问:
First Axiom of Countability and Convergent Sequences
I am reading through part of my real analysis book about topology, and it asserts the following theorem:

If a topological space $T$ satisfies the first axiom of countability, then every contact point of a set $M \subset T$ is the limit of a convergent sequence of points in $M$.

The proof then goes on to say that if we can select a sequence of $O_n \in \mathcal{O}$, where $\mathcal{O}$ is a countable neighborhood base at some point $x \in [M]$, such that $O_{n+1} \subset O_n$, then we can choose some $x_n$ for each $O_n$ that results in a sequence converging to our desired point $x$. The part that I don't understand is the assumption that we can choose $O_{n+1} \subset O_n$ for all $n \in \mathbb{N}$. Why is this the case? The obvious solution might seem to be to choose some other $G'$ such that there exists some $O' \in \mathcal{O}$ and $O' \subset O$, but what if no such $G'$ exists in the set of open sets of $T$? If the first axiom of countability were to fail because of this, could someone explain it to me?

Any help with this question would be greatly appreciated! Thanks!

答: Suppose that $\{U_n:n\in\Bbb N\}$ is any countable base at $x$. For each $n\in\Bbb N$ let $V_n=\bigcap_{k=0}^nU_k$; then $\{V_n:n\in\Bbb N\}$ is a countable base at $x$ such that $V_{n+1}\subseteq V_n$ for each $n\in\Bbb N$.


问:
Differential equation in Complex numbers
How do I solve this differential equation
$$Z'=Z^*$$
such that $Z$ is complex numbers and $Z^*$ is the conjugate of $Z$?

Sketch the phase portraits of solutions.

答:
Writing

$z = x + iy, \tag 1$

where $x$, $y$ and $z$ are functions of the independent variable $t$, we have

$\dot z = \dot x + i \dot y, \tag 2$

and

$z^\ast = x - iy; \tag 3$

then the given equation

$\dot z = z^\ast \tag 4$

becomes

$\dot x + i \dot y = x - iy; \tag 5$

equating the real and imaginary parts we find

$\dot x = x \tag 6$

and

$\dot y = -y; \tag 7$

to proceed further, we assign $z$ the initial condition

$z(t_0) = x(t_0) + iy(t_0), \tag 8$

which in fact initializes $x$ and $y$; now (6) and (7) have the well-known unique solutions

$x(t)= x(t_0)e^{t - t_0}, \tag{9}$

$y(t) = y(t_0)e^{t_0 - t}, \tag{10}$

whence

$z(t)  = x(t_0)e^{t - t_0} + iy(t_0)e^{t_0 - t}. \tag{11}$

In order to sketch the phase portrait of the system (4), we first take note of the fact that (9) and (10) together yield

$x(t)y(t) = x(t_0)y(t_0), \tag{12}$

which whenever

$x(t_0) \ne 0 \ne y(t_0) \tag{13}$

shows that the solution curves lie in he hyperbolas

$xy = x(t_0)y(t_0) \tag{14}$

in the $xy$-plane $\Bbb R^2$; if precisely one of $x(t_0)$, $y(t_0)$ does not vanish, then the corresponding phase curve is either the positive or negative $x$ or $y$ half-axis.  For example, if

$x(t_0) > 0, \; y(t_0) = 0, \tag{15}$

the solution curve traverses the entire positive $x$-axis as $-\infty \to t \to \infty$.  Similar results apply when

$x(t_0) = 0, \; y(t_0) \ne 0 \tag{16}$

etc; the reader may easily complete the necessary details; finally when

$x(t_0) = 0 = y(t_0) \tag{17}$

the phase curve is the single point $(0, 0)$.  With these facts in mind, the phase portrait of (4) may be sketched with little additional effort.

ContentLicense: CC BY-SA 4.0


问:
$\mathcal{A}\perp_\mathcal{G}\mathcal{B}\wedge\mathcal{H}\subseteq\mathcal{G}\implies\mathcal{A}\perp_\mathcal{H}\mathcal{B}$?
If $\mathcal{A}\perp_\mathcal{G}\mathcal{B}$ and $\mathcal{H}\subseteq\mathcal{G}$, is it the case that $\mathcal{A}\perp_\mathcal{H}\mathcal{B}$? Here $\mathcal{A}$, $\mathcal{B}$, $\mathcal{G}$ and $\mathcal{H}$ are sub-$\sigma$-algebras of some common probability space $\left(\Omega,\mathcal{F},P\right)$ and $\cdot\perp_\square\cdot$ is the conditional independence relation.</p>

答: Consider a complete probability space for all the following underlying $\sigma$-fields. Then $\mathcal A\cap \mathcal B\subset\mathcal G$ (take $A$ in the intersection, then $P(A\mid\mathcal G)\in\{0,1\}$, and completeness ensures that $A\in\mathcal G$). To get a counter-example, take $\mathcal H:=\mathcal A\cap\mathcal B=\{\emptyset,X\}$ and $\mathcal A$ and $\mathcal B$ are not independent. </p>

ContentLicense: CC BY-SA 3.0


问:
How is this second order difference equation solved?
I have and want to solve the following difference equation:

$ x_{n+2} - 4X_n = 27n^2, x_0 = 1, x_1 = 3 $.

First, I find a solution to the homogeneous case (by solving the auxiliary equation $ (\lambda -2)(\lambda + 2) = 0 $), which is:

$ x^c_{n} = A2^n+B(-2)^n $

Then I have to "guess", what's called the particular integral on the ODE world. And as $27n^2$ is a polynomial, I guess with the generic same degree polynomial $ x^p_{n} = Cn^2+Dn+E $.

Now, I substitute $ x^p_{n} $ in the initial equation to find $ C=-9, D=-12 and E=-20 $.

Therefore, as the solution $ x_n = x^c_{n} + x^p_{n} $, we have:

$ x^n = A2^n+B(-2)^n-9n^2-12n-20 $.

We substitute in the initial conditions and get:

$ x_0 = A + B - 20 = 1 $

$ x_1 = 2A - 2B - 9 - 12 - 20 = 3 $

Which means $ A = \frac{43}{2} $ and $ B = -\frac{1}{2} $.

Therefore, the solution to this difference equation should be:

$ x_n = \frac{43*2^n}{2} - \frac{(-2)^n}{2} - 9n^2 - 12n - 20 $

But, when substituting into the initial equation it's clearly not a solution.

Where did I go wrong?

答:
Here's the theory to justify your guess:

Denote $\mathbb{E}$ the forward shift operator such that $\mathbb{E} X_n=X_{n+1}, \mathbb{E}^2 X_n=X_{n+2}$, etc.

Note that $(\mathbb{E}-1)^3 n^2=0$ (the third difference of a second order polynomial is zero, just like third derivative of a second order polynomial is zero), then

$$(\mathbb{E}^2-4)X_n=27n^2
\implies (\mathbb{E}-1)^3 (\mathbb{E}^2-4) X_n = 0\\
$$

Therefore $$X_n = c_1 2^n + c_2 (-2)^n + c_3 + c_4 n + c_5 n^2.$$

Also see this example.


问:
Family of linear independent vectors
I have $K \in \{\mathbb{R},\mathbb{Z_{2}} \}$ (I want to show both cases), and $V$ vectorspace. I want to proof:
If the family of vectors $(u,v,w)$ is linear independent, then $(u+v,u+w,v+w)$  is linear independent.

Alright let's start for $\mathbb{R}$

Consider $a,b,c \in \mathbb{R}$.

$u,v,w$ are linear independent if and only if $au+bv+cw=0$ means that $a=b=c=0$.

$(u+v)\cdot a + (u+w) \cdot b +(v+w) \cdot c = 0$.

$\leftrightarrow ua+va+ub+wb+vc+wc = 0$

$\leftrightarrow u(a+b)+v(a+c)+w(b+c) = 0$

Since $u,v,w$ linear independent $(a+b)+(a+c)+(b+c) =0$, therefore  $(u+v,u+w,v+w)$ linear independent.

Now I want to look at the second case.

Consider $a,b,c \in \mathbb{Z_2}$.

In my opinion, nothing changes since $\mathbb{Z_2}$ and $\mathbb{R}$ are both fields. Am I missing something or is this proof valid if I just say that both of these are fields?

答:
What you have so far is
$$a+b=b+c=c+a=0\quad\ (1)$$
To obtain the result, you need to conclude $a=b=c=0$, which is easy in the case of $\Bbb R$:

Adding all three equations we get
$$2(a+b+c)=0\quad\ (2)$$
Therefore, dividing by $2$, we get $a+b+c=0$ and we can conclude.

In the case of characteristic $2$, however, division by $2$ is not allowed, as $2=1+1=0$.

And, indeed, $a=b=c=1$ is a solution for $(1)$, proving that $u+v,\,u+w,\,v+w$ are linearly dependent in this case.

Title: Family of linear independent vectors


问:
Identity type and its negation
In type theories, an intensional identity type $\textsf{Id}(a,b,A)$ is well-formed iff $a,b:A$ for $A:\textsf{Type}_i$. I have two small questions about the negation of intensional identity types:

Let $a$ and $b$ be different, does $\neg\textsf{Id}(a,b,A)$ mean that $a$ and $b$ are non-equal terms of $A$?

Let $a$ and $b$ be equal, then we have $\textsf{Id}(a,a,A)$. In this case, what does $\neg\textsf{Id}(a,a,A)$ mean? It appears to me that this negation is nonsense since the well-formedness of $\neg\textsf{Id}(a,a,A)$ already presupposes that $a:A$ and trivially $a$ and $a$ are equal terms of $A$.

I am not sure if I understood correctly the basics of intensional identity. Could anyone help with this? Thanks in advance!

答:
Let $a$ and $b$ be different, does $\neg\textsf{Id}(a,b,A)$ mean that
$a$ and $b$ are non-equal terms of $A$?

First, let's fix that by equality of terms we mean definitional equality, i.e. two terms are equal if they are $\beta\eta$-convertible.

The answer depends on the typing context. It's possible that $\neg\textsf{Id}(a,a,A)$ is an assumption in context, in which case it does not follow that $a$ is not definitionally equal to $a$. So in arbitrary contexts the answer is no.

If we're in the empty context, then the answer is yes. In this case we can use the consistency of the system (which holds for usual type theories). Consistency says that the empty type has no proof in the empty context. Assume that $\neg\textsf{Id}(a,b,A)$ is provable in the empty context. Now, $a$ cannot be definitionally equal to $b$, because in that case $\textsf{Id}(a,b,A)$ would be provable by $\textsf{refl}$, which would make it possible to get an inhabitant of the empty type, contradicting consistency.

Let $a$ and $b$ be equal, then we have $\textsf{Id}(a,a,A)$. In this
case, what does $\neg\textsf{Id}(a,a,A)$ mean?

As pointed out in the comments, $\neg\textsf{Id}(a,a,A)$ is just a type which is equivalent to the empty type, as it immediately implies the empty type.

Title: Identity type and its negation


问:
Explanation of graphical meaning of derivative
today my lecturer taught us derivatives. So I am a complete newbie at this and I am struggling with the graphic of a derivative. To be more specific, I want to know how we can "read" a graphic of a derivative and conclude some stuff about the given function?

Example:

What we can "read" from the blue graph about the red one and how are they connected with each other?

I know this might be a stupid question but I honestly can't get it.

答:
The derivative of a function gives the slope of the tangent line to that function at the same $x$ co-ordinate, so looking at a graph of the derivative tells you something about how the function is moving at various points. In particular:

If the derivative is 0, then the tangent line is horizontal, so the function must go completely flat at that point (either a local maximum or minimum, or a point of inflexion). For example, on your graph, the derivative is zero at -1, 0 and 1, which are all points where the function itself is completely flat.

If the derivative is positive, then the tangent line is diagonal going up-right, so the function is increasing at that point. For example, the derivative in your graph is positive between -1 and 0, and from 1 onwards, and that's where the graph slopes up.

If the derivative is negative, then the tangent line is diagonal down-right, so the function is decreasing at that point.

If you want to get fancier, then since bigger values of the derivative (either positive or negative) mean more significant slopes, then the fact that your derivative trends to infinity in both directions means that your graph will get more extremely sloping as it goes along. By comparison, if the derivative stayed very small, then the function would be much more gentle even as it trended wherever it was going.


问:
How do I calculate the annihilator of series with fraction?
How do I calculate the annihilator of this:
$$s_n = s_{n-1} + \frac{1}{(n + 1)(n + 2)(n + 3)}$$

I know that $E-1$ is the annihilator part for $s_n - s_{n-1}$ but I don't know how to get the part for the fraction.

I need to find the closed form of $s_n$ and the first thought I had was to use annihilators.

答:
I don't know what an annialtor is but if you want a closed form of $s_n$ you can telescope as follows:

$$\frac{1}{(n+1)(n+2)(n+3)}=\frac{1}{n+1} \left( \frac{1}{n+2} - \frac{1}{n+3} \right) = \frac{1}{(n+1)(n+2)} - \frac{1}{(n+1)(n+3)}\\
=\frac{1}{n+1} - \frac{1}{n+2} - \frac 12\left( \frac{1}{n+1} - \frac{1}{n+3}\right)\\
=\frac12 \left(\frac{1}{n+1} - \frac{2}{n+2} + \frac{1}{n+3}\right).
$$

Can you start from here?


问:
Is this proposition incomplete?
In mendelson logic textbook , there is this proposition,(I will replace some letters to make it easier to read)

"If $P$ is a tautology containing as statement letters $A_1,A_2,...,A_n$ and $Q$ arises from $T$ by substituting statement forms $B_1,B_2,...,B_n$ for $A_1,A_2,...,A_n$ ,respectively, then $Q$ is a tautology;that is ,substituting in a tautology yields a tautology"

For me , this seems incomplete.Shouldn't there be this line also that ,"$A_1 \equiv B_1 , A_2 \equiv B_2 , ... , A_n \equiv B_n$ "?

答:
As Mauro showed: No, this is not necessary.

In fact, the fact that the $B_i$'s need not be equivalent to the $A_i$'s reveals the real 'power' of this proposition. It shows that once you know that a certain statement is a tautology, then anything with that same abstract 'form' will be a tautology as well. And, since the $A_i$'s are statement letters, it doesn't matter what you substitute them with: the abstract form will be retained, and hence the result is still a tautology.

To illustrate this with Mauro's example: we know that $p \lor \neg p$ is a tautology, and indeed anything of the form $[something] \lor \neg [something]$. So, no matter what $[something]$ you substitute for $p$, the result is a tautology.

And finally, to demonstrate the importance that we are subsitutiong statements for statemwnt letters, note that $p \lor \neg p$ could also be seen as being if the form $[some statement]$, as of course every statement is ... but clearly just substituting justanything  for $p \lor \neg p$ as a whole does not mean we end up with a tautology.

And if that latter example is a bit too abstract for you:

Consider the statement $(p \lor \neg p) \land (q \to q)$

This is a tautology, as of course anything of the form $[something1] \lor [something1] \land ([something2] \to [something2])$ is. This is why you can systematically substitute anything for $p$ and $q$ and still have a tautology.

However, this statement is also of the form $[something1] \land [something2]$. But that is not a tautologous form: $r \land s$ is not a tautology. Hence, substituting just anything for the complex statements $p \lor \neg p $ and for $q \to q$ does not necessarily result in a tautology. It is only when you work with the atomic variables that you can be guaranteed to retain 'tautologousness'.

Title: Is this proposition incomplete?


问:
How to calculate the complex integral
How to calculate: $\int_{1}^{i}{\log(z)}\,dz$? I am beginner on this topic.

答:
Suppose the integration is done along the segment joining $1$ and $i$ whose parametric equation is $\gamma(t)=t+i (1-t),\;0\le t\le 1$. So we have (look here)

$$\int_{\gamma} \log z \, dz=\int_0^1 (1-i) \log (t+i (1-t)) \, dt=\frac{\pi }{2}-1+i$$

Title: How to calculate the complex integral


问:
Finding $\int_{0}^{\infty}\frac{\cos(ax)}{(x^2 + 1)^2}\,dx $
I have a contour integral problem I need to solve, but I don't know the answer, so I wanted to verify that my work is correct.

$$ \int_{0}^{\infty}{\frac{\cos(ax)}{(x^2 + 1)^2}dx} $$

For this one, the function being integrated is even, so I can just take the integral over the entire real line and multiply by $ \dfrac{1}{2} $. That is $ \int_{0}^{\infty}{\dfrac{\cos(ax)}{(x^2 + 1)^2}dx} = \dfrac{1}{2}\int_{-\infty}^{\infty}{\dfrac{\cos(ax)}{(x^2 + 1)^2}dx} $. In the upper half-plane, the function being integrated has a double pole at $ i $. Therefore, I want to say that this is true:

$$ \int_{-\infty}^{\infty}{\frac{\cos(ax)}{(x^2 + 1)^2}dx} = \operatorname{Re} [2 \pi i\ \operatorname{Res}(\dfrac{e^{iz}}{(x^2 + 1)^2}, i)] $$

My solution yields: $ \int_{0}^{\infty}{\dfrac{\cos(ax)}{(x^2 + 1)^2}dx} = \dfrac{\pi}{4e} $

I have no way to verify the correctness of my answer, so is this correct or have I made a mistake somewhere?

答:
You have:
$$\int_{-\infty}^{\infty}{\frac{\cos(ax)}{(x^2 + 1)^2}dx} = \operatorname{Re} \left[2 \pi i\ \operatorname{Res}\left(\dfrac{e^{i a z}}{(z^2 + 1)^2}, i\right)\right]$$
And
$$\operatorname{Res}\left(\dfrac{e^{i a z}}{(z^2 + 1)^2}, i\right)=\operatorname{Res}\left(\dfrac{e^{i a z}}{(z +i)^2(z -i)^2}, i\right)=\lim_{z \to i} \frac{d}{dz}\left( (z-i)^{2}f(z) \right)$$
where  $f(z)=\dfrac{e^{i a z}}{(z +i)^2(z -i)^2}$.

 So
$$\frac{d}{dz}\left( (z-i)^{2}\dfrac{e^{i a z}}{(z +i)^2(z -i)^2} \right)=\frac{i a e^{i a z}}{(z+i)^2}-\frac{2 e^{i a z}}{(z+i)^3}$$
Taking the limit, multiplying by $2\pi i$, taking the real part and finally multiplying by $\frac{1}{2}$ will give you $\frac{1}{4} \pi  (a+1) e^{-a}$.


问:
$A,B$ subsets of a metric space, subset $ U=\{x\in X: d(x,A) < d(x,B) \}$ is open.
Let $X,d$ be a metric space with non-empty subsets $A,B$, we define:
$$ d(x,A):= \inf_{a \in A}d(x,a).$$

We also define the set:
$$ U=\{x\in X: d(x,A) < d(x,B) \}.$$

In a sense this seems to be the set of points that are "closer" to $A$ than to $B$? (any thoughts on this, or a good way to visualise this?)

The claim is that this set $U$ is open.

So given some $x\in U$, we prove that $\exists R$, such that $B(x,R) \subseteq U$.

What would be a good way to find such an $R$?
I tried it out with $R=\frac{1}{2}(d(x,A) +d(x,B))$. Observe that for $y\in B(x,R)=\{b \in X | d(b,x) <R\}$ we have:
$$ d(y, A) \leq d(y,x) + d(x,A) $$
By the triangle inequality (proven in an earlier exercise for this distance function). We then take the property from $U$ that says $d(x,A) < d(x,B)$ and get:
$$ d(y, A) < d(y,x) + d(x,B)< R + d(x,B)= \frac{1}{2} d (x,A) + \frac{3}{2} d(x,B) \dots $$
And as you can see I got kind of stuck. I am a far way from proving $d(y,A) < d(y,B)$ it seems.

答:
Here is an argument just using the triangle inequality.

Suppose that $d(x,A)<d(x,B)$, let $\epsilon=\frac13\big(d(x,B)-d(x,A)\big)$ and suppose that $d(x,y)<\epsilon$. Then for each $b\in B$ we have $d(x,b)\le d(x,y)+d(y,b)$ and hence

$$d(y,b)\ge d(x,b)-d(x,y)>d(x,B)-\epsilon\,,$$

so that $d(y,B)\ge d(x,B)-\epsilon$.

There is an $a\in A$ such that $d(x,a)<d(x,A)+\epsilon$, and for that $a$ we have

$$d(y,a)\le d(y,x)+d(x,a)<d(x,A)+2\epsilon\,,$$

so $d(y,A)<d(x,A)+2\epsilon$.

Thus,

$$\begin{align*}
d(y,B)-d(y,A)&>d(x,B)-\epsilon-\big(d(x,A)+2\epsilon\big)\\
&=d(x,B)-d(x,A)-3\epsilon\\
&=0\,,
\end{align*}$$

and $y\in U$. In other words, the $\epsilon$-ball centred at $x$ is contained in $U$, and $U$ is open.


问:
Recursive sequence with first term defined in interval
I have to prove that the recursive sequence $$a_{n+1}=2a_n(a_n+1)$$ converges for $-1 \le a_0 \lt 0$.

I began by trying to prove that the sequence is decreasing:
$$ a_n - a_{n+1}>0, $$
$$ a_n-2a_n(a_n+1)>0, $$
$$ a_n-2a_n^2 - 2a_n>0. $$
I don't know how to prove that inequality since $a_0$ is defined on an interval.
What should I do next?

答:
Hint 1: Try a few cases in Excel and see where $a_n$ goes.

Hint 2: complete the square on your RHS.

 Next $$a_{n+1} + \frac 12 = 2a_n^2+2a_n+\frac 12 = 2\left(a_n+\frac 12\right)^2.$$ Can you end it now?

 Next: $$2a_{n+1}+1 = (2a_n+1)^2 \implies 2a_n+1 = \left(2a_0+1\right)^{2^n} \to 0, \text{ if } a_0 > -1;\\ 2a_n+1 \equiv  1, \forall n>0, \text { if } a_0=-1.$$


问:
Limit with variable in integral boundary
I have the following integral

$$\int_{\vert x\vert\leq\varepsilon}\frac{1}{\varepsilon}\frac{1}{\vert x\vert^2}\,\mathrm{d}x,$$

where the integral should be understood as a Lebesgue integral and where $x\in\mathbb{R}^{n}$. My question is now, can we say what happens if $\varepsilon\to 0$? I don't know how to approach this question in a mathematical way. The problem here is that on the one hand our integral region becomes smaller and smaller, which means that in the end we will end up with a Lebesgue-zero set, but in the same time our integrand becomes larger and larger.  Any ideas?

答:
Of course if $n\le2$ all the integrals are $+\infty$. Assume $n>2$.

Say $I(\epsilon)$ is the integral in question. We can use a change of variable, in particular a dilation, to rewrite $I(\epsilon)$ in terms of $I(1)$. (Similar tricks are often useful; here it's perhaps motivated by an attempt to fix the difficulty you point out, that the domain and the function are both changing.)

A simple way to get straight how an integral in $\Bbb R^n$ transforms under dilations is this:

If $\lambda<0$ then $\int f(\lambda x)\frac{dx}{|x|^n}=\int f(x)\frac{dx}{|x|^n}.$

We apply this with $f(x)=|x|^{n-2}\chi_{B_1}(x)$: $$\begin{align}I(\epsilon)&=\epsilon^{n-3}\int_{|x/\epsilon|<1}|x/\epsilon|^{n-2}\frac{dx}{|x|^n}
\\&=\epsilon^{n-3}\int f(x/\epsilon)\frac{dx}{|x|^n}
\\&=\epsilon^{n-3}\int f(x)\frac{dx}{|x|^n}
\\&=\epsilon^{n-3}I(1).\end{align}$$

So the limit as $\epsilon\to0$ is this or that, depending on the value of $n$.


问:
Find the components of a vector in the standard basis given a vector in the general basis
The vector $\vec{v}_B$ has the following components in the general basis $B$.

$\vec{v}=\begin{bmatrix}1\\0\end{bmatrix}_{B}$

$B=\left\langle \begin{bmatrix}1\\-1\end{bmatrix},\begin{bmatrix}2\\4\end{bmatrix}\right\rangle$

Find the components of the vector $\vec{v}_\cal{E}$ in the standard basis $\cal{E}$.

$\vec{v}=\begin{bmatrix}x\\y\end{bmatrix}_{{\cal E}}$

${\cal E}=\left\langle \begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix}\right\rangle$

I honestly have no idea. All help is appreciated.

Body: $$\vec v=\begin{bmatrix}\color{blue}1\\\color{blue}0\end{bmatrix}_\mathcal{B}=\color{blue}1 \begin{bmatrix}1\\-1\end{bmatrix}_\mathcal{E}+\;\color{blue}0 \begin{bmatrix}2\\4\end{bmatrix}_\mathcal{E} = \begin{bmatrix}1\\-1\end{bmatrix}_\mathcal{E}$$


问:
Evaluating $\int_0^4 \int_{\sqrt {x}}^2 \frac{x^2 e^{y^2}}{y^5}\mathrm dy\,\mathrm dx$
I need help starting with this. I can't find an example like this anywhere in my book

$$\int_0^4 \int_{\sqrt {x}}^2 \frac{x^2 e^{y^2}}{y^5}\mathrm dy\,\mathrm dx$$

答: Hint: Try the substitution $y^2=u$, then $2ydy=du$. Hence you get $$\int_{\sqrt{x}}^2\frac{e^{y^2}}{y^5}dy=\int_{x}^4 \frac{e^{u}}{2u^3}du$$


问:
Ergodicity under transformation
Suppose $\Omega := [0,1]^{\mathbb Z}$ is equipped with the product topology and endowed with the Borel $\sigma$-algebra $\mathcal B(\Omega)$ and there is a probability measure $\mathbb P$ on $(\Omega,\mathcal B(\Omega))$ such that the shift $T:\Omega \to \Omega$, $$T(\omega)(k) := \omega(k+1),\quad \omega\in\Omega,k\in \mathbb Z$$ is measure preserving, i.e. $\mathbb P = \mathbb P \circ T^{-1}$ on $\mathcal B(\Omega)$, and ergodic, i.e. $A=T^{-1}(A)$ implies $\mathbb P (A)\in\{0,1\}$ for any $A\in\mathcal B(\Omega)$.
Now let $f:[0,1]^3\to[0,1]$ a measurable function and $U:\Omega \to \Omega$ the transformation defined by $$ U(\omega)(k) := f(\omega(2k-1),\omega(2k),\omega(2k+1)),\quad \omega\in\Omega,k\in\mathbb Z.$$ We consider the probability measure $\widetilde {\mathbb P}:= \mathbb P\circ U^{-1}$ where $U^{-1}$ denotes the preimage.

Then, by $T\circ U= U\circ T^2$, it holds that $(\Omega,\mathcal B(\Omega), \widetilde {\mathbb P},T)$ is still a measure-preserving dynamical system. Is it also ergodic?

Edit: What are examples of probability measures $\mathbb P$ on $\mathcal B(\Omega)$ and sets $A\in\mathcal B(\Omega)$ such that $T^{-2}(A)=A$ but $\mathbb P(A)\notin \{0,1\}$ (and hence necessarily $T^{-1}(A)\neq A$)?

答:
The answer is negative: Let \begin{align*} \mathbb P &:= \frac 1 2 \left(\delta_{\left(\mathbb 1_{2 \mathbb Z}(k)\right)_{k\in \mathbb Z}} + \delta_{\left(\mathbb 1_{2 \mathbb Z+1}(k)\right)_{k\in \mathbb Z}} \right), \\ A &:= \left\lbrace (1)_{k\in\mathbb Z} \right\rbrace ,\\ f(x,y,z) &:= y. \end{align*}

The probability $\mathbb P$ corresponds to the irreducible Markov chain on state space $\{0,1\}$ with transition matrix $P = \begin{pmatrix} 0 & 1\\ 1& 0\end{pmatrix}$ which has unique stationary distribution $(\frac 1 2 \,\,\, \frac 1 2)$. In light of the answer to this math.SE question the dynamical system $(\Omega,\mathcal B(\Omega),\mathbb P, T)$ is measure-preserving and ergodic (but not mixing). Now, $T^{-1}(A)=A$ but $$U^{-1}(A) = \prod_{k\in \mathbb Z} \begin{cases} \{1\},&\quad k\in 2 \mathbb Z \\ [0,1],&\quad k\in 2\mathbb Z+1\end{cases}, $$ whence $\widetilde{\mathbb P}(A) = \frac 1 2 $.

Title: Ergodicity under transformation


问:
Solving linear congruence (modular inverse or fraction) via gcd Bezout equation
Isn't finding the inverse of $a$, that is, $a'$ in $aa'\equiv1\pmod{m}$ equivalent to solving the diophantine equation $aa'-mb=1$, where the unknowns are $a'$ and $b$? I have seem some answers on this site (where the extended Euclidean Algorithm is mentioned mainly) as well as looked up some books but there is no mention of this. Am I going wrong somewhere or is this a correct method of finding modular inverses? Also can't we find the Bézout's coefficients by solving the corresponding diophantine equation instead of using the extended Euclidean Algorithm?

答:
Yes, you can find the Bézout's coefficients without using the Euclidean algorithm.

An algorithm can be constructed by 'lifting' some algebra/logic from the Wikipedia existence proof.

Our algorithm operates on the input

$\tag 1 ax+by=z$

as follows:

If $z \mid a$ and $z \mid b$ the algorithm terminates and $\text{(1)}$ is the solution.

If $z \not\mid a$ the algorithm transforms the equation to

$\tag 2 au+bv=r, \text{ where } a = zq + r \;\text{(euclidean division)} \land \big[u = 1-qx \big] \land \big[v = -qy\big]$

and repeats with $u \text{ replacing } x$ and $v \text{ replacing } y$ and and $r \text{ replacing } z$.

If $z \not\mid b$ the algorithm transforms the equation to

$\tag 3 au+bv=r, \text{ where } b = zq + r \;\text{(euclidean division)} \land \big[u = -qx \big] \land \big[v = 1-qy\big]$

and repeats with $u \text{ replacing } x$ and $v \text{ replacing } y$ and and $r \text{ replacing } z$.

You can assume that $a \gt 0$ and initialize the input to the algorithm as the equation

$\quad a(1) + b(0) = (a) ::: a(x) + b(y) = (z)$

Observe that the algorithm iterates on $\text3{-tuples}$ until the desired form is obtained.

Example: Construct the Bézout's identity for $a = 12$ and $b=42$.

$\quad 12\,(1) + 42\,(0) = (12)$ and with $42$, $q = 3$ and $r=6$

$\quad 12\,(-3) + 42\,(1) = (6)$ and with $42$, $q = 6$ and $r=6$

$\quad 12\,(18) + 42\,(-5) = (6)$ and the algorithm stops on the Bézout's identity


问:
Proof that if $|X_n|>n $ i.o. then $|S_n|>n $ i.o.
I’m having trouble with this problem: given $(X_n)$ a sequence of independent and identically distributed random variables such that $E|X_n|=\infty$. First it asks to prove that $|X_n|>n$ i.o., which I did using

$$E|X_1|=\infty \iff \sum_{n=1}^{\infty}P(|X_1|>n)=\infty \iff \sum_{n=1}^{\infty}P(|X_n|>n)=\infty \iff P(\limsup(|X_n|>n))=1$$

So $|X_n|>n$ i.o. happens almost surely. Then it asks to show that $|S_n|>n$ i.o. where $S_n = \sum_{i=1}^n X_i$. I think the argument could be using Borell-Cantelli, but since the $|S_n|$ might not be independent I don’t know how to continue.

The last step of the exercise is to prove that $\limsup |S_n|/n = \infty$ almost surely. Again I think I can play with the series and use Borell-Cantelli but I’m not sure how.

答:
By the same argument of yours, we can see that $|X_n|>2n$ i.o

So if  for a $n$ we have $X_n>2n$, then either $S_{n-1}+X_n>n$ or $S_{n-1} < -n$. Hence either $|S_n|  >n$ or $|S_{n-1}|>n$

Likewise for the case $X_n<-2n$.

Both of them imply that $$|S_n| >n$$ infinitely often.

I owe this answer to this link:
https://mathoverflow.net/questions/376007/random-variables-with-no-first-moment


问:
Convergence of $\sum ( 1 - \frac{2}{\sqrt{n}} )^n$
How can I prove that $\sum ( 1 - \frac{2}{\sqrt{n}} )^n$ converges?

I've tried root test, and ratio test, but it was inconclusive.

Wolfram Alpha also says that this sum is convergent by comparison test.

What series can I use for comparison test?

答:
Another way to see this series converges is by Schlömilch's Test, a generalization of the Cauchy Condensation Test. Basically, if $u(n)$ is a positive increasing sequence with a uniformly bounded ratio of successive differences, then $\sum_{n} f(n)$ converges iff $\sum_{n} (u(n+1)-u(n)) f(u(n))$ converges. In our case, put $u(n)=n^2$: we have $u(n+1)/u(n)<5$ and the new series is
$$
\sum (2n+1)\left(1-\frac{2}{n}\right)^{n^2}
$$Now we can use the Root Test:
$$
\lim_{n\to\infty}\left|\sqrt[n]{(2n+1)\left(1-\frac{2}{n}\right)^{n^2}}\right|
$$We will split this into two limits.
$$
=\lim_{n\to\infty}\left|\sqrt[n]{2n+1}\right|\cdot \lim_{n\to\infty}\left|\left(1-\frac{2}{n}\right)^{n}\right|
$$These limits are both well-known: the first is $1$ and the second is $e^{-2}$. Their product is less than $1$, so we have convergence.


问:
Finding the image for the fundamental group of a covering of $S^1 \vee S^1$
I am representing $S^1 \vee S^1$ via the following diagram:

I have the following connected covering of $S^1 \vee S^1$, call it $p: \tilde{X} \rightarrow X$, for $X=S^1 \vee S^1$.

I am having difficulty in computing $p_{*}(\pi_1(\tilde{X}, \tilde{x}_0))$, where $p_{*}$ indicates the image of the map $p$ under the $\pi_1$ functor.

I know that $\pi_1(\tilde{X},\tilde{x}_0) \cong <\alpha, \beta, \gamma, \sigma, \theta, \mu>$ , but my issue lies in finding the images $\alpha, \dots, \mu$ under $p_*$. My naive guess would be that  $\alpha = b, \beta = a^2, \gamma = b^2,  \sigma = a^2,  \theta = b^2,  \mu = a$ but I believe this doesn't respect the basepoint $\tilde{x}_0$.

答:
The concern you mentioned at the end of your post is pretty accurate. It may be helpful to recall how you arrived at the fundamental group of your covering space, this should give you an answer to your question. The computation can be done via the Seifert-Van-Kampten theorem, that inductively gives you the fundamental group your wedge sum. But to make life easier you will change the basepoint at each step, which is possible for path connected spaces. The change of basepoint also has an explicit isomorphism of fundamental groups by picking a path and conjugating your loop by it (so of the form $f \mapsto sfs^{-1}$). For you that means remembering the path you had to chose to get to the loop you want, I think this is the idea you had, that each loop represents a generator. So $p(\alpha )=  b^{-1}a^{-1}bab$, $p(\beta) = b^{-1}aab$ ... . Algebraic Topology by A.Hatcher also gives those Covering spaces as an example.

Edit: forgot to take inverses


问:
Prove recursion for sum
One should show, that
$$S(n,x,y):= (y-n)\sum_{k=0}^n \binom{n}{k} (x+k)^{k+1} (y-k)^{n-k-1}$$
satisties the recursion
$$S(n,x,y)=x(x+y)^n +n S(n-1,x+1,y-1),\quad \text{for} ~n\geq1,\quad S(0,x,y)=x.$$
I know that we should use Abels' generalization of the binomial theorem, that is
$$\begin{equation} (x + y)^n = \sum_{k=0}^n \binom{n}{k} x (x-kz)^{k-1} (y + kz)^{n-k}\end{equation}$$
for any $x, y, z \in\mathbb{R}$.

STARTING SOLUTION:

If I start with the right side, I get
$$\begin{align*}x (x+y)^n &+ n S(n-1,x+1,y-1) \\
         &= x (x+y)^n + n (y-n+1) \sum_{k=0}^{n-1} \binom{n-1}{k} (x+1+k)^{k+1} (y-1-k)^{n-k-2}\\
         &= x\sum_{k=0}^n \binom{n}{k} x (x-kz)^{k-1} (y + kz)^{n-k} \\
         &\hspace{2.1cm}+ n (y-n+1) \sum_{k=0}^{n-1} \binom{n-1}{k} (x+1+k)^{k+1} (y-1-k)^{n-k-2}\\
         &=\cdots\\
         &= (y-n) \sum_{k=0}^n \binom{n}{k} (x+k)^{k+1} (y-k)^{n-k-1} = S(n,x,y),\end{align*}$$
I do not really know how to continue from here, those anyone have a clue what could be inserted for z (and how one could shift the indices of the second sum to get the right term)?

答:
We start with the right-hand side and keep at first the focus on $nS(n-1,x+1,y+1)$.

We obtain
\begin{align*}
&\color{blue}{nS(n-1,x+1,y-1)}\\
&\qquad =n(y-n)\sum_{k=0}^{n-1}\binom{n-1}{k}(x+1+k)^{k+1}(y-1-k)^{n-k-2}\tag{1}\\
&\qquad= n(y-n)\sum_{k=1}^{n}\binom{n-1}{k-1}(x+k)^k(y-k)^{n-k-1}\tag{2}\\
&\qquad=(y-n)\sum_{k=0}^n\binom{n}{k}k(x+k)^k(y-k)^{n-k-1}\tag{3}\\
&\qquad\,\,\color{blue}{=S(n,x,y)-(y-n)\sum_{k=0}\binom{n}{k}x(x+k)^k(y-k)^{n-k-1}}\tag{4}
\end{align*}

Comment:

In (1) we use the definition of $S(n,x,y)$.

In (2) we shift the index $k$ by one.

In (3) we use the binomial identity $\binom{n}{k}=\binom{n-1}{k-1}\frac{n}{k}$. We also start with index $k=0$ which doesn't change anything.

In (4) we use $k=(x+k)-x$ and obtain a representation with $S(n,x,y)$.

Looking at (4) we observe the following is left to show:
\begin{align*}
(x+y)^n=\color{blue}{(y-n)}\sum_{k=0}\binom{n}{k}(x+k)^k(y-k)^{n-k-1}\tag{5}
\end{align*}

We use Abel's generalisation with $z=-1$ which gives
\begin{align*}
(x + y)^n = \color{blue}{x}\sum_{k=0}^n \binom{n}{k} (x+k)^{k-1} (y - k)^{n-k}\tag{6}
\end{align*}
and we see (6) is close to the wanted representation (5) indicating we could give the substitution $\color{blue}{x\to y-n}$ a try.

This substitution finally does the job.

We obtain
\begin{align*}
\color{blue}{(x+y)^n}&=((y-n)+(x+n))^n\tag{7}\\
&=\sum_{k=0}^n\binom{n}{k}(y-n)((y-n)+k)^{k-1}((x+n)-k)^{n-k}\tag{8}\\
&=\sum_{k=0}^n\binom{n}{k}(y-n)(y-(n-k))^{k-1}(x+(n-k))^{n-k}\\
&\,\,\color{blue}{=(y-n)\sum_{k=0}^n\binom{n}{k}(y-k)^{n-k-1}(x+k)^{k}}\tag{9}\\
\end{align*}
which is besides a factor $x$ the sum in (4) and the claim follows.

Comment:

In (7) we use the substitution $x\to y-n$ and $y\to x+n$.

In (8) we use Abel's generalised binomial theorem from (6).

In (9) we change the order of summation $k\to n-k$.

Title: Prove recursion for sum


问:
Limit in two variables of $x e^{-y^2/x}$

I'm trying to compute the limit

$$\lim_{(x,y)\rightarrow (0,0)} x e^{-y^2/x}$$

In order to compute it, I would just use the following bound:

$$e^{-\frac{-y^2}{x}} \leq 1$$ and hence

$$|x e^{-y^2/x}|$$ and hence the limit is $0$. Is it right?

Using polar coordinates I obtain

$\lim_{r \rightarrow 0} r cos(\theta) e^{- r \frac{sin^2(\theta)}{cos(\theta)}}$

why isn't this $0$?

答:
If $$f(x,y)=xe^{-\frac{y^2}{x}}$$

then

$$F(y)=f(y^3,y)=y^3e^{-\frac 1y}$$

and

$$\lim_{y\to 0^-}F(y)=-\infty$$

Your limit does not exist.

Observe also that
$$\lim_{y\to 0}f(y^2,y)=0$$


问:
A difficulty with Fourier sine transform
I'm trying to work through an exercise on Fourier sine transform. The function given is
$$ f(x) = \frac{\sin(\omega x)}{x^2 + 1}\quad \omega > 0 $$
with Fourier sine transform pair defined as
$$ f_s(x) = \sqrt{2}f(x),$$
$$\hat{F}_s(k) = \int_{0}^{\infty}f_s(x) \sin(kx)\mathrm{d}x.$$

I'm told I need to consider two cases: $\omega > k$ and $\omega < k$ and I'm not sure why, I think it is something to do with Jordan's lemma when performing the contour integration.

答:
You have
$$
\hat{F}_s(k)
= \int_{0}^{\infty}f_s(x) \sin(kx)\mathrm{d}x
= \int_{0}^{\infty} \sqrt{2} f(x) \sin(kx)\mathrm{d}x
= \sqrt{2} \int_{0}^{\infty} \frac{\sin(\omega x)}{x^2+1} \sin(kx)\mathrm{d}x
\\
= \sqrt{2} \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin(\omega x)}{x^2+1} \sin(kx)\mathrm{d}x
= \sqrt{2} \frac{1}{4} \int_{-\infty}^{\infty} \frac{\cos(\omega x-k x) - \cos(\omega x+k x)}{x^2+1} \mathrm{d}x
\\
= \operatorname{Re} \sqrt{2} \frac{1}{4} \int_{-\infty}^{\infty} \frac{e^{i(\omega-k)x} - e^{i(\omega+k)}}{x^2+1} \mathrm{d}x
$$

Now calculate
$$
I_\pm := \int_{-\infty}^{\infty} \frac{e^{i(\omega\pm k)x}}{x^2+1} \mathrm{d}x
$$
by a standard contour integration along $[-R,R]$ on the real axis and a semicircle. Be careful with whether the semicircle should be in the upper or lower half-plane to make the integral over it vanish when $R\to\infty.$ For that you will have to look at the sign of $\omega\pm k.$


问:
If X is normal, A closed in X, U an open set in X that contains A, find a set V open in X that contains A whose closure is contained in U
My attempt at a solution: since X is normal, it is regular, and this implies U is regular as a subspace of X.  Now, since A is closed in X, it is closed in U.  If A=U, we're done; U is closed since A is closed, so  U itself is the set V we are looking for (closure of U is just U in this case).  If A is not equal to U, we can find some point x in U that isn't in A, and apply regularity: we can find open, disjoint sets U', V such that U' contains x and V contains A.  Then since U is closed in itself, the closure of V must be a subset of U, as desired, so this is the set we're looking for.

The problem with this solution that was pointed out to me is that I made the mistake of conflating the closure of V in U with the closure of V in X.  But is there a way to salvage this proof, or do we need to do something completely different?

答:
You need something completely different (and much simpler). Let $B=X\setminus U$; then $A$ and $B$ are disjoint closed subsets of the normal space $X$, so there are are disjoint open sets $V$ and $W$ such that $A\subseteq V$ and $B\subseteq W$. $V\cap W=\varnothing$, so $V\subseteq X\setminus W$. Finally $X\setminus W$ is closed, so

$$\operatorname{cl}V\subseteq X\setminus W\subseteq X\setminus B=U\,.$$


问:
Uniform Convergence on $[-b,b]$.
Suppose that $f_0 = 1$ and for $n \geq 1$, $$f_n(x)  = \sum_{k=0}^{n-1} \frac{x^{2k+1}}{\prod\limits_{i=0}^k(2i+1)} + \frac{x^{2n}}{\prod\limits_{i=1}^n (2i)}.$$

I'm wondering whether this sequence $(f_n)$ converges uniformly on $[-b,b]$ where $b > 0$. I'm trying to show that it's Cauchy in $C[-n,n]$, but this seems to be really difficult. Is there any hint or suggestion?

答: Yes, it does. All that you need to prove is that the series$$\sum_{k=0}^\infty\frac{x^{2k+1}}{\prod_{i=0}^k(2i+1)}$$and the sequence$$\left(\frac{x^{2n}}{\prod_{i=0}^n(2i)}\right)_{n\in\Bbb N}$$both converge uniformly on $[-b,b]$. The first assertion is true by the Weierstrass $M$-test: $\sup_{x\in[-b,b]}\left|\frac{x^{2k+1}}{\prod_{i=0}^k(2i+1)}\right|=\frac{b^{2k+1}}{\prod_{i=0}^k(2i+1)}$ and the series $\sum_{k=0}^\infty\frac{b^{2k+1}}{\prod_{i=0}^k(2i+1)}$ converges (by, say, the ratio test). And, on the other hand, $\sup_{x\in[-b,b]}\left|\frac{x^{2n}}{\prod_{i=0}^n(2i)}\right|\leqslant\frac{b^{2n}}{\prod_{i=0}^n(2i)}$ and $\lim_{n\to\infty}\frac{b^{2n}}{\prod_{i=0}^n(2i)}=0$.

Title: Uniform Convergence on $[-b,b]$.


问:
I want to figure out my mistake in a prime factorization problem.
Here is the problem.
"How many ordered pairs $(x,y)$ of positive integers satisfy the equation $xy = 144$?"

I would consider this to be a fairly easy problem as you just have to find the number of factors. I did the prime factorization of $144$ and got $2^4 \cdot{3^2}$. With that, the number of factors is $(4 + 1) x (2 + 1) = 15$. Since it's odd, that means $144$ is a perfect square and therefore involves $7$ pairs of factors and another pair to account for the perfect square. So I got $8$ as my answer.

However the answer that is given is $15$. I'm not sure where I went wrong here.

答:
Because $x$ and $y$ are separate variables, the ordered pairs $(1,144)$ and $(144,1)$ are different solutions.

So, there are $15$ solutions.


问:
Proving a matrix is not nilpotent

Let $A$ be an $n \times n$ matrix with real entries. Prove that no matrix $A^2 + A + I$ is nilpotent.

I tried to approach by contradiction and suppose that such an $A$ did exist so $(A^2 + A + I)^n = [0]$. I then need to figure $A^2 + A + I$, but the function $x^2 + x + 1$ doesn't factor over $\mathbb{R}$, and the factorization over $\mathbb{C}$ isn't very helpful. Is the fact that it doesn't factor over $\mathbb{R}$ enough?

答:
This statement is false. We find that the companion matrix associated with the polynomial $(x^2 + x + 1)^2 = x^4 + 2x^3 + 3x^2 + 2x + 1$ is given by
$$
A = \pmatrix{0&0&0&-2\\1&0&0&-3\\0&1&0&-2\\0&0&1&-1}.
$$
This matrix has real entries, but it satisfies $(A^2 + A + I)^2 = 0$. In fact, we find that $A^2 + A + I \neq 0$, which means that $A^2 + A + I$ is a non-zero nilpotent matrix.

For posterity: as discussed in the comments to this answer, you can get an answer in "real Jordan form", namely
$$
A = \pmatrix{C&I\\0 & C}, \quad C = \pmatrix{-1/2 & -\sqrt{3}/2\\\sqrt{3}/2& -1/2}.
$$
We could also take
$$
C = \pmatrix{0&-1\\1&-1},
$$
which is an interesting blend of approaches.

If you use another version of the real Jordan form, you might instead end up with the example
$$
A = \left[\begin{array}{cc|cc}
-1/2 & 1 & -\sqrt{3}/2 & 0\\
0 & -1/2 & 0 & -\sqrt{3}/2\\
\hline
\sqrt{3}/2 & 0 & -1/2 & 1\\
0 & \sqrt{3}/2 & 0 & -1/2
\end{array}\right].
$$

Title: Proving a matrix is not nilpotent


Title: Recursive sequence with first term defined in interval

Body:
I have to prove that the recursive sequence $$a_{n+1}=2a_n(a_n+1)$$ converges for $-1 \le a_0 \lt 0$.

I began by trying to prove that the sequence is decreasing:
$$ a_n - a_{n+1}>0, $$
$$ a_n-2a_n(a_n+1)>0, $$
$$ a_n-2a_n^2 - 2a_n>0. $$
I don't know how to prove that inequality since $a_0$ is defined on an interval.
What should I do next?

Body:
If
$a_{n+1}=2a_n(a_n+1)$
then
$2a_{n+1}
=4a_n(a_n+1)
=(2a_n+1)^2-1
$
so
$2a_{n+1}+1
=(2a_n+1)^2
$.

Let
$b_n = 2a_n+1
$.
Then
$b_{n+1}
=b_n^2
$
so
$b_{n+k}
=b_n^{2^k}
$.

Since
$-1 \le a_0 < 0$,
$-1 \le b_n < 1$.

If $a_0 = -1, a_n = 0$ for $n \ge 1$.

If $a_0 >-1$,
then
$-1 < b_0 < 1$
so
$b_n
=b_0^{2^n}
\to 0$
so
$a_n
=\dfrac{b_n-1}{2}
\to -\dfrac12
$.


问:
BA diagonalisable show that AB diagonalisable ( A and B are not square)
let A $\in \mathcal{M}_{4,3}(\mathbb{R})$ and B $\in \mathcal{M}_{3,4}(\mathbb{R})$ such as BA$ = \begin{pmatrix} 0&1&1 \\ 1&0&1 \\ 1&1&0 \end{pmatrix}$

show that AB is diagonalizable

答:
we have $\:\:\:\:p_{BA}(t) = \begin{vmatrix} t&-1&-1 \\ -1&t&-1 \\ -1&-1&t \end{vmatrix}  = \begin{vmatrix} t-2&-1&-1 \\ t-2&t&-1 \\ t-2&-1&t \end{vmatrix} =\:(t-2) \begin{vmatrix} 1&0&0 \\ 1&t+1&0 \\ 1&-2&t+1 \end{vmatrix} =(t-2)(t+1)^{2}$
$BA+I_3= \begin{pmatrix} 1&1&1 \\ 1&1&1 \\ 1&1&1 \end{pmatrix}$ we have clearly $\:\:\:\:rg(BA+I_3)=1$ thus dim$(Ker((BA+I_3))=2 $ and BA is diagonalisable.

using the relation $t^4p_{BA}(t)=t^3p_{AB}(t)\:\:\:$we have $p_{AB}(t)=t(t-2)(t+1)^{2}$ thus $0,(-1),(-1)\:$and$\:2$ are the four eigenvalues of AB
thus $\mathbb{R^4} = Ker(AB) ⊕ Ker(AB-2I_4)⊕Ker((AB+I_4)^2)\\
dim(ker(AB)) = dim(Ker(AB-2I_4))=1\\ (the\:dimention\:of\: the\: eigenspace \:is\: majorised\: by\: the\: multiplicity\: of \:the \:associated \:eigenvalue) \:thus\: dim(Ker(AB+I_4)^2)=2$

I would like to show that dim$(Ker(AB+I_4)) =2$

$edit: \\ $ On one site, we have  $dim(Ker(AB+I_4)) \le 2 (same \: reason\: cf\: above)$

On the other side
$\\ \begin{array}{ccccc}
Φ & : & Ker(BA+I_3) & \to & Ker(AB+I_4) \\
 & & v & \mapsto & Av \\
\end{array}$

Φ  is well defined and clearly linear $\:\:\:\:\:\:let \:v \in Ker(BA+I_3) \\ ABAv =A*-1*v=-1*A*v$

Φ is injective $let \:x \in Ker(Φ) \\ Ax = 0 \:\: ie \:\: x \in Ker(A)\cap Ker(BA+I_3) \\(BA+I_3)x = BAx +I_3x =0 = I_3x\\ so \: x = 0 \:since \:I_3 \: is \:invertible \:thus\: ker(Φ)= {0}$

As a consequence $dim(Ker(BA+I_3))\leq dim(Ker(AB+I_4)) \:ie\: 2\leq dim(Ker(AB+I_4))$

We have dim$(Ker(AB+I_4)) =2$ thus $Ker(AB+I_4) = Ker((AB+I_4)^2)$ thus $ \mathbb{R^4} = Ker(AB) ⊕ Ker(AB-2I_4)⊕Ker(AB+I_4)$ thus AB is diagonalisable.


问:
How to prove $a \ln(b) \sqrt{b} > b \ln(a) \sqrt{a}$ for $a,b > 0$ and $a>b$
Visually, it is easy to see that the inequality holds for $a \neq b$ and $a,b > 0$. Do note that I'm only considering $a,b \in \mathbb{N}$ so I may miss something if they were real numbers.

I did think earlier about the inequality $\ln(x) < \sqrt{x}$ but I don't know how to prove $\ln(x)\sqrt{x} < x$, which I think would help.

答: The inequality doesn't hold. If $b=1$, then LHS=0, RHS > 0.


问:
How do you denote and define the slope of the complex tangent plane to a function from $\Bbb C$ to $\Bbb R$?
Consider the function

\begin{align}
f: \Bbb C &\to \Bbb R\\
f(x) &\mapsto \frac{\lvert x\rvert^2}2
\end{align}

What is the complex slope of the tangent plane at $(x, f(x)$.  If this isn't $f'(x)$, does it have a name?

Also, what is $f'(x)$?  What is its range and how is it defined in terms of limits?  WolframAlpha seems to be defining this somehow.

答:
For a continuous function $f:\Bbb{R}^n \to \Bbb{R}^k$ with continuous partial derivatives, we look at the directional derivative $$D_v f(a) = \lim_{h\to 0,h\in \Bbb{R}}\frac{f(a+tv)-f(a)}{t}, \qquad a,v\in \Bbb{R}^n$$
For each fixed $a$ then $v\to D_v f(a)$ is a $\Bbb{R}$-linear map, which is just $$D_v f(a)=\sum_{j=1}^n v_j \partial_{x_j} f(a)$$
It works the same way for your function under the identification $\Bbb{C}\cong \Bbb{R}^2$, $n=2,k=1$.

Your 'tangent plane' is $(a,f(a))+\Bbb{R}(1,D_1(a))+\Bbb{R}(i,D_i(a))$.

As you see no complex slope appears.

When $f:\Bbb{C\to C}$, again with the identification $\Bbb{C}\cong \Bbb{R}^2$, if $v\to D_vf(a)$ is $\Bbb{C}$-linear then $f$ is said complex differentiable at $a$, and holomorphic when it is so at every $a$ on an open.


问:
Degree of a splitting field over $\mathbb{Q}$
How to calculate the degree of the splitting field $X^3+2$ over $\mathbb{Q}$.

I know that I need to first find the minimal polynomial which will give me what the splitting field is equal to but I don't know how to factorise it.

答:
The splitting field is $\mathbb{Q}(r_1,r_2,r_3)$, where $r_1,r_2,r_3$ are the three roots of the polynomial $X^3+2$.

This is just the definition, but it's the starting point anyway. How do we know that the polynomial has distinct roots? The derivative test, of course: the derivative is $3X^2$ which has no nonconstant common factor with $X^3+2$.

Now let's look at the polynomial again: it has no rational roots, because $\pm1$ and $\pm2$ aren't roots. Since it has degree three, it is irreducible over $\mathbb{Q}$, but this is not that important in order to find the splitting field.

Suppose we are in an extension where all roots exist. Thus we can factor $X^3+2=(X-r_1)(X-r_2)(X-r_3)$; however, $r_1^3=-2$, so we also get the factorization
$$
(X-r_1)(X^2+r_1X+r_1^2)
$$
What does this entail? That $(X-r_2)(X-r_3)=X^2+r_1X+r_1^2$ and therefore
$$
r_2+r_3=-r_1,\quad r_2r_3=r_1^2
$$
This is the same as saying that $a+b=-1$, $ab=1$, where $a=r_2/r_1$, $b=r_3/r_1$. In particular $a$ and $b$ are roots of $X^2+X+1$, so they're the nonreal cube roots of $1$. If we call $\omega$ one of them, the other one is $\omega^2$.

Hence $r_2=\omega r_1$ and $r_3=\omega^2 r_1$. Now we can make a choice: take $r_1=r=-\sqrt[3]{2}$ (the real cube root). Then the splitting field is
$$
\mathbb{Q}(r,\omega r,\omega^2r)=\mathbb{Q}(r,\omega)
$$
You know that $[\mathbb{Q}(r):\mathbb{Q}]=3$ and $[\mathbb{Q}(\omega):\mathbb{Q}]=2$.

Can you compute $[\mathbb{Q}(r,\omega):\mathbb{Q}(r)]$?

You're starting with the wrong foot. The given polynomial is irreducible, so it is the minimal polynomial of any of its roots.

You also seem to believe that if $f(X)$ is the minimal polynomial for some algebraic number, then $\mathbb{Q}[X]/(f(X))$ is the splitting field for $f(X)$. This is not true, in general, and the example shows it: the minimal polynomial for $r=-\sqrt[3]{2}$ is $X^3+2$, but $\mathbb{Q}[X]/(X^3+2)\cong\mathbb{Q}(r)$ is not the splitting field because it doesn't contain $\omega$.


问:
Homomorphism between $Q_8$ and $S_4$
Prove that there does not exist a homomorphism $\phi :Q_8 \rightarrow S_4$ such that $\phi (i) = (1 2 3 4), \phi (j) = (1 2 4 3) $.

Using homomorphism properties, we obtain $\phi(i) \phi(j) = \phi(ij) = \phi(k) = (1 2 3 4)(1 2 4 3) = (1 3 2)$, then $\phi(j) \phi(k) = \phi(jk) = \phi(i) = (1 2 4 3)(1 3 2) = (3 4)\neq (1 3 2)$, hence $\phi$ is not even a well-defined function. Is what I did right? I am not completely convinced. Thanks!

答:
Your calculations are good and your proof is right, you could perhaps word it the following way to make the logic crystal clear:

Let $\phi$ be some arbitrary function from $Q_8$ to $S_4$ that happens to take $i$ to $(1234)$ and $j$ to $(1243)$.

Let's show that $\phi$ is not a homomorphism.

If $\phi$ were a homomorphism, then $\phi(i) = \phi(jij) = \phi(j)\phi(i)\phi(j)$. But this implies
$$
(1234) = (1243)(1234)(1243)
$$
which is false: the right hand side fixes $1$ and the left hand side doesn't.

Title: Homomorphism between $Q_8$ and $S_4$


问:
Convergence of $\int_{x=0}^\infty x^8 e^{-\sqrt x} dx$

Does $\int_{x=0}^\infty x^8 e^{-\sqrt x} dx$ converge?

I've tried splitting it into $\int_{x=0}^1 x^8 e^{-\sqrt x} dx$ + $\int_{x=1}^\infty x^8 e^{-\sqrt x} dx$. The first one is a continuous function over a bounded interval so that converges however I'm struggling with the second one. I have tried using limit comparison with $e^{- \frac{\sqrt x}{2}}$ as a test function but ended up going in circles. I have also tried direct comparison and integration by parts.

Thanks

Morgzy

答:
Put $t=\sqrt{x}$ or $t^2=x$, which yields $2t\,dt=dx$ and $x^8=t^{16}$. Then we have
$$
\int_0^{\infty}x^8 e^{-\sqrt{x}}\,dx = 2\int_0^{\infty} t^{17}e^{-t}\,dt
$$This is the well-known Gamma function. It converges as long as the power of $t$ is non-negative. You can figure out the exact value yourself by doing integration by parts once or twice until you see the pattern; you should get $711374856192000$.


答:
A limit $L$ of a multivariate function $f$ exists at a point $P$ if any only the limits taken on any path at $P$ is equal to $L$.

This is quite technical, so I'll simplify it. Take $P=(x_0, y_0)$ (In your case, $P=(0,0)$). Then, a path that meets $P$ is any continuous function $y(x)$ where $y(x_0)=y_0$. (I.e. $P$ is a point on the path) Likewise, we might also use $x(y)$ s.t. $x(y_0)=x_0$.

Then we may take: $$\lim_{(x,y)\to (x_0, y_0)}f(x,y)=\lim_{x\to x_0}f(x,y(x))$$

If the right-hand limit is the same no matter what $y(x)$ we pick (subject to the above constraint that $y(x_0)=y_0$), then it's limit $L$ is the limit of the left-hand side. If there is any function where they are unequal or the limit doesn't exist, the multivariate function has no limit.

hamam_Abdallah pointed out in his answer that taking $x(y)=y^3$, we have that $x(0)=0$ so the constraint is met, and then $$\lim_{(x,y)\to (0,0)}f(x,y)=\lim_{y\to 0}y^3e^{-\frac 1y}$$
which doesn't exist (The LHS limit is $-\infty$ and the RHS limit is $0$), and so the general limit doesn't exist.

Title: Limit in two variables of $x e^{-y^2/x}$

Body:
I'm trying to compute the limit

$$\lim_{(x,y)\rightarrow (0,0)} x e^{-y^2/x}$$

In order to compute it, I would just use the following bound:

$$e^{-\frac{-y^2}{x}} \leq 1$$ and hence

$$|x e^{-y^2/x}|$$ and hence the limit is $0$. Is it right?

Using polar coordinates I obtain

$\lim_{r \rightarrow 0} r cos(\theta) e^{- r \frac{sin^2(\theta)}{cos(\theta)}}$

why isn't this $0$?


答:
Your argument is correct. There is another, perhaps simpler argument:
$$
(1234)^2=(13)(24),\qquad (1243)^2=(14)(23)
$$
so $\phi(i^2)=(\phi(i))^2=(13)(24)$. On the other hand $\phi(j^2)=(14)(23)$. A contradiction, because $i^2=j^2=-1$.

Title: Homomorphism between $Q_8$ and $S_4$

Body:
Prove that there does not exist a homomorphism $\phi :Q_8 \rightarrow S_4$ such that $\phi (i) = (1 2 3 4), \phi (j) = (1 2 4 3) $.

Using homomorphism properties, we obtain $\phi(i) \phi(j) = \phi(ij) = \phi(k) = (1 2 3 4)(1 2 4 3) = (1 3 2)$, then $\phi(j) \phi(k) = \phi(jk) = \phi(i) = (1 2 4 3)(1 3 2) = (3 4)\neq (1 3 2)$, hence $\phi$ is not even a well-defined function. Is what I did right? I am not completely convinced. Thanks!


问:
Equivalence of the two expression for dual averaging
I have a question about two versions of the algorithm for dual averaging.
The first one is
$$x_{t+1}=x_t+\nabla f(w_t)\to w_{t+1}=P_C\left(-\frac{1}{\sqrt{t+1}}x_{t+1}\right)=\arg\min_{w\in C}\left\|w+\frac{1}{\sqrt{t+1}}x_{t+1}\right\|_2$$
The second one is

What should we define the $d$ and $\beta$ and $p$ so that we would have the equivalent algorithm?
The objective function to solve is $\min_{w\in C}f(w)$ where $C$ is compact convex and $f$ is closed convex. I have been given a smoothed dual $\min_sf^*(s)-\min_{w\in C}\langle w,s\rangle+\frac{1}{2\mu}\|w\|_2^2$ where $f^*$ is Fenchel Conjugate, but I do not know where it comes from and will this help building the equivalence.

Source: Here

答:
The first algorithm is the vanilla projected subgradient method (with the standard diminishing step size $\beta_t = \frac{1}{\sqrt{t+1}}$) for minimizing a set-constrained minimization problem: $\boxed{\min_{w\in C} f(w)}$.

The second algorithm is a more advanced variant of the same method, but for solving an unconstrained maximization problem: $\boxed{\max_{\lambda\in\mathcal{R}^n} d(\lambda)}$.

Note that the second algorithm was presented as a solution to a dual problem to another constrained minimization problem, and this dual problem turns out to be unconstrained. However, it can handle constrained problems $\boxed{\max_{\lambda\in C} d(\lambda)}$ just as well, with a small adjustment of the projection step:
\begin{equation}
\pi_{\beta}(x) = \arg\min_{\lambda\in C} \{\beta p(\lambda) - \langle x, \lambda\rangle\}.
\end{equation}
Indeed, this is equivalent to the unconstrained projection: it suffices to replace $p(\lambda)$ by $p(\lambda) + \mathcal{I}_{C}(\lambda)$ where $\mathcal{I}_{C}$ is the indicator of the set $C$. If $C$ is closed convex and $p$ is strongly convex then the new proximity function $p + \mathcal{I}_{C}$ is also strongly convex, thus the assumptions of the second (unconstrained) algorithm still hold.

Now comes your main question: How to obtain the first algorithm from the second one?

The answer is simple, just make the following change of notation:

\begin{align}
\lambda &\to w \\
d &\to -f \\
x &\to -x \\
k &\to t \\
\gamma_k &\to 1 \\
\beta_k &\to \frac{1}{\sqrt{k+1}} = \frac{1}{\sqrt{t+1}} \\
p(\lambda) &\to p(w) = \frac{1}{2}\|w\|_2^2.
\end{align}

Note that I have used the constrained variant of the second algorithm. If you want to use the unconstrained one, just replace the last step with $p(\lambda) \to p(w) = \frac{1}{2}\|w\|_2^2 + \mathcal{I}_{C}(w)$.


问:
Probability, Lebesgue Stieltjes measure
Assume $ F $ is a non-decreasing, right continuous function $ F: \mathbb{R} \to [0,1] $ such that

$F(\infty)=1 $ and $F(-\infty) = 0 $.

By Lebesgue Stieltjas measure theorem, there exists a unique probability measure denoted by $ dF $, such that $ (\mathbb{R},\mathcal(B),dF) $ is a probability space.

Now let for any $ D \in \mathcal{B} $ , let $ \mathcal{B} (D) = \{A\cap D :A\in \mathcal{B}\}$.

For any $ A\in \mathcal{B} $ such that $ A\subseteq [0,1] $ define $ \mathbb{P}(A)=dF(A) $.

Is it true that $ ([0,1],\mathcal{B}([0,1]),\mathbb{P} $ is a probability space?

答:
No, not in general, since you cannot conclude that $dF([0,1])=1$. You can actually compute the total mass as
$$dF([0,1]) = F(1) - \lim_{x\uparrow 0}F(x)$$
and thus $\mathbb{P}$ is a probability measure if and only if $F(1)=1$ and $\lim_{x\uparrow 0}F(x)=0$, where the last condition is equivalent to $F(0)=0$.

If you want to secure that $\mathbb{P}$ becomes a probability measure, you can instead define it as
$$\mathbb{P}(A) = \frac{dF(A)}{dF([0,1])} \quad \quad (A\subseteq [0,1]).$$
The last formula should look familiar and can be interpreted as the conditional probability of the event $A$ given that we know that the outcome is inside $[0,1]$.

Title: Probability, Lebesgue Stieltjes measure


问:
This finite sum involving roots of unity is bounded, but why?
For $N \in \mathbb{N}$, let $\omega := \exp(\frac{2 \pi}{N} \sqrt{-1} )$. For $k,l \in \{1, \dots, N \}$, define

$$d_{k,l}^{(N)} := \begin{cases}
 \frac{\sin\left(\frac{2\pi}{N}(k-1)\right)}{\sin\left(\frac{2\pi}{N}(k-1)\right)^2 + \sin\left(\frac{2\pi}{N}(l-1)\right)^2}&\mbox{ if } \sin\left(\frac{2\pi}{N}(k-1)\right) \neq 0\\
 0 & \mbox{ otherwise }
 \end{cases}$$

For $i,j,m,n \in \{1, \dots, N\}$, let

$$a_{i,j,m,n}^{(N)} :=\frac{1}{N^2}  \sum_{l=1}^N \sum_{k=1}^N (\omega^{i-j})^{k-1} (\omega^{m-n})^{l-1} d^{(N)}_{k,l} $$

Numerical experiments very strongly suggest that there exists $C>0$ independent of $N$ ($C \approx 0.25$) such that $\forall N$,
$$
 \max_{1 \le i,j,m,n \le N} | a^{(N)}_{i,j,m,n} | \le C
$$

How can one formally show that this is true?

答:
I eventually found a (fairly lengthy) proof of this fact. I will not write it in detail, but I sketch the ideas, in hopes somebody might eventually find them helpful -- I would be happy to clarify any question in the comments. Let for convenience $p := i-j$, $q := m-n$.

Using Euler's formula $e^{ix} = \cos(x) + i\sin(x)$, expand the terms $\omega^{p(k-1)}$ and $\omega^{q(l-1)}$ as sines and cosines.

The sum can then be identified with four Riemann sums, for integrals
$$\int_{y=0}^{2\pi}\int_{x=0}^{2\pi} \frac{\sin(x)}{\sin(x)^2 + \sin(y)^2 }h^x(px)h^y(qy) \,dx\, dy $$
for $h^x, h^y \in \{ \sin, \cos\}$, up to multiplicative constants.

A simple argument of parity shows that three of these four integrals are zero, and the only nonzero is:
$$I_{p,q} := \int_{y=0}^{2\pi}\int_{x=0}^{2\pi} \frac{\sin(x)}{\sin(x)^2 + \sin(y)^2 }\sin(px)\cos(qy) \,dx\, dy $$

I refer to user TheSimpliFire's excellent proof here that this integral tends to zero as $q\to\infty$. A completely analogous argument shows that it also tends to zero as $p\to\infty$.

A careful application of Riemann-Lebesgue's lemma to the integrals in the proof above determines that the rate of convergence to zero is $I_{p,q} = O(p^{-1}q^{-2})$, and $I_{p,0} = O(p^{-1})$. The asymmetry with respect to $p$ and $q$ is caused by the following fact: if one takes the innermost integral to be the one with respect to $x$, as in the aforementioned proof, then eventually one arrives to an integral
of the form $\int_0^{2\pi} g(y)\cos(qy)\,dy $, with $g$ smooth and such that $g(0) = g(2\pi)$. Riemann-Lebesgue thus gives a bound of $O(q^{-2})$. Conversely, integrating first by $y$, one reaches an integrand of the form $\int_0^{2\pi} f(x)\sin(px) \,dx$ with $f$ smooth but $f(0)\neq f(2\pi)$; Riemann-Lebesgue only gives a $O(p^{-1})$ bound.

The integrands $\frac{\sin(x)}{\sin(x)^2 + \sin(y)^2 }h^x(px)h^y(qy)$ are (smooth) periodic functions, integrated over one period. Periodic functions display accelerated convergence of its Riemann sums (e.g. see Guillemin and Stroock's "Some Riemann sums are better than others"); so the error of the Riemann sums, $\epsilon_N$, decays faster than any polynomial of $N (\ge |p|,|q|)$, and can effectively be ignored.

In conclusion, if $m\neq n$,
$$a_{i,j,m,n}^{(N)} = \begin{cases}\epsilon_N + O(|i-j|^{-1}|m-n|^{-2}) &\text{ if }m\neq n \\ \epsilon_N + O(|i-j|^{-1}) &\text{ if } m=n,\,i\neq j\\
\epsilon_N &\text{ if } m=n,\,i=j\end{cases}$$

Which is obviously bounded for all $i,j,m,n$.


问:
Connection between the predual of a von Neumann algenra and $L^{\infty}(\mu)$
I saw the following conclusion from Miguel Martin and Yoshimichi Ueda's paper.

Let $M$ be an arbitrary von Neuamnn algebra and denote its unique predual by $M_{*}$, and they are known to be non-commutative counterparts to $L_{\infty}(\mu)$ and $L_1(\mu)$, respectively.

How can we view $M_{*}$ as non-commutative counterparts to $L_{\infty}(\mu)$ and $L_1(\mu)$, respectively

答:
The category of commutative von Neumann algebras is
contravariantly equivalent
to the category of compact strictly localizable enhanced measurable spaces
with morphisms being equivalence classes of measurable measure-zero reflecting
maps modulo the equivalence relation of weak equality almost everywhere.
(Here “compact” is an abstract measure-theoretic analogue of Radon measures,
and “strictly localizable” generalizes σ-finiteness.)

Under this correspondence, commutative von Neumann algebras
are precisely algebras of bounded measurable functions
on compact strictly localizable enhanced measurable spaces,
modulo equality almost everywhere.
Preduals of commutative von Neumann algebras
are precisely the vector spaces of finite complex-valued measures on such spaces,
or, equivalently, L^1-spaces with respect to any faithful measure.

Thus, it is completely natural to see the predual of a noncommutative
von Neumann algebra as the vector space of measures
on a noncommutative measurable space.


问:
Spectrum of fourier transform of $\delta(t-t_0)$
We know that $$\mathcal{F}(\delta(t))=1.$$ In frequency domain, it is of magnitude $1$ for all frequency $\omega = 2\pi f$.

However $$\mathcal{F}(\delta(t-t_0))=e^{i\omega t_0}.$$

It looks like in frequency domain, it oscillates due to $e^{i \theta} $. However, this is not a real number.

So my question is how the frequency domain looks like for $$\mathcal{F}(\delta(t-t_0))=e^{i\omega t_0}$$ how the imaginary part in  frequency domain (spectrum).

Thanks!

答:
The magnitude is everywhere unit,
$
|\mathcal{F}(\delta(t-t_0))| = |e^{i\omega t_0}| = 1,
$
but the angle (complex argument) varies linearly.

Using $e^{ix}=\cos x + i \sin x$ we see that
$$
\mathcal{F}(\delta(t-t_0))
= e^{i\omega t_0}
= \cos\omega t_0 + i \sin\omega t_0
$$
so the graph $\{ (\omega, x, y) \mid \mathcal{F}(\delta(t-t_0))(\omega) = x+iy \}$ is a helix.


问:
How to prove that $\sum_{i=1}^{m} 2^{i} B_{i}\binom{m}{i} \frac{1}{m-i+1}=\frac{2 m+1}{m+1}$ when $m$ is an even positive integer?
I want to prove the identity below $$
\sum_{i=1}^{m} 2^{i} B_{i}\binom{m}{i} \frac{1}{m-i+1}=\frac{2 m+1}{m+1}
$$
I've carried out some computations and verified it but it only seems to hold when $m$ is an even number. Does someone have a proof of this? Preferably without induction? Maybe  multiplying $\frac{z}{1-e^{-z}}$ by some $f(z)(e^z-1)$ and then comparing coefficients of $z^n$? I tried starting off with $$\frac{2z}{1-e^{-2z}}  = \sum_{k=0}^{\infty}  \frac{2^kz^k}{k!} B_k^+ $$ but I didn't end up anywhere useful as I just got
$$
\sum_{m=0}^{\infty} \frac{2^{m+1} z^{m+1}}{m !}=\sum_{m=0}^{\infty} z^{m+1} \sum_{k=0}^{m} \frac{2^{m+1} B_{k}^{+}}{k !(m-k+1) !}
$$
and comparing coefficients just gave the classic identity
$$ \sum_{m=0}^{k} B_m^+ \binom{m}{k}\frac{1}{m-k+1} = 1  $$
This does look similar to
$$ \sum_{i=1}^{m} 2^{i} B_{i}\left(\begin{array}{c}
m \\
i
\end{array}\right) \frac{1}{m-i+1}=\frac{2 m+1}{m+1} $$
Or rather, since $m$ is even (so hence $m = 2k$ for some $ k \in \mathbb{N}$)
$$ \sum_{i=1}^{2k} 2^{i} B_{i}\left(\begin{array}{c}
2k \\
i
\end{array}\right) \frac{1}{2k-i+1}=\frac{4k+1}{2k+1} $$
Any help would be greatly appreciated, I tried using induction but $\binom{2k+2}{i}$ causes problems and I can't prove the inductive step using my induction hypothesis. Also, I wonder why the formula only holds for even $m$. Is there a related one for odd $m$?

答:
Here is a simple proof that works when $ n $ is even :

Let $ n\in 2\mathbb{N}^{*} :$

\begin{aligned} \sum_{k=0}^{n}{B_{k}\binom{n}{k}\frac{2^{k}}{n-k+1}}&=2^{n°1}\sum_{k=0}^{n}{B_{k}\binom{n}{k}\int_{0}^{\frac{1}{2}}{x^{n-k}\,\mathrm{d}x}}\\&=2^{n+1}\int_{0}^{\frac{1}{2}}{B_{n}\left(x\right)\mathrm{d}x} \end{aligned}

Using the substitution $ \left\lbrace\begin{aligned}y&=1-x\\ \mathrm{d}y &=-\,\mathrm{d}x\end{aligned}\right.$, we get : $$ \int_{0}^{\frac{1}{2}}{B_{n}\left(x\right)\mathrm{d}x}=\int_{\frac{1}{2}}^{1}{B_{n}\left(1-x\right)\mathrm{d}x}=\left(-1\right)^{n}\int_{\frac{1}{2}}^{1}{B_{n}\left(x\right)\mathrm{d}x}=\int_{\frac{1}{2}}^{1}{B_{n}\left(x\right)\mathrm{d}x} $$

Thus : $$ \int_{0}^{\frac{1}{2}}{B_{n}\left(x\right)\mathrm{d}x}=\frac{1}{2}\int_{0}^{1}{B_{n}\left(x\right)\mathrm{d}x} = 0 $$

That means $$ \sum_{k=0}^{n}{B_{k}\binom{n}{k}\frac{2^{k}}{n-k+1}}=0 $$

Now let's try to get the general result using a different method.

Denoting $ \left(\forall n\in\mathbb{N}\right),\ u_{n}=\sum\limits_{k=0}^{n}{B_{k}\binom{n}{k}\frac{2^{k}}{n-k+1}} $ we can observe that if $ x $ is such that $ \left|x\right|<2\pi $, $ \sum\limits_{n\geq 0}{\frac{B_{n}}{n!}x^{n}} $ and $ \sum\limits_{n\geq 0}{\frac{2^{-n}}{\left(n+1\right)!}x^{n}} $ both converge. Thus the series $ \sum\limits_{n\geq 0}{\frac{u_{n}}{2^{n}n!}x^{n}} $ also converges and we have : \begin{aligned} \sum_{n=0}^{+\infty}{\frac{u_{n}}{2^{n}n!}x^{n}}&=\left(\sum_{n=0}^{+\infty}{\frac{B_{n}}{n!}x^{n}}\right)\left(\sum_{n=0}^{+\infty}{\frac{2^{-n}}{\left(n+1\right)!}x^{n}}\right)\\ &=\frac{x}{\mathrm{e}^{x}-1}\times\frac{\mathrm{e}^{\frac{x}{2}}-1}{\frac{x}{2}}\\ &=\frac{1}{\sinh{\left(\frac{x}{2}\right)}}-\frac{2}{\mathrm{e}^{x}-1}\\ &=\frac{2}{x}-\sum_{n=1}^{+\infty}{B_{2n}\frac{2\left(1-2^{1-2n}\right)}{\left(2n\right)!}x^{2n-1}}-\frac{2}{x}\sum_{n=0}^{+\infty}{\frac{B_{n}}{n!}x^{n}}\\ &=1-\sum_{n=1}^{+\infty}{B_{2n}\frac{4\left(1-2^{-2n}\right)}{\left(2n\right)!}x^{2n-1}}\end{aligned}

From that formula we get the following : $$\fbox{$\begin{array}{rcl}\begin{aligned} u_{0} &=1\\ \left(\forall n\geq 1\right),\ u_{2n} &= 0\\ \left(\forall n\geq 1\right),\ u_{2n-1}&=B_{2n}\frac{1-2^{2n}}{n} \end{aligned} \end{array}$}$$

From these expressions you can get expressions for the sum $ \sum\limits_{k=1}^{n}{B_{k}^{+}\binom{n}{k}\frac{2^{k}}{n-k+1}} $ : $$ \fbox{$\begin{array}{rcl}\left(\forall n\in\mathbb{N}^{*}\right),\ \displaystyle\sum_{k=1}^{n}{B_{k}^{+}\binom{n}{k}\frac{2^{k}}{n-k+1}}=\frac{2n+1}{n+1} + u_{n} \end{array}$}$$


问:
Successor in Peano Axioms
I am an engineer who is learning real analysis myself. I read Peano axioms which are about the set of natural numbers $\mathbb{N}$. However, I find the statements, especially the word "successor" is not well defined, and cannot be converted into well-formulated formulae in first-order logic. Could anyone explain how to accept the word "successor"?

答: In the context of Peano, we use the language of first-order logic with equality, enhanced by two symbols: $0$ and $S$, which are intended to stand for the number zero and for the successor operation. It is the addition of these symbols to the language that makes the respective axioms (e.g., $\forall n\colon \neg(Sn=0)$ well-formed. There is nothing to be defined about $S$ (or $0$). The meaning of these symbols is comprised by the axioms.

Title: Successor in Peano Axioms


问:
Order of complexity proofs. Show that $n^3-5n+1=O(n^3)$.
Show that $n^3-5n+1=O(n^3)$.
I did the following:
Using the definition of big-Oh, we need to show that there exist $n_0$  and $C$ such that $n^3-5n+1 \leq Cn^3$  is valid for some constant $C > 0$ and for all $n > n_0$.
First, choose $n_0=1$. Then $1^3-5 \cdot 1+1=C\cdot 1^3$=> $C=-3$. Since by definition of big-Oh, $C$ should be greater than $0$. $C= -3$ should not work. How can I prove this order of complexity?

答:
Hint: The condition is really
$$\bigl|n^3-5n+1\bigr| \leq Cn^3$$
so you can use the triangle inequality.

Another approach:  a polynomial is asymptotically equivalent to its leading term, so $n^3-5n+1\sim_\infty n^3$, which is stronger than being $O\bigl(n^3\bigr)$.


问:
Generating function of the squared Riemann zeta function
It's a well known fact that
$$\sum_{k=2}^{\infty} \zeta(k) x^k=-x \psi(1-x)-x\gamma \space (|x|<1) $$
but I didn't meet yet a version for squared Riemann zeta function
$$\sum_{k=2}^{\infty}\zeta(k)^2 x^k$$

Do you know such a generating function? If yes, what is this one and how to derive it?

答:
There are really no miracles here, the best we can do is to use a double integral:

$$\zeta(k+1)^2=\frac{1}{(k!)^2}\int _0^1\int _0^1\frac{(\log(u)\log(v))^k}{(1-u)(1-v)}\,du\,dv$$

which stems from one of the polylogarithm formulae valid for the positive integers (source is Wikipedia).

Now, let's set the generating function to the below form for convenience:

$$\sum _{k=1}^{\infty} x^{2k}\zeta(k+1)^2$$

A series with squared factorials involves the Bessel function, and the one which is useful for our case has order 0:

$$J_0(x)-1=\sum_{k=1}^{\infty}\frac{(-1)^k}{(k!)^2}\left(\frac{x}{2}\right)^{2k}$$

So, after we put it all together, we find the below not so simple formula, which should be valid for $|x|<$1:

$$\sum _{k=1}^{\infty} x^{2k}\zeta(k+1)^2=\int _0^{1}\int _0^{1}\frac{-1+J_0\left(2i x\sqrt{\log(u)\log(v)}\right)}{(1-u)(1-v)}\,du\,dv$$

I've tested it numerically and it's matching.

It's possible that this formula reduces to a simpler, more human-like, form if the zeta at the odd integers are not part of the series.


问:
A hyperplane separates a euclidean space into two half spaces
In this Wikipedia article
It is written the following:

A hyperplane in a Euclidean space separates that space into two half spaces.

Is this precise? I mean a hyperplane in
$\mathbb C^n$ does not separate $\mathbb C^n$ into two half spaces. Indeed, the complement of a hyperplane in $\mathbb C^n$ is connected, it is even path connected.

答: This is a notational issue in a sense. What people usually mean by Euclidean space is something of the form $\mathbb{R}^n$. While it is true that $\mathbb{C}^n$ viewed as a vector space over $\mathbb{R}$ is isomorphic to $\mathbb{R}^{2n}$, we never call $n$-dimensional complex space a Euclidean space. The property you mentioned holds true for real $n$-dimensional space.


答:
There might be some ambiguity with the definition that you are working with, but at least for me I am used to $f(x)=O(g(x))$ if there is $x_0\in\mathbb{R}$ and $C\geq 0$ such that $x\geq x_0$ implies $|f(x)|\leq Cg(x)$ (notice the absolute values). Then in your case, choose $n_0=5$ and $C=1$. We have for $n\geq n_0$

$$0<n(n^2-5)=n^3-5n<n^3-5n+1<n^3$$

Then

$$|n^3-5n+1|\leq 1\cdot n^3$$

implies $n^3-5n+1=O(n^3)$.

Title: Order of complexity proofs. Show that $n^3-5n+1=O(n^3)$.

Body:
Show that $n^3-5n+1=O(n^3)$.
I did the following:
Using the definition of big-Oh, we need to show that there exist $n_0$  and $C$ such that $n^3-5n+1 \leq Cn^3$  is valid for some constant $C > 0$ and for all $n > n_0$.
First, choose $n_0=1$. Then $1^3-5 \cdot 1+1=C\cdot 1^3$=> $C=-3$. Since by definition of big-Oh, $C$ should be greater than $0$. $C= -3$ should not work. How can I prove this order of complexity?


问:
Quantum field theory free particle field operators
I am reading the book Abrikosov methods of quantum field theory in statistical physics.
I got stuck at some derivation of the field operators for the free particle.
The reasoning in the book is as follows. First particle field operators are defined as

$
\Psi(\xi)=\sum_{i}\varphi_{i}(\xi)a_{i}
$

$
\Psi^{+}(\xi)=\sum_{i}\varphi^{*}_{i}(\xi)a^{+}_{i}
$

where $\varphi_{i}(\xi)$ are wavefunctions of a particle in state i. $\xi$ is a combination of
spin and position coordinates. $a_{i}$ and $a^{+}_{i}$ are the ladder operators.
The field operators satisfy the same commutation/anticommutation relations as the ladder operators

$
[\Psi(\xi),\Psi^{+}(\xi^{'})]=\delta(\xi-\xi^{'})
$
and $[\Psi(\xi),\Psi(\xi^{'})]=0$ and $[\Psi^{+}(\xi),\Psi^{+}(\xi^{'})]=0$.

Then a free particle wavefunction (without spin) is introdcued

$
\frac{1}{\sqrt{V}}e^{i\mathbf{p}\mathbf{r}}
$

where $\mathbf{p}$ is the momentum and $\mathbf{r}$ is the position.

The field operator $\Psi(\xi)$ in the Schroedinger picture for the free spinless particle is

$
\Psi(\mathbf{r})=\sum_{\mathbf{p}}e^{i\mathbf{pr}}a_{\mathbf{p}}
$

The Hamiltonian is
$
H = \sum_{\mathbf{p}}n_{\mathbf{p}}\epsilon_{0}(\mathbf{p})
$

where $n_{\mathbf{p}}=a^{+}(\mathbf{p})a(\mathbf{p})$ is the number of particle operator and $\epsilon_{0}(\mathbf{p})$ is
the energy of a particle with momentum $\mathbf{p}$.

Now the field operator in the Heisenberg picture is introduced as
$a(t)=e^{iHt}ae^{-iHt}$.

This is done for the free particle field operator as

$
\tilde{\Psi}(\mathbf{r},t)=\frac{1}{\sqrt{V}}
\sum_{\mathbf{p}}e^{i\sum_{\mathbf{p}^{'}}\epsilon_{0}(\mathbf{p}^{'})n_{\mathbf{p}^{'}}t}
a_{\mathbf{p}}e^{-i\sum_{\mathbf{p}^{''}}\epsilon_{0}(\mathbf{p}^{''})n_{\mathbf{p}^{''}}t}
e^{i\mathbf{pr}}=\frac{1}{\sqrt{V}}\sum_{\mathbf{p}}a_{\mathbf{p}}e^{i(\mathbf{pr}-\epsilon_{0}(\mathbf{p})t)}
$

This is exactly where I got stuck. Because I don't understand the
mathematical identity or what "trick" was used to obtain the result after the second equal sign in the last equation.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Use Heisenberg Equation of Motion:
\begin{align}
\ic\,\totald{a_{\bf p}\pars{t}}{t} & = \bracks{a_{\bf p}\pars{t},H} =
\expo{\ic Ht}\
\underbrace{\bracks{a_{\bf p},H}}
_{\ds{\epsilon_{\bf p}\,a_{\bf p}}}\expo{-\ic Ht} =
\epsilon_{\bf p}\,a_{\bf p}\pars{t}
\\[5mm] \implies &
a_{\bf p}\pars{t} = a_{\bf p}\expo{-\ic\epsilon_{\bf p}\,t}
\end{align}


问:
Prove $P\left( A_n\right) = 0 \Longrightarrow P\left(\bigcup\limits_{n=1}^{\infty} A^c_n\right)=1 $
If $ (\Omega,F,P)$ is  a  probability  space  and $A_1, A_2..\in F$ Show   that
$$\ P\left( A_n\right) = 0 \Longrightarrow P\bigg(\bigcup\limits_{n=1}^{\infty} A^c_n\bigg)=1$$

I don't know how to approach the problem

答: If $P(A_m) = 0$, then $P(A_m^c)=1$. But $P(\cup_n A_n^c) \geq P(A_m^c)$, so the former is equal to $1$.


问:
$\overline{\mathrm{span} \{e_{2n} \ \mid \ n \}} \cap \overline{\mathrm{span} \{ e_{2n} + \frac{1}{n+1}e_{2n+1} \} } = 0$, but their sum is dense

Regarding this question:Show that the intersection of $\overline{\text{span }\{e_{2n}\}}$ and $\overline{\text{span }\{e_{2n}+\frac{1}{n+1}e_{2n+1}\}}$ is zero..

The above has an accepted answer, but there is no proof of the result from the title. I am looking for a proof for that result and also that
$$\overline{\overline{\mathrm{span} \{e_{2n} \ \mid \ n \}} + \overline{\mathrm{span} \{ e_{2n} + \frac{1}{n+1}e_{2n+1} \} }} = H. $$
I have not been able to do it myself.

For the first part, I have started by picking $x \in \overline{\mathrm{span} \{e_{2n} \}} \cap \overline{\mathrm{span} \{e_{2n} + \frac{1}{n+1} e_{2n+1} \}} .$

Then, we know that $$x = \sum_{n=1}^\infty \langle x, e_n \rangle e_n. $$
Moreover, there is a sequence $(x_k)_{k}$ with $x_k \in \mathrm{span} \{e_{2n} \}$ for every $k$ and such that $x_k \to x$.
Using the same reasoning, there is a sequence $(y_k)_k$ with $y_k \in \mathrm{span} \{e_{2n} + \frac{1}{n+1} e_{2n+1} \}$ and such that $y_k \to x$. However, I do not know how to proceed from here.

I do not know how to start the second part either.

答:
You just need some basic properties of orthogonal sequences. First verify that $(e_{2n}+\frac  1 {n+1} e_{2n+1})$ is an orthogonal sequence.

If  $(f_n)$ is an orthogonal sequence then  any vector in the closed subspace spanned by this sequence can be written as a sum $\sum a_n x_n$.

Suppose $x$ belongs to the intersection of the closed subspace spanned by the sequences $(e_{2n})$ and $(e_{2n}+\frac  1 {n+1} e_{2n+1})$. Then $x=\sum a_n e_{2n}$ and $x=\sum b_n (e_{2n}+\frac  1 {n+1} e_{2n+1})$. This gives $\sum (a_n-b_n)e_{2n}=\sum b_n \frac  1 {n+1} e_{2n+1}$. This implies that $a_n=b_n$ for all $n$ and $b_n=0$ for al $n$. Hence $x=0$.

To show that the sum of the two spaces is dense it is enough to show that if vector $x$ is orthogonnal to the sum then it is $0$. But this is obvious since $x$ must be orthogoanal to $e_{2n}$ as well as $e_{2n+1}$ for every $n$.


Title: How to prove that $\sum_{i=1}^{m} 2^{i} B_{i}\binom{m}{i} \frac{1}{m-i+1}=\frac{2 m+1}{m+1}$ when $m$ is an even positive integer?

Body:
I want to prove the identity below $$
\sum_{i=1}^{m} 2^{i} B_{i}\binom{m}{i} \frac{1}{m-i+1}=\frac{2 m+1}{m+1}
$$
I've carried out some computations and verified it but it only seems to hold when $m$ is an even number. Does someone have a proof of this? Preferably without induction? Maybe  multiplying $\frac{z}{1-e^{-z}}$ by some $f(z)(e^z-1)$ and then comparing coefficients of $z^n$? I tried starting off with $$\frac{2z}{1-e^{-2z}}  = \sum_{k=0}^{\infty}  \frac{2^kz^k}{k!} B_k^+ $$ but I didn't end up anywhere useful as I just got
$$
\sum_{m=0}^{\infty} \frac{2^{m+1} z^{m+1}}{m !}=\sum_{m=0}^{\infty} z^{m+1} \sum_{k=0}^{m} \frac{2^{m+1} B_{k}^{+}}{k !(m-k+1) !}
$$
and comparing coefficients just gave the classic identity
$$ \sum_{m=0}^{k} B_m^+ \binom{m}{k}\frac{1}{m-k+1} = 1  $$
This does look similar to
$$ \sum_{i=1}^{m} 2^{i} B_{i}\left(\begin{array}{c}
m \\
i
\end{array}\right) \frac{1}{m-i+1}=\frac{2 m+1}{m+1} $$
Or rather, since $m$ is even (so hence $m = 2k$ for some $ k \in \mathbb{N}$)
$$ \sum_{i=1}^{2k} 2^{i} B_{i}\left(\begin{array}{c}
2k \\
i
\end{array}\right) \frac{1}{2k-i+1}=\frac{4k+1}{2k+1} $$
Any help would be greatly appreciated, I tried using induction but $\binom{2k+2}{i}$ causes problems and I can't prove the inductive step using my induction hypothesis. Also, I wonder why the formula only holds for even $m$. Is there a related one for odd $m$?

Body:
We seek to show that for $m$ even

$$\sum_{q=1}^m 2^q B^+_q {m\choose q} \frac{1}{m-q+1} = \frac{2m+1}{m+1}$$

or alternatively for $m$ even

$$\sum_{q=1}^m 2^q B^+_q {m+1\choose q} = 2m+1$$

where

$$B^+_q = q! [z^q] \frac{z}{1-\exp(-z)}.$$

We get for the LHS

$$-B^+_0 - 2^{m+1} B^+_{m+1}
+  \sum_{q=0}^{m+1} 2^q B^+_q {m+1\choose q}
\\ = - 1  - 2^{m+1} B^+_{m+1} +
(m+1)! [z^{m+1}] \exp(z) \frac{2z}{1-\exp(-2z)}
\\ = - 1 - (m+1)! [z^{m+1}] \frac{2z}{1-\exp(-2z)}
\\ + (m+1)! [z^{m+1}] \exp(z) \frac{2z}{1-\exp(-2z)}
\\ = - 1 + (m+1)! [z^{m+1}]
\frac{2z (\exp(z)-1)}{1-\exp(-z)^2}
\\ = - 1 + (m+1)! [z^{m+1}]
\frac{2z \exp(z) (1-\exp(-z))}{1-\exp(-z)^2}
\\ = - 1 + (m+1)! [z^{m+1}]
\frac{2z \exp(z)}{1+\exp(-z)}
\\ = - 1 + (m+1)! [z^{m+1}]
\left(\frac{2z \exp(z) + 2z}{1+\exp(-z)}
- \frac{2z}{1+\exp(-z)} \right)
\\ = - 1 + (m+1)! [z^{m+1}]
\left(2z \exp(z)
- \frac{2z}{1+\exp(-z)} \right)
\\ = - 1 + 2 (m+1)! [z^{m}]
\left(\exp(z)
- \frac{1}{1+\exp(-z)} \right)
\\ = - 1 + 2 (m+1)! \frac{1}{m!}
- 2 (m+1)! [z^{m}] \frac{1}{1+\exp(-z)}
\\ = 2m+1 - 2 (m+1)! [z^{m}] \frac{1}{1+\exp(-z)}.$$

We observe  that this evaluates to zero when $m=0$ as claimed and
henceforth assume $m\ge 1.$ Continuing,

$$2m+1 - 2 (m+1)!
[z^{m}]
\left( \frac{1/2-\exp(-z)/2}{1+\exp(-z)} + \frac{1}{2}\right).$$

Now note that with $f(z) = \frac{1/2-\exp(-z)/2}{1+\exp(-z)}$ we have

$$f(-z) = \frac{1/2-\exp(z)/2}{1+\exp(z)}
= \frac{\exp(-z)/2-1/2}{\exp(-z)+1}
= - f(z)$$

so the function $f(z)$ is odd and hence the even order coefficients of
its Taylor series vanish. Since the constant $1/2$ does not contribute
to $[z^m]$  when $m  \ge 1$ that  leaves just $2m+1$  and we  have the
desired result. For the odd order coefficients we have to evaluate the
remaining coefficient extractor.

$$2 (m+1)! [z^{m}] \frac{1}{1+\exp(-z)}
= 2 (m+1)! (-1)^m [z^{m}] \frac{1}{1+\exp(z)}
\\ = 2 (m+1)! (-1)^m [z^{m}] \frac{1}{2+\exp(z)-1}
\\ = (m+1)! (-1)^m [z^{m}] \frac{1}{1+(\exp(z)-1)/2}
\\ = (m+1)! (-1)^m [z^{m}]
\sum_{q=0}^m (-1)^q 2^{-q} (\exp(z)-1)^q
\\ = (m+1) (-1)^m
\sum_{q=0}^m (-1)^q 2^{-q} q! {m\brace q}.$$

We thus obtain the closed form

$$\bbox[5px,border:2px solid #00A000]{
2m+1 - [[m \;\text{odd}]]
(m+1) (-1)^m
\sum_{q=1}^m (-1)^q 2^{-q} q! {m\brace q}.}$$

The Iverson bracket is not essential here as the sum evaluates to zero
anyway when $m$  is even. This yields for the  original as proposed by
OP that it is

$$\bbox[5px,border:2px solid #00A000]{
\frac{2m+1}{m+1} - [[m \;\text{odd}]]
(-1)^m \sum_{q=1}^m (-1)^q 2^{-q} q! {m\brace q}.}$$


问:
If $f$ is bounded and continuous and $X_n \xrightarrow{\mathbb{P}} X$, do we have $f(X_n) \xrightarrow{a.s.} f(X)$?
Basically the title, if $f: \mathbb{R} \to \mathbb{R}$ is bounded and continuous and the sequence of random variables $X_1, X_2, ...$ converges to $X$ in probability, do we have almost sure convergence of $f(X_1), f(X_2), ...$ to $f(X)$?

答: No, as a counterexample take $f(x) = x$.  Then you're asking whether convergence in probability implies almost sure convergence, which it doesn't.


问:
Is there a deeper connection among $\int_0^1 x^{e-2} dx$, $\int_0^{\infty}\{x\}e^{-x}\,dx$, and $\sum_{n=1}^{\infty} e^{-n}$?
Consider the following integrals $I_1,I_2$:
$$
I_1=\int _1^{\infty} \lfloor x\rfloor e^{-x}\,dx = \sum_{n=1}^{\infty}n \int_{n}^{n+1}e^{-x}\,dx = (e-1)\sum_{n=1}^{\infty} n e^{-n-1} = \frac{e-1}{(e-1)^2}= \frac{1}{e-1}
$$
$$
I_2 = \int_0^1 \sqrt{x\sqrt[3]{x\sqrt[4]x{\sqrt[5]{x\cdots }}}}\,dx = \int_0^1 x^{e-2}\,dx = \frac{1}{e-1}\left. x^{e-1} \right|_0^1= \frac{1}{e-1}
$$Note that $I_1$ is an infinite series of definite integrals and $I_2$ is a definite integral of an infinite series, and they both evaluate to the same number. Millions of definite integrals evaluate to the same value but these are striking enough that I wonder if there is a deeper connection between them, or if one could possibly be transformed into the other. This might be a wild-goose chase but I'm curious if there is a transformation that takes one to the other, possibly using the closed-form of the geometric series $\sum_{n=1}^{\infty} e^{-n} = \frac{1}{e-1}$ as an intermediate step.

答:
Here's a fun connection that takes $I_2$ to something like $I_1$ without passing through the exact result, but uses advanced tools.

Consider $\int_{0}^1 x^a dx$ and Taylor expand the integrand around $x=1$. The following string of equalities ensues

$$\int_0^1 x^a dx=\sum_{n=0}^{\infty}\frac{(a)_n}{n!}\int_0^1 (x-1)^n dx=\sum_{n=0}^{\infty}\frac{(-1)^n (a)_n}{(n+1)!}$$

We want to expand the falling factorials $(a)^n=a(a-1)...(a-n+1)$ into powers of $n$ (they are polynomials in it after all) and for that we use their connection to the Stirling numbers of 1st kind $(a)_n=\sum_{k=0}^ns(n,k)a^k$. After reversing the order of summation we obtain

$$\int_0^1 x^a dx=\sum_{k=0}^{\infty}a^k\sum_{n=k}^{\infty}\frac{(-1)^n}{(n+1)!}s(n,k)$$

Finally by manipulating the generating function we obtain the relation

$$\sum_{n=k}^{\infty}\frac{(z)^n}{(n+1)!}s(n,k)=\frac{1}{zk!}\int_{0}^zdt \log^k(1+t)$$

After putting $z=-1$ and performing the integral we finally obtain that

$$\sum_{n=k}^{\infty}\frac{(-1)^n}{(n+1)!}s(n,k)=(-1)^k$$

and hence

$$\int_0^1 x^{e-2}dx=\sum_{n=0}^{\infty}(2-e)^n$$

which is almost a surprise (at least from this point of view). I tried to connect $I_1$ and $I_2$ directly using this method and various cheap manipulations, and while it is possible to get a geometric series in $1/e$ (alas not quite the right one above) out of an appropriate expansion after a substition, I found the manipulation unenlightening. Hope this is interesting!


问:
Calculating $P(Y>1)$, Y follows an random uniform distribution on $[0,X]$ and X follows an exp. distribution.
I need to calculate $P(Y>1)$ with Y being uniformly distributed on the random interval $[0,X]$ and X being exponentially distributed with parameter $\lambda = 1$.

So I need to condition on X to find $P(Y>1)$ but I do not know to continue. Appreciate some help

Body:
$$
\mathbb{P}(Y>1) = \int \mathbb{P}(Y>1 | X=x) \,dx = \int_1^{\infty} \frac{x-1}{x} e^{-x} \,dx \approx 0.15
$$


问:
Plus construction of a presheave factors every sheaf-valued morphism.
I'm having some trouble understanding the correctness of some proof in Sheaves in Geometry and Logic (Mac Lane, Moerdijk). It concerns the lemma III.5.3 :

If $F$ is a sheaf and $P$ a presheaf, then any map $\phi \colon P \to F$ of
  presheaves factors uniquely through $\eta$ as $\phi = \tilde\phi \circ \eta$.

$$ \begin{matrix} P & \stackrel \eta \to & P^+ \\ & \!\!\!\!\!\!_\phi\searrow & \downarrow \small{\tilde\phi} \\ & & \!\!\!\!F\end{matrix}$$

Recall the plus construction : $P^+(C)$ is the equivalent class of matching families under the relation $(x_f)_{f\in S} \sim (y_g)_{g \in R}$ ($R,S$ covering sieves of $C$) iff there is some covering sieve $T \subseteq S \cap R$ on which $x_f = y_f$ forall $f \in T$.

Recall the morphism $\eta$ : $$\eta_C \colon P(C) \to P^+(C),\, x \to [(Pf(x))_{f \colon D \to C}].$$

I easily understand the definition of $\tilde\phi_C(\mathbf x)$ for some equivalent class $\mathbf x = (x_f)_{f\in S}$ : we push the matching family $(x_f)$ by $\phi_C$ into a matching family of $F$ ; being a sheaf, $F$ admit a unique amalgamation in $F(C)$ that we define to be $\tilde\phi_C(\mathbf x)$. The well-definition is no problem. What bother me is to check that $\tilde\phi = (\tilde\phi_C)_C$ is actually natural in C.

We want a commutative diagram
$$ \begin{matrix} P^+C & \stackrel {Ph} \to & P^+D \\
\small{\tilde\phi_C}\downarrow \ \ \ & & \ \ \ \downarrow\small{\tilde\phi_D} \\
FC & \stackrel{Fh} \to & FD \end{matrix}$$
for all $h \colon D \to C$. For an element $\mathbf x = (x_f)_{f\in S}$, and $h \in S$, the stability axiom of Grothendieck topologies assure $h^\ast S = \hom(-, D)$, and so the commutativity is immediat from the definition of $\tilde\phi_C(\mathbf x)$ as amalgamation.

But I'm stuck with the case where $h \notin S$, and (it seems to me that) it is not treated in the proof of Mac Lane. Maybe can we always find $\mathbf y = (y_g)_{g\in R}$ with $\mathbf x \sim \mathbf y$ and $h \in R$, but I can't see it.

答:
Let me write $w$ for the equivalence class in $P^{+}C$ of a matching family $(x_{f})_{f\in R}$ for a covering sieve $R$ of $C$. Now the down-right path sends $w$ to $Fh(y)$, where $y$ is the only amalgamation for the matching family $(\phi_{dom (f)}(x_{f}))_{f\in R}$. The right-down path maps $w$ to $z\in FD$, where $z$ is the only amalgamation for $(\phi_{dom (hf')}(x_{hf'}))_{f'\in h^{\ast}(R)}=:k$ (recall the definition of $P^{+}h$ for an arrow $h$). In order to show that going the two way gives the same result it suffices to show that $Fh(y)$ is an amalgamation of $k$: if $f'\in h^{\ast}(R)$
$$Ff'(Fh(y))=(F(hf'))(y)=\phi_{dom (hf')}(x_{hf'})$$ where the last equality is by definition of $y$ and we are done.
I hope it's clear enough (fill in the details!) and that I didn't make some mistakes. </p>


问:
Confusion in solving this recursion
I am trying to solve the following recursion
$$
\sum _{i=1}^{n-1} -\mu (n-i)+\sum _{i=1}^{n-1} 2 i \mu (n-i)=(n-1)^2 (\mu (n)-1) \tag{1}
$$

By increasing $n$ to $n+1$, we get
$$
\sum _{i=1}^n -\mu (-i+n+1)+\sum _{i=1}^n 2 i \mu (-i+n+1)=n^2 (\mu (n+1)-1) \tag{2}
$$
and then by shifting index, we get
$$
\sum _{i=1}^{n-1} -\mu (n-i)+\sum _{i=1}^{n-1} 2 (i+1) \mu (n-i)+\mu (n)=n^2 (\mu (n+1)-1)\tag{3}
$$
Taking the difference of (1) and (3), we arrive at
$$
-2 \sum _{i=1}^{n-1} \mu (n-i)-\mu (n)=(n-1)^2 (\mu (n)-1)-n^2 (\mu (n+1)-1)\tag{4}
$$

Replace $n$ by $n+1$ and shift index again, we get
$$
(n+1)^2 \mu (n+2)+1= \\2 \sum _{i=1}^{n-1} \mu (n-i)+n^2 \mu (n+1)+2 \mu (n)+\mu (n+1)+2 n+2 \tag{5}
$$
Then taking the difference between (4) and (5), we get
$$
(n+1)^2 \mu (n+2)+(n-2) n \mu (n)=\left(2 n^2+1\right) \mu (n+1)+2 \tag{6}$$
This has a solution
$$
\mu(n)=\frac{2}{3} H_{n-1}:=\frac{2}{3} \sum_{i=1}^{n-1} \frac{1}{i}
$$
What bothers me is that this solution does not fit in the original equation (1)! How is this possible? I don't think I have made a mistake in the computations.

答:
The sums are
$$
\sum _{i=1}^{n-1} -\mu (n-i)+\sum _{i=1}^{n-1} 2 i \mu (n-i)=(n-1)^2 (\mu (n)-1) \tag{1}
$$
Your equation you have arrived
$$
(n+1)^2\mu(n+2)-(2n^2+1)\mu(n+1)+n(n-2)\mu(n)-2=0\tag{eq}
$$
Have indeed general solution
$$
\mu(n)=-1+\frac{2}{3(n-1)}+\frac{2}{3}H_{n-2}+C_1+C_{2}\left(-\frac{1}{2}\psi^{(1)}(n-2)+\frac{\pi^2}{12}+\frac{-12+43n-46n^2+20n^3-3n^4}{4(2-3n+n^2)^2}\right)\tag{sol}
$$
as long as $n\geq 3$. That is because for $n=1,2$ the solution has singularities at $n=1$ and $n=2$ ($n=2$ is also a curious point since $H_n=\sum_{1\leq n\leq n}k^{-1}$ and by convection we use $H_0=0$). It is like beeing solving a two order differential equation with no initial conditions. The solution it can be far away from what you might expecting. The Mathematica do its best. Its gives you $-1+\frac{2}{3(n-1)}+\frac{2}{3}H_{n-2}+C_1$ and not $C_1+\frac{2}{3}H_{n-1}$, to tell you that $n=1$ and $n=2$ are pathological cases. Hence for $n\geq 3$ the solution goes well. But for $n=1,2$ you have to define or give more input. The values at $n=1$ and $n=2$ are usefull for your computation of the sums you have. By demanding your sums to be in accordance to each other you can define
$$
\mu(n)=\left\{\begin{array}{cc}
c_1\textrm{ , for } n=1\\
c_2\textrm{ , for }n=2\\
2/3H_{n-1}-1\textrm{ , for }n\geq 3
\end{array}\right\}\tag 2
$$

where $H_0=0$. Setting these into (eq) and (1), you take a solution: $c_1=-5/4$, $c_2=-1/4$.

Title: Confusion in solving this recursion


问:
Mathematical problem related to spectral method for time evolution
I have a mathematical problem related to spectral method for time evolution in QM.

Considering the standard evolution for a generic quantum state $\psi(t) \in \mathbb{C}^N $ which is not express in a basis where U is diagonal, setting $\hbar=1$ we have:
$$| \psi(t) \rangle = U |\psi(0) \rangle \hspace{5em} \text{where} \hspace{2em} U^\dagger U  = UU^\dagger  = 1_N \hspace{1em} \text{and} \hspace{1em} U=\exp[-iH(t-t_0)]  $$
it is known that in the basis of the eigenvectors it can be written as:
$$| \psi(t) \rangle =\sum_{n}c_n \exp[-iE_n(t-t_0)] | \phi_n \rangle  \hspace{5em} (1)$$
I'm stuck in derive this simply expression, supposing to store the eigenvectors in the columns of P:
$$| \psi(t) \rangle = U |\psi(0) \rangle = P D P^{\dagger} |\psi(0) \rangle $$
because of $ P^{\dagger} |\psi(0) \rangle=\vec{c} \hspace{2em} \text{where} \hspace{2em}  c_n=\langle \phi_n|\psi(0)  \rangle
$
$$| \psi(t) \rangle =P D \vec{c} \neq \vec{c} D P
$$
The commutation relation beacuse of the elements on the diagonal are different does not seem to me to be valid. I'm stuck

答:
So indeed, we have
$$
| \psi(t) \rangle = PD \vec c.
$$
Note that the diagonal entries of $D$ are $\exp(-iE_n(t-t_0))$. So, $D\vec c$ is the vector $$
D\vec c = \pmatrix{c_1\exp(-iE_1 (t-t_0))\\ \vdots \\ c_N \exp(-iE_N(t-t_0))}.
$$
Since the columns of $P$ are $|\phi_1\rangle,\dots,|\phi_N\rangle$, it follows that
$$
PD\vec c = \sum_{n=1}^N c_n \exp(-iE_n(t-t_0)) \cdot |\phi_n\rangle,
$$
which is what we wanted.


问:
$p:\tilde{X}\to X$ a covering map, $\tilde{X}$ path connected, show a deck transformation $\tau$ fixes a point in $\tilde{X}$ iff $\tau=id_\tilde{X}$
The backwards direction is clear, and for the forwards direction I think I need to use the homotopy lifting property for paths since $\tilde{X}$ is path connected. So I start with for some $x_0\in\tilde{X}$, $\tau(x_0)=x_0$. Now I think I need to use $x_0$ as a base for a path $\gamma$ and somehow show that $\tau(\gamma)=\gamma$. Not really sure if this is right or how to proceed if it is.

答:
Let $p : \tilde X \to X$ be any covering map and $f : Y \to X$ be any map. It is a well-known theorem in the theory of covering maps that for any two lifts $\tilde {f_1}, \tilde {f_2}: Y \to \tilde X$ of $f$ the set of points of $Y$ where $\tilde {f_1}$ and $\tilde {f_2}$ agree is open and closed in $Y$. See for example Hatcher's book on Algebraic Topology. Also see Unique Lifting Property - Algebraic Topology.

Thus, if $Y$ is connected, then $\tilde {f_1} = \tilde {f_2}$ if and only if $\tilde {f_1}(y_0) = \tilde {f_2}(y_0)$ for a single $y_0 \in Y$.

Therefore in your question it suffices to assume that $\tilde X$ is connected. Now simply observe that any deck transformation $\tau$ is a lift of $p$.


问:
Sheaf of $\mathscr{O}_X$-modules generated by global sections
In the Hartshorne book (Algebraic Geometry) we have the definition of sheaf of $\mathscr{O}_X$-modules generated by global sections, which is given as follows:

Definition. Let $X$ be a scheme, and let $\mathscr{F}$ be a sheaf of $\mathscr{O}_X$-modules. We say
that $\mathscr{F}$ is generated by global sections if there is a family of global sections
$\{s_i\}_{i\in I}$, $s_i \in  \Gamma(X, \mathscr{F})$, such that for each $x \in X$, the images of $s_i$ in the stalk $\mathscr{F}_x$ generate that stalk as an $\mathscr{O}_X$-module.

I do not know what means "the images of $s_i$ in the stalk $\mathscr{F}_x$".

I always like to look at the stalk as follows:

$\mathscr{F}_x = \{(U,s)| x\in U \in Ouv(X), s \in \mathscr{F}(U)\}/ \sim$

In that case "the images of $s_i$ in the stalk $\mathscr{F}_x$" would be $(U, s_i)\in \mathscr{F}_x$???

答: You are almost correct: the image of $s_i$ in $\mathscr{F}_x$ is the equivalence class of $(X,s_i)$.


问:
Derivation of 2D Korteweg-de-Vries equation
Coming from engineering rather than mathematics, I am recently dealing with non-linear partial differential equations e.g. like the well known Korteweg-de-Vries equation:

$$u_{t} + uu_x + u_{xxx} = 0$$

In the literature I found the two- and three-dimensional versions of the KdV-equation:

$$(u_{t} + uu_x + u_{xxx})_x + u_{yy} = 0 $$
$$(u_{t} + uu_x + u_{xxx})_x + u_{yy} + u_{zz} = 0$$

What bugs me now is how one derives the two- respectively three-dimensional versions from the original version? In all of the literature I have found so far, the multi-dimensional version seems to fall from heaven, nowhere I could find a complete derivation or any intuition behind it. This is unfortunately not the first time I can not grasp how one author jumps from the 1-D to a n-D version.

So generally asked: How does one derive a multi-dimensional version of a (non-linear)PDE given a 1 + time-dimensional form like the first equation? It would be more clear to me if the original version would be written in vector notation (with $\nabla$, $\cdot$, $\times$, $\Delta$, etc. operators). Would be great if someone with a more profound mathematical background could educate me.

Thanks a lot!

Edit:

The 2D version is also called Kadomtsev-Petviashvili equation, if that rings a bell for anyone.

答:
I'm not sure how one would derive the analog, but my gut feeling is that one would generalize the KdV hamiltonian (which is the second (aka "energy") integral of motion on the wikipedia page).  I'm not sure if Hamiltonian mechanics is something you've covered in engineering courses, as it can quickly turn into very advanced/technical pure mathematics, but the legend Jerrold Marsden was an absolute pioneer of not only developing this advanced pure math, but also applying it to real engineering problems (he was a Professor of Engineering and Control & Dynamical Systems at CalTech).  There is a great lecture of his on YouTube that highlights applications of his research to problems in engineering.

There is a "2D analog" of KdV, it's called the "Kadomtsev–Petviashvili Equation" (KP).  Like KdV, it is integrable (it has countably infinitely many conserved quantities/integrals of motion).  I wonder if you took the Hamiltonian for the KP equation and only kept one spatial dimension, if it would be equal to the KdV Hamiltonian.  Perhaps I shall try it!


问:
Integration of a rational function from +/- infinity
I am trying to calculate the integral

$$\int_{-\infty}^{\infty}{\frac{a+x}{b^2 + (a+x)^2}\frac{1}{1+c(a-x)^2}}dx$$

where $\{a, b, c\}\in \mathbb{R}$. I have looked in a table of integrals for rational functions, but with no luck. Is there a smart trick I can utilize?

答:
Potential Hint:

$$\int_{-\infty}^{\infty}{\frac{a}{(b^2 + (a+x)^2)(1+c(a-x)^2)}} - {\frac{x}{(b^2 + (a+x)^2)(1+c(a-x)^2)}}dx$$

The first part can be solved using a chain rule like method. The second half can be solved using integration by parts with $u = x$ i.e $u\prime = 1$.

Hope this helps!


问:
Why isn't the Fourier transform self-inverse?
I'm reading the Wikipedia article on Pontryagin duality.

(This is a fuzzy question about intuition for the difference between the forward and inverse Fourier transforms in this context.)

It defines the Fourier transform for all $\chi$ in $\widehat{G}$:
$$\widehat f(\chi) = \int_G f(x) \overline{\chi(x)}\ d\mu(x)$$

And the inverse transform
$$ \check{g} (x) = \int_{\widehat{G}} g(\chi) \chi(x)\ d\nu(\chi)$$

The transform on $\widehat{G}$ is given by:
$$
\begin{aligned}
\widehat{g}(\operatorname{ev}_G(x)) &= \int g(\chi) \overline { \operatorname{ev}_G(x)(\chi) } d\nu(\chi) \\
&= \int g(\chi) \overline { \chi(x) } d\nu(\chi)
\end{aligned}
$$
where $\operatorname{ev}_G(x)$ for $x$ in $G$ is a generic element of $\widehat{\widehat{G}}$.

What is the intuitive reason for the minus sign in
$$ \check{g}(x) = \widehat{g} ( \operatorname{ev}_G(-x)) $$

Is the isomorphism $\psi : G \to \widehat{\widehat{G}}$ given by $x \mapsto \operatorname{ev}_G(-x)$ more "natural" than $\operatorname{ev}_G$?

答:
The Fourier inversion theorem is saying that the Fourier transform $W: L^2(G)\to L^2(\hat{G})$ is unitary
$$\|W(f)\|_{L^2(\hat{G})} = \|f\|_{L^2(G)}$$ so that its inverse is its adjoint $W^*$, which is defined by $$\langle Wf,\hat{h} \rangle=\langle f,W^*\hat{h} \rangle,\qquad \forall f\in L^2(G),\hat{h}\in L^2(\hat{G})$$
When everything is $L^1$ we can change the order of integration to get

$$\int_G f(x)\overline{\int_{\hat{G}} \hat{h}(\chi) \chi(x)}d\chi dx=\int_{\hat{G}} \overline{\hat{h}(\chi)} \int_G f(x)\overline{\chi(x)}dx d\chi = \langle Wf,\hat{h} \rangle$$ whence $W^* \hat{h}(x) = \int_{\hat{G}} \hat{h}(\chi) \chi(x)d\chi$.

That $(WW f)\circ ev_G(x)= f(-x)$ is then a 'coincidence' coming from $G=\hat{\hat{G}}\circ ev_G$ and $\overline{\psi(t)}=\psi(-t)$.

That said proving the Pontryagin Fourier inversion theorem seems the only way to really see what is happening.


答:
Let $k\in\mathbb N$, such that $\mathbb P(A_k)=0$, then $\mathbb P(A^c_k)=1$. So, taking complements one gets,

$$\mathbb P\big[\bigcup_n A^c_n\big] = \mathbb P\Big[\Omega\setminus\bigcap_n A_n\Big]=\mathbb P\big[\bigcup_n(\Omega\setminus A_n)\Big]\geq \mathbb P\big[\Omega\setminus A_k\big] =1$$

But $A_1, A_2,\ldots\in\mathcal F$, is in the probability space, so $\mathbb P\big[\bigcup_n A^c_n\big]=1$.

Title: Prove $P\left( A_n\right) = 0 \Longrightarrow P\left(\bigcup\limits_{n=1}^{\infty} A^c_n\right)=1 $

Body:
If $ (\Omega,F,P)$ is  a  probability  space  and $A_1, A_2..\in F$ Show   that
$$\ P\left( A_n\right) = 0 \Longrightarrow P\bigg(\bigcup\limits_{n=1}^{\infty} A^c_n\bigg)=1$$

I don't know how to approach the problem