问:
Integral for Biot-Savart Law on a box
I'm trying to make a basic computer model of a bar magnet. In the process I came across this question and answer that appears to have an appropriate equation for me to use. I say appears because my maths knowledge is only barely past high-school level, and the answerer stops here:

you can simply use the Biot-Savart law to calculate the magnetic field:
$$\mathbf B(\mathbf x) = \frac{\mu_0}{4\pi}\int_{\mathbb S}d\mathbf a' \ \mathbf K(\mathbf x') \times \frac{\mathbf{x-x'}}{|\mathbf{x-x'}|^3}$$
I believe you can take it from here.

So I understand that in this case I can use the fact that the net magnetic field (the thing I want to model) is composed of the sum of all the magnetic fields produced in this situation. Which is to say I need to add up the field produced by each of the faces with respect to each point I want to model.

I also figure that that is what the part of the equation that I don't understand is trying to express, namely the section $\int_{\mathbb S}d\mathbf a' \ \mathbf K(\mathbf x')$

So in the above example a bar magnet is being modeled like so:

You can model a bar magnet by a rectangular box with a constant magnetization in one direction. Let's take the box $[0,a]\times[0,b]\times[0,c]$, with a constant magnetization $\mathbf M(\mathbf x) = M_0 \ \hat{\mathbf k}$, where $\hat{\mathbf k}$ is the unit vector in the $z$ direction. The bound volume and surface current densities are:
$$\mathbf J_b(\mathbf x) = \boldsymbol{\nabla}\times\mathbf M(\mathbf x)$$
$$\mathbf K_b(\mathbf x) = \mathbf M(\mathbf x) \times \hat {\mathbf n}$$
The volume current density is zero because $\mathbf M$ is constant. For the surface current density, the top and bottom faces don't contribute since $M_0 \hat{\mathbf k}\times\hat {\mathbf k}=0$. For the other four faces we have:
$$\mathrm{x=0 \ face:} \ \mathbf K_1 = M_0 \ \hat{\mathbf k}\times (-\hat{\mathbf i}) = -M_0 \ \hat{\mathbf j}$$
$$\mathrm{x=a \ face:} \ \mathbf K_2 = M_0 \ \hat{\mathbf k}\times \hat{\mathbf i} = M_0 \ \hat{\mathbf j}$$
$$\mathrm{y=0 \ face:} \ \mathbf K_3 = M_0 \ \hat{\mathbf k}\times (-\hat{\mathbf j}) = M_0 \ \hat{\mathbf i}$$
$$\mathrm{y=b \ face:} \ \mathbf K_4 = M_0 \ \hat{\mathbf k}\times \hat{\mathbf j} = -M_0 \ \hat{\mathbf i}$$
Now that you know the bound current distribution, you can simply use the Biot-Savart law to calculate the magnetic field:
$$\mathbf B(\mathbf x) = \frac{\mu_0}{4\pi}\int_{\mathbb S}d\mathbf a' \ \mathbf K(\mathbf x') \times \frac{\mathbf{x-x'}}{|\mathbf{x-x'}|^3}$$
I believe you can take it from here.

My question is how do I evaluate the integral portion of this equation? I'm looking to turn this into a piece of computer code, and my background is pretty shallow when it comes to this level of maths.

Edit: I understand Matlab has an integrate function, but I would prefer not to buy a license for that if possible.

Edit2:

After thinking about this some more, and with the help of Ian's comments I have determined what I think I need to do, which is best expressed graphically by the diagram I have just drawn:

答:
So it seams to me the specific translation you were seeking looks something like this, and probably would have better been asked on physics or stack overflow.</p>
```<code>
for each position (x) in space:

    for each face in the set of faces:

        #create a function for this faces surface current
        Kface(dx) = M_0 cross surface_normal ??? This function feels more like a simple vector

        for each point (df) on current face:

             B_x_df = mu_0 / 4 * pi * Kface(dx) cross x-dx / mag(x-dx)^3

    B_x = sum of intermediate B_x_df values
<span class="math-container">```</span>
</code>```

Title: Integral for Biot-Savart Law on a box

Body:
So it seams to me the specific translation you were seeking looks something like this, and probably would have better been asked on physics or stack overflow.

for each position (x) in space:

    for each face in the set of faces:

        #create a function for this faces surface current
        Kface(dx) = M_0 cross surface_normal ??? This function feels more like a simple vector

        for each point (df) on current face:

             B_x_df = mu_0 / 4 * pi * Kface(dx) cross x-dx / mag(x-dx)^3

    B_x = sum of intermediate B_x_df values
```


问:
Summing a recurrence to infinity
A friend sent this recreational maths problem:

Let $a_0=0$ and $a_1=1$. If $a_n=n(a_{n-1}+3a_{n+1})$, compute $\sum_{n\ge 0}a_n$.

My current work:
Since this is not your classic linear recurrence, you can't use the characteristic polynomial method. I tried Engineer's induction, but the terms didn't seem to follow a pattern of any sort (I listed values for both $a_n$ and $\sum_{n=0}^{i}a_n$ and got nowhere).

It would be great if someone could give me a hint as to solving this.

答:
Let $f(x) = \displaystyle \sum_{n=0}^\infty a_n x^n$. Then:
$$\begin{array}{rcl}
f(x) &=& \displaystyle \sum_{n=0}^\infty a_n x^n \\
x f(x) &=& \displaystyle \sum_{n=0}^\infty a_{n-1} x^n \\
x^{-1} f(x) &=& \displaystyle \sum_{n=0}^\infty a_{n+1} x^n \\
(x + 3x^{-1}) f(x) &=& \displaystyle \sum_{n=0}^\infty (a_{n-1} + 3a_{n+1}) x^n \\
(x + 3x^{-1}) f'(x) + (1-3x^{-2}) f(x) &=& \displaystyle \sum_{n=0}^\infty n (a_{n-1} + 3a_{n+1}) x^{n-1} \\
(x^2 + 3) f'(x) + (x-3x^{-1}) f(x) &=& \displaystyle \sum_{n=0}^\infty n (a_{n-1} + 3a_{n+1}) x^n \\
\end{array}$$
So we obtain the differential equation:
$$f(x) = (x^2 + 3) f'(x) + (x-3x^{-1}) f(x)$$
Which we now solve:
$$\begin{array}{rcl}
(x^2 + 3) f'(x) + (x-3x^{-1}) f(x) &=& f(x) \\
(x^2 + 3) f'(x) &=& (1-x+3x^{-1}) f(x) \\
\dfrac{f'(x)}{f(x)} &=& \dfrac{1-x+3x^{-1}}{x^2+3} \\
(\ln f)' &=& \dfrac{x-x^2+3}{x(x^2+3)} \\
\ln f &=& -\log(x^2+3) + \log(x) + \dfrac1{\sqrt3} \arctan\left(\dfrac{x}{\sqrt3}\right) + C \\
f(x) &=& e^C \dfrac{x}{x^2+3} \exp\left( \dfrac1{\sqrt3} \arctan\left(\dfrac{x}{\sqrt3}\right) \right)
\end{array}$$

Given the constraint $a_1 = 1$ we know $e^C = 3$, so:
$$f(x) = \dfrac{3x}{x^2+3} \exp\left( \dfrac1{\sqrt3} \arctan\left(\dfrac{x}{\sqrt3}\right) \right)$$

So $a_n$ can be found from the Taylor series expansion of $f$.

ContentLicense: CC BY-SA 4.0

Title: Summing a recurrence to infinity


问:
Do discrete geometric objects exist in the same space as non-discrete geometric object?
Do discrete geometric objects exist in the same space as non-discrete geometric object? I am wondering if the spaces are defined differently in discrete geometry. After reading a little about it, I am still confused as to what purpose discrete geometry serves exactly. It seems to be about the study of graphs, but I don't see how it relates to geometry.</p>

答:
I'd say discrete geometry mostly takes place in $d$-dimensional Euclidean space for some $d$, often denoted $\mathbb{E}^d$, which is the same thing as $d$-dimensional real space $\mathbb{R}^d$ with the usual Euclidean metric. Plenty of work is done in other settings, though (hyperbolic space, toroids, etc.)

It includes the study of polytopes, like the regular polyhedra (the Platonic solids), Archimedean solids, and so on. This is the focus of Euclid's Elements, making it probably the oldest and longest-studied branch of mathematics there is.

By contrast, geometry which isn't discrete, which I think of as "swoopy" geometry, has only been developed much more recently.

Whenever you're assembling discrete components, like making polyhedra by attaching polygons, or tiling a surface, or designing a space-filling honeycomb or weight-supporting truss, you're doing discrete geometry. The study of crystals, quasi-crystals, and other periodic or aperiodic systems crop up here.

As you mention, embedding graphs in space is another example, like planar graphs, or with distance constraints in three dimensions, which gives us protein folding. (And global positioning, although the geometric aspects aren't the most challenging part of the GPS system.)

Computer graphics are another "purpose" served by discrete geometry. Objects are modeled by a very large number of polyhedra, to which various Euclidean transformations are applied as the viewpoint moves.


答:
We can prove that $0< a_{n}< 1$ inductively by making use of the graph$:\quad y= x\left ( 1- \ln x \right ).$ We let
$$b_{n}:= 1- a_{n}, \left \{ b_{n} \right \}_{n= 1}^{\infty}\Leftrightarrow b_{1}= \frac{1}{2}, b_{n+ 1}= b_{n}+ \left ( 1- b_{n} \right )\ln\left ( 1- b_{n} \right )$$
Well$,\quad a_{n+ 1}- a_{n}= -a_{n}\ln n> 0,$ then $b_{n+ 1}< b_{n}\Rightarrow 0< b_{n}< b_{1}= \frac{1}{2},$ according to inequality
$$\ln\left ( 1- x \right )< -x,\quad x\in\left ( 0, 1 \right )$$
So$,\quad n> 2$
$$\Rightarrow b_{n}= b_{n- 1}+ \left ( 1- b_{n- 1} \right )\ln\left ( 1- b_{n- 1} \right )< b_{n- 1}- \left ( 1- b_{n- 1} \right )b_{n- 1}= b_{n- 1}^{2}\Rightarrow b_{n}< b_{2}^{2^{n- 2}}=$$
$$= \left ( \frac{1- \ln 2}{2} \right )^{2^{n- 2}}\Rightarrow\sum_{k= 1}^{n}b_{k}< \sum_{k= 1}^{n}\left ( \frac{1- \ln 2}{2} \right )^{2^{k- 2}}< \sum_{k= 1}^{\infty}\left ( \frac{1- \ln 2}{2} \right )^{2^{k- 2}}= 0.56\cdots< \frac{2}{3}$$

ContentLicense: CC BY-SA 3.0

Title: show that $\sum_{k=1}^{n}(1-a_{k})<\frac{2}{3}$

Body:
Let $a_{1}=\dfrac{1}{2}$, and such $a_{n+1}=a_{n}-a_{n}\ln{a_{n}}$,show that
$$\sum_{k=1}^{n}(1-a_{k})<\dfrac{2}{3}$$

My attemp: let $1-a_{n}=b_{n}$,then we have
$$b_{n+1}=b_{n}+(1-b_{n})\ln{(1-b_{n})}<b^2_{n}<\cdots<(b_{1})^{2^{n}}=\dfrac{1}{2^{2^n}}$$
where use $\ln{(1+x)}<x,x>-1$
so
$$\sum_{k=1}^{n}(1-a_{k})<\sum_{k=1}^{n}\dfrac{1}{2^{2^{k-1}}}?$$

But $$\sum_{k=1}^{+\infty}\dfrac{1}{2^{2^{k-1}}}=0.816\cdots$$big than$\frac{2}{3}$,so this inequality How to prove it?


问:
How do we know that PDE solutions obtained via separation of variables are the only ones?
You can find solutions to, for example, the 1D Schrödinger equation $-\frac{\hbar^2}{2m}\Psi_{xx}(x,t) + V(x, t)\Psi(x, t) = i\hbar\Psi_{t}(x,t)$ by assuming solutions of the form $\Psi(x,t) = X(x)T(t)$. How do we know that there aren't other (potentially unphysical, but still mathematically valid) solutions where $x$ and $t$ aren't separate? Does the fact that you can arbitrary linear combinations of the separable solutions to construct a new solution "make up" for the fact that we didn't go looking for the non-separable ones?   </p>

答:
We are indeed using separable solutions to build non-separable ones.

Suppose $\psi_1 = e^{-\frac{i}{\hbar}E_1t}\phi_1(x)$ and  $\psi_2 = e^{-\frac{i}{\hbar}E_2 t}\phi_2(x)$ are two solution for the Schrödinger equation, where $\phi_i$'s are the eigenvector of eigenvalue $E_i$ for the operator $H = -\frac{\hbar}{2m}\Delta + V$, then the superposition
$$\psi_{12} = \alpha_1 \psi_1 + \alpha_2 \psi_2$$
is non-separable.

The separability comes from the heuristic that for a state $\psi(x,0)$ "evolves" with time, and using the evolution operator we can get the solution at time $t$:
$$
\psi(x,t) = e^{-\frac{i}{\hbar}H t}\psi(x,0),
$$
which is intrinsically separable for each term in the expansion in the eigenvalues of $H$:
$$
e^{-\frac{i}{\hbar}H t} = \sum_{n} |{\phi_n}\rangle e^{-\frac{i}{\hbar}E_n t}\langle{\phi_n}|,
$$
translated into mathematics notations:
$$
\psi(x,t) = e^{-\frac{i}{\hbar}H t} |\psi_{t=0}\rangle = \sum_{n} |{\phi_n}\rangle e^{-\frac{i}{\hbar}E_n t}\langle{\phi_n}|\psi_{t=0}\rangle = \sum_{n} \alpha_n e^{-\frac{i}{\hbar}E_n t}\phi_n(x),
$$
where $\alpha_n$ is the inner product $\langle{\phi_n}|\psi_{t=0}\rangle$. Each one term is separable, but not $\psi(x,t)$.

In general, we can get non-separable solution from the evolution of the separable eigen-solution in different energy level.

Aside from Christopher A. Wong said about existence and uniqueness, another reason why we are more interested in the separable solution is because of its physical meaning, for example time harmonic wave $\psi(x,t) = e^{-i\omega t} \phi(x)$.


问:
Cohomology of line bundles on the blowup of $\mathbb P^2$
Let $X$ denote the blowup of $\mathbb P^2$, $E$ the exceptional divisor, and $H$ the pullback of the hyperplane class.  How can I compute $H^0(X,mE+nH)$, $H^1(X,mE+nH)$, and $H^2(X,mE+nH)$ for $m,n \in \mathbb Z$?  If I'm working analytically, how can I think of these geometrically (e.g. "an element of $H^1(X,2E)$ is equivalent to a $1$-form on $\mathbb P^2$ such that...")?</p>

答:
Let $X$ be a smooth surface and $\pi: \widetilde{X}\rightarrow X$ be the blow up of $X$ at one point. By Hartshorne, Algebraic Geometry, Proposition V.3.4,
$$R^{i}\pi_{*}\mathcal{O}_{\widetilde{X}}=0$$ for all $i>0$ and $\pi_*\mathcal{O}_\widetilde{X}\cong\mathcal{O}_X$.
By the projection formula, for any divisor $D\subset X$, we have
$$H^{i}(\widetilde{X},\pi^{*}D)=H^{i}(X,D).$$

Now we come back to your question. If $X$ is the blow up of $\mathbb{P}^{2}$, $H^{i}(X,nH)=H^{i}(\mathbb{P}^{2},\mathcal{O}(n))$.
$$0\rightarrow \mathcal{O}_{X}(nH+(m-1)E)\rightarrow \mathcal{O}_{X}(nH+mE)\rightarrow \mathcal{O}_{E}(nH+mE)\rightarrow 0$$
Since $E(nH+mE)=mE^2=-m$, $\mathcal{O}_{X}(nH+mH)\cong \mathcal{O}_{\mathbb{P}^2}(-m)$. Then by taking the long exact sequence, we may relate the cohomology of $nH+(m-1)E$ and $nH+mE$.
Since we already know the cohomology of $nH$, we can inductively get the cohomology of $nH+mE$ for any $m,n\in \mathbb{Z}$.

OwnerDisplayName: Anonymous

ContentLicense: CC BY-SA 2.5


问:
Set with a property
This was a mock math olympiad problem sent to me by a friend.

Let $P$ be a set of prime numbers. Then, create a set $S$ of positive integers that satisfies the following property:

For every element $p \in P$, $p$ is a factor of at least three elements in $S$.

Prove that for all sets $P$ and $S$, it is possible to divide $S$ into 4 nonempty subsets such that each $p$ is a factor of at least three integers.

I tried to consider the element $a \in P$ that was a factor of the least number of elements in $S$ and see what I could discover, but got nowhere. I also thought that perhaps some sort of algorithm would work too, but I didn't make any progress. Overall, I'm just not quite sure how to even start.

答:
Not entirely an answer, but too long to be a comment: note that primes and factoring here are a little bit of a red herring. It's possible to rewrite this question in such a way that they don't come into it at all. To see this, let the set of primes be labeled $\{p_1, p_2, \ldots, p_N\}$ (where possibly there is no last $N$, in which case we'll say that $N=\omega$ just for convenience). Then label each number $i\in S$ by the set $s_i$ of numbers $j$ such that $p_j$ is a factor of $S$. For instance, if $P$ is the set $\{2, 3, 11\}$ and $S$ is the set $\{12, 22, 99\}$, then we would have $p_1=2, p_2=3, p_3=11$ and $s_{12}=\{1,2\}$ since $p_1$ and $p_2$ are the factors of $12$; similarly $s_{22}=\{1,3\}$ and $s_{99}=\{2,3\}$. But now we can write the condition '$p_j$ divides $i$' as '$j$ is a member of $s_i$' and just write $S$ in terms of the $s_i$ (e.g., here $S$ would be $\left\{\{1,2\}, \{1,3\}, \{2,3\}\right\}$) and your initial properties become:

for every $n \leq N$, there are at least two sets $s\in S$ such that $n\in s$

for every $i, j\leq N$ there is at least one set $s\in S$ such that $i\in s$ and $j\in s$ (or equivalently, $\{i,j\}\subseteq s$)

And your result becomes that it's possible to divide $S$ into three sets $S_1, S_2, S_3$ such that for every $n\leq N$ at least two $S_i$ contain a set which has $n$ as an element.

Title: Set with a property


问:
Expected hitting time in a Markov chain
Consider a Markov chain on $\{0, 1, \ldots\}$ with $p_{i, i-1}=q$ and $p_{i,i+1}=p$.

If $p = q = 0.5$, find the expected hitting time to hit State $0$ from all states $k \geq 1$.

If $q>p$, find the expected hitting time from each state $k\geq 1$.

I tried to set up systems of equations:

$E(X_0) = 0$

$E(X_n) = \frac{1}{2}(1 + E(X_{n-1})) + \frac{1}{2}(1 + E(X_{n+1}))$ for $n \geq 1$.

I need help analyzing this system in the two cases though. I think the answer in the first case is just $\infty$, so I'm guessing I get a divergent series or something. I'm not entirely sure.

Any help is greatly appreciated. I know it's related to the Gambler's ruin problem, but I can't find anything online.

答:
Let $y_n=E[X_n]$ then you have a linear recurrence relation with constant coefficients:

$$y_{n+1}-2y_n+y_{n-1}=-2.$$

This system needs two boundary conditions. You only really have one, which is $y_0=0$. To fix that, you can artificially introduce $y_N=0$ (so that you are solving for the expected time to hit $0$ or $N$ from inside) and then send $N \to \infty$ (so that you are asymptotically guaranteed to hit $0$ first since $N$ is being moved further and further away).

To solve the finite problem, the procedure is to split into a homogeneous and particular solution. For a particular solution, you may first guess that a constant might work, but you find you're wrong (you get $0=-2$). Next you may guess that a linear function might work, but in this special case with the probabilities being equal you are again wrong (you get $0=-2$). So you continue all the way to a quadratic function and get

$$c \left ( (n+1)^2-2n^2+(n-1)^2 \right ) = c \left ( n^2 + 2n + 1 - 2n^2 + n^2 - 2n + 1 \right ) = 2c = -2$$

so $c=-1$. Thus a particular solution is $y_n=-n^2$.

Next you need the general homogeneous solution. It turns out that you already found it in the course of doing the guesswork to find the particular solution: the general homogeneous solution is $y_n=c_1 + c_2 n$. So you have

$$y_n=c_1 + c_2 n - n^2$$

where $c_1,c_2$ are to be found. Plugging in $0$ you get $c_1=0$. Plugging in $N$ you get $c_2 N - N^2 = 0$ so $c_2=N$. Thus

$$y_n=N n - n^2.$$

Now what happens when you send $N \to \infty$ for $n>0$ fixed?

The case $q>p$ is fairly similar. The differences are:

In finding the particular solution, the linear function will actually work

To get the homogeneous solution you will need to look for exponential solutions, i.e. the homogeneous solution will be of the form $c_1 \lambda_1^n + c_2 \lambda_2^n$.

Title: Expected hitting time in a Markov chain


问:
Understanding subsitution by parts of $\int_0^{1/4}x\sin^{-1}(4x)dx$
I am doing an Integration by Parts exercise and can't wrap my head around one of the answers:

Consider the definite integral

$$\int_0^{1/4}x\sin^{-1}(4x)dx$$

We select $u = \sin^{-1}(4x)$.

We select $dv = xdx$.

Consider the known formula $$\int udv = uv - \int vdu$$

Then, after integrating by parts, we obtain the integral $$\int_0^{1/4}vdu = \int_0^{1/4}f(x)dx$$ on the right hand side where
$$f(x) = \frac{x^2}{2} \cdot \frac{1}{(\sqrt{1-(4x)^2})} \cdot 4$$

I get it all up to this point, but then the exercise says this:

The most appropriate subsitution to simplify this integral is $x = g(t)$ where $$g(t) = \frac{\sin(t)}{4}$$

It's not clear to me what do they mean by "simplify this integral". What's there to simplify and why?

And of course, why $\frac{\sin(t)}{4}$? I can almost see what's going on because that integral looks a lot like the differentiation of $\arcsin(x)$, but still failing to make the connection.

What is it that I'm missing? What was the train of thought behind that substitution?

答:
How do you deal with a problem like
$$\int\frac{\text{d}x}{1-x^2}$$
Most likely, you would substitute $x=\sin u$ so that you can use the trig identity for the denominator and everything simplifies.

In your regard, there is a $4x$ instead of $x$, so they just substituted $4x=\sin u$, (and used their fancy way to write the same thing!)

Hope this helps. Ask anything if not clear :)


问:
Unique questions about pentagon
Consider the (non-regular) pentagon with consecutive vertices at (-1,-1), (-1,1), (0,2), (1,1), and (1,-1).
a) Prove that there is no circle that is tangent to all 5 sides of the pentagon
b) Is there an ellipse that is tangent to all five sides of the pentagon? If so, find its equation.

Here is what I have done so far:
I have drawn out the pentagon. For part a) I am very bad at proving. For part b) I have some idea. I drew the ellipse and tried to match the ellipse inside of the pentagon and tried to see its tangency. My best idea is to take the derivative. I may be completely off from the true solution, I am not sure. Please click the solution button below and tell me the correct solution. This is a question that is not for homework, so if you posted the solution there would be no issue.

答:
For a), note that the diameter formed by the points of tangency of the vertical sides, must be horizontal, and therefore must be 2. In other words, the radius must be 1. This implies that the center of the circumference must have an X coordinate equal to 0. But then for it to be tangent to the horizontal side, it must have Y coordinate equal to 0. So the center is (0,0). Prove that that is impossible so that it is tangent to the crooked sides.

For b) use a general formula and try to clear the parameters, knowing that the foci must have an x coordinate equal to 0.

Title: Unique questions about pentagon


问:
Distributive Law in Boolean Algebra problem
Im practicing boolean algebra on the following problem

```<code>(A+B)(¬A+¬B)
</code>```

In my textbook they apply the distributive property to get

```<code>A(¬A+¬B)+B(¬A+¬B)
</code>```

Im not quite sure how this is being applied and would appreciate some clarification.

I know the distributive property is defined as followed

```<code>A(B+C)=(AB)+(AC)
</code>```

But in our problem we have (A+B)(¬A+¬B) not (AB)+(AC)

答:
They implicitly assumed $C=A'+B'$, and so
$$(A+B)(A'+B')=(A+B)C$$$$=AC+BC=A(A'+B')+B(A'+B')$$
Hope this helps. Ask anything if not clear :)


问:
Solve $y'-2=e^{x^2}y$
This is for some help. A differential equation $y'-1=e^{x^2}y$, and if $f(x)$ and $g(x)$ are solutions to the differential equation, then will $h(x)=f(x)+g(x)$ solve the equation $y'-2=e^{x^2}y$. So basically it's saying 2 solutions to the old equation added together, does it solve the equation $y'-2=e^{x^2}y$. I think it's no. It's 2 solutions from a totally different differential equation! It's combined, but it doesn't make a difference.

答:
Simply check it

We know that $f$ and $g$ are solutions to the differential equation

$$ y'-1=e^{x^2}y$$

Let's check wether $h(x)=f(x)+g(x)$ satisfies the following equation:

$$ y'- 2=e^{x^2}y$$
Then
$$ (f(x)+g(x))'-2
=e^{x^2}(f(x)+g(x))
$$
$$
f'(x)+g'(x)-1-1
=e^{x^2}f(x)+e^{x^2}g(x) \hspace{1cm} (1)
$$

Since $f$ is a solution, then

$$ f'(x) - 1 = e^{x^2}f(x)$$

We cancel the two parts from $(1)$ :

$$ g'(x) -1 = e^{x^2}g(x) $$

Which is also correct because $g$ is a solution to the first equation

So yes, $h(x)=f(x)+g(x)$ is a solution to your second equation.

It doesn't matter from where you got the solutions, as long as the sum satisfies the equation we're all good

Title: Solve $y'-2=e^{x^2}y$


问:
Prove $\int_{0}^{\frac{\pi}{2}}f(\cos(x))dx = \int_{0}^{\frac{\pi}{2}}f(\sin(x))dx$ with hint
Assume $f$ is continuous, prove $\int_{0}^{\frac{\pi}{2}}f(\cos(x))dx = \int_{0}^{\frac{\pi}{2}}f(\sin(x))dx$...

I was given a hint, $\sin x=\cos(\frac{\pi}{2}-x)$. I used this property to get $\int_{0}^{\frac{\pi}{2}}f(\sin x)dx=\int_{0}^{\frac{\pi}{2}}f(\cos(\frac{\pi}{2}-x))dx$ and $\int_{0}^{\frac{\pi}{2}}f(\cos \frac{\pi}{2})dx+\int_{0}^{\frac{\pi}{2}}f(\cos(x))dx$. I am a few steps away, and I know $\cos(\frac{\pi}{2})=0$, but I only have $f(0)$, which is not $0$. If I can make it $0$, then I'm done. What is the additional step needed here?

答:
Well, use the King's Rule of Integration:
$$\int_a^bg(x)\,\text{d}x=\int_a^bg(a+b-x)\,\text{d}x$$

 For proof, substitute $u=a+b-x$ so that $\text{d}u=-\text{d}x$. The limits become $b$ to $a$. So, our integral is
 $$\int_b^a-g(a+b-x)\,\text{d}x=\int_a^bg(a+b-x)\,\text{d}x$$

Here,
$$\int_{0}^{\pi/2}f(\cos x)\,\text{d}x$$$$=\int_{0}^{\pi/2}f\left(0+\frac{\pi}{2}-\cos x\right)\,\text{d}x$$
$$=\int_{0}^{\pi/2}f(\sin x)\,\text{d}x$$

Alternatively, just substitute $u=\dfrac{\pi}{2}-x$ so that $\text du=-\text dx$, and $0\rightarrow\frac{\pi}{2}$ and $\frac{\pi}{2}\rightarrow0$. So, the integral becomes
$$\int_{\frac{\pi}{2}}^0-f\Bigg(\cos\left(\frac{\pi}{2}-u\right)\Bigg)\,\text du$$
$$=\int_0^{\frac{\pi}{2}}f(\sin u)\,\text du$$
$$=\int_0^{\frac{\pi}{2}}f(\sin x)\,\text dx$$
Hope this helps. Ask anything if not clear :)


问:
Suppose that $\Sigma \vdash \exists x \theta$. Do we then have that $\Sigma \vdash \theta$?
This is a problem from A Friendly Introduction to Mathematical Logic by Christopher Leary and Lars Kristiansen which is in Page number 65.

Here's my attempt:

Consider a structure $\mathfrak{U}$ and suppose that its universe is $A=\{ 1,2 \}$. Consider $\Sigma$ be the singleton consisting of the formula $\exists x(x=y)$. Then $\Sigma \vdash \exists x(x=y)$ and suppose that we had $\Sigma \vdash x=y$. Then by Soundness Theorem, $\Sigma\models x=y$.

Consider the variable assignment function $s$ into $\mathfrak{U}$ which sends $x$ to $1$ and $y$ to $2$. Then clearly, $\mathfrak{U}$ satisfies $\exists x (x=y)$ with the assignment $s$ as $\mathfrak{U}$ satisfies $x=y$ with the substituition $s[x|2]$. But on the other hand, $\mathfrak{U}$ does not satisfy $x=y$ with the substituition $s$. This would contradict that $\Sigma\models x=y$. Hence, it cannot be that $\Sigma \vdash x=y$.

Is this counterexample correct? Is there a way to do without appealing to the Soundness Theorem?

答:
Yes, your counterexample works.

As to avoiding soundness: it's difficult, although of course not impossible, to avoid the soundness theorem when proving non-instances of $\vdash$. This is because the definition of $\vdash$ is a "positive inductive" one - it's set up to be very easy to prove instances of $\vdash$, but that has a cost. To see the issue, think about proving $\top\not\vdash\perp$; although blindingly obvious, this isn't actually easy (depending on what definition of "$\vdash$" you're using anyways).

Broadly speaking, in order to prove $\Phi\not\vdash\psi$ you need to produce a set of formulas $\Theta$ which is closed under $\vdash$, contains $\Phi$, and does not contain $\psi$ (and verify that of course). The soundness theorem is really just a way of repackaging this task: every structure $\mathfrak{M}$ has its associated theory $Th(\mathfrak{M})$, which is closed under $\vdash$ by soundness, and so we can think about structures instead of theories. However, we can totally "cut out the middleman:" you can produce an entirely syntactic description of $Th(\mathfrak{A})$, and prove the relevant properties of this, without ever actually referring to structures.

And this is in fact a worthwhile problem to tackle:

Can you give a "structure-free" description of $Th(\mathfrak{A})$?

Even the case of quantifier-free formulas is instructive here. (HINT: think about ways of splitting the set of free variables into two possibly empty pieces.)


问:
Wouldn't an element of a set always spans a ray of the cone of that set?
For context, I'm reading the paper A linear optimization oracle for zonotrope computation.

In algorithm 1, line 8, I don't understand the testing condition:

For a discrete set $G \subset \mathbb{R}^n$.

It is testing condition:  $\forall z \in G$  if $z$ spans a ray of $\operatorname{cone}(G)$.

Isn't it always true for all elements of $G$?

答:
Presumably, this paper is using "ray" to mean "extreme ray", that is (borrowing a definition from Fenchel 1951)

A ray (x) of a convex cone C is an extreme ray of C if x is not a
positive linear combination of two linearly independent vectors of C.

It seems clear this is what is meant in your paper, given this statement (part of proposition 2.4):

The half-line spanned by a point $x$ in $A$ is a ray of the cone spanned
by $A$ if and only if the line through 0 and $x$ is disjoint from the
convex hull of $A\setminus\{x\}$.


问:
Convergence of distance between Cauchy sequences
If $\{a_n\}$ and $\{b_n\}$ are Cauchy sequences in $(X,d)$, how do I show that $d(a_n , b_n)$ converges?</p>

答:
The key is that for all $m, n \in \mathbb{N}$,
$$d(a_m, b_m) \leq d(a_m, a_n) + d(a_n, b_n) + d(b_m, b_n)$$
from using triangle inequality twice.

It suffices to show that $\{d(a_n, b_n) \}$ is a Cauchy sequence in $\mathbb{R}$.

Fix $\varepsilon > 0$. $\{ a_n \}$ and $\{b_n\}$ are Cauchy in $(X, d)$, so $\exists N \in \mathbb{N}$ such that $m, n > N \implies d(a_m, a_n) < \varepsilon/2$ and $d(b_m, b_n) < \varepsilon/2$. So for $m, n > N$, if $d(a_m, b_m) \geq d(a_n, b_n)$ then
$$|d(a_m, b_m) - d(a_n, b_n)| = d(a_m, b_m) - d(a_n, b_n) \leq d(a_m, a_n) + d(b_m, b_n) < \varepsilon \text, $$
and the case $d(a_n, b_n) > d(a_m, b_m)$ is the same.


答:
Hint: consider ellipses with centre on the $y$ axis and axes parallel to the coordinate axes, tangent to the vertical and horizontal sides.  If the major axis of the ellipse is too small, it will look like this:

If too large, like this:

You want it just right so it is tangent to the slanted sides.

Title: Unique questions about pentagon

Body:
Consider the (non-regular) pentagon with consecutive vertices at (-1,-1), (-1,1), (0,2), (1,1), and (1,-1).
a) Prove that there is no circle that is tangent to all 5 sides of the pentagon
b) Is there an ellipse that is tangent to all five sides of the pentagon? If so, find its equation.

Here is what I have done so far:
I have drawn out the pentagon. For part a) I am very bad at proving. For part b) I have some idea. I drew the ellipse and tried to match the ellipse inside of the pentagon and tried to see its tangency. My best idea is to take the derivative. I may be completely off from the true solution, I am not sure. Please click the solution button below and tell me the correct solution. This is a question that is not for homework, so if you posted the solution there would be no issue.


问:
Is the cardinality of a set necessarily a natural number?
I've never seen phrases like "$\sqrt{5}$ people" or "a set with $\pi$ many elements". Are there sets with cardinality, say, $\frac{1}{2}$?

Edit: As Brian M. Scott pointed out, the only real numbers that are cardinalities of sets are the non-negative integers. Could you explain why is it so?

答:
I am copying Brian M. Scott's comment here, because I think it answers the question completely and correctly:

As the term cardinality is normally used, the only real numbers that are cardinalities of sets are the non-negative integers. The cardinalities of infinite sets, of course, are not natural numbers.


答:
I know the distributive property is defined as followed

```<code>A(B+C)=(AB)+(AC)
</code>```

It is also the case that conjunction is commutative, so

```<code> (A+B)(¬A+¬B) = (¬A+¬B)(A+B)          Commute
              = (¬A+¬B)A+(¬A+¬B)B     Distribute
              = A(¬A+¬B)+B(¬A+¬B)     Commute
</code>```

Title: Distributive Law in Boolean Algebra problem

Body:
Im practicing boolean algebra on the following problem

```<code>(A+B)(¬A+¬B)
</code>```

In my textbook they apply the distributive property to get

```<code>A(¬A+¬B)+B(¬A+¬B)
</code>```

Im not quite sure how this is being applied and would appreciate some clarification.

I know the distributive property is defined as followed

```<code>A(B+C)=(AB)+(AC)
</code>```

But in our problem we have (A+B)(¬A+¬B) not (AB)+(AC)

Body:
I know the distributive property is defined as followed

A(B+C)=(AB)+(AC)

It is also the case that conjunction is commutative, so

 (A+B)(¬A+¬B) = (¬A+¬B)(A+B)          Commute
              = (¬A+¬B)A+(¬A+¬B)B     Distribute
              = A(¬A+¬B)+B(¬A+¬B)     Commute


问:
If $g$ is bounded and measurable and $\mu$ is obtained by taking the lim inf of $g$ approaching each point, is $\mu$ Riemann integrable?
Let $$g:[a,b] \to \mathbb{R}$$ be bounded and measurable. Define $$\mu = \lim_{\delta \to 0} \mu_\delta(x),$$
where $$\mu_\delta(x) = \inf \{g(y):y \in b_\delta(x) \cap [a,b]\}.$$
Is $\mu$ always Riemann integrable?

I know that $\mu$ is Lebesgue integrable but I'm not sure whether it is always Riemann integrable. I'm suspecting not, but I can't seem to find a counterexample to this. Is there any suggestion?

答:
Let $K\subset[a,b]$ be a fat Cantor set and let $g$ be the characteristic function of $[a,b]\setminus K$.  Then $\mu=g$ is not Riemann integrable since it is discontinuous at every point of $K$ and $K$ has positive measure.

In fact, the assumption that $g$ is measurable is not really relevant.  For any function $g$, $\mu$ will be a lower semicontinuous function (and thus is measurable, even if $g$ was not!), and if $g$ was already lower semicontinuous, then $\mu=g$.  So you are really just asking whether every bounded lower semicontinous function is Riemann integrable.


问:
Cyclic Homology of a mixed complex as derived tensor product.
Let $k$ be a commutative ring and $A$ a $k$-algebra. Denote by $C_{\bullet}(A)$ the Hochschild chain complex, with $C(A)_n = A^{\otimes n+1}$. Let $B$ denote Connes' operator
$$B_n \colon A^{\otimes n+1} \to A^{\otimes n+2}\, .$$
This induces a bicomplex $\mathcal{B}_{\bullet\bullet}(A)$ with $\mathcal{B}(A)_{pq} = A^{\otimes q-p+1}$ if $q \geq p$ and $0$ otherwise, where the vertical differential $b$ is the Hochschild differential and the horizontal differential is given by $B$. Define the cyclic homology of $A$ as
$$HC_{\bullet}(A) := H_n(\text{Tot}\, \mathcal{B}_{\bullet\bullet}(A)) \, .$$

This definition makes sense for any mixed complex $(M, b, B)$ (i.e. $M$ is a graded $k$-module together with a homological differential $b$ and a cohomological differential $B$ such that $bB + Bb = 0$). Any mixed complex can be identified with a dg $\Lambda$-module over the dg-algebra $\Lambda$ given by
$$\Lambda_0 = k \, , \; \Lambda_1 = k[\epsilon]/\epsilon^2 \, \; \Lambda_i = 0 \quad \text{for $i >1$}\, ,$$
where $\epsilon$ has degree $1$. An alternative definition of cyclic homology that works for any bicomplex $(M, b, B)$ is
$$HC_{\bullet}(M) = H^n(k \otimes_{\Lambda}^{\mathbf{L}} M) \, ,$$
where $-\otimes_{\Lambda}^{\mathbf{L}}-$ stands for the derived tensor product. My question is

Why are these two definitions equivalent?

If I wanted to compute $HC_{\bullet}(M)$ as in the second definition, I need a resolution of $k$ by dg $\Lambda$-modules, and then somehow compute the derived tensor product. I don't know how to proceed, probably because my comprehension of the "derived world" is very limited. Any help would be very much appreciated.

答:
What you want is explained in this paper of Kassel, in Proposition 1.3.

A mixed module is a (non-negatively graded) dg-module $(M,b)$ for the algebra $\Lambda = k[t]/(t^2)$ where $t$ has degree $1$, so it comes equipped with an extra differential $B$ of degree $1$ that (anti)commutes with $b$.

The trivial module $k$, seen as a (dg) module over $\Lambda$ admits a very simple free resolution over $\Lambda$, where the maps are multiplication by $t$, where I use $s$ as a formal symbol to remember the degree where $\Lambda$ sits on:

$$C:\cdots\longrightarrow s^4\Lambda \longrightarrow s^3\Lambda \longrightarrow s^2\Lambda \longrightarrow s\Lambda \longrightarrow 0$$

If you tensor this with $M$ over $\Lambda$, you get a complex $C\otimes_\Lambda M$ representing the derived tensor product
such that $$(C\otimes_\Lambda M)_n = \bigoplus_{i+j=n} s^iM_j = M_n\oplus M_{n-1} \oplus \cdots$$

and where we both have an internal differential $b$ that preserves this homological grading (and simply goes from $M_n$ to the next summand, etc) and the external differential $B$ going from one "row of $M_i$s" to the next one above. This is precisely what your cyclic bar (double) complex looks like:

$$\begin{matrix}
\uparrow & & \uparrow &  &  \uparrow\\
M_2 &\to& M_1& \to &M_0 \\
\uparrow & &\uparrow \\
M_1 &\to &M_0 \\
\uparrow \\
M_0\end{matrix}$$

right? (You need to totalize at the end and look at diagonals, and this introduces the funny situation that as in Kassel's paper the total complex now goes $M_n\oplus M_{n-2}\oplus\cdots$.)


问:
Convergence in probability for a sequence of random variable
I'm currently stuck on the following problem which involves proving the convergence in probability for a sequence of independent random variables. Here's the full statement:

Let $X_1, X_2, X_3,\dots$ be a sequence of independent random variables such that $\mathbb E[X_i]=\mu$ for all $i$ and exist a constant $L$ satisfying $0\lt L\lt\infty$ such that $\mathbb V[X_i]\lt L$ for all $i$. Show that $n^{-1}\sum\limits_{i=1}^nX_kX_{k+1} \to \mu^2, \text{
in probability for $n\ge1$ }$

My attempt: Let $Y_n$= ${n^{-1}\sum\limits_{i=1}^nX_kX_{k+1}}$,

$P \left( \left | {Y_n-\mu  } \right | \ge \epsilon \right) \le \frac{Var \left(Y_n \right )}{\epsilon^2}$

$\mathbb V[Y_n]= \mathbb V[{n^{-1}\sum\limits_{i=1}^nX_kX_{k+1}}]$=$n^-2
 \{{\sum\limits_{k=1}^n\mathbb V[X_kX_{k+1}]}+2({\sum\limits_{1\le i\le j\le n}Cov(X_iX_{i+1}, X_jX_{j+1})}$}

I don't know how to proceed any further than that. I appreciate any help. Thank you in advance.

答: $X_iX_{i+1}$ and $X_jX_{j+1}$ are independent if $|i-j|>1$, so their covariance is zero if $|i-j|>1$.  There are only $O(n)$ non-zero terms in the covariance sum, so dividing it by $n^2$ gives an $O(n^{-1})$ contribution.


问:
Showing that a set is open with metric
Let $(A, d_1)$ and be a metric. Suppose $d_2: A \times A \rightarrow \mathbb{R}$ be defined in this way:

For $a, b \in A$, $d_2(a, b) = \frac{d_1(a, b)}{1 + d_1(a, b)}$.

Show that $E \subset A$ is open with respect to $d_1 \iff E$ is open with respect to $d_2$.

My attempt:

I have replaced $a, b$ with $x, y$*

Proof.

Assume $E$ is open inside $(A, d_2)$. Then there is a $r > 0$ s.t. the open ball $B_2(x, r) = \{y \in A: d_2(x, y) < r\}$ is contained in $E$.

We need to show that for any $x \in E$, there is an $\rho > 0$ so that the open ball $B_1(x, \rho)  = \{y \in A: d_1(x, y) < \rho\}$ is contained in $E$.

Let $r < 1$ and set $\rho = \frac{r}{1 - r}$ then for $y \in B_1(x, \rho)$ we have

$$d_2(x, y) = \frac{d_1(x, y)}{1 + d_1(x, y)} < \frac{\rho}{1 + \rho} = r$$

So $y \in B_2(x, r)$ and so $B_1(x, \rho) \subset E$. Hence $E$ is open inside $(A, d_1)$.

The other way.

Assume $E$ is open inside $(A, d_1)$. Then there is a $\rho > 0$ s.t. the open ball $B_1(x, \rho) = \{y \in A: d_1(x, y) < \rho\}$ is contained in $E$.

We need to show that for any $x \in E$, there is an $r > 0$ so that the open ball $B_2(x, r)  = \{y \in A: d_2(x, y) < r\}$ is contained in $E$.

Set $r = \rho$ then for $y \in B_2(x, r)$ we have:

$$d_2(x, y) = \frac{d_1(x, y)}{1 + d_1(x, y)} \leq d_1(x, y) < \rho = r$$

So $y \in B_1(x, \rho)$ and $B_2(x, r) \subset E$. We may conclude $E$ is open inside $(A, d_2)$.

QED.

Is this proof correct? If not, can somebody explain where I went wrong. Any assistance much appreciated.

答:
Nearly perfect.  Here's some quibbles (my changes in italic):

Line 1, Then for every $x\in E$ there is a $r>0$...

Line 3, if $r\geq1$, $B_2(x,1/2)\subseteq B_2(x,r)$, which is why your logic suffices.

Line 4, depending on your reader, it may not be clear that $\rho=\frac r{r-1}\implies r=\frac\rho{1+\rho}$, likewise for $\frac{d_1(x,y)}{1+d_1(x,y)}<\frac\rho{1+\rho}\implies d_1(x,y)<\rho$.  Only the algebra convinces me that the inequality is the direction it is.

Line 7, Then for every $x\in E$..., I'm being pedantic.

Title: Showing that a set is open with metric


问:
I need a limit definition for the Hessian, does this work?

Let $f: \mathbb{R}^n \to \mathbb{R}$ be of class $C^2$. Let $x$ be a non-degenerate critical point of $f$. Prove that there is an open neighborhood of $x$ which contains no other critical points of $f$.

$\textbf{Proof:}$  Suppose the contrary. Then for every $n \in \mathbb{Z}_+$ there is an $x_n \in B\left(x,\frac{1}{n}\right)$, which is a critical point of $f$ not equal to $x$. So $x_n$ converge to $x$. Now let $v_n = \frac{x_n - x}{||x_n-x||}$. Let us just assume $v_n$ converges to some $v \in \mathbb{S}^{n-1}$ (since $v_n$ is a bounded sequence, it has a convergent subsequence). So by the definition of the Frechet derivative, and since $x_n \to x$, we have that
\begin{align*}
0 &= \lim_{n\to\infty}  \frac{||Df(x_n) - Df(x) - D^2f(x)(x_n -x)||}{||x_n - x||}\\
  &=\lim_{n\to\infty}  \frac{||- D^2f(x)(x_n -x)||}{||x_n - x||}\\
  &=\lim_{n\to\infty}  ||D^2f(x)v_n||\\
  &= D^2f(x)v,
\end{align*}
where we use the fact that $Df(x) = Df(x_n) = 0$ for all $n \in \mathbb{Z}_+$ ($x$ and $x_n$ are critical points). This implies that $D^2f(x)v = 0$, but since $x$ is a nondegenerate point, $D^2f(x)$ is invertible, and since $v \ne 0$, we have a contradiction. Thus, there is an open neighborhood of $x$ that contains no other critical points of $f$.

Does this approach work? I am not sure if I used a correct definition for the Hessian. Also, I did not use that the Hessian is continuous, which worries me. Regardless of whether this is correct or not, is there a direct approach, or even maybe a more elegant one?

答:
Your proof, as it is, is correct! You do not need continuity of the second derivative, only that it is non-degenerate. Indeed, the following is universally true:

Lemma 1: If $g: R^n \to R^m$ (not necessarily $C^1$) has injective derivative at $x_0$ (as a linear map), then there is a neighborhood of $x_0$ on which no other point maps to $g(x_0)$.

In the above, you will apply this to $g=Df$, $n=m$, as people pointed out. Indeed, your proof is basically how this lemma is proved!

Last, probably your proof will benefit from justifying the very first line of calculations, which is equivalent to saying that Hessian is the Frechet derivative of $Df$:

Lemma 2: If $f\colon R^n \to R$ is $C^2$, then
$$
\lim_{h \to 0} \frac{\nabla f(x_0+h) - \nabla f(x_0) - Hf(x)h}{|h|} = 0 \, .
$$


问:
Over C, invertible operators have square roots?
We know that on finite-dimensional vector spaces, it is true. This is 8.33 in Sheldon Axler's Linear Algebra Done Right:

Suppose $V$ is a complex vector space and $\, T\in L(V) \,$ is
invertible. Then $T$  has a square root.

The question is that, if V is infinite-dimensional, does this result still hold?

I'd be much obliged if anyone could give a counterexample or prove that this result is true in general. Thanks!

答:
Be careful about the first example above. On the space $\ell^2$, the shift operator $T$ defined in the first comment above is a bounded invertible operator. Intuitively, a square root of $T$ should be a shift by one-half slot, which of course does not make sense. Thus one can easily believe that this invertible operator $T$ does not have a square root. However, this operator does have square root!! To see this, identify $\ell^2$ with $L^2$ of the unit circle and identify the shift with the operator of multiplication by $z$. Then the operator of multiplication by the square root of $z$ (choose any branch of the square root function) is a square root of $T$.

For an invertible operator that does not have a square root, consider the operator of multiplication by $z$ on the Hardy space of an annulus centered at the origin. This operator is invertible (its inverse is multiplication by $1/z$) but it does not have a square root.


问:
How to prove something combinatorially
I'm being asked to prove $ 2 \cdot 3^0 + 2 \cdot 3^1 + 2 \cdot 3^2 + \ldots + 2 \cdot 3^{n-1} = 3^n - 1. $ combinatorially using the question "How many length-$n$ lists can we form using the elements in $\{0, 1, 2\}$ in which the elements are not all 0?"

Frankly, I'm not even sure where to start and not entirely sure what a combinatorial proof is, to be honest. How would I go about approaching this problem?

I assume that the right side is $3^n-1$ because it represents the number of lists of length n minus the list of all 0's.

答: A combinatorial proof shows that an identity is true by demonstrating that the two sides count the same objects in two different ways.  For your given identity, you have already determined the objects and an explanation for the RHS.  To explain the LHS, condition on the first appearance of $1$ or $2$.

Title: How to prove something combinatorially


问:
Examples of a random variable $X$ and a mapping $f$ such that $f(X)$ is not a random variable
I am reading Probability and Random Process and in chapter 4.2 where the author says something like that given a random variable $X$, $f(X)$ is also a valid random variable if $f$ is sufficiently smooth or regular by being continuous or monotonic. This explanation is a little bit abstract for me. Can anybody explain this further or maybe provide a counter example? Thanks a lot!

答:
As defined by Grimmett and Stirzaker in sec. 2.1, a random variable $X$ on the sample space $\Omega$ with $\sigma$-algebra $\mathcal F$ is a real-valued function $X$ on $\Omega$ such that for every real $x$, $\{\omega \in \Omega : X(\omega) \le x \} \in \mathcal F$.  Equivalently, for every Borel set $B \subseteq \mathbb R$, $\{\omega \in \Omega: X(\omega) \in B\} \in \mathcal F$.

Now $\{\omega \in \Omega: f(X(\omega)) \in B\} = \{\omega \in \Omega: X(\omega) \in f^{-1}(B)\}$.  So in order for $f(X)$ to be a random variable (with the same $\mathcal F$), it suffices for $f^{-1}(B)$ to be Borel whenever $B$ is Borel.  That is true if
$f$ is a Borel measurable function.  In particular it is true if $f$ is continuous.

On the other hand,  for an example where $f(X)$ is not a random variable, you might take $\Omega = \mathbb R$ with Borel $\sigma$-algebra $\mathcal F$, and $X(\omega)=\omega$, and $f$ some function that is not Borel measurable.


答:
A combinatorial proof is just a proof that depends on counting something.  You're correct about the right-hand side.

As to the left hand side, how many lists can you make that don't have a $0$ in the first place?  There are $2$ choices for the first place, and $3^{n-1}$ for the remaining places.  What about that lists that have a $0$ in the first place but not the second?  It's $1\cdot2\cdot3^{n-2}$ and so on.

A neater proof, in my view, is that the left-hand side is $22\dots2_3$ where there are $n$ $2$'s.  If we add $1$, the usual addition algorithm gives $10\dots0_3$ where there are $n$ $0$'s.

Title: How to prove something combinatorially

Body:
I'm being asked to prove $ 2 \cdot 3^0 + 2 \cdot 3^1 + 2 \cdot 3^2 + \ldots + 2 \cdot 3^{n-1} = 3^n - 1. $ combinatorially using the question "How many length-$n$ lists can we form using the elements in $\{0, 1, 2\}$ in which the elements are not all 0?"

Frankly, I'm not even sure where to start and not entirely sure what a combinatorial proof is, to be honest. How would I go about approaching this problem?

I assume that the right side is $3^n-1$ because it represents the number of lists of length n minus the list of all 0's.


问:
Let $A$ and $B$ be two sets. Then, prove that $A - B = A$ iff $B - A = B$
Can someone help me with this question?

Question: Let $A$ and $B$ be two sets. Then, prove that

$$A - B = A \Longleftrightarrow B - A = B$$

答:
Welcome to Math SE!

Motivation

Given two sets, $A$ and $B$, if no part of $B$ is in $A$, then the two sets must be disjoint. The converse is also true i.e. if the two sets are disjoint, then no part of $B$ can be in $A$. If you have trouble seeing this, draw a simple Venn diagram.

Idea

Here, I will only prove the $\implies$ direction as the same argument can be used to prove the $\impliedby$ direction.

The proof is very short and sweet.

If $A - B = A$,

then $A \cap B = \emptyset = B \cap A$.

$\because B \cap A = \emptyset$

$\therefore B - A = B$


问:
Clarification regarding the space of continuous functions and its completeness (Banach)
A Banach space is a normed vector space with completeness, i.e. all Cauchy series converge to an element within the vector space.

In the case of continuous functions on an interval in the real line, $C[a,b],$ equipped with the uniform or infinity norm, it forms a Banach space.

However, the following Cauchy sequence in the space $C[-\infty,\infty]$ seems to be a counter-example, or much more likely, a clue that I misunderstood the whole thing:

$$f^{k}(x) = \frac 1 2 +\frac 1 \pi \tan^{-1}(kx)$$

The square function is not a continuous function, yet this Cauchy sequence clearly converges to it. Therefore, the vector space of continuous functions is not complete, and shouldn't be a Banach space (?).

What am I missing? Is the reason the fact that the chosen norm (the supremum of the function is always constant at $1$) renders this not a Cauchy sequence)? And if this is the case, is the step function convergence invalidating the convergence of a Cauchy sequence still valid for other norms?

The reason cannot be that there could be unbounded functions in the interval, since the argument will hold for the space of bounded functions $C[-1,1].$

答:
Note that $C(-\infty,\infty)$ is not  a Banach algebra with the supremum norm, as it has unbounded functions. But that's irrelevant to your example as it can be considered in any $C[-b,b]$.

In any case, your sequence is not Cauchy. For example you have
$$
|f_k(1/k^2)-f_{k^2}(1/k^2)|=\frac1\pi|\arctan1/k-\arctan1|=(\frac14-\pi\arctan\frac1k).
$$
This tells you that for all $k$ big enough such that $\arctan \tfrac1k<\tfrac1{8\pi}$,
$$
\|f_k-f_{k^2}\|_\infty\geq\frac18.
$$


问:
No continuous function switches $\mathbb{Q}$ and the irrationals
Is there a way to prove the following result using connectedness?

Result:

Let $J=\mathbb{R} \setminus \mathbb{Q}$ denote the set of irrational numbers. There is no continuous map $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(\mathbb{Q}) \subseteq J$ and $f(J) \subseteq \mathbb{Q}$.

Link

答: Suppose such a continuous function exists. $f(\Bbb R)$ is connected, and hence uncountable. Then, $f(\Bbb Q)$ is countable because $\Bbb Q$ is so. Moreover, $f(\Bbb R\setminus\Bbb Q)$ is countable since it is contained in $\Bbb Q$. This is a contradiction.


问:
Find the mistake (AM GM ineqality)
According to AM GM inequality

$$\frac{\left(x^3-\frac{1}{x}-\frac{1}{x}+\frac{2}{x}\right)}{4}\ge\left[(x^3)\left(-\frac{1}{x}\right)\left(-\frac{1}{x}\right)\left(\frac{2}{x}\right)\right]^\frac{1}{4}$$

$$\Rightarrow x^3 \ge 2^\frac{9}{4}\;\;\;\;;\;
\forall\; x\in \mathbb R^+ $$
which is not true

答: Your use of the AM-GM is equivalent to saying something like $$\frac{2 + (-1) + (-1)}{3} \ge \sqrt[3]{2(-1)(-1)}.$$  It doesn't work because we require each term to be a nonnegative real number.

Title: Find the mistake (AM GM ineqality)


问:
Set Theory element argument proof help please
Prove that $(A U B)$ = $A$ or $B$ - $A$ and $B$

答:
I think you have the right idea, but your proof can be improved. To show two sets have the same cardinality, we just need to show there is a bijection between them.

In this case, define $f:P(S)\to X$ by $f(R)=R\cup \{x\}$, where $R$ is any subset of $S$. We need to show $f$ is one-to-one and onto. The sentence "every element of $X$ is the union of a subset of $S$ with $\{x\}$" means $f$ is onto. Indeed we can check for any $A\in X$, $A$ contains $\{x\}$ by definition. Then $A-\{x\}\in P(S)$ and is mapped to $A$ by $f$. Similarly, the sentence "if you take different subsets of $S$ you get different elements of $X$" means $f$ is one-to-one. Indeed, if $S_1,S_2$ are in $P(S)$ and $S_1\neq S_2$, their images under $f$ are $S_1\cup \{x\}$ and $S_2\cup \{x\}$, which are not equal. Therefore, $f$ is a bijection between $P(S)$ and $X$.


答:
The $GM-AM$ said the following

Let $$a,b>0 \, \, \text{Then the following inequality holds} \sqrt{ab}< \frac{a+b}{2}$$

as @Benjamin Wang remark your RHS is non positive.

Title: Find the mistake (AM GM ineqality)

Body:
According to AM GM inequality

$$\frac{\left(x^3-\frac{1}{x}-\frac{1}{x}+\frac{2}{x}\right)}{4}\ge\left[(x^3)\left(-\frac{1}{x}\right)\left(-\frac{1}{x}\right)\left(\frac{2}{x}\right)\right]^\frac{1}{4}$$

$$\Rightarrow x^3 \ge 2^\frac{9}{4}\;\;\;\;;\;
\forall\; x\in \mathbb R^+ $$
which is not true


问:
What is the covariance of the exponential of two jointly distributed normal random variables?
Suppose I have the bivariate normal variable $[X \ \ Y]'$ which has mean $[\mu_X \ \ \mu_Y]'$ and covariance matrix $ \left[ \begin{array}{cc}
\sigma_X ^2 & \sigma_Y \ \sigma_X \ \\
\sigma_Y \ \sigma_X \ & \sigma_Y^2
\end{array} \right]$.

I am trying to figure out what is  $\text{Cov}(e^X, e^Y)$.

I'm aware of Stein's lemma which says $\text{Cov}(g(X), Y) = \text{Cov}(X,Y)E[g'(X)]$, but I don't know how it's derived, or how to extend it to functions on both $X$ and $Y$. Presumably this case should be simpler since it's the exponential function.

Otherwise I'm unsure what other steps to take. I don't need to prove this, but the answer is necessary for a calculation I am doing.

答: $$\operatorname{Cov}[e^X, e^Y] = \operatorname{E}[e^{X}e^{Y}] - \operatorname{E}[e^X]\operatorname{E}[e^Y] = \operatorname{E}[e^{X+Y}] - M_X(1) M_Y(1),$$ where $M_X, M_Y$ are the moment-generating functions of the marginal distributions of $X$ and $Y$.  The handling of $\operatorname{E}[e^{X+Y}]$ requires an additional step, which is to compute the mean and variance of $X+Y$, since their sum is normal whenever $(X,Y)$ are bivariate normal.  Specifically, we obviously have $$\mu = \operatorname{E}[X+Y] = \mu_X + \mu_Y$$ and $$\sigma^2 = \operatorname{Var}[X+Y] = \sigma_X^2 + 2\sigma_{XY} + \sigma_Y^2.$$ This results in a third MGF to compute.  These are all straightforward calculations if one is familiar with bivariate normal distributions.


问:
Inverse Rubik's Cube
If I am given five faces of a rubik's cube, is it possible to

a) Determine if these are five sides of an actually solvable cube

b) Extend this to the sixth face in a unique way

Assuming one eliminated the obvious issues: Colour occurring twice on an edge or corner, opposite colours on the same edge or corner, or a repeated edge or corner. Would this guarantee that the cube can be extended to a solvable 6 face cube?

Asking out of curiousity but also because I was making a Rubik's cube cake and want to know if it would theoretically be solvable.

答:
The answer to (b) is no. Start with a solved cube with yellow on the top. Then, flip all 4 edges on the yellow face. This is a solvable position. Call this position A.

Now swap the position of the two pairs of opposite edges on the yellow face. Call this position B.

Finally, remove the yellow face from A and B. The results are the same. So the extension is not unique.

Title: Inverse Rubik's Cube


问:
Does infinite variance imply infinite expectation?
If a random variable has infinite variance, does it necessarily have infinite expectation or is its expectation not well-defined?</p>

答:
No, the fact that the variance (or, rather, second moment) is infinite tells you nothing about the expectation.

It may not exist: e.g., take the standard Cauchy distribution, with pdf $f(x)=\frac{1}{\pi(1+x^2)}$ for $x\in\mathbb{R}$. [Note: this is more in line with "second moment $\mathbb{E}[X^2]$ infinite", not variance, since not having a well-defined expectation means the variance itself is tricky to define]

It may exist: e.g., take the distribution on $\mathbb{R}$ with pdf $f(x)=\frac{3\sqrt{3}}{4\pi}\cdot\frac{1}{1+|x|^3}$ for $x\in\mathbb{R}$. (The expectation exists and is 0.)

The examples above are for continuous distributions, but you have analogous examples for discrete distributions over $\mathbb{N}$ or $\mathbb{Z}$.


问:
Taking sheaf of union of algebraic open subsets is same as intersection of sections
Let $I(V) = \{ f \in k[x_1, \dots, x_n] : f(P) = 0, \forall p \in V\}$ be the vanishing ideal of $V$, and let $k[V] = k[x_1, \dots, x_n]/I(V)$ be the coordinate ring of $V$ (ring of regular functions). Let $(V, \mathcal{O}_V)$ be a sheaf of regular functions, that is for each $f \in \mathcal{O}_V(U)$ and $P \in U$ there exists a neighborhood of $P$ such that on that neighborhood $f = \cfrac{g}{h}$ for $h(P) \neq 0$ and $g,h \in k[V]$, where $V$ is irreducible and $U$ is an open (wrt Zariski topology) algebraic subset of $V$. My question, if we have an open cover $ \bigcup\limits_{i=1}^n U_i = U$, is it true that $\mathcal{O}_V(U) = \bigcap\limits_{i=1}^n \mathcal{O}_V(U_i)$? If we define principal open subsets $ U_i = D(h_i) = \{P \in V: h_i(P) \neq 0\}$ (any open algebraic subset can be written as a finite union of principal open subsets), then is it true for them, i.e. in this case does $\mathcal{O}_V(\bigcup\limits_{i=1}^nU_i) = \bigcap\limits_{i=1}^n \mathcal{O}_V(U_i)$?

答:
As I mentioned in the comments, for this intersection to make sense I want to explicitly state that we are dealing with subrings of $k(V)$. In case you're unfamiliar, $k(V)$ will be defined as the set of all pairs $(U, f)$ where $U \subseteq V$ is nonempty and open and $f$ is regular on $U$, i.e. $f \in \mathcal O_V(U)$. We furthermore insist that $(U, f) = (W, g)$ if $f, g$ are equal on $U \cap W$. This defines a field $k(V)$ and all of the $\mathcal O_V(U)$ naturally embed into $k(V)$ via $f \mapsto (U, f)$. You can check that this is injective by density of nomempty open subsets of $V$. See Hartshorne 1.3 for a reference.

By the way, we can effectively forget the $U$ in a pair $(U, f)$ by considering the domain $dom(f)$ to be the largest open set on which $f$ is regular. More formally, this is the union over all $W$ such that $(W, f) = (U, f)$, which yields $(dom(f), f) = (U, f)$. I'm abusing notation a bit here but I hope you'll believe that such a representive for the class of $(U, f)$ can be found.

With this aside, given $f \in k(V)$ what does it mean for $f$ to be in $\mathcal O_V(U)$? As you wrote, it means $f$ must be regular on $U$, so using the notation I had above, $U \subseteq dom(f)$. This immediately yields one inclusion. As $U_i \subseteq U$, $U \subseteq dom(f)$ implies $U_i \subseteq dom(f)$. In english, as $f$ is regular on $U$ it is regular on its subset $U_i$. Symbolically, $\mathcal O_V(U) \subseteq \mathcal O_V(U_i)$. This therefore holds for the intersection, so $\mathcal O_V(U) \subseteq \bigcap \mathcal O_V(U_i)$.

On the other hand, suppose $f \in \bigcap \mathcal O_V(U_i)$ Then $f$ is in every $\mathcal O_V(U_i)$, i.e. it is regular at every point in every $U_i$. Formally, this means $U_i \subseteq dom(f)$ for all $i$. Hence, the union $U = \bigcup U_i \subseteq dom(f)$ and we this have $f \in \mathcal O_V(U)$. Hence, $\mathcal O_V(U) \subseteq \bigcap \mathcal O_V(U_i) \subseteq \mathcal O_V(U)$ and we arrive at equality.

I'd like to explain a bit more of the equalizer sequence I mentioned in the comments, as it gives you a similar result without worrying about making sure everything is a subring of $k(V)$. Indeed, what I mean is that the following diagram is an equalizer sequence:

$$
\mathcal O_V(U) \rightarrow \prod_i \mathcal O_V(U_i) \rightrightarrows \prod_{i, j} \mathcal O_V(U_i \cap U_j)
$$

The first map takes $f \mapsto (f|_{U_1}, \dots, f|_{U_n})$. Also, we write an element of $\prod_{i, j} \mathcal O_V(U_i \cap U_j)$ as a matrix $(f_{ij})$ where $f_{ij} \in \mathcal O_V(U_i \cap U_j)$. The top map $\mathcal O_V(U_i) \rightarrow \prod_{i, j} \mathcal O_V(U_i \cap U_j)$ takes $(f_1, \dots, f_n) \mapsto (f_i|_{U_i \cap U_j})_{ij}$ and the bottom map is the transpose of this, i.e. $(f_j|_{U_i \cap U_j})_{ij}$.

If you're unfamiliar with category theory, this basically states that the map $\mathcal O_V(U) \longrightarrow \prod \mathcal O_V(U_i)$ is injective and that the image of this map is the set of tuples $(f_1, \dots, f_n)$ which map to the same thing under both maps $\prod \mathcal O_V(U_i) \rightrightarrows \prod_{i, j} \mathcal O_V(U_i \cap U_j)$, i.e. $f_i|_{U_i \cap U_j} = f_j|_{U_i \cap U_j}$ for all $i, j$. These are formulations of the identity and gluong axioms of a sheaf. So this equalizer diagram means that $\mathcal O_V(U)$ is isomorphic to the subring of $\prod \mathcal O_V(U_i)$ consisting of compatible intersections in the sense I just defined. This yields an understanding of $\mathcal O_V(U)$ in terms of all of the $\mathcal O_V(U_i)$ similar to what you were asking for.


问:
For all $f: D_1(0) \to D_1(0)$ analytic with $f(\frac{i}{3}) = 0$, find $\displaystyle \sup_f\{\operatorname{Im} f(0) \}$
Let $\mathcal{F}$ denote the family of all analytic functions $f$ that map the unit disc onto itself with $f(\frac{i}{3}) = 0$. Find $M \equiv\sup\{\operatorname{Im} f(0) : f \in \mathcal{F}\}$.

I am preparing for my qualifying exam in complex analysis and I'm pretty stuck on this question. I will admit that I have not gotten very far and I'm just looking for a few hints to get me started. Hopefully then I will be able to post an answer to my own question.

A few thoughts. I know that that, for each $f \in \mathcal{F}$, $\operatorname{Im} f$ is a harmonic function and thus satisfies a maximum principle on each $\overline{D_r(0)} \subset D_1(0)$, $0 < r <1$. Could this be useful?

Also, I do know that the automorphism $\phi : D_1(0) \to D_1(0)$ given by $\phi(z) = \frac{\frac{i}{3} - z}{1 + \frac{i}{3}z}$ is an element of $\mathcal{F}$, with $\operatorname{Im} \phi(0) = \frac{1}{3}$, so $M \ge \frac{1}{3}$.

Any hints, or even a solution, are greatly appreciated.

答: Take arbitrary $f \in \mathcal{F}$ and look at $f \circ \phi : D_1(0) \to D_1(0)$. Since $f \circ \phi (0) = 0$, the Schwarz Lemma applies. In particular, we get $|f\circ \phi (\frac{i}{3})| \le |\frac{i}{3}| = \frac{1}{3}$. But then $\operatorname{Im} f (0) \le |f(0)| = |f\circ \phi (\frac{i}{3})| \le \frac{1}{3} \implies M \le \frac{1}{3}$. Since we already mentioned that $M \ge \frac{1}{3}$ above, the proof is complete.</p>


问:
Set with a property
This was a mock math olympiad problem sent to me by a friend.

Let $P$ be a set of prime numbers. Then, create a set $S$ of positive integers that satisfies the following property:

For every element $p \in P$, $p$ is a factor of at least three elements in $S$.

Prove that for all sets $P$ and $S$, it is possible to divide $S$ into 4 nonempty subsets such that each $p$ is a factor of at least three integers.

I tried to consider the element $a \in P$ that was a factor of the least number of elements in $S$ and see what I could discover, but got nowhere. I also thought that perhaps some sort of algorithm would work too, but I didn't make any progress. Overall, I'm just not quite sure how to even start.

答:
Consider a partition $\{S_1,S_2,\ldots,S_k\}$ such that

($\ast$) each $p \in P$ has multiples in at least two different sets $S_i$

for which $k$ is as small as possible. (Note that there exist partitions for which ($\ast$) is true, such as the partition with only singletons, because of (1).) We intend to show that $k \le 3$.

Indeed, suppose that $k \ge 4$. By minimality of $k$, there exists a $p \in P$ that only divides numbers in $S_1$ and $S_2$; otherwise, we could merge $S_1$ and $S_2$ and ($\ast$) would still hold. Similarly, there is a $q \in P$ that only divides numbers in $S_3$ and $S_4$. But now $S$ does not contain a multiple of $pq$, contradicting (2).

Title: Set with a property


问:
Particle moving along an ellipse; negative/positive displacement
I have the following problem.

A particle moves along an ellipse  $$3x^2 + y^2 = 1$$ with a position vector $$\vec{r} (t)= (f(t), g(t))$$ the displacement is such that the horizontal component of the velocity vector in each  instant $ t $ is  $ -g(t)$.

Is the particle having a negative or positive displacement?

Show that the vertical component of the velocity vector in instant  $ t $ is proportional to $ f(t)$. Find the constant of proportionality.

This is what I have:

$$\vec{r} (t)= ( \cos(t)/\sqrt{3} , \sin(t))$$

$$\vec{v} (t) = ( -\sin(t)/\sqrt{3} , \cos(t))$$

Velocity vector with horizontal component  $ -g(t)$:

$$\vec{v} (t) = ( -\sin(-sin(t))/\sqrt{3} , \cos(t))$$
This is equal to:
$$\vec{v} (t) = ( \sin(sin(t))/\sqrt{3} , \cos(t))$$

My conclusion, if I am not wrong, is that since both components of the velocity vector are positive then the particle displacement is positive.

I am still not sure about what to do regarding if the vertical component of the velocity vector in instant  $ t $ is proportional to $ f(t)$, and about finding the constant of proportionality.

Any tips on how to solve that problem?

Edit:

Constant of proportionality

Please excuse my english. Thanks.

答:
You claim that
$$ \vec r(t) = (f(t), g(t))
   \stackrel?= \left( \tfrac1{\sqrt3} \cos(t), \sin(t)\right). \tag1$$

Therefore you claim that at $t = 0,$ the particle is at
$\left( \tfrac1{\sqrt3}, 0\right)$.
You claim that at time $t = \frac\pi2,$ the particle is at $(0, 1).$
You claim that the particle continues around the ellipse counterclockwise and that if we note the position of the particle at any instant, it takes exactly $2\pi$ units of time for the particle to return to that position.

Very specifically, you claim that $f(t) \stackrel?= \tfrac1{\sqrt3} \cos(t)$
and that $g(t) \stackrel?= \sin(t).$
This is exactly what Equation $(1)$ says.

None of this can be deduced from the statement that
a particle moves along the ellipse  $3x^2 + y^2 = 1$ with a position vector
$\vec{r} (t)= (f(t), g(t))$.
You cannot even deduce it from the statement that the displacement is such that the horizontal component of the velocity vector in each  instant $ t $ is  $ -g(t)$,
because in fact that statement implies that Equation $(1)$ is not the correct equation of motion of the particle.

If Equation $(1)$ were correct, then the horizontal component of the velocity of the particle at any time $t$ would be
$-\tfrac1{\sqrt3} \sin(t).$
But since you have claimed that $g(t) \stackrel?= \sin(t),$ it follows that
$-g(t) \stackrel?= -\sin(t) \neq -\tfrac1{\sqrt3} \sin(t).$
So the horizontal component of the velocity would not be $-g(t).$

I have no idea why you would think the horizontal component of $\vec v(t)$
is $-{\sin(-\sin(t))}/{\sqrt3}.$
According to your other claims, that's $f'(-g(t)),$ not $-g(t).$

What you do know is that the position of the particle at any instant can be described by the formula
$$ \left( \tfrac1{\sqrt3} \cos(\theta), \sin(\theta)\right) $$
for some number $\theta$. If you knew the position of the particle at time $t$ you could find a value of $\theta$ that makes the formula above produce the coordinates of that position, and then you could declare that this defines $\theta$ as a function of $t$. That is, you could conclude that
$$ \vec r(t) = \left( \tfrac1{\sqrt3} \cos(\theta(t)), \sin(\theta(t))\right). $$

You could even reasonably conclude that this should be true for some continuous function, that is, you can suppose that $\theta(t)$ is a continuous function of $t$.
(You could just as easily make it discontinuous by randomly adding multiples of $2\pi$ to some values of $\theta(t)$, but you can choose not to do such a silly thing.)

You could even reasonably assume $\theta(t)$ is differentiable.
There's actually nothing in the problem statement that requires $\theta(t)$
to be differentiable everywhere, but it's a reasonable assumption and it does need to be true in enough places to answer the question.

Assuming that $\theta(t)$ is differentiable, you should be able to apply the chain rule to find components of the particle's velocity,
and setting the horizontal component equal to $-g(t)$ you may then be able to draw some conclusions about $\theta(t)$ which will tell you in which direction the particle travels around the ellipse.


问:
Show an isometry is unitarily equivalent to a shift (Wold decomposition).
This is a part of Wold's Decomposition Theorem as in Theorem 1.1 of https://arxiv.org/pdf/1704.04200.pdf

Let $H$ be an infinite dimensional, complex Hilbert space. Suppose $T: H \rightarrow H$ is an isometry, i.e. $\| Tx\| = \| x\|$ for all $x\in H$.

Show that there is a unitary operator $W$ and a cardinal $\alpha$ such that $U \bigoplus V^{(\alpha)} =W T W^*$, where $V$ is the unilateral forward shift $Ve_n=e_{n+1}$.

I'm looking for the proof and\or constructing a proof by myself.

My biggest struggles are to construct such unitary operator $U$ and find the cardinal $\alpha$.

Any help will be greatly appreciated!

答:
Suppose $TH \ne H$. (Otherwise $T$ is unitary.) Then there exists a unit vector $x\in H$ such that $x\perp TH$. It follows that $\{ x, Tx, T^2x, \cdots \}$ is an orthonormal sequence because
$$  \langle T^{n+k}x,T^{n}x\rangle = \langle T^n(T^k x),T^n x\rangle = \langle T^kx,x\rangle=0,\;\; n=0,1,2,3,\cdots,\;\; k=1,2,3,\cdots. $$
The restriction of $T$ to the closed linear span $H_x$ of $\{ x,Tx,T^2x,\cdots \}$ is the unilateral shift. There is one copy of the unilateral shift for every vector $e_{\alpha}$ in an orthonormal basis $\{ e_{\alpha} \}_{\alpha\in\Lambda}$ of $(TH)^{\perp}$, and the corresponding spaces $H_{e_{\alpha}}$ are mutually orthogonal. The number of such mutually orthogonal representations is the same as the cardinality $|\Lambda|$ of the orthonormal basis $\{ e_{\alpha} \}_{\alpha\in\Lambda}$ of $(TH)^{\perp}$.

$T$ is unitary on $\bigcap_{\alpha\in\Lambda}H_{e_{\alpha}}$, unless the intersection is $\{0\}$.


问:
Prove $\mathbb{R}^2$ is regular
Prove $\mathbb{R}^2$ is regular. I got stuck on this proof even though the answer was almost given to me Infimum problem lemma to prove $\mathbb{R}^2$ is regular.

Let $p\in \mathbb{R}^2$, $p \notin A$, where $A$ is closed. Then there is some $r>0$ with $B_r(p) \cap A =\varnothing.$ Now I know I need to find an open set containing $A$ disjoint from $B_r(p)$, and this was my issue. I just guessed to take for each $a \in A$, $U_a=\text{inf}\{d(a,p)|a \in A\}$, the infimum of this set. Then would $\cup_{a \in A}U_a$ be an open set containing $A$ disjoint from $B_r(p)$? Now I realize my answer does not make sense without defining the open set containing $A$ dependent on $r$, any hints?

答:
The easiest way is to let $s\in(0,r)$ and use the open sets $B_s(p)$ and $\Bbb R^2\setminus\operatorname{cl}B_s(p)$: $\operatorname{cl}B_s(p)\subseteq B_r(p)$, so $\Bbb R^2\setminus\operatorname{cl}B_s(p)\supseteq\Bbb R\setminus B_r(p)\supseteq A$.

Your definition of $U_a$ makes it the infimum of a set of non-negative real numbers, so it is simply a real number, not a subset of $\Bbb R$. I suspect that you wanted to do something like this: let $s=\inf\{d(a,p):a\in A\}$, and consider $\bigcup_{a\in A}B_s(a)$. That gives you an open nbhd of $A$ that does not contain $p$, but it does not give you one that is disjoint from $B_r(p)$. You could, however, let $s=\frac{r}2$ and set $U=\bigcup_{a\in A}B_s(a)$; then it’s an easy exercise in the triangle inequality to show that $B_s(p)\cap U=\varnothing$.

Title: Prove $\mathbb{R}^2$ is regular


问:
Prove $\int_{0}^{\frac{\pi}{2}}f(\cos(x))dx = \int_{0}^{\frac{\pi}{2}}f(\sin(x))dx$ with hint
Assume $f$ is continuous, prove $\int_{0}^{\frac{\pi}{2}}f(\cos(x))dx = \int_{0}^{\frac{\pi}{2}}f(\sin(x))dx$...

I was given a hint, $\sin x=\cos(\frac{\pi}{2}-x)$. I used this property to get $\int_{0}^{\frac{\pi}{2}}f(\sin x)dx=\int_{0}^{\frac{\pi}{2}}f(\cos(\frac{\pi}{2}-x))dx$ and $\int_{0}^{\frac{\pi}{2}}f(\cos \frac{\pi}{2})dx+\int_{0}^{\frac{\pi}{2}}f(\cos(x))dx$. I am a few steps away, and I know $\cos(\frac{\pi}{2})=0$, but I only have $f(0)$, which is not $0$. If I can make it $0$, then I'm done. What is the additional step needed here?

答:
Hint: Use King's property

Proof. If $x=a+b−y$ then
$$
\int_{x=a}^{x=b}f(x)dx=\int_{y=b}^{y=a}f(a+b-y)d(a+b-y)=\\
=\int_{y=b}^{y=a}f(a+b-y)(-dy)=\\
=-\int_{y=b}^{y=a}f(a+b-y)dy=\\
=\int_{y=a}^{y=b}f(a+b-y)dy
$$
Now again you can change the dummy variable and the property is proven to be true.


问:
How is the solution of an ODE identically zero?
I am reading Sturmian theory on Edward Ince's classic Ordinary Differential Equation and I again come across the statement that's been baffling me:

Consider $\frac{d}{dx}\left(K\frac{dy}{dx}\right)-Gy=0$ with $K>0$ and $G$ both being continuous on the interval $a\leq x\leq b$, and let $c\in[a, b]$. Then the initial value problem with initial values $y(c)=y'(c)=0$ has no solution not identically zero. (Page 224.)

I don't see why the initial values would imply that the solution is identically zero. What I can conclude from the information at hand is the fact that $y(c)=y'(c)=\frac{d}{dx}\left(K\frac{dy}{dx}\right)\bigg\rvert_{x=c}=0$, which, if $K$ is a nonzero constant, also means that $y''(c)=0$. Continued differentiation of the differential equation also shows that $y^{(n)}(c)=0$ for all $n\in\mathbb{N}$. But how can we conclude that $y$ is identically zero?

I am aware that there are works of literature out there that say that $y$ is identical zero because the Taylor series around $x=c$ is $0$. But this can't be true unless we know that the Taylor series exists and is convergent at $x=c$ and even that is not enough as we will need solutions $y$ to the differential equation to be always analytic (which I doubt so); it is even not enough to assume that the solution is smooth ($C^\infty([a, b])$).

Am I missing something obvious?

答: There is a unique solution $y$ given the requirement that $y(c)=A$, $y'(c)=B$. This is because the ODE is a regular second-order linear ODE on $[a,b]$.


问:
Determine the number of integer solutions of the equation $ x_{1}+x_{2}+x_{3}+x_{4}=20 $
Determine the number of integer solutions of the equation
$$
x_{1}+x_{2}+x_{3}+x_{4}=20
$$
under the restrictions:
$$
\begin{array}{l}
2 \leq x_{1} \leq 6 \\
0 \leq x_{2} \leq 5 \\
2 \leq x_{3} \leq 8 \\
2 \leq x_{4} \leq 6
\end{array}
$$
I did $x_1=y_1+2$ then $0\leq y_1 \leq4$

$x_2=y_2$ then $0\leq y_2 \leq5$

$x_3=y_3+2$ then $0\leq y_3 \leq6$

$x_4=y_4+2$ then $0\leq y_4 \leq4$

So $y_1+y_2+y_3+y_4 \leq 19$
So the number of solutions is

$\sum_{k=0}^{19}\left(\begin{array}{c}
k+3 \\
k
\end{array}\right)=\sum_{k=0}^{19}\left(\begin{array}{c}
k+3 \\
3
\end{array}\right)$
is right?

答:
You can try the standard generating function approach which I explain first and then apply it to current question.

Consider a simpler equation where $i, j$ are non-negative and $i+j=10$. And next consider the following product $$(1+x+x^2+\dots+x^i+\dots)(1+x+x^2+\dots+x^j+\dots)$$ After multiplying one obtains terms of the form $x^{i+j} $ where $i, j$ are non-negative integers. And all such like terms need to be added together to get the final product. Thus the coefficient of $x^{10}$ in the above product will equal the number of pairs $(i, j) $ such that $i+j=10$.

One can create similar product expressions to deal with equation $i+j+k=12$. And we can get more creative to deal with equations like $2i+j=20$ by forming the product $$(1+x^2+x^4+\dots+x^{2i}+\dots)(1+x+x^2+\dots+x^j+\dots)$$ And if the variables have got constraints like $1\leq i\leq 10$ then we can keep the corresponding factor as $$(x+x^2+\dots+x^{10})$$ Now what do we get out of all this? Multiplying such factors is a challenge if we do it term by term. Luckily we have the formula to sum a geometric progression to simplify each factor. And further we make use of the standard formula $$(1-x)^{-n}=\sum_{r=0}^{\infty} \binom{n+r-1}{r}x^r$$

For the current problem the desired number of solutions is equal to the coefficient of $x^{20}$ in the product $$\sum_{i=2}^{6}x^i\cdot\sum_{i=0}^5 x^i\cdot\sum_{i=2}^{8}x^i\cdot \sum_{i=2}^{6}x^i$$ The above expression equals $$x^6\cdot\frac{1-x^5}{1-x}\cdot\frac{1-x^6}{1-x}\cdot\frac{1-x^7}{1-x}\cdot\frac{1-x^5}{1-x}$$ or $$x^6(1-x^5)^2(1-x^6)(1-x^7)(1-x)^{-4}$$ Thus we need the coefficient of $x^{14}$ in $$(1-x^5)^2(1-x^6)(1-x^7)(1-x)^{-4}$$ The above expression equals $$(1-2x^5+x^{10})(1-x^6-x^7+x^{13})(1-x)^{-4}=(1-2x^5-x^6-x^7+x^{13}+2x^{11}+2x^{12}-2x^{18}+x^{10}-x^{16}-x^{17}+x^{23})(1-x)^{-4}$$ We can get rid of terms with powers higher than $x^{14}$ in first factor and the desired number of solutions is thus equal to the coefficient of $x^{14}$ in $$(1-2x^5-x^6-x^7+x^{10}+2x^{11}+2x^{12}+x^{13})(1-x)^{-4}=(1-2x^5-x^6-x^7+x^{10}+2x^{11}+2x^{12}+x^{13})\sum_{r=0}^{\infty} \binom{4+r-1}{r}x^r$$ This coefficient clearly equals $$\binom{4+14-1}{14}-2\binom{4+9-1}{9}-\binom{4+8-1}{8}-\binom{4+7-1}{7}+\binom{4+4-1}{4}+2\binom{4+3-1}{3}+2\binom{4+2-1}{2}+\binom{4+1-1}{1}$$ The above simplifies to $$\binom{17}{3}-2\binom{12}{3}-\binom{11}{3}-\binom{10}{3}+\binom{7}{3}+2\binom{6}{3}+2\binom{5}{2}+\binom{4}{1}=680-440-165-120+35+40+20+4=54$$ The generating function approach is well suited to such problems and is intuitive and easy to understand. Some care however needs to be taken while performing the computations so that no term is left out.


问:
Is the following Statement always True?
Let $\sum\limits_{n=1}^{\infty}a_n$ be a series such that $a_n\to 0$. Suppose that for any $ n\geq 1$ we have that $a_n = \frac{1}{2^{k_n}}$, where $k_n$ is some positive integer. Then the series is always convergent. Is the statement True or Flase?

答: There is a very simple counterexample: take $k_n = 1$ for all $n$.

Title: Is the following Statement always True?


问:
continuous onto function from irrationals in [0,1] onto rationals in [0,1]
Give a continuous surjective function from the irrationals in $[0,1]$ onto the rationals in $[0,1]$. Can we at least prove the existence of such a function? I couldn't see a function off the top of my head. Here we assume $[0,1]\setminus\mathbb Q$ with its usual metric.</p>

答: The function $x\mapsto\left\lfloor\frac1x\right\rfloor$ maps the irrationals in $[0,1]$ continuously onto the countable discrete space $\mathbb N=\{1,2,3,\dots\}$, which can be mapped continuously onto any countable (nonempty) topological space, such as the space $\mathbb Q\cap[0,1]$.


问:
Uniform continuity general case
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function defined on (all of) the real axis. Assume that $$f(x) \rightarrow a, \quad \text{for } x \rightarrow -\infty$$ $$f(x) \rightarrow b, \quad \text{for } x \rightarrow \infty$$ for some real numbers $a,b$. Show/explain that for every $\epsilon >0$ there exists a $M >0$ such that $$\left | f(x)-f(y) \right |<\varepsilon$$ given $x,y \geq M$. Furthermore, show that $g$ is uniformly continuous.

Attempt (or rather thoughts)

I am fairly sure that I have gotten the solution down for the first part of the question as it can be solved directly form the definition of a limit at infinity.

Now, I have absolutely no idea on how to show that the function is uniformly continuous as I am not confident at all in the approach of these sort of problems. I would have to use my the fact given above that I have shown but, again, no idea how to approach such a problem as I seem to be missing a general approach and lack of properties. I really need some explanation as to what is going on.

答:
You have to consider different cases. Let's take an example: $f:[-2,2] \rightarrow \mathbb{R}$ defined by $f(x)=x^2$, now consider $f:[-1,1]\rightarrow \mathbb{R}$ and choose $\epsilon=1$, then if we take $\delta=\frac{1}{2}$ see that whenever $x,y \in [-1,1]$ and $|x-y|<\frac{1}{2}$, $|f(x)-f(y)|<1$. But if you take $x=0.9,y=1.35$, then also $|x-y|=0.45<\frac{1}{2}$ but $|f(x)-f(y)|=1.0125>1$. So one $\delta$ doesn't work for all $|x-y|<\delta$.

For given $\epsilon>0~\exists~M_1,M_2 >0$ such that $|f(x)-a|<\frac{\epsilon}{2}$ whenever $x<-M_1$ and $|f(x)-b|<\frac{\epsilon}{2}$ whenever $x>M_2$. Let $M=\max\{M_1,M_2\}$.
So there will be the following cases:

(i) $x,y<-M$, (ii) $x,y>M$, (iii) $x,y \in [-M,M]$, (iv) $x \in [-M,M],~y<-M$, (v) $x \in [-M,M],~y>M$ and (vi) $x<-M,~y>M$, (where $M>0$ is what we got from the definition of limit at infinity).

(i) Let $x,y<-M(\leq-M_1)$, then $|f(x)-f(y)|=|f(x)-a-f(y)+a|\leq |f(x)-a|+|f(y)-a|<\epsilon$. So no specific role of $\delta $ here.

(ii) is done similarly.

From (iii) for chosen $\epsilon>0$ we get a $\delta_1>0$ such that $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta_1$ (using the fact that continuous function on compact set is uniformly continuous).

For (iv) we have to the continuity of $f$ at $-M$, for chosen $\epsilon>0~~~\exists~\delta_2>0$ such that $|f(x)-f(-M)|<\frac{\epsilon}{2}$ whenever $|x-(-M)|<\delta_2$. For $|f(x)-f(y)|$ to be small enough $x,y$ can't be too far apart and $x,y$ should belong to $(-M-\frac{\delta_2}{2},-M+\frac{\delta_2}{2})$. Now if we take $|x-y|<\frac{\delta_2}{2}$ then since $|x-(-M)|\leq |x-y|+|y+M|<\frac{\delta_2}{2}+\frac{\delta_2}{2}=\delta_2$ and similarly $|y-(-M)|<\delta_2$ we get $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\frac{\delta_2}{2}$.

For (v) similarly using continuity of $f$ at $M$ we get a $\frac{\delta_3}{2}$.

Note that we have to find a $\delta>0$ for which the definition of uniform continuity is satisfied. So we take $\delta =\frac{1}{2}\min\{\delta_1, \frac{\delta_2}{2}, \frac{\delta_3}{2},2M\}$ (if certain $\delta$ works, then so will anything less than that $\delta$, so multiplying by 0.5 doesn't violate anything! It just helps with the strict inequalities. You may not take it if you wish. Actually we could have also simply taken $\epsilon$ instead of $\frac{\epsilon}{2}$ in the beginning.). Because in the case of (vi) $|x-y|\geq 2M$ and $\delta<2M$ (why strict inequality?!), we don't have to actually consider (vi).
All in all now $|x-y|<\delta\Rightarrow |f(x)-f(y)|<\epsilon$.

Title: Uniform continuity general case


问:
$\int_{0}^{1}\frac{1}{x^p} \not= 1$?
Is the following statement true: $\int_{0}^{1}\frac{1}{x^p} \not= 1$. This is for any $p \in \mathbb{R}$. I gave an counterexample with $p=-2$ then the integral is $\int_{0}^{1}\frac{1}{x^-2}=1$ exactly. But I'm bad at math, so I want to make sure is right.

答:
The Power Rule of Integration states
$$\int x^p\,\text dx=\frac{x^{p+1}}{p+1}+C$$
So,
$$\int_0^1\frac{\text dx}{x^p}=\left[\frac{x^{-p+1}}{-p+1}\right]_0^1$$
$$=\frac{1^{1-p}}{1-p}-0=\frac{1}{1-p}$$
If this is to be one,
$$1-p=1\implies \boxed{p=0}$$
Hence, the statement is false for $p=0$.

Hope this helps. Ask anything if not clear :)

Title: $\int_{0}^{1}\frac{1}{x^p} \not= 1$?


问:
Given a subgroup $H\le G$ and an element $g\in G$, is $H\cap gHg^{-1}$ normalized by $g$?
Let $G$ be a group, $H\le G$ a subgroup, and $g\in G$ an element. Must $g$ normalize $H\cap gHg^{-1}$?

This is true if $H\le G$ is cyclic and finite index, but I don't see if the proof can be extended to a more general case.

Proof in the case that $H = \langle h\rangle$ is cyclic and finite index in $G$: =
In this case, $H\cap gHg^{-1}$ is also cyclic, with the same (finite) index in both $H$ and $gHg^{-1}$, so let $x$ be a generator. It suffices to show that $gxg^{-1}\in H$.

For some integers $i,j$, we have $x = h^i = gh^jg^{-1}$. On the other hand, $gxg^{-1} = gh^ig^{-1}$, but
$$[gHg^{-1} : \langle gh^ig^{-1}\rangle] = [H : \langle x\rangle] = [H : H\cap gHg^{-1}] = [gHg^{-1}:H\cap gHg^{-1}] = [gHg^{-1}:\langle x\rangle] = [gHg^{-1} : \langle gh^jg^{-1}\rangle]$$
which implies that $\langle gh^ig^{-1}\rangle = \langle gh^jg^{-1}\rangle$ by uniqueness of subgroups of cyclic groups of a given index. This implies that $gh^ig^{-1}\in\langle x\rangle\in H$, as desired.

答:
Here is a concrete example.
Take as $G$ the group $S_5$ and as $H$ the subgroup $S_4$ of $S_5$. Let x=(125).
Then we obtain $H^x\cap H=\{e,(23),(24),(34),(234),(243)\}=K$ and
$(152)(23)(125)=(13)\notin K$.


问:
Why does Alon's combinatorial Nullstellensatz require working over a field.
In Alon's Nullstellensatz theorems (theorems 1.1 and 1.2 here) why is it necessary for $F$ to be a field? As far as I can tell, all the arguments in the proofs should work when $F$ is, say, an integral domain.

答:
Here $ F $ is just simply a integral domain. You can use any domain.  Like Ali commented if you take any part of $ F $ or any other integral domain then also the theorem will work.

Like example you have been asked to find the integer whose successor is 6. Many places you will have the equation $ x+1 = 6 $ but we know even if we take the equation as $ y+1=6$ then also nothing will be wrong.

Edit: by any part I mean substring


问:
Two maximum values for $|iz+3-4i|$ what went wrong?
I need to find the maximum value of $|iz+3-4i|$ given that $|z|\leq4$

$|iz+3-4i|\leq |iz|+|3-4i|$

$|iz+3-4i|\leq 4+5$

But I could write an inequality for $|3-4i|$ whose maximum value would be 7, not 5 so I get the maximum value to be 11 when doing so rather than the original answer of 9.

What went wrong here? , thanks in advance.

答: Nothing seems to have gone wrong. $9 = 4 + 5 \leq |iz| + |3 - 4i| \leq |iz| + |3| + |4i| \leq 4 + 3 + 4 = 11$. And, well, $9 < 11$.


答: A minor simplification is $|iz+3-4i|=|i(z-4-3i)|=|z-4-3i|$, which dispenses of the distracting factor of $i$. Moreover, it allows us to interpret $|z-4-3i|$ as the distance between $z$ and $4+3i$. That suggests a tidy geometric interpretation and a simple answer.

Title: Two maximum values for $|iz+3-4i|$ what went wrong?

Body:
I need to find the maximum value of $|iz+3-4i|$ given that $|z|\leq4$

$|iz+3-4i|\leq |iz|+|3-4i|$

$|iz+3-4i|\leq 4+5$

But I could write an inequality for $|3-4i|$ whose maximum value would be 7, not 5 so I get the maximum value to be 11 when doing so rather than the original answer of 9.

What went wrong here? , thanks in advance.


答: For a point in a disk of radius $4$ centered at the origin, what is the maximum distance of that point from $(3,-4)$?  Clearly, the farthest (and closest!) point away is the one that is on the line that joins $(3,-4)$ and the circle center $(0,0)$.  Since the distance between $z$ and the center is at most $4$, and the distance from the center to $(3,-4)$ is $5$, the maximum distance is $4+5 = 9$ and the minimum is $5-4 = 1$.  The fact that $z$ is multiplied by $i$ is of no consequence, because we can set $iz = w$ and then $|z| \le 4$ is equivalent to $|w/i| = |w| \le 4$.


答:
Particle moves along $3x^2+y^2=1$

So we define $\vec r(t) = (\frac{1}{\sqrt3} \cos (at), \sin (at))$ where $a$ is a constant. For $a = 1$, it will take $2\pi$ unit of time for the particle to go along the ellipse and return to its starting position.

$\vec v(t) = (- \frac{a}{\sqrt3} \sin (at), a \cos (at))$

We also know that $\vec r(t) = (f(t), g(t))$ such that $x$ component of velocity vector at any given $t$ is $-g(t)$.

So what is the value of $a$? Can you take it from here?

Title: Particle moving along an ellipse; negative/positive displacement

Body:
I have the following problem.

A particle moves along an ellipse  $$3x^2 + y^2 = 1$$ with a position vector $$\vec{r} (t)= (f(t), g(t))$$ the displacement is such that the horizontal component of the velocity vector in each  instant $ t $ is  $ -g(t)$.

Is the particle having a negative or positive displacement?

Show that the vertical component of the velocity vector in instant  $ t $ is proportional to $ f(t)$. Find the constant of proportionality.

This is what I have:

$$\vec{r} (t)= ( \cos(t)/\sqrt{3} , \sin(t))$$

$$\vec{v} (t) = ( -\sin(t)/\sqrt{3} , \cos(t))$$

Velocity vector with horizontal component  $ -g(t)$:

$$\vec{v} (t) = ( -\sin(-sin(t))/\sqrt{3} , \cos(t))$$
This is equal to:
$$\vec{v} (t) = ( \sin(sin(t))/\sqrt{3} , \cos(t))$$

My conclusion, if I am not wrong, is that since both components of the velocity vector are positive then the particle displacement is positive.

I am still not sure about what to do regarding if the vertical component of the velocity vector in instant  $ t $ is proportional to $ f(t)$, and about finding the constant of proportionality.

Any tips on how to solve that problem?

Edit:

Constant of proportionality

Please excuse my english. Thanks.


问:
Differential form of Conservation of Momentum related to the Navier-stokes equation
In the context of the Navier-Stokes equation the Conservation of Momentum is given in my textbook as $$\frac{d}{dt}\int_{W_t} \rho \mathbf{u}dV = -\int_{W_t}\rho \cdot \mathbf{n} - \mathbf{\sigma}(\mathbf{x},t)dA$$ where $\sigma(\mathbf{x},t)$ is called the $\textit{stress tensor}$ and is written as $$\sigma(\mathbf{x},t) = 2\mu[D - \frac{1}{3}(div(u))Id]+\xi (div(u))Id$$ where $\mu$ is called the first coefficient of viscosity and $\xi = \lambda + \frac{2}{3}\mu$ is called the second coefficient of viscosity, and $D$ called the deformation densor.

Plugging in the above $\sigma$ into the integral version of the conservation of momentum apparently yields the differential form of the conservation of momentum below:

$$\rho \frac{D \mathbf{u}}{Dt} = -\nabla \rho + (\lambda + \mu )\nabla (div(\mathbf{u}))+\mu\Delta \mathbf{u}$$

However, I cannot get the same result. Could someone please show me how this is done? Here $\frac{D}{Dt} = \partial_t + u\cdot \nabla$ denotes the material derivative. I can plug in $\sigma$, but how are the integrals removed?

答:
First, you need to recognize that $W_t$ represents a material control volume that is time-dependent as it moves and deforms with the fluid.  Then you pass the derivative under the integral on the right-hand side by applying Reynolds transport theorem to obtain,

$$\frac{d}{dt} \int_{W_t} \rho \mathbf{u} \, dV = \int_{W_t} \frac{\partial (\rho \mathbf{u})}{\partial t} \, dV + \int_{\partial W_t}\rho \mathbf{u} \mathbf{u} \cdot \mathbf{n} \, dS,$$

where $\mathbf{n}$ is the outer unit normal vector field at the surface $\partial W_t$ bounding the control volume and the integral on the far right-hand side is a surface integral.

By the divergence theorem, it follows that

$$\tag{1}\frac{d}{dt} \int_{W_t} \rho \mathbf{u} \, dV = \int_{W_t} \frac{\partial (\rho \mathbf{u})}{\partial t} \, dV + \int_{W_t}\nabla \cdot(\rho \mathbf{u} \mathbf{u}) \,dV = \int_{W_t} \left[\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u})\right] \,dV $$

The right-hand side you have written is incorrect.  This should represent the net force acting on the control volume due to pressure and stress at the surface $\partial W_t$. Correctly written it is

$$\int_{\partial W_t}(-p\mathbf{n} + \mathbf{\sigma} \cdot \mathbf{n}) \, dS,$$

where $p$ is the pressure field and $\sigma $ is the (deviatoric) stress tensor.  Applying the divergence theorem, we get

$$\tag{2}\int_{\partial W_t}(-p\mathbf{n} + \mathbf{\sigma} \cdot \mathbf{n}) \, dS = \int_{W_t} (- \nabla p + \nabla \cdot \sigma) \, dV$$

Equating (1) and (2), we have

$$\int_{W_t} \left[\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u})\right] \,dV = \int_{W_t} (- \nabla p + \nabla \cdot \sigma) \, dV$$

Since this is true for any control volume $W_t$ we can eqaute the integrands, to obtain

$$\tag{3}\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u}) =- \nabla p + \nabla \cdot \sigma $$

Now it is just  a matter of simplifying both sides.

For the left-hand side we have

$$\tag{4}\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u}) = \rho \left[\frac{\partial \mathbf{u}}{\partial t}+ \mathbf{u} \cdot \nabla \mathbf{u}\right] + \left[\frac{\partial \rho}{\partial t}+ \nabla \cdot (\rho\mathbf{u}) \right]\mathbf{u}$$

Conservation of mass is expressed in terms of the equation of continuity

$$\frac{\partial \rho}{\partial t}+ \nabla \cdot (\rho\mathbf{u}) =0,$$

which can be derived in a similar way. Substituting into (4) , we get

$$
\frac{\partial (\rho \mathbf{u})}{\partial t} +\nabla \cdot(\rho \mathbf{u} \mathbf{u}) = \rho \left[\frac{\partial \mathbf{u}}{\partial t}+ \mathbf{u} \cdot \nabla \mathbf{u}\right]= \rho \frac{D\mathbf{u}}{Dt}, $$

and substituting into (3) with this result,

$$ \rho \frac{D\mathbf{u}}{Dt} = -\nabla p + \nabla \cdot \sigma$$

It remains for you to expand $\nabla \cdot \sigma$ using the expression you gave for $\sigma$ in terms of the deformation tensor which should be $D = \nabla \mathbf{u} + [\nabla \mathbf{u}]^T.$


问:
Group of units in a finite field
Let $f(x) \in \mathbb{Z}/3\mathbb{Z}[x]$ be a cubic irreducible polynomial and let $F = \mathbb{Z}/3\mathbb{Z}[x]/(f(x))$. I want to show that either $x$ or $2x$ generates the group of units of $F$.

My only thoughts so far are that $F$ is a finite field of order $27$, so its group of units has order $26$. Clearly both $x,2x$ do not have order $2$, so they must have order $13$ or $26$. But how can I narrow it down more?

I could try to find all cubic irreducibles, or I could use that $F$ is a splitting field of $x^{27}-x$, or use that the group of units is cyclic, but I'm not sure. Those strategies all seem to computational or don't lead anywhere.

答:
The $f$ is a red herring. The procedure is equivalent to picking $x \in \Bbb F_{27} \setminus \Bbb F_3$.

Let $\zeta$ be a primitive $26$th root of unity, so $\Bbb F_3 = \{0, 1, \zeta^{13}\}$. Let $x = \zeta^e$ with $0 < e < 26$, so $e \ne 13$, and $2x = \zeta^{e+13}$.

It suffices to show that one of $e$ or $e+13$ is coprime to $26 = 2 \times 13$.

Since $0 < e < 26$ and $e \ne 13$, we know that $e$ is not divisible by $13$, so $e+13$ is also not divisible by $13$.

One of $e$ or $e+13$ is odd, i.e. not divisible by $2$ either.

So one of $e$ or $e+13$ is coprime to $26 = 2 \times 13$.

So one of $x = \zeta^e$ or $2x = \zeta^{e+13}$ generates the unit group.

Title: Group of units in a finite field


问:
A bilateral weighted shift $T$ with $\sigma(T) =$ unit circle
I am looking for an example of a bilateral weighted shift $T\in B(\ell^2(\mathbb{Z})) $ with $\sigma(T) = \mathbb{T}$, which is the  unit circle.

NOTE: what I'm looking for a bilateral weighted shift such that $Te_n=t_ne_{n+1}$ with $t_n$ a non-constant.

I know that a bilateral shift $S$ with $Se_n=e_{n+1}$, $n\in \mathbb{Z}$,  has spectrum $\sigma(S)=\mathbb{T}$.

So I started trying out things.

Here are some ideas:

Let $T$ be defined by $Te_n =t_ne_{n+1}$, where

$ t_n=
\begin{cases}
    \frac{1}{1+n},& \text{if } n\geq 0\\
    \frac{1}{1-n},              & \text{if } n <0
\end{cases}
$

First we know that the spectrum is always closed.

But I'm not sure how to find out if the spectrum of this is $\mathbb{T}$.

To show $\sigma(T) \subset \mathbb{T}$, some facts that might be needed are:

The spectrum is always closed.

Could someone please let me know if my example works?

If it does, why $\sigma(T) = \mathbb{T}$?

If not, could I get some suggestions on the other possible weighted shifts? Thank you!

答:
Any bilateral shift with weights in $\mathbb T$ will have spectrum $\mathbb T$, so you can choose $t_n$ to be whatever you want, as long as $|t_n|=1$.

This is because if you define $Ve_n=\lambda^ne_n$ for some $\lambda\in\mathbb T$, then
$$
VTe_n=t_nVe_{n+1}=t_n\lambda^{n+1}e_{n+1},\qquad\qquad TVe_n=\lambda^nTe_n=t_n\lambda^ne_n.
$$
So $TV=\lambda VT$. As $V$ is a unitary, $V^*TV=\lambda T$. Then
$$
\sigma(T)=\sigma(V^*TV)=\lambda\sigma(T).
$$
That is, $\sigma(T)$ is invariant for rotation, as we can do this for any $\lambda\in\mathbb T$. So $\sigma(T)=\mathbb T$.


答:
You’ve noted that $x$ has order $13$ or $26$. If it has order $26$ you are done. If $x$ has order $13$, then just note that since $2=-1$ in $F$, then $2$ has order $2$. So $\gcd(o(2),o(x))=1$, and since the multiplicative group is abelian, it follows that $o(2x)=\mathrm{lcm}(o(2),o(x)) = 26$.

So if $x$ does not generate, then $2x$ certainly does.

Title: Group of units in a finite field

Body:
Let $f(x) \in \mathbb{Z}/3\mathbb{Z}[x]$ be a cubic irreducible polynomial and let $F = \mathbb{Z}/3\mathbb{Z}[x]/(f(x))$. I want to show that either $x$ or $2x$ generates the group of units of $F$.

My only thoughts so far are that $F$ is a finite field of order $27$, so its group of units has order $26$. Clearly both $x,2x$ do not have order $2$, so they must have order $13$ or $26$. But how can I narrow it down more?

I could try to find all cubic irreducibles, or I could use that $F$ is a splitting field of $x^{27}-x$, or use that the group of units is cyclic, but I'm not sure. Those strategies all seem to computational or don't lead anywhere.


问:
Suggest a method to solve $y + y' + y'' = ry - ry^2$
I'm stuck with the following differential equation:

$$y+\frac{d^2y}{dx^2} + \frac{dy}{dx} = ry - ry^2$$

where $r$ is a real positive constant and $x,y \in \mathbb{R}$.

Since it is for a small research purpose, I wish to get a proper analytic solution that is not a series expansion.

I tried multiplying with $\frac{dy}{dx}$ and simplifying. This gave:

$$\frac{d}{dx} {\left(\frac{1}{2}{y'}^2-\frac{r-1}{2}y^2+ \frac{r}{3}y^3\right)}=-{y'}^2$$

But I'm unable to continue. Just to mention I also tried solving the first order equation:

$$y+ \frac{dy}{dx} = ry - ry^2$$

and tried similar solutions. That didn't work too.

答:
Substituting $\mathrm dy/\mathrm dx=-u$ brings your ODE to an Abel equation of the second kind in normal (canonical?) form
\begin{align}
u\frac{\mathrm du}{\mathrm dy}=u+(r-1)y-ry^2.\tag{1}
\end{align}
Via my cheat sheet http://eqworld.ipmnet.ru/en/solutions/ode/ode0124.pdf I can tell you that you'll be hard pressed to get an analytical solution to Eqn. (1). I've also tried my hand at solving it to no avail.