问:
Logarithmic differentiation of a complicated product
In this answer, it says that one may take the logarithmic derivative w.r.t. $z$ on both sides of the equation
$$(q^4;q^4)_\infty\left\{z(-z^4q^3;q^4)_\infty(-z^{-4}q;q^4)_\infty-z^{-1}(-z^4q;q^4)_\infty(-z^{-4}q^3;q^4)_\infty\right\}\\=(z-z^{-1})(q;q)_\infty(z^2q;q)_\infty(z^{-2}q;q)_\infty,\tag{1}$$
to get
$$1+4\sum_{n\ge1}(-1)^n\frac{q^{2n-1}}{1-q^{2n-1}}=\frac{(q;q)_\infty^3}{(q^4;q^4)_\infty(-q^3;q^4)_\infty(-q;q^4)_\infty},\tag{2}$$
after having set $z=1$.

I am having some trouble with doing this.

I decided to simplify things a bit so I came up with
$$z^2(-z^4q^3;q^4)_\infty(-z^{-4}q;q^4)_\infty-(-z^4q;q^4)_\infty(-z^{-4}q^3;q^4)_\infty=\frac{(q;q)_\infty}{(q^4;q^4)_\infty}(z^2-1)(z^2q;q)_\infty(z^{-2}q;q)_\infty.\tag{3}$$
The logarithmic derivative of the R.H.S. of $(3)$ is easy, and it evaluates to
$$\frac{\partial}{\partial z}\log\left\{\frac{(q;q)_\infty}{(q^4;q^4)_\infty}(z^2-1)(z^2q;q)_\infty(z^{-2}q;q)_\infty\right\}=\frac{2z}{z^2-1}+\frac{2z}{(z^2q;q)_\infty}A(1,z^2;q)-\frac{2z^{-3}}{(z^{-2}q;q)_\infty}A(1,z^{-2};q),$$
where
$$A(a,b;q)=\sum_{n\ge1}\frac{(aq)^n}{1-bq^n}.$$
But, simplifying the logarithm of the left hand side of $(3)$ is proving to be a difficult task. Is there a better way to go about this? Thanks.

Body:
Let \begin{align*}
f(z)&=z\prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{4n - 1} z^4)(1 + q^{4n - 3} z^{-4})\\
g(z)&=z^{-1}\prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{4n - 3} z^4)(1 + q^{4n - 1} z^{-4})\\
h(z)&= (z - z^{-1})\prod_{n = 1}^\infty (1 - q^n)(1 - q^n z^2)(1 - q^n z^{-2}).
\end{align*} Then \begin{align*}
\log f(z)&=\log z+\sum_{n = 1}^\infty\left[\log(1 - q^{4n})+\log(1 + q^{4n - 1} z^4)+\log(1 + q^{4n - 3} z^{-4})\right]\\
\log g(z)&=-\log z+\sum_{n = 1}^\infty\left[\log(1 - q^{4n})+\log(1 + q^{4n - 3} z^4)+\log(1 + q^{4n - 1} z^{-4})\right]\\
\log h(z)&=\log(z - z^{-1})+\sum_{n = 1}^\infty\left[\log(1 - q^n)+\log(1 - q^n z^2)+\log(1 - q^n z^{-2})\right]
\end{align*} so \begin{align*}
\frac{f'(z)}{f(z)}&=\frac1z+\sum_{n = 1}^\infty\left[\frac{4q^{n-1}z^3}{1 + q^{4n - 1} z^4}-\frac{4q^{4n-3}z^{-5}}{1 + q^{4n - 3} z^{-4}}\right]\\
\frac{g'(z)}{g(z)}&=-\frac1z+\sum_{n = 1}^\infty\left[\frac{4q^{4n-3}z^3}{1 + q^{4n - 3} z^4}-\frac{4q^{4n-1}z^{-5}}{1 + q^{4n - 1} z^{-4}}\right]\\
\frac{h'(z)}{h(z)}&=\frac{1+z^{-2}}{z - z^{-1}}+\sum_{n = 1}^\infty\left[\frac{-2q^nz}{1 - q^n z^2}+\frac{2q^nz^{-3}}{1 - q^n z^{-2}}\right].
\end{align*} This gives \begin{align*}
f'(1)&=f(1)\left(1+\sum_{n = 1}^\infty\left[\frac{4q^{n-1}}{1 + q^{4n - 1}}-\frac{4q^{4n-3}}{1 + q^{4n - 3}}\right]\right)\\&=\left(\prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{4n - 1})(1 + q^{4n - 3})\right)\left(1 + 4\sum_{n = 1}^\infty (-1)^n \frac{q^{2n - 1}}{1 + q^{2n - 1}}\right)\\
g'(1)&=g(1)\left(-1+\sum_{n = 1}^\infty\left[\frac{4q^{4n-3}}{1 + q^{4n - 3}}-\frac{4q^{4n-1}}{1 + q^{4n - 1}}\right]\right)\\&=\left(\prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{4n - 1})(1 + q^{4n - 3})\right)\left(-1-4\sum_{n = 1}^\infty (-1)^n \frac{q^{2n - 1}}{1 + q^{2n - 1}}\right)\\
h'(1)&=\left(\lim_{z\to1}\frac{h(z)}{z-z^{-1}}\right)\times2=2\prod_{n = 1}^\infty (1 - q^n)^3
\end{align*} so $f'(1)-g'(1)=h'(1)$ is equivalent to $$1 + 4\sum_{n = 1}^\infty (-1)^n \frac{q^{2n - 1}}{1 + q^{2n - 1}}=\frac{\prod\limits_{n = 1}^\infty (1 - q^n)^3}{\prod\limits_{n = 1}^\infty (1 - q^{4n})(1 + q^{4n - 1})(1 + q^{4n - 3})}.$$


问:
Finding intervals for a two variable function
I had a similar question about 1 year ago, and I find more properties about my question. Now I want to ask it again and so thankful for new  solutions.
Suppose
$$ H(x,y)=x e^{\pi y}-\frac{x(\pi-x)}{\pi}e^{xy}+(\pi-x)(\frac{\pi^2}{12}-1)e^{xy}-\frac{\pi^2}{12}x \sinh(\pi y)\\ +(1-\frac{\pi^2}{12})(\pi-x)-x$$.

I guess  there exist $\alpha\geq 0$ and $\beta\geq 0$ such that
$$ H(x,y)\geq 0, \quad for \quad (x,y) \in [\alpha,\pi)\times [\beta, \infty).$$
I am looking for the best choices for  $\alpha$ and $\beta$.
With several calculations and plot graphs, I guess that if $\alpha=0$then $0.4<\beta<0.5$.

Also we have,
$$1)\,H(0,y)=0,
\qquad 2)\,H(\pi,y)= 0,
\qquad 3)H(x,0)=\frac{1}{\pi}x^2-1.$$
Furthermore $H(x,0)\geq 0$  if $x\geq \sqrt{\pi}$.

I appreciate any solutions, comments and hints.

答:
A substitution $u=x/\pi$, $v=e^{\pi y}$, and $\rho=12-\pi^2$ simplifies $0\le H(x,y)$ to

$$0\le G(u,v)=2(u-1)(12u+\rho)v^u+(12+\rho)vu+(12-\rho)v^{-1}u-(24+2\rho )u+2\rho,$$

$(x,y) \in [\alpha,\pi)\times [\beta, \infty)$ iff $(u,v)\in [\alpha\pi,1)\times [e^{\pi\beta},\infty)$.

At the horizontal axis is $v$ from $1$ to $4$, at the vertical axis is $u$  from $0$ to $1$, the points with non-negative value of $g$ are red, and with negative value of $g$ are blue, the graph of a function $u=\tfrac{6-v}5$ is black.

The graph suggests that $H(x,y)\ge 0$ provided $v\ge 3.63$ ($\beta\ge 0.41$) or $u\ge  \tfrac{6-v}5$
($\alpha\ge \frac{\pi}5 \left(6-e^{\pi\beta}\right)$).

ContentLicense: CC BY-SA 4.0


问:
What's the solution to $ 2|2x - a| < |x + 3a|$ (found critical values)
$$ 2|2x - a| < |x + 3a|$$

I found the critical values to be $x = \frac{5a}{3}$ and $x = \frac{-a}{5}$ by removing the abs val signs and equation the left side to the +/- right side.

Now this is an inequality, and I have an arbitrary constant $a$. I'm not sure how to find the range.

P.S.: I'd prefer to learn how to solve it algebraically in particular. Also, if I knew a, I'd take those critical values and test inputs between and beyond those values. Not sure what to do here.

答:
An inequality $|P(x)|<|Q(x)|$ is the same as
$$
(P(x))^2<(Q(x))^2
$$
which becomes
$$
(Q(x)-P(x))(Q(x)+P(x))>0
$$
In your case you get
$$
(x+3a-4x+2a)(x+3a+4x-2a)>0
$$
and, in shorter form,
$$
(3x-5a)(5x+a)<0
$$
Now it’s easier: for $a=0$ you have no solution. For $a>0$, the roots are $-a/5<5a/3$, so
$$
-\frac{1}{5}a<x<\frac{5}{3}a
$$
You can do the $a<0$ case.


问:
Proving that $(\sqrt 3+\sqrt 2)/(\sqrt 2)$ is irrational
Can you please help me prove this ? I can prove that $\sqrt p$, where p is prime is irrational, also that the sum $\sqrt 3 + \sqrt 2$ is irational, bud dont know how to prove that the whole fraction is irrational. Thanks for any answer.

答:
Consider $r=(\sqrt{3}+\sqrt{2})/\sqrt{2}$. Then
$$
3=2(r-1)^2
$$
The possible positive rational roots of
$$
2x^2-4x+1
$$
are $1$ and $1/2$, neither of which is actually a root.


问:
Problem on conditional probability
We have a set of $s$ elements and we choose $n$ elements, where $n \le s$, from it randomly. There are $M$ special numbers (including $M'$). What is the probability of choosing $n$ elements from $s$, which contains at least $k$ special numbers given that $M'$ is in selected elements?</p>

答:
I'll make the following assumptions about the intent of the question: The $M$ "special numbers" are elements of the set of $s$ elements. (It would have been clearer to call them "special elements", as the remaining elements are not limited to numbers and the restriction of the special elements to numbers isn't being used anywhere.) The name $M'$ refers to one of the $M$ special numbers. (It would have been clearer to use two different letters here, since one is a count of elements and the other is an element.)

Under these assumptions, the question can be reduced to this question by first removing the element $M'$ known to be among the selected elements; i.e., substitute $n-1$ for $C$, $s-1$ for $T$, $k-1$ for $K$ and $M-1$ for $G$.

ContentLicense: CC BY-SA 3.0

Title: Problem on conditional probability


问:
Why can I find Fixed points in a Dynamical System by Setting a map F(x) = x, and with a O.D.E by setting $\dfrac{dx}{dt} = 0$, what is the connection?
In elementary differential equations, one can find the fixed points of a single variable system by setting the derivative equal to zero and solving.

$$\dfrac{dx}{dt} = \mu\,x - x^{3} = 0$$

In Dynamical systems and Chaos - there is a similar concept with regard to maps ...

$$ x_{n} = \mu\,x_{n+1} - x_{n}^{3} \rightarrow x_{n} - \mu x_{n+1} - x_{n}^{3} = 0$$

What is the connection between these two Ways of finding fixed points - and how would you convert from one to the other if possible?

This answer explores this question - but I'm not following the connection:
Strange attractors: what is the difference between a map and differential equation system?

答:
The connection is constant solutions. If in a discrete dynamic $x_{n+1}=g(x_n)$ you get $x_n=c$ for all $n$, then obviously $c=g(c)$. In a continuous time dynamic $\dot x=f(x)$ the constant solutions $x(t)=c$ have derivative $\dot x=0$, so to satisfy the dynamic you need $f(c)=0$.

After that one starts to ask about stability. That is, starting close to a constant solution, does the dynamic lead further towards the constant solution or away from it.

Then there are periodic solutions that then have their own kinds of stability. If the dynamical system depends on a parameter, then usually such structures are generally stable towards small changes of the parameter, for example, if there is a periodic orbit, then with the changed parameter there will usually also be a periodic orbit close-by with a similar period. The interesting parameters are those where this breaks down, these are the bifurcation points (even for non-fork bifurcations).


问:
Bisection method complex roots
Can the bisection method of root finding be used to find the complex roots of a polynomial, i.e, complex values of $x$ such that $f\left(x\right)\:=\:0$?

In the bisection method we arbitrarily choose two starting points with opposite signs and then see if the value of the function at the midpoint of these two points is a root. If it is, we stop, if it is not we halve the interval depending on which side of the interval gives us a closer result.

Link here: https://en.wikipedia.org/wiki/Bisection_method

My gut feeling is to say that it can't but I am not entirely sure how to justify it.

I know Newton's method can but am not sure about this one.

Any help would be highly appreciated!

答:
Hint.

If $f(z) = 0$ can be represented as

$$
\cases{
f_r(x,y) = 0\\
f_i(x,y) = 0
}
$$

we can use a bintree to determine into a plane region the trace of $f_r=0, f_i=0$ and then we will have within a required precision the set of points which verify the equation. The binthree for $\mathbb{R}^2$ or octree for $\mathbb{R}^3
$ can be processed recursively in a fashion very akin to the binary search process to $\mathbb{R}$ problems

Title: Bisection method complex roots


问:
$f(x)$ is monotonically increasing function for all real $x$, $f’’(x)>0$ and $f^{-1}$ exists, then prove that [CONT]...

$f(x)$ is monotonically increasing function for all real $x$, $f’’(x)>0$ and $f^{-1}$ exists, then prove that $\frac{f^{-1}(x_1)+f^{-1}(x_2)}{2}<f^{-1}(\frac{x_1+x_2}{2})$

The conclusions that can be drawn from the given data is that $f(x)$ is monotonous and increasing, so $f^{-1}$ is also monotonous and increasing. Also $f(x)$ is concaved upwards, so $f^{-1} (x)$ is concaved downwards. How do I solve from here?

答:
Visualisation helps.$g(x)=f^{-1}(x)$.

Imgine $g(x)$ concave downward.Let P be the point $(x_1,g(x_1))$ and Q be $(x_2,g(x_2))$.

Thus the point $(\frac{x_1+x_2}{2},\frac{g(x_1)+g(x_2)}{2})$ is  the mid point of of segment $PQ$.

Also a perpendicular line  line drawn from $(\frac{x_1+x_2}{2},0)$ meets the curve in $(\frac{x_1+x_2}{2},g(\frac{x_1+x_2}{2}))$.

It is easy to see that this point lies above midpoint of $PQ$


问:
PGM and Derivation some inference?
I study my notes on Probabilistic Theory and notation and d-separation topics. we know PGM's are a rich framework for encoding probability distributions over complex domains. I couldn't get the point why these two ones are True according to this fig. how we can derived these two are conducted from this fig.

答:
d-separation allows us to infer conditional independences using the concept of active an inactive paths.

A path between two variables $C_1 - ... - C_n$ is active in a context $S$ if:

there is no fork $\leftarrow C_i \rightarrow $, chain $\rightarrow C_i \rightarrow$ or reverse chain $\leftarrow C_i\leftarrow$ such that $C_i \in S$

for every collider $\rightarrow C_i \leftarrow$ either $C_i \in S$ or there exists a descendent $C_i \to D_1 \to ... \to D_n$ such that $D_n \in S$.

We say that two variables $A,B$ are d-separated in a context $S$ if there is no active path in $S$ between them.

d-separation is a useful concept because in a Bayesian network it implies conditional independence (though the reverse is not always true).

Now we are prepared to answer the question.

There are two paths between $X$ and $J$ in the diagram. But both paths include the collider $\rightarrow R \leftarrow$. Hence both paths will be inactive unless we control for $R$. In particular, $X\perp J$.

Between $L$ and $J$ there are also two paths. Let's focus on the path $L \to H \leftarrow W \rightarrow V \leftarrow J$. Here, $H,V$ are colliders. Thus, controlling for them without controlling for the fork $W$ opens up the path. Thus it is not necessarily true that $L\perp J | H,V$.

Title: PGM and Derivation some inference?


问:
How to solve for $x$ in trigonometric (sine wave) equation with $x$ passing through the floor function?
I have the following equation:

$\displaystyle 1=\sin\left(2\pi\left(\frac{2^{\operatorname{floor}\left(x\right)-1}-1}{\operatorname{floor}\left(x\right)}+\frac{1}{4}\right)\right)$

Is it possible to solve for $x$, here? When removing the floor function, the sine wave changes entirely (so this is not ideal when solving for roots). I would grateful for any help.

答:
Warning : I am not totally sure that my process is fully correct.

The solution will be obtained when the argument of the sine will be equal to $(4k+1)\frac \pi 2$.

For the time being, let $y=\lfloor x\rfloor$ and the equation is
$$2 \pi  \left(\frac{2^{y-1}-1}{y}+\frac{1}{4}\right)=(4k+1)\frac \pi 2\implies 2^{y-1}=ky+1$$ the solution of which being
$$y=\lfloor x\rfloor=-\frac{W_{-1}\left(-\frac{2^{-\frac{1}{k}-1} \log (2)}{k}\right)}{\log
   (2)}-\frac{1}{k}$$ where $W(.)$ is Lambert function.

And $x  \leq \lfloor x\rfloor \leq x+1$.


问:
Are there any valid continuous Sudoku grids?
A standard Sudoku is a $9\times 9$ grid filled with digits such that every row, column, and $3\times 3$ box contains all the integers from $1$ to $9$.

I am thinking about a generalization of Sudoku which I call "continuous Sudoku", which consists of a unit square where every point on that square corresponds to a real number. The rules for continuous Sudoku are designed to be analogous to the rules for standard Sudoku, and I've devised two different rulesets:

The first ruleset I call "weak" continuous Sudoku. In weak continuous Sudoku, the only restriction is that every row and column of the square contains every real number in the interval $[0,1]$ exactly once.

The second ruleset I call "strong" continuous Sudoku. In strong continuous Sudoku, the rules of weak continuous Sudoku apply, and, in addition, every square sub-region of the unit square contains every real number in the interval $[0,1]$ at least once. This is analogous to the $3\times 3$ box restriction in standard Sudoku.

Let $U = [0,1]$ and $U^2 = U\times U$. More precisely, a weak continuous Sudoku is essentially a function $f:U^2\to U$, which satisfies the following four properties:

If $x,y_1,y_2\in U$ and $y_1\neq y_2$, then $f(x,y_1)\neq f(x,y_2)$.

If $x_1,x_2,y\in U$ and $x_1\neq x_2$, then $f(x_1,y)\neq f(x_2,y)$.

If $x\in U$ then $\{z: f(x,y)=z,y\in U\} = U$.

If $y\in U$ then $\{z: f(x,y)=z,x\in U\} = U$.

Now, strong continuous Sudoku is a bit harder to define precisely. A set $S$ is a square sub-region of $U^2$ iff $S\subseteq U^2$ and there exists $z = (z_1,z_2)\in U^2$ and $r>0$ such that $S = \{(x,y)\in U^2:z_1\leq x\leq z_1+r,z_2\leq y\leq z_2+r\}$. Thus, using this definition, a strong continuous Sudoku is a weak continuous Sudoku which satisfies the following additional property:

If $S$ is a square sub-region of $U^2$, then $f(S) = U$.

I've been trying to look for specific examples of both strong and weak continuous Sudoku grids, but have so far been unsucessful.

I'm not sure whether any weak continuous Sudoku exists. My first attempt:
$$
f(x,y)=\begin{cases} x+y &\text{if }x+y\leq 1 \\
x+y-1 & \text{if }x+y>1\end{cases}
$$
almost works. It satisfies properties $3$ and $4$, and almost, but not quite, satisfies $1$ and $2$. The issue occurs only at boundaries of the square, for example, $f(0.5,0) = 0.5$ and $f(0.5,1)=0.5$.

Any example of a strong continuous Sudoku will likely need to be an extremely discontinuous pathological function, similar to the Conway base 13 function. Obviously, if there are no weak continuous Sudoku grids, then there are no strong continuous Sudoku grids. Even if there are no weak Sudoku grids, it may be possible to slightly modify the definitions to permit small exceptions such as in the above example.

The main question I'm asking is: Do any weak continuous Sudoku grids exist, and if they do, do any strong continuous Sudoku grids exist?

答:
Weak continuous Sudoku:

A weak continuous Sudoku can be constructed based on the ideas that you already provided.

First, we construct a weak continuous Sudoku for the set $U=(0,1]$ instead of $U=[0,1]$.
Here, a weak continuous Sudoku can be constructed by using the function
$f$ from your attempt but as a function $f:(0,1]^2\to (0,1]$ (since one boundary is gone, the problems that you observed are now gone, too).
Then, choose a bijection $h:[0,1]\to (0,1]$
(an explicit bijection can be constructed if you prefer a constructive soution).
Then we define
$$
g:[0,1]^2\to [0,1],
\qquad
(x,y)\mapsto h^{-1} (f(h(x),h(y))).
$$
This function $g$ then can be shown to be a weak continuous Sudoku.

Strong continuous Sudoku:

As for strong continuous Sudoku, things get more complicated
and it would be a lot of work to explain my construction in full detail,
but I can provide a sketch.

First, the bijection $h$ above should be chosen such that
each interval in $[0,1]$ contains a subinterval $[ a,b ]$ such that $h(x)=x$
for all $x\in[a,b]$, see the comments below for such a construction.
Furthermore, it uses a bijection $j:[0,1]\to [0,1]$
such that $j((c,d))$ is dense in $[0,1]$ for all intervals $(c,d)$,
see the comments below for such a construction for $j$.

Then one can mix the rows or columns of the previous weak Sudoku according to $j$,
i.e. $\tilde g(x,y)=g(j(x),y)$.
This function $\tilde g$ should then be a strong continuous Sudoku.
Let me provide a rough sketch how this can be done.

Let $S$ be a square sub-region of $[0,1]^2$.
Let $S_2=[a,b]\times [c,d]\subset S$ be a smaller square sub-region,
where $a<b,c<d$ are such that
$h(x)=x$ holds for all $x\in[a,b]\cup[c,d]$
(such a sub-region exists due to the comments above on the choice of $h$).
It suffices to show that $\tilde g(S_2)=[0,1]$ instead of $\tilde g(S)=[0,1]$.

Let $t\in [0,1]$ be given.
Let $m:=(c+d)/2$.
Since $j([a,b])$ is dense in $[0,1]$,
the function values $\{\tilde g(x,m)| x\in[a,b]\}$ are also dense in $[0,1]$.
Let $s\in[a,b]$ be such that $\tilde g(s,m)$ is close to $t$ in the sense that
$$
t-\frac{d-c}{2} < \tilde g(s,m) < t+\frac{d-c}{2}.
$$
By exploiting the definitions of $\tilde g,g,f$ we have
$\tilde g(s,m+x)=\tilde g(s,m)+x$
for $x\in (-\frac{d-c}{2},\frac{d-c}{2})$
(with the exception that the values wrap around at $1$).
By setting $x=\tilde g(s,m)-t$,
we get $t=\tilde g(s,m+x)$ and $(s,m+x)\in S_2 = [a,b]\times [c,d]$.
Thus $t$ can be reached and the condition (5.) for strong continuous Sudoku
is satisfied.

on the existence of a function $h$:

We can define $h:[0,1]\to (0,1]$ by setting $h(0)=1/2$, $h(1/2)=1/3$, $h(1/3)=1/4$,
etc., and $h(x)=x$ for all other $x$.
Then for each interval one can find a sufficiently small subinterval
$[a,b]$ such that $h(x)=x$ for all $x\in[a,b]$.

on the existence of a function $j$:

This is more complicated, so let me provide a rough sketch.
Let $(q_k)_k$ be an enumeration of the rational numbers in $[0,1]$
and let $I_k$ be an interval of length $2^{3-2k}$ centered at $q_k$.
We define the sets
$$ A_k := I_k\setminus \bigcup_{l>k} I_l.$$
These sets form a partition of $[0,1]$ and each set $A_k$ has cardinality
equal to $[0,1]$.

Let $(B_k)_k$ be another sequence of subsets of $[0,1]$ which
form a partition of $[0,1]$ such that each $B_k$ is dense and has
cardinality equal to $[0,1]$
(such a partition exists, one can append dense countable sets with enough other elements to form sets $B_k$,
but I think this requires the axiom of choice).
Then we construct $j$ by (bijectively) mapping
$A_k$ to $B_k$.

Since the lengths of the sets $A_k$ get smaller and smaller
and the rationals $q_k$ are dense,
each interval has a subinterval of the form $I_k$.
Since $I_k$ contains $A_k$ and $A_k$ is mapped to a dense set $B_k$,
we obtain the desired property that $j(I_k)$ is dense in $[0,1]$.


问:
Conditional probability calculation
Given $P(B \cap A') = 0.25$, $ P(A \cap B') = 0.2$ and $P(A' \mid B') = 0.5$, how do we find $P( A \cap B)$, and $P( B' \cap A')$?

(Here $A', B'$ are the complements of $A$ and $B$ respectively)

答:
From venn diagrams

$P(only~ B)=0.25$, $P(only~A)=0.2$, $P(\overline{A\cup B})/P(\overline{B})=0.5$

Let $P[A\cap B]= x$

$$\implies \frac{0.55-x}{0.75-x}=0.5 \implies x=0.35$$

Now you can find the rest.

Title: Conditional probability calculation


问:
A closed set is intersection of countable collection of open sets
I am currently reading Apostol Mathematical Analysis. There was this question

Question : Prove that a closed set in $\mathbb{R}^1$ is the intersection of countable collection of open sets.

[Here $N_a(\varepsilon) =$ the open set $(a-\varepsilon, a+\varepsilon)$ ]

My Attempt : Let $A$ be a closed set. Let $A^c$ denote the complement of $A$. Take $G$ be a collection of neighborhoods where for each $N_a(\varepsilon)\in G $ where $a\in \mathbb{Q}\cap A^c=A'$(Let) and $\varepsilon$ is the minimum real number such that $A\subseteq N_a(\varepsilon) $

As the set of rational numbers are countable hence the set $G$ is also countable. Hence $$A=\bigcap_{S\in G}S$$

Can we write the last statement ?

答:
Here's an approach that works more generally in any metric space. As in the statement of the question, we will let $N_r(x)$ be the set of points whose distance to $x$ is less than $r$. For all $n$, let
$$U_n = \bigcup_{a \in A} N_{1/n}(a).$$
This is a union of open sets, and hence is open itself. I claim that
$$\bigcap_{n \in \Bbb{N}} U_n = A.$$
It's clear that $A \subseteq U_n$ for each $n$, simply because each $a \in A$ belongs in $N_{1/n}(a) \subseteq U_n$.

On the other hand, suppose that $x \notin A$. Then some open neighbourhood of $x$ exists that fails to intersect $A$, i.e. there is some $\varepsilon > 0$ such that $N_{\varepsilon}(x) \cap A = \emptyset$. Fix some $n$ such that $1/n < \varepsilon$. If we had $x \in U_n$, then some $a \in A$ exists such that $x \in N_{1/n}(a)$. That is, the distance from $x$ to $a$ is no larger than $1/n$, which is less than $\varepsilon$. But then $a \in N_\varepsilon(x) \cap A = \emptyset$, a contradiction. Therefore, $x \notin U_n$, and hence $x \notin \bigcap_{n \in \Bbb{N}} U_n$, completing the proof.


问:
A simple doubt about rigor.
Using the rigor of analysis, how do I argue that a sequence $a_{n}$, tends to be "fast" than another sequence $b_{n}$?

For exemple, considering the case where, $a_n \to 0$ and $b_n \to \infty$, I want to proof that $a_n b_n \to 0$

I have a intuitive idea of calculus, but I can't express it on paper.

答:
The concept you are looking for is Little-o notation

For instance $n^2\to \infty$ and $e^{-n}\to 0$ as $n\to\infty$.

As $$\underset{n\to \infty }{\text{lim}}\frac{n^2}{e^n}=0$$
we say that $n^2 =o(e^n)$.

In your example if $a_n\to 0$ and $b_n\to\infty;\;\frac{1}{b_n}\to 0$ if $a_n=o(b_n)$ then $$\underset{n\to \infty }{\text{lim}}\frac{a_n}{b_n}=\underset{n\to \infty }{\text{lim}}a_n\cdot b_n=0$$
Intuitively, $a_n$ goes to $0$ faster than $b_n$ goes to $\infty$

Title: A simple doubt about rigor.


答: Here's a weak solution. Using your favorite bijection, replace $[0,1]$ with the Cantor group $2^\mathbb N$ of infinite binary sequences. Then let $f(x,y)=x+y$. That is, just use the group operation: pointwise XOR.

Title: Are there any valid continuous Sudoku grids?

Body:
A standard Sudoku is a $9\times 9$ grid filled with digits such that every row, column, and $3\times 3$ box contains all the integers from $1$ to $9$.

I am thinking about a generalization of Sudoku which I call "continuous Sudoku", which consists of a unit square where every point on that square corresponds to a real number. The rules for continuous Sudoku are designed to be analogous to the rules for standard Sudoku, and I've devised two different rulesets:

The first ruleset I call "weak" continuous Sudoku. In weak continuous Sudoku, the only restriction is that every row and column of the square contains every real number in the interval $[0,1]$ exactly once.

The second ruleset I call "strong" continuous Sudoku. In strong continuous Sudoku, the rules of weak continuous Sudoku apply, and, in addition, every square sub-region of the unit square contains every real number in the interval $[0,1]$ at least once. This is analogous to the $3\times 3$ box restriction in standard Sudoku.

Let $U = [0,1]$ and $U^2 = U\times U$. More precisely, a weak continuous Sudoku is essentially a function $f:U^2\to U$, which satisfies the following four properties:

If $x,y_1,y_2\in U$ and $y_1\neq y_2$, then $f(x,y_1)\neq f(x,y_2)$.

If $x_1,x_2,y\in U$ and $x_1\neq x_2$, then $f(x_1,y)\neq f(x_2,y)$.

If $x\in U$ then $\{z: f(x,y)=z,y\in U\} = U$.

If $y\in U$ then $\{z: f(x,y)=z,x\in U\} = U$.

Now, strong continuous Sudoku is a bit harder to define precisely. A set $S$ is a square sub-region of $U^2$ iff $S\subseteq U^2$ and there exists $z = (z_1,z_2)\in U^2$ and $r>0$ such that $S = \{(x,y)\in U^2:z_1\leq x\leq z_1+r,z_2\leq y\leq z_2+r\}$. Thus, using this definition, a strong continuous Sudoku is a weak continuous Sudoku which satisfies the following additional property:

If $S$ is a square sub-region of $U^2$, then $f(S) = U$.

I've been trying to look for specific examples of both strong and weak continuous Sudoku grids, but have so far been unsucessful.

I'm not sure whether any weak continuous Sudoku exists. My first attempt:
$$
f(x,y)=\begin{cases} x+y &\text{if }x+y\leq 1 \\
x+y-1 & \text{if }x+y>1\end{cases}
$$
almost works. It satisfies properties $3$ and $4$, and almost, but not quite, satisfies $1$ and $2$. The issue occurs only at boundaries of the square, for example, $f(0.5,0) = 0.5$ and $f(0.5,1)=0.5$.

Any example of a strong continuous Sudoku will likely need to be an extremely discontinuous pathological function, similar to the Conway base 13 function. Obviously, if there are no weak continuous Sudoku grids, then there are no strong continuous Sudoku grids. Even if there are no weak Sudoku grids, it may be possible to slightly modify the definitions to permit small exceptions such as in the above example.

The main question I'm asking is: Do any weak continuous Sudoku grids exist, and if they do, do any strong continuous Sudoku grids exist?


问:
Calculus & Mathematical functions to design rollercoaster.
I am designing a roller coaster using functions (ie. linear, cubic, logarithmic).
I need the roller coaster to have no sudden changes in gradient - smooth transitions between say a linear function and a parabola.

If I had y = 30 (over domain {0<x<20}) & y = -((x-20)/8)^2)+30 (over domain {20<x<60}; how would I prove there is a smooth transition mathmatically? If I plot this I can tell visually but I would like to know how to prove this mathmatically if possible?

Also I would like to determine that gradient of the parabola function above (so I can work out the angle it drops at). Can someone explain a method of doing this mathmatically?

All assistance is greatly appreciated.

答:
This is precisely the field of application of the Bezier's curve.

They are widely used in designing car bodies and similar, because they respond to the need you are telling.


问:
Seems that I just proved $2=4$.
Solving $x^{x^{x^{.^{.^.}}}}=2\Rightarrow x^2=2\Rightarrow x=\sqrt 2$.

Solving $x^{x^{x^{.^{.^.}}}}=4\Rightarrow x^4=4\Rightarrow x=\sqrt 2$.

Therefore, $\sqrt 2^{\sqrt 2^{\sqrt 2^{.^{.^.}}}}=2$ and $\sqrt 2^{\sqrt 2^{\sqrt 2^{.^{.^.}}}}=4\Rightarrow\bf{2=4}$.

What's happening!?

答:
Let's add the hypothesis that $x>0$ to the problem, so that it's clear your derivations are correct.

Pay attention to what you've proven:

If $x^{x^{\cdot^\cdot}} = 2$, then $x = \sqrt{2}$

If $x^{x^{\cdot^\cdot}} = 4$, then $x = \sqrt{2}$

This is very different from

If $x = \sqrt{2}$, then $x^{x^{\cdot^\cdot}} = 2$

If $x = \sqrt{2}$, then $x^{x^{\cdot^\cdot}} = 4$

Your argument that $2=4$ requires this latter pair of statements, but you haven't proven either of them; instead, what you've proven are the first pair of statements!

It's easy to get in the habit of forgetting about the direction you've argued a problem, and in many situations, arguments are reversible, making it hard to see why direction matters. But this is an example of the dangers of getting things wrong!

Incidentally, if $x = \sqrt{2}$ then $x^{x^{\cdot^\cdot}} = 2$ is correct, if we assume the usual meaning of infinite power towers as a limit of finite ones. If you're familiar with limits of sequences, then you can use an inductive proof to show that the sequence

$$ a_0 = \sqrt{2} \qquad \qquad a_{n+1} = \sqrt{2}^{a_n} $$

is strictly increasing and bounded above by $2$, and so the limit converges. And if $L$ is the limit, then because exponentiation is continuous, we can take the limit of the recursive relation to see that

$$L = \sqrt{2}^L$$

letting you complete the proof.

Title: Seems that I just proved $2=4$.


问:
Nonnegative graphic integers proof
Let $d = (d_{1}, d_{2},...,d_{n})$ be a sequence of nonnegative integers that is graphic. Prove that for each integer k = $1,2,...,n$,
$$
\sum_{i=1}^kd_{i} \leq k(k-1) + \sum_{i=k+1}^n \min \{k,d_{i}\}.
$$

I am not exactly sure on how to proceed. I know that a graphic sequence is a sequence of numbers which can be the degree sequence of some graph (from Wolfram). I am still quite new to this topic, so I thought of using mathematical induction, but the base case $\sum_{i=1}^1d_{i} = d_{1} \leq 1(1-1) + \sum_{i=2}^n \min \{1,d_{i}\}$ already seems complicated. Please help me, thank you!

答: Hint: fix a graph with those degrees. $\sum_{i=1}^kd_i$ counts the number of edge-ends meeting the first $k$ vertices. Every edge-end counted is either from an edge between two vertices in the first $k$, or between one of these vertices and one of the other vertices.

Title: Nonnegative graphic integers proof


问:
What is the size of a multinomial?
The answer to this question uses the phrase "multinomial of size".

What is the definition of the size of a multinomial? They are using a negative multinomial.

答: The size is a metric on on an elementary event, which is defined as a sequence of multinomial events before a stopping event occurs. Each multinomial event can have $n$ outcomes where $n \geq 2$. It is not a metric on the number of outcomes of each trial.

Title: What is the size of a multinomial?


问:
a problem related to continuous functions
Let $f$ be a function defined in $[0,1]$ such that $f(0) > 0 > f(1)$. Suppose there is a continuous function $g$ on $[0,1]$ such that $f + g$ is increasing. Prove that the equation $f(x)=0$ has at least one solution in $[0,1]$. I tried to consider another continuous function in order to use the theorem of intermediate values but I failed. Thanks a lot.

答:
An increasing function has only jump discontinuities, with left and right limits at every point:
$$
\forall a \in (0, 1] : \lim_{x \to a-} (f(x) + g(x)) \le f(a) + g(a) \, ,\\
\forall a \in [0, 1) : f(a) + g(a) \le \lim_{x \to a+} (f(x) + g(x)) \, .
$$
But $g$ is continuous, so that
$$
\forall a \in (0, 1] : \lim_{x \to a-} f(x) \le f(a) \, , \\
\forall a \in [0, 1) : f(a)  \le \lim_{x \to a+} f(x) \, .
$$
Now define
$$
 c = \sup \{ x \mid f(x) > 0 \} \, .
$$
Then
$$
c > 0 \implies 0 \le f(c) \, ,\\
c < 1 \implies f(c) \le 0 \, .
$$
Since $f(0) > 0 > f(1)$ is given, the only possibility is that $0 < c < 1$ and $f(c) = 0$.


问:
Finding stable, unstable and centre manifold

Find the stable, unstable, and center manifolds of the following system
\begin{align}
\dot{x}& = -x,\\
 \dot{y}& = -2y + 2z^2(1-z^2), \\
 \dot{z} &= -z^3
\end{align}
at the origin. Prove that origin is the local attractor.

Solving the above system I got: $$x = x_0 \ e^{-t}, \ y(t) = \frac{1}{2}\bigg[\frac{1}{t+z_0} + \bigg(\frac{2y_0  z_0 -1}{z_0}\bigg)e^{-2t} \bigg], \ z(t) = \sqrt{\frac{1}{2(t+z_0)}}$$

Stable manifold: $$W^{s}(0,0,0) := \{(x,y,z):x(t), y(t) \text{ and } z(t) \to 0 \text{ as } t \to +\infty\} =\{(x,0,z)\}$$

Unstable manifold: $$W^{u}(0,0,0) := \{(x,y,z):x(t), y(t) \text{ and } z(t) \to 0 \text{ as } t \to -\infty\} = \{(x,y,z):
x=0, yz=\frac{1}{2} \} $$

I do not whether my calculations are right or not. I do not know how to prove the second part.

答:
First note that the third equation has the solution
$$
z(t)^{-2}-z_0^{-2}=2t\implies z(t)=\frac{z_0}{\sqrt{1+2z_0^2t}}
$$
where now indeed $z(0)=z_0$.

Next, you can be a little creative in solving the second equation by inserting the third to reduce the 4th degree term,
$$
\dot y+2y=2z^2-2z^4=2z^2+2z\dot z=\frac{d}{dt}(z^2)+2z^2.
$$
This means that "luckily" the terms turn out to be that the same differential operator $(D+2)$ is applied to both sides. The solution then is
$$
y(t)-z(t)^2=(y_0-z_0^2)e^{-2t},\\
y(t)=\frac{z_0^2}{1+2z_0^2t}+(y_0-z_0^2)e^{-2t}.
$$
I do not see how there can be any unstable manifold. From points outside the set with $x_0=0$ and $y_0=z_0^2$ the solutions will converge exponentially fast towards points within that set, and then slowly along that set towards the origin. In my mind this was the characteristic of a center manifold?


问:
Why is the curl of the gradient of the polar angle phi zero, counter to my intuition?
I was trying to test my intuition about curl and divergence by constructing vector functions whose curl and divergence I think I have intuitive expectations about. An idea I came up with was to think of a scalar function with some shape whose gradient I would intuitively expect to have the desired property. I thought, for example, the function $\phi=\arctan(y/x)$ can be geometrically thought of as a staircase of linearly increasing angle winding around the origin. Therefore I thought that $\vec{\nabla}\phi$ would be a vector field whirlpooling around the z-axis (i.e. in the x-y plane), and therefore would have a nonzero curl in the z direction at (0,0). However in this case $\vec{\nabla}\times\vec{\nabla}\phi = \vec{0}$ for all (x,y), contrary to my expectation. What is wrong about my intuition here?

答:
If $\phi=arctan(y/x)$ is electric scalar potential, I believe that $\phi$ is single valued.
If $\phi$ is single valued, the vector $\nabla\phi$ directs in reverse within the small region $(2\pi-\epsilon)<\theta<2\pi$, where $\epsilon$ is some positive small number.
Thus the vector $\nabla\phi$ field is not whirlpooling around the z-axis.


问:
Prove $f$ is injective if and only if $g$ is surjiective
Problem: Give mapping $f,g: X \to X$ and $f \circ g$ is bijective. Prove that
$$f \text{ is injective} \iff g \text{ is surjective}.$$
My attempt:

($\Leftarrow$)Give $y,y' \in X$ satisfy $f(y)=f(y')$. Since $g$ is surjective then exists $x,x' \in X$ that $y=g(x),y=g(x')$. This yields
$$f[g(x)]=f[g(x')].$$
Thus, $f\circ g(x)=f\circ g(x')$. And then, we have $x=x'$ since $f\circ g$ is bijiective. Therefore, $g(x)=g(x')$ or $y=y'$.

So, $f$ is injective.

($\Rightarrow$) I am stucking here now.

Loof forward to your support. Thanks

答:
$f\circ g$ is bijective implies that $f$ is onto and $g$ is one-one.

Given $f$ is one-one, so $f$ is a bijection, so is $f^{-1}$.

So $f^{-1}\circ(f\circ g)=(f^{-1}\circ f)\circ g=g$ is a composition of bijections and hence a bijection, i.e. $g$ is onto.

You can use similar reasoning for reverse implication.


问:
Bounded linear operator from orthonormal sequence
I'm working on the following problem:

Suppose $\{e_n\}$ is a complete orthonormal sequence in a Hilbert space $H$. Let $\{c_n\}$ be a sequence of complex numbers.

Prove that there is a bounded linear operator $T$ on $H$ such that $Te_n=c_ne_n$ for each $n$, if and only if $\{c_n\}$ is bounded.

Determine $\|T\|$ when $\{c_n\}$ is bounded.

Determine $T^*$, the adjoint of $T$, when $\{c_n\}$ is bounded.

What I've done so far: I know that if $\{e_n\}$ is a complete orthonormal sequence then any $x\in H$ can be written as $\sum_{1}^\infty\langle x,e_n\rangle e_n$ and $\|x\|^2=\sum_1^\infty|\langle x,e_n\rangle|^2$. Then if $T$ is linear then $Tx=T(\sum\langle x,e_n\rangle e_n)=\sum\langle x,e_n\rangle c_ne_n$. But I'm not sure how to proceed.

Any help on this problem would be greatly appreciated!

答:
Suppose $(c_n)$ is bounded. $\|\sum c_n\langle x, e_n \rangle e_n\|^{2}=\sum |c_n\langle x, e_n \rangle |^{2} \leq M \sum |\langle x, e_n \rangle |^{2}=M^{2}\|x\|^{2}$ where $M =\sup_m |c_n|$. Hence   $T$ is bounded with$\|T\| \leq \sup_m |c_n|$.

Also $\|Te_n\| =|c_n|$ and this implies (by definition of norm of an operator) that $\|T\| \geq |c_n|$. This is true for each $n$ so we conclude that $\|T\|=\sup_m |c_n|$.

Conversely,  $Te_n=c_n e_n$ show that $|c_n| \leq \|T\|$ for each $n$ so boundedness of $T$ implies that $(c_n)$ is bounded.

$T^{*}$ is given by $T^{*}x=\sum \overline {c_n} \langle x, e_n \rangle e_n $ and I will let you verify it using definitiom of the adjoint.


答:
Write $h:=f\circ g$ and $h^{-1}$ for its inverse.

Assume $f$ is injective. Let $y\in X$ be given. Then $$f(g(h^{-1}(f(y))))=h(h^{-1}(f(y))= f(y)$$
so that by injectivity of $f$,
$$ g(h^{-1}(f(y))) = y,$$
which shows that $y=g(x)$ for $x=h^{-1}(f(y))$.

Title: Prove $f$ is injective if and only if $g$ is surjiective

Body:
Problem: Give mapping $f,g: X \to X$ and $f \circ g$ is bijective. Prove that
$$f \text{ is injective} \iff g \text{ is surjective}.$$
My attempt:

($\Leftarrow$)Give $y,y' \in X$ satisfy $f(y)=f(y')$. Since $g$ is surjective then exists $x,x' \in X$ that $y=g(x),y=g(x')$. This yields
$$f[g(x)]=f[g(x')].$$
Thus, $f\circ g(x)=f\circ g(x')$. And then, we have $x=x'$ since $f\circ g$ is bijiective. Therefore, $g(x)=g(x')$ or $y=y'$.

So, $f$ is injective.

($\Rightarrow$) I am stucking here now.

Loof forward to your support. Thanks


问:
Calculate distance from plane to parallel plane in O using vector and normal
I'm trying to figure out what's the best method to get the distance between two planes where i have the normalized vector of the plane and a point in the plane. What I want to do is to create a parallel plane where 0,0,0 is in and then calculate the distance between the two planes.
My thought was that it's actually the same like the distance between the plane and 0,0,0 which I tried just to calculate as the dot product between the normalized vector and the given point.

e.g.: (The values are that strange because they are from a 3D camera system)

```<code>{ 0.00102464424908972; -1.5806471007608384E-4; -0.0014898099362794807} // normalized vector (plane normal)
{ 1.2168973210867913; -0.2862541104934054; -0.39146720489920095} //given point
0.0018753452463538794  //calculated distance  ?
</code>```

I'm not sure but this looks not like the result I expected. Is it the right formula? I tried to use the Hessesche Normalform

Update: somehow I droped my normalize function and the normal was not normalized. now my results look much better. Thankf for all answers.

答:
Call $\pi$ the plane, $p$ a point in $\pi$, and $n$ a vector normal to the plane. Let $\pi'$ be the plane through $(0,0,0)$ parallel to $\pi$. Then the distance between $\pi$ and $\pi'$ is indeed given by $\frac{\lvert n\cdot p\rvert}{\lVert n\rVert}$.

To see this, recall that the distance between two parallel planes is given by the length of the segment between any couple of points (one for each plane) lying on a line orthogonal to both. To get the Cartesian equations of $\pi$, let $d=-n\cdot p$ and $n=(a,b,c)$, so that
$$
\pi\colon\; ax+by+cz+d=0
$$
and the line through $(0,0,0)$ orthogonal to $\pi$ has parametric equations
$$ r\colon \;
\begin{cases}
x=at\\
y=bt\\
z=ct
\end{cases}
$$
Intersecting $\pi$ and $r$ gives
$$
(a^2+b^2+c^2)\bar{t}+d = 0 \;\Rightarrow\; \bar{t}=\frac{-d}{a^2+b^2+c^2} = \frac{\lvert n\cdot p\rvert}{\lVert n\rVert^2}
$$
Finally, the distance between $\pi$ and $\pi'$ is the norm of the vector $(a\bar{t},b\bar{t},c\bar{t})$, i.e.
$$
\sqrt{(a\bar t)^2+(b\bar t)^2+(c\bar t)^2}=\lvert \bar{t}\rvert\sqrt{a^2+b^2+c^2}=\frac{\lvert n\cdot p\rvert}{\lVert n\rVert^2}\lVert n\rVert=\frac{\lvert n\cdot p\rvert}{\lVert n\rVert}
$$


问:
Global contact form defining contact structure
Let $M$ be a $(2n+1)$-manifold and $\xi$ a smooth hyperplane field. It's easy to see that for each point $p$ exists a neighborhood $U$ of $p$ such that $\xi_{|U} = \ker \alpha_{U}$, where $\alpha_U$ is a smooth 1-form. I have read that if $M$ and $\xi$ are both oriented, we can find $\alpha$ $\textbf{globally}$ defined such that $\xi = \ker \alpha$ in all $M$. I can't see why, I mean, if $M$ is oriented by a volume form $\omega \in \Omega^{2n+1}(M)$, without supposing any kind of regularity in $\xi$ I don't know how to construct this globally defined 1-form. This exercise is proposed in $\textit{Introductory Lectures on Contact Geometry}$ by John B. Etnyre.

EDITED: In this post have a nice proof of this statement. I am going to introduce another proof based on it without using Riemannian metrics. When we say that $\xi$ is orientable we refer to $TM/\xi$ is the trivial bundle (by definition). We will say that $\xi$ is $\textit{corientable}$. In fact, since $\xi$ is a field of hyperplanes the previous quotient is a line bundle, that is, a 1-dimensional bundle.

Suposse $\xi$ is coorientable. If $TM/\xi$ is trivial, there exists a $\textbf{global}$ section throughout $TM/\xi$ and therefore a global section $\overline{\alpha}$ in $(TM/\xi)^*$. Consider $\pi : TM \to (TM/\xi)^*$ and the pullback $\alpha := \pi^*\overline{\alpha}$. In fact, this $\alpha$ satisfies $\ker \alpha = \xi$ by definition, and it's globally defined. Therefore we have proved that

$$
\xi \text{ coorientable} \Longrightarrow \alpha \text{ globally defined}
$$

Reciprocally, assume $\alpha$ globally defining $\xi = \ker \alpha$ throughout $M$. Then this $\alpha$ induces a non-zero section in $TM/\xi$, that's mean, $TM/\xi$ is trivial.

答:
Your exercice does not use the usual co-orientability property of globally defined contact $1$-form, but use the orientability of $M$ and $\xi$ instead.
Here is an idea. $M$ is orientable, that means that $\Lambda^{2n+1}TM$ is a trivial line bundle. $\xi$ is orientable, so $\Lambda^{2n}\xi$ is a trivial line bundle. Given a riemannian metric $g$ on $M$, you can define $\xi^{\perp}$ as the $g$-orthogonal supplementary of $\xi$ in $TM$. Thus, you can show that

\begin{align}
\Lambda^{2n+1}TM = \Lambda^{2n}\xi\otimes \Lambda^1 \xi^{\perp}
\end{align}
And triviality of the two first line bundles implies the triviality of $\Lambda^1\xi^{\perp}$ as a line bundle. Thus you can find a non-vanishing section of $\Lambda^1\xi^{\perp}$, that is a non-vanishing $1$-form in $\Gamma(\Lambda^1\xi^{\perp})$, which will be a globally defined contact $1$-form.


问:
Is this relation true for orthonormal matrices elements?
Let $R$ be a $3 \times 3$ proper orthogonal and normal matrix such that $RR^T = I$ and $det(R) = 1$, so basically a rotation matrix.

$R = \begin{bmatrix} r_{11} & r_{12} & r_{13} \\ r_{21} & r_{22} & r_{23} \\ r_{31} & r_{32} & r_{33} \\  \end{bmatrix}$

Is the following relation true?

$r_{13} = r_{32}r_{21} - r_{22}r_{31}$

答:
We have:

$$\mathbf{RR^T}=\mathbf{I} \ \iff \ \mathbf{R^{-1}}=\mathbf{R^T}\tag{1}$$

Remember now the classical formula

$$\mathbf {A} ^{-1}=\frac{1}{\det(\mathbf {A})}\operatorname {adj} (\mathbf {A} )\tag{2}$$

where $\operatorname {adj} (\mathbf {A})$ is the adjugate matrix of square matrix $\mathbf {A}$, defined as the transpose of its matrix of cofactors.

Taking $\mathbf {A}=\mathbf {R}$ in (2), we obtain, using (1), the identity:

$$\mathbf {R^{-1}}= \operatorname {adj} (\mathbf {R} )=\mathbf {R^T}
 \implies \mathbf {R}= (\operatorname {adj} \mathbf {R)^T} $$

Otherwise said, $\mathbf {R} $ is equal to the matrix of its cofactors (without need to transpose it), explaining in particular your identity.


问:
Question about polar decomposition. Is the book wrong?
An introduction to the classification of amenable C*-algebra.

page 140 Lemma 3.5.1 Let $x\in A$ with the polar decomposition $x=u|x|$ in $A''$ and $B=\overline{x^*Ax}$. Then $ub\in B$ for every $b\in B$.

$A''$ refers to the enveloping C*-algebra of $A$, the weak closure of $A$ in $B(H)$ where $A$ is universally represented. But these are not important.

Let $A$ be $B(l^2(\mathbb N))$ and let $x$ be the shift operator such that $x(e_j)=e_{j-1}$. Then $\overline {x^*Ax}=B(0\oplus l^2(\mathbb N^+))$ (regarding $B(0\oplus l^2(\mathbb N^+))$ as a subalgebra of $B(l^2(\mathbb N))$). Since $x$ is a partial isometry itself so $x=x|x|$ and $|x|$ is the projection onto $0\oplus l^2(\mathbb N^+)$.
However, $x=x|x|\not\in \overline{x^*Ax}$ since no element in $\overline{x^*Ax}=\overline{|x|A|x|}$ has range larger than $0\oplus l^2(\mathbb N^+)$.

Am I wrong, or is the book wrong?

答:
Yes, the book is wrong. It was supposed to be $ub \in A$ for all $b \in B$.

See also Farah's book Corollary 1.6.13.


问:
Understanding equation manipulation on manifolds
I cannot understand a specific equation manipulation of the paper:

C. Forster, L. Carlone, F. Dellaert and D. Scaramuzza, "On-Manifold Preintegration for Real-Time Visual--Inertial Odometry," in IEEE Transactions on Robotics, vol. 33, no. 1, pp. 1-21, Feb. 2017, doi: 10.1109/TRO.2016.2597321

Specifically, I do not understand the equation manipulation from line 1 to line 2 of equation (35) in the paper. In this question i stick to the same equation numbering as in the paper and simply copy the equations. They define the following properties of the exponential maps of the $\operatorname{SO}(3)$ manifold:

\begin{align}
 R \operatorname{Exp} (\phi) R^\top &= \operatorname{exp} (R \mathbf{\phi}^\wedge R^\top ) = \operatorname{Exp} (R\phi) \tag{10}\label{eq10} \\
\Leftrightarrow \operatorname{Exp}(\mathbf{\phi}) R &= R \operatorname{Exp} (R^\top \phi)
\tag{11}\label{eq11}
\end{align}

They say that they rearrange line 1 of equation \eqref{eq35} by using the relation \eqref{eq11}:

\begin{align}
&\dotsc \Pi_{k=i}^{j-1}
\big[
\operatorname{Exp} ( ( \mathbf{\tilde{\omega}}_k - \mathbf{b}_i^g ) \Delta t)
\operatorname{Exp} \big( -J_r^k\mathbf{\eta}^{gd}_k  \Delta t \big)
\big] \\
&\stackrel{\eqref{eq11}}{=} \Delta \tilde{R}_{ij} \Pi_{k=i}^{j-1}
\big[
\operatorname{Exp} \big( - \color{red}{\Delta \tilde{R}_{k+1 j}^\top } J_r^k \mathbf{\eta}^{gd}_k \Delta t \big)
\big] \tag{35}\label{eq35}  ,
\end{align}

where
$\Delta \tilde{R}_{ij} \doteq \Pi_{k=i}^{j-1} \operatorname{Exp} (( \mathbf{\tilde{\omega}} - \mathbf{b}_i^g ) \Delta t ) $.

But I do not understand where the red marked term
$ \color{red}{\Delta \tilde{R}_{k+1 j}^\top } $
comes from. As far as I am concerned the red term appears magically since they summarize all factors

$\Delta \tilde{R}_{ij} \doteq \Pi_{k=i}^{j-1} \operatorname{Exp} (( \mathbf{\tilde{\omega}} - \mathbf{b}_i^g ) \Delta t ) $
into one term and it simply remains
$ \Pi_{k=i}^{j-1}
\big[
\operatorname{Exp} \big( -J_r^k\mathbf{\eta}^{gd}_k  \Delta t \big)
\big] $
without $ \color{red}{\Delta \tilde{R}_{k+1 j}^\top } $ . What am I missing here?

答:
I found the answer after working on it for a while. Using

\begin{align}
\operatorname{Exp} ( \mathbf{\phi} ) R = R \operatorname{Exp} (R^\top \phi) \tag{1}\label{eq1}
\end{align}

Weird the above formula is displayed correctly in editing mode but is wrongly displayed in the answer for me.

and defining
$(\mathbf{\tilde{\omega}}_k- \mathbf{b}_i^g)\Delta t = a_k$
and
$-J_r^k\mathbf{\eta}_k^{gd} \Delta t = b_k $ we have:

\begin{align}
&\dotsc \Pi_{k=i}^{j-1}
\big[
\operatorname{Exp} ( ( \mathbf{\tilde{\omega}}_k - \mathbf{b}_i^g ) \Delta t)
\operatorname{Exp} \big( -J_r^k\mathbf{\eta}^{gd}_k  \Delta t \big) \\
&= \Pi_{k=i}^{j-1}
\big[ \operatorname{Exp} (a_k) \operatorname{Exp} (b_k) \big] \\
&=  \operatorname{Exp} (a_i) \operatorname{Exp} (b_i) \operatorname{Exp} (a_{i+1}) \operatorname{Exp} (b_{i+1}) .. \operatorname{Exp} (a_{j-2}) \operatorname{Exp} (b_{j-2}) \operatorname{Exp} (a_{j-1}) \operatorname{Exp} (b_{j-1}) \\
&= R_i \operatorname{Exp} (b_i) R_{i+1} \operatorname{Exp} (b_{i+1}) .. R_{j-2} \operatorname{Exp} (b_{j-2}) R_{j-1} \operatorname{Exp} (b_{j-1}) \\
&\stackrel{\eqref{eq1}}{=} R_i \operatorname{Exp} (b_i) R_{i+1} \operatorname{Exp} (b_{i+1}) .. R_{j-2} R_{j-1} \operatorname{Exp} (R_{j-1}^\top b_{j-2})  \operatorname{Exp} (b_{j-1}) \\
&\stackrel{\eqref{eq1}}{=} R_i \operatorname{Exp} (b_i) R_{i+1}..R_{j-1} \operatorname{Exp} (R_{i+2}^\top..R_{j-1}^\top b_{i+1}) .. \operatorname{Exp} (R_{j-1}^\top b_{j-2})  \operatorname{Exp} (b_{j-1}) \\
&\stackrel{\eqref{eq1}}{=} R_i..R_{j-1} \operatorname{Exp} (R_{i+1}^\top..R_{j-1}^\top b_i)\operatorname{Exp} (R_{i+2}^\top..R_{j-1}^\top b_{i+1}) .. \operatorname{Exp} (R_{j-1}^\top b_{j-2})  \operatorname{Exp} (b_{j-1}) \\
&= \Delta \tilde{R}_{ij} \Pi_{k=i}^{j-1} \operatorname{Exp}
\big( \Delta \tilde{R}_{k+1 j}^\top b_k \big) \\
&= \Delta \tilde{R}_{ij} \Pi_{k=i}^{j-1} \operatorname{Exp}
\big( -\Delta \tilde{R}_{k+1 j}^\top J_r^k\mathbf{\eta}_k^{gd} \Delta t \big)
\end{align}

The last simplification is done using $\Delta \tilde{R}_{ij} \doteq \Pi_{k=i}^{j-1} \operatorname{Exp} (( \mathbf{\tilde{\omega}} - \mathbf{b}_i^g ) \Delta t ) = \Pi_{k=i}^{j-1} \operatorname{Exp} (a_k)  $.


问:
Finding the eigenvectors of a matrix.
Find the normalized eigenvectors of this matrix:

$A=\begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix}$

My work:

$$\det(A-\lambda I) =(\lambda-1)^2(\lambda-4)$$

$$\lambda_1=1, \lambda_2=1,\lambda_3=4$$

$$\lambda_1\iff x + y +z=0$$
$$\lambda_2\iff x + y +z=0$$
$$\lambda_3\iff -2x + y +z=0$$
$$\lambda_3\iff x + -2y +z=0$$
$$\lambda_3\iff x + y +-2z=0$$

When my lecturer did this the equations were easy to solve... How do I find the normalized eigevectors from these equations?

Please Help :)

答:
For the matrix:

$$A=\begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix}$$

The CP is given by:

$$|A - \lambda I| = 0$$

For the CP, we get:

$$-\lambda^3+6 \lambda^2-9 \lambda+4 = -(\lambda-4) (\lambda-1)^2 = 0$$

This leads to three eigenvalues as $\lambda_1 = 4, \lambda_2 = 1, \lambda_3 = 1$.

We have a repeated eigenvalue.

To find an eigenvector, for each eigenvalue, we solve:

$$[A - \lambda_i I]v_i = 0$$

When we have repeated eigenvalues, we may need to resort to generalized eigenvectors, which I assumed you learned in class.

Lets find an eigenvector as an example for $\lambda_2 = 1$:

$$[A - \lambda_2 I]v_2 = 0$$

$$[A - \lambda_2 I]v_2 = \begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}v_2 = 0$$

The row-reduced-echelon-form (RREF) is given by:

$$A=\begin{bmatrix}1&1&1\\0&0&0\\0&0&0\end{bmatrix}v_2 = 0$$

So, we can choose an eigenvector to satisfy the RREF for:

$$x + y + z = 0 \rightarrow (-1, 0 , 1)$$

Since we have a repeated eigenvalue, we can choose a second, linearly independent, eigenvector as:

$$(-1, 1, 0)$$

Notice that both of these satisfy the RREF equation.

Repeat the above process for the other eigenvector so you can practice.

This leads to the eigenvalues and eigenvectors as:

$$\lambda_1 = 4, v_1 = (1, 1, 1)$$

$$\lambda_2 = 1, v_2 = (-1, 0, 1)$$

$$\lambda_3 = 1, v_3 = (-1, 1, 0)$$

However, we are being asked to find the normalized eigenvectors, so we have to do an additional step.

The way we normalize is to divide each eigenvector by it's length.

$v_2 = (-1,0, 1)$

$|v_2| = \sqrt{-1^2 + 0^2 + 1^2} = \sqrt{2}$

Thus, the normalized eigenvector is:

$$\tilde v_2 = \frac{v_2}{|v_2|} = \frac{1}{\sqrt{2}}(-1,0, 1)$$

You should verify that the normalized eigenvector has length one.

Repeat this for the other two eigenvectors.

Title: Finding the eigenvectors of a matrix.


问:
Does it make sense to ask what the probability that a coin is biased (in favour of heads) is?
A coin is tossed $5$ times and $5$ heads are observed. What is the probability the coin is biased in favour of heads? Does this question even have any meaning without using a significance level?

Let's have a look at the usual Binomial test.

Let $X$ be the random variable: the number of heads out of $5$ tosses. Then $X \sim B(5,p).\ H_0: p = 0.5;\ H_1: p > 0.5.$

Suppose $H_0$ is true, that is,  $X \sim B(5,0.5)$. Then $P( X \geq 5) = P(X=5) = 0.5^5 = 0.03125.$ This is the probability that the coin actually is fair, and we just happened to get $5$ heads in a row. My question is: is the probability that the coin is not biased in favour of heads equal to $0.03125$? Or like I said at the beginning, is this question meaningless?

答:
Yes, this question makes sense. However, to answer it you need first a prior distribution for the probability of rolling heads (i.e. your initial suppositions on possible values of $p$ presented in form of probability distribution). Because you decide between $p = \frac{1}{2}$ and $p > \frac{1}{2}$ (no matter how much), I, personally, think that the following prior will be most natural in this case: $p$ equals to $\frac{1}{2}$ with probability $\frac{1}{2}$ and is uniformly distributed on $[\frac{1}{2};1]$ otherwise.

Now, you need to find the posterior probability the following way (using Bayes theorem):

$$P(p=\frac{1}{2}|\text{ 5 heads in 5 tosses}) = \frac{P(\text{ 5 heads in 5 tosses}|p=\frac{1}{2})P(p=\frac{1}{2})}{P(\text{ 5 heads in 5 tosses})}$$

Note, that $P(\text{ 5 heads in 5 tosses}|p=p_0) = p_0^5$ for any $p_0 \in [\frac{1}{2}; 1]$ and thus for our prior we have:

$$P(\text{ 5 heads in 5 tosses}) = \frac{1}{2}(P(\text{ 5 heads in 5 tosses}|p=\frac{1}{2}) + \int_{\frac{1}{2}}^{1} P(\text{ 5 heads in 5 tosses}|p=t)dt) = \frac{1}{2}(\frac{1}{32} + \int_{\frac{1}{2}}^{1} t^5 dt) = \frac{1}{64} + \frac{1}{12}(1 - \frac{1}{64}) = \frac{25}{256}$$

Thus the posterior probability of the coin being unbiased is $$P(p=\frac{1}{2}|\text{ 5 heads in 5 tosses}) = \frac{\frac{1}{64}}{\frac{25}{256}} = \frac{4}{25}$$

And from that we can conclude, that the posterior probability of the coin being biased towards heads is:

$$P(p>\frac{1}{2}|\text{ 5 heads in 5 tosses}) = 1 - P(p=\frac{1}{2}|\text{ 5 heads in 5 tosses}) = \frac{21}{25}$$

Note, that you can get a different result from different prior (but the procedure of obtaining it will still be the same).


问:
$\frac{1}{t}e^{F(t)} \to \infty \Rightarrow t F'(t) \to \infty$?
Let $F:[0,\infty)\to[0,\infty)$ be a non-decreasing continuously differentiable function satisfying $F(t)\to\infty$ as $t\to\infty$. Consider the following statements:

$t F'(t) \to \infty$, as $t\to\infty$.

$\frac{F(t)}{\log t} \to \infty$, as $t\to\infty$.

For each $\epsilon>0$, $\frac{1}{t}e^{\epsilon F(t)} \to \infty$, as $t\to\infty$.

For each $\epsilon>0$, $F'(t)e^{\epsilon F(t)} \to \infty$, as $t\to\infty$.

I can derive that
1 $\Rightarrow$ 2 $\Rightarrow$ 3 and 1 $\Rightarrow$ 4 $\Rightarrow$ 3, as follows.

1 $\Rightarrow$ 2: Straightforward application of L'Hôpital's rule.

2 $\Rightarrow$ 3: $\frac{1}{t}e^{\epsilon F(t)} = \exp( \epsilon F(t) - \log t ) = \exp\left[ \left(\epsilon \frac{F(t)}{\log t} - 1 \right) \log t \right]$.

1 $\Rightarrow$ 4: Using 1 and 3, $F'(t)e^{\epsilon F(t)} = t F'(t) \cdot \frac{1}{t}e^{\epsilon F(t)}$.

4 $\Rightarrow$ 3: Straightforward application of L'Hôpital's rule.

So here are my questions:

Are the above derivations correct?

Can we prove that 3 $\Rightarrow$ 1? If so, then we get two closed loops 1 $\Rightarrow$ 2 $\Rightarrow$ 3 $\Rightarrow$ 1 and 1 $\Rightarrow$ 4 $\Rightarrow$ 3 $\Rightarrow$ 1, so that all the four statements are equivalent. If not, any counterexamples?

TIA...

答:
Here is a counterexample of 2 $\Rightarrow$ 1:

Let $F(t) = t + \sin(t)$, then $F(t)$ is non-decreasing since $F'(t) = 1 + \cos(t) \ge 0$.

In addition, $\lim_\limits{t\to +\infty}\dfrac{F(t)}{\log t} = +\infty$, but $\lim\limits_{t\to +\infty}tF'(t)$ does not exist since $F'((2n+1)\pi) = 0$.

I believe this is also a counterexample of 3 $\Rightarrow$ 1.


问:
Extension of a maximal ideal
Let $R$ be a commutative ring, with the natural ring homomorphism $f:R \to R\left[ x \right]$, let $M$ be a maximal ideal of $R$.

For each ideal $I$ of $R$, the ideal $f(I)R[x]$ of $R[x]$ generated by $f(I)$ is called the extension of $I$ to $R[x]$, denoted by ${I^e}$.

Is ${M^e}$ a maximal ideal of $R\left[ x \right]$?

答:
No , because
$$R[X]/M^{\text e}=R[X]/MR[X]\simeq(R/M)[X],$$
which is a P.I.D., not a field. All you can say is that $M^{\text e}$ is a prime ideal.

A maximal ideal containing $M^{\text e}$ would be, for instance, $(M^{\text e}, X-r)$, where $r\in R$.

Title: Extension of a maximal ideal


问:
What does $^t$ mean in $x={}^t(x_1,x_2,...,x_n)$
I'm reading this paper and came across some notation I don't know and which is pretty difficult to search for. The relevant section is the first paragraph of section 2.1 in the paper, copied below:

Let $q$ be a power of prime and $k$ be a finite field of order $q$. For integers $n, m\ge1$, denoted by $f_1(x),f_2(x),...,f_m(x)$ quadratic polynomials of $x={}^t(x_1,x_2,...,x_n)$ over $k$.

Question: What is the meaning & name of the $^t$ notation before the brackets? It is not defined and is obviously not an exponent due to its placement. They use it in a similar fashion later in section 2.2:

$$
f_i(x_1,x_2,...,x_n)={}^txF_ix+ \text{(linear.)}
$$

where $F_1,...,F_m$ are $n×n$ matrices over $k$

答:
It should be the transpose. It is an aesthetic notation.

Instead of writing a vector as column, you would write it as a row.

So $(1,2,3)^t=\begin{pmatrix}1\\2\\3\end{pmatrix}$


问:
How to calculate the determinant of a matrix using Laplace?
How to calculate the determinant using Laplace?
$$
\det \begin{bmatrix}
       0 & \dots & 0 & 0 &   a_{1n}           \\[0.3em]
       0 & \dots & 0 & a_{2,n-1} &   a_{2n}   \\[0.3em]
       \dots & \dots & \dots & \dots & \dots             \\[0.3em]
       a_{n1} & \dots & a_{n,n-2} & a_{n,n-1} & a_{nn}  \\[0.3em]
     \end{bmatrix}
$$
I think it's something like:
$$
(a_{n1} * ... *a_{1n}) * (-1)^{n(n+1)}
$$
But I'm not sure about it.</p>

答:
You can swap rows, or just develop with respect to the first row:
$$
\det A=
\det\, \begin{bmatrix}
       0 & \dots & 0 & 0 &   a_{1n}           \\[0.3em]
       0 & \dots & 0 & a_{2,n-1} &   a_{2n}   \\[0.3em]
       \dots & \dots & \dots & \dots & \dots             \\[0.3em]
       a_{n1} & \dots & a_{n,n-2} & a_{n,n-1} & a_{nn}  \\[0.3em]
     \end{bmatrix}
=(-1)^{1+n}a_{1n}
\det\, \begin{bmatrix}
       0 & \dots & 0 & a_{2,n-1} \\[0.3em]
       \dots & \dots & \dots & \dots \\[0.3em]
       a_{n1} & \dots & a_{n,n-2} & a_{n,n-1} \\[0.3em]
     \end{bmatrix}
$$
which has the same form; hence, in the end, you get
$$
\det A=(-1)^{n+1}(-1)^n(-1)^{n-1}\dots(-1)^2(a_{1n}a_{2,n-1}\dots a_{n1})
$$
and the final exponent of $-1$ is
$$
(n+1)+n+\dots+2=\frac{n(n+3)}{2}
$$
Thus
$$
\det A=(a_{1n}a_{2,n-1}\dots a_{n1})(-1)^{n(n+3)/2}.
$$</p>


问:
How to calculate the remaining video length if watched in different speeds?
I'm searching for a formula which can calculate the remaining video speed if selected different speeds.

So if I have a video and watch it twice as fast, then it's half the time. That's easy.
But what should I calculate if I use 1.25, 1.5 or 1.75 times speed?

And how can I use the formula the opposite so when watching with slower speed like 0.25 or 0.5?

答:
$\frac{T_1}{T_2} = \frac{S_2}{S_1}$, where $T_1, T_2$ refer to times (lengths or durations of time) and $S_1, S_2$ refer to speeds of the videos.

Note the reversed ratios, because the speed and time length are inversely proportional.

Example of application: you have a video length 5 min at normal (1x) speed. What is the length (in time) at 1.75x speed?

$\frac{T_1}{T_2} = \frac{S_2}{S_1}$

$\frac{5}{T_2} = \frac{1.75}{1}$

So $T_2 = 2.857...$ minutes, which is approximately $2$ minutes $51$ seconds.

For slower than normal speeds (like 0.5x), just put that in place of $S_2$.


问:
Find $G$ and $H$ such that $dF$ has given form
Let $F$ be a cumulative distribution function and $$dF(x)=\begin{cases}dx/3&x\in (0,1)\cup(2,3)\\1/6&x \in \{1,2\}\\0&\mathrm{elsewhere}\end{cases}$$

Find a continuous cdf $H$, a discrete cdf $G$ and a real constant $c$ such that $$F(x)=cG(x)+(1-c)H(x)\ \forall x$$

What is the best method of solving such tasks?

$c=1/3$,  $H(x)=x/2,\ x \in (0,1)\cup (2,3)$ but that's just a guess. How can I solve the problem instead of trial and error?

答:
$$cG+(1-c)H=F=\begin{cases}0,&x<0\\x/3,&0\le x<1\\1/2,&1\le x<2\\x/3,&2\le x<3\\1,&x\ge3\end{cases}$$

$G(x)$ is a staircase function and $H(x)$ is continuous. From the above equation, for $c\ne0$, you know two things:

If $F$ has a finite jump $\Delta$ at $y$, then $G$ has a jump at $y$ equal to $\Delta/c$.

If $G$ has a finite jump $\delta$ at $y$, then $F$ has a jump at $y$ equal to $c\delta$.

Thus, the sets of points of discontinuity of $F$ and $G$ are identical. So $G(x)$ has a jump at $1,2$ and these jumps are equal to $\Delta/c=1/6c$. If you take$$G(x)=\begin{cases}0,&x<1\\1/6c,&1\le x<2\\1/3c,&x\ge2\end{cases}$$Then $x_2=1$ gives $c=1/3$.

Now you can find $H(x)$ by solving $H(x)=[F-cG]/(1-c)$.


问:
Is it true that every connected graph (except $K1$) with no perfect matching $\textbf{must}$ have more than one maximum matching?
I have been trying to find a counterexample for such statement,
namely - a unique maximum matching (for a graph that is not K1 - 1 vertex, and has no perfect matching)
and cannot find one at all. Is it time to prove it formally, or does such a graph exist?

答:
If the graph is not required to be connected the answer is no:
Consider a graph with 3 vertices and 1 edge, then clearly this graph has no perfect matching but its maximum matching (the matching containing the single edge) is unique.

However, for connected graphs the answer is yes:
Suppose we are given a graph $G$ which has no perfect matching and let $M$ be a maximum matching of $G$.
As $M$ is not perfect, there must exist some vertex $v$ not covered by $M$ which is connected to some vertex $w$ which is covered by some edge $e \in M$ and hence $(M \setminus \{e\}) \cup \{vw\}$ is maximal matching as well, so in connected graphs the answer is positive.


问:
asymptotic behavior of a matrix multiplying with a vector
I want to find the asympototic behaviors of $||M^nv||$ with some rate. [Just claiming that it's divergent is not enough - it requires finding some parallel bounds.]

We are given that M is 2x2 with characteristic polynomial $t^2-6t+9$.

(1) How does the asymptotic behavior of $||M^nv||$ have anything to do with diagonalizability of a matrix?

(2) I only saw that it simulates asymptotically to $3^nv$, but I was told that there is another situation involving $n-1$ somewhere. Is this the situation when it's diagonalizable or not? What is the other situation and how is it derived?

This is not a homework question and I don't really know which topic this problem belongs to. Any help with two questions above will be appreciated!

答:
You can have two slightly different cases, as you suspected.

In the diagonalizable case, in some basis you get $M=\pmatrix{3&0\\0&3}$ so that indeed $M^nv=3^nv$.

In the non-diagonalizable case, there is a basis where you get the Jacobi normal form $M=\pmatrix{3&1\\0&3}$ so that $M^n=3^n\pmatrix{1&\frac{n}{3}\\0&1}$. So indeed, the norm (in the original basis) of the matrix-power-vector product can have components that behave like $n3^{n-1}$.


问:
Show that sigma algebra is coarser than other sigma algebra
Consider the probability space $(\Omega,\mathcal F, \mathbb P)$ and the measure space $(S,\mathcal S)$. $I\neq \emptyset$.  Let $X=(X_i)_{i\in I}$ be $\mathcal F-\mathcal S$ measurable random variables with values in $S$.

Let \begin{equation}\mathcal E:=\Big\{\{X_{i_1}\in A_1,\dots,X_{i_n}\in A_n\}\Big|n \in \mathbb N, i_1,\dots,i_n \in I, A_1,\dots,A_n \in \mathcal S\Big\}\end{equation}
What is the argument for
$$\mathcal E\subseteq \sigma(X_i:i \in I)$$

or does this follow from definition, i.e. $\sigma(X_i:i \in I)$ is the sigma-algebra generated by $X_i, i \in I$?

Body: $X_{i_1},X_{i_2},...,X_{i_n}$ are all measurable w.r.t. $\sigma (X_i: i \in I)$ and hence $X_{i_1}^{-1}(A_1) \cap X_{i_2}^{-1}(A_2)... \cap X_{i_n}^{-1}(A_n) $ belongs to $\sigma (X_i: i \in I)$.


问:
Finding a unique definition of eigenspace for degenerate eigenvalues
I am diagonalizing an Hermitian matrix numerically, and this results in a set of eigenvalues $\varepsilon_1, ..., \varepsilon_n$ and a set of eigenvectors $v_1, ..., v_n$.

If the eigenvalues $\varepsilon_i = \varepsilon_{i+1} = \varepsilon_{i+2}$ are degenerate this results in an eigenspace, spanned by $v_i, v_{i+1}, v_{i+2}$.

The Problem is, that unlike the eigenvalues, $v_i, v_{i+1}, v_{i+2}$ are not uniquely defined and they differ between different Lapack and ScaLapack implementations, which makes debugging very hard.

Is there an established algorithm to transform the eigenspace, so that it always results in a
presentation that is:

still orthonormal

unique regardless of the original state

I have been toying with different ideas, but I would prefer to use something established, that then is also numerically stable etc.

答:
I came up with a similar solution. First I find the vector in the space which is "closest" to some target vector using a least squares and the I perform a gram-schmidt on it, which gives me an orthonormal basis. Here I use a set of sinus-functions as target vectors:</p>
```<code>def t(n):
    tar = np.sin(np.arange(old_eig_vecs.shape[0])*(n+1) * old_eig_vecs.shape[0]/2.0)
    return tar

def unify_eigvec(old):
    new = np.zeros(old.shape)
    for i in range(3):
        target_vec = t(i)
        res = np.linalg.lstsq(old, target_vec, rcond=None)
        new[:,i] = old@res[0]
    return gram_schmidt_columns(new)

def gram_schmidt_columns(X):
    Q, _ = np.linalg.qr(X)
    return Q
<span class="math-container">```</span>
</code>```

Body:
I came up with a similar solution. First I find the vector in the space which is "closest" to some target vector using a least squares and the I perform a gram-schmidt on it, which gives me an orthonormal basis. Here I use a set of sinus-functions as target vectors:

def t(n):
    tar = np.sin(np.arange(old_eig_vecs.shape[0])*(n+1) * old_eig_vecs.shape[0]/2.0)
    return tar

def unify_eigvec(old):
    new = np.zeros(old.shape)
    for i in range(3):
        target_vec = t(i)
        res = np.linalg.lstsq(old, target_vec, rcond=None)
        new[:,i] = old@res[0]
    return gram_schmidt_columns(new)

def gram_schmidt_columns(X):
    Q, _ = np.linalg.qr(X)
    return Q
```


问:
How to factorize $2x^2-9x+9$ by completing the square?
I know that $x^2-bx+c=(x-k)^2=x^2-2kx+k^2$ if it is a complete square. If not we create one by adding and subtracting $\left(\frac{b}{2}\right)^2$

I tried $$2\left(x^2-\frac{9}{2}x+\frac{9}{2}\right)=2\left(x^2-\frac{9}{2}x+\left(\frac{9/2}{2}\right)^2-\left(\frac{9/2}{2}\right)^2+\frac{9}{2}\right)$$

What am I missing and what is the best way to factorize when we have coefficients?(Is this the right term, ($a x^2$)?)

答:
Make the polynomial equal to $0$ to find the roots so that you can factorise.

$$2x^2-9x+9=0$$

$$x^2-\frac{9x}{2}+\frac{9}{2}=0$$

$$x^2-\frac{9x}{2}=\frac{-9}{2}$$

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of b.

$$(\frac{b}{2})^2=(\frac{-9}{4})^2$$

Add the term to each side of the equation.

You get:

$$(x-\frac{9}{4})^2=\frac{9}{16}$$

$$x=\frac{9}{4} \pm \frac{3}{4}$$

$$\implies x=3,\frac{3}{2}$$

Hence, ($x$-$\frac{3}{2}$)($x$-$3$) is your answer.


问:
How to show that ${\omega \in \Omega : X(\omega) \neq Y(\omega)}$ is a measurable set, $X,Y$ being $\overline{\mathbb{R}}$-valued.
How to show that $\{\omega \in \Omega : X(\omega) \neq Y(\omega)\}$ is a measurable set, $X,Y$ being $\overline{\mathbb{R}}$-valued. I mean, how to fix the case when $X$ and $Y$ = $\infty$? Here i can't pass to the $\{\omega \in \Omega : X(\omega) - Y(\omega) \neq 0\}$ because $\infty - \infty$ isn't defined. Thank you!

答: $\{\omega: X(\omega)=\infty, Y(\omega)<\infty\} =\bigcup_n  \{\omega: X(\omega)=\infty\}\bigcap \{\omega: Y(\omega)<n\}$ which is measurable. Simialrly, $\{\omega: Y(\omega)=\infty, X(\omega)<\infty\}$ is measurable. Can you finish?


答:
I'm assuming $X,Y$ are measurable functions.

We need to show that $\{\omega : X(\omega)>Y(\omega)\} = \bigcup_{q\in\mathbb Q}(\{\omega: X(\omega) > q\}\cap\{\omega: Y(\omega) < q\})$ is measurable.

To show this, we only need to notice $\{\omega: X(\omega) > q\}$ and $\{\omega: Y(\omega) < q\}$ are measurable for every $q$, and the set $\mathbb Q$ is countable.

Title: How to show that ${\omega \in \Omega : X(\omega) \neq Y(\omega)}$ is a measurable set, $X,Y$ being $\overline{\mathbb{R}}$-valued.

Body: How to show that $\{\omega \in \Omega : X(\omega) \neq Y(\omega)\}$ is a measurable set, $X,Y$ being $\overline{\mathbb{R}}$-valued. I mean, how to fix the case when $X$ and $Y$ = $\infty$? Here i can't pass to the $\{\omega \in \Omega : X(\omega) - Y(\omega) \neq 0\}$ because $\infty - \infty$ isn't defined. Thank you!


问:
Representation of recurrence as a polynomial
I'm studying linear-feedback shift register generators. In those kinds of generators, there's this recurrence:
$$
b_i\equiv(a_{p}b_{i-p}+a_{p-1}b_{i-p+1}+\ldots+a_{1}b_{i-1})\,\, mod\,\,2
$$
In a book, it says that as the module is prime, then we can express that recurrence as the following polynomial:
$$
f(z)=z^p-(a_{1}z^{p-1}+\ldots+a_{p-1}z+a)
$$
over the Galois filed $\mathcal{G}(2)$ defined over the integers 0 and 1 with the addition and multiplication being defined in the usual way followed by a reduction modulo 2.

Can someone explain to me why the module needs to be prime?

In addition, it says that "an important result from the theory developed for such polynomials is that, as long as the initial vector of bs is not all 0s, the period of the recurrence is $2^p-1$ if and only if the polynomial is irreducible over $\mathcal{G}(2)$.

Note: the book I'm following is 'Random Number Generation and Monte Carlo Methods, Second Edition, by James E. Gentle, page 38,39.

答:
This theory heavily relies on the theory of Galois fields (fields on a finite set), which enjoy suitable arithmetic properties. It turns out that the only possibly Galois fields have $m^p$ elements, where $m$ is a prime. $G(2)$ is the smallest Galois field.

The result 2. is a known theorem is the theory of Galois fields. It is invoked because a long period is desired, and $2^p-1$ has maximum length.


问:
Why is the derivative operator never bounded?
Consider $ X= C^\infty ([0,1],\mathbb{R} )$ and the operator
$$ D: (X,\|\cdot\| ) \to (X,\|\cdot\|).$$
given by
$$D(x)=x'.$$ (derivation operator).

Why is unbounded, independently of the choice of norm? I can prove this for cases where I know the norm, but why does this hold in the general case?

答:
Because the exponential functions $e^{cx}$ with $c\in\mathbb{R}$ exist in $C^\infty([0,1],\mathbb{R})$:
$$\|D(e^{cx})\|=\|ce^{cx}\|=|c|\cdot\|e^{cx}\|,$$
and there is no constant $M\geqslant0$ such that this norm is $\leqslant M\cdot\|e^{cx}\|$ for all $c \in \mathbb{R}$.


问:
Seems that I just proved $2=4$.
Solving $x^{x^{x^{.^{.^.}}}}=2\Rightarrow x^2=2\Rightarrow x=\sqrt 2$.

Solving $x^{x^{x^{.^{.^.}}}}=4\Rightarrow x^4=4\Rightarrow x=\sqrt 2$.

Therefore, $\sqrt 2^{\sqrt 2^{\sqrt 2^{.^{.^.}}}}=2$ and $\sqrt 2^{\sqrt 2^{\sqrt 2^{.^{.^.}}}}=4\Rightarrow\bf{2=4}$.

What's happening!?

答: Your reasoning does not make sense because you did not specified what does $x^{x^{x\ldots}}$ mean. It is not a finitary operation so it is not clear what does that term denote. If it stands for an outcome of a certain limiting procedure then you are in trouble since that limit can be $1$ or $\infty$ for positive $x$. Thus with this interpretation your premises are false and you can deduce from them anything you want, for instance that $0=1$.</p>

Title: Seems that I just proved $2=4$.


答:
Joining the two reasoning, Classical and Bayesian, I suggest to assume an uniform prior in $(0;1)$ thus we can get focus on the likelihood (I think this is better for the O.P as he said he didn't study Bayesian Inference yet.

The observed likelihood  (5 H on 5 Tosses) is the following

$$L(\theta) =\theta^5$$

If we look at this function as a function of $\theta$ and we want to consider it  as a density, we have to normalize it setting

$$C\int_0^1 \theta^5 d \theta=1$$

finding $C=6$

Thus we can  use the following density (whithout explicitly introducing Beta distribution)

$$f_{\Theta}(\theta)=6\theta^5$$

$\theta \in (0;1)$

...and now we can compute the probability that the coin is biased in favour of head in the following way

$$\int_{0.5}^{1}6\theta^5 d \theta=\frac{63}{64}$$

Graphically, the probability the coin is biased in favour of H is the purple area below

@Adam Rubinson: For a basic review of Bayesian thinking, if interested, you can read this paper

Title: Does it make sense to ask what the probability that a coin is biased (in favour of heads) is?

Body:
A coin is tossed $5$ times and $5$ heads are observed. What is the probability the coin is biased in favour of heads? Does this question even have any meaning without using a significance level?

Let's have a look at the usual Binomial test.

Let $X$ be the random variable: the number of heads out of $5$ tosses. Then $X \sim B(5,p).\ H_0: p = 0.5;\ H_1: p > 0.5.$

Suppose $H_0$ is true, that is,  $X \sim B(5,0.5)$. Then $P( X \geq 5) = P(X=5) = 0.5^5 = 0.03125.$ This is the probability that the coin actually is fair, and we just happened to get $5$ heads in a row. My question is: is the probability that the coin is not biased in favour of heads equal to $0.03125$? Or like I said at the beginning, is this question meaningless?


问:
Probability of picking at least $3$ green balls and atmost $1$ red ball from $4$ different balls in $6$ tries

I have four balls in the bag {, , ,}. What's the probability of picking (with replacement) $3$ or more green balls and $1$ or less red ball out of $6$ tries?

When I am running experiment programmatically, result is $0.13$. $4096$ all possible combinations and $539$ combinations with $3$ or more greens and $1$ or less red. $539/4096\approx0.13$.

Now I am wondering what would be a math/formulaic way to derive same result?

I have tried multiplication rule but $P(\text{green}\ge3)\times P(\text{red}\le1) = 0.09$ which makes no sense :(

Any help would be much appreciated

答:
$$1Red(3 Green+4Green+5Green)+\\
0Red(3Green+4Green+5Green+6Green)$$
For example, 1Red gives a factor of ${6\choose1}=6$.  For that case, 3Green has a count of ${5\choose3}2^2=40$ because there are two Other balls.


答:
Note that out of 6 tries we have that
$$P(\text{green$=3$ and red$=1$})=\frac{6!}{3!1!2!}\left(\frac{1}{4}\right)^3\cdot \left(\frac{1}{4}\right)^1\cdot \left(\frac{2}{4}\right)^2.$$
Your probability is equal to
$$p=\sum_{j=3}^6 P(\text{green$=j$ and red$=0$})+\sum_{j=3}^5 P(\text{green$=j$ and red$=1$}).$$

Can you take it from here and compute the exact value of $p$?

P.S. I confirm that $p\approx 0.13$,

Title: Probability of picking at least $3$ green balls and atmost $1$ red ball from $4$ different balls in $6$ tries

Body:
I have four balls in the bag {, , ,}. What's the probability of picking (with replacement) $3$ or more green balls and $1$ or less red ball out of $6$ tries?

When I am running experiment programmatically, result is $0.13$. $4096$ all possible combinations and $539$ combinations with $3$ or more greens and $1$ or less red. $539/4096\approx0.13$.

Now I am wondering what would be a math/formulaic way to derive same result?

I have tried multiplication rule but $P(\text{green}\ge3)\times P(\text{red}\le1) = 0.09$ which makes no sense :(

Any help would be much appreciated


答:
Split the event into two disjoint events:

$E_1$: When you get atleast $3G$ and exactly $0R$,

$E_2$: When you get atleast $3G$ and exactly $1R$.

$P(E_1)=\sum_{k=3}^6\binom 6k(1/4)^k(2/4)^{6-k}$ since you pick $k$ out of $6$ tries to get $k\ge3$ green balls and on the remaining $6-k$ tries, you get either orange or yellow ball.

$P(E_2)=\sum_{k=3}^5\binom 6k(1/4)^k\binom{6-k}1(1/4)(2/4)^{6-k-1}$ since you pick $k$ out of $6$ tries to get $k\ge3$ green balls, $1$ try out of remaining $6-k$ tries to get a red ball and on the remaining $6-k-1$ tries, you get either orange or yellow.

The required probability is $P=P(E_1)+P(E_2)$ which indeed turns out to be $539/4096$.


问:
Probability density function of $W = X + Y$ where $X \sim \mathrm{Unif}[0,1]$ and $Y \sim \mathrm{Exp}(\lambda)$ are independent random variables?
This is the question that I was given:

And this was the provided solution:

I can't seem to make sense of it - firstly, what exactly is the question asking you to do? How did they know to divide the range of $w$ in that way? And why did they use those particular limits for the integrals?

答: $X$ can take values in $[0,1]$, and $Y$ in $[0,\infty)$, and so $W=X+Y$ can take values in $[0,\infty)$.  But if $W<1$, then there are limitations on the value that $X$ can take.  For example, if $W=0.5$, how could $X>0.5$ when $Y$ cannot be less than zero?  This is precisely what Case 1 is saying: when $W<1$, then $X<W$ as well.  But if $W>1$, then all values of $X$ in its range $[0,1]$ are permissible.  So the convolution is taken by integrating over the permissible ranges of $X$ as defined by the cases.  Finally, the question is asking you to find the complete density function, which can only be found in cases conditioned on the value of $W$.</p>


问:
When is a halfspace a subset of another halfspace?
Let's say we have the following halfspaces:$$H_1=\left \{x\mid a^tx\leq b\right \}$$and$$H_2=\left \{x\mid \tilde a^tx\leq \tilde b\right \}.$$I want to find the conditions that need to hold such that $H_1\subseteq H_2$. So these halfspaces are defined by the hyperplanes, say,$$A_1=\left \{x\mid a^tx=b\right \}$$and$$A_2=\left \{x\mid \tilde a^tx=\tilde b\right \}.$$Intuitively I understand that $H_1\subseteq H_2$ demands the hyperplanes to be parallel or equivalently the normal vectors $a$ and $\tilde a$ to be parallel, so:\begin{align*}A_1\parallel A_2 & \iff a\parallel \tilde a \\
& \iff \exists t\in \mathbb{R}\text{ such that }\tilde a=ta.
\end{align*}The hypothesis $H_1\subseteq H_2$ also requires the $H_2$ to lie above $H_1$ which restricts us to consider $\tilde b\geq tb$ for positive values of $t$ ($t>0$). But how can I formally prove that these two conditions are enough and necessary?

答:
For the general case, to appear $t> 0$, let us normalize the vectors $a, b$. Then we have,
$H_1=\{x\mid a^Tx \le b\} = \{x\mid\frac{a^T}{\|a\|} \le \frac{b}{\|a\|}\}$.
Do the same for $H_2$ and do the same as above. We will show $t =\frac{\|a\|}{\|a'\|}$.


问:
proving the infimum of a set
I am trying to find the supremum and infimum of the set $A = \{1/x +(-1)^x|x \in \mathbb{N}
\}$.

Obviously, the supremum is $3/2$, however, I am not sure about the infimum. In the limit as $x \rightarrow \infty$, $1/n \rightarrow 0$ and since $(-1)^x$ is always either $-1$ or $1$, I would think the infimum should be $-1$, but I may be confusing the definition of a lower bound and an infimum.

答:
You are correct, the infimum is indeed $-1$. $-1$ is also a lower bound of $A$.

You can prove that $-1=\inf(A)$ by proving two things:

That $-1$ is a lower bound for $A$ (this one should be easy)

That $-1$ is the largest lower bound for $A$, in other words, that for all $\epsilon > 0$, the value $-1+\epsilon$ is not a lower bound for $A$.

Title: proving the infimum of a set


问:
Solve $\displaystyle\lim_{x \to 0}{(1 - \sqrt{\cos 4x})\cot^2x}$
I just use  L'Hospital's Rule (by changing into $\dfrac{1}{\tan^2x}$ first) for this, but it makes me even more confusing to solve. Could anyone help me to figure this out please?

答:
HINT

$$\lim_{x\to 0} \left(1-\sqrt{\cos(4x)}\right)\cot^2{x}\\=\lim_{x\to 0}\frac{1-\cos^2{4x}}{\tan^2{x}\cdot \left(1+\sqrt{\cos(4x)}\right)\cdot \left(1+\cos(4x)\right)}\\=\lim_{x\to 0}\frac{\sin^2{4x}}{\tan^2{x}}\cdot \lim_{x\to 0}\frac{1}{\left(1+\sqrt{\cos(4x)}\right)}\cdot\lim_{x\to 0}\frac{1} {\left(1+\cos(4x)\right)}$$


问:
Not sure is my conclusions are good in set-theory exercise?
I have to prove equality of sets:

 $f^{-1}(C\cup D)=f^{-1}(C)\cup f^{-1}(D)$ where $f \subseteq X\times Y$ is function and $C,D\subseteq Y$.

My proof goes like this:

Let $x \in f^{-1}(C\cup D) \iff f(x) \in C\cup D \iff f(x) \in C \vee f(x) \in D \iff x \in f^{-1}(C) \vee x \in f^{-1}(D) \iff x \in f^{-1}(C)\cup f^{-1}(D)  $

Is it all ok? Are all these equivalence signs good or I have just implication somewhere. This one is supicious to me but I can't find counterexample why it wouldn't be good?
$f(x) \in C \vee f(x) \in D \iff x \in f^{-1}(C) \vee x \in f^{-1}(D)$

答:
Well, aside from the fact that you have $D$ in some spots and $U$ in others, it is just fine.

By definition $f(x)\in C$ if and only if $x\in f^{-1}(C)$ (similarly with $D$). Thus, that "suspicious" equivalence is okay. Basically, all you've done is rewritten the two statements in your disjunction. This is no more problematic than saying something like $$(1+x\ge0)\vee(1+y\ge0)\Longleftrightarrow(x\ge-1)\vee(y\ge-1)$$ for real numbers $x,y$.


问:
Stalk functor is adjoint to the skyscraper functor.
In Proof that sheafification induces isomorphism on stalks using adjoints, Zhen Lin affirms that we can use adjointness to prove that the sheafification preserves stalks in the following way.

Concatenating the sheafification / inclusion (on the right) and skyscraper / stalk (on the left) adjuntions

we get that the functor $\sf{C}_X^\text{pre}\to\sf{C}$ which takes a presheaf, sheafifies and then takes the stalk is left adjoint to the skyscraper sheaf functor $\sf{C}\to\sf{C}_X^\text{pre}$. By unicity, it suffices to show that stalk functor $\sf{C}_X^\text{pre}\to\sf{C}$ is also a left adjoint of the skyscraper functor.

I would like to know why this adjunction holds.

Little discussion: let's recall how one could prove that the adjunction on the left of my diagram holds. Taking the adjunction between direct and inverse images in the case of the inclusion $i_p:\{p\}\to X$ of a point in $X$, we get

Now, the category $\sf{C}_{\{p\}}$ is equivalent to $\sf{C}$ since defining a sheaf on $\{p\}$ is the same thing as giving an object of $\sf{C}$ (as the image of $\varnothing$ is automatically defined). This proves the adjunction on the left above. Nevertheless, this same proof doesn't seems to work when working with presheaves since $\sf{C}_X^\text{pre}$ is not equivalent to $\sf{C}$. (The image of $\varnothing$ is not necessarily a final object of $\sf{C}$ in this case.)

答:
As requested, the two comments above are turned into an answer.

You need to prove the adjunction directly. Note that a morphism of presheaves $A\to {i_p}_*B$ is the data for all U of a morphisms $A(U)\to{i_p}_*B(U)$ such that some diagrams commute. For all $U\not\ni p$, there is no data here since ${i_p}_*B(U)$ is the terminal object of your category and thus there is a single possibility for $A(U)\to{i_p}_*B(U)$. It follows that the morphism $A\to {i_p}_*B$ is equivalent to the data for all $U\ni p$ of $A(U)\to {i_p}_*B(U)$ such that some diagrams commute which in turn is equivalent to a morphism $A_p\to B$ by definition of the colimit.

This is basically the same proof that the stalk functor and the skyscraper functor are adjoints in the category of sheaves, the proof does not uses the sheaf condition. In fact, the proof for presheaves implies the proof for sheaves :
$$Hom_{Sh}(A,{i_p}_*B)=Hom_{PSh}(A,{i_p}_*B)=Hom(A_p,B)$$
And you have that the sheaffification commutes with stalks :
$$Hom(A^a_p,B)=Hom_{Sh}(A^a,{i_p}_*B)=Hom_{PSh}(A,{i_p}_*B)=Hom(A_p,B)$$
which implies that $A^a_p=A_p$ by Yoneda.


答:
HINT

By standard limits

$$(1 - \sqrt{\cos 4x})(\cot^2x)=\frac{1 - \sqrt{\cos 4x}}{\tan^2x}=\frac{1 - \cos 4x}{(4x)^2}\cdot \frac{x^2}{\tan^2x}\cdot\frac{16}{1 + \sqrt{\cos 4x}}$$

Title: Solve $\displaystyle\lim_{x \to 0}{(1 - \sqrt{\cos 4x})\cot^2x}$

Body: I just use  L'Hospital's Rule (by changing into $\dfrac{1}{\tan^2x}$ first) for this, but it makes me even more confusing to solve. Could anyone help me to figure this out please?


问:
help solve inequality
help solve inequality:

$\sin 3x > \cos 3x$

答:
Hint:

On a full period, $$\sin\theta>\cos\theta$$ for $$\theta\in\left(\frac\pi4,\frac{5\pi}4\right).$$

You can also use

$$\sin\theta-\cos\theta\propto\sin\left(\theta-\frac\pi4\right).$$

Title: help solve inequality


答:
Note,

$$1-\sqrt{\cos4x}={1-\cos4x\over1+\sqrt{\cos4x}}={1-(\cos^22x-\sin^22x)\over1+\sqrt{\cos4x}}={2\sin^22x\over1+\sqrt{\cos4x}}={8\sin^2x\cos^2x\over1+\sqrt{\cos4x}}$$

Consequently, for the limit as $x\to0$, we have

$$(1-\sqrt{\cos4x})\cot^2x={8\sin^2x\cos^2x\over1+\sqrt{\cos4x}}\cdot{\cos^2x\over\sin^2x}={8\cos^4x\over1+\sqrt{\cos4x}}\to{8\cdot1^4\over1+\sqrt1}={8\over2}=4$$

The key thing here, really, is that multiplying by the conjugate $1+\sqrt{\cos4x}$  eliminates any difficulty with the problematic square root, since $1+\sqrt{\cos4x}$ tends to a nice non-zero limit.


问:
Is the set of all rational sequences with constant tail countable?
I am unable to prove or disprove that the set of all rational sequences with constant tail is countable. Precisely the set in question is

\begin{equation}
\{(x_n)_{n\ge1}: x_n \in \mathbb{Q}, \exists N \in \mathbb{N}, \, c \in \mathbb{Q} : \forall n > N : x_n = c\}
\end{equation}

What about when the tail is the same for all sequences? That is

\begin{equation}
\{(x_n)_{n\ge1}: x_n \in \mathbb{Q}, \exists N \in \mathbb{N} : \forall n > N : x_n = 0\}
\end{equation}

答:
Here are some rough steps, with details for you to fill in.

First consider the set $S_0$ of sequences with a constant $0$ tail. Find a bijection between $S_0$ and the set of finite sequences of rationals. Then prove that the set of finite sequences of rationals is countable.

Now, for a rational $c$, consider the set $S_c$ of sequences with constant tail $c$. Find a bijection between $S_c$ and $S_0$. Finally, the set of all sequences with constant tail is the union of $S_c$ for all rational $c$. Argue that this union is countable.


答:
Let $\mathscr{A}$ be the set of rational-valued sequences which are eventually constant and, for each $n\in \mathbf{N}$, let $\mathscr{A}_n$ be the set of all $x \in \mathscr{A}$ such that $x_{k}=x_n$ for all $k\ge n$. Then $\mathscr{A}=\bigcup_n \mathscr{A}_n$ and it is enough to show that each $\mathscr{A}_n$ is countable. To this aim, it is sufficient to see that
$$
\mathscr{A}_n=\bigcup_{q_1 \in \mathbf{Q}} \cdots \bigcup_{q_n \in \mathbf{Q}} \{x \in \mathbf{R}^{\mathbf{N}}: x_1=q_1, \ldots, x_{n-1}=q_{n-1}, x_k=q_n \text{ for all }k\ge n\}.
$$
for all $n \in \mathbf{N}$. Therefore each $\mathscr{A}_n$ is countable.

Basically, this is the reason why $(c, \|\cdot\|_\infty)$ is separable.

Title: Is the set of all rational sequences with constant tail countable?

Body:
I am unable to prove or disprove that the set of all rational sequences with constant tail is countable. Precisely the set in question is

\begin{equation}
\{(x_n)_{n\ge1}: x_n \in \mathbb{Q}, \exists N \in \mathbb{N}, \, c \in \mathbb{Q} : \forall n > N : x_n = c\}
\end{equation}

What about when the tail is the same for all sequences? That is

\begin{equation}
\{(x_n)_{n\ge1}: x_n \in \mathbb{Q}, \exists N \in \mathbb{N} : \forall n > N : x_n = 0\}
\end{equation}


答:
Note that since $\mathbb Q$ is countable we may rewrite the question in terms of sequences over $\mathbb N$ and additionally it is then easy to give a bijection between the first set you mention and the set $\mathbb{N}^\ast$ of finite tuples over $\mathbb N$ by simply mapping a sequence $(x_1, ..., x_N, c, ...)$ to the tuple $(x_1, ..., x_N, c)$.

Now, to show that $\mathbb N^\ast$ is countable one can simply note that it is the countable union of countable sets.


问:
Show that there are infinitely many primes $p$ with $p^5 \equiv 5 \pmod 6$
Show that there are infinitely many primes $p$ with $$p^5 \equiv 5 \pmod 6$$

I am very confused by this question.

I am familiar with Euclid's proof of there being infinitely many primes and can see that most proofs of this nature run along a similar path.

For instance, most usually involve assuming a limited number of primes of this type and then multiplying them all together and adding a number to this and this leads to a contradiction later on.

For this problem, I have begun by also assuming there are only a finite number of primes with the above property. However, due to the index of 5, I don't see how I can simply multiply all the primes in the finite set together to figure this out. Multiplying two to more numbers with a remainder of 5 on division by six together always leads to a remainder of either 1 if the index is odd and 5 if it is even.

I do not know where I can go any further with my analysis.

Any hints at how I can tackle this question would be greatly appreciated.

答:
We know that
$$p \equiv \pm1 \pmod{6}$$
for every prime $p>3$.

Also, one can verify (see below) that, there are infinitely many primes such that
$$p \equiv -1 \pmod {6} \tag 1$$

Raising to the $5^\text{th}$ power, we observe that prime such that

$$p^5 \equiv -1 \equiv 5 \pmod{6}$$
are infinite, as well.

Proof of $(1)$: Assume there are only finitely many primes such that
$$p \equiv -1 \pmod {6}$$
Let $P = \{p_1,p_2,\dots,p_n\}$ be all such primes. Now, we will look at the following number:
$$Q = 6p_1p_2\dots p_n - 1$$
Since, it's asserted not to be prime and since $\forall p\in P: p \nmid Q$, and $2\nmid Q$, $3\nmid Q$, we have that all prime factors of $Q$ are of the form $6k + 1$. Then, $Q$ must be of the form $6k+1$, as well. However, this contradicts the fact that $$Q \equiv -1 \pmod{6}$$ by definition. $\square$


答:
Both of those sets are countable, and both of them can be almost identified with the set of all finite sequences of rational numbers.

In fact:

In the first case look at the map which maps a finite sequence of length $n\ge 1$ to the infinite sequence where the last element is repeated ad infinitum.

In the second case, look at the map that maps a finite sequence of length $n\ge 0$ to the infinite sequence made up by adding repeated zeros.

Both those maps are onto but not necessarily 1-1, so your sets will be smaller than, or equal, to the size of the set of all finite sequences of rational numbers.

However, the latter is the union (for $n\ge 1$ or $n\ge 0$) of all sequences of rational numbers of size exactly $n$. Each of those sets has the same cardinality as $\mathbb Q\times\mathbb Q\times\cdots\mathbb Q$ ($n$ times) and so is countable. (For $n=0$ it is an $1$-element set.) The resulting countable union of countable sets is therefore countable.


问:
Why does every limited succession have a convergent subsuccession?
Working on some real analysis homework, I faced this question:

Let $(a_n)\ n ∈ N$ a real number succession. Prove that $(a_n)\ n ∈ N$ does not have a converging subsuccession if and only if $\lim_{n → +∞} |a_n| = + \infty $

I solved it being a sufficient condition, but I couldn't prove it being a necessary one. I came to the conclusion that any succession which has no limit but doesn't go to infinity is limited, but nothing beyond.
I asked my professor and he said that the result that every limited succession have a convergent subsuccession is a given, but didn't prove it.
How does one go about proving it?

答:
$\lim_n |a_n|\neq \infty$ can be rewritten as $$\kappa:=\liminf_{n\to \infty}|a_n| \in [0,\infty).$$
Hence, there exists a subsequence $(|a_{n_k}|: k\ge 1)$ which is bounded, i.e., such that its image is contained in the compact set $[0,\kappa]$. Therefore by the Bolzano--Weierstrass theorem, there exists a sub-subsequence $(|a_{n_{k_t}}|: t\ge 1)$ which is convergent to some limit  $\ell \in [0,\kappa]$. It follows that there is an infinite set $T\subseteq \mathbf{N}$ such that $a_{n_{k_t}} \ge 0$ for all $t \in T$ or $a_{n_{k_t}} \le 0$ for all $t \in T$. Therefore $(a_{n_{k_t}}: t\in T)$ is convergent (to $\ell$ or to $-\ell$).


问:
Is $\frac{1}{x}$ integrable?
I am taking a course on theory of integrals and measure, where I have learned a very fundamental definition of integral. One of the things I've learned is that $f$ is integrable iff $\int|f|<\infty$. Then, according to this definition, $f(x)=\frac{1}{x}$ wouldn't be integrable, since the integral of $|f(x)|=|\frac{1}{x}|$ diverges. However, I can calculate $\int_a^b\frac{1}{x}dx$ for any $a, b \in (-\infty, +\infty)$ and this isn't problematic at all. How do I deal with this?

答:
You can deal with it using Cauchy principal value taking symmetric boundaries in limit. When $a<0<b$
$$\int\limits_{a}^{b}\frac{dx}{x}=\lim\limits_{\varepsilon \to 0+}\left(\int\limits_{a}^{-\varepsilon}+\int\limits_{\varepsilon}^{b}\right)\frac{dx}{x}$$

Title: Is $\frac{1}{x}$ integrable?


问:
How can I identify the graph of a function and the graphs of its first derivative and second derivative?

Identify the graphs $A$ (blue), $B$( red) and $C$ (green) as the graphs of a function and its derivatives:

_____ is the graph of the function

_____ is the graph of the function's first derivative

_____is the graph of the function's second derivative

答: Find where one is always increasing, and the other always corresponds to be positive, and find where the function is decreasing, to find the corresponding negative function.  The graph that's positive and negative while the other is increasing and decreasing is a derivative function.  The graph that is increasing and decreasing would be the antiderivative function.</p>


答:
Let me rephrase this in a more standard way: you want to show that a bounded sequence in $\mathbb{R}$ has a convergent subsequence. This is a sketch of the proof (it should be a exercise to make this rigorous): say that $$|a_n|\leq R\quad \forall\:n\in\mathbb{N},$$ which means that $$a_n\in[-R,R]\quad \forall\:n\in\mathbb{N}.$$

Now, consider the half intervals $[-R,0]$ and $[0,R]$. One of them must have infinitely many terms of the sequence, right? Say that it's $[0,R]$ and consider $[0,\frac{R}{2}]$ and $[\frac{R}{2},R]$; again one of them has to have infinitely many terms of the sequence.

You can continue this process indefinitely and get a sequence of nested intervals whose length goes to $0$. Take $a$ to be the intersection of all this intervals. Then $x$ is an accumulation point of $\{a_n\}_n$, in other words a subsequence of $\{a_n\}_n$ converges to $a$.

Hope it helps.

PS: this is a classical result from real analysis called the Bolzano–Weierstrass theorem.

Title: Why does every limited succession have a convergent subsuccession?

Body:
Working on some real analysis homework, I faced this question:

Let $(a_n)\ n ∈ N$ a real number succession. Prove that $(a_n)\ n ∈ N$ does not have a converging subsuccession if and only if $\lim_{n → +∞} |a_n| = + \infty $

I solved it being a sufficient condition, but I couldn't prove it being a necessary one. I came to the conclusion that any succession which has no limit but doesn't go to infinity is limited, but nothing beyond.
I asked my professor and he said that the result that every limited succession have a convergent subsuccession is a given, but didn't prove it.
How does one go about proving it?


答:
If $\lim\limits_{n\to\infty}|a_n| \neq \infty$, then it has a bounded subsequence: if it did not, then for any $C$, there would be some $n$ such that for all $m > n$, $|a_n| > C$, so $\lim\limits_{n\to\infty}|a_n| = \infty$, a contradiction. Choose some $C$ at which this fails.

We now proceed by "lion hunting":

Define $b_0 = -C$ and $c_0 = C$. We shall then construct sequences $(b_n)$ and $(c_n)$ such that there are always infinitely many terms of $a_n$ between them. We shall do this inductively as follows:

Let $m_n$ be the midpoint of $b_n$ and $c_n$. Then at least one of $[b_n, m_n]$ and $[m_n, c_n]$ contains infinitely many terms of $a_n$ (we know that there are infinitely many in $[b_0,c_0]$, and by our inductive hypothesis, in $[b_n, c_n]$). Choose $b_{n+1}$ and $c_{n+1}$ to be the endpoints of that interval.

Now, $(b_n)$ and $(c_n)$ are monotone and bounded. Define $b = \limsup b_n$. We shall show that $(b_n) \to b$.

For the former, note that, by definition of $b$, for any $\varepsilon > 0$, there is some $n$ such that $0 < b - b_n < \varepsilon$. But then, for any $m > n$, we have $0 < b - b_m < b - b_n < \varepsilon$. Thus, $(b_n) \to b$.

We note also that $b_n < c_n$ for all $n$, so $b$ lies in $[b_n, c_n]$ for all $n$.

Let $n_k$ be the smallest natural number such that $a_{n_k} \in [b_k, c_k]$. Then we have $|a_{n_k} - c| \leq |c_k - b_k| = 2^{k-1}C \to 0$, and so $(a_{n_k}) \to c$, and in particular, that $(a_{n_k})$ is a convergent subsequence of $(a_n)$.


问:
Upper and Lower semicontinuity and right-left continuity.
Let $f: [0,1]\rightarrow\mathbb{R}$ be a function. Is there any relation between upper-lower semicontinuity of $f$ and right-left continuity of $f$?

I mean, is there any implication?

Thanks in advance.

答:
The only implication is that both left + right continuity and upper + lower semicontinuity are equivalent (and, of course, are equivalent to continuity). Otherwise, we can create a table of counterexamples. Here, we define several real functions of a variable $x$, which possess the given continuity properties at $0$:

$$\begin{array}{|c|c|}
\hline & \text{Left-continuous} & \text{Right-continuous} & \text{Both} & \text{Neither} \\
\hline \text{USC} & \begin{cases} 1 & \text{if }x \le 0 \\ 0 & \text{if } x > 0 \end{cases} & \begin{cases} 0 & \text{if }x < 0 \\ 1 & \text{if } x \ge 0 \end{cases} & - &\begin{cases} 0 & \text{if }x \neq 0 \\ 1 & \text{if } x = 0 \end{cases} \\
\text{LSC} & \begin{cases} 0 & \text{if }x \le 0 \\ 1 & \text{if } x > 0 \end{cases} & \begin{cases} 1 & \text{if }x < 0 \\ 0 & \text{if } x \ge 0 \end{cases} & - &\begin{cases} 1 & \text{if }x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \\
\text{Both} & - & - & 0 & - \\
\text{Neither} & \begin{cases} 0 & \text{if }x \le 0 \\ \sin\left(\frac{1}{x}\right) & \text{if } x > 0 \end{cases} & \begin{cases} \sin\left(\frac{1}{x}\right) & \text{if }x < 0 \\ 0 & \text{if } x \ge 0 \end{cases} & - & \begin{cases} \sin\left(\frac{1}{x}\right) & \text{if }x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \\
\hline
\end{array}$$