问:
Motivation behind the Canonical Model in Modal Logic
I am working through Blackburn et al.'s Modal Logic (ML), and I am currently on the fourth chapter, i.e. Completeness. The idea of completeness is reduced to saying that consistency implies satisfiability. In propositional logic (PL), we have seen a similar notion, namely that of maximally consistent extensions.

In PL, the idea was:

Start with a consistent set of formulae $\Gamma$, want to show $\Gamma$ is satisfiable

Extend $\Gamma$ to its maximally consistent extension, $\Gamma'$.

Turns out that $\Gamma'$ is satisfiable, hence so is $\Gamma$. In PL, this is pretty simple to digest because valuations are all we care about and there are no modalities in the picture.

However, I'm very confused about the working process to achieve something similar in ML. The author introduces maximally consistent sets (calls them MCSs), and then brings up a canonical model. The canonical model is defined as follows, the motivation behind which I strongly lack:

$W^\Lambda$ is the set of all MCSs.

$R^\Lambda$ is the binary relation on $W^\Lambda$ defined by $R^\Lambda wu$ if for all formulas $\psi$, $\psi\in u\implies \diamond\psi \in w$.

$V^\Lambda$ is given by $V^\Lambda(p) = \{w\in W^\Lambda:p\in w\}$

I understand the construction, but why are we doing this? I did see the author proving the Existence Lemma and the Truth Lemma right after this construction, and stating the Canonical Model Theorem, i.e. "Any normal modal logic is strongly complete with respect to its canonical model."

What does this mean? I'd appreciate it if someone can help me understand the bigger picture here since I do understand the tiny details (and that's it). Thank you so much for reading so far!

答:
So, you want to prove the completeness of a modal logic ($\mathit{K}$, $S5$, etc. ), which is formally can be stated as follows:

Completeness: For all $\varphi \in \mathcal{K}$, if $\varphi$ is valid, then $\varphi$ is derivable in the proof system $\mathbb{K}$.

In the above, $\mathcal{K}$ is the language of $K$, and $\mathbb{K}$ is the axiomatisation of $K$. Validity means that for all pointed models $M_s$ we have $M_s \models \varphi$.

The classic way to prove Completeness is by proving the contrapositive:

Completeness (Contraposition): For all $\varphi \in \mathcal{K}$, if $\varphi$ is not derivable, then, then $\varphi$ is not valid.

The trick here is, similar to your PL example, to establish the connection between the notion of satisfiability, and the set theoretic notion of membership. In particular, again, similar to your PL example, we provide deductively closed maximal consistent sets that, intuitively, depict all possible consistent valuations of all formulas of your logic. Naturally, there are infinitely many ways to satisfy formulas of $\mathcal{K}$, and, hence, there are infinitely many maximal consistent sets.

The canonical model is that crucial link in the construction that allows us to go back and forth between formula belonging to a MCS $x$, and being true in a model, where $x$ is one of the states. Having this in mind, it is, hopefully, a bit more clear why $V^\Lambda$ and $R^\Lambda$ are defined that way (we mimic the semantics of modal logic with the set-theoretic membership).

After having established this, the proof is usually straightforward. You assume that $\varphi$ is not derivable, and hence $\{\lnot \varphi\}$ is a consistent set, then you extend the set to a MCS (Lindenbaum) $x$ such that $\lnot \varphi \in x$, then 'jump' to modal logic semantics $M^\Lambda_x \models \lnot \varphi$, and get the result.

Going to your question about the meaning of "Any normal modal logic is strongly complete with respect to its canonical model", remember that some modal logics require models to have certain properties (e.g. relations in $S5$ should be equivalences). Hence, there is usually an additional step of showing that your canonical model satisfies this properties.

Of course, there are also other interesting questions that could be asked about the canonical model. For example, does it have to be infinite? And I think it is answered in further chapters of Modal Logic.


问:
Newton Binomial for number of integer solutions to $|x|+|y|+|z|=30$
Following my post here: Newton Binomial to find $(x_1+x_2+x_3)(x_4+x_5+x_6+x_7)=77$

I would like to know the number of integer solutions to |x|+|y|+|z| = 30. Without the absolute value I get 32 choose 30 (496). With the absolutes, I get 3602 (by computer simulation) and I don't know why...

答:
Hint Consider the equation
$$
a+b+c=30
$$
with $0 \leq a,b,c$.

For each solution with $a,b,c \geq 1$ you get $8$ solutions $x= \pm a, y= \pm b, z=\pm c$ to your equation.

For each solution with exactly one of $a,b,c$ equals zero you get $4$ solutions to your equation.

For each solution with exactly one of $a,b,c$ equals zero you get $2$ solutions to your equation.

ContentLicense: CC BY-SA 4.0


问:
Evaluating $\int_{0}^{1} \frac{ \ln x \ln (1-x)}{\sqrt{x} \sqrt{1-x}} dx$
I have the following integral:
$$\int_{0}^{1} \frac{ \ln x \ln (1-x)}{\sqrt{x} \sqrt{1-x}} dx$$

I think I may be to evaluate this with beta and gamma functions but I am not quite sure how. Any help?

答:
An alternative way, using elementary methods, which begins with the same substitution as sos440's solution:

$$\int_0^1 \frac{\ln x\ln (1-x)}{\sqrt{x(1-x)}}\,dx=8\int_0^{\frac{\pi}{2}}\ln \sin x\ln \cos x\,dx$$

Note that:

$$\begin{aligned}2\int_0^{\frac{\pi}{2}}\ln \sin x\ln \cos x\,dx &=\int_0^{\frac{\pi}{2}}(\ln \sin x+\ln \cos x)^2\,dx-2\int_0^{\frac{\pi}{2}}\ln^2\sin x\,dx\\&=\int_0^{\frac{\pi}{2}}(\ln \sin 2x-\ln 2)^2\,dx-2\int_0^{\frac{\pi}{2}}\ln^2\sin x\,dx\\&=\int_0^{\frac{\pi}{2}}(\ln \sin x-\ln 2)^2\,dx-2\int_0^{\frac{\pi}{2}}\ln^2\sin x\,dx\\&=\ln^2 2\int_0^{\frac{\pi}{2}}\,dx-2\ln 2\int_0^{\frac{\pi}{2}}\ln \sin x\,dx-\int_0^{\frac{\pi}{2}}\ln^2\sin x\,dx\\&=\frac{3\pi\ln^2 2}{2}-\int_0^{\frac{\pi}{2}}\ln^2\sin x\,dx\end{aligned} $$

Now we are left to evaluate the latter integral, for which a little preliminary work must be done:

$$ \sum_{k=1}^n \sin 2kz =\frac{1}{2\sin z} \Big( \cos z-\cos (2n+1)z\Big)$$ which upon integration with $0<x<\frac{\pi}{2}$ gives:

$$ \int_x^{\frac{\pi}{2}} \frac{\cos (2n+1) z}{\sin z}\,dz=\sum_{k=1}^n \frac{\cos 2k x}{k}+\sum_{k=1}^n \frac{(-1)^{k+1}}{k}+\ln \sin x$$

Furthermore, $$\int_x^{\frac{\pi}{2}} \frac{\cos (2n+1) z}{\sin z}\,dz=\frac{(-1)^{n+1}}{2n+1}+\frac{\sin (2n+1) x}{(2n+1)\sin x}- \int_x^{\frac{\pi}{2}} \frac{\sin (2n+1) z\cos z}{(2n+1)\sin^2 z}\,dz$$

Which $\to 0$ as $n\to\infty$. Hence we obtain the relation: $$ \ln \sin x= -\sum_{k=1}^{\infty} \frac{\cos 2k x}{k}-\ln 2$$

$$ \ln^2 \sin x= \ln^2 2+\left(\sum_{a=1}^{\infty} \frac{\cos 2a x}{a}\right)\left(\sum_{b=1}^{\infty} \frac{\cos 2b x}{b}\right)+2\ln 2\sum_{k=1}^{\infty} \frac{\cos 2k x}{k}$$

Integrating over $(0,\frac{\pi}{2})$ the latter sum clearly vanishes and all the paired cosines for which $a\neq b$ as well. Thus we have:

$$\begin{aligned}\int_0^{\frac{\pi}{2}}\ln^2 \sin x\,dx &=\ln^2 2\int_0^{\frac{\pi}{2}}dx+\sum_{a=1}^{\infty}\frac{1}{a^2}\int_0^{\frac{\pi}{2}}\cos^2 2a x\,dx\\&=\frac{\pi\ln^2 2}{2}+\sum_{a=1}^{\infty}\frac{\pi}{4a^2}\\&=\frac{\pi\ln^2 2}{2}+\frac{\pi^3}{24}\end{aligned}$$

Combining everything and multiplying by $4:$

$$\int_0^1 \frac{\ln x\ln (1-x)}{\sqrt{x(1-x)}}\,dx=4\pi\ln^2 2-\frac{\pi^3}{6}$$

ContentLicense: CC BY-SA 3.0


问:
Finding correlation between U and W

Two investments $X$ and $Y$ give returns as follows (expectation and variance):
$E(X)=0.5$,$E(Y)=0.4$ and $V(X)=2$, $V(Y)=1$. The correlation between X and Y is ρ (X,Y) = 0.6.

(a)

Find the expectation and variance of the "combined" investments $U = X + Y$ and $W = 2Y$, and finally find the correlation between $U$ and $W$. Hint: First find $V (U + W)$.

\begin{align}\\
& E(U) = E(X+Y) = E(X) + E(Y) = 0,5+0,4 = 0,9 \\
& E(W) = E(2Y) = (2)E(Y) = (2)(0,4) = 0,8 \\
& \\
& V(U) = V(X+Y) = V(X)+V(Y)+2(p(X,Y))(\sigma_X)\\
& V(U) = 2 + 1 + 2(0,6)(\sqrt{2}) = 3 + 1,2(\sqrt{2}) \\
&\\
& V(W) = V(2Y) = (2^2)(V(Y)) = 2^2(1) = 4 = \sigma_W^2\\
\end{align}

Correlation:
\begin{align}\\
& V(U+W) = V(U) + V(W) = 3 + 4 = 7 = \sigma_{U+W}^2
\end{align}

Definition:
The correlation coefficient to $X$ and $Y$ $(Corr(X,Y)/p_{X,Y}/p)$ is defined as: $p_{X,Y}=\frac{Cov(X,Y)}{{\sigma_X}*{\sigma_Y}}$

So we have:
\begin{align}\\
& \sigma_U = \sqrt{\sigma_U^2} = \sqrt{3} \\
& \sigma_W = \sqrt{\sigma_W^2} = \sqrt{4} = 2
\end{align}

and:

$Cov(U,W) = E[ (U-\mu_U)(W-\mu_W)] = E(UW) - E(U)*E(W)$

This is where I'm stuck, because I don't know how to calculate $Cov(U,W)$ without $E(UW)$, and I don't know if U and W are independent or not. (If U and W are independent, then $E(UW)=E(U)*E(W)$)

答:
First error:

$$\mathbb{V}[X+Y]=2+1+2\cdot0.6\sqrt{2}=3+1.2\sqrt{2}$$

because $X,Y$ are not independent but correlated via $\rho=0.6$

Covariance between $(U,W)$, the following definition will help

$$\mathbb{Cov}[U,W]=\mathbb{E}[UW]-\mathbb{E}[U]\cdot\mathbb{E}[W]$$

substitute $U=X+Y$ and $W=2Y$ and find all the expectations w.r.t X,Y

so  you can find

$$\mathbb{E}[UW]=\mathbb{E}[(X+Y)2Y]=2\mathbb{E}[XY]+2\mathbb{E}[Y^2]$$

Title: Finding correlation between U and W


问:
Infinite subset of natural numbers
I want to show that if the set $A$ is an infinite subset of natural numbers $N$, then $\text{card}(A)=\text{card}(N)$. To do so, it suffices to find an injective function $f:N\rightarrow A$.

One natural way to construct $f$ is as follows: we assign to $f(1)$ the smallest element of $A$. Then we omit the smallest element of $A$ and we assign to $f(2)$ the smallest element of the remaining numbers and so on.

I do not want to use the well ordering principle. I argue as follows. I assign to $f(1)$ an arbitary element of A like $a_1$.Then I remove $a_1$ from $A$ and assign to $f(2)$ an arbitrary element of $A/a_1$. Since $A$ is infinite, I can assign to $f(n)$ a number from $A$ for any $n$ because $A$ never gets empty.

Do I need to use the axiom of choice here??

答:
No. You don't need the axiom of choice. $\Bbb N$ is well-ordered without relying on the axiom of choice, indeed it is a defining property of $\Bbb N$: it is the unique non-empty linear order that is well-ordered, without a maximal, and every element has a successor.

The thing about the term "arbitrary" is that you're effectively hiding a choice function from the non-empty subsets of $A$ (or at least the co-finite subsets). So the question is why such a function exists, and in order to prove this you'll need to resort to proving that $A$ is either countable, or at least there is an injection from $\Bbb N$ into $A$.

So you run into a slight circularity. Of course, the axiom of choice might be used for the argument a priori, and then removed in the a postriori by arguing "actually since there is a well-ordering of $A$ as a subset of $\Bbb N$ here's a choice function".

Title: Infinite subset of natural numbers


答:
Yes, you do need some amount of the axiom of choice to run the argument you're describing (although of course the result doesn't need choice at all, and there's absolutely no reason to avoid using the well-orderedness of $\mathbb{N}$). Specifically, you're using dependent choice.

It's worth noting that the following statement is not provable in $\mathsf{ZF}$ (= set theory without choice) alone:

If $X$ is infinite, then there is an injection $\mathbb{N}\rightarrow X$.

(Sets not admitting an injection from $\mathbb{N}$, or equivalently sets with no non-surjective self-injection, are called Dedekind-finite; in $\mathsf{ZFC}$ every Dedekind-finite set is finite, but this is not provable in $\mathsf{ZF}$ alone. In fact, it is consistent with $\mathsf{ZF}$ that there is an infinite Dedekind-finite set of real numbers!)

So if you want to prove "Every infinite set of natural numbers admits a surjection from $\mathbb{N}$" without using the axiom of choice, you need to at some point use something particular about the natural numbers - such as their well-orderedness. Given that the well-orderedness of $\mathbb{N}$ is basically its most fundamental property, this isn't exactly a high cost.

Title: Infinite subset of natural numbers

Body:
I want to show that if the set $A$ is an infinite subset of natural numbers $N$, then $\text{card}(A)=\text{card}(N)$. To do so, it suffices to find an injective function $f:N\rightarrow A$.

One natural way to construct $f$ is as follows: we assign to $f(1)$ the smallest element of $A$. Then we omit the smallest element of $A$ and we assign to $f(2)$ the smallest element of the remaining numbers and so on.

I do not want to use the well ordering principle. I argue as follows. I assign to $f(1)$ an arbitary element of A like $a_1$.Then I remove $a_1$ from $A$ and assign to $f(2)$ an arbitrary element of $A/a_1$. Since $A$ is infinite, I can assign to $f(n)$ a number from $A$ for any $n$ because $A$ never gets empty.

Do I need to use the axiom of choice here??


问:
Calculate the sum of the coefficients
If f(x) = $x^3-(a-2)x^2+bx+c$
calculate a+b+c (Answer:4)

I try: $f(x)= x^3-(a-2)x^2 +bx+c\\
0 = 2^3-(a-2)2^2+2b+c\\
0 = 8-4a+8+2b+c\rightarrow \boxed{4a-2b-c=16}\\
x=0\rightarrow y=0-0+0=c\rightarrow y = c$

But i stopped here..I don't see another equation

答:
Since $2$ is a repeated root, $(x-2)^2$ divides $f(x)$, i.e. $$f(x) = (x^2-4x+4)( x +d)= x^3+(d-4)x^2+(4-4d)x +4d$$

Comparing the constant term, $d=\frac c4$ and comparing the rest gives $$ a-2=4-\frac c4 \\ b=4-c $$ Adding up, $$a+b+c = 10-\frac c4 $$

This means the answer is dependent on $c=f(0)$. If you can provide that, the question is complete.

Title: Calculate the sum of the coefficients


问:
Does weak convergence in $H^1$ already imply weak convergence in $L^6$ to the same limit?
Let $\Omega\subset \mathbb{R}^n$ be bounded and let $(f_n)_{n\in \mathbb{N}}\subset H^1(\Omega)$ be weak convergent to $f$ in $H^1(\Omega)$. Is it true that $(f_n)_{n\in \mathbb{N}}$ admits a subsequence which converges weakly to $f$ in $L^6(\Omega)$ ?

By the compact embedding of $H^1(\Omega)\hookrightarrow L^p(\Omega)$ for $p\in [1,6)$ we have for example a subsequence $(f_{n_k})_{k\in \mathbb{N}}$ so that $f_{n_k}\to f$ in $L^5(\Omega)$. On the other hand we know that there exists $h\in L^6(\Omega)$ so that $f_n\rightharpoonup h$ in $L^6(\Omega)$. This implies $f_n \rightharpoonup h$ in $L^5(\Omega)$ as $L^{\frac{5}{4}}(\Omega)\subset L^{\frac{6}{5}}(\Omega)$. By the strong convergence we conclude $f=h$ but only in $L^5(\Omega)$. Is there a way to extend my argument or maybe a completely different approach to show $f_n\rightharpoonup f$ in $L^6(\Omega)$?

I would be grateful for any advice.

答: If $n\le 3$ then the embedding $H^1\hookrightarrow L^6$ is continuous, this implies weak convergence in $L^6$ directly, without using compact embeddings.


问:
Monty Hall Problem with unknown probabilities
Does someone know a solution to the following generalization of the Monty Hall Problem:

The Problem: Assume you are on Let's Make a Deal and are presented with the regular dilemma of the Monty Hall Problem. Unlike normally, however, the car isn't placed randomly behind a door, but according to a certain probability. So the probability that the game show host puts the car behind door $1$ is $A_1$, behind door $2$ is $A_2$ and behind door $3$ is $A_3$. (With $A_1+A_2+A_3=1$, of course)

Additionally, you have a preference for choosing some of the doors over others, meaning that the probability of you choosing a random one of the three doors on your first try is not necessarily $1/3$. So the probability of you choosing door $1$ is some $B_1$, the probability for choosing door $2$ is $B_2$ and the door $3$ has probability $B_3$ of being chosen by you. (Once again with $B_1+B_2+B_3=1$)

Now what is the probability of you winning the car?

As the normal Monty Hall Problem is already tough enough for me, I thought I might ask you all for help. If my notations are bad please do not hesitate to change them. Nonetheless, my progress so far:

My Progress: Obviously the probability of choosing the $1st$ door and the car being behind it is $A_1\cdot B_1$ if you do not switch. The probability of you choosing the $2$nd door and finding the car behind it if you do not switch is $A_2\cdot B_2$ and the accordingly the probability of finding the car behind door $3$ without switching is $A_3 \cdot B_3$.

And this is right about where my head starts spinning, because I don't know where to go from here on

答:
Since you don't know the values of $A_1, A_2, A_3$ then the safe play is to always switch.  Suppose that if you were assigned door $k$, that your win factor was $p(k)$.

Then, your overall expectation will be $\sum_{i=1}^3 B_k \times p(k).$

Therefore, the problem is reduced to calculating $p(k)$ based on the OP's query and response to my comment/question.

I will calculate $p(1)$, with the idea that the calculation of $p(2)$ and $p(3)$ should result in a symmetrical formula.

Assuming that the participant was assigned door 1:

$A_2$ of the time, door 2 is correct and door 3 is shown.

$A_3$ of the time, door 3 is correct and door 2 is shown.

$\frac{A_1}{2}$ of the time, door 1 is correct and door 2 is shown.

$\frac{A_1}{2}$ of the time, door 1 is correct and door 3 is shown.

Therefore, the chance that door 2 will be shown is
$A_3 + \frac{A_1}{2}.$  When that happens, the participant's chances of winning will be $\frac{A_3}{A_3 + (A_1/2)}.$

Similarly, the chance that door 3 will be shown is
$A_2 + \frac{A_1}{2}.$  When that happens, the participant's chances of winning will be $\frac{A_2}{A_2 + (A_1/2)}.$

Therefore,

$$p(1) = \left\{[A_3 + (A_1/2)] \times \frac{A_3}{A_3 + (A_1/2)}\right\}$$
$$+
\left\{[A_2 + (A_1/2)] \times \frac{A_2}{A_2 + (A_1/2)}\right\}$$
$$ = A_3 + A_2 = (1 - A_1).$$

Addendum

Examine the above computation.

Suppose that you are assigned door 1, you are computing $p(1)$, and you are focusing on what happens when Door 1 is correct.

It is actually totally irrelevant what percentage of the time Monty Hall will show Door 2, rather than Door 3. This is demonstrated by the fact that in each of the two terms used to compute $p(1)$, the denominator is always exactly cancelled by the applied scalar.


问:
Let $X_1, X_2, X_3$ be the numbers that appear when three balanced dice are rolled. Find the expected value of the largest number that you will see.
Let $X_1, X_2, X_3$ be the numbers that appear when three balanced dice are rolled. Find
the expected value of the largest number that you will see.

My attempt:
$P[X_i=x] = 1/6$ for $i=1,2,3$.

$E[X] = \sum{xp(x)} = 1×1/6 + 2×1/6 + 3×1/6 + 4×1/6 + 5×1/6 + 6×1/6$

Is my attempt correct?

答:
No, six has nothing to do with this. What is the probability that the largest element equals $1$ - it is $\frac16^3.$ What is the probability that it is equal to $2?$ It is $\left(\frac26\right)^3 -\left(\frac16\right)^3.$ And so on. So, the expectation is:

$$\sum_{i=1}^6 i \left(\left(\frac{i}6\right)^3 - \left(\frac{i-1}6\right)^3\right).$$


问:
Average integral of a function over a ball
How to show that the average integral of a continuous function over a ball converges to the value at the center as the radius goes to zero , i.e,

答:
Note that the closed ball $B(x,\epsilon)$ is compact, so $f$ restricted to $B(x,\epsilon)$ is uniformly continuous. This means that for any $\delta > 0$, there exists an $\epsilon > 0$ such that,

$$\sup_{y \in B(x,\epsilon)} \left|f(x) - f(y)\right| < \delta.$$

Letting $|\partial B(x,\epsilon)| = \int_{\partial B(x,\epsilon)}\,dS(x)$, this implies that,

$$\int_{\partial B(x,\epsilon)} f(x) - \delta\,dS(x) = (f(x) - \delta)|\partial B(x,\epsilon)| \leq \int_{\partial B(x,\epsilon)} f(x)\,dS(x) \leq (f(x)+\delta)|\partial B(x,\epsilon)|.$$

Dividing by $|\partial B(x,\epsilon)|$ yields,

$$f(x) - \delta \leq \frac{1}{|\partial B(x,\epsilon)|}\int_{\partial B(x,\epsilon)} f(x)\,dS(x) \leq f(x) + \delta.$$

Since our choice of $\delta > 0$ was arbitrary, it follows that

$$\lim_{\epsilon\to 0} \frac{1}{|\partial B(x,\epsilon)|}\int_{\partial B(x,\epsilon)} f(x)\,dS(x) = f(x).$$

Just multiply by $-1$ to get your answer.


问:
Evaluating $\int_0^{\pi/2} \log(1 - x \cot x) \, dx\;$ (Is there a closed form?)
I am interested in knowing if it is possible to find a closed-form solution to the following challenging log-cotangent integral

$$\int_0^{\frac{\pi}{2}} \log(1 - x \cot x) \, dx$$

I very much doubt a closed form in terms of known mathematical constants can be found and have no reason to suspect one exists.

In the absence of a simple closed-form value being found, maybe the integral can be evaluated in terms of an infinite series. In this direction, since $0 < x \cot x < 1$ for all $x \in (0,\frac{\pi}{2})$ the log term appearing in the integrand can be expanded. Doing so produces
$$\int_0^{\frac{\pi}{2}} \log(1 - x \cot x) \, dx = -\sum_{n = 1}^\infty \frac{1}{n} \int_0^{\frac{\pi}{2}} (x \cot x)^n \, dx = -\sum_{n = 1}^\infty \frac{a_n}{n},$$
where
$$a_n = \int_0^{\frac{\pi}{2}} (x \cot x)^n \, dx. \tag1$$
Noting that
\begin{align*}
a_1 &= \int_0^{\frac{\pi}{2}} x \cot x \, dx = \frac{\pi}{2} \log (2)\\
a_2 &= \int_0^{\frac{\pi}{2}} (x \cot x)^2 \, dx = \pi \log (2) - \frac{\pi^3}{24}\\
a_3 &= \int_0^{\frac{\pi}{2}} (x \cot x)^3 \, dx = \frac{9\pi}{16} \zeta (3) - \frac{\pi^3}{16} + \frac{3\pi}{2} \log (2) - \frac{\pi^3}{8} \log (2),
\end{align*}
perhaps it is possible to find a general expression for $a_n$. Indeed, an attempt at finding (1) can be found here.

Any other approaches or suggestions to this challenging integral would be welcome.

答:
Just based on approximations.

First of all, a Taylor expansion
$$\log (1-x \cot (x))=2\log(x)-\log(3)+\frac 13\sum_{n=1}^\infty \frac {a_n}{5^n\,b_n} x^{2n}$$ where the first $a_n$'s are
$$\{1,13,22,907,55282,298183,738788,11537791247,\cdots\}$$ the first $b_n$'s are
$$\{1,42,189,19404,2837835,35756721,202621419,7114443263928,\cdots\}$$  makes the series to converge very fast. Truncating and using $p$ terms give the following numerical results
$$\left(
\begin{array}{cc}
 p &  I_p\\
 1 & -3.362471345 \\
 2 & -3.354577981 \\
 3 & -3.353531646 \\
 4 & -3.353370377 \\
 5 & -3.353343240 \\
 6 & -3.353338390 \\
 7 & -3.353337483 \\
 8 & -3.353337307 \\
 9 & -3.353337272 \\
 10 & -3.353337265
\end{array}
\right)$$
Another,almost equivalent, possibility  could be to use the $[2p,4]$ Padé approximant of the integrand and use partial fraction decomposition

$$\left(
\begin{array}{cc}
p & I_p \\
 1 & -3.353356609 \\
 2 & -3.353338184 \\
 3 & -3.353337337 \\
 4 & -3.353337269 \\
 5 & -3.353337263
\end{array}
\right)$$

Just for the fun of it, the result is "close" to
$$-\frac{1}{10} \zeta \left(\frac{1}{2}\right) \left(\frac{1}{2}-\Gamma
   \left(\frac{1}{24}\right)\right)$$ which is in a relative error of $5.75 \times 10^{-7}$%.

Edit

In a comment, @James Arathoon suggested to work instead
$$\int_0^{\frac{\pi }{2}} \left(\log \left(\frac{x^3}{3}\right)-\log (\cos (x))\right) \, dx+\int_0^{\frac{\pi }{2}} \log \left(-\frac{3 (x \cos (x)-\sin (x))}{x^3}\right) \, dx$$

The first integral does not make any problem
$$\int_0^{\frac{\pi }{2}} \left(\log \left(\frac{x^3}{3}\right)-\log (\cos (x))\right) \, dx=-\frac{\pi}{2}   (3+\log (12)-3 \log (\pi ))$$

The second integrand can write
$$\log \left(-\frac{3 (x \cos (x)-\sin (x))}{x^3}\right)=-\sum_{n=1}^\infty \frac {c_n}{5^n\,d_n} x^{2n}$$ where the first $c_n$'s are
$$\{1,1,1,37,59,2753,1654,\cdots\}$$ and the first $d_n$'s are
$$\{2,28,189,38808,315315,71513442,202621419,\cdots\}$$ and the convergence is quite fast.
$$\int_0^{\frac{\pi}{2}} \log(1 - x \cot (x)) \, dx=
-\frac{\pi}{2}   (3+\log (12)-3 \log (\pi ))-$$ $$\frac{\pi ^3}{240}\Bigg[1+\frac{3 \pi ^2}{1400}+\frac{\pi ^4}{88200}+\frac{37 \pi ^6}{465696000}+\frac{59 \pi ^8}{92492400000}+\cdots \Bigg]$$ Using the above terms leads to an absolute error equal to $7.39 \times 10^{-8}$.

Using the $[8,4]$ Padé approximant and partial fraction decomposition leads to an absolute error equal to $5.72 \times 10^{-11}$.

Update

Using the same approach as @River Li
$$I = \int_0^{\frac \pi 2} \log(1 - x\cot (x))\,dx
= - \frac{\pi^3}{24} + \frac{\pi}{2}\log( 2) - \int_0^{\pi/2} \frac{x^3}{\tan (x) - x} \,dx$$ we can easily build the $[2p,4]$ Padé approximant of the left integrand which write
$$\frac{x^3}{\tan (x) - x} =\frac {3+\sum_{k=1}^p a_k\,x^{2k} } {1+b_1 x^2+b_2 x^4 }$$ the integration of which leading to a polynomial of degree $(2p-1)$ plus two hyperbolic arctangents.

As a function of $p$ the result of this integration is given below
$$\left(
\begin{array}{cc}
 2 &  \color{red}{3.150202}313227377435283622 \\
 3 &  \color{red}{3.15020211}4234142271205176 \\
 4 &  \color{red}{3.1502021130}45512153905225 \\
 5 &  \color{red}{3.15020211302}9044077891702 \\
 6 &  \color{red}{3.15020211302878}5049324311 \\
 7 &  \color{red}{3.150202113028780}655331848 \\
 8 &  \color{red}{3.15020211302878057}7184507 \\
 9 &  \color{red}{3.1502021130287805757}51197 \\
 10 & \color{red}{3.15020211302878057572}4358 \\
 11 & \color{red}{3.1502021130287805757238}48 \\
 12 & \color{red}{3.150202113028780575723839}
\end{array}
\right)$$


问:
arrange k identical robots in 15 chairs, with limitations
I ran into this question, want to challenge you.

in how many ways can you arrange a number of identical robots on $15$ chairs?

limitations:
1) $2$ robots cannot sit next to each other.
2) each empty chair has at least one neighbour with a robot.

what i got so far: i know that it has something to do with $\sum_{5}^{8}$ because the minimum amount of robots is $5$ and maximum is $8$.

i'm not sure how to proceed from here though.

Thank you very much in advance,

Yaron

答:
Consider $xRExRExRExRExRx$.  The $R$'s stand for robots, the $E$'s stand for empty chairs.  All the $x$'s can be zero or one empty chairs.   However there are 15 chairs altogether, so in fact each $x$ must be one empty chair.

Consider now one more robot.  $xRExRExRExRExRExRx$.  We have six robots and five empty chairs accounted for, of 15 chairs.  That means there are four empty chairs to distribute among the seven possible $x$'s, which can be done in ${7\choose 4}$ ways.

I leave the cases of 7 or 8 robots for you to consider.


问:
How to obtain the probability density function of a function of random variable?
Assume we have three random variables $x, y, z$, and they follow joint Gaussian distribution. We define $W$ as a new random variable, which is the function of $x, y, z$, denoted as $f(x, y, z)$. The question is how to obtain the probability density function of $W$ in the following two conditions:

1、$f$ is a closed-form function;

2、$f$ is not a closed-form and the value of $f(x, y, z)$ can be determined by simulation methods such as Monte Carlo simulation when $x, y, z$ are fixed.

The question is how to obtain the density function of $W$ in the above two conditions. Does such a method exist? Any comment will be thankful.

答:
These settings are quite general. In the case of 1, it might be possible to calculate $F_W(x) = P(f(X,Y,Z) \le x) $, and differentiate it to get the density function, but this could be tough if $f$ is not easy to work with. In the setting 2, it is at least easy to get an approximation of the density if one is able to simulate the random variable $W=f(X,Y,Z)$. In this case one would simply simulate many, many replicates of $W$, and look at the histogram. One could also apply proper density estimation techniques to "smooth" the histogram. See e.g.

https://machinelearningmastery.com/probability-density-estimation/


问:
Compact onto linear operator from Banach space imply space is finite-dimensional
I am trying to prove this question, which from the book Foundations of Modern Analysis by Avner Friedman pp189 5.1.7. the question is below.

Let T be a compact linear operator from a Banach space X onto itself. If $\mathbf{T^{-1}}$ is a bounded operator, then X is finite-dimensional.

I think this question is not right, because if T is a compact operator then T is a bounded operator and due to the open-mapping theorem,  $\mathbf{T^{-1}}$ must be bounded. It seems that this question just tells us that T is one-one and onto.

答:
Two key facts here.

$(1)$ The compact operators form an ideal.

$(2)$ The identity operator on a space $X$ is compact if and only if $X$ has finite-dimension.

By your assumption, $T \circ T^{-1}= 1_X$ and $T$ is compact. Because of $(1)$, we see that $1_X$ is compact. By $(2)$, $X$ has finite-dimension.


问:
Calculate $\int_{-\infty}^{\infty}\frac{\cos^2(x)}{x^2}e^{-ikx}dx$ for $k\in[0,2]$
After some alegbra, we find that the integral equals:

$$\frac{1}{4}\int_{-\infty}^{\infty}\frac{2e^{-ikx}+e^{i(2-k)x}+e^{-i(2+k)x}}{x^2}dx$$.

Now since $k\in[0,2]$, for contour integration, we consider the LHP for the first and third term in the integrand and then the UHP for the second term.

If we call $C_{\epsilon,U}$ the semi-circle of radius $\epsilon$ in the UHP, $C_{\epsilon,L}$ the semi-circle of radius $\epsilon$ in the LHP, $C_{R,U}$ the semi-circle of radius $R$ in the UHP, $C_{R,L}$ the semi-circle of radius $R$ in the UHP, with closed contours $\Gamma_{U}$ and $\Gamma_{L}$, we have:

$$\int_{-\infty}^{\infty}=\lim_{R\to\infty}\lim_{\epsilon\to 0}\Bigg(\int_{\Gamma_U}-\int_{C_{\epsilon,U}} -\int_{C_{R,U}}\Bigg) \text{ for the second term}$$

$$\int_{-\infty}^{\infty}=\lim_{R\to\infty}\lim_{\epsilon\to 0}\Bigg(-\int_{\Gamma_L}-\int_{C_{\epsilon,L}} -\int_{C_{R,L}}\Bigg) \text{ for the remaining terms}$$

(Note the negative sign in line 2 due to negative orientation of $\Gamma_L$)

No poles lie inside the closed $\Gamma_U$ and $\Gamma_L$ so the first integral on the RHS of each line is zero, and the last integral on the RHS each line $\to 0$ as $R\to \infty$.

So, for the top line, we're left with:

$$ \frac{1}{4}\int_{-\infty}^{\infty}\frac{e^{i(2-k)x}}{x^2}dx=-\frac{1}{4}\lim_{\epsilon\to 0}\int_{C_{\epsilon,U}}\frac{e^{i(2-k)z}}{z^2}dz=-\frac{1}{4}\lim_{\epsilon\to 0}\int_{\pi}^{0}\frac{e^{i(2-k)\epsilon e^{i\theta}}}{\epsilon^2e^{2i\theta}}i\epsilon e^{i\theta}d\theta$$.

Expanding $e^{i(2-k)\epsilon e^{i\theta}}=1+i(2-k)\epsilon e^{i\theta}+...$ (can ignore higher order terms as they tend to $0$ as $\epsilon \to 0$).

So we get (for the second term in the original integrand): $$\frac{1}{4}\lim_{\epsilon\to 0}\int_{0}^{\pi}\frac{1+i(2-k)\epsilon e^{i\theta}}{\epsilon^2e^{2i\theta}}i\epsilon e^{i\theta}d\theta=\frac{1}{4}\lim_{\epsilon\to 0}\int_{0}^{\pi}\frac{ie^{-i\theta}}{\epsilon}d\theta-\frac{1}{4}\int_{0}^{\pi}(2-k)d\theta=\frac{1}{4}\lim_{\epsilon\to 0}\int_{0}^{\pi}\frac{ie^{-i\theta}}{\epsilon}d\theta-\frac{1}{4}(2-k)\pi$$

But surely this is divergent? Have I gone wrong somewhere with this? When I consider the remaining terms (the 1st and 3rd terms in the original integrand, using the LHP), using the same method I get: $$=-\frac{3}{4}\lim_{\epsilon\to 0}\int_{-\pi}^{0}\frac{ie^{-i\theta}}{\epsilon}d\theta-\frac{1}{4}(2k+2)\pi$$ which again is divergent. So summing these two results would mean that the original integral in my question is divergent (assuming my method is correct - which I don't think it is)?

答:
Around zero

$$\frac{\cos^2(x)}{x^2}e^{-ikx} \simeq \frac{1}{x^2}$$ so the integral is surely divergent.


问:
Relationship between the expected value of the determinant and the determinant of the expected value?
Let $X$ be an $n \times n$ matrix of random variables. Is $E[\det(X)] = \det(E[X])$?

If it doesn't hold in general, under what assumptions on $X$ does it hold?

答: It is not true in general. Let $X= [A, B ; C , D] \in \mathbb{R}^{2\times 2}$. Then the statement is equivalent with $E[ AD - BC ] = E[A]E[D]- E[B]E[C]$. If $B$ and $C$ are uncorrelated, but $A$ and $D$ are correlated, this is false. @yotam maoz is right though that it does hold in the case when the entries are independent. As suggested by @Lorenzo Najt, if only rather the rows or columns are independent, then the same holds, since when calculating the determinant, one can expand along the rows or the columns, calculating the determinant by multiplying each entry in a row/column by the determinant of the corresponding minorant matrix, which will be comprised of sums/products of uncorrelated terms.


问:
Symbol for very small variable
Is there any representation to state that a variable is close to (not equal) zero? Let me give you an example. Consider the function

$u(x)=\alpha (e^{i\omega \delta t}-1) f(x)$

I am interested in the function $u(x)$ when $\delta t$ is very small. For this case, it should be easy to see that

$u(x)|_{\text{small }\delta t} \approx i \alpha \omega \delta t f(x)$

Is there any "nice" notation to represent such an equation (without having to write small)? I thought that I could use the limit notation for that [for example, $\lim_{\delta t \to 0} u(t)$], but then I realized that if $\delta t$ goes to zero, then $u(x)=0$. Therefore, it is not what I need.

I found this link in the same forum, but it did not help.

答:
The issue with taking limits is that as $\delta t\to 0$ both sides go to zero, but what you want to say is stronger than that.

I imagine what you actually mean by $u(x)\approx i\alpha\omega\delta tf(x)$ can be formalised as
$$\lim_{\delta t\to 0}\frac{u(x)}{\delta t}=i\alpha\omega f(x).$$

This is simply a matter of rearranging so that the limit captures the relative error of the approximation, not just the absolute error.

Title: Symbol for very small variable


问:
Inequality in the eigenvalues of $B$
Suppose $B \in M_{n \times n} (\mathbb C)$ is Hermitian. Prove that $$\lambda_{min}\leq b_{kk} \leq \lambda_{max}, \; \text{for}\; 1 \leq k \leq n.$$ With the equality in both sides for some $k$ only if $b_{kj}=b_{jk}=0,$ for all $k\neq j.$ Here $\lambda_{min} \;\& \; \lambda_{max}$ are the minimum and maximum eigenvalues of $B$ and $b_{kj}$ is the $(k,j)$- entry of $B$.

 Attempt: Using Rayleigh-Ritz theorem we achieved the inequalities above. How can I show the equality in both sides if $b_{kj}=b_{jk}=0,$ does it follow from Rayleigh-Ritz? Also suppose we take the matrix $\text{diag}(1,2,3)$
does the condition $b_{kj}=b_{jk}=0,$ for all $k\neq j$ imply that $b_{kk}$ is equal to either $\lambda_{min} \;\vee \; \lambda_{max}?$

Any help and hint would be much appreciated.

答: It follows from Rayleigh Ritz. In particular: part of the Rayleigh Ritz theorem states that if $x$ is a unit vector with $x^*Bx = \lambda_{\max}$, then it must hold that $x$ is an eigenvector of $B$ associated with $\lambda_\max$. Now, note that if $b_{kk} = \lambda_{\max}$, then then the vector $x = e_k$ (the $k$th standard basis vector) satisfies $x^*Bx = \lambda_\max$. Similar logic can be applied in the case that $b_{kk} = \lambda_\min$.

Title: Inequality in the eigenvalues of $B$


答:
I think you are asking about the first order (linear) approximation using the derivative.

You might say
$$
u(x)= i \alpha \omega \delta t  f(x) + o(\delta t ).
$$

The fact that this is an approximation for small $\delta t$ should be clear to your reader. If not you can say so.

Title: Symbol for very small variable

Body:
Is there any representation to state that a variable is close to (not equal) zero? Let me give you an example. Consider the function

$u(x)=\alpha (e^{i\omega \delta t}-1) f(x)$

I am interested in the function $u(x)$ when $\delta t$ is very small. For this case, it should be easy to see that

$u(x)|_{\text{small }\delta t} \approx i \alpha \omega \delta t f(x)$

Is there any "nice" notation to represent such an equation (without having to write small)? I thought that I could use the limit notation for that [for example, $\lim_{\delta t \to 0} u(t)$], but then I realized that if $\delta t$ goes to zero, then $u(x)=0$. Therefore, it is not what I need.

I found this link in the same forum, but it did not help.


问:
Calculate KKT point - simple example
Problem: $\min φ(x)$ subject to $l ≤ x ≤ u$ - how do I find the KKT point?

I reformulated the problem so that we now min φ(x) subject to $x-l ≥ 0$ and $-x+u ≥ 0$, which means the lagrangian $L = φ(x) - λ_1(x - l) - λ_2(-x + u)$. How do I proceed from here?

答:
Once you have formed your Lagrangian $\mathcal{L}(x, \lambda)$, you need to write the KKT conditions:

$\frac{\partial \mathcal{L}}{\partial x} = 0$ (stationarity)

$x - l \ge 0$, $u - x \ge 0$ (primal feasibility)

$\lambda_1 (x - l) = 0$, $\lambda_2 (u - x) = 0$ (complementarity)

$\lambda \ge 0$ (dual feasibility)

Title: Calculate KKT point - simple example


问:
Intuition for spherical coordinates of a highly symmetric point
Consider the sphere $\{(x,y,z)\in{\mathbb{R}^3}:x^2+y^2+z^2=1\}$ and the point on the sphere $(x,y,z)=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$. Intuitively, I thought "oh, that point lies on the "dead center" of the sphere as far the first octant is concerned", so I thought in spherical coordinates the same point should be $\left(r,\theta,\varphi)=(1,\frac{\pi}{4},\frac{\pi}{4}\right)$, but later when I was getting some inconsistent calculations I came back to this statement and when I applied the formal conversions:
$$x(r,\theta,\varphi)=r\sin(\theta)\cos(\varphi)$$
$$y(r,\theta,\varphi)=r\sin(\theta)\sin(\varphi)$$
$$z(r,\theta)=r\cos(\theta)$$
I got that the point $(r,\theta,\varphi)=\left(1,\frac{\pi}{4},\frac{\pi}{4}\right)$ should actually be $(x,y,z)=\left(\frac{1}{2},\frac{1}{2},\frac{1}{\sqrt{2}}\right)$, and this has been driving me insane! Either there's some really silly arithmetic error I'm making or my geometric intuition is failing me. Which one is it?

答: Your intuition is failing you.  If you do the conversion, the point $(\frac 1{\sqrt 3},\frac 1{\sqrt 3},\frac 1{\sqrt 3})$ you first need $\cos (\theta) =\frac 1{\sqrt 3}$, which says $\theta\approx 0.955$.  You do have $\phi=\frac \pi 4$


答:
Your intuition is wrong. $\cos\left(\frac{\pi}4\right) = \frac{1}{\sqrt{2}}$.

The correct angle here is $\arccos\left(\frac 1{\sqrt{3}}\right)$ which does not have a closed form solution I am aware of. If $\cos(\theta) = \frac 1 {\sqrt{3}}$, then $\sin{\theta} = \sqrt{\frac{2}{3}}$. Thus $\sin(\varphi )= \frac 1{\sqrt{2}}$. Thus $\varphi = \frac \pi4$.

Title: Intuition for spherical coordinates of a highly symmetric point

Body:
Consider the sphere $\{(x,y,z)\in{\mathbb{R}^3}:x^2+y^2+z^2=1\}$ and the point on the sphere $(x,y,z)=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$. Intuitively, I thought "oh, that point lies on the "dead center" of the sphere as far the first octant is concerned", so I thought in spherical coordinates the same point should be $\left(r,\theta,\varphi)=(1,\frac{\pi}{4},\frac{\pi}{4}\right)$, but later when I was getting some inconsistent calculations I came back to this statement and when I applied the formal conversions:
$$x(r,\theta,\varphi)=r\sin(\theta)\cos(\varphi)$$
$$y(r,\theta,\varphi)=r\sin(\theta)\sin(\varphi)$$
$$z(r,\theta)=r\cos(\theta)$$
I got that the point $(r,\theta,\varphi)=\left(1,\frac{\pi}{4},\frac{\pi}{4}\right)$ should actually be $(x,y,z)=\left(\frac{1}{2},\frac{1}{2},\frac{1}{\sqrt{2}}\right)$, and this has been driving me insane! Either there's some really silly arithmetic error I'm making or my geometric intuition is failing me. Which one is it?


问:
Change of variable in Bertsekas nonlinear programming chapter 1.3
I'm trying to understand how a very simple change of variables works for the following gradient descent equation.
\begin{align}
x^{k+1} = x^k - \alpha^kD^k \nabla f(x^k)
\end{align}
This is supposedly viewed as a scaled version of steepest descent and they prove it by setting:
\begin{align}
S = (D^k)^{1/2}
\end{align}
They then use this to consider the transformation of variables
\begin{align}
x = Sy
\end{align}
The minimization problem gets rewritten as
\begin{align}
&\text{minimize } h(y) \equiv f(Sy) \\
&\text{subject to } y \in \mathbb{R}^n
\end{align}
And finally they say they can write this as follows:
\begin{align}
y^{k+1} = y^k - \alpha^k \nabla h(y^k)
\end{align}

Where does $D^k$ go? My work so far is simply plugging in $Sy$ for $x$ and then multiplying by the inverse as follows which gets me the following:
\begin{align}
Sy^{k+1} &= Sy^k - \alpha^k D^k \nabla f(Sy^k) \\
S^{-1}(Sy^{k+1}) &= S^{-1}(Sy^k - \alpha^k D^k \nabla h(y^k)) \\
y^{k+1} &= y^k - \alpha^k S^{-1}D^k \nabla h(y^k) \\
y^{k+1} &= y^k - \alpha^k D^{-1/2}D^k \nabla h(y^k) \\
\end{align}
Where do I find my other $D^{-1/2}$ to make $D^k$ go to 1?

Edit: Here is the solution for anyone who needs it:

We need the $D = S^2$ and $\nabla h(y^k) = S\nabla f(Sy^k)$

\begin{align}
Sy^{k+1} &= Sy^k - \alpha^k S^2 \nabla f(Sy^k) \\
Sy^{k+1} &= Sy^k - \alpha^k S \nabla h(y^k) \\
S^{-1}(Sy^{k+1}) &= S^{-1}(Sy^k - \alpha^k S \nabla h(y^k)) \\
y^{k+1} &= y^k - \alpha^k \nabla h(y^k) \\
\end{align}

答:
You need to write the gradient of $h(y) = f(Sy)$ wrt $y$:

$\nabla_y h(y) = S^T \nabla_x f(Sy) = S \nabla_x f(Sy)$ ($S$ seems to be symmetric?)

Could be where you had a missing $S$.


问:
Can you find a formula for the nth term of this particular sequence?
While I was solving a question, I had to find out the values in specific cases. I found the individual values as: 0,0,1,6,35 for n = 1,2,3,4,5 respectively. Is there anyway to represent this as a function f(n)   $\forall$ n $\in \mathbb{Z}$ that uses (n!)

答:
One way, out of many ways, to represent your function is
$$
f(n)=(n-1)!- (-\frac{1}{3}n^4 + \frac{19}{6}n^3 - \frac{32}{3}n^2+ \frac{89}{6}n - 6).
$$
So $f(1)=0!-1=0$, $f(2)=0$, $f(3)=1$, $f(4)=6$ and $f(5)=13$.

Furthermore $f(6)=169$.

Of course, we cannot know what you could do with an arbitrary formula.


问:
Does $\sum_{k=0}^\infty t^{ak} $ converges if $0<t<1$ and $a>0$?
Let $t\in(0,1)$ and $a>0$. Does it holds that
$$\sum_{k=0}^\infty t^{ak} <\infty?$$

If $a>1$, then each $t^{ak}<t^k$. And so $\sum_{k=0}^\infty t^{ak}<\sum_{k=0}^\infty t^{k}=\frac{1}{1-t}<\infty$. What if $0<a<1$?

Body:
$0<t<1$ and $a>0$ so $t^{a} < 1$. Hence,

$$\sum_{k=0}^{\infty}t^{ak}={1 \over 1-t^a} < \infty$$


问:
Prove by induction that for all $n \geq 3$: $n^{n+1} > (n+1)^n$
I am currently helping a friend of mine with his preperations for his next exam. A big topic of the exam will be induction, thus I told him he should practice this a lot. As at the beginning he had no idea how induction worked, I showed him some typical examples.

Now he showed me an exercise he was having trouble with, which states that one should prove
$$n^{n+1} > (n+1)^n$$
for all $n \geq 3$. I have to admit that I too have trouble showing this inequality, as in all of my attempts, my lower bound is too low. Also, I have not yet figured out, how one gets the $(n+2)^{n+1}$, primarely the number $2$ is a problem. I think this might be solved using the binomial theorem, however, I don't think they have already seen the binomial theorem in school.

Is there an easy method to show this inequality by induction not using the binomial theorem? If no: How can one show it using the binomial theorem?

Thanks for answers in advance.

答:
The premise of the question is incorrect. This is not a problem where integer induction is useful for seeing or proving the truth of the statement.

In one form, the problem is to show that $n^{1/n} > {(n+1)}^{1/{(n+1)}}$.  This has the inductive $n \to (n+1)$ pattern, but is better understood as the statement that $f(x) = \frac{\log x}{x}$ is decreasing for all real $x \geq 3$ (where the meaning of $3$ is "anything $\geq e$") which is most apparent by differentiation, $f' = \frac{1- \log x}{x^2}$.

In another form, $n > {(1 + \frac{1}{n})}^{n}$, the problem is essentially a request to prove that $n > e = 2.718... $ for all $n \geq 3$, if you allow the fact that $e_n = {(1 + \frac{1}{n})}^{n}$ is an increasing sequence converging to $\hskip2pt e \hskip2pt$ from below.  The fact does not require induction, it is another statement about real $n$ and can be proved most easily by computing $\frac{d}{dn}\log e_n$.

A variant of the second form avoids the need to know an accurate value for $e$, or to mention $e$ at all, by using the decreasing sequence of upper bounds $ E_k = {(1 + \frac{1}{k})}^{k+1}$, and selecting a value of $k$ such that $E_k < 3$  (see the recently posted solution by robjohn with $k=6$).  The decrease of $E_k$ is again shown by differentiation of $\log E_k$ with respect to real $k$.  In this view the problem is (essentially) asking to prove that $e < 3$, which is not fundamentally an inductive statement about any function of $n$.


问:
Taylor series for $\sqrt[3]{x+1}$
I'm trying to find the Taylor series for $\sqrt[3]{x}$, but since the n-th derivative of the function $$f(x)=\sqrt[3]{x}$$
not definded at $x=0$, I've swithched to $f(x)=\sqrt[3]{x+1}$, This is my Attempt:

$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)x^n}{n!}$$
$$\sqrt[3]{x+1}=1+\frac{x}{3}-\frac{2x^2}{9\cdot2!}+\frac{10x^3}{27\cdot 3!}-\frac{80x^4}{81\cdot4!}+...$$

but when i plug $x=125 \iff \sqrt[3]{x+1}=\sqrt[3]{126}$ i get something way bigger than the acttual value which is about $5.01...$, you can check it here.

But I can't find the mistake in the equation.

答:
For $x\gt 1$, expand in terms of $\frac1x<1$ to ensure convergence

$$\sqrt[3]{x+1}= \sqrt[3]{x} \sqrt[3]{1+\frac1x}
= \sqrt[3]{x}  (1+\frac{1}{3x}-\frac{2}{9\cdot2!}\frac1{x^2}+\frac{10}{27\cdot 3!}\frac1{x^3}-\frac{80}{81\cdot4!}\frac1{x^4}+...)
$$

Then, set $ x=125$ to obtain the approximation
$$\sqrt[3]{126} =5(1+\frac1{3\cdot 125}-\cdots )=5.0133$$

Title: Taylor series for $\sqrt[3]{x+1}$


问:
Finding least-norm solutions of a linear Diophantine equation
I am looking for an algorithm solving the following problem.

Let $\Vert x\Vert_{*} := \sqrt{x^\mathrm{T} D \,x}$, where $D$ is a diagonal matrix with positive diagonal entries.
Given $a\in \mathbb{N}^n$, $b \in \mathbb{N}$, find
$$\arg \min_{x \in \mathbb{N}^n} {\vert a^{\mathrm{T}} x - b\vert}$$
with minimal $\Vert x \Vert_{*}$.

Motivation

I am creating a tools website for tabletop role-playing game (TTRPG) players. In these games, players fight together against monsters, and one player (called "GM") has to ensure these fights are of appropriate difficulty. This is done using "experience points" (XP). XP is used to estimate the strength of monsters. Each monster awards a set amount of XP (tougher monsters award more XP) and each difficulty has a target XP number. When the players defeat the monsters they gain these XP, becoming stronger as a result. If the GM selects a few monsters, how many of each do we need to ensure a good battle?

For example, the GM selects goblins, hobgoblins, and bugbears (who award 1,2, and 4 XP) to be his monsters. As threshold they select 12 XP. We would have $a^T=[1,2,4], b=12.$ $x$ describes the number of monsters picked. It is clear that there are many ways to select monsters, but most of them are undesirable. We don't want a fight against just 12 goblins, or only 3 bugbears. My idea is giving each monster a somewhat arbitrary weight: $D=diag(1,6,12)$. With this we obtain the optimal $x^T=[4,2,1]$ (4 goblins, 2 hobgoblins and 1 bugbear), a well balanced encounter.
If the GM selects creatures such that we can't meet the threshold, we want to get as close as possible to it.

Why don't I use a greedy algorithm? Because I thought it would be cool and interesting not to.

答: You seem to be juggling two criteria (getting close to the threshold and minimizing the weighted norm of $x$). There are several ways you could approach this via an optimization model. One would be to minimize a weighted combination of $\Vert x \Vert_*^2$ and $\vert a^Tx - b\vert$. (I squared the norm so that the objective function is quadratic and convex.) Another would be to minimize the squared norm subject to an arbitrary limit on the absolute deviation. Alternatively, you could minimize the absolute deviation subject to a limit on the squared norm. All three approaches should lead to tractible problems.


问:
How to solve $\int\frac{dx}{(x^2 - 8x + 1)}$?
How to solve  $\int\frac{dx}{(x^2 - 8x + 1)}$?

I can't spread it into partial fractions. I can't use u-substitution. And you can't do integration by parts. So, I'm wondering how to do it. Thanks for the help.

答:
You can use partial fractions.  It isn't necessary that the constants be integers for it to work.
$$\frac{1}{x^2-8x+1} = \frac{1}{(x-4-\sqrt{15})(x-4+\sqrt{15})}$$
or if you prefer to substitute $u = x-4$ first:
$$\frac{1}{u^2-15} = \frac{A}{u-\sqrt{15}} + \frac{B}{u+\sqrt{15}}$$


答:
When there are no obvious roots, you can complete the square to arrive either

to $\displaystyle\int \dfrac {du}{1+u^2}=\arctan(u)$

or to $\displaystyle\int \dfrac{du}{1-u^2}=\operatorname{argth}(u)$

In the present case $x^2-8x+1=(x-4)^2-15$, so set $u=\dfrac{x-4}{\sqrt{15}}$ and check you get the second form above.

Body:
How to solve  $\int\frac{dx}{(x^2 - 8x + 1)}$?

I can't spread it into partial fractions. I can't use u-substitution. And you can't do integration by parts. So, I'm wondering how to do it. Thanks for the help.


问:
How to understand cluster points?
We talked about cluster points yesterday in class and I am now very confused. I know the formal definition of the limit: $a$ is a cluster point, if for every $\epsilon > 0$, there exist infinitely many natural numbers $n ∈ ℕ$ such that $|a_n -a|<\epsilon$

Okay. I think I sort of understand this intuitively. If we have some sort of open interval around $a$, then a is a cluster point because we can fit infinitely many elements from a set or sequence into that interval no matter how small it is ($\epsilon$ is). The integer set does not contain any cluster points because it every element is isolated. Okay cool.

But I still cannot do this:

$1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, ... .$

Shouldn't there be no cluster points? If I draw an open interval around 1, there are no other elements from the sequence except 1 that I can fit inside it since every elements of the sequence are integers.

Consider:

In this case, I think $1$ and $-1$ are the cluster points. I don't think $0$ can be a cluster point for that I can't construct a small interval to include elements from the sequence.

and what happens if we have a something like this

my teacher gave an explanation of the definiton of cluster point and that's it :(

答:
Some of what partially answers the question is given in the comments, and I might later try to expand what follows to include concerns discussed there. However, for now I will only discuss cluster points for the third sequence. I will show in great detail (for expository purposes) that the set of all cluster points of the third sequence is $\{-\frac{1}{2},\,0\}.$

For cluster points we don't care what $a_1$ is, so let's ignore $a_1.$ Letting $k=1,2,3,\ldots$ we get

$$a_2 = 0 \;\; \text{and} \;\; a_4 = -\frac{1}{4} \;\; \text{and} \;\; a_6 = -\frac{2}{6} \;\; \text{and} \;\; a_8 = -\frac{3}{8} \;\; \text{and} \;\; a_{10} = -\frac{4}{10} \;\; \text{and} \;\; \cdots $$

(fractions not reduced so that the pattern is easy to see), and we get

$$a_3 = \frac{\sin 3}{\sqrt 3} \;\; \text{and} \;\; a_5 = \frac{\sin 5}{\sqrt 5} \;\; \text{and} \;\; a_7 = \frac{\sin 7}{\sqrt 7} \;\; \text{and} \;\; a_9 = \frac{\sin 9}{\sqrt 9} \;\; \text{and} \;\; \cdots $$

It is clear that $-\frac{1}{2}$ is a cluster point, because given any open interval about $-\frac{1}{2},$ we have infinitely many (in fact, all but finitely many) of the even-numbered terms in that interval. The assertion about open intervals is because the limit of the even-numbered exists and is equal to $-\frac{1}{2}.$

It is clear that $0$ is a cluster point, since the limit of the odd-numbered terms exists and is equal to $0.$ The limit is $0$ because the odd-numbered terms are fractions with numerators between $-1$ and $1$ and denominators that become (and remain) arbitrarily large.

Therefore, $\{-\frac{1}{2},\,0\}$ is a SUBSET of the set of all cluster points of the third sequence. To show that $\{-\frac{1}{2},\,0\}$ is EQUAL to the set of all cluster points of the third sequence, I will show that no other real number is a cluster point of the third sequence.

To this end, let $x$ be a real number such that $x \neq -\frac{1}{2}$ and $x \neq 0.$ I will show that $x$ is NOT a cluster point of the third sequence. To show that a “for all $\epsilon > 0$ something” statement is false (note that the definition of “$x$ is a cluster point” is a “for all $\epsilon > 0$ something” statement), it suffices to show that the corresponding “there exists $\epsilon > 0$ not-something” statement is true. Specifically, I will show there exists $\epsilon > 0$ such that it is false that infinitely many terms of the sequence are a distance of at most $\epsilon$ from $x.$

Intuitive idea for what follows: Because the even-numbered terms converge to $-\frac{1}{2}$ and the odd-numbered terms converge to $0,$ it follows that the terms will get closer and closer to these two numbers (and no other numbers; I'll show this below), which implies that the terms will NOT get (and remain) arbitrarily close to $x,$ because the minimum of the distances of $x$ from these two numbers is positive. Thus, even we imagine $x$ to be very close to one of these two numbers, what happens for terms far enough out in the sequence is that each of the terms will be super-close to $-\frac{1}{2}$ or $0,$ so close in fact that $x$ will seem to be far off in the distance like a distant star compared to a person standing next to you on your left or a person standing next to you on your right. (In this analogy, the distant star is $x,$ the person who could be next to you is $-\frac{1}{2}$ or $0,$ and you is any term that is far enough out in the sequence.)

Let $d$ be the minimum of the distances of $x$ from $-\frac{1}{2}$ and $0.$ That is, let $d = \min \left\{\left|x - \left(-\frac{1}{2}\right)\right|,\;\left|x - 0\right|\right\}.$ Since $x$ is not equal to $-\frac{1}{2}$ or $0,$ it follows that $d > 0.$ Now let $\epsilon = \frac{d}{2}$ and note that $\epsilon > 0.$ Since the even-numbered terms converge to $-\frac{1}{2},$ there exists a positive integer $N_1$ such that at most $N_1$ many of the even-numbered terms are NOT less than a distance of $\epsilon$ from $-\frac{1}{2}.$ Since the odd-numbered terms converge to $0,$ there exists a positive integer $N_2$ such that at most $N_2$ many of the odd-numbered terms are NOT less than a distance of $\epsilon$ from $0.$ It follows that all but at most $N_1 + N_2$ many of the terms (whether even-numbered or odd-numbered) are a distance of less than $\epsilon$ from one of the two numbers $-\frac{1}{2}$ and $0.$

Therefore, all but at most $N_1 + N_2$ many of the terms are a distance of more than $\epsilon$ from $x,$ since anything a distance less than $\frac{d}{2}$ from one of the two numbers will be a distance of more than $\frac{d}{2}$ from $x.$ To see this geometrically, draw a number line, mark points for $-\frac{1}{2}$ and $0,$ draw two open intervals of radii $\frac{d}{2}$ whose centers are these two points (these two open intervals may overlap, but that won't matter), and notice that $x$ is a distance of more than $\frac{d}{2}$ from any point within either of these two intervals. To see this algebraically, let $z$ represent either $-\frac{1}{2}$ or $0,$ and let $y$ be a number less than a distance of $\frac{d}{2}$ from $z$ $(y$ represents an appropriate term of the sequence, and we have $|y - z| < \frac{d}{2}),$ and note that if we assume (for a later contradiction) $|x - y| \leq \frac{d}{2},$ then by the triangle inequality we have

$$|x - z| \; \leq \; |x - y| + |y - z| \; \leq \; \frac{d}{2} + |y - z| \; < \; \frac{d}{2} + \frac{d}{2} \; = \; d,$$

which contradicts the fact that $|x - z| \geq d$ (i.e. the distance from $x$ to either of the numbers $-\frac{1}{2}$ or $0$ is at least $d),$ and hence $|x - y| \leq \frac{d}{2}$ must be false, and thus we have $|x - y| > \frac{d}{2}$ (i.e. the sequence term $y$ is a distance of more than $\frac{d}{2}$ from $x).$

Thus, to repeat what the stuff just above shows, we know that all but at most $N_1 + N_2$ many of the terms are a distance of more than $\epsilon$ from $x.$ That is, only finitely many of the terms are a distance of less than $\epsilon$ from $x,$ and hence it is false that infinitely many of the terms are a distance of less than $\epsilon$ from $x,$ which implies that $x$ cannot be a cluster point of the third sequence.

Generalization: The argument above can be adapted to prove that if a sequence is a finite union of sequences, each of which converges, then the set of all cluster points of the original sequence is the set whose elements are the limits of those finitely many sequences.

Title: How to understand cluster points?


Title: How to solve $\int\frac{dx}{(x^2 - 8x + 1)}$?


问:
Integral inequality (Divergence theorem)
I'm trying to prove the following inequality:

$$2 \int_U |\nabla \phi|^2 dx \leq \int_U \phi^2 dx + \int_U |\Delta \phi|^2 dx$$

where $U \subset \mathbb{R}^n$ is bounded and open and $\phi \in C^\infty_c(U)$. I actually think I have managed to prove this just using one of Green's identities $$\int_UD \phi \cdot D \phi dx = -\int_U \phi \Delta \phi dx + \int_{\partial U}\frac{\partial \phi}{\partial \nu} \phi dx$$
(which comes from the divergence theorem), and then using the fact that $\phi = 0$ on $\partial U$. This then gives:

$$\int_U |\nabla \phi|^2 dx = -\int_U \phi \Delta \phi dx \leq \int_U |\phi| |\Delta \phi| dx \leq \int_U \frac{\phi^2}{2}+\frac{|\Delta \phi|^2}{2}$$ where the last inequality comes from Cauchy's inequality $ab \leq a^2/2 + b^2/2$.

Does this seem correct? The problem is that I have been given a hint which says to use the fact that $\nabla \cdot(\phi \nabla \phi) = |\nabla \phi|^2 + \phi \Delta \phi$ and I'm not sure how to use this hint. Would this just lead to an alternative proof of the inequality?

答: You already used it :) That's where the Green's identity came from! </p>


问:
Find expressions for the mean and variance of X.
We say that $X$ is a discrete uniform random variable if $X$ takes any of the values
$1, 2, . . . , n$ equally likely. That is, $P[X = i] = 1/n$ for $i = 1, . . . , n$.

Find expressions for
the mean and variance of X.

My attempt:

$E[X] = \sum{xp(x)} =\sum_{i=1}^n{ i×1/n} = \sum_{i=1}^n{i/n}$

$Var[X] = \frac{\sum_{i=1}^n(i-\sum_{i=1}^n{i/n})^2}{n}$

Is my attempt correct, how do I continue expressing $Var[X]$?

Body:
$$\mathbb{E}[X]=\frac{1}{n}\sum_{i=1}^n i=\frac{n+1}{2}$$

$$\mathbb{V}[X]=\mathbb{E}[X^2]-\mathbb{E}^2[X]$$

that is

$$\mathbb{V}[X]=\frac{1}{n}\sum_{i=1}^n i^2-\left(\frac{n+1}{2}\right)^2=$$

$$=\frac{n(n+1)(2n+1)}{6n}-\left(\frac{n+1}{2}\right)^2=...$$


问:
Manageable project to learn some arithmetic geometry
OK, this may be an unusual question, since I'm asking you for advice, the way I would ask a supervisor were I a grad student.

I am a thirty-something maths teacher with an unresolved attraction to arithmetic geometry, and would like to know more about themes I find fascinating when I read texts about arithmetic geometry, such as:

the local-global principle

the way geometry influences the arithmetic behaviour of Diophantine equations, like in Faltings's theorem

the way scheme theory helps number theory by making a lot of geometric constructions possible, starting from base change

One of the reasons this attraction is unresolved is that I never managed to seriously learn enough algebraic number theory, algebraic geometry, etc. in a "bottom-to-top" way: I have often started to read for instance R. Vakil's The rising sea, D. Eisenbud & J. Harris The geometry of schemes or Q. Liu's Algebraic geometry and arithmetic curves  and find them to be very good texts, but I don't have the stamina to ingest hundreds of pages without a more specific goal.

Because of this there are a number of subjects I'm vaguely familiar with, but don't know in any depth. For instance, I know what a scheme is, I even think I understand in part their raison d'être, but am probably unable to pass an scheme theory exam (and the same goes for elliptic curves, derived categories, class field theory...).

To change this, I would like to try and understand a specific result and working my way from the top down to acquire the needed skills. The result has to be interesting enough to preserve my motivation in the long run, but "small" enough so that I could realistically understand it in a few months. I would love it to really mix number theory and geometry.

So the question is:

what theorem would constitute a reasonable-size project?

Personal facts that might be relevant: my PhD is in low-dimensional topology, so I'm quite comfortable with things like cohomology and Riemann surfaces, and I like it when geometric intuition is useful. Large doses of homological algebra or category theory don't frighten me.

答:
You could have a look at Lenstra's book Galois Theory for Schemes. It's only a bit over 100 pages, is very well-written, and contains tons of nice exercises. It builds up to the beautiful and useful result that any connected scheme $X$ has an etale fundamental group $\pi = \pi_1^{et}(X,x)$, and the category of finite etale covers of $X$ is equivalent to finite sets on which $\pi$ acts continuously. This is about as arithmetic geometric as I can imagine. Many examples are given. There is a deep relation to the topological fundamental group/algebraic topology, which is explained in the first chapter. The prerequesites are clearly laid out at the end of the introduction.

If this doesn't convince you, maybe Lenstra will, as the very first sentence of the book states, "One of the most pleasant ways to familiarize oneself with the basic language of abstract algebraic geometry is to study Galois theory for schemes."


问:
Finding a volume integral in an ellipsoid
I am trying to find the volume integral of $\rho=\rho_{0}\left(\frac{R^{2}-r^{2}}{R^{2}}\right)$ inside an ellipsoid given by
$\frac{x^{2}}{(3 R)^{2}}+\frac{y^{2}}{(4 R)^{2}}+\frac{z^{2}}{(5 R)^{2}}=1$

I've tried using jacobian to move from an ellipsoid to an unit ball by these mapping relationships

$x=3 R u$,
$y=4 R v$,
$z=5 R w$

But the resulting integral is still heavy

$\int\rho_0\left(1-\left(9 u^{2}+16 w^{2}+25 w^{2}\right)\right) 60 dudvdw$

Does anyone have any insight to a more elegant way.

Thank you

答:
The resulting integral in this case is actually very simple due to symmetry:
\begin{align}
\int_Bu^2\,dV = \int_Bv^2\,dV = \int_Bw^2\,dV=\frac{1}{3}\int_B(u^2+v^2+w^2)\,dV = \frac{1}{3}\int_0^1 r^2\,4\pi r^2\,dr =\frac{4\pi}{15}.
\end{align}

This symmetry should be intuitive enough. Otherwise use the change of variables theorem to permute $(u,v,w)$ and the fact that the unit ball remains invariant under this change of coordinates, and that the absolute value of the determinant of this is $1$ if you want a really rigorous proof.


答: The sphere intersects the $x$, $y$, and $z$ axis at $(1, 0, 0)$, $(0, 1, 0)$, and $(0, 0, 1)$. To see if a point is in the "dead center" you can find the distance to each of these points and see if they are the same. For $(\frac{1}{2}, \frac{1}{2}, \frac{1}{\sqrt{2}})$, the distances are $1$, $1$, $\sqrt{2-\sqrt{2}}$, so it isn't in the "dead center". However, for $(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$, the distances are all $\sqrt{2-\frac{2}{\sqrt{3}}}$.


答:
You're right of course that the point $\left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$ is symmetric with respect to permutations of the $x,y,z$ axes, and so e.g. lies on the line of symmetry about which you can rotate $x,y,z$. But you're neglecting the fact that $(r,\theta,\varphi)$ coordinates do not transform nicely under rotations that move the $z$-axis. So you cannot reason that the angle $\theta$ must take some special value by symmetry.

You can reason this way for $\varphi$; considering the fact the point is unchanges by reflection in the plane $x = y$ guarantees that $\varphi \equiv \pi/2 - \varphi$ is either $\pi/4$ or $5\pi/4$. Clearly it's the former as it lies in the first $(x,y)$ quadrant.

But $\theta$ must be expressed differently. Note that $\theta$ is the angle made with the $z$ axis, which can equivalently be expressed in terms of the angle with the $(x,y)$ plane. Since $$\sqrt{x^2+y^2} = \sqrt{2} \times \frac{1}{\sqrt{3}} \qquad \text{ but }\qquad  \sqrt{z^2} = \frac{1}{\sqrt{3}}$$
we see that in an appropriate sense you should think of your point as being "$\sqrt{2}$ times more in the $(x,y)$ plane than in the $z$ direction". This is made precise by noting
$$ \theta = \arctan\frac{\sqrt{x^2+y^2}}{\sqrt{z^2}} = \arctan \sqrt{2} $$
rather than $\arctan 1 = \pi/4$. (This value of $\theta \approx 54.7^\circ$ is the so-called magic angle.)

So to correct your intuition, I think you should internalize the idea that we are in 3 dimensions; the point has equal magnitude in each direction; and $\theta$ reflects how close you are to one of three possible directions; hence, since only one third of the magnitude [squared] is concentrated in that particular direction, the angle made $\theta$ will be larger than $\pi/4$.


答:
I recommend Silverman's The Arithmetic of Elliptic Curves, especially since you already have an understanding of compact Riemann surfaces.

(Or, phrased in terms of results: you should try to learn the complete proof of the Mordell-Weil theorem for Elliptic Curves.)

One caveat is that he avoids scheme language, and so some definitions are rather ad hoc in terms of Weierstrass equations. This is an advantage in that it makes the heart of the research on elliptic curves accessible more quickly, but a disadvantage for your particular use case of motivating scheme theory.

Title: Manageable project to learn some arithmetic geometry

Body:
OK, this may be an unusual question, since I'm asking you for advice, the way I would ask a supervisor were I a grad student.

I am a thirty-something maths teacher with an unresolved attraction to arithmetic geometry, and would like to know more about themes I find fascinating when I read texts about arithmetic geometry, such as:

the local-global principle

the way geometry influences the arithmetic behaviour of Diophantine equations, like in Faltings's theorem

the way scheme theory helps number theory by making a lot of geometric constructions possible, starting from base change

One of the reasons this attraction is unresolved is that I never managed to seriously learn enough algebraic number theory, algebraic geometry, etc. in a "bottom-to-top" way: I have often started to read for instance R. Vakil's The rising sea, D. Eisenbud & J. Harris The geometry of schemes or Q. Liu's Algebraic geometry and arithmetic curves  and find them to be very good texts, but I don't have the stamina to ingest hundreds of pages without a more specific goal.

Because of this there are a number of subjects I'm vaguely familiar with, but don't know in any depth. For instance, I know what a scheme is, I even think I understand in part their raison d'être, but am probably unable to pass an scheme theory exam (and the same goes for elliptic curves, derived categories, class field theory...).

To change this, I would like to try and understand a specific result and working my way from the top down to acquire the needed skills. The result has to be interesting enough to preserve my motivation in the long run, but "small" enough so that I could realistically understand it in a few months. I would love it to really mix number theory and geometry.

So the question is:

what theorem would constitute a reasonable-size project?

Personal facts that might be relevant: my PhD is in low-dimensional topology, so I'm quite comfortable with things like cohomology and Riemann surfaces, and I like it when geometric intuition is useful. Large doses of homological algebra or category theory don't frighten me.


问:
Newton Binomial for number of integer solutions to $|x|+|y|+|z|=30$
Following my post here: Newton Binomial to find $(x_1+x_2+x_3)(x_4+x_5+x_6+x_7)=77$

I would like to know the number of integer solutions to |x|+|y|+|z| = 30. Without the absolute value I get 32 choose 30 (496). With the absolutes, I get 3602 (by computer simulation) and I don't know why...

答:
The formula for the number of positive solutions to the equation
$$
x_1+x_2+\dots+x_k=n
$$
reads
$$
\binom {n-1}{k-1}.
$$
Therefore the number your are looking for is:
$$
2^3\binom{29}2+\binom312^2\binom{29}1+\binom322^1\binom{29}0=3602,
$$
where the terms count contributions with the number of zeros in the set $\{a,b,c\}$ being $0,1$ and $2$, respectively.


问:
Prove that the sum $2^0+2+2^2+\cdots+2^{p-1}(2^p-1)$ is $2^{p-1}(2^p-1)$
Attempt: In an attempt to find the sum of the series given that $2^{p-1}$ is prime,

$$1+2+2^2+\cdots+2^{p-2}+2^{p-1}+(2^p-1)+2(2^p-1)+2^2(2^p-1)+\cdots+2^{p-1}(2^p-1)$$

A teacher grouped elements as follows,

$$(1+2^p-1)+2(1+2^p-1)+2^2(1+2^p-1)+\cdots+2^{p-2}(1+2^p-1)+2^{p-1}$$

Question: I am confused a little about the last two terms in the summation above as what I think it should be is as follows,

$$(1+2^p-1)+2(1+2^p-1)+2^2(1+2^p-1)+\cdots+2^{p-2}(1+2^p-1)+2^{p-1}(1+2^p-1)$$

Edit:
Question: I should get the sum $2^{p−1}(2^p − 1)$, but with the way I wrote it I did not get the sum wanted, but I got $2^p(2^p-1)$ for,

$$(1+2^p-1)+2(1+2^p-1)+2^2(1+2^p-1)+\cdots+2^{p-2}(1+2^p-1)+2^{p-1}(1+2^p-1)$$

Edit 2: The only solution to this is that the sum is $2^p(2^p-1)$.

答:
Your grouping method is correct. Your teacher seems to have made a mistake.

From your sum you get
$$(1+2^p-1)+2(1+2^p-1)+2^2(1+2^p-1)+\cdots+2^{p-2}(1+2^p-1)+2^{p-1}(1+2^p-1)$$

which is equal to
$$(2^p)+2(2^p)+2^2(2^p)+\cdots+2^{p-2}(2^p)+2^{p-1}(2^p)=$$
$$2^p(2+2^2+...+2^{p-1})$$

Now, you only have to solve

$$2+2^2+...+2^{p-1}$$

and multiply by $2^p$ and get your answer. Is that clear?


答:
Your last integral is not that heavy.

The constant $1$ yields the volume of the unit sphere. Then using spherical coordinates, the integral of $w^2$ is

$$\int_0^{2\pi}\int_0^\pi\int_0^1 r^2\cos^2\phi\,r^2\sin\phi\,dr\,d\phi\,d\theta=2\pi\frac23\frac15.$$

By symmetry, the integrals of $u^2$ and $v^2$ are equal.

Title: Finding a volume integral in an ellipsoid

Body:
I am trying to find the volume integral of $\rho=\rho_{0}\left(\frac{R^{2}-r^{2}}{R^{2}}\right)$ inside an ellipsoid given by
$\frac{x^{2}}{(3 R)^{2}}+\frac{y^{2}}{(4 R)^{2}}+\frac{z^{2}}{(5 R)^{2}}=1$

I've tried using jacobian to move from an ellipsoid to an unit ball by these mapping relationships

$x=3 R u$,
$y=4 R v$,
$z=5 R w$

But the resulting integral is still heavy

$\int\rho_0\left(1-\left(9 u^{2}+16 w^{2}+25 w^{2}\right)\right) 60 dudvdw$

Does anyone have any insight to a more elegant way.

Thank you


问:
Evaluating $\frac{\Gamma(m+b)}{\Gamma(m+a)}\sum_{k=0}^{\infty}\frac{\Gamma(k+m +a)}{\Gamma(k+m+b)}(k+m)$
I'm trying to solve the following sum for a project. Many thanks in advance for answering the question!

$$\frac{\Gamma(m+b)}{\Gamma(m+a)}\sum_{k=0}^{\infty}\frac{\Gamma(k+m+a)}{\Gamma(k+m+b)}(k+m)$$

where $m$ is an integer but $a$ and $b$ can be any real numbers.

答:
Mathematica says (after slight massaging):

$$m \, _2F_1(1,a+m;b+m;1)+(a+m) \Gamma (b+m) \,
   _2\tilde{F}_1(2,a+m+1;b+m+1;1)$$


问:
On the derivative of $\phi(u) = {}^tu'u'$
I am reading Pseudo-differential Operators and the Nash-Moser Theorem by Gerard and Alinhac. I am confused with one of the results stated (but not proven) in the book.

On page 133, given $u : M \to \mathbb{R}^n$ (where $M$ is a $C^\infty$ compact Riemannian manifold), the book defines (for a quadratic form $g = \sum g_{ij}(x) \mathrm{d}x_i \mathrm{d}x_j$):
$$
g_{ij} = \frac{\partial u}{\partial x_i} \cdot \frac{\partial u}{\partial x_j}
$$
and abbreviates $g = {}^tu'u'$ (alternatively, I believe one may view $g$ as an $n \times n$ matrix). On page 140, the book then defines:
$$
\phi(u) = {}^tu'u'
$$
I have two questions here.

Below the definition, the book writes $\phi'(u)v = {}^tu'v' + {}^tv'u'$, where . I'm not sure how they obtained this result. I presume they are computing the derivative of $\phi(u)$ then apply it to the vector $v$. One may refer to page 135 for the precise definition of $\phi'$.

I would like to compute the second derivative $\phi''(u)(v,w)$, but I'm not sure how to proceed.

Any help is appreciated.

答:
This turns out to be rather simple calculations, with the fact I didn't realise the workings is because I didn't realise the author stated a different definition of the derivative (on page 135), which is:
\begin{align*}
    \Phi^{(k)}(u) \cdot (v_1,\dots,v_k) = \left.\frac{\partial^k}{\partial t_1 \cdots \partial t_k}\Phi(u + t_1v_1 + \cdots + t_kv_k)\right|_{t_1 = \cdots = t_k = 0}
\end{align*}
Then:
\begin{align*}
    \phi'(u)v &= \left.\frac{\partial}{\partial t}\phi(u + tv)\right|_{t=0} \\
    &= \left.\frac{\partial}{\partial t}({}^t(u + tv)'(u + tv)')\right|_{t=0} \\
    &= \left.\frac{\partial}{\partial t}({}^tu'u' + t({}^tu'v' + {}^tv'u') + t^2({}^tv'v'))\right|_{t=0} \\
    &= {}^tu'v' + {}^tv'u'
\end{align*}
and for the second derivative:
\begin{align*}
    \phi''(u)(v,w) &= \left.\frac{\partial^2}{\partial t_1\partial t_2}\phi(u + t_1v + t_2w)\right|_{t_1 = t_2 = 0} \\
    &= \left.\frac{\partial^2}{\partial t_1\partial t_2}{}^t(u + t_1v + t_2w)'(u + t_1v + t_2w)'\right|_{t_1 = t_2 = 0} \\
    &= \left.\frac{\partial^2}{\partial t_1\partial t_2}{}^t(u' + t_1v' + t_2w')(u' + t_1v' + t_2w')\right|_{t_1 = t_2 = 0} \\
    &= {}^tv'w' + {}^tw'v'
\end{align*}


问:
Continuously differentiable $LU$ factorization matrix
Hello I have the following problem and I hope you can help me:

Suppose the entries of $A(\epsilon)\in\mathbb R^{n\times n}$ are continuously differentiable functions of the scalar $\epsilon$. Assume that $A\equiv A(0)$ and all its principal sub matrices are nonsingular. Show that for sufficiently small $\epsilon$ the matrix $A(\epsilon)$ has an $LU$ factorization $A(\epsilon)=L(\epsilon)U(\epsilon)$ and that $L(\epsilon)$ and $U(\epsilon)$ are both continuously differentiable.

My attempt at reasoning has been as follows:

The fact that the matrix is continuously differentiable implies that each of its inputs are continuously differentiable, the fact that $ A \equiv A(0)$ I don't see what I can contribute, but the fact that each of its of its matrices is non-singular implies that the matrix has an $LU$ factorization and since to obtain the $LU$ factorization it is necessary to perform elementary operations on the matrix, the only thing that is done is to combine continuously differentiable functions, which is continuously differentiable.

But the problem is that if my reasoning were true it would work for every $\epsilon$, so I think that I am wrong. I would appreciate any help you could give me and in advance, thank you.

This is problem P3.2.2 from Golub's book matrix computations

答: Your method is correct. The reason for the small $\epsilon$ is that the non-singularity assumption will eventually fail. If a matrix is nonsingular, small perturbations of it are too, but once you drop the small, it's the Wild West.


问:
How to solve equation of motion of cosmological scalar field
I am told that equation for a scalar field is:

$$\ddot{\phi}+3 H \dot{\phi}+V^{\prime}(\phi)=0$$

where $H$ is defined as: $H=\dot{R} / R$

For a potential $V(\phi)=V_{0} \exp (-\lambda \phi)$, I am supposed to show that:

$$R(t)=R_{0}\left(\frac{t}{t_{0}}\right)^{\frac{2}{\lambda^{2}}}$$

and that

$$\phi(t)=\frac{2}{\lambda} \ln \left[\left(\frac{V_{0}}{2\left(6-\lambda^{2}\right)}\right)^{1 / 2} \lambda^{2} t\right]$$

I can verify that these solutions indeed satisfy the equation. How would I go about solving it directly though? (I am happy with complete solutions, but hints are also appreciated.)

Verification of the provided solution

Consider:

$$\dot{\phi}(t)=\frac{2}{\lambda} t^{-1}$$

$$\ddot{\phi}(t)=-\frac{2}{\lambda} t^{-2}$$

$$\dot{R}(t)=R_{0}\left(\frac{1}{t_{0}}\right)^{\frac{2}{\lambda^{2}}} \frac{2}{\lambda^{2}} t^{\frac{2}{\lambda^{2}}-1}$$

$$\frac{\dot{R}}{R} \dot{\phi}=\frac{R_{0}\left(\frac{1}{t_{0}}\right)^{\frac{2}{\lambda^2}} \frac{2}{\lambda^{2}} t^{\frac{2}{\lambda^{2}}-1}}{R_{0}\left(\frac{t}{t_{0}}\right)^{\frac{2}{\lambda^{2}}}} \frac{2}{\lambda} t^{-1}=\frac{4}{\lambda^{3}} t^{-2}$$

Then we can substitute into original equation:

$$\ddot{\phi}+\frac{3 \dot{R}}{R} \dot{\phi}-\lambda V_{0} e^{-\lambda \phi}=$$
$$=-\frac{2}{\lambda} t^{-2}+3 \frac{4}{\lambda^{3}} t^{-2}-\lambda \operatorname{V_0exp}\left(-\lambda \frac{2}{\lambda} \ln \left[\left(\frac{V_{0}}{2\left(6-\lambda^{2}\right)}\right)^{1 / 2} \lambda^{2} t\right]\right)$$

$$=-\frac{2}{\lambda} t^{-2}+\frac{12}{\lambda^{3}} t^{-2}-\lambda V_{0}\left[\left(\frac{V_{0}}{2\left(6-\lambda^{2}\right)}
\right)^{\frac{1}{2}} \lambda^{2} t\right]^{-2}$$

$$=-\frac{2}{\lambda} t^{-2}+\frac{12}{\lambda^{3}} t^{-2}-\lambda V_{0}\left(
\frac{V_{0}}{2\left(6-\lambda^{2}\right)}
\right)^{-1} \lambda^{-4} t^{-2}$$

$$=-\frac{2}{\lambda} t^{-2}+\frac{2 \lambda^{2}}{\lambda^{3}} t^{-2}=0$$

as wanted.

My thoughts:

I think I would be able to solve for $\phi$ if $H$ was constant. Would probably resort to writing $\phi$ as $\phi=\sum a_n x^n$ & then find $a_n$ coefficients, which is probably not the best way but I think it would work.

I am not used to solve ODEs where I have two functions to worry about ($\phi$ & $R$). Is this ODE a certain class of ODE which I should look up?

答:
You are missing one key equation to close the system.

A scalar field carries energy and its this energy that determines how fast the Universe expands. For a canonical scalar field the kinetic energy is $\frac{1}{2}\partial_\mu \phi\partial^\mu\phi$ which is just $\frac{1}{2}\dot{\phi}^2$ in a smooth Friedmann universe. The potential energy is $V(\phi)$ so the Friedmann equation (in Planck units: $8\pi G = 1$) is
$$H^2 = \left(\frac{d\log R}{dt}\right)^2 = \frac{1}{3} \rho_\phi = \frac{1}{3}\left(\frac{1}{2}\dot{\phi}^2 + V(\phi)\right)$$
when the scalar field is the only source of energy. Thus the full system reads
$$H^2 = \frac{1}{3} \rho_\phi = \frac{1}{3}\left(\frac{1}{2}\dot{\phi}^2 + V(\phi)\right)$$
$$\ddot{\phi} + 3H\dot{\phi} + V'(\phi) = 0$$
Only for very special classes of potentials do we have analytical solutions. A general method of studying such systems is to write them as a coupled set of first order equations - a so-called dynamical system.

If we introduce $x_1 = \frac{\dot{\phi}}{\sqrt{6}H}$ and $x_2 = \frac{\sqrt{V}}{\sqrt{3}H}$ then the Hubble equation says that $x_1^2 + x_2^2 = 1$. We further find with $x = \log R$ as our time-coordinate:
$$\frac{d x_1}{dx} = -3x_1 + \frac{\sqrt{6}}{2}\lambda_\phi x_2^2 - x_1 \frac{d\log H}{dx}$$
$$\frac{d x_2}{dx} = -\frac{\sqrt{6}}{2}\lambda_\phi x_1x_2 - x_2 \frac{d\log H}{dx}$$
$$\frac{d\lambda_\phi}{dx} = -\sqrt{6}\lambda_\phi^2(\Gamma-1)x_1$$
where
$$\frac{d\log H}{dx} = -\frac{3}{2}(1+x_1^2 - x_2^2)$$
$$\lambda_\phi = -\frac{V'}{V},\,\,\,\Gamma = \frac{V''V}{V'^2}$$
Now for an exponential potential $\lambda_\phi = \lambda$ stays constant and $\Gamma = 1$ and the system simplifies greatly down to
$$\frac{d x_1}{dx} = -3x_2^2[x_1 - \frac{\lambda}{\sqrt{6}}]$$
$$\frac{d x_2}{dx} = 3x_1x_2[x_1 - \frac{\lambda}{\sqrt{6}}]$$
or simply
$$\frac{d x_1}{dx} = -3(1-x_1^2)[x_1 - \frac{\lambda}{\sqrt{6}}]$$
This is possible to integrate though its hard to invert the resulting function, but we see right away that if $x_1 \equiv \frac{\lambda}{\sqrt{6}}$ then we have one solution. This is a so-called fixpoint of this system (put derivatives to zero and solve the resulting algebraic system). Finding and classifying such fixpoints (i.e. are they attractive or repellent) is super useful (much more than trying to find analytical solutions that might not exist) for understanding the dynamics of the system and the given analytical solution is such a fixpoint. We can also see that the fixpoint is attractive so even if we did not start with this initial value of $x_1$ we would be driven towards it and the system would settle close to it.

If you now unwrap this solution you will find the given solution. For example $\frac{d\log H}{dx} = -3x_1^2 = -\frac{\lambda^2}{2}$ gives us $H \propto R^{-\frac{\lambda^2}{2}}$ and $H = \frac{d\log R}{dt}$ gives us $R(t)$ and from the definition of $x_1$ we get $\dot{\phi} \propto H \implies \phi \propto \log R$.


问:
2-color version of Sylvester-Gallai theorem
I was reading Sylvester-Gallai theorem and thought about the following question.

Let there be some finite number of points on a plane.  Each point is either red or blue.  Every straight line passing through two points of the same color must also pass through another point of the other color.  Is it true that all points are on the same straight line?

My thinking was : if they are not on a single straight line then by Sylvester-Gallai, we know that there is a line passing through exactly two points.  Then these points can't have the same color, so I'll call the red one $R_0$ and the blue one $B_0$.  Now there has to be another point not on $R_0B_0$, so let this point be red (wlog) and call it $R_1$.  Now $R_0R_1$ has to have a blue point $B_1\ne B_0$.

Then inductively $R_n$ is a blue point on $B_{n-2}B_{n-1}$ and $B_n$ is a blue point on $R_{n}R_{n-1}$.  I think all these points are different but not sure how to prove it.

答:
This is known as the Motzkin-Rabin theorem. Based on the existing proofs in the literature, I do not think this can be easily proved using the Sylvester-Gallai theorem as a lemma.

The earliest proofs of this theorem actually prove the dual version: given several lines in the plane colored red and blue, there exists a point where two or more lines meet, and all lines meeting there are the same color. This is equivalent since the entire problem can be formulated in the projective plane, and because of projective duality.

Motzkin's original proof (see page 130 in https://doi.org/10.1007/BF02112289) used the area of triangles in the projective plane as a monovariant. This can be formulated in terms of non-projective triangles alone, see below.

Chakerian published a proof (https://doi.org/10.2307/2317330) which relies on Euler's formula.

There also exists a proof, due to Pretorius and Swanepoel (https://doi.org/10.2307/4145133), which sticks to the original formulation of colored points, though it is quite subtle.

Here is a sketch of the version of Motzkin's proof for red and blue lines without projective geometry. Consider two intersecting blue lines, $B_1$ and $B_2$. We are done unless there exists a red line $R_1$ lying on their intersection, so assume this is so. Furthermore, assume there is some red line $R_2$ not passing through this intersection, to avoid trivial cases.

The entire configuration is either equivalent to situation 1 or 2 in the top row of the image below, defining a certain gray triangle $T$. If there is no blue line $B_3$ at the intersection of $R_1$ and $R_2$, we are done, so assume $B_3$ exists. In situation $1$, this creates a smaller version of situation 1, with the colors swapped, such that the corresponding triangle $T'$ has a smaller area. Iterating this, we cannot have smaller areas forever, so eventually the process stops.

In situation 2, there is a small hiccup where $T'$ can be larger than $T$. Iterating the process, $T'$ can only grow finitely many times, and once it shrinks, we go back to situation 1 where it will continue to shrink, so this process still must stop.


问:
Japanese theorem for cyclic quadrilaterals Proof Inversion
Two weeks ago our professor taught us the Japanese theorem for cyclic quadrilaterals. It states that the inscribed centres of the four triangles formed by two sides and a diagonal of a cyclic quadrilateral always form a rectangle.

After the proof he mentioned that he was convinced that the reverse also holds, that is, that whenever the incircle centres form a rectangle the outer quadrilateral is a cyclic quadrilateral, but he never managed to prove it. After finding absolutely nothing on the internet about the inversion of the Japanese theorem, I set out to prove it myself.

The big problem is that the construction of the rectangle is not unique, many different chordal quadrilaterals have the same rectangle of incircle centres. Since there is thus no construction rule in the opposite direction, all my elementary geometric attempts have failed.

So I came up with the idea of a purely arithmetic proof for the Japanese theorem. In the hope that, unlike the geometric proof, it works in both directions. Without limiting generality, I have given the four vertices of the chord quadrilateral the following coordinates: $A (0, 0), B (1, 0), C(x_c, y_c), D(x_d, y_d)$. This is possible because by moving and scaling in the coordinate system alone, any quadrilateral can be brought into this shape. The simple values for A and B simplify the calculation enormously. After much effort I have found formulas to represent all vectors between the incircle centres only dependent from $x_c, y_c, x_d, y_d$:

$$\overrightarrow{M_1M_2} =    \left(\begin{array}{c}      \frac{\sqrt{x_c^2 + y_c^2} + x_c}{\sqrt{1 + x_c^2 - 2x_c + y_c^2} + \sqrt{x_c^2 + y_c^2} + 1} - \frac{\sqrt{x_d^2 + y_d^2}+ x_d}{1 + \sqrt{1 + x_d^2 - 2x_d + y_d^2} + \sqrt{x_d^2 + y_d^2}}\\      \frac{y_c}{\sqrt{1 + x_c^2 - 2x_c + y_c^2} + \sqrt{x_c^2 + y_c^2} + 1} - \frac{y_d}{1 + \sqrt{1 + x_d^2 - 2x_d + y_d^2} + \sqrt{x_d^2 + y_d^2}}    \end{array}\right)$$

$$\overrightarrow{M_2M_3} =     \left(\begin{array}{c}      \frac{\sqrt{x_c^2 + y_c^2 + x_d^2 + y_d^2 - 2x_cx_d - 2y_cy_d} + \sqrt{1 + x_d^2 - 2x_d + y_d^2} \cdot x_c + \sqrt{1 + x_c^2 - 2x_c + y_c^2} \cdot x_d}{\sqrt{x_c^2 + y_c^2 + x_d^2 + y_d^2 - 2x_cx_d - 2y_cy_d} + \sqrt{1 + x_d^2 - 2x_d + y_d^2} + \sqrt{1 + x_c^2 - 2x_c + y_c^2}} - \frac{\sqrt{x_c^2 + y_c^2} + x_c}{\sqrt{1 + x_c^2 - 2x_c + y_c^2} + \sqrt{x_c^2 + y_c^2} + 1}\\      \frac{\sqrt{1 + x_d^2 - 2x_d + y_d^2} \cdot y_c + \sqrt{1 + x_c^2 - 2x_c + y_c^2} \cdot y_d}{\sqrt{x_c^2 + y_c^2 + x_d^2 + y_d^2 - 2x_cx_d - 2y_cy_d} + \sqrt{1 + x_d^2 - 2x_d + y_d^2} + \sqrt{1 + x_c^2 - 2x_c + y_c^2}} - \frac{y_c}{\sqrt{1 + x_c^2 - 2x_c + y_c^2} + \sqrt{x_c^2 + y_c^2} + 1}    \end{array}\right)$$

$$\overrightarrow{M_3M_4} =    \left(\begin{array}{c}       \frac{\sqrt{x_d^2 + y_d^2} \cdot x_c + \sqrt{x_c^2 + y_c^2} \cdot x_d}{\sqrt{x_c^2 + y_c^2 + x_d^2 + y_d^2 - 2x_cx_d - 2y_cy_d} + \sqrt{x_d^2 + y_d^2} + \sqrt{x_c^2 + y_c^2}} - \frac{\sqrt{x_c^2 + y_c^2 + x_d^2 + y_d^2 - 2x_cx_d - 2y_cy_d} + \sqrt{1 + x_d^2 - 2x_d + y_d^2} \cdot x_c + \sqrt{1 + x_c^2 - 2x_c + y_c^2} \cdot x_d}{\sqrt{x_c^2 + y_c^2 + x_d^2 + y_d^2 - 2x_cx_d - 2y_cy_d} + \sqrt{1 + x_d^2 - 2x_d + y_d^2} + \sqrt{1 + x_c^2 - 2x_c + y_c^2}}\\       \frac{\sqrt{x_d^2 + y_d^2} \cdot y_c + \sqrt{x_c^2 + y_c^2} \cdot y_d}{\sqrt{x_c^2 + y_c^2 + x_d^2 + y_d^2 - 2x_cx_d - 2y_cy_d} + \sqrt{x_d^2 + y_d^2} + \sqrt{x_c^2 + y_c^2}} - \frac{\sqrt{1 + x_d^2 - 2x_d + y_d^2} \cdot y_c + \sqrt{1 + x_c^2 - 2x_c + y_c^2} \cdot y_d}{\sqrt{x_c^2 + y_c^2 + x_d^2 + y_d^2 - 2x_cx_d - 2y_cy_d} + \sqrt{1 + x_d^2 - 2x_d + y_d^2} + \sqrt{1 + x_c^2 - 2x_c + y_c^2}}     \end{array}\right)$$

$$\overrightarrow{M_4M_1} =     \left(\begin{array}{c}       \frac{\sqrt{x_d^2 + y_d^2}+ x_d}{1 + \sqrt{1 + x_d^2 - 2x_d + y_d^2} + \sqrt{x_d^2 + y_d^2}} - \frac{\sqrt{x_d^2 + y_d^2} \cdot x_c + \sqrt{x_c^2 + y_c^2} \cdot x_d}{\sqrt{x_c^2 + y_c^2 + x_d^2 + y_d^2 - 2x_cx_d - 2y_cy_d} + \sqrt{x_d^2 + y_d^2} + \sqrt{x_c^2 + y_c^2}}\\        \frac{y_d}{1 + \sqrt{1 + x_d^2 - 2x_d + y_d^2} + \sqrt{x_d^2 + y_d^2}} - \frac{\sqrt{x_d^2 + y_d^2} \cdot y_c + \sqrt{x_c^2 + y_c^2} \cdot y_d}{\sqrt{x_c^2 + y_c^2 + x_d^2 + y_d^2 - 2x_cx_d - 2y_cy_d} + \sqrt{x_d^2 + y_d^2} + \sqrt{x_c^2 + y_c^2}}    \end{array}\right)$$

As I feared, the formulae that result from the geometric construction are so complicated that I could not work with them even with the help of Mathematica and Matlab. So I discarded the arithmetic approach. After many more failed attempts, I wanted to rule out the possibility that a counterexample exists and that the reverse of the Japanese theorem is not true in the first place. I have written a programme in C++ which tries out millions and millions of constellations and indeed all points where the centres of the incircles form a rectangle always lie on a circle.

But I've run out of ideas how I could still attempt the proof. And apparently no one has ever cared about the validity of the reversion, I've only found one old StackExchange post. But it is very sparse and the counterexample presented in the only answer is invalid because the vertices of the rectangle do not coincide with the inscribed centres.

答:
Easy angle chasing shows that lines $CM_3, CM_4$ are isogonal in angle $DCM_2$ and lines $DM_3, DM_4$ are isogonal in angle $M_1DC$. Hence, letting $X$ be the intersection of $CM_2$ and $DM_1$, we get that $M_3$ and $M_4$ are isogonal conjugates in triangle $CDX$. Hence $\angle M_2XM_3 = \angle M_4XM_1$.

Since $M_1M_2M_3M_4$ is a rectangle and $\angle M_2XM_3 = \angle M_4XM_1$, an easy symmetry argument shows that $X$ lies on the perpendicular bisector $\ell$ of $M_1M_2$ (which, of course, coincides with the perpendicular bisector of $M_3M_4$).

Similarly, letting $Y$ be the intersection of $AM_4$ and $BM_3$, we get that $Y$ lies on $\ell$.

Let $P$ be the intersection of $AC$ and $BD$. Let $Z$ be the intersection of lines $BY, CX$. Let $T$ be the intersection of lines $DX, AY$. Then $Z$ and $T$ are incenters of $BCP$ and $DAP$, respectively. Again, a symmetry argument shows that $Z$ and $T$ are symmetric wrt to $\ell$. In particular $\angle PZX = \angle XTP$. Therefore $\angle CZP = \angle PTD$, so $\frac\pi 2 + \frac 12\angle CBD = \frac \pi 2 + \frac 12 \angle CAD$, hence $\angle CBD = \angle CAD$, hence $ABCD$ is cyclic.


问:
nth term of $2,2+\frac{1}{2},2+\frac{1}{2+\frac{1}{2}},2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}},2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}}...$
What is the nth term of the sequence: $$2,2+\frac{1}{2},2+\frac{1}{2+\frac{1}{2}},2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}},2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}}...$$

in terms of $s_{n-1}$.

I have tried for some time to calculate this but to no avail. Hopefully somebody with a better understanding of the world of sequences could help/point me in the right direction!

答:
Assuming you already know the sequence converges, and that is not trivial as the subsequence $\;\left\{x_{2n}\right\}_{n=0}^\infty\;$ is monotonic descending whereas the subsequence $\;\left\{x_{2n+1}\right\}_{n=0}^\infty\;$ is monotonic ascending ( yet both sequences are appropiatedly bounded and converge to the same non-zero finite limit ), and if we put $\;\alpha=\lim\limits_{n\to\infty} x_n\;$  , we can then use arithmetic of limits and get

$$\alpha=\lim_{n\to\infty} x_{n+1}=\lim_{n\to\infty}\left(2+\frac1{x_n}\right)=2+\frac1\alpha\implies\alpha^2-2\alpha-1=0$$

and solving the above quadratic we get that the only plausible limit is $\;1+\sqrt2\;$ ,since the other root of the quadratic is negative: $\;1-\sqrt2<0\;$ .


问:
How to prove that using law of cosines?
I'm reading a book about Cartesian Tensors. The author said that is possible to obtain the following, applying the law of cosines.
\begin{equation}
\sum_{k=1}^{3}\lambda_{ik}\lambda_{jk} = 0\qquad i,j = 1,2,3 \qquad\ \ \ \ \ \ \ \ \ \ \ \ \ (1)
\label{eq1}
\end{equation}
where $\lambda$ is a rotation matrix. I don't know how to obtain (1) using just the law of cosines. I can obtain (1) by applying three consecutive rotations in regarding to $x_1$, $x_2$ and $x_3$ axes, yielding a rotation matrix
$$R = R(\alpha)R(\beta)R(\gamma) =
 \left(
 \begin{array}{ccc}
  \cos (\beta ) \cos (\gamma ) & \cos (\beta ) \sin (\gamma ) & \sin (\beta ) \\
  -\sin (\alpha ) \sin (\beta )\cos (\gamma )-\cos (\alpha ) \sin (\gamma ) & \cos (\alpha ) \cos (\gamma )-\sin (\alpha ) \sin (\beta ) \sin (\gamma ) & \sin (\alpha ) \cos (\beta ) \\
  \sin (\alpha ) \sin (\gamma )-\cos (\alpha ) \sin (\beta ) \cos (\gamma ) & -\sin (\alpha )\cos (\gamma )-\cos (\alpha ) \sin (\beta ) \sin (\gamma ) & \cos (\alpha ) \cos (\beta ) \\
 \end{array}
 \right)
$$
then applying (1) above I can obtain the results equal to zero. But I'm curious to know how can I get it using a geometrical approach, e.g. law of cosines or another geometrical approach.

答:
The approach I find most intuitive is the following: the key point is that a rotation is an isometric (equivalently "orthogonal") linear transformation: the length of a vector $x$ is equal to the length of the transformed vector $\lambda x$.

Consider any $x = (x_1,x_2,x_3) \in \Bbb R^3$. We have
\begin{align}
\|\lambda x\|^2 &=
\sum_{i=1}^3 (\lambda x)_i^2 = \sum_{i=1}^3 \left(\sum_{j=1}^3 \lambda_{ij} x_j\right)^2
\\ & = \sum_{i=1}^3 \left(\sum_{j=1}^3 \lambda_{ij} x_j\right)\left(\sum_{k=1}^3 \lambda_{ik} x_k\right)
= \sum_{i,j,k=1}^3 \lambda_{ij} \lambda_{ik} x_j x_k.
\end{align}
Let $M$ denote the matrix with $M_{jk} = \sum_{i} \lambda_{ij}\lambda_{ik}$ (that is, $M = \lambda^T\lambda$). Note that this matrix is symmetric, which is to say that $M_{jk} = M_{kj}$. From the above, we have $\|\lambda x\|^2 = \sum_{jk} M_{jk} x_jx_k$. On the other hand, from the isometric property we have
$$
\|\lambda x\|^2 = \|x\|^3 = \sum_{j}x_j^2 = \sum_{jk} \delta_{jk} x_j x_k,
$$
where $\delta_{jk}$ denote the Kronecker delta. That is,
$$
\sum_{jk} M_{jk}x_jx_k = \sum_{jk}\delta_{jk} x_jx_k.
$$
Now, argue that because $M$ is symmetric this holds for all choices of $x \in \Bbb R^3$, it must be that $M_{jk} = \delta_{jk}$. That is, we have
$$
\lambda^T \lambda = I \implies \lambda^T = \lambda^{-1} \implies \lambda\lambda^T = I \implies\\
\sum_k \lambda_{ij}\lambda_{jk} = \delta_{ij},
$$
which is what we wanted to show.

Title: How to prove that using law of cosines?


问:
Sufficient condition for complete local ring to be the completion of a certain subring
The context is two statements from this paper:

A quasi-local ring $(R, M \cap R)$ contained in a complete local ring $(T,M)$ is Noetherian and has completion $T$ provided the map $R \to T/M^2$ is surjective and finitely generated ideals of $R$ are closed in the topology induced from $T$.

This fact is stated without explanation, and I couldn't track it down in Matsumura or the other references. It appears to be used in the final line of the main theorem (Theorem 8), which simply concludes

Finally, $A \to T/M^2$ onto implies $\widehat{A} \to T$ is surjective.

(Here, $A$ is playing the same role as $R$ above).

I haven't been able to figure out why this is true: in particular, why does the map $R \to T/M^2$ give us information about the $\widehat R \to \widehat T$? But my understanding of completions is fairly shallow, so it's possible this is a simple enough fact that the author felt it was self-evident.

答:
I tracked down a reference in a later paper by the author, Completions of Local Rings with an Isolated Singularity.

Proposition 1 is exactly the statement above, and is presented with proof this time.


问:
Please help me clear confusion over principal roots and identities for n-th radicals

From my old high school math textbook:

If ${a{\geq }0}$ and $n\in \mathbb{N} ^{\ast }$, then ${\sqrt[{n}] {a}}$ is the non-negative solution of ${{x}^{n}}=a$.

It then goes on to infer a number of properties and identities in the case that $\alpha \geq 0$ and $n\in \mathbb{N} ^{\ast }$.

However the choice of defining the n-th root of something only as the positive solution of this equation leaves me quite confused. For example, we could define $\sqrt[3] {-8}=-2$ but we don't.

Instead the book goes on to break down the solution of ${{x}^{n}}=a$ based on $\pm \sqrt[n] {\left| a\right| }$ instead of just $\sqrt[n] {a}$.

Why do we do this?

The second part is please tell me if these expressions are in fact correct because the book doesn't mention them (although it might be using them) and I've seen them elsewhere online.

If $x \geq 0$ and $a\geq 0$ and $n\in \mathbb{N} ^{\ast }$, then $\ x^{n}=a\Leftrightarrow x=\sqrt[n] {a}$

$\left( \sqrt[n] {a}\right) ^{n}=a$ holds for every $a\in \mathbb{R}$, while $\sqrt[n] {a^{n}}$ holds only for $a\geq 0$

答:
It is quite likely that your textbook deals with it that way in order not to have to deal with case case in which $n$ is even (in which case no negative number has a $n$th root) and the case in which $n$ is odd (in which case each number has one and only one $n$th root) separately.

And, yes, those assertions are correct.


问:
character of a group is constant on each conjugacy class in the group
I'm trying to solve this problem and I'm not quite sure what I need to prove here. Can you guys please help?. Here is the question.

Definition:
Character of a group $G$ is defined to be a group homomorphism  $\chi: G \rightarrow \mathbb{C^*}$. Prove that the character of a group $G$ is constant on each conjugacy class in $G$.

My answer: Let $x\in G$ be fixed. Then $\chi(gxg^{-1})=\chi(g) \chi(x) \chi(g) ^{-1}$ since $\chi$ is a homomorphism. But $\mathbb{C^*}$ is abelian so $\chi(gxg^{-1})=\chi(x)$.

Have I proved what I needed to prove? :-). I guess my question is what exactly do they mean by "constant on each conjugacy class in $G$".

Thanks for all your help.

答: You are absolutely right in your work. Conjugacy class of an element $x$ in $G$ is precisely the set $\{gxg^{-1} | g \in G\}$. Thus, you have accomplished what you wanted.</p>


问:
If $n-1$ integers are so that no partial sums are divisible by $n$, then the integers are congruent mod $n$ to one number coprime to $n$.
Let $a_1,a_2,...,a_{n-1}$ be integers.  Suppose that there is no subset $S\ne \emptyset$ of $\{1,2,...,n-1\}$ such that $\sum_{k\in S}a_k\equiv 0\pmod{n}$.  Show that there exists an integer $a$, $0< a< n$ st $\gcd(a,n)=1$ and $a_1,a_2,...,a_n\equiv a\pmod{n}$.

Source : https://artofproblemsolving.com/community/c6h626708p3759215 (mavropnevma's comment)

My attempt : Let $s_k=a_1+a_2+...+a_k$.  Then $s_1,s_2,...,s_{n-1}$ form a permutation of $1,2,...,n-1$ (mod n).  Let $s_0=0$ and consider the subscripts mod $n$.  Then the condition that $\sum_{k\in S}a_k\ne 0\pmod{n}$ for all $S\subseteq\{1,2,...,n-1\}$ st $S\ne \emptyset$ implies $\sum_{k\in S}s_k \ne \sum_{k\in S}s_{k-1}\pmod{n}$.  This is where I'm stuck.

答:
(This is essentially identical to reuns' solution, but has an easier presentation / more reasonable to come up with the solution.)

If there exists 2 element that are different from each other, WLOG $a_1 \neq a_2$, then consider the $n$ terms

$$ a_1, a_2, a_1 + a_2, a_1 + a_2 + a_3, \ldots, \sum a_i .$$

Consider their remainder mod $n$.

Since none of them are 0, there are $n-1$ possibilities.

By the PP, 2 of them have the same remainder mod $n$.

This pair of terms cannot be $a_1, a_2$.

Hence, we may taking the difference of these terms, which gives us a sequence that sums to 0 mod n.

Contradiction.

Note:

The solution can be arrived by thinking that it likely has a PP solution, so how can we apply PP?

Clearly we (likely) have the $n-1$ remainder as holes.

However, taking the standard $\sum_{i=1}^k a_i$, we seem to only have $n-1$ pigeons.
Thus, we need to "force" out another pigeon.