问:
Quadratic equation where constant term is an integral function
Consider a quadratic equation ${x^2} + 2x = k + \int\limits_0^1 {\left| {t + k} \right|dt}$. Then choose the correct option(s),

(A) Roots are Real

(B) Roots are Imaginary

(C) Roots are Distinct

(D) Roots are complex number

My approach is as follow

Let $-c=k + \int\limits_0^1 {\left| {t + k} \right|dt}$

$a=1$ & $b=2$

$b^2-4ac$

$4-4(k + \int\limits_0^1 {\left| {t + k} \right|dt})$

How do I check the nature of the roots

答:
The discriminant $b^2 - 4ac$ is actually

$$d = 4 + 4(k + \int_{0}^{1}\left| t + k \right| dt) \tag{1}\label{eq1A}$$

Regarding calculating $k + \int_{0}^{1}\left| t + k \right| dt$, there are $3$ cases to consider:

Case $1$: $\; k \lt -1$

Since $t + k \lt 0$ for $t$ from $0$ to $1$, this gives

$$\begin{equation}\begin{aligned}
k + \int_{0}^{1}\left| t + k \right| dt & = k + \int_{0}^{1}(-t - k) dt \\
& = k + \left. \left(-\frac{t^2}{2} - kt\right)\right|_{0}^{1} \\
& = k + \left(-\frac{1}{2} - k\right) \\
& = -\frac{1}{2}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$

In \eqref{eq1A}, this gives $d = 2$.

Case $2$: $\; -1 \le k \lt 0$

This requires splitting the integration into $2$ parts, the initial one for $t$ from $0$ to $-k$ where $t + k \le 0 \implies \left|t + k\right| = -t -k$ and a second part for $t$ from $-k$ to $1$ where $t + k \ge 0 \implies \left|t + k\right| = t + k$, giving

$$\begin{equation}\begin{aligned}
k + \int_{0}^{1}\left| t + k \right| dt & = k + \int_{0}^{-k}(-t - k) dt + \int_{-k}^{1}(t + k) dt \\
& = k + \left. \left(-\frac{t^2}{2} - kt\right)\right|_{0}^{-k} + \left. \left(\frac{t^2}{2} + kt\right)\right|_{-k}^{1} \\
& = k - \frac{k^2}{2} + k^2 + \frac{1}{2} + k - \frac{k^2}{2} + k^2 \\
& = 2k + k^2 + \frac{1}{2} \\
& = k^2 + 2k + 1 - 1 + \frac{1}{2} \\
& = (k + 1)^2 - \frac{1}{2}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$

For $k \ge -1$, the value above is an increasing function, going from $-\frac{1}{2}$ to $\frac{1}{2}$. Thus, $d$ in \eqref{eq1A} goes from $2$ to $6$.

Case $3$: $\; k \ge 0$

This obviously gives $d \gt 0$, but doing the calculations results in

$$\begin{equation}\begin{aligned}
k + \int_{0}^{1}\left| t + k \right| dt & = k + \int_{0}^{1}(t + k) dt \\
& = k + \left. \left(\frac{t^2}{2} + kt\right)\right|_{0}^{1} \\
& = k + \left(\frac{1}{2} + k\right) \\
& = 2k + \frac{1}{2}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$

Thus, $d$ in \eqref{eq1A} is $6$ at $k = 0$, and increases without bound as $k$ goes up.

This shows all $3$ cases gives $d \gt 0$, so the roots are real and distinct. Since the roots are also always complex numbers, this means options (A), (C) and (D) are true, with only (B) being false since there are no imaginary roots.


问:
Understanding Hecke Characters as Extension of Dirichlet Characters
I understand the concept of a Dirichlet character, and am interested in its generalizations to arbitrary number fields. I have heard that this generalization is called a Hecke character. However, I am not familiar with adeles or ideles, so I don't understand that definition. I know there is a "classical" definition of Hecke characters, but I am having trouble finding an easy to understand source about this definition. Could anyone provide a reference on the classical definition of Hecke characters that covers some of the preliminary knowledge needed to understand this definition?</p>

答:
The original papers, in German, are quite well written, and certainly do not use ideles or adeles. Yes, identical titles:

E. Hecke, Eine neue Art von Zetafunktionen und ihre
Beziehungen der Verteilung der Primzahlen}, Math. Z. 1 no. 4
(1918), 357-376.

E. Hecke, Eine neue Art von Zetafunktionen und ihre
Beziehungen der Verteilung der Primzahlen, Math. Z. 6 no. 1-2
(1920), 11-51.

ContentLicense: CC BY-SA 4.0


问:
Logical Conjunction of Binary Decision Diagrams
Compute a Binary Decision Diagram for $B1∧B2$. Furthermore, for an arbitrary BDD B you can use the equations $B∧F=F$, $F∧B=F$, $B∧T=B$ and $T∧B=B$.

To construct the BDD i start from the leaves and work myself to the top. So i have:

but here is the problem:

Now i do not know how to proceed from here on. Has someone experience with Binary Decision Diagrams ? I would like to understand this problem and solve it.

答: This is the solution step by step :

ContentLicense: CC BY-SA 3.0

Body: This is the solution step by step :


问:
Matrix representation of trace
I got trace is $4$ for a $2\times2$ matrix. I know trace is a linear operation which transforms a matrix to a field. So the matrix representation of a trace should be like $1×n$, where $n$ is the dimension of that square matrix.

How I can represent number $4$ as a $1\times4$ matrix?
Thanks in advance!

答:
You can represent a $2\times2$ matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ by the $4\times1$ vector $\begin{bmatrix}a\\b\\c\\d\end{bmatrix}$, then the trace matrix is $\begin{bmatrix}1&0&0&1\end{bmatrix}$. We have$$\begin{bmatrix}1&0&0&1\end{bmatrix}\begin{bmatrix}a\\b\\c\\d\end{bmatrix}=a+d=\text{Tr}\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

We can get to this matrix easily: the trace of $e_1=\begin{bmatrix}1\\0\\0\\0\end{bmatrix}$ is $1$, which is the first column of trace matrix. The trace of $e_2=\begin{bmatrix}0\\1\\0\\0\end{bmatrix}$ is $0$, which is the second column of trace matrix. Similarly the traces of $e_3,e_4$ are $0,1$ respectively.

Title: Matrix representation of trace


问:
How can I solve $y'' + 8y' + 15y = 35 \theta(2-x)e^{2x}$ using Laplace transform?
I have the following homework problem. Using Laplace transform, I need to solve the following differential equation
$$
y'' + 8y' + 15y = 35 \theta(2-x)e^{2x}
$$
where $\theta(x)$ is the Heaviside step function. I evaluated this differential equation using WolframAlpha, but the solution it presents is for the homogeneous equation, which is not the same as the equation I have since for $2-x>0$ the R.H.S. is $35 e^{2x}\neq 0$. Since I believe there's no general formula for the Laplace transform of a product, I explicitly calculated the Laplace transform of the R.H.S. which gives
$$
\mathcal{L}\left\{35 \theta(2-x)e^{2x} \right\} = 35 \int_{0}^{2}e^{-px}e^{2x} \ dx = \frac{35\left(1-e^{4-2s}\right)}{s-2} \tag{1}
$$
Which would then imply, by taking the Laplace transform on both sides of the original equation, that
$$
Y = \frac{1}{(s+3)(s+5)}\left[\frac{35\left(1-e^{4-2s}\right)}{s-2} +y(0)(s+8) +y'(0) \right]\tag{2}
$$
Now, for the last $2$ summands I know I can apply partial fractions to get the inverse Laplace transform, but for the first term, I have no idea how to inverse Laplace it. I plugged it into WA and got a really long expression for the answer. This last part seemed odd to me and I started doubting if my approach was correct.

I then had the idea that maybe I should try to solve the equation piece-wise, getting $2$ "simpler" equations for $x<2$ and $x>2$, the latter being the homogeneous case. The problem with this is that I don't think this results in a solution to the original problem since in both cases I would get different functions, and I need a function that satisfies both conditions ($x>2$ and $x<2$) simultaneously.

Am I on the right path to solving this problem? Or is my approach wrong? And in any case, is there a simple way to obtain the inverse Laplace transform of equation $(2)$? Thank you!

答:
Yes, you are on the right path.

It will be easier to separate the domain of this ODE, solve them respectively and then glue the solutions together using boundary conditions than trying to obtain a "single" solution.

Generally there is no simpler way to do the inverse Laplace transform of equation (2), and the answer provided by  WA seems nothing suspicious. The Heaviside function corresponds to the exponential term in (2), and the three exponential functions of $t$ come from the three poles in (2).


问:
Product of absolute values nonzero implies sum of one minus absolute value is bounded
I'm reading a proof that claims a well known equivalence: $\Pi_{k=1}^\infty |\alpha|>M>0$ implies $\sum_{k=1}^\infty 1-|\alpha|<\infty$ for $|\alpha_k|<1$ in the complex numbers. It seems to suggest writing $1-(1-|\alpha|)=\alpha$, but I don't really see how this helps. Expanding the product, many more terms appear than just our desired sum.

答: Hints: $\frac {\ln x} {x-1} \to 1$ as $ x \to 1$ by L'Hopital's Rule. The hypothesis implies that  $|\alpha_k| \to 1$ as $ k \to \infty$ and $\sum \ln |\alpha_k|$ is absolutely convergent. By Comparison test $\sum |(|\alpha_k| -1)|$ is convergent.


问:
How to show if the complex series $\sum_{n=2}^\infty \frac{(-i)^n}{\ln n}$ converges or diverges?
I have the series
$$\sum_{n=2}^\infty \frac{(-i)^n}{\ln n}$$
and I'm trying to determine whether the series will converge or diverge.The Ratio test gives me $L=1$, so it doesn't tell me anything.

I tried to use the comparison test, so I used $\ln n<n$ and thus $\frac{1}{n}<\frac{1}{\ln n}$ for all $n>2$. So I have:
$$|z_n|=\left|\frac{(-i)^n}{\ln n}\right|=\frac{1}{\ln n}\Rightarrow \frac{1}{n}<|z_n|$$ and since $\sum_{n=2}^\infty \frac{1}{n}$ diverges, thus $\sum_{n=2}^\infty |z_n|$ diverges, but I can't say that the series of $z_n$ diverges if the series of absolute values |$z_n|$ diverges. So, how can I show if this series converges or diverges?

答:
Here’s how I would argue:

The real terms and the imaginary terms of the series have nothing to do with each other. The real part is $\sum_{k=1}^\infty\frac{(-1)^k}{\log(2k)}$, while the imaginary part is $\bigl(\sum_{k=0}^\infty\frac{(-1)^k}{\log(2k+3)}\bigr)i$.

More precisely, if $N\equiv1\pmod4$, that is, if $N=4K+1$, then
$$
\sum_{n=2}^N\frac{(-i)^n}{\log n}=\sum_{k=1}^{2K}\frac{(-1)^k}{\log(2k)}
\>+\>\biggl(\sum_{k=0}^{2K-1}\frac{(-1)^k}{\log(2k+3)}\biggr)i\>,
$$
in which I do hope that I have gotten the running indices right. At any rate, the real part and the imaginary part are, separately, alternating series, and there you are.


问:
Is there an educational platform like Webassign or OLI that you don't need to be in a class to use and offers upper level and/or grad level math?
(Please let me know if I am posting this in the right place, or if there are better tags I could use.)

Is there an educational platform like Webassign or OLI (Open Learning Initiative) that you don't need to be in a class to use?  Also, are there any that offer upper level and/or grad level math?  The first question is more important than the second, but please answer the second if you can.

These platforms give a bunch of multiple choice and/or free response problems.  When you are wrong, some of them tell you what is wrong about your answer without giving away the correct answer.  They also have multiple versions of the same type of question or set of questions to practice with.

So, if you know of any websight or app (fee required or not) that does any of this where you don't need to be in a college class to do the exercises, can you please let me know?

Although, if you know of one where you only have to enroll in a MOOC (Massive Open Online Course), (like EdX, Coursera, Kahn Academy, Udacity, MITOpenCourseware) can you please let me know?

Edit: In case there is any confusion about what types of classes I'm talking about, the most common class that would indicate the level I'm talking about would be abstract algebra (AKA modern algebra). It doesn't necessarily have to include abstract algebra, but it would be the most likely indicator.

The ambiguous ones (as indicators) would be linear algebra, discrete math, and number theory (as these are commonly taught at community colleges, meaning lower division).  But, please share any site that offers these.

There is a long list of (not so ambiguous) upper level (and especially grad level) classes, e.g. modern analysis (AKA real analysis), complex analysis, functional analysis, numerical analysis, numerical linear algebra, differential geometry, algebraic geometry, topology, algebraic topology, differential topology, logic, set theory, combinatorics, graph theory, probability theory, and I'll just include category theory for the sake of wishful thinking

答:
I would reccomend you register for MathXL under self-study.

A wide array of Pearson published Math textbooks have MathXL content.

This includes textbooks for post-secondary level courses up to Differential Equations and Linear Algebra. No higher level courses, unfortunately.

You can register here: https://www.pearsonmylabandmastering.com/northamerica/mathxl/students/get-registered/index.html

You will have to pay for access though. Just like enrolled students need to beyond a trial period.

I personally find MathXL far more comprehensive than WebAssign and am using it for self-study while taking Calculus II.

Another reccomendation is Zybooks online textbooks, if you are looking for an online interactive textbook. However they do not provide a drill evironment like MathXL. Again nothing for higher level courses.

https://www.zybooks.com/

Edit:

I thought you might be interested in reasorces for the basic set of undergraduate math courses (Calculus I-III + Differential Equations and Linear Algebra). If you want similar reasouces beyond that level I have no idea.


问:
Determine $[1] =\{n \in\Bbb Z | 1 R n\}$
Let the relation $R$ be defined on the set $\Bbb Z$ of integers by $a R b$ if and only if $a^2 − b^2$ is an integer multiple of $3$.

a) Determine whether $R$ is an equivalence relation. Either prove that $R$
is an equivalence relation or specifically demonstrate which properties of an
equivalence relation it fails to have.

b) Determine $[1] =\{n \in\Bbb Z | 1 R n\}$.

I have already done part a (I believe it is reflexive, symmetric, and transitive, therefore it is an equivalence relation), however the second part is crazy.
My professor's lecture notes on equivalence class was viciously underwhelming and he explained nothing in a way that was easy to understand. He is also not taking questions. I am not sure what "Determine $[1] =\{n \in\Bbb Z | 1 R n\}$. even means, let alone how to solve the problem. Can someone please explain and walk me through this in such a way that is easy to comprehend?

Thank you.

答:
An equivalence relation partitions the domain into disjoint sets called equivalence classes. Every element of an equivalence class is related to every element of the same class but to no element outside the class. The union of the disjoint equivalence classes is the domain, i.e. each element of the domain belongs to exactly one equivalence class.

The equivalence class containing $x$ is denoted by $[x]$. It is the set of all members of the domain $\Bbb Z$ that are related to $x$. The expression $[1]=\{n\in\Bbb Z:1Rn\}$ says exactly that. Note that two integers may belong to the same equivalence class, but the distinct equivalence classes must be disjoint.

Can you reason why every element in $[1]$ defined in this manner is related to every other element in $[1]$, and not to any element in $\Bbb Z-[1]$?

 If $x,y\in[1]$, then by definition $1Rx$ and $1Ry$. By transitivity of $R$ we have $xRy$. Further, if $\exists z\in\Bbb Z-[1]$ such that $xRz$, then by transitivity, $1Rx$ implies $1Rz$, a contradiction.

We know from the definition of the relation that $1Rn\iff nR1\iff3\mid(n^2-1)$. We can take $n=3k+i,i\in\{0,1,2\},k\in\Bbb Z$. Then $n^2-1=9k^2+i^2+6ki-1$.

Thus $3\mid n^2-1\iff3\mid i^2-1\iff i=1,2$.

Thus, $[1]=\{3k+1,3k+2:k\in\Bbb Z\}$.


问:
Efficient Computation of the expression
I have been working on a problem for quite some time now and looking for a better approach.

Problem :

You are given two integers $K$ and $N$.
 For each m ($1 \le m \le N)$, you have to find the answer of the below expression

$$X=\sum_{i=1}^K i^m$$
Since answer can be large we have to return $X$ modulo $10^9+7$ for each m.

Constrainsts:
 $1 \le K \lt 10^9+7 $
 $1 \le N \le 10^5$

An obvious solution to the problem is a brute force approach of complexity $O(N*K)$.

A slightly better approach is the use of
Faulhaber's forumula which would reduce the complexity to $O(N^2)$ but still is quite slower.

I am looking for a better solution. So is there a way to compute Faulhaber's formula faster or is there a different approach for the above problem?

Any sort of help is appreciated!!

答:
If you look for approximations.

Using generalized harmonic numbers$$X=\sum_{i=1}^k i^m=H_k^{(-m)}$$ If $k$ is large, you can use the asymptotics
$$\frac{H_k^{(-m)}-\zeta(-m)}{k^m}=\frac{k}{m+1}+\frac{1}{2}+\frac{m}{12 k}-\frac{(m-2) (m-1) m}{720
   k^3}+\frac{(m-4) (m-3) (m-2) (m-1) m}{30240
   k^5}+O\left(\frac{1}{k^7}\right)$$

Title: Efficient Computation of the expression


问:
Projection onto the convex hull of a sequence
For context, I am trying to show that a certain orthogonal projection operator is not directionally differentiable at a point in direction $d$. The set in question is the convex hull of the set $S = \{(\cos x_n,\sin x_n) : n \geq 1 \} \cup \{(0,0),(1,0)\}$, where $(x_n)$ satisfies $x_1 = \frac{\pi}{2}$, and $x_n$ decreases monotonically to $0$ after that.

I am told that the projection onto this set of the point $\left(2, \sin x_n + (2-\cos x_n)\tan\left(\frac{x_n + x_{n-1}}{2}\right)\right)$ is $(\cos x_n, \sin x_n)$ but I am not sure how to prove this. What I have tried so far is working with the convex constrained optimality conditions and the gradient of the norm to obtain that this is equivalent to proving the inequality $$\tan\left(\frac{x_n + x_{n-1}}{2}\right)(u_2 - \sin x_n) \leq \cos x_n - u_1$$
for all $u = (u_1,u_2) \in \text{conv}(S)$.

However I have been unable to prove this inequality, and maybe there is a better way to do it.
Any help is much appreciated.

答:
Let $p_n = (\cos x_n, \sin x_n)$.

Orange polygon is the convex hull of $S$.

Point $C$ is constructed in the following way: it is the intersection of the line $y=2$ and line, that is perpendicular to the segment $p_{n-1}p_n$ and passes through the point $p_n$. So by construction $p_n$ is the projection of $C$ on the segment $p_{n-1}p_n$.

Calculating coordinates of $C$ we find, that $C = \left(2, \sin x_n + (2-\cos x_n)\tan\left(\frac{x_n + x_{n-1}}{2}\right)\right)$.


问:
Posterior Probability Distribution from Geometric Distribution
Suppose an experimenter believes that $p \sim Unif([0, 1])$ is the chance of success on any given experiment.  The experimenter then tries the experiment over and over until the experiment is a success.  This number of trials will be $[G \mid p] \sim Geo(p)$.

How can you find the distribution of $[p \mid G]?$
Do you use Bayes' Theorem?

答:
Your prior is a $Beta(1;1)$

$$\pi(\theta)=1$$

$\theta \in[0;1]$

Your likelihood is

$$p(x|\theta)=(1-\theta)^{x-1}\theta$$

$x=1,2,3...$

Considering $x$ as a given number (after seeing the observations), the posterior is the same expression as a function of the parameter, say

$$\pi(\theta|x)\propto \theta(1-\theta)^{x-1}$$

$\theta \in[0;1]$

That is

$$\pi(\theta|x)=Beta(2;x)$$

The complete expression of your posterior is

$$\pi(\theta|x)=(x+1)x \theta(1-\theta)^{x-1}$$

To derive this expression not any integral is needed as you immediately recognize it is a Beta distribution


问:
How to show that $\exp(\mathbb{C}) $ is open subset of $\mathbb{C}^*$, the set of non zero complex numbers
How to show that $\exp(\mathbb{C}) $ is open subset of $\mathbb{C}^*$, the set of non zero complex numbers?

The Hint given to me says that, for $w\in \exp({\mathbb{C}})$

$D_{|w|}(w) = wD_1(1)\subset w\exp(\mathbb{C})= \exp(\mathbb{C})$

I don't understand how we got $D_{|w|}(w) = wD_1(1)$

Also, how is $wD_1(1) \subset w\exp(\mathbb{C})?$

答:
I can answer the question.  However, I am unfamiliar with the syntax that the query
is using.  Therefore, I have to use my own syntax.

Let the set $S \equiv ~$ exp$(\mathbb{C})$.

I have to show three things.

(1) $~0 ~\not\in ~S.$

(2) $~z \neq 0 \implies ~z \,\in \,S.$

(3) $~\forall ~z ~\in ~S,~$ there exists a neighborhood $N$ around $z$ such that
$N \subseteq S.$

For any $~(x + iy) ~\in \mathbb{C},$ and
for any $~z ~\in \mathbb{C}$

$e^{(x + iy)} = z \iff \{x = \text{ln}|z| ~\text{and}
~y \equiv \text{Arg}(z) \pmod{2\pi} ~\}.$

(1)

As $z \to 0,~ $ ln$|z| \to -\infty.$

For any $~(x + iy) ~\in ~\mathbb{C}, ~x~$ is finite.

Therefore, for any $~(x + iy) ~\in ~\mathbb{C}, ~e^{(x + iy)} \neq 0.$

(2)

For $z \neq 0,~$ let $~x = \text{ln}|z|~$ and let $~y = \text{Arg}(z).$

Then $~(x + iy) \,\in \,\mathbb{C}~$ and $~e^{(x + iy)} = z.$

Therefore $~z \,\in \,S.$

(3)

Given any $~z \,\in \,S,~$ from (1) above, $~z \neq 0.$

Thus:

Given any $~z \,\in \,S,~$
define the neighborhood $N$ to be the open circle of radius
$~\frac{|z|}{2}~$ around $~z$.

Take any $~n \,\in \,N$.

Then, $~n \neq 0.$
Therefore, by (2) above, $~n ~\in ~S.$


问:
Simulating bernoulli trials with unknown p
Assume I have access to independent Bernoulli trials with bias $p$ where $p$ is unknown and $p < 0.01$ (or any other small positive number).

Is it possible to simulate Bernoulli trials with bias $cp$ for some $c > 1$ independent of $p$? Any single value of $c$ would be good, it is not necessary to solve it for all $c$. By simulation, I mean an algorithm that uses the Bernoulli trials as an oracle and returns a Bernoulli trial with the larger $p$. I have (unsuccessfully) tried the following:

Get two samples and take OR. This gives me probability of success of $2p - p^2$, so there is a $p^2$ that I do not want. Now, if I instead take AND, it gives me bias $p^2$. Flipping an unbiased coin and taking AND/OR depending on the outcome gives me average of those two success probabilities, thus removing the unwanted $p^2$. However, it also decreases the coefficient in front of $p$ to 1 so this does not work.

Edit: as noted in a comment, I am interested in an exact solution. As can be seen from my failed attemtp, it is easy to get a relatively good approximation for small $p$.

答:
This is possible in your scenario only if you know that p is such that cp is not more than 1 − ε. In three different papers, Huber has presented algorithms that simulate the probability cp in this particular case.

However, if p is totally unknown, this is impossible; once cp touches 1 somewhere in (0, 1), it's impossible to simulate the probability cp using coins of unknown probability p of heads (Keane and O'Brien 1994).

REFERENCES:

Keane, M. S., and O'Brien, G. L., "A Bernoulli factory", ACM Transactions on Modeling and Computer Simulation 4(2), 1994.

Huber, M., "Optimal linear Bernoulli factories for small mean problems", arXiv:1507.00843v2 [math.PR], 2016.

Huber, M., "Designing perfect simulation algorithms using local correctness", arXiv:1907.06748v1 [cs.DS], 2019.

Huber, M., "Nearly optimal Bernoulli factories for linear functions", arXiv:1308.1562v2 [math.PR], 2014.


问:
Proof of SVD (Singular Value Decomposition)
I'm reading this article.

I don't understand the last sentence of the section Proof (optional):

Now, we just solve U, V and S for   $A = USV^T$   and prove the
theorem.

It seems to me that this article suddenly jumped to the conclusion.

Do we have to calculate further?

I think I understand both $A^TAV = VS^2$ and $AA^TU = US^2$ well because I calculated them with a concrete example, like:
$$
A =
    \begin{pmatrix}
    3 & 3 & 2 \\
    2 & 3 & -2 \\
    \end{pmatrix}
$$

How can we utilize the information from $A^TAV = VS^2$ and $AA^TU = US^2$ to prove the theorem?

Please help me. Thank you in advance.

答: That last sentence is poorly worded but is actually a summary. The few lines before have demonstrated how to find $V$, by solving for the eigen-vectors of $A^TA$, find $U$ from the eigen-vectors of $AA^T$. You know $S$ up to a sign ( since you have actually have $S^2$) and these signs are the only unknown, which you can decide by demanding $A=USV^T$


问:
Understanding Algebras With Alteratives to the Distributive Law
If we want to quantify how much an operation $*$ associates under repeated application, we can consider the associator $(a,b,c)_* := (a*b)*c-a*(b*c)$. I understand the distributive law as a bit like associativity, but looking at how two different binary operations associate with each other, instead of how one operations associates with itself.

So, consider a set $X$ with a binary operation, $*$, and addition of elements, $+$, so that $X$ is a ring but without the distributive law, and that that $*$ may or may not be associative.  Is there a name for this structure?

Is there some similar definition for a 'distributator' for my structure $X$ which quantifies how much the binary operation distributes over addition in the same way that associator quantifies how much a binary operator associates with itself?
It seems natural to define $(a,b,c)_{*+} := (a*b)+(a*c)-a*(b+c)$; then we can see that $(a,b,c)_{*+} =0$ when the binary operation distributes over addition.
Is this definition used to analyse non-distributive algebras?

Secondly, consider adding in an additional binary operation $\cdot$, and allowing both binary operations to distribute over addition. Is it also possible to define a distributator which quantifies how much one of the binary operations distributes over the other? Again, something along the lines of $(a,b,c)_{*\cdot} := (a*b)\cdot(a* c)-a*(b\cdot c)$.  Although in this case it also seems natural to look at $(a,b,c)_{*\cdot} := (a*b)\cdot c-a*(b\cdot c)$, just taking the associator and replacing the second $*$ operation with $\cdot$

I'm also interested in any reading associated with this topic, and any examples of algebras which have the forms that I've described.

答:
(Note: For simplicity, when I say "is missing" I really mean "we do not require".)

Also I misinterpreted your question slightly. The examples I give below all assume that we have multiplicative inverses.

If it has distributivity but is missing associativity it is a semifield.

If it is missing associativity and missing one side of distributivity (so distributes from right but not from left), it is a quasifield.

If it has associativity but is missing one side of distributivity, it is a near-field.

I believe if it is missing both associativity and distributivity, it is called a "Cartesian group".

Dembowski's book Finite Geometries gives a good reference for this (he discusses these as examples of "planar ternary rings" which are used to coordinative projective planes).

I don't know that you will find a lot of material on structures that are missing both associativity and distributivity. But there is a lot of active research on finite semifields (and I believe the finite near-fields are classified). In response to the second part of your question, for example with semifields, we can consider the left, right, and middle nucleus, which are the elements that associate with everything when they are in the left, right, and middle position, respectively; ie the left nucleus of a semifield $\mathbb{S}$ would be
$$N_{L}(\mathbb{S}) = \{a : a \in \mathbb{S} \mid (ab)c = a(bc) \mbox{ for all } b,c \in \mathbb{S} \}$$


问:
definite integral. closed form?
Does this definite integral have a closed form $?$:
$$
\int_{0}^{1}
\frac{\mathrm{e}^{1/\log\left(x\right)}}
{x^{1/5}\,
\left[\log\left(x\right)\right]^{1/5}}
\,\mathrm{d}x
$$
Although it may not, I'm hopeful for a closed form. The troublesome part is the fractional exponents.

For example if the denominator were $x^{1}$ times $\log^{n}\left(x\right)$ then the integral would turn out to be related to $\Gamma\left(n\right).$

However when both exponents go fractional, it becomes much more diffiucult for me.

I have a hunch that the closed form is in terms of a modified Bessel function of the second kind due to having evaluated similar integrals.

Although this integral looks daunting I think it can be solved $!$.

Thanks so much.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{m \in
\braces{\pm 1,\,\pm 3,\,\pm 5,\ldots}}$:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{1}
{\expo{1/\ln\pars{x}} \over x^{1/5}\,\,\bracks{\vphantom{\large A}\ln\pars{x}}^{1/5}}\,\dd x} =
\int_{0}^{1}
{\expo{1/\ln\pars{x}} \over x^{1/5}\,\,\bracks{\vphantom{\large A}\expo{m\pi\ic}\verts{\ln\pars{x}}}^{1/5}}\,\dd x
\end{align}
Lets $\ds{x \equiv \expo{-1/t} \iff t = -\,{1 \over \ln\pars{x}}}$:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{1}
{\expo{1/\ln\pars{x}} \over x^{1/5}\,\,\bracks{\vphantom{\large A}\ln\pars{x}}^{1/5}}\,\dd x}
\\[5mm] = &\
\expo{-m\pi\ic/5}\int_{0}^{\infty}
{\expo{-t} \over \expo{-1/\pars{5t}}\,\,t^{-1/5}}\,
\pars{\expo{-1/t}\,{1 \over t^{2}}\,\dd t}
\\[5mm] = &\
\expo{-m\pi\ic/5}\int_{0}^{\infty}t^{-9/5}\,\,
\exp\pars{-t - {4 \over 5}\,t}\,\dd t
\\[5mm] = &
\expo{-m\pi\ic/5}
\bracks{50^{1/5}\,\on{K}_{4/5}\pars{4\root{5} \over 5}}
\label{1}\tag{1}
\end{align}

where $\ds{\on{K}_{\nu}}$ is a Modified Bessel Function. The
last result is from DLMF.

Since $\ds{\quad 50^{1/5}\,\on{K}_{4/5}\pars{4\root{5} \over 5} \approx 0.3740}$, the $\ds{\underline{\mbox{initial integral value}}}$ is
$$
\approx 0.302595 + 0.219848\ic
$$
when $\ds{m = \pm 1}$ besides, of course, other values which arise from the $\ds{\expo{-m\pi\ic/5}}$ periodicity.

The final result, which agrees with
$\ds{\tt Mathematica}$, is given by
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{1}
{\expo{1/\ln\pars{x}} \over x^{1/5}\,\,\bracks{\vphantom{\large A}\ln\pars{x}}^{1/5}}\,\dd x} =
\expo{\pi\ic/5}
\bracks{50^{1/5}\,\on{K}_{4/5}\pars{4\root{5} \over 5}}
\\[5mm] = &
\pars{{1 + \root{5} \over 4} +
\root{5 - \root{5} \over 8}\ic}50^{1/5}
\,\on{K}_{4/5}\pars{4\root{5} \over 5}
\\[5mm] \approx &\ \bbx{0.302595 + 0.219848\,\ic} \\ &
\end{align}

Title: definite integral. closed form?


问:
Invertible matrices and zero rows
I have a doubt related to proof of a theorem:

Theorem:
Suppose that $A$ and $B$ are matrices such that the product $AB$ is an identity matrix. Then the reduced
row-echelon form of $A$ does not have a row of zeros.

Proof. Let $R$ be the reduced row-echelon form of $A$. Then $R = EA$ for some invertible square matrix $E$.
By hypothesis $AB = I$ where $I$ is an identity matrix, so we have a chain of equalities
$R(BE^{-1}) = (EA)(BE^{-1}) = E(AB)E^{-1} = EIE^{-1} = EE^{-1} = I$.
If $R$ would have a row of zeros, then so would the product $R(BE^{-1})$. But since the identity matrix $I$ does
not have a row of zeros, neither can $R$ have one.

Why the statement "But since the identity matrix $I$ does
not have a row of zeros, neither can $R$ have one." is correct?

I have found a similar question here

This theorem is taken from the text

答:
Let $BE^{-1}=M$. The $(i,j)^{\text{th}}$ entry of $RM$ is given by $\sum_{k}r_{ik}m_{kj}$.

If $R$ has a zero row, then for some $i=i',r_{i'k}=0\forall k$. Thus the $(i',j)^\text{th}$ entry of $RM$ is $0$ for all $j$ i.e. $RM$ has a zero row (row $i'$).

Title: Invertible matrices and zero rows


问:
Variance of a random vector with entries either 0 or normal distribution
I have a random vector $x$ of length $N$, which is constructed as follows:

I know up to $k$ of the entries are non-zero.

The exact number of non-zero entries is based on a poisson distribution with parameter $\lambda$. i.e., Let $N_{nz}$ denote the number of non-zero entries, then

$\mathbb{P}(N_{nz}=i)=\frac{e^{-\lambda}\lambda^{i}/i!}{\sum_{j=0}^{k}{}e^{-\lambda}\lambda^{j}/j!}$

Once $N_{nz}=n_{nz}$ is realized, we uniformly pick $n_{nz}$ positions in  $x$ for the non-zero entries . Moreover, each of the non-zero entry in $x$ follows an independent normal distribution with a different mean but same variance $N(h_i,\sigma^{2})$.

My aim is to compute the variance of $x$ defined as $\mathbb{E}(xx^{T})-(\mathbb{E}x)(\mathbb{E}x)^{T}.$

答:
Before presenting my answer, let's establish some conventions.

Given $Z$ and $W$  random variables and $f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ a functión, then $\mathbb{E}_{Z}[f(Z,W)]$ denote the expected value respect to $Z$. Note that with this convention $\mathbb{E}_{Z}[f(Z,W)]$ is also a random variable.

Given $M\leq N$ we define the set $\mathcal{P}_{M}$ given  by $$\mathcal{P}_{M}:=\mbox{All subsets of } \{1,2,\ldots,N\}\mbox{ of size }M.$$

In my answer, I will repeatedly use the Law of Total Probability for expected values.
We are assuming that $\mathrm{Var}(x):=\mathbb{E}[xx^{T}]-(\mathbb{E}[x])(\mathbb{E}[x])^{T}$, So we just have to calculate $\mathbb{E}[xx^{T}]$ and $\mathbb{E}[x]$. Let's start with $\mathbb{E}[xx^{T}]$.

\begin{align}
\mathbb{E}[xx^{T}]&=\mathbb{E}[xx^{T}\mathbf{1}_{\{N_{nz}>0\}}+xx^{T}\mathbf{1}_{\{N_{nz}=0\}}] \\
&=\mathbb{E}[\mathbb{E}[xx^{T}\mathbf{1}_{\{N_{nz}>0\}}+xx^{T}\mathbf{1}_{\{N_{nz}=0\}}|N_{nz}]] \\
&=\mathbb{E}\left[\mathbb{E}[xx^{T}\mathbf{1}_{\{N_{nz}>0\}}|N_{nz}]+\mathbb{E}[xx^{T}\mathbf{1}_{\{N_{nz}=0\}}|N_{nz}]\right] \\
&\qquad\qquad \mbox{Nore that if }N_{nz}=0\mbox{ then }x=\mathbf{0}\mbox{, then }\mathbb{E}[xx^{T}\mathbf{1}_{\{N_{nz}=0\}}|N_{nz}]=0.\\
&= \mathbb{E}\left[\mathbb{E}[xx^{T}\mathbf{1}_{\{N_{nz}>0\}}|N_{nz}]\right]  \\
&=\mathbb{E}_{\color{red}{N_{nz}}}\left[\mathbf{1}_{\{N_{nz}>0\}}\mathbb{E}_{\color{red}{x,U}}\left[\sum_{i=1}^{N_{nz}}x_{u_{i}}^{2}\right]\right]\\
& \qquad\qquad\mbox{where }U:=(u_{1},\dots,u_{N_{nz}})\sim\mathrm{Uniform}(\mathcal{P}_{N_{nz}})\\
& \qquad\qquad\mbox{and } x_{u_{i}}\neq 0 \mbox{for each }i=1,\ldots,N_{nz}.\\
& \qquad\qquad \mbox{Also, note that for $u_{i}$ fixed } x_{ui}\sim N(h_{u^{i},\sigma^{2}}), \mbox{ then} \\
& \qquad\qquad x_{u_{i}}=\sigma Y_{u_{i}}+h_{i}\mbox{ where  }\{Y_{i}\}_{i=1}^{N} \mbox{ are iid }N(0,1) \mbox{ random variables.}\\
& \qquad\qquad \mbox{Therefore, } x_{u_{i}}^{2}=\sigma^{2}Y_{u_{i}}^{2}+2h_{u_{i}}\sigma Y_{u_{i}}+h_{u_{i}}^{2}\\
&= \mathbb{E}_{\color{red}{N_{nz}}}\left[\mathbf{1}_{\{N_{nz}>0\}}\mathbb{E}_{\color{red}{Y,U}}\left[\sum_{i=1}^{N_{nz}}\sigma^{2}Y_{u_{i}}^{2}+2h_{u_{i}}\sigma Y_{u_{i}}+h_{u_{i}}^{2}\right]\right]\\
 &= \mathbb{E}_{\color{red}{N_{nz}}}\left[\mathbf{1}_{\{N_{nz}>0\}}\mathbb{E}_{\color{red}{U}}\left[\mathbb{E}\left[\left.\sum_{i=1}^{N_{nz}}\sigma^{2}Y_{u_{i}}^{2}+2h_{u_{i}}\sigma Y_{u_{i}}+h_{u_{i}}^{2}\right|U\right]\right]\right]\\
&= \mathbb{E}_{\color{red}{N_{nz}}}\left[\mathbf{1}_{\{N_{nz}>0\}}\mathbb{E}_{\color{red}{U}}\left[\mathbb{E}_{\color{red}{Y}}\left[\sum_{i=1}^{N_{nz}}\sigma^{2}Y_{u_{i}}^{2}+2h_{u_{i}}\sigma Y_{u_{i}}+h_{u_{i}}^{2}\right]\right]\right]\\
&= \mathbb{E}_{\color{red}{N_{nz}}}\left[\mathbf{1}_{\{N_{nz}>0\}}\mathbb{E}_{\color{red}{U}}\left[\sum_{i=1}^{N_{nz}}\sigma^{2}\mathbb{E}_{\color{red}{Y}}\left[Y_{u_{i}}^{2}\right]+2h_{u_{i}}\sigma \mathbb{E}_{\color{red}{Y}}\left[Y_{u_{i}}\right]+h_{u_{i}}^{2}\right]\right]\\
& \qquad\qquad \mbox{Note that for $u_{i}$ fixed we have }Y_{u_{i}}^{2}\sim\mathcal{X}^{2}_{1} \mbox{ where}\\
&\qquad\qquad \mathcal{X}_{1}^{2} \mbox{ is  Chi-squared distribution with 1 degree of freedom.}\\
&\qquad\qquad \mbox{Therefore, }\mathbb{E}_{\color{red}{Y}}\left[Y_{u_{i}}^{2}\right]=1. \mbox{In addition, we know that } \mathbb{E}_{\color{red}{Y}}\left[Y_{u_{i}}\right]=0.\\
&= \mathbb{E}_{\color{red}{N_{nz}}}\left[\mathbf{1}_{\{N_{nz}>0\}}\mathbb{E}_{\color{red}{U}}\left[\sum_{i=1}^{N_{nz}}(\sigma^{2}+h_{u_{i}}^{2})\right]\right]\\
&= \mathbb{E}_{\color{red}{N_{nz}}}\left[\mathbf{1}_{\{N_{nz}>0\}}\sum_{(u_{1},\ldots,u_{N_{nz}})\in \mathcal{P}_{N_{nz}}}\frac{1}{\binom{N}{N_{nz}}}\left(\sum_{i=1}^{N_{nz}}(\sigma^{2}+h_{u_{i}}^{2})\right)  \right]\\
&= \mathbb{E}_{\color{red}{N_{nz}}}\left[\mathbf{1}_{\{N_{nz}>0\}}\left(N_{nz}\sigma^{2}+\frac{1}{\binom{N}{N_{nz}}}\sum_{(u_{1},\ldots,u_{N_{nz}})\in \mathcal{P}_{N_{nz}}}\left(\sum_{i=1}^{N_{nz}}h_{u_{i}}^{2}\right)\right)  \right]\\
&= \mathbb{E}_{\color{red}{N_{nz}}}\left[ \mathbf{1}_{\{N_{nz}>0\}}\left( N_{nz}\sigma^{2}+\frac{1}{\binom{N}{N_{nz}}}\sum_{j=1}^{N}h_{j}\binom{N-1}{N_{nz}-1}  \right)\right]\\
\end{align}
\begin{align}
&= \mathbb{E}_{\color{red}{N_{nz}}}\left[ \mathbf{1}_{\{N_{nz}>0\}}\left(N_{nz}\sigma^{2}+\frac{\binom{N-1}{N_{nz}-1} }{\binom{N}{N_{nz}}}\sum_{j=1}^{N}h_{j}^{2} \right)\right] \color{white}{--------.}\\
&= \mathbb{E}_{\color{red}{N_{nz}}}\left[\mathbf{1}_{\{N_{nz}>0\}}\left(N_{nz}\sigma^{2}+\frac{N_{nz} }{N}\sum_{j=1}^{N}h_{j}^{2} \right)\right]\\
&= \sum_{n_{nz}=0}^{k}\mathbf{1}_{\{n_{nz}>0\}}\left(n_{nz}\sigma^{2}+\frac{n_{nz} }{N}\sum_{j=1}^{N}h_{j}^{2} \right)\mathbb{P}[N_{nz}=n_{nz}]\\
&= \sum_{n_{nz}=1}^{k}\left(n_{nz}\sigma^{2}+\frac{n_{nz} }{N}\sum_{j=1}^{N}h_{j}^{2} \right)\mathbb{P}[N_{nz}=n_{nz}]\\
&= \sum_{n_{nz}=1}^{k}n_{nz}\left(\sigma^{2}+\frac{1 }{N}\sum_{j=1}^{N}h_{j}^{2} \right)\frac{e^{-\lambda}\lambda^{n_{nz}}/(n_{nz}!)}{\sum_{i=0}^{k}e^{-\lambda}\lambda^{i}/i!}\\
&= \left(\sigma^{2}+\frac{1 }{N}\sum_{j=1}^{N}h_{j}^{2} \right)\sum_{n_{nz}=1}^{k}n_{nz}\frac{e^{-\lambda}\lambda^{n_{nz}}/(n_{nz}!)}{\sum_{i=1}^{k}e^{-\lambda}\lambda^{i}/i!}\\
&=\left(\sigma^{2}+\frac{1 }{N}\sum_{j=1}^{N}h_{j}^{2} \right)\left(\frac{1}{\sum_{i=0}^{k}e^{-\lambda}\lambda^{i}/i!}\sum_{n_{nz}=1}^{k}n_{nz}\frac{e^{-\lambda}\lambda^{n_{nz}}}{n_{nz}!}\right)\\
&=\left(\sigma^{2}+\frac{1 }{N}\sum_{j=1}^{N}h_{j}^{2} \right)\left(\frac{1}{\sum_{i=0}^{k}e^{-\lambda}\lambda^{i}/i!}\sum_{n_{nz}=1}^{k}\frac{e^{-\lambda}\lambda^{n_{nz}}}{(n_{nz}-1)!}\right)\\
&=\left(\sigma^{2}+\frac{1 }{N}\sum_{j=1}^{N}h_{j}^{2} \right)\left(\frac{\lambda}{\sum_{i=0}^{k}e^{-\lambda}\lambda^{i}/i!}\sum_{\ell=0}^{k-1}\frac{e^{-\lambda}\lambda^{\ell}}{\ell!}\right)\\
&=\left(\sigma^{2}+\frac{1 }{N}\sum_{j=1}^{N}h_{j}^{2} \right)\left(\lambda -\lambda\frac{e^{-\lambda}\lambda^{k}/k!}{\sum_{i=0}^{k}e^{-\lambda}\lambda^{i}/i!}\right)\\
&=\left(\sigma^{2}+\frac{1 }{N}\sum_{j=1}^{N}h_{j}^{2} \right)\lambda\left(1-\mathbb{P}[N_{nz}=k]\right)\\
\end{align}

Now we are going to calculate $\mathbb{E}[x]:=[\mathbb{E}[x_{1}],\ldots,\mathbb{E}[x_{N}]]$, note that we only have to calculate $\mathbb{E}[x_{i}]$ for each $i=1,2,\ldots,N$.

\begin{align}
\mathbb{E}[x_{i}] &= \mathbb{E}[x_{i}\mathbf{1}_{\{N_{nz}>0\}}+x_{i}\mathbf{1}_{\{N_{nz}=0\}}] \\
&= \mathbb{E}\left[\mathbb{E}[x_{i}\mathbf{1}_{\{N_{nz}>0\}}+x_{i}\mathbf{1}_{\{N_{nz}=0\}}|N_{nz}]\right] \\
&= \mathbb{E}\left[\mathbb{E}[x_{i}\mathbf{1}_{\{N_{nz}>0\}}|N_{nz}]+\mathbb{E}[x_{i}\mathbf{1}_{\{N_{nz}=0\}}|N_{nz}]\right] \\
&\qquad\qquad \mbox{Nore that if }N_{nz}=0\mbox{ then }x=\mathbf{0}\mbox{, then }\mathbb{E}[x_{i}\mathbf{1}_{\{N_{nz}=0\}}|N_{nz}]=0.\\
&=\mathbb{E}\left[\mathbb{E}\left[\left. x_{i}\mathbf{1}_{\{N_{nz}>0\}}\right|N_{nz}\right]\right]\\
&= \mathbb{E}_{\color{red}{N_{nz}}}\left[\mathbf{1}_{\{N_{nz}>0\}}\mathbb{E}_{\color{red}{X,U}}\left[X_{i}\mathbf{1}_{i\in U}+X_{i}\mathbf{1}_{i\notin U}\right]\right]\\
&\qquad\qquad \mbox{where } U:=(u_{1},\dots,u_{N_{nz}})\sim\mathrm{Uniform}(\mathcal{P}_{N_{nz}})\\
&= \mathbb{E}_{\color{red}{N_{nz}}}\left[\mathbf{1}_{\{N_{nz}>0\}}\mathbb{E}\left[\mathbb{E}\left[\left.X_{i}\mathbf{1}_{i\in U}+X_{i}\mathbf{1}_{i\notin U}\right|U\right]\right]\right]\\
&=\mathbb{E}_{\color{red}{N_{nz}}}\left[\mathbf{1}_{\{N_{nz}>0\}}\mathbb{E}_{\color{red}{U}}\left[\mathbb{E}_{\color{red}{X}}\left[X_{i}\mathbf{1}_{i\in U}\right]+\mathbb{E}_{\color{red}{X}}\left[X_{i}\mathbf{1}_{i\notin U}\right]\right]\right]\\
&\qquad\qquad \mbox{Note that if }i\in U \mbox{ then }x_{i}\sim N(h_{i},\sigma^{2}), \mbox{therefore}, \mathbb{E}_{\color{red}{X}}\left[X_{i}\mathbf{1}_{i\in U}\right]=h_{i}.\\
&\qquad\qquad \mbox{Note that if }i\notin U \mbox{ then }x_{i}=0, \mbox{therefore}, \mathbb{E}_{\color{red}{X}}\left[X_{i}\mathbf{1}_{i\notin U}\right]=0.\\
&=\mathbb{E}_{\color{red}{N_{nz}}}\left[\mathbf{1}_{\{N_{nz}>0\}}\mathbb{E}_{\color{red}{U}}\left[h_{i}\mathbf{1}_{i\in U}\right]\right]\\
&=\mathbb{E}_{\color{red}{N_{nz}}}\left[\mathbf{1}_{\{N_{nz}>0\}}h_{i}\mathbb{P}\left[i\in U\right]\right]\\
&=\mathbb{E}_{\color{red}{N_{nz}}}\left[\mathbf{1}_{\{N_{nz}>0\}}h_{i}\frac{\binom{N-1}{N_{nz}-1}}{\binom{N}{N_{nz}}}\right]\\
&=\frac{h_{i}}{N}\mathbb{E}_{\color{red}{N_{nz}}}\left[\mathbf{1}_{\{N_{nz}>0\}}N_{nz}\right]\\
&=\frac{h_{i}}{N}\sum_{n_{nz}=0}^{k}\mathbf{1}_{\{n_{nz}>0\}}n_{nz}\mathbb{P}[N_{nz}=n_{nz}]\\
&=\frac{h_{i}}{N}\sum_{n_{nz}=1}^{k}n_{nz}\frac{e^{-\lambda}\lambda^{n_{nz}}/(n_{nz}!)}{\sum_{j=0}^{k}e^{-\lambda}\lambda^{j}/j!}\\
&=\frac{h_{i}}{N\sum_{j=0}^{k}e^{-\lambda}\lambda^{j}/j!}\sum_{n_{nz}=1}^{k}\frac{n_{nz}e^{-\lambda}\lambda^{n_{nz}}}{n_{nz}!}\\
&=\frac{h_{i}}{N\sum_{j=0}^{k}e^{-\lambda}\lambda^{j}/j!}\sum_{n_{nz}=1}^{k}\frac{e^{-\lambda}\lambda^{n_{nz}}}{(n_{nz}-1)!}\\
&=\frac{h_{i}\lambda}{N\sum_{j=0}^{k}e^{-\lambda}\lambda^{j}/j!}\sum_{n_{nz}=1}^{k}\frac{e^{-\lambda}\lambda^{n_{nz}-1}}{(n_{nz}-1)!}\\
&=\frac{h_{i}\lambda}{N\sum_{j=0}^{k}e^{-\lambda}\lambda^{j}/j!}\sum_{\ell=0}^{k-1}\frac{e^{-\lambda}\lambda^{\ell}}{\ell!}\\
&=\frac{h_{i}\lambda}{N\sum_{j=0}^{k}e^{-\lambda}\lambda^{j}/j!}\left(\sum_{\ell=0}^{k}\frac{e^{-\lambda}\lambda^{\ell}}{\ell!}-\frac{e^{-\lambda}\lambda^{k}}{k!}\right)\\
&=\frac{h_{i}\lambda}{N}\left(1-\mathbb{P}[N_{nz}=k]\right)
\end{align}

Therefore, we have
\begin{align}
(\mathbb{E}[x])(\mathbb{E}[x])^{T}&=\sum_{i=1}^{N} (\mathbb{E}[x_{i}])^{2}\\
&= \sum_{i=1}^{N} \frac{h_{i}^{2}\lambda^{2}}{N^2}\left(1-\mathbb{P}[N_{nz}=k]\right)^{2}\\
&= \lambda^{2}\left(\frac{1-\mathbb{P}[N_{nz}=k]}{N}\right)^{2}\sum_{i=1}^{N} h_{i}^{2}.
\end{align}

So, we get the following

\begin{align}
\mathrm{Var}(x)&:=\mathbb{E}[xx^{T}]-(\mathbb{E}[x])(\mathbb{E}[x])^{T}\\
&= \left(\sigma^{2}+\frac{1 }{N}\sum_{j=1}^{N}h_{j}^{2} \right)\lambda\left(1 -\mathbb{P}[N_{nz}=k]\right)-\lambda^{2}\left(\frac{1-\mathbb{P}[N_{nz}=k]}{N}\right)^{2}\sum_{i=1}^{N} h_{i}^{2}\\
&=\lambda(1-\mathbb{P}[N_{nz}=k])\left(\sigma^{2}+\frac{(N-\lambda(1-\mathbb{P}[N_{nz}=k]))}{N^{2}}\sum_{i=1}^{N} h_{i}^{2}\right).
\end{align}


问:
How can I deduce when this equation is an integer?
The equation is $\displaystyle\frac{k(2-k)}{\sqrt{4k-3}}$. I have determined that it's integer when $k\in \{2,3,7,57\}$. Are there any other possibilities? How do I find all of them (and convince myself that I've found them all)?

Thank you!

答:
Assuming you want $k \in \mathbb N$, you missed the case $k=1$.

Let $x =\dfrac {k(2-k)}{\sqrt{4k-3}}$. Then if $x \in \mathbb Z$, $256 x^2 \in \mathbb Z$.

We have:

$$256x^2 = \frac {256 (2 - k)^2 k^2}{4 k - 3} = 64 k^3 - 208 k^2 + 100 k + \frac {225}{4 k - 3} + 75$$

so if $x\ne 0$, $4k-3$ must be a divisor of $225$, and since $\sqrt{4k-3}$ is in the denominator, $4k-3$ must also be square. This gives the cases:

$$(k=2) \text{ or } (4k-3 = 1,9,25,225)$$


问:
Showing that $\lfloor\frac{x-1}3\rfloor=\lfloor\frac{x}3+\frac23\rfloor-1$ and $\lfloor\frac{x+1}3\rfloor=\lfloor\frac{x}3+\frac13\rfloor$.
I have 2 questions about the floor functions:

1) $\left\lfloor \frac{x-1}{3}\right\rfloor =\left\lfloor \frac{x}{3}+\frac{2}{3}\right\rfloor -1$

2) $\left\lfloor \frac{x+1}{3}\right\rfloor =\left\lfloor \frac{x}{3}+\frac{1}{3}\right\rfloor$

As we know that the definitions and properties of floor functions are:

1) $\lfloor x\rfloor =m$ if $m\leq x<m+1$ and

2) $\lfloor m+x\rfloor =\lfloor x\rfloor +m$ if $m$ is an integer.

Questions:

1) Why the first floor function above has to +1 inside the floor brackets and -1 outside the floor brackets:
  $\left\lfloor \frac{x-1}{3}\right\rfloor$
= $\left\lfloor \frac{x-1}{3}+1\right\rfloor -1$
= $\left\lfloor \frac{x}{3}+\frac{2}{3}\right\rfloor -1$

2)Why the second floor function above doesn't need to add or minus 1 inside or outside the floor brackets:
  $\left\lfloor \frac{x+1}{3}\right\rfloor$
= $\left\lfloor \frac{x}{3}+\frac{1}{3}\right\rfloor$

Does anyone here know the reason? Thank you.

答:
If n is an integer, then $\lfloor x + n \rfloor = \lfloor x \rfloor + n$.

\begin{align}
   \left\lfloor\frac{x-1}3 \right\rfloor
      &= \left\lfloor\frac{x-1}3 \right\rfloor + 1 - 1 \\
      &= \left\lfloor\frac{x-1}3 + 1\right\rfloor - 1 \\
      &= \left\lfloor\frac x3 + \frac 23 \right\rfloor - 1
\end{align}

The second is true because $\dfrac{x+1}3 = \dfrac x3 + \dfrac 13$.


问:
Equation of three lines passing through the origin
If $dx^3+7x^2y+bxy^2+y^3=0$ represent three straight lines of which two of them makes complementary angles with x-axis then the value of $|d^2-bd|$=_______

My approach is as follows

From the above equation, presume that line passes through origin

The equation of lines are $y=m_1x$, $y=m_2x$ & $ y=m_3x$

For complementary angle the angle between them is $180^o$

$(y-m_1x)(y-m_2x)(y-m_3x)=0$

$(y^2-(m_1+m_2)xy+m_1m_2x^2)(y-m_3x)=0$

$(y^3+(m_1m_2+m_2m_3+m_3m_1)x^2y-(m_1+m_2+m_3)xy^2+m_1m_2m_3x^3)=0$

How do I proceed from here. For supplementary angle $m_1m_2=-1$  then what is the condition for complementary angles?

Body:
$m_1m_2=1$. Let $m_1 + m_2 = S$.

Comparing coefficients,

$$y^3 - xy^2\sum m_1  + x^2y \sum m_1m_2 \color{red}{-x^3}m_1m_2m_3  = 0 =$$$$ y^3 +  bxy^2 + 7x^2y + dx^3$$

we get

$$ d=-m_3$$

$$-b = S+m_3$$

$$ 7 = 1 + m_3S$$

so that $$|d(d-b)| = |-m_3(S)| $$

$$ = 6 $$


问:
Finding differential of inverse?
Let $f(x)=6x^3-11x^2-2$, $x\geqslant 1.
5$. Find $\frac{\text{d}f^{-1}}{\text{d}x}$ at the point $x=1517=f(7)$.

I thought you should first find the inverse and then differentiate but from looking at the inverse, that's clearly not what I should do. Any ideas? Also, why did they give me x>=1.5?

答:
Just to show you what you did avoid to face.

If $y>0$ the inverse is given by
$$x=\frac{11}{18} \left(1+2 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{486y+2303}{1331}\right)\right)\right)$$
$$\frac {dx}{dy}=\frac{11 \sinh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{486
   y+2303}{1331}\right)\right)}{\sqrt{729 y^2+6909 y+10902}}$$ If $x=1517$, $y=20921018297$ making at this point
$$\frac {dx}{dy}=\frac{11 \sinh \left(\frac{1}{3} \cosh
   ^{-1}\left(\frac{10167614894645}{1331}\right)\right)}{20694914
   \sqrt{745016541}}$$ while,as given in the answers,
$$\frac {dy}{dx}=18 x^2-22 x$$ which is for $x=1517$
$$\frac {dy}{dx}=41389828 \implies \frac {dx}{dy}=\frac 1 {41389828}$$

Which one do you prefer ?

Title: Finding differential of inverse?


问:
Trace conserves singularity for certain case
Suppose $A, B$ are matrices and $A\sim D$.

Then does $\text{tr}(ABA^{-1}B^{-1})=\text{tr}(DBD^{-1}B^{-1})$ hold?

I think this works because $\text{tr}(AB)=\text{tr}(BA)$, but it isn't easy to prove.

答:
No. It should be easy to construct a counterexample. E.g. when $B$ commutes with $A$ but not $D$, the commutator $ABA^{-1}B^{-1}$ is always equal to $I$ but $DBD^{-1}B^{-1}$ is more arbitrary and the equality in question will likely fail. Here is a concrete counterexample on any field of characteristic $\ne2$:
\begin{aligned}
&A=B=B^{-1}=\pmatrix{0&1\\ 1&0},\ D=\pmatrix{1&0\\ 0&-1},\\
&ABA^{-1}B^{-1}=I,\\
&DBD^{-1}B^{-1}=-I,\\
&\operatorname{tr}(ABA^{-1}B^{-1})=2\ne-2=\operatorname{tr}(DBD^{-1}B^{-1}).
\end{aligned}
On a field of characteristic $2$, replace the $D$ above by $D_1=\pmatrix{1&1\\ 0&1}$. Then $A\sim D_1$ and $\operatorname{tr}(ABA^{-1}B^{-1})=0\ne1=\operatorname{tr}(D_1BD_1^{-1}B^{-1})$.


问:
Show that if a vector $v$ is orthogonal to $u$ then $Pv=0$ being $P=uu^T$
Show that if a vector $v$ is orthogonal to $u$ then $Pv=0$ being $P=uu^T$ and $\displaystyle u=(\frac{1}{6},\frac{1}{6},\frac{3}{6},\frac{5}{6})$.

First I showed that $uu^T=1$, and the hypothesis of this problem tells me that $<v,u>=0$ (since they're orthogonal).

Note:  is an inner product.

If $P=1$, why would $Pv=0$ and not lead to $v$?

答:
Note that vectors are denoted by column matrices, not row matrices. So $u=\begin{bmatrix}1/6\\1/6\\3/6\\5/6\end{bmatrix}=\begin{bmatrix}1/6&1/6&3/6&5/6\end{bmatrix}^T$, so your question is missing the transpose.

Now $uu^T$ is not $1$ but a $4\times4$ matrix. Given $\langle u,v\rangle=u^Tv=0$, we have $Pv=(uu^T)v=u(u^Tv)=\vec0$ by associativity of matrix product.


问:
Jacobson radical and Frattini group
The Frattini subgroup of $\Bbb{Z}_p \times\Bbb Z _{p^2}.$

According to this answer, the Frattini subgroup is equivalently its Jacobson radical as a ring (this
is actually true for any ring that is a finite direct product of
quotient rings of $\mathbf{Z}$).

How can I formally prove this statement?

I want to use this statement to find the Frattini subgroup of a given ring like product of quotient groups of $\mathbf{Z}$. Thank you.

答:
Lemma 1: if $M$ and $N$ are any abelian groups, $C<M\times N$ is a maximal subgroup, and $D\leqslant M\times N$ is any subgroup, then $C\cap D\leqslant D$ is either all of $D$ or is a maximal subgroup of $D$.

Proof: Note that, since $C$ is maximal in $M\times N$, $M\times N\big/C$ is simple. But also, $$D\big/C\cap D\cong (C+D)\big/C$$ is a subgroup of $M\times N\big/C$, so by simplicity either $D\big/C\cap D$ is trivial or $D\big/C\cap D\cong M\times N\big/C$. In the first case, $C\cap D=D$, and in the second case, $D\big/C\cap D$ is simple, whence $C\cap D$ is maximal in $D$.

Lemma 2: if $M,N$ are any abelian groups, then $\Phi(M\times N)=\Phi(M)\times\Phi(N)$. (Note that we do not need a finiteness condition here, nor any condition on whether $M$ or $N$ have maximal subgroups. So the Frattini subgroups are much better behaved for abelian groups than for non-abelian ones. This lemma does not hold for non-abelian groups in general.)

Proof: If $A$ (resp. $B$) is a maximal subgroup of $M$ (resp. $N$), then $A\times N$ (resp. $M\times B$) is a maximal subgroup of $M\times N$. Hence we have

\begin{align}
\Phi(M\times N)&\leqslant(\bigcap_{\substack{A<M \\{ \text{max}}}}A\times N)\space\cap\space(\bigcap_{\substack{B<N \\{ \text{max}}}}M\times B)
\\&=(\Phi(M)\times N)\cap(M\times\Phi(N))
\\&=\Phi(M)\times\Phi(N)
\end{align}

For the other direction, it suffices to show the two inclusions $\Phi(M)\times 0\leqslant\Phi(M\times N)$ and $0\times\Phi(N)\leqslant\Phi(M\times N)$. Hence let $C<M\times N$ be any maximal subgroup. By lemma 1, $C\cap(M\times 0)$ is either all of $M\times 0$ or is maximal in $M\times 0$, and in both cases $\Phi(M)\times 0=\Phi(M\times 0)\leqslant C$. Since $C$ was arbitrary, we hence have $\Phi(M)\times 0\leqslant\Phi(M\times N)$, and exactly the same argument shows $0\times\Phi(N)\leqslant\Phi(M\times N)$, so we are done.

Now let $R$ be a finite direct product of quotient rings of $\mathbb{Z}$, say $R=(\mathbb{Z}\big/n_1\mathbb{Z})\times\dots\times(\mathbb{Z}\big/n_k\mathbb{Z})$. By induction and lemma 2 we have $\Phi(R)=\Phi(\mathbb{Z}\big/n_1\mathbb{Z})\times\dots\times\Phi(\mathbb{Z}\big/n_k\mathbb{Z})$. On the other hand, every ideal of a finite product of rings is a product of the constituent rings' ideals, so every maximal ideal of $R$ is of the form $(\mathbb{Z}\big/n_1\mathbb{Z})\times\dots \times M\times\dots\times(\mathbb{Z}\big/n_k\mathbb{Z})$, where $M$ is a maximal ideal of some $\mathbb{Z}\big/n_i\mathbb{Z}$. Hence $J(R)=J(\mathbb{Z}\big/n_1\mathbb{Z})\times\dots\times J(\mathbb{Z}\big/n_k\mathbb{Z})$.

(Looking at your previous questions it looks like you have a bit of confusion about this latter point, so try filling in the details yourself and show that $J(R_1\times\dots\times R_k)=J(R_1)\times\dots\times J(R_k)$ for any commutative rings $R_i$.)

Thus to show $J(R)=\Phi(R)$ all we need to show is that $J(\mathbb{Z}\big/n\mathbb{Z})=\Phi(\mathbb{Z}\big/n\mathbb{Z})$ for any $n\in\mathbb{Z}$, but this is clear, as the ideals of $\mathbb{Z}\big/n\mathbb{Z}$ are precisely the subgroups of $\mathbb{Z}\big/n\mathbb{Z}$.

Title: Jacobson radical and Frattini group


问:
Is there a mathematical formula for the nearest-square function?
Let $x$ be a positive integer.

Is there a mathematical formula for
$$f(x)=\text{nearest square to } x \text{ }(\text{in terms of } x)?$$

I tried searching for related questions in MSE and found this one, but my inquiry is not covered there.

I also tried searching via Google and found this closely related question in StackOverflow.

I used the search string "nearest-square function".

答:
By virtue of the accepted answer in your linked question, you can take the nearest integer function of $\sqrt x$ and square it. However it may not be obvious (or seemingly too obvious to be true) so here are more details:

The nearest integer function (for $x \notin \{k+\frac12| k \in \mathbb Z\}$ to avoid ambiguities) is equal to $\left\lfloor x + \frac12\right\rfloor$. Fortunately for the square root, it is never an integer plus half. Hence we can also write

$$f(x) = \left\lfloor \sqrt x + \frac12\right\rfloor^2$$

To prove that this formula works, we consider the cases (where $n \in \mathbb N \cup \{0\}$):

Case 1: $n^2 \le x \le n^2+n < (n+\frac12)^2$

Case 2: $(n+\frac12)^2< n^2+n+1 \le x < (n+1)^2$

and we see that both cases work perfectly fine, given that $x$ is a positive integer.


问:
Probability mass function of a, random variable considered on a probability mass function
This is a proof that I developed for myself and I'm asking whether it is true.

Let X be a random variable that take integer values with probability mass function of PX(x). Let Y be a random variable that also take integer values.

Consider the below discrete random variable,

PX(Y)

We can describe this as follows,

Let b be an event such that Y(b) = z. The random variable PX(Y) describes the probability of event a  that corresponds to X(a) = z.

Now I tried to solve the probability mass function of PX(Y) for a given value.

Which is PPX(Y)(0.1).

My argument is, PPX(Y)(0.1) tells us to find the probability of such event c.

To find this event c, I thought it as the event which gave PX(Y) = 0.1. That means there is such an event b in Y's sample space and an event a in X's sample such that X(a) = Y(b). And P(a) = 0.1.

Therefore the event that gave the random variable 0.1 value is a. Hence c = a.

Therefore PPX(Y)(0.1) = 0.1.

Am I correct. If you can point out what I've done is correct or wrong I'm very glad.

答:
I went through some further thinking on the problem and found an elegant solution.

Given above random variables $X$ and $Y$.
Since ${P_X(Y)}$ is a random variable, it has a PMF. So my interpretation for $P_ X(Y)$ is follows.

Let $b$ be an event s.t $Y(b) = z$, for this event $P_ X(Y)$ takes the value of $P(a)$, where $a$ is an event in $X$'s sample space s.t $X(a) = z$

Now lets consider the PMF of $P_ X(Y)$, which is $P_{P_X(Y)}$.

If we calculate the $P_{P_X(Y)}(0.1)$, my argument is,

$P_{P_X(Y)}(0.1)$ tells us to find the probability of such event $c$.

To find this event $c$, I thought it as the event which gave $P_ X(Y)$ = 0.1. That means there is such an event $a$ in $X$'s sample space and $P(a) = 0.1$. This event $a$ is found  for $P_ X(Y(b))$ such that $X(a) = Y(b)$ for an event $b$ in $Y$'s sample space.
Therefore the event of interest that gave the random variable a value of 0.1, is $b$. Hence $c = b$.

Therefore $P_{P_X(Y)}(0.1)$= $P(b)$.


问:
Compute the desired area
I solved it on my own but I need some feedback on whether I solved it correctly or not.

Consider the function $$f(x)=x^3-2x^2-x+2$$

Let $h(x)$ be the function that results when $f(x)$ is shifted by $−1$ in the $x$-direction.

a) Sketch the graph of the function $f(x)$.

I sketched it.

b) Sketch the graph of the function $h(x)$ and give an algebraic definition
for $h(x)$.

I sketched it and I formulated the formula $$h(x)=f(x+1)=(x+1)^3-2(x+1)^2-(x+1)+2= x^3-2x^2-2x$$

Is this correct?

c) Find the intersection points of $f(x)$ and $h(x)$.

I solved this simultaneously and got $x=-2$
Then I plugged this value in at $f(x)$ and got $y=-12$

d) Compute the desired area.

this is the part i am confused about. At what interval do I need to do this? -2 to 2? I used -2 to 2 and got the Area= 16.

答:
[F(x)-H(x)] from -2/3 to 1, where H'=h and F'=f
u can see the desired area in graph-picture</p>

Title: Compute the desired area


问:
A Group Isomorphic to $D_8$
How to find a group isomorphic to $D_8$ consisting of elements of $GL(2,\Bbb C)$?

How about $GL(2,\Bbb R)$?

Answer:
As we know $D_8=\{e, a, a^2, a^3, b, ab, a^2b, a^3b\}$ and $a^4=b^2=e, ba= a^3b$.

Obviously:
$$e=\left[\begin{matrix}
1&0\\
0&1\\
\end{matrix}\right]$$

If $$a=\left[\begin{matrix}
0&1\\
-1&0\\
\end{matrix}\right]$$ and $$b=\left[\begin{matrix}
0&1\\
1&0\\
\end{matrix}\right]$$
Is it true?

We should prove $\phi: D_8 \rightarrow G$ is $1-1$ and onto and $$\phi(a*b)=\phi(a)\cdot\phi(b)$$

Thank you for your answers.

答: The isometries of a square in $\mathbb{R}^2$ form a group isomorphic to $D_8$. If you center the square at the origin, each isometry can be represented by a linear transformation (like a rotation or a reflection), which can be written as a $2 \times 2$ real matrix.

Title: A Group Isomorphic to $D_8$


问:
Binomial coefficient sum over top index
I am trying to evaluate a sum over binomial coefficients which is giving me some problems. Specifically I want to calculate:

$$\sum_{r=0}^{c-1}\binom{r+n}{n}\frac{1}{c-r}$$

My main thought was to convert the fraction here into:

$$\int_0^1 x^{c-r}dx,$$
move the integral out of the sum, alongside the $x^c$ and then attempt to rewrite as some closed form function. I however, cannot see what the generating function should be.

Note, my aim here is to avoid having a sum - some product of binomial coefficients would be ideal but I obviously do not know if this exists!

Any help on summing this would be greatly appreciated.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\sum_{r = 0}^{c - 1}{r + n \choose n}{1 \over c - r}}
\\[2mm] = &\
\bracks{z^{c}}\sum_{\ell = 0}^{\infty}z^{\ell}
\sum_{r = 0}^{\ell - 1}{r + n \choose n}
{1 \over \ell - r}
\\[5mm] = &\
\bracks{z^{c}}\sum_{r = 0}^{\infty}{r + n \choose n}
\sum_{\ell = r + 1}^{\infty}\,\,{z^{\ell} \over
\ell - r}
\\[5mm] = &\
\bracks{z^{c}}\sum_{r = 0}^{\infty}
{r + n \choose n}z^{r}\
\underbrace{\sum_{\ell = 1}^{\infty}\,\,
{z^{\ell} \over \ell}}_{\ds{-\ln\pars{1 - z}}}
\\[5mm] = &\
-\bracks{z^{c}}\ln\pars{1 - z}\ \times
\\[2mm] &\
\sum_{r = 0}^{\infty}
{\bracks{-r - n} + r - 1 \choose r}
\pars{-1}^{r}\,z^{r}
\\[5mm] = &\
-\bracks{z^{c}}\ln\pars{1 - z}
\sum_{r = 0}^{\infty}
{- n  - 1 \choose r}\pars{-z}^{r}
\\[5mm] = &\
-\bracks{z^{c}}\ln\pars{1 - z}\pars{1 - z}^{-n  - 1}
\\[5mm] = &\
-\bracks{z^{c}}\bracks{\nu^{1}}
\pars{1 - z}^{\nu -n  - 1}
\\[5mm] = &\
-\bracks{\nu^{1}}
{\nu - n - 1 \choose c}\pars{-1}^{c}
\\[5mm] = &\
-\bracks{\nu^{1}}{-\nu + n + 1 + c - 1 \choose c}
\pars{-1}^{c}\pars{-1}^{c}
\\[5mm] = &\
-\bracks{\nu^{1}}{c + n - \nu \choose c}
\\[5mm] = &\
\left.{n + c - \nu \choose c}
\pars{H_{n + c - \nu}\ -\ H_{n - \nu}}
\right\vert_{\ \nu\ =\ 0}
\\[5mm] = &\
\bbx{{n + c \choose c}
\pars{H_{n + c}\ -\ H_{n}}} \\ &
\end{align}

Title: Binomial coefficient sum over top index


问:
Show that the harmonic conjugates are collinear - Menelaus and Ceva's Theorems with homogeneous coordinates
Suppose we have a triangle $\triangle ABC$ such that $D$ is an arbitrary point on $BC$, $E$ is an arbitrary point on $AB$ and $F$ is an arbitrary point on $AC$. Let $G$ be the harmonic conjugate of $E$ with respect to $AB$ (the cross ratio $(A,B,E,G)=-1$), $I$ the harmonic conjugate of $D$ with respect to $BC$ (the cross ratio $(B,C,D,I)=-1$) and $H$ the harmonic conjugate of $F$ with respect to $AC$ (the cross ratio $(A,C,F,H)=-1$). Show that $G,I,H$ are collinear if and only if $AD$, $CE$ and $BF$ have a point in common.

The picture would look like this (I amended the pic as pointed out in the comments).

My idea is to convert $A,B,C$ to homogeneous coordinates, for example, $A=(1:0:0), B=(0:0:1)$ and $C=(0:1:0)$, but I am not sure where to go from there. I read that these are equivalent statements from Menelaus and Ceva's theorems but I cannot find anything on the internet with homogeneous coordinates, so any help is appreciated.

答:
I finally worked out the solution and I'd like to share it below.
Let $A=(1:0:0)$, $B=(0:0:1)$, $C=(0:1:0)$, and denote $M:=AD\cap CE$. Then let $M=(1:1:1)$ since $A,B,C,M$ are in general position so we can choose this labelling. Since $F\in AC$ we can write it as $F=(1:a:0)$ for some $a\neq 0$ by homogeneity. We use harmonic conjugacy as follows.

We know that two points $F'$ and $G'$ are harmonic conjugates with respect to $D',E'$ if $(D',E',F',G')=-1$ so if $F=\lambda_1D'+ \mu_1E'$ then $G=-\lambda_1D'+\mu_1E\sim \lambda_1D'-\mu_1E'$ since the cross-ratio is then $\frac{\lambda_1}{\mu_1}:\frac{-\lambda_1}{\mu_1}=-1$.

Then $(1:a:0)=F=1\cdot A + a\cdot C$ so $H=-1\cdot A +a \cdot C$ because they are harmonic conjugates. So $H=(-1:a:0)$. Similarly $I=(0:-1:1)$ and $G=(-1:0:1)$.

The three lines are concurrent if and only if $M\in BF$. The equation for $BF$ is $X_1=aX_0$ so $M\in BF$ if and only if $a=1$.

$G,H,I$ are collinear if and only if the determinant

$$\begin{vmatrix}-1 & a & 0 \\ 0 & -1 & -1 \\ -1 & 0 & 1\end{vmatrix}=0,$$
which happens if and only if $a=1$.

So we have equivalent statements.


问:
How to find the intersection of two given lines
How to find intersection of two lines below:
$$
\left \{ \begin{array}{ccc}
3x − 2y & = & 5\\
7x + 3y & = & 4
\end{array} \right .
$$
I know the answer is $(1,-1)$, but I cannot find it with the methods. I have already tried to solve pairs of linear equations.</p>

答:
try to equal any variable's coefficient in both eqn,either x or y.I take y to equal the coefficient.multiply 1st eqn by 3(2nd eqns y variable) and 2nd eqn by 2(1st eqns y variable)
$$
(3x − 2y  = 5)\cdot3$$
$$(7x + 3y  =  4)\cdot2
$$
now eqn will be:
$$
9x − 6y  = 15$$
$$14x + 6y  =  8
$$

now add both eqn side by side
$$9x − 6y +14x + 6y  =  15+8$$
$$23x=23$$
$$x=1$$
put this value of x in any eqn .I am taking 2nd eqn.
$$7\cdot(1)+3y=4$$
$$7+3y=4$$
$$3y=4-7$$
$$3y=-3$$
$$y=-1$$
so answer will be x=1 and y=-1 this is called elimination method


问:
Linear dependence and independence of vectors
Let $v_1 = (−1,−1,1,0)$ and $v_2 = (3,0,3,1)$ be two vectors in $\Bbb R^4$. (a) Find a vector $v$ such that the set $\{v_1,v_2,v\}$ is linearly independent. (b) Find real numbers $s,t \in\Bbb R$ such that the vector $v = (−2,1,s,t)$ is a linear combination of $v_1$ and $v_2$.

The solution I think (a) $v=(n,n,0,0),n\in\Bbb Z$

(b) $s,t = (-3,-1)$.

Please let me know if I'm wrong

答:
Your solution to part $(a)$ is almost correct. In order for $v_1,v_2,(n,n,0,0)$ to be linearly independent, $n\ne0$ otherwise $v=0$. You can check that$$\begin{bmatrix}-1&-1&1&0\\3&0&3&1\\n&n&0&0&\end{bmatrix}$$is a full rank matrix for $n\ne0$ by performing the row operations $R_2\to R_2+3R_1,R_3\to R_3+nR_1$ to get

$$\begin{bmatrix}-1&-1&1&0\\0&-3&6&1\\0&0&n&0&\end{bmatrix}$$

Let $(-2,1,s,t)=av_1+bv_2=a(-1,-1,1,0)+b(3,0,3,1)$.

So we have $-a+3b=-2,-a+0b=1$. We get $a=b=-1$.

So $s=a+3b=-4,t=b=-1$.


问:
Convergence in uniform convergence in probability to Brownian motion for left continuous riemann type sum of Brownian motion
Consider a standard Brownian motion $B_t$ with $B_0 = 0$.
I would like to show that if $\pi_n$ is a refining sequence of partitions of $[0,\infty)$ whose mesh tends to $0$, then $B_t^n = \sum_{t_k \in \pi_n} B_{t_k} 1_{(t_k, t_{k+1}]}$ converges to $B_t$ in ucp, i.e.
for all $t \ge 0$, $$ \lim_{n\to \infty}P(\sup_{0 \le s \le t}|B_s^n - B_s| >\epsilon) \to 0$$ for all $\epsilon>0$.

I know that on compact intervals, the Brownian motion is uniformly continuous, so we can choose the mesh small enough that for each fixed $\omega$, we can take the inner event to be empty. I.e. we could get $P(\lim_n \sup_{0 \le s \le t} |B_s^n - B_s|>\epsilon)=0$. However, I cannot figure out how to take this limit outside of the probability.

How can I prove this result?

答: All you need is $\lim \sup P(E_n) \leq P( \lim \sup E_n)$ which follows by applying Fatou's Lemma to then sequence $(I_{E_n^{c}})$.


问:
How do you show Tonelli’s theorem can fail in the σ-nonfinite case, and also that product measure does not need to be unique.
For this, I showed that
If we have the set $I=[0,1]$ equipped on one side with the Borel set and Lebesgue measure λ for X and on the other side the same set $I=[0,1]$ equipped with the power set and the counting measure m for Y.

The diagonal set Δ={$(x,x) : x∈[0,1]$} is a closed subset of $I_2$. Hence
Δ$\in(I_2)=\mathcal{B}(I)×\mathcal{B}(I)⊂\mathcal{B}(I)×\mathcal{B}(I)$.
Therefore, the characteristic map $1_Δ$ is measurable for the measure induced on $X×Y$. However, we have on one side
$\int_{I}(\int_{I}1_Δ(x,y)dm(y))dλ(x)=\int_{I}(\int_{I}1_{x}(y)dm(y))dλ(x)=\int_{I}m({x})dλ(x)=λ(I)=1$

and on the other side

$\int_{I}(\int_{I}1_Δ(x,y)dλ(x))dm(y)=\int_{I}(
\int_{I}1_{y}(x)dλ(x))dm(y)=\int_{I}λ({y})dm(y)=\int_{I}0 dm(y)=0$

These are not equal and Tonelli's Theorem failed. Is this right?

Also, in the case for uniqueness. I need to show that there is more than one measure $\mu$ on $\mathcal{L}[0,1]\times 2^{[0,1]}$ with the property that $\mu(E\times F)=\mu(E)$#$(F)$ (where # is the counting measure of F) for all $E\in \mathcal{L}([0,1])$ and $F\in 2^{[0,1]}$. How should I start with this?

答:
The measurability of $\Delta$ is actually due to the fact that one can write
$$\Delta = \bigcap_{n=1}^\infty \left(\bigcup_{k=1}^{2^n}\left[\frac{k-1}{2^n},\frac{k}{2^n}\right]^2\right)$$
and each of the $\left[\frac{k-1}{2^n},\frac{k}{2^n}\right]^2\in \mathcal{B}(I)\times 2^I$ (note that this also works in the case that you have $\mathcal{B}(I)$ as your $\sigma$-algebra for $Y$). The reason the closedness argument you gave is a bit tricky is that it requires one to show that the product $\sigma$-algebra on $I^2$ is equivalent to the Borel $\sigma$-algebra on $I^2$ that's generated by open sets.

Your iterated integrals are correct. Further, notice that (I'll adopt your notation and set $\mu = \lambda\times m$, where $\lambda$ is the Lebesgue measure and $m$ is the counting measure)
$$\int_{X\times Y}1_{\Delta}\ \mathrm{d}\mu = \mu(\Delta) = \infty$$
with the last equality coming from the fact that any covering of the diagonal will include infinitely many elements in $Y$.

For uniqueness of the measure, you should take a look here. I'll briefly go over the idea behind it. Let $\varpi$ be a measure on $\mathcal{B}(I)\otimes2^I$ such that
$$\varpi(E) = \sup\{\mu(E\cap (M\times N)):M\in \mathcal{B}(I),N\in 2^I, \lambda(M),m(N)<\infty\}.$$
The link I provided shows that such definition indeed gives us a measure on $\mathcal{B}(I)\otimes2^I$, the product $\sigma$-algebra.

We first show that $\varpi=\mu$ on $\mathcal{B}(I)\times2^I$. Let $A\in\mathcal{B}(I)$ and $B\in 2^I$. If $\#B<\infty$, then $\varpi(A\times B) = \mu(A\times B)$ is straightforward. On the other hand, if $\#B=\infty$, we know that $\mu(A\times B)=\infty$. Further, observe that, by our above definition, we have
$$\varpi(A\times B)\geq \lambda(A)(\sup\{m(N):N\subseteq B,m(N)<\infty\}) = \infty$$
since $m(B) = \#B = \infty$.

We will now show that $\varpi(\Delta)=0$ and thus $\varpi\ne\mu$ on $\mathcal{B}(I)\otimes2^I$. Consider some $M\in\mathcal{B}(I)$ and $N\in 2^I$ such that $m(N)=\#N<\infty$. To show that $\varpi(\Delta)=0$, note that it's sufficient to consider $M=[0,1]$ since $\lambda([0,1]) = 1<\infty$. Further, let $N=\{n_1,\ldots,n_k\}$ so we have
$$\mu(\Delta\cap ([0,1]\times\{n_1,\ldots,n_k\})) \leq \sum_{i=1}^k\mu(\Delta\cap ([0,1]\times\{n_i\})) = \sum_{i=1}^k\mu(\{n_i\}^2) = 0. $$
Since $N$ was chosen arbitrarily, we then have $\varpi(\Delta)=0$ as it's a supremum of a set of zeros.


问:
Why can't this be a coset?
Let $H$ be a subgroup of $G$ and H is not normal, there are left  cosets $aH$ and $bH$ whose product isn't a coset.

My attempt:

$ab H\subset aHbH$ and if H is not normal, if $ah_1bh_2=abh_3$ for all $h_1$, $h_2$, and $b$ then $h_1b=bh'$ for all $b$ and $H$ is normal a contradiction. So,$ab H\neq aHbH$ for some  $a,b$ and as cosets don't overlap $aHbH$ is not a coset of $H$. However, I don't know how to show that this isn't a coset of some other subgroup.

答: You can do this: What you have to prove is that in general it's not true that $aHbH=abH$. If it was true, then, for all $h_{1},h_{2},h_{3}\in H$ you'd have $ah_{1}bh_{2}=abh_{3}$, which implies, by cancellation rule, $h_{1}bh_{2}=bh_{3}$. Now take $h_{2}=1$, you have $h_{1}b=bh_{3}$, valid for all $h_{1}$ and $h_{3}$, so $H$ is normal, absurd.</p>

Title: Why can't this be a coset?


问:
About matrix exponentials
Suppose $M$ is a $n$-dimensional matrix and $v$ and $w$ are two vectors in $n-$dimensions.  I want to be able to say that $\vert \langle e^{-M \cdot t} v, w \rangle \vert$ or $e^{-M \cdot t} v$ itself, is function  monotonically decreasing to $0$ with $t$.

What is a necessary and/or sufficient condition for the above conclusion to hold?

Likewise, is $M$ being a positive definite matrix enough?

答:
For $e^{-Mt}v$ to diminish to zero, what is required is that $v\in E^+(M)$, the subspace generated by eigenspaces of eigenvalues having strictly positive real part. If $M$ is positive definite self-adjoint, then $E^+(M)$ is the whole vector space.

If $v\in E^+(M)$, then $v=\sum_i\alpha_ie_i$ with $Me_i=\lambda_ie_i$, $\mathrm{Re}(\lambda_i)>0$. $$e^{-Mt}v=\sum_i\alpha_ie^{-\mathrm{Re}(\lambda_i)t}e^{\mathrm{i Im}(\lambda_i)t}e_i\to0$$ This is not true when $v$ has a component in $E^0(M)$ or $E^-(M)$.

For $\langle w,e^{-Mt}v\rangle$ to go to zero, the condition can be relaxed to $$w\in(E^0(M)+E^-(M))^\perp$$ because then any component of $v$ in those spaces will be dotted to zero.

Title: About matrix exponentials


问:
Sum of cube roots of two conjugate quadratic integers makes an integer.
Consider the following expression:

$$(20-\sqrt{392})^{1/3}+(20+\sqrt{392})^{1/3}.$$

This equals $4$, but how can I show this?

Note that I do not want to make use of the following line of reasoning: 4 is a solution to $x^3-6x-40=0$, that this cubic has exactly one solution, and that all solutions to the cubic
$x^3+px +q = 0$ are given by
$$(-q/2-((q/2)^2+(p/3)^3)^{0.5})^{1/3}+(-q/2+((q/2)^2+(p/3)^3)^{0.5})^{1/3}.$$

I have noted the method set out here, How can I show that this complicated expression with square and cube roots reduces to the value 7? , but implementing it does not work very well.

I attempted this and got more complicated expressions than I started off with! It doesn't result in a unique solution for $a$ and $b$.

I do think that this is the right approach, though, i.e. making use of the fact that if $\sqrt c$ is irrational and $a+b\sqrt c=d+e \sqrt f$ then $a=d$ and $b=e$. (How do you prove this fact, by the way?)

答:
kludginess and meta-cheating say hello.

alternative approach.

Let $~a ~\equiv ~\left[20 - \sqrt{392}\right]^{(1/3)}.$

Let $~b ~\equiv ~\left[20 + \sqrt{392}\right]^{(1/3)}.$

Let $~x ~\equiv a + b.$

The problem is to solve for $x$.

It is immediate that $~(ab) = (400 - 392)^{(1/3)} = 2.$

Therefore,

$$x^3 = (a+b)^3 = a^3 + b^3 + 3ab(x) = 40 + 6x.$$

Examining $~f(x) = x^3 - 6x - 40~$ gives the following:

Clearly $~a > 0~$ and $~b > 0.~$  Therefore $x > 0.$

$f(0) = -40, ~f(\sqrt{2}) < 0~,$ and by the OP's own observation, $f(4) = 0.$

$f'(x) = 3x^2 - 6 = 3(x^2 - 2).~$ This means that in the interval $~[0, \sqrt{2}]~, f(x)~$ is strictly decreasing and in the interval $~[\sqrt{2}, +\infty), f(x)~$ is strictly increasing.

Therefore, $~f(x)~$ can have at most 1 positive real root.

This proves that $x = 4.$


问:
Twisting with limits
We know,$\lim_{x \to 0} \frac{\sin x}{x}=1$,but can we write $\lim_{x \to 0} \sin x=x$?
Another question,can we apply limits on both sides of an inequality?For example when we prove $\lim_{x \to 0}\frac{\sin x}{x}=1$,we deduce it from
$\cos x \le \frac{\sin x}{x} \le 1$ and then apply limits on all the inequalities.Is it justified?If so,then what's the proof?

答:
We know,$\lim_{x \to 0} \frac{\sin x}{x}=1$,but can we write $\lim_{x \to 0} \sin x=x$

No. On the right hand side, the expression $x$ is undefined, so the equation $\lim_{x\to 0}\sin x = x$ is meaningless.

Technically speaking, the expression is correct so long as the value of $x$ is $0$. However, in that case, the equality $\lim_{x\to 0} \sin x = \sin x$ is also technically correct.

For all practical purposes, the expression is meaningless, and dangerously so, because you use the same symbol, $x$, for two different purposes. On the left hand side, $x$ is a bound variable (meaning you cannot replace it with a fixed value and still get a valid expression), while on the right hand side, it is unbound (meaning you can replace it with a fixed value and get a valid expression, one that can be either correct or incorrect depending on what you replace $x$ with). Note the difference between a valid expression and a correct expression.

As for your second question, the conclusion is justified because we apply a well known theorem called the Squeeze theorem.

Title: Twisting with limits


问:
Analytically solve $\Phi(\frac{d-\mu}{\sigma}) = c$ for $\mu$
If $\Phi(\frac{d-\mu}{\sigma}) = c$, for known parameters $\sigma$, $c$ and $d$, but unknown $\mu$, is there a way to analytically solve for $\mu$? I know that the CDF is strictly decreasing in $\mu$, so finding $\mu$ numerically is easy, but I am looking for an analytical way.

Body: $\mu = d - \sigma \Phi^{-1}(c)$..


答:
Note that $\lim_{x \rightarrow 0} \sin(x) = 0$, so the equality "$\lim_{x \rightarrow 0} \sin(x)=x$" has the same meaning as "$0=x$".

When you learn about asymptotic expansions, you will be able to write something like $\sin(x)=x+o(1)$, which is the rigorous way of write what you want.

Title: Twisting with limits

Body:
We know,$\lim_{x \to 0} \frac{\sin x}{x}=1$,but can we write $\lim_{x \to 0} \sin x=x$?
Another question,can we apply limits on both sides of an inequality?For example when we prove $\lim_{x \to 0}\frac{\sin x}{x}=1$,we deduce it from
$\cos x \le \frac{\sin x}{x} \le 1$ and then apply limits on all the inequalities.Is it justified?If so,then what's the proof?


问:
Question about Bayes' theorem
A box contains $2$ coins and $p_i (I = 1, 2)$ denotes the probability of a head
when the $I^{th}$ coin is tossed. Suppose $p_1 = 1/4, p_2 = 1/2$

a) Suppose that one coin is selected at random from the box and when it is tossed three
times, two head are obtained. Work out the probability that it is the $I^{th}$  coin that has been selected $(I = 1, 2).$

b) If a tail has been obtained on the first toss of the selected coin and the same coin is tossed again, what would be the probability of obtaining a head this time?

答:
...the teacher decided to give a suprise online test so i dont even know where to start

The exercise is a standard application of Bayes' theorem

a)

$$\mathbb{P}[2H]=\frac{1}{2}\cdot\binom{3}{2}\cdot\Big(\frac{1}{4}\Big)^2\cdot\frac{3}{4}+\frac{1}{2}\cdot\binom{3}{2}\cdot\Big(\frac{1}{2}\Big)^3=A+B$$

Thus

$$\mathbb{P}[1^{st}|2H]=\frac{A}{A+B}=\frac{3}{11}$$

b) I think you can try to solve it with similar reasoning

Title: Question about Bayes' theorem


答: You're right about how we deduce the first limit. But limits, where they exist, are constants. In particular, $\lim_{x\to0}\sin x=0$. You may be more interested in the asymptotic notation $\sin x\sim x$.

Title: Twisting with limits


问:
Average, Instantaneous Change And Linear Approximation
I have these 2 question:

A function f(x), is any function, so that
$L_{a}(x)=L_{b}(x), a \not= b$. Show $\frac{f(b)-f(a)}{b-a}=\frac{f'(b)+f'(a)}{2}$.

Assume that $a \not= b$ and that $\frac{f(b)-f(a)}{b-a}=\frac{f'(b)+f'(a)}{2}$, prove or disprove $L_{a}(x)=L_{b}(x)$.

For 1. I find that $\frac{f(b)-f(a)}{b-a}=\frac{f'(b)+f'(a)}{2}$ this is asking when the average rate of change is equal to the instantaneous rate of change, of a and b. But, that is all I find. How do I show this equation is true using this information?

For 2. I know $L_{a}(x)=f(a)$ and $L_{b}(x)=f(b)$. They are differentiable. at x=a, x=b, for f(x). I think, the linear approximation is a linear equation. And, a linear equation is a polynomial, so if $L_{a}(x)=L_{b}(x)$, then every term in the linear equation is equal.

Thank you for the help.

答:
So it I understand well $L_a(x)$ being the linear approximation of $f$ at $a$, we have

$$L_a(x) = f^\prime(a)(x-a) +f(a)$$

and in a similar way
$$L_b(x) = f^\prime(b)(x-b) +f(b).$$

If those two coincide, we have in particular
$$\begin{cases}
L_a(b)&=f(b)&=f^\prime(a)(b-a) +f(a)\\
L_b(a)&=f(a)&=f^\prime(b)(a-b) +f(b)
\end{cases}$$

From which we get the requested equality by substracting the two previous equations.

The converse is not true. For a counterexample, take $f(x)=x^2$ and $a=-b=1$. We have

$$\frac{f(b)-f(a)}{a-b} = \frac{f^\prime(a)+f^\prime(b)}{2}=0$$ however the linear approximations are different.


问:
Minimal number of objects, each containing $n$ features, s.t. in the set of all objects each feature is represented at least $k$ times.
Can anyone please help frame this problem theoretically and/or guide me to its solution?

I think it might be partly related to 'set cover' type of problems, but the requirements are a bit different.

Suppose you have a total of $M$ features ($M \ge 3$), and you have the set of all objects possessing (each) exactly $n$ of these features ($2 \le n \le M-1$).

E.g. the features could be the first $M=5$ uppercase letters of the Latin alphabet: $\{A,B,C,D,E\}$.

For $n=2$, you have a set of objects $\{(A,B); (A,C); (A,D); (A,E); (B,C); ...; (D,E)\}$.

The number of such objects is of course $M \choose n$; in this example, ${5 \choose 2} =10$.

I need to calculate the minimal number of objects from this set that I am required to select, such that in their union each of the $M$ features is represented at least $k$ times ($1 \le k \le M-1)$.

E.g. in the above example ($M=5,n=2$), if I want each feature to be represented at least twice ($k=2$), I need at least $5$ objects, which could be $\{(A,B);(C,D);(A,E);(C,E);(B,D)\}$.

Note that I don't particularly need to know the objects themselves (as there are probably several degenerate solutions): just how many they are.

Supposing that we want to have each feature repeated $k=3$ times, the required number of objects is of course larger ($8$ for $M=5,n=2,k=3$).

Conversely, if we increase the number of features each object can have, the required number of objects decreases (e.g. $4$ for $M=5,n=3,k=2$)

I have tried framing this in a matrix format, to figure out a formula, meaning a function of $M,n,k$ that gives me the desired number, but I can't see it.

Any suggestions?

Thanks

PS

Find below an R script that calculates the number using linear programming

```<code># calculate the minimal number of objects, each containing n features out of a total of M features
# such that in the set of all objects each feature is represented at least k times

M = 5
n = 2
k = 2

# make the sparse matrix of all objects containing n features out of M
# objects are in columns, and are as many as choose(M,n)
# features are in rows, and are M

N.objects <- choose(M,n)
obj.mat <- sparseMatrix(j = rep(1:N.objects,each = n), i = unlist(combn(1:M, m = n, simplify = F)), x = 1)

# the constraint is that each row of the matrix sums to at least k

dir <- rep(">=",M)
rhs <- rep(k,M)

# the objective is all 1's (we want to count and minimise the number of objects used)

obj <- rep(1,N.objects)

# solve the task by linear programming

if (length(find.package(package="Rsymphony",quiet=TRUE))==0) install.packages("Rsymphony")
require(Rsymphony)
lp.out <- Rsymphony_solve_LP(obj,obj.mat,dir,rhs,types="B",max=FALSE)

cat("\nN objects needed =",lp.out<span class="math-container">$objval)
cat("\n\n")
print(obj.mat[,as.logical(lp.out$</span>solution)])
</code>```

答:
I am going to attempt a justification of the result we worked out with Paul's help.

Suppose the solution is a binary matrix with the $N$ objects in columns and the $M$ features in rows.

E.g. for $M=7, n=4, k=3$, where $6$ objects are necessary:

```<code>     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    1    1    1    1    0    0
[2,]    1    1    0    0    1    0
[3,]    1    0    1    0    1    0
[4,]    0    1    0    1    0    1
[5,]    1    0    1    1    0    1
[6,]    0    0    0    1    1    1
[7,]    0    1    1    0    1    1
</code>```

It is required to represent each of the $M$ features at least $k$ times, thus each row must contain at least $k$ 1's.

Each object has exactly $n$ distinct features, thus each column contains exactly $n$ 1's.

Therefore the total number of 1's ($N_1$) in the matrix, counting by rows, is $N_1 \ge M \cdot k$; counting by columns, it is $N_1=N \cdot n$.

Substituting $N_1$ from the equation into the inequality, we have:

$N \cdot n \ge M \cdot k$

i.e.:

$N \ge \frac {M \cdot k} {n}$

As all variables are positive integers, and we want the minimal $N$:

$N = \lceil \frac {M \cdot k} {n} \rceil$

I hope this makes sense and I haven't overlooked anything.

Body:
I am going to attempt a justification of the result we worked out with Paul's help.

Suppose the solution is a binary matrix with the $N$ objects in columns and the $M$ features in rows.

E.g. for $M=7, n=4, k=3$, where $6$ objects are necessary:

     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    1    1    1    1    0    0
[2,]    1    1    0    0    1    0
[3,]    1    0    1    0    1    0
[4,]    0    1    0    1    0    1
[5,]    1    0    1    1    0    1
[6,]    0    0    0    1    1    1
[7,]    0    1    1    0    1    1

It is required to represent each of the $M$ features at least $k$ times, thus each row must contain at least $k$ 1's.

Each object has exactly $n$ distinct features, thus each column contains exactly $n$ 1's.

Therefore the total number of 1's ($N_1$) in the matrix, counting by rows, is $N_1 \ge M \cdot k$; counting by columns, it is $N_1=N \cdot n$.

Substituting $N_1$ from the equation into the inequality, we have:

$N \cdot n \ge M \cdot k$

i.e.:

$N \ge \frac {M \cdot k} {n}$

As all variables are positive integers, and we want the minimal $N$:

$N = \lceil \frac {M \cdot k} {n} \rceil$

I hope this makes sense and I haven't overlooked anything.


问:
Trivariate normal distribution with mean 0 and covariance matrix $\Sigma$
Consider a trivariate normal vector (X, Y, Z) with mean 0 and covariance matrix $\Sigma$. How to construct a covariance matrix that X, Y  are conditionally independent given Z but X, Y are not marginally independent.

I think  $\Sigma = \begin{bmatrix}
    1  & 1 & 0 \\
    1  & 1 & 0 \\
    0  & 0 & 1
\end{bmatrix}$ can fulfill the requirement but I don't know how to show it conditionally independent.

答: Take any example where $cov(X,Y) \neq 0$ and take $Z=X$.


答:
Express your system in matrix notation
$$\left(
\begin{array}{cc}
3&-2\\
7&3
\end{array}
\right)\cdot\left(
\begin{array}{c}
x\\
y\end{array}
\right)=\left(
\begin{array}{c}
5\\
4
\end{array}
\right)$$
Now mutliply both sides by the inverse of the matrix, which is the matrix
$$\frac{1}{23}\left(
\begin{array}{cc}
3&2\\
-7&3
\end{array}
\right)$$
You finally get
$$\left(
\begin{array}{c}
x\\
y
\end{array}
\right)=\left(
\begin{array}{c}
1\\
-1
\end{array}
\right)$$</p>

Title: How to find the intersection of two given lines

Body:
How to find intersection of two lines below:
$$
\left \{ \begin{array}{ccc}
3x − 2y & = & 5\\
7x + 3y & = & 4
\end{array} \right .
$$
I know the answer is $(1,-1)$, but I cannot find it with the methods. I have already tried to solve pairs of linear equations.</p>


问:
Confusion over semantics behind ∃-elimination
$\exists-$ states that: if $\Sigma, A(u) \vdash B$ where u occurs nowhere else, then $\Sigma, \exists x A(x) \vdash B$.

Why does this translate to $\exists$ and not $\forall$? Intuitively, $A(u)$ where $u$ occurs nowhere else should mean that $u$ can be any arbitrary thing, so shouldn't $\Sigma, A(u) \vdash B$ be valid for any conceivable $u$ (a.k.a, all $x$) meaning $\Sigma, \exists xA(x) \vdash B$?

This is giving me trouble mostly because of the below theorem I've already learned:

Why is it that in this case $A(u)$ can translate in to a "for all" while in $\exists-$ it only gives us "there exists"? In both cases $u$ is arbitrary, so it should refer to anything$-$"for all"$-$in my  head.

答:
Intuitively, $A(u)$ where $u$ occurs nowhere else should mean that $u$ can be any arbitrary thing.

No, $u$ is not arbitrary.

"Arbitrary" means: whatever term $u$ we may choose...

This is not so: we have to choose a “fresh” term, in the sense of having no other occurrences in the derivation.

We have (as per the title of your question) to consider the semantics.

With $(\forall \text I)$ rule we may, under the condition that $u$ is "fresh", derive from $A(u)$ the "generalization" $\forall x A(x)$.

Here every possible choice of a term $u$ not already used will do, and this is consistent with the intuitive meaning of the rule: if $A$ holds of an "object" $u$ whatever, then it holds of every "object".

A simple arithmetical example is: we have proved $\mathsf {PA} \vdash (x \ge 0)$, where $\mathsf {PA}$ is the collection of Peano's axioms.

Then we may apply $(\forall \text I)$ to conclude with: $\mathsf {PA} \vdash \forall x(x \ge 0)$.

Consider now the formula $\exists x (x > 0)$.

If we want to use it in a proof with $(\exists \text E)$ rule, we have to choose a suitable term $u$ for the assumption $(u > 0)$: the formula $\exists x (x > 0)$ says there must be (at least) one.

But we cannot choose an "arbitrary" term, i.e. a term whatever: for $u=0$ we get $(0 > 0)$, which is false.


问:
Infinitely many solutions vs one solution vs no solution in systems involving an unknown constant
Just need a little clarification in case my assumptions are incorrect. If I were to have the matrix

{{1, 3, 3}, {2, 7, 6}, {1, 4, k^2 - 18}} * {x1, x2, x3} = {1, 3, k+1}

which has a row echelon form of:

 {{1, 3, 3}, {0, 1, -12}, {0, 0, 1}} * {x1, x2, x3} = {1, 3, (k - 3)/(k^2 - 9)} , didn't want to reduce any further to avoid dealing with that ugly fraction.

would it be the case where k = 3, -3 that would be no solution? I mean logically I suppose so but in terms of linear algebra I thought no solution required an inconsistency and not an undefined point. Or would an undefined k value make for an inconsistency?

for one solution is it right to assume any unique result? suppose I chose k = 0, then would the unique solution be whatever I get from plugging in? And if so is it safe to assume that we have infinitely many unique solutions so long k does not equal 3, -3?

and for infinitely many solutions is it again the case where any result so long k does not equal 3, -3?

I would really appreciate a response with an explanation that can be understood. Thank you.

oh and how would k = 3 play into the solutions? since if we were to simplify (k -3)/(k^2 - 9) we would end up with 1/(k+3) in which case k = 3 fails for the former but is valid for the latter.

答:
This is a case where complete row reduction is preferred, and leads to different values of $k$ to avoid: The complete row reduction leads to
$$x_1=\frac{3-3k}{k^2-21}-2,\\
x_2=1,\\
x_3=\frac{k-1}{k^2-21}.$$
So the only cases where the solution might not be unique are when $k = \pm \sqrt{21}.$ A good way to deal with these, which avoids the "zero divide" in the above equations, and issues of undefined answers, is just to substitute $k=\pm \sqrt{21}$ into the original matrix having the formulas in it, and row reduce each. Neither has any solution, i.e. both particular systems are inconsistent, in this case.

ADDED: I checked, and found your not-completely reduced matrix is off. It should have row 1 as $[1,3,3,1]$, row 2 as $[0,1,0,1]$, and the crucial row 3 as
$$[0,0,1,(k-1)/(k^2-21)].$$
Thus you would be led to the same question about the only exceptions possible being $k=\pm \sqrt{21}$ which come from the completely reduced matrix. So the same course of action would be suggested, namely to substitute $k=\pm \sqrt{21}$ into the original system, obtaining matrices with only numbers, which would then be row reduced in the usual way to see if consistent. Note that before division the last row reads as
$$[0,0,k^2-21,k-1].$$
So even here, without going back to the original system, you can see directly that if $k=\pm \sqrt{21}$ the last row reads $[0,0,0,(\pm \sqrt{21}-1]$ so the last entry is nonzero, making the system inconsistent. This avoids the plugging in of the radical expression for $k$ into the original system.


问:
Cauchy sequence of rationals converging to a non zero limit
I'm trying to show that if $(x_n)_n∈N $ is a Cauchy sequence of rational numbers with $\lim_{n \to \infty} \neq 0$, that there exists  $b \in \mathbb{Q}$ with $b> 0$ and $N \in \mathbb{N}$ so that $|x_n| > b$ for all $n \geq N$

So far, my proof is as follows:

Given that $x_n$ is a Cautchy sequence it is convergent. Hence, $x_n$ is a convergent sequence. Also, since  $\lim_{n \to \infty} \neq 0  $,  say $\lim_{n \to \infty} = L$, where $L \neq 0$. Then, given $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that

$|x_n - L| < \epsilon \forall \; n \geq N$

$ - \epsilon < x_n - L < \epsilon  \forall \; n \geq N$

$L - \epsilon < x_n < L + \epsilon   \; \; \; \forall \; n \geq N$

Let $b =  L - \epsilon$. It follows then that $|x_n| > b$.

However, I realize I have not shown that $b>0$. How could I do so to complete my proof? Also, are there any changes you would recommend to make my proof that it is better?

答: Choose $N$ such that $|x_n-L| <\frac {|L|} 2$ for $n \geq N$. Then $|x_n|=|L+(x_n-L)| \geq |L| -|x_n-L| >\frac {|L|} 2$ for $n \geq N$. Take $b$ to be any rational number in $(0, \frac {|L|} 2)$.


答:
As noted by Jyrki Lahtonen, it might well be that the product $a H b H$, for a particular choice of $a, b$, is a coset of another subgroup $K$. Here is a general argument, and another particular example, also based on the dihedral group of order $8$.

Suppose $a H b H = c K$. Multiplying by $a^{-1}$ on the left, we have $H b H = a^{-1} c K$. Now $b \in H b H = a^{-1} c K$, so we have $a^{-1} c K = b K$ and multiplying $H b H = b K$ on the left by $b^{-1}$ we have
$$
(b^{-1} H b) H = K.\tag{prod}
$$

So the point is to construct a group $G$, subgroups $H \le K \le G$, and an element $b \in G$ such that (prod) holds.

Consider the dihedral group $K$ of order $8$, whose elements are
$$
1, r, r^2, r^3, s, s r, s r^2, s r^3.
$$
(Here $r$ is a rotation of $\pi/ 2$, and $s$ is a reflection. A presentation is given by $\langle r, s : r^4 = 1, s^2 = 1, s^{-1} r s = r^{-1} \rangle$.)

Consider the subgroup $H = \{ 1, r^2, s, s r^2 \}$ of $K$.

Now $K$ has an automorphism $b$ of order $2$ that maps $r \mapsto r$ and $s \mapsto s r$. Consider the semidirect product of $G$ by $\langle b \rangle$.

In $G$ we have $b^{-1} H b = \{ 1, r^2, s r , s r^3 \}$, and $b^{-1} H b H = K$.

Jyrki Lahtonen's example is simpler than mine, as it relies on $H = \{ 1, s \}$, $b = r$, so that $b^{-1} H b = \{ 1, s r^{2} \}$, and $K = \{ 1, s, r^{2}, s r^{2} \}$.

Title: Why can't this be a coset?

Body:
Let $H$ be a subgroup of $G$ and H is not normal, there are left  cosets $aH$ and $bH$ whose product isn't a coset.

My attempt:

$ab H\subset aHbH$ and if H is not normal, if $ah_1bh_2=abh_3$ for all $h_1$, $h_2$, and $b$ then $h_1b=bh'$ for all $b$ and $H$ is normal a contradiction. So,$ab H\neq aHbH$ for some  $a,b$ and as cosets don't overlap $aHbH$ is not a coset of $H$. However, I don't know how to show that this isn't a coset of some other subgroup.


问:
For anypair of $(x,y)$ is $x+y + xy$ always unique?
I was solving a computer science problem and was thinking of a way to encrypt an unordered pair of $(x,y)$ to a single value, I tried playing with bits a bit but did not get any conclusions But I think for any unordered pair of $(x,y)$ the Value of $x+y + xy$ is always unique for it, It is purely intuition-based, I am looking for a valid proof, If it is wrong Please present a counterexample for it.
For eg a pair of $(2,5)$ no other pair of numbers can give me the value $17$.
If anyone has any other way to encrypt two numbers Please suggest
The numbers are always Positive Numbers.

答:
If you look at the graph of $f(x,y) = x+y+xy$, you see that different pairs may result in the same value.

You say $f(2,5)=17$. Now let's fix $x=1$ and we will find the $y$- co0rdinate such that $f(1,y)=17$
$$f(1,y) = 1+y+1\cdot y=17 \implies y = 8$$
So, below is the counter-example:
$$f(2,5)=f(1,8)=17$$.


答:
$1 + 25 + 1\cdot 25 = 51 = 3 + 12 + 3\cdot 12$.

If you want an injective mapping from $\mathbb N^2$ to $\mathbb N$, then you could take any two prime numbers $p,q$ and define the mapping

$$(x,y)\mapsto p^x\cdot q^y$$

or, if you want something a little less explosive, take

$$(x, y) \mapsto 2^{x-1}\cdot (2y - 1)$$

This mapping is not only injective, it is bijective, i.e. it hits all integers, and each one exactly once.

Title: For anypair of $(x,y)$ is $x+y + xy$ always unique?

Body:
I was solving a computer science problem and was thinking of a way to encrypt an unordered pair of $(x,y)$ to a single value, I tried playing with bits a bit but did not get any conclusions But I think for any unordered pair of $(x,y)$ the Value of $x+y + xy$ is always unique for it, It is purely intuition-based, I am looking for a valid proof, If it is wrong Please present a counterexample for it.
For eg a pair of $(2,5)$ no other pair of numbers can give me the value $17$.
If anyone has any other way to encrypt two numbers Please suggest
The numbers are always Positive Numbers.


问:
A problem related to $\int_{0}^{1} f(x)(x-f(x))dx=1/12$

Consider  a differentiable  function satisfying $$\int_{0}^{1} f(x)(x-f(x))dx=1/12$$ Then find the nearest integer less than or equal to $\frac{1}{f'(1)}$.

Let $$F(x)=\int_{0}^{x}f(x)(x-f(x))$$ we notice $F(1)=1/12$ , $F(0)=0$ and $$F'(1)=f(1)-{f(1)}^2$$
What do i do next?
I have tried using LMVT and Rolle's but not getting anything nice.

Body:
We can manipulate $$\int_{0}^{1} (4xf(x)-4f^2(x)) \, dx=1/3$$

$$\int_{0}^{1} (4xf(x)-4f^2(x)-x^2) \, dx=1/3 - \int_0^1 x^2 dx$$

$$\int_{0}^{1} (2f(x)-x)^2 \, dx=0 $$

$$\Rightarrow f(x)=x/2 $$


问:
How to prove $a x^{2a} -2 a x^a \geq x^{a+1}-(a+1)x$
For $x \in [0, 1]$ and $a \geq 1$, how to prove
$$a x^{2a} -2 a x^a \geq x^{a+1}-(a+1)x?$$

Because the inequality holds with equality at $x = 0$ and $x = 1$, we can not show that for each $a$, the function LHS - RHS is increasing / decreasing in $x$.

Because the inequality holds with equality at $a = 1$ for any $x$, this suggests that we can show, for each $x$, the LHS - RHS is increasing in $a$. But it is not easy to show this because the derivative is messy.

Is there any easy, elegant way to show this inequality?

Thanks

答:
HINT: You plugged $x=0$ and checked it works. So exclude this value and work on $(0,1]$; in this way you can divide by $x$ to get something easier to handle; at this point, study the sign of
$$
f(x)=ax^{2a-1}-2ax^{a-1}-x^{a}+(a+1)\;.
$$
taking the derivative:
$$
f'(x)=a(2a-1)x^{2(a-1)}-2ax^{a-2}-x^{a-1}
=x^{a-1}\underbrace{\left[a(2a-1)x^{a-1}-2ax^{-1}-1\right]}_{=:g(x)}
$$
and since $x^{a-1}$ is always positive, you can just study the sign of the functions I have named $g$:
\begin{align*}
g(x)\ge0
&\Longleftrightarrow a(2a-1)x^{a-1}\ge2ax^{-1}+1\\
&\Longleftrightarrow a(2a-1)x^{a}\ge2a+x\\
\end{align*}
where in the last step I have multiplied both sides by $x$ since it is strictly positive on $(0,1]$.


问:
Let A be a UFD and $A_{P}$ be the localisation of $A$ at prime ideal $P$. Then prove that $A_{P}$ is principal
I have proved that $A_{P}$ is a local ring that is, it has unique maximal ideal containing elements of type $\frac{a}{r}$ where $a\in P$ and $r\in A\backslash P$. But, I have no idea how to prove that $A_{P}$ is principal.

答:
This is not true. Recall that the prime ideals of $S^{-1} A$ are in a one to one order preserving correspondence with the prime ideals of $A$ that do not intersect $S$. In the case $S=A-\mathfrak p$, this means that the primes of $A_{\mathfrak p}$ correspond to the primes contained in $\mathfrak p$. Now let $A=k[x, y]$ for $k$ a field. This is a UFD and we can then consider the chain of primes $0 \subsetneq (x) \subsetneq (x, y) = \mathfrak p$. By correspondence, this yields a chain of primes in $A_{\mathfrak p}$: $0 \subsetneq (x) A_{\mathfrak p} \subsetneq (x, y) A_{\mathfrak p}$. In a PID all nonzero prime ideals are maximal, but we see here that $(x) A_{\mathfrak p}$ is a prime ideal which is strictly contained in $(x, y) A_{\mathfrak p}$, another prime ideal. Hence, $A_{\mathfrak p}$ is not a PID.

We can phrase this proof a bit more succinctly with dimension theory (although it's really the exact same proof). The chain of primes I listed above proves that $\mathrm{ht}(\mathfrak p) \geq 2$. Thus, $\dim(A_{\mathfrak p}) = \mathrm{ht}(\mathfrak p) \geq 2$ but a PID must have dimension 1 (or 0 if you allow fields to be PIDs).


问:
An existence of a certain continuous map $\Delta^{n+1}\to \Delta^n$
I wonder if there is a continuous map $f\colon \Delta^{n+1}\to \Delta^n$ such that $f(x_0,x_1,...,x_n) = \left( \dfrac{x_1}{1 - x_0}, \dots, \dfrac{x_n}{1 - x_0}\right)$ if $x_0 \neq 1$.

This is an auxiliary result needed here.

I know short questions are not recieved well on this site, but I don't know what else to add.

答:
Such a map cannot exist for $n\ge1$. The sequences $(1-1/n,1/n,0,...,0)_n$ and $(1-1/n,0,...,0,1/n)_n$ both converge to $(1,0,...,0)$, so, if such a continuous $f$ existed, we would have (since $\Delta^n$ is Hausdorff)
$$f(1,\overbrace{0,...,0}^{n+1\text{ zeroes}})=\lim_{n\rightarrow\infty}f(1-1/n,1/n,\overbrace{0,...,0}^{n\text{ zeros}})=(1,\overbrace{0,...,0}^{n\text{ zeros}}),\\
f(1,\overbrace{0,...,0}^{n+1\text{ zeros}})=\lim_{n\rightarrow\infty}f(1-1/n,\overbrace{0,...,0}^{n\text{ zeros}},1/n)=(\overbrace{0,...,0}^{n\text{ zeros}},1).$$
This cannot happen unless $n=0$.