问:
Showing that $\det(J_f(x))\det(J_{f^{-1}}(x))=1$

Let $f: U \to V$ be a diffeomorphism, where $U$ and $V$ are both subsets  of $\Bbb R^n$. Show that $\det(J_f(x))\det(J_{f^{-1}}(x))=1$ for all $x \in U.$

So I have that $
\det(J_f(x)) = \det
\begin{bmatrix}
\frac{\partial f_1}{x_1} & \frac{\partial f_1}{x_2} & \dots & \frac{\partial f_1}{x_n} \\[1ex]
\frac{\partial f_2}{x_1} & \frac{\partial f_2}{x_2} & \dots & \frac{\partial f_2}{x_n} \\
\vdots & \vdots & \ddots & \vdots \\[1ex]
\frac{\partial f_n}{x_1} & \frac{\partial f_n}{x_2} & \dots & \frac{\partial f_n}{x_n}
\end{bmatrix}
$ and similarly $\det(J_{f^{-1}}(x)) = \det
\begin{bmatrix}
\frac{\partial f^{-1}_1}{x_1} & \frac{\partial f^{-1}_1}{x_2} & \dots & \frac{\partial f^{-1}_1}{x_n} \\[1ex]
\frac{\partial f^{-1}_2}{x_1} & \frac{\partial f^{-1}_2}{x_2} & \dots & \frac{\partial f^{-1}_2}{x_n} \\
\vdots & \vdots & \ddots & \vdots \\[1ex]
\frac{\partial f^{-1}_n}{x_1} & \frac{\partial f^{-1}_n}{x_2} & \dots & \frac{\partial f^{-1}_n}{x_n}
\end{bmatrix}$

Now from the fact that $\det(AB) = \det(A) \det(B)$ I have that

$\det
\begin{bmatrix}
\frac{\partial f_1}{x_1} & \frac{\partial f_1}{x_2} & \dots & \frac{\partial f_1}{x_n} \\[1ex]
\frac{\partial f_2}{x_1} & \frac{\partial f_2}{x_2} & \dots & \frac{\partial f_2}{x_n} \\
\vdots & \vdots & \ddots & \vdots \\[1ex]
\frac{\partial f_n}{x_1} & \frac{\partial f_n}{x_2} & \dots & \frac{\partial f_n}{x_n}
\end{bmatrix} \cdot \det
\begin{bmatrix}
\frac{\partial f^{-1}_1}{x_1} & \frac{\partial f^{-1}_1}{x_2} & \dots & \frac{\partial f^{-1}_1}{x_n} \\[1ex]
\frac{\partial f^{-1}_2}{x_1} & \frac{\partial f^{-1}_2}{x_2} & \dots & \frac{\partial f^{-1}_2}{x_n} \\
\vdots & \vdots & \ddots & \vdots \\[1ex]
\frac{\partial f^{-1}_n}{x_1} & \frac{\partial f^{-1}_n}{x_2} & \dots & \frac{\partial f^{-1}_n}{x_n}
\end{bmatrix} = \det (\begin{bmatrix}
\frac{\partial f_1}{x_1} & \frac{\partial f_1}{x_2} & \dots & \frac{\partial f_1}{x_n} \\[1ex]
\frac{\partial f_2}{x_1} & \frac{\partial f_2}{x_2} & \dots & \frac{\partial f_2}{x_n} \\
\vdots & \vdots & \ddots & \vdots \\[1ex]
\frac{\partial f_n}{x_1} & \frac{\partial f_n}{x_2} & \dots & \frac{\partial f_n}{x_n}
\end{bmatrix} \begin{bmatrix}
\frac{\partial f^{-1}_1}{x_1} & \frac{\partial f^{-1}_1}{x_2} & \dots & \frac{\partial f^{-1}_1}{x_n} \\[1ex]
\frac{\partial f^{-1}_2}{x_1} & \frac{\partial f^{-1}_2}{x_2} & \dots & \frac{\partial f^{-1}_2}{x_n} \\
\vdots & \vdots & \ddots & \vdots \\[1ex]
\frac{\partial f^{-1}_n}{x_1} & \frac{\partial f^{-1}_n}{x_2} & \dots & \frac{\partial f^{-1}_n}{x_n}
\end{bmatrix})$

But this feels already very messy and I'm not sure how the multiplication turns out... Is there another way I should approach this?

答:
Let $f$ map a neighborhood $U$ of $p$ diffeomorphically onto a neighborhood $V$ of $q:=f(p)$. Then the map $g:=f^{-1}:\>V\to U$ is again differentiable, and we have
$$g\bigl(f(x)\bigr)=x\qquad(x\in U)\ .$$
The chain rule then says that
$$dg(q)\circ df(p)=d(g\circ f)(p)=d\,{\rm id}_U(p)={\rm id}_{T_p}\ .$$
In terms of matrices this means that
$$J_g(q) J_f(p)=I_n\ ,$$
so that
$$\det\bigl(J_{f^{-1}}(f(p))\bigr)\det\bigl(J_f(p)\bigr)=1\ .$$
Note that you have a slight typo in the stated formula: The $J_{f^{-1}}$ is taken at $q=f(p)$, not at $p$.


问:
Number of non-negative integer solutions of linear Diophantine equation

Given $n \ge 0$, find the number of non-negative integer solutions of the following equation $$2x_1+x_2+x_3+x_4+x_5=n$$

I tried putting one $x_1$ on the right-hand side, as follows

$$x_1+x_2+x_3+x_4+x_5=n-x_1.$$

Then I am not sure what to do. On the left-hand side, we now have non-negative numbers, but not sure how to write this sum because of the right-hand side.

ContentLicense: CC BY-SA 4.0

Body:
The number of nonnegative integer solutions of $2 x_1 + x_2 + x_3 + x_4 + x_5 = n$ is the coefficient of $t^n$ in the following generating function

$$\frac{1}{(1-t^2) (1-t)^4} = \frac{1}{(1+t) (1-t)^5}$$

which is the generating function for integer sequence OEIS A001752

$$a(n) = \left\lfloor \frac{((n+3)^2 - 1)((n+3)^2 - 3)}{48} \right\rfloor$$

Code

In Python:

>>> from sympy import *
>>> t = Symbol('t', real=True)
>>> f = 1 / ( (1 - t**2) * (1 - t)**4 )
>>> f.series(t,0,13)
1 + 4*t + 11*t**2 + 24*t**3 + 46*t**4 + 80*t**5 + 130*t**6 + 200*t**7 + 295*t**8 + 420*t**9 + 581*t**10 + 784*t**11 + 1036*t**12 + O(t**13)

In Haskell:

λ> map (\n -> floor(((n+3)^2 - 1)*((n+3)^2 - 3)/48)) [0..12]
[1,4,11,24,46,80,130,200,295,420,581,784,1036]

generating-functions


问:
Survival analysis: probability of dying between two given times
If the probability of surviving $1$ month is $0.5$ and the probability of surviving $2$ months is $0.3$, I understand that the conditional probability of surviving $2$ months given having survived one month is $0.3 / 0.5 = 0.6$.

However, I'm trying to calculate the probability at the outset of dying between $1$ and $2$ months. This must mean the probability of both surviving $1$ month and surviving less than $2$ months given having survived $1$ month. How do I set up the math for this? If they were independent events, we could multiply: $0.5 \times 0.3 = 0.15$. But they're not independent, are they?

答:
Let $A_i$ be the event of surviving at the end of month $i$.
So $A_1$ is the event of surviving at the end of first month.
Similarly $A_2$.

Then $A_1,A_2$ are dependent because$$P(A_1\wedge A_2)=P(\text{surviving at the end of first and second month})=P(A_2)\\\ne P(A_1)\times P(A_2)$$

In fact $A_2$ can happen only if $A_1$ does.

You want to find $P(\text{surviving at the end of first month and not surviving at the end of second month})=P(A_1\wedge A_2^C).$

First remember for any two events $P,Q,P(P|Q)+P(P^C|Q)=1$. So$$\begin{align*}&P(A_2^C|A_1)=1-\underbrace{P(A_2|A_1)}_\color{red}{\text{found by you as }0.6}=0.4\\\implies&P(A_2^C\wedge A_1)=0.4P(A_1)=0.4\times0.5=0.2.\end{align*}$$


问:
Find the area of the next polygon
The problem didn't give me any angles.
I was approaching the angle of 640 and 650 as a right angle to calculate the hypotenuse and thus have an edge, but I don't think it is the most correct

答:
$Area\approx 344,430\text{m}^2$

I used GeoGebra. Imported the picture, drew a polygon over the given one, made a proportion and got the result. There is no other way to solve without  angles.

Title: Find the area of the next polygon


问:
counterexample for cycles in a graph
I'm trying to find a counterexample to the statement that a graph that has cycles containing x and y and a cycle with all 3
I know this is a false statement</p>

答: Cycle with $(x,y)$, cycle with $(y,z)$, cycle with $(x,z)$ but no cycle with all 3.

Title: counterexample for cycles in a graph


问:
How many times harder is to think in 3D as compared with 2D?
I'm a chemist, currently going through a course about molecular symmetry and group theory applied to Chemistry. This subject is very demanding in terms of visualization in 3D space. To really grasp the subject, one must be able to see, for example, how the orbitals around a given nucleus transform when subjected to the different symmetry operations in space.

I'm stumped by the sheer difficulty I have at it, specially given I deem myself good at reasoning with geometric concepts in the 2D plane. Add just one dimension and now it feels like to be in a quagmire.

That makes me wonder, can you put hard numbers in this increased overhead when going from 2D to 3D? How much more processing power is necessary to reason about 3D space? A linear scaling can be surely ruled out as, at least for me, 3D doesn't feel like just 50% more difficult. It feels like a several fold increase in complexity. Could be the case it scales like area or volume, as a power of the number of dimensions, say, n³? What about a generalization to higher dimensions? How much more difficult a 4D being would have reasoning about 4D as compared with 3D?

I would not be surprised if grasping something like a 4D were very difficult or even impossible for us, as we don't experience 4D spatial dimensions. But we do live in 3D space, and yet thinking in 3D can be very hard.

答:
It is often with examples that one can approach a subject.

I give here 5 counter-examples where 3D can be a "blessing" for understanding a 2D situation...

A) First, an answer of mine some time ago about Monge and Desargues theorems.

B) A different situation where a family of curves can be more globally interpreted as a set of level curves on a surface. Consider for example the following representation of a double family of hyperbolas situated "on" the parabolic hyperboloid with equation $z=x^2-y^2$:

Remark: The horizontal sections are hyperbolas whereas the vertical sections are parabolas.

C) Still with a level curve interpretation, but with a physical situation: the fact that sound waves emitted from one of the foci of an elliptical mirror are reflected on the mirror in such a way that they are eventually re-focalized into the other focus. You will find a description in this answer of mine (again !): begin by the last figure and consider the figure before this one as a oblique "lifting" in 3D of the 2D situation: just an issue of (shifted) level curves...

Edit (Nov 16, 2020)

D) Let me present here, still another case, with some common features with example C). Consider the first figure below:

It describes (in a classical way) a conical curve by a point (a focus), a line (its directrix) and a number $e$ (its eccentricity) as the set of points $M$ such that the distances' ratio (from the point to the focus and the shortest distance from the point to the directrix) is constant and equal to $e:

$$MF/MH = e\tag{1}$$

This process encompasses the three types of conic curves : ellipse if $e<1$, parabola if $e=1$, hyperbola if $e>1$.

Could this picture be seen in a different manner ? Yes, in the following way (second figure): we can transform  in a certain vertical plane our different types of curves in a unified type, i.e.,  circles, moreover concentric... with a well placed light source $S$:

In fact, afterwards, one can think to the different curves of the first figure as the limits of the light projected on the ground by a torchlight having an adjustable conical beam width...

Remarks:

The "mathematical machinery" behind the transformation seen in the second figure is "projective geometry" which has become a recognized branch of geometry in the 19th century, whereas the definition of conics we just saw using foci, directrix and eccentricity has $2300$ years...

Squaring relationship (1) gives a 2nd degree equation common to all conics:

$$x^2+y^2=e^2(1-x)^2.$$

E) See this recent answer as well...


问:
how to prove $f$ is an arithmetic function with this property $\sum_{d\mid n} f(d)=n^2$
how to prove $f$ is an arithmetic function with this property
$$\sum_{d\mid n} f(d)=n^2$$

Arithmetic function

答:
I think by saying arithmetic function,OP means that $f(1)=1$ and for all $a,b,$
$$(a,b)=1 \Rightarrow f(a)f(b)=f(ab).\tag1$$
From  Möbius inversion formula as Frank Science commented,
$$f(n)=n^2\sum_{d|n}{\dfrac{u(d)}{d^2}}=n^2\prod_{p|n}{(1-\frac{1}{p^2})},\tag2$$
for every $n\geq1.$

Now it's obvious that $f(1)=1$,and $(1)$ is hold.Hence $f(n)$ is an arithmetic function.

ContentLicense: CC BY-SA 3.0


问:
Radius of limit cycle for van der Pol oscillator in the limit of $\varepsilon\ll1$ using Green's Theorem

We want to determine the radius of the nearly circular limit cycle of the van der Pol oscillator $$\ddot x + \varepsilon \dot x \left( x^2 - 1 \right) + x = 0$$ in the limit $\varepsilon\ll1$. Assume that the limit cycle is a circle of unknown radius $a$ about the origin, and invoke the normal form of Green’s theorem (i.e., the $2$-dimensional divergence theorem) $$\oint_C\mathbf{v}\cdot\mathbf{n} \, dl=\iint_A \nabla\mathbf{v} \, dA$$ where $C$ is the cycle and $A$ is the region enclosed. By substituting $\mathbf{v}=\dot x =(\dot x,\dot y)$and evaluating the integrals, show that $a\approx2$.

I first rewrote the van der Pol equation in terms of $x$ and $y$ to get
$$\dot x=y \\ \dot y=-x-\varepsilon y(x^2-1)$$
The double integral then becomes
$$\iint_A \nabla\mathbf{v} \, dA=\iint_A -\varepsilon (x^2-1) \, dA$$
rewriting this in polar coordinates, we get that
$$\iint_A -\varepsilon (x^2-1) \, dA=\int_0^{2\pi}\int_0^a -\varepsilon r(r^2\cos^2(\theta)-1) \, dr \, d\theta \\=\int_o^{2\pi}-\frac{\varepsilon}{4}r^4\cos^2(\theta)+\frac{\varepsilon}{2}r^2\biggr|_0^ad\theta \\=\int_o^{2\pi}-\frac{\varepsilon}{4}a^4\cos^2(\theta)+\frac{\varepsilon}{2}a^2d\theta \\=-\frac{\varepsilon}{4}a^4\left(\frac{\sin(\theta)\cos(\theta)+\theta}{2} \right) + \frac{\varepsilon}{2}a^2\theta\biggr|_0^{2\pi} \\=\varepsilon a^2\pi-\frac{\varepsilon a^4\pi}{4} \\ =\varepsilon a^2\pi(1-\frac{1}{4}a^2)$$

I'm not exactly sure how to get this answer when doing the line integral. It seems the $\varepsilon$ term disappears. My work is as follows:

Using $r(t)=\langle a\cos(t), a\sin(t) \rangle$ such that $r'(t)=\langle -a\sin(t), a\cos(t) \rangle$, we have
$$\oint_C\mathbf{v}\cdot\mathbf{n} \, dl=\int_0^{2\pi}\langle a\sin(t),-a\cos(t)-\varepsilon a\sin(t)(a^2\cos^2(t)-1)\rangle \cdot \langle -a\sin(t), a\cos(t) \rangle \, dt \\\int_0^{2\pi}=-a^2-\varepsilon a^4 \sin(t)\cos^3(t)+\varepsilon a^3\sin(t)\cos(t) \,dt \\=-a^2t+\frac{a^4}{4} \varepsilon \cos^4(t)-\frac{a^2}{2}\varepsilon \cos^2(t)\biggr|_0^{2\pi} \\=-a^22\pi$$
If anyone could point out where I went out in either integral I'd be super appreciative!

答:
You should use instead $\mathbf{n}=\langle \cos(t), \sin(t) \rangle$ such that $||\mathbf{n}||=1$, and we have
$$\oint_C\mathbf{v}\cdot\mathbf{n} \, dl=\int_0^{2\pi}\langle a\sin(t),-a\cos(t)-\varepsilon a\sin(t)(a^2\cos^2(t)-1)\rangle \cdot \langle \cos(t), \sin(t) \rangle \, dt \\\int_0^{2\pi}=-\varepsilon a^4 \sin(t)\cos^3(t)+\varepsilon a^3\sin(t)\cos(t) \,dt \\=\frac{a^4}{4} \varepsilon \cos^4(t)-\frac{a^2}{2}\varepsilon \cos^2(t)\biggr|_0^{2\pi} \\=0$$
Now you can see that these two are only equivalent if $a=2$


问:
If $|x|^2u=0$ for $u\in\mathcal D'(\mathbb R^d)$, then $\psi\in\mathcal D(\mathbb R^d)$ and $\operatorname{supp}(\psi)\cap B(0,10)$ imply $u(\psi)=0$
Problem: Show that if $\vert x\vert^2u=0$ for $u\in\mathcal D'(\mathbb R^d)$, then for all $\psi\in\mathcal C_c^\infty(\mathbb R^d)$ such that $\operatorname{supp}(\psi)\cap B(0,10)=\varnothing$ we have $u(\psi)=0.$

My Thoughts: I think that the problem statement is not clear, as there seems to be nothing special about $B(0,10)$. For example, fix $\varepsilon>0$ and consider $B(0,\varepsilon)$. Note that $x\mapsto \vert x\vert^2$ is smooth on $\mathbb R^d\setminus B(0,\varepsilon)$. Therefore, if $\psi\in\mathcal D(\mathbb R^d)$ has $\operatorname{supp}(\psi)\cap B(0,\varepsilon)=\varnothing$, then
$$u(\psi)=\vert x\vert^2u\left(\frac{\psi}{\vert x\vert^2}\right)=0.$$
I justify the above calculation by noticing that $\psi/\vert x\vert^2\in\mathcal D(\mathbb R^d\setminus B(0,\varepsilon))$ and $\operatorname{supp}(\psi/\vert x\vert^2)=\operatorname{supp}(\psi)$. Also if $u\in\mathcal D'(\mathbb R^d)$, then $u\vert_{\mathbb R^d\setminus B(0,\varepsilon)}\in\mathcal D'(\mathbb R^d\setminus B(0,\varepsilon)).$

Therefore, it seems that the result holds for any ball centered at the origin.

Could anyone tell me if I am on the right track or if I have made a horrible mistake at the beginning in my understanding of the definitions?

Thank you for your time and appreciate the valuable feedback.

答:
You are correct that there is nothing special about radius $10$. You can as well take radius $R>0$ or $\epsilon>0.$ That doesn't make the problem statement non-clear, though. But I understand that it might be confusing.

I was about to write that I do not understand how you justify $u(\psi)=|x|^2u(\psi/|x|^2)$ as $|x|$ is not a constant. But then I realized that you do not mean $|x|^2$ multiplied with $u(\psi/|x|^2)$ but the distribution $|x|^2 u$ applied to $\psi/|x|^2$. In the "bracket" notation that I'm used to it would be $\langle u, \psi \rangle = \langle |x|^2 u, |x|^{-2}\psi \rangle.$ So, yes, you are on the right track; $|x|^{-2}$ is smooth outside of the ball.


问:
Why is $\sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{k}=\frac{1/3}{2/3}$
$\sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{k}=\frac{1/3}{2/3}$ what theorem or algebra leads to this equality?

EDIT: The sum should have been infinite.

Body:
Sum should be infinite, because for finite sum you should get:
$$\sum_{k=0}^{n}(\frac{1}{3})^k=1+\frac{1}{3}+\dots+(\frac{1}{3})^n=\frac{1-(\frac{1}{3})^{n+1}}{1-\frac{1}{3}},$$ where we used formula $1+q+\dots+q^n=\frac{1-q^{n+1}}{1-q}$ for $q=\frac{1}{3}$.
Therefore,
$$\sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{k}=\frac{1}{3}\sum_{k=0}^\infty \left(\frac{1}{3}\right)^{k}=\frac{1}{3}\lim_{k\to\infty}\sum_{i=0}^k (\frac{1}{3})^i=\frac{1}{3}\lim_{k\to\infty}\frac{1-(\frac{1}{3})^{k+1}}{1-\frac{1}{3}}=\frac{1}{3}\frac{1}{\frac{2}{3}}=\frac{1}{2},$$
since
$$\lim_{k\to\infty} (\frac{1}{3})^{k+1}=0.$$


问:
How to represent the following in predicate logic
I'm struggling to represent the following in logic:

"When he was watching people in game, Jason only seen people on 2 teams."

The relations include:
watching(x,y), IsOnTeam(x,y).

From my interpretation of the sentence, I got:

∃a∃b∀x( (-(a=b) ^ watching(Jason,x)) -> ( (IsOnTeam(x,a) | IsOnTeam(x,b)) ^ ∀y(IsOnTeam(x,y) -> (y=a|y=b))) ) )

Which in my head translates to:
There Exists a team A and Exists a team B for All players such that team A!=team B, and if Jason is watching player X, then player X must be on team A or B, and if there is a team C, then C must be A or B.

I believe the last part is necessary to show that he could only see exactly 2 teams.

The second one I'm struggling with is:

"Everybody in game was playing"

For this I got:

∀x(Playing(x) -> ∃a(IsOnTeam(x,a))

Meaning for all people, if they were playing then there must be a team which they are playing on.

答:
Let $\cal{T}$ be the set of all teams and $\cal{P}$ the set of all people in the game. Then the sentence

When he was watching people in game, Jason only seen people on 2 teams.

can be written as

$$\exists A,B \in \mathcal{T}, \forall x \in \mathcal{P} : \text{watching}(\text{Jason},x) \wedge \big(\text{IsOnTeam}(A) \vee \text{IsOnTeam}(B))$$

I believe this is the right answer. Translating back to English would mean that

There exists teams $B$ and $C$ such that for all person $x$, Jason was watching $x$ and $x$ was on team $A$ or team $B$.

Note that you don’t have to guarantee that $A \neq B,$ because if it is the same team then

$$\begin{align} \text{IsOnTeam}(A) \vee \text{IsOnTeam}(A) \iff \text{IsOnTeam}(A) && (\text{Idempotent Law})\end{align}$$

Also, note that I am assuming that everybody is necessarily on a team.


问:
Is a modular element of maximal ideal unique if it exists?
Let $A$ be an algebra over the complex numbers (not necessary unital or commutative). An ideal $I$ in $A$ is called modular if there exists a non-zero $e\in A$ such that $ea-a\in I$ and $ae-a\in I$ for all $a\in A$. Equivalently, $e+I$ is a unit for the quotient algebra $A/I$. In this case, we call $e$ a modular element for $I$. Also, it is easy to see that $e\in I$ if and only if $I=A$.

If $A$ is unital, then $1$ is a modular element for every ideal in $A$.

Although I have not seen a counterexample yet, I am quite sure that a modular element is  not unique in general. Namely, if $e,f\in A$ are both modular elements for $I$, then we only know that $e+I=f+I$ by uniqueness of the unit and this does not imply that $e=f$.

Now I was wondering if we can make assumptions on $I$ that give uniqueness of $e$.

My first guess was maximality: Can we conclude that $e=f$ when $I$ is maximal?

(If no, are there other conditions on $I$ that guarantee uniqueness?)

EDIT: My second guess is the assumption that we can write $I=\ker(\phi)$ for some non-zero homomorphism $\phi\colon A\to\mathbb{C}$. Then $I$ is automatically a modular maximal ideal (so this case is included in my other guess). I can prove that $\phi(e)=1=\phi(f)$...

答:
Although I have not seen a counterexample yet, I am quite sure that a modular element is not unique in general.

It looks like you didn't notice that if $i$ is any element of $I$ at all,

$(e+i)a-a=(ea-a)+ia\in I$ for every $a\in A$, making every element in the coset $e+I$ a modular element for $I$.

That means there's only one condition to make it unique: $I=\{0\}$.


问:
What does this definition of the upper and lower integral mean?
I'm struggling to understand this definition for a class project on Riemann integration.

Let $\mathcal{P}$ represent all possible partitions over $[a,b]$, where $f:[a,b]\rightarrow\mathbb{R}$, then the upper integral of $f$ is given by $U(f)=$ inf$(\{U(f,P):P\in\mathcal{P}\})$ and the lower integral of $f$ is given by $L(f)=$ sup$(\{L(f,P):P\in\mathcal{P}\})$.

I understand what it's saying, but I have no idea how to apply this. The example I'm illustrating is $f(x)=x$ over the interval $[0,2]$ with four subintervals split off by halves.

I calculated the lower sum as 1.5 and the upper sum as 2.5 by summation. The kicker here is Abbott chose to use the concepts of infimum and supremum with the Axiom of Completeness rather than Riemann sums and limits (so I'm really lost since I'm only familiar with the latter method).

How do I determine the upper and lower integrals for this function?

答:
One approach to all this is to find a sequence of partitions $P_n$ for which $\lim_{n \to \infty} L(f,P_n) = \lim_{n \to \infty}U(f,P_n) = L$.  If you do this, then we can say that
$$
L = \lim_{n \to \infty} L(f,P_n) \leq \sup_{P \in \mathcal P} L(f,P) \leq \inf_{P \in \mathcal P} U(f,P) \leq \lim_{n \to \infty} U(f,P_n) = L,
$$
which means that these four quantities must all be the same.

In fact, you should find that if you take any "common sense" choice of $P_n$ (e.g. breaking the interval into $n$ equal subintervals), then these two limits are indeed equal. All the "math" that needs to be done here is straightforward; the trick from there is to just use the argument I present to say that $L(f) = U(f) = 2$.

TL;DR: use the usual "Riemann sums and limits" method (for both the left endpoint and right endpoint sums), but provide a logical argument that this leads to the correct answer.


问:
Searching for a neat proof
I'm searching for a neat solution for the following exercise: << Prove that:
$$\forall a,n\in \mathbb{N^*}:\,((a\leq n \, \, \wedge 0\leq50-a\leq100-n)\Rightarrow \, \frac{a}{n}+\frac{50-a}{100-n}\leq \frac{148}{99})  $$

and determine for which couples $(a,n)$ equality holds.>>

Actually I solved the exercise, but I didn't manage to find a solution which does not use derivatives, so an elementary one.

My solution is the following:

Let's call $f(a,n)$ the expression contained in the LHS of the inequality. Since $f(a,n)=f(50-a,100-b)$ with $50-a,\, 100-n$ respecting the same conditions on $a$ and $n$, we can say WLOG $\,\,\,n\leq 50$.

Let's consider the function $g: \, [0,1]\times \{n\in \mathbb{N^*}\vert \,\, n\leq 50\}\rightarrow \mathbb{R}$ such that
$$g(x,n)=x+\frac{50-nx}{100-n}$$

Then

$$\frac{\partial g(x,n)}{\partial x}=1-\frac{n}{100-n}=\frac{100-2n}{100-n}\geq 0$$

and so, fixed the value of $n$, the maximum value of $g(x,n)$ is reached only when $x=1$. So the maximum value of $g(x,n)$ is reached when $$1+\frac{50-n}{100-n}$$ is maximum, and so only when $n=1$. So the maximum value of $g(x,n)$ is $1+49/99=148/99$. Noticing that $g(a/n,n)=f(a,n)$, we obtain that
$$f(a,n)\leq \frac{148}{99}$$
with the equality if and only if $(a,n)=\{(1,1),(99,99)\}$.

Is there a simpler solution?

答:
It's equivalent to finding the maximum of $\frac an + \frac bm$ with the constraint $a+b=50, n+m=100, a\leqslant n, b\leqslant m$.

BTW I believe your condition should be $1\leqslant n \leqslant 99$.

Since $100=2\times 50$, $\min (\frac an, \frac bm) \leqslant \frac{1}{2} \leqslant \max (\frac an, \frac bm)$. WLOG assume $\frac an \leqslant \frac bm$.

If $\frac an = \frac bm$ then both are exactly one half, the sum is 1.

If $\frac an < \frac 12 < \frac bm$, $2a < n, n\geqslant 2a+1, a\leqslant 49$, then

$$
\frac an \leqslant \frac{a}{2a+1}=\frac 12 \left( 1 - \frac{1}{2a+1}\right) \leqslant \frac 12 \left( 1- \frac{1}{99}\right)=\frac{49}{99}.
$$

(basically it means the largest integer fraction less than one half, with denominator not exceeding 100, is 49/99.)

and
$$
\frac bm \leqslant 1.
$$

When $a=49, n=99$, both inequalities become equality simultaneously.

Therefore $\frac an + \frac bm \leqslant \frac{49}{99}+1 = \frac{148}{99}. \blacksquare$

Title: Searching for a neat proof


问:
Does the existence of a function in this proof require the axiom of choice?
I have the following proof which I think might use the axiom of choice implicitly:

Let $\mathcal{A}$ be a pairwise disjoint cover of a set $S\neq \emptyset$ such that $A\cap S\neq \emptyset$ for all $A\in \mathcal{A}$. Then $\vert \mathcal{A}\vert \leq \vert S\vert$, since we have an surjective function $\Phi:S\to \mathcal{A}$ by choosing for all $s\in S$ the unique set $A\in\mathcal{A}$  for which $s\in A$. It seems to me that this map is indeed well defined since the assumptions on $\mathcal{A}$ ensure such an $A$, but this feels like there is a choice being made here.

I would appreciate if someone could point out if I'm overlooking something, or just overcomplicating the argument.

答:
Yes, of course there is a use of AC here. You are proving, in effect, that if there is a surjection from $X$ onto $Y$, then there is an injection from $Y$ into $X$.

This is known as "The Partition Principle", and I have minced many a word on that topic over this site and the interwebs. It implies, for example, that every infinite set has a countably infinite subset, or that every well-ordered family of non-empty sets admits a choice function. Both of these are known to be unprovable from $\sf ZF$.

You might ask whether or not The Partition Principle implies the Axiom of Choice, that is, do we need the full power of AC to prove this statement. And the answer is that we do not know yet. This has been open for over a century. Very recently a proof was suggested using a new technique, but there are several significant flaws and gap in that paper, so the problem is still open.


问:
Simplify using laws of indices: $\frac{\sqrt{1-x}\,\frac12(1+x)^{1/2} + \frac12(1-x)^{-1/2} \sqrt{1+x} }{1-x}$

Simplify using the laws of indices:

$$\frac{\sqrt{1-x}\,\frac{1}{2} (1+x)^{1/2} + \frac{1}{2}(1-x)^{-1/2} \sqrt{1+x} }{1-x}$$

I tried factorizing it using surds and I am stuck

Body: Split into two terms,$$\begin{align}\tfrac12(1-x)^{-1/2}(1+x)^{-1/2}+\tfrac12(1-x)^{-3/2}(1+x)^{1/2}&=\tfrac12(1-x)^{-3/2}(1+x)^{-1/2}(1-x+1+x)\\&=(1-x)^{-3/2}(1+x)^{-1/2}.\end{align}$$You're free to write that in other ways of your choice, e.g. $(1-x^2)^{-3/2}(1+x)$ or $(1-x^2)^{-1/2}(1-x)^{-1}$.


问:
Let $A\in M_{5×5}(\mathbb{R})$ be a matrix such that $\operatorname{rank}(A)=2$ and $A^3 = 0$. Is A guaranteed to be diagonalizable over R?
Let $A\in M_{5×5}(\mathbb{R})$ be a matrix such that $\operatorname{rank}(A)=2$ and $A^3 = 0$. Is A guaranteed to be diagonalizable over R?

I've been searching my notes for some sort of theorem that would help me with this problem, but I have had no luck.

Without the characteristic polynomial or actual eigenvalues, I can't seem to use the two step test for diagonalizability or any other theorem I know of for that matter.

Any thoughts?

答: If $A$ was diagonalizable, then $D^3=0,$ so $D=0,$ so $A=0,$ contradicting the fact that the rank is $2.$


问:
Closed form of geometric series $\sum_{i=1}^n p^{i+1}$
I know that $\sum_{i=1}^n p^i = \frac{p-p^{n+1}}{1-p}$, but I am not sure how the i+1 factors into the closed form for $\sum_{i=1}^n p^{i+1}$, what is the closed form for the second sum?

答:
You basically answered your question yourself.

You know,
$\sum_{i=1}^n p^i = \frac{p \left(p^n-1\right)}{p-1}$

Now

$$\sum_{i=1}^n p^{i+1} = p \sum_{i=1}^n p^i$$

So you get
$$\sum_{i=1}^n p^{i+1} =  \frac{p^2 \left(p^n-1\right)}{p-1}$$


问:
If $\int {yy''dx} = 3xy$ is it possible to find y'?
So one of the people I know on discord gave me this problem to solve for fun.
I've tried using integration by parts on the left side and it didn't do anything.
The integral transforms into the ODE $yy''- (3xy)'=0$ but it's non-linear and I have no idea how to solve that. Would a series solution work?
Notice that I'm looking for y' not y so maybe there's some trick there?
Any help is appreciated. Thank you for reading.
Here, $y'$ and $y''$ denote the first and second order derivatives of $y$ w.r.t. $x$.

答:
$$3 x y'+3 y=y y'';\;y(0)=1,y'(0)=0$$
Approximated graph of $y'$

Approximated graph of $y$


问:
Exercise in Peano arithmetic and order relation
Show that if m<n. and k$\ne 0$ that km<kn

Working assumptions:

6.22 Theorem Let m $\leqslant$  n denote the fact that m ∈ n or m = n. Then the relation   is an order relation in ω.

6.19 Theorem (Associative Law for Multiplication). (mn)k = m(nk).

Attempted proof

If m<n;by 6.22 m=n or m$\in$n

If m=n we are done

If m$\in$n and k$\ne 0$ by 6.19 mk$\in$nk
Then if mk$\in$nk; mk$\subseteq$nk

Is it possible to do this theorem inductively on n?
Help

答:
Welcome to MSE!

Hint: induct on $k$. At some point you'll need to know that $<$ is compatible with addition. By this I mean

$$m < n \quad \& \quad a < b \implies m+a < n+b$$

If you've already proven this, great. If not, it is an easy induction on $a$.

I hope this helps ^_^


问:
Construction of QR decomposition for a singular matrix
Consider a matrix $A \in \mathbb{C}^{n \times n}$. Suppose it's non-singular. Then it's columns $a_1, a_2, ..., a_n$ form a basis in $\mathbb{C}^n$.

Let's apply Gram-Schmidt process to them. We will obtain some basis $q_1, q_2, ..., q_n$. The transition matrix $S$ from the basis $a$ to the basis $q$ is upper-triangular(by construction).
Now, consider the matrix $Q=\big[ q_1|q_2|...|q_n \big]$. It's columns are orthonormal $\implies$ $Q$ is unitary.

And now we know that the matrix $A$ can be represented as:
$$
AS=Q \\
A=QS^{-1}
$$

$S$ is invertible since it is a transition matrix. $S$ is upper-triangular $\implies$ $S^{-1}$ is upper-triangular as well.

E.g. we have shown that each non-degenerate matrix $A$ can be represented as a product of unitary and upper-triangular matrices.

But what if $A$ is singular? I've tried the following:

Let's select a basis of column space of $A$: $a_1, ..., a_r$ (they are columns of $A$). And also let's reorder columns of $A$ s.t. these linearly independent vectors $a_1, ..., a_r$ are the first $r$ vectors of the matrix. So, we've done permutation of columns of $A$. This operation can be represented by right-multiplication of A with some matrix $P$: $A \rightarrow AP$.

Now we apply Gram-Schmidt to the basis of column space(vectors $a_1,...,a_r$), and after extend the resulting basis to an ONB of $\mathbb{C}^n$:

$$q_1,q_2, ..., q_n$$

We have constructed matrix $Q=\big[ q_1 | q_2 | ... | q_n \big]$ that is unitary(since it's columns form an ONB).
The matrix $S$ s.t. $AP=QS$ is upper-triangular. Moreover, it looks like:
$$
\begin{bmatrix}
s_{11} & s_{12} & s_{13} & ... & s_{1r} & ... & s_{1n} \\
0 & s_{22} & s_{23} & ... & s_{2r} & ... & s_{2n} \\
0 & 0 & s_{33} & ... & s_{3r} & ... & s_{3n} \\
... & ... & ... & ... & .... & ... & ... \\
0 & 0 & 0 & ... & s_{rr} & ... & s_{rn} \\
0 & 0 & 0 & ... & 0 & ...  & 0\\
0 & 0 & 0 & ... & 0 & ...  & 0\\
... & ... & ... & ... & .... & ... & ... \\
0 & 0 & 0 & ... & 0 & ...  & 0\\
\end{bmatrix}
$$

But how to show now that the matrix $SP^{-1}$ is upper-triangular? For me, it seems like it's not. Can you see a mistake in my thoughts? Maybe, you have other suggestions for construction of $QR$-decomposition in case of singular matrix? I'm stuck with this issue for a week now and I have no ideas what to do. Thanks!

答:
conceptually the simplest approach using Gram Schmidt, is to lean on what you already know.  Suppose $\text{rank}\big(A\big)=r$.

1.) run Gram Schmidt on $A$ as you normally do, and each time you run into a linearly dependent column, discard it.  The end result is

$$A' = Q'R'=\bigg[\begin{array}{c|c|c|c} \mathbf q_1' & \mathbf q_2' &\cdots & \mathbf q_{r}'\end{array}\bigg]R'$$

where $A'$ is $n\times r$ and $R$ is $r\times r$ and upper triangular.  This is equivalent to running QR facotrization on a tall skinny matrix with full column rank.

2.) Now extend $Q'$  to an orthonormal basis (i.e. 'find' $n-r$ linearly independent vectors not in the span of $A'$ and run more Gram Schmidt. Common places to look -- left nullspace of $A$ or use a random number generator, ref here: In $A = QR$ factorization do we need basis vectors from the left null space?)

3.) Now one at a time re-insert the deleted columns to recreate $A$.  E.g. suppose $\mathbf a_1$ was non-zero but $\mathbf a_2$ was linearly dependent (and hence deleted when constructing $A'$).  So re-insert $\mathbf a_2$  on the LHS

$$A'=\bigg[\begin{array}{c|c|c|c} \mathbf a_1' & \mathbf a_2' &\cdots & \mathbf a_{r}'\end{array}\bigg]=\bigg[\begin{array}{c|c|c|c} \mathbf a_1 & \mathbf a_2' &\cdots & \mathbf a_{r}'\end{array}\bigg]\rightarrow \bigg[\begin{array}{c|c|c|c|c} \mathbf a_1 & \mathbf a_2& \mathbf a_2' &\cdots & \mathbf a_{r}'\end{array}\bigg]$$

Now to make dimensions match on the RHS, first insert some $\mathbf q_j$ that you found when doing step (2) in the appropriate column (i.e. 2nd column)

$$\bigg[\begin{array}{c|c|c|c} \mathbf q_1' & \mathbf q_2' &\cdots & \mathbf q_{r}'\end{array}\bigg]\rightarrow \bigg[\begin{array}{c|c|c|c} \mathbf q_1'& \mathbf q_j & \mathbf q_2' &\cdots & \mathbf q_{r}'\end{array}\bigg]$$

and you need to insert a row and column into $R'$ to make dimensions match (here $k=1$)

$$R'=\begin{bmatrix} R_{k\times k}' & * \\ \mathbf 0 & R_{r-k\times r-k}'\end{bmatrix}\rightarrow \begin{bmatrix} R_{k+1\times k+1}' & * \\ \mathbf 0 & R_{r-k\times r-k}'\end{bmatrix}$$
and the RHS is still upper triangular because its two diagonal blocks are upper triangular, i.e. one of them is unchanged (bottom right) and the other (top right) had a row and column inserted on the bottom and right respectively, but this is still upper triangular because $\mathbf a_2$ is written a linear combination of $\mathbf a_i$ with lower indices (i.e. it is linearly dependent).

Thus the structure holds one step at a time and after finitely many steps you recover $A=QR$

remark 1:

QR factorization is not unique for non-invertible $A$

remark 2:

a much cleaner way of getting to this result involves using Householder matrices.  The proof is essentially is the same as when using Householder matrices for QR factorization of invertible A, except you skip the step at iteration $j$ if the block submatrix you are looking at has a zero in column j (which occurs iff $\mathbf a_j = \sum_{k=1}^{j-1}x_k \mathbf a_k$).


问:
Is there an object such that its unique existence follows from choice axiom, while its existence cannot be proven without choice axiom?
Let $\varphi$ be a formula, and suppose $\text{ZFC}\vdash \exists x\varphi(x)\wedge\forall x\forall y(\varphi(x)\wedge\varphi(y)\to x=y)$

Is $\text{ZF}\vdash \exists x\varphi(x)$ true?

答:
I posted this originally as a comment to Alex's answer but, at his suggestion, I am expanding it into a proper answer.

This situation actually occurs in practice in infinitary combinatorics: we use the axiom of choice to establish the existence of an object, but its uniqueness then follows without further appeals to choice. I point this out to emphasize that this is actually a common and interesting phenomenon, rather than the result of metamathematical tricks, as one might erroneously conclude from the examples in the other answers.

For a simple example, from choice one can prove that there are uncountable regular cardinals. Once we know that they exist, it is obvious that the least such cardinal is unique. However, $\mathsf{ZF}$ is not enough to prove its existence. This example is perhaps a bit misleading in that, under $\mathsf{ZFC}$, $\aleph_1$ is the least uncountable regular cardinal, but $\mathsf{ZF}$ suffices to prove the existence of $\aleph_1$. What it cannot prove is its regularity.

More interesting examples identify objects whose existence requires the axiom of choice and have no counterpart in the choiceless setting in general. This was studied in some detail, in the context of pcf theory, in

MR2078366 (2005k:03105). Cummings, James; Foreman, Matthew; Magidor, Menachem. Canonical structure in the universe of set theory. I. Ann. Pure Appl. Logic 129 (2004), no. 1-3, 211–243.

From the introduction: "Of particular interest are invariants which are canonical, in the sense that the Axiom of Choice is needed to show that they exist, but once shown to exist they are independent of the choices made. For example the uncountable regular cardinals are canonical in this sense."

The examples studied in the paper have to do with the combinatorics at the successor of a singular cardinal $\mu$. The simplest kind of objects considered in this setting were introduced by Shelah: these are certain nice collections of points in $\mu^+$, that he dubbed good and approachable. These collections are unique modulo the nonstationary ideal.


问:
If the order of a finite abelian group is not divisible by a square, show that the group must be cyclic.
If the order of a finite abelian group is square free, show that the group is cyclic.

This is a question from "basic abstract algebra" by bhattacharya

答:
The structure theorem for finitely generated abelian groups says that if $G$ is finitely generated and abelian, then it is a product of finitely many groups of the form $\mathbb{Z}$ and $\mathbb{Z}_q = \mathbb{Z}/q\mathbb{Z} $ where $q$ is power of a prime. If $G$ is finite, then surely it is finitely generated and has no factor isomorphic to $\mathbb{Z}$; thus it is of the form:
$$ G \simeq \mathbb{Z}_{q_1} \times \mathbb{Z}_{q_2} \times \dots \times \mathbb{Z}_{q_n} $$
where $q_i$ are (not necessarily distinct) powers of primes. It is clear that $|G| = \prod_{i=1}^n q_i$. Because $|G|$ is square-free, $q_i$ have to be all primes (i.e. not some higher powers) and distinct. In particular, $q_i$ and $q_j$ are compime for $i \neq j$. The Chinese Remainder Theorem now yields:
$$ \mathbb{Z}_{q_1} \times \mathbb{Z}_{q_2} \times \dots \times \mathbb{Z}_{q_n} \simeq \mathbb{Z}_{q_1 q_2 \dots q_n} $$
This is a cyclic group.</p>


问:
What does it mean for the word 'propagate' in the context of method of characteristics?
Perhaps I am being pedantic but I saw the word 'propagate' used in the method of characteristics multiple times but I am not sure what it means. Here are a few examples I encountered.

$1)$ The boundary conditions and initial conditions propagates along the characteristics.

$2)$ Consider the ODE $U_x+U_y=0$ subject to $U(0,y)=1_{\{y\geq 0\}}.$ Then the discontinuity of the initial condition propagates along the characteristics.

I am not sure what the meaning of these sentences are trying to convey, could someone please kindly explain to me?

答:
This is the idea behind the method of characteristics. The dynamic of the PDE reduces to ODE systems along the characteristic curves. The initial conditions of the ODE are given by the boundary conditions of the PDE, so that these get "propagated" along the characteristic curves to the area where these are regular, do not intersect.

In the example the characteristic curves are described by $y-x=c$. Along these the value of $U$ stays constant, this constant is thus propagated to the whole line, in total $U(x,y)=U(0,y-x)={\bf 1}_{\{y\ge x\}}$.


问:
negation of statement with "at least"
The negation of the following sentence:

At least one girl has bought her friend a book, which she hasn't read herself.

I have tried to write it with logical symbols

p = at least one girl has bought her friend a book

q = she has not read the book herself

so would the negation be $\lnot( p \land q)$, and this would be $\lnot p \lor \lnot q$?

but how could I say it in english?

would the negation of "at least" be "All girls....?"

答:
To answer the first part of your question, yes.
$\lnot(p \land q) = \lnot p \lor \lnot q$.

To answer the second, think when "at least one girl has blah" is true. Is it true when $0$ girls have blah? Is it true when $1$ girl has blah? Is it true when $2$ girls have blah?

Since "at least one girl has blah" is true for $\geq 1$ girls, the only time it fails is when $0$ girls have blah. So that's the negation.

As mentioned in the comments, you may notice that "at least one" is the same thing as "there exists". So if you know that
$\lnot \exists x \varphi = \forall x \lnot \varphi$, you can do this slightly quicker.

I hope this helps ^_^

Title: negation of statement with "at least"


答: First part of your questions is by De Morgan’s law. And second part is also correct. “At least one girl did something” is equivalent to saying that “there exists a girl who did something”. The negation of “there exists” is “for all”.</p>

Title: negation of statement with "at least"

Body:
The negation of the following sentence:

At least one girl has bought her friend a book, which she hasn't read herself.

I have tried to write it with logical symbols

p = at least one girl has bought her friend a book

q = she has not read the book herself

so would the negation be $\lnot( p \land q)$, and this would be $\lnot p \lor \lnot q$?

but how could I say it in english?

would the negation of "at least" be "All girls....?"


问:
Evaluating the integral of $(\csc(x))^5$ with the reduction formula of $(\sin(x))^n$
I have found the reduction formula for the integral of $\sin^nx$

$$\int \sin^{n}x\, dx = -\frac{1}{n}\sin^{n-1}x \cos x  + \frac{n-1}{n}\int \sin^{n-2}x \, dx$$

Using this, I am tasked to find the integral $\int \csc^{5}x\,dx$.

What I have now is simply to convert $\csc^{5}x$ to $\sin^{-5}x$ and substitute $n = -5$ in the reduction formula. However, I found that the last integral will simply go into loops of power $-7, -9,\dotsc$ with no end.

Does anyone have tips on how I can solve something like this?

答:
Use the reduction formula

$\int \csc ^{m}(x) d x=-\frac{\cos (x) \csc ^{m-1}(x)}{m-1}+\frac{m-2}{m-1} \int \csc ^{-2+m}(x) d x$

When m=5, you get

$-\frac{1}{4} \cot (x) \csc ^{3}(x)+\frac{3}{4} \int \csc ^{3}(x) d x$

using the formula for m=3 to carry out the integral,
you get

$-\frac{3}{8} \cot (x) \csc (x)-\frac{1}{4} \cot (x) \csc ^{3}(x)+\frac{3}{8} \int \csc (x) d x$

\begin{aligned}
&\text { Using the fact that integral of } \csc (x) \text { is }-\log (\cot (x)+\csc (x))\\
&-\frac{1}{4} \cot (x) \csc ^{3}(x)-\frac{3}{8} \cot (x) \csc (x)-\frac{3}{8} \log (\cot (x)+\csc (x))+\text { constant }
\end{aligned}

Which can be simplified to

$$-\frac{1}{64} \csc ^4\left(\frac{x}{2}\right)-\frac{3}{32} \csc ^2\left(\frac{x}{2}\right)+\frac{1}{64} \sec ^4\left(\frac{x}{2}\right)+\frac{3}{32} \sec ^2\left(\frac{x}{2}\right)+\frac{3}{8} \log \left(\sin \left(\frac{x}{2}\right)\right)-\frac{3}{8} \log \left(\cos \left(\frac{x}{2}\right)\right)+constant$$


问:
Is $10^{-3}$ error or tolerance? .Find a solution accurate within $10^{-3}$ for $() =  ^{3} + 3 − 5$ on$ [1, 2].$

According to the algorithm,$f(c)=0$ 0r $(a-b)/2<TOL$ is stop procedure.

My problem is in this question, Is $10^{-3}$ error or tolerance. If it is an error, should I check $|aₙ-bₙ|<10^{-3}$ to stop the iteration? and if it is TOL should I check $(a-b)/2<TOL$ to stop iteration?
I have confusion about the differences between tolerance and error and the relationship between them.

答: Tolerance is usually synonymous for error. As stated in the procedure, you want to stop when $(b-a)/2<10^{-3}$. Note that $|a-b|=b-a>0$ by the initial condition and that the distance the actual root is from $c$ is at most $(b-a)/2$ when the root is at $a$ or $b$ (the farthest points from $c$, the midpoint).


问:
Kolmogorov‘s Maximal inequality and convergence a.s.
I have a question about Durrett exercise 2.5.12 (i).

where the (*) is given in 2.5.9

When I try to use Maximal Inequality, it is in the form of $|\frac{S_N}{a(n)}-\frac{S_m}{a(m)}|$. So that, I am not able to apply Maximal Inequality directly. Could anyone help me with this?

答:
It suffices to show that
$$
M_n:=\frac1{a(2^n)}\max_{2^n+1\leqslant k\leqslant 2^{n+1}}\lvert S_k-S_{2^n}\rvert\to 0 \mbox{ a.s.}
$$
Indeed,

$$
\max_{2^n+1\leqslant k\leqslant 2^{n+1}}\frac{\lvert S_k\rvert}{a(k)}
\leqslant\max_{2^n+1\leqslant k\leqslant 2^{n+1}}\frac{\lvert S_k\rvert}{a(2^n)}\leqslant
\frac{\lvert S_{2^n}\rvert}{a(2^n)}+M_n
$$

Noticing that $S_k-S_{2^n}=S_{2^n,k}$ we have by 2,5,9:
$$
\mathbb P\left(M_n>2\varepsilon\right)\leqslant \frac{\mathbb P\left(\lvert S_{2^{n+1}}-S_{2^n}\rvert>\varepsilon a(2^n)\right)}{\min_{2^n+1\leqslant k\leqslant 2^{n+1}}\mathbb P\left(\lvert  S_k-S_{2^n}\rvert \leqslant \varepsilon a(2^n)\right)}.
$$
Using boundedness of $a(2^n)/a(2^{n-1})$, we know that $\lvert S_{2^{n+1}}-S_{2^n}\rvert/ a\left(2^n\right)\to 0$ in probability.


答: If you are familiar with minimal polynomials, then it is easy to see that the only eigenvalue of $A$ is $0$ and the system $(A-0I)v=0$ has only $3$ linearly independent solutions (eigenvectors) by virtue of $A$ having rank $2$. So $A$ is guaranteed to be non-diagonalizable.

Body:
Let $A\in M_{5×5}(\mathbb{R})$ be a matrix such that $\operatorname{rank}(A)=2$ and $A^3 = 0$. Is A guaranteed to be diagonalizable over R?

I've been searching my notes for some sort of theorem that would help me with this problem, but I have had no luck.

Without the characteristic polynomial or actual eigenvalues, I can't seem to use the two step test for diagonalizability or any other theorem I know of for that matter.

Any thoughts?


问:
How to find all points for which the tangent line of a parametric equation (x, y, z) passes through a point

For the function
$$ \vec x(t) = \begin{pmatrix}
    2t + 3 \\
    2 - t \\
    t^3 - 2t^2 + t \\
    \end{pmatrix}
 t ≥ 0$$
Find all points $\vec x(t_0)$ for which the tangent line passes through the
point (1, 3, 0).

I understand how I would go about solving the problem if I had a normal polynomial function or only x and y values.  However, with x, y, and z values I am not sure how to approach the question.

I know that the derivative of the function is

$$ \vec x'(t) = \begin{pmatrix}
    2 \\
    -1 \\
    3t^2 - 4t + 1 \\
    \end{pmatrix} $$

Would I then have to find the equation of the tangent line and use that with the given point to find the other points?

Body:
Hint:

Just write the vectors $\vec x(t)-(1,3,0)$ and $\vec x'(t)$ are collinear:
$$\frac{2t+3-1}2=\frac{-t-1}{-1}=\frac{t^3-2t^2+t}{3t^2-4t+1}\iff t+1=\frac{t^3-2t^2+t}{3t^2-4t+1}=\frac{t(t-1)^2}{(3t-1)(t-1)}=\dotsm$$


问:
Name or notation for a function f(n) that equals 1 if n = 1 and 0 if n > 1
While solving number theory problems I sometimes I have to use a function that can be defined as

$$
f(n) =
\begin{cases}
1 & \text{ if } n = 1, \\
0 & \text{ if } n > 1.
\end{cases}
$$

where $ n $ is a positive integer.

For example, using this function, we can define Euler's totient function as

$$ \varphi(n) = \sum_{k=1}^n f(\gcd(k, n))$$

where $ n $ is a positive integer.

Is there a name or notation for such a function already? I just want to make sure that I do not create my own notation for something that already has a popular name or notation in mathematics.

答:
In Apostol's Introduction to Analytic Number Theory, he calls this function $I$.

Part of Apostol's rationale is that $I$ acts like the identity element in the group of arithmetic functions (where Dirichlet multiplication is the operator), and so something like $I$ is a good name.

Elsewhere I've seen it called many things. There is no standard name, though there are lots of unambiguous names. If I were king of the notational universe, I might use Kronecker-delta based names like $\delta(\cdot)$, $\delta_1(\cdot)$, or $\delta_{[n=1]}(\cdot)$ --- but (fortunately) I am not king of the notational universe.


问:
Show that $ | S_{2n + 2} | - | S_{2n + 1} | $ follows the same law as the $ X_k $.
We consider a sequence $ (X_n)_{n \leq 1} $ of mutually independent random variables defined on a probability space $ (\Omega, \mathcal{A}, \mathbb{P}) $  with value in $\{-1,1\}$ and such that, for all $ k \geq 1$ $$ \mathbb{P} (X_k = -1) = \mathbb{P} (X_k = 1) = \frac{1}{2} $$ For all $ n \geq 1 $, we set $ S_n = X_1 + \cdots + X_n $

Problem

Show that $ | S_{2n + 2} | - | S_{2n + 1} | $ follows the same law as the $ X_k $.

For all $w\in\Omega $, we have $ | S_{2n + 2}(w) | - | S_{2n + 1}(w) |\in\{-1,1\} $.  How to show that

$$
\mathbb{P} (| S_{2n + 2}(w) | - | S_{2n + 1}(w) | = -1) = \mathbb{P} (| S_{2n + 2}(w) | - | S_{2n + 1}(w) | = 1) = \frac{1}{2}
$$

An idea please

答:
Condition on $S_{2n+1}$.

$$P(|S_{2n+2}| - |S_{2n+1}| = 1 \mid S_{2n+1})
= P(|S_{2n+1} + X_{2n+2}| - |S_{2n+1}| = 1 \mid S_{2n+1})$$

You can check (with a little bit of casework on whether $S_{2n+1}$ is positive or negative) that no matter what value $S_{2n+1}$ is, this conditional probability is always $1/2$. (It is important to note that it is impossible for $S_{2n+1}$ to be zero!) This then implies that the unconditional probability is also $1/2$.


答: It is usually called an indicator function in signal processing. It is denoted as either $\delta(1)$ or $\mathbb{1}_{n=1}$.

Title: Name or notation for a function f(n) that equals 1 if n = 1 and 0 if n > 1

Body:
While solving number theory problems I sometimes I have to use a function that can be defined as

$$
f(n) =
\begin{cases}
1 & \text{ if } n = 1, \\
0 & \text{ if } n > 1.
\end{cases}
$$

where $ n $ is a positive integer.

For example, using this function, we can define Euler's totient function as

$$ \varphi(n) = \sum_{k=1}^n f(\gcd(k, n))$$

where $ n $ is a positive integer.

Is there a name or notation for such a function already? I just want to make sure that I do not create my own notation for something that already has a popular name or notation in mathematics.


问:
What is the sequence represented by the exponential generating function $e^{x^4}$?
Problem

For the exponential generating function $e^{x^{4}}$, give a formula in closed form for the sequence $\{a_n:n \geq 0\}$ it represents.

My Attempt

I have that
$$e^{x^{4}}=\sum_{n=0}^{\infty}\frac{(4n)!\,x^{4n}}{n!\,(4n)!},$$

which leads me to believe that $a_n=\frac{(4n)!}{n!}$. However, the series representation is not in the correct form, because it is not of the form $\sum_{n=0}^{\infty}a_n\frac{x^n}{n!}$. Is my answer still correct?

答:
Define:
$$a_n=\frac{1}{(n/4)!},\ \ {\rm for} \ \ n\equiv0\pmod 4\ {\rm or}\  n=0$$
and
$$a_n=0\ \ {\rm otherwise}$$


问:
Find the extreme values of $f(x,y)=x^2y$ in $D=\{x^2+8y^2\leq24\}$
Find the extreme values of $f(x,y)=x^2y$ in $D=\{x^2+8y^2\leq24\}$

It was easy to find using Lagrange multipliers the local extreme values on $\partial{D}$ since we have the condition $x^2+8y^2=24.$ I found that $(-4,1),(-4,-1),(4,1)$ are critical points.

Now I have to find the extreme values inside $D$. $\nabla f(x,y)=(2xy,x^2)=(0,0)$ gives me that every $(0,y)$ is a critical point. However, I'm not sure how to get extremes from here, since I cannot use the determinant test. Would be a possibility to use the monotony of $x^2$ depending the sign of $y$?

答: $24\geq x^2+8y^2=\frac{x^2}{2}+\frac{x^2}{2}+8y^2\geq 3\sqrt[3]{2x^4y^2}=3\sqrt[3]{2}(x^2y)^{\frac{2}{3}}$ by AM-GM inequality. Maximal value for $x^2y$ is attained for $x=4y=4$ and it is $16$.


答:
While these formal statements are made, here are some real life examples,

"The cardinality of a Hamel basis of $\Bbb R$ over $\Bbb Q$", is a unique and well-defined object once some Hamel basis exists (we can prove that a Hamel basis must always have the cardinality of the continuum). Nevertheless, it is consistent with $\sf ZF$ that no such basis exists.

Similarly we can talk about "the cardinal of a free ultrafilter on $\omega$".

In general, any type of object whose cardinality in $\sf ZFC$ is unique (indeed the above cases are even better: the uniqueness of cardinality follows from $\sf ZF+$"The object exists") but $\sf AC$ is necessary for the existence of the objects, is a good target.

Title: Is there an object such that its unique existence follows from choice axiom, while its existence cannot be proven without choice axiom?

Body:
Let $\varphi$ be a formula, and suppose $\text{ZFC}\vdash \exists x\varphi(x)\wedge\forall x\forall y(\varphi(x)\wedge\varphi(y)\to x=y)$

Is $\text{ZF}\vdash \exists x\varphi(x)$ true?


问:
Understanding Formula for the Pair Correlation for the Farey Sequence
I'm reading this paper https://arxiv.org/pdf/math/0404114.pdf, and I'm unsure how to interpret the formula in Theorem 2 (1.5), regarding the pair correlation function of the sequence of Farey Fractions. Specifically, I don't understand the $\lambda$ that they used.

In (1.6), they wrote as $\lambda \rightarrow \infty$,  $g_2(λ) = 1 + O(λ^{−1})$. So this means the function approaches 1 as $\lambda$ goes to infinity. This seems reasonable as I could see the function asymptotically approaching 1, the problem is I'm not sure what $\lambda$ means in this context.

The only time they mentioned $\lambda$ beforehand is on page 2 when they defined

$R_F^{(v)}(\lambda_1,...,\lambda_{v-1})=2^{-v+1}R_F^{(v)}(\prod_{j=1}^{v-1}[-\lambda_j,\lambda_j]).$

My guess right now is that $\lambda$ refers to the interval for which the correlation measures are taken. Thus, as the interval of normalized Farey Fractions become arbitrarily large, the function converges to 1. Is that reasonable? Then $\lambda$ would represent a box in $\mathbb{R}^{v-1}$ for $v \geq3$.

Could someone give me some clarity on what this $\lambda$ represents and how this would change for $v\geq 3$? Thanks a lot.

答:
It might be a better idea not to look at the (free) parameter $\lambda$, but instead to look at $g_2$. On page 2, $g_2$ is defined as a function such that the $2$-level correlation measure is given by

$$ \mathcal{R}^{2}(\mathcal{B}) = \int_{\mathcal{B}} g_2(x_1) dx_1.$$

So $\lambda$ (in their Theorem 2) is just a number. I suppose they could have called it $x_1$, but that seems pretty odd.

Note that this is strongly analogous to the pair correlation function for $\zeta(s)$.

An analogous function for higher level correlation would necessarily be higher-dimensional. This is sometimes studied along with random-matrix theory, and called "$n$-level density". See for example this preprint by Chandee and Lee.


问:
Can a D.E. still be defined even if terms are unbounded?
$$(1 - x^2)y'' - 2xy' + l(l + 1)y = 0, -1 \leq x \leq 1$$

I want to go ahead and use the Sturm-Liouville theorem to prove that this equation's eigenstates are orthogonal, but it's not a given that $y'$ is bounded at $a$ or $b$, only that the D.E. is defined over $[-1,1]$. Is that sufficient?

答:
It turns out that imposing boundedness on the solutions is equivalent to requiring that certain continuous endpoint functionals vanish on these functions. So the formalism works, and gives you discrete eigenvalues where there are eigenfunctions that are bounded near both endpoints of the interval. And the operator is self-adjoint on this domain, which gives you completeness and orthogonality of the eigenfunctions. That said, there are other conditions you can impose at the endpoints instead, and the eigenfunctions will not be bounded, but will be orthogonal and complete. The eigenvalues will be different than for the Legendre polynomials.

This problem and solution that I posed may be helpful to you: Selfadjoint Restrictions of Legendre Operator $-\frac{d}{dx}(1-x^{2})\frac{d}{dx}$


问:
Is the endomorphism ring of $\mathbb{R}$ self-opposite?
Is the endomorphism ring $End\mathbb{R}$ of the Abelian group $\mathbb{R}$ isomorphic to its opposite ring?

All subrings of a self-opposite ring are self-opposite. By choosing an isomorphism of $\mathbb{R}$ with $\mathbb{R}^n$ and observing that linear endomorphisms are also endomorphisms of an Abelian group, we see that $End\mathbb{R}$ contains a copy of the linear endomorphism ring of $\mathbb{R}^n$ for every $n$. However, all of these subrings are self-opposite (by transposition).

My motivation for this is to help me decide on notation for a problem in geometry, which I will not elaborate upon here.

答:
We know that $\mathbb R$ is a direct sum of subgroups isomorphic to $\mathbb Q$; suppose that $I$ is the indexing set of this direct sum. It follows that the endomorphism ring of $\mathbb R$ is the ring of matrices whose columns and rows are indexed by $I$ and whose columns have finitely many non-zero items.

Then you are asking if the ring of column-finite matrices with entries in $\mathbb Q$ is isomorphic to its opposite ring. It seems it isn't: this has been proved in J. J. Simón, There is no analogue for the transpose map for infinite matrices, Publicacions Matemàtiques, Vol 41 (1997), 653–657.


问:
show that a language is context-free by giving a pushdown automaton that accepts the language
Given the language $K$ : the words x$y where x and y have odd lengths on {a, b} and the median letter of x is equal to the median letter of y.

Example:

abb$b and

aaaaa$bab are in K

aba$abaab isn't in K

So i am kinda lost here, i am trying to figure out how to do this problem. I struggle to understand how will it recognize that the numbers are odd and how can it know that the median of both words are the same letter.

Could anyone help please, thanks a lot.

I started this drawing but i am so unsure on what to put on state 2 and 3 to make it work.
( I used : http://madebyevan.com/fsm/ if u wanna draw).

EDIT
I wrote a new solution this seem ok, if someone could confirm it would be nice.

答:
Its been a little while since I've done this stuff, so sorry if my description isn't great. But here is an approach you can flesh out.

Start in $x_L$. This state scans the input and pushes onto the stack until it randomly decides to switch to $x_?$ state.

In $x_?$ state it looks to see if the current character is $a$ or $b$ and moves to $x_a$ or $x_b$ depending on which one it sees.

In either $x_a$ or $x_b$ it starts popping off the stack. If it gets to $\#$ before the stack is empty or it empties the stack before it gets to $\#$ it fails. Otherwise, if it gets to $\#$ exactly when the stack is empty, it moves to $y^a_L$ or $y^b_L$ depending on if it is in $x_a$ or $x_b$.

At this point, if it hasn't failed, it knows that $x$ is of odd length and if the middle character is $a$ or $b$. Then, you can do something similar for $y$ and only succeed if both middle characters are the same.

Hope this helps.


问:
If $\int {yy''dx} = 3xy$ is it possible to find y'?
So one of the people I know on discord gave me this problem to solve for fun.
I've tried using integration by parts on the left side and it didn't do anything.
The integral transforms into the ODE $yy''- (3xy)'=0$ but it's non-linear and I have no idea how to solve that. Would a series solution work?
Notice that I'm looking for y' not y so maybe there's some trick there?
Any help is appreciated. Thank you for reading.
Here, $y'$ and $y''$ denote the first and second order derivatives of $y$ w.r.t. $x$.

答:
Disclaimer: I managed to get a recursive definition of the coefficients of the Maclaurin series, but not a closed form.

From the ODE $$yy'' - (3xy)' = 0$$

This expands to $$yy'' - 3xy' -3y = 0$$

Let $y = \sum_{n=0}^{\infty} a_nx^n$. Then the relation is $$\left(\sum_{n=0}^{\infty} a_nx^n\right)\left(\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}\right) + \sum_{n=0}^{\infty} (-3na_n-3a_n)x^{n} = 0$$

Expanding this, it becomes

$$\sum_{n=0}^{\infty} x^n \left(-3na_n - 3a_n + \sum_{m=0}^n (m+2)(m+1)a_{m+2}a_{n-m}\right) = 0$$

For it to be $0$, each of the coefficients of $x^n$ must be $0$, so it is $$(-3n-3)a_n + \sum_{m=0}^n (m+2)(m+1)a_{m+2}a_{n-m} = 0$$

for all nonnegative integer $n$. Assume that $y(0) = a_0 \not = 0$ and $y'(0) = a_1$ are fixed at some values. Then solving for $a_{n+2}$, it is $$a_{n+2} = \frac{(3n+3)a_n - \sum_{m=0}^{n-1} (m+2)(m+1)a_{m+2}a_{n-m}}{(n+2)(n+1)a_0}$$

For the first few values, this comes out to $a_2 = \frac{3}{2}$, $a_3 = \frac{a_1(3-a_2)}{3a_0} = \frac{a_1}{2a_0}$, $a_4 = \frac{9a_{2}-2a_{2}a_{2}-6a_{3}a_{1}}{12a_{0}} = \frac{3a_{0}-a_{1}^{2}}{4a_{0}^{2}}$.


问:
Example for non-Lebesgue integrable function on $(0,1)$ which is integrable on $(1/n,1)$
Q. Let $\lambda$ denote Lebesgue measure on $\mathbf{R}$. Give an example of a continuous function $f:(0,1) \rightarrow \mathbf{R}$ such that $\lim _{n \rightarrow \infty} \int_{\left(\frac{1}{n}, 1\right)} f d \lambda$ exists (in $\left.\mathbf{R}\right)$ but $\int_{(0,1)} f d \lambda$ is not
defined.

I hope the area under $f$ from $1/n$ to $1$ need to be constant.

Any role for conditionally convergent series?

答:
$f(x)=\dfrac{\sin(2\pi/x)}{x^2}$ is an example. Proof: Let $x=1/y$ to see

$$\tag 1\int_{[1/n,1]} \frac{\sin(2\pi/x)}{x^2}\,dx = \int_{[1,n]} \sin(2\pi y)\,dy.$$

Now the integral of $\sin(2\pi y)$ over any interval of length $1$ is $0.$ This implies the right side of $(1)$ is $0$ for any $n\in \mathbb N.$ But for $\int_0^1 f$ to exist as a real number, we must have $\int_0^1 |f|<\infty.$ You can check, using the same substitution $x=1/y,$ that $\int_0^1 |f|=\infty.$


问:
Estimate on complex path integral
I came across the question in this post, where I wrote a partial answer. The question is to show that, for some constant $k>0$, we have
$$
\left|\int_{[-1,i]}(1+z+z^2+z^3)^n dz\right| \leq  k \left(\frac{4}{3 \sqrt{3}} \right)^n.
$$

What I did:

We can easily  compute the integral up to a certain point, obtaining
\begin{align*}
\int_{[-1,i]} (1+z+z^2+z^3)^n dz= &\int_{[-1,i]}(1+z)^n(1+z^2)^n \\
= & \int_0^1 (1+i) (t(1+i)^n)(2(1-t)-2t(1-t)i)^ndt\\
=& (1+i)^{n+1} 2^n \int_0^1 t^n(1-t)^n(1-ti)^n dt
\end{align*}

and so,
$$
\left| \int_{[-1,i]} f(z)^n dz \right|\leq \sqrt{2} (2\sqrt{2})^n \int_0^1 t^n (1-t)^n \left(\sqrt{1+t^2}\right)^n dt
$$

Although it is true that this upper bound is smaller than $\left(\frac{4}{3 \sqrt{3}}\right)^n$, and the result follows with $k=1$, I feel that there must be some simpler approach and would appreciate some ideas. In particular, nothing in my calculations gave any hint on the origin or motivation for the ratio $\frac{4}{3 \sqrt{3}}$.

答:
This has much to do with saddle points of $|f(z)|$ where $f(z)=1+z+z^2+z^3$.

Namely, the solutions of $f'(z)=0$ are $z=(-1\pm i\sqrt2)/3$, with $$f(z)=\frac{20\pm4i\sqrt2}{27},\quad|f(z)|=\color{blue}{\frac{4}{3\sqrt3}}.$$ Observe also that $f(-1)=f(i)=0$, so that we can choose a path $\gamma$ of integration from $-1$ to $i$, passing through $(-1+i\sqrt2)/3$, and such that $|f(z)|\leqslant 4/(3\sqrt3)$ for $z\in\gamma$ (apart from "general theory", it is actually easy to construct $\gamma$ explicitly).

This gives the desired estimate $\left|\int_\gamma f(z)^n\,dz\right|\leqslant\left(\frac{4}{3\sqrt3}\right)^n\int_\gamma|dz|$ immediately.

Title: Estimate on complex path integral


问:
Nonexistence of spherical objects
This is example 8.10(vi) (on page 170-171) in Huybreachts' Fourier-Mukai transform in algebraic geometry. Let $X$ be a variety of dimension at least three with nontrivial $H^2(X,O_X)$. I want to show $X$ has no spherical object with nontrivial rank. In the book he said that this is because $Ext^2(E,E)\rightarrow H^2(X,O_X)$ is surjective for $rk(E)\neq0$ and $E$ is a complex of coherent sheaves (which is induced by the trace map) and thus it dos not exist. Howenver, I do not know why the given map above is surjective.

答: $\operatorname{Ext}^2(E,E)=H^2(E^*\otimes E)$ for any vector bundle. The trace map gives $E^*\otimes E\to O$ which splits ($E\neq 0$) and thus the above map is surjective.

Title: Nonexistence of spherical objects


问:
Topology: Show, that if $A\cap\mathcal{O}=\emptyset$ then $\overline{A}\cap\mathcal{O}=\emptyset$
Within a proof I need the following step.

Let $(X,\mathcal{T}$) be a topological space and $A\subset X$. If $\mathcal{O}\in\mathcal{T}$ such that $A\cap\mathcal{O}=\emptyset$, then also $\overline{A}\cap\mathcal{O}=\emptyset$.

答:
$A \cap \mathcal{O} = \emptyset \overset{(1)}{\iff} A \subset \mathcal{O}^{c} \overset{(2)}{\iff} \overline{A} \subset \mathcal{O}^{c} \overset{(1)}{\iff} \overline{A} \cap \mathcal{O} = \emptyset$

$(1)$ Property of sets.

$(2)$ Because $\mathcal{O}^{c}$ is closed, and because by definition  closure is intersection of all closed supersets.


问:
what is the value of this trigonometric expression
I want to find out value of this expression
$$\cos^2 48°-\sin^2 12°$$
Just hint the starting step.Is there any any formula regarding $\cos^2 A-\sin^2 B$?</p>

答:
I've got a formula :
$$\cos(A+B).\cos(A-B)=\cos^2A-\sin^2B$$
so from this formula this question is now easy
$$\cos^248-\sin^212$$
$$\cos60.\cos36$$
$$\frac{\sqrt{5}+1}{8}$$</p>


问:
dist$(x,A)=0$ if and only if $x\in \overline{A}$ (closure of $A$)
Let $(X,d)$ be  a metric space and non-empty $A\subset X$. The distance from $x$ to $A$ is defined as $$\mbox{dist}(x,A)=\inf\{d(x,a):a\in A\}$$

I want to show that dist$(x,A)=0$ if and only if $x$ is in the closure of $A$.

(=>): dist$(x,A)=0$ implies $\inf \{d(x,a):a\in A\}=0$. Suppose $x\not\in \overline{A}$, then $x\in X\setminus \overline{A}$, which is an open set. Thus $\exists\epsilon >0$ such that punctured open ball $B(x;\epsilon)\setminus\{x\}= \{y\in X:0<d(x,y)<\epsilon\} \subset X\setminus\overline{A}$. But then dist$(x,A)=\inf\{d(x,a):a\in A\}\ge \inf\{d(x,y):y\in B(x,\epsilon)\setminus\{x\}\}>0$. This is a contradiction! Hence, $x\in \overline{A}$.

(<=): Suppose $x\in \overline{A}$. If $x\in A^\circ$ (the interior of $A$) then $\inf\{d(x,a):a\in A\}=d(x,x)=0$.
  If $x\not\in A^\circ$ then $\exists $ a sequence $(a_k)_k\subset A$ such that $a_k$ converges to $x$. Then, $\forall \epsilon > 0, \exists N\in\mathbb{N}$ such that $k\ge N\implies d(a_k,x)<\epsilon$.

So that dist$(x,A)=\inf\{d(x,a_k)\}=\lim\limits_{k\to\infty} d(x,a_n) = 0$.

Please let me know if my proof is rigorous / correct. I'm not completely sure, think it's better to check.

答:
Here is another proof, hopefully it is correct. Assume $x$ is a limit point of $A$ (otherwise the result is achieved in a straightforward manner).

Let $x\in \overline{A}$ and $d = \operatorname{dist}(x,A)>0$. Choose $0<r<d$. The ball $B_r(x)$ is a neighborhood of $x$ such that $a\notin B_r(x)$ for all $a\in A$. This shows that $x$ is not a limit point of $A$, contrary to the hypothesis. Thus $\operatorname{dist}(x,A)=0$.

Conversely, let $\operatorname{dist}(x,A)=0$ and let $U$ be any neighborhood of $x$. Then there exists some $B_r(x), r>0$, such that $B_r(x) \subseteq U$. Since $\operatorname{dist}(x,A)=0$, there is at least an $a\in A$ such that $\rho(x,a)<r$ (this is the alternative characterization of the infimum). Thus $a\in B_r(x)$ by definition and also $a\in U$. Since $U$ was chosen arbitrary and there is at least one element of $A$ in $U$, $x$ is a limit point of $A$.


问:
determine the smallest n for which an in the interval
Let $(a_n)_{n\in \mathbb{N}}$ be defined by
$$ a_n = (-1)^{n+2} \left(\frac{90}{100}\right)^n.$$

Determine $N$ such that $a_n \in (0,\frac{2}{1000})$,

Body:
$$(a_n)=\{\frac{9}{10},-\frac{81}{100},\frac{729}{1000},-\frac{6561}{10000},\frac{59049}{100000},-\frac{531441}{1000000},\ldots\}$$
There is no way to find an $N$ such that for $n\ge N$ we have $0<a_n<\frac{1}{100}$ because it is an alternating sequence.

If you mean $|a_n|<\frac{1}{100}$ then we solve the inequality
$$\left(\frac{9}{10}\right)^n<\frac{1}{100}$$
$$n\log\frac{9}{10}<\log\frac{1}{100}$$
$$n\log\frac{10}{9}>\log 100\to n>43.7$$
So taking $N=44$ we have that for any $n\ge N$, $|a_n|<\frac{1}{100}$


问:
(A,B) regular => there is a scalar s such that A+s*B is regular ??
Given two matrices $A,B \in \mathbb{C}^{n \times n}$, is it true that $rank([A,B])=n \implies \exists s\in \mathbb{C}: rank(A+sB)=n$

It seems to me this could be easily proved by writing both in Jordan normal form. Am I mistaken? Is there a result I could use, or can someone suggest a name for the problem so I could google it?

If this is true, then does

Given two matrices $A,B \in \mathbb{R}^{n \times n}$, is it true that $rank([A,B])=n \implies \exists s\in \mathbb{R}: rank(A+siB)=n$

also hold?

答:
In general, you can not Jordan normalize two matrices simultaneously.

Hint: what if $\det(A+sB)$ as $s$ goes to infinity.


问:
Check on differentiability of a function
I need a check on the following exercise:

Consider the following function
$$
f(x,y)=
\begin{cases}
\frac{x^3 y}{x^2 + y^4} \quad (x,y) \ne (0,0) \\
0\quad \quad (x,y)=(0,0)
\end{cases}
$$

determine where the function is differentiable.

I checked the function is continuous over the whole $\mathbb{R}^2$ and that the derivatives $\partial_x f(0,0)=\partial_y f(0,0)=0$.

Differentiability: first of all, for any point $(x,y)\ne (0,0)$ the partial derivatives are continuous at $(x,y)$ and they exist in a neigbourhood of $(x,y)$. The tricky part is for $(x,y)=(0,0)$. There I know only the value of the derivatives at $(0,0)$ but I have no other information, therefore I have to study the following limit:

$$\lim_{(h,k) \rightarrow (0,0)} \frac{f(h,k)}{\sqrt{h^2+k^2}}$$

I note that $$|\frac{h^3 k}{(h^2 + k^4) \sqrt{h^2+k^2}}| = |\frac{h^2 \cdot h k}{(h^2 + k^4) \sqrt{h^2+k^2}}| $$ and since $$|\frac{h^2}{h^2 + k^4}|<1$$ I have
$$|\frac{h^2 \cdot h k}{(h^2 + k^4) \sqrt{h^2+k^2}}|<|\frac{hk}{\sqrt{h^2+k^2}}| < |\frac{hk}{|h|}| =_{\text{h} \ne 0} |k|$$ (if $h=0$ then the limit is identically $0$)

Now, as $(h,k) \rightarrow (0,0)$ the r.h.s goes to $0$ and hence the limit is $0$ and the function is differentiable.

Therefore the function is differentiable on the whole $\mathbb{R}^2$, right?

is everything correct

答:
Here is a proof that
$$\lim_{(x,y) \rightarrow (0,0)} \frac{f(x,y) - f(0,0)}{\sqrt{x^2+y^2}}=0.$$

Let $\varepsilon>0$. Take $\delta=\varepsilon$, let $(x,y) \in \mathbb{R}^2\backslash \{(0,0)\}$ such that $||(x,y) - (0,0)|| \leq \delta$.

\begin{align}
\left|\frac{f(x,y) - f(0,0)}{\sqrt{x^2+y^2}} - 0\right| &= \left|\frac{x^3y}{(x^2 + y^4) \sqrt{x^2+y^2}}\right| \\
&= \left| \frac{x^2}{x^2 + y^4} \right| \left| \frac{x}{\sqrt{x^2+y^2}} \right| \left| y \right| \\
&\leq \left| y \right| \leq \delta = \varepsilon.
\end{align}

So, $f$ is differentiable at $(0,0)$.


问:
Simplify using laws of indices: $\frac{\sqrt{1-x}\,\frac12(1+x)^{1/2} + \frac12(1-x)^{-1/2} \sqrt{1+x} }{1-x}$

Simplify using the laws of indices:

$$\frac{\sqrt{1-x}\,\frac{1}{2} (1+x)^{1/2} + \frac{1}{2}(1-x)^{-1/2} \sqrt{1+x} }{1-x}$$

I tried factorizing it using surds and I am stuck

Body: $\frac{\sqrt{1-x} \frac{1} {2} (1+x)^{\frac{1} {2}} + \frac{1} {2}\sqrt{1+x} (1-x)^{\frac{-1} {2}} }{1-x}=\frac{1}{2} ((1-x)^{-\frac{1} {2}} (1+x)^{\frac{1} {2}} +\sqrt{1+x}(1-x)^{\frac{-3}{2}}) .  =\frac {1} {2} \sqrt{1+x}(1-x)^{\frac {-1}{2}}((1+(1-x)^{-1})=\frac {1} {2} \sqrt{\frac{1+x} {1-x}} ((\frac{2 -x}{1-x})$


问:
Prove that planar graph with girth at least 6 contains a vertex of degree at most 2.
I'm trying to prove that planar graph with girth at least 6 contains a vertex of degree at most 2.

However, I'm not sure how would I do so. Could you please help me?

答:
By Euler's formula for plane graphs,
$$ v+f=e+2.$$
If each vertex has degree $\ge 3$, then $2e\ge 3v$. By the assumption about the girth, each face has at least $6$ edges, so $2e\ge 6f$. Thus
$$ 12=6v+6f-6e\le 4e+2e-6e=0,$$
contradiction.


问:
Prove if a power series is zero on an interval then all coefficients are zero
Suppose the power series $P(x) =  \sum_{n=1}^\infty b_n x^n$  converges for $|x| \leq 1$ and that for some $0<c\le1$ it is given that $$P(x)=0 \quad \forall x \;\text{such that}\;|x| < c$$ Show that $b_n = 0$ for all $n \geq 1$.

What I was able to do was:

$$\lim_{n\to\infty} |b_n x^n|^{1/n} < 1$$ (by the root theorem for convergence)

and as $P(x)=0$ for $|x| < c$, we have:

$$\sum_{n=1}^\infty b_n x^n = 0$$

Now how to get what is asked in the question from these known facts? I am not able to proceed beyond this. Thanks.

答: A power series has the same radius of convergence as the series of its derivatives. In particular, we can take the derivative term by term for $|x|<c$. We have $P^{(k)}(0)=0$, and translating this in term of $\sum_k a_kx^k$, this means that $a_k=0$.


问:
Prove that a relation is Equivalent where: $7\mid x+6y$ where x and y both come from a set of numbers (1,2,3.....10)
Like I've mentioned, I need to prove that it is equivalent, and get all of its classes.
Proving that it is Reflexive is super easy and I did it myself.
Proving symmetry and transitivity is what bothers me.
I don't even want to talk about classes, which I definitely do not know how to define.

Thanks for your help in advance

答:
Hint

$\frac{x+6y}{7}\in \mathbb{Z}$ if and only if $\pm\frac{x+6y}{7}+k\in \mathbb{Z}$ for all $k\in \mathbb{Z}$. The right hand side is easier to use to show symmetry or transitivity.


问:
Why using Napierian logarithm gives me the same result of doing properties?
I wanted to change the base of this logarithm $\log_432$ from 4 to 2 so it could be easier to solve. Using some properties I get 2.5 as a result. However I was told that if I use $\frac{\ln 32}{\ln 4}$ I get the same result. Can someone explain me how this ln works?

答: In general$$y:=\log_ax\implies b^{y\log_b a}=a^y=x\implies y\log_b a=\log_bx\implies y=\tfrac{\log_bx}{\log_ba},$$whether you use $b=2$ or $b=e$.


问:
Nikish choosing 5 games question.
Nikish wants to install 5 new games. 2 are word games, 5 quiz and 16 saga. She needs to have at least one of each. How many different selections of games could she choose?

Thanks for any help, the textbook I was using only has the answer at 9080, which I cannot get to.

答:
There are ${2+5+16}\choose{5}$ ways to choose some 5 games.
But we need to exclude ${5+16}\choose{5}$ options when we didn't take any word game, ${2+16}\choose{5}$ options when we didn't take any quiz game and ${2+5}\choose{5}$ options when we didn't take any saga games. But now we excluded twice ${5}\choose{5}$ options, when we took quiz games only and ${16}\choose{5}$ options, when we took saga games only. Let's add them back in, and we get the answer:

$$
{{2+5+16}\choose{5}} - {{5+16}\choose{5}} - {{2+16}\choose{5}} - {{2+5}\choose{5}} + {{5}\choose{5}} + {{16}\choose{5}} \\
= 33649 - 20349 - 8568 - 21 + 1 + 4368 \\
= 9080
$$

Title: Nikish choosing 5 games question.


答:
$\log_432$ is the number $x$ such that $4^x=32$.

Taking the natural logarithm at both sides, we get

$$\ln(4^x)=x·\ln(4)=\ln(32)$$

Finally,
$$x=\frac{\ln(32)}{\ln(4)}$$

Title: Why using Napierian logarithm gives me the same result of doing properties?

Body: I wanted to change the base of this logarithm $\log_432$ from 4 to 2 so it could be easier to solve. Using some properties I get 2.5 as a result. However I was told that if I use $\frac{\ln 32}{\ln 4}$ I get the same result. Can someone explain me how this ln works?


问:
Help understanding Casella & Berger's explanation of a sufficient statistic
This is from Casella and Berger's Statistical Inference:

Definition: A statistic $T(\mathbf{X})$  is a sufficient statistic for $\theta$ if the conditional distribution of the sample $\mathbf{X}$ given the value of $T(\mathbf{X})$ does not depend on $\theta$.

In the discrete case,

Let $t$ be a possible value of $T(\mathbf{X})$ , that is, a value such that $P_\theta(T(\mathbf{X})  = t) > 0$. We wish to consider the conditional probability $P_\theta(\mathbf{X} = \mathbf{x}|T(\mathbf{X}) = t)$. If $\mathbf{x}$ is a sample point such that $T(\mathbf{x}) \neq t$, then clearly, $P_\theta(\mathbf{X} = \mathbf{x}|T(\mathbf{X}) = t) = 0$. Thus, we are interested in $P(\mathbf{X} = \mathbf{x}|T(\mathbf{X}) = T(\mathbf{x}))$. By the definition, if $T(\mathbf{X})$ is a sufficient statistic, this conditional probability is the same for all values of $\theta$ so we have omitted the subscript.

This is the part I'm having trouble with:

A sufficient statistic captures all the information about $\theta$ in this sense.  Consider Experimenter 1, who observes $\mathbf{X} = \mathbf{x}$ and, of course, can compute $T(\mathbf{X} = T(\mathbf{x})$. To make an inference about $\theta$, he can use the information that $\mathbf{X} = \mathbf{x}$ and $T(\mathbf{X}) = T(\mathbf{x})$. Now consider Experimenter 2, who is not told the value of $\mathbf{X}$ but only that $T(\mathbf{X}) = T(\mathbf{x})$. Experimenter 2 knows $P(\mathbf{X} = \mathbf{y}|T(\mathbf{X}) = T(\mathbf{x}))$, a probability distribution on $A_{T(\mathbf{x})} = \{\mathbf{y}: T(\mathbf{y}) = T(\mathbf{x})\}$, because this can be computed from the model with knowledge of the true value of $\theta$.

So far, so good. But below, what exactly is this random variable $\mathbf{Y}$? I'm having trouble unraveling why exactly this conclusion means that Experimenter 2 has the same information that Experimenter 1 has regarding the parameter $\theta$. I apologize for not framing my question better -- I'm just quite confused by the point the author is trying to make in the paragraph below. I will update with an edit if I can clarify my question further.

Thus, Experimenter 2 can use this distribution and a randomization device, such as a random number table, to generate an observation $\mathbf{Y}$ satisfying $P(\mathbf{Y} = \mathbf{y}|T(\mathbf{X}) = T(\mathbf{x})) = P(\mathbf{X} = \mathbf{y}|T(\mathbf{X}) = T(\mathbf{x}))$. It turns out that, for each value of $\theta$, $\mathbf{X}$ and $\mathbf{Y}$ have the same unconditional probability distribution, as we shall see below. So Experimenter 1, who knows $\mathbf{X}$, and Experimenter 2, who knows $\mathbf{Y}$ have equivalent information about $\theta$, but surely the use of the random number table to generate $\mathbf{Y}$ has not added to Experimenter 2's knowledge of $\theta$. All his knowledge about $\theta$ is contained in the knowledge that $T(\mathbf{X}) = T(\mathbf{x})$. So Experimenter 2, who knows only $T(\mathbf{X}) = T(\mathbf{x})$, has as much information about $\theta$ as does Experimenter 1, who knows the entire sample $\mathbf{X} = \mathbf{x}$.

To complete the above argument, we need to show that $\mathbf{X}$ and $\mathbf{Y}$ have the same unconditional distribution, that is, $P_\theta(\mathbf{X} = \mathbf{x}) = P_\theta(\mathbf{Y} = \mathbf{x})$ for all $\mathbf{x}$ and $\theta$. Note that the events $\{\mathbf{X} = \mathbf{x}\}$ and $\{\mathbf{Y} = \mathbf{x}\}$ are both subsets of the event $\{T(\mathbf{X}) = T(\mathbf{x})\}$

Also recall that
  $$ P(\mathbf{X} = \mathbf{x}|T(\mathbf{X}) = T(\mathbf{x})) = (\mathbf{Y} = \mathbf{x}|T(\mathbf{X}) = T(\mathbf{x})) $$
  and these conditional probabilities do not depend on $\theta$. Thus, we have
  $$ P_\theta(\mathbf{X} = \mathbf{x}) = P_\theta(\mathbf{X} = \mathbf{x} \text{ and } T(\mathbf{X}) = T(\mathbf{x})) \\
=  P(\mathbf{X} = \mathbf{x}|T(\mathbf{X}) = T(\mathbf{x}))P_\theta(T(\mathbf{X}) = T(\mathbf{x})) \\
=  P(\mathbf{Y} = \mathbf{x}|T(\mathbf{X}) = T(\mathbf{x}))P_\theta(T(\mathbf{X}) = T(\mathbf{x})) \\
= P_\theta(\mathbf{Y} = \mathbf{x} \text{ and } T(\mathbf{X}) = T(\mathbf{x}))\\
= P_\theta(\mathbf{Y} = \mathbf{x})$$

答:
Some further intuition on how sufficiency preserves the information contained in the sample about a parameter of interest: given a random sample $X_1,\dots,X_n$ from a distribution family having densities $f_X(x; \theta)$. Let $\mathcal{X}$ denote the sample space of the random vector $(X_1,\dots,X_n)$, then we consider a statistic
$$
T= t(X_1,\dots,X_n)
$$
can be viewed as a partitioning of $\mathcal{X}$. To see this, let $\mathcal{T}$ be the sample space of $T$, and consider the sets
$$
\mathcal{X}_t = \{ (x_1,\dots,x_n)\in\mathcal{X} : t(x_1,\dots,x_n) = t\}
$$
for each $t \in \mathcal{T}$, then it is clear that the collection fo sets $\{\mathcal{X}_t\}_{t \in \mathcal{T}}$ forms a partition of $\mathcal{X}$.

Now, a statistic $T$ is useful if it has good data reduction properties. This can be judged by how good the corresponding partition is at reducing the number of possible values to be considered, as well as the degree to which all relevant information regarding the parameter $\theta$ is retained. So if decisions are to be made based on the value of the statistic instead of the observed data, then this decision will be the same for any dataset within the same partition $\mathcal{X}_t$. Therefore, for a statistic to be sufficient, the information which distinguishes the individual elements of each $\mathcal{X}_t$ should have no bearing on the value of $\theta$ (i.e. if an observed sample is known to be in a given $\mathcal{X}_t$, the probability of a sample taking any of the values within this member of the sample space partition should be independent of $\theta$.

Consider the example in which we have a sample of size $3$, $X_1,X_2,X_3$ from a Bernoulli distribution with parameter $p = P(X_i=1)$. The sample space for $(X_1,X_2,X_3)$ is
$$
\mathcal{X} = \{(0,0,0), (1,0,0), \dots, (1,1,1)\}
$$
with $|\mathcal{X}| = 2^3=8$. Consider two statistics $T_1 = X_1X_2+X_3$ and $T_2 = X_1+X_2+X_3$. We can see that $\mathcal{T}_1 = \{0,1,2\}$ and $\mathcal{T}_2 = \{0,1,2,3\}$. $T_1$ induces the sample space partitions:
\begin{align*}
\mathcal{X}^1_0 &= \{(0,0,0), (0,1,0), (1,0,0) \} \\
\mathcal{X}^1_1 &= \{(0,0,1), (0,1,1), (1,0,1),(1,1,0) \}\\
\mathcal{X}^1_2 &= \{(1,1,1) \},
\end{align*}
and similarly, $T_2$ induces the partitions
\begin{align*}
\mathcal{X}^2_0 &= \{(0,0,0) \} \\
\mathcal{X}^2_1 &= \{(0,0,1), (0,1,0), (1,0,0) \}\\
\mathcal{X}^2_2 &= \{(0,1,1), (1,0,1), (1,1,0) \}\\
\mathcal{X}^2_3 &= \{(1,1,1) \}
\end{align*}

Now, we can consider the distribution of the sample space values within each element of the two partitions. Suppose you are given that $T_1=0$, so the possible values for our original samples are those elements of $\mathcal{X}^1_0$. Then, consider the probability that the actual dataset was all zeroes:
\begin{align*}
P(X_1=0,X_2=0,X_3=0 | T_1=0)
&= \frac{P(X_1=0,X_2=0,X_3=0,T_1=0)}{P(T_1=0)}\\
&= \frac{P(X_1=0,X_2=0,X_3=0)}{P(X=(0,0,0) \text{ or } X=(0,1,0) \text{ or } X=(1,0,0) }\\
&=\frac{(1-p)^3}{(1-p)^3 + 2p(1-p)^2}\\
&=\frac{1-p}{1+p}
\end{align*}
and so the probability is not independent of $p$, in other words $T_1$ is not sufficient, as it does not induce an appropriate partition. In other words, if you based your decision on the value of $T_1=0$, it would have to be the same regardless of whether the actual sample had been $(0,0,0)$ or $(0,1,0)$, but these samples contain different information about the parameter of interest, $p$.

Now, suppose you are told that $T_2=1$, so that the possible values of the original sample are the elements of $\mathcal{X}^2_1$. You can easily show that $P(X_1=0,X_2=1,X_3=0|T_2=1) = \frac{1}{3}$, and you can actually show that for any value $T_2=t$, the chance that the actual dataset was one of the possible elements of $\mathcal{X}^2_t$ does not depend on $p$. So, $T_2$ is sufficient and basing estimates on its value retains all the relevant information in the sample $(X_1,X_2,X_3)$ regarding $p$.

For a formal definition: Let $X_1,\dots,X_n$ be a random sample from a distribution family with density function $f_X(x;\theta)$ where $\theta \in \mathbb{R}^k$ is a parameter vector. A vector valued statistic $T= t(X_1,\dots,X_n)$ is sufficient for $\theta$ if and only if the conditional distribution of $X_1,\dots,X_n$ given $T$ does not depend on $\theta$.


问:
Quaternions and special orthogonal transformations
I am asked to show that for $A,B\in\operatorname{SU}(2)$, the map $\phi\colon (A,B)\colon\mathbb H\to\mathbb H\colon h\mapsto AhB^{-1}$ is in $\operatorname{SO}(\mathbb H)$, where $\mathbb H$ are the quaternions (and we consider them as being embedded in $2$-by-$2$ complex matrices). Now, I know of a way to brute force this; I can consider a basis for $\mathbb H$, and then write out how $\phi(A,B)$ looks like. By means of Mathematica it can then be verified that indeed $\phi(A,B)\phi(A,B)^T=I$ and $\det(\phi(A,B))=1$. However... this method is unsatisfying. I'm hoping that if I were to choose a nice basis for $\mathbb H$, a quicker argument could be made. Would anyone have any idea?

答:
You need only to realize that for any two quaternions $u,w$,

$$|uv| = |u| |w|.$$
Thus
$$ | A hB^{-1} | = |A| |h| |B^{-1}|=|h|$$
and so $h\mapsto AhB^{-1}$ is in $O(4)$. Since $SU(2)$ is connected, we see that the determinant has to be one (by connecting to $A,B$ to $I$).


问:
Prove: If $f$ is bounded and uniformly continuous on $R$ and $g(x)=f(x)^2$ for all $x \in R$, then $g$ is uniformly continuous on $R$.
My initial approach was to use the definition of uniform continuity and altering the epsilon since there exists $M \in R$ such that $f(x)\leq M$

I'm confused about the selection of delta in this process, if using the epsilon-delta definition of uniform continuity even is the right way of doing this.
Thank you very much!

答:
Since $f$ is uniformly continuous and bounded, we know that:

$$\exists M \in \mathbb{R} \setminus \{0\}: \forall x \in \mathbb{R}: |f(x)| \leq M$$

$$\forall \epsilon > 0: \exists \delta > 0: \forall x,y \in \mathbb{R}: |x-y| < \delta \implies |f(x)-f(y)| < \epsilon$$

Let $\epsilon > 0$ be given. Let $x,y \in \mathbb{R}$. Then:

$$|g(x)-g(y)| = |(f(x))^2-(f(y))^2| = |(f(x)-f(y))(f(x)+f(y))| = |f(x)-f(y)| \cdot |f(x)+f(y)|$$

By the triangle inequality, wee can see that:

$$|g(x)-g(y)| \leq |f(x)-f(y)| \cdot (|f(x)|+|f(y)|) \leq |f(x)-f(y)| \cdot 2M$$

Then, we know that there is a $\delta_1 > 0$ such that:

$$|x-y| < \delta \implies |f(x)-f(y)| < \frac{\epsilon}{2M}$$

Hence, let $\delta = \delta_1$. Then:

$$|x-y| < \delta \implies |g(x)-g(y)| \leq |f(x)-f(y)| \cdot 2M < \frac{\epsilon}{2M} \cdot 2M = \epsilon$$

which is exactly what we wanted. Hence, $g$ is uniformly continuous on $\mathbb{R}$. Does that make sense?


Title: (A,B) regular => there is a scalar s such that A+s*B is regular ??

Body:
Given two matrices $A,B \in \mathbb{C}^{n \times n}$, is it true that $rank([A,B])=n \implies \exists s\in \mathbb{C}: rank(A+sB)=n$

It seems to me this could be easily proved by writing both in Jordan normal form. Am I mistaken? Is there a result I could use, or can someone suggest a name for the problem so I could google it?

If this is true, then does

Given two matrices $A,B \in \mathbb{R}^{n \times n}$, is it true that $rank([A,B])=n \implies \exists s\in \mathbb{R}: rank(A+siB)=n$

also hold?

Body: Consider $$A=\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}, \qquad B=\begin{pmatrix}1&0&0\\0&0&0\\0&1&0\end{pmatrix}.$$


问:
Finding a value using Runge-Kutta 4th order
I'm having trouble comprehending the following question.

Using Runge-Kutta 4th order method and a step size of 0.1, find the value of y=0.4

$$y\frac{dy}{dx}-y^{2}+x=0, y(0)=2$$

I am stuck in understanding what is supposed to mean by $y = 0.4$. Suppose that I make it $x(2)=0$ so that I am now looking for the value of $x(0.4)=0$. It still doesn't make sense for me since $y$ is initially $2$, unless a negative step size is valid.

What is your understanding of the problem? Should a negative step size be considered? Does the question actually mean $x = 0.4$?

答:
In the equation you presented, we are looking for a function $y(x)$ satisfying the differential equation and the initial condition $y(0)=2$. I believe there is a typo in the question and you are just being asked to provide an estimate of $y(0.4)$, using RK4. If you were meant yo find a value $x = x^*$ such that $y(x^*)=0.4$, the question would have surely been written in a different way.

Regarding the application of the Runge-Kutta method, you just need to look up the formula use it instead of the "$\cdots$" below
\begin{align*}
y(0) &= 2\\
y(0.1) &\approx y_1 = \cdots\\
y(0.2) &\approx y_2 = \cdots\\
y(0.3) &\approx y_4 = \cdots\\
y(0.4) &\approx y_4 = \cdots\\
\end{align*}


问:
tr$A$=tr$AB$ Can you say anything about B?
How to prove that if
$tr{A}=tr(AB),$ then $B=I$ if A and B are positive semidefinite?

My attempt: $\langle A, I\rangle =\langle A, B\rangle=\langle A, I-B \rangle$.

I do not know where to start. I feel that if I knew that eigenvalues of $B$ are at most 1, then $I-B$ would have been positive semidefinite and so the inner product of positive semidefinite matrices is 0 iff one of the matirces is 0. But I do not have that...

From my problem it follows that $A$ and $B$ are positive semidefinite, but I do not know anything about the eigenvalues of $B$. Is it necessary that $A,B\ge 0$ or is it enough that at least one of $A,B$ is psd?

Thank you!

答:
Even if $A$ and $B$ are assumed to be positive definite, it's not true.  Try
$$ A = \pmatrix{1 & 0\cr 0 & 1\cr},\ B = \pmatrix{t & 0\cr 0 & 2-t}$$
for $0 < t < 2$.


问:
Proving $\sum_{k=0}^{n}\left(\sum_{l=0}^{k}\binom{n}{k}\binom{k}{l} l\right)=n \times 3^{n-1}$ with a combinatoric argument

I want to prove
$$\sum_{k=0}^{n}\left(\sum_{l=0}^{k}\binom{n}{k}\binom{k}{l} l\right)=n \times 3^{n-1}$$
with combinatoric argumentation.

I have tried to understand it by comparing it with similiar problems. One such problem would be to prove the following equation with combinatoric argumentation:

$$\sum_{k=1}^n {{n}\choose{k}}*k = n* 2^{n-1}$$  The proof would be

https://math.stackexchange.com/a/7767/788271

How can i apply that solution to the given problem?

答:
We have $n$ people, One of them will be the Captain & all the others will be one of $3$ types (A,B or C).

Choose $n-k$ to be of type $A$, now choose $k-l$ out of the remaining $k$ to be of type $B$, now choose one of the remainig $l$ to be the captain and let the final $l-1$ be of type $C$.


答:
hint For transitivity

Assume that
$$7\mid(x+6y)$$
and
$$7\mid(y+6z)$$

then

$$x+6y=7k$$
and
$$y+6z=7k'$$

So, the sum gives
$$x+6z=7(k+k'-y)=7k''$$

Other approach

$$x+6y\equiv 0 \mod 7$$
$$y+6z\equiv 0 \mod 7$$
$$-7y\equiv 0 \mod 7$$
the sum gives
$$x+6z\equiv 0 \mod 7$$
Done.

Title: Prove that a relation is Equivalent where: $7\mid x+6y$ where x and y both come from a set of numbers (1,2,3.....10)

Body:
Like I've mentioned, I need to prove that it is equivalent, and get all of its classes.
Proving that it is Reflexive is super easy and I did it myself.
Proving symmetry and transitivity is what bothers me.
I don't even want to talk about classes, which I definitely do not know how to define.

Thanks for your help in advance


问:
Subtractive cancellation (same sign)
I know that subtractive cancellation occurs when subtracting two numbers close to each other. But if you have two numbers close to each other, one being negative and the other postive. You subtract the postive from the negative. Will there be subtractive cancellation here?</p>

答:
Summary: If we consider the relative error then the subtraction of two real numbers is ill-conditioned if the relative error is small. If we consider the absolute error, then the subtraction of two real number is well-conditioned.

Analysis: By design, condition numbers measure the sensitivity of the output to small changes in the input. It is critical that we carefully consider how we measure size. Let $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ be given by $$f(x) = x_1-x_2.$$ and consider the auxiliary function $$\kappa_f^{\text{abs}}(x,\delta) = \sup \left\{ \frac{|f(x)-f(y)|}{\|x-y\|_\infty} \: : \:  0 < \|x-y\|_\infty \leq \delta \right\}.$$
It is clear that $\delta \rightarrow \kappa_f^{\text{abs}}(x,\delta)$ is a monotone increasing and nonnegative function. This shows that the limit $$ \underset{\delta \rightarrow 0_+}{\lim}\kappa_f^{\text{abs}}(x,\delta)$$
exists. By definition,
$$
\kappa_f^{\text{abs}}(x) = \underset{\delta \rightarrow 0_+}{\lim}\kappa_f^{\text{abs}}(x,\delta)
$$
is the normwise absolute condition number of $f$ at the point $x$.
It is easy to see that $\kappa_f^{\text{abs}}(x) = 2$. Simply choose $y_1 = x_1 - \delta$ and $y_2 = x_2 + \delta$, so that $$f(x)-f(y) = 2\delta, \quad \|x-y\|_\infty = \delta.$$ This shows that $$\kappa_f^{\text{abs}}(x,\delta) = 2.$$
It follows that subtraction is well-conditioned with respect to absolute errors.

We now consider the relative error and define a new auxiliary function
$$\kappa_f^{\text{rel}}(x,\delta) = \sup \left\{ \frac{|f(x)-f(y)|}{|f(x)|} \big{/} \frac{\|x-y\|_\infty}{\|x\|_\infty} \: : \:  0 < \|x-y\|_\infty \leq \delta \|x\|_\infty \right\}.$$
It is clear that $\delta \rightarrow \kappa_f^{\text{rel}}(x,\delta)$ is a monotone increasing nonnegative function. By definition, $$\kappa_f(x) = \underset{\delta \rightarrow 0_+}{\lim}\kappa_f^{\text{rel}}(x,\delta)$$
is the normwise relative condition number of $f$ at the point $x$. However, this number is not necessarily small. The choice of $$y_1 = x - \delta |x_1|, \quad y_2 = x_2 + \delta |x_2|$$ ensures that $$\frac{f(x)- f(y)}{f(x)} = \delta \frac{|x_1|+|x_2|}{|x_1-x_2|}, \quad \|x-y\|_\infty = \delta \|x\|_\infty.$$
This implies that $$\kappa_f^{\text{rel}}(x,\delta) \geq \frac{|x_1|+|x_2|}{|x_1-x_2|}$$ and equality is not hard to verify. We conclude that $$\kappa_f^{\text{rel}}(x) = \frac{|x_1|+|x_2|}{|x_1-x_2|}$$
It follows that subtraction of two real numbers is ill-conditioned in the relative sense if $$|x_1-x_2| \ll |x_1|+|x_2|.$$ In particular, this happens when then relative difference between $x_1$ and $x_2$ is tiny.

Emphasis: Subtractive cancellation cannot happen if $x_1$ and $x_2$ have different sign, because in that case $$|x_1-x_2| = |x_1| + |x_2|.$$ Subtractive cancellation is unrelated to the absolute size of the numbers.

Intuition. Suppose that multi-billion dollar company is almost breaking even. A small relative change in the earning can easily cause a net profit. The switch from a negative to positive result is relative change which is larger than unity. This illustrates why the subtraction of near equals is ill-conditioned in the relative sense.

When I initially commented on your question I had not identified what could be the root cause of the confusion. It is the distinction between absolute and relative errors that must be emphasized.

Title: Subtractive cancellation (same sign)


问:
What is the geometric meaning of this vector equality? $\vec{BC}\cdot\vec{AD}+\vec{CA}\cdot\vec{BD}+\vec{AB}\cdot\vec{CD}=0$
I was doing some exercises for linear algebra. One of them was to prove that for any four points $A, B, C, D \in \mathbb{R}^3$ the following equality holds:
$$\overrightarrow{BC} \cdot \overrightarrow{AD}\ +\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}\ = 0$$
The proof is easy; you just make three vectors starting in $A$ and then see that all the terms cancel out.

My question is: what is the geometric interpretation of this equality? How can I visualize it or understand its deeper meaning? Does this equality have a name or where can I read more about it?

I'm asking this because it turns out that it is not just a random equality and is rather useful. For example, if we want to prove the existence of orthocenter, we can do it surprisingly easily and quickly using this equality.

答:
Here is a geometrical interpretation

beeing $H$ the projection of $D$ onto the plane containing $A$, $B$ and $C$ such that

$\overrightarrow{AD}=\overrightarrow{AH}+\overrightarrow{HD} $

$\overrightarrow{BD}=\overrightarrow{BH}+\overrightarrow{HD} $

$\overrightarrow{CD}=\overrightarrow{CH}+\overrightarrow{HD} $

and since $\overrightarrow{HD}$ is orthogonal to the plane containing $A$, $B$ and $C$, the given identity is equivalent to

$$\overrightarrow{BC} \cdot \overrightarrow{AH}\ +\ \overrightarrow{CA} \cdot \overrightarrow{BH}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CH}\ = 0$$

which is trivially true indeed by $\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}$ we obtain

$$(\overrightarrow{BA}+\overrightarrow{AC}) \cdot \overrightarrow{AH}\ +\ \overrightarrow{CA} \cdot \overrightarrow{BH}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CH}\ = $$

$$=\overrightarrow{BA}\cdot \overrightarrow{AH}+\overrightarrow{AC}\cdot \overrightarrow{AH}+\ \overrightarrow{CA} \cdot \overrightarrow{BH}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CH}\ = $$

$$=\overrightarrow{AB}\cdot \overrightarrow{HA}+\overrightarrow{CA}\cdot \overrightarrow{HA}+\ \overrightarrow{CA} \cdot \overrightarrow{BH}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CH}\ = $$

$$=\overrightarrow{AB}\cdot (\overrightarrow{CH}+\overrightarrow{HA})+\overrightarrow{CA}\cdot (\overrightarrow{BH}+\overrightarrow{HA})=$$

$$=\overrightarrow{AB}\cdot\overrightarrow{CA}+\overrightarrow{CA}\cdot\overrightarrow{BA}=0$$


问:
Show that $\bf (a^\top Q b)^2 \le a^\top Q a \ b^\top Q b$!
where $Q \in \mathbb{R}^{n\times n}$ is a positive definite matrix and $a, b \in \mathbb{R}^n$.

The actual problem is, that $\bf ||x||_Q = \sqrt{x^\top Qx}$ should be a norm, but this is as far as I've gotten trying to show that $\bf||a+b||_Q \le ||a||_Q + ||b||_Q$.

答:
Factor $Q = M^{\mathsf T}M$ for some matrix $M$; let $\mathbf x = M\mathbf a$ and $\mathbf y = M \mathbf b$.

(One of the characterizations of positive semidefinite matrices is that they have such a characterization. If you haven't seen it, it follows from the existence of an orthogonal basis of eigenvectors. This gives us a factorization $Q = P^{-1}D P = P^{\mathsf T} D P$, and then we can let $M = D^{\frac12} P$.)

Then your statement becomes $(\mathbf x^{\mathsf T}\mathbf y)^2 \le \mathbf x^{\mathsf T}\mathbf x \mathbf y^{\mathsf T}\mathbf y$, which is the Cauchy-Schwarz inequality.

Also, $\|\mathbf a\|_Q, \|\mathbf b\|_Q, \|\mathbf a + \mathbf b\|_Q$ become $\|\mathbf x\|, \|\mathbf y\|, \|\mathbf x+\mathbf y\|$, so the inequality you want to show turns into the triangle inequality for the standard norm.