问:
How does a single $d$ in Leibniz notation work? Eg, $y\,\mathrm{d}x-x\,\mathrm{d}y\equiv-x^2\,\mathrm{d}\left(\frac{y}{x}\right)$
Recently a lecturer used this notation, which I assume is a sort of twisted form of Leibniz notation:

$$y\,\mathrm{d}x - x\,\mathrm{d}y \equiv -x^2\,\mathrm{d}\left(\frac{y}{x}\right)$$

The logic here was that this could be used as:

$$\begin{align}
-x^2\,\mathrm{d}\left(\frac{y}{x}\right) &\equiv -x^2\,\left(\frac{\mathrm{d}y}{x} -\frac{y}{x^2}\,\mathrm{d}x\right)\\
&\equiv y\mathrm{d}x - x\mathrm{d}y
\end{align}
$$

Why is this legal?

I can see some kind of differentiation going on with the second term in the above equivalence, producing the $\frac{1}{x^2}$, but having the single $\mathrm{d}$ seems like a really weird abuse of notation, and I don't quite follow why it splits the single $\frac{y}{x}$ fraction into two parts.

答:
You should know that the differential at a point $\mathbf x_0$ of a function $\;\mathbf R^m\longrightarrow \mathbf R^n$ is the linear map $\:\ell:\mathbf R^m\longrightarrow \mathbf R^n$, that yields the best linear approximation of $f(\mathbf x_0)$ in a neighbourhood of $\mathbf x_0$, in the sense that we have
$$f(\mathbf x_0+\mathbf h)=f(\mathbf x_0)+\ell(\mathbf h)+o\bigl(\|\mathbf h\|\bigr).$$
This differential is denoted $\:\mathrm d f_{\mathbf x_0}$ (or simply $\mathrm df$ for the differential at a generic point). A linear function is of course its own differential.

With the usual  abuse of language that denotes a function $f$ by its value at a given point, we obtain that the differential of the $i$-th projection $\:p_i:\mathbf x=(x_1,x_2,\dots,x_m)\longmapsto x_i$ is denoted $\mathrm dx_i$.

As an example, in the case of a function of a single variable $x$, the linear map defining the differential simply corresponds to the equation of the tangent to the graph of $f$ with abscissa $x_0$:
$$f(x_0+h)=f(x_0)+f'(x_0)h,\enspace\text{i.e. }\quad \mathrm df_{x_0}:h\longmapsto f'(x_0)h,$$
that we may write as $\enspace\mathrm df=f'(x)\,\mathrm dx$. This notation is generalised to functions of $m$ variables under the form
$$\mathrm df=\frac{\partial f}{\partial x_1}\,\mathrm dx_1+\frac{\partial f}{\partial x_2}\,\mathrm dx_2+\dots+\frac{\partial f}{\partial x_m}\,\mathrm dx_m. $$
The usual formulæ for the derivatives have a ‘differential version’:

$\mathrm d(f+g)=\mathrm df+\mathrm dg$,

$\mathrm d(fg)=f\,\mathrm dg+g\,\mathrm df$,

$\mathrm d\Bigl(\dfrac fg\Bigr)=\dfrac1{g^2}\bigl(g\,\mathrm df-f\,\mathrm dg\big),$

$\mathrm d(g\circ f)=\mathrm dg_{f(x)}\circ\mathrm df_x$.


问:
If $B$ is a $C^*$-subalgebra of $A$, do we have $AB= B?$
Let $B$ be a $C^*$-subalgebra of $A$. Do we have $AB= A?$ In other words, is $\text{span}\{ab \mid a \in A, b \in B\}$ norm-dense in $A$?

The usual way to approach these questions is by using approximate units, so that's how my attempt will go. Consider the canonical approximate unit of $A$, that is
$$(u_i)_{i \in I}$$
where $I$ is the set of all positive elements in the closed unit ball of $A$ and $u_i= i$.

Then if $a \in A$, we have $$a = \lim_i au_i$$

Now, consider the subnet $(u_j)_{j \in J}$ where $J$ is the set of all positive elements in the closed unit ball of $B$. Then we still have
$$a= \lim_j a u_j \in AB$$

Is this correct?

答:
This is not true.  Consider $A=M_2(\mathbb C)$, and
$$B=\left\{\begin{bmatrix}b&0\\0&0\end{bmatrix}:b\in\mathbb C\right\}.$$
Note that
$$\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\begin{bmatrix}b&0\\0&0\end{bmatrix}=\begin{bmatrix}a_{11}b&0\\a_{21}b&0\end{bmatrix},$$
so clearly $AB$ cannot be all of $A$.

ContentLicense: CC BY-SA 4.0


问:
General solution and intervals (over which the general solution is defined)
So I did a homework and I got x^2/(x+1)e^x + C/(x+1)e^x for all R (except -1), but they say it's for -1 < x < oo+. I don't understand how that makes any sense. Is there something I am not getting? e^x can never equal 0, so I think my solution makes more sense than their solution.</p>

答:
The interval of existence of a solution is an interval. The set $\mathbb R\setminus \{-1\}$ is not an interval, it consists of two disjoint intervals $(-\infty,-1)$ and $(-1,\infty)$. So, $\mathbb R\setminus \{-1\}$ can never be the correct answer to "what is the interval of existence?"

Which of the two intervals will be the correct answer depends on the context. If you solve an initial value problem which specifies the value of $y(0)$, or $y(1)$, or $y(42)$, then the interval of existence is $(-1,\infty)$. If you solve an initial value problem which specifies the value of $y(-10)$  or $y(-2)$, then the interval of existence is $(-\infty,-1)$.

Simply put, the solutions of this differential equations don't jump over $-1$: they explode upon reaching it.

ContentLicense: CC BY-SA 3.0


问:
Nonconstructive proof of the irrationality of a power
I'm given some rational $x$ and irrational $y$, I want to show nonconstructively that that $\exists x \exists y(x^y)$ is irrational.

I'm given the example $4^{\sqrt 2}$. It's just a one-liner. $4$ is rational, $\sqrt 2$ is irrational, and so $4^\sqrt{2}$ is irrational.

I intuitively understand this, but is there a slightly more detailed explanation I can give to show that $4^\sqrt 2$ is irrational?

答:
As has been remarked, it is not so easy to show that $4^{\sqrt 2}$ is irrational.

If the goal here is to find a rational $x$ and an irrational $y$ such that $x^y$ is irrational then take $$10^{\log_{10} \sqrt 2}=\sqrt 2$$

To see that $\log_{10} \sqrt 2=\frac 12 \times \log _{10}2$ is irrational note that it suffices to prove that $\log_{10}2 $ is irrational.  But were it otherwise then $$\log_{10}2=\frac ab\implies 2^b=10^a$$ and, since $a,b$ are intended to be natural numbers, this is impossible ($b$ can not be $0$ and $5$ does not divide $2^a$ for any $a$).


问:
Units in a ring between an integral domain $R$ and its fraction field.
Let $R$ be an integral domain with fraction field $K$, and let $\mathcal{T}$ be the set of subrings $T$ of $K$ such that $R\subseteq T$.

Call $R\;$unit-complete$\;$if for all $T\in \mathcal{T}$, either $T=R$ or $T$ has a unit element which is not in $R$.

Call $R\;$strongly unit-complete$\;$if for all $T\in \mathcal{T}$, either $T=R$ or $T$ contains ${\large{\frac{1}{r}}}$ for some non-unit element $r\in R$.

It's immediate that strongly unit-complete implies unit-complete.

It's easily shown that if $R$ is a $\text{PID}$, then $R$ is strongly unit-complete.

Proof:

Let $R$ be a PID with fraction field $K$, and let $T$ be a subring of $K$ with $R\subset T$ (proper inclusion).

Fix $t\in T{\setminus}R$, and write $t={\large{\frac{a}{b}}}$, where $a,b\in R$ are such that
$b$ is a non-unit of $R$ and $\gcd(a,b)=1$.

From ${\large{\frac{a}{b}}}\in T$, we get  $1-{\large{\frac{a}{b}}}\in T$, so ${\large{\frac{b-a}{b}}}\in T$.

From $\gcd(a,b)=1$, we get  $\gcd(a,b-a)=1$.

Then since $R$ is a $\text{PID}$, there exist $u,v\in R$ such that $ua+v(b-a)=1$, so
$$
\frac{a}{b},\frac{b-a}{b}\in T
\implies
u\Bigl(\frac{a}{b}\Bigr)+v\Bigl(\frac{b-a}{b}\Bigr)\in T
\implies
\frac{1}{b}\in T
$$
hence $R$ is strongly unit-complete.

Some questions:

$(1)\;$If a noetherian integral domain $R$ is strongly unit-complete, must $R$ be a $\text{UFD}$?

$(2)\;$Is there an integral domain $R$ which is unit-complete but not strongly unit-complete?

答:
Here's a proof that if $R$ is a Dedekind domain with torsion class group, then $R$ is strongly unit-complete. This shows that the answer to question (1) is no, as there are Dedekind domains with torsion class group which are not UFDs, e.g. $\Bbb Z[\sqrt{-5}]$.

Let $K$ be the fraction field of $R$.
Let $K \supset T \supsetneq R$ be an overring, then we can find $\frac{a}{b} \in T \setminus R, a,b \in R$ so that $b$ is not a unit in $R$. As the class group of $R$ is torsion, we can find some $n$ such that $(a,b)^n=(c)$. Write $c=\sum_{i=0}^nc_ia^ib^{n-i}$ for some $c_i \in R$, divide by $b^n$ and we get $$\frac{c}{b^n}=\sum_{i=0}^n c_i (\frac{a}{b})^i \in T.$$
Because $b^n \in (a,b)^n=(c)$, we get $b^n=tc$ for some $t \in R$, so we get
$$T \ni \frac{c}{b^n}=\frac{1}{t}$$

So it remains to show that $t$ is not a unit in $R$. Suppose $t\in R^\times$, then $(b)^n=(c)=(a,b)^n$. By unique factorization of ideals into prime ideals in $R$, this implies $(a,b)=(b)$, i.e. $b \mid a$, contradicting $\frac{a}{b} \notin R$.


问:
ito formula apply to integral
I am a beginner of SDE, and I am working on an exercise, the problem ask us to find the $d Y_{t}$ for different $Y_t$. The setting of problem is:

Suppose $B_{t}$ is a standard Brownian motion and $X_{t}$ satisfies
$$
d X_{t}=X_{t}^{2} d t+X_{t} d B_{t}
$$
Find $d Y_{t}$ for different $Y_t$

$Y_{t}=B_{t}^{2}$

$Y_{t}=X_{t}^{3}$

$Y_{t}=\exp \left\{\int_{0}^{t}\left(X_{s}^{2}+1\right) d s\right\}$

I know how to use Ito's formula to find first and second answer, but don't how to proceed the third one, can anyone help or give some suggestion.

答: It sounds like the main difficulty is finding the differential of $\int_0^t (X_s^2+1)ds$, and for that I would give the hint: For a deterministic function $f$, what is $\frac{d}{dt} \int_0^t f(s)ds$?  Did you need the chain rule for that, or even to know what $\frac{df}{dt}$ is?

Title: ito formula apply to integral


问:
Sum of two squares theorem
I am working on the problem: I need to quickly check if positive number $n$ can be expressed as $n^2=a^2+b^2$. I found this theorem: https://en.wikipedia.org/wiki/Sum_of_two_squares_theorem

But it seems I am missing something. For example number $4$ falls under theorem condition: its prime decomposition doesn't have any prime $p\equiv 3{\pmod {4}}$. But $4$ cannot be expressed as sum of 2 squares.

答:
In the theorem $a,b$ have to be integers and don't have to be nonzero. So in your example: $4=2^2+0^2$.

So the theorem doesn't really help in your case. $n^2=n^2+0^2$ but that doesn't help you much, since I assume that you have additional assumptions on $a,b$, like for example that they should both be nonzero.

Title: Sum of two squares theorem


问:
Why is one of the solutions to a equality-constrained minimization problem independent of the constraint
I am trying to understand the interpretation of an equality constraint minimization problem which I posted in a previous question here.

The function being constrained is $C(x,y)=tx+wy$ where the constants $t, w > 0$.
The constraint function is $P=ay^2 + ax$ where the constants $P, a > 0$.

In the accepted solution to the question linked to above, the Lagrangian multiplier technique is used to get the following solution:

$P_c = (\frac{P}{a} - \frac{w^2}{4t^2}, \frac{w}{2t}, \frac{t}{a})$

I understand how the process to reach this result was done, but I don't understand why it is that the second term, ie. $y = \frac{w}{2t}$, does not have a dependence on $P$ as the first term ($x$) does. Intuitively, I would expect changing the value of $P$ would result in a some change in $y$, but it seems not to be the case as both $w$ and $t$ are constants.

Why is this so?

答:
This is a result of the special nature of your equations. By imposing your constraint you are restricting the set of points you are considering to a parabola $x=P/a-y^2$ (this parabola is "horizontal"). Minimizing $C$ involves finding the leftmost (or rightmost, depending on the signs of some of the coefficients) of a family of parallel lines that intersects this parabola. If you change $P$ you will move the parabola left or right, and the leftmost curve intersecting it will also move up by the same amount. The tangency point (which is where your extremal point is) will correspondingly move left or right (but not up or down), meaning that the $y$ coordinate of the optimal point is independent of $P$.  Another way to say it, is that changing $P$ in the constraint function happens to be the same as replacing $y$ by $y+const$, and this happens to also move lines $tx+wy=const$ to each other. The reason $\lambda$ is unchanged is a bit trickier, but it can be explained in similar way  -- both the constraint function and the family of level sets of the cost function are simply shifted, so the proportionality between the constraint and level set gradients (which is what $\lambda$ is) is preserved.

BTW, your original problem does not need Lagrange multipliers. Just solve the constraint equation to get $x=P/a-y^2$ and plug into the cost to get $C=t(P/a-y^2)+wy=-ty^2+wy+tP/a$, which is a parabola. A parabola $az^2+bz+c$ has extremal value at $z=-b/2a$, so in your case the extremum is attained indeed at $y=w/2t$ (if $t>0$ this is a maximum, if $t<0$ a minimum). Correspondingly $x=P/a-y^2=P/a-w^2/4t^2$, as before.


问:
Find the smallest positive integer, whose last $4-digit$ part is $2010$ and is divisible by $2011$.
Find the smallest positive integer, whose last $4-digit$ part is $2010$ and is divisible by $2011$.

I solved it in the following way:

I state that $a$ is the number in front of $2010$, in other words if the number we're looking for is $A$, then $A=10^4\cdot a+2010$.

Hence, $a\cdot10^4\equiv1\bmod2011$.

$1956a\equiv1\bmod2011$

$55a\equiv2010\bmod2011$

So $55a=2010+2011k$ for some positive integer $k$.

So $30+31k\equiv 0 \bmod55$

After doing some trial and error for $k\in[0,54]$

$k_{min}=15$

So $55a=32175$, $a=585$.

Hence the smallest number possible is $5850000$

I solved this question using classical number theory and trial-and-error. However, I realize that this question can be solved using the euclidean algorithm. When I looked it up on the internet the only thing I could find was this: https://en.wikipedia.org/wiki/Euclidean_algorithm which is how to find the gcd using euclid's algorithm. Could you please explain to me how to solve the question using the euclidean algorithm, and also please explain to me what the euclidean algorithm is, how it can be applied and some resources to study from?

答:
Update: just found another way. Not the simplest but very safe and calculator-free because dividing by 10 is easy.

$$
\pmod{2011}: \frac{1}{10^4} \equiv \frac{-2010}{10^4}  \equiv \frac{-201+2011}{10^3} \equiv \frac{181-2011}{100} \equiv \frac{-183+2011\cdot 3}{10} = 585.
$$

I recently learned Gauss's Algorithm from Bill Dubuque (How to find the inverse modulo m?). Let's try:

$$
\frac{1}{10^4}\equiv \frac{1}{-55} \equiv \frac{36}{-55\cdot 36} \equiv \frac{36}{31} \equiv \frac{36\cdot 65}{31\cdot 65} \equiv \frac{329}{4} \equiv \frac{329+2011}{4}=\frac{2340}{4} =585 \pmod{2011}.\blacksquare
$$

Or, even simpler
$$
\frac{1}{10^4}\equiv \frac{-2010}{-55} = \frac{402}{11} \equiv \frac{402\cdot 183}{11\cdot 183} \equiv \frac{1170}{2} = 585 \pmod{2011}.
$$


问:
Integrate $\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx$
Please help me to solve this integral:
$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx.$$

I managed to calculate an indefinite integral of the left part:
$$\int\frac{\cos x}{\sin x}x\ \mathrm dx=\ x\log(2\sin x)+\frac{1}{2} \Im\ \text{Li}_2(e^{2\ x\ i}),$$
where $\Im\ \text{Li}_2(z)$ denotes the imaginary part of the dilogarithm. The corresponding definite integral $$\int_0^\pi\frac{\cos x}{\sin x}x\ \mathrm dx$$ diverges. So, it looks like in the original integral summands compensate each other's singularities to avoid divergence.

I tried a numerical integration and it looks plausible that
$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx\stackrel{?}{=}\pi \log 54,$$
but I have no idea how to prove it.

答:
Begin with $u$-substitution using
$$\begin{align}
u & =x\\
dv & = \frac{3\cos(x)+\sqrt{8+\cos^2(x)}}{\sin(x)}\,dx\end{align}$$
so that $du=dx$, and my CAS tells me (which I suppose could be verified through differentiation and identities) that
$$\begin{align}
v & = \sinh^{-1}\left(\frac{\cos(x)}{\sqrt{8}}\right)+\frac{3}{2}\ln\left(\frac{3\sqrt{\cos^2(x)+8}-\cos(x)+8}{3\sqrt{\cos^2(x)+8}+\cos(x)+8}\right)+3\ln(1-\cos(x))
\end{align}$$

Now we have
$$\begin{align}
\left[x\left(\sinh^{-1}\left(\frac{\cos(x)}{\sqrt{8}}\right)+\frac{3}{2}\ln\left(\frac{3\sqrt{\cos^2(x)+8}-\cos(x)+8}{3\sqrt{\cos^2(x)+8}+\cos(x)+8}\right)+3\ln(1-\cos(x))\right)\right]_0^\pi\\
-\int_0^\pi\left(\sinh^{-1}\left(\frac{\cos(x)}{\sqrt{8}}\right)+\frac{3}{2}\ln\left(\frac{3\sqrt{\cos^2(x)+8}-\cos(x)+8}{3\sqrt{\cos^2(x)+8}+\cos(x)+8}\right)+3\ln(1-\cos(x))\right)dx
\end{align}$$

and most of the integral part can be evaluated by taking advantage of symmetry about $\pi/2$:

$$\begin{align}
\left[x\left(\sinh^{-1}\left(\frac{\cos(x)}{\sqrt{8}}\right)+\frac{3}{2}\ln\left(\frac{3\sqrt{\cos^2(x)+8}-\cos(x)+8}{3\sqrt{\cos^2(x)+8}+\cos(x)+8}\right)+3\ln(1-\cos(x))\right)\right]_0^\pi\\
-3\int_0^\pi\ln(1-\cos(x))dx
\end{align}$$

($\sinh^{-1}$ is odd and $\cos(x)$ has odd symmetry about $\pi/2$. For the logarithmic term, the input to $\ln()$ at $x$ is the reciprocal of the input at $\pi/2-x$.)

Some of the nonintegral-part can be cleanly evaluated:

$$\begin{align}
\pi\sinh^{-1}\left(\frac{-1}{\sqrt{8}}\right)+\frac{3\pi}{2}\ln\left(\frac{9}{8}\right)+\left[3x\ln(1-\cos(x))\right]_0^\pi\\
-3\int_0^\pi\ln(1-\cos(x))dx
\end{align}$$

and now moving the "unclean" part back into an integral:
$$\begin{align}
\pi\sinh^{-1}\left(\frac{-1}{\sqrt{8}}\right)+\frac{3\pi}{2}\ln\left(\frac{9}{8}\right)+\int_0^\pi\left(3\ln(1-\cos(x))+\frac{3x\sin(x)}{1-\cos(x)}\right)\,dx\\
-3\int_0^\pi\ln(1-\cos(x))dx\\
=\pi\sinh^{-1}\left(\frac{-1}{\sqrt{8}}\right)+\frac{3\pi}{2}\ln\left(\frac{9}{8}\right)+\int_0^\pi\frac{3x\sin(x)}{1-\cos(x)}\,dx
\end{align}$$

My CAS says this is

$$\begin{align}
\pi\sinh^{-1}\left(\frac{-1}{\sqrt{8}}\right)+\frac{3\pi}{2}\ln\left(\frac{9}{8}\right)+\pi\ln(64)
\end{align}$$

which is the only thing the CAS does that I don't quite get. But it's nothing special about endpoints: even WA can give an antiderivative if we can use the dilogarithm. It looks like an integral that might even appear somewhere on this site. A conversion of the arcsinh and logarithm rules yields

$$\begin{align}
\pi\ln(2^{-\frac{1}{2}})+\pi\ln\left(\frac{27}{8^{3/2}}\right)+\pi\ln(64)=\pi\ln(54)
\end{align}$$


问:
$\lim\limits_{n\to\infty}\sqrt[n]{1^k +2^k+\cdots+ n^k},$
What is
$$\lim\limits_{n\to\infty}\sqrt[n]{1^k +2^k+\cdots+ n^k},$$
where $ k\in \mathbb{N}$?

Body: Hint: $$1\le 1^k +2^k+\cdots+ n^k\le n^{k+1}$$ and $$\lim_{n\to\infty}n^{1/n}=1.$$


问:
distortion to virtually rotate an orthographic sphere image
How, exactly do I produce a 2D distortion that takes an orthographic image of a sphere and make it look like it's rotating?

I have this image of an eyeball  What I'd like to do is come up with a way to distort the image so that it looks as if the eyeball is looking around (rotating).  So, as the iris gets closer to the edge it becomes more compressed and warped to the curve of the sphere.

I've tried all sorts of stuff having to do with polar and Cartesian coordinates and I can get it close but nothing quite gives the effect that an actual sphere is rotating.

答:
There are two distinct parts in your question 1) and 2).

Let us consider a rotation with angle $R$ to the right for the  owner of the eye, therefore to the left for the observer. let us call $A$ the Azimuth and $E$ the Elevation.

Fig. 1: A small feature and its image by a rotation with $R=5 \pi/12$ using formulas (1).

We have to make first a mathematical computation from cartesian coordinates $(x,y) \to (x',y')=(x',y)$.
$$\begin{cases}x&=&\cos(A)\cos(E)\\y&=&\sin(E)\end{cases} \ \ \to \ \  \begin{cases}x'&=&\cos(A')\cos(E)\\y'&=&y\end{cases}\tag{1}$$

where $A$ and $A'$ are the old and new azimuth, $E$ the common elevation, and  $A'=A+R$ where $R$ is the rotation angle. (1) can be done by using the following computations:

$$A=\arccos\left(\dfrac{x}{\sqrt{1-y^2}}\right) \ \ \to \ \ x'=\cos(A+R)\sqrt{1-y^2}\tag{2}$$

Please note that we have taken a radius equal to $1$.

But this is enough for the contour of shapes only. If one wants to take into account  the image itself, with its pixels and their colors, it is not sufficient. We have to use the fact that the image of a group of $N$ horizontal pixels undergoing a rotation will give a group of $N'$ horizontal pixels with, in general, $N \ne N'$. The idea is to work backwards. Let us take the example of pixels in the iris ; all of them will be more or less compressed by a $75°$ rotation. The pixel which is for example right in the center of the iris in its rotated position will correspond to, say, $5$ pixels in the initial position of the center. Therefore you will have to find them, make an average color and attribute this color to the new pixel. Saying that for $75°$, we must take a ratio $1:5$, that for $45°$ the ratio is $2:3$ depends in general from the initial position.
The precise rule of compression (or dilatation according to the case at hand) can be established as $N'/N = $

$$\text{relative distorsion coefficient: } \ \ \dfrac{\sin(A)}{\sin(A')}\tag{3}$$

Please note that (3) is independent from Elevation.


答:
Very often it works to just think of $\mathrm{d}f$ as an alternative notation to $f'$ or the derivative of $f$ with respect to some unnamed variable.

For example, assume that $x$ and $y$ are functions of some common variable. Then we have
$$
y\,x' - x\,y' = -x^2 (y/x)'
$$
Just using the notation $\mathrm{d}f$ instead of $f'$ gives
$$
y\,\mathrm{d}x - x\,\mathrm{d}y = -x^2 \mathrm{d}(y/x).
$$

Title: How does a single $d$ in Leibniz notation work? Eg, $y\,\mathrm{d}x-x\,\mathrm{d}y\equiv-x^2\,\mathrm{d}\left(\frac{y}{x}\right)$

Body:
Recently a lecturer used this notation, which I assume is a sort of twisted form of Leibniz notation:

$$y\,\mathrm{d}x - x\,\mathrm{d}y \equiv -x^2\,\mathrm{d}\left(\frac{y}{x}\right)$$

The logic here was that this could be used as:

$$\begin{align}
-x^2\,\mathrm{d}\left(\frac{y}{x}\right) &\equiv -x^2\,\left(\frac{\mathrm{d}y}{x} -\frac{y}{x^2}\,\mathrm{d}x\right)\\
&\equiv y\mathrm{d}x - x\mathrm{d}y
\end{align}
$$

Why is this legal?

I can see some kind of differentiation going on with the second term in the above equivalence, producing the $\frac{1}{x^2}$, but having the single $\mathrm{d}$ seems like a really weird abuse of notation, and I don't quite follow why it splits the single $\frac{y}{x}$ fraction into two parts.


问:
Question in the proof of Clifford Theorem (representation theory)

Let $A$ be a normal subgroup of $G$. Let $\rho:G\to GL(V)$ be a
irreducible representation. Write $Res^G_A\rho$ as a direct sum of
isotypic representations of $A$;$V=\bigoplus\limits_{i=1}^k W_i $. Since
$A$ is normal subgroup of $G$, $\rho(g)W_i$ is stable under $A$. Since
$W_i$ is isotypic, $\rho(G)W_i$ is also isotypic and conjugate to
$W_i$ so $G$ permutes $W_i$'s. Since $V$ is irreducible representation
of $G$, we know that the action of $G$ on $W_i$'s is transitive and
hence all $W_i$'s are conjugate.

I understand that $\rho(g)W_i$ is stable under A. But I don't understand the next two sentences. Can you please explain what is going on here? I would really appreciate any help.

答:
Any group element $g \in G$ defines an automorphism on the normal subgroup $A$ by conjugation: $c_g(a) = g^{-1} a g$. For any representation $\pi \colon A \to \operatorname{GL}(W)$ of $A$, we can pull the representation back along this automorphism to get another representation $\pi \circ c_g$ of $A$, namely the representation where we have $a \cdot w = \pi(g^{-1} a g) w$. If $\pi$ is irreducible then so is $\pi \circ c_g$, so precomposing by $c_g$ permutes the isomorphism classes of irreps of $A$. Accordingly if $\pi$ is isotypic then so is $\pi \circ c_g$.

Now suppose we have a representation $\rho \colon G \to \operatorname{GL}(V)$. We can treat $\rho$ as a representation of $A$, yielding new representations $\rho \circ c_g$ as before, but now we can factorise $\rho(g^{-1} a g) = \rho(g^{-1}) \rho(a) \rho(g)$ because $\rho$ is defined on $G$ rather than on $A$. The normality of $A$ implies that for any $A$-subrepresentation $W \subseteq V$, the subspace $\rho(g) W$ is again $A$-stable, and hence $(\rho(g) W, \rho)$ is another $A$-subrepresentation of $V$. In fact, left multiplication by $\rho(g)$ gives an isomorphism $(W, \rho \circ c_g) \to (\rho(g) W, \rho)$ which is $A$-equivariant:
$$ \rho(g)(\rho \circ c_g)(a) w = \rho(g g^{-1} a g) w = \rho(a) \rho(g) w.$$
So the conjugated representation $(W, \rho \circ c_g)$ of $A$ is isomorphic to the subrepresentation $(\rho(g) W, \rho)$. So something that used to be quite abstract (changing the action on a space by precomposition with an automorphism) becomes quite concrete (another subrepresentation in the same ambient space).

Now specialising to your question: by "$\rho(g) W_i$ is conjugate to $W_i$" they mean in the sense above: $(\rho(g) W_i, \rho)$ is isomorphic to the conjugated representation $(W_i, \rho \circ c_g)$. Since $W_i$ is isotypic it is of the form $W_i \cong X^{\oplus k}$ for some particular irrep $X$ of $A$. Since conjugation permutes the irreps of $A$, it must permute the isotypic components of $V$, hence $\rho(g) W_i = W_j$ for some $j$ depending on $i$ and $g$.

Lastly, for each $W_i$ and $W_j$ there must exist a $g \in G$ such that $\rho(g) W_i = W_j$, since if this were not the case $V$ would not be irreducible as a representation of $G$, as $\sum_{g \in G} \rho(g) W_i$ would be a proper $G$-stable subspace.


问:
Systems of Differential Equations and higher order Differential Equations.
I've seen how one can transform a higher order ordinary differential equation into a system of first-order differential equations, but I haven't been able to find the converse. Is it true that one can transform any system into a higher-order differential equation? If so, is there a general method to do so?</p>

答:
If I am understanding your question, you just would reverse the process on the last equation from the system.

An $n^{th}$ order diﬀerential equation can be converted into an $n$-dimensional system of ﬁrst order equations.

There are various reasons for doing this, one being that a ﬁrst order system is
much easier to solve numerically (using computer software) and most diﬀerential equations you encounter in “real life” (physics, engineering etc) don’t have nice exact solutions.

If the equation is of order $n$ and the unknown function is $y$, then set:

$$x_1 = y, x_2 = y', \ldots , x_n = y^{n−1}.$$

Note (and then note again) that we only go up to the $(n − 1)^{st}$ derivative in this process. Lets do an example in both directions (practice some where a known system has been converted to such a system and make sure you can work backward).

Forward Approach

$$\tag 1 y^{(4)} - 3y' y'' + \sin(t y'') -7ty^2 = e^t$$

Let: $x_1 = y, x_2 = y', x_3 = y'', x_4 = y'''$ and substitute into $(1)$, yielding:

$x_1' = y' = x_2$

$x_2' = y'' = x_3$

$x_3' = y''' = x_4$

$x_4' = y^{(4)} = 3yy''-\sin(ty'')+7ty^2+e^t = 3x_2x_4-\sin(tx_3)+tx_1^2+e^t$

Backward Approach

Looking at the last equation from the system, we let: $y = x_1, y' = x_2, y''=x_3, y'''=x_4$ and substitute into the system's last equation above, yielding:

$y^{(4)} = 3y'y''-\sin(ty'')+7ty^2+e^t$


问:
Prove $\int_{0}^{\pi/2} \int_{0}^{\pi/2} \frac{\theta\cot\theta-\varphi\cot\varphi}{\cos\theta-\cos\varphi} \text{d}\varphi\text{d}\theta = \pi\ln2$
Month ago I encounter a nice result numerically checked by Mathematica
$$
\int_{0}^{\pi/2} \int_{0}^{\pi/2} \frac{\theta\cot\theta-\varphi\cot\varphi}{\cos\theta-\cos\varphi} \mathrm{d}\varphi\mathrm{d}\theta = \pi\ln2
$$
where the integrated function is actually well-defined even around its singularity $\theta=\varphi=0$.

At my first sight, I thought it might be a trivial conclusion derived from a kind of typical integral like
$$
\int_{0}^{\pi} \frac{\cos n\theta}{\cos\theta-\cos\varphi} \mathrm{d}\theta = \pi\frac{\sin n\varphi}{\sin\varphi}
$$
just using a proper series expansion. However, when I review it in detail, the result over $(0,\pi/2)$ will be awkwardly complicated. I realize this double integrals may not be done directly, or I may lack some essential insight to solve it.

So I question it here for some further discussion, and thanks in advance for any suggestion.

答:
The following illustrates an approach that looks promising, but
that I could not yet complete: wish you and the community here might help
to finalize it.

Starting from the integrand
$$
f(\theta ,\varphi )
 = {{\theta \cot \theta  - \varphi \cot \varphi } \over {\cos \theta  - \cos \varphi }}
 = {{\theta {{\cos \theta } \over {\sin \theta }} - \varphi {{\cos \varphi } \over {\sin \varphi }}}
 \over {\cos \theta  - \cos \varphi }}
 = f(\varphi ,\theta )
$$
it's structure suggests the idea that Green's Theorem might be applied.

Formulated as above, it is difficult to find the needed potential function.

But applying the transformation
$$
\left\{ \matrix{
  \theta  = x + y \hfill \cr
  \varphi  = y - x \hfill \cr}  \right.\quad  \Leftrightarrow \quad \left\{ \matrix{
  x = {{\theta  - \varphi } \over 2} \hfill \cr
  y = {{\theta  + \varphi } \over 2} \hfill \cr}  \right.
$$
we get
$$
\eqalign{
  & g(x,y) = f(x + y,\;y - x) =   \cr
  &  = {{\left( {y - x} \right){{\cos \left( {y - x} \right)} \over {\sin \left( {y - x} \right)}}
 - \left( {y + x} \right){{\cos \left( {y + x} \right)}
 \over {\sin \left( {y + x} \right)}}}
 \over {\cos \left( {y - x} \right) - \cos \left( {y + x} \right)}} =   \cr
  &  = {{y\left( {{{\cos \left( {y - x} \right)} \over {\sin \left( {y - x} \right)}}
 - {{\cos \left( {y + x} \right)} \over {\sin \left( {y + x} \right)}}} \right)
 - x\left( {{{\cos \left( {y - x} \right)} \over {\sin \left( {y - x} \right)}}
 + {{\cos \left( {y + x} \right)} \over {\sin \left( {y + x} \right)}}} \right)}
 \over {2\sin x\sin y}} =   \cr
  &  = {{y\sin \left( {2x} \right) - x\sin \left( {2y} \right)}
 \over {2\sin x\sin y\sin \left( {y - x} \right)\sin \left( {y + x} \right)}} =   \cr
  &  = y{{\sin \left( {2x} \right)} \over {2\sin x\sin y\sin \left( {y - x} \right)\sin \left( {y + x} \right)}}
 - x{{\sin \left( {2y} \right)} \over {2\sin x\sin y\sin \left( {y - x} \right)\sin \left( {y + x} \right)}} =   \cr
  &  = y{{\cos x} \over {\sin y\left( {\sin ^2 y\cos ^2 x - \cos ^2 y\sin ^2 x} \right)}} - x{{\cos y}
 \over {\sin x\left( {\sin ^2 y\cos ^2 x - \cos ^2 y\sin ^2 x} \right)}} =   \cr
  &  = y{{\cos x} \over {\sin y\left( {\cos ^2 x - \cos ^2 y} \right)}}
 - x{{\cos y} \over {\sin x\left( {\cos ^2 x - \cos ^2 y} \right)}} =   \cr
  &  = g( - x,y) \cr}
$$

Putting:
$$
\eqalign{
  & M(x,y) =  - {y \over {2\sin ^2 y}}\ln \left( {{{ - \sin x + \sin y}
 \over {\sin x + \sin y}}} \right)  \cr
  & L(x,y) = {x \over {2\sin ^2 x}}\ln \left( {{{ - \sin x + \sin y}
 \over {\sin x + \sin y}}} \right) \cr}
$$
then we have the expression required for applying the Green's theorem
$$
g(x,y) = {\partial  \over {\partial x}}M(x,y) - {\partial  \over {\partial y}}L(x,y)
$$

With the help of this sketch

we have
$$
\eqalign{
  & I = \int_0^{\pi /2} {\int_0^{\pi /2} {f(\theta ,\varphi )\,d\varphi \,d\theta } }  =   \cr
  &  = \int\!\!\!\int\limits_{square} {f(\varphi ,\theta )\,d\varphi \,d\theta }
  = 2\int\!\!\!\int\limits_{triang} {f(\varphi ,\theta )\,d\varphi \,d\theta }  =   \cr
  &  = 4\int\!\!\!\int\limits_{triang} {g(x,y)\,dx\,dy}
  = 4\int\!\!\!\int\limits_{triang} {\left( {{\partial  \over {\partial x}}M
 - {\partial  \over {\partial y}}L} \right)\,dx\,dy}  =   \cr
  &  = 4\oint\limits_{triang} {Ldx + Mdy}  \cr}
$$
that is
$$
\eqalign{
  & {I \over 4} = \oint\limits_{triang} {Ldx + Mdy}  =   \cr
  &  = \left( \matrix{
  \int_{y = 0}^{\pi /4} {\left( {L\left( {y,y} \right) + M(y,y)} \right)dy}  +  \hfill \cr
   + \int_{y = \pi /4}^{\pi /2} {\left( { - L\left( {\pi /2 - y,y} \right) + M(\pi /2 - y,y)} \right)dy}
  +  \hfill \cr
   - \int_{y = 0}^{\pi /2} {M(0,y)dy}  \hfill \cr}  \right) =   \cr
  &  = I_{\,1}  + I_2  + I_3  \cr
}
$$

Now
$$
I_3  =  - \int_{y = 0}^{\pi /2} {M(0,y)dy}
  = \int_{y = 0}^{\pi /2} {{y \over {2\sin ^2 y}}\ln \left( {{{\sin y} \over {\sin y}}} \right)dy}
  = 0
$$

$I_{\,1}$ instead,  is not null as it might appear.

If the path is taken to approach $y=x$ while maintaining positive the difference $y-x$ then a numerical
computation gives a value of $\approx -1.233688 \ldots$, which together with the computation of $I_{\, 2}$ returns the
expected value for $I$.

I am blocked at this point:  analytically establish and justify the path for $I_{\, 1}$ and find a close expression for it and for $I_{\, 2}$.


问:
How to find $P(N=3\mid X_4=1,X_5=2)$ where $X_i$'s are i.i.d $\text{Exp}(1)$ and $N=\min\{n\ge 1:X_n\in [1,2]\}$?
This is a rather simple exercise I stumbled upon:

Let $X_1,X_2,\ldots$ be i.i.d $\text{Exp}(1)$ random variables and $N=\min\{n\ge 1:X_n\in [1,2]\}$.

How can we find the conditional probability $P(N=3\mid X_4=1,X_5=2)$?

For every $n\ge 1$, I have $$P(N>n)=P(X_1\notin [1,2],X_2\notin [1,2],\ldots,X_n\notin [1,2])=(1-p)^n\,,$$  where $$p=P(X_1\in[1,2])=\frac1e-\frac1{e^2}$$

In other words $N$ has a geometric distribution with pmf $$P(N=n)=p(1-p)^{n-1}\mathbf1_{n\in \{1,2,\ldots\}}$$

But how to argue regarding the value of the conditional probability? I am not sure how to express the event $\{N=3\}$ in terms of the $X_i$'s or the event $\{X_4=1,X_5=2\}$ in terms of $N$ so that I can work with either the distribution of $X_i$'s or $N$.

Does $\{N=3\}$ imply $\{X_1\notin [1,2],X_2\notin [1,2],X_3 \in [1,2]\}$? Any suggestion would be great.

答:
Does $\{N=3\}$ imply $\{X_1\notin [1,2],X_2\notin [1,2],X_3 \in [1,2]\}$?

Yes, that is correct.   Furthermore, it is an equivalence.   $N$ is the value of the index for the first sample from the series to occur in the interval.

$${~~~~N=\min\{n: n\geq 1, X_n\in[1..2]\}\\\Updownarrow\\~~~~ X_N\in[1..2]\wedge\forall k\in[\![1..N-1]\!]~.X_k\notin[1..2]}$$

So then, $\mathsf P(N=3)~{=\mathsf P(X_1\notin[1..2])\mathsf P(X_2\notin[1..2])\mathsf P(X_3\in[1..2])\\=(1+\mathrm e^{-2}-\mathrm e^{-1})^2\,(\mathrm e^{-1}-\mathrm e^{-2})}$

Now use  Bayes' Rule .$$\mathsf P(N=3\mid X_4=1, X_5=2)=\dfrac{\mathsf P(N=3)f_{X_4,X_5\mid N=3}(1,2)}{f_{X_4}(1)f_{X_5}(2)}$$.

And  $f_{X_4,X_5\mid N=3}(1,2)=f_{X_4,X_5\mid X_1\notin[1..2],X_2\notin[1..2],X_3\in[1..2]}(1,2) = ....$


问:
Finding Number of Zeros in the First Quadrant of a Complex Polymomial
I am trying to determine the number of roots of $p(z) = z^6 + 4z^4 + z^3 + 2z^2 + z + 5$ in the first quadrant. I I understand that one does this by finding the change in argument of $p(z)$ over the boundary of the quarter disc of radius $R$ (for large $R$) centered at the origin. I found the change over the real axis and the circular arc, but I don't understand how to find the change over the imaginary axis from $iR$ to $0$.I consulted my book but found the discussion very confusing and not very rigorous. I cou use some help.

Note that $p(it) = -t^6 + 4t^4 - 2t^2 + 5 + i(-t^3 + t)$, where $0 \le t \le R$. Because $-t^6$ dominates the real part, and $-t^3$ dominates the imaginary part, we see that $p(iR)$ lies in the third quadrant.
Not exactly sure where in the 3rd quadrant it is, so I can't say exactly what $\arg p(iR)$ is, but whatever I guess. Then from my understanding I am suppose to find the nonnegative zeros of the real and imaginary part and track where it crosses the real and imaginary axis, and then look at it's final position. But this how can I calculate the argument change if I don't know exactly where these points are?
$p(0) = 5$, so it ends up in the first quadrant. According to a solution I found, the change in argument over the imaginary axis is $\cong \pi$, but I don't see it for some reason. I don't like these estimates.

答: There is not too much more to say, and what little it takes is basically Calculus I curve-sketching analysis. Observe that when $t>1$, $p(it)$ is in the lower half-plane and, as $t\to\infty$, $\arg(p(it))\to -\pi$, since the slope for large $t$ is approaching $0$. On the interval $[0,1]$, $p(it)$ starts and ends on the positive real axis, staying in the first quadrant between. Thus, $\Delta\arg_{[0,i]} p(z)=0$. Since $p(z)$ is in the second and third quadrants for $t>1$, we conclude that $|\arg(p(it)) - \arg(p(i))|<\pi$ for all $t>1$. Thus $\Delta_{[0,R]}\arg f(it)$ approaches $-\pi$ as $R\to\infty$. Now just turn everything upside-down.


问:
Finding the R-minimal and R-maximal
For any two integers $a,b$, we say that $a$ divides $b$ (written as $a\mid b$) if $b=ak$ for some $k∈\Bbb Z$.
Let $A=\{3,6,7,9,12,14,21,42,252\}$. Consider the partial order relation $R$ on $A$ given by $aRb↔a∣b$.
Find all the $R$-minimal and $R$-maximal elements of A.
How would I go about finding the $R$-minimal and $R$-maximal elements of $A$ in this question? Would it just be the smallest number (which is $3$) for the $R$-minimal and the  biggest number (which is $252$) for the $R$-maximal? If not, how would I go about solving this question?

答:
Yes, $3$ is a $R$-minimal element, but not the only one. Another $R$-minimal element is $7$, since $a\mathrel R7$ never occurs, unless $a=7$. Actually, $3$ and $7$ are the only $R$-minimal elements of $A$: if $n\in A\setminus\{3,7\}$, then $3\mathrel Ra$ or $7\mathrel Ra$.

Can you find the $R$-maximal elements now?

Title: Finding the R-minimal and R-maximal


问:
How do you visualize polynomials rings and their quotient?
When working in the framework of group theory I could easily visualize the group by imagining its elements and I could see the group "collapse" when forming the quotient group by a normal subgroup by simply imagining the elements of the subgroup all being collapsed onto the identity, same thing for the other cosets.

But for polynomial rings I have a very hard thing visualizing this and it has lead to a poor understanding of ideals and quotient rings since I cannot rely on my visualize intuition anymore. I cannot make sense of $\mathbb{R}[x]$, its ideal $x^2+1$ and its quotient ring by that ideal.

I was wondering if you had any help in this regard because I am struggling here.

答:
I visualize elements of the polynomial ring $\mathbb R[x]$ as, well, polynomials, although perhaps it helps to keep one's focus on the $0$ degree term:
\begin{align*}
    & a_0 \\
a_1 x + & a_0 \\
a_2 x^2 + a_1 x + &a_0 \\
a_3 x^3 + a_2 x^2 + a_1 x + &a_0
\end{align*}
and so on.

And then, when you mod out by $x^2+1$, welp, that's just substituting every $x^2$ by $-1$, and every $x^3$ by $-x$,
\begin{align*}
    & a_0 \\
a_1 x + & a_0 \\
a_2 (-1) + a_1 x + &a_0 \\
a_3 (-x) + a_2 (-1) + a_1 x + &a_0
\end{align*}
and so on... and then you have to collect terms of course, which shifts the terms formerly known as the $x^3$ and $x^2$ terms two to the right, and shifts the terms formerly known as the $x^5$ and $x^4$ terms four to the right, and so on.


问:
Proving by contradiction check if correct
Consider the statement: For all x, y, z ∈ Z. At least one of x− y, x− z
and y − z is even

I want to check if my proof is correct via contradiction:

Suppose(for a contradiction) that all of x-y, x-z, and y-z are odd:

x - y = 2k+1 for some k ∈ Z

rearranging for x:

x=2k+1+y

y - z  = 2n+1  for some n ∈ Z

rearranging for z:

z=y-2n-1

taking x-z and subbing in

x - z  =2k+1+y-(y-2n-1)

= 2k+2n+2

=2(k+1+n)

therefore x-y is even which is a contradiction therefore the original statement is true

答:
Yes your proof looks fine.

Another way to think about this is to notice that the sum
$$ ( x - y) + (y - z) + (x-z) = 2 (x-z) $$
is even, so the numbers $(x-y), (y-z), (x-z)$ can't all be odd.

If you want to continue posting here you should consider learning LaTeX. It will make your equations easier to read.


问:
Visualization of branched covering map between Riemann surface
I was having a hard time imagine the picture of a branched covering, the locally $z^k$ description is more "real" or "1-dimension" to me. Anybody have a good picture/example to visualize this?

答: This picture (https://upload.wikimedia.org/wikipedia/commons/9/9c/Riemann_sqrt.svg) gives you a pretty good idea of what $z \mapsto z^2$ looks like near $z=0$. For the general case $z \mapsto z^k$, there will be $k$ distinct sheets away from the origin, but they will all come together at $0$.


问:
What are the eigenvalues of the matrix $A+2I$ when $A$ is given?
Question: What are the eigenvalues of the matrix $A+2I$ if
$A =\begin{bmatrix}3 & 2\\5 & 0\end{bmatrix}$?

My attempted solution: I'm actually not quite sure how I should approach this problem. I know that that the augumenting matrix $A+2I$ is $\begin{bmatrix}5 & 2\\5 & 2\end{bmatrix}$, but other than that I'm not quite sure how to go about solving this problem. I'd be much appreciated if someone could explain this to me.

Thanks in advance!

答:
If $\lambda$ is an eigenvalue of $A$, then $Av=\lambda v$ for some $v\ne0$. Therefore
$$
(A+2I)v=Av+2v=\lambda v+2v=(\lambda+2)v
$$
and so $\lambda+2$ is an eigenvalue of $A+2I$.

The converse is also true: if $\mu$ is an eigenvalue of $A+2I$, then $\mu-2$ is an eigenvalue of $A$ (similar proof).

Thus you find all the eigenvalues of $A+2I$ by adding $2$ to the eigenvalues of $A$. Note that $A$ can be of any order.

The characteristic polynomial of $A$ is $x^2-3x-10=(x+2)(x-5)$, so the eigenvalues are $-2$ and $5$. The eigenvalues of $A+2I$ are $0$ and $7$.


答:
The theorem is not at all useless.
Note that if $n = x^2 + y^2$, $n^2 = (x^2-y^2)^2 + (2xy)^2$.  Thus if $n$ is the sum of two squares, neither $0$ and not both equal, then $n^2$ is the sum of two nonzero squares.

Title: Sum of two squares theorem

Body:
I am working on the problem: I need to quickly check if positive number $n$ can be expressed as $n^2=a^2+b^2$. I found this theorem: https://en.wikipedia.org/wiki/Sum_of_two_squares_theorem

But it seems I am missing something. For example number $4$ falls under theorem condition: its prime decomposition doesn't have any prime $p\equiv 3{\pmod {4}}$. But $4$ cannot be expressed as sum of 2 squares.


问:
How to prove every closed interval in R is compact?
Let $[a,b]\subseteq \mathbb R$. As we know, it is compact. This is a very important result. However, the proof for the result may be not familiar to us. Here I want to collect the ways to prove $[a,b]$ is compact.

Thanks for your help and any link.

答:
Here's a proof using a non-standard approach.

With the help of transfer principle, we could reformulate compactness as:

$A \subset \mathbb R$ is compact, iff for each $y \in ^\ast A$, there is $x \in A$ such that $x$ is infinitely close to $y$.

We could prove a bounded closed interval is compact by showing that every closed and bounded subset of $\Bbb R$ is compact.

Proof: Let $B$ be a closed and bounded subset of $\mathbb{R}$. It suffice to show that $\operatorname{st}(^\ast B) \subseteq B$. Since $B$ is closed, we only need to show that $\operatorname{st}(^\ast B) \subseteq \overline B$. For each $x \in \operatorname{st}(^\ast B)$, there is $y \in ^\ast B$ such that $\operatorname{st}(y) = x$. Appealing to the ultrapower construction, $y$ is actually a sequence $\{r_i\}_{i \in \Bbb{N}}$ such that $r_i \in B$ for all $i \in \Bbb{N}$. We can construct a new sequence of elements in $B$ that converge to $x$, which implies $x \in \overline B$. This construction is borrowed from an answer by Brian M. Scott. Given the free ultrafilter $\mathscr{U}$. Define $A_n = \{r_i:|r_i -x |<\frac{1}{n+1}\} \in \mathscr{U}$ for all $n \in \mathbb{N} $. With $\bf AC$, we have a choice function $f$ such that $A_n \mapsto a_n$ satisfying $a_n \in A_n$. Then the sequence $\{a_n\}_{n \in \Bbb N}$ is just what we need.


问:
Inequality for an linear operator
Consider in $X=\left ( C\left [ 0,1 \right ],\left \| \cdot  \right \|_{\infty} \right )$ the operator $T$ defined by $\left ( Tf \right )\left ( t \right )=\int_{0}^{t}f\left ( s \right )ds$, where $t \in\left [ 0,1 \right ]$. Show that $\left \| T^n \right \| \leq \frac{1}{n!}$  that for all $n \in \mathbb{N}$. Is $\left \| T^n \right \|=\left \| T \right \|^n$?

My attempt:
First, I found $\left \| Tf \right \|_{\infty} \leq 1$ through some inequalities, and then I fixed $f \equiv 1$ in order to get $\frac{\left \| Tf \right \|_{\infty} }{\left \|f \right \|_{\infty} }=1$, therefore $\left \| T \right \|=1$. Now I'm stuck with the inequality above, can anyone give me a hint? Thanks in advance.

答: Hint: Let $g(t)=1$ for $t \geq 0$ and $g(t)=0$ for $t <1$. Then $Tf=f*g$. Show  that $T^{n}f=f*g^{(n)}$ where $g^{(n)}$ stands for $g*g*...*g$ ($n-$ fold convolution). Show by direct computation that $g^{(n)}(t)=\frac {t^{n}} {n!}$ for $t >0$ and $0$ for $t <1$. This gives $\|T^{n}\| \leq \frac 1  {n!}$. So $\|T^{n}\|<\|T\|^{n}$ for $n >1$.

Title: Inequality for an linear operator


问:
For any two positive operator $a,b$ with norm $<1$ there is a $c$ such that $a,b\leq c$ and $\|c\|<1$. Does this hold for $\|a\|=\|b\|=1$?
I mean, for two positive operator $a,b$ with norm $1$, is there still always a positive $c$ such that $a,b\leq c$ and $\|c\|=1$?
Suppose it is in a non-unital C*algebra.

And $a\leq b$ means $b-a$ is positive.

答:
The answer is negative and here is a counter-example.   Let us begin with the following:

Lemma. If $T$ and $S$ are positive operators on a Hilbert space, with $0\leq T\leq S\leq 1$,
and if $\xi $ is any vector such that $T\xi =\xi $,  then  $S\xi =\xi $,  as well.

Proof.
We have
$$
  \|\xi \|^2 = \langle \xi , \xi \rangle = \langle T\xi , \xi \rangle \leq \langle S\xi , \xi \rangle \leq   \|S\xi \|\|\xi \| \leq \|\xi \|^2,
  $$
so equality holds throughout.   Cauchy-Scwartz inequality,  namely the second inequality above, being an equality, we deduce
that $S\xi =\xi $, as desired.  QED

Let us introduce  the subalgebra
$$
  A\subseteq  C([0, 1])\otimes M_2(\mathbb C) = C\big ([0, 1], M_2(\mathbb C)\big )
  $$
formed by all continuous functions
$$
  f: [0, 1]\to  M_2(\mathbb C),
  $$
such that
$$
  f(1) = \pmatrix{z&0\cr 0&0},
  $$
for some $z$ in $\mathbb C$.
Consider the elements $a$ and $b$ in $A$ given, for every $t$ in $[0,1]$, by
$$
  a(t) = \pmatrix{1&0\cr 0&0},
  $$
and
$$
  b(t) = \pmatrix{t&\sqrt{t-t^2}\cr \sqrt{t-t^2}&1-t}.
  $$

Notice that both $a$ and $b$ are projections, hence positive elements with norm 1.

Theorem.  There exists a unique element $c$ in $C([0, 1])\otimes M_2(\mathbb C)$, such that  $a, b\leq c$, and $\|c\|\leq 1$.

Proof.  The existence is easily verified by taking $c=1$.  Next
suppose that $c\in C([0, 1])\otimes M_2(\mathbb C)$ is such that $a, b\leq c\leq 1$.  It then follows that
$$
  a(t), b(t)\leq c(t)\leq 1,
  $$
for every $t\in [0, 1]$.

Setting $\xi =(1,0)$, we have that $a(t)\xi =\xi $, so $c(t)\xi =\xi $, by the Lemma.

On the other hand,  setting $\eta _t=\left(t,\sqrt{t-t^2}\right)$, we have that $b(t)\eta _t=\eta _t$, so again $c(t)\eta _t=\eta _t$.

Observing that $\{\xi , \eta _t\}$ spans $\mathbb C^2$,  for $t\in (0,1)$, we deduce that $c(t)$ is the identity $2\times 2$ matrix for
every such $t$, and hence also for $t=0$ and $t=1$, by continuity.  QED

Observing that the element $c$ referred to above is not in $A$, we see that there is no element $c$ in $A$  doing the job!


问:
Let $\mathcal{G}$ be a $\sigma$-algebra. Show that if $\mathbb{E}((\mathbb{E}(X|\mathcal{G}))^2)=\mathbb{E}(X^2)$ then $\mathbb{E}(X|\mathcal{G})=X$?
$X$ has finite mean.

Of course, in general $\mathbb{E}(X^2)=\mathbb{E}(Y^2)$ does not imply $X=Y$.

I tried using the fact that  $(\mathbb{E}(X|\mathcal{G}))^2=\mathbb{E}(X^2|\mathcal{G})-\text{Var}(X|\mathcal{G})$ and also $\mathbb{E}(X^2)=(\mathbb{E}(X))^2+\text{Var}(X)$, hence:

$\mathbb{E}((\mathbb{E}(X|\mathcal{G}))^2)=\mathbb{E}(\mathbb{E}(X^2|\mathcal{G})-\text{Var}(X|\mathcal{G}))  = \mathbb{E}((\mathbb{E}(X))^2+\text{Var}(X)) = \mathbb{E}(\mathbb{E}(X^2)) $

So we can apply the Tower rule and linearity of expectation to get:

$\mathbb{E}(X^2)-\mathbb{E}(\text{Var}(X|\mathcal{G}))=   (\mathbb{E}(X))^2+\mathbb{E}(\text{Var}(X)) $.

This is where I'm stuck. We could rearrange to get $\text{Var}(X)= \mathbb{E}(\text{Var}(X|\mathcal{G})) + \mathbb{E}(\text{Var}(X)) $

And so this implies that $\mathbb{E}(\text{Var}(X)) = \text{Var}(\mathbb{E}(X|\mathcal{G})) $ by the Law of  Total Variance. I.e. $ \text{Var}(X)) = \text{Var}(\mathbb{E}(X|\mathcal{G}))$, but I don't think that this implies $X=\mathbb{E}(X|\mathcal{G})$?

答:
We also need $X$ to be square-integrable in order for $\mathbb E[X^2]$ to make sense. I assume this is the case.

First, note that
\begin{align*}
\mathbb E\big[\mathbb E[X|\mathcal G]^2\big]=\mathbb E\big[\mathbb E[X|\mathcal G]\mathbb E[X|\mathcal G]\big]\underset{\spadesuit}{=}\mathbb E\bigg[\mathbb E\big[\mathbb E[X|\mathcal G]X\big|\mathcal G\big]\bigg]\underset{\heartsuit}{=}\mathbb E\big[\mathbb E[X|\mathcal G]X\big],\tag{$\star$}
\end{align*}
where $\spadesuit$ is due to the fact that $\mathbb E[X|\mathcal G]$ is $\mathcal G$-measurable (so it can be pulled inside the other conditional expectation) and $\heartsuit$ is the law of total expectation.

Now define $Y\equiv\mathbb E[X|\mathcal G]-X$. Clearly, $\mathbb E[Y]=0$, so $\operatorname{Var}[Y]=\mathbb E[Y^2]$. But
\begin{align*}
\mathbb E[Y^2]&=\mathbb E\big[\mathbb E[X|\mathcal G]^2\big]+\mathbb E[X^2]-2\mathbb E\big[\mathbb E[X|\mathcal G]X\big]\\
&=\mathbb E\big[\mathbb E[X|\mathcal G]^2\big]+\mathbb E[X^2]-2\mathbb E\big[\mathbb E[X|\mathcal G]^2\big]\\
&=\mathbb E[X^2]-\mathbb E\big[\mathbb E[X|\mathcal G]^2\big]\\
&=0,
\end{align*}
where the second equality comes from $(\star)$ and the last one from your leading assumption. Consequently, $Y$ (almost surely) equals $0$.


问:
Does L'Hopital's rule imply that $\lim_{x\to a}\frac{f'(x)}{g'(x)} = \lim_{x\to a}\frac{f(x)}{g(x)}$ always?
My question is summarised as follows:

Is $$\lim_{x\to a}\frac{f'(x)}{g'(x)} $$
always equal to $$\lim_{x\to a}\frac{f(x)}{g(x)} $$
or can we only deduce values of limits using differentiation in the forward direction, ie we can't deduce that the value of $\lim_{x\to a}\frac{f'(x)}{g'(x)} $ is the same as $\lim_{x\to a}\frac{f(x)}{g(x)} $?

Logically I cannot see why this shouldn't be true (ie I can't see why we can't also integrate the numerator and denominator). HOWEVER, the following result seems to throw some doubt on this:

Consider the identity
$$f(x)^2\frac{d}{dx}\frac{g(x)}{f(x)}+g(x)^2\frac{d}{dx}\frac{f(x)}{g(x)}\equiv0$$
$$\implies\frac{\frac{d}{dx}\frac{f(x)}{g(x)}}{\frac{d}{dx}\frac{g(x)}{f(x)}}\equiv-\frac{f(x)^2}{g(x)^2}$$
On applying L'Hopital's rule we obtain
$$\lim_{x\to a}\frac{\frac{d}{dx}\frac{f(x)}{g(x)}}{\frac{d}{dx}\frac{g(x)}{f(x)}}\equiv\lim_{x\to a}-\frac{f(x)^2}{g(x)^2}\implies\lim_{x\to a}\frac{\frac{f(x)}{g(x)}}{\frac{g(x)}{f(x)}}\equiv\lim_{x\to a}-\frac{f(x)^2}{g(x)^2}$$
$$\implies\lim_{x\to a}\frac{f(x)^2}{g(x)^2}\equiv\lim_{x\to a}-\frac{f(x)^2}{g(x)^2}$$
which is obviously false.

Any help explaining this apparent proof would be appreciated.

答:
For clarity, let's let $F(x)=f(x)/g(x)$ and $G(x)=g(x)/f(x)$. The identity you call attention to can be written as

$${F'(x)\over G'(x)}=-{f(x)^2\over g(x)^2}$$

and we also, of course, have

$${f(x)^2\over g(x)^2}={f(x)/g(x)\over g(x)/f(x)}={F(x)\over G(x)}$$

Now if L'Hopital's rule applied to the limit for $F/G$, then we would have

$$\lim{f(x)^2\over g(x)^2}=\lim{F(x)\over G(x)}\color{red}{=}\lim{F'(x)\over G'(x)}=-\lim{f(x)^2\over g(x)^2}$$

(where the red equal sign is the L'Hopital step).  But in order to apply L'Hopital to the ratio $F/G$, we must have either $\lim F(x)=\lim G(x)=0$ or $\lim F(x)=\lim G(x)=\pm\infty$, and this cannot happen, since $F(x)$ and $G(x)$ are reciprocals (i.e., $G(x)=1/F(x)$). In short, there is no paradox, because L'Hopital does not apply to the ratio $F/G$.


问:
Limits of a vector proof
If $x_n, x, y$ are vectors in $R^m$ and $x_n \rightarrow x$, prove that $\langle x_n, y\rangle \rightarrow \langle x, y \rangle$.

I am assuming one can simply say that $x_n$ $\rightarrow$ $x$ so we get the result but I don’t understand how vectors work and an explanation of how this proof works would go a long way.

答:
By Cauchy-Schwarz inequality and $\|x_n - x\| \to 0$ as $n \to \infty$:
\begin{align*}
|\langle x_n, y \rangle - \langle x, y \rangle |
= |\langle x_n - x, y \rangle| \leq \|x_n - x\|\|y\| \to 0
\end{align*}
as $n \to \infty$.

Title: Limits of a vector proof


问:
Solving an invertible block lower triangular matrix with zeros on the right hand side
Suppose we're given the following (physics-based) system of two matrix equations, involving an invertible block lower triangular matrix:
\begin{align}
    \begin{bmatrix}
        \mathbf{A}_{1,1} & \mathbf{0}\\
        \mathbf{A}_{2,1} & \mathbf{A}_{2,2}
    \end{bmatrix}
    \begin{bmatrix}
        \mathbf{x}_1 \\ \mathbf{x}_2
    \end{bmatrix}
    =
    \begin{bmatrix}
        \mathbf{0} \\ \mathbf{b}_2
    \end{bmatrix},
\end{align}
where $\mathbf{A}_{1,1}$ has size $m \times m$, $\mathbf{A}_{2,2}$ has size $n \times n$.
The above equation represents two systems of equations which were discretized via the boundary element method.

Since we know that $\mathbf{A}_{1,1}$ must be invertible (because we're told the entire system matrix is invertible, e.g.), then it means $\mathbf{A}_{1,1}$ should not have a null space, therefore the only solution to this system should be
\begin{align}
    \mathbf{x}_1 &= \mathbf{0},\\
    \mathbf{x}_2 &= \mathbf{A}_{2,2}^{-1}\,\mathbf{b}_2.
\end{align}
Assuming the above makes sense, my issue is that the physics dictates that $\mathbf{x}_1 \neq \mathbf{0}$.
Does this mean that I have set up my system wrong, and am missing some terms? Or is it possible somehow to impose a constraint on $\mathbf{x}_1$ so that it isn't $\mathbf{0}$? I'm not very experienced in linear algebra, unfortunately.

Thanks.

答:
A block matrix like yours
\begin{bmatrix}
A & 0 \\
B & C
\end{bmatrix}
is invertible if and only if $A$ and $C$ are. Its inverse is
\begin{bmatrix}
A^{-1} & 0 \\
X & C^{-1}
\end{bmatrix}
provided
$$
\begin{bmatrix}
A & 0 \\
B & C
\end{bmatrix}
\begin{bmatrix}
A^{-1} & 0 \\
X & C^{-1}
\end{bmatrix}
=
\begin{bmatrix}
I & 0 \\
BA^{-1}+CX & I
\end{bmatrix}
$$
is the identity matrix, which means $X=-C^{-1}BA^{-1}$.

Thus the unique solution to
$$
\begin{bmatrix}
A & 0 \\
B & C
\end{bmatrix}
\begin{bmatrix} x \\ y \end{bmatrix}
=
\begin{bmatrix} a \\ b \end{bmatrix}
$$
is
$$
\begin{bmatrix} x \\ y \end{bmatrix}=
\begin{bmatrix}
A^{-1} & 0 \\
-C^{-1}BA^{-1} & C^{-1}
\end{bmatrix}
\begin{bmatrix} a \\ b \end{bmatrix}
=
\begin{bmatrix} A^{-1}a \\ -C^{-1}BA^{-1}a + C^{-1}b \end{bmatrix}
$$
If $a=0$, then $x=0$. So if you want $x\ne0$, the upper block of $\left[\begin{smallmatrix} a \\ b \end{smallmatrix}\right]$ has to be nonzero.


问:
What exactly is the relationship between the concepts of conjugate complex vector space and conjugations/real structures?
I started studying the book of Daniel Huybrechts, Complex Geometry An Introduction. I tried studying backwards as much as possible, but I have been stuck on the concepts of almost complex structures and complexification. I have studied several books and articles on the matter including ones by Keith Conrad, Jordan Bell, Gregory W. Moore, Steven Roman, Suetin, Kostrikin and Mainin, Gauthier

I have several questions on the concepts of almost complex structures and complexification. Here is one:

Assumptions, notations and what I understand so far: Let $V$ be a $\mathbb C$-vector space. Let $W$ be an $\mathbb R$-vector space.

Let $V_{\mathbb R}$ be the realification of $V$. For any almost complex structure $I$ on $V_{\mathbb R}$, denote by $(V_{\mathbb R},I)$ as the unique $\mathbb C$-vector space whose complex structure is given $(a+bi) \cdot v := av + bI(v)$. Let $i^{\sharp}$ be the unique almost complex structure on $V_{\mathbb R}$ such that $V=(V_{\mathbb R},i^{\sharp})$.

Let $W^{\mathbb C}$ denote the complexification of $W$ given by $W^{\mathbb C} := (W^2,J)$, where $J$ is the 'canonical' almost complex structure on $W^2$ given by $J(v,w):=(-w,v)$. The map $\chi: W^2 \to W^2$, $\chi(v,w):=(v,-w)$ is such that $\chi^J: W^{\mathbb C} \to W^{\mathbb C}$, which is $\chi$ now viewed as a map on $W^{\mathbb C}$ instead of $W^2$, is the 'canonical' conjugation/real structure. Here, 'canonical' is meant in the sense that we would use $J$ and $\chi$ to define complexifications of $W$ and of elements of $End_{\mathbb R}(W)$. (See here.)

Then the complex conjugate of $V$ is defined $\overline V := (V_{\mathbb R},-i^{\sharp})$.

Question: What exactly is  the relationship between the concept of $\overline V$, the conjugation of $V$ and the concept of conjugations/real structures on $V$?

答:
A common misconception I've run across is to conflate the conjugate space $\overline V$ with $V$ in too strong a fashion. I will attempt to describe what I mean by this below. (Disclaimer: I am not a complex geometer. Far from it. This answer is just sharing  what I have understood from attempts to resolve my previous and lasting confusions. ;-;)

a) For a complex vector space $(V,+,\cdot)$, the conjugate space $\big(\overline V,+,\star\big)$ is defined to have the same underlying set and additive structure, but scalar multiplication is conjugated: $\lambda \star v=\bar\lambda\cdot v.$ Since $V$ and $\overline V$ have the same dimension, there is a $\mathbb C$-linear isomorphism $V\cong \overline V,$ but it is not canonical. The $\mathbb C$-linear isomorphisms $V\cong \overline V$ are in natural bijection with $\mathbb C$-antilinear bijections $V\rightarrow V$ (which does not include $\text{Id}_V$).

b) Now what is a real structure on $V$? There are several ways to look at it, but one is as a $\mathbb C$-antilinear involution $\varphi:V\rightarrow V.$ That means that it has the extra condition that $\varphi^2=\text{Id}_V$, which is stricter than just having some $\mathbb C$-antilinear bijection $V\rightarrow V$. Thus identifying the space $V$ with its conjugation $\overline V$ is a strictly weaker notion than choosing a real structure on $V$.

To see this in action, check out this question that I had a while back. Constructions tend to only work on bundles if you can do them canonically (perhaps up to homotopy), because otherwise things may twist up and clash globally. The answer to the linked question gives a complex vector bundle that is isomorphic to its conjugate bundle, but is not a complexification of a real vector bundle. In other words, you can do (a) globally, but you can't do (b) globally.

This "weakness" isn't to say that conjugate vector spaces aren't useful; in fact, they will probably occur far more than real structures in your reading of Huybrechts. For example, everything that you do in terms of $(p,q)$-forms is based on a decomposition $V\otimes_{\mathbb R}\mathbb C\cong V\oplus \overline V$.

Update: Here's another "false resemblance." A real structure on a $\mathbb C$-vector space $(V,J)$ is a way of identifying $V$ as the complexification of an $\mathbb R$-vector space. But what if we complexify something that already has a complex structure? By this, I mean that we form $(V_\mathbb{R})^\mathbb{C}=V\otimes_\mathbb{R}\mathbb C$. We can write this as $V\otimes_\mathbb{R}\mathbb C=V\oplus iV$ or $V\otimes_\mathbb{R}\mathbb C=V\oplus \overline V$. This seems to imply that we can identify $iV$ and $\overline V$, but this is not the case. If we are careful, we actually see that the two copies of $V$ are different:

In the decomposition $V\otimes_\mathbb{R}\mathbb C=V\oplus iV$, we embed $V\hookrightarrow V\otimes_\mathbb{R}\mathbb C$ and $iV\hookrightarrow V\otimes_\mathbb{R}\mathbb C$ by the maps $v\mapsto v\otimes 1$ and $v\mapsto v\otimes i$, respectively. This "remembers" how $V\otimes_\mathbb{R}\mathbb C$ was formed as a complexification, but it doesn't make $J$ and $i$ work in a particularly compatible way. Note that $V$ and $iV$ are both closed under $J$, but they are interchanged when we multiply by $i$. Thus, we have two complex structures, but these are only complex subspaces with respect to one.

In the decomposition $V\otimes_\mathbb{R}\mathbb C=V\oplus \overline V$, we embed $V\hookrightarrow V\otimes_\mathbb{R}\mathbb C$ and $\overline V\hookrightarrow V\otimes_\mathbb{R}\mathbb C$ by the maps $v\mapsto v\otimes 1-Jv\otimes i$ and $v\mapsto v\otimes 1+Jv\otimes i$, respectively. This is different from the above, although perhaps not super enlightening. More useful is the description $$V=\{w\in V\otimes_\mathbb{R}\mathbb C:Jv=iv\}\quad\text{and}\quad \overline V=\{w\in V\otimes_\mathbb{R}\mathbb C:Jv=-iv\}.$$ This shows us that $V$ and $\overline V$ are complex subspaces under both complex structures. These two structures are equal on $V$ and opposite on $\overline V,$ which allows us to talk about "holomorphic" and "anti-holomorphic" things on a complex (or almost complex) manifold.


问:
What are some relationships between a matrix and its transpose?
All I can think of are that
If symmetric, they're equivalent

If A is orthogonal, then its transpose is equivalent to its inverse.

They have the same rank and determinant.

Is there any relationship between their images/kernels or even eigenvalues?

答:
Fix a ring $R$, and let $A \in M_n(R)$.  The characteristic polynomial for $A$ is $$\chi_A(x)=\det (xI-A),$$
so that $\chi_{A^T}(x) = \det (x I -A^T)= \det ((xI-A)^T)=\det(xI-A)$.  Since the eigenvalues of $A$ and $A^T$ are the roots of their respective characteristic polynomials, $A$ and $A^T$ have the same eigenvalues.  (Moreover, they have the same characteristic polynomial.)

It follows that $A$ and $A^T$ have the same minimal polynomial as well, because the minimal polynomial arises as the radical of the characteristic polynomial.


问:
Suppose that $\sec(\theta)=\frac{5}{3}$ and $\sin(\theta)<0$ find the following:
I am looking for some help with the following questions I am a little lost. It has been sometime before I worked with trig.

Suppose that $\sec(\theta)=\frac{5}{3}$ and $\sin(\theta)<0$ find the following:

$\tan(2\theta)$

$\tan(\frac{\theta}{2})$

$\cos\frac{\theta}{2}$

答:
$\sec \theta =
\frac 5 3 \iff \cos \theta = \frac 3 5$

$\sin \theta =-\sqrt{1-\cos^2 \theta}=-\frac4 5 $

You can find $\tan \theta=\frac{\sin \theta}{\cos \theta}$, and hence $\tan 2 \theta = \frac{2\tan \theta}{1-\tan^2 \theta}$.

$\tan \frac \theta 2 = \frac{\sin \theta}{1+\cos \theta}$

$\cos \theta = 2\cos^2 \frac \theta 2 -1$


问:
Why do we study irreducibility of polynomials? What are some real-life applications?
Irreducibility of polynomial functions seems like one of the major topics in introductory abstract algebra, and there are many theorems (such as Eisenstein's Criterion, etc) on testing whether a polynomial is reducible in $\mathbb{Q}[x], \mathbb{R}[x],$ and $\mathbb{C}[x]$, etc.

I wonder, why do we care so much about irreducibility? What are some of the applications (preferably in engineering or physics, but other areas are also great)?

Thanks!

答:
Polynomial codes are error detection and correction codes that start by choosing a finite field of prime power order, $GF(p^n)$.  The easiest way to explicitly construct such a field in which you can actually compute is to construct a non-prime field.  For instance, if $P(X)$ is an irreducible polynomial of degree $n> 1$, then $GF(p^n) = GF(p)[X]/P(X)$ is a finite field of order $p^n$.  (It doesn't matter which irreducible polynomial of degree $n$ you use -- you get the same field, but the names of some of the elements seem different.)

Polynomial codes include

cyclic codes, which give cyclic redundancy checks(CRCs).  iSCSI and SCTP have incorporated CRCs.;

BCH codes and Reed-Solomon codes, which are used in radio communication, CDs, DVDs, HDDs, SSDs, some 2-d barcodes, e.g. QR codes, data transmission and reconstruction, and other forms of data storage and retrieval.


问:
If $U$ is a subspace of a normed space $X$, must points outside $U$ be a nonzero distance from $U$?
Let $X$ be a normed space over $\mathbb K=\mathbb R$ or $\mathbb C$ and $U\subsetneq X$ be a proper subspace. Suppose further that $X\ni x \notin U$.

Must we have $d(x,U)>0$ in general?

Must we have $d(x,U)>0$ if $\dim U<\infty$?

Note that $d(x,U):=\inf_{u\in U} \| x-u \|$.

I'm pretty sure the answer to (2) is "yes," since in this case $X\simeq\mathbb K^{\dim U}$, and then we can use topological theorems about $\mathbb K^n$ (as in this related question). Even so, I'd like to see a simpler proof that doesn't rely on the isomorphism with $\mathbb K^n$, since this seems like unnecessary baggage.

答:
Hints:

take $X$ to be a space of convergent sequences and $U$ to be the subspace of $X$ comprising sequences that are finitely non-zero. This shows that claim (1) is false.

choose a basis $\langle u_1, \ldots u_k \rangle$ for $U$ and extend it to the basis $\langle u_1, \ldots u_k, x \rangle$ for $\mathrm{span}(U, x)$. If $d(x, U) = 0$, then $x \in U$. (This is essentially the argument that you don't like because of what you describe as excess baggage, but there is no way of avoiding it.)


答: $d(x,U)=0$ if and only if $x \in \overline U$. So a dense porper subspace gives  a counter-example to the first part. For a specific example take $U=\ell^{0}$ in $X=\ell^{2}$.  Since finite dimensional subspaces are closed we have $d(x,U) >0$ when $U$ is finite dimensional.

Title: If $U$ is a subspace of a normed space $X$, must points outside $U$ be a nonzero distance from $U$?

Body:
Let $X$ be a normed space over $\mathbb K=\mathbb R$ or $\mathbb C$ and $U\subsetneq X$ be a proper subspace. Suppose further that $X\ni x \notin U$.

Must we have $d(x,U)>0$ in general?

Must we have $d(x,U)>0$ if $\dim U<\infty$?

Note that $d(x,U):=\inf_{u\in U} \| x-u \|$.

I'm pretty sure the answer to (2) is "yes," since in this case $X\simeq\mathbb K^{\dim U}$, and then we can use topological theorems about $\mathbb K^n$ (as in this related question). Even so, I'd like to see a simpler proof that doesn't rely on the isomorphism with $\mathbb K^n$, since this seems like unnecessary baggage.


问:
What exactly is $\mathbb{Z_n}$?
My impression has been that $\mathbb{Z_n}$ is the set $\{0,1,...,n-1\}$ under binary operation addition modulo $n$. However I'm also coming across this notion that $\mathbb{Z_n}$ is actually a set of equivalence classes of equivalence relation $x\sim y \iff x \equiv y$ mod $n$ and the addition here is actually addition of equivalence classes rather than simply addition of integers modulo $n$. Is this correct? So would it be correct to say $\mathbb{Z_n} = \{n,n+1,...,2n-1\}$? if we are considering these elements as being equivalence classes?

答:
Below I explain the general idea, which works not only for rings but also for any algebraic structure definable by $\rm\color{#c00}{universal}$ equational axioms ("identities") such as $\,\color{#c00}{\forall x,y,z\!:}\,x(y+z) = xy+xz\,$.

$a\equiv b\pmod{\! n}\iff n\mid a-b\,$ is a congruence relation (on a ring) i.e. it is  an equivalence relation that is furthermore compatible with all of the ring operations addition and multiplication, i.e. congruence satisfies the following Congruence Sum & Product Rules

$$\begin{align} a_1\equiv b_1\\ a_2\equiv b_2\end{align}\ \Rightarrow\ \begin{array}{}a_1 + a_2\equiv b_1+b_2\\ a_1 \times a_2\equiv b_1 \times b_2 \end{array}\qquad$$

This implies that the ring operations descend to well-defined operations on the equivalence classes $\,[a] = a+n\Bbb Z\,$ via $\,[a]+[b] := [a+b],\ [a]\times [b] := [a\times b],\,$ and the map $\, a\mapsto [a]\,$ is a surjective ring hom, which immediately implies that all the ring laws persist in the image $\,\Bbb Z_n,\,$ so $\,\Bbb Z_n\,$ has associative and commutative addition and multiplication, connected via the distributive law, so arithmetic in $\,\Bbb Z_n\,$ is essentially the same as in $\,\Bbb Z,\,$ except that some elements are forced equal.

For computational purposes it is often convenient to map the classes to normal (canonical) representatives $\,h\,:\, [a]\mapsto \bar a.\,$ The most common choice is its least nonnegative element $\,\bar a := a\bmod n\,$ (remainder reps), but also convenient are least magnitude reps, e.g. $\,0,\pm1,\pm2\pmod{\!5}.\,$ Generally we can use any complete system of residues, i.e. any $\,n\,$ integers such that every integer is congruent to exactly one in our set of complete residue reps.

Then we can transport the ring structure to our normal forms by pulling (back) the ring operations along $h$ to obtain the induced ring operations on the normal forms as follows:

$$\color{#c00}{\bar a} + \bar b = h([a])+h([b]) = h([a]+[b]) = h([a+b]) = \color{#0a0}{\overline{a+b}}$$

e.g. this becomes   $\ \color{#c00}{a\bmod n} + b\bmod n\, = \, \color{#0a0}{(a+b)\bmod n}\ $ using common normal forms.

Said equivalently, to perform an operation on normal forms $\,\bar a,\,\bar b,\,$ we apply $h^{-1}$ to map them to their associated classes $\,[a],[b],\,$ then we perform the operation on the classes yielding $\,[a+b],\,$ then finally we apply $h$ to map that result to its normal form $\,\overline{a+b}.\,$  So the normal forms are essentially canonical  "labels" or "names" for their congruence classes. We could instead use any set of $\,n\,$ elements as labels, but using elements from the original ring $\,\Bbb Z\,$ makes it more intuitive how the normal rep corresponds to the class.

As above, in Euclidean domains like $\,\Bbb Z\,$ and $\,F[x]\,$, which enjoy Euclidean division with smaller remainder, it is convenient to use the the remainder ("least Euclidean size") as the normal rep, which is discussed further here, showing how Hamilton's pair representation for complex numbers is just a special case of this, viz. $\Bbb R[x]\bmod x^2\!+\!1\cong \Bbb R[i]\cong \Bbb C$.

Title: What exactly is $\mathbb{Z_n}$?


问:
draw the pushdown automaton
How can I draw the pushdown automaton for language $L = \{ bin(n) \text{#} bin(3n)^{R} | n \in  \mathbb{Z}, n \geq 0 \}$ when bin(n) is binary notation of $n$ without leading zeros. Alphabet is $\{ 0, 1, \text{#} \}$

答:
Design it to mimic ordinary long multiplication by $3$ in binary. I’ll outline what’s needed.

The key is to recognize that multiplying a binary number by $3$ is just adding the number to the result of shifting it one place to the left, as in this example of $185\cdot3$:

```<code>            10111001
                  11
            --------
            10111001
           10111001
           ---------
          1000101011
</code>```

The automaton should stack the input until it reads a $\#$. At that point the top element of the stack will be the least significant bit of $n$. If the input is in $L$, that bit should agree with the next bit of input. The basic processing cycle from this point on is:

Use the bit at the top of the stack and the current state to determine what the next bit of input ought to be and whether there should be a carry into the next ‘column’;

If the next input bit is correct, pop one bit off the stack and go to one of four states, depending on the popped bit, the present state, and whether or not there is a carry to the next ‘column’. The states keep track of whether the popped bit was $0$ or $1$ and whether there is an outgoing carry.

Of course if the next input bit is not correct, the input is not in $L$.

Title: draw the pushdown automaton

Body:
Design it to mimic ordinary long multiplication by $3$ in binary. I’ll outline what’s needed.

The key is to recognize that multiplying a binary number by $3$ is just adding the number to the result of shifting it one place to the left, as in this example of $185\cdot3$:

            10111001
                  11
            --------
            10111001
           10111001
           ---------
          1000101011

The automaton should stack the input until it reads a $\#$. At that point the top element of the stack will be the least significant bit of $n$. If the input is in $L$, that bit should agree with the next bit of input. The basic processing cycle from this point on is:

Use the bit at the top of the stack and the current state to determine what the next bit of input ought to be and whether there should be a carry into the next ‘column’;

If the next input bit is correct, pop one bit off the stack and go to one of four states, depending on the popped bit, the present state, and whether or not there is a carry to the next ‘column’. The states keep track of whether the popped bit was $0$ or $1$ and whether there is an outgoing carry.

Of course if the next input bit is not correct, the input is not in $L$.


问:
Proving by contradiction check if correct
Consider the statement: For all x, y, z ∈ Z. At least one of x− y, x− z
and y − z is even

I want to check if my proof is correct via contradiction:

Suppose(for a contradiction) that all of x-y, x-z, and y-z are odd:

x - y = 2k+1 for some k ∈ Z

rearranging for x:

x=2k+1+y

y - z  = 2n+1  for some n ∈ Z

rearranging for z:

z=y-2n-1

taking x-z and subbing in

x - z  =2k+1+y-(y-2n-1)

= 2k+2n+2

=2(k+1+n)

therefore x-y is even which is a contradiction therefore the original statement is true

答:
That's fine.  ANd you proof is good.

Alternatively:

By pigeon hole principle, if we are given three items, each with two possible options, then at least two of the items must have the same option, as for them all to be different options would require at least three options.  That's why if you flip three coins two of them must agree on heads or tails or if you pick three socks from a drawer of red and blue two must match.

Likewise given $x,y,z$ each either odd of even, then two of them are the same parity. (Because there you can't have all three be three different parities!)

And lemma: If $a,b$ are the same parity then $a-b$ is even.

Pf:  If $a=2m$ and $b =2n$ are even then $a-b = 2(m-n)$ is even.  If $a=2m-1$ and $b=2n-1$ are odd then $a-b = (2m-1)-(2n-1)= 2(m-n)$ is even.

And that's it. Of $x,y,z$ two of them are the same parity and depending on which of the two $x-y$ or $y-z$ or $x-z$ is even. (If $x,y$ are same parity $x-y$ is even; if $x,z$ are same parity then $x-z$ is even; if $y,z$ are same parity then $y-z$ is even.)


问:
Let a, b and c be natural numbers. If a^2 = bc and gcd(b, c) = 1, then there exist natural numbers s and t such that b = s^2 and c = t^2 .
We were given this proof under our Fundamental Theorem of Arithmetic section and it is supposed to involve prime factorization.
I'm totally lost so far and could use any help you can offer.</p>

答:
Hint.

Write the prime factorization of $bc$.

What can you say about the parity of the power to which each prime is raised?

Can $b$ and $c$ share any of the prime factors of their product?


问:
$AA^T =A$, what can you say about the rank of $A$?
$AA^T =A$, what can you say about the rank of $A$?

Interview question, and I have no idea. The closest thing I can relate to is the inverse. But that's not the case here.

答:
Robert Israel has already given the answer to the question asked—the rank can be any nonnegative integer up to the dimension of the matrix. However I wanted to point out that one can say a lot more: note that
$$
A^T = (AA^T)^T = (A^T)^TA^T = AA^T = A,
$$
and so $A$ symmetric. In particular, we now know that $A^2=A$ and so $A^2-A=0$, which means that the minimal polynomial of $A$ divides $t^2-t$. This tells a lot about the matrix $A$ (that it's diagonalizable and that its eiegnvalues can only be $0$ or $1$), and ultimately says that it is a projection matrix.


问:
Punctured torus with antipodal boundary points identified
Given a torus with an open disk removed and its antipodal boundary points identified (note that the boundary refers that of the removed disk, which is a $S^1$), what space is it?

At first I viewed the torus as a square with opposite edges identified, cut along a diagonal, and tried to glue the two parts together in a nice way, but it ended up more confusing. Then I tried to view the torus as a square glued with two strips (consider torus as two cylinders glued together over a square patch, [shown by the following picture from Basic Topology by Armstrong], I simply cut off the square and get three pieces), and the procedure amounts to replacing the two strips with two Mobius strips and identifying several points. With some struggle, I got the result: a real projective plane with two tangent disks replaced by two Mobius strips, which is not satisfactory.

I've been stuck since then. Please help. I'm not sure what kind of description can be considered good and clear. (Maybe the cell structure will do, but I couldn't determine it either.)

答:
Here is a third answer, to complement the two that Kevin has given above.

Imagine you have a cylinder $S^1\times I$. On the bottom $S^1\times 0$, you glue on a disk by identifying boundary circles. On the top $S^1\times 1$, you identify opposite points of the circle. What do you get? A projective plane! Because once you glue a disk on the bottom, what you have left is homeomorphic to a disk. Then you identify opposite points along the boundary circle, which is precisely how (in one approach) the projective plane is constructed.

Now let's go back in the other direction. Cut the bottom disk back off of your cylinder (so you've only glued opposite points on the top $S^1\times 1$). Now you have a projective plane with a disk removed. Gluing this to your torus with a disk removed (along boundary circles) is precisely how we construct the connected sum $\mathbb T\#\,\mathbb P.$

Now it remains to see that this is the same as directly gluing opposite points on your punctured torus. To see this, choose a bigger (closed) disk on your torus and drill out a small (open) disk from near its center. Now you have a cylinder/annulus, where one boundary circle is adjacent to the rest of the torus and the other boundary circle is undergoing this gluing of opposite points. Hence, we see that this is the cylinder-with-a-wonky-end glued onto the torus, from which we got $\mathbb T\#\,\mathbb P$ above. But only the smaller disk underwent any sort of change. So forgetting about the  cylinder and the larger disk, this is the same as the construction that you initially described.


问:
Abstract Algebra - The Equality of Numbers - 0 in the Reals equals 0 in Zero Ring?
I've been learning a bit about Abstract Algebra and it has me thinking a lot about numbers. One thing that has occurred to me is that if we group numbers into sets based on legal operations (the definitions of groups, rings, and fields for example define the properties of the numbers in their respective sets). Are numbers themselves unique? Or are they only defined within their own classes? Can we say for example that the number $0$ in the Zero-Ring is equal to (or even identical to) the number $0$ in say the real numbers: $$0 \in \left\{0\right\} = 0 \in \mathbb{R}$$

Are they the same number? Is there an equality that can be defined through some kind of morphism? Or are these numbers different from each other because they belong to different classes, and so are defined differently?

答:
The answer (to the more general question of whether the zeroes of different structures are equal) is going to unavoidably depend on your formalism, and tell you essentially nothing, regardless of what it is.

For example, if we define $\mathbb{Q}$ as the set of all equivalence classes of pairs $(a,b)$ where $a$ and $b$ are integers with $b$ non-zero under the equivalence relation generated by $(a,b) \sim (c,d)$ if $bc = ad$, and use ZFC as our formalism (and also define ordered pairs by $(a,b) = \{\{a\},\{a,b\}\}$ as is standard), then we cannot have $0_\mathbb{Q} = 0_\mathbb{Z}$, as we have $0_\mathbb{Z} \in \{0_\mathbb{Z}\} \in \{\{0_{\mathbb{Z}}\},\{0_\mathbb{Z},1_\mathbb{Z}\}\} = (0,1) \in [(0,1)] = 0_\mathbb{Q}$, so $0_\mathbb{Q} = 0_\mathbb{Z}$ would give a loop of set membership, violating the axiom of constructibility.

However, if we instead define $\mathbb{Q}$ first, then define $\mathbb{Z}$ as the subring generated by $1_\mathbb{Q}$, then we obviously have $0_\mathbb{Z} = 0_\mathbb{Q}$ in the most literal of senses.


问:
Integrate $\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx$
Please help me to solve this integral:
$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx.$$

I managed to calculate an indefinite integral of the left part:
$$\int\frac{\cos x}{\sin x}x\ \mathrm dx=\ x\log(2\sin x)+\frac{1}{2} \Im\ \text{Li}_2(e^{2\ x\ i}),$$
where $\Im\ \text{Li}_2(z)$ denotes the imaginary part of the dilogarithm. The corresponding definite integral $$\int_0^\pi\frac{\cos x}{\sin x}x\ \mathrm dx$$ diverges. So, it looks like in the original integral summands compensate each other's singularities to avoid divergence.

I tried a numerical integration and it looks plausible that
$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx\stackrel{?}{=}\pi \log 54,$$
but I have no idea how to prove it.

答:
Here's one way to go.

First, note that
$$\begin{eqnarray*}
\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx
&=& \int_0^\pi\frac{3x(1+\cos x)}{\sin x} \mathrm dx
+\int_0^\pi\frac{3x}{\sin x} \left(-1+\sqrt{1-\frac{\sin^2x}{9}}\right)\ \mathrm dx.
\end{eqnarray*}$$
For now I'll simply claim that
\begin{equation*}
\int_0^\pi\frac{3x(1+\cos x)}{\sin x} \mathrm dx = \pi\log 64.\tag{1}
\end{equation*}
(I would be surprised if this integral has not been handled somewhere on this site.)
But
$$\begin{eqnarray*}
\int_0^\pi\frac{3x}{\sin x} \left(-1+\sqrt{1-\frac{\sin^2x}{9}}\right)\ \mathrm dx
&=& \int_0^\pi\frac{3x}{\sin x} \sum_{k=1}^\infty {1/2\choose k} \frac{(-1)^k}{3^{2k}} \sin^{2k}x \ \mathrm dx \\
&=& \sum_{k=1}^\infty {1/2\choose k} \frac{(-1)^k}{3^{2k-1}}
\int_0^\pi x  \sin^{2k-1}x \ \mathrm dx \\
&=& \sum_{k=1}^\infty {1/2\choose k} \frac{(-1)^k}{3^{2k-1}}
\frac{\pi^{3/2}\Gamma(k)}{2\Gamma(k+1/2)} \\
&=& -\pi \sum_{k=1}^\infty \frac{1}{3^{2k-1}2k(2k-1)} \\
&=& -\pi \log \frac{32}{27}.
\end{eqnarray*}$$
(The last sum can be found by standard methods.
Schematically, $\sum \frac{a^{2k-1}}{2k(2k-1)} = \sum \int {\mathrm da} \frac{a^{2k-2}}{2k}$.)
Thus, the integral is $\pi \log 54$ as claimed.

Proof of (1):
We have
$$\begin{eqnarray*}
\int_0^\pi \frac{3x(1+\cos x)}{\sin x} \ \mathrm dx
&=& \int_{0^+}^\pi \frac{3x(1+\cos x)}{\sin x} \ \mathrm dx \\
&=& 3\int_{0^+}^\pi x \cot\frac{x}{2} \ \mathrm dx
\hspace{5ex}\textrm{(double angle formulas)} \\
&=& 12 \int_{0^+}^{\pi/2} t\cot t \ \mathrm dt
\hspace{5ex} (t = x/2) \\
&=& -12\int_{0^+}^{\pi/2} \log\sin t \ \mathrm dt
\hspace{5ex}\textrm{(integrate by parts)} \\
&=& -6\int_{0^+}^{\pi/2} \log\sin^2 t \ \mathrm dt \\
&=& -6\int_{0^+}^{\pi/2} \log(1-\cos^2 t) \ \mathrm dt \\
&=& 6 \int_{0^+}^{\pi/2} \sum_{k=1}^\infty \frac{1}{k}\cos^{2k}t \ \mathrm dt
\hspace{5ex}\textrm{(series for log)} \\
&=& 6\sum_{k=1}^\infty \frac{1}{k} \int_{0^+}^{\pi/2} \cos^{2k}t \ \mathrm dt
\hspace{5ex} \textrm{(Tonelli's theorem)}\\
&=& 6\sum_{k=1}^\infty \frac{1}{k}
    \frac{\sqrt{\pi}\Gamma(k+1/2)}{2\Gamma(k+1)}  \\
&=& 3\pi \sum_{k=1}^\infty {1/2 \choose k}(-1)^{k+1}\frac{2k-1}{k} \\
&=& \pi \log 64.
\end{eqnarray*}$$
Note that
$$\begin{eqnarray*}
6\pi \sum_{k=1}^\infty {1/2 \choose k}(-1)^{k+1}
&=& -6\pi \left[\sum_{k=0}^\infty {1/2 \choose k}(-1)^{k} - 1\right] \\
&=& -6\pi[(1-1)^{1/2} - 1] \\
&=& 6\pi
\end{eqnarray*}$$
and
$$\begin{eqnarray*}
-3\pi \sum_{k=1}^\infty {1/2 \choose k}(-1)^{k+1} \frac{1}{k}
&=& 3\pi \sum_{k=1}^\infty {1/2\choose k}(-1)^k  \int_0^1 x^{k-1} \ \mathrm dx \\
&=& 3\pi \int_0^1 \frac{1}{x}  \left[
\sum_{k=0}^\infty {1/2\choose k}(-1)^k x^{k} -1
\right] \ \mathrm dx \\
&=& 3\pi \int_0^1 \frac{1}{x}  \left(
\sqrt{1-x} -1
\right) \ \mathrm dx \\
&=& 3\pi(-2+\log 4) \\
&=& -6\pi + \pi\log 64.
\end{eqnarray*}$$


答:
Every ring has an additive identity - that's part of the definition of a ring.

We (almost) always call that element "$0$" because it's convenient. It helps us remember how it behaves in that ring's arithmetic.

But those $0$'s aren't the same thing at all (except when one ring is a subring of another). They are different things that happen to have the same name. Context disambiguates, so you know what each $0$ "really is".

Title: Abstract Algebra - The Equality of Numbers - 0 in the Reals equals 0 in Zero Ring?

Body:
I've been learning a bit about Abstract Algebra and it has me thinking a lot about numbers. One thing that has occurred to me is that if we group numbers into sets based on legal operations (the definitions of groups, rings, and fields for example define the properties of the numbers in their respective sets). Are numbers themselves unique? Or are they only defined within their own classes? Can we say for example that the number $0$ in the Zero-Ring is equal to (or even identical to) the number $0$ in say the real numbers: $$0 \in \left\{0\right\} = 0 \in \mathbb{R}$$

Are they the same number? Is there an equality that can be defined through some kind of morphism? Or are these numbers different from each other because they belong to different classes, and so are defined differently?


问:
for every $v$ vector If $Av = Bv$ then $A=B$
I just can't grasp the idea, I mean I can give you billion vectors and matrices for which this
sentence is not correct.
take vector which build with multiple zeros and whatever different $A,B$ matrices you want
Can someone shed some light

答:
Two maps $A,B$ are equal if they have the same domain, and the same action on each element in the domain

Sorry I cannot comment yet!


问:
Solution verification: Divergence of gradient descent (recursive sequence)
Consider the recursion:
$$x_{k+1} = x_k - \frac{\alpha_0}{k^p} \left( e^{x_k} - e^{-x_k} \right)$$

for finite $p$ and $\alpha_0 > 0$.

Show that when $x_1$ is large enough, $x_k$ grows super-exponentially, for all $k$:

$$\log\left(\left| \frac{x_{k+1}}{x_k} \right |\right) \geq 2^k$$

What I've Tried

It's actually not clear to me that the argument in the log above is always positive, for large $x_1$ we have:

$$x_2 = x_1 - \alpha_0\left(e^{x_1} - e^{-x_1}\right)$$

For large enough $x_1$ I'd expect the second term to be larger than the first term, so how can $x_1$ and $x_2$ have the same sign?

EDIT: I updated the above to include an absolute value in the argument of the logarithm.

I initially tried bounding the exponentials with linear terms, but that is too coarse.

First Ill try to show that $x_k$ is an increasing sequence.

Checking the condition for $|x_2|$:

$$|x_2| = |x_1 - \alpha_0(e^{x_1}-e^{-x_1})|$$

is larger than $|x_1|$ when $x_1 \leq 0$ or $\alpha_0/1^p(e^{x_1}-e^{-x_1}) \geq 2x_1$ or $\alpha_0/1^p(e^{x_1}-e^{-x_1}) \leq 2x_1$. These conditions can be met when $x_1$ is large enough in magnitude. For $|x_k|$ we get the same expression in terms of $|x_{k-1}|$ instead of in terms of $x_1$. Since our condition is met when $|x_{k-1}|$ is large enough, we can ignore the signs and simply choose a $x_1$ such that $|x_k|$ is large enough. This can be done for any $k$.

Since $x_k = O(e^{x_{k-1}})$ we can iterate this to arrive at $x_k = O(e^{e^{x_{k-2}}})$ which shows that $|x_k|$ grows superexponentially.

I could use some feedback on my arguments.

This is example 3 in [1]

References

[1] Asi, H., Duchi, J. (2019). The importance of better models in stochastic optimization arXiv https://arxiv.org/abs/1903.08619

答:
$\color{brown}{\textbf{Preliminary notes.}}$

Since the derivative
$$(z\ln a - c\ln z)' = \ln a -\dfrac cx$$
has the single root $\;z_m = \dfrac c{\ln a},\;$ then the function $\;a^z - z^c\;$ increases when $z\in(z_m, \infty).$

Let
$$b=4\left(\ln\,\frac{11}{10}\right)^{-1}\,\max\left(1,\,\ln\,\dfrac2{\alpha_0},\,p \right),\tag1$$
then under the condition $\;y \ge \max(b,k)\;$ are valid the inequalities

$\left(\frac{11}{10}\right)^y > \dfrac2{\alpha_0};$

$\left(\frac{11}{10}\right)^y > y;$

$\left(\frac{11}{10}\right)^y > k^p;$

$\;e^y - e^{-y} > \left(\dfrac83\right)^y > 2^y\left(\dfrac{11}{10}\right)^{3y}
 > \dfrac{2y\, k^p}{\alpha_0}\,2^y.\tag2$

$\color{brown}{\textbf{Recurrence relations for the majorants.}}$

Assuming $\;x_1 = b,\;$ one can get
$$x_{2} = x_1 - \dfrac{\alpha_0}{1^p}\left(e^{x_1}-e^{-x_1}\right)
 < x_1 - 2x_1\,2^{x_1},$$
$$|x_2| > |x_1|\,2^{|x_1|} > \max(b,2),\tag3$$
$$x_{3} = x_2 + \dfrac{\alpha_0}{2^p}\left(e^{-x_2}-e^{x_2}\right)
 > -|x_2| + 2|x_2|\,2^{|x_2|},$$
$$|x_3| > |x_2|\,2^{|x_2|} > \max(b,3),\dots,\tag4$$
$$|x_{k+1}| > |x_k|\,2^{|x_k|} > \max(b,k+1).\tag5$$

From $(5)$ should
$$\ln|x_{k+1}| - \ln |x_k| > |x_k| \ln2 > |x_{k-1}|\,2^{|x_{k-1}|} \ln2 > k\,2^k\,\ln2,$$

$$\color{brown}{\mathbf{\ln\left|\dfrac{x_{k+1}}{x_k}\right| > 2^k.}}$$


答:
If $\{e_1, \ldots, e_n\}$ is the standard basis for $\Bbb{R}^n$, then notice that
$$Ae_j = j\text{-th column of $A$}, \quad Be_j = j\text{-th column of $B$}$$

If $Av = Bv$ for every vector $v$, then in particular $Ae_j = Be_j$ for all $1 \le j \le n$ so $A$ and $B$ have equal columns i.e. $A=B$.

Title: for every $v$ vector If $Av = Bv$ then $A=B$

Body:
I just can't grasp the idea, I mean I can give you billion vectors and matrices for which this
sentence is not correct.
take vector which build with multiple zeros and whatever different $A,B$ matrices you want
Can someone shed some light


答: If $A$ and $B$ agree on any vector, then in particular they agree on a basis.  This implies that they agree everywhere, as the values on a basis determine a linear transformation.


问:
Ehresmann's lemma equivalent etale topology
In complex algebraic geometry we have Ehresmann's lemma which states that given $f:X \to Y$ a smooth surjective proper map between complex manifold this is locally a fibration in the analytic topology.

I was wondering whether there was a generalization of that in the context of etale topology. Lots of statement which are true analytic locally in the context of $\mathbb{C}$ can also be carried in the etale topology for smooth varieties over any algebraically closed field $k$.

In the (beautiful) notes by Conrad https://www.math.ru.nl/personal/bmoonen/Seminars/EtCohConrad.pdf at page 53, he states what he calls etale Ehresmann's lemma: given $f:X \to S$ smooth and proper, the sheaf $R^if_*F$ are locally free whenever $F$ is a torsion etale sheaf on $X$.

The hypothesis are however very general:assuming $X,S$ to be both smooth varieties over an algebraically closed field, it is true the stronger statement that $f$ is a fibration etale locally?

答:
No. I think you might be misinterpretting what 'locally trivial fibration' means.

Namely, what Ehresmann's theorem implies is the following:

Corollary to Ehresmann's theorem: Let $f:X\to Y$ be a smooth proper surjection of complex manifolds. Then, there is an open cover
$\{U_i\}$ of $Y$ such that $f^{-1}(U_i)$ is diffeomorphic (as a manifold over $U_i$) to
$U_i\times B$ for some smooth manifold $B$.

One cannot replace diffeomorphic by 'biholomorphic' or, if $X$ and $Y$ are algebraic and the $U_i$ is a Zariski cover, 'algebraic'.

Here's a silly counterexample to your literal claim.

Consider $D(\Delta)\subseteq \mathbb{A}^2_k=\mathrm{Spec}(k[a,b])$ where, for simplicity, $k$ is an algebraically closed field of characteristic greater than $3$ where

$$\Delta(a,b)=-16(4a^3+27b^2)$$

We then have the universal family of Weierstrass equations living over $D(\Delta)$

$$\mathcal{W}^\text{univ}=V(y^3-axz^2-bz^3)\subseteq \mathbb{P}^2_k\times D(\Delta)$$

We then note that

$$\mathcal{W}^\text{univ}\to D(\Delta)$$

is a family of elliptic curves and so, in particular, smooth proper and surjective. Moreover, each $\mathcal{W}^\text{univ}$ and $D(\Delta)$ is smooth.

That said it's not an etale local fibration since this, in particular, would imply that the fibers of $\mathcal{W}^\text{univ}\to D(\Delta)$ are Zariski locally isomorphic. Indeed, let $\{U_i\to D(\Delta)\}$ be an etale cover such that $\mathcal{W}^\text{univ}\times_{D(\Delta)}U_i\cong U_i\times B_i$ for some $k$-space $B_i$. Then, in particular we see that for all $x\in V_i:=\text{im}(U_i\to D(\Delta))$ (this is an open set) we have that

$$\mathcal{W}^\text{univ}\times_{D(\Delta)}U_i\times_{V_i}x\cong \bigsqcup_{y\mapsto x}B_i$$

which since $k(x)\to k(y)$ is an isomorphism implies that $\mathcal{W}^\text{univ}_x\cong B_i$. Thus, $\mathcal{W}^\text{univ}_x\cong B_i$ for all $x\in V_i$.

Note that we have an algebraic map $j:D(\Delta)\to \mathbb{A}^1_k$, the $j$-function map, which sends, in particular, $x$ to the $j$-invariant $\mathcal{W}^\text{univ}_x$. Since $\mathcal{W}^\text{univ}_x$ is locally isomorphic this implies that $j$ is locally constant which implies, since $D(\Delta)$ is connected, that $j$ is constant and thus $\mathcal{W}^\text{univ}_x$ is actually constant for $x$ in $D(\Delta)$.

Of course, this is ludicrous since $\mathcal{W}^\text{univ}_x$ contains isomorphism classes of every elliptic curve over $k$.

Of course, it is true that the fibration $W^\text{univ}\to D(\Delta)$, when $k=\mathbb{C}$, is analytically locally on the target DIFFEOMORPHIC to the trivial fiber bundle. The operative point being that any two elliptic curves are diffeomorphic but are rarely isomorphic/biholomorphic.


问:
List of Local to Global principles
What are some of the local to global principles in different areas of mathematics?</p>

答:
Number Theory

If an integer $n \not \equiv 0 \pmod m$ (for any $m$) then $n \not = 0$.

Title: List of Local to Global principles


问:
Quartic equation $ x^4-mx^2-4=0$ has root $\sqrt[4]{3-2\sqrt2}+\sqrt[4] {3+2\sqrt2}$. Find $m$
Find $m$ knowing that
$$x^4-mx^2-4=0$$ has
$$ \sqrt[4]{3-2\sqrt2}+\sqrt[4] {3+2\sqrt2}$$
as root.

I was thinking that if it has this root it must have  $ \sqrt[4]{3-2\sqrt2}-\sqrt[4] {3+2\sqrt2}$ too. Am I wrong? I tried solving by replacing $x^2$ with t too but it doesn't lead anywhere cause the roots are kind of weird looking.

答:
Rewrite $x^4-mx^2-4=0$ as
$$(x^2-\frac m2)^2=4+\frac {m^2}4$$
which leads to the solution
$$x =\sqrt{\sqrt{4+\frac{m^2}4}+\frac m2 }$$

Then, compare with the given root
$$ \sqrt[4]{3-2\sqrt2}+\sqrt[4] {3+2\sqrt2}
= \sqrt{\sqrt2-1}+\sqrt {\sqrt2+1}= \sqrt{\sqrt8+2}
$$
to arrive at $m=4$.


问:
It is possible to prove that these two collections generate the same topology on $ \mathbb{X} $?
Let $ \mathbb{Y} $ a topological space whose topology $ \tau $ is generated
by a collection of subsets $ \mathcal{B} \subset 2^\mathbb{Y} $.

In other words the topology $ \tau $ of $ \mathbb {Y} $ is the smallest topology
("Smaller" in partial order on $ \subset $ sub-collections in the class of $ 2^\mathbb{Y} $)
containing the collection $ \mathcal{B} $ of subsets of $ \mathbb{Y}$.

Let $ \{\varphi_i\}_{i \in \Lambda} $ a family of functions defined in a set $ \mathbb{X} $ and
taking values in topological space $ \mathbb{Y}$. Put $ \mathcal{A}_{\tau} $ as the collection of all
subsets of $ \mathbb{X}$ in the form
$$
\{\varphi_i^{-1}(V): V \in \tau \mbox{ and } i\in \Lambda \}.
$$
and $ \mathcal{A}_\mathcal{B}$ as
$$
\{\varphi_i^{-1}(B): B\in\mathcal{B}\mbox{ and } i \in \Lambda \}.
$$
It is possible to prove that these two collections generate the same topology on $ \mathbb{X} $?
If not what assumptions are sufficient for this to be true?
I am particularly interested in the case where $ \mathbb{Y} = \mathbb{R} $.

答:
(I know the OP said to disregard this question, but I discovered some nice things that I'd like to share in this answer.)

Notations. Let me start by introducing a slightly different notation, writing $D^\#$ instead of $\mathcal{A}_D$: $$ D^\# = \{ V,f : V \in D : f^{-1}[V] \} $$
Read this as "the set which, for each $V$ and $f$ such that $V \in D$, contains $f^{-1}[V]$".  Here, and in the rest of the post, we let $f$ range implicitly over $\{\varphi_i\}_{i \in \Lambda}$.

We use $D^\cup$ and $D^\cap$ for the set of all arbitrary unions and of finite intersections of elements of $D$, respectively:
$$
\begin{align}
& D^\cup = \left\{ \mathscr{F} : \mathscr{F} \subseteq D : \bigcup \mathscr{F} \right\} \\
& D^\cap = \left\{ F,G : F,G \in D : F \cap G \right\} \\
\end{align}
$$
(See also another answer of mine.)  This means that we can write "the topology generated by $B$" as $B^{\cap\cup}$.

Problem statement. Using these notations, we are asked whether, for all $B$ and $\tau$ with $\tau = B^{\cap\cup}$, we have $\tau^{\#\cap\cup} = B^{\#\cap\cup}$; or equivalently, whether for all $B$
$$
(0) \;\;\; B^{\cap\cup\#\cap\cup} = B^{\#\cap\cup}
$$
Proof.  The only thing we know about $^\#$, $^\cap$, and $^\cup$ are their definitions: these apply $f^{-1}[\cdot]$, $\cap$, and $\cup$ many times.  And looking at the latter operators, we know that the order of $f^{-1}[\cdot]$ and $\cap$ doesn't matter, and the same for $f^{-1}[\cdot]$ and $\cup$.  (I will not prove those distribution properties here.)  Therefore it seems likely that for all $D$
$$
(1) \;\;\; D^{\cup\#} = D^{\#\cup} \\
(2) \;\;\; D^{\cap\#} = D^{\#\cap} \\
$$
If we would be able to prove these, then $(0)$ directly follows from a simple calculation:
$$
\begin{align}
& B^{\cap\cup\#\cap\cup} \\
= & \;\;\;\;\;\text{"using $(1)$ with $D := B^\cap$"} \\
& B^{\cap\#\cup\cap\cup} \\
= & \;\;\;\;\;\text{"using $(2)$ with $D := B$"} \\
& B^{\#\cap\cup\cap\cup} \\
= & \;\;\;\;\;\text{"$^{\cap\cup}$ is idempotent"} \\
& B^{\#\cap\cup} \\
\end{align}
$$
(I won't prove here that $^{\cap\cup}$ is idempotent.)

It turns out we can prove $(1)$, as follows:
$$
\begin{align}
& D^{\cup\#} \\
= & \;\;\;\;\;\text{"definition of $^\#$"} \\
& \{ V,f : V \in D^\cup : f^{-1}[V] \} \\
= & \;\;\;\;\;\text{"write $V$ as a union using the definition of $^\cup$"} \\
& \{ \mathscr{F},f : \mathscr{F} \subseteq D : f^{-1}\left[\bigcup \mathscr{F} \right] \} \\
= & \;\;\;\;\;\text{"distribution property of $^{-1}$"} \\
& \{ \mathscr{F},f : \mathscr{F} \subseteq D : \bigcup \{ V : V \in \mathscr{F} : f^{-1}[V]\} \} \\
= & \;\;\;\;\;\text{"rewrite using the definition of $^{\cup}$"} \\
& \{ V,f : V \in D : f^{-1}[V] \}^{\cup} \\
= & \;\;\;\;\;\text{"definition of $^\#$"} \\
& D^{\#\cup} \\
\end{align}
$$
The proof for $(2)$ is very similar.

Observations. As a first observation, I quite like how the $^\#$ notation works out, and interacts with the $^\cap$ and $^\cup$ notations I'd invented earlier: $(0)$ brings out the structure of the problem in a way that is much clearer than writing $(\mathcal{A}_{B^{\cap\cup}})^{\cap\cup} = (\mathcal{A}_B)^{\cap\cup}$: after making the latter explicit, I suddenly realized that $\mathcal{A}$ is essentially a unary operator.

Also, I like how we've been able to avoid $\subseteq$, and instead have an equality proof.  That is not what I expected when I started to work on this proof.

Finally, I dislike the set comprehension manipulation: the two steps above "using the definition of $^\cup$" are not really easy to follow.  I don't yet see how to fix that.