问:
A nonlinear differential equation
We are to solve

$$y=(y'-1)\cdot e^{y'}$$

Let $p=y'$, so $$y= (p-1)\cdot e^p$$
Differentiate: $$dy=(e^p + pe^p-e^p)dp=pe^p\, dp$$
From $$dx=\dfrac{dy}{p}=e^p\,dp$$
I find $$x=\int(e^p)dp=e^p$$.but the answer on my book is $$y=[\ln(x-c)-1]\cdot(x-c)$$
Help?

答:
$$x=e^p \implies p=\mathrm d y/\mathrm d x = \log x$$

You need to integrate this for $y=y(x)$.

Note that to get the constant shift, you should use $x =\int e^p\mathrm d p = e^p+c$. You should also go back and check the original equation to make sure you haven't got any incorrect solutions by accident. This is always a good idea with nonlinear equations. (The point is that you differentiated a first order equation with one arbitrary constant so you introduced an extra fictitious one.)

Title: A nonlinear differential equation


问:
Path-connected is connected (proof verification)
Assume that $A$ is path-connected, and $U\subseteq A$ which is clopen in respect of $A$. Assume that $U\neq\emptyset \land U\neq A$/ then there are $x\in U \;\; y\in A-U$. From path connected there is $\gamma:[0,1]\to A$ which is continous. and as $[0,1] = \gamma^{-1}(A)=\gamma^{-1}(U)\uplus \gamma^{-1}(A-U)$ , and U,A-U are open

(A-U is open because U is also closed.)

then we have the connected interval $[0,1]$ written as a disjoint union of open sets which is contradictory to the connectives of the interval $[0,1]$.

Does this make sense?
I also want to prove that the interval is connected, but I couldn't do it: Tried to state that if it wasn't connected then by taking $x$ in one of the sets and $y$ in the second set then the interval $(x,y)$ satisfies that for every $z$ in the interval: $z$ belongs to just one of the sets, and also that there is a ball around $z$ that is contained in that subspace. Hence, if the two sets are not empty then at least one of them must have a boundary point that is contained in the set and that leads to a contradictory because we wrote the interval as a union of the open sets without the boundary.

but I wasn't able to formalise that.. (maybe there is a need to build a sequence of points that leads to a contradictory?)

Thanks.

答:
I can't decipher the symbols you're using, so I can't really say if you're correct tor not. Here is the standard answer;

If $X=A \cup B$  with $A,B$ disjoint and nonempty and open in $X$. then pick $a \in A$ and $b \in B$. There is a cts map, $f:[0,1] \rightarrow X$ with $f(0)=a,f(1)=b$. Then:
$$f^{-1}(X)=f^{-1}(A \cup B)=f^{-1}(A)\cup f^{-1}(B)=[0,1]$$

This implies $[0,1]$ is disconnected (why?), a contradiction.

ContentLicense: CC BY-SA 4.0


问:
Remainder When Divided By 70 : Remainder Theorem Problem
If $x$ = $16^3$ + $17^3$ + $18^3$ + $19^3$ , then $x$ divided by $70$ leaves a remainder of?

I tried to solve this problem by using the remainder theorem which states that remainder of $$Rem[\frac{a+b+c+....}{x}] = Rem[\frac{a}{x}] + Rem[\frac{b}{x}] + Rem[\frac{c}{x}] + .... $$ where 'Rem' means 'Remainder Of'.

Using this same logic I can write $$Rem[\frac{16^3+17^3+18^3+19^3}{70}] = Rem[\frac{16^3}{70}] + Rem[\frac{17^3}{70}] + Rem[\frac{18^3}{70}] + Rem[\frac{19^3}{70}] = 18 +13+11+69 = 111 \Rightarrow Rem[\frac{111}{70}] = 41$$

So my answer comes up as 41. Can someone please explain to me what I might be doing wrong or what I have understood wrong?

Body: Hint: $\,a+b\mid a^3+b^3\Rightarrow 35\mid 16^3\!+\!19^3,\,17^3\!+\!18^3\Rightarrow 35\mid x.\,$  $x$ is even so  $\,2\cdot 35\mid x$


问:
Find the minimum of $\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$
Find the minimum of $$f(x,y)=\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$$ It seems that $f_x=f_y=0$ is very hard to compute. Is there any easier idea?

答: Complete the square in the first two terms and let $Y=2y-3$ and $X=3x-3/2$ to get $$f(X,Y)=\sqrt{Y^2+1}+\sqrt{2X^2+\frac12}+\sqrt{2\left(X+\frac12\right)^2+\left(Y+\frac12\right)^2-2XY+\frac74}$$ so that \begin{align}f_X&=\frac{2X}{\sqrt{2X^2+1/2}}+\frac{2X-Y+1}{\sqrt{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}}=0\\f_Y&=\frac Y{\sqrt{Y^2+1}}+\frac{Y-X+1/2}{\sqrt{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}}=0\end{align} which can be rewritten as \begin{align}\frac{4X^2}{2X^2+1/2}&=\frac{(2X-Y+1)^2}{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}\\\frac{Y^2}{Y^2+1}&=\frac{(Y-X+1/2)^2}{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}\end{align} and extensively simplified to give the symmetrical forms \begin{align}X^2(4Y^2+16Y+12)+4(Y-1)X-(Y-1)^2&=0\\Y^2(4X^2+12X+5)+4(2X-1)Y-(2X-1)^2&=0\end{align} on clearing denominators. Equating the two expressions on the left gives $$8X^2Y-6XY^2+8X^2-3Y^2-2XY-4X+3Y=0$$ which factorises to $(4X-3Y)(2XY+2X+Y-1)=0$, where the two interaction terms significantly aid this observation. Substituting $X=3Y/4$ gives the polynomial $$9Y^4+36Y^3+35Y^2-4Y-4=0$$ which has roots at $Y=-2,\pm1/3$ after using the rational root theorem. Hence $X=-3/2,\pm1/4$ and a minimum of $\sqrt{10}$ is obtained at $$(X,Y)=\left(-\frac14,-\frac13\right)\implies(x,y)=\left(\frac5{12},\frac43\right).$$ In fact, it is the global minimum as the expression $2XY+2X+Y-1=0$ is an anomaly due to squaring. The square root in the third term of $f(X,Y)$ means that the condition $f_X=0$ only holds if $X\le-3/2$ or $(X,Y)=(0,1)$ and $f_Y=0$ only holds if $-3/2\le X<0$ or $(X,Y)=(1/2,0)$.


问:
An exact equation to calculate the extrema of $\dfrac{\sin(x)}{x}$
A couple of days ago I calculated that the $m^{th}$ extrema of $\dfrac{\sin(x)}{x}$ denoted by $y_m$ is given by this equation below:

$$ \sqrt{1-y_m^2} +y_m \sin^{-1}(y_m)-y_m(-1)^m \left(2m+1\right)\dfrac{\pi}{2}=0$$

which can be simplified to the following :

$$ \sin^{-1}(y_m)+\cot(\sin^{-1}(y_m))-(-1)^m \left(2m+1\right)\dfrac{\pi}{2}=0$$

I cannot find any reference to this expression I calculated anywhere online in any published work or in general. Could anyone tell me whether it is an already known equation?

References:

Locations and amplitudes of the extrema of the sinx/x function

Edit: Since this is confusing some people. I'll clarify that I do not want to solve for $x_k$ (the location of the extrema) but the extrema itself, there are a lot of papers which already have done that in different manners. I instead was looking for a function that directly gives you the extrema. The function which I mentioned above directly gives you the extrema $y(x_m)$ (and not its location $x_m$) when numerically solved for its root. I wanted to know whether this equation which I found was already known/used before. I cannot find any published work where they have given a equation that directly gives you the extremum values for $\dfrac{\sin(x)}{x}$.

答:
Here is the derivation of the OP equation.

The extrema $x_m$ of the function
$$
f(x)=\frac{\sin x}{x}\tag1
$$
are given by the fixed point of $\tan x$:
$$
x_m=\tan x_m\tag2
$$
and lie in the interval $(m\pi,(m+\frac12)\pi)$.

Substituting $x$ in the denominator of (1) with $\tan x_m$ one obtains:
$$
y_m=\frac{\sin x_m}{\tan x_m}=\cos x_m
$$
or
$$
x_m=s_m\arccos y_m+k_m\pi.\tag3
$$
More detailed analysis reveals:
$$
s_m=(-1)^m;\quad k_m=\begin{cases}m,&m\text{ even}\\
m+1,&m\text{ odd}\\
\end{cases}.$$

Substitution of (3) in (1) gives rise to the equation
$$
y_m=\frac{\sin((-1)^m\arccos y_m+k_m\pi)}{(-1)^m\arccos y_m+k_m\pi}=
\frac{(-1)^m\sqrt{1-y_m^2}}{(-1)^m\arccos y_m+k_m\pi}
$$
or
$$
\frac{\sqrt{1-y_m^2}}{y_m}-\arccos y_m=(-1)^mk_m\pi,\tag4
$$
which should be equivalent to the OP equation.

The equation (4) can be cast in a simpler form:
$$
\frac{\sqrt{1-y_m^2}}{|y_m|}-\arccos |y_m|=m\pi.\tag5
$$
However it delivers only the absolute values of $y_m$ which should be additionally multiplied by $(-1)^m$.


问:
Write $\sum\limits_{i=0}^n 2^i + 2^{n+1}$ as a single summation: any guidance?
I'm working on an assignment for Discrete Math, and sort of stuck on this question. I'm finding it a bit challenging to wrap my head around some of these concepts doing it remotely, and being removed from other students doing it.

Anyway: I'm kind of confused - $\sum\limits_{i=0}^n 2^i + 2^{n+1}$ seems to be a single summation already.

The direction I was going was that by Theorem 1 in my textbook for a geometric sum
$\sum\limits_{i=0}^n 2^i = 2^{n+1} - 1$

So I was thinking the entire statement would be equal to $2^{n+2} - 1$ (edited from a silly mistake: $4^{n+1} - 1$), but that doesn't really seem to line up with what the question is asking.

Thanks in advance for your patience and advice.

答:
Assuming it's only $2^i$, not $2^i+2^{n+1}$, that's being summed, you can rewrite $\sum_{i=0}^n2^i+2^{n+1}$ as a single summation as follows:

$$\sum_{i=0}^n2^i+2^{n+1}=\sum_{i=0}^{n+1}2^i$$

Presumably this is an exercise in manipulating summation notation, which can be a useful thing to know how to do.


问:
Cube root of $-2+i$
Edit:

My question comes from finding the solutions of this equation using Cardano's method(because our teacher said :D ):

$$x^3-6x+4=0$$

And finally I got:

$$x=(\sqrt[3]{2})\sqrt[3]{-2+\sqrt{-1}}⠀+(\sqrt[3]2)\sqrt[3]{-2-\sqrt{-1}}$$

I want to denest and find the cube roots of this two radicals.

According to this links answers:

First link

Second link

Third link

I tried to find this using norm(actually for dinest):

$$\sqrt[3]{-2+i}$$

So the problem's norm is $5$ and I must find a root with norm = $\sqrt[3]{5}$  .

And I found this :

$$(\sqrt{\sqrt[3]{5}⠀-1}⠀+i)$$

But:

$$(\sqrt{\sqrt[3]{5}⠀-1}⠀+i)^3\neq {-2+i}$$

Anybody can tell me whats my wrong?

And how to correct that?

答:
Applying Cardano's method to $x^3- 6x+4 =0$ I get:

set $x=u+v$, then

$$x^3 = u^3 + v^3 +3uv(u+v)$$

$$-6x = -6(u+v)$$

so in order  to cancel the last terms we set $uv=2$.

So the equation transforms into

$$u^3 + v^3 + 4 = 0; uv=2 $$

Using $v = \frac{2}{u}$ we get

$$u^3 + \frac{8}{u^3} + 4 =0$$

and we set $t = u^3$ and multiply both sides by $t$ to get

$$t^2 + 4t + 8 = 0 \to (t+2)^2 + 4  =0$$

And so $$t+2 = 2i \to t = 2i-2 \text{ or } t+2 = -2i \to t = -2i -2$$

This doesn't seem to lead to your solution?

Added $x=2$ is a solution (the only rational roots are integer divisors of $4$), and the equation factorises to $$(x-2)(x^2 +2x-2)=0$$

Title: Cube root of $-2+i$


问:
Question about initial topology and dual vector space
Consider the following fragment from the book "Lectures on von Neumann algebras".

Why is the line
$\varphi$ is  $\sigma(\mathcal{E}, \mathcal{F})$-continuous $\implies$ there exist $\psi_1, \dots, \psi_n \in \mathcal{F}$ such that $|\varphi| \leq \sum_k p_{\psi_k}$ true? This is probably an easy consequence of the initial topology definition, but I can't see it.

答:
By definition of initial topology and using that $\mathscr{F}$ is a subspace, we can find $\psi_1, \dots, \psi_n \in \mathscr{F}$ such that
$$\bigcap_{k=1}^n \{x \in \mathscr{E}:|\psi_k(x)| < 2\}\subseteq \{x \in \mathscr{E} :|\varphi(x)| < 1\}.$$

In particular, consider the vector $z:=x/\sum_{k}|\psi_k(x)|$ when $\sum_k |\psi_k(x)| \neq 0$. Then
$$|\psi_k(z)| = |\psi_k(x)|/\sum_k|\psi_k(x)| \le 1 < 2$$
and thus
$$|\varphi(z)| < 1$$
I.e. $$|\varphi(x)| < \sum_k |\psi_k(x)| = \sum_k p_{\psi_k}(x)$$

If $\sum_k |\psi_k(x)|= 0$, then $x \in \bigcap_k \ker \psi_k$ and thus $x \in \ker \varphi$ so that we always have the desired inequality.


问:
Is the Fibonacci lattice the very best way to evenly distribute N points on a sphere? So far it seems that it is the best?
Over in the thread "Evenly distributing n points on a sphere" this topic is touched upon:
https://stackoverflow.com/questions/9600801/evenly-distributing-n-points-on-a-sphere.

But what I would like to know is: "Is the Fibonacci lattice the very best way to evenly distribute N points on a sphere?  So far it seems that it is the best.  Does anyone know of a better method?"

I have a Ph.D. in physics and may have an application for some of this research in physics.

I came across this wonderful paper:

http://arxiv.org/pdf/0912.4540.pdf "Measurement of areas on a sphere using
Fibonacci and latitude–longitude lattices"

The paper states, "The Fibonacci lattice is a particularly appealing alternative [15, 16, 17, 23, 65, 42, 66, 67, 68, 76, 52, 28, 56, 55]. Being easy to construct, it can have any odd number of points [68], and these are evenly distributed (Fig. 1) with each point representing almost the same area. For the numerical integration of continuous functions on a sphere, it has distinct advantages over other lattices [28, 56]."

It the Fibonacci lattice the very best way to distribute N points on a sphere so that they are evenly distributed?  Is there any way that is better?

As seen above, the paper states, "with each point representing almost the same area. "

Is it impossible, in principle (except for special rare cases of N such as 4, etc.), to exactly evenly distribute N points on a sphere so that each point/region has the exact same are?

So far it seems to me that the Fibonacci lattice the very best way to distribute N points on a sphere so that they are evenly distributed.  Do you feel this to be correct?

Thanks so much!

答:
The Fibonacci lattice is not the best way to evenly distribute points on a sphere.
The problem of distributing N points evenly on a unit sphere is only known for specific N.

Moreover, the vertices of Platonic solids are not always optimal. This is succinctly described on the Wolfram Mathworld site:

“For two points, the points should be at opposite ends of a diameter.
For four points, they should be placed at the polyhedron vertices of
an inscribed regular tetrahedron. There is no unique best solution for
five points since the distance cannot be reduced below that for six
points. For six points, they should be placed at the polyhedron
vertices of an inscribed regular octahedron. For seven points, the
best solution is four equilateral spherical triangles with angles of
80 degrees. For eight points, the best dispersal is not the polyhedron
vertices of the inscribed cube, but of a square antiprism with equal
polyhedron edges. The solution for nine points is eight equilateral
spherical triangles with angles of arcos(1/4). For 12 points, the
solution is an inscribed regular icosahedron.”

There are many approximate solvers for this ( SpherePoints[] and Offset Lattice ).
The Fibonacci Spiral is easy to program, but is not optimal.

ContentLicense: CC BY-SA 3.0


问:
Truth value of a statement
I have a problem:

Let P be the statement " x $\in$ A and x $\in$ $\mathbb{Z}$ "

statement: ($\forall$x)P $\Longrightarrow$ ($\exists$x)P

Is there a set A for which the truth value of the above statement is false? Explain.

My approach is:

The statement is only false when the antecedent is true and the consequent is false.

This is not possible because ...

Can someone help, please?

NOTE: Please forgive me : our assumption is that x is in the universal set R

答:
Yes, when $A \cap \mathbb Z$ is the empty set.

Then there are no such $x$ and so $(\forall x) P$ is true vacuously.

But as there exist no such $x$, there can exist no $x$ such that $P$ holds, so $(\exists x)P$ is false.

Title: Truth value of a statement


问:
Towards a consistent notation for entropy and cross-entropy
I am learning about the cross entropy, defined by Wikipedia as
$$H(P,Q)=-\text{E}_P[\log Q]$$
for distributions $P,Q$.

I'm not happy with that notation, because it implies symmetry, $H(X,Y)$ is often used for the joint entropy and lastly, I want to use a notation which is consistent with the notation for entropy:
$$H(X)=-\text{E}_P[\log P(X)]$$

When dealing with multiple distributions, I like to write $H_P(X)$ so it's clear with respect to which distribution I'm taking the entropy. When dealing with multiple random variables, it thinks it's sensible to make precise the random variable with respect to which the expectation is taken by using the subscript $_{X\sim P}$. My notation for entropy thus becomes
$$H_{X\sim P}(X)=-\text{E}_{X\sim P}[\log P(X)]$$

Now comes the point I don't understand about the definition of cross entropy: Why doesn't it reference a random variable $X$? Applying analogous reasoning as above, I would assume that cross entropy has the form \begin{equation}H_{X\sim P}(Q(X))=-\text{E}_{X\sim P}[\log Q(X)]\tag{1}\end{equation}
however, Wikipedia makes no mention of any such random variable $X$ in the article on cross entropy. It speaks of

the cross-entropy between two probability distributions $p$ and $q$

which, like the notation $H(P,Q)$, implies a function whose argument is a pair of distributions, whereas entropy $H(X)$ is said to be a function of a random variable. In any case, to take an expected value I need a (function of) a random variable, which $P$ and $Q$ are not.

Comparing the definitions for the discrete case:
$$H(p,q)=-\sum_{x\in\mathcal{X}}p(x)\log q(x)$$
and
$$H(X)=-\sum_{i=1}^n P(x_i)\log P(x_i)$$

where $\mathcal{X}$ is the support of $P$ and $Q$, there would only be a qualitative difference if the events $x_i$ didn't cover the whole support (though I could just choose an $X$ which does).

My questions boil down to the following:

Where is the random variable necessary to take the expected value which is used to define the cross entropy $H(P,Q)=-\text{E}_{P}[\log Q]$

If I am correct in my assumption that one needs to choose a random variable $X$ to compute the cross entropy, is the notation I used for (1) free of ambiguities.

答:
Personally, I feel that $H(X)$ is already a (very convenient) abuse of notation. After all, entropy is a function of purely the law of $X$ (and not the values it takes). So the IMO fundamental object is (for discrete laws), $H(p) := - \sum p_x \log p_x$ and writing $H(X)$ for the entropy of the law of a random variable $X$ is a convenient shorthand - more formally I'd write this as $H(X\mu)$, where $\mu$ is the measure of the underlying probability space, and $X\mu$ is its pushforward under $X$ (which is a fancy way of saying the law of $X$). There's an exact parallel for differential entropy, but with the catch that you need to have a base measure on the outcome space for differential entropy to make sense.

With this view, $H(p,q) := -\sum p_x \log q_x$ is a natural definition and notation -you don't really need to invoke a random variable, it's about the laws. In fact, I'd maybe go one step further - cross entropy, much like KL divergence, is not really about random variables, but it's a measure(in the non-mathematical sense) of how well aligned two laws are.

If you want to highlight the asymmetry, there's a couple of options that i've seen floating around -

$H(p\|q)$, which highlights the connection between cross-entropy and KL-divergence - $H(p\|q) = D(p\|q) + H(p)$. The notation is nice because the asymmetry of cross entropy is entirely the same as that of KL divergence.

$H_p(q)$ which is sort of expressing that the 'base measure' is $p$. I find this notation quite natural when I consider operational meanings of cross entropy - e.g. if one designs a good compression scheme for an iid $q$-source, but the actual law on which this code is used were iid $p$, the average code length per symbol one would get is (asymptotically) $H_p(q).$ So in this case the base measure $p$ is what's 'truly happening', while $q$ is what you're trying to 'impose'.


问:
Linear Independence of $v$, $Av$, and $A^2v$
Let A be a 3x3 matrix and $ v \in \Re^3 $ with $ A^3v = 0 $ but $ A^2v \neq 0 $.
Show that the vectors $ v, Av $ and $ A^2v $ are linearly independent.
(From Alan MacDonald's Linear and Geometric Algebra, Problem 3.1.5).</p>

答:
First, you might try to see what would happen if they were linearly dependent?

Certainly, you could write an equation down from that assumption.  That equation would involve a linear combination and, once you've got a linear combination, you can multiply through by $A^2$ to get the result.

Edit

Per your request, here's a more geometric way to think of it.

First, $v$ cannot be an eigenvector of $A$.  (Can you see why?)  Thus, $v$ and $Av$ together span a 2D subspace, say $S$, of $\mathbb R^3$.

Next, we need to show that $A^2v$ does not lie in $S$.  Well, to say that $A^3v=0$, while $A^2v\neq 0$ is to say that $A^2 v$ lies in the null space of $A$. Thus, if $A^2v$ did lie in $S$, there would be some linear combination of $v$ and $Av$ that lied in the null-space of $A$.  But since $Av$ already maps to the null-space under application of $A$, we would get that $v$ must also map to the null-space of $A$, which is a contradiction.

More generally, you might think of multiplication by $A$ as folding up space, even in higher dimensions.  Multiplication by $A$ once sends the one-dimensional null-space to zero and something also maps to that null-space.  Multiplication again does it, well, again.


问:
Sum of complex exponential which gives cosine
I have the following expression: let be $\omega = \exp(\frac{2\pi i}{n})$. I have to calculate

$$\omega^j+ \omega^{(n-1)j}$$ with $0\le j \le n-1$. I have to find a cosine from the expression above. How can I do this?

答:
\begin{align*}
\omega^j + \omega^{(n-1)j} &= \exp(\frac{2\pi i j}{n}) + \exp(\frac{2\pi i (n-1)j}{n}) = \exp(\frac{2\pi i j}{n}) + \exp(-\frac{2\pi i j}{n} + 2\pi ij) \\
&= \exp(\frac{2\pi i j}{n}) + \exp(-\frac{2\pi i j}{n}) = 2 \cos(\frac{2 \pi j}{n})
\end{align*}


问:
If unit quaternions $q_1,q_2$ satisfy $q_1^2=q_2^2=-1$, then the map $S^3\to S^3$, $x\mapsto q_1xq_2^{-1}$ has a fixed point
Let $\Bbb H$ denote the quaternion algebra. We can identify $S^3$ as the subset of unit quaternions.
For fixed $q_1,q_2\in S^3$, consider the map $\phi:S^3\to S^3$ defined by $\phi(x)=q_1xq_2^{-1}$. How can we show that if $q_1^2=q_2^2=-1$ (so that $\phi^2=\text{id}$) then $\phi$ has a fixed point?

(This is claimed in a lecture note that I was reading, contained in the sketch of proof of the statement that if $G$ is a subgroup of $SO(4)$ of order $2$ acting freely on $S^3$, then $G=\{\pm \text{id}\}$.)

答:
The map $\phi:x\mapsto q_1x\overline q_2$ is ($\Bbb R$-)linear and as $\phi^2={\rm id}$, its minimal polynomial $m_\phi$ divides $(X^2-1)$.

If $m_\phi=(X-1)$ or $(X^2-1)$, then we're ready because $1$ is an eigenvalue of $\phi$.

The only remaining case is $m_\phi=(X+1)$, i.e. $\phi(x)=-x$.

But in that case $q_1\overline q_2=\phi(1)=-1$ implying $q_2=-q_1$, so
$$q_1x\overline q_1=-\phi(x)=x$$
for all $x\in\Bbb H$, or, $q_1x=xq_1$, that is, $q_1$ is in the center of $\Bbb H$, therefore $q_1\in\Bbb R$, namely $q_1=\pm1$, contradicting to $q_1^2=-1$.


问:
What are a , b and c?
$$y = ax^2 + bx + c$$

which is tangent at the origin with the line $y=x$, It is also tangential with the line $y=2x + 3$. Determine the function! Draw a figure!

My main question is this solvable? I am doubtful?

答:
This is the Graph of $f(x)= -\frac{1}{5}x^{2}+x$ which I graphed using KmPlot. The figure should give you an intuitive idea of how to go about solving.

The Green line is $y=2x+3$.

The Blue line is $y=x$.

If the line $y=2x+3$ and the parabola $y=ax^{2}+bx+c$ are going to be tangent at a given point then their slopes are equal. Let's find that out. Slope of line $y=2x+3$ is $2$ and we have $$2 = \frac{dy}{dx} = 2ax+1$$ So you have $x=\frac{1}{2a}$. Also we have
\begin{align*}
2x+3 & = ax^{2} + x
\end{align*}
which says that $$2 \times \frac{1}{2a} + 3 = a \times \frac{1}{4a^{2}} + \frac{1}{2a}=\frac{3}{4a}$$ From this we have  $$\frac{1}{a} -\frac{3}{4a} = -3 \Longrightarrow a=-\frac{1}{12}$$

This is for the value $a=-\frac{1}{3}$

This is for the value $a=-\frac{1}{7}$.

Title: What are a , b and c?


问:
Improper Integral Convergence Properties
Let $f,g:(0,\infty) \to \mathbb{R}$ be continuous function with $f,g>0$ on $(0,\infty)$.

a) If $\int_1^{\infty}f(x) dx$ converges, then $\int_0^{1}\frac{1}{f(x)} dx$ converges.

b) If $\int_1^{\infty}f(x) dx$ converges and $\lim_{x \to \infty}g(x)=L>0$, then $\int_1^{\infty}f(x)g(x) dx$ converges.

Are these two statements true? I'm not sure where to begin to prove them or how to think of counterexamples. Any hints are appreciated.

答:
I am gonna answer the question b)

Suppose $\int_1^{+\infty}{f(x)dx} = K$.

Since $f, g > 0$, we have that the function $\int_1^t{f(x)g(x)dx}$ is strictly increasing. Then $\int_1^t{f(x)g(x)dx} < K$.

From the fact that $g(x) \to L > 0$ when $x\to +\infty$ and the definition of infinite limit we have that
$$\forall ε > 0, \exists M>0:\forall x>M \implies |g(x) - L|< ε$$
Take $ε_0$ such that $L - ε_0 > 0$(you can always find one since $L>0$), so there exists $M_0$ such that $\forall x>M_0$ we have
$$0<L - ε_0<g(x)<L+ε_0 \implies$$
$$0<g(x)<L+ε_0 \implies $$
$$0<f(x)g(x)<f(x)(L+ε_0)\implies$$
$$0<\int_{M_0}^t{f(x)g(x)dx}<(L+ε_0)\int_{M_0}^t{f(x)}dx<(L+ε_0)K\implies$$
$$\int_{M_0}^t{f(x)g(x)dx}<(L+ε_0)K \implies$$
$$\int_{1}^{M_0}{f(x)g(x)dx}+\int_{M_0}^t{f(x)g(x)dx}<\int_{1}^{M_0}{f(x)g(x)dx}+(L+ε_0)K\implies$$
$$\int_{1}^{t}{f(x)g(x)dx}<\int_{1}^{M_0}{f(x)g(x)dx}+(L+ε_0)K$$

where $\int_{1}^{M_0}{f(x)g(x)dx}$ is constant. So $\int_1^t{f(x)g(x)dx}$ is strictly increasing and bounded, the limit when $t \to +\infty$ exists. So the integral $\int_1^{+\infty}{f(x)g(x)dx}$ converges.


答:
It’s much easier. Suppose $X$ were disconnected and $\{A,B\}$ a disconnection by open sets.

As both sets are non-empty pick $a\in A$ and $b \in B$. Let $f: I \to X$ be a continuous path from $a$ to $b$ ( so $I = [0,1]$ in the usual topology which makes it a connected space).

Then $f[I]$ is connected by continuity of $f$ but $\{A,B\}$ disconnects it, contradiction.

Title: Path-connected is connected (proof verification)

Body:
Assume that $A$ is path-connected, and $U\subseteq A$ which is clopen in respect of $A$. Assume that $U\neq\emptyset \land U\neq A$/ then there are $x\in U \;\; y\in A-U$. From path connected there is $\gamma:[0,1]\to A$ which is continous. and as $[0,1] = \gamma^{-1}(A)=\gamma^{-1}(U)\uplus \gamma^{-1}(A-U)$ , and U,A-U are open

(A-U is open because U is also closed.)

then we have the connected interval $[0,1]$ written as a disjoint union of open sets which is contradictory to the connectives of the interval $[0,1]$.

Does this make sense?
I also want to prove that the interval is connected, but I couldn't do it: Tried to state that if it wasn't connected then by taking $x$ in one of the sets and $y$ in the second set then the interval $(x,y)$ satisfies that for every $z$ in the interval: $z$ belongs to just one of the sets, and also that there is a ball around $z$ that is contained in that subspace. Hence, if the two sets are not empty then at least one of them must have a boundary point that is contained in the set and that leads to a contradictory because we wrote the interval as a union of the open sets without the boundary.

but I wasn't able to formalise that.. (maybe there is a need to build a sequence of points that leads to a contradictory?)

Thanks.


问:
How to remember which direction Jensen's inequality is in?
For a real valued random variable and a convex function $f$, Jensen's inequality is given by:
$$f(\mathbb E X) \leq \mathbb E f(X)$$

How do you remember which direction the inequality is in? I always have to think about how for convex functions, the chord connecting two points on the graph is always above the graph itself.

答: For this and other variants of this inequality, I memorize it by asking in which side do you evaluate more times the function $f$, then that side is greater (or equal).


答: Hint: write down the equation : $a_1v+a_2Av+a_3A^2v=0$  .Now the idea is to  multiply $A$ on both sides. What must the coefficient $a_1,a_2,a_3$ be if you multiply several times $A$ on both sides?</p>

Title: Linear Independence of $v$, $Av$, and $A^2v$

Body:
Let A be a 3x3 matrix and $ v \in \Re^3 $ with $ A^3v = 0 $ but $ A^2v \neq 0 $.
Show that the vectors $ v, Av $ and $ A^2v $ are linearly independent.
(From Alan MacDonald's Linear and Geometric Algebra, Problem 3.1.5).</p>


问:
Confused about open and closed sets
I am slightly confused about part of a problem about open and closed sets, from Stephen Abbott's Understanding Analysis. Consider the set $A = \{ 1 + \frac{1}{4} + \frac{1}{9} + \cdots + \frac{1}{n^2} : n \in \mathbb{N} \}.$ Obviously the set is not closed, because $\sum_{n=1}^\infty \frac{1}{n^2}$ converges. What I'm confused about is the fact that $A^c = (-\infty) \cup (1 , 1+\frac 14) \cup (1+\frac 14, 1+\frac 14 + \frac 19)\cup ......\cup (\sum_{k=1}^n \frac 1{k^2}, \sum_{k=1}^{n+1} \frac 1{k^2})$. This seems to be a union of open sets, which should be open. But clearly, if $A^{\mathsf{c}}$ was open, then $A$ would be closed, which it is not. What am I missing here?

答:
But $A^{\mathsf{c}} \ne (-\infty, 1) \cup (1, 1+\frac{1}{4}) \cup (1 + \frac{1}{4}, 1 + \frac{1}{9}) \cup \cdots \cup (1+\frac{1}{n^2}, 1 + \frac{1}{(n+1)^2}) \, \forall n \in \mathbb{N}$[*]

You are omitting that $\{x| x > \sum_{k=1}^n \frac 1{n^2} \forall n \in \mathbb N\}\subset A^c$.

And $\{x| x > \sum_{k=1}^n \frac 1{n^2} \forall n \in \mathbb N\}\subset A^c = [\lim_{n\to \infty} \sum_{k=1}^n \frac 1{n^2}, \infty)$ is a closed set.

Title: Confused about open and closed sets


问:
How to write a constraint for time bound binary integers?
In these constraints $u_{g,t}^G$ is on at t=1, then $u_{g,t}^{G,start}$ should become 1 after  $t_g^{G,start}$ time-steps then after it stays 1 for $t_g^{G,cr}$ time-steps then $u_{g,t}^{G,on}$ becomes 1. T is the complete time horizon. Is there any simple or alternate approach to these constraints?

$\sum\limits_{t \in{T} } {\left( {1 - u_{g,t}^{G,start}} \right)}  = \left[ {\sum\limits_{t \in{T}} {\left( {1 - u_{g,t}^G} \right)} } \right] + t_g^{G,start},g \in {G}$, where $G=1,2,3,...N$

$\sum\limits_{t \in{T} } {\left( {1 - u_{g,t}^{G,on}} \right)}  = \left[ {\sum\limits_{t \in{T}} {\left( {1 - u_{g,t}^G} \right)} } \right] + t_g^{G,start} + t_g^{G,cr},g \in{G} $ , where $G=1,2,3,...N$

For every g, there are two times, for example $t_g^{G,start}=2$ and $t_g^{G,cr}=3,$ and time $t=1,2,...,T$.
For example in the MIP, at $t=3,$ the variable $u_{g,t}^G$ becomes $u_{g,t}^G=1,$ then $u_{g,t}^{G,start}$ becomes $u_{g,t}^{G,start}=1,$ at $t=5,$ and remains $1$ for the rest of the time interval $T$. After that $u_{g,t}^{G,on}$ becomes $u_{g,t}^{G,on}=1$ at $t=8$ and remains $1$ for the rest of the time interval $T$. I am trying to formulate a single constraint for this formulation.

答:
If I understand correctly, you want to enforce a logical implication $x \implies \bigwedge_{i=1}^n y_i$, which you can do with linear constraints $x \le y_i$ or a single (weaker) aggregated linear constraint $n x \le \sum_{i=1}^n y_i$.  The derivation via conjunctive normal form is:
\begin{align}
&x \implies \bigwedge_{i=1}^n y_i \\
&\lnot x \lor \bigwedge_{i=1}^n y_i \\
&\bigwedge_{i=1}^n (\lnot x \lor y_i) \\
&(1 - x) + y_i \ge 1 && \text{for $i \in \{1,\dots,n\}$} \\
&x \le y_i && \text{for $i \in \{1,\dots,n\}$} \\
\end{align}


问:
Convergence in probability as well as almost surely
If $(X_n)_n$ converges to T in probability and also $(X_n)_n$ converges almost surely to Y can we say that $$P(T=Y)=1$$?

答:
$X_n$ converges to $Y$ almost surely implies that $X_n$ converges to $Y$ in probability. You already know that $X_n$ converges to $T$ in probability. Therefore $Y=T$ almost surely.
Note that
$$
|Y-T|\leq |Y-X_n|+|T-X_n.|
$$
Therefore $|Y-T|>\varepsilon$ implies $|Y-X_n|>\varepsilon/2$ or $|T-X_n|>\varepsilon/2$. Thus, for any fixed $\varepsilon>0$
$$
\mathbb{P}(|Y-T|>\varepsilon) \leq \mathbb{P}(|X_n-T|>\varepsilon/2) + \mathbb{P}(|X_n-Y|>\varepsilon/2)
$$
Take the limit as $n \to \infty$ and you get for every $\varepsilon>0$
$$
\mathbb{P}(|T-Y|>\varepsilon) =0.
$$


答:
Yes, because you can have $T\ne Y$ only with probability zero.

Convergence in probability is implied by a.s. convergence

Title: Convergence in probability as well as almost surely

Body: If $(X_n)_n$ converges to T in probability and also $(X_n)_n$ converges almost surely to Y can we say that $$P(T=Y)=1$$?


问:
showing map $\mathbb{RP}^2 \rightarrow \mathbb{RP}^5$ is smooth embedding
Condider $f:\mathbb{RP}^2 \rightarrow \mathbb{RP}^5$ given by
$$f[x,y,z]=[x^2,y^2,z^2,yz,xz,xy]$$
I would like to show that $f$ is a smooth embedding, meaning that it is a smooth immersion and a topological embedding. For $f$ to be a smooth immersion, I need the differential to be injective. This is given by
$$df=\begin{pmatrix}
2x & 0 & 0\\
0 & 2y & 0\\
0 & 0 & 2z\\
0 & z & y\\
z & 0 & x\\
y & x & 0\\
\end{pmatrix}.$$
However, I am not sure how homogeneous coordinates influence this. Do I need to use charts rather?

答:
Yes, you need to use charts. What you have written is not $df$, it is the differential of a map $\mathbb{R}^3 \to \mathbb{R}^6$. But we can restrict $f$ to an affine chart
$$ f[x,y,1] = [x^2,y^2,1,y,x,xy] $$
and compute its differential as a map $\mathbb{R}^2 \to \mathbb{R}^5$
$$ df = \begin{bmatrix} 2x & 0\\ 0 & 2y\\ 0 & 1\\ 1 & 0\\ y & x \end{bmatrix} $$

Edit: Here's how to finish the argument. By computing the differential you can show that $f$ is a smooth immersion. Next we show that $f$ is injective. Suppose that $f[x,y,z] = f[a,b,c]$. Then $x^2=a^2,y^2=b^2,$ and $z^2=c^2$ (since e.g. $x^2=-a^2$ is impossible) so $x=\pm a, y=\pm b, z=\pm c$. Now use the equations $xy=ab,yz=bc,xz=ac$ to see that either $x=a,y=b,$ and $z=c$ or $x=-a, y=-b$, and $z=-c$. In either case $[x,y,z]=[a,b,c]$ so $f$ is injective.

Now $f$ is an injective smooth immersion defined on compact manifold, so it is an embedding.


问:
An automorphism acting on the sheaf of differentials
I am trying to do the following problem, taken from Iitaka's "Algebraic Geometry".

Let C be a smooth, geometrically connected curve of genus g over a field k. Assume that $g \geq 2$. If $f \in Aut(C)$ satisfies that $f^\ast w = w$ for all $w \in \Omega^1_{C/k}$, show that $f=id$.

I have no idea so far, so any hint would be nice (or solution).

答:
I am not sure if the following method works, but for the case when C is non-hyperelliptic, consider the following:
The canonical map $g:C \rightarrow \mathbb{P}^{g-1}_k$ is a closed immersion since C is not hyperelliptic, so assume that $f \in Aut(C)$ acts as the identity on $\Omega^1_{C/k}$. Then we have a commutative diagram
$$\require{AMScd}
\begin{CD}
C @>{g}>> \mathbb{P}^{g-1}_k;\\
@VfVV @VidVV \\
C @>{g}>> \mathbb{P}^{g-1}_k;
\end{CD}.$$

This gives us $g= g \circ f$, and since g is a monomorphism (closed immersion) we get that $f=id_C$. However, I am a bit unsure if this is true, since the result seem a bit too strong and I would like people to tell me if it is wrong, so don't trust me all too much!


问:
Absolute value of sum of absolute values
Wondering if it's generally true that:

$$||a|+| b||=|a|+|b|$$

If I assumme the contrary, that is $||a|+| b|| = m$, $|a|+|b|=k$, and $k \neq m$,

then $|a|+| b| = m$ or $|a|+| b| = -m$. But since $k \neq m$, this is a contradiction. So it has to be the case that they're equal.

Does this reasoning make sense ?

答:
In the most naive sense, you could think of an absolute value as being the first square root of the square of a number. For example, $$|a| = \sqrt{a^2}$$

Then you could substitute and solve as follows:
$$||a|+|b||=|a|+|b|$$
$$\sqrt{(|a|+|b|)^2}=|a|+|b|$$
$$\sqrt{(\sqrt{a^2}+\sqrt{b^2})^2}=\sqrt{a^2}+\sqrt{b^2}$$
$$(\sqrt{a^2}+\sqrt{b^2})^2=(\sqrt{a^2}+\sqrt{b^2})^2$$
$$\sqrt{a^2}+\sqrt{b^2}=\sqrt{a^2}+\sqrt{b^2}$$

So I would say this generally holds true. I can't think of circumstance when it wouldn't -- even in the case of imaginary numbers, we seem to make out OK:

$$|a| = ||a||$$
$$\sqrt{a^2} = \sqrt{\sqrt{a^2}^2}$$
$$\sqrt{(1 + i)^2} = \sqrt{\sqrt{(1 + i)^2}^2}$$
$$\sqrt{1 + 2i - 1} = \sqrt{\sqrt{1 + 2i - 1}^2}$$
$$\sqrt{2i} = \sqrt{\sqrt{2i}^2}$$
$$\sqrt{2i} = \sqrt{(\sqrt{2}\sqrt{i})^2}$$
$$\sqrt{2i} = \sqrt{(\sqrt{2}\sqrt{i})(\sqrt{2}\sqrt{i})}$$
$$\sqrt{2i} = \sqrt{2i}$$


问:
What is the heuristic 'integration by parts' in computing $\int_{\mathbb R} x e^{ix\xi} d\xi = 0$?
This question is about heuristics to understand with the tempered distribution given by the oscillatory integral $\int xe^{ix\xi} d\xi$ (all integrals are over $\mathbb R$). Source is M. Zworski's Semiclassical Analysis book, page 54.

Maybe it bears mentioning: the author is "not a crank" and this was supposed to be a simple example as more calculations are yet to come in later chapters. I am aware how to compute things rigourously, but it seems that the author has a way to guess the right answer beforehand that I cannot understand.

PS $\int x e^{ix\xi} d\xi= x\int e^{ix\xi} d\xi = 2\pi x\delta = 0$ as tempered distributions (obviously just heuristically but do it with Fourier inversion starting from \eqref{B} below). In contrast with the previous form of reuns' answer below which computed $\int i\xi e^{ix\xi}d\xi =D_x\int e^{ix\xi}d\xi =2\pi \delta'$.

First, its alleged that "integration by parts" (quotation marks are from the book)
shows $$ \int x e^{ix\xi} d\xi = \int D_\xi(e^{ix\xi})d\xi = 0.$$
(Nevermind the missing $-i$ since things should be zero)
I suppose the author intended me to envision the following?
$$ \int D_\xi (e^{ix\xi}) d\xi =  e^{ix\xi}|_{-\infty}^{+\infty} -\int(D_\xi1)e^{ix\xi}d\xi\tag{IBP}\label{IBP}$$
but I don't have the 'feeling' that this 'should' evaluate to zero.

Second, the author tries to justify this as follows: for some large $k$,
$$ u(\psi) := \lim_{\epsilon\to 0+} \iint xe^{ix\xi} \psi(x) \chi(\epsilon\xi) dxd\xi = \iint \frac x{(1+\xi^2)^k} e^{ix\xi} (1-\partial_x^2)^k \psi(x) dxd\xi\tag{A}\label{A}$$
From here, it is said that integration by parts in $\xi$ gives the result. But I think that there's an error here and I cannot commute $x$ with the $x$ derivatives, $[(1-\partial_x^2), x]\psi = 2\psi'$?
Instead what I get is either
$$ u(\psi) = \iint \frac 1{(1+\xi^2)^k} e^{ix\xi} (1-\partial_x^2)^k \Big(x\psi(x)\Big) dxd\xi\tag{B}\label{B}$$
or
$$ u(\psi) = \iint \frac x{(1+\xi^2)^k} e^{ix\xi} L\psi(x) dxd\xi\tag{C}\label{C}$$
where $L\neq(1-\partial_x^2)^k$ is some other differential operator. From \eqref{B}, I can conclude with the Fourier inversion formula, but I see no integration by parts. Whereas if I start from \eqref{C} and try to follow the steps in \eqref{IBP}, I'm led to
$$ \int L\psi(x) \left(\int (1+\xi^2)^{-k} D_\xi e^{ix\xi}d\xi \right) dx = \int L\psi(x) \left( \left.\frac{e^{ix\xi}}{(1+\xi^2)^k}\right|_{\xi=-\infty}^\infty - \int D_\xi((1+\xi^2)^{-k}) e^{ix\xi} d\xi\right)d x$$
and while there is now decay to say $\left.\dfrac{e^{ix\xi}}{(1+\xi^2)^k}\right|_{\xi=-\infty}^\infty=0$, its not immediately obvious to me that the other term should be zero.

Questions:

Should \eqref{IBP} be intuitive? If it should, how far is this intuition correct, is it e.g. correct only if I see power law growth in $\xi$?

Did I make a mistake?

What is the intended calculation?

答:
There is no heuristic and you should not try to "evaluate" divergent integrals that converge only in the sense of distributions.

$$h_A(x)=x\int_{-A}^A e^{ixt}dt =2\sin(A x)$$
For $\phi \in C^\infty_c$
$$\lim_{A\to \infty}\int_{-\infty}^\infty h_A(x)\phi(x)dx=
\lim_{A\to \infty}\int_{-\infty}^\infty \frac{2\cos(Ax)}{A} \phi'(x)dx=0$$
ie. $$\lim_{A\to \infty} h_A = 0 \text{ in the sense of distributions}$$


答:
Very informally, a function $f$ is convex if the value of $f$ at the average of two points lies below the average of the values at the two points.  In symbols, $f\left(\frac{x + y}{2}\right) \leq \frac{1}{2} \left(f(x) + f(y)\right)$.  However, it's very good to experiment with drawing what a continuous function satisfying this inequality can possibly look like.  The version you have above is a direct generalization, where "average of two points" is replaced by "expectation."  Both are notions of averaging, hence it's not so hard to build a mental association between the midpoint and expectation versions.

In fact, $f(\mathbb{E}(X)) \leq \mathbb{E}(f(X))$ for all random variables $X$ --- on a sufficiently rich probability space (if you want, this is "legalese") --- is equivalent to convexity of $f$.

Now you could ask: how to remember $f\left(\frac{x + y}{2}\right) \leq \frac{1}{2} \left(f(x) + f(y)\right)$ as opposed to $f\left(\frac{x + y}{2}\right) \geq \frac{1}{2} \left(f(x) + f(y)\right)$?  Well, in a way, there is no way to remember which one is convex since the other one is concave and the distinction is, after all, arbitrary --- as naming conventions always are.  (Here it's unfortunate that most American students have it beaten into them that these notions are "concave up" and "concave down," only to see "convex" and "concave" later in life and have to rethink these concepts.)

At any rate, with some experience, you learn that convex means "bending upwards" or I just think "convex functions have local minima, but not local maxima" and that helps with distinguishing between convex and concave.  If you think "concave = bending down" and "convex = bending up" (or "convex = upwards parabola" versus "concave = downwards parabola"), only one of these mental pictures coincides with $f\left(\frac{x + y}{2}\right) \leq \frac{1}{2} \left(f(x) + f(y)\right)$.

Title: How to remember which direction Jensen's inequality is in?

Body:
For a real valued random variable and a convex function $f$, Jensen's inequality is given by:
$$f(\mathbb E X) \leq \mathbb E f(X)$$

How do you remember which direction the inequality is in? I always have to think about how for convex functions, the chord connecting two points on the graph is always above the graph itself.


问:
About a proof of Abel's theorem by Kazuo Matsuzaka in "A Course in Analysis vol.2" (in Japanese). I cannot derive an inequality in his proof.
I am reading "A Course in Analysis vol.2" (in Japanese) by Kazuo Matsuzaka.

There is the following theorem (Abel's theorem) in this book:

Let $\sum a_n x^n$ be a power series whose radius of convergence is $1$.

Let $f(x) := \sum_{n=0}^{\infty} a_n x^n$ for $|x|<1$.

Suppose that $\sum_{n=0}^\infty a_n$ converges.

Then, $\lim_{x\to 1} f(x)=\sum_{n=0}^\infty a_n$ holds.

And the author's proof of this theorem is here:

Let $s_n := a_0+a_1+\dots+a_n$.

Let $s:=\lim_{n\to\infty}s_n=\sum_{n=0}^\infty a_n$.

Let $|x|<1$.

$$a_0 + a_1 x+a_2 x^2+\dots+a_n x^n=s_0+(s_1-s_0)x+(s_2-s_1)x^2+\dots+(s_n-s_{n-1})x^n=(1-x)(s_0+s_1 x+\dots+s_{n-1} x^{n-1})+s_n x^n.$$
So, $f(x)=(1-x)\sum_{n=0}^\infty s_n x^n.$

Since $1=(1-x)\sum_{n=0}^\infty x^n$, $s=(1-x)\sum_{n=0}^\infty s x^n$.

$f(x)-s=(1-x)\sum_{n=0}^\infty(s_n-s)x^n$.

Let $\epsilon$ be any positive real number.

Then there exists an positive integer $N$ such that if $n>N$, then $|s_n-s|<\frac{\epsilon}{2}$.

If $|x|<1$, then $$|f(x)-s|\leq(1-x)\sum_{n=0}^N|s_n-s||x|^n+\frac{\epsilon}{2}|(1-x)\sum_{n=N+1}^\infty x^n|<(1-x)\sum_{n=0}^{N}|s_n-s|+\frac{\epsilon}{2}.$$

Since $\sum_{n=0}^{N}|s_n-s|$ is a constant real number, if $0<1-x<\delta$ for sufficiently small positive real number $\delta$, then $$(1-x)\sum_{n=0}^N|s_n-s|<\frac{\epsilon}{2}$$ holds.

So, if $0<1-x<\delta$, then $$|f(x)-s|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$

This means $$\lim_{x\to 1}f(x)=s.$$

I cannot derive the following inequality:

If $|x|<1$, then
$$|f(x)-s|\leq(1-x)\sum_{n=0}^N|s_n-s||x|^n+\frac{\epsilon}{2}|(1-x)\sum_{n=N+1}^\infty x^n|$$ holds.

By the way, we are interested in $x$ which is very close to $1$, so we can assume that $x$ is positive.

If $0<x<1$, then
$$|f(x)-s|\leq(1-x)\sum_{n=0}^N|s_n-s|x^n+(1-x)|\sum_{n=N+1}^\infty (s_n-s)x^n|\\\leq (1-x)\sum_{n=0}^N|s_n-s|x^n+(1-x)\sum_{n=N+1}^\infty |s_n-s|x^n\\\leq (1-x)\sum_{n=0}^N|s_n-s|x^n+(1-x)\sum_{n=N+1}^\infty \frac{\epsilon}{2}x^n\\= (1-x)\sum_{n=0}^N|s_n-s|x^n+\frac{\epsilon}{2}(1-x)\sum_{n=N+1}^\infty x^n$$ holds.

答:
For $\lvert x \rvert < 1$ we have
$$\frac{f(x)-s}{1-x}=\sum_{n=0}^\infty(s_n-s)x^n .$$
I have chosen this form to remove the factor $(1-x)$ from further transformation of the RHS. But you can also do everything with the original equation. We get
$$\left\lvert \frac{f(x)-s}{1-x} \right\rvert = \left\lvert\sum_{n=0}^\infty(s_n-s)x^n \right\rvert \le \left\lvert\sum_{n=0}^N(s_n-s)x^n \right\rvert + \left\lvert\sum_{n=N+1}^\infty(s_n-s)x^n \right\rvert \\ \le  \sum_{n=0}^N\lvert s_n-s\rvert \lvert x \rvert^n  + \left\lvert\sum_{n=N+1}^\infty(s_n-s)x^n \right\rvert \le  \sum_{n=0}^N\lvert s_n-s\rvert  + \left\lvert\sum_{n=N+1}^\infty(s_n-s)x^n \right\rvert .$$
But $\sum_{n=N+1}^\infty(s_n-s)x^n$ is absolutely convergent because $\lvert (s_n-s)x^n \rvert \le \frac{\epsilon}{2}\lvert x \rvert^n$ and we have
$$\left\lvert\sum_{n=N+1}^\infty(s_n-s)x^n \right\rvert \le \sum_{n=N+1}^\infty\frac{\epsilon}{2}\lvert x \rvert^n = \frac{\epsilon}{2}\sum_{n=N+1}^\infty\lvert x \rvert^n \le \frac{\epsilon}{2}\sum_{n=0}^\infty\lvert x \rvert^n = \frac{\epsilon}{2} \frac{1}{1 - \lvert x \rvert} .$$
Therefore for $0 \le x < 1$ we get
$$\lvert f(x)-s \rvert \le (1 - x)\rvert\sum_{n=0}^N\lvert s_n-s\rvert + \frac{\epsilon}{2} .$$
Note, however, that your text claims that for $\lvert x \rvert < 1$
$$\left\lvert(1-x)\sum_{n=N+1}^\infty(s_n-s)x^n \right\rvert \le \frac{\epsilon}{2}\left\lvert (1-x)\sum_{n=N+1}^\infty x^n \right\rvert ,$$
i.e.
$$\left\lvert\sum_{n=N+1}^\infty(s_n-s)x^n \right\rvert \le \frac{\epsilon}{2}\left\lvert \sum_{n=N+1}^\infty x^n \right\rvert .$$
Here I have some doubts: If $-1 < x < 0$ and the sequence $(s_n-s)$ has alternating signs, the all terms in $\sum_{n=N+1}^\infty(s_n-s)x^n$ have the same sign whereas $\sum_{n=N+1}^\infty x^n$ is an alternating sequence. Anyway, the range $-1 < x < 0$ is irrelevant.


问:
A step in proof of existence for SDE: handling locally Lipschitz function
On page 56 of Mao's SDE book, the author considers an existence result for SDE under the local Lipschitz assumption:

Here $a\bigvee b:=\max\{a,b\}$.
What he does is to reduce the argument to the global Lipschitz case by considering

$$
f_{n}(x, t)=\left\{\begin{array}{ll}
f(x, t) & \text { if }|x| \leq n \\
f(n x /|x|, t) & \text { if }|x|>n,
\end{array}\right.
$$
also the same for $g$. It is claimed that $f_n$ and $g_n$ satisfies the global Lipschitz and linear growth condition, but I do not see how.

Let us omit $t$. For $|x|\leq n$ and $|y|\leq n$, this is clear, and for $|x|>n$ and $|y|>n$, we have
\begin{align}
|f_n(x)-f_n(y)|&=|f(nx/|x|)-f(ny/|y|)|\leq \sqrt{K_n} |nx/|x|-ny/|y||\\
&=\sqrt{K_n}\, n|x/|x|-y/|y||\leq 2n\, \sqrt{K_n} |x-y|,
\end{align}
where in the last inequality, $|x|,|y|\geq 1\implies |x/|x|-y/|y||\leq 2|x-y|$ is used; cf. $\left|\frac{x}{|x|}-\frac{y}{|y|}\right|\leq |x-y|$, for $|x|, |y|\geq 1$?.

But I do not see how to show the global Lipschitz continuity for other cases; when, WOLG, we have $|x|\leq n$ and $|y|>n$.

A related question is Cutting off local Lipschitz gives global Lipschitz, which is on another way of cooking up a function like this.

答:
If you have points on a line, then you can apply the triangle identity to get
$$
|f_n(x_M)-f_n(x_0)|\le\sum_{k=0}^{M-1}|f_n(x_{k+1})-f_n(x_k)|\le L\sum_{k=0}^{M-1}| x_{k+1} - x_k |=L|x_m-x_0|
$$
if all the parts satisfy the Lipschitz condition. Thus you do not need to consider mixed cases as you can take the inner points on the sphere $|x_k|=n$ so that the segments are either completely outside or completely inside.

Note that the link for the outside inequaltity was for $|x|\land|y|>1$. Here you have $|x|\land|y|>n$, so that the inequality scales as
$$
|\hat y-\hat x|=\frac{\Bigl||x|\,y-|y|\,x\Bigr|}{|x|\,|y|}
\le\frac{|x|\,|y-x|+\Bigl|(|y|-|x|)\,x\Bigr|}{|x|\,|y|}
=2\frac{|y-x|}{|y|}\le\frac2n|y-x|.
$$

This above inequality is true for all vector norms. Specifically for the Euclidean norm one can use Euclidean geometry to get
$$
|\hat y-\hat x|=2d = \frac2{|x|+|y|}\left( |x| d + |y|d\right)\le\frac{2}{|x|+|y|}(|y-z|+|z-x|)=\frac{2}{|x|+|y|}(|y-x|)
$$
where $z$ is the intersection point of the angle bisector of the rays through $x$ and $y$ and the segment $[x,y]$, see the sketch below

Along these lines more directly
$$
|\hat y-\hat x|
=\frac{\Bigl|y-x+|x|\hat y-|y|\hat x\Bigr|}{|x|+|y|}
\le\frac{\Bigl|y-x\Bigr|+\Bigl||x|\hat y-|y|\hat x\Bigr|}{|x|+|y|}
=\frac{2|y-x|}{|x|+|y|}
$$
as $|x|\hat y-|y|\hat x$ is the reflection of $y-x$ on the angle bisector between $x$ and $y$.


问:
What is the square root of complex number i?
Square root of number -1 defined as i, then what is the square root of complex number i?, I would say it should be j as logic suggests but it's not defined in quaternion theory in that way, am I wrong?

EDIT: my question is rather related to nomenclature of definition, while square root of -1 defined as i, why not j defined as square root of i and k square root of j and  if those numbers have deeper meanings and usage as in quaternions theory.

答:
To get rid of real numbers, x^2=-y^2, because if you add a number to it's negative the result is 0, and what's left is an imaginary number. If x^2=-y^2, then x=iy. So lets plug in x=1, and y=i. (1+i)^2=1^2+2i+i^2=1+2i-1=2i. Not quite, but almost there. (sqrt1/2+sqrt1/2i)^2=(sqrt1/2)^2+2sqrt1/2sqrt1/2i+(sqrt1/2i)^2=1/2+21/2i-1/2=0+21/2i=i
Therefore, the answer is sqrt1/2+sqrt1/2i or -(sqrt1/2+sqrt1/2i) or sqrt1/2-sqrt1/2i ((-i)^2=-1) or sqrt1/2i-sqrt1/2.


问:
urn problem with replacement drawing only two different numbers in 4 tuple
I have an urn with 7 distinct balls which have the number 1 - 7 written on each.
I draw 4 times with replacement and want to find the probability of having only two distinct numbers in each drawn 4 tuple.

I came up with arranging my result space as $(a_1,a_2,a_3,a_4)$ but exactly one time for $0 \leq i < j \leq 4 $a_i ≠ a_j$ within each tuple.

for example for $a_1 = 1$ this would be the following outcome: $(1,2,2,2),...,(1,1,1,2),(1,3,3,3),...,(1,1,1,3),...,(1,1,1,7)$ for every possible number $a_4 \in \{2,...,7\}$

This would lead to having $7 \cdot 6 \cdot 3$ outcomes. $7$ since every number 1 to 7 can be chosen as $a_1$, 6 since I can choose 6 remaining numbers as $a_4$ and sample them on 3 different positions of $j$ since I only have 3 place to choose $j$.

For the sample space I would choose having $7^4$ possibilities since I choose with replacement and have 7 to choose for every pick. This would lead to a probability of $7 \cdot 6 \cdot 3 \over 7^4$

I am actually not sure if this is correct and if there might be a more general way to approach this maybe even for different numbers.

答: I think you're making this too hard.  There are $\binom72$ choices for the two numbers that appear.  Having chosen them, there are $2^4$ ways to make $4$ choices from them, but we have to exclude the two cases where we choose the same number four times.  That gives $$14\binom72=7\cdot7\cdot6$$ choices, and a probability of $$\frac6{49}$$


问:
Does the following integral converge or diverge and can it be solved?
I have been stuck solving the following integral for some time now: $$\int_0^{2R} \frac{\sin^2\left(\frac{x}{4R}\pi\right)R^2}{\left(\sqrt{\left(d+Z\right)^2+\sin^2\left(\frac{x}{4R}\pi\right)}\right)^3}\,\mathrm{d}x$$
If it isn't possible to solve, is there a way to show whether this integral converges or diverges for $\lim_\limits{R\to\infty}$?

Any help or suggestions on how to solve this are appreciated.

答:
If you are only interested in convergence when $R\rightarrow\infty$, you can observe that
$$
0 \le K\sin^2(\frac{x}{4R} \pi)\le \frac{\sin^2(\frac{x}{4R} \pi)R^2}{\left(\sqrt{(d+Z)^2+\sin^2(\frac{x}{4R} \pi)}\right)^3} = I
$$
with
$$
K=\frac{R^2}{\left(\sqrt{(d+Z)^2+1)}\right)^3}
$$

By consequence
$$
\int_R^{2R}  K\sin^2(\frac{x}{4R} \pi)dx \le \int_0^{2R}  K\sin^2(\frac{x}{4R} \pi)dx \le I
$$

But for $x\in[R,2R]$, $\frac{1}{2}=\sin^2(\frac{\pi}{4})\le \sin^2(\frac{x}{4R} \pi)$

By consequence
$$
\frac{KR}{2}=\int_{R}^{2R}\frac{K}{2}dx \le \int_R^{2R}  K\sin^2(\frac{x}{4R} \pi)dx
$$
and your integral $\rightarrow \infty$ when $R\rightarrow \infty$

Note: If you want an analyitcal expression you will have to use elliptic integrals: https://en.wikipedia.org/wiki/Elliptic_integral


问:
A Question on Integers and Fractions

Given that $a,b,c,d,e,f$ are all distinct positive integers, prove that there are no solutions for the following knowing their fractions are non-integers and completely simplified:
$$\left(\frac{a}{c}\right)^2 + \left(\frac{b}{d}\right)^2 = e^2 \tag{1}$$

I coulden't find a proof for this one but I managed to prove this for the case of
$$\left(\frac{a}{c}\right)^2 + \left(\frac{b}{d}\right) ^2 = \left(\frac{e}{f}\right)^2 \tag{2}$$

by simplifying:

$$\frac{(ac)^2 + (bd)^2}{(cd)^2}  = (\frac{e}{f})^2 $$

Since $(ac)^2 + (bd)^2$  must satisfy a Pythagorean triple, we know that each side length(exculding the hypotenuse) must be the product of two distinct integers, which can be proven through Euclid's formula :

$$ac = (2m)(n)$$
$$bd = (m -n)(m + n)$$

Q.E.D.

I also noted that $(1)$ can be written as:

$$(ac)^2 + (bd)^2 = (ecd)^2$$

This lead me to conjecture that there are no Pythagorean triples where the hypotenuse has 2 distinct integer factors where one of each is shared with its side lengths. Furthermore, I also conjecture the hypotenuse's remainder must be a prime such that those conditions can never be satisfied.

Can anyone prove or disprove this?

答:
Suppose the equation $\left(\frac ab\right)^2+\left(\frac cd\right)^2 = e^2$ holds, where the fractions are in their simplest form, i.e. $\gcd(a,b) = \gcd(c,d) = 1$. Then $$\frac {a^2}{b^2} = \frac {e^2d^2 - c^2}{d^2}$$

Both sides are in their simplest form since
$$\gcd(a^2,b^2) = 1 = \gcd(c^2,d^2) = \gcd(e^2d^2 - c^2, d^2)$$

so $a^2 = e^2d^2-c^2, b^2 = d^2$.

Given $b,d > 0$, $b=d$. Therefore no solutions exists, and this fact is unrelated to Pythagorean triples.

Title: A Question on Integers and Fractions


问:
Eisenbud Exercise 2.4 (Practice with $\text{Hom}$ and $\otimes$)
I finished doing Exercise 2.4 from Eisenbud: Commutative Algebra, with a View Toward Algebraic Geometry, and I was wondering if I could have someone look over my solutions to check if everything is correct. Thanks!

(If someone can just tell me if my answers are correct or not, without checking the proofs, I would appreciate that too!)

I will write my solution to each problem in separate bullets, with their justifications immediately below.

$\text{Hom}_\mathbf{Z}(\mathbf{Z}/(n),\mathbf{Z}/(m)) \cong (\mathbf{Z}/(m))[n]$ as $\mathbf{Z}$-modules.

If $\phi:\mathbf{Z}/(n)\to\mathbf{Z}/(m)$ is a $\mathbf{Z}$-module homomorphism, then it is determined by $\phi(1)$ whence it is necessary that $n\cdot \phi(1) = \phi([n]) = \phi([0])\equiv [0]$
. So $\phi(1)$ is an $n$-torsion element of $\mathbf{Z}/(m)$. Moreover, it is not difficult to check that every map which sends $\phi(1)$ to an $n$-torsion element in $\mathbf{Z}/(m)$ is a well-defined $\mathbf{Z}$-module homomorphism. Therefore, the map sending $\phi\to \phi(1)$ defines an isomorphism of $\mathbf{Z}$-modules.

$\text{Hom}_{k[x]}(k[x]/(x^n),k[x]/(x^m)) \cong k[x]/(x^{n})$ as $k[x]$-modules.

The maps $k[x]/(x^n)\to k[x]/(x^m)$ are in bijection with maps $k[x]\to k[x]/(x^m)$ for which $(x^n)$ is contained in its kernel.

Let $\phi:k[x]\to k[x]/(x^m)$ be a module homomorphism. Then $\phi([f]) = f\cdot \phi([1])$ whence $\phi$ is determined by $\phi([1])$. Then $(x^n)$ is contained in $\ker\phi$ if and only if $ x^n\cdot \phi([1]) = \phi([x^n]) = \phi([0]) = [0] $ so that if $\phi([1]) = [a_{m-1}x^{m-1} + \dots a_1x + a_0]$ then $a_i = 0$ for all $i < m-n$. This is a necessary condition if $\phi$ exists. I will show that $\phi$ exists for every such choice of $\phi([1])$.

Let $\phi_g$ be the multiplication-by-$g$ map defined as $\phi_g([f]) = [fg]$. It is clearly well-defined and a module homomorphism. Indeed, $\phi([1]) = [g]$. We just have to choose $g$ so that if $[g] = [a_{m-1}x^{m-1} + \dots a_1x + a_0]$ then $a_i = 0$ for all $i < m-n$. This will ensure that $\phi_g$ sends $(x^n)$ to zero, and hence our homomorphisms are exactly these multiplication by $g$ maps.

It is not difficult to see that the map $\text{Hom}_{k[x]}(k[x]/(x^n),k[x]/(x^m)) \to k[x]/(x^{n})$ by sending the element $\phi_g$ to the polynomial $g$ after shifting its degrees down by $m-n$ steps is a module isomorphism.

So far we have assumed that $m > n$. If $m \leq n$ then $g$ is allowed to be any polynomial, and hence the isomorphism still holds.

$\mathbf{Z}/(n)\otimes_{\mathbf{Z}}\mathbf{Z}/(m) \cong \mathbf{Z}/(\gcd(n,m))$

First, notice that the tensor product is generated by $1\otimes 1$ and is therefore a cyclic module.
\begin{align*}
 n(1\otimes 1) &= n\otimes 1 = 0 \\
 m(1\otimes 1) &= 1\otimes m = 0
\end{align*}
Therefore, $\gcd(m,n)(1\otimes 1) = 0$ and thus the following map is well-defined:
\begin{align*}
 \mathbf{Z}/(\gcd(n,m))&\to \mathbf{Z}/(n)\otimes_{\mathbf{Z}}\mathbf{Z}/(m) \\
 [k] &\mapsto k (1\otimes 1)
\end{align*}
To show that it is an isomorphism, we will construct its inverse by invoking the universal property of the tensor product. Basically, it suffices to see that the following map is $\mathbf{Z}$-bilinear.
\begin{align*}
 \mathbf{Z}/(n)\times\mathbf{Z}/(m) &\to \mathbf{Z}/(\gcd(n,m)) \\
 (k,l) &\mapsto [kl]
\end{align*}
Let us just say we are done, because I do not want to belabor this point.

$k[x]/(x^n) \otimes_{k[x]} k[x]/(x^m) \cong k[x]/(x^{\min(n,m)})$ as $k[x]$-modules.

First, we notice that $k[x]/(x^n) \otimes_{k[x]} k[x]/(x^m)$ is a cyclic module generated by $[1]\otimes [1]$.

In one direction, we send $[f]\in k[x]/(x^{\min(n,m)})$ to $[f]([1]\otimes [1])$ which is well-defined by the bilinearity of the tensor product. In the other direction, we define a $k[x]$-bilinear map as follows.
\begin{align*}
 k[x]/(x^n) \times k[x]/(x^m) &\to k[x]/(x^{\min(n,m)}) \\
 [f]\times [g] &\mapsto [fg] \ (\text{mod } (x^{\min(n,m)}))
\end{align*}
This is well-defined, and everything works (I'm pretty sure) so we finish by just calling the universal property.

$k[x]\otimes_k k[x] \cong k[x,y]$ as a $k$-algebra.

We write $k[x]\otimes_k k[y]$ for better clarity. As modules, we see that:
\begin{align*}
 k[x]\otimes_k k[y] &\cong k[x]\otimes_k \oplus_{n=1}^\infty ky^n \\
 &\cong \oplus_{n=1}^\infty (k[x]\otimes_k ky^n) \tag{commute the sum and tensor} \\
 &\cong \oplus_{n=1}^\infty (k[x]\otimes_k k) \\
 &\cong \oplus_{n=1}^\infty k[x]
\end{align*}
However, for each factor in the direct sum, the multiplicative structure is basically $(\ast)$ given by adding the monomial $y^n$ and hence as $k$-algebras we have that $k[x]\otimes_k k[x]\cong (k[x])[y]$

$(\ast)$ Is there an easy way to make this rigorous, without writing out elements explicitly? Is this even correct?

Thanks!

答:
I think everything is okay.

For the first item, note also that $\mathbb Z/(m) [n]$ consists of the classes $[k]$ for which $m \mid nk$. Writing $d = \gcd(m,n)$ and $m = dx, n = dy$, this is equivalent to having $x \mid yk$. But since $x$ and $y$ are coprime, moreover this it the same as $x \mid k$.

Hence $\mathbb Z/(m)[n] = \{[k] \colon k \in (m/\gcd(n,m))\mathbb{Z}\} = (m/\gcd(n,m))\mathbb{Z}/m\mathbb{Z} = \mathbb{Z}/(\gcd(n,m))$.

Next; if $I,J$ are ideals of a commutative ring $R$, then $R/I \otimes R/J = R/I+J$, see here. In PIDs such as $\mathbb{Z}$ and $k[x]$ we get $(a)+(b) = (\gcd(a,b))$ which coincides with what you have done.

For the latter exercise, here is a nonsensical argument: in general if $A$ and $B$ are commutative $k$-algebras, then

$$
\hom(A \otimes_k B, C) \simeq \hom(A,C) \times \hom(B,C).
$$
naturally in $A,B$ and $C$.

The $\hom$ is the $k$-algebra one. On the other hand, if $S$ is a finite set then the polynomial algebra on $S$ satisfies
$$
\hom(k[S],C) = \prod_{s\in S}C
$$
and thus we have a natural isomorphism

$$
\hom(k[x]\otimes_k k[y],C) \simeq \hom(k[x],C)\times \hom(k[y],C) \simeq C \times C \simeq \hom(k[x,y],C).
$$
We can then apply the Yoneda lemma.


问:
Show $f(x)=x^2+1$ is irreducible over $\mathbb{Q} (i\sqrt{2})$
I am wanting to show $x^2+1$ is irreducible over $\mathbb{Q} (i\sqrt{2})$, however, I'm a little stuck.

Here's my attempt: Suppose, for contradiction's sake, that $x^2+1$ is reducible, and factors as $(x-\alpha)(x-\beta)$. We can expand this as $x^2-(\alpha + \beta)x + \alpha\beta$, or in other words, obtain the system:

$$\alpha\beta = 1, \alpha + \beta = 0$$

Is this the right way to go? Is there a contradiction reached, and why?

答:
Let $a+bi\sqrt2$ be a root of $x^2+1$, so $(a+bi\sqrt2)^2 = -1$, i.e. $(a^2-2b^2)+2abi\sqrt2 = -1$.

Equate coefficients and derive a contradiction.


答:
Hint: $x^2+1$ has a unique factorization in an appropriate extension field, say $\Bbb Q(i)$. Then $x^2+1 = (x+i)(x-i)$.

You have to show that $i\not\in \Bbb Q(\sqrt 2 i)$.

Title: Show $f(x)=x^2+1$ is irreducible over $\mathbb{Q} (i\sqrt{2})$

Body:
I am wanting to show $x^2+1$ is irreducible over $\mathbb{Q} (i\sqrt{2})$, however, I'm a little stuck.

Here's my attempt: Suppose, for contradiction's sake, that $x^2+1$ is reducible, and factors as $(x-\alpha)(x-\beta)$. We can expand this as $x^2-(\alpha + \beta)x + \alpha\beta$, or in other words, obtain the system:

$$\alpha\beta = 1, \alpha + \beta = 0$$

Is this the right way to go? Is there a contradiction reached, and why?


问:
Possible mistake in book Indra's Pearls regarding setup for finding Hausdorff dimension of Schottky limit sets
I'm reading the book Indra's Pearls and on page 138 it says:

"Let $W$ be a reduced word which
we suppose for example doesn't end with $B$. If we let $L_w$ be that part
of the limit set (dust) contained in the disk $D_w$, then we can express
the splitting into finer levels by saying that
$L_W = L_{Wa} \cup L_{WA} \cup L_{WB}$. Let's suppose also that $W$ doesn't begin with $A$. Then we can expand everything using $a$ to get
$a(L_w) =L_{aw} = a(L_{wa}) \cup a(L_{wA}) \cup a(L_{wB})$.
"

I think that there is a mistake here, namely that it should be that the reduced word $W$ DOES end with $B$. Below I'll give some more context for people who are unfamiliar with the book and explain my reasoning for why I think this is a mistake. I do have to note that the book is very consistent with this mistake if it is one, as I have seen this again on page 156.

We have four disjoint Schottky disks $D_a, D_b, D_A$ and $D_B$ arranged like this: http://prntscr.com/10drwmt and four möbius transformations $a, b, A=a^{-1}$ and $B=b^{-1}$ which each map the outside of the disk with label the inverse of the corresponding map to the inside of the disk with as label the corresponding map; and they map the inside of the disk with as label the inverse of the corresponding map to the outside of the disk with as label the corresponding map. So for example: the outside of the disk $D_A$ is mapped by $a$ to the inside of the disk $D_a$ and the inside of the disk $D_A$ is mapped to the outside of the disk $D_a$. The inverses $A$ and $B$ behave just like the maps $a$ and $b$ as for example the outside of the disk $D_a$ is mapped by $A$ to the inside of the disk $D_A$ and the inside of the disk $D_a$ is mapped to the outside of the disk $D_A$. Applying these maps after one another and focusing on what happens with certain starting points we see that points get mapped into smaller disks inside the original disks $D_a, D_b, D_A$ and $D_B$. For example the map $aB=a(B)$ first maps a point outside of disk $D_b$ into disk $D_B$, and then maps this point to the inside of disk $D_a$, so we see that this map sends points into a smaller disk which in the book is denoted as $D_{aB}\subset D_a$.

Combinations of the maps $a, b, A$ and $B$ are called words in the book, so for example $a(b(a(B(B(b)))))=abaBBb$ is a word. A reduced word is a word where all operations which 'do nothing' are filtered out. So $abaB$ is a reduced word of $abaBBb$ as the $Bb=Id$ has been filtered out.

So to come back to my confusion: if $W$ is a reduced word which does not end with $B$ then it could end with $a$ for example. So we could consider the word $W=a$. Filling this into what is stated in the book we get that for the part of the limit set contained in disk $D_a$ holds that $L_a=L_{aa}\cup L_{aA}\cup L_{aB}$, but $L_{aA}=a(L_A)$ is the image set of the part of the limit set which is contained in disk $D_A$ obtained by applying the map $a$ on it, which means that this image set is in the EXTERIOR of $D_a$. The image sets $a(L_B)$ and $a(L_a)$ on the other hand are in the interior of $D_a$. I think the authors were trying to say that for some word $W$ the parts of the limit set in disk $D_W$ equals the part of the limit set which is inside the three disks which are inside this disk $D_W$ (note that in the fragment above they state "splitting into finer levels"). This way it is possible to set up some way to find the Hausdorff dimension (this is not done explicitly in the book but they stated the Hausdorff dimension and claimed they found it via an algorithm by McMullen's which is based on the idea they did explain). Either way, I think this word $W$ should end explicitly in $B$ to make sure that this idea works and we don't get any cases where parts of the limit set not in $D_a$, like $a(L_A)$, are part of the given equation $a(L_w) =L_{aw} = a(L_{wa}) \cup a(L_{wA}) \cup a(L_{wB})$. I'm also thinking that this is a mistake as they didn't add an $L_{Wb}$ term and only for that term we would have the same problem if the word $W$ would end explicitly in $B$, namely we would then have that $B(L_b)$ is in the exterior of $D_B$ while the other parts of the equation are in the interior of $D_B$. The only logical conclusion why the authors excluded this case I can come up with is that the word $W$ does end in $B$.

Is my reasoning correct and is this indeed a mistake? Or is the book correct and have I made some mistake in my reasoning? Thanks in advance!

答: Yes, it should be "$W$ does end with $B$".  The decomposition of $L_W$ consists of $L_{Wa}$, $L_{WA}$, $L_{WB}$, but not $L_{Wb}$. That is exactly right if $W$ ends with $B$, because $Wb$ would then end with $Bb$ so it would not be in reduced form, so we disallow it.


问:
Is $\frac{dx(t-T)}{d(t-T)}=\frac{dx(t-T)}{dt}$? Where T is a constant and t is an independent variable.
Given a function x(t) and a constant T, is the following relation true?:
$$\frac{dx(t-T)}{d(t-T)}=\frac{dx(t-T)}{dt}$$

If it is true, what is the proof?

Also, if this is true, does this mean that this statement is wrong?:
$$\frac{dx(t-T)}{d(t-T)}=\frac{dx(z)}{dz}=\frac{dx(t)}{dt}$$ where $z=t-T$.

Here is a snippet from Example 1.11 of Linear Systems and Signals by B. P. Lathi and Roger Green using this relation:

Applying input $x(t)$ to the system produces output $y(t) = \frac{d}{dt}x(t)$; delaying
this output by T produces $y(t − T) = \frac{d}{d(t−T)}x(t − T) = \frac{d}{dt}x(t − T)$. This is just the output of
the system to a delayed input $x(t−T)$. Since the T-delayed output of the system to input $x(t)$
equals the output of the system to the T-delayed input $x(t−T)$, the system is time invariant.

This part refers to proving the system $y(t)=\frac{d}{dt}x(t)$ to be time invariant.

答:
The proof is the chain rule, using $$\frac{d(t-T)}{dt} = 1\qquad\text{or its consequence}\qquad \frac{dt}{d(t-T)} = 1$$

Yes,
$$
\frac{dx(z)}{dz}=\frac{dx(t)}{dt}
$$
is wrong, where $z=t-T$.  Unless $x(t)$ is periodic with period $T$, then it is right.


问:
Is $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2}+\sqrt{3})$?

Is $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2}+\sqrt{3})$ ?

$$\mathbb{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \mid a,b,c,d\in\mathbb{Q}\}$$

$$\mathbb{Q}(\sqrt{2}+\sqrt{3}) = \lbrace a+b(\sqrt{2}+\sqrt{3}) \mid a,b \in \mathbb{Q} \rbrace  $$

So if an element is in $\mathbb Q(\sqrt{2},\sqrt{3})$, then it is in $\mathbb{Q}(\sqrt{2}+\sqrt{3})$, because $\sqrt{6} = \sqrt{2}\sqrt{3}$.

How to conclude from there?

答:
To recap the notation: $\mathbb{Q}[x]$ denotes the ring of polynomials with rational coefficients. The square bracket notation $\mathbb{Q}[\sqrt{2}]$ means $\{p(\sqrt{2}) : p \in \mathbb{Q}[x]\}$. It's easy to show that $\mathbb{Q}[\sqrt{2}] = \{a+b\sqrt{2}:a,b,\in \mathbb{Q}\}.$

A really nice fact is that $\mathbb{Q}[\sqrt{2},\sqrt{3}] = \mathbb{Q}[\sqrt{2}][\sqrt{3}],$ where
\begin{array}{ccc}
\mathbb{Q}[\sqrt{2}][\sqrt{3}] &=& \{a+b\sqrt{3} : a,b \in \mathbb{Q}[\sqrt{2}] \} \\ \\
&=& \{p + q\sqrt{2} + r\sqrt{3} + s\sqrt{6} : p,q,r,s \in \mathbb{Q}\}.
\end{array}
These all use square brackets because they are considered as rings. The round brackets give us the set of rational expressions, which are fields, e.g.

$$\mathbb{Q}(\sqrt{2},\sqrt{3}) = \left\{ \frac{\alpha}{\beta} : \alpha,\beta \in \mathbb{Q}[\sqrt{2},\sqrt{3}]\right\}$$

It turns out that, as sets, $\mathbb{Q}[\sqrt{2},\sqrt{3}] = \mathbb{Q}(\sqrt{2},\sqrt{3})$.

In turns of the representation of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ we have seen that, as a set, we have $\{p + q\sqrt{2} + r\sqrt{3} + s\sqrt{6}:p,q,r,s \in \mathbb{Q}\}$. There are many representations for this fiels, e.g. $\mathbb{Q}(1,\sqrt{2},\sqrt{3},\sqrt{6})$ or $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{6})$ or $\mathbb{Q}(1,\sqrt{2},\sqrt{3})$ or $\mathbb{Q}(\sqrt{2},\sqrt{3})$, etc. We can show that $\mathbb{Q}(\sqrt{2}+\sqrt{3})$ is also a representation of the same field too.

Think of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ as a $\mathbb{Q}$-vector space with $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ as a basis. Let $\gamma := \sqrt{2}+\sqrt{3}.$ We have $\gamma^2 = 5+2\sqrt{6},$ $\gamma^3 = 11\sqrt{2}+9\sqrt{3}$ and $\gamma^4 = 49 + 20\sqrt{6}$. Putting this together:

$$\left[\begin{array}{cccc} 0 & 1 & 1 & 0 \\ 5 & 0 & 0 & 2 \\ 0 & 11 & 9 & 0 \\ 49 & 0 & 0 & 20 \end{array}\right]\left[\begin{array}{c} 1 \\ \sqrt{2} \\ \sqrt{3} \\ \sqrt{6} \end{array}\right] = \left[\begin{array}{c} \gamma \\ \gamma^2 \\ \gamma^3 \\ \gamma^4 \end{array}\right]$$

The 4-by-4 matrix on the left is non-singular, and so we can invert:

$$\left[\begin{array}{c} 1 \\ \sqrt{2} \\ \sqrt{3} \\ \sqrt{6} \end{array}\right] = \frac{1}{2}\!\left[\begin{array}{cccc} 0 & 20 & 0 & -2 \\ -9 & 0 & 1 & 0 \\ 11 & 0 & -1 & 0 \\ 0 & -49 & 0 & 5 \end{array}\right]\left[\begin{array}{c} \gamma \\ \gamma^2 \\ \gamma^3 \\ \gamma^4 \end{array}\right]$$

This tells us that $1$, $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{6}$ can all be expressed as rational polynomials in $\gamma = \sqrt{2}+\sqrt{3}$.

\begin{array}{ccc}
10\gamma^2-\gamma^4 &=& 1 \\
\tfrac{1}{2}(\gamma^3-9\gamma) &=& \sqrt{2} \\
\tfrac{1}{2}(11\gamma - \gamma^3) &=& \sqrt{3} \\
\tfrac{1}{2}(5\gamma^4-49\gamma^2) &=& \sqrt{6}
\end{array}

It follows that $\mathbb{Q}(1,\sqrt{2},\sqrt{3},\sqrt{6}) \cong \mathbb{Q}(\gamma) = \mathbb{Q}(\sqrt{2}+\sqrt{3}).$


问:
$\bigwedge^{2}T\cong\mathbb{Z}/2\mathbb{Z}$
Hello I have the next issue

Consider $T$ the subgroup of diagonal matrices of $GL(2, 3) $ (group of invertible matrices of $2$ by $2$ over a field of three elements)

It is well known that $T$ is isomorphic to the Klein four group, therefore

$\mathbb{Z}/2\mathbb{Z} \cong H_{2}(T, \mathbb{Z})\cong\bigwedge^{2}T $

I would like to give isomorphisms from $\mathbb{Z}/2\mathbb{Z}$ to $\bigwedge^{2}T$

Any hint would be helpful, thank you!

答:
Fix your well known iso $T \simeq \mathbb{Z}_2 \oplus \mathbb{Z}_2$. You get for free that this is a $\mathbb{Z}_2$-vector space isomorphism; hence you obtain a basis $\{\alpha, \beta\}$ for $T$.

Now map

$$
1 \in \mathbb{Z}_2 \mapsto \alpha \wedge \beta.
$$

This map is a $\mathbb{Z}_2$-linear monomorphism (the wedge on the RHS is a basis element of the exterior algebra) and by a dimension argument it must be an iso.


答:
You are on the right track assuming it's irreducible, but you have not reached a contradiction yet. This is because you need to use what elements in $\mathbb{Q}(i\sqrt{2})$ are. Notice that any element in $\mathbb{Q}(i\sqrt{2})$ can be written as $a + ib\sqrt{2} $ for $a,b \in \mathbb{Q}$ (why?).

So for an element in $a + ib\sqrt{2} \in \mathbb{Q}(i\sqrt{2})$ to be a solution to the equation, we must have

$-1= (a + ib\sqrt{2})^2= a^2-2b^2+iab\sqrt{2}$

From this equation, we first get that $ab = 0$. Secondly, if $b=0$, we have $-1=a^2$, which is impossible. So $a=0$. Therefore we get

$-1=-2b^2 \implies b^2=\frac{1}{2}$

which is impossible.


问:
If $x,y,z>0$ with $x+y+z=\sqrt{3}$, prove that: $\sqrt{x^2+1-yz}+\sqrt{y^2+1-zx}+\sqrt{z^2+1-xy}\ge 3$ (Venezuela 1960)

If $x,y,z>0$ with $x+y+z=\sqrt{3}$, prove that: $$\sqrt{x^2+1-yz}+\sqrt{y^2+1-zx}+\sqrt{z^2+1-xy}\ge 3$$

I tried solving it as follows:

The condition we want proved is: $\sqrt{3x^2+3-3yz}+\sqrt{3y^2+3-3zx}+\sqrt{3z^2+3-3xy}\ge 3(x+y+z)$

$\sqrt{3x^2+(x+y+z)^2-3yz}+\sqrt{3y^2+(x+y+z)^2-3xz}+\sqrt{3z^2+(x+y+z)^2-3xy}\ge 3(x+y+z)$

And this is where I got stuck. I couldn't continue it. Could you please explain to me how to solve the question?

答:
The inequality ia actually very easy with just one hidden point kept in mind

Note that $$\sum \sqrt{x^2+1-yz}\ge \sum \sqrt{x^2+1-\frac{{(y+z)}^2}{4}}$$ Now use the fact $$x^2+1-\frac{{(y+z)}^2}{4}=x^2+\frac{{(x+y+z)}^2}{3}-\frac{{(y+z)}^2}{4}=\frac{1}{12} {(4x+y+z)}^2$$


问:
Find $x_{n}=\frac{5(-1)^{n}-3}{4(-1)^{\left \lfloor \frac{n}{2} \right \rfloor}n+2}, n\in \mathbb{N}$ subsequential limits
Given sequence $x_{n}=\frac{10(-1)^{n}*n-4}{4(-1)^{\left \lfloor \frac{n}{2} \right \rfloor}n+2}, n\in \mathbb{N}$. How to find all of the subsequential limits of this sequence?

答:
Answer

The answer is that the elements

$$x_{n}=\frac{5 (-1)^n-3}{4 n (-1)^{\left\lfloor \frac{n}{2}\right\rfloor }+2}, n=0,1,2,...\tag{1}$$

create 4 different sequences, and all of them have the limit $0$ for $n\to\infty$.

But the problem deserves a more thorough study.

Revealing the 4 sequences

The first 10 terms of the sequence are

$$\{x_{n}\}_{n=0}^{n=10} = \left\{1,-\frac{4}{3},-\frac{1}{3},\frac{4}{5},\frac{1}{9},-\frac{4}{11},-\frac{1}{11},\frac{4}{13},\frac{1}{17},-\frac{4}{19},-\frac{1}{19}\right\}\tag{2}$$

These look rather confusing, a pattern would be hard to guess.

Inspecting the formula for $x_{n}$ we focus on the terms $c(n)=(-1)^{n}$ and $d(n) = (-1)^{\left \lfloor \frac{n}{2} \right \rfloor}$ and study the possible parities of the index $n$.

$n$ can be even or odd, for $n=2m$ we have $c(2m)=+1$ and $d(2m)=(-1)^{m}$, and for $n=2m-1$ we have $c(n)=(-1)$ and again $d(2m-1)=(-1)^{m}$. Now in $m$ in turn can be even or odd, leading to 4 cases for $n$.

These cases can be parametrized by letting $n=4m+k$, with $m=0,1,2,...$ and $k=0,1,2,3$.

Inserting this we find

$$x_{n} =x_{m,k}:= \frac{5(-1)^{4 m+k}-3}{4(-1)^{\left \lfloor \frac{4m+k}{2} \right \rfloor}(4m+k+2)}=\frac{5(-1)^{k}-3}{4(-1)^{\left \lfloor \frac{k}{2} \right \rfloor}(4m+k+2)}\tag{3}$$

explcitly we find for the four sequences

$$\begin{align}\{x_{m,k}\}_{k=0}^{3}
& =\left\{\frac{1}{8 m+1},-\frac{4}{8 m+3},\frac{1}{-8 m-3},\frac{4}{8 m+5}\right\}, m=0,1,2,...\\
& =\left\{\frac{1}{2 n+1}|_{n=4m},\frac{-4}{2n+1}|_{n=4m+1},
\frac{-1}{2n-1}|_{n=4m+2},\frac{-4}{2n-1}|_{n=4m+3}\right\}, \\& n=0,1,2,...
\end{align}\tag{4}$$

Obvioulsly, the limit for $m\to\infty$ is zero for all sequences.

Plot of the 4 sequences

Generating function

The generating function of the sequence $x_{n}$ is defined as

$$g(z) = \sum_{n=0}^{\infty} z^n x_{n}\tag{5}$$

Taking care of the four sequences we have here

$\begin{align}
g(z) =& \sum _{k=0}^3 \left(\sum _{m=0}^{\infty } x(k+4 m) z^{k+4 m}\right)\\
& =\, _2F_1\left(\frac{1}{8},1;\frac{9}{8};z^4\right)\\
& -\frac{4}{3} z \, _2F_1\left(\frac{3}{8},1;\frac{11}{8};z^4\right)\\
& -\frac{1}{3} z^2 \, _2F_1\left(\frac{3}{8},1;\frac{11}{8};z^4\right)\\
& +\frac{4}{5} z^3 \, _2F_1\left(\frac{5}{8},1;\frac{13}{8};z^4\right)
\end{align}\tag{6}$

here $\, _2F_1\left(a,b;c;z\right)$ is a hypergeometric function.

Expanding $g(z)$ in a power series around $z=0$ gives

$$g(z) = 1-\frac{4}{3}z-\frac{1}{3}z^2+\frac{4}{5}z^3+\frac{1}{9}z^4-\frac{4}{11}z^5-\frac{1}{11}z^6+\frac{4}{13}z^7+\frac{1}{17}z^8-\frac{4}{19}z^9-\frac{1}{19}z^{10}+...$$

and reproduces the sequence $(2)$.

Here is a plot of the generating function

The plot shows that $g(z)$ is finite in the range $-1\le z \le 1$

The boundary values which are at the same time the values of the series  and the alternating series of the $x_{n}$, respectively, are

$$\begin{align}
g_1 = g(z=1) &= \sum_{n=0}^{\infty}x_{n}\\
& =\frac{1}{8} \left(5 \pi +2 \sqrt{2} \left(\coth ^{-1}\left(\sqrt{2}\right)-2 \pi \right)\right)\\
&\simeq 0.0536666
\end{align}\tag{7a}$$

$$\begin{align}
g_{-1}=g(z=-1) &= \sum_{n=0}^{\infty}(-1)^n x_{n}\\
& = \frac{1}{8} \left(2 \sqrt{2} \left(2 \pi +\coth ^{-1}\left(\sqrt{2}\right)\right)-3 \pi \right)\\
& \simeq 1.35496
\end{align}\tag{7b}$$

Finally, we can also calculate the integral over the generating function

$$i= \int_{-1}^{1} g(z)\,dz = \frac{\pi }{3 \sqrt{2}}+\frac{\log (2)}{3}+\frac{\log \left(17+12 \sqrt{2}\right)}{3 \sqrt{2}} \simeq 1.8025\tag{8}$$

Summarising, we have discovered several interesting relations and rather smooth functions and unexpected explicit results from the bizarre looking sequence term $x_{n}$.


问:
secant method in maple

secant method in maple. Find a root of the statement $x^3-3x^2+4x-1=0$ with the initial value $x_0=0$ and $x_1=1$ with 5 digits point approximation.

Title: secant method in maple

Body:
another one:

eps_step := 0.00001;

eps_abs := 0.00001;

f := x -> x^3-3*x^2+4*x-1;

x[0] := 1.0;

x[1] := 1.5;

for i from 2 to 100 do

x[i] := (x[i - 2]*f(x[i - 1]) - x[i - 1]*f(x[i - 2]))/(f(x[i - 1]) - f(x[i - 2]));

if abs( x[i] - x[i - 1] ) < eps_step and abs( f( x[i] ) )  < eps_abs then

eps_abs then break;

elif i = 100 then

end if;

end do:

evalf(x[i],5);

eps_step := 0.00001

eps_abs := 0.00001

                 f := x -> x^3-3*x^2+4*x-1

                     x[0] := 1.0

                     x[1] := 1.5

                        0.31767


问:
Solve for $x,y$ : $x^4+y^4+2 = 4xy$

Solve for $x,y$ : $x^4+y^4+2 = 4xy$

Obviously, the very obvious solutions are $(1,1)$ and $(-1,-1)$.

But I don't know how to reach the answer. I got the answer through hit and trial.

I tried adding and subtracting $2x^2y^2$ in the $LHS$:

$x^4+y^4+2 +2x^2y^2-2x^2y^2 = 4xy$

$(x^4+y^4)^2 = 4xy + 2x^2y^2 -2$

Giving us:

$(x^4+y^4)^2 =2(2xy + x^2y^2 -1)$

How to proceed further? I don't really think this helps. Is there a better method? Any hints or solutions would be appreciated!

答:
Hint:

Using AM-GM  inequality for real $x^2,y^2\ge0$

$$\dfrac{x^4+y^4+1+1}4\ge\sqrt[4]{x^4y^4}$$

the equality occurs if $x^4=y^4=1$

Alternatively,

$$x^4+y^4+2-4xy=(x^2-y^2)^2+2(xy-1)^2$$

Title: Solve for $x,y$ : $x^4+y^4+2 = 4xy$


问:
Group object in the category of formal schemes vs. formal group law
I've been trying to see how both definitions of a formal group coincide:

Take a ring $ R $, a group object in the category of formal schemes over $ R $ is then given by two morphims $ e : R \rightarrow X $, $ \mu : X \times X \rightarrow X $.

For $ X = \widehat{\mathbb{A}}^1_R $, this amounts to two morphisms of $R$-adic algebras $ e : R[[x]] \rightarrow R $ and $ \mu : R[[x]] \rightarrow R[[x,y]] $, but it is easy to see that $ \mathrm{Hom}_R(R[[x]], R) \simeq \mathrm{Nil}(R) $, so there might be more than $1$ possible choice for $ e $. Yet, in the definition of a formal group law, unitality is then stated as $ \mu(x) = x + y + \text{higher terms} $, but if for example $ R = \mathbb{Z}/4\mathbb{Z} $, take $ e : X \mapsto 2 $, then unitality reduces to $ \mu(x)(2, Y) = Y, \mu(x)(X, 2) = X $, but then $ \mu(x) = x + y -2 $ satisfies unitality and associativity.

In the case of $ R $ a field, there is of course no longer any problem, but I've seen the formal group law definition stated for $ R $ in this way.

So is it that we require formal group laws to actually have a fixed unit morphism?

答: I think in the context of formal group laws, one defines adic algebras over $R$ to be augmented $R$-algebra $\eta:A \to R.$ The topology on $A$ is the adic topology generated by $\text{Ker}(\eta),$ which is sometimes also assumed to be complete and separated. The maps between such adic algebras are required to be compatible with this data of the augmentation which fixes the problem you mentioned.


问:
Inverse number using gcd
If I want to find the inverse number of $5$ $mod448$ using $gcd$.
I try to do like that:

$448=89\cdot5+3$

$gcd(5,448)=gcd(89,3)$

$89=3\cdot29+2$

$gcd(89,3)=gcd(29,2)$

$1=29-2\cdot14$

$1=29-14\cdot(89-29\cdot3)$

$1=43\cdot29-14\cdot89$

But here I don't know how to continue, maybe I have mistake...

答:
The second row is $5 = 1\cdot 3 + 2$ and the third one, $3 = 1\cdot 2 + 1$.

From here you have to calculate backwards.

$1 = 3 - 2 = 3 - (5-3) = 2\cdot 3 - 5 = 2\cdot (448 - 89\cdot 5) - 5 = 2\cdot 448 - 179\cdot 5$.

From here you see that $179\cdot 5\equiv 1\mod 448$. So $5^{-1}= 179$ in $\Bbb Z_{448}$.

Title: Inverse number using gcd


问:
Please correct me if the following Relations are wrong.
If $A=\{1,2,3,4\},~B=\{4,5\}$ and $C=\{5,6\}$ find:

$A\times (B-C)$

$(A\times B)-(A\times C)$

$(A\times B)\cap (A\times C)$

My answers that I came up with were:

$\{(1,4),(2,4),(3,4),(4,4),(1,5),(1,5),(3,5),(4,5)\}$

$\{(1,4),(2,4),(3,4),(4,4),(1,6),(2,6),(36),(4,6)\}$

$\{(1,5),(1,5),(3,5),(4,5)\}$

Please correct me if I am wrong in any of these answers.

Editor's note: originals found here https://i.stack.imgur.com/sPMvp.png
https://i.stack.imgur.com/POZon.png

答:
$A \times (B -C)$:

First $B-C = \{4,\color{red}5\} - \{\color{red}5, \color{purple}6\} = \{4\}$

So $A \times (B-C) = \{1,2,3,4\} \times \{4\}= \{(1,4),(2,4),(3,4),(4,4)\}$

Why did you include any with $5$ in the second term?

$(A\times B) - (A\times C)$:

This means we can have any $(x,y)$ so that $x\in A;y\in B$ so long as we do not have both $x\in A$ and $y\in C$.  As $C$ contains $5$ and $6$ we can't have any $x\in A; y = 5$ or $y= 6$.  The only $y \in B$ that are not $5$ or $6$ are $y=4$ so we can only have $(x,y)$ there $x \in A$ and $y = 4$.  And we can have all of those.

..... or more directly

$$A \times B = \{(1,4),(2,4),(3,4),(4,4),\\
(1,\color{red} 5),(2,\color{red} 5),(3,\color{red} 5),(4,\color{red} 5)\}$$

but we must exclude $$A\times C = \{(1,\color{red} 5),(2,\color{red} 5),(3,\color{red} 5),(4,\color{red} 5),\\
(1,\color{purple} 6),(2,\color{purple} 6),(3,\color{purple} 6),(4,\color{purple} 6)\}$$

So that leaves $A\times B - A\times C = \{(1,4),(2,4),(3,4),(4,4)\}$

$(A\times B)\cap (A\times C)$:

Again $$A \times B = \{(1,4),(2,4),(3,4),(4,4),\\
(1,\color{red} 5),(2,\color{red} 5),(3,\color{red} 5),(4,\color{red} 5)\}$$

and $$A\times C = \{(1,\color{red} 5),(2,\color{red} 5),(3,\color{red} 5),(4,\color{red} 5),\\
(1,\color{purple} 6),(2,\color{purple} 6),(3,\color{purple} 6),(4,\color{purple} 6)\}$$

so we must take what the have in common:

$(A\times B)\cap (A\times C)=\{(1,\color{red} 5),(2,\color{red} 5),(3,\color{red} 5),(4,\color{red} 5)\}$


答:
From your equation $x^4+y^4+2x^2y^2-2x^2y^2+2=4xy$ it follows
$$x^4+y^4-2x^2y^2=4xy-2-2x^2y^2 \Longleftrightarrow
(x^2-y^2)^2=-2(xy-1)^2$$

Over the real numbers this is possible if and only if
$$x^2=y^2 \text{ and } xy=1$$
which has only $(1,1)$ and $(-1,-1)$ as solutions.

Title: Solve for $x,y$ : $x^4+y^4+2 = 4xy$

Body:
Solve for $x,y$ : $x^4+y^4+2 = 4xy$

Obviously, the very obvious solutions are $(1,1)$ and $(-1,-1)$.

But I don't know how to reach the answer. I got the answer through hit and trial.

I tried adding and subtracting $2x^2y^2$ in the $LHS$:

$x^4+y^4+2 +2x^2y^2-2x^2y^2 = 4xy$

$(x^4+y^4)^2 = 4xy + 2x^2y^2 -2$

Giving us:

$(x^4+y^4)^2 =2(2xy + x^2y^2 -1)$

How to proceed further? I don't really think this helps. Is there a better method? Any hints or solutions would be appreciated!


Title: Solve for $x,y$ : $x^4+y^4+2 = 4xy$

Body:
After realizing that $x^2, y^2≥0$, then

$$x^4+1≥2x^2$$

$$y^4+1≥2y^2$$

$$x^4+y^4+2≥2(x^2+y^2)$$

$$x^4+y^4+2-4xy≥2(x^2+y^2)-4xy=2(x-y)^2≥0$$

So, the solution occurs for only $x=y$, which follows $x=y=±1.$


问:
Is uniform convergence equivalent to Cauchy sequences for the space of continuous functions on an interval?
Having received a perfect answer to this question, I am left trying to revisit what I thought I grasped at an intuitive level.

The present question asks for clarification on the comment:

The example is not using uniform convergence. A uniform limit of continuous functions is continuous. Because of that $C[a,b]$ is a Banach algebra when equipped with the infinity norm.

The broader concept for me to come to terms with is whether this example, which is clearly not a Cauchy sequence as per the answer to my prior question, is still a valid example for non-completeness, and if so, in what respect does it fail to contradict the idea that $C[a,b]$ is a Banach space.

This is the example (to avoid having to cross-reference the prior question):

$$f^{k}(x) = \frac 1 2 +\frac 1 \pi \tan^{-1}(kx)$$

答:
The example is pretty bad. The lecturer wants to describe non-completeness and tries to use this example. From the little I saw in the video, he doesn't discuss what topology he is using. It looks like he is using the vector space of all continuous functions on the real line, which makes no sense as the "problem" with his example only  occurs at zero. And it looks like he is thinking of pointwise convergence.

With pointwise convergence, the same pathology can be found on $C[0,1]$ if one uses the usual example of $f_n(x)=x^n$.


问:
Monty hall with $n$ doors and Monty opening $m$ doors
In Monty Hall problem generalized to $n$ doors, the Monty hall's problem was generalized to $n$ doors with Monty opening $m$ of them, where $m < n - 1$.

The probability of winning by switching is $\frac{n - 1}{n(n - m - 1)}$. I am confused about what the probability of winning by not switching is in this case. Is it $\frac{1}{n}$, $\frac{1}{n - m}$, or $1- \frac{n - 1}{n(n - m - 1)}$?
I think it's $\frac{1}{n}$, but this doesn't seem to incorporate monty opening $M$ doors.

答:
The probability of winning by not switching is just the probability of picking the door with the car at your initial guess.

Since there are $n$ doors, the car is behind one unique door and you only choose one door, the probability of winning without switching is just
$$\frac{1}{n}.$$


问:
Conjugacy Class of $(2 1)$ in $S_4$
What are the conjugacy classes of $(21)$ in $S_4$?

I think they are $\{(12),(13),(14),(23),(24),(34)\}$

But I'm not sure and the centralizer of $(12)$ is $\{(21),(34)\}$. Is this right?

答:
Conjugation of permutations preserves cycle type, so, yes, the conjugacy class is correct.

Note that the identity is in the centraliser too. It is also closed (as it is a subgroup of $S_4$), so the product $(12)(34)$ is in there as well.

Title: Conjugacy Class of $(2 1)$ in $S_4$


问:
Finite difference central scheme equivalent for non centered point
I need to use a central difference scheme:

$$\frac{dy}{dx}=\frac{y_{i+1}-y_{i-1}}{2 \Delta x}$$

equivalent for non centered point:

Unfortunately I cannot find a book in which I have this formula. Could anybody give me this in answer?

答:
Actually you can use Taylor expansion to derive the formula
$$y_{i-1}=y(x-\Delta x_i)=y(x)-\frac{dy}{dx}\Delta x_i+O(\Delta x^2)$$
$$y_{i+1}=y(x+\Delta x_{i+1})=y(x)+\frac{dy}{dx}\Delta x_{i+1}+O(\Delta x^2)$$
By neglecting higher order terms $O(\Delta x^2)$
$$y_{i+1}-y_{i-1}=\frac{dy}{dx}\Delta x_{i+1}+\frac{dy}{dx}\Delta x_i\Rightarrow \frac{dy}{dx}=\frac{y_{i+1}-y_{i-1}}{\Delta x_{i+1}+\Delta x_i}$$</p>


问:
Test whether a given function is polynomial
You have a black box function to which you can give real number inputs and from which you can receive real number outputs. How would you test whether it is likely to be a polynomial?

One expensive idea is to use finite differences:

Choose a maximum degree n of the "polynomial" you are testing.

Choose a consecutive sequence with random step size, and evaluate the function there to get an output sequence. E.g., $$[ 2, 2.3, 2.6, 2.9,\dots] \to [ 4.81, 5.02, 5.05, 4.90,\dots]$$

Using the output sequence as S[0], define S[n] so that its k^th entry S[n][k] = S[n-1][k+1]-S[n-1][k]. E.g. S[1] = [5.02-4.81,5.05-5.02,4.90-5.05,...] = [0.21,0.03,-0.15,...]

If the function is a polynomial (of degree at most n), then the sequence S[n+1] should be all zeros.

Some issues about programming this method:

Would be expensive for large n

If S[0] has large values, computer arithmetic will produce bad results for S[1] and beyond.

答:
I think the largest problem here is that given things like Taylor series, every function can eventually be closely approximated as a polynomial. As such, even a finite range of measured outputs of a blackbox representing a sine function will might yield a polynomial. Measuring a large range is computationally difficult as higher powers tend to yield very large values for increasing inputs. Anyway, this is what I would do:
Suppose you an input set $x\in \mathbb{R}^{n+1\times1}$ and the corresponding output set $S\in \mathbb{R}^{n+1\times1}$, where $n$ is the estimated polynomial order (or better, larger than). Then the polynomial itself can be estimated as follows:

create a square matrix $X = \begin{bmatrix}x^0 & x^1 & x^2 & ... & x^{n}\end{bmatrix} \in \mathbb{R}^{n\times n}$ containing numerical powers of the input vector.

Suppose we construct $A = \begin{bmatrix} a_0 & a_1 & ... & a_{n} \end{bmatrix}^T$ representing the black box polynomial parameters such that $f(x) = a_0 + a_1x+a_2x^2 + ... + a_{n}x^{n}$. Then $f(x) = XA$.

use the output vector $S$ to compute $A$ as follows: $A = X^{-1}S$, provided that $X$ is invertible (which should be the case if all entries of $x$ are unique).

If the black box does represent a perfect polynomial, the resulting $A$ vector should be unique and every entry with a higher order than this polynomial should be a flat zero (or approaching floating precision). The uniqueness can be verified by plugging in various different input sets.

This method can become just as troublesome for very large orders of $n$ and can yield just an estimated Taylor series of the function (at least I tested it with a sine function and it did return the Taylor series expansion of it, where the even powers had a parameters smaller than $1e^{-7}$).


问:
monodromy on kernel of polarization for abelian surfaces
Let $(A,{\mathcal L})$ be a $(1,n)$-polarized complex abelian surface. By general theory, the polarization corresponds to a map $A \to A^\vee$, and let $K({\mathcal L})$ be its kernel. Then $K({\mathcal L}) \simeq {\mathbb Z}/n \times {\mathbb Z}/n$.

Now let $M$ be the moduli space of $(1,n)$-polarized complex abelian surfaces and $a \in M$ corresponding to $(A,{\mathcal L})$. It looks like there should be a monodromy map
$$ \psi:\pi_1(M,a) \longrightarrow \mathrm{Aut}_{gp}( K({\mathcal L}) ). $$

Question. What is the image of $\psi$?

I presume that all the automorphisms in $Im(\psi)$ preserve the skew symmetric pairing (Mumford, Abelian Varieties, p. 227)
$$ K({\mathcal L}) \times K({\mathcal L}) \to {\mathbb C}^\times. $$
Is the converse true?

A reference would be greatly appreciated, though even a confirmation that this is correct is useful. Also, this clearly connects somewhat to the issue of monodromy on cohomology, and a reference for that would also be appreciated.

答:
I am going to discuss the case that $n=p$ is a prime number, the squarefree case can be reduced to this case.

Let $d$ and $N$ be coprime integers (where $d$ is either $p$ or $1$). We will adopt the (fairly standard) notation that $\mathcal{A}_{2,d,N}$ is the moduli (stack) over $\mathbb{C}$ of abelian surfaces $A$ with a polarisation $\lambda$ of degree $d$ and a symplectic level $N$-structure, that is, an isomorphism $ \alpha: A[N] \to (\mathbb{Z}/N \mathbb{Z})^{\oplus 4}$, preserving the symplectic form coming from the Weil-pairing.
The moduli stack of $(1,p)$-polarised abelian surfaces is just $\mathcal{A}_{2,p,1}$.

For $N \ge 3$ the stack $\mathcal{A}_{2,d,N}$ is actually a scheme, and it sometimes useful to work with it instead, the natural map $\mathcal{A}_{2,d,N} \to \mathcal{A}_{2,d,1}$ is a $\operatorname{Sp}_{4}(\mathbb{Z}/N\mathbb{Z})$-torsor. Moreover, $\mathcal{A}_{2,d,N}$ is (geometrically) connected for all choices of $d$ and $N$ (again $d$ is either $1$ or $p$, just to be safe).

Let $K(\lambda)$ be the kernel of $\lambda$ over $\mathcal{A}_{2,p,1}$, this is an étale group scheme over $\mathcal{A}_{2,p,1}$ equipped with a symplectic form. There is a $\operatorname{GL}_{2}(\mathbb{Z}/p\mathbb{Z})$-torsor $\mathcal{E}$ over $\mathcal{A}_{2,1,p}$ parametrising trivialisations of $K(\lambda)$, but it is clear that there is an $\operatorname{SL}_{2}(\mathbb{Z}/p\mathbb{Z})$ torsor $\mathcal{E}' \subset \mathcal{E}$, consisting of trivialisations of $K(\lambda)$ that preserve the symplectic form. Thus for a point $x=(A, \lambda) \in \mathcal{A}_{2,1,p}(\mathbb{C})$ we get a monodromy representation $$\psi:\pi_1^{\text{ét}}(\mathcal{A}_{2,1,p},x) \to \operatorname{SL}_{2}(\mathbb{Z}/p\mathbb{Z}).$$
Claim: The map $\psi$ is surjective.

Let $V$ be a one-dimensional subspace of $(\mathbb{Z}/p \mathbb{Z})^{\oplus 4}$, then there is a morphism
$$f_V:\mathcal{A}_{2,1,p} \to \mathcal{A}_{2,p,1} $$
which takes a principally polarised abelian surface $(B, \mu)$ together with an isomorphism $B[p] \simeq (\mathbb{Z}/p \mathbb{Z})^{\oplus 4}$ to the abelian surface $A=B/V$, together with the polarisation $\lambda$ of degree $p$ induced from $p \cdot \mu$. This polarisation descends because $V \subset \ker p \cdot \mu$ is an isotropic subspace. The kernel of the polarisation $\lambda$ is canonically identified with $V^{\perp}/V$, and so the map $f_V$ factors through $\mathcal{E}' \to \mathcal{A}_{2,p,1}$. It now suffices to show that the map $\mathcal{A}_{2,1,p} \to \mathcal{E}'$ is surjective.


问:
Why locally convex topological vector spaces Hausdorff?
In A Course in Functional Analysis, John B. Conway, 100p, it is written that

Definition. A locally convex topological vector space (LCTVS) is a TVS whose topology is defined by a family of seminorms $\mathcal{P}$ such that $\cap_{p\in\mathcal{P}}\{x:p(x)=0\}=\{0\}$.

This attitude that has been adopted in this book is that all topological spaces are Hausdorff. The condition in Definition above that $\cap_{p\in\mathcal{P}}\{x:p(x)=0\}=\{0\}$ is imposed precisely so that the topology defined by $\mathcal{P}$ be Hausdorff. In fact, suppose that $x\neq y$. Then there is a $p\in \mathcal{P}$ such that $p(x-y)\neq 0$; let $p(x-y)>\epsilon>0$. If $U=\{z:p(x-z)<\epsilon/2\}$ and $V=\{z:p(y-z)<\epsilon/2\}$, then $U\cap V=\emptyset$ and $U$ and $V$ are neighborhoods of $x$ and $y$, respectively.

Here I cannot agree with "$x\neq y\Rightarrow \exists p\in\mathcal{P},\,p(x-y)>0$". Isn't $\mathcal{P}$ the set of seminorms that is a priori given? Thank you in advance!

答: The condition on $\mathcal{P}$ says that the only vector $x$ all of whose seminorms (coming from $\mathcal{P}$) are zero is the zero vector itself. So if $x\neq y$, and hence $x - y \neq 0$, there must be some $p \in \mathcal{P}$ witnessing that.


问:
Are there nontrivial rational solutions to $\sqrt{1-x^2} + \sqrt{1-y^2} = \sqrt{1-z^2}$?
Obviously $(a,0,a)$,$(-a,0,a)$,$(0,a,a)$,$(0,-a,a)$ are solutions.

I tried finding solutions brute-forcing this problem, but I discovered there are no solutions with numerator and denominator smaller than 100'000.

I can prove that the question is equivalent to finding a bunch of natural numbers that satisfy:

$x'^2+ka^2 = w^2\\y'^2+kb^2 = w^2\\z'^2+kc^2 = w^2\\a+b=c$

(here is $x=x'/w$, $y=y'/w$, $z=z'/w$)

It can be proven they are equivalent as follows:

To go from the title to the 4 equations:

Call $w$ the lcm of the denominators of x, y, z

Define $x'=xw$, $y'=yw$, $z'=zw$

The equation is now $\sqrt{w^2-x^2} + \sqrt{w^2-y^2} = \sqrt{w^2-z^2}$

This equation can only hold if all sqrt are natural numbers, up to a common factor $k$ by this theorem

Now call $a=\sqrt{w^2-x^2}/\sqrt{k}$, $b=\sqrt{w^2-y^2}/\sqrt{k}$, $c=\sqrt{w^2-z^2}/\sqrt{k}$

The original equation becomes $a+b=c$, the other 3 can be calculated by the definitions of the previous step.

To go from the 4 equations to the title:

Rewrite the first 3 equations to $a=\sqrt{w^2-x^2}/\sqrt{k}$, $b=\sqrt{w^2-y^2}/\sqrt{k}$, $c=\sqrt{w^2-z^2}/\sqrt{k}$

Plug in in equation 4.

Substitude $x=x'/w$, $y=y'/w$, $z=z'/w$

But apparently being able to split a number into two squares in at least three different ways, is quite a rare property, so I would suspect that finding three with the extra property that $a+b=c$ is probably impossible, which makes me believe that there are no solutions, or that solutions are very rare.

But is there a way to find a solution, or prove that non exists except for the trivial case?

Edit: Added proof of equivalence.

答:
Here is a way to generate solutions for $k=3$.

It is well known that every number whose prime factors are all of the form $6m+1$, is also of the form $n=a^2+3b^2$.  Then
$$(2n)^2=4a^4+24a^2b^2+36b^4\\
=(2a^2-6b^2)^2+3(4ab)^2\\
=(a^2+6ab-3b^2)^2+3(a^2-2ab-3b^2)^2\\
=(a^2-6ab-3b^2)^2+3(a^2+2ab-3b^2)^2$$
Note two of the unsquared left-hand terms add up to the third, and two of the unsquared right-hand terms add up to the third.

There are also solutions for $k$ not equal to 3.


答:
Given two distinct vectors $a,b$, consider the difference vector $a-b$. This is nonzero, so the condition $$\bigcap_{p\in \mathcal{P}}\{x: p(x)=0\}=\{0\}$$ implies that for some $p\in\mathcal{P}$ we have $p(a-b)\not=0$.

Since seminorms take on only nonnegative values, this means $p(a-b)>0$.

Title: Why locally convex topological vector spaces Hausdorff?

Body:
In A Course in Functional Analysis, John B. Conway, 100p, it is written that

Definition. A locally convex topological vector space (LCTVS) is a TVS whose topology is defined by a family of seminorms $\mathcal{P}$ such that $\cap_{p\in\mathcal{P}}\{x:p(x)=0\}=\{0\}$.

This attitude that has been adopted in this book is that all topological spaces are Hausdorff. The condition in Definition above that $\cap_{p\in\mathcal{P}}\{x:p(x)=0\}=\{0\}$ is imposed precisely so that the topology defined by $\mathcal{P}$ be Hausdorff. In fact, suppose that $x\neq y$. Then there is a $p\in \mathcal{P}$ such that $p(x-y)\neq 0$; let $p(x-y)>\epsilon>0$. If $U=\{z:p(x-z)<\epsilon/2\}$ and $V=\{z:p(y-z)<\epsilon/2\}$, then $U\cap V=\emptyset$ and $U$ and $V$ are neighborhoods of $x$ and $y$, respectively.

Here I cannot agree with "$x\neq y\Rightarrow \exists p\in\mathcal{P},\,p(x-y)>0$". Isn't $\mathcal{P}$ the set of seminorms that is a priori given? Thank you in advance!