问:
Integrate the given function around the unit circle using Cauchy's

Integrate the given function around the unit circle

$\frac{z^3}{2z-i}$

I'm a bit confused here. So since the function is undefined at $z = \frac{1}{2}$, what does this imply? It implies that the function is not analytic everywhere does the integral does not equal 0 and we have to evaluate this by hand right? So...

Since the unit circle can be expressed as $z(t) = e^{it}$ and $\frac{dz}{dt} = ie^{it}$

we have:

$$f(z) = \frac{e^{3it}}{2e^{it} - i}$$

So the integral is:

$$\int_0^{2\pi} \frac{e^{3it}}{2e^{it} - i} \cdot ie^{it} $$

Is this the right setup? Where do I go from here?

答:
Here is how you could continue what you started:
$$
\begin{align}
\oint\frac{z^3}{2z-i}\,\mathrm{d}z
&=\int_0^{2\pi}\frac{i\,e^{4it}}{2e^{it}-i}\,\mathrm{d}t\tag1\\
&=\frac i2\int_0^{2\pi}\left[-\frac i8+\sum_{\substack{k=0\\k\ne3}}^\infty\frac{i^ke^{(3-k)it}}{2^k}\right]\,\mathrm{d}t\tag2\\
&=\frac\pi8+\frac i2\sum_{\substack{k=0\\k\ne3}}^\infty\left[\frac{i^ke^{(3-k)it}}{(3-k)2^k}\right]_0^{2\pi}\tag3\\[3pt]
&=\frac\pi8\tag4
\end{align}
$$
Explanation:

$(1)$: $z=e^{it}$

$(2)$: use the Taylor series for $\frac{iz^4}{2z-i}$, but separate out the $k=3$ term

$(3)$: integrate term by term

$(4)$: simplify

However, if you want to use Cauchy's Integral Formula, let $f(z)=z^3/2$:
$$
\begin{align}
f\!\left(\tfrac i2\right)&=\frac1{2\pi i}\oint\frac{f(z)}{z-\frac i2}\,\mathrm{d}z\tag5\\
-\tfrac i{16}&=\frac1{2\pi i}\oint\frac{z^3/2}{z-\frac i2}\,\mathrm{d}z\tag6\\
\frac\pi8&=\oint\frac{z^3}{2z-i}\,\mathrm{d}z\tag7
\end{align}
$$
Explanation:

$(5)$: Cauchy's Integral Formula

$(6)$: apply $f(z)=z^3/2$

$(7)$: simplify


问:
Orthogonality between geodesics and normal spheres in a Riemannian manifold
In do Carmo's Riemannian Geometry, right after proving Gauss' Lemma and defining normal neighborhoods and normal balls, the author writes:

By Gauss' Lemma, the boundary of a normal ball $B_\varepsilon(p)$ is a hypersurface (a submanifold of codimension $1$) in $M$ orthogonal to the geodesics leaving $p$, which is denoted by $S_\varepsilon (p)$ and called the normal (or geodesic sphere).

As I understand, the orthogonality refers to ortogonality between the tangent vector of the geodesic and the tangent plane to the said hypersurface. Is it correct?

Anyway, how to prove the claim about the sphere being a submanifold and the claim about orthogonality?

Thanks in advance.

答:
In short, the boundary of the normal ball is a hypersurface of codimension one because it just the image of a sphere in $T_pM$ under the diffeomorphism $\exp_p$ defined on some neighbourhood of zero vector. The orthogonality is pretty straightforward from Gauss' Lemma. Here's the details.

Let $V \subset T_pM$ is a normal neighbourhood of $p$ and $B_{\varepsilon}(p)= \text{exp}_p \big(B_{\varepsilon}(0)\big)$ be a normal ball. We know that the sphere $S_{\varepsilon}(0) = \{ v \in V \mid \|v\|=\varepsilon \}$ is a hypersurface of codimension one of $V$ since it is a regular level set of the smooth function $r : V \to \mathbb{R}$ defined as $r(x)=\|x\|$. More precisely $S_{\varepsilon}(0) = r^{-1}(\varepsilon)$. Since $\text{exp}_p : V \to U\subseteq M$ is a diffeomorphism, we have the normal sphere
$$
S_{\varepsilon}(p) := \partial B_{\varepsilon}(p) = \text{exp}_p \partial B_{\varepsilon}(0) =  \text{exp}_p (S_{\varepsilon}(0)) = \text{exp}_p \circ r^{-1}(\varepsilon) = (r \circ \exp_p^{-1})^{-1}(\varepsilon).
$$
Therefore $S_{\varepsilon}(p) \subseteq U$ is a regular level set of the smooth map $R := r \circ \exp_p^{-1} : U \to \mathbb{R}$. This establish that $S_{\varepsilon}(p)$ is a hypersurface of codimension one.

To show the orthogonality claim, let $\gamma(t) =\exp_p(tv)$ is a geodesic emanating from $p$ and passing through $S_{\varepsilon}(p)$, say $\|v\| = \varepsilon$ and $t$ defined in some open interval contain $[0,1]$, so $q:=\gamma(1) \in S_{\varepsilon}(p)$. For any $w \in T_q S_{\varepsilon} = \text{ker } dR_q$, we have
$$
0=dR_q(w) =dr_v \circ d(\exp_p^{-1})_q(w) = dr_v(w_0),
$$
with $ w_0:= d(\exp_p^{-1})_q(w) \in T_vV \approx T_v(T_pM) \approx T_pM$.
This means that
$$
w_0 \in \text{ker }dr_v = \text{tangent space of }S_{\varepsilon}(0) \text{ at }v.
$$
Therefore,
\begin{align}
\langle \gamma'(1),w\rangle &= \langle d(\exp_p)_v(v),d(\exp_p)_v(w_0) \rangle \\
&=\langle v,w_0 \rangle, \qquad \text{by Gauss' Lemma} \\
&= 0.
\end{align}

ContentLicense: CC BY-SA 4.0


问:
If $X$ is a Lévy process, can we show $\frac1tX_t\xrightarrow{t\to0+}\operatorname E\left[X_1\right]$?
Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $E$ be a $\mathbb R$-Banach space, $(X_t)_{t\ge0}$ be an $E$-valued Lévy$^1$ process on $(\Omega,\mathcal A,\operatorname P)$ and $\mu_t:=\mathcal L(X_t)$ for $t\ge0$.

Remember that $X$ is time-homogeneous and Markov with transition semigroup $$\kappa_t(x,B):=\mu_t(B-x)\;\;\;\text{for }(x,B)\in E\times\mathcal B(E)\text{ and }t\ge0.$$

Since $$Y_n:=X_n-X_{n-1}\;\;\;\text{for }n\in\mathbb N$$ is an independent identically $\mu_1$-distributed process, we know that $$\frac1n\sum_{i=1}^nf(Y_i)\xrightarrow{n\to\infty}\int f\:{\rm d}\mu_1\;\;\;\text{a.s. for all }f\in\mathcal L^p(\mu_1)\text{ and }p\ge1\tag1$$ by a strong law of large numbers.

Question: Is there an analogue of this result in continuous time?

I first thought the natural generalization would be to assume that $X$ is càdlàg and to consider $$Z_t:=\Delta X_t\;\;\;\text{for }t\ge0$$ and whether $$\frac1t\int_0^tZ_s\:{\rm d}s\xrightarrow{t\to\infty}\int f\:{\rm d}\mu_1\;\;\;\text{a.s. for all }f\in\mathcal L^p(\mu_1)\text{ and }p\ge1\tag2,$$ but on a second thought, this will fail whenenver $X$ is continuous, since then $Z_t=0$ for all $t\ge0$.

On the other hand, if $\mu_1$ has a finite first moment, we can choose $f=\operatorname{id}_E$ in $(1)$. For this choice of $f$, the natural generalization of $(1)$ would be $$\frac1tX_t\xrightarrow{t\to\infty}\operatorname E\left[X_1\right]\;\;\;\text{almost surely}\tag3.$$

Can we show this?

Remark: It might be useful to note that $\mu_t=\mu_1^{\ast t}$ (the rhs denoting the $t$th convolution power) for all $t\ge0$.

i.e.

$X_0=0$;

$X_{s+t}-X_s$ and $(X_r)_{0\le r\le s}$ are independent for all $s,t\ge0$;

$X_{s+t}-X_s\sim X_t$ for all $s,t\ge0$.

For the moment, I'm not imposing any continuity assumption, but we might need to in order to obtain a positive answer to this question.

答:
My understanding of Levy process is limited, so if I state anything trivial. Please kindly correct me.

(i)  Assume the cadlag property for $(X_t)$

(ii) Assume further that $X_1 \in L^{p}$ for a $p>1$  , that is $\exists M>0$ such that:$$ \mathbb{E}(|X_1|^p)< M$$

For simplicity of the proof, let $ \mathbb{E}(X_1)=0$.

For a $r<1$ that we choose so that $rp>1$, by Doobs inequality we have:
$$ \mathbb{P}\left( \max_{0 \le t \le 1} |X_t| \ge n^r \right) \le \dfrac{q^p}{n^{pr}}\mathbb{E}\left(|X_1|^p\right) < \dfrac{q^pM}{n^{pr}}$$
where $q$ is the positive real number such that $ \dfrac{1}{q}+\dfrac{1}{p}=1$

Hence,
$$ \mathbb{P}\left( \max_{0 \le t \le 1} |X_{n+t}-X_n| \ge n^r \right) \le \dfrac{q^p}{n^{pr}}\mathbb{E}\left(|X_1|^p\right) < \dfrac{q^pM}{n^{pr}}$$
As $rp>1$, by Borel-Catelli, we imply that:
$$\limsup_{n \rightarrow \infty}\left( \dfrac{ \max_{0 \le t \le 1} |X_{n+t}-X_n| }{n^r} \right) \le 1 \text{ a.s (1)} $$
On the other hand, LLN tells us that:
$$ \lim_{n \rightarrow +\infty} \dfrac{X_n}{n} \longrightarrow 0 \text{ as (2)}$$
Combining two previous results (1) and (2), and by noting that $\forall t >0$:
$$ \left| \frac{X_t}{t} \right| \le \left| \frac{X_n}{n} \right|+\dfrac{\max_{0\le s\le 1} |X_{n+s}-X_n|}{n} $$
Where $n=[t]$

We get our desired result :
$$ \lim_{t \rightarrow \infty} \frac{X_t}{t} =0 \text{ as}$$

Discussion:

The cadlag constraint is a usual thing we look for when dealing with stochastics processes. I don't know how to deal with one stochastic that we cannot prove that property.

The second assumption (ii), is not really a big thing because if $X$ is of bounded jumps, $X_1$ is also guaranteed to have moments at any order.

In fact, this problem looks like some kind of applications of the theory of semigroups. However, I don't have much knowledge of the semigroups of Levy process to be able to conduct such a solution.


问:
Relationship between matrix equations and determinants?
What is the relationship between a matrix equation and the determinants of the matrix?

Say if you have the matrix equation $A^3 - 4A = 0$, what can I learn about the matrix $A$'s determinants?

Or even if you have $A - 3B = 0$, does that imply $\det(A) = 3*\det(B)?$

答:
More generally, the eigenvalues of $p(A)$ are the images of the eigenvalues of $A$ under the polynomial $p$.  So if $A$ has eigenvalues $\lambda_1, \ldots, \lambda_n$ (counted by algebraic multipicity), $p(A)$ has eigenvalues $p(\lambda_1), \ldots,
p(\lambda_n)$ (again counted by algebraic multiplicity), and
in particular $\det p(A) = p(\lambda_1) \ldots p(\lambda_n)$.</p>

ContentLicense: CC BY-SA 3.0


问:
Confocal ellipses
The foci of an ellipse are S,S’ and P,P’ are two points on the curve on opposite sides of the major axis.SP’ meets S’P at Q’, and S’P’ meets SP at Q. To prove that Q and Q’ lie on another ellipse with foci S,S’. I am uncertain whether the best approach is geometric,or analytic. Any advice would be appreciated.</p>

答:
HINT1:

How to manipulate between ellipses of constant inter-focal distance $SS'= 2c$?

$$\dfrac{x_1^2}{a_1^2}+\dfrac{y_1^2}{a_1^2-c^2}\tag1$$
$$\dfrac{x_2^2}{a_2^2}+\dfrac{y_2^2}{a_2^2-c^2}\tag2$$

HINT2:

Imagine a laser shot in the direction $SQP.$ How does it get reflected/ricocheted at $Q$ and $P$?

Title: Confocal ellipses


问:
Collection of numbers to a single output, single output to the collection.
Is there a function or is it possible to make a function that will take in a collection of numbers (say 1, 2, 3, 5, 6, 7) and gives an output(say 56 for example) and another function that takes 56 and converts them back (to 1, 2, 3, 5, 6, 7) without disrupting the order. ?</p>

答:
If the numbers in your collection are all non-negative integers and bounded by a power of $10$ (say, by $10^b$), then a possible function would be:

$$\begin{align}
f(x_1,x_2,...,x_{n-1},x_n) &= x_1 \left[ \left( 10^b \right)^{n-1} \right] + x_2 \left[ \left( 10^b \right)^{n-2} \right] + \ldots + x_{n-1} \left( 10^b \right) + x_{n} \\
\\
&= \sum_{i=1}^n x_i \left[ \left( 10^b \right)^{n-i} \right] \\
\end{align}$$

The "return" function would be $g(N)=\{x_1,x_2,...,x_{n-1},x_n\}$, where

$$x_i = \left\lfloor \frac{N}{\left(10^b\right)^{n-i}} \right\rfloor - 10^b \left\lfloor \frac{N}{\left(10^b\right)^{n-(i-1)}} \right\rfloor$$

As an example, suppose your collection of numbers is $\{37,82,4,51\}$ (bounded by $10^2$) then

$$\begin{align}
f(37,82,4,51) &= 37 \left[ \left(10^2\right)^{4-1} \right] + 82 \left[ \left(10^2\right)^{4-2} \right] + 4 \left[ \left(10^2\right)^{4-3} \right] + 51 \left[ \left(10^2\right)^{4-4} \right] \\
\\
&= 37 \left(10^6\right) + 82 \left(10^4\right) + 4 \left(10^2\right) + 51 \left(10^0\right) \\
\\
&= 37820451
\end{align}$$

Then to return to the original collection, $g(x)$ would calculate, as one example,

$$\begin{align}
x_2 &= \left\lfloor \frac{37820451}{\left(10^2\right)^{4-2}} \right\rfloor - 10^2 \left\lfloor \frac{37820451}{\left(10^2\right)^{4-(2-1)}} \right\rfloor \\
\\
&= \left\lfloor \frac{37820451}{10^4} \right\rfloor - 10^2 \left\lfloor \frac{37820451}{10^6} \right\rfloor \\
\\
&= \lfloor 3782.0451 \rfloor - 100 \lfloor 37.820451 \rfloor \\
\\
&= 3782 - 3700 \\
\\
&= 82
\end{align}$$


问:
Defining projection maps in Affine geometry and Thales' theorem and related sources
I am from Physics background and have not much experience in dealing with rigorous mathematical language and proofs. Out of interest, I am trying to read the book Spacetime. The book, which seems advanced, is very hard for me to read due to my inexperience and lack of the coverage of its prerequisites. Here's an excerpt from the book on proving Thales' theorem,

As you can see, the author uses projection maps $\pi$ and $\vec{\pi}$ but no definitions of these maps have been given. The author assumes the reader is familiar with these definitions/concepts. And I am lost. Can someone kindly explain how these maps are defined (both rigorous and intuitive) and what are their properties, that are crucial for this proof?

Moreover, if you think I should consult other resources on affine and Euclidean geometry first before venturing into this book, kindly recommend those sources. (Please do not recommend Berger or Audin. Those seem even more difficult than this book)

Edit:

As I realized that, people are unlikely to have a copy of this book, I have decided to to add images of the first five pages of the book so that people can understand better the level of this book and that would help recommending resources.

答:
An answer to the unedited version of the question.

I think in order to understand the quoted fragment of a proof  is required basic knowledge on vector and affine spaces and their subspaces and quotient spaces. I expect it is sufficient to read Wikipedia articles devoted to these subjects. Concerning affine spaces, you can also start from my recent answer to the question “What is the affine space and what is it for?” and possible other stuff from that thread.

In the proof $\Bbb A^n$ should be the affine space associated to the vector space $\Bbb R^n$. Maps to equivalence classes usually are called quotient maps, not projections.  The map $\pi$ is defined via the map $\vec\pi$, $\vec x$ is a coset $x+\vec H$, that is an set $\{x+y:y\in\vec H\}$, which is an element of $\Bbb R^n/\vec H$.

As I understood the context, by a hypersuface in the book is meant a hyperplane, that is a set $\{y\in\Bbb R^n: (y,x)=c\}$, where $x\in\Bbb R^n$  is a fixed non-zero vector, $c\in\Bbb R$ is a constant and $(\cdot,\cdot)$ is the inner product on $\Bbb R^n$.

Then the proof via quotient spaces looks overcomplicated. Indeed, since the hyperplanes $H_i$ are parallel, there exists a non-zero vector $x\in\Bbb R^n$ and constants $c_i\in\Bbb R$ such that $H_i=\{y\in\Bbb R^n: (y,x)=c_i\}$ for each $i$. Remark that we
need a restriction $H_1\ne H_2$ missed in the lemma, because otherwise $x_2(l)-x_1(l)=0$. The the collinearity of points $x_i(l)$ implies that there exists $\lambda\in\Bbb R$ such that $x_3(l)-x_2(l)=\lambda(x_2(l)-x_1(l))$. Taking inner product with $x$ of both sides of this equality, we obtain that $c_3-c_2=\lambda (c_2-c_1)$, that is $$\frac{x_3(l)-x_2(l)}{ x_2(l)-x_1(l)}=\lambda=\frac {c_3-c_2}{c_2-c_1},$$which does not depend on $l$.

Update. I read the first paragraph of the first page of the book and briefly looked at the next pages and I guess that the affine space $\Bbb A^n$ is provided as a degenerated case of a manifold with a connection and its more general case corresponding to the general relativity theory will be considered later in the book. This subject is much more advanced than a basic knowledge of affine spaces and need a much more advanced background from differential geometry and tensor analysis.

Remark. I disagree with the author’s claims that a description of space by $\Bbb R^3$ has no physical counterpart. The origin $(0,0,0)$ of $\Bbb R^3$ naturally corresponds to an origin of a chosen  reference frame. Also elements of $\Bbb R^3$ can be considered as vectors with coordinatewise addition and scalar multiplication, so $\Bbb R^3$ has a natural vector space structure.


问:
What will be a circle look like considering this distance function?
I am working on some exercises in the book Geometry: A Metric Approach with Models by R.S. Millman. He defines the following map: $$d_S(P,Q):\mathbb R^2\times\mathbb R^2\to\mathbb R\\\ d_S(P,Q)=\max\{~|x_2-x_1|,~|y_2-y_1|~\}$$ where in $P(x_1,y_1),~Q(x_2,y_2)$ are points in $\mathbb R^2$ and then wants the reader to show that $d_S$ is a distance function on $\mathbb R^2$.

I was thinking to myself of what would be a circle looked like regarding this function in Cartesian plane? I fixed a point, for example the origin, as a centre and then probed the case with a given positive constant $r$ as a radius. I just got two horizon $|y|=r$ or two vertical $|x|=r$ lines. the results didn't have good taste. Any help? Was I right? Thanks

答:
We set

$$\max\{|x_2 - x_1|, |y_2 - y_1|\} = r$$

with $r > 0$ being a "radius".

Take a center for the "circle" -- say $(0, 0)$, to make this

$$\max\{|x|, |y|\} = r$$.

Now, this means at least one of $|x|$ and $|y|$ is $r$, and the other less. So there is a set of all $x$ for which $y = r$ and $|x| < r$, which is a horizontal line segment. For $y = -r$, we have another such segment, on the other side of the x-axis. Repeating for $x = r$ and $|y| < r$ and $x = -r$ and $|y| < r$, we get two vertical line segments. Altogether, we get a square with side lengths $2r$.


问:
Nested intervals theorem - a special case on open intervals
I learned the nested intervals theorem in the class: If $I_n\ (n\in\Bbb N)$ is a sequence of bounded closed intervals, i.e. $[a_n,b_n]$, then $\bigcap I_n\ (n\in\Bbb N)\neq\varnothing$. In the proof of the theorem, we used another theorem: monotone non-decreasing sequence (in this case $a_n$) is convergent, if bounded above.

And we discussed in the class that the theorem does not hold for open intervals, i.e. $(a_n,b_n)$. We made a counterexample $I_n=(0,\frac{1}{n}),\ n\in\Bbb N$, where $\bigcap I_n=\varnothing$. It also makes sense to me. There is no real staying in the intersection of nested intervals, because the candidate $0$ is ruled out by the open intervals.

Then, the professor gave us an extended question: What if we have nested $I_n$ as open intervals $(a_n,b_n)$, but this time, let $a_n$ be a strictly increasing sequence ($\forall n\in\Bbb N,\ a_n\lt a_{n+1}$) and $b_n$ a strictly decreasing sequence? It seems to me that there would be a real finally staying in the nested interval. However, I cannot prove it. Appreciated if anyone can provide me a hint. Thanks.

答:
Let $$A = \lim_{n\to \infty}
{a_n}$$$$B=\lim_{n\to \infty}{b_n}$$For all n, $ a_n \lt A \le B \lt b_n $according to Sandwich Theorem, thus $[A,B] \subset I_n$ for all n. So $$\bigcap_{n=1}^\infty{I_n} = [A,B] \neq \emptyset$$ Besides, for every $m \in \mathbb N$ and every $ n \in \mathbb N$, $a_m \lt b_n$.


问:
Question about derivation in Jordan algebra
Let $(G,\circ)$ be a Jordan algebra, then $\sigma:G\to G$ given by $$c\mapsto a\circ(b\circ c)-b\circ(a\circ c),\quad\forall c\in G,$$ is a derivation, where $a$ and $b$ are two fixed elements of $G$.</p>

答:
Just for the record, this is how I checked my claim that this works in special Jordan algebras:

The following Mathematica code makes the operator ** multilinear and defines $\circ$ to be its symmetrization.

```<code>Unprotect[NonCommutativeMultiply];
NonCommutativeMultiply[a___, b_Plus, c___] := NonCommutativeMultiply[a, #, c] & /@ b;
NonCommutativeMultiply[a___, \[Lambda]_?NumberQ b_, c___] := \[Lambda] NonCommutativeMultiply[a, b, c];

a_ \[SmallCircle] b_ := (a ** b + b ** a)/2;
</code>```

This is your map $\sigma$:

```<code>\[Sigma][c_] := a\[SmallCircle](b\[SmallCircle]c) - b\[SmallCircle](a\[SmallCircle]c);
</code>```

And this is the check that it is a derivation:

```<code>\[Sigma][x\[SmallCircle]y] - x\[SmallCircle]\[Sigma][y] - \[Sigma][x]\[SmallCircle]y // Simplify
</code>```

which gives zero.

(All this looks nicer in Mathematica...)

Body:
Just for the record, this is how I checked my claim that this works in special Jordan algebras:

The following Mathematica code makes the operator ** multilinear and defines $\circ$ to be its symmetrization.

Unprotect[NonCommutativeMultiply];
NonCommutativeMultiply[a___, b_Plus, c___] := NonCommutativeMultiply[a, #, c] & /@ b;
NonCommutativeMultiply[a___, \[Lambda]_?NumberQ b_, c___] := \[Lambda] NonCommutativeMultiply[a, b, c];

a_ \[SmallCircle] b_ := (a ** b + b ** a)/2;

This is your map $\sigma$:

\[Sigma][c_] := a\[SmallCircle](b\[SmallCircle]c) - b\[SmallCircle](a\[SmallCircle]c);

And this is the check that it is a derivation:

\[Sigma][x\[SmallCircle]y] - x\[SmallCircle]\[Sigma][y] - \[Sigma][x]\[SmallCircle]y // Simplify

which gives zero.

(All this looks nicer in Mathematica...)


问:
Show that the cone of the open interval (0, 1) can not be embedded in any Euclidean space
I've been trying to tackle this problem for some while now, but don't know how to start correctly. I know that the cone on $(0,1)$ is given by $$\text{Cone}((0,1)) = (0,1) \times [0,1]/((0,1)\times\{1\}).$$ But how do I show that it can not be embedded in an Euclidean space? Cause for me it looks like it is possible.(Open cylinder with the "ceiling" collapsed to one point. I'm guessing that the problem  for me also lies in what a quotient really is, cause I can't really get a good feeling for it.

I don't want the answer, I just want a push in the right direction so that I can think about how to solve it.

Edit:
New insight, when thinking about the cone, it should be something like this (I guess) but this would mean that it can be embedded in $\mathbb{R}^2$ I think, which contradicts the question.

Thanks

答:
Let me add a little bit of context and change the terminology such that
‘automatic thinking’ gets it right.

Let $\mathrm{C}(X)$ denote the (real, geometric, true...) cone on $X$.
Its points are the pairs $(\lambda, x) \in \mathbf R_+^* \times X$ and
a special point $0$.

An open $U$ of the real cone is either an open of $\mathbf R_+^* \times X$ if it does not contain $0$ and if it does contains $0$, it is additionally required to
contain an open of the form $(0, \varepsilon) \times X$.

Can you show that if $X$ is embeddable in $\mathbf R^n$, then the real cone
$\mathrm{C}(X)$ is embeddable in $\mathbf R^{n+1}$?

Let $\mathrm{CR}(X) = \mathbf R_+ \times X/0 \times X$ be the
collapsed rectangle.
Now, when $X$ is compact and separated, it happens to be true that
$$
\mathrm{C}(X) \simeq \mathrm{CR}(X)
$$
but not in general. Can you show that the two are different
in the case $X = (0,1)$?


问:
Prove Cauchy Sequence and limit
Below the following question is my solution but I am not sure whether the proofs are correct and rigorously exhaustive so, if present, kindly point out any errors.

Let $\{x_n\}$ be recursively defined as $   \left\{
\begin{array}{ll}
      x_{n+1} =\frac{1-x_n}{4} \\
      x_1 = \frac{1}{2}
\end{array}
\right. $

$(a)$ Prove that $|x_{n+1}-x_n| \le \frac{1}{4}|x_n-x_{n-1}|$

$(b)$ Use $(a)$ to prove that ${x_n}$ is a cauchy sequence

$(c)$ Compute $\underset{n \rightarrow \infty}{\rm lim} x_n$

Answer:

$(a)$
$x_{n+1} - x_n=\frac{1-x_n}{4}-x_n$ and $ x_n = \frac{1-x_{n-1}}{4}$. Thus, $x_{n+1}-x_n = (\frac{1-x_n}{4})-(\frac{1-x_{n-1}}{4}) = \frac{1}{4} \{1-x_{n}-1+x_{n-1}\}$. Therefore, $|x_{n+1} -x_n| \le |\frac{1}{4}(1-x_n-1+x_{n-1})| \Rightarrow |x_{n+1} -x_n| \le \frac{1}{4} |x_{n-1}-x_n|$ and $\frac{1}{4} |x_{n-1}-x_n| = \frac{1}{4} |x_{n}-x_{n-1}|$

$(b)$ From $(a)$, it is observable that $|x_{n+1}-x_n| \le \frac{1}{4}|x_{n-1}-x_n|$. Now $|x_{n-1}-x_{n-2}| \le \frac{1}{4} |x_{n-2} - x_{n-3}|$. Continuing this way, we get $|x_{n+1}-x_n| \le \frac{1}{4^n} |x_2-x_1| \Rightarrow |x_{n+1}-x_n| \le \frac{1}{4^n}(\frac{+3}{8})$. As $x \rightarrow \infty , \frac{1}{4^n} \rightarrow 0$. Hence, $\{x_n\}$ becomes a Cauchy sequence. Since every sequence in $\mathbb{R}$ is convergent, $\{x_n\}$ is convergent.

$(c)$ In $(b)$ we have shown $\{x_n\}$ is convergent. Therefore, we assume $l \in \mathbb{R}$ and $ lim\hspace{3pt} x_n = l$. We know $x_{n+1} = \frac{1}{4} (1-x_n)$;
$lim \hspace{3pt} x_{n+1} = \frac{1}{4}(1-lim\hspace{3pt} x_n) \Rightarrow l=(\frac{1}{4})(1-l) \Rightarrow l=\frac{1}{5}$

答: For (a) and (c) you proofs are correct But your argument for (b) is wrong. $x_{n+1}-x_n \to 0$ does not imply that $(x_n)$ is a Cauchy sequence. Use the fact that $|x_{n+m}-x_n| \leq |x_{n+m}-x_{n+m-1}|+|x_{n+m-1}-x_{n+m-2}|+\cdots +|x_{n+1}-x_{n}|$ to show that $x_{n+m}-x_n \to 0$ as $n,m \to \infty$. [You will have to compute a geometric sum for this].

Title: Prove Cauchy Sequence and limit


问:
Units on a ring
Let $A$ be a ring such that there exists units $u_1, \dots , u_p : u_1 + \dots + u_p=0$
and $p$ a prime. Did $A$ have a subring isomorphic to $\mathbb{Z}/p \mathbb{Z}$ in general

Thanks in advance.

答: The answer is clearly no in general. The $3$ non-zero elements of a field of order $4$ have sum zero, but there is no subring $\mathbb{Z}/3 \mathbb{Z}.$</p>

Title: Units on a ring


问:
How to calculate $I=\iint\limits_{\left[ 0,1 \right] \times \left[ 0,1 \right]}{e^{-\frac{u}{v^2}}\mathrm{d}v\mathrm{d}u} $?
I try to calculate by fixing $v$ first:
$$
I=\iint\limits_{\left[ 0,1 \right] \times \left[ 0,1 \right]}{e^{-\frac{u}{v^2}}\mathrm{d}v\mathrm{d}u}=\int_0^1{\left( \int_0^1{\left( e^{-\frac{1}{v^2}} \right) ^u\mathrm{d}u} \right) \mathrm{d}v}=\int_0^1{\left( -v^2\left( e^{-\frac{1}{v^2}} \right) ^u \right)\Bigg|_{u=0}^{u=1}\mathrm{d}v}=\int_0^1{-v^2\left( e^{-\frac{1}{v^2}}-1 \right) \mathrm{d}v}.
$$

Now the problem is $\int_0^1 v^2 e^{-(1/v^2)}\mathrm{d}v$. I've tried to make a substitution $s=1/v$, but it simply doen't work. It seems to me that the integral is not calculable... Does anyone have an idea about this? Thank you in advance!

答:
\begin{align}
I&=\int_1^{\infty} t^{-2} \int_0^1 e^{-ut^2}\,dudt\tag{$t=v^{-1}$}\\
&=\int_1^{\infty} t^{-4} \int_0^{t^2} e^{-s}\,dsdt\tag{$s=ut^{2}$}\\
&=\int_1^{\infty} t^{-4}\left(1-e^{-t^2}\right)\,dt\\
&=\frac13 - \frac{1}{3e}+\frac23\int_1^{\infty} t^{-2}e^{-t^2}\,dt\tag{Integration by parts}\\
&=\frac13 - \frac{1}{3e}+\frac2{3}\left(\frac1e -\sqrt\pi\operatorname{erfc}(1)\right)\tag{Integration by parts}\\
&=\frac13 + \frac{1}{3e}-\frac{2\sqrt\pi}{3}\operatorname{erfc}(1),
\end{align}
where $\operatorname{erfc}$ is the complementary error function.


问:
Finding limit of a function with root
I need to find the limit of this function and I'm not sure how to proceed
$$ \lim_{x\to 1}\frac{\sqrt {x+3} -\sqrt x}{1-x} $$
I multiply by the conjugate
$$ \lim_{x\to 1}\frac{\sqrt {x+3} -\sqrt x}{1-x}*\frac{\sqrt {x+3} +\sqrt x}{\sqrt {x+3} +\sqrt x} $$
and I am left with
$$ \frac{ {x+3} - x}{(1-x)(\sqrt {x+3} +\sqrt x)} = \frac{3}{(1-x)(\sqrt {x+3} +\sqrt x)} $$
but after this I am stuck there is nothing to cancel and pluging in 1 will give me $\frac{3}{0}$

Thank you everybody it is much clearer now !

答:
This limit does not exist, since when $x\to 1$ from the left we get result $\infty $ and from the right $-\infty$.

Limit of the factor $\frac{3}{\sqrt {x+3} +\sqrt x} $ is $1$, but not exists for ${1\over 1-x}$ (draw this function) when $x\to 1$.

Title: Finding limit of a function with root


答:
The function is positive when $1-x>0\to x<1$ therefore
$$\underset{x\to 1^-}{\text{lim}}f(x)=+\infty;\;\underset{x\to 1^+}{\text{lim}}f(x)=-\infty$$
The two limits are different so the limit doesn't exist.

Title: Finding limit of a function with root

Body:
I need to find the limit of this function and I'm not sure how to proceed
$$ \lim_{x\to 1}\frac{\sqrt {x+3} -\sqrt x}{1-x} $$
I multiply by the conjugate
$$ \lim_{x\to 1}\frac{\sqrt {x+3} -\sqrt x}{1-x}*\frac{\sqrt {x+3} +\sqrt x}{\sqrt {x+3} +\sqrt x} $$
and I am left with
$$ \frac{ {x+3} - x}{(1-x)(\sqrt {x+3} +\sqrt x)} = \frac{3}{(1-x)(\sqrt {x+3} +\sqrt x)} $$
but after this I am stuck there is nothing to cancel and pluging in 1 will give me $\frac{3}{0}$

Thank you everybody it is much clearer now !


答: An easy counterexample occurs with $A=\mathbb{Z}$, $p=2$, and $u_1=1$, $u_2=-1$.</p>

Title: Units on a ring

Body:
Let $A$ be a ring such that there exists units $u_1, \dots , u_p : u_1 + \dots + u_p=0$
and $p$ a prime. Did $A$ have a subring isomorphic to $\mathbb{Z}/p \mathbb{Z}$ in general

Thanks in advance.


答:
When you plug in $x=1$ in the beginning, you have the limit in the form $\frac{1}0$, which directly implies that the limit doesn't exist.

$$\lim_{x\to1^+}\frac{\sqrt{x+3}-\sqrt{x}}{1-x}=-\infty\quad\mbox{and}\quad\lim_{x\to1^-}\frac{\sqrt{x+3}-\sqrt{x}}{1-x}=\infty$$

Here's a simple graph of the function $f(x)=\frac{\sqrt{x+3}-\sqrt{x}}{1-x}$ drawn on Desmos.

Title: Finding limit of a function with root


问:
Argand Digram when arguments are equal
I am helping a student to draw the following equation on complex plane:
$$\arg\left[z-\left(3+i\right)\right]=\arg\left[z-\left(1+3\,i\right)\right]$$

Can anyone explain why the dashed line to be neglected.

答:
This equation gives two rays up and down on the line $x+y=4$, where points between (1,3) and (3,1) will not lie on this locus. For instance the points (2,2), (3/2,1/2), (1,1,3/2),... do not satisfy the equation that
$$Arg[z-(3+i)]=Arg[z-(1+3i)].$$ For the point $2+2i$, we get LHS: $Arg(-1+i)=-3\pi/4$ but RHS: $Arg(1-i)=-\pi/4.$

Title: Argand Digram when arguments are equal


问:
Convergence of Cauchy sample mean
If $X_i$'s are iid Cauchy random variables, does the sample mean $\frac{\sum^{n}_{i=1}X_i}{n}$ converge in distribution to any limit? If yes, can someone explain to me how the limit can be computed step by step? I don't think it would, since the variance of a Cauchy random variable is infinity, but I am not so sure about this.

答:
For every $n$ $\frac 1 n \sum\limits_{i=1}^{n} X_i$ also has the Cauchy distribution (!) so the Cauchy means certainly converge in distribution to the Cauchy distribution. The easiest way to prove this is to use characteristic functions: $Ee^{it \frac 1 n \sum\limits_{i=1}^{n} X_i}=(Ee^{itX_1/n})^{n}=(e^{-|t/n|})^{n}=e^{-|t|}$.

For the additional question in the comment below I will give a hint: If $S_n =\frac 1 n \sum\limits_{i=1}^{n} X_i$ converges in probability then $S_{2n}-S_n \to 0$ in probability, hence in distribution. So its characteristic function must tend to $1$ at every point. But the characteristic function converges to $e^{-|t|}$.

Title: Convergence of Cauchy sample mean


问:
Logarithm inequality including square root:minimum of $\sqrt{\log_2^2a+1}+\sqrt{\log_2^2b+1}+\sqrt{\log_2^2c+1}$
I just encountered this problem and tried a few ways but seems like it didn't work. Can some one help me?

Edited

I forgot to including what i have tried.

I tried to take the logarithm of both side of the hypothesis and then replace them in P, but it can not reduce the variables.

I also tried to calculate $P^2$, but did not seem to work out well though.

Let $a,b,c>0$ such that $abc=8$, find
\begin{align*}
\min P= \sqrt{\log_2^2a+1}+\sqrt{\log_2^2b+1}+\sqrt{\log_2^2c+1}.
\end{align*}
Any helps is appreciated, thanks in advance!

答:
let $\log_2a=x,\log_2b=y,\log_2c=z$

we have $x+y+z=3$

we have to minimise $$P=\sum_{cyc}\sqrt{x^2+1}$$
by C-S $$\sqrt{x^2+1}\ge \frac{|x+1|}{\sqrt{2}}$$
Thus by C-S along with triangle ineqality$$\sum_{cyc}\sqrt{x^2+1}\ge \sum \frac{1}{\sqrt{2}}|x+1|\ge \frac{1}{\sqrt{2}}|x+1+y+1+z+1|=?$$


问:
Find the limiting circle of a Hopf Bifurcation point of a non linear oscillator
This problem deals with the nonlinear oscillator
$$
y''+(y^2-\lambda)y'+(y+1)(y^2+y-\lambda)=0
$$Find the steady states and determine the values of $\lambda$ at which they are stable. Sketch the bifurcation diagram and classify the three bifurcation points.

Find a first-term approximation of the limit cycle that appears at Hopf bifurcation point.

My attemps:

the steady-states are
$$y=-1$$
$$y=-\frac12+\frac12\sqrt{1+4\lambda}$$
$$y=-\frac12-\frac12\sqrt{1+4\lambda}$$

we can draw out the bifurcation diagram as below

According to the solution key in the textbook i am reading, there are three bifurcation points. I can work out that the point $(0,-1)$ as a transcritical point. the details are:

Let
$$y\sim-1+\delta y_1+\cdots$$
substitute into the above differential equation, we have
$$y_1''+(1-\lambda)y_1'-\lambda y_1=0$$
and therefore, the two roots of the corresponding auxiliary equation is $r=\lambda$ and $r=-1$. And therefore, when $\lambda>0$, it is stable while when $\lambda<0$, it is unstable. We conclude that $(0,-1)$ is a transcritical point. Am i correct?
For the second steady state, i tried let
$$y\sim -\frac12+\frac12\sqrt{1+4\lambda}+\delta y_1$$
Substitute into the above differential equation, we have
$$y_1''+\left(\frac12-\frac12\sqrt{1+4\lambda}\right)y_1'+\left(\frac12\sqrt{1+4\lambda}+\frac12+2\lambda\right)y_1=0$$
From here, i worked out the two corresponding roots as well, it turns out that the real part of the two roots are
$$Re(r_-)=Re(r_+)=-\frac14+\frac14\sqrt{1+4\lambda}$$
Therefore, it is unstable when $\lambda>0$ while stable when $\lambda<0$. Am i correct?

Since when $\lambda=0$, the two roots are purely imaginary. Therefore, $(0,0)$ should be a Hobf bifurcation point.

I don't know why $\left(-\frac14,\frac12\right)$ is a saddle point. And I don't know how to work out the limiting circle as well. Anyone can help me work out?

答:
For the third steady state, let
$$y\sim -\frac12-\frac12\sqrt{1+4\lambda}+\delta y_1$$
Substitute into the above differential equation, we have
$$y_1''+\left(\frac12+\frac12\sqrt{1+4\lambda}\right)y_1'-\left(\frac12\sqrt{1+4\lambda}-\frac12-2\lambda\right)y_1=0$$
The two roots are $$r_\pm=-\frac14-\frac14\sqrt{1+4\lambda}\pm\frac12\sqrt{\frac52\sqrt{1+4\lambda}-\frac32-7\lambda}$$ we see that $r_+\ge0$ if $\lambda\in\left[-\frac14,0\right]$ and $r_+<0$ if $\lambda\in\left(0,\infty\right)$, while $r_-<0$ for $\lambda\in\left[-\frac14,\infty\right)$. Therefore, this steady state is unstable if $\lambda\in\left[-\frac14,0\right]$ and stable if $\lambda\in\left(0,\infty\right)$. From here we conclude that $\left(-\frac14,-\frac12\right)$ is a saddle-node and $\left(0,-1\right)$ is a transcritical point.

For the limiting circle, let $\lambda=\varepsilon>0$, and $y\sim -\frac12+\frac12\sqrt{1+4\lambda}+\delta y_1$ After subtituting into the orginal differential equation we have
$$y_1''+\left(\frac12-\frac12\sqrt{1+4\varepsilon}\right)y_1'+\left(\frac12\sqrt{1+4\varepsilon}+\frac12+2\varepsilon\right)y_1=0$$ By binomial expansion we have the following differential equation,
$$y_1''-\varepsilon y_1'+\left(1+3\varepsilon\right)y_1=0$$This equation is weakly damped for small $\varepsilon$. Based on experience, a multi-scale expansion with time scale $t_1=t$ and $t_2=\varepsilon t$ are used. and by assuming that $y_1\sim Y_1\left(t_1,t_2\right)+\varepsilon Y_2\left(t_1,t_2\right)+\cdots$
The final solution for this after some tedious calculation is as below
$$y_1\sim \frac2{\sqrt{1+ce^{-\lambda t}}} \cos\left( t+\phi_0 \right)$$
$$y_1'\sim -\frac2{\sqrt{1+ce^{-\lambda t}}} \sin\left( t+\phi_0 \right)$$ where $c,\phi_0$ are constants that depend on the initial conditions.

Final note to take so as to make the analysis complete, for the second steady state, the two roots of the auxiliary equations are
$$r_\pm=-\frac14+\frac14\sqrt{1+4\lambda}\pm\frac12\sqrt{-\frac52\sqrt{1+4\lambda}-\frac32-7\lambda}$$ When $\lambda>0$, the two roots are complex conjugates with positive real part, thus unstable focus. When $\lambda=0$, the two roots are conjugate and purely imaginary, this is the Hopf bifurcation point. When $\lambda\in\left(-0.2478,0\right)$, the two roots are conjugate with negative real parts, thus stable focus. Lastly, when $\lambda\in\left(-\frac14,-0.2478\right)$, the two roots are real negative, thus stable node.


问:
Proving an Integral domain is a field.
Let $R$ be an integral domain containing a field $F$ as a subring. Show that if $R$ is a finite dimensional vector space over $F$, then $R$ is a field.

This is a Ph.D. entrance question, I recently appeared.

Somehow I need to prove that every element in $R$ has an inverse, but can't figure out how.

答:
Hints:

$$(1)\;\;\;\forall\, r\in R\,\;\exists\,p(t)=a_0+a_1t+...+a_{n-1}t^{n-1}+t^n\in F[t]\;\;s.t.\;\;p(r)=0\;;$$

$$(2)\;\;\text{Lemma: let $\,A\subset B\,$ be two integral domains. If $\,B\,$ is integral over $\,A\,$ then} $$

$$\,A\, \text{ is a field iff $\,B\,$ is a field .}$$

Title: Proving an Integral domain is a field.


问:
solve for the Laurent series of the function $f(z) = \frac{z-12}{z^2+z-6}$ that is valid for $1<|z-1|<4$.
I am about to solve for the Laurent series of the function $f(z) = \frac{z-12}{z^2+z-6}$ that is valid for $1<|z-1|<4$.

What I did is I use the partial fraction decomposition to rewrite the $f(z)$ into $f(z) = \frac{3}{z+3}+\frac{2}{z-2}$.  Now based on the region to where the function $f$ should be valid on is that $1< |z-1|< 4$, I'm not sure if I did it right when I assumed that $\left|\frac{2}{z}\right|<1$ and $\left|\frac{3}{z}\right|<1$. (I just did it with brute force to arrive on the two inequalities). Can you help me with that?

Assuming my two inequalities are correct, I have now

$\frac{3}{z+3} = \frac{3/z}{1+3/z} = 3/z \sum_{n=0}^{\infty} \left(- \frac{3}{z}\right)^n = \sum_{n=0}^{\infty} (-1)^{n-1}3^nz^{-n} = \sum_{n=1}^{\infty} (-1)^n3^{n+1}z^{-(n+1)}$,

and

$\frac{2}{z-2} = \frac{2/z}{1-2/z} = 2/z \sum_{n=0}^{\infty}  \left(\frac{2}{z}\right)^n = \sum_{n=0}^{\infty} \frac{2^{n+1}}{z^{n+1}}.$

Then after that, I combined which I have the result as

$f(z) = \frac{z-12}{z^2+z-6} = \sum_{n=0}^{\infty} \frac{2^{n+1}}{z^{n+1}} + \sum_{n=1}^{\infty} (-1)^n(3)^{n+1}z^{-(n+1)}$.

Now, did I do it correctly? or I did wrong on the inequalities of the region to where the laurent should be valud to?  Thanks for those who can help.

答:
Your answer cannot possibly be an answer to the given question, since the answer should be a series of the form $\sum_{n=-\infty}^\infty a_n(z-1)^n$.

Note that\begin{align}\frac{z-12}{z^2+z-6}&=\frac3{z+3}-\frac2{z-2}\\&=\frac3{z-1+4}-\frac2{z-1-1}\\&=\frac34\frac1{1+\frac{z-1}4}+2\frac1{1-(z-1)}.\end{align}Now, since $1<|z-1|$ we have$$\frac1{1-(z-1)}=-\sum_{n=-\infty}^{-1}(z-1)^n$$and, since $|z-1|<4$ we have$$\frac1{1+\frac{z-1}4}=\sum_{n=0}^\infty\frac{(-1)^n}{4^n}(z-1)^n.$$Can you take it from here?


问:
Checking if a function is 3-linear?
I was doing some problems from Hoffman/Kunze, and came across one which asks to show check if a function, $D$ on the set of real 3x3 matrices is 3-linear or not. $D$ is defined as:

$$D(A) = A_{11}A_{12}A_{33}$$

I tried using counter-examples to show that it's not, however, so far, addition seems to work. However, a solution I found shows the following:

Let $A_{11} = 2, A_{12} = 2$ and $A_{33} = 1$ and all other $A_{ij} = 0$. Also, let $B_{11} = B_{12} = B_{33} = 1$ and all other $B_{ij} = 0$. Then $D(A) = D(B) + D(B)$ by n-linearity. But, $D(A) \neq D(B) + D(B)$. So, $D$ is not 3-linear.

However, isn't n-linearity focused on a specific entry rather than 2 or more entries? So, why does the solution above suggest that $D(A) = D(B) + D(B)$ when there are two instances with an entry of value 2 being compared i.e. $A_{11} = 2$ = $B_{11} + B_{11}$ and $A_{12} = 2$ = $B_{12} + B_{12}$, instead of just one or the other. That is, why are they using a matrix, $B$, that has $B_{11} = B_{12} = B_{33} = 1$ when n-linearity involves keeping all other rows constant; shouldn't they use a $B$ where only one entry of $B$ is written in terms of one entry of $A$? An example of such a $B$ being $B_{11} = 2$ and $B_{12} = B_{33} = 1$ or $B_{11} = 1$, $B_{12} = 2$ and $B_{33} = 1$. This would imply that D is, in fact, 3-linear.

I hope someone can explain whether they made a mistake or if I'm missing something. Any assistance is appreciated.

答: A function can be said to be associated with a multi-linear map when it is of the form $f(x)=g(x,\ldots,x)$ where $g$ is linear in each variable. In this case, $$D(A)=g(A,A,A),\qquad g(A,B,C)=A_{11}B_{12}C_{33}$$ It is not hard to show that $g$ is linear in each variable, e.g. $$g(A+\lambda A',B,C)=(A_{11}+\lambda A'_{11})B_{12}C_{33}=g(A,B,C)+\lambda g(A',B,C)$$

Title: Checking if a function is 3-linear?


问:
There are complex numbers $z_{1}, \cdots z_{n}$ such that $z^{n} + \cdots a_{0} = \prod_{i=1}^{n}(z-z_{i})$ - Final induction step
The following is a question from Spivak's Calculus - Ch 26 (3a)

If $a_{0}, \cdots a_{n-1}$ are any complex numbers, then there are complex numbers $z_{1}, \cdots z_{n}$ such that $$z^{n} + \cdots a_{0} = \prod_{i=1}^{n}(z-z_{i})$$

I had a question on obtaining the final conclusion from my inductive step.

This statement can be proved by induction using the fundamental theorem of algebra along with a fact that was established earlier in the book that stated

For any polynomial function $f$ and any $a$, there exists a polynomial $g$ and a constant $b$ s.t. $f(x) = (x-a)g(x) + b$

Base Case:

By the FTA we know there exists a root $z_{1}$ such that

$$(z-z_{1})(z^{n-1} + b_{n-2}z^{n-2}+ \cdots b_{1}z + b_{0})$$

Induction Hypothesis:

Suppose this is true for $n=k$, thus there exists a polynomial g(z) and constant $\gamma$ such that

$$\prod_{i=1}^{k}(z-z_{i})g(z) + \gamma$$

Induction Step:

Here is where I'm stuck. I know by the FTA there exists a root call it $z_{k+1}$ and constant $\phi$ which would allow me to write my expression as

$$\prod_{i=1}^{k+1}(z-z_{i})g(z) + \phi$$

But how would I state that this would be the end of the division algorithm that got me to this point and as such should be

$$\prod_{i=1}^{k+1}(z-z_{i})$$

?

答:
Here is how you can set out the induction if you can use the fundamental theorem of algebra and the division algorithm.

Base case

Let $\mathscr S_k$ be the statement that for any polynomial $f(z) = z^k+ a_{k-1}z^{k-1}+\cdots a_0 $ of degree $k > 0$ there exists roots $z_1, z_2, \cdots z_k$ such that
$$
f(z) = (z-z_1)(z-z_2)\cdots(z-z_k)
$$
Plainly $\mathscr S_1$ is true.

Inductive step

Suppose you have proved $\mathscr S_k$ for some $ k >0$.  Let $p(z) = z^{k+1} + b_k z^k +\cdots b_0$, a polynomial of degree $k+1$.  Then by the fundamental theorem of algebra there exists $z_{k+1}$ such that $p(z_{k+1}) = 0$.  Divide $p(z)$ by $(z-z_{k+1})$.  The division rule gives,
\begin{aligned}
p(z) = (z-z_{k+1})q(z) + \gamma
\end{aligned}
where the degree of $q(z)$ is $k$, being one less than the degree of $p(z)$.  But $p(z_{k+1}) = 0$, and substituting $z_{k+1}$ into this equation shows that $\gamma=0$.  Thus
$$p(z) = (z-z_{k+1})q(z).$$
We can now apply $\mathscr S_k$ to $q(z)$ to find $z_1, \cdots z_k$ such that $$q(z)=(z-z_1)(z-z_2)\cdots(z-z_k),$$ whence
$$p(z) = (z-z_1)(z-z_2)\cdots(z-z_{k+1})
$$
and we have proved $\mathscr S_{k+1}$.  It follows by the principle of induction that $\mathscr S_k$ is true for all $ k \geqslant 1$.

In your proof, you failed to spot that $\gamma=0$, so then couldn't easily proceed.  I also note that the four formulae at each step are not written as equations, which leaves their meaning unclear.


问:
Involution in $L^{1}(G)$ is isometric.
(Sorry for asking so many questions of the same type.  There is an underlying issue that I think once resolved will allow me to understand them all at once.)

Let $G$ be a locally compact group, and $f\in L^{1}(G)$.
I'm trying to verify that $\|f^{*}\|_{p} = \|f\|_{p}$, where the involution $*$ is defined by $f^{*}(x) = \delta(x)^{-1}\overline{f(x^{-1})}$.

So my computations so far look like:

\begin{eqnarray*}
\int_{G}f^{*}(y)\mu(dy) &=& \int_{G}\left|\delta(y^{-1})\overline{f(y^{-1})}\right|^{p}\mu(dy)\\
                        &=& \int_{G}\delta(y^{-1})^{p}\left|\overline{f(y^{-1})}\right|^{p}\mu(dy)\\
\end{eqnarray*}

Here I am stuck at the central issue I am facing where the calculations stop making sense and I feel like I'm just throwing things around hoping it's true.

How do I use the definition of the $\delta$ map:

$$\text{the unique group homomorphism }\delta:G\to(0,\infty)\text{ such that }\\\delta(x)\mu(E) = \mu(Ex)\text{ for all Borel sets }E$$

when its argument is not independant of the variable of integration?

Ted has already confirmed in his answer to:

How to use the modular function of a locally compact group?

that the definition of $\delta$ implies:

$$\int_{G}\delta(x)f(y)\mu(dy) = \int_{G}f(y)\mu(dy\cdot x)$$

If I try to use that step in my current problem, I end up with

\begin{eqnarray*}
\int_{G}\delta(y^{-1})^{p}\left|\overline{f(y^{-1})}\right|^{p}\mu(dy) &=& \int_{G}\left|\overline{f(y^{-1})}\right|^{p}\mu(dy\cdot (y^{-1})^{p})
\end{eqnarray*}
which I don't have any clue how to interpret, and I suspect is wrong.  What is the correct way to handle this function?  I dont know what measure to associate with the notation $\mu(dy\cdot (y^{-1})^{p})$.

I think the answer to this question is what I really need to be able to finish the verifications I'm working on.

Thanks to whoever can help me!

答:
I think you would find the last chapter of Real Analysis by Folland helpful (it covers the fundamentals of Haar measures).  The way I think of it, $\delta^{-1}$ is a factor attached to $\mu$ (a left Haar measure), not to $f$. To begin with, the measure $\delta^{-1}\mu$ is right-invariant:
$$\int_{E\,y}\delta(x)^{-1}\,d\mu(x)=\int_{E}\delta(xy)^{-1}\,\delta(y)\,d\mu(x)= \int_{E}\delta(x)^{-1} \,d\mu(x) $$
And here is another way to get a right invariant measure from $\mu$.
Let $\iota:G\to G$ be the inversion, $\iota(x)=x^{-1}$. This is an anti-homomorphism, it changes the order of group multiplication. Therefore,  the pushforward $\iota^*\mu$ is a right Haar measure.

Fact: $\delta^{-1}\mu =  \iota^*\mu$.

Proof. They are both right Haar measures, therefore one is a constant multiple of each other. To see that the multiple must be $1$, consider a symmetric neighborhood of identity, called $U$, which is small enough so that $|\delta(x)^{-1}-1|<\epsilon$ for $x\in U$. (Such $U$ exists because the modular function is continuous.) Now, $\iota^*\mu(U)=\mu(U)$ by symmetry, and also $|\delta^{-1}\mu(U)-\mu(U)|<\epsilon \,\mu(U) $. Since the ratio of $\iota^*\mu(U)$ to $\delta^{-1}\mu(U)$ can be made arbitrarily close to $1$, but is also a constant (independent ot $U$), the ratio is in fact $1$. $\Box$

The rest is a tautology from abstract measure theory: $$\int (f\circ \iota) \,d\mu = \int f \,d (\iota^*\mu) \tag1$$
hence $$\int f(x^{-1})\,d\mu(x) = \int f(x)\,\delta(x)^{-1} \,d\mu(x) \tag2$$
Actually,   you want to put   $f\circ \iota$ instead of $f$ here, so that the result is exactly as required:
$$\int f(x )\,d\mu(x) = \int f(x^{-1})\,\delta(x)^{-1} \,d\mu(x) \tag3$$

But I don't think your construction produces an isometry on $L^p$, at least not as stated. The identity (3) immediately generalizes to
$$\int |f(x )|^p\,d\mu(x) = \int |f(x^{-1})|^{p}\delta(x)^{-1} \,d\mu(x) \tag4$$ but if you rewrite (4) by attaching $\delta^{-1}$ to $f$, it has to be raised to power $1/p$.

Title: Involution in $L^{1}(G)$ is isometric.


问:
Occurrence rate $20/h$, $t=0.25h$ What is the chance to exactly $5$ Occurrences in that $t$?
I have a question in my homework:

Time interval: $1$ Hour

Divide that time to $4$ Equal periods, without overlap, each period $15$ Mins long.

If in that interval, there are $20$ Occurrences of the event, what is the probability that in each $15$ Mins occurs exactly $5$ Events.

My try:
It seems to me like poisson distribution, with $\lambda = 20$ And $t = 0.25$ Therefore: $\lambda t = 5$.

Therefore i should get:

$$
P(i = 5) = exp(-\lambda t) \frac{(\lambda t)^i}{i!} = exp(-5) \frac{5^5}{5!} = 0.175
$$

Yet, i have options to the answer, none of them is my answer.

What did i do wrong?

Thanks.

答:
Straight combinatorics:

Probability

$$= \frac{N\text{(umerator)}}{D\text{(enominator)}} $$

with

$$D = 4^{(20)}.$$

How many ways are there of selecting 5 events for each time period.

$$N = \binom{20}{5}\binom{15}{5}\binom{10}{5}.$$

Final answer:

$$\frac{N}{D}.$$


问:
Defining projection maps in Affine geometry and Thales' theorem and related sources
I am from Physics background and have not much experience in dealing with rigorous mathematical language and proofs. Out of interest, I am trying to read the book Spacetime. The book, which seems advanced, is very hard for me to read due to my inexperience and lack of the coverage of its prerequisites. Here's an excerpt from the book on proving Thales' theorem,

As you can see, the author uses projection maps $\pi$ and $\vec{\pi}$ but no definitions of these maps have been given. The author assumes the reader is familiar with these definitions/concepts. And I am lost. Can someone kindly explain how these maps are defined (both rigorous and intuitive) and what are their properties, that are crucial for this proof?

Moreover, if you think I should consult other resources on affine and Euclidean geometry first before venturing into this book, kindly recommend those sources. (Please do not recommend Berger or Audin. Those seem even more difficult than this book)

Edit:

As I realized that, people are unlikely to have a copy of this book, I have decided to to add images of the first five pages of the book so that people can understand better the level of this book and that would help recommending resources.

答:
First let me sketch an alternative proof of the lemma, which may provide some insight:

Given any pair of lines $l_1$ and $l_2$, there is a line $l_0$ that meets both lines and is not parallel to the planes $H_i$. For example, you could take $l_0:=x_1(l_1)x_2(l_2)$, assuming that $H_1\neq H_2$ as otherwise the lemma fails. Then $l_0$ and $l_1$ are coplanar, and $l_0$ and $l_2$ are coplanar, and so it suffices to show that the ratio does not depend on $l$ in these planes. That is, it suffices to prove the claim in dimension 2, for which we can draw a picture:

This simple case is of course an elementary exercise.

As for the proof in the book: First observe that the definition of $\frac{x_3-x_1}{x_2-x_1}$ makes no sense if  $x_1=x_2$, and hence that the statement of the lemma makes no sense if $H_1=H_2$. The notation $\frac{x_3-x_1}{x_2-x_1}$ is also very misleading as it suggests some algebraic properties of (differences of) vectors, which they don't have. This is illustrated by some confusion in the proof itself, which contains the expression
$$\frac{\pi(x_3(l))-\pi(x_1(l))}{\pi(x_2(l))-\pi(x_1(l))}.$$
This fails to make sense for some line $l$ even if $x_1(l)\neq x_2(l)$ and $H_1\neq H_2$.

As for your questions about the notation:

As you can see, the author uses projection maps $\pi$ and $\vec\pi$  but no definitions of these maps have been given.

This is not true; the author defines these maps in the same sentence where he first mentions them (emphasis mine):

This space has a natural affine structure with associated vector space $\Bbb{R}^n/\vec H$ given by $\pi(x)-\pi(z)=\vec\pi(x-z)$, where $\pi$, $\vec\pi$ denote the projections to the equivalence classes.

It is slightly ambiguous which projection map is which, but given that the expression
$$\pi(x)-\pi(z)=\vec\pi(x-z),$$
is supposed to be a definition of an affine structure on $\Bbb{A}^n/\vec H$, the sensible interpretation is that $\pi$ is the quotient map on $\Bbb{A}^n$ and $\vec\pi$ is the quotient map on $\Bbb{R}^n$. Explicitly, if $[x]\in\Bbb{A}^n/\vec H$ denotes the equivalence class of $x\in\Bbb{A}^n$, the quotient maps are given by
\begin{eqnarray*}
&&\pi:\ \Bbb{A}^n \longrightarrow \Bbb{A}^n/\vec H:\ x\ \longmapsto\ [x],\\
&&\vec\pi:\ \Bbb{R}^n \longrightarrow \Bbb{R}^n/\vec H:\ x\ \longmapsto\ x+\vec H.\\
\end{eqnarray*}
The latter is the quotient map of $\Bbb{R}^n/H$, and it should be familiar to you if you have taken linear algebra. See the Wikipedia page for basic properties of this map. The only property used is that it an $\Bbb{R}$-linear map.

As the question is tagged book-recommendation, if any of the above is still unclear (either my answer or the fragment you posted), I would recommend to pick up a book on linear algebra that takes an abstract approach. I can't recommend any particular book, but in my humble opinion you should be able to flip through Linear Algebra done wrong comfortably before attempting to read your book.


问:
If $A = \text{blkdiag}(G_1DG_1^T, G_2DG_2^T)$ is a block diagonal matrix and $B = C^TC$ is symmetric, how to manipulate $\text{tr}(AB)$?
$
\newcommand{\blkdiag}{\operatorname{blkdiag}}
\newcommand{\tr}{\operatorname{tr}}
$Suppose $G_1$ is a $m_1 \times n$ matrix, and $G_2$ is a $m_2 \times n$ matrix. $D$ is a $n \times n$ diagonal matrix with $\{d_1, d_2, \ldots, d_n\}$ along the diagonal such that $d_1 \neq d_2 \neq \ldots \neq d_n$. And $C$ is a $n \times m$ matrix.

Let $m = m_1 + m_2$. Then $$A = \blkdiag(G_1DG_1^T, G_2DG_2^T)$$ is a $m \times m$ matrix, and

$$B = C^TC$$ is a $m \times m$ matrix.

I'm interested in manipulating the expression for $\tr(AB)$ such that I can pull out $\sum_{i=1}^n d_i$ outside of the trace operator somehow.

I know this can be done for the following: suppose $A^* = G_1DG_1^T$ is a $m_1 \times m_1$ matrix, and $B^* = C^{*T}C^*$ is a $m_1 \times m_1$ matrix where $C^*$ is $n \times m_1$. Then we have
\begin{align*}
\tr(A^*B^*) &= \tr(G_1DG_1^TC^{*T}C^*)\\
&= \tr(DG_1^TC^{*T}C^*G_1)\\
& \sum_{i=1}^n d_i \tr(e_i^TG_1^TC^{*T}C^*G_1e_i)
\end{align*}
where $e_i$ is $i$th unit vector in $\mathbb{R}^n$.

Can something similar be done for $\tr(AB)$ where $A$ is now a block diagonal matrix? I know that $\tr(A) = \tr(G_1DG_1^T) +\ tr(G_2DG_2^T)$, but I'm not sure how to work with $\tr(AB)$ so that I might be able to pull $\sum_{i=1}^n d_i$ outside of the trace operator.

答: I assume that “$\operatorname{blkdiag}(\dots)$” denotes a block diagonal matrix with the respective block cells. Put $G=\operatorname{blkdiag}(G_1, G_2)$,  and $D’=\operatorname{blkdiag} (D,D)$. Then $A = GD’G^T$ so $$\operatorname{tr}(AB)= \operatorname{tr}(GD’G^TB)= \operatorname{tr}(D’G^TBG).$$


问:
Calculate limsup of $\frac{1}{n^{p/n}}$
I am trying to find-

$\displaystyle\limsup_{n\to\infty}\Big|\frac{1}{n^{p/n}}\Big|$ where $p\in\mathbb{Z}$.

I am not sure how to calculate that. All I have proved is that $\displaystyle\lim_{n\to\infty}n^{1/n}=1$. I am not sure how to proceed.

答:
Well, if you know $\lim_{n \to \infty} n^{1/n} = 1$ then for $p \in \mathbb Z$,
$$
\lim_{n \to \infty} \frac{1}{n^{p/n}} = \lim_{n \to \infty} (n^{1/n})^{-p} = \big(\lim_{n \to \infty} n^{1/n}\big)^{-p} = 1^{-p} = 1
$$ So, the sequence $(1/n^{p/n})_{n = 1}^\infty$ converges to $1$ and so the $\limsup$ must also be this same value.

Title: Calculate limsup of $\frac{1}{n^{p/n}}$


问:
Expectations of Indicator functions and Random Variables
Let $ X_1, X_2, ... $ be i.i.d with $P(X_i>x)=\frac{e}{x\log(x)}$ for $x\geq e$.  If I were to try to find the expected value of  the

$$ EX_k1_{\{|X_k|\leq n\}}$$ for $e \leq x \le n$,   what exactly would that be? I know the answer is supposed to be $e\log\log n$ but I can't figure out how? Will we not take the integral of  $$\int _e^n \frac{ex}{x\log(x)}dx$$

答:
You are given $P(X_k>x)$, so the CDF of $X_k$ is $P(X_k\le x)=1-P(X_k>x)=1-\frac e{x\log x}$.

The PDF of $X_k$ is$$\frac{dP(X_k\le x)}{dx}=\frac{e(1+\log x)}{x^2\log^2x}$$For $n>e$, the required expectation is $$\int_e^nx\frac{e(1+\log x)}{x^2\log^2x}dx=e\int_1^{\log n}\frac{(1+m)}{m^2}dm$$where $m=\log x$. This gives$$e\left[-\frac1m+\log|m|\right]_1^{\log n}=e\log\log n+e(1-1/\log n)$$


问:
How many ways do we have to count $1$ to $100$ by adding $1$ or $2$ each time?
We count $1$ to $20$ and each time we add $1$ or $2$. For example, the numbers can be $1,2,4,6,8,10,12,14,15,17,19$ and $20$. How many ways do we have to count $1$ to $100$ by adding $1$ or $2$ each time?

答:
This hint assumes that you must land squarely on 100.

You must have an even # of 1's to go with your 2's.

Therefore, as $k$ goes from 0 thru 50, you have $2k$ 1's.

You will have to do a summation.

Suppose that you had $2k$ 1's.

Then you must have how many 2's?

Suppose that you have $a$ 1's and $b$ 2's.

How many ways are there of arranging the $(a+b)$ steps?


问:
Understanding meromorphic/holomorphic forms on Riemann surface
I refer to Rick Miranda - Algebraic curves and Riemann surfaces chapter IV.1 (p. 105, p. 106, p. 107). I think I understand the regular Euclidean $\mathbb C$ case:

the idea of meromorphic/holomorphic $1$-form on open set $V_1$ of $\mathbb C$: $\omega_1 = f(z)dz$, for $f$ mero/holo function on $V$ and

the idea of the transformation rule: for $\omega_2 = g(w)dw$ on open set $V_2$ of $\mathbb C$ with $g$ mero/holo on $V$, we say that $\omega_1$ transforms to $\omega_2$ under $T$  if $g(w)=f(T(w))T'(w)$ for some holo $T: V_2 \to V_1$

Where it gets fuzzy for me is the case of Riemann surfaces. I wish Miranda would have 1st defined for charts on Riemann surface, but Miranda instead goes straight to Riemann surfaces. Apparently $\omega$, a mero/holo $1$-form on Riemann surface $X$ (in this book, all Riemann surfaces are connected), is a 'collection' (see (A1)) of mero/holo $$\{\omega_{\phi} | \phi: U \to V \ \text{is a chart in, I think, the max atlas of X}\} \tag{see (A2)}$$ such that for all charts $\phi_1: U_1 \to V_1$, $\phi_2: U_2 \to V_2$, with overlapping domains, we have that $\omega_{\phi_1}$ transforms to $\omega_{\phi_2}$ under $T=\phi_1 \circ \phi_2^{-1}$. I guess this is $T: \phi_2(U_1 \cap U_2) \to \phi_1 (U_1 \cap U_2)$.

Ostensibly, we have that for, say, $\omega_{\phi_1}$, the expression for $\omega_{\phi_1}$ is like '$\omega_{\phi_1} = f_1(z) dz$', for coordinate $z = \phi_1(x)$ and some mero/holo $f_1=f_1(z)$ on open subset $V_1$ of $\mathbb C$. But what I expected was something an expression involving some mero/holo $h_1=h_1(x)$ on the chart $U_1$ of $X$, like

$\omega$ is some map $$\omega: X \to \{\text{probably some bundle thing in complex geometry that I didn't learn yet}\},$$ where

the restriction $\omega|_{U_1}$ is a well-defined (because of the transformation rule for overlapping domains) mero/holo $1$-form on the chart domain $U_1$,  given as $\omega|_{U_1} = h_1(x) dx$, where the '$|_{U_1}$', is just omitted. And then

we can map this from $X$ to $\mathbb C$ like maybe there's some correspondence to the mero/holo $1$-form '$\omega|_{V_1}$' on the chart image $V_1$, given as something like $\omega|_{V_1} = (h_1 \circ \phi_1^{-1})(z) dz$ or even like $(h_1 \circ \phi_1^{-1})(z) d(\phi_1^{-1}(z))$. This way $f_1 = h_1 \circ \phi_1^{-1}: \phi_1(U_1) =V_1 \to U_1 \to \mathbb C$.

Question 1: Are $\omega$'s indeed locally like $\omega|_U = h(x) dx$ and then converted from $X$'s local coordinate $x$ on $U$ into $\mathbb C$'s local coordinate $z$ on $V$ into '$\omega|_{V}$' = $(h \circ \phi^{-1})(z) dz$?

Question 2: Later on, there's a definition for order. How should I understand the definition for order in terms of the above? In particular, is my definition as follows correct?

The definition is given as '$ord_p(\omega) := ord_0(f)$', for '$\omega = f(z) dz$', where $z=\phi(x)$, for chart $\phi: (U,p) \to (V,0)$, centred at $p \in U$. I understand this as $ord_p(\omega)$ $:= ord_{\{\phi(p)=0\}}(f \circ \phi^{-1}(z))$, for $\omega|_V = (f \circ \phi_1^{-1})(z) dz$, which in turn is from $\omega|_U = f(x) dx$.

Therefore, I can make this kind of definition chain: $ord_p(\omega) := ord_p(\omega|_U)$ and then $ord_p(\omega|_U) := ord_p(f)$ (and then finally $ord_p(f) := ord_{\{\phi(p)=0\}} (f \circ \phi^{-1})$).

In particular, this is why I was hoping we would 1st have a definition for $1$-forms on charts: like if a Riemann surface $X$ is covered by a single chart $\phi: U = X \to V$ then we can do for its 1-forms $\omega$ like $ord_p(\omega|_U) := ord_p(f)$ (where $\omega$ = $\omega|_U$ since $U=X$).

Question 2.1: Btw, for the original definition of '$ord_p(\omega) := ord_0(f)$', for '$\omega = f(z) dz$', can I just instead of any chart, that's not necessarily centred at $p$? This way, I would define $ord_p(\omega) := ord_{\phi(p)}(f)$, whether or not the chart $\phi: U \to V$, that gives us the local coordinate $z=\phi(x)$, is centred at $p$. Of course, it's more convenient to have Laurent series about 0, but just wondering if there's anything particular about the number 0.

Edit: Btw, there's also this thing in the text (but this is on 2-forms now) i noticed that goes like $$\int \int_{T} \eta = \int \int_{\phi(T)} f(z, \overline z) dz \wedge d \overline z,$$ where '$\eta = f(z, \overline z) dz \wedge d \overline z$'. I mean, if '$\eta = f(z, \overline z) dz \wedge d \overline z$', then one might think you wouldn't have to change the region of integration when replacing $\eta$ with $f(z, \overline z) dz \wedge d \overline z$. If this were 1-form, like $\eta = f(z) dz$, I'd think '$f(z)$' is actually like $f \circ \phi^{-1}(z)$

(A1): I guess similar to how a holo function on a non-connected open set is a 'collection' of holo functions on connected open sets.

(A2): I think initially mero/holo $1$-form is defined in Def IV.1.7/3 for every chart in max atlas and then later it's defined for every chart in an atlas in Lemma IV.1.8/4.

答:
Here is an attempt to answer your questions. Recall that a Riemann surface is a manifold $M$ of dimension $2$ such that there exists a complex atlas, that is a collection of charts $(U,\phi_U)$, with $\phi_U : U \to \phi(U)\subset \mathbb{C}$, with transition functions that are holomorphic.

Question 1 The definition of a holomorphic / meromorphic $1$-form on a Riemann surface is the following. Let $\omega$ be a $1$-form on $M$. It is holomorphic / meromorphic if for any $p\in M$, there exists a complex chart $(\phi,U)$ with $p\in U$, such that the push-forward $1$-form $\phi_*\omega$ on $\phi(U) \subset \mathbb{C}$ is a holomorphic / meromorphic one form. As a meromorphic $1$-form is defined to be of the form $f(z)\mathrm{d}z$ on open subsets of $\mathbb{C}$, $\omega$ is a meromorphic $1$-form if for any $p\in M$, there exists a complex chart $(\phi,U)$ with $p\in U$, such that there exists a meromorhic function $f$ on $\phi(U)\subset \mathbb{C}$ with $\phi_*\omega = f\mathrm{d}z$, that is $\omega = \phi^*\left(f(z)\mathrm{d}z\right)$. One can show that in an holomorphic atlas, the meromorphic functions $f$ behave very-well under change of charts.

Question 2 In a manifold $M^n$, the definition of a chart centered at $p\in M$ is a chart $(U,\phi)$ with $\phi : U \to \mathbb{R}^n$, with $\phi(p) = 0$. This is so that anything at $p$ can be read (in the charts centered at $p$) at the origin. Thus, for the definition of order, we define the order of a meromorphic $1$-form $\omega$ at $p$ to be the order of the meromorphic $1$-form $\phi_* \omega$ on $\phi(U)\subset \mathbb{C}$ at $0$, for $(U,\phi)$ centered charts at $p$. If one requires the chart not to be centered, one could define it to be the order at $\phi(p)$ of the meromorphic $1$-form $\phi_*\omega$. Note that any chart can be translated to a chart centered at $p$, so this is not a restrictive definition. The fact that transition functions are holomorphic shows that the order of a meromorphic $1$-form is a well-defined notion and does not depend on the chart, and is an intrinsic definition.

As order is a pointwise notion that depends on the local behavior, you are right when saying that the order of $\omega$ at $p$ is the same as the order of $\omega|_U$ at $p$.

If $M$ is covered by a single chart, then $M$ is an open subset of $\mathbb{C}$! Therefore, any local holomorphic / meromorphic function can be written as a global holomorphic / meromorphic function (this is a complex analysis result), and thus any holomorphic / meromorphic $1$-form is globally of the form $f(z)\mathrm{d}z$.

For the integration part of your question. A complex manifold is canonically oriented by its complex structure. In the case of a Riemann surface, there is a canonical volume form, defined on open subset $U$ by $\phi^* \left(\frac{i}{2}\mathrm{d}z\wedge \mathrm{d}\overline{z} \right)$. This is because if $z = x+iy$, $\mathrm{d}z\wedge \mathrm{d}\overline{z} = -2i\mathrm{d}x\wedge\mathrm{d}y$. One can choose the volume form to be $\mathrm{d}z\wedge \mathrm{d}\overline{z}$, it does not really matter (it seems it is what Miranda has chosen). Thus, if $\eta$ is a holomorphic / meromorphic $2$-form on $M$, in a chart $(U,\phi)$, there exists a holomorphic / meromorphic function $f$ such that $\eta = \phi^* \left( f(z) i\mathrm{d}z\wedge \mathrm{d}\overline{z} \right)$. This is because $\Lambda^2(\phi(U))$ is a rank $1$ trivial bundle with global section $\mathrm{d}z\wedge \mathrm{d}\overline{z}$ nowhere vanishing, so every $2$-form on $\phi(U)$ can be written $f \times \mathrm{d}z\wedge \mathrm{d}\overline{z}$. By the theory of integration of $n$-form over an oriented $n$ dimensional manifold $M$, the definition of $\int_U\eta$ is
\begin{align}
\int_{U}\eta = \int_{\phi(U)} \phi_*\eta = \int_{\phi(U)}f\mathrm{d}z\wedge\mathrm{d}\overline{z}
\end{align}
the $f(z,\bar z)$ part is just a notation in complex geometry. The function $f$ just depends on $z$ as a complex coordinate (and therefore on two real variables $x$ and $y$) but for theoretical purpose, considering $f$ as a function of $z$ and $\bar z$ is convenient. For example, a smooth function $f$ is holomorphic of and only if $\dfrac{\partial}{\partial \bar z}f = 0$ in this notations.


问:
Closed form of a recurrence relation using generating functions
It's been awhile since I have done this. The sequence is $\displaystyle a_n = a_{n-1} + 5~a_{n-2}$ with $a_{0}=0$ and $a_{1}=1$. </p>

<p>I found the generating function to be $\displaystyle G(x) = \frac{x}{(1-x-5x^2)}$ but I am lost on where to go from here. </p>

答:
The Characteristic equation of the Recurrence relation is $r^2-r-5=0$

So, $a_n=A\cdot r_1^n+B\cdot r_2^n$ where $r-1,r_2$ are the roots of the above Quadratic Equation

So, $r_1,r_2$ are $\frac{1\pm\sqrt{1^2-4\cdot1(-5)}}{2\cdot1}=\frac{1\pm\sqrt{21}}2$

$0=a_0=A+B\implies B=-A\implies a_n=A(r_1^n-r_2^n)$

$1=a_1=A(r_1-r_2)\implies A=\frac1{(r_1-r_2)}=\sqrt{21}$

$\implies a_n=\sqrt{21}\left(\left(\frac{1+\sqrt{21}}2\right)^n-\left(\frac{1-\sqrt{21}}2\right)^n\right)$


问:
Concerning the $\eta$-rule
In Martin Lof's type theory, one has a term

$$
x:A, y:A, z: Id_A(x,y) \vdash J(r(x);x,y,z): Id_A(x,y)
$$

can one derive judgemental equality $J(r(x);x,y,z)=z$ without the $\eta$-rule?

答: No. $J(r(x);x, y, z)$ is in $\beta$-normal form. The only judgmental equality for $J$ is $\mathit{J}(pr;x, x, r(x)) = pr$ (using your notation) which does not apply here. If instead of judgmental equality, we consider $\mathit{Id}(J(r(x);x, y, z), z)$, that's provable by induction on $z$.

Title: Concerning the $\eta$-rule


问:
What would be the appropriate way to calculate the probability $aX \leqslant Z \leqslant bY + cX$?
What would be the appropriate way the calculate the probability $aX \leqslant Z \leqslant bY + cX$ where $X,Y,Z$ are all independent exponential variable with parameter ${\lambda _1},{\lambda _2},{\lambda _3}$ and $a,b,c$ are all positive number.

Since $Z$ has been squeeze between $aX$ and $bY + cX$, I guess we could use some the difference between CDF but I cannot wrap my mind around the case where $a>c$ ?

Please help me with this

Thank you for your enthusiasm !

答: $$P(t \leq Z \leq s)=e^{-\lambda_3 t}-e^{-\lambda_3 s}.$$ So $$P(aX \leq Z \leq bY+cX)=E(e^{-\lambda_3 aX}-e^{-\lambda_3 (bY+cX)}).$$ This is equal to $$\int_0^{\infty} \int_0^{\infty} (e^{-\lambda_3 ax}-e^{-\lambda_3 (by+cx)}) \lambda_1 e^{-\lambda_1 x} \lambda_2 e^{-\lambda_2 x} dxdy.$$


问:
$16$ people, $8$ men, $8$ women, divide the group to $8$ couples, what is the chance for exactly $3$ male couples?
We have a group of $16$ people, $8$ men, $8$ women.

We divide them to $8$ couples.

Let $X$ Be the number of couples make of $2$ men.

Calculate: $P(X = 3)$

I am not sure how to approach this.

I thought that this has an hypergeometric distribution.

Therefore i chose:

$$
N = 16, n = 16, K = 8, k = 6
$$

For:

$$
\frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}
$$

But its wont gonna work.

Another way i thought is to choose the couples like this:

$$
\frac{\binom{8}{2} \binom{6}{2} \binom{4}{2} \binom{8}{1} \binom{7}{1} \binom{6}{2} \binom{4}{2}}{\frac{16!}{2^8}}
$$

The logic wass:

But again its not working.

I would like some help

答:
Choose 6 men to be paired with one another. This can be done in $\binom 86=28$ ways. Then pair them up. This can be done in $5!!=15$ ways. Then take the remaining two men, and pair them up with two women. This can be done in $_8P_2=56$ ways. Finally, pair up the remaining six women. This can be done in $5!!=15$ ways.

In total, there are
$$
28\cdot 15\cdot56\cdot 15=352\,800
$$
ways to make exactly three male couples. Finally, to get the probability, divide by the $15!!=2\,027\,025$ ways to make 8 couples.


问:
Find $\lim_{n \to \infty}\frac{\sin 1+2\sin \frac{1}{2}+\cdots+n\sin \frac{1}{n}}{n}$

Find $$\lim_{n \to \infty}\frac{\sin 1+2\sin \frac{1}{2}+\cdots+n\sin \frac{1}{n}}{n}.$$

This is a recent exam question, which I couldn't figure out in the exam. My guess is it doesn't exist.

答:
Another more general approach:

$$a_n\xrightarrow[n\to\infty]{} a\implies \frac{a_1+...+a_n}n\xrightarrow[n\to\infty]{}a$$

And since

$$\lim_{n\to\infty}\,n\,\sin\frac1n=\lim_{n\to\infty}\frac{\sin\frac1n}{\frac1n}=1\;\;\ldots\ldots$$


问:
$\sum_{n = 1}^{\infty}\frac{a^n}{n^\alpha}$, with $a>0$, $\alpha$ is a real number
After some calculations I found that the limit of this series is $a$.

Then, I've learnt that if $a$ is in $(0,1)$ the series is convergent and if it is in $(1, +\infty)$ it is divergent and I don't really understand why.

Can anybody explain this to me?
Thanks.

答: Hint: When $a>1$, then $\frac{a^n}{n^\alpha} \to \infty$, which contradicts with necessary condition convergence of the series.


问:
How two line segments are not equal if function of one line is ($f(x)=x$) and of another ($g(x)=cx))$
Line $c=g(x)=cx$;

Line $b=f(x)=x$

If for every point $(x,0)$ on line $(b)$ ,there exists a point  $(x,cx)$ on line $(c)$ and vice versa,then how two given  line segments are not equal (considering this reason).

答:
They are not congruent!

In geometry "congruent" means that they can be overlapped and clearly $b=3$ and $c=5$ can't.

The bijection you found is all another concept. It proves that any point of $b$ is related to one and only one point of $c$ and viceversa. The set of points of segment $b$ and the set of points of $c$ have the same cardinality.

Shocking fact: it can be proved that $b$ has the same cardinality of $\mathbb{R}^3$.


问:
Generalized optimisation question about open topped boxes made from regular polygons
This is based on the question about maximizing the volume of an open box being formed from a square with corners cut.

Original question is here
Optimisation question

edit: added above picture for extra clarity about the case of a square base.

I also solved this for the base being a triangle, pentagon and hexagon, the trig gets a little hectic. Spoiler alert - the answer in all cases ended up with $x$ being $1/6th$ the length of the original shape's side.

My question is how to prove if is this the case for all regular polygons.

In this low quality sketch, the cuts are along the green lines, making the rectangular flaps that fold up to make the triangular based box. Similar kite shapes need to be cut for the other shapes.

triangle base

答:
Notice that the base of the box is shrinked version of the original polygon. Let the scale factor be $\alpha$. If $S$ is the area of the original polygon, then the area of shrinked polygon is $\alpha^2 S$. If $h$ is the length of a perpendicular from the center of the polygon to one of it's sides, then the length of the shrinked perpendicular is $\alpha h$. Hopefully, it's clear that the height of the box is $(h - \alpha h) = (1-\alpha)h$.

Now the volume of the box is $\alpha^2 S (1 - \alpha) h$. It reaches it's maximum when $\alpha = \frac23$.

If $L$ is the length of a side of the polygon, then it's shrinked length is $\alpha L$, which means $x = (L - \alpha L) / 2 = L / 6$.


问:
Find a side of smaller triangle

$\triangle ABC$ has side $\overline{AB}=160 cm$, $\overline{BC}=190 cm$, and $\overline{CA}=190 cm$.
Point $D$ is along side $\overline{AB}$ and $\overline{AD} = 100 cm$. Point $E$ is along side $\overline{CA}$.

Determine the length of $\overline{AE}$ if the area of $\triangle ADE$ is three-fifths the area of $\triangle ABC$.

Is my solution correct? Thanks in advance!

答:
As pointed out by Jaap in the comments, $\triangle AED$ may not be isosceles so $h_2$ may not be $\sqrt{|ED|^2-(\color{red}{100/2})^2}$.

Label the base of $h_1$ as $N$. Let $EM$ be the perpendicular on $AD$ from $E$. Then $|EM|=h_2$. Then$$\text{ar}\triangle ADE=\frac12\times h_2\times |AD|=\frac35(13786.95)$$giving $h_2=165.44$. Note that $\triangle AEM\sim\triangle ACN$, thus$$\frac{h_2}{h_1}=\frac{|AE|}{|AC|}$$giving $|AE|=182.4$.

Title: Find a side of smaller triangle


答:
As indicated in comment, you're assuming $\triangle AED$ is isosceles (i.e., $DE$ || $BC$) which need not be true.

If you know formula for area of a $\triangle$ is also given as

$$ \dfrac{1}{2}AB\cdot AC\sin A$$

then using this one can directly solve

$$ \dfrac{1}{2}AD\cdot AE\sin A = \dfrac{3}{5} \cdot\dfrac{1}{2}AB\cdot AC\sin A$$

$$ \Rightarrow AE = \dfrac{3}{5}\dfrac{AB\cdot AC}{AD}$$

without any use of trigonometry.

Otherwise,

$$ \dfrac{1}{2}AD\cdot h_2 = \dfrac{3}{5} \cdot\dfrac{1}{2}AB\cdot h_1$$

$$ \Rightarrow h_2 = \dfrac{24}{25}h_1$$

and since the two right triangles formed by $h_1$, $h_2$ and containing $\angle A$ are similar

$$ \dfrac{AE}{AC} = \dfrac{h_2}{h_1}$$
$$ \Rightarrow AE = \dfrac{24}{25}AC$$

Either should give $AE = 182.4 $

Title: Find a side of smaller triangle

Body:
$\triangle ABC$ has side $\overline{AB}=160 cm$, $\overline{BC}=190 cm$, and $\overline{CA}=190 cm$.
Point $D$ is along side $\overline{AB}$ and $\overline{AD} = 100 cm$. Point $E$ is along side $\overline{CA}$.

Determine the length of $\overline{AE}$ if the area of $\triangle ADE$ is three-fifths the area of $\triangle ABC$.

Is my solution correct? Thanks in advance!


答:
Complicated problem.

Answer will be

$$\frac{N\text{(umerator)}}{D\text{(enominator)}}.$$

For the denominator, not only do you have to consider how many ways there are of choosing the 1st couple, then the 2nd, ...

You then have to adjust for over-counting, re each grouping into 8 couples will be counted $8!$ ways.

Therefore,

$$D = \frac{\binom{16}{2}\binom{14}{2}\binom{12}{2}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}}{8!}.$$

For the numerator, first you have to create exactly 3 male couples.

So

$$N_1 = \frac{\binom{8}{2}\binom{6}{2}\binom{4}{2}}{3!}.$$

Then, you have to select among 8 women for the 1st odd man, and then among 7 women for the 2nd odd man.

$$N_2 = 8 \times 7.$$

Then, the 6 remaining women must pair up.

$$N_3 = \frac{\binom{6}{2}\binom{4}{2}\binom{2}{2}}{3!}.$$

Final answer is

$$\frac{N_1 \times N_2 \times N_3}{D}.$$

Title: $16$ people, $8$ men, $8$ women, divide the group to $8$ couples, what is the chance for exactly $3$ male couples?

Body:
We have a group of $16$ people, $8$ men, $8$ women.

We divide them to $8$ couples.

Let $X$ Be the number of couples make of $2$ men.

Calculate: $P(X = 3)$

I am not sure how to approach this.

I thought that this has an hypergeometric distribution.

Therefore i chose:

$$
N = 16, n = 16, K = 8, k = 6
$$

For:

$$
\frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}
$$

But its wont gonna work.

Another way i thought is to choose the couples like this:

$$
\frac{\binom{8}{2} \binom{6}{2} \binom{4}{2} \binom{8}{1} \binom{7}{1} \binom{6}{2} \binom{4}{2}}{\frac{16!}{2^8}}
$$

The logic wass:

But again its not working.

I would like some help


问:
Find the flux of the vector field out of the region $T$ bounded by $z = x^2 + y^2$ and $z = 4$. Please help
Find the flux of the vector field $$\vec{F}(x,y,z)= <x,y,3>$$
out of the region $T$ bounded by $z = x^2 + y^2$ and $z = 4$.

It says the unit vector $n$ is $<0,0,1>$, but how do you find it?

答: You are asking how to find the unit vector? I expect by $(0,0,1)$, they mean the unit normal for the plane $z=4$, and we know that for any implicitly defined surface such as this, the gradient of the implicit defining function is normal to the surface. In this case, $\nabla(z-4)=(0,0,1)$. Seems we would then also consider the other part of the bounding surface, $x^2+y^2-z=0$, which has normal vector $\nabla(x^2+y^2-z)=(2x,2y,1)$. Divide by the magnitude to create a unit length vector. </p>


问:
Convergence of $\zeta(k)$ to arbitrary rational number
I have a question about Riemann's hypothesis for $s>1$, we know that $\displaystyle\zeta(s)=\sum_{n\ge 1}\dfrac{1}{n^s}$ converges, yet $\zeta(2)=\dfrac{\pi^2}{6}$ and $\zeta(4)$ and so forth are irrational. I wonder that there exists $s>1$ for which $\zeta(s)\in\mathbb{Q}$? Or even more general, is there exists $s>1$ for arbitrary rational number $q$ that $\displaystyle q=\zeta(s)$.

答: $\zeta$ is continuous on $(1,+\infty)$, strictly decreasing, $\lim\limits_{s\rightarrow 1}\zeta(s)=+\infty$ and $\lim\limits_{s\rightarrow +\infty}\zeta(s)=1$, thus for all rationnals $q>1$, there exists a unique $s_q>1$ such that $\zeta(s_q)\in\mathbb{Q}$.


答:
Let the edge length of the original polygon be $1$. The base of the polygonal box will then have edge length $1-2x$.

The height of the polygonal box is proportional to $x$ (because the polygon angle is fixed, the edges of the kites you cut off are proportional to $x$).

Therefore, the volume is proportional to $x(1-2x)^2$, which reaches its maximum at $x=\frac{1}{6}$.

Title: Generalized optimisation question about open topped boxes made from regular polygons

Body:
This is based on the question about maximizing the volume of an open box being formed from a square with corners cut.

Original question is here
Optimisation question

edit: added above picture for extra clarity about the case of a square base.

I also solved this for the base being a triangle, pentagon and hexagon, the trig gets a little hectic. Spoiler alert - the answer in all cases ended up with $x$ being $1/6th$ the length of the original shape's side.

My question is how to prove if is this the case for all regular polygons.

In this low quality sketch, the cuts are along the green lines, making the rectangular flaps that fold up to make the triangular based box. Similar kite shapes need to be cut for the other shapes.

triangle base


问:
If $f$ is bounded and $\lim_{x\to c}g(x)=\infty$, then $\lim_{x\to c}f(x)g(x)=\infty.$
I need to prove or give counter example of the following claim:

If $f$ is bounded and $\lim\limits_{x\to c}g(x)=\infty$, then $\lim\limits_{x\to c}f(x)g(x)=\infty.$

Would it be enough to have $f(x)$ with a limit of zero for a counter example?

Thanks for any input.

答: Really late counter example, but the function $f,g:[-1,1]\setminus\{0\}\to\mathbb{R}$ where $f(x)=x^2$ and $g(x)=\frac{1}{x^2}$ are good to look at. It can be shown that $f$ is bounded and $g$ is unbounded. Since each function is not, zero, then we can say $$\lim_{x\to 0}f(x)g(x)=\lim_{x\to 0}x^2\frac{1}{x^2}=\lim_{x\to 0}1=1 $$


问:
line bundles over the circle
I read in various places that up to isomorphism there are only two line bundles ( 1-d vector bundles) over a circle, the trivial one and the mobius strip. On the other hand, when I make a mobius strip from a strip of paper, I note that I can give the paper as many rotations as I want. No rotation of the paper and simply gluing the end pieces of the strip gives a cylinder without tops (which we can think as the trivial line bundle), one rotation gives the mobius strip. But more rotations of the paper are possible and they are not isomorphic to each other; you cannot make one change  to the other without tearing the paper.

Now I understand  that these are not line bundles as they are not extended to infinity and they are "bounded" together in a way that infinite fibers or vectors would not be.

I am just wondering: Is there a notion of "bundles" here over a circle which extends to more than the trivial and mobius strip? After all, no rotation and one rotation of my strip do match up with the two possible vector bundles (trivial and mobius). Can we define a notion of "bundle" for more rotations of the strip?

EDIT:  The comments below basically mean that I have not understood the situation correctly. I will not delete the question since I think others might run into the same misunderstanding.

答:
Not a PROOF but a non-sufficient simple algebraic reason for twice twisted is same as trivial for further readers:
Define an equivalence relation on $\Bbb R^2$ by
declaring that $(x', y')\sim(x,y)\iff (x',y')=(x+n,(-1)^ny)$ for some
$n \in\Bbb Z$. This defines the Mobius bundle. To see that if we twist the band twice is same as trivial bundle (that is the identification $(x,y)\sim (x+m,y)$ for some $m$), just apply this identification once again, i.e.: (If I am not mistaken by notations)
$$\text{twisted }\Bbb R^2\ni(a', b')\sim(a,b)\in\text{twisted }\Bbb R^2\iff (a',b')=(a+n,(-1)^nb)=(x+2n,(-1)^{2n}y)=(x+2n,y)\iff (a', b')\sim(x,y).$$
that says $(z,w)\sim (z',w')\iff (z',w')=(z+2n,w)$ that is the trivial bundle. (first factor is a circle and second is a line)

$\qquad\qquad\qquad\qquad\qquad$Source: Wikipedia

Title: line bundles over the circle


答:
The number of ways to make $n$ couples from $2n$ people is

$$\frac{(2n)!}{n!\cdot 2^n}$$

This is derived by starting like you did, multiplying

$$\begin{align}
&\quad\binom{2n}{2} \cdot \binom{2n-2}{2} \cdot \binom{2n-4}{2} \cdots \binom{4}{2} \cdot \binom{2}{2} \\
\\
=& \quad\frac{2n(2n-1)}{2} \cdot \frac{(2n-2)(2n-3)}{2} \cdot \frac{(2n-4)(2n-5)}{2} \cdots   \frac{(2)(1)}{2} \\
\\
=&\quad \frac{(2n)!}{2^n}
\end{align}$$

However, the above includes an order for the couples chosen which we don't want, so the above has to be divided by $n!$

Now to count the number of ways of choosing exactly three male couples, we start by choosing the males to form the three couples, then apply the above formula once, then choose the two females for the two lone males, then apply the above formula a second time to pair up the remaining six females.

$$
\quad \binom{8}{6} \cdot \frac{6!}{3!\cdot 2^3} \cdot 8 \cdot 7 \cdot \frac{6!}{3!\cdot 2^3} = 352,800
$$

The number of ways to make eight couples with no restriction is

$$\frac{16!}{8!\cdot 2^8} = 2,027,025$$

Thus the required probability is

$$\frac{352,800}{2,027,025} = \boxed{\frac{224}{1287}}$$


问:
Why is it given that an odd polynomial can be written in the form $ax^3 +bx$?
So a question give that P(x) is an odd polynomial of degree three. So now I know that $P(x) = ax^3 + bx^2
 +cx +d $. And $P(-x) = -P(x) $since it is odd. However, the solutions say that given this informations, it can actually be written simply in the form $ax^3 +bx$. Why is this so?

答:
The equation $P(-x)=-P(x)$ says that the polynomial $P(x)$ is pointsymmetric to the origin.

If you plug in, $P(-x)= -ax^3+bx^2-cx+d$ and $-P(x)=-ax^3-bx^2-cx-d$. Then the equation gives $P(x) = ax^3+cx$.


答:
For odd polynomial $P(x)$, we have
$P(x)+P(-x)=0$ for all real values of $x$, so if $P(x)=ax^3+bx^2+cx+d$, then
we get $$ax^3+bx^2+cx+d-ax^3+bx^2-cx+d=0 \implies 2bx^2+2d=0=0x^2+0,$$ for all  real values of $x$. This implies $b=0=d$. So finally we have $P(x)=ax^3+cx$

Title: Why is it given that an odd polynomial can be written in the form $ax^3 +bx$?

Body:
So a question give that P(x) is an odd polynomial of degree three. So now I know that $P(x) = ax^3 + bx^2
 +cx +d $. And $P(-x) = -P(x) $since it is odd. However, the solutions say that given this informations, it can actually be written simply in the form $ax^3 +bx$. Why is this so?


问:
Unilateral Laplace Transform vs Bilateral Fourier Transform
I would like to know why when we find the Laplace transform we use the one-sided (unilateral) version (all Laplace transform tables I can find are one-sided, like this one http://people.seas.harvard.edu/~jones/es154/Laplace/Table_pairs.html)

$$F(s)=\int^\infty_0e^{-st}f(t)dt$$

but when we find the Continuous Time Fourier Transform (CTFT) we use the two-sided (bilateral) version

$$F(\omega)=\int^\infty_{-\infty}e^{-j\omega t}f(t)dt$$

(a typical table http://www.mechmat.ethz.ch/Lectures/tables.pdf). What is the intuition or reasoning behind this? Since the analogy is often made that the CTFT is simply the Laplace transform with the real part of $s$ set to $0$, it would be natural to conclude that the limits of integration should not change.

答:
TL;DR

In short, If the system is not IAR (iff LTI) you can only use the unilateral transform. Otherwise, it is possible to, also, use the bilateral transform. If the system is not BIBO stable, you can only use the Laplace transform. Otherwise, you can use the Fourier transform as well.

With respect to the excellent answer by @Noir, I would like to add a few details and remarks.

First, it is important to understand that both Fourier and Laplace bilateral transforms are tools aimed at analyzing systems easily. None of them can be used without some assumptions. While Noir's assumption of a black-box model can describe any system, a system can only be defined as a convolution system if it is both linear and time-invariant (LTI).  Otherwise, it has no impulse response kernel $h(t)$, which can describe it as a convolutional system. In such a case, if there is no $h(t)$ there is no bilateral $H^L(s)$ or $H^F(\omega)$ to speak of. An example of such a system is simply the system $y(t)=\Psi \left\{ x \right\} (t)=x^2(t)$ which is obviously not linear or $y(t)=\Psi \left\{ x \right\} (t)=tx(t)$ which is clearly not time-invariant.

The simplest and most common way to define a system (if possible) is by an ordinary diffraction equation (ODE) and it is an LTI system if it is initially at rest (IAR). This condition is met if the output of the system is $0$ as long as the input is $0$. In a mathematical representation:
$$x(t)=0\forall t<t_0 \Rightarrow y(t)=0\forall t<t_0$$
IAR is "stronger" than the initial conditions of an ODE in the sense that we don't get a value for a specific time point, but a set of values prior to a specific time. If the system is not an ODE the discussion of an LTI system is by the specific linearity and time-invariancy definitions and everything formerly discussed does not apply including the existence of transforms.

The IAR LTI state is an assumption on the system which conditions the use of bilateral Laplace and\or Fourier transforms. The unilateral Laplace transform, on the other hand, considers the initial conditions of the signal in the form:
$$\mathcal{L}_+\left\{ x' \right\} (t)=sX_+^L(s)-x(0^-)$$
$$\mathcal{L}_+\left\{ x^{(n)} \right\} (t)=s^nX_+^L(s)-\sum_{k=1}^n x^{(k-1)}(0^-)s^{n-k}$$
In that sense, the unilateral transform can analyze non-LTI (none IAR) systems and is, therefore, very useful.

A second remark on the assumptions is in the difference between Fourier and Laplace transform. The Fourier transform is a private case of the Laplace transform given in the projection $H^F(\omega)=H^L(s=j\omega)$. This assumption implies that the imaginary axis $s=j\omega$ is included in the region of conversion (ROC) of the Laplace transform. If we analyze a system, the inclusion of the $j\omega$ axis is an iff for the bounded input bounded output (BIBO) stability of the system. If the system is not BIBO stable the axis is not in the ROC and only the Laplace transform is defined. There is no Fourier transform for such a case.


问:
Equivalence classes of the similarity relation on complex 2x2 matrices?
Let $\mathbb{K}$ be a field. Two matrices $A,B \in \mathbb{K}^{n \times n} $ are said to be similar if there exists an invertible matrix $S \in \mathbb{K}^{n \times n}$ such that $A = S^{-1}BS$. It can be shown the this relation is an equivalence relation.

My question is: How many equivalence classes of the similarity relation on the set of complex 2x2 matrices, $\mathbb{C}^{n \times n}$, are there? And which ones are they?

My idea would be to use the fact that matrices are similar if and only if they have the same Jordan normal form (up to a permutation of the Jordan blocks). Or is there any another way?

答:
As you said, a similarity class of a matrix $A \in \mathbb C^{n \times n}$ is given by its Jordan normal form, up to a permutation of the Jordan blocks.

As the cardinality of the Jordan normal blocks is the one of the continuum $\mathfrak c$ (i.e. the one of $\mathbb C$), the cardinality you're looking for is $\mathfrak c$.


问:
Steps in solving the equivalence of the set statements
$$ \overline{(B \setminus A) \cup (B\bigoplus A)} \equiv \overline{A\bigoplus B} $$

Hi guys how to solve or to prove the equivalence of this statement?

答:
This equivalence can be shown using properties of the operators involved, as well as De Morgan's laws. The main properties we'll need are commutativity, associativity, and idempotence of set intersection, as well as the fact that the symmetric difference of two sets is equal to the union of pairwise set differences.

Here is a purely symbolic manipulation showing the two are equal:

$$\begin{eqnarray}
\overline{(B \setminus A) \cup (B \oplus A)}
&=& \overline{(B \setminus A)} \cap \overline{(B \oplus A)} \\
&=& \overline{(B \setminus A)} \cap \overline{(B \setminus A) \cup (A \setminus B)} \\
&=& \overline{(B \setminus A)} \cap \overline{(B \setminus A)} \cap \overline{(A \setminus B)} \\
&=& \overline{(B \setminus A)} \cap \overline{(A \setminus B)} \\
&=& \overline{(A \setminus B)} \cap \overline{(B \setminus A)} \\
&=& \overline{(A \setminus B)\cup(B \setminus A)} \\
&=& \overline{(A \oplus B)}
\end{eqnarray}$$

The first, third, and sixth equalities follow from De Morgan's laws. The second and seventh equalities follow from the property of symmetric difference stated above. The fourth and fifth equalities follow from the idempotence and commutativity of set intersection, respectively.


问:
Finding Partial Derivative ($n$-dimensional) using implicit differentiation vs explicitly solving
This is a book example (not a homework question) about implicit differentiation on a composite of functions in $n$-dimensional space.

But my book explains this example in a very unclear manner.  So I really appreciate your help in clarifications, since I'm very lost now >_<

The task is as follow:

Given:

$F(x,y,u,v) = x^2 + ux + y^2 + v$

$G(x,y,u,v) = x + yu + v^2 + x^2  v$

Goal:  Find $\frac{\partial u}{\partial x}$ by 2 methods:

(1) by implicit differentiation

(2) explicitly solve for $u$

So far, I used the Jacobian to get the partial derivative, where:

$$\frac{\partial u}{\partial x} = \frac{-\left|\frac{\partial (F,G)}{\partial (x,v)}\right|}{\left|\frac{\partial (F,G)}{\partial (u,v)}\right|}$$

after some prep-calculations for partials of $F$ with respect to $x, u, v$; and partials of $G$ with respect to $x, u, v$

So my question is, is this method considered (1) or (2)?  What does it really mean to "explicitly solve" for $u$? Just treat $x,y,v$ as constants and solve for $u$ in $F$ and in $G$?

My friend said if I do method (2), I shall get a different answer from my answer using Jacobian, but I don't know what to do to get the 2nd answer.

Would someone please help me on this question?

Thank you in advance

答:
I rewrite the equations for readability:

$F(x,y,u,v)=x^2+ux+y^2+v$,

$G(x,y,u,v)=x+yu+v^2+x^2v$.

Maybe for method (2), the idea is to eliminate $v$, then solve for $u$. So we can take $v=F-x^2-ux-y^2$, then substitute that into the second equations, expand/simplify, and try to solve for $x$. I suppose it is possible you will have a quartic equation in $x$, so if you don't see a trick for solving it, you may try to have some software solve that algebraic equation.


问:
An Inconsistent Set of Formulas Syntactically Entails Anything
Here's what I'm trying to prove:

Let $\Delta$ be an inconsistent set of formulas. Then, for any formula $\psi$, $\Delta \vdash \psi$.

Proof Attempt:

We have previously shown that $\{(\alpha \land \beta) \} \vdash \alpha$. Let $\psi$ be any formula. Since $\Delta$ is inconsistent, it follows that:

$$\Delta \vdash (\lnot(\psi \to \psi))$$

However, $(\lnot(\psi \to \psi))$ is just $(\psi \land (\lnot \psi))$. Then, there is a deduction from $(\psi \land (\lnot \psi))$ to $\psi$. Hence, there is a deduction from $\Delta$ to $\psi$ and that is what we wanted to show. $\Box$

Does the argument above work? If it doesn't, then why? How can I fix it?

Would there be any alternative argument that doesn't actually rely on the fact that I proved previously?

答:
We assume as axioms those of the linked proof system.

With Ax.1 and Ax.2 we can prove (many copy of the proof on this site): $\vdash (\alpha \to \alpha)$.

Using the following instance of Ax.1:

$(\psi \to \psi) \to (\lnot \psi \to (\psi \to \psi))$.

and the previous result we have:

$\text { (1)   } \vdash (\lnot \psi \to (\psi \to \psi))$.

Using the following instance of Ax.1:

$\lnot (\psi \to \psi) \to (\lnot \psi \to \lnot (\psi \to \psi))$

we have:

$\text { (2)   } \lnot (\psi \to \psi) \vdash (\lnot \psi \to \lnot (\psi \to \psi))$.

Using the following instance of Ax.3:

$[(\lnot \psi \to \lnot (\psi \to \psi)) \to ((\lnot \psi \to (\psi \to \psi)) \to \psi)]$

we have, form (1) and (2) by MP:

$\text { (3)   } \lnot (\psi \to \psi) \vdash \psi$.

Thus, if $\text {Incons}(\Delta)$ is defined as: $\Delta \vdash ¬(α → α)$ for any formula $α$, using (3) above we have:

if $Δ$ is an inconsistent set of formulas, then $Δ \vdash ψ$, for any formula $\psi$.


问:
How do I prove this statement about greatest common divider?
$$2\gcd(a, b) \le \gcd(2a,2b)$$
I'm pretty sure it has something to do with Bezout's lemma or it could use prime factorization, but unsure on how to go exactly about it. Please help!

Full Disclosure: This came up on a midterm and something like this may be on the final so I want to make sure I have it down for next time.

答: Normally you would prove the equality, as suggested by the link provided in comments. In this case however, you only need to prove one inequality. For this notice $2\gcd(a,b)|2a$ (because $2|2$ and $\gcd(a,b)|a$ by definition) and similarly $2\gcd(a,b)|2b$. So by definition $2\gcd(a,b)|\gcd(2a,2b)$, which implies $2\gcd(a,b)\leq \gcd(2a,2b)$.


问:
solve for the Laurent series of the function $f(z) = \frac{z-12}{z^2+z-6}$ that is valid for $1<|z-1|<4$.
I am about to solve for the Laurent series of the function $f(z) = \frac{z-12}{z^2+z-6}$ that is valid for $1<|z-1|<4$.

What I did is I use the partial fraction decomposition to rewrite the $f(z)$ into $f(z) = \frac{3}{z+3}+\frac{2}{z-2}$.  Now based on the region to where the function $f$ should be valid on is that $1< |z-1|< 4$, I'm not sure if I did it right when I assumed that $\left|\frac{2}{z}\right|<1$ and $\left|\frac{3}{z}\right|<1$. (I just did it with brute force to arrive on the two inequalities). Can you help me with that?

Assuming my two inequalities are correct, I have now

$\frac{3}{z+3} = \frac{3/z}{1+3/z} = 3/z \sum_{n=0}^{\infty} \left(- \frac{3}{z}\right)^n = \sum_{n=0}^{\infty} (-1)^{n-1}3^nz^{-n} = \sum_{n=1}^{\infty} (-1)^n3^{n+1}z^{-(n+1)}$,

and

$\frac{2}{z-2} = \frac{2/z}{1-2/z} = 2/z \sum_{n=0}^{\infty}  \left(\frac{2}{z}\right)^n = \sum_{n=0}^{\infty} \frac{2^{n+1}}{z^{n+1}}.$

Then after that, I combined which I have the result as

$f(z) = \frac{z-12}{z^2+z-6} = \sum_{n=0}^{\infty} \frac{2^{n+1}}{z^{n+1}} + \sum_{n=1}^{\infty} (-1)^n(3)^{n+1}z^{-(n+1)}$.

Now, did I do it correctly? or I did wrong on the inequalities of the region to where the laurent should be valud to?  Thanks for those who can help.

答:
$$
\begin{align}
\frac{z-12}{z^2+z-6}
&=\frac3{z+3}-\frac2{z-2}\\
&=\frac3{(z-1)+4}-\frac2{(z-1)-1}\\
&=\frac34\underbrace{\sum_{k=0}^\infty\left(\frac{1-z}4\right)^k}_\text{converges for $|z-1|\lt4$}-2\underbrace{\sum_{k=1}^\infty\left(\frac1{z-1}\right)^k}_\text{converges for $|z-1|\gt1$}
\end{align}
$$


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

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

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

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

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

Thanks so much.

答:
With CAS help and Mellin Transform:

$$\int_0^1 \frac{\exp \left(\frac{1}{\log (x)}\right)}{\sqrt[5]{x} \sqrt[5]{\log (x)}} \, dx=\\\mathcal{M}_a^{-1}\left[\int_0^1 \mathcal{M}_a\left[\frac{\exp \left(\frac{a}{\log (x)}\right)}{\sqrt[5]{x}
   \sqrt[5]{\log (x)}}\right](s) \, dx\right](1)=\\\mathcal{M}_a^{-1}\left[\int_0^1 \frac{(-1)^{-s} \Gamma (s) \log ^{-\frac{1}{5}+s}(x)}{\sqrt[5]{x}} \, dx\right](1)=\\\mathcal{M}_s^{-1}\left[-(-1)^{4/5}
   \left(\frac{4}{5}\right)^{-\frac{4}{5}-s} \Gamma (s) \Gamma \left(\frac{4}{5}+s\right)\right](1)=\\-(-1)^{4/5} *\sqrt[5]{2}* 5^{2/5} *K_{\frac{4}{5}}\left(\frac{4}{\sqrt{5}}\right)\approx0.302595\, -0.219848 i$$

Where:

$\mathcal{M}_a[f(a)](s)$ is Mellin Transform,

$\mathcal{M}_s^{-1}[f(s)](a)$ is Inverse Mellin Transform,

$K_{\frac{4}{5}}\left(\frac{4}{\sqrt{5}}\right)$ is modified Bessel function of the second kind.

Title: definite integral. closed form?


问:
Compact absolute retracts have the universal extension property. Where's the mistake in this proof?
First, the definitions:

$Y$ has the universal extension property if given any normal $X,$ closed subspace $A \subseteq X,$ and continuous function $f: A \to Y,$ there exists a continuous extension of $f$ to $X.$

A normal space $Y$ is an absolute retract if for any normal $Z$ and closed subspace $Y_0 \subseteq Z$ homeomorphic to $Y,$ the space $Y_0$ is a retract of $Z.$

Here is the problem:

Show that if $Y$ has the universal extension property, $Y$ is an absolute retract. Show that if $Y$ is an absolute retract and compact, then $Y$ has the universal extension property.

Proof: The first part is easy. Given $Y_0, Z$ as above, let $g: Y_0 \to Y$ be a homeomorphism. Then $g$ has a continuous extension $f: Z \to Y$ and $g^{-1} \circ f$ is a continuous retraction. Indeed, for any $y \in Y_0,$ we have $g^{-1}(f(y)) = g^{-1}(g(y)) = y.$

For the second part, let $A \subseteq X, f : A \to Y$ be what we're given so that we wish to extend $f$ to $X.$ Since $Y$ is normal, $Y$ is completely regular, so there exists an embedding $g : Y \to [0,1]^J$ for some set $J.$ But $[0,1]^J$ has the universal property and $g \circ f : A \to [0,1]^J$ is continuous, so there exists a continuous extension $h: X \to [0,1]^J.$ Then $g^{-1} \circ h$ is the continuous extension of $f$ that we seek. Indeed, $a \in A \Rightarrow g^{-1}(h(a)) = g^{-1}(g(f(a))) = f(a).$

My proof of the 2nd part did not use the fact that $Y$ is compact, and did not use the fact $Y$ is an absolute retract in any way besides $Y$ being normal. But if it's wrong, where is the mistake?

答: The mistake was easier to spot than expected. $g^{-1} \circ g \ne \text{id}$ because $g$ is not a bijection.