问:
Finding a better bound for $c$ such that: $\sum_{i=1}^n{w_i}<\sum_{i=1}^n{w_i'}\implies\sum_{i=1}^n{w_i^2}<c\cdot \sum_{i=1}^n{w_i'^2}$
I am trying to solve the assignment problem between 2 sets of $n$ points in $\mathbb{R}^d$ minimizing the sum of square distances, i.e. finding a pairing between the two sets such that each point is mapped to exactly one point from the other set and minimize the sum of square distances.

There are many fast solutions for the problem for the case of minimizing distances instead of squared distances, unfortunately there aren't such fast solutions for the squared distances case (at least that I know of) so I was wondering if I'll just use the sum of distances algorithm would that give me a constant approximation for the squared distances case. I.e, if the weights in the optimal sum distances case are $w_1,...w_n$ and in the squared distances case are $w_1',...,w_n'$ is it true that:

$$\sum_{i=1}^n{w_i}<\sum_{i=1}^n{w_i'}\implies\sum_{i=1}^n{w_i^2}<c\cdot \sum_{i=1}^n{w_i'^2},$$

where $c<<n$.

Edit:

I was able to prove this for $c=n$ using Equivalent norm properties (see https://en.wikipedia.org/wiki/Norm_(mathematics)#Equivalence):
$$\sum_{i=1}^{n}{w_i^2}\leq\left(\sum_{i=1}^{n}{w_i}\right)^2\leq\left(\sum_{i=1}^{n}{w_i'}\right)^2=\left\lVert w'\right\rVert_1^2\leq n\left\lVert w'\right\rVert_2^2=n\cdot \sum_{i=1}^{n}{w_i'^2},$$

but I wanted to know if there's something better than $n$.

答:
This might not be the best result, but at least it puts some bound on things......

I claim that with all $w_i, w_i' \ge 0$

$\sum_{i=1}^n{w_i}<\sum_{i=1}^n{w_i'}\implies\sum_{i=1}^n{w_i^2}  \le n^3\cdot \sum_{i=1}^n{w_i'^2}$

Proof:

For each $i$ we have $w_i \le \sum_{i=1}^n{w_i'} \le n.W'$ where $W' = max\{w_i'\}$

So with $w_i \ge 0$, we have $w_i^2 \le n^2W'^2$

Then $\sum_{i=1}^n{w_i^2} \le n.n^2 W'^2  = n^3 W'^2 $

And $W'^2 \in \{w_i'^2\}$ so $ W'^2 \le \sum_{i=1}^n{w_i'^2}$

Then $\sum_{i=1}^n{w_i^2} \le  n^3 \sum_{i=1}^n{w_i'^2}$

(to get a strict inequality, add some small $\epsilon$ to $n^3$)


问:
What is the symmetry between the definitions of the bounded universal/existential quantifiers?
What is the symmetry between the definitions of the bounded universal/existential quantifiers?

$\forall x \in A, B(x)$ means $\forall x (x \in A \rightarrow B(x))$

$\exists x \in A, B(x)$ means $\exists x (x \in A \land B(x))$

These make intuitive sense, but I would expect there to be some kind of symmetry between how the definitions of the bounded quantifiers work, and I can't see one. $A \rightarrow B$ means $\lnot A \lor B$ which doesn't seem to have a direct relationship with $A \land B$. What am I missing?

答:
As Asaf Karagila says, those two connectives, $\forall x \in A, \ldots$ and $\exists x \in A, \ldots$ are dual of each other.

Just like the conjunction / disjunction pair or the universal / existential pair, we have that $\lnot (\forall x \in A, \phi(x)) \Leftrightarrow \exists x \in A, \lnot \phi(x)$.

Also, if $A$ is a singleton $\{a\}$, those two connective are logically equivalent, and $\forall x \in A, \phi(x) \Leftrightarrow \exists x \in A, \phi(x) \Leftrightarrow \phi(a)$

ContentLicense: CC BY-SA 3.0


问:
Joint density function problem with integration
So the question is: $f(x,y)= (2/81)x^3$ for $0\leq x \leq3, 0 \leq y \leq 2$ and $0$ otherwise. Find the probability that (x,y) satisfies $x+y \leq 3$. My thought on this question is to find the intersection between f(x,y) and $y=-x+3$. And they intersect at $x=2.577$. Then I went to integrate two parts: $0-2.577$ under equation $f(x,y)$ with respect to dx and $2.577-3$ under $y=-x+3$ with respect to dx.

However, the answer appears to be incorrect. What is wrong with my thought process here?

答:
You're on the wrong track.  You have a joint probability density, so the graph is 3D.  Probability is represented by volume.  The condition $x+y \leq 3$ is a constraint on the domain; it is telling you where to integrate.  Since the full domain is $[0,3]\times [0,2]$, you want to do a double integral over the region where $y$ ranges over $0$ to $2$ and $x$ ranges from $0$ to $3-y$.
$$\int_{0}^{2} \int_{0}^{3-y} \frac{2}{81} x^3 dx dy$$
Draw the region in the x-y plane if this is unclear, or you can integrate in the other order but then you have to split it into 0-1 and 1-3 for the outer integral.

ContentLicense: CC BY-SA 4.0


问:
Non-trivial example of algebraically closed fields
I'm beginning an introductory course on Galois Theory and we've just started to talk about algebraic closed fields and extensions.

The typical example of algebraically closed fields is $\mathbb{C}$ and the typical non-examples are $\mathbb{R}, \mathbb{Q}$ and arbitrary finite fields.

I'm trying to find some explicit, non-typical example of algebraically closed fields, but it seems like a complicated task. Any ideas?

答: One interesting thing to note is that the first-order theory of algebraically closed fields of characteristic $0$ is $\kappa$-categorical for $\kappa > \aleph_0$. That means that there other algebraically closed field of characteristic $0$ of the same cardinality as $\mathbb{C}$! Maybe not the most helpful response, but I think quite interesting.


问:
Can an everywhere continuous real function $f:\mathbb{R}\to\mathbb{R}$ be strictly increasing at only one point?
My definition of "strictly increasing at one point" is Did's definition here. I'll type it out here anyway:

A function $f$ is increasing at $x$ if $f(t)\gt f(x)$ for every $t\gt x$ close enough to $x$ and $f(t)\lt f(x)$ for every $t\lt x$ close enough to $x$. More rigorously, one asks that there exists $\varepsilon\gt0$ such that, for every $(t,s)$ such that $x-\varepsilon\lt t\lt x\lt s\lt x+\varepsilon$, $f(t)\lt f(x)\lt f(s)$.

My question is:

Can an everywhere continuous real function be strictly increasing at
only one point?

I think this is harder to prove than first meets the eye: for example we cannot use the fact that a monotone function has limits from the left and right at every point of it's domain, because to use it we would need to assume our function is monotone which I don't think is closely related to what we are trying to show. Also, trying to use this theorem on an interval of the domain (either including $0$ or excluding $0$) doesn't really help, as far as I can see.

My attempt, which I think is correct, but would like verification (of one step in particular):

Suppose without loss of generality that $f$ is strictly increasing at the point $x=0$ only. Then $f$ is decreasing everywhere else, that is, when $x\neq 0.$

$f$ is strictly increasing at $x=0,\ $ therefore $\exists \varepsilon>0\ $ such that for every $\ t,s\ $ such that $-\varepsilon<t<0<s<\varepsilon,\quad f(t)<f(0)<f(s).\quad $ Therefore $\frac{-\varepsilon}{2}<0\quad$ with $\ u = f(-\frac{\varepsilon}{2})<f(0).\quad$ But $f$ is decreasing for all $x< 0,\ $ so therefore $\displaystyle\lim_{x\to 0^-} f(x) \leq u <f(0)\ \ (*),\ $ contradicting the continuity at $x=0.$

Is $(*)$ obviously true? Is it good enough, or do I need to prove it with definitions? Or is this a matter of taste?

Edit: Clearly the above proof is faulty, as I found out in the comments. My statement at the beginning of the proof, "Then $f$ is decreasing everywhere else, that is, when $x\neq 0.$" is unjustified. It seems the problem remains open.

答:
Let $W(x)$ be the Weierstrass function, which is continuous, but is nowhere strictly monotone. Wikipedia: "Rather between any two points no matter how close, the function will not be monotone."

Then I believe $f(x) = x|W(x)|$ is continuous (product of continuous functions is continuous), and strictly increasing only at $x=0$, although I don't know how to prove that this function is not strictly increasing at $x\neq 0,$ though if true, it is probably similar to proving that the Weierstrass function itself is nowhere strictly monotone.

I can see that the Weierstrass function is defined differently in wikipedia and wolframalpha. I think my answer will work either way.


问:
Does $\sin (a_n)$ converge?
Given that an arbitrary $a_n > 0$, and $\sum a_n$ converges, then does $\sum \sin (a_n)$ converge also? Honestly I don't know where to start with this question. I tried some examples like $\frac{1}{n^2}$, $\frac{n^2+n}{n^4+5}$, and they all seem to follow this statement. I am beginning to think that this statement is correct. But how?

答:
Welcome to MSE ^_^

Hint:

$$\sin(x) \leq x$$

Do you see how to use this to solve your problem? Try to compare your $\sin(a_n)$ series to your original one.

I hope this helps ^_^

Title: Does $\sin (a_n)$ converge?


答:
Let $ x>0$.

$$t\mapsto \sin(t)\text{ is differentiable at } [0,x]$$

So, by MVT,

$$\sin(x)-\sin(0)=x\cos(c)\le x$$

$$a_n>0\text{ and } \sum a_n \text{ convergent}$$

$$\implies \lim_{n\to +\infty}a_n=0$$

$$\implies 0<a_n<\color{red}{\pi} \text{ for large enough  } \; n$$

$$\implies 0<\sin(a_n)\le a_n$$
$$\implies \sum \sin(a_n)\text{ converges}$$
by the test of comparison.

Title: Does $\sin (a_n)$ converge?

Body: Given that an arbitrary $a_n > 0$, and $\sum a_n$ converges, then does $\sum \sin (a_n)$ converge also? Honestly I don't know where to start with this question. I tried some examples like $\frac{1}{n^2}$, $\frac{n^2+n}{n^4+5}$, and they all seem to follow this statement. I am beginning to think that this statement is correct. But how?


问:
Does $a=\frac{\langle x,y\rangle}{\| y \|^2}$ imply $\| a \| =\frac{\| x \|}{\| y \|}$?
As the title, um, $\overset{?}{\text{suggests}}$ I'm wondering if $a=\frac{\langle x,y\rangle}{\lVert y \rVert^2}\overset{\color{pink} \heartsuit}{\implies}\lvert a \rvert=\frac{\lVert x \rVert}{\lVert y \rVert}$. So what do you think?

EDIT$^1$:

Where $z=x-ay$ and $y$ and $z$ are orthogonal, with $y\neq 0$ and assuming that
$$\lvert \langle x,y\rangle \rvert = \lVert x \rVert \cdot \lVert y \rVert.$$

Perhaps this literature may be of help:

EDIT$^2$:

OK, I think I've got it know. Thank you for everyone's kind and considerate help.

答:
No. If $x\perp y$, then $a=0$ even if $x\neq 0$.

If $\lvert\langle x,y\rangle\rvert=\lVert x\rVert \lVert y\rVert$ (that is, $x,y$ are colinear), then certainly $\lvert a\rvert=\frac{\lVert x\rVert}{\lVert y\rVert}$.

Not sure about the loving part, though.


问:
Finding size of test based on Poisson random variables
Problem:

Let $(X,Y)$ be a random sample from the Poisson distribution with unknown mean $\theta$. In order to test $H_0:\theta=1$ against $H_1:\theta >1$, the following procedure is suggested:

If $X\leq 1$, then accept $H_0$.

If $X\geq 4$, then reject $H_0$.

If $X$ equals $2$ or $3$, then reject $H_0$ if $X+Y\geq 4$, otherwise accept $H_0$.

Find the size of the test.

My attempt:

$X$ and $Y$ follow $Poisson(1)$ distribution independently.

Size of the test

$=P_{H_0}(X\geq 4)+P_{H_0}(X=2, X+Y\geq 4)+P_{H_0}(X=3, X+Y\geq 4)$

$=P_{H_0}(X\geq 4)+P_{H_0}(X=2, Y\geq 2)+P_{H_0}(X=3, Y\geq 1)$

$=P_{H_0}(X\geq 4)+P_{H_0}(X=2)(Y\geq 2)+P_{H_0}(X=3)(Y\geq 1)$

$=[1-\sum_{x=0}^{3}e^{-1}\frac{1}{x!}]+\frac{e^{-1}}{2}[1-\sum_{y=0}^{1}e^{-1}\frac{1}{x!}]+\frac{e^{-1}}{6}[1-e^{-1}]$

$=0.10635$

I want to know if I have done it correctly, or otherwise, I want to know where the mistake is. Thanks in advance.

答:
Yes, your answer is correct.  The size of the test is the conditional probability of rejecting $H_0$ given that $H_0$ is true.  Since $H_0$ is rejected in exactly two mutually exclusive cases:

$X \ge 4$

$X \in \{2, 3\}$ and $X + Y \ge 4$

the latter of which can be further broken apart into the two sub-cases you used, then the desired conditional probability is $$\Pr[X \ge 4 \mid H_0] + \Pr[(X = 2) \cap (Y \ge 2) \mid H_0] + \Pr[(X = 3) \cap (Y \ge 1) \mid H_0]$$ which is, apart from minor notational differences, is the same as what you wrote.

As an exercise, how would you calculate the power curve for this test?  In other words, given $H_1 : \theta = \theta_1$ for some $\theta_1 \ne 1$, what is the power of this test to reject $H_0$ as a function of $\theta_1$?


问:
Upper bound on first $k$ terms of binomial expansion where the variable is probability
If $p$ is some probability, meaning $0<p<1$, is there a good upper bound for $ \sum^{k}_{i=0} \binom{n}{i} \left(\frac{p}{1-p}\right)^{i}$ in terms of $n,k,p$ ?

There is of course the trivial bound without capturing the dependence on $k$, $  \left(1+\frac{p}{1-p}\right)^{n} = \left(\frac{1}{1-p}\right)^{n} \geq  \sum^{k}_{i=0} \binom{n}{i} \left(\frac{p}{1-p}\right)^{i}$.

答:
Let $x=p/(1-p)$. Assuming $k\leq n$, your sum can be evaluated in terms of the hypergeometric function as
$$
S:=\sum_{i=0}^k\binom{n}{i}x^i=(x+1)^n-x^{k+1} \binom{n}{k+1} F\left({1,k-n+1\atop k+2};-x\right).
$$
We can see that when $k=n$ the second term vanishes leaving your trivial upper bound; thus our goal will be to make the second term smaller without making it zero. Using transformation formula for the hypergeometric function we can write
$$
F\left({1,k-n+1\atop k+2};-x\right)=(1+x)^{-1}F\left({1,n+1\atop k+2};\frac{x}{x+1}\right)>(1+x)^{-1},
$$
since $F(1,n+1; k+2;z)$ is increasing on $z\in(0,1)$ and $F(1,n+1; k+2;0)=1$. Hence,
$$
S\leq (x+1)^n-x^{k+1}(1+x)^{-1} \binom{n}{k+1},
$$
with equality when $k=n$. Your application will determine if this upper bound is good or not. Using some trial values in Mathematica shows that this bound is almost identical to $S$ for some parameters and can be much larger for others.

For example, denoting $S^\ast$ as the upper bound above we have
$$
(p,n,k)=(0.0662009, 5, 1)\implies (S,S^\ast)=(1.35447, 1.36149),
$$
while
$$
(p,n,k)=(0.628947, 5, 2)\implies (S,S^\ast)=(38.2065, 124.103).
$$
At the very least, this upper bound does as good or better than the trivial bound.


问:
Prove that if $A \setminus B = \emptyset$, then $A \subseteq B$
Prove that if $A \setminus B =  \emptyset$, then $A \subseteq B$.

The Venn Diagram helped me to visualize what I'm trying to show (thanks @GA316), but the book asks for a written proof (step by step) by contradiction. Sorry if I wasn't more specific at first, is just that I've had many troubles in the past with proofs, somehow I have many ideas but I can't seem to connect them to get to the final proof.

This is what I have so far:
$P \rightarrow Q$ is equivalent to $\neg Q \rightarrow \neg P$ contraposition (thanks @The Chaz 2.0)

With P: $ A \setminus B = \emptyset$  and Q: $ A \subseteq B$

so $ \neg Q \equiv A \not\subseteq B\ ,   \exists x \in A : x \notin B $

be $ t: t \in A \wedge  t \notin B $ ...is this right?

as this is the definition for $A \setminus B \ne \emptyset$ ...is this right?

$\therefore \neg Q \rightarrow \neg P \equiv A \not\subseteq B\ \rightarrow A \setminus B  \ne\emptyset$

I have many concerns regarding if I'm using the correct notation. I am trying to learn this by myself and have nobody else to ask.

Also, sorry if it took me too long to update, I just started learning about this LaTEX notation.

Thank you very much in advance, you guys are so nice and helpful. You made me feel very welcomed and sure I need to read more about the rules and instructions for using this site.

答:
We assume $A \subset E$, $B \subset E$ and $A \setminus B = \varnothing$

\begin{align*}% Utiliser & pour aligner en colonne. Enlever * pour numéroter chaque ligne et \nonumber en fin de ligne pour enlever la numérotation
 x \in A & \implies x \in A \wedge x \in E                                                                              \\
   & \implies x \in A \wedge x \in B \cup \complement_E B                                                         \\
   & \implies x \in A \wedge \left(x \in B \vee x \in \complement_E B\right)                           \\
   & \implies x \in A \wedge \left(x \in B \vee x \notin B\right)                                      \\
   & \implies \left(x \in A \wedge x \in B\right) \vee \left(x \in A \wedge x \notin B\right) \\
   & \implies x \in A \cap B \vee x \in A\setminus B                                                            \\
   & \implies x \in \left(A \cap B\right) \cup \left(A\setminus B\right)                                                   \\
   & \implies x \in \left(A \cap B\right) \cup \varnothing                                                                 \\
   & \implies x \in \left(A \cap B\right)                                                                                \\
   & \implies x \in A \wedge x \in B                                                                              \\
   & \implies x \in B
\end{align*}
Then $A \subset B$


问:
Can we define the set of reals as the set containing all the convergences of the sequences?
My analysis teacher defined the set of reals as, and I quote, a collection of finite numbers $\{A_0, A_1, ..., A_n\}$ and a collection of finite or infinite numbers $\{d_1, d_2, ..., d_m, ...\}$ all in the range $0$ to $9$; we match this collection with

$x_+=+A_n...A_1A_0.d_1d_2...d_m$ a positive real

$x_-=-A_n...A_1A_0.d_1d_2...d_m$ a negative real

For example, $\pi=3.141...$ is defined such as $A_0=3$, $d_1=1$, $d_2=4$, $d_3=1$, ...

I find this definition very strange, as it seems to be based on syntax alone (or almost). In the rest of the course, we went on to discuss Cauchy sequences and the Cauchy criterion and what it implies:

The sequence $(U_n)_{n\in\mathbb{N}}$ converges if and only if it is a Cauchy one. This equivalence gives $\mathbb{R}$ the property of being a complete set, i.e. any Cauchy sequence of real numbers converges in $\mathbb{R}$.

My intuition then led me to think that the set of reals could be defined as the set containing all the limits of the real Cauchy suites. However, this is only an intuition and I don't know how to show this, I haven't found any result on the internet.

I wonder therefore if this intuition is good, if not (or if it is, because it is always interesting), I would like to know in what other ways we could define the reals, other than by the definition I gave above.

PS: obviously, we could define $\mathbb{R}$ in ensemblistic terms, and say that it contains all naturals, integers, rationals and irrationals, but I was looking for another definition than in terms of sets.

答:
Your teacher seems be taking the "naive" and preliminary school definition of real numbers as any number can be written as decimals but the decimals might need to go on forever.  But then wants to make it formal and rigorous.

It seems to me your teacher thinks this is the best of both worlds but IMO it is the worst of both worlds.

It does work.  We can get to any value with a $\frac 1{10^k}$ accuracy by taking an number with a $k$ decimal place and we can hone in to finer degree by taking a $k+1$ decimal places.  So if we take an infinite sequence $a_k$ where $a_k$ has $k$ decimal places and $a_{k+1}$ has $k+1$ decimal places and $|a_{k+1} - a_k| < \frac 1{10^k}$ that is all the $k$ decimal places of $a_{k+1}$ are the same as $a_k$'s decimal places. then $a_k$ is a cauchy sequence and $a_k \to a_\infty$ an decimal with an infinite number of decimals defined to be $\lim\limits_{k\to \infty}a_k$.

(In other words if $a_0 = 3; a_1 = 3.1; a_2 = 3.12; a_3=3.141; a_4 = 3.1415$ and $a_k=\pi$ rounded down to $k$ decimals, the $\{a_k\}$ is a cauchy sequence and $\pi = \lim a_k$.)

The subtle and difficult thing about define the reals is you don't have a "place" in which to define them.  The intuitive reason for the decimal definition as that it is taken for granted that numbers "live" in a continuous infinitely precise space and since decimals can hone into everywhere it must be that every point in in the continuous space can be honed into be an infinite decimal expansion.

That sort of begs the question though.

Grown up mathematicians make an abstract conception.  The no that the rationals aren't good enough-- they aren't complete as the "jump" from below an irrational such as $\sqrt 2$ or $\pi$ to above without ever being able to hit $\sqrt 2$ or $\pi$. And they can't say $\lim a_k =\pi$ because that $\pi$ doesn't "live" anywhere? Where does $\pi$ live?

They say.  If we can invent a space that isn't the numbers but behaves exactly like the numbers then it is equivalent and for all intents and purposes is.

And they do exactlY what you suggest.

The take the set of all cauchy sequences.  And the say two cauchy sequences are essentially the same as each other if their terms eventually always get arbitrarily close to each other.  And thus they define the real numbers as basically being identified by a cauche sequence.

So we'd say $\pi$ is essentially the same thing as $\{3,3.1,3.14,3.141, 3.1415.....\}$ and that is essentially the same thing as $\{4, 3.2, 3.14, 3.142,3.1416,.....$.

And yes, that does work.  But there are a lot if details I glossed over.


答:
Lets see why we cannot in general take e.g. $c  < \frac{n}{1.21}$.

Let $w_1',\ldots, w_n'=\frac{1.1}{n}$ and $w_1 =1, w_2=\ldots = w_n=0$. Then $\sum w_i = 1  < \sum w'_i =1.1$ but $\sum w_i^2 = 1$ with $\sum w_i'^2 = \frac{1.21}{n}$. Thus $c> \frac{n}{1.21}$, and similarly $c > \gamma n$ for all $\gamma < 1$.

So without further knowledge, such a general $c$ must be at least $n$, otherwise the norm property that you reference could also just be improved in a general setting.

Title: Finding a better bound for $c$ such that: $\sum_{i=1}^n{w_i}<\sum_{i=1}^n{w_i'}\implies\sum_{i=1}^n{w_i^2}<c\cdot \sum_{i=1}^n{w_i'^2}$

Body:
I am trying to solve the assignment problem between 2 sets of $n$ points in $\mathbb{R}^d$ minimizing the sum of square distances, i.e. finding a pairing between the two sets such that each point is mapped to exactly one point from the other set and minimize the sum of square distances.

There are many fast solutions for the problem for the case of minimizing distances instead of squared distances, unfortunately there aren't such fast solutions for the squared distances case (at least that I know of) so I was wondering if I'll just use the sum of distances algorithm would that give me a constant approximation for the squared distances case. I.e, if the weights in the optimal sum distances case are $w_1,...w_n$ and in the squared distances case are $w_1',...,w_n'$ is it true that:

$$\sum_{i=1}^n{w_i}<\sum_{i=1}^n{w_i'}\implies\sum_{i=1}^n{w_i^2}<c\cdot \sum_{i=1}^n{w_i'^2},$$

where $c<<n$.

Edit:

I was able to prove this for $c=n$ using Equivalent norm properties (see https://en.wikipedia.org/wiki/Norm_(mathematics)#Equivalence):
$$\sum_{i=1}^{n}{w_i^2}\leq\left(\sum_{i=1}^{n}{w_i}\right)^2\leq\left(\sum_{i=1}^{n}{w_i'}\right)^2=\left\lVert w'\right\rVert_1^2\leq n\left\lVert w'\right\rVert_2^2=n\cdot \sum_{i=1}^{n}{w_i'^2},$$

but I wanted to know if there's something better than $n$.


问:
Every topology for which every singleton subset is dense must be trivial
The whole exercise is:

Prove that a topological space is equipped with the trivial topology if and only if every singleton subset is dense.

$(\Rightarrow)$ Let $X\neq \emptyset$ be a topological space equipped with the trivial topology  $\tau = ${$\emptyset,X$}, then the closed sets are $\emptyset$ and $X$.

Let $A = ${$x_0$} with $x_0 \in X$. Since the only closed set containing $A$ is $X$, then we must have $\overline{A} = X$. So $A$ is dense in $X$.

$(\Leftarrow)$
To prove the topology is the trivial one, then we must prove that if $G \in \tau$, $G \neq \emptyset$, then $G=X$.

Since $G \neq \emptyset$, then there exists $x_0 \in G \subseteq X$. Let $A = ${$x_0$}$ \subseteq X$. Since $A$ is a singleton subset, then $\overline{A} = X$.

Also, there's this property: $\overline{\overline{A} \cap G} = \overline{A \cap G}$, which implies $\overline{G} = \overline{A}$ (note that $A \subseteq G$), then $\overline{G} = X$.

I don't know how to continue from here, or if there's another way. Any help is welcome.

答: Prove the contrapositive instead: suppose that $\varnothing\ne G\ne X$, and find a singleton that is not dense in $X$. HINT: Let $x_0\in X\setminus G$, and let $A=\{x_0\}$.


问:
Interchanging liminf and supremum under certain conditions?
Suppose a situation whereby I know for some bounded collection $a_{m,n}$ of real numbers the following:

For any $n \geq 1$, $\displaystyle\liminf_{m\to\infty} a_{m,n} \leq B_{n}$ and we can choose the $B_{n}$ such that $\sup_{n\geq 1}B_{n} <\infty$.

Can I conclude that $\displaystyle\liminf_{m \to \infty} \left(\sup_{n\geq 1}a_{m,n}\right) \leq \sup_{n\geq 1} B_{n}$?

Counter-examples also appreciated.

答:
Here's a counter-example: let $a_{m, n} = \begin{cases} 1 & m \equiv n \pmod{2} \\ 0 & m \not \equiv n \pmod{2} \end{cases}$. This is clearly bounded. Then for a fixed $n$ we have $\liminf\limits_{m \to \infty} a_{m, n} = 0$ and so we can pick $B_n = 0$ and so $\sup\limits_{n \geq 1} B_n = 0$. However, for any fixed $m$ we have $\sup\limits_{n \geq 1} a_{m, n} = 1$ and so
$$
\liminf\limits_{m \to \infty} \left ( \sup\limits_{n \geq 1} a_{m, n} \right ) = \liminf\limits_{m \to \infty} 1  = 1 > 0 = \sup\limits_{n \geq 1} B_n.
$$.


问:
Probability Distribution of Rolling Multiple Dice
What is the function for the probability distrabution of rolling multiple (3+) dice. The function is a bell curve but I can't find the actual function for the situation. Example, what is the function for rolling 50 6 sided dice?

EDIT: A six sided die returns an integer value from 1 to 6 inclusive. I am trying to find a function to add multiple dice together.

答:
In this answer, there is a section titled "Summing Dice". It describes how convolution of the discrete function that is $1$ for each integer from $1$ through $6$, and $0$ otherwise, yields the distribution for the sum of $n$ six-sided dice.

Rolling $50$ six-sided dice will yield an approximately Normal Distribution whose mean is $\mu=50\times\frac72$ and whose variance is $50\times\frac{35}{12}$; thus, a standard deviation of $\sigma=\sqrt{50\times\frac{35}{12}}$.

Mean and Variance of a Single Die

The mean of a single die whose faces vary from $1$ to $n$ is $\frac{n+1}{2}$. For $n=6$, this gives $\frac72$.

The variance of a single die whose faces vary from $1$ to $n$ is "the mean of the squares minus the square of the mean." The sum of the squares from $1$ to $n$ is $\frac{2n^3+3n^2+n}{6}$, so the mean is $\frac{2n^2+3n+1}{6}$. Subtracting $\frac{n^2+2n+1}{4}$ yields $\frac{n^2-1}{12}$. For $n=6$, this gives $\frac{35}{12}$.


答:
The trivial topology on a space $X$ can be written as $\{\emptyset, X\}$. We can also cleverly write this topology as $\{U \in \mathscr{P}(X)$ : $(\exists x \in X (x \in U)) \to U = X\}$.

In fact, the second definition is required if you're using constructive logic. But I digress.

We also note that a useful definition of "$A$ is dense" is the statement: for all open $U$, if $U$ has an element then $A \cap U$ has an element.

Suppose that every singleton subset of $X$ is dense. Now let $U$ be an open subset of $X$. Suppose we have some $x \in U$. Then consider an arbitrary $y \in X$. We see that $\{y\}$ is dense. And we see that $U$ is an open set, and $U$ has an element. So it must be the case that there is an element of $\{y\} \cap U$; that is, we must have $y \in U$. Then $U = X$.

This shows that $X$ has the trivial topology.

Title: Every topology for which every singleton subset is dense must be trivial

Body:
The whole exercise is:

Prove that a topological space is equipped with the trivial topology if and only if every singleton subset is dense.

$(\Rightarrow)$ Let $X\neq \emptyset$ be a topological space equipped with the trivial topology  $\tau = ${$\emptyset,X$}, then the closed sets are $\emptyset$ and $X$.

Let $A = ${$x_0$} with $x_0 \in X$. Since the only closed set containing $A$ is $X$, then we must have $\overline{A} = X$. So $A$ is dense in $X$.

$(\Leftarrow)$
To prove the topology is the trivial one, then we must prove that if $G \in \tau$, $G \neq \emptyset$, then $G=X$.

Since $G \neq \emptyset$, then there exists $x_0 \in G \subseteq X$. Let $A = ${$x_0$}$ \subseteq X$. Since $A$ is a singleton subset, then $\overline{A} = X$.

Also, there's this property: $\overline{\overline{A} \cap G} = \overline{A \cap G}$, which implies $\overline{G} = \overline{A}$ (note that $A \subseteq G$), then $\overline{G} = X$.

I don't know how to continue from here, or if there's another way. Any help is welcome.


答:
We want to prove that the only nonempty closed set $G$ in $X$ is $X$ itself (if this is the case then the same will also go for open sets, since they are just the complements of the closed sets).

As you already reasoned, since $G$ is nonempty it includes a singleton $\{x\}$, whose closure will have to be $X$. To complete the proof you just need two more lemmas:

The closure operator is increasing. So, since $\{x\} \subseteq G$, we have $\overline{\{x\}} \subseteq \overline{G}$ and hence $\overline{G} = X$.

Since $G$ is closed, it is equal to its closure. Hence $G = X$.


问:
Is there any online resource which can calculate the equation for the curve with given points?
I have several sets of Cartesian coordinates and I am looking for a resource which can generate the formula/equation/function for a curve (or line) which passes through these points.

The first set is:

1,1 - 2,2 - 3,5 - 4,7, - 5,12 - 6,15 - 7,22 - 8,35 - 9,40

All help is greatly appreciated. Thanks.

答:
A search found OEIS sequence A024702 "a(n) = (prime(n)^2 - 1)/24". For your sequence of values $\,(1,1),\,(2,2),\,(3,5),\dots\,$ The formula is
a(n) = (prime(n+2)^2-1)/24. The first few values are:

1, 2, 5, 7, 12, 15, 22, 35, 40, 57, 70, 77, 92, 117, 145, 155, 187, 210, 222

You are interested OEIS sequence A000040 "prime numbers" with approximation

```<code>a(n) = n log n + n log log n + (n/log n)*(log log n - log n - 2) + O( n (log log n)^2/ (log n)^2).
</code>```

which is probably what you are looking for. Consult the OEIS entry for more
information.

Body:
A search found OEIS sequence A024702 "a(n) = (prime(n)^2 - 1)/24". For your sequence of values $\,(1,1),\,(2,2),\,(3,5),\dots\,$ The formula is
a(n) = (prime(n+2)^2-1)/24. The first few values are:

1, 2, 5, 7, 12, 15, 22, 35, 40, 57, 70, 77, 92, 117, 145, 155, 187, 210, 222

You are interested OEIS sequence A000040 "prime numbers" with approximation

a(n) = n log n + n log log n + (n/log n)*(log log n - log n - 2) + O( n (log log n)^2/ (log n)^2).

which is probably what you are looking for. Consult the OEIS entry for more
information.


问:
Haar measure on $\mathbb{R}\backslash\{0\}$
I do not see how $\frac{dx}{|x|}$ can be a Haar measure on $\mathbb{R}\backslash\{0\}$ given that a Haar measure $\mu$ is supposed to have the property that $\int f\mu=\int L_yf \mu$ for all $y\in G$ where $L_{y}f(x)=f(y^{-1}x)$ and $f$ is a positive, compactly supported function on G.  Take $f(x)=x^{2}$, for example (ignoring the fact that this is not compactly supported).  Then $\int \frac{(y^{-1}x)^{2}}{y^{-1}x}dx=\int \frac{x}{y}dx\neq\int xdx=\int \frac{x^{2}}{x}dx$ for any $y\neq 1$.  Heuristically, this would seem to suggest that this definition doesn't work.  I suppose $dx$ should be thought of as a $1$-form so that $d(y^{-1}x)=y^{-1}dx$, but this is not the meaning of symbols like $\frac{dx}{x}$ in measure theory, so what is the right way to describe this situation?

EDIT:  More generally, let $G$ be an open subset of $\mathbb{R}^{n}$ with left multiplication given by $xy=A(x)y+b(x)$ where $A$ is an invertible linear transformation.  When it is said that "$|$det$A(x)|^{-1}dx$ is a Haar measure on $G$", should this be interpreted as "$|$det$A(x)|^{-1}dx$ is a left-invariant $n$-form" with the understanding that integration is with respect to differential forms as opposed to the integration developed in measure theory?

My confusion is that in measure theory when you say "$d\mu=f(x)d\lambda$ is a measure" this means $\int g(x) d\mu=\int g(x)f(x) d\lambda$.  This is purely a conceptual question: how do you think of these "Haar measures" in a measure theoretic way?  In other words, given a left-invariant differential $n$-form $\omega$ on an $n$-dimensional Lie group $G$, how can you describe the Haar measure arising from the positive linear functional $f \mapsto \int f\omega$?

答:
I think this is the right way to view it.  Start with
$$\int f(x)\frac{dx}{x}$$
For some $a \ne 0$, change variables $y = a^{-1}x$, so that $dy = a^{-1}\;dx$,
$$
\int f(x)\frac{dx}{x} = \int f(ay) \frac{a dy}{ay} = \int f(ay)\frac{dy}{y} .
$$


问:
The Scalar-to-Matrix Derivative of $\frac 1 2 \| V \sigma \left( Wx \right) - y \|_2^2$ w.r.t. $W$
I'd appreciate your help in explaining the steps of calculating $\frac {\partial L} {\partial W}$, where

$$
L := \frac  1 2 \| V \sigma \left( Wx \right) - y \|_2^2
$$

and $
x \in \mathbb{R}^{(d \times 1)},
W \in \mathbb{R}^{(k \times d)},
V \in \mathbb{R}^{(m \times k)},
y \in \mathbb{R}^{(m \times 1)}
$, for $\sigma \left( u \right) := \max \left( 0, u \right)$ element-wise.

答:
In the scalar case, the derivative of the ReLu function
is the Heaviside step function, i.e.
$$\frac{d\sigma(x)}{dx} = \theta(x) \quad\implies\quad d\sigma = \theta\,dx$$
When applied element-wise on a vector $(z)$ these functions produce vectors
$$\eqalign{
s &= \sigma(z) \quad\quad h &= \theta(z) \\
}$$
whose gradient can be calculated using the elementwise/Hadamard product
$$\eqalign{
ds &= h\odot dz = H\,dz \quad\implies\quad
\frac{\partial s}{\partial z} &= H \\
}$$
where $\odot$ denotes the Hadamard product and $H={\rm Diag}(h)$

For your specific problem let $\,z = Wx,\;p=(Vs-y)\;$ and
$$\eqalign{
 L &= \tfrac 12\;p:p \\
dL &= p:dp \\
 &= p:V\,ds \\
 &= p:VH\,dW\,x \\
 &= HV^Tpx^T:dW \\
 &= {\rm Diag}\big(\theta(Wx)\big)\,V^T\big(V\sigma(Wx)-y\big)x^T:dW \\
\frac{\partial L}{\partial W}
 &= {\rm Diag}\big(\theta(Wx)\big)\,V^T\big(V\sigma(Wx)-y\big)x^T \\\\
}$$

In the above, a colon is used to denote the trace/Frobenius product
$$\eqalign{
A:B &= {\rm Tr}(A^TB) \\
}$$
which happens to commute with itself and with the Hadamard product
$$\eqalign{
A:B &= B:A \\
A:(B\odot C) &= (A\odot B):C \\
}$$
The product rule for differentials
$$\eqalign{
d(A\star B) = dA\star B + A\star dB \\
}$$
is quite general since $(\star)$ can denote any product (Kronecker, Frobenius, Hadamard, Dyadic, Tensor, etc) and $(A,B)$ can be any two matrices (or scalars, vectors, tensors) whose dimensions are compatible with the underlying product.

If a product is commutative then this rule is analogous to the one for scalars, e.g.
$$\eqalign{
d\big(A\star A\big)
 \;=\; \big(dA\star A + A\star dA\big)
 \;=\; 2A\star dA \\
}$$


问:
Introduction to fractional calculus: problem with identity
I can't see the next step:

$D^\alpha e^{ix} = i^{\alpha}e^{ix} = e^{i\alpha \frac \pi2}e^{ix}$

答:
Using Euler's identity $$e^{ix} = \cos x + i \sin x$$
A complex number must be in the form:
$z = x+iy$, then $e^z = e^{x+iy} = e^x+e^{iy} = e^x(\cos y + i \sin y)$ and I want that this last equality be $i$, then, $e^x(\cos y + i \sin y) = i$ only if
$x = 0$ and $y = \frac \pi2$.</p>


问:
Question about the algebraic structure of function space
So, on the linear algebra class it is mentioned that function space is an example of the vector space.

By the definition of the vector space, any vector space $(V,+,\cdot)$ both an abelian group and a module. That means the function space has two operators working on a set. So far is correct, right?

One is the scalar multiplication $c\cdot f$ and one is the addition of functions $f+g$.

But in calculus we have seen that there are also many other operators that can work on function, for instance, the multiplication of functions $f\times g$ and the function composition $f\circ g$, also there are unary operators such as differential operator $\nabla$ and integral operator $\int$ and integral transformations $T f$, those are all linear, right?

[Main question] Does this means there are some more complex algebra structure for the function space? If not, how does the function composition enrolled into the discussion.

Decomposing the main question:

Part 1: If I obtain a function space with certain basis. It is closed under scalar multiplication and function addition because it is a vector space. Now, if I attach function composition to this structure, make it
$$
(F, \cdot, +, \circ)=(\text{set of function, scalar product, addition, function composition}),
$$
is the set still closed? Is there a proof or counter example for this.

[The question is explained by @RyanHowe in comment] Also, in wiki it is mentioned that function space is also a topological space idea. What's the difference when we treat the function space differently? Is the definition changed?

Does that means we only care scalar multiplication $c\cdot f$ and addition of functions $f+g$ in linear algebra. And, differential operator $\nabla$ and integral operator $\int$ are considered in Calculus?

Thank you guys very much.

答:
In general, if we have some algebraic structure $R$, the set of functions $f : A \to R$ inherit all the algebraic structure that $R$ has. For example, if $R$ is an Abelian group, the function space $\{f : A \to R\}$ inherits the group operations from $R$.

So $\{f : A \to \mathbb{R}\}$ in particular will inherit the structure of $\mathbb{R}$ as an $\mathbb{R}$-algebra. This means that not only can we multiply a function by a scalar, but we can also multiply a function by another function.

As for function composition, this is only a useful operation when $A = \mathbb{R}$. And it doesn't play very nice with the other operations for the most part. So you could come up with an "algebraic structure" by saying things like $(f \cdot g) \circ h = (f \circ h) \cdot (g \circ h)$, for instance. But there isn't a nice name for it.

Finally, consider that differentiation (and even integration) aren't total operators. They rely on the function being continuous. But if you're interested in looking into the algebraic structure of differentiation, you should look into differential algebra.


问:
Constructing formulas in first order logic

Let $\sigma = \{ +, R \}$ be a signature with a binary functional symbol $+$ and a binary relation symbol $R$,  and let  $\mathcal{N}(\mathbb{N}, +^\mathcal{N}, R^\mathcal{N})$ be a $\sigma-$strucutre where $+^\mathcal{N}$ is the normal addition of numbers and $R^\mathcal{N}:= \{ (x,y) \vert x \text{ divides } y \text{ and } x\neq 0\}$. Give for each of the following a formula $\varphi_i$:

$(i)$ $\varphi_1(\mathcal{N}) = \{ 1 \}.$

$(ii)$ $\varphi_2(\mathcal{N})$ is the set of all odd numbers.

$(iii)$ $\varphi_3(\mathcal{N}) = \{ (x, y) \vert gcd(x, y) = 1 \}$ (where gcd is the greatest common divisor)

For $(i)$ I have:

$$\varphi_1(x) = \forall y (R(x, y))$$

Since $R$ guarantees that $x$ cannot be $0$.

For $(ii)$ I have first defined $\varphi_2'$ such that $\varphi_2'(\mathcal{N}) = \{2\}:$

$$\varphi_2'(x):= \exists z (\varphi_1(z) \land x = z + z)$$

Here I'm forcing $z$ to be $1$. (Question: am I allowed to use $y$ instead of $z$ or would that change the meaning of $\varphi_1$?)

I have then defined $\varphi_2''$ such that $\varphi_2''(\mathcal{N})$ is the set of all even numbers:

$$\varphi_2''(x):= \exists y (R(y, x) \land \varphi'_2(y))$$

And finally

$$\varphi_2(x) := \lnot \varphi_2''(x)$$

For $(iii)$:

$$\varphi'_3(x, y) := (\varphi_2''(x) \land \varphi_2(y)) \lor (\varphi_2''(y) \land \varphi_2(x))$$

Here I'm forcing $x$ to be even and $y$ odd or vice-versa. Can someone please check my work?

答: (i) looks fine, (ii) looks correct but is overly complicated (and I'm not sure I understand your question about $y$) and (iii) is wrong. For (ii) it's enough to say $\varphi_2(x) \equiv \lnot \exists y (x = y + y)$. For (iii) an easy counterexample is that every two prime numbers have a gcd of 1. Maybe remember that if 1 is the gcd of two numbers then it is the $\textit{only}$ number that divides both?


问:
Is $((\frac{1}{a})^{\frac{1}{b}})^{\frac{1}{c}}=\frac{1}{\sqrt[bc]{a}}$?
Is $((\frac{1}{a})^{\frac{1}{b}})^{\frac{1}{c}}=\frac{1}{\sqrt[bc]{a}}$?

I believe it is since $((\frac{1}{a})^{\frac{1}{b}})^{\frac{1}{c}}=\frac{1}{\sqrt[b]{a}}^\frac{1}{c}=\frac{1}{\sqrt[bc]{a}}$.

Can someone please confirm my suspicions as I just came across a question that states in the answer that $((\frac{1}{a})^{\frac{1}{b}})^{\frac{1}{c}}=\sqrt[bc]{a}$ and I believe that this is a typo.

答:
assuming you are working with positive real numbers and assuming you are using the definition that if $m \ne 1$ and $m > 0$ and $k \in \mathbb N$ the $m^{\frac 1k}$ is defined to be (and you have proven there is such) the unique one and only positive number $w$ so that $w^k = m$ then

$((\frac 1a)^{\frac 1b})^{\frac 1c}$ (assuming $a\ne 1$) is the unique

$\beta$ so that $\beta^c = (\frac 1a)^{\frac 1b}$.

And $(\frac 1a)^{\frac 1b}$ is the unique $\gamma$ so that $\gamma^b = \frac 1a$.

But then $\beta^{bc} = (\beta^c)^b = ((\frac 1a)^{\frac 1b})^b = \gamma^b = \frac 1a$.

So $beta$ is the unique value so that $\beta^{bc} = \frac 1a$ so $\beta = (\frac 1a)^{\frac 1{bc}$.

SO $((\frac 1a)^{\frac 1b})^{\frac 1c}=\beta = (\frac 1a)^{\frac 1{bc}}$


问:
Is there an inequality for the product of two sequences, where one contains the term $a_i$ and the other its reciprocal $1/a_i$?
In my work I have come across the following product of summations:

$$
\left( \sum_{i=1}^n a_i b_i c_i \right) \left( \sum_{j=1}^n \frac{b_j}{a_j} \right),
\qquad\qquad (1)
$$
where $a_i, c_i \in \mathbb{R}_{> 0}, b_i \in \mathbb{R}$.

Is there an inequality I can use to get a better lower bound than
$$
\left( \sum_{i=1}^n a_i b_i c_i \right) \left( \sum_{j=1}^n \frac{b_j}{a_j} \right)
\geq
\frac{\min(a_i)}{\max(a_i)} \left( \sum_{i=1}^n b_i c_i \right) \left( \sum_{j=1}^n b_j \right)?
$$
I've been wrangling with this problem for a few days, and Google hasn't been much help. I am not a mathematician by training so this may be because I simply don't know how to phrase the right question.

Many thanks in advance!

(Something in my gut tells me this should be true:
$$
\left( \sum_{i=1}^n a_i b_i c_i \right) \left( \sum_{j=1}^n \frac{b_j}{a_j} \right)
\geq
\left( \sum_{i=1}^n b_i c_i \right) \left( \sum_{j=1}^n b_j \right);
$$
is it?)

Edit

Thanks to commenters for pointing out that both of my inequalities were false. I'll add some more details to hopefully clarify where my question comes from.

This question derives from trying to find the sign of the following expression:
$$
- \sum_{i=1}^n x_i y_i z_i^2 + \left( \sum_{i=1}^n x_i z_i \right)\left(
\sum_{j=1}^n x_j y_j z_j \right)
\qquad\qquad (2)
$$
where $x_i, y_i \in \mathbb{R}_{\geq 0}, z_i \in \mathbb{R}$, and $\sum_{i=1}^n x_i = 1$.

I used the Cauchy-Schwartz inequality on the first term to get:
$$
\left(\sum_{j=1}^n x_j\right) \left( \sum_{i=1}^n x_i y_i z_i \right)
\geq
\left( \sum_{i=1}^n x_i \sqrt{y_i} z_i \right)^2,
\qquad\qquad (3)
$$
so if $y_i$ is the same for all $i$, the expression $(2)$ is negative semidefinite. (To get the original expression (1) I assumed that $y_i \neq 0$ and multiplied (3) through by $\tfrac{\Pi_{i=1}^{n}\sqrt{y_i}}{\Pi_{i=1}^{n}\sqrt{y_i}}$, defining $a_i = \Pi_{j \neq i}\sqrt{y_i}, b_i = y_i, c_i = x_i z_i$.)

My question is: can something be said about the sign of $(2)$ for general $y_i \in \mathbb{R}_{\geq 0}$?

Thank you!

答:
Here are my two cents, I suspect that you cannot in general do much better. Here is a cooked up toy example: Let $a_1 = 1, a_2 = 100$, $b_1=1, b_2=100$ and $c_1=100^2, c_2=1$. Then
$$ \left(\sum a_ib_ic_i\right)\left(\sum b_i/a_i \right) =(100^2 + 100^2) \cdot (1+1) = 40000, $$
and
$$ \left( \sum b_ic_i \right) \left( \sum b_i\right) = (100^2+100)\cdot(1+100) = 1020100. $$
This shows in particular that your gut feeling in this case is incorrect, as the second sum is (quite a bit) larger than the first. Moreover,
$$\frac{\min (a_i)}{\max (a_i)} \left( \sum b_ic_i \right) \left( \sum b_i\right) = \frac{1}{100}1020100 = 10201.$$
This is roughly $1/4$th of the first sum, so we conclude that
$$ \left(\sum a_ib_ic_i\right)\left(\sum b_i/a_i \right)  \geq 4 \frac{\min (a_i)}{\max (a_i)} \left( \sum b_ic_i \right) \left( \sum b_i\right)  $$
does not hold in general.

This is just an exploratoty toy example, I think it is likely that by choosing the numbers in a smarter manner we can reduce the $4$ down, but not necessarily all the way to $1$.


问:
What is the name of a fully-twisted strip?
I know that an odd number of twists generate a Möbius strip, and that an even number of twists generate a shape with two sides. Now, as far as I could find, for the strip with an even number of twists it's only said that the figure is topologically homeomorphic to an ordinary cylinder.

What is the name of this figure? And if there is no name, why is that so?

答:
As you say, a strip with an odd number of half-twists is a Möbius up to homeomorphism, and one with an even number of strips is an annulus up to homeomorphism. So from that point of view, we only have the two cases. I mention this because you seem happy with the same term for $1$ and $3$ half-twists, but want different terms for $0$ an $2$ half-twists.

When embedded in $\mathbb R^3$, different numbers of twists of course give non-equivalent surfaces in another sense (specifacally, they are not ambient isotopic, or even isotopic). But in this case, I've always only heard it as a strip/band/annulus with a given number of half-twists or full twists. So your shape would just be a "band with a full twist", or similar.

Or maybe someone knows of a specific term for it.


问:
Finitely many Lefschetz fixed points
The questions is

Show that if $X$ is compact and all fixed points of $X$ are Lefschetz, then $f$ has only finitely many fixed points.

n.b. Let $f: X \rightarrow X$. We say $x$ is a fixed point of $f$ if $f(x) = x$. If $1$ is not an eigenvalue of $df_x: TX_x \rightarrow TX_x$, we say $x$ is a Lefschetz fixed point.

I have proved that $x$ is a Lefschetz fixed point of $f$ if and only if $\mathrm{graph}(f)$ and $\Delta = \mathrm{graph}(\mathrm{identity})$ intersect transversally at $(x,x) \in X \times X$, but not sure how to proceed.

Thank you!

答:
Some key points for this answer is contributed by @tessellation:

Suppose $x_0$ is a Lefschetz fixed point. Take a chart $(U,\phi)$ around $x_0.$ Then in this coordinate neighborhood (after composing with proper coordinate functions) I can think of $f$ as a map from open ball in $\mathbb{R}^n,\,\,$ (say $B$) to itself with $f(0)=0.$ Now consider we have a function $f:B\rightarrow B$  such that $f(0)=0.$ Consider the function $g=f-id.$ Then $g(0)=0$ and by lefschetz condition $det(dg)(0)\neq 0$ (as 1 is not an eigenvalue of $df$). Hence $g$ is a local diffeomorphism by inverse function theorem and we are done.

If $g$ is a diffeomorphism then $g$ is bijective in that small neighborhood. If $f$ has another fixed point a in that neighborhood then $g(a)=0$. This will contradict the bijcetivity of $g$. Therefore the Lefschetz fixed point is a 0-manifold. In other words, isolated.

Then we consider such an open cover of $X$: the open set around each fixed points, and the complement of the set of fixed poinots. Since $X$ is a compact manifold, so there must be finitely many of subcovers. Therefore, there are finitely many Lefschetz fixed points.

Title: Finitely many Lefschetz fixed points


问:
Expected value and variance of $X-Y$, where $X$ is the number of boys and $Y$ is the number of girls in a family with $N$ children
In a given family there are $N$ children, $X$ of them are boys and $Y$ are girls. We define a random variable $T$ as $T=X-Y$. Assuming that the probability of birth of males and females are equal, find the expectation and variance of $T$.

My efforts:

first try is by finding the PDF of $T$:

$P(T=t)=P(X-Y=t)=\sum\limits_{n=0}^{N}P(X-Y=t,Y=n)=\sum\limits_{n=0}^{N}P(X=n+t)=\sum\limits_{n=0}^{N} {N \choose n+t} (\frac{1}{2})^{n+t}(\frac{1}{2})^{N-(n+t)}=\frac{1}{2^N}\sum\limits_{n=0}^{N} {N \choose {n+t}}$
now that's seems a little off because $n+t$ exceeds the summation limit. Also, I can't get a nice closed formula from this expression.

second attempt:

$\mathbb E[T]=\mathbb E[X-Y]= \mathbb E[X]-\mathbb E[Y] = 0$, for simmetry reasons (I would like for a better explantion for that, because I'm practicing for an exam. Is it sufficient to say that $X,Y$ has the same range and the same probability?)

$\mathbb V[T]=\mathbb E[(X-Y)^2]-\mathbb E[X-Y]^2=\mathbb E[X^2]-2\mathbb E[XY]-\mathbb E[Y^2]=-2\mathbb E[XY]$

that's seem way more off because $X,Y$ and the variance is always non negative, so no reason for $\mathbb E[XY]$ to be negative.

Edit: as stated by lulu on the comments, I miscalculated the variance (actually on the sign of $\mathbb E[Y^2]$)

Any help would be appreciated.

Also keep in mind that I'm interested on finding the PDF, although it's not a part of the original question.

Thanks

答:
This is much too complicated.  Rather, note that because $X + Y = N$, then $$T = X - Y = X - (N - X) = 2X - N.$$ Since $N$ is not a random variable but a fixed parameter, we can compute the expectation and variance as
$$\operatorname{E}[T] = \operatorname{E}[2X - N] = 2\operatorname{E}[X] - N = 2(N/2) - N = 0,$$

and $$\operatorname{Var}[T] = \operatorname{Var}[2X - N] = \operatorname{Var}[2X] = 4\operatorname{Var}[X] = 4N(1/2)(1-1/2) = N.$$

I should also point out that the expectation calculation in your second attempt is correct:  you can justify $\operatorname{E}[X - Y] = \operatorname{E}[X] - \operatorname{E}[Y] = N/2 - N/2 = 0$ because $X$ and $Y$ need not be independent in order for the linearity of expectation to apply.  However, in your variance calculation, you must account for the fact that $X$ and $Y$ are dependent--in fact, they are perfectly correlated because $X + Y = N$.

The PDF of $T = 2X - N$ is simply $$\Pr[T = t] = \Pr[X = (t+N)/2] = \binom{N}{(t+N)/2} 2^{-N}, \\ t \in \{-N, -N+2, \ldots, N-2, N\}.$$  This is not a binomial distribution, but a location-scale transformation of one.


答:
You are probably familiar with the language of knots/links. A knot is a circle embedded in $S^3$ (or $\mathbb{R}^3$), considered up to isotopy of the ambient space. Now, for each integer $n$, there is a distinct framing of the unknot, which you can picture as embedding of a ribbon $([0,1] \times S^1 \hookrightarrow S^3)$ rather than a string $(S^1 \hookrightarrow S^3)$.

As you correctly observed, each framed unknot is homeomorphic to either the cylinder or Möbius strip, depending on parity. Thus, any pair of these are homeomorphic to one another if they have the same parity. But they are not ambient isotopic (i.e. you can't deform one into another without cutting and gluing, which is of course discontinuous).

Title: What is the name of a fully-twisted strip?

Body:
I know that an odd number of twists generate a Möbius strip, and that an even number of twists generate a shape with two sides. Now, as far as I could find, for the strip with an even number of twists it's only said that the figure is topologically homeomorphic to an ordinary cylinder.

What is the name of this figure? And if there is no name, why is that so?


问:
Logic: trying to prove function uniqueness using induction principle
I'm new to this topic, so I couldn't solve the problem although it seems to be easy:

```<code>prove using the induction principle that the following function +: N×N -> N is unique:

m+0=m

m+s(n)=s(m+n)

where s is successor operation.
</code>```

I defined the set A as follow:

m s.t +1 (m, s(n))=+2 (m, sn))

And tried to prove that A is inductive (and then add is unique). but I couldn't do that.

Any help would be appreciated.

答: Most times you have better luck proceeding by induction on the second term (in your case that's $n$). This has to do with the fact that both addition and multiplication are defined in most cases recursively on the second argument. Let $P(x)\equiv\forall m (m+_1x=m+_2x)$. If you don't like this notation just read $P(x)$ as $x\in P$. We have $P(0)$ by the first property of $+_1$ and $+_2$. If $P(n)$ is true we have $\forall m(m+_1n=m+_2n)$ so given any $m$ we have $m+_1S(n)=S(m+_1n)=S(m+_2n)=m+_2S(n)$. We used in these steps the second property of $+_1$ and $+_2$ and the induction hypothesis from which we got $m+_1S(n)=m+_2S(n)\equiv P(S(n))$. So by induction $\forall n P(n)$


问:
Implications of $\text{Spin}(3)\times \text{Spin}(3) \cong \text{Spin}(4)$
I recently learned that $\text{Spin}(3)\times \text{Spin}(3) \cong \text{Spin}(4)$, and find it intriguing: it seems like the sort of thing that would have lots of interesting consequences. What are the implications of this isomorphism? I'd be interested in connections between this and any other area of math (or physics!).

答:
I'm guessing an answer that best discusses the "implications" of this accidental isomorphism from a physical perspective is one that discusses the implications in spin geometry specifically. But that area is deeper than my expertise so I unfortunately can't comment on it. Instead, I can mention a bunch of ideas and facts that, are not really implications per se but go hand-in-hand with this isomorphism.

First, though, there is one elementary implication, and that is in classifying simplie Lie groups and algebras. The characterization of Lie groups splits into covering space theory and Lie algebras, and the characterization of Lie algebras splits into solvable and semisimple ones, and the characterization of semisimple ones is simply a matter of classifying simple Lie algebras. (I am talking about complex ones, there is some more to talk about going between real and complex algebras/groups.) The classification of simple Lie algebras is known, and simple enough to cover in a semester or two of graduate class. The case of $\mathfrak{so}(4)$ is the only $\mathfrak{so}(n)$ that is not simple. There are other accidental isomorphisms as well between simple lie algebras.

Another fact is how quaternions are used to model 3D and 4D rotations. Conjugating vectors by versors yields rotations (and there is an angle-doubling effect which highlights the spinorial nature of this homomorphism), and left/right-regular multiplications by pairs of versors yields arbitrary 4D rotations. The former is useful in graphics and computational models of physical orientation. I am not sure how useful the latter is - even ignoring the gravitational curvature of general relativity, the spacetime of special relativity has three spatial and one temporal dimension (rather than four spatial dimensions), and in this context there is an accidental isomorphism between $\mathrm{SL}_2\mathbb{C}$ and $\mathrm{Spin}(3,1)$ (which extends the one between $\mathrm{SU}(2)$ and $\mathrm{Spin}(3)$).

This is also useful for describing left and right isoclinic rotations. All rotations in higher dimensions restrict to plane rotations in a bunch of orthogonal (2D) planes. (Basically, the real version of the spectral theorem.) The orientations of the plane rotations either "add up" to the orientation of the whole space, or the opposite orientation - this creates a distinction between left/right isoclinic rotations in even dimensions (there is no such distinction in odd dimensions since the planes can't "add up" to the whole space anyway). In 4D, the left/right isoclinic rotations come from left/right multiplication by versors. In higher dimensions, the left/right isoclinic rotations don't commute and don't form subgroups (although they are unions of one-parameter subgroups, so are images of the exponential map).

One way to continue the progression $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$ is with octonions $\mathbb{O}$. These are Euclidean / Hurwitz / composition / normed division algebras (there are lots of names). But another way is with Clifford algebras. The Clifford algebra $\mathrm{C}\ell(p,q)$ is generated by $p$ and $q$ anticommuting square roots of $+1$ and $-1$ (respectively - although sign conventions differ). So $\mathbb{R},\mathbb{C},\mathbb{H}$ are $\mathrm{C}\ell(n,0)$ for $n=0,1,2$. We define the spin groups $\mathrm{Spin}(p,q)$ as generated by (products of evenly many) unit vectors in $\mathrm{C}\ell(p,q)$, these are double covers of $\mathrm{SO}(p,q)$ (mod $\pi_0$, with a couple exceptions). Spin representations come from modules over Clifford algebras. We can classify Clifford algebras recursively - by tensoring with $\mathrm{C}\ell(p,q)$ with $(p,q)=(2,0),(1,1),(0,2)$, which are $\mathbb{H}$, $\mathbb{R}(2)$, $\mathbb{R}(2)$ (the latter two are $2\times2$ matrix algebras). A microcosm of this is the $\otimes$ table

$$ \begin{array}{c|c|c|c}
\otimes & \mathbb{R} & \mathbb{C} & \mathbb{H} \\ \hline
\mathbb{R} & \mathbb{R} & \mathbb{C} & \mathbb{H} \\
\mathbb{C} & & \mathbb{C}^2 & \mathbb{C}(2) \\
\mathbb{H} & & & \mathbb{R}(4)
\end{array} $$

Note $\mathbb{H}\otimes\mathbb{H}\cong\mathbb{R}(4)$ is a "linearization" of $S^3\times_{\mathbb{Z}_2} S^3\cong \mathrm{SO}(4)$. So you could say this isomorphism (linearized) has implications for the classification of Clifford algebras, hence construction of spin representations (and also subsequently, for constructing maximal trivial subbundles of sphere's tangent bundles, IOW linearly independent vector fields on spheres).

The (real, oriented) Grassmanian $\mathrm{Gr}(n,k)$ is the space of all (oriented) $k$-dimensional subspaces of $\mathbb{R}^n$. It can be considered a subspace of $\Lambda^k\mathbb{R}^n$. (The unoriented Grassmanian is a quotient by $\mathbb{Z}_2$, and may be considered a subspace of $\mathbb{P}(\Lambda^k\mathbb{R}^n)$.) The first nontrivial Grassmanian would be $\mathrm{Gr}(4,2)$. In the case of $k=2$, we can identify $\Lambda^2\mathbb{R}^n\cong\mathfrak{so}(n)$ (as representations). The fact $\mathrm{so}(4)\cong\mathfrak{so}(3)\times\mathfrak{so}(3)$ corresponds to a decomposition $\Lambda^2=\Lambda^2_+\oplus\Lambda^2_-$. The Hodge star $\ast:\Lambda^k\mathbb{R}^n\to\Lambda^{n-k}\mathbb{R}^n$ (which linearly extends the orthogonal complement function $\mathrm{Gr}(n,k)\to\mathrm{Gr}(n,n-k)$) acts as a linear operator on $\Lambda^2\mathbb{R}^4$, whose $\pm1$ eigenspaces are $\Lambda^2_\pm$. The decomposition $\mathrm{Gr}(4,2)=S^2\times S^2$ exists within $\Lambda^2=\Lambda^2_+\oplus\Lambda^2_-$. All of this left/right-handedness and pairedness goes hand-in-hand with $\mathrm{Spin}(4)\cong\mathrm{Spin}(3)\times\mathrm{Spin}(3)$.


问:
Find polynomial with inequality for derivative
I would like to ask a question which is related to Find function with inequality for derivative

Let $C\geq 2$ and $L>0$ be fixed. Let $n \in \{1, 2, \ldots\}$. Does there exist a polynomial $g$ of degree $n$ such that
\begin{equation*}
0 < g(0) < g(L), \qquad g'(x) > (g(L) - g(x))C, \ \ \text{for all }x \in [0,L]
\end{equation*}
holds at least for $n$ big enough?

In Find function with inequality for derivative, a function $g$ fulfilling such conditions was found, but its expression involved the exponential function. Here, I would like to know if this problem can also be solved with polynomials.

You can see already that for $n = 1$, one has $g(x) = ax + b$, $g'(x) = a$, and consequently the conditions become
\begin{equation*}
0 < b < aL+ b, \qquad (L-x)C < 1, \ \ \text{for all }x \in [0,L],
\end{equation*}
which adds a constrain on $C$ and $L$.

答:
For any fixed $L,C>0$, pick $n$ sufficient large so that $L<\frac{n}{C}$. Now define the polynomial, $$p(x):=-\left(-\frac{C}{2n}x+\frac{1}{2}\right)^{2n}+1$$ Note that $p(x)$ satisfies the points $(-n/C,0)$, $(0,1-(1/2)^{2n})$, $(n/C,1)$, $(2n/C,1-(1/2)^{2n})$, which implies that $p(x)$ has a vertex at $n/C$ (the absolute maximum of $p$), and that $p(x)$ is increasing from $(-\infty,n/C]$ (moreover, $p(x)$ has positive derivative and is in fact concave down on that same region); additionally we see that $0<p(x)< 1$ for $x\in[0,L]$. Thus $0 < -\frac{C}{2n}x+\frac{1}{2} < 1$ for $x\in[0,L]$. Additionally, we clearly see that $0<p(0)<p(L)<1$. So for any $x\in[0,L]$ we have $$\begin{array}\
p'(x) &=& -2n\left(-\frac{C}{2n}x+\frac{1}{2}\right)^{2n-1}\cdot\frac{-C}{2n}  \\
      &=& C\left(-\frac{C}{2n}x+\frac{1}{2}\right)^{2n-1}    \\
      &>& C\left(-\frac{C}{2n}x+\frac{1}{2}\right)^{2n}      \\
      &>& C\left(\left(-\frac{C}{2n}x+\frac{1}{2}\right)^{2n}-\left(-\frac{C}{2n}L+\frac{1}{2}\right)^{2n}\right)  \\
      &=& C\left(p(L)-p(x)\right)  \\
\end{array}$$


问:
Can Ambient Isotopies be Deformed to be Constant Away from a Compact Set?
So I have this statement that I know (I guess I should say 99% sure until I actually see a proof):

Let $X, Y \subset \mathbb{R}^2$ be compact and suppose $F$ is an ambient isotopy carrying $X$ to $Y$.  Then there is an ambient isotopy $F'$ carrying $X$ to $Y$ such that $F'$ is constant in a neighborhood of $\infty$.

In other words, if two compact spaces are ambient-isotopic, then you don't need to radically deform the entire plane just to get one to the other.  I think it's true for general $\mathbb{R}^n$ with the same argument I'll outline below.

I am having trouble finding a reference for this or proving rigorously that it's true.  Can anyone help?

I'm trying to prove it using the Annulus Theorem.  My idea for the proof goes as follows.  Let $J_1$ be a circle large enough that the trace of $X$ and $Y$ under $F$ are contained in the inside (bounded complementary region, as opposed to outside) of $J_1$, and let $J_2$ be a circle such that the trace of $J_1$ under $F$ is contained in the inside of $J_2$.  I want to make it so $F'$ is constant on $J_2$ and its outside, and agrees with $F$ on $J_1$ and its inside.

So basically $F'$ should slide $J_1$ around inside $J_2$.  Let $A_t$ denote the annulus between $F_t(J_1)$ and $J_2$.  Then $A_0$ is the annulus between $J_1$ and $J_2$ and is homeomorphic to a 'typical' smooth annulus, so has a decomposition into 'transverse arcs'.  I want to drag the endpoints of these transverse arcs along $J_1$ according to the action of $F$, and have them stay fixed along $J_2$.

Visually it SEEMS clear that this process gives a continuous extension (it will be bijective by construction), since you can't get "infinite, large-scale winding" in "a compact amount of time," and that's "the only thing that can go wrong."  But that's a ton of handwaving.

Please help!

答:
Here is how to do this in the smooth category. Let $F_t(z)$ denote a smooth isotopy, $t\in [0,1]$, $z\in {\mathbb R}^2$; let $K$ denote
$$
\{F_t(z): z\in X, t\in [0,1]\},
$$
a compact in the plane. ($F_0=id$, $F_1(X)=Y$.)

Let $B(0,R)$ denote a ball containing $K$.
Define the tangent vector field $Z(z,t)=\frac{d}{dt}F_t(z)$ of the flow. Let $\eta(z)$ be a (smooth) cut-off function, which is identically $1$ on $B(0,R)$ and identically $0$ on a  ${\mathbb R}^2\setminus B(0,2R)$. Now, define a new vector field $W(z,t)= \eta(z) Z(z,t)$. Since it is identically zero outside of $B(0,2R)$, this vector field defines a flow $G_t(z), t\in [0,1]$ which is constant outside of $B(0,2R)$. By the uniqueness theorem, $G_t(z)=F_t(z)$ for all $z\in B(0,R)$, in particular, all $z\in X$. Hence, $G_t$ is an isotopy taking $X$ to $Y$, which is the identity  outside of $B(0,2R)$.

I am sure one can also find an isotopy in the topological category, but that would take more work and I am not sure about the motivation.


问:
What's wrong with the following argument showing that $L^p$ is Banach.
I want to show that $L^p$ is complete (assume I have everything else done). Here's my approach which is slightly different from the usual approach.

If $f_n$ is Cauchy then let $g_k$ be subsequence such that $\|g_{k+1}-g_k\|_p<\dfrac{1}{2^k}$. Then let $$f=g_1+\sum_{k=1}^{\infty}g_{k+1}-g_k$$

As the name suggest, I want to show that $f_n\rightarrow f$. Now suppose that I already know that $f$ is in $L^p$, then I only need to show that $f_n$ converges to $f$.

To show that I say, for $g_n$ we have

$$\| f-g_n\|=\|(g_1+\sum_{k=1}^{\infty}g_{k+1}-g_k)-g_n\|$$

But the sum is telescoping and it is equal to $g_n+\sum_{k=n}^{\infty}g_{k+1}-g_k$

and so
$$\| f-g_n\|=\|\sum_{k=n}^{\infty}g_{k+1}-g_k)\|\le \sum_{k=n}^{\infty}\|g_{k+1}-g_k)\|\le \dfrac{1}{2^n}$$

the second to last inequality is the Minkowski's inequality and continuity of the norm. So $g_n\rightarrow f$. But since $f_n$ is Cauchy, we conclude that $f_n\rightarrow f$.
I can't tell what's wrong with the argument, but I have my suspicions. First, the book does it differently and the difference is significant. Second, I can apply this argument to conclude that any functional space over any space (not necessary banach) is Banach, which sounds dubious. I would appreciate your feed back. Even more dubiously, the same argument tells me that $\|f\|= 0$

答: In a  complete normed linear space (i.e. a Banach sapce) $\sum \|x_n\| <\infty$ implies that $\sum x_n$ converges in the norm. Completeness of the space is essential for this. Your definition of $f$ already assumes that the space is complete since you are assuming that the series defining $f$ is convergent.


问:
$x_n\in\mathbb{N}$ such that $\cos(x_n)\to 1$
I'm trying to prove if it exists some sequence $x_n\in\mathbb{N}$ such that $\cos(x_n)\to 1$.

I've considered $x_n= 2\lfloor \pi\ 10^{n-1} \rfloor$, which is $x_1=2\cdot 3, x_2=2\cdot 31, x_3=2\cdot 314, x_4=2\cdot 3141$ and so on.

I don't think this initial approach verifies $\cos(x_n)\to 1$. Maybe it doesn't exist.

Any help would be appreciated.

答:
Yes, such a sequence exists. Define $x_n$ as follows. Consider the sequence $\{2\pi\}, \{4\pi\}, \{6\pi\},\ldots$ (where $\{x\}$ is the fractional part of $x$). As this is a list of numbers is the interval $[0,1)$, by the pigeonhole principle we can find two positive integers, say $m_1 < m_2$ such that $|\{2m_2\pi\}-\{2m_1\pi\}|<1/n$, hence $2(m_2-m_1)\pi\in (x_n-1/n,x_n+1/n)$ for some $x_n\in \mathbb{N}$.

Set $y_n=x_n-2(m_2-m_1)\pi$, then $y_n\in (-1/n,1/n) \quad \forall n\in \mathbb{N}$. By periodicity of $\cos$, $\cos(x_n)=\cos(y_n)\to 1$.


答:
All calculations are in $GF(2).$ I use $+$ instead of $\oplus$ for "xor".

You can make use of the special structure of this system of equations. If $R$ is the $32\times 32$ matrix that describes the circular shift:
$$
R = \begin{pmatrix}
0      &   &        & \cdots &   & 0 & 1      \\
1      & 0 &        &        &   &   & 0      \\
0      & 1 & 0      &        &   &   &        \\
       & 0 & 1      & \ddots &   &   & \vdots \\
\vdots &   & 0      & \ddots & 0 &   &        \\
       &   &        & \ddots & 1 & 0 &        \\
0      &   & \cdots &        & 0 & 1 & 0
\end{pmatrix}
$$
then your matrix $A$ can be written as $A=R^3+R^7+R^{18}.$ Futhermore, we know that $R^{32}+I=0.$ You can apply the Extended Euclidean Algorithm to the polynomials $x^{18}+x^{7}+x^{3}$ and $x^{32}+1$ to find two polynomials $p$ and $q$, such that
$$
\left(x^{18}+x^{7}+x^{3}\right)p(x) + \left(x^{32}+1\right)q(x)=1
$$
Then
$$
\left(R^{18}+R^{7}+R^{3}\right)p(R) + \left(R^{32}+I\right)q(R)=I
$$
or, as $R^{32}+I=0$,
$$
A\,p(R) = I
$$
which means $A^{-1} = p(R).$ Therefore, the solution of $Ax=b$ can be obtained by using $x=p(R)b.$

Note that the matrix $p(R)$ does not have to be explicitly calculated.
$$
x=p(R)b = \left(R^{a_0}+R^{a_1}+R^{a_2}+\ldots \right)b
  = R^{a_0}b+R^{a_1}b+R^{a_2}b+\ldots
$$
As $R$ is the circular shift, this means: Initialize your result with $0$. Then take $b$ and rotate it by $a_0$ bits and add/xor this to the result, then take $b$ and rotate it by $a_1$ bits and add/xor this to the result etc.

Edit

There is also a way to get $p$ without the Extended Euclidean Algorithm. In $GF(2)$, we have $(x+y)^{2^n}=x^{2^n}+y^{2^n},$ which means
$$
\left(R^{18}+R^{7}+R^{3}\right)^{32} = R^{18\cdot 32}+R^{7\cdot 32}+R^{3\cdot 32} = I+I+I = I
$$
If we use exponentiation by squaring and always reduce exponents as much as possible due to $R^{32}=I,$ we get
$$
\left(R^{18}+R^{7}+R^{3}\right)^{-1} = \left(R^{18}+R^{7}+R^{3}\right)^{31} \\
=\left(R^{18}+R^{7}+R^{3}\right)^{16}\left(R^{18}+R^{7}+R^{3}\right)^{8}\left(R^{18}+R^{7}+R^{3}\right)^{4}\left(R^{18}+R^{7}+R^{3}\right)^{2}\left(R^{18}+R^{7}+R^{3}\right)^{1} \\
=(I)(R^{16})(R^{8}+R^{28}+R^{12})(R^{4}+R^{14}+R^{6})(R^{18}+R^7+R^3)
$$
This product does not have to be expanded. It can easily be used directly to calculate $x.$

Title: How to solve a system of 32 XOR equations via Gaussian Elimination?

Body:
In the beginning, I needed to solve a system of linear XOR equations. I was inspired by the answer to this question:

how to solve system of linear equations of XOR operation?

, and started solving my problem with Gaussian Elimination. Now I have a system of XOR equations which is represented like this:

```<code>0:  0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 | 0
1:  0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 | 1
2:  0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 | 1
3:  1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 | 1
4:  0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 | 0
5:  0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 | 0
6:  0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 | 0
7:  1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 | 1
8:  0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 | 1
9:  0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 | 0
10: 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 | 1
11: 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 | 0
12: 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 | 1
13: 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 | 0
14: 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 | 0
15: 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 | 0
16: 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 | 0
17: 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 | 0
18: 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | 1
19: 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | 0
20: 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | 0
21: 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 | 1
22: 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 | 0
23: 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 | 0
24: 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 | 1
25: 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 | 1
26: 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 | 1
27: 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 | 0
28: 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 | 0
29: 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 | 1
30: 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 | 1
31: 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 | 0
</code>```

Every column is the variable (32 in total), and every row is another equation (32 in total too). So, since the amount of equations is equal to the amount of variables, it's 100% solvable. I'm just not experienced with Gaussian Elimination, and maybe you could give me at least basic ideas on how to solve that amount of equations.

Disclaimer

This question is part of my original question I asked on Stack Overflow, in case you're wondering why do I even need this in the first place:

https://stackoverflow.com/questions/66607696/reverse-sha-256-sigma0-function-within-complexity-of-on


问:
What if we take step functions instead of simple functions in the Lebesgue integral
When we define the Lebesgue integral, we first define it for simple functions $s(x) = \sum\limits_{j=1}^n c_j\chi_{A_j}(x)$ (where $A_j$ are measurable) as $\int sd\mu = \sum\limits_{i=j}^n c_j \mu(A_j)$ and then for $f\ge 0$ as $\int fd\mu = \sup\{\int sd\mu$ : s simple and $0\le s\le f\}$. But I was wondering what could go wrong if instead of taking simple functions in the supremum, we would take step functions, i.e. $s(x)=\sum\limits_{j=1}^nc_i\chi_{I_j}(x)$ where $I_j$ are intervals (any type, like $(a,b), (a,b], [a,b), [a,b])$).

答: It is easy to approximate measurable functions by simple functions: There is  a 'simple'construction of a sequence of simple functions increasing to a non-negative measuarble function. But it is not even true that any measuarble funciton is the ponit-wise limit of a sequence of step functions. However, it can be shown that if $f$ is integrable and $\epsilon >0$ then there exists a step function $g$ with $\int |f-g|d\mu <\epsilon$.


问:
Integral and periodic function
Let $f\colon\mathbb{R}\to\mathbb{C}$ be $2\pi$-periodic function. For all $a\in\mathbb{R}$ we consider the interval $I_a=[a-\pi, a+\pi]$. If $f\in L^1(I_0)$, then $$\int_{I_0}f\;d\lambda=\int_{I_a}\;f\;d\lambda,$$ where $\lambda$ is the Lebesgue measure.

I know the this result follows from the fact that the function $f$ is $2\pi$- periodic and traslation invariance of Lebesgue measure, but I don't know how to prove it explicitly.

Could anyone give me a suggestion? Thanks!

答: Hint: Write $I_a$ as the (disjoint) union $[a - \pi, (2k+1)\pi]\cup [(2k+1)\pi, a + \pi]$ for some $k\in\mathbb Z$. Can you now separately "shift" these two parts onto (disjoint) subsets of $I_0$ using the periodicity of $f$?

Title: Integral and periodic function


问:
Does $a=\frac{\langle x,y\rangle}{\| y \|^2}$ imply $\| a \| =\frac{\| x \|}{\| y \|}$?
As the title, um, $\overset{?}{\text{suggests}}$ I'm wondering if $a=\frac{\langle x,y\rangle}{\lVert y \rVert^2}\overset{\color{pink} \heartsuit}{\implies}\lvert a \rvert=\frac{\lVert x \rVert}{\lVert y \rVert}$. So what do you think?

EDIT$^1$:

Where $z=x-ay$ and $y$ and $z$ are orthogonal, with $y\neq 0$ and assuming that
$$\lvert \langle x,y\rangle \rvert = \lVert x \rVert \cdot \lVert y \rVert.$$

Perhaps this literature may be of help:

EDIT$^2$:

OK, I think I've got it know. Thank you for everyone's kind and considerate help.

答: The correct formula is $\dfrac{\langle x,y\rangle}{\|y\|^2}=\dfrac{\|x\|}{\|y\|}\cos\theta$, where $\theta$ is the angle between $x$ and $y$.</p>


问:
matrix and eigenvalues
Please help me solve this question. This is not a homework assignment. If $\lambda_M$ is the eigenvalue of the matrix M, please prove that $max|\lambda_M|<1$
$$M_{2n\times2n}=\begin{vmatrix} A_{n\times n} & I-A\\I_{n\times n} & 0_{n\times n}\end{vmatrix}$$
NOTE: $max|\lambda_A|<2$

I easily solved this problem in scalar mode but I have no idea about its matrix mode

答:
We write elements of $\mathbb{R}^{2n}$ as ordered pairs $(a, b)$ where $a, b \in \mathbb{R}^n$.

Suppose that $(a, b)$ is an eigenvector of $M$ with eigenvalue $\lambda$. Then we have $M (a, b) = (Aa + (I - A)b, Ia) = (Aa + (I - A)b, a) = \lambda(a, b) = (\lambda a, \lambda b)$.

Then in particular, we have $a = \lambda b$. So we have $M (a, b) = (A \lambda b + (I - A) b, \lambda b)$.

Then we have $A \lambda b + (I - A) b = \lambda a = \lambda^2 b$.

Then we have $(\lambda - 1) A b = (\lambda^2 - 1)b$.

We pause to note that every vector of the form $(b, b)$ is an eigenvector with eigenvalue 1.

Now we suppose that $\lambda \neq 1$. Then in that case, we have $Ab = (\lambda + 1) b$. Then $\lambda + 1$ is an eigenvalue of $A$.

On the other hand, let $\lambda_A$ be an eigenvalue of $A$ with eigenvector $b$. Then we see that, setting $\lambda = \lambda_A - 1$, we have

$\begin{equation}
  \begin{split}
  M (\lambda b, b) &= (A \lambda b + (I - A) b, I \lambda b) \\
&= (\lambda (\lambda + 1) b + (1 - (\lambda + 1)) b, \lambda b) \\
&= (\lambda (\lambda + 1) b - \lambda b, \lambda b) \\
&= \lambda ((\lambda + 1) b - b, \lambda b) \\
&= \lambda (\lambda b, b)
  \end{split}
\end{equation}$

So we see that $\{\lambda \in \mathbb{R}$ : $\lambda$ is an eigenvalue of $M\} = \{1\} \cup \{\lambda_A - 1$ : $\lambda_A$ is an eigenvalue of $A\}$.

Now, let's consider the constraint that $\max |\lambda_A| < 2$. This means for all eigenvalues $\lambda_A$, we have $-2 < \lambda_A < 2$. Thus, the strongest constraint we can get on the eigenvalues of $M$ is that for all eigenvalues $\lambda_M$, we have $-3 < \lambda_M \leq 1$.

Title: matrix and eigenvalues


问:
Usage of a basis
I've been learning how to determine and find a basis (which I do understand). For a set to be a basis, it must span the space and be linearly independent. However, I'm not acknowledging why I'm doing this.

Why do we need a basis? And how do I know I need to find a basis? I'm missing the basic concepts in discrete mathematics because I can't seem to connect the dots between all the material I've learned.

Any help would be greatly appreciated!

答:
The reason why we care about a basis is because many interesting functions between vector spaces are linear transformations, and linear transformations interact very nicely with bases. Let's say you have two vector spaces $V, W$ and you have a basis $(v_1, v_2, ..., v_n)$ for $V$. Let's also say that you have a linear transformation $T:V \to W$. Consider any element $v$ of the domain. $v$ can be written as $v = c_1 v_1 + \cdots + c_n v_n$. Now
$$T(v) = T( c_1 v_1 + \cdots + c_n v_n ) = c_1 T(v_1) + \cdots + c_n T(v_n)$$
The punchline is that if you can evaluate easily $T(v_i)$ for all the basis vectors, then you can easily evaluate $T(v)$ for any vector $v$. Here's an example of where this is useful.

Imagine the vector $(3, 6)$ in $\mathbb{R}^2$. If we rotate this vector by $45^{\circ}$ counterclockwise, where will it land? You might be tempted to draw a complicated diagram and break out your trig functions, but there is a better way to do this.

The function $T: \mathbb{R}^2 \to \mathbb{R}^2$ that rotates vectors by $45^{\circ}$ is a linear transformation (should be somewhat intuitive if you picture it). Now $e_1 = (1, 0)$ and $e_2 = (0, 1)$ form a basis for $\mathbb{R}^2$. And we can easily compute $T(e_1) = (\sqrt{2}/2, \sqrt{2}/2)$ and $T(e_2) = (-\sqrt{2}/2, \sqrt{2}/2)$. Thus
$$T(3, 6) = T(3 e_1 + 6e_2) = 3(\sqrt{2}/2, \sqrt{2}/2) + 6(-\sqrt{2}/2, \sqrt{2}/2)$$

$$= (- 3 \sqrt{2}/2, 9 \sqrt{2}/2).$$

Now, you may wonder "why do we need any other base than $e_1$ and $e_2$? Well, if $T$ is a rotation like above, it is easy to calculate $T(e_1)$ and $T(e_2)$. But for other transformations, this may not be so. For example: $T$ is projection on the line $y = \frac13 x$. Then it's much easier to calculate what $T$ does to the basis $\{(3, 1), (-1, 3) \}$ than to $e_1, e_2$.

Title: Usage of a basis


问:
How do i check if a polynomial is irreducible?
I have this excercise. let ${ f=2x \in R[x] }$. Determine the irreducibility of the polynomial ${f}$ if ${ R= \mathbb{Z} }$ or ${ R= \mathbb{Q} }$. Im not really sure how to do it and how it would change in the two domains that theyre giving me .

答:
In $\mathbb{Z}[x]$, $f$ is reducible, as $f(x)=g(x)h(x)$ for $g(x)=2$, $h(x)=x$ and both these polynomials are non-zero and non-units in $\mathbb{Z}[x]$.

In $\mathbb{Q}[x]$, $f$ is irreducible, as if we write $f(x)=g(x)h(x)$, one of the two polynomials must be a non-zero constant, which is a unit in $\mathbb{Q}[x]$, as $\mathbb{Q}$ is a field.


问:
Derivative of $\zeta=\partial_1f ⋅dx_2 ∧ dx_3-\partial_2f ⋅dx_1 ∧ dx_3+\partial_3f ⋅ dx_1 ∧ dx_2 \in \mathcal{A}^2(H)$
Let $b \in \mathbb{R}^3$ and $H= \mathbb{R}^3$ \ $\{b\}$. Let $f:H \to \mathbb{R}, f(x)=-|x-b|^{-1}$, and

$\zeta=\partial_1f ⋅dx_2 ∧ dx_3-\partial_2f ⋅dx_1 ∧ dx_3+\partial_3f ⋅ dx_1 ∧ dx_2 \in \mathcal{A}^2(H)$

I feel a bit stupid for asking this but how do I calculate the derivative of this? I know that the answer should be $0$ but the notation is a bit confusing for me.

答:
We compute using the exterior derivative.
The expression
$$ \zeta = \partial_1 f \ dx_2\wedge dx_3 - \partial_2 f \ dx_1\wedge dx_3 + \partial_3 f \ dx_1\wedge dx_2 $$

is a differential $2$-form on $H$ of smoothness class $C^1$, because $f$ is a $C^2$ function $H\to \mathbb{R}$.

So we compute
$$
d\zeta = d (\partial_1 f \ dx_2\wedge dx_3) - d(\partial_2 f \ dx_1\wedge dx_3) + d(\partial_3 f \ dx_1\wedge dx_2) \quad (\star)
$$
by linearity of the exterior derivative. Let's simplify the first term in $(\star)$. See that
\begin{align*}
 d (\partial_1 f \ dx_2\wedge dx_3) &= \partial_1\partial_1f \ dx_1\wedge dx_2 \wedge dx_3 + \partial_2 \partial_1 f\ dx_2\wedge dx_2 \wedge dx_3 + \partial_3 \partial_1 f\ dx_3\wedge dx_2 \wedge dx_3 \quad (*) \\
&= \partial_1\partial_1f \ dx_1\wedge dx_2 \wedge dx_3
\end{align*}

where the second and third terms in the right hand side of $(*)$ vanish by using the alternating property of the wedge product.
Similarly the second and third terms in $(\star)$ simplify to
$$ -d(\partial_2 f \ dx_1\wedge dx_3) = -\partial_2\partial_2 f\ dx_2\wedge dx_1\wedge dx_3 = \partial_2\partial_2 f\ dx_1\wedge dx_2\wedge dx_3  $$
and
$$ d(\partial_3 f \ dx_1\wedge dx_2) = \partial_3\partial_3f\  dx_3 \wedge dx_1\wedge dx_2 = \partial_3\partial_3f\  dx_1 \wedge dx_2\wedge dx_3  $$
by using the alternating property of the wedge product.

Therefore we use $(\star)$ to tell us
\begin{align*}
 d\zeta &= \partial_1\partial_1 f\  dx_1\wedge dx_2\wedge dx_3 + \partial_2\partial_2 f\  dx_1\wedge dx_2\wedge dx_3 + \partial_3\partial_3 f\  dx_1\wedge dx_2\wedge dx_3 \\
  &= (\partial_1\partial_1 f + \partial_2\partial_2 f + \partial_3\partial_3 f)\ dx_1\wedge dx_2\wedge dx_3 \\
  &= \Delta f\  dx_1\wedge dx_2\wedge dx_3
\end{align*}

where $\Delta f$ denotes the Laplacian of $f$.

I leave it to you the show that $f$ is a harmonic function on $H$, i.e. show that $\Delta f(p)=0$ for all points $p\in H$.


问:
On the number of ways to draw kissing circles
So I was watching Numberphile with Neil Sloane of OEIS fame on the number of ways to make circles intersect. During the intro, he explicitly forbid kissing (touching, tangent) circles. This made me think about the number of ways to draw circles that must all kiss, where intersections are not allowed. Kind of the dual problem, if you will. So in a nutshell:

Circles in the affine plane can be of any finite size

Circles must kiss at least one other (no separate circles, all must be connected)

Solutions that can be transformed into each other through mirroring, rotation, stretching, moving circles without making new tangent points, are considered the same

For $n=2$ circles there are two ways: touching on the outside, and touching on the inside of the larger circle.

For $n=3$ I believe the answer is 9. Try it, should be easy.

For $n=4$ I started drawing and came up with, so far, and with help from you guys, 38, 39, 42, 43, 44, 45, 47, 44 ways. The colored dots represent the number of kissing circles at the tangent points.

Questions:

Did I miss any? I'm sure I did.

Do I have inadvertent duplicates?

Does an integer sequence exist, or even a formula mapping $n = 2, 3, 4, ...$ to $2, 9, 44, ...$?

答:
This is not a complete answer, but a look at ways of describing these drawings and forming larger drawings from smaller ones.

Slightly abusing notation, let's say $A \sim B$ if circles $A$ and $B$ touch externally, or $A < B$ if circle $A$ is inside and tangent to circle $B$. Square brackets around a "formula" of three or more circles mean that they all meet at a common tangent point; otherwise, each relation symbol is a tangent point for only two circles.

Then for $n=2$, the two ways are $C_1 \sim C_2$ and $C_1 < C_2$.

For $n=3$, there are $9$ ways:

$[C_1 < C_2 < C_3]$

$C_1 < C_2 < C_3$

$C_1 \sim C_2 < C_3$

$C_1 < C_3$ and $C_2 < C_3$

$C_1 \sim C_2 < C_3$ and $C_1 < C_3$

$[C_1 < C_2 \sim C_3]$

$C_1 < C_2 \sim C_3$

$C_1 \sim C_2 \sim C_3$

$C_1 \sim C_2 \sim C_3 \sim C_1$

Note for any connected diagram of this sort with three or more circles, it's always possible to remove some circle with empty interior so that the remaining diagram is still connected. Proof:

Given a circle $C$, call its "inner depth" the largest integer $n$ such that there is a sequence of circles $C_i$ with $C_1 < C_2 < \cdots < C_n < C$. A circle with empty interior has inner depth zero.

If all circles in a diagram of at least three circles have inner depth zero, consider the graph defined by identifying the circles as vertices, with edges between vertices where the circles are tangent. A finite graph with at least two vertices always has at least two vertices which are either a leaf or within a graph cycle, and removing such a vertex from the graph leaves the remainder of the graph connected. So a corresponding circle can be removed and leave the remaining diagram connected, and in this case all the circles have empty interiors.

If not all circles in a diagram of at least three circles have inner depth zero, then there is at least one circle $C_0$ with inner depth $1$. Consider the graph whose vertices are $C_0$ and the circles it contains. There is at least one contained circle, and all these contained circles have inner depth $0$. This graph again has at least two vertices which can be removed without disconnecting the graph. Remove a circle which is not $C_0$ from the diagram. This circle has inner depth zero and empty interior. Since the graph remains connected, all the other remaining circles contained in $C_0$ are connected to $C_0$. All the diagram's circles outside $C_0$ are connected to $C_0$ via paths entirely outside $C_0$ (up to a tangent point), and those paths have not been affected. So the diagram is still connected.

The other way around, then, when looking for diagrams of $n+1$ circles, we can simplify by considering where to add a circle to a diagram of $n$ circles so that the added circle does not enclose any other circle. In the above list of diagrams for $n=3$, the first $5$ can be found from adding the minimal $C_1$ to a diagram $C_2 < C_3$, the next $2$ from adding to a diagram $C_1 < C_2$ OR $C_2 \sim C_3$, and the last $2$ from adding to a diagram $C_2 \sim C_3$.

Without looking back at the OP's drawings, I came up with 52 drawings for $n=4$. They're ordered by the nature of the diagram for $C_2,C_3,C_4$. A parenthesized color code matches the OP's convention of "r" for a point with two circles, "g" a point with three, or "b" a point with 4. A number after that indicates the equivalent drawing in the OP, if any.

(b 45): $[C_1 < C_2 < C_3 < C_4]$

(gr 37): $[C_2 < C_3 < C_4]$ and $C_1 < C_2$

(gr): $[C_2 < C_3 < C_4]$ and $C_1 \sim C_2$

(grr 43): $[C_2 < C_3 < C_4]$ and $C_2 \sim C_1 < C_3$

(gr): $[C_2 < C_3 < C_4]$ and $C_1 < C_3$

(gr 36): $[C_2 < C_3 < C_4]$ and $C_1 \sim C_3$

(grr 41): $[C_2 < C_3 < C_4]$ and $C_3 \sim C_1 < C_4$

(gr 38): $[C_2 < C_3 < C_4]$ and $C_1 < C_4$

(b 46): $[C_2 < C_3 < C_4 \sim C_1]$

(gr): $[C_2 < C_3 < C_4]$ and $C_1 \sim C_4$

(gr 35): $[C_1 < C_2 < C_3]$ and $C_3 < C_4$

(rrr 5): $C_1 < C_2 < C_3 < C_4$

(rrr 9): $C_1 \sim C_2 < C_3 < C_4$

(rrrr): $C_2 < C_3 < C_4$ and $C_2 \sim C_1 < C_3$

(rrr): $C_2 < C_3 < C_4$ and $C_1 < C_3$

(gr): $[C_2 < C_3 \sim C_1]$ and $C_3 < C_4$

(grr 44): $[C_2 < C_3 \sim C_1]$ and $C_1 < C_4$ and $C_3 < C_4$

(rrr): $C_2 < C_3 < C_4$ and $C_1 \sim C_3$

(rrrr 19): $C_2 < C_3 < C_4$ and $C_3 \sim C_1 < C_4$

(rrr 4): $C_2 < C_3 < C_4$ and $C_1 < C_4$

$C_2 < C_3$ and $[C_3 < C_4 \sim C_1]$

(rrr 12): $C_2 < C_3 < C_4 \sim C_1$

(rrrr 15): $C_4 > C_2 \sim C_3 < C_4 \sim C_1$

(grr 39=40): $[C_3 < C_4 \sim C_1]$ and $C_3 \sim C_2 < C_4$

(rrrrrr 30): $C_1 \sim C_2 \sim C_3 \sim C_1$ and $C_1 < C_4$ and $C_2 < C_4$ and $C_3 < C_4$

(rrrrr 26): $C_1 \sim C_2 \sim C_3 \sim C_1$ and $C_2 < C_4$ and $C_3 < C_4$

(rrrrr 25): $C_1 \sim C_2 \sim C_3$ and $C_1 < C_4$ and $C_2 < C_4$ and $C_3 < C_4$

(rrrr 22): $C_1 \sim C_2 \sim C_3$ and $C_2 < C_4$ and $C_3 < C_4$

(rrrr 20): $C_4 > C_2 \sim C_3 < C_4$ and $C_1 < C_4$

(gr): $[C_1 < C_2 \sim C_3]$ and $C_3 < C_4$

(rrr): $C_1 < C_2 \sim C_3 < C_4$

(rrr): $C_1 \sim C_2 \sim C_3 < C_4$

(rrrr 17): $C_4 > C_1 \sim C_2 \sim C_3 < C_4$

(rrrr): $C_3 \sim C_1 \sim C_2 \sim C_3 < C_4$

(rrr 7): $C_1 < C_4 > C_3 \sim C_2$

(gr): $[C_3 < C_4 \sim C_1]$ and $C_2 \sim C_3$

(rrr 11): $C_2 \sim C_3 < C_4 \sim C_1$

(rrr 6): $C_1 < C_4$ and $C_2 < C_4$ and $C_3 < C_4$

(gr 34): $[C_3 < C_4 \sim C_1]$ and $C_2 < C_4$

(rrr 13): $C_2 < C_4$ and $C_3 < C_4$ and $C_1 \sim C_4$

(grr 42): $[C_2 < C_3 \sim C_4]$ and $C_3 \sim C_1 \sim C_4$

(gr 32): $[C_2 < C_3 \sim C_4]$ and $C_1 \sim C_3$

(gr 33): $[C_2 < C_3 \sim C_4]$ and $C_1 \sim C_4$

(rrrr 21): $C_1 \sim C_3 \sim C_4 \sim C_1$ and $C_2 < C_3$

(rrr 8): $C_1 \sim C_3 \sim C_4$ and $C_2 < C_3$

(rrr 2): $C_3 \sim C_4 \sim C_1$ and $C_2 < C_3$

(rrr 1): $C_1 \sim C_2 \sim C_3 \sim C_4$

(rrr 10): $C_2 \sim C_3 \sim C_4$ and $C_1 \sim C_3$

(rrrr 14): $C_1 \sim C_2 \sim C_3 \sim C_4 \sim C_1$

(rrrr 18 (and 23)): $C_2 \sim C_3 \sim C_4 \sim C_2$ and $C_1 \sim C_4$

(rrrrr 24 (and 27)): $C_2 \sim C_3 \sim C_4 \sim C_2$ and $C_1 \sim C_3$ and $C_1 \sim C_4$

(rrrrrr 29): $C_2 \sim C_3 \sim C_4 \sim C_2$ and $C_1 \sim C_2$ and $C_1 \sim C_3$ and $C_1 \sim C_4$

Comparing to the OP's drawings, I see that my descriptions for numbers 18 and 24 also describe 23 and 27, respectively. Among the OP's drawings, I also missed

(rrr 3): $C_2 < C_3 \sim C_4 > C_1$

(gr 31): $[C_2 < C_3 \sim C_4]$ and $C_1 < C_4$

(b 47): $[C_1 < C_2 \sim C_3 > C_4]$

So this is at least 57 drawings for $n=4$. I think the 58th mentioned by @DanUznanski's answer is the variation of (rrrr 21: $C_1 \sim C_3 \sim C_4 \sim C_1$ and $C_2 < C_3$) where $C_2$ touches $C_3$ inside the triangle formed by the other three tangent points.

I don't think this notation is sufficient alone for higher numbers. With a loss of symmetry, it can matter which order the tangent points come in for a circle with three or more tangent points. We already see this sort of trouble for $n=4$ with pairs $18, 23$ and $24, 27$.


问:
Odds of getting at least one triple consecutive wins.
Given P(winning game) = p and P(losing game) = 1-p.

If you play 10 rounds of games only, what is the probability of getting at least one triple consecutive wins? in terms of p.

Where WWWLLLLLLL = one triple consecutive win.

答:
It is simpler to calculate the complimentary probability of no triples in a string of length $n$. Let the number of losses in the string be $k$.  There are $k+1$ gaps where we should place $n-k$ wins (no more than 2 wins per gap). Applying stars and bars and inclusion-exclusion principle one finds that the number of ways to arrange wins and losses is:
$$
N(n,k)=\sum_{j=0}^{k+1}(-1)^j\binom{k+1}j\binom{n-3j}k,
$$
so that the complimentary probability can be computed as
$$
P(n)=\sum_{k=0}^n N(n,k)p^{n-k}(1-p)^k.
$$


问:
Why the substitution is not working even though its bijective?
The following integral i was trying to evaluate:
$$\int_0^\infty \left( \sin(1/x) - \frac{\sin(\pi/x)}{\pi} \right) \,dx.$$ ,what i did was substituting pi/x = 1/t in the second integral , which converts to first integral which tells integral is zero ,but answer is ln(pi) , whats the fault that i wanted to know .

答:
It sounds like what you did was to write
$$
\int_0^\infty \left( \sin(1/x) - \frac{\sin(\pi/x)}{\pi} \right) \,dx = \int_0^\infty \sin(1/x)\, dx - \int_0^\infty \frac{\sin(\pi/x)}{\pi} \,dx
$$
You then made a $u$-substitution on the second term $u = x/\pi$, in which case the integrals are the same and "should" cancel.

The problem is that both of these improper integrals are divergent.  Technically what we have here is
$$
\lim_{R \to \infty} \left[\int_0^R \sin(1/x)\, dx - \int_0^R \frac{\sin(\pi/x)}{\pi} \,dx \right]
$$
but since both integrals approach $\infty$, this is a limit of the form $\infty - \infty$ and is not necessarily zero.

If we do the $u$-substitution for the proper integrals, it becomes instead
$$
\lim_{R \to \infty} \left[ \int_0^R \sin(1/x)\, dx - \int_0^{R/\pi} \sin(1/u) \,du \right] = \lim_{R \to \infty} \int_{R/\pi}^R \sin(1/x) \, dx
$$
which is more amenable to limit-taking.  (In particular, you might try substituting $w = 1/x$ at this point.)


答:
Let $a$ denote first $n$ coordinates and $b$ the last $n$ coordinates of a potential eigenvector, that is,
$$\lambda\pmatrix{a\\b}=\pmatrix{A&I-A\\I&0}\pmatrix{a\\b}=\pmatrix{Aa+b-Ab\\a}\,.$$
It follows that $\lambda b=a$ and $\lambda a=Aa+b-Ab$.

Substituting, we get
$$(\lambda^2-1)b=(\lambda-1)Ab$$
so in case $\lambda\ne 1$, $b$ must be an eigenvector of $A$ with eigenvalue $\lambda+1$.

Can you take it from here?

Title: matrix and eigenvalues

Body:
Please help me solve this question. This is not a homework assignment. If $\lambda_M$ is the eigenvalue of the matrix M, please prove that $max|\lambda_M|<1$
$$M_{2n\times2n}=\begin{vmatrix} A_{n\times n} & I-A\\I_{n\times n} & 0_{n\times n}\end{vmatrix}$$
NOTE: $max|\lambda_A|<2$

I easily solved this problem in scalar mode but I have no idea about its matrix mode


问:
Probability that XOR of an arbitrary number of random bits is 1
How can we say than the XOR between $n$ (uniform, independent) random bits is $1$ with probability $1/2$?
For example if we have 4 random bits, we know that the XOR of them will be $1$ with half probability. The same happens if we have 3, or 4 random bits. But if we have $n$ random bits?

答:
The last bit is $0$ or $1$, each with probability $\frac12$, independently of the XOR of the previous bits,

so leaves that result the same or changes it, each with probability $\frac12$,

meaning the overall XOR is $0$ or $1$, each with probability $\frac12$.


问:
Morphisms between locally ringed spaces and affine schemes
I need some hints to understand the conclusion of the proof of the following lemma from the Stacks Project:

Lemma $\mathbf{6.1.}\,$ Let $X$ be a locally ringed space. Let $Y$ be an affine scheme. Let $f\in\operatorname{Mor}(X,Y)$ be a morphism of locally ringed spaces. Given a point $x\in X$ consider the ring maps $$\Gamma(Y,\mathcal O_Y)\xrightarrow{f^{\#}}\Gamma(X,\mathcal O_X)\to\mathcal O_{X,x}$$ Let $\mathfrak p\subset\Gamma(Y,\mathcal O_Y)$ denote the inverse image of $\mathfrak m_x$. Let $y\in Y$ be the corresponding point. Then $f(x)=y$.

Proof. Consider the commutative diagram

$$\require{AMScd}\begin{CD}\Gamma(X,\mathcal O_X) @>>> \mathcal O_{X,x}\\ @AAA @AAA \\ \Gamma(Y,\mathcal O_Y) @>>> \mathcal O_{Y,f(x)}\end{CD}$$ (see the discussion of $f$-maps below Sheaves, Definition $21.7$). Since the right vertical arrow is local we see that $\mathbf m_{f(x)}$ is the inverse image of $\mathbf m_x$. The result follows.$\qquad\qquad\qquad\qquad\quad\square$

I don't understand  the last sentence, i.e.  "The result follows". If $m_{f(x)}$ (the maximal ideal of $\mathcal O_{Y,f(x)}$) is the inverse image of $m_x$, why should be $f(x)=y$?

Thanks in advice.

答: $Y$ is an affine scheme, so $f(x)$ is a prime ideal of $\mathscr{O}_Y(Y)$ and $\mathscr{O}_{Y,f(x)}$ is the localization of $\mathscr{O}_Y(Y)$ at $f(x)$. A basic fact about localization is that the inverse image in $\mathscr{O}_Y(Y)$ of the unique maximal ideal $\mathfrak{m}_{f(x)}$ of $\mathscr{O}_{Y,f(x)}$ is $f(x)$. Now it follows from the commutativity of the diagram, together with the fact you mentioned, which amounts to the map on stalks being local, that $f(x)$ is the inverse image in $\mathscr{O}_Y(Y)$ of $\mathfrak{m}_x$ under the map $\mathscr{O}_Y(Y)\rightarrow\mathscr{O}_X(X)\rightarrow\mathscr{O}_{X,x}$.</p>


问:
Solving Complex Quadratics containing Conjugate Imaginary Numbers
I've been trying to solve the equation: $z^2 + 2(z*) + 1 = 0$, where $(z*)$ is the complex conjugate of the complex number $z$, where $z \in \mathbb C$.

How do I interpret, and then solve this equation? Can I factorise for the real roots, and how do I find the complex roots?

Thanks all!!

答:
The direct way would be to write $z=x+iy$ for reals $x,y$. Given equation becomes
$$(x^2-y^2+2ixy)+2(x-iy)+1=0$$
$$\Rightarrow (x^2-y^2+2x+1)+i(2xy-2y)=0$$

On equating real and imaginary parts,
$$2y(x-1)=0$$

and

$$(x+1)^2-y^2=0$$

Can you solve these two simultaneously?


问:
What is an equation connecting $w$ and $r$ when $r=60$ and $W=4320$ when $w$ and $r$ are directly proportional
So this question is written weight w and Newton’s r is directly proportional to the cube of its radius r cm I thought I would just divide $4320$ by $60$ $= 72$ and the equation would be $w=r·72$, but according to the book it’s wrong and the answer is $w=0.02r^3$ But I don’t know how they got to this answer and help is much appreciated.thank you.

答:
so i assume the question says the weight is  directly proportional to the cube of its radius.

so writing it mathematically,

$w∝r^3$

This mean w is some function of $r_3$, that could look like $w=Ar^3$, where A is some constant. Now you can use the values given,

$4320=A \times 60^3$

making $A=0.02$

TBH if that all the question has given you its not worded very well, if w is proportional $r^3$ that could make the equation a lot of things i.e. $w=Ar^3+B$, where A&B are constants, and many others...


问:
ratio of exponential independent random variables (with limitations on the soltion)
Let $X,Y$ be indpendent exponential randonm variables with the same parameter $\lambda$.

Define $Z=\frac{Y}{X}$, find $f_z(z)$ , the PDF of $Z$.

Now here's the deal: this question was asked previously over here, however - in my solution i'm limited to a solution not including the use of an expected value or any theorems involving change of variables (I really don't know the right term because we didn't learned such theorem at my class on university).

Check how I solved a previous question to get an idea of the desired solution type over here.

My efforts:

$F_{_Z}(z) = P(\frac{Y}{X}\le z) = P(Y\le Xz) = \int\limits_0^\infty P(Y\le xz, X\color{red}=x)dx$

but that fails since $P(Y\le xz, X=x) = P(Y\le xz)P(X=x)=0$

$F_{_Z}(z) = \int\limits_0^\infty P(Y\le xz, X \color{red}\le x)dx = \int\limits_0^\infty F_{_Y}(xz)F_{_X}(x)dx = \int\limits_0^\infty(1-e^{-\lambda xz})(1-e^{-\lambda x})dx$

but that will go to infinity.

Any help would be appreciated,

Thanks

答:
This answer summarizes my comments. There are two basic ways:

1. Integrate the joint PDF:

$$ P[(X,Y) \in D] = \int \int_D f_{X,Y}(x,y)dxdy$$
In this case, by independence, we know $f_{X,Y}(x,y) = f_X(x)f_Y(y)$. Thus
\begin{align*}
P[Y \leq Xz] &= \int_{x=-\infty}^{\infty} \int_{y=-\infty}^{xz} f_X(x)f_Y(y)dydx \\
&= \int_{x=-\infty}^{\infty} \left[\int_{y=-\infty}^{xz} f_Y(y)dy  \right]f_X(x)dx
\end{align*}

2. Law of total probability in integral form:

\begin{align*}
P[Y\leq Xz] &= \int_{x=-\infty}^{\infty} P[Y \leq Xz|X=x] f_X(x)dx \\
&= \int_{x=-\infty}^{\infty} P[Y \leq xz|X=x] f_X(x)dx \\
&= \int_{x=-\infty}^{\infty} P[Y\leq xz]f_X(x)dx
\end{align*}
where the final equality holds by independence.  Since
$$ P[Y \leq xz] = \int_{y=-\infty}^{xz} f_Y(y)dy$$
we can see the two methods give the same answer.