问:
Determination of complex logarithm on spiral set
So, I was trying to understand the determination of the complex logarithm for sets 'trickier' than $\mathbb{C}$ minus a half line, and took the spiral $V=\{r\cdot e^{ir}\colon r\geq 0\}$ ( which begins at $0$ and rotates counterclockwise), and defined $S=\mathbb{C}\backslash V$ the set on which to find the logarithm determination. My first idea, since our teacher had done this in other examples, was to find a  continuous determination of the argument $\theta:S\rightarrow\mathbb{C}$ and from it take $g(z)=log(|z|)+i\theta(z)$ to be the new determination of the logarithm, but I can't see how to get neither the $\theta$ to  be continuous on $S$ nor an immediate new determination for the logarithm. Can you give me any clues on how to proceed? Thanks beforehand.

答:
Given a point $z\in S$ there is a unique $\tau\in[0,2\pi[\>$ such that $z=|z|\,e^{i\tau}$. Draw the ray from $0$ through $z$. This ray intersects the spiral $V$ in two points $a$, $b$ such that $z\in[a,b]$, and there is  an $n\geq0$ such that
$$a=(2n\pi+\tau)e^{i\tau},\qquad b=\bigl(2(n+1)\pi+\tau\bigr)e^{i\tau}\ .$$One has $|a|<|z|<|b|$, so that
$$2n\pi+\tau<|z|<2(n+1)\pi+\tau\ ,$$
or $$2n\pi<|z|-\tau<2(n+1)\pi\ .$$ This implies
$$n=\left\lfloor{|z|-\tau\over2\pi}\right\rfloor\ ,$$
so that your desired "spiral logarithm" $\ell$ can be written as
$$\ell(z)=\log|z|+\left(2\pi\left\lfloor{|z|-\tau\over2\pi}\right\rfloor+\tau\right)i\qquad(z\in S)\ .$$You can easily check that the discontinuity along the positive real axis is only apparent. In reality the function $\ell$ is analytic in $S$,
and satisfies
$$\ell'(z)={1\over z}\qquad(z\in S)\ .$$


问:
If $\text{dim}(X)< \infty$ and $\text{dim}(E)=\infty$, then $\text{dim}(E')=\infty$ and $E' \subset X'$?
Let $E=(E,\|\cdot \|_E)$ be a Banach space and $X \subset E$ a closed subspace of $E$. Then $X=(X,\|\cdot \|_E)$ is also a Banach space. Moreover, I know that, the dual of $E$ is cointaned in $X$, that is, $E' \subset X'$ (right?). And if $\text{dim}(X)< \infty$ then $\text{dim}(X')=\text{dim}(X)$. Here, $\text{dim}(X')$ and $\text{dim}(X)$ denote the dimension of $X$ and $X'$, respectively.

Question. If $\text{dim}(X)< \infty$  and $\text{dim}(E)=\infty$, then $\text{dim}(E')=\infty$ and $E' \subset X'$?

This seems strange to me, because in this case I would be saying that a space of finite dimension is contained in a space of finite dimension. Or am I thinking something wrong?

答:
The fact that you can think, via restriction, of a functional in $E'$ as a functional on $X$, doesn't mean that there is an inclusion $E'\subset X'$. This can be very easily seen in examples.

For instance let $E=\mathbb C^2$, and $X=\{(a,0):\ a\in\mathbb C\}$. Then $E'=\mathbb C^2$,  where you think of $(c,d)$ as the functional $\langle (c,d),(a,b)\rangle=ac+bd$, and $X'=\mathbb C$. So here $\dim E'=2$, $\dim X'=1$.

Even when both $X\subset E$  are infinite-dimensional, the inclusion $E'\subset X'$ often makes no sense. For instance let $E=\ell^2(\mathbb N)$, and $$X=\{f\in E:\ f(1)=0\}.$$ Now consider $\phi,\psi\in E'$ be given by
$$
\phi(f)=f(1),\qquad \psi(f)=2f(1).
$$
Both $\phi$ and $\psi$, which are distinct, are zero when restricted to $X$. In what sense do you want to see them as distinct elements in $X'$? Of course one can create an artificial inclusion (after all, in this example $E=X=E'=X'=\ell^2(\mathbb N)$), but the point is to have an inclusion that is canonical, in which functionals are considered as functionals.

For a more brutal example, let $E=\ell^\infty(\mathbb N)$ and $X=c_0$, the space of sequences that converge to zero. Then $X'=\ell^1(\mathbb N)$, which is separable, while $E'$ is not separable since $E$ isn't. So there is no possible inclusion $E'\subset X'$ as normed spaces, not even an artificial one.

ContentLicense: CC BY-SA 4.0


问:
How would this (x,y) function look like?
let $g:B=[0,1]^2 \rightarrow \mathbb{R}$ be

$g(x,y)=\begin{cases} \frac{1}{y^2} & 0 <x<y\leq 1\\
                      -\frac{1}{x^2}  & 0<y<x\leq 1 \\
                         0 &else \end{cases}$

Could someone please show me how this function would look like? I am having difficulties imaging a two dimensional piecewise function. I do not know how to draw it with a plotter

答:
Geogebra 3D formula:
a(x,y)=If(0<x ∧ x<y ∧ y≤1, (1)/(y^(2)), 1≥x ∧ x>y ∧ y>0, -((1)/(x^(2))), 0)

The input space is the blue/grey plane (more specifically, the square $[0,1] \times [0,1]$ on the grey plane), the output space is the blue axis. Your function is in 3 parts. The line $x=y$ maps to 0, the part above that line on the input plane maps to $1/y^2$ (the part above the grey plane), the part below that line on the input plane maps to $-1/x^2$ (the part below the grey plane)

Screenshots:


问:
Prove $ \sum_{n=1}^{\infty} \frac{2}{4n^2-1} = 1$
I want to prove that

$ \sum_{n=1}^{\infty} \frac{2}{4n^2-1} = 1$

My approach is the most logical one, rewrite as follows:

$ \sum_{n=1}^{\infty} \frac{2}{4n^2-1} = \sum_{n=1}^{\infty} \frac{1}{2n-1} - \sum_{n=1}^{\infty} \frac{1}{2n+1} $

But the remaining series are both divergent so I get to a sort of "$ \infty - \infty = 1 $".

答:
$$\lim_{n\to\infty}\left(\sum_{r=1}^{n} \frac{1}{2r-1} - \sum_{r=1}^{n} \frac{1}{2r+1}\right) $$
$\require{cancel}$
$$=\lim_{n\to\infty}\left(\frac11-
 \cancel{\frac13}+\cancel{\frac13}-\cancel{\frac15}+\cancel{\frac15}-\cdots+\cancel{\frac{1}{2n-1}} -  \frac{1}{2n+1}\right)$$

$$=\lim_{n\to\infty}\frac{2n}{2n+1}$$

I'll leave it to you to take the limit

This is known as the method of differences


答:
A concrete example that $X\subseteq E$ with $\text{dim}(X)<\infty$ and $X^*\subsetneq E^*$ is the following:

Take $E=l_2$ with the usual norm $||.||_2$ given by
$||x||_2=\biggl(\sum_{k=1}^{\infty}|x_k|^2\biggr)^{1/2}$, where $x=(x_k)_{k\in \mathbb{N}}$. Then for $X_m=\{x=(x_k)_{k\in \mathbb{N}}\in l_2:\ x_k=0\ \text{for all}\ k\geq m+1\}$ we have that $X_m$ is isometrically isomorphic to $\mathbb{R^m}$ hence, $X_m^*=\mathbb{R^m}$ whereas $E^*=l_2$.

Body:
Let $E=(E,\|\cdot \|_E)$ be a Banach space and $X \subset E$ a closed subspace of $E$. Then $X=(X,\|\cdot \|_E)$ is also a Banach space. Moreover, I know that, the dual of $E$ is cointaned in $X$, that is, $E' \subset X'$ (right?). And if $\text{dim}(X)< \infty$ then $\text{dim}(X')=\text{dim}(X)$. Here, $\text{dim}(X')$ and $\text{dim}(X)$ denote the dimension of $X$ and $X'$, respectively.

Question. If $\text{dim}(X)< \infty$  and $\text{dim}(E)=\infty$, then $\text{dim}(E')=\infty$ and $E' \subset X'$?

This seems strange to me, because in this case I would be saying that a space of finite dimension is contained in a space of finite dimension. Or am I thinking something wrong?


问:
relationship of polar unit vectors to rectangular
I'm looking at page 16 of Fleisch's Student's Guide to Vectors and Tensors.  The author is talking about the relationship between the unit vector in 2D rectangular vs polar coordinate systems.  They give these equations:

\begin{align}\hat{r} &= \cos(\theta)\hat{i} + \sin(\theta)\hat{j}\\
\hat{\theta} &= -\sin(\theta)\hat{i} + \cos(\theta)\hat{j}\end{align}

I'm just not getting it.  I understand how, in rectangular coordinates, $x = r \cos(\theta)$, but the unit vectors are just not computing.

答:
The symbols on the left side of those equations don't make any sense.  If you wanted to change to a new pair of coordinates $(\hat{u}, \hat{v})$ by rotating through an angle $\theta$, then you would have
$$
\left\{\begin{align}
\hat{u} &= (\cos \theta) \hat{\imath} + (\sin \theta)\hat{\jmath} \\
\hat{v} &= (-\sin \theta) \hat{\imath} + (\cos \theta)\hat{\jmath}.
\end{align}\right.
$$</p>

ContentLicense: CC BY-SA 3.0


问:
Unique solution to a system of n equations
$$ \begin{cases}
\sigma_1\left(\alpha x_1+\beta y_1\right)+\sigma_2\left(\gamma x_1+\delta y_1\right)=0&\\
\sigma_1\left(\alpha x_2+\beta y_2\right)+\sigma_2\left(\gamma x_2+\delta y_2\right)=0&\\
\vdots&\\
\sigma_1\left(\alpha x_n+\beta y_n\right)+\sigma_2\left(\gamma x_n+\delta y_n\right)=0&
\end{cases} , \ n \in \mathbb{Z}_+$$
How does one have to choose coefficients$$ \alpha, \ \beta, \gamma \ \mathrm{and} \ \delta$$ for this system of equations to have a unique solution $$ \begin{cases}
\sigma_1=0&\\
\sigma_2=0&
\end{cases}$$?

答:
COMMENT.-Put $A_i=ax_i+by_i$ and $B_i=cx_i+dy_i$ with $i=1,2,\cdots,n$. Since $\sigma_1=-\dfrac{B_i}{A_i}\sigma_2$ and $\sigma_2=-\dfrac{A_i}{B_i}\sigma_1$ because of  $\sigma_1=\sigma_2=0$ as only solution we need $A_i\ne0$ and $B_i\ne0$ for all $i$.

It follows for all $i$ the system
$$\begin{cases}ax_i+by_i=h_i\ne0 \\ cx_i+dy_i=k_i\ne0\end{cases}$$ then

$$x_i=\frac{\left|\begin{matrix}h_i&b\\k_i&d\end{matrix}\right|}{\left|\begin{matrix}a&b\\c&d\end{matrix}\right|}\hspace3cm
y_i=\frac{\left|\begin{matrix}a&h_i\\c&k_i\end{matrix}\right|}{\left|\begin{matrix}a&b\\c&d\end{matrix}\right|}$$ Since the point $P_i=(x_i,y_i)$ is unique we should have $\left|\begin{matrix}a&b\\c&d\end{matrix}\right|\ne0$ which is $$\color{red}{ad-bc\ne0}$$ as a first condition.

Another condition could be that no point $P_i$ is on the coordinate axes from which more restrictions could be deduced for the coefficients $a, b, c, d$. I leave this as an exercise for the $O.P.$ and for those who want to solve it.


问:
Prove that there exist a set $ S⊆V $ for a graph G which contains a matching M, such that there are $ frac{|E|+m}{2} $ between S and V\S
Let $ G =(V,E)$ be a graph.
Suppose that G contains a Matching M consisting of m edges. I have to prove that there exists a set $ S⊆V $ such that there are at least: $  \frac{|E|+m}{2} $ edges between S and V\S.

I know that it can be proven by determining the expectation of a suitable random variable, but if I choose S randomly among all subsets of V it is only suffice to obtain $\frac{|E|}{2}  $ edges.

Any tips how to determine such a suitable random variable?

Thank you in advance :)

答:
For every edge $vw$ of the matching, flip a coin to put either $v$ or $w$ in $S$.

For all vertices not in the matching, proceed as usual and flip a coin to decide if they're in $S$ or not.


问:
Is there a categorification of (infinite) ordinal arithmetic?

Background for the curious reader:

An ordinal $\beta$ is a transitive set in the sense that $\alpha\in\beta$ implies $\alpha\subset\beta$. Any ordinal is naturally well-ordered under $\in$ (so any subset of it has a least element), and any well-order is isomorphic to an ordinal. In fact, any class (naively, collection) of ordinals is itself well-ordered under the relation $\in$. This fact allows for the usage of transfinite induction and transfinite recursion.

We have three types of ordinals: the empty set $0=\emptyset=\{\; \}$, successor ordinals $S(\alpha)=\alpha\cup\{\alpha\}$ where $\alpha$ is an ordinal, and limit ordinals, which are all the other ones. Finite ordinals are either $0$ or successors, the set $\omega=\{0,1,2,\dots\}$ is a limit ordinal. A limit ordinal $\alpha$ has the property that $\alpha=\sup_{\beta<\alpha}\{\beta\}=\bigcup_{\beta<\alpha}\beta$.

My question

Ordinal arithmetic can be defined recursively as follows:

$\alpha+0=\alpha$, $\alpha+S(\beta)=S(\alpha+\beta)$, $\alpha+\sup_{\gamma<\beta}\{\gamma\}=\sup_{\gamma<\beta}\{\alpha+\gamma\}$;

$\alpha\cdot0=\alpha$, $\alpha\cdot S(\beta)=\alpha\cdot\beta+\alpha$, $\alpha\cdot\sup_{\gamma<\beta}\{\gamma\}=\sup_{\gamma<\beta}\{\alpha\cdot\gamma\}$;

$\alpha^0=1$, $\alpha^{S(\beta)}=\alpha^\beta\cdot\alpha$, $\alpha^{\sup_{\gamma<\beta}\{\gamma\}}=\sup_{\gamma<\beta}\{\alpha^\gamma\}$.

Alternatively, one can define:

$\alpha+\beta$ is the unique ordinal isomorphic to the disjoint union $\{0\}\times\alpha\cup\{1\}\times\beta$ given the lexicographic order.

$\alpha\cdot\beta$ is the unique ordinal isomorphic to the Cartesian product $\beta\times\alpha$ given the lexicographic order.

As the disjoint union and Cartesian product are simply the categorical coproduct and the categorical product, I wonder if there is some way to actually categorify these alternate definitions. Additionally, I am not aware of any non-recursive version of exponentiation, so I would be curious if a categorical formulation of addition and product of ordinals also allows for a categorical (hence non-recursive) formulation of exponentiation.

答:
Your question is quite vague, therefore I don't know if this is what you want. But I think this is a neat "all-in-one" description of ordinal operations, which in fact come down to operations on well-orderings.

The category of well-orderings has no coproducts (basically because there is no way how to decide the relations between the summands). But there are "directed coproducts" (I don't know if this construction has a canonical name):

Let $S$ be a well-ordering. For every $s \in S$ let $X_s$ be a well-ordering. Consider their underlying sets and take their disjoint union $X = \coprod_{s \in S} X_s = \bigcup_{s \in S} \{s\} \times X_s$. Endow this set with the following well-order: We have $(s,x) < (s',x')$ when $s < s'$ or ($s=s'$ and $x < x'$ in $X_s$). We may write
$$X = \coprod\limits_{s \in S}^{\longrightarrow} X_s$$
for this well-ordering in order to indicate the direction induced by $S$. It is not a colimit in the usual sense (but perhaps the category theorists can tell you if it is a weighted colimit?). But it enjoys the obvious universal property within the category of well-orderings and increasing maps:  Morphisms $X \to Y$ correspond 1:1 to families of morphisms $f_s : X_s \to Y$ that satisfy the property

$\forall s,s' \in S (s < s' \Rightarrow \forall x \in X_s \forall x' \in X_{s'} \,(f_s(x) < f_{s'}(x'))).$

Here are two examples:

If $S=\{0<1\}$, then we have just two well-orderings $X_0,X_1$, and we get $X_0 \stackrel{\longrightarrow}{\sqcup} X_1$. This categorifies the ordinal sum.

If $X_s=Y$ is constant, then we get $S \times Y$ with the lexicographic well-ordering. This categorifies the ordinal product.

Similarily, one can define and construct "directed products", which categorifies ordinal exponentiation.


问:
Is the identity $a^n \equiv a^{n \mod m} \mod m$ true if m is a power of 10
I stumbled on this code, which finds the last digits of a tetration: https://blog.dreamshire.com/project-euler-188-solution/.
The code seems to imply that $a^n \equiv a^{n \mod m} \mod m$ and takes advantage of this property to avoid calulation very large numbers.
This implies that if you are working with powers (mod 10^m) you only need to care about the last m digits of the exponent.

I tried a few examples with pythons pow() and it seems to hold, but i cant find a proof.
*with m > 1.

答:
Suppose $\,p = 2^{\large 2^{\Large k}}\!\!+1\,$ is a (Fermat) prime, e.g. $\,p=5\ (k=1)\,$ for the OP.

Then $\ a^{\large (2p)^{\large n}}\!\equiv 1\pmod{(2p)^n}\,$ if $\,\color{#0a0}{n\ge 2^k}\! = \log_2(p\!-\!1)\,$ and $\,(a,2p)=1\ $  since

$\!\bmod 2^n$ it's true by $\,\phi(2^n)\:=\: 2^{n-1}\mid\, (2p)^n,\,$ by Euler $\phi\,$ & $ $ MOR = modular order reduction.

$\!\bmod p^n$ it's true by $\,\phi(p^n)\! =\! 2^{2^{\large k}}\!p^{n-1}\mid (2p)^n\,$ by $\,\color{#0a0}{n\ge 2^k}\,$ with same justification as above.

Thus it's true mod their lcm $= (2p)^n$ by CCRT, so for $\,m\! =\! (2p)^n\,$ we have by MOR

$$\bbox[12px,border:1px solid #c00]{\bmod m\!:\,\ a^{m}\equiv 1\,\Rightarrow\, a^N\!\equiv a^{N\bmod m}}\qquad\qquad$$

Remark $ $ More generally see Carmichael's generalization of Euler's Theorem.


问:
Proving a necessary condition for an egyptian fraction with odd denominators
In the question posted in  Proving an equality involving cyclic sums, I realized that all the possible solutions to the following egyptian fraction with denominators from a set of odd positive integers $S=\{x_1,x_2,x_3,x_4\}$ $$1 = \frac{2}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}$$ have the following property:

Property

There is one $x_j\in S$ such that for every $x_i\in S$, $$\gcd(x_j,x_i)=x_i$$

I copy from the question mentioned the possible solutions of the above egyptian fraction, which have motivated my  present question, for you to check:

$$[3, 5, 9, 45]
[3, 5, 15, 15]
[3, 7, 7, 21]
[3, 9, 9, 9]
[3, 5, 5, 15]
[5, 5, 5, 5]$$

However, I am stuck when looking for a proof of this property as a necessary condition for every solution, other than the fact that the solutions listed are all the possible solutions. I am able to prove that every $x_k$ divides $\prod _{i\neq k} x_i$, but nothing more. Any help on this proof would be welcomed!

Body:
Too long to comment.

The solutions listed are not all the possible solutions.

In the following, I'm going to prove that all the possible solutions satisfying $x_2\le x_3\le x_4$ are

$$(x_1,x_2,x_3,x_4)=\color{red}{(9,3,3,9)},\color{red}{(15,3,3,5)},\color{red}{(7,3,3,21)},(3, 5, 9, 45),$$
$$(3, 5, 15, 15),(3, 7, 7, 21),(3, 9, 9, 9),(5, 3, 5, 15),(5, 5, 5, 5).$$

(The solutions in red are not listed in the question.)

Proof :

We have
$$3\le x_1,\quad (x_1,x_2)\not=(3,3),\quad 3\le x_2\le x_3\le x_4$$
and
$$1 = \frac{2}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}\le\frac{5}{\min(x_1,x_2)}\implies \min(x_1,x_2)\le 5$$

Now, let us separate it into three cases :

Case 1 : $x_1=x_2=5$

Suppose that $x_3\ge 7$. Then, we have
$$1=\frac 25+\frac{1}{5}+\frac{1}{x_3}+\frac{1}{x_4}\le\frac 25+\frac{1}{5}+\frac{1}{7}+\frac 17=\frac{31}{35}$$
which is impossible. So, we get $x_3=5$ and $x_4=5$.

Case 2 : $x_2\lt x_1$

Suppose that $x_2=5$. Then, we have
$$1 = \frac{2}{x_1}+\frac{1}{5}+\frac{1}{x_3}+\frac{1}{x_4}\le \frac{2}{7}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{31}{35}$$which is impossible. So, we get $x_2=3$.

Suppose that $x_1\ge 7$ and $x_3\ge 7$. Then, we have
$$1 = \frac{2}{x_1}+\frac 13+\frac{1}{x_3}+\frac{1}{x_4}\le \frac{2}{7}+\frac 13+\frac{1}{7}+\frac{1}{7}=\frac{19}{21}$$
which is impossible. So, we get either $x_1=5$ or $3\le x_3\le 5$.

If $x_1=5$, then we have$$1=\frac 25+\frac 13+\frac{1}{x_3}+\frac{1}{x_4}\implies \frac{4}{15}=\frac{1}{x_3}+\frac{1}{x_4}$$Multiplying the both sides by $4\times 15\times x_3x_4$ gives$$16x_3x_4-60x_3-60x_4=0$$Adding $15^2$ to the both sides gives $$ (4x_3-15)(4x_4-15)=225$$Since $4x_3-15\equiv 4x_4-15\equiv 5\pmod 8$ and $-3\le 4x_3-15\le 4x_4-15$, we get$$(4x_3-15,4x_4-15)=(5,45)\implies (x_3,x_4)=(5,15)$$

If $x_3=3$, then we have$$1 = \frac{2}{x_1}+\frac 13+\frac{1}{3}+\frac{1}{x_4}\implies (x_1-6)(x_4-3)=18$$Since $x_1-6$ is odd with $x_1-6\ge -1$ and $x_4-3\ge 2$, we have
$$(x_1-6,x_4-3)=(1,18),(3,6),(9,2)\implies (x_1,x_4)=(7,21),(9,9),(15,5)$$

If $x_3=5$, then we have$$1 = \frac{2}{x_1}+\frac 13+\frac{1}{5}+\frac{1}{x_4}\implies (7x_1-30)(7x_4-15)=450$$Since $7x_1-30\equiv 5\pmod{14}, 7x_4-15\equiv 6\pmod{14}, 7x_1-30\ge 5$ and $7x_4-15\ge 20$, we get $$(7x_1-30,7x_4-15)=(5,90)
\implies (x_1,x_4)=(5,15)$$

Case 3 : $x_1\lt x_2$

Suppose that $x_1=5$. Then, we have
$$1 = \frac{2}{5}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}\le \frac{2}{5}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}=\frac{29}{35}$$which is impossible. So, we get $x_1=3$.

Suppose that $x_2\ge 11$. Then, we have
$$1 = \frac{2}{3}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}\le \frac{2}{3}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}=\frac{31}{33}$$which is impossible. So, we get $x_2\le 9$.

If $x_2=5$, then we have$$1=\frac 23+\frac 15+\frac{1}{x_3}+\frac{1}{x_4}\implies (2x_3-15)(2x_4-15)=225$$Since $2x_3-15\equiv 2x_4-15\equiv 3\pmod 4$ and $-5\le 2x_3-15\le 2x_4-15$, we get $$(2x_3-15,2x_4-15)=(3,75),(15,15)\implies (x_3,x_4)=(9,45),(15,15)$$

If $x_2=7$, then we have$$1=\frac 23+\frac 17+\frac{1}{x_3}+\frac{1}{x_4}\implies (4x_3-21)(4x_4-21)=441$$Since $4x_3-21\equiv 4x_4-21\equiv 7\pmod 8$ and $7\le 4x_3-21\le 4x_4-21$, we get $$(4x_3-21,4x_4-21)=(7,63)\implies (x_3,x_4)=(7,21)$$

If $x_2=9$, then we have $$1=\frac 23+\frac 19+\frac{1}{x_3}+\frac{1}{x_4}\implies (2x_3-9)(2x_4-9)=81$$Since $2x_3-9\equiv 2x_4-9\equiv 1\pmod 4$ and $9\le 2x_3-9\le 2x_4-9$, we get $$(2x_3-9,2x_4-9)=(9,9)\implies (x_3,x_4)=(9,9).\quad\blacksquare$$


问:
Why is Banach-Alaoglu theorem so important?

According to Lawrence Narici and Edward Beckenstein, the Alaoglu
theorem is a "very important result - maybe the most important fact
about the weak-* topology - [that] echos throughout functional
analysis." (Source: Wikipedia)

It is a well-known fact (by Riesz) that the compactness of the unit ball with respect to the norm topology characterizes finite dimensional vector spaces. In a infinite dimensional setting, Banach-Alaoglu recovers the compactness of the unit ball in the weak*-topology which seems to come to relief of a lot of analyst.
I have come across a thread recently about the importance of Hahn-Banach which I found very illuminating. I would be interested about the different takes people have on Banach-Alaoglu.
Why is it so important? What if we were not to recover the compactness of the unit ball?

答:
I recall that during my undergraduate degree being told that compactness was important but we'll only realise why later.

It's important in the same way that finiteness is important in arithmetic. We can't add infinite numbers (well, until Cantor came along), but we can add finite numbers.

In the same way, compactness is a replacement for finiteness in the topological context: from any open covering we can extract a finite cover. It's best to see this in action in a major theorem such as integrating over a manifold where we use a partition of unity.


问:
Misconception about evaluating limits
Question

Find the limit of $$\lim\limits_{x\to\infty}\ \left[\frac 1 3 (3^{\frac 1 x} + 8^{\frac 1 x} + 9^{\frac 1 x})\right]^x\ .$$

My working

I know that the limit is $6$, which can be found through exponentiation. However, I would like to know why the following method (which I tried initially) gives the wrong answer of $1$. So, I thought that, since $x \rightarrow \infty\ $, $\frac 1 x \rightarrow 0\ $. Thus, $\lim\limits_{x\to\infty}\ \left[\frac 1 3 (3^{\frac 1 x} + 8^{\frac 1 x} + 9^{\frac 1 x})\right]^x = [\frac 1 3 (3^0 + 8^0 + 9^0)]^{\infty} = 1^\infty = 1\ $.

Any explanations as to why my original method is incorrect will be greatly appreciated :)

Edit

Thanks to some helpful comments/answers, I now know that $1^\infty$ is actually not $1$ but indeterminate!

答:
hint

$ 1^\infty $ is an inderminate form.

It is equivalent to compute

$$\lim_{x\to 0^+}(\frac 13(3^x+8^x+9^x))^{\frac 1x}$$

or, after taking logarithm,

$$\lim_{x\to 0^+}\frac 1x\ln(1+\frac 13(3^x-1+8^x-1+9^x-1))$$

$$=\lim_{x\to 0}\frac{1}{3x}(3^x-1+8^x-1+9^x-1)$$

Now, use the fact that

$$\lim_{x\to 0^+}\frac{a^x-1}{x}=\ln(a)$$

The result you will find is $e^{\ln(2.3)}=6.$

Title: Misconception about evaluating limits


问:
Generalizing Topological Spaces
I was studying topological spaces having studied metric spaces. I guess the properties of open sets in metric space led into (in terms of open set) definition of Topological spaces. Although there are many equivalent definitions such as Kuratowski closure axioms, the most used definition is in terms of open sets.

I know that we can extend these topological spaces into vector spaces or any other algebraic structures by defining appropriate operations in the given topological space. I'm not aware of any generalization of topological spaces which further broadens the definition of topological spaces.

For example, if we replace union and intersection by any other two operations (let's call it op1 and op2) such that arbitrary op1 and finite op2 are closed in the given class of subsets of a set and conventionally we can call them as open$^*$ sets(I'm calling them open$^*$ to distinguish from the open sets in the normal sense). This idea is itself is worth reflecting because if we consider op1 as intersection and op2 as union then closed sets(in the normal sense) are open$^*$ sets and open sets(in the normal sense) are closed$^*$ sets. If we consider some other operations then the properties will differ entirely. In general, it may not lead us into metric spaces or may not have pleasant properties which always is the aim of mathematicians.

Does this kind of generalization of topological space exist? Did anyone work in this direction?

答:
There are plenty of generalisations of topological spaces:

For example, NLab have pages on proximity spaces, pseudo-topological spaces, pretopologies and also Locales and Topoi. And the latter generalise to n-Topoi upto n=infinity.

There are also uniform spaces which generalise the notion of a space where uniform convergence applies; whilst Cauchy spaces generalise those spaces where completeness applies.

Title: Generalizing Topological Spaces


问:
Corollary: Let X be a connected and completely regular space with more than one point, then $| X | \geq | \mathbb R |$.
Demonstration: Let $x, y  \in  X$ be different. Because $X$ is completely regular, there is $f: X  \to  [0,1]$ continuous such that $f (x) = 0$ and $f (y) = 1$. Since $X$ is connected, $f [X]$ is an interval in $\Bbb R$, but $1,0  \in  f [X]$, so $f [X] = [0,1]$.

I found this corollary, but I am unable to understand the proof. Can anyone help me with this? Thanks in advance.

答:
Suppose there is some $a \in (0,1)$ that is not in $f[X]$. Then $X = f^{-1}[[0,a)] \cup f^{-1}[(a,1]]$ (if $x \in X$, then $f(x) \neq a$ by assumption so either $x$ is in the left hand or the right hand set). Both of these sets are open by continuity of $f$ and they are disjoint and non-empty ($x$, which maps to $0$ is in the left set, $y$ in the right). These facts combined imply that $X$ is disconnected which is a contradiction. It follows that $f[X]= [0,1]$.

The argument as given is summarising this argument by using two (presumably earlier) theorems: the continuous image of a connected set is connected, and a connected subset of $\Bbb R$ (or any space in an order topology) is order convex.


问:
Show that if some nontrivial linear combination of vectors $\vec{u}$ and $\vec{v}$ is $\vec{0}$, then $\vec{u}$ and $\vec{v}$ are parallel.
I've never been that great at writing proofs, but I'm getting a bit better. I think I have the answer correct, but I don't know if I'm missing anything. My logic seems right but there may be some minute detail that I'm leaving out. Can anybody give any feedback on this? Thanks.

$\vec{0}$ being a nontrivial linear combination of $\vec{u}$ and $\vec{v}$ implies that there exists a non-zero $a$ or $b$ such that $a\vec{u}=-b\vec{v}$. Without loss of generality, assume $a\neq 0$. Then divide by $a$ and the equality holds: $\vec{u}=-\frac{b}{a}\vec{v}$. And since $-\frac{b}{a}\vec{v}$ is a scalar multiple of $\vec{u}$, it remains that $\vec{u}$ and $\vec{v}$ are parallel.

More rigorous proof:
\begin{align*}
\vec{0}=a\vec{u}+b\vec{v}&\Longrightarrow a\neq 0\vee b\neq 0&&\text{Given}\\
&\Longrightarrow a\vec{u}=-b\vec{v}\\
&\Longrightarrow \vec{u}=-\frac{b}{a}\vec{v}&&\text{WLOG assume $a\neq 0$}\\
&\Longrightarrow \vec{u}\text{ and }\vec{v}\text{ are parallel.}
\end{align*}

答:
Almost perfect.

But, we don't have $a\ne 0$ and $b\ne 0$, only $a\ne 0$ or $b\ne 0$. So, one side might be $0$, but then the other is again parallel to it.


问:
A sufficient condition for $\omega$-categoricity
I am trying to prove that if a complete theory $T$ on a countable language $L$ has the following property:

$$\forall n<\omega \exists M\models T\quad\quad\text{s.t.}\ M\ \text{realizes only a finite number of complete n-types}\ S_n(T)$$
then $T$ is $\omega$-categorical.

What I am trying to show is that, if there was a non isolated type $p^*(x)$ for some $x$ of some length $n$ the for sure:
$$p^*(x)\longrightarrow\bigvee_{q_k\in S_n(T),q_k\neq p_1,...,p_n} \  q_k(x)$$
where $p_1,...,p_n$ are the types realised, and hence isolated by $M$.

Then one can find a formula $\varphi(x)$ such that
$$p^*(x)\wedge\varphi(x)\longrightarrow\  q_m(x)$$
for some $m$ indexing the disjunction, as shown here An excercise about omitting types theorem

Anyway I cannot reach a contradiction.

I aslo tried to reason with orbits but with no sucess.

Any hint or help would be much appreciated.

答:
I assume you know the Ryll-Nardzewski theorem, at least in the form which states that for a countable and complete theory $T$ with infinite models, $T$ is $\aleph_0$-categorical if and only if $S_n(T)$ is finite for all $n$.

What we know is that for all $n$, there is a model $M$ which only realizes finitely many types in $S_n(T)$.

Let's fix an arbitrary $n$ and show that $S_n(T)$ is finite. Let $M$ be a model of $T$ which only realizes finitely many complete types in $S_n(T)$, say $p_1,\dots,p_k$. For each $1\leq i \leq k$, let $\varphi_i$ be a formula in $p_i$ such that $\lnot \varphi_i\in p_j$ for all $j\neq i$. You can find $\varphi_i$ by picking, for each $j\neq i$, some formula $\psi_j\in p_i$ such that $\lnot \psi_j\in p_j$, and defining $\varphi_i = \bigwedge_{j\neq i} \psi_j$.

Now every $n$-tuple from $M$ realizes one of the types $p_i$, so $$M\models \forall x_1\dots\forall x_n\, \bigvee_{i=1}^k \varphi_i.$$
And for each $i$, if an $n$-tuple from $M$ satisfies $\varphi_i$, then the tuple must realize $p_i$, so for all $\theta\in p_i$, $$M\models \forall x_1\dots\forall x_n\, (\varphi_i\rightarrow \theta).$$

Now $T$ is complete, so any sentence satisfied in any model is actually satisfied in every model. So we have
$$T\models \forall x_1\dots\forall x_n\, \bigvee_{i=1}^k \varphi_i.$$
And for each $i$ and each $\theta\in p_i$, $$T\models \forall x_1\dots\forall x_n\, (\varphi_i\rightarrow \theta).$$

This shows that every model of $T$ only realizes the types $p_1,\dots,p_k$, since any $n$-tuple from any model of $T$ must satisfy $\varphi_i$ for some $1\leq i \leq k$, and thus must realize $p_i$. So in fact $S_n(T) = \{p_1,\dots,p_k\}$ is finite.

Since $n$ was arbitrary, $S_n(T)$ is finite for all $n$, so $T$ is $\aleph_0$-categorical.


问:
Difficulty understanding Measure Proof
I'm reading through Bass' "Real Analysis for Graduate Students" and came across the following leap in a proof for a measure $\mu$.

Theorem: $\text { If } A_{i} \in \mathcal{A} \text { and } A=\cup_{i=1}^{\infty} A_{i}, \text { then } \mu(A) \leq \sum_{i=1}^{\infty} \mu\left(A_{i}\right)$.

Beginning of proof: $\text { Let } B_{1}=A_{1}, B_{2}=A_{2}-A_{1}, B_{3}=A_{3}-\left(A_{1} \cup A_{2}\right), \dots$. The $B_i$ are pairwise disjoint, $B_{i} \subset A_{i} \text { for each } i, \cup_{i=1}^{n} B_{i}=\cup_{i=1}^{n} A_{i} \text { for each } n$ and $\cup_{i=1}^{\infty} B_{i}=\cup_{i=1}^{\infty} A_{i}$.

My question is - how does the author make the leap from showing $\cup A_i = \cup B_i$ for the finite $n$ case to the infinite case?

答: By $B_i\subset A_i$, we know that $\bigcup_{i=1}^\infty B_i\subset\bigcup_{i=1}^\infty A_i$. It then suffices to show that $\bigcup_{i=1}^\infty A_i\subset\bigcup_{i=1}^\infty B_i$. Now, if $x\in\bigcup_{i=1}^\infty A_i$, let $i^*$ be least such that $x\in A_{i^*}$ (exists by the well-ordering of the natural numbers). Then, $x\in A_{i^*}-\left(\bigcup_{i=1}^{i^*-1}A_i\right)=B_{i^*}\subset\bigcup_{i=1}^\infty B_i$.

Title: Difficulty understanding Measure Proof


问:
How do you find the number of solutions to $|(2x-\pi)\csc2x|=2$?
What are the solutions for the equation $|(2x-\pi)\csc2x|=2$?

My approach was to draw the graph of $|(2x-\pi)\csc2x|$ but it seemed to be too complicated

答:
$$\left| (2 x-\pi ) \csc 2x\right| =2$$
is verified when
$$(2 x-\pi ) \csc 2x=2\text{ or }(2 x-\pi ) \csc 2x=-2$$
This can be rewritten as
$$x-\frac{\pi}{2}=\sin 2x\text{ or }\frac{\pi}{2}-x=\sin 2x$$
The first equation has the solution $\frac{\pi}{2}$ discarded as $\csc 2x$ fail to exist at that value.

The second equation can be graphed as below and the equation has only two solutions $x_1\approx 0.623049, x_2\approx 2.51854$.


问:
How to see that $(p\rightarrow q)\leftrightarrow(\lnot q\rightarrow\lnot p)$ is a tautology, without using truth table?

Consider:
Statement-I: $(p\land\lnot q)\land(\lnot p\land q)$ is a fallacy.
Statement-II: $(p\rightarrow q)\leftrightarrow(\lnot q\rightarrow\lnot p)$ is a tautology. Which of these statements is true? If both are true then is statement-II a correct explanation of statement-I?

My attempt:

Statement-I: $(pq')(p'q)=0$. So, true.

Statement-II: $(p'+q)(q+p')+(pq')(q'p)$ [using the rules $p\rightarrow q=p'+q$ and $p\leftrightarrow q=pq+p'q'$]

So, statement-II becomes $p'q+q+qp'+pq'=p'q+q+pq'=q+pq'$, which is not a tautology. But the answer key says it is a tautology.

What's my mistake here?

答:
$$(p\implies q) \iff  (\neg q \implies \neg p)$$

$$((p\implies q) \implies  (\neg q \implies \neg p)) \land ((\neg q \implies \neg p)\implies (p\implies q))$$

$$((\neg p \lor q) \implies  (q \lor \neg p)) \land ((q \lor \neg p)\implies (\neg p\lor q))$$

$$(\neg(\neg p \lor q) \lor  (q \lor \neg p)) \land (\neg(q \lor \neg p)\lor (\neg p\lor q))$$

$$(\neg(\neg p \lor q) \lor  (q \lor \neg p)) $$

$$\text {tautology}$$


答: Note that $\lim\limits_{x\to \infty}a^x$ is $\infty$ when $a >1$; $1$ when $a=1$ and $0$ when $0<a<1$. In your question, $f(y)=\frac{1}{3}(3^y+8^y+9^y)$ (take $y=\frac{1}{x}$), $f(y)>1$ till $y \neq 0$. Hence, you can't just put $0$ in the expression.

Title: Misconception about evaluating limits

Body:
Question

Find the limit of $$\lim\limits_{x\to\infty}\ \left[\frac 1 3 (3^{\frac 1 x} + 8^{\frac 1 x} + 9^{\frac 1 x})\right]^x\ .$$

My working

I know that the limit is $6$, which can be found through exponentiation. However, I would like to know why the following method (which I tried initially) gives the wrong answer of $1$. So, I thought that, since $x \rightarrow \infty\ $, $\frac 1 x \rightarrow 0\ $. Thus, $\lim\limits_{x\to\infty}\ \left[\frac 1 3 (3^{\frac 1 x} + 8^{\frac 1 x} + 9^{\frac 1 x})\right]^x = [\frac 1 3 (3^0 + 8^0 + 9^0)]^{\infty} = 1^\infty = 1\ $.

Any explanations as to why my original method is incorrect will be greatly appreciated :)

Edit

Thanks to some helpful comments/answers, I now know that $1^\infty$ is actually not $1$ but indeterminate!


问:
Geometrically construct conic through four points and a tangent
I'm working on a class project about conic surfaces, and I'm reading the book: "History of the conic sections and quadric surfaces" by Julian Lowell Coolidge, and while talking about Newton, it mentions how to get a conic from four points and a tangent, with the following argument, but I can't manage to understand and apply it.
Here's the book explanation.

Does somebody has a complete explanation on the topic? Can you help me?

Thanks a lot!

答:
You can see below the figure provided by Newton himself (Book I of Principia, Sec. V Prop. XXIII - Case 2). We want to construct the conic passing through points $CPBD$ and tangent to line $HI$. To this end we need to find tangency point $A$ on line $HI$.

Suppose lines $CP$, $BD$ meet at $G$ and intersect the tangent at $I$, $H$ respectively. Construct the line through $H$ parallel to $PC$ and suppose it intersects the conic at $X$ and $Y$.

We can now use what is called "Newton's product theorem" in your book: "the ratio of the products of the distances from a point to two pairs of points of the curve, lying in given directions therefrom, is independent of the point chosen" (note that Newton doesn't mention this theorem, he just writes "by the properties of the conic sections").
If we consider lines $ICP$, $IAH$ from $I$ and their parallels $HYX$, $HAI$ from $H$, we obtain then:
$$
{IA^2\over IC\cdot IP}={HA^2\over HY\cdot HX},
$$
that is:
$$
{HA^2\over IA^2}={HY\cdot HX\over IC\cdot IP}
={HY\cdot HX\over HD\cdot HB}\ {HD\cdot HB\over IC\cdot IP}.
$$
But, by the same theorem:
$$
{HY\cdot HX\over HD\cdot HB}={GP\cdot GC\over GB\cdot GD}
$$
and substituting this into the previous equality we finally get:
$$
{HA^2\over IA^2}
={GP\cdot GC\over GB\cdot GD}\ {HD\cdot HB\over IC\cdot IP}.
$$
From there, the position of $A$ can be computed: note that two solutions are possible, depending on whether point $A$ lies inside or outside segment $HI$.

The recourse to points $X$, $Y$ is a dirty trick, as we have no guarantee that line $HXY$ will intersect the curve. However, Newton knew very well that the relation thus found for $HA/IA$ would retain its validity in any case.

Once tangency point $A$ has been found, the conic can be constructed (see Case 1 of the same proposition in Newton's Principia).


答: It allows you to cook up a subsequences almost anywhere and it very often turns out to be nice to have such a subsequence. For example in probability theory, since the set of probability measures(since they have measure equal to 1) it is weak$^*$-compact. So you can always get a convergent subsequence!

Title: Why is Banach-Alaoglu theorem so important?

Body:
According to Lawrence Narici and Edward Beckenstein, the Alaoglu
theorem is a "very important result - maybe the most important fact
about the weak-* topology - [that] echos throughout functional
analysis." (Source: Wikipedia)

It is a well-known fact (by Riesz) that the compactness of the unit ball with respect to the norm topology characterizes finite dimensional vector spaces. In a infinite dimensional setting, Banach-Alaoglu recovers the compactness of the unit ball in the weak*-topology which seems to come to relief of a lot of analyst.
I have come across a thread recently about the importance of Hahn-Banach which I found very illuminating. I would be interested about the different takes people have on Banach-Alaoglu.
Why is it so important? What if we were not to recover the compactness of the unit ball?


问:
Inverses where $f(g(x))=x\neq g(f(x))$
If $f: \mathbb{Z\to Z}$ and $G: \mathbb{Z\to Z}$, find $f$, and $g$ such that $f(g(x))=x\neq g(f(x))$.

I can find lots of $f$ and $g$ that aren't equal when composed with each other, but I have no idea how to proceed. A hint would be helpful.

Thanks in advance.

答:
E.g.
 $\ g(n):=2n$, $f(2n):=n$ and you can do anything on the odds...

Note that we won't have $x\ne g(f(x))$ for all $x$ as, if $x=g(y)$ for some $y$, then $g(f(x))=g(f(g(y)))=g(\,y\,)=x$.

Title: Inverses where $f(g(x))=x\neq g(f(x))$


答:
If you take Venn diagrams, $p \to q$ can be represented as having the set for $p$ being included in the set for $q$. It is geometrically obvious that this is equivalent to the set for $\neg q$ being included in the set for $\neg p$.

As for your error, it's at the very beginning. If we write $P := p \to q$ and $Q := \neg q \to \neg p$:

$$(P \Leftrightarrow Q) \Leftrightarrow$$
$$((P \cap Q) \cup (\neg P \cap \neg Q)) \Leftrightarrow$$
$$(((p \to q) \cap (\neg q \to \neg p)) \cup (\neg (p \to q) \cap \neg (\neg q \to \neg p))) \Leftrightarrow$$
$$(((\neg p \cup q) \cap (q \cup \neg p)) \cup (\neg (\neg p \cup q) \cap \neg (q \cup \neg p))) \Leftrightarrow$$
$$(((\neg p \cup q) \cap (q \cup \neg p)) \cup ((p \cap \neg q) \cap (\neg q \cap p))) \Leftrightarrow$$
$$((\neg p \cup q) \cup (p \cap \neg q)) \Leftrightarrow$$
$$((\neg p \cup q) \cup \neg \neg(p \cap \neg q)) \Leftrightarrow$$
$$((\neg p \cup q) \cup \neg (\neg p \cup q)) \Leftrightarrow$$
$$A \cup \neg A \Leftrightarrow$$
$$True$$

Since De Morgan's laws of negation invert internal conjunction/disjunction operators.

PS: you can use \Leftrightarrow, \neg, \to, \cup, and \cap  in your LaTeX.

Title: How to see that $(p\rightarrow q)\leftrightarrow(\lnot q\rightarrow\lnot p)$ is a tautology, without using truth table?

Body:
Consider:
Statement-I: $(p\land\lnot q)\land(\lnot p\land q)$ is a fallacy.
Statement-II: $(p\rightarrow q)\leftrightarrow(\lnot q\rightarrow\lnot p)$ is a tautology. Which of these statements is true? If both are true then is statement-II a correct explanation of statement-I?

My attempt:

Statement-I: $(pq')(p'q)=0$. So, true.

Statement-II: $(p'+q)(q+p')+(pq')(q'p)$ [using the rules $p\rightarrow q=p'+q$ and $p\leftrightarrow q=pq+p'q'$]

So, statement-II becomes $p'q+q+qp'+pq'=p'q+q+pq'=q+pq'$, which is not a tautology. But the answer key says it is a tautology.

What's my mistake here?


问:
Is this property of complex numbers true: $(a^* \times b)^*=(a\times b^*) $
I was wondering if this property of complex numbers is true:

$(a^* \times b)^*=(a\times b^*) $

$a^*$ is the complex conjugate of a

If anyone could tell me where I can find more about the properties of complex numbers, it would be helpful.

Do complex numbers follow all the same rules as normal numbers when it comes to addition, subtraction, multiplication and division?

答:
The product of conjugates is the conjugate of the product,

and conjugating a complex number twice gives the original number,

so indeed $(a^\ast\times b)^\ast=(a^\ast)^\ast\times b^\ast=a\times b^\ast$.

Complex numbers follow the commutative, associative, and distributive rules, like real numbers.


答: Hint: Applying $g$ to the equality gives $g(f(g(x)))=g(x)$ so if $y=g(x)$ then we must have $g(f(y))=y$. So the result could only fail for $y$ which are not $g(x)$ for any $x$. What is a simple way to construct a function which, say, misses out one such $y$?

Title: Inverses where $f(g(x))=x\neq g(f(x))$

Body:
If $f: \mathbb{Z\to Z}$ and $G: \mathbb{Z\to Z}$, find $f$, and $g$ such that $f(g(x))=x\neq g(f(x))$.

I can find lots of $f$ and $g$ that aren't equal when composed with each other, but I have no idea how to proceed. A hint would be helpful.

Thanks in advance.


问:
Show that $\left\lfloor\dfrac{\lfloor x \rfloor}{n} \right\rfloor = \left\lfloor\dfrac{x}{n}\right\rfloor$

Let $x \in \Bbb R$ and $n \in \Bbb Z$ with $n \gt 0$. Show that $\left\lfloor\dfrac{\lfloor x \rfloor}{n} \right\rfloor = \left\lfloor\dfrac{x}{n}\right\rfloor$;

in particular, $\left\lfloor\dfrac{\left\lfloor\dfrac {a}{b} \right\rfloor}{c} \right\rfloor = \left\lfloor\dfrac{a}{bc}\right\rfloor$

I don't know how to approach this problem, any hints to solve this would be greatly appreciated.

Thanks for your help!

答:
Let $m=\lfloor x\rfloor$. Then $m\leq x < m+1$. So $\dfrac{m}{n}\leq \dfrac{x}{n}< \dfrac{m+1}{n}$. By the division algorithm, $m =qn+r$ for some integers $q,r$ with $0\leq r\leq n-1$. Thus, $$q\leq\dfrac{m}{n}\leq\dfrac{x}{n}<\dfrac{m+1}{n}=\dfrac{qn+r+1}{n}\leq q+1.$$
So $\lfloor \tfrac{m}{n} \rfloor=\lfloor\tfrac{x}{n}\rfloor=q$.

Note: $\lfloor x\rfloor$ is the unique integer $N$ such that $N\leq x<N+1$.


问:
Recursively generating all partitions
Given a set $\{1,2,\ldots,n\}$, I would like to recursively generate all partition of size $r$ and $n-r$. For instance, we start with the partitions
$$\{1,2,\ldots, r\} \quad \{r+1,\ldots, n\}$$
and then iteratively swap elements.  We can write the solution in terms of all the collections of allowed swaps between the two sets. We do this by considering a swap operator $S_{ij}$, which swaps elements $i$, $j$ between these sets, for instance:

$$S_{1,r+1} \{\{1,2,\ldots, r\},\{r+1,\ldots, n\}\} = \{\{r+1,2,3,\ldots, r\},\{1, r+2,\ldots, n\}\}.$$

Then we can write the solution in terms of all $\binom{n}{r}$ strings of swap operators acting on the two sets.

I know that one way to do this is with Heap's algorithm, which would generate all the permutations, which we then just have to partition. However, this algorithm is inefficient for this problem since many permutations correspond to the same subsets. As such I am essentially looking for a simplification to Heap's algorithm for this specific case.

答:
I believe the answer to the question is merely all sets
$$\{\{i_1,i_2,\ldots i_r\}, \{\text{rest of the set}\}\}$$
where $i_r \in \{r,r+1,\ldots, n\}$, $i_{r-1} \in \{r-1,\ldots,i_r - 1\}$,..., $i_1 \in \{1,\ldots, i_2-1\}$. One way to easily generate this is by swapping elements of an ordered list (tuples). All tuples are generated by permutations $$\forall_{i_r = r}^{n} \forall_{i_{r-1}}^{i_r-1} \cdots \forall_{i_1 = 1}^{i_2 - 1} P_{r,i_r} P_{r-1,i_{r-1}}\cdots P_{2,i_2} P_{1,i_1} (1,\ldots r,r+1,\ldots n),$$
where $P_{ij}$ swaps the $i$ and $j$th element. Splitting the all these tuples then at the $r$th element yields all $\binom{n}{r}$ partitions exactly once.

I have checked numerically that this indeed works. A more formal mathematical prove would just use the fact that the way we choose the $i$'s, generates all ordered lists of size $r$ from $n$ elements. Combining this with the "rest" of the set, yields all partitions.

Title: Recursively generating all partitions


问:
prove the identity $\sum_{k=0}^n {s+k \choose k}{n-k \choose m} = {s+n+1 \choose s+m+1} $
I am not able to understand how to apply the vandemonde identity to prove this. Please help me proving this.</p>

Body:
$$\sum_{k=0}^{n}\binom{s+k}{k}\binom{n-k}{m}$$

Use the fact that $\binom{n}{k}=\binom{n}{n-k}$
$$\sum_{k=0}^{n}\binom{s+k}{k}\binom{n-k}{n-k-m}$$

Use $\binom{n}{k}=(-1)^k\binom{k-n-1}{k}$

$$\sum_{k=0}^{n}(-1)^{k}\binom{-s-1}{k}(-1)^{n-k-m}\binom{-m-1}{n-k-m}$$  $$(-1)^{n-m}\sum_{k=0}^{n}\binom{-s-1}{k}\binom{-m-1}{n-k-m}$$

Now apply Vandermonde's identity

$$(-1)^{n-m}\binom{-s-m-2}{n-m}$$$$\binom{n+s+1}{n-m}=\binom{n+s+1}{m+s+1}$$


问:
Invertiblity of ST and TS for linear maps
Could somebody explain why for any given linear maps S and T over a finite dimensional vector space V, ST is invertible if and only if TS is? Why is it so important that V is finite dimensional for this to hold true?

Edit: I am looking for an approach without using matrices or determinants.

答:
If $T : V \rightarrow V$ is a linear map on a finite dimensional vector space $V$. Then $T$ is injective iff $\text{dim ker }T = 0$. And $T$ is surjective iff $\text{dim Im }T = \text{dim }V$ (this fails if $V$ has infinite dimension). Furthermore we have the equation $\dim V = \dim \text{Im }T + \dim \text{ker }T$. Hence $T$ is surjective iff $ \dim \text{Im }T = \dim V $ iff $\dim \text{ker }T = 0$ iff $T$ is injective. So for a finite dimensional vector space we have:

$$T \text{ injective} \iff T \text{ surjective}$$

If $ST$ is invertible then $ST$ must be injective and surjective (I think of $ST$ as the composition of first applying $S$ and then applying $T$). This requires $T$ to be surjective and $S$ to be injective. Hence $T$ is also injective and $S$ is also surjective. So both $S$ and $T$ are invertible. Hence so is $TS$.


问:
Prove that this summation has a surprising result! (Or prove me wrong, it is possible that the pattern does not hold)
Essentially, prove the following summation:

$$\sum_{i=0}^{2x} {{2x \choose 2i}{2i \choose i}{2x-2i \choose x-i}} = {2x \choose x}^2$$

I came across this while attempting question 6 of this paper (BMO1 2019) and while I'm sure there are easier solutions to the problem, this summation seemed interesting in itself. Very similar to this summation is the following:

$$\sum_{i=0}^{x} {{2i \choose i}{2x-2i \choose x-i}} = {2}^{2x}$$

Which I also do not have a proof for.  The second one can also be thought of as going down the middle of Pascal's triangle and adding the products of opposite ends of the triangle.

Proofs of either summation, ideally using mathematical ideas which the BMO1 is based around (so basically pre-calculus), would be much appreciated.

答:
This is combinatorial proof. You have $2x$ people whom you will divide into four groups; A, B, C, and D. A and B has equal number of members and C and D has equal number of members.

The first way to do it:

Choose an even number of people who will join either A or B. $\binom{2x}{2i}$

Half of these $2i$ people join A while the rest join B. $\binom{2i}{i}$

Half of $2x-2i$ remaining people join C while the rest join D. $\binom{2x-2i}{x-i}$

Total ways of forming such four groups is $\sum_{i=0}^{x}{\binom{2x}{2i}\binom{2i}{i}\binom{2x-2i}{x-i}}$

The second way to do it:

Choose $x$ people. $\binom{2x}{x}$

Choose $x$ people again, does not have to be different people. $\binom{2x}{x}$

People who got chosen twice join A

People who never got chosen join B

People who got chosen only in first selection join C

People who got chosen only in second selection join D

Total ways of forming such four groups is $\binom{2x}{x}\binom{2x}{x}$


问:
Find a monic polynomial over $\mathbb{Q}$ whose Galois group is $V_4$
I need to find a polynomial over $\mathbb{Q}$ whose Galois group is $V_4=\langle(12)(34),(13)(24)\rangle\subset S_4$. I can find examples of such polynomials, but I wonder if one can construct such a polynomial by hand.

Can anyone offer some ideas? Thanks!

答: Try $x^4 - 10x^2 +1$, which as a splitting field of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ and Galois group isomorphic to $V_4$.</p>


问:
Solving $e^x + x = 5$ for $x$ without using a numerical method?
Canadian economist Mike Moffat asks on Twitter:

Math nerd Q: Is there a way to solve $e^x + x = 5$ for $x$, without using a numerical method?

Body:
If you do not want to use Lambert function or any numerical method, you can have explicit approximations using Newton-like methods of high order $n$ starting at $x=1$.

For example
$$x_{(2)}=\frac{5}{1+e}$$
$$x_{(3)}=\frac{10+14 e-e^2}{2+8 e+e^2}$$
$$x_{(4)}=\frac{60+296 e+32 e^2-4 e^3}{12+116 e+56 e^2+2 e^3}$$
$$x_{(5)}=\frac{720+8016 e+4344 e^2-120 e^3-18 e^4}{144+2592 e+3048 e^2+456 e^3+6 e^4}$$
$$x_{(6)}=\frac{14400+286464 e+381696 e^2+45504 e^3-4224 e^4-96 e^5}{2880+81984 e+189696
   e^2+69984 e^3+4416 e^4+24 e^5}$$ I stop here because the formulae start to be too large for the page.

Converted to decimals, the above approximations are
$$\{1.3447071,1.3061341,1.3064721,1.3065586,1.3065589\}$$


问:
S(n) properties
Recently, while reading a number theory textbook for Olympiads, i came across the following property;

$S(n_1+n_2) \le S(n_1) + S(n_2)$

Where S(A) is the sum of digits of A in base 10. In my textbook, there was a proof given but i couldn't follow it at all.

So I am looking for a proof of the following property and hopefully some reasoning behind it.

Any help would be appreciated, thanks!

答:
Assume
$$a = 10^0a_0 + 10^1a_1+10^2a_2+... \\
b = 10^0b_0 + 10^1b_1+10^2b_2+... $$then$$
S(a) = a_0+a_1+... \\
S(b) = b_0+b_1+... \\
\implies S(a)+S(b) = (a_0+b_0)+(a_1+b_1)+...$$
Also, $$a+b = 10^0(a_0+b_0) + 10^1(a_1+b_1)+... \\ a+b =:c = 10^0c_0 + 10^1c_1+10^2c_2+...$$ then $S(a+b) = c_0+c_1+...$

Now

if $a_k + b_k \ge 10$ for some $k$, then $c_k = a_k + b_k - 10 \le a_k + b_k$, but then $c_{k+1} = a_{k+1} + b_{k+1} + \color{red}1$. However this $\color{red}1$ cannot compensate $-10$ in the previous term. So, the inequality still holds overall.

else if $a_k + b_k < 10$ for some $k$ and the previous terms don't exceed $10$ (as the above case), then $c_k = a_k + b_k$. But if the previous terms do exceed, then by the same argument above, the inequality still holds.

Therefore,
$$c_0 + c_1 +... \le (a_0+b_0)+(a_1+b_1)+... $$
or
$$S(a+b) \le S(a) + S(b)$$  as desired.

Title: S(n) properties


答: Just figured out that $(\lnot q\rightarrow\lnot p)$ is a contrapositive of $(p\rightarrow q)$ i.e. both are just the same thing. So, $(p\rightarrow q)\leftrightarrow(\lnot q\rightarrow\lnot p)$ is a tautology.


问:
Induction Outset
I am referring to this older question, which I found here for preparation:

Proof by induction that $\sum_{j=0}^n 2^j = 2^{n+1} - 1$

Could somebody outline the Induction outset and the goal?
I am unsure where to put in my (n + 1).
My approach would be:
2⁰ - 1 + 2^(n+1) -1

答:
Base case $n=0$:  $\sum\limits_{j=0}^0 2^j=2^{1}-1$.

Inductive step:  assuming $\sum\limits_{j=0}^n 2^j=2^{n+1}-1$, show that $\sum\limits_{j=0}^{n\color{blue}{+1}} 2^j=2^{(n\color{blue}{+1})+1}-1$.

(Put $n\color{blue}{+1}$ wherever $n$ appeared.)

Title: Induction Outset


问:
If $A,B\in M(2,\mathbb{F})$ and $AB=I$, then $BA=I$
This is Exercise 7, page 21, from Hoffman and Kunze's book.

Let $A$ and $B$ be $2\times 2$ matrices such that $AB=I$. Prove that
  $BA=I.$

I wrote $BA=C$ and I tried to prove that $C=I$, but I got stuck on that. I am supposed to use only elementary matrices to solve this question.

I know that there is this question, but in those answers they use more than I am allowed to use here.

I would appreciate your help.

答:
I know this is old, but I think I have found the answer that was intended. I also struggled with this one for a while because, as spohreis mentioned, you don't have much to go on at the time this is asked (no determinants, no transposes, no inverses even).

That being said, in problem 3 of section 1.4 you prove that all $2\times 2$ row-reduced echelon matrices are of the following form:

$$ \left[ \begin{array}{cc}
0 & 0 \\
0 & 0 \end{array} \right]\quad,\quad
\left[ \begin{array}{cc}
0 & 1 \\
0 & 0 \end{array} \right]\quad,\quad
\left[ \begin{array}{cc}
1 & c \\
0 & 0 \end{array} \right]\quad,\quad
\left[ \begin{array}{cc}
1 & 0 \\
0 & 1 \end{array} \right] \,.$$

Now assume that $A$ and $B$ are $2 \times 2$ matrices such that $AB=I$. By theorem 5 (pg 12) we have that $B$ is row-equivalent to a row-reduced echelon matrix $R$, and by the corollary to theorem  9 (pg 20) this implies that $B=PR$ where $P$ is a product of elementary matrices. Similarly we have that $A=QT$ (where $Q$ is a product of elementary matrices and $T$ is in row-reduced echelon form).

Now we have that $AB=I \implies QTPR=I$, but now clearly $T=I$ because if the bottom row of $T$ were all zeros then the bottom row of $TPR$ would be zero, and this implies that the product $QTPR$ would have the form
$$\left[ \begin{array}{cc}
aQ_{11} & bQ_{11} \\
aQ_{21} & bQ_{21} \end{array} \right]$$
for some $a,b\in F$ and thus clearly could not be $I$. A similar argument shows that $R=I$. Thus $A$ and $B$ are both actually products of elementary matrices. By theorem 2 (pg 7) each elementary row operation has an inverse and using theorem 9 (pg 20) each elementary matrix therefore has an inverse, now we can write

$$\begin{align}
AB=QP=E_{q_1}E_{q_2} \cdots E_{q_t}E_{p_1} \cdots E_{p_s}&=I\\
E_{q_1}^{-1}E_{q_1}E_{q_2} \cdots E_{q_t}E_{p_1} \cdots E_{p_s}E_{q_1}&=E_{q_1}^{-1}E_{q_1}=I\\
&\vdots\\
E_{p_1} \cdots E_{p_s}E_{q_1} \cdots E_{q_t}&=I\\
PQ&=I\\
BA&=I\,.
\end{align}$$

(Note that at the end here, although I chose to use the standard inverse notation, it really is enough that such a matrix exists, which follows from theorem 2 and theorem 9 alone - no need to really "know" about inverse matrices yet. You could, if you so chose, just use theorem 9 to rewrite $QP=I$ in terms of elementary row operations, and then just use theorem 2 directly without ever mentioning inverse matrices.)


问:
Difference between integral and summation.
Well I am not a  math expert as most people are on this site but I feltb this site would help me.
$$\int_0^n x dx = \frac{n^2}{2}$$
also
$$\sum_{x=0}^{n} x=\frac{n(n+1)}{2}$$
(it is suggested that it may also be written as)
$$\lim_{n\to\infty}\sum_{0}^n x \triangle x
$$
What are we doing in the above two.Both sum up so in this case if $$n>0$$ why is $$\int<\sum$$?
I have read Difference between summation and integration but still it is not clear to me why integration is less than summation in this case for natural numbers.

答:
Draw a picture of the area under the parabola between $x=0$ and $x=n$ and superimpose rectangles with unit base corresponding to the terms in the sum. You will see the inequality.

Here is what it looks like (with intervals shorter that $1$).

Whenever the function you are integrating is increasing, this sum will overestimate the integral.

The corresponding sum using the left endpoints of the subdivision will reverse the inequality.

Source:
https://web.northeastern.edu/dummit/teaching_fa19_1341/calc1_4_introduction_to_integration_v2.00.pdf


答: The red line ist your function f(x) = x, in the sum, you essentially compute the area of the rectangles and add them up, thats why the integral is smaller.

Title: Difference between integral and summation.

Body:
Well I am not a  math expert as most people are on this site but I feltb this site would help me.
$$\int_0^n x dx = \frac{n^2}{2}$$
also
$$\sum_{x=0}^{n} x=\frac{n(n+1)}{2}$$
(it is suggested that it may also be written as)
$$\lim_{n\to\infty}\sum_{0}^n x \triangle x
$$
What are we doing in the above two.Both sum up so in this case if $$n>0$$ why is $$\int<\sum$$?
I have read Difference between summation and integration but still it is not clear to me why integration is less than summation in this case for natural numbers.


问:
How many words of length $10$ are there in which not all of the vowels A, E, I, O, U appear?
The word can have every letter as long as A,E,I,O,U aren't all in the same word.

Can this be the number of all words of length $10$ without all vowels?

$$ 21^{10}+ \sum_{i=1}^4  {10\choose i} 5^i \cdot 21^{10-i} + \sum_{i=5}^{10} {5\choose 4}{10\choose i} 4^i $$

(all consonant words + all words with less than $4$ vowels + all words with $5$ nonrepeated vowels or more)

Is this correct? If there is a simpler way, I would appreciate.

答:
The number of words in which not all the vowels appear is equal to the number of words in which at least one vowel does not appear.

Let $A, E, I, O, U$ be the set of ten-letter words in which, respectively, the letter $a, e, i, o, u$ does not appear.  The set of words in which at least one vowel does not appear is $A \cup E \cup I \cup O \cup U$.   By the Inclusion-Exclusion Principle,
\begin{align*}
& |A \cup E \cup I \cup O \cup U|\\
& = |A| + |E| + |I| + |O| + |U|\\
& \qquad - (|A \cup E| + |A \cup I| + |A \cup O| + |A \cup U| + |E \cup I| + |E \cup O| + |E \cup U| + |I \cup O| + |I \cup U| + |O \cup U|)\\
& \qquad + (|A \cap E \cap I| + |A \cap E \cap O| + |A \cap E \cap U| + |A \cap I \cap O| + |A \cap I \cap U| + |A \cap O \cap U| + |E \cap I \cap O| + |E \cap I \cap U| + |E \cap O \cap U| + |I \cap O \cap U|\\
& \qquad - (|A \cap E \cap I \cap O| + |A \cap E \cap I \cap U| + |A \cap E \cap O \cap U| + |A \cap I \cap O \cap U| + |E \cap I \cap O \cap U|)\\
& \qquad + |A \cap E \cap I \cap O \cap U|
\end{align*}

$|A|$:  If the vowel $a$ does not appear, then each of the ten positions can be filled in $25$ ways.  Thus, $|A| = 25^{10}$.

By symmetry, $$|A| = |E| = |I| = |O| = |U|$$

$|A \cap E|$:  If neither of the vowels $a$ nor $e$ appears, then each of the ten positions can be filled in $24$ ways.  Thus, $|A \cap E| = 24^{10}$.

By symmetry, $$|A \cap E| = |A \cap I| = |A \cap O| = |A \cap U| = |E \cap I| = |E \cap O| = |E \cap U| = |I \cap O| = |I \cap U| = |O \cap U|$$

$|A \cap E \cap I|$:  If none of the vowels $a, e, i$ appears, then each of the ten positions can be filled in $23$ ways.  Thus, $|A \cap E \cap I| = 23^{10}$.

By symmetry,
$$|A \cap E \cap I| = |A \cap E \cap O| = |A \cap E \cap U| = |A \cap I \cap O| = |A \cap I \cap U| = |A \cap O \cap U| = |E \cap I \cap O| = |E \cap I \cap U| = |E \cap O \cap U| = |I \cap O \cap U|$$

$|A \cap E \cap I \cap O|$:  If none of the vowels $a, e, i, o$ appears, there are $22$ ways to fill each of the ten positions.  Hence, $|A \cap E \cap I \cap O| = 22^{10}$.

By symmetry,
$$|A \cap E \cap I \cap O| = |A \cap E \cap I \cap U| = |A \cap E \cap O \cap U| = |A \cap I \cap O \cap U| = |E \cap I \cap O \cap U|$$

$|A \cap E \cap I \cap O \cap U|$:  If none of the vowels $a, e, i, o, u$ appears, there are $21$ ways to fill each of the ten positions.  Hence, $|A \cap E \cap I \cap O \cap U| = 21^{10}$.

Thus, by the Inclusion-Exclusion Principle, the number of ten-letter words in which at least one vowel does not appear is
\begin{align*}
|A \cup E \cup I \cap O \cap U| & = 5 \cdot 25^{10} - 10 \cdot 24^{10} + 10 \cdot 23^{10} - 5 \cdot 22^{10} + 21^{10}\\
& = \binom{5}{1}25^{10} - \binom{5}{2}24^{10} + \binom{5}{3}23^{10} - \binom{5}{4}22^{10} + \binom{5}{5}21^{10}\\
& = \sum_{k = 1}^{5} (-1)^{k - 1}\binom{5}{k}(26 - k)^{10}
\end{align*}
where $\binom{5}{k}$ is the number of ways of excluding $k$ of the $5$ vowels and $(26 - k)^{10}$ is the number of ways to form a ten-letter word with the remaining $26 - k$ letters of the alphabet.


答: Here a picture, in red we see the sum as described above. And the blue area is the integral analogue.


问:
Vertex $E$ of equilateral $\Delta ABE$ is in interior of $\square ABCD.$ $F$ is the midpoint of intersection of $AE$ and diagonal $BD$.

Vertex $E$ of equilateral $\Delta ABE$ is in interior of $\square ABCD.$ $F$ is the midpoint of intersection of $AE$ and diagonal $BD$. If $AB = \sqrt{1 + \sqrt{3}}$ and $[\Delta ABF] = A$, find $(4A)^2$.

What I Tried: Here is a picture :-

From a bit of angle chasing, we have the angles of $\Delta ABF$ .

Also $AB = \sqrt{1 + \sqrt{3}}$, let $AF = y$. We have :-
$$\frac{y}{\sin B} = \frac{AB}{\sin C}$$

We have $\sin 75^\circ = \frac{1 + \sqrt{3}}{2\sqrt{2}}$ , so :-

$$\rightarrow y = \frac{\sqrt{1 + \sqrt{3}} * \sqrt{2}}{\frac{1 + \sqrt{3}}{2\sqrt{2}}*2}$$

Let $\sqrt{1 + \sqrt{3}} = k$. After continuous solving we will have :-
$$y = \frac{2\sqrt{k}}{k}$$
From here I can get $$[\Delta ABF] = \frac{1}{2}(AB)(AF)\sin 60^\circ$$
$$\rightarrow \sqrt{k} * \frac{2\sqrt{k}}{k} * \frac{\sqrt{3}}{2}$$
$$\rightarrow [\Delta ABF] = \sqrt{3}$$

Hence $(4A)^2 = 48$ .

The question is, am I correct? Was it a coincidence that the $k$'s got cancelled out?

I am probably not correct, because right now I found out that $\sqrt{1 + \sqrt{3}} < \sqrt{3}$ , and this is impossible if my answer would have correct.

Can anyone let me know where I went wrong? Thank You.

答:
Let's use $AB=k=\sqrt{1+\sqrt 3}$. Then $$\sin 75^\circ=\frac{1+\sqrt 3}{2\sqrt 2}=\frac {k^2}{2\sqrt 2}$$
Area of triangle $ABF$ is then:
$$[\triangle ABF]=\frac12 AB\cdot AF\sin60^\circ=\frac12 k\frac {k\sin 45^\circ}{\sin 75^\circ}\sin 60^\circ\\=\frac 12 k^2\frac{\frac{\sqrt 2}2}{\frac{k^2}{2\sqrt 2}}\frac{\sqrt 3}2=\frac{\sqrt 3}2$$


答:
A much more general result is available in Donald Knuth's Concrete Mathematics and it is attributed to "R. J. McEliece when he was an undergrad". He proved it for ceiling functions, but I included a proof for the floor function in a book that I am writing. Here is a variation of what I wrote in my book:

Let $f$ be a continuous and monotonically increasing function, defined on an interval $I$ in $\mathbb{R},$ with the property that $$f(x)\in\mathbb{Z}\implies x\in\mathbb{Z}.$$ Then, for all $x\in I,$ $$ \lfloor f(x) \rfloor =\left\lfloor f(\lfloor x\rfloor) \right\rfloor \text{ and } \lceil f(x) \rceil =\left\lceil f(\lceil x\rceil) \right\rceil .$$ Note that each identity holds only if the expression on either side of it is defined, as $x$ being in $I$ does not necessarily mean that $\lfloor x\rfloor$ and $\lceil x\rceil$ are in $I.$

I'll leave it to you to check out Knuth's ceiling proof and adapt it to the floor scenario. Do you see why this applies to your problem? Roots, logs and linear functions with positive slope all work here (linear in particular).

Body:
Let $x \in \Bbb R$ and $n \in \Bbb Z$ with $n \gt 0$. Show that $\left\lfloor\dfrac{\lfloor x \rfloor}{n} \right\rfloor = \left\lfloor\dfrac{x}{n}\right\rfloor$;

in particular, $\left\lfloor\dfrac{\left\lfloor\dfrac {a}{b} \right\rfloor}{c} \right\rfloor = \left\lfloor\dfrac{a}{bc}\right\rfloor$

I don't know how to approach this problem, any hints to solve this would be greatly appreciated.

Thanks for your help!


答:
When you add two numbers, the sum of two facing digits is preserved, unless there is a carry. In this case, the sum is diminished by $9$.

$$\begin{align}\\1234\\+5678\\\hline6912
\end{align}$$

$$4+8\to 2+1,\\3+7\to 0+1,\\2+6\to8,\\1+5\to6.$$

Title: S(n) properties

Body:
Recently, while reading a number theory textbook for Olympiads, i came across the following property;

$S(n_1+n_2) \le S(n_1) + S(n_2)$

Where S(A) is the sum of digits of A in base 10. In my textbook, there was a proof given but i couldn't follow it at all.

So I am looking for a proof of the following property and hopefully some reasoning behind it.

Any help would be appreciated, thanks!


问:
Representation of Lindenbaum-Tarski algebras as a field of sets
I don't know if this changes anything, but for simplicity let's say we are only talking about propositional calculus.

Here are some facts that I've gathered when there are finite variables and I'm looking for their equivalent ones with infinite variables. Assume there are only $n$ variables, then:

The Lindenbaum-Tarski algebra $L$ is isomorphic to $2^{2^n}$

The atoms are of the form $(\lnot) p_1 \wedge (\lnot) p_2 \wedge \dots \wedge(\lnot) p_n$, where $(\lnot$) means that there may or may not be a $\lnot$. I think of this as only one row being true in a truth table.

For any formula $\phi \in L$, if the length of any chain from $\phi$ to $0$ is $k$, then it's the join of $k$ atoms. I think of this as only $k$ rows being true in a truth table.

The only ultrafilters $U$ are the principal ones. Hence the realizations $L/U \to \bf 2$ must take exactly one of the atoms to $1$. I interpret this as picking one row in a truth table.

I have tried to find some equivalences in the infinite case, but have not been able to do so. Assume there are infinite variables, then:

The Lindenbaum-Tarski algebra $L$ is isomorphic to $?$ By Stone's representation theorem there should be some subset of $f(\alpha)$, where $\alpha$ is the set of variables and $f$ is some function.

It is atomless and coatomless.

$-$? How do I think of the truth tables?

The only ultrafilters $U$ are the non-principal ones. How do these look? How do their realizations look as truth tables?

I am sorry these are many questions but I thought they were too related to make a question for each.

If you know some more basic facts like these, I'd appreciate them for building my intuition.

答:
Let $L(S)$ denote the Lindenbaum-Tarski algebra of propositional calculus with a set $S$ of variables.  Note that $L(S)$ is just the free Boolean algebra on the set $S$ (since the axioms of Boolean algebra are equivalent to imposing all propositional tautologies; this is a consequence of Stone's representation theorem).  Its Stone space is the Cantor space $\{0,1\}^S$: an ultrafilter on $L(S)$ is just an assignment of a truth value to each variable, i.e. a function $S\to\{0,1\}$.  So, elements of $L(S)$ can be thought of as clopen subsets of $\{0,1\}^S$.  This is just like the finite case--an element of $\{0,1\}^S$ is a row of an infinite "truth table" where you assign a truth value to each variable and use that to determine the truth value of any proposition.  The difference is that when $S$ is finite, $\{0,1\}^S$ is just a finite discrete space, so every set of rows in the truth table can be realized by some proposition, whereas if $S$ is infinite, then $\{0,1\}^S$ is not discrete, and only certain sets of rows are realized by propositions (those which are clopen subsets).

To get a more concrete understanding of what elements of $L(S)$ look like, it is helpful to observe that $L(S)$ is just the direct limit of $L(F)$ where $F$ ranges over all finite subsets of $S$ (and the inclusions $L(F)\to L(S)$ are all injective).  So given an element $x\in L(S)$, it comes from $L(F)$ for some finite subset of $S$, and so you can understand it in terms of the atoms of $L(F)$.  If you add a new element $s\in S$ to $F$, this just splits each of the atoms in two pieces, one obtained by taking a meet with $s$ and one obtained by taking a meet with $\neg s$.  To compute a join or meet of two elements of $L(S)$, you just find an $L(F)$ that contains both of them and compute the join or meet in $L(F)$.

If $S=\{s_1,s_2,\dots\}$ is countable, you can visualize this in terms of an infinite descending binary tree.  At the top of the tree is $1$, which then splits into the two atoms $s_1$ and $\neg s_1$ of $L(\{s_1\})$, which then split into the four atoms $s_1\wedge s_2,s_1\wedge\neg s_2,\neg s_1\wedge s_2,\neg s_1\wedge\neg s_2$ of $L(\{s_1,s_2\})$, and so on.  Each level of the tree consists of the atoms of $L(\{s_1,\dots,s_n\})$ for some $n$, and each element of $L(S)$ can be written as a join of atoms at some level of the tree (and thus also at each lower level, by splitting the atoms down into the lower level atoms).  To compute a join or meet of two elements of $L(S)$, you write both of them in terms of the atoms at some level of the tree and take the union or intersection of those sets of atoms.


问:
$n^\text{th}$ derivative of a function
I'm trying to show that $f^{n+1}(x)=0$. I've calculated the derivatives of $f(x)=2x^{5}+3x^{4}+4$.

$f'(x)=10x^{4}+12x^{3}$

$f''(x)=40x^{3}+36x^{2}$

$f'''(x)=120x^{2}+72$

$f^{4}(x)=240x+72$

$f^{5}(x)=240$

But I'm not sure how to proceed from there?

答:
Generally, assume
$$P(x) = \sum_{k=0}^n a_kx^k$$
then
$$\begin{align}
P'(x) &= \sum_{k=0}^n ka_kx^{k-1} \\
P''(x) &= \sum_{k=0}^nk(k-1)a_kx^{k-2} \\
...\\
P^{(n)}(x)&=\sum_{k=0}^nn!a_kx^0 \end{align}$$Finally,$$
P^{(n+1)}(x) = \sum_{k=0}^n(n!a_k)' = \sum_{k=0}^n0 = 0$$
Hence, for any polynomial of degree $\leqslant n$, we have $P^{(n+1)}(x) \equiv 0$.

Title: $n^\text{th}$ derivative of a function


问:
Convolution Integral - Shift property
I am trying to prove $f_1(t-T_1)*f_2(t-T_2)=c(t-T_1-T_2)$ given that $f_1(t)*f_2(t)=c(t)$. Here $f_1(t)*f_2(t)=\int_{-\infty}^{\infty}f_1(x)f_2(t-x)dx$.

I can only deduce that $f_1(t-T_1)*f_2(t)=c(t-T_1)$ and $f_1(t)*f_2(t-T_2)=c(t-T_2)$ by directly evaluating the integral and using the commutative property. According to my textbook, the conclusion should follows directly from these two relationships but I am not able to prove it.

If the 'left' function is delayed by $T_1$ seconds, so is $c(t)$. The same thing happens when the 'right' function is delayed by $T_2$ seconds. So it is quite 'logical' for $c(t)$ to be delayed by the sum of the delayed periods of time in $f_1$ and $f_2$. But this verbal explanation does not satisfy me.

I am also having trouble to express $f_1(t-T_1)*f_2(t-T_2)$ in integral form. I cannot only express it because the definition of the convolution integral seems limited. Thanks in advance.

答:
The primary problem is the notation: One should write (f*g)(t) (= the c(t) above) instead of f(t)*g(t).
To shift a function, we define the shift operator S as follows:

S[a,f](t)=f(t-a)

S[a,f] means: the function f with the argument shifted everywhere by a constant -a. That means f shifted everywhere by a. For example, if f(0)=0, then
S[a,f](a)=f(a-a)=0.
Obviously, as shifting by b=2 and then by a=3 is shifting by a+b=5 altogether:

S[a,S[b,f]](t)=S[b,f](t-a)=f(t-a-b)=S[a+b,f](t)

What has already been proved is:

S[a,f*g]=S[a,f]*g

i.e., the shift can be pulled into one of the factors. Now, it's easy to see:

S[a+b, f*g]=S[b+a,f*g]=S[b,S[a,f*g]]=S[b,S[a,f]*g]=S[b,g*S[a,f]]=S[b,g]*S[a,f]=S[a,f]*S[b,g]  QED

Title: Convolution Integral - Shift property


答:
Generalise:
$$1+x+x^2+\dots +x^n=\frac{x^{n+1}-1}{x-1}.$$
This formula is the simplest nontrivial model when one teaches proofs by induction.

The inductive step begins with
\begin{align}
1+x+x^2+\dots+x^{n+1}&=(1+x+x^2+\dots +x^n)+x^{n+1}\\
&=\frac{x^{n+1}-1}{x-1}+x^{n+1}=\dots
\end{align}

Title: Induction Outset

Body:
I am referring to this older question, which I found here for preparation:

Proof by induction that $\sum_{j=0}^n 2^j = 2^{n+1} - 1$

Could somebody outline the Induction outset and the goal?
I am unsure where to put in my (n + 1).
My approach would be:
2⁰ - 1 + 2^(n+1) -1


问:
Angle chasing in the triangle with incenter

In a triangle $ABC$ if $AD$, $BE$, $CF$ are the angle bisectors of $\angle A$, $\angle B$ $\angle C$ respectively with incenter $I$ and let $\angle BAC=2x$, $\angle ABC=2y$, $\angle ACB=2z$. Find all the possible angle in the triangle $DEF$ in term of $x,y,z$

I was able to find $\angle CID=\angle AIF=x+z$, $\angle BIF=\angle CIE=y+z$ and $\angle BID=\angle AIE=x+y$

I wasn't able to find $\angle IFD$,$\angle IDF$, $\angle IDE$, $\angle IED$, $\angle IEF$, $\angle IFE$. Please help

Body:
This is not a solution, but a 100% way to calculate angles of $EFD$ with a bit of trig.

Let $AB=c$, $BC=a$, $CA=b$. $\frac{BD}{DC}= \frac{AB}{AC}=\frac{c}{b}$, and $BD+DC = a \implies$ $$CD = \frac{ab}{b+c}, BD =\frac{ca}{b+c}.$$
Similarly, we can see that $$BF = \frac{bc}{a+b}, FA =\frac{ac}{a+b}.$$ $$AE = \frac{ab}{a+c}, EC =\frac{bc}{a+c}.$$ Now, we can apply cos theorem in $\triangle AFE: FE^2 = AF^2+AE^2-2\cdot AF \cdot AE \cdot cos(\angle A) = (\frac{ac}{a+b})^2+(\frac{ab}{a+c})^2-2\cdot \frac{ab}{a+c} \cdot \frac{ac}{a+b} \cdot \frac{b^2+c^2-a^2}{2bc} = a^2(\frac{c^2}{(a+b)^2}+\frac{b^2}{(a+c)^2}+\frac{b^2+c^2-a^2}{(a+c)(a+b)})= \frac{a^2(a + b + c) (-a^3 + 2 a b^2 + b^3 - a b c + 2 a c^2 + c^3)}{(a+b)^2(a+c)^2}\implies $ $$FE^2 = \frac{a^2(a + b + c)(b^3+c^3-a^3-abc+2ab^2+2ac^2)}{(a+b)^2(a+c)^2}$$ $$FD^2 = \frac{b^2(a + b + c)(a^3+c^3-b^3-abc+2ba^2+2bc^2)}{(b+a)^2(b+c)^2}$$ $$ED^2 = \frac{c^2(a + b + c)(a^3+b^3-c^3-abc+2ca^2+2cb^2)}{(c+a)^2(c+b)^2}.$$ Now you can see that $$EF^2=FD^2+DE^2-2\cdot FD \cdot FE \cdot cos(\angle FDE)\implies cos(\angle FDE)=\frac{FD^2+DE^2-EF^2}{2 \cdot FD \cdot FE}=\frac{b^2(a^3+c^3-b^3-abc+2ba^2+2bc^2)(a+c)^2+c^2(a^3+b^3-c^3-abc+2ca^2+2cb^2)(a+c)^2-a^2(b^3+c^3-a^3-abc+2ab^2+2ac^2)(b+c)^2}{2ab\sqrt{(b^3+c^3-a^3-abc+2ab^2+2ac^2)(a^3+c^3-b^3-abc+2ba^2+2bc^2)}(a+b)(b+c)} = \frac{2 a^5 b^2 + 2 a^5 b c + 2 a^5 c^2 + 2 a^4 b^3 + 2 a^4 b^2 c + 4 a^4 c^3 - 2 a^3 b^4 - a^3 b^2 c^2 - 4 a^3 b c^3 + 3 a^3 c^4 - 2 a^2 b^5 - 2 a^2 b^4 c + 2 a^2 b^3 c^2 + 2 a^2 b^2 c^3 - 4 a^2 b c^4 - 2 a b^5 c + 5 a b^3 c^3 + 6 a b^2 c^4 - a b c^5 - 2 a c^6 - b^5 c^2 + 3 b^3 c^4 + 3 b^2 c^5 - c^7}{2ab\sqrt{(b^3+c^3-a^3-abc+2ab^2+2ac^2)(a^3+c^3-b^3-abc+2ba^2+2bc^2)}(a+b)(b+c)}$$.

Note, that $a:b:c = sin(\angle A): sin(\angle B): sin(\angle C) = sin(2x):sin(2y):sin(2z)$. Plug this back you will find $\angle EDF$ in terms of $x,y,z$


问:
Multiply circulant matrix by usual vector
I found nice way to multiply circulant matrix by a usual vector (page 2):

http://web.mit.edu/18.06/www/Spring17/Circulant-Matrices.pdf

It means that I can find $\vec{y} = C\vec{x}$ in the following way (example in Python):

```<code>"""
I want to multiply the following matrix
            [[1, 2, 3],
             [3, 1, 2],
             [2, 3, 1]]
by vector [1, 2, 3]
"""
import numpy as np

# first raw of the circulant
c = [1, 2, 3]
# vector
x = [1, 2, 3]

# circular convolution
y = np.fft.ifft(np.fft.fft(c) * np.fft.fft(x))

</code>```

The result I obtained in this way is array([13.+0.j, 13.+0.j, 10.+0.j]) but the actual result that can be obtained by straight matrix-vector multiplication is array([14, 11, 11]). Why these results are different?

答:
So guys, I found the answer. These results are different becase the multiplication of circulant matrix by vector is not circular convolution.

It's easy to prove that
$$
\mathbf{y}=\hat{C}\mathbf{x}=DFT[DFT[\mathbf{c}]*IDFT[\mathbf{x}]]
$$
where $\mathbf{c}$ is the first row of the circulant matrix, $DFT[\cdot]$ denotes discrete Fourier transform, and $*$ is component-wise multiplication (Hadamart product).

To prove that, use properties of diagonalizable matrices and DFT-related properties of circulants matrices.

Let it remain as an exercise :)


问:
Numerical method to calculate sum of infinite series?
for example: I have a series

is there numerical computation method to find it ? thanks

Body:
Yoy can write
$$(1 - 4 n^2)^2 (36 n^2 + 1)=(2 n-1)^2 (2 n+1)^2 (6 n-i) (6 n+i)$$ Using partiai fractions, the summand is
$$\frac{7}{100 (2 n+1)}-\frac{7}{100 (2 n-1)}+\frac{81 i}{200 (6
   n+i)}-\frac{81 i}{200 (6 n-i)}+$$ $$\frac{1}{40 (2 n+1)^2}+\frac{1}{40 (2 n-1)^2}$$

Computing the partial sums up to $p$, we have at the end for $800 S_p$
$$\frac{56}{2 p+1}+\frac{5}{\left(p+\frac{1}{2}\right)^2}-54 i \psi
   \left(p+\left(1-\frac{i}{6}\right)\right)+54 i \psi
   \left(p+\left(1+\frac{i}{6}\right)\right)-10 \psi
   ^{(1)}\left(p+\frac{1}{2}\right)+5 \pi ^2-400+54 \pi  \coth \left(\frac{\pi
   }{6}\right)$$

Now, using asymptotics and Taylor expansions
$$S_p=\frac{1}{800} \left(5 \pi ^2+54 \pi  \coth \left(\frac{\pi
   }{6}\right)-400\right)-\frac{1}{2880 p^5}+\frac{1}{1152
   p^6}+O\left(\frac{1}{p^7}\right)$$

Computing for $p=10$
$$S_{10}=\frac{89936259827237312633515635583570094}{29615662049611209795735649868626520625}$$ which is
$0.00303678032$ while the above truncated expansion gives

$$S_{10}\sim -\frac{192000001}{384000000}+\frac{\pi ^2}{160}+\frac{27}{400} \pi  \coth
   \left(\frac{\pi }{6}\right)$$ which is $0.00303678042$


问:
How many commutative rings of order $p^2$
This question was part of my algebra quiz and I was unable to solve 1 option out of 4. So, I am asking it here.

There exists exactly 2 commutative rings of order $p^2$ are there?

I have a intuition that it's true as only two groups exists and as we are adding a multiplicative operation so that must be true but it's given false.

So, kindly tell what mistake I am making. I studied algebra from Joseph Gallian.

答:
More fun if $R$ is not assumed unital. $p$ is prime. I think we get the following cases:

If $char(R) = p^2$ (the least integer such that $\forall r\in R,nr=0$) we get one of $\Bbb{Z}/p^2\Bbb{Z},p\Bbb{Z}/p^3\Bbb{Z},p^2\Bbb{Z}/p^4\Bbb{Z}$.

Otherwise $char(R)=p$ so $R$ is a two dimensional vector space $R=a\Bbb{F}_p+b\Bbb{F}_p$. Let $Ann(R) = \{ r\in R, \forall s\in R, rs=0\}$.

If $Ann(R)=\{0\}$ then looking at how the multiplication by $a$ and $b$ acts on this basis we get an embedding $R\to M_2(\Bbb{F}_p)$ where the image is generated by two commuting matrices. If the characteristic polynomial of one of them is irreducible then $R\cong \Bbb{F}_{p^2}$.

If the characteristic polynomial of one of them has two distinct roots then in the basis diagonalizing it the two matrices are diagonal and $R=\pmatrix{u&0\\0&v}=\Bbb{F}_p\times \Bbb{F}_p$.

Otherwise in some basis the two matrices are of the form $\pmatrix{c&d\\0&c}$ so that  $R =\pmatrix{u&v\\0&u}=\Bbb{F}_p[t]/(t^2)$.

If $Ann(R)$ has $p^2$ elements then $R =p\Bbb{Z}/p^2\Bbb{Z}\times p\Bbb{Z}/p^2\Bbb{Z}$.

Otherwise $Ann(R)$ has $p$ elements, wlog $Ann(R)= b\Bbb{F}_p,ab=b^2=0,a\not \in Ann(R)$, so $a^2\ne 0$, the ideal $aR=a^2\Bbb{F}_p$ is a subring.

If $a^2\not \in Ann(R)$ then $R =a^2\Bbb{F}_p+b\Bbb{F}_p=\Bbb{F}_p\times p\Bbb{Z}/p^2\Bbb{Z}$.

The last case is $a^2 \in Ann(R)$, wlog $a^2=b$ together with $ab=b^2=0$ this defines the ring structure of $R$.


问:
Solving two algebraic equations
I'm trying to prove an identity from physics. I have the following two equations ($M$ is a constant):

$$e^2 = \left(1-\frac{2M}{r}\right)\left(1+\frac{l^2}{r^2}\right)$$ and $$r = \frac{l^2}{2M}\left[1 - \sqrt{1 - 12\left(\frac{M}{l}\right)^2}\right]$$  and I need to show that $$\frac{l}{e} = \sqrt{Mr}\left(1 - \frac{2M}{r}\right)^{-1}$$

I've tried rearranging the second eqation to get $l^2$ on its own and then dividing this by $e^2$ from equation 1 to get an expression for $l^2/e^2$ but this is quite messy and I can't see a clear way to simplify it. Can anyone see a straightforward way of doing this?

答: I'd let $\alpha =l/e$  so $ e=l/\alpha$ can be substituted in. Then one has $\alpha,l$ as coordinates. Eliminate $l$ and one is left with an equation for just $\alpha$. There might be neater ways of doing it, but this sort of method is at least free of guesswork.</p>

Title: Solving two algebraic equations


问:
Another question about "all odd moments vanish"
[Question inspired by Example of non-degenerate random variable with odd moments = 0  ]

Suppose $X$ is a real random variable such that all odd momenents vanish.
That is $\mathbb E[X^{2n+1}]=0$ for $n=0,1,2,3\cdots$.  Does it follow
that $X$ is symmetrically distributed about $0$?  That is, $X$ and $-X$ have the same
distribution.

Note: the case where $X$ is bounded is found here: Proof that $\mathbb{E} X^k = 0$ for all odd $k$ implies $X$ symmetric for bounded $X$ without characteristic functions

答:
Let $X$ have density
$$
f(x) = \frac1{48}\left(1-\mathsf{sign}(x)\sin\left(|x|^{\frac14}\right)\right) e^{-|x|^{\frac14}},\ x\in\mathbb R.
$$
Then for each positive integer $n$ we have
$$
\mathbb E[X^n] = \int_{-\infty}^\infty x^n f(x)\ \mathsf dx = \frac{(1+(-1)^n)(4(n+1))!}{12}.
$$
It follows immediately that $\mathbb E[X^{2n+1}=0]$ for all nonnegative integers $n$. It is clear though that $f$ is not an even function, so $X$ is not symmetric about zero.


问:
Expected time until random walk on hexagonal grid exceeds distance N from start
A particle starts in a cell in an infinite hexagonal grid, and every second, jumps to an adjacent cell uniformly randomly. What is the expected amount of time until the particle is $N$ cell-jumps away from its starting point? With some linear algebra, for example, one finds values of $1$, then $10/3$, then $213/29$, for cases $N=1,2,3$ respectively. Computer simulation shows growth is approximately $4N^2/5$.

I expected to be able to solve this problem with similar methods (using polynomials in barycentric coordinates, constrained by dihedral symmetries) as to my recent Puzzling question, but to no avail so far. Curiously, by a coupling argument, this problem is equivalent to computing the expected value of the variable $\text{min}\{X_1,X_2\}$ where $X_i$ are iid variables representing the escape time of the honeybee from the center of its triangle in the linked problem, but that observation doesn't seem to help much.

Some rambling about my current attempts: In barycentric coordinates $(\alpha, \beta, \gamma)$ whereby we always have $\alpha + \beta + \gamma = 3N$, it would seem reasonable to demand that—in order to find the average escape time at $(\alpha, \beta, \gamma)$ from the $N-1$-hexagon centered at $(N,N,N)$—we find a function $H(\alpha, \beta, \gamma)$ algebraically satisfying the "average-of-6-neighbors-plus-1" property everywhere, which also satisfies $H = 0$ whenever $\alpha = 0, 2N$ or $\beta = 0, 2N$ or $\gamma = 0, 2N$.

After all, this approach is exactly how the triangular escape time problem is solved, just leaving out the $2N$ constraints. In that case, we think of the elementary symmetric polynomials in $\alpha, \beta, \gamma$, and realize $\alpha\beta\gamma$ is a good candidate. It doesn't quite satisfy the averaging-plus-one law—its difference from its nearby-average function is $3N$ and not $1$—so we tweak it to $\frac{3\alpha\beta\gamma}{\alpha+\beta+\gamma}$ to solve the problem.

So that is how I proceeded here, examining the obvious candidate $H=\alpha \beta \gamma (\alpha-2\beta-2\gamma)(\beta-2\alpha-2\gamma)(\gamma-2\alpha-2\beta)$. But its difference from its nearby-average function is gnarly, and not susceptible to obvious tweaks. With some thought, one realizes the field of rational functions invariant up to angular and mirror symmetry are generated by $H$ as well as $e_1 = \alpha+\beta+\gamma$ and $e_2 = \alpha \beta + \alpha \gamma + \beta\gamma$. Especially considering the empirical evidence that our formula will be of degree $2$, one might try candidate tweaks like $\frac{H}{e_1^4}$ or $\frac{H}{e_1^2 e_2}$ or $\frac{H}{e_2^2}$ or $\frac{H^2}{e_1^4 e_2^3}$... but some time spent in Mathematica proved fruitless.

It's become clear to me now that no rational function of the form $\frac{F}{e_1^n e_2^m}$ will satisfy the criteria of the first paragraph, because such a function will still be defined on and inside the full triangular region, thus restricting to a solution of the honeybee escape time problem. By standard Markov chain reasoning, this solution is unique, and obviously not the solution to the problem at hand. So, either an even more complicated denominator is needed (one giving poles outside the hexagon but inside the triangle), or we need to allow possibilities like $H \neq 0$ even if $\alpha = 0$ as long as we're outside the hexagonal boundary, or we need some even more radical change to our techniques.

答:
Let us encode the hexagonal grid using the hexagonal lattice

$$ \mathsf{G} = \{ a + b \omega : a, b \in \mathbb{Z} \}, \qquad \omega = e^{i\pi/3},$$

where each $z \in \mathsf{G}$ represents the center of a hexagonal cell. Then two cells $z_1$ and $z_2$ are adjacent precisely when $\left| z_1 - z_2 \right| = 1$.

We also write $\mathsf{C}_n$ for the set of all cells with are precisely $n$ cells away from the origin.

Now let $(X_n)_{n\geq0}$ denote the simple random walk on $\mathsf{G}$, started at $X_0 = 0$. Denote by $\tau_n$ the hitting time of $\mathsf{C}_n$. Then by the second Wald's identity, the expectation of $\tau_n$ is

$$ \mathbb{E}[\tau_n] = \mathbb{E}\bigl[ \left| X_{\tau_n} \right|^2 \bigr]. $$

Now, if we define the continuous-time process $\tilde{X}^{(n)}_t = \frac{1}{n} X_{\lfloor n^2 t\rfloor}$ by the diffusive scaling of $X$, then by the invariance principle, $\tilde{X}^{(n)}$ converges to the complex Brownian motion $W$ started at $0$. So, if $\ell$ denotes the constant factor appearing in the asymptotic formula for $\mathbb{E}[\tau_n]$, then

$$ \ell
= \lim_{n\to\infty} \frac{1}{n^2}\mathbb{E}[\tau_n]
= \mathbb{E}\bigl[ \left| W_{\tau} \right|^2 \bigr]
= \int_{\mathsf{C}} \left| z \right|^2 \, \mathbb{P}(W_{\tau_{\mathsf{C}}} \in \mathrm{d}z), $$

where $\mathsf{C}$ is the regular hexagon with vertices $e^{ik\pi/3}$ for $k = 0, 1, \dots, 5$, which arises as the "limit" of the rescaled set $n^{-1}\mathsf{C}_n$, and $\tau_{\mathsf{C}}$ is the hitting time of $\mathsf{C}$.

In order to compute the last integral, consider the Schwarz–Christoffel mapping

$$ \phi(z) = K \int_{0}^{z} \frac{1}{(1-\zeta^6)^{1/3}} \, \mathrm{d}\zeta $$

over the unit disc $\mathbb{D}$, and the normalizing factor $K$ is chosen as

$$ K = \left( \int_{0}^{1} \frac{1}{(1-x^6)^{1/3}} \, \mathrm{d}x \right)^{-1} = \frac{6 \cdot 2^{1/3} \pi^{1/2}}{\Gamma(\frac{1}{6})\Gamma(\frac{1}{3})} $$

so that $\phi(1) = 1$ holds. It is well-known that $\phi$ maps $\partial\mathbb{D}$ to $\mathsf{C}$, and $\phi$ is a conformal mapping from $\mathbb{D}$ to the interior of $\mathsf{C}$. So by the conformal invariance of $W$, we obtain

\begin{align*}
\ell
&= \int_{\partial\mathbb{D}} \left| \phi(w) \right|^2 \, \mathbb{P}(W_{\tau_{\partial\mathbb{D}}} \in \mathrm{d}w)
= \frac{1}{2\pi} \int_{0}^{2\pi} \bigl| \phi(e^{i\theta}) \bigr|^2 \, \mathrm{d}\theta \\
&= K^2 \sum_{n=0}^{\infty} \binom{-1/3}{n}^2 \frac{1}{(6n+1)^2}
\approx 0.80957626278006891494.
\end{align*}


问:
Random coloring of a graph with $c^2$ edges using c colors
Suppose that $|| ≤ ^2$ for some fixed $ ∈ N$. Prove that it is possible to assign a
color to each vertex of  so that at most 2 different colors are used in total and
for each edge, its end vertices have different colors.

First, I need to consider a random coloring using only c colors but I am not sure how to continue..

答: Consider a random colouring using $c$ colours.  For each edge, the probability that both vertices are the same colour is $1/c$.  The expected number of monochromatic edges is then $\le c^2/c = c$.  The probability that there are at most $c$ monochromatic edges is nonzero, so there exist colourings with $c$ colours that have at most $c$ monochromatic edges.  Now take such a colouring, and using each of the remaining $c$ colours at most once, recolour one vertex of each monochromatic edge.


问:
How can I prove that the extremes of the interval of the solutions of this differential equation are equilibrium points?
I'm trying to proof if $x:I\to \mathbb R$ a maximal regular solution of $x'=f(x)$, such that the image $x(I)\subset \mathbb R$ is bounded and $f:\mathbb R\to \mathbb R$ is $C^1$, then the extremes $a$ and $b$ of x(I) are equilibrium points of $x'=f(x)$, i.e., $f(a)=f(b)=0$.

I need help to begin to solve this question.

Thanks in advance

答:
I'll leave it to you to formally prove it, but think about this;

Define the interval $I$ as $a<0<b$ (wlog).

Suppose $f(a)<0$, then the trajectory of $x(0)=a$ decreases and the image of $x(I)$ is unbounded.

Suppose $f(a)>0$, then trajectory stays inside the interval (at least initially) but it's not maximal as by continuity of $f$, there exists an $a'<a$ such that $f(a')=0$.

Same argument for the upper bound.


问:
Assume $G$ is a finite group such that every maximal subgroup of $G$ is normal in $G$ and for $H \leq G$ we have that $HG’=G$,then show that $H=G$.
Assume $G$ is a finite group such that every maximal subgroup of $G$ is normal in $G$ and for $H \leq G$ we have that $HG’=G$,then show  that $H=G$.

Where $G^\prime$ is the commutator subgroup (AKA derived subgroup) of $G$ defined as $$G’=\langle \left\{\left[a,b\right]:a,b\in\ G\right\}\rangle.$$  I know that Nilpotent iff every maximal subgroup is normal,but I’m not sure if that is helpful.  It would be nice if someone helps me,thanks in advance.

答:
If $M$ is maximal and hence normal, then $[G:M]=p$ is prime (since $G/M$ has no proper nontrivial subgroups and so is cyclic of prime order), and hence $G/M$ is abelian. Therefore, $G'\subseteq M$. Thus, $G'$ is contained in each maximal subgroup.

If $H\neq G$, then let $M$ be a maximal subgroup containing $H$. Then $HG'\leq MG' = M\neq G$. Thus, if $HG'=G$, then $H=G$.


问:
How can I perform local averaging on any manifold?
Say that I want to do local averaging on a circle.

I have values for angles $\phi \in [0,1]$ where 1 "spins around" to 0. Let us call it $c[0,1]$

If I just do normal averaging $$\frac{1}{N}\sum_{i=1}^N \phi_i$$

I will get trouble because of the cut at 0,1.

I can instead remap $\phi \to \exp(2\pi i \phi)$ and change sum to product:  $$\prod_{i=1}^{N} \exp(2\pi i \phi_i/N)$$

This will create a smooth behaviour around the angular jump.

How can I generalize this to more complicated surfaces ?

Own work:

Say for example I have an ordered pair on $\{c[0,1],\mathbb R^+\}$ which I want to smooth.

We can use the same construction above but encode the $\mathbb R^+$ into the radius part of the complex number. But then our complex numbers are kind of "exhausted" when comes to degrees of freedom of what to be able to represent.

答:
Not an answer, too long for a comment.

tl;dr To get a useful answer I think you will have to edit the question to provide more context. I don't understand your "Say for example" example.

To compute averages on a surface (or manifold of any dimension) you need a way to measure areas on that surface. Your question seems to assume some kind of underlying "uniform distribution".

Your "average angle" example is in fact more subtle than you realize. See

Average of angles

and the wikipedia page referred to there.


问:
Asymptotic Variance of MLE Exponential
Suppose we have a random sample $(X_{1},.....,X_{n})$, where $X_{i}$ follows an Exponential Distribution with parameter $\lambda$, hence:

$F(x) = 1 - exp(-\lambda x)$

$\mathbb{E}(X_{i}) = \dfrac{1}{\lambda}$

$ Var(X_{i}) =\dfrac{1}{\lambda^{2}}$

I know that the MLE estimator $\hat{\lambda} = \dfrac{n}{\sum_{i=1}^{n}X_{i}}$, asymptotically follows a normal distribution, but I'm interested in his variance.
So, since $\sqrt n (\hat{\lambda} - \lambda)\stackrel{D}{\rightarrow} \mathcal{N}(0, \sigma^{2}) $

I was thinking about dealing with:

\begin{align}
\sigma^{2} &= Var\left[\sqrt n (\hat{\lambda} - \lambda)\right]\\
&= n Var (\hat{\lambda})\\
&= n Var \left[ \dfrac{n}{\sum_{i=1}^{n}X_{i}} \right] \\
&= n^{3} Var \left[ \dfrac{1}{\sum_{i=1}^{n}X_{i}}\right]
\end{align}

But I don't know how to proceed from here. I think the sum of exponentials follows a gamma with parameters $(n,\lambda)$, and that $\dfrac{1}{\sum_{i=1}^{n}X_{i}}$ then follows an inverse gamma?

答:
Yes  you are almost there.

You can calculate the variance of $\hat{\lambda}=\frac{n}{Y}$ remembering that

$$Y=\Sigma_i X_i \sim \text{Inverse Gamma}$$

and thus you immediately solve the problem as its variance is a known parameter

...but you can also solve the problem using the integration, and calculating

$$\mathbb{V}\Bigg[\frac{1}{Y}\Bigg]=\mathbb{E}\Bigg[\frac{1}{Y^2}\Bigg]-\mathbb{E}^2\Bigg[\frac{1}{Y}\Bigg]$$

It is not difficult:

$$\mathbb{E}\Bigg[\frac{1}{Y}\Bigg]=\int_0^{\infty}\frac{1}{y}\frac{\lambda^n}{\Gamma(n)}y^{n-1}e^{-\lambda y}dy=\frac{\lambda}{n-1}\underbrace{\int_0^{\infty}\frac{\lambda^{n-1}}{\Gamma(n-1)}y^{(n-1)-1}e^{-\lambda y}dy}_{=1}=\frac{\lambda}{n-1}$$

and similarly for the second simple moment

Anyway this is not the asymptotic variance but it is the exact variance. To calculate the asymptotic variance you can use Delta Method

After simple calculations you will find that the asymptotic variance is $\frac{\lambda^2}{n}$ while the exact one is $\lambda^2\frac{n^2}{(n-1)^2(n-2)}$

Title: Asymptotic Variance of MLE Exponential


问:
Contemporary Mathematical Columns in Magazines
In good old days, Scientific American was host to some legendary mathematical (and computer science) oriented columns that inspired generations of scientists and engineers. Douglas Hofstadter, Martin Gardner, A. K. Dewdney... just to name a few.

Where are the contemporary (XXI century) equivalent of these masterpieces? Which magazines can I point out to a layman and get him learning group theory? Hacking on Core Wars? Figuring out some Quine in LISP?

P.S. There are a lot of good websites for dissemination and self-study, but I feel there is some difference between those old SA columns, and what I find nowadays. For the lack of better words, allow me to make an analogy between an iPad and a Raspberry Pi; while the former appeals to the masses, the later gives you the opportunity to tinker. It's the difference between an ephemeral (if somewhat polished) entertainment (or good read), and the life-lasting recreational experience of hacking.

Heck, I've learned more on AI from 1990 Dewdney's "The Magic Machine", then most MOOCs out there.

答: There are some good columns at the MAA (Mathematical Association of America) website. (Websites are the 21st century equivalent of magazines.)</p>


问:
How to decompose a vector regarding complementary subspaces
How to show that $U_1:=\langle(1,1,1)^T\rangle$ and $U_2:=\langle(1,0,0)^T,(0,0,1)^T\rangle$ are complementary subspaces and how to decompose $(1,2,3)^T$ regarding these ? Best Regards, Thomas</p>

答:
You need to show that every $x \in \mathtt{R}^3$ (or whatever vector space you are interested in, your question doesn't specify one) can be uniquely represented as $$
  x = u_1 + u_2 \quad\text{where $u_1 \in U_1$ and $u_2 \in U_2$.}
$$

Since you already have a basis for $U_1$ and $U_2$, this is equivalent to showing that combining those basis vectors produces a basis for $\mathbb{R}^3$, which in turn is equivalent to showing that the matrix built from those basis vectors has determinant $\neq 0$. In your case, that means showing that $$
  \left|\begin{array}{lll}
    1 & 1 & 0 \\
    1 & 0 & 0 \\
    1 & 0 & 1
  \end{array}\right| \neq 0 \text{.}
$$

To decompose a vector into $u_1, u_2$, one way is to find it's coordinates relative to the basis created from the basis vectors of the subspaces. You can then easily combine one vector per subspace, whose sum will produce the original vector.


问:
Why can you not factorialize all Real numbers??
I am a high school student and I learnt Factorials today. The Professor told that you can only factorialize natural numbers. Why is it so??

Is there a way to get the factorial of all Real numbers???

答:
To see why, consider this: Take $\pi$ things. In how many different ways can you order them? Does that make sense?

If you discard the combinatorial aspect, and focus on the algebraic aspect (namely $0!=1$ and $n!\cdot(n+1)=(n+1)!$), then there are many ways to generalize this to be valid for (almost) all real numbers, but the most common is the so-called $\Gamma$ function:
$$
\Gamma(x)=\int_0^\infty t^{x-1}e^{-t}dt
$$
with $n!=\Gamma(n+1)$.

The $\Gamma$ function is undefined on $\{0,-1,-2,\ldots\}$. This is basically unavoidable if you want the defining algebraic property above to hold. On the other hand, we have $\Gamma(\pi+1)\approx 7.188$, giving something like an answer to the question in the first paragraph.

But note that the $\Gamma$ function is not the factorial. The factorial works on the natural numbers and only the natural numbers.


问:
Is there a mathematical notation for the region enclosed by a set of curves
I want to define the region $S$ as the area enclosed by the curves $C_1, C_2, C_3, C_4$ as in the picture below, but do not know what notation to use.
Any help would be appreciated.

答: AFAIK there is no standard notation for this.  Best to say it in words.</p>


问:
Partial derivative confusion in lagrangian mechanics
An excerpt from an answer:

Consider the Lagrangian:

$L = (0.5)m(\dot{x}^2+\dot{y}^2 +\dot{z}^2)-U$.

The lagrange equation :

$\dfrac{\partial{L}}{\partial{x}}=\dfrac{d}{dt}\dfrac{\partial{L}}{\partial{\dot{x}}}$.

Evaluating this, we get:

$ -\frac{\partial{U}}{\partial{x}} = F_{x} $.

My issue with the evaluation:

The evaluation suggests $\frac{\partial{\dot x}}{\partial{x}}=0$ and $\frac{d}{dt}\left(\frac{\partial{U}}{\partial{\dot x}}\right)=0$, which unfortunately doesn't make sense to me.

How can we say in general that $\dot{x}$ is always not a function of $x$? We might have a situation where, say,$\dot{x} =ax+b$. How do we justify $\frac{\partial{\dot x}}{\partial{x}}=0$ for such a scenario?

If the answer is somehow "treat everything except $x$ as constant", then I will need some justification on what comes under "everything",especially $\dot{x}$, which might have a dependence on $x$ as I point out.

答:
The whole point of lagrangian mechanics is to start rather more abstractly from a configuration space of two variables, without any attached trajectory. At the moment, we can vary $x(t)$ and $x’(t)$ independently, and this independent variation of these quantities just corresponds to different paths between two events. Now Lagrange discovered (and this is where the physics comes in), that if you draw all the possible paths between two events in this configuration space, the one that extremizes the action is the one that nature follows.

So to answer your question: the $x(t)$ and $x’(t)$ are unrelated because they are the configuration space variables first and foremost, and are completely independent. Only when you fix a path do they have any correlation.


问:
Primary School fractions question
My daughter asked me for help with the following question. While it's fairly easy to work out with a bit of common sense and trial and error, I was wondering if there is a simple way to work mathematically that you'd expect an 8 year old to understand?

答:
We want to calculate $\frac{2}{3}$ of something. To make it easy, that "something" should be a multiple of 3 because it's easy to divide those into thirds. If that something is 3, 6 or 9 (and so on), we get 2, 4 and 6 (and so on) as results (so all the even numbers)

At the same time, we want to calculate $\frac{3}{10}$ of something. For simplicity that something should be a multiple of 10. If that something is 10, 20 or 30 (and so on), we get 3, 6 and 9 (and so on) as results. (so all the multiples of 3)

We now see that 6 appeared as a result in both: Two thirds of 6 is the same as three tenths of 20. Because 6 is a multiple of both 2 and 3.

Title: Primary School fractions question