问:
Graph theory problem dealing with the Laplace operator(?).

Let $\Gamma=(V,E)$ where $V=\{v_1,v_2,v_3,v_4,v_5,v_6\}$ and $E=\{\{v_1,v_2\}, \{v_2,v_3\},\{v_2,v_4\}, \{v_3,v_5\}, \{v_4,v_6\} \}$. Let $f: V \to \mathbb{R}, f(v_j)=(-1)^j.$ Define $$(\Delta f)(x) = \begin{cases} \sum_{\{x,y\} \in E(x)} (f(x)-f(y)), E(x)\ne\emptyset \\ 0, E(x) = \emptyset\end{cases},$$ where $E(x)=\{\{x,y\} \in E \mid y \in V\}$. Compute the value of $\Delta f$ at each vertex. Show that $\Delta : F(V) \to F(V)$, $f\mapsto \Delta f$ is a linear transformation, where $F(V) = f:V \to \mathbb{R}.$

Wouldn't $E(x)$ be here exactly the same as $E$? If so I would have that $$\Delta f(v_1) = 0 \\ \Delta f(v_2) = (-1)^1-(-1)^2 = -2\\ \Delta f(v_3) =(-1)^2-(-1)^3 = 2 \\ \Delta f(v_4) =(-1)^2-(-1)^4 = 0 \\ \Delta f(v_5)=(-1)^3-(-1)^5 = 1 \\ \Delta f(v_6)=(-1)^4-(-1)^6 = 0$$

I'm not entirely sure I understood this corrrectly. Any feedback/hints would be appreciated.

答:
$E(x)$ is defined in the text of the exercise to be the set of edges with one vertex being $x$. Judging from your notation for the set of edges $E$ (to be precise: from writing edges as sets $\{u,v\}$ and not ordered tuples $(u,v)$) I guess we are dealing with undirected graphs. So in your example we get
$$\begin{align*}
E(v_1)&=\{\{v_1,v_2\}\}\\
E(v_2)&=\{\{v_2,v_1\}, \{v_2,v_3\},\{v_2,v_4\}\}\\
E(v_3)&=\{\{v_3,v_2\}, \{v_3,v_5\}\}\\
&...
\end{align*}$$
Hence we get by definition of $\Delta f$
$$\begin{align*}
\Delta f (v_1) &= (-1)^1 - (-1)^2 = -2\\
\Delta f(v_2) &= ((-1)^2-(-1)^1) + ((-1)^2 - (-1)^3) + ((-1)^2 - (-1)^4)\\&= 2 + 2 + 0 = 4\\
\Delta f(v_3) &= ((-1)^3 - (-1)^2) + ((-1)^3 - (-1)^5) = -2 + 0 = -2\\
&...
\end{align*}$$


问:
Help in two improper integrals
For a physics problem, I need to evaluate two integrals:
$$I_n=\int_{0}^{\infty} f(x) dx=\int_{0}^{\infty} \frac{x^2 \sin^2x}{(n^2\pi^2-x^2)^2} dx,$$ and
$$J_n=\int_{0}^{\infty} \frac{x^2 \cos^2x}{((n+1/2)^2\pi^2-x^2)^2} dx, n=1,2,3..$$
At Mathematica the value of these integral turns out to be $\pi/4$ which is independent of $n$.

$I_n$ has got a pole of order 2 at $x=n\pi$ and when I calculate the residue at $x=n \pi$  is zero. But on the other hand I find that  $\lim_{x \to n\pi} f(x)=0,$  so $x=n\pi$ may not be a pole of $f(x)$. I face the same problem in the case of $J_n$. Also $I_0=\pi/2$ is a standard integral. In any case, I want to evaluate both $I_n$ and $J_n$.  Please help me.

ContentLicense: CC BY-SA 4.0

Title: Help in two improper integrals

Body:
Let is use $$\int_{n\pi}^{\infty} \frac{\sin^2 t}{t^2} dt=\frac{\pi}{2}-\text{Ei}(2n \pi)$$
Then
$$I_n=\int_{0}^{\infty} \frac{x^2 \sin^2 x}{x^2-n^2\pi^2} dx$$
$$=\frac{1}{4}\int_{0}^{\infty} dx \left(\frac{\sin^2 x}{(x-n\pi)^2}+\frac{\sin^2 x}{(x+n\pi)^2}+\frac{\sin^2 x}{4n\pi(x-n\pi)}-\frac{\sin^2 x}{4n\pi(x+n\pi)}\right)$$
$$I_n=\frac{1}{4}[\pi/2-\text{Ei}(2n\pi)+\pi/2+\text{Ei}(2n\pi)]+\int_{-n\pi}^{n\pi} \frac{\sin^2 t}{t}dt=\frac{\pi}{4} $$
Similarly
$$J_n=\int_{0}^{\infty} \frac{x^2 \cos^2 x}{x^2-m^2\pi^2} dx, m=n+1/2$$
$$=\frac{1}{4}\int_{0}^{\infty} dx \left(\frac{\cos^2 x}{(x-m\pi)^2}+\frac{\cos^2 x}{(x+m\pi)^2}+\frac{\cos^2 x}{4m\pi(x-m\pi)}-\frac{\cos^2 x}{4m\pi(x+m\pi)}\right)$$
$$I_n=\frac{1}{4}[\pi/2-\text{Ei}(2m\pi)+\pi/2+\text{Ei}(2m\pi)]+\int_{-m\pi}^{m\pi} \frac{\sin^2 t}{t}dt=\frac{\pi}{4} $$


问:
Find Coordinates of Intersection Between Hypotenuse of Right Triangle and Circle
Example 1: Big Triangle Intersecting Circle

Example 2: Small Triangle Intersecting Circle

1.
I am trying to get the coordinates of an intersection between a hypotenuse of a right triangle and a circle. I need an algorithm or equation that can give me the intersection coordinates using the radius of the circle and length of the 2 sides of the right triangle. I am completely stumped, but if I figure out the equation/algorithm I need, I will share it here.

2.
It is very difficult to explain exactly what I need so I have provided 2 pictures as examples. I need to find the equation/algorithm for both of these examples.

The 1st example shows a triangle that is bigger than the circle and intersects the circle. I need the coordinates for the intersection of the hypotenuse and the circle. Radius of the circle is 1. Sides of the triangle are 2 and 3.

The 2nd example shows a triangle that is smaller than the circle and I have drawn a dashed line that intersects with the circle. I need the coordinates for the intersection of the dashed line and the circle. Radius of the circle is 1. Sides of the triangle are 0.4 and 0.2.

3.
Below is unnecessary information that describes the project I'm working on that requires this solution:

I am trying to generate a common x,y value for an angle from a center point. The only values I can get are the x and y distance between point A and point B. This is for a game I'm developing. I am trying to add a force that has to be expressed as x,y values. The issue is that the magnitude is inversely proportional to distance, with exponential loss of force over distance. I basically need the common x,y coordinates, then I can apply the magnitude that I already have calculated. This is the best way I've come up with for solving this issue.

THANKS FOR ANY AND ALL HELP! ITS GREATLY APPRECIATED! i wish i paid more attention in school XD

EDIT:

Thanks to A-Level Student, I was able to get the equations I need for the intersection. These equations apply to both example 1 and 2.

$x_2$ = x of intercept, $y_2$ = y of intercept, $x_1$ = side width of triangle, $y_1$ = side length of triangle

Get X of the intercept with the following:

$$x_2 = \sqrt{1 \div ((\frac{y_1}{x_1})^2 + 1)}$$

Get Y of the intercept with the following:

$$y_2 = \sqrt{1 \div ((\frac{x_1}{y_1})^2 + 1)}$$

Please note that these equations only work for a circle with a radius of 1, since $x^2 + y^2 = r^2$.

Also, I would like to note that this formula was used in Unity (C# scripts) to essentially apply a force to a 2D object opposite to where a location was pressed on the screen, relative to the distance between the object and the touched position. (You can think of this as a 2D force that is applied to an object by a blast wave of a bomb. Essentially what I'm doing.) Also, this will only ever return positive values, so I had to add a conditional statement to apply the correct negative values. The if statement looks something like this: if (x_1 > 0) { //perform x_2 equation multiplied by negative-1}, and same for y_1 and also some logic for if x_1 or y_1 is equal to 0, so you don't get a null value. But everything worked. Hope this helps anyone else who comes across this problem as I did.

答:
Begin by finding the equations for each relevant curve and solving the simultaneous equations.

Question 1

The circle has equation
$$x^2+y^2=1$$
and the equation of the line that the hypotenuse is part of is
$$y=\frac{3}{2}x$$
Hence,
$$x^2+\left(\frac{3}{2}x\right)^2=1\implies \frac{13}{4}x^2=1$$
so $$x^2=\frac{4}{13}$$
Can you solve it from there? Watch out though: this equation gives you two solutions, but you are only interested in one of them.

Question 2

The circle has equation
$$x^2+y^2=1$$
and the line that includes the hypotenuse has equation
$$y=0.5x$$
Again, try solving these simultaneous equations.

If you need any more help please don't hesitate to ask :)


问:
Is a Cauchy sequence bounded in a metric space if the space is not complete?
I don't think a metric space needs to be complete for a Cauchy sequence to be bounded. Being complete means the sequence converges to a point in the space, but the standard proof for Cauchy sequence being bounded does not require you to know what the sequence converges to. For that reason, I think it doesn't matter whether the metric space is complete or not.</p>

答: A more conceptual way to see it: Let $X$ be a metric space, $x_n$ a Cauchy sequence in $X$, and let $\overline X$ be its completion. We can isometrically embed $X$ into $\overline X$, and the image $\overline x_n$ of the sequence $x_n$ will still be a Cauchy sequence, since its elements still have the same distance from each other. But this sequence converges, since we're in a complete space, so it must be bounded. And since the distances between the $x_n$ are the same as the ones between the $\overline x_n$, $x_n$ must also be bounded.


问:
Help with combinations problem?

Initially there are $m$ balls in one bag, and $n$ in the other, where $m,n>0$. Two different operations are allowed:

a) Remove an equal number of balls from each bag;

  b) Double the number of balls in one bag.

Is it always possible to empty both bags after a finite sequence of operations?

Operation b) is now replaced with

b') Triple the number of balls in one bag.

Is it now always possible to empty both bags after a finite sequence of operations?

This is question 4 on Round $1$ of the $2011/2012$ British Mathematical Olympiad.

I suck at combinatorics and the like but need to practise to try and improve my competition mathematics. If anyone could give me a hint on where to start I'd be most grateful :D

EDIT: Never mind guys, I just completely mis-read the question. I thought it said you had to double the numbers of balls in both bags. Thanks for the help!

答:
Regarding the first part...

Let $m>n$

Remove $n-1$ balls from each bag so that you have $m-n+1$ balls in one bag and $1$ ball in the other bag.

Now repeat the algorithm of doubling the balls in the bag which has $1$ ball and then taking away $1$ from each bag till you have $1$ ball in each bag. Finally remove $1$ ball from each bag and you have emptied them in finite number of steps.

For the second part

Again suppose $m>n$

Remove $n-1$ balls from each bag so that you have $m-n+1$ balls in one bag and $1$ ball in the other bag.

Now repeat the algorithm of tripling the balls in the bag which has $1$ ball and then taking away $2$ from each bag. If $2|(m-n)$, then you'll end up with $1$ ball in each bag in some steps. Otherwise you'll end up with $2$ balls in one bag and $1$ ball in the other bag and it seems like no matter what you do from here, you'll always end up with this arrangement.

Hence, in the second part, it may be conjectured that the bags cannot be emptied in finite steps if $m-n\equiv 1\pmod2$.

ContentLicense: CC BY-SA 3.0

Title: Help with combinations problem?


问:
ِIs it correct reasoning?
I am reading my friend explanation for a question , in proof of one theorem he writes that A1,A2,..⊆X
then closure of ∪ An
n⩾1 = ∪ closure of An
n⩾1.
I thing that its not correct but I cant find any counterexample ? Am I right ?does anybody can give me a counter example ? thanks

答: Here is a counterexample: Take $X=\Bbb R$, and $A_n=[0,1-\frac{1}{n}]$. Then $$\overline{\bigcup_{n\geq1} A_n}=\overline{[0,1)}=[0,1]$$ but $$\bigcup_{n\geq1} \overline{A_n}=\bigcup_{n\geq1} {A_n}=[0,1)$$

Title: ِIs it correct reasoning?


问:
linear combination of two factors of minimal polynomial is $1$

Let $K$ be a field and $V$ a finite dimensional $K$-vector space. Let
$A$ be a linear transformation of $V$ with minimal polynomial
$\phi_A$. Let $\phi_A = \phi_1 \cdot \phi_2$ with $\phi_1, \phi_2 \in K[X]$ monic and relative prime.

There exist polynomials $g,h \in K[X]$ so that $g \phi_1 + h \phi_2 = 1$.

I do not see the logic behind this? Why is this the case? I suppose it is because the product of $\phi_1$ and $\phi_2$ is the minimal polynomial for $A$, but I do not see entirely why.

答:
As you tagged abstract-algebra I hope the following is understandable to you. Otherwise feel free to ask for clarification!

In a principal ideal domain $R$ elements $a,b\in R$ are relatively prime, iff their gcd is 1. As ideals we have the identity $(a)+(b)=(\gcd(a,b))$. This is equivalent to the condition that $\gcd(a,b)$ is a unit, wlog 1. By definition of $(a)+(b)$ we thus have $ka + lb = 1$ for some $k,l \in R$.

If $K$ is a field then $K[X]$ is a principal ideal domain. So having polynomials $\phi_1,\phi_2$ being relatively prime, we find polynomials $g,h$ such that $g\phi_1 + h\phi_2=1$.


答:
For the first problem, if $m=n$, just take everything. Otherwise, apply operation (a) to reduce the smaller bag to a single ball, and let $a$ be the number of balls in the other bag at that point. Double the $1$ and take $1$ ball from each bag, so that you now have $a-1$ balls in one bag and $1$ in the other. Repeat until you have one ball in each bag and then empty the bags with operation (a).

For the second question I conjecture that $m=1$, $n=2$ is a position from which the bags cannot be emptied, though I don’t immediately see an argument.

Added: The conjecture is correct, and the other answer was on the right track. Suppose that $m$ and $n$ have opposite parity.Subtracting the same amount from each leaves two numbers of opposite parity, and multiplying a number by $3$ does not change its parity, so you always have two numbers of opposite parity. In particular, you cannot empty both bags.

Title: Help with combinations problem?

Body:
Initially there are $m$ balls in one bag, and $n$ in the other, where $m,n>0$. Two different operations are allowed:

a) Remove an equal number of balls from each bag;

  b) Double the number of balls in one bag.

Is it always possible to empty both bags after a finite sequence of operations?

Operation b) is now replaced with

b') Triple the number of balls in one bag.

Is it now always possible to empty both bags after a finite sequence of operations?

This is question 4 on Round $1$ of the $2011/2012$ British Mathematical Olympiad.

I suck at combinatorics and the like but need to practise to try and improve my competition mathematics. If anyone could give me a hint on where to start I'd be most grateful :D

EDIT: Never mind guys, I just completely mis-read the question. I thought it said you had to double the numbers of balls in both bags. Thanks for the help!


问:
Intersection between two parametric cubic functions?
If I have two parametric cubic functions in 2D space defined by
$$\vec r(t) = \vec A_1t^3 + \vec B_1t^2 + \vec C_1t + \vec D_1$$
and
$$\vec q(t) = \vec A_2t^3 + \vec B_2t^2 + \vec C_2t + \vec D_2$$
how can I find all the points of intersection $$\vec P_0, \vec P_1, ..., \vec P_n$$ between them?

答:
In principles, if $r_x,r_y, q_x, q_y$ are the coordinates of $\vec r, \vec q$, you have to find $(t_1, t_2) \in \mathbb R^2$ such that

$$\begin{cases}
r_x(t_1)&= q_x(t_2)\\
r_y(t_1)&= q_y(t_2)
\end{cases}$$

You have two variables and two equations... which looks good. This is the theory in a couple of lines. Now, if you're interested in solving the problem with a computer, I suggest that you Google

planar curves intersection algorithm computer aided design

as it is a very common topic in computer aided design.

You may have a look to open source package named OPEN CASCADE.


答:
Method 1:

You can convert one of the cubics to implicit form (a cubic equation) and plug the parametric equation of the second. Solve the resulting nonic polynomial equation and check that $t$ is in the useful range.

If you survive, you get a medal.

Method 2:

Convert the cubics to Bezier form and use the hull property to design a recursive split algorithm: if the two bounding boxes* of the control points overlap, split both curves (using the De Casteljau formula) and repeat with the four combinations. Do this recursively until the boxes are small enough (of finish with Newton).

*In principle you should use the convex hulls, but the extra complexity of constructing and intersecting the hulls is not worth the effort.

Title: Intersection between two parametric cubic functions?

Body:
If I have two parametric cubic functions in 2D space defined by
$$\vec r(t) = \vec A_1t^3 + \vec B_1t^2 + \vec C_1t + \vec D_1$$
and
$$\vec q(t) = \vec A_2t^3 + \vec B_2t^2 + \vec C_2t + \vec D_2$$
how can I find all the points of intersection $$\vec P_0, \vec P_1, ..., \vec P_n$$ between them?


问:
Continuous manifold dimension
First time encountering manifolds in general.

A smooth manifold $M$ is said to be of dimension $n$ if its (complete) atlas is of dimension $n$.

Now, what if we replace smooth by continuous ? Does the dimension is even well defined ?

I know from this that in case of smooth manifolds, there exists invariance of domain which means that there does not exist two U,V open sets of $M$ such that $U$ is homeomorphic to $\mathbb{R}^n$ and $V$ is homeomorphic to $\mathbb{R}^m$ with $m<n$.
It all boils down to the fact that there does not exist an homeomorphism between an open subset of $\mathbb{R}^n$ and an open subset of $\mathbb{R}^m$.

I do not know if this holds in the case of continuous maps.

答: The proof is exactly the same as in the smooth case. Invariance of domain doesn't use smooth maps, but continuous ones, and smooth maps happen to be continuous (in fact, smooth charts are defined to be homeomorphisms), which is why they also work. The real reason why invariance of domain can be applied is that smooth charts are homeomorphisms. But continuous charts are too, by definition, so the same reasoning applies.

Title: Continuous manifold dimension


问:
How to use Hall's theorem to show a necessary and sufficient condition for some problem (modified marriages problem)?
I undestand Hall's theorem but I'm not sure how to apply it to problems. I have such problem which I believe needs to be solved using Halls' theorem.

Problem:

We have a group of n girls and m boys. Show that a necessary and sufficient condition for k girls to find a husband (inside the group) is that each r girls know minimum k+r-n boys.

This almost screams to use Hall's theorem but I'm not sure how to apply it.

答:
Hall's theorem tells you something about finding a matching of size $n$ (i.e. a husband for all $n$ girls), while here you want to consider matchings of size at least $k$. The trick for applying Hall's theorem in this situation is to add $n-k$ new "boy" vertices, each of them connected to every "girl" vertex.

You should first check that a matching of size at least $k$ in the original graph can be extended to a matching of size $n$ in this new graph, and conversely, a matching of size $n$ in this graph gives a matching of size at least $k$ in the original graph. Then you can verify that the necessary and sufficient condition given is equivalent to Hall's condition in this "augmented" graph.

Do tell me if you need more help!


问:
Understanding how different groupings of terms in an Infinite series can lead to different answers.
One of the most, conceptually speaking, for me to understand is the topic of infinite series. I have always had a hard time proving that an infinite series diverges or even finding a solution for the sum of infinite series. So when I came across a video by Mathologer titled "Riemann's paradox: pi = infinity minus infinity", my mind had a mini-meltdown. In the video he mentions Reimann's Rearrangement Theorem, whereby the sum of the infinite series $1- \frac{1}{2}+\frac{1}{1}-\frac{1}{4}+...$ can equal whatever you want it to, depending on how you group the terms.

Even though in the video he explains how that is possible, I cannot for the life of me understand how different groupings lead to different answers. All my life I thought that no matter how you arrange the terms, because the series is infinite the sum will eventually be the same - no matter how you group them. Obviously, that's wrong but how can I understand this phenomenon better?

答: I don't know if this is exactly what you are looking for but it is a "mistake" to see an infinite series as an infinite addition. If this were the case, then such behaviors would be problematic since the addition is commutative as you know. So, you should better remember that an infinite series is in fact the limit of the sequence of partial sums. Now, if you group differently the terms, then the sequence of partial sums will be very different from the previous one and a priori, there is no reason that it will even converge  and if it does, there is no reason for it to converge to the same value.</p>


问:
Is log-likelihood bounded for continuous variables?
Let $(x_i,y_i)$, $i=1,...,N$, be an i.i.d. dataset. Suppose we have a  parametric distribution $f(y|x,\omega)$, parameterized by $\omega$, and we want to find the value $\omega^*$ such that $f(y|x,\omega^*)$ is the best approximation of the true distribution $p(y|x)$.

A usual criterion is the maximization of the likelihood of the $y_i$'s conditioned on the $x_i$'s, i.e.:

$\newcommand{\argmax}{\mathop{\mathrm{arg\,max}}}$
$\omega^* = \argmax_{\omega} f(y_1,...,y_N|x_1,...,x_N,\omega) =\argmax_{\omega} \prod_{i=1}^{N} f(y_i|x_i,\omega)$.

(the equality above holds due to the assumption that the examples in the dataset are independent)

For practical reasons, it is usual to work instead with the following (equivalent) formulation:

$\newcommand{\argmin}{\mathop{\mathrm{arg\,min}}}$
$\omega^* = \argmin_{\omega} \sum_{i=1}^{N} -\log(f(y_i|x_i,\omega))$

(these formulations are equivalent because the $\log(\cdot)$ function is monotonic)

For convenience, let me define $L(\omega)=\sum_{i=1}^{N} -\log(f(y_i|x_i,\omega))$ and call it a loss function.

If the $y_i$'s are discrete random variables, then $f(y|x,\omega) \leq 1$ and, therefore, $L(\omega) \geq 0$ for all $\omega$, so our loss function is always non-negative and so it is lower bounded.

If, instead, the $y_i$'s are continuous random variables, then $f(y|x,\omega)$ is a pdf and so it is not necessarily less than $1$ (although it integrates to $1$ w.r.t. $y$). Under these circumstances, is it possible to ensure that $L(\omega)$ is lower bounded (i.e. that the optimization problem actually has a solution regardless of the model family)? Is there any obvious lower bound for it?

Thank you in advance.

答:
There is no lower bound in general. Here is a counter-example !

Consider a single point $x_0$ and choose to model it using a Gaussian distribution with mean $x_0$ and an unknown variance $\sigma^2$.

The negative log-likelihood writes
\begin{equation}
\ell(\sigma,x_0) = \log\left(\sqrt{2\pi}\sigma\right)
\end{equation}
and has no lower bound.

This is the reason why it is important to choose appropriate models and to regularize your model using priors. For example, using a Frechet prior with $m=0$, $s=1$ and $\alpha=2$ for the standard deviation then the negative log-posterior writes
\begin{equation}
\ell_p(\sigma,x_0) = \log\left(\sqrt{2\pi}\sigma\right) + \log\left(\frac{\sigma^3}{2}\right) + \frac{1}{\sigma^2}.
\end{equation}
One can show that $\ell_p$ has a lower bound ($\sigma > 0$).


问:
Ordinary Differential Equation Solution.
I can't for the life of me figure out how to solve this ODE I found. It doesn't seem complicated and should have a solution but I have tried most online ODE solvers but they always return a "cannot compute solution". I know this is a dumb place to ask but I'm gonna have a hard time sleeping if I can't figure it out. It doesn't seem to be separable either.

The problem is:

$$2yy'= x-y+y^3$$

Even just an answer would be helpful,
Cheers.

答:
Consider first the equation
$$2yy'= -y+y^3$$ Switch variables to make
$$2\frac y{x'}=-y+y^3 \implies x'=\frac{2}{y^2-1}\implies x+C=\log \left(\frac{1-y}{1+y}\right)$$ that is to say
$$y=\frac{1-K e^x}{1+K e^x}$$ Now, trying the variation of parameters, this would give $\color{red}{\text{No hope}}$


问:
Linear algebra classic, Farkas lemma application
$A \in M_{m \times n}(\mathbb{R})$ and $b \in \mathbb{R}^m$.
Farkas' lemma says exactly one of the following holds:

(a) there exists some $x \in \mathbb{R}^n$, $x \geq 0$, such that $Ax = b$

(b) there exists some vector $p \in \mathbb{R}^m$ such that $p^TA \geq 0$ and $p^Tb < 0$.

where $0$ means all-zero vector of dimension $n$ and comparing two vectors means comparing all components of the vectors.

Let $x_1,x_2,\dots,x_m \in \mathbb{R}^n$. My task is to derive from Farkas’ Lemma that exactly one of the following holds:

(a) $0 \in$ convex.hull$\{ x_1,x_2,\dots,x_m \}$

(b) there is a row vector $y \in \mathbb{R}^n$ such that $yx_i > 0$ for $i = 1, . . . , m$.

I tried but cannot proceed. From the first you can deduce that $x_i$ are linearly dependent  but then i am stuck since i don't see how to apply the lemma :(

答:
$\def\Mat{\mathrm{Mat}}\def\R{\mathbb R}$Let $A \in \Mat_{n+1,m}(\R)$ given by
$$A = \begin{pmatrix} x_1 & x_2 & \cdots & x_m\\ 1 & 1 & \cdots & 1 \end{pmatrix} $$
and $b \in \R^{n+1}$ by $b = \binom{0_n}1$. Then, by Farkas,

either (a) for some $\lambda\in\R_{\ge 0}^m$, we have $A\lambda = b$, which is equivalent to $\sum_{i=1}^m \lambda_i x_i = 0$, $\sum_i \lambda_i = 1$, that is $0 \in \mathrm{conv}\{x_1, \ldots, x_m\}$.

or (b) for some $y = (\bar y, y_{n+1})^t \in \R^{n+1}$ we have $y^TA \ge 0$ that is $\bar y^t x_i + y_{n+1} \ge 0$ for each $i$, and $y^t b = y_{n+1} < 0$. So this holds, if for each $i$ $\bar y^t x_i \ge -y_{n+1} > 0$.


答:
A proof of the fact you talk about can be found here https://en.wikipedia.org/wiki/Riemann_series_theorem#Existence_of_a_rearrangement_that_sums_to_any_positive_real_M.

The essential idea is that if the series converges but not absolutely, then its positive terms add up to $\infty$, and the negative ones add to $-\infty$. Thus, at any point you can increase or decrease the sum as much as you want if you add the terms in the correct order, making it converge to any number you want.

Title: Understanding how different groupings of terms in an Infinite series can lead to different answers.

Body:
One of the most, conceptually speaking, for me to understand is the topic of infinite series. I have always had a hard time proving that an infinite series diverges or even finding a solution for the sum of infinite series. So when I came across a video by Mathologer titled "Riemann's paradox: pi = infinity minus infinity", my mind had a mini-meltdown. In the video he mentions Reimann's Rearrangement Theorem, whereby the sum of the infinite series $1- \frac{1}{2}+\frac{1}{1}-\frac{1}{4}+...$ can equal whatever you want it to, depending on how you group the terms.

Even though in the video he explains how that is possible, I cannot for the life of me understand how different groupings lead to different answers. All my life I thought that no matter how you arrange the terms, because the series is infinite the sum will eventually be the same - no matter how you group them. Obviously, that's wrong but how can I understand this phenomenon better?


问:
Lipschitz implies bounded gradient with any norm?
Assume I have a differentiable function $f: \mathbb{R}^n \rightarrow \mathbb{R}$ that is $L$-Lipschitz with respect to a norm $\|\cdot\|$ on $\mathbb{R}^n$:
$$ \forall x, y \in \mathbb{R}^n, |f(x) - f(y)| \le L\|x-y\|.  $$
I know that if $f$ is further assumed to be convex, and $\|\cdot\|$ is the euclidian norm, then convexity implies
$$ L\|\nabla f(x)\|_2 \ge f(x + \nabla f(x)) - f(x) \ge \langle \nabla f(x), \nabla f(x)\rangle = \| \nabla f(x)\|_2^2, $$
thus $\|\nabla f(x)\|_2 \le L$.

But does $\|\nabla f(x)\| \le L$ still hold if $\|\cdot\|$ is any norm on $\mathbb{R}^n$? And if $f$ is only assumed differentiable?

答:
Recall that for a differentiable function $f : \mathbb{R}^n \to \mathbb{R}$, its gradient is defined thanks to the euclidean structure of $\mathbb{R}^n$: it is the only vector field such that for all $x$ and $u$
$$
\mathrm{d}_xf\cdot u = \langle \nabla f(x),u\rangle.
$$

In what follows, all gradients $\nabla f$ are the classic euclidean gradients defined thanks to the usual euclidean structure.

An important result is that if $f$ is Lipschitz for some norm, then it is almost everywhere differentiable, and is Lipschitz for any norm. Moreover, the euclidean gradient is then defined almost everywhere.

Notice that for the euclidean norm, one has $\|\nabla f(x)\|_2 = \|\mathrm{d}_xf\|_2$, where this second norm is the operator norm subordinate to the euclidean one: it is because by Cauchy-Schwarz inequality:
$$
|\mathrm{d}_xf\cdot u| = |\langle \nabla f (x),u\rangle| \leqslant \|\nabla f (x) \|_2 \|u\|_2,
$$
hence $\|\mathrm{d}_xf\| \leqslant \|\nabla f(x)\|_2$, and equality is reached for $u = \nabla f(x)$. So saying $f$ is $L$-Lipschitz show that $\|\nabla f\|_2 \leqslant L$ almost everywhere.

But if you consider another norm on $\mathbb{R}^n$, there is no reason for (the above vector field) $\nabla f$ to stay bounded with the same constant: indeed, the equivalence of norms in a finite dimensional vector space assures that for any norm $\|\cdot\|$, there exists a constant $C>0$ such that
$$
\|\nabla f(x)\| \leqslant C\cdot \|\nabla f (x) \|_2.
$$
A counter example to having the same constant $L$ is just the following: suppose $f(x) = \langle u,x \rangle$ for a given $u \in \mathbb{R}$ and for $\langle\cdot,\cdot\rangle$ the usual inner product. Then $\nabla f \equiv u$ on $\mathbb{R}^n$, hence $\|\nabla f \|_2 = \|u\|_2$. But if one measures $\nabla f$ thanks to $\|\cdot\|_{\lambda}=\lambda \|\cdot \|_2$ with $\lambda >0$, one has
$$
\|\nabla f \|_{\lambda} = \lambda \|u\|_2
$$
which is strictly greater than $L = \|u\|_2$ if $\lambda >1$.

To conclude, note that that I never used any convexity assumption, but I used only the Lipschitz one.


问:
Is "find all extreme values" the same as finding all the critical points?
The instructions state to "find all extreme values and their type" for the function $f(x, y) = xy(5x + y - 15)$. So I went through the steps using partial derivatives, and the resulting critical points are $(0, 0)$, $(3, 0)$, $(0, 15)$ and $(1, 5)$. But when I re-read the question, am I even answering it correctly or am I incorrectly assuming that finding all extreme values refers to finding all the critical points?

If I go to Wolfram Alpha and type in extrema xy(5x + y - 15), the result it gives is "Local Minimum": min{x y (5 x + y - 15)} = -25 at (x, y) = (1, 5).

So in other words, it only shows one of the critical points that I found when I use the command extrema. But my example definitely asks for all extreme values, so now I'm just confused about what the question wants. When the question says "find their type", I'm guessing they mean whether it is a local minimum, maximum etc? Thanks for your ideas.

答:
The critical points of a differentiable function $\mathbb R^n\to\mathbb R$ are all points with vanishing gradient, while extreme points only include local maxima and local minima.

For example the map $\mathbb R^1\to\mathbb R$ given by $x\mapsto x^3$ has a critical point at $x=0$ that does not yield a local maximum or minimum.


答:
The intuitive reasoning is this: if a (real) series is convergent, but not absolutely so, then both the series consisting only of its negative terms and the series consisting only of its positive terms must diverge. If only one of them diverged, then the whole thing would have to diverge, too, and if none of them diverged, then the series of the absolute values would just converge to the sum of the absolute values of the two "subseries".

Now choose whatever value $x$ you want the rearrangement to converge to. For this, take the first few positive terms until you're above $x$. Then take the first few negative terms until you are below $x$. Take the next few positive terms until you are again above $x$, then the next few negative terms until you're again below $x$. Repeat ad infinitum, so you kinda oscillate around $x$, but getting closer to $x$ with every swing. This works because the negative and positive parts diverge, so you can always find enough terms to get you above or below the treshold, but the series itself converges, so the terms get small enough that you "overshoot" less and less with each swing.


问:
Prove or confute the follow proposition
I have to prove or confute the follow proposition:

Let $F$ a field and $ f: $$\mathbb Z \rightarrow F$ an homomorphism of rings such that $f(1_\mathbb Z ) =1_F$.

If $f$ isn't injective $\Rightarrow$ $F$ is finite

This is my work :

$f$ is an homomorphism of rings $\Rightarrow Ker f $ is an ideal of $\mathbb Z \Rightarrow $ $Ker f = n \mathbb Z $

$n \in \mathbb N$ is uniquely determined and is the characteristic of the field $F$

Such that $f$ is not injective $\Rightarrow Ker f \neq  (0)$ then $n \neq 0$

Now I can't conclude that $F$ is infinite neither that is finite.

If I take $\mathbb Z_2(x)$ I know that is a field infinite with characteristic 2, so I found an infinite field with finite characteristic.

Is it correct? I forgot something important to say?

How can I prove that $\mathbb Z_2 (x) $ is an infinite field?

答: The field $\Bbb Z_2(x)$ is an infinite field since $1,x,x^2,x^3\ldots\in\Bbb Z_2(x)$. And this proves that the statement is false.

Title: Prove or confute the follow proposition


问:
Constructing a conformal map from $\mathbb{D}$ to a cut plane
Source: Oxford Exam $A2 \ 1999$

We want to construct a conformal map $F$ from the unit disc $\mathbb{D}=\{z:|z|<1\}$ to $\mathbb{C} \setminus S$ where $S$ is the half-line $\{x+i:x \in (-\infty,0] \}$ with the additional property that $F(0)=0$.

Here are my thoughts, can someone see if they are correct please:

Use the Möbius transformation $f_1:z \mapsto \frac{1-z}{z+1}$ to send $\mathbb{D}$ to the right half plane.

Use the map $f_2:z \mapsto z+ \alpha$ where $\alpha \in \mathbb{C}$ we will determine later. This simply takes the right half plane to the right half plane.

Use the map $f_3:z \mapsto z^2$ to map the right half plane to $\mathbb{C}$ with a cut along the negative reals.

Use the map $f_4:z \mapsto z+i$ to move the cut up and map to $\mathbb{C} \setminus S$.

Now the composition of these four maps is conformal so we certainly have a conformal map between the sets.

We now need $F(0)=f_4 \circ f_3 \circ f_2 \circ f_1(0)=0$.

$f_1(0)=1$, $f_2(1)=1+\alpha$, $f_3(1+\alpha)=\alpha^2+2\alpha+1$, $f_4(\alpha^2+2\alpha+1)=\alpha^2+2\alpha+1+i$

So we need to solve $\alpha^2+2\alpha+1+i=0$ and this is the $\alpha$ required to ensure $F(0)=0$.

We can find that $\alpha=\sqrt{-i}-1$.

Thus we are done.

Is this correct?

答:
Your choice of $f_2$ is bad. We need a conformal map from the right-half plane to the right-half plane, that sends $f_1(0) = 1$ to $f_3^{-1}\circ f_4^{-1}(0) = f_3^{-1}(-i) = (1-i)/\sqrt 2$.

Your choice of a translation sends the right-half plane to one of its translate to the left, which is no longer the right-half plane, and then the square map messes up everything.

Since the automorphisms of the top-half plane are of the form $z \mapsto (az+b)/(cz+d)$ with $ad-bc > 0$, the conformal map we need is of the form $z \mapsto (az-ib)/(icz+d)$ with $ad-bc > 0$.
For example, $f_2(z) = (z-i)/\sqrt 2 $ is one map that works, and it is very easy to see that it does map the right-half plane to itself.


问:
Olympiad functional equations proof verification: $f\bigl(2f(x)+f(y)\bigr)=2x+y$ on $\mathbb Q$ and $f\bigl(xf(x)+f(y)\bigr)=f(x)^2+y$ on $\mathbb R$
First, let me be clear: I have absolutely no experience with olympiad functional equations, but I will start with a book in the very near future.

OK, but often, when I'm bored, I read solutions to functional equations on AOPS and/or YouTube, and often I find them interesting. This one time, I found a couple of problems (I don't exactly remember where, but they were from some early 2000's national olympiads), and I ran into the same issue with both these problems.

Find all functions $ f : \mathbb Q \to \mathbb Q $ such that
$$ f \bigl ( 2 f ( x ) + f ( y ) \bigr ) = 2 x + y \text . $$

Find all functions $ f : \mathbb R \to \mathbb R $ such that
$$ f \bigl ( x f ( x ) + f ( y ) \bigr ) = f ( x ) ^ 2 + y \text . $$

Here are my solutions:

Let $ P ( x , y ) $ denote the assertion that $ f \bigl ( 2 f ( x ) + f ( y ) \bigr ) = 2 x + y $.
$$ P ( x , x ) \implies f \bigl ( 3 f ( x ) \bigr ) = 3 x \label i \tag i \text . $$
$$ P \bigl ( 3 f ( x ) , 3 f ( y ) \bigr ) , \eqref{i} \implies f \bigl ( 6 f ( x ) + 3 f ( y ) \bigr ) = 6 f ( x ) + 3 f ( y ) \text . $$
Now letting $ a = 6 f ( x ) + 3 f ( y ) $, we have $ f ( a ) = a $.

Let $ Q ( x , y ) $ denote the assertion that $ f \bigl ( x f ( x ) + f ( y ) \bigr ) = f ( x ) ^ 2 + y $.
$$ Q ( 0 , 0 ) \implies f ^ 2 ( 0 ) = f ( 0 ) ^ 2 \text , $$
where $ f ^ 2 ( t ) $ is $ f \bigl ( f ( t ) \bigr ) $.
$$ Q ( 0 , y ) \implies f ^ 2 ( y ) = f ( 0 ) ^ 2 + y \text . $$
$$ Q \bigl ( 0 , f ( y ) \bigr ) \implies f \left ( f ( 0 ) ^ 2 + f ( y ) \right ) = f ( 0 ) ^ 2 + f ( y ) \text , $$
using the previous two relations.

Now letting $ t = f ( 0 ) ^ 2 + f ( y ) $, we have $ f ( t ) = t $.

For both of the questions $ f ( x ) = x $ was indeed a valid solution, but by just messing around I saw $ f ( x ) = - x $ is also a valid solution for both.

I'm not sure if my solution is wrong or correct but I think one possible reason it may be wrong is that letting $ a = \dots $ and $ t = \dots $ step because not all numbers in the domain can be written as that, that is my guess.

Please let me know where my solution is wrong, and also I would highly appreciate it if you could tell me some resource a novice like myself can start from.

Thanks!

Body:
Just after (i),  $P(3f(x),3f(y))=P(6x+3y)$, not $P(6f(x)+3f(y))$

$f(f(0)^2+f(y))$ should be $f(f(0)^2+y)$


问:
Work team sitting at a table question

A question  came up recently:

There is a team sitting at a round table. They all leave the table
for a lunch break, and after returning to the table, each person finds that
both their neighbours on the right and on the left do not sit next to
them anymore. How many people may be sitting around the table?
Describe all the possibilities.

Personally, I have no idea how to prove this mathematically -- all I know, after trial and error, that 1, 2, 3, and 4 don't work and 5 does. Is there a way I can prove this, and show (if they are) that greater numbers are possible?

答:
The question boils down to this: is there a Hamiltonian cycle through the complement of the cycle graph of n vertices?

For $n < 5$ the answer is obviously no: the complements here are the empty graph for $n = 2$ and $3$, and two separate $K_2$ graphs for $n = 4$.

For $5$, the answer is yes, and there is essentially only one way to do it: if our people are $ABCDE$ around the table originally, $ACEBD$ is the only layout that works.

For $6$ and higher, we can determine the yes-no answer as simply this: is it always true that the $n$th person can push a chair in in a place that isn't next to one of their former neighbors?  And the answer to this is obviously yes: the two neighbors only block off the four spots immediately next to them, and at more than $4$ seats already placed, there are more than $4$ spots to push in a new chair!

Title: Work team sitting at a table question


问:
Find Maclaurin series and radius of convergence for $f(x)=\int_{0}^{x} \frac{tdt}{(1+t)^2}$
Given $f(x)=\int_{0}^{x} \frac{tdt}{(1+t)^2}$, find  the Maclaurin series and radius of convergence.

My attempt:

By considerably simple computation we get that

$f(x)=\int_{0}^{x} \frac{tdt}{(1+t)^2} = \ln(1+x) -\frac{1}{(x+1)}$, for $t \ge 0$,

Now what I'm confused about is that the Maclaurin series of $\displaystyle \ln(x+1)=\sum_{n=o}^\infty (-1)^{n+1} \frac{x^n}{n}$, and $\frac{1}{(t+1)} = \left( \ln(t+1) \right) ' $

does that means that: $\displaystyle f(x) = \sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n} - (\frac{1}{x+1})$ but it seems to me that the convergence radius is $R=1$, but I'm missing the algebraic step to achieve that

答:
You don't have to explicitly compute the integral, nor the successive derivatives: by the 1st fundamental theorem of integral calculus, $f(x)$ is the function such that
$$f'(x)=\frac x{(1+x)^2},\qquad f(0)=0.$$
Therefore, we only have to find the expansion of $\dfrac x{(1+x)^2}$ and integrate term by term. The radius of convergence will be the same for $f(x)$ and $f'(x)$.

Now, $\dfrac 1{(1+x)^2}=-\Bigl(\dfrac1{1+x}\Bigr)'=\displaystyle\biggl(\sum_{n\ge0}(-1)^{n+1} x^n\biggr)'$, so
$$f'(x)=x\sum_{n\ge0}(-1)^{n+1}n\, x^{n-1}=\sum_{n\ge0}(-1)^{n+1}n\, x^n,$$
whence
$$f(x)=\sum_{n\ge0}(-1)^{n+1}\frac{n\,x^{n+1}}{n+1}.$$


问:
Trig Mountain Problem

I have a math problem.

Someone is on a football field and they are looking at a mountain. The angle of elevation from the floor to the mountain is 16 degrees. That person walks 127 meters towards the mountain and now it's 23 degrees.
 How high is the mountain?

I got 177m.

The way I solved it was by finding the difference in the angles and the difference in the distance away from the mountain.

X:16

X+127:23

Using that I was able to get

127:7

And with that ended up with

18.1428571429

I then multiplied that by 23 to get the bottom length of the triangle.

417.285714286

Which I put into tan

tan23 = y/417.285714286

417.285714286*tan23 = y

y = 177.127276878

It is never stated how far away they are at the start either

答:
Let $h = AB$ be the height of the mountain.

Let the person be initially at $C$, then moved $127$ m  to $D$ such that $BD = x$. So $BC = x+127$ m .

From the triangle $ABD$
$\cot(23^\circ) = \frac{x}{h} \Rightarrow \boxed{x = h\cot(23^\circ)}$

Similarly from triangle $ABC$

$x+127 = h\cot(16^\circ) \Rightarrow h\cot(23^\circ)  +127 = h\cot(16^\circ) \Rightarrow \boxed{h  = \frac{127}{\cot(16^\circ) - \cot(23^\circ)} \approx 112.234 \text{ m}} $

As mentioned in the comments below, $\cot$ is reciprocal of $\tan$. If you still have doubts, then in terms of $\tan$,

$\tan(23^\circ) = \frac h x \Rightarrow x = \frac{h}{\tan 23^\circ}$

and $x + 127 = \frac{h}{\tan 16 ^ \circ} \Rightarrow h = \frac{127}{\frac1{\tan 16^\circ} - \frac1{\tan23^\circ}} \approx 112.234 \text{ m}$

Title: Trig Mountain Problem


问:
Use of Chinese Remainder Theorem in proof of RSA correctness
RSA

$\gcd(p,q) = 1$

$N = pq$

$ed = 1 \bmod{\phi(N)}$

The proof of correctness of RSA involves 2 cases

Case 1)  $\gcd(m, N) = 1$

I understand the proof of correctness for this case using Euler's Theorem

Case 2) $\gcd(m, N) \neq 1$

For proving this, the Chinese Remainder Theorem is used

All the proofs say that as per CRT

If

$x = y \pmod{p}$   - 1
and

$x = y \pmod{q}$   - 2

then

$x = y \pmod{pq}$  ====> 3

I know that Line 1 is true & it's easily proven, but how does Chinese Remainder theorem come in here? How do you use CRT to get from (1)&(2) to (3)?

For e.g. check the Wikipedia article (it's not just Wikipedia, I saw it in several other places also)

https://en.wikipedia.org/wiki/RSA_(cryptosystem)#Proof_using_Fermat's_little_theorem

In that article check the line "To check whether two numbers, such as med and m, are congruent mod pq, it suffices (and in fact is equivalent) to check that they are congruent mod p and mod q separately" - they have [note2] for it where it says "This is part of the Chinese remainder theorem, although it is not the significant part of that theorem."

答:
Let $p$ and $q$ be distinct prime numbers, and let $a$ be an integer that is divisible by both $p$ and $q$. Then $a$ is divisible by $pq$. (Do you agree?)

In other words, if $a\equiv 0 \bmod p$ and $a\equiv 0\bmod q$, then $a\equiv 0\bmod pq$.

Now just let $a=x-y$.


问:
Why don't we consider non-units as quadratic residues?
Is there any specific reason in not including non-units of $\mathbb{Z}_n$ as quadratic residues? As an examples, we say that in $\mathbb{Z}_8$, the set of quadratic residues is just {1} and not {1,4}.</p>

答:
It is a nice idea to associate with unit quadratic residues the number $1$ and with unit quadratic non-residues the number $-1$ (this is the logic behind the Legendre and Jacobi symbols).

The reason this is a nice idea is that you have the following properties of unit residues and non-residues:

residue $\cdot$ residue $=$ residue

residue $\cdot$ non-residue $=$ non-residue

non-residue $\cdot$ non-residue $=$ residue

If you replace "residue" by $1$ and "non-residue" by $-1$ above, you get truthful statements about integers.

When you include non-unit residues, the above formulas are no longer correct. For example (taken from Wikipedia), if you're working modulo $15$, then $3\cdot 5 = 9 \cdot 5 = 9 \cdot 10 = 0$. But $3, 5$ are non-residues and $9, 10$ are residues.


问:
A transformation for independent variables to be dependent
I ran into a confusing question.

If two variables are independent, maybe they will be dependent after linear transformation.

How it can happen? Is it possible for independent variables?

What is the operation that it makes variables to be dependent?

In my opinion the transformation that maps all variables to a point is true for this fact.

What is wrong with my answer?

The mean of Variables are features of data.

答:
The converse statement is even more interesting: sometimes it is possible to transform dependent random variables into independent ones. For example, let $X$ and $Y$ be jointly normal, zero-mean, unit variance, with correlation coefficient $\rho$. Define $V=X+Y$ and $W=X-Y$. $V$ and $W$, are also jointly normal and zero-mean. As for correlation:
$$E\{VW\} = E\{X^2-Y^2\}=0$$
It is a simple and quite practical (in communication systems for instance) example of how a rotation can make it easier to deal with independent random variables instead of dealing with dependencies. Once you are done with the analysis, they can easily be reverted to the original variables. You can find many other examples like this, many of them with dramatic impacts. For instance, the main reason that the 4G cellular access links are way faster than 3G relies on a rather simple linear transform of signals, known as OFDMA. As fancy and complicated they might seem, the math behind them is quite simple: rotate the compound signals coming from hundreds of users so that they become independent, so you could deal with them separately. I am sure there are many abstract and practical examples that others can mention here. The bottom line is: statistical/probabilistic independence is subject to the representation of the phenomena and it is not necessarily inherent characteristics of the phenomena.


问:
Nice definition of Grothendieck topology?
In Scholze's masterclass on condensed mathematics, there was a discussion whether there is a "good definition of a site". It seems, that the definition of Grothendieck topology using sieves is the most general. If one works with Grothendieck pretopologies one has to worry about existence of certain pullbacks. So I am not interested in pretopologies, instead I want to conceptualize the sieve-definition. This is what I have:

The usual definition of Grothendieck topology based on sieves (see e.g. https://en.wikipedia.org/wiki/Grothendieck_topology, or Alternative formulation of Grothendieck topology) can be rewritten as follows:

Def. A Grothendieck topology on a small category $\mathcal C$ is a functor $J : \mathcal C^{\mathrm{op}} \to \mathrm{Set}$ with the following three properties:

$J$ is a subfunctor of $\mathrm{Sv}$. Here $\mathrm{Sv}(X)$ is the set of sieves on $X$ and for any morphism $f : Y \to X$ the map $f^* : \mathrm{Sv}(X) \to \mathrm{Sv}(Y)$ is given by pullback of sieves (see https://en.wikipedia.org/wiki/Grothendieck_topology)

$J_{\mathrm{ind}}$ is a subfunctor of $J$, where $J_{\mathrm{ind}}(X) = \{\mathrm{Hom}_{\mathcal C}(-,X)\}$.

For all $X \in \mathcal C$ and all $S \in J(X)$, we have
$$ J(X) = \{T \in \mathrm{Sv}(X) \mid \forall (f : Y \to X) \in S : f^*T \in J(Y)\}$$

1 and 2 just say, that $J$ is somewhere between the indiscrete and the discrete Grothendieck topology. I am completely happy with that.

The way I read 3 is: Being a $J$-covering sieve is a $J$-local property. This looks like a "circular" sheaf condition.

If $X \in \mathcal C$ and $T$ is a sieve on $X$, then we define a presheaf on $\mathcal C \downarrow X$ by
$$ \mathcal F_{T}(f : Y \to X) = \begin{cases} \{*\}, &f^*T \in J(Y), \\ \emptyset, &\text{else.} \end{cases} $$
This is a $J$-sheaf if and only if for all $Y$ and all covering sieves $S \in J(Y)$, the map
$$ \mathcal F_T(Y) \to \prod_{(f:V \to Y)\in S} \mathcal F_T(V) $$
is bijective. This for all $X$ and $T$ is equivalent to 3.

So we can take:

For all $X \in \mathcal C$ and all $T \in \mathrm{Sv}(X)$, the presheaf $\mathcal F_T$ on $\mathcal C \downarrow X$ is a $J$-sheaf.

Grothendieck topologies are often advertised as the "right generalization" of topological spaces for the definition of sheaves. Since 1 and 2 don't say much about $J$, all interesting topological features hide inside of 3. So it might be interesting to ask:

Is there a nice(r) way to express 3?
(or the entire definition?)

Creative answers are highly appreciated.

PS: I don't know the details of the discussion in the chat in the masterclass, because I watched it on Youtube (https://www.youtube.com/watch?v=OT65JC3gKPY, at 3:00). Comments about the content of this discussion are also appreciated.

答:
Rather than having $J: \mathcal{C}^{\mathrm{op}} \to \mathbf{Set}$, have the codomain of $J$ being the category $\mathbf{DistLat}_1$ of distributive lattices with a maximal element (the notation is ad hoc), since the collection of sieves is trivially closed under unions, and it is a consequence of the ordinary definition of Grothendieck topologies that it is closed under intersections. Moreover, pulling back sieves along morphisms in $\mathcal{C}$ preserves intersections, unions and maximal sieves. The maximal element eliminates the need for your axiom 2, since the maximal sieve is the maximal element of $\mathrm{Sv}(X)$ for every $X$, so being a subfunctor via an inclusion that preserves maximal elements ensures that the maximal sieve is always present.

Your axiom 3 in terms of the sheaf condition isn't as circular as it seems at first glance, since we can define the sheaf condition for any collection of sieves. However, all that's left to add to the axiomatisation is multicomposition (since stability under pullback is implied by functoriality of $J$), and this seems far less complicated to check than either the sheaf condition version of axiom 3 that you wrote. In summary, suppressing the subsidiary definitions of $\mathrm{Sv}$ and $\mathbf{DistLat}_1$, we have:

Def. A Grothendieck topology $J$ on a small category $\mathcal{C}$ is a subfunctor of the functor $\mathrm{Sv}: \mathcal{C}^{\mathrm{op}} \to \mathbf{DistLat}_1$, subject to the condition that for each $S \in J(X)$ and $T \in \mathrm{Sv}(X)$,
$$(\forall f:Y \to X \in S) (f^*(T) \in J(Y)) \implies T \in J(X).$$


问:
Definition of convergence in $C^\infty(\Omega)$
I am not convinced or may be don't understand, the way they define convergence and then topology as a consequence of convergence.

$\Omega$ is open subset of $\Bbb R^n.$Define standard topology on $\mathcal C^\infty(\Omega)$ through the following notion of convergence:

Which says, that a sequence $h^k $, $h^k \in \mathcal C^\infty(\Omega)$ converges to $h$ in $\mathcal C^\infty(\Omega)$ if $h^k$ converges to h uniformly with all partial derivatives of any order on any compact subset of $\Omega$.

I am wondering, why did they define convergence in such a way and how does it give birth to a topology?

[In above definition they are talking about the convergence of the sequence of derivatives of a function $h$ i.e. ($h^1, h^2....\to h)$. Now, if we want to discuss the convergence of any random sequence of functions (I mean, a sequence which is not made of derivatives of a particular function), how will we consider that sequence and the convergence....?]

Then writer says:

If $K$ is a compact subset of $\Omega$, we use the standard norm i.e $||\mathcal u||_{C^k(K)}= \sum _{|\alpha|\le k} sup_{x\in K}|D^\alpha\mathcal u(x)|$

Here, I've a basic question - why does he call it a standard norm and why does he consider only this specific norm...??

I will appreciate, if you can give me few minutes out of your precious time. Thank you so much.

答:
Julien has already given the construction for a topology. Let me give you some more motivation:

I am wondering, why did they define convergence in such a way and how does it give birth to a topology?

This is a very natural convergence: If you have a sequence of differentiable functions $h_n$ converging uniformly on compact sets to a function $h$ and the derivatives converge uniformly too, then you know from classical analysis that $h$ is differentiable and $h_n'$ converges to $h'$ uniformly on compact sets. And there are well-know examples that all these assumptions are necessary for the limit function $h$ being differentiable. So essentially this type of convergence guarantees that the limit function is again $C^\infty$.

Here, I've a basic question - why does he call it a standard norm and why does he consider only this specific norm...??

This is rather sloppy from the author: $\|u\|_{C^k(K)}$ is a norm on $C^k(K)$ but it is only a seminorm on $C^\infty$. And the family of seminorms $\{ \|\cdot\|_{C^k(K)}: K\subset \Omega \text{ compact }, k\in\mathbb{N}\}$ defines in a natural way a locally convex topology described here, in fact the coarsest locally convex topology such that all seminorms are continuous. And "luckily", this topology coincides with the topology described in the construction of julien. This is why these (semi-)norms are called standard.


问:
Moment generating function and expected value of $Y=\left|X-2\right|$
Random variable of probability density function of X is

${f}_{X}(x)=\frac{1}{2},0\le x\le 1\phantom{\rule{0ex}{0ex}} $

${f}_{X}(x)=\frac{1}{4},3\le x\le 5$

Find moment generating function ${M}_{Y}(t)$ of $Y=\left|X-2\right|$ and from here find $E(Y)$.

Here is my solution: ${M}_{Y}(t)=E\left[{e}^{tY}\right]=E\left[{e}^{t\left|X-2\right|}\right]=\int _{0}^{1}\frac{1}{2}{e}^{t(x-2)}dx+\int _{3}^{5}\frac{1}{4}{e}^{t(x-2)}dx=\frac{1}{4t}(2{e}^{-t}-2{e}^{-2t}+{e}^{3t}-{e}^{t})$

I want to use formula $E\left[Y\right]=\frac{d({M}_{Y}(t))}{dt}(t=0)$ to find expected value $E\left[Y\right]$ , but ${M}_{Y}(t)$ will be undefined when  $t=0$.

How can I calculate that? Any help will be appreciated.

答:
Your formula for the MGF is valid only for $t \neq 0$. For $t=0$ we have $M_Y(t)=Ee^{0X}=e^{0}=1$. With this value at $0$ $M_Y$ is actually continuous and even continuously differentiable at $0$.

Now $M_Y'(t)=\frac {(4t)(-2e^{-t}+4e^{-2t}+3e^{3t}-e^{t})-4 (2e^{-t}-2e^{-2t}+e^{3t}-e^{t})} {16t^{2}}$ and $M_Y'(0)$ is the limiting value of this at $0$. Using the fact that $e^{ct}\sim 1+ct+c^{2}t^{2}/2$ for $t$ near $0$ you can compute this limit.


问:
Infinite Limit Involving a Factorial
$$\lim_{x\to\infty} \left(\frac{x+1}{(x!)^{\frac{1}{x-1}}}\right) $$

I'm pretty sure that this approaches infinity, yet I'm having trouble proving it. Hope someone can help, thanks.

答:
$$a=\frac{x+1}{(x!)^{\frac{1}{x-1}}}\implies \log(a)=\log(x+1)-\frac{1}{x-1}\log(x!)$$

Using Stirling approximation anf then Taylor
$$\log(a)=1+\frac{\log \left(\frac{1}{x^{3/2}}\right)+2-\frac{1}{2} \log (2 \pi  )}{x}+O\left(\frac{1}{x^2}\right)$$ Then $\log(a)\to 1$

Title: Infinite Limit Involving a Factorial


问:
Upper bound of a function within an interval
I am trying to find the upper bound of the magnitude of a function within a given interval.

Here is the problem:

Find an upper bound on $|\partial f / \partial y|$ for the following function on the given domain:
$$
f(t, y)=\operatorname{ty} \sin (\beta y), \quad t \in[0,1], \quad y \in[-\pi, \pi]
$$
Here, $\beta$ is a fixed parameter. Your upper bound will depend on $\beta$.

I have currently gotten to the inequality: $|\partial f / \partial y|\leq |\sin(\beta y)+\beta y \cos(\beta y)|$

But from here I cannot see how I can further simplify this in order to remove the y dependence in the inequality. I don't see what else can be done without knowing any more information about $\beta$. Have I overlooked something that can allow me to strengthen the inequality?

More precisely I am trying to find a Lipschitz constant for those who are familiar with this procedure

答:
One bound, possibly not the best, can be obtained by
$$|\sin(\beta y)+\beta y\cos(\beta y)|\leq |\sin(\beta y)|+|\beta y||\cos(\beta y)|\leq 1+\pi|\beta|$$
where the assumption on $y$, namely, $y\in[-\pi,\pi]$, was used.


问:
Show that for all $a,b,c>0$, $\frac 1 {\sqrt[3]{(a+b)(b+c)(c+a)}}\geq\frac 3 {2(a+b+c)}$.

Show that for all $a,b,c>0$, $\displaystyle\frac 1 {\sqrt[3]{(a+b)(b+c)(c+a)}}\geq\frac 3 {2(a+b+c)}$.

I tried to cube the both sides, and expand it, but that'll be too troublesome, is there simpler ways? Thasnk you.

答:
HINT:

Using AM-GM inequality,

$$\frac{a+b+b+c+c+a}3\ge \sqrt[3]{(a+b)(b+c)(c+a)}$$


问:
system of equations calculator
I have a problem that is very simple in concept but very difficult to actually solve.

I need to find the best fit polynomial of degree at most 4 for f(x)=ln(x) on the interval [1,4].

So far I have gotten to the point where I need to solve the system of equations: $0=-3a_0-\frac{15a_1}{2}-21a_2-\frac{255a_3}{4}-\frac{1023a_4}{5}-3+ln\left(256\right),\\0=-\frac{15a_0}{2}-21a_1-\frac{255a_2}{4}-\frac{1023a_3}{5}-\frac{1365a_4}{2}-\frac{15}{4}+8ln\left(4\right),\\0=-21a_0-\frac{255a_1}{4}-\frac{1023a_2}{5}-\frac{1365a_3}{2}-\frac{16383a_4}{7}-7+\frac{64}{9}ln\left(64\right),\\0=-\frac{255a_0}{4}-\frac{1023a_1}{5}-\frac{1365a_2}{2}-\frac{16383a_3}{7}-\frac{65535a_4}{8}-\frac{255}{16}+64ln\left(4\right),\\0=-\frac{1023a_0}{5}-\frac{1365a_1}{2}-\frac{16383a_2}{7}-\frac{65535a_3}{8}-29127a_4-\frac{1023}{25}+\frac{1024ln\left(4\right)}{5}$

I know how to do this in theory, but the execution is incredibly tedious and errors easily find their way in. I was wondering if anyone knew a good system of equation calculator, symbolab seems to only be able to solve up to 4 terms, or code for solving a 5 term system of equations on matlab.

答:
Problem : Fitting to $\ln(x)$  a 4-th degree polynomial
$\quad a_0+a_1x+a_2x^2+a_3 x^3+a_4 x^4\quad $
on the range $1\leq x\leq 4$.

The sum of squared errors is :
$$\epsilon^2=\int_1^{4}\left(a_0+a_1x+a_2x^2+a_3x^3+a_4x^4-\ln(x) \right)^2dx$$
The minimum is obtained when the partial derivatives (wrt the coefficients) are all null.
$$\frac{\partial\epsilon^2}{\partial a_k}=2\int_1^{4}\left(a_0+a_1x+a_2x^2+a_3x^3+a_4x^4-\ln(x) \right)x^k dx =0 $$
This is for $k=0$ to $k=4$.

So we have five linear equations to be solved for the five unknowns $a_0,a_1,a_2,a_3,a_4$.

$$a_0\int_1^{4}x^{k} dx+a_1\int_1^{4}x^{k+1} dx+a_2\int_1^{4}x^{k+2} dx+a_3\int_1^{4}x^{k+3} dx+a_4\int_1^{4}x^{k+4} dx =$$
$$=\int_1^{4}\ln(x)\,x^k dx $$

After computing the integrals and writting the system of equations on matrix form :

On the range $\quad 0\leq x\leq 4\quad$ the Root Least Mean Square Error is $=0.001005$

On a graphical representation it is quite impossible to see a difference between the curves of the logarithm and the polynomial.

Following the integral method shown above it is easy to solve similar problems with higher polynomials and on different ranges of $x$.

For example with a polynomial of degree 5 :

Title: system of equations calculator


问:
Identification of problem in graph theory
Suppose a directed graph G=(V,E). I am interested in computing the smallest subset of nodes V' such that:

Every node n in G is reachable from at least one node in V', with a directed path

Every edge e in G is reachable from at least one node in V', with a directed path

Again, V' is the smallest subset that satisfies the above properties

comment 1: Nodes in V' are considered reachable  by definition.

comment 2: I suppose property 3. is already implied by property 2., but I added it to make sure there are no exceptions.

comment 3: Nodes with no links at all are also in V' for the same reason

comment 4: Nodes with no incoming links are in V' as they are not reachable by any other means

Questions:

Is there a name for this problem in the literature of graph theory?

Are there efficient algorithms to solve this problem?

答: An algorithm for solving this problem is given here. To wit: find the strongly connected components of the directed graph and pick one vertex from each "source" component. If you're not familiar with strongly connected components, the Wikipedia page might be helpful. I'm not aware of a name for this problem.</p>


问:
Help in two improper integrals
For a physics problem, I need to evaluate two integrals:
$$I_n=\int_{0}^{\infty} f(x) dx=\int_{0}^{\infty} \frac{x^2 \sin^2x}{(n^2\pi^2-x^2)^2} dx,$$ and
$$J_n=\int_{0}^{\infty} \frac{x^2 \cos^2x}{((n+1/2)^2\pi^2-x^2)^2} dx, n=1,2,3..$$
At Mathematica the value of these integral turns out to be $\pi/4$ which is independent of $n$.

$I_n$ has got a pole of order 2 at $x=n\pi$ and when I calculate the residue at $x=n \pi$  is zero. But on the other hand I find that  $\lim_{x \to n\pi} f(x)=0,$  so $x=n\pi$ may not be a pole of $f(x)$. I face the same problem in the case of $J_n$. Also $I_0=\pi/2$ is a standard integral. In any case, I want to evaluate both $I_n$ and $J_n$.  Please help me.

答:
$I_n=\int_{0}^{\infty} \frac{x^2 \sin^2x}{(n^2\pi^2-x^2)^2} dx$. Integrating by part end expanding integration to the whole axis (the integrand is even) we get

$I_n=-\frac{1}{2}\int_{0}^{\infty} \frac{\sin^2x+x\sin2x}{(n^2\pi^2-x^2)} dx=-\frac{1}{4}\int_{-\infty}^{\infty} \frac{x\sin2x}{(n^2\pi^2-x^2)}dx-\frac{1}{8}\int_{-\infty}^{\infty} \frac{1-\cos2x}{(n^2\pi^2-x^2)}dx=I_1+I_2$.

Then we choose the closed contour $C$ in upper half of the complex plane:

To close the contour, we added two small half-circles $C_1$ and $C_2$ of radius $r$ - around $x=-n\pi$ and $x=n\pi$ $(r\to0)$, and a big circle of radius $R$ ($R\to\infty$). The required integral $I_n$ is equal to the principal value of the integral along the axis. Then,  $\sin2x=Im \exp(i2x)$ and $\cos2x=Re\exp(2ix)$.

The integral over the closed contour $\oint_C=0$ (no singular points inside the contour), therefore$I_n+I_{C1}+I_{C2}+I_R=0$. It is easy to evaluate that $I_R\to0$ at $R\to\infty$ (using Jordan's lemma),
so $I_n=-I_{C1}-I_{C2}$

$I_1=-\frac{1}{4}Im\int_{-\infty}^{\infty} \frac{xe^{2ix}}{(n^2\pi^2-x^2)}dx=Im(-I_{C1}-I_{C2})=\frac{1}{4}Im\left((-\pi{i})[\frac{-\pi{n}\exp(-2\pi{in})}{2\pi{n}}+\frac{\pi{n}\exp(2\pi{in})}{-2\pi{n}}]\right)$

$$I_1=\frac{\pi}{4}$$

Making similar calculations it is easy to show that $I_2=0$
Finally, $$I_n=I_1+I_2=\frac{\pi}{4}$$

$J_n$ can be evaluated in the same way.

Title: Help in two improper integrals


问:
Prove subspace and subsets or R are polish space
"Let a,b∈R, with a < b. Prove that each discrete subspace of R and each of the spaces [a,b], (a,b), [a,b),(a,b], (−∞,a), (−∞,a], (a,∞), [a,∞), and {a}, with the topology induced as a subspace of R, is a Polish Space."

This question is a bit confusing. What exactly are discrete subspace of R? Is it just any subset or R with the discrete topology?

Also, Polish space means separable and completely metrizable, but just because R is complete under the usual Euclidean metric doesn't mean a subspace is complete so I must be confusing something.

答:
Each discrete subspace $D$ of $\Bbb R$ is at most countable, and has the discrete topology as its subspace topology (that's the definition of being discrete), and this topology can be induced by the discrete metric $d_1(x,y)=1$ whenever $x \neq y$, and this metric is complete (every $d_1$-Cauchy sequence is eventually constant, and hence converges). So $D$ is Polish (separable, due to its size) and completely metrisable.

Indeed $\Bbb R$ itself is Polish (it's usual metric will show that, and $\Bbb Q$ is countable and dense), and hence so is every closed susbet of $\Bbb R$: we just inherit the metric and the subspace is still sepaarble and compleet in the inherited metric, by standard facts. This handles sets like $[a,b]$, $[a,\infty)$ and $(-\infty,a]$ e.g. and also $\{a\}$ (it's discrete too BTW).

$[a,b) \simeq [a,\infty)$ so if one is Polish so is the other (being Polish only dpends on the topology; it's preserved by homeomorphism), etc. $(a,b) \simeq \Bbb R$ is Polish for the same reason.


问:
How to prove $\int_0^{2r} 2h \sqrt{r^2 -(x-r)^2}dx = \pi r^2 h$ without geometry?
I wanted to find the volume of a cylinder, radius r, height h by slicing it in to rectangles:

I placed the cylinder on the x-axis, one corner of the base diameter at (0,0) the opposite at (2r, 0). I have found that an area of a cross-section perpendicular to the x-axis is $A(x) = 2h \sqrt{r^2 -(x-r)^2}$ so:

$V=\int_0^{2r} 2h \sqrt{r^2 -(x-r)^2}dx$

I have tested this and it gives the volume correctly for various r and h. But how to show:

$\int_0^{2r} 2h \sqrt{r^2 -(x-r)^2}dx = \pi r^2 h$ without just saying "we know $V= \pi r^2 h$"?

This problem was my own device, perhaps it is not possible to do this.

答:
Put $(x-r) = r \sin\theta$ then you have $x = r + r \sin\theta$. Which says $\mathrm{dx}=r\cos\theta \rm{d}\theta$. When $x= 0$ we have $-r = r\sin\theta$ which says that $\theta = -\frac{\pi}{2}$. And when $x =2r$ we have $r=r\sin\theta$ which says that $\theta = \frac{\pi}{2}$. So your integral is now,

\begin{align*}
\int\limits_{0}^{2r}\sqrt{r^{2}-(x-r)^{2}} \ dx &=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{r^{2}-r^{2}\sin^{2}\theta} \cdot r\cos\theta \ \text{d}\theta \\ &= r^{2} \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2}\theta \ \text{d}\theta
\end{align*}


问:
How translation affects vectors angle?
Let $\vec a$ and $\vec b$ be vectors in finite-dimensional Euclidean space, s.t. $||\vec a|| > 0,$ $||\vec b|| > 0$.

What is the value of $cos(\beta)$ if $\beta = \angle(\vec v, \vec w)$ and  $\vec v = \vec a + [1, 1, \ldots, 1]$, $\vec w = \vec b + [1, 1, \ldots, 1]$?

答:
The cosine of the angle $\alpha$ between the non-zero vectors $\vec v$ and $\vec w$ is
$$
\cos(\alpha)=\frac{{\vec v}\cdot{\vec w}}{|\vec v||\vec w|}
$$
where the dot denotes the scalar product and $|\cdot|$ the norm.

Upon translation by a vector $\vec x$ and under the assumption that $\vec v+\vec x$ and $\vec w+\vec x$ are still non-zero the angle changes so that
$$
\cos(\alpha^\prime)=\frac{({\vec v+\vec x})\cdot({\vec w+\vec x})}{|\vec v+\vec x||\vec w+\vec x|}.
$$
While one can certainly make the latter expression explicit in terms of the components of the three vectors involved, the relation between the two expression is by no means simple except maybe for a few special cases.

Title: How translation affects vectors angle?


问:
How to prove the product rule in multivariable without chain rule?
It's easy to prove in one variable functions $f:I\subset \Bbb R \to \Bbb R$ but I tried in multivariable and I'm stuck.

Let $A$ be an open set in $\Bbb R^n$ and $a\in A$.
If $f,g:A\to \Bbb R$ are differentiable in $a$, then $f·g$ is differentiable in $a$ and
$D(f·g)(a)=g(a)·D(f)(a)+f(a)·D(g)(a)$

$f,g$ differentiable in $a$: $\lim\limits_{x\to a} \frac{f(x)-f(a)-Df(a)(x-a)}{||x-a||}=0$ and $\lim\limits_{x\to a} \frac{g(x)-g(a)-Dg(a)(x-a)}{||x-a||}=0$.

I started this way trying to prove it similar to one variable case:

$\lim\limits_{x\to a} \frac{f(x)g(x)-f(a)g(a)-g(a)·Df(a)(x-a)-f(a)·Dg(a)(x-a)}{||x-a||}=\lim\limits_{x\to a} \frac{f(x)g(x)-f(a)g(x)+f(a)g(x)-f(a)g(a)-g(a)·Df(a)(x-a)-f(a)·Dg(a)(x-a)}{||x-a||}=\lim\limits_{x\to a} f(a)\frac{g(x)-g(a)-Dg(a)(x-a)}{||x-a||}+\lim\limits_{x\to a} \frac{f(x)g(x)-f(a)g(x)-g(a)·Df(a)(x-a)}{||x-a||}.$

In this last step, I know the first limit is zero but I don't know what to do with the other one.

答:
Ok, now two days later I've already solved it. The key is adding in $f(x)g(x)-f(a)g(a)-g(a)·Df(a)(x-a)-f(a)·Dg(a)(x-a)$ the terms

$-f(a)g(x)+f(a)g(x)$

and also

$-g(x)·Df(a)(x-a)+g(x)·Df(a)(x-a)$

so that the limit can be separated this way

$\lim\limits_{x\to a} \frac{{\color{red}{f(x)g(x)-f(a)g(x)}}{\color{blue}{+f(a)g(x)-f(a)g(a)}}{\color{red}{-g(x)·Df(a)(x-a)}}+g(x)·Df(a)(x-a)-g(a)·Df(a)(x-a){\color{blue}{-f(a)·Dg(a)(x-a)}}}{||x-a||}=\lim\limits_{x\to a} \frac{f(x)-f(a)-Df(a)(x-a)}{||x-a||}g(x)+\lim\limits_{x\to a} f(a)\frac{g(x)-g(a)-Dg(a)(x-a)}{||x-a||}+\lim\limits_{x\to a} (g(x)-g(a))\frac{Df(a)(x-a)}{||x-a||}=$

$=0·g(a)+f(a)·0+\lim\limits_{x\to a} (g(x)-g(a))\frac{Df(a)(x-a)}{||x-a||}$

Since $\frac{Df(a)(x-a)}{||x-a||}$ is bounded by $|D_1f(a)|+...+|D_nf(a)|$, the limit is zero.

So $f·g$ is differentiable in $a$.


问:
Determine if $\|f\|=|\int_{0}^1 f(x) \ dx |$ defines a norm in $C([0,1]).$

Determine if $$\|f\|=|\int_{0}^1 f(x) \ dx |$$ defines a norm in $C([0,1]).$

Let $f,g \in C([0,1])$ such that $f\geqslant g$. Now $$\int_{0}^{1}f(x) \ dx \geqslant \int_{0}^{1}g(x) \ dx \Longrightarrow  |\int_{0}^{1}f(x) \ dx| \geqslant |\int_{0}^{1}g(x) \ dx| \geqslant 0$$ which gives positivity. Now take again $f,g \in C([0,1])$ $$\|f+g\| = |\int_{0}^1f(x)+g(x)  \ dx| \leqslant |\int_{0}^{1}f(x) \ dx| + |\int_{0}^{1}g(x) \ dx| = \|f\| +\|g\|.$$

is this enough to prove the result or am I missing some properties?

答: No.  Take $f:x \mapsto x - 0.5$ as an example.  $\int_0^1 f = 0$, but $f \not\equiv 0$.


问:
$T \cdot W=0$ for every skew symmetric tensor $T$ implies that $T$ is a symmetric tensor
I'm trying to prove by myself the following easy fact. The context is continuum mechanics (in particular I'm using Gurtin's book):

Let $T$ a tensor and $W%$ a skew-symmetric tensor. If $T \cdot W =0 $ for every $ W \in \text{Skw}$, then $T$ is symmetric.

I can't obtain $T=T^t$ by using the definition, but I can't.

$T \cdot W = \text{tr}(T^t W)=\text{tr}(T^t (W^t) ^t) =- \text{tr}(T^t W^t) = -\text{tr}(W^t T^t)=-W \cdot T^t$
so I have $$T \cdot W + W \cdot T^t =0$$

and I don't know how to get the thesis $T=T^t$. How should I move?

答:
Every linear operator $T$ is uniquely expressible as a combination $T=T_\mathrm{sym}+T_\mathrm{skew}$ of a symmetric operator and a skew-symmetric operator. Indeed, we can apply the adjoint to both sides and solve the system

$$ \begin{cases}
T & =T_\mathrm{sym}+T_\mathrm{skew} \\ T^\dagger &=T_\mathrm{sym}-T_\mathrm{skew}
\end{cases} $$

to yield the formulas

$$ T_\mathrm{sym}=\frac{1}{2}(T+T^\dagger), \qquad T_\mathrm{skew}=\frac{1}{2}(T-T^\dagger). $$

Not only is the vector space $\mathrm{End}(V)$ a direct sum of the subspaces of symmetric and skew-symmetric operators, this is an orthogonal direct sum. We verify orthogonality using the definition of the Frobenius-Schmidt inner product of two linear operators an an inner product space: $\langle T,W\rangle=\mathrm{tr}(T^\dagger W)$.

Suppose $T$ is symmetric and $W$ is antisymmetric. Then we may use $\mathrm{tr}(AB)=\mathrm{tr}(BA)$ and $\mathrm{tr}(X^\dagger)=\mathrm{tr}(X)$ (if we're talking about real inner product spaces) to reason that $\langle T,W\rangle=0$. (Can you do this?)

If $T$ were not symmetric, it would have nonzero skew-symmetric part. What would $T\cdot W$ be then if we were to set $W=T_\mathrm{skew}$? Can it be $0$?


问:
find the number of possible positive integer solutions when the inequality is $a \times b \times c \lt 180$
As most of you know there is a classical question in elementary combinatorics such that if $a \times b  \times c = 180$ , then how many possible positive integer solution are there for the equation $?$

The solution is easy such that $180=2^2 \times 3^2 \times 5^1$ and so , for $a=2^{x_1} \times 3^{y_1} \times 5^{z_1}$ ,   $b=2^{x_2} \times 3^{y_2} \times 5^{z_2}$ ,   $c=2^{x_3} \times 3^{y_3} \times 5^{z_3}$ .

Then: $x_1+x_2+x_3=2$ where $x_i \geq0$ ,  and $y_1+y_2+y_3=2$ where $y_i \geq0$ and $z_1+z_2+z_3=1$ where $z_i \geq0$.

So , $C (4,2) \times C(4,2) \times C(3,1)=108$.

Everything is clear up to now.However , i thought that how can i find that possible positive integer solutions when the equation is $a \times b  \times c \lt 180$  instead of $a \times b  \times c = 180$

After , i started to think about it. Firstly , i thought that if i can calculute the possible solutions for $x_1+x_2+x_3 \lt2$ where $x_i \geq0$ ,  and $y_1+y_2+y_3 \lt 2$ where $y_i \geq0$ and $z_1+z_2+z_3 \lt1$ where $z_i \geq0$ , then i can find the solution.However , there is a problem such that when i calculate the solution , i do not include the prime numbers and their multiplicites which is in $180$.

For example , my solution does not contain $1 \times 1 \times 179 \lt 180$

My question is that how can we solve these types of question . Is there any $\color{blue}  {\text{TRICK}}  $ for include all possible ways ? Moreover ,this question can be generalized for  $a \times b  \times c \leq 180$ , then what would happen for it ?

Thanks for helps..

答:
Addendums just added that moderately refine the enumeration of (for example) all positive integer solutions to $(xyz) \leq 180$.

The positive integer solutions to $(xyz) \leq 180$ can be partitioned into the $180$ mutually exclusive sets $(xyz) = a$, where $a \in \{1,2,\dots, 180\}.$

Then, using the same method that you used in your query, you will examine each value of $a$ separately, examining its prime factorization.  Although this approach dispenses with any attempt at elegance, the approach is certainly straightforward.

Addendum-1 : Overview

Additions to the answer are provided in sections, Addendum-1, Addendum-2, ... that discuss an alternative approach to the overall problem and then try to connect the two approaches.

The overall problem is:

Enumerate the number of positive integer solutions to $(xyz) \leq M \in \mathbb{Z^+}$.

Ideas to note:

Let $S_M$ denote the number of positive integer solutions to $(xyz) \leq M \in \mathbb{Z^+}$.

Let $T_M$ denote the number of positive integer solutions to $(xyz) = M \in \mathbb{Z^+}$.  Then clearly, $T_M = [S_M - S_{(M-1)}].$

The ordered triple $(a,b,c)$ will be used to denote the solution $(x=a, y=b, z=c).$

When (for example) computing $T_{(179)}$ the solutions $(1,1,179), (1,179,1),$ and $(179,1,1)$ will not be considered distinct.  To prevent overcounting, the constraint of $x \leq y \leq z$ will be enforced.

For $r \in \mathbb{R}, \lfloor r\rfloor$ will be used to denote the floor of $r$ (i.e. the largest integer $\leq r)$.

The alternative algorithm discards any ideas involving prime factorizations, and therefore supposedly renders the OP's analysis obselete.  In fact, the tail end of these addendums will be a (possibly laughable) alternative approach to computing prime factorizations.

Addendum-2 : Computing $S_{(180)}$

I think that the clearest demonstration of the alternative approach is to begin with an example. Given the constraint that $x \leq y \leq z$, the first consideration is that

$$\left\lfloor \left(\frac{180}{1}\right)^{(1/3)} \right\rfloor = 5.\tag1$$

Therefore, $x$ must be an element in $\{1,2,3,4,5\}.$  As a further illustration of the algorithm, suppose that you are enumerating all positive integer solutions $(x,y,z)$ where $x=3$.  Consider that

$$\left\lfloor \left(\frac{180}{3}\right)^{(1/2)} \right\rfloor = 7.\tag2$$

Therefore, when $x=3, y$ must be an element in $\{3,4,5,6,7\}.$  Continuing the illustration of the algorithm, suppose that you are enumerating all positive integer solutions $(x,y,z)$ where $x=3$ and $y=5$.  Consider that

$$\left\lfloor \left(\frac{180}{3 \times 5}\right)^{(1/1)} \right\rfloor = 12.\tag3$$

Therefore, when $x=3$ and $y = 5,$ $z$ must be an element in $\{5,6,7, \cdots, 12\}.$  There are therefore $[(12 + 1) - 5] = 8$ distinct solutions associated with $x=3$ and $y=5$.

Let $f_k(M,a) : ~k,M,a \in \mathbb{Z^+}, ~a \leq M~$ denote :

$$\left\lfloor \left(\frac{M}{a}\right)^{(1/k)} \right\rfloor.$$

Then $$S_{(180)} = \sum_{x=1}^{f_3(180,1)}~
\sum_{y=x}^{f_2(180,x)}~
\sum_{z=y}^{f_1(180,[xy])}~\{1\} $$
$$=~ \sum_{x=1}^{f_3(180,1)}~
\sum_{y=x}^{f_2(180,x)}~\{1 + f_1(180,[xy]) - y\}.\tag4$$

Addendum-3 : Computing $S_{M}$

The analysis inherent in equations (1) through (4) of the previous section will be unchanged.  Therefore:
$$S_{M} ~=~ \sum_{x=1}^{f_3(M,1)}~
\sum_{y=x}^{f_2(M,x)}~\{1 + f_1(M,[xy]) - y\}.\tag5$$

In the original answer, I speculated that employing a computer program on a PC to compute (for example) $S_{(100,000)}$ via prime factorizations should run okay.  I now speculate that employing a computer program on a PC to compute (for example) $S_{1,000,000,000}$ via the alternative algorithm should also be okay.

Further, if $L$ is any random number such that
$(10)^{(9)} \leq L < (10)^{(10)}$, then using a PC to compute
$T_L = S_L - S_{L-1}$ should also be okay.  It is unknown how large $L$ can be to allow $T_L$ to be readily computable on a modern super computer.

The remainder of the addendums discuss using the computation of $T_L$ to determine the prime factorization of $L$.

Addendum-4 : Using the enumeration of $T_{180}$ to compute the prime factorization of $(180)$.

In fact, $T_{180} = 20$, rather than $18$, as computed by the OP.  This is explained as follows.

$180 = 2^2 \times 3^2 \times 5^1.$

Setting:

$X = 2^{x_1} \times 3^{x_2} \times 5^{x_3}$

$Y = 2^{y_1} \times 3^{y_2} \times 5^{y_3}$

$Z = 2^{z_1} \times 3^{z_2} \times 5^{z_3}$

and then using Stars and Bars analysis to compute the number of non-negative integer solutions to

$(x_1 + y_1 + z_1 = 2) ~: \binom{4}{2} = 6.$

$(x_2 + y_2 + z_2 = 2) ~: \binom{4}{2} = 6.$

$(x_3 + y_3 + z_3 = 1) ~: \binom{3}{1} = 3.$

Then the initial estimate of $T_{180}$ is $6 \times 6 \times 3 = 108.$

The second estimate of $T_{180}$ as $\frac{108}{3!} = 18$ is closer, but also wrong.  This second estimate assumes that each solution generated by the previous paragraph occurs $(3!)$ times, among the solutions $(x,y,z)$.  This is wrong, because $180$ is divisible by $4$ perfect squares, $\{1,4,9,36\}.$  Therefore, the initial estimate of 108 solutions must be partitioned into two groups:

The 12 solutions that constitute the 3 permutations each of $(1,1,180), (2,2,45), (3,3,20), (6,6,5)$ and the other 96 solutions.  These other 96 solutions each involve 3 distinct factors which thus generates (3!) repetitions each.

Therefore, the correct enumeration is
$$\frac{96}{3!} + \frac{12}{3} = 20.$$

So the (?? laughable ??) question becomes : how can you use the computation of $T_{180} = 20$ to compute the prime factorization of $(180)$.

Suppose that $(180) = (p_1)^{a_1} \times (p_2)^{a_2} \times \cdots (p_r)^{a_r}$, where

$p_1, \cdots, p_r$ are distinct primes in ascending order and $a_1, \cdots, a_r \in \mathbb{Z^+}$.

Then you want to enumerate all distinct solutions $(X,Y,Z)$, where $(XYZ) = 180$, and

$X$ has form $p_1^{x_1} \times \cdots \times p_r^{x_r}$

$Y$ has form $p_1^{y_1} \times \cdots \times p_r^{y_r}$

$Z$ has form $p_1^{z_1} \times \cdots \times p_r^{z_r}.$

The first thing to do is compute the number of distinct solutions to
$$S_1 : x_1 + y_1 + z_1 = a_1 : \binom{a_1 + [3-1]}{3-1} = \binom{a_1 + 2}{2}$$
$$S_2 : x_2 + y_2 + z_2 = a_2 : \binom{a_2 + [3-1]}{3-1} = \binom{a_2 + 2}{2}$$
$$~~~\cdots~~~$$
$$S_r : x_r + y_r + z_r = a_r : \binom{a_r + [3-1]}{3-1} = \binom{a_r + 2}{2}.$$

Then, you must compute the number of solutions to $(xyz) = (180)$ where all three numbers are the same : $[0]$, and the number of solutions to $(xyz) = (180)$ where two of the three numbers are the same $[4]$.

Further, since $180 < (2 \times 3 \times 5 \times 7)$, you know immediately that $r < 4.$  Therefore, you have the following constraints:

$r \in \{1,2,3\}.$

$S_1 \times \cdots \times S_r = [(3!)d + (3)e]$ where $e = 4,$ and

$d + e = T_{180} = 20 \implies d = 16 \implies$

$(S_1 \times \cdots \times S_r) = [(3!)(16) + (3)(4) = 108].$

At this point, you are looking for no more than 3 factors $S_1, \cdots, S_r$ such that $S_1 \times \cdots \times S_r = 108$ and $S_1, \cdots, S_r$ are (not necessarily distinct) elements from

$$\left\{\binom{1 + 2}{2} = 3, \binom{2 + 2}{2} = 6, \binom{3 + 2}{2} = 10, \cdots\right\}.$$

Since the whole point of illustrating this section is to facilitate computing the prime factorization of $L$, for a very large $L$, when $T_L$ is known, this is a reasonable stopping point for this section.

Addendum-5 : Using the enumeration of $T_{L}$ to compute the prime factorization of $L$, for very large $L$.

This is a convenient place to emphasize that my understanding of Number Theory is at the undergraduate
level (e.g. my involvement with quadratic reciprocity has cobwebs on it), and (for example) I have zero knowledge of computer resources needed to compute all primes less than (large) $L$.

For all I know, all of the ideas that I will mention in this section have already been considered.

First of all, for large $n$,

$$n ~\text{is prime}~ \iff T_n = 1.$$

Next, instead of defining $S_n = $ the number of distinct positive integer solutions to $(xyz) \leq n$  you could define it to be ${}_3S_n$.  Similarly, you could redefine $T_n$ as ${}_3T_n.$  This suggests (perhaps wrongly) that for (relevantly) large $L$, it might be both feasible and helpful to compute (for example)

$$\{{}_{(10)}T_L, {}_9T_L, \cdots {}_2T_L\}.$$

I have glossed over a point that may be critical:

for large $L$, for $n,m \in \{1,2,\cdots, 10\} ~: m \leq n,$ it is unclear how feasible it will be to compute how many (non-distinct) solutions to $(f_1 \times \cdots \times f_n) = L$ will have exactly $m$ identical factors.


问:
Consider a net and its associated filter, then the associated net of the filter is the original net?
Take a net $s: D \rightarrow (X,T)$ and then its associated filter, $\mathcal{F}_s$. Then we take the associated net of this filter, $s_{\mathcal{F}_s}$. Do $s$ and $s_{\mathcal{F}_s}$ coincide?

It makes sense that they should, otherwise our definitions of filters and nets leave something to be desired, however, $s_{\mathcal{F}_s}$is dependant on the containment of sets in $\mathcal{D_\mathcal{F}}$ which isn't much to do with the original ordering.

Is the statement true and how can you prove it?

Thanks :)

答:
It's true, in essence. The nets cannot be equal literally, because $s$ has $D$ as its indexing set (with some directed preorder) while $s_{\mathcal{F}_s}$ has an entirely different indexing set (usually some product of the filter and the set of points), not $D$. And a net is essentially a function and so these functions with different domains are not equal.

What is true is that they are subnets (specifically so-called AA-subnets) of each other. This is sometimes called being equivalent nets. A cluster point (or limit) of one is a a cluster point (limit) of the other and vice versa. So for "practical", "theorem proving" purposes they can often be treated as the same thing. See the book by Schechter on the foundations of analysis for more on such matters.


问:
If $f(x) = 2x^4-15x^3+ax^2+bx+c$, then if $(x-5)^2$ is a factor of $f$, then $f^{'}(5)=0$.
Let $f(x) = 2x^4-15x^3+ax^2+bx+c$, then if $(x-5)^2$ is a factor of $f$, then $f^{'}(5)=0$.

It is not immediately obvious why this is so, since this question is only awarded 1 mark. I actually have to let $f(x) = (x-5)^2(x-k_1)(x-k_2)$ and differentiate to find the answer. The product rule guarantees that the derivative will contain $x-5$, hence the answer.

but given it is 1 mark in an IB exam, is there any easier way?

答:
Yes, if $f(x)=0$ has one double root say $r$ $\implies f(r)=0=f'(r)$ and $(x-r)^2$ is a factor of $f(x)$.
Geometrically, at $x=r$, $y=f(x)$ touches the $x$-axis at $x=r$.


问:
how to prove $f^\prime(x)$ is not bounded?
$f$ is a differentiable function in $(0, 1)$. The limit $\lim_{x \to 0^+}f(x)$ doesn't exist. Prove that $f^\prime(x)$ is not bounded in $(0, 1)$.

I can see why it's true, but I am struggling proving it.
I hope you guys can help me. Thank you!

答:
Put $\alpha=\limsup_{x \to 0^{+}} f(x), \beta = \liminf_{x \to 0^{+}} f(x)$.

We may choose strictly decreasing sequences $a_n, b_n \to 0^{+}$ with $a_n \neq b_n$ for each $n$, and with $f(a_n) \to \alpha, f(b_n) \to \beta$. Note the former condition implies $a_n - b_n \to 0$. Hence $$\limsup_{n \to \infty} \left|\frac{f(a_n) - f(b_n)}{a_n - b_n}\right| = \limsup_{n \to \infty} \overbrace{|\alpha - \beta|}^{>0}\left|\frac{1}{a_n - b_n}\right| = +\infty$$

and the result follows from the MVT.


问:
Can $3p^4-3p^2+1$ be square number?
I know $4p^4+1$ can't be square number, cause $4p^4<4p^4+1<4p^4+4p^2+1$ for all natural number p.
But I don't know the way to prove $3p^4-3p^2+1$ can(not) be square number. Is there a well known way to prove it?

答:
For the equation $s^2=3p^4-3p^2+1$, substitute
$$u=\frac{1}{p},v= \frac{s}{p^2}.$$
Then $v^2=u^4-3u^2+3$ and therefore $4v^2=4u^4-12u^2+12$. Let $2v=2u^2-X$ and then
$$4(X-3)u^2=X^2-12.$$
Then $(X^2-12)(X-3)$ is a rational square and the original equation has been transformed into $$Y^2=X^3-3X^2-12X+36.$$
In this form the equation can be solved e.g. by Maple or Sage.

This elliptic curve has rank $0$ and solutions $(0,\pm 6),( 3,0),(4,\pm2)$. These give the only solution as $$p=\pm 1.$$

Title: Can $3p^4-3p^2+1$ be square number?


问:
Intuitive Question about Klein Bottles and Torsion in Surface Groups
This is just a geometric sort of question about how to picture/intuit on homology, hope it's considered attractive and not just "vague" haha

So if you take a typical generator for the torsion element in $H_1(K)$, where $K$ is the Klein Bottle, you can picture deforming the loop down the 'length of the tube' and getting the orientation reversed when you get back around.

However, I'm wondering if there is another way to visualize the torsion.  If you take a stretchy piece of string and wind it around twice the torsion element, then tie the ends off, is there a way to deform the string so that it simply "falls off" the Klein Bottle?  Or is this asking too much?

Obviously "falls off" is sort of a 3-d notion.  I guess what I'm asking is that if your Klein bottle is sitting in a high-dimensional Euclidean Space (I guess 5+ should suffice to avoid any complications at all, but maybe 4 is fine as well), is the map that winds a circle around (a tubular neighborhood around) the torsion element twice homotopy-equivalent to a free loop rel $K$?  Or is this not quite the right question?

Thanks!

答:
What you're asking sounds more like torsion in the fundamental group. The intuition works for instance for $\mathbb{RP}^2$. Wind your loop around twice and suddenly you can continuously deform your loop into a point. Since $H_1$ is the abelianisation of $\pi_1$, part of this intuition is reflected onto $H_1$ as well.

The case of the Klein bottle would be a bit more subtle because although H1 has torsion, it doesn't come from torsion in $\pi_1$ (I think $\pi_1$ is torsionfree). Somewhat tautologically, you should still be able to say that, upon going around your loop twice, it becomes the boundary of some $2$-cycle, and perhaps if you squint hard enough you can see this geometrically.

OK so I said that if you squint you get to see the torsion so let's squint. We have two $2$-simplices $A$ and $B$, and we can make sense of their boundaries. We have $\partial A = a + b - c$, and $\partial B = -a + b + c$. That means that $\partial(A + B) = 2b$, which means that by adding $b$ up to itself we suddenly get to realise it as the boundary of a $2$-simplex.

This makes sense from the picture. $A$ and $B$ together can be thought of as the entire square, and tracing along its boundary we see that we pass through $b$ twice with the correct orientation.


问:
Closed level sets of a harmonic function
Suppose that $U$ is an open, connected subset of complex plane. Are there any non-constant harmonic functions on $U$ which have at least one closed isocurve?

My results so far:

If U is simply connected, this is impossible by maximum principle.

If U is not simply connected then we can construct harmonic functions which don't have a harmonic cojugate, i.e. harmonic $u$ for which there is no harmonic $v$ such that $u+iv$ is holomorphic. An example is $u(x,y)=\mathrm{Re} \log (x+i y)$ defined on $U = \mathbb C - \{0 \}$ (imaginary part is discontinuous on this set). Its isocurves are circles. From now I restrict attention to the case where this doesn't happen, so $u=\mathrm{Re} f(z)$ for some $f$ holomorphic in $U$.

If a closed isocurve $C$ exists, it contains two critical points of $f=u+iv$. That's because we have
$$ 0 = \int_C \mathrm{d}z f'(z) = \int_C \mathrm d s \vec n \cdot \vec \nabla v, $$
where $\vec n$ is unit tangent vector to $C$ and $\mathrm d s$ is arc-length element. However it follows from Cauchy-Riemann equations and the fact that $u$ is constant on $C$ that $\vec \nabla v$ is parallel to $\vec n$. Therefore if $\vec \nabla v$ doesn't vanish anywhere on $C$ then $\vec n \cdot \vec \nabla v$ has constant sign on $C$, so we have a contradiction.

答: Examples are provided by functions $e^z-e^{\frac{1}{z}}$ as well as $z - \frac{1}{z}$. This was pointed out in comments and the question was never officially answered, so I decided to post this as community wiki.


问:
$\int_{0}^{r} \sqrt{\frac{1}{r^2-x^2}} \ dx$ with improper integration or trigonometric substitution?
I've always done this limit with improper integration because the denominator goes to zero, but is a trigonometric substitution valid as well? (edit, I know of the trigonometric substitution $x=r \sin\left(\theta\right)$ but because I'd have the limits of integration going from $0$ to $\frac{\pi}{2}$ then wouldn't I still have an improper integral?)

For example, my denominator would be $\sqrt{r^2-r^2\sin\left(\theta\right)}$ and that would become $r\cos\left(\theta\right)$ but that would still be division by zero

答:
It is a basic formula that the antiderivative is
$$\int\frac{1}{\sqrt{r^2-x^2}} \,\mathrm dx=\arcsin\frac xr +C,$$
so just apply the 2nd fundamental theorem of integral calculus.


问:
Can anyone help me find the infimum of the sequence $A=\{1/(1+x^2) : x \in \mathbb{Q}\}$
Can anyone help me with this problem? I'am sorry, I don't know how to write math formulas in stack exchange.

I'am trying to prove that infA=0.

Thank you very much.

答:
Guide:

Clearly $\forall x \in A, x \ge 0$.

Given a small positive number $\epsilon>0$, can you find $x \in \mathbb{Q}$ such that

$$\frac1{1+x^2} < \epsilon$$

Try to convince yourself that you can do so by solving the inequality for $x$.


问:
Being inside or outside of an ellipse
Let $A$ be a point $A$ not belonging to an ellipse $E$. We say that $A$ lies inside
$E$ if every line passing trough $A$ intersects $E$. We say that $A$ lies otside $E$
if some line passing trough $A$ does not intersect $E$. Let $E$ be the ellipse
with semi-axes $a$ and $b$. Show that

(a) a point $A=(x,y)$ is inside $E$ if $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}< 1$

(b) a point $A=(x,y)$ is outside $E$ if $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}> 1$.

I wrote down the equation of an arbitrary line $L$ passing through $A$. If $A$ is inside $E$ then clearly there is another point, say $B$, that belongs to $A\cap L$. I ended up with a bunch of big equations which didn't give any result.

答: Set the line $L$ be passing through the origin and a point $A(x_0,y_0),x_0\neq0$ inside the ellipse. So we have $$L: y=\frac{y_0}{x_0}x$$ This line should intersect the ellipse in another point(s), so we should consider two equations below simultaneously: $$y=\frac{y_0}{x_0}x,~~~ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ Then we get $$\left(\frac{y_0^2}{b^2x_0^2}+\frac{1}{a^2}\right)x^2=1~~~~\text{or}~~~~~\frac{y_0^2}{b^2}+\frac{x_0^2}{a^2}=\frac{x_0^2}{x^2}$$ But $\frac{x_0^2}{x^2}<1$.</p>

Title: Being inside or outside of an ellipse


答:
Your integral is $$I=\int_{0}^{r} \frac{dx}{\sqrt{r^2-x^2}}$$
Note that $J=\int_{a}^{b} \frac{1}{(x-a)^p} dx$ is finite if $p<1$.
It is in this sense that your integral $I$ is finite as $p=1/2$ here. Now take $x=r \sin t$, then
$$I=\int_{0}^{\pi/2} 1 dt=\pi/2.$$

Title: $\int_{0}^{r} \sqrt{\frac{1}{r^2-x^2}} \ dx$ with improper integration or trigonometric substitution?

Body:
I've always done this limit with improper integration because the denominator goes to zero, but is a trigonometric substitution valid as well? (edit, I know of the trigonometric substitution $x=r \sin\left(\theta\right)$ but because I'd have the limits of integration going from $0$ to $\frac{\pi}{2}$ then wouldn't I still have an improper integral?)

For example, my denominator would be $\sqrt{r^2-r^2\sin\left(\theta\right)}$ and that would become $r\cos\left(\theta\right)$ but that would still be division by zero


问:
Meaning of the space $W_0^{1,p}$
I don't understand why the space $W_0^{1,p}(\Omega)$ is defined like the closure of $C^{\infty}_c(\Omega)$ functions and not like the space of $W^{1,p}(\Omega)$ functions with compact support in $\Omega$.

I don't understand if this is a useful definition, and in this case, why this is a useful definition?

Or if there is another reason for this definition.

答:
Let $\Omega \subset \mathbb{R}^N$ be a bounded domain with sufficiently smooth boundary. I will use $W_c^{1,p}(\Omega)$ to denote the Sobolev functions with compact support. First of all, you would have
$$
W_c^{1,p}(\Omega) \subsetneq W_0^{1,p}(\Omega)
$$
One function, which is in $W_0^{1,p}(\Omega)$ but not in $W_c^{1,p}(\Omega)$ is e.g. for $\Omega=B_1(0) \subset \mathbb{R}^2$

$$
f(x)=\frac{1}{4}(1-|x|^2)
$$
You can show that $f \in W_0^{1,p}(\Omega)$ by noting that it is continuous up to the boundary and use the trace operator.

Set $p=2$ for simplicity. Then this function is the weak solution to the BVP
\begin{equation}
\begin{cases}
\Delta f=1 &  \textrm{in } \Omega \\
f=0 &  \textrm{in } \partial \Omega
\end{cases}
\end{equation}
By regularity theory and the maximum principle, this solution is unique in $W_0^{1,p}(\Omega)$. However, we also have
$$
f \notin W_c^{1,p}  (\Omega)
$$
In other words, the BVP would have no solution in $W_c^{1,p}(\Omega)$, but in $W_0^{1,p}(\Omega)$.
You can also construct the same type of argument for variational problems, just find problems which have no compatly supported function as a solution.

Title: Meaning of the space $W_0^{1,p}$


答:
There are several definitions for the infimum of a set of real numbers. Depending on what you are expected to know, some will be easier to manipulate in different circumstances, but I will go with the most basic one: the infimum of $A \subseteq \mathbb R$ is the greatest lower bound of $A$.

So you have two points to prove:

0 is a lower bound of your set, i.e. $\forall a \in A, 0 \leq a$.
This should be easy to prove.

If you take any real number $\varepsilon > 0$, then it is not a lower bound of $A$. Said differently, for any $\varepsilon > 0$, you can find an element $a \in A$ smaller than $\varepsilon$. This amounts to solving an easy inequality.

Another method, using the sequential characterization of the infimum, would be to find a sequence $(a_n) \in A^{\mathbb N}$ converging to $0$. This is a quicker, easier way to prove that $\inf A = 0$, but it is a bit more advanced (i.e. the fact that this gives an equivalent defintion of the infimum is not completely immediate).

Title: Can anyone help me find the infimum of the sequence $A=\{1/(1+x^2) : x \in \mathbb{Q}\}$

Body:
Can anyone help me with this problem? I'am sorry, I don't know how to write math formulas in stack exchange.

I'am trying to prove that infA=0.

Thank you very much.


问:
Let $A$ be a $4\times 4$ matrix such that: $r(A+I)=3$ and $r(A-3I)=2$, then which of these is true?
a) $A$ has $0$ as an eigenvalue.

b) $A$ is not diagonalizable.

c) $A$ has at least $3$ different eigenvalues.

d) $A$ has at least $3$ eigen vectors that are linearly independent.

e) we can't infer any of (a-d) about $A$.

My Attempt :

$r(A+I)=3$, $(A+I)$ is a $4\times 4$ matrix, with a rank of $3$ then it's not invertible and $det(A+I)=0$, which means $-1$ is an eigen value of $A$ with geometric multiplicity of $n-r(A+I)=4-3=1$.

Same way with $r(A-3I)$, and got that $|A-3I|=0$, which means $3$ is an eigen value with geometric multiplicity of $2$.

So I looked back and said (a) isn't true since $0$ isn't necessary the last eigenvalue. and couldn't infer any more information so I went with (e) as an answer, but the answer was (d).

Would appreciate it if someone can point out my mistakes or what I have missed and led me to the wrong answer, thanks in advance.

答:
To disprove $(a)$ and $(b)$ consider
$$\begin{bmatrix} -1 & & & \\ & 3 & & \\ & & 3 & \\ & & & 1\end{bmatrix}$$
and to disprove $(c)$ consider
$$\begin{bmatrix} -1 & & & \\ & 3 & 1 & \\ & & 3 & \\ & & & 3\end{bmatrix}.$$


问:
General solution for a linear differential equation of second order (Weber differential equation)
$$R''(z)+\lambda^2*(\frac{-z^2}{2}+\frac{1}{2})*R(z)=0$$

The differential equation is only one part of the bigger differential equation. I have solved the other part which was only an ODE of 2 order. Is there a way to solve this?

lambda here is just a constant!
Also, according to Wolfram Alpha this is a Weber differential equation.

答:
When $\lambda=\sqrt{2}$, you may check that $R(z)=e^{-z^2/2}$ is one solution. Otherwise this equation is parabolic Cylinder ODE whose solution is discussed here:

https://en.wikipedia.org/wiki/Parabolic_cylinder_function