答:
On open set $S$ you can say that elements $x,y\in S$ are related if the closed interval (or singleton) determined by them is a subset of $S$.

(So $x\sim y$ if $x=y$ or $x<y$ and $[x,y]\subseteq S$ or $y<x$ and $[y,x]\subseteq S$)

This relation can be shown to be an equivalence relation and the equivalence classes can be shown to be open.

Now the density of $\mathbb Q$ comes in in order to prove that there are at most countable equivalence classes.

Title: Prove that every open set in $\mathbb{R}$ is the union of an at most countable collection of disjoint segments.

Body:
My attempt:

We know that $\mathbb{Q}$ is dense in $\mathbb{R}$.Now, we consider an open set $S$ in $\mathbb{R}$ then let $x \in S$, $x$ is an interior point of $S$ and we can get an open ball $(x-r,x+r)$ ,$r>0$ contained in $S$. Since $\mathbb{Q}$ is dense, let $q_1 \in (x-r,x+r)$.Now, $q_1 \in (x-r,x+r)$ and let $r_1=|q_1-x|$.We consider the rational number $ > r_1$ (which is possible to choose as $\mathbb{Q}$ is dense ) and let us call it $d_1$ and we consider the open ball $(q_1-d_1,q_1 +d_1)$ which will contain $x$.

We pick $x' \in S$ , and consider $r_2>0$ such that $(x'-r_2,x'+r_2) \cap (q_1-d_1,q_1+d_1) = \phi$ , if such an $r_2$ doesn't exist then we can conclude that $x' \in (q_1-d_1,q_1+d_1)$. Repeating this process we can conclude that the open set $S $ can be written as the union of disjoint segments.

Since $\mathbb{Q}$ is countable then it can be shown that union of this disjoint segment is also countable.

I am trying to do Rudin on my own and need someone to cross check my proofs.I know this might be a duplicate but if someone goes through my proof and picks up my mistake and a probable way out that would be helpful.


问:
RSA: what is $k$ in the formula to calculate $d$?
In RSA, to calculate $d$, when given $\phi(n)$ and $e$, I stumbled upon this formula:

$$d = \dfrac{k \phi(n) + 1}{e}$$

But what does $k$ stand for? How to obtain the value for $k$?

Thank you,

Chris

答:
We can rearrange your equation to obtain:

$$de = k\phi(n)+1$$

or:

$$de \equiv 1 \pmod {\phi(n)}$$

Therefore $d$ is the inverse of $e$ in $\mathbb Z_{\phi(n)}$. (This only makes sense if $e$ and $\phi(n)$ are coprime, by Bezout's Lemma.)

To find this $d$ (and the less important $k$), we can use the Euclidean Algorithm.

ContentLicense: CC BY-SA 4.0


问:
Prove a subring of rationals is a DVR
I am currently stuck on a problem being:

Prove that the ring of rational numbers $\frac{a}{b}$ such that $p \nmid b$ for a prime integer p is a discrete valuation ring.

Any help would be appreciated.

答:
We use the definition of discrete valuation ring in Atiyah and Macdonald.

The ring you mentioned is called $\mathbb{Z}_{(p)}$, and it has a map to $\mathbb{Z}_{\geq0}\cup\infty$ given by mapping a non-zero $\frac{a}{b}$ to $k$ where $k$ is the maximal interger such that $p^k|a$ and mapping $0$ to $\infty$.

Hence you can defined a discrete valuation on its fraction field $\mathbb{Q}$, namely $v(\frac{x}{y})=v(x)-v(y)$ where $x,y\in\mathbb{Z}_{(p)}-\{0\}$ and $v(0)=\infty$. You should check that this valuation is well-defined, and it is easy to see $\mathbb{Z}_{(p)}$ is the valuation ring of this discrete valuation.

Edit. In fact, the valuation over $\mathbb{Q}$ constructed above is exactly the $p$-adic valuation, and you get a norm $|x|_p=p^{-v_p(x)}$ over $\mathbb{Q}$. The only valuation over $\mathbb{Q}$ are $|\cdot|_p$ for some prime $p$ and the usual absolute value, this is proved in any book of algebraic number theory. The completion of $\mathbb{Q}$ associated to $p$-adic norm is the $p$-adic field $\mathbb{Q}_p$ and the completion of $\mathbb{Q}$ associated to  is the usual absolute value is just $\mathbb{R}$. Studying the completion of number field at some prime is called $p$-adic number theory.

Title: Prove a subring of rationals is a DVR


问:
A neighbor city should receive the double amount of refugees
So i have to solve a program in C, which is already a challenge, but it's likely finished. The only thing that i do wrong is the algorithm.
I have cities and one city will be destroyed while all the neighbor cities get the population from the destroyed city.

The task says a neighbor city that is twice as big as the destroyed city, should get twice as many refugees.

As an example:

```<code>Population: 149355 of the destroyed City A
Population: 189069  of the first neighbor City B.
</code>```

Now, City B should change from 189069 to 205057 citizens. So if i calculated that correct, City B receives 15988 citizens from City A. Why?

What is the mathematical algorithm here? I tried to divide City B trough City A and get around 1.26 for that. However, no matter what i try, i don't know what i need to do in a mathematical way to get to the 205057 population size of the first neighbor city.

Any ideas?

Edit:

All Neighbor cities in this example:

```<code>CityB 189069 135
CityC 119250 138
CityD 1365052 144
CityE 92807 162
</code>```

the number in the middle is the population, the right outer column the distance to the destroyed city.

CityC should change to 129334 in this example, after CityB got its new population.

答:
You just need to compare the populations of all the neighboring cities, the logic is: the larger city will receive the largest fraction of refugees, the second largest will receive the second largest fraction ...

In python this looks like

```<code>import numpy as np

destroyed = 149355

neighbors = np.array([189069, 119250, 1365052, 92807])

refugees = destroyed * neighbors / neighbors.sum()

new_populations = refugees + neighbors

</code>```

You should get

```<code>array([ 205057.42273825,  129334.25184211, 1480486.19828579,
        100655.12713384])
</code>```

Body:
You just need to compare the populations of all the neighboring cities, the logic is: the larger city will receive the largest fraction of refugees, the second largest will receive the second largest fraction ...

In python this looks like

import numpy as np

destroyed = 149355

neighbors = np.array([189069, 119250, 1365052, 92807])

refugees = destroyed * neighbors / neighbors.sum()

new_populations = refugees + neighbors

You should get

array([ 205057.42273825,  129334.25184211, 1480486.19828579,
        100655.12713384])


问:
Chinese remaining theorem, help with surjectivity
Let $R$ be a ring and let $I_1, \dots, I_n \subset R$ be ideals, which are pairwise coprime. Let furthermore $\pi_j:R \to R/I_j,\; r \mapsto r + I_j$, $j=1, \dots, n$ be the projection to the $jth$ quotient ring.

I know that for all $j \in \{1, \dots, n\}$, the ideals $I_j$ and $\cap_{i \neq j}I_i$ are coprime. For all $j \in \{1, \dots, n\}$ we take $d_j \in I_j$ and $e_j \in \cap_{i \neq j}I_i$ s.t. $d_j + e_j=1$.
$$\Rightarrow \pi_i(e_j)=\pi_i(1-d_j)$$
This is now equal to $0$, if $i \neq j$ and equal to $1+I_j$, if $ i=j$. I don't see why we get this.

Many thanks for any help!

答:
Ideals $I_j$ and $\bigcap_{i\ne j} I_i$ are coprime, which means that $I_j + \bigcap_{i\ne j} I_i = R$. Thus for $1\in R$ there are $d_j\in I_j$ and $e_j\in \bigcap_{i\ne j} I_i$ with $d_j+e_j=1.$

Then $\pi_i(e_j) = 0$, since $e_j\in I_i$ for $i\ne j$.


问:
Find probability with the help of Poisson distribution
There are $3$ cyclone on average in a year in russia .

What is the probability that there will be  cyclone  in the next TWO years  ?

I just want to know the value of $\lambda$ in Poisson distribution! As we know $\lambda$ actually works on the same time interval.
My calculations: as it's Two years, $\lambda = 3\cdot 2 = 6$.

But my confusion is should we calculate the present year too? I mean if we don't calculate this year we can calculate as time = $2$ years otherwise $3$ ...

Is my calculation ok?

答:
I suppose that your question means that there is some misunderstanding in the question about taking $2 \lambda$: otherwise there would be no questions.

Let $\xi$ be the number of cyclones during the first year and $\eta$ is the number of cyclones during the second year. Thus $\xi \sim Pois(\lambda)$,  $\eta \sim Pois(\lambda)$. We implicitly suppose that $\xi$ is independent of $\eta$ - otherwise this approach will not work.

$E \xi = \lambda =3$.

We use the next fact: the sum of independent r.v. with distributions $Pois(\lambda_1)$ and $Pois(\lambda_2)$ has distribution $Pois(\lambda_1 + \lambda_2)$. Hence $\xi+ \eta \sim Pois(2 \lambda)$.

Hence $P(\xi + \eta > 0) = 1 - e^{-6}$ is the probability of at least one cyclone and $P(\xi + \eta = 1) = 3e^{-6}$ is the probability that there will be exactly one cyclone.


问:
Find the number of integers $n$ of the form $30a0b03$ that are divisible by $13$.

Find the number of integers $n$ of the form $30a0b03$ that are divisible by $13$.

What I Tried: This problem looks simple but I cannot find it's solution. I recalled all the divisible rules of $13$ and tried to put them here, but none of them really worked. One thing which I got was :-
$$\rightarrow 13 | bo3 - 0a0 + 3$$
I only got this by applying one divisibility rule of $13$.
I might try and test for each values of $a$ and $b$ but there will be $100$ configurations, and I don't want to do that.

Can anyone help me?

Edit: After the hints @Daniel gave me, I figured out the solution and found $7$ solutions, can anyone tell me if this is correct or not?

答:
You can crunch it out like this:

$$3\cdot 10^6+a\cdot 10^4 + b\cdot 100 + 3 \equiv 3\cdot(-3)^6+a\cdot (-3)^4 + b\cdot 9 + 3 \pmod{13}.$$

Note that $3^3 = 27 \equiv 1 \pmod{13}$ so the above is $\equiv 3+3a+9b+3 \pmod{13},$ which you want to be congruent to $0$.  That reduced to

$$3a+9b \equiv 7 \equiv 33 \pmod{13}$$

or

$$a+3b \equiv 11 \pmod{13}.$$

You can solve this congruence for $b$ for each value of $a = 0,1,\ldots 9.$  Three of the solutions for $b$ are $10$, $11$, and $12$.  The other seven solutions work.


问:
Two numbers whose difference is a full square are called friendly numbers. For example,....
Two numbers whose difference is a full square are called friendly numbers. For example, the numbers $25$ and $16$ are friendly because their difference is $9$ full squares. Find the smallest natural number $n$ so that when we randomly paint the numbers $1, 2, 3, ..., n$ in three different colors, there are two friendly numbers of the same color.

I could not solve this question

答:
For $1\le k\le n-25$, numbers $k+9,k+16$ must have a different colour than $k$ and also different than $k+25$ and also $k, k+25$ must differ. We conclude that $k+9,k+16$ have same colour..
Then $k+8$ differs from $k+9$, $k+7$ differs from both $k+8$ and $k+16~k+9$, i.e. the three consecutive numbers $k+7,k+8,k+9$ use the three colours. If $n\ge33$, this holds for $k=1,2,\ldots,8$, so that numbers $8,9,\ldots,17$ cycle through the colours. In particular, $8$ and $17$ have sisme colour - contradiction!
Can you solve the puzzle with $n=32$?


问:
Can we prove this using Archimedean Property.

Question: let $s\in(0,1)$ then, is the following statement is true?

For every $m\in\mathbb{N}$ there exists $n\in\mathbb{N}$ such that
$s<m/n$

How can use Archimedean property to prove above?

I know by Archimedean property, for any two positive real numbers $m, s$ there exists $n\in\mathbb{N}$ such that, $ns>m$ (i.e. $s>m/n$). But inequality, in above statement is reversed.

答:
It is not necessary to use the Archimedean property to prove what you want, indeed

$s<1\leqslant\dfrac mn\quad\forall n\in\mathbb{N}\;$ such that $\;n\leqslant m\;.$


问:
Modular Congruences
I have the following exercise.

If $ abc \equiv b \pmod{m} $ and $ \gcd{(a,m)} = 1 $, then $ ac \equiv 1 \pmod{m} $.

My attempt.

From $ abc \equiv b \pmod{m} $, by definition, $ abc - b = b ( ac - 1 ) = mh $.

Now: if $ b \mid h $, then $ ac - 1 = mk \implies ac \equiv 1 \pmod {m} $; if $ b \mid m $... I don't know what to do...

Ideas? Other solutions?
Thanks.

答: This is incorrect.  Take $(a,b,c,m) = (1,2,3,4)$ as a counterexample: We have $\gcd(a,m) = \gcd(1,4) = 1$ and $$abc \equiv 3! \equiv 6 \equiv 2 \equiv b \pmod4,$$ but $$ac \equiv 3 \cdot 1 \equiv 3 \not\equiv 1 \pmod 4.$$

Title: Modular Congruences


问:
Prove that $\frac1{1+a}+\frac1{1+b}+\frac1{1+c}\leq\frac14(a+b+c+3)$ if $abc=1$
I found the following exercise in a problem book (with no solutions):

Given $a,b,c>0$ such that $abc=1$ prove that
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\leq\frac{a+b+c+3}{4}$$

I tried AM-GM for the fraction on the LHS but got stuck from there.

答:
Let $f(x) = \frac{1}{1 + x} - \frac{x+1}{4} + \frac{1}{2}\ln x$.
We have $f'(x) = \frac{(1-x)(x^2+x+2)}{4x(1+x)^2}$.
Thus, $f'(x) > 0$ on $(0, 1)$, and $f'(x) < 0$ on $(1, \infty)$.
Thus, $f(x)$ is strictly increasing on $(0, 1)$, and strictly decreasing on $(1, \infty)$.
Also $f(1) = 0$. Thus, $f(x) \le 0$ on $(0, \infty)$.

Thus, $f(a) + f(b) + f(c) \le 0$ which results in $\frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c}
\le \frac{a+b+c+3}{4}$ (using $\ln (abc) = 0$).

We are done.


答:
By Schur's inequality and AM-GM $$\begin{align*}x^3+y^3+z^3+3xyz\ge xy(x+y)+yz(y+z)+zx(x+z)&\\ \ge 2( {(xy)}^{3/2}+{(yz)}^{3/2}+{(zx)}^{3/2})\end{align*}\tag S$$

Now as $$\frac{1}{1+x}\le \frac{1}{2\sqrt{x}}$$ we have to prove $$a+b+c+3\sqrt[3]{abc}\ge 4\left(\frac{1}{2\sqrt{a}}+\frac{1}{2\sqrt{b}}+\frac{1}{2\sqrt{c}}\right)$$ or it suffices to prove $$a+b+c+3\sqrt[3]{abc}\ge 2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}$$ which is just a direct consequence of inequality (S) and so we are done!

Note here $\sqrt[3]{abc}=\sqrt{abc}=1$ is used to normalize inequality

Title: Prove that $\frac1{1+a}+\frac1{1+b}+\frac1{1+c}\leq\frac14(a+b+c+3)$ if $abc=1$

Body:
I found the following exercise in a problem book (with no solutions):

Given $a,b,c>0$ such that $abc=1$ prove that
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\leq\frac{a+b+c+3}{4}$$

I tried AM-GM for the fraction on the LHS but got stuck from there.


问:
Finding the equation of hyperbola that passes through $(2,0)$ and has foci points $(2,3)$ and $(1,0)$
If the hyperbola is given by $$(x-x_0)^2/a^2-(y-y_0)^2/b^2=1$$
and I know that if $F(c,0)$ is a foci point then  $c^2=a^2+b^2$ I can say that in this particular problem $a^2+b^2=1$.Then I plug in the given point and I get $$(2-x_0)^2/a^2-(y_0)^2/1-a^2=1$$
Now I have one equation with two unknowns. Could you please explain how to solve this problem.
Thank you.

答:
The center of the hyperbola is the midpoint of the segment joining $F_1(2,3),F_2(1,0)$ that is $M\left(\frac32,\frac32\right)\equiv(x_0,y_0)$.

Thus $c^2=MF_1^2$

This is a rotated hyperbola.

Edit

$(x_0,y_0)=\left(\frac32,\frac32\right)$

$c^2=MF_1^2=5/2$

$$\frac{\left(2-\frac{3}{2}\right)^2}{a^2}-\frac{\left(0-\frac{3}{2}\right)^2}{\frac{5}{2}-a^2}=1$$
solve for $a$


答:
We'll try to prevent friends being monochromatic, and see how big $n$ has to be to force a contradiction. Adjacent numbers are friendly, so say $1$ is red, $2$ is blue and the third colour is green. A computer-aided search finds the longest solutions have $n=28$, e.g. $rbgrbgrgrbgrgrbgbgrbgrbrbgrb$.

For $1\le i\le n-25$, the values $i,\,i+9,\,i+25$ must permute the colours, while $i+4,\,i+16$ must each differ in colour from $i$, and have the same colours as $i+25,\,i+9$ respectively. So $i,\,i+1,\,i+4,\,i+9,\,i+16,\,i+25$ have colours of the form $xyyzzy$ or $xyzyyz$. But $1$ to $4$ have possible colorations $rbgb,\,rbgr,\,rbrb,\,rbrg$. For example, the $n=28$ example above actually matches $rbgr$ and $xyyzzy$ for each of $i\in\{1,\,2,\,3\}$. A hypothetical $n=29$ solution is specified by which of $4\times2^4=64$ options it chooses for its start and the adding-squares patterns on $i\le4$, and presumably we can refute them in turn.

Title: Two numbers whose difference is a full square are called friendly numbers. For example,....

Body:
Two numbers whose difference is a full square are called friendly numbers. For example, the numbers $25$ and $16$ are friendly because their difference is $9$ full squares. Find the smallest natural number $n$ so that when we randomly paint the numbers $1, 2, 3, ..., n$ in three different colors, there are two friendly numbers of the same color.

I could not solve this question


问:
Let $f:\mathbb R\to \mathbb R$ be continuous satisfying specific condition is uniformly continuous.
Let $f:\mathbb R\to \mathbb R$ be continuous satisfying that $\forall \epsilon>0$, $\exists M>0$ such that if $|x|\ge M$ then $|f(x)|<\epsilon$

Then $f$ is uniformly continuous.

Attempt let $\epsilon>0$ be given then there exists $M> 0$ such that if $|x|\ge M$ then $\epsilon/2>|f(x)|$.

If $|x|\le M$ then since $[-M,M]$ is compact and $f$ is continuous then $f$ is uniformly continuous.

If $|x|\ge M$ then choose $\delta>0$ arbitrarily then $\forall |x|,|y|\ge M$ if $\delta >|x-y|$ then $|f(x)-f(y)|\le |f(x)|+|f(y)|<\epsilon$

I am not sure, because I didnot prove for single $\delta$ for given $\epsilon$, and not sure the proposition is true?

答:
You need to be a bit more careful with $\delta$ and how $x$ and $y$ are situated. What about the possibility that $x < M < y$?

You were off to a good start. $\epsilon > 0$ is fixed and you chose $M > 0$ accordingly. Since $f$ is continuous it is uniformly continuous on (say) $[-3M,3M]$ so there exists $\delta_M$ with the property that $x,y \in [-3M,3M]$ and $|x-y| < \delta_M$ implies $|f(x) - f(y)| < \epsilon$.

Define $\delta = \min\{\delta_M,M\}$.

Suppose $x,y \in \mathbf R$ and $|x-y| < \delta$.

In case $|x| \ge 2M$ then $|x| \ge M$, and in addition $$|y| \ge |x| - |x-y| > 2M - \delta \ge 2M - M = M$$ so that $|x|,|y| \ge M$ and
$$|f(x) - f(y)| \le |f(x)| + |f(y)| < \epsilon.$$

In case $|x| < 2M$ then $|x| < 3M$, and in addition $$|y| \le |x| + |x-y| < 2M + \delta < 3M$$ so that $x,y \in [-3M,3M]$ and $|f(x) - f(y)| < \epsilon$.


答:
Using A hyperbola as a constant difference of distances if $P(h,k)$ is an arbitrary point of the required hyperbola,

$$2a=|\sqrt{(h-2)^2+(k-3)^2}-\sqrt{(h-1)^2+(k-0)^2}|, a>0$$

As the hyperbola passes through $(2,0);$

$$2a=|\sqrt{(2-2)^2+(0-3)^2}-\sqrt{(2-1)^2+(0-0)^2}|=?$$

Title: Finding the equation of hyperbola that passes through $(2,0)$ and has foci points $(2,3)$ and $(1,0)$

Body:
If the hyperbola is given by $$(x-x_0)^2/a^2-(y-y_0)^2/b^2=1$$
and I know that if $F(c,0)$ is a foci point then  $c^2=a^2+b^2$ I can say that in this particular problem $a^2+b^2=1$.Then I plug in the given point and I get $$(2-x_0)^2/a^2-(y_0)^2/1-a^2=1$$
Now I have one equation with two unknowns. Could you please explain how to solve this problem.
Thank you.


问:
Evaluate $\lim\limits_{n\to +\infty}\frac{1}{n^2}\sum_{k=1}^{n} \cot^2\left(\frac{\pi k}{2n+1}\right)$
How to evaluate:
$$\lim\limits_{n\rightarrow +\infty}\frac{1}{n^2}\sum_{k=1}^{n} \cot^2\left(\frac{\pi k}{2n+1}\right)$$

Personally, i always have trouble to evaluate sum of trignometric series, if you know a paper to recommend, or part of a book, about this, please let me know.

Anyway

$$\frac{1}{n^2}\sum_{k=1}^{n} \cot^2\left(\frac{\pi k}{2n+1}\right) = \frac{1}{n^2}\left(\sum_{k=1}^{n} \frac{1}{\sin^2\left(\frac{\pi k}{2n+1}\right)} - \sum_{k=1}^{n} 1\right)$$

What now?

答:
It is possible to prove, by studying the function $ f: x\mapsto 1-x^{2}\cot^{2}{x}-\frac{2 x^{2}}{3} $ on $ \mathcal{D}=\left]-\frac{\pi}{2},\frac{\pi}{2}\right[\setminus\left\lbrace 0\right\rbrace $, that for any $ x\in\mathcal{D} $, we have : $$ \left|1-x^{2}\cot^{2}{x}\right|\leq\frac{2x^{2}}{3} $$

Let $ n \in\mathbb{N}^{*} $, and $ k\leq n $. Setting $ x\leftarrow \frac{k\pi}{2n+1} $, we get : $$ \left|1-\frac{k^{2}\pi^{2}}{\left(2n+1\right)^{2}}\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}\right|\leq\frac{2k^{2}\pi^{2}}{3\left(2n+1\right)^{2}} $$

Using the previous inequality, we can write the following : \begin{aligned} \left|\frac{1}{\left(2n+1\right)^{2}}\sum_{k=1}^{n}{\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}}-\frac{1}{\pi^{2}}\sum_{k=1}^{n}{\frac{1}{k^{2}}}\right|&\leq\frac{1}{\pi^{2}}\sum_{k=1}^{n}{\frac{1}{k^{2}}\left|1-\frac{k^{2}\pi^{2}}{\left(2n+1\right)^{2}}\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}\right|}\\ &\leq \frac{2n}{3\left(2n+1\right)^{2}}\underset{n\to +\infty}{\longrightarrow}0\end{aligned}

Thus, the sequence $ \left(\frac{1}{\left(2n+1\right)^{2}}\sum\limits_{k=1}^{n}{\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}}\right)_{n\in\mathbb{N}^{*}} $ does converge and : $$ \lim_{n\to +\infty}{\frac{1}{\left(2n+1\right)^{2}}\sum_{k=1}^{n}{\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}}}=\lim_{n\to +\infty}{\frac{1}{\pi^{2}}\sum_{k=1}^{n}{\frac{1}{k^{2}}}}=\frac{1}{6} $$

Hence : \begin{aligned}\lim_{n\to +\infty}{\frac{1}{n^{2}}\sum_{k=1}^{n}{\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}}}&=\lim_{n\to +\infty}{\left(\frac{2n+1}{n}\right)^{2}\times\frac{1}{\left(2n+1\right)^{2}}\sum_{k=1}^{n}{\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}}}\\ &=4\times\frac{1}{6}\\ \lim_{n\to +\infty}{\frac{1}{n^{2}}\sum_{k=1}^{n}{\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}}}&=\frac{2}{3}\end{aligned}


问:
How Many Possible Addition Equations Can Make a Certain Sum?
To jump right into the rawness of my topic, how many addition expressions create a certain sum?

Like, for example, if $n = 3$, a formula would return $4$, because the possible addition equations with a sum of $n$ are (with only whole numbers) $1+2$, $2+1$, $3+0$, and $0+3$, which means there are four possible equations with a sum of $n$. Is there a formula for this?

答: If order matters and $0$ is allowed, so $0+3$ and $3+0$ both count, you are talking about weak compositions.  For weak compositions into two parts there are $n+1$ of them.  The first number ranges from $0$ through $n$ and the other is what is left.


问:
Convergence of sum $\sum_{n=1}^\infty \frac{x^n}{(1-x^n)(1-x^{n+1})}$
I'm having difficulties showing that $\sum_{n=1}^\infty \frac{x^n}{(1-x^n)(1-x^{n+1})}$ converges to $\frac{x}{(1-x)^2}$ for $|x|<1$.

It's easy to see that:

$$\sum_{n=1}^\infty \frac{x^n}{1-x} = \frac{x}{(1-x)^2}$$

by using the geometric series. It seems I can't use that fact because clearly $(1-x^n)(1-x^{n+1})\neq 1-x$. Also, decomposing the fractions into several fractions with simple denominators seems too convoluted because I would have to work with cyclotomic polynomials. Is this the way to go or am I missing something?

Body:
Hint:

$$1-x^n-(1-x^{n+1})=x^n(x-1)$$

$$(x-1)\cdot\dfrac{x^n}{(1-x^n)(1-x^{n+1})}=f(n+1)-f(n)$$ where $f(m)=\dfrac1{1-x^m}$

For partial sum  using Telescoping Series,
$$(x-1)\sum_{n=1}^m\dfrac{x^n}{(1-x^n)(1-x^{n+1})}=\sum_{n=1}^m(f(n+1)-f(n))=f(m+1)-f(1)$$

Now set $m\to\infty$


Title: Convergence of sum $\sum_{n=1}^\infty \frac{x^n}{(1-x^n)(1-x^{n+1})}$

Body:
I'm having difficulties showing that $\sum_{n=1}^\infty \frac{x^n}{(1-x^n)(1-x^{n+1})}$ converges to $\frac{x}{(1-x)^2}$ for $|x|<1$.

It's easy to see that:

$$\sum_{n=1}^\infty \frac{x^n}{1-x} = \frac{x}{(1-x)^2}$$

by using the geometric series. It seems I can't use that fact because clearly $(1-x^n)(1-x^{n+1})\neq 1-x$. Also, decomposing the fractions into several fractions with simple denominators seems too convoluted because I would have to work with cyclotomic polynomials. Is this the way to go or am I missing something?

Body:
Let $u=x^n,$ then
$$ \frac{x^n}{(1-x^n)(1-x^{n+1})} = \frac{u}{(1-u)(1-xu)} = \frac{1}{1-x} \left( \frac{u}{1-u} - \frac{xu}{1-xu}\right) = \frac{1}{1-x} \left( \frac{x^n}{1-x^n} - \frac{x^{n+1}}{1-x^{n+1}}\right).$$

Now the series is telescoping.


问:
Prove that $f$ has limit in $\mathbb{\overline{R}}$
Let $f$ be a continuous function, that takes each value at most ten
times. Show that $f$ has a limit as $x \to \infty$.

It's easy to show that $$\forall y \ \exists u \ : \ (\forall x > u \ f(x) > y \text{ or }\forall x > u \ f(x) < y)$$

If for some $y$ we couldn't find $y'$ such that $\forall y' > y \ \exists u' : \forall x > u' f(x) < y'$, the limit is $=+\infty$ (or $-\infty$ in mirror situation).

But what I can do in other situations? How can we prove that limit exists?

答:
Let's try the contraposition. If $f$ does not have a limit for $x \to \infty$, then
$$ \varliminf_{x \to \infty} f(x) < \varlimsup_{x \to \infty} f(x) $$
Choose $a,b \in \mathbb R$ such that
$$ \varliminf_{x \to \infty} f(x) < a < b < \varlimsup_{x \to \infty} f(x)$$
As $b < \varlimsup f$, there is $x_0$ such that $f(x_0) > b$. As $a > \varliminf f$, we find $y_0 > x_0$ such that $f(y_0) < a$. By induction, we find sequences $(x_n)$, $(y_n)$ such that

$f(x_n) > b$, all $n \in \mathbb N$

$f(y_n) < a$, all $n \in \mathbb N$

$x_n < y_n < x_{n+1}$, all $n \in \mathbb N$.

Now let $m := \frac{a+b}2$. As $f(x_n)< a < m < b < f(y_n)$, by the intermediate value theorem there is $\xi_n \in (x_n, y_n)$ such that $f(\xi_n) = m$. Hence, $m$ is taken at infinitely many points.


答:
Let $ n\in\mathbb{N}^{*} $.

Observe that for any $ x\in\mathbb{R} $, and $ k\in\mathbb{N}^{*} $, we have : $$ \frac{x^{k}}{\left(1-x^{k}\right)\left(1-x^{k+1}\right)}=\frac{1}{1-x}\left(\frac{1}{1-x^{k}}-\frac{1}{1-x^{k+1}}\right) $$

Thus : \begin{aligned} \sum_{k=1}^{n-1}{\frac{x^{k}}{\left(1-x^{k}\right)\left(1-x^{k+1}\right)}}&=\frac{1}{1-x}\sum_{k=1}^{n-1}{\left(\frac{1}{1-x^{k}}-\frac{1}{1-x^{k+1}}\right)}\\ &=\frac{1}{1-x}\left(\frac{1}{1-x}-\frac{1}{1-x^{n}}\right) \end{aligned}

Now when $ n $ goes to $ +\infty $ :

If $ \left|x\right|<1 $, our series converges to $ \frac{1}{1-x}\left(\frac{1}{1-x}-1\right)=\frac{x}{\left(1-x\right)^{2}} $.

If $ \left|x\right|>1 $, our series does also converge, but this time to $ \frac{1}{\left(1-x\right)^{2}} $.


问:
Expected values normal order statistics
Given this equation pertaining to algorithm AS 177 (from Royston, 1982):

In a sample of size $n$ the expected value of the $r$th largest order statistic is given by
$$E(r,\,n)=\frac{n!}{(r-1)!(n-r)!}\int_{-\infty}^\infty x\{1-\Phi(x)\}^{r-1}\{\Phi(x)\}^{n-r}\phi(x)dx,$$
where $\phi(x)=\frac{1}{\sqrt{2\pi}}\exp(-\tfrac12x^2)$ and $\Phi(x)=\int_{-\infty}^z\phi(z)dz$.

I have been trying to code this (really the expected normal order statistic of the Terry-Hoeffding Test) into Python with some difficulty.
I can code everything up to the integral and I can also get the CDF and PDF functions.
Where I am confused is the $x$ (the actual value?) before the curly brackets and the $dx$ that follows the PDF.

Knowing little of mathematical notation, I am guessing I multiply the first fraction by $x$ times everything in the 2 sets of curly brackets and PDF but that doesn't seem correct. Also everything after "where" is Greek to me (literally and figuratively).

Any help in helping me understand this equation so that I can convert it to code would be greatly appreciated.

答:
This should do it, once you install scipy:

```<code>from scipy.integrate import quad
from scipy.special import binom
from scipy.stats import norm

inf, phi, Phi = float('inf'), norm.pdf, norm.cdf

def E(r, n):
    def f(x):
        F = Phi(x)
        return x*(1-F)**(r-1)*F**(n-r)*phi(x)
    return r*binom(n, r)*quad(f, -inf, inf)[0]
</code>```

See here, here & here.

Title: Expected values normal order statistics

Body:
This should do it, once you install scipy:

from scipy.integrate import quad
from scipy.special import binom
from scipy.stats import norm

inf, phi, Phi = float('inf'), norm.pdf, norm.cdf

def E(r, n):
    def f(x):
        F = Phi(x)
        return x*(1-F)**(r-1)*F**(n-r)*phi(x)
    return r*binom(n, r)*quad(f, -inf, inf)[0]

See here, here & here.


问:
Round Table Combinatorics Problem
I have an issue with two problems with round table.
I have $n$ families, each family have father, mother, kid and a dog.
First problem: I need to sit the families that every kid sits between his parents.
Second problem: I need to sit the families that the two parents sit together and no dogs near each other.

In the first problem I tried to stick the parents and the kid together like this:
I have total of $2n$ places in the table, so to sit the parents and kid between them I have $n!$ options (and multiplies by $2$ because the side the parents sit) and $(n-1)!$ for the dogs to sit.

$$n!\cdot 2\cdot (n-1)!$$
but I'm not really sure about it.

And in the second problem I thought to stick the parents together and the kid and the dog together, but it seems to be wrong because I need the dog the be alone because there can be a possibility that the dog can be between two couples of parents.

Thank you very much for your help!

答:
For the first problem, the factor $2$ applies to each family, so that would give $2^n$.  Presumably the dogs need to go between the families, but dogs can be together so you are ordering $2n$ (families plus dogs).  The result is $2^n(2n-1)!$

For the second, you have $n$ couples, $n$ kids, and $n$ dogs to seat.  You again get a factor $2^n$ for the couples' order.  Then attach a couple or kid to the right of each dog.  You have $2n$ choices for the first dog, $2n-1$ for the second, and so on.  This gives a factor $\frac {(2n)!}{n!}$.  Now you have $2n$ groups to arrange around the circle, which you can do in $(2n-1)!$ ways, for a final result of $2^n\frac {(2n)!(2n-1)!}{n!}$

Title: Round Table Combinatorics Problem


问:
Portfolio choice to maximize expectation of first order statistic among binary random variables
$X_1, X_2, \dots, X_n$ are independently distributed binary random variables. A portfolio is a set of $m < n$ of these.

The value of a portfolio $\{X_{i_1},\dots, X_{i_m}\}$ is $\max(X_{i_1},\dots, X_{i_m})$ (i.e. the first order statistic of the variables in the portfolio).

Is there an efficient algorithm to select from the $n \choose m$ possibilities a  portfolio with the highest expected value (i.e. to maximize the expectation of the first order statistic)?

I've only been able to think of the brute force solution to this optimization problem. It doesn't seem that a greedy approach works. Most of what I've been able to find on first order statistics deals with continuous distributions. And all of the work on portfolio choice that I've found deals with maximizing the expected sum.

答:
This article addresses exactly the question I had posted about:
https://onlinelibrary.wiley.com/doi/abs/10.1111/j.1468-0262.2006.00705.x</p>


问:
Converging sequence of irrationals

Let $x_{0}\in\mathbb{R}$ and $\eta>0$. Find a sequence $\left(x_{n}\right)_{n=1}^{\infty}$ that satisfies the following:

$\forall n\in\mathbb{N} \quad x_{n}\in (x_{0}-\eta,x_{0}+\eta)\setminus\{x_{0}\}$

$\forall n\in\mathbb{N} \quad x_{n}\in \mathbb{R}\setminus\mathbb{Q}$

$\underset{n\rightarrow\infty}{\lim}x_{n}=x_{0}$

Any idea on how to even begin to create such a sequence? I've thought about creating something recursively, but I couldn't find something that assures that I won't get a rational member of the sequence.

答:
Hint:

Create a rational sequence that converges to $x_0$, and add an irrational vanishing term like

$$\frac{1}{\sqrt2\,n\left\lceil\dfrac1\eta\right\rceil}.$$

Title: Converging sequence of irrationals


问:
What is the derivative of $y = \int_{0}^x t f(x^2-t^2)dt$?
Let $y = \int_{0}^x t f(x^2-t^2)dt$, where $f$ is differentiable. How to compute $\frac{dy}{dx}$?

My solution: I think that
$$
\frac{dy}{dx} = \int_{0}^x 2x t f'(x^2 - t^2)dt + tf(0).
$$
Is this correct? Thank you very much.

ContentLicense: CC BY-SA 3.0

Body:
Let $x^2-t^2=m$.

And  $-2tdt=dm$.

$\int_0^x tf(x^2-t^2) dt=\int_{x^2}^0 tf(m) \frac{dm}{-2t}$
$=-\frac{1}{2} \int_{x^2}^0 f(m) dm$

Hence,

$\frac{d}{dx}\int_0^xtf(x^2-t^2) dt=-\frac{1}{2} \frac{d}{dx}  \int_{x^2}^0 f(m) dm=xf(x^2)$


问:
Limit of a converging series
I want to calculate
$$\lim_{n\to\infty}\frac{x_n^3-1}{x_n-1}$$
whereas $$\lim_{n\to\infty}x_n=1.$$ and $x_n≠1$

My idea:
$$\lim_{n\to\infty}\frac{x_n^3-1}{x_n-1}=\lim_{n\to\infty}\frac{(x_n^2+x_n+1)(x_n-1)}{x_n-1}=\lim_{n\to\infty}x_n^2+x_n+1=1^2+1+1=3$$

Does this make any sense? Thanks for any advice!

答:
Your answer is correct but you should add facts like $$\lim_{n \to \infty}(1+x_n+x_n^2) = \lim_{n \to \infty}x_n^2+\lim_{n \to \infty}x_n+\lim_{n \to \infty}1 = (\lim_{n \to \infty}x_n) \cdot(\lim_{n \to \infty}x_n) + \lim_{n \to \infty} x_n + \lim_{n \to \infty} 1
$$

and as $\lim_{n \to \infty}x_n = 1$ the limit is 3

Title: Limit of a converging series


问:
How to calculate the correlation between the number of heads of 100 toss of coin and the number of heads of the first 10 toss of those 100 tosses?
It is a bit cumbersome to explain:

Toss a coin is a Bernoulli distribution, with the probability of seeing a head is p

if we toss this coin 100 times, we should expect $X_{1}$ times of head. Within that 100 toss (this is important, we are NOT tossing another 10 times), we should see $X_{2}$ heads from the first 10 toss.

How to calculate $corr(X_{1}, X_{2})$ ?

The only thing I can think of is, $X_{1} > X_{2}$, practially, we are doing two sets:

toss a coin 10 times, we see $X_{2}$ heads

independently toss a coin 90 times, we see $X_{3}$ heads

we want to calculate $corr(X_{2}, X_{2} + X_{3})$

答:
Let $Y_i$ take value $1$ if toss $i$ gives heads and let it take value $0$ otherwise.

Then to be found is:$$\mathsf{Corr}\left(\sum_{i=1}^{10}Y_i,\sum_{j=1}^{100}Y_j\right)=\frac{\mathsf{Cov}\left(\sum_{i=1}^{10}Y_i,\sum_{j=1}^{100}Y_j\right)}{\sqrt{\mathsf{Var}(\sum_{i=1}^{10}Y_i)}\sqrt{\mathsf{Var}(\sum_{i=1}^{100}Y_i)}}$$

Note that on base of bilinearity of covariance, independence and symmetry: $$\mathsf{Cov}\left(\sum_{i=1}^{10}Y_i,\sum_{j=1}^{100}Y_j\right)=\sum_{i=1}^{10}\sum_{j=1}^{100}\mathsf{Cov}(Y_i,Y_j)=10\mathsf{Cov}(Y_1,Y_1)=10\mathsf{Var}Y_1$$

Further, also on base of independence and symmetry: $$\mathsf{Var}\left(\sum_{i=1}^{10}Y_i\right)=10\mathsf{Var}Y_1$$and: $$\mathsf{Var}\left(\sum_{i=1}^{100}Y_i\right)=100\mathsf{Var}Y_1$$

Leading to answer: $$\frac1{\sqrt{10}}$$

It is not even necessary to calculate $\mathsf{Var}Y_1$.


问:
Notation in Steele's Stochastic Calculus
Steele, Stochastic Calculus and Financial Applications, p. 108, Proposition 7.13:

What does the notation $\tau_{u-}(\omega)$ mean?

答:
$$\tau_{u-}(\omega) := \lim_{t \uparrow u} \tau_t(\omega)$$
where the limit exists since $u \mapsto \tau_u(\omega)$ is increasing.


问:
Show that there's no continuous function that takes each of its values $f(x)$ exactly twice.
I need to prove the following:

There's no continuous function $f:[a,b]\to \mathbb{R}$ that takes each of its values $f(x)$, $x\in [a,b]$  exactly twice.

First of all, I didn't understand the question. For example $x^2$ takes $1$ twice, in the interval $[-1,1]$. Is it saying that it does not occur for all $x$ in the interval? But what about $f(x) = c$? Is it saying that it does not occur only exactly $2$ times, then? I have no idea about how to prove it. I know that for $f(x)$ such that $f(a)<f(x)<f(b)$, if $f$ is continuous then there is a $c\in [a,b]$ such that $f(c) = f(x)$.

Now, there's the following proof in my book and I really wanted to understand it, instead of just getting a new proof

Since the interval $[a,b]$ has only $2$ extreme points, then the maximum or minimum of $f$ must be in a point $c\in int([a,b])$ and and in another point $d\in [a,b]$. Then, there exists $\delta>0$ such that in the intervals $[c-\delta, c), (c,c+\delta)$ (and if $d$ is not extreme of $[a,b]$, $[d-\delta, d]$) the function takes values that are less than $f(c) = f(d)$. Let $A$ be the greatest of the numbers $f(c-\delta), f(c+\delta), f(d-\delta)$. By the intermediate value theorem, there are $x\in [c-\delta, c), y\in (c, c+\delta]$ and $z\in [d-\delta, d)$ such that $f(x)=f(y)=f(z)=A$. Contradiction.

Well, why the last part? Why is it that I can apply the intermediate value theorem to these values? For example, $<f(c-\delta)<p<f(c)$, then by the theorem I know that there exists $m\in [c-\delta, c)$ such that $f(m) = p$. Same for the other intervals. But what guarantees thhat the greatest of the values between  $x\in [c-\delta, c), y\in (c, c+\delta]$ will be inside the intervals $[c-\delta), c), (c,c+\delta), [d-\delta, d)$?

答:
I think the following way may be simpler.

Translation of the function doesnot change the generality. So assume that by translating f(x) we make it crosses the $x$ axis. Let denote it by $\tilde f$

By the hypothesis, $\exists x_1,x_2\in \mathbb R$ such that $\tilde f(x_1)=0=\tilde f(x_2)$ with $x_1<x_2$. Moreover there is no another point $x\in\mathbb R$ that $\tilde f(x)=0$.

So either $\tilde f\le 0$ or $\tilde f\ge 0$  on $[x_1,x_2]$ by extreme value theorem $\tilde f$ takes one of its extreme values at interior of $[x_1,x_2]$, according to its sign. And it is unique so we get contradiction.

Why is one of the extremas on $[x_1,x_2]$ unique?

i) Inductive geometrically: If we were to draw horizontal lines from $x$ axis since the hyphothesis states graph should intersect with these lines exactly twice, until to maxima or minima, there may be no problem but at these extremas the horizontal line can only intersects once. (This explanation is only intuitional, see the figure below)

ii) More formally (we emphasize the idea in formal way to elaborate the idea), WLOG assume $\tilde f\ge 0$ and we have minimum at boundaries, denote first maximum $c_1$. Assume we have another maximum at $x=c_2$. Since $\tilde f$  is continuous at $c_1,c_2$ there exist two closed neighborhood $\delta_1,\delta_2$, now consider $\delta=min\{\delta_1,\delta_2\}$. Observe $\tilde f(\delta_{c_1})=[c^{*}_1, \tilde f(c_1)=M]$ and $\tilde f(\delta_{c_2})=[c^{*}_2,\tilde f(c_2)=M]$

So in the $y$ axis closed interval $[min\{c^{*}_1,c^{*}_2\},M]$ every point corresponds to $4$ different value in $x$ axis.

Edit: About geometric interpretation of uniqueness of one of the extremas.


问:
Symmetric Matrices of $I_{2}$

Find $10$ symmetric matrices $ A = \begin{pmatrix}
a &b \\
 c&d
\end{pmatrix}$ such that $A^{2}=I_{2}$

(I'm going to call matrix A the "square root" of $A^{2}$. If this is the incorrect name for it, may someone please tell me what it is actually called?)

My professor posed this question in class and told us there was an infinite amount of square roots. (Assuming I understood him correctly). However I don't see how there would be many of these, as I was under the impression that a matrix only has one inverse, for $A  A^{-1}=I_{n}$. If someone could tell me if I either misunderstood the professor or if I'm thinking about something incorrectly, please correct me.

My other question is other than blatant guess and check, is there a method to think of these symmetric square roots?

Thanks in advance.

答:
Regarding your question about the inverse, of course $A^2=I$ tells you that $A=A^{-1}$. But for different $A$, no contradiction arises.

Edit: in the first answer I didn't take care to produce symmetric matrices, and was using arbitrary invertible matrices. What one needs here is unitaries (i.e. orthogonal matrices).

Here is one to construct all symmetric real matrices with square $I_2$. Let
$$
A_0=\begin{bmatrix}1&0\\0&-1\end{bmatrix}.
$$
Then $A_0^2=I_2$. Now define
$$
B_t=\begin{bmatrix}t&\sqrt{1-t^2}\\\sqrt{1-t^2}&-t\end{bmatrix}, \ \ \ t\in(-1,1).
$$ and $$
A_t=B_tA_0B_t^{-1}
%=\begin{bmatrix}t&-\sqrt{1-t^2}\\\sqrt{1-t^2}&t\end{bmatrix}\begin{bmatrix}t&\sqrt{1-t^2}\\\sqrt{1-t^2}&-t\end{bmatrix}
=\begin{bmatrix}2t^2-1&2t\sqrt{1-t^2}\\2t\sqrt{1-t^2}&1-2t^2 \end{bmatrix}
$$
Then $A_t\ne A_s$ for any $t,s\in(-1,1)$, and $A_t^2=I_2$ for all $t\in(-1,1)$ (since $A_t^2=B_tA_0B_t^{-1}B_tA_0B_t^{-1}=B_tA_0^2B_t^{-1}=B_tB_t^{-1}=I_2$).

Note that the matrices $B_t$ are all the real rotation matrices. Any symmetric matrix $A$ with $A^2=I_2$ is either the identity or has eigenvalues $1,-1$, and is thus unitarily equivalent to $A_0$.

Title: Symmetric Matrices of $I_{2}$


问:
Cauchy Criterion for Uniform Convergence of Functions
I have proved the following and I would like to know if I have made any mistakes:

A sequence of functions $(f_n)$ defined on a set $A\subseteq\mathbb{R}$ converges uniformly on $A$ if and only if for every $\varepsilon>0$ there exists an $N\in\mathbb{N}$ such that $|f_n(x)-f_m(x)|<\varepsilon$ whenever $m, n\geq N$ and $x\in A$.

My proof (NOTE: edited leftward implication according to answer below):

$\Rightarrow$ Let $\varepsilon>0$: then by hypothesis there exists a limit function $f$ and $N\in\mathbb{N}$ such that $|f_n(x)-f(x)|<\frac{\varepsilon}{2}$ for all $n\geq N$ and $x\in A$ so if we pick $m\geq N$ we have also that $|f_m(x)-f(x)|<\frac{\varepsilon}{2}$ for all $m\geq N$ and $x\in A$ so $|f_n(x)-f_m(x)|=|f_n(x)-f(x)+f(x)-f_m(x)|\leq |f_n(x)-f(x)|+|f(x)-f_m(x)|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$ for all $m,n\geq N$ and $x\in A$, as desired.

$\Leftarrow$ Fix $x\in A$: then $(f_n(x))_{n\in\mathbb{N}}$ is a Cauchy sequence of real numbers which is convergent by the Cauchy Criterion for sequences so the function $f:A\to\mathbb{R}, f(x):= \lim_{n\to\infty} f_n(x)$ is well defined for every $x\in A$.
Now, let $\varepsilon>0$: by hypothesis we have that there is $N\in\mathbb{N}$ such that $|f_n(x)-f_m(x)|<\frac{\varepsilon}{2}$ for all $m,n\geq N$ and $x\in A$ so $|f(x)-f_n(x)|=\lim_{m\to\infty}|f_m(x)-f_n(x)|\leq\frac{\varepsilon}{2}<\varepsilon$ for $n\geq N$ and $x\in A$, as desired.

Thank you.

答:
In the $\Leftarrow$ direction, note that your $N$ may depend on $x$. You write (red additions by me):

So, if $x\in A$ and $\varepsilon>0$, there exists $N\color{red}{(x,\varepsilon)}\in\mathbb{N}$ such that $|f_n(x)-f(x)|<\frac{\varepsilon}{2}$ for all $n\geq N$ (since $\lim_{n\to\infty} f_n(x)=f(x)$ for $x\in A$).

You have to show that there is an $N \in \mathbb N$ that does the trick for any $x \in A$. Let $\varepsilon > 0$, by the uniform Cauchy condition there is $N \in \mathbb N$ such that
$$ \left|f_n(x)- f_m(x) \right| < \varepsilon, \qquad n, m \ge N, x \in A.  $$
(Note that $N$ is independent of $x$). Now let $x \in A$. As $f_m(x) \to f(x)$ for $m \to \infty$, we have
$$ \left|f_n(x)- f(x) \right| = \lim_{m \to \infty} \left|f_n(x)- f_m(x) \right|\le  \varepsilon $$
That is
$$ \left|f_n(x)- f(x) \right| \le \varepsilon, \qquad n \ge N, x \in A.$$


问:
Condition for a hypergraph to be $2$-colorable
In a city, a bus company has several bus lines. Each bus line contains at least $4$ stops and $2$ different bus lines always have exactly $1$ stop in common. Show that all the bus stops in the city can be divided into $2$ groups such that each line has at least $1$ stop in each group.

So the rather obvious way to attack this problem is to consider a graph $G$ where each node corresponds to one bus stop and where each bus line is represented as a connected sub-graph of $G$. Then, I thought it might be interesting to look for "isolated" bus stops (i.e. bus stops that only belong to one line). If a bus line contains at least $2$ such stops then we can put one in each group and we are done for this particular bus line. The problem is that nothing guarantees the existence of such bus stops in any bus line, and even if we did have some it doesn't really solve the problem in the vast majority of cases. I think the way to proceed is to look at the total degree of the graph and isolated degrees of some vertices,  but I don't really know how to proceed.

答:
Let me call the bus lines lines and the bus stops points. The assumptions are:

(1) Each line contains at least $4$ points;

(2) any two distinct lines have exactly one point in common.

We want to prove that the hypergraph formed by the points and lines is $2$-colorable, i.e., we can color each point red or green so that every line contains at least one point of each color.

Condition (1) can be weakened slightly. It won't do to replace it with "every line contains $3$ points"; the Fano plane, which is not $2$-colorable, is a counterexample. However, we can weaken (1) to the following:

(1') Each line contains at least $3$ points; moreover, there is a point $a$ such that every line containg $a$ (and there is at least one) contains at least $4$ points.

Now, if every line contains our special point $a$, we can simply color $a$ red and the rest green. Therefore we may assume that there is a line $L$ which does not contain $a$, as well as a line $M$ which contains $a$. Let $b$ be the unique point commom to $L$ and $M$. Color $a$ red and $b$ green; then color the remaining points in $L$ red and the remaining points in $M$ green.

Now every line which contains neither $a$ nor $b$ already contains a point of each color, since it must meet $L$ and $M$ in some points other than $a$ and $b$. Hence we only have to worry about the lines, other than $L$ and $M$, which contain $a$ or $b$.

Let $V$ be the set of all uncolored points. Let $\mathcal A$ be the set of all lines cther than $M$ which contain $a$; and let $\mathcal B$ be the set of all lines, other than $L$ and $M$, which contain $b$. Note that $\mathcal A\cap\mathcal B=\emptyset$, and that each member of $\mathcal A\cup\mathcal B$ contains at least $2$ uncolored points. (Here we need the assumption that a line containing $a$ contains at least $4$ points.) For each line $K\in\mathcal A\cup\mathcal B$ draw an edge $e_K$ joining two uncolored points in $K$, and let $E$ be the set of all those edges $e_K$. Now the graph $G=(V,E)$ is bipartite (since $E$ is the union of two matchings), so we can color each vertex in $V$ red or green so that each edge $e_k$ joins a red point to a green point.

P.S. Nothing is assumed about finiteness; the number of lines may be finite or infinite, and a line may contain a finite or infinite number of points.


问:
Right Hand Limit of function consisting of inverse trignometric functions
I'm trying to evaluate

$$
\lim \limits_{x \to 0^+} \frac{\sin^{-1}(1 - \{x\}) \times \cos^{-1}(1 - \{x\})}{\sqrt{2\{x\}}\times(1 - \{x\})}
$$

where $\{x\}$ represents the fractional part of the number.

Clearly, for this problem $\{x\} = x$. We can also split up the limit into two parts

$$
\lim \limits_{x \to 0^+} \frac{\sin^{-1}(1 - x)}{1 -x}\times \lim \limits_{x \to 0^+} \frac{\cos^{-1}(1 - x)}{\sqrt{2x}}
$$

The limit on the left is $\frac{\pi}{2}$ which we get upon substituing the $x$ for a zero. The limit on the right is where I'm struggling.

I'm looking for a way to evaluate the limit on the right and/or a different way to approach this problem altogether.

答:
For the limit on right convert the $\cos^{-1}$ to $\sin^{-1}$ to get
$$\lim_{x \to 0^+}\dfrac{\sin^{-1}(\sqrt{1-(1-x)^2})}{\sqrt{2}x}$$

which can be evaluated easily


问:
$z^2 +z|z|+|2z|^2=0$

Find all complex solutions of the equation  $z^2 +z|z|+|2z|^2=0.$

My approach is as follows,

Let $z=rw$ , where $r=|z|$ and $w$ is a unimodular complex number.

The above equation could now be written as $r^2(w^2+w+4)=0$.

If $r=0$ we get $z=0$. Now let $r≠0$. The above equation simplifies to $w^2+w+4=0$.

After solving I got $|w|=2$ thus contadicting the fact that $w$ is unimodular.

Is $z=0$ the only solution or are there more solutions? Please correct me if I made mistakes anywhere.

答:
You could also write
\begin{equation}
4 |z|^2 = |2z|^2= |z^2+z|z||\le|z^2|+|z||z| = 2 |z|^2
\end{equation}
hence $2|z|^2 \le 0$, hence $z=0$.

Title: $z^2 +z|z|+|2z|^2=0$


问:
How to integrate $\int_0^1\frac{dx}{1+x+x^2+\cdots+x^n}$
I am interested in finding a solution to the integral
$$I_n=\int_0^1\frac{dx}{\sum_{k=0}^nx^k}$$
Since the denominator is a geometric series with $a=1$ and $r=x$ and it is within the radius of convergence, we should be able to say
$$\sum_{k=0}^nx^k=\frac{1-x^{n+1}}{1-x}=\frac{x^{n+1}-1}{x-1}$$
and
$$I_n=\int_0^1\frac{x-1}{x^{n+1}-1}dx$$
It makes sense to me that, for all values of $n$, $I_n$ is convergent since the bottom of the function is always above zero and the integral exists for $n\to\infty$ however I cannot seem to find a nice closed form for this.

One thought I did have was using:
$$\sum\ln(x_i)=\ln\left(\prod x_i\right)$$
but I cannot seem to make it work. Does anyone have any hints for this type of problem as I would like to try and complete it myself. Thanks :)

答:
Computing the Integral
$$
\begin{align}
&\int_0^1\frac{1-x}{1-x^{n+1}}\,\mathrm{d}x\\
&=\int_0^1\sum_{k=0}^\infty\left[x^{(n+1)k}-x^{(n+1)k+1}\right]\tag1\\
&=\sum_{k=0}^\infty\left(\frac1{(n+1)k+1}-\frac1{(n+1)k+2}\right)\tag2\\
&=\frac12+\sum_{k=1}^\infty\left(\frac1{(n+1)k+1}-\frac1{(n+1)k+2}\right)\tag3\\
&=\frac12+\frac1{n+1}\sum_{k=1}^\infty{\scriptsize\left(\frac1k-\frac1{k+\frac2{n+1}}\right)}-\frac1{n+1}\sum_{k=1}^\infty{\scriptsize\left(\frac1k-\frac1{k+\frac1{n+1}}\right)}\tag4\\[3pt]
&=\frac12+\frac1{n+1}\left(H\!\left(\frac2{n+1}\right)-H\!\left(\frac1{n+1}\right)\right)\tag5\\[3pt]
&=\frac1{n+1}\sum_{k=1}^n\log\left(2\sin\left(\frac{\pi k}{n+1}\right)\right)\left(\cos\left(\frac{4\pi k}{n+1}\right)-\cos\left(\frac{2\pi k}{n+1}\right)\right)\\
&+\frac1{n+1}\sum_{k=1}^n\left(\frac{\pi}2-\frac{\pi k}{n+1}\right)\left(\sin\left(\frac{2\pi k}{n+1}\right)-\sin\left(\frac{4\pi k}{n+1}\right)\right)\tag6
\end{align}
$$
Explanation:

$(1)$: expand the Taylor series

$(2)$: integrate

$(3)$: pull the $k=0$ term out front

$(4)$: rearrange into two series

$(5)$: Write as extended Harmonic numbers

$(6)$: apply $(7)$ from this answer

Although $(5)$ is always valid, since $(7)$ from this answer requires $p\le q$, $(6)$ is only valid for $n\gt0$. For $n=0$, the integral is $1$.

Mathematica Implementation

Here is a Mathematica implementation of $(6)$:

```<code>how[n_]:=1/(n + 1)Sum[
    Log[2 Sin[Pi k/(n+1)]](Cos[4 Pi k/(n+1)]-Cos[2 Pi k/(n+1)])
    + (Pi/2-Pi k/(n+1))(Sin[2 Pi k/(n+1)]-Sin[4 Pi k/(n+1)]),
    {k,1,n}]
</code>```

Body:
Computing the Integral
$$
\begin{align}
&\int_0^1\frac{1-x}{1-x^{n+1}}\,\mathrm{d}x\\
&=\int_0^1\sum_{k=0}^\infty\left[x^{(n+1)k}-x^{(n+1)k+1}\right]\tag1\\
&=\sum_{k=0}^\infty\left(\frac1{(n+1)k+1}-\frac1{(n+1)k+2}\right)\tag2\\
&=\frac12+\sum_{k=1}^\infty\left(\frac1{(n+1)k+1}-\frac1{(n+1)k+2}\right)\tag3\\
&=\frac12+\frac1{n+1}\sum_{k=1}^\infty{\scriptsize\left(\frac1k-\frac1{k+\frac2{n+1}}\right)}-\frac1{n+1}\sum_{k=1}^\infty{\scriptsize\left(\frac1k-\frac1{k+\frac1{n+1}}\right)}\tag4\\[3pt]
&=\frac12+\frac1{n+1}\left(H\!\left(\frac2{n+1}\right)-H\!\left(\frac1{n+1}\right)\right)\tag5\\[3pt]
&=\frac1{n+1}\sum_{k=1}^n\log\left(2\sin\left(\frac{\pi k}{n+1}\right)\right)\left(\cos\left(\frac{4\pi k}{n+1}\right)-\cos\left(\frac{2\pi k}{n+1}\right)\right)\\
&+\frac1{n+1}\sum_{k=1}^n\left(\frac{\pi}2-\frac{\pi k}{n+1}\right)\left(\sin\left(\frac{2\pi k}{n+1}\right)-\sin\left(\frac{4\pi k}{n+1}\right)\right)\tag6
\end{align}
$$
Explanation:

$(1)$: expand the Taylor series

$(2)$: integrate

$(3)$: pull the $k=0$ term out front

$(4)$: rearrange into two series

$(5)$: Write as extended Harmonic numbers

$(6)$: apply $(7)$ from this answer

Although $(5)$ is always valid, since $(7)$ from this answer requires $p\le q$, $(6)$ is only valid for $n\gt0$. For $n=0$, the integral is $1$.

Mathematica Implementation

Here is a Mathematica implementation of $(6)$:

how[n_]:=1/(n + 1)Sum[
    Log[2 Sin[Pi k/(n+1)]](Cos[4 Pi k/(n+1)]-Cos[2 Pi k/(n+1)])
    + (Pi/2-Pi k/(n+1))(Sin[2 Pi k/(n+1)]-Sin[4 Pi k/(n+1)]),
    {k,1,n}]


答:
Let $z=x+iy$, $|z|=r=\sqrt{x^2+y^2}$ then we get $(x+iy)*2+(xr+iry)+4r^2=0$, separating real and imaginary parts we get $x^2-y^2+rx+4r^2=0$ and $2xy=ry \implies y=0$ or $y=\pm x\sqrt{3}.$

Case 1: $y=0$ gives $x^2-x^2+4x^2=0 \implies x=0$ so $z=0$ is one root.

Case 2: $y=+\sqrt{3}x$ gives $x=0$ so again $z=0$

Case 3: Will again give $z=0$.

So the given equation has just one solution $z=0$.

Title: $z^2 +z|z|+|2z|^2=0$

Body:
Find all complex solutions of the equation  $z^2 +z|z|+|2z|^2=0.$

My approach is as follows,

Let $z=rw$ , where $r=|z|$ and $w$ is a unimodular complex number.

The above equation could now be written as $r^2(w^2+w+4)=0$.

If $r=0$ we get $z=0$. Now let $r≠0$. The above equation simplifies to $w^2+w+4=0$.

After solving I got $|w|=2$ thus contadicting the fact that $w$ is unimodular.

Is $z=0$ the only solution or are there more solutions? Please correct me if I made mistakes anywhere.


问:
Finding The Value Of Infinite Nested Radicals
The problem is to find the value of,

$\sqrt{-1+1\sqrt{-2+2\sqrt{-3+3\sqrt{-4+4\sqrt{...}}}}}$

Even though I have solved the problems which have a definite pattern which repeats itself and we make some substitution to get a quadratic which gives the final answer but I can't think of a way to follow that over here. Can anyone share the approach for this problem?

Note: I just came up with this problem (maybe it already exists) but then I wanted to know how could one solve such problems seeing it for the first time. I have verified it for some values and I believe the problem statement is correct. The method of generating this expression has been provided in the comments.

答:
OK.  My reaction based on your explanation in the comments.

Do not say
$$
1 = \sqrt{-1+1\sqrt{-2+2\sqrt{-3+3\sqrt{-4+4\sqrt{\dots}}}}}
\tag1$$
Instead say: one solution of the recurrence
$$
a_n^2 = -n+na_{n+1}
\tag2$$
is $a_n =n$.  The reason for saying it this way is that any other solution of the recurrence has just as much claim to be called the "value" of the infinite nested radical as this solution has.

Note that given any number $r$, there is a solution $(a_n)$ of the recurrence $(2)$ with $a_1 = r$.

Another, more conventional, interpretation of the problem would ask for the limit (if it exists) for the sequence
\begin{align}
a_1 &= \sqrt{-1}
= 0.0000000000 + 1.0000000000 i
\\
a_2 &=\sqrt{-1+1\sqrt{-2}}
=0.6050003336 + 1.168770894 i
\\
a_3 &= \sqrt{-1+1\sqrt{-2+2\sqrt{-3}}}
=0.9306048592 + 0.9306048592 i
\\
&\qquad\vdots
\end{align}
Always taking the principal square root.  Numerically, it seems this converges to $1$.  Can this be proved?


问:
$\chi(A\cup B)=\chi(A)+\chi(B)-\chi(A\cap B)$
I am trying to do an exercise that is the following :

Suppose that $A\cup B$ is a compact $n-$dimensional manifold such that $A,B\subset A\cup B $ are compact $n$-dimensional submanifolds and $A\cap B$ is a compact $n-1$ dimensional submanifold. Show that $$\chi(A\cup B)=\chi(A)+\chi(B)-\chi(A\cap B)$$

Now first since there are many definitions for $\chi(A\cup B)$, I will say with which one I am working with . Let $f:M\rightarrow TM$ be a section transverse to the zero section. Then for each $x\in M$ we have a canonical identification $T_{(x,0)}(TM)\cong T_x M$ and so $Tf$ induces an automorphism $\phi_x$ from $T_xM$ to $T_x M$ and we define $Ind_x f=\frac{Det \phi_x}{|Det \phi_x|}$.

The way we compute $\chi(A\cup B)$ is just to take a transversal section to the zero section and compute the $Ind_x f$, and then we can use this $f$ also to compute $\chi(A)$ and $\chi(B)$. Now for $\chi(A\cap B)$ I am not interely sure what section to use that is related to $f$.

Any help is aprecciated.

(I didn't include the definition for manifolds with boundary because I am trying to treat this simpler case fist without having to worry about outward pointing vector fields).

答:
As written, your question is ill-posed since you did not define what you mean by index for manifolds with boundary. One such definition can be found in the paper

Benoît Jubin, A generalized Poincaré-Hopf index theorem,     arXiv:0903.0697, 2009.

With the definition (and basic results) given there, the proof of the equality
$$
\chi(A\cup B)= \chi(A)+ \chi(B) - \chi(A\cap B)
$$
is straightforward, you use the same arguments as in the proof of Theorem 12 in that paper. (Isotope the vector fields on $A, B$ to make them equal on the common boundary and tangent to the common boundary and then count zeroes with sign.)


问:
How to solve the equation $x^2-2y=z^2$?
Consider the following equation
$$x^2-2y=z^2,$$
according some theorems in my research, I found that the only integer solution of the equation such that $xy\neq 0$ is $$(x,y,z)=(2,2,0).$$ Now my question is: how to solve the equation (or what is the way or method)? Anyone can help me. Thanks in advance(
I tried to rewrite the equation as follows
\begin{align}
2y=x^2-z^2 \Rightarrow 2y=(x-z)(x+z)\Rightarrow y=\frac{(x-z)(x+z)}{2}
\end{align}
but it did not work).

Title: How to solve the equation $x^2-2y=z^2$?

Body:
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did only a quick search with the following bounds: $-10\le x\le10$, $-10\le y\le10$, $-10\le z\le10$, and $x$, $y$, $z$ are all integers with $xy\ne0$.

I wrote and ran some Mathematica-code:

In[1]:=Clear["Global`*"];
FullSimplify[
 Solve[{x^2 - 2*y == z^2,
   x*y != 0 && -10 <= x <= 10 && -10 <= y <= 10 && -10 <= z <=
     10}, {x, y, z}, Integers]]

Running the code gives:

Out[1]={{x -> -6, y -> 10, z -> -4}, {x -> -6, y -> 10, z -> 4}, {x -> -5,
  y -> 8, z -> -3}, {x -> -5, y -> 8, z -> 3}, {x -> -4, y -> -10,
  z -> -6}, {x -> -4, y -> -10, z -> 6}, {x -> -4, y -> 6,
  z -> -2}, {x -> -4, y -> 6, z -> 2}, {x -> -4, y -> 8,
  z -> 0}, {x -> -3, y -> -8, z -> -5}, {x -> -3, y -> -8,
  z -> 5}, {x -> -3, y -> 4, z -> -1}, {x -> -3, y -> 4,
  z -> 1}, {x -> -2, y -> -6, z -> -4}, {x -> -2, y -> -6,
  z -> 4}, {x -> -2, y -> 2, z -> 0}, {x -> -1, y -> -4,
  z -> -3}, {x -> -1, y -> -4, z -> 3}, {x -> 1, y -> -4,
  z -> -3}, {x -> 1, y -> -4, z -> 3}, {x -> 2, y -> -6,
  z -> -4}, {x -> 2, y -> -6, z -> 4}, {x -> 2, y -> 2,
  z -> 0}, {x -> 3, y -> -8, z -> -5}, {x -> 3, y -> -8,
  z -> 5}, {x -> 3, y -> 4, z -> -1}, {x -> 3, y -> 4,
  z -> 1}, {x -> 4, y -> -10, z -> -6}, {x -> 4, y -> -10,
  z -> 6}, {x -> 4, y -> 6, z -> -2}, {x -> 4, y -> 6,
  z -> 2}, {x -> 4, y -> 8, z -> 0}, {x -> 5, y -> 8,
  z -> -3}, {x -> 5, y -> 8, z -> 3}, {x -> 6, y -> 10,
  z -> -4}, {x -> 6, y -> 10, z -> 4}}

So, using these bounds I get $36$ number of solutions. Extending the bounds to $-10^3$ and $10^3$ gives $12716$ number of solutions. Extending the bounds, again, to $-10^4$ and $10^4$ gives $173364$ number of solutions.


Title: Right Hand Limit of function consisting of inverse trignometric functions

Body:
I'm trying to evaluate

$$
\lim \limits_{x \to 0^+} \frac{\sin^{-1}(1 - \{x\}) \times \cos^{-1}(1 - \{x\})}{\sqrt{2\{x\}}\times(1 - \{x\})}
$$

where $\{x\}$ represents the fractional part of the number.

Clearly, for this problem $\{x\} = x$. We can also split up the limit into two parts

$$
\lim \limits_{x \to 0^+} \frac{\sin^{-1}(1 - x)}{1 -x}\times \lim \limits_{x \to 0^+} \frac{\cos^{-1}(1 - x)}{\sqrt{2x}}
$$

The limit on the left is $\frac{\pi}{2}$ which we get upon substituing the $x$ for a zero. The limit on the right is where I'm struggling.

I'm looking for a way to evaluate the limit on the right and/or a different way to approach this problem altogether.

Body:
You can use L'Hopital:$$\lim_{x\to0^+}\frac{\cos^{-1}(1-x)}{\sqrt{2x}}=\sqrt2\lim_{x\to0^+}\frac{\sqrt{x}}{\sqrt{1-(1-x)^2}}=1$$

You can keep $\cos^{-1}(1-x)=m$. Then $\cos m=1-x$,$$\lim_{m\to0^+}\frac{m}{\sqrt{2(1-\cos m)}}=\lim_{m\to0^+}\frac{m}{\sqrt{4\sin^2(m/2)}}=1$$


问:
Evalutate $\int\frac{dx}{x\sqrt{1-\frac{4}{3}x^4}}$
How to integrate
$$\int\frac{dx}{x\sqrt{1-\frac{4}{3}x^4}}$$
Mathematica found
$$\ln x-\frac{1}{2}\ln\left(1+\sqrt{1-\frac{4}{3}x^4}\right)$$
but I can't find a method to arrive at this solution.</p>

答: Let $ x^2 = \frac{\sqrt{3}\sin y}{2} $ and hence the integral is equivalent to $$\begin{align*} \displaystyle \int \frac{\cos y}{2\sin y \cos y} &= \frac{1}{2}\left(\ln(1 - \cos y) - \ln \sin y\right) + C\\&= \frac{1}{2}\left(\ln\left(1 - \sqrt{1 - \frac{4}{3}x^4}\right) - \ln\left(\frac{2}{\sqrt3}x^2\right)\right) + C \\&=\frac{1}{2}\left(\ln\left(\frac{1 - \sqrt{1 - \frac{4}{3}x^4}}{x^2}\right)\right) + C - \frac{1}{2}\ln \frac{2}{\sqrt{3}} \\&= \frac{1}{2}\left(\ln\left(\frac{\frac{4}{3}x^4}{x^2\left(1 + \sqrt{1 - \frac{4}{3}x^4}\right)}\right)\right) + C - \frac{1}{2}\ln \frac{2}{\sqrt{3}} \\&=\ln x - \frac{1}{2}\ln\left(1 + \sqrt{1 - \frac{4}{3}x^4}\right) + C + \frac{1}{2} \ln \frac{2}{\sqrt{3}}  \end{align*}$$</p>


答:
The generalized delta is just an antisymmetrizer. You just need to use the commutation identity that you wrote at the end of the first displayed equation to slide the $A$'s through the antisymmetrizer.

Edit with a few more details:

I said the generalized delta
$$
\delta^{i_1,\ldots, i_k}_{j_1,\ldots, j_k}=\sum_{\sigma\in S_k} {\rm sgn}(\sigma)
\delta^{i_{\sigma(1)}}_{j_1}\cdots \delta^{i_{\sigma(k)}}_{j_k}
$$
is an antisymmetrizer because if you contract it to a general tensor $T_{i_1,\ldots,i_k}$
and this way define a new tensor
$$
W_{i_1,\ldots,i_k}:= \delta^{i_1,\ldots, i_k}_{j_1,\ldots, j_k}\ T_{j_1,\ldots,j_k}
$$
(I used Einstein's convention, but I don't care about upstairs/downstairs)
what you get is an antisymmetric tensor, i.e., $W_{i_1,\ldots,i_k}$ changes sign if any two indices are switched.

Now the commutation, or sliding of $A$'s through the antisymmetrizer is the identity
$$
\delta^{i_1,\ldots, i_k}_{j_1,\ldots, j_k}A^{j_1}_{h_1}\cdots A^{j_k}_{h_k}=
A^{i_1}_{j_1}\cdots A^{i_k}_{j_k}\delta^{j_1,\ldots, j_k}_{h_1,\ldots, h_k}\ .
$$

As for the proof of the wanted identity, one can start as in the OP (a bad move that needs fixing)
$$
\frac{1}{k!} A^{i_1,\ldots,i_k}_{h_1,\ldots,h_k} (A^{-1})^{h_1,\ldots,h_k}_{j_1,\ldots,j_k}=
\frac{1}{k!} \left(
A^{i_1}_{p_1}\cdots A^{i_1}_{p_k}\delta^{p_1,\ldots, p_k}_{h_1,\ldots, h_k}\right)
\left(\delta^{h_1,\ldots, h_k}_{q_1,\ldots, q_k}(A^{-1})^{q_1}_{j_1}\cdots (A^{-1})^{q_k}_{j_k}\right)
$$
but the $A$'s and $A^{-1}$'s are on the wrong side of their attached generalized delta. So use the commutation twice to rewrite this as
$$
\frac{1}{k!}
\left(\delta^{i_1,\ldots, i_k}_{p_1,\ldots, p_k}A^{p_1}_{h_1}\cdots A^{p_k}_{h_k}\right)
\left((A^{-1})^{h_1}_{q_1}\cdots (A^{-1})^{h_k}_{q_k}\delta^{q_1,\ldots, q_k}_{j_1,\ldots, j_k}\right)
$$
$$
=\frac{1}{k!}\delta^{i_1,\ldots, i_k}_{p_1,\ldots, p_k}
(A^{p_1}_{h_1}(A^{-1})^{h_1}_{q_1})\cdots
(A^{p_k}_{h_k}(A^{-1})^{h_k}_{q_k})
\delta^{q_1,\ldots, q_k}_{j_1,\ldots, j_k}
$$
$$
=\frac{1}{k!}\delta^{i_1,\ldots, i_k}_{p_1,\ldots, p_k}
\delta^{p_1}_{q_1}\cdots\delta^{p_k}_{q_k}
\delta^{q_1,\ldots, q_k}_{j_1,\ldots, j_k}
=\frac{1}{k!}\delta^{i_1,\ldots, i_k}_{p_1,\ldots, p_k}
\delta^{p_1,\ldots, p_k}_{j_1,\ldots, j_k}
=\delta^{i_1,\ldots, i_k}_{j_1,\ldots, j_k}\ .
$$

Edit with even more details:

For completeness, let me prove the identity about the last two contracted generalized deltas giving a single one.

By definition, and summing over permutation $\sigma,\tau$ in the symmetric group $S_k$,
$$
\delta^{i_1,\ldots,i_k}_{p_1,\ldots,p_k}\delta^{p_1,\ldots,p_k}_{j_1,\ldots,j_k}=
\sum_{\sigma,\tau}{\rm sgn}(\sigma){\rm sgn}(\tau)
\delta^{i_{\sigma(1)}}_{p_1}\cdots\delta^{i_{\sigma(k)}}_{p_k}
\ \delta^{p_{\tau(1)}}_{j_1}\cdots\delta^{p_{\tau(k)}}_{j_k}
$$
But, by permuting factors,
$$
\delta^{p_{\tau(1)}}_{j_1}\cdots\delta^{p_{\tau(k)}}_{j_k}=
\delta^{p_1}_{j_{\tau^{-1}(1)}}\cdots\delta^{p_k}_{j_{\tau^{-1}(k)}}\ .
$$
We insert this equality and do the summation over the $p$ indices and get
$$
\delta^{i_1,\ldots,i_k}_{p_1,\ldots,p_k}\delta^{p_1,\ldots,p_k}_{j_1,\ldots,j_k}=
\sum_{\sigma,\tau}{\rm sgn}(\sigma){\rm sgn}(\tau)
\delta^{i_{\sigma(1)}}_{j_{\tau^{-1}(1)}}\cdots\delta^{i_{\sigma(k)}}_{j_{\tau^{-1}(k)}}
$$
$$
=\sum_{\sigma,\tau}{\rm sgn}(\sigma){\rm sgn}(\tau)
\delta^{i_{\sigma(\tau(1))}}_{j_1}\cdots\delta^{i_{\sigma(\tau(k))}}_{j_k}
$$
by reordering the factors again. To finish, use
${\rm sgn}(\sigma){\rm sgn}(\tau)={\rm sgn}(\sigma\tau)$ and notice that each permutation $\rho=\sigma\tau$ appears exactly $k!$ times.

Title: $\frac{1}{k!}\sum_{h_1,...,h_k=1}^NA^{i_1,...,i_k}_{h_1,...,h_k}\big(A^{-1}\big)^{h_1,...,h_k}_{j_1,...,j_k}=\delta^{i_1,...,i_k}_{j_1,...,j_k}$

Body:
ACHTUNG

TO FOLLOWS I USE ENSTEIN CONVENTION TO WROTE DETERMINANTS!

Definition

If $A$ is an $n\times n$ matrix we define its $k\times k$ minotr to be the determinant of $k\times k$ submatrix of $A$, e.g.
$$
A^{i_1,...,i_k}_{j_1,...,j_k}:=\det\begin{bmatrix}A_{j_1}^{i_1}\ &\cdots&A_{j_k}^{i_1}\ \\\vdots&\ddots&\vdots\\A_{j_1}^{i_k}&\cdots&A_{j_k}^{i_k}\end{bmatrix}=\delta^{h_1,...,h_k}_{j_1,...,j_k}A^{i_1}_{h_1}...A^{i_k}_{h_k}=\delta^{i_1,...,i_k}_{h_1,...,h_k}A^{h_1}_{j_1}...A^{h_k}_{j_k}
$$
where the $\delta$-symbol is the generalised Kronecker Delta as here defined.

So I ask to prove that
$$
\frac{1}{k!}\sum_{h_1,...,h_k=1}^NA^{i_1,...,i_k}_{h_1,...,h_k}\big(A^{-1}\big)^{h_1,...,h_k}_{j_1,...,j_k}=\delta^{i_1,...,i_k}_{j_1,...,j_k}
$$
for any non-singular matrix. In particular using the above definition I calculate that
$$
\frac{1}{k!}\sum_{h_1,...,h_k=1}^NA^{i_1,...,i_k}_{h_1,...,h_k}\big(A^{-1}\big)^{h_1,...,h_k}_{j_1,...,j_k}=\\
\frac{1}{k!}\sum_{h_1,...,h_k=1}^N\Big(\delta^{l_1,...,l_k}_{h_1,...,h_k}A^{i_1}_{l_1}...A^{i_k}_{l_k}\Big)\Big(\delta^{h_1,...,h_k}_{m_1,...,m_k}\big(A^{-1}\big)^{m_1}_{j_1}...\big(A^{-1}\big)^{m_k}_{j_k}\Big)=\\
\frac{1}{k!}\sum_{h_1,...,h_k=1}^N\delta^{l_1,...,l_k}_{h_1,...,h_k}\delta^{h_1,...,h_k}_{m_1,...,m_k}A^{i_1}_{l_1}...A^{i_k}_{l_k}\big(A^{-1}\big)^{m_1}_{j_1}...\big(A^{-1}\big)^{m_k}_{j_k}=\\
\frac{1}{k!}\sum_{h_1,...,h_k=1}^N\delta^{l_1,...,l_k}_{m_1,...,m_k}A^{i_1}_{l_1}\big(A^{-1}\big)^{m_1}_{j_1}...A^{i_k}_{l_k}\big(A^{-1}\big)^{m_k}_{j_k}
$$
but unfortunately I do not be able to conclude anything. So could someone help me, please?


问:
Prove that: if $\pi(m)\pi(m-2)-\pi^2(m-1)=0$, then $m \ne prime$
This is not a great way of testing for a prime number, $P_n$.

Where $\pi(m)$ is the prime counting function.

Let $m=2k+1$ where $k=1,2,3,4,...$

if $\pi(m)\pi(m-2)-\pi^2(m-1)=0$, then $m$ is not a prime number.$\tag1 $
if $\pi(m)\pi(m-2)-\pi^2(m-1)=n$, then $m$ is a prime number. $\tag2$
where $P_{n+1}=m$

How can you show that this conjecture, $(1)$ is correct?

examples:

$\pi(m)\pi(m-2)-\pi^2(m-1)=0$, $m \ne prime$

$\pi(15)\pi(13)-\pi^2(14)=0$, $15 \ne prime$

$\pi(25)\pi(23)-\pi^2(24)=0$, $25 \ne prime$

$\pi(123)\pi(123)-\pi^2(122)=0$, $123 \ne prime$

Of course if this conjecture $(1)$ is correct, we wouldn't use it to test for prime number, that would be madness!

答:
Proof by contradiction: Per your starting conditions, $m$ is odd. Assume $m=p_n$ is the $n$th prime. Let's look first at the case $m>3$. Then $\pi(m)=n\ge 3$ but $\pi(m-1)=\pi(m-2)=n-1$ because there is one fewer prime ($p_n$ itself), but no more than one fewer prime. The next smaller prime must be $\le p_n-2$, which is not eliminated from the count $\pi(m-2)$.

$n\ge 3 \Rightarrow n(n-1)-(n-1)^2\ne 0$.

So if $m>3$ is prime, the difference of products cannot be $0$

The case $m=3$ is readily excluded by examination, since $\pi(1)=0$.


答: Your rewrite to $y=\frac 12(x-z)(x+z)$ is exactly what you want.  You need $x$ and $z$ to have the same parity (both even or both odd) so the factors are even and the division by $2$ works.  Then you can choose any $x,z$ pair and compute $y$.  If you want positive integers, you must have $x \gt z$.  For example $(3,4,1)$ works.

Title: How to solve the equation $x^2-2y=z^2$?

Body:
Consider the following equation
$$x^2-2y=z^2,$$
according some theorems in my research, I found that the only integer solution of the equation such that $xy\neq 0$ is $$(x,y,z)=(2,2,0).$$ Now my question is: how to solve the equation (or what is the way or method)? Anyone can help me. Thanks in advance(
I tried to rewrite the equation as follows
\begin{align}
2y=x^2-z^2 \Rightarrow 2y=(x-z)(x+z)\Rightarrow y=\frac{(x-z)(x+z)}{2}
\end{align}
but it did not work).


问:
I don't imagine how to make a cartesian equation of the parametric one: $x=2 \sin (t), y=-\frac{4\cos (t)^2 (2 + 1.3\cos(t))}{3 + \sin (t)^2}$
I know it's a downvoting question but I'm not good at math. I know that $x=a \sin(t)$ , $y=\frac{a \cos^2(t)(2+ \cos(t))}{(3- \sin^2(t))}$ is $y^2(a^2-x^2)=(x^2+2ay-a^2)^2$ but what it could be as a cartesian equation with the negative $y$ part I can't even imagine. Sorry. Could you help me?

答:
$$x=2 \sin (t), ~~~~y=-\frac{4\cos (t)^2 (2 + 1.3\cos(t))}{3 + \sin (t)^2}$$
There's a very useful trigonometric identity:
$$\sin^2\theta+\cos^2\theta\equiv1$$
where the $\equiv$ symbol (the identity symbol) means the equation is true for all values of $\theta$; that's why it's called an identity. In your case, we know that
$$\sin t=\frac{x}{2}$$
Using the above identity,
$$\cos ^2t=1-\sin^2 t=1-\frac{x^2}{4}=\frac{4-x^2}{4}$$
Also,
$$\cos t=\sqrt{\frac{4-x^2}{4}}=\frac{\sqrt{4-x^2}}{2}$$
(The final equality assumes that $\cos t$ is $+$ve. In fact to be completely correct you should use $\cos t=\pm\frac{\sqrt{4-x^2}}{2}$. Thanks for spotting that @GEdgar.)

Try substituting these values into your equation. If you need more help, don;t hesitate to ask!


问:
Group homomorphisms into a field
Let $G$ be a finite group, and let $k$ be a field, which should be algebraically closed, I think. How to describe all homomorphisms $G\rightarrow k^*$ (i.e. one-dimensional representations: $k^*=\mathrm{GL}_1$) or just find the number of them?

I've got the following:

First, we assume $G$ is abelian. Since each finitely generated abelian group is a product of cyclic groups ($C_{k}$ or $\mathbb{Z}$), $G=C_{k_1}\times C_{k_2}\times\cdots \times C_{k_l}$. Let $n=|G|=k_1 k_2\cdots k_l$. So, for each homomorphism $\phi: G\rightarrow k^*$  we have $\phi(G)\subset Z(t^n-1)=H$ (I mean zeros of $t^n-1\in k[t]$).

Since each finite subgroup of $k^*$ is cyclic (let $H=C_m$) we are to investigate homomorphisms $$C_{k_1}\times C_{k_2}\times\cdots\times C_{k_l}\rightarrow C_m.$$ Is it necessarily the case that $m=n$? I know that in $\mathbb{Z}/p\mathbb{Z}$, we have $Z(x^p-1)=\{1\}$, so $m\ne n$. But how about algebraically closed fields?

However that may be, are there any other ways to describe such homomorphisms? To describe homomorphisms $C_{k_1}\times C_{k_2}\times\cdots\times C_{k_l}\rightarrow C_m$?

In conclusion, let me note that the case of nonabelian $G$ reduces to abelian one.

答:
Every homomorphism $C_{k_1}\times \cdots \times C_{k_l}\to k^\times$ factors as a product of homomorphism $C_{k_{\large i}}\to k^\times$, since the original is determined by where it sends the coordinates' generators. Each such coordinate homomorphism may be chosen independently. In particular the image of the full homomorphism is at most the $\ell={\rm lcm}(k_1,\cdots,k_l)$th roots of unity in $k^\times$ (which exist because $k$ is alg. closed), and it is indeed possible to obtain this as a surjective image.

To see that $\mu_\ell$, the $\ell$th roots of unity in $k$, are an "upper bound" for the image, note that everything in the image must be $\ell$-torsion since everything in $G$ is $\ell$-torsion itself. To see that this bound can be reached, it is simply a matter to see that the subgroups of $C_{\large \ell}$ of orders $k_i$ ($1\le i\le l$) are a generating set for the entire cyclic group $C_{\large\ell}$ (nice elementary number theory exercise).

The full collection of homomorphisms into $k^\times$ is itself a group under pointwise multiplication, called the dual group, and it is (noncanonically) isomorphic to the original finite abelian group.

If the original group $G$ is nonabelian then the homomorphisms factor through the abelianization $G^{\rm ab}:=G/[G,G]$, as you allude to, and there is no nice dual group for the full nonabelian group.

Title: Group homomorphisms into a field


问:
Treating Differentials like Fractions
Why does treating differential elements like fractions work so well? I studied calculus from a pure math perspective first, and have only basic knowledge of it, so can someone explain with a rigorous proof or a simple explanation why this hack works? Some people told me that treating differentials like fractions is a sin. Can anyone give me an example where treating differentials like fractions fails to work?</p>

答:
Physicists do this in more than one context, so a single answer might not be possible that covers all uses. I think you were poorly served by having this migrated to Math SE as some of these uses are related purely to how the physics is modeled and not to the math, although some are mathematical.  Some important cases:

Physics ultimately has to connect to something you can measure in a laboratory or in nature, and you can never measure an infinitesimal $dx$ directly, nor do you typically measure something that you represent by a derivative ("ratio of infinitesimals") directly.  What you actually measure is one or more finite values and then interpolate or extrapolate.  So an expression like $dx = v\ dt$ is short-hand for $\Delta x \approx v \Delta t$ for finite $\Delta x$ and finite $\Delta t$ and that approximation gets better and better as you make finer and finer resolution measurements.  The finite values $\Delta x$ and $\Delta t$ are regular numbers that you can move around using regular arithmetic.  It's essentially part of the theory that it doesn't matter a which point you take the limits because the theory asserts something about the extrapolated behavior of finite measurements.

The physicist accepts that theory may break down as some parameters of the model get big, get small, or take special values, but the physicist keeps writing the equal sign anyway.  For example, the physicist doesn't refuse to write $F = dp/dt$ because we know there are higher order terms from relativity or that there are issues with this representation of momentum in quantum mechanics.  The physicist separately keeps track of the fact that the theory will break down and where the bounds for that break might be.

For more purely mathematical operations, a lot of times whatever is done with infinitesimal symbols is ultimately going to be either used as a "ratio" to form the derivative or going to be put under an integral sign.  There, although it abuses notation, the chain rule or integration by parts saves many of the manipulations that look like they are treating the infinitesimal symbols like numbers and, e.g., canceling parts of fractions. (See more here: https://physics.stackexchange.com/questions/572956/the-usage-of-chain-rule-in-physics/573232#573232)  The chain rule and integration by parts, are, of course rigorously understood.

Title: Treating Differentials like Fractions


问:
Triple summation verification $\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}$
I am attempting the following triple summation. It would be great if someone would verify whether what I've done is correct.

$$\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}$$

$$\begin{aligned}\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}&=\sum_{i=1}^{\infty}\sum_{j=i+1}^{\infty}\sum_{k=j+1}^{\infty}\frac{1}{2^i3^j5^k}\\&^= \sum_{i=1}^{\infty}\sum_{j=i+1}^{\infty}\frac{1}{2^i3^j}\frac{1/5^{j+1}}{4/5}\\&=\frac{1}{4}\sum_{i=1}^{\infty}\frac{1}{2^i}\frac{1/15^{i+1}}{14/15}\\&=\frac{1}{56}\sum_{i=1}^{\infty}\frac{1}{30^{i}}=\frac{1}{56}\cdot\frac{1}{29}\end{aligned}$$

答: Wolfram Alpha yields $\dfrac{1}{1624}$, sounds like you are correct.


问:
$\lim_{x \to 0}{\frac{\sinh(x)-\sin(x)}{x(\cosh(x)-\cos(x))}}$
As stated in the title.

My attempt, begin with L'Hopital:
$$L=\lim_{x \to 0}{\frac{\sinh(x)-\sin(x)}{x(\cosh(x)-\cos(x))}}=\lim_{x \to 0}{\frac{\cosh(x)-\cos(x)}{(\cosh(x)-\cos(x))+x(\sinh(x)+\sin(x))}}$$
Dividing by the numerator
$$\lim_{x \to 0}\frac{1}{1+\left(\frac{x\left(\sinh(x)+\sin(x)\right)}{\cosh(x)-\cos(x)}\right)}=\frac{1}{1+\lim_{x \to 0}{\left(\frac{x\left(\sinh(x)+\sin(x)\right)}{\cosh(x)-\cos(x)}\right)}}$$
L'Hopital again
$$\frac{1}{1+\lim_{x \to 0}{\left(\frac{x(\cosh(x)+\cos(x))+(\sinh(x)+\sin(x))}{\sinh(x)+\sin(x)}\right)}}$$
Diving through
$$\frac{1}{1+\lim_{x \to 0}{\left(1+\frac{\cosh(x)+\cos(x)}{\left(\frac{\sinh(x)}{x}+\frac{\sin(x)}{x} \right)}\right)}}=\frac{1}{1+(1+\frac{1+1}{1+1})}=\frac{1}{3}$$

Is this correct, and is there a more elegant way of doing it?

答:
Hint: In fact,
\begin{eqnarray}
L&=&\lim_{x \to 0}{\frac{\sinh(x)-\sin(x)}{x(\cosh(x)-\cos(x))}}\\
&=&\lim_{x \to 0}{\frac{\sinh(x)-\sin(x)}{x^3}}\cdot\lim_{x \to 0}{\frac{x^2}{\cosh(x)-\cos(x)}}\\
&=&\lim_{x \to 0}{\frac{\cosh(x)-\cos(x)}{3x^2}}\cdot\lim_{x \to 0}{\frac{2x}{\sinh(x)+\sin(x)}}
\end{eqnarray}
and you can continue.


问:
Let $a,b \in \mathbb{R}$ such that $a0, f'(b)>0. $ Prove that $\exists c \in (a,b) $ such that $f(c) =0, f'(c)\leq0$
Preliminaries: $f : \mathbb{R} \to \mathbb{R} $ is differentiable;

Proved before that if $f'(p) > 0 \to \exists \delta > 0 : f(x)<f(p)<
f(y) \ \forall x,y \in \mathbb{R} : p-\delta < x < p < y < p+ \delta$

I wonder if there is an easier and more direct proof that this:

I proved the first part using Rolle's Theorem:
for $f(a)=f(b), \exists p: f(p)=0$
for the second part of the proof, I let $C={p \in \mathbb{R} : f(p)=0} $

Then, $c=\min(P)$

Since $f$ is differentiable,
$\lim G_c(h) _{h \to 0} = \lim G_c(h) _ {h \uparrow 0} $ where
$G_c(h) = \frac{f(c+h)-f(c)}{h}=\frac{f(c+h)}{h}$ and $h \in (-\delta,0) $

Claim: $\forall h, G_c(h) \leq 0$
Proof:

Want to show that $f(c+h) \geq 0 \because h < 0 $

Suppose not.
But $ \exists \delta_1 :$ for $ a+ \delta_1 < c, f(a+ \delta_1) > f(a)=0$ and thus by IVT, $\exists p' \in (a+\delta_1, c) : f(p')=0$ contradicting the minimality of $c$ as a root.
So $f(c+h) \geq 0$
Hence lim $G_c(h) _ {h \uparrow 0} \leq 0 $ i.e. $f'(c) \leq 0$.

Edit: $f'(c) \leq 0 $ copy-pasted the question wrongly, my bad.

答: Apart from @ancientmathematician's answer, let me give one more counterexample: take $f = \sin$, $a = 0$, $b = 2\pi$.  Note that $f' = \cos$, so $f'(a) = f'(b) = 1 > 0$, $f(a) = f(b) = 0$, but $f$ (sine) and $f'$ (cosine) never vanishes at the same time due to the basic identity $\sin^2 x + \cos^2 x = 1$ for all real number $x$.

Title: Let $a,b \in \mathbb{R}$ such that $a<b$ and $f(a)=f(b)=0$ with $f'(a)>0, f'(b)>0. $ Prove that $\exists c \in (a,b) $ such that $f(c) =0, f'(c)\leq0$


Body:
For $\lim_{x \to 0^+}$ we write $x=1+h, h>0$
then
$$
\lim \limits_{x \to 0^+} \frac{\sin^{-1}(1 - \{x\}) \times \cos^{-1}(1 - \{x\})}{\sqrt{2\{x\}}\times(1 - \{x\})}$$
$$ L=\lim_{h\to 0} \frac{\sin^{-1}(1-h) \cos^{-1}(1-h)}{\sqrt{2h} (1-h)}$$
$$L=\lim_{h\to 0}\frac{\sin^{-1}(1-h)}{(1-h)} \frac{\cos^{-1}(1-h)}{\sqrt{2h}}$$
$$L=\frac{\pi}{2} \lim_{h \to 0}\frac{\cos^{-1}(1-h)}{\sqrt{2h}}$$
Use L=Hosp. now
$$L=\frac{\pi}{2} \lim_{h \to 0} \frac{2/\sqrt{1-(1-h)^2}~\sqrt{h}}{\sqrt{2}}$$
Use $(1-h)^2\to 1-2h$, then
$$L=\frac{\pi}{2} \lim_{h \to 0} \frac{2\sqrt{h}}{2\sqrt{h}}=\frac{\pi}{2}.$$


答: Since $\sinh x-\sin x\sim\tfrac13x^3$ while $\cosh x-\cos x\sim x^2$, the limit is $\tfrac13$.

Title: $\lim_{x \to 0}{\frac{\sinh(x)-\sin(x)}{x(\cosh(x)-\cos(x))}}$

Body:
As stated in the title.

My attempt, begin with L'Hopital:
$$L=\lim_{x \to 0}{\frac{\sinh(x)-\sin(x)}{x(\cosh(x)-\cos(x))}}=\lim_{x \to 0}{\frac{\cosh(x)-\cos(x)}{(\cosh(x)-\cos(x))+x(\sinh(x)+\sin(x))}}$$
Dividing by the numerator
$$\lim_{x \to 0}\frac{1}{1+\left(\frac{x\left(\sinh(x)+\sin(x)\right)}{\cosh(x)-\cos(x)}\right)}=\frac{1}{1+\lim_{x \to 0}{\left(\frac{x\left(\sinh(x)+\sin(x)\right)}{\cosh(x)-\cos(x)}\right)}}$$
L'Hopital again
$$\frac{1}{1+\lim_{x \to 0}{\left(\frac{x(\cosh(x)+\cos(x))+(\sinh(x)+\sin(x))}{\sinh(x)+\sin(x)}\right)}}$$
Diving through
$$\frac{1}{1+\lim_{x \to 0}{\left(1+\frac{\cosh(x)+\cos(x)}{\left(\frac{\sinh(x)}{x}+\frac{\sin(x)}{x} \right)}\right)}}=\frac{1}{1+(1+\frac{1+1}{1+1})}=\frac{1}{3}$$

Is this correct, and is there a more elegant way of doing it?


问:
Finding side of rectangle using given information
Really simple question but I am stuck. The following information is given:
$$BD=8,\quad AB = 6,\quad  ED  =5,\quad EF = EC$$
and we want to find $AF$.

If we have three $90^\circ$, what does that really mean, and how I can find $AF$?

答:
Hint:
$$\begin{align*}
AB=6,\;BD=8,\;\text{Pythagorean theorem}&\implies AD=\mathbin{?}\\\\
ED=5&\implies AE=\mathbin{?}\\\\
EF+EC=AB=6,\;EF=EC&\implies EF=\mathbin{?}\\\\
AE=\mathbin{?},\;EF=\mathbin{?},\;\text{Pythagorean theorem}&\implies AF=\mathbin{?}\\\\
\end{align*}$$


Body: It looks fine, but I would have done it as follows\begin{align}\lim_{x\to0}\frac{\sinh(x)-\sin(x)}{x\bigl(\cosh(x)-\cos(x)\bigr)}&=\lim_{x\to0}\frac{\left(x+\frac{x^3}6+o(x^4)\right)-\left(x-\frac{x^3}6+o(x^4)\right)}{x\left(\left(1+\frac{x^2}2+o(x^3)\right)-\left(1-\frac{x^2}2+o(x^3)\right)\right)}\\&=\lim_{x\to0}\frac{\frac{x^3}3+o(x^4)}{x+o(x^4)}\\&=\lim_{x\to0}\frac{\frac13+o(x^3)}{1+o(x^3)}\\&=\frac13.\end{align}


问:
A pigeonhole riddle

John offers Dean the following challenge:

Dean would choose 8 natural numbers so that $10\leq n \leq 36$.
If John manages to create two equal sums using only numbers that Dean chose, at most once each, he wins.

Assuming John has adequate time to check all options, will he always win? and why?

My thoughts were counting the number of options that Dean has for choosing 8 numbers, which is $27 \choose 8$, and also counting the number of options John has of creating sums, which is $\binom{27}{8}\cdot\sum^8_{i=2} \binom{8}{i}$. But that doesn't bring me anywhere.

I feel like it's true that John would always win, and I know I need to use the pigeonhole principle somehow, which means showing that for sure there are more possible ways to arrange sums than the number of sums possible. So, there are at least two arrangements for the same sum. (By arrangement for the same sum I mean two different combinations of numbers that summed give the same answer).
But I'm kind of stuck there.

Would appreciate assistance!

答:
You can compute the maximum sum $S_M=36+35+34+33+32+31+30+29=260$ and the minimum sum $S_m=10+11=21$ (assuming we always have to sum at least two numbers). This means that any solution of the sum $S$ of any combination of the eight numbers will be $S_m \leq S \leq S_M$. From $S_M-S_m=239$ we get the number of different sums we can obtain from the combination of those numbers.

On the other hand, there are $\sum^8_{i=2} \binom{8}{i}$ possible sums from the 8 numbers chosen. It is enough to show that $\sum^8_{i=2} \binom{8}{i}>239$ in order to prove, by the Pigeonhole Principle, that there will always be a sum $S$ which can be computed in at least two different ways using different combinations of those 8 numbers.

Title: A pigeonhole riddle


问:
Define a $4 \times 4$ operator matrix so that pre-multiplying any matrix $X$ with this matrix will result in...
addition of $c_i$ times the $i$th row of $X$ to the $2$nd row of $X$ for each $i \in \{1, 2, 3, 4\}$ in one shot. Additionaly, Show that this matrix can be expressed as the product of three elementary addition matrices and a single elementary multiplication matrix.

My attempt:

$\begin{bmatrix}1 & 0 & 0 & 0\\ c & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 0 & 0 & 0\\ 0 & 1 & c & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & c\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix}1 & 0 & 0 & 0\\ c & 1 & c & c\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$

I am unsure about the part for $i \in 2$ (would it just be a $1+c$ in the $(2,2)$ position of the matrix?) or how to express it as a product of three elementary addition matrices and a multiplication matrix.

SOURCE: Linear Algebra and Optimization for Machine Learning (page 23, problem 1.3.1)

答:
I think you misunderstood the problem. You need to find $M(c_1,c_2,c_3,c_4)$ such that for any matrix $X=x_{i,j},\ i,j\in\{1,2,3,4\}$ you get $MX=Y$, where $Y=y_{i,j}$. If $i\ne 2$ then $y_{i,j}=x_{i,j}$. If $i=2$ then $$y_{2,j}=c_1x_{1,j}+(1+c_2)x_{2,j}+c_3x_{3,j}+c_4x_{4,j}$$
Once you write explicitly the $M$, $X$, and $Y$ matrices you can proceed in two ways. First (the easy one) is to write the $MX$ product, to check what properties does $M$ have. The more complicated way is to invert $X$, then write $M=YX^{-1}$.

Can you take it from here?


问:
Efficient way to find the remainder when $2001+ 2002+ 2003+ … + 2015+ 2016$ is divided by $2017$?
I can think of a couple of ways

Notice that unit digit of the first $8$ terms can be added in the last $8$ terms to make them $2017$. Now add the first $8$ terms without their unit digits (i.e. $2000*8$) and find a remainder on that.

Sum of AP series = $4017*8$, now find the remainder

But both would take a lot of computation.
Is there any more efficient way of doing it manually?

Body:
Welcome to MSE:

$2001 \equiv -16 \mod (2017)$

$2002 \equiv -15 \mod (2017)$

$2003 \equiv -14 \mod (2017)$

$\cdot$

$\cdot$

$\cdot$

$2016 \equiv -1 \mod (2017)$

Summing these relations we get:

$2001+2002+2002+2003+ . . . 2016 \equiv-\frac{16(1+16)}2=-136 \mod(2017)\equiv 1881 \mod(2017)$


Title: Efficient way to find the remainder when $2001+ 2002+ 2003+ … + 2015+ 2016$ is divided by $2017$?

Body:
I can think of a couple of ways

Notice that unit digit of the first $8$ terms can be added in the last $8$ terms to make them $2017$. Now add the first $8$ terms without their unit digits (i.e. $2000*8$) and find a remainder on that.

Sum of AP series = $4017*8$, now find the remainder

But both would take a lot of computation.
Is there any more efficient way of doing it manually?

Body: It's $-16-15-14-...-1=-16\times17/2=-8\times17=-136\equiv1881\bmod2017$.


问:
Finding the area of the intersection of two circles in terms of distance between centres of circles
I study maths as a hobby and have come across this problem.
Two circles with centres A and B intersect at points P and Q, such that $\angle APB$ is a right angle. If AB = xcm and $\angle PAQ = \frac{1}{3}\pi$ radians, find in terms of x the length of the perimeter and the area of the region common to the two circles.

I calculate the area of the segment APQ to be $\frac{\pi}{6}r^2$, where r = radius.
The area of $\triangle APQ$ I calculate as $\frac{1}{2}r^2\sin \frac{\pi}{3} = \frac{1}{2}r^2\sqrt3$

I also know that to find the right hand side of the central segment I need to subtract the area of the triangle APQ from the area of the segment APQ.

But I cannot proceed any further and certainly not in terms of the length x.

This is the diagram as I visualise it:

答:
HINT

We know that $\angle PAB=\frac{\pi}{6}$, hence the radius of the larger circle is $x\cos{\frac{\pi}{6}}=\frac{x\sqrt3}{2}$. Similarly, the radius of the smaller circle is
$x\sin{\frac{\pi}{6}}=\frac{x}{2}$.

Does that help? If you need any more help, please don't hesitate to ask. I have a full solution ready to post if you need it.


问:
Prove that every open set in $\mathbb{R}$ is the union of an at most countable collection of disjoint segments.
My attempt:

We know that $\mathbb{Q}$ is dense in $\mathbb{R}$.Now, we consider an open set $S$ in $\mathbb{R}$ then let $x \in S$, $x$ is an interior point of $S$ and we can get an open ball $(x-r,x+r)$ ,$r>0$ contained in $S$. Since $\mathbb{Q}$ is dense, let $q_1 \in (x-r,x+r)$.Now, $q_1 \in (x-r,x+r)$ and let $r_1=|q_1-x|$.We consider the rational number $ > r_1$ (which is possible to choose as $\mathbb{Q}$ is dense ) and let us call it $d_1$ and we consider the open ball $(q_1-d_1,q_1 +d_1)$ which will contain $x$.

We pick $x' \in S$ , and consider $r_2>0$ such that $(x'-r_2,x'+r_2) \cap (q_1-d_1,q_1+d_1) = \phi$ , if such an $r_2$ doesn't exist then we can conclude that $x' \in (q_1-d_1,q_1+d_1)$. Repeating this process we can conclude that the open set $S $ can be written as the union of disjoint segments.

Since $\mathbb{Q}$ is countable then it can be shown that union of this disjoint segment is also countable.

I am trying to do Rudin on my own and need someone to cross check my proofs.I know this might be a duplicate but if someone goes through my proof and picks up my mistake and a probable way out that would be helpful.

答:
We use the following facts: the connected open subsets of $\mathbb R$ are the open intervals. The union of two connected sets with a point in common is connected, so if two intervals have a point in common, then their union is an interval.

A careful proof, in steps:

$1).\ $ Every open set $U$ is a disjoint union of intervals:

Fix $x\in U$ and let $C_x$ be the union of all intervals in $U$ that contain $x.\ C_x$ is not empty because $U$ is open.

Claim: $C_x$ is an interval. It is clearly open. So now suppose $y,z\in C_x$ and $y<w<z.$ Then, $y$ and $z$ are contained in intervals $I_y$ and $I_z$, each of which is contained in $U$ and contains $x$. And since they are connected and have the point $x$ in common, their union $I:=I_y\cup I_z$ is connected and so is an interval. But now we are done because $I$ contains $y$ and $z$ and so must contain $w$. Hence, $C_x$ is an interval.

$2).\ C_x\cap C_y=\emptyset$ whenever $x\neq y.$ For, if $z\in C_x\cap C_y$ then $C_x\cup C_y$ is a union of two intervals with the point $z$ in common, and so is an interval. It follows that  $C_x\cup C_y\subseteq C_x$ and $C_x\cup C_y\subseteq C_y$ so $C_x\cup C_y\subseteq C_x\cap C_y.$ Then, $C_x\cup C_y=C_x\cap C_y.$ That is, $x=y$.

$1).\ $ and $2).\ $ show that $U=\bigsqcup_{x\in U} C_x$

$3).\ $ The union is actually countable: this follows immediately, for we may choose a rational $r_x$ in each $C_x$, and note that since the $C_x$ are disjoint, $r_x\mapsto C_x$ injects a subset of the rationals into $\{C_x\}_{x\in U}.$


问:
If $\langle\;\cdot\;,\;\cdot\;\rangle$ is a duality pairing satisfying the Cauchy-Schwarz inequality, is $\|x\|=\sup_{\|y\|\le1}|\langle x,y\rangle|$
Let $\langle\;\cdot\;,\;\cdot\;\rangle$ be a duality pairing between a Banach space $X$ and a normed space $Y$ with $$\left|\langle x,y\rangle\right|\le\left\|x\right\|_X\left\|y\right\|_Y\;\;\;\text{for all }(x,y)\in X\times Y\tag1.$$

Are we able to show that $$\left\|x\right\|_X=\sup_{\left\|y\right\|_Y\le1}\left|\langle x,y\rangle\right|\tag2$$ for all $x\in X$?

Let $x\in X\setminus\{0\}$. By $(1)$, $$\varphi:=\langle x,\;\cdot\;\rangle\in Y'\tag3$$ and $$c:=\sup_{\left\|y\right\|_Y\le1}\left|\langle x,y\rangle\right|=\left\|\varphi\right\|_{Y'}\le\left\|x\right\|_X\tag4.$$

On the other hand, since $\langle\;\cdot\;,\;\cdot\;\rangle$ is a duality pairing, there is a $y\in Y$ with $$\varphi(y)\ne0;\tag5$$ but at this point I'm stuck.

Motivation: If $E$ is a normed space, it is an immediate consequence of the Hahn-Banch theorem that, for all $x\in E$, there is a $\varphi\in E'$ with $\varphi(x)=\left\|x\right\|_E$ and $\left\|\varphi\right\|_{E'}\le1$. This yields that $$\langle x,\varphi\rangle:=\varphi(x)\;\;\;\text{for }(x,\varphi)\in E\times E'$$ is a duality pairing between $E$ and $E'$ satisfying $$\left\|x\right\|_E=\sup_{\|\varphi\|_{E'}\le1}|\langle x,\varphi\rangle|\;\;\;\text{for all }x\in E.\tag6$$

答:
As stated in the comments, $\frac{1}{2}\Phi$ is a duality pairing which satisfies $(1)$ but not $(2)$. Although, the norm is bounded above by the duality pairing in the sense that
$$||x||_{X}=\sup_{||y||\leq 2}\frac{1}{2}|\Phi(x,y)|$$
There are some duality pairings $\Phi :X\times Y\to \mathbb{R}$ for which
$$\tag{*}\bigl|\Phi(x,y)\bigr|\leq ||x||_{X}\cdot ||y||_{Y}$$
But there is no constant $C>0$ such that
$$\tag{**} ||x||_{X}\leq C\sup_{||y||\leq 1}\bigl|\Phi(x,y)\bigr|$$
For example, if $X=Y=\ell_2$ and $\Phi:X\times Y\to \mathbb{R}$ is given by
$$\Phi(x,y)=\sum_{n=1}^{\infty}\frac{1}{n}x(n) y(n)$$
$x=(x(n))_{n=1}^{\infty},\,y=(y(n))_{n=1}^{\infty}\in \ell_2$. Then, by the Cauchy-Schwarz inequality
\begin{align}
\bigl|\Phi(x,y)\bigr|&=\biggl|\sum_{n=1}^{\infty}\frac{1}{n}x(n)y(n)\biggr|\leq \sum_{n=1}^{\infty}\biggl|\frac{1}{n}x(n)y(n)\biggr|\\
&\leq \biggl(\sum_{n=1}^{\infty}\frac{1}{n^2}(x(n))^2\biggr)^{1/2}\cdot \biggl(\sum_{n=1}^{\infty}(y(n))^2\biggr)^{1/2}\\
&\leq \biggl(\sum_{n=1}^{\infty}(y(n))^2\biggr)^{1/2}\cdot \biggl(\sum_{n=1}^{\infty}(x(n))^2\biggr)^{1/2}=||x||_2\cdot ||y||_2
\end{align}
Hence, $(1)$ is satisfied. Now, consider the sequence $me_m$ where $e_m$ is the standard basis of $\ell_2$. I.e $e_m(n)=1$ if $n=m$ and $e_m(n)=0$ for every $n\neq m$. Then, for every $||y||_{2}\leq 1$
$$\Phi(me_m,y)=\frac{1}{m}m e_m(m)\cdot y(m)=y(m)$$
Hence, $\bigl|\Phi(me_m,y)\bigr|=|y(m)|\leq ||y||_{2}=1$ which means that $\sup_{||y||_2\leq 1}\bigl|\Phi(me_m,y)\bigr|\leq 1$. If there exists $C>0$ such that $(**)$ holds then we end up with
\begin{align}
m&=||me_m||_{2}\leq C\sup_{||y||_{2}\leq 1}\bigl|\Phi(me_m,y)\bigr|\leq C
\end{align}
for all $m$, which is obviously false.


Title: I don't imagine how to make a cartesian equation of the parametric one: $x=2 \sin (t), y=-\frac{4\cos (t)^2 (2 + 1.3\cos(t))}{3 + \sin (t)^2}$

Body: I know it's a downvoting question but I'm not good at math. I know that $x=a \sin(t)$ , $y=\frac{a \cos^2(t)(2+ \cos(t))}{(3- \sin^2(t))}$ is $y^2(a^2-x^2)=(x^2+2ay-a^2)^2$ but what it could be as a cartesian equation with the negative $y$ part I can't even imagine. Sorry. Could you help me?

Body:
$$y=-\frac{4 \cos ^2(t) \left(\frac{13 \cos (t)}{10}+2\right)}{\sin ^2(t)+3}$$
$$\sin t=\frac{x}{2}$$
thus
$$y=-\frac{4 \left(1-\frac{x^2}{4}\right) \left(\frac{13 \cos (t)}{10}+2\right)}{\left(\frac{x}{2}\right)^2+3}$$
with some algebra we get
$$\cos t=\frac{5 \left(x^2 y-8 x^2+12 y+32\right)}{26 \left(x^2-4\right)}$$
and then
$$\cos^2 t=\left(\frac{5 \left(x^2 y-8 x^2+12 y+32\right)}{26 \left(x^2-4\right)}\right)^2$$
as $\cos^2 t= 1-\sin^2 t$ we get
$$1-\frac{x^2}{4}=\left(\frac{5 \left(x^2 y-8 x^2+12 y+32\right)}{26 \left(x^2-4\right)}\right)^2$$
and finally
$$676 \left(x^2-4\right)^2=25 \left(x^2 y-8 x^2+12 y+32\right)^2+169 x^2 \left(x^2-4\right)^2$$


答:
Hint. For the common area,
$$\text{area of sector PBQ}+\text{area of sector PAQ}=\text{area of }\square\text{PBQA}+\text{common area of circles}$$

where $$\begin{align*}&\text{area of sector PAQ}=\frac16\pi r_1^2\\
&\text{area of sector PBQ}=\frac13\pi r_2^2\\
&\text{area of }\square\text{PBQA}=AB\times PD=xr_1\sin\angle PAD\end{align*}$$

Can you find $r_1,r_2$ in terms of $x$? Use the fact that $\triangle APB$ is right-angled.

Title: Finding the area of the intersection of two circles in terms of distance between centres of circles

Body:
I study maths as a hobby and have come across this problem.
Two circles with centres A and B intersect at points P and Q, such that $\angle APB$ is a right angle. If AB = xcm and $\angle PAQ = \frac{1}{3}\pi$ radians, find in terms of x the length of the perimeter and the area of the region common to the two circles.

I calculate the area of the segment APQ to be $\frac{\pi}{6}r^2$, where r = radius.
The area of $\triangle APQ$ I calculate as $\frac{1}{2}r^2\sin \frac{\pi}{3} = \frac{1}{2}r^2\sqrt3$

I also know that to find the right hand side of the central segment I need to subtract the area of the triangle APQ from the area of the segment APQ.

But I cannot proceed any further and certainly not in terms of the length x.

This is the diagram as I visualise it:


问:
Functionally, what does a symmetric matrix say about the linear transformation it represents?
I understand the definition of a symmetric matrix in terms of how it’s components are related. But functionally, what does it entail about the linear transformation it represents?
For example, block tri-diagonal matrices have special relations between entries but they also, functionally, tell us that some non-trivial vector subspace is invariant under the linear transformation with respect to a particular basis.
Incidentally, what do skew-symmetric matrices represent, functionally ?</p>

答:
In the comments (and in the linked discussion) on the question, I make the following claim:

$M$ is symmetric relative to at least one choice of (possibly oblique) basis if and only if $M$ is diagonalizable with real eigenvalues. $M$ is skew-symmetric relative to at least one choice of basis if and only if $M$ is a direct sum of scaled $90^\circ $ rotations and zero transformations.

First, the symmetric case. If $M$ is symmetric, then the spectral theorem states that $M$ is diagonalizable with real eigenvalues. Conversely, if $M$ is diagonalizable with real eigenvalues, then there is a basis relative to which the matrix of $M$ is diagonal with real diagonal entries. Since this diagonal matrix is symmetric, $M$ is symmetric relative to this choice of basis.

For case where $M$ is skew-symmetric, there are two common approaches. For the easy direction: if $M$ is a direct sum of $90^\circ$ rotations and zero transformations, then there is a basis relative to which the matrix of $M$ is the block-diagonal skew-symmetric matrix
$$
\pmatrix{0 & \kappa_1 \\-\kappa_1 & 0 \\ && \ddots \\ &&& 0 & \kappa_p\\ &&&-\kappa_p & 0
\\ &&&&&0 \\ &&&&&&\ddots\\ &&&&&&& 0}.
$$
There are two approaches for the converse. One is essentially to apply the spectral theorem for Hermitian matrices, noting that if $M$ is skew-symmetric then the complex matrix $iM$ is Hermitian. Alternatively, we can systematically construct a basis relative to which the matrix of $M$ has the above block-diagonal form as is outlined in this post and the proof linked therein.


问:
Polynomials - The sum of two roots
If the sum of two roots of
$$x^4  + 2x^3 - 8x^2 - 18x - 9 = 0$$ is
$0$, find
the roots of the equation</p>

答:
Let those two roots be $a,-a$ and the other two roots are $b,c$

$$\text{Now, }(x-a)(x+a)(x-b)(x-c)=(x^2-a^2)\{x^2-(b+c)x+bc\}$$
$$=x^4-x^3(b+c)+(bc-a^2)x^2+a^2(b+c)x-a^2bc$$

Comparing the coefficients of $x^3$ of this with that of $x^4  + 2x^3 - 8x^2 - 18x - 9 = 0,$

$b+c=-2$

Comparing the coefficients of $x,a^2(b+c)=-18\implies 2a^2=18\implies a=\pm3$

Comparing the  constants, $a^2bc=9\implies bc=1$

So, $b,c$ are the roots of the equation $t^2-(-2)t+(1)=0\implies t=-1,-1$

Title: Polynomials - The sum of two roots


问:
How best to explain partial order over vectors?
I have the following notational issue:

Given a vector space $\mathbf{V}$, I have a partial order such that for all $\mathbf{v},\mathbf{w} \in \mathbf{V}$:

iff for all components $v_i$ of $\mathbf{v}$ and  $w_i$ of $\mathbf{w}$
$$v_i\leq w_i$$

then

$$\mathbf{v} \leq \mathbf{w}$$

Is there a  well known way of stating this partial order ?

答:
I had not heard of this concept before, but searching for "componentwise partial order" brought up the "product order", which seems to be what you want.

I'm not an order theorist, but if you didn't know of a name for this offhand, your audience may not know this definition of "product order", so it would be good to give the definition in whatever you are writing, anyway.


问:
How to deduce that $\arctan\left(\frac{a}{i}\right)=-i\cdot \text{ arctanh}(a)$
I was doing some calculations in wolframalpha and I found the following equality:

$$\arctan\left(\frac{a}{i}\right)=-i\cdot \text{ arctanh}(a)$$

This is the first time I've seen this equality. How is this deduced? Is this just a specific case for $a/i$ or is it true also that:

$$\arctan\left(\frac{a}{b}\right)=-b\cdot \text{ arctanh}(a)?$$

答: First of all, note that $\frac ai=-ai$. So, the question becomes: is $\arctan(ai)=i\operatorname{arctanh}(a)$? Yes, because $\tan\bigl(\arctan(ai\bigr)=ai$ and$$\tan\bigl(i\operatorname{arctanh}(a)\bigr)=i\tanh\bigl(\operatorname{arctanh}(a)\bigr)=ia.\tag1$$The first equality from $(1)$ comes from$$\tan(ix)=\frac{\sin(ix)}{\cos(ix)}=\frac{i\sinh(x)}{\cosh(x)}=i\tanh(x).$$


问:
Are the Satake parameters of an automorphic representation pairwise distinct?
I have a rather naive question whose answer could turn out to be pretty useful to me: given an automorphic representation $\pi$ of $\operatorname{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$, are its Satake parameters at an unramified prime $p$ necessarily pairwise distinct?

答:
In this question in contrary to the others you probably meant primitive L-function, ie. irreducible representation, otherwise $\zeta(s)L(s,\chi_4)$ is a counter-example.

Then look at https://www.lmfdb.org/L/ArtinRepresentation/2.23.3t2.b.a/#moreep

($59$ splits completely in the ring of integers of the splitting field of $x^3-x^2+1\in \Bbb{Q}[x]$)