap the left and right parts of this expression, it will not be correct any more.

UPDATE.

Well, in this particular example it IS possible to swap left and right parts of expression and it still will be correct. (Thanks to J.G. - author of another answer to this question).

In this example it is possible to interpret the expression as "set of functions described by the left part and the set of functions described by the right side are equal".

But I am quite sure I encountered expressions like $o(x^2) = o(x)$ meaning "our function is infinitesimal compared to $x^2$, so it is also infinitesimal compared to $x$"

Title: Big o notation on both sides


问:
Proving that an ideal in a PID is maximal if and only if it is generated by an irreducible

Let $R$ be a PID (Principal Ideal Domain) and $x$ is an element R. Prove that the ideal $\langle x\rangle$ is maximal if and only if $x$ is irreducible.

Ok, so I know what an irreducible is. I'm thinking that this problem is asking us to set up a proof by contradiction but I can't see how. No one in my study group has any clue.

答:
Hint $ $ Recall for principal ideals: $\ \rm\color{#0a0}{contains} = \color{#c00}{divides}$, $ $ i.e.  $\,\color{#0a0}{(a)\supseteq (b)}\iff \color{#c00}{a\mid b},\,$   thus having no proper $\rm\color{#0a0}{containing}$ ideal (maximal) is the same as having no proper $\rm\color{#c00}{divisor}$ (irreducible), $ $ i.e.

$\qquad\quad\begin{eqnarray} (p)\,\text{ is maximal}
&\iff&\!\!\ (p)\, \text{ has no proper } \,{\rm\color{#0a0}{container}}\,\ (d)\\
&\iff&\  p\ \ \text{ has no proper}\,\ {\rm\color{#c00}{divisor}}\,\ d\\
&\iff&\  p\ \ \text{ is irreducible}\\
\end{eqnarray}$


问:
Formula for the sum of the value of a rational function over roots of unity
Let $n,k$ be integers, and let $U$ be the set of all $n$-ths roots of unity (so there are exactly $n$ elements in $U$). Let $U'=U \setminus \lbrace 1 \rbrace$. Are there simple formulas (in terms of $n$ and $k$) for the sums

$$
S_k=\sum_{\varepsilon \in U'} \frac{\varepsilon^k}{\varepsilon-1}, \
T_k=\sum_{\varepsilon \in U'} \frac{\varepsilon^k}{(\varepsilon-1)^2}
$$

答:
Note that the sums $S_k$ and $T_k$ obviously only depend on $k$ modulo
$n$. So for simplicity we may assume wlog that $1\leq k\leq n$.

If $\varepsilon$ is any complex number, we have the algebraic identity

$$
\bigg(\sum_{j=0}^{n-2} (j+1-n)\varepsilon^j\bigg)
(\varepsilon-1)-n=n-\bigg(\sum_{j=0}^{n-1} \varepsilon^j\bigg) \tag{1}
$$

If, in addition, $\varepsilon\in U'$ then we deduce

$$
\frac{1}{\varepsilon-1}=\frac{1}{n} \sum_{j=0}^{n-2} (j+1-n)\varepsilon^j
$$

and hence

$$
S_k=\frac{1}{n}
\sum_{j=0}^{n-2} (j+1-n)\sum_{\varepsilon\in U'}\varepsilon^{j+k}
$$

and the sum $\sum_{\varepsilon\in U'}\varepsilon^{j+k}$ is $-1$, except when $j+k$ is divisible by
$n$, i.e. when $j=n-k$. So

$$
\begin{eqnarray}
nS_k &=& (n-1)(1-k)+\sum_{j\neq n-k}^{} (n-(j+1)) \\
&=& (n-1)(1-k)-(k-1)+\sum_{j=0}^{n-2} (n-(j+1)) \\
&=& n(1-k)+\sum_{j=0}^{n-2} (n-(j+1)) \\
&=& n(1-k)+\sum_{j'=0}^{n-1} j’ \\
&=& n(1-k)+n\frac{n-1}{2}=\frac{n(n+1-2k)}{2}.
\end{eqnarray}
$$

and hence $S_k=\frac{n+1-2k}{2}$. Similarly, if $\varepsilon$ is any complex number, we have the algebraic identity

$$
\bigg(\sum_{j=0}^{n-2} (j+1)(n-(j+1))\varepsilon^j\bigg)
(\varepsilon-1)^2-2n=\bigg(\sum_{j=0}^{n-1} \varepsilon^j\bigg)\big(
(n-1)\varepsilon-(n+1)
\big)
$$

If, in addition, $\varepsilon\in U'$ then we deduce

$$
\frac{1}{(\varepsilon-1)^2}=\frac{1}{2n} \sum_{j=0}^{n-2} (j+1)(n-(j+1))\varepsilon^j
$$

and hence

$$
T_k=\frac{1}{2n}
\sum_{j=0}^{n-2} (j+1)(n-(j+1))\sum_{\varepsilon\in U'}\varepsilon^{j+k}
$$

and the sum $\sum_{\varepsilon\in U'}\varepsilon^{j+k}$ is $-1$, except when $j+k$ is divisible by $n$, i.e. when $j=n-k$. So

$$
\begin{eqnarray}
(2n)T_k &=& (n-k+1)(k-1)(n-1)+\sum_{j\neq n-k}^{} (j+1)(n-(j+1))(-1) \\
&=& (n-k+1)(k-1)(n-1)+(n-k+1)(k-1) -\sum_{j=0}^{n-2} (j+1)(n-(j+1)) \\
&=& (n-k+1)(k-1)n -\sum_{j=0}^{n-2} (j+1)(n-(j+1)) \\
&=& (n-k+1)(k-1)n -\frac{n^3-n}{6}
\end{eqnarray}
$$

So we finally have
$$
T_k=\frac{6(n-k+1)(k-1)-(n^2-1)}{12}
$$


问:
Is preimage of interval by concave function a convex set?
I have a concave function $F:A \rightarrow \mathbb R$ and an interval $I=[\underline y,\bar y ]$ for which I consider the preimage

$$f^{-1}(I) := \{ x \in A \lvert f(x) \in I\} = \{ x \in A\lvert \ \underline  y \leq f(x) \leq \bar y \}$$

my question is how to show that $f^{-1}(I)$ is convex if that is indeed the case?

Obviously for $x_1$ and $x_2$ in $f^{-1}(I)$ it must be the case that for $x_3 = \alpha x_1 + (1-\alpha)x_2$ that

$$f(x_3) = f(\alpha x_1 + (1-\alpha)x_2) \geq \alpha f(x_1) + (1-\alpha)f(x_2)\geq \underline y$$

but whether $f(x_3) \leq \bar y$ eludes me.

答:
The answer is no.  Consider $f : \mathbb{R} \to \mathbb{R}$ given by $f(x) = -x^{2}$.  We know that $f^{-1}([-4,-16]) = [-4,-2]\cup[2,4]$, which is disconnected and so not convex.

On the other hand, if $f$ is concave, then its super-level sets are convex.  That is, $f^{-1}([c,\infty))$ is always convex, as you already noted.


问:
Vector Subspaces and Linear Maps
Let $E$ be the subspace of $\mathbb{R}^5$ defined by the equations

$x_1 + 2x_2 + 5x_5 = 0$

$x_2 + 3x_3 = 0$

Calculate a basis of $E$ and expand it to a basis of $\mathbb{R}^5$. Now we define a linear map $f: E \to \mathbb{R}^2$ such that $f(x_1, x_2, x_3, x_4, x_5) = (x_1 + x_5, x_2 - x_3)$.

I am asked to find the matrix $M$ associated to $f$ with respect to the basis of $E$ obtained in the last question and the standard basis of $\mathbb{R}^2$.

I have found that a basis for $E$ is $\{(-5,0,0,0,1), (0,0,0,1,0), (6,-3,1,0,0)\}$, but now i don't know how to find that matrix $M$, can somebody help me please.

Title: Vector Subspaces and Linear Maps

Body:
Since:

$f(-5,0,0,0,1)=(-4,0)$;

$f(0,0,0,1,0)=(0,0)$;

$f(6,-3,1,0,0)=(6,-4)$,

the matrix that you're after is$$\begin{bmatrix}-4&0&6\\0&0&-4\end{bmatrix}.$$


问:
Which distance gives rise to the geometric mean?
The arithmetic mean has the nice property of minimising the sum of squares, or in other words, minimising the sum of quadratic-euclidean distances. Formally, given a set of points $x_0, \dots, x_n \in \mathbb{R}^d$, the arithmetic mean, $\mu = \frac{1}{n} \sum_{i=1}^n x_i$, is the unique solution to the equation
$$\mu = \underset{x \in \mathbb{R}^d}{\operatorname{argmin}} \sum_{i=1}^n d^2(x_i, x) \, ,$$
where $d$ denotes the Euclidean distance. I was wondering if there exists a distance measure on $\mathbb{R}$ that has a similar property for the geometric mean, i.e.

Question

Is there a distance measure $d$ on $\mathbb{R}$ such that for any set of points $x_1, \dots, x_n \in \mathbb{R}$ we have
$$ \sqrt[n]{x_1 \cdot \dots \cdot x_n} = \underset{x \in \mathbb{R}^d}{\operatorname{argmin}} \sum_{i=1}^n d^2(x_i, x) \; \textbf{?}$$

答: Consider the metric $d:(0,\infty)^2 \rightarrow [0,\infty)$ defined by $$d(x,y)=\Bigg|\ln\Big(\frac{y}{x}\Big)\Bigg|$$ This is the metric you're looking for. Given a dataset $\{x_1,\ldots ,x_n\}\subseteq (0,\infty)$ define a function $f$ on $(0,\infty)$ by $$f(x)=\sum_{i=1}^nd^2(x,x_i)=\sum_{i=1}^n(\ln(x)-\ln(x_i))^2$$ Taking a derivative yields $$f'(x)=\frac{2}{x}\sum_{i=1}^n(\ln(x)-\ln(x_i))=0 \iff \ln(x)=\frac{1}{n}\sum_{i=1}^n\ln(x_i)$$ This shows $x=(x_1 \times \dots \times x_n)^{1/n}$ is a critical point of $f$. Now since $f$ is continuous on $(0,\infty)$, is bounded below by $0$, and $f(x)\rightarrow \infty$ as $x \rightarrow 0^+$ or as $x \rightarrow \infty$, we see that $f$ attains a global minimum at its critical point.


问:
Finding arc lenght of a curve in $R^3$
I'm trying to find the arc lenght of a curve in $R^3$. I know that that the formula to find this is $\int_a^b \sqrt{f'(t)^2 + g'(t)^2+h'(t)^2}dt$.

But I want to find the arc lenght between the points $(\frac{1}{3},1,5)$ and $(9,5,13)$ of a curve $\vec{C} = \frac{t^3}{3}, 2t-1,t^2+4)$

So I am confused as to what my $a$ and $b$ would be in thi case? as I have a curve in $R^3$. so the interval [a,b] is not a single value? as I have three values, x,y z.

Do you get what I mean?

答: I believe a and b are the value of the parameter at the initial point and the final point since the integral integrates over the "speeds" (the magnitude of the derivative) at every value of the parameter between a and b. At $(\frac{1}{3}, 1, 5)$ $t=1$ and at $(9,5,13)$ $t=3$, so $a=1$ and $b=3$.

Title: Finding arc lenght of a curve in $R^3$


答: You need to find the $t$-values for which $C$ passes through the given points. It is easy to see that $$C(1)=\left(\frac{1}{3},1,5\right)\quad\mbox{and}\quad C(3)=(9,5,13),$$and since $C$ is injective (because of the first component), there are no ambiguities or other possibilities. The arclength is $$\int_1^3((t^2)^2+2^2+(2t)^2)^{1/2}\,{\rm d}t.$$

Title: Finding arc lenght of a curve in $R^3$

Body:
I'm trying to find the arc lenght of a curve in $R^3$. I know that that the formula to find this is $\int_a^b \sqrt{f'(t)^2 + g'(t)^2+h'(t)^2}dt$.

But I want to find the arc lenght between the points $(\frac{1}{3},1,5)$ and $(9,5,13)$ of a curve $\vec{C} = \frac{t^3}{3}, 2t-1,t^2+4)$

So I am confused as to what my $a$ and $b$ would be in thi case? as I have a curve in $R^3$. so the interval [a,b] is not a single value? as I have three values, x,y z.

Do you get what I mean?


问:
Factorial as a Integration over a Gaussian
I'm seeking a proof for a formula that I've affirmed through numerical checks. Let $n$ be a natural number. For

$$I_n = \frac{(-1)^n}{\sqrt{2\pi n}} \int\limits^{\infty}_{-\infty} dx\, x^n D^n (e^{-x^2/2n}),$$

where $D = d/dx$, how can we show that $I_n = n!$?

The formula can be checked for first few $n$'s with Wolfram alpha. n=6 case for identity

As an alternative question, what is $K_n$ for
$$K_n = \frac{(-1)^n}{\sqrt{2\pi n}} \int\limits^{\infty}_{-\infty} dx\, \left(\frac{d}{d\ln x}\right)^n e^{-x^2/2n},$$
where $(d/d\ln x)^nf(x) = x(d/dx)x(d/dx) \cdots x(d/dx) f(x)$. For example,
\begin{align} \left(\frac{d}{d\ln x}\right)^{3}x^{3} & = x\frac{d}{dx} \left[x\frac{d}{dx} \left[ x\frac{d}{dx}\left[ x^{3} \right] \right]\right]  \\
& = 27 x^3.
\end{align}

答:
Thanks @Professor Vector and @rrogers for the suggestions. I forgot that Hermite polynomials even exist! Integration by parts seems to be the best approach since you'd still need to use it for the Hermite approach, but I'll write out the two suggested solutions for posterity. Could probably be used as a fun calculus exercise for an integration by parts lesson!

Integration by parts

We rescale our $x$ variable by $\sqrt{n}$ so that $I_n$ is expressed as
$$ I_n = \frac{(-1)^n}{\sqrt{2\pi}} \int^{\infty}_{-\infty} dx\, x^n D^n(e^{-x^2/2}).$$

Applying integration by parts $n$ times on the RHS gives
$$ I_n = \frac{(-1)^{n+n}}{\sqrt{2\pi}} \int^{\infty}_{-\infty} dx\, D^n(x^n) e^{-x^2/2} = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty} dx\, n! e^{-x^2/2} = n!\qquad \checkmark$$

Hermite Polynomials

For the Hermite polynomials approach, we note that for the Hermite polynomial definition
$$ \mathit{He}_n(x) = (-1)^n e^{{x^2}/{2}}D^n(e^{-{x^2}/{2}})$$
we have the ortgonolaity relation
$$ \int_{-\infty}^\infty \mathit{He}_m(x) \mathit{He}_n(x)\, e^{-{x^2}/{2}} \,dx = \sqrt{2 \pi}\, n!\, \delta_{nm}.$$
Thus we find
$$(-1)^n\int_{-\infty}^\infty \mathit{He}_n(x) D^n(e^{-{x^2}/{2}})\,dx = \sqrt{2 \pi}\, n!.$$
We note that the highest power term in $\mathit{He}_n(x)$ is $x^n$ (i.e., $He_n(x) = x^n + o(x^{n-1})$). All other terms are of lower powers and vanish when we apply $n$ integrations by parts on the integral. Thus we have the starting relation.


答:
Your formula is incorrect, since $n$ is the dummy variable, so let's find a correct one.

Let $S=ar^0+ar^1+...$. Then $rS=ar^1+ar^2+...$. Thus $S-rS=ar^0=a\implies S(1-r)=a\implies S=\boxed{\frac{a}{1-r}}$. This only holds if $|r|<1$, otherwise the series diverges (goes to $\infty$).

For $ar^{n-1}$, the result is similar, we just need to add on the $ar^{-1}=\frac{a}{r}$ term.

Thus $\sum_{n=0}^{\infty}ar^n=\frac{a}{1-r}, \sum_{n=0}^{\infty}ar^{n-1}=\frac{a}{r}+\frac{a}{1-r}$, with $|r|<1$.

Title: Geometric Sums - Equivalency

Body:
Where $ar^0+ar^1+ar^2+ar^3+ar^4...=\sum_{n=0}^\infty ar^{n-1}=\frac{a(1-r^n)}{(1-r)}$

My question is: by this same definition, is it also true that $\sum_{n=0}^\infty ar^{n}=\frac{a(1-r^{n+1})}{(1-r)}$?


问:
Distribution of a division of two absolutely continuous random variables
Assume $Z,Y $ are independent standard normal random variables.

Find the distribution of $ Z/Y $.

The answer is that $ Z/Y $ is absolutely continuous random variable with density

$$ f_{\frac{Z}{Y}}(x)=\frac{1}{\pi(1+x^2)} $$

Here's what I don't understand:

I tried to write:

\begin{align}
F_{\frac{Z}{Y}}(t) & =\mathbb{P}\left(\frac{Z}{Y}\leq t\right)=\mathbb{P}\left(\frac{Z}{Y}\leq t\cap Y<0\right)+\mathbb{P}\left(\frac{Z}{Y}\leq t\cap Y>0\right) \\[8pt]
& =\mathbb{P} (Z\geq tY\cap Y<0)+\mathbb{P}\left(Z\leq tY\cap Y>0\right) \\[8pt]
& =\mathbb{P}(Z\geq tY) \mathbb{P}(Y<0)+\mathbb{P}(Z\leq tY) \mathbb{P}(Y>0) \\[8pt]
& =\frac{1}{2}\mathbb{P}(Z\geq tY)+\frac{1}{2}\mathbb{P} (Z\leq tY)
\end{align}

And then to calculate each term. for example :

$$ \mathbb{P}(Z\leq tY)=\mathbb{P} ((Z,Y)\in \{ (z,y):z\leq ty \} ) = \intop_{-\infty}^\infty \intop_{-\infty}^{ty} \frac{1}{2\pi} e^{-\frac{z^2+y^2}{2}} \, dz\, dy $$

But this leads me to a wrong answer, and I'm pretty sure that the problem is somewhere in what I wrote before.

Any help would be appreciated, thanks in advance.

答:
When you are separating two conditions (here: ${Y>0}$ and ${Y<0}$) you have to take it into account. For instance:

$ \Pr\{Z\leq tY \cap Y>0\} = \Pr\{Z\leq tY | Y>0\} \Pr\{Y>0\}$.

The condition effects the limits of integral that follows, i.e., where you have conditioned $Y$ to be positive, $y$ starts at $0$, not $-\infty$.


答: Since $o(f)$ is a set of functions, $o(o(f))$ is an abuse of notation for $\bigcup_{g\in o(f)}o(g)$. In particular, you can prove $h\in o(f) \iff\exists g\in o(f)(h\in o(g))$ using this definition. In fact, you can choose $g=\sqrt{|fh|}$.

Title: Big o notation on both sides

Body:
Can someone explain what exactly means $o(o(f(x))) = o(f(x))$?
I do understand what little oh means by definition(fraction limit of two functions converges to zero) , but somehow i don't get the idea of Big o or little o on both sides.


问:
Lie algebra of SO(V) is isomorphic to spin(V)

where $a\in Cl^2(V)$, and $\tau(a)(v)=[a,v]$ for $v\in V$. I was wondering is the first equality correct as written? Jost earlier defined the clifford algebra as the quotient of the tensor algebra by ideal generated by the relations $v\otimes v+\|v\|^2=0$, which yields
$$-2\langle v,w\rangle=vw+wv$$
upon polarizing. Back to the computation, I'm computing
$$\langle \tau(a)v,w\rangle=-\frac{1}{2}(\tau(a)vw+w\tau(a)v)$$
while
$$-\frac{1}{2}[[a,v],w]=-\frac{1}{2}[\tau(a)v,w]=-\frac{1}{2}(\tau(a)vw-w\tau(a)v)$$

These differ by a sign. Did I make an error somewhere?

答:
This is a follow up on the comment sequence above.

The following are taken from  Heat Kernels and Dirac Operators.

On page 39,

(2) is "the Jacobi identity." To verify it (in the case of a the bracket being the [super]commutator of multiplication), use that the parity of a product is the sum $\pmod 2$ of that of the individual terms:

$$ |a\cdot b| = |a| + |b| $$ (assuming that parity makes sense - i.e., $a$ and $b$ are 'pure').

Next, on page 103,

Finally, on page 105,

The reason that $[a,[v,w]] = 0$ is that $[v,w] = vw +wv = -2 Q(v,w)\in {\mathbb R}$, and so in the center.


问:
Find the least value of x which when divided by 3 leaves remainder 1, ...
A number when divided by 3 gives a remainder of 1; when divided by 4, gives a remainder of 2; when divided by 5, gives a remainder of 3; and when divided by 6, gives a remainder of 4. Find the smallest such number.

How to solve this question in 1 min?

Any help would be appreciated. :)

答:
Notice that this number is:
-2(mod 3)
-2(mod 4)
-2(mod 5)
-2(mod 6)
Thus, we need to calculate the LCM of 3,4,5,6 and subtract 2. The LCM is 60. Thus, the answer is 58.</p>


问:
How do I integrate the integral for the independent Poisson random variable?

I don't know how to derive this (my red pen mark)

答:
Possible explanation:

Since $\left| -\frac{1}{\sqrt n} \right | \approx 0,$ the interval of integration is very close to zero, and so $x^2 \approx 0 \implies e^{-x^2/2} \approx 1$.


答: When $x \approx 0$, we have $e^{-x^2/2} \approx 1$. Since the text states that we are doing a slightly less than rigorous proof, replace $e^{x^2/2}$ with $1$ and the answer follows immediately.

Title: How do I integrate the integral for the independent Poisson random variable?

Body: I don't know how to derive this (my red pen mark)


问:
Equivalence between continuity of $f$ and continuity of the $f_i$ in a countable product of metric spaces.
Context

We consider $(E, D)$ a metric space, and $F=\prod_{i=1}^{+\infty} E_i$ a countable product of metric spaces $(E_i, d_i)$.

We know we can choose a product-distance $d$ for $F$.

I choose to define the distances $d'_i$ on $E_i$ by $d'_i (x) = \min \{1, d_i(x)\}$ for $x\in E_i$. And then I defined $d$ by, for $(x_i)_{i\geqslant 1} \in F$ and $(y_i)_{i\geqslant 1} \in F$ : $$ d((x_i), (y_i)) = \sum\limits_{i = 1}^{+\infty} \frac{d'_i (x_i, y_i)}{2^i}$$

Let's consider $f_i$ applications from $E$ to $E_i$, and $f:x\mapsto (f_i (x))_{i\geqslant 1}$ from $E$ to $F$.

Question

The queston is : is $f$ continuous in $a \in E$ if and only if the $f_i$ are continuous in $a$ ? Is it true for any distance chosen for $F$?

(The result is of course true if the product is finite.)

My attempt

If $f$ is continuous then the $f_i$ are continuous too.

In fact let's take $\varepsilon > 0$. By noting $\varepsilon ' = \varepsilon / 2^i$, we get the existence of $\alpha >0$ such that for all $x\in E$, $D(x,a)\leqslant \alpha \Rightarrow d(f(x), f(a))\leqslant \varepsilon '$. By looking at $d$'s definition, we can see that $d_i(f_i(x), f_i(a))\leqslant 2^i d(f(x), f(a))$. So we have that for all $x\in E$, $D(x,a)\leqslant \alpha \Rightarrow d_i (f_i (x), f_i (a))\leqslant \varepsilon$. So $f_i$ is continuous in $a$.

If the $f_i$ are continuous, I think there's going to be a problem.

In fact, if, for $\varepsilon > 0$, and for every $i$, we note $\alpha_i >0$ such that for all $x\in E$, $D(x,a)\leqslant \alpha \Rightarrow d_i (f_i (x), f_i (a))\leqslant \varepsilon$, then $\underset{i\geqslant 1}{\inf} \alpha_i$ could be equal to $0$, so that there would be a problem.

But I can't find a counterexample.

Could someone help?

Thank you in advance.

答:
You can’t find a counterexample because the result is in fact true.

HINT: Let $\epsilon>0$. There is an $n_0\in\Bbb Z^+$ such that $\sum_{k>n_0}\frac1{2^k}<\frac{\epsilon}2$. Now just choose $\alpha$ small enough so that

$$\sum_{k=1}^{n_0}\frac{d_k'\big(f_k(x),f_k(a)\big)}{2^k}<\frac{\epsilon}2$$

whenever $D(x,a)<\alpha$.


问:
Why is it important to have a discrepancy between image and codomain?
A function $f:\mathbb{R}\rightarrow \mathbb{R}$ given by $f(x)=x^2$ has $\mathbb{R}_{\geq0}$ as its image and $\mathbb{R}$ as its codomain.

What's the need for this discrepancy? Why don't we just write $f:\mathbb{R}\rightarrow \mathbb{R}_{\geq0}$?

I believe the reason for this is that there may be some functions in which stating the image this way may be impractical - if this hypothesis is true, what is this class of functions? If yes or no, is there any other reason for this?

答: Codomain and Image of a function are two completely different concepts. The codomain of a function often has a structure, like being a topological space or something like this. It will be very inconvenient to put all that structure in the image, which is only a set, and often not such a nice one. Taking your example, one could say that for $x \to \infty$ the function converges to $0$; when you just look at the image of the function it can't converge to $0$ at all because $0$ is not in the image. And definitions like the pointwise addition of two functions will be much broader as it is sometimes hard to say what is the image of the sum of two functions.</p>


答:
Let $\alpha = (x_1x_2...x_n)^\frac{1}{n} $.  Then, $\ln(\alpha) = \frac{1}{n} \sum{\ln(x_i)}$ so because of the property you stated, $$\ln(\alpha) = \underset{x \in \mathbb{R}^d}{\operatorname{argmin}} \sum_{i=1}^n d^2(\ln(x_i), x)$$
Which implies that
$$\alpha = \underset{x \in \mathbb{R}^d}{\operatorname{argmin}} \sum_{i=1}^n d^2(\ln(x_i), \ln(x))$$

Then you can substitute to find the new metric
$$\delta^2(x, y) = d^2(\ln(x), \ln(y)) = (\ln(x) - \ln(y))^2$$
$$\delta(x, y) = \displaystyle\left\lvert \ln{\frac{x}{y}} \right\rvert$$

Title: Which distance gives rise to the geometric mean?

Body:
The arithmetic mean has the nice property of minimising the sum of squares, or in other words, minimising the sum of quadratic-euclidean distances. Formally, given a set of points $x_0, \dots, x_n \in \mathbb{R}^d$, the arithmetic mean, $\mu = \frac{1}{n} \sum_{i=1}^n x_i$, is the unique solution to the equation
$$\mu = \underset{x \in \mathbb{R}^d}{\operatorname{argmin}} \sum_{i=1}^n d^2(x_i, x) \, ,$$
where $d$ denotes the Euclidean distance. I was wondering if there exists a distance measure on $\mathbb{R}$ that has a similar property for the geometric mean, i.e.

Question

Is there a distance measure $d$ on $\mathbb{R}$ such that for any set of points $x_1, \dots, x_n \in \mathbb{R}$ we have
$$ \sqrt[n]{x_1 \cdot \dots \cdot x_n} = \underset{x \in \mathbb{R}^d}{\operatorname{argmin}} \sum_{i=1}^n d^2(x_i, x) \; \textbf{?}$$


问:
Showing that all points inside an enclosed region converge to origin
I have the dynamical system

$\dot{x_1}=x_2+x_1^2-\frac{1}{4} x_1(x_2-1+2x_1^2)$

$\dot{x_2}=-2(1+x_2)x_1$

and I have shown that the curves $x_2=1-2x_1^2$ and $x_2=-1$ are invariant.

Let $D$ be the region enclosed by the two curves just described. I am trying to make use of the function $H=x_1^2(1+x_2)+\frac{x_2^2}{2}$ to show that trajectories inside this region $D$ converge to the origin as $t\rightarrow-\infty$.

I am not sure what to do here because the only problems of this type that I have seen have to do with convergence as $t\rightarrow\infty$, and it is suitable to use Lyapunov's stability theorem in such a case. So how can I approach this problem?

答:
Convergence with $t\to -\infty$ is essentially the same as with $t\to\infty$, because you can reverse time. Let
$$\dot{x} = f(x).$$
Then the function $y(t)=x(-t)$ would satisfy
$$\dot{y}=-\dot{x}(-t)=-f(x(-t))=-f(y)$$
and $\lim_{t\to-\infty}x(t) = \lim_{t\to\infty}y(t)$. You can now examine the stability of $y$ with $t\to\infty$ in the usual way.

In consequence, you can see that to examine the convergence as $t\to-\infty$ you can apply the Lyapunov theorem with the opposite signs. In particular if $\frac{d}{dt}H(x(t))>0$ for $x\neq 0$, then from Lyapunov theorem we have instability as $t\to\infty$ and asymptotic stability as $t\to-\infty$.


问:
Limits in the localization of a category of fibrant objects
Suppose we have category $\mathcal{C}$ which has the structure of a category of fibrant objects, and suppose we have a functor $F:I\to \mathcal{C}$ with a limit $\lim F$ in $\mathcal{C}$.

If we have two maps $t_1,t_2:T\to \lim F$ such that for every $\pi_i:\lim F \to F_i$ satisfies $\pi_i\circ t_1\simeq\pi_i\circ t_2$, can we show that the two maps $t_1$ and $t_2$ are homotopic?

My first idea was to consider the localisation functor $\gamma:\mathcal{C}\to Ho(\mathcal{C})$, in the naive hope of it preserving finite limits, which it does not necessarily do unfortunately. However, intuitively it should be possible to use the homotopies $\eta_i:\pi_i\circ t_1\to \pi_i\circ t_2$, using the notation of Definition 4.16 on the nCatLab page, to assemble this into a homotopy $\eta:t_1\to t_2$.

Does anyone have any idea if this is true, or not? And, possibly give a reference to existing literature, or give a small argument on why it (does not) holds? Or, maybe a characterization of cases in which this does hold? For example; if all maps in the diagrams in $\mathcal{C}$ are weak equivalences.

Thanks!

答:
The category of topological spaces with the Quillen Model Structure is a category of fibrant objects (it really doesn't matter which structure you chose, as long as all objects are fibrant). Consider the (limit) pullback diagram [

Both $\mathbb{R}^2$ and the point $\ast$ are contractible so that any two maps $S^1 \rightarrow \mathbb{R}^2, \ast$ are homotopic. Since $\pi_1(S^1) = \mathbb{Z}$ we know that there are non homotopic maps $t_1$ and $t_2$ such that the composition with the legs of the cone are homotopic.

I don't know much about categories of fibrant objects, but in a model category weak equivalences are ismomorphisms in the homotopy category, so if every map and object in the diagram were fibrant-cofibrant and and least one leg $\pi_i$ were a weak equivalence we would have the following:

$\pi_i \circ t_1 \cong \pi_1 \circ t_2$ then $[\pi_i \circ t_1] = [\pi_1 \circ t_2]$, here [-] denote homotopy classes of morphisms.
Rewriting,
$[\pi_i] \circ [t_1] = [\pi_i] \circ [t_2]  $ and since $[\pi_i]$ is an isomorphism we would have that $[t_1]=[t_2]$


问:
Is this property true for all $p$ or for all primes?
In my textbook the following property is stated:

Let $p$ be a prime number. For any $a_1,...,a_k\in \mathbb Z$ we have that:
$$p|a_1...a_k \implies \exists i\in\{1,...,k\}:p|a_i$$
In particular, for any $a,b \in \mathbb Z$:

$p|ab \implies (p|a \vee p|b)$

$p|a^k \implies p|a$

I was able to prove this for but, In my proof, I never used the fact that $p$ is prime. Because of this, I don't know if my proof is valid but this got me wondering: Is this property also valid for any $p$ or only if $p$ is prime?

Edit: I'll show the proof I made for this.

Let $p$ be a prime number and $a_1,...,a_k \in \mathbb Z$ such that:

$$p|\prod_{n=1}^k a_n$$

The proof is by contradiction. Let's assume that $\forall i \in \{1,...,k\}, p \not | a_i$

Now, let's use mathematical induction:

$p \not | a_1$

Now let's assume that $p \not | \prod_{n=1}^{k-1} a_n$:

We have that $p \not | a_k \iff \gcd(p,a_k) = 1$ and $p \not | \prod_{n=1}^{k-1} a_n \iff \gcd(p,\prod_{n=1}^{k-1} a_n)=1$.

Since $\gcd(p,\prod_{n=1}^{k-1} a_n)=1$ and $\gcd(p,a_k) = 1$ we conclude that: $\gcd(p,a_k \cdot \prod_{n=1}^{k-1} a_n)=1 \iff p \not | \prod_{n=1}^{k} a_n$

This is a contradiction, hence $\exists i \in \{1,...,k\}: p | a_i$

答:
You make the claim that $p\not\mid a_k \iff \gcd(p, a_k) = 1$.

That's clearly false.  Consider $6\not \mid 14$ but $\gcd(6, 14) =2$.

.... The gyst of the statement (which is called Euclid's lemma) is that all numbers are composed of indivisible prime "atoms".  If $n|a\cdot b\cdot c$ then the prime atoms of $n$ may be spread across $a,b,$ and $c$ but each individual atom must belong to one of $a,b$ or $c$.

So the statement isn't true.  $6|24 = 3\cdot 8$ but $6\not \mid 3$ and $6\not \mid 8$.  But the prime atoms of $6$ are $2$ and $3$.  The $2$ atom is contained in the $8$.  And the $3$ atom is contained in the $3$.

So the question is to prove that if $p$ is prime and $p|\prod a_k$ prove that because $p$ can't "be broken up" that it can't be "spread" between $a$ and $b$ that it must divide and be "contained" entirely in either the $a$ or the $b$.

.....

Note:  $n\not \mid b \iff \gcd(n,b)=1$ is clearly false because $n$ can have some $\gcd(n,b) = d$ and $d|b$ but $n$ could have some "extra atoms" that $b$ doesn't share so we could have $\frac nd\not \mid b$.

But if $p$ is prime then $p\not \mid b \iff \gcd(p,b)$ is true because if $\gcd(p, b) \ne 1$ then $p$ and $b$ share some prime atoms in common.  But $p$ itself is a prime atom so the only atom they have in common is $p$ and $p|b$.

So.... formalize and prove that as a Lemma:

Lemma: If $p$ is prime then $p\not \mid b \iff \gcd(p,b) = 1$.

Then your proof will be better. (there's actually a problem in that you proof assumes that there is only one set of $a_1,....a_k$ whose product is the number.  You must assume it is true for every possible set of $k$ numbers and every possible product.)

Hint:  It might be easier to prove a stronger more detailed lemma:

Strong Lemma:  If $p$ is prime then for any integer $b$ either $\gcd(p,b) = 1$ and $p\not \mid b$, or $\gcd(p,b) = p$ and $p \mid b$.  No other option exists.


问:
If a normalized sequence of finite measures converges weakly, does the same hold true for the nonnormalized one?
Let $E$ be a metric space, $(\mu_n)_{n\in\mathbb N}$ be a sequence of finite nonnegative measures on $\mathcal B(E)$ and $\mu$ be a probability measure on $\mathcal B(E)$ s.t. $$\frac{\mu_n}{\mu_n(E)}\xrightarrow{n\to\infty}\mu\tag1$$ with respect to the to the topology of weak convergence of measures.

Assuming that $c:=\sup_{n\in\mathbb N}\mu_n(E)<\infty$, are we able to deduce that the nonnormalized sequence $(\mu_n)_{n\in\mathbb N}$ is convergent as well?

If not, does the other direction of this assertion hold, i.e. can we infer that the normalized sequence (i.e. the left-hand side of $(1)$) is convergent from knowing that the nonnormalized sequence is convergent? It's easy to see that $(\mu_n(E))_{n\in\mathbb N}$ is convergent in that case (since the function constantly equal to $1$ is bounded and continuous). So, this should be possible to obtain.

Remark: Without assuming $c<\infty$, the first implication is clearly wrong. We simply can consider any finite nonnegative measure $\nu$ on $\mathcal B(E)$ and set $\nu_n:=n\nu$ for $n\in\mathbb N$.

答:
Put $E = $ real line. Consider $\mu_{2n} = 2\delta_{ \{0 \} }$ and  $\mu_{2n-1} = \delta_{ \{0 \} }$, where $\delta_{ \{x \} }$  is a Dirac measure.

So the answer is negative.

The opposite is true if $lim_n \mu_n (E)$ (it exists, as you noticed) is not equal to $0$.

But if  $lim_n \mu_n (E) = 0$ we may have smth. like that: $\mu_{n} = \frac{(-1)^n + 1}{n}\delta_{ \{0 \} }$, that is counterexample, because normalized version of $\mu_{2n+1}$ doesn't exist.

Addition. Suppose that normalized sequence of measures converges to some measure $\nu$ and that $\mu_n(E) \to a \ne 0$, (hence $c < \infty$ and $a = \nu(E)$).
Statement: $\mu_n \to a \nu$ and $\nu$ is normalized (i.e. a probability measure).

Indeed, as $\lim_n \mu_n(E) \ne 0$ we have $\mu_n(E) \ne 0$ for all $n\ge n_0$. Thus $$\mu_n = \mu_n(E) \cdot \frac{\mu_n}{\mu_n(E)}.$$ Put $a_n =  \mu_n(E)$, $\nu_n = \frac{\mu_n}{\mu_n(E)}$. We want to show that $a_n \nu_n \to a \nu$. We know that  $\nu_n \to \nu$ and $a_n \to a \ne 0$.
By definition $\int f(x) d\nu_n \to \int f(x) d\nu$ for all bounded continuous $f$. Thus $$\int f(x) d(a_n \nu_n) = a_n \int f(x) d\nu_n \to a \int f(x) d\nu = \int f(x) d(a \nu)$$ and hence $a_n \nu_n \to a \nu$.

Further, $\nu(E) = \lim_n \nu_n(E) = 1$.

Moreover, under the condition $\lim_n \mu_n(E) \ne 0$ convergence of $\nu_n$ and convergence of $\mu_n$ are equivalent (because we may multiply and devide convergent sequence of measures by $ \mu_n(E)$).

Conclusion: if normalized measures converge then $\mu_n$ may not converge. If $\mu_n$ converges then in general case it doesn't mean that normalized measures converge (al least because normalized measures  may not exist). But  if $\mu_n$ converge and there is an additional condition $lim_n \mu_n(E) \ne 0$ then normalized measures converge.


问:
Make all element of a sequence equal
Some tricky questions from maths. I wonder it to be puzzling question. suppose we have given a sequence consisting of n numbers. An operation is given that we can choose three  consecutive elements and add 1 to all three of them. We can perform the above operation any number of times. Is it really possible to make all the array elements equal after performing operation any number of times? Is it possible to declare yes or no just after analyzing the numbers?

Note: elements of sequence can be $10^-9  to 10^9$.
Someone please tell me which algorithm will use here.Thanks

example: suppose given sequence is   5 4 1 -1 0 3 5. We can make all elements equal by performing above operation:

One operation on index [2,3,4] and the resultant sequence will be {5 5 2 0 0 3 5}.

Three operations on index [3,4,5] and the resultant sequence will be {5 5 5 3 3 3 5}.

Two operations on index [4,5,6] and the resultant sequence will be {5 5 5 5 5 5 5}.

答:
There is a straightforward algorithmic approach that will either make all the values on an $n$-tuple equal or show that it's impossible to do so.

Suppose we start with an $n$-tuple $(x_1,x_2,\ldots,x_n)$. Since the only allowable operation is to add $1$ to three consecutive numbers, we may as well limit the discussion to $n$-tuples of integers. (The numbers cannot possibly be made equal if they have fractional parts that are not all equal.) Now the only operation that increases $x_1$ also increases $x_2$ and $x_3$, and likewise the two operations that increase $x_2$ also increase $x_3$. Consequently, it will be impossible to make the first three numbers equal unless $x_1\ge x_2\ge x_3$.  (Similar considerations apply at the other end as well, but we'll work from left to right.)

If $n=3$, there is really nothing to do; we either have $x_1=x_2=x_3$ or we don't. The case $n=4$ is not much more difficult: we either have $x_1\ge x_2=x_3=x_4$ or $x_1=x_2=x_3\le x_4$, in which case it's possible to make all the numbers equal, or we don't, in which case it's not possible to make the numbers all equal.

For $n\ge5$, with $x_1\ge x_2\ge x_3$, it possible to make the first three numbers equal, by adding $1$ to $(x_3,x_4,x_5)$ enough times to make $x_3=x_2$, and then adding $1$ to $(x_2,x_3,x_4)$ enough times to make $x_2=x_1$. Once we have done this, we can add $1$ to $(x_1,x_2,x_3)$ as many times as we like, but we can no longer add any more $1$'s to $(x_2,x_3,x_4)$ or $(x_3,x_4,x_5)$, because doing so would leave us with $x_1\lt x_3$, which would produce an impossible situation already proscribed in the first paragraph.

The effect of this is to "detach" the first three numbers from the rest of the problem, reducing the rest of the problem to operations on the $(n-3)$-tuple $(x_4,x_5,\ldots,x_n)$.  In short, as long as we have at least $5$ numbers to work with, we can make the first three numbers equal and then deal with the remainder. Working systematically, the final remainder will have either $2$, $3$, or $4$ numbers. In the case of $2$ and $3$ remaining numbers, those numbers will either be equal or not and there's nothing one can do about it; in the case of $4$ remaining numbers, they can either be made equal or not as described earlier (in the second paragraph).

Finally, in the case of $2$ remaining numbers that are equal or $4$ remaining numbers that have been made equal, that value is either less than the value of one (or more) of the earlier split-off triples' values, or not. If it is (less than the greatest of the split-off values), nothing can be done; the numbers cannot all be made equal. But if it's greater than or equal to those values, then any triples who values are less can be brought up to equal it; and, clearly, in the case of $3$ remaining values, if they're equal, then all the triples can be made equal.

Demonstrating this on the OP's sequence $(5,4,1,-1,0,3,5)$, we have

$$(5,4,1,-1,0,3,5)\to(5,4,2,0,1,3,5)\to(5,4,3,1,2,3,5)\to(5,4,4,2,3,3,5)\to(5,5,5;3,3,3,5)\to(5,5,5;4,4,4,5)\to(5,5,5;5,5,5,5)$$

(where a semicolon replaces the comma to indicate the split-off triple).

Remark: Something that might be of interest here is whether there is any quick way, short of running the algorithm itself, to tell what the answer will be.

Title: Make all element of a sequence equal


问:
Sample $x$ from $g(x)$
I got confused with all this randomness and probability functions. I was trying to implement the rejection sampling method which (apparently) is really simple. I was reading from Rejection Sampling in Wikipedia and the first step says sample $x$ from $g(x)$ and $u$ from $U(0,1)$ what does this mean?

If I have a uniform distribution, when someone says "Generate/Sample/Draw a sample from $U(0,1)$" means that I have to take a random value from the uniform distribution right? In matlab terms, using the function rand will generate a sample from $U(0,1)$ right?

Now, for any pdf $g(x)$ defined over $(a, b)$. If I want to sample from $g(x)$ I have to pick a value of $x$ between $a$ and $b$ and evaluate the function $g(x)$ ? Is this correct ?
Then to pick the value of $x$ I can use a uniform distribution $U(a,b)$ to sample from, is this true?

I got confused easily. Thanks in advance.

UPDATE

I appreciate your help. But this is still not clear to me. Say I have a probability density function $g(x)=3x+1$. What would be the result of draw a sample from g(x) or pick any $x_0$ distributed according to $g(x)$?

UPDATE

I think I got it. I was confused with the graph of the PDF. When someone asks "draw a sample from a PDF, say $g(x)$" it should return a value for $x$ and this value has to be distributed as the function, that means that more samples will be drawn where the value of $g(x)$ is bigger. Does this sounds about right?

Finally, if I want to use the function I said before $g(x)=3x+1$ I have to use for instance the inverse method to draw samples form that distribution properly.

I think the confusion arose because I thought of the value of $g(x)$ as the probability but this was wrong, the value of $g(x)$ is some kind of density (is this correct?). Hope you can correct me or confirm my hypotheses.

答:
I think the Wikipedia article is misleading here.

Rather than saying sample from $g(x)$ which doesn't strictly speaking make sense it should read sample from a distribution whose probability density function is $g$.

A distribution is a way of describing a random number.  To specify the distribution of $X$ I need to be able to calculate the probability $\mathbb P(a < X \leq b)$ for every pair $a \le b$.  For a continuous random variable it's enough to specify its probability density function $g$.

So if $X$ has probability density function $g$ then
$$\mathbb P(a < x \leq b) = \int_a^b g(x) dx$$

Alternatively I can specify the cumulative distribution function
$$G(x) = \mathbb P(X\leq x) = \int_{-\infty}^x g(y)dy.$$

Now I can specify $\mathbb P(a < x \leq b) = G(b) - G(a)$.

The reason we use PDF's is that sometimes we can't write down the cumulative distribution function, because not all integrals have nice results.

Now if $X$ has cumulative distribution function $G$ I can sample from $X$ simply by taking a uniform $(0,1)$ random variable $U$ and setting $X = G^{-1}(U)$.  Because $G$ must be increasing we have
$$\begin{array}{rl}\mathbb P(X<x) &= \mathbb P (G(X) <G(x)) \\&=\mathbb P(U < G(x)) \\&=G(x)\end{array}$$

So if $X$ has a probability density function $g$ we need to integrate $g$ and find the inverse. Then I can sample using the uniform distribution.

Title: Sample $x$ from $g(x)$


问:
Formula for estimating n-sided regular polygon corner distance
I've encountered a situation where I'm trying to replicate the corner to corner distance of a regular polygon.  For instance, I know for a square, that the distance is roughly 1.414 times the inscribed diameter.  For a hexagon, it is 1.155 times the inscribed diameter.  Octagon, it is 1.082 times the inscribed diameter.

But if I want a general estimation or formula for determining what this value will be for any given regular polygon, what formula can be used?  It is clearly known and used within the metallurgical industry, but I am only able to find tables of the resulting output, not what formula is used to determine these constants.

Is there a general formula?

答:
I think I figured this out after going down a wrong path many times.  Essentially, the task is to find the hypotenuse of the triangle formed by the triangle of opposite corners when the apothem is 0.5 (and thus the diameter of the inscribed circle is 1).

Then it simply becomes a calculation of:
$H = \sqrt{1+(\tan({\pi \over n}))^2}$

This is because with a diameter of 1, we just have to find the side length, and use it as a term in the Pythagorean Theorem.


问:
Understanding integration by substitution
The integration by substitution technique is dervied from the following statement:
$$\int _{a}^{b}f(\varphi (x))\varphi '(x)\,dx=\int _{\varphi (a)}^{\varphi (b)}f(u)\,du$$
Now almost all the explanations I saw about the actual technique start playing around with the Leibniz notation and $dx$, $du$ to get to the new form, e.g stating $du=\varphi '(x)\,dx$ from ${\frac {du}{dx}}=\varphi '(x)$.

Now the problem is that $dx$ and $du$ are said to have no actual meaning, at least in elementary calculus, and yet we're playing around with them as if they were actual variables forming actual quotients. As far as I am aware this wouldn't be a problem in nonstandard analysis, but how should one make sense of it in the elementary settings? How can we apply this technique without "heuristic" methods (which actually seems to be used almost all the time for different techniques)?

Note: My problem is actually applying the method, not the proof, as there are many proofs available not using the Leibniz's notation.

答:
Here is one way to look at it. Let $F$ be an anti-derivative of $f$. Then $F(\phi(x))$ is an anti-derivative of $f(\phi(x)) \phi'(x)$ (by the chain rule). So by the fundamental theorem of calculus,
$$
\int_a^b f(\phi(x)) \phi'(x) \, dx = F(\phi(x)) \bigg \rvert_a^b = F(\phi(b)) - F(\phi(a)).
$$
But also, by the fundamental theorem of calculus,
$$
\int_{\phi(a)}^{\phi(b)} f(u) \, du = F(u) \bigg \rvert_{\phi(a)}^{\phi(b)} = F(\phi(b)) - F(\phi(a)).
$$
This reveals that
$$
\int_a^b f(\phi(x)) \phi'(x) \, dx = \int_{\phi(a)}^{\phi(b)} f(u) \, du.
$$

We should not throw "infinitesimal intuition" out the window, though. The following intuitive argument is quite illuminating. Suppose we chop up the interval $[a,b]$ into tiny pieces and the $i$th piece is $[x_i, x_{i+1}]$. Then correspondingly the interval $[\phi(a), \phi(b)]$ is chopped up into tiny pieces and the $i$th piece is $[\phi(x_i), \phi(x_{i+1}]$. To evaluate $\int_{\phi(a)}^{\phi(b)} f(u) \, du$, we must add up the contributions of each tiny piece:
$$
\int_{\phi(a)}^{\phi(b)} f(u) \, du \approx \sum_i f(\phi(x_i)) (\phi(x_{i+1}) - \phi(x_i)).
$$
But notice that $\phi(x_{i+1}) - \phi(x_i)) \approx \phi'(x_i)(x_{i+1} - x_i)$.
So,
$$
\int_{\phi(a)}^{\phi(b)} f(u) \, du \approx \sum_i f(\phi(x_i)) \phi'(x_i)(x_{i+1} - x_i) \approx \int_a^b f(\phi(x)) \phi'(x) \, dx.
$$


问:
Orthogonal Projection onto the Intersection of Convex Sets
The general method of Projection on Convex Sets (POCS) can be used to find a point in the intersection of a number of convex sets i.e.

$$ \text{find } x \in \mathbb{R}^N \text{ s.t. } x \in \bigcap_i C_i $$

This method can find any feasible point in the intersection of the convex sets. Now my question is: Is there a similar method that can find a point $x^*$ that has minimum norm instead i.e. solve

$$ x^* = \arg \min_{x \in \mathbb{R}^N} \Vert x\Vert \text{ s.t. } x \in \bigcap_i C_i$$

or more generally find the projection of a point $a \in \mathbb{R}^N$ onto the intersection of the convex sets i.e. solve

$$ x^* = \arg \min_{x \in \mathbb{R}^N} \Vert x - a\Vert \text{ s.t. } x \in \bigcap_i C_i$$  for some $a \in \mathbb{R}^N$?

Edit

The problem we are trying to solve is a 3D tomography reconstruction problem, and thus the variable $x$ takes gigabytes of RAM. There is already a POCS algorithm (A variant of Algebraic Reconstruction Technique (ART)) that finds a feasible point. So is there a way to use it as a black-box or adapt it to find a minimum-norm solution instead?

答:
ADMM Consensus Trick and Orthogonal Projection onto an Intersection of Convex Sets

The ADMM Consensus Optimization

Given a problem in the form:

$$\begin{aligned}
\arg \min_{ \boldsymbol{x} } \quad & \sum_{i}^{n} {f}_{i} \left( \boldsymbol{x} \right) \\
\end{aligned}$$

By invoking auxiliary variables one could rewrite it as:

$$\begin{aligned}
\arg \min_{ \boldsymbol{x} } \quad & \sum_{i}^{n} {f}_{i} \left( \boldsymbol{x}_{i} \right) \\
\text{subject to} \quad & \boldsymbol{x}_{i} = \boldsymbol{z}, \; i = 1, 2, \ldots, n \\
\end{aligned}$$

Namely, it is a separable form with equality constraint on each variable.

In matrix form it can be written as:

$$\begin{aligned}
\arg \min_{ \boldsymbol{x} } \quad & \sum_{i}^{n} {f}_{i} \left( \boldsymbol{x}_{i} \right) \\
\text{subject to} \quad & \boldsymbol{u} := \begin{bmatrix} \boldsymbol{x}_{1} \\ \boldsymbol{x}_{2} \\ \vdots \\ \boldsymbol{x}_{n} \end{bmatrix} = \begin{bmatrix} I \\ I \\ \vdots \\ I \end{bmatrix} \boldsymbol{z}
\end{aligned}$$

By defining $ {E}_{i} $ to be the matrix such that $ \boldsymbol{x}_{i} = {E}_{i} \boldsymbol{u} $, namely the selector of the appropriate sub vector form $ \boldsymbol{u} $ we can write the problem in the ADMM form (The Scaled ADMM form):

$$\begin{aligned}
\boldsymbol{u}^{k + 1} & = \arg \min_{ \boldsymbol{u} } \sum_{i}^{n} {f}_{i} \left( {E}_{i} \boldsymbol{u} \right) + \frac{\rho}{2} {\left\| {E}_{i} \boldsymbol{u} - \boldsymbol{z}^{k} + \boldsymbol{\lambda}^{k} \right\|}_{2}^{2} \\
\boldsymbol{z}^{k + 1} & = \arg \min_{ \boldsymbol{z} } \frac{\rho}{2} {\left\| \boldsymbol{u}^{k + 1} - \begin{bmatrix} I \\ I \\ \vdots \\ I \end{bmatrix} \boldsymbol{z} + \boldsymbol{\lambda}^{k} \right\|}_{2}^{2} \\
\boldsymbol{\lambda}^{k + 1} & = \boldsymbol{\lambda}^{k} + \boldsymbol{u}^{k + 1} - \begin{bmatrix} I \\ I \\ \vdots \\ I \end{bmatrix} \boldsymbol{z}^{k + 1} \\
\end{aligned}$$

Since the form is block separable it can be written in an element form:

$$\begin{aligned}
\boldsymbol{x}_{i}^{k + 1} & = \arg \min_{ \boldsymbol{x}_{i} } {f}_{i} \left( \boldsymbol{x}_{i} \right) + \frac{\rho}{2} {\left\| \boldsymbol{x}_{i} - \boldsymbol{z}^{k} + \boldsymbol{\lambda}_{i}^{k} \right\|}_{2}^{2}, & i = 1, 2, \ldots, n \\
\boldsymbol{z}^{k + 1} & = \arg \min_{ \boldsymbol{z} } \frac{\rho}{2} \sum_{i = 1}^{n} {\left\| \boldsymbol{x}_{i}^{k + 1} - \boldsymbol{z} + \boldsymbol{\lambda}_{i}^{k} \right\|}_{2}^{2} \\
\boldsymbol{\lambda}_{i}^{k + 1} & = \boldsymbol{\lambda}_{i}^{k} + \boldsymbol{x}_{i}^{k + 1} - \boldsymbol{z}^{k + 1}, & i = 1, 2, \ldots, n \\
\end{aligned}$$

Remark: Pay attention that $ \boldsymbol{z}_{k + 1} = \frac{1}{n} \sum_{i = 1}^{n} \boldsymbol{x}_{i}^{k + 1} + \boldsymbol{\lambda}_{i}^{k} $. Namely the mean value of the set $ { \left\{ \boldsymbol{x}_{i}^{k + 1} + \boldsymbol{\lambda}_{i}^{k} \right\} }_{i = 1}^{n} $.

The Proximal Operator is given by $ \operatorname{Prox}_{\lambda f \left( \cdot \right)} \left( \boldsymbol{y} \right) = \arg \min_{\boldsymbol{x}} \frac{1}{2} {\left\| \boldsymbol{x} - \boldsymbol{y} \right\|}_{2}^{2} + \lambda f \left( \boldsymbol{x} \right) $ and simplifying the optimization for $ \boldsymbol{z}_{k + 1} $ one can write the above as:

$$\begin{aligned}
\boldsymbol{x}_{i}^{k + 1} & = \operatorname{Prox}_{\frac{1}{\rho} {f}_{i} \left( \cdot \right)} \left( \boldsymbol{z}^{k} - \boldsymbol{\lambda}_{i}^{k} \right), & i = 1, 2, \ldots, n \\
\boldsymbol{z}^{k + 1} & = \frac{1}{n} \sum_{i = 1}^{n} \boldsymbol{x}_{i}^{k + 1} + \boldsymbol{\lambda}_{i}^{k} \\
\boldsymbol{\lambda}_{i}^{k + 1} & = \boldsymbol{\lambda}_{i}^{k} + \boldsymbol{x}_{i}^{k + 1} - \boldsymbol{z}^{k + 1}, & i = 1, 2, \ldots, n \\
\end{aligned}$$

Orthogonal Projection onto an Intersection of Convex Sets

The problem is given by:

$$\begin{aligned}
\arg \min_{ \boldsymbol{x} } \quad & \frac{1}{2} {\left\| \boldsymbol{x} - \boldsymbol{y} \right\|}_{2}^{2} \\
\text{subject to} \quad & \boldsymbol{x} \in \bigcap_i \mathcal{C}_{i}, \; i = 1, 2, \ldots n \\
\end{aligned}$$

Namely we're looking for orthogonal projection of $ \boldsymbol{y} $ onto the intersection of the sets $ {\left\{ \mathcal{C}_{i} \right\}}_{i = 1}^{n} $.

The problem could be rewritten as:

$$\begin{aligned}
\arg \min_{ \boldsymbol{x} } \quad & \frac{1}{2} {\left\| \boldsymbol{x} - \boldsymbol{y} \right\|}_{2}^{2} + \sum_{i}^{n} {\delta}_{\mathcal{C}_{i}} \left( \boldsymbol{x} \right)
\end{aligned}$$

Where $ {\delta}_{\mathcal{C}_{i}} (x) = \begin{cases} 0 & x \in \mathcal{C}_{i} \\ \infty & x \notin \mathcal{C}_{i}
\end{cases} $.

So we can use the ADMM Consensus Optimization by setting $ {f}_{1} \left( \boldsymbol{x} \right) = \frac{1}{2} {\left\| \boldsymbol{x} - \boldsymbol{y} \right\|}_{2}^{2} $ and $ {f}_{i} \left( \boldsymbol{x} \right) = {\delta}_{\mathcal{C}_{i - 1}} \left( \boldsymbol{x} \right), \; i = 2, 3, \ldots, n + 1 $.

In order to solve it we need the Proximal Operator of each function. For $ {f}_{1} \left( \cdot \right) $ it is given by:

$$ \operatorname{Prox}_{ \frac{1}{\rho} {f}_{1} \left( \cdot \right)} \left( \boldsymbol{v} \right) = \arg \min_{ \boldsymbol{x} } \frac{1}{2} {\left\| \boldsymbol{x} - \boldsymbol{y} \right\|}_{2}^{2} + \frac{\rho}{2} {\left\| \boldsymbol{x} - \boldsymbol{v} \right\|}_{2}^{2} = \frac{\rho \boldsymbol{v} + \boldsymbol{y}}{1 + \rho} $$

For the rest the Proximal Operator is the orthogonal projection:

$$ \operatorname{Prox}_{ \frac{1}{\rho} {f}_{i} \left( \cdot \right)} \left( \boldsymbol{v} \right) = \operatorname{Proj}_{ \mathcal{C}_{i - 1} } \left( \boldsymbol{v} \right), \; i = 2, 3, \ldots, n + 1 $$


问:
Which is the largest three digit number which when divided by 6 leaves the remainder 5 and when divided by 5 leaves the remainder 3?

Which is the largest three digit number which when divided by 6 leaves the remainder 5 and when divided by 5 leaves the remainder 3?

What I did:

On division by 6 leaves remainder 5

So numbers are 6, 11, 17, 23

On division by 5 leaves remainder 3

3,8,13,18,23

So common number between the two is 23. Largest three digit number divisible by 23 is 989

But answer is 983. How?

This is a gmat exam question.

答:
We can write this as:

6a+5=5b+3

5b-6a=2

5b(mod 6)≡2(mod 6), thus b≡4(mod 6). We can write b as 6x+4.

Our three digit number is then 5(6x+4)+3=30x+23. 30x+23<=999. The largest integer x that satisfies this is 32. Thus, our three digit number is 30(32)+23=983