问:
Elementary confusion about determinacy

In ZF we can prove Borel determinacy.

In ZF we can prove that the determinacy for closed sets is equivalent to the Axiom of Choice.

Since closed sets are Borel, it follows that the Axiom of Choice is a theorem of ZF. I am aware this is wrong and very likely an entirely trivial confusion emerging from my lack of knowledge on these matters. However, I'd appreciate someone to shed some light on this, probably with a line pointing out what is the confusion.

Thanks!

答:
Closed, or Borel, in what space?

$\mathsf{ZF}$ proves that the axiom of choice is equivalent to the following statement:

$(*)\quad$ For every set $X$, equip $X^\omega$ with the product topology coming from the discrete topology on $X$. Then every game on $X$ of length $\omega$, whose payoff set is closed as a subset of $X^\omega$, is determined.

But $(*)$ is an incredibly general principle! Every set $X$ has to be taken into account here. By contrast, when we talk about Borel determinacy we're (unless otherwise stated!) referring to specifically Borel subsets of $\omega^\omega$ (or equivalently for this context, $2^\omega$). While $\mathsf{ZF}$ proves "Every Borel subset of $\omega^\omega$ is determined," this is not enough to imply $(*)$ above.

Indeed, in general the phrase "$\Gamma$-determinacy" refers either to $\omega^\omega$ or $2^\omega$ specifically (and even this needs further clarification for $\Gamma$ sufficiently simple - closed games on $2$ are fairly boring but closed games on $\omega$ are extremely complicated). So the phrase "Over $\mathsf{ZF}$, choice is equivalent to closed determinacy" is misleadingly phrased; it should be something like "Over $\mathsf{ZF}$, choice is equivalent to the determinacy of all closed games on all sets."

Title: Elementary confusion about determinacy


问:
Suppose $n \geq a_1>a_2>...>a_k\geq1$ and $lcm(a_i,a_j) \leq n$ for $i,j=1,2,...,k$. Show that $ia_i \leq n$.
Could I receive some hints on the problem. I do not know how to touch it since index $i$ seems like it does not depend on any of the other conditions.

答:
HINT  Consider the case $k=2$. (It's always a good idea if you're stuck to try an easy case.)

If the result you are trying to prove is false then you have two unequal numbers, both greater than  $\frac {n}{2}$. Their LCM is a multiple of each of the two numbers and, since they are unequal, it cannot be the same multiple of each. So the LCM must be at least twice one of the numbers and is therefore greater than $n$, a contradiction.

Can you now do the cases $k=3, 4, ... $?

ContentLicense: CC BY-SA 4.0


问:
Taylor series of $(1+x)^{\frac{\ln x}{x}}$ at $0^{+}$
I want to find the Taylor series of $(1+x)^{\frac{\ln x}{x}}$ at $0^{+}$.

I already found that $(1+x)^{\large\frac{\ln x}x}=x-\frac{x^2 \ln x}{2} +o(x^2 \ln x)$ but I'm struggling to get a better precision.

Here is what I did:

using the classic Taylor series of $\ln(1+x)$ at $0$ we have that

$\ln x\:\frac{\ln (1+x)}x=\ln x-\frac{x\ln x}2+o(x\ln x)$.

As $(1+x)^{\large\frac{\ln x}x}=e^{\ln x-\frac{x\ln x}2++o(x\ln x)}$, we find

$(1+x)^{\large\frac{\ln x}x}=e^{\large \ln x} \cdot e^{-\frac{x\ln x}2+o(x\ln x)}=x (1- \frac{x \ln x}{2}+o(x\ln x))$ and the result follows.

Thanks in advance.

答:
For a more precision, this expansion can help:

$$x-\frac{1}{2} x^2 \ln (x)+\frac{1}{24} x^3 \left(3 \ln ^2(x)+8 \ln (x)\right)+\frac{1}{48} x^4 \left(-\ln ^3(x)-8 \ln ^2(x)-12 \ln (x)\right)+\frac{x^5 \left(15 \ln ^4(x)+240 \ln ^3(x)+1040 \ln ^2(x)+1152 \ln (x)\right)}{5760}+O\left(x^6\right)$$

I will add details in a while, after getting off the bus.


问:
How to calculate the circumference of a circle, excluding the part intersecting another circle?

I have one circle of $11.63$ inch bricks that is $21.25$ ft in radius (green in the picture). I already purchased the bricks for this circle already.

I need to add another circle to the side (marked in red) of bricks. The size is scaled by the golden ratio of $1.618$, or $13.133$ ft in radius. The center of this red circle falls directly on the edge of the green circle.

I need to purchase bricks to form the second (red) circle, but I don't need bricks to go inside the green circle.

How do I find the circumference of circle B, minus the part intersected with circle A, so I can calculate how many bricks to buy?

答:
One possibility is to think about the intersection of the two circles. Assuming that the red circle is centered at $(0,0)$, its ''northeast'' quadrant is described by $y=\sqrt{r^2-x^2}$ (where $r$ is its radius). The green circle is centered at $(R,0)$, $R$ being its radius, and its ''northwest'' quadrant is described by $y=\sqrt{R^2-(x-R)^2}$. If you solve

$$\sqrt{r^2-x^2}=\sqrt{R^2-(x-R)^2}$$

you obtain the points at which the circles intersect: $\left(\frac{r^2}{2R},r\sqrt{1-\frac{r^2}{4R^2}}\right)$  and, by symmetry, $\left(\frac{r^2}{2R},-r\sqrt{1-\frac{r^2}{4R^2}}\right)$. These two points define an arc that spans $2\arctan\frac{r\sqrt{1-\frac{r^2}{4R^2}}}{\frac{r^2}{2R}}$ of the $360º$.


问:
Prove converse Thales theorem, proportional sides imply parallel lines
I'm going through John Stillwell's Four Pillar's of Geometry and trying to follow the book's structure when doing the exercises.  Generally, a 'pillar' is divided into two chapters; the first chapter states useful results and motivation while the second chapter brings the machinery to do the proofs.

Thales' Theorem: Suppose that the line EF is parallel to BC. Then AE/EB=AF/FC.

Question

Prove Converse Thales' Theorem: Suppose that EG is not parallel to BC. Then AE/EB not equal to AG/GC.  Equivalently, AE/EB=AG/GC is sufficient for EG parallel BC.

The issue is that I don't think I'm 'allowed' to use much.  The ideas of similar triangles, SAS, etc. have not been developed.  I know that without having the book in front of you it's hard to judge what can be used but essentially I'm looking for the dirt simplest proof.  From what I can see the only known facts are Thales theorem and that was only stated.

If clarification is needed I'll be glad to answer in the comments.

答: Try for an indirect proof and use the (forwards) theorem on the parallel line in your linked picture.</p>

ContentLicense: CC BY-SA 3.0


问:
Given specific real Fourier coefficients, does there have to be a function that matches that?
We were taught this in class:

Given the real numbers b1, b2,..., bn, there exists a cyclic function such that its non-zero Fourier coefficients are b1, b2, ... bn.

Can someone please explain why this is true?

答:
One can just put formula for this function
$$f(t) = \sum_{k=1}^{n} b_{k}e^{-k*t*i}$$


问:
Scholarly references for definition and discussion of left and right inverses of functions
Can someone share some good scholarly references where the left and right inverse of functions are defined, and their properties analysed?

Even better if they are references that also discuss various different definitions in circulation, their advantages and drawbacks, and their different usefulness in different branches of mathematics.

答:
The google search "mathematics teacher" "function" "codomain" produces several items of possible interest to you, such as:

The mistakes that are made by students with regard to functions: evidence from Erzincan by Kani Başibüyük, et al. (2016 math education paper; 10 pages)

Undergraduate mathematics students’ understanding of the concept of function by Caroline Bardini, et al. (2014 math education paper; 23 pages)

The Function Concept and University Mathematics Teaching by Olov Viirman (2014 Ph.D. dissertation; 99 pages)

Teaching Students to Communicate with the Precise Language of Mathematics: A Focus on the Concept of Function in Calculus Courses by Derrick S. Harkness (2020 Ph.D. dissertation; xii + 137 pages)

Also of possible interest are some of the references I give in this answer, especially The rise of functions by Salomon Bochner as it focuses more on modern ideas than the other papers.


问:
An Euler-Lagrange Equation
I have an action with a Lagrangian which I would like to apply the Euler-Lagrange equations to https://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation but have spent hours really struggling with it. That is I define

$$
L\Big(\begin{pmatrix}x \\ y\end{pmatrix},\begin{pmatrix} \dot{x} \\ \dot{y}\end{pmatrix} \Big):= \Big\| A \Big( \begin{pmatrix}x \\ y\end{pmatrix} +  \begin{pmatrix} -\dot{y} \\ \beta\dot{y}\end{pmatrix}  \Big) \Big\|^2
$$

where

$$
A=\begin{pmatrix} c_1 & c_2 \\ c_2 & c_3 \end{pmatrix}, \text{for some }~c_i~\text{for which $A$ is invertible}.
$$

Which $x,y:[0,T]\to \mathbb{R}$ solve the Euler -Lagrange equation

$$
\nabla_{\begin{pmatrix}x \\ y\end{pmatrix}} L-\partial_t\nabla_{\begin{pmatrix} \dot{x} \\ \dot{y}\end{pmatrix} }L=0
$$
with initial and terminal conditions $x(0)=q,x(T)=q',y(0)=p,y(T)=p'$. ? Please help !

$\Big($would more information on how the constants $c_1,c_2,c_3$ relate to eachother be useful? because infact $c_1=\sigma^2+\gamma^2, c_2=-\gamma(\sigma+\alpha),c_3=\alpha^2+\gamma^2$, where these constants $\alpha,\sigma,\gamma$ are positive and $\alpha\sigma>\gamma^2$ $\Big)$.

EDIT : I got the E.L as with the above choices for $c_1,c_2,c_3$

\begin{equation}
    \nabla_{(x,y)} L = (\frac{1}{\alpha\sigma-\gamma^2})^2\begin{pmatrix}
    0 \\
    \nabla_{y} B_1+\nabla_{y} B_2
    \end{pmatrix}
\end{equation}

with

\begin{equation}
    \nabla_{y} B_1=2(\sigma+\gamma \beta) y + \frac{1}{\sigma+\gamma\beta}(\gamma \dot{y}-\sigma \dot{x})
\end{equation}

and

\begin{equation}
    \nabla_{y} B_2=2(\gamma + \alpha \beta) y + \frac{1}{\gamma+\alpha \beta}(\alpha \dot{x}-\gamma \dot{y}).
\end{equation}

Also

\begin{equation}
    \partial_t \nabla_{(\dot{x},\dot{y})} L=2 (A^{-1})^2(\begin{pmatrix}\ddot{x}\\ \ddot{y} \end{pmatrix}-\begin{pmatrix}
    \dot{y}
    \\
    -\beta \dot{y}
    \end{pmatrix})
\end{equation}

Which is a coupled ODE

答:
The Lagrangian doesn't depends on $t$  so it obeys Betrami's identity. With $p=(x,y)$

$$
L - \dot p\nabla_{\dot p}L = c_0
$$

or

$$
c_2^2 \left(x^2+y^2-\left(\beta ^2+1\right) \dot y^2\right)+c_3^2 \left(y^2-\beta ^2 \dot y^2\right)+2 c_2 c_1 \left(\beta  \dot y^2+x
   y\right)+2 c_2 c_3 \left(\beta  \dot y^2+x y\right)+c_1^2 \left(x^2-\dot y^2\right)=c_0
$$

From Euler-Lagrange's equations we obtain

$$
x = \frac{\left(c_2 \left(c_2-\beta  c_3\right)+c_1^2-\beta  c_2 c_1\right) \dot y-c_2 \left(c_1+c_3\right) y}{c_1^2+c_2^2}
$$

now substituting into the former ODE we obtain a new ODE now depending only on $y,\dot y$. Concluding, we have two independent constants to fix: $c_0$ an one additional boundary from the last obtained ODE, and four independent boundary conditions. This is not feasible.

NOTE

The lagrangian is kind of degenerate concerning the kinetic energy because

$$
\frac 12\nabla_{\dot p}\left(\nabla_{\dot p}L\right) = \left(
\begin{array}{cc}
 0 & 0 \\
 0 & \left(c_1-\beta  c_2\right){}^2+\left(c_2-\beta  c_3\right){}^2 \\
\end{array}
\right)
$$

which is not positive definite.

EDIT

Corrected some equations.

Title: An Euler-Lagrange Equation


问:
Prove that for $ f: \mathbb R \to \mathbb R , f(x)=\frac {x-1} {x^2 + x + 2}, \lim _{x \to 0} f(x) =- \frac12 $ by epsilon delta

As you can see from the image, I am not too sure where to go from after this. Could anyone give me a (not too ambiguous) hint?

答: Hint: Note that $x^2+x+2=\left(x+\frac12\right)^2+\frac74\geqslant\frac74$. On the other hand, you proved that $|x|<1\implies|x+3|<4$. Therefore$$|x|<1\implies\left|\frac{x(x+3)}{x^2+x+2}\right|<\frac{16}7|x|.$$


问:
How to know is every circle is on the same side of a line
How can I know if every circle is on the same side of a line, like this:

答:
enter image description here

For $n$ circles,

$S =\dfrac{1}{2n}\sum [sign(x-R)+sign(x+R)]$

if $S=+1$ then all the circles are on the right side of (D)

else if $S=-1$ then all the circles are on the left side of (D).


问:
Problem in the theory of finding the unique solution of $-u'' = \delta_0$ and $-u'' + cu= \delta_0$
I have theese two problems:

(I) Find the unique distribution $u \in H^1(-1,1)$ such that $-u'' = \delta_0$ and $u(-1) = u(1) = 0$.

(II) Let $c > 0$, find the unique distribution $u \in H^1(\mathbb{R})$ such that $-u'' + cu= \delta_0$

Because we want to use Lax-Milgram's theorem, solving theese problem we began like this:

$(I)$ $u \in H^1(-1,1)$ satisfies $-u'' = \delta_0$ in the distributional way if and only if for all $\varphi \in C_c^\infty(-1,1)$

$$\begin{cases}
u \in H^1_0(-1,1)\\[8pt]
\int_{-1}^1 u' \varphi' \ dx = \varphi(0)
\end{cases}$$

$(II)$ $u \in H^1(\mathbb{R})$ satisfies $-u'' +cu= \delta_0$ in the distributional way if and only if for all $\varphi \in H^1(\mathbb{R})$
$$
\begin{cases}
u \in H^1(\mathbb{R}) \\[8pt]
\int_{-1}^1 u'\varphi' +cu\varphi \ dx = \varphi(0)
\end{cases}$$

My questions are:

Why in $(I)$ we took $\color{red} {u \in H^1_0(-1,1)}$ and not $u \in H^1(-1,1)$ ?

Why in $(II)$ we took $\color{red}{\varphi \in H^1(\mathbb{R})}$ and not $\varphi \in C_c^\infty(\mathbb{R})$ ?

What is the general approach to be sure to choose "the right space for the right function" in this kind of problems?

I'm studying a course of a master's degree (Italy) in mathematics and as you may observe in my last questions on this site, this course was really linked to the solutions of the Dirichlet problem.

Introducing the "weak derivateive" of a function we mentioned that distributions live in spaces where there is not a normal topology to define continuity (taken for example from a norm) but we did not go deeper.

We defined the Sobolev space $W^{k,p}(\Omega)$ as follows:
$$W^{k,p}(\Omega) := \{u\in L^p(\Omega) \ : \ D^\alpha u \in L^p(\Omega) \ \text{for all}\  \alpha : |\alpha| \le k\}$$
where $D^\alpha u$ is used as a notion of the distributional derivative of $u$, that in our definition is the distribution that acts on every $\varphi \in C^\infty_c(\Omega)$ as
$$\langle D^\alpha u , \varphi \rangle := (-1)^{|\alpha|}\langle u , D^\alpha \varphi \rangle$$

We define $H^k(\Omega) := W^{k,2}(\Omega)$ and $W^{k,p}_0(\Omega)$ is the closure of $C^\infty_c(\Omega)$ inside $W^{k,p}(\Omega)$

Here I state Lax-Milgram theorem:

Let $H$ be an Hilbert real space and let $a:H\times H \to \mathbb{R}$ be a bilinear, continous and coerced form. Then for all $F\in H^*$ there exists one and one only $\bar{u} \in H$ such that
$$a(\bar{u},v) = F(v) \qquad \forall v \in H$$
If $a$ is also symmetric then $\bar{u}$ satisfies
$$\frac{1}{2}a(\bar{u},\bar{u}) - F(\bar{u}) = \min_{v\in H} \Big\{\frac{1}{2}a(v,v) + F(v)\Big\}$$

答:
On picking the right space for the right function

Equality in distribution is interpreted in terms of pointwise equality for functions in $C_c^{\infty}$. The "right space" for a function is picked by seeing its role in the resulting equation phrased for $C_c^\infty$ functions, suitably changed so as to facilitate the use of Lax-Milgram lemma. Suitable changes are performed via transferring derivatives to elements of $C_c^{\infty}$. However, note that the transfer of derivative should be done , while keeping in mind that we need a bilinear form on a space $H$. So the transfer of derivative is done in such a way that a bilinear form is definable and we can use the Lax-Milgram lemma. Usually that space $H$ is a Sobolev space. So the transfer of derivative is (usually) done so that we get the same amount of derivatives on both the test function and the variable function.

Let's take an example to illustrate this. Suppose you had a function $f$, and you wanted to find a $u$ such that $\Delta u = f$. Let's say all this is on a bounded open set $\Omega$, and for boundary conditions we want $u = 0$ on $\partial \Omega$. Classically, you'd demand that $f \in C(\Omega)$ and $u \in C^2(\Omega) \cap C(\overline{\Omega})$.

For the weak formulation : we will have $\int_{\Omega} \phi \Delta u = \int_{\Omega} \phi f$ for all $\phi \in C_c^{\infty}(\Omega)$. Using Green's theorem (an integration by parts for general domains), you can change the RHS to $-\int_{\Omega} \nabla \phi \nabla u = \int_{\Omega} \phi f$ for all $\phi \in C_c^{\infty}(\Omega)$. So now there's one derivative on both $\phi$ and $u$.

We stop now to think about what Lax-Milgram would look like : let's first start with $f$. We usually insist that $f \in L^2(\Omega)$ : this almost never changes for two reasons. One is that there's no derivative on $f$ so it doesn't need to be constrained, and second , the right hand side becomes a continuous functional on $C_c^\infty(\Omega)$ (continuous with respect to the topology you have not discussed, so I won't go into why it's continuous), so we can extend using density to $H_0^1(\Omega)$ if we need.

And we do need to : by continuity, if $\int_{\Omega} \nabla u \nabla \phi = \int_{\Omega} f \phi$  for all $\phi \in C_c^\infty(\Omega)$, then density gives that the same holds for all $\phi \in H_0^1(\Omega)$. But now, $\phi \in H_0^1(\Omega)$, and we need a bilinear form involving $u$ and $\phi$, so it is natural to take $u \in H_0^1(\Omega)$ as well, just so that the bilinear form $B(f,g) = \int_{\Omega} \nabla f \nabla g$ is well-defined on $H_0^1(\Omega) \times H_0^1(\Omega)$ and we can think about Lax-Milgram.

Note that if $u \not \equiv 0$ on $\partial \Omega$, then we can't take $u \in H_0^1(\Omega)$, because a necessary condition for membership in $H_0^1(\Omega)$ is that it should vanish on the boundary, because it is a (a.e. pointwise) limit of functions that have compact support so vanish on the boundary (This particular assertion is actually beyond your reach, since you don't know the topology on $C_c^{\infty}(\Omega)$. Take it at face value). Then we cannot use the Lax-Milgram lemma. We'd have to use something else, but most likely the bilinear form in use will be the same.

So that's how we choose our spaces : make sure that we are transferring derivatives to create the same Sobolev space in which both $u$ and $\phi$ can freely reside, and then we can think about Lax-Milgram.

The first question

In our example, much the same happens.

Indeed, note that we want $-u'' = \delta_0$. Following integration with test functions, we land up with :
$$
\forall \phi \in C_c^\infty(-1,1) \quad ; \quad \int_{-1}^1 u'\phi' = \phi(0)
$$

(I omit the $dx$ at the end of the integral, to focus on the integrand. This is quite common in PDE) Now , by density of $C_c^\infty$ we get that the same holds for $\phi \in H_0^1(\Omega)$. Now, to ensure that Lax-Milgram comes into play, we insist on $u \in H_0^1(\Omega)$, which doesn't conflict because $u(-1)=u(-1) = 0$ so $u$ is $0$ on the boundary. (Refer to second last paragraph of previous section)

Our bilinear form is then quite obviously $B(f,g) = \int_{-1}^1 f'g'$, and the $F \in (H_0^1(\Omega))^*$ will be given by $F(\phi) = \phi(0)$. Then we can frame our question as : does there exist a unique $u \in H_0^1(-1,1)$ such that $B(u,\phi) = F(\phi)$ for every $\phi \in H_0^1(\Omega)$?

We are in Lax-Milgram territory! All we need is that $B$ be a bilinear, continuous, coerced form. I won't explain why this happens in detail : continuity follows since $|B(u,v)| \leq \|u\|_{H_0^1} \|v\|_{H_0^1}$ (follows from the definition of the norm in $H_0^1$), and coercivity follows from the Poincare inequality, from which one gets that there is a constant $C>0$ such that $B(u,u) \geq C\|u\|^2_{H_0^1}$. Thus, by using the Lax-Milgram lemma, one sees that a $u$ satisfying the given conditions exists and is unique.

EDIT(UNRIGOROUSLY DONE) : To find $u'' = \delta_0$, we need to "integrate" $\delta_0$ twice, keeping track of the initial conditions. Let $v$ be such that $v(x) = v(-1) + \int_{-1}^x \delta_0(x)dx$. Note that $\int_{-1}^x \delta_0(x)dx = 0$ if $x<0$ and $1$ if $x \geq 0$. Therefore, we get that $v(x) = v(-1)$ for $x<0$ and $v(-1)+1$ for $x \geq 0$.

Finally, $u(x) = u(-1) + \int_{-1}^x v(x)dx$ so $u(x) = u(-1) + v(-1)(x + 1)$ for $x <0$ and $u(x) = u(-1) + v(-1) + xv(-1)+x$ for $x \geq 0$. We need to ensure that $u(-1)=u(1)=0$. From that, setting $x=1$ we get $u(1) = 2v(-1) +1 = 0$ so $v(-1) = \frac{-1}{2}$.

Thus, $u(x) = \frac{-(1+x)}{2}$ for $x<0$ and $\frac{x-1}{2}$ for $x \geq 0$.

So that's how solve problems using the Lax-Milgram lemma. Guessing solutions requires "physical intuition" on how $\delta$ functions are integrated.

The second question

Here things get a little thicker at the weak formulation stage : we have
$$
\int_{-1}^1 [u'\phi' + cu\phi] = \phi(0) \tag{1}
$$

for every $\phi \in C_c^\infty(\mathbb R)$. Now we want to apply Lax-Milgram, so we want to ensure that $u$ and $\phi$ are picked from the same space so we can let the LHS of the above equation be a bilinear form.

What is that Hilbert space? At first, it's a toss up between $H_0^1(\mathbb R)$ and $H^1(\mathbb R)$. But the point is, that $\mathbb R$ is unbounded, so it has no boundary. Therefore, we expect that $H_0^1(\mathbb R) = H^1(\mathbb R)$ i.e. that $C_c^\infty(\mathbb R)$ is in fact dense in $H^1(\mathbb R)$. This is a remarkable fact that is true!

From that fact and continuity, if equation $(1)$ was true for all $\phi \in C_c^\infty(\mathbb R)$ it automatically is true for all $\phi \in H^1(\mathbb R)$. Then we realize that $u \in H^1(\mathbb R)$ is also the case (once again, because the only two possible spaces we could have are in fact equal to each other, there's no doubt about this one).

Thus, our Hilbert space here is $H^1(\mathbb R)$, on which the bilinear form $B(f,g) = \int_{-1}^1 [f'g' + cfg]$ is well defined. We then define a functional $F$ on $H^1(\mathbb R)$ by $F(h) = h(0)$. Lax-Milgram is set when we reframe our problem as : we want a $u$ such that $B(u,\phi) = F(\phi)$ for every $\phi \in H^1(\mathbb R)$.

We need to check that $B$ is continuous and coercive. Continuity is pretty clear from elementary bounds. Coercivity is also clear since $B(u,u) \geq c\|u\|^2_{H_0^1}$ and $c>0$. Thus Lax-Milgram tells us there's a unique $u$ satisfying the conditions, and we are done.

We can find an explicit $u$ which satisfies the equation, using cues from classical treatment of the equation $u''+cu = f$. However, I don't wish to go into this because the priority of the post was understanding the Lax-Milgram lemma usage and the various Sobolev spaces in play. I hope I have done some justice.


问:
Determine where the following function is continuous
$f(z) = z$ for $\leq 1$, and $f(z) = |z|$ for $> 1$. I'm trying to find where $f$ is continuous. I know that $f$ is continuous where $z \leq 1$, since $f$ just maps elements to itself.

答:
When $|z_0|<1$ or $|z_0|>1$, $f$ is continuous at $z_0$, because the sets $D(0,1)$ and $\{z\in\Bbb C\mid |z|>1\}$ are open sets and the restrictions of $f$ any of these sets is continuous.

When $|z_0|=1$, there are $z$ points arbitrarily close to $z_0$ where $f(z)=z$ and there are $z$ points arbitrarily close to $z_0$ where $f(z)=|z|^2$. So, $f$ is continuous at $z_0$ if and only if $z_0=|z_0|^2$, which happens if and only if $z_0=1$.


答:
Given line
$$a x + b y + c = 0$$
the signed distance $d_i$ between the line and point $(x_i, y_i)$ is
$$d_i = \frac{a x_i + b y_i + c}{\sqrt{a^2 + b^2}}$$
In a computer program you might wish to use form
$$d_i = A x_i + B y_i + C$$
where
$$\begin{aligned}
A &= \displaystyle \frac{a}{\sqrt{a^2 + b^2}} \\
B &= \displaystyle \frac{b}{\sqrt{a^2 + b^2}} \\
C &= \displaystyle \frac{c}{\sqrt{a^2 + b^2}} \\
\end{aligned}$$

If the center of a circle $i$ is at $(x_i, y_i)$, and its distance $\lvert d_i \rvert$ to the line is less or equal to its radius $r_i$, i.e. $\lvert d_i \rvert \le r_i$, the circle intersects the line.

If all circles fulfill $d_i \lt -r_i$, or if all circles fulfill $d_i \gt r_i$, they are all on the same side of the line.  Otherwise, some of the circles either intersect with the line or are on different sides of the line.

Title: How to know is every circle is on the same side of a line

Body: How can I know if every circle is on the same side of a line, like this:


问:
I can't seem to find a solid proof for this question for every number , please help me out
There are 25 stones in a heap. The heap is divided into two parts, then one of the parts is divided in two again, et cetera, until we have 25 separate stones. After each division of one of the heaps into two smaller heaps we write the product of the numbers of stones in these two heaps on a blackboard. Prove that at the end the sum of all the numbers on the blackboard is 300.</p>

答:
Hint Prove by strong induction that if you have $n$ stones, at the end you get
$$\frac{n(n-1)}{2}$$

The inductive step is easy: if you have $n+1$ stones and at the first step you divide them into $k$ and $n+1-k$ then the first number you get is
$$k(n+1-k)$$

Next, apply $P(k)$ and $P(n+1-k)$.


问:
Comultiplication arising as pullback for group representations
Let $V$ and $W$ be $k[G]$-modules for an algebraically closed field, $k$ and $G$ a group. Now I know that $V\otimes W$ has a $k[G]\otimes k[G]$-module structure, and because $k[G]$ is in fact a Hopf algebra, I can give $V\otimes W$ a $k[G]$-module structure using,

$$k[G] \xrightarrow{\Delta} k[G]\otimes k[G] \xrightarrow{\rho_V\otimes \rho_W} \mathrm{End}(V)\otimes\mathrm{End}(W)\xrightarrow{}\mathrm{End}(V\otimes W)$$

where $\Delta$ is $g \mapsto g\otimes g$ and $\rho_{V}$ is the map $k[G]\to\mathrm{End}(V)$ or identically for $W$.

I have heard this described as taking a pullback. In terms of diagrams, what is this pullback? I.e. we are taking the limit of which diagram of the form $A \xrightarrow{} B \xleftarrow{} C$?

答:
The word "pullback" is sometimes used to refer to the map $f^* : \hom(Y,Z)\to \hom(X,Z)$ given by precomposition with $f: X\to Y$.

So in your case "pullback" just means $\Delta^*$, i.e. precomposition by the comultiplication.

The reason (or at least a reason) is that if you think of a function $g:Y\to Z$ as a sort of bundle over $Y$ (the fiber over $y$ being $g(y)$, so the elements of $Z$ are to be thought of as geometric or algebraic objets for instance) then the pullback bundle corresponds to $g\circ f$. There are ways in which this analogy is precise, others in which it is not.

In your situation, you're thinking of your tensor module as living "over $k[G]\otimes_k k[G]$" and you "pull it back" to $k[G]$ along the diagonal.

(Note : This conflicts with the geometric notion of pullback in the case of quasi-coherent sheaves over schemes which corresponds to an extension of scalars rather than a restriction such as here )


问:
Latitude given origin longitude, destination coordinates and distance
Given one origin coordinate (latitude or longitude), destination coordinates (both latitude and longitude) and the distance between both sets of coordinates in km, is it possible to compute the other origin coordinate?

Edit: title updated to reflect that bearing is not known.

答:
I assume we treat the Earth as a sphere; the math for an ellipsoid gets much more complicated.

For a given destination point with known latitude and longitude, and a known distance from origin to destination, the origin must be somewhere on the circle whose center is at the destination and whose geodetic radius (distance along the surface from the center to the circumference) is the desired distance. This circle is a small circle unless the geodetic radius is exactly $1/4$ of the circumference of the Earth.

If you know the longitude of the origin, that information places the origin on one half of a particular great circle. (The other half of that great circle has a longitude $180$ degrees different from the desired longitude.)
But if you know the latitude of the origin instead, that places the origin on the circle at that latitude (which is a small circle unless it is the equator).

So the origin will have to be at an intersection of the small circle around the destination with either the longitude circle or the latitude circle.

The intersection between two circles is one point if the circles are exactly tangent; two points if they are closer, and no intersection at all if they are farther apart.
In the case of a known longitude you also have to ask whether the intersection points are on the half of the great circle at the correct longitude.

Origin at a known latitude

For a known latitude, you can use the haversine distance formula:
$$
\operatorname{hav}\left(\Theta\right) =
  \operatorname{hav}\left(\varphi_2 - \varphi_1\right) + \cos\left(\varphi_1\right)\cos\left(\varphi_2\right)\operatorname{hav}\left(\lambda_2 - \lambda_1\right)
$$
where $\phi_1,\lambda_1$ are the latitude and longitude (respectively) of the origin, $\phi_2,\lambda_2$ are the latitude and longitude (respectively) of the destination, and $\Theta = \frac dR$ where $d$ is the geodetic distance from origin to destination and $R$ is the radius of the Earth.
You can also use the formula $\operatorname{hav}(\alpha) = \sin^2(\alpha/2).$

You then have an equation in which everything is known except for $\lambda_1,$ which occurs only once. Solve for $\lambda_1.$

For some input values there will be no solution for $\lambda_1,$ indicating that it is not possible to reach the desired destination traveling the desired distance from any point at the desired latitude.

Origin at a known longitude

For a known longitude, identify the great circle on which the line of longitude of the origin lies, and find the point $C$ on that great circle closest to the destination.
If $A$ is the destination and $B$ is a geographic pole of the Earth then $ABC$ is a right spherical triangle with right angle at $C.$
For simplicity in explanation let's suppose the destination is in the northern hemisphere and $B$ is the north pole; the same method works in mirror image in the southern hemisphere using the south pole.

Knowing the longitude of the destination you can determine the angle $B$ in the spherical triangle $ABC$; choose angle $B$ so it is no greater than a right angle, even if this means $C$ will be on the "wrong half" of the great circle.
Let $\gamma$ be the angle between $A$ and $B$;
this is the co-latitude of $A$, easily found be subtracting the destination latitude from a right angle.

You can now apply the right spherical trigonometric identity
$$ \sin(B) = \frac{\sin(\beta)}{\sin(\gamma)} $$
where $\beta$ is the angular distance from $A$ to $C$.
You know $B$ and $\gamma$; solve for $\sin(\beta).$
The spherical law of cosines then says that
$$ \cos(\gamma) = \cos(\alpha)\cos(\beta) $$
where $\alpha$ is the distance between $B$ and $C$.
You know $\gamma$ and $\sin(\beta)$; find $\cos(\beta)$ and solve for $\alpha.$

Knowing $\alpha,$ you know the latitude of $C.$

Now let $\gamma' = \frac dR$ where $d$ is the desired distance between origin and destination and $R$ is the radius of the Earth.
Then you can define a spherical right triangle $AB'C$ with right angle at $C$, where $B'$ is the desired origin point
and $\gamma$ is the angular distance between $A$ and $B'$.
If $\alpha'$ is the angular distance between $B'$ and $C,$
the spherical law of cosines says that
$$ \cos(\gamma') = \cos(\alpha')\cos(\beta). $$
You know $\gamma$ and $\sin(\beta)$; find $\cos(\beta)$ and solve for $\alpha'.$

Starting at the (now known) point $C$ on the great circle of the longitude of the origin point, travel an angle $\alpha'$ around the circle. Do this once in each direction to find two candidates for the point $B'.$ (Usually this means adding or subtracting $\alpha'$ from the latitude of $C$, but you have to do some extra arithmetic if the path crosses a pole.)

If either of your candidates for $B'$ is on the half of the great circle that has the desired longitude of the origin, it is a possible origin point.

You may find that at some point during this algorithm you are unable to solve for the next required step, in which case you know it is not possible to reach the desired destination traveling the desired distance from any point at the desired longitude.


问:
If $V$ is an $n$-dimensional vector space, $V= \operatorname{span}(W)$ and $|W|=n$ then $W$ is a basis for $V$
If $V$ is an $n$-dimensional vector space, $V= \operatorname{span}(W)$ and $W=\{w_1,...,w_n\}$ then $W$ is a basis for $V$.

Since $V= \operatorname{span}(W)$ in order to prove $W$ is a basis for $V$ the only thing left to prove is that $W$ is linearly independent, I decided to proceed by contradiction:

I supposed that $W$ is linearly dependent, so $\exists w_i \in W$ such that $w_i \in  \operatorname{span}(W\setminus \{w_i\})$, because $W$ spans $V$, then so does $W\setminus \{w_i\}$. Therefore, there exists a subset $S$ of $W\setminus \{w_i\}$ that is a basis for $V$ (I've already proved this property, so I can use it in good conscience).

Now, because $S \subset W\setminus \{w_i\}$ and $|W\setminus \{w_i\}|=n-1$, then $|S|\leq n-1$, then, because S is a basis for V, $\dim(V)\leq n-1$ and there's my contradiction.

I was just wondering if perhaps there's an easier way of proving this property, or if I made a mistake somewhere along the way.

答:
Your answer looks okay, but you took the long way round. A more direct argument is the following (same idea as Arturo's):

Since the generating set $W$ must contain a basis, and $W$ has $n$ elements, then $W$ already forms a basis, since all bases have the same cardinality $n$ (the dimension is defined precisely as the cardinality of any basis).

PS. A basis is not a mere set of vectors but rather an ordered sequence (or set). Order matters!


问:
Find n where $GCD(a_{n}, 14) = 7$
The question was:

Find $n$ where $GCD(a_{n}, 14) = 7$ where $n$ is natural if you knew that $a_{n} = n + 3$

The book solved it by saying that means $a_{n}$ is a multiple of $7$ but not $14$ so $a_{n} = 7a$ and by putting it in the equation

$GCD(7a, 14) = 7$ means $GCD(a,2) = 1$ which means $a = 2k + 1$ and by putting it in $a_{n}$ we find that n = 14a + 4$

The first, I understand that because if it was a multiple of 14 then we would take $14$ as the $GCD$ not $7$

What I didn't understand is that what putting $a_{n} = 7a$ has anything to do with not being a multiple of 14? 14 can be written as $7(2)$ which means $a_{n}$ can be clearly a multiple of 14 too.

答:
The argument you quote that $GCD(7a, 14) = 7$ means the same as  $GCD(a, 2) = 1$ and therefore $a=2k+1$ is correct.

You therefore know that $$n+3=7a=7(2k+1)=14k+7.$$

Then $n=14k+4$, as required

Title: Find n where $GCD(a_{n}, 14) = 7$


问:
Prove that connected graph $G$ with $\delta (G)\geq 3$ contains a cycle $C$ such that $G-E(C)$ is still connected.

In a connected simple graph every vertex has a degree at least $3$. Prove that the graph contains a cycle such that the graph remains connected when the edges of this cycle are deleted.

Source: https://www.komal.hu/verseny/2000-02/mat.e.shtml

I've tried this: Pull out of $G$ a cycle $C$ with smallest perimeter $p$.

Claim: No two nonconsecutive vertices in the circle $C$ are adjacent (else we could make smaller cycle then we already have).

Let $C_1,C_2,...,C_k$ be all components in $G-C$ and I want to prove $k=1$.

If $n_i = |V(C_i)|$ and $e_i= |E(C_i)|$, then we have
$$e_i = {3n_i\over 2}-b_i$$ where $b_i$ is a number of edges between $C_i$ and $C$. Because $G$ is connected we have $b_i\geq 1$ for each $i$. Because of the claim we have:  $$b_1+b_2+\cdots +b_k = p$$

I have no idea what to do now.

Edit: As Misha Lavrov pointed in a comment, this attemt is not correct.

答:
NOTE (2/6/2021): This answer was posted today and is a redo from the one posted on Wednesday. I came up with this independently of S. Dolan.

Proof SKETCH:
Let $G$ be the graph on $n$ vertices where every
vertex has degree at least 3. We assume that every simple graph of
minimum degree 3 and on at most $n-1$ vertices, has a cycle such
that the graph remains connected even after removing the edges of
the cycle.

Case A: $G$ has at least one triangle $T$. Let $T$ be a triangle in $G$. Let us write $T=y_1y_2y_3$.

Subcase A.1: $G \setminus E(T)$ has 2 components with $y_1 \in C_1$
and $y_2,y_3 \in C_2$.

A.1.1 If there is only one vertex $w_2 \in C_2$ that is adjacent to either
$y_2$ or $y_3$, AND there is only one vertex $w_1 \in C_1$:

A.1.1.1 If $w_2$ has 2 neighbors in $C_2 \setminus \{y_2,y_3\}$
then add the edge $w_1w_2$ and remove vertices $y_1,y_2,y_3$
Then the resulting graph $G'_{n-1}$ is simple and has minimum degree 3.
Also, a cycle $C'_{n-1}$ in $G'_{n-1}$ so that
$G'_{n-1} \setminus E(C'_{n-1})$ is connected, translates naturally
to a cycle $C_{n}$ in $G$ so that $G\setminus E(C_n)$ is
connected.

A.1.1.2 If $w_2$ has only one neighbor $w'_2$
in $C_2 \setminus \{y_2,y_3\}$ then add the edge $w_1w'_2$
and remove vertices $y_1,y_2,y_3,w_2$.
Then the resulting graph $G'_{n-1}$ is simple and has minimum degree 3.
Also, a cycle $C'_{n-1}$ in $G'_{n-1}$ so that
$G'_{n-1} \setminus E(C'_{n-1})$ is connected, translates naturally
to a cycle $C_{n}$ in $G$ so that $G\setminus E(C_n)$ is
connected.

A.1.2 If there is either more than one vertex in $C_2$ adjacent
to either $y_2$ or $y_3$ or there is more than one vertex in $C_1$
adjacent to $w_1$ collapse $T$ into a vertex $v_T$ and add the edges
$v_Tw$; $w \not \in T$; $w$ adjacent in $G$ to a vertex in $T$.
Then the resulting graph $G'_{n-1}$ is simple and has minimum degree 3.
Also, a cycle $C'_{n-1}$ in $G'_{n-1}$ so that
$G'_{n-1} \setminus E(C'_{n-1})$ is connected, translates naturally
to a cycle $C_{n}$ in $G$ so that $G\setminus E(C_n)$ is
connected.

Subcase A.2: $G \setminus E(T)$ has 3 components.  Collapse $T$ into a
vertex $v_T$ and add the edges $v_Tw$; $w \not \in T$; $w$
adjacent in $G$ to a vertex in $T$. Call the resulting graph $G'_{n-1}$.
Then the resulting graph $G'_{n-1}$ is simple and has minimum degree 3.
Also, a cycle $C'_{n-1}$ in $G'_{n-1}$ so that
$G'_{n-1} \setminus E(C'_{n-1})$ is connected, translates naturally
to a cycle $C_{n}$ in $G$ so that $G\setminus E(C_n)$ is
connected.

Subcase A.3: $G \setminus E(T)$ has one component. Then we would be done as here we have a cycle namely $T$ so that $G \setminus E(T)$ is
connected.

Case B: $G$ has no triangles. If $G$ has no triangles,
then pick a vertex $v_n$ and write $G_{n-1} \doteq G \setminus \{v_n\}$, and write $u_1,u_2,\ldots, u_k$ as the neighbors of $v_n$ in $G$. Add to
$G_{n-1}$ a matching $M$ of $\lfloor \frac{k}{2} \rfloor$ edges between the $u_i$s.

(C) If there remains [at most] 1 vertex $u_3$ of degree 2 in
the resulting graph, then contract $u_3$ to an edge $e_{u_3}$ and
call the resulting graph $G'_{n-1}$. As $G$ is triangle free
the resulting $G'_{n-1}$ is simple. Let $C'_{n-1}$ be a
cycle so that $G'_{n-1}\setminus E(C'_{n-1})$
is connected.

We consider 2 possibilities:

Subcase B.1: $C'_{n-1}$ contains 2 or more arcs in $M$.
Then let $u_1$ and $u_2$
be such that (i) $u_1$ and $u_2$ are incident to distinct edges in
$M \cap C'$, and (ii) there is a path $P'_{n-1}$ from
$u_1$ to $u_2$ in $C'_{n-1}$
that contains no arcs in $M$ and does not contain $e_{u_3}$.
One can observe that there exists such a $P'_{n-1}$. Then
let $C_n=vu_1P'_{n-1}u_2v$. Then
$G \setminus E(C_n)$ is connected. [Indeed let us suppose that there
is a vertex $u_3$ as in (C) above. Then let $u$ and $w$ be any 2 vertices
in $V(G) \setminus \{u_3,v_n\}$, and let $P'_{uw}$
be a path in $G'_{n-1} \setminus E(C'_{n-1})$. Then if $P'_{uw}$
contains any arcs $u_ju_{j+1}$
in $M$ or $e_{u_3}$ then as $M$ is
a matching $\{u_j,u_{j+1}\} \cap \{u_1,u_2\}$ is empty. So
replace $u_ju_{j+1}$ with $u_jv_nu_{j+1}$ and $e_{u_3}$ with $xu_3y$ where $xy \doteq e_{u_3}$ and so the resulting walk $W_{uw}$
is in $G \setminus E(C_n)$ so $u$ and $w$ are connected to each other
in $G \setminus E(C_n)$ for each $u,w \in V(G) \setminus \{u_3,v_n\}$.
If $P'_{uw}$ does not contain any arc in $M$ or $e_{u_3}$
then $P'_{uw}$ is in $G \setminus E(C_n)$ and
so $u$ and $w$ are connected to each other for each
in $G \setminus E(C_n)$ for each $u,w \in V(G) \setminus \{u_3,v_n\}$.
But then as $e_{u_3} \not \in P'_{n-1}$ it follows that $xe_{u_3}y$ is
in $G \setminus E(C_n)$ so $u_3$ is in the same component as

every other vertex in $V(G) \setminus \{u_3,v_n\}$, and as $v_n$
has degree 3, there is an edge in $G \setminus E(C_n)$ between $v_n$
and another vertex, so $v_n$ is in this component as well.
And so $G \setminus E(C_n)$ indeed has only 1 component. ]

Subcase B.2: $C'_{n-1}$ contains exactly 1 arc $e=u_1u_2$ in $M$ and $e_{u_3}$ as in (C) above exists and is in $C'_{n-1}$ as well.
Then write $e_{u_3} \doteq xy$; $x$ and $y$ the endpoints of $e_{u_3}$
in $G'_{n-1}$; so that the path $P'_{n-1}=xx_2x_3 \ldots x_lu_1$
in $C'_{n-1}$ contains neither $u_2$ nor $y$. Then let
$P_n=u_3xx_2x_3 \ldots x_lu_1$ and let $C_n=v_nu_3xx_2 \ldots x_lu_1v_n$.
Then $G \setminus E(C_n)$ is connected. [Indeed let $u$ and $w$ be any 2 vertices
in $V(G) \setminus \{u_3,v_n\}$, and let $P'_{uw}$
be a path in $G'_{n-1} \setminus E(C'_{n-1})$. Then if $P'_{uw}$
contains any arcs $u_ju_{j+1}$
in $M$ then $\{u_j,u_{j+1}\} \cap \{u_1,u_2\}$ is empty [as $u_1u_2$ is in $C'_{n-1}$ so it cannot be in $P'_{uw}$ and also $u_1u_2 \in M$ a matching], so
replace with $u_jv_nu_{j-1}$ and so the resulting walk $W'_{uw}$
is in $G \setminus E(C_n)$ so $u$ and $w$ are connected to each other
in $G \setminus E(C_n)$ for all $V(G) \setminus \{u_3,v_n\}$.
But then the edge $v_nu_2$ is in
$G \setminus E(C_n)$, and so is the edge $u_3y$.]

The remaining cases are easier and won't be covered here.


问:
Conditional logic in the definition of ordered field
In Rudin's "Principles of Mathematical Analysis", he defines an ordered field as field $F$ which is also an ordered set, such that:

$x + y < x + z$ if $x$, $y$, $z$ $\in$ F and $y < z$,

$xy > 0$ if $x \in F$, $y \in F$, $x > 0$, and $y > 0$.

I'm confused about the second condition. Its contrapositive is:

$xy \le 0$ if one of the following is true:

$x \le 0$, $y \le 0$

$x \le 0$, $y \ge 0$

$x \ge 0$, $y \le 0$

However, #1 is false. Can you help me find a flaw in my reasoning.

答: The contrapositive of$$x>0\wedge y>0\implies xy>0$$is$$xy\leqslant0\implies x\leqslant0\vee y\leqslant0.$$


问:
Using circles to figure out shortest distance between two curves
Based on this answer for the previous question I had asked on august of last year, the first difficulty for me to understand is why it is a condition that the circle must 'touch' the other curve?

Leaving the above point aside, it is clear to me that if we have two curves $C_1$ and $C_2$ , we can use the circle argument for a point on $C_2$ to find that nearest point's line segment is along the normal direction to curve $C_1$ and vice versa for $C_2$, but how is it possible that we can show that for the 'shortest distance' between two curves it must be that it is along mutual normal of two curves?

As in, we know shortest distance from one curve to a point is along it's normal, but , how can we guarantee that shortest distance between two curves is along mutual normal?

答: Look at this figure and just think about it:


问:
How do I reverse the order of integration for this double integral?
$$\int_0^{\pi/2}\int_{\sin(x)}^2dydx$$ This is the integral in question. My closest was $$\int_0^{2}\int^{\sin^{-1}(y)}_2dxdy$$
but I believe this is wrong.
Sorry for the bad formatting, It's my first time using it properly

Body:
Actually,
$$\int_0^{\pi/2}\int_{\sin(x)}^2\,dy\,dx = \int_0^{\pi/2}\left(2-\sin(x)\right)\,dx = \pi-1$$


问:
$\int_{0}^{1}\int_{0}^{\sqrt{1-x^2}} \sqrt{1-x^2-y^2} dy dx$
I need $\int_{0}^{1}\int_{0}^{\sqrt{1-x^2}} \sqrt{1-x^2-y^2}dy\;dx$

the first integral is very difficult, I plan to change the variable to obtain $r\sqrt{1-r^2}\;dr\;d\theta$

but the new limits of integration confuse me

答:
Note that the region you are integrating over is the part of the unit circle in the first quadrant. Thus, the new limits of integration are:

$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{1}r\sqrt{1 - r^{2}}\ dr\ d\theta=\int_{0}^{\frac{\pi}{2}}-\frac{(1-r^{2})^{\frac{3}{2}}}{3}\bigg\vert_{0}^{1}\ d\theta = \int_{0}^{\frac{\pi}{2}}\frac{1}{3}\ d\theta = \frac{\theta}{3}\bigg\vert_{0}^{\frac{\pi}{2}} = \boxed{\frac{\pi}{6}}$$


问:
Show that the set $\{A\textbf{x} \mid \|\textbf{x}\|_2=1\}$ is an ellipse, and describe its two semi-axes (direction and length) in terms of $A$.
Let $A$ be a $2 \times 2$ symmetric matrix with all positive eigenvalues. Show that the set
$$\{A\textbf{x} \mid \|\textbf{x}\|_2=1\}$$
is an ellipse, and describe its two semi-axes (direction and length) in terms of $A$.

How can we find two vectors $\bf{v_1}$, $\bf{v_2}$ and two positive numbers $\lambda_1$, $\lambda_2$ such that

$A\bf{v_1} = \lambda_1 f_1$, $A\bf{v_2} = \lambda_2 f_2$, $(\bf{v_i}, \bf{v_j})=\delta_{ij}$, where $\delta_{ij}$ denotes Kronecker's symbol?

答:
Note that since $A$ is symmetric, it is diagonalizable and we can find an orthonormal basis $\{ v_1, v_2 \}$ of eigenvectors with corresponding eigenvalues $\lambda_1$ and $\lambda_2$. For $x \in \mathbb{R}$ such that $\|x \| = 1$ we can find $\theta$ such that

$$x = \cos(\theta) v_1 + \sin(\theta) v_2.$$

Make sure you understand why this is!

Then we have

$$Ax =A(\cos(\theta) v_1 + \sin(\theta) v_2)=\lambda_1 \cos(\theta)v_1+\lambda_2 \sin(\theta)v_2,$$

which traces out an ellipse with axes along $v_1$ and $v_2$ with lengths $\lambda_1$ and $\lambda_2$ as $\theta$ goes from $0$ to $2\pi$.


答:
The contrapositive of "if x>0 and y> 0 then xy> 0" is
"if $xy\le 0$ then either $x\le 0$ or $y\le 0$."

Your mistake is treating that as if it were "if and only if" rather than just "if".  The fact that BOTH x and y negative gives xy positive does NOT contradict that statement.

Title: Conditional logic in the definition of ordered field

Body:
In Rudin's "Principles of Mathematical Analysis", he defines an ordered field as field $F$ which is also an ordered set, such that:

$x + y < x + z$ if $x$, $y$, $z$ $\in$ F and $y < z$,

$xy > 0$ if $x \in F$, $y \in F$, $x > 0$, and $y > 0$.

I'm confused about the second condition. Its contrapositive is:

$xy \le 0$ if one of the following is true:

$x \le 0$, $y \le 0$

$x \le 0$, $y \ge 0$

$x \ge 0$, $y \le 0$

However, #1 is false. Can you help me find a flaw in my reasoning.


问:
Derive the Hajek pojection of $T_n$.
Let $X_1, \dots , X_n$ i.i.d. copies of $X$ with distribution $F$ and density $f$. Let $(X_{1:n}, \dots , X_{i:n}, \dots , X_{n:n})$ be the order statistic. For a given $p \in (0, 1)$ consider the Harrell-Davis estimator defined by

$$T_n = \sum_{i=1}^n c_{ni}X_{n:i}$$

$$c_{ni} = \frac{\Gamma(n + 1)}{\Gamma(k)\Gamma(n − k + 1)} \int^{i/n}_{(i−1)/n} u^{k−1}(1−u)^{n−k} du, i = 1,\dots , n, k = [np].$$

I have three questions regarding this. I cannot seem to find any solid clear answers to them in my textbook

1st question: Which parameter $\theta$ is estimated? I'm not sure if I have completely misinterpreted but it is rather unclear what we are estimating?

2nd question: What is the definition of the Hajek projection? I get that the Hajek projection has to do with the Hoeffding Decomposition, but I would really like a definition of the Hajek projection without delving into Hoeffding Decomposition.

3rd question: how does one derive the Hajek projection of $T_n$ from this example?

答:
Answering in an itemized list:

$\theta$ is the expectation of your estimator, $T_n$. I'm not familiar with the Harrel-Davis estimator, but a quick google search mentions it estimates the quantiles of a distribution.

Hajek Projection is a projection of a random variable onto the class $\mathcal{S}$, which is the set of all variables of the form $\sum_{i=1}^n g_i(X_i)$, where $g_i \in L^2$, $X_i$ are iid. If you've had a class in Linear Modeling, then you would have been exposed to the "Projection Theorem," working with vectors and matrices. Hajek is a more general form of this.

To calculate directly, the Hajek Projection Principle can be used:
$$\hat{S} = \sum_{i=1}^n E \left[ T_n | X_i \right] - \left(n-1\right) E\left[ T_n \right].$$ Altneratively, but probably not suggested here, would be to express $T_n$ as a U-statistic, and use the projection formula specific to U-statistics.

Title: Derive the Hajek pojection of $T_n$.


问:
Show that $H$ is a group, and that it is isomorphic to $G$

Let $G$ be any group. Let $H$ consist of the same set of elements as $G$, but with
a new operation given by $a ∗ b = ba$, for all $a$ and $b$. Show that $H$ is a group, and
that it is isomorphic to $G$.

I am having trouble proving what is being asked of me. First, I have no information about how the operation is in $G$. I suppose I could define the operation in $G$ as is usually done for group definition, this is if $a, b \in G$, then its product is $ab$. Based on that, $H$ with the new operation would be a group as well, since they have the same elements. The truth seems like nonsense to me but I can't think of anything else.

Also I have tried to establish an isomorphism and let the rest follow from this fact, but so far it has been unsuccessful. Any hint would be helpful.

答:
You should be able to check the axioms of group for $H$. For instance if $e$ is the neutral element of $G$ then so is for $H$ since $e * g = ge=g$ and if $g^{-1}$ is the inverse of $g$ in $G$ then so is it in $H$ since $g * g^{-1}= g^{-1} g =e$.

For the isomorphism define $f: G \to H$, $f(g):= g^{-1}$. It is a group homomorphism since $$f(ab)=(ab)^{-1}= b^{-1}a^{-1} = f(b)f(a)= f(a)*f(b).$$ Its inverse $f^{-1}: H \to G$ is defined exactly in the same way, $f^{-1}(g):=g^{-1}$, and it is indeed the inverse of $f$ since $(g^{-1})^{-1}=g$.


问:
compute $\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}$

The question is:
$$
\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}
\quad \gamma:r(t)=(t,2t^2-2), \quad -1\leq t\leq 1
$$

I have tried to solve it like this:
since $Q_x=P_y$ it's potential vector field but the singularity is at origin, first i thought that i could evaluate it from $(-1,0)$ to $(1,0)$ along $x$ axis but the singularity is the problem, how should i proceed in this case?

Any suggestion would be great Thanks

答:
You may also note that by changing variable $y\to{y'}=y^2$ you get:
$\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}=\int\frac{-y'dx+x\ dy'}{x^2+y'^2}$.
We know that the integrand is a potential vector field, so the integral depends only on the starting and finishing points and does not depend on the way - we just have to avoid the singularity at $(0,0)$.

The starting point for $(x,y')$ is $(-1,0)$ and finishing $(1,0)$. Now we can introduce a parametrization $x=\cos(t)$ and $y'=\sin(t)$, where $t$ is changing from $\pi$ to $0$ - to meet the requirements for the starting and finishing points.

After that the integration is trivial: $\int_{(-1.0)}^{(1,0)}\frac{-y'dx+x\ dy}{x^2+y'^2}=\int_{\pi}^{0}\frac{-\sin(s)(-\sin(s))ds+\cos(s)\cos(s) ds}{\cos^2(s)+\sin^2(s)}=\int_\pi^0ds=-\pi$


问:
Number of functions
Let $F$ denote the set of all functions from $\{1,2,3\}$ to $\{1,2,3,4,5\}$

a) Find and simplify the number of functions $f$ in $F$ so that $f(1)=4$.

b) Find and simplify the number of one-to-one functions $f$ in $F$ so that $f(1)$ is greater than or equal to $4$.

c) Find and simplify the number of functions $f$ in $F$ so that $f(1)$ does not equal $f(2)$.

For (a) I have no idea how to write it out, and for (c) can't I make $f(1)=1$ and $f(2)=2,3,4,5$ that way they don't equal each other?

答:
a) We introduce some notation: Let $A,B$ be abitrary, non-empty (finite) sets and let $A^B := \{f \subseteq A \times B \ | \ f: A \mapsto B \}$. Let (for a given set $A$) $|A|$ be the number of elements of $A$.
Now, a nice way to solve your problem is the following:

Show that $|A^B| = |A|^{|B|}$

Show that $F_a := \{ f \in F \ | \ f(1) = 4 \} \overset{\text{bijective}}\cong \{2,3\}^{\{1,2,3,4,5\}}$

b) Assuming "one-to-one functions" refers to injective functions:
Let $F_b := \{ f \in F \ | \ f \text{ is injective }, f(1) \ge 4 \}$. Then  $F_b$ is the disjoint union $F_b = \{ f \in F \ | \ f \text{ is injective }, f(1) = 4\} \cup \{ f \in F \ | \ f \text{ is injective }, f(1) = 5\}$.
Hence, it is sufficient to prove that

$X := \{ f \in F \ | \ f \text{ is injective }, f(1) = 4\} \overset{\text{bijective}}\cong \{ f \in F \ | \ f \text{ is injective }, f(1) = 5\} \overset{\text{bijective}}\cong \{ (x,y) \in \{1,2,3,4\} \times \{1,2,3,4\} \ | \ x \neq y \}$

$|\{ (x,y) \in \{1,2,3,4\} \times \{1,2,3,4\} \ | \ x \neq y \}| = |\{1,2,3,4\}\times\{1,2,3,4\}| - |\{(1,1),(2,2),(3,3),(4,4)\}| = 4^2-4$

c) Let $F_c := \{ f \in F \ | \ f(1) \neq f(2)\}$. Then $F_c = F \setminus F_{1,2}$, where $F_{1,2} := \{ f \in F \ | \ f(1) = f(2) \}$ and show that $F_{1,2} \overset{\text{bijective}}\cong \{2,3\}^{\{1,2,3,4,5\}}$.
Finally use the technique developed in a) to calculate $|F|$ and $|F_{1,2}|$ which allows you to get $|F_c| \overset{!}= |F| - |F_{1,2}|$.

Title: Number of functions


答:
I will present two options to avoid doing the integral directly.

$\textbf{Option 1}$: Fundamental Theorem of Line Integrals

Since the vector field is conservative on any domain that doesn't contain the origin, we can find a potential function

$$f(x,y) = -\arctan\left(\frac{x}{y^2}\right)$$

There are many options for the potential function here, but this one is continuous on the lower half plane $y<0$. Being careful how we take limits, we approach the endpoints of the integral $(1,0)$ and $(-1,0)$ from below

$$I = \lim_{(x,y)\to(1,0^-)}-\arctan\left(\frac{x}{y^2}\right)+\lim_{(x,y)\to(-1,0^-)}\arctan\left(\frac{x}{y^2}\right) = -\frac{\pi}{2}-\frac{\pi}{2}=-\pi$$

$\textbf{Option 2}$: Green's Theorem/Partial Path Independence

As long we have two paths that (when taken together) do not enclose the origin, then we have path independence. This can be proven via Green's theorem. In this case we will automatically have path independence if we restrict our attention to paths only in the lower half plane. In this case consider the curve

$$x^2+y^4=1 \hspace{20 pt} y\leq 0$$

This curve is chosen because

(a) it simplifies the denominator to a constant and

(b) it contains the start and end points of the original curve.

We can parametrize this curve by

$$\begin{cases}x(t) = \cos t \\ y(t) = -\sqrt{-\sin t} \end{cases} \hspace{24 pt} t\in[-\pi,0] \hspace{18 pt}\implies \hspace{18 pt} \begin{cases}x'(t) = -\sin t \\ y'(t) = \frac{\cos t}{2\sqrt{-\sin t}} \end{cases}$$

Plugging into the integral gives us

$$\int_{-\pi}^0-\sin^2t-\cos^2t\:dt = -\pi$$

Title: compute $\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}$

Body:
The question is:
$$
\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}
\quad \gamma:r(t)=(t,2t^2-2), \quad -1\leq t\leq 1
$$

I have tried to solve it like this:
since $Q_x=P_y$ it's potential vector field but the singularity is at origin, first i thought that i could evaluate it from $(-1,0)$ to $(1,0)$ along $x$ axis but the singularity is the problem, how should i proceed in this case?

Any suggestion would be great Thanks


问:
Are there any plausible arguments for the infinity of right shifted prime numbers?
By "right shifted prime numbers" I mean prime numbers of the form:

$p_r \equiv$ $ 1 $ $mod $ $6$.

$p_l \equiv$ $5$ $mod$ $6$ on the other side would be a left shifted prime number.

Since all prime numbers $p_n>3$ are either right shifted or left shifted it seems to be a generalization of a huge amount of much more specific conjectures when simply asking if there are infinitely many right shifted prime numbers at all.

So before bothering about any folow up conjectures like "infinity of Twin primes", "infinity of Mersenne primes" and such, I'd like to understand why there should be infinitely many right shifted prime numbers, while some simple thought experiments don't suggest it:

Any amount of right shifted prime factors always results in a right shifted composite number.

Any even amount of left shifted prime factors always results in a right shifted composite number.

While on the other hand only an odd amount of left shifted prime factors results in a left shifted composite number.

Primorial -         intervals
$30<p<210$
$210<p<2310$
$2310<p<30030$
$30030<p<510510$
$510510<p<9699690$
$9699690<p<9699690*23$

All prime numbers
36
297
2905
39083
603698
11637502

Left shifted prime numbers
18
150
1462
19560
301928
5819042

Right shifted prime numbers
18
147
1443
19523
301770
5818460

The gap of right shifts
0
3
19
37
158
582

While at a first glance the ratios of Right-shifted-primes/All-primes might seem to converge to $0.5$, I highly doubt it, I'd rather expect to see an exponential increase of the right shift gap on further primorial intervals, when all the previous right shift gaps begin to really matter.

The exponential grwoth of gaps between the Mersenne-primes also highly suggests that there aren't infinitely many right shifted primes, since all Mersenne-primes are right shifted. This would then suggest that there exists some final right shifted prime number and no other prime number from there on can ever happen to be right shifted.

答:
Let $\pi(r,m;x)$ denote the count of primes $p$ less than $x$ such that $p\equiv r\,\,\mathrm{mod}\,\,{m}$. It is clear that if $r$ and $m$ are not relatively prime, i.e $\gcd(r,m)>1$, then there can't be infinitely many primes in $\pi(r,m;x)$ since any number $p\equiv r\,\, \mathrm{mod}\,\,m$ factors as $p=r+mk=\gcd(r,m)(r/\gcd(r,m)+m/\gcd(r,m)k)$.

Let $\varphi(m)$ be the count of relatively prime integers less than $m$. A natural generalization of your question is to assume whether or not the primes are in a sense equally distributed across the $\varphi(m)$ possibilities, meaning that

$$\lim_{x\to\infty}\frac{\pi(r,m;x)}{\pi(x)}=\frac{1}{\varphi(m)}$$

Where $\pi(x)$ is the total number of primes less than $x$ (i.e the percentage of primes less than $x$ is $\frac{1}{\varphi(m)}$). Your question is the case $m=6$, where we note that $\varphi(6)=2$ since $1$ and $5$ are the only numbers less than $6$ that are relatively prime to $6$.

This turns out to be very difficult to prove, but it was shown to be true in 1837 and it now known as Dirichlet's Theorem for Primes in Arithmetic Progressions. Research in this direction continues however, since people have notices than the quantities $\frac{\pi(r,m;x)}{\pi(x)}-\frac{1}{\varphi(m)}$ seem to have "preferences" for being positive or negative, meaning that for a given value of $r$ we generally see slightly less primes than expected.

This phenomenon is very interesting, and I highly recommend you check it out and learn more if you are interested.


问:
Let $X$ be a metric space and $A \subset X$ a bounded subset. If $d(A)$ equals $r$, show that $A$ is a subset of some ball $\bar{B}(a,r)$.

Let $X$ be a metric space and $A \subset X$ a bounded subset. If the diameter of $A$ denoted $d(A)$  equals $r$, show that $A$ is a subset of some ball $\bar{B}(a,r)$. Give an example of a scenario where $X=\mathbb{R}^2$ and $A$ doesn't belong to any $\bar{B}(a,r/2)$.

Let $a \in A$, then I have that $\bar{B}(a,r) = \{x \in X \mid d(x,a) \leqslant r = d(A) \}$. So essentially I would need to show that for any $x \in X$ I have that the distance between $x$ and $a$ is less than or equal to the diameter of $A$? I'm slightly confused about this I want to show that $A$ belong to an closed ball that has it's radius equal to the diameter of $A$?

答:
With metric $\rho.$ Definition: $d(A)=\sup \{\rho (a,b):a,b\in A\}$ if $A$ is not empty. And $d(\emptyset)=0.$ And $A$ is bounded iff $d(A)\in [0,\infty).$

We must distinguish between $\overline B(a,r)=\{b: \rho (a,b)\le r\}$ and the closure $\overline {B(a,r)}$ of the open ball $B(a,r).$ For example if $X=\{a,b\}$ and $\rho (a,b)=r\ne 0$ then $r=d(A)$ but $\overline{B(a,r)}=B(a,r)=\{a\}$ and $\overline{B(b,r)}=B(b,r)=\{b\}.$

If $a\in  A$ then $\forall b\in A\,(\rho(a,b)\le d(A)=r)$ so by definition of $\overline B(a,r)$ we have $\forall b\in A\,(b\in \overline B(a,r)\,).$ That is, $A\subseteq \overline B(a,r).$

If $A$ is empty (trivial special case) then  $A$ is a subset of every set.


问:
Does $f(x)\mid f'(x)$, where $f'(x)$ is the derivative of $f(x)$, and $f(x), f'(x)\in \mathbb{F[x]}$?
Herstein says "Let $f(x)$ and $f'(x)$ have a nontrivial common factor. Then $f(x)\mid f'(x)$."

$f'(x)$ is the derivative of $f(x)$, and $f(x), f'(x)\in \mathbb{F[x]}$.

How is this possible?

Let the common factor be $q(x)$. Then $f(x)=q(x)h(x)$, and $f'(x)=q(x)h'(x)+q'(x)h(x)=q(x)i(x)$, where $q(x), h(x), i(x)\in \mathbb{F[x]}$.

If $f(x)\mid f'(x)$, then the quotient will be $\dfrac{i(x)}{h(x)}$, and it is not obvious that this element should be there in the ring $F[x]$, as that would imply the multiplicative inverse of $i(x)$ is present in the ring.

答:
Indeed, you're right, based on what you've written. Consider for example $f(x)=x^3-x^2\in\Bbb R[x]$. Then $f(x)=x(x^2-x)$ and $f'(x)=3x^2-2x=x(3x-2)$, but we certainly can't say that $f$ divides $f'$ in $\Bbb R[x]$.

Now, if there are some further conditions on $f$ that you didn't mention, then this might work out.


问:
The number of ordered pairs $(p, q)$ of positive integers, whose $\operatorname{lcm}$ is $8100$
The number of ordered pairs $(p, q)$ of positive integers, whose $\operatorname{lcm}$ is $8100$, is $K$.
Then find the number of ways of expressing $K$ as a product of two distinct positive integers.

Now I have taken the $\text{lcm} (p,q)=8100$ which can be expressed as $2^2 \times 5^2 \times 3^4$.

After this, I have taken as $p$ to be $2^a 5^b3^c$ and $q$ to be $2^x5^y3^z$ where $a,b,c,x,y,z$ here represent non negative integers. From there I have taken
$\max\{a, x\}= 2$, $\max\{b, y\} = 2$, $\max\{c, z\} = 4$.
However from here, I am unable to proceed any further.

答:
I'm using "ordered" (as you clarify in the comments) to mean that the order of $(p,q)$ is important and makes a distinct pair, unless $p=q$.

Let's work up to this by solving for $4=2^2$ then $36=2^2\cdot 3^2$.

So the number of $(p,q)$ for $\text{lcm}(p,q)=4$ relies on having one of $p,q=4$, and the other can be $\in \{1,2,4\}$ - that is, $2+1=3$ options from the exponent going from $0$ to $2$. This gives us $3\times 2 = 6$ solutions (the ordering giving the $\times 2$), but then we remove the double counted $(4,4)$ giving $5$.

Now for $36$: We can solve independently as above for the powers of $2$ and $3$ but the only duplicate will be $(36,36)$ since no other repeated number will give lcm of $36$. So we have $6\times6-1 = 35$ solutions. (If we wanted "unordered" $(p,q)$, meaning $(18,4)\equiv (4,18)$, we would divide by $2$ rather than subtract $1$.)

Now it's clear that the number of solutions for $8100=2^2\cdot3^4\cdot5^2$ is $6\times 10\times 6 - 1 = 359$ options for $(p,q)$.

And, for the follow-on part of the question, $K=359$ is a prime number.


问:
Truncation and fixed finite domain
Let $P$ be a type, $||P||$ denotes a mere proposition obtained by truncating $P$. Let $D$ be a type and $A:D\rightarrow\textsf{U}$, then $\Pi x:D.A(x)$ is a well-formed type. Assuming that we are working with intuitionistic logic, we have $(*)$ but not its converse.
$$||\Pi x:D.A(x)||\rightarrow\Pi x:D.||A(x)||\tag{$*$}$$
My question is: let $D$ denote a finite domain such that $D=\{d_1,...,d_n\}:\textsf{U}$, can we have the converse of $(*)$, namely:
$$\Pi x:D.||A(x)||\rightarrow||\Pi x:D.A(x)||\tag{$**$}$$
It seems to me that $(**)$ should hold. Given $D=\{d_1,...,d_n\}:\textsf{U}$, we shall then write $\Pi x:D.A(x)$ as $A(d_1)\times...\times A(d_n)$. Since we have $||P\times Q||\leftrightarrow||P||\times||Q||$ (for $P,Q:\textsf{U}$), we shall generalize it to $A(d_1)\times...\times A(d_n)$ such that we have $(***)$:
$$||A(d_1)||\times...\times||A(d_n)||\rightarrow||A(d_1)\times...\times A(d_n)||\tag{$***$}$$

I find this very similar to the case of double negation shift (DNS). It is known that in intuitionistic logic, the converse of DNS is a theorem, but DNS itself is not. But if we require that the domain be finite, then writing $\Pi x:D.A(x)$ as a generalized conjunction $A(d_1)\times...\times A(d_n)$, DNS is valid by virtue of the fact that $\neg\neg$ distributes over $\rightarrow$ (that is, $\neg\neg(P\rightarrow Q)\leftrightarrow(\neg\neg P\rightarrow \neg\neg Q)$ for $P,Q:\textsf{U}$).
$$\Pi x:D.\neg\neg A(x)\rightarrow\neg\neg\Pi x:D.A(x)\tag{DNS}$$

I think it's something very basic, but since I have not read it in any introductory book I know, I am not sure if my reasoning is correct. Can anyone help me?

答: You are correct. Here is a cubical Agda proof. It essentially carries out your product-of-truncations = truncation-of-product equivalence by induction on the (size of the) finite set. The rest is just dealing with transporting along a (truncated) witness that a type is equivalent to the inductively defined finite sets.</p>

Title: Truncation and fixed finite domain


问:
Show that subspace is maximal ideal in commutative Banach algebra

Show that the subspace $c_{0} = \{ x = (x_{1}, x_{2}, \ldots ) \in \ell^{\infty} : \lim x_n = 0 \}$ is a maximal ideal in $c = \{ x = (x_{1}, x_{2}, \ldots ) \in \ell^{\infty} : \lim x_n \text{ exists } \}$ by finding a linear multiplicative functional with $c_{0}$ as its kernel.

I already showed that $c_{0}$ is a proper, closed ideal. I am stuck on showing that it is a maximal ideal by finding a multiplicative linear functional $\phi$ with $\ker(\phi) = c_{0}$.

答:
Consider $\phi(x_1,x_2,\dots):=\lim_n x_n$.

It's clearly linear and maps to the scalar field, its kernel is just $c_0$, and thus it's a one codimensional subspace, hence must be maximal.


问:
Weak* convergence equivalent to boundness, Banach-Steinhaus theorem
Let $(a_n)$ be a sequence of real numbers.
How to prove that weak* convergence to $0$ of $(a_ne_n)\in l_\infty(\mathbb{R})$ is equivalent to $(a_n)$ being bounded?

I'm asking for help, how to relate this to Banach-Steinhaus theorem of uniform boundness?

答:
Recall the canonical identification $(\ell^1)^* = \ell^\infty$ given by
$$\Psi: \ell^\infty \to (\ell^1)^*: f \mapsto \left(g \mapsto \sum_n f(n)g(n)\right)$$

Thus, by definition:

$a_ne_n$ converges weak$^*$-to $0$
$\iff$
$\Psi(a_n e_n)(g) \to 0$ for all $g \in \ell^1$.

Note that $$\Psi(a_n e_n)(g) = \sum_k (a_n e_n)(k) g(k) = a_ng(n) $$
so the latter condition is equivalent to $a_n g(n) \to 0$ for every $g \in \ell^1$.

Now, the claim follows from the following lemma which I leave as an exercise for you (see the other answer for a solution):

$$(a_n)_n \mathrm{\ bounded}\iff \forall g \in \ell^1: \lim_n a_n g(n) = 0$$


答:
Why there should be infinitely many right shifted prime numbers?

Suppose there were only finitely many and consider the number $X$ which is the product of all these primes together with $6$.

Let $p$ be a prime which divides $X^2+X+1$. Then $p$ would have to be of the form $6k+5$ since all other primes divide $X$.

By Fermat's little Theorem, $X^{6k+4}=X^{p-1}\equiv 1\pmod p$.

Since $X^3-1=(X^2+X+1)(X-1)$ we also have $X^3\equiv 1\pmod p$.

Therefore $X=X^{(6k+4)-3(2k+1)}\equiv 1\pmod p$.

But then  $X^2+X+1\equiv 3\pmod p$ and is not divisible by $p$ after all.

This contradiction proves that there are infinitely many right shifted primes.

Title: Are there any plausible arguments for the infinity of right shifted prime numbers?

Body:
By "right shifted prime numbers" I mean prime numbers of the form:

$p_r \equiv$ $ 1 $ $mod $ $6$.

$p_l \equiv$ $5$ $mod$ $6$ on the other side would be a left shifted prime number.

Since all prime numbers $p_n>3$ are either right shifted or left shifted it seems to be a generalization of a huge amount of much more specific conjectures when simply asking if there are infinitely many right shifted prime numbers at all.

So before bothering about any folow up conjectures like "infinity of Twin primes", "infinity of Mersenne primes" and such, I'd like to understand why there should be infinitely many right shifted prime numbers, while some simple thought experiments don't suggest it:

Any amount of right shifted prime factors always results in a right shifted composite number.

Any even amount of left shifted prime factors always results in a right shifted composite number.

While on the other hand only an odd amount of left shifted prime factors results in a left shifted composite number.

Primorial -         intervals
$30<p<210$
$210<p<2310$
$2310<p<30030$
$30030<p<510510$
$510510<p<9699690$
$9699690<p<9699690*23$

All prime numbers
36
297
2905
39083
603698
11637502

Left shifted prime numbers
18
150
1462
19560
301928
5819042

Right shifted prime numbers
18
147
1443
19523
301770
5818460

The gap of right shifts
0
3
19
37
158
582

While at a first glance the ratios of Right-shifted-primes/All-primes might seem to converge to $0.5$, I highly doubt it, I'd rather expect to see an exponential increase of the right shift gap on further primorial intervals, when all the previous right shift gaps begin to really matter.

The exponential grwoth of gaps between the Mersenne-primes also highly suggests that there aren't infinitely many right shifted primes, since all Mersenne-primes are right shifted. This would then suggest that there exists some final right shifted prime number and no other prime number from there on can ever happen to be right shifted.


问:
Pdf of the difference of random variables X and Y, both continuously, uniformly distributed on [0,1]
I tried to compute the pdf of the difference with the convolution formula and found that the pdf is x+1 for x ∈ [-1 ,0] and null otherwise, but my intuition tells me that I am missing the part where x ∈ [0,1]. Can anyone give me a hint what I got wrong?</p>

答:
It makes a difference whether you want the pdf for the absolute difference $|u_1-u_2|$ for two uniform independent vars $u_k$ ($k=1,2$) or their actual difference $u_1-u_2$ (or $u_2-u_1$.)

For the absolute difference consider the unit square $[0,1] \times [0,1]$ and draw a picture to see that the cumulative $|u_1-u_2| \le t$ has area $t^2/2$ and derivative of that is $t$ for $0\le t \le 1$ which is one of results you mention. If considering the difference it would be a piecewise function as you hint at.

By OP comment, it is $u_1-u_2$ whose density is desired, where each of the $u_k$ is uniform on $[0,1]$ and they are independent. So if one considers the "unit square" $S=[0,1] \times [0,1]$ it follows that probabilities of regions correspond to their areas. Now the cumulative distribution of the difference is by definition $F(t)=P(u_1-u_2 \le t)$ which (viewing $u_1$ as the horizontal and $u_2$ the vertical) is the area in $S$ lying above the line $u_1-u_2=t.$ This area is zero if $t \le -1.$ Then for $-1 \le t \le 0$ it is $(1/2)(1+t)^2.$ Using symmetry of the square $S$ we see that for $0 \le t \le 1$ the required area is $1-(1/2)(1-t)^2.$ and finally is constant at $1$ for $t \ge 1.$

Taking the derivative of $F(t)$ to get the density (ignoring as we may a few finite values) gives the function $f(t)=1+t$ if $t \in [-1,0]$ and $f(t)=1-t$ if $t \in [0,1].$ Of course $f(t)=0$ when $t$ is not in $[-1,1].$


问:
write five digits numbers using $0,0,0,1,1,2,2,2,2,3,3,3,4,4$.
It is given that we want to write five digits numbers using $0,0,0,1,1,2,2,2,2,3,3,3,4,4$.

How many ways are there to write it $?$

My thought In the first place, it seems cumbersome question a little bit , because i think that we should write them separately in terms of the leading term such that firstly i check over $1$ as leading term , then $2$ so on. However , this process is exhaustive and i hope to find  elegant approach for this question.

答:
Here is a method which still requires a bit of work, but which is still more efficient then naive brute force.

Start with a simpler question:

For each $k=1,2,\dots,5$, how many $k$ digit numbers can you make using $3, 3, 4, 4, 4$?

With a little work, the answers are

k
1
2
3
4
5

# numberss with k digits using 3s and 4s
2
4
7
10
10

Next....

For each $k=1,2,\dots,5$, how many $k$ digit numbers can you make using $2,2,2,2,3,3,4,4,4$?

The key is that we can leverage or work in the previous step to make this one easier. Take the summation of all ways to place the $2$'s, times the number of ways to fill in the rest of the empty spaces with $3$'s and $4$'s, the latter of which can be looked up in the previous table. For example,

\begin{align}
\text{# $5$-digit nums with $2$'s, $3$'s and $4$'s}
&= \binom{5}0\cdot \hspace{-1cm}
\overbrace{10}^\text{# 5 digit nums with 3,4}\hspace{-.1cm}
+ \binom{5}1\cdot \hspace{-1cm}\overbrace{10}^\text{# 4 digit nums with 3,4}\hspace{-1cm}
\\
&+ \binom{5}2\cdot \hspace{-1cm}\underbrace{7}_{\substack{\;\\\\\text{# 3 digit nums with 3,4}}}\hspace{-.1cm}
+ \binom{5}3\cdot \hspace{-1cm}\underbrace{4}_\text{# 2 digit nums with 3,4}\hspace{-.1cm}
\\&+ \binom{5}4\cdot \hspace{-1cm}\underbrace{2}_{\substack{\;\\\\\text{# 1 digit nums with 3,4}}}\hspace{-1cm}
\hspace{1.5cm}=\boxed{180}
\end{align}
You then have to do the same for the $1,2,\dots,4$ digit numbers, giving another row to the table. Then, use that row to calculate the number of numbers where you can use digits $1$ to $4$.

k
0
1
2
3
4
5

# numbers with k digits using 3s and 4s
1
2
4
7
10
10

# numbers with k digits using 2, 3, and 4
1
3
9
26
71
180

# numbers with k digits using 1, 2, 3 and 4
1
4
16
62
229
795

Finally, the number of five digit numbers using all available digits is found by summing over all ways to place the zeroes, times the number of ways to fill what remains with $1$ to $4$, which is in the last row of the table:
$$
\binom{4}0\cdot 795+\binom{4}1\cdot 229+\binom{4}2\cdot 62+\binom{4}3\cdot 16=\boxed{2147}
$$


答:
As saulspatz pinted out in the comment, you will have to split the integral into two if you switch the order:
$$
\int_0^1\int_0^{\sin^{-1}(y)}dxdy+\int_{1}^2\int_0^{\pi/2}dxdy
$$

Title: How do I reverse the order of integration for this double integral?

Body:
$$\int_0^{\pi/2}\int_{\sin(x)}^2dydx$$ This is the integral in question. My closest was $$\int_0^{2}\int^{\sin^{-1}(y)}_2dxdy$$
but I believe this is wrong.
Sorry for the bad formatting, It's my first time using it properly


问:
Probability of at least one empty subway car
I have the following task: there is a subway train with 4 empty cars and 15 passengers getting in it. What is the probability that at least one car will be empty?</p>

答:
Hint:

Let $E_i$ denote the event that car $i$ stays empty.

Then to be found is:$$P(E_1\cup E_2\cup E_3\cup E_4)$$
Apply the principle of inclusion/exclusion.


问:
Does $\sum^\infty_{n=3}\frac {\ln^2(n)}{n^{3/2}}$ converge or diverge?
I am trying to find a way to decide if this sum converges or diverges, I'm trying to use the integral test and bound the integrand or use the limit comparison test, but I am not finding a good function to compare my integrand with at infinity.
 I would appreciate any help or tips in how to find a function that behaves like $\frac {\ln^2(x)}{x^{3/2}}$ at $x\to \infty$ that I can use it for limit comparison test.

Thanks in advance.

答:
Integral test for the convergence. The integral is pretty doable
$$\int_3^{\infty } \frac{\log ^2(x)}{x^{3/2}} \, dx=\lim_{M\to\infty}\left[-\frac{2 \left(\log ^2(x)+4 \log (x)+8\right)}{\sqrt{x}}\right]_3^M=\frac{2 \left(8+\log ^2(3)+\log (81)\right)}{\sqrt{3}}$$
As the integral converges, the series converges as well.


问:
polynomials associated with generating function problem
Here is the question;

If there is an unlimited number (or at least 24 of each color) of red, green, white, and black jelly beans, in how many ways can Douglas select 24 of these candies so that he has an even number of white beans and at least six black ones?

This problem is solved using generating functions in my textbook (I am learning this for tje first time). I tried to understand it. But wasn't able to get it clearly. Here are the steps mentioned in book:

The polynomials associated with the jelly bean colors are as follows:

Red (green): $$1 + x + x^2 + .... + x^{24}$$
White: $$1 + x^2 + x^4 +.... +x^{24}$$
Black: $$x^6+x^7 + ... x^{24}$$

So the answer to the problem is the coefficient of $x^{24}$ in the generating function $$f(x) = (1 + x + x^2 + ... + x^{24})^2 (1 + x^2 + x^4 +...+x^{24})(x^6 + x^7+ ... + x^{24})$$

I am not worried about the answer. My problem is about writing these polynomials. What is the idea behind taking series for red and green upto $x^{24}$? Since 6 black beans are in every selection why isn't it just upto $x^{18}$? (Which is 24 -6)

My teacher told me it is because collection of beans is infinity. But I didn't get it. Even if the collection is infinity, the selection is limited to 24 with atleast 6 black beans and even number of white beans. This is my simple 'stupid' logic. I can't get rid of that logic. I need some conceptual clarity. Can you help me?

Question ref: Discret and Combinatorial Mathematics by Ralph Grimaldy 5 th edn example 9.2

答: You'll get the same answer either way (or even if you make each factor an infinite sum). Note that every term in the result will include a contribution of $x^6$ or larger from the last factor, so the $x^{19}$ and higher terms in the other factors will only contribute to $x^{25}$ and higher terms of the result - they won't affect the $x^{24}$ coefficient.


问:
Show $\int f \, \mathrm{d}\left( \sum_n \mu_n \right) = \sum_n \int f \, \mathrm{d} \mu_n$
Let $\mu_1,\mu_2,\dots$ measures on the measurable space $(E,\mathcal{E})$. Then $\mu = \sum_n \mu_n$ is a measure on $(E,\mathcal{E})$.

I want to show, that $$\mu f = \sum_n \mu_n f$$ for every positiv measurable function $f \in \mathcal{E}_+$ where $\mu f = \int_E f \, \mathrm{d}\mu$.

My strategy is to show for the function $L \colon \mathcal{E}_+ \to [0,\infty]$, $L(f)=\sum_n \mu_n f$ that the following properties hold:

$f=0 \Rightarrow L(f) = 0$

$f,g \in \mathcal{E}_+$, $a,b \in \mathbb{R}_+$  $\Rightarrow$ $L(af+bg)=aL(f)+bL(f)$

$(f_k) \subset \mathcal{E}_+$, $f_k \nearrow f$ $\Rightarrow$ $L(f_k) \nearrow L(f)$

Then there exists a unique measure $\nu$ on $(E,\mathcal{E})$ such that $L(f)=\nu f$ for every $f \in \mathcal{E}_+$ and because
$ \nu(B) = L(1_B) = \mu(B), \ B \in \mathcal{E}  $ the statement
$\nu = \mu$ is valid.

My problem with this is the third point:

For every $N \in \mathbb{N}$

$$ \sup_k \sum_{n=1}^{N} \mu_n(f_k) = \sum_{n=1}^{N} \sup_k \mu_n(f_k) = \sum_{n=1}^{N} \mu_n(f) $$ from the monotone convergence theorem.

Now I wan't to interchange $\lim$ and $\sup$ like this:

$$ \sum_n \mu_n (f) =  \lim_{N \to \infty}\sup_k \sum_{n=1}^{N} \mu_n(f_k)= \sup_k \lim_{N \to \infty} \sum_{n=1}^{N} \mu_n(f_k) = \sup_k \sum_{n} \mu_n(f_k) $$

Is this possible and why?

答:
Concerning the issue with the sup, don't forget that the supremum in $k$ there is also the limit in $k$.  Hence the monotone convergence theorem applies.  To see this, define functions $a_{k} : \mathbb{N} \to [0,\infty]$ by
\begin{equation*}
a_{k}(n) = \int_{E} f_{k} \, d \mu_{n}
\end{equation*}
Notice that $a_{k}(n) \leq a_{k+ 1}(n) \leq \int_{E} f \, d \mu_{n}$ and $\lim_{k \to \infty} a_{k}(n) = \int_{E} f \, d \mu_{n}$.  Therefore, by the monotone convergence theorem (applied to $\mathbb{N}$ with the counting measure),
\begin{equation*}
\lim_{k \to \infty} \sum_{n = 1}^{\infty} a_{k}(n) = \sum_{n = 1}^{\infty} \int_{E} f \, d \mu_{n}.
\end{equation*}

On the other hand, one can check directly that the formula $\mu(A) = \sum_{n = 1}^{\infty} \mu_{n}(A)$ defines a measure on $\mathcal{E}$.  Further, if $f$ is a simple function, then $\int_{E} f \, d \mu = \sum_{n = 1}^{\infty} f \, d \mu_{n}$ follows immediately.  Arguing using the monotone convergence theorem (as above) and simple approximation, one then finds that the same identity holds when $f$ is bounded and non-negative.  From there, one can generalize to arbitrary non-negative or integrable $f$.


问:
Prime Number Sums
The prime numbers $a,b$ and $c$ are such that $a+b^2=4c^2.$ Determine the sum of all possible values of $a+b+c.$

My Attempt
$a+b^2=4c^2$.
$a=4c^2-b^2$.
$a=(2c+b)(2c-b)$.

After this, I tried testing cases but I'm not totally sure how to account for everything.
Thanks!

答:
Note that $c=2,3$ yield solutions: $a=7,b=3,c=2 \text{ and }a=11,b=5,c=3$

As has been noted in the comments, $2c-b=1$ is necessary for $a$ to be prime.

For prime numbers $c\ge 5$, $c$ must have the form $6k\pm 1$. Note that for $c=6k-1$ we have $2c-b=1=12k-2-b \Rightarrow b=12k-3$. Since $3\mid 12k-3$, $b$ cannot be prime and primes $c$ of the form $6k-1$ can never give rise to solutions.

Prime numbers $c=6k+1$ yield $2c-b=1=12k+2-b \Rightarrow b=12k+1$, which may or may not be prime. In cases where $b$ is in fact prime, $a=2c+b=12k+2+12k+1=24k+3$. Since $3\mid 24k+3$, $a$ cannot be prime and primes $c$ of the form $6k+1$ can never give rise to solutions.

So the only solutions are those given first, in which the asked for sums are:  $a+b+c=12,19$

Title: Prime Number Sums


问:
How do I determine if $f(x)$ has a continuous derivative on $[a,b]$?
When finding the length of a curve using calculus, the first step is to determine whether $f(x)$ has a continuous derivative on the curve (i.e. on $[a,b]$). But how do we determine this?

My understanding is that you need to find the second derivative, i.e. $f''(x)$. If $f''(x)$ exists, this implies that $f'(x)$ is a continuous derivative along the curve of $f(x)$. Is that accurate?

I apologize if this is an obvious question. I just feel that my textbook/online resources give much more convoluted answers than this, when the solution is pretty simple. My concern is that I may be overlooking something.

答: Okay, it looks like my idea is wrong. Consider $y = 2x + 1$ on $[1,5]$, for example. This function has a continuous first derivative ($y' =2$) but does not have a meaningful second derivative ($y'' = 0$).


问:
Prove that $E[X|X]=X$
In my statistics notes , they used the following result in an exercise
$E[X|X]=X$ , where $X$ is a normal random variable. They simply stated that if I know $X$, the expected value is $X$. How can I prove that more rigorously?

答:
The solutions linked here already use the uniqueness, but another way to show this is to look at the sets

$$ E_n^+:=\Big \{ \omega: X(\omega)-\mathbb{E}[X\vert X] > \frac{1}{n} \Big\}  \quad \text{and} \quad  E_n^-:=\Big \{ \omega: X(\omega)-\mathbb{E}[X\vert X] < \frac{1}{n} \Big\}, $$

which are both measurable w.r.t $\sigma(X)$. Note that $ \{ X\neq \mathbb{E}[X\vert X] \}=E^+\cup E^- $, while

$$ E^+=\cup_{n}E_n^+ \quad \text{and} \quad E^-=\cup_{n}E_n^-. $$

Using condition $(2)$ of your definition, you can show that both $E^+$ and $E^-$ occur with probability $0$.

Title: Prove that $E[X|X]=X$


问:
Divisors of n! including 1 which sum to n!
I believe that given any positive integer $n$, we can find a set of divisors of $n!$, including $1$, which sum to $n!$ I checked for the first 6 positive integers, and tried to prove it by induction on $n$. However I was unable to formalise the proof, can anybody present a method of proving this? Or maybe without induction?

Body: $\frac{n!}{2}+\frac{n!}{3}+\frac{n!}{6}=n!$ works for all $n\ge 6$


答:
Digits given: $0,0,0,1,1,2,2,2,2,3,3,3,4,4$ and we need to make five digit numbers from them. Removing bad cases seems to be a good choice here, even though there is a bit more work in the end for removing numbers with leading zeroes. The point to note is that you do not have any overlap between bad cases of two numbers or more, on other words, we cannot simultaneously have bad cases for two of them.

Number of bad cases for $0$ or $3 = \underbrace{5 \cdot 4}_{\text{Four 0's}} + \underbrace{1}_{\text{Five 0's}} = 21$

Number of bad cases for $1$ or $4 = {5 \choose 3} \cdot 4^2 + 5 \cdot 4 + 1 = 181$

Number of bad cases for $2 = 1$

Total number of bad cases $ = 2 \times 21 + 2 \times 181 + 1 = 405$

Subtract this from total number of unrestricted cases $ = 5^5 = 3125$ (if we had $5$ choices for each digit). That gives us an answer of $2720$.

Now as we need proper $5$ digit numbers where leading zeroes are not allowed,

Numbers with leading zeroes $ = 5^4$. Out of which below are already counted in above bad cases -

In bad cases for $0$, only $4$ start with another digit so $17$ duplicates.

In bad cases for $3$, numbers with leading zeros $ = 1$ (fix first digit as $0$, then remaining four digits are all $3$).

In bad cases for $1$ (or $4$), numbers with leading zeroes $ = 4 \cdot 4 + 1 = 17$ (first digit $0$, choose three out of four remaining digits as $1$ and the last remaining digit other than $1$ or all four digits as $1$).

So total five digit numbers without leading zeroes $ = 2720 - 5^4 + 52 = 2147$

Body:
It is given that we want to write five digits numbers using $0,0,0,1,1,2,2,2,2,3,3,3,4,4$.

How many ways are there to write it $?$

My thought In the first place, it seems cumbersome question a little bit , because i think that we should write them separately in terms of the leading term such that firstly i check over $1$ as leading term , then $2$ so on. However , this process is exhaustive and i hope to find  elegant approach for this question.


问:
Complex function with values on the unit circle copied everywhere
If $f:\mathbb{C}\setminus \{0\}\to \mathbb{C}$ is a function such that $f(z)=f(\frac z{|z|})$ and its restriction to unit circle is continous,then

$(1)\lim _{z\to 0} f(z)$ exist.

$(2)f$ is analytic but not necessarily a constant function

$(3)f$ is continous but not necessarily analytic.

$(4)f$ is a constant function.

The title can be justified in the following way:-

Let $z=re^{i\theta}$, Then $f(re^{i\theta})=f(e^{i\theta})$

Hence, I can imagine rays of equal values radiating out from zero in every direction .

Now ,if $z=re^{i\theta}$, then as $r\to 0$, then $z\to 0$

$\lim_{z\to 0}f(z)=
\lim_{r\to 0}f(e^{i\theta})=f(e^{i\theta})$, which depends on $\theta$ and hence the limit does not exist.

I am quite confident that there exist no such non-constant analytic function but it is getting difficult for me to write an analytical proof. (I tried to arrive at a contradiction assuming $f$ is a analytic on  a bounded domain and using maximum modulus theorem but I failed.)

Since the restriction of $f$ to the unit circle is continous, that means for any fixed $\theta_0$, $\lim_{\theta\to \theta_0}f(e^{i\theta})=f(e^{i\theta_0})$

Now let $z_0=r_0e^{i\theta_0}, z=re^{i\theta}$ and $\epsilon \gt 0$ be given, then $|f(z_0)-f(z)| =|f(e^{i\theta_0})-f(e^{i\theta})|\lt \epsilon $,
if $| \theta-\theta_0|\lt \delta$, where the existence of $\delta$ for $\epsilon$ is garaunteed by continuity of $f$ on the unit circle.

So $f$ is continous.

$f$ may not be constant as suggested by above proof.

Please go through my work and help me complete it.

Thanks for your valuable time.

答:
Let $f(z)=\frac{z}{|z|}$, it is continuous in $\Bbb{C}\setminus \{0\}$.
Therefore,
$f(f(z))=f(\frac{z}{|z|})=f(z).$
composition of two continuous function is continuous. Thus $f$ is continuous.
$f$ is not analytic. one can check it by Cauchy-Riemann equation.

clearly, $f(z)=\frac{z}{|z|}$ is not constant.

Limit does not exist as $z\to 0$, one can check easily.
$f(x+iy)=\lim_{(x,y)\to (0,0)}\frac{x+iy}{x ^2+y^2}$.


问:
Is $f$ Lebesgue measurable?
I am having trouble with proof for this statement:
Let $V$ be Vitali set, is function $f : \Re \rightarrow \Re$  Lebesgue measurable?

$$
f(x) =
\left\{
\begin{array}{ll}
-7x-4 &x \in V \\
0 &x \notin V
\end{array}
\right. $$

I'm completely lost and I don't know where to begin, if $f$ would be continuous function it would be measurable, but it isn't continuous. It seems to me that this function is in fact measurable, but I don't know how to proof that and how to start. Thank you for any advice.

EDIT: I misplaced in sign, sorry

答: If $f$ were a measurable function then $\Bbb R\setminus [\, (f^{-1}\{0\})\cap ( \Bbb R\setminus \{-4/7\})\,]=V$ would be a measurable set.

Title: Is $f$ Lebesgue measurable?


问:
Prove that $ \sum_{i=1}^{\infty}\frac{f_{i}}{2^{i}} $ is integrable function.
Let $ I\subseteq\mathbb{R}^{n} $ be a box.Let $ f_{i}:I\to[0,1] $ be integrable functions.

Prove that $ \sum_{i=1}^{\infty}\frac{f_{i}}{2^{i}} $ is integrable function.

My first intuition was to use Weierstrass M-test, but we never proved it for multivariable functions and I cant see why would it hold.

The second intuition was to show that the set of discontinuities of $ \sum_{i=1}^{\infty}\frac{f_{i}}{2^{i}} $ is of measure zero, but im not sure how to show it.

Any help would be appreaciated. Thanks in advance.

答:
Let $F_k:=\sum_{i=1}^k \frac{f_i}{2^i}$ and let $F=\lim_k F_k=\sum_{i=1}^\infty \frac{f_i}{2^i}$.

Since $f_i\ge 0$ for all $i$, the sequence $F_k$ is non-decrasing, so by Monotone Convergence Theorem
$$\int_I F=\lim_k \int_I F_k.$$

But $f_i\le 1$ for all $i$, so (denoting $\lambda_n$ the $n$.dimensional Lebesgue measure)
$$\int_I F_k=\sum_{i=1}^k\frac{1}{2^i}\int_If_i =\sum_{i=1}^k \frac{1}{2^i} \lambda_n(I)=\lambda_n(I)\sum_{i=1}^k\frac{1}{2^i}\overset{k\to\infty}{\longrightarrow}\lambda_n(I)<+\infty.$$
So, $F$ is integrable.


问:
Does $f(x)\mid f'(x)$, where $f'(x)$ is the derivative of $f(x)$, and $f(x), f'(x)\in \mathbb{F[x]}$?
Herstein says "Let $f(x)$ and $f'(x)$ have a nontrivial common factor. Then $f(x)\mid f'(x)$."

$f'(x)$ is the derivative of $f(x)$, and $f(x), f'(x)\in \mathbb{F[x]}$.

How is this possible?

Let the common factor be $q(x)$. Then $f(x)=q(x)h(x)$, and $f'(x)=q(x)h'(x)+q'(x)h(x)=q(x)i(x)$, where $q(x), h(x), i(x)\in \mathbb{F[x]}$.

If $f(x)\mid f'(x)$, then the quotient will be $\dfrac{i(x)}{h(x)}$, and it is not obvious that this element should be there in the ring $F[x]$, as that would imply the multiplicative inverse of $i(x)$ is present in the ring.

答:
This statement isn't true in general. Take for example $f(x)=(x-1)^2$ so that $f'(x)=2(x-1)$. In this case, clearly $f(x)\nmid f'(x)$.

On the other hand, if we know $f(x)$ is the irreducible polynomial of some element, say $\alpha$, then $f(x)$ divides every polynomial that has $\alpha$ as a root. In particular, if $\alpha$ is a repeated root, then $f(x)\mid f'(x)$.


问:
Trick/Fake proof of 2x2=5 ; algebraic form
Just ran into "algebraic proof" that 2x2=5 on youtube. They didn't hide that it was a trick. Can't find where is the mistake.

Let a=4, b=5, c=1

c=b-a                   /multiply by (b-a)

c(b-a) = (b-a)(b-a)     /distribute

cb-ca = b^2-2ab+a^2     /subtract a^2

cb-ca-a^2 = b^2-2ab     /add ab

ab+cb-ca-a^2 = b^2-ab   /subtract cb

ab-ca-a^2 = b^2-ab-cb   /factor a and b respectively

a(b-c-a)=b(b-a-c)       /divide by (b-a-c)

a=b                     /substitute with given value

4=5

答: You can't divide by $b−a−c=5−4−1=0$. Theese kind of proofs always end up with a $0$-division and you can say that this is a good motivation you can't divide a number by $0$!


问:
Curl in spherical coordinates without Del operator
Trying to get practice with differential forms and related topics I had tried to calculate curl in spherical coordinates without mystic "del" operator.

Let $rot\: F = (\star d (F^\sharp))^\flat$;

$x = r \cos \phi \sin \theta$

$y = r \sin \phi \sin \theta$

$z = r \cos \theta $

Non normalized basis of tangent space $T$:

$\bar e_r = (\frac{dx}{dr}, \frac{dy}{dr}, \frac{dz}{dr}) = (\cos \phi \sin \theta, \sin \phi \sin \theta, \cos \theta)$

$\bar e_\phi = (\frac{dx}{d\phi},\frac{dx}{d\phi},\frac{dz}{d\phi}) = (-r\sin \phi \sin \theta, r\cos \phi \sin \theta, 0)$

$\bar e_\theta = (\frac{dx}{d\theta},\frac{dx}{d\theta},\frac{dz}{d\theta}) = (r \cos \phi \cos \theta, r \sin \phi \cos \theta, -r\sin \theta)$

Norms:

$\|\bar e_r\| = 1$

$\|\bar e_\phi\| = r \sin \theta$

$\|\bar e_\theta\| = r$

Orthonormal basis of tangent space $T$:

$ e_r = \frac{\bar e_r}{\| \bar e_r\|} = (\cos \phi \sin \theta, \sin \phi \sin \theta, \cos \theta)$

$ e_\phi = \frac{\bar e_r}{\| \bar e_r\|} = (-r\sin \phi \sin \theta, r\cos \phi \sin \theta, 0)$

$ e_\theta = \frac{\bar e_r}{\| \bar e_r\|} = (r \cos \phi \cos \theta, r \sin \phi \cos \theta, -r\sin \theta)$

Let $w_r, w_\phi , w_\theta$ be corresponding basis in $T^*$. Then

$dr = w_r$

$d\phi = \frac{1}{r\sin \theta} w_\phi$

$d\theta = \frac{1}{r} w_\theta$

So, going back to the $rot$:

$F = F^1 e_r + F^2 e_\phi + F^3 e_\theta$

$F^\sharp = F^1 w_r + F^2 w_\phi + F^3 w_\theta$

$d (F^\sharp) = $
$$ = F^1_\theta d\theta \wedge w_r + F^1_\phi d\phi \wedge w_r $$
$$ + F^2_\theta d\theta \wedge w_\phi + F^2_r dr \wedge w_\phi$$
$$ + F^3_\phi d\phi \wedge w_\theta + F^3_r dr \wedge w_\theta$$

$$ = F^1_\theta \frac{1}{r} w_\theta \wedge w_r + F^1_\phi \frac{1}{r\sin \theta} w_\phi \wedge w_r $$
$$ + F^2_\theta \frac{1}{r} w_\theta \wedge w_\phi + F^2_r w_r \wedge w_\phi$$
$$ + F^3_\phi \frac{1}{r\sin \theta} w_\phi \wedge w_\theta + F^3_r w_r \wedge w_\theta$$

$$ = (F^2_r - F^1_\phi \frac{1}{r\sin \theta}) w_r \wedge w_\phi$$
$$ + (F^3_r - F^1_\theta \frac{1}{r}) w_r \wedge w_\theta$$
$$ + (F^3_\phi \frac{1}{r\sin \theta} - F^2_\theta \frac{1}{r}) w_\phi \wedge w_\theta$$

$\star d (F^\sharp) = $
$$ = (F^2_r - F^1_\phi \frac{1}{r\sin \theta}) w_\theta$$
$$ - (F^3_r - F^1_\theta \frac{1}{r}) w_\phi$$
$$ + (F^3_\phi \frac{1}{r\sin \theta} - F^2_\theta \frac{1}{r}) w_r$$

$rot F  = (\star d (F^\sharp))^\flat = $
$$= (F^3_\phi \frac{1}{r\sin \theta} - F^2_\theta \frac{1}{r}) e_r$$
$$- (F^3_r - F^1_\theta \frac{1}{r}) e_\phi$$
$$+ (F^2_r - F^1_\phi \frac{1}{r\sin \theta}) e_\theta$$

But in wikipedia the answer is completely different. Please, help me to find which operations i had understood wrongly.

答:
I don't have any knowledge of differential forms, but the curl of a three dimensional vector field with coordinates expressed in some normalized basis can always be computed as a determinant
$$\operatorname{curl}\mathbf{V}=\frac{1}{h_1 h_2 h_3}\det\left(\begin{bmatrix} h_1 \mathbf{e}_1 & h_2 \mathbf{e}_2 & h_3 \mathbf{e}_3 \\ \partial_1 & \partial_2 & \partial_3 \\ h_1 V^1 & h_2 V^2 & h_3 V^3 \end{bmatrix}\right)$$
Here $h_i=\sqrt{g_{ii}}$.


问:
Conjecture a formula for the nth term of the sequence
I am struggling with finding any formula that fits this sequence, please help.

an=1,1,0,1,1,0,1,1,0,1

答:
Well, after thinking and trying different combinations. I got this:

$\boxed{a_n = ( \omega^n + \omega^{2n} ) - (-1)^{\omega^n + \omega^{2n}}\times2}$

Where $\omega$ is the imaginary cube root of unity.

This is based on the fact that,

$\omega^n + \omega^{2n} = \begin{cases}
 2&\text{if n is a multiple of 3}\, \\
 -1&\text{otherwise}\ \\
\end{cases}$

$\implies a_n = \begin{cases}
 (2) - (-1)^{2}\times 2&\text{if n is a multiple of 3}\, \\
 (-1) - (-1)^{-1}\times 2&\text{otherwise}\ \\
\end{cases}$

$\implies a_n = \begin{cases}
 (2) - 2&\text{if n is a multiple of 3}\, \\
 (-1) + 2&\text{otherwise}\ \\
\end{cases}$

$\implies a_n = \begin{cases}
 0&\text{if n is a multiple of 3}\, \\
 1&\text{otherwise}\ \\
\end{cases}$


问:
Union of generating sets is the generating set of the sum of the generated sets?
Let $(W_i)_{i\in I}$ be a (not necessarily finite) family of submodules of a $K$-module $E$. For each $i\in I$, let $S_i$ be the set that generates $W_i$. Does $\bigcup_{i\in I}S_i$ generate $\sum_{i\in I}W_i$?

Take $x\in\bigoplus_{i\in I}W_i$. Then $x_i=\sum_{s\in S_i}\alpha_{i,s}s$ for some $\alpha_i\in K^{(S_i)}$, for each $i\in I$. Thus $\sum_{i\in I}x_i=\sum_{i\in I}\sum_{s\in S_i}\alpha_{i,s}s$. But it's not clear how I can rewrite $\sum_{i\in I}\sum_{s\in S_i}\alpha_{i,s}s$ as a linear combination of elements of $\bigcup_{i\in I}S_i$. Is this something that's even possible? Does anything change if I replace "module" by "vector space"?

Edit:

In order to show that $\bigcup_{i\in I}S_i$ generates $\sum_{i\in I}W_i$, we have to demonstrate a $\xi\in K^{(\bigcup_{i\in I}S_i)}$ such that $y=\sum_{s\in\bigcup_{i\in I}S_i}\xi_s s$, for each $y\in\sum_{i\in I}W_i$.

Let $I:=\{1,2,3\}$. Set $S_1:=\{a,b\}$, $S_2:=\{a,b,c\}$ and $S_3:=\{a,d\}$. Write $x_1:=\eta_{1a}a+\eta_{1b}b$, $x_2:=\eta_{2a}a+\eta_{2b}b+\eta_{2c}c$ and $x_3:=\eta_{3a}a+\eta_{3d}d$. Take $y:=x_1+x_2+x_3$. Then
$$y=\eta_{1a}a+\eta_{1b}b+\eta_{2a}a+\eta_{2b}b+\eta_{2c}c+\eta_{3a}a+\eta_{3d}d.$$
This shows that $\eta$ can never be transformed into an element of $K^{S_1\cup S_2\cup S_3}$ because its domain has more elements than $S_1\cup S_2\cup S_3=\{a,b,c,d\}$. One solution is to use
$$S_1\times\{1\}\cup S_2\times\{2\}\cup S_3\times\{3\}$$
instead of $S_1\cup S_2\cup S_3$ but then that won't give us an element of $K^{S_1\cup S_2\cup S_3}$. What can be done?

答:
Yes, and you're on the right track for showing it!

You're completely right that we write $x$ as a (finite!) sum of $x_i \in W_i$. Then we write each $x_i$ as a (finite) linear combination of the generators in $S_i$.

But now $x$ is a linear combination of (linear combinations of the generators). But that's the same thing as one large (but finite) linear combination of the generators!

It might be helpful to write out this argument writing

$$
x =
(\alpha^1_1 s^1_1 + \ldots \alpha^1_n s^1_n) +
(\alpha^2_1 s^2_1 + \ldots \alpha^2_m s^2_m) +
\ldots +
(\alpha^k_1 s^k_1 + \ldots \alpha^k_l s^k_l)
$$

or something similar (importantly, not using $\sum$s). Here the $s^i$ are the generators for $W_i$.

Maybe that makes it more apparent that the $\{ s^i_j \}$ generate $\bigoplus W_i$?

I hope this helps ^_^


问:
MGF bound for non-centered random variables?
So I was trying to prove the following fact: if $|X| \leq 1$ is a random variable with mean $\mu$, then $\mathbb{E} e^{t (X - \mu)} \leq e^{t^2/2}$, for any real $t$.

I can show this in the special case $\mu = 0$:
for $x \in [-1, 1]$ we note by convexity
$$
e^{tx} \leq \frac{1-x}{2} e^{-t} + \frac{1 + x}{2} e^{t}.
$$
Hence, taking expectations: $\mathbb{E} e^{t (X -\mu)} \leq \cosh(t) \leq e^{t^2/2}$.

Any ideas how to extend this argument to general $\mu$?

答:
It suffices to show that for a random variable $Y \in [a,b]$ with $\mathbb E Y =0$, we have that $\mathbb E e^{tY} \leq e^{t^2(b-a)^2/8}$. Why? Let $Y = X - \mu$, then $\mathbb E Y = 0$, $b = 1 - \mu$, $a = -1 - \mu$, $(b-a)^2 = 4$, and the bound becomes $e^{t^2/2}$.

We use the same argument as you outlined. Fix a $t$. Using the convexity of $e^{ty}$, for any $y \in [a,b]$:
\begin{aligned}
e^{ty} \leq \frac{b-y}{b-a} e^{ta} + \frac{y-a}{b-a} e^{tb}.
\end{aligned}
Therefore, using the linearity of expectation and $\mathbb E Y = 0$, we get
\begin{aligned}
\mathbb E e^{tY} \leq \frac{be^{ta} - ae^{tb}}{b-a}  =: \varphi(t).
\end{aligned}
Make the following substitution: Let $\phi(t):= \log \varphi(\frac{t}{b-a})$ and $p = -a/(b-a)$.  We need to show that $\phi(t) \leq t^2/8$.

We have that $\phi(t) = -tp + \log(1 - p + pe^{t})$.

Notice that $\phi(0) = 0$, $\phi'(0) = 0$, and
\begin{aligned}\phi''(t) = \frac{(1-p)(pe^t)}{(1-p+pe^t)^2} \leq \frac{1}{4} ,\end{aligned}
where the last inequality holds for all $t$ by the AM-GM inequality.
Applying Taylor's theorem, we get that $\phi(t) \leq t^2/8$.

Remark: This is a very famous result called Hoeffding's Lemma. I followed the proof given on Wikipedia as it was the closet to your original approach. For alternate proofs, see, for example, Lemma 2.2 in the book Concentration Inequalities: A Nonasymptotic Theory of Independence
.