问:
Universal enveloping algebra vs algebra of continuous functions
I'm a physicist studying quantum groups, but this question is about the usual classical Lie groups (though it is related to the language that is used to describe the deformation to quantum groups, thus the quantum-group tag).

Suppose I have a finite-dimensional Lie algebra $\mathfrak{g}$ (you can assume it to be compact and/or semisimple, whatever suits your needs) and its universal cover Lie group $G$.

I can always make the universal enveloping algebra $U(\mathfrak{g})$ into a Hopf algebra by defining:

The co-product is defined as:
$$\Delta(1) = 1 \otimes 1$$
$$\Delta(t_a) = t_a \otimes 1 + 1 \otimes t_a$$
where $\{t_a\}$ are the set of generators of $\mathfrak{g}$, or, equivalently, the basis of the order-1 subspace of $U(\mathfrak{g})$.
This definition is then extended to the rest of $U(\mathfrak{g})$ through bialgebraic properties.

The trace is defined as
$$ \varepsilon(1) = 1$$
$$ \varepsilon(t_a) = 0$$
and extended to the rest of $U(\mathfrak{g})$ through bialgebraic properties.

The antipode is defined as
$$ S(1) = 1$$
$$ S(t_a) = - t_a $$
and extended to the rest of $U(\mathfrak{g})$ through bialgebraic properties.

Question: please mark the following claims as "true" or "false", and for those which are true, please provide a draft of the proof.

The group-like elements of the Hopf algebra $U(\mathfrak{g})$ (those which satisfy $\Delta(g) = g \otimes g$) form a Lie group, which is exactly the universal cover Lie group $G$.

The Hopf algebra $U(\mathfrak{g})$ is to some extent (which?) equivalent to the commutative algebra of functions over $G$. (I've seen this claim made many times in one form or another, see e.g. here, but I never actually understood this. How can a commutative algebra be equivalent to a noncommutative one? Anyway, looks like this duality is taken as the definition of the quantum group as a deformed Hopf algebra, so I assume it has to be valid in the classical case, too).

答:
In the meantime, I have posted an answer to the question Duality between universal enveloping algebra and algebras of functions you are citing. So, for the second question I suggest to check the corresponding literature. But the idea is that there is a pairing $\langle\cdot,\cdot\rangle\colon U({\frak g})\otimes O(G)\to\mathbb C$, where $O(G)$ is the algebra generated by the coordinate functions on $G$. If you interpret elements of $\frak g$ as tangent vectors of $G$ at identity, then elements of $U({\frak g})$ indeed act on $O(G)$. Namely by
$$\langle x_1\cdots x_n,f\rangle=\left.{\partial^n\over \partial t_1\cdots\partial t_n}\right|_{t_1,\dots,t_n=0}f(\exp(t_1x_1)\cdots\exp(t_nx_n)),$$
where $x_1,\dots,x_n\in{\frak g}$.

As for the first question, the answer is no. More precisely, the only group-like element in $U({\frak g})$ is the identity. To gain some insight, let's maybe look on the motivation behind group-like elements.

Let $\Gamma$ be a finite discrete group, then the group algebra $\mathbb C\Gamma$ forms a Hopf algebra with respect to the coproduct $\Delta(g)=g\otimes g$. In addition, for any Hopf algebra $A$, it holds that, if $g,h\in A$ are group-like, then also the product $gh$ is group-like. So, the set of all group-like elements forms a group – it somewhat cuts of exactly the part of $A$, which looks like $\mathbb C\Gamma$.

For instance, if $G$ is a Lie group or a compact group, so $O(G)$ forms a Hopf algebra, then the group-like elements of $O(G)$ are exactly all the one-dimensional representations of $G$. We know, that those indeed form a group. By the way, if $G$ is compact and abelian, then all the irreducible representations are one-dimensional and the group they form $\hat G$ is exactly the Pontryagin dual of $G$.

But what you are suggesting is that you would like to discover somehow the group algebra $\mathbb CG$ inside $U({\frak g})$. But in order to do that, you would have to somehow exponentiate the elements of ${\frak g}$ to get the elements of $G$, which is not possible inside $U({\frak g})$ (since you have only finite products and linear combinations here).

There is even a classification statement by Cartier–Gabriel (see Theorem 3.8.2 here) saying that every cocommutative Hopf algebra is of the form $A=U({\frak g})\rtimes \mathbb C\Gamma$. So, the property of being primitive (i.e. $\Delta(x)=1\otimes x+x\otimes 1$) and being group-like (i.e. $\Delta(x)=x\otimes x$) are rather complementary than compatible. Every cocommutative Hopf algebra consists of a part that looks like group algebra $\mathbb C\Gamma$ (generated by group-likes) and of a part that looks like an enveloping algebra $U({\frak g})$ (generated by primitives).


问:
Gödel's incompleteness wrt weakend versions of ZFC
Suppose for the sake of argument that we look at ZFC with the axiom of infinity removed.

http://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems#First_incompleteness_theorem

http://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems#Second_incompleteness_theorem

We would then be in a position where the hypotheses of Gödel's theorems are not satisfied, correct? Basically, I want to remove, for the sake of argument, a minimal amount of axioms of ZFC so that ZFC minus some axiom(s) leads to a theory that does not include arithmetical truths and is not capable of expressing elementary arithmetic.

Is it possible that ZFC- (my shorthand for ZFC with some axiom(s) removed) is consistent and complete?

答:
The standard answer is slightly incomplete. The true state of things is beautifully explained in Emil Jerabek's comments at https://mathoverflow.net/questions/63887/non-standard-models-of-finite-set-theory/ .  Just to record it in one place, and to bring out a surprising point from that MO thread:

It is well known that, in a suitable sense, the ZF theory of non-infinite sets is equivalent to PA.

Less well known, but familiar to experts, is that there is a subtle point about how to define ZF minus infinity for the equivalence to PA to hold.

But the story does not end there, and is not only about this PA/ZF- equivalence!  As revealed in the comments by Blass and Jerabek, the ZF axiom of foundation is oversimplified.  In more general situations, one wants not foundation but $\in$ induction. In full ZF one can do without it, but this is not an indication of how to axiomatize the same idea in more finitistic or constructive forms of set theory, including variants of ZF.

ContentLicense: CC BY-SA 3.0


问:
Use of directed lengths
I was studying Menelaus' Theorem which can be found here https://en.wikipedia.org/wiki/Menelaus%27s_theorem.

 I noticed they use directed lengths which specifies that points $E,D,F$(in reference to above link) are collinear. My doubt is when do we use this kind of notation, where the magnitude of a side is taken to be negative.

 Can someone please provide clarity on this issue or recommend a source. Thanks a lot for any help :)

答:
Signed lengths are a way to reduce the number of variations of a diagram that can be drawn. Consider the following basic use case:

If points $A$, $B$, and $C$ are collinear, then $AB + BC = AC$.

This is definitely true when point $B$ is between point $A$ and point $C$. We introduce signed lengths to make this statement true no matter what order $A, B, C$ appear in.

Signed lengths can be used in any statement of affine geometry: statements that only use properties which are unaffected by affine transformations. For example, in affine geometry:

we can talk about points being collinear, and lines being concurrent or parallel;

we can talk about ratios of areas, or ratios of parallel line segments;

we cannot compare line segments that are not parallel;

we cannot talk about angles;

we cannot even measure absolute lengths of line segments, only their ratios. (Statements like $AB + BC = AC$ are okay if they are understood as $\frac{AB}{AC} + \frac{BC}{AC}=1$, which is valid.)

In this setting, the ratio $\frac{AB}{CD}$ is positive if $AB$ and $CD$ point in the same direction (they are required to be parallel). We do not make statements about signed lengths that do not reduce to ratios in this way.

But we could define a signed length by picking a reference line segment $PQ$ in each class of parallel lines, which we treat as having length $+1$, and letting $AB$ have the signed length $\frac{AB}{PQ}$.

In general, any statement of affine geometry is automatically compatible with signed lengths. If we prove it for ordinary lengths when the points are in one configuration (with respect to order, points being inside a triangle, etc.) then we can deduce it for signed lengths and all configurations. For example, Ceva's theorem is often stated for ordinary lengths, with points $D, E, F$ lying on segments $BC, CA, AB$. But if we allow the points to lie anywhere on the lines $BC, CA, AB$, then it is true for signed lengths.

The handwavy reason for that is that theorems in affine geometry using signed lengths can be restated as polynomial equations in the coordinates of points. Checking a theorem for ordinary lengths in a large "region" of cases tells us that a polynomial equation holds in that region. But if a polynomial equation in $n$ variables holds over an open subset of $\mathbb R^n$, then it holds over all of $\mathbb R^n$, giving us the theorem for signed lengths. (Here, $n$ is the number of degrees of freedom in the diagram.)

ContentLicense: CC BY-SA 4.0

Title: Use of directed lengths


问:
Showing that a Krylov subspace is invariant under A
A is a symmetric positive definite matrix and I am trying to show that $S_K$(A,v)=span{v,Av,...,$A^{k-1}$v}.

To do this I have said let z belong to $S_K$(A,v). Then, z=$a_1$v+$a_2A$v+$a_1$$A^2$v+...+$a_k$$A^{k-1}$v

So I need to show Az belongs to $S_K$(A,v), so:
Az=$a_1A$v+$a_2$$A^2$v++...+$a_k$$A^{k}$v. Clearly the first k-1 terms belong to $S_K$(A,v) so it suffices to show that $a_k$$A^{k}$v belongs to $S_K$(A,v).

I am assuming I need to express $A^{k}$ in terms of lower powers, but I am unsure how to do this. Thank you for any help in advance.

答: If $A\in \mathbb{R}^{k\times k}$, you can use the Cayley-Hamilton theorem to express $A^k$ as a linear combination of $I$, $A$, ..., $A^{k-1}$, which, in turn, implies that $A^k v \in S_K(A,v)$.


问:
Creating a function similar to sin(x) with steeper downturns but equal peaks and lows
As the title states, I've been trying to create a function that is identical to $\sin(x)$, except that it has a steeper slope in the middle of the decreasing portion of the curve, but mellows out such that this function and $\sin(x)$ have the same turning point for both peak and bottom.

I've tried to describe it in the following image, if we pretend the top function is $\sin(x)$, I'd like to create something akin to the bottom image. Currently I am trying some variation of $\sin(x-0.5\sin(x))$, which gives me the steeper slope, but this shortens the actual duration and the peaks and lows don't line up.

Any hints on functional forms I could try would be appreciated!

Pardon my extremely poor paint skills:

答:
Here's a fairly straightforward solution:

$ f_k(x) = \arctan(k \cdot \sin x) / \arctan k $

Steep sine

We are using a non-uniform scaling function

$ f_k(x) = \arctan(k \cdot x) / \arctan k $

where $k$ determines the degree of deformation:

with small $k$ (e.g. $0.001$) it's almost identity transform.

with larger $k$ (e.g. $4$ or more) there's more upward/downward deformation

Because the domain is the result of $\sin x$, we don't worry what happens outside $[-1..1]$, so we don't need to make it piecewise.

Steep deformation

Other non-uniform scaling functions in the Sigmoid function family are applicable.

An application in image processing is for producing high contrast transformation.


问:
Prove $\lim_{t\to 0} \frac{P(N(t)>0)}{t}=E(N(1))$
A process $\{N(t)/ t\in [0, \infty]\}$ with $N (0)=0$ is called crudely stationary if $p_k(s, t) = P(N(t + s) - N (t) = k)$ depends only on the length $t-s$ and not on
$s$. It is called simple if, almost certainly, only if it has jump discontinuities of
size $1$. Prove that, for a simple crudely stationary process $N$, we have
what  $$\lim_{t\to 0} \frac{P(N(t)>0)}{t}=E(N(1)).$$

I have been looking for information to solve this problem and I have found that I can apply the Korolyuk-Khinchin theorem, but I don't know how to apply it, also before using it I should demonstrate it and I cannot find any clear proof on the web. I also think that maybe the problem can be solved in another way, that's why I go to this place. I would greatly appreciate any suggestions and even more if they solve the problem.

答:
(Do you acturally mean by crudely stationary that the distribution of $N(t+s)-N(t)$ depends only on $s$, not on $t$?)

If so, then here's an approach. Because $N$ has only unit jumps
$$
N(1) =\lim_{n\to\infty}\sum_{k=1}^n 1_{\{N(k/n)-N((k-1)/n)>0\}}.
$$
In effect, once $n$ is large enough, each interval $((k-1)/n,k/n]$ contains either precisely one jump or none. A little thought will reveal that the sum displayed above increases with $n$, so the monotone convergence theorem can be invoked. By crude stationarity, $E(1_{\{N(k/n)-N((k-1)/n)>0\}}) = E(1_{\{N(1/n)>0\}}) = P(N(1/n)>0)$.


问:
Integral over a ball
Let $a=(1,2)\in\mathbb{R}^{2}$ and $B(a,3)$ denote a ball in $\mathbb{R}^{2}$ centered at $a$ and of radius equal to $3$.

Evaluate the following integral:
$$\int_{B(a,3)}y^{3}-3x^{2}y \ dx dy$$

Should I use polar coordinates? Or is there any tricky solution to this?

答:
Note that $\Delta(y^3-3x^2y)=0$, so that $y^3-3x^2y$ is harmonic. By the mean value property, we get that the mean value over the ball is the value at the center. Since the area of the ball is $9\pi$ and the value at the center is $2$, we get
$$
\int_{B(a,3)}\left(y^3-3x^2y\right)\mathrm{d}x\,\mathrm{d}y=18\pi
$$

Title: Integral over a ball


问:
Check for values of $x$ that are not a basis for $V$ with basis $B$
I have a vector space $V$ with basis $B = \{v_1, v_2, v_3\}$.

I have a set $Z = \{v_1 + v_2, v_2 + v_3, v_3 + av_1\}$.

What value of $a$ makes $Z$ not a basis for $V$?

I know that $Z$ needs to be linearly independent to be a basis. But I'm lost on what to do after this...

答:
We know that $dimV=3$ also $|Z|=3$ so $Z$ forms a basis if and only if elements of $Z$ are linearly independent

if $B = \{v_1, v_2, v_3\}$ is a basis then

$c_1v_1+c_2v_2+c_3v_3=0 \Rightarrow c_1=c_2=c_3$

$a_1(v_1+v_2)+a_2(v_2+v_3)+a_3(v_3+av_1)=0 \Rightarrow (a_1+aa_3)v_1+(a_1+a_2)v_2+(a_2+a_3)v_3=0$

$(a_1+aa_3)v_1+(a_1+a_2)v_2+(a_2+a_3)v_3=0 \Rightarrow (a_1+aa_3)=(a_1+a_2)=(a_2+a_3)=0 $

$(a_1+a_2)=(a_2+a_3) \Rightarrow a_1=a_3$

$a_1=a_3 \Rightarrow a_1+aa_3=(a+1)a_1=0$

$(a+1)a_1=0 \Rightarrow a_1=0$ or $a=-1$

$a_1=0 \Rightarrow a_1=a_2=a_3=0$

$a=-1 \Rightarrow -a_2=a_3=a_1$ is a solution which means they are linearly dependent


问:
A question about an increasing bijection and dense sets.
Let $A, B$ be a dense subsets of intervals $[a,b], [c,d]$ respectively.
Consider an nondecreasing bijective function
$$f:A\rightarrow B$$

Is this a continous function? If so, then prove it. Otherwise give a counterexample.

I suspect that the answer is "yes" but i can't prove it.

I have thought that topological definition of continuity may be helpful, however i didn't achieve any result.

Thank you in advance for help

答:
The function would have to be continuous.

Let $p \in A$ be arbitrary. WLOG, assume that $p \neq b$. (The argument for the right end-point is similar.)

Since $A$ is dense, the limit $f(p^+) := \displaystyle\lim_{x \to p^+}f(x)$ makes sense. Since $f$ is increasing, the limit exists and we have $f(p) \le f(p^+)$.

Also note that $c \le f(p) \le f(p^+) \le d$.

Suppose that $f(p) < f(p^+)$. By density, there exists $q \in B$ such that $f(p) < q < f(p^+)$. Since $f$ is a bijection, there exists $x \in A$ such that $f(x) = q$. This is a contradiction. (Why?)

Thus, $f(p^+) = f(p)$. Similarly, if $p \neq a$, then $f(p) = f(p^-)$. This proves continuity at $p$.


问:
How do you describe the set solutions to $y = x + 1 + \frac{2}{x+3}$ in the ring $\mathbb{Z}[\sqrt{-2}$]?

How do you describe the set of solutions to
$$y = x + 1 + \frac{2}{x+3}$$ in the ring $\mathbb{Z}[\sqrt{-2}$] ?

$\mathbb{Z}[\sqrt{-2}] := \{ a + b\sqrt{-2} : a,b \in \mathbb{Z} \}$ is an integral domain and
I have also shown that $\mathbb{Z}[\sqrt{-2}]$ is a Euclidean domain so for every $x,y \in \mathbb{Z}[\sqrt{-2}]$ theres exists $q,r \in \mathbb{Z}[\sqrt{-2}]$ such that $x = qy + r$.

How do you use this fact to help you find solutions to the equation? Do I need to show $\frac{2}{x+3}$ has to be in form $a + b\sqrt{-2}$ ?

Any help would be appreciated thanks.

答: Hint: We need $x+3\mid 2$ in $\Bbb Z[\sqrt{-2}]$. In particular we have $N(x+3)\mid N(2)=4$. Let $x=a+b\sqrt{-2}$. Then the norm is given by $N(x)=a^2+2b^2$. Hence we have $((a+3)^2+2b^2 )\mid 4$. This has only very few integer solutions in $a$ and $b$.


问:
Is $C(\Bbb R)$ homeomorphic to $C([a,b])$?
Let $C(\Bbb R)$ be the space of all continuous self-maps of $\Bbb R$ in the compact convergence topology (which is same as compact-open topology) and $C([a,b])$ be the space of all continuous self-maps of [a,b] in the uniform topology. Is $C(\Bbb R)$ homeomorphic to $C([a,b])$?

答: Both are separable completely metrisable locally convex linear spaces, so yes, by "Anderson's theorem". See section 3 in this survey where the result is attributed to Dobrowolski and Toruńczyk. I've seen both attributions used. Added: not valid for self-maps of $[a,b]$ as this is not a linear space. I assumed the standard notion of real-valued maps instead. But in Bessaga and Pelzcynski, Topics in Infinite-dimensional topology, Thm 6.2 we find that a closed convex body (so it has non-empty interior) in an infinite-dimensional Fréchet space is homeomorphic to the whole space (and to $\ell^2$ and $\Bbb R^{\omega}$ etc.). And $C([a,b])$ is surely that. So it does indeed hold, by such general facts: there is very little topological variety in such function spaces (when Polish).


答:
Since $A$ is dense in $[a,b]$ and $B$ is dense in $[c,d]$, define the following function $g\colon[a,b]\to[c,d]$:
$$g(x) = \sup\{f(y)\mid y<x\},$$

Since $f$ is increasing and $A$ is dense, $g$ is well-defined. It is clear that $g$ is also increasing. Since $B$ is dense in $[c,d]$, and $f$ is a bijection, given any $y\in [c,d]$, $A_y=\{x\in A\mid f(x)<y\}$ is bounded in $[a,b]$. Take $x=\sup A_y$, then $$g(x)=\sup\{f(z)\mid z<x\}=\sup\{f(x)\mid x\in A_y\}=y.$$

Therefore $g\colon[a,b]\to[c,d]$ is an increasing bijection, and therefore it is continuous. By extension, $f$ must be continuous as well.

Title: A question about an increasing bijection and dense sets.

Body:
Let $A, B$ be a dense subsets of intervals $[a,b], [c,d]$ respectively.
Consider an nondecreasing bijective function
$$f:A\rightarrow B$$

Is this a continous function? If so, then prove it. Otherwise give a counterexample.

I suspect that the answer is "yes" but i can't prove it.

I have thought that topological definition of continuity may be helpful, however i didn't achieve any result.

Thank you in advance for help


问:
Show that $n \ge \sqrt{n+1}+\sqrt{n}$
(how) Can I show that:

$n \ge \sqrt{n+1}+\sqrt{n}$ ?

It should be true for all $n \ge 5$.

Tried it via induction:

$n=5$: $5 \ge \sqrt{5} + \sqrt{6} $ is true.

$n\implies n+1$:
I need to show that $n+1 \ge \sqrt{n+1} + \sqrt{n+2}$

Starting with $n+1 \ge \sqrt{n} + \sqrt{n+1} + 1 $ .. (now??)

Is this the right way?

答: You can replace the $\sqrt{n}$ on the RHS with another $\sqrt{n+1}$. Therefore you have $n\ge2\sqrt{n+1}$, or $n^2\ge4n+4$. $n^2-4n+4\ge8$, or $(n-2)^2\ge8$. The lowest integer solution to this is $5$, so $n\ge5.$</p>

Title: Show that $n \ge \sqrt{n+1}+\sqrt{n}$


问:
$\lim_{n\to\infty} \frac{2^n(n^4-1)}{4\cdot 3^n + n^7}$
How do I solve this example? I tried to point out the fastest growing term $2 ^ n$ and $3 ^ n$, but that doesn't seem to lead to the result. I know the limit is $0$ that's obvious, but I don't know how to work on it.

$$\lim_{n\to\infty} \frac{2^n(n^4-1)}{4\cdot 3^n + n^7}$$

Or can I make a power estimate of a theorem on a tightened sequence, for example, that $2 ^ n$ / $13 ^ n$ (smaller) goes to zero and the other side  $2 ^ n$ / $3 ^ n$ (bigger) also goes to zero?

$\lim_{n\to\infty} 2 ^ n$ / $13 ^ n$< $\lim_{n\to\infty} \frac{2^n(n^4-1)}{4\cdot 3^n + n^7}$ < $\lim_{n\to\infty} 2 ^ n$ / $3 ^ n$

答:
Observe that
$$
f(n)=\frac{(\frac23)^n\cdot (n^4-1)}{4+(\frac13)^n\cdot n^7}
$$

If you know the following limits, you are done:
$$
\lim_{n\to\infty} (\frac23)^nn^4=0,\quad \lim_{n\to\infty} (\frac23)^n=0,
\quad
\lim_{n\to\infty}(\frac13)^n\cdot n^7=0
$$

All you need is the following general statement:

If $0<r<1$ and $\alpha$ is a real number, then
$$
\lim_{n\to\infty}{n^\alpha}r^n
$$

To see why, let $k>0$ be an integer such that $k>\alpha$. Then for $n>2k$,
$$
(1+p)^n>\binom{n}{k}p^k=\frac{n(n-1)\cdots(n-k+1)}{k!}p^k>\frac{n^kp^k}{2^kk!},\qquad r=:\frac{1}{1+p}
$$
So
$$
0<n^\alpha r^n<\frac{2^kk!}{p^k}n^{\alpha-k},\quad (n>2k)
$$
Since $\alpha-k>0$, $n^{\alpha-k}\to 0$.


Body:
How do I solve this example? I tried to point out the fastest growing term $2 ^ n$ and $3 ^ n$, but that doesn't seem to lead to the result. I know the limit is $0$ that's obvious, but I don't know how to work on it.

$$\lim_{n\to\infty} \frac{2^n(n^4-1)}{4\cdot 3^n + n^7}$$

Or can I make a power estimate of a theorem on a tightened sequence, for example, that $2 ^ n$ / $13 ^ n$ (smaller) goes to zero and the other side  $2 ^ n$ / $3 ^ n$ (bigger) also goes to zero?

$\lim_{n\to\infty} 2 ^ n$ / $13 ^ n$< $\lim_{n\to\infty} \frac{2^n(n^4-1)}{4\cdot 3^n + n^7}$ < $\lim_{n\to\infty} 2 ^ n$ / $3 ^ n$

Body:
It follows from the ratio test.

If $\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = a < 1 \ $, then $\lim a_n = 0$.

$$ \dfrac{\frac{2^{n+1} \left((n+1)^4-1\right)}{4\cdot3^{n+1} + (n+1)^7}}{\frac{2^n(n^4-1)}{4\cdot3^n + n^7}} = \dfrac{2^{n+1} \left((n+1)^4-1\right)(4\cdot3^n + n^7)}{2^n(n^4-1)\left(4\cdot3^{n+1} + (n+1)^7\right)} \\ =\frac{2}{3} \left(\frac{n+1}{n}\right)^4 \left(\dfrac{1-\frac{1}{(n+1)^4}}{1-\frac{1}{n^4}}\right) \left(\dfrac{1+\frac{n^7}{3^n}}{1+\frac{(n+1)^7}{3^{n+1}}} \right) \longrightarrow \frac{2}{3} < 1 $$


问:
Meaning of derivative of a function with respect to another function
In chain rule we use,$\frac{dy}{dx}=\frac{dy}{dz} \times \frac{dz}{dx}$. My question is what is meant by $\frac{dy}{dz}$.When we find derivative,we find it with respect to an independent vairable meaning if $x$ changes by some amount,how much does $f(x)$ changes.But how much $g(f(x))$ changes by a change in $f(x)$ doesn't seem to make sense to me.Also derivatives are all about slopes of a graph.Now we will we draw the graph of $g(f(x))$ vs $f(x)$ since we can't place $f(x)$ on real axis as it can't achieve all real numbers for it being a dependent variable.So could someone please shed some light on what we are actually doing?

答:
The geometric idea: for $h,k$ small we have
$$x = x_0 + h\implies f(x)\approx f(x_0) + f'(x_0)h.$$
$$y = y_0 + k\implies g(y)\approx g(y_0) + g'(y_0)k.$$
Then, if $y = f(x), y_0 = f(x_0)$:
$$
g(f(x))\approx g(f(x_0) + f'(x_0)h)\approx
g(f(x_0)) + g'(f(x_0))f'(x_0)h.
$$


答: This is only a matter of considering the infinite hierachy: at the numerator the term $2^n$ is the fastest to go to $+\infty$ and at the denominator is $3^n$. So the resulting limit to determine is: $\lim_{n\to\infty}\frac{2^n}{4\cdot 3^n}=\lim_{n\to\infty}\frac{1}{4}(\frac{2}{3})^n=0$, (this is $0$ since $\lim_{n\to\infty}a^n=0$ if $|a|<1$).

Title: $\lim_{n\to\infty} \frac{2^n(n^4-1)}{4\cdot 3^n + n^7}$


问:
How to show that $\lim_{(x,y) \to (0,0)}\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}=0$
How to show that $$\lim_{(x,y) \to (0,0)}\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}=0$$
I know that

$$\left|\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}\right|\le\frac{x^{2}y^{2}}{\left|x\right|^{3}}=\frac{y^{2}}{\left|x\right|}$$

But now we have a problem with $\left|x\right|$ and I cannot find a lower bound other than $0$.

Also$$\left|\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}\right|\le\frac{x^{2}y^{2}}{\left|x^{3}+y^{3}\right|}=\frac{\left|x+y\right|}{\left|x^{2}-xy+y^{2}\right|}\le\frac{\left|x\right|+\left|y\right|}{\left|x^{2}-xy+y^{2}\right|}$$

But I'm not sure if that helps.

If we show that $\lim_{(x,y) \to (0,0)}\frac{y^{2}}{\left|x\right|}=0$ then we are done.

Body:
AM-GM:

$|x|^3 +|y|^3 \ge 2|x|^{3/2}|y|^{3/2};$

$\dfrac {x^2y^2}{|x|^3+|y|^3} \le \dfrac {x^2y^2}{2|x|^{3/2}|y|^{3/2}}=$

$(1/2)|x|^{1/2}|y|^{1/2} \rightarrow 0.$


问:
Derivative and inverse
Is there a function whos derivative of its inverse equals the inverse of its derivative? This may also hold on a specific interval only.
$$\frac{d}{dx}f^{\langle-1\rangle}(x)=\left(\frac{d}{dx}f(x)\right)^{\langle-1\rangle}$$
Considering that $f(x)=x$ is its own inverse and $e^x$ is its own derivative, neither of them satisfies the equation, but both are the only functions with the respective properties (over the reels). Is this already a proof that there is no solution to the above equation?

Also, I dont know much about complex analysis but if a complex-valued function meets the criteria I'd be happy to know too :)

答:
Edits: I leave here the preliminary lines of thoughts/failed attempts. The answer based on monotonicity of the functions is probably correct and one may try to give a more abstract justification, cf. part 2.

To begin with it is clear that the OP denotes $f^{\langle-1\rangle}$ the "reciprocal" function and not the inverse as in $\frac{1}{f}$. Hence if ever one wants to formulate the problem in algebraic terms, we need to introduce an algebra $\mathcal{A}$ of functions with two products: $\times$ the pointwise product and $\circ$ the composition. The notion of derivation is readily generalized as a linear map $D:\mathcal{A} \to \mathcal{A}$ which satisfies the Leibniz rule w.r.t. $\times$. One can define the group $G$ of $\circ$-invertible elements and study already its image by $D$, or stable subsets etc...

The immediate relation we get for $\mathcal{A}:=
   \mathcal{C}^{\infty}(\mathbb{R})$ are (to avoid confusion $g$ is the
inverse of $f$ for $\circ$, while $h$ is the inverse of $f'$ for that
same law, Id unit for $\circ$ while the constant $1$ function is the
unit fo $\times$)
$$ f\circ g=\mathrm{Id}_{\mathbb{R}}\quad \Longrightarrow\quad g' \times f'\circ g = 1\quad \label{1}\tag{1}$$
$$ f'\circ h=\mathrm{Id}_{\mathbb{R}}="x"\quad \Longrightarrow\quad \frac{1}{x} \times f'\circ h = 1 \quad \label{2}\tag{2}$$
with these notations, OP's equation reads $g'\overset{!}{=} h$ and by (\ref{1}): $\quad g' = \frac{1}{f'\circ g}=h$. (I assume the $\circ$-inverse are left and right inverse, as in the usual definition). What Ninad Munshi wrote on his first comment is the composition on the right of this equality with $f'$: $\quad \frac{1}{f'\circ g \circ f} = h\circ f =\mathrm{Id}_{\mathbb{R}}$. I did play with similar equalities but to no avail unfortunately... (@Christian by the way your "Feb 18" comment is correct, but for clarity you should for example use the variable $x$ for the domain and $y$ for the target space...)

Let us examine more abstractly the behavior of $\circ$ and of $f\mapsto f^{\langle -1\rangle}$ w.r.t. $+$ and $\times$. For the derivation of a composition there is that Faà di Bruno formula for derivation of composition but I was thinking more about a characterize the set $\mathcal{S}$ of solutions of the equation: does it define a subgroup/subvector space/ideal of $\mathcal{A}$. Unfortunately no, although (if not empty) $\mathcal{S}$ is stable under taking the inverse which we may denote $\mathrm{Inv}$: indeed if $f\in \mathcal{S}$ then so is $ g:=\mathrm{Inv} f$
$$\frac{d}{dx}\ g = \mathrm{Inv} \circ \frac{d}{dx}\circ\mathrm{Inv}\ g \quad\Longrightarrow\quad   \mathrm{Inv} \circ \frac{d}{dx}\ g = \mathrm{Inv} \circ \mathrm{Inv} \circ \frac{d}{dx}\circ\mathrm{Inv}\ g = \frac{d}{dx}\circ\mathrm{Inv}\ g $$
The equation can in fact be understood as commutation of the derivation operator $\frac{d}{dx}$ and that of taking the inverse
$$\frac{d}{dx}\circ \mathrm{Inv}\ f = \mathrm{Inv} \circ \frac{d}{dx}\ f  \label{Eq}\tag{Eq}$$
The contradiction I had hoped to get if $(\mathcal{S},\circ)$ were a group, was to exhibit a stable subvector space, and show that a group for $\circ$ could not contain a vector space, but $\mathcal{S}$ is not stable under composition!!! To do so, I assumed $\mathcal{A}:=\mathcal{C}^{\infty}_c(\mathbb{R}) \hookrightarrow H^1(\mathbb{R})$ (Sobolev space, or any space on which one could "diagonalize" $\frac{d}{dx}$, by Fourier transform usually). As a consequence of (\ref{Eq}), $\mathrm{Inv}$ should preserve the eigenspaces of $\frac{d}{dx}$. If we really want to continue along this line, I think we have to define a new kind of algebraic structure...

"Reciprocal" function also appear when one does a change of variable in integration, so one could also get an integral version of the equation $g'=h$ in the case $\mathcal{A}:= \mathcal{C}^{\infty}(\mathbb{R})$
$$\begin{split}
g(y)-g(0) &=\int_0^y g'(w)\, dw = \int_0^y h(w)\, dw =\genfrac{[}{]}{0pt}{0}{z:=h(w)}{dz = h'(w)\, dw = \frac{1}{f''\circ h(w)}\, dw}\\
&= \int_0^{h(y)} f''(z)\, z\, dz = \left[z\, f'(z) \right]^{h(y)}_0 - \int_0^{h(y)} f'(z)\, dz\\
&= h(y)\times y -\big(f\circ h(y) - f(0) \big)
\end{split}$$

Could there be power series solution? From the comments, the attempt with $f(x):= a x^{b} $ involved functions of the form $\sqrt[b]{x}$ and $\sqrt[b-1]{x}$ so maybe instead of looking at power series, one may look at sums of functions of this form $\sqrt[n]{x}$.

Can't we adapt some fixed point techniques? (very explicitly, start with some function $f_0,\ \Delta(f_0):=\frac{d}{dx}f_0^{\langle-1\rangle} - \left(\frac{d}{dx}f_0\right)^{\langle-1\rangle}$ not necessarily $0$ and look for some transformation $T$ s.t. $\Delta \big(T(f_0)\big)$ is closer to $0$ and so on). In particular for the case $\mathcal{A}:=\mathcal{C}^{\infty}(\mathbb{R})$ where inversion is symmetry w.r.t the diagonal in the plane $\mathbb{R}\times \mathbb{R}$ where we represent the graph of the function.

Trick to give an algebraic formulation of the argument based on monotonicity: define the action of translation on functions. If we want to break it into small steps, we can define translation by a constant $h\in \mathbb{R}$ acting on $\mathcal{A}:= \mathcal{C}^{\infty}(\mathbb{R})$
$$\tau_h: \left\lbrace \begin{aligned} \mathcal{A}\quad & \longrightarrow \quad \mathcal{A}\\ "f(t)" & \longmapsto "f(t-h) "\end{aligned} \right. \in GL(\mathcal{A}) $$
More generally the action of the translation group is defined by the group morphism
$$\left\lbrace \begin{aligned} \big(\mathbb{R},+\big)  & \longrightarrow  \big(GL(\mathcal{A}),\circ \big)\\ h\quad & \longmapsto\quad \tau_h \end{aligned} \right. \label{A}\tag{A}$$
To imitate the other reasoning, based on the formula for the derivative of a composition (\ref{1}-\ref{2}), we need to show that left hand side and r.h.s. of (\ref{Eq}) do not behave in the same way under (\ref{A}).

Title: Derivative and inverse


问:
Stable equilibrium
Consider the system:

$$ x'=(\epsilon x+2y)(z+1) $$
$$ y'=(-x+\epsilon y)(z+1) $$
$$ z'=-z^3 $$
where $\epsilon$ is a parameter. Find the parameters $\epsilon$ for which the system has a stable equilibrium at the origin.

I'm not sure about the definition of "stable equilibrium", and don't know how to start.

答:
$$
\cases{
x'=(\epsilon x+2y)(z+1) \\
y'=(-x+\epsilon y)(z+1) \\
z'=-z^3
}\Rightarrow \cases{
x x'=(\epsilon x^2+2x y)(z+1) \\
2y y'=(-2xy+2\epsilon y^2)(z+1) \\
z'=-z^3
}
$$

Adding the two first equations

$$
\cases{
\frac 12(x^2+2y^2)'-\epsilon(x^2+2y^2)(z+1)=0\\
\dot z+z^3=0
}
$$

and calling $u = x^2+2y^2$ we have

$$
\cases{
\frac 12 u'-\epsilon u(z+1)=0\\
z'+z^3=0
}
$$

here $z(t)\rightarrow 0$ and $u(t)\rightarrow 0$ as long as $\epsilon < 0$

Title: Stable equilibrium


答:
Let $U\subseteq B$ be a basic (relatively) open subset.
Then there exist real numbers $\alpha,\beta$ with $U=\{\,y\in B\mid \alpha<y<\beta\,\}$. Let $V=f^{-1}(U)\subseteq A$. We want to show that $V$ is (relatively) open. We need only consider the case $V\ne\emptyset$. Let $\gamma=\inf V\in\Bbb R$ and $\delta=\sup V\in \Bbb R$. Clearly, $V\subseteq [\gamma,\delta]\cap A$.

I claim that usually $V=(\gamma,\delta)\cap A$.
Indeed, assume $x\in (\gamma,\delta)\cap A$. Then there exist $x_1,x_2\in V$ with $\gamma < x_1<x<x_2<\delta$ and it follows that $f(x_1)<f(x)<f(x_2)$, i.e., $f(x)\in B$ and so $x\in V$. This shows $(\gamma,\delta)\cap A\subseteq V$.

Assume $\gamma\in V$ with $\gamma>a$. Then there exists $x_1\in A$ with $a<x_1<\gamma$, hence $c\le f(x_1)\le \alpha<f(\gamma)$. Then there exists $y_2\in U$ with $\alpha < y_2<f(\gamma)$ and so for $x_2:=f^{-1}(y_2)\in V$, we have $x_1<x_2<\gamma$, contradiction. This we can have $\gamma\in V$ only if $\gamma=a$. Similarly, $\delta\in V$ only if $\delta=b$.

We conclude that $V=A\cap(\gamma,\delta)$ or $V=A\cap(a-1,\delta)$  or $V=A\cap(\gamma,b+1)$ or $V=A\cap(a+1,b+1)$. At any rate, $V$ is (relatively) open.

This shows that $f$ is continuous.


问:
Proof verification for Rotman Exercise 6.7 (barycentric subdivision when $n=1$)
I was working on Exercise 6.7 of Rotman's book on algebraic topology, and would like to know if the following solution is correct. I'm a bit uncertain because my answer feels less "symmetric" than I would expect.

Definition. Let $E$ be a convex set. Then barycentric subdivision is a homomorphism $\textrm{Sd}_n:S_n(E)\to S_n(E)$ defined inductively on generators $\tau:\Delta^n\to E$ as follows:

If $n=0$, then $\textrm Sd_0(\tau)=\tau$;

If $n>0$, then $\textrm Sd_n(\tau)=\tau(b_n).\textrm{Sd}_{n-1}(\partial\tau)$, where $b_n$ is the barycenter of $\Delta^n$ and the $.$ represents the cone. In particular, if $\sigma:\Delta^n\to X$ is an $n$-simplex, then $b.\sigma$ is

$$b.\sigma(t_0,t_1,\dots,t_{n+1})=\begin{cases}b&\text{if}~t_0=1\\t_0b+(1-t_0)\sigma\left(\frac{t_1}{1-t_0},\dots,\frac{t_{n+1}}{1-t_0}\right)&\text{if}~t_0\ne1\end{cases}.$$

Then for any space $X$, we define $\textrm{Sd}_n:S_n(X)\to S_n(X)$ as $\textrm{Sd}_n(\sigma)=\sigma_\#\textrm{Sd}_n(\delta^n)$, where $\delta^n:\Delta^n\to\Delta^n$ is the identity.

The problem asks me to find $\textrm{Sd}_n(\delta^n)$ and $\textrm{Sd}_n(\sigma)$ for $n=1,2$. I'll just present my solution for $n=1$, and if anyone could just let me know if it looks correct or not, that'd be great.

My solution for $n=1$.
Since $\textrm{Sd}_0$ is the identity, we know that $\textrm{Sd}_1(\delta^1)$ is $b_1.(\partial\delta^1)$.
But $\partial\delta^1=e_1-e_0$, while $b_1=\frac{1}{2}(e_0+e_1)$, so
\begin{align*}
\textrm{Sd}_1(\delta^1)(t)&=
\begin{cases}
\frac12(e_0+e_1)&\text{if}~t=1\\
t\cdot\frac12(e_0+e_1)+(1-t)(e_1-e_0)&\text{if}~t\ne1.
\end{cases}\\
&= \frac t2(e_0+e_1)+(1-t)(e_1-e_0)\\
&=\left(1-\frac t2\right)e_1+\left(\frac{3t}2-1\right)e_0.
\end{align*}
For general $\sigma$, note that
$$\textrm{Sd}_1(\sigma)=\sigma_\#\textrm{Sd}(\delta)^n=\left(1-\frac t2\right)\sigma(e_1)+\left(\frac{3t}2-1\right)\sigma(e_0).$$

As I briefly mentioned in the beginning, I feel like the formula I end up with looks a little bit asymmetric, so I just wanted to double check my answer.

答:
There is a small conceptual error here to do with the notation:

when you write $b_1 = \frac{1}{2} (e_0 + e_1)$, you mean the $0$-chain consisting of a single map $\Delta^0 \to \Delta^1$ with image $\{\frac{1}{2} (e_0 + e_1)\}$, but

when you write $\partial \delta^1 = e_1 - e_0$, you mean the $0$-chain consisting of the formal difference of two maps $\Delta^0 \to \Delta^1$ with images $\{e_0\}$ and $\{e_1\}$ respectively.

Therefore, the correct context to use the formula for the cone should be $$\operatorname{Sd}_1(\delta^1) = \delta^1(b_1) . \operatorname{Sd}_0(e_1 - e_0) = b_1 . (e_1 - e_0) = (b_1 . e_1) - (b_1 . e_0).$$  Applying that formula, we get:
$$\begin{align}
(b_1 . e_0)(t) = t b_1 + (1-t) e_0, \\
(b_1 . e_1)(t) = t b_1 + (1-t) e_1,
\end{align}$$
so that $\operatorname{Sd}_1(\delta^1)$ is the $1$-chain given by the formal difference of two maps $\Delta^1 \to \Delta^1$, one from one endpoint of the interval to the midpoint and the other from the other endpoint to the midpoint.  This should agree with the intuition you have about barycentric subdivision.


问:
Combinatorics questions about balls in boxes
20 indistinguishable balls are to be placed in 5 distinguishable boxes. In how many ways can this be done if exactly one box should have no balls an at least one box should have at least 11 balls?

For context, this is what I've tried but it's been wrong.

5C1 x (9+4-1)C(4-1)

答:
Choose which box gets no balls. Recognize that only one box can receive at least 11 balls since there are so few balls to place. Choose which box that is and put 10 balls in it for now. Now, distribute the remaining 10 balls across the 4 not-designated-to-be-empty boxes subject to the constraint that each receive at least one. This last step is a standard problem-type solved by stars-and-bars.

So, first, we choose which box is empty.  We have $5$ choices.  (Note... we don't bother writing this as $\binom{5}{1}$ since $\binom{n}{1}$ is always just simply equal to $n$.  Don't use excessive notation when there is no benefit to doing so.)

Next, we choose which box will be greatly filled.  We have $4$ remaining choices.  We go ahead and place $10$ balls in this box for now.  We only place $10$ because we will soon be placing at least one more here and it is convenient for maintaining the symmetry that all remaining not-designated-to-be-empty boxes all require at least one more ball.

Now, we are in the position of having four boxes each needing at least one more ball and $10$ balls to place.  We distribute these balls according to stars-and-bars in $\binom{10-1}{4-1}=\binom{9}{3}$ ways

The final answer is then:

$$5\times 4\times \binom{9}{3}$$

Alternatively, if you preferred, rather than giving $10$ balls to the box designated to receive many, go ahead and give $11$ balls to it ahead of time.  Further, give one ball to each of the other remaining three boxes.  Now we are in the pattern where each of the four not-designated-to-be-empty boxes need $0$ or more additional balls and $6$ additional balls to place, the other standard problem type for stars-and-bars.  In such an explanation we have $\binom{6+4-1}{4-1}=\binom{9}{3}$ ways to do this, same as before, going to the same final answer.


问:
Quadratic variation of a square integrable continuous martingale
Is the total variation of the quadratic variation process of a square integrable right continuous martingale bounded?</p>

答: Since the quadratic variation process is non-decreasing, its total variation is equal to its value and the quadratic variation does not need to be bounded, although it will have finite expectation and hence be a.s. finite.  For example, let $B$ be a standard Brownian motion, $\tau := \inf\{t : |B_t| = 1\}$, and $M_t := B_{t \wedge \tau}$.  Then $M$ is square integrable (in fact $M$ is bounded) and continuous, but $\langle M,M \rangle_t = t \wedge \tau$ and $\tau$ is not bounded.


问:
Variable substitution in double integral

Let $(a,b)\in (0,1)$ and let $T>0$. Consider the following integral:
$$
\iint_{\Gamma }f(x)g(x+t)\,dt\,dx,
$$
where
$
\Gamma =\left\{ (t,x)\in (0,T)\times (0,1):t+x\in (0,1)\right\} .
$

Let $s=x+t$, then $s\in (0,1)$ and $s-t\in (0,1)$, $t\in (0,T)$.
Am I right?
$$
\iint_{\Gamma }f(x)g(x+t)\,dt\,dx={\int_{0}^{1}\int_{0}^{\min(s,T)}}f(s-t)g(s)\,dt\,ds,
$$
Thank you.

答:
I think you're right!

Your domain $\Gamma$ is defined by the inequalities:
$$\left\lbrace\begin{matrix} 0 & < & x & < & 1 \\ 0 & < & t & < & T \\ 0 & < & x+t & < & 1 \end{matrix}\right. $$
You can translate these into inequalities with $(s,t)$, they then read:
$$\left\lbrace \begin{matrix} 0 & < & s-t & < & 1 \\ 0 & < & t & < & T \\ 0 & < & s & < & 1 \end{matrix} \right.$$
Now the first inequality can be decomposed as $t < s$ and $s<1+t$. Assuming the last two inequalities hold, $s<1+t$ is always true so it can be discarded without loosing any information. Formally:
$$\left\lbrace \begin{matrix} 0 & < & s-t & < & 1 \\ 0 & < & t & < & T \\ 0 & < & s & < & 1 \end{matrix} \right. \Longleftrightarrow \left\lbrace \begin{matrix}
 & & t & < & s \\ 0 & < & t & < & T \\ 0 & < & s & < & 1 \end{matrix}\right.$$
The last system can be rewritten as:
$$ \left\lbrace \begin{matrix} 0 & < & t & < & \min (s,T) \\ 0 & < & s & < & 1 \end{matrix}\right.$$
and this gives you your new domain, as we have shown:
$$ \left\lbrace\begin{matrix} 0 & < & t & < & \min (1-x,T) \\ 0 & < & x & < & 1 \end{matrix} \right. \Longleftrightarrow \left\lbrace \begin{matrix} 0 & < & t & < & \min (s,T) \\ 0 & < & s & < & 1 \end{matrix}\right.$$


问:
Proving that either $2^n-1 $ or $ 2^n+1$ is not prime
Not sure what approach to take with this:

Prove that at least $2^n-1 $ or $ 2^n+1$ is composite $\forall$ $n>2$

答: If n is composit, so is $ 2^n -1 $. If n is prime or has an odd factor, $ 2^n +1 $ is composit.</p>


问:
Exercise Baby Rudin 4.8
Let $f$ be a real uniformly continuous function on the bounded set $E$ in $\mathbb{R}^1$. Prove that $f$ is bounded on $E.$

I am struggling to show that $E$ is compact (I know how to resolve the problem after that). I was originally going to say that it is a subset of a 1-cell since it's bounded but I know that $(0,1)$ isn't compact when $[0,1]$ is. Is it possible to make an argument about limit points?

答:
As pointed out, compactness of $E$ is not needed.

Let $\delta > 0$ be such that $|f(x) - f(y)| < 1$ whenever $x, y \in E$ and $|x - y| < \delta$.

Since $E$ is bounded, you can cover $E$ with finitely many (say, $N$) intervals of length $\delta$. We may also assume that all these intervals actually intersect $E$.

Pick $x_1, \ldots, x_N \in E$ from the each interval. Put $M = \max\{|f(x_1)|, \ldots, |f(x_N)|\} + 2.$

Now, if $y \in E$, then $y$ is in the $\delta$ neighbourhood of some $x_k$. In that case $$|f(y)| \le |f(y) - f(x_k)| + |f(x_k)| < 1 + |f(x_k)| < M.$$
Thus, $M$ is an upper bound.

Title: Exercise Baby Rudin 4.8


问:
The $2019$th strictly positive integer $n$ such that $\binom{2n}{n}$ is not divisible by $5$
This is problem $6$ from the $2019$ Pan African Mathematical Olympiad.

Here's my attempt:

We're looking for all integers $n$ such that $5$ doesn't divide $\binom{2n}{n}$.

Which means we need $v_5((2n)!) - 2 v_5(n!) = 0$

By Legendre's formula, this is equivalent to:

$$
\sum_{k = 1}^\infty \left\lfloor \frac{2n}{5^k} \right\rfloor - 2\sum_{k = 1}^\infty \left\lfloor \frac{n}{5^k} \right\rfloor = 0 \\
\iff \sum_{k = 1}^\infty \left(\left\lfloor \frac{2n}{5^k} \right\rfloor - 2\left\lfloor \frac{n}{5^k} \right\rfloor\right) = 0
$$
Which is true if and only if $\left\lfloor \frac{2n}{5^k} \right\rfloor = 2\left\lfloor \frac{n}{5^k} \right\rfloor$ for all strictly positive integers $k$.

Let's find the positive real solutions to the equation $\lfloor2x\rfloor = 2 \lfloor x \rfloor$

$$
\lfloor2x\rfloor = 2 \lfloor x \rfloor \\
\iff 2x - \{2x\} = 2x - 2 \{x\} \\
\iff \{2x\} = 2 \{x\}
$$
Which is true if $\{x\} < 1/2$.

Inversely, if $\{x\} > 1/2$ then $\{2x\} > 1$ which is absurd.

It's also trivially false for $x = 1/2$.

As such, for every strictly positive integer $k$, there exists a positive integer $m_k$ such that:
$$
m_k \le \frac{n}{5^k} < m_k + \frac1{2} \\
\iff 5^k m_k \le n < 5^k m_k + \frac{5^k}{2}
$$
I can see that this can be used to eliminate many possible remainders of $n$ modulo any power of $5$.

For example, we have:
$$
5 m_1 \le n < 5 m_1 + \frac{5}{2} \\
\iff 5 m_1 \le n \le 5m_1 + 2
$$
Which means $n \mod 5 \in \{0 , 1, 2 \}$

And similarly,
$$
25 m_2 \le n \le 25m_2 + 12
$$
So $n \mod 25 \in [0,12]$

But, since $n \mod 5 \in \{0,1,2\}$, we have $n \mod 25 \in \{0,1,2,5,6,7,10,11,12\}$

We can carry on with this for higher powers of $5$.

However, I don't know how exactly to proceed from here.

I don't just want an answer.

I want the reasoning and intuition behind your answer in order to know how I should approach such problems in the future so I would really appreciate it if you include a little bit about how you found the solution if it's not really obvious. I don't mind if you also include failed attempts on the way to finding the solution.

答:
A possible way to find the answer:

Exploring the numbers for which $\left\lfloor \frac{2n}{5} \right\rfloor - 2 \left\lfloor \frac{n}{5} \right\rfloor =0$ shows the $3$ out of every $5$ integers satisfy this, with the $3$th, $6$th etc. cases being multiples of $5$.

Similarly  exploring the numbers for which $\left\lfloor \frac{2n}{5^2} \right\rfloor - 2 \left\lfloor \frac{n}{5^2} \right\rfloor =0$ shows that $13$ out of every $25$ integers satisfy this, with the $13$th, $26$th etc. cases being multiples of $25=5^2$. Overlapping this with the previous result, $9$ out of every $25$ integers satisfy these two.

You can then prove that $\left\lfloor \frac{2n}{5^k} \right\rfloor - 2 \left\lfloor \frac{n}{5^k} \right\rfloor =0$ if and only if $n \in \{0,1,2,\ldots,\frac{5^k-1}{2}\} \mod 5^k$

and by induction that $\sum\limits_{k = 1}^t \left\lfloor \frac{2n}{5^k} \right\rfloor - 2\sum\limits_{k = 1}^t \left\lfloor \frac{n}{5^k} \right\rfloor = 0  $ for $3^t$ out of every $5^t$ integers with the $3^t$th, $2\times 3^t$th etc. being multiples of $5^t$.

To find the solution using this information:

we can find $3^6=729\le 2019$ and $3^7=2187>2019$.  So $t=7$ is too big.

$2019 = 2\times 729+2\times243 + 2\times 27 + 2\times 9 +1\times 3$ which is $2\times 3^6+2\times3^5 + 2\times 3^3 + 2\times 3^2 +1\times 3^1$.

The desired $n=2\times 5^6+2\times5^5 + 2\times 5^3 + 2\times 5^2 +1\times 5 ^1 = 37805$


问:
How to express the probability $P(X \text{ mod } 2 = 0)$ that an integer-valued $X$ is even using characteristic function?
Suppose $X$ is an integer-valued random variable. My question is, how can I express the probability $P(X \text{ mod } 2 = 0)$ that $X$ is even knowing the characteristic function $\phi_X(t)$? Can I do it without using a general inversion formula by summing up $P(X=2k)$ over $k \in \mathbb{Z}$? Maybe, I could use some sort of simpler argument and end up with some nice expression?

I almost don't have any idea. I tried to use $\phi_X(2t) = \sum\limits_{k \in \mathbb{Z}} e^{2ikt}P(X=k)$, but it just doesn't work. Do I miss something trivial here?

Actually, this is only a part of the problem that I'm trying to solve. Another part is the following:

Let $\{X_n\}_{n=1}^{+\infty}$ is a sequence of i.i.d random variables, and $P(X_1 \text{ mod } 2 = 0) = p \in (0, 1)$. Let $Y_n = X_1 + X_2 + ... + X_n$. Prove that $P(Y_n \text{ mod } 2 = 0) = 1/2$.

And I want to solve this problem using characteristic functions. I suspect, for this I would need to find $\phi_{Y_n}(t) = (\phi_{Z_1}(t))^n$ with $Z_i = X_i \text{ mod } 2$.

答:
Ok let's try something else. We'll start with the probability generating function

$$f_X(q) = \sum_{k \in \mathbf Z} \mathrm{P}(X = k) q^k. $$

So the characteristic function is where $q = e^{it}$.

Notice:

$f_{Y_n} = f_X^n$

$\displaystyle \tfrac12(f_X(q) + f_X(-q)) = \sum_{k \text{ even}} \mathrm{P}(X = k) q^k$

This is called "bisecting" the series.

So the probability that $Y$ is even is

$$\mathrm{P}(Y \text{ is even}) = \frac{f_X(1)^n + f_X(-1)^n}{2} = \frac12 + \frac{f_X(-1)^n}{2}.$$

You can write this in terms of the characteristic function by choosing appropriate values for $q = e^{it}$ (namely $t = 0, t = \pi$).

Also notice that

$$\mathrm{P}(X \text{ is even}) = p = \frac12 + \frac{f_X(-1)}{2}. $$

So combining these gives the limit of $1/2$.


问:
Proof of Sylow's Theorem (Herstein) - why is $no(H) = o(G)$?
The theorem is:

(Sylow's theorem): If $p$ is a prime number, and $p^\alpha |o(G)$, then $G$ has a subgroup of order $p^\alpha$.

Right before the proof, the author has established that if $n = p^\alpha m$ (where $p$ is prime), and if $p^r|m$ but $p^{r+1}\nmid m$, then $$p^r | {p^\alpha m\choose p^\alpha} \text{ but } p^{r+1} \nmid {p^\alpha m\choose p^\alpha}$$
This notation is used in the proof too.

I'm posting an image here, and the highlighted part is what bothers me. I'm sorry I didn't type it all up in MathJax, but as you can imagine it is a lot to type!

Question: How do I prove that $no(H) = o(G)$?

It's probably some class equation or such but I am not very comfortable with those yet and I would appreciate any help! Thanks a lot.

Clarification: $no(H)$ means $n$ multiplied with $o(H)$, the order of $H$.

Update (based on comments):
Herstein's proof never explicitly introduced group actions, and therefore I'm having trouble connecting it to the orbit-stabilizer theorem. Could someone please spell it out? In terms of group actions, what does the equivalence relation $\sim$ (and equivalence classes thereof) in Herstein's proof correspond to? What does $H$ mean? $H$ looks very similar to what a stabilizer of an element is defined as, but it's definitely not the same concept.

答: $G$ acts (from right) on $\mathcal{M}$ by right multiplication (in fact, $|Mg|=|M|$ for every $M\in\mathcal{M}$ and $g\in G$). $H$ is $\operatorname{Stab}(M_1)$ and $\{M_1,\dots,M_n\}$ is $\operatorname{Orb}(M_1)$. By the orbit-stabilizer theorem, then, $|\operatorname{Orb}(M_1)||\operatorname{Stab}(M_1)|=|G|$, i.e, in book's notation, "$no(H)=o(G)$".


问:
Using the Saddle point method (or Laplace method) for a multiple integral over a large number of variables
I am trying to understand the saddle point method used in the large N limit of matrix models.

First, for the case of the integral of a single variable I found this notes

There they say that you can approximate the integral

$$I(A)=\int_{-\infty}^\infty e^{Ag(x)} dx$$
by expanding $A g(x)$ in powers of $y=(x-x_0)\sqrt{A}$

$$Ag(x)=Ag(x_0)+\frac{1}{2}g''(x_0)y^2+\frac{g'''(x_0)}{6\sqrt{A}}y^3+\dots$$

where $x_0$ is a maximum of $g(x)$. Then we expand the exponential $e^{Ag(x)}$

$$e^{Ag(x)}=e^{Ag(x_0)}e^{\frac{1}{2}g''(x_0)y^2}(1+\frac{g'''(x_0)}{6\sqrt{A}}y^3+\frac{g''''(x_0)}{72A}y^4+\dots)$$

so now

$$I(A)=e^{Ag(x_0)}\int_{-\infty}^\infty \frac{dy}{\sqrt{A}}e^{\frac{1}{2}g''(x_0)y^2}(1+\frac{g'''(x_0)}{6\sqrt{A}}y^3+\frac{g''''(x_0)}{72A}y^4+\dots) $$

The first integral is a gaussian, the others are known, so this gives an asymptotic series

$$I(A)=e^{Ag(x_0)}\sqrt{\frac{2\pi}{-Ag''(x_0)}}(1+\frac{C_2}{A}+\frac{C_4}{A^2}+\dots) $$

Now, I want to do the same for the large N limit of partition function of a matrix model. The partition function is a multiple integral over a large number of variables

$$Z=\int_{-\infty}^\infty [dm] e^{-N^2 S(m)}$$

$$[dm]=\prod_{j=1}^{N}dm_j$$

Here, the action is a function of $N$ variables. For the gaussian model for example

$$S(m)=\frac{2}{\lambda N}\sum_{j'=1}^N m_{j'}^2-\frac{2}{N^2}\sum_{j'=2}^N\sum_{i=1}^{j'-1}\ln|m_i-m_{j'} |$$

Let's say $m_{(0)}$ is the minimun of $S(m)$. Using Taylor expansion of several variables we have

$$e^{-N^2S(m)}=e^{-N^2S(m_{(0)})} \ e^{-\frac{1}{2}(u\cdot\nabla)^2S(m_{(0)})} \ e^{-\frac{1}{3!N}(u\cdot\nabla)^3S(m_{(0)})-\frac{1}{4!N^2}(u\cdot\nabla)^4S(m_{(0)})+\dots}$$

where the vector $u$ is

$$u=N(m-m_{(0)})$$

$$e^{-N^2S(m)}=e^{-N^2S(m_{(0)})} \ e^{-\frac{1}{2}(u\cdot\nabla)^2S(m_{(0)})} \left(1 -\frac{1}{3!N}(u\cdot\nabla)^3S(m_{(0)})-\frac{1}{4!N^2}(u\cdot\nabla)^4S(m_{(0)})+\dots\right)$$

so now

$$Z=N^{-N}e^{-N^2S(m_{(0)})}\times\int [du]e^{-\frac{1}{2}(u\cdot\nabla)^2S(m_{(0)})} \left(1 -\frac{1}{3!N}(u\cdot\nabla)^3S(m_{(0)})-\frac{1}{4!N^2}(u\cdot\nabla)^4S(m_{(0)})+\dots\right)$$

Now, the first integral is a gaussian. The second integral

$$\int [du]e^{-\frac{1}{2}(u\cdot\nabla)^2S(m_{(0)})}\frac{1}{3!N}(u\cdot\nabla)^3S(m_{(0)})=0$$

The problem is in the third integral

$$C=\int [du]e^{-\frac{1}{2}(u\cdot\nabla)^2S(m_{(0)})}\frac{1}{4!N^2}(u\cdot\nabla)^4S(m_{(0)})$$

Since

$$(u\cdot\nabla)^4S(m_{(0)})=\sum u_{i_1}u_{i_2}u_{i_3}u_{i_4}\partial_{i_1}\partial_{i_2}\partial_{i_3}\partial_{i_4}S(m_{(0)})$$

we have

$$C=\frac{1}{4!N^2}\sum \partial_{i_1}\partial_{i_2}\partial_{i_3}\partial_{i_4}S(m_{(0)})\int [du]u_{i_1}u_{i_2}u_{i_3}u_{i_4}e^{-\frac{1}{2}(u\cdot\nabla)^2S(m_{(0)})}$$

The sum over indices will give an order $N^4$ so it seems that $C$ is bigger than the gaussian integral, the opposite of what you expect.

What is the problem here? Have I done the power counting properly? I want to get an asymptotic series where the terms get smaller and smaller.

答:
Provided that $(u\cdot\nabla)^2S(m_{(0)})$ is diagonal in u, something I do not know if it is true in your case, you only have to take into account the terms with even powers in the components of u, because the integral of a Gaussian with any odd power cancels out.

A quick numerical check with Mathematica shows that you have something close to $N^{2}$ non zero elements.


答:
Following on from your assessment, consider writing $n$ in base $5$; then every digit must be $\in \{0,1,2\}$, so valid values for $n$ in base $5$ are only those which look like numbers in base $3$.

Then we have $2019=[2202210]_3$ so  $n=[2202210]_5 = 37805$

Title: The $2019$th strictly positive integer $n$ such that $\binom{2n}{n}$ is not divisible by $5$

Body:
This is problem $6$ from the $2019$ Pan African Mathematical Olympiad.

Here's my attempt:

We're looking for all integers $n$ such that $5$ doesn't divide $\binom{2n}{n}$.

Which means we need $v_5((2n)!) - 2 v_5(n!) = 0$

By Legendre's formula, this is equivalent to:

$$
\sum_{k = 1}^\infty \left\lfloor \frac{2n}{5^k} \right\rfloor - 2\sum_{k = 1}^\infty \left\lfloor \frac{n}{5^k} \right\rfloor = 0 \\
\iff \sum_{k = 1}^\infty \left(\left\lfloor \frac{2n}{5^k} \right\rfloor - 2\left\lfloor \frac{n}{5^k} \right\rfloor\right) = 0
$$
Which is true if and only if $\left\lfloor \frac{2n}{5^k} \right\rfloor = 2\left\lfloor \frac{n}{5^k} \right\rfloor$ for all strictly positive integers $k$.

Let's find the positive real solutions to the equation $\lfloor2x\rfloor = 2 \lfloor x \rfloor$

$$
\lfloor2x\rfloor = 2 \lfloor x \rfloor \\
\iff 2x - \{2x\} = 2x - 2 \{x\} \\
\iff \{2x\} = 2 \{x\}
$$
Which is true if $\{x\} < 1/2$.

Inversely, if $\{x\} > 1/2$ then $\{2x\} > 1$ which is absurd.

It's also trivially false for $x = 1/2$.

As such, for every strictly positive integer $k$, there exists a positive integer $m_k$ such that:
$$
m_k \le \frac{n}{5^k} < m_k + \frac1{2} \\
\iff 5^k m_k \le n < 5^k m_k + \frac{5^k}{2}
$$
I can see that this can be used to eliminate many possible remainders of $n$ modulo any power of $5$.

For example, we have:
$$
5 m_1 \le n < 5 m_1 + \frac{5}{2} \\
\iff 5 m_1 \le n \le 5m_1 + 2
$$
Which means $n \mod 5 \in \{0 , 1, 2 \}$

And similarly,
$$
25 m_2 \le n \le 25m_2 + 12
$$
So $n \mod 25 \in [0,12]$

But, since $n \mod 5 \in \{0,1,2\}$, we have $n \mod 25 \in \{0,1,2,5,6,7,10,11,12\}$

We can carry on with this for higher powers of $5$.

However, I don't know how exactly to proceed from here.

I don't just want an answer.

I want the reasoning and intuition behind your answer in order to know how I should approach such problems in the future so I would really appreciate it if you include a little bit about how you found the solution if it's not really obvious. I don't mind if you also include failed attempts on the way to finding the solution.


问:
Smallest prime of the form 41033333333...?
What is the smallest prime of the form $410333333333...$ ?
 It should have more than $10,000$ digits.

[added from answer posted 2013 May 26 at 20:52 by Peter]

I thought it would be clear, but it seems not to be.
 Of course, after $410$, only $3$'s should follow.
 Otherwise, it would be very easy to find primes.
 I checked the numbers up to about $10,000$ digits,
 but of course, I would be glad if someone checks
 this, too.

I do not understand the question, WHY this number
 is interesting for me. Mersenne primes are not so
 much more interesting, but recently a prize of
 $100,000\$ $was payed for a community finding a
 $17$-million-digit mersenne prime. I would have
 better ideas what to do with all this money ...

Body:
41033333333323 = 41033333333300 + 23 is prime.
So is 4103333333333333333333000159.

Set $x= 4103333333333333333333...$ ($k$ times a 3). Then $\log x \approx 6 + 2.3 \cdot k$. By the Prime Number Theorem, you'll find a prime of the form $10^nx + r, \, r < 10^n$ with high probability (let's say 0.999999) if $10^n > 100 + 30k$ or so. That is, $n \ge 3 + \log_{10} k$ should be enough.


问:
How to Decompose a matrix (in cases rank=1) into two row vectors?
Suppose I have matrix A. rank=1. how can I decompose it into to row vectors? meaning that A=u*v(transpose)
I have found a way to do it when the matrix only has 1 row or 1 column, but how can I do it in other cases as well? (I am trying to do it on python but I need to understand the math behind it as well).
Thanks!</p>

答:
The rank of a matrix  $A$ is the dimension of the vector space spanned by its columns. So, if a matrix has rank one, the vector space spanned by its columns is spanned by one vector $\mathbf v = \begin{pmatrix}v_1\\ \vdots\\ v_n \end{pmatrix}$. We can always take $\mathbf v$ to be the first column of $A$.

By the definition of span, this means that the $i^{th}$ column of the matrix can be written as $a_i \mathbf v$, where $a_i$ is a scalar.
Writing this out, we have that the $i^{th}$ column is $\begin{pmatrix}a_iv_1\\ \vdots\\ a_iv_n \end{pmatrix}$. Thus, the $(j, i)$ entry of the matrix is $a_i  v_j$.

Now, we can put all of the scalars $a_1, ..., a_m$ into a vector $\mathbf a = \begin{pmatrix}a_1\\ \vdots\\ a_m\end{pmatrix}$. Then we can write $A$ as $\mathbf v \cdot \mathbf a^T $, where  $\mathbf a^T$ is the transpose of $\mathbf a$.

Here's an example: let $A = \begin{pmatrix} 2 & 3 \\ 4 & 6 \end{pmatrix}$. We take $\mathbf v = \begin{pmatrix} 2 \\ 4 \end{pmatrix}$. Now, we need to find $\mathbf a = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}$. Since we choose $\mathbf v$ to be the first column, $a_1 = 1$. Now, we need to find $a_2$ such that $a_2 \begin{pmatrix} 2 \\ 4 \end{pmatrix} = \begin{pmatrix} 3 \\ 6 \end{pmatrix}$. We can compute this by making the first entries match up by solving $2a_2 = 3$, $a_2 = 3/2$. This must work to make the second entry match up as well--otherwise, the second column wouldn't really be in the span of the first column, and the matrix wouldn't really be rank one.  Thus, we have $\mathbf a = \begin{pmatrix} 1 \\ 3/2\end{pmatrix}$. Thus, we have $$A = \mathbf v \cdot \mathbf a^T =\begin{pmatrix} 2 \\ 4 \end{pmatrix}\begin{pmatrix} 1 &3/2\end{pmatrix}. $$


问:
Application of Dynkin's formula
Let $x\in\mathbb{R}^d, (W_t)_{t\geq0}$ be a $d$-dim. Brownian motion. I have the following processes
$$
Z_{t, x} = b(Z_{t, x})dt + \sigma(Z_{t, x})dW_t, \qquad Z_{0, x} = x \\
V_{t, v} = \nabla b(Z_{t,x})V_{t,v}dt + \nabla \sigma(Z_{t,x})V_{t,v}dW_t, \qquad V_{0, v} = v\\
U_{t, v, v'} = (\nabla b(Z_{t,x})U_{t, v, v'} +  \nabla^2b(Z_{t, x})[V_{t, v'}]V_{t,v})dt \\ + (\nabla \sigma(Z_{t,x})U_{t, v, v'} +  \nabla^2\sigma(Z_{t, x})[V_{t, v'}]V_{t,v})dW_t, \qquad U_{0, v, v'}=0,
$$
i.e. a stochastic process and its first and second variation process, formally derived by differentiating $Z_{t, x}$ in direction $v\in\mathbb{R}^d$, and then $v'\in\mathbb{R}^d$.

Let $(P_t)_{t\geq0}$ denote the transition semigroup, i.e. $(P_tf)(x)=\mathbb{E}[f(Z_{t,x})]$. Can someone explain, why Dynkin's formula would lead to

$$
\mathbb{E}\left[ (P_tf)(x) \int_0^t\langle \sigma^{-1}(Z_{s,x})U_{s, v, v'}, dW_s \rangle \right] = 0?
$$
To my knowledge, letting $\mathcal{A}$ denote the infinitesimal generator, Dynkin's formula reads
$$
(P_tf)(x) = \mathbb{E}[f(Z_{t,x})] = f(x) + \mathbb{E}\left[\int_0^t(\mathcal{A}f)(Z_{s,x})ds\right] \\
= f(x) + \mathbb{E}\left[\int_0^t \langle b(Z_{s,x}), \nabla f(Z_{s,x}) \rangle +  \langle \sigma(Z_{s,x})\sigma^T(Z_{s,x}), \nabla^2 f(Z_{s,x}) \rangle_{\mathrm{F}} ds\right],
$$
where $\langle \cdot, \cdot \rangle_F$ denotes the Frobenius scalar product. I don't see how that helps here...

答:
Introduction to the paper

This answer focuses on the paper attached here. The paper measures sample quality using diffusions. The main idea for this is the idea of Stein's discrepancy method, a powerful idea in probability theory which allows the upper bounding of usual probabilistic distances by an appropriate Stein discrepancy. The Stein discrepancy involves the use of Ito diffusions, and in order to ensure that the Stein operator is well behaved, one needs good bounds on the behaviour of the underlying semigroup of the Ito diffusion, which we call $P_t$. The Ito diffusion will be called $Z_{t,x}$, the initial point being $x$ and time being $t$. We have $P_t f(x) = \mathbb E[f(Z_{t,x})]$.

On the use of Dynkin's formula

The main reference for stochastic calculus of the paper is Avner Friedman, Stochastic Differential Equations And Applications. As per the paper , equation (7.10) of the text has the following statement :

Dynkin's formula : Under suitable integrability conditions ,
$$
\mathbb Eu(\xi(t),t) - \mathbb Eu(\xi(0),0) = \mathbb E\int_0^\tau Lu(\xi(s),s)ds
$$
where $u(\psi(t),t)$ is a suitable function, $d\xi(t) = A(t)dt + B(t)dw(t)$ is an Ito diffusion and $L = \frac{1}{2}\sum_{i,j} (B^TB)_{ij} \frac{\partial^2}{\partial x_i\partial x_j} + \sum_{i} A_i\frac{\partial}{\partial x_i} + \frac{\partial}{\partial t}$ is the associated semigroup operator of $\xi$.

It is clear, that the use of Dynkin's formula must involve differentiation of a certain kind, since $L$ involves the same.

For a clear use of Dynkin's formula, we witness page number $32$ of the paper, where we see that in both applications of Dynkin's formula, derivatives clearly appear on the functions $\sigma$ and $b$.

The story of $\mathbb E[J_3(x)]$

To compute $\mathbb E[J_3(x)]$, nowhere is Dynkin's formula used, unlike what the author says. Indeed, what is used is equation $(29)$ at the end of page number $30$.

$$
f(Z_{t,x}) = (P_tf)(x) + \int_0^t \langle \Delta (P_{t-s}f)(Z_{s,x}) ,\sigma(Z_{s,x}) dW_s\rangle
$$

which is derived by applying the Ito formula to $(s,x) \to P_{t-s}f(x)$, a $C^2$ function.

From there, we have :
\begin{align}
\mathbb E[J_3(x)] & = \frac{1}{t} \mathbb E\left[f(Z_{t,x}) \int_0^t \left\langle \sigma^{-1}(Z_{s,x})U_{s,v,v'},dW_s\right\rangle\right]  \\ &= \frac{1}{t} \mathbb E\left[\left((P_tf)(x) + \int_0^t \langle \Delta (P_{t-s}f)(Z_{s,x}) ,\sigma(Z_{s,x}) dW_s\rangle\right) \int_0^t \left\langle \sigma^{-1}(Z_{s,x})U_{s,v,v'},dW_s\right\rangle\right] \\ &= \frac 1t \mathbb E\left[P_tf(x)\int_0^t \left\langle \sigma^{-1}(Z_{s,x})U_{s,v,v'},dW_s\right\rangle\right] \\ &+ \mathbb E\left[\int_0^t \langle \Delta (P_{t-s}f)(Z_{s,x}) ,\sigma(Z_{s,x}) dW_s\rangle\int_0^t \left\langle \sigma^{-1}(Z_{s,x})U_{s,v,v'},dW_s\right\rangle\right]
\end{align}

Of these, the first term vanishes since the $P_tf(x)$ comes outside the integral and we are left with the expectation of a stochastic integral, which is a martingale starting at $0$ hence has constant expectation equal to $0$. For the second term, we observe the form of Friedman's Ito theorem statement and how it was translated here, and obtain by expansion a result which I find bewildering but definitely true :
$$
\left\langle\Delta (P_{t-s}f)(Z_{s,x}) ,\sigma(Z_{s,x}) dW_s \right\rangle= \sum_{l=1}^n \sum_{i=1}^m [\Delta (P_{t-s}f)(Z_{s,x})]_i  \sigma(Z_{s,x})_{il} (dW_s)_l \\ = \sum_{l=1}^n (\Delta P_{t-s}f(Z_{s,x})\sigma(Z_{s,x}))_l (dW_s)_l =\left\langle \Delta (P_{t-s}f)(Z_{s,x}) \sigma(Z_{s,x}), dW_s   \right\rangle
$$

which is kind of saying that the $\sigma$ term can come from right to left side. I don't know why it was kept on the right in the first place, this could have even been a typo for all I know.

Anyway, now we come to :
$$
\mathbb E\left[\int_0^t \langle \Delta (P_{t-s}f)(Z_{s,x})\sigma(Z_{s,x}), dW_s\rangle\int_0^t \left\langle \sigma^{-1}(Z_{s,x})U_{s,v,v'},dW_s\right\rangle\right]
$$

and are in the "covariance of Ito-Integrals" territory. Indeed, by equation (7.1) of Friedman applied coordinatewise, we get :
$$
\mathbb E\left[\int_0^t \langle \Delta (P_{t-s}f)(Z_{s,x})\sigma(Z_{s,x}), dW_s\rangle\int_0^t \left\langle \sigma^{-1}(Z_{s,x})U_{s,v,v'},dW_s\right\rangle\right]
 \\ = \mathbb E\left[\int_0^t \left\langle \Delta (P_{t-s}f)(Z_{s,x})\sigma(Z_{s,x}), \sigma^{-1}(Z_{s,x})U_{s,v,v'}\right\rangle ds\right]$$

whence we can again use the same coordinatewise expansion logic we applied last time in the Ito integral form, for this expression to bring the $\sigma^{-1}$ over to the left of the $\sigma$ and cancel out, to get :
$$
\mathbb E\left[\int_0^t \left\langle \Delta (P_{t-s}f)(Z_{s,x}), U_{s,v,v'}\right\rangle ds \right]
$$

which is the final expression obtained by the authors of the paper as well.

Final thoughts

I feel that the paper skips many steps, and struggles to integrate its notation with that of the standard reference. However, the use of the Stein discrepancy in this particular instance is very nice, and I would suggest going through the paper without really looking at the details and just understanding the flow and what is happening in general, because it should not take too much effort for an expert to fill in details if and when required.

Title: Application of Dynkin's formula


问:
Recurrence relation related to integer partition
I have to prove a recurrence relation which is given by:

$p_{(n,S)} = \displaystyle\sum_{t=0}^{\lfloor \frac{n}{s} \rfloor}$${p_{(n − st , S-\{s\})}}$

assuming that $\{s\}$ is an arbitrary fixed element of $S$.

(We assume that $p_{(0,S')}$ is to be equal to $1$ for any $S'$ and  $p_{(m,S')}$ is  equal to $0$ for any
negative $m$ and every $S'$.)

so at last $p_{(n,S)}$ is the integer partition of $n$ from the set of allowed elements $S$. I read about integer partition and young's diagram and even went through some videos reagrading how we can construct a generating function for $p_{(n)}$ but I am unable to prove this recurrence relation. Any help will be appreciated.

Thanks in advance.

答:
This is an elaboration on Mike Earnest's comment and a slightly more formal statement of your claim.

For $S \subseteq \mathbb{N}$ and any $s \in S$,
$$ p(n,S) = \sum_{t=0}^{\lfloor n/s \rfloor} p(n-ts,S \setminus \{s\}).$$

Proof: A partition of $n$ included in the count $p(n,S)$ has some number of parts $s$ and the other parts come from $S \setminus \{s\}$.  More explicitly, if $s$ occurs exactly $t$ times in $\lambda \vdash n$, then removing those $t$ copies of $s$ from $\lambda$ leaves a partition of $n-ts$ with parts from $S \setminus \{s\}$.  There are $p(n-ts,S \setminus \{s\})$ of the "leftover" partitions.  Note that $t$ can vary from 0 (i.e., $\lambda$ contains no part $s$) to $\lfloor n/s \rfloor$ (the maximal possible number of parts $s$).


问:
Conjugate of negative normalized entropy
Given$$f(x) = \sum_{i=1}^n x_i \log \left( \frac{x_i}{\sum_{j=1}^n x_j} \right), \qquad dom f = R^n_{++}$$
Find $f^*(y)$

$$f^*(y) =
\sup_x ( y^T x - f(x) ) =$$
$$
\sup_x \left( y^T x - \sum_{i=1}^n x_i \log \left( \frac{x_i}{\sum_{j=1}^n x_j} \right) \right) =$$
$$
\sup_x \sum_{i=1}^n x_i \left(y_i - \log \left( \frac{x_i}{\sum_{j=1}^n x_j} \right) \right) =$$
$$
\sup_x \sum_{i=1}^n x_i \left(\log e^{y_i} - \log \left( \frac{x_i}{\sum_{j=1}^n x_j} \right) \right) =$$
$$
\sup_x \sum_{i=1}^n x_i \log \left( \frac{e^{y_i}\sum_{j=1}^n x_j}{x_i} \right)$$
Unfortunately, I can't come up what to do next. This is problem 3.41 from Convex Optimization by Boyd and Vandenberghe, and there is an answer to this problem available:

$$
\begin{equation*}
f^*(y) =
 \begin{cases}
   0 & \sum_{i=1}^n e^{y_i} \leq 1 \\
   +\infty & \text{otherwise}
 \end{cases}
\end{equation*}$$

答:
In case $\sum_{i=1}^n e^{y_i} > 1$, setting $x_i = r  \cdot e^{y_i}$ yields $r \cdot \sum_{i=1}^n e^{y_i} \cdot \log \sum_{i=1}^n e^{y_i}$, which goes to infinity for $r \rightarrow \infty$.

Write $S(x) = \sum_{j=1}^n x_j$.
The first derivatives of the $\sup$ objective are
\begin{align}
 \frac{\partial}{\partial x_i} &\left( y^T x - x^T \log x + \sum_{i} x_i \log S(x) \right) = y_i - \log x_i + \log S(x)   \ ;
\end{align}
the gradient being zero thus means that $\forall i: \frac{x_i}{S(x)} = e^{y_i}$. This can be fulfilled if and only if $\sum_i e^{y_i} = 1$, and then with $x_i = r \cdot e^{y_i}$ for some $r>0$.
The second derivative of the objective is
\begin{align}
   \frac{\partial^2 \left( y^T x - x^T \log x + \sum_{i} x_i \log S(x) \right)}{\partial x_i \partial x_j}(x) = -\frac{\delta_{ij}}{x_j} + \frac{1}{S(x)}
\end{align}
which is negative definite:
For $v \in \mathbb{R}^n$,
\begin{align}
 \sum_{ij} v_i  \left(-\frac{\delta_{ij}}{x_j} + \frac{1}{S(x)}\right) v_j
 &=  - \sum_{i}\frac{v_i^2}{x_i} + \frac{\left(\sum_{i} v_i\right)^2}{S(x)} \leq 0
\end{align}
where we used Jensen's inequality on the squaring function:
\begin{align}
 \sum_{i}\frac{v_i^2}{x_i}
 &= \frac{1}{S(x)}\sum_{i}\frac{v_i^2S(x)}{x_i} \\
 &= \frac{1}{S(x)}\sum_{i}\frac{x_i}{S(x)}\left(\frac{v_iS(x)}{x_i}\right)^2 \geq
  \frac{1}{S(x)}\left(\sum_{i}\frac{x_i}{S(x)}\frac{v_i S(x)}{x_i}\right)^2
  = \frac{\left(\sum_{i} v_i\right)^2}{S(x)} \ .
\end{align}

Thus, in case $\sum_i e^{y_i} = 1$ we have a global maximum with value  $h^*(y) = \sum_{j=1}^n r \cdot e^{y_j} y_j - \sum_{i=1}^n r \cdot e^{y_i}\log e^{y_i} = 0$.

In the last case $\sum_i e^{y_i} < 1$, there is $\hat{y} \succ y$ with $\sum_i e^{\hat{y}_i} = 1$. For any $x\succ0$ the term $x^T y - \sum_{i=1}^n x_i\log \frac{x_i}{S(X)}$ is increasing in $y_i$.
Therefore, $h^*(y) \leq h^*(\hat{y}) = 0$. Again taking $x_i = r \cdot e^{y_i}$ (also possible: simply $x_i = r$), we see that $0$ is realized as the supremum with $r \rightarrow 0$.


问:
Show that $v + iu$ is the reflection of the point $u + iv$ across the line $u = v$
Show that $v + iu$ is the reflection of the point $u + iv$ across the line $u = v$

My attempt :I know that the reflection of the point $(x,y)$ across
the line $y = x $ is the point $(y, x)$

i,e  line of equation will be  $x-y=0$

But here im confused that how to show $v + iu$ is the reflection of the point $u + iv$

答: You pretty much have it. If you know that the reflection of $(x,y)$ across $x=y$ is $(y,x)$, then just note that we identify $a+bi$ with $(a,b)$ when we identify $\Bbb{C}\cong \Bbb{R}^2$. In particular, the reflection across $x=y$ sends $a+bi$ to $b+ai$.


问:
Surjective function on product space
I know that, if $U$ and $V$ are closed subspaces of a Hilbert $(X,\langle\cdot,\cdot\rangle)$, then these statements are equivalents:
$$i)\ U^{\perp}\subseteq U+V\quad\quad\quad ii)\ X = U+V\quad\quad\quad iii)\ U^{\perp}+V^{\perp} \subseteq U+V.$$

Now, I want to prove this:

Let $X,Y_1, Y_2$ be Hilbert spaces, and $Y = Y_1\times Y_2$. Let $\mathcal{B}\in\mathcal{L}(X,Y)$ such that $\mathcal{B}(x) = \left(\mathcal{B}_1(x), \mathcal{B}_2(x)\right)^T$, $\forall x\in X$, where $\mathcal{B}_i\in\mathcal{L}(X,Y_i)$, $\forall i\in\{1,2\}$. Show that $\mathcal{B}$ is onto if and only if
$$i)\ \mathcal{B}_1 \mbox{ and } \mathcal{B}_2 \mbox{ are onto}.\quad\quad\quad ii)\ X = \mbox{Ker}(\mathcal{B}_1) + \mbox{Ker}(\mathcal{B}_2).$$

Please can sombody help me with this problem?

Thanks in advance

答:
First, suppose that $\mathcal{B}$ is onto. Let $\pi_1: Y_1 \times Y_2 \to Y_1$ and $\pi_2: Y_1 \times Y_2 \to Y_2$ be the orthogonal projections onto the factors $Y_1$ and $Y_2$ of $Y_1 \times Y_2$, which are surjective. Then $\mathcal{B}_1 = \pi_1 \circ \mathcal{B}$ and $\mathcal{B}_2 = \pi_2 \circ \mathcal{B}$ are compositions of surjections, and so... On the other hand, let $x \in X$, and by surjectivity of $\mathcal{B}$, let $x^\prime \in X$ be such that $\mathcal{B}(x^\prime) = (0,\mathcal{B}_2(x))$. Then $\mathcal{B}_1(x^\prime) = 0$ and $\mathcal{B}_2(x-x^\prime) = \mathcal{B}_2(x)-\mathcal{B}_2(x^\prime) = 0$, so that $x = x^\prime + (x-x^\prime) \in \ker \mathcal{B}_1 + \ker \mathcal{B}_2$.

Now, suppose that $\mathcal{B}_1$ and $\mathcal{B}_2$ are onto and that $X = \ker\mathcal{B}_1 + \ker\mathcal{B}_2$. Let $(y_1,y_2) \in Y_1 \times Y_2$, so that $y_1 = \mathcal{B}_1 x_1$ and $y_2 = \mathcal{B}_2 x_2$ for some $x_1$ and $x_2 \in X$. Since $X = \ker \mathcal{B}_1 + \ker \mathcal{B}_2$, write,
$$
 x_1 = x_1^\prime + x_1^0, \quad x_1^\prime \in \ker \mathcal{B}_2, \; x_1^0 \in \ker \mathcal{B}_1,\\
 x_2 = x_2^\prime + x_2^0, \quad x_2^\prime \in \ker \mathcal{B}_1, \; x_2^0 \in \ker \mathcal{B}_2,\\
$$
and observe that
$$
 y_k = \mathcal{B}_k(x_k) = \mathcal{B}_k(x_k^\prime+x_k^0) = \mathcal{B}_k(x_k^\prime) + \mathcal{B}_k(x_k^0) = \mathcal{B}_k(x_k^\prime), \quad k=1,2.
$$
What, then, is $\mathcal{B}(x_1^\prime+x_2^\prime)$?

Title: Surjective function on product space


问:
If $K, K' \subset \mathbb{S}^2$ are compact and $K \simeq K'$, is $\mathbb{S}^2 \setminus K \simeq \mathbb{S}^2 \setminus K'$?
One of the most immediate consequences of the Alexander-Pontryagin Duality Theorem is that if $X$ is compact and $f, g: X \rightarrow \mathbb{R}^n$ are two embeddings, then $\mathbb{R}^n \setminus f(X)$ and $\mathbb{R}^n \setminus g(X)$ have the same number of components.  The same applies for $\mathbb{S}^2$.

But in the plane, there are no 'wild sets'.  So is it in fact the case that $\mathbb{R}^2 \setminus f(X) \simeq \mathbb{R}^2 \setminus g(X)$ when $X$ is connected?  A union of two disjoint circles are a counterexample if we don't assume $X$ is connected, depending on whether one is in the interior region of the other.

I thought I had seen some counterexample to this in the plane, some sort of graph, but now I can't think of it and it wasn't in the book I thought I saw it in so now I'm feeling uncertain.  Note that $f$ and $g$ would have to be inequivalent embeddings, because every isotopy of compact subsets in the plane extends to an ambient isotopy.

EDIT: As pointed out in a comment below, it's true.  The proof outlined here uses shape theory: Homotopy type of the complement of a subspace

So now my question is, is there a more elementary proof?  It doesn't have to be for the full-strength "only homotopy type matters" of the above theorem, just when $K, K'$ are homeomorphic would be satisfactory to me.

答:
Here is a proof via the Alexander duality (using the form of duality with the Chech cohomology of $K$):

If $K\subset S^2$ is compact and connected, then for $\Omega= S^2\setminus K$ we have $H_1(\Omega)=0$.

Thus, each component $\Omega_i$ of $\Omega$ is an oriented surface with $H_1(\Omega_i)=0$. One then proves that such $\Omega_i$ is simply-connected. One proof is that for each noncompact connected surface $S$, $\pi_1(S)$ is free. Then apply Hurewicz theorem in conjunction with the fact that $H_1(\Omega_i)=0$.

Thus, assuming that $K$ is nonempty, Hurewicz theorem applied one more time implies that each $\Omega_i$ is contractible.

Applying the Alexander duality again, we see that ${H}_0(\Omega)\cong \check{H}^1(K)$. In other words, the homotopy type of $K$ also determines the number of connected components of $\Omega$. Hence, the homotopy type of $K$ determines the homotopy type of $\Omega=S^2\setminus K$.

One can do better, of course, and note that each connected noncompact simply-connected surface $S$ is homeomorphic to the plane. Proving this is not hard once you know the classification of surfaces; in fact, it is enough to know that $S$ is orientable and has $b_1(S)=0$ (thus, simplifying the proof above and avoiding most of the Algebraic Topology arguments):

Exhaust $S$ by compact subsurfaces with (smooth) boundary $S_k$ (find a smooth proper function on $S$ and consider its regular sublevel sets). Use the assumption that $b_1(S)=0$ to conclude that each boundary component of $S_k$ bounds a compact subsurface in $S$. Using this, enlarge each subsurface $S_k$ to a compact subsurface $S'_k$ which has exactly one boundary component and no handles (since $b_1=0$). WLOG, the surfaces $S'_k$ still form an exhaustion:
$$
S'_k\subset int(S'_{k+1}).
$$
Now, using the classification of surfaces conclude that $S'_k$ is homeomorphic to the disk and each $S'_{k+1} \setminus int(S'_k)$ is homeomorphic to the annulus. Using this information, construct a homeomorphism $S\to R^2$.

See also my answer here.

With this exhaustion argument in mind, you only need algebraic topology to relate the number of connected components of $R^2\setminus K$ to the topology of $K$. One can do this without algebraic topology but it will be long and complicated (since, among other things, you will be reproving Jordan curve theorem) and, as far as I am concerned, pointless.

To conclude: The homotopy type of a connected and compact subset $K\subset S^2$ determines the topology of $S^2 \setminus K$.


问:
Can a multiset be a subset of a set?
This is a bit of a silly question but its bothering me.

Given a multiset say $p=\{a,a,g,h,h\}$
And another set $t=\{a,g,h\}$
Can I say that $p\subset t$.

In other words is p a subset of t?.

答:
The problem is that there are several different, more or less equivalent, ways to present a multiset. Here are two of them.

A multiset is a pair $(X, \sim)$, with $X$ a set and $\sim$ an equivalence relation on $X$. In your case, one can make $p=(P, \sim_{p})$, with
$$P=\{a, a^{*}, g, h, h^{*}\},$$
and $a\sim_{p} a^{*}$ and $h\sim_{p} h^{*}$ (meaning, in a way, that $a$ and $a^{*}$, and $h$ and $h^{*}$, are the same elements, $a$ and $h$ respectively). In that case, a set is a multiset where $\sim$ is the identity, and we say, for two multisets $m_{1}=(X_{1}, \sim_{1})$ and $m_{2}=(x_{2}, \sim_{2})$, that $m_{1}$ is contained in $m_{2}$ if
$$X_{1}\subseteq X_{2}\quad\text{and}\quad \forall x_{1}, x_{2}\in X_{1} (x_{1}\sim_{1}x_{2}\Rightarrow x_{1}\sim_{2}x_{2})$$
(you could, instead of $X_{1}\subseteq X_{2}$, use that there is an injective function, but that would probably be considered more general). Then it is clear that, under this definition, $p$ is not a submultiset of $t=(T, \sim_{t})$, for $T=\{a, g, h\}$ and $\sim_{t}$ the identity on $T$; BUT, $t$ is a submultiset of $p$, and $T$ is a subset of $P$.

A multiset is a pair $(X, m)$, with $X$ a set and $m$ a function mapping elements of $X$ to cardinals. That way, we can take $p=(P, m_{p})$, with $P=\{a, g, h\}$, and $m_{p}(a)=2$, $m_{p}(g)=1$ and $m_{p}(h)=2$ (meaning $a$ and $h$ appear twice in $p$, while $g$ appears once). We say $(X_{1}, m_{1})$ is a submultiset of $(X_{2}, m_{2})$ if
$$X_{1}\subseteq X_{2}\quad\text{and}\quad \forall x\in X_{1}(m_{1}(x)\leq m_{2}(x)).$$
If we define $t$ as $(T, m_{t})$, for $T=\{a, g, h\}$ and $m_{t}(a)=m_{t}(g)=m_{t}(h)=1$ (a set is, with this definition, a multiset whose multiplicity function is always $1$), then we have $t$ is still a submultiset of $p$; but, and perhaps this definition is better regarding this intuitive aspect, $P$ and $T$ are EQUAL.

So, to summarize, you can not usually say $p$ is a submultiset of $t$; however,  if you consider the underlying set of $p$ (the $P$ in my second definition), it indeed equals $t$.

Title: Can a multiset be a subset of a set?


问:
Does it matter whether we consider field extensions as true supersets of a field?
I am trying to wrap my head around field extensions, and am using Pinter's abstract algebra. He presents a basic theorem on field extensions, namely that given any field $F$, and any polynomial $a(x)$ over $F$, we can always find a field $E$ containing $F$, and some element $c \in E$ s.t. $c$ is a root of $a(x)$ w.r.t. the operations in $E$.

As presented, the theorem seems to imply that $F$ is a 'genuine' subset of $E$. Also prior to stating and proving this theorem, we developed some theory where we assume a priori that we already have some extension $E$, with $F \subset E$.

The reason I am wondering, is that the proof we are presented with doesn't quite show that given any $F$ we have an extension $E$ with $F \subset E$. Rather, we show that actually, $F \cong H$, where $H$ is actually the subfield of $F[x]/(p(x))$ containing all the constant polynomials, and $p(x)$ is an irreducible factor of $a(x)$ in $F[x]$. So we actually prove that $F$ is isomorphic to another field $H$, and that this field $H$ has an extension, $F[x]/(p(x))$.

I assumed that since $F$ and $H$ are isomorphic, then defining a field $E_2$ such that $F \subset E_2$ and $E_2 \cong F[x]/(p(x))$ would be very easy. However I ran into problems actually coming up with this set and isomorphism. I took $E_2 = (E_1 \setminus H) \cup F$, and defined addition and multiplication on $E_2$ in a 'natural' way, but there were so many cases to consider that I struggled to actually prove it was even a field, specifically, with proving additive associativity in the most general case.

This leaves me with two questions: (i) can we simply consider $F$ and $H$ to be the same? That is, that $F[x]/(p(x))$ really is just an extension of $F$?
(ii) Is there an easy way to get a 'true' extension of $F$ easily from the extension of $H$?

答:
This is a very good question. Look, $F$ and $H$ are not the same by the rules of set theory. However, they are so naturally isomorphic that mathematicians usually just say they are the same thing. The only difference is that we call the elements $\alpha+(p(x))$ instead of $\alpha$. This is the only difference, this change of names respects all the operations of $F$. So in some sense, yes, they are the same object.

That being said, I like when things are as formal as possible. Such extensions are used a lot, and I don't always feel very comfortable knowing that it is not a "real" extension. But there is an easy way to fix it.

Note that the cardinality of $F[x]/(p)$ is $\leq\max\{\aleph_0, |F|\}$. So take any larger cardinal $\kappa$ and consider the set $X:=F\cup\kappa$. We have the natural isomorphism $H\to F$, and also can define any injective function $(F[x]/(p))\setminus H\to X\setminus F$. So together we obtain an injective function $f:F[x]/(p)\to X$ which satisfies $f(\alpha+(p))=\alpha$ for all $\alpha\in F$.

Let $L=Im(f)$, this is a set which really contains $F$. And now we can naturally give it a field structure by:

$f(\alpha)+f(\beta)=f(\alpha+\beta)$

$f(\alpha)f(\beta)=f(\alpha\beta)$

You can easily check that this  turns $L$ into a field, and it extends the usual operations we had on $F$ before. And $f$ is now an isomorphism of fields. So now we have a real extension of $F$, and the polynomial $p$ indeed has a root in it. (it is the image of the "root" from $F[x]/(p)$)


问:
Prove that the diameter of the incircle belongs and equals the height of the tangential trapezoid
I know it's probably obvious, but how can I prove that the diameter of any incircle between two parallel lines equals the distance between two lines? And in case it's tangential trapezoid, how to prove that the diameter of the incircle belongs and equals the height of the tangential trapezoid?

I see it and I know it's true, I just need to justify it. Everywhere in textbooks it is stated as a fact, so does it mean I can use this fact without proving it in any way when I'm solving a problem? I know that the angle between a tangent and radius is always 90 degrees, I know that only one perpendicular can be dropped from a point to the line and I also know that if two lines are parallel, they are perpendicular to the same line. Please help me combine these facts.

答:
First, we need to analyze the general scenario depicted in the diagram, the special case of which OP has described in his problem statement. Let us consider two straight lines $AD$ and $BE$, which are not parallel to each other. Therefore, they intersect each other at some point, say $C$. All points, which are at equidistance from two intersecting lines, are to be found on the angle bisector of the angle formed by the lines.

Now, we are going to draw some circle, which is tangential to the two line segments $AC$ and $BC$ and lies in the space delimited by them. We know that the center of this inscribed circle is located equidistant from $AC$ and $BC$, because the lines drawn from the center of this circle to the contact points between the circle and the lines are two of its radii. Therefore, the center of this circle must lie on the aforementioned angle bisector. Furthermore, since both lines are tangents of the circle, each of the two radii in question is perpendicular to the line, with which it shares a contact point (i.e. $G$ or $H$).

In the special case, where $AD$ and $BE$ are parallel to each other, their point of intersection $C$ is at infinity making the construction of the relevant angle bisector impossible. However, the fact, that all points on this line lie at equidistant from the two parallel lines running on either side of it, comes to our rescue. According to that, the equivalent angle bisector in this instance is another line parallel to the two parallel lines. The center of the inscribed circle is still located on this line. Furthermore, each of the two radii drawn from the center of the circle to the contact points between the circle and the straddling parallel lines is still perpendicular to the line, with which it shares a contact point. Since the two lines are parallel to each other, each radius is also perpendicular to the line, with which it does not share a contact point. Therefore, two radii are coinciding to make a diameter. Since this particular diameter is perpendicular to the two lines, it is also the distance between them.

If you are agreeing with my reasoning, you can use them to prove that the distance between the two parallel sides of a general tangential trapezoid is the decisive factor for dimensioning the diameter of its incircle. If you are unable to do so, let me know. I will extend my answer to address that part of your question.


问:
How to find the position vector at time t, when given plane curve equation?
I have a question about the following:
An object moves to the right along the plane curve $y=x^2$ with constant speed $v=5$. find the velocity and acceleration of the object when it is at the point (1,1).

It is concluded from this that the position of the object at time t is: $$\textbf{r}=x\textbf{i}+x^2\textbf{j}$$
But why is this exactly, how can we conclude this from the given info of this question?

If you know the answer and would like to share it with me, it would be very much appreciated :)

thanks in advance

答:
Let's say we denote the time as $t$ and the position of the object is
$$
r(t) = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}
$$
or in your notation
$$
r(t) = x(t) \mathbf{i} + y(t) \mathbf{j} = x(t) \begin{pmatrix}1 \\ 0 \end{pmatrix} + y(t) \begin{pmatrix}0 \\ 1 \end{pmatrix}.
$$

Since it has to be on the curve $y = x^2$, we know that $y(t) = x(t)^2$, hence
$$
r(t) = \begin{pmatrix} x(t) \\ x(t)^2 \end{pmatrix}
$$

It is important to notice that $r(t) \neq \begin{pmatrix}t \\ t^2 \end{pmatrix}$! Since then it would not have the correct speed! So, in your question, the variable $x$ depends on time!

It's velocity should be $v = 5$, so let's compute the velocity:
$$
v = \Vert \dot r(t) \Vert = \left\Vert \begin{pmatrix} \dot x(t) \\ 2 x(t) \dot x(t) \end{pmatrix} \right\Vert = \sqrt{ (1+4 x(t)^2) } \Vert \dot x(t) \Vert.
$$

From this equation, you can see that
$$
\dot x(t) = \pm \frac{5}{\sqrt{1+4x(t)^2}}.
$$

I leave the rest to you. You can use the above equation to determine $\dot x(t)$ and with that, you can find $\dot y(t)$ as well.


问:
Using the word or in proofs
When using the word "or" in proofs, what if one of the statements is true and provides the correct justification that finishes the proof, and the other thing you state is either false or true but not always true? Would this mean that the proof is incorrect? For example if you arrive at a contradiction and say this is false because otherwise $p$ or $q$ would happen leading to a contradiction, but $p$ is the statement that actually leads to the correct contradiction, whereas $q$ is not necessarily true, would this mean that the proof is incorrect?

I would think it would be incorrect because the statements $P \implies Q \vee R$ is logically equivalent to $P \wedge \sim Q \implies R$.

答:
For "$p$ or $q$" to be true, it suffices that one of $p$, $q$ be true. Both can also be true.

E.g.

$n$ is an arbitrary integer iff $n$ is even or $n$ is odd.

the natural $n$ is different of $2$ iff $n$ is odd or $n$ is composite.

if $n$ is even then $n=2m$ or $n=2m+1$.

For a slightly less trivial example, $n$ is a perfect number implies that $n$ is even or $n$ is odd.

Title: Using the word or in proofs


问:
Probability of car starting first time

The Question

One a given day, Claude's car has an 80% chance to starting first time
and Andre's car has a 70% chance of the same, Given that one of the
cars has started first time, what is the chance that Andre's car
started the first time?

My Understanding

Event A: Claude's car starting first time

Event B: Andre's car starting first time

We want at least one car started first time, also event B occurs, so there are only these $2$ cases: AB and A'B. I calculated the probability of these $2$ separately and multiply them together, which is $0.8\times 0.7\ {+}\ 0.2 \times 0.7=0.7$, which makes sense since event A and B are independent events, the occurrence of A should not affect the probability of B. I have to try something else because this is not the answer.

Then I tried to calculate the probability of B', which is when Andre's car is not starting first time.
there's only one case for this to be happening while at least one car is starting first time. A B'.
the probability for this is $0.8 \times 0.3 =0.24$. then I subtracted this from $1$ to get $76\%$.
the answer for this is $\frac{35}{47}$

It's close but not really. Can someone help me with this? I can't think of more ways of solving it.

答:
Let me rename the events, because I keep thinking event A refers to Andre.

Let A be the event that Andre's car starts the first time and let C be the event that Claude's car starts the first time.  We are asked for $$\Pr(A|A\cup C)=\frac{\Pr(A\cap(A\cup C))}{\Pr(A\cup C)}=\frac{\Pr(A)}{\Pr(A\cup C)}$$

Now, $\Pr(A\cup C)=\Pr(A)+\Pr(C)-\Pr(A\cap C) = .7+.8-.7\cdot.8=.94$, so the required probability is $$\frac{.7}{.94}=\frac{35}{47}$$

Title: Probability of car starting first time


问:
Find the general solution to the ODE
I would like to find the general solution to the ODE:
$$\frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{(1+x^2)^3(4x+1)}{4-x}y^5$$

and the tips given is "let $z:=\frac{1}{y^4}$. What is the differential equation that z satisfies?"

I tried by sub $$z=\frac{1}{y^4}$$ $$dz=-\frac{4}{y^5}dy$$

but the equation doesn't seem familiar to me. May I know how should I proceed? Thanks.

答:
HINT

Multiply both sides by $1 + x^{2}$ in order to obtain
\begin{align*}
(1+x^{2})y' + 2xy = \frac{(1+x^{2})^{4}(4x+1)}{4-x}y^{5} \Longleftrightarrow ((1+x^{2})y)' = \frac{4x+1}{(1+x^{2})(4-x)}((1+x^{2})y)^{5}
\end{align*}

Then make the substitution $u = (1+x^{2})y$ in order to obtain the following ODE
\begin{align*}
u' = \frac{(4x+1)u^{5}}{(1+x^{2})(4-x)}
\end{align*}
which is separable.

Can you take it from here?

Title: Find the general solution to the ODE


Title: Find the general solution to the ODE

Body:
I would like to find the general solution to the ODE:
$$\frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{(1+x^2)^3(4x+1)}{4-x}y^5$$

and the tips given is "let $z:=\frac{1}{y^4}$. What is the differential equation that z satisfies?"

I tried by sub $$z=\frac{1}{y^4}$$ $$dz=-\frac{4}{y^5}dy$$

but the equation doesn't seem familiar to me. May I know how should I proceed? Thanks.

Body:
$$\frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{(1+x^2)^3(4x+1)}{4-x}y^5$$
$$(1+x^2)\frac{dy}{dx}+{2x}y=\frac{(1+x^2)^4(4x+1)}{4-x}y^5$$
$$((1+x^2)y)'=\frac{(1+x^2)^4(4x+1)}{4-x}y^5$$
Try to separate now.
$$\dfrac {((1+x^2)y)'}{(y(1+x^2))^5}=\frac{(4x+1)}{(4-x)(1+x^2)}$$
Integrate both sides.
$$\int \dfrac {d(y(1+x^2))}{(y(1+x^2))^5}=\int \frac{(4x+1)}{(4-x)(1+x^2)} \, dx$$


问:
Cardinality summation
How do I prove that, for set $X$,

$$\sum_{S\subseteq X, S\neq \emptyset}\frac{(-1)^{|S|}}{|X|+|S|} = \frac{|X|!(|X|-1)!}{(2|X|)!}$$

I have been around this exercise all day and would much appreciate your help.

Title: Cardinality summation

Body:
Let us denote $n = |X|$. Then

\begin{align*}
\sum_{S\subset X} \frac{(-1)^{|S|}}{|X|+|S|}
&= \sum_{k=0}^{n} \binom{n}{k} \frac{(-1)^{k}}{n+k}
 = \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \int_{0}^{1} x^{n+k-1} \, dx \\
&= \int_{0}^{1} \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} x^{n+k-1} \, dx
 = \int_{0}^{1} x^{n-1} (1 - x)^{n} \, dx \\
&= \beta(n, n+1) = \frac{\Gamma(n)\Gamma(n+1)}{\Gamma(2n+1)} = \frac{(n-1)!n!}{(2n)!}
\end{align*}

as desired.


问:
Suppose $(X_n)_{n\geq 0}$ is a Markov Chain, would it be the case that $(Y_n)_{n\geq 0}:=X_{n+1}-X_n$ also a Markov Chain?
Suppose $(X_n)_{n\geq 0}$ is a Markov Chain, would it be the case that $(Y_n)_{n\geq 0}:=X_{n+1}-X_n$ is also a Markov Chain? To avoid any potential confusion, what I meant was $Y_0=X_1-X_0$, $Y_1=X_2-X_1,\cdots .$

I am really not sure how to prove this or disprove this. My intuition tells me that since $X_{n+1}$ only depends on $X_n$ (loosely speaking) and thus $X_{n+1}-X_n$ should also just only depends on $X_n-X_{n-1}$ (loosely speaking). However I could not convert this intuition to a proper, rigorous argument.

Could someone please help me with this rather simple-looking problem? Many many thanks in advance!

答: Hint: think about the gambler's ruin problem where the gambler stops at certain money thresholds. Does knowing whether the gambler won his last game tell you whether he plays the next one?</p>


问:
Smallest Image of an endomorphism?
Let $\sigma = \{E\}$ be a signature with a binary relation symbol $E$ and let $\mathcal{G}$ be a $\sigma$-structure given by the following graph:

I'm wondering what the size of the smallest image of an endomorphism $g : \mathcal{G} \rightarrow_{hom} \mathcal{G}$ could be. My initial thought was "cutting" it in half i.e.
$$ g(v_0) := v_0 \quad g(v_1) := v_1 \quad g(v_2) := v_2 \quad g(v_3) := v_1 \quad g(v_4) := v_4 \quad g(v_5) := v_5 \quad g(v_6) := v_6 \quad g(v_7) := v_5 $$
until it resulted in
$$ g(v_0) := v_5 \quad g(v_1) := v_1 \quad g(v_2) := v_5 \quad g(v_3) := v_1 \quad g(v_4) := v_1 \quad g(v_5) := v_5 \quad g(v_6) := v_1 \quad g(v_7) := v_5 $$
Since I (guess I) can't project everything on one vertex the smallest image would be $|\{v_1, v_5\}| = 2$, right? Is it still an endomorphism?

答:
A simpler way to see this is that $\mathcal G$ is bipartite as the subgraphs induced by $P_1 = \{v_1, v_3, v_4, v_6\}$ and $P_2 = \{v_0, v_2, v_5, v_7\}$ are independent sets and hence, the map $h$ defined as
$$h \colon \mathcal G \to \mathcal G, v \mapsto
\begin{cases}
v_1, \text{ if } v \in P_1 \\
v_5, \text{ if } v \in P_2
\end{cases}
$$
is an endomorphism of $\mathcal G$ of minimal image size.

Title: Smallest Image of an endomorphism?


问:
Is a finite flat module which is free modulo radical free?
Let $R$ be a commutative ring and $M$ is an $R$-module. The following statement is well-known.

If $M$ is finitely presented flat module, $M/J(R)M$ is free over $R/J(R)$ then $M$ is free.

Here $J(R)$ is the (Jacobson) radical of $R$. Is the statement true for finite $M$? I expect that it is not, but was not able to construct a counter-examples. Clearly, in a counter-example the ring $R$ is not Noetherian. Also, local $R$ won't work, since over a local ring any finite flat module is free.

答:
Set $I := J(R)$. Suppose $M/IM$ has rank $n$ as a free $A/I$-module; let $x_{1},\dotsc,x_{n} \in M$ be elements whose images in $M/IM$ form a basis for it as a free $A/I$-module, let $N \subseteq M$ be the $A$-submodule generated by the $x_{i}$. Then $M = N$ by the usual Nakayama argument ($M/IM = (N+IM)/IM$ implies $M = N+IM$ implies $M/N = (N+IM)/N = I(M/N)$ so $M/N = 0$) so we have a surjection $\varphi : A^{\oplus n} \to M$. For all maximal ideals $\mathfrak{m}$ of $A$, the localization $M_{\mathfrak{m}}$ is a free $A_{\mathfrak{m}}$-module (by e.g. the local case) of rank $n$ (since $(M/I)_{\mathfrak{m}}$ is a free $(A/I)_{\mathfrak{m}}$-module of rank $n$), so the localization $\varphi_{\mathfrak{m}}$ is an isomorphism (the "determinant trick", see Corollary (10.4) here), hence $\varphi$ is an isomorphism.

I guess more-or-less equivalently you could also use a limit argument to say that $M$ is locally finitely presented, then use e.g. Tag 00EO to conclude that $M$ is (globally) finitely presented.


问:
prove that an entire function is bounded
Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be a entire function and suppose that $\lim\limits_{z\rightarrow\infty}f(z)=0$. Does it impliy that $f$ is bounded? If so, how can I prove it?

答:
From the definition of a limit we know there is some $R>0$ such that $|f(z)|<1$ when $|z|> R$. Also, the set $A:=\{z\in\mathbb{C}: |z|\leq R\}$ is compact, and so $|f|$ is bounded on $A$, because it is a continuous function defined on a compact set.

So $f$ is bounded in both $A$ and $\mathbb{C}\setminus A$, thus it is bounded in all $\mathbb{C}$.


答: The statement that $\lim_{z \to \infty} f(z) = 0$ implies that $f$ is bounded in the set $\{z: |z| > r\}$ for some $r > 0$.  On the other hand, $\{z: |z| \le r\}$ is a compact set, on which any continuous function must be bounded.

Title: prove that an entire function is bounded

Body: Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be a entire function and suppose that $\lim\limits_{z\rightarrow\infty}f(z)=0$. Does it impliy that $f$ is bounded? If so, how can I prove it?