问:
Pair formation: Find max number of disjoint pairs with distance less than $K$
Here is an instance of the problem I am trying to solve:

In this graph there are two types of nodes:

$S$ stands for "start" nodes. You can think at them as the current locations of some postmen.

$E$ stands for "end" nodes. You can think at them as houses where a letter has to be delivered (people waiting for a letter to be delivered).

So basically there are some postmen that need to deliver letters to some locations (there can be any number of $E$ and $S$ edges, and the number of $E$ nodes does not necessarily match the number of $S$ nodes). Edge labels represent the known distance between two nodes.

Each postman cannot travel more than $K$. Each postman can deliver at most 1 letter. And each person $E$ is waiting exactly for 1 letter.

I need to find how many letters I can deliver. More formally, I am trying to find how many disjoint pairs $(s, e)$ with distance less than $K$ are there in my graph. In the example:

if $K=6$ then 3 letters can be delivered ($(s1, e1)$, $(s2, e3)$, $(s3, e4)$)

if $K=4$ then 2 letters can be delivered ($(s1, e2)$, $(s3, e3)$)

Note that the form of the graph is always like the one in the example, i.e. the nodes are "sorted" and form a "line".

Clearly, I have already tried with some brute force algorithm, but the graph I am dealing with can be very large and the complexity of such approach is not acceptable in my case. Do you have any idea? How can I efficiently find how many letters can be delivered?

答:
Maximising the number of letters delivered can be done in polynomial time:

For each postman determine the houses they can reach. Since the graph is a line, this can be easily done by walking both ways, up to a distance of $K$, and noting which houses were encountered.

Make a bipartite graph where the vertices on each side represent postmen and houses respectivelty. Edges represent the ability of a particular postman to reach a particular house.

Find the maximum-cardinality matching in this bipartite graph (this can be done in several ways, e.g. Hopcroft–Karp). The result is an optimal allocation of postmen to houses, in terms of the number of letters delivered.

The above algorithm can be generalised to arbitrary graphs, whereupon the linear scan in step 1 becomes an all-pairs shortest path determination.


问:
How do I evaluate a definite integral involving trigonometric functions?
Evaluate $$\int_0^{8\sqrt{2}} \dfrac{1}{\sqrt{256-s^2}}ds$$

I know that the antiderivative of $\dfrac{1}{\sqrt{1-x^2}}$ is $\sin^{-1}x + C$ but I am a little unsure how I would change the integrand to get hat form. Could someone give me a hint or explanation on how to evaluate this integral?

答:
The standard trick when faced with an integral like this is to take the terms that are being squared together with the term combining them, and to form a right-triangle out of them.  You have in the denominator of your original integral $256=16^2$ and $s^2$.  The term combining them is $\sqrt{256 - s^2}$. A right triangle with sides $16, s, \sqrt{256 - s^2}$ has legs $s$ and $\sqrt{256 - s^2},$ and hypotenuse $16$.

Let's arrange the sides so we can do a trigonometric substitution that keeps the lower limit $0$. We can do it by  taking the triangle $\triangle ABC$, where $\angle A = \theta$, $\angle B = \pi/2 - \theta$, and $\angle C = \pi/2$ and having the side lengths be $BC = s, AC = \sqrt{256 - s^2}, \text{and } AB = 16$. (I'd draw this if I could here.)

Then,
$$
\begin{aligned}
s/16 &= \sin\theta,\\
\frac{\sqrt{256 - s^2}}{16} &= \cos\theta,\\
s &= 16 \sin\theta,\\
\sqrt{256 - s^2} &= 16 \cos\theta,\\
ds &= 16 \cos\theta\,d\theta.\\
\end{aligned}$$
Also, when $s=0, \theta=0,$ and when $s=8\sqrt2, \sin\theta = 1/\sqrt2, \text{and } \theta=\pi/4.$

So,
$$
\int_0^{8\sqrt{2}} \dfrac{1}{\sqrt{256-s^2}}ds = \int_0^{\pi/4} \frac{16 \cos\theta\,d\theta}{16\cos\theta} = \frac{\pi}{4}.
$$

The typical trigonometric substitutions are $s=c\sin\theta$ when one has a constant minus $s^2$ or similar, $s = c\tan\theta$ when one has a constant plus $s^2$ or similar, and $s=c\sec\theta$ when one has $s^2$ minus a constant or similar.

ContentLicense: CC BY-SA 3.0


问:
Proof of the existence of a solution on IVP with a given interval for $x$
Prove that the solution $y(x)$ exists on the given interval for $x$:
$$y'=y^2 + \cos x^2$$
on $0\leq x\leq 1/2$ with $y(0)=0$.
I think that I have to use Picard's Theorem. But I don't know why I am given an interval for $x$. And the left bound of $x$ is the initial value. Any help?

答:
The task asks if $[0,\frac12]$ is a subset of the domain of the (or a) maximal solution of the IVP. This is a non-trivial question as the solution will diverge to infinity in finite time as soon as it reaches a value larger than $1$, as $y'>y^2-1$ gives an exploding lower bound.

First variant, $y'\le y^2+1$ gives that, as long as the solution exists, $y(x)\le \tan(x)$. In the consequence, the solution remains bounded above for $x<\frac\pi2$. As long as $\cos(x^2)\ge 0$, that is, $x\le\sqrt{\frac\pi2}$, it is also clear that $y$ is increasing and thus positive. Both together give that, for instance, $[0,1]$ is a subset of the maximal domain.

Second variant, if you want to go by the Picard iteration, you first fix a rectangle $R$ via $|x|\le a$, $|y|\le b$. Then the right side function is bounded by $|f(x,y)|\le M=b^2+1$ on $R$. Solutions are guaranteed by this bound to stay inside this rectangle for $|x|\le h$ if $Mh\le b$. Next for the convergence one needs that $q=Lh<1$ for the contraction factor of the Picard iteration, where $L=2b$ is an $y$-Lipschitz constant. Both conditions can be satisfied if $h$ is chosen small enough. As the bounds do not depend on $a$, one can set $a=h$. With $b=1$ these conditions are satisfied for any $h<\frac12$, it follows that the solution exists at least on $[0,1]$, with a limit at $x=\frac12$ due to boundedness.

ContentLicense: CC BY-SA 4.0


问:
How do I solve $\overline{(A \cap B) \cup (\overline{A} \cap C)} = (A \cap \overline{B}) \cup (\overline{A} \cap \overline{C})$?
The question is to prove  $\overline{(A \cap B) \cup (\overline{A} \cap C)} = (A \cap \overline{B}) \cup (\overline{A} \cap \overline{C})$ using laws of set theory.

I got stuck after the following steps:
$$
\overline{(A \cap B) \cup (\overline{A} \cap C)} = \overline{(A \cap B)} \cap \overline{(\overline{A} \cap C)}
\\ = (\overline{A} \cup \overline{B}) \cap (A \cup \overline{C})
\\ = ((\overline{A} \cup \overline{B}) \cap A) \cup ((\overline{A} \cup \overline{B}) \cap \overline{C})
\\ = (\overline{A} \cap A) \cup (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C})
\\ = (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C})
$$

Not sure what to do next. I somehow feel that the absorption law could be used here, but don't know how.

答:
Taking what you have done so far

$$\begin{align}
\overline{(A \cap B) \cup (\overline{A} \cap C)} & = \overline{(A \cap B)} \cap \overline{(\overline{A} \cap C)}\\
& = (\overline{A} \cup \overline{B}) \cap (A \cup \overline{C})\\
& = ((\overline{A} \cup \overline{B}) \cap A) \cup ((\overline{A} \cup \overline{B}) \cap \overline{C})\\
& = (\overline{A} \cap A) \cup (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C})\\
& = (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C}) & \tag{1}
\end{align}$$

it looks fine.

Next you go as follows. We know from definition of union that

$$(\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \subseteq (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C})$$

Although

$$\begin{align}
x \in (\overline{B} \cap \overline {C}) & \implies x \in \overline{B} \wedge x \in \overline{C}\\
& \implies \big(x \in \overline{B} \wedge x \in \overline{C} \big) \wedge(\overline{A} \vee A)\\
& \implies \big(x \in \overline{B} \wedge x \in \overline{C} \wedge x \in \overline{A} \big) \vee \big( x \in \overline{B} \wedge x \in \overline{C} \wedge x \in A \big)\\
& \implies (x \in \overline{C} \wedge x \in \overline{A}) \vee (x \in \overline{B} \wedge x \in A)\\
& \implies (x \in \overline{C} \cap \overline{A}) \vee (x \in \overline{B}\cap A)\\
& \implies x \in (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C})
\end{align}$$

which means that $(\overline{B} \cap \overline{C}) \subseteq (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}).$ So we conclude that

$$\begin{align}
(\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C}) & \subseteq (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C})\\
& \subseteq (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C})
\end{align}$$

Continuing from $(1)$ we get the following final conclusion

$$\therefore (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C}) = (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \quad \square$$


问:
System of infinite equations
While working I encountered the following system of equations:

$\binom{0}{0} x_0+\binom{1}{0} x_1+\binom{2}{0} x_2+\binom{3}{0} x_3+...=\alpha x_0$

$\binom{1}{1} x_1+\binom{2}{1} x_2+\binom{3}{1} x_3+...=\alpha x_1$

$\binom{2}{2} x_2+\binom{3}{2} x_3+...=\alpha x_2$

$\binom{3}{3} x_3+...=\alpha x_3$

$\vdots$

where $\alpha$ is a fixed real number. I want to find the values of $x_1,$ $x_2,$ $\ldots$ in terms of $x_0.$ The commonly used substitution method is not working in this case. Binomial coefficient is not allowing us to use the substitution method. Is there anyway to find the values of $x_1,$ $x_2,$ $\ldots$? Please help.

答:
The answer we are going to establish is:

$$\forall k, \ x_k=x_0\dfrac{1}{2^k} \ \ \ \text{and} \ \ \ \alpha=2. \tag{1}$$

In fact, your issue is equivalent to:

$$\underbrace{\begin{pmatrix}1&1&1&1&1&\cdots \\
0&1&2&3&4&\cdots \\
0&0&1&3&6&\cdots \\
0&0&0&1&4&\cdots \\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots
\end{pmatrix}}_P \underbrace{\begin{pmatrix}x_0 \\
x_1\\
x_2\\
x_3\\
\vdots
\end{pmatrix}}_X=\alpha \underbrace{\begin{pmatrix}x_0 \\
x_1\\
x_2\\
x_3\\
\vdots
\end{pmatrix}}_X\tag{2}$$

expressing that $X$ is an eigenvector of a so-called infinite Pascal matrix associated with eigenvalue $\alpha$.

You will find in this nice MIT paper by Strang and Edelman a description of the finite and infinite Pascal matrices and some of their properties.

In particular, one finds in this paper the following formula (numbered (12) page 9):

$$\begin{pmatrix}1&1&1&1&1&\cdots \\
0&1&2&3&4&\cdots \\
0&0&1&3&6&\cdots \\
0&0&0&1&4&\cdots \\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots
\end{pmatrix}\begin{pmatrix}1 \\
x\\
x^2\\
x^3\\
\vdots
\end{pmatrix}=\begin{pmatrix}1/(1-x) \\
x/(1-x)^2\\
x^2/(1-x)^3\\
x^3/(1-x)^4\\
\vdots
\end{pmatrix}\tag{3}$$

Let us remark that the different identities in (3) aren't at all mysterious:

the first entry expresses the sum of a geometric series,

the second one expresses the derivative of this sum multiplied by $x$ in order to generate a  shift, etc.

Otherwise said:

$$\begin{pmatrix}1&1&1&1&1&\cdots \\
0&1&2&3&4&\cdots \\
0&0&1&3&6&\cdots \\
0&0&0&1&4&\cdots \\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots
\end{pmatrix}\begin{pmatrix}1 \\
x\\
x^2\\
x^3\\
\vdots
\end{pmatrix}=\dfrac{1}{1-x}\begin{pmatrix}
1\\(x/(1-x))\\
(x/(1-x))^2\\
(x/(1-x))^3\\
\vdots
\end{pmatrix}\tag{4}$$

If we take $x$ such that

$$1=\dfrac{1}{1-x} \ \iff \ x=\dfrac12$$

(2) expresses the eigenvalues-eigenvector we were waiting for... with

$$x_k=\dfrac{1}{2^k} \ \ \ \text{and} \ \ \ \alpha=\dfrac{1}{x}=2.\tag{5}$$

But, as the multiple of an eigenvector is still an eigenvector, we can deduce from (5) the general solution (as a function of $x_0$) as given in (1).

Remark: A paradoxical fact takes place here (in fact not so paradoxical...).

If one imagine the infinite matrix in (1) as the limit of successive russian dolls matrices as depicted here:

These "nested" matrices have in particular all their eigenvalues equal to $1$... associated for example with the all-ones vector. No continuity argument can explain why "suddenly at infinity", eigenvalue $2$ suddenly appears. But in fact the image by $P$ of the "all-ones" infinite vector is indefinite (the vectors have to be in $\ell_2$ to be eligible)... This is typical of difficulties when dealing with infinite matrices.

Let us give another difference between finite and infinite matrices. Consider

$$A=\begin{pmatrix}0&1&0&0&0&\cdots \\
0&0&1&0&0&\cdots \\
0&0&0&1&0&\cdots \\
0&0&0&0&1&\cdots \\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots
\end{pmatrix} \ \text{giving} \ AA^T-A^TA=\begin{pmatrix}1&0&0&0&0&\cdots \\
0&0&0&0&0&\cdots \\
0&0&0&0&0&\cdots \\
0&0&0&0&0&\cdots \\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots
\end{pmatrix}$$

As a consequence: $$\operatorname{trace}(AA^T-A^TA)=1$$

which is false for finite dimensional matrices.

Title: System of infinite equations


问:
Help with a covariance inequality
Let $(X_i)_{i\in\mathbb{N}}$ be a sequence of real-valued random variables. In a proof of a theorem, the author starts with

$$\lvert Cov(X_i,X_j)\rvert\leq \int_0^{\alpha_{\lvert i-j\rvert}}[Q_i(u)]^2+[Q_j(u)]^2du\quad (a)$$

where $(\alpha_{k})_{k\geq 0}$ is a nonnegative and nonincreasing sequence with $\alpha_0=1/2$ and $Q_k(u)=\inf\{x\in\mathbb{R}:P(\lvert X_k\rvert>x)\leq u\}$, the quantile function of $\lvert X_k\rvert$ (the quantile function of a random variable is nonincreasing). This inequality $(a)$ is true, and assume it holds. But from $(a)$, he inferred that

$$\sum_{i=1}^n\sum_{j=1}^n\lvert Cov(X_i,X_j)\rvert\leq 2\sum_{i=1}^n\int_0^1 \sum_{j=1}^n I({u<\alpha_{\lvert i-j\rvert}})[Q_i(u)]^2du \quad (b)$$

with $I(\cdot)$ denoting the indicator function.

I'm struggling to see $(a)\to (b)$. Can you see the reasoning of this?

My attempt

It's clear that $\int_0^{\alpha_{\lvert i-j\rvert}}[Q_i^2(u)+Q_j^2(u)]du=\int_0^1 I(u<{\alpha_{\lvert i-j\rvert}})[Q_i^2(u)+Q_j^2(u)]du$. Then
$$\sum_{i=1}^n\sum_{j=1}^n\lvert Cov(X_i,X_j)\rvert\leq \sum_{i=1}^n\int_0^1 \sum_{j=1}^n I({u<\alpha_{\lvert i-j\rvert}})[Q_i^2(u)+Q_j^2(u)]du.$$

From this, I need to show that, in the summand, $Q_j^2(u)\leq Q_i^2(u)$ for each $1\leq i,j\leq n$, which is not clear to me. I see that the region of integration is maximized when $i=j$, which is equal to $[0,1/2]$. But I don't know the behavior of the height $Q_i^2(u)+Q_j^2(u)$ along $i,j$.

Reference: Rio's Book, pp. 9, Proof of Corollary 1.1

Title: Help with a covariance inequality

Body:
\begin{align}
\sum_{i,j}|\operatorname{Cov}(X_i,X_j)| &\le \sum_{i,j}\int_0^1 1\{u< \alpha_{|i-j|}\}Q_i^2(u)\, du \\
&\quad+\sum_{i,j}\int_0^1 1\{u< \alpha_{|i-j|}\}Q_j^2(u)\, du \\
&=\sum_i\int_0^1 \sum_j 1\{u< \alpha_{|i-j|}\}Q_i^2(u)\, du \\
&\quad+\sum_j\int_0^1 \sum_i 1\{u< \alpha_{|i-j|}\}Q_j^2(u)\, du \\
&=2\sum_i\int_0^1 \sum_j 1\{u< \alpha_{|i-j|}\}Q_i^2(u)\, du.
\end{align}


问:
Analogies between Prime Ideals and Knots
While reading this question posted at this link:

Subjects studied in number theory

i interestingly landed up on this Wikipedia page, and was quite amazed to see the variety of branches opening up. That's how I came to know about the subject Arithmetic Topology. Wikipedia describes this subject as:

Arithmetic topology is an area of mathematics that is a combination of algebraic number theory and topology. In the 1960s topological interpretations of class field theory were given by John Tate based on Galois cohomology, and also by Michael Artin and Jean-Louis Verdier based on Étale cohomology. In subsequent years, Barry Mazur and Yuri Manin pointed out a series of analogies between prime ideals and knots.

Can anyone explain me as to how Prime Ideals and Knot's are analogous (elementary explanation would be helpful) and why are we interested in studying these analogies.

答: I will try to write more later.  In the meantime, here is a link to a manuscript of Mazur on this topic.  (See the paper Remarks on the Alexander polynomial.)


问:
Show that $12n+5$ and $5n-2$ are relatively prime for all $n$ (in $\mathbb{Z}$)
Show that $12n+5$ and $5n-2$ are relatively prime for all $n$ (in $\mathbb{Z}$)

I've seen similar questions on here, I tried following their answers and doing it myself, but it's not working out for me. I tried the Euclidean Algo a few times and it's never ending.
I'd be happy if someone could post the way to do this.

答:
The Euclidean algorithm might not reach $0$, but it will help greatly:
$$
\begin{array}{cc}
12n+5&5n-2\\
12n+5-2(5n-2)&5n-2\\
2n+9&5n-2\\
2n+9&5n-2-2(2n+9)\\
2n+9&n-20\\
2n+9-2(n-20)&n-20\\
49&n-20
\end{array}
$$
So basically, if $n-20$ has a factor of $7$, then $12n+5$ and $5n-2$ will both have a factor of $7$ (and similarly for a factor of $49$). For instance, for $n=-1$ we get
$$
12n+5=-7\\
5n-2=-7
$$


问:
remove direction of non-standard basis from matrix
Given:

G1. C is $n\times n$ positive semidefinite matrix

G2. For positive semidefinite matrix SVD decomposition is equal to eigen value decomposition.

G3. Singular Value Decomposition of C $SVD(C) = U, S, V$, where U is eigen vectors (orthonormal basis) and S is diagonal matrix with eigen values.

G4. I decides to use 2 (for example) first values $s_0, s_1$ corresponding to  $u_0, u_1$ eigen vectors directions (columns in U)

G5. I receive another matrix $D\ne C$, D is $n\times n$ positive semidefinite matrix

Question:
Q1. I want to subtract directions of $u_0, u_1$ from D

The solution:

S1. Calculate the projection of D to U: $D_u = U^{-1}\times D\times U$

S2. Note: in order to return back $D = U\times D_u\times U^{-1}$

S3. U - orthonormal, $U^T=U^{-1}$

S4. Denote $U_2 = [u_0, u_1, 0...], U_{n-2} = [0, 0, u_2, ...u_n], U = U_2 + U_{n-2}$

S5. $U^T = U_2^T + U_{n-2}^T$

S6.  Get D for n-2 directions: $D_{n-2}=U_{n-2}\times D_u\times U_{n-2}^T$

Is it correct ?
Thanks

答:
Yes, that is correct. A note: D is the matrix of eigenvalues - it is defined on G2. What is D in G5?
You can also look at low-rank matrix approximation, or separable models here</p>


答:
$$(12n+5,5n-2)=(2n+9,5n-2)=(2n+9,n-20)=(n+29,n-20)=(n+29,49)$$

for $n=7t+6$ where $t\in Z$ both can divisible by 7. They are not relatively prime.

Body:
Show that $12n+5$ and $5n-2$ are relatively prime for all $n$ (in $\mathbb{Z}$)

I've seen similar questions on here, I tried following their answers and doing it myself, but it's not working out for me. I tried the Euclidean Algo a few times and it's never ending.
I'd be happy if someone could post the way to do this.


问:
Order of this group
Given $k\in{\mathbb{N}}$, we denote $\Gamma _2(p^k)$ the multiplicative group of all matrix $\begin{bmatrix}{a}&{b}\\{c}&{d}\end{bmatrix}$ with $a,b,c,d\in\mathbb{Z}$, $ad-bc = 1$, $a$ and $d$ are equal to $1$ module $p^k$ and $b$ and $c$ are multiples of $p^k$.

How can I show that $|\Gamma _2(p)/\Gamma _2(p^k)|\le p^{4k}$?

答:
By exhibiting a set of at most $N=p^{4k}$ matrices $A_1,\ldots,A_N\in\Gamma_2(p)$ such that for each $B\in\Gamma_2(p)$ we have $A_iB\in\Gamma_2(p^k)$ for some $i$.
Matrices of the form $A_i=I+pM_i$ suggest themselves.

Or much simpler: by counting how many matrices in $\Gamma_2(p)$ and $\Gamma_2(p^k)$ map to the same element of $SL_2(\mathbb Z/p^k\mathbb Z)$ under the canonical projection.

Title: Order of this group


问:
Limit of Sequence $x_n = (x_{n-1} +x_{n-2})/2$
I was unable to solve this particular question in my real analysis assignment and I am looking for help here.

Question: Let $x_1 =0$ and $x_2 =1$ and for n>2 define $x_n = (x_{n-1} +x_{n-2})/2$. Then which of following are true?

A ${x_n}$ is monotone

B ${x_n}$ is Cauchy

C Lim $n \to \infty $  $x_n=2/3$.

I was unable to prove $x_n$ is Cauchy and monotone. All I could prove was that $|x_{n+1} -x_n|$=$ 1/(2)^{n-2}$  and I put n tends to infinity in recurrence relation to get limit but I am only getting that x=2x/2 .

I am unable to prove/ doisprove any of the options.

Can you please help?

答: You can write an explicit formula for $x_n$: write characteristic equation$$ 2x^2-x-1=0$$ It solutions are $1$ and $-1/2$ so it is of form $$x_n = a+b({-1\over 2})^n$$ where you can find $a$ and $b$ from initial conditions. Clearly it is convergent with $a$ as a limit and it is not monotone.


问:
Question on proofs for order of elements in a non-abelian group
Question:

Let $H$ be a group, and $x, y, z \in H$.

Prove that $\mathrm{ord}(xyx^{-1}) = \mathrm{ord}(y)$.

Prove that $\mathrm{ord}(xy) = \mathrm{ord}(yx)$

Assume that $H$ is also abelian, show that $Q_m = \{x \in H: \mathrm{ord}(x) \mid m\}$
is a subgroup of $H$ for any positive integer $m$.

My attempts are as follows:

For part (1), I am stuck, because from logic, I know that the operations done after $x$, $y$ and then inverse of $x$ (namely $x^{-1}$),
is essentially doing the operation of $y$ alone, which is the right hand side of the statement. But I don't know how to write that out explicitly.

Or is there any formulas for ordering, which was not included in my lecture notes, will there be something like, i.e.,
$\mathrm{ord}(ab) = \mathrm{ord}(a)\mathrm{ord}(b) = \mathrm{ord}(b)\mathrm{ord}(a)$?

For part (2), as $H$ is not an abelian group in general, which means $xy$ not equal $yx$ in general, but from logic, I know that the $\mathrm{ord}(xy) = \mathrm{ord}(yx)$.

Because assume that $\mathrm{ord}(x) = 2$ and $\mathrm{ord}(y) = 3$, then $\mathrm{ord}(xy) = 6$, and also $\mathrm{ord}(yx) = 6$ as well, so $\mathrm{ord}(xy)=\mathrm{ord}(yx)$,
but I don't know how to write that out explicitly.

For part (3), to show $Q_m$ is subgroup of $H$, so I need to check on $2$ things, closure and inverse.

For check of closure, I need to show that for all $x, y \in Q_m$, $x\cdot y \in Q_m$, but I have no clue of what operation, i.e. $\cdot $ is for group $H$ or group $Q_m$.

For check of inverse, I need to show that for all $x \in Q_m$ , the inverse of $x$ is also in $Q_m$, but I have no clue how to find the inverse of $x$ as well,
or simply state it exists?

Please give me a little help, thank you!

答:
See my comments above for 1 and 2.

As for 3, I will use the one-step subgroup test.

Fix $m\in\Bbb N$.

Since $e\in H$, ${\rm ord}(e)=1$ and $1\mid m$, we have $e\in Q_m$. Thus $Q_m$ is nonempty.

By definition, $Q_m=\{\color{red}{x\in H}\; :\; {\rm ord}(x)\mid m\},$ so we have $Q_m\subseteq H$.

Let $x,y\in Q_m$. Then ${\rm ord}(x), {\rm ord}(y)\mid m$. Since the order of the inverse of an element is the order of the element, we have

$${\rm ord}(xy^{-1})\mid \frac{{\rm ord}(x){\rm ord}(y)}{{\rm gcd}({\rm ord}(x),{\rm ord}(y))}={\rm lcm}({\rm ord}(x),{\rm ord}(y))\mid m,$$

since $H$ is abelian. But $xy^{-1}\in H$ as $H$ is a group. Hence $xy^{-1}\in Q_m$.

Hence $Q_m\le H$.


答:
Observe that the unique solution of the IVP
$$
y'=1+y^2, \quad y(0)=0,
$$
is $y(x)=\tan x$, and its maximal interval is $(-\pi/2,\pi/2)$.

Next, define the Picard sequence $y_n(x)=\int_0^x \big(\cos^2t+y_{n-1}^2(t)\big)\,dt$ of
$$
y'=\cos^2(x)+y^2, \quad y(0)=0,
$$
and prove inductively that,
$$
0= y_0(x)\le y_1(x)\le \cdots\le y_n(x)\le y_{n+1}(x)\le \tan x, \quad x\in [0,\pi).
$$
Hence, $y_n$ converges to some $y(x)\le \tan x$, for $x\in [0,\pi/2)$.

Also, for $x\in [0,1/2]$, we have $½<\pi/6$, and hence $\tan(1/2)<\tan(\pi/6)=\frac{\sqrt{3}}{3}$.
$$
0\le y_{n+1}(x)-y_n(x)=\int_0^x \big(y_n^2(t)-y_{n-1}^2(t)\big)\,dt
=\int_0^x \big(y_n(t)+y_{n-1}(t)\big)\big(y_n(t)-y_{n-1}(t)\big)\,dt
\\ \le 2\tan x \int_0^x \big(y_n(t)-y_{n-1}(t)\big)\,dt \le
\frac{2\sqrt{3}}{3}\int_0^x \big(y_n(t)-y_{n-1}(t)\big)\,dt
$$
In particular, $y_1(x)-y_0(x)=\int_0^x \cos^2t\,dt \le x$,
$$
y_2(x)-y_1(x)\le \frac{2\sqrt{3}}{3}\int_0^x \big(y_1(t)-y_0(t)\big)\,dt \le
\frac{2\sqrt{3}}{3}\int_0^x t\,dt \le \frac{2\sqrt{3}}{3}\cdot \frac{x^2}{2}
$$
$$
y_3(x)-y_2(x)\le \frac{2\sqrt{3}}{3}\int_0^x \big(y_2(t)-y_1(t)\big)\,dt \le
\frac{2\sqrt{3}}{3}\int_0^x \frac{2\sqrt{3}}{3}\cdot \frac{t^2}{2}\,dt \le \left(\frac{2\sqrt{3}}{3}\right)^2\cdot \frac{x^3}{3!}
$$
and inductively
$$
0 \le y_{n}(x)-y_{n-1}(x)\le  \left(\frac{2\sqrt{3}}{3}\right)^{n-1}\cdot \frac{x^{n}}{n!} \le \frac{1}{n!} , \quad x\in [0,1/2],
$$
Hence $\{y_n(x)\}$ converges uniformly in $[0,1/2]$ (Weierstrass M-test) and the limit $y(x)$ satisfies
$$
y(x)=\int_0^x \big(\cos^2t+y^2(t)\big)\,dt, \quad x\in [0,1/2]
$$
and hence $y$ satisfies the IVP.

Note. The interval $[0,1/2]$ could be replaced by $[-a,a]$, for any $0<a<\pi/2$.

Title: Proof of the existence of a solution on IVP with a given interval for $x$

Body:
Prove that the solution $y(x)$ exists on the given interval for $x$:
$$y'=y^2 + \cos x^2$$
on $0\leq x\leq 1/2$ with $y(0)=0$.
I think that I have to use Picard's Theorem. But I don't know why I am given an interval for $x$. And the left bound of $x$ is the initial value. Any help?


问:
Determine all pairs of perfect powers $2^m$, $3^n$ such that their difference is a perfect square.
Determine all pairs of perfect powers $2^m$, $3^n$ such that their difference is a perfect square.

I found these pairs work by mod 3 and mode 4:

Case 1, $2^m \ge 3^n$, $(0,0), (2,1), (1,0)$

Case 2, $2^m \le 3^n$, $(0,0), (1,1), (1,3), (3, 2), (5, 4)$

How to prove there's no others?

I also found this post: Pairs forming perfect square. There's good discussions. Do we have any complete solution?

答:
You can solve the problem in three cases:

Case 1: $m\ge2$ and $2^m>3^n$.

Here, looking modulos $3$ and $4$ demands that $m$ is even and $n$ is odd. The problem then reduces to the form
$$(2^{m/2})^2=x^2+3(3^{(n-1)/2})^2\,~\,~\,~\,~ (1)$$
for some integer $x$; thus we are looking at solutions to the Diophantine equation
$$z^2=x^2+3y^2\,.$$
Checking modulo $8$ shows that the only possibility for both $x,y$ to be odd is when $z^2\equiv 4\mod 8$, which thus demands that the only solutions to Eq (1) must be $$(m,n)=(2,1)\,.$$

Case 2: $m\ge2$ and $2^m<3^n$.

Here also, looking modulos $3$ and $4$ rather demands that $m$ is odd and $n$ is even. Now, the problem reduces to the form
$$(3^{n/2})^2=x^2+2(2^{(m-1)/2})^2\,~\,~\,~\,~ (2)$$
for some integer $x$; thus, this time, we are looking at solutions to the Diophantine equation
$$z^2=x^2+2y^2\,.$$
The general solution to this equation (in the manner similar to deriving Pythagorean Triples) is given by $$z=a^2+2b^2\,,~\,~\,~\,x=a^2-2b^2\,,~\,~\,~\,~y=2ab$$ for some integers $a,b$. Comparing to Eq (2) forces $a=\pm 1$ and $b=\pm 2^{(m-3)/2}$, which implies that $3^{n/2}=z=(\pm 1)^2+2(\pm 2^{(m-3)/2})^2= 1+2^{m-2}$——that is, $3^{n/2}-2^{m-2}=1$——which has the only solutions $$(m,n)\in\{(3,2),(5,4)\}$$ thanks Catalan-Mihailescu’s Theorem.

Case: $m\in\{0,1\}$.

For $n=0$, we have the obvious solutions $$(m,n)\in\{0,0),(1,0)\}\,.$$ Hence suppose $n\ge 1$. Here, we have $3^n-2^m=y^2$ for some integer $y$, which can be rewritten as
$$y^2=3^{e_n}\left(3^{(n-e_n)/3)}\right)^3-2^m\,,$$ where $e_n\in\{0,1,2\}$ and $n\equiv e_n\mod 3$,. Multiplying through by $3^{2e_n}$, we see that we are actually looking at integer solutions to the Mordell elliptic curve $$y^2=x^3-2^m\cdot 3^{2e_n}\,,$$ and a solution to your problem corresponds to $x=3^{(n+2e_n)/3}$. Mordell (1920) proved that there are only finitely many integral solutions to such elliptic curves (you can find the integer solutions listed here https://web.archive.org/web/20110618033146/http://tnt.math.se.tmu.ac.jp/simath/MORDELL/MORDELL-); in particular, for $m=1$ and $e_n=0$, it is a theorem or Fermat that the only integral solutions to the elliptic are $(x,y)\in\{3,\pm 5\}$, which thus gives the only solution to your problem as $$(m,n)=(1,3)\,.$$ When $m=e_n=0$, the only integral solution is $(x,y)=(1,0)$, which doesn’t correspond to a solution to your problem.
For $e_n=1$, using the linked address above, the only integral solutions are no solutions for $m=0$ but $(x,y)=(3,\pm 3)$ for $m=1$, which forces the only solution to your problem in this case $$(m,n)=(1,1)\,.$$ Finally, for $e_n=2$, there are no integral solutions to the elliptic curve for $m=1$, but for $m=0$, the only solution is $(x,y)=(13,46)$, which does not correspond to a solution to your problem.


Title: Limit of Sequence $x_n = (x_{n-1} +x_{n-2})/2$

Body:
I was unable to solve this particular question in my real analysis assignment and I am looking for help here.

Question: Let $x_1 =0$ and $x_2 =1$ and for n>2 define $x_n = (x_{n-1} +x_{n-2})/2$. Then which of following are true?

A ${x_n}$ is monotone

B ${x_n}$ is Cauchy

C Lim $n \to \infty $  $x_n=2/3$.

I was unable to prove $x_n$ is Cauchy and monotone. All I could prove was that $|x_{n+1} -x_n|$=$ 1/(2)^{n-2}$  and I put n tends to infinity in recurrence relation to get limit but I am only getting that x=2x/2 .

I am unable to prove/ doisprove any of the options.

Can you please help?

Body:
$$
x_{n+1}=\frac{1}{2}(x_{n}+x_{n-1})
\quad\Longrightarrow\quad
x_{n+1}-x_n=-\frac{1}{2}(x_{n}-x_{n-1})=\cdots
=\frac{(-1)^{n-1}}{2^{n-1}}(x_2-x_1)=\frac{(-1)^{n-1}}{2^{n-1}}.
$$
and
$$
x_{n+1}=\frac{1}{2}(x_{n}+x_{n-1})
\quad\Longrightarrow\quad
x_{n+1}+\frac{x_n}{2}=x_n+\frac{x_{n-1}}{2}=\cdots=x_2+\frac{x_1}{2}=1
$$
Subtracting the above we obtain
$$
x_n=\frac{2}{3}+\frac{(-1)^{n-1}}{3\cdot 2^{n-1}}\to \frac{2}{3}
$$


答:
Since $x_1=0$, $x_2=1$, and $x_3=\frac12$, it is clear that $(x_n)_{n\in\Bbb N}$ is not monotonic.

For each $n\in\Bbb N$, you have$$\begin{bmatrix}x_{n+2}\\x_{n+1}\end{bmatrix}=\begin{bmatrix}\frac12&\frac12\\1&0\end{bmatrix}\begin{bmatrix}x_{n+1}\\x_n\end{bmatrix}$$
and therefore$$\begin{bmatrix}x_{n+1}\\x_n\end{bmatrix}=\begin{bmatrix}\frac12&\frac12\\1&0\end{bmatrix}^n\begin{bmatrix}1\\0\end{bmatrix}.$$But then, since$$\lim_{n\to\infty}\begin{bmatrix}\frac12&\frac12\\1&0\end{bmatrix}^n=\begin{bmatrix}\frac23&\frac13\\\frac23&\frac13\end{bmatrix},$$and since$$\begin{bmatrix}\frac23&\frac13\\\frac23&\frac13\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}\frac23\\\frac23\end{bmatrix},$$the limit of your sequence is indeed $\frac23$. And, since the sequence converges, it is a Cauchy sequence.


问:
If $f$ is differentiable on $[x_1,x_2]$, $x_1x_2>0$, prove that $\frac{x_1f(x_2)-x_2f(x_1)}{x_1-x_2}=f(c)-cf'(c)$, where $c\in(x_1,x_2)$
So I thought of making up a function $\phi(x)$ (just as how the mean value theorem was proven) and by using rolle's theorem($\phi(x_1)=\phi(x_2)=0$), I could get the answer-- $f(c)-cf'(c)$ reminds me of....something related to $e^{-x}f(x)$? but is that how its done? At least, I couldn't devise the function; couldn't link the dots together... any help would be appreciated! Thanks!

答:
You can write the LHS as $$ \frac{\displaystyle \frac{f(x_2)}{x_2} - \frac{f(x_1)}{x_1}}{\displaystyle \frac{1}{x_2} - \frac{1}{x_2}} $$

Which reminds us of cauchy mean value theorom which just proves the result as $x_1x_2 > 0$ so both $x_1 , x_2$ are at the same side of zero so both the functions are derivable.


问:
Normal basis of $GF(2^4)$ over $GF(2^2)$
I'm reading this paper: Scrutinizing the Tower Field Implementation of the $F(2^8)$
Inverter – with Applications to AES, Camellia, and SM4. It mentioned a sequence of field extensions $F_2 \subseteq F_{2^2} \subseteq F_{2^4} \subseteq F_{2^8}$. Let $r(y) \in F_{2^4}[y], s(z) \in F_{2^2}[z], t(w) \in F_2[w]$ be irreducible polynomials over their respective fields, and let $Y \in F_{2^8}, Z \in F_{2^4}, W \in F_{2^2}$ be roots of $r(y),s(z),t(w)$ over the corresponding fields respectively. Then it obtain a set of normal basis: $[Y^{16},Y]$ is a basis of $F_{2^8}$ over $F_{2^4}$, $[Z^4,Z]$ is a basis of $F_{2^4}$ over $F_{2^2}$, $[W^2,W]$ is a basis of $F_{2^2}$ over $F_{2}$.

But I really don't understand how to get $[Y^{16},Y], [Z^4,Z], [W^2,W]$. HELP!

答:
$F_{p^{2n}} = F_{p^n}[x]/(x^2+ax+b)$ for any $x^2+ax+b\in F_{p^n}[x]$ irreducible.

$F_{p^n}$ is the splitting field of $x^{p^n}-x\in F_p[x]$.

The $F_{p^n}$-conjugate of $x$ is $x^{p^n}$ and $x,x^{p^n}$ is a normal basis of $F_{p^{2n}}/F_{p^n}$ iff $x,x^{p^n}$ are $F_{p^n}$-linearly independent iff $x^{p^n-1}$ is not in $F_{p^n}$.

If $x^{p^n-1}\in F_{p^n}$,

Then $(x^{p^n-1})^{p^n-1}=1$.

On the other hand $x\in F_{p^{2n}}$ implies that $x^{p^{2n}-1}=1$.

Together this would imply that $x^{\gcd(p^{2n}-1,(p^n-1)^2)}=1$.

Now we assume that $p=2$, then $\gcd(2^{2n}-1,(2^n-1)^2)=2^n-1$ so $x^{2^n-1}=1$ ie. $x\in F_{2^n}$ which is a contradiction.

Whence $x,x^{2^n}$ is always a normal basis.


问:
Weird Identities with Scalar Product & Transpose: $\vec{a}\cdot\vec{b} = \vec{b}^T \cdot {a}^T$, $\vec{a}^T \cdot \vec{b} = \vec{b}^T \cdot \vec{a} $?
Let's say I have a row vector $\vec{a}$ and a column vector $\vec{b}$:

\begin{align}
\vec{a}= \begin{pmatrix}4 & 5 & 6\end{pmatrix} \qquad \vec{b} = \begin{pmatrix}1\\2\\3\end{pmatrix}
\end{align}

Now if I want to calculate a scalar product, it is easy for $\vec{a} \cdot \vec{b}= \begin{pmatrix}4 & 5 & 6\end{pmatrix}\begin{pmatrix}1\\2\\3\end{pmatrix} = \begin{pmatrix}4\cdot1+ 5\cdot 2 + 6\cdot 3\end{pmatrix} = 32.$

but it is not as easy for $\vec{b}\cdot \vec{a}$ where the scalar product is undefined:

\begin{align}
\vec{b}\cdot \vec{a} = \begin{pmatrix}1\\2\\3\end{pmatrix} \begin{pmatrix}4 & 5 & 6\end{pmatrix} \qquad{\scriptsize\text{Not defined!}}
\end{align}

So if I take a transpose of both of the vectors I get:

\begin{align}
\vec{b}{\hspace{0.4mm}}^\mathsf{T}\cdot \vec{a}{}^\mathsf{T} = \begin{pmatrix}1&2&3\end{pmatrix} \begin{pmatrix}4 \\ 5 \\ 6\end{pmatrix} = \begin{pmatrix}1\cdot4 + 2 \cdot 5 + 3 \cdot 6\end{pmatrix} = 32
\end{align}

So I get a relation that $\vec{a}\cdot\vec{b} = \vec{b}^T \cdot {a}^{T} = 32$. But why do I find a different equation on Wikipedia which says: $(\vec{a}\cdot\vec{b} ){}^\mathsf{T}= \vec{b}{}^\mathsf{T} \cdot \vec{a}{\hspace{0.4mm}}^\mathsf{T}$?

EDIT: If i do it like @Samuel says i define original vectors differently:

\begin{align}
\vec{a}= \begin{pmatrix} 4\\ 5 \\ 6 \end{pmatrix} \qquad \vec{b}= \begin{pmatrix} 1\\ 2 \\ 3 \end{pmatrix}
\end{align}

And now i try to calculate $\vec{a} \cdot \vec{b}$ which is not defined. I can calculate ${\vec{a}}^T \vec{b}$ though:

\begin{align}
{\vec{a}}^T \cdot \vec{b}= \begin{pmatrix} 4& 5 &6 \end{pmatrix} \begin{pmatrix} 1\\ 2 \\ 3 \end{pmatrix} = 4\cdot 1 + 5\cdot 2 + 6 \cdot 3 = 32
\end{align}

Now i try to calculate scalar product $\vec{b} \cdot \vec{a}$ and neither this one is defined. But i can calculate ${\vec{b}}^T\cdot \vec{a}$ though:

\begin{align}
{\vec{b}}^T \cdot \vec{a}= \begin{pmatrix} 1& 2 &3 \end{pmatrix} \begin{pmatrix} 4\\ 5 \\ 6 \end{pmatrix} = 1\cdot 4 + 2\cdot 5 + 3 \cdot 6 = 32
\end{align}

So now i can say ${\vec{a}}^T \cdot \vec{b} = {\vec{b}}^T \cdot \vec{a} $. So this now is a lot wierder and i cant find it on Wikipedia.

答:
A vector should always be a column vector. If you want to talk about a "row vector", you should write it as the transpose of some vector, or as a matrix with just one row.

I will add that, technically, the scalar product can not be written as a matrix product the way you are doing it. If $u,v$ are vectors, then the matrix product $u^T v$ is a $1\times 1$-matrix, whereas the scalar product $u\cdot v$ is a scalar. It is common to ignore this and consider them the same, but one should sometimes be careful when using this, for example when one wants to multiply the scalar product with a matrix.

Added: You cannot say that $a$ is a vector and then write $a=(4,5,6)$ as a row matrix. You can however say that $a$ is a $1\times 3$-matrix $(4,5,6)$, or that $a$ is a vector and $a^T=(4,5,6)$ (and in the latter case, $a$ is a column vector). Let us suppose that you defined the matrices $a=(4,5,6)$ and $b=\begin{pmatrix}1\\2\\3\end{pmatrix}$. Now, the matrix product $a\cdot b$ is equal to the $1\times 1$-matrix $[32]$, not the number 32, which is the scalar product of the vectors $\begin{pmatrix}4\\5\\6\end{pmatrix}$ and $\begin{pmatrix}1\\2\\3\end{pmatrix}$. Likewise, the matrix product $b^T\cdot a^T$ is equal to the $1\times 1$-matrix $[32]$. Then we can conclude that $a\cdot b=b^T\cdot a^T$ (note that you made a typo in your post and wrote $a^T\cdot b^T$ for the right-hand-side). This is consistent with Wikipedia, which states that $(a\cdot b)^T=b^T\cdot a^T$. However, your typo is also correct in this particular case, namely $(a\cdot b)^T=a^T\cdot b^T$, but this is only because both matrices are $1\times 1$-matrices, so they are equal to their own transposes.

Added after the edit: You get the identity $a^T\cdot b=b^T\cdot a$. This equation is not true for general matrices $a,b$, so you will not find it on Wikipedia. What is generally true, however, is the identity $(a^T\cdot b)^T=b^T\cdot a$. The reason your identity is true is because both matrices are $1\times 1$-matrices, and the transpose of a $1\times 1$-matrix does not change the matrix.


问:
Intuition for divergence of $\arctan \left(\frac{1}{\sqrt n}\right)$, the angle in the Spiral of Theodorus
A (UK sixth form; final year of high school) student of mine raised the interesting question of how to prove that the total angle in the Spiral of Theodorus (formed by constructing successive right-angled triangles with hypotenuses of $\sqrt{n}$), diverges.

He identified that this is equivalent to proving the divergence of the series $$\sum_{r=1}^\infty \arctan \left(\frac{1}{\sqrt r}\right)$$ and came up with an interesting proof attempt which didn't conceptually work (although it was very nicely thought of).

The best I could offer by way of intuition is that $\arctan\left(\frac{1}{\sqrt n}\right) \approx \frac{1}{\sqrt n}$, and the latter series diverges by comparison with $\frac{1}{n}$. But the $\arctan$ value is strictly lesser, so that doesn't convert into a precise proof as far as I can see.

He hasn't been taught formal convergence tests (and it's a while since I was taught them!) although I'm sure he'd be very open to learning. However, I can't shake the feeling there ought to be a nice geometrical demonstration that the spiral does indeed keep winding around the starting point.

答:
It is not "intuition". By MacLaurin expansion at $x=0$ we have

$$\arctan\sqrt{x}= \sqrt{x}+O\left(x^{3/2}\right)$$
Therefore
$$\arctan\sqrt{\frac{1}{n}}\sim \sqrt{\frac{1}{n}};\quad n\to\infty$$
So the given series diverges.

edit

This is a rigorous proof of the divergence


问:
$\lim_{n \to \infty} \frac{\cal P(n)}{n}$
In an exercise I'm asked to evaluate the following limit:

$$\lim_{n \to \infty} \frac{\cal P(n)}{n}$$
Where $\cal P(n)$ is the number of prime numbers that devide $n$.

I have no idea how to approach this. I know that $1< \cal P(n) <n$ So this must mean that

$$\lim_{n \to \infty} \frac{\cal P(n)}{n} \in [0,1]$$

if the limit exists.

But I'm not able to find the exact value.

答: Even if $\mathcal P(n)$ counts prime factors with multiplicity, it's easy to show that $\mathcal P(n)\le\log_2n$. Take the prime factorisation $n=\prod_ip_i^{m_i}$ and replace all $p_i$ by $2$; since all primes are at least $2$ the product cannot increase, yielding $n\ge2^{\mathcal P(n)}$. Thus the limit expression is upper-bounded by $\frac{\log_2n}n$ and the limit is $0$.


问:
The normal mapping of a continuous function with a weak derivative
Let $\Omega\subset\mathbb{R}^n$ be a bounded domain and suppose $u\in W^{2,1}_{\operatorname{loc}}(\Omega)\cap C(\Omega)$. The book I am reading claims that the normal mapping of $u$ satisfies

\begin{equation}
\chi_u(y) := \{p\in\mathbb{R}^n:u(x)\leq u(y) + p\cdot(x-y)~\forall x\in\Omega\} = \{Du(y)\} \quad \text{for a.e. }y\in\Gamma_u := \{z\in\Omega:\chi_u(z)\not=\emptyset\}.
\end{equation}

The given argument is as follows. We know that $w(x):= u(y)+p\cdot(x-y) - u(x)$ takes its minimum value (of zero) on $\Gamma_u\subset\Omega$. It is a fact that for all $\xi\in\mathbb{R}^n$ with $|\xi|=1$, we have
\begin{equation}
\frac{w(y+h\xi)-w(y)}{h} \rightarrow \frac{\partial w}{\partial \xi} \quad\text{in }L_{\operatorname{loc}}^1(\Omega),
\end{equation}
where $\frac{\partial w}{\partial \xi}$ denotes the weak derivative of $w$ in the direction $\xi$. We therefore know that, up to a subsequence, this convergence holds a.e. in $\Omega$. The authors then say -- and this is the step I do not understand -- that by taking $h\rightarrow 0^+$ and $h\rightarrow 0^-$ along a suitable subsequence, we conclude that
\begin{equation}
\frac{\partial w}{\partial \xi}=0 \quad\text{a.e. in }\Gamma_u.
\end{equation}
(Of course, the definition of $w$ and taking $\xi$ to be the coordinate directions then gives the desired result).

How do the authors conclude that the weak derivative of $w$ in the direction $\xi$ is zero a.e. in $\Gamma_u$? Is it a general fact that if the weak derivative of a continuous function exists, then the weak derivative is zero almost everywhere on the set where the function attains its minimum? Thanks in advance!

答:
$\newcommand{\R}{\mathbb{R}}$
Fix $\xi \in \R^n$ of length one and denote
$$ w_h(y) := \frac{w(y+h\xi)-w(y)}{h} \quad \text{for } h \in \R. $$
If the rest of the argument is fine, we have $w_h \to \frac{\partial w}{\partial \xi}$ in $L_{\operatorname{loc}}^1(\Omega)$ as $h \to 0$.

Focus first on $h > 0$. Since $w = 0$ on $\Gamma_u$ and $w \ge 0$ in general, we have $w_h \ge 0$ on $\Gamma_u$. This last property is preserved by $L^1$-convergence (one can choose an a.e. convergent subsequence), so also $\frac{\partial w}{\partial \xi} \ge 0$ a.e. on $\Gamma_u$. Considering $h < 0$, we obtain the opposite inequality.


问:
Solve the differential equation with variable separable method
I am not able to figure out how to integrate after separating

Equation is:
$$y\sqrt{1-x^2}dy+x\sqrt{1-y^2}dx=0$$

In my book there is no solution but have answer:
$$\sqrt{1-x^2}+\sqrt{1-y^2}=c$$

Body:
$$y\sqrt{1-x^2}dy+x\sqrt{1-y^2}dx=0$$
$$-2y\sqrt{1-x^2}dy-2x\sqrt{1-y^2}dx=0$$
$$\sqrt{1-x^2}d(1-y^2)+\sqrt{1-y^2}d(1-x^2)=0$$
Substitute $u=1-x^2 \,\,, v=1-y^2$:
$$\sqrt{u}dv+\sqrt v du=0$$
$$\dfrac {dv} {\sqrt{v}}+\dfrac {du} {\sqrt u}=0$$
Integrate.
$$ {\sqrt{v}}+ {\sqrt u}=C$$
$$ {\sqrt{1-y^2}}+ \sqrt {1-x^2}=C$$


Title: $ X \sim B(200,1/9) $ Calculate $ E[2^{x+3}] $

Body:
$$
X \sim B(200,1/9)
$$

Calculate

$$
E[2^{x+3}]
$$

What i did:

$$
E[2^{x+3}] = E[2^x2^3] = 2^3E[2^x] = 8E[2^x] = 8(\sum_{x=0}^{\infty}2^xPr(X=x)) = 8(\sum_{x=0}^{\infty}2^x\binom{200}{x}(1/9)^x(8/9)^{200-x}) =
$$
$$
= 8(\sum_{x=0}^{\infty}\binom{200}{x}(2/9)^x(8/9)^{200-x})
$$

Here I'm stuck. How can i calculate this serie?

I solved it using MGF, but i am not allowed to use it.

$$
8E[(e^{\ln 2})^X] = 8M_X(t=\ln 2) = 8(8/9+(1/9)e^{ln2})^{200} = 8(10/9)^{200}
$$

Thanks.

Body:
$X$ take value only up to $200$.

$$
= 8(\sum_{x=0}^{200}\binom{200}{x}(2/9)^x(8/9)^{200-x}) = 8(\frac  29 +\frac 8 9)^{200}
$$ by Bimomial Theorem.


问:
''Commutative'' 2-cocycles
Let ba  $G$ an abelian group and $L$ is $G$-module.

If $f$ is a 2-cocycle in $Z(G,L)$, is it true that $f(g,h)=f(h,g)$ for all $g,h \in G$? Or even for $\bar{f} \in H^{2}(G,L)$ is $\overline{f(g,h)}=\overline{f(h,g)}$, here $\bar{f}$ is the cohomology class of $f$.

Thank you

答:
The answer is no, and counterexamples are easy to construct using the isomorphism from Schreier's theorem, which says that $H^2(G,A)$ is isomorphic as a pointed set $0\in H^2(G,A)$ is the point) to $EXT(G,A)$ which is the set of short exact sequences $1\to A\to\hat G\to G\to 1$ up to equivalence, with the point being the semi-direct product $\hat G=A\rtimes G$.

The construction of the cocylces involved is easy. Suppose you have a short exact sequence $1\to A\to\hat G\to G\to 1$ (i.e. identify $A$ as a normal subgroup of some $\widehat G$ so that $\widehat G/A\cong G$). Then pick a system of representatives $\widehat g\in G$ for every $g\in G$. Then $\widehat g\widehat h=f(g,h)\widehat{gh}$ defines a function $f\colon G^2\to A$. The condition that $f$ is a $2$-cocycle falls out from the equality $(\widehat g\widehat h)\widehat k=\widehat g(\widehat h\widehat k)$, $f$.

If $f$ is "commutative", then $\widehat g\widehat h=f(h,g)\widehat{h,g}=f(g,h)\widehat{g,h}=\widehat g\widehat h$ where the middle equality is from commutativity of $G$. So all you need for a counterexample is to find a non-abelian group with quotient $G$, kernel $A$, and a system of representatives that do not commute with each other.

What is the smallest non-abelian group? $S^3$ of order $6$! So let $A=\left<(123)\right>\cong\mathbb Z/3$ and $G=S^3/A\cong\mathbb Z/2$. Then $(12)$ and $(123)$ are coset representatives of $A$ in $S^3$, but $(12)(123)=(13)$ while $(123)(12)=(23)$. So $f((123),(12))\neq f((12),(123))$.

What about the second question: if this is necessarily the case for the homology classes? Well, the different cocycles come from different choices of coset representatives, but the cohomology class of the cocycle is determined by the choice of short exact sequence $1\to A\to \widehat G\to G\to 1$. So an example when $\bar f(g,h)\neq\bar f(h,g)$ would be one in which there is no choice of representatives that commute with each other.

I suspect such an example is furnished by the next smallest non-abelian group: the quaternions of order $8$, which are $\{1,-1,i,-i,j,-j,k,-k\}$ with multiplication table $i^2=j^2=k^2=ijk=-1$, and the requirement that $A=\{1,-1\}$ is the center subgroup (then $G\cong\mathbb Z/2\times\mathbb Z/2$). It would be a fun exercise to think through the cocycles that various coset representatives give you.

P.S. Not every $2$-cocycle arises as a system of representatives in a short exact sequence! But every cohomology class is represented by such a cocycle in some (unique up to isomorphism) short exact sequence. That is the content of Schreier's theorem.

Title: ''Commutative'' 2-cocycles


问:
Closest point to the origin
Using Lagrange multipliers to find the closest points to the origin on the surface
$$x^2 -z^2 =1$$

My attempt:

The distance $f(x,y,z) = \sqrt{x^2 +y^2 +z^2}$

Then
$\frac{2x}{ \sqrt{x^2 +y^2 +z^2}}$=$kx$

$\frac{z}{ \sqrt{x^2 +y^2 +z^2}}$=$-kz$
$y=0$

Then
$\frac{1}{ \sqrt{x^2 +y^2 +z^2}} = \frac{-1}{ \sqrt{x^2 +y^2 +z^2}}$

But the last equation has no solution !
What the mistake that i have done ?

答:
Distance to origin $f(x,y,z) = \sqrt{x^2 +y^2 +z^2}$

Constraint $g(x,z) = x^2 - z^2 -1 = 0$

Applying Lagrange Multiplier method as asked in the question -

$\sqrt{x^2 +y^2 +z^2} = \lambda (x^2 - z^2 - 1)$

Differentiating wrt $x, y, z, \lambda$ we get

$\frac{x}{\sqrt{x^2 +y^2 +z^2}} = 2\lambda x$

$\frac{y}{\sqrt{x^2 +y^2 +z^2}} = 0$

$\frac{z}{\sqrt{x^2 +y^2 +z^2}} = -2 \lambda z$

$x^2 - z^2 - 1 = 0$

Based on first equation, $x = 0$ is a possibility but it does not meet our constraint.

So we have $\lambda = \frac{1}{2\sqrt{x^2 +y^2 +z^2}}$

Second equation gives us $y = 0$. Third equation for above value of $\lambda$ gives $z = 0$.

Substituting in our constraint, we have $x = \pm 1$.

So $x = \pm 1, y = 0, z = 0$ are two points closest to the origin and the distance is $1$.

As you can see from the equation of the constraint, it is a hyperbola with vertices at $(\pm 1, 0, 0)$ and of course vertices of the hyperbola will have the min distance from the origin.

Title: Closest point to the origin


问:
Probability of choosing 4 sick people from 100 where there is a probability of 8.6% for each of them to have a disease.
I have the following task:
There is 4% of sick people in a population.
There is 5% chance for a positive test for a healthy person.
There is 2% chance for a negative test for a a sick person.
We tested 100 people and 4 had a positive test.
What is the probability of such an outcome?

So far I found that the probability for a test to be positive is 0.086, but I am not sure what to do next.

答: Assuming your overall positivity rate is correct: You then have a binomial probability situation where you want to know the probability of 4 successes in 100 trials.</p>


问:
Prove the inequality $x-4x \frac{x}{\sqrt{x}+x} +3x \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 <0$.
I want to prove $$x-4x \frac{x}{\sqrt{x}+x} +3x  \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 <0$$
for $x>0$ real number.
I tried supposing that  $$x-4x \frac{x}{\sqrt{x}+x} +3x  \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 \geq 0$$ holds to get a contradiction but I couldn't find the solution.
Can you help me proceed?

答:
EDIT: The question was updated, so here's my new answer.

Notice the LHS is

and since $x>0$, it's clear that it is $<0$ for all $x$.

Old answer: This is false. Indeed, since $x>0$, we have

\begin{align*}
   &x-4x \frac{1}{\sqrt{x}+x} +3x \left( \frac{1}{\sqrt{x}+x} \right)^2 -3 \left(\frac{1}{\sqrt{x}+x} \right)^2 >0\\
  \iff & (x+\sqrt x)^2\left(x-4x \frac{1}{\sqrt{x}+x} +3x \left( \frac{1}{\sqrt{x}+x} \right)^2 -3 \left(\frac{1}{\sqrt{x}+x} \right)^2 \right) >0\\
  \iff & x^3 + 2x^{5/2}-3x^2-4x^{3/2}+3x-3 >0\\
 \iff & (\sqrt x+1)(x^{5/2}+x^2-4x^{3/2}+3x^{1/2}-3)>0,
\end{align*}
but this isn't true for all $x$, e.g. put $x=1$ and the quintic becomes $-2<0$.


答:
Rather than concentrating on the angle, consider the arc length of the spiral.

At the $n$th step, we add a right triangle of base $\sqrt n$ and hypotenuse $\sqrt{n+1}$, with the outer side (of course) of length 1 perpendicular to the base. So we add a new arc to the spiral which must be greater than 1. Hence the arc length at $n$ steps must be greater than $n$. Thus the arc length diverges, and so must the angle.

(FWIW, a decade or so ago, I spent some time trying to come up with a good approximation of $n$ given a total angle of $\theta$. It gets rather messy...).

Wikipedia has some info, including an analytic continuation of the spiral published by Davis in 2001. It also has this nice diagram.

And here's a link to their SVG version. (I'm sure it'd be easy to make a much smaller SVG).

Title: Intuition for divergence of $\arctan \left(\frac{1}{\sqrt n}\right)$, the angle in the Spiral of Theodorus

Body:
A (UK sixth form; final year of high school) student of mine raised the interesting question of how to prove that the total angle in the Spiral of Theodorus (formed by constructing successive right-angled triangles with hypotenuses of $\sqrt{n}$), diverges.

He identified that this is equivalent to proving the divergence of the series $$\sum_{r=1}^\infty \arctan \left(\frac{1}{\sqrt r}\right)$$ and came up with an interesting proof attempt which didn't conceptually work (although it was very nicely thought of).

The best I could offer by way of intuition is that $\arctan\left(\frac{1}{\sqrt n}\right) \approx \frac{1}{\sqrt n}$, and the latter series diverges by comparison with $\frac{1}{n}$. But the $\arctan$ value is strictly lesser, so that doesn't convert into a precise proof as far as I can see.

He hasn't been taught formal convergence tests (and it's a while since I was taught them!) although I'm sure he'd be very open to learning. However, I can't shake the feeling there ought to be a nice geometrical demonstration that the spiral does indeed keep winding around the starting point.


问:
Image of sup/inf under Galois connection between lattices
This is Lemma 2.4 in Pete Clark's notes on commutative algebra.

Given two posets $(X,\le)$ and $(Y,\le)$, an (antitone) Galois connection between $X$ and $Y$ is a pair of maps $F:X\to Y$ and $G:Y\to X$ such that $F$ and $G$ are antitone (= order-reversing) and for all $x\in X$ and $y\in Y$
$$x\le G(y)\iff y\le F(x)\quad\quad(*).$$

Lemma 2.4: Given a Galois connection as above, if $X$ and $Y$ are lattices, then for all $x_1,x_2\in X$

$F(x_1\lor x_2)=F(x_1)\land F(x_2)$

$F(x_1\land x_2)=F(x_1)\lor F(x_2)$

I can prove the first formula (proof below), but I don't see how to show the second one.  Can anyone show how?

Proof of $F(x_1\lor x_2)=F(x_1)\land F(x_2)$: Given $x_1, x_2$, for each $y\in Y$ we have
$$
\begin{align*}
& y\le F(x_1\lor x_2) & \\
\iff\;& x_1\lor x_2\le G(y) &\text{ from (*)} \\
\iff\;& x_1\le G(y)\quad\text{and}\quad x_2\le G(y) & \text{ from definition of }\lor\\
\iff\;& y\le F(x_1)\quad\text{and}\quad y\le F(x_2) & \text{ from (*)} \\
\iff\;& y\le F(x_1)\land F(x_2) & \text{ from definition of }\land \\
\end{align*}
$$
The desired formula follows.

答:
That's not right.
Maybe you're overlooking some underlying hypothesis.

To see that that's not true (the second condition), consider the lattice $\mathbf L$ in the picture below.

We will define a Galois connection $\varphi : \mathbf L \to \mathbf L$.

(In this case, it's the same map in both ways, having also the same poset in both sides.)

Notice that in order to show that
$$x \leq \varphi(y) \quad\text{iff}\quad y \leq \varphi(x),$$
it is enough to see that $x \leq y$ implies that $\varphi(y) \leq \varphi(x)$ and that $x \leq \varphi(\varphi(x))$, for all $x,y \in L$ (this is a well known alternative definition—or characterization—of Galois connections).

These conditions are immediate if we define
$$\varphi(0)=1,\; \varphi(a) = b,\; \varphi(b) = a,\;\varphi(c) = \varphi(1) = 0.$$
However,
$$\varphi(a) \vee \varphi(b) = b \vee a = c < 1 = \varphi(0) = \varphi(a \wedge b).$$


问:
Tensor algebra functor and coproducts.
Let $V_i$ be a family of vector spaces for $i\in I$. I'm trying to show that
$$T\bigoplus_{i\in I} V_i\cong \coprod_{i\in I}TV_i$$
where the RHS is the free product of algebras. My idea is to show that the LHS has the right universal property for the free product. The natural linear maps $V_i\rightarrow \bigoplus_{i\in I}V_i$ give algebra homomorphisms $TV_i \rightarrow T\bigoplus_{i\in I}V_i$, and I need to show these make the latter a free product of the $TV_i$s, but from here I'm not sure what to do.

答: Hint: What's the universal property of $T$ ?

Title: Tensor algebra functor and coproducts.


问:
What does it mean to "Decide to which group $G$ is isomorphic" for a given group $G$?
I have a homework question which is

Decide to which group $(\mathbb{Z}_n^*,\,\cdot\,)$ is isomorphic (classification of finite abelian groups), for

    (i) $n = 9$,

    (ii) $n = 15$.

But I don't understand what it means. Should I decide whether $\mathbb{Z}_9^*$ and $\mathbb{Z}_{15}^*$ is isomorphic or I need to do something else?

答:
I'm assuming you mean $\mathbb Z_n^*$.

Your question means that for

$(i)$ Find the group that is isomorphic to $\mathbb Z_9^*$: you'll want to determine how to represent it as the decomposition into the direct product of cyclic groups (one or more factors): Express this isomorphism as the decomposition.

$(ii)$ Find the group that is isomorphic to $\mathbb Z_{15}^*$. Again, use decomposition to express this as the direct product of cyclic groups, and express this isomorphism as its decomposition.

For any group denoted by $\mathbb Z_n^*,\,$ alternatively denoted by $(\mathbb Z/n\mathbb Z, \cdot),\,$ there are $\phi(n)$ elements in this group, the elements being those integers (each representing a congruence class) strictly between $0$ and $n$ and which are coprime to $n$. $\;\phi(n) = n − 1\,$ if and only if $n$ is itself a prime. Once you determine the order of each group, decomposition into the direct product of cyclic products of the same order should easily fall out. (If you have further questions, feel free to comment below.)


问:
In $\Delta ABC$, $AB = 14, BC = 16, AC = 26$. $M$ is the midpoint of $BC$ and $D$ is the point on $BC$ such that $AD$ bisects $\angle BAC$.

In $\Delta ABC$, $AB = 14, BC = 16, AC = 26$. $M$ is the midpoint of $BC$ and $D$ is the point on $BC$ such that $AD$ bisects $\angle BAC$. Let $P$ be the foot of the perpendicular from $B$ to $AD$ . Find $PM$ .

What I Tried: Here is a picture :-

The first thing which I noticed is that, $AC \parallel PM$ from Geogebra. I tried everything to prove that from the diagram but without success. I got no idea as I don't think checking for areas will help, and I tried angle-chasing to the most. And obviously, I am in $50$ steps away from solving the problem.

Can someone show me how $AC \parallel PM$ ? Thank You.

答:
Hint:

PM||AC because it connects midpoints of BC and BK.P is mid point of BK, do you know why?


答:
$P$ is the midpoint of $BK$. $M$ is the midpoint of $BC$.

So $\triangle BPM \sim \triangle BKC$ (by A-S-S).

$KC = AC - AK = 26 - 14 = 12$

So $PM = 6$.

Body:
In $\Delta ABC$, $AB = 14, BC = 16, AC = 26$. $M$ is the midpoint of $BC$ and $D$ is the point on $BC$ such that $AD$ bisects $\angle BAC$. Let $P$ be the foot of the perpendicular from $B$ to $AD$ . Find $PM$ .

What I Tried: Here is a picture :-

The first thing which I noticed is that, $AC \parallel PM$ from Geogebra. I tried everything to prove that from the diagram but without success. I got no idea as I don't think checking for areas will help, and I tried angle-chasing to the most. And obviously, I am in $50$ steps away from solving the problem.

Can someone show me how $AC \parallel PM$ ? Thank You.


问:
Prove exists $c\in(0,1)$ such that $f''(c)=0$.
$y=f(x)$ has one intersection with the straight line that passes thorugh $(0,f(0))$ and $(1,f(1))$ at $(x_0,f(x_0))$, where $0<x_0<1$,
Where $f$ is continuous on $[0,1]$ and twice differentiable on $(0,1)$，
Prove exists $c\in(0,1)$ such that $f''(c)=0$.

since it involved the line $y=x$, I thought of let $g(x)=f(x)-x$, but this gets me nowhere-- can it be proved by contradiction? suppose there does not exist $f''(c)=0$ then by Darboux $f'$ is strictly monotonic-- I don't know how to continue next! Any help would be appreciated! Thanks!

答:
Consider the function $g : [0,1] \rightarrow \mathbb{R}$ defined by $$g(x)=f(x)-(f(1)-f(0))x -f(0)$$ By hypothesis, $$g(0)=g(x_0)=g(1)=0$$

Because $g$ is differentiable, you can apply Rolle's theorem : there exists $y_0 \in (0,x_0)$ such that $g'(y_0)=0$, and $y_1 \in (x_0,1)$ such that $g'(y_1)=0$. And again by Rolle's theorem, there exists $c \in (y_0, y_1)$ such that $g''(c)=0$. But $g''=f''$, so you have $f''(c)=0$.


问:
Laplacian of viscosity kernel function (Müller SPH)
I'm studying Müller's paper about Smoothed Particle Hydrodynamics (SPH).

To compute fluid viscosity, the following kernel function is proposed (paragraph 3.5, eq. 22):

$$W_{viscosity}(\mathbf{r}, h)=\frac{15}{2\pi h^3}
\begin{cases}
-\frac{r^3}{2h^3} + \frac{r^2}{h^2} + \frac{h}{2r} - 1,  & 0 \le r \le h \\
0 & \text{otherwise}
\end{cases}
$$

I'm interested in the 2D case, so I'm assuming that $$ h \in \mathbb{R}^+, \mathbf{r} \in \mathbb{R}^2, r=||\mathbf{r}||_2$$

Immediately after the kernel definition, it is stated that
$$\nabla^2 W(\mathbf{r}, h)=\frac{45}{\pi h^6} (h - r)$$

but I get a different result when I calculate the Laplacian on my own.

I start with the linearity of the Laplacian operator:

$$ \nabla^2 (a \cdot f(\mathbf{x}) + b \cdot g(\mathbf{x})) = a\cdot\nabla^2 f(\mathbf{x}) + b \cdot \nabla^2 g(\mathbf{x})$$

Assuming $ 0 \le r \le h $ I have:

$$
\nabla^2 W_{viscosity}(\mathbf{r}, h) = \frac{15}{2\pi h^3}  \left[ -\frac{ \nabla^2  ||\mathbf{r}||_2^3}{2h^3}  + \frac{\nabla^2  ||\mathbf{r}||_2^2}{h^2}  + \frac{h}{2}\nabla^2 \frac{1}{||\mathbf{r}||_2}\right]
$$
where on the last term I dropped the $-1$ because it's an additive constant.

Given the following general rule:

$$\nabla^2 ||\mathbf{r}||_2^n = n^2 ||\mathbf{r}||_2^{n-2}$$

I need in particular:
$$\nabla^2 ||\mathbf{r}||_2^3 = 9||\mathbf{r}||_2$$
$$\nabla^2 ||\mathbf{r}||_2^2 = 4$$
$$\nabla^2 \frac{1}{||\mathbf{r}||_2} = \frac{1}{||\mathbf{r}||_2^3}$$

Thus

$$
\frac{15}{2\pi h^3}  \left[ -\frac{ \nabla^2  ||\mathbf{r}||_2^3}{2h^3}  + \frac{\nabla^2  ||\mathbf{r}||_2^2}{h^2}  + \frac{h}{2}\nabla^2 \frac{1}{||\mathbf{r}||_2}\right]$$
$$= \frac{15}{2\pi h^3}  \left[ -\frac{ 9||\mathbf{r}||_2}{2h^3}  + \frac{4}{h^2}  +\frac{h}{2||\mathbf{r}||_2^3}\right]$$
$$= \frac{15}{2\pi h^3}  \frac{-9||\mathbf{r}||_2^4 + 8h||\mathbf{r}||_2^3 + h^4}{2 h^3 ||\mathbf{r}||_2^3}$$
$$=\frac{15\left( -9||\mathbf{r}||_2^4 + 8h||\mathbf{r}||_2^3 + h^4 \right)}{4\pi h^6 ||\mathbf{r}||_2^3}$$

which is different from the Laplacian presented in the paper.

Surprisingly, Mathematica seems to agree with my derivation:

```<code>Wviscosity[rx_, ry_, h_] := With[
   {
       r = Sqrt[rx^2 + ry^2]
   },
   15/(2*\[Pi]*h^3)*
   Piecewise[{{-r^3/(2*h^3) + r^2/h^2 + h/(2*r) - 1, 0 <= r <= h}}, 0]
];

Laplacian[Wviscosity[rx, ry, h] , {rx, ry}] // FullSimplify
</code>```

yields

Mathematica Laplacian result

I think it's unlikely that the paper is wrong, so what am I missing?

Thanks

答:
I finally found the error: I'm considering $\mathbf{r} \in \mathbb{R}^2$,  while the paper assumes $\mathbf{r} \in \mathbb{R}^3$, which affects the laplacian formulas as well as the definition of $W_{viscosity}$.

I've found here a very good derivation of the smoothing kernels, which gives some insight on why the kernels are what they are.
The presented 2D kernels agree with the ones reported by Muller himself here.

On a final note, it seems that many 2D SPH implementations just use the 3D kernel formulas whereas the 2D kernels should be used instead.


问:
Given an equation with coefficients a and b, how can I find the values a and b for which the equation has 1 solution?
Given system of equations: {x+ay=1, x-ay=b} where x,y are variables and a,b parameters.

My idea was that I should write the system of equations as a reduced echelon matrix, but it doesn't seem to work instead I did this.

$$
    \begin{bmatrix}
    1 & a & 1 \\
    1 & -a & b \\
    \end{bmatrix}
$$

~ $$
    \begin{bmatrix}
    1 & 0 & (1+b)/2 \\
    0 & a & (b-1) \\
    \end{bmatrix}
$$

And this is where I get uncertain. My idea is that if a = 0 and b-1≠0 we should not have any solution to the system. if a = 0 and b-1=0 we should have an infinite amount of solutions and else we should have one solution.

答:
There is a simple criterion for the existence of solutions:

A non-homogeneous linear system of equations $A \mathbf x=\mathbf b$ has solutions if & only if  the matrix $A$ and the augmented matrix $A\mathbf b$have the same rank.

Furthermore, this common rank is the codimension of the affine subspace of solutions.


问:
Exponent of prime $p$ in $n!$

For how many positive integral values of $n$ does $n!$ end with precisely $25$ zeroes?

My work

Number of zeroes at the end of $n!$=$25$

Using Legendre's Theorem,

$$E_5(n!)=25$$

$$\lfloor \frac{n}{5} \rfloor+\lfloor \frac{n}{25} \rfloor +...=25.$$

How to solve this? Can you give me hint?

答:
You solved upto this step.

$$\lfloor \frac{n}{5} \rfloor+\lfloor \frac{n}{25}\rfloor+...=25$$

First put $n$=$100$.

$$\lfloor \frac{100}{5} \rfloor+\lfloor \frac{100}{25}\rfloor+...=20+4=24$$

You need one more $0$.

Now put $n$=$105$

$$\lfloor \frac{105}{5} \rfloor+\lfloor \frac{105}{25}\rfloor+...=21+4=25$$

This continues upto $n$=$109$.

So, there are 5 such integers.

Title: Exponent of prime $p$ in $n!$


问:
calculating an infinte sum
given that : $\alpha_{i}$ = $\frac1{2^{i+1}}$  and  $\beta_{i}$ = (1+$\frac{{(i-1)i}}{2} +(2i))$ 2

How can we calculate this infinite  sum :

$$\sum_{k=1}^\infty{ \alpha_{i}\beta_{i}}$$

I calculate 1200 terms of this sum using python , and it is seems that it converges to 8 .

How can we proove that mathematically ?

This was my try :

$$\sum_{k=1}^\infty{ \alpha_{i}\beta_{i}}$$ $$=\sum_{k=1}^\infty{ \left(\frac1{2^{i}} + \frac{{(i-1)}i}{2^{i+1} } + \frac{{2}i}{2^{i} }\right)}$$

The first term is easy ( geometric serie ), but what the second and the third term ?

答:
You seem to indicate from your question that you understand how to calculate the first term. As you mentioned, it's a geometric series, the formula for which is well known. The last term is a little more complicated. It's an arithmetico-geometric sequence. The formulas for these are known, and can be found in the linked article:

$$\Sigma i(r)^i = \frac{r}{(1 - r)^2}$$

So for $r = \frac12$ we have:

$$\Sigma i(\frac12)^i = \frac{\frac12}{(\frac12)^2} = \frac{\frac12}{(\frac14)} = 2$$

The middle term is yet again slightly more complicated. Let's work out a general formula first that'll help us. Suppose we want to calculate the infinite sum $S$ given below:

$$S = \Sigma i^2r^i$$

Assume that $r$ is such that the sum converges. Then we have can algebraically manipulate it to get an identity as follows:

$$S(1 - r) = S - rS = \Sigma i^2r^i - \Sigma i^2r^{i+1} = \\
\Sigma i^2r^i - \Sigma (i - 1)^2r^i = \\
\Sigma i^2r^i - (i - 1)^2r^i = \Sigma (2i - 1)r^i = \\
2\Sigma ir^i - \Sigma r^i$$

Now we've reduced to terms we already know: a geometric series plus an arithmetico-geometric series. Applying the formulas we've already seen for those:

$$ = 2\left(\frac{r}{(1 - r)^2}\right) - \frac{r}{1 - r}$$

Remember this is the term for $S(1 - r)$ so to get $S$ we divide by $1 - r$ to get:

$$S = 2\left(\frac{r}{(1 - r)^3}\right) - \frac{r}{(1 - r)^2}$$

In particular, when $r = \frac12$ then $S$ is:

$$S = 2\left(\frac{\frac12}{(\frac12)^3}\right) - \frac{\frac12}{( \frac12)^2} = \\
2\left(\frac{\frac12}{(\frac18)}\right) - \frac{\frac12}{(\frac14)} = \\
2(\frac82) - \frac42 = 8 - 2 = 6$$

You now have mathematical formulae for all the terms in your expression. Let's plug them into your expression and just use some algebra/rearrangement of sums to work out the result. First, though, I think your formula is slightly incorrect. You have:

$$\sum{ \left(\frac1{2^{i}} + \frac{{(i-1)}i}{2^{i+1} } + \frac{{2}i}{2^{i} }\right)}$$

But I believe you should start with:

$$\sum{ \left(\frac2{2^{i}} + \frac{{(i-1)}i}{2^{i+1} } + \frac{{2}i}{2^{i} }\right)}$$

Notice the $2$ multiplier on the first term, which I think you forgot. From here, you can derive your convergence to $8$:

$$\sum{ \left(\frac2{2^{i}} + \frac{{(i-1)}i}{2^{i+1} } + \frac{{2}i}{2^{i} }\right)} = \\
2\sum\frac{1}{2^{i}} + \sum\frac{{(i-1)}i}{2^{i+1} } + \sum\frac{{2}i}{2^{i} } = \\ 2(1) + \sum\frac{i^2}{2^{i+1} } - \sum\frac{i}{2^{i+1} } + 2\sum\frac{i}{2^{i} }
= \\ 2 + \frac12\sum\frac{i^2}{2^i } - \frac12\sum\frac{i}{2^i} + 2\sum\frac{i}{2^{i} }
= \\ 2 + \frac12\sum\frac{i^2}{2^i } + \frac32\sum\frac{i}{2^{i} } \\
= \\ 2 + \frac12\sum i^2(\frac{1}{2})^i + \frac32\sum i(\frac{1}{2})^i \\
= \\ 2 + \frac12(6) + \frac32(2) = \\
2 + 3 + 3 = 8$$

Title: calculating an infinte sum


答: Hint:  Simplify that LHS to  $\displaystyle -\frac{2x}{1+\sqrt x}< 0$, which is obvious for $x>0$.

Title: Prove the inequality $x-4x \frac{x}{\sqrt{x}+x} +3x \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 <0$.

Body:
I want to prove $$x-4x \frac{x}{\sqrt{x}+x} +3x  \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 <0$$
for $x>0$ real number.
I tried supposing that  $$x-4x \frac{x}{\sqrt{x}+x} +3x  \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 \geq 0$$ holds to get a contradiction but I couldn't find the solution.
Can you help me proceed?


问:
prove the ratio of the line in the question

the book's solution uses $(a+b)/a= a/b$. I got my answer using b and c but want to understand the book's method.

Body:
$$\frac{a}{b}=\frac{b}{c}$$

$$\frac{b}{a}=\frac{c}{b}$$

$$\frac{b}{a}+1=\frac{c}{b}+1$$

$$\frac{a+b}{a}=\frac{c+b}{b}=\frac{a}{b}$$

$$a^2-ba-b^2=0$$

$$a=\frac{b(1+\sqrt {5})}{2}$$

but

$$a=\frac{b(1-\sqrt {5})}{2}<0$$


问:
Convergence of the improper integral $\int_0^2\frac x{(16-x^4)^{1/3}}\,\mathrm dx$
Need your help in the checking convergence of this improper integral :$$\int_0^2\frac{xdx}{(16-x^4)^\frac{1}{3}}$$I tried my luck and somehow was able to establish a relationship between  $(1-t^4)$  and $t^4$ i.e $$(1-t^4)\geq t^4$$
$$\frac{1}{1-t^4} \leq \frac{1}{t^4}$$
$$\Rightarrow \frac{1}{(1-t^4)\frac{1}{3}} \leq \frac{1}{t^\frac{4}{3}}$$ $$\Rightarrow \frac{t}{(1-t^4)\frac{1}{3}} \leq \frac{t}{t^\frac{4}{3}}$$
for some $$t\in\left[0,\frac{1}{2^\frac{1}{4}}\right]$$ It can be shown that$$\int_0^2\frac{xdx}{(16-x^4)^\frac{1}{3}}=2^\frac{2}{3}\int_0^1\frac{tdt}{(1-t^4)^\frac{1}3} $$
Therefore ,
$$F(x) =\int_0^\frac{1}{2^\frac{1}{4}}\frac{tdt}{(1-t^4)^\frac{1}3} \leq \int_0^\frac{1}{2^\frac{1}{4}}\frac{tdt}{t^\frac{4}{3}} = g(x)$$ For this particular interval I was able to prove its convergence as $g(x)$ converges but I am unable to comment on the remaining interval. Please help and also provide some insight which might help me in future. Thanks in advance...

Body: Since$$16-x^4=(2-x)(8+4x+2x^2+x^3),$$you have$$\frac x{(16-x^4)^{1/3}}=\frac x{\sqrt[3]{2-x}\sqrt[3]{8+4x+2x^2+x^3}}$$and therefore$$\lim_{x\to2^-}\frac{\frac x{(16-x^4)^{1/3}}}{\frac1{\sqrt[3]{2-x}}}=\frac1{\sqrt[3]4}\ne0.$$So, since the integral$$\int_0^2\frac1{\sqrt[3]{2-x}}\,\mathrm dx$$converges, then so does your integral.


问:
Find a general solution
$$(y+e^x)\frac{dy}{dx}=-\frac{y^2}{2}-2ye^x$$

I attempted to turn this inexact eq. to exact:

I multiply by IF = $e^t$ and got:

$$(ye^x+e^{2x})dy+(\frac{y^2}{2}e^x+2ye^{2x})dx = 0$$

which is exact.

Then I integrate, and I get $y^2e^x+2ye^{2x} + c = 0$.

My question is if this is considerate a general solution? or does this expression need to be expressed in terms of y?

If so, how can I do this?

Thanks

答:
A general solution of an ODE is an explicit expression of the form
$$
y=\varphi(x,c),
$$
which contains all the solutions of the ODE for different values of $c$.

The $c$ is a scalar if the equation is scalar of first order. In general, however, it is a vector in $\mathbb R^n,\,$ where $\,n=$order of ODE $\times$ dimension of system.

Nevertheless, such an expression is not always easy to find. Very often the general solution is obtained implicitly, i.e.,
$$
\varPhi(x,y,c)=0,
$$
and this still is acceptable as an implicitly expressed general solution.

Title: Find a general solution


问:
Domain of function $\frac{1}{x + \sqrt{x+1}}$
I need to find the domain of the fuction: $$\frac{1}{x + \sqrt{x+1}}$$

I can see that the conditions for the domain are $x+\sqrt{x+1} \neq 0$ and $x + 1\geq 0$. Hence:
$$x+\sqrt{x+1} \neq 0 \quad \wedge \quad x + 1\geq 0$$
$$x \neq -\sqrt{x+1} \quad \wedge \quad x\geq -1$$
$$x^2 \neq |x+1| \quad \wedge \quad x\geq -1$$
Since $x \geq -1$ the expression $x+1$ is never negative, therefore:

$$x^2 \neq x+1 \quad \wedge \quad x\geq -1$$
$$x^2 - x - 1 \neq 0 \quad \wedge \quad x\geq -1$$
$$x \neq \frac{1\pm \sqrt{5}}{2} \quad \wedge \quad x\geq -1$$

Now both $\frac{1 + \sqrt{5}}{2}$ and $\frac{1 - \sqrt{5}}{2}$ are greater than $-1$. So why isn't my domain $$\mathbb{R} \setminus \{\frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2}\}$$

When I plot the graph I can see that $\frac{1 + \sqrt{5}}{2}$ is part of the domain, but analitically I can't see why or derive from these equations that conclusion.

答:
From
$$x\ne-\sqrt{x+1}$$
you are deducing
$$x^2\ne|x+1|$$
But this is invalid! You can't deduce $a^2\ne b^2$ from $a\ne -b$, because $a$ may be equal to $b$. (And in fact $x$ is equal to $\sqrt{x+1}$ when $x=\frac{1+\sqrt 5}{2}$.)


答:
All the side length $a,b,c$ are known,
hence, we can get all the angles $\alpha,\ \beta,\ \gamma$;

$|BM|=\tfrac12\,a$,

$|BP|$ can be found from $\triangle ABP$,

$\angle PBM=\tfrac12\,\alpha+\beta-90^\circ$,

use the cosine rule for $\triangle BMP$ to find $|MP|$:

\begin{align}
|MP|^2=|BM|^2+|BP|^2-2\cdot|BM|\cdot|BP|\cdot\cos\angle PBM
.
\end{align}

Title: In $\Delta ABC$, $AB = 14, BC = 16, AC = 26$. $M$ is the midpoint of $BC$ and $D$ is the point on $BC$ such that $AD$ bisects $\angle BAC$.


问:
System of linear equations with complex parameters
This system has to be solved depending on (in relation to) the value of the complex parameters p and q

$$x_1  +x_2  +x_3  +x_4  = -1$$
$$x_1  -x_2  +x_3  -x_4  =  1$$
$$x_1  +x_2  -x_3  +x_4  =  1$$
$$x_1  -x_2  +px_3 +qx_4 =  2$$

I would like to show some work I've done but I'm stuck how to start because I don't think I understand properly the problem. I would like someone to explain how these type of problems are approached, are there steps I could follow when solving? I couldn't find similar problems for reference.

答:
$$A=\left(
\begin{array}{cccc}
 1 & 1 & 1 & 1 \\
 1 & -1 & 1 & -1 \\
 1 & 1 & -1 & 1 \\
 1 & -1 & p & q \\
\end{array}
\right)$$
For Cramer's rule the system has one unique solution is and only if $\det A\ne 0\to =4+4q\ne0 \to q\ne-1$

Thanks to Rouché–Capelli theorem

If $q=-1$ then $\text{rank}(A)=3$. the system has solution only if $\text{rank}(A|B)=\text{rank}(A)=3$.

This happens if all the $4$th order determinants of $A|B$ are zero, in particulat if
$$\det\left|
\begin{array}{cccc}
 1 & 1 & 1 & -1 \\
 1 & -1 & 1 & 1 \\
 1 & 1 & -1 & 1 \\
 1 & -1 & p & 2 \\
\end{array}
\right|=4p=0\to p=0$$
The system has infinite solution if $q=-1$ and $p=0$.

From the augmented matrix
$$A|B=\left(
\begin{array}{cccc|c}
 1 & 1 & 1 & 1 & -1 \\
 1 & -1 & 1 & -1 & 1 \\
 1 & 1 & -1 & 1 & 1 \\
 1 & -1 & 0 & -1 & 2 \\
\end{array}
\right)$$
we choose a order $3$ submatrix with nonzero determinant, for instance
$$A'=\left(
\begin{array}{ccc}
 1 & 1 & 1 \\
 1 & -1 & 1 \\
 1 & 1 & -1 \\
\end{array}
\right)$$
and solve
$$\begin{cases}
x_1  +x_2  +x_3    = -1-x_4\\
x_1  -x_2  +x_3    =  1+x_4\\
x_1  +x_2  -x_3    =  1-x_4\\
\end{cases}
$$
which gives
$$x_1= 1,x_2= -x_4-1,x_3= -1$$


问:
Linearize objective function for intlinprog
This is the predecessor to the following question:

I have some constraints based on which I have written a Matlab program that provides me some feasible binary bits for Finite state machines(FSM). However, there is no objective function.

My objective function should be to minimize the Hamming distance (HD) between all of the transitions between the obtained binary bits.

It should be Minimize $\sum_{1 \leq i < j \leq k}   p_{i,j}\sum_{l=1}^{n} |x_{il} - x_{jl}|$ , $i\neq j $

$\sum_{l=1}^{n} |x_{il} - x_{jl}|$ is the HD between the binary bits that I would like to minimize and then sum all of them, considering each transition. Notations- $n$ binary bits, $p_{i,j}$ is some value I would like to multiply with each of the HD, $k$ state FSM.

How do I linearize this?

答:
Every time you have an objective of the form
$$\mathrm{minimize} \sum_i p_i|u_i|$$
where $u_i$ are some variables or linear expressions of other variables, and $p_i$ are nonnegative constants, you can linearize it as
$$\begin{array}{rl}
   \mathrm{minimize}  & \sum_i p_it_i\\
   \mathrm{subject\ to}& t_i\geq u_i, \\
                      & t_i\geq -u_i,
\end{array}$$
where $t_i$ are additional auxiliary variables you introduce.


问:
Question about a puzzle in to mock a mockingbird
In to mock a mockingbird, we have the following puzzle: There are four people:

A is an accurate truth teller

B is an inaccurate truth teller

C is an accurate liar

D is an inaccurate liar

Smullyan claims that only c will claim to be b when asked "who are you?" But it seems to me like
D would also claim to be b - he thinks he's anyone but d and then lies about it, so he could claim to be anyone.

What am I missing?

Here is the full puzzle:

答:
Note: in this answer, I have taken "inaccurate" to mean inconsistent with whatever philosophy governs their behavior.

Well, going through them, we see that, when asked "Who are you?":

A will answer A, as he is an accurate truth teller.

B, in order to be inconsistent with the fact that he is a truth teller, can answer with anyone other than B, as he is inaccurate and cannot think he is B, but will tell you whom he truly (but inaccurately) thinks he is.

C is an accurate liar, and so when asked who he is, he will lie, and lie accurately, i.e, he can answer someone other than C. Thus, he can consistently answer that he is B.

D is an inaccurate liar, i.e., he will try to lie about who he is, but because he is inaccurate, in order to be inconsistent with his philosophy, he must name himself when he is asked who he is. If he were to name anyone other than D, then he would be an accurate liar. As the problem statement says, he is both deluded and dishonest, and so he will try to give you inaccurate information but can't - in other words, he tells the truth (without meaning to!)


问:
On Rouche's theorem: Find the value of $c$ such that $z^5 -3z^2 +1= 0$ has no zeros in $|z| > c$
According to Fundamental theorem of algebra it must have 5 zeros in some region.

If we chose

$f(z) = z^5$ and

$g(z)= -3z^2 -1$

Then $|f(z)| > |g(z)|$

This condition is satisfied for the positive integers such that $|z|≥2$.

We chose $|z| = 2$

Then according to Rouché's theorem

$f(z)$ and $f(z)+g(z)$ will have same number of zeros inside $|z|=2$

It means that all the five zeros will lie inside $|z| = 2$

And there will be no zeros outside for $|z| > 2$

This was my solution

But according its correct ans is $|z|>2^⅔$

I did not get.

How this could be get.

Body:
Observe that
$$
(2^{2/3})^5-3\cdot(2^{2/3})^2-1=8\cdot 2^{1/3}-6\cdot 2^{1/3}-1=2\cdot 2^{2/3}-1>0.
$$
Thus $|z|^5>3|z^2|+1\ge |3z^2-1|$ on the circle $|z|=2^{2/3}$ and
hence $c=2^{2/3}$ works.

You can improve it a little bit, if you replace $c=2^{2/3}$ with $c=1.51<2^{2/3}$, and the same argument works.

In fact, the optimal $c$ is a root of the equation $z^5-3z^2-1=0$, which is equal to
$$
c=1.50941045653627123243833773286186\ldots
$$


问:
Prove it when $ P(X) = a_1 x + a_2 x^2 + \cdots +a_n x^n $

If $ P(X) = a_1 x + a_2 x^2 + \cdots +a_n x^n $ and $ m \in \mathbb{N} $, prove that
$$\lim_{x \to 0} \frac{\sqrt[m]{1 + P(x)} - 1}{x} = \frac{a_1}{m} $$

I tried to solve it using induction but nothing good happend please help me

答: Hint: If $f(x)=\sqrt[m]{P(x)+1}$, then all you have to do is to compute $f'(0)$.


答:
A value of $x$ satisfying the equation
$$
x=-\sqrt{x+1}
$$
must be nonpositive since $\sqrt{x+1}$ is nonnegative.

Squaring both sides is legal but of the two solutions you found, one of them is positive so doesn't satisfy the given equation, hence should not excluded from the domain.

Thus the domain is $[-1,r)\cup(r,\infty)\;$where $r={\large{\frac{1 - \sqrt{5}}{2}}}$.

Title: Domain of function $\frac{1}{x + \sqrt{x+1}}$

Body:
I need to find the domain of the fuction: $$\frac{1}{x + \sqrt{x+1}}$$

I can see that the conditions for the domain are $x+\sqrt{x+1} \neq 0$ and $x + 1\geq 0$. Hence:
$$x+\sqrt{x+1} \neq 0 \quad \wedge \quad x + 1\geq 0$$
$$x \neq -\sqrt{x+1} \quad \wedge \quad x\geq -1$$
$$x^2 \neq |x+1| \quad \wedge \quad x\geq -1$$
Since $x \geq -1$ the expression $x+1$ is never negative, therefore:

$$x^2 \neq x+1 \quad \wedge \quad x\geq -1$$
$$x^2 - x - 1 \neq 0 \quad \wedge \quad x\geq -1$$
$$x \neq \frac{1\pm \sqrt{5}}{2} \quad \wedge \quad x\geq -1$$

Now both $\frac{1 + \sqrt{5}}{2}$ and $\frac{1 - \sqrt{5}}{2}$ are greater than $-1$. So why isn't my domain $$\mathbb{R} \setminus \{\frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2}\}$$

When I plot the graph I can see that $\frac{1 + \sqrt{5}}{2}$ is part of the domain, but analitically I can't see why or derive from these equations that conclusion.


问:
If the absolute values of all the eigenvalues of $C$ are smaller than $1$, then $I-C$ is non-singular.
I am reading "Matrix Calculus"(in Japanese) by Masahiko Saito.

There is the following proposition in this book:

Let $I$ be an identity matrix.

Let $C$ be a square matrix.

Let $A=\begin{pmatrix} I&B \\ O&C \end{pmatrix}$.

If the absolute values of all the eigenvalues of $C$ are smaller than $1$, then $I-C$ is non-singular and $$\lim_{p\to\infty} A^p = \begin{pmatrix}I&B(I-C)^{-1}\\O&O\end{pmatrix}.$$

Proof:

By assumption, $I-C$ is non-singular.

By the previous proposition, $$A^p=\begin{pmatrix} I&B(I-C)^{-1}(I-C^p) \\ O&C^p\end{pmatrix}.$$

By theorem 3.3.8, $$\lim_{p\to\infty} C^p = O.$$

Therefore, $$\lim_{p\to\infty} A^p = \begin{pmatrix}I&B(I-C)^{-1}\\O&O\end{pmatrix}.$$

We must show $I-C$ is non-singular, but by mistake, the author wrote it is non-singular by assumption.

I wanna know a proof of the following proposition:

If the absolute values of all the eigenvalues of $C$ are smaller than $1$, then $I-C$ is non-singular.

My attempt:

$(I-C)(I+C+\dots+C^{p-1}) = I - C^p$.

I know $\lim_{p\to\infty} C^p = O$.

So, $\lim_{p\to\infty} (I - C^p) = I$.

答: If $I-C$ is singular it has a zero eigenvalue. But the eigenvalues of $I-C$ are $1-\lambda$ as $\lambda$ runs through the eigenvalues of $C$. So $C$ has an eigenvalue $1$ - a contradiction.


问:
Finite cyclic groups are isomorphic to their product with $\Bbb Z$?
I'm currently making a start on group theory and have hit a roadblock with a relatively basic theorem on finite cyclic groups. The specific relation killing me is:
$$\mathbb{Z}_n \times \mathbb{Z}_m \cong \mathbb{Z}_{mn} \Leftrightarrow \text{gcd}(m,n) = 1$$
So, the most straightforward result I see there is
$$\mathbb{Z}_n \times \mathbb{Z} \cong \mathbb{Z}_n$$
For some reason this doesn't sit right with me. Why should a cyclic group be unchanged (up to isomorphism) by a direct product with $\Bbb Z$?

Does anybody have a nice example to ease my mind?

Thanks!

答:
The given biconditional, taken as a whole, extends to the infinite cyclic group $\mathbb Z$ when the latter is treated as $\mathbb Z_0$. Let $m=0$:
$$\mathbb Z×\mathbb Z_n\iff\gcd(n,0)=1\iff n=1$$
So your conclusion is false except if $n=1$, in which case it's trivial.


问:
Lowest degree polynomial vanishing on the integers mod $n$?
This problem comes from D.J. Newman's A Problem Seminar.

Problem: What is the lowest degree monic polynomial $p(x)$ such that the value of $p(x)$ is divisible by $100$ whenever $x$ is an integer?

(The number $100$ is not particularly special. You may want to try solving it when $100$ is replaced by an arbitrary integer $n$.)

答:
HINT:
We have $x^5\equiv x\pmod 5$ and $x^5\equiv x\pmod 2$, therefore
$$x^{10}-2x^6+x^2=(x^5-x)^2$$
is a multiple of $100$ for all integers $x$.

Remains to show that no degree $\le 9$ works.
Here's how to reduce this to $d\le 5$:

Let $p$ be monic of minimal degree $d$ with $p(x)\equiv0\pmod{100}$ for all $x\in\mathbb Z$.
Wlog. $p(0)=0$.
Then the first difference $\Delta p(x)=p(x+1)-p(x)$ is also a multiple of $100$ for all integers, but is not monic. In fact, it starts with $d\cdot x^{d-1}$ (from $(x+1)^d-x^d=dx^{d-1}+\ldots$). Assume $5<d<10$. As $\gcd(d,25)=1$, find $u,v$ with $ud+25v=1$. Then $u\Delta p + 25vx^{d-1}$ is monic, divisible by $25$ at integers and has degree $d-1$. Modulo $4$, all monomials are congruent at interger arguments to one of $1, x, x^2, x^3$ (because $x^4\equiv x^2\pmod 4$), hence divisibility by $4$ can be achieved by adding a degree $3$ polynomial $q$. Thus $u\Delta p + 25vx^{d-1}+q$ is monic, of degree $d-1$ and a multiple of $100$ at integers - contradiction.
We conclude that $d\ge 10$ (and we are done) or $d\le 5$.


答:
Alternatively to José Carlos Santos answer you can use the Taylor expansion
$$(1+x)^a=1+ax+o(x^2)$$
with $p(x)$ instead of $x$ (both tend to zero as $x\to 0$) and $a=\frac{1}{m}$

Or even simpler, Bernoulli de l'Hopital:

$$ \lim_{x \to 0} \frac{\sqrt[m]{1 + P(x)} - 1}{x} = \lim_{x \to 0} \left(\sqrt[m]{1 + P(x)}\right)'=\lim_{x \to 0} \frac{P'(x)}{m}({1 + P(x)})^{(1-m)/m}$$

Title: Prove it when $ P(X) = a_1 x + a_2 x^2 + \cdots +a_n x^n $

Body:
If $ P(X) = a_1 x + a_2 x^2 + \cdots +a_n x^n $ and $ m \in \mathbb{N} $, prove that
$$\lim_{x \to 0} \frac{\sqrt[m]{1 + P(x)} - 1}{x} = \frac{a_1}{m} $$

I tried to solve it using induction but nothing good happend please help me


问:
How to prove that there's at least one of them to be multiple 5?
how do we prove that if $a^2+b^2=c^2$, and they are all integers, there would be at least one of them to a multiple of 5.

What I did is to assume that none of them to be multiple of 5. So$$a=5p_1+r_1, b=5p_2+r_2, c=5p_3+r_3, 0<r<5$$.

By direct substitution back to the equation, we would have:

$$5(5p_1^2+2p_1r_1+5p_2^2+2p_2r_2-5p_3^2-2P_3r_3)=r_3^2-r_1^2-r_2^2$$.
Then, we can list all possible scennarios to show that for this equality to be held, there must be at least 1 r to be 0. I want to ask is there any quick way we can show $$r_3^2-r_1^2-r_2^2$$is a multipler of 5 unless one of them to be 0. Thank you.

答:
Modulo $5$, all squares are in $\{-1,0,1\}$. The only ways two squares can add up to a third square modulo $5$ are
$$0+0=0$$
$$1+0=1$$
$$-1+0=-1$$
$$1+(-1)=0$$
There is a zero in all four possibilities. So in $a^2+b^2=c^2$, at least one variable is a multiple of $5$.


答:
First, I will use that $\lim_{n\to \infty} \mathbf{C}^n = \mathbf{O}$, proven in If modulus of each one of eigenvalues of $B$ is less than $1$, then $B^k\rightarrow 0$

Now, we consider the homogeneous system
\begin{equation*}
(\mathbf{I}-\mathbf{C})\mathbf{x}=\mathbf{0},
\end{equation*}
where $\mathbf{0}$ is a column of $0$'s with the same length as the order of $\mathbf{C}$. Then, we can rewrite it as
\begin{equation*}
\mathbf{C}\mathbf{x}=\mathbf{x}.
\end{equation*}
However, this results in the equation
\begin{equation*}
\mathbf{C}^n\mathbf{x}=\mathbf{x}.
\end{equation*}
that holds for every $n\in \mathbb{N}\cup \{0\}$. Provided that $\mathbf{C}^n\to \mathbf{0}$ as $n\to \infty$, taking limits at both side, we get
\begin{equation*}
    \lim_{n\to \infty}\mathbf{x}=\lim_{n\to \infty}\mathbf{C}^n\mathbf{x}=\mathbf{0}.
\end{equation*}
Hence, the unique solution to this system is the trivial solution, so the rank of the square matrix $(\mathbf{I}-\mathbf{A})$ is maximum (by Rouché-Frobenius' theorem) and we can define its inverse.

Body:
I am reading "Matrix Calculus"(in Japanese) by Masahiko Saito.

There is the following proposition in this book:

Let $I$ be an identity matrix.

Let $C$ be a square matrix.

Let $A=\begin{pmatrix} I&B \\ O&C \end{pmatrix}$.

If the absolute values of all the eigenvalues of $C$ are smaller than $1$, then $I-C$ is non-singular and $$\lim_{p\to\infty} A^p = \begin{pmatrix}I&B(I-C)^{-1}\\O&O\end{pmatrix}.$$

Proof:

By assumption, $I-C$ is non-singular.

By the previous proposition, $$A^p=\begin{pmatrix} I&B(I-C)^{-1}(I-C^p) \\ O&C^p\end{pmatrix}.$$

By theorem 3.3.8, $$\lim_{p\to\infty} C^p = O.$$

Therefore, $$\lim_{p\to\infty} A^p = \begin{pmatrix}I&B(I-C)^{-1}\\O&O\end{pmatrix}.$$

We must show $I-C$ is non-singular, but by mistake, the author wrote it is non-singular by assumption.

I wanna know a proof of the following proposition:

If the absolute values of all the eigenvalues of $C$ are smaller than $1$, then $I-C$ is non-singular.

My attempt:

$(I-C)(I+C+\dots+C^{p-1}) = I - C^p$.

I know $\lim_{p\to\infty} C^p = O$.

So, $\lim_{p\to\infty} (I - C^p) = I$.


答:
You can identify $\mathbb Z_0 = \mathbb Z / 0 \mathbb Z$ with $\mathbb Z$. And then, it is of course true that when $n$ is coprime to $0$, then
$$\mathbb Z_n \times \mathbb Z \cong \mathbb Z \,.$$
But the only $n$ that are coprime to $0$ are $\pm 1$, and the above isomomorphism is just
$$\{0\} \times \mathbb Z \cong \mathbb Z \,.$$
It seems you forgot the condition that $\gcd(n, 0) = 1$.

Title: Finite cyclic groups are isomorphic to their product with $\Bbb Z$?

Body:
I'm currently making a start on group theory and have hit a roadblock with a relatively basic theorem on finite cyclic groups. The specific relation killing me is:
$$\mathbb{Z}_n \times \mathbb{Z}_m \cong \mathbb{Z}_{mn} \Leftrightarrow \text{gcd}(m,n) = 1$$
So, the most straightforward result I see there is
$$\mathbb{Z}_n \times \mathbb{Z} \cong \mathbb{Z}_n$$
For some reason this doesn't sit right with me. Why should a cyclic group be unchanged (up to isomorphism) by a direct product with $\Bbb Z$?

Does anybody have a nice example to ease my mind?

Thanks!


问:
Find the limit of the series $6^n/n!$ as $n$ tends to infinity.
I need to find the limit of the following serie:

$\lim_\limits{n\to \infty}$$\frac{6^n}{n!}$

I was thinking of the following solution, but i'm not sure it's correct, please let me know your thoughts :)

$0$ $<$ $\frac{6^n}{n!}$ $<$ $\frac{6}{1}\cdot$$\frac{6}{2}\cdot$ $\frac{6}{3}\cdot$$\frac{6}{4}\cdot$$\frac{6}{5}\cdot$$\frac{6}{6}\cdot$ $1\cdot$ $1\cdot$ $1\cdot$ .... $\cdot1\cdot$ $\frac{6}{n}$ $=$ $\frac{1944}{5n}$

$\lim_\limits{n\to \infty}$$0$ $= 0$

$\lim_\limits{n\to \infty}$$\frac{1944}{5n}$ $= 0$

Hence, by using the sandwich theorem, $\lim_\limits{n\to \infty}$$\frac{6^n}{n!}$ $= 0$.

答: Yes, this is a valid solution. The upper bound you provide works for $n > 6$, but this is okay for $n \rightarrow \infty$.


问:
Plot any irrational number on number line.
I have a basic question:

Can we plot any irrational number on a number line? I can plot all integers and rational numbers on it but how can I plot any irrational number, such as $\sqrt2$,$\sqrt3$ etc.?

答:
Choosing a unit length, you can use the spiral of Theodorus to find any length of the form $\sqrt{n}$ in which n is an integer:

As a matter of fact, as shown in the picture below, using a similarity argument we can deduce that
$CH^2=AH\times BH$, which means by choosing AH to be the unit length, we can construct $\sqrt{BH}$. You just have to find a way to construct such a triangel (Hint: Use a semicircle!).

For other irrational numbers, it depends on the number AND the tools you are allowed to use. For example, it has been proved that these numbers are not constructible  using just a straight edge and compass: $2^{\frac{1}{3}}$, $\pi$, $e$. But, if you're allowed to use a tool to construct a hyperbola and another for a parabola, then you can construct $2^{\frac{1}{3}}$.


答:
$$x\ge-1$$ is obvious.

Then the denominator is zero when

$$x+\sqrt{x+1}=0$$ or by squaring $$x^2=x+1$$
and there is a single valid root
$$x=1-\phi\ge-1.$$


问:
Describe class of rational functions such as $ \left|f(z) \right| \leq M(1 + |z|^\pi),\:\: z\in \mathbb{C} $
I'm trying to describe class of rational functions with the following property: $\exists M = M(f) > 0$ and in the complex plane there is an estimate
$$
\left|f(z) \right| \leq M(1 + |z|^\pi),\:\: z\in \mathbb{C}
$$

What have i try to do?

I tried to estimate the derivatives of the order higher than $\pi$ using the Cauchy integral formula
$$
f^{m} = \frac{m}{2\pi i}\int_{\Gamma}\frac{f(\zeta)}{(\zeta - z)^{m+1}}d\zeta, \: \: z\in \mathbb{C}
$$
and then i need to expand a function $f$ in a power series but i don't know how

UPD: Now I know the answer. It's class of polynomials of degree at most k, but how to get an answer

答:
Hint

If $f = \frac{h}{g}$ where $h, g$ are coprimes then $\deg g = 0$. If not, $g$ has a root. What is the limit of $f$ at such a root?

Therefore $f$ is a polynomial. The inequality $$\left|f(x) \right| \leq M(1 + |z|^\pi),\:\: z\in \mathbb{C}$$

implies that $\deg f \le 3 < \pi$. Conversely, any polynomial of degree at most equal to $3$ satisfies such an inequality.

Finally the class $\mathcal C$ of requested functions is the one of polynomial of degree at most equal to $3$.


问:
Do continuous injections preserve open sets?
Do continuous injections preserve open sets?

I'm pretty sure that's true in euclidean space.

If we let the singleton sets of integers generate the topology of the domain, and then identity map it to the real with standard topology, is that a counterexample?

If they don't, what combination of injective, surjective, continuous, and inverse continuous is the minimum to be an open map?

Edit:  Either I was really tired and distracted on the bus when I typed this into my phone and somehow forgot to say continuous (entirely possible), or whoever put "from $\mathbb{R}^m$ to $\mathbb{R}^n$" in my title deleted it.  Should I start a new one?

答:
Isn't the case $\mathbb R^n \to \mathbb R^n$ a result of L.E.J. Brouwer?  In any case: $\mathbb R^1 \to \mathbb R^1$ is easy from calculus facts, while $\mathbb R^2 \to \mathbb R^2$ is harder, but can be proved from the Jordan curve theorem.

added

I remembered the name, "Invariance of domain"

See https://en.wikipedia.org/wiki/Invariance_of_domain


问:
Find all natural $n$ numbers.
Find all natural $n$ numbers such that
$15(n!)^2+1$ is divisible by $2n-3$.
My try: First I assumed $2n-3$ is not prime number. Let $a$ be divisor of $2n-3$. It's clear that $a<n-1$, so $15(n!)^2$ is divisible by $a$. Which means $15(n!)^2+1$ is not divisble by $a$. But it is given that $15(n!)^2+1$is divisble by $n$ which means it is divisible by $a$ too. But we have already proved it is not. Contradiction!
So $2n-3$ must be prime number. Now if we change $2n-3$ as $p$. We can say $((p+3)/2)!*((p+3)/2)!$ is congruent to $-1$ by $p$ module. By Wilson's theorem $((p+3)/2)!*((p+3)/2)!$ is congruent to $(p-1)!$ by $p$ module

答: it has been checked by a computer up to n = 40000. the set now contains the numbers {1,2,10,77}. I do not have a proof that there are no new numbers that satisfy the conditions for the set above n = 40000. I can see why people thought this question was impossible. it may actually be a nice problem that is solvable if you can prove that the statement is false for n > 77. the rest are checked by computer.</p>

Title: Find all natural $n$ numbers.


问:
Continuity of function with parameter
I have this function. I need to find all the parameters a (real number) so that the function is continuous.

$${f(x)=\begin{cases}
sin(ax-a)/(x-1) & \text{, } x< 1 \\
(a/(x+2))+ax & \text{, } x\geq1
\end{cases}} \\
$$

When I substitute first I get a limit of type 0/0. If I extend the expression by a/a and make adjustments I get as a result a. For the second limit I get a/3 + a. So I get a = a / 3 + a which is probably not good. How should I work with the first limit?

答:
The language you use to describe what you're doing isn't the most accurate, but it is true that

$$\lim_{x \rightarrow 1^-} f(x) = a,$$
and
$$\lim_{x \rightarrow 1^+} f(x) = (a/3) +a,$$

and continuity at $1$ requires that
$\lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1^+} f(x) = f(1)$.

So the restriction you found on $a$ is correct. And it is a possibility that there are no values of $a$ that will make $f$ continuous, or maybe there is a value but it would make $f$ very boring.

(This is more a point about what kind of questions teachers and text book writers tend to create for us, than it is a point about what kind of math problems exist in the real world. Our classes seem to give us problems where finding solutions involves integers or small denominator fractions, and answers are of middle complexity, not super complicated and not boringly simple. But hopefully we build up our skills so that when the answer is boring we are confident enough to say "Yep, the answer here looks suspicious but it's right.".)

Title: Continuity of function with parameter


问:
Are these two writings equivalent?
Is saying $A$ an element of a power set of the power set of $F$ the same as saying $A$ is a subset of the power set $F?$

Mathematically, $A\subset \mathcal P(F) \Leftrightarrow A\in \mathcal P(\mathcal P(F))$

答:
Note that the definition of power set says the following.

Let $A$ be a set. We define the power set of $A,$ denoted by $\mathcal{P}(A)$ to be the set defined as
$$\mathcal{P}(A) = \{X \mid X \subseteq A\}.$$

Therefore the elements of $\mathcal{P}(A)$ are the sets $X$ such that $X$ is a subset of $A,$ i.e., $X \subseteq A.$

Hence, saying that a certain object $X \in \mathcal{P}(A)$ is equivalent to saying that $X \subseteq A,$ i.e.,

$$X \in \mathcal{P}(A) \iff X \subseteq A.$$

Hence, if you let $A = \mathcal{P}(B)$ for some set $B$ you get

$$X \in \mathcal{P}(\mathcal{P}(B)) \iff X \subseteq \mathcal{P}(B)$$

Title: Are these two writings equivalent?


问:
Probability of choosing 4 sick people from 100 where there is a probability of 8.6% for each of them to have a disease.
I have the following task:
There is 4% of sick people in a population.
There is 5% chance for a positive test for a healthy person.
There is 2% chance for a negative test for a a sick person.
We tested 100 people and 4 had a positive test.
What is the probability of such an outcome?

So far I found that the probability for a test to be positive is 0.086, but I am not sure what to do next.

答: Take this answer with a grain of salt, and call me in the morning. The probability of getting a positive result is .0872, as they commented above. IF the probability follows the binomial PMF, then (nCk)(p^k)(q^n-k) gives .0356 or 3.5%. If there is a different distribution, then we may be off the mark.</p>


问:
Let $P$ be a point on $DE$ and let the extensions of $BP$ and $CP$ meet the sides $AC$ and $AB$ at $G$ and $F$ respectively.

In $\Delta ABC$ , $AB = AC = 1$ and $D,E$ are midpoints of $AB$ and $AC$ respectively. Let $P$ be a point on $DE$ and let the extensions of $BP$ and $CP$ meet the sides $AC$ and $AB$ at $G$ and $F$ respectively. Find $\frac{1}{BF} + \frac{1}{CG}$ .

What I Tried: Here is a picture :-

If we take $P$ at either ends like $D$ or $E$ then we get $BF = 0.5$ cm and $CG = 1$ cm . From here I get that $\frac{1}{BF} + \frac{1}{CG} = 3$ . But the surprising thing is wherever I choose $P$ the answer is always coming $3$ in Geogebra. I cannot understand how that is coming.

Can someone explain why the value is always constant? Thank You.

答:
$\frac{GE}{GC} = \frac{PE}{BC}$ (by similar triangles $\triangle GPE, \triangle GBC$)

Similarly,

$\frac{FD}{FB} = \frac{PD}{BC}$

Adding both,

$\frac{GE}{GC} + \frac{FD}{FB} = \frac{1}{2}$

$\frac{GC-EC}{GC} + \frac{FB-BD}{FB} = \frac{1}{2}$

$2 - \frac{EC}{GC} - \frac{BD}{FB} = \frac{1}{2}$

As $EC = BD = \frac{1}{2}$

We get $\frac{1}{GC} + \frac{1}{FB} = 3$