问:
Progress on a conjecture of Burnside...
Given a group $G $, the set of automorphisms of $G $ also forms a group, $\rm {Aut}(G) $,with composition as the operation.

An inner automorphism is one determined by conjugation by some element $g\in G $.  That's we have the automorphism  $i_g $ given by $i_g (h)=ghg^{-1}\,\forall h $.

I learned from a comment by @LeeMosher on this site that Burnside once conjectured that any class-preserving automorphism is inner.  Does anyone know about any progress on this?

Of course the converse is trivial.

(Btw, the reference here is to conjugacy classes.  Of course the class equation gives the sizes of these classes.  For example,  in the case of an abelian group,  the class equation consists in all ones.)

For reference,  this would have been early in the  $20$-th century.

答:
It is true for finite simple groups. This is a result of Walter Feit and Gary Seitz (1984). Seitz communicated this to me in a letter and I am not sure if it was formally published afterwards. I needed the result for classifying finite groups whose complex irreducible characters are all primitive, see here (in particular $(6.1)$).

Note (added April 20th 2021): the result of Feit and Seitz appeared in On finite rational groups and related topics, Illinois Jour. of Math., vol. 33 (1989), pages 103-131.

Title: Progress on a conjecture of Burnside...


问:
Combinatorics Problem: Building numbers from the difference of the first 2021 numbers
I recently found this problem, it was part of a regional qualifier in Southern America (Venezuela- I believe) in January 2021. As I can’t find the solution anywhere and it is very different from the problems I have tackled so far I thought I might ask you all for some help. I have tried finding cases where $(a)$ or $(b)$ is true, but I wasn’t able to do that.

THE PROBLEM:

On a blackboard are the numbers $1, 2, . . . , 2020$ and $2021$. The following operation is carried out:
You choose two numbers, write the amount of their difference on the board and delete the two numbers chosen.
Repeat this until there is only one number left on the board.

$(a)$ Show that $2021$ can be the last number on the board.

$(b)$ Show that $2020$ cannot be the last number on the board.

答:
The given operation preserves the parity of the number of odd numbers on the board -- that number always drops either by $2$ or $0$.  Since you started with an odd number of odd numbers, it's impossible for $2020$ to be the last number on the board.

As for the first part, just pair up numbers in order:  $1$ and $2$, $3$ and $4,
\ldots$.   You'll get a string of an even number of $1$s followed by $2021$.  Now pair up the $1$s and you'll get a string of $0$s followed by $2021$.  Now eliminate $0$s until you're left with $0$ and $2021$.

ContentLicense: CC BY-SA 4.0


答:
$(a)$ First step, choosing $(1,2), (3,4),...,(2019,2020)$. After these operations, the remaining numbers on the board are 1010 times of ones and $2021$.
Second step, choosing $(1,1),...(1,1)$  ($\frac{1010}{2} = 505$ couple $(1,1)$), the remaining numbers on the board are 505 times of zeros and $2021$.
Now, it suffices to choose whatever two numbers on the board. $2021$ will always be the last number on the board.

$(b)$
We notice that there are an odd numbers of odd numbers ($1011$ odd numbers). After every operation, the number of odd numbers must always be odd, because:
\begin{cases}
odd-odd =even \qquad \#OddNumber \text{ reduces 2}\\
odd-even =odd \qquad \#OddNumber \text{ is unchanged}\\
even-even =even \qquad \#OddNumber \text{ is unchanged}\\
\end{cases}

If the end, $2020$ is the remaining number, then the number of odd number is even (there is $0$ odd numbers) $\implies$ contracdiction.

So,  $2020$ cannot be the last number on the board.

Title: Combinatorics Problem: Building numbers from the difference of the first 2021 numbers

Body:
I recently found this problem, it was part of a regional qualifier in Southern America (Venezuela- I believe) in January 2021. As I can’t find the solution anywhere and it is very different from the problems I have tackled so far I thought I might ask you all for some help. I have tried finding cases where $(a)$ or $(b)$ is true, but I wasn’t able to do that.

THE PROBLEM:

On a blackboard are the numbers $1, 2, . . . , 2020$ and $2021$. The following operation is carried out:
You choose two numbers, write the amount of their difference on the board and delete the two numbers chosen.
Repeat this until there is only one number left on the board.

$(a)$ Show that $2021$ can be the last number on the board.

$(b)$ Show that $2020$ cannot be the last number on the board.


问:
Conditional expectation of exponential distribution
We were learning about the memoryless property of the exponential distribution, when I came across these seemingly easy problems:

If $X$ ~ $Exp(\lambda)$, find:

$E[X|X>t]$

For this, I used the tail sum formula. $$P(X>s+t|X>t) = P(X>s)$$ $$\int_0^{\infty}P(X>s+t|X>t)ds = \int_0^{\infty}P(X>s)ds$$

I'm not too sure about this step, because I'm not entirely sure of tail sum expectation working for conditional expectations the way I've written it.

If it is correct, it gives the unsurprising result that $E[X|X>t] = E[X]$

$E[X^2|X>t]$

This seemed much harder to tackle for me, since I couldn't easily use the memoryless properly as above. I feel like this should also maybe come out cleanly as $E[X^2]$, but I'm not sure how to go about this.

Thanks in advance!

答:
The memoryless property works in deriving this conditional expectation like so:

$$\begin{align}\mathsf E(X\mid X>t)&=\int_0^\infty (t+s)\,f_X(t+s\mid X>t)\,\mathrm ds\\&=\int_0^\infty (t+s)\,f_X(s)\,\mathrm d s&&\star\\&=t\int_0^\infty f_X(s)\,\mathsf ds+\int_0^\infty s\,f_X(s)\,\mathrm d s\\&=t+\int_0^\infty\int_0^sf_X(s)\,\mathrm d u\,\mathrm d s\\&=t+\int_0^\infty \int_u^\infty f_X(s)\,\mathrm d s\,\mathrm d u\\&=t+\int_0^\infty \mathsf P(X>u)\,\mathrm du\\[2ex]\therefore\quad\mathsf E(X\mid X>t)&=t+\mathsf E(X)\end{align}$$


问:
Is this true for integers matrices $A=XA’Y$
Let be $A$ a matrix with integer entries. Prove that there are some matrices $X, Y$ with integer entries and having $\det (X), \det(Y)$ equal to $1$ or $-1$ such that $$A=XA’Y$$ where

$$ A’=\begin{pmatrix} a_1 & 0 & \ldots & 0\\ 0& a_2& \ldots & 0\\ \vdots & \vdots & \ddots & 0 &\\ 0& 0 &0 &a_n \end{pmatrix} $$
and $a_1$, $a_2$, ... $a_n$ are integers.

I came across this while solving some mathematics problems but only for the case when $n=2$ . Is this true for the general case? I would be extremely grateful if somebody can provide a complete proof for this problem.

答:
This is the Smith normal form of a matrix over a principal ideal domain. The ring of the integers $(\mathbb Z, + , \cdot)$ is a principal ideal domain.

Wikipedia entry provides an algorithm to compute the Smith normal form of a matrix and therefore the proof you're looking for. See also this paper to get some elements on algorithms performance.


问:
Evaluating the series $\sum\limits_{n=1}^\infty \frac1{4n^2+2n}$
How do we evaluate the following series:

$$\sum_{n=1}^\infty \frac1{4n^2+2n}$$

I know that it converges by the comparison test. Wolfram Alpha gives the answer $1 - \ln(2)$, but I cannot see how to get it. The Taylor series of logarithm is nowhere near this series.

答:
Hint:
Another approach is to Complete the square in the denominator and use the Fourier transform and Poison summation formula approach.

ContentLicense: CC BY-SA 3.0


问:
Definition of horizontal subspace using horizontal lift
Let $\pi: E \longrightarrow M$ be a vector bundle, $e \in E$ and  $h: T_{\pi(e)}M \longrightarrow T_eE$, $X \mapsto \tilde{X}=\sigma_*(X)$, where $\sigma$ is any local section of $E$ which is parallel in the direction of $X$ at $p$ and satisfies $\sigma(p)=e$.

In "Lectures on Kähler Geometry" by Andrei Moroianu the horizontal subspace $T^h_{e}E$ is defined as $\text{im}(h)$. Furthermore it is claimed that $\pi_* \circ h = Id$. Can you help me understand, why this is true?

答:
In a vector bundle $\pi \circ \sigma = Id_M$ for all $\sigma \in \Gamma(E)$.

$(\pi_* \circ h)(X) = \pi_*(h(X)) = \pi_*(\sigma_*(X)) = (\pi_* \circ \sigma_*)(X) = (\pi \circ \sigma)_*(X) = (Id_M)_*(X) = X$


问:
Explicit computation Legendre transform random variable
Let $X$ be uniformly distributed in $\{-1,1\}$. Show that the Legendre transform is
$$
\psi^*(x) = \frac{(1+x)\log(1+x)+(1-x)\log(1-x)}{2}
$$

I computed the moment generated function
$$
\psi(x) = \frac{(e^{-t}+e^t)}{2}
$$
and tried to apply the definition

$$
\psi^*(x) = \sup_{t \in \mathbb{R}} tx-\log \psi(t)
$$
but can't get the required form.

答:
Using the linked section, we see that $\dfrac{d}{dx} \log(\psi(x)) = \tanh(x)$ is invertible.

Hence, $\dfrac{d}{dx} \phi^*(x) = \tanh^{-1}(x)$.

This is an easy exercise in integration by parts.

Indeed, it is easy to see the solution from the well-known identity $\tanh^{-1}(x) \equiv \dfrac{\log(1+x) - \log(1-x)}{2}$.


问:
Proving that free modules are flat (without appealing projective modules)
Suppose $R\neq 0$ is a commutative ring with $1$. Let $M$ be a free $R$-module. I would like to prove that $M$ is a flat $R$-module. Everywhere I have looked (mostly online) this is proved by first proving that every free module is projective, and then proving that every projective module is flat. Unfortunately, Atiyah & Macdonald's "Introduction to Commutative Algebra" (Chapter 2) does not discuss projective modules. But the result that every free module is flat comes very handy in the exercises.

So my question is,

Is it possible to prove that every free module is flat just by
  definitions and without appealing to projective modules?

Thanks!

答:
The identity functor is exact.

If $\{F_i\}$ is a family of exact functors, then $\oplus_i F_i$ is also exact.

Now apply this to $M \otimes - $ for a free module $M$. 1. deals with the case $M=R$, and 2. generalizes this to $M = \oplus_i R$.


问:
How do we know the Newton-Raphson method produces a better approximation than the initial point.

I know this method works by finding the root of the tangent line to the initial point, which is the best linear approximation of the function at that point. And I also know the closer the initial point to the root, the better the approximation, since a tangent line approximation is more accurate for nearby values.

But how do we know the root of the tangent line always produces a value that is closer the the root than the initial point? I haven't seen any author provide an explanation on this other than providing some diagrams. Is it possible that sometimes it gives an value that is slightly further away from the initial point (provided the initial guess is fair)?

答:
The short answer is that it doesn't.  You don't know, because it depends on the function and the initial guess.

Refer to the following question and answer:  Examples of when Newton's Method will fail?  In that answer, you can see that for certain functions, you could have nonconvergent behavior, even repelling fixed points.


问:
For positive integers $x$, $y$, and positive real $k$, if $\frac{x+1}{y+k}=\frac{y}{x}$, then $k\geq1$
I saw this problem in the prep book for the USAMO in the Algebra section, but I don’t know how to tackle it. As the problems before were all solvable by factoring/Cauchy-Schwarz, I have tried finding a way to factor/rearrange this into a C.S.-form , but I have made no progress at all. As I’m getting quite frustrated with this problem I would appreciate all help!

Let $x$ and $y$ be positive integers and $k$ be a positive real number for which $$\frac{x + 1}{y + k} = \frac{y}{x}$$
is satisfied.

Prove that $k\geq1$ holds.

My thoughts: After rearranging we arrive at $(x+1)x$ which must at least be 2. And $y(y+k)$ where $y$ is at least $1$. So we have
$$x^2+x=y^2+yk$$
and $$\frac{x^2+x-y^2}{y}=k$$
This obviously holds for $x>y$, but for $y<x$ I am running into difficulties.

Body:
\begin{align*}
&
\frac{a^2+a-b^2}{b}=c
\\[4pt]
\implies\;&
\frac{a^2+a-b^2}{b} > 0
\\[4pt]
\implies\;&
a^2+a-b^2 > 0
\\[4pt]
\implies\;&
a^2+a > b^2
\\[4pt]
\implies\;&
(a+1)^2 > b^2
\\[4pt]
\implies\;&
a+1 > b
\\[4pt]
\implies\;&
a\ge b\;\;\;\text{$\bigl($for integers $x,y$,$\;$if $x+1 >y\;$then $x\ge y\bigr)$}
\\[4pt]
\implies\;&
c=\frac{a^2+a-b^2}{b}\ge \frac{b^2+b-b^2}{b}=1
\\[4pt]
\end{align*}


问:
Continuity of a function proof approach
Given a positive number b and a set E ⊆ R, let us say that a function f : E → R has
property X(b) on E provided that for each pair x, y ∈ E, |f(y) − f(x)| ≤ b|y − x|

How do I go about showing that if f has property X(b) on E then f is continuous on E?

what is the intuition here?

答:
If $f(x)$ has property $X(b)$ on $E$ , then we can get that for any $\epsilon > 0$ , we can get a $0 <\delta < \frac \epsilon b $ set $|y-x| < \delta$ , follows
$$
|f(y) - f(x)| \le b|y-x| <\epsilon
$$
that's the definition of Uniform convergence

Title: Continuity of a function proof approach


问:
Basis of a basis
I'm having troubles to understand the concept of coordinates in Linear Algebra.

Let me give an example:

Consider the following basis of $\mathbb R^2$:

$S_1=\{u_1=(1,-2),u_2=(3,-4)\}$ and $S_2=\{v_1=(1,3),v_2=(3,8)\}$

Let $w=(2,3)$ be a vector with coordinates in $S_1$, then $w=2u_1+3u_2=2(1,-2)+3(3,-4)=(11,-16)$.

When I tried to found the coordinates of $w$ in $S_2$, I found the following problem:

Which basis $(11,-16)$ belongs to? I suppose the same of $u_1$ and $u_2$, but which basis $u_1$ and $u_2$ belongs to? and if I discover the basis of $u_1$ and $u_2$, what's the basis of the basis of $u_1$ and $u_2$?

I found an infinite recurrence problem and I was stuck there.

Maybe I'm seeing things more complicated than it is, but it seems that there is a deeper and philosophical question inside of this doubt, I couldn't see what a coordinate really is.

I would be very grateful if anyone help me with this doubt.

答: The basis for everything, unless specified, is the standard basis $\{\textbf{e}_1=(1,0),\textbf{e}_2=(0,1)\}$</p>

Title: Basis of a basis


问:
Estimate involving the exponential function
Context: I've just got into reading papers but sometimes struggle a little bit with the "basics" i.e. the things that the authors take for granted.

Problem: I'd like to show that $$e^{-x} \leq 1-x+x^2 \textrm{ for all } x\geq 0. $$

Ideas: I tried using the taylor series of $e^{-x}$  which led me to $$e^{-x}=1-x+\frac {x^2} 2 +\sum_{k=3}^\infty \frac{(-x)^k}{k!} $$ and now it remains to show that $$\sum_{k=3}^\infty \frac{(-x)^k}{k!} \leq \frac {x^2}{2}.$$

This is where I'm stuck. Any help or hint is appreciated!

答:
First do a drawing and you realize that

$$e^{-x}<1-x+x^2$$

$\forall x >0$

to prove it simply compare the derivatives of LHS vs RHS

If you compare the two derivatives you see that for all $x_0 \in (0;0.5)$ the derivative of $e^{-x}$ is greater, in absolute value, w.r.t. the other one. when $x>0.5$ the parabola increases....


问:
Prove $\pi=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}$
How could it be proved that
$$\pi=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}?$$

What I tried

Let
$$L=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}.$$
Unwinding $\Gamma (n+3/4)$ into a product gives
$$\Gamma \left(n+\frac{3}{4}\right)=\Gamma\left(\frac{3}{4}\right)\prod_{k=0}^{n-1}\left(k+\frac{3}{4}\right).$$
Then
$$\lim_{n\to\infty}\frac{(2n)!}{4^n}\prod_{k=0}^{n-1}\frac{16}{(3+4k)^2}=\frac{\Gamma ^2(3/4)}{\sqrt{L}}.$$
Since
$$\frac{(2n)!}{4^n}\prod_{k=0}^{n-1}\frac{16}{(3+4k)^2}=\prod_{k=1}^n \frac{4k(4k-2)}{(4k-1)^2}$$
for all $n\in\mathbb{N}$, it follows that
$$\prod_{k=1}^\infty \frac{4k(4k-2)}{(4k-1)^2}=\frac{\Gamma ^2(3/4)}{\sqrt{L}}.$$
But note that this actually gives an interesting Wallis-like product:
$$\frac{2\cdot 4\cdot 6\cdot 8\cdot 10\cdot 12\cdots}{3\cdot 3\cdot 7\cdot 7\cdot 11\cdot 11\cdots}=\frac{\Gamma ^2(3/4)}{\sqrt{L}}.$$

I'm stuck at the Wallis-like product, though.

答:
You can make a shift $n\to m-1/4$ (the limit is the same along real numbers as well). Thus,
\begin{align*}
& \mathop {\lim }\limits_{n \to  + \infty } 2^{4n} \frac{{\Gamma ^4 (n + 3/4)}}{{\Gamma ^2 (2n + 1)}} = \mathop {\lim }\limits_{m \to  + \infty } 2^{4m - 1} \frac{{\Gamma ^4 (m + 1/2)}}{{\Gamma ^2 (2m + 1/2)}}
\\ & = \mathop {\lim }\limits_{m \to  + \infty } \pi \left( {  \frac{\sqrt {\pi 2m}}{{4^{2m} }}\binom{4m}{2m}} \right)^{ - 2} \left( { \frac{\sqrt {\pi m }}{{4^m }}\binom{2m}{m}} \right)^2  = \pi .
\end{align*}
Here I made use of the well-known asymptotics for the central binomial coefficients:
$$
\binom{2k}{k} \sim \frac{{4^k }}{{\sqrt {\pi k} }}, \quad k\to +\infty.
$$
In fact, the only thing I used was the fact that the limit
$$
\mathop {\lim }\limits_{k \to  + \infty } 4^{-k}\sqrt{k}\binom{2k}{k}
$$
exists and is finite.


问:
Prove $(AB)^{k} = A^{k}B^{k}$
Apologies if this question was posed already.

I need to prove that $(AB)^{k} = A^{k}B^{k}$ holds if $AB=BA$. After trying it myself, I looked at the solution and found it quite strange.

"Use induction: $(AB)^{k} = A^{k}B^{k}$ is true for $k=1$ since since $AB=BA$. Assume $(AB)^{k-1}=(BA)^{k-1}$ and prove it true for k:
$$ (AB)^{k} = (AB)^{k-1} AB $$
$$  = A^{k-1}B^{k-1} AB $$
$$  = A^{k-1}A B^{k-1} B $$
$$  = A^{k} B^{k} $$"

I don't understand how this really proves anything. We had to assume $(AB)^{k-1}=(BA)^{k-1}$, how does if follow then that $(AB)^{k} = A^{k}B^{k}$  holds?

答:
In regards to "I don't understand how this really proves anything":

I think it should be easy to see that the first case for $k=1$ holds because $(AB)^1=AB=A^1B^1$

for $k=2$, $(AB)^2 = (AB)(AB)= ABAB = A(BA)B=_{\text{because }AB=BA} A(AB)B=A^2B^2$

assuming that $(AB)^{k-1}=A^{k-1}B^{k-1}$ holds for some $k\geq 2$ uses the case $k=2$ as jumping point, because we know it is true for $k=2$.

The objective is to prove that it holds for all $k \in \mathbb{N}$, so far we only know it holds for $k=1,2$.

What about $k=3$, could we use the knowledge we have already gathered? Yes:

$(AB)^3 = (AB)(AB)(AB) = (AB)^2(AB)\color{red}{=_*}A^2B^2AB=A^2BBAB=A^2B(BA)B=A^2B(AB)B=A^2(BA)B^2=A^2(AB)B^2 = A^3B^3$

therefore we were able to prove that $(AB)^3 =A^3B^3$ by knowing that $(AB)^2=A^2B^2(\color{red}{*})$.

Instead of doing this ad infinitum and checking for $k=4,5,6,7,...$ individually, which would be impossible, tedious and boring, instead we notice a pattern:

If we know an arbitrary previous case held, like say for the case $k-1$: $(AB)^{k-1} = A^{k-1}B^{k-1}$, as in say you checked all the cases up to $k-1$ $(1,...,k-1, \text{in our case } k-1=2)$ could we show that the next case $k$ held as well?

Well, $(AB)^k=(AB)^{k-1}(AB)=A^{k-1}B^{k-1}(AB)= A^{k-1}(B^{k-1}A)B$

and because we know that $AB=BA$, given $B^{k-1}A =B^{k-2}(BA)=B^{k-2}(AB)=B^{k-3}(BA)B=B^{k-3}AB^2 = \cdots =AB^{k-1}$ we can always shift $A$ to the far left for powers of $B$.

Thus $A^{k-1}(B^{k-1}A)B=A^{k-1}(AB^{k-1})B=A^{k}B^{k}$.

In essence this is checking for $k=4,5,6,7,...$ without checking each one individually, because we just proved that given $2$ holds, which it does, $3$ holds as well, and given $3$ holds, $4$ holds as well... and so on and so on:

If $k=2$ holds, then $k=3$ holds,

If $k=3$ holds, then $k=4$ holds,

$\vdots$

If $k=k'-1$ holds, then $k=k'$ holds,

$\vdots$

So it might be weird proving if the case $k=3$ holds, then $k=4$ holds, without knowing anything about the case $k=3$ itself at first, but we know that it holds true because we proved that if the case $k=2$ held then the case $k=3$ held true, as $k=2$ was our base case, we know it's true because it was proven in the first line. All the pieces fall one after another like an infinite chain reaction...

Hope this helps!

Title: Prove $(AB)^{k} = A^{k}B^{k}$


问:
Difficulty in solving the following integral
How to solve the following definite integral:

$$I= \int_{\pi/4}^{3\pi/4} e^{-e^{\frac{1}{x}}}dx$$

I have tried solving this using Feynman's trick but to me, it turns out there is no obvious way to parametrize this function. I am very weak in contour integration and hence couldn't apply it here. I couldn't think of any obvious substitution as well.

The only thing I can see is that $1/x$ is the derivative of $log$ $x$ which is the inverse of exponential. So probably this integral requires exploitation of that but I can't see how.

Also, I prefer answers that do not go beyond undergraduate knowledge in mathematics (if possible) since I encountered it in an undergraduate exam.

答:
Just for the fun.

It would be amuzing to approximate the integral using the $[2,2]$ Padé approximant built around $\frac \pi 2$ to make
$$ e^{-e^{\frac{1}{x}}}=\frac {a_0+a_1 \left(x-\frac{\pi }{2}\right)+a_2 \left(x-\frac{\pi
   }{2}\right)^2} {1+b_1 \left(x-\frac{\pi }{2}\right)+b_2 \left(x-\frac{\pi
   }{2}\right)^2 }$$ The problem is that the constants are quite ugly but perfectly usable.

As usual, integrating will give two arctangents and two logarithms.

I shall not write any formula but, numerically, the result is $0.221184$ to be compared with the "exact" $0.221115$.

The inverse symbolic calculator provides, as an approximation,
$$\frac 1{10}\left(\log (\pi ) \,\,\Gamma \left(\frac{13}{24}\right)\right)^{\sqrt[3]{2}}=0.221115422\cdots$$to be compared with the "exact" $      0.221115434\cdots$


答:
The vector $(11,-16)$ is written in the standard basis - it means $11\times(1,0)+(-16)\times(0,1)$. This is exactly the computation that you do when you write:

$$2u_1+3u_2=2(1,-2)+3(3,-4)=(11,-16)$$

If we write $e_1=(1,0)$ and $e_2=(0,1)$, then what you're writing is:

$$2u_1+3u_2=2(e_1-2e_2)+3(3e_1-4e_2)=11e_1-16e_2$$

so you've changed coordinates from $u_1$ and $u_2$ to $e_1$ and $e_2$.

It's not strictly correct to say that a vector $(a,b)$ "belongs to" a basis, but you do need to know what basis you're working in to be able to interpret it.

More about vectors vs. coordinates

Normally we take $\mathbb{R}^2:=\{(a,b):a,b\in\mathbb{R}\}$ to be the set of pairs of real numbers, with vector space structure given by:

$$(a,b)+(c,d)=(a+c,b+d)$$
$$\lambda(a,b)=(\lambda a,\lambda b)$$

So its points (vectors) are pairs $(a,b)$. We also have that $(a,b)=a(1,0)+b(0,1)$, so the vectors $e_1=(1,0)$ and $e_1=(0,1)$ span (I'm not claiming they're a basis yet, but of course they are), but I can still understand $(a,b)$ without knowing about $e_1$ and $e_2$.

Now pick a basis $v_1,v_2$ of $\mathbb{R}^2$ (not necessarily the standard one). Then for any $u\in\mathbb{R}^2$, we have $u=xv_1+xv_2$, so we could write $u=(x,y)$ in the coordinates $v_1$ and $v_2$. However, because $u$ is a point of $\mathbb{R}^2$, it is a pair $(a,b)$ of real numbers. But unless $v_1=e_1$ and $v_2=e_2$, we won't have $a=x$ and $b=y$. This is confusing (now we can truthfully say $u=(a,b)$ and $u=(x,y)$, but the pair $(a,b)$ isn't equal to the pair $(x,y)$), so instead I'll write $u=[x,y]_v$ to mean that $u$ has coordinates $x$ and $y$ in the basis $v_1$ and $v_2$. Now we can say $(a,b)=[x,y]_v$ without getting confused, and our notation separates a vector in $\mathbb{R}^2$, which is just a pair of numbers $(a,b)$ that doesn't depend on any basis, from a coordinate representation $[x,y]_v$, which requires the basis $v$ to be understood. (Note that we do have $(a,b)=[a,b]_e$).

Now to return to your example, $u_1$ really is the vector $(1,-2)$, and $u_2$ really is the vector $(3,-4)$ - I don't need any basis to understand this. In our new coordinate-emphasising notation, your calculation is now:

$$[2,3]_u=2u_1+3u_2=2(1,-2)+3(3,-4)=(11,-16)$$

where $(11,-16)$ is interpreted as just being a vector in $\mathbb{R}^2$, with no chosen basis (although we could think of it as $(11,-16)=[11,-16]_e$ if we wanted). Now to write it in terms of the basis $v_1$ and $v_2$ you need to find $x,y$ such that $[x,y]_v=(11,-16)$, or rather such that:

$(11,-16)=[x,y]_v=xv_1+yv_2=x(1,3)+y(3,8)=(x+3y,3x+8y)$

Title: Basis of a basis

Body:
I'm having troubles to understand the concept of coordinates in Linear Algebra.

Let me give an example:

Consider the following basis of $\mathbb R^2$:

$S_1=\{u_1=(1,-2),u_2=(3,-4)\}$ and $S_2=\{v_1=(1,3),v_2=(3,8)\}$

Let $w=(2,3)$ be a vector with coordinates in $S_1$, then $w=2u_1+3u_2=2(1,-2)+3(3,-4)=(11,-16)$.

When I tried to found the coordinates of $w$ in $S_2$, I found the following problem:

Which basis $(11,-16)$ belongs to? I suppose the same of $u_1$ and $u_2$, but which basis $u_1$ and $u_2$ belongs to? and if I discover the basis of $u_1$ and $u_2$, what's the basis of the basis of $u_1$ and $u_2$?

I found an infinite recurrence problem and I was stuck there.

Maybe I'm seeing things more complicated than it is, but it seems that there is a deeper and philosophical question inside of this doubt, I couldn't see what a coordinate really is.

I would be very grateful if anyone help me with this doubt.


问:
What is $\arcsin(\cos{x})$ (Spivak Calculus exercise 15-18)?
In the first part of the exercise I proved that $\sin(x+\frac{\pi}2)=\cos{x}$.

The second part is asking what $\arcsin(\cos{x})$ is. Presumably, they want a rule for unrestricted $x$.

We have $\arcsin(\cos{x})=\arcsin(\sin(x+\frac{\pi}2))$. It was said that the domain of $\sin$ usually restricted to $\left[-\frac{\pi}2, \frac{\pi}2\right]$, hence I assume that $x+\frac{\pi}2 \in \left[-\frac{\pi}2, \frac{\pi}2\right]$, and this set is also the co-domain of the $\arcsin$ function.

Since $\sin$ is periodic with period $2\pi$, $\sin(x+\frac{\pi}2)=\sin(2k\pi+x+\frac{\pi}2)$, $2k\pi+x+\frac{\pi}2 \in \left[2k\pi-\frac{\pi}2, 2k\pi+\frac{\pi}2\right], k \in \mathbb{Z}$.

And we have that $\arcsin(\sin(2k\pi+x+\frac{\pi}2))=2k\pi+x+\frac{\pi}2$.

But the answer in the solutions manual is $\begin{cases} x-2k\pi+\frac{\pi}2, x \in \left[2k\pi-\pi, 2k\pi\right] \\ 2k\pi+\frac{\pi}2-x, x \in \left[2k\pi, 2k\pi+\pi\right] \end{cases}$

I can plot the graph and see how $2k\pi+x+\frac{\pi}2$ doesn't work, but where did they get this piecewise function?

答:
Let $\arcsin(\cos x)=y$

$\implies\sin y=\cos x=\sin\left(\dfrac\pi2+x\right)$

$y=n\pi+(-1)^n\left(\dfrac\pi2+x\right)$ where $n$ is any integer

If $n$ is even $=2m$(say) $$y=2m\pi+\dfrac\pi2+x$$

What if $n$ is odd $=2m+1$(say)?

Now in your solution we need $$x-2k_1\pi+\dfrac\pi2=2k_2\pi+x+\dfrac\pi2\implies k_2=-k_1$$

As $k_1$ is any integer,  $k_2$ also has the same domain!


问:
Perturbation Problem (ODE)

Need answers to question 5a(iv), I'm confused whether it is regular perturbation or singular perturbation.

Also, 5b also.

Thank you.

答:
If setting $\epsilon=0$ changes the character of the equation, then it is a singular perturbation. Thus a(iv) is singular, the order of the ODE changes from 2 to 1, while (iii) is regular, the ODE stays at order 2.

for b) one would need to know what techniques you know of. You can solve this via characteristic polynomial, via inner and outer solutions, via WKB approximation,...

The boundary layer is at $x=0$, the outer solution is $e^{(x-1)/2}$ which has value $e^{-1/2}$ at $x=0$, the inner solution would have to deliver this jump, the proposed formula does not seem to give that,...

Title: Perturbation Problem (ODE)


答:
This sort of problems can be proved by construct a function as follows:
$$
g(x) = e^{-x} - 1 + x - x^2
$$
Just to prove $g(x) \le 0 $ for $x \ge 0$

you can discover that usually $g(0) = 0 , g'(0) = 0, g''(x) < 0,x \ge 0 $
by Monotonicity you can get what you want to prove

Title: Estimate involving the exponential function

Body:
Context: I've just got into reading papers but sometimes struggle a little bit with the "basics" i.e. the things that the authors take for granted.

Problem: I'd like to show that $$e^{-x} \leq 1-x+x^2 \textrm{ for all } x\geq 0. $$

Ideas: I tried using the taylor series of $e^{-x}$  which led me to $$e^{-x}=1-x+\frac {x^2} 2 +\sum_{k=3}^\infty \frac{(-x)^k}{k!} $$ and now it remains to show that $$\sum_{k=3}^\infty \frac{(-x)^k}{k!} \leq \frac {x^2}{2}.$$

This is where I'm stuck. Any help or hint is appreciated!


问:
Artin section 1.4: permutation matrices
Proposition 4.7(a) in Artin's book states:

Let $p,q$ be two permutations, with associated permutation matrices $P$, $Q$. Then the matrix associated to the permutation $pq$ is the product $PQ$.

The proof he gives is then:

Since $P$ operates by permuting rows according to $p$ and $q$ operates by permuting entries according to $q$, the associative law for matrix multiplication tells us that $PQ$ permutes according to $pq$:
$$(PQ)X = P(QX).$$
Thus $PQ$ is the permutation matrix associated to $pq$.

I do not understand how associativity comes into play. Is the only idea that I want to apply the matrix $PQ$ (the product) to $X$, but this is exactly the same thing as first applying $Q$ and then applying $P$? Do I need to use anywhere that a composition of permutations is a permutation? In essence, that seems to be what we're using, so this is in effect a homomorphism between $S_n$ and the set of permutation matrices, if I am not mistaken.

答:
You are right about the role of associativity; the idea is: lets write $X = \begin{pmatrix}x_1\\\vdots\\x_n\end{pmatrix}$ with rows $x_1^t , \ldots , x_n^t$. Then $QX = \begin{pmatrix}x_{q(1)}\\\vdots\\x_{q(n)}\end{pmatrix}$ and therefore $$(PQ)X=P(QX) = \begin{pmatrix}x_{p(q(1))}\\\vdots\\x_{p(q(n))}\end{pmatrix}\overset{(*)}=\begin{pmatrix}x_{(pq)(1)}\\\vdots\\x_{(pq)(n)}\end{pmatrix},$$(In $(*)$ you need, that $p \circ q$ is again a permutation.) which shows that $PQ$ is the associated permutation matrix to the permutation $pq$.

You are right that this is a homomorphism $S_n \to GL_n(\mathbb R)$ which is probably what should be constructed here, as it is a representation of $S_n$.

Title: Artin section 1.4: permutation matrices


问:
Vector Calculus Question
My question is: How can I solve the following problem?

$C$ is the curve resulting from the intersection of $x+y+z=0$  with  $x^2+y^2+z^2=4$,
parameterized counterclockwise as seen from the positive $x$-direction.

Evaluate $\int { ydx + zdy + xdz }$

Thanks.

答:
Hint:

You have the vector field $F(x,y,z)=(y,z,x)$, and you want to evaluate $$\int_C F\cdot ds.$$  What is the Curl of your vector field?  Notice that it must be a constant since $F$ is linear.  Let $K=\nabla \times F$ denote this constant.

Now, apply Stokes Theorem.  What is the area of the enclosed curve?  (Hint: it is a circle of radius $2$).  If we call that area $A$, then the line integral is equal to $$K A.$$

Hope that helps,

Title: Vector Calculus Question


问:
Why this term becomes zero when integrating?
I'm reading this text for Boltzmann's equation derivation, which works with a 6N dimensional system, being N the number of particles of the system (each particle has a 3D position vector r and a 3D momentum vector p). At Eq. 2.10 (page 5 of pdf https://www.damtp.cam.ac.uk/user/tong/kintheory/two.pdf) we have the solution of the previous integral, but I don't see this step. Let's take only the first term, so we come from:

$$-N\int\prod_{i=2}^Nd^3r_id^3p_i\sum^N_{j=1}\frac{\vec{p}_j}{m}\cdot\frac{\partial f}{\partial \vec{r}_j}$$

Here f depends on the 6N dimensions. The text says that all the summs except for j=1 are zero, because if you integrate by parts the derivatives which depend on r are with respect p and vice-versa, but I cannot see it. For example, let's take j=2:

$$-N\int\prod_{i=2}^Nd^3r_id^3p_i\frac{\vec{p}_2}{m}\cdot\frac{\partial f}{\partial \vec{r}_2}=-N\int\prod_{i=3}^Nd^3r_id^3p_i\int d^3p_2\int d^3r_2\frac{\vec{p}_2}{m}\cdot\frac{\partial f}{\partial \vec{r}_2}$$

If we do by parts the last integrand we have:

$$\int d^3r_2\frac{\vec{p}_2}{m}\cdot\frac{\partial f}{\partial \vec{r}_2}=\frac{\vec{p}_2}{m}\cdot f-\int d^3r_2\frac{1}{m}\frac{d^3\vec{p}_2}{dr^3}\cdot f$$

The second term is zero. However, why the first term becomes zero?

答: At the end I got an answer. f is a probability distribution and to be normalizable it must be 0 at infinity. Hence that term is also zero.</p>


答:
The statement is clearly true for $x=0$. If $x>0$, we have, by Taylor's theorem with Lagrange remainder,
$$
e^{ - x}  = 1 - x + \frac{{x^2 }}{2}e^{ - \xi }  < 1 - x + \frac{{x^2 }}{2}<1+x+x^2
$$
with a suitable $0<\xi<x$.


答:
For $x\ge0$, $$1-e^{-x}\ge0$$

Integrate twice from $0$ to $x$:

$$x+e^{-x}-1\ge0$$

$$\frac{x^2}2-e^{-x}+1-x\ge0.$$

You can continue forever.

Justification:

If $f(x)\ge0$ from $x=a$, $\displaystyle\int f(x)\,dx$ is a growing function from $x=a$ and $\displaystyle\int_a^x f(t)\,dt\ge0$.


问:
Endomorphism ring of $p$-adic integer $Z_{p}$
I would like to figure out the endomorphism ring of p-adic integer $Z_{p}$.

I know $End（Z）$ is isomorphic to $Z$,

And I guess $End（Zp）$ is also $Zp$.

My image : End$（Z_{p}）$$＝End（limZ/p^nZ）＝lim （End（Z/p^nZ））$$＝limZ/p^nZ＝$$Z_{p}$

My image might be wrong, I want to formally prove that the endomorphsm ring of $Zp$ is $Zp$.
Thank you in advance.

答:
There is an inclusion $\mathbb Z_p \to \operatorname{End}_{\mathbb Z_p}(\mathbb Z_p)$, which sends $a$ to multiplication by $a$.

You are correct that this map is surjective: Let $f$ be an endomorphism and let $a = f(1)$. By $\mathbb Z_p$-linearity, we then have $f(b \cdot 1) = b \cdot f(1) = ab$ for all $b \in \mathbb Z_p$. Thus $f$ is multiplication by $a$.

The same proof works for any commutative ring (with 1): every $A$-module endomorphism of $A$ is multiplication by some $a \in A$.


问:
Induced topology on subset are equal
got a huge dubt on a fact that was presented today to me in class. I would really appreciate if you could help me out with this!

Today in class the teacher mentioned the following:

Lets consider three subsets $W \subset Y \subset X$. Let $\tau$ be a topology on X and let $\hat{\tau}$  be a topology induced by $\tau$ in Y. Then the topologies induced by $\tau$ and $\hat{\tau}$ on $W$ are equal.

As I said, we are just starting the course on some kind of intruductory topology and I've searched for this fact online but couldnt find anything. Any help would be great.

Thanks in advance!

答:
$U  \subseteq W$ is open in the $X$ induced topology if and only if there is an open set $V \subseteq X$ such that
$$
U = V \cap W
$$
Note that
$$
V \cap W = (V \cap Y) \cap W
$$
since $W \subseteq Y$, and so if $U$ is open in the $X$ induced topology, it is open in the $Y$ induced topology and vice versa, since every open set in $Y$ can be written in the form $V \cap Y$ for some $V$ open in $X$.

Title: Induced topology on subset are equal


问:
Find the running speed of the student.
$60$ students are standing in a line with with a distance a $1$ $m$ between each other . The line is moving forward with a speed of $3$ $m/s$ . The last student in the line wanted to convey a message to the first student . So, he ran towards the first student, conveyed the message and returned back to his place . In total he covered $240$ $m$ . Find the running speed of the student .

Since the line is moving forward with a certain speed, the last student must have covered more distance while going towards the first student and less distance while returning back. Which quantity should I consider as the variable?

答:
Hint:

Let the speed, time spent running forward, and time spent running backward be $s, t_1,$ and $t_2$ respectively.

Then

$(s-3)t_1 = 59.$

$(s+3)t_2 = 59.$

$s(t_1 + t_2) = 240.$

Edit

Thanks to Toby Mak for indicating that initially, my first two equations, which specified $60$ on the RHS, were wrong.

Title: Find the running speed of the student.


问:
Determining planarity??

Sorry i couldnt think of any way to ask this so i took a picture of it and pasted it below.
I'll start with what little i do know, firstly this graph is not homemorphic to  $K_5$ simply because that would require 5 vertices of degree 4 which we clearly don't have ( only 3) i also can't seem to figure out hot make it homemorphic to $k,3,3$ though i think i could label it as a bipartite graph in some was where everything was degree 3 i cant find a way to do that with only 6 vertices.

Lastly i don't think its planar why? if you cut out b1b4, b2b5 and a6b3 it becomes obvious by looking at it that you can't untwist it in two dimensions.

Please help thx.

答:
Being non-planar doesn't mean being homeomorphic to $K_5$ or $K_{3,3}$.  Rather, a graph is non-planar if and only if it contains either $K_5$ or $K_{3,3}$ as a minor.  Here, a graph $H$ is a minor of a graph $G$ if $H$ can be obtained from $G$ by deleting vertices, deleting edges, and/or contracting edges.  (I.e., $G$ has a subgraph homeomorphic to $H$.)

Here's a hint for this problem: show that your graph contains $K_{3,3}$ as a minor.

Title: Determining planarity??


问:
On integration over distribution.
When I read various journal articles related to machine learning, I often face integrals over distribution.

In an article I am reading now, for example, a risk function associated with distribution $\phi$ is defined by $$ R_i(\theta) = \int f_L(f_\theta(x),y) \, d\phi(x,y),$$ where

$\mathcal{X}$ and $\mathcal{Y}$ are a feature space and a label space, respectively.

$f_\theta:\mathcal{X}\to\mathcal{Y}$ is a given model parameterized by $\theta\in\Theta$,

$f_L:\mathcal{X}\times\mathcal{Y}\to\mathbb{R}_{\ge0}$ is a loss function, and

$\phi$ is the data generating distributions.

In addition to the above case and the others (even not related to this field), I have seen many times integration formulas over distributions. However, whenever I encountered them, I couldn't grasp what it is.

Rather, I am familiar with the following equation:
$$
\int f_L(x) p_X(x) \, dx,
$$
where $f_L(x)$ is a cost (or reward) function achieved by an event $x$, and $p_X(x)$ is a probability that an event $x$ occurred.

Can someone please let me know what the integral over a distribution means?

答:
I think you lack relevant knowledge on Riemann–Stieltjes integral

In the Probability theory，the Expectation of Discrete distribution and continuous distribution can be written uniformly as Riemann–Stieltjes integral form, that is
$$ E_X(f(x)) = \int f(x)dF(x)$$

where $F(x)$ is the  cumulative distribution function of random variable $X$,
you can roughly think that $dF(x) = p(x)dx$

I hope it is useful to you.
Sorry, if it didn't help you.

Title: On integration over distribution.


问:
Prove if $x > 0$, $1+\frac{1}{2}x-\frac{1}{8}x^2 \le \sqrt{1+x}$
Prove if $x > 0$,  $1+\frac{1}{2}x-\frac{1}{8}x^2 \le \sqrt{1+x}$.

I find online, one person suggested using Taylor Theorem to expand the right-hand side, and apply Bernoulli's inequality.

So, if $x_0 = 0$, $\sqrt{1+x} = 1+\frac{1}{2}x-\frac{1}{4(2!)}x^2+R$, where R is larger than $0$, this makes sense to me, but I'm trying to find another way to prove the equality. Like Mean Value Theorem for inequality $\sqrt{1+x} \le 1+\frac{1}{2}x$.

I see the part where $1+\frac{1}{2}x$ is the same, but get trouble to put $\frac{1}{8}x^2$ into the equation.

答:
Well putting $x=\sinh^2(a)$

We have to show (using $\cosh^2(a)-\sinh^2(a)=1$):

$$\cosh(a)\geq 1+\frac{1}{2}\sinh^2(a)-\frac{1}{8}\sinh^4(a)$$

Or :

$$\sinh^6\Big(\frac{a}{2}\Big)(\cosh(a)+3)\geq 0$$

Wich is clearly obvious !


问:
Corollary of convergence in distribution
Can I say that if $\xi_n \xrightarrow {d} \xi$ and $\xi_n, \xi$ are non-negative random variables with finite expected value then
$$\mathbb{E}\xi \le \lim{\inf{}_{n\to\infty}\mathbb{E}\xi_n}?$$
I have problems to disprove or prove it. May somebody can help me? I would be grateful.

答: Write each expectation as integral of the tail on the positive real line, and apply Fatou's lemma to the positive real line (with Lebesgue measure) and the sequence $f_n(t)=\mathbb P(X_n>t)$.


问:
What is another more direct proof of this exercise?
The exercise states that " If we know that $(A \cup B)-(A \cap B)=(A \cup C)-(A \cap C)$, then $B=C$ ". At my first sight, I don't know how to do that. However, when I see another exercise in the later of the book, I find a way to work out. That exercise states that we can definite $+$ and $.$, such that, $A+B:=(A \cup B)-(A \cap B), A.B=A \cap B$. Then it let me prove that :

$1.A+B=B+A$

$2.A+\emptyset=A$

$3.A+A=\emptyset$

$4.A.A=A$

$5.A+(B+C)=(A+B)+C$

These are very easy to prove. And from these property, I start my proof of the former exercise. I use property 3 to prove the former. Because now I can translate the  $(A \cup B)-(A \cap B)=(A \cup C)-(A \cap C)$ to $A+B=A+C$. By using property 3, $A+A+B=A+A+C$, using property 5, I can conclude that $(A+A)+B=(A+A)+C$, so by using property 3, $B=C$.

Though I work out the exercise finally, I think that there must be some simpler and more direct proof, since this exercise is situated before than those five properties. Can anyone help with a more direct and more intuitive proof of this?

答:
Sidenote: a common way to denote $(A\cup B)-(A\cap B)$ is $A\Delta B$.

It is the symmetric difference of $A$ and $B$.

Using indicator functions we find that:$$1_{A\Delta B}=1_A+1_B-2\cdot1_A1_B$$

From: $$1_A+1_B-2\cdot1_A1_B=1_A+1_C-2\cdot1_A1_C$$it follows that:
$$(1_B-1_C)(1-2\cdot1_A)=0$$

This implies that $1_B=1_C$ (since we cannot have $1_A=\frac12$) or equivalently: $$B=C$$


答:
Proof by contradiction: suppose there were some $A, B,$ and $C$ such that $(A \cup B) - (A \cap B) = (A \cup C) - (A \cap C)$ but $B \neq C$.

Because $B \neq C$, there must be some element $x$ which is in either $B$ or $C$ but not the other. Otherwise, if every element of $B$ was in $C$ and vice versa, clearly the sets must be equal by mutual containment.

Suppose that $x$ is in $B$ and not in $C$. If $x$ is also in $A$, then $x$ is in both $A \cup B$ and $A \cap B$, so $x$ is not in $(A \cup B) - (A \cap B)$. However, $x$ will be in $A \cup C$ and not in $A \cap C$, so $x$ is in $(A \cup C) - (A \cap C)$, violating the equation and causing a contradiction.

If $x$ is instead not in $A$, then $x$ is in $A \cup B$ but not $A \cap B$, so $x$ is in $(A \cup B) - (A \cap B)$, but $x$ is in neither $A \cup C$ nor $A \cap C$ so it is not in $(A \cup C) - (A \cap C)$, also violating the equation.

Because of the symmetry of the question, you can apply the same argument to show that we reach a contradiction if $x$ is in $C$ but not $B$. Therefore, our assumption has led us to a contradiction and it must be false.

Title: What is another more direct proof of this exercise?

Body:
The exercise states that " If we know that $(A \cup B)-(A \cap B)=(A \cup C)-(A \cap C)$, then $B=C$ ". At my first sight, I don't know how to do that. However, when I see another exercise in the later of the book, I find a way to work out. That exercise states that we can definite $+$ and $.$, such that, $A+B:=(A \cup B)-(A \cap B), A.B=A \cap B$. Then it let me prove that :

$1.A+B=B+A$

$2.A+\emptyset=A$

$3.A+A=\emptyset$

$4.A.A=A$

$5.A+(B+C)=(A+B)+C$

These are very easy to prove. And from these property, I start my proof of the former exercise. I use property 3 to prove the former. Because now I can translate the  $(A \cup B)-(A \cap B)=(A \cup C)-(A \cap C)$ to $A+B=A+C$. By using property 3, $A+A+B=A+A+C$, using property 5, I can conclude that $(A+A)+B=(A+A)+C$, so by using property 3, $B=C$.

Though I work out the exercise finally, I think that there must be some simpler and more direct proof, since this exercise is situated before than those five properties. Can anyone help with a more direct and more intuitive proof of this?


问:
Unitary Operators are Connected in a $C^*$-algebra
Let $A$ be a unital $C^*$- algebra. Let $U = \{ u \in A : u^*u=uu^*=1\}$ be the unitary group of $A$.

Let $U'= \{ e^{ia_1}e^{ia_2} \cdots e^{ia_n} : a_k = a^*_k \in A, \text{for } 1\leq k \leq n \}$.

Show that $U'$ is the connected component of the identity in $U$.

If $A$ is commutative, show that $U' = \{ e^{ia} : a = a ^* \in A\}$.

I thought of using the theorems below from Banach Algebra Techniques in Operator Theory by Douglas.

Following the proof of 2.14, if $f = e^{ia} \in A$. Then $f\in U$. Consider $\phi: [0,1 ] \rightarrow e^(A)$ defined by $\phi(\lambda) = e^{i \lambda a}$, I don't see why $f \in U'$, and $e^A$ is contained in $U'$...

I also showed that if $\| u-1\| <2 $ , then we have $-1 \ne spec(u) $ and we can write $u = e^{ia}$ for some self-adjoint $a \in A$

Any help or suggestions will be appreciated.

Thank you!

答:
First of all, in the case that $A$ is commutative the result easily follows: Note that, when $ab=ba$, then $e^{b}\cdot e^{a}=e^{a}\cdot e^{b}=e^{a+b}$. Using this elementary result (which follows easily from the identity $e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$), we have that
$$U'=\{e^{ia_1}\cdots e^{ia_n}: n\geq1,a_k\in A_{sa}\}=\{e^{i(a_1+\dots+a_n)}: n\geq1, a_k\in A_{sa}\}=\{e^{ia}:a\in A_{sa}\}$$
where $A_{sa}$ denotes the self-adjoint part of $A$.

Now forget about the commutative case and let $A$ be any unital $C^*$-algebra. Let's denote by $C$ the path connected component of $1_A$ in $U(A)$, i.e.
$$C=\{u\in U(A): \text{ there exists a continuous path from }u\text{ to }1_A\} $$
First of all, if $a\in A$ is a self-adjoint element, then the unitary $e^{ia}$ is path-homotopic to $1_A$: simply define $\alpha:[0,1]\to U(A)$ by setting $\alpha(t)=e^{ita}$. Then $\alpha(0)=1_A$, $\alpha(1)=e^{ia}$ and it is easy to verify that $\alpha$ is continuous. Therefore $e^{ia}\in C$ for all $a\in A_{sa}$. Moreover, if $u,v\in C$, then $uv\in C$. Indeed, if $\alpha,\beta:[0,1]\to U(A)$ are continuous paths from $u$ to $1_A$ and from $v$ to $1_A$ respectively, then $t\mapsto\alpha(t)\cdot\beta(t)$ is a continuous path from $uv$ to $1_A$, so $uv\in C$. These two results combined allow us to infer that, finite products of elemenets of the form $e^{ia}$ where $a\in A_{sa}$ belong to $C$. In other words, $U'\subset C$. Also, observe that $U'$ is closed under multiplication (obviously).

Now for the other inclusion, which is the hard one. Since $C$ is a path-connected component, it is connected, so if we show that $U'$ is a clopen subset of $C$, we are done.

$U'$ is open in $C$: We show that $D(u,2)\cap C\subset U'$ for any $u\in U'$. Indeed, if $u\in U'$ and $v\in C$ satisfies $\|u-v\|<2$, then $\|u^*v-1_A\|<2$, so $-2\not\in\sigma(u^*v-1_A)$, so $-1\not\in\sigma(u^*v)$, therefore, as you explain in your post, we can define a logarithm which will give us $u^*v=e^{ih}$ for some self-adjoint element $h$, thus $v=ue^{ih}$, so $v\in U'$, (because $U'$ is closed under multiplication).

$U'$ is closed in $C$: It suffices to show that $U'$ is closed in $U(A)$ (why?). So we will show that $U(A)\setminus U'$ is open. Note that, $(U(A), \cdot)$ is a group and $U'$ is a subgroup of $U(A)$. Let's pause for a moment and have a group-theory throwback:

Assume that $G$ is a group and $H\leq G$ is a subgroup. We have the relation $\sim$ on $G$ given by $x\sim y$ iff $y^{-1}x\in H$. This is an equivalence relation and $[x]=xH=\{xh: h\in H\}$. We take the quotient set $\{[x]: x\in G\}$ and we make a choice from each class (using of course the axiom of choice) and in this way we obtain a set $P\subset G$ so that, if $x,y\in P$ are distinct then $x\not\sim y$ and if $z\in G$ there exists a unique $x\in P$ so that $x\sim z$. From this we can conclude that
$$G=\bigcup_{x\in P}xH $$
Now if $x_0\in P$ is the element that satisfies $[x_0]=H$, we have that
$$G\setminus H=\bigcup_{x\in P-\{x_0\}}xH$$
So we have written the set $G\setminus H$ as a disjoint union of "copies" of $H$. But this will solve the problem, as we shall see:

Okay, back to our situation. We want to show that $U(A)\setminus U'$ is open; by the preceding remark, we have a set $P\subset U(A)$ so that
$$U(A)\setminus U'=\bigcup_{v\in P}vU'.$$

It is elementary to verify that the map $U'\mapsto vU'$, $u\mapsto vu$ is a topological homeomorphism for all $v\in P$. But, as we did in bullet 1, the set $U'$ is open in $U(A)$ (the exact same proof will work). Therefore, since every set $vU'$ is topologically homeomorphic to $U'$, every set $vU'$ is open in $U(A)$. Unions of open sets are open, so we have shown that $U(A)\setminus U'$ is open in $U(A)$, as we wanted.

I hope this is detailed enough. The idea of the proof is demonstrated in M. Rordam's book on K-theory of $C^*$-algebras, you might find Chapter 2 of this book very helpful.


问:
Translation an orthogonal transformation?
Translation is when we move an object in an $n$-dimensional space, to
another point in the same space. It can be used to remove the mean of data points from a dataset. Mathematically we can define by a transformation matrix equation as follows:

If we have a dataset $\mathbf{A}$, with $m$ rows (instances), and $n$ columns (features), then we mean centre it:

$$
\mathbf{A}_{mc}= \begin{bmatrix}
\mathbf{A} & \mathbf{1}
\end{bmatrix}
\mathbf{W}_{mc}.
$$
The translation matrix is
$$
\begin{bmatrix}
1 & 0 & \dotsb & \dotsb & 0 \\
0 & 1 & \dotsb & \dotsb & 0 \\
\vdots & \vdots  & \ddots  & \ddots & \vdots \\
\vdots & \vdots  & \ddots  & \ddots  & \vdots \\
- \mu_1 & - \mu_2 & \dotsb & \dotsb & 1 \\
\end{bmatrix}
$$

So now is it an orthogonal transformation or not ? I think, it is not an orthognoal transformation since the translation matrix is not orthogonal, but I am not sure, could some one clarify this ?

答:
By definition, an orthogonal transformation of $A$ is multiplication of $A$ by an orthogonal matrix, and nothing else. You have (cleverly) included an additive term alongside $A$. That disqualifies the map from being orthogonal.

Translation is an isometry of Euclidean space (see: Euclidean group). These are precisely the maps of the form $x\mapsto Mx+b$ where $M$ is an orthogonal matrix and $b$ is any vector. Translation is a special isometry, with $M$ being the identity map.


问:
Question on the restricted product topology
Let $I$ be an index set and $G_i$ a collection of compact topological groups.  Suppose that for each $i \in I$ there are compact (proper) subgroups $H_i \subseteq G_i$.  We say that $G =\{(a_i) \in \prod G_i \,:\, a_i \in H_i \text{ for almost all $i \in I$}\}$ is the restricted product of $G_i$ with respect to the $H_i$.

We can give $G$ a topology by declaring the collection $\mathcal{B} = \{B \,:\, B = \prod U_i \text{ with $U_i$ open in $G_i$ and $U_i \neq H_i$ only finitely many times}\}$ as a basis.

I must be missing something really basic here, but I can't see why this is closed under arbitrary unions.  For example, consider the sets
\begin{align*}
W_i = G_i \times \prod_{j\neq i} H_j.
\end{align*}
Then $W_i \in \mathcal{B}$, but $W = \bigcup_{i\in I} W_i = \prod_iG_i \not\subseteq G$.  What is going wrong here?

答: You're just using the standard base of the product topology on $\prod_i G_i$, restricted to the subspace $G$. So $G$ has the subspace topology wrt that product topology, nothing more or less.


问:
Is this random variable continuous or discrete?
Here's this random variable with a given $CDF_X$:

$x<1$
$1\le x<2$
$2\le x<3$
$3\le x<4$
$4\le x$

$0  $
$0.2$
$0.35$
$0.9$
$1$

A friend of mine says that $P\left(X=1\right)=0.2$, $P\left(X=2\right)=0.35-0.2=0.15$, $P\left(X=3\right)=0.9-0.35=0.55$, $P\left(X=4\right)=1-0.9=0.1$.

I'm ok with all that but at the same time he says that the random variable is continuous and that puzzles me.

If a variable is continuous, shouldn't we have zero probability for every point?

So is this random variable continuous or discrete?

答:
The CDF is the function where you put in a value $x$ and get out the probability $\Pr(X\leq x)$.

If this function jumps at a point, there is a positive probability of taking that exact value. So we know $\Pr(X\leq 1)=0.2$, but $\Pr(X\leq x)=0$ for every $x<1$ - the only way this can happen is if all of the probability $0.2$ corresponds to taking a value of exactly $1$.

Similarly, if the CDF is constant on an interval, that means the probability of a value strictly inside that interval is $0$. Since $\Pr(X\leq x)=\Pr(X\leq 1)$ for any $1\leq x< 2$, we know that $\Pr(1<X<x)=0$.

Since this CDF consists only of jumps between constant sections, it is the CDF of a purely discrete random variable.


问:
What's wrong with my solution for this complex integral?
Here I got a complex integral given by$$\frac{1}{2\pi}\int_{0}^{2\pi}{\frac{1}{z_0+2a\cos{k}}dk}$$where $z_0$ is a arbitrary complex number and $a$ is a real number. I try to solve this by taking $z=e^{ik}$ then I can substitute $\cos k=(z+\frac{1}{z})/2$, $dk=dz/iz$ in the original integral, to obtain$$\frac{1}{2\pi i}\oint{\frac{1}{az^2+z_0z+a}dz}$$The integral contour is a unit circle centered at the origin. I can evaluate this integral to obtain$$\frac{1}{2\pi i}\oint{\frac{1}{az^2+z_0z+a}dz}=\sum_i{Res[f(\lambda_i)]}$$where $\lambda_i$ denotes the singularity which lies inside the integral contour. So the integral results in$$\frac{\Theta(1-|\lambda_1|)}{a(\lambda_1-\lambda_2)}+\frac{\Theta(1-|\lambda_2|)}{a(\lambda_2-\lambda_1)}$$ where $\lambda_{1,2}=\frac{-z_0\pm\sqrt{z_0^2-4a^2}}{2a}$ and $\Theta$ denotes the HeavisideTheta function. In my solution, the result of this integral will be $0$ if both of singularities lie outside the contour, but it's totally different with the result from MMA( which couldn't be $0$ ). What's wrong with my solution?

答:
For simplicity I will rescale the integral as
$$
\frac1{2\pi a}\int_{0}^{2\pi}\frac{dx}{c+\cos x}
=\frac1{2\pi a i}\oint_{|z|=1}\frac{dz}{z^2+2cz+1}
$$
with $c=\frac{z_0}{2a}$.

The integrand has poles at:
$$
z_{1,2}=-c\pm\sqrt{c^2-1}.
$$

In view of $z_1\cdot z_2=1$ there are two options:

$|z_1|=|z_2|=1$

$|z_1|\ne |z_2|$.

It is not hard to demonstrate that the former case is equivalent to $c\in [-1,1]$. This corresponds exactly to the case when the original integral diverges.

In the general (second) case one of the poles is inside the integration contour and the other is outside of it. Therefore the integral is non-zero.


问:
How many triplets (a, b, p) satisfy $a^3 - 2ba^2 + 16ap^2 + 32p^2b = 0 $ where $p$ is a prime and $a,b$ $\epsilon$ $N$
How many triplets (a, b, p) satifsy $a^3 - 2ba^2 + 16ap^2 + 32p^2b = 0 $ where $p$ is a prime and $a,b$ $\epsilon$ $N$

I tried solving this question using the quadratic formula since this equation is quadratic in $p$. But I wasn't successful in solving this question.

答:
From your eq we have $$a^2(2b-a)=16p^2(2b+a).$$
If $p\nmid a$ then $p^2\mid 2b-a$.

Write $2b-a=kp^2$.  So $2b+a=kp^2+2a$.  That is
$ka^2=16(kp^2+2a)$ or
$$ka^2-32a-16kp^2=0.$$
The discriminant has to be square so
$$32^2+4\cdot 16k^2p^2=n^2.$$
We must have $n=8m$ and so
$$m^2=4^2+k^2p^2.$$
The only possible value of $kp$ is $3$ so $k=1$ and $p=3$.  Then
$$a^2-32a-16\cdot 9=0$$
or $a=36$ (since $a\in \Bbb N$).  But $2b-a=kp^2=9$ we get $2b=45$ but this is impossible.

Therefore $p \mid a$.  That is $a=px$.  We get
$$x^2(2b-px)=16(2b+px).$$
Then
$$2b(x^2-16)=px(x^2+16).$$

If $p\mid 2b$, then write $2b=py$ so that
$$y(x^2-16)=x(x^2+16)=x(x^2-16)+32x.$$  That is $x^2-16\mid 32 x$.  That is $x-4\mid x^2-16 \mid 32x$.  So $x-4\mid 32\cdot 4=128$.  Therefore $x-4=2^i$ for some $i=0,1,..,7$.  But $x+4\mid x^2-16\mid 32 x$ too, so $x+4=2^j$ for some $j=0,1,...,7$.  That is $2^j-2^i=8$ so $i=3$ and $j=4$, so $x=12$.  Therefore $a=px=24$ and
$$y=\frac{12(12^2+16)}{12^2-16}=15.$$
That is $2b=15p$ so $p=2$. Then we have an answer $(a,b,p)=(24,15,2)$.

If $p\nmid 2b$, then $p$ is odd.  But $2b(x^2-16)=px(x^2+16)$, so $x$ is even.  Write $x=2z$.  We have
$$b(z^2-4)=pz(z^2+4).$$
If $z$ is odd, then $\gcd(z,z^2-4)=1$ and so $z\mid b$.  Write $b=zu$.  We get
$$u(z^2-4)=p(z^2+4).$$
Since $p\nmid 2b$ and $b=zu$, we conclude that $p$ is a factor of $z^2-4=(z-2)(z+2)$.  Therefore after dividing by $p$, either $z-2$ or $z+2$ divides $z^2+4$.  Thus, $z\pm 2\mid 2^2+4=8$.  But $z$ is odd, so $z\pm 2=1$.  Hence, $z=3$ is the only possible value.  Then
$u(3^2-4)=p(3^2+4)$ so $5u=13p$.  That is, $p=5$ and $u=13$.  Hence $b=zu=39$.  Also $x=2z=6$ and $a=px=30$.  We have one more answer $(a,b,p)=(30,39,5)$.

If $z$ is even, then $z=2v$.  So $$b(v^2-1)=2pv(v^2+1).$$  Because $\gcd(v,v^2-1)=1$, we get $v\mid b$.  Write $b=vw$.  So
$$w(v^2-1)=2p(v^2+1).$$
Thus $2(v^2+1)\mid w(v^2-1)$.  Note that $\gcd(v^2+1,v^2-1)=1,2$ depending on whether $v$ is even or odd.

If $v$ is even, then $2(v^2+1)\mid w$.  So
$$p=\frac{w}{2(v^2+1)}(v-1)(v+1).$$
But $p$ is prime, we get $\frac{w}{2(v^2+1)}=1$ and $v-1=1$.  So $v=2$, $w=10$, $b=vw=20$, $z=2v=4$, $x=2z=8$, $p=v+1=3$, and $a=px=24$.  We get another solution $(a,b,p)=(24,20,3)$.

If $v$ is odd, then $8\mid v^2-1$ but $8\nmid 2p(v^2+1)$ because $p$ is odd and $v^2+1\equiv 2\pmod{4}$.

Therefore, all $(a,b,p)$ are $(24,15,2)$, $(30,39,5)$, $(24,20,3)$.  This problem is so long.


问:
Calculate volume of intersection two non-aligned cuboids
I'm on the lookout to finding an algorithm that calculates the volume of the intersection between two cuboids. Most answers so far assume all the axes of the two cuboids in question are nicely aligned. Not the case in my problem. See figure below

Finding out what corners of the two cuboid are in each others space is not that difficult. Also determining some boundary points on the cuboids faces shouldn't be problematic. The hard part where I get lost is to determine what kind of shape the intersection actually is and how to calculate it's volume. I think it's safe to assume the intersection will be convex. Any suggestions?

答:
Here is an alternative approach:

Define each of your cuboids or parallelepipeds as six opposing faces.

For each face of the target convex polyhedron, define both the face plane (unit normal vector $\hat{n}$ and signed distance from origin $d$, so that point $\vec{p}$ is inside the halfspace if and only if $\hat{n}\cdot\vec{p} - d \ge 0$),  and the vertices of the face polygon $\vec{v}_i$, $i = 1 \dots N$, in counterclockwise order so that
$$\hat{n}\cdot\bigl((\vec{v}_{i-1} - \vec{v}_i) \times (\vec{v}_{i+1} - \vec{v}_i)\bigr) \gt 0 \quad \forall i$$
where the index wraps around, i.e. $\vec{v}_0 = \vec{v}_N$ and $\vec{v}_{N+1} = \vec{v}_1$.

Then, apply each halfspace $\hat{n}$, $d$ of the other convex polyhedron, cutting out the vertices $\vec{v}$ in the target polyhedron that are outside the cutting halfspace, $\hat{n}\cdot\vec{v} - d \lt 0$, one at a time.

Because you "cut" the convex polyhedron left with halfspaces, the result is either a convex polyhedron, or nothing.  Note that the centroid is always within the convex polyhedron, so the total volume can be calculated as the sum of the volumes of pyramids with the apex at the centroid, and the base the face polygon.

If the halfspace includes all existing vertices, it includes the entire polyhedron, and the halfspace has no effect.

Otherwise, the halfspace removes at least one vertex, and adds a new face.

If the halfspace removes all vertices of a face, that face is completely removed.

On each face where at least one vertex, but not all vertices, are removed by the halfspace, a section of the face polygon is replaced by a single edge, corresponding to the intersection with the face plane and the new halfspace (plane), so that all removed vertices are in the replaced segment; and the removed vertices are continuous in the face polygon.

So, by identifying first the vertices excluded by the halfspace, then finding the face edges between an excluded vertex and not-excluded vertex (line-plane intersection – very fast to compute), we can quickly update each partially cut face.  The added new edges across all existing faces form the new face, too.

If you describe each convex polyhedron with a data structure that contains

All $N$ vertices $\vec{v}_i$, $i = 1 \dots N$, on the convex polyhedron

All $F$ face unit normal vectors $\hat{n}_f$ and signed distances $d_f$ from origin, $f = 1 \dots F$

For each face $f$, the vertices (vertex indexes) in counterclockwise order, defining the face polygon

you can quickly construct the set of edges that will be modified (for example, as triplets from vertex, to vertex, face), and their replacements.

Let $\vec{v}_k$ are the $K$ vertices of a convex polyhedron, $k = 1 \dots K$.  The centroid $\vec{C}$ is then
$$\vec{C} = \frac{1}{K} \sum_{k=1}^{K} \vec{v}_k$$

Let $\vec{p}_i$ are the $N$ vertices of one planar face of the convex polyhedron, $i = 1 \dots N$, with cyclic indexing so $\vec{p}_0 = \vec{p}_N$ and $\vec{p}_{N+1} = \vec{p}_1$.  Let $\hat{n}$ be the face plane unit normal vector, pointing inwards.  Because the polyhedron is convex, each face polygon is convex also, and therefore the face centroid $\vec{c}$ is
$$\vec{c} = \frac{1}{N} \sum_{i=1}^{N} \vec{p}_i$$
and the area of this face is
$$A = \frac{1}{2} \left\lVert \sum_{i=1}^{N} (\vec{p}_i - \vec{c}) \times (\vec{p}_{i+1} - \vec{c}) \right\rVert = \frac{1}{2} \sum_{i=1}^{N} \left\lvert \hat{n} \cdot (\vec{p}_{i} - \vec{c}) \times (\vec{p}_{i+1} - \vec{c}) \right\rvert$$
(because each cross product summand is always $\lambda \hat{n}$, $\lambda \in \mathbb{R}$, by definition – otherwise the face is not planar! – and $\lambda \ge 0$ or $\lambda \le 0$ depending on whether the vertices are in counterclockwise or clockwise order in the convex planar polygon) and the volume contribution of this face is
$$V_f = \frac{A}{3} \hat{n} \cdot \left(\vec{C} - \vec{c}\right) = \frac{\hat{n} \cdot (\vec{C} - \vec{c})}{6} \sum_{i=1}^{N} \left\lvert \hat{n} \cdot (\vec{p}_{i} - \vec{c}) \times (\vec{p}_{i+1} - \vec{c}) \right\rvert$$
Summing the volume contributions of all faces yields the total volume,
$$V = \sum_{f=1}^{F} V_f$$


问:
In the sequence, if a term is less than $\frac{3}{4}$ and some term after it is greater than $\frac{3}{4}$, can $\frac{3}{4}$ exists in between?
I need help verifying the solution to the following problem from the book Reading, writing, and proving: A closer look at mathematics. In Reading, writing, and proving: A closer look at mathematics 2nd ed on page 10, Problem 1.12.

Shlomo Sureshot started the basketball season with a free throw shooting percentage of below 75%. By the end of the season he brought it up to above 75%. Must there have been a time in the season (after a free throw attempt) when his free throw percentage was exactly 75%?

After having solved this problem and looking back at your solution, are there questions that you would like to answer? Can you answer them?

The following is my attempted solution:

The free throw percentage of a basketball player is calculated by dividing the number of free throws the player has connected by the number of free throw the player has attempted in a given time in the season after a free throw.

Consider a sequence of non-negative rational numbers which represented the player's free throw percentage after each attempted free throw from the beginning of the season to the end of the season.

$...,\frac{x}{y}, \frac{x+1}{y+1},...$ where $\frac{x}{y} < \frac{3}{4}$ and $\frac{x+1}{y+1} > \frac{3}{4}$ for some $x,y \in \mathbb{Z^*}$ and $y\ne 0$

If there exist two consecutive terms in the sequence which the first term in the pair is less than $\frac{3}{4}$ and the term after it is more than $\frac{3}{4}$ in which the latter term's numerator is equal to the first term's numerator added by $1$ and it's denominator is equal to the first term's denominator added by $1$, then there exist a sequence which does not contain the term $\frac{3}{4}$.

However, no such $x,y$ satisfy the sequence.

If $\frac{x}{y} < \frac{3}{4}$, then for some term after it to become greater than $\frac{3}{4}$, there must exist at least two terms after $\frac{x}{y}$ which some term before the last term in the sequence is exactly $\frac{3}{4}$.

Therefore, if a basketball player starts a season with a free throw percentage less than 75%, then for the player to achieve the free throw percentage above 75% by the end of the season, there must be a moment after a free throw during the season which the player's free throw percentage is exactly 75%.

Reference:

Daepp, U., & Gorkin, P. (2011). Reading, writing, and proving: A closer look at mathematics. In Reading, writing, and proving: A closer look at mathematics (2nd ed., p. 10). New York: Springer.

答:
I would say that you have asserted without justification that the player at some point must have had a free-throw percentage of exactly $75\%$.

It may be worth noting that at least for some free-throw percentages you can go from below to above without landing exactly.  For example a player could go from 1 for 4 ($25\%$) to 2 for 5 ($40\%$). They never hit $30\%$ exactly; they just passed over it.


问:
What is $\arcsin(\cos{x})$ (Spivak Calculus exercise 15-18)?
In the first part of the exercise I proved that $\sin(x+\frac{\pi}2)=\cos{x}$.

The second part is asking what $\arcsin(\cos{x})$ is. Presumably, they want a rule for unrestricted $x$.

We have $\arcsin(\cos{x})=\arcsin(\sin(x+\frac{\pi}2))$. It was said that the domain of $\sin$ usually restricted to $\left[-\frac{\pi}2, \frac{\pi}2\right]$, hence I assume that $x+\frac{\pi}2 \in \left[-\frac{\pi}2, \frac{\pi}2\right]$, and this set is also the co-domain of the $\arcsin$ function.

Since $\sin$ is periodic with period $2\pi$, $\sin(x+\frac{\pi}2)=\sin(2k\pi+x+\frac{\pi}2)$, $2k\pi+x+\frac{\pi}2 \in \left[2k\pi-\frac{\pi}2, 2k\pi+\frac{\pi}2\right], k \in \mathbb{Z}$.

And we have that $\arcsin(\sin(2k\pi+x+\frac{\pi}2))=2k\pi+x+\frac{\pi}2$.

But the answer in the solutions manual is $\begin{cases} x-2k\pi+\frac{\pi}2, x \in \left[2k\pi-\pi, 2k\pi\right] \\ 2k\pi+\frac{\pi}2-x, x \in \left[2k\pi, 2k\pi+\pi\right] \end{cases}$

I can plot the graph and see how $2k\pi+x+\frac{\pi}2$ doesn't work, but where did they get this piecewise function?

答:
Let $y=\arcsin (\cos x) $. Note that since $|\cos x|\le 1,y$ is defined on all real $x$.

It is true for all $x\in [-1,1]$ that $\arcsin x+\arccos x=\pi/2$

Hence $y=\pi/2-\arccos (\cos x)$

$\arccos(\cos x)=\begin{cases}x; x\in [0,\pi]\\2\pi-x; x\in [\pi,2\pi]\\ -2\pi+x;x\in[2\pi,3\pi]\\ 2(2\pi)-x; x\in[3\pi,4\pi]\end{cases}$ etc.

In general,

$\arccos(\cos x)=2(-1)^{n+1}n\pi-(-1)^{n+1}x$  when $x\in [n\pi, (n+1)\pi], n\in \mathbb Z^+\cup\{0\}$

$y=\pi/2+(-1)^{n+1}x-2(-1)^{n+1}n\pi$ when $\in [n\pi, (n+1)\pi]$.

Can you take it from here by taking $n=$ even and then taking $n=$ odd number?

Also note that $\cos x$ is even function so above definitions are valid for negative $x$ values as well.


答:
A direct proof would be to take $x\in B$ and then to deduce $x\in C$, that is $B\subset C$.
If we can do that then we may exchange $B$ and $C$ and reach $B=C$.

So, let us assume $x\in B$.

Now either  $$x\in (A\cup B) - (A\cap B)\tag{1}$$ or
$$x\notin (A\cup B) - (A\cap B) \tag{2}$$
Our assumption $x\in B$ certainly imply $x\in A\cup B$, hence in case (1) $x\notin A$.

But by the assumption on $C$ we have
$$(A\cup C) - (A\cap C) = (A\cup B) - (A\cap B)\ni x$$
and since $x\notin A$, $x$ must belong to $C$.

The case (2) can be handled similarly, I leave the fun for you.


答:
I don't follow your reasoning, which reads to me as a mere assertion that it must be true, not as a proof.

The proof is simple however given that you have already set up the inequalities that the variables $x$ and $y$ must satisfy.
$$\frac{x}y<\frac34 \implies 4x<3y$$
$$\frac{x+1}{y+1}>\frac34 \implies 4x+4>3y+3$$
Put those together to find that $4x<3y<4x+1$, which is impossible in whole numbers - there cannot be an integer $3y$ in between the successive integers $4x$ and $4x+1$.

Title: In the sequence, if a term is less than $\frac{3}{4}$ and some term after it is greater than $\frac{3}{4}$, can $\frac{3}{4}$ exists in between?

Body:
I need help verifying the solution to the following problem from the book Reading, writing, and proving: A closer look at mathematics. In Reading, writing, and proving: A closer look at mathematics 2nd ed on page 10, Problem 1.12.

Shlomo Sureshot started the basketball season with a free throw shooting percentage of below 75%. By the end of the season he brought it up to above 75%. Must there have been a time in the season (after a free throw attempt) when his free throw percentage was exactly 75%?

After having solved this problem and looking back at your solution, are there questions that you would like to answer? Can you answer them?

The following is my attempted solution:

The free throw percentage of a basketball player is calculated by dividing the number of free throws the player has connected by the number of free throw the player has attempted in a given time in the season after a free throw.

Consider a sequence of non-negative rational numbers which represented the player's free throw percentage after each attempted free throw from the beginning of the season to the end of the season.

$...,\frac{x}{y}, \frac{x+1}{y+1},...$ where $\frac{x}{y} < \frac{3}{4}$ and $\frac{x+1}{y+1} > \frac{3}{4}$ for some $x,y \in \mathbb{Z^*}$ and $y\ne 0$

If there exist two consecutive terms in the sequence which the first term in the pair is less than $\frac{3}{4}$ and the term after it is more than $\frac{3}{4}$ in which the latter term's numerator is equal to the first term's numerator added by $1$ and it's denominator is equal to the first term's denominator added by $1$, then there exist a sequence which does not contain the term $\frac{3}{4}$.

However, no such $x,y$ satisfy the sequence.

If $\frac{x}{y} < \frac{3}{4}$, then for some term after it to become greater than $\frac{3}{4}$, there must exist at least two terms after $\frac{x}{y}$ which some term before the last term in the sequence is exactly $\frac{3}{4}$.

Therefore, if a basketball player starts a season with a free throw percentage less than 75%, then for the player to achieve the free throw percentage above 75% by the end of the season, there must be a moment after a free throw during the season which the player's free throw percentage is exactly 75%.

Reference:

Daepp, U., & Gorkin, P. (2011). Reading, writing, and proving: A closer look at mathematics. In Reading, writing, and proving: A closer look at mathematics (2nd ed., p. 10). New York: Springer.


问:
When is $\int |f| =\int f$
Sorry if this is a duplicate, I couldn't find a similar thread.

I'm currently trying to prove a fact regarding the total variation of a function and the integral of it's derivative.

If $f$ is non-decreasing is it true that $\int^b_a |f|=\int^b_a f$?

答: No, take for example $f(x)=x$ and $a=-2$ and $b=-1$. Nevertheless, your statement holds whenever $f\geq 0$.

Title: When is $\int |f| =\int f$


答:
Take $f(x)=x$, $a=-1$, $b=1$.

In general, call $f^+=\max\{f,0\}$ and $f^-=\max\{-f,0\}$. In the language of Lebesgue integration (id est, if we discuss measurable functions such that either $\int \lvert f\rvert<\infty$ or at least one of $\int f^+$ and $\int f^-$ is finite), notice that these are non-negative functions and that $$\lvert f\rvert=f^++f^-\\ f=f^+-f^-\\ \int \lvert f\rvert=\int f^++\int f^-\\ \int f=\int f^+-\int f^-$$

And therefore $\int \lvert f\rvert=\int f$ if and only if either $\int f^+=\infty$ and $\int f^-<\infty$ or $\int f^-=0$. Non-negativity of $f^-$ implies that the second case is equivalent to $f$ being non-negative almost everywhere.

With similar argument, $\int \lvert f\rvert=\lvert\int f\rvert$ if and only if either $f$ has almost everywhere the same sign, or one of $\int f^\pm$ is infinite and the other is finite.

Title: When is $\int |f| =\int f$

Body:
Sorry if this is a duplicate, I couldn't find a similar thread.

I'm currently trying to prove a fact regarding the total variation of a function and the integral of it's derivative.

If $f$ is non-decreasing is it true that $\int^b_a |f|=\int^b_a f$?


问:
Horizontal asymptotes of y = 1/f(x)
If $f(x)$ has a horizontal asymptote of $y=a$ then the graph of $y = 1/f(x)$ has a horizontal asymptote of $y=1/a$.

However what happens if $y=f(x)$ has a horizontal asymptote of $y=0$. I'm thinking there's no horizontal asymptote for $y = 1/f(x)$, as $y=1/0$ is undefined.

Is this thinking correct or am I missing something?

答: Your thinking is correct iff we have $$\lim_{x\to\infty}f(x)=\lim_{x\to-\infty}f(x)=L$$ in which case if $L=0$ then $y=1/f(x)$ will have no horizontal asymptote. However, there exist functions other than rational functions that have horizontal asymptotes, like exponential functions and the arctangent function;  notice that you cannot make the same assumption since the statement above does not apply to these functions.

Title: Horizontal asymptotes of y = 1/f(x)


问:
Minimize height, above a hole, needed for getting a rectangle out of the hole.
I am looking for a solution to a particular task of manipulating a rectangle out of a hole. I will be happy for any suggestions. I wasn't able to find any simplifications (e.g.: transferring the problem to a problem of manipulating a rod), which would lead to an equivalent optimum.

Problem: Minimize height, above a hole, needed for getting a rectangle out of the hole.

Illustration picture:

Having rectangle with known sides
$$
w, h
$$
and having a hole with known width and height:
$$
a, b
$$
Minimize height, above the hole, needed for pulling the rectangle out of the hole
$$
min(x)
$$

EDIT
Two answers (one of them was edited, I will consult it later) suggested placing bottom left/right corner of a rectangle to upper left/right point of the hole and rotating rectangle as much as possible to the right/left. I think this is correct for a "short" rectangle.

I suggest following for a "long" rectangle:

When I take "long" rectangle, I think the height needed (x) is what the picture below shows:

答:
Hint.

A figure worth a thousand words.

NOTE

Attached a MATHEMATICA script which calculates the maximum height clearance needed to free the frame

```<code>parms1 = {w -> 1, a -> 1.2, h -> 1.5, b -> 1.9};
parms2 = {w -> 1, a -> 1.6, h -> 1.7, b -> 2.1};
frame = {{-w/2, -h/2}, {w/2, -h/2}, {w/2, h/2}, {-w/2, h/2}};
s = (a - w)/2;
r = (b - h)/2;
g = {gx, gy};
R = RotationMatrix[theta];
ftg = Table[R.frame[[k]] + g, {k, 1, 4}];
lw[lambda_] :=  lambda {-w/2 - s, -h/2 - r} + (1 - lambda) {-w/2 - s, h/2 + r}
rw[lambda_] :=  lambda ftg[[2]] + (1 - lambda) ftg[[3]]
restr1 = (ftg[[1]] - lw[lambda]).(ftg[[1]] - lw[lambda]);
restr2 = ({w/2 + s, h/2 + r} - rw[mu]).({w/2 + s, h/2 + r} - rw[mu]);
restr3 = 0 < lambda < 1;
restr4 = 0 < mu < 1;

obj = ftg[[4]].{0, 1} /. parms1;
restrs = Flatten[{restr1 == 0, restr2 == 0, restr3, restr4}] /. parms1;
sol = NMaximize[Join[{obj}, restrs], {theta, gx, gy, lambda, mu}]
ftg0 = Join[ftg, {ftg[[1]]}] /. parms1 /. sol[[2]];
grfr = ListLinePlot[ftg0];
lw0 = lw[u] /. parms1;
glw = ParametricPlot[lw0, {u, 0, 1}];
grw = ParametricPlot[{w/2 + s, -h/2 - r} u + (1 - u) {w/2 + s, h/2 + r} /. parms1, {u, 0, 1}];
Show[grfr, glw, grw, PlotRange -> All, AspectRatio -> 1.8]
x = obj - b/2 /. parms1 /. sol[[2]]
</code>```

Follows two plots relative to parms1 and parms2 respectively

With $p_1 = (-w/2,-h/2), p_2 = (w/2,-h/2), p_3 = (w/2,h/2), p_4 = (-w/2,h/2)$ we form a frame $F_0 = \{p_1,p_2,p_3,p_4\}$ This frame rotated by $R(\theta)$ and translated by $g = (g_x,g_y)$ is now $F(\theta,g_x,g_y) = R(\theta)\cdot F_0 + g$. The restrictions are $p_1(\theta,g)\in l_w(\lambda),\ 0\le \lambda\le 1$ and $p_0=(a/2,b/2)\in l_r(\mu),\ \ 0\le \mu\le 1$. Here $l_r(\lambda)$ is the left hole wall and $l_r(\mu)$ is the rectangle side with extremum points $p_2(\theta,g)$ and $p_3(\theta,g)$. The clearance is given by $x = p_4(\theta,g)\cdot(0,1)-b/2$

Body:
Hint.

A figure worth a thousand words.

NOTE

Attached a MATHEMATICA script which calculates the maximum height clearance needed to free the frame

parms1 = {w -> 1, a -> 1.2, h -> 1.5, b -> 1.9};
parms2 = {w -> 1, a -> 1.6, h -> 1.7, b -> 2.1};
frame = {{-w/2, -h/2}, {w/2, -h/2}, {w/2, h/2}, {-w/2, h/2}};
s = (a - w)/2;
r = (b - h)/2;
g = {gx, gy};
R = RotationMatrix[theta];
ftg = Table[R.frame[[k]] + g, {k, 1, 4}];
lw[lambda_] :=  lambda {-w/2 - s, -h/2 - r} + (1 - lambda) {-w/2 - s, h/2 + r}
rw[lambda_] :=  lambda ftg[[2]] + (1 - lambda) ftg[[3]]
restr1 = (ftg[[1]] - lw[lambda]).(ftg[[1]] - lw[lambda]);
restr2 = ({w/2 + s, h/2 + r} - rw[mu]).({w/2 + s, h/2 + r} - rw[mu]);
restr3 = 0 < lambda < 1;
restr4 = 0 < mu < 1;

obj = ftg[[4]].{0, 1} /. parms1;
restrs = Flatten[{restr1 == 0, restr2 == 0, restr3, restr4}] /. parms1;
sol = NMaximize[Join[{obj}, restrs], {theta, gx, gy, lambda, mu}]
ftg0 = Join[ftg, {ftg[[1]]}] /. parms1 /. sol[[2]];
grfr = ListLinePlot[ftg0];
lw0 = lw[u] /. parms1;
glw = ParametricPlot[lw0, {u, 0, 1}];
grw = ParametricPlot[{w/2 + s, -h/2 - r} u + (1 - u) {w/2 + s, h/2 + r} /. parms1, {u, 0, 1}];
Show[grfr, glw, grw, PlotRange -> All, AspectRatio -> 1.8]
x = obj - b/2 /. parms1 /. sol[[2]]

Follows two plots relative to parms1 and parms2 respectively

With $p_1 = (-w/2,-h/2), p_2 = (w/2,-h/2), p_3 = (w/2,h/2), p_4 = (-w/2,h/2)$ we form a frame $F_0 = \{p_1,p_2,p_3,p_4\}$ This frame rotated by $R(\theta)$ and translated by $g = (g_x,g_y)$ is now $F(\theta,g_x,g_y) = R(\theta)\cdot F_0 + g$. The restrictions are $p_1(\theta,g)\in l_w(\lambda),\ 0\le \lambda\le 1$ and $p_0=(a/2,b/2)\in l_r(\mu),\ \ 0\le \mu\le 1$. Here $l_r(\lambda)$ is the left hole wall and $l_r(\mu)$ is the rectangle side with extremum points $p_2(\theta,g)$ and $p_3(\theta,g)$. The clearance is given by $x = p_4(\theta,g)\cdot(0,1)-b/2$


问:
How to understand transition probabilities from Durrett.

I understand his definition of transition probability but I do not understand his integral notation. I know that if we fix the first component of the transition probability we get a probability measure. However, how does he pick the $x_0,x_1,..,x_{n-1}$ points used in determining the measures.

The second issue I have is more measure theoretic. He defined the finite dimensional distributions on the cylinder cells only but how is that sufficient to determine a finite dimensional distribution on the product sigma-algebra. Are we using Caratheodary extension theorem here. i.e. cylinder cells can form an algebra over which the definition given is a pre-measure which can be extended. If so, is the definition given just the product measure given by $\mu \otimes p(x_0,.)\otimes...\otimes p(x_{n-1},.)$?

答:
The $x_n$ are chosen by $X_n$. Plug $\omega$ into $X_n$ at time n.  This gives you $x_n$ which you use “choose” your probability measure for the next time.

It’s a little weird because the image of $X_n$ might be a proper subset of S.  So there could be $x_n$ that give probability measures that have nothing to do with $\Omega$.  But we will never use those.  They are a technicality.

What you propose for your second question seems reasonable, but is complicated by the fact that we need a possibly unique probability measure for each sample path of the random variable.  Will have to think about that bit. I usually just think of it as “conditioning” and the multiplication rule.


问:
Simple explanation of difference between $\cos(x)$ and $\cos(x^2)$ integrals.
I'm not looking for the solved integrals, I'm just looking for a simple explanation for why

$\displaystyle\int\cos(x^2)dx$

is so much more complicated than

$\displaystyle\int\cos(x)dx$

With simple I mean something I can say to explain the difference to a person just starting to learn about integration. Perhaps something visual?

答: For one thing, look at their graphs. Since $x$ increases at a constant rate, $\cos(x)$ has a constant period but $x^2$ increases faster and faster for larger $x$ so the "period" of $\cos(x^2)$ keeps getting smaller and smaller.</p>


问:
Covariant derivative of the parallel transport operator.
I know that it is possible to define a covariant derivative of a tensor field given the parallel transport operator, but I wonder whether the covariant derivative of the parallel transport operator itself can be defined in any meaninful way? Or at least the covariant derivative along the curve along which  the transport is happening.

I'm working on a problem where using such derivative is tempting, but I don't know if it has any sense.

答:
First, note that the parallel transport is not an operator on $M$, neither on a curve: it is an operator from one fixed vector space to a set of sections of a vector bundle.

Given a curve $\gamma \colon I \to M$, one can define the parallel transport along $\gamma$ with basepoint $t_0\in I$. Given a point $t_0 \in I$, the parallel transport is the operator
$$
P_{t_0} \colon T_{\gamma(t_0)}M  \longrightarrow \Gamma\left(\gamma^*TM \right)
$$
assigning to any vector $v \in T_{\gamma(t_0)}M$ the unique vector field along $\gamma$, say $V = P_{t_0}v$,  such that $D_{\gamma'}V = 0$ and $V(t_0) = v$.

Also note that $\gamma$ can intersect itself for two different values $t_0 \neq t_1$, but that $P_{t_0}$ and $P_{t_1}$ are two distinct operators (if it is not clear for you, just look at the figure-eight curve).

Now, here is a question for you: how would you define for $P_{t_0}$ to be parallel? As it is not a tensor on $M$, neither along $\gamma$, this seems impossible (to me) to define this notion.


答: $$ \int |f| = \int f \Longleftrightarrow \int(|f|-f) = 0 \Longleftrightarrow f=|f|~~a.e. \Longleftrightarrow f \geq 0 ~~a.e. $$

Title: When is $\int |f| =\int f$


问:
Finding all triples of numbers with the following divisibility
Could somebody tell me how to tackle this problem? I don’t know where to start with it. The problem was part of the Math Olympiad in Kazhakstan (or Kyrgistan, I keep forgetting…) this February, but there are no solutions online.

Find all triples $(x, y, z)$ of positive integers for which $$x|(y + 1)$$ $$y|(z + 1)$$ $$z|(x + 1)$$
holds.

答:
The $3$ divisibility requirements, and that $(x,y,z)$ are positive integers, means there are positive integers $m_1$, $m_2$ and $m_3$ such that

$$y + 1 = m_1 x \tag{1}\label{eq1A}$$

$$z + 1 = m_2 y \tag{2}\label{eq2A}$$

$$x + 1 = m_3 z \tag{3}\label{eq3A}$$

There is a symmetry with each equation above being of the form of a variable plus $1$ equaling a multiple of another variable. With some $m_1 = a_1$ and $x = v_1$, then \eqref{eq1A} gives $y = a_1 v_1 - 1 = v_2$. Then with some $m_2 = a_2$, \eqref{eq2A} gives $z = a_2 v_2 - 1 = v_3$. Finally, with some $m_3 = a_3$, \eqref{eq3A} gives $x = a_3 v_3 - 1$, which leads back to \eqref{eq1A}.

Note we could have started with $a_1$ and $v_1$ in \eqref{eq2A} (or \eqref{eq3A}) instead, and then used \eqref{eq3A} and \eqref{eq1A} (or \eqref{eq1A} and \eqref{eq2A}) with the next $2$ multiplier values of $a_2$ and $a_3$. This indicates there's a rotational symmetry, i.e., if $(x_1, y_1, z_1)$ is a solution, then so is moving each value's position to the right (or left), and "rotating" the position at the end (or start) to the start (or end), i.e., $(z_1, x_1, y_1)$ and $(y_1, z_1, x_1)$ are also solutions.

Next, adding the $3$ equations above gives

$$\begin{equation}\begin{aligned}
x + y + z + 3 & = m_1 x + m_2 y + m_3 z \\
3 & = (m_1 - 1)x + (m_2 - 1)y + (m_3 - 1)z
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$

Note the $3$ terms on the right are all non-negative integers. There are $3$ cases regarding the number of $0$ terms on the right side to consider:

Case #$1$:

None of the terms are $0$. Thus, they each must be $1$, i.e., $m_1 = m_2 = m_3 = 2$ and $x = y = z = 1$. This means the only the solution here is $(1,1,1)$.

Case #$2$:

Just one of the terms on the right of \eqref{eq4A} is $0$, say the first one. This gives $m_1 = 1$. Then one of the other terms must be $1$ and the final term must be $2$. Consider first $(m_2 - 1)y = 1$, which means $m_2 = 2$ and $y = 1$. From \eqref{eq2A}, this gives $z = 1$. Note we will also get $y = z = 1$ by having $(m_3 - 1)z = 1$ instead. With $m_1 = y = 1$, then \eqref{eq1A} gives $x = 2$. Thus, a solution is $(2,1,1)$, with the other $2$ associated ones being $(1,2,1)$ and $(1,1,2)$. Note this actually covers all of the possibilities for this case.

Case #$3$:

Here, $2$ of the terms on the right of \eqref{eq4A} are $0$, say the first $2$. This gives

$$m_1 = m_2 = 1, \; 3 = (m_3 - 1)z \tag{5}\label{eq5A}$$

This results in $2$ possibilities. First, $z = 1$ and $m_3 - 1 = 3 \implies m_3 = 4$. From \eqref{eq3A}, this gives $x = 3$. Also, $m_1 = 1$ gives in \eqref{eq1A} that $y = 2$. Thus, a solution is $(3, 2, 1)$. Due to the rotational symmetry I mentioned (with this coming from the cases where either $m_2$ or $m_3$ are not $1$), the other $2$ related solutions are $(2, 1, 3)$ and $(1, 3, 2)$.

The other possibility from \eqref{eq5A} is $z = 3$ and $m_3 - 1 = 1 \implies m_3 = 2$. From \eqref{eq3A}, this gives $x = 5$. Also, $m_1 = 1$ gives from \eqref{eq1A} that $y = 4$. Thus, a solution is $(5, 4, 3)$. The other rotational related solutions are $(3, 5, 4)$ and $(4, 3, 5)$.

Altogether, the set of $(x,y,z)$ solutions are $\{(1,1,1)$, $(1,1,2)$, $(1,2,1)$, $(2,1,1)$, $(1,3,2)$, $(2,1,3)$, $(3,2,1)$, $(3,5,4)$, $(4,3,5)$, $(5,4,3)\}$.


问:
Taylor series to solve limit equation
I came across this problem while studying taylor series and limits.

$(1+1/n)^{(n+x)}=e$

calculate $x$ when $n$ tends toward infinity.

Looking at the question the obvious answer would be $0$, but through taylor series the answer becomes $x=1/2$. I can calculate it, but I have no intuition of why it is not $0$. Is my though of

$(1+1/n)^n = e \Rightarrow $

$(1+1/n)^{(n+x)}=e \Rightarrow $

$(1+1/n)^x=1$ wrong?

Thank you.

Edit:
I should probably call it Maclaurin expansion, and this is the solution I came to.

$y=\frac{1}n$,
${y\to0}$

$(1+y)^{\frac{1}y+x}=e$ $<=>$

$(\frac{1}y+x)ln(1+y)=1$ <=>

$x=\frac{y-ln(1+y)}{yln(1+y)}$

$ln(1+y)=y-\frac{y^2}2+y^3B(y)$,  where B(y) is bounded close to $y=0$

$=> x=\frac{y-y+\frac{y^2}2-y^3B(y)}{y^2-\frac{y^3}2+y^4B(y)}=\frac{\frac{1}2-yB(y)}{1-\frac{y}2+y^2B(y)}\to\frac{1}2, y\to0$

This however is not my initial intuition. Therefore my assumption that x should be 0 is incorrect, but I don't know where my thinking is wrong.

Sorry for the clumsy math notation. It is a first time for me.

Title: Taylor series to solve limit equation

Body:
$$(1+1/n)^{n+x}=(1+1/n)^n \cdot (1+1/n)^x  \to e \cdot1^x=e$$

as $n \to \infty$ for all $x$.


问:
About a 5-fold cyclic branched cover of $\Bbb CP^1$ along 3 points
Suppose $F$ is a smooth connected complex curve, and we have a cyclic 5-fold branched cover $\pi:F\to \Bbb CP^1$ along 3 points. Let $\gamma$ denote the corresponding $\Bbb Z_5$-action on $F$. We have $\chi(F)=5(\chi(\Bbb CP^1)-3)+3=-2$, so the genus of $F$ must be $2$. Let $D$ be the diagonal in $\Bbb CP^1\times \Bbb CP^1$ and let $D'=(\pi\times \pi)^{-1}(D)\subset F\times F$. Then we can decompose $D'$ into union of $5$-curves: $D_i:=\{(x,y)\in D': y=\gamma^i x\}$ $(i=1,\dots,5)$. On the other hand, note that $H_2(\Bbb CP^1\times \Bbb CP^1;\Bbb Z)=\Bbb Z^2$ with basis $\alpha:=[\Bbb CP^1\times x], \beta:=[y\times \Bbb CP^1]$. Since $[D]\cdot \alpha=1=[D]\cdot \beta
$, we see that $[D]=\alpha+\beta \in H_2(\Bbb CP^1\times \Bbb CP^1)$.

In this situation, how can we show that

$[D']=5([F\times x]+[y\times F]) \in H_2(F\times F;\Bbb Z)$, and that

$[D_i]^2= [D_j]^2$ (self-intersection) for $i,j=1,\dots,5$?

答:
Throughout I use Poincare duality without much explanation.

1.
The important fact here is that the map $\pi: F \rightarrow \mathbb{CP}^1$ is degree $5$. I.e the map $\pi^*: H^{2}(\mathbb{CP}^1,\mathbb{Z}) \rightarrow  H^{2}(F,\mathbb{Z})$ is $x \mapsto 5x$, with respect to the positive generators of these groups (both are $\mathbb{Z}$). The degree can be computed by taking the number of preimages of a generic point which is 5, which can be made rigorous with a little bit of work.

Then it follows that $(\pi \times \pi)^*: H^{2}(\mathbb{CP}^1  \times \mathbb{CP}^1 , \mathbb{Z}) \rightarrow H^{2}(F \times F , \mathbb{Z}) $ is the map $(x,y) \mapsto (5x,5y)$ in the obvious bases (i.e. the Poincare dual of the two diagonal factors). To prove this just restrict the map to the Poincare dual of the two factors.  Now 1. can be proved using the fact that the class of the diagonal in $H^{2}(\mathbb{CP}^1 \times \mathbb{CP}^1,\mathbb{Z})$ is $(1,1)$ (i.e. $\alpha + \beta$ in your notation, I am taking these as an integral basis of $H^2$).

2.

For this, note that $D'$ consists precisely of points $(x,y)$ such that $\pi(x) = \pi(y)$. In other words, this may be restated as: $D'$ consists of exactly the points $(x,y)$ such that $\exists g \in \mathbb{Z}_5 $ such that $gx=y$.

It is not hard to see that for any $g$, the subset $U_{g} = \{(x,y): gx=y\} \subset F \times F$ is a holomorphic curve, isomorphic to $F$. To see this note that the map $F \rightarrow U_g$ , $p \mapsto (p,gp)$ is an isomorphism of Riemann surfaces.

Moreover, there is a global biholomorphism of $F \times F$ taking $U_{g}$ to $U_{g'}$ for any $g,g'$, namely the map $(p,q) \mapsto (p, g' g^{-1} q)$. So in particular these holomorphic curves have the same self-intersection.


问:
Inverse of a function $B_r(0) \to \mathbb{R}^n$

Definition. Let $B_r(0) := \{x \in \mathbb{R}^n \mid \|x\| < r\}$ be the open ball in $\mathbb{R}^n$ with radius $r>0$ and center $0:=(0,\ldots,0) \in \mathbb{R}^n$.

Let $s_r : B_r(0) \to \mathbb{R}^n$ be the map, defined by $$s_r(x) = \frac{r}{\sqrt{r^2-\|x\|^2}} \cdot x $$

Does this function have an inverse? If so, how does $s_r^{-1} : \mathbb{R}^n \to B_r(0)$ look like?

答: Let $y=s_r(x)$. Take norm on both sides and solve the equation for $\|x\|$  to get $\|x\|=\frac {r\|y\|} {\sqrt {r^{2}+|\|y\|^{2}}}$. Can you write down $x$ in terms of $y$ now?


问:
Induced topology on subset are equal
got a huge dubt on a fact that was presented today to me in class. I would really appreciate if you could help me out with this!

Today in class the teacher mentioned the following:

Lets consider three subsets $W \subset Y \subset X$. Let $\tau$ be a topology on X and let $\hat{\tau}$  be a topology induced by $\tau$ in Y. Then the topologies induced by $\tau$ and $\hat{\tau}$ on $W$ are equal.

As I said, we are just starting the course on some kind of intruductory topology and I've searched for this fact online but couldnt find anything. Any help would be great.

Thanks in advance!

答:
This is a special case of the "transitive law of initial topologies", that I prove and describe here.

But this special case is also easy to see without abstract machinery if you prefer:

Let $U \subseteq W$ be open in the subspace topology on $W$ wrt $X$. Then $U = O \cap W$ for some $O$ open in $X$.
Then $O \cap Y$ is open in $Y$ by definition so $O \cap Y \cap W$ is open in $W$ when we give $W$ the subspace topology wrt $Y$. But $Y \cap W= W$ so $O \cap Y \cap W = O \cap W=U$. SO $U$ is also open wrt $Y$.

Now let $U \subseteq W$ be open in the subspace topology on $W$ wrt $Y$. Then $U= V \cap W$ where $V$ is open wrt $Y$. But as $Y$ has the subspace topology wrt $X$ we know that $V = O \cap Y$ where $O$ is open in $X$. But then $U = V \cap W = (O \cap Y) \cap W = O \cap (Y \cap W) = O \cap W$ and $U$ is also open in the subspace topology on $W$ wrt $X$.

So "subspace topology on $W$ wrt $X$" is the same topology as "subspace topology on $W$ wrt $Y$".

Title: Induced topology on subset are equal


问:
Proving that $T^{-1}(V)$ is a subspace of $X$ when $V$ is a subspace of $X$?
I just came across a question which is as follows:

Let $X$ and $Y$ be normed linear spaces and let $ T : X \rightarrow Y$ be a linear operator with domain $D(T) \subset X$ and range $R(T) \subset Y$.

The first part of the question asks us to show that $T^{-1}$ is linear as well, which I have been able to do.

However I'm confused with the second part. It says:

Let $V$ be a subspace of $X$ and $W$ be a subspace of $Y$. Show that $T^{-1}(V)$ is a subspace of $X$ and $T^{-1}(W)$ is a subspace of $Y$.

I simply don't understand how $T^{-1}(V)$ can be a subspace of $X$. Shouldn't it have been:

Show that $T^{-1}(W)$ is a subspace of  $X$ and $T(V)$ is a subspace of $Y$?

答:
Yes, I agree with your diagnosis. The notation $T^{-1}(V)$ isn't justified since $V$ is not necessarily in the range of $T$. It's much more likely that $T(V)$ and $T^{-1}(W)$ were meant.

By the way you mentioned you were able to show "$T^{-1}$ is linear." Whenever $T$ isn't one to one, it isn't even a homomorphism, so it will often fail to be linear. When it's 1-1 though it can be regarded as a homomorphism from a subspace of $Y$ to a subspace of $X$, yes.

One good thing to keep in mind here is that $T^{-1}(W):=\{v\in D(T)\mid T(v)\in W\}$ is defined even if $T^{-1}$ isn't a function.


问:
How to generate a set of points that are equidistant from each other and lie on a normal distribution
I am trying to generate n points on a normal distribution(say, mean \mu and standard deviation \sigma) that are equidistant(Let's say the distance between pints is d) from each other in C. Points at normal distribution are from (-3\sigma + \mu) to (3\sigma + \mu). I am trying to have beads on the Gaussian curve, where beads are at equal distance (curved length).

Thank you in advance for the answer.

答:
Curve length of a function in $x$ can be found as

$$\ell_f(a, b) =\int_a^b \sqrt{1+\left(\frac{df(x)}{dx}\right)^2}dx$$

Which is nasty enough that in general we don't get an integrable result out of it; in our case we need to perform the integral and then invert it, and this is a recipe for sadness.

So what we'll do instead of this is do it numerically, replacing integration with the Riemann sum and differentiation with the difference quotient and doing a little simplifying.

$$\ell_f(a, b) = \lim_{\Delta x \rightarrow 0} \sum_{t=\frac{a}{\Delta x} + 1}^{\frac{b}{\Delta x}}\sqrt{\Delta x^2+\left(f(t\Delta x)-f((t-1)\Delta x)\right)^2}$$

Which still looks kinda mean but is possible to plug & chug by picking a reasonable value for $\Delta x$.

Of course, that only works forward, and we have to work backwards.  Our best bet, since this is a series, is to do linear interpolation: draw straight lines between the points that make up the series.  This gives a monotonic piecewise linear function, which is trivial to invert.

The procedure

We have our function $f(x)$, a low and high end of the interval in question $p$ and $q$, and a number of intervals to break the function into $n$.

Calculate $\Delta x = \frac{q-p}{n}$

For each integer $k$, $x_k = a + k\Delta x$, $y_k = f(x_k)$.

$d_k = \sqrt{(x_k - x_{k-1})^2 + (y_k - y_{k-1})^2} = \sqrt{\Delta x^2 + (y_k - y_{k-1})^2} $

$\ell_k = \sum_{i=1}^{k} d_k$

$m_k = \frac{\Delta x}{d_k}$

$b_k = x_k - \ell_k m_k$

And now we have a piecewise linear function: $x_f(\ell) = m_k \ell + b_k \text{ for } x_{k-1} \le \ell < x_k$

Now we wish to place $m$ spaces between points, so we take values $\frac{\ell_ni}{m}$ for $i$ in $0 \le i \le m$, find their $x$ values from the piecewise linear function above, and then from there the $f(x)$ values.

Example

Our star today is the standard normal distribution density function, $$f(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$$

over the interval $-3 \le x \le 3$. I'll use $n = 6$ here for relative brevity and for relatively nice $x$ values.

$x$
$y$
$d$
$\ell$
$m$
$b$
formula
interval

$-3$
$0.004432$

$-2$
$0.053991$
$1.001227$
$1.001227$
$0.998774$
$-3.000000$
$0.998774\times \ell -3.000000$
$0.000000\le \ell < 1.001227$

$-1$
$0.241971$
$1.017515$
$2.018742$
$0.982787$
$-2.983993$
$0.982787\times \ell -2.983993$
$1.001227\le \ell < 2.018742$

$0$
$0.398942$
$1.012245$
$3.030987$
$0.987903$
$-2.994322$
$0.987903\times \ell -2.994322$
$2.018742\le \ell < 3.030987$

$1$
$0.241971$
$1.012245$
$4.043232$
$0.987903$
$-2.994322$
$0.987903\times \ell -2.994322$
$3.030987\le \ell < 4.043232$

$2$
$0.053991$
$1.017515$
$5.060747$
$0.982787$
$-2.973635$
$0.982787\times \ell -2.973635$
$4.043232\le \ell < 5.060747$

$3$
$0.004432$
$1.001227$
$6.061974$
$0.998774$
$-3.054544$
$0.998774\times \ell -3.054544$
$5.060747\le \ell < 6.061974$

at this point I want to point something out: the curve length of this function is only about 1% more than the width of our interval in $x$.  There isn't going to be much difference between evenly spaced on $x$ and evenly spaced along the curve!  In fact when we get our points, the largest difference between the two is about $0.008$.  Still, the point here is a reasonably general procedure, so let's continue.

Now that we've got our piecewise linear function, we'll just plug in evenly-spaced $\ell$ values.

$\ell$
$x$
$y$

$\ell$
$x$
$y$

$0.000000$
$-3.000000$
$0.004432$

$3.030987$
$0.000000$
$0.398942$

$0.303099$
$-2.697273$
$0.010498$

$3.334086$
$0.299432$
$0.381453$

$0.606197$
$-2.394546$
$0.022689$

$3.637185$
$0.598864$
$0.333452$

$0.909296$
$-2.091818$
$0.044744$

$3.940283$
$0.898296$
$0.266493$

$1.212395$
$-1.792467$
$0.080026$

$4.243382$
$1.196705$
$0.194954$

$1.515494$
$-1.494586$
$0.130572$

$4.546481$
$1.494586$
$0.130572$

$1.818592$
$-1.196705$
$0.194954$

$4.849579$
$1.792467$
$0.080026$

$2.121691$
$-0.898296$
$0.266493$

$5.152678$
$2.091818$
$0.044744$

$2.424790$
$-0.598864$
$0.333452$

$5.455777$
$2.394546$
$0.022689$

$2.727888$
$-0.299432$
$0.381453$

$5.758876$
$2.697273$
$0.010498$

$6.061974$
$3.000000$
$0.004432$

And we can plot these.


Title: Taylor series to solve limit equation

Body:
I came across this problem while studying taylor series and limits.

$(1+1/n)^{(n+x)}=e$

calculate $x$ when $n$ tends toward infinity.

Looking at the question the obvious answer would be $0$, but through taylor series the answer becomes $x=1/2$. I can calculate it, but I have no intuition of why it is not $0$. Is my though of

$(1+1/n)^n = e \Rightarrow $

$(1+1/n)^{(n+x)}=e \Rightarrow $

$(1+1/n)^x=1$ wrong?

Thank you.

Edit:
I should probably call it Maclaurin expansion, and this is the solution I came to.

$y=\frac{1}n$,
${y\to0}$

$(1+y)^{\frac{1}y+x}=e$ $<=>$

$(\frac{1}y+x)ln(1+y)=1$ <=>

$x=\frac{y-ln(1+y)}{yln(1+y)}$

$ln(1+y)=y-\frac{y^2}2+y^3B(y)$,  where B(y) is bounded close to $y=0$

$=> x=\frac{y-y+\frac{y^2}2-y^3B(y)}{y^2-\frac{y^3}2+y^4B(y)}=\frac{\frac{1}2-yB(y)}{1-\frac{y}2+y^2B(y)}\to\frac{1}2, y\to0$

This however is not my initial intuition. Therefore my assumption that x should be 0 is incorrect, but I don't know where my thinking is wrong.

Sorry for the clumsy math notation. It is a first time for me.

Body:
$$a_n=\left(1+\frac{1}{n}\right)^{n+x}\implies \log(a_n)=(n+x)\log\left(1+\frac{1}{n}\right)$$

$$\log(a_n)=(n+x)\Bigg[\frac{1}{n}-\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right) \Bigg]=1+\frac {2x-1}{2n}+O\left(\frac{1}{n^2}\right)$$
$$a_n=e^{\log(a_n)}=e+\frac {e(2x-1)}{2n}+O\left(\frac{1}{n^2}\right)$$