问:
The Birthday Paradox with combinatorics
Assume a year has $365$ days, how many are required to have a $50%$ chance of $2$ people having the same birthday?

According to Scientific American, there are $23$ people needed to achieve the goal.

$${\begin{pmatrix} 23 \\ 2\end{pmatrix}}=253$$
$$1-(1-\frac{1}{365})^{253}\approx 0.50048$$

However, I have a different approach but I'm not sure if this is correct.

One could be any day in a year. And $23$ people would be $365^{23}$ possibilities.

Suppose no one in $23$ people has the same birthday.
That would be :- ${\begin{pmatrix} 365 \\ 23 \end{pmatrix}} \cdot23!$

The possibility of having at least 2 people having the same birthday is :-

$$1-\frac{{\begin{pmatrix} 365 \\ 23 \end{pmatrix}} \cdot23!}{365^{23}}\approx0.5072972$$

Although $0.50048\approx 0.5072972$, these two numbers are not equal to each other.

Can anyone explain this difference and the reason behind this phenomenon?

答:
It's easy to see that SA's argument is incorrect. If we have 366 people, then according to SA's argument, we should take $1- \left(1 - \frac 1 {365}\right) ^{\begin{pmatrix} 366 \\ 2\end{pmatrix}} $. We don't need to calculate the exact value to know this won't be exactly equal to $1$. But the probability of having 366 different people all with different birthdays is clearly zero, because there aren't 366 different birthdays to have (this is all ignoring leap days).

The problem with SA's formula is that it calculates the number of pair, and calculates the probability of a particular pair sharing a birthday (364/365), and treats all of those events as being independent. But the probability of Alice and Bob having different birthdays and Bob and Charles having different birthdays isn't (364/365)^2. If Alice and Bob don't have the same birthday, then there are 363 birthdays left for Charles to have that's different from them, not 364.

Title: The Birthday Paradox with combinatorics


问:
Lie algebras of GL(n,R) and differentials
This question comes from a proof in John Lee's Introduction to Smooth Manifolds, page 194. I am questioning a line in the proof of the following proposition:

The composition of the maps

$\text{Lie}(GL(n,\mathbb{R}))\rightarrow T_{I}(GL(n,\mathbb{R}))\rightarrow\mathfrak{gl}(n,\mathbb{R})$

gives a Lie algebra isomorphism between $\text{Lie}(GL(n,\mathbb{R}))$ and the matrix algebra $\mathfrak{gl}(n,\mathbb{R})$

He uses the standard coordinates, $X^i_j$, on $GL(n,\mathbb{R})\subset \mathfrak{gl}(n,\mathbb{R})$. As I understand it, these $n^2$ coordinate functions take $A\in GL(n,\mathbb{R})$ to the $ij$-th entry of the matrix representation of $A$.

He writes any $A=(A^i_j)\in \mathfrak{gl}(n,\mathbb{R})$ determines a left-invariant vector field $A^l\in \mathfrak{g}$, given by

$A^L|_X=(dL_X)_I (A)=(dL_X)_I\left(A^i_j\frac{\partial}{\partial X^i_j}\bigg|_I\right)$.

This is fine. It is his next few lines which confuse me. He says

Since $L_X$ is the restriction to $GL(n,\mathbb{R})$ of the linear map $A\mapsto XA$ on $\mathfrak{gl}(n,\mathbb{R})$, its differential is represented in coordinates by exactly the same linear map. In other words, the left-invariant vector field $A^L$ determined by $A$ is the one whose value at $X\in GL(n,\mathbb{R})$ is

$
\begin{align}A^L|_X=X^i_j A^j_k \frac{\partial}{\partial X^i_k}\bigg|_X\end{align}$.

I understand the fact that the matrix representation of the differential of a linear map is just the matrix representation of the linear map itself. And I do believe this is essentially what is going on here. The reason I am confused, is because I feel he is using two different meanings for the same notation $X^{\alpha}_{\beta}$.

As I understand it, the $X^i_k$ in the $\partial/\partial X^i_k$ are the global coordinates defined on $GL(n,\mathbb{R})$. Whereas, I feel that the $X^i_j$ in the coefficient of each basis vector is the $ij$-th entry of the matrix representation of $X$. Wouldn't it be more appropriate to write,

$A^L|_Y = (dL_y)_I(A)=Y^i_jA^j_k\frac{\partial}{\partial X^I_k}\bigg|_Y$, where $Y^i_j$ is the $ij$-th entry of the matrix representation of $Y\in GL(n,\mathbb{R})$?

I don't know if I am being stupid or pedantic. Probably both. But if someone is able to clear this up for me it would be much appreciated!

Thanks.

答:
The computation of $[A^L, B^L]$ that follows on the same page shows that the choice of notation is intentional, as Lee uses the fact that $\partial X^p_q/\partial X^i_k$ is equal to $1$ whenever $p=i$ and $q=k$, and is equal to $0$ otherwise. For better or worse, differential geometry is full of (mostly) harmless abuses of notation like this.

It’s worth thinking about Lee’s remarks on page 63, where he discusses change of coordinates:

Here we are indulging in a typical abuse of notation: in the expression $\tilde x^i(x)$, we think of $\tilde x^i$ as a coordinate function (whose domain is an open subset of $M$, identified with an open subset of ${\bf R}^n$ or ${\bf H}^n$); but we think of $x$ as representing a point (in this case, in $\varphi(U\cap V)$).

(For reference, other treatments of this computation can be found in Loring Tu’s An Introduction to Manifolds, page 184,  or Michael Spivak’s A Comprehensive Introduction to Differential Geometry,  volume 1, page 377.)

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
For each natural number $n \ge 5$ give a simple graph with $n$ vertices like $G$ so that: $\kappa(G) < \delta(G)\ ,\ \delta() = n−3$
For each natural number $n \ge 5$ give a simple graph with $n$ vertices like $G$ so that:

$\kappa(G) < \delta(G)\ ,\ \delta(G) = n−3$

$\kappa(G)$ is the minimum size of vertex cut and a vertex cut of a graph $G$ is a subset of vertices $V$ such that $G-{V}$ is disconnected.

答: This is the most natural construction I can come up with. Here there are two disjoint copies of $K_2$ and one copy of $K_{n-4}$, and we add every possible edge between a $K_2$ and the $K_{n-4}$.


问:
Determine no. elements in $ \operatorname{Aut}(H)$ where $H$ is the 6 point/5line graph in the shape of H
Determine the amount of automorphisms in the group $\operatorname{Aut}(H)$ where $H$ is the graph with 6 points and five lines in the shape of a capital 'H'. Here is what it should look like, I labeled the points:

Now I can clearly see that rotation through the line $12$ and rotation through a vertical axis through the middle will be automorphisms. These can be described by permutation in $S_6$ in the following way:
$$\sigma_1 = (1\ 3)(4\ 6)$$
and
$$\sigma_2 = (1\ 4)(2\ 5)(3\ 6)$$
then we also have point reflection through the middle which is given by:
$$\sigma_3 = \sigma_1 \sigma_2 = (1\ 6)(2\ 5)(3\ 4)$$
Is this all the symmetries of $H$? I am confused because I do not think the answer should be this simple. If I for instance have the permutation $(2\ 5)$ then will the points $2$ and $1$ still be connected meaning the line piece from $2$ to $1$ will be diagonal implying $(2\ 5)$ is not an automorphism or am I missing something? Thanks in advance

答:
First of all, what's a graph homomorphism? In my definition, it's a map $f: G\to H$ (by which I mean it sends vertices to vertices and edges to edges) such that, if $(a,b)$ is an edge in $G$, then $(f(a),f(b))$ is an edge in $H$.

First of all, you should convince yourself that it's enough to work out where the vertices go: if $f: H\to H$ is an automorphism and you know what $f(a)$ is for all vertices $a$ in $H$, you know where all the edges are. (Why is this?) Having done this, you can just specify automorphisms as permutations in $S_6$ like you've been doing.

Secondly, to answer you rather obliquely: the two reflections and one point rotation you've found are automorphisms; but are there more? (What about the map given by $(1\; 3)$? What about the identity map? Importantly, automorphisms form a group; what group is it?) If you apply the permutation $f = (2\;5)$, the edge between $1$ and $2$ will turn into an edge appearing between $f(1) = 1$ and $f(2) = 5$, which you don't want, so this doesn't give you an automorphism.


问:
Can D'Alembert's functional be derived from Cauchy functionas?
Is it possible to derive D'Alembert's functional equation from Cauchy's functional equations? If so, can somebody kindly point me to a reference?

Edit:
Can
$f(x + y) + f(x − y) = 2f(x)f(y)$

be derived from either of

$f(x+y)=f(x)+f(y)$,

$f(xy)=f(x)f(y)$,

$f(x+y)=f(x)f(y)$,

$f(xy)=f(x)+f(y)$.

PS: I am not a mathematician.

答:
Not in a meaningful way.

You cannot derive d'Alembert's functional equation from the single Cauchy functional equation $f(x+y)=f(x)+f(y)$, because this latter equation has the solution $f(x)=x$, and if you make algebraic manipulations to the equation $f(x+y)=f(x)+f(y)$ you can only end up with equations which have the solution $f(x)=x$, and possibly more solutions which the original equation did not have, but $f(x)=x$ is not a solution of d'Alembert's equation (which is easy to check), so no matter which manipulations you make, you can not end up with an equation which only has the solutions which d'Alembert's equation has. Likewise none of the other Cauchy functional equations can individually imply d'Alembert's functional equation, since they have solutions which the latter does not.

It can be done in a silly way however. The only continuous solutions to $f(x+y)=f(x)+f(y)$ are those of the form $ax$ for different $a$, and the only continuous solutions to $f(x+y)=f(x)f(y)$ are those of the form $b^x$ for different $b$. (See http://www.jstor.org/stable/2972575 .) If we assume both of these equations to be true, then the only common continuous solution is $f(x)=0$, and from the functional equation $f(x)=0$ it of course follows that $f(x+y)+f(x-y)=2f(x)f(y)$ since $0+0=2\cdot0\cdot0$.


问:
Constructing Finite Separable Field Extensions From Sets Equipped With a Continuous Galois Group Action
This question is a follow-up to the following question:

Connection Between Finite Separable Extensions and Galois Group Actions

I would like to understand Theorem 1.5.2 in the book "Galois groups and Fundamental Groups" by Tamás Szamuely.

Definitions and Previous Work

Let $k$ be a field and $k_s$ a separable and algebraic closure. Let furthermore $\mathbf{FinSep}_k$ be the category of finite separable extensions of $k$. Let furthermore $\operatorname{Gal}(k):=\operatorname{Gal}(k_s|k)$, be the absolute Galois group. We define $\operatorname{Gal}(k)-\mathbf{FinSetCont}$ denote the finite sets equipped with a continuous group action by the absolute Galois group.

I would like to show that we have an anti-equivalence
$$\mathbf{FinSep}_k\simeq \operatorname{Gal}(k)-\mathbf{FinSetCont}.$$
I have shown that we have a functor
$$\Phi:\mathbf{FinSep}_k\to \operatorname{Gal}(k)-\mathbf{FinSetCont},$$
by assigning $$L\mapsto \operatorname{Hom}(L,k_s)\\ \Big(M\xrightarrow{f} L\Big )\mapsto\phi\circ f,$$
where $\phi:L\to k_s$.

Constructing Inverse

In the book, they show that the functor is essentially surjective and fully faithful.

However, by doing it like that, it doesn't seem to give me a clue about how to map a finite set equipped with a  continuous Galois action to a a finite separable extension of $k$ (I may be wrong).

I would like to have a functor
$$\Psi:\operatorname{Gal}(k)-\mathbf{FinSetCont}\to \mathbf{FinSep}_k,$$
and then show that it is an inverse to $\Phi$.

I think it should be possible to do by the Fundamental Theorem of Galois - I am not sure though.

If $S$ is a set equipped with a continuous action by $\operatorname{Gal}(k)$, where should I send it? I would like to have a map like the following one, I think
$$\operatorname{Gal}(k_s|L)\mapsto L=k_s^{\operatorname{Gal}(k_s|L)}.$$

But in order to do this, it feels like I need to recover the Galois group from the category of $\operatorname{Gal}(k)-\mathbf{Sets}$.

I did some work on the étale fundamental group of varieties defined over the complex numbers earlier. I showed the connection to the classical fundamental group by taking the profinite completion. In order to do this connection, I showed the following:

Let $\widehat{\mathbf{G}}$ be the profinite completion of a group $\mathbf{G}$, considered as groupoids, and define the category $\mathbf{\widehat{G}-FinSets}$ to be the category of functors $q:\mathbf{\widehat{G}}\to\mathbf{FinSets}$, from the profinite completion to the category of finite sets. Then I showed that
$$\operatorname{Aut}(q)\cong\mathbf{\widehat{G}}.$$
Maybe I can use this somehow? Maybe a variation of it? Notice that I didn't use anything about continuity in this construction, I think.

I talked with a professor earlier about this. He said that it may be good to understand the following equivalence of categories:
$$
\mathbf{\widehat{G}-FinSetCont}\simeq \mathbf{G-FinSet}.
$$

I am not sure how to use that though. Or well, he didn't mention in which direction the above result might be good, and I don't think I used it to construct $\Phi$. So I suppose he meant I might need it to construct $\Psi$.

Questions

1. Can I use $\operatorname{Aut}(q)$ somehow?

2. Or should I prove the equivalence the professor recommended: $\mathbf{\widehat{G}-FinSetCont}\simeq \mathbf{G-FinSet}$? Maybe this lets me use the classical fundamental theorem of Galois somehow?

Or maybe use a combination of 1. and 2.?

Do you have any idea where I should send the objects and morphisms to define $\Psi$, and what I need to do so? I really have no idea, and I would be really happy if you could help me to figure this out.

Best wishes,

Joel

I always try to mention some useful results, this time I wasn't sure if I should mention the fundamental theorem of Galois for finite extension or the Krull variant - but I did choose to mention the latter in the end.

Theorem 1.3.11 (Krull) Let $L$ be a subextension of the Galois extension $K|k$. Then $\operatorname{Gal}(K|L)$ is a closed subgroup of $\operatorname{Gal}(K|k)$. Moreover, the maps
$$
L\mapsto H:=\operatorname{Gal}(K|L)\qquad \text{ and }\qquad H\mapsto L:=K^H
$$
yield an inclusion-reversing bijection between subfields $K\supset L\supset k$ and closed subgroups $H\subset G$. A subextension $L|k$ is Galois over $k$ if and only if $\operatorname{Gal}(K|L)$ is normal in $\operatorname{Gal}(K|k)$; In this case there is a natural isomorphism $$\operatorname{Gal}(L|k)\cong \operatorname{Gal}(K|k)/\operatorname{Gal}(K|L).$$

答:
For $L/k$ finite separable then
$G_k$ acts on $Hom_k(L,k_s)$ and $L =k_s^H$ where $H$ is the stabilizer of $Id$. Since the action is transitive then the other stabilizers are conjugate to $H$ and $k_s^{\sigma H \sigma^{-1}} = \sigma(L)$. Thus the correspondence is between isomorphism classes of finite separable extensions of $k$ and transitive continuous actions of $G_k$ on finite sets.

For a class of separable extension given by $L/k$ there is only one finite continuous transitive action $\rho$, up to a permutation of the finite set. This is because $\rho$ defines an action of $G_k$ on $G_k/H\cong Hom_k(L,k_s)$.


问:
Find the general solution to the differential equation $y' + 12 x^{11} y = x^{12}$
$y' + 12 x^{11} y = x^{12}$

Use the variable $I = \int e^{x^{12}} dx$ where it occurs in your answer.

I attempted this one and found the variable $I$ also works as the integrating factor. Then I arrived at $ ye^{x^{12}} = \int x^{12} e^{x^{12}}dx $. I'm stuck here.

Body:
This is linear ODE, its integrating factor is
$e^{x^{12}}$, then $$y=e^{-x^{12}}\int x^{12} e^{x^{12}} dx+Ce^{-x^{-12}}$$
$$y=e^{-x^{12}}\int x. x^{11} e^{x^{12}}dx+Ce^{-x^{-12}}$$
Integrate by parts
$$y=e^{-x^{12}}[x \frac{e^{x^{12}}}{12}-\int e^{x^{12}} dx]+C e^{-x^{12}}$$
$$y=\frac{x}{12}-e^{-x^{12}} ~\frac{x \Gamma[1/12,-x^{12}]}{{144(-x^{12})}^{1/12}}+C e^{-x^{12}}$$


问:
How to prove these 7 inequation about characteristic function?
In Problems in Probability by Shiryaev, Many problems about characteristic function are based on these inequalities? Does anybody have hints to prove some of them? Here's the image. Thank you.

Prove that any characteristic function $\varphi=\varphi(t)$ must satisfy the following inequalities:

$$1-\operatorname{Re} \varphi(n t) \leq n\left[1-(\operatorname{Re} \varphi(t))^{n}\right] \leq n^{2}[1-\operatorname{Re} \varphi(t)], n=0,1,2, \ldots ;\tag{*}$$

$$|\operatorname{Im} \varphi(t)|^{2} \leq \frac{1}{2}[1-\operatorname{Re} \varphi(2 t)] ; \quad 1-\operatorname{Re} \varphi(2 t) \geq 2(\operatorname{Re} \varphi(t))^{2}$$

$$|\varphi(t)-\varphi(s)|^{2} \leq 4 \varphi(0)|1-\varphi(t-s)| ; \quad 1-|\varphi(2 t)|^{2} \leq 4\left[1-|\varphi(t)|^{2}\right]$$

$$|\varphi(t)-\varphi(s)|^{2} \leq 2[1-\operatorname{Re} \varphi(t-s)]$$

$$\frac{1}{2 h} \int_{t-h}^{t+h} \varphi(u) d u \leq(1+\operatorname{Re} \varphi(h))^{1 / 2}, t>0$$

答: For $|\varphi(t)-\varphi(s)|^{2} \leq 2[1-\operatorname{Re} \varphi(t-s)]$, apply Schwarz inequality to the LHS, should be straight forward afterward.


问:
Limit of two variables function by defintion
I have to prove by definition that
$$\lim_{(x,y)\to(3,\pi)}x^{2}\sin(y)=0$$

Since
$$|f(x,y)-0|=|x^{2}|\cdot|\sin(y)|$$
and
$$|\sin(y)|\leq1$$
then
$$|f(x,y)-0|\leq|x^{2}|$$

Can I say that since
$$|x|\leq\sqrt{x^{2}+y^{2}}$$
so
$$|x^{2}|=|x|^{2}\leq x^{2}+y^{2}$$
and therefore
$$|x|^{2}\leq(x-3)^{2}+(y-\pi)^{2}?$$

I'm not sure about the last inequality, but it's the only way I found to try to write $\delta$ using $\epsilon$.

EDIT

I tried to change the variables:
$$x=\bar{x}+3,\quad y=\bar{y}+\pi$$
so, I have the limit
$$\lim_{(\bar{x},\bar{y})\to(0,0)}(\bar{x}+3)^{2}\sin(\bar{y}+\pi)$$
and, still
$$|f(\bar{x},\bar{y})-0|\leq|(\bar{x}+3)^2|$$
but now I must to find $\delta$ such that
$$||(\bar{x},\bar{y})-(0,0)||<\delta$$
implies on
$$|f(\bar{x},\bar{y})-0|<\epsilon$$
Since
$$||(\bar{x},\bar{y})-(0,0)||=\sqrt{\bar{x}^{2}+\bar{y}^{2}}$$
then
$$\sqrt{\bar{x}^{2}+\bar{y}^{2}}<\delta\Rightarrow \bar{x}^{2}+\bar{y}^{2}<\delta^{2}$$
and, therefore
$$|\bar{x}|^{2}=\bar{x}^{2}\leq \bar{x}^{2}+\bar{y}^{2}<\delta^{2}\Rightarrow |\bar{x}|^{2}<\delta^{2}$$
and so
$$|\bar{x}|<\delta$$
Hence,
$$|(\bar{x}+3)^{2}|=|\bar{x}^{2}+6\bar{x}+9|\leq|\bar{x}^{2}|+6|\bar{x}|+9<\delta^{2}+6\delta+9=(\delta+3)^{2}$$
Then, if I take
$$\epsilon=(\delta+3)^{2}$$
the limit is proved.

I want to know if it's right.

Body:
I'd do it this way:

We want:

$|x^2\sin(y)-0|<\epsilon$

We know that

$|x^2\sin(y)-0|$

$=|x^2\sin(y)|$

$= |x^2||\sin(y)|$

$= |(x-3+3)^2||\sin(y-\pi+\pi)|$

$ = |(x-3)+3|^2|\sin(y-\pi+\pi)|$

$ \le (|x-3|+|3|)^2|\sin((y-\pi)+\pi)|$

$\sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b)$

so

$ = (|x-3|+3)^2|\sin(y-\pi)|$

So here I use that $|sin(a)|\le|a|$

$ \le (|x-3|+3)^2|y-\pi|$

We force $|x-3|<1$.

So then

$ (|x-3|+3)^2|y-\pi|$

$ < (1+3)^2|y-\pi|$

$ = 16|y-\pi|$

We want $16|y-\pi|$ to be $< \epsilon$

ie: we want $|y-\pi| < \frac{\epsilon}{16}$

Now we can do our proof.

Give an $\epsilon>0$ we choose $\delta = \min\{1,\frac{\epsilon}{16}\}$

We've already shown that

$|x^2\sin(y)-0|$

$\le (|x-3|+3)^2|y-\pi|$

Now since $|x-3|<\delta$ and $|y-\pi|<\delta$

$(|x-3|+3)^2|y-\pi|$

$< (\delta+3)^2\delta$

$\le (1+3)^2\delta$

$\le 16\delta$

$\le 16(\frac{\epsilon}{16})$

$= \epsilon$


问:
Without calculating limit directly show that it is equal to zero
$$\lim_{n\rightarrow\infty}\left(\frac{n+1}{n}\right)^{n^2}\frac{1}{3^n}=0$$
I am not really sure what it means by "without calculating limit" and I don't really have ideas how to do it.</p>

答:
By Cauchy test for the series we have :

$$\sqrt[n]{(\frac{n+1}{n})^{n^2}\frac{1}{3^n}}=(\frac{n+1}{n})^{n}\frac{1}{3}\sim_\infty \frac{e}{3}<1$$
hence the series
$$\sum_{n=1}^\infty (\frac{n+1}{n})^{n^2}\frac{1}{3^n}$$
is convergent and then the general term is convergent to $0$.


问:
Pushforward Lorentz metric: $ds^2=dxdy $
Consider an open embedding $\varphi:\Bbb R^{1,1}_- \hookrightarrow (0,1)^2$ with $\varphi(x_1,x_2)=(e^{x_1},e^{x_2})$

I need to put a metric on $(0,1)^2$.

Here’s what I know so far. I need to transport the Lorentz metric: $ds^2=dxdy. $

Since my map is a diffeomorphism I can try to push forward the Lorentz metric to $(0,1)^2$.

I know that $n=\log(x)\log(y)$ is a preserved metric of a transformation on $(0,1)^2$ that is related to the classical Lorentz transformation.

I’ve tried to get a metric from this preserved quantity but have not been successful.

How can I push the Lorentz metric forward?

Thanks so much.

答:
The new metric in $(0,1)^2$ is $ds^2=\frac{1}{u}du\frac{1}{v}dv.$

This is done by pulling back the Lorentz metric $ds^2=dxdy.$

$\begin{align}
 (\varphi^{-1})^* (dxdy) &=( d((\varphi^{-1})^*x))( d((\varphi^{-1})^*y)) \\
&= (d (x\circ \varphi^{-1}))(d (y\circ \varphi^{-1})) \\
&= (d \log u)(d\log v) \\
&= \left( \frac{1}{u} du\right)\left( \frac{1}{v} dv\right) \\
&= \frac{1}{u} du\frac{1}{v} dv.
\end{align}$


问:
Push-forward of vector fields over smooth manifolds
Let $M, N$ be smooth manifolds with appropriate dimensions, $X$ be a (smooth) vector-field on $M$.

Let $f: M \rightarrow N$ be a map. Now, what I know is that to pushforward $X$ on $N$, $f$ is required to be an injection; and that for 'pushed' $X$ on $N$ to be smooth, $f$ is required to be smooth. (given $X$ is a smooth vector field)

But I've seen people say that $f$ should be a diffeomorphism (i.e. $M\cong_{diff} f(M)$).

Does $f$ have to be a diffeomorphism?
This naturally poses another question - Does a bijection that is smooth in one way, necessarily smooth the other way? (If yes, then $f$ is already a diffeomorphism)

答:
Let $X\in\mathfrak{X}(M)$ be a smooth vector field on $M$. If all you have is a smooth bijection $F\colon M\to N$ with an inverse $F^{-1}$ that isn’t smooth, there is no way to guarantee that the resulting vector field $F_*X$ is smooth. That said, a smooth bijection is enough to define a unique rough (that is, not necessarily smooth) vector field.

Following the suggestion of @Tom Ariel in the comments, consider the (constant) coordinate vector field $X$ defined by $X_p=\partial/\partial x|_p$  on $\mathbf{R}$. Its pushforward under the map $F\colon\mathbf{R}\to\mathbf{R}$ defined by $F\colon x\mapsto x^3$ gives us a rough vector field $F_*X$ defined by
$$\begin{align*}
(F_*X)_p
&=dF_{F^{-1}(p)}(X_{F^{-1}(p)}) \\
&=dF_{p^{1/3}}(\partial/\partial x|_{p^{1/3}}) \\
&=3p^{2/3}(\partial/\partial x|_p).
\end{align*}$$
This vector field is not smooth; the map $x\mapsto 3x^{2/3}$ isn’t even differentiable at $x=0$.

However, if we have a diffeomorphism $G\colon M\to N$, then we are guaranteed that $G_*$ takes smooth vector fields to smooth vector fields, since $G_*X\colon N\to TN$ is then the composition of smooth maps
$$N
\overset{G^{-1}}{\longrightarrow}
M
\overset{X}{\longrightarrow}
TM
\overset{dG}{\longrightarrow}
TN.$$

(For references, see for example Propositions 8.1 and 8.19 in John Lee’s Introduction to Smooth Manifolds.)


问:
Combinatorial Proof of $|T| \leq 2 \binom{n}{\lfloor n/2 \rfloor}$
I need some help proving that if $T$ is a collection of subsets of $\{1,2,3,...,n\}$ such that there doesn't exists $A, B, C \in T$ with $A \subset B \subset C$ then $|T| \leq 2 \binom{n}{\lfloor n/2 \rfloor}$

答:
Not sure if this proof counts as combinatorial.

Let $T_1=\{B\in T\mid \exists A\in T:B\subsetneq A\}$ and show that $T_1$ and $T\setminus T_1$ are anti-chains, which lets us apply Sperner’s Theorem to both.


问:
Adjacency Matrix vs Incidence Matrix
Let $A$ be the adjacency matrix of a simple graph $G=(X,U)$ in relation to the marking $(x_1,x_2,\ldots,x_n)$ of its vertices and $B$ the matrix of incidences of $G$ in relation to the same marking $(u_1,u_2,\ldots,u_m)$ of your edges.

How can I show that $A=BB^T-D$ where $D=[d_{ij}]$ is the diagonal matrix, of order $n$ and $d_{ii}=d_G(x_i)$, $ i = 1,\ldots,n$? Does someone have any hint on how can I start the proof?

答:
Let $b_{ij}$ be the entry in row $i$, column $j$ of $B$. $b_{ij}=1$ if $x_i$ is incident to $u_j$, otherwise zero.

The $(i,j)$-entry of $BB^t$ is $\sum_kb_{ik}b_{jk}$.

If $i\ne j$, then this sum is one if $x_i$ and $x_j$ are adjacent (since in that case there's a unique $k$ such that $x_i$ and $x_j$ are incident to $u_k$), and zero otherwise, so it equals the $(i,j)$-entry of $A$.

If $i=j$, so the sum is $\sum_k(b_{ik})^2$, this counts one for each edge incident to $x_i$, that is, it counts the degree of $x_i$, and we're done.

Title: Adjacency Matrix vs Incidence Matrix


问:
Looking to vectorize P2 Penrose tiling
I'm looking for a tool to vectorize a P2 Penrose tiling.

I've been able to find tools to vectorize P3 tilings, but not P2.

Goal is to put p2 tiling on cycling kit

答:
The part of the English Wikipedia page on Penrose tilings describing the P2 tiling has a vector image of the pieces and a vector image of the vertex figures, both under the Attribution-ShareAlike 3.0 Unported Creative Comments license.

Edit: For images of a significant P2 tiling, rather than just the pieces or vertex figures, here are some resources. Except for the first point, I found them by searching Google for penrose kite and dart vector or penrose kite and dart svg:

If you don't need too much of a tiling, Wikipedia has a "sun" svg (mentioned on the vertex figures page linked above) and a "star" svg.

Sarah Marsh has a Python program PenroseGenerator on GitHub which generates svg files for tilings.

Jeff Preshing's blog has a tutorial and Python code for generating tilings.

Rosetta Code's page for Penrose tiling has code for tilings in a variety of languages, most or all of which are for P2 tilings.

Jonas Arnfred has a convenient tiling generator on the web that lets you download svg files and which has a corresponding GitHub page.

Title: Looking to vectorize P2 Penrose tiling


问:
Finding the next point when the angle is give
I'm trying to create a code for a function which evaluates the next point when the initial point and the angle  is given...

Here, if the angle $\theta$ is give as follows (That is the angle with respect to the standard x-axis):

And if the length between $(x_0,y_0)$ and $(x_1,y_1)$ is known then I can use simple trigonometric relationships and evaluate the position of $(x_1,y_1)$.

But what if the angle is given along the vector as below:

Here $(x_0,y_0),(x_1,y_1),L_1,L_2,\theta$ are known. Then is there a method that I can use to evaluate the point $(x_2,y_2)$

Appreciate your help

答:
Draw a line through $(x_1,y_1)$, parallel to the $x$ axis. The angle between the dashed pink line and the horizontal is given by the arctan2 function.$$\alpha=\arctan2(y_1-y_0,x_1-x_0)$$
Then the angle $L_2$ makes with the horizontal is $\alpha+\theta$. The problem reduces to the case you already solved.


问:
Cubic Formula Doesn't Seem to Work for $x^3+3x^2-50x-52=0$
So I'll jump right into it. I've derived the following formula set for solving cubics of the form $ax^3+bx^2+cx+d=0$:

$$z_{k\pm}=\sqrt[3]{\frac{9abc-2b^3-27a^2d}{54a^3}\pm\sqrt{\frac{4ac^3+27a^2d^2-18abcd-b^2c^2+4b^3d}{108a^4}}}$$

$$x=z_{k\pm}-\frac{\left(\frac{-b^2}{3a^2}+\frac{c}{a}\right)}{3z_{k\pm}}-\frac{b}{3a}$$

The idea is that you solve for the 6 $z$-values defined by $k=0, 1, 2$ and each of these $k$-values has a positive and negative result based upon the selection for the $\pm$ sign. Also note that this derivation is heavily based upon this article (http://math.sfsu.edu/smith/Documents/Cubic&Quartic.pdf). The article dictates that these 6 $z$-values translate to only 3 distinct $x$-values (since cubics only have 3 solutions).

This all went smoothly and I was able to successfully show the viability of the formula set by solving the cubic $-2x^3+3x^2-x+5=0$. The formula was successful in capturing all 3 $x$-values without any issues.

However, upon further inspection of the formula, I have noticed an issue when the radicand of the square root is negative. This would make the radicand of the cube root imaginary, thus leaving $z$ with no real solutions; hence, $x$ would have no real solutions when $\frac{4ac^3+27a^2d^2-18abcd-b^2c^2+b^3d}{108a^4}\lt0$. Of course this cannot be true since every cubic has at least one real solution.

To test this theory I tried to solve
$$x^3+3x^2-50x-52=0$$
using the formula. This cubic does indeed make the radicand of the square root negative and happens to have three real roots: $-1, -1-\sqrt{52}, -1+\sqrt{52}$. However, upon solving the formula for these values, $z_{0+}$ and $z_{0-}$ were complex and $z_{1+}$ gave a domain error on my TI-84+ calculator. My reference article provides some insight into dealing with a negative radicand, but it doesn't make sense to me. Any help is greatly appreciated!

答:
For negative radicands, the 6 $z$-values translates into 3 distinctive $x$-values as well.

Take the example $x^3+3x^2-50x-52=0$ given in the post, $z$‘s are evaluated as

$$z_{k\pm} =\sqrt[3]{\pm \sqrt{-\frac{53^3}{27}}}= \pm\sqrt{\frac{53}3}e^{i \frac{1+4\pi k}6},\>\>\>k=0,1,2$$
and the corresponding $x$’s are
$$x_{k\pm} = z_{k\pm} +\frac{53}{3z_{k\pm}}-1
= z_{k\pm} +{z_{k\mp}}-1=2Re(z_{k\pm})-1
$$
Explicitly, the 6 $x$-values are
\begin{align}
x_{0+} & = 2\sqrt{\frac{53}3}\cos\frac\pi6-1=\sqrt{53}-1\\
x_{0-} & = -2\sqrt{\frac{53}3}\cos\frac\pi6-1=-\sqrt{53}-1\\
x_{1+} & = 2\sqrt{\frac{53}3}\cos\frac{5\pi}6-1=-\sqrt{53}-1\\
x_{1-} & = -2\sqrt{\frac{53}3}\cos\frac{5\pi}6-1=\sqrt{53}-1\\
x_{2+} & = 2\sqrt{\frac{53}3}\cos\frac{3\pi}2-1=-1\\
 x_{2-} & =- 2\sqrt{\frac{53}3}\cos\frac{3\pi}2-1=-1\\
\end{align}
of which, only 3 values, $-1$, $\sqrt{53}-1$ and $-\sqrt{53}-1$, are distinctive, as expected.


问:
What does $\sum \mu(d)\lfloor{\frac{x}{d}}\rfloor$ count?
Where the sum is over all $d$, where $d$ only has prime factors $\leq \sqrt{x}$. I'm trying to determine, in plain English, precisely what this sum is counting.

I know the Möbius function will be 0 if $p^2|x$ when $p|x$, and $(-1)^{k}$ when $x$ is the product of $k$ distinct primes. Further, I understand that $\lfloor{\frac{x}{d}}\rfloor$ determines the number of multiples of $d\leq x$, I'm just struggling to put these functions together in my head. I know that the only such $d$'s that will contribute to the sum will be the product of distinct primes, to the exponent 1, where the primes are less than or equal to $\sqrt{x}$.

I know that my question is related to Legendre's formula ($\pi (x) = -1 + \pi(\sqrt x) + \sum \mu (d)\lfloor \frac {x}{d}\rfloor$,), but I'd like to understand it more combinatorially.

答:
Since Mobius function is zero as long as the input number is divisible by some squares. Hence, that summation sums over the $M$, product of all primes less $\le\sqrt x$. Hence, the original sum in the question is equivalent to finding

$$
\begin{aligned}
\sum_{d|M}\mu(d)\left\lfloor\frac xd\right\rfloor
&=\sum_{r=0}^{\pi(\sqrt x)}(-1)^r\sum_{1\le i_1<\cdots<i_r\le\pi(\sqrt x)}\left\lfloor x\over p_{i_1}\cdots p_{i_r}\right\rfloor
\end{aligned}
$$

wherein $\left\lfloor\frac xm\right\rfloor$ denotes number of multiples of $m$ (including itself) within $x$, and let's denote the set of integer-multiple of $m$ within $x$ by $S_{x,m}$, then for any $(a,b)=1$ we have

$$
\left\lfloor x\over ab\right\rfloor=|S_a\cap S_b|
$$

As a result, we can rewrite the sum above by

$$
\sum_{d|M}\mu(d)\left\lfloor\frac xd\right\rfloor=1+\sum_{r=0}^{\pi(x)}(-1)^{r-1}\sum_{1\le i_1<i_r\le\pi(\sqrt x)}\left|\bigcap_{k=1}^rS_{p_{i_r}}\right|
$$

wherein the last sum, by inclusion-exclusion principle, represents the number of primes within $x$ but greater than $\sqrt x$. Therefore,

$$
\sum_{d|M}\mu(d)\left\lfloor\frac xd\right\rfloor=1+\pi(x)-\pi(\sqrt x)
$$

Rewriting this identity gives a formula for the prime-counting function:

$$
\pi(x)=\pi(\sqrt x)+1+\sum_{d|M}\mu(d)\left\lfloor\frac xd\right\rfloor
$$


答:
Does f need to be a diffeomorphism?

It's not always possible to push-forward tangent fields on a manifold. This is why we have the restriction to diffeomorphisms, or more generally, tangent fields that are f-related.

However, it's an easy observation that when we work in the category of tangent fields over an auxiliary manifold, that is parametrised tangent fields, we can always do so.

Title: Push-forward of vector fields over smooth manifolds

Body:
Let $M, N$ be smooth manifolds with appropriate dimensions, $X$ be a (smooth) vector-field on $M$.

Let $f: M \rightarrow N$ be a map. Now, what I know is that to pushforward $X$ on $N$, $f$ is required to be an injection; and that for 'pushed' $X$ on $N$ to be smooth, $f$ is required to be smooth. (given $X$ is a smooth vector field)

But I've seen people say that $f$ should be a diffeomorphism (i.e. $M\cong_{diff} f(M)$).

Does $f$ have to be a diffeomorphism?
This naturally poses another question - Does a bijection that is smooth in one way, necessarily smooth the other way? (If yes, then $f$ is already a diffeomorphism)


问:
Show that this language is context-free by giving a pushdown automaton
I have to show that the language L is context-free by drawing the pushdown automaton.

L = $x\$y^r$: where x is the binary representation of an integer nx and y represents the integer nx+ 1

$y^r$ represents the word y reversed.

So example : $10010\$11001$ is in T because  $10010 = 18$ and $y^r = 10011$ which is equal to 19

So I have started a drawing which is not finished because I am stuck:

From there I am a bit stuck on how to handle the part after the $.

I understand that $y^r$ is $x^r + 1$ but I can't seem to find a way to illustrate it in the pushdown automaton

Thanks for your help. Here's the tool I use to draw if u wanna help out: http://madebyevan.com/fsm/

答:
Actually, $y^r$ is not necessarily equal to $x^r+1$. Think about $10 = 1 + 1$ but $01 \neq 1 + 1$ as a counterexample.

The reverse of $y$ really gives a helpful hand.

Aside: $x\$y$ where $x$ and $y$ has the same meaning as above is not a CFL.

Hint: apply pumping lemma to this string: $\underbrace{11\cdots 1}_p0\underbrace{11\cdots 1}_p\$\underbrace{11\cdots1}_{p+1}\underbrace{00\cdots0}_p$.

In this circumstance, it is a good manner to remember whether the PDA has read through the string of $'1'$ in the front when disposing of $y^r$, which is represented by the self-looping edge of the node $D$. For simplicity, we confine $x$ to a positive integer and require that $0$ should not be the prefix either in $x$ or $y$.

The reason why there are two ways out of $D$ is that if $x$ is a string full of $1$, then the length of $y$ is one more than $x$. In other cases, $x$ and $y$ should always be at the same length.


问:
Help proving that $\sum_{d^k|n}\mu(d) =$ 1 if is k-free and 0 otherwise
In class we saw a similar result when $k=2$, and now I'm trying to extend this to an arbitrary $k\in \mathbb{Z}^+$. When I plug in values this identity seems to hold, however I'm unsure how to tackle the proof.

This is how I started:
Suppose that $p_{1},...,p_{q}$ are the distinct prime factors in the prime factorization of $n$ that have exponent $\geq 2$ in the prime factorization (it is possible that $q=0$. Other than when $d=1$, the only integers that will contribute to the sum are products of these primes with powers $\leq k-1$. Furthermore we see that any $d$ in which some exponent of a prime is $\geq k$ makes no contribution. I'm trying to write,
$\sum_{d^k|n} \mu(d)=\sum_{d|p_{1}^{\alpha_{1}}\cdot...\cdot p_{q}^{\alpha_{q}}} \mu(d),$
where $\alpha_{i} \leq k-1$. This is where I get stuck and I'm not sure how to make a generalization about when this will evaluate to $0$ versus $1$. Any help or guidance would be greatly appreciated!

答:
Let $(a,b)=1$, then

$$
\sum_{d^k|a}\mu(d)\sum_{r^k|b}\mu(r)=\sum_{(dr)^k|ab}\mu(dr)=\sum_{m|ab}\mu(m)
$$

Hence, this function is multiplicative. As a result, let's consider prime powers $n=p^r$:

$$
\sum_{d^k|p^r}\mu(d)=\mu(1)+\mu(p)[k\le r]
$$

If $k\le r$, the number $n$ is not $k$-free, so we conclude

$$
\sum_{d^k|n}\mu(d)=\begin{cases}
1 & \text{$n$ is $k$-free} \\
0 & \text{otherwise}
\end{cases}
$$


问:
Prove function under subspace topology
In a topological space $X$, $Y\subset X$ is the subspace topology and their Borel $\sigma$-algebras. Show that $f \upharpoonright_Y$ is measurable if $f:X\rightarrow Z$ is ($Z$ is a measurable space).

答:
Welcome to MSE ^_^

Hint:

To show that $f \upharpoonright_Y$ is measurable, it suffices to show that

$$(f \upharpoonright_Y)^{-1} (E) = \{ y \in Y ~|~ (f \upharpoonright_Y)(y) \in E \}$$

is borel (in $Y$) for every $E \subseteq Z$ measurable.

But we know that

$$f^{-1}(E) = \{ x \in X ~|~ f(x) \in E \}$$

is measurable because that's the definition of $f : X \to Z$ being measurable.

Can you show that $(f \upharpoonright_Y)^{-1} (E) = f^{-1}(E) \cap Y$?

Can you use this to show that $(f \upharpoonright_Y)^{-1} (E)$ is indeed borel in $Y$?

I hope this helps ^_^

Title: Prove function under subspace topology


问:
Using Ito calculus to prove that $\int_0^t W_s^2dW_s = \frac{1}{3} W_t^3 - \int_0^t W_s d_s$
I am busy trying to teach myself some stochastic calculus and have come across a statement that I am trying to prove.

How can I prove that
\begin{align}
\int_0^t W_s^2dW_s = \frac{1}{3} W_t^3 - \int_0^t W_s d_s
\end{align}
where $W \in \mathbb{R} $ is Brownian motion, using Ito's rule?

答:
Use the other author's hint, $f(w,t)=w^3$. Then, compute the partial derivatives needed for Ito's lemma:
$$\partial_{w}f=3w^2$$
$$\partial_{w^2}f=6w$$
$$\partial_{t}f=0$$
Plug in Ito's formula,
$$d(W_t^3)=3W_t^2dW_t+3W_tdt$$
Then integrate both sides from $0$ to $t$ (introduce dummy variable $s$),
$$W_t^3=3\int_{0}^{t}W_s^2dW_s+3\int_{0}^{t}W_sds$$
Reaarange, done. Of course, there are other $f(w,t)$'s that you can plug in.


问:
How to remember elements of Quaternion group $Q_8$
Quaternion group acts a great example. But how to remember this? Any shortcut? i know this is silly question but just wanted to know. Each 2x2 matrix element is represented by $\{ 1,-1,i,-i,j,-j,k,-k\}$ What i am asking is how to remember matrix representation correspondence with 1, i, j, k

答:
The common mnemonic for remembering quaternion multiplication is to write $i,j,k$ clockwise around a circle.

To multiply two consecutive elements in the clockwise order, you get the third element not chosen. If you multiply consecutive elements in counterclockwise order, you get $-1$ times the third.

to remember matrix representation

Well, the matrix representation isn’t unique.  Find one you like, and you only have to remember three matrices for $i,j,k$.  Remembering three things does not really merit a special trick...  You could even get away with just memorizing $i,j$ and multiply them to get $k$.

The standard representation of quaternions as 2 by 2 matrices looks like this

$\begin{bmatrix}a&b\\-\bar{b}&\bar{a}\end{bmatrix}$.  The three matrices can be all that form.


答: Let me reveal you my secret: The quaternion group $Q_8$ is the four dimensional analogue of $2$-dimensional complex unit group $$\{\pm1, \pm i : i^2=-1\}$$ with two additional complex directions $j$ and $k.$ Moreover it is not commutative but anti-commutative. Altogether it is $$\{\pm1, \pm i, \pm j, \pm k : i^2=j^2=k^2=ijk=-1\}.$$ Note that using $ijk=-1$ with other relations you can prove that $$ij=-ji=k,\qquad jk=-kj=i,\qquad ki=-ik=j.$$ Put them in a clockwise cycle make an easier to memorize all these at onece.

Title: How to remember elements of Quaternion group $Q_8$

Body: Quaternion group acts a great example. But how to remember this? Any shortcut? i know this is silly question but just wanted to know. Each 2x2 matrix element is represented by $\{ 1,-1,i,-i,j,-j,k,-k\}$ What i am asking is how to remember matrix representation correspondence with 1, i, j, k


答:
As per the request in the previous answer, let's assume all points are in 3D, in the $z=0$ plane. Then let's use the following notations:$$\vec v_1=(x_1,y_1,0)-(x_0,y_0,0)\\\vec v_2=(x_2,y_2,0)-(x_1,y_1,0)$$
We can decompose $\vec v_2$ into two components, one along $v_1$ and one along a direction perpendicular to $\vec v_1$. To get $\theta$ to increase counterclockwise, we need this axis to point towards the upper left side in the figure. To create such a perpendicular vector, we use the cross product of a vector with $\vec v_1$. And to keep it in the horizontal plane, we need the other vector in the cross product to be perpendicular to the $z=0$ plane. So my two directions are going to be $$\hat v_{||}=\frac{\vec v_1}{|\vec v_1|}\\\hat v_\perp=\hat z\times \hat v_1$$
Here $\hat z=(0,0,1)$. Now we know that the length of $\vec v_2$ is $L_2$, so the component along $\hat v_{||}$ is $L_2\cos\theta$ and along $\hat v_\perp$ is $L\sin\theta$, so $$\vec v_2=\hat v_{||}L_2\cos\theta+\hat v_\perp L_2\sin\theta$$

Title: Finding the next point when the angle is give

Body:
I'm trying to create a code for a function which evaluates the next point when the initial point and the angle  is given...

Here, if the angle $\theta$ is give as follows (That is the angle with respect to the standard x-axis):

And if the length between $(x_0,y_0)$ and $(x_1,y_1)$ is known then I can use simple trigonometric relationships and evaluate the position of $(x_1,y_1)$.

But what if the angle is given along the vector as below:

Here $(x_0,y_0),(x_1,y_1),L_1,L_2,\theta$ are known. Then is there a method that I can use to evaluate the point $(x_2,y_2)$

Appreciate your help


答:
Here's a trick you can use to remember the matrix correspondence you give.

On the one hand, multiplying $\mathbb{C}^2$ by $i$ has matrix $[\begin{smallmatrix} i&0\\0&i\end{smallmatrix}]$ ("obviously").

On the other hand, multiplying by $i$ is an anticlockwise $90^{\circ}$ rotation in the complex plane $\mathbb{C}\cong\mathbb{R}^2$, and you should memorize the matrix of such a rotation is $[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}]$ (seriously, very useful); you can extend this to a linear operator on $\mathbb{C}^2$ and call it $j$ instead of $i$.

Then you can get $k$ from $k=ij$.


问:
Is my understanding of how not to use the ($\rightarrow$ Introduction) rule correct?
Prove $ A \vee B, \neg A \vdash B $

$A \vee B \quad\quad\quad$ data

$\neg A \quad\quad\quad\quad$ data

$\neg A \rightarrow B \quad\quad$ $\rightarrow $ Introduction, Subcomputation below

$\quad\quad 3.1 \space \neg A \quad $ Assumption

$\quad\quad \color{red}{3.2} \space B \quad$ from (2)(3), $\rightarrow$ Elimination (Modus Ponens)

Am I correct in saying that line 3.2 is incorrect as you can't use line 3 to prove itself?

答:
From

Yes, 3.2 is incorrect since you cannot use what is to be derived as part of the derivation. - @md2perpe

I can indeed see that 3.2 is not a correct derivation.


问:
How do you determine if a formula is satisfiable in Predicate Logic?
For example:

$ (\forall x)(P(x) \rightarrow Q(x))$

Are you suppose to invent your own Interpretation (domain, and giving the meaning to the predicates), and make it satisfiable under that Interpretation?

Attempt:

Interpretation I:

Domain: $ \mathbb{N} $

P(x) - "x is a natural number"

Q(x) - "x is an integer"

I is a model of $ (\forall x)(P(x) \rightarrow Q(x)) $

Therefore, $ (\forall x)(P(x) \rightarrow Q(x)) $ is satisfiable.

答:
From

Yes. More precisely, you have to construct your interpretation such
that the formula becomes satisfied under it. - @lemontree

It is clear how a formula can be satisfied.


问:
Real Projective Space

Corollary $\bf7.15.$ The real projective space $\mathbb{R}P^n$ is second countable.

How I can prove this corollary 7.15.I consider that ı can apply the corollary 7.10.But I could not can you help me.

Corollary $\bf7.10.$ If $\,\sim\,$ is an open equivalence relation on a second-countable space $S$, then the quotient space $S/\sim$ is second countable.

答: Recall that $S^n\subset\mathbb{R}^{n+1}$ is second countable as it is a subspace of a second countable space. Can you see how to use your corollary 7.10, along with this fact to show that $\mathbb{R}P^n$ is second countable?</p>

Title: Real Projective Space


问:
Uniform boundedness of the partial Fourier sums of the sawtooth function
This question is about exercise 19 in the third chapter of Stein and Shakarchi's Fourier analysis.
We are asked to prove that
$$
\label{eq:1}
\sum_{n=1}^N \frac{\sin nx}{n}=\frac{1}{2}\int_0^x(D_N(t)-1)dt,  \tag{*}
$$
where $D_N(t) = \sum_{|n|<=N} \exp(int) = 1 + 2\sum_{n=1}^N \cos nt$ is the Dirichlet kernel. And then as a consequence of the identity in $\eqref{eq:1}$ to prove that $\sum_{n=1}^N \frac{\sin nx}{n}$ is uniformly bounded in $N$ and $x$ for $0 \leq x \leq \pi$ (and hence for all $x$ given that the expression in the LHS above is odd and $2\pi$ periodic).

The identity in  $\eqref{eq:1}$ easily follows from the fact that
$$
\frac{d}{dt} \sum_{n=1}^N \frac{\sin nt}{n} = \frac{1}{2}(D_N(t) - 1),
$$
but I am unable to see how the integral representation implies uniform boundedness in $N$ and $x$.

答:
Let $$F_N(x) = \int_0^x (D_N(t) - 1)dt = \int_0^x \left( \frac{ \sin (N+1/2)t}{\sin(t/2)} -1 \right) dt$$ and let $$G_N(t) = \int_0^x \left( \frac{\sin (N+1/2)t}{t/2} - 1 \right)dt.$$

We have to show $$ \sup_{ \substack{ 0 \leq x \leq \pi \\ N \geq 1} } |F_N(x)| < \infty.$$

It is sufficient to show $ \sup_{ \substack{ 0 \leq x \leq \pi \\ N \geq 1} } |F_N(x) - G_N(x)| < \infty$ and  $ \sup_{ \substack{ 0 \leq x \leq \pi \\ N \geq 1} } |G_N(x)| < \infty$.

Note that $$g(x) = \frac{1}{\sin(x/2)} - \frac{1}{x/2}$$ is a continuously differentiable function for $x \in [0,\pi].$

Now $$F_N(x) - G_N(x) = \int_{0}^x \sin( (N+1/2) t) g(t) dt = \int_{0}^x \frac{d}{dt} \left(\frac{1 -\cos( (N+1/2) t)}{(N+1/2)}\right) g(t)dt . $$

Integrating by parts we get,
$$ F_N(x) - G_N(x) = \frac{1-\cos( (N+\frac{1}{2})x)}{(N+\frac{1}{2})}g(x) - \frac{1}{(N+\frac{1}{2})}\int_{0}^x (1 - \cos( (N+\frac{1}{2})t )) g'(t) dt.$$

Let $M_1 = \sup_{0 \leq x \leq \pi} |g(x)|$ and let $M_2 = \sup_{0 \leq x \leq\pi} |g'(x)|,$ then
$$
\label{e:1}
|F_N(x) - G_N(x)| \leq \frac{2M_1}{N+\frac{1}{2}} + \frac{2M_2\pi}{N+\frac{1}{2}} \leq 4M_1 + 4M_2\pi \tag{*}$$ for all $N \geq 1$ and $0 \leq x \leq \pi.$

We have,
$$
G_N(x) = 2\int_0^x \frac{\sin(N+\frac{1}{2})t}{t}dt - x = 2 \int_0^{ (N+1/2)x} \frac{\sin t}{t}dt - x.
$$

From the convergence of the improper Riemann integral $ \int_0^\infty \frac{\sin t}{t}dt$ we have $ \sup_{0 \leq x < \infty} | \int_0^x \frac{\sin t}{t}dt |\, \colon= M_3 < \infty$.

From which it follows $\label{e:2}|G_N(x)| \leq 2 M_3 + \pi \tag{+}$ for all $N \geq 1$ and all $0 \leq x \leq \pi$.

From $\eqref{e:1}$ and $\eqref{e:2}$ the result follows.


答:
These are the representations you obtain if you consider elements $g \in Q_8$ as linear maps $g : \mathbb{C} \oplus \mathbb{C}j \to \mathbb{C} \oplus \mathbb{C}j$ with $g(a + bj) = ag + bgj$, where we have identified $1 \in \mathbb{C}$ with $e \in Q_8$ and $i \in \mathbb{C}$ with $i \in Q_8$ (and likewise for $\overline{e}$ and $\overline{i}$).

To determine the matrix of $j$ in this representation, it suffices to look at its action on basis elements.
\begin{align}
j(1 + 0j) &= 1j + 0j^2 = 1j + 0(\overline{e}) = 1j + 0 = 0 + 1j\\
j(0 + 1j) &= 0j + 1j^2 = 0j + 1(-1) = -1 + 0j
\end{align}
This means
$$
j \mapsto \begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}
$$

This all might seem very complicated, but I'm just being pedantic. Another more compressed example might make things clearer.

We have $k(1) = 1k = k = ij$ and $kj = -i$. Thus
$$
k \mapsto \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}
$$

This is the same trick that is used to represent complex numbers as matrices in $M_2(\mathbb{R})$. A trick worth remembering if you ever study complex analysis.


问:
Let $X$ be the disjoint union of copies of a subspace of $\mathbb{R}$. Is the quotient of $X$ by identifying accumulation points first countable?
Sorry the title isn't super clear, I ran out of characters.

From Willard's General Topology, 16A.5:

For each $n \in \mathbb{N}$, let $X_n$ be a copy of the subspace
$\{0\} \cup \{\frac{1}{m}: m = 1, 2, \ldots \}$ of $\mathbb{R}$. Let $X$ be the disjoint union of the $X_n$. Is the quotient $Y$ of $X$ obtained by identifying all accumulation points of $X$ first
countable?

I'm really having a hard time with this question. I'm struggling more generally with both the notions of the disjoint union topology, and the quotient topology, so this is really throwing me for a loop.

The relevant information that I know is:

If $X$ is a topological space and $a \in X$, then we say $a$ is an accumulation point of a set $A$ iff each nbhood of $a$ meets $A$ in some point other than $a$,

The set $X_n'$ of all accumulation points of $X_n$ (defined above) is $X_n' = \{0\}$.

But I'm not really sure what $X$, the disjoint union of these $X_n$, "looks" like. Is the set of accumulation points of $X$ the union of the accumulation points of the $X_n$? I'm really just not sure how to proceed.

Any help is appreciated.

答:
The quotient topology on $Y$ is defined as the strongest topology such that the quotient map $q:X\to Y$ is continuous. Let $T_X$ be the topology on $X.$

Let $T_Y=\{S\subseteq Y: q^{-1}S\in T_X\}.$ Then $T_Y$ is a topology on Y, merely because $q$ is surjective and because $T_X$ is a topology on $X$.

(1). If $T$ is a topology on $Y$ with $T\supsetneqq T_Y$ then  there exists $S'\in T$ \ $T_Y,$ which by def'n of $T_Y$ implies $q^{-1}S'\not \in T_X,$ so $q$ is not continuous  with the topology $T$ on $Y.$

So $T_Y$ is the quotient topology on $Y.$

Let $E$ be an equivalence relation on $X.$ The quotient map $q$ induced by $E$ is $q(x)=[x]_E=\{x'\in X:x'Ex\}.$ This can be notationally and conceptually inconvenient. When $\phi\ne C\subset X$ and $[x]_E=\{x\}$ for $x\in X$ \ $C$ while $[x]_E=C$ for $x\in C,$ we can take some (any) $p\not \in X,$ and let $q(x)=x$ for $x\in X$ \ $C$ while $q(x)=p$ for $x\in C.$ The space $Y=\{p\}\cup (X \setminus C)$ with the quotient topology is called the identification of C to a point.

(2). For some set $I$, suppose $T_i$ is a topology on $X_i$ for each $i\in I.$ The topological sum  (topological disjoint union) is $(X, T)$ where $X=\cup_{i\in I} (X_i\times \{i\})$ and $T$ is the topology on $X$ generated by the base (basis) $\cup_{i\in I}\{t\times \{i\}:t\in T_i\}.$ Note that each $X_i\times \{i\}$ is an open-and-closed subspace of $X$ and its subspace topology is $\{t\times \{i\}:t\in T_i\},$ so it is homeomorphic to the $T_i$ topology on $X_i.$ In particular, if $x=(x_i,i)\in X$ then $x$ is an isolated point of $X$ iff $x_i$ is an isolated point of $X_i.$


问:
Double sum: $\sum_{i=1}^3 \sum_{j=1}^i j$
I have been trying to solve for $\sum_{i=1}^3 \sum_{j=1}^i j$

I was given the answer = 10 but I am not sure how to even begin.

Tried  $\sum_{i=1}^3 (1+2+3+\cdots+i)$ so far but it doesn't seem right. Could someone provide some tips please? Thanks!

答:
First, we have $\sum_{j = 1}^{i} = 1 + 2 + ... + i = \frac{i(i + 1)}{2}$. Then:

$$\sum_{i=1}^{3}\sum_{j=1}^{i} j = \sum_{i=1}^{3}\frac{i(i + 1)}{2} = \frac{1(1 + 1)}{2} + \frac{2(2 + 1)}{2} + \frac{3(3 + 1)}{2} = 1 + 3 + 6 = \boxed{10}$$

This approach is most useful when large values of $i$, $j$ are present. A more efficient way to calculate it here is simply $1 + (1 + 2) + (1 + 2 + 3) = 10$.


Title: Double sum: $\sum_{i=1}^3 \sum_{j=1}^i j$

Body:
I have been trying to solve for $\sum_{i=1}^3 \sum_{j=1}^i j$

I was given the answer = 10 but I am not sure how to even begin.

Tried  $\sum_{i=1}^3 (1+2+3+\cdots+i)$ so far but it doesn't seem right. Could someone provide some tips please? Thanks!

Body: It's $\sum\limits_{j=1}^1 j+\sum\limits_{j=1}^2 j+\sum\limits_{j=1}^3 j = 1+(1+2)+(1+2+3)=10$.


答:
"Solve for" is the wrong term. If you're given $3x+5=9$ and you're trying to find $x$ then you are solving that equation for $x.$ But if you're wondering what number $17\times43$ is, then you're evaluating that expression.

When $i=1$ then $\sum_{j=1}^i j = 1.$

When $i=2$ then $\sum_{j=1}^i j = 1 + 2.$

When $i=3$ then $\sum_{j=1}^i j = 1 + 2 + 3.$

So $\sum_{i=1}^3 \sum_{j=1}^i j = 1 + (1+2) + (1+2+3). $


问:
Pushforward Lorentz metric: $ds^2=dxdy $
Consider an open embedding $\varphi:\Bbb R^{1,1}_- \hookrightarrow (0,1)^2$ with $\varphi(x_1,x_2)=(e^{x_1},e^{x_2})$

I need to put a metric on $(0,1)^2$.

Here’s what I know so far. I need to transport the Lorentz metric: $ds^2=dxdy. $

Since my map is a diffeomorphism I can try to push forward the Lorentz metric to $(0,1)^2$.

I know that $n=\log(x)\log(y)$ is a preserved metric of a transformation on $(0,1)^2$ that is related to the classical Lorentz transformation.

I’ve tried to get a metric from this preserved quantity but have not been successful.

How can I push the Lorentz metric forward?

Thanks so much.

答:
I would like to explain my comments to the OP's post here. As this is too long to be in the comments, I have decided to include them in an "answer format".

The metric on $2$-dimensional Minkowski spacetime is

$$ g = dt^2 - dx^2.$$

Of course, some people prefer to write it as $-g$. In the end, this is just a matter of convention. Now, let

$u = f_1(t,x) := t+x \quad \text{ and } \quad v = f_2(t,x) := t-x.$

Denote by $f$ the resulting map which maps $(t,x)$ to $(u,v)$, and thus has as components $f_1$ and $f_2$.

Therefore, if we let $h = dudv$, then

$$f^*(h) = dt^2 - dx^2 = g.$$

In other words, $f$ is an isometry from $\mathbb{R}^2$ with coordinates $(t,x)$ and Minkowski metric $g$ onto $\mathbb{R}^2$ with coordinates $(u,v)$ and metric $h = dudv$.

Finally, note that both $du$ and $dv$ are null, meaning that

$h^{-1}(du,du) = 0 \quad \text{ and } h^{-1}(dv,dv) = 0.$

This is what I meant by the statement that $u$ and $v$ are null coordinates.


问:
What is 'Operator valued function is holomorphic' ? Does Neumann series imply holomorphic?
Let $H_1$ and $H_2$  be Hilbert spaces and $K(w):H_1 \rightarrow H_2$ be an operator for given $w\in\Omega\subset\mathbb{C}$.(like a resolvent. It would not be resolvent.)

In complex analysis, a function $f:\Omega \rightarrow \mathbb{C}$ is holomorphic if
$$f'(w_0):=\lim_{w\rightarrow w_0} \frac{f(w)-f(w_0)}{w-w_0}$$
exists for every $w_0\in\Omega$. And its limit takes modulus.
However, since an operator valued function $w \mapsto K(w) $ is not complex value, $\left|\frac{K(w)-K(w_0)}{w-w_0}\right|$ cannot be defined.
So, I can GUESS its limit is defined in operator norm, i.e. if $K'(w_0)$ is defined such that
$$\left\|\frac{K(w)-K(w_0)}{w-w_0} - K'(w_0)\right\|_{H_1 \rightarrow H_2} \rightarrow 0 \quad as \quad w \rightarrow w_0,$$ $K(w_0)$ is holomorphic for $w_0 \in \Omega$. In my book, there is no mention about this detail.

By the way, consider the Neumann series,
$$(I-K(w))^{-1} = \sum_k^\infty (K(w))^k.$$
In order to show $(I-K(w))^{-1}$ is holomorphic, I can think about 'analytic' first (since analytic and holomorphic is equivalent in $\mathbb{C}$) and Neumann series say '$K(w)$ is analytic for each $w$'. But, again,  it is not complex number.

Questions

What is definition of 'Operator valued function is holomorphic' ? Is my guess right ?

Does it suffice to show existence of its Neumann series ? If then, why ?

答:
Let $F : \mathcal{O}\subset\mathbb{C}\rightarrow \mathcal{L}(X,Y)$ where $\mathcal{O}$ is an open subset of $\mathbb{C}$, and $\mathcal{L}(X,Y)$ denotes the bounded linear operators from a complex Banach space $X$ to another complex Banach space $Y$. Then $F$ is holomorphic iff the following limits exist for all $z\in\mathcal{O}$:
$$
             \lim_{w\rightarrow z}\frac{1}{w-z}(F(w)-F(z)).
$$
In that case, the limit is denoted by $F'(z)$.

It can be proved that $F$ is holomorphic iff $z\mapsto F(z)x$ is a holomorphic vector function for all $x\in X$. And $F(z)x$ is holomorphic iff $z\mapsto y^*(F(z)x)$ is a complex holomorphic function for all $y^*\in Y^*$. The proof follows from the uniform boundedness principle, and a power series argument. This remarkable bootstrapping is due to the Uniform Boundedness Principle and a power series argument, where the existence of a complex derivative of a complex function $f : \mathcal{O}\rightarrow\mathbb{C}$ amounts to boundedness of the following for all $w$ in a punctured neighborhood of $z$:
$$
              \frac{1}{w-z}\left[\frac{f(w)-f(z)}{w-z}-f'(z)\right],\;\;\;w\in B_r(z)\setminus \{z\}.
$$


问:
Combinatorics Equation
I am studying for a test on combinatorics and I recently came across this equation I had lost in my notes.

I do not really have a clue as to why this equation holds or its higher-level meaning. If anyone can offer some explanation as to why this equation is true for any positive integer k and non-negative integer n it would be greatly appreciated. It would even better if someone could give any insights on how I can prove this to be true so that it will be more mathematically intuitive.

答:
I think this is best shown with a simple example. The equation is rather scary in full generality, even though the concept it describes is simple.

Let's start with the simplest example: $(x+y)^2$. We can multiply this out to see

$$(x+y)^2 = x^2 + 2xy + y^2.$$

If I prime you by saying "pascal's triangle", you might notice that the coefficients of this expression $(1 \quad 2 \quad 1)$ form a row of pascal's triangle. Let's try the next one:

$$(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$$

The coefficients of this $(1 \quad 3 \quad 3 \quad 1)$ are also a row of pascal's triangle. You may or may not be aware of this (shocking!) connection, but in general

$$(x+y)^n = \text{something } x^n + \text{something } x^{n-1}y + \ldots + \text{something } xy^{n-1} + \text{something } y^n$$

It turns out that these "somethings" will always be the $n$th row of pascal's triangle!

Let's take a moment to think about that. This is one of the places where mathematics exists to let us be lazy! If you wanted to expand out $(x+y)^{100}$, you would spend all day doing it if you wanted to distribute everything out. Moreover you would almost certainly make a mistake somewhere along the way. But this gives us a quick trick™ for computing these easily! Simply find the $100$th row of pascal's triangle (which is easy, if tedious, to do) and we're done.

Of course, this still sounds tedious... Is there a way for us to be lazier?

Always!

The next step comes from combinatorics, which is why you're being taught this. Let's think about what the coefficients really mean in these expressions. Why do we have $1 \quad 3 \quad 3 \quad 1$ showing up here?

Well, let's do some distribution and see what happens:

$$(x+y)^2 = xx + \color{green}{xy + yx} + yy$$

$$(x+y)^3 = xxx + \color{blue}{xyy + yxy + yyx} + \color{red}{xxy + xyx + yxx} + yyy$$

Notice what's happening! When we do the distribution, we get every possible choice of $x$s and $y$s. Then, since $xy = yx$, we can reorder them so that they all look like $x^iy^{n-i}$. So our coefficient on $x^i y^{n-i}$ is exactly the number of ways to have $i$ xs and $n-i$ ys in a term!

But we know what that coefficient is! We want to count the number of ways to have $i$ xs and $n-i$ ys in a string of length $n$. This is what binomial coefficients are made for!

$$\text{The coefficient of } x^iy^{n-i} \text{ is } \frac{n!}{i!(n-i)!}$$

We write this as $\binom{n}{i}$, but this is kind of a special case. When we only have $x$ and $y$, things are easy. If we know where the xs go, we have no choice where to put the ys! More generaly we could write this as $\binom{n}{i, n-i}$ to indicate we're choosing $i$ xs and $n-i$ ys.

Ok, last step! What are these multinomial coefficients floating around? Well, they exist to solve this same problem for more variables!

Say you want to calculate $(x+y+z)^{10}$, but you're too lazy to expand it out. Keeping in mind what we've done so far, you might come up with a clever strategy:

You notice the terms of $(x+y+z)^{10}$ are going to be all of the strings of $10$ xs, ys, and zs. Then the coefficient of $x^i y^j z^{10 - i - j}$ is going to be exactly the number of ways you can choose that string.

Again, we know the answer to this problem! How many ways are there to split $10$ things into $3$ groups of size $i$ $j$ and $10-i-j$? There's exactly

$$\binom{10}{i, j, 10-i-j} = \frac{10!}{i! j! (10 - i - j)!}$$

many ways to do this. So that is the coefficient!

More generally, if you want the coefficient of $x^iy^jz^kw^l$ in $(x+y+z+w)^n$, you know what to do! It's $\binom{n}{i,j,k,l}$. Of course, we should have $i+j+k+l = n$, since each term will have $n$ letters in total.

And now, finally, we find ourselves with the formula you started with.

What is $(x_1 + x_2 + \ldots + x_k)^n$?

Well, the coefficient of $x_1^{i_1}x_2^{i_2}\ldots x_k^{i_k}$ is exactly $\binom{n}{i_1, i_2, \ldots, i_k}$. So if we want to express this as a sum, we see:

$$(x_1 + x_2 + \ldots + x_k)^n = \sum \binom{n}{i_1, i_2, \ldots, i_k} x_1^{i_1} x_2^{i_2} \ldots x_k^{i_k}$$

here the constraint on the sum is, again, that $i_1 + i_2 + \ldots + i_k = n$. Because there must still be $n$ letters in any given term.

But people don't like writing $x_1 \ldots x_n$, so we write it using $\prod$ instead. This gives us exactly the formula that you were initially asking about:

$$(x_1 + x_2 + \ldots x_k)^n = \sum_{i_1 + \ldots + i_k = n} \binom{n}{i_1, i_2, \ldots i_k} \prod x_j^{i_j}$$

As some quick exercises to see that you understand:

What is the coefficient of $x^3 y^2 z^4$ in $(x+y+z)^9$?

Can you expand out in full the polynomial $(w+x+y+z)^3$ without foiling?

I hope this helps ^_^

Title: Combinatorics Equation


问:
Polynomials and Limits
Let $P(x)$ be any polynomial with positive real coefficients. Determine the following limit with proof

$ \lim \limits_{x \to \infty} \dfrac{\lfloor P(x) \rfloor}{P(\lfloor x \rfloor)}$

I am unable to approach these kind of sums. Anyone please can guide me and help me with this solution?

答:
Here, we assume $P(x)$ is nonconstant. Your limit is equal to, for some $a,b\in[0, 1)$ ($a$ and $b$ are also functions of $x$):

$$\lim_{x\to\infty}\frac{P(x) - a}{P(x - b)}$$

Note that we can rearrange this as:

$$\lim_{x\to\infty}\frac{P(x) - a}{P(x)}\frac{P(x)}{P(x-b)} = \lim_{x\to\infty}\frac{P(x) - a}{P(x)}\lim_{x\to\infty}\frac{P(x)}{P(x - b)}$$

$$\lim_{x\to\infty}\bigg(1 - \frac{a}{P(x)}\bigg)\lim_{x\to\infty}\frac{P(x)}{P(x-b)}$$

Then, since $P(x)$ has positive coefficients and is nonconstant, $P(x)\to\infty$ as $x\to\infty$. Thus, since $a$ is bounded:

$$\lim_{x\to\infty}\bigg(1 - \frac{a}{P(x)}\bigg)\lim_{x\to\infty}\frac{P(x)}{P(x-b)} = \bigg(1 - \frac{a}{\infty}\bigg)\lim_{x\to\infty}\frac{P(x)}{P(x-b)} = \lim_{x\to\infty}\frac{P(x)}{P(x-b)}$$

Then, taking $x$ at the integers, which are also unbounded in the positive direction, we find that $b = 0$, so:

$$\lim_{x\to\infty}\frac{P(x)}{P(x-b)}=\lim_{x\to\infty, x\in\mathbb{Z}}\frac{P(x)}{P(x-b)} = \lim_{x\to\infty, x\in\mathbb{Z}}\frac{P(x)}{P(x)} = 1$$
Thus:

$$\boxed{\lim_{x\to\infty}\frac{\lfloor P(x)\rfloor}{P(\lfloor x\rfloor)} = 1}$$

In the case of the constant function $P(x) = p$, then $\lim_{x\to\infty}\frac{\lfloor P(x)\rfloor}{P(\lfloor x\rfloor)} = \frac{\lfloor p\rfloor}{p}$, which is not always equal to $1$.

Title: Polynomials and Limits


答:
ok lets consider a string x , we have 2 cases :

x ends with a 0 :

for ex , x = 1010 , x+1 = 1010 + 1 = 1011

so in this case , clearly y is the same as x , except for the last symbol

so , if the last symbol in x , that is the top of the stack after pushing all symbols of x , if it is 0 , we need to make sure that the first symbol in y^r (which is the last symbol in y ) is 1 , then we need to make sure all other symbols match x ,

x ends with a 1 :

for ex , x = 1011 , x+1 = 1011 + 1 = 1100 , what happened here ?

if the last symbol in x is 1 , then in y , starting from least significant bit (rightmost symbol) , all 1s become 0s , then the first 0 we encounter becomes a 1 , the rest of the string remains the same , this is due to adding 1 , then having the carry added and so on

so in case 2 , we make sure that all 1s in x are 0s in y , till we reach the first 0 in x , which is a 1 in y , then all symbols are the same

other cases :

when we reach the state same we expect both binary strings to be the same from this point on , but x or y may have different number of leading 0s

the leading 0s are found at beginning of y , thus at end of y^r , so if we reach the accept state , meaning y = x+1 , we allow reading as much 0s as we want since we know these are leading 0s that do not change the number

if x is the one with more leading 0s , then we guess non-deterministically that input has ended , and what remains on stack are all leading 0s , we head to the state x 0s where we pop all leading 0s then accept , of course in the process if we are face with a 1 , this means these 0s are not leading 0s and we reject

we mentioned how if y ends with a 1 , we need to check for the first 0 in x to change to 1 in y , but what if x has only 1s ? , ex x = 1 , y = 10 , we check that 1 has turned to 0 , then if we find no 0s , we will find the # , the # will tell us that the string ended , and thus when x = 1 , the stack is #1 but we treat the # as a leading 0 , then we push it back for the final check for empty string , so the original rule is 1,0,ε , the rule we add is 1,#,# , we pop the # instead of the 0 ,fair enough since we know this is the end of the string anyway so we can say that #1 is same as #01 , then we push the # again for the final test of emptiness of stack

Now here is the PDA :

Title: Show that this language is context-free by giving a pushdown automaton

Body:
I have to show that the language L is context-free by drawing the pushdown automaton.

L = $x\$y^r$: where x is the binary representation of an integer nx and y represents the integer nx+ 1

$y^r$ represents the word y reversed.

So example : $10010\$11001$ is in T because  $10010 = 18$ and $y^r = 10011$ which is equal to 19

So I have started a drawing which is not finished because I am stuck:

From there I am a bit stuck on how to handle the part after the $.

I understand that $y^r$ is $x^r + 1$ but I can't seem to find a way to illustrate it in the pushdown automaton

Thanks for your help. Here's the tool I use to draw if u wanna help out: http://madebyevan.com/fsm/


答:
First of all, I think your desire to remember those $2 \times 2$ matrices is misguided.  If you need to know it, then look it up. What should be learned well is the multiplication rules among $i$, $j$, and $k$. That is what everyone knows who has studied quaternions.

I think a better question than "how to remember" those $2 \times 2$ matrices is "where do such matrices comes from"? It is nearly the same question but gets at a more mathematically interesting issue: how would such a matrix representation be discovered? It just sounds like a more compelling issue than "how to remember?" What we are trying to do with the representation as matrices is turn quaternions into a subgroup of ${\rm GL}_2(\mathbf C)$, so we want to make quaternions look like $\mathbf C$-linear maps. Now that is tricky because $\mathbf C$, in its natural embedding into the quaternions, does not commute with all quaternions (unlike the real numbers inside the quaternions). But it can be done and there is a method involved that helps in other situations.

First, let's back up and explain why complex numbers can be viewed as $2 \times 2$ real matrices: associate to $z = a + bi$ in $\mathbf C$ (so $a, b$ are real) the mapping $m_z \colon \mathbf C \to \mathbf C$ given by multiplication by $z$: $m_z(w) = zw$.  The function $m_z$ is an $\mathbf R$-linear map, and using the $\mathbf R$-basis $\{1,i\}$ of $\mathbf C$ we can represent $m_z$ as a $2 \times 2$ real matrix by expressing $m_z(1)$ and $m_z(i)$ in terms of the basis $\{1,i\}$:
$$
m_z(1) = z = a + bi, \ \ m_z(i) = zi = -b + ai \Longrightarrow [m_z] = \begin{pmatrix}a&-b\\b&\ a\end{pmatrix},
$$
where $[m_z]$ is the matrix representation for $m_z$ with respect to the (ordered) $\mathbf R$-basis $\{1,i\}$ of $\mathbf C$.

Now let's try to do the same thing for the division ring of quaternions $\mathbf H = \mathbf R + \mathbf R i + \mathbf R j + \mathbf R k$. We will view $\mathbf H$ as a complex vector space, but we have to be careful to decide whether to make $\mathbf H$ a left $\mathbf C$-vector space or a right $\mathbf C$-vector space.

Method 1. Let's treat $\mathbf H$ as a left $\mathbf C$-vector space:
$$
a + bi + cj + dk = (a + bi) + (c+di)j \in \mathbf C + \mathbf C j.
$$
Should we associate to $q \in \mathbf H$ the map $\ell_q \colon \mathbf H \to \mathbf H$ given by left multiplication by $q$ ($\ell_q(s) = qs$) or the map $r_q \colon \mathbf H \to \mathbf H$ given by right multiplication by $q$ ($r_q(s) = sq$)?  Left multiplication is not good because it is not $\mathbf C$-linear when we view $\mathbf H$ as a left $\mathbf C$-vector space: $\ell_q(zs) = q(zs) \not= z(qs) = z\ell_q(s)$ in general for complex $z$.  But right multiplication by $q$ is $\mathbf C$-linear when $\mathbf H$ is a left $\mathbf C$-vector space thanks to associativity of multiplication: $r_q(zs) = (zs)q = z(sq) = zr_q(s)$.

If we associate to $q \in \mathbf H$ the mapping $r_q \colon \mathbf H \to \mathbf H$, then this is additive in $q$ ($r_{q+q'} = r_q + r_{q'}$), but it is not multiplicative: $r_q(r_{q'}(s)) = (sq')q = s(q'q) = r_{q'q}(s)$, so $r_q \circ r_{q'} = r_{q'q}$. The multiplication got swapped. So we will use conjugation on $\mathbf H$ on top of using right multiplication: associate to $q$ the map "right multiplication by $\overline{q}$": let $m_q \colon \mathbf H \to \mathbf H$ by $m_q(s) = s\overline{q}$.  Then $m_q$ is $\mathbf C$-linear since
$$
m_q(s + s') = (s+s')\overline{q} = s\overline{q} + s'\overline{q} = m_q(s) + m_q(s')
$$
and
$$
m_q(zs)= (zs)\overline{q} = z(s\overline{q}) = zm_q(s).
$$

Moreover, $m_q$ is additive and multiplicative in $q$: for $q, q' \in \mathbf H$ and $s \in \mathbf H$,
$$
(m_q + m_{q'})(s) = m_q(s) + m_{q'}(s) = s\overline{q} + s\overline{q'} = s(\overline{q} + \overline{q'}) = s\overline{q+q'} = m_{q+q'}(s)
$$
and
$$
(m_q \circ m_{q'})(s) = m_q(m_{q'}(s)) = m_q(s\overline{q'}) = (s\overline{q'})\overline{q} = s(\overline{q'} \ \overline{q}) = s\overline{qq'} = m_{qq'}(s).
$$
Therefore $m_{q+q'} = m_q + m_{q'}$ and $m_{qq'} = m_q \circ m_{q'}$. That proves $q \mapsto m_q$ is a ring homomorphism $\mathbf H \to {\rm End}_{\mathbf C}(\mathbf H)$. It is injective since $m_q(1) = \overline{q}$, so we can recover $q$ from $m_q$. Therefore $q \mapsto m_q$ is an embedding of $\mathbf H$ as a subring of ${\rm End}_{\mathbf C}(\mathbf H)$.

For $q = a + bi + cj + dk = (a + bi) + (c + di)j$, we can turn $m_q$ into  a $2 \times 2$ complex matrix by using a $\mathbf C$-basis of $\mathbf H$ as a (left) $\mathbf C$-vector space. We'll use the basis $\{1,j\}$:
$$
m_q(1) = \overline{q} = a - bi - cj - dk = (a-bi) + (-c - di)j
$$
and
$$
m_q(j) = j\overline{q} = aj + bk + c - di = c - di + (a + bi)j.
$$
Therefore the representation of $m_q$ as a matrix in ${\rm M}_2(\mathbf C)$ using the (ordered) basis $\{1,j\}$ is
$$
[m_q] =
\begin{pmatrix}
a-bi&c-di\\
-c-di&a+bi
\end{pmatrix}.
$$
For example,
$$
[m_i] =
\begin{pmatrix}
-i&0\\
0&i
\end{pmatrix}, \ \ \
[m_j] =
\begin{pmatrix}
\ 0&1\\
-1&0
\end{pmatrix}, \ \ \
[m_k] =
\begin{pmatrix}
\ 0 &-i\\
-i& \ 0
\end{pmatrix}.
$$
Comparing this $2 \times 2$ complex matrix representation of $\mathbf H$ to the one in your question, it is not the same.

Method 2. Let's view $\mathbf H$ as a right $\mathbf C$-vector space, so $z \cdot q := qz$ for quaternions $q$ and $z \in \mathbf C$.  Then
the standard notation $q = a + bi + cj + dk$ looks in the basis $\{1,j\}$ like
$$
q = (a + bi) + j(c - di).
$$
The left multiplication mapping $\ell_q(s) = qs$ is $\mathbf C$-linear in this new sense (multiplication by $\mathbf C$ is on the right!) and $\ell_{q + q'} = \ell_q + \ell_{q'}$ while $\ell_{qq'} = \ell_q\ell_{q'}$, and $q \mapsto \ell_q$ is injective since $\ell_q(1) = q$.

Since $\ell_q(1) = q = (a + bi) + j(c - di)$ and $\ell_q(j) = qj = -(c+di) + (a+bi)j = -(c+di) + j(a-bi)$, the representation of $\ell_q$ as a matrix in ${\rm M}_2(\mathbf C)$ using the (ordered) basis $\{1,j\}$ is
$$
[\ell_q] =
\begin{pmatrix}
a+bi&-c-di\\
c-di&a-bi
\end{pmatrix}.
$$
For example,
$$
[\ell_i] =
\begin{pmatrix}
i&\ 0\\
0&-i
\end{pmatrix}, \ \ \
[\ell_j] =
\begin{pmatrix}
0&-1\\
1&\ 0
\end{pmatrix}, \ \ \
[\ell_k] =
\begin{pmatrix}
\ 0 &-i\\
-i& \ 0
\end{pmatrix}.
$$
Comparing this $2 \times 2$ complex matrix representation of $\mathbf H$ to the one in your question, it is (again) not the same.

Method 3. Let's go back to viewing $\mathbf H$ as a left $\mathbf C$-vector space, so each quaternion is written as $z + wj$ with $z, w \in \mathbf C$. For $q \in \mathbf H$, let $f_q \colon \mathbf H \to \mathbf H$ by
$$
f_q(z + wj) = q(\overline{z} + \overline{w}j).
$$
This mapping $f_q$ is $\mathbf C$-linear: it is easily seen to be an additive function on $\mathbf H$, and for $\alpha \in \mathbf C$,
$$
f_q(\alpha(z + wj)) = f_q((\alpha z) + (\alpha w)j) = q(\overline{\alpha z} + \overline{\alpha w}j) = q(\overline{\alpha}\,\overline{z} + \overline{\alpha}\,\overline{w}j) = q\overline{\alpha}(\overline{z} + \overline{w}j).
$$
For $q \in \mathbf H$ and $\alpha \in \mathbf C$, $\alpha q = q\overline{\alpha}$, so
$$
f_q(\alpha(z + wj)) = \alpha q(\overline{z} + \overline{w}j) = \alpha f_q(z+wj).
$$
That completes the proof that $f_q$ is $\mathbf C$-linear.

It is left to you to check that $f_{q+q'} = f_q + f_q$ and $f_{qq'} = f_q \circ f_{q'}$ as functions $\mathbf H \to \mathbf H$. Therefore $q \mapsto f_q$ is a ring homomorphism $\mathbf H \to {\rm End}_{\mathbf C}(\mathbf H)$ and it is injective since $f_q(1) = q$.  By picking a $\mathbf C$-basis of $\mathbf H$ (as a left $\mathbf C$-vector space), we can turn this homomorphism into an embedding $\mathbf H \to {\rm M}_2(\mathbf C)$. Using the $\mathbf C$-basis $\{1,j\}$, write $q = a+bi + cj + dk = (a+bi) + (c+di)j$ and then
$$
[f_q] = \begin{pmatrix}
a+bi&-c-di\\
c+di&a+bi
\end{pmatrix}.
$$
For example,
$$
[f_i] =
\begin{pmatrix}
i&0\\
0&i
\end{pmatrix}, \ \ \
[f_j] =
\begin{pmatrix}
0&-1\\
1&\ 0
\end{pmatrix}, \ \ \
[f_k] =
\begin{pmatrix}
0 &-i\\
i& \ 0
\end{pmatrix}.
$$
This is the matrix realization of quaternions in your question. But notice how it corresponds to the rather "peculiar" $\mathbf C$-linear mapping $f_q(z+wj) = q(\overline{z} + \overline{w}j)$.
That is part of what makes it pointless to try to "remember" a specific $2 \times 2$ matrix representation of $\mathbf H$. There are many "natural" ways to do this and it's hopeless to try to remember what the results are.  If you need a matrix representation, look it up.


问:
Counting Problem: Pigeonhole principle
How many people are needed to guarantee that at least two were born on the same day of the week and in the same month (perhaps in different years) ?

I couldn't understand the wording of this question, so please explain the exact solution which is required.

Thanks!

答:
I will assume that the problem asks for both conditions to be satisfied (i.e, the two people must be born on both the same day of the week and in the same month). Then we have that there are $84 = 7\cdot 12$ combinations of days of the week and months ("holes"), so then the number of people ("pigeons") such that pigeons must be in the same hole is $\boxed{84 + 1 = 85}$

We don't have to deal with leap-years or anything like that because each month always contains more than $7$ days (and thus every day of the week).

Title: Counting Problem: Pigeonhole principle


问:
Calculate the convergence of MMSE
$$X(n) = \mathcal{N}(0, 1), n=1,2,...$$
$$N(n) = \mathcal{N}(0.3, 1), n=1,2,...$$
$$X(n) \text{ and } N(n) \text{ are both i.i.d.}$$
$$\hat{X}(n)=\frac{1}{5}(4X(n)+2N(n)-0.6)$$
$$f(k)= \frac{1}{k}\sum_{n=1}^{k}[(X(n)-\hat{X}(n))^2]$$
Prove:

1: $\lim_{k \rightarrow \infty}f(k)=0.2$

2: $\lim_{k \rightarrow \infty}f(k)$ is almost surely convergence

答:
This is a direct application of the strong law of large numbers for the random variable $(X(1) - \hat{X}(1))^2$.

$$E[(X(1)-\hat{X}(1))^2] = \text{Var}((X(1) - \hat{X}(1)) = (1/5)^2 + (2/5)^2 = 0.2.$$

Title: Calculate the convergence of MMSE


答:
We need to find the number of pigeonholes. Two birthdays are in the same pigeonhole if they occur in the same month on the same day. So the different pigeonholes are of the form $(\text{Month, Day})$ i.e. $(\text{January, Monday}),(\text{January, Tuesday}),...,(\text{January, Sunday}),(\text{February, Monday}),...,(\text{February, Sunday})...$

and so on, giving a total of $7\times 12 =84$ pigeonholes. So you can take $>84$ people.

Title: Counting Problem: Pigeonhole principle

Body:
How many people are needed to guarantee that at least two were born on the same day of the week and in the same month (perhaps in different years) ?

I couldn't understand the wording of this question, so please explain the exact solution which is required.

Thanks!


问:
Logarithmic convexity of incomplete gamma function.
The incomplete Gamma function $F(t)$ satisfies

$$
1 - F(t) \sim \int^\infty_t dx \, e^{-x} x^{\alpha - 1}
$$

for $t > 0$ and its derivative $F^\prime \sim e^{-t} t^{\alpha - 1}$ is the density function of the Gamma distribution.

It is claimed that the hazard rate, also known as the failure rate

$$h = -\frac{d}{dt} \ln (1 - F) = \frac{F^\prime}{1-F}$$

is increasing everywhere (i.e. $h^\prime (t) > 0$) for the case that $\alpha > 1$ and decreasing everywhere for the case that $\alpha < 1$.

In other words, the logarithmic convexity of $1 - F$ depends only on the sign of $\alpha - 1$.

This result is also given as a proof exercise in problem 5.22 of this book.

It's unclear to me how to prove this result. We have

$$ h^\prime = \frac{(1-F)F^{\prime \prime} + (F^\prime)^2}{(1-F)^2} = \frac{F^\prime}{(1-F)^2} \left[ (1-F)\left(\frac{\alpha - 1}{t} - 1 \right) + F^\prime \right], $$

but from this point the next step is unclear.

答:
Consider $t=0$. Then, $h(0)=0$ and since for any $\epsilon > 0$ we have $h(\epsilon) > 0$, the derivative (to the right) at zero is positive. (I assume we can omit to consider $t<0$).

Now, consider $t > 0$. On this set, $1/h$ is well defined and given by
$$1/h = \int_t^{\infty} dx e^{t-x} \left(\frac{x}{t}\right)^{\alpha - 1}$$
Doing a change of variable, we get
$$1/h = \int_0^{\infty} dx e^{-x} \left(1+\frac{x}{t}\right)^{\alpha - 1}$$

Now, note that the term
$$\left(1+\frac{x}{t}\right)^{\alpha - 1}$$
increases (decreases) in $t$ if $\alpha<1$ ($\alpha>1$).

The claim then follows.


问:
Show that the linear operator $(Tf)(x)=\frac{1}{\pi} \int_0^{\infty} \frac{f(y)}{(x+y)} dy$ satisfies $\|T\|\leq 1$.

Show that the linear operator $T$ given by $(Tf)(x) = \frac{1}{\pi} \int_0^{\infty} \frac{f(y)}{(x+y)} dy$ is bounded on $L^2(0, \infty)$ with norm $||T|| \leq 1$.

The professor also wrote, $``$In fact, you should observe that $||T|| = 1$, where $$||T|| = \sup\limits_{f\neq 0}\frac{||Tf||_2}{||f||_2}."$$

答:
Upper bound

For any $g\in L^2(0,\infty)$,
$$
\begin{align}
\int_0^\infty Tf(x)g(x)\,\mathrm{d}x
&=\frac1\pi\int_0^\infty\int_0^\infty\frac{f(y)g(x)}{x+y}\,\mathrm{d}y\,\mathrm{d}x\\
&=\frac1\pi\int_0^\infty\int_0^\infty\frac{f(xy)g(x)}{1+y}\,\mathrm{d}y\,\mathrm{d}x\\
&=\frac1\pi\int_0^\infty\int_0^\infty\frac{f(xy)g(x)}{1+y}\,\mathrm{d}x\,\mathrm{d}y\\
&\le\frac1\pi\int_0^\infty\|f\|_2\|g\|_2\frac{y^{-1/2}}{1+y}\mathrm{d}y\\
&=\frac1\pi\|f\|_2\|g\|_2\int_0^\infty\frac2{1+u^2}\,\mathrm{d}u\\
&=\|f\|_2\|g\|_2
\end{align}
$$
This says that $\|Tf\|_2\le\|f\|_2$ ; that is, $\|T\|_2\le1$.

Lower bound

Let
$$
f(y)=\left\{\begin{array}{}
\frac1{\sqrt{y}}&\text{for }1\le y\le\lambda\\
0&\text{otherwise}
\end{array}\right.
$$
Then $\|f\|_2=\sqrt{\log(\lambda)}$ . Furthermore,
$$
\begin{align}
Tf(x)
&=\frac1\pi\int_1^\lambda\frac{y^{-1/2}}{x+y}\,\mathrm{d}y\\
&=\frac1{\pi\sqrt{x}}\int_{1/x}^{\lambda/x}\frac{y^{-1/2}}{1+y}\,\mathrm{d}y\\
&=\frac2{\pi\sqrt{x}}\int_{\sqrt{1/x}}^{\sqrt{\lambda/x}}\frac1{1+u^2}\,\mathrm{d}u\\
&=\frac2{\pi\sqrt{x}}\left(\tan^{-1}\left(\sqrt{\lambda/x}\right)-\tan^{-1}\left(\sqrt{1/x}\right)\right)\\
&=\frac2{\pi\sqrt{x}}\tan^{-1}\left(\frac{1-1/\sqrt\lambda}{1+x/\sqrt\lambda}\sqrt{x}\right)
\end{align}
$$
For any $\alpha\gt0$,
$$
\begin{align}
\frac{\|Tf\|^2}{\|f\|^2}
&=\frac1{\log(\lambda)}\int_0^\infty|Tf(x)|^2\,\mathrm{d}x\\
&\ge\frac4{\pi^2\log(\lambda)}\int_{\lambda^\alpha}^{\lambda^{1-\alpha}}\frac1x\tan^{-1}\left(\frac{1-1/\sqrt{\lambda}}{2}\lambda^{\alpha/2}\right)^2\,\mathrm{d}x\\
&=(1-2\alpha)\frac4{\pi^2}\tan^{-1}\left(\frac{1-1/\sqrt{\lambda}}{2}\lambda^{\alpha/2}\right)^2\\
&\stackrel{\lambda\to\infty}{\to}1-2\alpha
\end{align}
$$
Since $\alpha\gt0$ was arbitrary, we get that $\|T\|_2\ge1$. Therefore, $\|T\|_2=1$.


问:
When is a bounded lattice isomorphic to some $(\mathbf{2}^S,\subseteq)$?
If I have a set $S$, I can turn it into a bounded lattice in any number of ways, the simplest of which is to designate an initial object and a terminal object, adding a single object if necessary. Such an extension is clearly not isomorphic to a power set unless $|S|\leq 4$.

On the other hand, if I have a bounded lattice $L$, is there some property strictly in terms of $\lor$, $\land$, and $\leq$ that guarantees that $L$ is isomorphic to a power set? In other words, is there a set of properties of $L$ implying $\exists S_L$ s.t. $L\simeq(\mathbf{2}^{S_L},\subseteq)$ without explicitly constructing $L\leftrightarrow(\mathbf{2}^{S_L},\subseteq)$ as a part of the proof for each specific $L$?

As a starting point, it seems like if $\exists S$ s.t. $L\simeq(\mathbf{2}^S,\subseteq)$ then we can define $S$ as the set of completely join prime objects of $L$. For fixed $x\in L$, we can then define the subset $S_x\equiv\{y\in S\mid y\leq x\}\in\mathbf{2}^S$.

Additionally, I'm fairly certain that $L$ must be atomic and that the atoms of $L$ must be exactly $S\setminus$, i.e., that $L$ can be expressed strictly in terms of joins on subsets of $S$.

The problem with this approach is that it requires showing that each object in $L$ can be constructed from $S$ and that $\mathbf{2}^S$ is fully represented in $L$. It would be nice if there was some existing proof for simple conditions on $L$ that guarantees that such an $S$ can be found without needing to find it.

(To put the question in context, I'm interested in a specific situation where I want to extend a posetal category by including all products and coproducts. I'd like to know what the uniqueness requirements are for this to be isomorphic to a power set.)

答:
One such characterization is that a poset $L$ is isomorphic to a power set iff $L$ is a complete lattice (every subset has a join and a meet), complemented (for each $x\in L$ there exists $\neg x\in L$ such that $x\wedge \neg x=0$ and $x\vee \neg x=1$), and completely distributive, meaning it satisfies the infinite distributive law $$\bigwedge_{i\in I}\bigvee_{j\in J}a_{ij}=\bigvee_{f:I\to J}\bigwedge_{i\in I}a_{if(i)}$$ for arbitrary index sets $I$ and $J$.

This characterization is well-known, and you can probably find it in any text that covers the basic theory of complete Boolean algebras (a Boolean algebra is a complemented distributive lattice, so the characterization above says $L$ is isomorphic to a power set iff it is a complete Boolean algebra that is completely distributive).  As a sketch of a proof, given such an $L$, note that by complete distributivity $$1=\bigwedge_{a\in A}(a\vee \neg a)$$ is equal to the join of all elements of the form $\bigwedge_{a\in A}a^*$, where each $a^*$ is either $a$ or $\neg a$.  But if such an element $b=\bigwedge_{a\in A}a^*$ is nonzero, then it is an atom, since if $0<c<b$ then $b\not\leq c$ and also $b\not\leq \neg c$ which is impossible since either $c$ or $\neg c$ is a term in the meet defining $b$.  So, $1$ is a join of atoms in $L$, and it follows from distributivity that every element of $L$ is the join of the atoms below it.  Now let $S$ be the set of atoms of $L$, let $f:L\to \mathbf{2}^S$ send each element of $L$ to the set of atoms below it, and let $g:\mathbf{2}^S\to L$ send each set of atoms to its join.  It is clear that $f$ and $g$ are both order-preserving, and we know $gf=1_L$.  Distributivity can be used to show that $f$ preserves arbitrary joins, and so $f$ is surjective since $L$ is complete and the image of $f$ contains all singletons.  So, $f$ is a bijection and $g=f^{-1}$, so $f$ is an order-isomorphism.


问:
How to find the values of $x,y,z$ to give a maximum value of $\min\{ax,by,cz\}/(x+y+z)$.
$a,b,c$ are all known positive rational numbers (e.g. $a=2.1,b=3.5,c=7.4$). They are also always $\geq1$ but I'm not sure if that helps. $\min\{ax,by,cz\}$ means the minimum out of those three products.

This feels like an optimisation problem but I'm not sure how to approach it. I'm hoping to solve this programmatically, so that it can be solved for several thousand sets of coefficients.

答:
Suppose that $ax$ is the minimum; then $y$ can be reduced to $ax/b$ without changing the numerator but decreasing the denominator; and similarly for $z$. And similarly if one of the other two products is the minimum.

So the maximum of the quotient is going to be achieved when $ax=by=cz$, that is, when $x=\lambda/a$, $y=\lambda/b$, $z=\lambda/c$ for some constant $\lambda$, so that the quotient equals $1/(\frac1a+\frac1b+\frac1c)$.


问:
Sufficient condition on $f:\mathbb R \to \mathbb R$ to ensure the function $F:x \mapsto f(x^\top a)$ is geodesically convex on $n$-sphere
Let $\mathbb S_{n-1}$ be the $(n-1)$-dimensional sphere in $\mathbb R^n$. Fix $a \in \mathbb R^n$ and  consider a function of the form $F(x) = f(x^\top a)$, for some one-dimensional function $f:\mathbb R \to \mathbb R$.

Question. What is a sufficient (and hopefully, necesssary) condition on $f$ for which
$F$ is geodesically convex on $\mathbb S_{n-1}$ ?

You may assume $f \in \mathcal C^2(\mathbb R)$.

Definition. Recall that if $\mathcal M$ is a manifold, then a function $G:\mathcal M \to \mathbb R$ is said to be geodescially convex iff $G\circ \gamma:[0,1] \to \mathbb R$ is convex for every geodesic curve $\gamma:[0,1] \to \mathcal M$. Geodesic strong-convexity, geodesic concavity, and geodesic strong-concavity are defined similarly.

Examples of functions that I have in mind

Linear function. $f(z) \equiv z$

Nonlinear functions. $f(z) \equiv e^{z}$ or $f(z) \equiv \log(1 + e^{z})$.

Note that all of the above functions are convex in the usual sense in $\mathbb R^n$.

Update: solution to case of linear function

Consider the case where $f(z) \equiv z$. One computes

Euclidean gradient: $\nabla_{\mathbb R^n} F(x) = a$.

Euclidean hessian: $\nabla_{\mathbb R^n}^2 F(x) = 0$.

Therefore, using standard formula, the riemannian hessian is
$$
\nabla_{\mathbb S_{n-1}}^2 f(x)[u] = P_{x}(\nabla_{\mathbb R^n}^2 f(x)[u]) - \mbox{trace}(x^\top \nabla_{\mathbb R^n}f(x))u = -F(x)u.
$$

Therefore if we define the spherical cap $H_a := \{x \in \mathbb S_{n-1} \mid x^Ta < 0\}$, then
$$
\nabla_{\mathbb S_{n-1}}^2 f(x)[u,u] = -F(x)\|u\|^2 = \begin{cases}\le 0,&\mbox{ if }x \in H_{a},\\\ge 0,&\mbox{ if }x \in H_{-a}\end{cases}
$$

and so $F$ is geodesically convex on $H_a$ and geodescially concave on $H_{-a}$.

答:
The resulting function must be constant, since all geodesically convex functions on $S^n$ are constant.

for any function $f:S^1\to\mathbb{R}$, there is a corresponding periodic function $\tilde{f}:\mathbb{R}\to\mathbb{R}$. $f$ is geodesically convex iff $\tilde{f}$ is convex on any closed interval, which, by periodicity, is only the case if $\widetilde{f}$ is constant.

A geodesically convex function $f:S^n\to\mathbb{R}$ must be constant along all great circles by the previous argument, and thus $f$ must be constant on $S^n$, since any two points are joined by a great circle.

This result isn't specific to $S^n$: If I'm not mistaken, all geodeiscially convex functions are constant on any compact, connected Riemannian manifold without boundary.


问:
Proving $\lim_{p\to 0} ||f||_p^p = \mu(\operatorname{supp}f)$
Q) Let $\mu$ be a counting measure on $\mathbb{N}$. Prove that

$$\lim_{p\to 0} ||f||_p^p = \mu(\operatorname{supp} f)$$

My attempt: I can prove this if the support of $f$ is a finite set. Because then,

$$\lim_{p\to 0} \sum_n |f(n)|^p = \sum_n \lim_{p\to 0} |f(n)|^p = \sum_{n:f(n)\neq 0} 1 = \mu(\operatorname{supp} f)$$

But if $\operatorname{supp} f$ is an infinite set, may I know how to show that $\lim_{p\to 0} ||f||_p^p = \infty$?

答: Suppose the support if $f$ is an infinite set, say $\{n_1,n_2,...\}$. Then $\sum |f(n)|^{p} \geq \sum\limits_{k=1}^{N} |f(n_k)|^{p} \to N$ as $p \to 0$ for each positive integer $N$. Hence the limit is $\infty$, as required.


答: You can assume that $|f(n)|<1$ for all $n$ because if there are infinitely many integers for which $|f(n)|\ge 1$ the result is clear. Now,  define $f_m(n)=|f(n)|^{1/m}.$ Then $f_m$ is positive increasing so the Monotone Convergence Theorem applies to show that $\underset{m\to\infty}\lim \sum_n |f(n)|^{1/m}=\sum_n \underset{m\to\infty}\lim|f(n)|^{1/m}=\infty.$

Title: Proving $\lim_{p\to 0} ||f||_p^p = \mu(\operatorname{supp}f)$

Body:
Q) Let $\mu$ be a counting measure on $\mathbb{N}$. Prove that

$$\lim_{p\to 0} ||f||_p^p = \mu(\operatorname{supp} f)$$

My attempt: I can prove this if the support of $f$ is a finite set. Because then,

$$\lim_{p\to 0} \sum_n |f(n)|^p = \sum_n \lim_{p\to 0} |f(n)|^p = \sum_{n:f(n)\neq 0} 1 = \mu(\operatorname{supp} f)$$

But if $\operatorname{supp} f$ is an infinite set, may I know how to show that $\lim_{p\to 0} ||f||_p^p = \infty$?


问:
Looking for Linear Algebra FREE course supplements (book and other)
I am well aware that the tag "free" limits the search quite a bit, but I'm having some issue finding free resources myself so I thought I might extend the question to this community to see what you all might think up.

Specifically, I am looking for some free material to help me understand my linear algebra course better as I am unable to meet with my professor due to a massively busy schedule and my current book is hardly any help. The book we use is offered by UCDavis and I find it quite lacking in the example department.

Anything helps! Thanks everyone.

答:
Linear Algebra by Jim Hefferon has been mentioned by Vikram in the comments, but I'd like to promote it to a full answer. This is a first-semester linear algebra textbook made available as a free PDF file (see the previous link) and as a relatively inexpensive paperback (see Google Books). Hefferon's Linear Algebra, now in its fourth edition, won the 2020 Daniel Solow Author's Award from the Mathematical Association of America; here one can read their reasons for the award, of which I quote an excerpt:

The nomination packet for Dr. Hefferon’s work spoke directly to how Linear Algebra works extraordinarily well for students. Not every student will have the same interests or abilities, but all can be encouraged and supported as they explore the parts of the subject that they can grasp and build on. As recent research has shown, active learning is demonstrably more effective than traditional lecture, and the book’s approach helps teachers to get students involved. This is especially true of the application sections that are highlighted in [a previously cited] review, as they give plenty of opportunity to have students put recently learned theory into context.

[…]

The clear writing style, tremendous variety of exercises, amenability to use with active learning strategies, and the careful attention to detail in preparation mean that the text is exceptionally adaptable.


问:
Is iteration process through equal sign possible?
Is iteration process through equal sign possible?

for an example: repeatedly pressing on the equal button of calculator

Repeatedly pressing on the equal button of calculator until full converging the Answer.

this is (Ahmad Reza Estakhr's) symbol of iteration of equal sign $\Re_{i=1}^{\infty}=$ which means repeatedly pressing on equal sign.

答:
That just depends on what the equal sign on your calculator is programmed to do.

Usually the equal sign is meant to compute as much as it can (in your terms, carry out as many "iterations" in one step as possible) - if you type in 5+(3*2)+8+9, it will return 28, instead of returning 5+6+8+9.

It's possible, yes, for certain cases (e.g. if your calculator is a polynomial solver), especially if your equal sign is programmed to carry out simplification; however, by and large, it will just land you returning the same value over and over again. In other words, the general answer to your question is no.