问:
34x+51y = 6z , we got z,y are Primary numbers , so we need to know how many numbers are true in that
34x+51y =  6z  , we got z,y are Primary numbers , so we need to Find all the integers , ,  ∈ ℤ which hold 34x+51y = 6z</p>

答:
you probably have this in comments,  the lattice of points is given by all
$$  (x,y,z) = u(-3,2,0) + v(0,2,17)  $$
with integers $u,v$

Apparently the requirement is $y,z$ prime
That is, $z = 17v$ and can only be a (positive) prime when $v = 1$

Next $y = 2(u+v)$ and can only be a positive prime when $u+v=1,$ so that $u=0$
$$  (x,y,z) = (0,2,17)  $$

$$  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc   \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc   \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  $$

For $x,z$ prime:
That is, $z = 17v$ and can only be a positive prime when $v = 1$

$x = -3u$ and can only be a positive prime when $u = -1$

Thsu
$$  (x,y,z) = -(-3,2,0) + (0,2,17)  = (3,-2,0) + (0,2,17)=( 3,0,17) $$
You cannot also have $y$ prime

$$  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc   \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc   \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  $$

```<code>parisize = 4000000, primelimit = 500000
? row = [ 34,51,-6]
%1 = [34, 51, -6]
? r1 = [ 1,0,0; 1,1,0; 0, 0, 1]
%2 =
[1 0 0]

[1 1 0]

[0 0 1]

? row1 = row * r1
%3 = [85, 51, -6]
? r1 = [ 1,0,0; 0,1,0; 6, 8, 1]
%4 =
[1 0 0]

[0 1 0]

[6 8 1]

? row1 = row * r1
%5 = [-2, 3, -6]
?
? r2 = [ 1,0,0; 1,1,0; 0, 0, 1]
%6 =
[1 0 0]

[1 1 0]

[0 0 1]

? row2 = row1 * r2
%7 = [1, 3, -6]
?
? r3 = [ 1,-3,6; 0,1,0; 0, 0, 1]
%8 =
[1 -3 6]

[0  1 0]

[0  0 1]

? row3 = row2 * r3
%9 = [1, 0, 0]
? r
%10 = r
? r = r1 * r2 * r3
%11 =
[ 1  -3  6]

[ 1  -2  6]

[14 -34 85]

? row
%12 = [34, 51, -6]
? row * r
%13 = [1, 0, 0]
? colselect = [ 0,0; 1,0;0,1]
%14 =
[0 0]

[1 0]

[0 1]

? r * colselect
%15 =
[ -3  6]

[ -2  6]

[-34 85]

? colbasis = r * colselect
%16 =
[ -3  6]

[ -2  6]

[-34 85]

? queeg = qflll(colbasis)
%17 =
[5 2]

[2 1]

? reduced_col_basis = colbasis * queeg
%18 =
[-3  0]

[ 2  2]

[ 0 17]

? redt= mattranspose(reduced_col_basis)
%19 =
[-3 2  0]

[ 0 2 17]

? redt * reduced_col_basis
%20 =
[13   4]

[ 4 293]

?
</code>```

$$  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc   \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc   \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  $$

Body:
you probably have this in comments,  the lattice of points is given by all
$$  (x,y,z) = u(-3,2,0) + v(0,2,17)  $$
with integers $u,v$

Apparently the requirement is $y,z$ prime
That is, $z = 17v$ and can only be a (positive) prime when $v = 1$

Next $y = 2(u+v)$ and can only be a positive prime when $u+v=1,$ so that $u=0$
$$  (x,y,z) = (0,2,17)  $$

$$  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc   \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc   \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  $$

For $x,z$ prime:
That is, $z = 17v$ and can only be a positive prime when $v = 1$

$x = -3u$ and can only be a positive prime when $u = -1$

Thsu
$$  (x,y,z) = -(-3,2,0) + (0,2,17)  = (3,-2,0) + (0,2,17)=( 3,0,17) $$
You cannot also have $y$ prime

$$  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc   \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc   \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  $$

parisize = 4000000, primelimit = 500000
? row = [ 34,51,-6]
%1 = [34, 51, -6]
? r1 = [ 1,0,0; 1,1,0; 0, 0, 1]
%2 =
[1 0 0]

[1 1 0]

[0 0 1]

? row1 = row * r1
%3 = [85, 51, -6]
? r1 = [ 1,0,0; 0,1,0; 6, 8, 1]
%4 =
[1 0 0]

[0 1 0]

[6 8 1]

? row1 = row * r1
%5 = [-2, 3, -6]
?
? r2 = [ 1,0,0; 1,1,0; 0, 0, 1]
%6 =
[1 0 0]

[1 1 0]

[0 0 1]

? row2 = row1 * r2
%7 = [1, 3, -6]
?
? r3 = [ 1,-3,6; 0,1,0; 0, 0, 1]
%8 =
[1 -3 6]

[0  1 0]

[0  0 1]

? row3 = row2 * r3
%9 = [1, 0, 0]
? r
%10 = r
? r = r1 * r2 * r3
%11 =
[ 1  -3  6]

[ 1  -2  6]

[14 -34 85]

? row
%12 = [34, 51, -6]
? row * r
%13 = [1, 0, 0]
? colselect = [ 0,0; 1,0;0,1]
%14 =
[0 0]

[1 0]

[0 1]

? r * colselect
%15 =
[ -3  6]

[ -2  6]

[-34 85]

? colbasis = r * colselect
%16 =
[ -3  6]

[ -2  6]

[-34 85]

? queeg = qflll(colbasis)
%17 =
[5 2]

[2 1]

? reduced_col_basis = colbasis * queeg
%18 =
[-3  0]

[ 2  2]

[ 0 17]

? redt= mattranspose(reduced_col_basis)
%19 =
[-3 2  0]

[ 0 2 17]

? redt * reduced_col_basis
%20 =
[13   4]

[ 4 293]

?

$$  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc   \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc   \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  $$


问:
Dependent or independent samples?
For the scenario given below, I am confused about if the samples are dependent or independent since the scenario does not mention anything about the samples being paired/related or vice versa.

I am aware if terms such as paired, repeated measurements, within-subject effects, matched pairs, and pretest/posttest are instructed in scenarios then it indicates that the samples are dependent and the opposite applies to independent samples, but I am clueless for the given scenario. Any help would be appreciated.

Alice and Bob work evening shifts in a supermarket. Alice has complained to
the manager that she works, on average, much more than Bob. The manager claims that on
average they both work the same amount of time, i.e. the competing claim is that the average
working hours are different. After a short discussion between the manager and Alice, the manager
randomly selected 50 evenings when Alice and Bob both worked.

答:
The scenario you are describing seems to be setting up a hypothesis test, where the null hypothesis is $\mu_A=\mu_B$ and the alternative is $\mu_A\ne\mu_B$ , where $\mu_A$ ($\mu_B$) is the average number of hours worked by A (B).

It is given that the 50 days are "randomly selected". This usually means that the days are selected independently at random from the population of all available days where A and B both worked (though you have to wonder if the manager is going to allow a given day to be selected more than once).

By construction the pairs $(x_i,y_i),\ldots,(x_{50},y_{50})$ of hours worked by A and B are IID, which means you can conduct a paired test. The two individual samples $(x_1,\ldots,x_{50})$ and $(y_1,\ldots,y_{50})$ are not independent, since A's behavior can affect B's behavior and vice-versa, so you shouldn't perform a two-sample test (which anyhow would be less powerful than the paired test).

This whole scenario seems quite artificial to me, and a bad setting for illustrating a hypothesis test. When the manager looks only at days where both Alice and Bob are working (which we can reasonably assume is the only basis for Alice's complaint), the hypothesis being tested is specifically "on the days where both A and B are working, the average hours worked by A equals the average hours worked by B". It is possible, as @user2661923 mentions, that average hours differ on the days where only one of A and B are working, so the manager should be performing that test as well, and/or testing the global average hours on all days worked by A versus B.

Worst of all, if the manager has the power to randomly select from the population of all work days, why doesn't the manager simply summarize everything and report the population statistics? There's no need to drag a bizarre sampling ritual and a hypothesis test into this scenario at all.

ContentLicense: CC BY-SA 4.0

Title: Dependent or independent samples?


问:
Shouldn't the derivative not exist
derrivativeProfessor has shown us this calculations but does the function have a partial derrivative with respect to x in the point (0,0)? the original function is $$u(x,y)=\frac{(x^3+y^3)}{x^2+y^2}$$

答: The partial derivative with respect to $x$ is obtained by holding $y$ constant and differentiating with respect to $x$.  Notice that $u(x,0)=x$ so $\frac{\partial u}{\partial x}=1$ at $(0,0)$.

Title: Shouldn't the derivative not exist


答:
First, my guess is that $u(0,0)$ is defined to be zero and not via the rule for $u(x,y)$ that you give above. Then, use the limit definition of the derivative so that

\begin{equation*} \frac{u(h,0)-u(0,0)}{h}=\frac{\frac{h^{3}}{h^{2}}-0}{h}=\frac{h^{3}}{h^{3}}=1\end{equation*}

so taking $h\to 0$ gives $u_{x}(0,0)=1$.

Title: Shouldn't the derivative not exist

Body: derrivativeProfessor has shown us this calculations but does the function have a partial derrivative with respect to x in the point (0,0)? the original function is $$u(x,y)=\frac{(x^3+y^3)}{x^2+y^2}$$


问:
$\lim\limits_{n \to \infty}{\lfloor x \rfloor + \lfloor x^2 \rfloor + ... + \lfloor x^n \rfloor\over x^n}, 1<x\in \mathbb R^+$
$$\lim\limits_{n \to \infty}{\lfloor x \rfloor + \lfloor x^2 \rfloor  + ... + \lfloor x^n \rfloor\over x^n}, 1<x\in \mathbb R^+$$
I have a problem solving this limit.Here is my solution. Let x be an integer, then we have:
$$\lim\limits_{n \to \infty}{ x + x^2 + ... + x^n\over x^n} = \lim\limits_{n \to \infty} {(x^n -1)x \over (x - 1) x^n } = \lim\limits_{n \to \infty} {\frac{x^{n+1}}{x^n}-\frac{x}{x^n}\over \frac{x^{n+1}}{x^n} -\frac{x^{n}}{x^n}} = \lim\limits_{n \to \infty} {x - \require{cancel} \cancelto{0}{\frac{x}{x^n}}\over x -1}= {x\over x-1}$$

But I don't know how to solve for non-integer x.

答:
Hint: Test the cases $0 \le x < 1$ and $x = 1$.

If $x > 1$, use that $\lfloor x \rfloor > x - 1$.


问:
Question on the right hand rule
Say I'm taking the  cross product of vectors $a$ and $b$. Say that $b$ is totally in the $z$ direction and has length $7$, so $b = 7k$. Say that $a$ is in the $xy$-plane with positive coefficients, $a = 3x + 4y$.

I want to understand the sign of the components of $a \times b$ using the right hand rule. Now, surele since $a \times b$ is orthogonal to both $a$ and $b$, it's $z$ component will be zero. But will the $x$ and $y$ components be positive or negative, and how can i see this with the right hand rule? Thank you for your time

答:
Take a piece of paper, and place a pencil on top of it. That pencil would represent the $z$-axis.

Your thumb should follow the direction of vector $a$, your index finger is pointing towards the $b$ direction and you will notice that your midddle finger should be pointing to the fourth quadrant, hence the $x$ coordinate is positive and the $y$ coordinate is negative.

Title: Question on the right hand rule


答:
You proved that if $x \in \mathbb{N}_{\geq 2}$ the limit is $\frac{x}{x-1}$, but we can actually prove that for any $x>1$ we have $\lim\limits_{n \to \infty} \frac{x+\ldots+x^n}{x^n} = \frac{x}{x-1}$.

$x>1$, then $(x+\ldots+x^n) -n\leq\lfloor x \rfloor + \ldots + \lfloor x^n \rfloor \leq x + \ldots +x^n$, which show us that the limit is again $\frac{x}{x-1}$ (Why?)

Body:
$$\lim\limits_{n \to \infty}{\lfloor x \rfloor + \lfloor x^2 \rfloor  + ... + \lfloor x^n \rfloor\over x^n}, 1<x\in \mathbb R^+$$
I have a problem solving this limit.Here is my solution. Let x be an integer, then we have:
$$\lim\limits_{n \to \infty}{ x + x^2 + ... + x^n\over x^n} = \lim\limits_{n \to \infty} {(x^n -1)x \over (x - 1) x^n } = \lim\limits_{n \to \infty} {\frac{x^{n+1}}{x^n}-\frac{x}{x^n}\over \frac{x^{n+1}}{x^n} -\frac{x^{n}}{x^n}} = \lim\limits_{n \to \infty} {x - \require{cancel} \cancelto{0}{\frac{x}{x^n}}\over x -1}= {x\over x-1}$$

But I don't know how to solve for non-integer x.


答:
Yep, the cross product will have no z component. The cross product of vector in the xy plane with a vector perpendicular to the plane has the effect of rotating the vector plus or minus 90 degrees then scaling it. If the perpendicular vector has a positive component, the plane's vector rotates clockwise, otherwise couter-clockwise, by the right hand rule.

This applies just as well to the unit vectors. So the positive $\hat{y}$ unit vector becomes the positive $\hat{x}$ unit vector.

Rotating a vector doesn't necessarily change the sine of its component. $<\frac{-\sqrt{2}}{2},\frac{\sqrt{2}}{2}>$ doesn't change the sign of its y component upon taking the cross product.

Title: Question on the right hand rule

Body:
Say I'm taking the  cross product of vectors $a$ and $b$. Say that $b$ is totally in the $z$ direction and has length $7$, so $b = 7k$. Say that $a$ is in the $xy$-plane with positive coefficients, $a = 3x + 4y$.

I want to understand the sign of the components of $a \times b$ using the right hand rule. Now, surele since $a \times b$ is orthogonal to both $a$ and $b$, it's $z$ component will be zero. But will the $x$ and $y$ components be positive or negative, and how can i see this with the right hand rule? Thank you for your time


答:
Remember that $a$ points in direction $(3,4)$ which is little above $y=x$ ($45^{\circ}$) line.

Open your palm so your fingers point in direction of a. Now curl your four fingers towards b. Your stretched out thumb will point in direction of $a \times b$.

Easy to see $x$-component is positive and $y$-component negative. More precisely, the components will be proportional to $(4,-3)$ so that $$(3,4)\cdot(4,-3)=0$$

Title: Question on the right hand rule


问:
Will a path in Cayley graphs end at unique vertices when started at distinct vertices and traversed along same edges
Let $G$ be a finite group and $S$ be a subset of $G$. Let the Cayley graph of $G$ with respect to $S$ be defined as follows, provided that $1 {\not\in} S$ and $S$ is inverse closed.

"The Cayley graph of $G$ with respect to $S$, $Cay(G,S)$ is the graph whose vertices are the elements of $G$ and $g$ is adjacent to $gs$ for all $g \in G, \, s \in S$."

Consider a Cayley graph $X$ of a non-abelian group generated by elements $\{s,t\}$. Let $P=sts^2t^3$ (as an example) represent a path in the Cayley graph expressed in terms of the generating elements.

Suppose,

$v_1 \cdot sts^2t^3 = w_1$

$v_2 \cdot sts^2t^3 = w_2$

$v_3 \cdot sts^2t^3 = w_3$ etc.

for vertices $v_1,v_2,v_3, \cdots, w_1, w_2, w_3, \cdots \in V(X)$. If $v_1 \neq v_2 \neq v_3 \neq \cdots \neq v_n$, then will $w_1 \neq w_2 \neq w_3 \neq \cdots \neq w_n$, for any path $P$? ($n$ is the order of the Cayley graph).

The Cayley graphs are regular and we know that each vertex in a Cayley graph has similar environments. Thus, starting from any vertex we can move along edges corresponding to a similar set of generating elements. So it is clear that when any vertex is multiplied by $P$ it will reach another vertex. So my question expressed in another way is that, if $v_1,v_2,v_3, \cdots, v_n$ are distinct, then will $w_1,w_2,w_3, \cdots, w_n$ be also distinct for any arbitrary path $P$?

Thanks a lot in advance.

答:
Yes, they will be distinct. This is a more basic and fundamental fact about groups that can be stated without mention of Cayley graphs.

Suppose $v_1 s = v_2 s$, where $v_1,v_2,s$ are any elements of a group. Then just multiply both sides by $s^{-1}$ to get that $v_1 = v_2$.

This shows that: "if $v_1 s = v_2 s$, then $v_1=v_2$".

What you are asking is the contrapositive of this statement, which is equivalent: "if $v_1 \neq v_2$, then $v_1 s \neq v_2 s$"


问:
Non cyclic Galois extension of degree 12
Are there examples of Galois extensions of degree 12 which are not cyclic(i.e the corresponding Galois group is non cyclic)?
If yes, please give me an hint on how to come up with such an extension?</p>

答: I would say that the easiest is just to adjoin the sixth root of some innocent prime like $7$, and make the extension normal by adjoining the sixth roots of unity, which is only a quadratic extension of the preceding. Thus $\Bbb Q(\sqrt{-3},\sqrt[6]7\,)$. Galois group is the dihedral group of order twelve.


问:
Setting up and solving differential equation with The Euler Method
I recently started this question and it gave me some insight into the world of differential equations.

However the solution was not fit for my goals as I wanted a general method for calculating the position of a particle travelling through a vector field where the vectors in the field represents a force pushing on the particle.

So this is kind of a follow-up from the information I got from my other question.

So what do I have?

A vector field that is defined by a function:

$\vec{F}(x,y)$

I have left out the definition of the function since I wish to find a solution that works regardless off the function definition. But one example could be:
$\vec{F}(x,y) = (\sin(x),\cos(y))$

A particle, which have some initial values for position and velocity

$\vec{p}(0)=P$

$\dot{\vec{p}}(0)=Q$

I know wish to find the position of the particle after a specified time $\vec{P}(t)$ and velocity $\dot{\vec{P}}(t)$.

From what I learned from the other question it's not possible to find a closed form solution to this, but you have to use some approximation with for example Euler's Method. However I don't know how to model this scenario as a differential equation and how to apply Euler's Method (or any other approximation technique) to my scenario. If someone could, step by step explain this process it would be much appreciated!

I should also add, the goal is to implement this in a program, so keep that in mind!

Thanks!

P.S.

I'm sure I missed stuff and you have questions about my logic here so I'd be happy to edit and answer any questions.

答: If the vector field represents the force acting on the particle, then by Newton's law you have $F(p(t)) = m \ddot p(t)$, where $m$ is the mass. Since you don't say anything about mass, let's take $m=1$. You can write Newton's equation as a first-order differential equation by setting $u(t)=(p(t),\dot p(t))$ and getting $\dot u = (\dot p, \ddot p) = (\dot p, F(p))$. In this form, Euler's method is simply $u^{k+1} = u^k + h (\dot p^k, F(p^k))$, which gives you two equations: $p^{k+1} = p^k + h \dot p^{k}$, $\dot p^{k+1} = \dot p^{k} + h F(p^k)$, with $p^0 = P$, $\dot p^0 = Q$.</p>

ContentLicense: CC BY-SA 3.0


问:
Do there exist polynomial rings with nonzero prime ideals that are not maximal?
I know that with $F$ a field, $F[x]/(f(x))$ is a field iff $f(x)$ is irreducible in $F$. Due to the fact that in a UFD irreducible elements are necessarily prime, we would have that $(f(x))$ is both prime and maximal.

Do there exist any rings $R$ such that $R[x]$ is not a UFD and contains nonzero ideals that are prime but not maximal?

答: Take $R$ an integral domain which is not a field. Since $R[X]/(X)\simeq R$, we find that $(X)$ is a prime and non-maximal ideal. Furthermore, if $R$ is not a UFD, then $R[X]$ is not a UFD either. (Now take $R=\mathbb Z[\sqrt{-3}]$, for example.)</p>


问:
Proof of recursive formula for "fusible numbers"
The set of fusible numbers is a fantastic set of rational numbers defined by a simple rule. The story is well told here but I'll repeat the definitions. It's the formula on slide 17 that I'm trying to understand.

Define $\displaystyle a \oplus b = \frac{a+b+1}{2}$. A number is fusible if it is $0$ or if it can be written as $a \oplus b$ where $a, b$ are fusible and $|a-b|<1$. Let $F$ be the set of fusible numbers. More formally, $F$ is the intersection of all sets of real numbers that are closed under $\oplus$ applied to arguments at a distance at most 1.

The set $F$ is a well-ordered set of non-negative rational numbers. The proof that it's well-ordered isn't included in the PDF file I linked to, but it's not hard to show this. (It wouldn't be true if we hadn't insisted on the condition $|a-b|<1$, by the way.)

Amazingly, the order type of $F$ is $\varepsilon_0$. It's also true that $F$ is closed under ordinary addition; this isn't hard to prove either but I don't know if it plays a part in what follows.

Because $F$ is well-ordered, we may define $f(x)$ to be the least fusible number greater than $x$, for any real $x$. Further, set $m(x) = f(x)-x$. We obviously have $m(x) = -x$ for $x<0$, while for $x \geq 0$, it is posited that
$$m(x) = \frac{1}{2}m(x-m(x-1))$$
The question is: why is this last formula true?

I'm able to show one of the necessary inequalities, namely that $\displaystyle m(x) \leq \frac{1}{2}m(x-m(x-1))$:

Given $x$, observe that
$$(x-1+t) \oplus (x-t+u) = x + u/2$$
Take $t = m(x-1)$, which guarantees that ($x-1+t$) is indeed fusible. Now set $u = m(x-t)$ which makes ($x-t+u$) fusible as well, and the distance between those two fusible numbers can't be greater than $1$. It follows that ($x+u/2$) is fusible, and so $m(x)$ is at most $u/2$ for that particular $u$, which is indeed $m(x-m(x-1))$.

The question, then, is:

How can we prove that no other choice of $t$ yields an even smaller value for $m(x)$?

It's not hard to show that there's no loss of generality in focusing on $x-1+t$ and $x-t+m(x-t)$, but greedily minimizing $t$ by setting $t=m(x-1)$ is not in any obvious way guaranteed to yield the minimum value for $m(x)$, as far as I can see.

What am I missing?

答: It may be worth noting that a new paper on Fusible Numbers 'snuck onto' the arXiv earlier this year; it proves the order type $\epsilon_0$ and shows (unsurprisingly, given that) unprovability in PA of the statement 'there exists a next fusible number after $n$ for any $n\in\mathbb{N}$'. It doesn't entirely solve the minimal-gap question but establishes some new bounds, as well.  The paper is "Fusible numbers and Peano Arithmetic" from Jeff Erickson, Gabriel Nivasch, and Junyan Xu, at https://arxiv.org/abs/2003.14342 .


问:
How do you determine a set that a series converges on absolutely uniformly?
Alright, so I know that there are a bunch of tests to figure out whether a series is convergent/the types of convergence, but how do you find the specific set that a particular series is convergent on? I was checking the Weierstrass M-test, but I don't think that specifies the interval?

Also, there is a specific series that I'm looking at right now, but I want to figure out if there's a general way to do this, so I'll add an image of the series later if anyone wants to look at it.

EDIT: This is the series that I'm looking at right now:

答:
$$\sum _{n=1}^{\infty } \frac{\left(\frac{x}{x+1}\right)^n}{\sqrt{n}}=\sum _{n=1}^{\infty }f_n(x)$$
Ratio test
$$\underset{n\to \infty }{\text{lim}}\left|\frac{f_{n+1}}{f_n}\right|=\underset{n\to \infty }{\text{lim}}\left|\frac{\sqrt{n} x}{\sqrt{n+1} (x+1)}\right|=\left|\frac{x}{x+1}\right|$$
converges if $$\left|\frac{x}{x+1}\right|<1\to x>-\frac12$$
In that interval it is absolutely convergent.

Edit

As $|f_n(x)|=\lim_{x\to+\infty} f_n(x)=\frac{1}{\sqrt n}$ Weierstrasss M-Test cannot be applied.
but the given series converges uniformly on any interval $[x_1,x_2];\; \left(x_1>-\frac12\right)$ because the
$$\frac{1}{\sqrt n} \left(\frac{x}{x+1}\right)^n$$
converges.

It converges because for $x>-\frac12$ we have $\left|\frac{x}{x+1}\right|<1$ that means that there exists a $z>1$ such that $\left|\frac{x}{x+1}\right|<\frac{1}{z}$, therefore

$$\frac{1}{\sqrt n} \left|\frac{x}{x+1}\right|^n<\frac{1}{\sqrt{n}z^n}<\frac{1}{z^n}$$
which converges


问:
To find the $n$-th power of a Jordan block such that there is no complex entries.
Let $A = \begin{pmatrix}7&3\\ \:-3&7\end{pmatrix}$ be a $2 \times 2$ matrix. We need to find the $n$-th power of $A$ such that there is no complex entries.

Note that $A$ is a Jordan block. I tried with the eigenvector method but it gives complex entries.

答:
$$A = \begin{pmatrix}7&3\\ \:-3&7\end{pmatrix}$$
$$A = 7\begin{pmatrix}1&0\\ \:0&1\end{pmatrix}+3 \begin{pmatrix}0&1\\ \:-1&0\end{pmatrix}$$
$$A=7I+3N$$
With $N^{2n}=(-1)^nI$ and $N^{2n+1}=(-1)^nN$. And $N$ and $I$ commute. Then use binomial theorem to find $A^n$


问:
Computing an Integral With Parameters
Let $p > -1$ and $n \in \Bbb{N} \cup \{0\}$. I am trying to compute the integral

$$\int_{0}^{1} x^{2n+1} (1-x^2)^p dx$$

I suspect that it equals something like $\frac{1}{(1+p)(n+1)^{p+1}}$ (after trying small $n$), but I don't know how to integrate it (I could also be wrong as to my conjecture, since I am awful at arithmetic). It seems like an annoying application of induction and integration by parts, but I could be completely wrong, as integrals are not my forte. Perhaps there is a simpler, more elegant way to calculate the integral. I could use some help.

Title: Computing an Integral With Parameters

Body:
With the substitution $1-x^2=t$, then$-xdx=dt$ and $x^{2n}=(x^2)^n=(1-t)^n$. So
$$\int_0^1x^{2n-1}(1-x^2)^p dx=-\int_1^0(1-t)^n t^pdt=\int_0^1(1-t)^nt^pdt=B(n+1,p+1)=\frac{n! \Gamma(p+1)}{\Gamma(p+n+2)}$$


问:
Noisy channel coding: feedback does not improve the channel capacity
In the context of noisy channel coding, a theorem by Shannon says that, by using suitable channel codes, communication with rate up to the channel capacity is possible.

I've seen numerous claims that this capacity (of one way communication) cannot be improved by adding a feedback channel, which tells the sender which sent symbols were corrupted by the noisy channel.

In a lecture by MacKay, he proves this fact for the binary erasure channel (BEC). This channel essentially loses a fraction $p$ of sent symbols and has a capacity $C=1-p$. I have doubts about the proof presented in the lecture, since it seems to assume that any lost symbol can be re-transmitted and will then not be lost a second time. Perhaps asymptotically this assumption can be justified but it's not obvious to me.

On the other hand, I've seen claims in literature saying that the feedback does not help for any channel. I'm not sure how the feedback channel is defined exactly for this claim to hold. Is the feedback channel itself also noisy? I think, with added noiseless feedback, the binary symmetric channel (BSC) and the BEC should have the same capacity, since it no longer matters if the bits are lost of flipped. This seems to contradict the claim, meaning that the feedback channel assumed there is not noiseless. Is it then the same channel added backwards?

Finally, how does one prove this result rigorously (perhaps for BSC as an example, or in general)?

答:
I have doubts about the proof presented in the lecture, since it seems to assume that any lost symbol can be re-transmitted and will then not be lost a second time.

No, it doesn't assume that. You send a symbol until the decoder tells you it was received ok. Hence, for a probability of channel error $p$, the number of tries (channel use) for each symbol follows a geometric distribution, and the average number of tries is $1/(1-p)$.

I've seen claims in literature saying that the feedback does not help for any channel. I'm not sure how the feedback channel is defined exactly for this claim to hold.

That holds only for memoryless channels. And "it does not help" only in the sense that the capacity is the same. This is all explained in Cover & Thomas textbook, sect 7.12.

Is the feedback channel itself also noisy?

It's assumed that the feedback channel is perfect (noiseless). If the above property is true with this assumption, then it's also true if the feedback is noisy.

I think, with added noiseless feedback, the binary symmetric channel (BSC) and the BEC should have the same capacity, since it no longer matters if the bits are lost of flipped

Not true. For a BEC, the receiver knows if a channel error occurred, for the BSC it doesn't.


问:
Let $A$ be a matrix of size $5\times 5$ with rank(A)=3 .Prove that there exists a matrix $B$ such that $AB=0$
I have to prove that the rank of the matrix $B=2$ and given that $B$ is a $5\times 5$ non zero matrix.

My attempt :
Let the columns of $B$ be $X_1,X_2,...,X_5$.Then $AX_1=0$,$AX_2=0$,...,$AX_5=0$.Now since the rank of the matrix $A$ is given as $3$ .So there are $2$ free variables and the possible dimensions of $X_1,..,X_5$ are $2$.

Hence combining we can find the possible rank of $B$ is 2 .

Where am I going wrong in the proof?Also I have not been taught linear transformation so I cannot use it here.

答: $B$ can be any $5 \times k$ matrix whose columns are in the null space of $A$ which is 2 dimensional. So $B$ has rank at most $2$.


问:
Interpretable Explanation of Conditional Probability Problem
I've given a question to students of mine, conditional probability.

Essential information is we have an email filter that filters 99.9% of junk emails to the spam folder, but it also filters 7% of legitimate emails there as well. If 50% of emails are junk and 50% are legit, how many emails in a spam folder containing 500 emails can we expect to be legitimate?

I did the work, it comes out to be roughly 32.74, using Bayes theorem, and the calculation speaks for itself, but I hate not being able to give an interpretable reason for why, beyond the formula, the answer makes sense. A few students, of course, answered 35, because that's an interpretable answer.

I gave them the classic Blue/Green taxi-cab witness problem already, which is easier to explain. Look at how many Blue cars he'd call Green vs how many Green cars he'd call Green, then put your "good" outcomes over the total outcomes, where the only valid ones are the ones where the witness says he saw a green taxi. But a similarly plain and interpretable result for the spam email folder is hard for me to find.

答:
Note sure that this is an improvement over Bayes but:

Say $2N$ was the total number of emails received.  Then we have (on average) $N$ junk mails and $N$ legitimate emails.  And the spam filter will contain $.999N+.07N$ emails.  Thus we want $$(.999+.07)N=500\implies N\approx 468$$

It follows that the spam folder contains about $.07\times 468=32.76$ legitimate emails.


问:
Partial fraction decomposition issue
I'm trying to do the partial fraction decomposition of the following rational expression:

$(x-4) / (x-2)(x-3)$

Here are the steps I preformed:

$ x-4 = A/(x-3) + B(x-2)$

$x-4 = Ax - 3A + Bx - 2A  $

$ x-4 = x(A+B) - (3A + 2B)$

Form a system of equations by equating the coefficients of like powers of x:

$A+B = 1$

$ -3A - 2B = 0$

Solve the system by substitution:

$ A = -B+1$

$ -3(-B+1) -2B = 0$

$ 3B -3 -2B = 0$

$3B - 2B = 3$

$     B = 3$

$    A + 3 = 1$

$  A = -2$

So, my final decomposition is $-2/(x-2) + 3(x-3)$

However, the answer in the back of my packet is: $2/(x-2) + -1(x-3)$

What have I done wrong?

答:
The Answer in your book is correct and your calculation is wrong. I am writting correct one for you now.

$$\dfrac{x-4}{(x-2)(x-3)}=\dfrac{A}{x-2}+\dfrac{B}{x-3}=\dfrac{Ax-3A+Bx-2B}{(x-2)(x-3)}=\dfrac{(A+B)x+(-3A-2B)}{(x-2)(x-3)}$$

so we should have

$$A+B=1,-3A-2B=-4$$

By multiping first equation by 2 and adding result by second one you will have $-A=-2$ so $A=2$ and by $A+B=1$ we have $B=-1$

Title: Partial fraction decomposition issue


问:
Estimating Number of Lottery Players Per Game
Below I have created an excel spreadsheet showing real data of the 6/49 national lottery in Canada. The "ODDS 1 IN:" column are the official odds of winning what is shown in the "Prize Category" column.

The "PLAYERS" column is one I created by multiplying the "Winners" by the "ODDS 1 IN:" column. Then, when I average all the numbers in the "PLAYERS" column, I get an estimate of the number of players in this particular Lotto 6/49 game.

Based on my calculations, there were roughly 6,034,755 players in this particular game on this particular day if each person bought only 1 ticket. Knowing some people buy more than one ticket, this means the number of players was actually smaller.

I wanted to know if my reverse engineering estimate is an accurate calculation of the number of players per game? If not, how do I calculate a better estimate?

Thank you!

答:
Take this answer with a grain of salt, as I'm not to familiar with statistics, but I think this is a good way of getting a much more accurate estimate:

The biggest issue here is that you are taking an average when you really should be doing something like a weighted average, since getting 6.055 million people as an estimate based on a sample of 729,000 people is much much more significant than getting 6.4 million from a sample of 116 people, or getting 4.6 million from a sample of 2.

You could disregard the 3 smallest samples, and average over the 3 largest ones instead, but you still have the issue that 700,000 is much bigger than 77,000.

So how do you do something like a weighted average? Well it will require something a bit more advanced than what you've done. For a given number of tickets sold, you can generate the probability of seeing the number of winning tickets that we've seen.
Using this, you can find the number of ticket sold which maximises this probability.


问:
Solve the recurrence $\ T(n) = \sqrt{n}\cdot T(\sqrt{n}) + n\cdot \log(n) $
Solve the following recurrence:

$$\ T(n) = \sqrt{n}\cdot T(\sqrt{n}) + n\cdot \log(n)$$

I tried to just unroll it to see if I can understand, but it is too complicated .

$\ T(n) = n^{\frac{1}{2}} \cdot T(n^{\frac{1}{2}}) +n\cdot \log(n)\\
= n^{\frac{1}{2}} \cdot \left( n^{\frac{1}{4}} \cdot T(n^{\frac{1}{4}} ) + \frac{n}{2} \cdot \log(\frac{n}{2}) \right) + n \cdot \log(n) \\
= n^{\frac{1}{2}} \left( n^{\frac{1}{4}} \cdot \left( n^{\frac{1}{8}} \cdot T(n^{\frac{1}{8}}) +  \frac{n}{4}\cdot \log(\frac{n}{4})  \right) + \frac{n}{2} \cdot \log(\frac{n}{2}) \right) + n\cdot \log(n) $

Not really clear how to proceed here? I also tried recurrence tree, but couldn't see a clear pattern. I've found a similar question here but I could not understand the answer

答:
Step 1 : get rid of square root

$$T(n^2)=nT(n)+2n^2\ln(n)$$

Step 2 : get rid of $n^2$ factor

Let set $U(n)=\frac {T(n)}n$

$\require{cancel}\cancel{n^2}U(n^2)=T(n^2)=nT(n)+2n^2\ln(n)=\cancel{n^2}U(n)+2\cancel{n^2}\ln(n)$

After simplification we get $$U(n^2)=U(n)+2\ln(n)$$

Step 3 : get rid of the logarithm

Let set $V(n)=U(n)+a\ln(n)$

The idea is that $\ln(n^2)$ will transform to $2\ln(n)$

$V(n^2)=U(n^2)+a\ln(n^2)=U(n)+2\ln(n)+2a\ln(n)=V(n)+(2+a)\ln(n)$

So for $a=-2$ the logarithm disappear and $$V(n^2)=V(n)$$

Step 4 : linearize

Notice that powers transform multiplications into additions so if we set $n=2^p$ then $n^2=2^p2^p=2^{p+p}=2^{2p}$

If we do it again by setting $p=2^k$ then $2p=2^{k+1}$

Therefore let set $W(k)=V(2^{2^k})=V(2^p)=V(n)$ then we get $$W(k+1)=W(k)$$

Step 5 : solve

In this case the solve is trivial, this is just (for $c$ a random constant) $$W(k)=c$$

Step 6 : go back the chain of substitutions

$T(n)=nU(n)=n\big(V(n)+2\ln(n)\big)=n\big(W(k)+2\ln(n)\big)$

Finally $$T(n)=c\,n+2n\ln(n)$$


答:
Your expression is $\dfrac{x-4}{(x-2)(x-3)}$

Now you have $A(x-3)+B(x-2)=x-4$

Plug in $x=3$ to cancel out $A$,

You get the value of $B$. Now you can calculate the value of $A$. You have to cancel out the multiple variables to make the calculation easier.

Title: Partial fraction decomposition issue

Body:
I'm trying to do the partial fraction decomposition of the following rational expression:

$(x-4) / (x-2)(x-3)$

Here are the steps I preformed:

$ x-4 = A/(x-3) + B(x-2)$

$x-4 = Ax - 3A + Bx - 2A  $

$ x-4 = x(A+B) - (3A + 2B)$

Form a system of equations by equating the coefficients of like powers of x:

$A+B = 1$

$ -3A - 2B = 0$

Solve the system by substitution:

$ A = -B+1$

$ -3(-B+1) -2B = 0$

$ 3B -3 -2B = 0$

$3B - 2B = 3$

$     B = 3$

$    A + 3 = 1$

$  A = -2$

So, my final decomposition is $-2/(x-2) + 3(x-3)$

However, the answer in the back of my packet is: $2/(x-2) + -1(x-3)$

What have I done wrong?


答:
Working with $10000$ mails for ease of keeping calculations in integers.

A) Expected number of junk mails $= 5000$

Filtered to spam folder $ = 4995$ $\, \, (99.9\%)$

Comes to Inbox $ = 5 \, \, (0.1\%)$

B) Expected number of legit mails $= 5000$

Filtered to spam folder $ = 350$ $\, (7\%)$

Came to Inbox $ = 4650$ $\, \, (93\%)$

So total number of mails in spam folder $ = 5345$

Legit mails out of those $ = 350$.

If $500$ mails in spam folder, expected number of legit mails = $\displaystyle \frac{350 \times 500}{5345} \approx 32.74$

Title: Interpretable Explanation of Conditional Probability Problem

Body:
I've given a question to students of mine, conditional probability.

Essential information is we have an email filter that filters 99.9% of junk emails to the spam folder, but it also filters 7% of legitimate emails there as well. If 50% of emails are junk and 50% are legit, how many emails in a spam folder containing 500 emails can we expect to be legitimate?

I did the work, it comes out to be roughly 32.74, using Bayes theorem, and the calculation speaks for itself, but I hate not being able to give an interpretable reason for why, beyond the formula, the answer makes sense. A few students, of course, answered 35, because that's an interpretable answer.

I gave them the classic Blue/Green taxi-cab witness problem already, which is easier to explain. Look at how many Blue cars he'd call Green vs how many Green cars he'd call Green, then put your "good" outcomes over the total outcomes, where the only valid ones are the ones where the witness says he saw a green taxi. But a similarly plain and interpretable result for the spam email folder is hard for me to find.


问:
Zero to the zero power – is $0^0=1$?
Could someone provide me with a good explanation of why $0^0=1$?

My train of thought:

$x>0$

$0^x=0^{x-0}=0^x/0^0$, so

$0^0=0^x/0^x=\,?$

Possible answers:

$0^0\cdot0^x=1\cdot0^0$, so $0^0=1$

$0^0=0^x/0^x=0/0$, which is undefined

PS. I've read the explanation on mathforum.org, but it isn't clear to me.

答: Knuth's answer is at least as good as any answer you're going to get here: http://arxiv.org/pdf/math/9205211v1.pdf See pp. 4-6, starting at the bottom of p. 4.</p>

Title: Zero to the zero power – is $0^0=1$?


问:
The Limit of a Wallis-like Integral
Let

$\displaystyle I_n=\int^{\pi}_{0} e^x \sin^n x dx$,

then, what number does

$\displaystyle \sqrt{n} I_n$

tend to?

According to wolfram alpha it tends to 12.05... but I want to know what is this number.

I posted similarly titled question about two months ago: The limit and Asymptotic Behavior Wallis-like Integral but the theme here is not that.

答:
Denoting $ \left(\forall n\in\mathbb{N}\right),\ I_{n}=\int_{0}^{\pi}{\mathrm{e}^{x}\sin^{n}{x}\,\mathrm{d}x} $.

Let $ n \in\mathbb{N} :$

Integrating by parts : \begin{aligned}\int_{0}^{\pi}{\mathrm{e}^{x}\sin^{n}{x}\,\mathrm{d}x}=-\frac{n-1}{2}\int_{0}^{\pi}{\mathrm{e}^{x}\left(\sin{x}-\cos{x}\right)\sin^{n-2}{x}\cos{x}\,\mathrm{d}x}\end{aligned} Reintegrating by parts : \begin{aligned} \int_{0}^{\pi}{\mathrm{e}^{x}\sin^{n}{x}\,\mathrm{d}x}=\frac{\left(n-1\right)^{2}}{4}\int_{0}^{\pi}{\mathrm{e}^{x}\left(\sin{x}+\cos{x}\right)\sin^{n-2}{x}\cos{x}\,\mathrm{d}x}+\frac{n-1}{2}\int_{0}^{\pi}{\mathrm{e}^{x}\sin^{n-2}{x}\cos^{2}{x}\,\mathrm{d}x}\end{aligned}

Thus : \begin{aligned}\left(1+\frac{n-1}{2}\right)\int_{0}^{\pi}{\mathrm{e}^{x}\sin^{n}{x}\,\mathrm{d}x}&=\frac{\left(n-1\right)^{2}}{2}\int_{0}^{\pi}{\mathrm{e}^{x}\sin^{n-2}{x}\cos^{2}{x}\,\mathrm{d}x}+\frac{n-1}{2}\int_{0}^{\pi}{\mathrm{e}^{x}\sin^{n-2}{x}\cos^{2}{x}\,\mathrm{d}x}\\ &=\frac{n\left(n-1\right)}{2}\int_{0}^{\pi}{\mathrm{e}^{x}\sin^{n-2}{x}\left(1-\sin^{2}{x}\right)\mathrm{d}x}\\ \frac{n+1}{2}I_{n}&=\frac{n\left(n-1\right)}{2}\left(I_{n-2}-I_{n}\right)\end{aligned}

Thus : $$ \fbox{$\begin{array}{rcl}\left(1+n^{2}\right)I_{n}=n\left(n-1\right)I_{n-2}\end{array}$} $$

Define a sequence $ \left(u_{n}\right)_{n\geq 1} $ as follows : $$ \left(\forall n\in\mathbb{N}^{*}\right),\ u_{n}=\frac{n}{\left(n!\right)^{2}}\prod_{k=1}^{n}{\left(1+k^{2}\right)}I_{n-1}I_{n} $$

Since $ \left(u_{n}\right) $ is a constant sequence and $ \frac{I_{n+1}}{I_{n}}\underset{n\to +\infty}{\longrightarrow}1 $, we get : $$ I_{n}\underset{n\to +\infty}{\sim}\frac{n!\sqrt{2\,\mathrm{e}^{\pi}\sinh{\pi}}}{\sqrt{n\prod\limits_{k=1}^{n}{\left(1+k^{2}\right)}}} $$

We have $ \prod\limits_{k=1}^{n}{\left(1+k^{2}\right)}=\frac{\sinh{\pi}\,\Gamma\left(n+1-\mathrm{i}\right)\,\Gamma\left(n+1+\mathrm{i}\right)}{\pi}\underset{n\to +\infty}{\sim}\frac{\sinh{\pi}\,\left(n!\right)^{2}}{\pi} $, which means : $$\fbox{$\begin{array}{rcl} I_{n}\underset{n\to +\infty}{\sim}\sqrt{\frac{2\pi\,\mathrm{e}^{\pi}}{n}} \end{array}$} $$

The limit would, thus, be equal to : $$ \sqrt{2\pi\,\mathrm{e}^{\pi}} $$

Title: The Limit of a Wallis-like Integral


问:
Does the eigenvalue that the Power Method converges to have to be "strictly" dominant?
In almost all the text I can find on the Power method, the authors say something to this effect:

If we assume A has an eigenvalue that is strictly greater in magnitude than its other eigenvalues and the starting vector b_0 has a nonzero component in the direction of an eigenvector associated with the dominant eigenvalue, then a subsequence (b_k) converges to an eigenvector associated with the dominant eigenvalue.
Without the two assumptions above, the sequence (b_k) does not necessarily converge.

source

I can intuitively understand the second condition. But can someone explain to me intuitively (or with an example) why an $n \times n$ matrix with, say, 2 (or 3 or maybe even n-1) equal eigenvalues which are greater than the remaining n-2(or n-3... or 1) eigenvalues could not converge?

What I understand is, if we have a single strictly dominant eigenvalue, then the initial guess $b_0$ (which has a component in the direction of the corresponding eigenvector) is bound to converge after some time to the eigenvector.

What I don't understand is, however, why this logic cannot be extended to more than 1 and written this way: "if we have dominant eigenvalue(s), then the initial guess $b_0$ (which has a component in the eigenspace corresponding to the dominant eigenvalue) is bound to converge after some time to a vector in the eigenspace." (Assuming of course that the matrix is diagonalisable which is a requirement for all cases anyway.)

答:
The answer is “yes”: take, for example, the basic two-circulant matrix
$$
C= \begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix},$$
which has two dominant eigenvalues ($\pm 1$).

If $$x =
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix},$$
then
$$Cx =
\begin{bmatrix}
x_2 \\
x_1
\end{bmatrix}$$
and

$$C^2x =
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}.$$

Thus, convergence fails in general.


问:
Show complex function is bijective
I'm asked to show that the function $f:\mathbb{C}\setminus\{1\}\rightarrow \mathbb{C}\setminus\{i\}$ given by
$f(z)=i\frac{z+1}{z-1}$ is bijective.

I get stuck very early on - I think I get thrown off by the fact that we're in $\mathbb{C}$ and not good old $\mathbb{R}$.

My attempt:

injective: assuming $f(z_1)=f(z_2)$, I need to show $z_1=z_2$. Usually I would just set it up like this: $$i\frac{z_1+1}{z_1-1}=i\frac{z_2+1}{z_2-1} $$
And then manipulate the expressios to get $z_1=z_2$, but I can't seem to do it here.

Am I choosing a poor strategy here?

In regards to surjectivity, I am just lost.

Any help is much appreciated.

答: If you solve the equation$$i\frac{z+1}{z-1}=w,$$you will get$$z=\frac{w+i}{w-i}.$$So, define $g\colon\Bbb C\setminus\{i\}\longrightarrow\Bbb C\setminus\{1\}$ by$$g(z)=\frac{z+i}{z-i}$$and check that it is the inverse of $f$. In other words, check that$$(\forall z\in\Bbb C\setminus\{i\}):f\bigl(g(z)\bigr)=z\quad\text{and that}\quad(\forall z\in\Bbb C\setminus\{1\}):g\bigl(f(z)\bigr)=z.$$

Title: Show complex function is bijective


问:
Is there a better way to visualize the midpoint formula?
I've been trying to teach this to my bro and I kept wondering of the best way to "see" why midpoint of $A=x_1,B=x_2$ is $(x_1+x_2)/2$.

One trick I learned recently is to use change of coordinates.

We know the midpoint of $OA$ is $OA/2$.

Move the origin to $A$. With this new origin, $B$ is represented as $(-x_1+x_2)$

Midpoint is then $(-x_1+x_2)/2$

Move the origin back to $O$. Then the midpoint is represented as $x_1 + (-x_1+x_2)/2$

But translation is not linear... still this change of coordinates trick seems to work. Am I doing anything wrong here? Also anyone has a more better/simpler way for the proof?

答:
If we are on the real line we have that the mid point of $x_1 < x_2$ is the point $x$ such that:
$$x-x_1=x_2-x \iff 2x=x_1+x_2$$
which is exactly what we want.


答:
Here's one way to visualize it with the help of some algebra:  Suppose $x_{1} < x_{2}$.  Then the midpoint means finding a point $x_{m}$ such that $x_{m} = x_{1} + a$ and $x_{m} = x_{2} - a$ as in the image below.

This means that $$x_{1} + a = x_{2} - a \implies a = \frac{x_{2} - x_{1}}{2}$$ and so $$x_{m} = x_{1} + a = x_{1} + \frac{x_{2} - x_{1}}{2} = \frac{x_{1} + x_{2}}{2}.$$

Title: Is there a better way to visualize the midpoint formula?

Body:
I've been trying to teach this to my bro and I kept wondering of the best way to "see" why midpoint of $A=x_1,B=x_2$ is $(x_1+x_2)/2$.

One trick I learned recently is to use change of coordinates.

We know the midpoint of $OA$ is $OA/2$.

Move the origin to $A$. With this new origin, $B$ is represented as $(-x_1+x_2)$

Midpoint is then $(-x_1+x_2)/2$

Move the origin back to $O$. Then the midpoint is represented as $x_1 + (-x_1+x_2)/2$

But translation is not linear... still this change of coordinates trick seems to work. Am I doing anything wrong here? Also anyone has a more better/simpler way for the proof?


问:
The degree-2 curves through three points is given by $\gamma L_1L_2+\lambda L_2L_3+\mu L_3L_1=0$, where the $L_i$ are the lines joining those points
I came across this equation while reading on conics and circles.

The family of degree-$2$ curves passing through three points is given by
  $$\gamma L_1L_2 + \lambda L_2L_3 + \mu L_3L_1 = 0$$
  where $L_1$, $L_2$, and $L_3$ are the equations of lines joining the three points, and $\lambda$, $\mu$ and $\gamma$ are parameters.

I can see that the three points definitely satisfy this equation , but I am unable to say much about rest of the points.

How would you prove this relation? Explain, please. :-)

答:
Conics can be expressed in terms of trilinear coordinates, which express points in terms of relative directed distances from three lines (which can be thought of as the sidelines of a triangle.  In these terms it is much easier to reason about the relations between a family of conics and a given triangle.  Variations of the formula given in the OP will, for example, give expressions for the conics that are tangential to a given triangle, or have that triangle as a self polar triangle.

A good reference for this was published over 150 years ago: Whitworth's Trilinear Coordinates.  The URL points to Chapter XV, which discusses the case in the OP. Using the key $\alpha\rightarrow L_1,
\beta\rightarrow L_2,
\gamma\rightarrow L_3,
l\rightarrow \lambda,
m\rightarrow \mu,
n\rightarrow \gamma,$ the equation $l\beta\gamma+m\gamma\alpha+n\alpha\beta=0$ translates directly into the OP equation.

But how do we get there from here, namely a conic (expressed in homogeneous coordinates)
$$Ax^2 + By^2 + 2Gxz + 2Fyz + 2Hxy + Cz^2 = 0 \tag{1}$$
that may or may not go through the three points?

The answer is a change of variables. We know that (again, using homogeneous coordinates)
$$
\begin{align}
L_1 &= a_1 x+b_1 y + c_1z \\
L_2 &= a_2 x+b_2 y + c_2z \\
L_3 &= a_3 x+b_3 y + c_3z, \\
\end{align}
$$

so we can solve for $x,y,z$ in terms of $L_1,L_2,L_3$ and then plug the result into (1) to get an equation of the form

$$
fL_1^2+gL_2^2+hL_3^2+2lL_2L_3+2mL_3L_1+2nL_1L_2=0,\tag{2}\
$$

which maps (using the key given earlier) to the second equation on pg 192.  Assuming that the conic (1) passes through the three points we can use the reasoning on that page to determine that $f=g=h=0,$ which reduces (2) to

$$
lL_2L_3+mL_3L_1+nL_1L_2=0.
$$

This answer could have been given without reference to Whitworth, but I think it's important to underline the use of trilinear coordinates. For a point $P=(x,y,z),L_n(P)$ gives a relative directed distance of point $P$ from line $L_n$, and can be thought of as a trilinear coordinate.  The cases of conic families that inscribe the triangle, or for whom the triangle is self-polar are treated in Chapters XVI and XIV of the reference.  Also note that trilinear coordinates are projective/homogeneous, so it's easier to make this argument using homogeneous coordinates $(x,y,z)$ rather than Cartesian coordinates $(x/z,y/z).$


问:
When do these functions equal each-other?
I'm looking for the $x$ such that

$$ -\frac{1}{q^x}+1=a^x -1$$

For any $q$ and $a$

A.K.A the $x$ at the intersection of the two functions. Am I asking the wrong question here?

One obvious solution is $x=0$ but there is another solution I can't seem to find in terms of $q$ and $a$

Graph for Visualization of Second Solution

答: $0$ is the only solution that works for all values of $q$ and $a$. Let $f(x) = - \frac{1}{q^x}+1$ and $g(x) = a^x -1$, and take $q=1$ and $a=2$ as a simple example. Then $f(x) = 0$ for all $x$, hence $h(x) = f(x) - g(x) = 1-2^x$. Note that for positive $x$, we have $2^x > 1$, so clearly $h(x)$ is always negative for positive $x$, then crosses the $y$-axis at zero. For negative $x$ it's equivalent to $1 - \frac{1}{2^x}$, which is always positive. No other zeroes apart from $h(0)$ are possible.


问:
Finding the limit for functions with two variables
I know that when we have a limit of a function with $2$ variables, the limit must be the same, regardless of the path we take. So this is useful for proving that a limit doesn not exist. But when you've tried this method for different "paths" (e.g., $(x,0),(0,y),(x,x),(y,y)$, etc...) and you think that the limit does exist, how you do show it?

For example here is a question from my textbook:

Evaluate the limit:
  $$\lim_{\large{(x,y) \to (0,0)}} \dfrac{xy \sin(xy)}{x^2+y^2}$$

The answer is supposed to be $0$ but I don't see how you can prove that it is $0$ for any direction you approach $(0,0)$ from. Can someone please help me? Thanks.

答: Hint: $\left|\dfrac{xy}{x^2+y^2}\right| \le \dfrac12$.</p>


答:
With the substitution $t=1-x^2$, the integral becomes
$$\frac12 \int_0^1(1-t)^nt^p\,\mathrm{d}t $$
This is the beta function but in case you're not familiar with it, you can derive a recurrence relation. Let
$$I_n=\int_0^1(1-t)^nt^p\,\mathrm{d}t \\
I_0=\int_0^1t^p\,\mathrm{d}t=\frac{1}{p+1} $$
$$\begin{align}
I_n&=\int_0^1(1-t)^nt^p\,\mathrm{d}t=\int_0^1(1-t)^{n-1}(1-t)t^p\,\mathrm{d}t\\
&=\int_0^1(1-t)^{n-1}\left(t^p-t^{p+1}\right)\,\mathrm{d}t=I_{n-1}-\int_0^1(1-t)^{n-1}t^{p+1}\,\mathrm{d}t
\end{align}$$
In the last integral in the previous line, use integration by parts with $u=t^{p+1},dv=\left(1-t\right)^{n-1}dt$ to find
$$\int_0^1(1-t)^{n-1}t^{p+1}\,\mathrm{d}t=\frac{p+1}{n}\int_0^1(1-t)^nt^p\,\mathrm{d}t=\frac{p+1}{n}I_n $$
Hence,
$$I_n=I_{n-1}-\frac{p+1}{n}I_n$$
$$I_n=\frac{n}{n+p+1}I_{n-1}=\frac{(n-1)n}{(n+p)(n+p+1)}I_{n-2}=\dots=\frac{1\cdot 2\cdots n}{(p+2)\cdots(n+p+1)}I_0\\
=\frac{n!}{(p+1)(p+2)\cdots(n+p+1)}
$$

The final answer is $\frac12 I_n$.

Title: Computing an Integral With Parameters

Body:
Let $p > -1$ and $n \in \Bbb{N} \cup \{0\}$. I am trying to compute the integral

$$\int_{0}^{1} x^{2n+1} (1-x^2)^p dx$$

I suspect that it equals something like $\frac{1}{(1+p)(n+1)^{p+1}}$ (after trying small $n$), but I don't know how to integrate it (I could also be wrong as to my conjecture, since I am awful at arithmetic). It seems like an annoying application of induction and integration by parts, but I could be completely wrong, as integrals are not my forte. Perhaps there is a simpler, more elegant way to calculate the integral. I could use some help.


问:
What is wrong with this proof there is no $\omega$-th worldly cardinal
Call a cardinal $\kappa$ worldly iff $V_\kappa\vDash ZFC$. Let $\kappa_\alpha$ be the $\alpha$th worldly cardinal, i.e. the least worldly cardinal such that $\{\beta\lt\kappa_\alpha|\beta\text{ is worldly}\}$ has order-type $\alpha$. Since the class of worldly cardinals are closed, then $\kappa_\omega=\sup\kappa_n$.

Therefore, $V_{\kappa_\omega}\vDash\forall\alpha(\exists\delta\gt\alpha(V_\delta\vDash ZFC))$. Now define $f: \omega\rightarrow\kappa$ by $f(n)=\kappa_n$. We have that $f(n)$ is definable and absoloute to $V_{\kappa_\omega}$. Therefore, $f"\omega\in V_{\kappa_\omega}$ by the axiom of replacement. But this is impossible.

答: The worldly cardinals are not closed, and you have proved this. You are probably remembering the similar-sounding fact that $ \{\alpha: V_\alpha\prec V_\kappa\}$ is a club in $\kappa$ for worldly $\kappa$ of uncountable cofinality. (And note that the argument for closure relies on an elementary chain argument.) Picking out this cofinal sequence requires $V_\kappa$ know what is elementary to it, which it doesn't. And nor does $V_\alpha$ for the $\omega$-th such $\alpha$ know which worldly cardinals beneath it are elementary submodels of $V_\kappa.$


问:
If $f$ is uniformly continuous, converges to $0$ at $\pm\infty$, is $\int_{-\infty}^\infty f^2 < \infty$?
Is it true that if $f$ is uniformly continuous and converges to $0$ at $\pm\infty$, that the improper integral $$\int_{-\infty}^\infty f^2 < \infty$$
is finite?

I am currently reading a probability theory proof that seems to take this for granted, but it is not obvious to me. How can this be argued?

答: No, it's not true.  Every continuous function $f$ on $\mathbb R$ such that $f \to 0$ at $\pm \infty$ is uniformly continuous.  Try e.g. $f(x) = 1/(1+\sqrt{|x|})$.


答:
This is false. Consider the function

$$f(x)=\begin{cases}
      \frac{1}{\sqrt{-x}} & x\leq -1 \\
      -\frac{1}{4}x^2+\frac{5}{4} & -1< x< 1 \\
      \frac{1}{\sqrt{x}} & x\geq 1
   \end{cases}$$

This function is differentiable for all $x\in\mathbb{R}$ and converges to $0$ at $\pm \infty$. The derivative is

$$f'(x)=\begin{cases}
      \frac{1}{2(-x)^{3/2}} & x\leq -1 \\
      -\frac{1}{2}x& -1< x< 1 \\
      -\frac{1}{2 x^{3/2}} & x\geq 1
   \end{cases}$$

It is easy to see that this is continuous for all $x\in\mathbb{R}$. Further, we see that $|f'(x)|\leq \frac{1}{2}$ for all $x\in\mathbb{R}$. Thus, $f(x)$ is Lipschitz Contiuous (see the first property in the link). Thus, $f(x)$ is uniformly continuous over the real numbers. However,

$$\int_1^\infty f(x)^2dx=\int_1^\infty \frac{1}{x}dx=\infty$$

Thus, the integral does not converge.

Title: If $f$ is uniformly continuous, converges to $0$ at $\pm\infty$, is $\int_{-\infty}^\infty f^2 < \infty$?

Body:
Is it true that if $f$ is uniformly continuous and converges to $0$ at $\pm\infty$, that the improper integral $$\int_{-\infty}^\infty f^2 < \infty$$
is finite?

I am currently reading a probability theory proof that seems to take this for granted, but it is not obvious to me. How can this be argued?


问:
Show that $\angle PAB = \angle CAQ$
Given a triangle $\triangle ABC$ with a point $P$ in the perpendicular bisector of $BC$. Take point $Q$ also in the bisector of $BC$ s.t. $\angle ABP + \angle ACQ =180^o$ and such that $P$ and $Q$ are in the interior of $\angle BAC$.

Prove that $\angle PAB = \angle CAQ $

I tried to find some circles in this configuration. Tried to find some interesting locus and I failed so I went for trigonometry. This problem is equivalent to show that $\frac{BP}{AP} = \frac{BQ}{AQ}$... wait this looks like an apolonian circle.

Ok it looks like $CQ$ is tangent to the apollonian circle that goes throught $P$ and $Q$ EDIT: it is not tangent.

Can we work it with coaxial systems? Tha apollonian circle through fixed points $AB$ form a coaxial system orthogonal to the system of circles that pass throught $A$ and $B$

答:
Drawing the circumcircle about triangle $ABC$, since $P$ is on the perpendicular bisector of chord $BC$, $P$ lies on the diameter.  Extend $BP$ to $F$, join $FC$ and extend it to meet the diameter's extension at $Q$. Join $QA$, crossing the circle at $D$. We want to show$$\angle PAB=\angle CAQ$$

Note that since $\angle ACF=\angle ABP$, the supplement of $\angle ACQ$, drawing and extending $FC$ from $F$ on the circumcircle determines point $Q$.

Join $DP$ extended to $K$, join $PC$, and $KQ$ crossing the circle at $E$.

Since $QP$, the perpendicular bisector of $BC$, bisects $\angle BQF$, then$$\frac{BP}{BQ}=\frac{FP}{FQ}$$And by congruent triangles$$\frac{BP}{BQ}=\frac{CP}{CQ}$$Therefore points $B$, $F$, $C$ lie on an Apollonian circle with $P$, $Q$ as "foci", and we have$$\frac{DP}{DQ}=\frac{KP}{KQ}$$and hence$$\angle DQP=\angle KQP$$Therefore $AH$, $KH$ are equal arcs and, by symmetry, $AP$ extended is concurrent with $KQ$ at $E$, and makes $EG$, $DG$ likewise equal arcs. And since $G$ is thus the midpoint of arcs $DE$ and $BC$, it follows that$$\angle BAE=\angle CAD$$that is$$\angle PAB=\angle CAQ$$

Title: Show that $\angle PAB = \angle CAQ$


问:
Relation between null space of a matrix and column space.
Let $A$ be an $m\times n$ matrix. Then under what conditions null space of $A$ and column space of $A$ may have common elements.

My attempt: I start by taking a vector $v\in \mathbb{R}^n$ such that $$v\in N(A)\cap Col(A)$$ where $N(A)$ is the column space of A and $Col(A)$ is column space of A. Then above means
$$v\in N(A)\implies Av=0, \ v\in Col(A)\implies v=Ay\ \text{for some}\ y.$$
Combining two we get $$A^2y=0.$$
Can I deduce anything from above? Please help.

答: Assume $A$ is square matrix and zero is an eigenvalue of $A$. If number of zero eigenvalues is $k$ and dimension of $N(A)$(also zero eigenspace) is strictly less than $k$, then we can directly say that $C(A) \cap N(A)$ has nontrivial element. It looks like the converse is true. But I could not prove yet.


答:
The partial fraction decomposition is given by:

$$\displaystyle \frac{x-4}{(x-3) (x-2)} = \frac{2}{x-2}-\frac{1}{x-3}$$

We would have written:

$$\displaystyle \frac{x-4}{(x-3) (x-2)} = \frac{A}{x-3}+ \frac{B}{x-2}$$

This yields:

$$x - 4 = A ( x- 2) + B(x-3)$$

Equating terms yields:

$$A+ B = 1$$

$$-2A -3B = -4$$

Solving for the two variables yields:

$$A = -1, B = 2$$

Title: Partial fraction decomposition issue


问:
Find a Lyapunov candidate for scalar system
The following scalar system:

$$\dot{x} = x^{2} - x^{3} $$

has two equilibrium points, one located at $x = 0$ and the other at $x = 1$. After looking at the phase portrait, setting xdot(2,1) = 0 and xdot(1,1) = x(1)^2 - x(1)^3) it is possible to say that both equilibrium points are asymptotic stable.

Knowing this, how can I obtain a valid Lyapunov function that can guarantee asymptotic stability, since just using:

\begin{align}
V(x) &= \frac{1}{2}x^{2}\\
\dot{V} &= x^{3} - x^{4}
\end{align}

Can only guarantee that $V$ is positive definite, but cannot be used to guarantee that $\dot{V}$ is negative definite.

答:
You have a scalar system, and there is no phase portrait for it, it lives on $\mathbb{R}$. And two equilibrium points connected by a feasible trajectory cannot be asymptotically stable simultaneously.

For your system, trajectories starting at $x_0<0$ converge to zero, and trajectories starting with $x_0>0$ converge to 1. The equilibrium point $x=1$ is locally asymptotically stable.

More precisely, let $z=x-1$ and consider $V = \frac{1}{2}z^2$. Then $\dot{z} = (z+1)^2-(z+1)^3 = -z(z+1)^2$ and $$\dot{V} = -z^2(z+1)^2.$$
You observe that $\dot V<0$ for all $z$ except for $\dot{V}=0$ at $z=-1$ and $z=0$. So trajectories always move towards $z=0$ (i.e., $x=1$), but they stuck at $z=-1$ (i.e., $x=0$).


问:
How can you prove that the square root of two is irrational?
I have read a few proofs that $\sqrt{2}$ is irrational.

I have never, however, been able to really grasp what they were talking about.

Is there a simplified proof that $\sqrt{2}$ is irrational?

答:
The irrationality of $\sqrt{2}$ is equivalent to that of $1+\sqrt{2}$, which is equivalent to there being no length $\ell$ such that the hypotenuse and perimeter of an isosceles right triangle are both whole-number multiples of $\ell$. Suppose there were such a length. We say that the hypotenuse (indicated schematically by the black path in the triangle on the left)  and the perimeter (indicated by the brown path in the triangle on the left) are measured by $\ell$. To find $\ell$, we apply the Euclidean algorithm, measuring off the perimeter in units of the hypotenuse, and observing that any remainder is measured by $\ell$. In any right triangle, the perimeter contains at least two hypotenuse-lengths (since the legs together are greater than the hypotenuse) but never three (since the hypotenuse is greater than each leg separately). (The isosceles right triangle, in fact maximizes the ratio of perimeter to hypotenuse, this ratio being $1+\sqrt{2}\approx 2.4$.) The two hypotenuse-lengths are indicated schematically by the black diagonal path and the green L-shaped path in the middle figure. The remainder is the short black vertical path. We claim that the brown triangular path in the middle figure is also a hypotenuse-length path, and therefore also measured by $\ell$.  But then this small isosceles right triangle, which is less than half as big (in linear dimensions) as the original isosceles right triangle, has hypotenuse and perimeter that are measured by $\ell$. Which is a contradiction: the Euclidean algorithm can be iterated, always producing a new isosceles right triangle of less than half the size of the previous one whose hypotenuse and perimeter are measured by $\ell$. Such repeated reductions to less than half must eventually result in a figure whose size is less than any conceivable $\ell$.

To establish the claim that the brown triangular path is hypotenuse-length, note that if three congruent isosceles right triangles are arranged like the gray triangles in the middle figure, and the rectangular gap in the middle filled in, the result is an isosceles right triangle. Then the black, green, and brown paths all have length $2s+d$, using the notation of the figure on the right.

For a more organic derivation, let $\triangle ABC$ be an isosceles right triangle. Draw a circle with center $C$ and radius $\overline{CB}$ intersecting $\overline{AC}$ at $G$. The length of  $\overline{AG}$ is the amount by which the length of the hypotenuse exceeds that of a leg, which is the length that the short segment of the green L in the middle figure needs to have. Now construct a segment perpendicular to $\overline{AC}$ at $G$, intersecting $\overline{AB}$ at $D$. Triangle $AGD$ is an isosceles right triangle, so $\overline{DG}\cong\overline{AG}$. But also $\overline{DB}\cong\overline{DG}$ as required, which can be seen by observing that the right triangles $CGD$ and $CBD$ share a hypotenuse and have corresponding legs $\overline{CG}$ and $\overline{CB}$ congruent, and are therefore congruent. The points $F$ and $E$ can be constructed by a symmetric procedure.

One final image: if the hypotenuse and perimeter of the largest triangle in the fractal below are both measured by $\ell$, then the hypotenuse and perimeter of every similar triangle in the fractal, and the perimeter of every similar rectangle in the fractal, are measured by $\ell$ as well.

Remarks: This proof is essentially the same as the continued fraction proof of Aryabhata and the algebraic proof of Thomas Andrews. The only added feature is the geometric interpretation and the pictures.

To relate it to continued fractions, let the hypotenuse be $1$, so that the perimeter is $1+\sqrt{2}$. Taking out two hypotenuse lengths gives
$$
1+\sqrt{2}=2+(\sqrt{2}-1).
$$
The quantity in parentheses is the length of the short black vertical segment in the middle diagram, which is the hypotenuse of the original triangle scaled down by a factor of $1+\sqrt{2}$. So
$$
1+\sqrt{2}=2+\frac{1}{1+\sqrt{2}}.
$$
Iterating gives
$$
1+\sqrt{2}=2+\frac{1}{2+\frac{1}{1+\sqrt{2}}}=2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\ldots}}}.
$$

To relate it to the algebraic proof of Thomas Andrews, refer to the rightmost diagram and let
$$
\sqrt{2}=\frac{p}{q}=\frac{2s+d}{s+d},
$$
where $p$ has been interpreted as the length of the hypotenuse and $q$ as the length of a leg. But the small right triangles in the diagram make it clear that also
$$
\sqrt{2}=\frac{d}{s}=\frac{2q-p}{p-q}.
$$
If $\frac{p}{q}$ was in lowest terms, this is a contradiction, which is another way of expressing the infinite descent idea of the geometric proof.

ContentLicense: CC BY-SA 2.5


答:
The claim $V_{\kappa_{\omega}} \models \forall\alpha(\exists\delta > \alpha)(V_{\delta} \models ZFC))$ presupposes that not only is $\kappa_{\omega}$ the $\omega$th worldly cardinal, it is in fact the limit of the $\kappa_n$ for $n < \omega$. Your argument in fact handily shows that the limit of the first $\omega$ worldly cardinals is not worldly, so the $\omega$th worldly cardinal will be much larger.

For a similar, sillier example of what's going on, consider the following "proof":

Let $\lambda_{\alpha}$ denote the $\alpha$th infinite regular cardinal.

Consider $\lambda_{\omega}$.

The sequence $\langle\lambda_n\mid n < \omega\rangle$ is a countable sequence cofinal in $\lambda_{\omega}$.

But $\lambda_1 = \omega_1$ is uncountable, so $\lambda_{\omega}$ is also uncountable.

Thus $\lambda_{\omega}$ is an uncountable cardinal with a countable cofinal sequence, so $\lambda_{\omega}$ is not regular. Contradiction.

This argument is essentially the same as the one you've proposed, but talking about regularity rather than worldliness hopefully makes the flaw (step 3) clearer.

Title: What is wrong with this proof there is no $\omega$-th worldly cardinal

Body:
Call a cardinal $\kappa$ worldly iff $V_\kappa\vDash ZFC$. Let $\kappa_\alpha$ be the $\alpha$th worldly cardinal, i.e. the least worldly cardinal such that $\{\beta\lt\kappa_\alpha|\beta\text{ is worldly}\}$ has order-type $\alpha$. Since the class of worldly cardinals are closed, then $\kappa_\omega=\sup\kappa_n$.

Therefore, $V_{\kappa_\omega}\vDash\forall\alpha(\exists\delta\gt\alpha(V_\delta\vDash ZFC))$. Now define $f: \omega\rightarrow\kappa$ by $f(n)=\kappa_n$. We have that $f(n)$ is definable and absoloute to $V_{\kappa_\omega}$. Therefore, $f"\omega\in V_{\kappa_\omega}$ by the axiom of replacement. But this is impossible.


问:
Example of a Ring that has nothing to do with numbers
What is an example of a ring that has nothing to do with numbers?
For example, for groups, we have dihedral groups, quaternions, etc.

I'm missing the analogue of the characteristic of a ring, when the ring is not related to numbers (complex, real, integers,...).

Part of why I want this example is because I want to see if the characteristic of a ring must be an element of the ring itself.

答:
Part of why I want this example, is because I want to see if the characteristic of a ring must be an element of the ring itself.

It is not really sensible to think of the characteristic of a ring as being an element of the ring, even when it literally is.

Characteristic is a sort of "dimension" on the ring, having a value in the natural numbers.

It's kind of like asking if the dimension of a vector space can be an element in the vector space or not.

Now, coincidentally the characteristic of $\mathbb Z$ is in $\mathbb Z$ if you like $\mathbb N$ being a subset of $\mathbb Z$, but the observation is not really useful.

The closest you can come to putting the characteristic "in" the ring is if the ring has identity so that you can see the homomorphic image of $\mathbb Z$ within your ring. But even then, the characteristic always maps to the $0$ element of your ring, so nothing is achieved.

An example, though?

I dunno, can you think of an abelian group $G$ that doesn't involve numbers? if so, the endomorphism ring $End(G_\mathbb Z)$ is a ring that apparently does not involve numbers.


答:
Let $S$ be any set. We can turn its powerset
$$P(S)=\{A\mid A\subseteq S\}$$
into a ring by defining the sum
$$
A+B:=(A\setminus B)\cup (B\setminus A)
$$
and the product
$$
A\cdot B:=A\cap B.
$$
The empty set is the zero element of this ring, and the full set $S$ plays the role of the multiplicative neutral element. Leaving it to you to verify all the ring axioms.

This ring is isomorphic to the ring of functions $f:S\to\Bbb{Z}_2$ with addition and product defined pointwise. I chose to write it as above, because then no numbers are present (as requested in the title).

Clearly the ring $(P(S),+,\cdot,\emptyset,S)$ has characteristic two.

Title: Example of a Ring that has nothing to do with numbers

Body:
What is an example of a ring that has nothing to do with numbers?
For example, for groups, we have dihedral groups, quaternions, etc.

I'm missing the analogue of the characteristic of a ring, when the ring is not related to numbers (complex, real, integers,...).

Part of why I want this example is because I want to see if the characteristic of a ring must be an element of the ring itself.


问:
Gradient of norm of gradient of scalar field
I have a function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ representing a surface $z=f(x,y)$ in euclidean space. Is there an elegant way to apply the chain rule to expand $\nabla||\nabla f||_2$, where $||\cdot||_2$ is the euclidean norm?

答:
The gradient of $f$ is a vector field (i.e. a function $\Bbb{R}^2 \to \Bbb{R}^2$). Then the norm is a scalar function $\Bbb{R}^2 \to \Bbb{R}$.

In general, if we have $G \circ F$, where $F \colon \Bbb{R}^2 \to \Bbb{R}^2$ and $G \colon \Bbb{R}^2 \to \Bbb{R}$, then the chain rule says:

$$ \nabla (G \circ F) = \nabla G \cdot DF $$

On the right-hand side, $\nabla G$ is evaluated at $F(x,y)$, and $DF$ means the Jacobian matrix of $F$.

In your example, $F$ is $\nabla f$, so the Jacobian matrix is the matrix of second partial derivatives of $f$:

$$ DF = \begin{pmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{pmatrix} $$
And for you, $G(x,y) = \sqrt{x^2+y^2}$, so the gradient is
$$ \nabla G = \left( \frac{x}{\sqrt{x^2+y^2}}, \; \frac{y}{\sqrt{x^2+y^2}} \right) $$
As I said above, we want to evaluate this at $F(x,y)$, meaning we want to plug in the derivatves $f_x$ and $f_y$ for $x$ and $y$, so we get:
$$ \nabla G (f_x,f_y) = \left( \frac{f_x}{\left| \nabla f \right|}, \; \frac{f_y}{\left| \nabla f \right|} \right) = \frac{\nabla f}{\left| \nabla f \right|} $$

Now just put it all together, and do the vector-matrix multiplication.


问:
Proving that this sum $\sum\limits_{0 < k < \frac{2p}{3}} { p \choose k}$ is divisible by $p^{2}$
How does one prove that for a prime $p \geq 5$ the sum :  $$\sum\limits_{0 < k < \frac{2p}{3}} { p \choose k}$$ is divisible by $p^{2}$.

Since each term of $\displaystyle \sum\limits_{0 < k < \frac{2p}{3}} { p \choose k}$ is divisible by $p$, only thing remains is to prove that the sum $$\sum\limits_{ 0 < k < \frac{2p}{3}} \frac{1}{p} { p \choose k}$$ is divisible by $p$.

How to evaluate this sum: $\displaystyle \frac{1}{p} { p \choose k} = \frac{(p-1)(p-2) \cdots (p-k+1)}{1 \cdot 2 \cdot 3 \cdots k}$

答:
Since we are working in the field $\mathbb{F}_p$ we can write

$$\frac{(p-1)(p-2) \cdots (p-k+1)}{1 \cdot 2 \cdots k}$$ as

$$\frac{(-1)(-2) \cdots (-(k-1))}{1 \cdot2 \cdots k}$$

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

Let $N = [\frac{2p}{3}]$ and $M = [\frac{N}{2}]$

Thus what we need is

$$ \sum_{k=1}^{N} \frac{(-1)^{k-1}}{k}$$  $$  = \sum_{k=1}^{N} \frac{1}{k} - 2 \sum_{k=1}^{M}\frac{1}{2k}$$  $$ =\sum_{k=M+1}^{N}\frac{1}{k}$$

Now $N+M+1 =p$ so we can rewrite as

$$\frac{1}{N} + \frac{1}{M+1} + \frac{1}{N-1} + \frac{1}{M+2} + \cdots = $$

$$\frac{p}{N(M+1)} + \frac{p}{(N-1)(M+2)} + \cdots $$

which is $0$.

There are $N-M$ terms, which is even, so each term gets paired off.

OwnerDisplayName: anonymous


问:
Elementary question about integral extensions
I'm reading page $59$ of Reid's "Undergraduate commutative algebra" book.

In example (ii) it says, $k[x^{2}] \subset k[x]$ is an integral extension.

How do we know this? I mean, in order to show this we must take a polynomial $f(x) \in k[x]$ and show there is a monic polynomial $g(x) \in k[x^{2}]$ such that $g(f(x))=0$, right? Why can we do this?

答: It's enough to prove that $x$ is integral over $k[x^2]$, which it clearly is, being a root of $T^2-x^2 \in k[x^2][T]$.</p>


问:
Showing that the area of a triangle is $\frac{1}{2}|Im[w_1\bar{w_2}]|$?
I'm reading Beardon's Algebra and Geometry.

Let $T$ be a triangle in $\mathbb{C}$ with vertices at $0$, $w_1$, $w_2$. By applying the mapping $z\mapsto \bar{w_2}z$, show that the area of $T$ is $\frac{1}{2}|Im[w_1\bar{w_2}]|$.

I am lost in this question mainly in two points:

The reason for the existence of the mapping $z\mapsto \bar{w_2}z$;

I believe I know the the reason for the existence of this mapping - it is there to ensure we have 3 points that aren't positioned in, say: $(0,0),(0,2),(0,3)$ because it wouldn't constitute a triangle. Is this correct?

I am aware that $\frac{1}{2}|Im[w_1\bar{w_2}]|$ is a form of $\frac{bh}{2}$.

But I can't understand why it constitute the area of the triangle, I've evaluated this expression and it yields $$\frac{1}{2} \sqrt{\left(a_1 b-a b_1\right){}^2}\tag{1}$$ for $w_1=a+bi$ and $w_2=a_1+b_1i$.

In a trial of understanding the problem, I started thinking on how I would do it myself, I would first get  the module of $0$ to $w_1$, then the module from $w_1$ to $w_2$ and then use $\frac{bh}{2}$:

$$\frac{|a+bi|\cdot|a_1+b_1i|}{2}$$

which yields

$$\frac{1}{2} \sqrt{a^2+b^2} \sqrt{a_1^2+b_1^2}$$

And is different of $(1)$, from here I don't know what else to think.

答:
I am not exactly sure answering this question rather than the former is the best course of action, but this question is the more fleshed-out one and is more amenable to the answer I gave you in chat, which I am expanding upon here.

No, the transformation $z\mapsto \bar{w}_2z$ has nothing to do with degenerate triangles. In fact, if you start with a degenerate triangle (where the three endpoints are collinear), applying the map to the triangle will just yield another degenerate triangle - it doesn't change anything. This should be more apparent once you see what multiplication does to the complex plane. But it's not like degenerate triangles pose any issues in the first place!

The idea is that we want to use the formula Area = Base x Height / 2, however it isn't entirely obvious what the "height" of the triangle will be for any choice of edge as "base." This can be fixed, however, if we rotate the triangle (around the origin) so that one of its edges is on the real axis - in this case, the base is the length of that edge, and the height is the imaginary part of the third point.

Let's review some basic facts of the complex plane and its geometry. Every nonzero number $\alpha\in\bf C$ can be written in polar form, $\alpha=re^{i\theta}$. Here, $r=|\alpha|$ is the modulus, or magnitude - the distance between $\alpha$ and $0$. And $\theta$ is the angle made between the line segment from $0$ to $\alpha$ and the positive real axis. Note that $e^{i\theta}$ is the rescaling of $\alpha$ that is on the unit circle in the same direction.

So if $\alpha=re^{i\theta}$ and $\beta=se^{i\phi}$, then $\alpha\beta=rse^{i(\theta+\phi)}$. In words, this means that the effect of multiplying anything by $\alpha$ is to (a) expand/shrink by a factor of $r=|\alpha|$, and (b) rotate about the origin by an angle of $\theta$. Rotations and dilations are achieved by multiplication.

Now, suppose we want to rotate $T$ so it has an edge on the real axis, which will necessitate multiplying by something. We already know that $u\bar{u}=|u|^2\in\bf R$ for any $u\in\bf C$, so we can achieve our goal of rotating the triangle by multiplying the vertices by $\bar{w}_2$. Denote the rotated triangle $T'$, so that the vertices of $T'$ are $\bar{w}_2\cdot0=0$, $\bar{w}_2w_2=|w_2|^2$, and $\bar{w}_2w_1$. The base of this triangle has length $|w_2|^2$, and the height of this triangle (the vertical distance of the third vertex from the real axis) is ${\rm Im}(\bar{w}_2w_1)$ Hence ${\rm Area}(T')=\frac{1}{2}|w_2|^2{\rm Im}(\bar{w}_2w_1)$.

The issue at this point is that multiplying by $\bar{w}_2$ didn't just rotate the triangle $T$, it also stretched it. Luckily, multiplication stretches the complex plane uniformly - all lengths are scaled by a given factor. In particular, since lengths are stretched by a factor of $|w_2|$, areas must be stretched by a factor of $|w_2|^2$. Hence ${\rm Area}(T')=|w_2|^2{\rm Area}(T)$. Solving for ${\rm Area}(T)$ yields the conclusion.


问:
Translating Polar Functions
How do I rewrite a polar function (expressed in a polar coordinate system $r = F(\theta)$ so the entire curve is shifted right or left $h$ units and up or down $k$ units?</p>

答:
If you are translating the pole by $\Delta x$ + $\Delta y$, it's simply:

$$
r_{new}= r + \Delta x.\cos(\theta) + \Delta y.\sin(\theta)
$$

Title: Translating Polar Functions


问:
If a homomorphism is surjective it cannot be trivial
My group theory is a bit rusty, but I recall such a property existing. Is this true, if so, can anyone give me some hints as to how to prove this?</p>

答: You need an additional hypothesis. If a homomorphism whose codomain is a nontrivial group is surjective then it is nontrivial (this is immediate from the fact that the trivial homomorphism by definition takes on only one value), but of course the trivial homomorphism to the trivial group is surjective.


问:
Where did 33 come from in this pigeon hole principle problem?
I have thought about this for some time and I can't justify how they got the number 33.
For the  $1\leq x_1<x_2<...<x_{20}\leq 33$, it should be $1 \leq x_1<x_2<...<x_{20} = 35$ because the total number of calls at the end of the last day should be 35.

However, if we continue the in the same way, then we arrive at 40 integers under 41 in which the pigeonhole principle does not help. Where did 33 come from, or is there a different way to solve this problem?

答:
Here's how you can adapt the existing solution to work for 35 calls, which is a deeper application.

Include $ x_0 = 0$ (which makes sense because otherwise we're ignoring what Susan does on the first day).

As per before, we have

$0 \leq x_0 < x_1 <  \ldots < x_{20} \leq 35.$

$6 \leq x_0 + 6 < x_1 + 6 < \ldots < x_{20} + 6 \leq 35+6 = 41.   $

There are 42 integers (pigeons), which range from $0$ to $41$ (holes).

If the integers are distinct, then

$ \sum_{i=0}^{41} i = \sum x_i + \sum (x_i + 6) = \sum ( 2 x_i + 6)$.

However, observe that the LHS is odd and the RHS is even, so this is a contradiction.

Hence, (at least) 2 of the integers are not distinct, which means we have $x_i = x_j + 6$.

Notes

The inclusion of $x_0$ would have allowed us to make the conclusion for 34 calls using PP directly.

36 and 37 calls are not possible. See below.

For 38 calls, I have the counter example of 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1

For 36 calls, we have 42 integers (pigeons) which range from 0 to 42 (holes).

If the integers are distinct, then consider their values mod 6.

Holes: Mod 6, we have 8/7/7/7/7/7 holes.

Taken mod 6, $x_i$, $x_i + 6$ pigeons are in the same mod 6 hole.

Hence, each hole has an even number of pigeons.

Hence, there are at most $8 + 6 + 6 + 6 + 6 + 6 = 38  $ pigeons, which is a contradiction.

For 37 calls, we have 42 integers (pigeons) which range from 0 to 43 (holes).

If the integers are distinct, then consider their values mod 6.

Holes: Mod 6, we have 8/8/7/7/7/7 holes.

Taken mod 6, $x_i$, $x_i + 6$ pigeons are in the same mod 6 hole.

Hence, each hole has an even number of pigeons.

Hence, there are at most $8 + 8 + 6 + 6 + 6 + 6 = 40  $ pigeons, which is a contradiction.

35 calls using the above formulation.

If the integers are distinct, then consider their values mod 2.

Holes: Mod 2, there are 21/21 holes.

Taken mod 2, the $x_i$ and $x_i + 6$ pigeons are in the same mod 2 hole.

Hence, each hole has an even number of pigeons.

Hence, there are at most $20 + 20 = 40$ pigeons, which is a contradiction.


答:
Kind of. But we want to do it very rigorously (at least one time). Furthermore we will assume that $f \in C^2(\mathbb{R}^2)$, i.e. that it is twice continuously differentiable. You will later see why.

Now let $g(x) := \lVert x \rVert_2$, $g:\mathbb{R}^2 \rightarrow \mathbb{R}$. Note that we have
$$
\partial_{x_1} g(x) = \partial_{x_1} \sqrt{x_1^2 + x_2^2} = \frac{x_1}{\sqrt{x_1^2 + x_2^2}}
$$
by the chain rule for one variable. You can do the same computation to get $\partial_{x_2} g$. So, (we set $Dg := \nabla g^\top$)
$$
\nabla g(x) = \frac{x}{g(x)} \implies Dg(x) = \frac{x^\top}{g(x)}
$$
which is a formula that can be very useful if you know it by heart. But this also means that $g \in C^1(\mathbb{R} \setminus \lbrace (0, 0)^\top \rbrace)$.

Furthermore let $h: \mathbb{R}^2 \rightarrow \mathbb{R}^2 $, $h(x) := \nabla f(x)$. Note that $Dh(x) = Hf (x)$. So, for $x \in Z$ where $Z := \lbrace y \in \mathbb{R}^2: \nabla f(x) \neq 0 \rbrace $:
$$
D (\lVert \nabla f(x) \rVert_2) = D(g \circ h)(x) = Dg(y)\vert_{y = h(x)}Dh(x) = \frac{1}{\lVert \nabla f(x) \rVert_2}\nabla f(x)^\top Hf(x)
$$
We conclude:
$$
\nabla \lVert \nabla f(x) \rVert_2 = \left( \frac{1}{\lVert \nabla f(x) \rVert_2}\nabla f(x)^\top Hf(x) \right)^\top = \frac{1}{\lVert \nabla f(x) \rVert}_2 Hf(x) \nabla f(x)
$$
I think the latter is the expression you looked for.

Some nice facts:
The assumption $f \in C^2(\mathbb{R})$ allowed us to write $Hf(x)^\top = Hf(x)$ in the very last step. This is calles Schwartz's theorem.
Also, the set $Z$ is open: We can write
$$
Z = \nabla f^{-1}((-\infty, 0) \cup (0, \infty)).
$$
So $Z$ is the preimage of an open set under a continuous function, hence open. (The topology is always the euclidian one...)

Title: Gradient of norm of gradient of scalar field

Body: I have a function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ representing a surface $z=f(x,y)$ in euclidean space. Is there an elegant way to apply the chain rule to expand $\nabla||\nabla f||_2$, where $||\cdot||_2$ is the euclidean norm?


答:
If you change to polar coordinates, you get
$$
\lim_{r \rightarrow 0} \sin(\theta) \cos(\theta) \sin(r^2 \sin(\theta) \cos(\theta)).
$$

It might look a bit confusing, but the essential thing is that $\sin(x)$ goes to zero as $x$ goes to zero, and that $\sin(\theta)$ and $\cos(\theta)$ are at most 1.

Title: Finding the limit for functions with two variables

Body:
I know that when we have a limit of a function with $2$ variables, the limit must be the same, regardless of the path we take. So this is useful for proving that a limit doesn not exist. But when you've tried this method for different "paths" (e.g., $(x,0),(0,y),(x,x),(y,y)$, etc...) and you think that the limit does exist, how you do show it?

For example here is a question from my textbook:

Evaluate the limit:
  $$\lim_{\large{(x,y) \to (0,0)}} \dfrac{xy \sin(xy)}{x^2+y^2}$$

The answer is supposed to be $0$ but I don't see how you can prove that it is $0$ for any direction you approach $(0,0)$ from. Can someone please help me? Thanks.


问:
Is there any simple formula for $x \sum\limits_{k=1}^n \frac{p_{n-k}(x)}{k}$?
Is there any simple formula?

$$
\begin{align}
p_0(x) &= 1 \\
p_n(x) &= x \sum\limits_{k=1}^n \frac{p_{n-k}(x)}{k}
\end{align}
$$

I am thankful for any help or hint.

Edit: The first few polynomials are

$$
\begin{align}
p_0(x) &= 1 \\
p_1(x) &= x \\
p_2(x) &= x^2 + \frac{1}{2}x \\
p_3(x) &= x^3 + x^2 + \frac{1}{3}x \\
p_4(x) &= x^4 + \frac{3}{2}x^3 + \frac{11}{12}x^2 + \frac{1}{4}x \\
p_5(x) &= x^5 + 2x^4 + \frac{7}{4}x^3 + \frac{5}{6}x^2 + \frac{1}{5}x
\end{align}
$$

I know that the first coefficient is always $1$, the second is $\frac{n-1}{2}$ and the last is $\frac{1}{n}$.

答:
This is not an answer!

Just too long for a comment.

A closed form could be quite complicated...

Using Mathematica FindSequenceFunction powerful feature, I found
$$\frac{x}{n}+\frac{ (2 (\psi ^{(0)}n+\gamma ))}{n}x^2+\frac{\left(6 \psi ^{(0)}n^2+12 \gamma  \psi ^{(0)}n+6 \psi ^{(1)}n+6 \gamma ^2-\pi ^2\right)}{2 n}x^3+\ldots$$
where $\gamma$ is the Euler constant, and $\psi ^{(0)},\psi ^{(1)}$ are polygamma and its derivative.
Indeed we get
$$
\begin{array}{cr}
 1 & x \\
 2 & x^2+\frac{x}{2} \\
 3 & x^3+x^2+\frac{x}{3} \\
 4 & \ldots+\frac{3 x^3}{2}+\frac{11 x^2}{12}+\frac{x}{4} \\
 5 & \ldots\frac{7 x^3}{4}+\frac{5 x^2}{6}+\frac{x}{5} \\
\end{array}
$$


问:
Question about Slutsky's Theorem: How to Show Convergence for $Y_n$ if Knowing a $E(X_n-Y_n)^2 \longrightarrow 0$ Term between Two Distributions
If $X_n \longrightarrow \theta$ in probability, where $\theta$ is a constant and $E(X_n - Y_n)^2  \longrightarrow 0$ in probability, can we say $Y_n  \longrightarrow \theta$ in probability?

My first thought is to write the MSE term into$$
E(X_n - Y_n)^2 = \frac{1}{n}\sum_n [(X_n)^2 - 2(X_n)(Y_n) + (Y_n)^2],
$$ and then figure out each term in the expansion.

$\frac{1}{n}\sum_n (X_n)^2$: Since $X_n \longrightarrow \theta$ in probability, $g(X_n) \longrightarrow g(\theta)$ in probability. Thus, $g(X_n) \longrightarrow g(\theta)=\theta^2$. Plug into the above form, we can have $\frac{1}{n} * n * \theta^2 = \theta^2$.

$\frac{1}{n} \sum_n 2(X_n)(Y_n)$: Assuming $Y_n \longrightarrow y$ in probability, then by the Slutsky's theorem, it can be expressed as $\frac{1}{n}*n*y\theta = y\theta$.

$\frac{1}{n} \sum_n (Y_n)^2$: Following the similar calculation above, we can have $\frac{1}{n} * n * y^2 = y^2$.

Since $E(X_n - Y_n)^2  \longrightarrow 0$ in probability, thus $y$ must be equal to $\theta$.

However, I am not quite sure the above calculation is correct or not, since if we take limit for deterministic sequence such as $\frac{1}{n}$, it goes to 0 as n goes to infinity, which causes the above calculation doesn't work anymore I think.

答:
Since $$\{|Y_n-X_n|+|X_n-\theta|>\varepsilon\}\subset\{|Y_n-X_n|>\varepsilon/2\}\cup\{|X_n-\theta|>\varepsilon/2\},$$
\begin{align*}
P(|Y_n-\theta|>\varepsilon)
&=P(|Y_n-X_n+X_n-\theta|>\varepsilon)\\
&\le P(|Y_n-X_n|+|X_n-\theta|>\varepsilon)\\
&\le P(|Y_n-X_n|>\varepsilon/2)+P(|X_n-\theta|>\varepsilon/2)
\end{align*}
using the triangle inequality. Using Markov's inequality,
$$
P(|Y_n-X_n|>\varepsilon/2)\le\frac{\operatorname E|Y_n-X_n|^2}{(\varepsilon/2)^2}
$$
and since $\operatorname E|Y_n-X_n|^2\to0$ as $n\to\infty$, we conclude that $Y_n\to\theta$ in probability as $n\to\infty$.

I hope this helps.


问:
Let $(10)$ be the subgroup of $\langle R^+,*\rangle $ generated by 10,Prove that $R^+/(10)$ $\cong$ $\mathbb{R}/Z$
Let $(10)$ be the subgroup of $\langle\mathbb{R}^+,*\rangle$ generated by 10,Prove that $\mathbb{R}^+/(10)$ $\cong$ $\mathbb{R}/\mathbb{Z}$

In this question ,I don't how to find the isomorphism which take the 1 to 0 without other number over the size.
Please give me some help .

答: Try $x \mapsto \log_{10}(x)$.  It takes $1$ to $0$, turns multiplication into addition, and takes the powers of $10$ to the integers, which is what you want.


问:
Laplace transform of time until absorption of Brownian motion
Let $B_t$ be a one dimensional Brownian motion. For $a, b > 0$ let $T_x = \inf\{t \in \mathbb{R}_+: B_t = x\}$ and $T_{-a \wedge b} = T_{-a} \wedge T_b$.

I now want to show the Laplace transform of $T_{a, b}$ is

$\mathbb{E}[\exp(-\lambda T_{a, b})] = \frac{cosh(\frac{a-b}{2}\sqrt{2\lambda})}{cosh(\frac{a+b}{2}\sqrt{2\lambda})}$

I know the usual way to this is to take a variant of the exponential martingale $M_t$, then use optional stopping to show that for some stopping time $T$

$E[M_T] = E[M_0] = c$

where $c \in \mathbb{R}$. This is the way it was presented in this answer for example.

My main problem is that I'm having trouble dealing with the two different values $-a$ and $b$. I tried working with the martingale

$M_t^{\lambda, c} = \sinh(\lambda (B_t + c))\exp(-\frac{\lambda^2}{2}t)$

but because the $c$ is part of it, the expected value at $t = 0$ is not the same for $-a$ and $b$. Then I tried using the fact, that I know that

$P(T_b < T_a) = \frac{-a}{b-a} \quad P(T_a < T_b) = \frac{b}{b-a}$

as well as

$\mathbb{E}[\exp(-\lambda T_{a})] = \exp(a\sqrt(2\lambda))$

to calculate something similar, but I feel like I keep getting stuck.

答:
For simplicity, let $T:=T_{-a,b}$ and let :

$A := \mathbb{E} ( e^{-\frac{\lambda^2}{2} T} \mathbb{1}_{ T_{-a}<T_b} )$

$B := \mathbb{E} ( e^{-\frac{\lambda^2}{2} T} \mathbb{1}_{ T_{-a}>T_b} )$

By the same exponential martingale method, you imply that:
$$1= e^{-\lambda a} A+e^{\lambda b}B$$
and
$$1= e^{\lambda a} A+e^{-\lambda b}B$$
While the transform Laplace of $T$ at $t= \lambda^2/2$ is, by definition:
$$\mathbb{E}( e^{-\frac{\lambda^2}{2} T} )=A+B $$

So if our formula is correct, the work is done by now. $\square$


问:
Show complex function is bijective
I'm asked to show that the function $f:\mathbb{C}\setminus\{1\}\rightarrow \mathbb{C}\setminus\{i\}$ given by
$f(z)=i\frac{z+1}{z-1}$ is bijective.

I get stuck very early on - I think I get thrown off by the fact that we're in $\mathbb{C}$ and not good old $\mathbb{R}$.

My attempt:

injective: assuming $f(z_1)=f(z_2)$, I need to show $z_1=z_2$. Usually I would just set it up like this: $$i\frac{z_1+1}{z_1-1}=i\frac{z_2+1}{z_2-1} $$
And then manipulate the expressios to get $z_1=z_2$, but I can't seem to do it here.

Am I choosing a poor strategy here?

In regards to surjectivity, I am just lost.

Any help is much appreciated.

答: A broader context: your function is a Möbius transformation $\frac{az+b}{cz+d}$ with$$a=b=c=-d=i\to ad-bc=2\ne0.$$These transformations are famously isomorphic to invertible $\left(\begin{array}{cc}a & b\\c & d\end{array}\right)$.

Title: Show complex function is bijective


问:
Notation for the set of edges of a directed graph whose tails belong to vertex set $S_1$ and heads belong to $S_2$
For a directed graph $D$, and sets of vertices $S_1$, $S_2$, is there a standard notation to refer to the set of directed edges whose tail belong to $S_1$ and head belong to $S_2$? If so what text specifically uses this notation?

I have been told that a notation is defined in Bollobas's Modern Graph Theory, but have not been able to get my hands on it yet.

答: Bollobas uses the notation $\vec{E}(S_1,S_2)$.


答: Consider $\phi : \mathbb R \to \mathbb R^+/(10)$ where $\phi(x)$ is the equivalence class of $10^x$.

Title: Let $(10)$ be the subgroup of $\langle R^+,*\rangle $ generated by 10,Prove that $R^+/(10)$ $\cong$ $\mathbb{R}/Z$

Body:
Let $(10)$ be the subgroup of $\langle\mathbb{R}^+,*\rangle$ generated by 10,Prove that $\mathbb{R}^+/(10)$ $\cong$ $\mathbb{R}/\mathbb{Z}$

In this question ,I don't how to find the isomorphism which take the 1 to 0 without other number over the size.
Please give me some help .


答:
Let $G$ be a group. Consider

$$\begin{align}
\varphi: G &\to \{e\}\\
g & \mapsto e.
\end{align}$$

Then for any $g,h\in G$, we have $\varphi (gh)=e=ee=\varphi(g)\varphi(h)$, so $\varphi$ is a homomorphism.

Also, we have, for $e\in \{e\}$, there exists a $k\in G$ such that $e=\varphi(k)$ by definition of $\varphi$; namely, we can take $e_G=k$. So $\varphi$ is surjective.

But, clearly, $\varphi$ is trivial.

Title: If a homomorphism is surjective it cannot be trivial

Body: My group theory is a bit rusty, but I recall such a property existing. Is this true, if so, can anyone give me some hints as to how to prove this?</p>


问:
Show the set $S=\{(x,0): a <x < b\} \subset \mathbb{R}^{2}$ is not open
I'm working through Shifrin's Multivariable Mathemtics and I'm working on an example from the book:

Claim: the set $S=\{(x,0): a <x < b\} \subset \mathbb{R}^{2}$ is not open.

The only concept I have to this point is a subset, $U \subset \mathbb{R}^{n}$, is open if for every $\textbf{a} \in U$ there is some ball centered at $\textbf{a}$ that is completely contained in $U$; that is, there is $\delta >0$ so that $B(\textbf{a}, \delta) \subset U$.

I understand why it is not open and the picture of the object also illustrated this. Essentially the idea being that any ball drawn around a point in $S$, for example $B(\textbf{a}, \delta)$ will contain points whose $y$ component will not be $y = 0$.  But I was trying to construct a formal argument using what I have learned up to this point, which  in terms of multivariable objects is not much. Can I do it using anything simplistic? Because any idea I attempt I'm running into the trouble of stating the idea above.

答:
Take any $a<x<b$. Then $\xi=(x,0)\in S$. Now take any $\delta>0$ and we will show that the ball $B(\xi,\delta)$ is not contained in $S$. Indeed, the point $(x,\frac{\delta}{2})$ belongs to this ball, but it is not in $S$. It belongs to the ball because:

$||(x,0)-(x,\frac{\delta}{2})||=||(0,\frac{\delta}{2})||=\frac{\delta}{2}<\delta$

So the set is not just not open, but even more so - it doesn't have any interior points.


问:
Logic: Show that {∧, ↔, +} is complete, but every proper subset of it, isn't. (Symbol "+" is exclusive or)
I know that, for example, to show {¬,→} is not complete, if φ is a propositional type that only includes ¬,→ has truth value T, then every propositional variable in φ must have a truth value T. But ¬p has not this property (if p has a truth value T then ¬p has a t.v. F).

But when the case is {∧, ↔, +}, I really don't know how to prove this.

答:
You probably already have an example of a minimal complete set of operations, like $S=\{\wedge,\neg\}$.  So proving that a set of operations is complete is as easy as showing that you can derive all of the operations in $S$.

After that, all you need to do is show that none of the three doubleton subsets of $\{\wedge,\leftrightarrow,+\}$ are complete.  You could do that like the example you cited where you describe a feature that all of the derivable operations have and then show that there is a logical operation that doesn't have that feature.  The usual suspects for counterexamples are $\bot$, $\neg$, and $\to$ because a set of operators either can't return false if all the arguments are true, can't be contrary, or can't be asymmetric".


答:
You describe it fairly correctly.  TO do it formally.

Let $P\in S$.  Then $P = (w,0)$ for some real $w: a <x < b$.

Consider any radius $r > 0$ and consider the open ball $B_r(P)$.

Let $Q = (w,\frac 12 r)\in \mathbb R^2$.  We can easily show that $Q\in B_r(P)$ (because $d(P,Q) = \sqrt{(w-w)^2 +(\frac 12r-0)^2} = \frac 12r < r$)  and we can easily see that $Q \not \in S$ (because $Q = (w, \frac 12r)$ and $\frac 12r \ne 0$).

This is true for any open ball around $P$ of any radius, so there is no open ball around $P$ completely contained is $S$ so $P$ is not an interior point of $S$.  So $S$ contains a point that is not an interior point (indeed none of the points of $S$ are interior points) so $S$ is not open.

Title: Show the set $S=\{(x,0): a <x < b\} \subset \mathbb{R}^{2}$ is not open

Body:
I'm working through Shifrin's Multivariable Mathemtics and I'm working on an example from the book:

Claim: the set $S=\{(x,0): a <x < b\} \subset \mathbb{R}^{2}$ is not open.

The only concept I have to this point is a subset, $U \subset \mathbb{R}^{n}$, is open if for every $\textbf{a} \in U$ there is some ball centered at $\textbf{a}$ that is completely contained in $U$; that is, there is $\delta >0$ so that $B(\textbf{a}, \delta) \subset U$.

I understand why it is not open and the picture of the object also illustrated this. Essentially the idea being that any ball drawn around a point in $S$, for example $B(\textbf{a}, \delta)$ will contain points whose $y$ component will not be $y = 0$.  But I was trying to construct a formal argument using what I have learned up to this point, which  in terms of multivariable objects is not much. Can I do it using anything simplistic? Because any idea I attempt I'm running into the trouble of stating the idea above.


问:
Is it "legal" to switch $\forall x$ with $\forall s\forall t$?
In the book The Foundations of Mathematics by Kenneth Kunen, in a chapter about set theory formalizations, the following first order logic sentence is provided (referred to as the Axiom of Comprehension/Separation Schema):

$\forall z \Big (\exists y \color{red}{\forall x} \big ( \color{red}{x} \in y \leftrightarrow x \in z \land \varphi(\color{red}{x}) \big ) \Big )$. I have colored in $\color{red}{\text{red}}$ the variable that I will be focusing on for the remainder of this question.

Consider the following two sets:

$A = \{ x \in \mathbb N \ |\  x \lt 5\}$.

$B = \{ \langle s, t \rangle \in \mathbb N \times \mathbb N \ |\ s \lt 5 \land t\lt 4 \}$

Now, let's say I wanted to look at set $A$ and confirm that it exists through the perspective of the Axiom of Comprehension Schema:

$\exists y \forall x (x \in y \ \leftrightarrow \ x \in \mathbb N \land x \lt 5)$ where $\varphi(x):= x \lt 5$. Cool. Set A exists.

To confirm set $B$'s existence, however, I run into an issue when it comes to inserting the predicate $\varphi$:

$\exists y \forall x (x \in y \ \leftrightarrow \ x \in \mathbb N \times \mathbb N \ \land \text{something} )$

I use the phrase "$\text {something}$" because "$\varphi (x)$" does not appear to decompose correctly into a statement about two variables...i.e. $\varphi(x)$ cannot be used to represent "$s \lt 5 \land t \lt 4$" because $s$ and $t$ are not necessarily equal, and therefore do not represent one object $x$.

A quick fix seems to be to switch $\forall x$ with $\forall s \forall t$ in order to generate the following statement:

$\exists y \forall s \forall t (\langle s,t \rangle \in y \ \leftrightarrow \ \langle s,t \rangle \in \mathbb N \times \mathbb N \ \land s \lt 5 \land t \lt 4  )$

I am uncertain why this is something that is allowed. Specifically, this seems to suggest that $\forall x  \ \varphi(x)$ is 'semantically' (regular English definition) equivalent to $\forall s \forall t \ \varphi (s,t)$.

From what I understand about first order logic, $\forall x \ \varphi (x)$ essentially means "For all singular objects in the domain of discourse, the following sentence about a singular object is true". Comparatively, I interpret $\forall s \forall t \ \varphi(s,t)$ as saying "For any two objects in the domain of discourse, the following sentence is true about two objects".

Is the syntax "$\ s,t\ $" a singular object in the domain of discourse? It doesn't strike me as one...hence my confusion.

Any input is greatly appreciated! Thank you.

答:
You use the projection functions - or rather, the projection relations. These are the binary relations $\pi_0(x,y)$ and $\pi_1(x,y)$, which we read as "$x$ is an ordered pair with left coordinate $y$" and "$x$ is an ordered pair with right coordinate $y$." Note that the expression "$u=\langle v,w\rangle$" is really shorthand for "$\pi_0(u,v)\wedge\pi_1(u,w)$."

Writing out the projection relations in terms of $\in$ alone is a good exercise. Once you've done this, you can produce $B$ by applying Separation to the formula

$$\varphi(x)\equiv \forall s,t(\pi_0(x,s)\wedge \pi_1(x,t)\rightarrow s<5\wedge t<4).$$

Separation applied to this formula, with "container set" $z=\mathbb{N}\times \mathbb{N}$, says that there is a set $B$ consisting of exactly those $a$ which are elements of $z$ and which satisfy $\varphi$ - and this $B$ is precisely what we want.