问:
Equivalence Between Category of Covering Spaces and Category of Sets with an Action By The Fundamental Groupoid
Let $\Pi_1(X)$ be the fundamental groupoid of a locally path-connected topological space $X$ and define  $\Pi_1(X)-\mathbf{Sets}$ to be the category of sets equipped with an action by the fundamental groupoid. Let furthermore $\mathbf{Cov}(X)$ be the category of covering spaces over $X$. I want to show that we have the following equivalence of categories
$$\mathbf{Cov}(X)\simeq \Pi_1(X)-\mathbf{Sets}.$$

So far, I have constructed two functors
$$\Phi:\mathbf{Cov}(X)\to\Pi_1(X)-\mathbf{Sets}$$
and (I think and hope this is correct)
$$\Psi:\Pi_1(X)-\mathbf{Sets}\to\mathbf{Cov}(X).$$
I now want to show that they give me the equivalence between the categories. However, I find it quite difficult to show that we have natural isomorphisms $\Phi\Psi\Longrightarrow \mathbf{1}_{\Pi_1(X)-\mathbf{Sets}}$ and $\Psi\Phi\Longrightarrow \mathbf{1}_{\mathbf{Cov}_{/X}}$.

In the first section, I will briefly go through the construction of the functors $\Phi$ and $\Psi$. After that, I will explain my concerns about the equivalence.

This post will only concern one of the directions about the equivalence. Just to make the post more concentrated and not "too much". I will (maybe) post another question later, depending on if I don't manage to figure out the other direction by myself.

Construction of Functors

We begin with an explanation of the functor $\Phi$, then with $\Psi$.

The Construction of the Functor $\Phi$

In the category $\Pi_1(X)-\mathbf{Sets}$, an object is a functor $F:\Pi_1(X)\to\mathbf{Sets}$. So, we need to construct such a functor using a covering space.

Objects. Let $p:Y\to X$ be a covering space, we define a functor
$$F:\Pi_1(X)\to\mathbf{Sets}$$
as follows.

Objects of the fundamental groupoid are points of $X$, so we define
$$F(x)=p^{-1}(x),$$
where $x\in X$.

A morphism in $\Pi_1(X)$, between two points $x_0$ and $x_1$, is a path $\alpha:x_0\to x_1$ modulo homotopy equivalence. Given such a path, we construct a set-theoretical map $\alpha_*:p^{-1}(x_0)\to p^{-1}(x_1)$ (from a point $y_0\in p^{-1}(x_0)$ to a point $y_1\in p^{-1}(x_1)$) as follows:

We have a covering space $p:Y\to X$, which makes it possible to apply the path-lifting property. That is, we lift $\alpha:x_0\to x_1$ to a path $\widetilde{\alpha}:y_0\to y_1$, such that $\beta(0)=y_0$ and $p_*\widetilde{\alpha}=\alpha$. Letting $\widetilde{\alpha}(1)=y_1$, one can check that $p_*\widetilde{\alpha}(1)=p(y_1)=\alpha(1)=x_1$, which gives a well-defined map
$$\alpha_*:p^{-1}(x_0)\to p^{-1}(x_1),$$
which is the morphism in $\mathbf{Sets}$. So $F$ is, indeed, a functor.

Morphisms. A morphism in $\Pi_1(X)-\mathbf{Sets}$ between two functors $F,G:\Pi_1(X)\to\mathbf{Sets}$ is a natural transformation $\eta:F\Longrightarrow G$.

One can construct a natural transformation $\eta$ using a covering space homomorphism $\varphi:(Y_0,p_0)\to (Y_1,p_1)$, by letting $\eta_x:F(x)\to G(x)$ be defined by $\eta_{x}(y)=\varphi_{|p_0^{-1}(x)}(y)$. This gives us a commutative diagram
$\require{AMScd}$
\begin{CD}
p_0^{-1}(x_0) @>\eta_{x_0}>> p_1^{-1}(x_0)\\
@VVF(\alpha)V @VVG(\alpha)V\\
p_0^{-1}(x_1) @>\eta_{x_1}>> p_1^{-1}(x_1),
\end{CD}
with $\alpha:x_0\to x_1$ a homotopy path.

Hence, we know where to map objects through the functor $\Phi$.

The Construction of the Functor $\Psi$

Objects. Factor $X$ as follows
$$
X=\bigcup_i V_i,
$$
where $V_i$ has the subspace topology and such that $\Pi_1(V_i)=\pi_1(V_i,x_i)=0$ (with a basepoint $x_i$), which can be done since $X$ is locally path-connected.

An object in $\Pi_1(X)-\mathbf{Sets}$ is a functor $G$, say. The functor $G$ is constant at each $V_i$. That is, for $v_i\in V_i$, we have
$$
G(v_i)=S_i,
$$
where $S_i$ is some set, equipped with the discrete topology.

To build a covering space of $X$, let us define an equivalence relation $\sim_a$ on points in $\bigsqcup_i V_i$, we will do it through the collection of inclusions
$$
h_i:V_i\hookrightarrow X.
$$
We say that two points $x\in V_i$ and $y\in V_j$ are equivalent under $\sim_a$ if and only if
$$
h_i(x)=h_j(y).
$$
We then paste it together and express $X$ as the quotient space
$$
X=\bigsqcup_i V_i / \sim_a
$$
Furthermore, let $V_j\times S_j$ be the topological space, where $V_j$ has the subspace topology, $S_j$ the discrete topology and their product the product topology. We define another equivalence relation $\sim_b$ on the space
$$
\bigsqcup_i \Big [V_i\times S_i \Big ]
$$
as follows
$$
V_i\times S_i\ni(x,s)\sim_b (y,t)\in V_j\times S_j \iff \\
h_i(x)=h_j(y)\\
s\in G(v_i)\cong G(x)=G(y)\cong G(v_j)\ni t
$$
This gives us the space $\widetilde{X}$, which covers $X$
$$
\widetilde{X}=\bigsqcup_i \Big [V_i\times S_i \Big ]\Big /\sim_b.
$$

Morphisms. A morphism in $\Pi_1(X)-\mathbf{Sets}$, is a natural transformations $\mu:G_1\Longrightarrow G_2$ between $G_1,G_2:\Pi_1(X)-\mathbf{Sets}$. That is, given a point $x\in\Pi_1(X)$, we assign a morphism $\mu_x:G_1(x)\to G_2(x)$. And given a path $\gamma:x\to y$ in $X$, such that we have the following commutativity $\mu_y\circ G_1(\gamma)=G_2(\gamma)\circ\mu_x$.

We use that natural transformation to construct a covering space homomorphism.

Let $f: \bigsqcup_i\Big [ V_i\times G_1(V_i)\Big ] \to  \bigsqcup_i\Big [ V_i\times G_2(V_i)\Big ]$ be defined by $f(v,s)=(v,\mu_{x}(s))$, where $x\in V_j$ for some $j$. This gives us the commutativity of the diagram follows, and hence, we have a covering homomorphism
$$\require{AMScd}
\require{cancel}
\def\diaguparrow#1{\smash{\raise.6em\rlap{\ \ \scriptstyle #1}
   \lower.6em{\cancelto{}{\Space{2em}{1.7em}{0px}}}}}
\begin{CD}
&& \bigsqcup_i\Big [ V_i\times G_2(V_i)\Big ]\\
& \diaguparrow{f} @VVp_2V \\
\bigsqcup_i\Big [ V_i\times G_1(V_i)\Big ] @>>p_1> \bigsqcup_i\Big [ V_i\Big ]
\end{CD}$$
which induces a covering space homomorphism from $\widetilde{X}$ onto $X$.

Equivalence of Categories

I now want to show that the categories are equivalent
$$\mathbf{Cov}(X)\simeq \Pi_1(X)-\mathbf{Sets}.$$
This means that we have to construct natural isomorphisms of functors
$$
\Psi\circ \Phi\cong \operatorname{Id}_{\mathbf{Cov}(X)}\qquad \text{and}\qquad \Phi\circ\Psi\cong \operatorname{Id}_{\Pi_1(X)-\mathbf{Sets}}
$$

Constructing a Natural Isomorphism $\eta:\Psi\circ\Phi\Longrightarrow \operatorname{Id}_{\mathbf{Cov}(X)}$

What we need is to construct an isomorphism $\eta_{(Y,p)}:\Psi\circ\Phi(Y,p)\to \mathbf{1}_{\mathbf{Cov}(X)}(Y,p)$, for every covering space $(Y,p)$ of $X$, such that the following diagram commutes
$\require{AMScd}$
\begin{CD}
\Psi\circ\Phi(Y_0,p_0) @>\eta_{(Y_0,p_0)}>> \operatorname{Id}_{\mathbf{Cov}(X)}(Y_0,p_0)\\
@VV\Psi\circ\Phi(\varphi)V @VV\operatorname{Id}_{\mathbf{Cov}(X)}(\varphi)V\\
\Psi\circ\Phi(Y_1,p_1) @>\eta_{(Y_1,p_1)}>> \operatorname{Id}_{\mathbf{Cov}(X)}(Y_1,p_1)
\end{CD}

The above diagram reduces to

$\require{AMScd}$
\begin{CD}
\Psi\circ\Phi(Y_0,p_0) @>\eta_{(Y_0,p_0)}>> (Y_0,p_0)\\
@VV\Psi\circ\Phi(\varphi)V @VV\varphi V\\
\Psi\circ\Phi(Y_1,p_1) @>\eta_{(Y_1,p_1)}>> (Y_1,p_1)
\end{CD}

Question 1. Maybe this is obvious, but how should I define $\eta_{(Y,p)}$ to make it both an isomorphism and such that the diagram commutes?

Question 2. It is difficult for me to understand what $\Psi\circ\Phi(Y,p)$ and $\Psi\circ\Phi(\varphi)$ looks like. Applying $\Phi$ to $(Y,p)$ gives a functor $F:\Pi_1(X)-\mathbf{Sets}$, such that $F(x)=p^{-1}(x)$ and which maps paths according to rule we gave earlier.

Given the functor $F$, what does $\Psi(F)$ look like? Do I need to understand it in order to prove the commutativity of the diagram and/or construct $\eta_{(Y_0,p_0)}$?

Likewise, given a covering space homomorphism $\varphi:(Y_0,p_0)\to (Y_1,p_1)$, what does $\Psi\circ\Phi(\varphi)$ give me? First, I map $\varphi$ to a natural equivalence $\mu:F_0\Longrightarrow F_1$ such that we have morphisms $\mu_x:F_0(x)\to F_1(x)$ and a diagram which commutes
$\require{AMScd}$
\begin{CD}
p_0^{-1}(x_0) @>\mu_{x_0}>> p_1^{-1}(x_0)\\
@VVF_0(\alpha)V @VVF_1(\alpha)V\\
p_0^{-1}(x_1) @>\mu_{x_1}>> p_1^{-1}(x_1),
\end{CD}
where does $\Psi$ map this natural transformation?

This is probably easy. But there are a lot of abstract things going on at the same time, which makes me lose myself.

Best wishes,

Joel

答:
In simple terms, both a covering map $p$ (by looking at preimages) and a $\Pi_1(X)$-set are assigning discrete sets to the points of $X$, in a path compatible way.

For a given cover $(Y,p)$, we first construct the functor that assigns each point of $x\in X$ the fibre of $Y$ above it $\Phi(Y,p)$, then we construct a covering $(Z,q):=\Psi\circ\Phi\,(Y,p)$ such that its fibres are $q^{-1}(x)\cong\Phi(Y,p)\,(x)=p^{-1}(x)$.

Note that, by construction of $\Psi$, for every $x$ an index $i$ can be chosen such that $x\in V_i$ and we should also fix a path $w_x^i:v_i\leadsto x$ in $V_i$, so that
$$s\in G(v_i)\cong G(x)=G(y)\cong G(v_j)\ni t$$
should really mean $x=y$ and $G(w_x^i)\,(s)=G(w_x^j)\,(t)$, so that when we will lift the path $w_x^i$, point $s$ over $v_i$ goes to the same place where point $t$ over $v_j$.

(Since $V_i$ was chosen to be simply connected, this construction is independent of the actual choices $w_x^i$.)

Now, let $(x,s)\in Z$ be a point, then $s$ actually implicitly refers to an index $i$ such that $x\in V_i$ and $s\in \Phi(Y,p)\,(v_i)$, but $\Phi(Y,p)\,(v_i)=p^{-1}(v_i)$, so you receive the correspondent of $(x,s)$ by travelling $s\in p^{-1}(v_i)$ along $w_x^i$.

This correspondence is indeed bijective, for its inverse, travel a point $y\in p^{-1}(x)$ along $(w_x^i)^{-1}$ for some $i$ to obtain an element of $p^{-1}(v_i)$.

There are still lot of details to check, e.g. that these maps are well defined, continuous and natural.

I hope it helps. Good luck!


问:
Inequality proof 2
How to prove the inequality : for real numbers $\alpha_1, \ldots \alpha_n, \beta_1, \ldots \beta_n$:

$$\sqrt{(\alpha_1 + \beta_1)^2+\cdots+(\alpha_n + \beta_n)^2} \leq \sqrt{\alpha_1^2 + \cdots + \alpha_n^2}+\sqrt{\beta_1^2+\cdots+\beta_n^2}.$$

Thanks!

答: One easy way is to square both sides. Then use Cauchy-Schwarz to show the inequailty between the terms that don't vanish on both sides.</p>

ContentLicense: CC BY-SA 3.0

Title: Inequality proof 2


答:
(Edited to add details)

Due to the symmetry of vertical parabola, for distinct $x_1, x_2,$ $f(x_1)=f(x_2) \Rightarrow x_1+x_2=-b/a$ and $f'(x_1)+f'(x_2)=0$

For a quadratic, w cannot have $f(x_1)=f(x_2)=f(x_3)$ for distinct three $x_i$ since this would imply, $f(x)−f(x_1)$ has three distinct roots. There arise three conditions of arguments being pairwise equal.

$f(f(1))=f(f(2))=f(f(3)) \Rightarrow$ following cases :

$f(1)=f(2) \Rightarrow 1+2=-b/a$.

Now $f(1) \neq f(3)$. But $f(f(1))=f(f(3))$.  $\Rightarrow f(1)+f(3)=-b/a=3$

Here $f(1)+f(2) \neq -b/a$ since $f(1),f(2)$ are identical. Similarly,

$f(2)=f(3) \Rightarrow 2+3=-b/a$$f(1) \neq f(3) \Rightarrow f(1)+f(3)=-b/a=5$

$f(1)=f(3) \Rightarrow 1+3=-b/a$$f(1) \neq f(2) \Rightarrow f(1)+f(2)=-b/a=4$

But $f(1)+f(3)=10a+4b+2c$ is even. So only the third case holds: $f(1)=f(3)$ and $f'(1)+f'(3)=0$.

Thus $b=-4a$. $f(x)=ax^2-4ax+c \Rightarrow f'(x)=2a(x-2)$

Also $f'(f(x)) = (ay^2-4ay+c)'=2a(y-2)y'=2af'(x)(f(x)-2)$

Evaluating, $f'(f(1))+f(f'(2)+f'(f(3))=c$

where $a,c$ have been computed as below.

Remark :

Inspired from the other answer, even the numerical value of $c$ can be calculated. Subbing $f(x)=ax^2-4ax+c$ into $f(f(1))=f(f(2))$, one obtains $$c=\dfrac{7a+4}{2}$$

Quickly checking for $a\in \{2,4,6,8\}$, $c=23$ a prime, only for $a=6$.

ContentLicense: CC BY-SA 4.0

Body:
Let $a,b,c\in\mathbb{Z}$, $1<a<10$, $c$ is a prime number and $f(x)=ax^2+bx+c$. If $f(f(1))=f(f(2))=f(f(3))$, find $f'(f(1))+f(f'(2))+f'(f(3))$

My attempt:
\begin{align*}
f'(x)&=2ax+b\\
(f(f(x)))'&=f'(f(x))f'(x)\\
f'(f(x))&=\frac{(f(f(x)))'}{f'(x)}\\
\end{align*}


问:
Prove that the system has a single limit cycle
We have the following system:
\begin{cases}
      x'=x(x^2+y^2-2x-3)-y\\
      y'=y(x^2+y^2-2x-3)+x
   \end{cases}
We want to determine that it has a single limit cycle.
To prove this we had to use the Poincare Bendixson theorem.

We have the linearized system:
\begin{cases}
   x'=-y\\
   y'=x
\end{cases}
$A=\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}$. We can see that we have a center point.

After that I found the stationary points. My problem arose when I had to determine the set A, which does not contain stationary points and is invariant to flow. I tried to solve this problem, but without any success.

答:
As was established, for the dynamic of the radius you get
$$
\dot r=r(r^2-2x-3).
$$
This can be weakened to a scalar differential inequality
$$
r((r-1)^2-4)\le\dot r\le r((r+1)^2-4)\\
r(r-3)(r+1)\le r\le r(r-1)(r+3).
$$
The lower bound is positive for $r>3$, the upper bound negative for $r<1$. This means that the annulus $1\le r\le 3$ is a trapping region in negative time direction. As the angular velocity around the origin is always $1$, there are no equilibrium points except the origin. This establishes the existence of a repelling limit cycle inside the annulus. However not that it is single, there could in theory also be a triple of repelling, attracting and again repelling limit cycles.

Numerical experiments suggest that the limit cycle is close to the circle of radius $2$ around $(1,0)$, that is the root locus of the factor $x^2+y^2-2x-3$. Let $R=\sqrt{(x-1)^2+y^2}$, then
$$\begin{align}
\dot RR&=(x-1)\dot x+y\dot y=((x-1)x+y^2)(R^2-4)-(x-1)y+yx\\
&=R^2(R^2-4)+(x-1)(R^2-4)+y
\end{align}$$
By Cauchy-Schwarz
$$
|(x-1)(R^2-4)+y|\le R\sqrt{(R^2-4)^2+1}
$$
so that
$$
R(R^2-4)-\sqrt{(R^2-4)^2+1}\le \dot R\le \underbrace{R(R^2-4)}_{=f(R)}+\underbrace{\sqrt{(R^2-4)^2+1}}_{=g(R)}
$$

The upper bound is negative on $[1.1,1.8]$, the lower bound positive for $R>2.2$, so that the annulus $1.8\le R\le 2.2$ is a trapping region in negative time direction. The plot of the numerical solution shows that this bound is rather tight.


答:
To prove the equivalence of the categories, it suffices to take one of the above functors (e.g. $\Phi:\text{Cov}(X)→\Pi1(X)−Sets$) and show that it is:

Full

Faithful

Essentially Surjective

In other words, you don't need to explicitly construct the inverse functor and the natural isomorphisms. Hope this helps shorten the proof.

Title: Equivalence Between Category of Covering Spaces and Category of Sets with an Action By The Fundamental Groupoid

Body:
Let $\Pi_1(X)$ be the fundamental groupoid of a locally path-connected topological space $X$ and define  $\Pi_1(X)-\mathbf{Sets}$ to be the category of sets equipped with an action by the fundamental groupoid. Let furthermore $\mathbf{Cov}(X)$ be the category of covering spaces over $X$. I want to show that we have the following equivalence of categories
$$\mathbf{Cov}(X)\simeq \Pi_1(X)-\mathbf{Sets}.$$

So far, I have constructed two functors
$$\Phi:\mathbf{Cov}(X)\to\Pi_1(X)-\mathbf{Sets}$$
and (I think and hope this is correct)
$$\Psi:\Pi_1(X)-\mathbf{Sets}\to\mathbf{Cov}(X).$$
I now want to show that they give me the equivalence between the categories. However, I find it quite difficult to show that we have natural isomorphisms $\Phi\Psi\Longrightarrow \mathbf{1}_{\Pi_1(X)-\mathbf{Sets}}$ and $\Psi\Phi\Longrightarrow \mathbf{1}_{\mathbf{Cov}_{/X}}$.

In the first section, I will briefly go through the construction of the functors $\Phi$ and $\Psi$. After that, I will explain my concerns about the equivalence.

This post will only concern one of the directions about the equivalence. Just to make the post more concentrated and not "too much". I will (maybe) post another question later, depending on if I don't manage to figure out the other direction by myself.

Construction of Functors

We begin with an explanation of the functor $\Phi$, then with $\Psi$.

The Construction of the Functor $\Phi$

In the category $\Pi_1(X)-\mathbf{Sets}$, an object is a functor $F:\Pi_1(X)\to\mathbf{Sets}$. So, we need to construct such a functor using a covering space.

Objects. Let $p:Y\to X$ be a covering space, we define a functor
$$F:\Pi_1(X)\to\mathbf{Sets}$$
as follows.

Objects of the fundamental groupoid are points of $X$, so we define
$$F(x)=p^{-1}(x),$$
where $x\in X$.

A morphism in $\Pi_1(X)$, between two points $x_0$ and $x_1$, is a path $\alpha:x_0\to x_1$ modulo homotopy equivalence. Given such a path, we construct a set-theoretical map $\alpha_*:p^{-1}(x_0)\to p^{-1}(x_1)$ (from a point $y_0\in p^{-1}(x_0)$ to a point $y_1\in p^{-1}(x_1)$) as follows:

We have a covering space $p:Y\to X$, which makes it possible to apply the path-lifting property. That is, we lift $\alpha:x_0\to x_1$ to a path $\widetilde{\alpha}:y_0\to y_1$, such that $\beta(0)=y_0$ and $p_*\widetilde{\alpha}=\alpha$. Letting $\widetilde{\alpha}(1)=y_1$, one can check that $p_*\widetilde{\alpha}(1)=p(y_1)=\alpha(1)=x_1$, which gives a well-defined map
$$\alpha_*:p^{-1}(x_0)\to p^{-1}(x_1),$$
which is the morphism in $\mathbf{Sets}$. So $F$ is, indeed, a functor.

Morphisms. A morphism in $\Pi_1(X)-\mathbf{Sets}$ between two functors $F,G:\Pi_1(X)\to\mathbf{Sets}$ is a natural transformation $\eta:F\Longrightarrow G$.

One can construct a natural transformation $\eta$ using a covering space homomorphism $\varphi:(Y_0,p_0)\to (Y_1,p_1)$, by letting $\eta_x:F(x)\to G(x)$ be defined by $\eta_{x}(y)=\varphi_{|p_0^{-1}(x)}(y)$. This gives us a commutative diagram
$\require{AMScd}$
\begin{CD}
p_0^{-1}(x_0) @>\eta_{x_0}>> p_1^{-1}(x_0)\\
@VVF(\alpha)V @VVG(\alpha)V\\
p_0^{-1}(x_1) @>\eta_{x_1}>> p_1^{-1}(x_1),
\end{CD}
with $\alpha:x_0\to x_1$ a homotopy path.

Hence, we know where to map objects through the functor $\Phi$.

The Construction of the Functor $\Psi$

Objects. Factor $X$ as follows
$$
X=\bigcup_i V_i,
$$
where $V_i$ has the subspace topology and such that $\Pi_1(V_i)=\pi_1(V_i,x_i)=0$ (with a basepoint $x_i$), which can be done since $X$ is locally path-connected.

An object in $\Pi_1(X)-\mathbf{Sets}$ is a functor $G$, say. The functor $G$ is constant at each $V_i$. That is, for $v_i\in V_i$, we have
$$
G(v_i)=S_i,
$$
where $S_i$ is some set, equipped with the discrete topology.

To build a covering space of $X$, let us define an equivalence relation $\sim_a$ on points in $\bigsqcup_i V_i$, we will do it through the collection of inclusions
$$
h_i:V_i\hookrightarrow X.
$$
We say that two points $x\in V_i$ and $y\in V_j$ are equivalent under $\sim_a$ if and only if
$$
h_i(x)=h_j(y).
$$
We then paste it together and express $X$ as the quotient space
$$
X=\bigsqcup_i V_i / \sim_a
$$
Furthermore, let $V_j\times S_j$ be the topological space, where $V_j$ has the subspace topology, $S_j$ the discrete topology and their product the product topology. We define another equivalence relation $\sim_b$ on the space
$$
\bigsqcup_i \Big [V_i\times S_i \Big ]
$$
as follows
$$
V_i\times S_i\ni(x,s)\sim_b (y,t)\in V_j\times S_j \iff \\
h_i(x)=h_j(y)\\
s\in G(v_i)\cong G(x)=G(y)\cong G(v_j)\ni t
$$
This gives us the space $\widetilde{X}$, which covers $X$
$$
\widetilde{X}=\bigsqcup_i \Big [V_i\times S_i \Big ]\Big /\sim_b.
$$

Morphisms. A morphism in $\Pi_1(X)-\mathbf{Sets}$, is a natural transformations $\mu:G_1\Longrightarrow G_2$ between $G_1,G_2:\Pi_1(X)-\mathbf{Sets}$. That is, given a point $x\in\Pi_1(X)$, we assign a morphism $\mu_x:G_1(x)\to G_2(x)$. And given a path $\gamma:x\to y$ in $X$, such that we have the following commutativity $\mu_y\circ G_1(\gamma)=G_2(\gamma)\circ\mu_x$.

We use that natural transformation to construct a covering space homomorphism.

Let $f: \bigsqcup_i\Big [ V_i\times G_1(V_i)\Big ] \to  \bigsqcup_i\Big [ V_i\times G_2(V_i)\Big ]$ be defined by $f(v,s)=(v,\mu_{x}(s))$, where $x\in V_j$ for some $j$. This gives us the commutativity of the diagram follows, and hence, we have a covering homomorphism
$$\require{AMScd}
\require{cancel}
\def\diaguparrow#1{\smash{\raise.6em\rlap{\ \ \scriptstyle #1}
   \lower.6em{\cancelto{}{\Space{2em}{1.7em}{0px}}}}}
\begin{CD}
&& \bigsqcup_i\Big [ V_i\times G_2(V_i)\Big ]\\
& \diaguparrow{f} @VVp_2V \\
\bigsqcup_i\Big [ V_i\times G_1(V_i)\Big ] @>>p_1> \bigsqcup_i\Big [ V_i\Big ]
\end{CD}$$
which induces a covering space homomorphism from $\widetilde{X}$ onto $X$.

Equivalence of Categories

I now want to show that the categories are equivalent
$$\mathbf{Cov}(X)\simeq \Pi_1(X)-\mathbf{Sets}.$$
This means that we have to construct natural isomorphisms of functors
$$
\Psi\circ \Phi\cong \operatorname{Id}_{\mathbf{Cov}(X)}\qquad \text{and}\qquad \Phi\circ\Psi\cong \operatorname{Id}_{\Pi_1(X)-\mathbf{Sets}}
$$

Constructing a Natural Isomorphism $\eta:\Psi\circ\Phi\Longrightarrow \operatorname{Id}_{\mathbf{Cov}(X)}$

What we need is to construct an isomorphism $\eta_{(Y,p)}:\Psi\circ\Phi(Y,p)\to \mathbf{1}_{\mathbf{Cov}(X)}(Y,p)$, for every covering space $(Y,p)$ of $X$, such that the following diagram commutes
$\require{AMScd}$
\begin{CD}
\Psi\circ\Phi(Y_0,p_0) @>\eta_{(Y_0,p_0)}>> \operatorname{Id}_{\mathbf{Cov}(X)}(Y_0,p_0)\\
@VV\Psi\circ\Phi(\varphi)V @VV\operatorname{Id}_{\mathbf{Cov}(X)}(\varphi)V\\
\Psi\circ\Phi(Y_1,p_1) @>\eta_{(Y_1,p_1)}>> \operatorname{Id}_{\mathbf{Cov}(X)}(Y_1,p_1)
\end{CD}

The above diagram reduces to

$\require{AMScd}$
\begin{CD}
\Psi\circ\Phi(Y_0,p_0) @>\eta_{(Y_0,p_0)}>> (Y_0,p_0)\\
@VV\Psi\circ\Phi(\varphi)V @VV\varphi V\\
\Psi\circ\Phi(Y_1,p_1) @>\eta_{(Y_1,p_1)}>> (Y_1,p_1)
\end{CD}

Question 1. Maybe this is obvious, but how should I define $\eta_{(Y,p)}$ to make it both an isomorphism and such that the diagram commutes?

Question 2. It is difficult for me to understand what $\Psi\circ\Phi(Y,p)$ and $\Psi\circ\Phi(\varphi)$ looks like. Applying $\Phi$ to $(Y,p)$ gives a functor $F:\Pi_1(X)-\mathbf{Sets}$, such that $F(x)=p^{-1}(x)$ and which maps paths according to rule we gave earlier.

Given the functor $F$, what does $\Psi(F)$ look like? Do I need to understand it in order to prove the commutativity of the diagram and/or construct $\eta_{(Y_0,p_0)}$?

Likewise, given a covering space homomorphism $\varphi:(Y_0,p_0)\to (Y_1,p_1)$, what does $\Psi\circ\Phi(\varphi)$ give me? First, I map $\varphi$ to a natural equivalence $\mu:F_0\Longrightarrow F_1$ such that we have morphisms $\mu_x:F_0(x)\to F_1(x)$ and a diagram which commutes
$\require{AMScd}$
\begin{CD}
p_0^{-1}(x_0) @>\mu_{x_0}>> p_1^{-1}(x_0)\\
@VVF_0(\alpha)V @VVF_1(\alpha)V\\
p_0^{-1}(x_1) @>\mu_{x_1}>> p_1^{-1}(x_1),
\end{CD}
where does $\Psi$ map this natural transformation?

This is probably easy. But there are a lot of abstract things going on at the same time, which makes me lose myself.

Best wishes,

Joel


问:
Are there any practical implications of Infinite Monkey Theorem?
I've recently come across the Infinite Monkey Theorem which loosely states that if you gave a monkey a typewriter and an infinite amount of time, it would almost surely type out any given finite text (say, a Shakesperian sonnet).

I understand also that the "monkey" is a placeholder for some notion of an abstract machine that continually produces random strings of symbols (say, letters and numbers).

What are the implications of this theorem? I find the theorem fascinating, but am really not sure how (if at all) it has contributed to other branches of mathematics, real world applications, or academia in general. Any insight would be appreciated!

答:
Given an alphabet of size $n$, and a text with $k$ characters in it, the probability of typing this text in $k$ characters is:

$$\frac{1}{n^k}$$

If we allow $t$ attempts to write the text, the probability becomes

$$\frac{1}{n^k}\cdot\left(1+\left(1-\frac{1}{n^k}\right)+\left(1-\frac{1}{n^k}\right)^2+\dots+\left(1-\frac{1}{n^k}\right)^{t-1}\right)$$

and as $t\to\infty$, the probability tends to:

$$\frac{1}{n^k}\cdot\frac{1}{1-(1-\frac{1}{n^k})}$$

which equals $1$.

This can be used to prove Murphy's Law:

"Anything that can go wrong will go wrong."


答:
Hint: Square both sides

$(\alpha_1 + \beta_1)^2+\cdots+(\alpha_n + \beta_n)^2 \leq \alpha_1^2 + \cdots + \alpha_n^2+\beta_1^2+\cdots+\beta_n^2+ 2\sqrt{\left(\alpha_1^2 + \cdots + \alpha_n^2\right) \left(\beta_1^2+\cdots+\beta_n^2\right)}$

$\alpha_1 \beta_1+\cdots+\alpha_n \beta_n \leq  \sqrt{\left(\alpha_1^2 + \cdots + \alpha_n^2\right) \left(\beta_1^2+\cdots+\beta_n^2\right)}$

Title: Inequality proof 2

Body:
How to prove the inequality : for real numbers $\alpha_1, \ldots \alpha_n, \beta_1, \ldots \beta_n$:

$$\sqrt{(\alpha_1 + \beta_1)^2+\cdots+(\alpha_n + \beta_n)^2} \leq \sqrt{\alpha_1^2 + \cdots + \alpha_n^2}+\sqrt{\beta_1^2+\cdots+\beta_n^2}.$$

Thanks!


问:
Determining if certain sets are open or closed.
I would like someone to verify my answers to the below questions are correct.

Exercise 3.2.8 from Understanding Analysis by Stephen Abbott.

Assume that $A$ is an open set, and $B$ is a closed set. Determine if the following sets are definitely open, definitely closed, both, or neither.

(a) $\overline{A \cup B}$

(b) $A-B = \{x \in A: x \notin B\}$

(c) $(A^C \cup B)^C$

(d) $(A \cap B) \cup (A^C \cap B)$

(e) $(\overline{A})^C \cap \overline{A^C}$

My Attempt.

(a) $\overline{A}$ is definitely closed for any set $A$. So, $\overline{A \cup B}$ is definitely closed.

(b) $A - B$ is definitely open. For all $x$ remaining in $A - B$, there exists an $\epsilon$-neighbourhood, that is contained entirely in $A - B$.

Consider $A = (0,1)$, $B=\{1/n:n \in \mathbf{N}\} \cup \{0\}$.
$A-B$ is still open, because for the difference set is
\begin{align*}
 \bigcup_{n=1}^{\infty}\left(\frac{1}{n},\frac{1}{n+1}\right)
\end{align*}
which is a countable union of open sets.

(c) $(A^C \cup B)^C$ is definitely open. $A^C$ is closed, so $A^C \cup B$ is closed and therefore $(A^C \cup B)^C$ is open.

(d) $(A \cap B) \cup (A^C \cap B) = (A \cup A^C) \cap B = \mathbb{R} \cap B = B$. This set is definitely closed.

(e) $(\overline{A})^C \cap \overline{A^C}$. I can't tell this one for sure, because its the intersection of an open set with a closed set.

答:
Your answers for a)-d) are correct.

For e): Since $A$ is open its complement is closed. So $(\overline {A^{c})}=A^{c}$. Now $(\overline A)^{c} \cap A^{c}=(\overline A)^{c}$ which is open.


答:
Applying L'Hopital's rule you get
$$
\lim_{x\to 2}\dfrac{2^{x+2^x} \log ^2 2-2^{x^2+1} x \log (2)}{2^x \log 2-2 x} = 16 \log 2.
$$

Title: Find $\lim_{k\to 2}\frac{2^{2^k}-2^{k^2}}{2^k-k^2}$

Body:
Find $\displaystyle \lim_{k\to 2}\frac{2^{2^k}-2^{k^2}}{2^k-k^2}$

My attempt:
\begin{align*}
A&=\lim_{k\to 2}\frac{2^{2^k}-2^{k^2}}{2^k-k^2}\\
\ln A&=\lim_{k\to 2}\ln\frac{2^{2^k}-2^{k^2}}{2^k-k^2}\\
&=\lim_{k\to 2}(\ln(2^{2^k}-2^{k^2})-\ln(2^k-k^2))
\end{align*}
I stuck at $\ln(0)$.


Body:
$$\lim_{k\to2}\dfrac{2^{2^k}-2^{k^2}}{2^k-k^2}=\lim_{k\to2}2^{k^2}\cdot\lim_{k\to2}\dfrac{2^{2^k-k^2}-1}{2^{k}-k^2}$$

Now use $2^{k}-k^2=y$ as $2=e^{\ln2},$

$$\implies\lim_{y\to0}\dfrac{2^y-1}y=\lim_{y\to0}\dfrac{e^{y\ln2}-1}y=\ln2$$


答:
Okay here's the outline of how you solve this.

first, you write down the system of equations
$$
0=f(f(2))-f(f(1))=(3a+b)*(b+5a^2+3ab+2ac) \\
0=f(f(3))-f(f(2))=(5a+b)*(b+13a^2+5ab+2ac)
$$
and this leaves you with four possibilities:

(a) $(3a+b=0)$ and $(5a+b=0)$ which is impossible since $a \ne 0$

(b) $(3a+b=0)$ and $(b+13a^2+5ab+2ac=0)$
in which case we obtain $b=-3a$, then our second condition turns into $2ac-2a^2-3a=0$, which further yields $c=a+\frac{3}{2}$, and we see no acceptable solutions here since both $a$ and $c$ are asked to be integers.

(c) $(b+13a^2+5ab+2ac=0)$ and $(b+5a^2+3ab+2ac=0)$ which, if we subtract one of the equations from the other, gives us $b=-4a$ and, shortly after, $p=\frac{7a+4}{2}$, and we can just check the values of $a$ from $2$ to $9$ to find the only solution at $a=6, b=-24, c=23$.

(d) $(5a+b=0)$ and $(b+5a^2+3ab+2ac=0)$, in which case we get $b=-5a$, the second condition simplifies to $10a-2c+5$, which gives us $c=5a+\frac{5}{2}$, and we also see that $c$ will never be integer if $a$ is integer, and $a$ is always integer.

This leaves us with the only acceptable solution found in (c). As I already mentioned above, putting $(a,b,c)=(6,-24,23)$ into $f'(f(1))+f(f'(2))+f'(f(3))$ gives you $95$.

Title: Let $a,b,c\in\mathbb{Z}$, $1<a<10$, $c$ is a prime number and $f(x)=ax^2+bx+c$. If $f(f(1))=f(f(2))=f(f(3))$, find $f'(f(1))+f(f'(2))+f'(f(3))$


问:
Revisiting questions 3888565 and 3894925
The topic of odd perfect numbers likely needs no introduction.

I would like to revisit these two questions:

Is it possible to improve on the bound $D(q^k) < \varphi(q^k)$ if $k>1$?

On the quantity $I(q^k) + I(n^2)$ where $q^k n^2$ is an odd perfect number with special prime $q$

Here, $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$, $D(x)=2x-\sigma(x)$ is the deficiency of $x$, $\varphi(x)$ is the Euler totient function of $x$, and $I(x)=\sigma(x)/x$ is the abundancy index of $x$.

In the OP of the second question, the following lower bound is derived:
$$3 - \bigg(\frac{q-1}{q^2}\bigg) < I(q^k) + I(n^2).$$

In the accepted answer to the first question, MSE user mathlove gives the following improved lower bound, under the assumption that $k>1$:
$$3 - \bigg(\frac{q-1}{q^2}\bigg) + \bigg(\frac{q^4 - 1}{q^{k+1} (q - 1)}\bigg) = 3 - \frac{1}{q^{k+1}}\bigg(\varphi(q^k) - \frac{q^4 - 1}{q - 1}\bigg) < I(q^k) + I(n^2).$$

Here are my two questions:

(1) How does one prove the improved lower bound given by mathlove?

(2) Is the improved lower bound given by mathlove already "best-possible", in the sense of being obtainable via the $D(q^k)$-route?

(Note that I am already aware of the lower bound
$$3 - \bigg(\frac{q-2}{q(q-1)}\bigg) < I(q^k) + I(n^2),$$
which does not answer Question (2), because while this latter lower bound improves on mathlove's result, to the best of my knowledge it is not obtainable via the $D(q^k)$-route.)

答:
This is an answer to the question (1).

Let
$$f(k):=\varphi(q^k)-D(q^k)=q^{k-1}(q-1)-\bigg(2q^k-\frac{q^{k+1}-1}{q-1}\bigg)=\frac{q^{k-1}-1}{q-1}$$
Then, we see that $f(k)$ is increasing.

So, if $k\gt 1$ and $k\equiv 1\pmod 4$, then we have $f(k)\ge f(5)=\dfrac{q^{4}-1}{q-1}$ which is equivalent to
$$D(q^k)\le\varphi(q^k)-\frac{q^4-1}{q-1}$$
Using
$$D(n^2) < \frac{2n^2}{q}$$
we get
$$D(q^k)D(n^2) < \frac{2n^2}{q}\bigg(\varphi(q^k)-\frac{q^4-1}{q-1}\bigg)$$
Dividing the both sides by $2q^k n^2$, we obtain
$$\frac{D(q^k)D(n^2)}{2q^k n^2} < \frac{1}{q^{k+1}}\bigg(\varphi(q^k)-\frac{q^4-1}{q-1}\bigg)$$
which can be written as
$$3 - \bigg(I(q^k) + I(n^2)\bigg) < \frac{1}{q^{k+1}}\bigg(\varphi(q^k)-\frac{q^4-1}{q-1}\bigg)$$

We therefore finally have
$$3 - \frac{1}{q^{k+1}}\bigg(\varphi(q^k)-\frac{q^4-1}{q-1}\bigg) < I(q^k) + I(n^2) $$


问:
How is the depth of a parabolic mirror calculated
How is the depth of a parabolic mirror calculated. I couldn't find a similar question/answer.

I have this specific problem to solve (not related to parabolic mirrors, but is a starting point, and not interested in a focal point):

L=6"

D=4.5"

What is d?

Thanks for your help

答:
Staring from @Ishraaq Parvez's answer, as said, the equation
$$L=\frac{1}{2 a}\sinh ^{-1}\left(\frac{9 }{2}a\right)+\frac{9}{8} \sqrt{81 a^2+4}$$ requires some numerical method. However we can make nice approximations.

Let $x=\frac{9 }{2}a$ and $k=\frac{4 }{9}L$ to obtain
$$k=\sqrt{x^2+1}+\frac{\sinh ^{-1}(x)}{x}$$ and expand the rhs as a Taylor series
$$k=2+\frac{x^2}{3}-\frac{x^4}{20}+\frac{x^6}{56}-\frac{5 x^8}{576}+\frac{7
   x^{10}}{1408}+O\left(x^{12}\right)$$ Now, let $y=x^2$ and use series reversion to obtain
$$y=t+\frac{3 t^2}{20}-\frac{3 t^3}{350}+\frac{23 t^4}{8400}-\frac{5889
   t^5}{5390000}+O\left(t^6\right)$$ where $t=3(k-2)$.

Using $L=6$, $k=\frac{8}{3}$ then $t=2$, this would give
$$y\sim \frac{2567266}{1010625}\implies x\sim \frac 1 {175}\sqrt{\frac{2567266}{33}}\implies a\sim \frac 2 {1575}\sqrt{\frac{2567266}{33}}$$ Converted to decimals, this gives
$a=0.3542$ while the solution is $a=0.3553$.


问:
Find all positive integer solutions for the following equation $5^a + 4^b = 3^c$:

Find all positive integer solutions for the following equation:
$$5^a + 4^b = 3^c$$

My first guess would be to study the equation in mod, but I tried modulo 3, 4, 5, and 9 and I can't find anything.

答:
Hint: $$1^a + 0 \equiv _4 (-1)^c \implies c \equiv_2 0$$
so $c=2d$, $d\in \mathbb{N}$.

So $$ 5^a = (3^d-2^b)(3^d+2^b)$$

so $3^d-2^b = 5^x$ and $3^d+2^b = 5^y$. Now add/substract both equations...

...we see that $x=0$. Now if $b>1$ then we have $$(-1)^d -0 \equiv _41 $$ so $d$ is even. So we have $$(3^m-1)(3^m+1)= 2^b$$ Now $(3^m-1)$ and $(3^m+1)$ are two consecutive even numbers so exactly one is divisible by $4$ so the other number is exactly $2$.


问:
If $V$ is a vector space of dimension $n$ and $T: V → V$ is a transformation of V with $T ^ 2 = 0$, then $\text{rank }(T) ≤n / 2$.
Let $V$ be a vector space of dimension $n$ and  $T: V → V$ be a linear transformation of $V$ with $T ^ 2 = 0$.  Prove that $\text{rank }(T) ≤\frac{n }{ 2}$.

That's my homework and really I'm awful at this.
Can anyone help me here?

答:
Hint: Let $\alpha\in V.$ Then $T(\alpha)\in \text{range}( T).$
Since $$T^2(\alpha)=T(T(\alpha))=0 \\
\implies T(\alpha)\in \ker (T).$$ Thus, $\text{range} (T)\subseteq \ker (T).$ So that $\text{rank}(T)\leq\text{nullity}(T)$.

Now, by Rank-Nullity Theorem, we have
$$\text{rank}(T)+\text{rank}(T)\leq\text{rank}( T) + \text{nullity}( T)=n$$
Hence, we must have $\text{rank}( T)\leq n/2.$


问:
Logarithm approximation
I saw on https://math.stackexchange.com/a/2382470/432488 the following approximation: $\ln(x) \approx a x^{1/a}-a$, which is good for large value of $a$. Where does it come from?

答: This is the same thing as asserting that, if $a>0$ and $a$ is very small, then$$\log(x)\approx\frac{x^a-1}a,$$which is true, since\begin{align}\lim_{a\to0^+}\frac{x^a-1}a&=\lim_{a\to0^+}\frac{e^{a\log x}-1}a\\&=\log(x);\end{align}in other words, the derivative at $0$ of $a\mapsto x^a$ is $\log x$.

Title: Logarithm approximation


答:
$T$ has a non trivial kernel.  Lets say the dimension of the kernel of $T = m$

That means that there exist $m$ independent vectors in $V$ that are getting flattened  (mapped to the zero vector) when we apply $T(V)$.  When we apply the transformation a second time, no more than $m$ vectors in the image of $T(V)$ get mapped to the zero vector.

The rank of $T = n-m.$

The rank of $T^2 \ge n-2m.$

The rank of $T^k \ge n-km.$

It is given that the rank of $T^2 = 0.$  What does that say about $m$ relative to $n$?

Title: If $V$ is a vector space of dimension $n$ and $T: V → V$ is a transformation of V with $T ^ 2 = 0$, then $\text{rank }(T) ≤n / 2$.

Body:
Let $V$ be a vector space of dimension $n$ and  $T: V → V$ be a linear transformation of $V$ with $T ^ 2 = 0$.  Prove that $\text{rank }(T) ≤\frac{n }{ 2}$.

That's my homework and really I'm awful at this.
Can anyone help me here?


答:
For large $a$ we have

$$x^{ \frac{1}{a} } = e^{\frac{\log x}{a}} = 1 + \frac{\log x}{a} + O \left( \left( \frac{\log x}{a} \right)^2 \right)$$

(by Taylor's theorem) which gives

$$ax^{\frac{1}{a}} - a = \log x + O \left( \frac{(\log x)^2}{a} \right)$$

so the approximation is accurate once $a$ is large relative to $(\log x)^2$. By the way, I would describe this in exactly the opposite way: that $\log x$ is a good approximation to $ax^{ \frac{1}{a} } - a$! It's not as if $ax^{ \frac{1}{a} } - a$ is particularly easy to compute for large $a$.

Title: Logarithm approximation

Body: I saw on https://math.stackexchange.com/a/2382470/432488 the following approximation: $\ln(x) \approx a x^{1/a}-a$, which is good for large value of $a$. Where does it come from?


问:
Computing matrix from its eigenvectors
Could someone help me with solving this eigenvector question?
Thanks!

Complete the $2\times2$ matrix $A=\begin{bmatrix}x&y\\1&0\end{bmatrix}$ so that $A$ has eigenvectors $v_1=[3~1]^T$ and $v_2=[-2~1]^T$?

答:
Let $$D=\begin{bmatrix} 3 & 2 \\-2 & 1\end{bmatrix}$$ and let $$A=\begin{bmatrix} x & y \\ 1 & 0 \end{bmatrix}$$
Then $$B=D^{-1}AD= \frac{1}{5}\begin{bmatrix} 3(2+x)-y &  -2(2+x)+y \\ 3(3-x)-y  &  -2(3-x)-y \end{bmatrix}$$
Now set the off diagonal elements of $A$ as zero to make $B$ as adiagonal matrix, get
$y=2(2+x), y=3(3-x)  \implies x=1, y=6$, this gives
$$A=\begin{bmatrix} 1 & 6 \\ 1 & 0 \end{bmatrix}, B=\begin{bmatrix} 3 & 0 \\ 0 & -2 \end{bmatrix}$$

Title: Computing matrix from its eigenvectors


问:
Proving that a function in two variables is (or is not) differentiable in (0,0)
The Problem:

Consider the function:
$$ f : \mathbb{R^2} \to \mathbb{R} : (x,y) \mapsto \begin{cases} 2 \sin x + x^2 y \sin\Big(\frac{1}{x^2 + y^4}\Big) \quad &\text{if } (x,y) \neq (0,0) \\ 0 & \text{if } (x,y) = (0,0) \end{cases}$$
In which points of $\mathbb{R^2}$ is $f$ totally differentiable? What is the total derivative in those points?

Attempt:

We can use elementary methods of calculus to find the partial derivatives of $f$ for all $(x,y) \in \mathbb{R^2} \setminus \{(0,0)\}$. In $(0,0)$ we can use the limit definition of partial derivatives and find $(\partial_x f)(0,0) = 2$ and $(\partial_y f)(0,0) = 0$. We can now construct a candidate total derivative for all $(x,y) \in \mathbb{R}^2$, which we name $(df)(x,y)$. It has the $(1 \times 2)$-matrix representation:
$$ (df)(x,y) = \begin{bmatrix} (\partial_x f)(x,y) & (\partial_y f)(x,y) \end{bmatrix} $$
After making the tedious (yet simple) calculations, it is easy to see that both partials are surely continuous everywhere except maybe in $(0,0)$. Because of this, we know that $f$ is differentiable everywhere except maybe in $(0,0)$ and the total derivative is given by $(df)(x,y)$.

It is the point $(0,0)$ that is giving me trouble. After creating some plots with computer software, I have made an educated guess that $f$ is indeed differentiable in $(0,0)$. I have thought of two ways in which I could show this. One way involves showing that $\lim_{(x,y) \to (0,0)} (\partial_x f)(x,y) = 2$ and $\lim_{(x,y)\to(0,0)} (\partial_y f)(x,y) = 0$, which would imply that both partial derivatives are also continuous in $(0,0)$ and hence $f$ is totally differentiable everywhere. The other way involves substituting $(df)(0,0)$ into the definition of the total derivative:
$$ f(x,y) = f(0,0) + (df)(0,0)(x-0,y-0) + o\big(||(x-0,y-0)||\big) \quad \text{as $(x,y) \to (0,0)$}.$$
Simplifying and applying the definition of Landau's little-$o$ notation would then imply that:
\begin{align*} \lim_{(x,y) \to (0,0)} &\frac{|f(x,y) - f(0,0) - (df)(0,0)(x,y)|}{||(x,y)||} \\
= \lim_{(x,y) \to (0,0)} &\frac{\big|2 \sin x + x^2 y \sin \Big(\frac{1}{x^2 + y^4}\Big) - 2x\big|}{\sqrt{x^2 + y^2}} = 0
\end{align*}
Verifying the differentiability of $f$ then comes down to verifying that this limit expression indeed equals $0$. But calculating a limit in two variables is anything but obvious. My attempts to show this limit is not $0$ or does not exist (which would imply $f$ is not differentiable in $(0,0)$) have so far been futile.

Am I on the right path? Have I made any mistakes in the first part of my attempt? Is there any way I could (relatively) easily see that a function is differentiable or not, so that I don't need to resort to plotting the graph? Any tips or extra information would be much appreciated.

答:
Going with the hints in comments...
$$\lim_{(x,y) \to (0,0)} \frac{\big|2 \sin x - 2x + x^2 y \sin \big(\frac{1}{x^2 + y^4}\big)\big|}{\sqrt{x^2 + y^4}} \leq \lim_{(x,y) \to (0,0)} \Big|\frac{2 \sin x - 2x}{x}\Big| + \lim_{(x,y) \to (0,0)} \big|xy \sin \big(\frac{1}{x^2 + y^4}\big)\big|$$.
By noting that $\frac{\sin x - x}{x} < 0$ for any $x \neq 0$, we can get rid of the absolute value in the first limit. It is then easily seen to be zero. We use the fact that $0 \leq |\sin( \cdot )| \leq 1$ to obtain the following upper bound to the second limit: $\lim_{(x,y) \to (0,0)} |xy|$. While it may be obvious that this bound is $0$, it is in my opinion more rigourous to use the $\varepsilon,\delta$-definition for a quick proof.

Let $\varepsilon > 0$ and take $\delta = \sqrt{\varepsilon} > 0$. Choose $(x,y) \in \mathbb{R}^2$ such that $||(x,y)|| = \sqrt{x^2 + y^2} < \delta$. Then we surely have $|x| < \delta$ and $|y| < \delta$ and, therefore, $|xy| < \delta^2 = \varepsilon$. This proves that $\lim_{(x,y) \to (0,0)} |xy| = 0$.

We have now bounded the desired limit above by 0. But since the expression inside the limit itself is always greater than or equal to 0, the limit must be 0. This proves that $f$ is totally differentiable in $(0,0)$.


问:
Application of Sylow Theorems to groups of order $pq^2r$

Suppose G is a group of order $pq^2r$, where $p,q,r$ are primes and $p  < q < r.$ Suppose further that $q^2 \not\equiv 1\pmod{r}$, $pq^2 \not\equiv 1\pmod{r}$, and $\gcd(q,r-1) = 1$. Show that either $G$ contains a unique Sylow $r$-subgroup, or that $Z(G)$ contains an element of order $q$.

Here is my attempt at the problem. Striving for a contradiction, suppose that neither $G$ has a unique Sylow $r$-subgroup, nor does $Z(G)$ have an element of order $q$. By the third part of Sylow's Theorem, $n_r \equiv 1\pmod{r}$ and $n_r \in \{1,p,q,q^2,pq,pq^2\}$. By our assumption that $G$ does not have a unique Sylow $r$-subgroup, we cannot have $n_r = 1$. We cannot have $n_r = p$, since $p \not\equiv 1\pmod{r}$, and similarly $n_r \neq q.$ Furthermore, by our hypotheses that $q^2, pq^2 \not\equiv 1\pmod{r}$, we deduce by exhaustion that $n_r = pq.$

Next, we look at possible values for $|Z(G)|.$ By Lagrange's Theorem, we have $|Z(G)| \in \{1,p,q,q^2,r,pq,pq^2,pr,qr,q^2r,pqr,pq^2r\}.$ By Cauchy's Theorem, combined with our assumption that $Z(G)$ contains no element of order q, we immediately reduce this set of possibilities to $\{1,p,r,pr\}.$ Now suppose $R \in \text{Syl}_r(G)$. Then $[G:N_G(R)] = n_r = pq$ implies $|N_G(R)| = qr$. Since $Z(G) \leq N_G(R)$, we must have $|Z(G)|$ divides $qr$. Thus $|Z(G)| \in \{1,r\}$. But if $|Z(G)| = r$, then $Z(G)$ is a normal Sylow $r$-subgroup, which would imply $n_r = 1$, a contradiction. We deduce that $G$ has trivial center.

Finally, if we look at $n_q$, we see that $n_q \in \{1,p,r,pr\}$. Since $n_q \equiv 1\pmod{q}$, we cannot have $n_q = p$. If $n_q = r$, then we may write $r = 1 + kq$ for some $k \in \mathbb{Z}$. But then $r-1 = kq$, which implies that $q = \gcd(q,r-q)$, a contradiction. Thus either $n_q = 1$ or $n_q = pr.$

From here I'm kind of stuck.

答: if $n_{q}=1$ then let $Q<G$ be a sylow q-subgroup of $G$ and consider $\phi :G \to Aut(Q)$ s.t. for all $g \in G$ and $x \in Q$ we have $\phi (g)(x) = gxg^{-1}$, since $Q$ is normal we get that for all $g\in G$ the function $\phi (g)$ is really an automorphism on $Q$ From the first isomorphism theroem we deduce that $|G| = pq^2r = |ker(\phi)||Im(\phi)|$, also $Q$ is a $q$ group of order $q^2$ so $Q\cong \Bbb Z_{q^2}$ or $Q \cong \Bbb Z_{q} \times \Bbb Z_{q}$, and so $|Aut(Q)| = q(q-1)$ or $|Aut(Q)| = q(q-1)^2(q+1)$. From this follows that necceceraly $|Im(\phi)| = q,|Ker(\phi)| = pqr$  or  $|Im(\phi)| = 1,|Ker(\phi)| = pq^2r$. Also notice that $Q$ is abelian (group of order $q^2$) so $Q<Ker(\phi)$ and so $|Im(\phi)| = 1,|Ker(\phi)| = pq^2r$ and from here $Q<Z(G)$ (since every element in $G$ commutes with every element in $Q$)and clearly $Z(G)$ has an element with order $q$ (caushy).


答:
$Av_1 = \lambda_1 v_1\\
\begin{bmatrix} 3x + y\\3\end{bmatrix} = \lambda_1\begin{bmatrix} 3\\1\end{bmatrix}\\
\lambda_1 = 3\\
3x + y = 9$

Using similar logic we find that $\lambda_2 = -2$ and $-2x+ y = 4$

Leaving you with a system of two equations that you should be able to solve.

There is also a theorem that says that the trace of $A = \lambda_1+\lambda_2$ and the determinant of $A = \lambda_1\lambda_2$

Title: Computing matrix from its eigenvectors

Body:
Could someone help me with solving this eigenvector question?
Thanks!

Complete the $2\times2$ matrix $A=\begin{bmatrix}x&y\\1&0\end{bmatrix}$ so that $A$ has eigenvectors $v_1=[3~1]^T$ and $v_2=[-2~1]^T$?


问:
Induction proof to prove inequality?
I'm quite comfortable proving by induction, but I've stumbled upon an exercise where you need to prove a inequality using induction, something I have never done before, and where I'm absolutely clueless how to start. It goes as follows:

The numbers $a_k$ with $k\in \mathbb{N}$ is given by $a_0=1, a_1=2$ and $$a_k=5a_{k-1}-4a_{k-2}$$ for $$k\leqq2$$

Use proof by induction to prove $$a_{k-1}\leqq a_{k} \leqq 4a_{k-1}$$

If someone knows the answer it would be very much appreciated if you would share your wisdom :)

Thanks in advance

答:
At least in most elementary cases, induction is always : base case, induction case. What are we trying to prove? That $a_{n} \leq a_{n+1} \leq 4a_n$ for all $n \geq 0$.

Base Case

The base case is to check that this is true for $n=0$. This gives $a_0 \leq a_1 \leq 4a_0$. This is true, since the inequality becomes $1 \leq 2 \leq 4$.

The base case is clear by checking that $1<2<4$.

Induction Case

For the induction case, we'll assume that $a_{k-1} \leq a_k \leq 4a_{k-1}$ holds true up till index $n$, and prove it for $n+1$. In fact, we only need to assume that $4a_{n-1} \leq a_n \leq 4a_{n-1}$.

Now, remember these simple facts, which we will be using (all variables below take values in the real numbers) :

If $y\geq z$ then for any $x$ we have $x+y > x+z$. (Hate $2$, so I'll skip it)

If $y \geq z$ and $x<0$ then $xy \leq xz$ (note the sign reverses)

That's what we need, these two facts.

We want to prove $a_{n+1} \geq a_{n}$ first. Remember that $a_{n+1} = 5a_{n} - 4a_{n-1}$. Now, from our induction hypothesis, we know that $a_n \geq a_{n-1}$. Multiply by $-4>0$ and use point $3$ to get $-4 a_n \leq -4a_{n-1}$. Now, add $5a_n$ to both sides, then use point $1$ to get $5a_n - 4a_n \leq 5_n - 4a_{n-1}$. The left hand side is $a_n$, the right hand side is $a_{n+1}$ so $a_n \leq a_{n+1}$.

Now let's prove that $a_{n+1} \leq 4a_n$. Once again, $a_{n+1} = 5a_{n} - 4a_{n-1}$. We know that $a_n \leq 4a_{n-1}$. Again, multiply by $-1$ and use point $3$ to get $-a_n \geq -4a_{n-1}$. Now, add $5a_{n}$ to both sides and use point $1$ to get $5a_n - a_n \geq 5a_n - 4a_{n-1}$. The left side is $4a_n$ , the right side is $a_{n+1}$ so $4a_n \geq a_{n+1}$.

Thus $a_n \leq a_{n+1} \leq 4a_n$ and the induction is complete!

Title: Induction proof to prove inequality?


问:
Is this proof of integrability of a random variable correct?
My textbook on advanced probability states that
$$\forall \epsilon >0 \exists \delta > 0 \forall A \in \mathbb{F}: P(A) < \delta \implies \int_A |X| dP < \epsilon$$
for a random variable $X$ on $(\Omega,\mathbb{F},P)$ such that $E|X| < \infty$.

I am trying to figure out whether the converse holds, that is that when this implication holds than $E|X| < \infty$. I came up with the following proof:

Let $A_i = (|X|)^{-1}(i)$ -- that is $A_i$ contains all $\omega \in \Omega$ s.t. $|X(\omega)| = i$. Let $\epsilon = 1$ (any other value would work) and let $\delta$ be some value that satisfies the formula above (we are assuming it exists). Now consider separately all $A_i$ with $P(A_i) \geq \delta$ and those with $P(A_i) < \delta$. There are finitely many of the first kind, meaning that they contribute a finite amount to $E|X|$. Now we partition all the $A_i$'s of the second kind into $B_j$'s such that each $B_j$ is a union of some number of $A_i$'s such that $\delta/2 \leq P(B_j) < \delta$ for each $j$ except one $j$ for which only $P(B_j) < \delta$. Note that such partition exists. The inequality $P(B_j) < \delta$ can be satisfied easily and then if we consider a partition that satisfies $P(B_j) < \delta$ for all $j$ and the inequality $\delta/2 \leq P(B_j)$ is violated for smallest possible number of $j$'s, it will only be violated for one as otherwise, we could merge two such sets of the partition. Now the number of $B_j$'s is clearly finite as all but one have probability at least $\delta/2$. Each contributes at most $1$ to $E|X|$ and, therefore, the first absolute moment is finite (one could write it out formally as splitting the integral that is $E|X|$ into two parts, estimating each as I described).

Is this idea correct? The thing is that ew have been told that the converse implication does not hold but it seems to me that this proof should work.

答:
Note that the set $\{\{i\}:i\in\mathbb{R}\}$ is uncountable, so you cannot get some conclusion from there. By example if $X:(0,1)\to (0,1),\, x\mapsto 1/x$ and the probability measure is the Lebesgue measure in $(0,1)$ then you will have that $\Pr [A_i]=0$ for all $i\in[0,1]$, however $\int_0^1 X\,dP =\infty $.

A counterexample is this: let the probability space $[0,1]$ with the probability measure given by $\Pr [A]=1$ if and only if $0\in A$ and zero otherwise. Now let $X(c)=\infty $ if and only if $c=0$ and zero otherwise, then $\int_{0}^1 X \mathop{}\!d P=\infty $ but for every $\epsilon \in(0,\infty )$ we have that $\int_{A}X\mathop{}\!d P<\epsilon $ if and only if $0\notin A$, so choosing $\delta \in(0,1)$ the condition holds.


问:
Transformation between two equations for hyperbola: $y=\frac1x$ and $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Back in school we were taught that the equation for hyperbola is:
$$y=\frac1x \tag1$$
The equation used in the university is different:
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \tag2$$

I can't understand how they are the same and how to get one from another.

What is the relation between these two forms?

答:
$$xy=1~~~~(1)$$ is rectangular hyperbola but
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1~~~~(2)$$ is more standard hyperbola.

Interestingly, (2) can also be written as
$$XY=1, X=\frac{x}{a}-\frac{y}{b}, Y=\frac{x}{a}+\frac{y}{b}~~~(3)$$
in the transformed co-ordinate system.

EDIT:
We can write $$\begin{bmatrix} X \\ Y \end{bmatrix}=(ab)^{-1} \begin{bmatrix} b & -a \\ b & a \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$$
This transformation matrix can be belated to the rotataion matrix:
$$\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin\theta     & \cos\theta\end{bmatrix}.$$ in a special case.


问:
Confused on definition of strong induction
I found the following statement in Munkres' Topology:

Theorem 4.2 (Strong induction principle). Let $A$ be a set of positive integers. Suppose that for each positive integer $n$, the statement $S_n \subset A$ [here $S_n = \{1, 2, \dots, n\}$] implies the statement $n \in A$. Then $A = \mathbb{Z}_+$.

Now, I think I understand strong induction. But what I don't understand here is, doesn't $S_n \subset A$ always imply $n \in A$? It's pretty much by definition of a subset, that if $\{1,2,\dots,n\} \subset A$ then $n \in A$. Did the author mean "Suppose that for each positive integer $n$ the statement $S_n \subset A$ is true"?

答: If you check the previous page, you should see that $S_n$ is the set of positive integers that are less than $n$. $S_n$ here denotes the section of $\Bbb Z_+$ by $n$. See also the more general definition immediately before Lemma 10.2.


答:
Those are both hyperbolas, though the second equation is more "general" in the sense that it describes many different hyperbolas (precisely those with their foci on the x-axis).
Actually the equation $y = \frac{1}{x}$ describes the graphic of the "homographic function" which is an hyperbola such that it is tangent to both axises. This geometric transformation is necessary to make it an actual fuction.

To obtain the first from the second, first choose $a=b = \sqrt{2}$ in the second equation (in this way you obtain a rectangular hyperbola with perpendicular tangents, then rotate your hyperbola by -45° degrees, which is apply the following transformation:
\begin{align}
x &= \frac{\zeta +\eta}{\sqrt{2}}\\
y &= \frac{-\zeta +\eta}{\sqrt{2}}
\end{align}
Thus the second equation becomes
\begin{align}
\frac{2\zeta\eta}{2} = 1
\end{align}
Solving for $\eta$ you obtain
\begin{align}
\eta = \frac{1}{\zeta}
\end{align}
Now changing name of your variables you get the first equation.
Can you see why now this is a function?

Title: Transformation between two equations for hyperbola: $y=\frac1x$ and $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Body:
Back in school we were taught that the equation for hyperbola is:
$$y=\frac1x \tag1$$
The equation used in the university is different:
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \tag2$$

I can't understand how they are the same and how to get one from another.

What is the relation between these two forms?


问:
Convergence of these series
$$\sum_{n=1}^\infty \frac{2nx^{n}}{(n+1)^{2}3^{n}} \tag{1}$$

$$\sum_{n=1}^\infty x^{n}\tan\frac{x}{4^{n}}\tag{2}$$

Is there any good article that describes an equivalents like if $$ \lim_{n\to\infty} \sin\frac{2\pi}{3^{n}} \sim \frac{2\pi}{3^{n}}\tag{3}$$

(am I right about $(3)$?)

答:
$\lim_{n\to\infty} \sin\frac{2\pi}{3^{n}} \sim \frac{2\pi}{3^{n}}\tag{3}$ Is correct since $\lim_{x\to 0 }  \dfrac{sinx}{x}=1$ and since sinus is continuous this holds for any sequence
 ${x_n\to 0}$</p>

Title: Convergence of these series


问:
Find the range of a given representing matrix of a linear transformation
Given the following:

Let $T:\mathbb R^3 \rightarrow \mathbb R^3$ be a linear transformation. And let $B = ((1,1,1),(1,1,0),(1,0,1))$ be a basis of $\mathbb R^3$.

The representing matrix of the transformation is as follows: $[T]_B = \begin{bmatrix}1&0&1\\3&2&1\\2&1&1\end{bmatrix}$

Find a basis to the range ($Im T$) and the Kernel ($ker T$)

I've found the kernel by taking a generic vector from $\mathbb R^3$, applying the coordinates of the basis on it and found that $ker T = Sp{(1,0,0)}$.

I'm just not sure how to go about finding the range.

As far as I know, the steps to find it are as follows:

Transpose the representing matrix.

Use elementary actions to bring it to a canonical form.

Restore the given matrix to a vectors via the basis coordinates.

I'm failing to understand the last part, here's what I've found:
$\begin{bmatrix}1&0&1/2\\0&1&1/2\\0&0&0\end{bmatrix}$

How do I restore these vectors via the basis coordinates?

EDIT: Fixed the wrong basis B.

答:
The coordinates $[x,y,z]$ in the ordered basis $\{v_1,v_2,v_3\}$ represent the vector $xv_1+yv_2+zv_3$ in the standard basis $\{e_1,e_2,e_3\}$ (assuming $v_i$ are written in standard basis).

For the kernel basis, take $v\in\Bbb R^3$ in basis $B$.

$\begin{bmatrix}1&0&1\\3&2&1\\2&1&1\end{bmatrix}v=0$ gives $v=\begin{bmatrix}-k\\k\\k\end{bmatrix}$ which is the linear span of $-[1,1,1]^T+[1,1,0]^T+[1,0,1]^T=(1,0,0)$ in the standard basis.

Note that the range-space of a linear transformation is its column space (subspace spanned by column vectors) or the row-space of its transpose. You have found the linearly independent rows of $[T]_B^T$ as $[1,0,1/2]^T,[0,1,1/2]^T$ which serve as the basis vectors of the range in basis $B$. In the standard basis,

$$~[1,0,1/2]^T_S=1[1,1,1]^T+0[1,1,0]^T+\frac12[1,0,1]^T\\~[0,1,1/2]^T_S=0[1,1,1]^T+1[1,1,0]^T+\frac12[1,0,1]^T$$


问:
How to prove that $(A×B) ∩ (B×A)=(A ∩ B) × (B ∩ A)$?
It is known that, to prove that two sets are equal, we need to show that they are subset of each other. Here we have, intersection of cartesian products on the left side and cartesian product of intersections of sets on the right side, I know the definitions here that is:

Cartesian Product: If $A$ and $B$ are any two sets, then $A\times B=$ {$ (a,b), a \in A, b \in B$ }.
Intersection of sets: $A \cap B=${$ x\colon x \in A, x \in B $}.

but don't know how to prove that.

答:
Let $(a,b) \in (A \times B) \cap(B \times A),$ then we have that $a \in B$ and $b \in A.$ This gives $a,b \in A \cap B.$ Since $A \cap B=B \cap A, $ we derive $(a,b) \in (A \cap B) \times (B \cap A).$

Thus we have shown that

$$(A \times B) \cap(B \times A) \subseteq (A \cap B) \times (B \cap A).$$

It is now your turn to show that

$$(A \times B) \cap(B \times A) \supseteq (A \cap B) \times (B \cap A).$$


问:
partial derivations and evaluation commute?
Let $F(x,y,z): \mathbb{R}^3 \to \mathbb{R}$ a smooth function in three variables.
I am trying to figure out taking derivative with respect to $x$ or $y$ commute with setting $z = 0$?
In other words does
$$
\frac{\partial F}{\partial x}(x, y, 0) = \frac{ \partial F(\cdot, \cdot, 0) }{\partial x} (x, y)
$$
for all $x,y \in \mathbb{R}$? I have tried some examples and they all seem to  satisfy this.. Any comments appreciated. Thank you!

答: That is correct, just write out the definitions:$$\frac{\partial F}{\partial x}(x,y,0):=\lim_{t\to0}\frac{F(x+t,y,0)-F(x,y,0)}{t}=:\frac{\partial F(\cdot,\cdot,0)}{\partial x}(x,y)$$


问:
Exponential formula to assign list items an equally distributed percentage
I am going to explain this as best as I can:

We have a number of list items, that dynamically changes (sometimes it has 5 items sometimes it has 7 items or 43 items, etc.) We are trying to find a relevant prioritypercentage to assign to each list item (starting from bottom to top) but cannot figure out a formula to do so based on the number of items.

For example (percentages are COMPLETE guesses):

Is there a formula that will take a number and distribute 100 points exponentially or any other way as shown below?

A list item with 2 items:

List item 1 (75%)

List item 2 (25%)

A list item with 3 items:

List item 1 (50%)

List item 2 (30%)

List item 3 (20%)

答: If you want each one to be smaller than the last by a ratio $r$, the sum of $1+r+r^2+\dots+r^{n-1}$ ($n$ terms) is $\frac{1-r^n}{1-r}$ so you could give $100\frac{r^{i-1}(1-r)}{1-r^n}$ to iterm $i$, where $i$ ranges from $1$ to $n$.  So for $r=\frac{2}{3}, n=4$, the sum is $\frac{65}{27}$ and the points are about $41.53, 27.69, 18.46, 12.30.$  Does this work for you?</p>


问:
What is the value of this repeated square root: $\sqrt{1\sqrt{2\sqrt {3 \sqrt{4\cdots}}}}$

Find the value of
  $$\sqrt{1\sqrt{2\sqrt {3 \sqrt{4\sqrt{5\sqrt{6\cdots\sqrt{\infty}}}}}}}$$

What is the absolute value of the root in below question and what does it represent geometrically, I had a few approaches leading to possible values to approximation and know the answers, but I also need what it represents geometrically.

答:
As indicated by Ron Gordon the logarithm of this is :
$$l= \sum_{k=1}^\infty \frac{\log{k}}{2^k}$$
Consider the polylogarithm function :
$$f(s):=\sum_{k=1}^\infty \frac{(1/2)^k}{k^s}=\operatorname{Li}_s\left(\frac 12\right)$$
then
$$f'(s)=-\sum_{k=1}^\infty \frac{\log(k)}{2^k\;k^s}$$
so that the final answer may be written as $\ \boxed{\displaystyle e^{-\operatorname{Li}_{0'}\left(1/2\right)}}$ with the meaning :
$e^{-\lim_{s\to 0^+}\frac d{ds}\operatorname{Li}_s\left(1/2\right)}$</p>


答:
$\Rightarrow: $ If $(x, y) \in A \times  B \cap B \times A$, then $x$ must be in both $A$ and $B$, and likewise for $y$. So $(x,y) $ must be in the right side as well.

$\Leftarrow$: The reasoning is nearly identical to the previous reasoning, just reversed.

Title: How to prove that $(A×B) ∩ (B×A)=(A ∩ B) × (B ∩ A)$?

Body:
It is known that, to prove that two sets are equal, we need to show that they are subset of each other. Here we have, intersection of cartesian products on the left side and cartesian product of intersections of sets on the right side, I know the definitions here that is:

Cartesian Product: If $A$ and $B$ are any two sets, then $A\times B=$ {$ (a,b), a \in A, b \in B$ }.
Intersection of sets: $A \cap B=${$ x\colon x \in A, x \in B $}.

but don't know how to prove that.


问:
Question on injective envelopes
Let $A$ be an Artin algebra and let $\text{mod}(A)$ denote the category of finitely generated left $A$-modules.

Let $S$ be a simple module in $\text{mod}(A)$ and let $\iota_S: S \rightarrow I(S)$ be the injective envelope of $S$. Then  the natural epimorphism $p: I(S) \rightarrow I(S)/\text{Im}(\iota_S)$ is irreducible.

Unfortunately I have no idea how to prove this. Can anybody help?

答:
Suppose not. Then $p$ factors as
$$I(S)\xrightarrow{s}X\xrightarrow{t}I(S)/\operatorname{Im}(\iota_S),$$
where $s$ is not a split monomorphism.

Since $I(S)$ is injective, $s$ can not be a monomorphism, or it would be a split monomorphism. Since $\operatorname{Im}(\iota_S)$ is the unique simple submodule of $I(S)$, it must be contained in the kernel of $s$, and so $s$ factors through $p$: say $s=rp$ for some $r:I(S)/\operatorname{Im}(\iota_S)\to X$. So $p=ts=trp$, and so $tr$ is the identity map of $I(S)/\operatorname{Im}(\iota_S)$, since $p$ is surjective.

So $t$ is a split epimorphism.

Title: Question on injective envelopes


答:
Let $(a,b)\in (A\times B)\cap (B\cap A).$

Then $$(a,b)\in (A\times B) \text{ and } (a,b)\in (B\times A)$$
iff $a\in A, b\in B \text{ and } a\in B, b\in A$

iff $a\in (A\cap B), b\in (B\cap A)$

iff $(a,b)\in (A\cap B)\times (B\cap A).$


问:
An Exercise on projective covers and injective envelopes of simple modules over an Artin algebra
Let $A$ be an artin algebra and let $\text{mod}(A)$ denote the category of finitely generated left $A$-modules.

Let $S$ be a simple module in $\text{mod}(A)$. Let $\iota_S: S \rightarrow I(S)$ and $\pi_S: P(S) \rightarrow S$ be its injective envelope and projective cover, respectively. If $f:M \rightarrow I(S)$ is a non-zero monomorphism in $\text{mod}(A)$, then there exists a morphism $g: P(S) \rightarrow M$ such that $fg \neq 0$.

My attempt:

If $f$ is injective then we have an exact sequence $0 \rightarrow M \xrightarrow{f} I(S) \xrightarrow{p} I(S)/\text{Im}(f) \rightarrow 0$. Hence if we show $p\iota_S \pi_S=0$ then there must exist a morphism $g: P(S) \rightarrow M$ with $fg=\iota_S \pi_S \neq 0$.

Now in order to show $p\iota_S \pi_S=0$ I think one needs to prove that $S$ is a submodule of $\text{Im}(f)$. That is the point where I have no idea how to continue. Does anybody have a clue?

答: $S$ is the unique simple submodule of $I(S)$ [how easy this is to see depends on the definition of "injective envelope" that you are using]. So every nonzero submodule of $I(S)$ contains $S$.


问:
Show that $f(x, y) = (x+y, x-y)$ is a bijection.

Show that the function $f: \mathbb{R}\times\mathbb{R}\to \mathbb{R}\times\mathbb{R}$ be defined by $f(x, y) = (x+y, x-y)$ is a bijection.

I tries this by matrix inverse method but i am not able to solve the problem.

Please someone guide me to the solution.

答:
Injectivity: $f(x_1,y_1)=f(x_2,y_2)\implies(x_1+y_1,x_1-y_1)=(x_2+y_2,x_2-y_2)$. So $(x_1+y_1)+(x_1-y_1)=(x_2+y_2)+(x_2-y_2)$ and $(x_1+y_1)-(x_1-y_1)=(x_2+y_2)-(x_2-y_2)$.

Surjectivity: For $(k,m)\in\Bbb R^2,f\left(\frac{k+m}2,\frac{k-m}2\right)=(k,m)$.

Alternatively note that $f$ is a linear transformation with the matrix \begin{bmatrix}1&1\\1&-1\end{bmatrix} which is invertible.


答:
You need to use a double inclusion demonstration.

I'll help you for the first one :

I note $C=(A×B) ∩ (B×A)$ and $D=(A ∩ B) × (B ∩ A)$.

Let's show that $C \subset D$.

To prove that, it is enough to prove that all elements $x$ in $C$ belongs to $D$.

Let $x$ be an element in $C$. By definition of $C$, it means that

$$x \in A \times B \text{ and } x \in B \times A.$$
The first inclusion means that there exists $a_1 \in A$ and $b_1 \in B$ such that $x=(a_1,b_1)$.

The second one means that there exists $a_2 \in A$ and $b_2 \in B$ such that $x=(b_2,a_2)$.

Therefore we have $x=(a_1,b_1)=(b_2,a_2)$, which gives that $\underbrace{a_1}_{\in A}=\underbrace{b_2}_{\in B}$ and $\underbrace{b_1}_{\in B}=\underbrace{a_2}_{\in A}$.

This gives you that $a_1 \in A$ and $a_1 \in B$, which means $a_1 \in A \cap B$. For the same reason, $b_1 \in A \cap B$.

Conclusion : we have proved that $x=(a_1,b_1)$ with $a_1$ and $b_1$ in $A \cap B$, which means exactly that $x \in (A \cap B) \times (A \cap B)=D$. Since we did this for any $x$ in $C$, it proves that $C \subset D$.

I hope this help, now try to do the other way around : $D \subset C$, which will give you $D=C$.


答:
Solution using linear algebra: I suggest the notation $\mathbb{R}^{2}$ instead of $\mathbb{R}\times \mathbb{R}$, but as Arthur pointed out, there is no need to change the notation.  Now returning to the problem in your question, note that the $f$ is a linear map and the $\color{blue}{\text{matrix representation}}$  is $$[f]=\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}$$ and we can see that $$\det([f])=\det \begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}=(-1\cdot 1)-(1\cdot 1)=-1-1=-2\not=0.$$
So, $[f]$ is an $\color{blue}{\text{invertible matrix}}$ and therefore $f$ is a $\color{blue}{\text{bijective function}}$.

Title: Show that $f(x, y) = (x+y, x-y)$ is a bijection.

Body:
Show that the function $f: \mathbb{R}\times\mathbb{R}\to \mathbb{R}\times\mathbb{R}$ be defined by $f(x, y) = (x+y, x-y)$ is a bijection.

I tries this by matrix inverse method but i am not able to solve the problem.

Please someone guide me to the solution.


答:
Apply those definitions like so:

$$\begin{align}&(A\times B)\cap(B\times A)\\[1ex]=~&\{z: z\in (A\times B)\land z\in(B\times A) \}&&\textsf{definition of }\cap\\[1ex]=~&\{\langle x,y\rangle: (x\in A\land y\in B)\land(x\in B\land y\in A) \}&&\textsf{definition of }\times\\[-1ex]\vdots~~~&&&\textsf{comutation and association}\\[-1ex]\vdots~~~&&&\textsf{definition of }\cap\\[0ex]=~&(A\cap B)\times (B\cap A)&&\textsf{definition of }\times\end{align}$$


问:
Let $x$ be a $p$-letter word, $f^{n}(x)=x \implies x$ must be a 'boring' word
Let $D$ be the set of permutation of $p$-letter word, where $p$ is prime number. And let $f$ be the function $f:D \rightarrow D$ that moves the last letter to the front. For example, if $p=7$,
$$ f(abcdefgh) = habcdefg  $$

We may call a word $x$ is a 'boring' word, only if $f(x) = x$. Notice that there exist only 26 boring words: all $a$s $\underbrace{aa...a}_{p}$, all $b$s, ..., and all $z$s $\underbrace{zz...z}_{p}$.

Proof that if $f^{n}(x) = x$, where $1 \le n \le p-1$, then $x$ is a 'boring' word.

My attempt, if $n=1$ is clear. Now let $p>2$, $n=2$ and $f^{2}(x)=x$. Define $\alpha_{i}$ to be the letter at position $i$th in $x$.

$$ x = \alpha_{1}, \alpha_{2}, ..., \alpha_{p} $$
$$ \implies f(x) = \alpha_{p}, \alpha_{1} , ..., \alpha_{p-1} $$
$$ \implies f^{2}(x) = \alpha_{p-1} , \alpha_{p}, ..., \alpha_{p-2} $$

This means that
$$ \alpha_{p} = \alpha_{p-2} = \alpha_{p-4} = ... = \alpha_{1} $$
but
$$ \alpha_{1} = \alpha_{p-1} = \alpha_{p-3} = ... = \alpha_{2} $$

so all the letters must be the same, which means a 'boring' word.

Now if $n > 2$ and $f^{n}(x)=x$. Notice that we always have $p = qn + r$, where $0<r<n$ is the remainder, then
$$ \alpha_{p} = \alpha_{p-n} =  ... = \alpha_{(p -qn) = r} $$

how to continue..?

答:
Let us look at a specific example where $p=13$, $n=4$, then
$\alpha_{13} = \alpha_{9} = \alpha_{5} = \alpha_{1}$.
Then $\alpha_{1} = \alpha_{(1+13)-4} = \alpha_{(1+13)-2 \cdot 4} = ... $.

So basically
$$ \alpha_{13} = \alpha_{(13 - kn) \mod 13} $$
for non-negative integer $k$.

And since $\gcd(n,13)=1$ then $ lcm (n,13)=13n$, this means
$$ 13 \bmod 13, \: 13-(4) \bmod 13, \: 13-(2 \cdot 4) \bmod 13, ..., \: 13-(12 \cdot 4) \bmod 13 $$
are all different, and then it starts again at $13 - (13 \cdot 4) \equiv 13 \bmod 13$.

Proof: assume that there are $1< k_{1}, k_{2} < 13$ with $k_{1} \ne k_{2}$, such that
$$ 13 - (k_{1}n) \equiv 13 - (k_{2}n) \bmod 13 $$
then $(k_{2} - k_{1})n$ is divisible by $13$ which is impossible because the smallest multiple of $n=4$ that is divisible by $13$ is $13 \cdot 4$. This means that $(13-kn)$ must be congruent with all $1,2,...,12 \bmod 13$. This means $\alpha_{1}=\alpha_{2}=...=\alpha_{13}$.

Now for general prime $p > 2$, and $1 \le n \le p-1$, then we will also have $lcm(p,n)=np$, so we will end up with $\alpha_{1}=\alpha_{2}=...=\alpha_{p}$.


问:
Matrix derivative: $\frac{\partial}{\partial A_{ij}}(\mathbb{A}(\mathbb{A}^T\Sigma\mathbb{A})^{-1}\mathbb{A}^T)$
Let $\mathbb{A} = (A_{ij})$ be a rectangular, $m\times n$ real matrix, full-rank and with $m \le n$. Let $\mathbb{C}$ be a real $n\times n$ symmetric and pos-def matrix. With these assumptions, $\mathbb{A}^T\mathbb{C}\mathbb{A}$ is invertible. Finally, let $\mathbf{u}$ be an $m$-vector.

I need to compute the following derivative:

$$\frac{\partial}{\partial \mathbb{A}}(\mathbf{u}^T\mathbb{A}(\mathbb{A}^T\mathbb{C}\mathbb{A})^{-1}\mathbb{A}^T\mathbf{u})$$

where $\Sigma$ is considered a constant.

But I keep struggling with the algebra which gets really messy.

Any tricks I can use?

答: Hint: first note$$\frac{\partial}{\partial A_{ij}}(A^T\Sigma A)_{kl}=\frac{\partial}{\partial A_{ij}}(A_{mk}\Sigma_{mn}A_{nl})=\delta_{jk}(\Sigma A)_{il}+(A^T\Sigma)_{ki}\delta_{jl}.$$Define $X_{ijlm}:=\frac{\partial}{\partial A_{ij}}(A^T\Sigma A)^{-1}_{lm}$ so$$0=\frac{\partial}{\partial A_{ij}}\delta_{km}=[\delta_{jk}(\Sigma A)_{il}+(A^T\Sigma)_{ki}\delta_{jl}](A^T\Sigma A)^{-1}_{lm}+(A^T\Sigma A)_{kl}X_{ijlm}.$$Solve this to get $X_{ijlm}$, then compute $u_ku_n\frac{\partial}{\partial A_{ij}}[A_{kl}(A^T\Sigma A)^{-1}_{lm}A_{nm}]$, which is a matrix.


问:
If $p$ and $q$ are solutions of the equation $x \tan x = 1$, show the integral of $\cos^2 px$ entirely in terms of $p$
I am working through a pure maths text book out of interest. I have finished the chapter on integration and differentiation of trigonometric functions and am  doing the end of chapter questions. This is causing me a problem:

Given that $p$ and $q$ are solutions of the equation  $x \tan x = 1$,

Find an expression for $\int^1_0 \cos^2 px \ dx$ entirely in terms of $p$, not involving any trigonometric functions.

This is my working so far:

if $p$ and $q$ are solutions of $x \tan x = 1$,

$$p\frac{\sin p}{\cos p} = 1$$

$$\cos^2px = \frac{1 + \cos 2px}{2}$$

$$\int \cos^2px ~dx = \frac{1}{2}(x + \frac{\sin 2px}{2p})$$

$$= \frac{1}{2}(\frac{2px + \sin 2px}{2p})$$

I need to find

$$\left[
    \frac{1}{2}((\frac{2px + \sin 2px}{2p})
   \right]_{0}^1$$

But I cannot arrive at the answer in the book which is:

$$\frac{2 + p^2}{2(1 + p^2)}$$

答:
Write

\begin{gather*}
\tan p=\frac{1}{p}\\
\
\end{gather*}
and then use the formula

\begin{equation*}
\sin 2x=\frac{2\tan x}{1+\tan^{2} x}
\end{equation*}
This should lead you right to your answer.


问:
Definition of "essential limit sup"?
In the article of Lions https://www.jstor.org/stable/2045002?seq=1 the following notion is used:
$$L=\underset{y \to x_0}{\text{lim sup ess }} f(y),$$
for a measurable function $f:\Omega \rightarrow \mathbb{R}$ but I have no idea what it means.

What is the precise definition of this $L$ with quantifiers ?

How to pronounce it, is it "essential supremum limit" ?

Where to find it in classical textbooks ?

I have also seen "$\text{ess lim sup }$" in another article, is it the same object ?

答:
$L$ is defined by the following properties:

a) For every $\epsilon >0$ ther exist $\delta >0$ such that $|y-x_0| <\delta$ implies $f(x) \leq L+\epsilon$ almost everywhere.

b) No smaller number has property a).

'ess lim  sup' is same as 'lim sup ess'.

You pronounce it as 'essential lim sup' (or eseential limit superior).

Title: Definition of "essential limit sup"?


问:
Theorems that rely on the fact that the atlas of a smooth manifold is maximal/complete
When I started to learn Differential geometry, I wondered why a smooth manifold is defined to have an atlas that is not only smooth, but also maximal/complete.

Do you know of any theorems that rely on the fact that the atlas of a smooth manifold is maximal?

Edit: I am aware of the fact that a smooth atlas is contained in exactly one maximal smooth atlas, but I would be interested in an answer nevertheless.

答:
The maximality condition is included so that, in effect, anything that should be a coordinate chart is actually a coordinate chart. Here is an excerpt from Loring Tu's An Introduction to Manifolds which formalizes this.

Proposition 6.11. Let $U$ be an open subset of a manifold $M$ of dimension $n$. If $F \colon U \to F(U) \subset \mathbb{R}^n$ is a diffeomorphism onto an open subset of $\mathbb{R}^n$, then $(U, F)$ is a chart in the differentiable structure of $M$.

The proof is a straightforward application of the maximality assumption on the differentiable structure of $F$; for completeness, I have included it. (Proposition 6.10, referenced therein, just says that the coordinate charts are diffeomorphisms.)

Proof. For any chart $(U_\alpha, \phi_\alpha)$ in the maximal atlas of $M$, both $\phi_\alpha$ and $\phi_\alpha^{-1}$ are $C^\infty$ by Proposition 6.10. As composites of $C^\infty$ maps, both $F \circ \phi_\alpha^{-1}$ and $\phi_\alpha \circ F^{-1}$ are $C^\infty$. Hence, $(U, F)$ is compatible with the maximal atlas. By the maximality of the atlas, the chart $(U, F)$ is in the atlas.

Here is an example of Proposition 6.11 at work, again from Tu. You can think of it as a version of the inverse function theorem, if you'd like.

Corollary 6.27. Let $N$ be a manifold of dimension $n$. A set of $n$ smooth functions $F^1, \dots, F^n$ defined on a coordinate neighborhood $(U, x^1, \dots, x^n)$ of a point $p \in N$ forms a coordinate system about $p$ if and only if the Jacobian determinant $\det [\partial F^i / \partial x^j(p)]$ is nonzero.

The proof requires (a version of) the inverse function theorem on manifolds, so in order not to stray too far off topic, I have omitted it.

Another example of this in action is in nearly any theorem that has to do with the existence of local coordinates near a point. For example, the constant rank theorem, which states that under certain conditions on a function at a point, you can find coordinate charts near that point and its image in which the function has a particularly nice "normal form." (See Tu, Theorem 11.1, or John M. Lee's Introduction to Smooth Manifolds, Theorem 4.12.) Or, perhaps, in the theorem that any set of non-zero commuting coordinate vector fields are actually the coordinate vector fields for some chart. (See Lee, Theorem 9.46.)


Title: If $p$ and $q$ are solutions of the equation $x \tan x = 1$, show the integral of $\cos^2 px$ entirely in terms of $p$

Body:
I am working through a pure maths text book out of interest. I have finished the chapter on integration and differentiation of trigonometric functions and am  doing the end of chapter questions. This is causing me a problem:

Given that $p$ and $q$ are solutions of the equation  $x \tan x = 1$,

Find an expression for $\int^1_0 \cos^2 px \ dx$ entirely in terms of $p$, not involving any trigonometric functions.

This is my working so far:

if $p$ and $q$ are solutions of $x \tan x = 1$,

$$p\frac{\sin p}{\cos p} = 1$$

$$\cos^2px = \frac{1 + \cos 2px}{2}$$

$$\int \cos^2px ~dx = \frac{1}{2}(x + \frac{\sin 2px}{2p})$$

$$= \frac{1}{2}(\frac{2px + \sin 2px}{2p})$$

I need to find

$$\left[
    \frac{1}{2}((\frac{2px + \sin 2px}{2p})
   \right]_{0}^1$$

But I cannot arrive at the answer in the book which is:

$$\frac{2 + p^2}{2(1 + p^2)}$$

Body:
$$I=\int_0^1 \cos ^2(p x) \, dx=\frac{\sin (2 p)}{4 p}+\frac{1}{2}$$
it is given that $\tan p=\frac{1}{p}$
From the formula
$$\sin \alpha=\frac{2t}{1+t^2};\;t=\tan\frac{\alpha}{2}$$
we get
$$\sin 2p=\frac{2\tan p}{1+\tan^2 p}=\frac{2\cdot\frac{1}{p}}{1+\left(\frac{1}{p}\right)^2}=\frac{2 p}{p^2+1}$$
and finally
$$I=\frac{\frac{2 p}{p^2+1}}{4p}+\frac{1}{2}=\frac{p^2+2}{2 p^2+2}$$


问:
What are the steps involved to get the following equation?
I'm just a bit confused on the math process to get from the left equation to the right
$$\left(\frac{s}{6}\frac{1-s^6}{1-s}\right)^2 = \frac{s^2}{36} (s^5 + s^4 +s^3+s^2+s+1)^2 $$

答:
The modified statement is now correct.

$$\Bigg(\frac{s(1-s^6)}{6(1-s)}\Bigg)^2=\frac{s^2}{36}[s^5+s^4+s^3+s^2+s+1]^2$$

To understand the algebraic passages let's focus on

$$(1-s^6)=(1+s^3)\underbrace{(1-s^3)}_{(1-s)(1+s+s^2)}$$

... simplify and get the correct result


问:
Let $a$, $b$, $c$ be positive real numbers. Prove the following inequality.
Let $a$, $b$, $c$ be positive real numbers. Prove the following inequality;
$$
(a+b+c)\Bigl(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Bigr)
\ge 9+3\sqrt[\scriptstyle 3]{\cfrac{(a-b)^2(b-c)^2(c-a)^2}{a^2b^2c^2}}
$$

The root is a cube root in case it isn't clear.

I'm trying to use the AM-GM, Cauchy-Schwarz, triangle and other inequalities.

答:
Using the AM-GM inequality, we have
$$(a+b+c) \cdot \Bigl(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Bigr) - 9  = \left(\frac{a}{b}+ \frac{b}{a}-2\right)+\left(\frac{b}{c}+ \frac{c}{b}-2\right)+\left(\frac{c}{a}+ \frac{a}{c}-2\right)$$
$$= \frac{(a-b)^2}{ab}+\frac{(b-c)^2}{bc}+\frac{(c-a)^2}{ac}\geqslant 3\sqrt[\scriptstyle 3]{\cfrac{(a-b)^2(b-c)^2(c-a)^2}{a^2b^2c^2}}.$$


问:
Calculate the length of a polar curve

Calculate the length of the polar curve $$\theta (r)=\frac{1}{2}\left( r+\frac{1}{r}\right)$$ from r = 1 to r = 3.

I understand mostly how to get the length of a polar curve by:

$$\int_{a}^{b} \sqrt[]{(f(\theta ))^{2}+(f'(\theta ))^{2}} \ d\theta $$

But in this exercise i dont get how to do it. Maybe i need to write the function $\theta (r)$ in terms of $\theta$

Any ideas or hints? Thanks

答: Building on my comment, since $rd\theta/dr=\tfrac12(r-\tfrac1r)$, $\int ds=\int_1^3\tfrac12(r+\tfrac1r)dr=2+\tfrac12\ln3$.

Title: Calculate the length of a polar curve


问:
conditioning of the monomial basis
It is well-known that the Vandermonde matrix is ill-conditioned when all nodes, i.e., the points that generate it, are real. I have read the following on a paper:

The ill-
conditioning of the Vandermonde basis has an elementary explanation
in cases where the points xj are unequal in size. If $|x_j |$ varies with j, then the powers
$|(x_j)^k|$ vary exponentially. This means that function information associated with smaller
values of $x_j$ will only be resolvable through exponentially large expansion coefficients,
and accuracy will quickly be lost in floating point arithmetic

I have not understood the explanation because he terms "function information" and "expansion coefficients" are not defined in this context on the paper. Could somebody provide an explanation on why the Vandermonde matrix is ill-conditioned?.

答:
The Vandermonde matrices are not necessarily ill-conditioned when the nodes are real.

Example: In the case of $n=2$ nodes, the Vandermonde matrix is
$$A = \begin{bmatrix} 1 & x_0 \\ 1 & x_1 \end{bmatrix}.$$
If $x_0 = -1$ and $x_1 = 1$, then columns are orthogonal and the 2-norm condition number of $A$ is given by
$$\kappa_2(A) = \|A\|_2 \|A^{-1}\|_2 = 1.$$
This is the smallest possible value.

In order to understand exactly why the Vandermonde matrices can be arbitrarily ill-conditioned we investigate the singular values. Let $$\sigma_1 \ge \sigma_2 \dots \ge \sigma_n \ge 0$$
denote the singular values of $A$. Then $\{\sigma_i^2\}_{i=1}^n$ are the eigenvalues of $A^TA$ and $$\kappa_2(A) = \frac{\sigma_1}{\sigma_n}.$$ We can establish simple bounds on the singular values as follows. For a general symmetric matrix $B = [b_{ij}]$, the smallest eigenvalue satisfies
$$ \lambda_{\min}(B) = \min \{ x^T B x \: : \: \|x\|_2 = 1 \} \leq b_{ii} $$
for all $i$ and the largest eigenvalue satisfies
$$ \lambda_{\max}(B) = \max \{ x^T B x \: : \: \|x\|_2 = 1 \} \ge b_{ii} $$
for all $i$.

Example: In the case of $n=2$ nodes, we have
$$ A^T A = \begin{bmatrix} 2 & x_0+x_1 \\ x_0 + x_1 & x_0^2 + x_1^2 \end{bmatrix}.$$
It follows that $$ \sigma_2^2 \leq 2, \quad x_0^2 + x_1^2 \leq \sigma_1^2$$
We conclude that $$\kappa_2(A) \ge \sqrt{ \frac{x_0^2 + x_1^2}{2} }.$$
The case of $x_1 = -x_0 = -x$ is particular clear. In this case, the relevant linear system is
$$ \begin{bmatrix} 1 & x \\ 1 & -x \end{bmatrix} \begin{bmatrix} c_0 \\ c_1 \end{bmatrix} = \begin{bmatrix} y_0 \\ y_1 \end{bmatrix}. $$
The solution is
$$ \begin{bmatrix} c_0 \\ c_1 \end{bmatrix} = \frac{1}{2x} \begin{bmatrix} x & x  \\ 1 & -1\end{bmatrix} \begin{bmatrix} y_0 \\ y_1 \end{bmatrix} = \begin{bmatrix} \frac{y_0+y_1}{2} \\ \frac{y_1-y_0}{2x} \end{bmatrix}.$$
Here we note that while $c_0 = \frac{y_0+y_1}{2}$ is insensitive to small relative changes in the right-hand side when $y_1$ and $y_2$ are nearly equal, the divided difference
$$c_1 = \frac{y_1 - y_0}{2x}$$ is extremely sensitive to small relative changes in the right-hand side when $y_1$ and $y_2$ are nearly equal. Specifically, a tiny relative change can flip the sign of $c_1$. The situation is reversed when $y_1 + y_2 \approx 0$. In both cases, we see that a small componentwise relative change in the right hand-side $y$, can cause a large componentwise relative change in the solution $c$.

The analysis presented here extends to the general case of $n>2$. Here we have $$\sigma_n^2 \leq n, \quad \sum_{j=0}^{n-1} (x_j^{n-1})^2 \leq \sigma_1^2$$
and
$$ \kappa_2(A) \ge \sqrt{\frac{\sum_{j=0}^{n-1} (x_j^{n-1})^2}{n }}.$$

Title: conditioning of the monomial basis


问:
Specific proof: $ span\{a_1,a_2\} \cap span\{b_1,b_2,b_3\} \subseteq \{ \vec0 \} $
Prove that:
$ span\{a_1,a_2\} \cap span\{b_1,b_2,b_3\} \subseteq \{ \vec0 \} $
where:

$ a_1 =\left[\begin{array}{cc} 1 \\ 1 \\ 1 \\ 3 \end{array}\right] $
, $ a_2 =\left[\begin{array}{cc} 1 \\ 2 \\ 1 \\ 3 \end{array}\right] $
;
$ b_1 =\left[\begin{array}{cc} 1 \\ 1 \\ 2 \\ 2 \end{array}\right] $
, $ b_2 =\left[\begin{array}{cc} 1 \\ 1 \\ 1 \\ 1 \end{array}\right] $
, $ b_3 =\left[\begin{array}{cc} 2 \\ 2 \\ 3 \\ 3 \end{array}\right] $

My attempt:

Let $ v$ be arbitrary. Suppose $ v \in span\{a_1,a_2\} \cap span\{b_1,b_2,b_3\} $ be arbitrary.
Therefore there exists scalars $ \alpha_1,\alpha_2,\beta_1,\beta_2,\beta_3 \in  \mathbb{R} $, such that

v = $ \alpha_1 \left[\begin{array}{cc} 1 \\ 1 \\ 1 \\ 3 \end{array}\right] $+ $ \alpha_2 \left[\begin{array}{cc} 1 \\ 2 \\ 1 \\ 3 \end{array}\right] $
and     v= $  \beta_1 \left[\begin{array}{cc} 1 \\ 1 \\ 2 \\ 2 \end{array}\right] $+ $
 \beta_2 \left[\begin{array}{cc} 1 \\ 1 \\ 1 \\ 1 \end{array}\right] $+ $ \beta_3 \left[\begin{array}{cc} 2 \\ 2 \\ 3 \\ 3 \end{array}\right] $

eventually we get the following matrix which describes the above system of equations:

$ \begin{bmatrix}
1 & 1 & -1 &  -1 & -2  | 0 \\
1 & 2 & -1 &  -1 & -2  | 0 \\
1 & 1 & -2 &  -1 & -3  | 0 \\
3 & 3 & -2 &  -1 & -3  | 0 \\  \end{bmatrix} $
but there are four equations and five variables, so there is one free variable and thus $ v $ is not necessarily zero.  [ Stopped here ]

Now I am stuck because the statement to be proven is a true one but in my solution I saw that $ v $ isn't necessarily zero, so it looks like there is a contradiction ( if so, I'd rephrase the proof to disproof ) but I'm not convinced.

The solution reduces $ span\{b_1,b_2,b_3\}  $ to $ span\{b_1,b_2\}   $, that is because  $ b_3 = b_1 + b_2 $ and I understood it, but I want to prove the statement with my way of not reducing the span a-priori , and essentially how do I deal with the apparent contradiction that v is not always zero vector?

答:
For your system of equations, there are non-zero solutions like

$$\begin{bmatrix}\alpha_1\\\alpha_2\\\beta_1\\\beta_2\\\beta_3\end{bmatrix}
= k\begin{bmatrix}0\\0\\1\\1\\-1\end{bmatrix}$$

But the $\alpha$s are still zero, which means the intersection of the spans is still $\{0a_1+0a_2\}= \left\{\vec 0\right\}$ when calculated from $a_1, a_2$.

Or when calculated from vectors $b_1, b_2, b_3$, the required intersection is ($k\in\mathbb R$)

$$\{kb_1+kb_2-kb_3\} = \left\{k\left(\begin{bmatrix}1\\1\\2\\2\end{bmatrix} + \begin{bmatrix}1\\1\\1\\1\end{bmatrix} - \begin{bmatrix}2\\2\\3\\3\end{bmatrix}\right)\right\} = \left\{k\vec0\right\} = \left\{\vec 0\right\}.$$


答: Your method is perfectly fine, just continue. Your solution will give a non-trivial combination of $(\alpha_1,\alpha_2,\beta_1,\beta_2,\beta_3)$, but if you plug it back into $v$, you will still get $v = 0$.

Title: Specific proof: $ span\{a_1,a_2\} \cap span\{b_1,b_2,b_3\} \subseteq \{ \vec0 \} $

Body:
Prove that:
$ span\{a_1,a_2\} \cap span\{b_1,b_2,b_3\} \subseteq \{ \vec0 \} $
where:

$ a_1 =\left[\begin{array}{cc} 1 \\ 1 \\ 1 \\ 3 \end{array}\right] $
, $ a_2 =\left[\begin{array}{cc} 1 \\ 2 \\ 1 \\ 3 \end{array}\right] $
;
$ b_1 =\left[\begin{array}{cc} 1 \\ 1 \\ 2 \\ 2 \end{array}\right] $
, $ b_2 =\left[\begin{array}{cc} 1 \\ 1 \\ 1 \\ 1 \end{array}\right] $
, $ b_3 =\left[\begin{array}{cc} 2 \\ 2 \\ 3 \\ 3 \end{array}\right] $

My attempt:

Let $ v$ be arbitrary. Suppose $ v \in span\{a_1,a_2\} \cap span\{b_1,b_2,b_3\} $ be arbitrary.
Therefore there exists scalars $ \alpha_1,\alpha_2,\beta_1,\beta_2,\beta_3 \in  \mathbb{R} $, such that

v = $ \alpha_1 \left[\begin{array}{cc} 1 \\ 1 \\ 1 \\ 3 \end{array}\right] $+ $ \alpha_2 \left[\begin{array}{cc} 1 \\ 2 \\ 1 \\ 3 \end{array}\right] $
and     v= $  \beta_1 \left[\begin{array}{cc} 1 \\ 1 \\ 2 \\ 2 \end{array}\right] $+ $
 \beta_2 \left[\begin{array}{cc} 1 \\ 1 \\ 1 \\ 1 \end{array}\right] $+ $ \beta_3 \left[\begin{array}{cc} 2 \\ 2 \\ 3 \\ 3 \end{array}\right] $

eventually we get the following matrix which describes the above system of equations:

$ \begin{bmatrix}
1 & 1 & -1 &  -1 & -2  | 0 \\
1 & 2 & -1 &  -1 & -2  | 0 \\
1 & 1 & -2 &  -1 & -3  | 0 \\
3 & 3 & -2 &  -1 & -3  | 0 \\  \end{bmatrix} $
but there are four equations and five variables, so there is one free variable and thus $ v $ is not necessarily zero.  [ Stopped here ]

Now I am stuck because the statement to be proven is a true one but in my solution I saw that $ v $ isn't necessarily zero, so it looks like there is a contradiction ( if so, I'd rephrase the proof to disproof ) but I'm not convinced.

The solution reduces $ span\{b_1,b_2,b_3\}  $ to $ span\{b_1,b_2\}   $, that is because  $ b_3 = b_1 + b_2 $ and I understood it, but I want to prove the statement with my way of not reducing the span a-priori , and essentially how do I deal with the apparent contradiction that v is not always zero vector?


问:
Expected value of sum of two dependent Binomial variables
Question: Assume we toss a coin in two trials. In the first trial we toss it n times. In the second trial we toss it as many as the number of tails observed in the first trial.

Calculate the expectation of total number of tails in both trails:

Solution: This is what I have tried but can't find any closed form for the answer when n is not given. I am not sure if I have taken the right approach:

$T_1$ = Number  of  tails in the 1st trial

$T_2$ =  Number of tails in the 2nd trial

T1 ~ Binomial(n,p)

T2 ~ Binomial(T1,p)

$E(T1 + T2) =
E(T1) + E(T2) = \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n p+\sum_{i=0}^{n} E\left(T_{2} \mid T_{1}=i\right) P\left(T_{1}=i\right) =\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n p+\sum_{i=0}^{n} i p \times\left(\begin{array}{l}n \\ i\end{array}\right) p^{i}(1-p)^{n-i}$

I couldn't find any closed form for the answer. Any advice on how to approach this?

答:
To calculate $E(T_2)$, consider using the law of total expectation.

$$E(X) = E\left[E(X|Y) \right]$$

Which in your case would give

\begin{align*}
E[T_2] &= E\left[E(T_2|T_1) \right] \\
&=E\left[p \cdot T_1 \right] = p E\left[T_1 \right] \\
&= np^2
\end{align*}


问:
Find the intersection point of a logarithmic and a quadratic functions.
Situation

I have following equation:

$|\log_{10}x| + (x-1)^2 = a$ or I think it will be better to write it in following form: $|\log_{10}x| = - (x-1)^2 + a$. I think with this, we can draw the parabola way better.

The question

Find out the the x when a = 82 and $x \in Z.$

What I've tried

Well... I thought maybe we can solve like a usual $|f(x)|=g(x)$ and find out what happens:

$|\log_{10}x|=-(x-1)^2+82$

$
  \begin{cases}
  -(x-1)^2+82 \geqslant 0 \\
  \log_{10}x = -(x-1)^2+82 \quad OR \quad \log_{10}x=(x-1)^2-82
  \end{cases}
 $

After simplifying, I found out that I can't do really much here.

Second option was to draw the graph. After doing it manually by hand(which wasn't that hard) I tried plot it with a program (called Desmos) to see if I was missing something.

What I have found out is that

when my a is 0, then I have only one x.

when my a is greater than 0, then I have 2 xs.

when my a is less than 0, then I have no solution.

But I wasn't able to find the intersection point.

How do you solve this kind of problems?

答: The RHS is an integer, so the LHS must also be an integer, i.e. $x$ is a power of ten. Eyeballing shows that $x=10$ works.


答: Your mistake lies in deducing from the fact that the system has necessarily a non-zero solution that there is a non-zero vector $v$. In fact,, and since $b_3=b_1+b_2$, if you take $\alpha_1=\alpha_2=0$, $\beta_1=\beta_2=1$, and $\beta_3=-1$, then $(\alpha_1,\alpha_2,\beta_1,\beta_2,\beta_3)$ is a solution of your system. But the $v$ that you et then is$$v=\alpha_1a_1+\alpha_2a_2=\beta_1b_1+\beta_2b_2+\beta_3b_3=0.$$


问:
Boundedness of a bilinear operator
Let $X, Y$ and $Z$ be Banach spaces, and $T: X \oplus Y \to Z$ be a bilinear operator, where $X \oplus Y$ denotes the direct sum of $X$ and $Y$.

Is the following true?

If $\| T(x,y) \|_Z \leq C_x\|y\|_Y$ and $\|T(x,y)\| \leq C_y \|x\|_X$, then $T$ is bounded, i.e. $\|T(x,y)\|_Z \leq C \|x\|_X \|y\|_Y$

答: For unit $y$, let $T_y(x):=T(x,y)$. Each $T_y$ is a bounded linear operator since $$\|T_y(x)\|\le C_y\|x\|$$ Also, it is given that for each $x$, $$\|T_y(x)\|\le C_x$$ By the Banach-Steinhaus theorem, it follows that $\|T_y\|\le C$, independent of $y$. But this means that for any $y$, not necessarily unit, $$\|T(x,y)\|=\|T_{y/\|y\|}(x)\|\|y\|\le C\|x\|\|y\|$$

Title: Boundedness of a bilinear operator


答:
First observe that $f^p(x) = x$ for all $x\in D$, i.e. $f^p = \mathrm{id}_D$. In particular, $f$ is invertible. Furthermore, notice that if $f^n(x) = x$ for some $x$, then $f^{kn}(x) = x$, for any integer $k$.

Now, let $1\leq n < p$ and $x$ be such that $f^n(x) = x$. Since $p$ is prime, $n$ and $p$ are relatively prime. By the Bézout's identity, there exist integers $a$ and $b$ such that $an+bp = 1$. Finally,
$$f(x) = f^{an+bp}(x) = (f^{an}\circ f^{bp})(x) = x.$$

Thus, $x$ is boring.

Title: Let $x$ be a $p$-letter word, $f^{n}(x)=x \implies x$ must be a 'boring' word

Body:
Let $D$ be the set of permutation of $p$-letter word, where $p$ is prime number. And let $f$ be the function $f:D \rightarrow D$ that moves the last letter to the front. For example, if $p=7$,
$$ f(abcdefgh) = habcdefg  $$

We may call a word $x$ is a 'boring' word, only if $f(x) = x$. Notice that there exist only 26 boring words: all $a$s $\underbrace{aa...a}_{p}$, all $b$s, ..., and all $z$s $\underbrace{zz...z}_{p}$.

Proof that if $f^{n}(x) = x$, where $1 \le n \le p-1$, then $x$ is a 'boring' word.

My attempt, if $n=1$ is clear. Now let $p>2$, $n=2$ and $f^{2}(x)=x$. Define $\alpha_{i}$ to be the letter at position $i$th in $x$.

$$ x = \alpha_{1}, \alpha_{2}, ..., \alpha_{p} $$
$$ \implies f(x) = \alpha_{p}, \alpha_{1} , ..., \alpha_{p-1} $$
$$ \implies f^{2}(x) = \alpha_{p-1} , \alpha_{p}, ..., \alpha_{p-2} $$

This means that
$$ \alpha_{p} = \alpha_{p-2} = \alpha_{p-4} = ... = \alpha_{1} $$
but
$$ \alpha_{1} = \alpha_{p-1} = \alpha_{p-3} = ... = \alpha_{2} $$

so all the letters must be the same, which means a 'boring' word.

Now if $n > 2$ and $f^{n}(x)=x$. Notice that we always have $p = qn + r$, where $0<r<n$ is the remainder, then
$$ \alpha_{p} = \alpha_{p-n} =  ... = \alpha_{(p -qn) = r} $$

how to continue..?


问:
What is the name of a game that cannot be won until it is over?
Consider the following game:

The game is to keep a friend's secret. If you ever tell the secret, you lose. As long as you don't you are winning. Clearly, it's a game that takes a lifetime to win.

Another example would be honor. Let's say, for our purposes, once you lose your honor you can't get it back. Call it a game. Then the only way to win is to never lose your honor.

Both of these games are highly unfavorable for the player. Because they are games that can never be won, until they are over. Is there a mathematical name for such a game?

答:
I would argue that, mathematically speaking, there is only one game of this form (or perhaps one distinct game for each number of players). It's essentially the color/country game, in which players take it in turn to either lose or not lose.

The differences between different forms of this game are significant, but non-mathematical; in general, there will be some reason why choosing not to lose is difficult. For example:

The games given in the question aren't really independent from the rest of your life. Presumably there's some reason why you might be tempted to spill a secret or give up your honor. So the right way to analyze them involves having some model of morality and temptation.

There are a number of video games of this form (e.g., Tetris), where "choosing not to lose" can only happen by performing progressively more difficult physical actions. So the right way to analyze this kind of game involves knowing something about human physiology.

In the color/country game itself, choosing not to lose is an exercise in willpower (or willingness not to declare that the game is stupid). So analyzing it boils down to knowing something about the attention span of the players.

In any case, you need some non-mathematical (and probably unmathematizable) knowledge to say anything interesting about the game. So mathematicians are unlikely to have come up with very much terminology for talking about them...


问:
diameter and radius in graph and disconnected graph
We have an undirected Graph $G$.

Def $1$: The diameter of a graph is defined as the maximum of shortest paths between two vertices of $G$.

Def $2$: we define $L(S)$ as maximum length of shortest paths from $S$ to other vertices.

Def $3$: we define the radius of a graph as minimum value of $L(S)$ between all vertices in $G$.

If $\mathrm{diam}$ and $\mathrm{rad}$ be diameter and radius of graph which of the following was correct always (choose the best option):

$\mathrm{rad} \geq \frac{\mathrm{diam}}{2}$

$\mathrm{rad} \leq \mathrm{diam}$

Answer: 1 is the best option.

My question is why the author choose the 1 while 2 is also
true. Then my idea is because in the question there is no assumption
for connected graph so on disconnected graph option 1 can handle
$\infty$ but option 2 cannot.

I think here by using best option words it means there is a case that we can support by one option and cannot support by another ones.

I know both of them is upper and lower bound but here there is a trick
by the words "best option". is my conclusion about supporting
disconnected graph is correct and reason for choosing $(1)$ is that support connect and disconnected graph at same time? if no is
there any reasonable logic here?

答:
Both are always true (with the convention that for a disconnected graph, radius and diameter are infinite).

Which is better really depends on context: what do you need the bound for? If you need a lower bound then the first is better, and if you need an upper bound the second is.

Without context, there's no reasonable way to say one bound is better than the other.

One thing which could apply, but doesn't here, is whether the bounds are best possible. For example, if the second said $\mathrm{rad}\leq 2\mathrm{diam}$, then that would be a worse bound, because although true it is weaker than it needs to be.

However, in this case both bounds are best possible (so equally good by this measure), since for any $k\geq 1$ there exist graphs $G,H$ with $\mathrm{rad}(G)=k,\mathrm{diam}(G)=2k$ and $\mathrm{rad}(H)=k,\mathrm{diam}(H)=k$ (e.g. $G$ is the path of length $2k$ and $H$ is the cycle of length $2k$).


问:
Why the operation hierarchy works?
I don't know if this is a silly question, but they never teached me this at school, i will have to trust my teacher and not doubt about this affirmation, but why we are completly sure that performing simple arithmetic operations in an specific order gives a correct answer?

For example:

2 + 4 × 2

Why we completly affirm that this is 10 and not 12?

Why this hierarchy is difined like this?

Why this works?

答:
This is a mere convention ... so it could've historically ended up differently, but it hasn't, we are where we are, it's been accepted by everyone, and so everyone needs to learn it.

If you write something like:

$$2+4\times 2$$

(or, let's make it worse:

$$2+2\times 2-2-(2^{2+2^2}+2):2+2+2\times 2:2$$

) ... what does this writing even mean? The convention is about deciding what it means, i.e. which operations will be done first, and which will follow. Over centuries, people settled with the following:

Brackets (Parentheses): Do those first. Whatever is in brackets needs to be calculated already (as a separate, standalone, expression), before you go forward.

Indices (Exponents): Calculate those next, right-to-left. If you have something like $2^2$ do it here. If you have something like $2^{3^4}$ you do $3^4(=81)$ first, then you do $2^{81}$

Division/Multiplication: Do both, from left to right, in the order they appear.

Addition/Subtraction: Do both of them, again, from left to right, in the order they appear.

Those rules are usually remembered using a mnemonic BIDMAS or PEDMAS (or PEMDAS).

The flipside is, if you want the operations in your expression to be done in a different order than prescribed, you then put brackets/parentheses to modify the order.

In your question: $2+4\times 2$, there are no parentheses and no indices, but we do the multiplication first ($4\times 2 = 8$), and then we do addition ($2+8=10$). If you want it to be calculated differently, i.e. to first add $2+4$, and then multiply by $2$, you put $2+4$ in brackets, i.e. the expression becomes: $(2+4)\times 2$.

Hope this helps.

Title: Why the operation hierarchy works?