问:
Cyclic Automorphism group

Show that no group can have its automorphism group cyclic of odd order.

I have shown it only if $G$ is cyclic, but I could not do that if $G$ is not cyclic. Can you help?

答:
Here are some hints:

Let $G$ be a group with a cyclic and odd group of automorphisms.

Since $G/Z$ (Z being the center) is a subgroup of $\operatorname{Aut} G$, it will be cyclic. Deduce that $G = Z$, i.e. $G$ is abelian.

Since $G$ is abelian, what can you say about $x\mapsto x^{-1}$ ? Deduce that $G \cong \bigoplus_{i} \mathbb Z/2\mathbb Z$, i.e. $G$ is an elementary abelian $2$-group.

Now you know a lot about $G$, so you may try to find automorphisms of $G$ that will contradict that $\operatorname{Aut} G$ is cyclic.

I can give more hints if you tell where you're stuck.

-- editted:  for step 3: For instance: If there are at least 3 factors involved in the direct product $G \cong \bigoplus_i (\mathbb Z/2\mathbb Z)$ then permuting these factors gives rise to an automorphism (for instance $(a,b,c,...)\mapsto (b,a,c,...)$). This implies $S_3$ appears as subgroup of the automorphism-group so it will surely not be cyclic. If there are $2$ factors $G\cong (\mathbb Z/2\mathbb Z)^2$ and it's easy to see that any permutation of the three involutions of this group is an automorphism.

I think there is a more beautiful way to derive the contradiction but I don't see it right now.

-- editted (much later): I just thought of the more beautiful way: If the direct sum has at least two terms, consider the automorphism that switches these terms $(a,b,c,\dots)\mapsto (b,a,c,\dots)$. This is an automorphism of order 2, a contradiction.

Title: Cyclic Automorphism group


问:
Show that $X :=$ {$0, 1$}$^{\mathbb R}$ is not first-countable.
Show that $X :=$ {$0, 1$}$^{\mathbb R}$ is not first-countable.

According to Munkres Topology:
Definition: A space $X$ is said to have a countable basis at $x$ if there is a countable collection $\beta$.
A space that has a countable basis at each of its points is said to be first countable.

I can also proceed indirectly by showing that there exists a real-valued function on some subspace of $[0,1]^{\mathbb R}$ that is sequentially continuous but not continuous. It is the case of $F(f) = \int_0^1 f(t) dt$ where $f$ is measurable. But I don't know how to prove that there exists such a function. Any help please?

答:
You can show that there is no countable basis at $0$ in the following way.

First, notice that if there is a countable basis, this implies that any basis has a countable subset which is also a basis, to it is enough to find a basis at $0$ whose countable subsets are all not bases.

Consider, then, for each finite $A\subseteq \mathbf R$ the set $U_A:=\{x\in X\mid (\forall a\in A)\, x(a)=0 \}$. This is clearly a basis at $0$.

The conclusion follows from the observation that if $\{A_n\mid n\in \mathbf N\}$ is a countable family of finite subsets of $\mathbf R$, then $\mathbf R\setminus \bigcup_n A_n$ is nonempty, so for any $r\in \mathbf R\setminus \bigcup_n A_n$, the set $U_{\{r\}}$ is a neighbourhood of $0$ which does not contain $\bigcap_n U_{A_n}$, so it certainly does not contain any of the sets $U_{A_n}$.
Hence, $\{U_{A_n}\mid n\}$ is not a basis.

ContentLicense: CC BY-SA 4.0


问:
When can you not use the second derivative test?
According to google, it's when $f'(x)$ doesn't exist. I was given the following functions:

\begin{align}
y & = -\tfrac 1 3x^3 -4x + 16x \\[6pt]
y & = xe^{-x/4} \\[6pt]
y & = -\cos(x-4) \\[6pt]
y & = -x^2 + 8x
\end{align}

I was able to automatically rule out the first and last because they're polynomials. Then I was stuck with the middle two. But they both have first derivatives. AND second derivatives. So, is there any other way to determine when I can't use the second derivative test?

答:
"I was able to automatically rule out the first and last because they're polynomials."

What does that mean? What question were you trying to answer?

What is usually called "the second derivative test" is used for identifying local extreme values – maxima and minima. Many polynomials do have local maxima and local minima. In particular, your fourth example has a local (and in this case, also global) maximum at a point where the first derivative is $0$ and the second derivative is negative.

And the same is true of your second example.

The third example has a global maximum that is attained at infinitely many points in the domain, all of them points where the first derivative is $0$ and the second is negative.

In the first example, I wonder whether there's a typo.


问:
Prove prime number p divides 1+...+n^{p-2}
I am working on a problem in number theory:

Problem: Let $p>2$ be prime. Let $n$ be an integer with $\gcd(n,p)=1$ and $n\not\equiv1\pmod{p}$. Prove that $p\mid(1+n+n^2+n^3+\dots+n^{p-2})$.

Right now, I am trying to solve this problem with Fermat's Little Theorem. The Theorem tells me $n^p\equiv n\pmod{p}$. Then I have $n(n^{p-1}-1)\equiv0\pmod p$, so I know $p\mid n(n^{p-1}-1)$. Because $p\not\mid n$, I know then that $p\mid(n^{p-1}-1)$. From here, I am stuck and cannot see how to proceed. Thank you in advance for assistance.

答:
Since $n\not\equiv1\pmod{p} \implies n \neq pk+1 \implies n-1 \neq pk \implies p \nmid n-1$.

As Greg Martin mentions, consider $(1+n+n^2+n^3+\dots+n^{p-2})(1-n)=1-n^{p-1}$.

By Fermat's Little Theorem, for any prime $p$ we have $n^{p-1} \equiv 1 \pmod p
\implies p\mid 1-n^{p-1} \implies p \mid (1+n+n^2+n^3+\dots+n^{p-2})(1-n).$

But $p \nmid n-1$, so $p\nmid 1-n$ and $p \mid(1+n+n^2+n^3+\dots+n^{p-2})$.


问:
Inverse function of $x+n^x$
I need to find the inverse function of $f(x)=x+n^x$, where $n$ is a variable in the range $(0, 1]$. I understand that the result would not  be a function because of the vertical-line test, so it would need  to be broken up.

By swapping $x$ for $y$ we get $x=y+n^y$, but I have no idea where to go from here to isolate $y$ on one side of the equation.

答:
Equations of the form $a^x+cx=b$ are in general solvable using the Lambert W function, the inverse of $xe^x$. In your case, if $y=x+n^x$ then
$$x=y-\frac{W(n^y\ln n)}{\ln n}$$
Choosing different branches of $W$ gives all possible values of $x$ given $y$.

Title: Inverse function of $x+n^x$


答:
If you consider the function $f(x)=x+n^x$ from $\mathbb{R} \to \mathbb{R}$ notice that since $f(x)=n^x+x$ is not surgetive for some  $n\in (0,1]$, the function don´t should be inverse function for all $n$.

Is clear that for $n=1$ the function have inverse, but for arbitrary $n$ you can´t afirm these.

Title: Inverse function of $x+n^x$

Body:
I need to find the inverse function of $f(x)=x+n^x$, where $n$ is a variable in the range $(0, 1]$. I understand that the result would not  be a function because of the vertical-line test, so it would need  to be broken up.

By swapping $x$ for $y$ we get $x=y+n^y$, but I have no idea where to go from here to isolate $y$ on one side of the equation.


问:
Topology on $\mathbb{K}\times E$
This is probably a really basic question, but here it goes. Let $E$ be a vector space over $\mathbb{K}$ and let $\tau$ be a topology on $E$. Then $E$ is called a topological vector space if the sum and scalar multiplications are continuous, i.e. if the maps $s: E\times E \to E$ and $m: \mathbb{K}\times E \to E$ defined by $s(x,y) := x+y$ and $m(\lambda, x) := \lambda x$ are continuous.

The topology on $E\times E$ is obviously the product topology but what is the topology on $\mathbb{K}\times E$?

答: Typically $\mathbb{K}$ would be $\mathbb{R}$ or $\mathbb{C}$, which have natural topologies already.  Then the product topology is what is meant for $\mathbb{K} \times E$. If $\mathbb{K}$ were something else, it would need a prescribed topology for this to make sense.

Title: Topology on $\mathbb{K}\times E$


问:
Finding Euler-Lagrange equations
Maybe you can help here.  There is kind of a lengthy setup to understand what the question is asking.  There is a paper I'm reading, and in one section of it I can't make heads or tails of the result.  The reference is "Global Carleman Estimates for Waves and Applications" by Baudouin, Buhan, Ervedoza.

The setup (taken from the paper) : Suppose $p \in L^{\infty}(\Omega \times (-T,T))$.   Given initial data $(y_0^{-T},y_1^{-T}) \in L^2(\Omega)\times H^{-1}(\Omega)$, find a function $u \in L^2(\Gamma_0 \times (-T,T))$ such that the solution $y$ of

\begin{eqnarray}
\partial_t^2y -\Delta y + py = 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ in  }  \Omega \times(-T,T)  \\
y = u|_{\Gamma_0} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \text{ on } \partial \Omega \times (-T,T)\\
y(-T) = y_0^{-T}, \partial_ty(-T) = y_1^{-T} \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ in } \Omega
\end{eqnarray}
 solves $y(T) = \partial_ty(T) = 0$.

There is a claim that we can get an explicit form for $u$ and $y$.  Let $\phi = e^{\lambda \psi}$, where $\psi(x,t) = |x-x_0|^2 - \beta t^2 +C$.  For $s$ a parameter, define the functional
$$K_{s,p}(z) = \frac{1}{2s}\int_{-T}^{T}\int_{\Omega}e^{2s\phi}|\partial_t^2z - \Delta z + pz|^2 dx \ dt + \frac{1}{2}\int_{-T}^{T}\int_{\Gamma_0}e^{2s\phi}|\partial_{\nu}z|^2 d \sigma dt $$ $$+<(y_0^{-T},y_1^{-T}),(z(-T), \partial_t z(-T))>_{(L^2 \times H^{-1}) \times (H_0^1 \times L^2)}$$

Here, $<(y_0^{-T},y_1^{-T}),(z(-T), \partial_t z(-T))>_{(L^2 \times H^{-1}) \times (H_0^1 \times L^2)} = \int_{\Omega}{y_0^{-T}z_1^{-T} dx} - <y_1^{-T},z_0^{-T}>_{H^{-1} \times H_0^1}$,
and
$<y_1^{-T},z_0^{-T}>_{H^{-1} \times H_0^1} = \int_{\Omega} \nabla(-\Delta_d)^{-1}y_1^{-T}\cdot \nabla z_0^{-T} dx$
where $\Delta_d$ is the Laplace operator with Dirichlet boundary conditions.

Part of the paper shows that $K_{s,p}$ has a unique minimizer $Z[s,p]$, for each $s,p$.

The setup is above.  Now comes the two parts I don't get.

(1).  The paper claims that the Euler-Lagrange equation given by the minimization of $K_{s,p}$ is
$$\frac{1}{s}\int_{-T}^{T}\int_{\Omega}e^{2s\phi}(\partial_t^2z - \Delta z + pz)(\partial_t^2Z -\Delta Z +pZ) dx \ dt + \int_{-T}^{T}\int_{\Gamma_0}e^{2s\phi}\partial_{\nu}z \partial_{\nu}Z d\sigma dt$$  $$+ <(y_0^{-T},y_1^{-T}),(z(-T), \partial_t z(-T))>_{(L^2 \times H^{-1}) \times (H_0^1 \times L^2)}$$

I don't understand how this result is obtained. From what I know, the Euler Lagrange equations are as follows (from Evans book).  If $I[w] = \int L(Dw(x),w(x),x)$, and we call these variables $p, z, x$ respectively, then the Euler Lagrange equations satisfy $-\sum{({L_{p_i}(Du,u,x)})_{x_i}} + L_z(Du,u,x) = 0$.  When I try to do this to $K_{s,p}$, I get a huge mess, because it seems like we need to use the product rule.  I don't get how it simplifies to this form, and why the third term $<\cdot,\cdot>$ stays the same.

(2)  Let $Y = \frac{1}{s}e^{2s\phi}(\partial_t^2 - \Delta + p)Z[s,p]$, and let $U[s,p] = e^{2s\phi}\partial_{\nu}Z[s,p]|_{\Gamma_0}$.

Then, we get
$$\int_{-T}^{T}\int_{\Omega}e^{2s\phi}(\partial_t^2z - \Delta z + pz)Y dx \ dt + \int_{-T}^{T}\int_{\Gamma_0}e^{2s\phi}\partial_{\nu}z U d\sigma dt$$  $$+ <(y_0^{-T},y_1^{-T}),(z(-T), \partial_t z(-T))>_{(L^2 \times H^{-1}) \times (H_0^1 \times L^2)} = 0$$

The paper claims that this is the dual formulation of the problem.  What does this mean exactly, and how does this help us show that Y,U works as a solution?

Any help is greatly appreciated.  Thanks in advance

答: Consider $g(\epsilon)_{z,v}=K_{s,p}(z+\epsilon v)$ for $v$ an arbitrary function in the appropriate space. Now take the derivative in $\epsilon$ and evaluate at $\epsilon =0$. This is the directional derivative of the functional along the line given by $v$.</p>

ContentLicense: CC BY-SA 3.0

Title: Finding Euler-Lagrange equations


答: We give $\mathbb{K}=\mathbb{R},\mathbb{C}$ the typical topology (induced for example by $|\cdot |$) and let $\mathbb{K}\times E$ have the product topology.

Title: Topology on $\mathbb{K}\times E$

Body:
This is probably a really basic question, but here it goes. Let $E$ be a vector space over $\mathbb{K}$ and let $\tau$ be a topology on $E$. Then $E$ is called a topological vector space if the sum and scalar multiplications are continuous, i.e. if the maps $s: E\times E \to E$ and $m: \mathbb{K}\times E \to E$ defined by $s(x,y) := x+y$ and $m(\lambda, x) := \lambda x$ are continuous.

The topology on $E\times E$ is obviously the product topology but what is the topology on $\mathbb{K}\times E$?


问:
If both $x+10$ and $x-79$ are perfect squares, what is $x$?

If both $x+10$ and $x-79$ are perfect squares, what is $x$?

That is:

$x+10=$perfect square  (or can be square such that the result of it squared is a positive integer)

$x-79=$perfect square  (or can be square such that the result of it squared is a positive integer)

I have no idea how to solve this.I tried inequality like

$$x+10>x-79$$
$$x-x>-79-10$$
$$0>-89$$

Note: You are not allowed to use trial and error or guess and check to solve the question.

答:
It is helpful to keep in mind that the difference between consecutive perfect square numbers gives us an arithmetic sequence. $$\left\{n^2\right\}_{n=1} ^{\infty},\space \space \space a_n = n^2  $$ $$a_{n+1}-a_n =(n+1)^2-n^2= 2n+1$$ Therefore, we have $x+10$ and $x-79$ as consecutive perfect square numbers. Assume that $a_{n+1}=x+10$, $a_n = x-79$. $$\begin{align} \ a_{n+1}-a_n & = 2n+1 \\
& = \ (x+10)-(x-79) \\
& = \ 89 \\ \end {align}$$ $$n=44 \space \Rightarrow \space \begin{align}
\ a_{45} & =\ 45^2 \\
& =\ 2025 \\
& =\ x+10 \\ \end{align}$$ $$x=2015$$


答:
Starting with $x=y+n^y$, and manipulating the equation in some non-obvious ways:
\begin{align*}
x&=y+n^y\\[5pt]
\implies x-y &=n^y\\[5pt]
\implies (x-y)\ln(n)n^{x} &= n^{y}\ln(n)n^{x}\\[5pt]
\implies(x-y)\ln(n)n^{x-y} &= n^{x}\ln(n)\\[5pt]
\implies \ln\left(n^{x-y}\right)n^{x-y} &= n^{x}\ln(n)\\[5pt]
\implies \ln\left(n^{x-y}\right) \exp\left(\ln\left(n^{x-y}\right)\right) &= n^{x}\ln(n)\\[5pt]
\implies W(n^{x}\ln(n)) &= \ln\left(n^{x-y}\right)\\[5pt]
\implies W(n^{x}\ln(n)) &=(x-y)\ln(n)\\[5pt]
\implies W(n^{x}\ln(n)) &=x\ln(n) - y\ln(n)\\[5pt]
\end{align*}
and so by solving the last equation for $y$ we get $$\boxed{y = x - \frac{W(n^{x}\ln(n))}{\ln(n)}}$$ where $W$ is the Lambert W function.

Title: Inverse function of $x+n^x$


问:
How to write $\pi$ as a set in ZF?
I know that from ZF we can construct some sets in a beautiful form obtaining the desired properties that we expect to have these sets. In ZF all is a set (including numbers, elements, functions, relations, etc...).

For example we can define a copy of $\mathbb{N}$, using the empty set axiom and others, called sometimes $\omega$ and defining,

\begin{align}
0_\mathbb{N}&=\{\}\\
1_\mathbb{N}&=\{\{\}\}\\
2_\mathbb{N}&=\{\{\},\{\{\}\}\}\\
3_\mathbb{N}&=\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}\\
4_\mathbb{N}&=\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}, \{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}\}\\
&\ \vdots
\end{align}

Well, ZF allow us to build this type of sets that are kind of arrangements of brackets and commas. We can follow with $\mathbb{Z}$ since we can define $(a,b)=\{\{a\},\{a,b\}\}$:

$\mathbb{Z}$ is defined as the set of equivalence classes $\mathbb{Z}=(\mathbb{N}\times\mathbb{N})\big/\sim$ where

$$\sim\, =\{\big((m,n),(h,k)\big)\in(\mathbb{N}\times\mathbb{N})\times (\mathbb{N}\times\mathbb{N}):(m+_\mathbb{N} k)= (h+_\mathbb{N} n)\}$$

here, the integers are sets more complicated than natural numbers. For example,

\begin{align}
-2_\mathbb{Z}&=\{(1_\mathbb{N},3_\mathbb{N}),(2_\mathbb{N},4_\mathbb{N}),(3_\mathbb{N},5_\mathbb{N}),\ldots,(n_\mathbb{N},(n+2)_\mathbb{N}),\ldots\}\\
&=\{\{\{1_\mathbb{N}\},\{1_\mathbb{N},3_\mathbb{N}\}\},\{\{2_\mathbb{N}\},\{2_\mathbb{N},4_\mathbb{N}\}\},\ldots\}\\
-2_\mathbb{Z}&=\{\{\{\{\{\}\}\},\{\{\{\}\},\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}\}\},\\
&\quad\{\{\{\{\},\{\{\}\}\}\},\{\{\{\},\{\{\}\}\},\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}, \{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}\}\}\},\ldots\}\\
\end{align}

Here we note the importance of notations. We continue with  $\mathbb{Q}=(\mathbb{Z}\times(\mathbb{Z\setminus\{0_\mathbb{Z}\})}\})\big/\sim$ where

$$\sim\, =\{\big((m,n),(h,k)\big)\in(\mathbb{Z}\times(\mathbb{Z\setminus\{0_\mathbb{Z}\})}\})\times (\mathbb{Z}\times(\mathbb{Z\setminus\{0_\mathbb{Z}\})}\}):m\odot_\mathbb{Z}k=h\odot_\mathbb{Z}n\}$$

And for example:

$$(0.2)_\mathbb{Q}=\{(1_\mathbb{Z},5_\mathbb{Z}),(2_\mathbb{Z},10_\mathbb{Z}),(3_\mathbb{Z},15_\mathbb{Z}),\ldots,(n_\mathbb{Z},(5n)_\mathbb{Z}),\ldots\}$$

Imagine if we write the integer numbers as before and write the ordered pairs in unabridged form ($(0.2)_\mathbb{Q}$ is a nice abbreviation for this monster). However we can.

Finally we define $\mathbb{R}$ as the set of all Dedekind cuts, for example:

$$(0.2)_\mathbb{R}=\{x\in\mathbb{Q}:x<_\mathbb{Q} (0.2)_\mathbb{Q}\}$$

Note that $(0.2)_\mathbb{R}$ is even more monstrous than $(0.2)_\mathbb{Q}$. Also I can write $(\sqrt{2})_\mathbb{R}$ showing its elements in a simple form,

$$(\sqrt{2})_\mathbb{R}=\{x\in\mathbb{Q}:(x^2<_\mathbb{Q} 2_\mathbb{Q}) \lor (x<_\mathbb{Q} 0_\mathbb{Q})\}$$

But I don't know how to do it with $\pi_\mathbb{R}$, since

$$\pi=\lim_{k\to\infty}\sum_{n=0}^{k}\cfrac{2^{n+1} n!^2}{(2n + 1)!}$$

I only know that

$$(\pi)_\mathbb{R}=\bigcup_{k=1}^{\infty} \left(\sum_{n=0}^{k}\cfrac{2^{n+1} n!^2}{(2n + 1)!}\right)_\mathbb{R}$$

Since monotonically converge to $\pi$ we have

$$(\pi)_\mathbb{R}=\bigcup_{k=1}^{\infty} \left\{x\in\mathbb{Q}:x<_\mathbb{Q} \left(\sum_{n=0}^{k}\cfrac{2^{n+1} n!^2}{(2n + 1)!}\right)_\mathbb{Q}\right\}$$

Is there any way to avoid infinite union (and
the choice of a particular convergent sequence
) as the case of $(\sqrt{2})_\mathbb{R}$? If not, why?

Can we write the set representing $\pi$ listing its elements as we do with integers or rationals (as $\mathbb{Q}$ is countable I guess that should be able to do, but I don't know how to do)?

If in ZF all is a set, is so surprising the fact that so many things can be defined, then my last question is

How many more things can be built using ZF?, ZF could define us what is a derivative, an integral, a limit or a measure?

Thanks in advance.

答:
Let me try this $$(\pi)_{\mathbb{R}}=\left\{x\in \mathbb{Q}:\;\exists k\in\mathbb{N}.\;x< \sum\limits_{n=0}^{k}\cfrac{2^{n+1} n!^2}{(2n + 1)!}\right\}$$ This would be the Dedekind construction of the reals, and note that there is no union here.

Or, I would complete the rationals first by taking the quotient of ring of fundamental sequences with maximal ideal of null sequences $\mathcal{F}_Q/\mathcal{N}_Q$ to obtain a complete field $\mathbb{R}$, then I can safely say that $$\pi=\lim\limits_{k\to\infty}\sum\limits_{n=0}^{k}\cfrac{2^{n+1} n!^2}{(2n + 1)!}$$ (or define exp first, then define sin, then define $\pi$)

Title: How to write $\pi$ as a set in ZF?


问:
Is it true that the cost of a closed walk is nonnegative if all its cycles have nonnegative costs?
Consider a directed graph $G$ and a closed (not necessarily directed) walk $\omega$ constructed from some edges of $G$. The term "not necessarily directed" means here that $\omega$ is a walk of the underlying undirected graph of $G$.
Let $\kappa^+(\omega)=\sum\limits_{i=1}^n(v_i + S_i u_i)$ and $\kappa^-(\omega)=\sum\limits_{i=1}^n(v_i-S_i u_i)$ be the cost and anti-cost of $\omega$, respectively, where $u_i$ and $v_i$ are two nonnegative parameters for the $i$th edge within $\omega$ and $S_i$ is a sign for the $i$th edge. Here $S_i=1$, if the $i$th edge points in the direction of the endpoint of $\omega$, otherwise $S_i=-1$.

My question is whether it is true that $\kappa^\pm(\omega) \ge 0$ if $\kappa^\pm(\chi) \ge 0$ for all $\chi \subseteq \omega$ (not necessarily directed) cycles.
If no, then what other $\psi \subseteq \omega$ subwalks should satisfy $\kappa^\pm(\psi) \ge 0$ to achieve $\kappa^\pm(\omega) \ge 0$?

答:
We can prove this by strong induction on the length of the walk. (When $\omega$ is a closed walk of length $0$, its cost is of course nonnegative.)

In general, suppose $\omega$ is a closed walk of length $n$ visiting vertices $v_0, v_1, v_2, v_3, \dots, v_n = v_0$. I am not requiring these vertices to be distinct, of course. Now pick the pair $(v_i, v_j)$ with $v_i = v_j$, $i<j$, and $j$ as small as possible: the first time that $\omega$ revisits a vertex.

The sub-walk $\chi$ visiting vertices $v_i, v_{i+1}, \dots, v_j$ is a cycle, because none of its vertices can be repeats, and so we know that the total cost of its edges $v_i v_{i+1}, \dots, v_{j-1} v_j$ is nonnegative. Deleting it from $\omega$, we get the sub-walk $\omega'$ visiting vertices $v_0, v_1, \dots, v_i, v_{j+1}, \dots, v_n$ which is also a closed walk. In fact, all cycles contained in $\omega'$ are nonnegative, because they were also contained in $\omega$.

So the total cost of $\omega'$ is nonnegative, by the inductive hypothesis (it's shorter). Since the total cost of $\omega$ is the sum of the costs of $\chi$ and $\omega'$, we know that the total cost of $\omega$ is nonnegative, proving the result we wanted.

We did not assume anything about the costs beyond the fact that edges have them, so this works for your cost and anti-cost equally well.

One caveat: you might want to allow your walks to backtrack (going from a vertex $v$ to another vertex $w$ back to the same vertex $v$) but not allow the same for a cycle. In that case, the argument above no longer quite works: the sub-walk $\chi$ might not be a cycle in that case.

In this case, $\chi$ will have length $2$ and take edge $vw$ followed by edge $wv$. Your formulas for $\kappa^+$ and $\kappa^-$ have a "constant" cost and a "direction-dependent" cost; let these be $c$ and $d$ for the edge in question. Then $\kappa^+(\chi)$ and $\kappa^-(\chi)$ will both have cost $(c+d) + (c-d) = (c-d) + (c+d) = 2c$ for traversing this edge once in each direction. As long as the "constant" cost $c$ is always nonnegative, which is assumed, the cost of $\chi$ will still be nonnegative.

So in this case, the result still holds, though we have to use a bit more information about the details of the costs.


问:
General solution of $\cos (2\arctan x) = \frac{{1 - x^2 }}{{1 + x^2 }}?$

How to find the general solution of $$\cos (2\arctan x) = \frac{{1 - x^2 }}{{1 + x^2 }}$$

答:
$t=\arctan x\iff \tan t=x$.

Recall that $1+x^2=1+\tan^2 t=\frac{1}{\cos^2 t}$, so $\cos^2 t=\frac{1}{1+x^2}$.

Now, $$\cos(2\arctan x)=\cos 2t=\cos^2t-\sin^2t=2\cos^2t-1=2\frac{1}{1+x^2}-1=\frac{1-x^2}{1+x^2}$$ and it is a tautology.


问:
$f: (a, b] \to \mathbb{R}$ continuous. Prove: $f$ continuously resumable in $a \iff f$ is uniformly continuous
Let $f: (a, b] \to \mathbb{R}$. Prove that: $f$ is continuously resumable in $a \iff f$ is uniformly continuous.

My approach:

"$\implies$:"

Since $a$ is continuously resumable, then $a$ has to be an accumulation point and $\lim_{x \to a} f(x)$ exists. I define another function $g$:

$$g: [a, b] \to \mathbb{R}, \left\{\begin{array}{lr}
        x \to f(x), & \text{for } x \in (a, b]\\
        \lim_{z \to x} f(z), & \text{for } x = a\\
        \end{array}\right\}$$

$g$ is continuous on $[a, b]$ so $g$ is also uniformly continuous on $[a, b]$ and therefore also uniformly continuous on $(a, b]$. This implies $f$ is also uniformly continuous on $(a, b]$.

"$\impliedby$:"

I think this direction is quite trivial to prove. Edit: This direction is not so trivial, as Danny Pak-Keung Chan's proof shows.

My question is: Is my proof correct? Did I made a too big jump from "$g$ is uniformly continuous in $[a, b]"$ to "$g$ is uniformly continuous in $(a, b]$"?

答:
Suppose $f$ extends to a continuous function $g$ on $[a, b]$.  Since $[a, b]$ is compact, the function $g$ is uniformly continuous.  Therefore, so is its restriction
$$
f = g|_{(a, b]}.
$$

If $f$ is uniformly continuous on $(a, b]$, suppose $\{x_{n}\}_{n}$ is an arbitrary sequence in $(a, b]$ that converges to $a$. The uniform continuity of $f$ implies that the sequence $\{f(x_{n})\}_n$ is Cauchy, hence has a limit $y$, and also that this limit is the same for all sequences $\{x'_{n}\}_{n}$ in $(a, b]$ that converge to $a$.  Therefore, letting $g$ agree with $f$ on $(a, b]$ and putting $g(a) = y$ gives us a continuous extension $g$ of $f$ to $[a, b]$.


问:
Find shortest distance between lines in 3D
Find shortest distance between lines given by

$$\frac{x-2}{3}=\frac{y-6}{4}=\frac{z+9}{-4}$$
and
$$\frac{x+1}{2}=\frac{y+2}{-6}=\frac{z-3}{1}$$
Is there any shortcut method for this problems?

答:
I'm using a method similar to marty cohen's but without matrix/linear system

We first parametrize the two lines as:
$$
\vec{L_0} = \vec{p_0}+s\vec{u}\\
\vec{L_1} = \vec{p_1}+t\vec{v}
$$
The distance between any points on line 1 to line 0 is
$$
d^2 = \dfrac{\left[(\vec{p_1}-\vec{p_0}+t\vec{v})\times\vec{u}\right]^2}{\vec{u}\cdot \vec{u}}
$$
To find the minimum distance, we need to minimize the numerator
$$
\dfrac{d\left[(\vec{p_1}-\vec{p_0})\times \vec{u}+t\vec{v}\times\vec{u}\right]^2}{dt}=\dfrac{d\left[2(\vec{p_1}-\vec{p_0})\times\vec{u}\cdot(t\vec{v}\times\vec{u})+t^2(\vec{v}\times\vec{u})^2\right]}{dt}\\
=(\vec{p_1}-\vec{p_0})\times\vec{u}\cdot(\vec{v}\times\vec{u})+t(\vec{v}\times\vec{u})^2=0
$$

Then we can get the solution of t
$$
t = -\dfrac{(\vec{p_1}-\vec{p_0})\times\vec{u}\cdot(\vec{v}\times\vec{u})}{(\vec{v}\times\vec{u})^2}
$$

Taking the value of $t$ back to the distance expression,
$$
d^2=\dfrac{\left[(\vec{p_1}-\vec{p_0})\times\vec{u}\right]^2(\vec{v}\times\vec{u})^2-\left[((\vec{p_1}-\vec{p_0})\times\vec{u})\cdot(\vec{v}\times\vec{u})\right]^2}{\vec{u}\cdot \vec{u}(\vec{v}\times\vec{u})^2}\\
= \dfrac{\left[((\vec{p_1}-\vec{p_0})\times\vec{u})\times(\vec{v}\times\vec{u})\right]^2}{\vec{u}\cdot \vec{u}(\vec{v}\times\vec{u}^2}\\
= \dfrac{\left[\vec{u}\cdot((\vec{p_1}-\vec{p_0})\times\vec{v})\vec{u}\right]^2}{\vec{u}\cdot \vec{u}(\vec{v}\times\vec{u})^2}\\
= \dfrac{\left[\vec{u}\cdot((\vec{p_1}-\vec{p_0})\times\vec{v})\right]^2}{(\vec{v}\times\vec{u})^2}\\
= \dfrac{\left[(\vec{p_1}-\vec{p_0})\cdot(\vec{v}\times\vec{u})\right]^2}{(\vec{v}\times\vec{u})^2}
$$
The final solution
$$
d=\dfrac{\left|(\vec{p_1}-\vec{p_0})\cdot(\vec{v}\times\vec{u})\right|}{|\vec{v}\times\vec{u}|}
$$
By simply switching the index, we can also get the solution of s
$$
s = -\dfrac{(\vec{p_1}-\vec{p_0})\times\vec{v}\cdot(\vec{v}\times\vec{u})}{(\vec{v}\times\vec{u})^2}\\
$$

Then we can also get the direction of the line connecting the two point:
$$
\vec{n}=\vec{p_0}+s\vec{u}-\vec{p_1}-t\vec{v}
$$


问:
Changing order of summation for simple integral

I want to change the summation or the following integral:

$ \intop_{0}^{2\pi}dx\intop_{0}^{\sin x}f\left(x,y\right)dy $.

Now I know that $ x=\left(-1\right)^{k}\arcsin\left(y\right)+\pi k $ and if we want to write $ sinx $ as combination of funcions of $ y $ it should be as described in the attached picture (first the green part, then the purple and the blue).

But I'm not sure that I understand why exactly it is correct. Can anyone provide a simple explanation? thanks in advance.

答:
The key thing here is that $$-\frac{\pi}{2} \le \arcsin y \le \frac{\pi}{2} $$

When $x\in [0, \frac{\pi}{2} ] $, $x$ is just $\arcsin y$.

When $x\in[\frac{\pi}{2}, \frac{3\pi}{2} ] $, we have to manipulate the expression so as to match up the ranges, while still making sure that taking the sine of it gives $y$. This is only achieved by the expression $$\frac{\pi}{2} \le\pi - \arcsin y  \le \frac{3\pi}{2}$$

Similarly, when $x\in [\frac{3\pi}{2} , \frac{5\pi}{2} ] $ we have to add $2\pi$. This makes sure that the value of its sine is conserved, and that the ranges of the new expression and $x$ are the same: $$\frac{3\pi}{2} \le \arcsin y +2\pi \le \frac{5\pi}{2} $$


Title: General solution of $\cos (2\arctan x) = \frac{{1 - x^2 }}{{1 + x^2 }}?$

Body: How to find the general solution of $$\cos (2\arctan x) = \frac{{1 - x^2 }}{{1 + x^2 }}$$

Body: If $t=\arctan x$ then $\cos(2t)=2\cos^2t-1=\tfrac{2}{1+x^2}-1=\tfrac{1-x^2}{1+x^2}$.


答:
It is hard to tell what "Avoid infinite union" means. Of course all dedekind cuts are infinite sets of elements of $\mathbb Q$, which are themselves infinite sets and so on.
If you want to write an expression for $\pi_{\mathbb R}$ that avoids blobs of ink looking like $\bigcup_{k=1}^\infty$, you can try to define $\pi$ as smallest positive root of any nonzero function that has a root at zero and is the negative of its second derivative. It may turn out as fun to formulate all this using only rationals.</p>

Title: How to write $\pi$ as a set in ZF?

Body:
I know that from ZF we can construct some sets in a beautiful form obtaining the desired properties that we expect to have these sets. In ZF all is a set (including numbers, elements, functions, relations, etc...).

For example we can define a copy of $\mathbb{N}$, using the empty set axiom and others, called sometimes $\omega$ and defining,

\begin{align}
0_\mathbb{N}&=\{\}\\
1_\mathbb{N}&=\{\{\}\}\\
2_\mathbb{N}&=\{\{\},\{\{\}\}\}\\
3_\mathbb{N}&=\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}\\
4_\mathbb{N}&=\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}, \{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}\}\\
&\ \vdots
\end{align}

Well, ZF allow us to build this type of sets that are kind of arrangements of brackets and commas. We can follow with $\mathbb{Z}$ since we can define $(a,b)=\{\{a\},\{a,b\}\}$:

$\mathbb{Z}$ is defined as the set of equivalence classes $\mathbb{Z}=(\mathbb{N}\times\mathbb{N})\big/\sim$ where

$$\sim\, =\{\big((m,n),(h,k)\big)\in(\mathbb{N}\times\mathbb{N})\times (\mathbb{N}\times\mathbb{N}):(m+_\mathbb{N} k)= (h+_\mathbb{N} n)\}$$

here, the integers are sets more complicated than natural numbers. For example,

\begin{align}
-2_\mathbb{Z}&=\{(1_\mathbb{N},3_\mathbb{N}),(2_\mathbb{N},4_\mathbb{N}),(3_\mathbb{N},5_\mathbb{N}),\ldots,(n_\mathbb{N},(n+2)_\mathbb{N}),\ldots\}\\
&=\{\{\{1_\mathbb{N}\},\{1_\mathbb{N},3_\mathbb{N}\}\},\{\{2_\mathbb{N}\},\{2_\mathbb{N},4_\mathbb{N}\}\},\ldots\}\\
-2_\mathbb{Z}&=\{\{\{\{\{\}\}\},\{\{\{\}\},\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}\}\},\\
&\quad\{\{\{\{\},\{\{\}\}\}\},\{\{\{\},\{\{\}\}\},\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}, \{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}\}\}\},\ldots\}\\
\end{align}

Here we note the importance of notations. We continue with  $\mathbb{Q}=(\mathbb{Z}\times(\mathbb{Z\setminus\{0_\mathbb{Z}\})}\})\big/\sim$ where

$$\sim\, =\{\big((m,n),(h,k)\big)\in(\mathbb{Z}\times(\mathbb{Z\setminus\{0_\mathbb{Z}\})}\})\times (\mathbb{Z}\times(\mathbb{Z\setminus\{0_\mathbb{Z}\})}\}):m\odot_\mathbb{Z}k=h\odot_\mathbb{Z}n\}$$

And for example:

$$(0.2)_\mathbb{Q}=\{(1_\mathbb{Z},5_\mathbb{Z}),(2_\mathbb{Z},10_\mathbb{Z}),(3_\mathbb{Z},15_\mathbb{Z}),\ldots,(n_\mathbb{Z},(5n)_\mathbb{Z}),\ldots\}$$

Imagine if we write the integer numbers as before and write the ordered pairs in unabridged form ($(0.2)_\mathbb{Q}$ is a nice abbreviation for this monster). However we can.

Finally we define $\mathbb{R}$ as the set of all Dedekind cuts, for example:

$$(0.2)_\mathbb{R}=\{x\in\mathbb{Q}:x<_\mathbb{Q} (0.2)_\mathbb{Q}\}$$

Note that $(0.2)_\mathbb{R}$ is even more monstrous than $(0.2)_\mathbb{Q}$. Also I can write $(\sqrt{2})_\mathbb{R}$ showing its elements in a simple form,

$$(\sqrt{2})_\mathbb{R}=\{x\in\mathbb{Q}:(x^2<_\mathbb{Q} 2_\mathbb{Q}) \lor (x<_\mathbb{Q} 0_\mathbb{Q})\}$$

But I don't know how to do it with $\pi_\mathbb{R}$, since

$$\pi=\lim_{k\to\infty}\sum_{n=0}^{k}\cfrac{2^{n+1} n!^2}{(2n + 1)!}$$

I only know that

$$(\pi)_\mathbb{R}=\bigcup_{k=1}^{\infty} \left(\sum_{n=0}^{k}\cfrac{2^{n+1} n!^2}{(2n + 1)!}\right)_\mathbb{R}$$

Since monotonically converge to $\pi$ we have

$$(\pi)_\mathbb{R}=\bigcup_{k=1}^{\infty} \left\{x\in\mathbb{Q}:x<_\mathbb{Q} \left(\sum_{n=0}^{k}\cfrac{2^{n+1} n!^2}{(2n + 1)!}\right)_\mathbb{Q}\right\}$$

Is there any way to avoid infinite union (and
the choice of a particular convergent sequence
) as the case of $(\sqrt{2})_\mathbb{R}$? If not, why?

Can we write the set representing $\pi$ listing its elements as we do with integers or rationals (as $\mathbb{Q}$ is countable I guess that should be able to do, but I don't know how to do)?

If in ZF all is a set, is so surprising the fact that so many things can be defined, then my last question is

How many more things can be built using ZF?, ZF could define us what is a derivative, an integral, a limit or a measure?

Thanks in advance.


问:
Median of the subsets of a finite set of real numbers
Starting from this$^1$ question, using Mathematica I did the following.

Generated a sequence of $n$ random distinct "real" numbers.

Computed the median of the $2^n$ subsets (conventionally $\emptyset$ has a median zero) and deleted duplicates.

The number $m$ of distinct medians I got, was constant for fixed $n$ and followed a simple rule
$$m=\frac{1}{2} \left(n^2+n+2\right)$$

The code I used is the following

```<code>Table[{n,
  Length[DeleteDuplicates[
    Median /@ Subsets[Table[RandomReal[10^12], {k, 1, n}]]]]}, {n, 1, 20}]
</code>```

Do you know why this happens?

Edit

There are exceptions

Consider $A=\{1,2,3,4\}$

The value predicted by the formula is $m=11$ but we have
$$\mathcal{P}(A)=\{\emptyset,\{1\},\{2\},\{3\},\{4\},\{1,2\},\{1,3\},\{1,4\},\\\{2,3\},\{2,4\},\{3,4\},\{1,2,3\},\{1,2,4\},\{1,3,4\},\{2,3,4\},\{1,2,3,4\}\}$$

Medians are $$\left\{0,1,2,3,4,\frac{3}{2},2,\frac{5}{2},\frac{5}{2},3,\frac{7}{2},2,2,3,3,\frac{5}{2}\right\}$$ and the distinct ones are only $8$:
$$\left\{0,1,2,3,4,\frac{3}{2},\frac{5}{2},\frac{7}{2}\right\}$$
[1]: Efficient way to get median of all subsets.

答:
Median has to be equal to either some number from $x_1,\ x_2,\ldots,x_n$, or arithmetic mean of two of these numbers (plus our case with zero). So number of different results is

$$ {{n}\choose{0}} + {{n}\choose{1}} + {{n}\choose{2}}  = \frac{n^2+n+2}{2} $$


问:
Linking number of torus links
Let $(p,q)$ be a pair of coprime integers; the torus knot $K(p,q)$ is the unique (up to isotopy) curve on the boundary of a solid torus which is homologous to $p\cdot \lambda + q \cdot \mu$, where $\lambda, \mu$ are standard longitude and meridian of the solid torus. One can define torus knots even when $p$ and $q$ are not coprime, and one gets $n=\text{gcd}(p,q)$ copies of the torus knot $K(p/n, q/n)$; these links are called torus links. Let us suppose that $n=2$, so our torus link has two components. I would like to say that the linking number of these two components is non zero. An example of such a link is the Hopf link, which has linking number 1.

Indeed, I've found a formula (https://discover.wooster.edu/jbowen/files/2013/10/Total-linking-numbers-of-torus-links-and-Klein-links.pdf, Theorem 4.2) for the total linking number of torus links, which in the case of two components is the linking number of the components (which is what I need). The formula is

$$lk(T(m,n))=m \big(n- \frac{n}{gcd(m,n)}\big).$$

Ok, great. But for the Hopf link, which is a $T(2,2)$ torus link, the formula gives

$$lk(T(2,2))=2 \big(2-1)=2,$$

but the linking number of the Hopf link is 1! Am I missing something obvious?

P.S. If you know a simple way to say that the linking number of two-components torus links is nonzero, which does not involve any general formula, for me it's okay.

答:
I believe the paper you referenced has a typo in Theorem 4.2. It should say that the total linking number $\operatorname{lk}(T(m,n))$ of the torus link $T(m,n)$ is
$$\operatorname{lk}(T(m,n)) = \frac{m}{2}\left(n - \frac{n}{\gcd(m,n)}\right).$$

The formula they give in Theorem 4.2 (without the extra $1/2$ factor) does not agree with their own computation in Figure 9. They compute $\operatorname{lk}(T(6,4)) = 6$ in Figure 9, but their own formula yields
$$\operatorname{lk}(T(6,4)) = 6\left(4-\frac{4}{\gcd(6,4)}\right) = 12.$$

If you work through their proof, you'll see that they accurately count that the total number of crossings between different components of the $T(m,n)$ torus link is $m\left(n-\frac{n}{\gcd{m,n}}\right)$. Since all of those crossings are positive, it follows that the linking number is
$$\operatorname{lk}(T(m,n)) = \frac{m}{2}\left(n - \frac{n}{\gcd(m,n)}\right).$$

Additional note: If all you are looking for is that the linking number is nonzero, then it suffices to know that torus links are positive, nonsplit links. The nonsplit part guarantees that there are crossings between different components, and the positive part tells you that all crossing count positively in the linking number sum.

Title: Linking number of torus links


问:
Banach Algebra exponentials
Let $\mathcal{A}$ be a unital Banach Algebra. Suppose $a,b \in \mathcal{A}$.

show that if $a$ and $b$ commute, i.e. ab=ba, then $\exp(a) \exp(b)= \exp(a+b)$.

Give a counterexample to show that $\exp(a) \exp(b)= \exp(a+b)$ is not necessarily true when $ab \ne ba$.

For 1, I know we have that $\exp(a)=\sum_{n=0}^{\infty}\frac{a^n}{n!}.$

So $\exp(a) \exp(b)=\sum_{n=0}^{\infty}\frac{a^n}{n!}\sum_{n=0}^{\infty}\frac{b^n}{n!}$

And $\exp(a+b)=\sum_{n=0}^{\infty}\frac{(a+b)^n}{n!}.$

But this is almost like proving in real numbers using Taylor series. If this is correct, what is the difference between the proof in the real numbers and in the unital Banach Algebra?

For 2, I couldn't come up with anything...

Thanks!

答:
For (1), as Cameron Williams says in the comment, mimic the proof of the real number case. This is a strategy that might appear a few times in the elementary theory, see for example the accepted answer in this post: Why is $f(t) = e^{ta}$ differentiable in a unital Banach algebra?

For (2), whenever you need a counter-example for the non-commuting setting, you always look in $M_2(\mathbb{C})$ first. Keep this in mind, as I suspect you are now starting Murphy's book.

Here is a counter example: try $a=e_{12}=\begin{pmatrix}0&1\\0&0\end{pmatrix}$, $b=e_{21}=\begin{pmatrix}0&0\\1&0\end{pmatrix}$. Note that these do not commute.

since $a+b=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, we have that $(a+b)^2=I_2$, so
$$e^{a+b}=\sum_{n=0}^\infty\frac{1}{n!}(a+b)^n=\sum_{n=0}^\infty\frac{1}{(2n)!}\cdot I_2+\sum_{n=0}^\infty\frac{1}{(2n+1)!}\cdot(a+b)=\begin{pmatrix}s_0&s_1\\s_1&s_0\end{pmatrix}$$
where $s_0=\sum_{n=0}^\infty\frac{1}{(2n)!}$ and $s_1=\sum_{n=0}^\infty\frac{1}{(2n+1)!}$.

On the other hand, since $a^2=0$ and $b^2=0$ we have that
$$e^a=I_2+a=\begin{pmatrix}1&1\\0&1\end{pmatrix}$$
and $$e^b=I_2+b=\begin{pmatrix}1&0\\1&1\end{pmatrix} $$
so $e^a\cdot e^b=\begin{pmatrix}2&1\\1&1\end{pmatrix}$, so $e^ae^b\neq e^{(a+b)}$ (unless $2=1$).

Title: Banach Algebra exponentials


问:
Let $z_1$, $z_2$ and $z_3$ be complex vertices of an equilateral triangle. Show $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$.
Prompt:

Let $z_1$, $z_2$ and $z_3$ represent vertices of an equilateral triangle in the complex plane. Show $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$.

Question:

I hope the following question isn't too soft. I know a solution to the above question by taking the sides of the triangle and equating them by rotations in the complex plane of $e^{i\pi/3}$.

What, if anything, is the significance of the result other than showing some clever algebraic manipulation? Does it tell us something unique about equilateral triangles defined by vertices in the complex plane?

Is there any intuition to be gained here?

答:
Does it tell us something unique about equilateral triangles?

In the complex plane the sides of any triangle are represented by numbers $a,b,c$ where $$a+b+c=0.$$

The equation you are interested in is equivalent to the equation $$a^2+b^2+c^2=0$$ and therefore tells us that equilateral triangles have the very special property that all quadratic symmetric polynomials of the sides are constant.

This gives us an elegant tool for investigating these triangles with minimal algebraic manipulation. I will give just two examples.

The sides of an equilateral triangle satisfy $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.

Simply divide the symmetric quadratic equation $ab+ba+cb=0$ by $abc$.

The equation $a^2+b^2+c^2=0$ only represents equilateral triangles

$abc-c^3=c(ab+bc+ca)-c^2(a+b+c)=0$. Then $a^3,b^3,c^3$ are all equal (to $abc$) and therefore $a,b,c$ all have the same length.


答:
For another approach to the problem of proving that if $f : (a, b] \to \mathbb{R}$ is uniformly continuous, then $\lim_{x \to a^+}$ exists, we can adapt a common proof that any Cauchy sequence in $\mathbb{R}$ has a limit.

First, observe that the uniform continuity implies that $f$ is "Cauchy as $x \to a^+$"; in other words, for every $\epsilon > 0$, there exists $\delta > 0$ such that whenever $x, y \in (a, a + \delta) \cap (a, b]$, then $|f(x) - f(y)| < \epsilon$.  To see this, apply the uniform continuity to get $\delta > 0$ such that whenever $|x - y| < \delta$, we have $|f(x) - f(y)| < \epsilon$; and then note that whenever $x, y \in (a, a + \delta) \cap (a, b]$ we certainly have $|x - y| < \delta$.

Now, the fact that $f$ is Cauchy as $x \to a^+$ implies that $$\limsup_{x\to a^+} f(x) \overset{def}{=} \inf_{\delta > 0} \sup \{ f(x) | x \in (a, a + \delta) \cap (a, b] \}$$ and $$\liminf_{x\to a^+} f(x) \overset{def}{=} \sup_{\delta > 0} \inf \{ f(x) | x \in (a, a + \delta) \cap (a, b] \}$$ are both finite, and that the two numbers are equal.  Furthermore, that implies that $\lim_{x\to a^+} f(x)$ exists and is equal to the common value of $\limsup_{x\to a^+} f(x) = \liminf_{x\to a^+} f(x)$.  Here, the proofs of these facts will be analogous to the proofs mentioned previously for sequences.

Title: $f: (a, b] \to \mathbb{R}$ continuous. Prove: $f$ continuously resumable in $a \iff f$ is uniformly continuous

Body:
Let $f: (a, b] \to \mathbb{R}$. Prove that: $f$ is continuously resumable in $a \iff f$ is uniformly continuous.

My approach:

"$\implies$:"

Since $a$ is continuously resumable, then $a$ has to be an accumulation point and $\lim_{x \to a} f(x)$ exists. I define another function $g$:

$$g: [a, b] \to \mathbb{R}, \left\{\begin{array}{lr}
        x \to f(x), & \text{for } x \in (a, b]\\
        \lim_{z \to x} f(z), & \text{for } x = a\\
        \end{array}\right\}$$

$g$ is continuous on $[a, b]$ so $g$ is also uniformly continuous on $[a, b]$ and therefore also uniformly continuous on $(a, b]$. This implies $f$ is also uniformly continuous on $(a, b]$.

"$\impliedby$:"

I think this direction is quite trivial to prove. Edit: This direction is not so trivial, as Danny Pak-Keung Chan's proof shows.

My question is: Is my proof correct? Did I made a too big jump from "$g$ is uniformly continuous in $[a, b]"$ to "$g$ is uniformly continuous in $(a, b]$"?


答:
$$y(x)=x+A^{\,x}\qquad A=\text{constant}$$
$$y-x=A^x=A^{x-y}A^y$$
$$(y-x)A^{y-x}=A^y$$
$$(y-x)e^{(y-x)\ln(A)}=e^{y\ln(A)}$$
$$(y-x)\ln(A)e^{(y-x)\ln(A)}=\ln(A)e^{y\ln(A)}$$
Let $\quad X=(y-x)\ln(A)\quad$ and $\quad Y=\ln(A)e^{y\ln(A)}$
$$Xe^X=Y$$
The inverse function $X(Y)$ cannot be written with a finite number of elementary functions. A special function is required, namely the Lambert W function.

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

$$X=W(Y)$$

Note that this function is multivalued, namely $W_0(Y)$ and $W_{-1}(Y)$ in reals and more determinations in complex.

$$(y-x)\ln(A)=W\left(\ln(A)e^{y\ln(A)} \right)$$
Thus the formal answer to your question is :
$$x=y-\frac{1}{\ln(A)}W\left(\ln(A)e^{y\ln(A)} \right)$$

$\boxed{\text{I see that Parcly Taxel and DMcMor gave the answer while I was still typing.}  }$

Title: Inverse function of $x+n^x$


问:
Cute problem: determinant of $I_n+(f_if_j)_{i,j}$
I thought of the following little problem.

Given numbers $f_1,\dots f_n$, what is the determinant of the symmetric matrix $I_n+(f_if_j)_{i,j}$?

I have found a cute combinatorial-style proof that it is $1+\Sigma_i f_i^2$. using the sum over permutations formula for the determinant. Here $F(\sigma)$ denotes the set of fixed points of $\sigma$.

Does anyone have a faster/more elegant method?

答: The eigenvalues of $(f_if_j)$ are $(\|f\|^2, 0,\dots,0)$ so those of your matrix are $(1+\|f\|^2, 1,\dots,1)$


答:
Note that when we define the natural numbers we have a good sense of addition and multiplication (ordinal arithmetics), and from those we can define the operations on $\Bbb Z$ and $\Bbb Q$ and then by using Dedekind cuts construction we can extend these to $\Bbb R$ as well.

So we have that $\Bbb R$ has the operations $+,\cdot$ and they all satisfy all the things we know they do from the times we did mathematics without writing all the sets explicitly.

Now we can use these things to start and define anything else that we desire using the $+$ and $\cdot$ and whatnot as our stones. For example you can define $\pi$ to be the length of the semi-circle of radius $1$.

How do we do that? We define what is an integral, and a path integral, and so on. All from the sets which are addition and multiplication and so on, and then we can define $\pi$ in a painfully tedious way.

The whole point of using set theory, and in this case $\sf ZF$, as our foundation is that we can do things, once we can define the real numbers with their basic properties we have formulas which define things from that structure, and we don't have to write everything in set-form explicitly.

Once we have the real numbers (with the order) it is easy to define the collection of open intervals, and then it is easy to define the standard topology (the smallest collection containing the intervals and having certain properties), from there we can define the Borel sets, the Lebesgue sets, and the Lebesgue measure (being the unique function from the Lebesgue sets into the real numbers which satisfies certain properties), then we can define integration and with respect to the measure, and we can define derivation.

All these things end up being immensely long and complicated formulas, but the point is that we can write them up. And all this with just $\in$ and the axioms of $\sf ZF$. (Although we may want to add $\sf DC$ or even $\sf AC$ if we discuss measure theory.)

But if you do want to insist on $\pi$ being written in set form:

$$\pi = \left\{x\in\Bbb Q\mathrel{}\middle|\mathrel{} x<_\Bbb Q0\lor \left(x\geq_\Bbb Q0\land\exists k\in\Bbb N:\frac{x^2}6<_\Bbb Q\sum_{n=1}^k\frac1{n^2}\right)\right\}$$

Title: How to write $\pi$ as a set in ZF?


问:
Can $n!$ start with the digits 2020...
Can anyone give me any hints on how to tackle this? I assume the answer is no otherwise why ask the question. I am aware that $k \mid n!$
for all $k\leq n$, and I am aware that divisibility by $3$, $9$ and $11$ in particular have nice and easy tests involving sums of digits etc. I know that $2020 = 2^2 \cdot 5 \cdot 101$ written as a product of primes. However I am not seeing how these things help..

Thanks

Edit: Plot twist... $384!$ begins with the digits $2020$ so now the question is how do we prove this can happen (let's pretend we don't have computers)?

答:
Let᾽s start with some definition.

Definition. Let $x\in(0,\infty)$. The decimal part of the decadic logarithm of $x$ is called mantissa of $x$ and we denote it as
$$\text{m}(x)=\log_{10}x-\lfloor\log_{10}x\rfloor.$$

Observe first that $n!$ starts with 2020 if and only if
$$
\text{m}(n!)=\log_{10}(n!)-\lfloor\log_{10}(n!)\rfloor\in [\log_{10}2.020,\log_{10}2.021).
$$

Hence, suffices to show that the values of the sequence
$$a_n=\text{m}(n!)=\log_{10}(n!)-\lfloor\log_{10}(n!)\rfloor,\,\,n\in\mathbb N,$$
are dense in $[0,1]$.

In particular, we shall prove that for any $c,d$, with $(c,d)\subset [0,1]$, there exist an $n\in\mathbb N$, such that $a_n\in (c,d)$.

This will be achieved in the following way. Will shall try to find $k,\ell\in\mathbb N$, such that the mantissas of
$$
N_j=10^k+j, \quad j=1,\ldots,\ell
$$
are all less that $d-c$, i.e., $\text{m}(N_j)\in (0,d-c)$, and their sum exceeds 1,
i.e., $\sum_{j=1}^\ell \text{m}(N_j)>1$.

Suppose that we have found such $k$ and $\ell$. If $\ell_0<\ell$, such that $\sum_{j=1}^{\ell_0} \text{m}(N_j)\le 1$, while $\sum_{j=1}^{\ell_0+1} \text{m}(N_j)> 1$, then the numbers
$$
b_j=\sum_{i=1}^j\text{m}(N_i)=\text{m}(N_1N_2\cdots N_j), \quad j=1,\ldots,\ell_0,
$$
are distributed in $[0,1)$ is a way that
$$
0<b_1<b_2<\ldots<b_{\ell_0}\le 1<b_{\ell_0+1}
$$
and $b_j-b_{j-1}<d-c$. Hence, at least of the numbers
$$
\text{m}(n!)\in(c,d), \quad \text{for at least one of the numbers}\,\,\,n=N_1,\ldots,N_{\ell_0},
$$

The existence of suitable $k,\ell$ is provided by the following estimate:

$$
\prod_{j=1}^{\lfloor 5\sqrt{10^k}\rfloor} \left(1+\frac{j}{10^k}\right)>\sum_{j=1}^{\lfloor 5\sqrt{10^k}\rfloor}\frac{j}{10^k}\ge
\frac{1}{10^k}\cdot \frac{1}{2}(\lfloor 5\sqrt{10^k}\rfloor)^2>10.
$$
Hence, $k$ nad $\ell$ should be picked so that
$$
\log\left(1+\frac{5}{\sqrt{10^k}}\right)<d-c\quad\text{and}\quad
\ell=\lfloor 5\sqrt{10^k}\rfloor.
$$

Title: Can $n!$ start with the digits 2020...


答:
Overkill but Sylvester's Theorem tells us:
$$\det(I_n + XY) = \det(I_m+YX)$$
for $X, Y$ of sizes of $n\times m$ and $m\times n$ respectively. Then for $F = (f_1,f_2,\dots f_n):$
$$\det(I_n + F^TF) = \det(I_1+FF^T) = 1+\sum_{i=1}^n f_i^2$$

Title: Cute problem: determinant of $I_n+(f_if_j)_{i,j}$

Body:
I thought of the following little problem.

Given numbers $f_1,\dots f_n$, what is the determinant of the symmetric matrix $I_n+(f_if_j)_{i,j}$?

I have found a cute combinatorial-style proof that it is $1+\Sigma_i f_i^2$. using the sum over permutations formula for the determinant. Here $F(\sigma)$ denotes the set of fixed points of $\sigma$.

Does anyone have a faster/more elegant method?


问:
Is there some kind of $0$-countable space?
Is there some kind of $0$-countable space?

That for every $x\in X$ there exists at least continuum many neighborhood. Is anyone interested in such kind of non-first countable spaces? If so are these spaces interesting?

答:
There is a whole theory of cardinal invariants in topology, with a large literature, that studies such questions. If you’re interested in local bases, you can define for a space $X$ and a point $p \in X$, the cardinal number $\chi(p,X)$ which is the minimal cardinal number of a local base at $p$. The character $\chi(X)$ is then defined as the supremum of all $\chi(p,X)$ over all $p \in X$. Elementary facts in set theory imply that these cardinals are always well-defined.

A first countable space is exactly one where we have $\chi(X) \le \aleph_0$.

It seems that what you want to call 0-countable is a space where $\min\{\chi(x,X)\mid x \in X\} \ge \mathfrak{c}$?


问:
How to find the minimum value of cos(2x)+cos(4x)+cos(6x)+cos(8x)+...+cos(20x)
I want to find the minimum value of the series cos(2x)+cos(4x)+cos(6x)+cos(8x)+...+cos(2nx). x could be 2pit. Anyone can share a method of how to determine the minimum value of the series?

答:
Comment:

Let $t=2x$ , you get:

$s=\cos (t)+\cos(2t)+\cos(3t)+\cdot\cdot\cdot +\cos(10t)$

Now use this formula:

$$s=\cos (t)+\cos(2t)+\cos(3t)+\cdot\cdot\cdot +\cos(nt)=\frac{\sin \frac n2 t\cos\frac {n+1}2 t}{\sin (\frac t 2)}$$

So you have to find minimum of this function:

$$y=\frac{\sin(10x)\cos(11x)}{\sin(x)}$$

By plotting in Wolfram minimum of y is about -2.8.


问:
How do you prove that only two 4-regular graphs on 7 vertices exist? (Up to isomorphism)
I am wondering how to determine all graphs that are 4-regular and have 7 vertices. I just attempted to draw different graphs until I found two, then verified online that those are indeed the (only) two graphs that fit the requirement.
My two questions are, how, if there is a way, can I determine those two graphs originally without just using guess-and-check, and most importantly how do I go about proving that no other graph works?
Would I have to prove that any adjacency matrix that is 4-regular is just a permutation of the two I already have? I don't quite know where to start.

Thanks!

答: The complement of $4$-regular graph on $7$ vertices is $2$-regular. Any $2$-regular graph is sum of disjoint cycles. Since every cycle has to have at least three vertices, $3+4$ and $7$ are the only options for these cycles, and they correspond to two $4$-regular graphs.


问:
Conditions for uniqueness of the solution of a linear system of equations
QUESTION

Consider the positive integers, $N,\{T_n\}_{n=1}^N,J$, where $N>\max\{T_1,...,T_N\}$, $N>J$. $1_{a}$ denotes the $a\times 1$ vector of ones.

Consider the system of equations
$$
\begin{pmatrix}
D'Y\\
F'Y
\end{pmatrix}= \underbrace{\begin{pmatrix}
D'D & D'F\\
F'D & F'F
\end{pmatrix}}_{\equiv W}
\beta
$$
where $Y$ is an $\sum_{n=1}^N T_n\times 1$ vector; $D$ is an $\sum_{n=1}^N T_n\times N$ matrix; $F$ is an $\sum_{n=1}^N T_n\times J$ matrix; $\beta$ is an $(N+J)\times 1$ vector.

Could you help me to find some necessary (and, if possible, also sufficient) conditions such that the system has a unique solution wrto $\beta$. That is, such that the matrix $W$ is invertible?

I report below some properties of the system that I analyse.

Note: a recent question of mine (here) seems very similar. However, there is a key difference that prevents me from implementing the same answer: it is the fact that here the matrices $D$ and $F$ can be decomposed in blocks that do not have necessarily the same dimension. If you believe that the same answer can be applied, please advise on how.

SOME PROPERTIES OF THE SYSTEM

$Y$ does not have elements equal to zero. The matrices $F$ and $D$ have specific structures.

With regards to the matrix $F$:

Let me decompose $F$ in $N$ submatrices, but cutting it every $T_1,...,T_N$ rows. Then, with some abuse of notation, let me write $F$ as the collection of such $N$ submatrices, $F\equiv \{F_1,F_2,...,F_N\}$, where each submatrix $F_i$ has size $T_i\times J$.

Property 1: For each $i\in \{1,...,N\}$, the submatrix $F_i$ is such that in each row, one and only one element takes value in $(0,1)$ and all the other elements takes value zero.

Property 2: If the submatrix $F_i$ has each of its nonzero elements positioned in the same columns as the submatrix $F_k$, then it should be that $F_i=F_k$.

Construct the bipartite graph such the nodes in one side (hereafter, side 1) represent the column indices $j\in \{1,...,J\}$ and the the nodes in the other side (hereafter, side 2) represent the submatrix indices $i\in \{1,...,N\}$. Draw an edge between nodes $j\in \{1,...,J\}$ and $i\in \{1,...,N\}$ if at least one element of the $j$-th column of $F_i$ has a nonzero element.

Property 3: The aforementioned bipartite graph should be such that each node in side 1 is indirectly connected to at least another node in side 1. Further, the bipartite graph should be such that each node in side 2 is indirectly connected to at least another node in side 2. This is called a connected bipartite graph.

For example, for $N=4$, $T_1=2$, $T_2=3$, $T_3=1$, $T_4=4$, and $J=3$, we could have
$$
F\equiv \Big\{\underbrace{\begin{pmatrix}
0.3 & 0 & 0\\
0 & 0.1 & 0\\
\end{pmatrix}}_{F_1},
\underbrace{\begin{pmatrix}
0.7 & 0 & 0\\
0.2 & 0 & 0\\
0 & 0 & 0.1
\end{pmatrix}}_{F_2},
\underbrace{\begin{pmatrix}
0 & 0 & 0.5\\
\end{pmatrix}}_{F_3},
\underbrace{\begin{pmatrix}
0 & 0 & 0.6\\
0 & 0 & 0.4\\
0.2 & 0 & 0\\
0 & 0.8 & 0\\
\end{pmatrix}}_{F_4}\Big\}
$$
with bipartite graph

With regards to the matrix $D$:

As done for $F$, let me write $D\equiv \{D_1,D_2,...,D_N\}$, where each submatrix $D_i$ has size $T_i\times N$.

Property 4:  For each $i\in \{1,...,N\}$, the submatrix $D_i$ has its $i$-th column equal to $1_{T_i}$ and all the other elements equal to zero.

For example, for $N=4$, $T_1=2$, $T_2=3$, $T_3=1$, and $T_4=4$, we have
$$
D\equiv \Big\{\underbrace{\begin{pmatrix}
1 & 0 & 0 & 0\\
1 & 0 & 0& 0\\
\end{pmatrix}}_{D_1},
\underbrace{\begin{pmatrix}
0 & 1 & 0& 0\\
0 & 1 & 0& 0\\
0 & 1 & 0& 0\\
\end{pmatrix}}_{D_2},
\underbrace{\begin{pmatrix}
0 & 0 & 1& 0\\
\end{pmatrix}}_{D_3},
\underbrace{\begin{pmatrix}
0 & 0 & 0&1\\
0 & 0 & 0&1\\
0 & 0 & 0&1\\
0 & 0 & 0&1\\
\end{pmatrix}}_{D_4}\Big\}
$$

Given, the properties above, $D'D$ and $F'F$ are diagonal matrices with strictly positive diagonal elements, where

$$
\underbrace{D'D}_{N\times N}=   \begin{pmatrix}
    T_1 & & \\
    & \ddots & \\
    & & T_N
\end{pmatrix}, \underbrace{F'F}_{J\times J}=\begin{pmatrix}
    \sum_{k=1}^{\sum_{n=1}^N T_n}F(k,1)^2 & & \\
    & \ddots & \\
    & &     \sum_{k=1}^{\sum_{n=1}^N T_n}F(k,J)^2
\end{pmatrix}
$$

Further,
$$
F'D=\begin{pmatrix}
\sum_{t=1}^{T_1} F(1,t) & \sum_{t=T_1+1}^{T_1+T_2} F(1,t) & ... \\
\vdots & \vdots & \vdots \\
\sum_{t=1}^{T_1} F(J,t) & \sum_{t=T_1+1}^{T_1+T_2} F(J,t) & ... \\
\end{pmatrix}
$$

答:
As in the earlier post, $W$ is invertible if and only if $[D\ \ F]$ has a trivial nullspace.

Extend the columns of $D$ to form an orthogonal basis for $\Bbb R^k$ (where $k = \sum_n T_n$). Let $P$ denote the matrix whose columns are these additional vectors. Note that we have
$$
D'P = 0, \quad P'D = 0,
$$
and both $D'D$ and $P'P$ are diagonal. Now, note that $[D\ \ F]$ has a trivial nullspace if and only if $[D\ \ P]'[D\ \ F]$ has a trivial nullspace. Compute
$$
\pmatrix{D & P}' \pmatrix{D & F} = \pmatrix{D'D & D'F\\P'D & P'F} = \pmatrix{D'D & D'F\\0 & P'F}.
$$
We can conclude that $W$ is invertible if and only if $P'F$ has a trivial nullspace.


问:
Example of a function $f:\mathbb{R}^n\to\mathbb{R}$ that is not a subversion for all open $U$
The definition of submersion is:

Let $f:U\to\mathbb{R}^n$ a differentiable function defined in the open $U\subset\mathbb{R}^m$, if for all $x\in U$, $f'(x)$ is an surjective linear transformation then $f$ is a submersion.

I have to find a open function $f:\mathbb{R}^2\to\mathbb{R}$ of class $C^\infty$, so that $f$ is not a submersion.

I thought in the function, $f(x,y)=x^3$.

If the open $U$ contains any point of the form $(0,y)$, we will have that $f'(0,y)$ is not surjective. But if $U=\{(a,b) : a>1 \}$ we will have that $f$ will be a submersion.

I was wondering if it would be possible to construct a open function $f:\mathbb{R}^2\to\mathbb{R}$ of class $C^\infty$ so that for every open $U\subset\mathbb{R}^2$, $f$ it would not be a submersion.

For another example see this link.

答:
This is not possible, as a consequence of Sard's theorem.  In fact, given a smooth open map $f:\mathbb{R}^2\rightarrow \mathbb{R}$, and given any non-empty $U\subseteq \mathbb{R}^2$, there is an open subset $V\subseteq U$ for which $f|_V$ is a submersion.

To see this, suppose $f:\mathbb{R}^2\rightarrow \mathbb{R}$ is a smooth open map and let $U\subseteq \mathbb{R}^2$ be any open subset.

Then $f(U)$ is open, so by Sard's theorem, it contains a regular value $r\in f(U)$.  Pick $x\in U$ with $f(x) = r$.  Saying $r$ is a regular value means that $(\nabla f)(x)\neq \langle 0,0\rangle$.  Since the partial derivatives $f_x$ and $f_y$ are continuous, there is some neighborhood $V$ of $x$ (which we can take to be a subset of $U$ if we wish) for which $\|(\nabla f)(y)\| \geq \|(\nabla f)(x)\|/2  > 0$ for any $y\in V$.

Because $\nabla f$ is non-zero on $V$, $f|_V$ is a submersion.


问:
How to find the amount to added every month or year to get the required amount after certain years?
I want to do a Java application for which after giving the current savings, and the rate of interest and  and required amount after specified  no of years, it has to show how much a person has to earn a month or year to achieve his target.
I'm aware of the formula $$\text{present value}={\text{future value} \over (1+\text{interest rate})^{\text{no. of years}}}$$
Now after
getting present value, should I divide present value by no. of months in those years, to get the amount he has to invest per month ?</p>

答:
The formula you need is the future value of an ordinary annuity:

$$F=R \times \frac{(1+i)^n-1}{i}$$where F is the value at the time of the last payment, R is the size of the regular payment, i is the interest rate per payment period and n is the number of regular payments.

The compounding period of the interest rate must match the payment period...

REVISED ANSWER:
Assume you start with B dollars, and wish to accumulate a total of C dollars after n years, making annual deposits of R dollars, and earning an annual rate of i per year as you go.

The original amount will grow, all by itself, to $B'$ at the end of the payments:

$$B'=B \times (1+i)^n $$Assuming that this is less than your target of C dollars, the regular payments must make up the difference of $C-B'$.  Re-arranging the formula for the future value of an annuity: $$R=(C-B') \times \frac{i}{(1+i)^n-1}$$


问:
Homomorphism of complex unital Banach algebra is continuous

Let $A$ be a unital Banach algebra, $n$ a positive integer, and $w : A \rightarrow \mathbb{C}^{n \times n}$ be a homomorphism of complex algebras such that $w(A) = \mathbb{C}^{n \times n}$ ($\mathbb{C}^{n \times n}$ is the algebra of complex $n \times n$ matrices).

Show that $w$ is continuous.

Also deduce that any linear functional $f : A \rightarrow \mathbb{C}$ satisfying $f(xy) = f(x)f(y)$ is continuous. HINT: Use that $\mathfrak{m} \subseteq A$ is a maximal ideal if and only if the quotient algebra $A / \mathfrak{m}$ is simple.

For the first part we have to show that for any open set $U_{n \times n}$ of $\mathbb{C}^{n \times n}$ we have that the pre-image $w^{-1}(U_{n \times n})$ is open in $A$; or equivalently for any closed set $C_{n \times n}$ of $\mathbb{C}^{n \times n}$, the pre-image $w^{-1}(C_{n \times n})$ is closed in $A$. I am kind of at a loss here.

I suppose the second question, namely, showing that any linear functional $f$ is continuous then follows from the first part by taking $n = 1$? Since the condition $f(xy) = f(x)f(y)$ makes $f$ into a homomorphism?

答: I will denote $\mathbb{C}^{n\times n}$ by $M_n(\mathbb{C})$. Since $w$ is an algebra homomorphism, we have that $\ker(w)$ is an ideal in $A$ and from the first isomorphism theorem one easily sees that $A/\ker(w)\cong M_n(\mathbb{C})$. Now any matrix algebra is simple, so $A/\ker(w)$ is simple. As the hint suggests, this implies that $\ker(w)$ is a maximal ideal. But this means that $\ker(w)$ is closed, since $\overline{\ker(w)}$ is also an ideal containing $\ker(w)$ (and of course if $\overline{\ker(w)}=A$ then $w=0$, a contradiction: indeed, if $\ker(w)$ is dense in $A$, then there exists an element $z\in B(1_A,1)$ with $w(z)=0$. But then $z$ is invertible, so $w(1)=w(zz^{-1})=w(z)w(z^{-1})=0$). But if the kernel is closed then $w$ is continuous.


问:
Polya theorem unequality
I need a little hint. I have a complex function $f(z) = z^2 - 2 $. I have to show that $(x^2+y^2)^2 \leq 4(x^2 - y^2)$ for all $z$ with $|f(z)| \leq 2$.

I would start like this:

$|f(z)| \leq 2$

$|z^2 -2| \leq 2$

$|(x + iy)^2 -2| \leq 2$

$|x^2 + 2ixy - y^2 -2| \leq 2$

But I don´t know further steps. Maybe someone has an hint for me?

Title: Polya theorem unequality

Body:
Square the whole thing:

$$|x^2 + 2ixy - y^2 -2| \leq 2\implies (x^2-y^2-2)^2+4x^2y^2<4\implies$$

$$(x^2-y^2)^2-4(x^2-y^2)+4x^2y^2<0\implies (x^2+y^2)^2<4(x^2-y^2)$$


问:
Is the dihedral group $D_n$ nilpotent? solvable?
Is the dihedral group $D_n$ nilpotent? solvable?

I'm trying to solve this problem but I've been trying to apply a couple of theorems but have been unsuccessful so far. Can anyone help me?

答:
A general proof that $D_n$ is solvable for all $n \geq 1$.

Here is the normal series with abelian quotients:
$$\{e\} \trianglelefteq \langle \sigma \rangle \trianglelefteq D_n$$

$$\langle \sigma \rangle/\{e\} \cong \mathbb{Z}_n,   D_n/\langle \sigma \rangle \cong \mathbb{Z}_2$$ both obviously abelian, therefore $D_n$ is solvable.


问:
Is $\sum\limits_{k=1}^{n-1}\binom{n}{k}x^{n-k}y^k$ always even?
Is
$$ f(n,x,y)=\sum^{n-1}_{k=1}{n\choose k}x^{n-k}y^k,\qquad\qquad\forall~n>0~\text{and}~x,y\in\mathbb{Z}$$
always divisible by $2$?</p>

答: Try setting $y=-x+a$ and expand the powers.</p>


问:
Find an expression for a double integral
Given the compact set with the boundary $ \left(x-y\right)^{2}+x^{2}=a^{2}\thinspace\thinspace\thinspace,a>0 $, I have to find an expression for $ \intop dx\int f\left(x,y\right)dy $ and $ \intop dy\intop f\left(x,y\right)dx $, where $ f $ is some function (dosent really matter what exactly the function is, because it is an exercise with finding bounderies of the integral and changing order of summation.

Now, I cannot really see what exactly the geometric shape of this equation is. Is it still possible to solve and find the bounderies of the integral, just by using the equation?

Thanks in advance.

答:
$(x-y)^2 + x^2 = a^2 \implies (x-y)^2 = a^2 - x^2 \implies x-y = \pm \sqrt{a^2-x^2}\quad  (1)$

Thus,
$y = x \pm \sqrt{a^2-x^2}$. But $(1)$ imposes a constraint on $x$, $a^2-x^2 \geq 0 \implies x\in [-a,a]$.

Therefore, the $\int dx~\int f(x,y)dy$ order is given by $$-a\leq x \leq a$$ $$x - \sqrt{a^2-x^2} \leq y \leq x+ \sqrt{a^2-x^2} $$

For the alternative order, the idea is pretty much the same.


问:
Maximal solutions of a generalized differential equation for logistic growth
The ordinary differential equation  for logistic growth is given by
$$\dot{x}(t)=\lambda(t)x(t)(1-x(t))\quad\quad(1)$$
If $\lambda\colon\Bbb R\to\Bbb R$ is continuous and if the initial condition $x(0)=x_0\in(0,1)$ is imposed one may use the standard method of separating the variables to get the solution
$$x\colon\Bbb R\to\Bbb R,\,x(t)=(1+c\exp(-\Lambda(t))^{-1},$$ where $\Lambda(t):=\int_0^t \lambda(s)\mathrm{d}s$ and $c=(1-x_0)/x_0$. Thus the maximal open interval $I$ containing $0$ such that (1) with $x(0)=x_0\in(0,1)$ has a solution on $I$ is $\Bbb R$.

Now we could think of a harvesting term $\mu\colon\Bbb R\to\Bbb R$ and replacing (1) by
$$\dot{x}(t)=\lambda(t)x(t)(1-x(t))+\mu(t)\quad\quad(2)$$
Since the right hand side $\lambda x(1-x)+\mu$ is continuous and differentiable with respect to $x$ there is also in this case a maximal interval $I$ and a solution $x$ defined on $I$ such that $x(0)=x_0\in(0,1)$.

Question: Is $I$ unbounded to the right, i.e., is $[0,\infty)\subseteq I$?

答:
Meanwhile I found an equation such that maximal solution of a corresponding initial value problem is bounded from above. The problem is
$$I'(t)=I(t)(1-I(t))-\frac{1}{4}, I(0)=\frac{1}{5}$$

Its solution is
$$I(t)=\frac{1}{t-\frac{10}{3}}+\frac{1}{2}.$$
The maximal interval of definition  for this function is $(-\infty,\frac{10}{3})$.


问:
Convergence of filters in top.
Let $X$ be any set equip with the dis. topology and $x\in X$. What filters $\mathscr{F}$ converge to $x$, written $\mathscr{F}\to x$?

A filter $\mathscr{F}$ is said to converge to $x$, written $\mathscr{F}\to x$, if $\mathscr{F}_x\subseteq\mathscr{F}$, where $\mathscr{F}_x$ is the nhbd filter of $x$.

Therefore, $\mathscr{F}\to x$, if,

$$\mathscr{A}:=\{A\subseteq X:\exists U\in\tau\text{ such that } x\in U\subseteq A\}\subseteq\mathscr{F}.$$

However, I don't see where to go from here. Some help would be appreciated.

答:
If $X$ is a non-empty set then for every point $x \in X$ there is the fixed filter $\mathcal{F}_x=\{A \subseteq X\mid x \in A\}$. It's clear that this is even a maximal (by inclusion) filter, i.e. an ultrafilter.

If $(X,\tau)$ is the discrete space and $\mathcal{F}$ is a (proper) filter on $X$:

$$\forall x: (\mathcal{F} \to x \iff \mathcal{F}=\mathcal{F}_x)$$

Left to right: $\{x\}$ is a neighbourhood of $x$ in $(X,\tau)$ so if $\mathcal{F}$ converges to $x$, $\{x\}$ must be an element of $\mathcal{F}$. But then by the filter axioms all $A \in \mathcal{F}_x$, so all $A$ with $\{x\} \subseteq A$, are also in $\mathcal{F}$, hence $\mathcal{F}_x \subseteq \mathcal{F}$ and if $A \in \mathcal{F}$ it must intersect $\{x\}$ so that $x \in A$ and so $\mathcal{F} \subseteq \mathcal{F}_x$ too. Hence $\mathcal{F} = \mathcal{F}_x$.

Right to left is trivial, as $\mathcal{F}_x$ is exactly the neighbourhood filter of $x$ in $(X,\tau)$.

So precisely the fixed filters converge, and no others.

Title: Convergence of filters in top.


问:
Let $Y\sim~\text{Exp}(\lambda)$. What is $P(|Y-\frac1\lambda|>a)$ with $a>\frac1\lambda$?
I think, that I need to find the density of $|Y-\frac1\lambda|$. I also know, that for independent random variables $X,Y$, the density of $X+Y$ is $\int f(y)\cdot g(x-y)\,dy$. However I am having troubles with actually applying that.

答: Hint: $$\begin{align}P\left(\left|Y-\frac1\lambda\right|>a\right) &= P\left(Y-\frac1\lambda>a\right) + P\left(Y-\frac1\lambda<-a\right)\\&=P\left(Y>a+\frac1\lambda\right) + P\left(Y<\frac1\lambda-a\right) \end{align}$$


问:
In the triangle shown, what is the degree measure of $\angle ADB$?

In the triangle shown, what is the degree measure of $\angle ADB$

I have mainly tried angle-chasing, but it turns out that through plain angle-chasing you cannot find $\angle ADB$ or $\angle EBD$. I explored extending $AE$ and $BD$ to their intersection, but that did not help much either. Can someone get me started on how to solve this problem? I have a feeling dropping a line from $B$ to $DC$ might help, but I cannot figure out which line it would be. EDIT: I realized that dropping a line from B to DC which is parallel to EA might work.

Thanks,

Mike

答:
In these problems there are two methods to pick from: Synthetic geometry or trigonometry. Although synthetic solutions are more beautiful, they are harder to come up with. To obtain a synthetic solution you need to create isosceles triangles until you end up with an equilateral triangle that can be used to solve the problem or can be used to obtain a congruency.

A more practical theorem is Trigonometric Ceva's Theorem. (The proof of this theorem is done by applying the law of sines to triangles $\bigtriangleup$EOD, $\bigtriangleup$EOA, $\bigtriangleup$AOB, $\bigtriangleup$BOD, and multiplying all the obtained expressions.) To apply the theorem just multiply the ratios of the sines of the angles at each vertex respectively.
$$\frac{\sin(x)}{\sin(50)}×\frac{\sin(20)}{\sin(60)}×\frac{\sin(50)}{\sin(30)}×\frac{\sin(40)}{\sin(150 - x)}= 1$$
Rearranging,
$$2\sin(x)×\frac{\sin(20)}{\sin(60)}×\frac{\sin(40)}{\sin(150 - x)}= 1$$
From here, the rest is simply trigonometry which I am too tired to finish up right now.


问:
What is larger -- the set of all positive even numbers, or the set of all positive integers?
We will call the set of all positive even numbers E and the set of all positive integers N.

At first glance, it seems obvious that E is smaller than N, because for E is basically N with half of its terms taken out.  The size of E is the size of N divided by two.

You could see this as, for every item in E, two items in N could be matched (the item x and x-1).  This implies that N is twice as large as E

On second glance though, it seems less obvious.  Each item in N could be mapped with one item in E (the item x*2).

Which is larger, then?  Or are they both equal in size?  Why?

(My background in Set theory is quite extremely scant)

答:
The word 'size' doesn't have a intuitive meaning for set of infinite items.

Mathematicians defined cardinality by one-to-one correspondence(bijection), and it's generally what it means by 'size'.

If there exist a bijection between A and B, then the two sets have the same cardinality. You have shown the existence of a bijection, therefore E and N have the same cardinality.

You might mean the 'density' of N is twice as large as E. density of A(a subset of natural number) is limit of |{a<=n|a\in A}|/n as n goes to infinity.

ContentLicense: CC BY-SA 2.5


问:
Confusion between operation and relation: Clarification needed
I'm doing some old exams and found following question:

Set $S={{1,2,3}}$ is given. Provide an
  example of binary operation in set S,
  binary relation in set S and a
  function $f:S\rightarrow R$.

So, I'm thinking about $+_4$ as operation, but wouldn't it be a relation too? I can take ordered pair from $S$ and for every ordered pair associate an element form $S$ with every pair.

For the third part, I'm thinking about $y=x*1.25$ or something similar, but that part isn't so problematic for me.

I did read the Wikipedia articles, but the difference between operations  and relations isn't clear to me.

答:
A binary operation is a function from $S\times S \to S$ such as addition, multiplication or anything really.

A binary relation is just a subset of $S^2$, that is not necessarily a function and it doesn't have to include all the elements of $S$ in one way or another.

A function $f\colon S\to R$ is a relation, this time it's a subset of $S\times R$ however it satisfies a certain property, if you take some $s \in S$ then there is only a unique ordered pair with $s$ in it, so if you have $\langle s,r_1\rangle$ as well $\langle s,r_2\rangle$ then you can say that $r_1 = r_2$.


Title: Polya theorem unequality

Body:
I need a little hint. I have a complex function $f(z) = z^2 - 2 $. I have to show that $(x^2+y^2)^2 \leq 4(x^2 - y^2)$ for all $z$ with $|f(z)| \leq 2$.

I would start like this:

$|f(z)| \leq 2$

$|z^2 -2| \leq 2$

$|(x + iy)^2 -2| \leq 2$

$|x^2 + 2ixy - y^2 -2| \leq 2$

But I don´t know further steps. Maybe someone has an hint for me?

Body:
A variant:
\begin{align}
|f(z)|^2=(z^2-2)(\bar z^2-2)\le 4&\iff (z\bar z)^2-2(z^2+\bar z^2)+4\le 4 \\
& \iff  (z\bar z)^2\le 2(z^2+\bar z^2),
\end{align}
which yields explicitly
$$x^2+y^2\le 2\bigl((x^2+2ixy-y^2)+(x^2-2ixy-y^2)\bigr)=4(x^2-y^2)$$


问:
Symmetric positive definitine matrix $I-X^TX$ => $X^TX<1$
If $I-X^{\top}X$ is symmetric positive definite, why is $\lVert X \rVert < 1$?
It seems logically to me but why does it follow?

It's
$v^{ \top} ( I - X^{ \top } X ) v
=
\lVert v \rVert^2 - \lVert X v \rVert^2      > 0$....

(An attempt was $- \lVert X v \rVert^2   \geq - \lVert X \rVert^2  \cdot  \lVert v \rVert^2 $ but it doesn't help..)

答:
Suppose $X\in \mathbb{R}^{m \times n}$ and write $X^TX=Q^TDQ$ where $Q\in \mathbb{R}^{n\times n}$ is orthogonal and $D\in \mathbb{R}^{n\times n}$ is the diagonal matrix $D=\text{diag}(\lambda_1 , \ldots ,\lambda _m)$ with $\lambda _j \geq 0$ for all $1\leq j \leq n$. Since $I-X^TX=Q^T\big(I-D\big)Q$ is positive definite and $I-D=\text{diag}(1-\lambda _1 ,\ldots ,1-\lambda _n )$ we have $1-\lambda_j>0$ for all $j$. Therefore $\lambda_j \in [0,1)$ for all $j$ which means so is $M=\max_{1 \leq j \leq n}\lambda _j$.

Now choose $v\in \mathbb{R}^n, ||v||=1$ and set $w=Qv=(w, \ldots ,w_n)^T$. The orthogonality of $Q$ implies $||w||=1$ too. We get $$||Xv||^2=w^TDw=\lambda_1 w_1^2 + \dots + \lambda _n w_n^2\leq M||w||^2<1$$ Hence $||Xv||<1$ for any $v\in \mathbb{R}^n$ with $||v||=1$ and by compactness of the unit sphere in $\mathbb{R}^n$ we have the result.


问:
$s$ is integral over $R$, if and only if $R[s]$ is a finitely generated $R$-module.
I am trying to prove this lemma for a bigger proof that I am working on.

Let $R$ be a subring of the commutative ring $S$ and let $s \in S$. Then $s$ is integral over $R$, if and only if $R[s]$ is a finitely generated $R$-module.

I am able to prove that if $s$ is integral over $R$, then $R[s]$ is a finitely generated $R$-module. My problem is showing the reverse implication. Can someone please help me with the reverse implication.

Here $R$ and $S$ are rings with unity.

答: Choose a basis for $R[s]$, and observe that multiplying each basis element by $s$ yields some linear combination of basis elements. Putting these equations together, you get a singular matrix, so its determinant is $0$. This determinant exhibits $s$ as an integral element over the base ring.


问:
Integral of harmonic functions on a smooth compact set
I need to show that the integral of the derivative of a harmonic function $h$ along the surface normal $\nu$ multiplied by the function over the surface of a smooth compact set G is equal to $\int_G || \nabla h ||^2 d \lambda$. I have some progress on both ends, but I am missing a key insight on the step in the middle. Does anybody have a tip how I could show that?

$$
\begin{align*}
\int_{\partial G} h \frac{\partial h}{\partial \nu} dS
&= \int_{\partial G} h  \left\langle \nabla h, \nu\right\rangle dS \\
&= \text{some steps and the usage of the divergence theorem} \\
&= \int_{G} \sum_i \left( \frac{\partial h}{\partial x_i} \right)^2 d \lambda \\
&= \int_{G} || \nabla h ||^2 d\lambda
\end{align*}
$$

答:
By a form of the product rule, $h\nabla h = \frac{1}{2}\nabla(h^2)$, so by the divergence theorem,

\begin{align}
\int_{\partial G} h\langle \nabla h,\nu\rangle\, dS = \frac{1}{2}\int_{\partial G}\left\langle \nabla (h^2), \nu\right\rangle\, dS = \frac{1}{2}\int_G \nabla \cdot [\nabla(h^2)]\, dV.
\end{align}
Now, you just have to calculate that carefully. $\frac{1}{2}\nabla(h^2) = \sum_{i=1}^nh\frac{\partial h}{\partial x^i}\, e_i$, where $e_i$ is the $i^{th}$ basis vector. Now, taking the divergence of this yields:
\begin{align}
\frac{1}{2}\nabla^2(h^2)&=\sum_{i=1}^n\frac{\partial}{\partial x^i}\left(h\frac{\partial h}{\partial x^i}\right) = \sum_{i=1}^n\left(\frac{\partial h}{\partial x^i}\right)^2 + h\frac{\partial^2h}{(\partial x^i)^2}
\end{align}
Since $h$ is harmonic, what can you now conclude?


问:
$\mathbb{R}^n$ minus a simple closed curve is simply connected for $n\geq 4$?
I would like to prove the statement in my question, at least for the case $n=4$. Here is some context for it; in the study of the general Jordan curve theorem, one confronts with the following result (36.2 in Munkres' book, 19.4 in Bredon's, 2B.1 b) in Hatcher's)

Theorem. For an embedding $h:\mathbb{S}^r\rightarrow\mathbb{S}^n$ with $r<n$, one has that
$$\tilde H_i(\mathbb{S}^n-h(\mathbb{S}^r))\cong\begin{cases}
\mathbb{Z},&\text{ if }i=n-r-1\\
0&\text{ otherwise}
\end{cases}$$

This completely chatacterizes the homology of the space $\mathbb{S}^n-h(\mathbb{S}^r)$, but it does not say anything about its fundamental group; a very well-known example of this wild behaviour is given by Alexander's Horned sphere, for it is an embedding of $\mathbb{S}^2$ in $\mathbb{S}^3$ whose complement is not simply connected.

However, I think this kind of behaviour does not happen when $r=1$ and $n\geq 4$; in other words, the complement of any simple closed curve in $\mathbb{R}^n$ for $n\geq 4$ must be simply connected.

The inuition is clear in simple cases, but how can we prove it from the previous facts, if possible?

答:
Your intuition is false unless the embedding is smooth, in which case this is a standard result of differential topology, using the tool of transversality. It is probably sufficient to assume that the sphere has codimension at least 3 and is locally flat but I do not know anything about locally flat embeddings.

The Double suspension theorem of Cannon (expanding on work of Edwards) shows that if $P$ is a homology sphere of dimension $n$, then $\Sigma(\Sigma(P)) = \Sigma^2(P) \cong S^{n+2}$.

Just as there is a natural map $S^0 \to \Sigma P$ whose complement is homeomorphic to $P \times \Bbb R$, there is a map $h: S^1 = \Sigma(S^0) \to \Sigma^2 P$ whose complement is homeomorphic to $P \times \Bbb R^2$.

Corollary. If $G$ is the fundamental group of a homology n-sphere, then $G$ is the fundamental group of the complement of a wild embedding of $S^1$ into $S^{n+2}$.

Thus you can get, for instance, a group of order 120 (called the binary icosahedral group) as the complement of a wild 1-knot in $S^5$.

In fact work or Kervaire shows that any finitely presented superperfect group, a group whose first and second homology is zero, arises as the fundamental group of a homology n-sphere for every $n \geq 5$. Possibly $n \geq 4$: I did not check.

So the situation is more or less as bad as it can be.


问:
Trying to understand why $\gamma(x) = \langle p : p \in Irr(L), p \leq x \rangle$ is a lattice isomorphism
A lattice $L$ is called distributive if the join and meet operations distribute, i.e., for all $x,y,z \in L$ we have that

$$x \vee (y \wedge z) = (x \wedge y) \vee (x \wedge z)$$

$$x \wedge (y \vee z) = (x \vee y) \wedge (x \vee z)$$

An element $x \in L$ is called join-irreducible if it can't be written as the join of two other elements. That is, if $x = y \vee z$ then either $x = y$ or $x=z$. The subposet (not sublattice!) of $L$ consisting of all join-irreducible elements is denoted by $Irr(L)$.

Now, let $J(Irr(L))$ denote the set of order ideals of $Irr(L)$.

For a finite distributive lattice $L$, i'm trying to show that the map $\gamma: L \rightarrow J(Irr(L))$ given by $$\gamma(x) = \langle p : p \in Irr(L), p \leq x \rangle$$ is a lattice isomorphism

Can somebody help me to see what is happening here? Thank you for your time

答:
This is called the Birkhoff's representation theorem for finite distributive lattices.

I'll use the following lemma, which I'll state without proof:

Lemma. Let $L$ be a finite lattice.
Then, for all $a \in L$,
$$a = \bigvee \{ p \in Irr(L) : p \leq a \}.$$

If $a \in L$, then let us see that $\gamma(a) \in J(Irr(L))$.

Now
$$\gamma(a) = \{p \in Irr(L) : p \leq a\},$$
and if $q \in Irr(L)$ is such that $q \leq p_1$, for some $p_1 \in \gamma(a)$, then, by transitivity (since $p_1 \leq a$), we have $q \leq a$;
thus, $q \in \gamma(a)$.
We conclude that $\gamma$ is an application.

If $a \leq b$ in $L$, then, for any $p \in Irr(L)$, if $p \leq a$ then $p \leq b$, by transitivity.
Hence $\gamma(a) \subseteq \gamma(b)$.

Conversely, if $\gamma(a) \subseteq \gamma(b)$, then, by the lemma
$$a = \bigvee \gamma(a) \leq \bigvee \gamma(b) = b.$$
So we have
$$a \leq b \quad\text{ iff }\quad \gamma(a) \leq \gamma(b),$$
that is, $\gamma$ is an order-embedding.
If $\gamma$ is also onto, then it's an isomorphism.

So let $A \in J(Irr(L))$, and let us see that there exists $a \in L$ with $\gamma(a) = A$.

If $A = \varnothing$, then $A = \gamma(0)$, as it is clear.
So suppose that $A \neq \varnothing$.

$A$ is certainly finite, so we can take $A = \{p_1, \ldots, p_n\}$.
Let $a = \bigvee_{i=1}^n p_i$.

We'll see that $\gamma(a) = A$.

Indeed, if $p \in \gamma(a)$, then $p \leq p_1 \vee \cdots \vee p_n$, and, since these are join-irreducible, $p \leq p_i$, for some $i$.

Since $A$ is an order-ideal, it follows that $p \in A$.
So $\gamma(a) \subseteq A$.

Conversely, if $p \in A$, then $p \in \gamma(a)$ because $p \in Irr(L)$ and $p \leq a$.


问:
Cech homology groups
Let X and Y two compact Hausdorff spaces. If f is a homeomorphism of X into Y, then for each covering $\mathfrak{U}$ in $\Sigma(X)$, the collection $f(\mathfrak{U})$ of all images of elements of $\mathfrak{U}$  is an open covering of Y and conversely.

How can I prove that the Cech homology groups $H_{p}(X, G)$ and $H_{p}(Y, G)$ are isomorphic?

Thank you

答:
Note that $H_p(-,G):Top\to Ab$ is a functor.

Recall that homeomorphisms are isomorphisms in $Top$.

Finally prove (it is immediate) that any functor maps isomorphisms to isomorphisms.

Title: Cech homology groups


问:
If $G$ is not complete, then $\kappa(G) \ge 2\delta(G) -n + 2$
Let $G$ be a graph with $n$ vertices. Show that if $G$ isn't complete, then $\kappa(G) \ge 2\delta(G) - n + 2$, where $\delta(G)$ is the minimum degree of the vertices of $G$, and $\kappa(G)$ is the connectivity of $G$.

Why is this true, if $G$ has only one vertex for example, then isn't $\kappa(G)= 0 < 2\cdot0 -1 + 2$, showing that the statement is false. Maybe this is just an edge case though. Still, I don't know how to approach this proof.

答:
Try to prove first, that if graph on $n$ vertices is disconnected, then $\delta(G) \leqslant \frac{n-2}{2}$.

So, suppose we deleted $k$ vertices and obtained disconnected graph $H$. Then $|H|=n-k$ and $\delta(H) \geqslant \delta(G)-k$. So

$$ \delta(G)-k \leqslant \frac{n-k-2}{2}. $$

After simple transformations, we arrive at thesis.


问:
List of books in which at the end I can understand most mathematical symbols
I am so sorry if my question seems selfish. There are already so many existing questions asking for a list of books to better learn math is some way or another.

This is a list of all the math symbols:

https://www.rapidtables.com/math/symbols/Basic_Math_Symbols.html

Can anyone recommend a short list, possibly of 10-20 books (for beginners or easy to read) in which through learning their subjects, at the end I will have a basic understanding with at least most symbols?

Update: My knowledge ends with high school symbols: Basic Algebra, Basic Geometry and Basic calculus. I was very good in math in high school, but my life took another direction. I am very good at understanding theorems, but in a very procedural fashion. I would like to close the gap with better understanding of symbols.

If it helps to understand what my goal is: my goal is to read these books and be able to come back to this forum and understand most questions in this forum (not necessarily their answers). From there I would be able to further my studies in the directions I want.

What I am trying to do is to get the most knowledge of most symbols in an elementary way, Instead of diving through topics. My assumption is that it will then be easier for me to dive into a specific topic.

I know there is Khan Academy, but I am a little bit old fashioned and my plan is to buy 10 books.

答:
Maybe you are interested:
A history of mathematical notations, Florian Cajori 1928</p>


问:
Diagonalizing a matrix where the pairwise difference of eigenvalues are units
Let $R$ be a commutative ring (with $1$), and $M$ be an $n\times n$ matrix over $R$ whose characteristic polynomial is
$$\det(tI - M) = (t-\alpha_1)\cdots(t-\alpha_n)$$
with $\alpha_i \in R$ for all $i$. Assume that $\alpha_i - \alpha_j \in R^\times$ whenever $i\ne j$. Is it true that there is some invertible $P \in GL_n(R)$ with
$$M = P\begin{pmatrix}\alpha_1 & & & \\ & \alpha_2 & & \\ & & \ddots & \\ & & & \alpha_n\end{pmatrix}P^{-1}?$$

This is true if $R$ is a field. The proof here shows that we must have $R^n = \oplus_{i=1}^n \ker(M - \alpha_i I_n)$, at least in the case when $R$ is a local ring (but also it seems to work for general $R$). But would it follow that we can write $M$ in the form above? It is not clear to me that each $\ker(M - \alpha_i I_n)$ is free (if $R$ is not local), so I am not sure if we can necessarily find a basis of $R^n$ consisting of eigenvectors of $M$.

If the above is false, might there be some conditions on $R$ that would make it true, e.g. if we assume $R$ is an algebra over a field?

答: Let $A_i$ be the product of all $M-\alpha_j$ $j\neq i$. Let $K_i\subset R^n$, $K_i=A_i(R^n)$. Then $M$ acts as $\alpha_i$ on $K_i$. Next show that $\sum K_i=R^n$ using your condition $\alpha_i-\alpha_j$ are units and $K_i\cap \sum_{j\neq i} K_j=0$, so $R^n=\oplus K_i$. This gives you more or less what you want, except, the $K_i$ are projective modules of rank one, but may not be free, so not sure you will get your diagonalization globally.


问:
Finding posterior mean from prior density
Suppose the probability p of success on an experiment is given a prior with density $f_p(s)=2(1−s) \mathbb{1}_{[0,1]}(s)$. The experiment is independently conducted three times and is a success $N=2$ times.

What is the posterior mean of p?

Does anyone the steps to solving this problem?

答:
$$\pi(s|\mathbf{x})\propto (1-s)s^2(1-s)=s^2(1-s)^2$$

That is a $Beta(3;3)$ thus the posterior mean is

$$\mathbb{E}[s|\mathbf{x}]=1/2$$


问:
Show that an unrestricted $3$-coloring of a plane triangulation $G$ has an even number of $3$-colored faces
Show that an unrestricted $3$-coloring of a plane triangulation $G$ has an even number of $3$-colored faces.

Here an unrestricted $3$-coloring means that adjacent vertices can receive the same color. And a $3$-colored face is a face where its frontier has distinct colors in each vertex.

答:
I think the idea should be something like this below.
Because I've seen somewhere 25+ years ago a competitive math problem of this kind whose solution impressed me quite a lot back then.

Let $s$ be any segment. Let also

$f(s) = 1$ if the two ends/vertices of the segment are colored differently;

$f(s) = 0$ if the two ends are of the same color.

Now for each triangular face $a$ denote $G(a)$ to be the sum of $f(s_{ai})$ for its three sides $s_{ai}$ where $i=1,2,3$.

Obviously $G$ can take only the values $0, 2, 3$.

You can make a few drawings and convince yourself in this.

Now sum $G(a)$ over all triangular faces $a$ and denote this sum by $S$.

On one side: $$S = n_0 \cdot 0 + n_2 \cdot 2 + n_3 \cdot 3$$

where $n_i$ is the number of triangular faces for which $G$ takes the value $i$.

On the other side each segment takes part in exactly two triangles. So in $S$ we  in fact sum up the $f(s)$ values for all segments, but actually each segment $s$ contributes two times its own $f(s)$ to $S$.

Therefore $S$ is even.

But then $n_3$ has to be even (from the above equation).

And this is what we wanted to prove.

Side note: I don't quite understand what you mean by triangulation of the plane.

Is this infinite? If so, how can we talk about even or odd number of the 3-colored faces?

Also I see in Wikipedia that different types of triangulations can be defined.

Triangulation (geometry)

But I am sure this problem (which you state) has something to do with this idea.


问:
Solve PDE $\rho_t + c\rho_x +\rho^2 = 0$ using method of characteristics
$$
\frac{\partial\rho}{\partial t} + c\frac{\partial\rho}{\partial x} + \rho^2 = 0
$$

I am not sure how to begin this problem, I have looked up how to use the method of characteristics but can find no example where $\rho^2$ so I am unsure of how one would approach this.

答:
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{dx}{ds}=c$ , letting $x(0)=x_0$ , we have $x=cs+x_0=ct+x_0$

$\dfrac{d\rho}{ds}=-\rho^2$ , we have $\rho(x,t)=\dfrac{1}{s+f(x_0)}=\dfrac{1}{t+f(x-ct)}$


问:
A physics approach to the Jordan curve theorem.
The Jordan curve theorem states that if $f:S^1\to \mathbb R^2$ is an injective continuous function then $\mathbb R^2\setminus \text{image}(f)$ has two connected components. I want to discuss an approach to proving this theorem which is as follows.

Let us instead try to prove the following. Let $f:S^1\to S^2$ be an injective continuous function.
Then $S^2\setminus \text{image}(f)$ has two connected components.

We have $f$ depicted in the following figure in pink.

Now thinking of $S^2$ as the surface of Earth, and thinking of the iamge of $f$ as a massive string.
We endow the image of $f$ with a uniform positive charge density so that the string starts moving (since it is repelling itself). To keep the string from flying off the planet let us also switch on gravity.

This should homotope the image of $f$ into a perfect circle. Now one would need to prove that the number of components that any intermediate curve (during gravity assisted homotopy) partitions the sphere into does not change with time, whence the Jordan curve theorem (for the sphere) can be proved since the number of components induced by the final curve is clearly $2$.

I guess one should be able to define an energy function of the system and show that the system has the tendency to go towards the lowest energy configuraton. This configuration shuld be the one where the pairwsie distances of the charges is the greatest, which happens when the string is a perfect circle.
Of course, there would be many technical problems along the way.

I do not know if a proof of the Jordan curve theorem already exists along these lines. If not, then I think a proper execution of this idea should make for a publication.

EDIT:
Thanks to copperhat for pointing out a flaw in the previous post where there was no mention of an electric field.

答:
This proposal appears to be similar to the idea of the gradient flow of a knot energy, see

Simon Blatt, The gradient flow of O’Hara’s knot energies, Math. Ann. 370, No. 3-4, 993-1061 (2018). ZBL1398.53071.

(O'Hara's knot energies are certain regularizations of the energy given by the electrostatic potential.)

In his paper Blatt proves that, given a smooth knot in $S^3=R^3\cup \{\infty\}$, the gradient flow converges to critical points of the energy functional. If I understand this correctly, in the case of an unknot, the only critical points are round circles. (The trouble is that there are so many different versions of the knot energy that I cannot keep track which of these satisfy this uniqueness property.) In particular, this applies to knots contained in round spheres $S^2\subset S^3$. In view of the reflection symmetry, if you apply the gradient flow to a knot contained in a round sphere, the flow will be contained in that sphere.

This is all interesting for its own sake and for the sake of potential applications of the knot theory (say, to study DNA and protein folding), but should not be used in order to prove Jordan Curve theorem for smooth curves, since the latter has a very easy proof using transversal intersections with rays.   None of these will work for topological Jordan curves.


问:
Discontinuity points in linearly ordered space
It is a simple well known result that given a monotonic function $f:\mathbb{R}\to \mathbb{R}$, its discontinuity points can only be jump discontinuities.  My question is probably a simple one. Given an order preserving map $f:(X,\mathcal{T}_X)\to (Y,\mathcal{T}_Y)$, where these space are linearly topological ordered spaces, can we say that a discontinuity point $x_0\in X$ of $f$ is a jump discontinuity in general?
i.e., can we say that
$$ \underset{x<x_0}{\sup} f(x)< \underset{x>x_0}{\inf} f(x)? $$

Do we need to assume further that $Y$ is a complete lattice, or is this true in the general case?

答:
I'll try to write an answer following the hint of Henno Brandsma, hoping that if it is wrong someone will point to a mistake.

Attempt:

We assume that

$$ \sup_{x < x_0} f(x) = f(x_0) = \inf_{x > x_0} f(x) $$

Basic open sets in a linearly ordered topological space $Z$ are of the form

$$ L_Z(z):=\{ z'\in X: z'<z \} \quad \text{and} \quad U_Z(z):=\{ z'\in Z: z'>z \}. $$

It is enough to check continuity against basic open sets in $Y$. Hence, it will suffice to show that:

$f^{-1}\big[ L_Y(u) \big]$ contains an open set of the form $L_X(x_u)$ for $u>f(x_0)$ when $\inf_{x>x_0}f(x)=f(x_0)$ and

$f^{-1}\big[ U_Y(\ell) \big]$ contains an open set of the form $U_X(x_\ell)$ for $\ell<f(x_0)$ when $\inf_{x>x_0}f(x)=f(x_0)$.

Since $u>f(x_0)$, there exists $x_u>x_0$ such that $u>f(x_u)\geq f(x_0)$. Since $f$ is order preserving, we conclude that

$$ x_0\in L_X(x_u) \subseteq f^{-1}\big[ L_Y(u) \big] . $$

By similar arguments, for all $\ell<f(x_0)$ there exists $x_\ell<x_0$ satisfying $\ell< f(x_\ell)\leq f(x_0)$. Using again the fact that $f$ is order preserving yields

$$  x_0\in U_X(x_\ell) \subseteq f^{-1}\big[ U_Y(\ell) \big]. $$

Does this argument seem flawed to anyone? Also I use the fact that sets bounded from above  have a supremum and sets bounded from below have an infimum. Does anyone think I can omit this condition?


问:
Having trouble understanding upper and lower limits.
I am seriously having trouble understanding the meaning of upper and lower limits. Can someone give me easy-to-follow examples and explanations for the following?

Def: Let $\{s_n\}$ be a sequence of real numbers. Let $E$ be the set of numbers $x$ such that $s_{n_k} \to x$ for some subsequence $\{s_{n_k}\}$. This set $E$ contains all subsequential limits plus possibly the numbers $+\infty$ and $-\infty$.

Putting $s^* =\sup(E)$ and $s_* = \inf(E)$, we call them the upper and the lower limits. We also use the notation $$\lim_{n \to \infty} \sup (s_n) = s^*$$ $$\lim_{n \to \infty} \inf (s_n) = s_*$$

This is my understanding.

a), $\{s_n\}$ is a sequence, and its subsequence (there are $\infty$ many patterns) $\{s_{n_k}\}$ has many different limits. That's why it's possible to have a set $E$ which may contain more than one limit.

b), The limit of  $\{s_n\}$ is not necessarily the limit of all subsequence of  $\{s_n\}$.

c), The "largest" number  $x$ in $E$ is the "upper limit." The analogue of the lower limit would be the "smallest".

Now this is what I'm getting confused.

What does $\sup (s_n)$ mean? Why do we have to take the limit as $n \to \infty$ to get the supremum? Isn't $\sup (s_n)$ already the supremum of $E$?

Another thing, can't there be a subsequence of $\{s_n\}$ that has a limit point greater than that of $\{s_n\}$?

I may be asking some weird questions, but that's just because this idea still doesn't click with me.

The reason I am asking this question in the first place is, because I thought I understood it but I couldn't uderstand the following.

Consider the series $$ \frac{1}{2}+\frac{1}{3}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^3} + \cdots $$

I was working on the ratio test and I miserably failed to understand the text.

My claim is that $$a_n = \frac{3^n+2^n}{6^n}$$ so the ratio test would give me $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{1}{6} \left({\frac{3+2(\frac{2}{3})^n}{1+(\frac{2}{3})^n}}\right)$$

But the book tells me that $$\lim_{n \to \infty} \inf {\frac{a_{n+1}}{a_n}} = \lim_{n \to \infty} (\frac{2}{3})^n$$ $$\lim_{n \to \infty} \sup{\frac{a_{n+1}}{a_n}} = \lim_{n \to \infty} (\frac{3}{2})^n$$  $$\lim_{n \to \infty} \inf{\sqrt[n]{a_n}} = \lim_{n \to \infty} \sqrt[2n]{\frac{1}{3^n}}$$ $$\lim_{n \to \infty} \sup {\sqrt[n]{a_n}} = \lim_{n \to \infty} \sqrt[2n]{\frac{1}{2^n}}$$

I've spent 4 hours trying to understand this but I have absolutely no idea how and why these numbers popped up. Especially the "2n th" roots for the root test.

I am asking a lot, but it's really driving me mad and I need help :P

答:
What does $\sup(s_n)$ mean ?

In this context, I believe that they mean $\sup_{k\geq n}(s_k)$ so that the expression for the upper limit is $\lim_{n\rightarrow \infty} \sup_{k\geq n}(s_k)$. Note that you can have some numbers in your sequence larger than the upper limit. In fact, every number in the sequence can be larger than the upper limit, take the sequence $\{\frac{1}{n}\}$. Its upper limit is 0, but every number in the sequence is larger than 0. Another important property is that a sequence converges in $\mathbb{R}$ if and only if the upper and lower limits are in $\mathbb{R}$ and coincide.

Edit: The supremum of the set $\{\frac{1}{n}\}$ is 1, but the upper limit is not 1. To compute the upper limit, the idea is this. Take your sequence $\{s_n\}$ and form a new sequence, call it $\{t_n\}$ (they all may be infinite). Define $t_1$ to be the sup of $\{s_1,s_2,s_3\ldots\}$. Define $t_2$ to be the sup of $\{s_2,s_3,s_4\ldots\}$. $t_3$ to be the sup of $\{s_3,s_4,s_5\ldots\}$. and so on. Now check that $\{t_n\}$ is a decreasing sequence (you are throwing away potentially larger terms). Therefore it has a limit (maybe infinite). This limit is the upper limit.

Edit 2: Now for the lower limit. Start with the given sequence $\{s_n\}$. The lower limit is $\lim_{n\rightarrow \infty}\inf_{k\geq n}(s_k)$. Let's unravel this definition as we did for the upper limit. Define a new sequence of numbers $\{t_n\}$ (possibly including $-\infty$) as follows.  Define $t_1$ to be the inf of $\{s_1,s_2,s_3\ldots\}$. Define $t_2$ to be the inf of $\{s_2,s_3,s_4\ldots\}$. $t_3$ to be the inf of $\{s_3,s_4,s_5\ldots\}$. and so on. Now check that $\{t_n\}$ is an increasing sequence, and therefore has a limit (possibly infinite). This limit is called the lower limit.