问:
Line bundles on $S^2$ and $\pi_2(\mathbb{R}P^2)$
Real line bundles on $S^2$ are all trivial, but what about the following way to think about a line bundle: we view a line bundle on $S^2$ (thought of as living in $3$d space) as providing a map $S^2 \to \mathbb{R}P^2$ the space of lines; and since $\pi_2(\mathbb{R}P^2) \simeq \mathbb{Z}$ we should have a $\mathbb{Z}$ worth of different line-bundles.

Clearly I've cheated very badly in the latter line of thought; but I perhaps am still a bit confused at what step I have cheated so badly; is it that the following intuition just makes no sense?

if I have a manifold $M \hookrightarrow \mathbb{R}^n$ and a line bundle $L \mapsto M$; can I always find a vector field $V$ on $\mathbb{R}^n$ that is not zero anywhere on $M$, such that the line bundle $L$ looks like what you get from 'integrating $V$ around $M$'? I think this is another way of saying my inuition that line bundles come from maps $M \to \mathbb{R}P^{n-1}$ which might be totally wrong; and which I'm not too sure how to formalise

EDIT : I just realised since $\pi_2(\mathbb{R}P^3)$ is trivial, even if my intuition above was correct, you'd still have just the trivial line bundle on $S^2$ since you can always "view the lines in a higher dimension" and rotate them away; if that makes any sense. Sorry if this is all just gibberish: just trying to learn

答:
The classifying space of real line bundles is $\mathbb{RP}^{\infty}$, not $\mathbb{RP}^2$; $\mathbb{RP}^2$ instead classifies line subbundles of the trivial $3$-dimensional real vector bundle $\mathbb{R}^3$ (and similarly $\mathbb{RP}^n$ classifies line subbundles of the trivial $n+1$-dimensional real vector bundle $\mathbb{R}^{n+1}$).

So what the calculation of $\pi_2(\mathbb{RP}^2)$ vs. $\pi_2(\mathbb{RP}^n), n \ge 3$ reveals is that there are a $\mathbb{Z}$'s worth of real line subbundles of $\mathbb{R}^3$ on $S^2$ but that these bundles all become isomorphic after adding an additional copy of $\mathbb{R}$. With a little effort it should be possible to write down these line subbundles and the resulting isomorphisms explicitly. Probably the normal bundle of the embedding $S^2 \to \mathbb{R}^3$ is a generator.

If $X$ is any compact Hausdorff space then every vector bundle on it is a direct summand of a trivial bundle, so for line bundles what this tells us is that every line bundle is represented by a map $X \to \mathbb{RP}^n$ for some $n$ (but isomorphisms of line bundles may require passing to a larger value of $n$ to define). This connects up nicely with the picture where $\mathbb{RP}^{\infty}$ is the filtered colimit of the $\mathbb{RP}^n$'s, because a map $X \to \mathbb{RP}^{\infty}$ has image contained in some $\mathbb{RP}^n$ by compactness.


问:
Let $M$ be a noetherian $R$-module and $I=\mathrm{Ann}_R(M)$. Then $R/I$ is a noetherian ring for all ring $R$?
Matsumura $p22$,thm$3.5$ says

Let $M$ be a noetherian $R$-module and $I=\mathrm{Ann}_R(M)$. Then $R/I$ is a noetherian ring.

My question is,; is this true for arbitrary ring $R$,or,is this true for non-commutative rings?

I would be appreciated if you could give me some examples.
Thank you for your kind help.

答:
It is not true that $R/\text{ann}_R(M)$ must be right Noetherian if $M$ is a Noetherian right $R$-module.  Any right primitive ring that is not right Noetherian is a counterexample: take $M$ to be a faithful simple module (which is certainly Noetherian).  Examples of such rings include endomorphism rings of infinite dimensional vector spaces, free algebras on more than one generator, and group algebras of free groups on more than one generator.  (See, e.g., ``The Algebraic Stucture of Group Rings'' by Passman, Corollary 9.2.11 for the group algebra case.)

If you want a noncommutative result, you have to assume $M$ is a bimodule.  For example, if $M$ is an $R$-$R$-bimodule that is Noetherian as both a left and a right $R$-module, then $R/\text{ann}_R(M)$ is right (or left) Noetherian, where the annihilator is the right (or left) annihilator.

To see this, suppose $m_1,\dots,m_n$ generate $M$ as a left $R$-module and define a right $R$-module map $\phi:R\to M^n$ by $\phi(r)=(m_1r,\dots,m_nr)$.  Because $m_1,\dots,m_n$ generate $M$ as a left $R$-module, the kernel of $\phi$ is $\text{ann}_R(M)$, where the annihilator is the right annihilator of $M$.  Thus $R/\text{ann}_R(M)$ emebeds in $M^n$ as a right module and so is right Noetherian.

We can weaken the hypotheses.  If we assume $M$ is an $S$-$R$-bimodule for some ring $S$ and assume that ${}_SM$ is finitely generated and $M_R$ is Noetherian, then the same proof shows $R/\text{ann}_R(M)$ is right Noetherian.

ContentLicense: CC BY-SA 4.0


问:
Expected number of pieces of a chessboard
If n squares are randomly removed from a $p \ \cdot \ q$ chessboard, what will be the expected number of pieces the chessboard is divided into?

Can anybody please provide how can I approach the problem? There are numerous cases and when I go through case consideration it becomes extremely complex.

答: I am posting this as an answer to make it more visible. This is currently a Brilliant.org problem, in which the specific case $ 2 \times 500 $ is asked for. While it is an interesting problem meriting discussion, that should wait until after 8:00 PM EST 5/19 when the problem set is closed.</p>

ContentLicense: CC BY-SA 3.0


问:
Given positive real numbers a and b, prove that $\frac{2}{\frac{a}{x}+\frac{b}{y}} \leq ax + by, x>0, y>0$
I am working through Problem Solving through Problems by Larsons. I am stuck on question 1.8.5b) which is as follows:

"
Given positive real numbers $a$ and $b$ such that $a + b = 1$, prove that

$$\frac{2}{\frac{a}{x}+\frac{b}{y}} \leq ax + by, \ x>0, y>0$$

"

It appears to me that this question has a typo. Given any $a$ and $b$, choose $x = 1$, $y = 1$. Then:

$$\frac{2}{\frac{a}{1}+\frac{b}{1}} \leq a + b \Leftrightarrow  \frac{2}{a+b} \leq a + b \Leftrightarrow \frac{2}{1} \leq 1 \Rightarrow \Leftarrow Contradiction$$

My long-shot hypothesis is that the "$2$" is supposed to be a "$1$". I think that I actually proved this version of the problem (it was how I first realized that something seemed weird in the problem -- won't get into that proof here). However, I am hugely unsure about all of this. It seems very bizarre to me that Problem Solving would have this sort of error, and I suspect that I am just getting something very simple wrong somewhere. Anyone agree with my reasoning or know where I went wrong? Thank you!

Edit: Thought I would drop a link to the original text where the problem appears as above: https://math.la.asu.edu/~ifulman/spring13/mat194/problem-solving.pdf

答:
You are right, the problem is wrong. Actually we have
$$
\left( \frac{a}{x}+\frac{b}{y} \right) \left( ax+by \right) =a^2+b^2+ab\left( \frac{y}{x}+\frac{x}{y} \right) \ge a^2+b^2+2ab=\left( a+b \right) ^2=1
$$
which indicates
$$
\frac{1}{\frac{a}{x}+\frac{b}{y}}\le ax+by
$$


答:
Note
$$\frac{1}{\frac{a}{x}+\frac{b}{y}} \leq ax + by\tag1$$
is equivalent to
$$ (ax+by)(\frac{a}{x}+\frac{b}{y})\ge1 $$
or
$$ a^2+b^2+ab(\frac{x}{y}+\frac{y}{x})\ge1. \tag2$$
Since
$$ \frac{x}{y}+\frac{y}{x}\ge 2 $$
one has
$$ a^2+b^2+ab(\frac{x}{y}+\frac{y}{x})\ge a^2+b^2+2ab=(a+b)^2=1 $$
which says (2) is true. So (1) is true.

Title: Given positive real numbers a and b, prove that $\frac{2}{\frac{a}{x}+\frac{b}{y}} \leq ax + by, x>0, y>0$

Body:
I am working through Problem Solving through Problems by Larsons. I am stuck on question 1.8.5b) which is as follows:

"
Given positive real numbers $a$ and $b$ such that $a + b = 1$, prove that

$$\frac{2}{\frac{a}{x}+\frac{b}{y}} \leq ax + by, \ x>0, y>0$$

"

It appears to me that this question has a typo. Given any $a$ and $b$, choose $x = 1$, $y = 1$. Then:

$$\frac{2}{\frac{a}{1}+\frac{b}{1}} \leq a + b \Leftrightarrow  \frac{2}{a+b} \leq a + b \Leftrightarrow \frac{2}{1} \leq 1 \Rightarrow \Leftarrow Contradiction$$

My long-shot hypothesis is that the "$2$" is supposed to be a "$1$". I think that I actually proved this version of the problem (it was how I first realized that something seemed weird in the problem -- won't get into that proof here). However, I am hugely unsure about all of this. It seems very bizarre to me that Problem Solving would have this sort of error, and I suspect that I am just getting something very simple wrong somewhere. Anyone agree with my reasoning or know where I went wrong? Thank you!

Edit: Thought I would drop a link to the original text where the problem appears as above: https://math.la.asu.edu/~ifulman/spring13/mat194/problem-solving.pdf


问:
Finding the distribution of the sum of independent random variables.
I cannot imagine that this has not been asked before, but the search function floods you with specific examples, so I cannot find it. The thing I want to prove is as follows:

Suppose $X, Y: (\Omega, \mathcal{F}, \mathbb{P}) \to \mathbb{R}$ are independent random variables, let $Z = X + Y$. Denote $F_X, F_Y, F_Z$ for their distributions and $\mu_Y$ for the law of $Y$ (so $\mu_Y(B) = \mathbb{P}(Y^{-1}(B))$. What I want to show is

$F_Z(z) = \int F_X(z - y) \mu_Y(dy)$

I basically have no idea on how to solve this.

The only way I can think of to use the independence is by noting that $\mathbb{P}(Z^{-1}(-\infty, z]) = \mathbb{P}\left((\langle X, Y \rangle)^{-1}(C)\right)$ where $C = +^{-1}((-\infty, z]) = \{(x, y) \ | \ x + y \leq z\}$. Now for rectangles $A \times B$ we have by independence that $\mathbb{P}\left((\langle X, Y \rangle)^{-1}(X \times Y)\right) = \mathbb{P}(X^{-1}(A)) \mathbb{P}(Y^{-1}(B))$, so if we can write $C$ as a suitable disjoint union of rectangles we may be able to do something with it.

答:
As $X$ and $Y$ are independent then the distribution of $W:=(X,Y)$ is just the product of the distributions, that is: $F_W=F_X\cdot F_Y$, this means that
$$
\Pr [W\in A]=\int_{A}\mathop{}\!d F_W=\int_{A}\mathop{}\!d (F_X\cdot F_Y) \tag1
$$
In your case we have that $A=\{(x,y)\in \mathbb{R}^2: x+y\leqslant c\}$, so
$$
\Pr [Z\leqslant c]=\Pr [W\in A]=\int_{A}d(F_X\cdot F_Y)=\int_{\mathbb{R}^2}\mathbf{1}_{A}(x,y)\mathop{}\!d (F_X(x)\cdot F_Y(y))\\
=\int_{\mathbb{R}}\int_{\mathbb{R}}\mathbf{1}_{A}(x,y)\mathop{}\!d F_X(x)\mathop{}\!d  F_Y(y)=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}\mathbf{1}_{A}(x,y)\mathop{}\!d F_X(x)\right)\mathop{}\!d F_Y(y)\tag2
$$
Now observe that for fixed $y$ we have that $\mathbf{1}_{A}(x,y)=1 \Leftrightarrow x+y\leqslant c \Leftrightarrow x\leqslant c-y$, so we have that
$$
\mathbf{1}_{A}(x,y)=\mathbf{1}_{\mathbb{R}}(y)\mathbf{1}_{(-\infty .c-y]}(x)\tag3
$$
Using (3) in (2) we finally find that
$$
\Pr [Z\leqslant c]=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}\mathbf{1}_{\mathbb{R}}(y)\mathbf{1}_{(-\infty ,c-y]}(x)\mathop{}\!d F_X(x)\right)\mathop{}\!d F_Y(y)\\
=\int_{\mathbb{R}}\left(\int_{(-\infty ,c-y]}\mathop{}\!d F_X(x)\right)\mathop{}\!d F_Y(y)=\int_{\mathbb{R}}F_X(c-y)\mathop{}\!d F_Y(y)\tag4
$$
∎


问:
Cannot understand why cosine similarity is calculated as $\cos(u,v) = 1 - \frac{u\cdot v}{\|u\|_2 \cdot \|v\|_2}$
I am reading this article (PDF via arXiv.org).

At page 3 of the PDF, right column, under section 3.2 "Quantifying Bias Removal", there is a formula to calculate the cosine distance between two vectors:

$$\cos(\mathbf{u},\mathbf{v}) = 1 - \dfrac{\mathbf{u} \cdot \mathbf{v}}{\|u\|_2 \cdot \|v\|_2} \tag1$$

But I thought that in an Euclidean space:

$$\mathbf{u} \cdot \mathbf{v} = \|u\|_2 \cdot \|v\|_2 \cdot \cos(\theta) \tag2$$

So the right hand side of $(1)$ would be $ 1-\cos(\theta) $ and not $\cos(\theta)$.

答: Notice that "cosine distance" is actually different from "cosine"(which is defined in analytical geometry textbooks). It is just a new concept lead into the context by the author himself for further descriptions.


问:
How to deduce the Vieta's substitution $y = z-\frac {p}{3z}$ for cubic equation $y^3+py +q=0$
In deriving the formula of cubic equations, Vieta substituted the following
$$y = z-\frac {p}{3z}$$

for the depressed cubic equation

$$y^3+py +q=0$$

and transformed it into a quadratic one.

My question: How did he get that substitution, or how did he know that by substituting $y = z-\frac {p}{3z}$, he could turn cubic into quadratic?

Please help me!

答:
This is a sligntly different way of looking at your problem and getting the same results.

You have an equation of the form $$t^3 + pt + q=0 \tag{A.}$$

Let $t=z-w$. Then

$\begin{align}
   t^3 &= z^3 - 3z^2V + 3zw^2 - w^3 \\
   t^3 &= 3zw(z - w) + (z^3 - w^3) \\
   t^3 &= 3zwt + (z^3 - w^3)
\end{align}$

Hence

$$t^3 - 3zwt - (z^3-w^3) = 0 \tag{B.}$$

Comparing (A.) and (B.), we get

\begin{align}
   p &= -3zw \\
   q &= -z^3 + w^3 \\
\end{align}

Rewriting the second equation as
$$z^3 + q - w^3 = 0$$

And substituting $w = -\frac{p}{3z}$, we get
$z^3 + q - \left(\dfrac{p}{3z}\right)^3 = 0$, which becomes

$$z^6 + qz^3 - \left(\dfrac{p}{3}\right)^3 = 0$$


问:
Introduction to p-adic numbers
I am a freshman and for a final project of a subject I have to give an introduction to p-adic numbers, I look for some sources (books, videos,  articles) to be able to do my work, the only bases I have is a very basic "number theory" (construction of $\mathbb{N}$,  $\mathbb{Z}$, $\mathbb{Q}$, through equivalence relations and the construction of $\mathbb{R}$ by Dedekind cuts. And dense sets in $\mathbb{R}$ and $\mathbb{R}^n$), also basic things of modular arithmetic and diophantic equations and more things that are seen in the first year (calculus I, basic set theory etc.)

Could you give me some recommendations?

Thanks in advance.

答: If you're interested in video format content, Richard Borcherds (a prominent algebraist, if you're not familiar) has a three part video series on the $p$-adic numbers which is aimed at advanced high school students (https://www.youtube.com/watch?v=VTtBDSWR1Ac).

Title: Introduction to p-adic numbers


问:
Solve this : $ x^{\frac 34 { \log_2^2 x} + \log_2 {(x-5/4) }} = \sqrt 2 $

The question : $$ x^{\frac 34 { \log_2^2 x}  + \log_2 {\left(x-\frac 54\right) }}  = \sqrt 2 $$

My try at it : I think that the whole equation should be converted to log with base 2.  What to do next?

答:
Taking logarithms of base $2$ we have after simplification
$$3\log_2^3(x)+4\log_2(x)\log_2\left(x-\frac54\right)=2$$ This is a trascendental equation for which we must apply some numerical approximation.

The function is increasing, and we get an approximation, for example, with the values
$$f(2.0501058)\approx1.9999998948\lt2\\f(2.0501059)\approx2.0000012306\gt2$$ so a good enough approximation is given by $x=2.0501058$


问:
Is this the right way to view infinity in real analysis?
So, I've lately been having confusion on how to understand infinity, but I think I have progress in my intuition. So, I'd appreciate if someone would let me know if I'm on the right track, and which parts I still don't have right.

This is my understanding so far. Originally I had looked at the infinite intersection: $$\bigcap_{n\in\mathbb{N}} \left(0,\frac{1}{n}\right)$$
I had thought that $\bigcap_{n\in\mathbb{N}} \left(0,\frac{1}{n}\right) \neq \emptyset$ since for an arbitrary $n \in \mathbb{N}$, the intersection is non-empty. Thereofre, since every element in the intersection is non-empty, it must be that the total intersection is non-empty since the intervals are nested.

My current understanding is this: that while the above information could be used to prove that any finite/arbitrarily long intersection is non-empty, an arbitrarily large number will never cover all of $\mathbb{N}$. I reasoned that for any $m \in \mathbb{N}$, $\bigcup_{n=1}^{m}n$ does not cover all the natural numbers, since $m+1 \notin \bigcup_{n=1}^{m}n$. So while it covers arbitrarily many natural numbers, it can't possibly cover all of them.

So the only way to cover all of the natural numbers is using some object which is not in $\mathbb{N}$ and is greater than all natural numbers. If I have an $M$ such that $\forall n \in \mathbb{N},M > n$, then $\mathbb{N} \subset \bigcup_{n=1}^{M}n$. So, even if the intersection is non-empty for each natural number, its still neccesary to check that $(0,\frac{1}{M})$ is non-empty. Checking this, gives that its non-empty only if $\exists n\in \mathbb{N} \text{ such that } 0<\frac{1}{n}<\frac{1}{M}$. Since this implies $M<n$, but $\forall n\in\mathbb{N}, M>n $, then this fails for all natural numbers.

While my explanation is very unrigorous, it seems like it can be formalized by using limit ordinals, and including limit ordinals as a requirement when doing induction. Is this correct? Is this the right way to view these problems? Also, does a value the same size as the limit oridinal occur infinite times, or does it only "occur" at the end? More formally, would $$\left|\bigcup_{n \in\mathbb{N}} \begin{cases} \emptyset, \text{if} \left(0,\frac{1}{n}\right) \neq \emptyset \\\{n\},\text{otherwise} \end{cases} \right|$$
be 0, or $\omega$?

答:
Infinite collections are very different from finite collections.

From the definition of $\Bbb R,$ there is no positive $r\in \Bbb R$ that is less than $1/n$ for all $n\in \Bbb N,$ so $\cap_{n\in\Bbb N}(0,1/n)=\emptyset.$

When you have an infinite sequence $(t_n)_{n\in \Bbb N}$ of things, nothing happens at the end because there is no end.


答: It's going to be pretty difficult to find a source appropriate for your level, so I really just recommend checking out lots of graduate textbooks on the subject and trying to take away the key ideas and calculations. One good text that I can think of is "p-adic Numbers, p-adic Analysis, and Zeta-Functions" by Neal Koblitz.</p>

Title: Introduction to p-adic numbers

Body:
I am a freshman and for a final project of a subject I have to give an introduction to p-adic numbers, I look for some sources (books, videos,  articles) to be able to do my work, the only bases I have is a very basic "number theory" (construction of $\mathbb{N}$,  $\mathbb{Z}$, $\mathbb{Q}$, through equivalence relations and the construction of $\mathbb{R}$ by Dedekind cuts. And dense sets in $\mathbb{R}$ and $\mathbb{R}^n$), also basic things of modular arithmetic and diophantic equations and more things that are seen in the first year (calculus I, basic set theory etc.)

Could you give me some recommendations?

Thanks in advance.


问:
How many permutations of the string "000011112222" contain the substring "2020”?
So far, this is what I know. There are 9 ways to place 2020 and $\frac{8!}{4!2!2!}$ ways to arrange the remaining numbers. Im having a problem tackling the overcounting case.. If there are 2 seprate 2020s then i need to subtract $6C4$? Thanks in advance!

Is there a solution using PIE or recursion?

答:
@Funaizhang... This is what I am thinking.. But I can't figure out the discrepancy. Start with 2200001111. For this string, consider 2020 as one character. There are 7!/(4!2!)=105 ten character strings that contain 2020. At this stage, we are missing two 2s. Case 1: the two 2s are together. There are 8 spaces, so 8C1105=840. Here, no repetition of 2020 occurs because the two 2s are together. Case 2: The two 2s are separated. (There are double counts here.) We have a total of 8C2105=2940 ways to do so. But creation of new 2020s are possible, so the cases with 202020 are double-counted. So treat 202020 as 1 string, there are then 7!/4!=210 ways of arranging it. But the strings with 20202020

(5 of them) will be double-counted. So for this case, total should be 2940-210+5=2735. Finally, 2735+840=3575.


答:
There's a lot to unpack here.

First, there seems to be an issue with a fact of typical induction that (in my experience) sometimes escapes students:
a statement being true for each $n\in\Bbb N$ does not mean it 'spills over' to infinity.

Typical induction goes as follows:
$P(n)$ is a proposition about $n\in\Bbb N$.
If $P(0)$ is true and if $P(n)$ being true implies $P(n+1)$ is true, then $P(n)$ is true for all $n\in\Bbb N$.

Notice how this is fundamentally a result about natural numbers, all of which are finite.
Infinity is not a natural number.
You yourself have pointed that out, and indeed in mentioning ordinals it seems you might be acquainted with (or would be interested in) transfinite induction. Transfinite induction is what's used when we want a proposition to 'spill over' to infinity, and it tells us precisely that one needs to take special care with this transition (more precisely, the case of limit ordinals needs its own inductive step).

Moreover, and somewhat related, is the fact that, outside the context of induction, but in the context of limits, there are properties that may hold in each term of a limit but fail to hold in the limit.

You might have heard of fluke proofs that $\pi = 4$, by starting from a square that is progressively made to approach the circle that inscribes it.
In each step the resulting polygon preserves the perimeter of the initial square, and yet in the limit, we get the circle.

This is an example where the kind of convergence matters.
In this case, the limiting process is a pointwise convergence, for which plenty of properties fail to hold over the limit.
There are other kinds of stronger convergence (like uniform convergence), for which more properties hold over the limit.

Now, perhaps this example was discrete, and therefore it still smells of typical induction.
Of course, rather than consider it to be done in steps, you could smooth out the transition between each step along the real numbers, so the whole thing becomes a continuous limit.

But you can observe this in much simpler cases too.
Consider $\lim_{x\to\infty}1/x$.
This is a continuous limit, and for every $x>0$ the expression $1/x$ is positive.
And yet, the limit is $0$ (non positive).

Finally, there seems to be just plain misunderstanding about what the symbol $\bigcap_{n\in\mathbb{N}}$ means.
Or rather, given what you state in your question, it seems your intuition fails to conform to the understanding of the symbol.

There are plenty of counterintuitive things in mathematics.
Most paradoxes are usually named so not because they're wrong, but because when they were found out, they were deemed counterintuitive.

It is important to be able to recognize when our intuition fails.
This is usually a moment of growth: if we understand a posteriori why something fails, we'll have learned something valuable from it.
This applies to both points I shared previously: we learned something about induction, and we learned something about limits.

In the case of $\bigcap_{n\in\mathbb{N}} A_n$, the meaning is the set of elements that belong to all $A_n$.
In your example question, $A_n = (0,1/n)$.
We have

$$\begin{align}
a \in \bigcap_{n\in\mathbb{N}} \left(0,\frac1n\right)
&\iff \forall n\in\Bbb N, a \in\left(0,\frac1n\right)
\\&\iff \forall n\in\Bbb N, 0 < a < \frac1n
\end{align}$$

so $a$ is positive, and yet, $n<1/a$ for all $n\in\Bbb N$.
We are forced to conclude that no such $a$ exists, so the intersection $\bigcap_{n\in\mathbb{N}} \left(0,\frac1n\right)$ must be empty.

Title: Is this the right way to view infinity in real analysis?

Body:
So, I've lately been having confusion on how to understand infinity, but I think I have progress in my intuition. So, I'd appreciate if someone would let me know if I'm on the right track, and which parts I still don't have right.

This is my understanding so far. Originally I had looked at the infinite intersection: $$\bigcap_{n\in\mathbb{N}} \left(0,\frac{1}{n}\right)$$
I had thought that $\bigcap_{n\in\mathbb{N}} \left(0,\frac{1}{n}\right) \neq \emptyset$ since for an arbitrary $n \in \mathbb{N}$, the intersection is non-empty. Thereofre, since every element in the intersection is non-empty, it must be that the total intersection is non-empty since the intervals are nested.

My current understanding is this: that while the above information could be used to prove that any finite/arbitrarily long intersection is non-empty, an arbitrarily large number will never cover all of $\mathbb{N}$. I reasoned that for any $m \in \mathbb{N}$, $\bigcup_{n=1}^{m}n$ does not cover all the natural numbers, since $m+1 \notin \bigcup_{n=1}^{m}n$. So while it covers arbitrarily many natural numbers, it can't possibly cover all of them.

So the only way to cover all of the natural numbers is using some object which is not in $\mathbb{N}$ and is greater than all natural numbers. If I have an $M$ such that $\forall n \in \mathbb{N},M > n$, then $\mathbb{N} \subset \bigcup_{n=1}^{M}n$. So, even if the intersection is non-empty for each natural number, its still neccesary to check that $(0,\frac{1}{M})$ is non-empty. Checking this, gives that its non-empty only if $\exists n\in \mathbb{N} \text{ such that } 0<\frac{1}{n}<\frac{1}{M}$. Since this implies $M<n$, but $\forall n\in\mathbb{N}, M>n $, then this fails for all natural numbers.

While my explanation is very unrigorous, it seems like it can be formalized by using limit ordinals, and including limit ordinals as a requirement when doing induction. Is this correct? Is this the right way to view these problems? Also, does a value the same size as the limit oridinal occur infinite times, or does it only "occur" at the end? More formally, would $$\left|\bigcup_{n \in\mathbb{N}} \begin{cases} \emptyset, \text{if} \left(0,\frac{1}{n}\right) \neq \emptyset \\\{n\},\text{otherwise} \end{cases} \right|$$
be 0, or $\omega$?


问:
A theoretical model of a flow network.
I believe I've got a correct understanding of how flow network works.

Yet, I've bumped into a theoretical question that I'm not even sure if interpreted right. Here it is:

Provide an example of a network with n nodes and with integer
capacities such that there is an exponential (i.e. greater than $c^n$
for some constant $c \gt 1$) number of different integer maximal flows
in it. Two flows are considered different if they differ on at least
one arc of the network

Unfortunately, I cannot post picture of a graph that I've constructed (due to the lack of 10 reputation), so I'll describe it formally

$V = \lbrace S, 1, 2, 3, T\rbrace$
$C(S,1) = s1$

$C(S,2) = s2$

$C(S,3) = s3$

$C(T,1) = t1$

$C(T,2) = t2$

$C(T,3) = t3$

$C(1,2) = ?$

$C(S,1) = ?$

So, one thing that I understood about that excercise, is that in order to have sufficient input flow and output capacity $$t1 + t2 + t3 \ge c^n \le s1 + s2 + s3$$

But what about $C(1,2)$ and $C(2,3)$? Maybe I've got totally wrong idea about question itself?

It would be a huge help even to paraphrase the task, as I think I'm reading it wrong.

答: Let $N$ be a network with nodes named $1,2,3,\dots,n$ and with an arc, of capacity one, from node $a$ to node $b$ precisely when $a<b$. Then node $1$ is the source, node $n$ is the sink, every flow is maximal, and the number of flows is $2^{n-2}$, because every subset of $\{\,2,3,\dots,n-1\,\}$ corresponds to a flow.

Title: A theoretical model of a flow network.


问:
How to compute the the last digit of $122^{122}$?
I try to use
$$122^{122} = 2^{122} (\mod 10)$$

But I use MATLAB that $122^{122}=2  (\mod 10)$ and $2^{122}=4  (\mod 10)$. Why they are not the same? The screenshot is as follows:

答: Actually, $ 122^{122} mod 10 = 4$. They are the same.


答:
You can use the Chinese remainder theorem.  Since $2$ and $10$ aren't relatively prime, Euler isn't directly applicable.

$10=2\cdot5$, and $2^{122}\equiv0\bmod2$.  We get $\varphi (5)=4\implies2^{122}\equiv2^2\equiv4\bmod5$, and the answer is $4$.

Title: How to compute the the last digit of $122^{122}$?

Body:
I try to use
$$122^{122} = 2^{122} (\mod 10)$$

But I use MATLAB that $122^{122}=2  (\mod 10)$ and $2^{122}=4  (\mod 10)$. Why they are not the same? The screenshot is as follows:


问:
How to deal with negative Z score problems
Given mean life time of 40 days and standard deviation of $1.25$ days. If $12\%$ of the cars produced are having a maximum lifespan of $n$ days, find the value of n.

My work::

$Z = \dfrac{X- 40}{1.25}$

$P (X<n) = 0.12$

Using normal table conversion,

$\dfrac{X- 40}{1.25} = -1.175$

$X = -1.175(1.25)+40$

$X = 38.5325 $ days

Is the $Z$ value correct? As per the statement, the maximum of $n$  days $(X<n)$ but it gives negative $Z$ score.

答: It is correct. Note that the variable is $n$, not $X$ in the last three lines. $n$ was expected to be less than the mean $40$ otherwise $P(X<n)$ would have been greater than or equal to $0.5$, which is not the case.


答:
Actually,
you can get an identity
rather than just
an inequality.

I'll start like FFjet.

$\begin{array}\\
\left( \frac{a}{x}+\frac{b}{y} \right) \left( ax+by \right)
&=a^2+b^2+ab\left( \frac{y}{x}+\frac{x}{y} \right)\\
&=a^2+b^2+2ab +ab\left( \frac{y}{x}-2+\frac{x}{y} \right)\\
&=(a+b)^2+ ab\left( \sqrt{\frac{y}{x}}-\sqrt{\frac{x}{y}} \right)^2\\
&=(a+b)^2+ \frac{ab}{xy}\left(y-x\right)^2\\
&\ge (a+b)^2
\qquad\text{with equality } \iff x=y\\
\end{array}
$

So it doesn't depend on
what $a+b$ is.


问:
How many ways can I arrange ROONEY KIM if R and K can't be next to each other?
Please help. Me and my friend are on a debate on who is right.

How many ways can I arrange ROONEY KIM if R and K can't be next to each other? My friend said it's $9!/2! -2!(8!)= 181440$. I said it's $9!/2! - 8!/2!$.

How many ways can you arrange ARRANGED If N and D can't be together?

Please help us resolve this conflict.

答:
$\frac{9!}{2!}$ would be the answer if the R&K rule did not apply, because of the two Os

The number of arrangements with R immediately followed by K would be $\frac{8!}{2!}$ treating RK as a single letter

The number of arrangements with K immediately followed by R would be $\frac{8!}{2!}$ treating KR as a single letter

So the number of arrangements if R and K can't be next to each other is  $\frac{9!}{2!}- 2\frac{8!}{2!}$

(ARRANGEMENT does not have a D in it)


问:
Be $\Omega \subset \mathbb{C}$ an open and connected set. If $f \in A\mathbb{(\Omega)}$ and $f \neq 0$ show that the set of all zeros is numerable

Be $\Omega \subset \mathbb{C}$ an open and connected set. If $f \in A\mathbb{(\Omega)}$ and $f \neq 0$ show that the set of all zeros is numerable.

I already demostrated by contradiction that $Z(f)$ is discrete, but I don't know how use it to demostrated that $Z(f)$ is numerable. Any hint please?.

答:
Let $f$ be analytic and suppose that $z_0$ is a zero of $f$.

Then $f$ has a power series centered at $z_0$, that is, there exists $r > 0$ such that $f(z) = \sum_{n=0}^\infty a_n(z-z_0)^n$ for all $z$ such that $|z-z_0|<r$.  Now there exist a minimal $N > 0$ such that $a_N \neq 0$ or else $f$ would be identically equal to $0$ by the uniqueness theorem for analytic functions.  Hence, $f(z) = \sum_{n=N}^\infty a_n(z-z_0)^n = (z-z_0)^N\sum_{n=N}^\infty a_n(z-z_0)^{n-N} = (z-z_0)^Ng(z)$.  Now $g(z)$ is continuous and $g(z_0) \neq 0$ so there exists a neighborhood of $z_0$ such that $g(z) \neq 0$ in this neighborhood. Hence each zero is isolated so just pick a rational in each such disk $\implies$ the zeros are countable.


问:
How to show that $k[x^2,xy,y^2] \cong k[u,v,w]/(uw - v^2)$
I'm working through the exercises of Eisenbud's commutative algebra. I want to prove the isomorphism stated in the question. Assume that $k$ is a field of characteristic not $2$.

Here's what I've done. Consider the map $\phi : k[u,v,w] \to k[x^2,xy,y^2]$ determined by $\phi(u) = x^2, \phi(v) = xy, \phi(w) = y^2$. Clearly $\phi$ is onto so we want to determine its kernel. Moreover, it is clear that $(uw - v^2) \subseteq \ker \phi$. To show that $(uw - v^2)$ is the whole kernel, consider $p \in k[u,v,w]$ such that $p(x^2,xy,y^2) = 0$. At this point, I would want to say something like "we can replace $x$ and $y$ with $\pm\sqrt{u}$ and $\pm\sqrt{w}$ respectively, and obtain that $\pm\sqrt{uw}$ are roots of $p$ viewed as a polynomial over the ring $k[u,w]$." Then $uw - v^2$ divides $p$ as desired.

The end of the argument doesn't feel rigorous to me. Another concern I have is how do I know over what ring $uw - v^2$ would divide $p$? How do I know that it is not over the ring of rational functions in $u,w$ instead of the ring of polynomials?

答:
For the ideal $(uw - v^2)$ let's take $v^2 - uw$ to be the preferred generator, since it is monic in $v$ as a polynomial in $k[u,v,w] = k[u,w][v]$.  In  $R[v]$ for an arbitrary commutative ring $R$, did you know that there is unique division with remainder by each monic polynomial in $v$ (or a polynomial whose leading coefficient in $v$ is a unit in $R$)? That means for each $p(u,v,w) \in k[u,v,w]$ we can (uniquely) write
$$
p(u,v,w) = (v^2 - uw)q(u,v,w) + r(u,v,w)
$$
where $q(u,v,w)$ and $r(u,v,w)$ are in $k[u,w][v]$ with $r(u,v,w) = 0$ or $\deg_v(r(u,v,w)) < 2$. That means
$$
r(u,v,w) = a(u,w)v + b(u,w)
$$
for some $a(u,w)$ and $b(u,w)$ in $k[u,w]$.  Now use the condition $p(x^2,xy,y^2) = 0$ in $k[x,y]$ to prove $a(u,w) = 0$ and $b(u,w) = 0$ in $k[u,w]$ (that is, all of their coefficients in $k$ are $0$).This part requires some careful considerations.  And from $r(u,v,w) = 0$, you get $p(u,v,w) = (v^2-uw)q(u,v,w) \in (v^2 - uw)$.

Why are you assuming the field $k$ doesn't have characteristic $2$?


Body:
$122^{122} \equiv 2^{122} (\text{mod $10$})$

As, $2^5\equiv 2 (\text{mod $10$})$
$\implies 2^{120}\equiv 2^{24} \equiv 2^{4} (\text{mod $10$})$
$\implies 2^{122} \equiv 2^{6} \equiv 4 (\text{mod $10$})$


问:
Conditional expectation property: independence versus conditional independence properties
My question is about those two properties of conditional expectation:

-If $\mathcal{H}$ is independent of $\sigma(\sigma(X), \mathcal{G}),$ then
$$
\mathrm{E}[X \mid \sigma(\mathcal{G}, \mathcal{H})]=\mathrm{E}(X \mid \mathcal{G}), \quad \text { a.s. }
$$
In particular, if $X$ is independent of $\mathcal{H},$ then $\mathrm{E}(X \mid \mathcal{H})=\mathrm{E}(X), \quad$ a.s.

-If $X, Y$ are conditionally independent given $Z$, then $P(X \in B \mid Y, Z)=P(X \in B \mid Z)$ a.s (equivalently, $\left.E\left(1_{\{X \in B\}} \mid Y, Z\right)=E\left(1_{\{X \in B\}} \mid Z\right) a.s \right)$

Is one of those properties more general than the other ? (in the sense does one imply the other)

答: The second statement is the definition of conditional independence. It is more general than "$Y$ is independent of $(X,Z)$."


问:
Find next point in ellipse given the chord length
I would like to draw a cloud programmatically. For this reason I need to know where to draw the next circle around the ellipse.

Given the chord (circle radius), how can I calculate the next point in the ellipse?

Thanks!

答:
Suppose the ellipse is centered at $(0,0)$ with equation $x^2/a^2+y^2/b^2$, with major and minor axes alligned with the $x$ and $y$ axes. This ellipse can be parametrized by $(x,y)=(a \cos t, b \sin t)$, and as the parameter $t$ increases, the point moves around the ellipse in a counterclockwise direction.

If your first center is at $(x_1,y_1)$ then first find the $t$ value (call this $t_1$) by use of $\cos t_1=x_1/a,\ \sin t_1 = y_1/b.$ Now let $t$ be the parameter value for your "next" circle center, which is to be the first point, going around the ellipse counterclockwise, which is at distance $r$ from $(x_1,y_1).$ Set up the squared distance to $(x_1,y_1)$ as
$$f(t)=a^2(\cos t - \cos t_1)^2 +b^2 (\sin t -\sin t_1)^2.$$
This starts out as $0$ when $t=t_1$, and you can (since it's a program) now increment $t$ and get the least $t$ beyond $t_1$ for which $f(t)=r^2.$ [This will likely involve some sort of bisection approach, once you find a value of $t$ for which $f(t)>r^2$, then back up, etc. (I tried getting an exact expression for this first $t$ value but it was a bit involved, maybe not possible in closed form...)

Anyway after the value of $t$ is found as above your "next center" say $(x_2,y_2)$ becomes $(a \cos t,b \sin t).$


问:
Simplification of rational function
I saw the following simplification done:

$$\frac{100}{\frac{900sL}{900+sL}+100}$$
is equal to
$$\frac{sL + 900}{10sL + 900}$$
and was wondering how this is done?

Thank you

Title: Simplification of rational function

Body:
Just simple algebraic transformations
$$
\frac{100}{\frac{900sL}{900+sL}+100}=\frac{100}{\frac{900sL}{900+sL}+\frac{100\left( 900+sL \right)}{900+sL}}\\=\frac{100}{\frac{1000sL+90000}{900+sL}}=100\times \frac{900+sL}{1000sL+90000}=\frac{sL+900}{10sL+900}
$$


问:
Change of variable to show $\int_0^{\infty}\frac{dx}{1+x^2}=2\int_0^1\frac{dx}{1+x^2}$
Which change of variable can be used to show
$$\int_0^{\infty}\frac{dx}{1+x^2}=2\int_0^1\frac{dx}{1+x^2}$$
(direct integration isn't allowed, must use change of variable)

Body:
$$I=\int_0^\infty\frac{1}{1+x^2}dx=\int_0^1\frac{1}{1+x^2}dx+\int_1^\infty\frac{1}{1+x^2}dx$$
now:
$$I_1=\int_1^\infty\frac{1}{1+x^2}dx$$
$$u=\frac1x,du=-1/x^2dx\Rightarrow dx=-\frac{1}{u^2}du$$
$$I_1=-\int_1^0\frac{1}{1+(1/u)^2}\frac{du}{u^2}=\int_0^1\frac{1}{1+u^2}$$
$$\therefore I=2\int_0^1\frac{1}{1+x^2}dx$$


答:
You could do it by first dividing numerator and denominator by $100$ to get $\dfrac{1}{\dfrac{9sL}{900+sL}+1}$

and then multiplying numerator and denominator by $900+sL$ to get $\dfrac{900+sL}{9sL+900+sL}.$

Can you take it from here?

Title: Simplification of rational function

Body:
I saw the following simplification done:

$$\frac{100}{\frac{900sL}{900+sL}+100}$$
is equal to
$$\frac{sL + 900}{10sL + 900}$$
and was wondering how this is done?

Thank you


问:
Show that if $m$, $n$, $m^2+n^3$, $m^3+n^2$ are all prime, then $(m,n)=(2,3)$ or $(3,2)$
if $m$, $n$, $m^2+n^3$, $m^3+n^2$ are all prime

then $(m,n)=(2,3)$ or $(3,2)$

I found out that one of the $m$ and $n$ must be two since $n^2+m^3$ will be even if $m$ and $n$ have the same parity.

However, how to prove that if $4+n^3$ and $8+n^2$ are both prime, then $n=3$?

答: If $n \neq 3$ and is also prime, then $n \equiv 1 \pmod{3} \implies 8 + n^2 \equiv 0 \pmod{3}$, or $n \equiv 2 \pmod{3} \implies 4 + n^3 \equiv 0 \pmod{3}$. Since both $4 + n^3$ and $8 + n^2$ are greater than $3$, this means at least one of them can't be prime.


问:
Given positive real numbers a and b, prove that $\frac{2}{\frac{a}{x}+\frac{b}{y}} \leq ax + by, x>0, y>0$
I am working through Problem Solving through Problems by Larsons. I am stuck on question 1.8.5b) which is as follows:

"
Given positive real numbers $a$ and $b$ such that $a + b = 1$, prove that

$$\frac{2}{\frac{a}{x}+\frac{b}{y}} \leq ax + by, \ x>0, y>0$$

"

It appears to me that this question has a typo. Given any $a$ and $b$, choose $x = 1$, $y = 1$. Then:

$$\frac{2}{\frac{a}{1}+\frac{b}{1}} \leq a + b \Leftrightarrow  \frac{2}{a+b} \leq a + b \Leftrightarrow \frac{2}{1} \leq 1 \Rightarrow \Leftarrow Contradiction$$

My long-shot hypothesis is that the "$2$" is supposed to be a "$1$". I think that I actually proved this version of the problem (it was how I first realized that something seemed weird in the problem -- won't get into that proof here). However, I am hugely unsure about all of this. It seems very bizarre to me that Problem Solving would have this sort of error, and I suspect that I am just getting something very simple wrong somewhere. Anyone agree with my reasoning or know where I went wrong? Thank you!

Edit: Thought I would drop a link to the original text where the problem appears as above: https://math.la.asu.edu/~ifulman/spring13/mat194/problem-solving.pdf

答:
This is just Cauchy-Schwarz:

$$1=(a+b)^2=\left(\sqrt{ax} \sqrt{\frac ax} + \sqrt{by} \sqrt{\frac by}\right)^2 \le (ax+by)\left(\frac ax + \frac by\right)$$


问:
Determinants of matrices defined by the minimum/maximum indices of their entries
$
\newcommand{\m}[1]{\left( \begin{matrix} #1 \end{matrix} \right)}
$
The Problem: Consider the matrices $A_n := \left(a^{(n)}_{i,j}\right)_{1 \le i,j \le n},B_n := \left(b^{(n)}_{i,j}\right)_{1 \le i,j \le n} \in M_{n \times n}(\Bbb R)$ with entries defined as follows:

$$a^{(n)}_{i,j} := \min \{i,j\} \qquad b^{(n)}_{i,j} := \max \{i,j\}$$

(Examples of these matrices follow momentarily, to show how the structure of each "evolves" and emerges as $n$ increases.)

What are $\det(A_n)$ and $\det(B_n)$ in the general $n \times n$ case, for each $n \in \Bbb Z^+$?

Background/Context: I happened to have a question similar to this on my linear algebra final a few weeks ago. (In full disclosure: final grades have already been determined and all that, so this isn't any longer an "active" test question.) However, that problem used a matrix similar to $A_n$, without specific numbers, and only the $4 \times 4$ case.

I then recently found a video by Dr. Peyam which considered the $5 \times 5$ cases for both $A_n$ and $B_n$, and hinted that these possibly came from a Putnam exam. I'm not sure if it was, though. I looked back through all of the Putnam problems back to $1985$ and didn't see this exact one, though one was similar: $A2$ from $2014$, where $A_n$ is instead defined by $a^{(n)}_{i,j} = 1/\min\{i,j\}$. (Maybe I'll look at that some other time.)

I imagine the definition for $B_n$ is just considered a natural generalization of this problem. In any event, I couldn't find this problem handled in its generality on MSE, so I figured I'd try to figure it out, and post my own solution.

Of course, if you have your own solutions, they're greatly welcomed! Each can provide their own insights.

Small Cases for the Matrices: To help show the structure of the matrices, $A_n$ and $B_n$ look like, in the cases for $n=1,2,\cdots,6$, as follows. Determinants are included, but calculated by Wolfram.

\begin{alignat*}{3}
&\boxed{n=1} &&\qquad A_1 = \m{
1  }  &&\qquad \det(A_1) = 1\\
& &&\qquad B_1 = \m{
1 }
&&\qquad \det(B_1) = 1\\
&\boxed{n=2} &&\qquad A_2 = \m{
1 & 1 \\
1 & 2  }
&&\qquad \det(A_2) = 1\\
& &&\qquad B_2 = \m{
1 & 2    \\
2 & 2   }
&&\qquad \det(B_2) = -2\\
&\boxed{n=3} &&\qquad A_3 = \m{
1 & 1 & 1 \\
1 & 2 & 2 \\
1 & 2 & 3 }
&&\qquad \det(A_3) = 1\\
& &&\qquad B_3 = \m{
1 & 2 & 3   \\
2 & 2 & 3  \\
3 & 3 & 3  }
&&\qquad \det(B_3) = 3\\
&\boxed{n=4} &&\qquad A_4 = \m{
1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 \\
1 & 2 & 3 & 4 }
&&\qquad \det(A_4) = 1\\
& &&\qquad B_4 = \m{
1 & 2 & 3 & 4  \\
2 & 2 & 3 & 4 \\
3 & 3 & 3 & 4   \\
4 & 4 & 4 & 4  }
&&\qquad \det(B_4) = -4\\
&\boxed{n=5} &&\qquad A_5 = \m{
1 & 1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 & 3 \\
1 & 2 & 3 & 4 & 4 \\
1 & 2 & 3 & 4 & 5 }
&&\qquad \det(A_5) = 1\\
& &&\qquad B_5 = \m{
1 & 2 & 3 & 4 & 5  \\
2 & 2 & 3 & 4 & 5 \\
3 & 3 & 3 & 4 & 5  \\
4 & 4 & 4 & 4 & 5 \\
5 & 5 & 5 & 5 & 5 }
&&\qquad \det(B_5) = 5\\
&\boxed{n=6} &&\qquad A_6 = \m{
1 & 1 & 1 & 1 & 1 & 1\\
1 & 2 & 2 & 2 & 2 & 2\\
1 & 2 & 3 & 3 & 3 & 3\\
1 & 2 & 3 & 4 & 4 & 4\\
1 & 2 & 3 & 4 & 5 & 5\\
1 & 2 & 3 & 4 & 5 & 6}
&&\qquad \det(A_6) = 1\\
& &&\qquad B_6 = \m{
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 2 & 3 & 4 & 5 & 6\\
3 & 3 & 3 & 4 & 5 & 6 \\
4 & 4 & 4 & 4 & 5 & 6 \\
5 & 5 & 5 & 5 & 5 & 6 \\
6 & 6 & 6 & 6 & 6 & 6}
&&\qquad \det(B_6) = -6
\end{alignat*}

One reasonably conjectures that $\det(A_n) = 1$ for all $n$ and $\det(B_n) = (-1)^{n-1} n$ for all $n$.

答:
$
\newcommand{\RR}{\mathcal{R}}
\newcommand{\m}[1]{\left( \begin{matrix} #1 \end{matrix} \right)}
\newcommand{\mdet}[1]{\left| \begin{matrix} #1 \end{matrix} \right|}
$
Finding $\det(A_n)$:

I will pursue this one in part by simple row reduction: adding multiples of one row to another will preserve the determinant. So, for now, note the form of $A_n$:

$$A_n =
\m{
1 & 1 & 1 & \cdots & 1 \\
1 & 2 & 2 & \cdots & 2 \\
1 & 2 & 3 & \cdots & 3 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 2 & 3 & \cdots & n}$$

Throughout the entirety of this answer, let $\RR_i$ denote the $i$th row of the matrix, at whatever stage in the row reduction process we're at. We'll let $\sim$ denote equivalence by row reduction. At this point, subtract $\RR_1$ from each of $\RR_k$ for $k \ge 2$. We see that

$$A_n  \sim
\m{
1 & 1 & 1 & \cdots & 1 \\
0 & 1 & 1 & \cdots & 1 \\
0 & 1 & 2 & \cdots & 2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 1 & 2 & \cdots & n-1}$$

Next, subtract $\RR_2$ from $\RR_1$. Then

$$A_n  \sim
\m{
1 & 0 & 0 & \cdots & 0 \\
0 & 1 & 1 & \cdots & 1 \\
0 & 1 & 2 & \cdots & 2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 1 & 2 & \cdots & n-1}$$

Expanding the determinant along the first row (or column) trivially yields that

$$\det(A_n) = 1 \cdot \mdet{
1 & 1 & 1 & \cdots & 1 \\
1 & 2 & 2 & \cdots & 2 \\
1 & 2 & 3 & \cdots & 3 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 2 & 3 & \cdots & n-1} = \det(A_{n-1})$$

This readily defines a linear homogenous recurrence relation, with initial condition $\det(A_1) = 1$, which sufficiently verifies that $\det(A_n) = 1$ for all $n \in \Bbb Z^+$. (This is particularly clear on back-iteration, which shows $\det(A_n) = \det(A_k)$ for $k \in \{1,2,\cdots,n-1\}$.)

Finding $\det(B_n)$:

I will similarly utilize row reduction for this, as I did for $A_n$. In general, $B_n$ looks like this:

$$B_n = \m{
1 & 2 & 3 & \cdots & n \\
2 & 2 & 3 & \cdots & n \\
3 & 3 & 3 & \cdots & n \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n & n & n & \cdots & n}$$

Subtract $\RR_2$ from $\RR_1$ and we obtain

$$B_n \sim \m{
-1 & 0 & 0 & \cdots & 0 \\
2 & 2 & 3 & \cdots & n \\
3 & 3 & 3 & \cdots & n \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n & n & n & \cdots & n}$$

Next, add $k \RR_1$ to each $\RR_k$ for $k \ge 2$:

$$B_n \sim \m{
-1 & 0 & 0 & \cdots & 0 \\
0 & 2 & 3 & \cdots & n \\
0 & 3 & 3 & \cdots & n \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & n & n & \cdots & n}$$

Similarly, subtract $\RR_3$ from $\RR_2$, and then add $k \RR_2$ to each $\RR_k$ for $k \ge 3$.

$$B_n \sim \m{
-1 & 0 & 0 & \cdots & 0 \\
0 & -1 & 0 & \cdots &0 \\
0 & 0 & 3 & \cdots & n \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & n & \cdots & n}$$

We proceed inductively. At the $r$th step, we subtract $\RR_{r+1}$ from $\RR_r$, turning $\RR_r$ into a row vector with all zeroes except $-1$ in its $r$th position. Then we add  $k$ times this new $\RR_r$ to each $\RR_i$ with $i > k$, to negate their $r$th column and turn it into a similar row vector.

At the end of this process, after the $(n-1)$th step, we have that

$$B_n \sim \m{
-1 & 0 & 0 & \cdots & 0 \\
0 & -1 & 0 & \cdots & 0 \\
0 & 0 & -1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & n} = \mathrm{diag}(\underbrace{-1,-1,-1,\cdots,-1,-1}_{n-1 \; (-1)'s},n)$$

The determinant of $B_n$ is that of this diagonal matrix, and the determinant of a diagonal matrix is the product of the entries on the diagonal. Thus,

$$\det(B_n) = (-1)^{n-1} n$$

Alternate Method for $\det(B_n)$:

This still utilizes row reduction, but more in line with how Dr. Peyam did it for $B_5$ in his video. So we instead subtract $\frac{n}{n-1} \RR_{n-1}$ from $\RR_n$. Then

\begin{align*}
B_n &= \m{
1 & 2 & 3 & \cdots & n-1 & n \\
2 & 2 & 3 & \cdots & n-1 & n \\
3 & 3 & 3 & \cdots & n-1 & n \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
n-1 & n-1  & n-1  & \cdots & n-1 & n \\
n & n & n & \cdots & n & n} \\
&\sim \m{
1 & 2 & 3 & \cdots & n-1 & n \\
2 & 2 & 3 & \cdots & n-1 & n \\
3 & 3 & 3 & \cdots & n-1 & n \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
n-1 & n-1  & n-1  & \cdots & n-1 & n \\
0 & 0 & 0 & \cdots & 0 & n - \frac{n^2}{n-1}  }
\end{align*}

We expand the determinant along the bottom row. Since the only nonzero entry is the bottom-right entry (i.e. $b_{n,n}^{(n)})$), the sign associated with this expansion is simply $+1$. (The sign of this in the cofactor expansion is more formally $(-1)^{n+n}=(-1)^{2n} = 1$, since the exponent is even.)

Thus,

\begin{align*}
\det{B_n} &= \left( n - \frac{n^2}{n-1} \right) \mdet{
1 & 2 & 3 & \cdots & n-1  \\
2 & 2 & 3 & \cdots & n-1  \\
3 & 3 & 3 & \cdots & n-1  \\
\vdots & \vdots & \vdots & \ddots & \vdots  \\
n-1 & n-1  & n-1  & \cdots & n-1  } \\
&= \left( n - \frac{n^2}{n-1} \right) \det(B_{n-1})
\end{align*}

Visibly, one may proceed inductively, iterating backwards to the initial condition of $\det(B_1) = 1$, and observe

$$\det(B_n) = \prod_{k=2}^{n}  \left( k - \frac{k^2}{k-1} \right)$$

One now notes that

$$ k - \frac{k^2}{k-1} = \frac{k(k-1) - k^2}{k-1} = \frac{k^2 - k - k^2}{k-1} = -\frac{k}{k-1}$$

and thus

$$\det(B_n) = \prod_{k=2}^{n} \left( -\frac{k}{k-1} \right)$$

This is a telescoping product, insofar as consecutive terms cancel nicely. Writing out the product, we have

\begin{align*}
\require{cancel}
\prod_{k=2}^{n} \frac{k}{k-1}
&= \frac 2 1 \cdot \frac 3 2 \cdot \frac 4 3 \cdot \frac 5 4 \cdot \cdots \cdot \frac{n-2}{n-3} \cdot \frac{n-1}{n-2} \cdot \frac{n}{n-1}\\
&= \frac{\cancel{2}} 1 \cdot \frac {\cancel{3}} {\cancel{2}} \cdot \frac {\cancel{4}} {\cancel{3}} \cdot \frac {\cancel{5}} {\cancel{4}} \cdot \cdots \cdot \frac{\cancel {n-2}}{\cancel {n-3}} \cdot \frac{\cancel{n-1}}{\cancel {n-2}} \cdot \frac{n}{\cancel{n-1}}\\
&= n
\end{align*}

Meanwhile, what remains now is the sign. Since it is constant with respect to $k$, we can pull it out of the product altogether. Note that the product has $n-1$ members. Thus,

$$\det(B_n) = (-1)^{n-1} \prod_{k=2}^{n} \frac{k}{k-1} = (-1)^{n-1} n$$

as desired.


问:
Finding sequence in a set $A$ that tends to $\sup A$
I have been reading the book at http://www.neunhaeuserer.de/short.pdf, and have noticed that in the proof of the intermediate value theorem (Theorem 5.8 in the book), it seems to be quietly assumed that you can always find a sequence of points in a set that tend to the supremum of the set (in $\mathbb{R}$).

I presume this is true, but how would one go about proving it? I can't seem to find the result easily, but maybe I'm searching the wrong. What does one call this property (of there being a sequence converging to you?). It seems "limit point" would be the natural term, but I've discovered that the definition doesn't quite mean that--e.g. an isolated point $a$ cannot be limit point, but the sequence $a,a,a,a,a,a \dots$ converges to it (or is that a matter of how we define convergence?).

答: Very astute question. Provided you discard the constant sequence (and, I guess, eventually constant sequences), limit point is the right notion in reasonable spaces. Assuming that $\sigma=\sup A$ is not an element of $A$, you are guaranteed for each $n$ an element $a_n\in A$ with $0<\sigma-a_n<1/n$.


答:
Look at the general problem of find the last digit of $n^m$.

Take $n=10N+h$ where $0\le h\le9$ and $m=4M+k$ where $0\le k\le3$.

Power of digits have a period of $4$ modulo $10$ the following way:
$$1\to1\to1\to1\\2\to4\to8\to6\\3\to9\to7\to1\\4\to6\to4\to6\\5\to5\to5\to5\\6\to6\to6\to6\\7\to9\to3\to1\\8\to4\to2\to6\\9\to1\to9\to1$$

Example: $797^{723}=(10N+7)^{4\cdot180+3}\equiv7^3\pmod{10}=3\pmod{10}$.

Apply this to find the answer to your problem.


问:
How many ways that we can distribute $4$ tasks to $6$ workers if no worker can perform all four tasks?
I have confusing problem about counting. I don't know how to start. Here is my problem.

How many ways that we can distribute $4$ tasks to $6$ workers?

Note: Not only distributing to all workers, the tasks can be distributed to only one, two, or three workers.

Answer:
I try solve by $6 \times 5 \times 4 \times 3$ to give the task to $4$ workers. But what's next? I am sorry for my confusion.

答:
If there were no restrictions, there would be six ways to assign each of the four tasks, so there would be $6^4$ ways to distribute the tasks.  However, these include the six distributions in which all four tasks are assigned to the same worker.  Hence, there are
$$6^4 - 6$$
ways to distribute the four tasks to six workers so that no worker is assigned all four tasks.


问:
Dividing $x^{81}+x^{48}+2x^{27}+x^{6}+3$ by $x^{3}+1$
Letting $f(x)=x^{81}+x^{48}+2x^{27}+x^{6}+3$, and we seek to divide $f(x)$ by $x^{3}+1$, or just any $x^3-a$. Is subbing $y=x^3$ allowed in this kind of division? I've experimented doing so and so far it seems correct.

答:
Of course it is allowed. Since the substitution doesn't change the roots of the polynomials, and thus the divisibility.

In fact, any substitution $y=g(x)$, where $g(x)$ is a polynomial can be applied.


答:
This is a consequence of the Axiom of Countable Choice, which the author is taking for granted (actually, the author is taking stronger assumptions for granted, based on section 6 of the book).

The idea is as follows:

If $\xi\in A$, then there is a constant sequence of points of $A$ converging to $\xi$. Otherwise, since $\xi$ is the supremum of $A$, then for each positive integer $n$, we have that $A\cap[\xi-\frac1n,\xi)$ is non-empty, and we simultaneously choose $x_n\in A\cap[\xi-\frac1n,\xi)$ for each such $n$. This gives us a sequence of points of $A$ converging to $\xi$.

Title: Finding sequence in a set $A$ that tends to $\sup A$

Body:
I have been reading the book at http://www.neunhaeuserer.de/short.pdf, and have noticed that in the proof of the intermediate value theorem (Theorem 5.8 in the book), it seems to be quietly assumed that you can always find a sequence of points in a set that tend to the supremum of the set (in $\mathbb{R}$).

I presume this is true, but how would one go about proving it? I can't seem to find the result easily, but maybe I'm searching the wrong. What does one call this property (of there being a sequence converging to you?). It seems "limit point" would be the natural term, but I've discovered that the definition doesn't quite mean that--e.g. an isolated point $a$ cannot be limit point, but the sequence $a,a,a,a,a,a \dots$ converges to it (or is that a matter of how we define convergence?).


答:
I no longer have access to Matlab so I can't reproduce this error exactly on my end. A similar mistake is observed when I use Octave online:

```<code>octave:2> mod(122^122, 10)
ans = 0
</code>```

You should use a function such as powermod. The trick is we do not want to compute $122^{122}$ explicitly.

Notice that $122^{122}$ is a very large number and it is working in double precision float point that exceeds flintmax. Above this value, double-precision format does not have integer precision, and not all integers can be represented exactly.

The answer is indeed $4$.

Here are the Python outcome:

```<code>>>> 122**122 % 10 # cool, it can be computed
4
>>> pow(122, 122, 10) # preferred.
4
</code>```

Body:
I no longer have access to Matlab so I can't reproduce this error exactly on my end. A similar mistake is observed when I use Octave online:

octave:2> mod(122^122, 10)
ans = 0

You should use a function such as powermod. The trick is we do not want to compute $122^{122}$ explicitly.

Notice that $122^{122}$ is a very large number and it is working in double precision float point that exceeds flintmax. Above this value, double-precision format does not have integer precision, and not all integers can be represented exactly.

The answer is indeed $4$.

Here are the Python outcome:

>>> 122**122 % 10 # cool, it can be computed
4
>>> pow(122, 122, 10) # preferred.
4


问:
convolution of two sums of shifted dirac functions
So I have this

$$
x=0.5\delta(t-1)+0.5\delta(t+2)\\
y=0.3\delta(t+1)+0.7\delta(t-2)
$$

and I have to find the convolution.  They give me the integral of $\delta(a-t)\delta(t-b)=\delta(a-b)$

I have searched on google and in the book but I have not found a similar problem anywhere else.

I tried to do the convolution but I must be missing a property or something because it doesn't get to the type I want and need.

Body:
$$(x*y)(t)=\int_{-\infty}^{+\infty}{x(\tau)y(t-\tau)d\tau}$$

$$(x*y)(t)=\int_{-\infty}^{+\infty}{[0.5\delta(\tau-1)+0.5\delta(\tau+2)][0.3\delta(t-\tau+1)+0.7\delta(t-\tau-2)]d\tau}$$

$$(x*y)(t)=\int_{-\infty}^{+\infty}{0.15\delta(\tau-1)\delta((t+1)-\tau)\,d\tau} + \int_{-\infty}^{+\infty}{0.35\delta(\tau-1)\delta((t-2)-\tau)\,d\tau} + \int_{-\infty}^{+\infty}{0.15\delta(\tau-(-2))\delta((t+1)-\tau)\,d\tau} + \int_{-\infty}^{+\infty}{0.35\delta(\tau-(-2))\delta((t-2)-\tau)\,d\tau}$$

Knowing that

$$ \int_{-\infty}^{+\infty}{\delta(a-x)\delta(x-b)\,dx} = \delta(b-a) $$

your convolution becomes

$$(x*y)(t)= 0.15\delta(1-(t+1)) + 0.35\delta(1-(t-2)) + 0.15\delta(-2-(t+1)) + 0.35\delta(-2-(t-2))$$

$$(x*y)(t)= 0.15\delta(-t) + 0.35\delta(3-t) + 0.15\delta(-3-t) + 0.35\delta(t)$$

$$(x*y)(t)= 0.5\delta(t) + 0.35\delta(t-3) + 0.15\delta(t+3)$$


问:
Power series of $\frac{z}{e^z-1}$
I have the function $ f(z)=\frac{z}{e^z-1} $. I'm looking for a power series.

My idea:

I define a power series $$ \frac{z}{e^z-1}=f(z):=\sum\limits_{k=0}^\infty c_k\cdot z^k. $$ Now I start equating the coefficients $ c_k $:

$$ \begin{align}z&=(e^z-1)\cdot \left(\sum\limits_{k=0}^\infty c_k\cdot z^k \right)=e^z\left( \sum\limits_{k=0}^\infty c_k\cdot z^k\right)-\sum\limits_{k=0}^\infty c_k\cdot z^k\\&=\left(\sum\limits_{k=0}^\infty \frac{1}{k!}\cdot z^k\right)\cdot \left(\sum\limits_{k=0}^\infty c_k\cdot z^k\right)-\left(\sum\limits_{k=0}^\infty c_k\cdot z^k\right)\\&=\sum\limits_{k=0}^\infty\left(\sum\limits_{l=0}^k \frac{1}{l!}\cdot z^l\cdot  c_{k-l}\cdot z^{k-l}\right)-\left(\sum\limits_{k=0}^\infty c_k\cdot z^k\right) \\&=\sum\limits_{k=0}^\infty\left(\sum\limits_{l=0}^k \frac{1}{l!}\cdot  c_{k-l}\cdot z^{k}\right)-\left(\sum\limits_{k=0}^\infty c_k\cdot z^k\right)\\&=\sum\limits_{k=0}^\infty\left(\sum\limits_{l=0}^k \frac{1}{l!}\cdot  c_{k-l}\right)\cdot z^{k}-\left(\sum\limits_{k=0}^\infty c_k\cdot z^k\right)\\&=\sum_{k=0}^\infty \left[\left(\sum\limits_{l=0}^k\frac{1}{l!}\cdot c_{k-l} \right)-c_k \right]\cdot z^k\\&=\sum_{k=0}^\infty \left(\sum\limits_{l=1}^k\frac{1}{l!}\cdot c_{k-l} \right)\cdot z^k\end{align}$$

$ \deg k=0: \quad 0=\sum\limits_{l=1}^0\frac{1}{l!}\cdot c_{k-l}=0\\\deg k=1: \quad 1=\sum\limits_{l=1}^1\frac{1}{l!}\cdot c_{k-l}=c_0\\\deg k\geq 2:\quad 0=\sum\limits_{l=1}^k\frac{1}{l!}\cdot c_{k-l}=\left(\sum\limits_{l=2}^k\frac{1}{l!}\cdot c_{k-l}\right) +c_{k-1}\\\qquad \qquad \ \ c_{k-1}=-\sum\limits_{l=2}^k\frac{1}{l!}\cdot c_{k-l}$.

I struggle to show that $c_{k-1}$ converges absolutely (I tried ratio test and root test) and I'm not to find an explizit formula of $c_k$ and taylor approximation is a catastrophy concerning of the derivatives of $ f $. How else could I show that this series converges absolutely?

答:
This is the generating function
for the Bernoulli numbers:

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

Title: Power series of $\frac{z}{e^z-1}$


问:
For every prime of the form $2^{4n}+1$, 7 is a primitive root.
What I want to show is the following statement.

For every prime of the form $2^{4n}+1$, 7 is a primitive root.

What I get is that

$$7^{2^{k}}\equiv1\pmod{p}$$
$$7^{2^{k-1}}\equiv-1\equiv2^{4n}\pmod{p}$$
$$7^{2^{k-2}}\equiv(2^{n})^2\pmod{p}$$
Thus $(\frac{2^n}{p})=(\frac{7^{2^{k-2}}}{p})=1$.

I think that $7$ is important because $7$ is a primitive root but I don't know how to use $7$.

答:
Recall that any  prime $p$ has $\varphi(p-1)$ primitive roots. In our case, $\varphi(p-1)=(p-1)/2$.

Any prime $p$ has $(p-1)/2$ quadratic non-residues. Any primitive root is a non-residue, so for primes $p$ of the form $2^w+1$, every NR is a primitive root.

It remains to show that $7$ is a NR of $p$. This is dealt with in the answer by Jyrki Lahtonen. Briefly, use Reciprocity.


问:
Prove by induction floor and ceiling
I couldn't solve below question ... I was able to start the solution .. but I couldn't turn left side to be same as write side in the inductive step:

答:
Hint: you only need to show

$$\left\lceil \frac{n+1}{m}\right\rceil - \left\lceil \frac nm \right\rceil = \left\lfloor \frac{n+m}{m} \right\rfloor - \left\lfloor \frac{n+m-1}{m}\right\rfloor$$

Now the LHS is either $0$ or $1$, usually $0$. So is the RHS.

When is LHS $1$? When is RHS $1$?

 When $\frac nm \in \mathbb N$, both LHS and RHS are equal to 1. Otherwise they are both zeros. Therefore, $$\left\lceil \frac{n+1}{m}\right\rceil - \left\lceil \frac nm \right\rceil = \left\lfloor \frac{n+m}{m} \right\rfloor - \left\lfloor \frac{n+m-1}{m}\right\rfloor \tag1$$ $$\left\lceil \frac nm \right\rceil = \left\lfloor \frac{n+m-1}{m}\right\rfloor \text{ (from last step)} \tag 2$$ $(1)+(2)$, then we have $$ \left\lceil \frac{n+1}{m}\right\rceil = \left\lfloor \frac{n+m}{m} \right\rfloor $$

Title: Prove by induction floor and ceiling


问:
How many ways can I arrange ROONEY KIM if R and K can't be next to each other?
Please help. Me and my friend are on a debate on who is right.

How many ways can I arrange ROONEY KIM if R and K can't be next to each other? My friend said it's $9!/2! -2!(8!)= 181440$. I said it's $9!/2! - 8!/2!$.

How many ways can you arrange ARRANGED If N and D can't be together?

Please help us resolve this conflict.

答:
You should learn more than one way of attacking a problem, especially if the methods are equally easy, so I am solving by a different method.

There are $(7!/2!)$ arrangements of the letters without the $R$ and $K$

The two letters taken out can be inserted back into the $8$ gaps between the letters (including gaps at the ends) in $8\cdot7$ places, thus $(7!/2!)\cdot8\cdot7$ ways

Similarly, for $N$ and $D$ to not be next to each other in $ARRANGED, [6!/(2!2!)]\cdot 7\cdot6$ ways


答:
Using the Schur complement,

$$\det \left( {\rm A}_{n+1} \right) = \det \begin{bmatrix} {\rm A}_n & {\rm A}_n \, {\rm e}_n\\ {\rm e}_n^\top {\rm A}_n &  n+1 \end{bmatrix} = \underbrace{\left( n + 1 - {\rm e}_n^\top {\rm A}_n {\rm A}_n^{-1} {\rm A}_n \, {\rm e}_n \right)}_{= n + 1 - n} \det \left( {\rm A}_n \right) = \det \left( {\rm A}_n \right)$$

and, since $\det \left( {\rm A}_1 \right) = 1$, we conclude that $\det \left( {\rm A}_n \right) = 1$ for all $n \geq 1$.

Title: Determinants of matrices defined by the minimum/maximum indices of their entries

Body:
$
\newcommand{\m}[1]{\left( \begin{matrix} #1 \end{matrix} \right)}
$
The Problem: Consider the matrices $A_n := \left(a^{(n)}_{i,j}\right)_{1 \le i,j \le n},B_n := \left(b^{(n)}_{i,j}\right)_{1 \le i,j \le n} \in M_{n \times n}(\Bbb R)$ with entries defined as follows:

$$a^{(n)}_{i,j} := \min \{i,j\} \qquad b^{(n)}_{i,j} := \max \{i,j\}$$

(Examples of these matrices follow momentarily, to show how the structure of each "evolves" and emerges as $n$ increases.)

What are $\det(A_n)$ and $\det(B_n)$ in the general $n \times n$ case, for each $n \in \Bbb Z^+$?

Background/Context: I happened to have a question similar to this on my linear algebra final a few weeks ago. (In full disclosure: final grades have already been determined and all that, so this isn't any longer an "active" test question.) However, that problem used a matrix similar to $A_n$, without specific numbers, and only the $4 \times 4$ case.

I then recently found a video by Dr. Peyam which considered the $5 \times 5$ cases for both $A_n$ and $B_n$, and hinted that these possibly came from a Putnam exam. I'm not sure if it was, though. I looked back through all of the Putnam problems back to $1985$ and didn't see this exact one, though one was similar: $A2$ from $2014$, where $A_n$ is instead defined by $a^{(n)}_{i,j} = 1/\min\{i,j\}$. (Maybe I'll look at that some other time.)

I imagine the definition for $B_n$ is just considered a natural generalization of this problem. In any event, I couldn't find this problem handled in its generality on MSE, so I figured I'd try to figure it out, and post my own solution.

Of course, if you have your own solutions, they're greatly welcomed! Each can provide their own insights.

Small Cases for the Matrices: To help show the structure of the matrices, $A_n$ and $B_n$ look like, in the cases for $n=1,2,\cdots,6$, as follows. Determinants are included, but calculated by Wolfram.

\begin{alignat*}{3}
&\boxed{n=1} &&\qquad A_1 = \m{
1  }  &&\qquad \det(A_1) = 1\\
& &&\qquad B_1 = \m{
1 }
&&\qquad \det(B_1) = 1\\
&\boxed{n=2} &&\qquad A_2 = \m{
1 & 1 \\
1 & 2  }
&&\qquad \det(A_2) = 1\\
& &&\qquad B_2 = \m{
1 & 2    \\
2 & 2   }
&&\qquad \det(B_2) = -2\\
&\boxed{n=3} &&\qquad A_3 = \m{
1 & 1 & 1 \\
1 & 2 & 2 \\
1 & 2 & 3 }
&&\qquad \det(A_3) = 1\\
& &&\qquad B_3 = \m{
1 & 2 & 3   \\
2 & 2 & 3  \\
3 & 3 & 3  }
&&\qquad \det(B_3) = 3\\
&\boxed{n=4} &&\qquad A_4 = \m{
1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 \\
1 & 2 & 3 & 4 }
&&\qquad \det(A_4) = 1\\
& &&\qquad B_4 = \m{
1 & 2 & 3 & 4  \\
2 & 2 & 3 & 4 \\
3 & 3 & 3 & 4   \\
4 & 4 & 4 & 4  }
&&\qquad \det(B_4) = -4\\
&\boxed{n=5} &&\qquad A_5 = \m{
1 & 1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 & 3 \\
1 & 2 & 3 & 4 & 4 \\
1 & 2 & 3 & 4 & 5 }
&&\qquad \det(A_5) = 1\\
& &&\qquad B_5 = \m{
1 & 2 & 3 & 4 & 5  \\
2 & 2 & 3 & 4 & 5 \\
3 & 3 & 3 & 4 & 5  \\
4 & 4 & 4 & 4 & 5 \\
5 & 5 & 5 & 5 & 5 }
&&\qquad \det(B_5) = 5\\
&\boxed{n=6} &&\qquad A_6 = \m{
1 & 1 & 1 & 1 & 1 & 1\\
1 & 2 & 2 & 2 & 2 & 2\\
1 & 2 & 3 & 3 & 3 & 3\\
1 & 2 & 3 & 4 & 4 & 4\\
1 & 2 & 3 & 4 & 5 & 5\\
1 & 2 & 3 & 4 & 5 & 6}
&&\qquad \det(A_6) = 1\\
& &&\qquad B_6 = \m{
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 2 & 3 & 4 & 5 & 6\\
3 & 3 & 3 & 4 & 5 & 6 \\
4 & 4 & 4 & 4 & 5 & 6 \\
5 & 5 & 5 & 5 & 5 & 6 \\
6 & 6 & 6 & 6 & 6 & 6}
&&\qquad \det(B_6) = -6
\end{alignat*}

One reasonably conjectures that $\det(A_n) = 1$ for all $n$ and $\det(B_n) = (-1)^{n-1} n$ for all $n$.


问:
A binary quadratic form: $nx^2-y^2=2$
For which $n\in\mathbb{N}$ are there $(x, y)\in\mathbb{N}^2$ such that $nx^2-y^2=2$ ?</p>

答:
This looks very similar to Pell's Equation and luckily it's already been studied, such as here.

Go through their listed transformations to arrive at $$ \left(y^2 + 1\right)^2 - n(xy)^2 = 1 $$ I'm not sure how to proceed from here, as Pell's Equation doesn't have a closed form for the fundamental solution and so it can't easily be said if one of the the solutions will be in the form $ y^2 + 1 $.

Title: A binary quadratic form: $nx^2-y^2=2$


问:
Optimising Graph Vertex Value Assignment for Minimum Weights
I have a real world problem that can be modelled as a graph. I'm looking for algorithms that can solve the general problem, probably as a graph or by linear programming.

I have a system that can be thought of as a highly connected <s>undirected</s> directed graph. Every vertex connects to every other vertex in both directions. I have a fixed set of values that can be assigned to a vertex. It is possible for the vertex value to be null. The number of values will always be <= to the number of vertices. There is a real-world physical distance between the locations of each vertex. The weight of an edge depends on the physical separation distance and the values assigned to the vertices at each end of the edge.

I am interested in finding the optimum assignment of values to vertices that minimises the weight of all the edges. I don't want to minimise the average of the edges in case the result is skewed by a wide variance of weights. I think I want to optimise for the minimum highest weight, but would be open to suggestions. The method for calculating the weights could probably be reworked if the problem were easier to solve as optimising for a maximum. If there are multiple optimum solutions, it is OK for my purposes to just find one of them.

In the real world case, it is possible for two vertices to have the same assigned value. In this case, some different constraints apply and we don't need to include it in the optimisation. I can ignore this complication if it simplifies solving the problem.

Are there any pre-existing algorithms or methodologies for solving this type of problem?

答:
Let binary decision variable $y_{i,v}$ indicate whether $x_i=v$.  The problem is to minimize $z$ subject to
\begin{align}
z &\ge P_i - a d_{i,j}^2 - b \sum_{v_i, v_j} |v_i-v_j| y_{i,v_i} y_{j,v_j} &&\text{for all $i,j$} \tag1 \\
\sum_v y_{i,v} &= 1 &&\text{for all $i$} \tag2 \\
\sum_v v y_{i,v} &= x_i &&\text{for all $i$} \tag3 \\
\end{align}
Constraint $(1)$ enforces $z \ge \max_{i,j} W_{i,j}$.
Constraint $(2)$ assigns exactly one value to each node.
Constraint $(3)$ enforces $y_{i,v}=1 \iff x_i=v$.

You can linearize the quadratic constraint $(1)$ by replacing the product $y_{i,v_i} y_{j,v_j}$ with binary variable $u_{i,v_i,j,v_j}$ and linear constraints
\begin{align}
\sum_{v_i} u_{i,v_i,j,v_j} &= y_{j,v_j} &&\text{for all $i,j,v_j$} \tag4 \\
u_{i,v_i,j,v_j} &\le y_{i,v_i} &&\text{for all $i,v_i,j,v_j$} \tag5 \\
\end{align}


问:
Prove that there exist infinitely many Gaussian integers with the following properties
Prove that there exist infinitely many Gaussian integers $\alpha$ with the following properties:

$N(\alpha)$ is the product of two prime numbers.

At least one of the potential factors of $\alpha$, as discovered by the “norm method” for determining factors, is not actually a factor.

[Hint: you may assume without proof the fact that there are infinitely many prime numbers that are congruent to 1 modulo 4.]

This is an exercise I'm trying to do. I've listed my attempt below. I get stuck at the final case though I'm not even certain my proof is correct up until that point.

Let $\alpha \in \mathbb{Z[i]}$ and $\beta = a + bi$ a factor of $\alpha$ in $\mathbb{Z[i]}$. Then $\alpha = \beta\gamma$ for some $\gamma$ in $\mathbb{Z[i]}$.  Taking norms, $N(\alpha) = N(\beta)N(\gamma)$. Suppose that $N(\alpha) = pq$ for primes $p,q$ in $\mathbb{Z}$. Then $N(\beta)N(\gamma) = pq$. We have 4 cases to obtain potential factors of $\alpha$ in $\mathbb{Z[i]}$.

We ignore two cases, $N(\beta) = 1$ and $N(\beta) = pq$.  In both of these cases, $\beta$ and $\gamma$ are units or associates of $\alpha$. These are always factors of $\alpha$ so are not interesting to the proposition.

Then we are left with two cases to consider, $N(\beta) = p$ and $N(\beta) = q$.

When both $p,q$ are congruent to $3$ modulo $4$, $p,q$ cannot be written as a sum of squares.  So $\alpha$ is irreducible in $\mathbb{Z[i]}$ and there are no factors to consider.  There are infinitely many primes congruent to $3$ modulo $4$ (by the hint) so there are infinitely many such $\alpha$. We are done.

Now suppose only one of $p,q$ are congruent to $1$ modulo $4$. WLOG, let $p$ be congruent to $1$ modulo $4$. Then $q$ is congruent to $3$ mod $4$ so $q$ cannot be written as a sum of squares. We are done similar to the previous case.

Finally, consider when both $p,q$ are congruent to $1$ modulo $4$.

答:
You are only asked to demonstrate that there are infinitely many numbers with this property.  You don't have to exhaust the cases of $N(\alpha)$, just to find an infinite set that have this property.  For an extreme example, if you could show that the norm of all the Fermat numbers $2^{2^n}+1$ for $n \gt 1000$ was the product of two primes and one was not a factor, you would be done.

$N(\alpha)$ is always a natural and you are looking for natural number factors of it, not members of $\Bbb Z[i]$.  You then look for members of $\Bbb Z[i]$ that have that norm and ask if they are factors of $\alpha$

When $p,q$ are equivalent to $3 \bmod 4$ it is true that they cannot be written as the sum  of two squares, but neither can $N(\alpha)$.  This comes from the sum of two squares theorem.  As an example, $p=3, q=7$ gives a proposed $N(\alpha)$ of $21$.  As $21$ cannot be written as a sum of two squares, there are no numbers in $\Bbb Z[i]$ with norm $21$.

On the other hand, consider $65=5 \cdot 13$, which is the product of two primes equivalent to $1 \pmod 4$.  We know there are two ways to write it as the sum of two squares, to wit $65=8^2+1^2=7^2+4^2$.  We then know that $8+i$ and $7+4i$ have norm $65$.  Some factor of $8+i$ must not be a factor of $7+4i$, but this factor would be suggested by the norm method.  This applies to any product of two primes equivalent to $1 \pmod 4$


问:
Integrating Dirac Delta Gives Two Different Answers
I am trying to solve the following differential equation

$$\frac{dx}{dt} = x \delta(t-\tfrac{1}{2})$$

from $t=0$ to $t=1$ where $\delta$ denotes the typical Dirac function. If I separate variables and integrate, I get

$$ \int_{x_0}^{x_t} \frac{1}{x} dx = \int_0^t \delta (\tau-\tfrac{1}{2}) d\tau$$

leading to

$$ \ln (x_t) - \ln(x_0) = \begin{cases} 0, & t \leq 1/2 \\ 1, & t > 1/2 \end{cases} $$

On the other hand, if I rewrite the original equation as

$$ \int_{x_0}^{x_t} dx = \int_0^t x(\tau) \delta (\tau-\tfrac{1}{2}) d \tau$$

and simply apply the definition of the Dirac function, I get

$$ x_t - x_0 = \begin{cases} 0, & t \leq 1/2 \\ x\left(\tfrac{1}{2}\right), &t > 1/2 \end{cases} $$
Why do I get two different answers? Is either of these right or is the original equation ill-defined in some sense ... perhaps the Radon-Nikodym derivative does not exist since the LHS is wrt to Lebesgue and the RHS is wrt Dirac? Thanks in advance!

答:
The second way is not very well defined, but there's not really a contradiction between them. The only problem is that $x$ is not really the same for $t < 1/2$ as it is for $t = 1/2$, it's tricky to define $x$ at this point, since there's a discontinuity in the function here.

If you look at the first equation, you have

$$ \ln (x_t) - \ln(x_0) = \begin{cases} 0, & t < 1/2 \\ 1, & t > 1/2 \end{cases} $$

Using logarithm rules to rewrite it,

$$ \ln (x_t/x_0) = \begin{cases} 0, & t < 1/2 \\ 1, & t > 1/2 \end{cases} $$
$$ x_t/x_0 = \begin{cases} e^0, & t < 1/2 \\ e^1, & t > 1/2 \end{cases} $$
$$ x_t = \begin{cases} x_0, & t < 1/2 \\ x_0 e, & t > 1/2 \end{cases} $$

And the second equation is

$$ x_t = \begin{cases} x_0, & t < 1/2 \\ x_0 + x(1/2), & t > 1/2 \end{cases} $$

So the definitions are the same for $t < 1/2$ and if you pretend $x(1/2) = x_0(e-1)$, then they are the same for $t > 1/2$ as well.


问:
How to show that there exists a scalar $\lambda$ so that $A = \lambda I$?
Let $V$ be a finite dimensional vector space, $V \ne \{ 0 \}$ and $A\in L(V)$ an operator which commutes with every operator from $L(V)$. Show that there is a scalar $\lambda$ so that the following applies: $A = \lambda I$.

$L(V)$ = The set of all linear mappings (linear operators) from V to V

If $B\in L(V)$ is arbitrary and if I understood the assignment correctly, this would mean $AB = BA \implies AB - BA = 0 $

I believe, a matrix representation of the linear operators would apply here but I don't have any idea what to do here.

EDIT (Hint by Jared)

I tried to examine it on 5x5 matrices:  example 1, example 2.

For example, for:

$\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} & a_{15}
\\ a_{21} & a_{22} & a_{23} & a_{24} & a_{25}
\\ a_{31} & a_{32} & a_{33} & a_{34} & a_{35}
\\ a_{41} & a_{42} & a_{43} & a_{44} & a_{45}
\\ a_{51} & a_{52} & a_{53} & a_{54} & a_{55}\end{bmatrix}
\begin{bmatrix} 0 & 0 & 0 & 0 & 0
\\ 0 & 0 & 0 & 1 & 0
\\ 0 & 0 & 0 & 0 & 0
\\ 0 & 0 & 0 & 0 & 0
\\ 0 & 0 & 0 & 0 & 0\end{bmatrix} =
\begin{bmatrix} 0 & 0 & 0 & a_{12} & 0
\\ 0 & 0 & 0 & a_{22} & 0
\\ 0 & 0 & 0 & a_{32} & 0
\\ 0 & 0 & 0 & a_{42} & 0
\\ 0 & 0 & 0 & a_{52} & 0\end{bmatrix}$

$\begin{bmatrix} 0 & 0 & 0 & 0 & 0
\\ 0 & 0 & 0 & 1 & 0
\\ 0 & 0 & 0 & 0 & 0
\\ 0 & 0 & 0 & 0 & 0
\\ 0 & 0 & 0 & 0 & 0\end{bmatrix}
\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} & a_{15}
\\ a_{21} & a_{22} & a_{23} & a_{24} & a_{25}
\\ a_{31} & a_{32} & a_{33} & a_{34} & a_{35}
\\ a_{41} & a_{42} & a_{43} & a_{44} & a_{45}
\\ a_{51} & a_{52} & a_{53} & a_{54} & a_{55}\end{bmatrix}
 =
\begin{bmatrix} 0 & 0 & 0 & 0 & 0
\\ a_{41} & a_{42} & a_{43} & a_{44} & a_{45}
\\ 0 & 0 & 0 & 0 & 0
\\ 0 & 0 & 0 & 0 & 0
\\ 0 & 0 & 0 & 0 & 0\end{bmatrix}$

which means $a_{22}=a_{44}$.
For a 1 on the $ij$th and $ji$th position, I get 2 equations (actually 4 but 2 of them are equal to the other 2).

But I still don't understand what I can conclude from that. I reviewed older homework assignments and found the same problem but with a hint:
"Show that A has at least one eigenvalue and observe the corresponding subspace of the eigenvalue. Before that, show that every subspace of an eigenvalue of the operator $T \in L(V)$ is invariant for every operator $S \in L(V)$ which commutes with $T$."

EDIT 2

For $B=E_{ij}$ I conclude that $ AB=BA \implies a_{ii}=a_{jj}, a_{ij} = 0$ if $i \ne j$

(In my example, i is 2 and j is 4 so it turned out $a_{ii}=a_{22}=a_{44}=a_{jj}$)

For $B = E_{ij} + E_{ji}, AB = BA \implies a_{ii} = a_{jj},  a_{ij} = a_{ji},  a_{kl} = 0$ for $ k=l \ne i,$ and $ k=l \ne j $ and $ k\ne i \space \& \space l\ne j$ and $ k\ne j \space \& \space l\ne i$

But we can write B in a different way:

$B = \sum_{i=1}^n\sum_{j=1}^n\beta_{ij}E_{ij}$

$AB = A(\sum_{i=1}^n\sum_{j=1}^n\beta_{ij}E_{ij}) =$ associativity of matrix and scalar multiplication & distributivity over matrix addition $=\sum_{i=1}^n\sum_{j=1}^n(\beta_{ij}A)E_{ij} = \sum_{i=1}^n\sum_{j=1}^n\beta_{ij}(AE_{ij})$

$BA = (\sum_{i=1}^n\sum_{j=1}^n\beta_{ij}E_{ij})A = \sum_{i=1}^n\sum_{j=1}^n\beta_{ij}E_{ij}A = \sum_{i=1}^n\sum_{j=1}^nE_{ij}(A\beta_{ij}) = \sum_{i=1}^n\sum_{j=1}^n(E_{ij}A)\beta_{ij}$

$AB = BA \implies \beta_{ii}a_{ii} = \beta_{jj}a_{jj},  \forall i,j\in \{1, ... , n\} \space \& \space \beta_{ij}a_{ij} = 0 \space , i \ne j$

Which would give us a system of $n^2$ linear equations:

$\beta_{11}a_{11} = \beta_{11}a_{11}$

$\beta_{11}a_{11} = \beta_{22}a_{22}$

...

$\beta_{11}a_{11} = \beta_{nn}a_{nn}$

$\beta_{22}a_{22} = \beta_{11}a_{11}$

$\beta_{22}a_{22} = \beta_{22}a_{22}$

...

$\beta_{22}a_{22} = \beta_{nn}a_{nn}$

...

$\beta_{nn}a_{nn} = \beta_{11}a_{11}$

$\beta_{nn}a_{nn} = \beta_{22}a_{22}$

...

$\beta_{nn}a_{nn} = \beta_{nn}a_{nn}$

$\implies \beta_{11}a_{11} = \beta_{22}a_{22} = ... = \beta_{nn}a_{nn} = \beta a$
$\implies A = \lambda I$ for $\lambda = \beta a$

答:
Let $E_{i,j}$ be the matrix with $1$ in $i,j$ position and 0 everywhere else.<br>
Use that $AE_{1,i}=E_{1,i}A$ to show that
$a_{1,1}=a_{i,i}$, $a_{j,1}$ for $j=2,3,\ldots,n$ and $a_{i,j}=0$ for $j\neq i$. </p>

Body:
Let $E_{i,j}$ be the matrix with $1$ in $i,j$ position and 0 everywhere else.

Use that $AE_{1,i}=E_{1,i}A$ to show that
$a_{1,1}=a_{i,i}$, $a_{j,1}$ for $j=2,3,\ldots,n$ and $a_{i,j}=0$ for $j\neq i$.


问:
Find minimal primary decompositions of ideals
(1) In $R = \mathbb{R}[x,y]/(x^2 + y^2 -1)$, find a minimal primary decomposition of $(\bar{x}^2)$.

(2) In $R = \mathbb{Z}[\sqrt{-5}]$, find a minimal primary decomposition of $(6)$.

For (1), I know that
$$(\bar{x}^2) = (1 - \bar{y}^2)=\left((1-\bar{y})(1+\bar{y})\right)$$

Then
$$((1-\bar{y})(1+\bar{y}))=(1-\bar{y})\cap(1+\bar{y}) $$
and $(1-\bar{y}), (1+\bar{y})$ are prime ideals in $R$,

since
$$R \cong \mathbb{R}[\bar{x}, \bar{y}] \text{ and } R/(1-\bar{y}) \cong \mathbb{R}[\bar{x}, \bar{y}]/(1-\bar{y})\cong \mathbb{R}[\bar{x}]$$ which is an integral domain and similarly for the case of $(1+\bar{y})$.

Then both of these ideals are primary ideals, too.

But for (2), I don’t know how to proceed.

All I’ve done is:

$$(6) = (2\cdot3)=((1+\sqrt{-5})\cdot(1-\sqrt{-5}))$$

And I’ve searched this question but all I’ve found was about ideal factorization.

It is written as a product of ideals, not as an intersection of primary ideals.

Is that factorization same with primary decomposition?

Any comment will be appreciated.

Thank you.

答:
For (1), I'm still working on.

For (2), I think I found a solution.

Here it is:

$Claim: (6) = (2) \cap (3, 1 + \sqrt{-5}) \cap (3 ,1 - \sqrt{-5})$

Consider $(2, 1+ \sqrt{-5}), (3, 1+ \sqrt{-5}$) and $(3, 1-\sqrt{-5})$.
$(2, 1+ \sqrt{-5}) = (2, 1-\sqrt{-5})$,
since $2 - (1+\sqrt{-5}) = 1 - \sqrt{-5}\\$

And $(3, 1+\sqrt{-5})(3, 1-\sqrt{-5}) = (3)$.

We claim that $(2, 1+\sqrt{-5})$ is maximal in $\mathbb{Z}[\sqrt{-5}]$.
Because $\Bbb{Z}[\sqrt{-5}] \simeq \mathbb{Z}[x]/(x^2 + 5)$,

\begin{align}
   \mathbb{Z}[\sqrt{-5}]/(2,1+\sqrt{-5}) &\simeq \mathbb{Z}[x]/(2,1+x,x^2+5) \\
   &\simeq \mathbb{Z}_2[x]/(1+x, 1+x^2) \\
   &\simeq \mathbb{Z}_2[x]/(1+x, (1+x)^2) \\
   &\simeq \mathbb{Z}_2[x]/(1+x) ≃ \mathbb{Z}_2
\end{align}
This shows that $(2, 1+ \sqrt{-5})$ is maximal in $\mathbb{Z}[\sqrt{-5}]$

Similarly, we can show that $(3, 1+\sqrt{-5}),\ (3, 1-\sqrt{-5})$ are both maximal in $\mathbb{Z}[\sqrt{-5}]$.

Since $(2) = (2, 1+\sqrt{-5})^2$, This implies that $(2)$ is $(2,1+\sqrt{-5})$-primary.

And $(3, 1+\sqrt{-5}) + (3, 1-\sqrt{-5}) = (1)$, so that
$$(3, 1+\sqrt{-5}) \cap  (3,1-\sqrt{-5}) = (3,1+\sqrt{-5})(3,1-\sqrt{-5})$$

Then we can prove our claim now.
\begin{align*}
   (6) = (2)\cdot(3) &= (2)\cap(3) \\
   &=(2) \cap \left((3,1+\sqrt{-5})\cdot(3, 1-\sqrt{-5})\right) \\
   &=(2) \cap  (3,1+\sqrt{-5}) \cap (3, 1-\sqrt{-5}) \\
   \end{align*}

$\sqrt{(2)}, \sqrt{(3,1+\sqrt{-5})}, \sqrt{(3, 1-\sqrt{-5})}$ are all distinct.
and none of three primary ideals $(2), (3, 1+\sqrt{-5}), (3, 1-\sqrt{-5})$ contains the intersection of the others.
This shows that this primary decomposition is indeed minimal.

Any comment will be appreciated.


问:
Factoring $a^4+2a^3 b-3a^2 b^2-4a b^3-b^4$
I have tried rearranging and grouping, considering the expression $f(a)$ and finding a root in terms of $b$ (which so far have worked for this sort of problems), but none was fruitful. In fact, this will factor into two trinomials, but I am absolutely clueless so as to how to go about factoring it intuitively (without knowing in the first place that the factors are trinomials).

答:
This solution is specific to this problem.

As others suggested, we look at $y=x^4+2x^3-3x^2-4x-1$.

We know $y'=4x^3+6x^2-6x-4$.

"Obviously" $x=1$ is a root of $y'=0$. Working out the rest, we find
$y'=2(x-1)(2x+1)(x+2)$ and the zeros are $-2, -\frac 12,$ and  $1$. We also notice that
$-\frac 12$ is exactly halfway between the other two roots. This suggests that the polynomial may be symmetric about $x = -\frac 12$.

Letting $x = t-\frac 12$, we get

\begin{align}
   y &= t^4-\frac 92t^2 + \frac{1}{16} \\
     &= \left(t^2 - \frac 94 \right)^2 - \frac{80}{16} \\
     &= \left(t^2 - \frac 94 \right)^2 - 5 \\
     &= \left(t^2 - \frac 94 - \sqrt 5 \right)\left(t^2 - \frac 94 + \sqrt 5 \right) \\
     &\phantom{=} \text{Letting $t = x + \frac 12$, we get} \\
     &= \left(x^2 + x - 2 - \sqrt 5 \right)\left(x^2 + x - 2 + \sqrt 5 \right) \\
\end{align}


问:
If $\lceil\frac{x}{3}\rceil < \lceil\frac{y}{3}\rceil$ what can you conclude about $x$ and $y$?
Certainly if there were no ceiling functions around I would know what that inequality implies. I'm just not sure what else I have to be worried about.

My intuition tells me that I should just be able to conclude that $x<y$, however I'm concerned there's more to the ceiling function than meets the eye. I know that if the input is an integer the function will return itself, if it's not an integer it will round up to the nearest integer.

Edit: Certainly if $x<y$ then it's not necessarily true that $\lceil\frac{x}{3}\rceil < \lceil\frac{y}{3}\rceil$. For example take $x=2$ and $y=3$, obviously $x<y$ but $\lceil\frac{x}{3}\rceil = \lceil\frac{y}{3}\rceil$.

答: Your edit gets the implication backwards.  It is true that if $\lceil \frac x3 \rceil \lt \lceil \frac y3 \rceil$ then $x \lt y$.  It is not true that if $x \lt y$ then $\lceil \frac x3 \rceil \lt \lceil \frac y3 \rceil$, which your example shows.  You can add $x \lt y-1$ to the conclusion when $x \equiv 0 \pmod 3$ or $y \equiv 1 \pmod 3$


问:
Differential equation about Complex functions.
$z\in \mathbb{C}$

$f(z)$ : meromorphic function in a connected domain

I cannot solve this differential equation.

\begin{equation}
f'(z)=f(z)^2 \cdots \ast
\end{equation}

My mathematic book says "Integrate the equation $\ast$ for $f$.

But I cannot know what I should do.

If I integrate $\ast$ for $f$,

\begin{equation}
f(z)=\int f(z)^2 dz
\end{equation}

I cannot proceed from this.

And I tried by another way.

I represent $f(z)=\sum_{n=0}^{\infty} c_n z^n$, and the equation $\ast$ is
\begin{equation}
\sum_{n=0}^{\infty} (n+1)c_{n+1} z^n
=\left(
\sum_{n=0}^{\infty} c_n z^n \right)^2
\end{equation}
But this approach doesn't seem to work.

So I want to solve by integrating the equation $\ast$ for $f$, as my book says.

What should I do?

答:
One might proceed as follows: in any region where

$f(z) \ne 0, \; \; (1) \tag 1$

we have

$\left (\dfrac{1}{f(z)} \right )' = (f^{-1}(z))' = -f^{-2}(z)f'(z) = -\dfrac{f'(z)}{f^2(z)}
;  \; \; (2) \tag 2$

now if

$f'(z) = f^2(z), \tag 3$

then

$\dfrac{f'(z)}{f^2(z)} = 1, \tag 4$

so in accord with (2),

$\left (\dfrac{1}{f(z)} \right )' = -1; \tag 5$

we may integrate this along a path 'twixt $z_0$ and $z$:

$\dfrac{1}{f(z)} - \dfrac{1}{f(z_0)} = \displaystyle \int_{z_0}^z \left (\dfrac{1}{f(w)} \right )'\; dw = \int_{z_0}^z (-1)\; dw = z_0 - z; \tag 6$

we now solve for $f(z)$ algebraically:

$\dfrac{1}{f(z)} = z_0 - z + \dfrac{1}{f(z_0)} = \dfrac{f(z_0)(z_0 - z) + 1}{f(z_0)}, \tag 7$

or

$f(z) = \dfrac{f(z_0)}{f(z_0)(z_0 - z) + 1} = f(z_0)(f(z_0)(z_0 - z) + 1)^{-1}, \tag 8$

which may easily be checked by direct differentiation; from (8),

$f'(z) = -f(z_0)(f(z_0)(z_0 - z) + 1)^{-2}(-f(z_0)) = f^2(z_0)(f(z_0)(z_0 - z) + 1)^{-2} = f^2(z), \tag 9$

which is (3).  We also see that (8) yields

$f(z_0) = f(z_0)(f(z_0)(z_0 - z_0) + 1)^{-1} = f(z_0)(1^{-1}) = f(z_0), \tag{10}$

which demonstrates that $f(z)$ satisfies the appropriate boundary/initial condition at $z_0$.

It is perhaps worth observing that the hypothesis (1) is satisfied by (8) for all $z \in \Bbb C$ as long as $f(z_0) \ne 0$, that is, the solution is non-trivial.


问:
maximin of difference between projections
I am wondering if the following problem is well studied. Given two matrices $A, B \in R^{n\times d}$, solve
$$\max_{x:\|x\|\leq 1}\min_{y:\|y\|\leq 1} \|Ax - By\|^2,$$
where $x, y \in R^d$. Assume $n\geq d$, and $A$, $B$ are full column-rank.

I have derived something, but I wonder if it's already been studied somewhere?

答:
I have a partial solution and am glad to see more discussions.

First, let's solve the inner minimization problem for any given $x$,
$$y^\ast(x)\triangleq \arg\min_{\|y\|\leq 1} \|Ax-By\|^2 = (B^\top B + \lambda I)^{-1}B^\top Ax,$$
where $\lambda=0$ if $(B^\top B)^{-1}B^\top Ax \leq 1$. Otherwise $\lambda$ is such that $(B^\top B + \lambda I)^{-1}B^\top Ax=1$. In general, we can write $\lambda=\lambda(x)$ to emphasize that the $\lambda$ value depends on $x$. This is not hard. A detailed derivation can be found at Least Squares with Euclidean ($ {L}_{2} $) Norm Constraint

Now substitute $y^\ast(x)$ to the outer maximization problem, solve $$\max_{\|x\|\leq 1} \|Ax-B(B^\top B + \lambda I)^{-1}B^\top Ax\|^2 \equiv \max\|[I-B(B^\top B + \lambda I)^{-1}B^\top] Ax\|^2.$$
If $\lambda$ were a constant, the maximizer is simply the leading right singular vector of $[I-B(B^\top B + \lambda I)^{-1}B^\top] A$.

However in reality, $\lambda$ also depends on $x$. So I'm not sure what we can do with this. Could anyone give more ideas?


问:
Under what conditions can we find a basis of a Lie algebra such that the adjoint representation acts by skew-symmetric matrices?
My question might be very silly as I've only recently started to learn about Lie algebras.

Given a Lie algebra $\mathfrak{g}$, and a vector space basis $\{e_i|i=1 \ldots \dim \mathfrak{g}\}$, we can write the adjoint action $\text{Ad}_x: \mathfrak{g} \to \mathfrak{g}: y \mapsto [x,y]$ in matrix form, where the $i,j$th entry of the matrix $[\text{Ad}_x]$ equals $\lambda_i$ if $[x,e_j] = \sum_{k}\lambda_k e_k$. I am looking for necessary and sufficient conditions on our Lie algebra $\mathfrak{g}$ such that there exists a basis in which all of these matrices $\{ [\text{Ad}_x]| x \in \mathfrak{g} \}$ are skew-symmetric.

If the condition turns out to be either simplicity of compactness of the associated simply connected Lie group, I would be very happy.

答:
Malkoun's beautiful answer shows that for a semisimple real Lie algebra, a sufficient criterion for such a basis to exist is that the Lie algebra is anisotropic, i.e. what in the real case is commonly called "compact", i.e. the corresponding Lie groups are compact.

I just want to point out that for semisimple real Lie algebras $\mathfrak g$, this is also necessary (as alluded to by Tsemo Aristide in the answer to this related question). Namely, if a real semisimple Lie algebra is not anisotropic (compact), then it contains elements $x \neq 0$ which are ad-diagonalisable, i.e. such that $ad(x)$ has non-zero real eigenvalues. However, all eigenvalues of real skew-symmetric matrices are purely imaginary, so there can be no basis of $\mathfrak g$ in which all $ad(x)$ are skew-symmetric.

Incidentally, Malkoun's answer can also be used to show that somewhat surprisingly, the result is true for all complex semisimple Lie algebras $\mathfrak g$. Namely, it is a classical cornerstone of the theory that for each such $\mathfrak g$ there exists a (unique) compact real form, i.e. an anisotropic real semisimple Lie algebra $\mathfrak g_0$ such that the scalar extension $\mathbb C \otimes\mathfrak g_0$ is isomorphic to $\mathfrak g$. Now take an orthonormal (real) basis $e_1, ..., e_n$ of $\mathfrak g_0$ w.r.t. its Killing form as per Malkoun's answer, and the matrices of all the adjoints to those elements will be skew-symmetric. Now all elements of $\mathfrak g$ are $\mathbb C$-linear combinations of elements of $\mathfrak g_0$, hence the matrices of all $ad(x), x \in \mathfrak g$ in the basis $1 \otimes e_1, ..., 1 \otimes e_n$ will be skew-symmetric (complex) matrices. (Note that what I wrote in my second paragraph does not contradict this, as complex skew-symmetric matrices can have all eigenvalues one could want, as long as they add up to $0$.)


问:
Dividing $x^{81}+x^{48}+2x^{27}+x^{6}+3$ by $x^{3}+1$
Letting $f(x)=x^{81}+x^{48}+2x^{27}+x^{6}+3$, and we seek to divide $f(x)$ by $x^{3}+1$, or just any $x^3-a$. Is subbing $y=x^3$ allowed in this kind of division? I've experimented doing so and so far it seems correct.

答:
Even without the substitution, the problem will succumb to

polynomial long division.
Therefore, the only issue is how much perspiration you can replace with inspiration.

As already discussed, setting $y = x^3$ reduces the problem to

$$\text{Divide} ~g(y) = (y^{27} + y^{16}  + 2y^9 + y^2 + 3) ~\text{by}~ (y+1). \tag1$$

In attacking equation (1), note that

$(16 - 9) = (9 - 2)$.

the middle terms have coefficients : $1,2,1$ which match the 2nd row of Pascal's triangle.

This suggests that the three middle terms of equation (1) above can be re-expressed as

$$(y^2)(y^7 + 1)^2.$$

Further, it is well settled that for $n$ odd, $(y + 1)$ divides $(y^n + 1).$

For example: $(y + 1) \times (y^2 -y + 1) = (y^3 + 1)$ and

$(y + 1) \times (y^4 -y^3 + y^2 - y + 1) = (y^5 + 1).$

Therefore, you now know (immediately) that $(y + 1)$ divides both $(y^{7} + 1)$ and $(y^{27} + 1).$

Therefore, you can immediately compute the division of $g(y)$ by $(y + 1)$, avoiding polynomial long
division, and immediately conclude that the remainder will be $2$.


问:
Reasoning about least common multiples and their relationship to binomial coefficients
I have been fascinated by this inequality which I believe is valid:

Let:

$x \ge n$ be integers

lcm$(a,b,c)$ be the least common multiple of integers $a, b, c$

$$\text{lcm}(x+1, x+2, \dots, x+n) \le {{x+n}\choose{n}}\text{lcm}(1,2,\dots,n)$$

Here's my reasoning:

lcm$(x+1,x+2,\dots,x+n) < \frac{(x+n)!}{x!}$

If a prime $p > n$ and $p | \dfrac{(x+n)!}{x!}$, then $p | {{x+n}\choose{n}}$ since $p \nmid n!$

If a prime $p \le n$ with $p^t > n$ and $p^t | \dfrac{{x+n}!}{x!}$, then there exists $c$ where $p^{t-c} | {{x+n}\choose{n}}$ where $p^c | \text{lcm}(1, 2, \dots, n)$ and $p^{c+1} \nmid \text{lcm}(1,2,\dots,n)$

It follows that lcm$(x+1, x+2, \dots, x+n) \le {{x+n}\choose{n}}\text{lcm}(1,2,\dots,n)$

I am wondering if the following extension holds:

$$\text{lcm}(\text{lcm}(x-n+1, x-n+2,\dots,x),\text{lcm}(x+1, x+2, \dots, x+n)) = \text{lcm}(x-n+1, x-n+2, \dots, x+n) \le {{x}\choose{n}}{{x+n}\choose{n}}\frac{(n!)^2}{(2n)!}\text{lcm}(1,2,\dots,2n)$$

Here's my thinking for this extension:

(1) lcm$(x-n+1, x-n+2, \dots, x+n) \le {{x+n}\choose{2n}}\text{lcm}(1,2,\dots,2n)$

(2) ${{x+n}\choose{2n}} = \dfrac{(x+n)!}{(x-n)!(2n)!} = \left(\dfrac{x!}{(x-n)!(n!)}\right)\left(\dfrac{(x+n)!}{(x!)(n!)}\right)\left(\dfrac{(n!)(n!)}{(2n)!}\right) = {{x}\choose{n}}{{x+n}\choose{n}}\frac{(n!)^2}{(2n)!}$

Does this extension hold?  Is there a standard way to represent this relationship?  I am especially interested if there is a related statement in terms of Stirling Approximation and the second Chebyshev function.

答:
Your inequality, i.e.,

$$\operatorname{lcm}(x + 1, x + 2, \, \ldots, \, x + n) \le {{x + n}\choose{n}}\operatorname{lcm}(1, 2, \, \ldots, \, n) \tag{1}\label{eq1A}$$

is true, actually for all integers $n \ge 1$ and $x \ge 0$, not just for where $x \ge n$.

However, there are several problems with your attempted proof. A fairly minor one is with your first bullet point. Note both sides of the inequality are actually always equal for $n = 1$ and $n = 2$, and equal for $n = 3$ when $x + 1$ is odd, so you should be using $\le$ instead of $\lt$.

Regarding your third bullet point, as pointed out in mathlove's question comment, there's a counter-example. In your comment reply, you suggested defining $t$ as being $p^t \mid \operatorname{lcm}(x + 1, \, \ldots \, x + n)$ instead, but you don't explain how to modify the rest of the bullet point accordingly.

With your extension inequality, if you were trying to use the result of your first inequality, due to its restriction of the first value in the $\operatorname{lcm}$ being $\ge n + 1$, then your extension would be limited to where $x - n + 1 \ge n + 1 \implies x \ge 2n$. Otherwise, your thinking is correct. Actually, your extension is true for all $x - n + 1 \ge 1 \implies x \ge n$.

Here is one way to prove \eqref{eq1A} is true for all integers $n \ge 1$ and $x \ge 0$. First, for somewhat simpler algebra, let

$$a = \operatorname{lcm}(x + 1, \ldots, \, x + n), \; b = {{x + n}\choose{n}}, \; c = \operatorname{lcm}(1, \, \ldots, \, n) \tag{2}\label{eq2A}$$

Note $b$ can also be expressed as

$$b = \frac{(x + 1)(x + 2)\, \ldots \, (x + n)}{n!} = \frac{\prod_{i=1}^{n}(x + i)}{n!} \tag{3}\label{eq3A}$$

Using the prime factorization exponents based on the definition of $\operatorname{lcm}$, and the $p$-adic order function, then for any prime $p$,

$$d = \nu_{p}(a) = \max_{1 \, \le \, i \, \le n}(\nu_{p}(x + i)) \tag{4}\label{eq4A}$$

$$e = \nu_{p}(b) \tag{5}\label{eq5A}$$

$$f = \nu_{p}(c) = \max_{1 \, \le \, i \, \le n}(\nu_{p}(i)) \tag{6}\label{eq6A}$$

By proving

$$d \le e + f \iff d - f \le e \tag{7}\label{eq7A}$$

for all primes $p$, then \eqref{eq1A} will be true.

Note \eqref{eq6A} gives

$$p^f \le n \lt p^{f+1} \tag{8}\label{eq8A}$$

Since $a$ is the product of at least $p^f$ consecutive integers, they represent a complete residue system of integers modulo $p^f$, so one or more of them must be $\equiv 0 \pmod{p^f}$, which means

$$d \ge f \tag{9}\label{eq9A}$$

If $d = f$, then \eqref{eq7A} gives $e \ge 0$, which is always true. Consider instead that $d \gt f$. Due to the upper bound in \eqref{eq8A}, there's only one $1 \le j \le n$ where $\nu_{p}(x + j) = d$. Also, let $1 \le k \le n$ be any of the one or more values where $\nu_{p}(k) = f$. Next, from \eqref{eq5A},

$$\begin{equation}\begin{aligned}
e & = \left(\sum_{i=1, \, i \neq j}^{n}\nu_{p}(x + i) + d\right) - \left(\sum_{i=1, \, i \neq k}^{n}\nu_{p}(i) + f\right) \\
e & = (d - f) + \sum_{i=1, \, i \neq j}^{n}\nu_{p}(x + i) - \sum_{i=1, \, i \neq k}^{n}\nu_{p}(i) \\
e - (d - f) & = \sum_{i=1, \, i \neq j}^{n}\nu_{p}(x + i) - \sum_{i=1, \, i \neq k}^{n}\nu_{p}(i)
\end{aligned}\end{equation}\tag{10}\label{eq10A}$$

Thus, we want to prove the right side of the last line above is non-negative. For some positive integer $m$, we have

$$x + j = mp^{d} \implies \color{blue}{x - mp^{d}} = j \tag{11}\label{eq11A}$$

Define a new positive starting value of $x$ as

$$x_{1} = p^{f+1} - (\color{blue}{x - mp^{d}}) \implies x_1 + j = p^{f
 + 1} \tag{12}\label{eq12A}$$

Note

$$\begin{equation}\begin{aligned}
x + i & = x + j + (i - j) = mp^d + (i - j) \\
x_1 + i & = x_1 + j + (i - j) = p^{f+1} + (i - j)
\end{aligned}\end{equation}\tag{13}\label{eq13A}$$

Since $\nu_p(x + i) \le f$ for all $1 \le i \le n, \; i \neq j$, then we also have for all such $i$ that

$$\nu_{p}(i - j) = \nu_{p}(x + i) = \nu_{p}(x_1 + i) \tag{14}\label{eq14A}$$

Thus, the value of the right side of the last line of \eqref{eq10A} is the same if $x$ is replaced by $x_1$. Next,

$$\begin{equation}\begin{aligned}
e_1 & = \nu_{p}\binom{x_1 + n}{n} \\
& = \left(\sum_{i=1, \, i \neq j}^{n}\nu_{p}(x_1 + i) + (f + 1)\right) - \left(\sum_{i=1, \, i \neq k}^{n}\nu_{p}(i) + f\right) \\
& = 1 + \sum_{i=1, \, i \neq j}^{n}\nu_{p}(x_1 + i) - \sum_{i=1, \, i \neq k}^{n}\nu_{p}(i)
\end{aligned}\end{equation}\tag{15}\label{eq15A}$$

Thus, we just need to prove that $e_1 \ge 1$. This can be shown using Kummer's theorem which states $e_1$ is the number of carries when $n$ is added to $x_1$ in base $p$. Since $2p^{f + 1} \gt x_1 + n \ge x_1 + j = p^{f+1}$, then $x_1 + n$ has a $1$ for the $p^{f+1}$ digit in base $p$. However, $p^{f+1} \gt x_1$ and $p^{f+1} \gt n$, so both have $0$ in that digit, which means there must be at least one carry in the addition, giving that $e_1 \ge 1$, so the left side of \eqref{eq10A} is $\ge 0$. Thus, this shows \eqref{eq7A} is true in this case. Since we've now covered all of the cases, then \eqref{eq1A} is also true.