答:
Let $w=\frac1{1-z}$. Then, $z=1-\frac1w$ and, with unit circle $|z|=1$
$$|z|^2=(1-\frac1w)(1-\frac1{\bar w})=1\implies w+\bar{w}=1
\implies Re(w)=\frac12$$
which is the vertical line $x=\frac12$ in the complex plane.

Title: The map $\frac1{1-z}$

Body:
I want to find the image of the unit circle under the map $\frac1{1-z}$. I first rewrote this as
$$
\frac{1+z}{1-z^2}
$$
But on the unit circle, $z^2 = 1$. So I get $\infty$ for the image of all points on the unit circle as the result. But I thought linear fractional always map circles to lines or circles. Can anyone point out what I did wrong?

EDIT: It turns out that I was confusing $z^2$ with $\bar{z}$, so not all points on the unit circle satisfy $z^2 = 1$. I should let $z = x + iy$ and plug that into the map instead.


问:
discrete groupoid
Here
they say:

First of all, since A is a discrete groupoid, the only morphisms are of the form $\text{id}:x\to x$
Also, here is the definition of discrete category in ncatlab:

```<code>A category is discrete if it is both a groupoid and a preorder.
</code>```

I would say that there might be a $x$ and a $y$ with $x\neq y$ and $\text{hom}(x,y)=\star$.

So in my view there are definitely more morphisms than $\text{id}_x:x\to x$!

答:
There's a mild disagreement on terminology here. Your first source wants a discrete category to satisfy the axiom that if there's a morphism $f:x\to y$, then $x=y$. The nLab, as usual, is a bit less demanding: their notion of a discrete category is one such that, if there's a morphism $f:x\to y$, then it is unique and an isomorphism. Thus, for instance, the category $a\leftrightarrows b$ generated by two objects with a unique isomorphism between them is discrete for the nLab.

The reason for this gap in terminology is that discrete categories in the sense of the nLab are equivalent to categories discrete in the first sense, and the nLab, along with many categorical thinkers, prefers definitions that are invariant under equivalence of categories.

ContentLicense: CC BY-SA 4.0

Title: discrete groupoid


问:
Can we always find $x_n\to\infty$ with $f_n(x_n)\to0$ if the $(f_n)$ are non-decreasing and pointwise convergent to zero?
Suppose we have a sequence of functions $(f_n)$ with each $f_n:[0, \infty) \to \mathbb{R}$, each $f_n$ is monotone non-decreasing and $f_n\to 0$ pointwise. Is it always possible to choose a sequence $(x_n)$ with $x_n\to \infty$ slow enough such that we still have $f_n(x_n) \to 0$? If not, is this possible with additional hypotheses on the $(f_n)$ (e.g. continuity perhaps)?

答:
There is no need for the $f_n$ to be monotone non-decreasing.

Given $n\in \Bbb Z_+$, let $$m(n):=\min\{m\in\Bbb Z_+:|f_k(j)|<1/n \;(j=1,...,n)\;\text{for}\, k\geqslant m\}.$$ Now define $M(0):=0$ and  $M(n):=\max\{n,M(n-1)+1,\,m(1),...,m(n)\}\;(n\in\Bbb Z_+).$ Then $$|f_{M(n)}(n)|<\frac1n\quad(n\in\Bbb Z_+).$$Now $M(\centerdot)$ is strictly increasing and so has an inverse $M^{-1}:M(\Bbb N)\to\Bbb N$ which we can extend to a map $L:\Bbb N\to\Bbb N$ by defining $L(0):=0$ and $$L(k):=M^{-1}(l)\quad\text{for}\;k=M(l)+1,...,M(l+1)\quad(l\in\Bbb N).$$It follows that $$|f_k(L(k))|<\frac1{L(k)}\quad(k\in\Bbb Z_+).$$Note that $L$ is increasing and unbounded. Thus, by taking $x_n:=L(n)\;(n\in\Bbb Z_+)$, we obtain a sequence with the required property that $f_n(x_n)\to0$ as $n\to\infty$.


问:
Why is $x = c$ an inflection point only if the lowest-order (above the second) non-zero derivative is of odd order (third, fifth, etc.).
So according to wikipedia inflection point,

If the second derivative, $f''(x)$ exists at $x_0$, and $x_0$ is an inflection point for $f$, then $f''(x_0) = 0$, but this condition is not sufficient for having a point of inflection, even if derivatives of any order exist. In this case, one also needs the lowest-order (above the second) non-zero derivative to be of odd order (third, fifth, etc.). If the lowest-order non-zero derivative is of even order, the point is not a point of inflection, but an undulation point.

It is also mentioned that:

the condition that the first nonzero derivative has an odd order implies that the sign of $f'(x)$ is the same on either side of $x$ in a neighborhood of $x$

My question is why is it the case? For example, if $f^{3}(c)$ is nonzero, while $f''(c) = 0$.  How does the first condition contribute to the fact that $(x, f(x))$ is not an undulation point?.  I found this non-obvious but I couldn't find any explanation online. So I am seeking a proof (or any kind of intuition) for this. Thanks in advance.

答:
Following the Taylor development, assuming WLOG the inflection point at $x=0$, we have

$$f(x)\approx f(0)+f'(0)x+f^{(n)}(0)\frac{x^n}{n!}$$

where $n$ is the order of the first nonzero derivative ($n>1$). For $x$ small enough, the higher order terms have no influence. To get an inflection, the last term must be an odd function, so that the curve can cross the straight line $y=f(0)+f'(0)x$.

Below, plots of $1-x, 1-x+x^2,1-x+x^3,1-x+x^4,1-x+x^5$:


问:
If $f$ is a function such that $f'(x)$ exists for all $x\in\mathbb{R}$. Then $f'$ has period $2\pi $.
If $f$ is a function such that $$ \int_{-\pi}^{\pi}f(x)dx=\int_c^{c+2\pi}f(x)dx \quad\forall c\in\mathbb{R}$$ and $f'(x)$exists for all $x\in\mathbb{R}$.

Then $f'$ has period $2\pi $.

I know I've to show that $f'(x)=f'(x+2\pi)$ for all  $\quad x\in\mathbb{R}$. Any suggestions would be great!

答:
By the second Fundamental Theorem of Calculus, after taking derivative for both sides w.r.t $c$, we have
$$ f(c+2\pi)-f(c) =0. $$
Changing $c$ into $x$ gives
$$ f(x+2\pi)=f(x). $$
Taking derivative for both sides w.r.t $x$, we have
$$ f'(x+2\pi)=f'(x). $$
Namely $f'(x)$ is a periodic function with period $2\pi$.


问:
Trying to understand a double summation formula
I am trying to understand a double summation mentioned in this paper: https://doi.org/10.1016/0024-3795(95)00696-6. How can we prove that this is true?

$$\sum_{l = 1}^n\;\; \sum_{k = l}^{n+l} = \sum_{k \,=\, 1}^n \;\; \sum_{l \,=\, 1}^{k} \,+\, \sum_{k \,=\, n+1}^{2n}\;\; \sum_{l \,=\, k-n}^{n}$$

答:
Here is another variation to show the identity for $n\geq 1$:
\begin{align*}
\sum_{l=1}^n\sum_{k=l}^{n+l}a_{k,l}
=\sum_{k=1}^n\sum_{l=1}^ka_{k,l}+\sum_{k=n+1}^{2n}&\sum_{l=k-n}^na_{k,l}\tag{1}
\end{align*}

We start with the right-hand side of (1) and consider at first the left sum.

We obtain
\begin{align*}
\sum_{k=1}^n\sum_{l=1}^ka_{k,l}=\sum_{\color{blue}{1\leq l\leq k\leq n}} a_{k,l}=\sum_{l=1}^n\sum_{k=l}^{n}a_{k,l}\tag{2}
\end{align*}

In the middle of (2) we represent the index range of summation as inequality chain. This might help to better see the region of validity and how to exchange the sums.

Now the other sum. We obtain
\begin{align*}
\sum_{k=n+1}^{2n}&\sum_{l=k-n}^na_{k,l}\\
&=\sum_{k=1}^{n}\sum_{l=k}^na_{k+n,l}
=\sum_{\color{blue}{1\leq k\leq l\leq n}} a_{k+n,l}=\sum_{l=1}^n\sum_{k=1}^la_{k+n,l}\tag{3}\\
&=\sum_{l=1}^n\sum_{k=n+1}^{n+l}a_{k,l}\tag{4}
\end{align*}

Comment:

In (3) we shift the index $k$ by $n$ to start with $k=1$. We are now in the same situation as in (2) and continue accordingly.

In (4) we shift the index back.

Adding (2) and (4) gives
\begin{align*}
\sum_{l=1}^n\sum_{k=l}^{n}a_{k,l}+\sum_{l=1}^n\sum_{k=n+1}^{n+l}a_{k,l}
\color{blue}{=\sum_{l=1}^n\sum_{k=l}^{n+l}a_{k,l}}
\end{align*}
and the claim (1) follows.


问:
Problem with the density of polynomials.
Problem: Suppose that $X \subset \mathbb{R}^n$ is compact. Let $Y = \{y_1,...,y_p\}$ be a finite subset of $X$. Let $\mathscr{P} = \{p \in \mathbb{P}[x_1,...,x_n]: p(y_i) = 0$ for all $p\}$ (the collection of polynomials that vanish on $Y$). Show that $\mathscr{P}$ is dense on $I = \{f \in C_\mathbb{R}(X): f|_Y = 0\}$.

I'm not quite sure how to proceed with this problem. Is there any suggestion or hint?

EDIT: Dense wrt to the sup norm.

答:
Hint: Suppose $Y$ is just one point, say $Y=\{a\}.$ By Stone-Weierstrass there exists a sequence $p_m$ of polynomials on $\mathbb R^n$ that converges uniformly to $f$ on $X.$ What can you say about

$$f(y)-[p_m(y)-p_m(a)]?$$


问:
Finding the indefinite integral $\int \frac{3x+2}{(6x^2+8x)^7}\,\mathrm dx$
I'm not too familiar with how to solve this. Could anyone present a step by step guide on how to get the answer?

$$\int \dfrac{3x+2}{(6x^2+8x)^7}\,\mathrm dx$$

答:
Put $6x^2+8x=u\implies (12x+8)dx=du \implies 4(3x+2)dx=du$

So, $$\int \dfrac{3x+2}{(6x^2+8x)^7}\,\mathrm dx=\int\frac{du}{4u^7}=\frac14\int u^{-7}du=\frac14\cdot \frac{u^{-7+1}}{-7+1}+C$$ where $C$ is an arbitrary constant of indefinite integration .

ContentLicense: CC BY-SA 3.0


问:
Limit of $\frac{1 - F(x)}{f(x)}$
Suppose $f: [0, 1] \rightarrow \mathbb{R}+$ is a continuous density function. Let $F$ be its c.d.f. Suppose $f > 0$ over $(0, 1)$ and $$\frac{1 - F(x)}{f(x)}$$ is decreasing. Is it true that $$\lim_{x \rightarrow 1} \frac{1 - F(x)}{f(x)} = 0?$$

Trivially, if $f(1) > 0$, then it is true. But how about $f(1) = 0$?

答:
The assertion is true. The reciprocal function to $\frac{1-F(x)}{f(x)}$ is the hazard function
$$h(x):=\frac{f(x)}{1-F(x)}.\tag1$$
Check that $h$ satisfies
$$
\int_0^x h(u)\,du=-\log[1-F(x)]\tag2
$$
for every $x$. In particular, as $x$ approaches $1$ from below,
$$
\lim_{x\to1}\int_0^x h(u)\,du=\infty.\tag3
$$
Reason: For $x\in(0,1)$ we know $f(x)>0$, so $F(x)$ increases to $1$ as $x$ approaches $1$, hence the RHS of (2) approaches infinity.

Note that $h$ is increasing (since by assumption its reciprocal is decreasing), so (3) implies that $h(x)\to\infty$ as $x\to1$; otherwise the integral $\int_0^1 h(u)du$ would be finite. As a consequence we see
the reciprocal function must decrease to $0$.

Title: Limit of $\frac{1 - F(x)}{f(x)}$


问:
Can we assume that this separation is formed of connected sets?
Suppose $Y$ is a subset of $\mathbb{R}^n$ and we want to prove that $Y$ is connected. By contradiction we assume this is not the case, and so there are sets $A$ and $B$ such that $Y=A\cup B$ and the closure of each set does not interset the other set. My question is: can we ASSUME that $A$ and $B$ are connected?

答:
The other two  answers are indeed conclusive but the OP's resistance to accepting any one of them makes me suspect  that
they indeed want something slightly different, namely what one gets upon replacing
"$Y=A\cup B$" by "$Y\subseteq A\cup B$", in the statement of the question,  requiring also that both $A\cap Y$ and $B\cap Y$ are nonempty.

To be 100% precise here is my interpretation of the question:

Question.  Given a disconnected subset $Y⊆\mathbb R^n$, can one find open connected subsets $A,B ⊆\mathbb R^n$,
such that both $A\cap Y$ and $B\cap Y$ are nonempty, and $Y\subseteq A\cup B$?

In case $Y$ is a subset of $\mathbb R^1$, this is  indeed possible  since a disconnected subset $Y\subseteq \mathbb R$ cannot
be an
interval,  so there are points $y_1<x<y_2$, with $y_1,y_2\in Y$, and $x\notin Y$, and then  one may take
$A=(-\infty ,x)$ and   $B=(x, +\infty )$.

In higher dimensions I believe this is a very interesting and challenging problem, to which I have no answer.
Nevertheless,  in case $Y$ is compact, this is not so difficult,  and here is my attempt at proving it:

Given that $Y$ is disconnected, choose open subsets $A$ and $B$ of $\mathbb R^n$, with $Y\subseteq A\cup B$,  and   $A\cap Y\neq \emptyset\neq B\cap Y$.

For each point $y\in A\cap Y$,  let us choose an open cube $C_y$ such that $y\in C_y\subseteq A$.  Since the $C_y$ cover the compact set
$A\cap Y$,  we may find $y_1,\ldots ,y_n$ such that
$$
  A\cap Y\subseteq A':=\bigcup_{i=1}^nC_y \subseteq A.
  $$
Doing the same  relative to $B\cap Y$ we may find a finite union of open cubes $B'$ such that
$B\cap Y\subseteq B'\subseteq B$.

This said we may assume WLOG that the originally chosen sets $A$ and $B$  are each a finite union of open cubes.

Denoting by $A_1, A_2, \ldots ,  A_n$ the connected components of $A$,  each of which is obviously again a finite union of open
cubes,  we then have that $Y\cap A\subseteq \bigcup_{i=1}^nA_n$.

Doing the same with $B$,  and relabling
$$
 (C_1, C_2, \ldots ,  C_p):=  (A_1, A_2, \ldots ,  A_n; B_1,  B_2,  \ldots ,  B_m),
  $$
we see that

each $C_i$ is a connected finite union of open cubes,

$Y\subseteq \bigcup_{i=1}^pC_i$.

If  we discard the $C_i$ having no intersection with $Y$, we may also assume that

$C_i\cap Y\neq \emptyset$, for all $i$,

$p\geq 2$.

To conclude we must show how to  arrange a situation in which $p$ is exactly equal to 2.  For this assume that $p$ is
minimal and, by contradiction, that $p>2$.

The final part of the argument might be a bit awkward  to describe in detail,  but intuitively  it is pretty obvious:  clearly
one may find two such sets,  say $C_i$ and $C_j$,  and a bridge between them formed by a finite union of open cubes, without
going over any one of the remaining cubes.

One may then replace $C_i$ and $C_j$ by the single connected set formed by the
union of $C_i$,  $C_j$,  and the bridge,  hence reducing $p$ by 1 and contradicting minimality!


问:
Why do we not consider the case where $u$ and $v$ in the factored form $(x+u)(x+v)$ are fractions?
When factoring the quadratic $x^2 + bx + c $, where $b$ and $c$ are integers, why do we not consider the case where $u$ and $v$ in the factored form $(x+u)(x+v)$ are fractions?

We wish to write this as $(x + u) (x + v)$. Where $u = \frac{p}{q}$ and $v =\frac{m}{n}$.

\begin{align}
\left(x + \frac{p}{q}\right)\left(x + \frac{m}{n}\right)&= x\left(x+\frac{m}{n}\right)+\frac{p}{q}\left(x + \frac{m}{n}\right)\\
&=x^2 + \frac{mx}{n} + \frac{px}{q} + \frac{pm}{qn}\\
&=qnx^2 + (mq + pn)x + pm.
\end{align}

This is quite different from the quadratic $x^2 + bx + c$.

Am I along the right path? What's the complete reasoning?

答:
Notice that if $u = \frac{p}{q}$ and $v = \frac{m}{n}$ for some integers $p,q,m,n$ then you obtain the equation $$qn\left(x + \frac{p}{q}\right)\left(x + \frac{m}{n}\right) = qnx^2 + (mq + pn)x + pm.$$  In the expression $qnx^2 + (mq + pn)x + pm$ we can conclude that $qn, mq+pn$, and $pm$ are all integers.  If we use the labels

\begin{align}a &= qn\\
b&=mq+pn\\
c&=pm\\
u &= \frac{p}{q}\\
v &= \frac{m}{n}\end{align} then we again end up with $$(x-u)(x-v) = ax^2 + bx + c,$$ so in reality this form covers your case as well.  Generally, though, we don't require $a,b,c,u,v$ to be integers.


问:
In how many ways can 5 objects be placed in 3 groups?
Consider this question:

In how many ways can $5$ tourists be placed in $3$ different hotels, such that there is at least one tourist in each hotel?

To start working on this, I removed the lower bound on the number of tourists in a hotel. I reasoned that any tourist can be placed in any of the $3$ hotels, resulting in $5\times 3=15$ permutations.

However, the correct answer with the lower bound is $150$, which is more than $15$.

Obviously, I am undercounting by a large amount. How?

答:
The number 150 was very logically solved by Phicar. For fun, I wrote a Python that confirms it.</p>
```<code>from random import seed
from random import randint
seed(1)
Number_of_Tourists = 5
Number_of_Hotels = 3

class Permutation():
    def __init__(self):
        self.make_permutation()

    def make_permutation(self):
        while True:
            self.tourists =[]
            for t in range(Number_of_Tourists):
                self.tourists.append(randint(1, Number_of_Hotels))
            self.hotels = {}
            for t in self.tourists:
                if t not in self.hotels:
                    self.hotels[t] = 1
            # Enforce lower bound, each hotel as at least one tourist
            if len(self.hotels) == Number_of_Hotels:
                break

def find_permutations():
    permutations = []
    for i in range(2000):
        permutations.append(Permutation())
    permutation_hash = {}
    for p1 in permutations:
        hash = str(p1.tourists)
        if hash not in permutation_hash:
            permutation_hash[hash] = 1
    print('Answer = '+ str(len(permutation_hash)))

find_permutations()
# Output: Answer = 150

<span class="math-container">```</span>
</code>```

Body:
The number 150 was very logically solved by Phicar. For fun, I wrote a Python that confirms it.

from random import seed
from random import randint
seed(1)
Number_of_Tourists = 5
Number_of_Hotels = 3

class Permutation():
    def __init__(self):
        self.make_permutation()

    def make_permutation(self):
        while True:
            self.tourists =[]
            for t in range(Number_of_Tourists):
                self.tourists.append(randint(1, Number_of_Hotels))
            self.hotels = {}
            for t in self.tourists:
                if t not in self.hotels:
                    self.hotels[t] = 1
            # Enforce lower bound, each hotel as at least one tourist
            if len(self.hotels) == Number_of_Hotels:
                break

def find_permutations():
    permutations = []
    for i in range(2000):
        permutations.append(Permutation())
    permutation_hash = {}
    for p1 in permutations:
        hash = str(p1.tourists)
        if hash not in permutation_hash:
            permutation_hash[hash] = 1
    print('Answer = '+ str(len(permutation_hash)))

find_permutations()
# Output: Answer = 150

```


问:
A problem on modular arithmetic.
How is $\left(\frac {p-1}{2}\right)! ^2 \equiv -1 \pmod p$. I was using this result for a proof ($x^2$ = -1 (mod p) ) and was stuck on this part. NOTE, this result is only when p is a prime of form 4k+1.

note: please don't delete my question without atleast informing me of the reason.

Title: A problem on modular arithmetic.

Body:
Your claim seems false:

Take $ p=7$.

$$(\frac{p-1}{2})!=3!=6$$

$$6^2=36\equiv 1 \mod 7$$


问:
Proof involving derivative product rule for transpose and gradient

Let $f,g: \mathbb{R} \rightarrow \mathbb{R}^n$ be differentiable
functions and let $h: \mathbb{R} \rightarrow \mathbb{R}$ be defined by
$h(t) = f(t) \cdot g(t)$.

Prove that $\frac{dh}{dt}(a) = (Df(a))^T \cdot g(a) + f(a) \cdot (Dg(a))^T$,

where $a \in \mathbb{R}$ and $M^T$ represents the transpose of a
matrix $M$ (we are considering, for example, the row matrix
$(Df(a))^T$ as an element of $\mathbb{R}^n$.

I know that this has something to do with gradient vectors and product rule, but I actually have no ideia of how to prove this. Any help, please?

答:
You can track the vector components.

Let $f_i, g_i$ be the vector components of the function values. Note that $(Df)_i=\frac{d}{dt}f_i$.

$$
\frac{dh}{dt}=\frac{d}{dt}\sum_i f_i(t) g_i(t)=\sum_i f_i(t)\frac{d}{dt}g_i(t)+g_i(t)\frac{d}{dt}f_i(t)
$$

which leads to the product rule formula you stated by definition.

The use of transpose in the original statement is superfluous. Transpose only matters if you phrase $h$ as the matrix-vector multiplication of $f^Tg$, where you can see the application of (the true $\mathbb{R}^m\rightarrow\mathbb{R}^n$) product rule simply retains the transpose on $f$; except that is not what your question asks.


问:
Equation on cross product
I have to solve the equation $\vec x\times\vec a=\vec b-\vec x$  ,   where $\vec x, \vec a,\vec b$ are vectors and the last two are known.

I have proven that $\vec a\cdot \vec x=\vec a\cdot\vec b$, but that is all.

Thank you in advance !

<< Symbol : $\vec a \vec b = \vec a \cdot \vec b$ >>

答:
Given:

$$\vec{x} = \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}, \vec{a} = \begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix} ~\text{and}~ \vec{b} = \begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix},$$
the equation
$$\vec x\times\vec a=\vec b-\vec x$$

corresponds to:

$$\begin{cases}
a_3 x_2 - a_2 x_3 +x_1 - b_1 = 0\\
a_1 x_3 - a_3 x_1 +x_2 - b_2 = 0\\
a_2 x_1 - a_1 x_2 +x_3 - b_3 = 0
\end{cases}.$$

The solution is:

$$x_1 = \frac{a_1(\vec{a} \cdot \vec{b}) + b_1 + a_2 b_3 - a_3 b_2}{|\vec{a}|^2 + 1},$$

$$x_2 = \frac{a_2(\vec{a} \cdot \vec{b}) + b_2 + a_3 b_1 - a_1 b_3}{|\vec{a}|^2 + 1},$$

$$x_2 = \frac{a_3(\vec{a} \cdot \vec{b}) + b_3 + a_1 b_2 - a_2 b_1}{|\vec{a}|^2 + 1}.$$

Note that $\vec{a} \cdot \vec{b}$ is the scalar product, while $|\vec{a}|^2 = \vec{a} \cdot \vec{a}$ is the squared norm of the vector $\vec{a}$.

Title: Equation on cross product


问:
Does hold the isomorphism$ (A \otimes_R B)/J(A \otimes_R B) \cong (A/J(A)) \otimes_R (B/J(B))$ where $R$ is a commutative ring?
while reading an article, I saw the isomorphism $(A \otimes_K B)/J(A \otimes_K B) \cong (A/J(A)) \otimes_K (B/J(B))$ where $A$ and $B$ are two algebras over the field $K$ and $J$ denotes Jacobson radical. Why is this isomorphism true? Also, if $A$ and $B$ are two algebras over the commutative ring $R$ then can we say this isomorphism true?

答: This is false over any field $K$ that isn't perfect. For example, if $K = \mathbb{F}_p(t)$ and $A = B$ are both the purely inseparable extension $\mathbb{F}_p(\sqrt[p]{t})$, then $A \otimes_k B \cong A[x]/x^p$ has a nontrivial nilpotent, so $J(A \otimes_k B) \cong (x)$ is nontrivial even though $J(A)$ and $J(B)$ are both trivial (since $A, B$ are fields).


问:
Probability recursion on normal distribution
I'm currently reading this paper. I have one question on some probability part.
In the paper they have the following equations.
$$ P(A)\sim \mathcal {N}(\Pi,\tau \Sigma), r_A \sim \mathcal{N}(P(A),\Sigma) \tag{11}\label{11}$$
$$ r_A \sim \mathcal{N}(\Pi,(1+\tau)\Sigma)  \tag{12}\label{12}$$
My question is that I have absolutely no idea how they pass from $\eqref{11}$ to $\eqref{12}$. It is of course some lack in probability theory for something like recursive functions maybe?. I don't know. Can somebody explain this relationship? Why $\mathcal{N}(\mathcal {N}(\Pi,\tau \Sigma),\Sigma)$ equals $\mathcal{N}(\Pi,(1+\tau)\Sigma)$? (if I'm not mistaken)

答:
We have
$$
f(x|\theta) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{1}{2}\left(\frac{x-\theta}{\sigma}\right)^2}
$$
Conjugating the prior distribution for $\theta$ which is a normal distribution as
$$
f(\theta)  = \frac{1}{r\sqrt{2\pi \sigma^2}}e^{-\frac{1}{2}\left(\frac{\theta-\mu}{r\sigma}\right)^2}
$$
So the joint g-pdf would be
$$
f(x,\theta) = f(x|\theta)f(\theta) = \frac{1}{\sqrt{2\pi\sigma^2}}\frac{1}{r\sqrt{2\pi\sigma^2}}e^{-\frac{1}{2}\left[(\frac{x-\theta}{\sigma})^2 + (\frac{\theta-\mu}{r\sigma})^2\right]}
$$
By expanding the quadratic inside the exponent,
$$
(\frac{x-\theta}{\sigma})^2 + (\frac{\theta-\mu}{r\sigma})^2 = \frac{x^2 + \theta^2-2x\theta}{\sigma^2} + \frac{\theta^2 + \mu^2 - 2\theta\mu}{r^2\sigma^2}
$$

$$
=\frac{x^2r^2 + \theta^2r^2-2x\theta r^2 + \theta^2 + \mu^2 - 2\theta\mu }{r^2\sigma^2}
$$

$$
=\frac{\theta^2(1+r^2) - 2\theta(xr^2 + \mu) + (x^2r^2 + \mu^2)}{r^2\sigma^2}
$$

Now the marginal probability of $f(x)$ could be calculated from integration of joint probability of $f(x,\theta)$,
$$
f(x) = \int f(x,\theta)d \theta = \frac{1}{\sqrt{2\pi\sigma^2}}\frac{1}{r\sqrt{2\pi\sigma^2}}\int e^{\frac{-\theta^2(1+r^2) + 2\theta(xr^2 + \mu) - (x^2r^2 + \mu^2)}{2r^2\sigma^2}} d\theta
= \frac{1}{\sqrt{2\pi\sigma^2}}\frac{1}{r\sqrt{2\pi\sigma^2}}\sqrt{\frac{\pi}{\frac{1+r^2}{2r^2\sigma^2}}}e^{\frac{(2\frac{(xr^2+\mu)}{2r^2\sigma^2})^2}{4\frac{1+r^2}{2r^2\sigma^2}}-\frac{x^2r^2+\mu^2}{2r^2\sigma^2}}
$$
We could simplify the terms inside the exponent as
$$
\frac{(xr^2+\mu)^2}{(2r^2\sigma^2)(1+r^2)}-\frac{x^2r^2+\mu^2}{2r^2\sigma^2} = \frac{(xr^2+\mu)^2-(x^2r^2+\mu^2)(1+r^2)}{(2r^2\sigma^2)(1+r^2)}
$$
$$
=\frac{x^2r^4+\mu^2+2xr^2\mu-x^2r^2-\mu^2-x^2r^4-r^2\mu^2}{(2r^2\sigma^2)(1+r^2)}
=\frac{2xr^2\mu-x^2r^2-r^2\mu^2}{(2r^2\sigma^2)(1+r^2)}
=\frac{-(x-\mu)^2}{(2\sigma^2)(1+r^2)}
$$
Thus, the pdf of $f(x)$ will be
$$
f(x) =  \frac{1}{\sqrt{2\pi\sigma^2(1+r^2)}}e^{\frac{-1}{2}\left[\frac{(x-\mu)^2}{(\sigma^2)(1+r^2)}\right]}
$$


问:
Is the area independent of $n$? $ A=\int_0^\infty \frac{e^{-\frac{1}{tn}}(e^{-tn}(tn+1))}{t}~dt $
Consider the parametric equation $x=\frac{1}{t}e^{-tn}$ and $y=te^{-\frac{1}{tn}}.$ To integrate under this curve I put it in the proper form:

$$ A=\int_0^\infty \frac{e^{-\frac{1}{tn}}(e^{-tn}(tn+1))}{t}~dt $$

I noticed that by changing $n$ from 1 to 2 to 3 that the area was the same!

Is the area $A$ independent of $n?$

Do integrals like these have applications?

答:
Letting $u=nt$ so that $du=ndt$ gives
$$A=\int_0^{\infty}\dfrac{e^{-\frac{1}{u}}(e^{-u}(u+1))}{u}du$$
which is precisely the integral when $n=1$.


问:
Zorn's lemma and maximal linearly ordered subsets
Let $T$ be a partially ordered set. We say $T$ is a tree if $\forall t\in T$ $\{r\in T\mid r < t\}$ is linearly ordered (such orders can be considered on connected graphs without cycles, i.e. on trees). By a branch we mean a maximal linearly ordered subset in $T$.

It is easy to prove that each tree has a branch using Zorn's lemma. However, the converse is also true (I read it recently in an article). Can anybody give a sketchy proof?

答:
Here is a nice way of proving the well-ordering principle from "Every tree has a branch":

Let $A$ be an infinite set, and let $\lambda$ be the least ordinal such that there is no injection from $\lambda$ into $A$. Consider the set $A^{<\lambda}$, that is all the functions from ordinals smaller than $\lambda$ into $A$, and order those by end-extension (which is exactly $\subseteq$ if you think about it for a moment).

It is not hard to verify that $(A^{<\lambda},\subseteq)$ is a tree. Therefore it has a branch. Take the union of that branch, and we have a "maximal" function $f$ whose domain is an ordinal $\alpha<\lambda$. Note that it has to be a maximal element, otherwise we can extend it and the branch was no branch.

If the range of $f$ is not all $A$ then there is some $a\in A$ which witnesses that, and we can extend $f\cup\{\langle\alpha,a\rangle\}$ to a strictly larger function. Since $f$ is maximal, it follows it has to be a surjection, and therefore $A$ can be well-ordered.

(Note: If we restrict to the sub-tree of the injective functions we can pull the same trick and end up with a bijection, but a surjection from an ordinal is just as good).

From this we have that every set can be well-ordered, and therefore the axiom of choice holds.


答: If $\vec{x}=\vec{y},\,\vec{x}=\vec{y}+\vec{z}$ both work, $\vec{z}\times\vec{a}=-\vec{z}$, so $\vec{z}$ is self-orthogonal and vanishes, and the solution $\vec{x}$ is unique. Now try the Ansatz $\vec{x}=A\vec{a}+B\vec{b}+C\vec{a}\times\vec{b}$. As I noted in a comment on @thecandyman's answer, you should find$$\vec{x}=\frac{(\vec{a}\cdot\vec{b})\vec{a}+\vec{b}+\vec{a}\times\vec{b}}{a^2+1}.$$

Title: Equation on cross product

Body:
I have to solve the equation $\vec x\times\vec a=\vec b-\vec x$  ,   where $\vec x, \vec a,\vec b$ are vectors and the last two are known.

I have proven that $\vec a\cdot \vec x=\vec a\cdot\vec b$, but that is all.

Thank you in advance !

<< Symbol : $\vec a \vec b = \vec a \cdot \vec b$ >>


问:
Newton tangent method (1D)
Newton tangent method for finding root of $f(x)=x^2-a \;$   is
$$x_{k+1} = x_k-\dfrac{x_k^2-a}{2x_k},\;\;(k=0,1,2,...)$$
and its error
$$|x_k-\sqrt{a}|\leq \dfrac{M}{2m}|x_k-x_{k-1}|^2$$
where $M=\max_{[c,d]}|f''(x)|$ and $m=\min_{[c,d]}|f'(x)|$. In one reference, I find special case $a=3$ and it needs to be found with precision $10^{-6}$. First $6$ approximations are

$$3.00000,\;2.0000000,\; 1.7500000,\; 1.7321429,\; 1.7320508,\; 1.7320508$$
They stopped when $x_k$ and $x_{k+1}$ got same first $6$ digits. Is it always case and does this guarantee that error will be less than $10^{-6}$ or is this some practical evaluation?
I ask this because of these two numbers $M$ and $m$, it seems to me that they are ignored. No matter how small $|x_k-x_{k-1}|$, $M$ and $m$ can produce some large number. Can someone explain me is this $100$% correct to stop when $x_k$ and $x_{k-1}$ get same first $6$ digits

答:
The formula give you the certainty that the error, the distance to the root, is indeed bounded by the step updates. Remember that if the step updates go to zero like $1/n$ or $1/n^2$ or even still with a geometric behavior with a factor close to $1$, then the step updates themselves have no or little information on the remaining distance to the limit.

Now here you have $|x_5-x_4|\le 10^{-7}$, which gives $|x_5-\sqrt3|\le \frac{M}{2m}10^{-14}$, so that if the derivatives behave not too extreme, the first factor is smaller than, say, $100$, the goal for the accuracy is reached.

If you take $x_{k+1}$ as proxy for the limit $\sqrt a$, then $\frac{|x_{k}-x_{k+1}|}{|x_k-x_{k-1}|^2}$ is a guess for the factor $\frac{M}{2m}$. For the gien values that gives
\begin{array}{c|c|c}
   k &   x_k  &  \frac{|x_{k}-x_{k+1}|}{|x_k-x_{k-1}|^2} \\
\hline
   0& 3.00000000&\\
   1& 2.00000000& 0.25 \\
   2& 1.75000000& 0.2857 \\
   3& 1.73214286& 0.2887 \\
   4& 1.73205081& 0.2887
\end{array}
so that indeed the amount of correct decimal digits doubles in every step after the second step.

Title: Newton tangent method (1D)


问:
Solving equation $a_n=1+\frac12a_{n-1}+\frac12a_{n+1}$
Let's consider sequence following for $n=0...,T$ for which $a_0=a_T=0$. I want to solve equation $$a_n=1+\frac12a_{n+1}+\frac12a_{n-1}$$

My work so far

$$a_1=1+\frac12a_2 \Leftrightarrow a_2=2a_1-2$$
$$a_2=1+\frac12a_3+\frac12a_1 \Leftrightarrow2a_1-2=1+\frac12a_3 \Leftrightarrow a_3=3a_1-6$$
Also I calculated $a_4$ to get $a_4=4a_1-12$. So my observation is that for any $n \in \{1,...,T-1\}$
we have $$a_n=na_1+n(n-1)$$

And here I stacked, I'm not sure what to do next. Could you please give me a hand solving this equation ?

答:
Hint 1: Prove by induction that your guess is correct.

Hint 2  If $a_T=0$ then
$$Ta_1+T(T-1)=0$$

Use this to find $a_1$.


问:
Dominant morphism $P^1\to P^1$ which is not an isomorphism
Let $P^1$ be over an algebraically closed base field.

My aim is to find a dominant morphism $P^1\to P^1$ which is not an isomorphism. I am disturbed by this since I am convinced it is not possible.

First, such a morphism $\phi$ must in fact be surjective. I also know that $\phi^*$ on the function fields would be injective. I believe that in fact $\phi^*$ is also necessarily surjective, although I cannot prove this. But then this would mean that $\phi^*$ is given by a linear polynomial, so $\phi$ would be an isomorphism. So if I am right, I simply need to show that $\phi^*$ is surjective, but if I am wrong, then there would be counterexample exploiting exactly this potential failure of $\phi^*$ is to be surjective.

I also have a similar, potentially related result in my text, which states that for any smooth variety of dimension $1$ and a dominant morphism $\phi:P^1\to X$, we must have that $X$ is isomorphic to $P^1$. The proof of this uses Luroth's theorem in a critical way, but I'm not sure how exactly to connect this to the problem at hand.

答:
First, such a morphism $\phi$ must in fact be surjective. I also know that $\phi^*$ on the function fields would be injective. I believe that in fact $\phi^*$ is also necessarily surjective, although I cannot prove this.

This is false. A simple counterexample is the map $k(x) \xrightarrow{x \mapsto x^2} k(x)$, which exhibits $k(x)$ as a subfield of itself of index $2$ (namely $k(x^2)$). This corresponds to a map $\mathbb{P}^1 \to \mathbb{P}^1$ of degree $2$ given in homogeneous coordinates by $(X : Y) \mapsto (X^2 : Y^2)$.

More generally dominant maps $\mathbb{P}^1 \to \mathbb{P}^1$ over a field $k$ correspond naturally to homomorphisms $k(x) \xrightarrow{f} k(x)$ (necessarily injective) over $k$, which in turn correspond to nonzero rational functions $f(x) = \frac{p(x)}{q(x)} \in k(x)$ (the image of $x$ under the map $f$). This map is an isomorphism iff $f$ is a Mobius transformation $f(x) = \frac{ax + b}{cx + d}, ad - bc \neq 0$.


问:
Why does the form of the normal distribution have different forms for the 2nd parameter?
Sometimes I see the normal distribution written as $N(\mu,\sigma)$ sometimes as $N(\mu,\sigma^2)$ and sometimes, when the Central Limit Theorem is involved, as $N(\mu,\sigma^2/n)$

As a software developer I would not write a function that overloads the meaning of a parameter.
It looks to me as if that is what is being done here.

How should I be thinking of the 2nd parameter here?

Are we specifying how the variance is to be created in a new normal distribution out of parameters of an old distribution?

答:
As a software developer you have totally nailed it, in that the second parameter is overloaded, and that there is a

$$N( \text{mean}, \text{std_dev}) $$
constructor and a
$$N( \text{mean}, \text{variance}) $$
constructor.

The mathematician parser is generally smart enough to be able to distinguish between the two when it encounters "$N(\cdot, \cdot)$". Sometimes it can tell from how it is written (i.e. $N(\mu,\sigma)$ versus $N(\mu,\sigma^2)$, other times it can tell from the units, e.g. if the mean is in $cm$, then it looks at whether the second parameter has units of $cm$ or $cm^2$. If it cannot disambiguate it throws an error and demands that the writer write it more clearly.

Your observation about the CLT, and the use of $N(\mu,\sigma^2/n)$, however,  points out that mathematicians can be pretty lazy about notation, relying on context, which can be confusing to the beginner, who may not know all of the same assumed context. And statisticians are even worse than your average mathematician in this regard. (I've taught stats, and have found myself apologizing to students for this.)

In the context of the CLT there is an underlying random variable (call it $X$), with mean $\mu$ and standard deviation $\sigma$. This underlying random variable may or may not be normally distributed, and so $\mu$ is shorthand for $E[X]$, and similarly for $\sigma$.

In the CLT we construct a new random variable, the "sample mean (from $n$ samples)", and call it $\overline{X}$. We want to compare its distribution with a particular Normal Distribution, $N(\mu,\sigma^2/n)$. The content of the CLT is that $\overline{X}$ looks more and more like $N(\mu,\sigma^2/n)$ as $n$ gets larger.

So yes, you are right, in the CLT "we [are] specifying how the variance is to be created in a new normal distribution out of parameters of an old distribution". The only thing to add is that there is an easy CLT, where the original distribution is normal, and a more general CLT, where the original distribution is fairly arbitrary (within some broad limits). In the easy CLT, $\mu$ and $\sigma$ are the parameters of the original distribution, but in the more general version they are "computed properties" of the distribution. For example, if the original distribution is Poisson (with parameter $\lambda$), then the CLT holds, but with $\mu = \lambda$, and $\sigma = \sqrt \lambda$.


问:
Highest and lowest, exercise.
Let $ A $ and $ B $ be non-empty sets of $ \mathbb R $ where $ A $ is upper bound and $ B $ is lower bound. Suppose that for every $ \epsilon> 0 $, there exist $ x \in A $ and $  y\in B $ such that $ 0 <y-x <\epsilon $. Is it true then that $ \sup A = \inf B $?

答: For a counterexample, you could take $A = B$

Title: Highest and lowest, exercise.


问:
If finer topology is second-countable then coarses is second-countable
Let $(X,\tau_1)$ and $(X,\tau_2)$ be topological spaces and $\tau_2$ finer than $\tau_1$. Prove if $\tau_2$ is second-countable then $\tau_1$ is also second-countable.

My try:

 Let $B=\{B_n,n\in \mathbb{N}\}$ be countable basis for $\tau_2$. Let's prove that $B_1=B\cap \tau_1$ is countable basis for $\tau_1$. It's countable because it's subset of countable set.

And now I'm stuck, because I don't know how to prove it's basis for $\tau_1$. I tried to prove it like this: Let $U\in\tau_1\subset\tau_2$ that means there is $B'\subseteq B$ so $U=\cup B'$. What next? How to show it's also union of elements of $B_1$ ?

答:
Will a counterexample do? Let $\tau_2$ be the discrete topology on $\mathbb{N}$, and suppose $\tau_1$ is a non-first-countable (equivalently, non-second-countable) topology on $\mathbb{N}$. Since $\tau_2$ is second-countable and finer than $\tau_1$, this will be a counterexample to what you wanted to prove. So all we have to do is find a non-first-countable topology on $\mathbb{N}$. Here are two ways to do that.

I. Let $\mathcal{U}$ be a uniform ultrafilter on $\mathbb{N}$. Call a set $X\subseteq\mathbb{N}$ open if either $1\notin X$ or else $X\in\mathcal{U}$.

II. For $X\subseteq\mathbb{N}$ let $d(X)$ be the asymptotic density of $X$. Call a set X open if either $1\notin X$ or else $d(X)=1$.

In words: we make the topology discrete except at one point. For the neighborhood system of the special point we take some filter which is not countably generated.


问:
Efficient way to assemble polynomial terms with more than one variable?
Assume that I have a scalar variable $x$. For this variable, I can write down a second degree polynomial:

$$f(x) = c_0 + c_1x + c_2x^2$$

where $c_0$, $c_1$, and $c_2$ are scalar coefficients. Now assume that my polynomial does not only depend on $x$, but also on two other variables $y$ and $z$. For simplicity, assume that we wish to limit the order of terms of each summand to 2, so we get:

$$f(x,y,z) = c_0 + c_{1}x + c_{2}y + c_{3}z + c_{4}x^2 + c_{5}y^2 + c_{6}z^2 + c_{7}xy + c_{8}yz + c_{9}xz$$

Obviously, the first seven terms are simply the independent polynomial terms of all three variables. The remaining terms are cross-variable terms, whose (summed) order I somewhat arbitrarily limited to 2 (different limitations would of course also be possible).

Now my question: Is there an easy version to find all combinations of variables for the remaining terms for arbitrary numbers of particles and upper order limits?

A brute-force solution would be to create all possible combinations, then discard the ones which violate the order limit:

$$\require{cancel}x+y+z+x^2+y^2+z^2+xy+xz+yz+\cancel{x^2y}+\cancel{x^2z}+\cancel{xy^2}+\cancel{y^2z}+\cancel{xz^2}+\cancel{yz^2}+\cancel{x^2y^2}+\cancel{x^2z^2}+\cancel{y^2z^2}$$

However, this quickly becomes infeasible for larger variable counts.

答:
I don't quite know what you mean by "find" in your question. Do you mean how to list or name them all? The general term of a polynomial in $k$ variables $x_1, \dots x_k$ has the form

$$c_{i_1, i_2, \dots i_k} x_1^{i_1} x_2^{i_2} \dots x_k^{i_k}$$

with total degree $\sum_{j=1}^k i_j$. If you want to bound the total degree then you just enforce a bound on this sum, so the general polynomial in $k$ variables of degree at most $n$ is

$$\sum_{i_1, i_2, \dots i_k \in \mathbb{Z}_{\ge 0} : \sum i_j \le n} c_{i_1, i_2, \dots i_k} x_1^{i_1} x_2^{i_2} \dots x_k^{i_k}.$$

Writing all those subscripts can get annoying so it's convenient to use multi-index notation instead: we package up the tuple $(i_1, i_2, \dots i_k)$ of indices into a single vector index $I$ and similarly package up the tuple $(x_1, x_2, \dots x_k)$ of variables into a single vector variable $x$, then define $x^I = x_1^{i_1} x_2^{i_2} \dots x_k^{i_k}$. If we also write $|I| = \sum_{j=1}^k i_j$ for the total degree then the above summation can be written in the simplified form

$$\sum_{I \in \mathbb{Z}_{\ge 0}^k : |I| \le n} c_I x^I.$$


问:
How does $\frac{\cos x}{1-\cos^2x}=\frac23$ become $\cos x=\frac12$?
I'm sure its just gaps in my trig knowledge, however I cant seem to find a similar example easily.

The circled part of the problem is the simplification I'm having trouble with.

I understand where the identity is coming from, what I'm having trouble with is how it is getting eliminated. Any help is appreciated. Thank you.

Body:
$$2 \cot x \csc x=\frac{4}{3}$$
$$\cot x \csc x=\frac{2}{3}$$
$$\frac{\cos x}{\sin x}\cdot \frac{1}{\sin x}=\frac{2}{3}$$
$$\frac{\cos x}{\sin^2 x}=\frac{2}{3}$$
$$\frac{\cos x}{1-\cos^2 x}=\frac{2}{3}$$
set $\cos x= y$
$$\frac{y}{1-y^2}=\frac{2}{3}$$
$$2y^2+3y-2=0\to y=\frac{1}{2};\;y=-2$$
$$\cos x=\frac{1}{2}\to x=\pm\frac{\pi}{3}+2k\pi$$
$\cos x=-2$ is discarded because has no real solutions.


问:
monic irreducible polynomial in $\mathbb Z[x]$ have a multiple root in $F_p$ over $F_p[x]$ for some prime $p$
Could you help me with the following problem which I cannot solve ?

Let $f\in \mathbb Z[x]$ be a monic irreducible polynomial with $deg(f)>1$.

Prove or disprove that there is always a prime $p$ satisfying the following condition:
$\overline{f}$ have a multiple root in $F_p$
where $\overline{f}$ is a polynomial in $F_{p}[x]$, whose coefficients are the reduction of those of $f$ modulo $p$.

From Minkowski's bound, |disc$(f)|>1$.

So we can take a prime $p$ dividing  |disc$(f)|$.

Then $\overline{f}$  have a multiple root in algebraic closure of $F_p$.

However, this does not mean $\overline{f}$ have a multiple root in $F_p$.

For example, $f=x^4 + 2x^3 + 3x^2 + 2x + 6$ and $p=5$.

In this case, disc$(f)$$=35600$ and $\overline{f}=(x^2+x+1)^2$ over $F_p$.

(On the other hand,$\overline{f}=x^2(x+1)^2$ over $F_2$)

答:
The statement is actually false.

Consider the polynomial $f=\Phi_{12}=X^4-X^2+1$. It's straightforward to see that $\Phi_{12}$ divides $X^{12}-1$. The latter polynomial is coprime with its derivative in $\mathbb{F}_p$ for $p > 3$. So we need to check that $\Phi_{12}$ has no double root in $\mathbb{F}_p$ for $p=2$ and $p=3$.

For $p=2$, $\overline{f}(x) = (x^2+x+1)^2$ so we're done. For $p=3$, $\overline{f}(x)=(x^2+1)^2$ and we're also done.


Title: Solving equation $a_n=1+\frac12a_{n-1}+\frac12a_{n+1}$

Body:
Let's consider sequence following for $n=0...,T$ for which $a_0=a_T=0$. I want to solve equation $$a_n=1+\frac12a_{n+1}+\frac12a_{n-1}$$

My work so far

$$a_1=1+\frac12a_2 \Leftrightarrow a_2=2a_1-2$$
$$a_2=1+\frac12a_3+\frac12a_1 \Leftrightarrow2a_1-2=1+\frac12a_3 \Leftrightarrow a_3=3a_1-6$$
Also I calculated $a_4$ to get $a_4=4a_1-12$. So my observation is that for any $n \in \{1,...,T-1\}$
we have $$a_n=na_1+n(n-1)$$

And here I stacked, I'm not sure what to do next. Could you please give me a hand solving this equation ?

Body:
$$a_{n+1} - a_n -(a_n-a_{n-1}) = -2 \implies a_n - a_{n-1} = (a_1-a_0) - 2(n-1).$$

Can you take it from here?

 Next $$a_n-a_0=\sum_{k=1}^n(a_k-a_{k-1}) = n(a_1-a_0) - 2\sum_{k=1}^n(k-1).$$

Lastly, plug in $n=T$ and solve for $a_1$.


问:
What is the point of computing for $dy$?
I can understand the significance of computing for a derivative $\frac{dy}{dx}$ because you can see the slope (change in y in relation to the change in x) at any given point of a function.

By that logic, computing for $dy$ can find the change in the y axis at a given point x? What is the point of this?

答:
In calculus, we generally define $\frac{d}{dx}$ as an operator that acts on a function $y$, meaning that $\frac{dy}{dx}$ is just the output of an operator and the top and bottom are not independently defined. Throughout history many people have had the same urge as you to defined "dy" and "dx" independently, which leads to a field known as infinitesimal calculus.

When Newton was first creating calculus, he originally thought of dy and dx as infinitesimals as well, and the math was only formalized with operators later.


问:
Why do we use open intervals in most proofs and definitions?
In my class we  usually use intervals and balls in many proofs and definitions, but we almost never use closed intervals (for example, in Stokes Theorem, etc). On the other hand, many books use closed intervals.

Why is this preference? What would happen if we substituted "open" by "closed"?

答:
My guess is that it's because of two related facts.

The advantage of open intervals is that, since every point in the interval has an open neighbourhood within the interval, there are no special points 'at the edge' like in closed intervals, which require being treated differently.

Lots of definitions rely on the existence of a neighbourhood in their most formal aspect, like differentiability for instance, so key properties within the result may require special formulation at the boundary.

In PDE/functional analysis contexts for example boundaries are very subtle and important objects which are treated separately.


问:
About finding the closure of a set.
I want to learn how should I find the closure of a set $S$. It's a long that I was not worked with topological concepts, and I am afraid if I am mistaken about some easy things. Here I will list some of the facts which will help me to solve my problems and pursue my purposes.

Suppose that a topological space $X$ is given, and let $R, S \subseteq X$ be two sets.

Is this true? The closure of $S$ is the union of $S$ with all limit points of $S$. (I think it should be the definition of the closure, or at least it should be equivalent to the definition, any other equivalent definitions are welcome.)

If $S$ is dense in $R$, then $R \subseteq \overline S$.

$\overline S$ is the largest set, in which $S$ is dense. (i.e. if $\overline S \subsetneq R'$, then any $x\in R'\backslash \overline S$ has a positive distance from $S$, therefore is not a limit point)

The next two points, are not related to the closure, but I have some doubts.

A function $f: X \rightarrow Y$ between two topological spaces is called continuous, iff for every open $V \subseteq Y$, $f^{-1}(V)$ is open in $X$. (I think this was a standard definition, any other equivalent definition are welcome.)

A function $f: X \rightarrow Y$ between two topological spaces, is continuous, iff for every close $V \subseteq Y$, $f^{-1}(V)$ is close in $X$.

My intuition says that, the first $4$ points are true. Am I right? If I am mistaken about these facts, please tell me, and if it is possible please give me a counter-example. Also, I think the last point is also equivalent to the $4^{th}$ point, because if a set is closed iff and only if its complement is open. Having this in mind it seems the last two points are equivalent to each other as the definition of a continuous function.

Any equivalent definitions to the $1^{st}$ point and $4^{th}$ point are welcome. Especially those equivalent definitions, which are useful for computing the closure. Using the first three points I am able to find the closure of some simple sets in $p$-adic topology.

答: They are all correct. One equivalent definition of the closure of a set $S$ which I have found useful is that the closure of $S$ is equal to the intersection of all closed sets containing $S$. In other words, $\overline{S}$ is the smallest closed set containing $S$.

Title: About finding the closure of a set.


问:
What is the difference between the first Chern class and integral first Chern class?
I'm reading a paper which says that the first Chern class of a manifold is $0$, but the integral first Chern class is not $0$.

What is the difference between the two? Does it have to do with taking integer coefficients?

答:
This is strange terminology; does it come from a physics paper or something? I think a topologist would think of the first Chern class $c_1 \in H^2(-, \mathbb{Z})$ as being integral by default.

I would guess that whoever wrote this is considering "the first Chern class" to refer to the real first Chern class $c_1 \in H^2(-, \mathbb{R})$, which has the advantage of being definable via Chern-Weil theory in terms of the curvature of a connection. There's a natural map $H^2(-, \mathbb{Z}) \to H^2(-, \mathbb{R})$ sending the first Chern class to the real first Chern class, whose kernel is (under mild hypotheses) the torsion subgroup of $H^2(-, \mathbb{Z})$, so if the real first Chern class vanishes this means the (integral) first Chern class is torsion.


问:
Mass matrix second order differential equation to first order ODEs
I want to simulate the behaviour of a 2-DOF robotic manipulator, which is described by the following model:

$$ M(q)\ddot{q} = -C(q,\dot{q})\dot{q}-G(q)+\tau \ \ \ \ \ (1) $$

Considering the fact that the $\text{2x2}$ mass matrix $M$ is positive definite, I could use the inverse matrix and break down the problem into the $4$ first order ordinary differential equations and simulate it:

$$ x_1 = q_1 \Rightarrow \dot{x_1} = x_2 $$
$$ x_2 = \dot{q_1} \Rightarrow \dot{x_2} = -M^{-1}(1,:)\cdot C\cdot \begin{bmatrix} x_2 & x_4 \end{bmatrix}^{T}-M^{-1}(1,:)\cdot G + M^{-1}(1,:)\cdot \tau $$
$$ x_3 = q_2 \Rightarrow \dot{x_3} = x_4 $$
$$ x_4 = \dot{q_2} \Rightarrow \dot{x_4} = -M^{-1}(2,:)\cdot C\cdot \begin{bmatrix} x_2 & x_4 \end{bmatrix}^{T}-M^{-1}(2,:)\cdot G + M^{-1}(2,:)\cdot \tau $$

Suppose I would like to use a solver that takes as an argument the mass matrix (a MATLAB ODE solver in particular) and don't use its inverse because this will also simplify the computation of the jacobian (I intend to simulate a 7-DOF robotic manipulator after that so providing the mass matrix would be great). How can I write the initial equation $(1)$ as a series of first order ordinary differential equations and be able to simulate its response by using some software solvers ?

答:
Let us choose $\eta = \begin{bmatrix} q & \dot{q} \end{bmatrix}^\top$. Then
$$\begin{bmatrix} I & 0\\ 0&M(q) \end{bmatrix}\dot{\eta} = \begin{bmatrix} 0 & I \\ 0 & -C(q,\dot{q})\end{bmatrix}\eta + \begin{bmatrix} 0 \\ -G(q) + \tau \end{bmatrix}.$$


问:
Infinity norm of the inverse of a matrix with integer coefficients
Let $A$ be a $k$-dimensional non singualar matrix with integer coefficients. Is it true that $\|A^{-1}\|_\infty \leq 1$? How can I show that? Could you give me a counterexample?It is clear that $\|A^{-1}\|_{\infty}=\frac{1}{\min\{\|Ax\|_{\infty}:\|x\|_{\infty}=1\}}$. My idea is to show that the minimum is obtained on an integer point so the denominator is bigger than $1$. Is mi idea right?

Thank you very much!

答:
It is not true.

Note that inf-norm on matrices is the max absolute sum of per row.

Thus

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

is a counter example.


答:
I presume that by $\|\cdot\|_\infty$ for matrices you mean the matrix norm induced by the $\|\cdot \|_\infty$ norm on vectors.  Then you are right that
$$ \|A^{-1}\|_\infty = \frac{1}{\min \{ \|Ax\|_\infty \; : \; \|x\|_\infty = 1\}}$$
But there is no reason to think that the minimum is obtained for an integer point, and
it is not true that $\|A^{-1}\|_\infty \le 1$.  For example, consider
$$ A = \pmatrix{3 & -4\cr 2 & -3\cr}$$
which has $A^{-1} = A$.

Title: Infinity norm of the inverse of a matrix with integer coefficients

Body:
Let $A$ be a $k$-dimensional non singualar matrix with integer coefficients. Is it true that $\|A^{-1}\|_\infty \leq 1$? How can I show that? Could you give me a counterexample?It is clear that $\|A^{-1}\|_{\infty}=\frac{1}{\min\{\|Ax\|_{\infty}:\|x\|_{\infty}=1\}}$. My idea is to show that the minimum is obtained on an integer point so the denominator is bigger than $1$. Is mi idea right?

Thank you very much!


问:
help to find SOR optimal parameter $w$
Please let following linear system as $Ax=b$:
$$\begin{array}{l}
6a{x_1} + {x_2} + {x_3} = 1\\
{x_1} - 3a{x_2} + 4{x_3} = 2\\
{x_1} + {x_2} - 2a{x_3} = 3
\end{array}$$
Help me to prove that the optimal value $w$ in SOR method is $\dfrac{6a}{1+6a}$.</p>

答: The SOR method is convergent if eigenvalues are positive. Therefore a should be greater than o.5 or smaller than -5/6. The optimal parameter w=2/(1+sqrt(lambda(min)*lambda(max))). See S.M.Grzegorski, On optimal parameter not only for the SOR method, in Applied and Computational Mathematics, November 2019</p>

Title: help to find SOR optimal parameter $w$


问:
Why does the graph look like that?
I am studying about Specific Energy in open channel hydraulic from a book.

The equation for that is: $E_s=y+\cfrac{\alpha q^2}{2gy^2}$ where $q$ and $\alpha$ are constant. therefore we have:

$$\left(E_s-y\right)y^2=\cfrac{\alpha q^2}{2g}=constant$$
Or:
$$\left(E_s-y\right)=constant/y^2$$
If we only consider positive solutions of $y$
Then according to the book the graph of that is :

But I don't understand why the graph is look like that. for example why there is it has two asymptotes and why for every $E_s$ we have two values of $y$ corresponding to it?

答:
$$E_s=\frac{k+y^3}{y^2}$$
$$E_s=y+\frac{k}{y^2}$$
As $y\to 0^+$ we have $E_s\to +\infty$, horizontal asymptote $y=0$.

As $y\to+\infty$ we have $E_s\to y$, slant asymptote $y=E_s$.

Furthermore $y=E_s-\frac{k}{y^2}$ ($y>0$) shows that the graph is always under the first quadrant bisector $y=E_s$.

Title: Why does the graph look like that?


问:
Is there a difference between 'inconsistent', 'contrary', and 'contradictory'
Is there a difference between 'inconsistent' 'contrary' and 'contradictory'? As far as I understand, two statements are inconsistent when they can not both be true; two statements are contradictory when they can not both bear the same truth value.

Does 'contrary' have a unique meaning?

答:
A contradiction is an pair of statements of the form $p$ and $\lnot p$.

A set of statements is contradictory if there is a pair of statements in it that form a contradiction.

A set of statements is inconsistent if you can derive a contradiction from them.

We will sometimes use the terms "contradiction" and "contradictory" more loosely to mean "obviously inconsistent."

"Contrary" is used informally to mean the negation of a statement. I don't think I've ever seen a set of statements called "contrary," for example, just because there is a contradiction. Perhaps there is a formal use of the term that I am missing.


问:
Geometric multiplicity for non zero eigen values of matrices $AB$ and $BA$.
As lot of information is given in this site about eigen values of $AB$ and $BA$ for square matrices $A$ and $B$. As characteristics polynomial of $AB$  and $BA$ are same so both have same set of eigen values with multiplicity. Now I want to know about geoemetric multiplicity and as one of $AB$ and $BA$ may become zero and other not even diagonalizable so i can conclude that geometric multiplicity of eigen value $0$ may not equal. Now what about geometric multiplicity of common non zero eigen values? Are they will be same? i.e. if $a\neq 0$ then can we say $$Geo.Mult_a(AB) =Geo.Mult_a(BA)? $$ please explain or give counter example. Thanks.

答: It's true. Let $x_1,x_2,\ldots,x_k$ be a basis of the eigenspace of $AB$ corresponding to a nonzero eigenvalue $\lambda$. Then $Bx_1,Bx_2,\ldots,Bx_k$ are linearly independent, for, if $\sum_ic_iBx_i=0$, then $\lambda\sum_ic_ix_i=A(\sum_ic_iBx_i)=0$ and hence all $c_i$s are zero. However, as $BA(Bx_i)=B(ABx_i)=\lambda Bx_i$, each $Bx_i$ is an eigenvector of $AB$ corresponding to the eigenvalue $\lambda$. Therefore the geometric multiplicity of $\lambda$ in $BA$ is greater than or equal to the geometric multiplicity of $\lambda$ in $AB$. The reverse inequality is also true if we interchange the roles of $A$ and $B$ in the above. Therefore the geometric multiplicities of $\lambda$ in $AB$ and $BA$ are the same.


问:
The projection $\mathbb{A}^n\ni V(I)\to \mathbb{A}^{n-1}$ is closed
During a lecture in algebraic geometry the following fact was mentioned without proof

If we consider an affine variety $X\subset \mathbb{A}^n$ and the map $f_n:\mathbb{A}^n\to \mathbb{A}^{n-1}$ $f(x_1,\dots,x_n)=(x_1,\dots, x_{n-1})$ then if $(0:0:\dots:1)$ is not in the projective closure of X then $f(X)$ is closed in $\mathbb{A}^{n-1}$.

The only suggestion was to look at the (to be proven to be) commutative diagram (where $\varphi_n(x_1,\dots x_n)=(1:x_1,\dots, x_n)$ and $\overline{f_n}(x_0:\dots:x_n)=(x_0:\dots :x_{n-1}$))

$$ \begin{matrix}
X& \rightarrow_{\varphi_n}& \overline{X}\\
\downarrow_{f_n} & &  \downarrow_{\overline{f}_n}\\
\mathbb{A}^{n-1} &\rightarrow_{\varphi_{n-1}}& \mathbb{P}^{n-1}
\end{matrix}$$

However I don't see how this makes thing easier, it looks to me that removing the last coordinate from an affine variety is as hard as doing it from a projective one (id est quite hard given that all my attempts failed), on the other hand proving this is a commutative diagram is quite easy.

Am I missing something obvious?

答:
This looks a lot like a homework or an exercise you have to solve, so I don't think is very usueful to spoon feed you the solution.

What I can suggest is to focus on the relationship between the two vertical functions and aftert you got that look at what does $\overline{f}_n$ to closed sets.


问:
Universal Generalisation ($\forall$ - I)
With this deduction rule, in the premise of the rule: the term to be substituted for a variable must be arbitrary (refer to an arbitrary d $\in$ D).

What constitutes arbitrary and not arbitrary?

$ P(a) \quad\quad Premise$

$ \forall x P(x) \quad (1), \forall-I: a/x$

Would term john not be considered arbitrary and thus line 2 be incorrect?

$ P(john) \quad Premise$

$ \forall x P(x) \quad (1), \forall-I: john/x$

答:
The $(\forall \text I)$ rule is:

if $\Gamma \vdash \varphi[x/a]$, then  $\Gamma \vdash \forall x \varphi$, provided that parameter $a$a is “fresh” in the sense of having no other occurrences in $\Gamma , \varphi$

The proviso is consistent with the intuitive meaning of the rule: if $\varphi$ holds of an object $a$ whatever, then it holds of every object.

The proviso is needed in order to avoid the fallacy: John is  a Philosopher, therefore everything is a Philospher.

In your wrong proof above, you have committed exactly these fallacy: the parameter $a$ [in your case: John] must not occur in $\Gamma$. In your case $\Gamma = \{ P(\text {John}) \}$.

In conclusion, the issue is: how can you prove $\vdash P(\text {John})$?

Example: consider the first-order language of arithmetic with individual constants $0$ and $1$ and let $\mathsf {PA}$ the collection of first-order Peano axiom.

We have: $\mathsf {PA} \vdash (0 \ne 1)$,

Now, applying $(\forall \text I)$ to it, using $0$ as $\text {John}$, we conclude with:  $\mathsf {PA} \vdash \forall x (x \ne 1)$.

Where is the mistake ?


问:
Generating functions for combinatorics
I have no formal education in generating functions, but, based on another question, I have seen that they can be powerful for combinatorics.

Are there a few general principles for using generating functions to construct counting arguments (e.g. counting the number of subsets or number of total orderings) that can be remixed to count different quantities?

If the answer is not amenable to this forum, where can I learn more?

I've been trying to read Concrete Mathematics (Graham, Knuth, Patashnik), but the topic spans several chapters and over 100 pages. At the moment, I just have patience to learn the gist of the combinatorics applications of generating functions. Maybe later I'll go back and get an extensive education from GKP.

答:
Flajolet and Sedgewick's Analytic Combinatorics is dense, but equips you with amazing tools to construct, manipulate, and extract information from generating functions. I also found the relevant chapters of Stanley's Enumerative Combinatorics (both volumes) extremely helpful.

Generating functions happen to be a favorite topic of mine, so I've written several posts on the subject on my blog. You might want to start at this post, which builds to an explanation of the Polya enumeration theorem, and a few years ago I wrote up these notes which might be useful.

There are in fact a few general principles, but I have enormous difficulty concisely describing them.

Title: Generating functions for combinatorics


问:
Finding the indefinite integral $\int \frac{3x+2}{(6x^2+8x)^7}\,\mathrm dx$
I'm not too familiar with how to solve this. Could anyone present a step by step guide on how to get the answer?

$$\int \dfrac{3x+2}{(6x^2+8x)^7}\,\mathrm dx$$

答:
Let $u = 6x^{2} + 8x$.  Then $du = (12x + 8) dx \rightarrow du = 4(3x + 2) dx$.  This is equivalent to $\frac{du}{4} = (3x + 2) dx$

Make the substitution for the integral, so we have:

$$\int \frac{((3x + 2) dx)}{u^{7}}$$
$$= \int \frac{1/4 du}{u^7}$$
$$= \frac{1}{4} \int \frac{du}{u^7}$$
$$= \frac{1}{4} \int u^{-7} du$$

Thus, by power rule, we have:

$$\frac{1}{4} \frac{u^{-7 + 1}}{-7 + 1} + c$$
$$= \frac{1}{4} \frac{u^{-6}}{-6} + c$$
$$= \frac{-1}{24} u^{-6} + c$$
$$= \frac{-1}{24} (6x^{2} + 8x)^{-6} + c \text{ where c is an arbitrary constant}$$

In terms of positive exponent, we obtain:

$$\frac{-1}{24(6x^{2} + 8x)^{6}} + c$$


答:
Hint:

If $\lambda \ne 0$ is an eigenvalue of $AB$ and $BA$, check that the linear maps $$\ker(AB-\lambda I) \to \ker (BA - \lambda I), \quad x \mapsto Bx$$
$$\ker(BA-\lambda I) \to \ker (AB - \lambda I), \quad x \mapsto Ax$$
are injective. It follows $\dim \ker (AB - \lambda I) = \dim \ker (BA - \lambda I)$.

Title: Geometric multiplicity for non zero eigen values of matrices $AB$ and $BA$.

Body: As lot of information is given in this site about eigen values of $AB$ and $BA$ for square matrices $A$ and $B$. As characteristics polynomial of $AB$  and $BA$ are same so both have same set of eigen values with multiplicity. Now I want to know about geoemetric multiplicity and as one of $AB$ and $BA$ may become zero and other not even diagonalizable so i can conclude that geometric multiplicity of eigen value $0$ may not equal. Now what about geometric multiplicity of common non zero eigen values? Are they will be same? i.e. if $a\neq 0$ then can we say $$Geo.Mult_a(AB) =Geo.Mult_a(BA)? $$ please explain or give counter example. Thanks.


问:
Let $K$ be a field. Show that $K[t]$ is finitely generated as a $K[t^2,t^3]$-module and find a generating set
Using the variables $t^2,t^3$ I think that $K[t^2,t^3]$ contains all polynomials of $K[t]$ except those of degree $1$ since $t$ cannot be made from products of $t^2,t^3$ so I think that the module is generated by the set $\{1,t\}$ since given any $f(t)=a_nt^n+...+a_1t+a_0\in K[t]$ we can write this as $(a_nt^n+...+a_2t^2+a_0)(1)+a_1(t)$ where $a_nt^n+...+a_2t^2+a_o$ and $a_1$ are in $K[t^2,t^3]$.Is this correct and how I can I write it more rigorously? Do I need to use the fact that $K$ is a field and hence $K[t]$ is a PID?

答: Yes, $K[t]$ as a $K[t^2, t^3]$-module can be generated by the two elements $1$ and $t$, just as you say.


问:
Find the number of ordered triples of subsets of $S$

Let $S = \{1, 2, 3,...., n\}$. Find the number of ordered triples $(A,B,C)$ of subsets of $S$ such that $A \subseteq B$ and $A \cup B \cup C = S$.

I'm not sure how to approach this question and saw a similar question where the solution involved something along the lines of dividing $S$ into three parts $A_1, A_2, A_3$ such that $A = A_1$, $B = A_1 \cup A_2$, $C = A_3$ and $S = A_1 \cup A_2 \cup A_3$ but I didn't really understand it. Please help.

答:
HINT: Notice that since $A\subseteq B$, saying that $A\cup B\cup C=S$ really just amounts to saying that $B\cup C=S$. In other words, $C$ has to contain all of $S\setminus B$, and in addition it can contain any subset of $B$. And $A$, of course, can be any subset of $B$.

If $|B|=k$, how many possibilities are there for $A$?

If $|B|=k$, how many possibilities are there for $C$?

How many $B\subseteq S$ are there with $|B|=k$?

Now put the those three pieces together to get an answer expressed as a summation. This summation has a nice closed form. If you don’t see what it is, calculate its value for $n=1,2$, and $3$; from those you should be able to guess the closed form, and that form may suggest the theorem that you can use to get it directly from the summation.


问:
Solving equation $a_n=1+\frac12a_{n-1}+\frac12a_{n+1}$
Let's consider sequence following for $n=0...,T$ for which $a_0=a_T=0$. I want to solve equation $$a_n=1+\frac12a_{n+1}+\frac12a_{n-1}$$

My work so far

$$a_1=1+\frac12a_2 \Leftrightarrow a_2=2a_1-2$$
$$a_2=1+\frac12a_3+\frac12a_1 \Leftrightarrow2a_1-2=1+\frac12a_3 \Leftrightarrow a_3=3a_1-6$$
Also I calculated $a_4$ to get $a_4=4a_1-12$. So my observation is that for any $n \in \{1,...,T-1\}$
we have $$a_n=na_1+n(n-1)$$

And here I stacked, I'm not sure what to do next. Could you please give me a hand solving this equation ?

答:
Consider the generating function $$f(x)=\sum_{n\ge0}a_nx^n,\qquad|x|<1.$$
We may write
$$2a_{n+1}x^{n+2}=2x^{n+2}+a_{n+2}x^{n+2}+a_{n}x^{n+2},$$
and sum over $n\ge 0$:
$$2x\sum_{n\ge0}a_{n+1}x^{n+1}=2x^2\sum_{n\ge0}x^n+\sum_{n\ge0}a_{n+2}x^{n+2}+x^2\sum_{n\ge0}a_nx^n,$$
which is
$$2x(f(x)-a_0)=\frac{2x^2}{1-x}+f(x)-a_0-a_1x+x^2f(x).$$
Using $a_0=0$, and treating $a_1$ as a constant,
$$(x^2-2x+1)f(x)=\frac{(a_1+2)x^2-a_1x}{x-1},$$
so that
$$f(x)=\frac{(a_1+2)x^2-a_1x}{(x-1)^3}.$$
Then let the sequences $p_n$ and $q_n$ be defined by
$$\begin{align}
p(x)=\frac{x^2}{(x-1)^3}&=\sum_{n\ge0}p_nx^n\\
q(x)=\frac{x}{(x-1)^3}&=\sum_{n\ge0}q_nx^n,
\end{align}$$
so that
$$f(x)=(a_1+2)p(x)-a_1q(x)=\sum_{n\ge0}((a_1+2)p_n-a_1q_n)x^n,$$
giving $a_n=(a_1+2)p_n-a_1q_n$.
Differentiating the series $\frac{1}{1-x}=\sum_{n\ge0}x^n$ three times gives
$$\frac{1}{(x-1)^3}=\sum_{n\ge0}x^n\left(-\frac{1}{2}(n+1)(n+2)\right),$$
so that
$$\begin{align}
p(x)=\frac{x^2}{(x-1)^3}&=\sum_{n\ge0}x^n\left(-\frac{1}{2}n(n-1)\right)\quad &\Rightarrow p_n=-\frac12n(n-1)\\
q(x)=\frac{x}{(x-1)^3}&=\sum_{n\ge0}x^n\left(-\frac{1}{2}n(n+1)\right)\quad &\Rightarrow q_n=-\frac12n(n+1).
\end{align}$$
Therefore
$$a_n=(a_1+2)p_n-a_1q_n=n(a_1+1-n).$$
To find $a_1$, we use $a_T=0$. Specifically we have
$$T(a_1+1-T)=0\Rightarrow a_1=T-1.$$

Thus we have $$a_n=n(T-n).$$


问:
How to find the time needed to fill a reservoir when one faucet changes its condition?
The problem is as follows:

Two faucets $A$ and $B$ together fill an empty tank in $30$ hours. If
tap $B$ were to drain, it would take $60$ hours to fill the tank. How
long would tap $A$ require to fill the tank when the latter is empty?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{44 hours}\\
2.&\textrm{50 hours}\\
3.&\textrm{40 hours}\\
4.&\textrm{42 hours}\\
4.&\textrm{54 hours}\\
\end{array}$

I attempted to solve this problem but I came stuck as it states that tap $B$ becomes a drain, how should this be translated as an equation?. Can someone help me with this?.

An answer which would help me the most is an algebraic approach. So far what I came up with was:

Adding speeds: Assuming the volume of the tank is $a$:

$\frac{a}{t_{1}}+\frac{a}{t_{2}}=a\frac{t_{1}+t_{2}}{t_{1}t_{2}}$

$\frac{1}{t_{1}}+\frac{1}{t_{2}}=\frac{t_{1}+t_{2}}{t_{1}t_{2}}$

$\frac{t_{1}+t_{2}}{t_{1}t_{2}}=\frac{30}{t_{1}t_{2}}$

But as mentioned, what it confuses me the most is how to assess the part where it mentions that faucet B is drain. Does this means such tap will have a negative speed? Can such thing happen?. Can someone help me with this part please?. Please try to use a step-by-step approach so I can understand. By the way, I'm aware of the existence of this question but it doesn't address my doubt regarding on how to understand negative velocities I don't understand why can arise a negative velocity?. Does this means the water is flowing in another direction or what?. Therefore I need this doubt to be resolved.

答:
If tap $A$ fills tank in $a$ hours and tap $B$ fills in $b$ hours individually,

then equating part of tank filled in an hour if both fill together,

$\displaystyle \frac{1}{a} + \frac{1}{b} = \frac{1}{30}$

When $B$ becomes a drain and assuming same rate of outflow,

$\displaystyle \frac{1}{a} - \frac{1}{b} = \frac{1}{60}$

Adding both, $\displaystyle \frac{2}{a} = \frac{3}{60} \implies a = 40 \,$ hours


答:
The phrasing is peculiar. "If tap B were to drain, it would take 60 hours to fill the tank."  Normally "draining" empties a tank!

I am going to interpret this as saying "Two faucets A and B together fill an empty tank in 30 hours. If tap B only were used, it would take 60 hours to fill the tank."

So tap B fills at the rate of 1 tank per 60 hours or 1/60 "tank per hours".  Call the number of hours it would take to fill the tank with faucet A only "x".  With A and B both working, it would take 1/x+ 1/60= 1/30.    Multiply both sides by 60x: 60+ x= 2x so A= 60.  It would take 60 hours using A  only.

Another way of looking at this is that B alone could fill half the tank in the 30 hours so A has to fill the other half in those same 30 hours so A is filling at the same speed as B.

Title: How to find the time needed to fill a reservoir when one faucet changes its condition?

Body:
The problem is as follows:

Two faucets $A$ and $B$ together fill an empty tank in $30$ hours. If
tap $B$ were to drain, it would take $60$ hours to fill the tank. How
long would tap $A$ require to fill the tank when the latter is empty?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{44 hours}\\
2.&\textrm{50 hours}\\
3.&\textrm{40 hours}\\
4.&\textrm{42 hours}\\
4.&\textrm{54 hours}\\
\end{array}$

I attempted to solve this problem but I came stuck as it states that tap $B$ becomes a drain, how should this be translated as an equation?. Can someone help me with this?.

An answer which would help me the most is an algebraic approach. So far what I came up with was:

Adding speeds: Assuming the volume of the tank is $a$:

$\frac{a}{t_{1}}+\frac{a}{t_{2}}=a\frac{t_{1}+t_{2}}{t_{1}t_{2}}$

$\frac{1}{t_{1}}+\frac{1}{t_{2}}=\frac{t_{1}+t_{2}}{t_{1}t_{2}}$

$\frac{t_{1}+t_{2}}{t_{1}t_{2}}=\frac{30}{t_{1}t_{2}}$

But as mentioned, what it confuses me the most is how to assess the part where it mentions that faucet B is drain. Does this means such tap will have a negative speed? Can such thing happen?. Can someone help me with this part please?. Please try to use a step-by-step approach so I can understand. By the way, I'm aware of the existence of this question but it doesn't address my doubt regarding on how to understand negative velocities I don't understand why can arise a negative velocity?. Does this means the water is flowing in another direction or what?. Therefore I need this doubt to be resolved.


问:
Show that $\int_0^t\!\!\left(t^2-x^2\right)^n\mathbb{d}x=\frac{\sqrt{\pi}}{2}t^{2n+1}\frac{\Gamma(n+1)}{\Gamma(n+\frac{3}{2})}$
The question asks to prove the identity:
$$\int_0^t\!\!\left(t^2-x^2\right)^n\mathbb{d}x=\frac{\sqrt{\pi}}{2}t^{2n+1}\frac{\Gamma(n+1)}{\Gamma(n+\frac{3}{2})}$$
where $n\in\mathbb{Z}$

I have no idea where to start. Thank you for your help!

答:
The key formula in this kind of problems is the beta function evaluation
$$B(p,q)=\int_0^1s^{p-1}(1-s)^{q-1}ds=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}.$$
To guess what kind of change of variables may reduce your integral to this one, consider zeros of the integrand and compare with the bounds of integration. In your case the change of variables is very easy to guess: setting $s=x^2/t^2$, we obtain
\begin{align}
\int_0^t\left(t^2-x^2\right)^n dx=\frac12 t^{2n+1}\int_0^1 s^{-\frac12}(1-s)^nds=
\frac12 t^{2n+1}B\left(\frac12,n+1\right)=\\
=\frac12 t^{2n+1}\frac{\Gamma\left(\frac12\right)\Gamma(n+1)}{\Gamma\left(n+\frac32\right)}=\frac{\sqrt{\pi}}{2} t^{2n+1}\frac{\Gamma(n+1)}{\Gamma\left(n+\frac32\right)},
\end{align}
where it is also used that $\Gamma\left(\frac12\right)=\sqrt{\pi}$.</p>


问:
What can be Jordan forms of matrix with charcterstic polynomial $(x-3)^2 (x-2)^4$ and minimal polynomial $(x-3)(x-2)^2$
The following question was asked in my linear algebra quiz and I was unable to think which result I should use to solve it.

Question: Let A be a $6 \times 6$ matrix over $\mathbb{R}$ with chacterstic polynomial =$(x-3)^2 (x-2)^4$ and minimal polynomial = $(x-3)(x-2)^2$ .Then Jordan canonical form of A can be? I have to find all the matrix which are possible.

Algebraic multiplicity of 2 and 3 are 4,2 respectively and Geometric multiplicity of 2 and 3 are 2 and 1 respectively.

But i don't know which result should be used to move foreward. I calculated these in case they are useful.

Can you please shed some light on this?

答:
If a factor $(x-\lambda)^k$ appears in the characteristic polynomial, then $k$ is total size of all Jordan $\lambda$-blocks.

If a factor $(x-\lambda)^k$ appears in the minimal polynomial, then $k$ is the size of the largest Jordan $\lambda$-block.

In our case we get:

size of the largest $3$-block is $1$ and the total size is $2$ so there are two $3$-blocks of size $1$

size of the largest $2$-block is $2$ and the total size is $4$ we can have two $2$-blocks of size $2$ or one $2$-block of size $2$ and two $2$-blocks of size $1$

$$\begin{bmatrix}
3 & & & & & \\
& 3 & & & & \\
& & 2 & 1 & & \\
& & & 2 &  & \\
& & & & 2 & 1 \\
& & & & & 2
\end{bmatrix}\quad\text{ or } \quad
\begin{bmatrix}
3 & & & & & \\
& 3 & & & & \\
& & 2 & 1 & & \\
& & & 2 &  & \\
& & & & 2 &  \\
& & & & & 2
\end{bmatrix}$$


问:
Rearrange to get subject
I am looking the rearrange the equation to get h2 as the subject.

$$8-3.5 = (0.028)(h2 - 2724.9)$$

Now i have been dividing h2 both sides to get

$$\frac{4.5}{h2} = (0.028(h2 - 2724.9)$$

But this doesnt seem right. Can anyone help please

Title: Rearrange to get subject

Body:
$$8-3.5 = (0.028)(h2 - 2724.9)$$
$$8-3.5 = 0.028h2 - 76.2972$$
$$8-3.5+76.2972=0.028h2$$
$$80.7972=0.028h2$$
$$h2=\frac{80.7972}{0.028}$$
$$h2=2885.61$$


问:
Joint density of $\chi^2$ distribution and t distribution from a normal distribution N(0,$\sigma^2$)
Let $X_1$ and $X_2$ be independent N(0,$\sigma^2$) random variables. Find the joint $Y_1 = X_1^2 + X_2^2$ and $Y_2 = X_1/\sqrt{X_1^2 + X_2^2}$. Show that they are independent.

Attempt:
Step 1: Find $x_1 = g(y_1,y_2) = y_2\sqrt{y_1}$ and $x_2 = h(y_1,y_2) = \sqrt{y_1(1-y_2^2)}$

Step 2: Apply the Jacobian Transformation to get $|J| = \frac{1}{2\sqrt{1-y_2^2}}$

Step 3: Plug in into $f_{X_1,X_2}(x_1,x_2)$ to get $f(y_1, y_2) = \frac{e^{\frac{y_1}{2\sigma^2}}}{4\pi\sigma^2\sqrt{1-y_2^2}}$

But I am not sure whether my calculation for Jacobian is correct or not, since getting the $g(y_1,y_2)$ and $h(y_1,y_2)$ involving the sqrt calculation. I don't know how to deal with it there.

答:
You are quite close. Fixing the missing minus sign in the exponent, we have
$$f(y_1, y_2) = \frac{1}{2 \sigma^2} e^{-\frac{y_1}{2\sigma^2}} \cdot \frac{1}{2\pi \sqrt{1-y_2^2}}$$
where $y_2 \in (-1, 1)$ and $y_1 > 0$.
Because this joint density can be written as $g(y_1) h(y_2)$, you have shown $Y_1$ and $Y_2$ are independent. You can check that the first factor is the distribution of $X_1^2+X_2^2$ by referring to the density of a chi-squared distribution of degree $2$ (scaled by $\sigma$). I'm not sure what the name of the distribution of $Y_2$ is, but it can't be a $t$-distribution since $-1 \le Y_2 \le 1$.


问:
The Probability of Getting X More Heads Than Tails in Y Tosses of a Coin
You toss a coin 30 times. A head is a win, a tails is a loss. What's the probability of getting two heads ahead of the cumulative tails in these 30 tosses. You stop tossing the coin once you have achieved two heads more than tails.

For example:

tails, heads, heads, heads

is a win because you have two more heads than tails.

tails, tails tails, tails, heads, tails, heads, heads, tails, heads, tails, heads, heads, heads, heads, heads

is a win because you have 7 tails, and 9 heads.

What is the probability of having x heads ahead of the cumulative tails in y number of tosses?

Hopefully that makes sense and thanks in advance, and please provide a formula for how this is calculated.

答:
Really interesting question! Thanks for posting. I don't think you have received an answer yet so I will take a bit of time to talk you through it all. Let us use a little formal notation and define some sets that will help us solve this problem. We are going to define and examine 3:

$\bf{\Omega:}$

Let us define  $\Omega$ as our sample space of $\textbf{all}$ outcomes. I.e $\Omega$ is the set of all different sequences of $30$ coin flips. Clearly the size of $\Omega$ is $2^{30} $

$\bf{\mathcal{S}}: $

We define $\mathcal{S} \subset \Omega$ As the set of $\bf{all} $ successful sequences. i.e $HHHHHHHTHHHHTHHHHHHHHHHHHHHHTTT \in \mathcal{S} $ Clearly the probability we are after is $\frac{|\mathcal{S}|}{|\Omega|}$ and as we already have $|\Omega|$ all we need is $|\mathcal{S}|$ and we are done. Eyes on the prize.

$\bf{\mathcal{U}:}$

Now call $\mathcal{U}$ this set of unique ways to reach the position where #H = #T $+2$ for the first time. Elements of $\mathcal{U}$ are a sort of truncation of element from $\mathcal{S}$

For example: $HH , THHH, $and the example you gave are all some elements of $\mathcal{U} $

Notice that $\mathcal{U} \not \subset \Omega $ as $\mathcal{U} $ is $\textbf{not}$ a set of $\textbf{full}$ sequences of 30 flips, it is just the opening moves to reach this net $+2$ heads. Elements of $\mathcal{U} $ will correspond somewhat to elements of $\mathcal{F}$ eventually. For example:

$ \mathcal{U} \ni HH \not \in \Omega  $ but $HHHHHHHTHHHHTHHHHHHHHHHHHHHHTTT \in \mathcal{S} \subset \Omega$ Every element of $\mathcal{S}$ has one unique corresponding element in $\mathcal{U}$

We are going to solve this problem in 5 distinct steps.

Firstly we will construct a partition of $\Omega$ into $3$ pieces

We will show that $2$ of these pieces belong to $\mathcal{S}$

We will then construct more elements of $\mathcal{S}$ that are not in our initial partitions of $\Omega$

We will then show that $\mathcal{S}$ contains only elements of these three sets, the two partitions of $\Omega$ and the final set we constructed in step 3

The sum of the size of these $3$ pieces make up the size of $\mathcal{S}$ (what we are after)

$\textbf{Step 1:}$

Call $\Omega_{16},  \Omega_{>16}  $ and $\Omega_{<16}$ respectively the elements of $\Omega$ that contain 16, more than 16 and less than 16 heads. This is clearly a mutually disjoint partition.

Examples: 16 heads followed by 14 tails is in $\Omega_{16}.$ All heads and all tails a respectively in  $\Omega_{>16}$ and $\Omega_{<16}$

$\textbf{Step 2:}$

Clearly elements of $\Omega_{16}$ and $  \Omega_{>16}  $ are all elements of $\mathcal{S}$ as these are sequences with at least 16 heads and thus at most 14 tails and as such there exists (at least one) point in which we have 2 more heads than tails. Hence  $\Omega_{16} \cup \Omega_{>16} \subseteq \mathcal{S}  $

$\textbf{Step 3:}$

What I intended to do now is, for $\textbf{every single element of} $ $\bf{\Omega_{>16}}$,  generate another element of $\mathcal{S} $ that is in $\Omega_{<16}$

For each element $x \in \Omega_{>16} \subset \mathcal{S} $ we shall call $u$ the corresponding element in $\mathcal{U}$ For example if $ x = HTHTHTHHHHHHHTTTHHHHHHHHHHHHHHH$ then $u = HTHTHTHH$ or if $ x = HHHHHHHHHHHHHHHHHHHHHHHHHHHHH$ then $u = HH$

Now call $l$ the "leftover" of $x$ when we take away $u$

Examples: if  $ x = HTHTHTHHHHHHHTTTHHHHHHHHHHHHHHH$ then $u = HTHTHTHH$ and $l = HHHHHTTTHHHHHHHHHHHHHHH$

Now define a function "flipped", $f:$ Sequences $\to$ Sequences by flipping each tail to a head and each head to a tail.

Example: $f(HTHTTT) = THTHHH $

Now finally for each $x \in \Omega_{>16} $ define $g(x): \mathcal{S} \to \mathcal{S} $ by $g(x) := u +f(l) $

Example: If $x = TTHHHHHHHHHHHTHTHHHHHHHHHHHHHH $ then $g(x) = TTHHHH + f(HHHHHHHTHTHHHHHHHHHHHHHH) =  TTHHHH + TTTTTTTHTHTTTTTTTTTTTTTT = TTHHHHTTTTTTTHTHTTTTTTTTTTTTTT$

Properties of g:

If $x \in \mathcal{S} \implies g(x) \in \mathcal{S} $ as g does not change the inital elements $u$ that result in a net $+2$ heads

Notice that $g$ is its own inverse as g never changes the beginning terms, it simply flips the last and the "flipping" function is clearly its own inverse. Hence $g(g(x)) = x $

g is a bijection.

$g : \Omega_{>16} \to \Omega_{< 16}$ or $g : \Omega_{<16} \to \Omega_{> 16}$

As if $x$ has $h$ heads and $t$ tails with $h_u$ heads in $u$ and $h_u - 2$ tails in $u$  with $h$ > 16 (i.e it is in $\Omega_{>16} $ ) then $g(x) $ has $h_u$ heads in $u$ and $t-(h^u-2) = t + 2 - h^u$ heads in $l$ Hence has $t + 2$ heads in total. And as $h = 30 - t > 16 $ we have $t +2 < 16 $ Hence $g(x) \in  \Omega_{<16} \iff x \in \Omega_{>16} $ and visa versa

Define $ \Omega_{<16} \supset \Omega_{G} := g(\Omega_{>16}) \subset \mathcal{S}  $ as everything in $ \Omega_{>16} $ is in $\mathcal{S} $ and property 1) of g shows that it will keep everything in $\mathcal{S} $

$\textbf{Step 4:}$

We now have $3$ disjoint sets: $\Omega_{16} , \Omega_{>16} , \Omega_{G} $

Define $\mathcal{B} := \Omega_{16} \cup \Omega_{>16} \cup \Omega_{G}$

We will prove $\mathcal{B} = \mathcal{S}$

We have already show that everything in $\Omega_{16} , \Omega_{>16} , \Omega_{G} $ is in $\mathcal{S}$ Hence $\mathcal{B} \subseteq \mathcal{S}$ Thus all we wish to show for equality is : $\mathcal{S} \subseteq \mathcal{B}$

Finally assume $ x \in  \mathcal{S}$ we have $3$ options as to which of $\Omega_{16} , \Omega_{>16} , \Omega_{<16} $ $x$ belongs to. If $x$ is in one of the first two then great! As this would imply $x \in \mathcal{B} $

If  $ x \in \Omega_{<16} \cap \mathcal{S} \implies g(x) \in \Omega_{>16} \implies g(g(x)) = x \in g(\Omega_{>16}) = \Omega_{G} \implies x \in \mathcal{B} $

Hence $\mathcal{S} \subseteq \mathcal{B}$ and as $\mathcal{B} \subseteq \mathcal{S}$ we have $\mathcal{B} = \mathcal{S}$

$\textbf{Step 5:}$

$|\mathcal{S}| = |\mathcal{B}| = |\Omega_{16} \cup \Omega_{>16} \cup \Omega_{G}| = |\Omega_{16}|+ |\Omega_{>16}|+|\Omega_{G}| $ (as they are disjoint sets.) $ = {30 \choose 16} + \sum\limits_{i=17}^{30}{30 \choose i} + |\Omega_G|$

However $\Omega_G$ is the image of an bijective map from a finite set and as such its size is equal to the size of its domain and as thus$ |\Omega_{>16}|=|\Omega_{G}| = \sum\limits_{i=17}^{30}{30 \choose i} $

And thus : $|\mathcal{S}| = {30 \choose 16} + 2\sum\limits_{i=17}^{30}{30 \choose i} $

And hence probability to have won this game, $\frac{|\mathcal{S}|}{|\Omega|} $ = $\frac{{30 \choose 16} + 2\sum\limits_{i=17}^{30}{30 \choose i}}{2^{30}} \approx 0.720100131817 $

I hope this helped.

Oskar


问:
Show that the action of $\mathfrak{so}(3)$ on $\mathbb R^3$, $M_a(x)=a\wedge x$, satisfies $M_{ga}=g M_a g^{-1}$
In the notes on the Lie groups $SU(2)$ and $SO(3)$ found here, the authors state (page 75, before Proposition 2.2) the following.

Consider the $\mathbb R$-linear map $\mathbb R^3\ni a\mapsto M_a\in\mathfrak{so}(3)$ defined by $e_k\mapsto \eta_k$, where $e_k\in\mathbb R^3$ are the elements of the canonical basis of $\mathbb R^3$, $\eta_k\in\mathfrak{so}(3)$ are the corresponding generators: $\exp(i\eta_k)=\mathrm{Rot}(e_k,t)$, and $\mathrm{Rot}(a,t)$ denotes the rotation by angle $t$ around the axis $a\in\mathbb R^3$.

We can then write $M_a(x)=a\wedge x$. This is easy to see, for example writing $\eta_i=\varepsilon_{ijk} (e_j e_k^* - e_k e_j^*)$.
Then, they say, given $g\in SO(3)$ and $a,x\in\mathbb R^3$, we have
$$M_{ga}(x) = (ga)\wedge x = g(a\wedge (g^{-1}x)) = (g M_ag^{-1})(x). \tag1$$
What's a good way to show (1)?

By direct computation, using $x\wedge y=\sum_{ijk}\varepsilon_{ijk}e_i x_j y_k$, I get
$$[(ga)\wedge x]_i = \sum_{jk\ell}\varepsilon_{ijk} g_{j\ell}a_\ell x_k, \\
[g(a\wedge (g^{-1}x))]_i = \sum g_{ij}\varepsilon_{jk\ell} a_k g^{-1}_{\ell m}x_m.
$$
Because $g\in SO(3)$ I know that $g^{-1}_{\ell m}=g_{m\ell}$, but even with this I'm not sure how to simplify the expressions to show that they are equal.

I could probably just get the identity from geometrical intuition, observing that it must be that $g(x\wedge y)=(gx)\wedge (gy)$, but I wanted to show it explicitly algebraically here.

答:
This would be one way of proving the statement: Fixing some $r$ and $s$, the expression $$\sum_{i,j,k} \epsilon_{ijk} \cdot g_{ir}g_{js}$$
is the vector product between the $r$-th and the $s$-th column of the matrix $g$. Now, since $g \in SO(3)$, the columns of $g$ form a positively oriented orthonormal basis of $\mathbb{R}^3$, or, in other words, the vector product of two columns gives, up to a sign, the third column, i.e. if we denote the columns of $g$ by $g_1,g_2,g_3$

$$g_1 \times g_2 = - g_2 \times g_1 = g_3, \quad g_2 \times g_3 = - g_3 \times g_2  = g_1, \dots$$
and so on. Using the $\epsilon$-tensor, we can summarize this as

$$\sum_{i,j,k} \epsilon_{ijk} \cdot g_{ir}g_{js} = [g_r \times g_s]_i = \left[\sum_k \epsilon_{rsk} \cdot g_{k}\right]_i = \sum_k \epsilon_{rsk} \cdot g_{ik}. $$

Now, using the identity that you mentioned, and playing around with the indices (recall for example that $\epsilon_{ijk} = \epsilon_{kij} = \epsilon_{jki}$), this will give you the identity you want.


问:
Proof by contradiction : $n^3 −n−6 = 0$ means $m^3 − m − 6 ≠ 0$ when $m≠n$
I have to prove by contradiction that if $n$ is a positive integer such as $n^3-n-6 = 0$, so for any positive integer $m$ such as $m ≠ n$, we have $m^3-m-6 ≠ 0$

I am new to proofs, but I tried and here is what I did.

I first assume the opposite of the conclusion, so
$m^3-m-6 = 0$

$(m-2)(m^2+2m+3) = 0$

m=2

If I do the same thing for the n function and find the solution for $n^3-n-6 = 0, n=2$.

This is a contradiction since $m=n$, but we said $m ≠ n$.

I feel like im doing something wrong here and I would like someone to tell me what it is ! (Again, I am a beginner)

答: It looks like your proof would be assuming $n^{3}-n-6=0, m\neq n$ (where $m,n \in \mathbb{Z}$) showing a contradiction if we had $m^{3}-m-6=0$ as well. You have shown by factoring and the fact that $m,n \in \mathbb{Z}$, that $m, n = 2$. Hence  $m=n$, a contradiction.


问:
What is the limit of $\sin(n^2\pi/2)$ as $n$ approaches $\infty$?
I said it does not exist because $\sin(n^2\pi/2)$ oscillates back and forth as $n$ approaches infinity. Am I correct?

答:
As suggested in the comments, we can construct two subsequences with two different limits. Let $s_n:=\sin(n^2 \pi/2)$, $a_n:=s_{2n}$, $b_{n}:=s_{2n+1}$ for all $n\geq 0$. Then $a_n=0$ for all $n$, and hence $a_n\to0$, while $b_n=1$ for all $n$, and hence $b_n\to1$. Therefore, there are two convergent subsequences with different limit, and thus the limit for $s_n$ does not exist.

On the why the limit does not exist, see here.


问:
Fun with distributivity of graph intersection over graph union
In algebra or, number theory, it is fun playing with distributivity of one operation over the other; it helps in simplifying the calculations; for instance, $5\times(106)=5\times 100+5\times 6=530.$ Or, $16\times 6+ 16\times 4=16\times(6+4)=16\times 10=160.$

We know that graph intersection $\cap$ distributes over graph union $\cup$. What is a fun or, an interesting fact about the distributivity of graph intersection $\cap$ over graph union $\cup$? I can't find any geometrical significance or important usage of distributivity of graph operations.

Note that the graphs are simple and undirected here.

答:
This is a very broad answer, but here are some things I can think of.

First of all, there are two common versions of 'graph union'. One is the general union of the vertex-sets and the edge-sets, and the other is the disjoint union where the operation remains the same but the graphs are seen to be disjoint (empty intersection).

Rough example uses

There are more interesting things you can do with disjoint union. For instance, any graph can be seen as the disjoint union of its connected components and that decomposition is unique among decompositions into the disjoint union of connected graphs. In this sense, anything that applies component-wise can be seen to distribute over the disjoint union. Many things can be done component-wise, in particular, the search for any connected subgraph such as paths, cycles, trees, cliques is such an example and graph intersection is one way to 'select' subgraphs from a graph. So if the connected components $G_1, \cdots, G_k$ of a graph $G$ have some special structure which makes intersection easier, than you can use that to compute $(G_1 \cap H) \cup (G_2 \cap H) \cdots (G_k \cap H)$ faster than $G \cap H$ vertex-wise and edge-wise. Direct computational advantages cannot seem to be generalizable to any class of graphs.

Problems

Unfortunately, a big hurdle against finding very many interesting things to do with graph union and intersection is that union is not invertible (there is such a thing as a $-5$ number, but there is no $-G$ graph) and graph intersection does not even have a unit (there is no '1 graph' $G$ such that $H \cap G = H$ for all graphs $H$ (at least not nice, finite graphs). To see why these cause problems, consider the case on integers where multiplication by 10 is easy. This multiplication is considered easy because the digits of 10 are 1, the multiplicative identity, so you don't change the digits and 0, multiplicative absorbing element, so the multiplication by 0 is 0 allowing you to say that $16 \times 10 = 16 0$ the same digits, but shifted.

Other operations

Since you appear to be searching for interesting uses of some properties of graph operations more broadly, I would like to direct you toward a generalization of these graph operations in category theory as products and sums. Here, union and intersection can be seen as categorical sums and products in the category where arrows are inclusion (in other words where functions can be seen as indicating the subgraph relation), but there are other distributive operations that might resemble those on numbers more, namely, the tensor product and the edge-disjoint union.

EDIT: Categorical products don't have to distribute over categorical sums Distributivity of categorical product and sum


问:
How to find $\lim_{n\to+\infty} \frac{n^2}{7^{\sqrt{n}}}$? I know that it exists and can't find it with Stolz theorem.
How do I find $\lim_{n\to+\infty} \frac{n^2}{7^{\sqrt{n}}}$?

I thought about Stolz theorem so I got:
$$ \frac{(n+1)^2-n^2}{7^{\sqrt{n+1}}-7^{\sqrt{n}}}, $$
but it doesn't seem to lead anywhere (at least not through $e$ and $\ln$).

And I know for fact that there is some limit because:
$$\lim_{n \to +\infty} \frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)^2}{7^{\sqrt{n+1}}}}{\frac{n^2}{7^{\sqrt{n}}}} = \frac{(n+1)^2*7^{\sqrt{n}}}{7^{\sqrt{n+1}}*n^2} = \frac{(n+1)^2*7^{\sqrt{n}}}{7^{\sqrt{n}\sqrt{\frac{1}{n}+1}}*n^2} \implies \frac{(n+1)^2}{n^2}\implies 1$$

答:
This is $\lim\limits_{n\to\infty}\exp\left(2\ln n-\sqrt n\ln 7\right)=0$, because $\lim\limits_{x\to\infty}2\ln x-\sqrt x\ln 7=-\infty$.

For the record, you should be warned that Riemann rearrangement theorem implies that for all $0\le x\le y\le\infty$ there is a sequence $a_n>0$ such that $$\begin{cases}\liminf\limits_{n\to\infty}a_n=x\\ \limsup\limits_{n\to\infty}a_n=y\\ \lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1\end{cases}$$

Therefore $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ is inconsequential. Also, while the value of this limit is certainly $1$, I think you made some mistake either of algebra or in manipulation of limits; I won't inspect it any further, because I am not a tutor.


问:
Connectivity of this set: $A=\{x: \operatorname{dist}(x,F)=\epsilon\}?$
Suppose $F$ is a closed subset of $\mathbb{R}^n$ ($n>1$). How to prove that if $F$ and the complement of $F$ are connected, then so is
$A=\{x: \operatorname{dist}(x,F)=\epsilon\}?$ Suppose $\epsilon>0$.

Intuitively we can say that if $A$ is disconnected and $C$ is one of its components, then there is an open set $V$ that contains $C$ (which is closed) and the boundary $b V$ doesn't intersect $A\setminus C$. So at each point on $b V$ the distance to $F$ must be either $>\epsilon$ or $<\epsilon$ (intermediate value theorem). But at the same time, if you take a point $x\in A$ and $y\in F$ such that $\operatorname{dist}(x,y)=\epsilon$, then the line $L$ through $x$ and $y$ intersects $b V$ at (at least) two points: $z_1$ and $z_2$ and the distance of both from $F$ cannot be neither $<\epsilon$ nor $>\epsilon$. Is this correct? If yes how can we write it down as a proof?

答:
This is not true.

In $\mathbb{R}^2 \simeq \mathbb{C}$, consider $B = \lbrace z \in \mathbb{C} \text{ }|\text{ } 2\varepsilon \leq z \leq 3\varepsilon \rbrace$, and
$$F = B \setminus \lbrace z \in \mathbb{C} \text{ }|\text{ } \mathrm{Im}(z)<0, |\mathrm{Re}(z)|< \varepsilon/2\rbrace$$

It is easy (just make a picture) to see that $F$ is connected, $\mathbb{C} \setminus F$ is connected, but $A = \lbrace z \in \mathbb{C} \text{ }|\text{ }d(z,F) = \varepsilon \rbrace$ looks like the union of the two circles of radii $\varepsilon$ and $4\varepsilon$, therefore $A$ is not connected.


答:
It's correct, but you're not actually using contradiction and it's indeed hard to write a real proof by contradiction of this particular statement.

What you proved is that if $n^3-n-6=0$, then $n=2$. Hence for no $m\ne2$ one can have $m^3-m-6=0$.

Title: Proof by contradiction : $n^3 −n−6 = 0$ means $m^3 − m − 6 ≠ 0$ when $m≠n$

Body:
I have to prove by contradiction that if $n$ is a positive integer such as $n^3-n-6 = 0$, so for any positive integer $m$ such as $m ≠ n$, we have $m^3-m-6 ≠ 0$

I am new to proofs, but I tried and here is what I did.

I first assume the opposite of the conclusion, so
$m^3-m-6 = 0$

$(m-2)(m^2+2m+3) = 0$

m=2

If I do the same thing for the n function and find the solution for $n^3-n-6 = 0, n=2$.

This is a contradiction since $m=n$, but we said $m ≠ n$.

I feel like im doing something wrong here and I would like someone to tell me what it is ! (Again, I am a beginner)


问:
If polynomial $P(x)$ has integer coefficients and at least three integer roots, then $P(x)+5^m$ has no more than one integer root for $m\geq 1$
I've been doing some polynomial excersises lately and in that one I got completly stuck.

Let $m \geqslant 1$ be natural number and $P(x)$ polynomial with integer coefficients which has at least three different integer roots. Prove that $P(x)+5^m$ has no more than one integer root.

At first I considered the easiest case: $(x-x_1)(x-x_2)(x-x_3)+5$, but it did not turned out in anything helpful, so I am seeking for some clues on how to crack that problem.

Also, I'd like to ask for as elementary hint/solution as possible since this question is from (inactive) high school contest.
https://om.mimuw.edu.pl/static/app_main/problems/om48_1.pdf

答:
Suppose $r_1$, $r_2$, $r_3$ are three distinct integer roots of $P(x)$, and $y_1$ and $y_2$ are integer roots of $P(x) + 5^m$.  Thus $P(x) = Q(x)(x-r_1)(x-r_2)(x-r_3)$ where $Q(x)$ also has integer coefficients.  Now
$P(y_i) = -5^m = Q(y_i) (y_i - r_1)(y_i-r_2)(y_i-r_3)$, so $y_i - r_j$ are integers that divide $5^m$, in particular they are $\pm$ powers of $5$.  And $$y_1 - y_2 = (y_1 - r_1)-(y_2-r_1) = (y_1-r_2)-(y_2-r_2) = (y_1-r_3)-(y_2-r_3)$$
is written as the difference between two of those in three different ways.  But that is impossible unless $y_1 - y_2 = 0$.


问:
What is the solution to $\arctan -i$?
I know that $\operatorname{Arctan} z = \frac{1}{2i} \operatorname{Ln} \frac{1+iz}{1-iz}$ but in this case we have $$\frac{1}{2i} \operatorname{Ln}\frac{1+1}{1-1}$$
Is it possible to deal with this formula or what other methods should I try?

答:
Thanks to all! That's what I've got. $\\$
The function $\operatorname{Ln} z$ is considered on the set $\mathbb C \setminus \{ 0 \}$, and the set $\mathbb C $ itself does not include an infinitely distant point. Therefore $\operatorname{Ln}(\infty)$ does not exist. That's why $\operatorname{Arctan} (-i)$ does not exist either.

Title: What is the solution to $\arctan -i$?


问:
Local ring if and only if $x$ or $1-x$ is unit

A commutative ring $R$ with identity is local if and only if for all $x
\in R$, $x$ or $1-x$ or both is unit.

I have solved the 'only if' part, but I have stuck on the 'if' part for a very long time$\ldots$ My idea is to prove that all non-units form an ideal. But I don't know how to check $rx$ is non-unit for any $r\in R$ and non-unit $x$$\ldots$

答:
In fact, the hard part is proving the nonunits are closed under addition.

Also, the ring can be noncommutative. The same proof will work.

Suppose the nonunits are closed under addition. Then it is impossible for $x$ and $1-x$ to both be nonunits, because their sum (which is $1$) would have to be a nonunit, and of course $1$ is a unit.

Now suppose $R$ has the property that for every $x\in R$, at least one of $x$ and $1-x$ is a unit.  Let $M$ be any maximal right ideal. We claim that every nonunit is in $M$ so that the sum of nonunits is again a nonunit.

First, a little Lemma:

Under the assumptions immediately above, any one-sided invertible element is a unit. Suppose $xy=1$. Then it satisfies $(1-yx)yx=0$.  By assumption, one of $yx$ or $1-yx$ is a unit: but in the former case, $1-yx=0$ is a contradition, and in the latter case $yx=0$ leads to a contradiction too.

If we assume to the contrary there is a nonunit $x$ outside of $M$, it would have to be the case that $xR+M=R$. Selecting $r\in R$ and $m\in M$ such that $xr+m=1$, since $x$ is not a unit, $xr$ is not a unit either (because of the lemma: if $xr$ were a unit, $x$ would be right invertible, hence a unit also.) By assumption again, $1-xr$ is a unit... but $1-xr=m\in M$, which is a proper ideal... that is a contradiction.

Hence all nonunits are contained in $M$, and so the nonunits are closed under addition.


问:
Is the area independent of $n$? $ A=\int_0^\infty \frac{e^{-\frac{1}{tn}}(e^{-tn}(tn+1))}{t}~dt $
Consider the parametric equation $x=\frac{1}{t}e^{-tn}$ and $y=te^{-\frac{1}{tn}}.$ To integrate under this curve I put it in the proper form:

$$ A=\int_0^\infty \frac{e^{-\frac{1}{tn}}(e^{-tn}(tn+1))}{t}~dt $$

I noticed that by changing $n$ from 1 to 2 to 3 that the area was the same!

Is the area $A$ independent of $n?$

Do integrals like these have applications?

答:
"Integrals like these" is pretty vague, but here is a remarkable fact:

We can split the integral up as
$$A=\int_0^\infty \exp\left(-t-\frac{1}{t}\right)\mathrm{d}t+\int_0^\infty \exp\left(-t-\frac{1}{t}\right)\frac{1}{t}\mathrm{d}t$$
Which, using an integral identity on DLMF can be expressed in terms of modified Bessel functions -
$$A=2(K_{-1}(2)+K_0(2))$$
The modified Bessel functions obey the recurrence
$$\frac{2\nu}{z}K_\nu(z)+K_{\nu+1}(z)=K_{\nu-1}(z)$$
From which we can conclude $K_{-1}(z)=K_1(z)$. So we have
$$A=2(K_{1-1}(2)+\frac{2\cdot1}{2}K_{1}(2))=2K_2(2)$$
The modified Bessel function has many applications in applied mathematics and physics.


问:
Let a, b and c be odd positive integers . Show that the quadratic equation ^2++=0 has no rational solution.
To prove this, I think the Δ should =$k^2$ so I let a=2p-1, b=2q-1, c=2r-1, where p, q, r are all positive integers, then I calculated $
b^2-4ac$ which is $-16 p r + 8 p + 4 q^2 - 4 q + 8 r - 3$
and find it hard to prove that $-16 p r + 8 p + 4 q^2 - 4 q + 8 r - 3 ≠ k^2$
so how to prove Δ ≠ $k^2$ and is it possible to use method of contradiction ( let a root $x_0$= p/q and $gcd(p,q)=1$)

答:
Step 1: We show that it either has zero or two rational solutions. Assume then contrary, that is $x_1 \in \mathbb{Q}, x_2 \in \mathbb{R}\setminus\mathbb{Q}$. Then $x_1x_2 \in  \mathbb{R}\setminus\mathbb{Q}$, or $-ac\in  \mathbb{R}\setminus\mathbb{Q}$. Contradiction.

Step 2: Assume that is has two rational solutions. So it can be written as:
\begin{align*} (x - \frac{n_1}{m_1})(x - \frac{n_2}{m_2}) &= 0\\ (m_1x - n_1)(m_2x - n_2) &= 0 \\
m_1m_2x^2 - (n_1m_2 + n_2m_1)x + n_1n_2 &= 0
\end{align*}

Now I will claim that we are done. Note that we can pick $n_i, m_i$ such that $\gcd(n_i, m_i) = 1$. We need the coefficients of $x$ to have the same parity. If $m_1$ is even then $n_2$ must be even which gives $n_1m_2 + n_2m_1$ odd. The symmetric argument applies when $m_2$ is even. Finally, if both $m$ are odd then $n$ are odd but now $(n_1m_2 + n_2m_1)$ is even so we reach the required contradiction.