问:
Given set of 1011 positive numbers prove there are 2 numbers that their sum or difference is a multiple of 2020
I'm struggling solving this problem :
I need to prove that given any set of 1011 positive integers, there are 2 numbers such that
$x_1 + x_2$  or $x_1 - x_2$ is a multiple of 2020.

basically I understand that I need to show that there are 2 numbers such that $ x_1 + x_2 = 0 \pmod{ 2020} $ or $ x_1 - x_2 = 0\pmod {2020}$ but can't find a way to solve that.

Hope to get a help.

edit - also able to make $x_1+x_1=0 (mod 2020)$ if necessary

答:
As stated, the problem is flawed; $1011$ numbers is not enough.

For example, if $x,y$ are two distinct elements of the $1011$-element set $
S=\{0,...,1010\}$, then
$$
\left\lbrace
\begin{align*}
&x+y\in\{1,...,2019\}\\[4pt]
&|x-y|\in\{1,...,1010\}\\[4pt]
\end{align*}
\right.
$$
hence neither of $x+y,x-y$ is divisible by $2020$.

But $1012$ numbers is enough.

Claim:

If $S$ is a set of integers with $|S|\ge1012$, then there exist two distinct elements of $S$ whose sum or difference is divisible by $2020$.

Proof:

If two distinct elements of $S$ are congruent to each other mod $2020$, then their difference is a multiple of $2020$, and we're done.

Thus assume no two distinct elements of $S$ are congruent to each other mod $2020$.

Then without loss of generality, we can reduce the elements of $S$ mod $2020$, hence we can assume $S\subseteq\{0,...,2019\}$.

Our goal now is to show that there are two distinct elements of $S$ whose sum is $2020$.

Let $A=S{\setminus}\{0,1010\}$ and let $B=\{2020-a{\,\mid\,}a\in A\}$.

Then $A,B$ are subsets of the $2018$-element set
$$
\{1,...,1009\}\cup\{1011,...,2019\}
$$
hence, since $|A|\ge 1010$ and $|B|=|A|$, it follows by the pigeonhole principle that $A\cap B\ne {\large{\varnothing}}$.

Let $x\in A\cap B$.

Since $x\in B$ we have $x=2020-y\;$for some $y\in A$.

Then $x,y\in A$ and $x+y=2020$.

If $x=y$ then $x+y=2020$ would imply $x=1010$ and $y=1010$, contradiction, since $1010\not\in A$.

Thus $x,y$ are two distinct elements of $S$ whose sum is $2020$.

This completes the proof,


问:
Proving $\lim_{x\to\infty} \frac{\Gamma(x+a)}{x^a\Gamma(x)}=1$
For any real $a>0$:
$$\lim_{x\to\infty} \frac{\Gamma(x+a)}{x^a\Gamma(x)}=1$$
I got the hint to use Stirling's approximation but I cant seem to solve it. Any tips would be very appreciated.

答:
I just solved it and for anyone interested the answer

$ \frac{\Gamma(x+a)}{x^a\Gamma(x)}=\frac{\sqrt{2\pi(x+a)}(\frac{x+a}{e})^{x+a}}{\sqrt{2\pi(x)}(\frac{x}{e})^{x}}=
 \frac{\sqrt{2\pi }\sqrt{x+ a}(x+a)^{x+a}e^x}{\sqrt{2\pi }\sqrt{x }(x)^{x}e^{x+a}}=
(\frac{x+a}{x})^{\frac{1}{2}ax}\cdot\frac{1}{e^a}  \to \frac{e^a}{e^a}=1 $

ContentLicense: CC BY-SA 4.0


问:
We put $15$ balls into $5$ boxes. No box can contain more than $4$ balls. Prove that there are at least $3$ boxes, containing $3$ or more balls.
We put $15$ balls into $5$ boxes. No box can contain more than $4$ balls. Prove that there are at least $3$ boxes, containing $3$ or more balls.

I know this problem is crying to be solved by the pigeonhole principle. However I couldn't figure how to apply it exactly and started looking into cases (which turned out to be a lot). How can one apply the pigeonhole principle properly to this problem? Thanks in advance.

答:
This method is tedious but straight forward:

Suppose there are NO boxes containing three or more balls.  Then every box contains no more then 2 balls.  There cannot be more than 2(5)= 10 balls, not 15.

If only one box contains 3 or more balls then we have four boxes containing at most 2 balls and one containing at most four balls.  There cannot be more than 2(4)+ 4= 12 balls, not 15.

If only one two boxes contains 3 or more balls then we have three boxes containing at most 2 balls and two containing at most four balls.  There cannot be more than 2(3)+ 2(4)= 14 balls, not 15.


问:
Uniform convergence sequence of positive reals
Suppose $f_n$ for $n ∈ \Bbb N$ and $f$ are functions on $J$. If there exists a sequence $(a_n)$ of positive reals satisfying $a_n→0$ as $n→∞$ and
$$|f_n(x)−f(x)|≤a_n, \quad  ∀n∈\Bbb N, ∀x∈J$$
then $(f_n)$ converges uniformly to $f$.

Can someone help me with the proof of this theorem?

答:
$$(\forall n\in \Bbb N)\;(\forall x\in J)\;$$
$$|f_n(x)-f(x)|\le a_n \implies $$

$$(\forall n\in \Bbb N)$$
$$ 0\le \sup_{x\in J}|f_n(x)-f(x)|\le a_n$$

and squeeze.


问:
Rudin Theorem 1.17
1.17 THEOREM

Let $f: X \to [0, \infty)$ be measurable. There exist simple measurable functions $s_n$ on $X$ such that

(a) $0 \le s_1 \le s_2 \le \dots \le f.$

(b) $s_n(x) \to f(x)$ as $n \to \infty \forall x$

PROOF Put $\delta_n = 2^{-n}$. To each positive integer $n$ and each real number $t$ corresponds a unique integer $k = k_n(t)$ that satisfies $k \delta_n \le t < (k + 1) \delta_n$. Define

$
\phi_n(t)=
\begin{cases}
k_n(t) \delta_n \space \space \text{if} \space 0 \le t <n \\
n \space \space \text{if} \space n \le t < \infty \\
\end{cases}
$

Each $\phi_n$ is then a Borel function on $[0,\infty)$,

$t~\delta_n< \phi_n(t)\le t \space \space \space \space \text{if} \space \space 0 \le t <n$

$0 \le \phi_1(t) \le \phi_2(t) \le \dots \le  t$ and $\phi_n(t) \to t$ as $n \to \infty$ It follows that the functions

$s_n = \phi_n (f)$

satisfy (a) and (b); they are measurable, by Theorem 1.12(d). $\square$

$\textbf{Question}$: Why can't we set $\phi_n(t)= n$ ? What would we lose?

$0 \le \phi_1(t) \le \phi_2(t) \le \dots \le  t$ is only true for $t<n$. I just don't see what we gain by having $k_n(t) \delta_n \space \space \text{if} \space 0 \le t <n $

答:
What would we lose? Well, we need to keep our eyes on the prize: approximating $f$ by simple functions. The $k_n(t)\delta_n$ part is what is actually doing it.

Imagine filling up a glass with some ice in the bottom, ripe for filling. As you pour water in, ignoring displacement there will be a level it would go up to, but in some spots it will be "stopped" by the ice.

That is whats going on here essentially, you are trying to fill up to level "n" but the function gets in the way (the $k_n(t)\delta_n$ part). But, as you fill up more and more, the degree to which you are approximating $f$ gets better.

A good picture of what is going on can be found in Folland, chapter 2.

ContentLicense: CC BY-SA 3.0

Title: Rudin Theorem 1.17


问:
Infinite strictly decreasing sequence of ordinals
I know that on a linearly ordered set $X$, $\leq$ is a well order if and only if $X$ contains no strictly decreasing sequence. However this conflicts in my head with the following problematic 'intuition' which I do not know where it fails.

Let $\alpha$ be a limit ordinal such that $\alpha\geq \omega_1$. All ordinals strictly less than $\alpha$ are a proper set and therefore we should not be able to find an infinite strictly decreasing sequence of ordinals. However since there uncountably many ordinals lesser than $\alpha$, it seems like we should be able to choose a strictly decreasing sequence in some complicated way. Aside from this proposition, can someone explain to me why a decreasing sequence must stablize?

I know there is probably some simple point which has not fully clicked for me.

答:
Well. Choose $\alpha_0$, then maybe it was finitely many points below $\alpha$, but maybe it was infinitely many points below $\alpha$, i.e. there was a limit ordinal between $\alpha_0$ and $\alpha$.

Repeat the process. After finitely many steps you must have reached $\omega_1$, or went below it. Otherwise we had an infinite set of ordinals without a minimal element (that would be $\{\alpha_n\mid n<\omega\}$). Say that $\alpha_5$ was $\omega_1$. Now you have to choose a point below it, $\alpha_6$. But that's a countable ordinal, so you've discarded "most" ordinals already.

Rinse and repeat, and we hit $\omega$, say at $\alpha_{52}$, and then you have to choose $\alpha_{53}$ which is a natural number, and then there's only finitely many points left anyway. So your sequence had to be finite.

You might read my answer and object that I didn't really use the fact that it was $\omega_1$. And that's true. Because it's true for every ordinal. And it is one of those situations where "uncountable" is a red herring. Start with $\alpha_0=\omega$, there are infinitely many points below it, so principle dictates that we can find a decreasing sequence. But we can't, since after choosing just one we have only finitely many points to work with.


问:
Can $S_n=a_1a_2+a_2a_3+\cdots+a_na_1=0$ if $a_i=±1$ for $n=28$ and $n=30$?

Let $$S_n=a_1a_2+a_2a_3+\cdots+a_na_1$$
If $a_i=\pm 1$, can $S_{28}=S_{30}=0$?

My approach: Start from small $n$ if we can see a pattern

For $n=1$:

$S_1=a_1=±1 \neq0$

For $n=2$:

$S_2=a_1a_2+a_2a_1=2a_1a_2=±2 \neq0$

For $n=3$:

$S_3=a_1a_2+a_2a_3+a_3a_1$

Because $a_1a_2$, $a_2a_3$ and $a_3a_1$ are odd numbers and $0$ is an even number, $S_3$ cannot be $0$. (this can be applied to other $S_n$ if $n$ is an odd number)

For $n=4$:

$S_4=a_1a_2+a_2a_3+a_3a_4+a_4a_1=(a_1+a_3)(a_2+a_4)$

However, I can't do the same with $n=6$ and above...

答:
I would like to suggest one more approach to show that existence of the solution implies $4|n$.
Define $x_i = a_i a_{i+1}$, obviously $x_i = \pm 1$.
For any choice of signs of $x_i,\dots,x_{n-1}$ you can always find appropriate $a_1,\dots,a_n$.
Now what about the last term $a_n a_1$?
See that $x_1 x_2 \dots x_{n-1} = a_1 a_2^2\dots a_{n-1}^2 a_n = a_1 a_n$ and obviously $x_1 \dots x_{n-1} = \pm 1$.
Now the problem reads
$$
x_1 + \dots + x_{n-1} + x_1 x_2 \dots x_{n-1} = 0.
$$
Solution exists if and only if ($x_1+ \dots + x_{n-1} = 1$ and $x_1 \dots x_{n-1} = -1$) or ($x_1+ \dots + x_{n-1} = -1$ and $x_1 \dots x_{n-1} = 1$).
Obviously, considering only one case suffices.
Suppose, $x_1 \dots x_{n-1} = -1$. This means, we have odd number of $-1$-s, say $2m+1$.
Then the number of $+1$-s is $n - 1 - (2m + 1)$.
The latter means
$$
1 = x_1+ \dots + x_{n-1} = [n - 1 - (2m + 1)] - (2m + 1) = n - 3 - 4m,
$$
thus
$$
n = 4m + 4 = 4(m+1).
$$


问:
Why is $\log_{-7}49$ undefined instead of $2$?
When you have $\log_{7}49$, you can tell that it is equal to $2$ because $7^2 = 49$. But, when you have $\log_{-7}49$, isn't this also equal to $2$ because $(-7)^2 = 49$ also? Instead, this is undefined.

Can someone explain why this is?

答: In general it is not possible to take a number $x$ and a negative number $n$ and find an exponent $y$ such that $n^y=x.$ For example, it won't work if $x=7$ and $n=-7.$ It works out in the specific case of $x=49$ and $n=-7$, but it's just not useful enough to define the logarithm function for a few special values where logarithm by a negative number actually works.


问:
Inverse sine simplifying problem
It's a simple but not popular problem and there isn't a lot to say here, the problem is:

simplify the following function:

\begin{align}
 \arcsin(b\sin(x))
\end{align}

And for example, if b=2 what is the resulting simplified function?

答:
Use power series to get the result
$$ \arcsin(b\sin(x)) = (b)\frac{x^1}{1!} +
 (b^3-b)\frac{x^3}{3!} + (9b^5-10b^3+b)\frac{x^5}{5!}
+\cdots. \tag{1} $$
This series converges for certain values of $\,b\,$ and $\,x.\,$
Except for $\,b=\pm 1\,$ the series is very unlikely to be
able to be simplified.

Note  that the (unsigned) polynomial coefficients in equation
$(1)$ are given by
OEIS sequence A008596.

Note the superficially similar result
$$ \sin(b\arcsin(x)) = (b)\frac{x^1}{1!} +
 (-b^3+b)\frac{x^3}{3!} + (b^5-10b^3+9b)\frac{x^5}{5!}
+\cdots. \tag{2} $$
Here, if $\,b\,$ is an integer, then this does simplify.
For example,
$$\sin(2\arcsin(x)) = 2x\sqrt{1-x^2},\;
\sin(3\arcsin(x)) = 3x-4x^3, \dots. \tag{3}$$
This case is closely related to the Chebyshev polynomials of the second kind.

Title: Inverse sine simplifying problem


答:
If you consider the function $f(x) = b^x$, the function logarithm in base $b$ is defined to be the inverse of $f$. In other words, if we denote $f^{-1}$ this function, we should have
$$(f \circ f^{-1})(x) = x \quad \text{and} \quad (f^{-1} \circ f)(x) = x.$$
However the notion of inverse function make sens only for bijective functions. Since $b^x$ in not bijective for $b = 0$ and not even well-defined for $b < 0$ (indeed, for $b = -1$ and $x = 1/2$ for example, $b^x$ is not a real number), the inverse function is only defined for $b > 0$ and we note $\log_b$ this function.

Title: Why is $\log_{-7}49$ undefined instead of $2$?

Body:
When you have $\log_{7}49$, you can tell that it is equal to $2$ because $7^2 = 49$. But, when you have $\log_{-7}49$, isn't this also equal to $2$ because $(-7)^2 = 49$ also? Instead, this is undefined.

Can someone explain why this is?


问:
A small query about the integration by substitution formula: $\int f(u) \, du= \int f(g(x))g'(x) \, dx, \ u=g(x) \, .$
In Michael Spivak's Calculus (p.368), he makes the following remark about integration by substitution:

The substitution formula is often written in the form $$\int f(u) \, du= \int f(g(x))g'(x) \, dx, \quad u=g(x) \, .$$This formula cannot be taken literally (after, all $\int f(u) \, du$ should mean a primitive of $f$ and $\int f(g(x))g'(x) \, dx$ should mean a primitive of $(f \circ g) \cdot g'$; these are certainly not equal). However, it may be regarded as a symbolic summary of the procedure which we have developed.

I don't understand what Spivak is referring to here. Say $F$ is a primitive of $f$. Then,
\begin{align}
\text{LHS} &= \int f(u) \, du = F(u)+C \\
\text{RHS} &= \int f(g(x))g'(x) = F(g(x))+C=F(u)+C
\end{align}
I'm also not quite on board with the view that the LHS is a primitive of $f$. If this were true, then wouldn't we simply write
$$
\int f(u) \, du = F \, ?
$$
I think if we were being precise then we would say the LHS is a primitive of $f$, evaluated at the point $u$. I'm probably misunderstanding something here, but I'm not what it exactly is.

I am aware that this question has been asked before, but I wasn't entirely satisfied with the answer, and I'm worried that I might have a deeper confusion about integration in general.

答:
To be precise, and under some conditions, you can write

$$\int_{g(a)}^{g(x)}f(u)du=$$

$$\int_a^xf(g(u))g'(u)du=$$

$$\int_a^xf(g(t))g'(t)dt$$

the substitution made is
$$u=g(t)$$


问:
Pettis integrability condition
Let $B$ be a Banach Space, $(X,\mu)$ a measure space and $f:X\longrightarrow B$ a weak measurable function, that is a function such that:
$$ \phi\circ f:X \longrightarrow \mathbb C $$
is measurable forall $\phi\in B^*$ (the topological dual).

If $||f(x)||\leq M(x)$ with $M\in L^1$, can we deduce that $f$ is Pettis integrable?

答:
$\newcommand{\bf}[1]{\mathbb #1}\newcommand{\sc}[1]{\mathscr #1}$The answer is affirmative for separable Banach spaces.

In order to prove it, consider the linear functional $\Lambda $ defined on the (topological) dual space $B'$  by
$$
  \Lambda (\varphi )=\int_X \varphi \big (f(x)\big )\, d\mu (x), \quad\forall \varphi \in B'.
  $$
Notice that
the integrand satisfies
$$
  |\varphi \big (f(x)\big )| \leq  \|\varphi \| \|f(x)\| \leq  \|\varphi \| M(x),
  $$
so
$$
  |\Lambda (\varphi )|\leq \|\varphi \|\|M\|_1,\quad\forall \varphi \in B',
  $$
and hence  $\Lambda $ is seen to be continuous.  We next claim that $\Lambda $ is also continuous relative to  the
weak-star topology on $B'$.

Using
V.12.8 in Conway's "A Course in Functional Analysis", a Corollary of Krein-Smulian's Theorem,  it is enough to show that $\Lambda $ is weak-star sequentially
continuous.   So let us suppose that $\{\varphi _n\}_n$ is a sequence in $B'$ converging to some $\varphi \in B'$ relatively to  weak-star topology.

If tollows that
$$
  \varphi _n\circ f \to  \varphi \circ f
  $$ pointwise everywhere on $X$.  Moreover, by the Banach-Steinhauss Theorem, we have that the $\varphi _n$ are uniformly
bounded, whence
$$
  K := \sup_{n\in {\bf N}}\|\varphi _n\| <\infty ,
  $$
and hence
$$
  |\varphi _n(f(x))| \leq  \|\varphi _n\| \|f(x)\| \leq   K M(x),\quad \forall x \in X.
  $$
We may then  use Lebesgue's Dominated Convergence Theorem to conclude that
$$
  \lim_{n\to \infty } \Lambda (\varphi _n) =
  \lim_{n\to \infty } \int_X \varphi _n(f(x))\, d\mu (x) = $$$$ =
  \int_X \varphi (f(x))\, d\mu (x) = \Lambda (\varphi ),
  $$
hence concluding the proof of the weak-star continuity of $\Lambda $.

As it is well known, every weak-star continuous linear functional on $B'$ must be a point-evaluation, so there exists
$b\in B$ such that
$$
  \Lambda (\varphi )= \varphi (b),\quad\forall \varphi \in B'.
  $$
Consequently
$$
  \int_X \varphi (f(x))\, d\mu (x) = \varphi (b),
  $$
so $f$ is Pettis integrable!

Thanks to MathOverflow user @Mikael de la Sale for pointing me towards Krein-Smulian's Theorem!

Title: Pettis integrability condition


问:
Confusion on definition of complete metric spaces
So I am a complete beginner to functional analysis and topology and I recently learned what a complete metric space is I understand that every cauchy sequence must converge to a limit in the set.
My confusion is why is the set of all natural numbers $\mathbb{N}$ not a complete metric space.
I also don't grasp why a complete metric space is automatically not a closed set because a closed set contains all of its limit points?.
This is probably very trivial to you all and I admit my inquiries on the matter may not seem clear but it has really confused me.
Thanks in advance

答:
I'm going to expand on Brian Scott's comment. First of all, the natural numbers $\mathbb{N}$ are in fact complete. You can see this because all Cauchy sequences of natural numbers must eventually be constant and therefore converge.

Secondly, you should be aware that while completeness is an internal/absolute property of a metric space, being closed is defined relative to some ambient space. For example, while $(0,1)$ is not closed when considered as a subset of $\mathbb{R}$, it is closed when considered as a subset of itself. That's because $0$ and $1$ are no longer limit points of $(0,1)$ as a subset of $(0,1).$ They can't be because they are not even points of $(0,1).$ Actually, all metric spaces are closed as subsets of themselves.

There is a sense in which your intuition holds. If $X\subseteq Y$ are metric spaces and $X$ is complete then $X$ must be closed. If $Y$ is assumed to be complete, then $X$ is complete if and only if it is closed.


问:
Understanding cup products for Tate cohomology group
I am reading about 'Cup products' from the book 'Algebraic Number Theory' by Cassels and Frohlich. The following theorem appears on page 105.

Theorem. $G$ is a finite group then there exists one and only one family of morphisms $$\hat H^p(G,A ) \otimes \hat H^q(G,B ) \rightarrow \hat H^{p+q}(G,A \otimes B ) $$  defined for integers $p, q$ and $G$ modules $A,B$ such that

*the usual properties of cup products are listed

The second of these listed properties is

(ii) For $p=q=0 $ they (the family of morphism) are induced by natural product  $$A^G \otimes B^G= (A \otimes B)^G $$

I suppose the natural map is $a\otimes b \mapsto a \otimes b$. While this map works for cohomology groups, when we are dealing with Tate cohomology groups $\hat H^0(G,A)=A^G/Nm_G A $. So the cup product morphism is $$A^G/Nm_G A \otimes B^G/Nm_G B \rightarrow (A\otimes B)^G/Nm_G(A \otimes B)$$ The natural induced map should be, to my mind, $\overline a \otimes \overline b \mapsto \overline{ a\otimes b }$.

But I don't see how this map is well defined? I know that in general the map  $A/A' \otimes B/B' \rightarrow A \otimes B/(A'\otimes B')$ is not well defined.

Suppose for example $a' \in Nm_GA, b' \in Nm_G B $ then is $\overline {a+a'\otimes b+b'} = \overline {a \otimes b} $.

I am stuck at  $ \overline {a+a'\otimes b+b'}=\overline {a \otimes b} + \overline {a \otimes b'} + \overline {a' \otimes b}$

Feel free to give any reference and any help is appreciated.

答:
To show that the map$\newcommand\Nm{\mathrm{Nm}}$
$$\frac{A^G}{\Nm_GA}\otimes\frac{B^G}{\Nm_GB}\to\frac{(A\otimes B)^G}{\Nm_G(A\otimes B)}$$
is well-defined, we need it to be induced from a bilinear map
$$\frac{A^G}{\Nm_GA}\times\frac{B^G}{\Nm_GB}\to\frac{(A\otimes B)^G}{\Nm_G(A\otimes B)}.$$
In other words, we need the map
$$A^G\times B^G\ni(a,b)\longmapsto\overline{a\otimes b}\in\frac{(A\otimes B)^G}{\Nm_G(A\otimes B)},$$
which is visibly bilinear, to vanish when $a\in\Nm_GA$ or $b\in\Nm_GB$ (so it descends to a bilinear map on the product of the quotient spaces). First suppose
$$a=\Nm_Gx=\sum_{g\in G}g\cdot x$$
for some $x\in A$. Then, for any $b\in B^G$,
$$(a,b)\mapsto\overline{\sum_{g\in G}(g\cdot x)\otimes b}=\overline{\sum_{g\in G}(g\cdot x)\otimes(g\cdot b)}=\overline{\Nm_G(x\otimes b)}=0\in\frac{(A\otimes B)^G}{\Nm_G(A\otimes B)},$$
as desired. One similarly shows that for any $a\in A^G$ and $b=\Nm_Gy$ ($y\in B$), $(a,b)\mapsto0$. Thus, the map $\overline a\otimes\overline b\mapsto\overline{a\otimes b}$ you wrote down is well-defined.


答:
When working in the real numbers, we require that the base of the logarithm is greater than $0$ (and not equal to $1$). While the expression
$$
\log_{-7}(49)=2
$$
does have some sense to it, it is formally incorrect for this simple reason.

This begs the question—why do we avoid negative bases? This is because the function $\log_{-7}$ would be ugly. It would only be defined for a limited subset of the real numbers, and it wouldn't be differentiable, unlike other logarithm functions. The exponential function $(-7)^x$ only makes sense for some rational $x$ (where the denominator is odd), and so exponentiation of negative numbers is a touchy subject in general. Note that the definition of $a^x$ is
$$
\exp(x\log(a))
$$
where $\log$ is the natural logarithm and $\exp$ is the exponential function ($\exp(x)=e^x$). Negative bases have to be dealt with separately, as $\log(a)$ is undefined for $a \leq 0$.


问:
Uniform convergence implies pointwise convergence
I am having trouble proving a simple proposition regarding uniform convergence and pointwise convergence in Real Analysis.

Problem:

Suppose that $\left(f_{n}\right)$ is a sequence of functions $f_{n}: A \rightarrow \mathbb{R}$ such that $\left(f_{n}\right)$ converges uniformly to $f: A \rightarrow \mathbb{R}$. Prove that $\left(f_{n}\right)$ also converges pointwise to $f: A \rightarrow \mathbb{R}$

Relevant definitions/Notations:

One says that $f_{n} \rightarrow f$ uniformly on $A$ if, for every $\epsilon>0,$ there exists $N \in \mathbb{N}$ such that $n>N$ implies that $$\left|f_{n}(x)-f(x)\right|<\epsilon$$ for all $x \in A$.

Finally, we say that $f_{n} \rightarrow f$ pointwise on $A$ if, given $\epsilon>0,$ for each $x \in A$ there exists $N \in \mathbb{N}$ such that $n>N$ implies that $$\left|f_{n}(x)-f(x)\right|<\epsilon$$

Attempts:

I tried to write down the definition of uniform convergence and then arguing that, in particular, since $N \in \mathbb{N}$ from uniform convergence works for any given point in $x \in A$, then it must work for a given point and from that conclude pointwise convergence.

I also checked some proofs like the one that states that uniform continuity implies continuity and write something similar, but i just dont know how to do it.

I would highly appreciate a detailed proof regarding this fact, i am trying to become proeficient at proof writing.

Thanks in advance, Lucas

答:
To begin with, it is useful to remind that uniform convergence is equivalent to convergence with respect to the sup norm.

Having said that, let $f_{n}:(X,d_{X})\to(Y,d_{Y})$ converges uniformly to $f:(X,d_{X})\to(Y,d_{Y})$.

Thus for every $\varepsilon > 0$ and every $x\in X$ there corresponds $n_{\varepsilon} \in \mathbb{N}$ such that we have that
\begin{align*}
n\geq n_{\varepsilon} & \Rightarrow d_{Y}(f_{n}(x),f(x)) \leq \sup_{x\in X}d_{Y}(f_{n}(x),f(x)) \leq \varepsilon
\end{align*}

Hence $f_{n}$ converges to $f$ point-wise.

Hopefully this helps!


问:
Monotonically decreasing sequence $(a_n) \to$ mo. dec. sequ. $(a_1+a_2+\cdots+a_n)/n$
I think this isn't quite difficult, however I don't get the point..

I have to prove:

$(a_n)$ is a monotonically decreasing sequence. Show, that the sequence $\frac{a_1 + a_2 + \cdots + a_n}{n}$ is monotonically decreasing, too

I thought about using something like Cauchy-Limit or so.. Since the sequence $\frac{a_1 + a_2 + \cdots + a_n}{n} = \frac{1}{n} (a_1 + a_2 + \cdots+ a_n)$ looks quite interesting so far..

If you have any hints for me, I'd be glad to use them :)

答:
In a sense, the proof writes itself. The details depend on whether we use $\lt$ or $\le$ in the definition of monotonically decreasing. Let's use $\le$. We want to know whether
$$\frac{a_1+a_2 +\cdots+a_{n+1}}{n+1} \overset{?}{\le} \frac{a_1+a_2+\cdots+a_n}{n}.\tag{$1$}$$
Algebraic manipulation shows that this is equivalent to
$$na_{n+1}\overset{?}{\le} a_1+a_2+\cdots+a_n.\tag{$2$}$$
Since $a_{n+1}\le a_i$ for $i=1,2,\dots, n$, Inequality $(2)$ is obvious.  </p>


问:
(New Kids on The) Block Matrix Multiplication
Block matrices are a great time-saver if one is tasked with multiplying two large matrices and one (or both) of them has an identity submatrix or, better still, a zero submatrix. However, in those cases when neither has any identity or zero submatrices, is there any advantage to dividing them into blocks before multiplying?</p>

答:
Block matrix multiplication is very useful in a practical sense. For instance, the Strassen algorithm for matrix multiplication speeds up computation for large matrix products by exploiting block-matrix multiplication. Parallel and distributed algorithms for matrix multiplication also rely on block matrix multiplication in order to properly "distribute" the necessary computation.

It is also very useful for theoretical purposes. For instance, consider the inductive proof of the existence of the Schur decomposition of a matrix.


答: Your attempt looks fine. Take $x\in A$; you want to prove that $\lim_{n\to\infty}f_n(x)=f(x)$. Now, take $\varepsilon>0$; you want to prove that there is some $N\in\Bbb N$ such that$$n\geqslant N\implies\bigl|f_n(x)-f(x)\bigr|<\varepsilon.\tag1$$So, take $N\in\Bbb N$ such that$$(\forall a\in A)(\forall n\in\Bbb N):n\geqslant N\implies\bigl|f_n(a)-f(a)\bigr|<\varepsilon.$$It follows from this that, for such a $N$, $(1)$ holds.

Title: Uniform convergence implies pointwise convergence

Body:
I am having trouble proving a simple proposition regarding uniform convergence and pointwise convergence in Real Analysis.

Problem:

Suppose that $\left(f_{n}\right)$ is a sequence of functions $f_{n}: A \rightarrow \mathbb{R}$ such that $\left(f_{n}\right)$ converges uniformly to $f: A \rightarrow \mathbb{R}$. Prove that $\left(f_{n}\right)$ also converges pointwise to $f: A \rightarrow \mathbb{R}$

Relevant definitions/Notations:

One says that $f_{n} \rightarrow f$ uniformly on $A$ if, for every $\epsilon>0,$ there exists $N \in \mathbb{N}$ such that $n>N$ implies that $$\left|f_{n}(x)-f(x)\right|<\epsilon$$ for all $x \in A$.

Finally, we say that $f_{n} \rightarrow f$ pointwise on $A$ if, given $\epsilon>0,$ for each $x \in A$ there exists $N \in \mathbb{N}$ such that $n>N$ implies that $$\left|f_{n}(x)-f(x)\right|<\epsilon$$

Attempts:

I tried to write down the definition of uniform convergence and then arguing that, in particular, since $N \in \mathbb{N}$ from uniform convergence works for any given point in $x \in A$, then it must work for a given point and from that conclude pointwise convergence.

I also checked some proofs like the one that states that uniform continuity implies continuity and write something similar, but i just dont know how to do it.

I would highly appreciate a detailed proof regarding this fact, i am trying to become proeficient at proof writing.

Thanks in advance, Lucas


问:
Question in using a set of ordered pairs and its union
The question became more verbose than I expected but it is really simple.

Let $A\subseteq X$, $A$ is non-empty, and $b \not\in A$.
Suppose that for all $x \in A$, there are two disjoint subsets of $X$ such that each of them has $x$ and $b$, respectively, that is
$$
\forall x \in A: \exists U, V \subseteq X: x\in U \land b \in V \land U\cap V = \varnothing
$$
Then, we can see that the union of all such $U$ contains $A$, while the intersection of all such $V$ has $b$. This style of proof stated in English is really what is done in many textbooks. But I want to utilize more mathematical tools to clarify for practice.

Now I am going to show this fact using a set of ordered pairs:
Let
$$
S = \{(U,V)\in \mathcal{P}(X)^2~|~U\cap A \neq \varnothing \land b\in V \land U \cap V = \varnothing\}
$$

where $\mathcal{P}$ denotes the power set. ($S$ is non-empty due to assumption.)

My question is whether the following expressions conform to the standard mathematical notation and the proofs are correct:

Claim 1:
$$ A \subseteq \bigcup_{(U,V)\in S}U$$

Proof: Let $x \in A$ be arbitrary. Then, there is $(U,V)\in S$ such that $x \in U$, which is obtained directly from the assumption.

Claim 2:
$$b \in \bigcap_{(U,V)\in S}V$$

It is a trivial result by definition.

To make the expressions more concise, I tried using a projection.

$$A \subseteq \bigcup_{B\in S}\pi_1 (B)$$

This is meant to be identical to Claim 1. And we can say that
$$
C =
\{\pi_1(B)~|~B\in S\}
=
\{N \subseteq X~|~ \exists B \in S : N = \pi_1(B)\}
$$
covers $A$.

Claim 3:

$$
\left(\bigcup_{B\in S}\pi_1(B) \right) \cap \left(\bigcap_{B\in S}\pi_2(B) \right) = \varnothing
$$

This is derived from the condition $U \cap V = \varnothing$ in the definition of $S$.

答:
The notations are correct, though I prefer the simpler first version to the second: the projection functions are unnecessary clutter. However, what you’ve done symbolically is not the same as what you did in words at the top. Initially you said only that for each $x\in A$ there is (at least) one pair of sets $U,V\subseteq X$ such that $x\in U$, $b\in V$, and $U\cap V=\varnothing$. To capture this in symbols you might say that for each $x\in A$ there are sets $U_x,V_x\subseteq X$ such that $x\in U_x$, $b\in V_x$, and $U_x\cap V_x=\varnothing$. Your claims are then simply that

$$A\subseteq\bigcup_{x\in A}U_x$$

and

$$b\in\bigcap_{x\in A}V_x\,.$$

The set $\mathcal{S}$ that you defined contains much more than your introduction describes: it contains every pair $\langle U,V\rangle$ of subsets of $X$ such that $U$ contains some member of $A$, $b\in V$, and $U\cap V=\varnothing$. That’s fine if that’s what you wanted, but it is definitely not what you described at the beginning of the question. And as drhab pointed out, it allows a much stronger conclusion: the pair $\big\langle X\setminus\{b\},\{b\}\big\rangle$ is in $\mathcal{S}$, so

$$\bigcup_{\langle U,V\rangle\in\mathcal{S}}U=X\setminus\{b\}\supseteq A\,,$$

and

$$\bigcap_{\langle U,V\rangle\in\mathcal{S}}V=\{b\}\,.$$


问:
Partial recursive set from values of recursive function
So I know that $E = \{n\ |\ \text{if } F(n,n) \text{ is defined}\}$, is recursively enumerable set which isn't recursive. But how i find a recursive function which values is that set?

答:
Fix some $m$ such that $F(m,m)$ is defined. Then define a function $g$ by setting
$$
g(n,s)=\begin{cases}
n&\text{if the computation of }F(n,n)\text{ produces an output in }<s\text{ steps.}\\
m&\text{otherwise.}
\end{cases}
$$
Then $g$ is recursive, and its range is your $E$.

Incidentally, I don't recall ever hearing "partial recursive set".  You probably meant "recursively enumerable set". Partial recursiveness is a property of functions, not of sets.


问:
Prove that $\mathbb{Z}$ with the finite complement topology is not Hausdorff.
Prove that $\mathbb{Z}$ with the finite complement topology is not Hausdorff.

What are the closures of the following two subsets of $\mathbb{Z}$ with the finite complement topology? (Fully justify your answers.)

i) $A=\{1,2,3\}$ $\,\,\,\,\,$ ii) $\mathbb{N}=\{1,2,3,4,...\}$

$\textbf{Proof:}$ By way of contradiction, assume $\mathbb{Z}$ with the finite complement topology is Hausdorff. Then $\forall a \neq b \,\, \exists U,V \subset \mathbb{Z}$ where $U \cap V = \emptyset$ with $a \in U$, $b \in V$, and $U,V$ are open sets.

Since $a,b \in \mathbb{Z}$, then pick $a=0$ and $b=1$. Since $U,V \subset \mathbb{Z}$ are open in the finite complement topology, then $U^c$,$V^c$ are finite.

Thus, $U,V$ are infinite. So $V \not \subset U^c$. Since $1 \in V$, then $1 \not \in V^c$, so $1 \in (U^c)^c=U$ by infinite/finite sets.

But, $U \cap V = \emptyset$. $\rightarrow \leftarrow$ (contradiction!)

$\therefore$ By way of contradiction, $\mathbb{Z}$ with the finite complement topology is not Hausdorff.

(i) $A=\{1,2,3\}$

$A \subset \mathbb{Z}$ with the finite complement topology. So, $\mathbb{Z}\backslash A$ is open $\implies A$ is closed.

$\therefore A=\overline{A}$.

(ii) $\overline{\mathbb{N}}=\mathbb{N} \cup \mathbb{N}^{'}=\mathbb{N}$ where $\mathbb{N}^{'}$ is the set of all limit points of $\mathbb{N}$.

I am checking if my proof is valid and could use any feedback on it.
Any help with part (ii) would be nice. I am unsure how to provide a further explanation.

答:
Your proof that the space is not Hausdorff is correct, but there is no need to use proof by contradiction. Let $m$ and $n$ be distinct integers, and let $U$ and $V$ be open nbhds of $m$ and $n$, respectively; there are finite sets $F,G\subseteq\Bbb Z$ such that $U=\Bbb Z\setminus F$ and $V=\Bbb Z\setminus G$. Then

$$U\cap V=(\Bbb Z\setminus F)\cap(\Bbb Z\setminus G)=\Bbb Z\setminus(F\cup G)\ne\varnothing\,,$$

since $F\cup G$ is finite. Thus, $m$ and $n$ do not have disjoint open nbhds. (I could just as well have worked with a specific pair of integers, but it’s just as easy to make the argument for an arbitrary pair and prove the stronger result that no two distinct points can be separated by disjoint open sets.)

For part (ii) you can prove the stronger result that every infinite subset of $\Bbb Z$ is dense in $\Bbb Z$, i.e., has closure equal to $\Bbb Z$. Suppose that $A\subseteq\Bbb Z$ is infinite. Let $n\in\Bbb Z$. If $U$ is an open nbhd of $n$, $\Bbb Z\setminus U$ is finite, so $A\nsubseteq\Bbb Z\setminus U$, and therefore $U\cap A\ne\varnothing$. Thus, $n\in\operatorname{cl}A$, and since $n$ was an arbitrary integer, $\operatorname{cl}A=\Bbb Z$.


问:
How does the coefficient ring influence the Euler characteristic?
Notation/background. Let $C_\bullet$ be a bounded chain complex of finitely generated abelian groups, with $C_i\cong\mathbb{Z}^{a_i}$. For any principal ideal domain $R$, let $C_\bullet\otimes R$ be the chain complex of finitely generated $R$-modules obtained by applying the functor $(-)\otimes_\mathbb{Z}R\colon \mathbf{Ab}\to R\mathbf{Mod}$ termwise. Furthermore, let $$\chi_R=\sum_{i\in\mathbb{Z}}(-1)^i\operatorname{rank} H_i(C_\bullet\otimes R)$$ be the Euler characteristic of $C_\bullet\otimes R$. It's a classical exercise in homological algebra (see e.g. Theorem 2.44 in Hatcher) to show that $$\chi_{R}=\sum_{i\in\mathbb{Z}} (-1)^i {a_i}$$ whenever $R$ is a field and whenever $R=\mathbb{Z}$.

Question. What happens when $R$ is some other PID? Do we still get $\chi_{R}=\sum_{i\in\mathbb{Z}} (-1)^i{a_i}$?

Attempt/thoughts. Suppose that $H_i(C_\bullet)\cong\mathbb{Z}^{b_i}\oplus T_i$, where $T_i$ is the torsion subgroup of $H_i(C_\bullet)$. The universal coefficient theorem then tells us that
$$H_i(C_\bullet\otimes R)\cong (H_i(C_\bullet)\otimes _{\mathbb{Z}}R)\oplus \mathrm{Tor}_1^{\mathbb{Z}}(H_{i-1}(C_\bullet),R)\\ \cong R^{b_i}\oplus (T_{i}\otimes_{\mathbb{Z}} R)\oplus \mathrm{Tor}_1^{\mathbb{Z}}(T_{i-1},R)\,.$$
When $R$ is a field (see this thread),  then all free copies of $R$ obtained from the tensor products with $T$ are cancelled by free copies of $R$ obtained from the Tor-terms when we compute the alternating sum. It intuitively doesn't feel very likely that things would work out as nicely when $R$ is a general PID, but I find it hard to come up with a counterexample...

Does anyone have any ideas or suggestions about how to approach this?

答:
I'll try to turn my comment into a proof. Let $F$ be the field of fractions of $R$. Since $R$ is a PID, the Universal Coefficient Theorem tells us that
$$H_i(C;F) \cong (H_i(C;R) \otimes_R F) \oplus \operatorname{Tor}_1^R(H_{i-1}(C;R),F)$$
(this is an isomorphism of $F$-vector spaces). Since $F$ is flat over $R$, the $\operatorname{Tor}$ term is $0$. Now
$$H_i(C;F) \cong H_i(C;R) \otimes_R F$$
implies
$$\operatorname{rk}_F H_i(C;F) = \operatorname{dim}_F H_i(C;F) = \dim_F (H_i(C;R) \otimes_R F) = \operatorname{rk}_R H_i(C;R).$$
Thus, $\chi_R = \chi_F = \chi_{\mathbb{Z}}$.

Edit: To be extra clear; the thing I called $H_i(C;F)$ here is technically $H_i((C_{\bullet} \otimes_{\mathbb{Z}} R) \otimes_R F)$. The precise definition of $H_i(C;F)$ is $H_i(C_{\bullet} \otimes_{\mathbb{Z}} F)$. However, this is the same thing, since the map $(A \otimes_{\mathbb{Z}} R) \otimes_R F \to A \otimes_{\mathbb{Z}} F$ defined by $(a \otimes r) \otimes f \mapsto a \otimes (rf)$ gives a natural isomorphism $({(-)} \otimes_{\mathbb{Z}} R) \otimes_R F \to {(-)} \otimes_{\mathbb{Z}} F$. This is easy to see by thinking about these tensor products as pullbacks of sheaves on affine schemes.


问:
Is rational number system the motivation behind the definition of field structure?
It might sound a very silly question but I didn't find satisfactory answer anywhere. When I read concept of order(I am reading baby Rudin now), I find it very intuitive in sense that it connects to real world arrangement of object. Meaning if we had collection of objects, what we naturally do is arrange them one after another. That's exactly what the definition of order does with set. Also I find definition of metric space very much connected to real world in sense of distance. So on basis of what I have read so far, I find that, we take a set and attach some notion to it and thus convert it to some structure. I am curious to know how that particular notion came, what's the idea behind it, like I said about notion of order and distance function in metric space. Now when I look back at the definition of field, I thought for a long time about how someone came up with notion of addition, multiplication, element 0, element 1, additive and multiplicative inverse.. there's so much. What was motivation behind defining this all notions. Then I felt that it might be the case that we were already familiar with the rational number system before the birth of 'field structure'. And taking intuition from the properties that rationals posses, we made precise definition of field. But this all are my thoughts. Unknowingly I am getting habit of finding intuition or motivation behind every definition, which I feel is not good always. Thanks.</p>

答:
You are asking about the intuition and motivation for the definition of
a field. It seems to be the case that finite rings and fields led to the
definition. Early in the 1800s Gauss developed the idea of the congruence
relation of integers. This implicitly defined the ring of integers
modulo $\,n\,$ which is a finite ring. When $\,n\,$ is a prime it is
also a finite field. A little later Galois extended this by introducing
finite fields with prime power number of elements. This led to the
idea of algebraic extensions of existing fields such as the real number
field. These important special cases motivated the definition
and study of fields.

The Wikipedia article Fields has more details about
the history and development of field theory going back
before 1800.


问:
Can I use control theory when the output is stochastic and has delayed reactions to the controller?
I am a complete novice in control theory. My understanding of control theory is that it can be used to adjust parameters of a system based on feedback to reach some desired state. It sounds like a useful set of tools for a problem that I have, but I'm struggling to formulate it.

My problem is (essentially) as follows: Users are allowed to click on one of two buttons, A or B, whenever they want to, over the course of an hour. My desired state is that I want the number of A-presses to be about the same as the number of B-presses ($A-B = 0$). There is an internal parameter $u$ that can make one of the buttons more desirable to click than the other to an arbitrary degree, which will cause more users to click on that button in the future.

Let's say that after 10 minutes, A has been pressed 300 times and B has been pressed only 100 times ($y = 300-100 = 200$). Clearly, for some unknown external reason, pressing A is more desirable to the users than pressing B. I would like to take this feedback and adjust the internal parameter $u$ to make B more enticing and encourage users to press B more, with the goal of bringing $y$ nearer to zero.

Can I use control theory techniques to help with this problem? It seems like a perfect fit based on the description of control theory, but every example that I've seen so far has had a well defined output based on a physical system such as the car's speed in cruise control, rather than something as stochastic as the users choices based on their desires. Additionally, the car's speed updates immediately, whereas the impact of changing $u$ on $y$ could be substantially delayed. I've tried implementing an ad-hoc solution for this problem with simulated users, but it regularly overshoots the desired solution and oscillates. Any tips are greatly appreciated!

Simulation Details

The simulation is coded up in Python and works as follows. At each timestep, one of the buttons is pressed by a "user". The probability of the pressed button being A, $P(A) = \sigma(u+v)$ where $\sigma(x) = \frac{1}{1 + e^{-x}}$ is the sigmoid function to produce a probability, $u$ is the internal parameter, and $v$ is some pre-determined value indicating how the users feel about the buttons from the beginning. $P(B) = 1 - P(A)$.

For example, in the case that $v = 0.5$, the users will favour button A. To perfectly counteract this, $u$ needs to be set to $-0.5$.

答:
Model

Let $A$ be the number of clicks on button A, and $B$ - on button B. Define $e:=A-B$, and the goal is to drive $e$ to zero.

At each step we have
$$
e(t+1) = \begin{cases}e(t)+1 \text{ with the probability }p(A), \\e(t)-1 \text{ with the probability }1-p(A). \\\end{cases}
$$
I assume that the next control action is taken after $N$ clicks are collected, and thus I consider expectations. Then we obtain $e(t+1) = e(t) + N(2p(A)-1) = e(t)+Ng(u(t))$,
where $g \in (-1,1)$
$$
g(u(t)) = 2\frac{1}{1+e^{-u(t)-v}}-1,
$$
$u$ is the control signal, and $v$ is a constant.

Control

Note that $g(u)$ is monotonic. Without any deep analysis, I would suggest trying a simple PI controller
$$
\begin{aligned}
z(t) &= z(t-1)+\alpha e(t),\\
u(t) &= -z(t) - \beta e(t).
\end{aligned}
$$
Here the gains $\alpha$ and $\beta$ are to be tuned.


问:
Integration of $2$-form on the sphere using the stereographic projection
Let $\omega$ be the $2$ form $\omega = x \, dy \wedge dz - y \, dx \wedge dz + z \, dx \wedge dy$ on $S^2$. I want to integrate $\int_{S^2} \omega$ using the definition, with the stereographic projection ${\varphi}^{- 1} : {\mathbb{R}}^2 \to S^2 \setminus \{(0 , 0 , 1)\}$ given by
$$
{\varphi}^{- 1}(u , v) = \left(x = \frac{2 u}{1 + u^2 + v^2} , y = \frac{2 v}{1 + u^2 + v^2} , z = \frac{u^2 + v^2 - 1}{1 + u^2 + v^2}\right).
$$
Then
$$
\int_{S^2} \omega = \int_{{\mathbb{R}}^2} {\left({\varphi}^{- 1}\right)}^*(\omega).
$$
I proceed to calculate ${({\varphi}^{- 1})}^*(\omega)$. It is
$$
{\left({\varphi}^{- 1}\right)}^*(\omega) = x {\left({\varphi}^{- 1}\right)}^*(dy) \wedge {\left({\varphi}^{- 1}\right)}^*(dz) - y {\left({\varphi}^{- 1}\right)}^*(dx) \wedge {\left({\varphi}^{- 1}\right)}^*(dz) + z {\left({\varphi}^{- 1}\right)}^*(dx) \wedge {\left({\varphi}^{- 1}\right)}^*(dy).
$$
For instance,
$$
{\left({\varphi}^{- 1}\right)}^*(dx) = \frac{\partial x}{\partial u} \, du + \frac{\partial x}{\partial v} \, dv = \frac{2 (1 + u^2 + v^2) - 4 u^2}{{(1 + u^2 + v^2)}^2} \, du - \frac{4 u v}{{(1 + u^2 + v^2)}^2} \, dv
$$
and similarly
$$
{\left({\varphi}^{- 1}\right)}^*(dy) = - \frac{4 u v}{{(1 + u^2 + v^2)}^2} \, du + \frac{2 (1 + u^2 + v^2) - 4 v^2}{{(1 + u^2 + v^2)}^2} \, dv
$$
and
$$
{\left({\varphi}^{- 1}\right)}^*(dz) = 4 \left(\frac{u}{{(1 + u^2 + v^2)}^2} \, du + \frac{v}{{(1 + u^2 + v^2)}^2} \, dv\right).
$$
We calculate now the exterior products:
$$
x {\left({\varphi}^{- 1}\right)}^*(dy) \wedge {\left({\varphi}^{- 1}\right)}^*(dz) = - \frac{16 u^2}{{(1 + u^2 + v^2)}^4} \, du \wedge dv,
$$
$$
- y {\left({\varphi}^{- 1}\right)}^*(dx) \wedge {\left({\varphi}^{- 1}\right)}^*(dz) = - \frac{16 v^2}{{(1 + u^2 + v^2)}^4} \, du \wedge dv,
$$
$$
z {\left({\varphi}^{- 1}\right)}^*(dx) \wedge {\left({\varphi}^{- 1}\right)}^*(dy) = - 4 \frac{{(u^2 + v^2)}^2 - 2 (u^2 + v^2) + 1}{{(1 + u^2 + v^2)}^4} \, du \wedge dv.
$$
Therefore
$$
{\left({\varphi}^{- 1}\right)}^*(\omega) = \frac{4}{{(1 + u^2 + v^2)}^4} (- 2 u^2 - 2 v^2 - 1 - u^4 - 2 u^2 v^2 - v^4) \, du \wedge dv
$$
if I did not have any mistake. But how can I proceed with this expression? On the other hand, I know that the integral should be $4 \pi$.

答:
Your results so far are actually all correct. To proceed, you just need to be a little less eager in expanding all the expressions, but opt for factoring more. In particular, the result for $(\phi^{-1})^* (z\, dx\wedge dy)$ can be factored. The numerator is actually just $4(u^2+v^2-1)^2$. When you add up the pullback of the other two terms, you are adding $16(u^2+v^2)$ to the numerator. Thus you get $4(u^2+v^2+1)^2$, which cancels neatly with the denominator.

Alternatively, if you are familiar enough with the expansion of $(x+y+z)^2$, you can recognize straight away that the numerator of your final result is $4(u^2+v^2+1)^2$.

To proceed with the integral, you can either convert the integral into polar coordinates in the $uv$-plane, or do it with trig substitutions. The former method is easier by far.


问:
Why does $\log^{100}(n)$ grow asymptotically slower than $n^{7/8}$
More formally:
How do I compute the following limit:

$$
\lim _{x \rightarrow \infty} \frac{\log ^{100}(x)}{x^{7 / 8}}
$$

I would start using l'Hôpital's rule and notice the pattern, is there some other procedure? (without knowing it)

答:
It is standard that, for any $\alpha,\beta>0$, $\;\ln^\alpha(x)=o\bigl(x^\beta\bigr)$ as $x\to\infty$.

This is very simple to prove: set $u=x^{\tfrac \beta\alpha}$. We have
$$\frac{\ln^\alpha(x)}{x^\beta}=\frac{\ln^\alpha\Bigl(u^{\tfrac\alpha\beta}\Bigr)}{u^\alpha}=\underset{\strut\text{constant}}{\biggl(\frac\alpha\beta\biggr)^{\mkern-5mu\alpha}} \biggl(\underbrace{\frac{\ln u}u}_{{}\to 0}\biggr)^{\mkern-5mu\alpha}.$$


问:
What is the fastest algorithm to solve the eigenvector of a transition matrix of a Markov Chain?
Given a transition matrix of a Markov chain, $P$, I want to solve the left eigenvector of $P$, namely a row vector $\alpha$ such that
$$
\alpha P = \alpha
$$

I know the algorithm to solve a linear equation takes $O(n^3)$, using LU decomposition.

I wonder if there is any faster algorithm?

By the way, I may not need the exact solution, approximate one is OK.

答:
Because this is a transition matrix, the stationary vector $\alpha$ can be found by merely computing successive powers of $P$ against a random starting vector; this is the power method. In practice, for a large $P$, the matrix is sparse and it is easy to calculate the map $x \mapsto Px$, so you can use the Arnoldi algorithm.

In general, you can never get an exact solution to the eigenvalue problem, in that you'll never be able to find a closed form algebraic expression; this is because solving for eigenvalues of an $n \times n$ matrix is equivalent to finding the roots of its $n$th degree characteristic polynomial.


问:
How many words can be created by using the letters of "MATHEMATIK" when no same letters can be next to each other?
The word MATHEMATIK (in german) has 10 letters.

I want to know how many different words can be created by using the letters of this word. However, the same letter can never be next to each other.

So for example, MAATHEMTIK is not allowed.

I did the following:

$n = 10$

$k_1 = 2$ (M)

$k_2 = 2$ (A)

$k_3 = 2$ (T)

$k_4 = 1$ (H)

$k_5 = 1$ (E)

$k_6 = 1$ (I)

$k_7 = 1$ (K)

I use the formula $\frac{n!}{k_1! \cdot k_2! ... \cdot k_n! }$ and get:

$$\frac{13!}{2! \cdot 2! \cdot 2! \cdot 1! \cdot 1! \cdot 1! \cdot 1!} = 778377600$$

This is the number of all possible words (permutations). Now we need to subtract the number of words which can be created by having the same letter next to each other, i.e.

MAATHEMTIK

AAMTHEMTIK

MTAAHEMTIK

etc.

But I don't know how to figure that out.

答:
Please apply Principle of Inclusion Exclusion. It is a more standard approach.

Your answer would simply be

$\displaystyle \frac{10!}{2! \ 2! \ 2!} - \binom{3}{1} \ \frac{9!}{2! \ 2!} + \binom{3}{2} \ \frac{8!}{2!} - \binom{3}{3} \ 7! = 236880$

Some details -

The first term is unrestricted number of arrangements. Now we subtract from it all arrangements where either of $AA, MM, TT$ are together. If you take $AA$ together as one letter, you now have $9$ letters and the number of arrangements are $\frac{9!}{2! \ 2!}$. Now you have same number of arrangements for $MM$ and $TT$. But these overcount arrangements where two identical letters (say AA, MM) are together. As we subtracted them, we need to add them back. The same follows for the next.


问:
Computing $\exp(tT)$ where $T(b)=\{\frac{1}{2m}p^2,b\}$
I've recently been reading Baez's paper on Noether's theorem and I was trying to test the simplest case I could think of: a mass moving in one dimension in a space with potential $0$. The Hamiltonian in this case would just be the kinetic energy $H(q,p)=\frac{1}{2m}p^2$.  The manifold is $M=\mathbb R$, the cotangent bundle is $T^\ast M =\mathbb R\times \mathbb R$.

I'm doing this to test out and reconcile the usual statement of Noether's theorem for Hamiltonians with Baez's exposition using observables (elements of $C^\infty(T^\ast M)$) which form a Poisson algebra under the bracket $\{a,b\}:=\frac{\partial a}{\partial p}\frac{\partial b}{\partial q}-\frac{\partial a}{\partial q}\frac{\partial b}{\partial p}$.

Now translation by $t$ is a one-parameter subgroup $\Phi_t(q)=q+t$ of diffeomorphisms of $M$. According to the classical statement of Noether's theorem, the conserved quantity is $I(q,p)=\left. p\cdot \frac{d\Phi_t(q)}{d t}\right|_{t=0}=p$, which makes complete sense: momentum is conserved.

In terms of Baez's exposition, $H$ and $I$ are two observables such that $\{H,I\}=\{0\}$, and according to his formulation of Noether's theorem, $H$ should generate symmetries of $I$ and $I$ should generate symmetries of $H$. In this context, given two observables $a,b$, the observable $a$ is said to generate symmetries of $b$ if there's a flow $F^a_t:C^\infty(T^\ast M)\to C^\infty(T^\ast M)$ such that $\frac{d}{dt}F^a_t(b)=\{a,F^a_t(b)\}$ and $F_t^a(b)=b$.  In this case, $F^a_t$ can be computed as $\exp(tT)$ where $T(c):=\{a,c\}$.

Now one quickly discovers that for $a=I$ above, $T=\frac{\partial}{\partial q}$. After a brief search I seemed to find something which I think says that $\exp(tT)(b)(q,p)=b(q+t, p)$.  This makes sense because $b(q+t, p)$ is the result of lifting the original one-parameter group to $T^\ast M$ results in $\Phi_t^\ast(q,p)=(q+t,p)$ and then lifting it to be on $C^\infty(T^\ast M)$ just uses precomposition.  So in short, $\exp(tT)$ corresponds exactly with my original flow $\Phi_t$.

Finally, to test the symmetry of the relationship between $H$ and $I$ fully, I wanted to compute the symmetries of $I$ that are generated by $H$. To do so, we'd need to look at $T(c):=\{H, c\}$ which I believe works out to $T(c)=\frac{p}{m}\frac{\partial c}{\partial q}$, so $T=g\frac{\partial}{\partial q}$ where $g$ is the function $g(q,p)=\frac{p}{m}$.

The problem is, I have no idea how to get a handle on what $\exp(tT)$ is in that case. At least I'm interested in knowing what the flow on $T^\ast M$ looks like.

Is there some way to sort it out to something recognizable? Or is it unexpectedly complicated?

Update

I started to wonder if it isn't just $\exp(tT)(h)=h(q+t\frac{p}{m}, p)$. It feels heuristically right, since $t\frac{p}{m}$ is as constant with respect to $q$ as $t$ is alone.  But more importantly, the flow $\Phi_t(q)=q+t\frac{p}{m}=q+t\dot q$ makes a ton of sense.  It's just the displacement after time $t$.

Also importantly, its cotangent lift is $(q,p)\mapsto (q+\frac{p}{m}, p)$ again, and that fixes the observable $I$ above, as hoped.

However, when I compute its conserved quantity, $I'(q,p)=\left. p\cdot \frac{d\Phi_t(q)}{d t}\right|_{t=0}=p\cdot \frac{p}{m}$ which is twice the original Hamiltonian.  I suppose that modifying an observable by scaling does not change the level sets, and so it tells you the same information.

I'm still sort of looking for confirmation of whether or not my conclusions are on solid ground.

答:
Your conclusions are essentially correct. Some useful operator identities to have around in this context are

$$\exp\left( a \frac{d}{dx} \right)(f)(x) = f(x+a), \qquad \exp\left( bx \frac{d}{dx} \right)(f)(x) = f(e^b x)$$

when $a, b$ do not depend on $x$, and their generalization to multiple variables

$$\exp\left( \mathbf{a} \cdot \nabla \right)(f)(\mathbf{x}) = f(\mathbf{x}+\mathbf{a}), \qquad \exp\left( \mathbf{x}^\top B^\top \nabla \right)(f)(\mathbf{x}) = f(e^B \mathbf{x}).$$

where $\mathbf{a}$ is a vector and $B$ is a matrix (and $e^B$ denotes a matrix exponential). The latter identity becomes particularly useful when dealing with rotational or dilational symmetries. In fact, an alternative way to arrive at the expression in your question is to note that

$$T = \frac{p}{m} \frac{\partial}{\partial q}=\mathbf{z}^\top \begin{pmatrix} 0 & 0\\ \frac{1}{m} & 0\end{pmatrix} \nabla_{\mathbf{z}},$$

where $\mathbf{z} = (q, p)^\top$, so that

$$\exp\left(tT \right)(h)(\mathbf{z}) = h\left(\exp\begin{pmatrix} 0 & \frac{1}{m}\\ 0 & 0\end{pmatrix}\: \mathbf{z} \right) = h\left(\begin{pmatrix} 1 & \frac{1}{m}\\ 0 & 1\end{pmatrix} \mathbf{z} \right) = h\left(q+\frac{p}{m},p\right)$$

using the identity above and the fact that $e^B =I+B$ for $B$ nilpotent. In physics this kind of transformation is called a Galilean transformation or Galilean boost, and can be thought of as the limiting case of a relativistic pseudo-rotation when we formally send the speed of light $c$ to $\infty$.


问:
Transformation to complex spherical basis.
According to this Wikipedia page about the complex spherical basis, it is stated that the spherical basis vectors are written in terms of the Cartesian basis vectors the following way:

$$
\left(
\begin{array}{c}
e_{+} \\
e_{-} \\
e_{0}
\end{array}
\right)
=
\left(
\begin{array}{ccc}
-\frac{1}{\sqrt{2}} & -\frac{i}{\sqrt{2}} & 0 \\
+\frac{1}{\sqrt{2}} & -\frac{i}{\sqrt{2}} & 0 \\
0 & 0 & 1
\end{array}
\right)
\left(
\begin{array}{c}
e_{x} \\
e_{y} \\
e_{z}
\end{array}
\right)
=
\textbf{U}
\left(
\begin{array}{c}
e_{x} \\
e_{y} \\
e_{z}
\end{array}
\right)
$$

However, the "components" of a vector transform by:

$$
\left(
\begin{array}{c}
A_{+} \\
A_{-} \\
A_{0}
\end{array}
\right)
=
\left(
\begin{array}{ccc}
-\frac{1}{\sqrt{2}} & +\frac{i}{\sqrt{2}} & 0 \\
+\frac{1}{\sqrt{2}} & +\frac{i}{\sqrt{2}} & 0 \\
0 & 0 & 1
\end{array}
\right)
\left(
\begin{array}{c}
A_{x} \\
A_{y} \\
A_{z}
\end{array}
\right)
=
\textbf{U*}
\left(
\begin{array}{c}
A_{x} \\
A_{y} \\
A_{z}
\end{array}
\right)
$$

I understand that this is what you get by substituting the components of the vectors in the basis definition but, is there a better way to explain why the transformation matrix is conjugated when I go from transforming the basis vectors to transforming the "components" of a vector?

答:
Let $B_c$ denote the Cartesian basis as a column vector. Let $B_s$ be the spherical basis. Then we have $B_s=U B_c.$ If $v$ is a vector, let $v_c$ and $v_s$ be the coordinate representation in each basis. Then
$$
{B_c}^T v_c={B_s}^T v_s=(UB_c)^T v_s={B_c}^T U^T v_s.
$$
But because $B_c$ is a basis this gives $v_c=U^T v_s$ or $v_s=(U^T)^{-1} v_c.$ But because $U$ is unitary,
$$
(U^T)^{-1}=(U^{-1})^T=((U^*)^T)^T=U^*.
$$


问:
Are there any non-casework proofs of the $18$-point problem?
There is a cute little problem stated as follows:

Choose a sequence $x_1,x_2,x_3,\ldots$ with $x_i\in[0,1)$ such that $x_1$ and $x_2$ are in different halves of the unit interval, $x_1,x_2,$ and $x_3$ are in different thirds, and so on. How long can such a sequence be?

It turns out the answer is exactly $17$; it is not possible to find a length-$18$ sequence obeying these constraints. See Wolfram Mathworld for more.

Berkelamp and Graham's original 1970 paper proves this with a rather involved casework argument on the relative positions of various terms in the sequence, though later in the paper they present a non-casework argument that implies the sequence can be at most $63$ terms long.

I am curious whether a more direct or "natural" argument has been developed in the 50 years hence; I wouldn't be surprised if getting an exact bound ultimately required a fairly casework-y approach, but it seems plausible that at least better upper bounds could be reached with some other method. Is anything more known about this problem?

答:
"A supplementary note on the irregularities of distributions" by M. Warmus (1976) gives a simpler proof. In fact, Warmus may have been the first to solve the problem, though not the first to publish. The paper states

Berlekamp and Graham [2] published a paper in which they considered the same problem, although in a generalized form, and they gave a proof that $17$ is the greatest possible $N$. They also mention my name as the author of this result. It is not true, however, that I solved the problem with the aid of a computer.

Since my proof is very simple and seems to be shorter than the one given by Berlekamp and Graham, I give it here. It has not been published before.

I think the paper should be accessible by everyone by the link above, since I can access it and I'm at home. The argument is still casework, but short casework: it considers only six cases of $x_9^5$, defined to be the unique element of $\{x_1, x_2, \dots, x_9\}$ which lies in the interval $[\frac49, \frac59)$. Without loss of generality, $x_9^5 \le \frac12$, so we split the possible values as $$x_9^5 \in [\tfrac49, \tfrac5{11}) \cup [\tfrac5{11}, \tfrac6{13}) \cup [\tfrac6{13},\tfrac7{15}) \cup [\tfrac7{15},\tfrac8{17}) \cup [\tfrac8{17}, \tfrac12) \cup \{\tfrac12\}.$$
Each of these cases tells us something about the other similarly-defined $x_n^k$ that ultimately leads to a contradiction, and in fact it seems like all intervals other than $[\frac5{11}, \frac6{13})$ lead to a contradiction even for $N=17$.


问:
Real rank of $C_0(R)$
I was trying to calculate Real rank of $C_0(\mathbb{R} \otimes \mathbb{K}(l^2))$. $\mathbb{K}$, compact operators algebra, is inductive limit of matrix algebras. So, $RR(C_0(\mathbb{R} \otimes \mathbb{K}(l^2)) = lim_{\rightarrow}RR (C_0(\mathbb{R} \otimes \mathbb{M_n})) = C_0(\mathbb{R})$. But what the real rank of $C_0(\mathbb{R})$?

答: I think I found answer: If $X$ is a locally compact Hausdorff space, then $RR(C_0(X)) = dim(X)$, where $dim$ is the Lebesgue covering dimension of $X$.

Title: Real rank of $C_0(R)$


答:
Here is an attempt to combine the classical proof for the case $R=\mathbb{Z}$ (see e.g. Thm. 2.44 in Hatcher) with @diracdeltafunk's suggestion to work with the field of fractions.

Definition. Let $R$ be an integral domain, and let $F=\operatorname{Frac}(R)$ be its field of fractions. The rank of an $R$-module $M$ is then defined as $\operatorname{rk} M=\dim_F (M\otimes_R F)$.

Remark. If $R$ is a PID and $M$ is a finitely generated $R$-module with $M\cong R^r\oplus R/(a_1)\oplus \cdots\oplus R/(a_m)$, then $M\otimes_R F\cong F^r$ and we get $\operatorname{rk} M=r$. In other words: this definition of rank coincides with the usual definition of rank for finitely generated modules over a PID.

Lemma. Let $R$ be an integral domain, and let $0\to A\to B\to C\to 0$ be a short exact sequence of finitely generated $R$-modules. Then $\operatorname{rk} B=\operatorname{rk} A+ \operatorname{rk} C$.

Proof. Note that $F$ is a flat $R$-module (this follows for example from the more general fact that localization is an exact functor). Hence, the induced sequence
$0\to A\otimes_R F\to B\otimes_R F\to C\otimes_R F\to 0$ of $F$-vector spaces is exact. Since every short exact sequence of vector spaces splits, this gives $B\otimes_R F\cong (A\otimes_R F)\oplus(C\otimes_R F)$. $\:\square$

Proposition. Let $\chi_R$ be defined as in the original post (but for an arbitrary integral domain $R$, with field of fractions $F$). Then $\chi_R=\sum_{i\in\mathbb{Z}} (-1)^i\,a_i$.

Proof. Let $d_i^R=d_i\otimes \mathrm{id}_R$ denote the differentials of the chain complex $C_{\bullet}\otimes_{\mathbb{Z}} R$. Note that we for every index $i\in\mathbb{Z}$ have short exact sequences
$$0\to\operatorname{im}(d_{i+1}^R)\hookrightarrow \ker(d_{i}^R)\to H_i(C_\bullet\otimes_\mathbb{Z} R)\to 0\,,$$
$$0\to\operatorname{ker}(d_{i}^R)\hookrightarrow C_i\otimes R \to \operatorname{im}(d_{i}^R)\to 0\,.$$
By the lemma, this gives rise to equalities
$$\operatorname{rk}\ker(d_{i}^R)=\operatorname{rk}\operatorname{im}(d_{i+1}^R)  + \operatorname{rk} H_i(C_\bullet\otimes_\mathbb{Z} R)\,,$$
$$\operatorname{rk}(C_i\otimes_\mathbb{Z} R)=\operatorname{rk}\operatorname{ker}(d_{i}^R)  + \operatorname{rk}\operatorname{im}(d_{i}^R)\,,$$
which implies
$$a_i=\operatorname{rk}(C_i\otimes_\mathbb{Z} R)=\operatorname{rk} H_i(C_\bullet\otimes_\mathbb{Z} R)+\operatorname{rk}\operatorname{im}(d_{i+1}^R)  +  \operatorname{rk}\operatorname{im}(d_{i}^R)\,.$$
The desired result now follows by computing the alternating sum over all $i\in\mathbb{Z}$. $\:\square$

Title: How does the coefficient ring influence the Euler characteristic?

Body:
Notation/background. Let $C_\bullet$ be a bounded chain complex of finitely generated abelian groups, with $C_i\cong\mathbb{Z}^{a_i}$. For any principal ideal domain $R$, let $C_\bullet\otimes R$ be the chain complex of finitely generated $R$-modules obtained by applying the functor $(-)\otimes_\mathbb{Z}R\colon \mathbf{Ab}\to R\mathbf{Mod}$ termwise. Furthermore, let $$\chi_R=\sum_{i\in\mathbb{Z}}(-1)^i\operatorname{rank} H_i(C_\bullet\otimes R)$$ be the Euler characteristic of $C_\bullet\otimes R$. It's a classical exercise in homological algebra (see e.g. Theorem 2.44 in Hatcher) to show that $$\chi_{R}=\sum_{i\in\mathbb{Z}} (-1)^i {a_i}$$ whenever $R$ is a field and whenever $R=\mathbb{Z}$.

Question. What happens when $R$ is some other PID? Do we still get $\chi_{R}=\sum_{i\in\mathbb{Z}} (-1)^i{a_i}$?

Attempt/thoughts. Suppose that $H_i(C_\bullet)\cong\mathbb{Z}^{b_i}\oplus T_i$, where $T_i$ is the torsion subgroup of $H_i(C_\bullet)$. The universal coefficient theorem then tells us that
$$H_i(C_\bullet\otimes R)\cong (H_i(C_\bullet)\otimes _{\mathbb{Z}}R)\oplus \mathrm{Tor}_1^{\mathbb{Z}}(H_{i-1}(C_\bullet),R)\\ \cong R^{b_i}\oplus (T_{i}\otimes_{\mathbb{Z}} R)\oplus \mathrm{Tor}_1^{\mathbb{Z}}(T_{i-1},R)\,.$$
When $R$ is a field (see this thread),  then all free copies of $R$ obtained from the tensor products with $T$ are cancelled by free copies of $R$ obtained from the Tor-terms when we compute the alternating sum. It intuitively doesn't feel very likely that things would work out as nicely when $R$ is a general PID, but I find it hard to come up with a counterexample...

Does anyone have any ideas or suggestions about how to approach this?


问:
Derivative of $(Ax)^T Ax$
This derivative comes up in the differentiation of the euclidean norm and I can't seem to find an appropriate rule to apply that the dimensions match. I know the solution is:

$$ \frac{\partial}{\partial x}\left((A x)^{\top} A  x\right)=2  A^{\top} A x $$

I tried applying the product rule and swapping the transpose with the derivative:

\begin{align}
 \frac{\partial}{\partial x}\left((A x)^{\top} A  x\right) &= \frac{\partial}{\partial x} \left( (Ax)^{\top} \right) Ax  + (Ax)^{\top} \frac{\partial}{\partial x}\left( Ax \right) \\&= A^{\top} Ax  + (Ax)^{\top} \frac{\partial}{\partial x}\left( Ax \right)
\end{align}

The left term seems to be correct but the right term is bothering me because the dimensions don't match.

I'm looking for some rule that I can apply and remember in case I ever see a similar derivative. I looked through the Matrix Cookbook and googled but couldn't find anything.

I also came across this but the answers didn't really explain it to my satisfaction.

Thanks in advance!

答: It's better to work componentwise with Einstein notation:$$\frac{\partial}{\partial x_i}(Ax)_j(Ax)_j=2(Ax)_j\frac{\partial}{\partial x_i}(A_{jk}x_k)=2(Ax)_jA_{ji}=2(A^TAx)_i,$$so the gradient is $2A^TAx$.

Title: Derivative of $(Ax)^T Ax$


问:
Inequality $\frac{1}{2028}<\frac{a}{2027+a}+\frac{b}{2026+b²}+\frac{c}{2024+c⁴}+\frac{d}{2020+d^8}<\frac{1}{2019}$
Given:

$a+b+c+d=1$

$a,b,c,d∈R_+$

I want to prove or refute the following inequality:

$\frac{1}{2028}<\frac{a}{2027+a}+\frac{b}{2026+b²}+\frac{c}{2024+c⁴}+\frac{d}{2020+d^8}<\frac{1}{2019}$

This is my first inequality proving, and I've not any idea how to do that. Could you give me some tip and advise book for the topic?

Body:
Since $0\leq a, b, c, d\leq 1$, we see
$$\begin{aligned}
\frac{a}{2027+a}+\frac{b}{2026+b²}+\frac{c}{2024+c⁴}+\frac{d}{2020+d^8}
> & \frac{a}{2027+1}+\frac{b}{2026+1}+\frac{c}{2024+1}+\frac{d}{2020+1}\\
\geq & \frac{a}{2028}+\frac{b}{2028}+\frac{c}{2028}+\frac{d}{2028}\\
=& \frac{a+b+c+d}{2028}=\frac{1}{2028},
\end{aligned}$$
and
$$\begin{aligned}
\frac{a}{2027+a}+\frac{b}{2026+b²}+\frac{c}{2024+c⁴}+\frac{d}{2020+d^8}
< & \frac{a}{2027}+\frac{b}{2026}+\frac{c}{2024}+\frac{d}{2020}\\
\leq & \frac{a}{2020}+\frac{b}{2020}+\frac{c}{2020}+\frac{d}{2020}\\
=& \frac{a+b+c+d}{2020}=\frac{1}{2020}<\frac{1}{2019}.
\end{aligned}$$


问:
Determine cartesian coordiantes of the points on the polar curve, furthest from the origin.
Let $K$ be the level curve $f(x,y)=0$ of $f(x,y)=(x^2+y^2)^2-6x^2y$

$K$ is given in polar coordinates: $r=3\sin(2\theta)\cos(\theta)$, $\theta \in[0,\pi]$

Determine the cartesian coordinates of the points on $K$, furthest from the origin.

I've been trying for quite some time and have no clue what direction to look in. I tried some things with the gradient of $f(x,y)$, the derivative of $r$. I found the critical points and plugged them into the polar equation, but when graphing, it was clear those weren't the points. I'm truly lost here.

Note: This was an ex-exam question so I do not have a solution for this question.

答:
$$r=3\sin(2t)\cos(t)$$
The maximum of $r$ is when $r'=0$ and $r''<0$

$r'=6 \cos (t) \cos (2 t)-3 \sin (t) \sin (2 t)=3 \cos (t) (3 \cos (2 t)-1)$

$r'=0\to 3 \cos (t) (3 \cos (2 t)-1)=0$

$\cos t=0\to t =\frac{\pi}{2}$

$3 \cos (2 t)-1=0\to \cos 2t=\frac{1}{3}\to t=\frac{1}{2}\arccos\left(\frac{1}{3}\right)\approx 35° $

To show it is an actual maximum compute second derivative

$r''=-12 \sin (t) \cos (2 t)-15 \sin (2 t) \cos (t)=-\frac{3}{2} (\sin (t)+9 \sin (3 t))$

At $t=\frac{\pi}{2}$ $r''=12$, $r=0$ minimum

At $t=\frac{1}{2}\arccos\left(\frac{1}{3}\right)$, $r''\approx -13.9$ thus it is a maximum.

$r=2 \sqrt{2} \cos \left(\frac{1}{2} \arccos\left(\frac{1}{3}\right)\right)$

$t=\frac{1}{2}\arccos\left(\frac{1}{3}\right)$

To compute cartesian coordinates we plug back the values we have found in the equations

$x=r\cos t;\;y=r\sin t$

$x=2 \sqrt{2} \cos ^2\left(\frac{1}{2} \arccos\left(\frac{1}{3}\right)\right),y=2 \sqrt{2} \cos \left(\frac{1}{2} \arccos\left(\frac{1}{3}\right)\right) \sin \left(\frac{1}{2} \arccos\left(\frac{1}{3}\right)\right)$

Using trig bisection formula  we arrive to

$$x=\frac{4 \sqrt{2}}{3},y=\frac{4}{3}$$


问:
Let $f : N → Y$ be a map having right inverse $g$. Prove that $Y$ is at most countable?
I know this implies $F$ is surjective, but not sure if this helps.

答: In this case, $g$ is an injection from $Y$ to $\mathbb N$, and then there is a bijection between $Y$ and $\operatorname{im} g \subseteq \mathbb N$. Since  every subset of $\mathbb N$ is at most countable, the result follows.


答: Show that all the denominators are bounded between $2027+a$ and $2020$.  For the left one, reduce the numbers in the middle by replacing the denominators by the larger $2027+a$.  For the right one, increase the numbers in the middle by replacing the denominators with the smaller $2020$.

Title: Inequality $\frac{1}{2028}<\frac{a}{2027+a}+\frac{b}{2026+b²}+\frac{c}{2024+c⁴}+\frac{d}{2020+d^8}<\frac{1}{2019}$

Body:
Given:

$a+b+c+d=1$

$a,b,c,d∈R_+$

I want to prove or refute the following inequality:

$\frac{1}{2028}<\frac{a}{2027+a}+\frac{b}{2026+b²}+\frac{c}{2024+c⁴}+\frac{d}{2020+d^8}<\frac{1}{2019}$

This is my first inequality proving, and I've not any idea how to do that. Could you give me some tip and advise book for the topic?


问:
How can we say later that x can be equal to zero when we took x as a common factor and hence divided by it?
To solve an equation like this:

$x^2+x=0$

We can take $x$ as a common factor

$\frac{x^2}{x}+\frac{x}{x}=0$

$x(x+1)=0$

However to do such a step we have to say that $x$ isn't equal to zero, because after all, if $x$ is equal to zero, then $\frac{x}{x} \neq1$, like we factorized. Instead our factorization should look like this:

$x(∞)=0$

And even though we stated clearly that $x$ can't be equal to zero above, later on to get the solutions we state:

$x=-1$ $or$ $x=0$

I know this is a very silly question and that's why I didn't want to ask my teacher about it. And I totally get that it doesn't matter in this equation anyway.

But shouldn't this be considered wrong, to state that $x$ shouldn't be a certain value to reach a certain step then when we are over that step and we reach another step we say that $x$ is that certain value we based our work on the fact that it shouldn't be it.

答:
The 4 biggest words in math are If, Then, Or, And.  For practice, write  whole math arguments  in complete grammatically correct sentences and do not omit the big 4. And do not omit any justifications.

Illogical: If $x^2+x=0$ then $x+1=(x^2+x)/x=0/x=0.$

Logical: If $x^2+x=0$ and if $x\ne 0$ then $x+1=(x^2+x)/x=0/x=0.$

"Illogical" includes an unwarranted hidden assumption that division by $x$ is possible. But "Logical" above still doesn't answer the whole  original Q, because it includes an extra "if" (i.e. if $x\ne 0$) that's not in the original Q.

Complete: $x^2+x=0\iff x(x+1)=0$ [by the Distributive Law] $\iff (x=0\lor x+1=0)$ [by a basic axiom of arithmetic] $\iff (x=0\lor x=-1)$ [Let's not get too tedious; you can skip the justification for this].

It is especially useful when you can manage to have all "$\iff$", as then the argument reads logically in reverse order.


Body:
Note that $\frac{a}{2027+a} > \frac{a}{2028}$, $\frac{b}{2026+b^2} > \frac{b}{2028}$,$\frac{b}{2024+c^4} > \frac{c}{2028}$, $\frac{b}{2020+d^4} > \frac{d}{2028}$ for all $a,b,c,d \in (0,1)$.

Likewise, note that $\frac{a}{2027+a} < \frac{a}{2019}$, $\frac{b}{2026+b^2} > \frac{b}{2019}$,$\frac{b}{2024+c^4} > \frac{c}{2019}$, $\frac{b}{2020+d^4} > \frac{d}{2019}$ for all $a,b,c,d \in (0,1)$.

Then use the fact that $a+b+c+d=1$ to conclude.


问:
Does the integral $\int_{2}^{\infty}\frac{dx}{\sqrt{1+x^{3}}}$ exists?
I need to prove this. I need your help to verify that my proof is correct (or not) please.

Prove that this integral exists: \begin{align}
 \int_{2}^{\infty}\frac{dx}{\sqrt{1+x^{3}}} \end{align}

My attempt:

Fist we need to observe that $\frac{1}{\sqrt{1+x^{3}}}<\frac{1}{\sqrt{x^{3}}}\Longrightarrow \int_{2}^{\infty}\frac{dx}{\sqrt{1+x^{3}}}<\int_{2}^{\infty}\frac{dx}{\sqrt{x^{3}}}$

Next, we note that $\lim_{x \rightarrow \infty} \frac{\frac{1}{\sqrt{1+x^{3}}}}{\frac{1}{\sqrt{x^{3}}}}=1 \Longrightarrow$ for $f(x)=\frac{1}{\sqrt{1+x^{3}}}$, $g(x)=\frac{1}{\sqrt{x^{3}}}$ both integrals converges or both diverges.

By last, integrals of the form $\int_{1}^{\infty}\frac{dx}{x^{p}}$ converges if $p>1$, $\Longrightarrow \int_{1}^{\infty}\frac{dx}{\sqrt{x^{3}}}$  converges $\Longrightarrow \int_{2}^{\infty}\frac{dx}{\sqrt{x^{3}}}$ converges

That implies that, $\int_{2}^{\infty}\frac{dx}{\sqrt{1+x^{3}}}$ converges, therefore it exists.

Is it correct? Is there another way to prove it? Thank you very much

答: Your solution is correct, though you didn't really need to use the limit comparison test. You could just stop after the first line. Since $\frac{1}{\sqrt{1+x^3}}\leq\frac{1}{\sqrt{x^3}}$ and the integral $\int_2^{\infty}\frac{1}{\sqrt{x^3}}dx$ converges, we know from the usual comparison test that $\int_2^{\infty}\frac{1}{\sqrt{1+x^3}}dx$ converges as well. Of course it is important to note that these are nonnegative functions in the solution.


问:
Prove that $\mathbb{Z}$ with the finite complement topology is not Hausdorff.
Prove that $\mathbb{Z}$ with the finite complement topology is not Hausdorff.

What are the closures of the following two subsets of $\mathbb{Z}$ with the finite complement topology? (Fully justify your answers.)

i) $A=\{1,2,3\}$ $\,\,\,\,\,$ ii) $\mathbb{N}=\{1,2,3,4,...\}$

$\textbf{Proof:}$ By way of contradiction, assume $\mathbb{Z}$ with the finite complement topology is Hausdorff. Then $\forall a \neq b \,\, \exists U,V \subset \mathbb{Z}$ where $U \cap V = \emptyset$ with $a \in U$, $b \in V$, and $U,V$ are open sets.

Since $a,b \in \mathbb{Z}$, then pick $a=0$ and $b=1$. Since $U,V \subset \mathbb{Z}$ are open in the finite complement topology, then $U^c$,$V^c$ are finite.

Thus, $U,V$ are infinite. So $V \not \subset U^c$. Since $1 \in V$, then $1 \not \in V^c$, so $1 \in (U^c)^c=U$ by infinite/finite sets.

But, $U \cap V = \emptyset$. $\rightarrow \leftarrow$ (contradiction!)

$\therefore$ By way of contradiction, $\mathbb{Z}$ with the finite complement topology is not Hausdorff.

(i) $A=\{1,2,3\}$

$A \subset \mathbb{Z}$ with the finite complement topology. So, $\mathbb{Z}\backslash A$ is open $\implies A$ is closed.

$\therefore A=\overline{A}$.

(ii) $\overline{\mathbb{N}}=\mathbb{N} \cup \mathbb{N}^{'}=\mathbb{N}$ where $\mathbb{N}^{'}$ is the set of all limit points of $\mathbb{N}$.

I am checking if my proof is valid and could use any feedback on it.
Any help with part (ii) would be nice. I am unsure how to provide a further explanation.

答:
Any infinite set with the cofinite topology is $T_1$ but not $T_2$.  The proof that it is not Hausdorff is given, in essence,  by @BrianM.Scott above. (The gist of the argument is that two cofinite subsets of an infinite set have to intersect nontrivially.)

For part $ii)$ you need a correction: $\bar{\Bbb N}=\Bbb Z$.  The only proper closed subsets are the finite ones.


答:
There are two ways to solve your equation. First, we could deal with the cases separately. If $x \neq 0$, then we can safely divide by $x$, meaning that
$$
x^2+x=0
$$
implies
$$
x+1=0
$$
which in turn implies that $x=-1$. Then, we have to deal with the case that $x=0$. This is also a solution to your equation—we can verify this by plugging it in.

There is a second, more elegant way of solving the equation, that you yourself have alluded to. If $x^2+x=0$, then
$$
x(x+1)=0 \, ,
$$
which implies $x=0$ or $x=1$. Note that there is no division by $x$ here at all. Finally, you appear to suggest at one point in your question that $0/0 = \infty$. This is not true. In fact, $0/0$ is undefined. It is as meaningless to talk about $0/0$ as it is to talk about $\text{yellow}/\text{blue}$.

Title: How can we say later that x can be equal to zero when we took x as a common factor and hence divided by it?

Body:
To solve an equation like this:

$x^2+x=0$

We can take $x$ as a common factor

$\frac{x^2}{x}+\frac{x}{x}=0$

$x(x+1)=0$

However to do such a step we have to say that $x$ isn't equal to zero, because after all, if $x$ is equal to zero, then $\frac{x}{x} \neq1$, like we factorized. Instead our factorization should look like this:

$x(∞)=0$

And even though we stated clearly that $x$ can't be equal to zero above, later on to get the solutions we state:

$x=-1$ $or$ $x=0$

I know this is a very silly question and that's why I didn't want to ask my teacher about it. And I totally get that it doesn't matter in this equation anyway.

But shouldn't this be considered wrong, to state that $x$ shouldn't be a certain value to reach a certain step then when we are over that step and we reach another step we say that $x$ is that certain value we based our work on the fact that it shouldn't be it.


问:
Why does convergence in distribution not imply convergence in mean square sense?
I know convergence in mean square sense implies convergence in distribution. But a distribution defines the moments of the random variable. Given the convergence in distribution, why don't we observe convergence in the moments? Should not the reverse also hold going by this logic?</p>

答:
Two things:

Convergence in mean square (also called convergence in $L^2$) doesn't mean that $\mathbb{E}X^2_n\rightarrow \mathbb{E}X^2$ but that $\mathbb{E}\left(X-X_n\right)^2\rightarrow 0$ (equivalently: $\lVert X-X_n\rVert_2=\sqrt{\mathbb{E}(X-X_n)^2}\rightarrow 0$). One implication of this is that moments converge (as a consequence of the reverse triangular inequality), but it is also stronger because it says that the expected square distance between the sequence and its limit goes to $0$.

Convergence in distribution doesn't imply convergence of the moments. A classic counterexample is to take for all $n\geq 0$ the random variable $X_n$ with $\mathbb{P}(X_n=2^n)=2^{-n}$ and $\mathbb{P}(X_n=0)=1-2^{-n}$. It can be shown that this sequence converges in probability (therefore in distribution) to $0$ but for all $n\geq 0$ we have $\mathbb{E}X_n=1\neq 0$. Note that, by the first point, this means that $(X_n)_{n\geq 0}$ does not converge to $0$ in $L^2$.


问:
Could exists a vector field on $\mathbb{S}^{2}$ with exactly $n$ zeroes?
I just started to learn index theory of tangent vector fields. I'm aware of two examples on the sphere $\mathbb{S}^{2}$ with exactly one zero, which, which are $F(x,y) = (1-x^2-y^2)\partial x$ thought on $\mathbb{D}^2$ and then indentify the boundary, I think it is

And one with two zeroes should be something like this

I was wondering, it exists a tangent vector field with $n$ zeroes on the sphere ? I thought that if I have $n$ zeroes (isolated), then thanks to the Poincarè-Hopf theorem the sum of the indices must be  $\chi(\mathbb{S}^{2}) = 2$. I thought that some combination of $-1$ and $1$ and a final zero of index $2$ could do the job, but can this vector field actually exist?
Something like defining a polynomial with $n$ distinct zeroes $p : \mathbb{C} \longmapsto \mathbb{C}$ can be extended to $\mathbb{S}^{2}$ to gain a tangent vector field on the sphere?

Thanks in advance, any help would be appreciated.

答: This is a somewhat boring answer, but I believe you can always construct such a vector field. Given a vector field on the plane, you can artificially put in zeros of index $0$ by multiplying it with a "dent" function, for example something like $1-e^{-r^2}$. Thus you can take the constant vector field, put in $n$ "dents" wherever you please, then stereographically wrap it back onto the sphere, which will give you an extra zero of index $2$ at the north pole.


答:
Yes, you compute a limit that is not necessary to compute. You just need to bound your integrand by something that is integrable (like you did):

$$
0\leq \int_2^{+\infty} \frac{1}{\sqrt{1+x^3}} dx \leq \int_2^{+\infty} \frac{1}{x^{3/2}} dx
$$
and the last integral is finite. Therefore also the one in the middle of the inequalities is finite.

Title: Does the integral $\int_{2}^{\infty}\frac{dx}{\sqrt{1+x^{3}}}$ exists?

Body:
I need to prove this. I need your help to verify that my proof is correct (or not) please.

Prove that this integral exists: \begin{align}
 \int_{2}^{\infty}\frac{dx}{\sqrt{1+x^{3}}} \end{align}

My attempt:

Fist we need to observe that $\frac{1}{\sqrt{1+x^{3}}}<\frac{1}{\sqrt{x^{3}}}\Longrightarrow \int_{2}^{\infty}\frac{dx}{\sqrt{1+x^{3}}}<\int_{2}^{\infty}\frac{dx}{\sqrt{x^{3}}}$

Next, we note that $\lim_{x \rightarrow \infty} \frac{\frac{1}{\sqrt{1+x^{3}}}}{\frac{1}{\sqrt{x^{3}}}}=1 \Longrightarrow$ for $f(x)=\frac{1}{\sqrt{1+x^{3}}}$, $g(x)=\frac{1}{\sqrt{x^{3}}}$ both integrals converges or both diverges.

By last, integrals of the form $\int_{1}^{\infty}\frac{dx}{x^{p}}$ converges if $p>1$, $\Longrightarrow \int_{1}^{\infty}\frac{dx}{\sqrt{x^{3}}}$  converges $\Longrightarrow \int_{2}^{\infty}\frac{dx}{\sqrt{x^{3}}}$ converges

That implies that, $\int_{2}^{\infty}\frac{dx}{\sqrt{1+x^{3}}}$ converges, therefore it exists.

Is it correct? Is there another way to prove it? Thank you very much


问:
It's $M$ the centroid of $\triangle ABC$?
Let $ABC$ a triangle and $M$ a point inside it. Find the minimum of $MA^2+MB^2+MC^2$.

It's $M$ the centroid of $\triangle ABC$?

答:
Consider a triangle on an $xy$-plane, where $A=(x_1, y_1)$, $B=(x_2, y_2)$, $C=(x_3, y_3)$.

Let $P = (s,t)$. The sum of squared distances from $P$ to vertices is

$$\begin{align*}
d(s,t) &= PA^2 + PB^2 + PC^2\\
&= \sum_{i=1}^3\left[(s-x_i)^2+(t-y_i)^2\right]\\
&= 3s^2+3t^2-2s\sum_{i=1}^3x_i-2t\sum_{i=1}^3y_i + \sum_{i=1}^3(x_i^2+y_i^2)
\end{align*}$$

As a quadratic function in $s$ with fixed $t, x_i,y_i$, the minimum of $d(s,t)$ happens independent of $t$ when

$$s = -\frac{-2\sum_{i=1}^3x_i}{2\cdot 3} = \frac{x_1+x_2+x_3}{3}$$

Similarly, as a quadratic function in $t$ with fixed $s, x_i, y_i$, the minimum of $d(s,t)$ happens independent of $s$ when

$$t = -\frac{-2\sum_{i=1}^3y_i}{2\cdot 3} = \frac{y_1+y_2+y_3}{3}$$

These $s,t$ are the same as the coordinates of the centroid $G$ of $\triangle ABC$, given by formula

$$G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$$


问:
Infinite minimal left ideals
Let $R$ be an infinite field and let $S$ be the ring

$$S = \begin{pmatrix}
R & R & R\\
0 & R & 0\\
0 & 0 & R
\end{pmatrix}$$

Show that there are infinite minimal left ideals of $S$.

I tried to consider the possible minimal left ideals of $S$. Given a nonzero matrix $A$ in one of these ideals, say $I$, I think I could multiply only once, or maybe twice in the left by some matrices of $S$ to get a matrix $B$ whose nonzero entries are in the same positions as $A$, but $B's$ nonzero entries are simply $1$'s, so that, since $B$ is in the ideal $I$, we can get any matrix with the nonzero entries in the same positions as $A$ but being any elements of $R$. Hence, since there are finitely many possible choices for the positions of the nonzero entries of a matrix $3$ x $3$, it seems to follow that $S$ has finitely many minimal left ideals... contradicting what I'm supposed to prove.

Any help would be appreciated.

答:
It appears that you are misinterpreting the use of the symbol $R$ in the definition of $S$.

Each instance of the symbol $R$ in that definition is intended as an independent, arbitrary choice of an element of $R$.

With that understanding, we can proceed as follows . . .

For each $r\in R$, let $I_r$ be the left ideal of $S$ generated by the matrix
$$
T_r
=
\pmatrix{
0&1&r\\
0&0&0\\
0&0&0\\
}
$$
Then for any $A\in S$ we have
$$
AT_r=
\pmatrix{
0&a&ar\\
0&0&0\\
0&0&0\\
}
$$
where $a=A_{1,1}$.

It follows that

Each of the left ideals $I_r$ is minimal.$\\[4pt]$

No two of the left ideals $I_r$ are equal.

Title: Infinite minimal left ideals


问:
Measure of the union of the sets of a numerable family of measurable sets
I'm working with some measure exercises . I've tried to get a formula to know the measure of the union of n sets ( finite number ), I think I have it, proved by induction, I wonder if there is also a formula to measure the infinite union of all the sets in a numerable family or if that can't be simply measured or if there doesn't exist a formula.
Thanks.</p>

答: If $A=\{A_n:n\in \Bbb N\}$ is a family of measurable sets then $\mu(\cup A)=\sum_{n\in \Bbb N}\mu (B_n)$ where $B_1=A_1$ and $B_{n+1}=A_{n+1}\setminus (\cup_{j=1}^n A_j).$


问:
Total expectation theorem on exponentially distributed random variables
The problem is that an absent minded professor schedules two student appointments for the same time. The appointment durations are independent and exponentially distributed with mean thirty minutes. Both students arrive at the same time, what is the expected value of the time between the arrival of the first student and the departure of the second student?

This is my variation of the problem from https://www.youtube.com/watch?v=X5ch4xZf3LE&ab_channel=MITOpenCourseWare. The original problem has the second student arriving five minutes late.

I decided to redefine a simpler problem (as stated in the first paragraph) where the students arrive at the same time. I think the solution should be trivial, as the average is 30 min, we can expect the total time to be 60 min. However, I decided to solve this with the total expectation theorem based on the original solution to see if I would get 60 min. Unfortunately I do not and I am not sure why. My solution is as follows:

Let $T$ be the the time between the arrival of the first student and the second student. Let $T_1$ be the appointment duration of the first student and $T_2$ be the appointment duration of the second student.

From the total expectation theorem:
$$ E[T] = P(T_1\leq t)E[T|T_1\leq t] + P(T_1 > t)E[T|T_1 > t]$$

Since $T_1$ and $T_2$ are exponentially distributed we have $P(T_1\leq t) = 1-e^{-\lambda t}$ and $P(T_1> t) = e^{-\lambda t}$

Next, by the definition of expectation we have $E[T|T_1\leq t] = E[T_1|T_1\leq t] + E[T_2|T_1\leq t] = \int_0^tx\lambda e^{-\lambda x} + 30 dx$

For the other scenario we have $E[T|T_1> t] = E[T_1|T_1> t] + E[T_2|T_1> t] = 5 + 30 + 30 = 65$, this is because of the memoryless property, given that $t$ minutes have passed for student 1's appointment it will take another 30 min to finish.

Finally,
$$ E[T] = (1-e^{-\lambda t}) (\int_0^tx\lambda e^{-\lambda x} + 30) + (e^{-\lambda t})65$$

Here $\lambda=\frac{1}{30}$ and suppose that we use $t=5$. Plugging this in WolframAlpha results in $E[T]=59.68$.

So by complete accident, I took the integral out of the brackets and found that this equals $60$... But I have no idea why:

$$ E[T] = \int_0^tx\lambda e^{-\lambda x} + (1-e^{-\lambda t}) (30) + (e^{-\lambda t})65 = 60$$

Does anyone know why my solution does not produce 60 min, where am I making a mistake?

答:
$\mathbb E[T_1 \mid T_1 \le t]$ is not equal to $\int_0^t x \lambda e^{-\lambda x}\,dx$.

The problem is that $\lambda e^{-\lambda x}$ is the PDF of $T_1$, not the PDF of $(T_1 \mid T_1 \le t)$. To find the PDF of $(T_1 \mid T_1 \le t)$ on $[0,t]$, we divide by $\Pr[T_1 \le t] = 1 - e^{-\lambda t}$. Then, we get
$$
   \mathbb E[T_1 \mid T_1 \le t] = \int_0^t x \cdot \frac{\lambda e^{-\lambda x}}{1 - e^{-\lambda t}} \,dx.
$$


问:
$\lim_{x\to 0} \frac{\sin(t\sqrt{x^2-k^2})}{\sqrt{x^2-k^2}}=?$
$\lim_{x\to 0} \frac{\sin(t\sqrt{x^2-k^2})}{\sqrt{x^2-k^2}}=?$

Anyone can help, please.
Does is equal to $t$?

答:
if $\lim x\to k$ then you can make the substitution $\sqrt{x^2-k^2 } =y$

you get : $\lim_{y\to 0} \dfrac{\sin(t{y})}{{y}}=t$ (this is a well known limit)


问:
Subset of countable set is countable proof
I'm self studying Understanding Analysis and there's an exercise that asks to prove that if $ A\subseteq B $ and $B$ is countable, then $A$ is either countable or finite. I want to be sure that my proof is not wrong or badly written.

Being finite is trivial, so let's suppose it's not.
In the exercise, we assume there exists $f:N \to B$ which is 1-1 and onto and $n_1=\min\{n\in N:f(n)\in A\}$. As a start to a definition of $g:N \to A$, we set $g(1)=f(n_1)$. These are given by the exercise and we're asked to continue the process inductively.

I construct sets $a_n=\{n_c \text{ for } c \le n\}$ for example $a_1=\{n_1\}$, $a_2=\{n_1, n_2\}$ etc.

Then I construct: $$n_c=\min\{n\in N:f(n)\in A\ and\ n\notin a_{n_{c-1}}\}$$
and thus $g(c)=f(n_c)$ for every $c\in N$ gives us a $1-1$ function from $N$ onto $A$.

I would love to hear any sort of feedback, thank you!

答: By definition, we have an injection $j:B\hookrightarrow\Bbb N$.  But if $A\subset B$, then $|A|\le |B|$.  That is, the inclusion $i:A\hookrightarrow B$ is an injection.  Then $j\circ i:A\hookrightarrow\Bbb N$ is injective.


问:
Determinant, same for transpose?
$$\det(A) = \det(A^\top)$$

Wiki confirms this.  How can I see this this is true?  I tested out a few examples and it seems to hold, however unsure the exact reasoning.

For instance:

$$\begin{vmatrix} a & b \\ c & d \\ \end{vmatrix} = ad - bc$$

Vectors $(a,b)$ and $(c,d)$ define the parallelogram, but how would the transpose define a similar shape with same area?

答:
Use the Leibniz definition of determinants, that $$\det(A) = \sum\limits_{\sigma\in S_n}\text{sgn}(\sigma)\prod\limits_{i=1}^na_{\sigma(i),i}$$

Recognize that the transpose of a permutation matrix is simply the inverse of the permutation matrix and so has the same sign.

Your sum then consists of exactly the same terms before and after the transposition with the same signs as originally had, just perhaps appearing in a slightly different order.

Title: Determinant, same for transpose?


问:
Alternative to Ceiling Function for Simple Sequence
I came across a problem where I was wondering if there is a simple expression to map a sequence of numbers to the same adjacent values without using a ceiling function.

For instance, given the sequence:

{1, 2, 3, 4, 5, 6} -> {1, 1, 2, 2, 3, 3}

My solution would be $
f(x) = \left \lceil \frac{x}{2} \right \rceil
$

Is there an alternative form that would not require using the ceiling or floor functions, perhaps using the mod operator instead?

答:
Since $f:\{1,2,3,4,5,6,\cdots\}\mapsto\{1,1,2,2,3,3,\cdots\}$

we see that

\begin{eqnarray}
f(n)+f(n+1)&=n+1\tag{1}\\
-f(n)+f(n+1)&=\frac{1+(-1)^n}{2}\tag{2}
\end{eqnarray}

Subtracting equation $(2)$ from equation $(1)$ and solving for $f(n)$ gives

$$ f(n)=\frac{2n+1-(-1)^n}{4} $$


问:
Find the missing angle in triangle
In the below triangle, we are looking for the value of angle $φ$.

We are given $α=30, β=18, γ=24$ and also that $CD=BD$.

I have solved it with trigonometry (sine law) and found the required angle to be 78 but I need to solve it with Geometry only.

What I have tried so far:

First of all, the angle is constructible, which means to me that there must be a geometrical solution. I first drew triangle ABC; easy, since we know 2 of its angles. We are not interested in the lengths of the sides. Then, with side AC as a base, and angle of 24 degrees, we can draw a ray from the point A.

Then, since $CD=BD$, triangle DCB is isosceles, therefore D must lie on the perpendicular bisector of CB, which we can draw. The point of intersection of the ray from A and the perpendicular bisector, is point D.

From triangle FEB we have that

angle AFD = 108.

From triangle AFD,

$ADC+CDE+54+108=180$ so $ADC+CDE=18$

We also have $24+ACD+ADC=180$

$ACB=132$

$132+φ+ACD=180$

$18+φ+54+ADC+2CDE=180$

I am always one equation short.

Any ideas?

Many thanks in anticipation!

EDIT:

Sine law in triangle ABD:

$\frac {sin (φ+18)}{AD} = \frac {sin (54)}{BD}$

Sine law in triangle ACD:

$\frac {sin (360-132-φ)}{AD} = \frac {sin (24)}{CD} = \frac {sin (24)}{BD}$

so

$\frac {sin (φ+18)}{sin (228-φ)} = \frac {sin (54)}{sin (24)}$

hence $φ=78$.

答:
Since $\angle DAB=54^o$, if we construct a regular pentagon on $AD$, then $AB$ bisects $\angle DAG=108^o$, and $AB$ extended to $K$ on the circumcircle passes through center $N$.

Extend $AC$ to $I$, $DB$ to $L$, and join $IK$, $KL$, $LA$, $IL$, and $DG$.

Since cyclic quadrilateral $AIKL$ has a right angle at $I$, it is a rectangle.
Therefore $\angle AIL=\angle IAK=30^o$, $\angle LAK=60^o$, and$$\angle LAG=\angle LAK-\angle GAK=60^o-54^o=6^o=\angle LDG$$And since in the regular pentagon $\angle ADG=36^o$, and as OP notes $\angle ADE=18^o$, then $\angle LDG=\angle ADC$.

Therefore$$\angle CDB=\angle ADG-2\angle LDG=36^o-2\cdot6^o=24^o$$and$$\angle DCB=\phi=\frac{180^o-24^o}{2}=78^o$$

Title: Find the missing angle in triangle


问:
Cech nerve and descent data
When generalizing from sheafs on a site to 2-sheafs or stacks, it is useful to first rephrase the descent data for ordinary (pre)sheafs in terms of the Cech nerve of a coverage (e.g. https://ncatlab.org/nlab/show/%C4%8Cech+groupoid): $$\text{Match}(\mathcal{U}, F)\cong[\mathbf{C}^{op}, \mathbf{Grpd}](C(\mathcal{U}), F)$$ where $\mathcal{U}$ is a covering family, $F$ is a presheaf (it takes values in $\mathbf{Grpd}$ after composition with the embedding $\mathbf{Set}\hookrightarrow\mathbf{Grpd}$) and $C(\mathcal{U})$ is the Cech groupoid of $\mathcal{U}$, i.e. the 2-coskeleton of the full Cech nerve.

Then, by replacing $F$ with a 2-presheaf (pseudofunctor) $\mathbf{C}^{op}\rightarrow\mathbf{Cat}$ and $C(\mathcal{U})$ with the 3-(co)skeleton of the Cech nerve we obtain the descent data for $F$.

However, I was wondering about the occurence of the category $\mathbf{Grpd}$ of groupoids in the definition of ordinary matching families and, by extension, $\mathbf{2Grpd}$ in the definition of descent data for stacks. Is there any intuition of why we have to regard our presheafs as taking values in these higher categories instead of the ordinary cosmos $\mathbf{Set}$ (and its higher versions)?

And partially related: The Cech nerve is defined as a simplicial object in $\mathbf{C}$ and hence it technically takes values in $\mathbf{C}$, not in $\mathbf{Set}$ (or higher versions). Do we turn this into a $\mathbf{Set}$-valued functor by extending with a forgetful functor $\mathbf{C}\rightarrow\mathbf{Set}$? (And in general by a functor $\mathbf{C}\rightarrow\mathcal{E}$ when working internal to an arbitrary topos $\mathcal{E}$?) I have seen how people turn the data in the nerve into a ($n$-)groupoid, but I haven't really heard them talk about how this happens in terms of functors.

答:
$\mathbf{Grpd}$ is actually the higher categorical analog of $\mathbf{Set}$. Indeed, recall that a set is a collection of object together with an equality, that is a relation which is

transitive (composition of morphisms)

reflexive (existence of identities)

symmetric (existence of inverses)

The parentheses above are the categorical interpretation of an equivalence relation and shows that the higer categorical version of $\mathbf{Set}$ is indeed $\mathbf{Grpd}$.

But even without this abstract nonsense, just recall what a stack is : $F$ is a stack if you can glue a family of objects along a descent datum : if you have objects $x_i\in F(U_i)$ defined on $U_i$, and want to glue them, you don't want to assume that $x_i=x_j$ on $U_i\cap U_j$, but rather that there is an isomorphism $x_i\simeq x_j$. These objects, together with the isomorphisms $x_i\simeq x_j$ forms the descent datum. This is where the groupoids appears in the formulation $[\mathbf{C}^{op},\mathbf{Grpd}](C(\mathcal{U}),F)$.

Title: Cech nerve and descent data


问:
How to prove that $f^{-1}(B)\subset A \Rightarrow B \subset f(A)$ if and only if $f$ is surjective?
so I'm trying to prove that $f^{-1}(B)\subset A \Rightarrow B \subset f(A)$ if and only if f is surjective. I'm used to proving proofs with just one implication, but now I'm confused because of the "if and only if" statement. I will show you the actual question and show you what I know. I hope someone can help me, because I'm clueless for the moment.

Question

$$\forall A \in P(X): \forall B \in P(Y): f^{-1}(B) \subset A \Longrightarrow B \subset f(A) \text{ if and only if $f$ is surjective}$$

What I know

$P(X)$ is the powerset = is the set of all subsets of $X$

Inverse image: If $K$ is a subset of $Y$ and $f^{-1}(K)$ is a subset of $X$ given by: $f^{-1}(K)=\{x\in
X|f(x)\in K\}$

If $A\subset B$ = "every element in A is also an element of B"

surjective =  $\forall y\in Y|\exists x\in X: f(x)=y $

If someone knows the answer it would be very kind to share your wisdom :)

thanks in advance

答:
Let's consider function $f\colon X \to Y$ and consider definition of surjective function as: $f(X)=Y$ i.e. $\forall y \in Y, \exists x \in X, y=f(x)\quad (1)$.

Now let's take your sentence $f^{-1}(B)\subset A \Rightarrow B \subset f(A)$ for $\forall A \subset X$, $\forall B \subset Y\quad (2)$.

If holds $(2)$, taking $B=Y$ and $A=X$, obviously gives surjectivity.

Now assume $(1)$ and let's take any such $A,B$, that $f^{-1}(B)\subset A$. If we take $y \in B$, then as $B \subset Y$ and function is surjective we have, that $\exists x \in X, y=f(x)\in B$. From last we have $x \in f^{-1}(B) \subset A$, so $x \in A$, so $y=f(x) \in f(A)$.


问:
When can the Killing form be put in the form $\kappa_{ij}=-\kappa\delta_{ij}$, and thus be treated as an Euclidean dot product?
Let $\mathfrak{g}$ be a Lie algebra and $\mathfrak{h}$ is a Cartan subalgebra. We know when the lie algebra is semi simple then the killing form, $\kappa(X,Y) = Tr(X^{ad}Y^{ad})$, is non degenerate in $\mathfrak{g}$ and more importantly in $\mathfrak{h}$. So it is a bilinear form analogous to the metric providing an isomorphism between $\mathfrak{h}$ and $\mathfrak{h}*$ and allowing us to take the inner product across these spaces.

One thing that isn't clear fromt the literature I've read (and from my Universities lecture notes) is when we are able to to put the killing form in the form $\kappa_{ij} = -\kappa \delta_{ij}$ and so we can treat the inner product  as the normal euclidean dot product. Some notes I have read say this is only true for compact groups, others I have read say you can always do it and some don't even mention it (my lecture notes).

答:
It is true for compact Lie algebra (Lie algebras of compact simply connected groups) for which Killing is negative definite.

https://en.wikipedia.org/wiki/Killing_form#Connection_with_real_forms