问:
Genus of a graph and planarity
In 'Introduction to graph theory' by Trudeau it is proved the theorem

Every graph has a genus

by first embedding the graph onto a sphere , and then adding handles, say we use $k$ of them, where the graph has edge-crossing , this is the example used :

Then, he writes that using a continuous deformation of the surface (stretching,shrinking or distorting the surface) it is possible to preserve planarity while deforming the $k$-handles sphere into a $k$-hole torus. So, the genus will be at most $k$.

I'm not able to visualize the bold part, how can I be sure that planarity will be preserved? Should it be obvious? Is there a simple explanation to understand clearly why this happens ? (I'm totally a beginner in algebraic topology, just visualization of trasformations are used in the book).

Another proof of a theorem stated earlier in the book : "every planar graph has genus $0$ (it can be embedded on a sphere)" was obscure to me but then searching online I found the explanation through the stereographic projection and all my doubts were gone , does something similar exist? Or even a good intuition is difficult to set up ?

EDIT :

For clarity I add the definition of genus given in the book :

The genus of a graph, denoted $g$, is the subscript of the first surface among the family $S_0,S_1,S_2,\dots$ on which the graph can be drawn without edge crossing .

I also add the explanation of the author to pass from the $k$-handle sphere to the $k$-hole thorus :

Think of the surface consisting of $S_0$ with $k$ handles as made of extremely pliable rubber which we are free to stretch, shrink or otherwise distort as much as we want , provided that we don't tear it off of fold it back onto itself. With this understanding we see that the surface consisting of $S_0$ with $k$ handles can be 'continuously deformed' into $S_k$ and the vertices and edges of $G$ can be carried along with this continuos deformation. Thus $G$ can be drawn on $S_n$ without edge-crossings.

The bold part is obscure to me in the sense explained above.

答:
If you have a mug (that has glaze on it) and a whiteboard marker, try drawing $K_5$ on it without any edge crossings.

Once you do that, it's not a far leap to imagine the handle expanding to make the mug more donut-like, and noticing that this deformation does not create any new edge crossings. Then try generalizing this to more holes.

Title: Genus of a graph and planarity


问:
Can the Stern-Brocot tree be employed for better convergence of $2^m/3^n$?
Prerequisite reading:

Can any positive real be approximated as $2^m/3^n$ with $(m,n)$ large enough?

Stern Brocot tree sequence

Something unsatisfactory is going on with convergence of $\,2^m/3^n\,$ towards a positive real $\,r\,$. As soon as we have reached sufficient approximation,
the next step in our current iteration procedure is to increase $\,m \to m+1\,$ if $\,2^m/3^n < r\,$ or to increase $\,n \to n+1\,$ if $\,2^m/3^n > r\,$.
But then we have actually destroyed our approximation so far, according to $\,2^m/3^n \to 2.2^m/3^n\,$ or $\,2^m/3^n \to 2^m/3^n/3\,$ respectively.
Thus it looks like we are starting all over every time again without making much progress. The number of iterations needed indeed is very large.

Reason why I have been looking for a procedure that does not have this drawback, i.e. where the next approximation is always closer to the desired outcome.
This is what I have tried so far.

According to question (2.),
for every positive real number $0 \lt g \lt 1$ , there exists an infinite sequence in the Stern Brocot tree [ .. ] that converges to the real number.
Meanwhile, this question has an answer,
and the main result in there reads as follows:
$$
- \frac{1}{n_1(n_1+n_2)} \lt g - \frac{m_1+m_2}{n_1+n_2} \lt \frac{1}{(n_1+n_2)n_2}
$$
In view of question (1.), we substitute $\ln(2)/\ln(3)$ for that number $g$. Then it follows that:
$$
- \frac{1}{n_1(n_1+n_2)} \lt \frac{\ln(2)}{\ln(3)} - \frac{m_1+m_2}{n_1+n_2} \lt \frac{1}{(n_1+n_2)n_2} \\
- \frac{\ln(3)}{n_1} \lt \ln(2)(n_1+n_2) - \ln(3)(m_1+m_2) \lt + \frac{\ln(3)}{n_2} \\
\ln\left(3^{-1/n_1}\right) \lt \ln\left(\frac{2^{n_1+n_2}}{3^{m_1+m_2}}\right) \lt \ln\left(3^{+1/n_2}\right) \\
3^{-1/n_1} \lt \frac{2^{n_1+n_2}}{3^{m_1+m_2}} \lt 3^{+1/n_2}
$$
The search through the Stern-Brocot tree can be pictured. The blue line is the function $\,\color{blue}{x\ln(2)-y\ln(3)=0}\,$, small circles are fractions,
mapped on a grid $\,m/n \to (m,n)\,$, massively black filled dots are the fractions in the Stern-Brocot tree. It is seen that searching through
the tree is much more efficient than increasing $m$ and $n$ with increments one at a time.

Now compare the expression at the second line of the above formulas with an analogous expression in reference (1.):
$$
\ln(2)(n_1+n_2) - \ln(3)(m_1+m_2) \quad \Longleftrightarrow \quad m\ln(2) - n\ln(3) - \ln(r)
$$
And be prepared for a disappointment: the logarithm of the arbitrary real $r$ is missing! Or alternatively: $\ln(r)=0$ or $r=1$.
This means that our "infinite search" through the Stern-Brocot tree, though highly efficient, finally arrives at an approximation only for the number one.
I find this weird, because - graphically - there doesn't seem to be a great difference between $\color{red}{2^m/3^n \to r}$ and $\color{blue}{2^m/3^n \to 1}$:

Hence the QUESTION: does there exist a means for adapting the Stern-Brocot procedure such that it works for reals other than one?

EDIT.

Here comes another graph that shows the astonishing convergence with the Stern-Brocot method, in comparison with analogous pictures in my Q & A
Can any positive real be approximated as $2^m/3^n$ with $(m,n)$ large enough? :

答:
I will give an approach which doesn't use the Stern-Brocot procedure.

It suffices to show that $\frac{2^{m}}{3^{n}}$ is dense in the interval [1,2]. Since taking $\alpha\in (0,\infty)$ outside this interval there is some $k\in Z$ so that $\alpha = 2^{k}\gamma $ for some $\gamma \in [1,2]$. Then we know there is a sequence in $\frac{2^{m}}{3^{n}}$ which approaches $\gamma$, multiplying the sequence termwise by $2^{k}$ (possibly taking a tail of the sequence), we get a sequence in $\frac{2^{m}}{3^{n}}$ which approaches $\alpha$.

Next consider that the map $f:[1,2] -> [0,1]$ with $f(x) = log_{2}(x)$ is a bijection.

The image of $\frac{2^{m}}{3^{n}}$ under the map is $N-Nlog_{2}(3)$. So it is enough to show that $N-Nlog_{2}(3)$ is dense in $[0,1]$.

This is a consequence of Weyl's Equidistribution theorem, which is a special case of the Ergodic theorem.

Consider $a=2-log_{2}(3) = log_{2}(\frac{4}{3})$, so $a$ is in the image of the set, So is $na = log_{2}(\frac{4^{n}}{3^{n}})$ and so is the fractional part of $na$.

The weyl Equidistribution theorem (which is not a trivial result) shows that for irrational a the fractional part of $na$ is uniformly distributed and hence dense on [0,1]. Since $2-log_{2}(3)$ is irrational you can use this theorem.

ContentLicense: CC BY-SA 4.0


问:
Finding example
Assume that $\Omega$ is a simple connected open set and $z_0 \in \Omega$. Let $f : \Omega \rightarrow \Omega$ be a holomorphic function such that $f(z_0) = z_0$ and $|f'(z_0)| = 1$.

If $\Omega \neq \mathbb{C}$, show that $f$ is a bijection.

Give an example that shows that $f$ must not be a bijection if $\Omega = \mathbb{C}$.

I was able to show the first part, but I do not find a counterexample for the second part.
I was looking for a simplest example possible, so I tried first $f(z) = z$, but that function is bijective. Next, I tried $f(z) = z^2$, but there are no points that statisfies the conditions.
From this, I thought that no polynomial will do the job. Therefore I tried something in the form of $e^{iz}$, but this function does not have fixed points.

From this, I am out of ideas, someone who can help me?

Thanks in advance!

答: Take $f(z)=\sin(z)$ and $z_0=0$. Then $f(0)=0$ and $f'(0)=\cos(0)=1$. But the sine function is indedeed not injective.

Title: Finding example


问:
Rank of coherent sheaf on complex manifolds
Let $X$ be a complex smooth connected manifold of dimension $n$ and let $\mathcal{O}_{X,x}$ be the ring of germs of holomorphic functions at $x$. Since the stalk of $\mathcal{O}_X$ at $x$ does not depend in which open subset is contained the point, then it can be identified with the ring of convergent power series in $n$ variables. Let $K(x)$ be the field of fractions of $\mathcal{O}_{X,x}$, i.e. it is the field of germs of meromorphic functions at $x$.

Definition: Let $\mathcal{F}$ be a sheaf over $X$, then we say that $\mathcal{F}$ is coherent if for each $x\in X$ there exist a neighborhood $U$ of $x$ and an exact sequence \begin{equation*}\mathcal{O}_U^{\oplus q}\rightarrow\mathcal{O}_U^{\oplus p}\rightarrow\mathcal{F}_{|U}\rightarrow 0\end{equation*}for some $p,q>0$ where $\mathcal{O}_U$ is the restriction to $U$ of the sheaf of holomorphic functions on $X$. In particular the sheaf $\mathcal{F}$ is locally finitely presented.

Definition: Let $\mathcal{F}$ be a coherent sheaf on $X$, since the stalk $\mathcal{F}_x$ is a finitely-generated $\mathcal{O}_{X,x}$-module then we can define the rank of $\mathcal{F}$ to be \begin{equation*}\text{Rank}(\mathcal{F}):=\text{dim}_{K(x)}\bigg(\mathcal{F}_x\bigotimes_{\mathcal{O}_{X,x}}K(x)\bigg)
\end{equation*}

Question 1

Since $X$ is assumed to be smooth and connected, is the rank of a
coherent sheaf $\mathcal{F}$ constant and equal to $r$ say?

Now suppose that $\mathcal{F}$ is coherent and torsion-free (i.e. $\mathcal{F}_x$ is a torsion-free $\mathcal{O}_{X,x}$-module for each $x\in X$), then it can be proved that there exist a closed complex manifold $Y\subset X$ of codimension at least $2$ sucht that $\mathcal{F}|_{X\setminus Y}$ is locally-free and in particular has finite constant rank (since over this open subset is a holomorphic vector bundle, am I right?).

Question 2

Adding the hypothesis of being torsion-free, what can we say about the rank over $Y$? Could it jump?

答:
Question: "Now suppose that $F$ is coherent and torsion-free (i.e. $F_x$ is a torsion-free $O_{X,x}$-module for each $x∈X$), then it can be proved that there exist a closed complex manifold $Y⊂X$ of codimension at least 2 sucht that $F|X∖Y$ is locally-free and in particular has finite constant rank (since over this open subset is a holomorphic vector bundle, am I right?)."

Answer: If $X^a\subseteq \mathbb{P}^n_{\mathbb{C}}$ is a complex projective manifold and $E^a$ is a coherent $\mathcal{O}_{X^a}$-module, it follows $X^a:=X$ is algebraic and to $E^a$ corresponds a coherent $\mathcal{O}_X$-module $E$ with the property that there is an open sub variety $U\subseteq X$ where the restriction $E_U$ is a locally trivial $\mathcal{O}_U$-module of finite rank $e$. This implies the same result for $E^a$ I believe: The $\mathcal{O}_{X^a}$-module $E^a$ when restricted to $U$ is locally trivial of finite rank $e$. I believe the functor $F$ (and its "inverse" $G$) that associates the coherent $\mathcal{O}_X$-module $E$ to the $\mathcal{O}_{X^a}$-module $E^a$ respects restrictions and isomorphisms. Hence when $E_U \cong \mathcal{O}_U^e$ it follows

$$E^a_U \cong G(E_U) \cong G(\mathcal{O}_U^e)\cong \mathcal{O}_{U^a}^e.$$

You must check the original GeoAlg-paper where this correspondence is proved to exist.

There is the "generic freeness" result in Matsumura (page 185, $24):

Theorem 1. Let $A$ be any noetherian ring and let $M$ be a fintely generated $A$-module.  It follows there is an element  $a\in A$ such that $M_a$ is a free $A_a$-module of finite rank.

Corollary. Let $Y \subseteq \mathbb{P}^n_{\mathbb{C}}$ be any projective sub scheme/variety and let $\mathcal{E}$ be a coherent $\mathcal{O}_Y$-module. It follows there is an open set $U \subseteq Y$ with $\mathcal{E}_U$ a locally trivial $\mathcal{O}_U$-module of finite rank.

Proof. Choose any open affine subset $U:=Spec(A) \subseteq Y$ and let $\mathcal{E}(U):=E$. It follows from Theorem 1 there is an open subset $D(a) \subseteq U$ where $E_a \cong \mathcal{E}(D(a))$ is a finite rank trivial $\mathcal{O}_{D(a)}$-module. Hence there is an open subscheme $D(a) \subseteq U \subseteq Y$ where $\mathcal{E}_U$ is locally trivial of finite rank. QED.

Hence it seems the result holds in greater generality if your complex manifold/space is projective.


问:
Visualizing infinitesimals
I am having problems visualizing infinitesimals. Like in a curve, when we integrate we can adjust in it a perfect infinitely thin rectangle. But how do I visualize it? Every time I try to visualize it, I always notice some gap area or error in the curve. Also in differentiation, $ dx $ is supposed to an infinitely close step. But every time I draw a line, I can cut it into smaller line and that becomes $ dx $. So I am facing bit problema visualizing infinitesimals.

答:
There's an easy workaround: forget infinitesimals, learn the definitions of limits, derivatives, integrals,... and work with them. I'll admit there may be some collateral damage: after several years of exercise and experience, you may start seeing those "infinitesimals", and which "gaps" you can safely ignore.

Alternative: you may study nonstandard analysis, making infinitesimals a rigorous notion. But there are several systems (Robinson, Nelson,...), none of them being trivial, all of them having their own pitfalls.

Title: Visualizing infinitesimals


问:
Minimum perimeter of a quadrilateral with an inscribed circle
Let us say a circle is given, and the circle is inscribed to a quadrilateral. How would you prove that the minimum perimeter is achieved when the quadrilateral is a square?</p>

答:
Suppose the inscribed circle has radius 1. In the picture we have a poligon $IHFG$, and we have drawn the perpendicular segments from the center of the circle to the sides of the poligon. The angle $\alpha_1$ is the angle $JOK$, and the rest of the angles are analogous. The segment $OF$ bisects $\alpha_1$, so using the right triangle $FJO$ you have that $\overline{FJ}=\overline{JO}\tan\left(\frac{\alpha_1}{2}\right)=\tan\left(\frac{\alpha_1}{2}\right)$, and also $FK=\tan\left(\frac{\alpha_1}{2}\right)$.

Doing the same for the rest of the segments of the perimeter, we have that the perimeter is $2\left(\tan\frac{\alpha_1}{2}+\tan\frac{\alpha_2}{2}+\tan\frac{\alpha_3}{2}+\tan\frac{\alpha_4}{2}\right)$. So, calling $\beta_i=\frac{\alpha_i}{2}$, you have to minimize $\tan\beta_1+\tan\beta_2+\tan\beta_3+\tan\beta_4$, where $\beta_i\in(0,\frac{\pi}{2})$ and $\sum\beta_i=\pi$.

But $\tan(x)$ is a convex function in $\left(0,\frac{\pi}{2}\right)$, so using Jensen´s inequality, we have that $\tan\beta_1+\tan\beta_2+\tan\beta_3+\tan\beta_4\geq4\tan\left(\frac{\beta_1+\beta_2+\beta_3+\beta_4}{4}\right)=4\tan\left(\frac{\pi}{4}\right)=4$. That is, the minimum value is achieved when $\beta_i=\frac{\pi}{4}$, which is precisely the case of the square.


问:
Why doesn't my derivation of the arcsec derivative formula not work?
We would like to find the derivative of $\operatorname{arcsec}(x) = y$. Rearranging this, we get $x = \sec(y)$. Taking the derivative of both sides, we get $1 = \sec(y)\tan(y)y^\prime$. Thus, $$y^\prime = \frac{1}{\sec(y)\tan(y)}.$$ Now, draw a right triangle with an acute angle $y$, hypotenuse $x$, and a side adjacent to angle $y$ with length $1$. By the Pythagorean theorem, the side opposite angle $y$ is equal to $\sqrt{x^2-1}$. Thus, we have $x = \sec(y)$, which we already knew and got from the triangle, and $$\tan(y) = \frac{1}{\sqrt{x^2-1}}$$ from the triangle we drew. We substitute these values in to our derivative expression to get $$y' = \frac{1}{x\sqrt{x^2-1}}.$$

However, $y^\prime$ should equal $$\frac1{|x|\sqrt{x^2-1}}$$ (note the absolute value). Since the difference occurred in the $|x|$ part, that means I must've done something wrong when I said $\sec(y) = x$, but I do not know why this assumption is wrong. Please correct my proof.

答:
Ted Shifrin has already explained the problem with your derivation, but if you're looking for a 'safer' way of finding the derivative then consider this: if $\DeclareMathOperator{\arcsec}{arcsec} y = \arcsec x$, then using the identity $\arcsec x = \arccos 1/x$, we see that
\begin{align}
\frac{dy}{dx}=\frac{d}{dx}\left(\arccos 1/x\right) &= -\frac{1}{\sqrt{1-\left(\frac{1}{x}\right)^2}} \cdot -\frac{1}{x^2} \\[4pt]
&= \frac{1}{x^2\sqrt{1-\frac{1}{x^2}}} \, .
\end{align}


问:
Need help with Ex 3.3.3 from Topology Without Tears
Let (X $\tau$) be any topological space

Prove that (X $\tau$) is not connected iff it has proper nonempty disjoint open subsets
A and B s.t $ A\cup B=X$

My ideas come from here

(If $X$ is connected then there're not $F$ and $G$ such that $F\cap G=\emptyset$ and $X=F\cup G$)

Proof
(l thought of proof by contradiction but
did not know how to proceed)

(<=)Suppose  we have proper nonempty disjoint
open subsets A an B s.t A$\cup $ B=X

As X is not connected ,$\exists  U \in\tau$
and.U=A B=X\U. So U is clopen

Then $\tau $ is nonconnected

How is this half of the proof

Any help would be appreciated.

答:
Suppose $X$ is not connected according to the definition you gave. So there is a clopen subset $C$ that is not euql to $X$ or $\emptyset$. Then note that $A=C$ and $B=X\setminus C$ are both open and disjoint and non-empty and $A \cup B=X$ is trivial.

If $A ,B$ are open disjoint and non-empty and $X=A\cup B$ then $A$ (and $B$ too) is clopen and unequal to $X$ or $\emptyset$. So $X$ is then not connected.


问:
Integrals of the type $\int\frac{dx}{\left(a^2+x^2\right) \sqrt{a^2+b^2+x^2}}$
I'm interested in understanding the sequence of substitutions needed to obtain
\begin{align}
\int\frac{dx}{\left(a^2+x^2\right) \sqrt{a^2+b^2+x^2}}
=\frac{\tan ^{-1}\left(\frac{b x}{a \sqrt{a^2+b^2+x^2}}\right)}{a b}\, ,
\end{align}
with $a,b>0$.

The obvious substitution
\begin{align}
x=\sqrt{a^2+b^2}\tan\xi
\end{align}
gets rid of the square root but complicates the $a^2+x^2$ factor:
\begin{align}
\int\frac{dx}{\left(a^2+x^2\right) \sqrt{a^2+b^2+x^2}}
= \int \frac{d\xi}{ (a^2+ (a^2+b^2)\tan^2\xi)\cos\xi} \tag{1}
\end{align}
which doesn't seem to be immediately helpful.

The substitution $a^2+x^2=b^2\tan^2\xi$ makes things worse since
$dx$ will convert to $b^2 \tan\xi d\xi/(\cos^2\xi\sqrt{b^2\tan^2\xi-a^2})$
with no realistic chance of eliminating the square root.

That would leave me with Euler's substitution
\begin{align}
\sqrt{a^2+b^2+x^2}=x+t
\end{align}
but before I get there I wonder if there's a more "obvious" way.

答: Your first choice is a better start than it seems, because Bioche's rules advise continuing with$$u=\sin\xi=\frac{\sqrt{a^2+b^2}\tan\xi}{\sqrt{a^2+b^2}\sec\xi}=\frac{x}{\sqrt{a^2+b^2+x^2}}.$$Then$$\frac{d\xi}{(a^2+ (a^2+b^2)\tan^2\xi)\cos\xi}=\frac{\cos\xi d\xi}{a^2\cos^2\xi+(a^2+b^2)\sin^2\xi}=\frac{du}{a^2+b^2u^2}.$$


问:
How to solve a system of nonlinear equations?
How can i solve this system of equations:
$$
\begin{align}
x(t) = a \cos(\omega t + \phi) + \dfrac{\beta a^3}{32} \cos(3(\omega t + \phi)), \\
p(t) = -a \omega \sin(\omega t + \phi) - 3 \omega \dfrac{\beta a^3}{32} \sin(3(\omega t + \phi)),\\
\omega = \omega_0 + \dfrac{3}{8} a^2,
\end{align}, \\
\text{if } x(0) = x_0, p(0) = p_0
$$
where $\omega_0, \beta$ are constants?

I can write:

$$
\begin{align}
x_0 = a \cos(\phi) + \dfrac{\beta a^3}{32} \cos(3\phi), \\
p_0 = -a \omega \sin(\phi) - 3 \omega \dfrac{\beta a^3}{32} \sin(3\phi)),\\
\omega = \omega_0 + \dfrac{3}{8} a^2,
\end{align}
$$
But i can't figure out how to solve it for $a$ and $\phi$. Could you help?

答:
Not a solution but possibly helpful. Make use of the triple angle formulas
\begin{align}
\sin(3\theta) &=3\sin\theta−4\sin^3\theta,\\
\cos(3\theta) &= 4 \cos^3\theta - 3 \cos \theta.
\end{align}
The first equation ($x_0 = \cdots$) will end up being third order in $\cos\phi$ and  the other ($p_0 = \cdots$) third order in $\sin\phi$. With a bit of luck you can solve the first for $\cos\phi$ and the second for $\sin\phi$, then take  it from there knowing that $\cos^2\phi + \sin^2\phi = 1$.


问:
Find the maximum volume for a rectangular solid in the first octant with one vertex at the origin and opposite vertex on the plane x + y + z = 1.
I have to find the maximum volume for a rectangular solid in the first octant (  $x \ge 0 , y \ge 0 , z \ge 0$ ) with one vertex at the origin and opposite vertex on the plane $x + y + z = 1$  .

I need to use the method of Lagrange multipliers to find a set of equations that when solved will give the max volume of the rectangle solid

答:
A point on the plane $x+y+z=1$ has coordinates $(u, v, 1 - u - v)$.

If one vertex of the cuboid is $(0,0,0)$ then the volume is

$W=uv(1-u-v)$

set partial derivatives $W_u=0;\;W_v=0$

that is
$$\begin{cases}
u (1-u-v)-u v=0\\
v (1-u-v)-u v=0\\
\end{cases}
$$


问:
Restriction of quotient map $\pi: \mathbb{C}^{n+1}\setminus \{0\} \to \mathbb{P}^n (\mathbb{C})$ to $S^{2n+1}$ is closed
My lecturer, to demonstrate that the projective complex space $\mathbb{P}^n(\mathbb{C})$ is Hausdorff, proves that the restriction of the quotient map to $S^{2n+1}$ is a closed map. Here is how the proof begins.

Let
$$ \pi_0 := \pi_{\mid S^{2n+1}} : S^{2n+1} \to \mathbb{P}^n(\mathbb{C})$$
be the restriction of the quotient map $\pi: \mathbb{C}^{n+1}\setminus \{0\} \to \mathbb{P}^n (\mathbb{C})$. We want to prove the closedness of $\pi_0$: $\forall C$ closed set in $S^{2n+1}$ $\pi_0(C)$ is a closed set in $\mathbb{P}^n(\mathbb{C})\iff \pi_0^{-1}\left(\pi_0(C)\right)$ is closed in $S^{2n+1}$.

However, being in a quotient topology with quotient map $\pi$, proving that $\pi_0(C)$ is a closed set in $\mathbb{P}^n(\mathbb{C})$ should be equivalent to prove that $\pi^{-1}\left(\pi_0(C)\right)$ is closed in $\mathbb{C}^{n+1}\setminus \{0\}$. Therefore, my question is: does $\pi^{-1}\left(\pi_0(C)\right)$ being closed in $\mathbb{C}^{n+1}\setminus \{0\}$ implies that $\pi_0^{-1}\left(\pi_0(C)\right)$ is closed in $S^{2n+1}$ and viceversa?

答:
This community wiki solution is intended to clear the question from the unanswered queue.

As shown in the comments, the answer is "yes". Also have a look at The restriction of a quotient map $p : X \to Y$ to a retract of $X$ is again a quotient map. This shows that $\pi_0 : S^{2n+1} \to  \mathbb{P}^n(\mathbb{C})$ is a quotient map because $\pi: \mathbb{C}^{n+1}\setminus \{0\} \to \mathbb{P}^n (\mathbb{C})$ is one. Therefore the following are equivalent for $A \subset \mathbb{P}^n (\mathbb{C})$:

$A$ is closed in $\mathbb{P}^n (\mathbb{C})$.

$\pi^{-1}(A)$ is closed in $\mathbb{C}^{n+1}\setminus \{0\}$.

$\pi_0^{-1}(A)$ is closed in $S^{2n+1}$.


问:
Which positive integers can NOT be written as a sum of consecutive positive integers
I have been wondering about this question for a while (I am almost sure that I read it in some contest mathematics book a while ago)

Determine which positive integers cannot be written as a sum of consecutive positive integers.

All numbers will be considered positive integers in the following unless stated otherwise.

I was thinking as follows: odd numbers will always work since if we take an odd $m$, we have

$$
m=2\cdot k+1=k+(k+1).
$$

Then I realised that all numbers having at least one odd prime factor will also work. Take $n$ as a number having an odd prime factor and let this odd prime factor be $m$ and let $n=m\cdot k$. Then
$$
n=\underbrace{k+k+\ldots+k}_{m \ \text{times}}
$$
and since $m$ is odd, say $m=2\cdot l+1$ we have
$$
n=\underbrace{k+k+\ldots+k}_{l \ \text{times}}+k+\underbrace{k+k+\ldots+k}_{l \ \text{times}}
$$
and we can start moving $1$s one by one from the left side of the middle "$k$" to the right side of it
$$
n=\underbrace{k+k+\ldots+(k-1)}_{l \ \text{times}}+k+\underbrace{(k+1)+k+\ldots+k}_{l \ \text{times}}
$$
$$
n=\underbrace{k+k+\ldots(k-2)+(k-1)}_{l \ \text{times}}+k+\underbrace{(k+1)+(k+2)+\ldots+k}_{l \ \text{times}}
$$
this way we end up in a sum of consecutive integers. It may happen that some integers on the left will be negative which is undesired but since these numbers are consecutive integers we always be able to cancel out the negative ones without disturbing the structure of the uncancelled ones. To see this in action let us take $22=11\cdot 2$. This would give
$$
2+2+2+2+2+2+2+2+2+2+2
$$
so
$$
2+2+2+2+1+2+3+2+2+2+2
$$
the next step gives
$$
2+2+2+0+1+2+3+4+2+2+2
$$
it seems that now we are out of numbers but we just continue moving an increasing number of $1$s to the right
$$
2+2+(-1)+0+1+2+3+4+5+2+2
$$
$$
2+(-2)+(-1)+0+1+2+3+4+5+6+2
$$
finally
$$
\underbrace{(-3)+(-2)+(-1)+0+1+2+3}_{=0}+4+5+6+7
$$
so we get
$$
22=4+5+6+7.
$$

Now we have a method to generate the desired expression for all odd numbers and all the ones having at least one odd prime factor. Which numbers are left? The ones having ONLY even primes in their prime factorisation. These are of the form $2^n$. I was thinking what do these have in common and I thought that they are separated by a factor of $2$. The first of these are $1$ for $n=0$ and $1$ cannot be expressed in the desired for so if I could show that
$$
m \ \text{cannot be expressed}\Rightarrow 2m \ \text{cannot be expressed}
$$
I am done. So I was trying to show the contrapositive
$$
2m \ \text{can be expressed}\Rightarrow m \ \text{can be expressed}.
$$
So assume that $2m$ can be expressed. Now, eihter $m$ is odd in which case we are done since odds were easy to express or it is even, that is $m=2l$. From here the argument repeats. Eihter $l$ is odd in which case we are done or it is even, so $l=2k$ and so on. After a while we factored out all of the $2$'s and we end up with and odd number which we can express and we are done.

So integers of the form $2^n$ cannot be expressed in this form.

Is my reasoning correct?

If not please point out where I made a mistake. If it turns out to be correct then we are all happy I can sleep well tonight but it would be nice to see a more rigorous proof of this so if someone can provide one I am more than happy to read it. Since this is a contest question I expect it to have a really elegant solution and I do not consider mine an elegant one.

答:
Yes, I think your reasoning is correct. I have also given a solution below.

Let's consider sum of $r$ consecutive positive integers with first term $a$ and last term $l$. Since the integers are consecutive thus the terms are in Arithmetic Progession and thus we can apply formula related to AP. Sum of AP is number of terms into its mean, where mean of whole arithmetic series is equal to mean of first and last term.

Sum of consecutive terms $=$ $r*(a+l)/2$ where $r>1$.

$Case 1$

If the number of terms($r$) is odd then we already have a odd factor in the sum. Please note that both $a$ and $l$ will have same parity, thus mean $(a+l)/2$ will be an integer.

$Case 2$

If number of terms($r$) is even then we can say that parity of $a$ and $l$ will be different, i.e., if $a$ is even then $l$ will be odd and vice-versa. Thus the sum will have an odd factor as $a+l$. Since we are talking about postive integers thus ${a+l}>1$. Here $r/2$ will be integer.

Both cases show that sum of positive consecutive integers will always have an odd factor in it. So sum of positive consecutive integers can only represent numbers having at least one odd factor, say $m$, which is $>1$.

Now to show that all positive integers of form $n = 2^{p}*m$, where $m>1$ is odd number, can be expressed as sum of consecutive integers. We can see two cases. We will use the fact that sum of consecutive numbers is number of terms into its mean.

$Case 1$ : $m>2^{p+1}$

Here the idea will be to take $2^{p+1}$ consecutive numbers with mean of numbers as $m/2$. Please note that $m/2$ is not an integer. So we will take equal number of integers from left and right side of $m/2$ on the number line till we get a total of $2^{p+1}$terms.

$Case 2$ : $m<2^{p+1}$

Here the idea will be to take $m$ consecutive numbers with mean of numbers as $2^{p}$. So starting form $2^p - ((m-1)/2)$ we will take all consecutive numbers till $2^p + ((m-1)/2)$


问:
Right versus left derivative by increasing one-sided derivatives
Condider a map $f:D\to\mathbb{R}$, $D$ open interval.
Let be $y,z\in D$, $y<z$. Suppose that $f$ has everywhere in $D$ both left and right derivative (this implies that $f$ is continuous), both increasing.
I want to show that
$$
f'(y^+)\le f'(z^-).
$$

I have found a way, but I'm not completely convinced that it is formally correct:
$$
f'(y^+)
=\lim_{\varepsilon\to 0^+}\frac{f(y+\varepsilon)-f(y)}{\varepsilon}
\overset{\forall\delta}{\le}
  f'((z-\delta)^+)
= \lim_{\varepsilon\to 0^+}\frac{f(z-\delta+\varepsilon)-f(z-\delta)}{\varepsilon}
\overset{\varepsilon=\delta}{=} \lim_{\varepsilon\to 0^+}\frac{f(z)-f(z-\varepsilon)}{\varepsilon}
=f'(z^-)
$$
where $\delta>0$ and $y<z-\delta$.

Is there a way to formalise my proof?

答:
We can prove the stronger statement  $$\lim_{x\to y^+}f'_-(x)=f'_+(y)$$

By the Monotone Convergence Theorem $\lim_{x\to y^+}f'_-(x)$ exists. Call that limit $A$.

Now, let $\epsilon>0$ be given and choose a value of $\delta>0$ such that both of these are true:

$$x\in(y,y+\delta] \implies f'_-(x)\in \left[A,A+\frac{\epsilon}2 \right]$$
and
$$\left|\frac{f(y+\delta)-f(y)}{\delta}-f'_+(y)\right|\le \frac{\epsilon}2$$
That is possible because of the definitions of the left hand derivative and the limit.

Similar to the answer from @MartinR, define the function $g(x)=f(x)-\frac{f(y+\delta)-f(y)}{\delta}(x-y)$ so that $g(y)=g(y+\delta)$ and both $g'_-(x)=f'_-(x)-\frac{f(y+\delta)-f(y)}{\delta}$ and $g'_+(x)=f'_+(x)-\frac{f(y+\delta)-f(y)}{\delta}$ are increasing.

Claim 1. $g'_+(y) \le 0$

This follows because if it were not true, then $g'_+(x)$ would have to be positive for all $x\in  (y,y+\delta)$. That would imply $g$ is increasing (proof here) in that interval and it would be impossible for $g(y+\delta)$ to equal $g(y)$.

Claim 2. $g'_-(y+\delta) \ge 0$

Same argument as Claim 1. If it were not true, then $g'_-$ would have to be negative in the whole interval which would make it impossible for the function to have the same value at both ends of the interval.

Now,

$$g'_-(y+\delta)\ge0$$
$$f'_-(y+\delta)-\frac{f(y+\delta)-f(y)}{\delta}\ge0$$
$$f'_-(y+\delta) \ge \frac{f(y+\delta)-f(y)}{\delta} \ge f'_+(y)-\frac{\epsilon}2$$
We also know that $f'_-(y+\delta)\in \left[A,A+\frac{\epsilon}2 \right]$

Thus,
$$A \ge f'_-(y+\delta)-\frac{\epsilon}2 \ge f'_+(y)-\epsilon \label{eq1}\tag{1}$$

On the other hand,

$$g'_-(y)\le0$$
$$f'_-(y)-\frac{f(y+\delta)-f(y)}{\delta}\le0$$
$$f'_-(y) \le \frac{f(y+\delta)-f(y)}{\delta} \le f'_+(y)+\frac{\epsilon}2$$

So,
$$A \le f'_-(y+\delta) \le f'_+(y)+\frac{\epsilon}2 \label{eq2}\tag{2}$$

Combining the two inequalities (1) and (2), since $\epsilon$ was arbitrary, we have $A=f'_+(y)$.

To answer the original question, if $z>y$ then by the monotonicity assumption $f'_-(z)>\lim_{x\to y^+}f'_-(x)=f'_+(y)$


问:
Conjecture: primes non-decreasing from $(n,2n)$ to $(2n,4n)$.
More generally, I assert:

$$\pi(na^k)-\pi(na^{k-1})\leq \pi(na^{k+1})-\pi(na^k)\quad\text{for }n,a,k \in \mathbb N.$$

The most useful case of this is probably to say that if there are $m$ primes in $[n,2n]$, then there will be at least $m$ primes in $[2n,4n]$. I thought it worth noting this can also be written as

$$\pi(2n) \leq \frac{\pi(4n)+\pi(n)}{2}.$$

I doubt there are counterexamples, but if someone finds one, please share. Barring that, since this seems like an intermediate-strength statement, I'm curious how easy or difficult this might be to prove, and whether there are known similar results.

Update

Looks like a much tighter bound can also be specified below as $ \pi(3n)+\pi(n) \leq 2\pi(2n)$  for $n>1323$.

Also, note this can be used in reverse.

E.g. suppose we've worked out that the only primes in $[8,16]$ are $11$ and $13$. Then, even knowing nothing whatsoever about primes other than this rule (and that they start at $2$), it follows that there could be at most $2$ more primes in $[4,8]$, and another $2$ in $[2,4]$, so we'd know that $\pi(13)\leq 6$.

答:
One readily tests that
$$ 2\pi(2n)\le\pi(4n)+\pi(n)$$
holds at least for $1\le n<60184$. Following [Pierre Dusart, 2010], we know that
$$ \frac x{\ln x-1}<\pi(x)<\frac x{\ln x -1.1}$$
for all $x\ge60184$.
Hence at least for $n\ge 60184$, we have
$$ \pi(2n)<\frac{2n}{\ln 2n -1.1}$$
and
$$ \pi(4n)+\pi(n)>\frac{4n}{\ln 4n-1}+\frac n{\ln n-1}.$$

For the claim it would be sufficient to have
$$ \frac{4n}{\ln 2n-1.1}<\frac{4n}{\ln 4n-1}+\frac n{\ln n-1}.$$
If we multiply by $\frac 1{4n}$ and substitute $\ln n=t+1$, our goal becomes
$$ \frac{1}{t +\ln 2-0.1}<{\frac{1}{t+\ln 4}+\frac 1{4t}}$$
for all $t\ge \ln 60184-1\approx 10$.
But this follows directly from
$$ \frac{1}{t +\ln 2-0.1}-\frac{1}{t+\ln 4}
=\frac{\ln 2 +0.1}{(t +\ln 2-0.1)(t+\ln 4)}<\frac1{t^2}<\frac1{4t}.$$

Replacing $2$ with an integer $a\ge2$, the same method boils down to
$$ \frac{2a}{\ln a+t-0.1}-\frac{a^2}{2\ln a+t}<\frac {1}{t}$$
which is readily shown. However, the "base" case for $n<60184$ but with arbitrary $a$ is not so trivial.


问:
Proving that if $f'$ has at most $n-1$ zeros, then $f$ has at most $n$ zeros
Is this proof correct?

The problem is the following.

Let $n$ be a natural number. Suppose that the function $f:\mathbb{R}\to\mathbb{R}$ is differentiable and that the following equation has at most $n-1$ solutions:
  $$f'(x)=0, \quad x \in \mathbb{R}.$$
  Prove that the following equation has at most $n$ solutions:
  $$f(x)=0,\quad x \in \mathbb{R}.$$

My proof is:

Let $f'(x)=0$ have solutions $x_1$, $x_2,\ldots ,x_{n-1}$.

since
$$f'(x_1)=\lim\limits_{{x}\to{x_1}}\frac{f(x)-f(x_1)}{x-x_1} = 0$$
and doing so, $f(x)=f(x_{n-1})$
Therefore $f(x)=f(x_1)=f(x_2)=\cdots=f(x_{n-1})$.

Let $x_n$ be one of solutions of $f(x)=0$.
$$0=f(x_n)=f(x_1)=f(x_2)=\cdots=f(x_{n-1}),$$
so $f(x)=0$ has solutions like $x_1$, $x_2,\ldots, x_n$.

If there is wrong part, please let me know.

答:
Suppose $f(x)=0$ has more than n solutions and then show that then $f'(x)=0$ would have more than $n-1$ solutions.

It's a proof by contrapositive.

Instead of showing that $a\implies b$, you show that $\neg b \implies \neg a$.

ContentLicense: CC BY-SA 3.0


问:
Local integrability of a Lebesgue integral
Fix a Banach space $X$, an open subset $U \subset X$, and a measure space $(\Omega, \mathcal A, \mu)$. Let $f: U \times \Omega \to \mathcal R$ be continuously Frechet-differentiable in the first argument and let it and its Frechet derivative be measurable in the second argument.

Now fix $u \in U$, and suppose that $f(u,\omega)$ and, for all $x \in U$, $Df(u,\omega)(x)$ are Lebesgue-integrable over $\omega$. Does it follow that $f(\tilde u,\omega)$ is integrable for all $\tilde u$ in some neighborhood of $u$?

I suspect not but I cannot construct a counterexample, and this would be a very useful fact for me.

答:
Nevermind! Here is a counter-example:

$X = \mathbb R$, $\Omega = (0,1]$ with Lebesgue measure.

$$f(x,\omega) = x^2 \frac{1}{\omega}$$


问:
Order of calculation about nested absolute values
In case of nested absolute values, is there a difference between working first on the outer abs value and the opposite?
$$ \frac{|2x^2-3|x|+1|}{|x|+1} = 1 $$
Should I consider first the inner one?

My real solutions for this equation are $\left \{0, \pm 2\right \}$, and I've found those working first with the inner one.

Body:
Yet another solution.
$$\begin{align}
&|2x^2-3|x|+1|=|x|+1\\
\iff\quad& (2|x|-1)\cdot(|x|-1)=|x|-1+2\\
\iff\quad &(|x|-1)(2|x|-1-1)=2\\
\iff\quad & (|x|-1)^2=1\\
\iff\quad &|x|-1=\pm1\\
\iff\quad &x\in\{0,\pm2\}
\end{align}
$$


问:
A smart Ox to solve math problems.
Found an interesting 1907 article and am curious about how and why the underlying math works.

On a livestock fair in late Victorian Plymouth, England, a
statistician with the name of Francis Galton asked around 800
attendees to guess the weight of an ox that was on display. He then
calculated the mean of all estimates, which ended up being 1208
pounds¹. To his surprise, the measured weight of said ox was 1197
pounds, which put the mean estimation at ~0.01% off from the real
weight.

Are similar "herd" concepts applicable to solve other math problems? Thanks.

答: The major factor here is that most people either guess too high or too low, so if you average them they'll cancel each other out. This is reproducible at a large scale. The specific case cited here is probably just one time it happened to be really close, and if it was done again it probably wouldn't work (The media has a knack for these things, ignoring the mass results and only looking at specific cases that are "impressive"), but if you use this method of estimation you'll get pretty good results.</p>

Title: A smart Ox to solve math problems.


问:
How to consider "and" between quantifiers?
Let $\{x_n\}$ be a sequence in $\mathbb{R}$ and $a \in \mathbb{R}$. How would I negate

$\exists \varepsilon>0$ and $\exists N \in\mathbb{N}$ such that $\forall n\ge N$, $|x_n - a| \ge \varepsilon$

?

I get really confused with the "and" between the quantifiers. It makes me want to change it to an "or". Also, this statement is equivalent to no subsequence of $x_n$ is convergent to $a$, but I am having a difficult time understanding why. If anyone could explain this, and give the negation for the statement above, that would be great.

答:
Formally, we would write "there exists $a$ and there exists $b$ such that for all $c$ we have $\phi$" as
$$ \exists a\,\exists b\,\forall c\,\phi$$
If we prepend a negation, it percolates as follows:
$$ \neg\exists a\,\exists b\,\forall c\,\phi\iff\forall a\,\neg \exists b\,\forall c\,\phi\iff \forall a\,\forall b\,\neg \forall c\,\phi
\iff \forall a\,\forall b\, \exists c\,\neg\phi$$

(A more proper wording would be "There exists $a$ for which there exists $b$ for which ...")


问:
Surjectivity of $f(z)=\frac{(1+z)^2}{(1-z)^2}$
I am trying to prove that the function $f:A\longrightarrow H^+$, with $f(z)=\frac{(1+z)^2}{(1-z)^2}$, $A=\{z\in\mathbb{C}:|z|<1, Imz>0\}$ and $H^+=\{z\in\mathbb{C}:Imz>0\}$, is bijective.

I have managed to prove that it is injective but I am struggling to prove that it is surjective. I thought about seeing the function as the composition of the functions $h(z)=\frac{1+z}{1-z}$ and $g(z)=z^2$, trying to prove they are both surjective, but it did not work because I do not know what the domain of $g$ would be.

Any suggestions on how to approach this problem will be very helpful.

Thanks in advance.

答: To achieve $w=f(z)=1+\frac{4z}{1-z^2}$, rearrange to $z^2+\frac{4z}{w-1}-1=0$. The roots are $\pm z_0^{\pm1}$ with $z_0:=\frac{2}{1-w}+\sqrt{\frac{4}{(1-w)^2}+1}$. They have product $-1$ but aren't $\pm1$ for finite $w$, so exactly one of them has modulus $<1$. So $z_0$ is a unit complex number, and it and $-z_0^{-1}$ are of the form $\pm a+ib$. We need only check $\Im w>0\implies\Im z>0$ by showing the roots' sum has positive imaginary part; indeed, this sum is $\frac{4(1-w^\ast)}{|1-w|^2}$, with imaginary part greater than $\frac{4}{|1-w|^2}$.


问:
Find some solution $w$ which is not divided by any divisor of $a^{2x}$ and $b^{2y}$
Let us consider the following conguence:
$$2^{x+z}b^{y}(2a^{x}+b^{y})w=0  \bmod 2^{x}a^{2x}$$
$$ 2^{y+z}a^{x}(a^{x}+2b^{y})w=0  \bmod 2^{y}b^{2y}$$

I am asking if it is possible to find some solution $w$ which is not divided by any divisor of $a^{2x}$ and $b^{2y}$.

答: Sorry if i'm misinterpreting here, but you can find a solution $w$ for any two integers $a$ and $b$ by breaking them down into their prime factors and choosing a prime number that is not one of them. As there are an infinite amount of prime numbers, you'll always be able to find one.


问:
Stuck on an exercise regarding orthonormal bases.
Given a basis $e_1, e_2, e_3$ for $\mathbb{R}^3$ such that $||e_1||=1$, $||e_1||=\sqrt{2}$, $||e_1||=\sqrt{3}$ and

$e_1\cdot  e_2 = 1$, $e_1\cdot  e_3 = 1$, $e_2\cdot  e_3 = 1$

I'm asked to find an orthonormal basis $B$ of $\mathbb{R}^3$ in terms of $e_1, e_2, e_3$. I try to use Gram-Schimdt but get stuck.

My attempt:

Let $v_2=e_2-\text{proj}_{e_1}e_2=e_2-(e_1\cdot e_2)e_1=e_2-e_1$.
Since the length of the projection is $1$, and the length of $e_2$ is $\sqrt{2}$, we have that $||e_2-e_1||=1$. Now let

$$v_3=e_3-\text{proj}_{e_1,e_2}e_3=e_3-(e_3e_1)e_1-(e_3e_2)e_2=e_3-e_1-2e_2$$

the length of $\text{proj}_{e_1,e_2}e_3$ is $$\sqrt{||\text{proj}_{e_1}e_3||^2+||\text{proj}_{e_2}e_3||^2}=\sqrt{(e_3e_1)^2e_1+(e_3e_2)^2e_2}=\sqrt{1+4}=\sqrt{5}$$

yet this certainly can't be since $||e_3||=\sqrt{3}$.


问:
Compound Poisson Distribution with Sum of Poisson Random Variables
I am trying to find the distribution of a compound Poisson distribution with sum of Poisson random variables.

Suppose that $N \sim \text{Poisson}(\lambda_1)$,

and that $X_1, X_2, ...$ are i.i.d. r.v.s. that are again $\sim \text{Poisson}(\lambda_2)$.

What is the distribution of $Y = \displaystyle \sum_{n=1}^{N}X_n$ ?

Is there any analytic solution for this distribution?

Any kind of help is thanked in advance.

答:
The characteristic function is derived here.

$$\exp(\lambda_1 ( \exp(\lambda_2 (e^{it}-1))-1))$$
You could differentiate this to find the mean, variance, and other moments- see here.

Mean is $\lambda_1 \lambda_2$, variance is $\lambda_1 \lambda_2(\lambda_2 +1)$.


问:
Identify a graph coloring problem in graph theory
Suppose a directed graph $G=(V,E)$ and an initial  subset of colored nodes $R_{0} \subseteq V$. We then produce the following subsets $R_{1} \subseteq R_{2} \subseteq ..\subseteq R_{n}$ with the following iterative procedure:

$$ R_{1} = R_{0} \cup \{v\;|\; \forall (w,v) \in E: w \in R_{0}\}$$

and generally

$$ R_{i+1} = R_{i} \cup \{v\;|\; \forall (w,v) \in E: w \in R_{i}\}$$

Essentially, at each step of the iteration, we color a node if all of it's incoming edges originate from nodes that have already been colored.

This procedure can have two outcomes, (a) all nodes eventually get colored at a step $j$ of the iteration such that $R_{j}=V$ or (b)  an iteration is reached where no other nodes can get colored and some remain uncolored ($R_{j} = R_{j+1} \subset V$).

Now, the problem is to compute the smallest possible inital set $R_{0}$ such that eventually all nodes will be colored. My questions are:

a) Does this problem have a name in the literature of graph theory?

b) If so, are there any efficient algorithmic solutions?

Thank you!

答:
If the graph is non-oriented (that is, $(u,v) \in E \Leftrightarrow (v,u) \in E$), then you're looking for a vertex cover of the graph. Indeed, if an edge $(u,v)$ has none of its ends in $R_0$, this will be a deadlock preventing both $u$ and $v$ from ever getting in one of the $R_i$ ; and if you have a vertex cover for $R_0$ then $R_1 = V$.

Even in this subclass of graphs, the problem already is NP-hard, so I guess the answer to b) is no. But for the vertex cover problem there are approximation algorithms, so maybe you could look there to find something similar for your problem...


问:
A transformation for independent variables to be dependent
I ran into a confusing question.

If two variables are independent, maybe they will be dependent after linear transformation.

How it can happen? Is it possible for independent variables?

What is the operation that it makes variables to be dependent?

In my opinion the transformation that maps all variables to a point is true for this fact.

What is wrong with my answer?

The mean of Variables are features of data.

答: If you are asking about taking linear transformations of each variable separately, then they will remain independent- see here. If you are asking about linear transformation of both variables together, then the transformed variables may or may not be independent.</p>


问:
Gradient with respect to x?
What is the gradient (with respect to $x \in \mathbb{R}^n$) of $f(\vec{x}) = e^{\vec{x}^TA\vec{x}+b}$ where $A \in \mathbb{R}^{n\times n}$ is a symmetric matrix and $b$ is a real number?

I know that the derivative of $\vec{x}^TA\vec{x}+b$ is $2\vec{x}^TA$, but I am not sure of how to get the gradient of $f(\vec{x}) = e^{\vec{x}^TA\vec{x}+b}$ with respect to $x$?

答:
This is just the chain rule.

Let:

$\alpha: \mathbb{R}^n \rightarrow \mathbb{R}$ such that $\alpha(x)=xAx+b$

$\exp: \mathbb{R} \rightarrow \mathbb{R} $ such that $exp(t)=e^t$

Then you have that

$$ f(x)=(exp \, \circ \, \alpha)(x)  $$

thus, buy chain rule for the gradient

$$ \nabla f (x) = \nabla exp(\alpha(x)) \, \cdot \, \nabla \alpha (x) $$

$exp$ being from the reals to the reals, the first gradient is simply a derivative: $ \nabla exp(t) = \frac{d}{dt}|_t e^t = e^t $, while $ \nabla \alpha (x) = \nabla (xAx+b)(x) = \nabla (xAx) +\nabla b = \nabla xAx = 2Ax $ (the latest term should be proven but you can also take it for granted).

In the end:

$$ \nabla f (x)= 2e^{xAx+b} (A\cdot x)$$ where $ 2e^{xAx+b} $ is a number and $ (A\cdot x) $ is a vector.

Title: Gradient with respect to x?


答:
Recall that adding a multiple or subtracting a multiple of one row does not change the value of the determinant, see, for example ProofWiki.

Using this fact and Laplace expansion you get

$$|A_4|=
\begin{vmatrix}
  1&  1&  1&  1\\
  1&  2&  2&  2\\
  1&  2&  3&  3\\
  1&  2&  3&  4
\end{vmatrix}=
\begin{vmatrix}
  1&  1&  1&  1\\
  0&  1&  1&  1\\
  0&  1&  2&  2\\
  0&  1&  2&  3
\end{vmatrix}=
\begin{vmatrix}
  1&  1&  1\\
  1&  2&  2\\
  1&  2&  3
\end{vmatrix}=|A_3|.$$
(We have subtracted the first row from the other rows.)

In the same way you get $|A_{n+1}|=|A_n|$.

Title: Evaluation of a specific determinant.

Body:
Evaluate $\det{A}$, where $A$ is the $n \times n$ matrix defined by $a_{ij} = \min\{i, j\}$, for all $i,j\in \{1, \ldots, n\}$.

$$A_2
\begin{pmatrix}  1&  1\\
  1&  2
\end{pmatrix};
A_3 = \begin{pmatrix}  1&  1&  1\\
  1&  2&  2\\
  1&  2&  3
\end{pmatrix};
A_4 = \begin{pmatrix}  1&  1&  1&  1\\
  1&  2&  2&  2\\
  1&  2&  3&  3\\
  1&  2&  3&  4
\end{pmatrix};
A_5 = \begin{pmatrix}  1&  1&  1&  1&  1\\
  1&  2&  2&  2&  2\\
  1&  2&  3&  3&  3\\
  1&  2&  3&  4&  4\\
  1&  2&  3&  4&  5
\end{pmatrix}$$

$$A_6 = \begin{pmatrix}  1&  1&  1&  1&  1&  1\\
  1&  2&  2&  2&  2&  2\\
  1&  2&  3&  3&  3&  3\\
  1&  2&  3&  4&  4&  4\\
  1&  2&  3&  4&  5&  5\\
  1&  2&  3&  4&  5&  6
\end{pmatrix};
A_7 = \begin{pmatrix}  1&  1&  1&  1&  1&  1&  1\\
  1&  2&  2&  2&  2&  2&  2\\
  1&  2&  3&  3&  3&  3&  3\\
  1&  2&  3&  4&  4&  4&  4\\
  1&  2&  3&  4&  5&  5&  5\\
  1&  2&  3&  4&  5&  6&  6\\
  1&  2&  3&  4&  5&  6&  7
\end{pmatrix}
$$


答:
(I'll use the notation $f'_+$ and $f'_-$ for the right and left derivative.)

Unfortunately your proof does not work. This
$$
\lim_{\varepsilon\to 0^+}\frac{f(z-\delta+\varepsilon)-f(z-\delta)}{\varepsilon}
\overset{\varepsilon=\delta}{=} \lim_{\varepsilon\to 0^+}\frac{f(z)-f(z-\varepsilon)}{\varepsilon}
$$
makes no sense because $\delta$ is fixed and the limits are taken for $\varepsilon \to 0 $. Also it would imply that $f'_+(z-\delta) = f'_-(z)$ for $0 < \delta < z-y$, which is true only if $f$ is linear between $y$ and $z$.

What we can show is that
$$ \tag{*}
 f'_+(y) \le \frac{f(z)-f(y)}{z-y} \le f'_-(z)
$$
for $y < z$, which implies the desired conclusion. (This is motivated by the fact that both an increasing right derivative and an increasing left derivative imply that $f$ is convex.)

The proof of $(*)$ mimics the proof of Rolle's theorem and the mean-value theorem. We consider the function
$$
 g(x) = f(x) - (x-y)\frac{f(z)-f(y)}{z-y}
$$
which is continuous and satisfies $g(y) = g(z)$. It follows that $g$ attains its maximum on the interval $[y, z]$ at some point $w \in [y, z)$. Then $g'_+(w) \le 0$ and it follows that
$$
f'_+(y) \le f'_+(w) = g'_+(w) + \frac{f(z)-f(y)}{z-y} \le \frac{f(z)-f(y)}{z-y} \, .
$$
This proves the left inequality in $(*)$, the proof of the right inequality works similarly.

Title: Right versus left derivative by increasing one-sided derivatives

Body:
Condider a map $f:D\to\mathbb{R}$, $D$ open interval.
Let be $y,z\in D$, $y<z$. Suppose that $f$ has everywhere in $D$ both left and right derivative (this implies that $f$ is continuous), both increasing.
I want to show that
$$
f'(y^+)\le f'(z^-).
$$

I have found a way, but I'm not completely convinced that it is formally correct:
$$
f'(y^+)
=\lim_{\varepsilon\to 0^+}\frac{f(y+\varepsilon)-f(y)}{\varepsilon}
\overset{\forall\delta}{\le}
  f'((z-\delta)^+)
= \lim_{\varepsilon\to 0^+}\frac{f(z-\delta+\varepsilon)-f(z-\delta)}{\varepsilon}
\overset{\varepsilon=\delta}{=} \lim_{\varepsilon\to 0^+}\frac{f(z)-f(z-\varepsilon)}{\varepsilon}
=f'(z^-)
$$
where $\delta>0$ and $y<z-\delta$.

Is there a way to formalise my proof?


问:
Modulus of continuity of Brownian Motion, independent of particular choice of Brownian Motion
Suppose I can show for one Brownian Motion $(B(t))$ on a probability space $(\Omega,\mathscr{A},\mathbb{P})$ that there exists a constant $C>0$ and a set $\mathcal{N}\in \mathscr{A}$ with $\mathbb{P}(\mathcal{N}) = 0$ such that $\omega\notin \mathcal{N}$ implies that we can find $\delta(\omega)$ such that $0<h<\delta(\omega)$ implies that
$$|B(t+h)-B(t)|\leq C\sqrt{h\log(1/h)}$$
for every $t\in [0,1-h]$. Then how can I show that this property and constant is independent of the particular choice of Brownian motion. I.e. if $(\tilde{B}(t))$ is another Brownian Motion then we can find a new set $\tilde{\mathcal{N}}$ and bound $\tilde{\delta}$  but such that the rest carries over.

My interest in this comes from the fact that it is easy to prove the continuity condition for a Brownian Motion stemming from the Levy construction however in order to prove it for all Brownian Motions it needs to be shown to be invariant.

My attempt: Suppose that $(B(t))$ is a Brownian Motion which satisfies the continuity condition. Let $a>1$. Then we know that
$$\mathbb{P}\left(\limsup_{h\rightarrow 0}\sup_{0\leq t\leq 1-h}\frac{|B(t+h)-B(t)|}{C\sqrt{h\log(1/h)}}>a\right)=0.$$

So what we need now is to show that

$$\mathbb{P}\left(\limsup_{h\rightarrow 0}\sup_{0\leq t\leq 1-h}\frac{|B(t+h)-B(t)|}{C\sqrt{h\log(1/h)}}>a\right) = \mathbb{P}\left(\limsup_{h\rightarrow 0}\sup_{0\leq t\leq 1-h}\frac{|\tilde B(t+h)-\tilde B(t)|}{C\sqrt{h\log(1/h)}}>a\right)$$
but this is where I get stuck. Any tip on how to proceed would be appreciated.

答:
The Brownian motion $B$ induces a measure $\mu$ ("Wiener measure") on the measurable space $(C[0,1],\mathcal C)$ via the map $\Phi:\Omega\to C[0,1]$ defined by $\Phi(\omega) := (B(t,\omega))_{0\le t\le 1}$. The event
$$
G:=\left\{x\in C[0,1]: \limsup_{h\to 0}\sup_{0\le t\le 1-h}{|x(t+h)-x(t)|\over C\sqrt{h\log(1/h)}}>a\right\}
$$
is an element of $\mathcal C$ and
$$
\mu(G)=\Bbb P(\Phi^{-1}(G))=\mathbb{P}\left(\limsup_{h\rightarrow 0}\sup_{0\leq t\leq 1-h}\frac{|B(t+h)-B(t)|}{C\sqrt{h\log(1/h)}}>a\right) = 0.
$$
Because $\tilde B$ is also Brownian motion, it induces the same measure $\mu$, and so
$$
\mathbb{P}\left(\limsup_{h\rightarrow 0}\sup_{0\leq t\leq 1-h}\frac{|\tilde B(t+h)-\tilde B(t)|}{C\sqrt{h\log(1/h)}}>a\right) = \mu(G) =0,
$$
as well.


问:
Is there something similar to a discriminant for algebraic curves of the form $y^2 = f(x)$ (similar to the discriminant of elliptic curves)?
Consider an algebraic curve $X$ defined by an equation $y^2 = f(x)$ where $f$ is a polynomial of degree $d \geq 3$ with coefficients in a field $k$.

For $m = 2$ and $d = 3$, we can get an equation of the form $f(x) = x^3 + Ax + B$ (if the characteristic of $k$ is not $2$ or $3$). Then for this case, a general result says that
$$
\Delta \neq 0 \Longleftrightarrow X \text{ is non-singular (i.e. an elliptic curve)}.
$$

Question: Is there something similar to a discriminant of $X$ for polynomials $f$ of arbitrary degrees? (i.e. $X$ is a superelliptic curve)

答:
If $k$ is any field and $F(x,y):=y^n-f(x)$ with $n\geq 2$ an integer and $deg(f)=d$, you may consider the algebraic curve $C:=Spec(A)$ where $A:=k[x,y]/(F(x,y))$. The jacobian ideal $J$ given by

$$J:=(ny^{n-1},-f'(x))$$

defines the singular subscheme $C_{sing}:=V(J) \subset C$. Assume for simplicity $char(k) \neq n$. Let $(a,b)\in C_{sing}(K)$ be a $K$ rational point of $C_{sing}$, where $k\subseteq K$ is a field. It follows $a,b\in K$ and $f(a)=f'(a)=0$.
Hence the curve $C$ has a $K$-rational singular point $(a,0)\in K^2$ iff

$$f(a)=f'(a)=0.$$

If $f(x):=a_0+a_1x+\cdots +a_dx^d$ we define the discriminant $\Delta(f)$ to be the resultant $\Delta(f):=Res(f(x),f'(x))$ of the polynomial $f(x)$ and it's derivative $f'(x)$. It follows $\Delta(f)=0$ iff $f(x)$ has a multiple root in $K$, where $K$ is a finite extension of $k$. Hence the discriminant $\Delta(f)=0$ iff the curve $C$ has a $K$-rational singular point, where $k\subseteq K$ is a finite extension.

Question: "Is there something similar to a discriminant of X for polynomials f of arbitrary degrees? (i.e. $X$ is a superelliptic curve)?"

Answer: Yes, it seems to me you may use the classical discriminant $\Delta(f)$ of the polynomial $f(x)$ as explained above. If $char(k) \neq n$ it follows $\Delta(f)=0$ iff $C_{sing}(K) \neq \emptyset$ for some finite extension $k\subseteq K$.

Note: The discriminant $\Delta(f):=Res(f(x),f'(x))$ is a polynomial in the coefficients of $f(x)$, hence you can calculate  $\Delta(f)$ without knowing anything on the roots of $f(x)$ or if the roots of $f(x)$ live in a finite extension $k \subseteq K$. There is a well known determinantal formula for the resultant $Res(f(x),g(x))$ of any pair of polynomials $f(x),g(x)$ - the "Sylvester determinant".

As remarked in the comments: If $k$ is not algebraically closed you may get singular points defined over non-trivial extensions $k \subseteq K$. The curve $C$ is finite type over $k$ and closed points $x\in C$ have residue field $k \subseteq \kappa(x)$ which is a finite extension in general. Hence points of $C_{sing}$ are not in general defined over the base field $k$.

Example: If $k=\mathbb{Q}$ is the field of rational numbers and $f(t):=g(t)^2$ where $g(t):=t^2-2$. Let $F(x,y):=y^n-f(t)$. It follows

$$f'(t)=2g(t)g'(t)=4t(t^2-2) \in k[t].$$

Hence $f(t)=0$ and $f'(t)=0$ does not have a solution in $k$. It has a solution in $k(\sqrt{2})$. This phenomenon also happens in characteristic $p>0$.

Note: If the polynomial $f(x)\in k[x^p]$ is a polynomial in $x^p$ it follows $f'(x)=0$ hence $\Delta(f)=Res(f(x), f'(x))$ is identically zero in characteristic $p>0$. Over the algebraic closure $\overline{k}$ it follows $f(x^p)=g(x)^p$, hence $f(x)$ has multiple roots and the curve is singular.


问:
Proving $f(x) = x^2 \sin(1/x)$, $f(0)=0$ is differentiable at $0$, with derivative $f'(0)= 0$ at zero
I need a solution for this question. I've been trying out this question for days and I haven't been able to find out its solution yet. And some explanation would help too.

Show that the function f defined by:
  $$f(x):= \begin{cases}
  x^2\sin(1/x) &:\text{if $x \ne 0$} \\
  0 &:\text{if $x=0$}
\end{cases}$$
  is differentiable at $x=0$, and that $f'(0)=0$.

答:
Hint: We have $-x^2 \le f(x) \le x^2$, hence
$$ -x \le \frac{f(x) - f(0)}{x-0} = \frac{f(x)}x \le x $$
for $x \ne 0$.


答: Sum over repeated indices: since$$\partial_if=f\partial_i\ln f=f\partial_i(x_jx_kA_{jk}+b)=(\delta_{ij}x_kA_{jk}+x_j\delta_{ik}A_{jk})f=(x_kA_{ik}+x_jA_{ji})f,$$it follows $\nabla f=f(A+A^T)x$. Since $A$ is symmetric, this simplifies to $\nabla f=2fAx$.

Title: Gradient with respect to x?

Body:
What is the gradient (with respect to $x \in \mathbb{R}^n$) of $f(\vec{x}) = e^{\vec{x}^TA\vec{x}+b}$ where $A \in \mathbb{R}^{n\times n}$ is a symmetric matrix and $b$ is a real number?

I know that the derivative of $\vec{x}^TA\vec{x}+b$ is $2\vec{x}^TA$, but I am not sure of how to get the gradient of $f(\vec{x}) = e^{\vec{x}^TA\vec{x}+b}$ with respect to $x$?


问:
Prove that the integral of a differential form on an open set is well defined.
What to follow is a summary of part of the sixth chapter of the text Analysis on Manifolds by James Munkres.

Definition

Given $x\in\Bbb R^n$ we define a tangent vector to $\Bbb R^n$ at $x$ to be a pair $(x;\vec v)$ where $\vec v\in\Bbb R^n$. The set of tangent vector to $\Bbb R^n$ at $x$ forms a vector space if we define
$$
(x;\vec v)+(x;\vec w):=(x;\vec v+\vec w)\\
c(x;v\vec):=(x;c\vec v)
$$
It is called the tangent space to $\Bbb R^n$ at $x$ and is denoted by the symbol $\mathcal T_x(\Bbb R^n)$.

Definition

Let $A$ be open set in $\Bbb R^n$. A $k$-tensor field in $A$ is a function $\omega$ assigning to each $x\in A$ a $k$-tensor defined on the vector space $\mathcal T_x(\Bbb R^n)$, that is
$$
\omega(x)\in\mathcal L^k\big(\mathcal T_x(\Bbb R^n)\big)
$$
for each $x\in A$. In particular if it happens that $\omega(x)$ is an alternanting $k$-tensor for each $x\in A$ then $\omega$ is called a differential form of order $k$ on $A$.

Now if the vectors $\vec v_1,...,\vec v_n$ form a basis of $\Bbb R^n$ then the vectors $(x;\vec v_1),...,(x;\vec v_n)$ form a bsis of $\mathcal T_x(\Bbb R^n)$ and thus we define a $1$-form $dx^i$ on $\Bbb R^n$ by letting that
$$
[dx^i(x)](x;\vec v_j)=\delta^i_j
$$
for any $x\in\Bbb R^n$ and for any $j=1,...,n$. Well it is possible to prove that if $\omega$ is a $k$-form defined in any open set $A$ then
$$
\sum_{i_1<...<i_k}f_{i_1,..,i_k}dx^{i_1}\curlywedge...\curlywedge dx^{i_k}
$$
where $f_{i_1,...,i_k}$ is for any ascending $k$-tuple of indices $i_1,...,i_k$ is a scalar function defined on $A$.

Definition

Let $A$ be open in $\Bbb R^k$; let be $\alpha:A\rightarrow\Bbb R^n$ be of class $C^\infty$; let be $B$ and open set of $\Bbb R^n$ containing $\alpha[A]$. We define a dual transformation of forms
$$
\alpha^*:\Omega^l(B)\rightarrow\Omega^l(A)
$$
as follows: given a $0$-form $f:B\rightarrow\Bbb R$ on $B$ we define a $0$-form $\alpha^* f$ on $A$ by setting $\alpha^* f:=f\circ\alpha$ and given an $l$-form on $B$ with $l>0$ we define an $l$-form $\alpha^*\omega$ on $A$ by the equation
$$
[\alpha^*\omega(x)]\big((x;v\vec v_1),...,(x;\vec v_l)\big):=[\omega(\alpha(x)]\big((\alpha(x);D\alpha(x)\cdot\vec v_1),...,(\alpha(x);D\alpha(x)\cdot\vec v_l)\big)
$$
for any $x\in A$ and for any $\vec v_1,...,\vec v_k\in\Bbb R^n$.

Definition

Let $A$ be an open set in $\Bbb R^k$ and let be $\eta$ a $k$-form defined in $A$ so that it can be written uniquely in the form
$$
\eta=fdx^1\curlywedge...\curlywedge dx^k
$$
for any scalar function $f$ defined on $A$. So we define the integral of $\eta$ over $A$ by the equation
$$
\int_A\eta:=\int_A f
$$
provided the latter integral exists.

This definition seems to be coordinate-dependent: indeed the function $f$ dependes to the choice of a basis in $\Bbb R^n$. One can, however, formulate the definition in a coordinate-free fashion: specifically if $\vec v_1,...,\vec v_k$ is any right-handed orthonormal basis for $\Bbb R^k$ tehn it is an lemenetary exercise to show that
$$
\int_A\eta=\int_{x\in A}[\eta(x)]\big((x;\vec v_1),...,(x;\vec v_k)\big)
$$
Thus the integral of $\eta$ does not depend on the choice of any orthonormal basis in $\Bbb R^k$ although it does depend on the orientation of $\Bbb R^k$.

Definition

Let $A$ be open in $\Bbb R^k$; let $\alpha A\rightarrow\Bbb R^n$ be a $k$-parametrized manifold. If $\omega$ is a $k$-form defined in an open set of $\Bbb R^n$ containing $Y_\alpha:=\alpha[A]$ then we define the integral of $\omega$ on $Y_\alpha$ by letting
$$
\int_{Y_\alpha}\omega:=\int_A\alpha^*\omega
$$
So now I would like to know if it is possible to define the integral of a differential form on an open set independently of any basis and not only for orthonormal bases.

So if $dx^{1}\curlywedge...\curlywedge dx^{k}$ and $dy^{1}\curlywedge...\curlywedge dy^k$ are the $k$-elementary form with respect two any different bases $\mathcal V:=\{\vec v_1,...,\vec v_k\}$ and $\mathcal W:=\{\vec w_1,...,\vec w_k\}$ then there exist two scalar functions $f$ and $g$ defined on $A$ such that
$$
f dx^1\curlywedge...\curlywedge dx^k=\eta=g dy^1\curlywedge...\curlywedge dy^k
$$
so that they are related by the equation
$$
g(x)=[\eta(x)]\big((x;\vec w_1),...,(x;\vec w_k)\big)=\\ [(f\cdot dx^1\curlywedge...\curlywedge dx^k)(x)]\big((x;\vec w_1),...,(x;\vec w_k)\big)=...=f(x)[dx^1(x)\wedge...\wedge dx^k(x)]\big((x;\vec w_1),...,(x;\vec w_k)\big)=\\
f(x)\det\begin{pmatrix}[dx^1(x)](x;\vec w_1)&&\cdots&&[dx^1(x)](x;\vec w_k)\\
\vdots&&\ddots&&\vdots\\
[dx^k(x)](x;\vec w_1)&&\cdots&&[dx^k(x)](x;\vec w_k)\end{pmatrix}=f(x)\det\begin{pmatrix}w^1_1&&\cdots&&w^1_k\\
\vdots&&\ddots&&\vdots\\
w^k_1&&\cdots&&w^k_k\end{pmatrix}
$$
where $w^i_j$ are the components of the basis $\mathcal W$ with respect the basis $\mathcal V$. So by the previous equation clearly we conclude that generally if the two bases are not orthonormal then
$$
\int_A g=\det\begin{pmatrix}w^1_1&&\cdots&&w^1_k\\
\vdots&&\ddots&&\vdots\\
w^k_1&&\cdots&&w^k_k\end{pmatrix}\cdot\int_A f\neq\int _A f
$$
and thus the integral of $\eta$ it is not uniquely determined. Anyway I tried to expalin this using the following theorem

Theorem

Let $g:A\rightarrow B$ be a diffeomorphism of open sets in $\Bbb R^k$. Assume $\det Dg$ does not change sign on $A$. Let $\beta: B\rightarrow\Bbb R^n$ be a map of class $C^\infty$; let $Y_\beta:=\beta[B]$. Let $\alpha:=\beta\circ g$ and thus $Y_\alpha:=\alpha[A]$. If $\omega$ is a $k$-form defined in an open set of $\Bbb R^n$ containing $Y$ then $\omega$ is integrable over $Y_\beta$ if and only if it is integrable over $Y_\alpha$ and in this case
$$
\int_{Y_\alpha}\omega=\pm\int_{Y_\beta}\omega
$$

but unfortunately I did not conclude anything. However I am sure that using the change of variables theorem it is possible to prove that the above two quantities are equal unless the sign because the above matrix is the matrix of the identity map with respect the basis $\mathcal V$ and $\mathcal W$. Indeed we suppose that $\mathcal W$ is the standard basis of $\Bbb R^k$ and thus we define the diffeormorphism $\phi:\Bbb R^k\rightarrow\Bbb R^k$ through the position
$$
\phi(x):=x^iw^j_i\vec v_j=x^i\vec w_i=x
$$
for any $x\in\Bbb R^n$ so that by the change of variable theorem it follows that
$$
\int_{x\in A} f=\int_{y\in\phi^{-1}[A]}(f\circ \phi)|\det D\phi|=\int_{y\in A} f|\det[w^i_j]|=\pm\int_{y\in A} g
$$
as we desired.
However, I fear that there is some error in the above identity. So could anyone give some clarification about this, please?

答:
Consider the function $f=1$ and the region $A\subset\Bbb R^2$ to be the rectangle spanned by $(1,0)$ and $(0,2)$. I consider the linear mapping $T(x^1,x^2)=(x^1,2x^2)=(y^1,y^2)$ mapping the unit square $Q$ to $A$. Then we have $\omega = dy^1\wedge dy^2 = 2\,dx^1\wedge dx^2$. What the second equals sign means, more carefully, is that $T^*\omega = 2\,dx^1\wedge dx^2$, as officially these $2$-forms are defined on different versions of $\Bbb R^2$. Now,
\begin{multline*}
2 = \int_A  dy^1\wedge dy^2 = \int_{T(Q)} dy^1\wedge dy^2 = \int_Q T^*(dy^1\wedge dy^2) \\= \int_Q 2\,dx^1\wedge dx^2 = 2\cdot 1.
\end{multline*}
This generalizes to any invertible linear map, of course.


问:
How to identify the independent and dependent variables in a differential equation?
I had doubt regarding the equation $y^4 = x^5$. Which one is the dependant and the independent variable here? Thanks in advance.

答:
Typically, we use $y$ as the dependent variable, but this doesn't always have to be the case. One could interpret the above equation as
$$y(x)=x^{5/4}$$
Or as
$$x(y)=y^{4/5}$$

Can't really say much more without context.


问:
Taylor Approximations
I'm learning about Taylor approximations and I must be missing something. The textbook I'm working with shows that a function $f(x)$ can be approximated near $z$ like this:

$\hat{f}(x) = f(z) + \nabla f(z)^T (x - z)$

where $\nabla f(z)$ is an $n$-vector, the gradient of $f$ at the point $z$.

My question is, if we know $f(x)$ and it's differentiable, why would we need to approximate it? Why can't we just work directly with $f$?

答:
Suppose you want to know how much is
$e^{0.35}$.

How to do? I don't know and nobody else knows.

With Taylor you expand $e^x$ around $x=0$ and get
$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+O\left(x^5\right)$$
Thus you can compute, with a little patience if you libe in 18th century,

$$e^{0.35}\approx 0.35\, +1+\frac{0.35^2}{2}+\frac{0.35^3}{6}+\frac{0.35^4}{24}$$

Title: Taylor Approximations


问:
Being inside or outside of an ellipse
Let $A$ be a point $A$ not belonging to an ellipse $E$. We say that $A$ lies inside
$E$ if every line passing trough $A$ intersects $E$. We say that $A$ lies otside $E$
if some line passing trough $A$ does not intersect $E$. Let $E$ be the ellipse
with semi-axes $a$ and $b$. Show that

(a) a point $A=(x,y)$ is inside $E$ if $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}< 1$

(b) a point $A=(x,y)$ is outside $E$ if $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}> 1$.

I wrote down the equation of an arbitrary line $L$ passing through $A$. If $A$ is inside $E$ then clearly there is another point, say $B$, that belongs to $A\cap L$. I ended up with a bunch of big equations which didn't give any result.

答:
Case $a:$

Let the point be $P(h,k)$ and any arbitrary line passing through $P$ be $y=mx+c$

$\implies k=m\cdot h+c\iff c=k-m\cdot h$

Let us find the intersection of the line with the given ellipse

$${So,}\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\implies \dfrac{x^2}{a^2}+\dfrac{(mx+c)^2}{b^2}$$

$$\text{On simplification,}x^2(b^2+m^2)+2a^2cm x+a^2(c^2-b^2)=0$$

For intersection, there will be $2$ distinct values of $m$

i.e., the discriminant $(2a^2cm)^2-4(b^2+m^2)a^2(c^2-b^2)=4a^2(a^2m^2+b^2-c^2)>0$

i.e., $a^2m^2+b^2-c^2>0$

Putting the value of $c, a^2m^2+b^2-(k-m\cdot h)^2>0$

Putting $m=0, b^2-k^2>0$

Put $m=\frac1n, \frac{a^2}{n^2}+b^2-(k- \frac hn)^2>0 \iff a^2+b^2n^2-(k\cdot n-h)^2>0$

Putting $n=0, a^2-h^2>0$

$$\implies (b^2-k^2)(a^2-h^2)>0 \implies a^2b^2> a^2k^2+ b^2h^2\implies 1> \frac{k^2}{b^2}+\frac{h^2}{a^2}$$

Case $b:$ there will be no intersection i.e.,  the discriminant $<0$

Title: Being inside or outside of an ellipse


问:
Hierarchy in logical systems
I had an informal conversation in which I was told that logical systems could be intuitively drawn in a hierarchy according to their expressive power, i.e. the amount of things we can prove with them. This hierarchy, would be infinitely dense, but some points have been studied and have special names. Here is a list of names in order of strength:

Presburger arithmetic

Presburger arithmetic with bitwise operators

-------- we end the decidable side

Robinson arithmetic

First-order peano arithmetic

System T

System F (maybe System F is not strictly stronger than System T?)

---------big gap: set theories are much more powerful

Zermelo's set theory (somewhat equivalent to higher-order logic in
power)

Fraenkel axioms allowing to speak about certain ordinals

Axioms of existence of large cardinals............

I want to know more about this hierarchy. Is there a comprehensive study or some sample analysis of the hierarchy?

In that respect, Is there an overview of several logical systems? does not count as a duplicate since the focus there is non-classical logics, here I think everything is fairly classical.

答:
I don't know of a good source which covers this whole spectrum. That said, there are a variety of sources covering different pieces of it. Below I'll describe how I view this hierarchy of systems, and say specifically which texts I have in mind - texts by Hajek/Pudlak, Simpson, Mathias, Jech, and Kanamori. (I've copied the links below as well, so you don't have to bounce around in this answer.)

I'm not an expert (to put it mildly) in decidable theories or even "ultraweak" theories like $\mathsf{Q}$, so I'll only go as low as theories which admit "reasonably simple" coding of finite sequences. Moreover, I'll only consider three languages for the theories in question: first-order arithmetic (e.g. $\mathsf{PA}$) possibly augmented with additional symbols for natural functions like exponentiation, second-order arithmetic (e.g. $\mathsf{RCA_0}$), and set theory (e.g. $\mathsf{ZFC}$).

Annoyingly, in this context second-order arithmetics are first-order theories. "Second-order" refers to treating both natural numbers and sets of natural numbers; "two-sorted arithmetic" or "theory of analysis" (which was used early on), would be better terms, but sadly that's not how the terminology wound up.

Here, then, is the hierarchy as I perceive it:

We begin with the "not-too-weak" first-order arithmetics. Personally I'd draw the line at $\mathsf{I\Delta_0+exp}$; this theory consists of the basic "$+/\times/<$" properties of $\mathbb{N}$ without induction, the induction scheme restricted to $\Delta_0$ formulas, and an axiom asserting that exponentiation is total and satisfies the usual recursive definition. This is the weakest natural theory of arithmetic which allows "easy" coding of finite sequences. Climbing up from here we get theories like $\mathsf{I\Sigma_n}$ (the basic properties plus $\Sigma_n$-induction) and $\mathsf{B\Sigma_n}$ (a more technical variant of $\mathsf{I\Sigma_n}$) which satisfy $$...<\mathsf{I\Sigma_n<B\Sigma_{n+1}<I\Sigma_{n+1}}<...$$ These climb all the way up to $\mathsf{PA}$, and $\mathsf{PA}$ proves "$\mathsf{I}\Sigma_n$ is consistent" for each $n$ (although of course $\mathsf{PA}$ does not prove "$\mathsf{I\Sigma_n}$ is consistent for each $n$"). The standard text here is Metamathematics of first-order logic.

$\mathsf{PA}$ is in my opinion the strongest "natural" theory of first-order arithmetic, so going above it I'll switch to the context of second-order arithmetic. A relevant term here is reverse mathematics. Typically the weakest theory considered here is $\mathsf{RCA_0}$, which is conservative over $\mathsf{I\Sigma_1}$; a couple theories up we get $\mathsf{ACA_0}$, which is conservative over $\mathsf{PA}$. Then higher up we get things like $\mathsf{ATR_0}$ which are much stronger than $\mathsf{PA}$. The strongest common theory in reverse mathematics is $\Pi^1_1$-$\mathsf{CA_0}$, and then we keep going through higher levels of the second-order comprehension hierarchy. At the top of these sits $\mathsf{Z_2}$, which has comprehension for all second-order formulas and bears essentially the same relation to the $\Pi^1_n$-$\mathsf{CA_0}$s as $\mathsf{PA}$ does to the $\mathsf{I\Sigma_n}$s. The standard text here is Subsystems of second-order arithmetic.

At this point we switch over to set theory. As with the transition between first- and second-order arithmetics, things "glue together" long before we need to make the shift: the weakest set theory I know much about here is $\mathsf{KP\omega}$ (Kripke-Platek + Infinity), which is weaker in consistency strength than $\mathsf{ATR_0}$ (which is itself weaker than $\Pi^1_1$-$\mathsf{CA_0}$). That said, there is also the surprisingly-weak set theory $\mathsf{Z}$ (= $\mathsf{ZFC}$ without $\mathsf{Replacement}$ or, less importantly, $\mathsf{Choice}$). Mathias has an epic paper about these systems, but it's quite technical.

Then there's a big gap.

The gulf between things like $\mathsf{Z+KP}$ and $\mathsf{ZF}$ is ridiculously huge, and I don't know a good source that covers that range. There are some famous results here, e.g. Harvey Friedman's analysis of the role of $\mathsf{Replacement}$ in Borel determinacy, but for the most part this is territory I don't know anything about. My knowledge kicks back in at $\mathsf{ZF}$. Of course here we have a huge amount of robustness, e.g. the agreement in consistency strength between $\mathsf{ZF}$, $\mathsf{ZFC+V=L}$, $\mathsf{ZF+DC}$ + "Every set of reals has the Baire property," and so on. The study of e.g. fragments of the axiom of choice over $\mathsf{ZF}$ doesn't play with consistency strength at all, although of course we see huge changes in what we can prove in general.

And then ...

Large cardinals.

Yippee!

There are two standard texts here: Kanamori's Higher infinite and Jech's Set theory. Very roughly, there are three "blocks" of large cardinal axioms:

The small large cardinals. Basically, these are the ones compatible with $\mathsf{V=L}$. They include inaccessibles, Mahlos, weakly compacts, and so on - more-or-less at the top of this are the remarkable cardinals.

The medium large cardinals. The natural starting point here is $\mathsf{0^\sharp}$, which technically isn't a large cardinal but fits the picture; a bit higher is the easier-to-explain measurable cardinal, and a fair way above are the Woodin cardinals. Strength-wise, the characterizing feature of these cardinals I have in mind is that, while incompatible with $\mathsf{V=L}$, they still have well-understood inner model theory. However, there's another important phase transition which happens here: large cardinals at this and the next level are connected with "large-scale" model-theoretic phenomena, even if they have "local" characterizations. E.g. we can prove (in an appropriate conservative extension of $\mathsf{ZFC}$) that there is a measurable cardinal iff there is a nontrivial elementary embedding of $V$ into some inner model, and this was in fact how Scott proved that measurables are incompatible with $\mathsf{V=L}$.

And then we get the large large cardinals, which are beyond the reach of current inner model theory. Supercompacts and rank-into-rank cardinals are perhaps the most famous, although I also want to plug Vopenka's principle here, if only because of its hilarious history.

Title: Hierarchy in logical systems


问:
Proof that Expression Involving Moment Generating Function is Expectation
Suppose we have a random variable (RV) $X$ defined on a measurable space $\mathcal{M} = (\Omega, \Sigma)$. Suppose we equip the measurable space with a probability measure $P$ and associated expectation operator $E$ such that for all $\theta \in \mathbb{R}$ we have $E[e^{\theta X}] < \infty$. Then, for all bounded, continuous $f : \mathbb{R} \rightarrow \mathbb{R}$ define $F_{\theta}$ as follows:

$$F_{\theta}(f) = \dfrac{E[f(X)e^{\theta X}]}{E[e^{\theta X}]}$$

Question: Show that there exists a probability measure $P_{\theta}$ defined on $\mathcal{M}$ such that for all bounded continuous functions $f$ the following identity holds:

$$F_{\theta}(f) = E_{\theta}[f(X)]$$

where $E_{\theta}[Y]$ is the expectation of any RV $Y$ w.r.t. $P_{\theta}$.

Background: Around 45 minutes into this video, the lecturer defines an operator $E_{\theta}$ w.r.t to a moment generating function and makes a brief argument as to why this operator is in fact the expectation of an implicitly defined probability measure $P_{\theta}$. I can't see how this conclusion can be made.

Note: If the above result is, or relies on, some known theorem I'd be delighted if someone could point me in the direction of that.

Note: in expressions such as $f(X)$ and $e^{\theta X}$ we are composing the RV $X$ with continuous functions to build new RVs. Also in the expression $f(X)e^{\theta X}$ we use both composition with $X$ and mutliplication of two RVs to get a new RV.

Note: I'm not sure if maybe we need to restrict $\theta > 0$. I'm thinking about it...

答:
Using the definition of expectation and the transfer theorem, we have that :

$$ F_{\theta}(f) := \frac{\mathbb{E}[f(X)e^{\theta X}]}{\mathbb{E}[e^{\theta X}]} = \frac{1}{\mathbb{E}[e^{\theta X}]} \int_{\omega \in \Omega} f(X(\omega))e^{\theta X(\omega)}d\mathbb{P}(\omega) \tag{1}$$

Similarly, we can compute $\mathbb{E}_{\theta}[f(X)]$ :

$$ \mathbb{E}_{\theta}[f(X)] = \int_{\omega \in \Omega} f(X(\omega)) d\mathbb{P}_{\theta}(\omega) \tag{2} $$

If (1) and (2) are equal, we have the equality :

$$\int_{\omega \in \Omega} f(X(\omega)) d\mathbb{P}_{\theta}(\omega) = \int_{\omega \in \Omega} f(X(\omega))\frac{e^{\theta X(\omega)}d\mathbb{P}(\omega)}{\mathbb{E}[e^{\theta X}]} \tag{3} $$

A sufficient condition for equality (3) to hold is to have

$$ d\mathbb{P}_{\theta}(\omega) := \frac{e^{\theta X(\omega)}}{\mathbb{E}[e^{\theta X}]}d\mathbb{P}(\omega) \quad \forall \omega \in \Omega \tag{4} $$

Or, written differently :

$$ \mathbb{P}_{\theta}(A) := \int_{\omega \in A} \frac{e^{\theta X(\omega)}}{\mathbb{E}[e^{\theta X}]}d\mathbb{P}(\omega) \quad  \forall A\in \Sigma \tag{4'}$$

Now if condition (4) is fulfilled, you can check that $\mathbb{P}_{\theta}$ defines a probability measure on $(\Omega,\Sigma) $ and that it is by construction such that $F_{\theta}(f) = \mathbb{E}_{\theta}[f(X)]$.

Intuitively, what we want is to find a measure $\mathbb{P}_{\theta}$ under which the expectation is such that $\mathbb{E}_{\theta}[f(X)] = \frac{\mathbb{E}[f(X)e^{\theta X}]}{\mathbb{E}[e^{\theta X}]} $. The naïve way one might try to find such a measure would be to simply "pull out" the $e^{\theta X}$ in the expectation term and set $``\mathbb{P}_{\theta} = \frac{e^{\theta X}}{\mathbb{E}[e^{\theta X}]} \times \mathbb{P}"$. The law of the unconscious statistician (which has quite a fitting name here) allows to write that intuition in a more formal way and shows that it is (somewhat) correct.


问:
Is $\bar{\mathbb{Q}}/\mathbb{Q}$ finitely generated field extension?
Sorry for my bad English.

I know $\bar{\mathbb{Q}}/\mathbb{Q}$ is infinitely extension. Namely, $[\bar{\mathbb{Q}}:\mathbb{Q}]=\infty.$

Now I want to know is $\bar{\mathbb{Q}}/\mathbb{Q}$  finitely extension as field extension? Namely, are there $\alpha_1,\dots,\alpha_n\in \bar{\mathbb{Q}}$ such that $\bar{\mathbb{Q}}=\mathbb{Q}(\alpha_1,\dots,\alpha_n)$?

Please tell me proof or hint or good book, thanks.

答: Hint: $\mathbb Q(\alpha_1, ... , \alpha_n) = \mathbb Q(\alpha_1, ... , \alpha_{n-1})(\alpha_n)$.  If $F$ is a field, and $\alpha$ is algebraic over $F$, what can you say about $[F(\alpha) : F]$?


问:
Evaluate $\int^{441}_0\frac{\pi\sin \pi \sqrt x}{\sqrt x} dx$
Evaluate this definite integral:
$$\int^{441}_0\frac{\pi\sin \pi \sqrt x}{\sqrt x} dx$$</p>

答:
This integral (even the indefinite one) can be easily solved by observing:

$$\frac{\mathrm d}{\mathrm dx}\pi\sqrt x = \frac{\pi}{2\sqrt x}$$

which implies that:

$$\frac{\mathrm d}{\mathrm dx}\cos\pi\sqrt x = -\frac{\pi \sin\pi\sqrt x}{2\sqrt x}$$

Finally, we obtain:

$$\int\frac{\pi\sin\pi\sqrt x}{\sqrt x}\,\mathrm dx = -2\cos\pi\sqrt x$$

whence the definite integral with bounds $0, n^2$ evaluates to $2(1-(-1)^n)$.


问:
Genus of a graph and planarity
In 'Introduction to graph theory' by Trudeau it is proved the theorem

Every graph has a genus

by first embedding the graph onto a sphere , and then adding handles, say we use $k$ of them, where the graph has edge-crossing , this is the example used :

Then, he writes that using a continuous deformation of the surface (stretching,shrinking or distorting the surface) it is possible to preserve planarity while deforming the $k$-handles sphere into a $k$-hole torus. So, the genus will be at most $k$.

I'm not able to visualize the bold part, how can I be sure that planarity will be preserved? Should it be obvious? Is there a simple explanation to understand clearly why this happens ? (I'm totally a beginner in algebraic topology, just visualization of trasformations are used in the book).

Another proof of a theorem stated earlier in the book : "every planar graph has genus $0$ (it can be embedded on a sphere)" was obscure to me but then searching online I found the explanation through the stereographic projection and all my doubts were gone , does something similar exist? Or even a good intuition is difficult to set up ?

EDIT :

For clarity I add the definition of genus given in the book :

The genus of a graph, denoted $g$, is the subscript of the first surface among the family $S_0,S_1,S_2,\dots$ on which the graph can be drawn without edge crossing .

I also add the explanation of the author to pass from the $k$-handle sphere to the $k$-hole thorus :

Think of the surface consisting of $S_0$ with $k$ handles as made of extremely pliable rubber which we are free to stretch, shrink or otherwise distort as much as we want , provided that we don't tear it off of fold it back onto itself. With this understanding we see that the surface consisting of $S_0$ with $k$ handles can be 'continuously deformed' into $S_k$ and the vertices and edges of $G$ can be carried along with this continuos deformation. Thus $G$ can be drawn on $S_n$ without edge-crossings.

The bold part is obscure to me in the sense explained above.

答:
Maybe I found a satisfactory not much mathematical way for myself to explain what's going on.

Take a nonplanar graph on a sphere with $1$ unavoidable intersection, then you can add a bridge to delete the intersection, so now if we are able to imagine the trasformation in a torus we are done in the case of one intersection. Take the bridge and make it say an half ring, this is simple enough for me to see that it preserves planarity, now we will have something like a kettlebell, now get bigger the half ring so that it incorporates the sphere, here we have the torus!

Now, say we have a graph with $k$ unavoidable intersections , then we can add $k$ bridges one by one (applying the trasformation above each time) so at most the genus is $k$.

Title: Genus of a graph and planarity


问:
How to find whether point is maxima/minima/saddle point for this case?
I read that we have a formula to check signs of

$f_{xx}*f_{yy}-(f_{xy})^2$      and       $f_{xx}$

And then we conclude whether the point is a maxima minima or saddle point for the given two variable function $f_{(x,y)}$

The text also states that for

$f_{xx}*f_{yy}-(f_{xy})^2$ $=0$  The test is inconclusive. So, how do we proceed for such points in the following example :

$f_{(x,y)}=2(x-y)^2-x^4-y^4$   at $(0,0)$ check maxima/minima/saddle point

答:
Well, if the second derivative test fails (i.e. the determinant of the Hessian matrix is zero at that point)

We need to check by definition, we take a neighborhood of the point $(0,0)$ using some variables $h,k$, if the point $f(0+h, 0+k) - f(0,0) $ has a certain sign we know it is an extremum, if it is positive for all $h,k$ then we know it's maxima, negative minima, and if the sign is not known and depends on the variables then it's a saddle point.

$$ f(0+h, 0+k) -f(0,0) = f(h,k) -0 = 2(h-k)^2 -h^4 - k^4 $$

We can clearly choose some values for $h,k$ that would return a positive value, and some that return negative values, thus it is a saddle point.


问:
Region of integration for double integral:
When tasked with reversing the order of integration in the following integral:

$$\int^1_0\int^{y^2}_{-y}x\ \mathrm{d}x\mathrm{d}y$$

It seems sensible to sketch out our situation:

we have that $-y\le x\le y^2$ and that $0 \le y \le 1$ which gives the following enclosed region:

However, where is the fallacy in writing:

$$\int^1_0\int^{y^2}_{-y}x\ \mathrm{d}x\mathrm{d}y = \int^1_0\left[\int^{0.1}_{-y}x\ \mathrm{d}x + \int^{y^2}_{0.1}x\ \mathrm{d}x\right]\mathrm{d}y$$

(Where 0.1 is chosen arbitrarily) which will then give an enlarged region and thus a larger answer:

This seems clearly wrong, but there's something I'm missing when we have both upper and lower limits of an "inner" integral being a function of $y$. If anyone can spot the exact assumption I'm making which is wrong, that would be appreciated.

答:
If you switch the order, the integral is given by
$$
\int_{-1}^0\left(\int_{-x}^1 x dy\right) dx
+\int_0^1\left(\int_{\sqrt{x}}^1xdy\right) dx
$$


答:
I think your second image is actually showing the integral of:

$$\int^1_0\left[\int^{0}_{-y}x\ \mathrm{d}x + \int^{y^2+0.1}_{0}x\ \mathrm{d}x\right]\mathrm{d} \neq \int^1_0\int^{y^2}_{-y}x\ \mathrm{d}x\mathrm{d}yy$$

I think rather than shifting $$\mathbb{x = y^2}$$ you need to add a line $$\mathbb{x}=0.1$$
which splits R vertically

Title: Region of integration for double integral:

Body:
When tasked with reversing the order of integration in the following integral:

$$\int^1_0\int^{y^2}_{-y}x\ \mathrm{d}x\mathrm{d}y$$

It seems sensible to sketch out our situation:

we have that $-y\le x\le y^2$ and that $0 \le y \le 1$ which gives the following enclosed region:

However, where is the fallacy in writing:

$$\int^1_0\int^{y^2}_{-y}x\ \mathrm{d}x\mathrm{d}y = \int^1_0\left[\int^{0.1}_{-y}x\ \mathrm{d}x + \int^{y^2}_{0.1}x\ \mathrm{d}x\right]\mathrm{d}y$$

(Where 0.1 is chosen arbitrarily) which will then give an enlarged region and thus a larger answer:

This seems clearly wrong, but there's something I'm missing when we have both upper and lower limits of an "inner" integral being a function of $y$. If anyone can spot the exact assumption I'm making which is wrong, that would be appreciated.


问:
Plane geometry tough question
$\triangle ABC$ is right angled at $A$. $AB=20, CA= 80/3, BC=100/3$ units. $D$ is a point between $B$ and $C$ such that the $\triangle ADB$ and $\triangle ADC$ have equal perimeters. Determine the length of $AD$.</p>

答:
Hint:

$AB+BD+AD=AD+CD+AC$

$AB+BD=CD+AC$

$AB+BD+DC=2CD+AC$

$20+\dfrac{100}{3}=2CD+\dfrac{80}{3}$

Solve for $CD$ .And then for $AD$.

Title: Plane geometry tough question


问:
How to solve this Riccati Differential Equation
I need to solve following Differential Equation

$\dfrac{dy}{dx} = \dfrac{1}{x^2} - y^2$, $x >0$

I know  this is a Riccati Equation, but I am unable to see the substitution required to convert this into a Linear Differential Equation.

Any hints will be helpful.

Thank You.

答:
Hint:

$$\dfrac{dy}{dx} = \dfrac{1}{x^2} - y^2$$
Try to find a particular solution like:
$$y_p=\dfrac {C}{x}$$
Then substtute  $y=y_p+u$


问:
Functoriality of the $hom$-functor?
(Originally posted on stackoverflow, but I got no satisfying (for me) answer there.)

I'm a bit lost in Section 8.8 from Category Theory for Programmers.

The book reads

[…] the mapping that takes a pair of objects $a$ and $b$ and assigns to it the set of morphisms between them, the $hom$-set $C(a, b)$, is a functor.

I don't understand it. If a functor is something that can be mapped over, what does it mean to map on the mapping […]?

However, I tried to keep reading.

It is a functor from the product category $C^{op} × C$ to the category of sets, $Set$.

My understanding is that, as long as we talk of objects and not of morphisms, $C^{op} × C$ is not different from $C × C$. Indeed the book addresses that right now:

Let's define its action on morphisms. […]

and the rest of the paragraph can be made in drawing as follows

```<code>              a     f    a'
┌┬──────┬┬┬── • <─────── • ──┬┬┬────────┬┐
││      │││                  │││        ││
││C(a,b)│││                  │││C(a',b')││
││  …   │││                  │││   …    ││
││      │││   b     g    b'  │││        ││
└┴──────┴┴┴─> • ───────> • <─┴┴┴────────┴┘
</code>```

where it's clear that $g \cdot h \cdot f$ is in $C(a',b')$ for each $h$ from $C(a,b)$.

However, I can't really match this objects/morphisms/compositions with the statement at the beginning of the section and the one at the very end, therefore I don't get what this $hom$-functor is.

答:
$\newcommand{\C}{\textsf{C}} \newcommand{\op}{\operatorname{op}} \newcommand{\Hom}{\operatorname{Hom}}$

Lets start with any category $\C$. We want to define the functor $\Hom$ as follows:
$$\Hom: \C^{\op}\times \C \to \textsf{Set}$$
We first need to define it on objects. That's easy: $\Hom(X,Y)$ is by the definition the set of all morphism from $X$ to $Y$.

Next we need to define it on morphisms. But what exactly are morphisms in $\C^{\op} \times \C$? Well for a given pair of objects $(X,Y)$ and $(X',Y')$, a morphism $(X, Y) \to (X',Y')$ is actually a pair of morphisms $(f,g)$ where $f:X'\to X$ and $g:Y\to Y'$. Note the direction of arrows due to $\op$. Also you need to understand how composition works in $\C^{\op} \times \C$. If $(f,g)$ and $(f',g')$ are two morphisms then $$(f,g)\circ (f',g'):=(f'\circ f, g\circ g').$$ The different direction again comes from the $\op$.

And so given $(f,g)$ morphism, or equivalently $f$ and $g$ morphisms, we are somehow supposed to create another morphism (which in case of $\textsf{Set}$ is just a function) by going through $\Hom$. Meaning $\Hom(f,g)$ is supposed to be a function $\Hom(X,Y) \to \Hom(X',Y')$. Remember that $\Hom(X,Y)$ is a set, and thus we can consider functions between such sets. There's nothing really deep here. For example if $C(\mathbb{R})$ denotes the set of all continuous functions $\mathbb{R} \to \mathbb{R}$, then we can consider $p:C(\mathbb{R})\to C(\mathbb{R})$ given by $$p(f)(x)=f(x)+1.$$ Meaning it takes any continuous function and applies "$+1$" to it. It is a similar idea here, we are transforming morphisms and such transformation is a function itself.

Anyway how do we define $\Hom(f,g)$? Note that we can evaluate $\Hom(f,g)$ on any $\alpha:X\to Y$. So for a given morphism $\alpha:X\to Y$ we get that $\Hom(f,g)(\alpha)$ is supposed to be an element of $\Hom(X',Y')$, i.e. it is supposed to be a morphism $X'\to Y'$. And so given those four pieces of data we have: $f$, $g$, $\alpha$ and the expected output, there is only one real way to construct such morphism:

$$\Hom(f,g)(\alpha):=g\circ\alpha\circ f.$$

I encourage you to write down every object, morphism and composition to see why exactly it is well defined, and why we need that $\op$ (otherwise the composition is invalid, and actually there's no way to fix that).

With this we have that $\Hom$ is a functor. For example composition preserving goes like this:
\begin{align}
\Hom\big((f,g)&\circ(f',g')\big)(\alpha) = \\
&= \Hom(f'\circ f,g\circ g')(\alpha) = \\
&=g\circ g'\circ\alpha\circ f'\circ f= \\
&=g\circ \Hom(f',g')(\alpha)\circ f= \\
&=\Hom(f,g)\big(\Hom(f',g')(\alpha)\big).
\end{align}

Note that $(f,g)\circ (f',g')=(f'\circ f,g\circ g')$ again due to $\op$. This gives us

$$\Hom\big((f,g)\circ(f',g')\big)=\Hom(f,g)\circ \Hom(f',g')$$

because $\alpha$ is arbitrary. Just like in the classical set theory if $f(g(x))=h(x)$ for all $x$ then we write $f\circ g=h$.

Also note that I use "$\circ$" symbol here in three different (but closely related) meanings: as a composition in $\C^{\op}\times \C$, as a composition in $\C$, and as a composition in $\textsf{Set}$.

Title: Functoriality of the $hom$-functor?


答:
The outer integral is evaluated for $y \in [0,1]$, and the inner integral's limits are on $x \in [-y, y^2]$, which you partition into $x \in [-y, 0.1] \cup (0.1, y^2]$.  So if $y \in [0, \sqrt{0.1})$, the integral over the second interval is negative, because the upper limit of integration is smaller than the lower limit of integration.  For example, if $y = 0.2$, then $y^2 = 0.04 < 0.1$, and the two inner integrals in your expression become $$\begin{align}
\int_{x=-y}^{0.1} x \, dx + \int_{x=0.1}^{y^2} x \, dx &= \int_{x=-0.2}^{0.1} x \, dx + \int_{x=0.1}^{0.04} x \, dx \\
&= \int_{x=-0.2}^{0.1} x \, dx - \int_{x=0.04}^{0.1} x \, dx \\
&= \int_{x=-0.2}^{0.04} x \, dx.
\end{align}$$
This resolves your contradiction.


问:
Is the Arabic "4" a value, numeral, or numerical expression?

What I understand:

A number is an abstract concept. We use numerals such as names (i.e. cuatro), symbols (i.e. Arabic "4"), and numerical expressions (i.e. "9 - 5") to represent numbers.

Where I get tripped up:

"Value" is defined as the number a numeral represents, where "4" is a value of "3 + 1". Doesn't this contradict the author's earlier definition of "4" being a symbol?

Is "4" a number or a numeral?

Bonus Question

"4" is then defined as a numerical expression.
Is a numerical expression a type of numeral, or are they synonymous to one another?

I value your insight!

答: Just as "four" and "quatro" and "$4$" are symbols for the value $4$, the words "one hundred and one" is a symbol for the value $101$. It has more "words" but it is just a symbol. Here, "three plus one" might be called "three and one" (think "four and twenty" blackbirds baked in a pie) and it is just another symbol as is "$3+1$" so the conflict is resolved.


问:
Could I use an average of one random process to estimate average in other random process?
I am designing a drawing game for a kid's class. A kid could draw a ball out of a box numbered 1, each ball has a number 1, 2, or 3 printed on it. After the kid gets a number, he or she then draws from another box N times (N is the number the kid obtained from the first box) to get N numbers. In the second drawing, each number drawn will be put back to the box for the next draw. And the sum of the numbers drawn from the second box is the number of candies the kid could have. I am trying to find what is the average number of candies each kid could have.

There is a fair chance to have 1,2, or 3 in the first box. The numbers available in the second box is 1,2,3,4,5 in the second box with chance $p_1, p_2, p_3, p_4, p_5$.

My gut feeling tells that the average number of candies should be given as

$$\overline{C} = \frac{1+2+3}{3}\times \sum_{n=1}^5 np_n = 2\sum_{n=1}^5 np_n$$

I did a simulation and it gives me something very close to the formula. However, I want to prove it in a rigorous way and thus I simplify the problem to the following. Let's assume the kid always gets a 2 in the first box and there are 2 numbers (1 and 2) in box 2 with a chance $p_1$ and $p_2$ to be drawn. The kid could get 1-1 (2 candies) or 1-2 (3 candies) or 2-1 (3 candies) or 2-2 (4 candies) from box 2 so the average number of candies in the case should be

$$
  \overline{C} = 2p_1^2 + 3p_1p_2 + 3p_2p_1 + 4p_2^2 = 2(1+p_2) = 2(p_1+2p_2)
$$
It seems to match the conclusion I have before. But how to generalize it to the case of the initial problem?

答:
The first part of your analysis is an application of the law of total expectation

In your simplified example, the two expressions $2p_1^2 + 3p_1p_2 + 3p_2p_1 + 4p_2^2$ and $2(1p_1+ 2p_2)$ are equal if you use $p_1+p_2=1$, as they are both equal to any of $2p_1+4p_2=2+2p_2 = 4-2p_1 = 3+p_2-p_1$.  You can see this with

$$2p_1^2 + 3p_1p_2 + 3p_2p_1 + 4p_2^2 \\= 2p_1^2 + 2p_1p_2 +4p_1p_2 +4p_2^2  \\= 2p_1(p_1+p_2) +4(p_1+p_2)p_2 \\= 2p_1+4p_2\\ = 2(1p_1+ 2p_2)$$


问:
How to show $\frac{P(x)}{Q(x)}=\sum_{k=1}^n \frac{Q(x_k)}{P'(x_k)(x-x_k)}$ when the following condition holds?
Let $P$ be a polynomial of degree $n$ with n different real roots
$x_1,x_2,....x_n$ and let $Q$ be a polynomial of degree at most $n -1$. How to  Show
that</p>

<p>$ \ \ \frac{P(x)}{Q(x)}=\sum_{k=1}^n \frac{Q(x_k)}{P'(x_k)(x-x_k)}$ </p>

答:
Your typo is that the LHS should be $ Q(x)/P(x) $, it is easy to observe that $ x_1,...,x_n $ being roots of $P$ you have always
$$ \lim_{x\rightarrow x_j} \frac{P(x)}{(x-x_k)} = P'(x_j)\delta_{jk} $$
Now look at the ploynomial $ F \in \mathbb{R}[X] $ given by the following (which is indeed a polynomial as $ (x-x_k)\  |\ P(x) $ )
$$ F(x) = \sum_k \frac{Q(x_k)P(x)}{P'(x_k)(x-x_k)} $$
Clearly from the behavior of limit as shown above you find that $ F(x_j) = Q(x_j) $, that is $ x_1, x_2,...,x_n $ are roots of $ F- Q $, but as $ \deg(P) = n $ so we have $ \deg(Q) \leq n-1 = \deg (F) $, so $\deg(F-Q)\leq n-1 $, but it has n roots, hence $ F = Q $ which implies
$$ \frac{Q(x)}{P(x)} = \sum_k \frac{Q(x_k)}{P'(x_k)(x-x_k)} $$</p>


问:
$G=\langle g_1,\ldots,g_n \rangle$. Proof that $x \in G \Rightarrow \exists k \in \mathbb{N}: x=x_1 \cdots x_k$ where $x_i \in \{g_1,\ldots,g_n\}$
Let $G$ be a finite group. Assume that $G=\langle g_1, \ldots, g_n \rangle$.

Show that each element in $G$ can be written as $x_1 \cdots x_k$ for a $k \in \mathbb{N}$ with $x_1, \ldots, x_k \in \{g_1, \ldots, g_n\}$

My attempt:

Let $g \in G$. Since $G=\langle g_1, \ldots, g_n \rangle=\overline{\{g_1,\ldots,g_n\}}$ we can write $g=x_1^{\epsilon_1} \cdots x_k^{\epsilon_k}$ where $x_i \in \{g_1, \ldots, g_n\}$ and $\epsilon_i= \pm 1$. Now I need to show that $\epsilon_i=1$ for all $i \in \{1,\ldots,k\}$, however I don't know how to do this.

答: If $G$ is finite, then so are the orders of all its elements, i.e. for each $g\in G$ there is an ord$(g)<\infty$ with $g^{\text{ord}(g)}=e$, the identity element. So if you need $g^{-1}$, you just need to use $$g^{-1}=g^{\text{ord}(g)-1}=\underbrace{gg\cdots g}_{\text{ord}(g)-1\text{ times}}.$$