答:
Might be easier to do $P(n) \Rightarrow P(n+2)$, i.e. prove it for odd and even numbers separately.

$P(1), P(2)$ are clear.

$P(n) \Rightarrow P(n+2)$:

$$10^{n+2}= 10^n*100=99*10^n+10^n \cong 10^n \cong (-1)^n \mod 11$$

Title: Complete induction of $10^n \equiv (-1)^n \pmod{11}$

Body:
To prove $10^n \equiv (-1)^n\pmod{11}$, $n\geq 0$, I started an induction.

It's $$11|((-1)^n - 10^n) \Longrightarrow (-1)^n -10^n = k*11,\quad k \in \mathbb{Z}. $$

For $n = 0$:
$$ (-1)^0 - (10)^0 = 0*11 $$

$n\Rightarrow n+1$
$$\begin{align*}
 (-1) ^{n+1} - (10) ^{n+1} &= k*11\\
(-1)*(-1)^n - 10*(10)^n &= k*11
\end{align*}$$
But I don't get the next step.


问:
Proving a function is quasi-concave
This is from economics, but I think there's a lot of math involved and I want to make sure I didn't mess anything up.  There is a utility function U = $x_1$ + $\ln(x_2)$ such that $x_1$ and $x_2$ are nonnegative.

How do I prove this function is quasi-concave?

I was looking online for any notes about this, and it tells me I need to create this

https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/qcc/t

"bordered Hessian matrix"

As in for this problem it seems to me I need to calculate

\begin{bmatrix}
0 & f'_{1}(x) & f'_{2}(x)\\
f'_{1}(x) & f''_{11}(x) & f''_{12}(x) \\
f'_{2}(x) & f''_{21}(x) & f''_{22}(x) \\
\end{bmatrix}

So I do that for the above function and I get

\begin{bmatrix}
0 & 1 & \frac{1}{x_2} \\
1 & 0 & 0 \\
\frac{1}{x_2} & 0 & -\frac{1}{x_{2}^{2}} \\
\end{bmatrix}

Then it says I need to find the determinant and if n = even, I need $D_{n} \geq 0$.  So here $D_{2}$ for the above matrix I got to be $\frac{1}{x_{2}^{2}}$.  Since $x_2$ is nonnegative, this means that the determinant is $\geq 0$.

Does this prove this function is quasi-concave?  Is my general approach right or am I totally missing the point of this question?

What about strictly quasi-concave?   The online notes I found never mentions this, but is that just when it is strictly greater than 0?  So in this case, this function is also strictly quasi-concave too?  If someone can verify if my guess is correct, that would be great.

Finally, on an unrelated note, how do I prove this function is continuous using math?  I know it is when I graph it, but is there a mathematical proof for this?

Thank you so much for the help!

答:
Eigenvalues of the Hessian matrix of the function $U=x_1+\mathrm{ln}(x_2) $ turns out to be $0$ and $-\frac{1}{{x_{2}}^2}$ which implies that $U$ is concave in $\mathbb{R}^2_{++}.$

Every function that is concave is also Quasi Concave too.

ContentLicense: CC BY-SA 4.0

Title: Proving a function is quasi-concave


问:
Consider a game with 2 stages. In the first stage, the player tosss a coin until a head is appears
Consider a game with 2 stages. In the first stage, the player tosss a coin until a head is appears. Say that it takes j flips to get a head. Then, in the second stage, he random draws an integer between 1 and $ 2^j $ to get his winnings (so for example if he draws 2,he gets 2 dollars).
a) What is the average amount of dollars he gets?
b)Given that he wins 4 dollars, what is the probability that the coin was flipped exactly 6 times,i.e,that j=6?

My work for question a is I have the probability $\frac{1}{2^j}$ until getting a head,and each time I have a head,i calculate the expected value of money in range from 1 to j,so:

\begin{equation}
 E(X) = \sum_{j=1}^\infty \frac{1}{2^j}\sum_{i=1}^{j}i\frac{1}{2^j}
\end{equation}
b) I call A is the event he gets 4 dollars and B is the event he flipped the coin exactly 6 times,so

\begin{equation}
 P(B|A) = \frac{P(B \cap A)}{P(A)}
\end{equation}
with

\begin{equation}
 P(B \cap A) = \frac{1}{2^6} \ and \ P(A)=\sum_{i=2}^\infty \frac{1}{2^i}
\end{equation}
is it a right way to do? if it's right, how can i calculate the sums?

答:
(a) Your inner summation should go from $1$ to $2^j$. You can also simplify the summations:

$$ \begin{equation}
 E(X) = \sum_{j=1}^\infty \frac{1}{2^j}\sum_{i=1}^{2^j}i\frac{1}{2^j} = \sum_{j=1}^\infty \left( \frac{1}{2^j}\cdot \frac{1+2^j}{2} \right)
=\frac{1}{2}\sum_{j=1}^\infty \left( \frac{1}{2^j} + 1 \right) = +\infty
\end{equation}
$$

(b) How did you get $P(B \cap A) = \frac{1}{2^6}$? Isn't that $P(B)$? Since $P(A)$ and $P(A \cap B)$ are hard to computer I suggest computing the conditional this way:

$$
\begin{equation}
P(flip=6|won=$4) = \frac{P(won=$4|flip=6)P(flip=6)}{\sum\limits_{i=1}^\infty P(won=$4|flip=i)P(flip=i)}
\end{equation}
$$


问:
Triangulation of 2-Manifolds
I am reading John Lee's "Introduction to Topological Manifolds" and in chapter 5 he states:

Theorem 5.36 (Triangulation Theorem for 2-Manifolds). Every 2-manifold
is homeomorphic to the polyhedron of a 2-dimensional simplicial
complex, in which every 1-simplex is a face of exactly two
2-simplices.

The proof is highly technical and beyond our scope, so we can only
describe some of the main ideas here (etc.)

This was a bit disappointing since this theorem is central in proving his classification theorem of compact 2-manifolds in chapter 6. Thus I decided to look up on the net if there were some proofs and came up across this link: https://www.maths.ed.ac.uk/~v1ranick/papers/doylemoran.pdf

Which seems to prove that all compact 2-manifolds are triangulable and the paper seems to date back to 1967 (I haven't read it yet, I want to confirm its legitimacy). Now I realize that Theorem 5.36 refers to all 2-manifolds not just compact ones but given that he's specifically classifying compact manifolds why did he not include the above (seemingly short) proof if it is indeed a legitimate one? Or is it because the second part of the claim in Theorem 5.36 (i.e. "in which every 1-simplex is a face of exactly two 2-simplices") that is technical and hard to prove?

答:
The second part is very easy. I recommend as an exercise that you show this is true of any polyhedron homeomorphic to a surface.

The point is that the paper you link is not a complete reference: it pushes the problem back to needing a proof of the Schoenflies theorem, which is technical no matter how you slice it.

Title: Triangulation of 2-Manifolds


问:
Proving that there are infinitely many prime numbers of the form $4k+3$

Anyone wanna help me solve this one? Been at it for a little bit but haven't really gotten anywhere..

答: Suppose there are finitely many primes $p_1, \dots, p_k$ of the form $4n + 3$. Let $m = 4p_1 \dots p_k -1$.  Note that $m \equiv 3 \bmod 4$, and in particular, $m$ is odd so $2 \nmid m$.  Thus all primes dividing $m$ have to be of the form $4n + 1$, but that implies $m \equiv 1 \bmod 4$, a contradiction.</p>

ContentLicense: CC BY-SA 3.0


问:
$(a+c)(b+d)=1$, $a,b,c,d \in R^{+}$. Prove that $\frac{a^3}{b+c+d} + \frac{b^3}{a+c+d} + \frac{c^3}{a+b+d} + \frac{d^3}{a+b+c} \geq \frac{1}{3}$
$(a+c)(b+d)=1$, $a,b,c,d \in R^{+}$. Prove that

$$\frac{a^3}{b+c+d} + \frac{b^3}{a+c+d} + \frac{c^3}{a+b+d} + \frac{d^3}{a+b+c} \geq \frac{1}{3}$$

rather annoying that the powers of denominators and numerators are not the same and can't think of easy way to flatten them..

答:
A direct application of C-S and AM-GM works here. We have:
$$\left(\sum_{\text{cyc}} \dfrac{a^3}{b+c+d} \right) \left(\sum_{\text{cyc}} a(b+c+d)\right) \geq \left(\sum_{\text{cyc} }a^2 \right)^2.$$

It thus suffices to prove that

$$\dfrac{\left(\displaystyle \sum_{\text{cyc} }a^2 \right)^2}{\left(\displaystyle \sum_{\text{cyc}} a(b+c+d)\right) } \geq \dfrac{1}{3}.$$

Note that $a^2+b^2 \geq 2ab, a^2+d^2 \geq 2ad, b^2+c^2 \geq 2bc, c^2+d^2 \geq 2cd.$ Thus, $2a^2+2b^2+2c^2+2d^2 \geq  2(ab+ad+bc+cd) = 2(a+c)(b+d)=2$. This gives us  $a^2+b^2+c^2+d^2 \geq 1.$ It remains for us to prove that

$$
3\left(\displaystyle \sum_{\text{cyc} }a^2 \right) \geq \left(\displaystyle \sum_{\text{cyc}} a(b+c+d)\right) = 2ab+2ac+2ad+2bc+2bd+2cd.
$$
But this is just a simple application of AM-GM, and we are done.


问:
Notation used in Russell's paradox
I am slightly confused by the notation used in Russell's paradox. I am following this text.

I understand that $\phi (x)$ is this boolean function, which outputs either True or False. I understand that $R$ is the set of all $x$ which has the image True when subjected to $\phi$. This is denoted by $$R = \{x:\phi (x)\}$$

Great. I am good so far.

Now, the author defines $\phi(x)$ such that $x \in x$. I don't understand this notation. How can an object belong to itself? Shouldn't it be '$=$' instead of '$\in$'?

Then the author goes ahead and defines $R$ is a set that contains all $x$ such that $x$ does not belong in $x$, or using symbols,
$$R = \{ x: \, \sim\phi(x)\}$$
I don't understand this $x\in x$ notation used. I believe this very argument is used to explain the paradox. Could someone explain this?

答:
It's not a typo: although the idea of a set containing itself as an element may be weird, and we may even want to include an axiom in our set theory forbidding such shenanigans (such as the axiom of regularity in the setting of $\mathsf{ZFC}$), it makes sense to ask whether or not a set is an element of itself, and so we can ask whether there is in fact a set consisting exactly of those sets which are not elements of themselves.

Whether or not self-elementhood actually happens doesn't affect the paradox. This may be made clearer by distinguishing between sets and classes (= a collection of sets which may or may not be a set itself). In a theory like $\mathsf{ZFC}$, which prohibits self-elementhood, the Russell class $\{x: x\not\in x\}$ is the whole universe of sets: every set satisfies $x\not\in x$. In theories like $\mathsf{NFU}$ or $\mathsf{GPK^+_\infty}$ which do allow self-elementhood, the Russell class is more complicated. But either way, the Russell class can never be a set, as Russell's argument shows.

Title: Notation used in Russell's paradox


问:
Negative Definiteness of a real symmetric block matrix
Here is a question I have in my mind. Would really appreciate it if anyone can help offer a clue.

Say we have a real symmetric $2n\times 2n$ matrix $M$, with the form of

$$M = \left[ \begin{array}{*{20}{c}}
A & B\\
B^T& C
\end{array} \right],$$

where $A$ and $C$ are symmetric and negative definite, with rank $n.$ Matrix $B$ is known to have zero trace.

Q: Is there anything we can say about the negative definiteness of the matrix $M$?

PS: In the specific problem I am looking at, matrix $C$ is a permutation of $A$. Therefore, you can say $C=P^T A P$, where $P$ is some permutation operation.

Update:
The B matrix has the following format:
$$B = \left[ \begin{array}{*{20}{c}}
0 &... &... &... \\
... & 0 & -\frac{3\sin((m+n)\theta)}{|\sin^3(\frac{m-n}{2}\theta)|^3} &...\\
... & -\frac{3\sin((n+m)\theta)}{|\sin^3(\frac{n-m}{2}\theta)|^3} & 0 &...\\
... &... &... & 0\\
\end{array}
\right],$$
Basically, the diagonal terms are zero and the off diagonal terms $B_{m,n}=-\frac{3\sin((m+n)\theta)}{|\sin^3(\frac{m-n}{2}\theta)|^3}$, and $\theta=2\pi/N$.

The A matrix has the following format:
$$A = \left[ \begin{array}{*{20}{c}}
-c &... &... &... \\
... & -c & \frac{3\cos^2(\frac{m+n}{2}\theta)-1}{|\sin^3(\frac{m-n}{2}\theta)|^3} &...\\
... & \frac{3\cos^2(\frac{m+n}{2}\theta)-1}{|\sin^3(\frac{m-n}{2}\theta)|^3} & -c &...\\
... &... &... & -c\\
\end{array}
\right],$$
where c is a positive constant value and the off diagonal terms $A_{m,n}=\frac{3\cos^2(\frac{m+n}{2}\theta)-1}{|\sin^3(\frac{m-n}{2}\theta)|^3} $.

答:
There's the counterexample of
\begin{align*}
A = C &=
\begin{bmatrix}
-1 & 0\\
0 & -1
\end{bmatrix}\\
B &=
\begin{bmatrix}
-2 & 0\\
0 & 2
\end{bmatrix}
\end{align*}
which results in an $M$ with eigenvalues $-3$ and $1$ ($[0, 1, 0, 1]^T$ is one of the eigenvectors with positive eigenvalue). More loosely, the eigenvectors of all the matrices can line up, and trace of $B$ being zero still allows for large positive eigenvalues, so for $M$ to be negative definite there has to be some constraint that prevents that from happening.

Update:

Since $\operatorname{Tr}(B)= 0$, and $B$ is not the zero matrix, there always is some eigenvector $v$ of $B$ with positive eigenvalue $\lambda>0$. Then,
\begin{align}
\begin{bmatrix} v^T&v^T\end{bmatrix} M\begin{bmatrix} v\\v\end{bmatrix}
&= v^TAv + v^TBv+v^TB^Tv + v^TCv\\
&= v^TAv + 2\lambda v^Tv + v^TCv
\end{align}
could still be positive if, for example, $v$ is an eigenvector of $A$ and $C$, and the corresponding eigenvalues have absolute value less than $\lambda$.

Propositions 2.1 and 2.2 involving the Schur complement described here might be useful. The fact that $B$ is symmetric simplifies the condition slightly to $A-B C^{-1} B$ being negative definite, but it appears some information about $A$ is still needed to conclude anything useful.


问:
Prove that this set (involving fractional part of any rational number) is a partition of the set of rationals.
For any rational number $x$, we can writte $x=q+\,n/m$ where $q$ is an integer and $0\le n/m<1$. Call $n/m$ the fractional part of $x$. For each rational $r\in \{x : 0\le x<1\}$ ,let $A_r = \{ x\in \Bbb Q: \text{the fractional part of $x$ is equal to}\; r\}$.

Prove :$\{A_r : 0\le r<1\}$ is partition of $\Bbb Q$

答:
HINT: To prove that $\{A_r:r\in\Bbb Q\cap[0,1)\}$ is a partition of $\Bbb Q$, you must show that each $A_r$ is a non-empty set of rational numbers, and that every rational number belongs to exactly one $A_r$. It’s clear from the definition that each $A_r\subseteq\Bbb Q$, and you should have no trouble exhibiting a particular rational number that belongs to $A_r$. That leaves you only two things to show:

each rational number belongs to some $A_r$ with $r\in\Bbb Q\cap[0,1)$, and

if $r,s\in\Bbb Q\cap[0,1)$, and $r\ne s$, then $A_r\cap A_s=\varnothing$.

The first sentence of your question gives the justification for (1), so really all that’s left is to prove (2). I suggest proving the contrapositive: if $x\in A_r\cap A_s$, then $r=s$.


答:
For the $x\in x$ part, you can just think $\phi(x)$ is false for every $x$. This will not affect the paradox. So now $R$ is literally the set of "everything," and the paradox is asking "does $R$ belong to $R$ itself?" If $R\in R$, i.e., the set of "everything" is one of "everything," then $\phi(R)$ is true, contradicting the construction of $R$; if $R\notin R$, then $\phi(R)$ is true, so $R\in R$, contradiction.

See this. ZFC says that the construction of $R$ is not allowed, for we have to specify a "bound" for $x$.

Title: Notation used in Russell's paradox

Body:
I am slightly confused by the notation used in Russell's paradox. I am following this text.

I understand that $\phi (x)$ is this boolean function, which outputs either True or False. I understand that $R$ is the set of all $x$ which has the image True when subjected to $\phi$. This is denoted by $$R = \{x:\phi (x)\}$$

Great. I am good so far.

Now, the author defines $\phi(x)$ such that $x \in x$. I don't understand this notation. How can an object belong to itself? Shouldn't it be '$=$' instead of '$\in$'?

Then the author goes ahead and defines $R$ is a set that contains all $x$ such that $x$ does not belong in $x$, or using symbols,
$$R = \{ x: \, \sim\phi(x)\}$$
I don't understand this $x\in x$ notation used. I believe this very argument is used to explain the paradox. Could someone explain this?


问:
Existence of Lebesgue integral on limit domain
Suppose that for some $a,b \in \mathbb R \cup \{ \infty \}$ with $a < b$, a function $f:(a,b) \to \mathbb R$ is continuous (but possibly unbounded). Moreover, there exists $L \in \mathbb R$ such that for all decreasing sequences $a_n \to a$ and increasing sequences $b_n \to b$,
$$\lim_{n \to \infty} \int\limits_{[a_n,b_n]} f(x) dx = L.$$

Can I conclude that the Lebesgue integral $\int\limits_{(a,b)} f(x) dx$ exists?

To me, the setup seems similar to Fatou's Lemma, but with a special domain, without non-negativity, and with this "limit exists" condition. But I can't see how / whether the pieces fit together. Thanks for your help.

答: No. Let $f(x)=\frac 1 x \sin (\frac 1 x)$ on $(0,1)$. Then the hypothesis holds with $L=\int_1^{\infty} \frac {\sin x} xdx$ but $f$ is not Lebesgue integrable. [Make the substitution $y=\frac 1 x $ for the proof].


问:
A polynomial divisible by primes of special form

Problem. Let $p$ be a prime number of the form $9k + 1$. Show that there exists an integer $n$ such that $p \mid n^3 - 3n + 1$.

Source. Here.

My only idea was that $n^3-3n+1$ might form a complete residue modulo such primes. So, I tried it out for $19$ and it turned out to be wrong. After that I could not find anything useful so I read the solution. But I don't understand it. Any elaboration on the solution given in the provided link or any new solution both are appreciated. Thanks in advance

答: Hint: $ x^3 - 3x + 1 $ is the minimal polynomial of $ \zeta_9 + \zeta_9^{-1} $ over $ \mathbf Q $, and if $ p \equiv 1 \pmod{9} $ then there is an element of order $ 9 $ in $ (\mathbf Z/p \mathbf Z)^{\times} $.


问:
Question about repertoire method example: performing linear combinations.
In @MarkusScheuer's excellent answer on Mathematical Explanation for the Repertoire Method the introduction gives a concrete example with two provided repertoire items, $x_n$ and $y_n$ which are then linearly combined to find the solution to $z_n$. Quoting that section below:

To make it more concrete, let $a_n = 3$ and $b_n = 5n^2 + 1$. Assuming we know the solutions of $x_n$ and $y_n$ of the recurrences

$$
\begin{align*}
x_0&=3&y_0&=1\\
x_n&=3+x_{n-1},\quad n>0&y_n&=5n^2+1+y_{n-1},\quad n>0
\end{align*}
$$

then we also know by linearity that the solution of the recurrence

$$
\begin{align*}
z_0&=7\\
z_n&=2n^2+7+z_{n-1}
\end{align*}
$$

is

$$
\begin{align*}
z_n=\frac{11}{5}x_n+\frac{2}{5}y_n
\end{align*}
$$

To find the eventual closed form cooefficient for the $n^2$ variable (say $\beta$) I think we do the following

$$
\begin{align}
2 &= \beta 5 \\
\frac25 &= \beta
\end{align}
$$

to use with $y_n$. However I have not been able to figure out how to arrive at $\frac{11}{5}$. I am not sure how to "map" the constant in recurrence $x_n = 3 + x_{n-1}$ to the $7$ in recurrence $z_n$ as well as take into account the $1$ in $y_n = 5n^2 + 1 + y_{n-1}$. I suspect I'm overthinking this.

答: Use $z_n=\alpha x_n+\beta y_n$ for $n=0, 1$ to get the equations $3\alpha+\beta=7$ and $6\alpha+7\beta=16$. Solving for $\alpha,\beta$ gives $\alpha=11/5, \beta=2/5$.


问:
About the property of $m$: if $n < m$ is co-prime to $m$, then $n$ is prime
The number $30$ has a curious property:

All numbers co-prime to it, which are between $1$ and $30$ (non-inclusive) are all prime numbers!

I tried searching(limited search, of course) for numbers $\gt 30$ that have this property, but could not find any.

Are there any such numbers $\gt 30$?

答:
There are no larger numbers, for which all the coprimes are primes.  While 30 itself is the largest such number, there are no multiples of 30 that contain eight primes, however, there are intervals of the form 30n+/-15, which do contain eight primes (such as around 1230).

The largest base that contains 'prime decades' is 18.  A prime decade is where all the co-primes exist between two multiples of the base are all prime.  For example, in decimal, the co-primes $101, 103, 107, 109$ are all prime.  Likewise, there are four primes $8.1, 8.5, 8.7, 8.11$ in base 12, are also prime.

In $base 18$, they are very rare, but do exist.  The second smallest starts with $43776$, and runs $7.9.2.1$, $7.9.2.5$, $7.9.2.7$, $7.9.2.11$, $7.9.2.13$, $7.9.2.17$ are all prime.  They are supprisingly rare.


问:
The equivalence of a 2 by 2 positive semidefinite matrix and a 3 by 3 positive semidefinite matrix?
I came across the following:
$$\begin{bmatrix}
-x^TAx-2b^Tx+c &-(Ax+b)^TR\\
-R(Ax+b) & \lambda I -RAR
\end{bmatrix}\geq0 \iff
\begin{bmatrix}
b^TA^{-1}b+c & 0 &(x+A^{-1}b)^T\\
0 & \lambda I &R\\
(x+A^{-1}b) & R &A^{-1}
\end{bmatrix}\geq0$$
where $A\in\mathbb{S}^n_+$ is a positive semidefinite matrix, and R can be either a symmetric matrix or a scalar.
This is not obvious to me. Can someone kindly teach me how is this equivalent?

答:
This is not a final answer, but the work may be useful regardless.
I'll denote the matrices as $M_2,M_3$ respectively. Then the Schur complement of the diagonal element $A^{-1}$ in $M_3$ is given by

\begin{align}
M_3\setminus A^{-1}&=
\begin{bmatrix} -b^\top A^{-1}b+c & 0\\ 0 & \lambda I\end{bmatrix}-\begin{bmatrix} (x+A^{-1}b)^T \\ R\end{bmatrix}A\begin{bmatrix} x+A^{-1}b & R\end{bmatrix}\\
&=\begin{bmatrix} -b^\top A^{-1}b+c & 0\\ 0 & \lambda I\end{bmatrix}
-\begin{bmatrix}(x+A^{-1}b)^T A(x+A^{-1}b) & (x+A^{-1}b)^TAR\\ RA (x+A^{-1}b) & RAR \end{bmatrix}\\
&=\begin{bmatrix} -b^\top A^{-1}b+c-(x^\top Ax +2b^\top x+b^\top A^{-1}b) & -(Ax +b)^TR\\ -R(Ax+b)  & \lambda I-RAR\end{bmatrix}
\end{align}

This is altogether close to
$$M_2=\begin{bmatrix}
x^TAx+2b^Tx+c &(Ax+b)^TR\\
R(Ax+b) & \lambda I -RAR
\end{bmatrix}.$$

But it isn't identical. One obvious difference is the sign of $R(Ax+b)$ in the corners. However, this may not be a big difference in context: If you're maximizing some $f(R)$ subject to $M_2\geq 0$, then relabeling $R\to -R$ isn't an issue. The more concerning problem is that the upper-left corner doesn't simplify the way one would want. For it to do so, one would first need to have $b^T A^{-1}b$ in the initial statement of $M_3$. But that would still only simplify to $$b^\top A^{-1}b+c-(x^T Ax +2b^\top x+b^\top A^{-1}b)=-x^T Ax-2b^T x+c$$
which is similar but not identical to the upper-left element of $M_2$.


问:
Non-principal Ultrafilter on Measurable Subsets of $\mathbb R$
Let $\Omega$ be the set of all measurable subsets of $\mathbb R$ ordered by inclusion. Is it possible to construct a non-principal ultrafilter on $\Omega$?

答:
I think that it is possible. Let $U$ be a non-principal ultrafilter on $\mathcal{P}(\mathbb{N})$.
Define $W := \{ A \in \Omega \mid  \exists X \in U,\ X\subset A\}$.
Then, $W$ is an ultrafilter on $\Omega$:
If $A,B \in W$, then there are $X_A , X_B \in U$ such that $X_A \subset A$ and $X_B \subset B$. Thus $X_A \cap X_B \subset A \cap B$.
If $A \in \Omega \setminus W$, then $A \cap \mathbb{N} \notin U$ so that $\mathbb{N}\setminus A \in U$. Since $\mathbb{N} \setminus A \subset \mathbb{R} \setminus A $, we have $\mathbb{R} \setminus A \in W$.


问:
Find surface area of cylinder $y^2 + z^2 = 1$ between two planes: $x + y - 2 = 0$ and $x-z+4 = 0$.
Find surface area of a cylinder $y^2 + z^2 = 1$ between two planes: $x + y - 2 = 0$,  $x-z+4 = 0$.

This was my approach:

I drew the picture and projected the figure onto $xOy$ plane, and after that I found partial detivatives $\frac{\partial z}{\partial x} $ and $ \frac{\partial z}{\partial y}$ where $z = + \sqrt{1-y^2}$. I choose the positive sqare root because of the symmetry, and that's why I will multiply following integral by $2$.

Eventually, my integral looks like this:

$P(S) = 2\begin{gather*}
    \iint_D \sqrt{1+ \frac{y^2}{1-y^2} + 0^2}\,dx\,dy =
 2\iint_D \frac{1}{\sqrt{1-y^2}}\,dx\,dy =
... = 9\pi
\end{gather*}
$ .

$D$ is $\{ -4 \leq x \leq 1 \land -1\leq y \leq 1  \} \cup \{ 1 \leq x \leq 3 \land  -1 \leq y \leq 2-x \}$

I wonder if this is correct and if anyone could tell me if I am wrong on this, I would really appreciate it.

答:
It is easier to integrate the surface in cylindrical coordinates, with $z=r\cos \theta$, $y=r\sin\theta$,  $x=x$ and around the circle of radius $1$

$$A=\int_0^{2\pi} (x_2-x_1)\cdot 1\cdot d\theta
= \int_0^{2\pi} (2-\sin\theta -(\cos\theta -4))d\theta=12\pi
$$


答: $\Omega$ contains the cofinite topology $\tau$ on $\mathbb R$, which $\tau$ can then be extended to a ultrafilter $U$ on $\mathbb R$ by the ultrafilter lemma. $U$ is non-principal as for any $a \in \mathbb R$, $I \setminus a \in \tau$. I believe $U \cap \Omega$ is then the desired non-principal ultrafilter.

Title: Non-principal Ultrafilter on Measurable Subsets of $\mathbb R$

Body: Let $\Omega$ be the set of all measurable subsets of $\mathbb R$ ordered by inclusion. Is it possible to construct a non-principal ultrafilter on $\Omega$?


问:
What do the elements of a ring $R/I$ look like?
I am a bit confused, as my class is currently on Section 9.2 of Dummit and Foote, and what elements of $R/I$ are of the form of, where $R$ is a polynomial ring.

For example, I am reading that $\mathbb{Z}[x,y]/(x^2,y^2,2)$ are of the form $a + bx + cy + dxy$, where $a,b,c,d \in \mathbb{Z}_2$. How is this deduced?

I suppose there is a definition $R/I = \{ r + I, r \in R \}$, but how does this form $a + bx + cy + dxy$, and can anyone give some other examples of how we can write elements of $R/I$ with a polynomial ring?

How should I be thinking about this more deeply?

答: You can think of quotienting by an ideal as a way of making the elements in the ideal become $0$. A simple example is $\mathbb{Z}/6\mathbb{Z}$ which is just the ring with elements $\{0,1,2,3,4,5\}$. We've made $6 = 0$ which in turn meant that $7 = 1, 8 = 2, ...$. In your example we are making $x^2, y^2$ and $2$ equal $0$. Having $x^2$ be $0$ means that any term of the form $ ax^ny^m = 0$ if $n \ge 2$. We have a similar situation for $y^2$. This means that your polynomial can only have the form $ a + bx + cy + dxy $ since the higher order terms will be $0$. The reason why the coefficients are in $Z_2$ is because we have quotiented by $2$ which means $2 = 0$. Hence all even numbers are actually $0$ since they can be written as $2n = 2\cdot 0 = 0$. All odd numbers are just equal to $1$ since they can be written as $2n + 1 = 2 \cdot 0 + 1 = 1$.


问:
find the condition on A for the summation to be convergent
The summation is:
$$\sum_{n=1}^\infty \frac{ \sqrt { n + 1 } - \sqrt n }{n^A}$$

I don't know how to even begin. Hints??

答:
In addition to both of neat answers, you could use the following facts to find proper values of $A$ as well:

Let $\lim_{n\to\infty}~n^pu_n=A$. Then:

If $p>1$ and $A$ is finite then $\sum u_n$ converges.

If $p\le1$ and $A\neq0$ or $A=\infty$ then $\sum u_n$ diverges.


问:
every ideal in a Noetherian ring is an intersection of ideals primary to maximal ideals?
I'm reading a lecture note Tight Closure of Huneke and he stated that "Every ideal in a Noetherian ring is
an intersection of ideals primary to maximal ideals". I don't see why this is the case.

I know that every ideal in a Noetherian ring has a primary decomposition into an intersection of ideals primary to prime ideals, but I don't see how I can refine this result.

Here the ring is of prime characteristic, but I don't think that will help.

答: Let $A$ be a ring, let $I$ be a proper ideal of $A$, let $\mathfrak{m}$ be a maximal ideal of $A$ which contains $I$. Then $I = \bigcap_{n \ge 1} (I + \mathfrak{m}^{n})$ by the Krull intersection theorem, and each $I + \mathfrak{m}^{n}$ is $\mathfrak{m}$-primary.


问:
How to perform vertical line test algebraically?
I understand how a function should have one value mapped to a single value in the range. Thus, when graphing the function I understand how a vertical line can determine if a point or multiple points intersect the vertical line. This is visual. However, how can we perform the same test algebraically across a domain of the function?

I am learning calculus from the start again since college, so in case my question is naive or obvious I apologize.

答:
A vertical line is the set of all points $(x_0,y)$ where $x_0$ is a fixed constant and $y$ is allowed to vary freely. Given an equation in the form $f(x,y)=0$, we say it passes the vertical line test if, for every fixed choice of $x_0$, the resulting equation $f(x_0,y)=0$ has either $0$ or $1$ solutions in terms of $y$.

For example, if we were checking the equation
$$ x = y^2 $$
we would define the function $f(x,y) = x - y^2$. If we choose $x_0=4$, notice that the equation
$$f(4,y) = 4 - y^2 = 0$$
fails the vertical line test since $y=2$ and $y=-2$ are both solutions.

As an exercise, see if you can prove that if $y$ is a function of $x$, such as $y=g(x)$, then the equation $f(x,y)=y-g(x)$ will always pass the vertical line test :)


问:
Prove that $x/(1 + \frac{2}{\pi} \cdot x) < \arctan(x)$ $\forall x > 0$
The first thing that I tried to do is to differentiate both functions and try to see if there establishes inequality that we want (considering that they are equal when $x = 0$). This attempt failed because firstly it really is true but after that we get opposing inequality. Also, I have noticed that $\lim \frac{x}{1 + x \cdot \frac{2}{\pi}} = \lim \arctan(x) = \frac{\pi}{2}$ as $x \rightarrow + \infty$ but it didn't lead me to solution. I also tried to apply Taylor's formula but it didn't help much either.

So, what are available ways to solve this problem?

答:
With the idea of @trancelocation:   the map $x\mapsto x/(1 + a x)= \frac{1}{a} ( 1 - \frac{1}{1+a x})$  is increasing on $(0, \infty)$  ( for $a>0$), so the inequality is equivalent to

$$x < \frac{ \arctan x}{ 1 - \frac{\arctan x}{\pi/2}} \textrm{ or } \tan t < \frac{t}{1- \frac{t}{\pi/2}}$$

We have the product expansions

$$\cos t = \prod_{n=1}^{\infty} \left(1- \frac{t^2}{(n-1/2)^2\pi^2} \right), \ \ \sin t = \prod_{n=1}^{\infty} \left(1- \frac{t^2}{n^2 \pi^2}\right)$$
so

$$\frac{\tan t }{t} = \frac{1}{1- (\frac{t}{\pi/2})^2} \cdot\prod_{n=1}^{\infty} \frac{\left(1- \frac{t^2}{n^2 \pi^2}\right)}{ \left(1- \frac{t^2}{(n+1/2)^2 \pi^2}\right)}< \frac{1}{1- (\frac{t}{\pi/2})^2}$$
for $0< t <\pi/2$. So we have in fact

\begin{equation}
\boxed{\color{indigo}{
 \frac{1-(\frac{t}{\pi})^2 }{1- (\frac{t}{\pi/2})^2}<\frac{\tan t}{t} < \frac{1}{1- (\frac{t}{\pi/2})^2}  < \sec t
}}
\end{equation}
for real $t$, $0<|t|< \frac{\pi}{2}$.

In the above inequalities, we have also the corresponding inequalities for the corresponding coefficients of the Taylor series, clear if we examine the product formulas.

From the second inequality we get
\begin{eqnarray}
\boxed{\color{indigo}{
\frac{\frac{\pi}{2} x}{ \sqrt{x^2 + (\frac{\pi}{4 })^2} + \frac{\pi}{4 }} < \arctan(x)
}}
\end{eqnarray}
for $x > 0$, an improvement over the inequality $\frac{\frac{ \pi }{2} x}{  x+\frac{\pi}{4}+\frac{\pi}{4}} < \arctan x$.


答: To say that a "relation" (or oftentimes "graph") is a "function" from A to B (A is the domain and B is the range) is to say that for any number a (in A) (so any number we take a "y value" at), there is exactly one number that this a gets mapped to. In practice, this typically amounts to checking how the "relation" is defined, and comparing it with this "exactly one" condition. Let's say our "relation" is f(x) = x (so the "identity function", sends numbers to themselves), we start with a number a, and f sends this a to f(a), which is just a in this case, so we start with a, and this a gets sent just to a, so where the function sends any number a (to itself) is definitely unique. $f(x) = x^2$ is another function, to check this is a function amounts to just checking that when you take any number x and square it, the output is unique (there is a single output, for example, -3 gets sent to the unique output $(-3)^2 = 9$) In this way, $x^2$ is a function. This process doesn't have to be "algebraic" it is just "clear" (or "rigorous") (more so than looking at a picture at least).

Title: How to perform vertical line test algebraically?

Body:
I understand how a function should have one value mapped to a single value in the range. Thus, when graphing the function I understand how a vertical line can determine if a point or multiple points intersect the vertical line. This is visual. However, how can we perform the same test algebraically across a domain of the function?

I am learning calculus from the start again since college, so in case my question is naive or obvious I apologize.


问:
How to write the negation of a biconditional?
I have two statements $p$ and $q$ such that:

Statement $p$: Ravi reads Mathematics

Statement $q$: Ravi reads Chemistry

Now, I am required to write the negation of $p \iff q$, i.e., I need to write $\lnot (p \iff q)$, which, if I understand correctly, is, by definition: $(p \land \lnot  q) \lor(\lnot p \land q )$

So, according to me, it should be as follows:

Ravi reads Mathematics and not Chemistry or Ravi doesn't read Mathematics and reads Chemistry.

But, the book states the answer to be as follows:

Ravi reads neither Mathematics nor Chemistry

Are the two statements equivalent? Or am I going wrong here?

Check question 3 (e) for the question asked here

The answer for question 3 (e)

答:
Your reasoning is correct. You can rewrite the negation of the biconditional as $ (p \land \lnot  q) \lor(\lnot p \land q )$ which is the same as your english sentence. The sentence your book has is equivalent to $\lnot p \land \lnot q$ which is certainly not equivalent to $\lnot (p \iff q)$ since they don't have the same truth value when $p,q$ are both false.

Either your book is wrong or you have misunderstood what the problem is asking. Maybe it was asking for the negation of the disjunction, $\lnot (p \lor q)?$

Edit: Now that you've posted pictures of the book I can definitely say that the book is wrong.


答:
How can an object belong to itself?

Well, it depends on what "belong" means, doesn't it? I'm inclined to agree with you, but this is a somewhat controversial point; some people have found uses for set theories where a set can belong to itself.

In a sane world, where things can't belong to themselves, Russell's paradox would take the simpler form:
$$R=\{x:x=x\}$$
Now $R=R$, so $R\in R$, which is absurd. Paradox!

But what if someone objects: "I'm not so sure. Normal sets can't belong to themselves, but maybe there are some abnormal sets that do belong to themselves? Maybe your $R$ is one of those?"

To satisfy those people, instead of defining $R$ to be the set of all sets, we define it to be the set of all normal sets:
$$R=\{x:x\not\in x\}$$
And now, as you know, we get a paradox, regardless of whether abnormal sets exist or don't.

Title: Notation used in Russell's paradox


问:
Evaluate $\int_{-\infty}^{-1}e^{\lambda t}\operatorname{Log}\left(\frac{\lambda+1}{\lambda}\right)\Bbb d\lambda$
This question is still related to my question 3 days ago about inverse laplace transform. Here is the link: enter link description here.
Now, i'm working on path $L_1$ which led me to this integral when i set the radius around branch point goes to $0$. Now forget about my question 3 days ago, and my question here is how to evaluate this improper integral:

$$\int_{-\infty}^{-1}e^{\lambda t}\operatorname{Log}\left(\frac{\lambda+1}{\lambda}\right) \Bbb d\lambda$$.

$\lambda$ is now real variable instead of complex variable since my path goes close to real axis. I need to compute this integral but the problem is $t$ is unkown and i can't approximate it using geogebra. How to process this? I haven't tried to taylor it then integrating term by term cz i'm not sure if the given answer will be as my desired result.

答:
Note that $\ln\left(\frac{\lambda+1}\lambda\right)=\ln(\lambda+1)-\ln\lambda$. Given this, we write

$$\int^{-1}_{-\infty} e^{\lambda t}(\ln(\lambda+1)-\ln\lambda)d\lambda=\int^{-1}_{-\infty} e^{\lambda t}\ln(\lambda+1)d\lambda-\int^{-1}_{-\infty} e^{\lambda t}\ln\lambda d\lambda$$

Let's first evaluate $\int^{-1}_{-\infty} e^{\lambda t}\ln\lambda d\lambda$ since it looks simpler.

We first substitute $u=\lambda t$, which means $\lambda=\frac ut$ and $d\lambda=\frac1t du$.

$$\int^{-1}_{-\infty} e^{\lambda t}\ln\lambda d\lambda=\frac1t\int^{-t}_{-\infty} e^u\ln\frac ut du=\frac1t\int^{-t}_{-\infty} e^u(\ln u-\ln t) du=\frac1t\left(\int^{-t}_{-\infty} e^u\ln u du-\int^{-t}_{-\infty}\ln te^udu\right)$$

$\int^{-t}_{-\infty}\ln te^udu$ is trivial. One can clearly see $\int^{-t}_{-\infty}\ln te^udu=\ln t\int^{-t}_{-\infty}e^udu=\ln t \left.e^u\right|^{-t}_{-\infty}=(\ln t\pm i\pi) e^{-t}$

$\int^{-t}_{-\infty} e^u\ln u du$ is trickier, but doable. We use integration by parts.

Let $v=\ln u$, $dw=e^u du$. Then $dv=\frac1u$, $w=e^u$.

Thus, we have

$$\int^{-t}_{-\infty} e^u\ln u du=\left. \ln u e^u\right|^{-t}_{-\infty}-\int^{-t}_{-\infty}\frac{e^u}u du$$

$\left.\ln u e^u\right|^{-t}_{-\infty}=\ln(-t)e^{-t}-\ln(-\infty)e^{-\infty}=(\ln t+i\pi)e^{-t}-\left(\ln\infty+i\pi\right) e^{-\infty}$. You can show that $\lim_{x\to\infty}\ln xe^{-x}=0$, so $(\ln t+i\pi)e^{-t}-\left(\ln\infty+i\pi\right) e^{-\infty}=(\ln t\pm i\pi)e^{-t}.$

$\int^{-t}_{-\infty}\frac{e^u}u du$ is the Exponential integral, which has no closed form. However, we can write $\int^{-t}_{-\infty}\frac{e^u}u du=\operatorname{Ei}(-t)$.

Combining the two integrals, we have

$$\int^{-1}_{-\infty} e^{\lambda t}\ln\lambda d\lambda=\frac1t\left((\ln t\pm i\pi)e^{-t}-\operatorname{Ei}(-t))-(\ln t\pm i\pi)e^{-t}\right)=-\frac1t\operatorname{Ei}(-t)$$

To evaluate $\int^{-1}_{-\infty} e^{\lambda t}\ln(\lambda+1)d\lambda$, we first write $u=\lambda+1$, which means $\lambda=u-1$ and $du=d\lambda$. Hence

$$\int^{-1}_{-\infty} e^{\lambda t}\ln(\lambda+1)d\lambda=\int^0_{-\infty} e^{(u-1)t}\ln udu=\int^0_{-\infty} e^{ut}e^{-t}\ln udu=e^{-t}\int^0_{-\infty} e^{ut}\ln udu$$

Does $\int^0_{-\infty} e^{ut}\ln udu$ look familiar to you? It should, since we evaluated this $\int^{-1}_{-\infty} e^{ut}\ln udu$ earlier. Using the same substitutions, but replacing $-1$ and $-t$ with $0$ in the upper bound, we arrive with

$$\int^0_{-\infty} e^{\lambda t}\ln(\lambda+1)d\lambda=\frac1t\left(\ln u \left.e^u\right|^0_{-\infty}-\operatorname{Ei}(0)-\ln t \left.e^u\right|^0_{-\infty}\right)=\frac1t(\ln0-\operatorname{Ei}(0)-\ln t)$$

Inserting this back in,

$$e^{-t}\int^{-1}_{-\infty} e^{ut}\ln udu=\frac{e^{-t}}t(\ln0-\operatorname{Ei}(0)-\ln t)$$

$\ln0-\operatorname{Ei}(0)$ is undetermined as $\ln0=\operatorname{Ei}(0)=-\infty$, but we can squeeze some meaning out of it by interpreting it as $\lim_{x\to0}\ln x-\operatorname{Ei}(x)$ instead. To evaluate this, we first use the integral definitions of $\ln x$ $\operatorname{Ei}(x)$:

$$\lim_{x\to0}\ln x-\operatorname{Ei}(x)=\lim_{x\to0}\int^x_1\frac1zdz-\int^x_{-\infty}\frac{e^z}zdz$$
$$=\int^x_1\frac1zdz-\int^x_1\frac{e^z}zdz-\int^1_{-\infty}\frac{e^z}zdz$$

You may recall that $\int^b_a [f(x)-g(x)]dx=\int^b_a f(x)dx-\int^b_a g(x)dx$. The opposite also applies, i.e $\int^b_a f(x)dx-\int^b_a g(x)dx=\int^b_a [f(x)-g(x)]dx$. Knowing this, we can combine $\int^x_1\frac1zdz-\int^x_1\frac{e^z}zdz$:

$$\lim_{x\to0}\int^x_1\frac1zdz-\int^x_1\frac{e^z}zdz-\int^1_{-\infty}\frac{e^z}zdz=\lim_{x\to0}\int^x_1\frac{1-e^z}zdz-\operatorname{Ei}(1)$$

Now apply the sum definition of $e^z$:

$$\lim_{x\to0}\int^x_1\frac{1-e^z}zdz-\operatorname{Ei}(1)=\lim_{x\to0}\int^x_1\frac{1-\sum^\infty_{n=0}\frac{z^n}{n!}}zdz-\operatorname{Ei}(1)$$

$$=\lim_{x\to0}\int^x_1\frac{1-1-\sum^\infty_{n=1}\frac{z^n}{n!}}zdz-\operatorname{Ei}(1)$$

$$=\lim_{x\to0}\int^x_1\sum^\infty_{n=1}\frac{z^{n-1}}{n!}dz-\operatorname{Ei}(1)$$

$$=\sum^\infty_{n=1}\frac1{n!}\int^1_0z^{n-1}dz-\operatorname{Ei}(1)=\sum^\infty_{n=1}\frac1{n!}\cdot\frac1n-\operatorname{Ei}(1)$$

Interestingly, according to MathWorld $\operatorname{Ei}(x)=\gamma+\ln x+\sum^\infty_{n=1}\frac{x^n}{n\cdot n!}$. This means $\operatorname{Ei}(1)=\gamma+\ln 1+\sum^\infty_{n=1}\frac{1^n}{n\cdot n!}\Rightarrow \sum^\infty_{n=1}\frac1{n\cdot n!}=\operatorname{Ei}(1)-\gamma$. An interesting coincidence. Inserting this back:

$$\sum^\infty_{n=1}\frac1{n!}\cdot\frac1n-\operatorname{Ei}(1)=\operatorname{Ei}(1)-\gamma+\operatorname{Ei}(1)=-\gamma$$

Hence,

$$\int^{-1}_{-\infty} e^{\lambda t}\ln(\lambda+1)d\lambda=\frac{e^{-t}}t(-\gamma-\ln t)=-\frac{e^{-t}}t(\gamma+\ln t)$$

Thus,

$$\int^{-1}_{-\infty} e^{\lambda t}\ln(\lambda+1)d\lambda-\int^{-1}_{-\infty} e^{\lambda t}\ln\lambda d\lambda=-\frac{e^{-t}}t(\gamma+\ln t)+\frac{\operatorname{Ei}(-t)}t$$

(Note 1: If $\Gamma(x,y)$ refers to the upper incomplete gamma function, $\Gamma(0,t)=\int^\infty_t s^{-1}e^{-s} ds=\int^\infty_t\frac{e^{-s}}s ds$. Let $z=-s$, $-dz=ds$. $\int^\infty_t\frac{e^{-s}}s ds=\int^{-\infty}_{-t}\frac{e^z}{-z}-dz=-\int^{-t}_{\infty}\frac{e^z}zdz=-\operatorname{Ei}(-t)$. Hence, $\Gamma(0,t)=-\operatorname{Ei}(-t)$, and the $-\frac{\Gamma(0,t)}t$ that appears in Mathematica's answer is equivalent to the $\frac{\operatorname{Ei}(-t)}t$ that appears in my answer.)

(Note 2: Note that $\ln -t=\ln t+i\pi$ or $\ln -t=\ln t-i\pi$ depending on which branch of $\ln t$ you use, but considering $\frac{\lambda+1}{\lambda}$ is always nonnegative, $\ln\left(\frac{\lambda+1}{\lambda}\right)$ is always real and hence there should be no complex terms.)


问:
Show that $\langle X\rangle$ is a subgroup of $G$
Let $G$ a group and $X\subseteq G$ show that $\langle X\rangle=\left\{x_{1}^{r_{1}} x_{2}^{r_{2}} \ldots x_{n}^{r_{n}} \mid n \geq 1, x_{i} \in X, r_{i} \in \mathbb{Z}\right\}$ is a subgroup of G

How $0 \in \mathbb{Z}$ then exist $x\in \langle X\rangle$ such that $x=x_{i}^0=e$ then $e\in\langle X\rangle$

Let $x=x_{1}^{r_1}...x_{n}^{r_n}$ and  $y=y_{1}^{s_1}...x_{k}^{s_k}$ elements in $x\in \langle X\rangle$

then $xy=x_{1}^{r_1}...x_{n}^{r_n}y_{1}^{s_1}...x_{k}^{s_k}$ but i dont know  how to justify xy is in $\langle X\rangle$

答:
This is somwhere between an answer and a comment: So are you assuming $x_1, \ldots, x_n$ are different elements of $X$? Or can the $x_i$ repeat. From your question I took it to mean the former.

Unless you know anything further about the structure of $G$, there is no guarantee that each $x\in \langle X \rangle$ can be written as a word of length no more than $|X|$, where the letters are $x^m$; $x \in X$ and $m$ a positive integer. Indeed, there are large groups $G$ with a set $X$ of 2 generators $a$ and $b$; $a$ and $b$ of low order; where most elements $y$ in $G$ require a word of length $\log |G|$ i.e., $y=a^{m_1}b^{m_2}a^{m_3}b^{m_4} a^{m_5} \ldots a^{m_{2r+1}}$ where $r$ is $\log |G|$, and there is no more succinct way to write $y$ from $a$ and $b$.

If you are allowing words of any length and $G$ is finite, note that for each $x \in X$, the element $x^{-1}$ is in $X$; indeed $x^{|G|} = e$ so $x^{|G|-1} = x^{-1}$. So you can use this to see that $\langle X \rangle$ is closed under inversion.


问:
Spanning trees of ladder graphs...
a. Draw the 1-ladder, 2-ladder, and 3-ladder graphs, and calculate the number of spanning
trees for each.
- I have completed this part and wanted to confirm that these numbers look accurate, I feel like i was able to get all of them.

b. A spanning tree of a connected graph $G$ is a a tree which contains all the vertices
of $G$ and which is a subgraph of $G$. If $T(n)$ is the number of spanning trees for the $n$-ladder, then
$T(n)=4 · T(n - 1) - T(n - 2)$. Solve this recurrence relation to obtain a closed formula the number of spanning trees for a $n$-ladder.
I don't really know how to do this part of the problem and would appreciate any help.

答:
Another interesting "ladder" graph is the so-called Mobius ladder $M_n$ of order $n$. Namely, take cycle graph with $2n$ vertices and then add $n$ edges that connect every vertex with its diametrically opposite counterpart. Another way to describe it is to take a regular ladder graph with $n+1$ levels and then glue the two vertices of level $1$ with two vertices of level $n+1$, flipping the sides (left vertex of level $1$ will become the same as the right vertex of level $n+1$, and vice versa).

Then there exists a very similar formula for $t(M_n)$, that is, for the number of spanning trees of $M_n$.

Alas, I never saw any "decent" formula for $t(Sq_{\,n,m})$ where $Sq_{\,n,m}$ is the graph of two-dimensional square grid with dimensions $m\times n$ (this would be the most obvious generalization of order $n$ ladder graph). I am sure it is possible to produce a fairly manageable formula for $t(Sq_{\,3,m})$ but it probably looks rather ugly... although I just might be wrong.

Title: Spanning trees of ladder graphs...


答:
Since $x$ and $y$ are finite products of elements in $X$, so is $xy$, so $xy\in\langle X\rangle$.

The identity and inverses are clearly in $\langle X\rangle$, being the empty product and elementwise inversion respectively, so $\langle X\rangle$ is a group.

Title: Show that $\langle X\rangle$ is a subgroup of $G$

Body:
Let $G$ a group and $X\subseteq G$ show that $\langle X\rangle=\left\{x_{1}^{r_{1}} x_{2}^{r_{2}} \ldots x_{n}^{r_{n}} \mid n \geq 1, x_{i} \in X, r_{i} \in \mathbb{Z}\right\}$ is a subgroup of G

How $0 \in \mathbb{Z}$ then exist $x\in \langle X\rangle$ such that $x=x_{i}^0=e$ then $e\in\langle X\rangle$

Let $x=x_{1}^{r_1}...x_{n}^{r_n}$ and  $y=y_{1}^{s_1}...x_{k}^{s_k}$ elements in $x\in \langle X\rangle$

then $xy=x_{1}^{r_1}...x_{n}^{r_n}y_{1}^{s_1}...x_{k}^{s_k}$ but i dont know  how to justify xy is in $\langle X\rangle$


问:
An area of a convex polyhedral surface homeomorphic to a sphere vs an area of its shadow
Consider a convex polyhedral surface $P$ in $\mathbb{R}^3$. If $\pi_v : \mathbb{R}^3\rightarrow
\{ x\in \mathbb{R}^3| x\perp v\}$ where $v\in \mathbb{S}^2$ is an orthogonal projection, then $$
\frac{1}{{\rm area}\ \mathbb{S}^2 }\int_{v\in \mathbb{S}^2} \ {\rm area}\ \pi_v(P)\ d{\rm Vol}_v = \frac{1}{4}\cdot {\rm area}\ (P) $$

This can be proved by using mutivariable calculus. But by using the fact that unit ball in $\mathbb{R}^2$ has an area, which is $\frac{1}{4}$ of area of $\mathbb{S}^2$, how can prove this ?

答:
This is true not just for polyhedral surfaces, but for any convex body in $\mathbb{R}^3$.

To reformulate the problem in a much more intuitive sense, this is saying that if we project a convex set onto a plane oriented in a random direction, the expected area of the "shadow" cast onto the plane by a perpendicular beam of light is $1/4$ of the surface area of $S$. (Equivalently, we can think of holding $S$ in a random orientation and projecting light onto a fixed horizontal plane.)

To prove this, we use linearity of expectation. Consider some very small part of the surface of $S$. We say that its contribution to the shadow is $0$ if the beam of light does not touch it, and is equal to the area of its projection if the beam of light does reach it. (This way, we don't overcount the shadow area from both the top and bottom parts.) Then obviously the expected area of the shadow is just the integral of the expected contributions from infinitesimal regions of the surface of $S$.

But because $S$ is convex, it's impossible for a portion of the surface to be in the shadow of another part of the surface - it has no "overhangs". So the contribution of a small patch of area $\epsilon$ depends only on the direction of the beam, and not on the rest of $S$ - this means that the expected contribution is just linear in $\epsilon$, because all small patches of $S$ act the same.

So the expected shadow area of a convex set $S$ is $C\cdot\text{Area}(S)$, for some universal constant $C$ independent of $S$. Taking $S$ to be the sphere, it is obvious that the shadow area is always $\pi$ and the surface area is $4\pi$, so we must have $C=1/4$.


问:
If $x\in L$ lies in exactly one Cartan subalgebra, then $x$ is regular

(Question 15.3 in Humphreys's book Introduction to Lie Algebras) Let $L$ be semisimple (char $\mathbb F = 0$), $x\in L$ semisimple. Prove that if $x$ lies in exactly one CSA (Cartan subalgebra), then $x$ is regular.

Definitions. An element $x$ is said regular if $C_L(x) = \{y\in L: [y,x] = 0\}$ is a maximal toral subalgebra. In particular, when $L$ is semisimple, being maximal toral is equivalent to being a CSA which is equivalent to being a minimal Engel subalgebra.
An Engel subalgebra is one of the form $L_0(\mbox{ad }x) = \{y \in L: (\mbox{ad }x)^ky =0 \mbox{ for some integer } k \}, x\in L$.

Remark. This question have been asked here Cartan Subalgebra and regular elements and here $x$ regular $\Leftrightarrow$ $x$ is in exactly one CSA. Both of them lack of an answer. Moreover, the suggestion given in the latter is in my opinion wrong, since taking "$y\in C_L(x)\setminus H$ such that $L_0(\mbox{ad }y)$ is minimal" (I'm assuming he meant minimal among all the $y\in C_L(x)\setminus H$) does not ensure that it is minimal among all Engel subalgebras. I've tried to work a little bit more with this argument as follows. Of course, if $L_0(\mbox{ad }y)$ is not minimal, it will contain a minimal Engel subalgebra, say $L_0(\mbox{ad }z)$ ($L$ has finite dimension). But I can't see how to argue that $x\in L_0(\mbox{ad } z)$ to obtain the absurd of $x$ lying both in $H$ and $L_0(\mbox{ad } z)$.

Some ideas. Suppose $x$ is not regular. On one hand, being semisimple implies that $C_L(x) = L_0(\mbox{ad } x)$. Furthermore, there exists a maximal toral $H\subseteq C_L(x),$ where $H$ is the subalgebra spanned by the semisimple elements of $L$. Since $H = C_L(H)$, it follows that $x\in H$. On the other hand, $x$ being irregular implies that $L_0(\mbox{ad }x) = C_L(x)$ is not miminal, so let us take $L_0(\mbox{ad }y)\subseteq C_L(x)$ properly and minimal. In particular, $y\in C_L(x)$ gives us that $x\in L_0(\mbox{ad }y).$  Both $H$ and $L_0(\mbox{ad }y)$ are CSA, so it remains to prove that they are not the same to obtain a contradiction. $H$ and $L_0(\mbox{ad }y)$ are too generic, I can't see a way to argue that. I'm stuck and out of ideas.

Any insight or help is very much appreciated.

答:
I've finally found a proof which works for any (finite-dimensional) semisimple Lie algebra $L$ over an arbitrary field of characteristic $0$ (and possibly even more). It rests on several "well-known" facts, which, however, when you don't accept them as well-known, are not exactly trivial. In the following, the reference "Bourbaki" is obviously to their books on Lie Groups and Algebras, actually all to chapter VII of that.

Remember that the general definition of a Cartan subalgebra of any Lie algebra is: a subalgebra which is nilpotent and self-normalising. For semisimple Lie algebras, this is equivalent to being a maximal toral subalgebra, where a subalgebra is called "toral" if it is abelian and consists of $ad$-semisimple elements. Cf. 1, 2, 3, 4.

Fact 1: If $x \in L$ is an $ad$-semisimple element, then
$$\lbrace \text{Cartan subalgebras of } L \text{ containing } x \rbrace = \lbrace \text{Cartan subalgebras of } C_L(x) \rbrace. $$

(This is a special case  of Bourbaki, chap. VII, §2 no. 3 Proposition 10, where indeed we don't need $L$ to be semisimple. The inclusion "$\subseteq$" is straightforward in particular in the case of semisimple $L$, from the known equivalence of the definitions in that case. For "$\supseteq$",  let $\mathfrak h$ be a CSA of $C_L(x)$. Clearly $\mathfrak h$ is nilpotent and contains $x$. Further, if $N:=N_L(\mathfrak h)$ is the normaliser of $\mathfrak h$ in $L$, then because $ad(x)(N) \subseteq \mathfrak h$ and $x$ is $ad$-semisimple, there is a decomposition into $ad(x)$-stable vector spaces $N = \mathfrak h \oplus M$ where further $ad(x)(M) = 0$, meaning that $M$ and hence all of $N$ is contained in $C_L(x)$. But then $N = N_{\color{red}{C_L(x)}}(\mathfrak h)$ which, by $\mathfrak h$ being a CSA in $C_L(x)$, is just $=\mathfrak h$.)

Fact 2: If $x \in L$ is a semisimple element, then its centraliser $C_L(x)$ is a reductive Lie algebra.

(See e.g. here, although the proof there uses an algebraically closed field / split semisimple $L$, but then it's really straightforward to reduce to that case: Because the centraliser in a scalar extension is the scalar extension of the centraliser, and a Lie algebra is reductive iff its scalar extension is. Actually, the stronger fact that for any toral subalgebra $\mathfrak a \subset L$, the centraliser $C_L(\mathfrak a)$ is a reductive Lie algebra, is Corollary 3.1.21 in my thesis, where it's also proved via scalar extension. It is also a special case of Bourbaki, chap. VII §1 no. 5 Proposition 13.)

Fact 3: The Cartan subalgebras of a reductive Lie algebra $\mathfrak r$, written as $\mathfrak z \oplus \mathfrak s$ with centre $\mathfrak z$ and semisimple $\mathfrak s$,  are exactly the subalgebras of the form $\mathfrak z \oplus \mathfrak c$ where $\mathfrak c$ is a Cartan subalgebra of $\mathfrak s$.

(This should be clear from the general definition of CSAs as subalgebras which are nilpotent and self-normalising; indeed, in Bourbaki chap VII §2 no. 1 it is mentioned right after that definition that the only CSA of any nilpotent Lie algebra (in particular, an abelian one like $\mathfrak z$) is itself, and it is Proposition 2 that the CSAs of a direct product are the direct products of the CSAs.)

Fact 4: Any semisimple Lie algebra $\mathfrak s \neq 0$ contains more than one CSA.

(Not as trivial as it sounds. But for sure there are many ways to see this in general: Indeed, Bourbaki shows that over any infinite field, every non-nilpotent Lie algebra contains infinitely many CSAs (see my answer to this question). Here is an ad hoc argument: W.l.o.g. $\mathfrak s$ is simple. In a semisimple Lie algebra, every element can be decomposed into a semisimple and a nilpotent part which commute. Basically, we have to find two semisimple elements $x_1, x_2$ which don't commute, because then $x_1$ is contained in a CSA which, by virtue of being abelian (toral!), cannot contain $x_2$, and vice versa, so there are at least two different CSAs. -- Now, if $\mathfrak s$ contains no nonzero nilpotent (this happens, e.g. compact forms over $\mathbb R$), then all its elements are semisimple, but since $\mathfrak s$ is not abelian we are done. So assume there is a nilpotent element $y$. Then, like in Levent's answer, we extend $y$ to an $\mathfrak sl_2$-triple $y,h,z$ and can explicitly find non-commuting semisimple elements from that, e.g. like in his answer, or we note that the inner automorphism $e^{ad(y)}$ of $\mathfrak s$ sends the semisimple $h$ to the semisimple $h \mp 2 y$ (sign depending on how you define $\mathfrak sl_2$-triples), and these two do not commute with each other.)

Now let's go, let $x$ be a semisimple element of $L$. By facts 2, 3, and 4, the Lie algebra $C_L(x)$, in and of itself, contains a unique CSA if and only if it is abelian, if and only if it is itself a CSA (in itself); otherwise it contains several CSAs.

Via fact 1, this translates to:

There is a unique CSA in $L$ containing $x$ if and only if $C_L(x)$ is a CSA in $L$; which by your definition is equivalent to $x$ being regular.


问:
Prove $\sin(\pi/2-x)\cot(x+\pi/2) = -\sin(x)$
I've managed to use cofunction identities to get the left side of the equation to (pi/2-x)-tanx. From here, I keep ending up at -cotx? I'm pretty sure that's wrong, and I have no idea how to to go about proving this in another way.</p>

答:
Just muck it out.

The following are basic identities.

$\sin(\frac \pi 2-x) =\cos x$

$\sin(x+\frac \pi 2) = \cos x$

$\cos(x+\frac \pi  2) = \sin x$.

We can know this by:

Rote memory

Considering that by considering angles $\frac \pi 2 -x$ or $x +\frac \pi 2$ is simply a matter of rotating a unit circle a quarter turn and possibly reflecting the circle on an axis and thus the $x,y$ cooridinates will "switch".  (It helps to draw a picture.

Considering $\sin (A+ B) = \sin A\cos B + \cos A \sin B$ and $\cos (A+B) =\cos A\cos B -\sin A\sin B$ and so

$\sin (\frac \pi 2 - x) = \sin \frac \pi 2\cos x + \cos \frac \pi 2 \sin(-x)=1\cdot \cos x + 0\cdot (-\sin x) = \cos x$.

$\sin ( x+ \frac \pi 2) = \sin x\cos \frac \pi 2 + \cos x \sin(\frac \pi 2)=\sin x \cdot 0 + \cos x\cdot 1 = \cos x$

$\cos(x+ \frac \pi 2) = \cos x\cos\frac \pi 2 - \sin x \sin \frac \pi 2=\cos x \cdot 0 -\sin x \cdot 1=-\sin x$.

ANd as a result of these three identities.

$\sin(\frac \pi -x)\cot (x+\frac \pi 2) = $

$\sin(\frac \pi -x)\frac {\cos (x+\frac \pi 2)}{\sin (x+\frac \pi 2)}=$

$\cos x\frac {-\sin x}{\cos x}= $

$-\sin x$.

That is ..... IF $\cos x\ne 0$.  Or in otherwords if $x \ne \pm\frac \pi 2$.

If $x =\pm \frac \pi 2$ then $\cot (x+\frac \pi 2)$ is not defined.


问:
How can we calculate $\int_{r\in \mathbb{S}^2}\ {\rm Area}\ \Delta p_0q_0r\ d{\rm Area}_r$
Define a function $f : \mathbb{S}^2\times \mathbb{S}^2\times \mathbb{S}^2\rightarrow \mathbb{R}$ by $f(p,q,r)$ to be a area of the geodesic triangle $pqr$ in the unit sphere $\mathbb{S}^2$.

(Here the area of the triangle $pqr$ is $\angle p + \angle q +\angle r -\pi$).

Question 1 : Here how can we find $$ \int_{\mathbb{S}^2\times \mathbb{S}^2\times \mathbb{S}^2 } \ f(p,q,r) d{\rm Vol}(p,q,r) $$

Question 2 : Furthermore, how can we find the following $$\int_{\mathbb{S}^2 } \ f(p_0,q_0,r) d{\rm Vol}(r) $$

Question 3 : Consider the triangle $\Delta pqr$ where the order of $p,\ q,\ r$ is
positively oriented. And define
$$A = \frac{ q\times p }{|
 q\times p| },\ B = \frac{r\times q}{|r\times q|},\ C =
 \frac{p\times r}{|p\times r|} $$

Note that $\angle \ (A,B) =\pi-\angle p $ so that $$ \angle (A,B) +
\angle (B,C)+\angle (A,C) = 2\pi - {\rm Area}\ \Delta pqr $$

Hence ${\rm perim}\ \Delta ABC$, the sum of all side lengths in
$\Delta ABC$, is equal to $2\pi - {\rm Area}\ \Delta pqr$.

Hence we want to calculate $$ \int_{(A,B,C)\in \mathbb{S}^2
 \times\mathbb{S}^2\times
\mathbb{S}^2}\ {\rm perim}\ \Delta ABC \ d{\rm Vol}_{ (A,B,C) }$$

答:
The area of a spherical triangle $ABC$ on a unit sphere with respective angles $\alpha,\beta,\gamma$ in radians has an extremely nice formula:

$$\text{Area}(\triangle ABC) = \alpha+\beta+\gamma-\pi$$

There is a very cute proof of this formula using some clever geometric arguments given here among other places.

Because this formula is linear in the angles, we can just use linearity of expectation: compute the average value $x$ of an angle in a spherical triangle, and our answer will be $x+x+x-\pi$.

Without loss of generality, let $A$ be at the north pole, and fix $B$ to be on the Prime Meridian. Then the longitude of $C$ will be uniformly distributed on $[-180,180]$, so the angle $\alpha$ is uniformly distributed between $0$ and $\pi$; in particular, its expectation is $\pi/2$. So the expected area is simply

$$3\cdot\frac{\pi}{2}-\pi = \frac{\pi}2$$

or $1/8$ of the sphere.


问:
Solve $(1+x) d y-y d x=0$
Here is the worked example by the author,

$$
\int \frac{d y}{y}=\int \frac{d x}{1+x}
$$

$$
\ln |y|=\ln |1+x|+c_{1}
$$

$$
|y|=e^{\ln |1+x|+c_{1}}=e^{\ln |1+x|} \cdot e^{c_{1}}=|1+x| e^{c_{1}}
$$

$$
y=\pm e^{c_{1}}(1+x)
$$

Relabeling $\pm e^{c_{1}}$ as $c$ then gives $y=c(1+x)$.

My question: Shouldn't you say $c=\pm e^{c_1},0$ because $y=0$ is a constant solution?

I guess that when $x=-1,y=0$ so I thought maybe that covers the solution $y=0$, but $(1+x) d y-y d x=0$ can be rearranged to $\frac{d y}{d x}=\frac{y}{1+x}$ and this implies that $x \neq-1$, so the constant solution $y=0$ cannot be achieved from $y=c(1+x)$ where $c=\pm e^{c_1}$, is what I think.. Thanks!

答: When dividing by $y$ you are assuming that $y$ is not the zero function – it may be zero at isolated points. So what happens at $x=-1$ does not matter for this problem. It turns out that the $y=0$ case can be subsumed into the general solution, since it fills the gap left by the expression $c=\pm e^{c_1}$ (which implies $c$ may be any real number except zero). Hence $c$ may be any real number.

Title: Solve $(1+x) d y-y d x=0$


问:
Partition $3n$ people into $n$ groups of $3$ people with conditions
There are $N$ number of sons, fathers and mothers. How to partition $3n$ of them into $n$ groups of $3$ people such that in each group is one son, one father and one mother but no son with his own father and mother?

Thanks in advance for giving me an approach or a hint?

答: I think the easiest approach would be to start with the sons, then square the number of derangements of $N$ because each father and each mother can be with anyone but their own son. For instance, for $N=2$ you have one solution (the parents swap children or vice versa) and for $N=3$ you have two options for each parent-type, hence four solutions in total.


问:
Unexpected result on the number of permutations with a restriction.
Let $p=(p_1,p_2,\dots,p_n)$ be a weak composition of a positive integer number $n$ into $n$ non-negative integer parts and let $k_i$ be the count of the part $i$ ($i=0,1,2,\dots$) in the composition.

A permutation $\tilde p$ of $p$ is allowed if its partial sums obey the inequality:
$$
\forall m (1\le m\le n):\ s_m\equiv\sum_{j=1}^m \tilde p_j\ge m.
$$

The simple but rather unexpected result valid by numerical evidence  is that the number of allowed permutations is
$$
\frac{n!}{\displaystyle(k_0+1)!\prod_{i>0}k_i!}.\tag1
$$

This means that the number is exactly $n+1$ times less than the number of all permutations of the composition supplemented by an additional 0. This simplicity suggests that there should be almost obvious
explanation of the fact but I could not find it.

Any hint is appreciated.

答:
Given any weak composition $p = (p_1, p_2, \ldots, p_n)$, append an extra $0$ to it, and consider the rearrangement $ p' = (p_1' p_2', \ldots , p_n', p_{n+1}' )$.

There are $\frac{ (n+1)! } { (k_0 + 1) ! \prod k_i ! } $ such rearrangements.

We can naturally pair them up into groups of size $n+1$, obtained by cyclically permuting the indices.

For such a rearrangement, we associate it to the "up and right path" of:

Start at $(0,0)$,

Go up $p'_1$, go right 1.

Go up $p'_2$, go right 1.

$\ldots $

Go up $p'_n$, go right 1.

Go up $p'_{n+1}$, which brings us to $(n, n)$.

We use the bijection as described in the third proof of Catalan's number. The following  is a section of the writeup which contains the necessary information, though I encourage you to read the full details on Wikipedia for more context.

The exceedance of the  path is defined to be the number of vertical edges which lie above the diagonal.

Given a path whose exceedance is not 0, then we may apply the following algorithm to construct a new path whose exceedance is one less than the one we started with.

Starting from the bottom left, follow the path until it first travels above the diagonal.

Continue to follow the path until it touches the diagonal again. Denote by X the first such edge that is reached.

Swap the portion of the path occurring before X with the portion occurring after X.

The algorithm will cause the exceedance to decrease by one, for any path that we feed it, because the first vertical step starting on the diagonal <s>(at the point marked with a black dot)</s> is the unique vertical edge that under the operation passes from above the diagonal to below it; all other vertical edges stay on the same side of the diagonal.

It is also not difficult to see that this process is reversible: given any path P whose exceedance is less than n, there is exactly one path which yields P when the algorithm is applied to it.

Using this algorithm, we conclude that the $n+1$ cyclic permutations each have a unique exceedance. Since this exceedance ranges from 0 to $n$, and there are $n+1$ of them, hence it's exactly the integers from 0 to $n$ (inclusive).

Here's the bijection for the case of $n = 3$.

The rows are cyclic permutations of $ (3,0,0,0), (2, 1, 0, 0), (1, 2, 0, 0), (2, 0, 1, 0), (1, 1, 1, 0)$.

The columns have exceedance $3, 2, 1, 0$.

(source: wikimedia.org)

Now, verify that an allowed permutation has an associated staircase diagram that stays on or above the line $ y = x$, hence has exceedence exactly $n$.

Finally, the number of allowed permutations is (the desired):
$$ \frac{1}{n+1} \times \frac{ (n+1)! } { (k_0 + 1) ! \prod k_i ! } = \frac{ (n)! } { (k_0 + 1) ! \prod k_i ! } $$

In the case of $n=3$, this corresponds to the first column, and the allowed permutations are $(3, 0, 0), (1, 2, 0), (2, 0, 1), (1, 1, 1)$.

Corollary: The total number of allowed permutations is the Catalan number $C_n$.


答:
The procedure followed to solve the equation assumes $y\neq 0$ and $x\neq -1$. Assuming that you get solutions
$$
y=\pm e^{c_1}(1+x)
$$
which is the same as $y=C(1+x)$ with $C\neq 0$. But you realize that $y=0$ is also a solution when the equation is written in the form
$$
\frac{dy}{dx}=\frac y{x+1}
$$
and $x=-1$ is a solution when it is written in the form
$$
\frac{dx}{dy}=\frac{x+1}{y}
$$
You can include the first special solution by allowing $C =0$ and the second one by allowing $C=\infty$ (or by shifting the constant to the other side and allowing it to be. zero).

The differential form of an ODE is more symmetrical and admits solutions which are not necessarily functional relations of the form $y=f(x)$ or $x=g(y)$, but are locally made up of those, except for the points where both coefficients in front of $dx$ and $dy$ vanish (no direction field defined).

Title: Solve $(1+x) d y-y d x=0$

Body:
Here is the worked example by the author,

$$
\int \frac{d y}{y}=\int \frac{d x}{1+x}
$$

$$
\ln |y|=\ln |1+x|+c_{1}
$$

$$
|y|=e^{\ln |1+x|+c_{1}}=e^{\ln |1+x|} \cdot e^{c_{1}}=|1+x| e^{c_{1}}
$$

$$
y=\pm e^{c_{1}}(1+x)
$$

Relabeling $\pm e^{c_{1}}$ as $c$ then gives $y=c(1+x)$.

My question: Shouldn't you say $c=\pm e^{c_1},0$ because $y=0$ is a constant solution?

I guess that when $x=-1,y=0$ so I thought maybe that covers the solution $y=0$, but $(1+x) d y-y d x=0$ can be rearranged to $\frac{d y}{d x}=\frac{y}{1+x}$ and this implies that $x \neq-1$, so the constant solution $y=0$ cannot be achieved from $y=c(1+x)$ where $c=\pm e^{c_1}$, is what I think.. Thanks!


问:
Classify all abelian groups of order $2^4 \cdot 5^2 \cdot 11^3$.
I know I can use the Fundamental Theorem of Finitely Generated Abelian Groups, but that would be a total of $4 \cdot 2\cdot 3 = 24$ decompositions and simplifications, which would leave a lot of room for error. I was wondering if there is a more direct way of doing this.

答: The group $\mathbb Z_{pqr\cdots}$ where $p,q,r\dots$ are pairwise coprime is isomorphic to $\mathbb Z_p×\mathbb Z_q×\mathbb Z_r×\cdots$ so we only need count the number of ways to partition each prime power into smaller prime powers. There are $5$ ways to decompose $2^4$ in this way, $2$ ways for $5^2$ and $3$ ways for $11^3$. Hence there are $5×2×3=30$ non-isomorphic groups of order $2^45^211^3$.


问:
If $f\left(x+\frac{5}{6}\right)+f(x)=f\left(x+\frac{1}{2}\right)+f\left(x+\frac{1}{3}\right)$ then check for periodicity
Let $f:\mathbb R\rightarrow \mathbb R$ be a function which satisfies the functional equation
$$f\left(x+\frac{5}{6}\right)+f(x)=f\left(x+\frac{1}{2}\right)+f\left(x+\frac{1}{3}\right)$$
Is the function periodic ?

I tried by replacing $x$ by $x+\frac{1}{2}$ and also $x+\frac{1}{3}$ and so on but couldn't reach any conclusion

答: No, it does not have to be periodic. A linear function $f(x)=ax$ satisfies this relation, as you can easily check.


问:
Gradient of dot product
If $\vec A$ and $\vec B$ are two vectors such that $\vec A=<A_x,A_y,A_z>$ and $\vec B=<B_x,B_y,B_z>$ then

$$\nabla({\vec A}.{\vec B})=A \times (\nabla \times B)+B \times (\nabla \times A)+(A.\nabla)B+(B.\nabla)A$$

In this relation what does this $(A.\nabla)B$ imply? How does this work on vector? Isn't gradient meant to be operated on scalar function?

答:
The operator $\vec A\cdot\nabla$ is the directional derivative operator in the direction $\vec A$. If you have some experience with analysis, you might know that if $f:\mathbb R^n\to\mathbb R^n$ is totally differentiable with Jacobian $\mathrm Jf(x)$ in $x$, then all its directional derivatives exist and can be calculated as $\mathrm Jf(x)v$, where $v$ is the direction.

Sometimes this is also written as $(\nabla f)^T v$, using the nabla operator. We can write this as

$$(\nabla f)^Tv=(\partial_1 f)v_1+\dots+(\partial_n f)v_n.$$

Keep in mind that $\partial_i f$ are vector valued functions. This expression we could also write as $(v_1\partial_1+\dots+v_n\partial_n)f$. Formally, the expression in parentheses is just $v\cdot\nabla$, so we can write the whole thing as

$$(\nabla f)^T v=(v\cdot\nabla)f.$$

The takeaway is: $v\cdot\nabla$ is the directional derivative operator in direction $v$ because $(v\cdot\nabla)f=(\nabla f)^T v$, and the expression on the right side is known to give you the directional derivative.

Title: Gradient of dot product


问:
An ellipse determined by circumcenters
Can you prove or disprove the following claim:

Claim. A convex hexagon $ABCDEF$ is circumscribed about an ellipse. Let $G$ be the point of concurrency of hexagon's principal diagonals , and let the points $O_1$ , $O_2$ , $O_3$ , $O_4$ , $O_5$ , $O_6$ be the circumcenters of $\triangle ABG$ , $\triangle BCG$  , $\triangle CDG$ , $\triangle DEG$ , $\triangle EFG$ and $\triangle FAG$ , respectively . Then, the points  $O_1$ , $O_2$ , $O_3$ , $O_4$ , $O_5$ , $O_6$ lie on a new common ellipse.

GeoGebra applet that demonstrates this claim can be found here.

Consider a hexagon $O_1O_2O_3O_4O_5O_6$ . My idea is to apply Braikenridge–Maclaurin theorem on it in order to prove that points  $O_1$ , $O_2$ , $O_3$ , $O_4$ , $O_5$ , $O_6$ are conelliptic. But, to do that I need to show that three intersection points of the three pairs of lines through opposite sides of a hexagon $O_1O_2O_3O_4O_5O_6$  lie on a line l. How this can be done?

答:
Since the circumcenter lies on the perpendicular bisector of the sides of each triangle, $O_{1}O_{6}\perp AD$ and $O_{3}O_{4}\perp AD$. This means that $O_{1}O_{6}\ \vert\vert\ O_{3}O_{4}$, and similarly for the other pairs of opposite sides. Thus, the hexagon is a parallelogon, which means that the lines through its opposite vertices concur and furthermore the hexagon is centrally symmetric through this point of concurrency (moved to the origin for then next part.)

Consider the parallelogram formed by $2$ pairs of opposite points, and note that any conic through them will have a center at the origin, which means that that conic will also be centrally symmetric. Next, we add in a $5$th point. Since a conic can go through any $5$ points, we can construct a conic through these $5$ points. This conic must still be centrally symmetric, which means that the $6$th point must also lie on this conic.

It remains to prove that this conic is an ellipse, which is trivial since each of the interior angles of the hexagon are between $0^{\circ}$ and $180^{\circ}$.

Thus, $\boxed{O_{1}O_{2}O_{3}O_{4}O_{5}O_{6}\text{ has a circumscribed ellipse.}}$

Title: An ellipse determined by circumcenters


问:
A polynomial divisible by primes of special form

Problem. Let $p$ be a prime number of the form $9k + 1$. Show that there exists an integer $n$ such that $p \mid n^3 - 3n + 1$.

Source. Here.

My only idea was that $n^3-3n+1$ might form a complete residue modulo such primes. So, I tried it out for $19$ and it turned out to be wrong. After that I could not find anything useful so I read the solution. But I don't understand it. Any elaboration on the solution given in the provided link or any new solution both are appreciated. Thanks in advance

答:
Elementary solution:

Suppose $n=t-1$, putting in we get:

$A=n^3-3n+1=t^3-3t^2+3$

For $t=3m$ we get:

$A=27m^3-27+3=3[9(m^3-m^2)+1]$

We can assume $m^3-m^2=k$ so $9k+1|A$

Then $n=3m-1$

For example:

$m=2$, $\rightarrow:$, $p=37$, $n=5$

$m=3$, $\rightarrow:$, $p=163$, $n=8$

$m=4$, $\rightarrow:$, $p=433$, $n=11$

Hence values of n make an arithmetic progression .


问:
prove or disprove $(x+1)^{2p^2}\equiv x^{2p^2}+\binom{2p^2}{p^2}x^{p^2}+1\pmod {p^2}$
The following question I read in a book, but the book does not give proof. I doubt the correctness of the result

let $p>3$ be prime number. prove or disprove
$$(x+1)^{2p^2}\equiv x^{2p^2}+\binom{2p^2}{p^2}x^{p^2}+1\pmod {p^2}\tag{1}$$

I think use binomial theorem
it maybe show $$\binom{2p^2}{k}\equiv 0,\pmod {p^2},k=1,2,\cdots,p^2-1,p^2+1,\cdots,2p^2-1\tag{2}$$
but  other hand I think (2) is not right.becuase let  $p=5,k=5$
we have $$\binom{2p^2}{k}=\binom{50}{5}=\dfrac{50\cdot 49\cdot 48\cdot 47\cdot 46}{5\cdot 4\cdot 3\cdot 2\cdot 1}\ne 0\pmod {25}$$,so I think $(1)$ is not right?

答:
You got it right. By Lucas correspondence the congruence (1) would be correct if it were modulo $p$ only. If we did this modulo $p$ we would be doing arithmetic in the ring $\Bbb{Z}_p[x]$ — a commutative ring of characteristic $p$. This means that we can apply Freshman's Dream twice, and get
$$
\begin{aligned}
(x+1)^{2p^2}&=\left((x+1)^2\right)^{p^2}\\
&=\left((x^2+2x+1)^p\right)^p\\
&\equiv\left(x^{2p}+2^px^p+1\right)^p\pmod{p}\\
&\equiv\left(x^{2p}+2x^p+1\right)^p\pmod{p}\\
&\equiv x^{2p^2}+2x^{p^2}+1\pmod{p}.
\end{aligned}
$$
But, this is only modulo $p$.

Your example is the first failure modulo $p^2$. Well, the same way you would see that $\binom{18}3$ is not divisible by $9$, but the question specified $p>3$. Well done.

Kummer's theorem (see Qiaochu's link) says the same thing. When you do grade school addition of $5+45$ in base five there will only be a single carry, from the $5^1$-digit to the $5^2$-digit. Hence $\binom{50}5$ is divisible by $5^1$ but not divisible by $5^2$. For the same reason $\binom{2p^2}p$ is not divisible by $p^2$ for any prime $p>2$.

As the last observation, when $p=2$ the congruence (1) will be correct modulo $4$. For example the binomial coefficient $\binom 82$ is divisible by $2^2$ as $8$ has a non-zero bit at position $2^3$, and we get the required two carries when calculating $2+6$ in base two.


问:
Mathematical philosophy
I have seen that they have already asked before in this forum for recommendations for books about mathematical philosophy, but I think my case is particular. I have never read anything about it and I would like to start with an easy-to-read book, since at the moment I am devoting my energy to studying other topics, but I would like to gradually delve into this topic that has caught my attention. Thanks in advance.</p>

答: It sounds like you would like to read The Mathematical Experience by Davis and Hersh.</p>

Title: Mathematical philosophy


答:
We obtain the abelian groups of order $2^4\cdot 5^2\cdot 11^3$ by
taking one abelian group from each of the three lists(right-hand
column below) and taking their direct product.

Title: Classify all abelian groups of order $2^4 \cdot 5^2 \cdot 11^3$.

Body: I know I can use the Fundamental Theorem of Finitely Generated Abelian Groups, but that would be a total of $4 \cdot 2\cdot 3 = 24$ decompositions and simplifications, which would leave a lot of room for error. I was wondering if there is a more direct way of doing this.


问:
Define the following residue for using the residue theorem
I wanna calculate the following integral
$$\oint_{\{\vert z \vert =\frac{5}{2}\}} \frac{1}{sin^2(z)cos(z)}dz$$
with the residuum theorem.
Well first I have to calculate the residue of this integral. We have that $cos(z)=0 \iff \frac{\pi}{2}+k \pi, k \in \Bbb Z$ and $sin(z)=0 \iff z=k \pi, k \in \Bbb Z$. This means, we get the following isolated zero:

$z_0=0$

$z_1=\frac{\pi}{2}$

$z_2=-\frac{\pi}{2}$

Now I have to calculate the residue of the different $z_i$ but for this I have to know, which kind of isolated zero we have. Here is where I got stucked.. How can I find this out?

EDIT: For the real case we have that $sin^2(x)cos(x)=cos(x)-cos^3(x)$. Can I maybe use it here?

答:
To find out which kind of isolated zero you have, you can have a look on the series expansion of $\sin$ and $\cos$ in the point $z_i$. For $z_0=0$ the terms of lowest order of $\sin$ is $z$ and of $\cos$ it is $1$. This means the term of lowest order of $\sin^2(z)\cos(z)$ is $z^2$ and thus you have a second-order pole. Then you can do the same for $z_1$ and $z_2$, where you will find out that you have simple poles.

Another way to do so is you try to find out for which $n$
$$\frac{(z-z_i)^n}{\sin^2(z)\cos(z)}
$$
is holomorphic and nonzero in a neighbourhood of $z_i$, what is actually the defintion of the order of a pole.

A third way is to calculate the laurent series, which is a very similar method to the first method, but delivers you directly your residue. Again for example for the point $z_0=0$, you have
$$ \sin z=z+\mathcal O(z^3)
$$and
$$ \cos z=1+\mathcal O(z^2)
$$
thus
\begin{align}\frac{1}{\sin^2(z)\cos(z)}&=\frac{1}{(z+\mathcal O(z^3))^2(1+\mathcal O(z^2))}=\frac{1}{z^2(1+\mathcal O(z^2))^2(1+\mathcal O(z^2))}\\
&=\frac{1}{z^2(1+\mathcal O(z^2))(1+\mathcal O(z^2))}=\frac{1}{z^2(1+\mathcal O(z^2))}\\
&=\frac{1}{z^2}(1+\mathcal O(z^2)).
\end{align}
In the last step we have used the geometric series. So your residue in $z_0=0$ is $0$.


问:
Find constants $a$, $b$, $c$, $d$ and $e$ such that $\cos4x=a\sin^4x+b\sin^3x+c\sin^2x+d\sin x+e$ for all angles $x$
Basically, write $\cos4x$ as a polynomial in $\sin x$.

I've tried the double angles theorem and $\cos2x=\cos^2x-\sin^2x$. I'm still having trouble right now though.

Please help! Thanks!

Body: $$\begin{align}\cos4x&=1-2\sin^22x\\&=1-2(1-\cos^22x)\\&=1-2(1-(1-2\sin^2x)^2)\\&=1-2(4\sin^2x-4\sin^4x)\\&=1-8\sin^2x+8\sin^4x\end{align}$$


答:
Let $p = 5$. For any integer $k$, we have

$$\begin{equation}\begin{aligned}
& ((5k + 1)(5k + 2))((5k + 3)(5k + 4)) \\
& \equiv (25k^2 + 15k + 2)(25k^2 + 35k + 12) \\
& \equiv (15k + 2)(10k + 12) \\
& \equiv 150k^2 + 200k + 24 \\
& \equiv -1 \pmod{25}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Using this for each group of a product of $4$ consecutive integers between multiples of $5$, we get

$$\begin{equation}\begin{aligned}
\binom{50}{25} & \equiv \frac{50(-1)45(-1)40(-1)35(-1)30(-1)}{25(-1)20(-1)15(-1)10(-1)5(-1)} \\
& \equiv \frac{2(45)(2)(35)(2)}{10(5)} \\
& \equiv 9(2)(7)(2) \\
& \equiv 2 \pmod{25}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Since $\varphi(25) = 20$, with $x = 1$, your congruence equation becomes

$$\begin{equation}\begin{aligned}
2^{50} \equiv 1 + 2 + 1 \pmod{25} \\
2^{10} \equiv 4 \pmod{25} \\
1024 \equiv 4 \pmod{25} \\
24 \equiv 4 \pmod{25}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Of course, this is not true, so this gives one specific counter-example to show the book's statement is not correct for all primes $p \gt 3$.

Title: prove or disprove $(x+1)^{2p^2}\equiv x^{2p^2}+\binom{2p^2}{p^2}x^{p^2}+1\pmod {p^2}$

Body:
The following question I read in a book, but the book does not give proof. I doubt the correctness of the result

let $p>3$ be prime number. prove or disprove
$$(x+1)^{2p^2}\equiv x^{2p^2}+\binom{2p^2}{p^2}x^{p^2}+1\pmod {p^2}\tag{1}$$

I think use binomial theorem
it maybe show $$\binom{2p^2}{k}\equiv 0,\pmod {p^2},k=1,2,\cdots,p^2-1,p^2+1,\cdots,2p^2-1\tag{2}$$
but  other hand I think (2) is not right.becuase let  $p=5,k=5$
we have $$\binom{2p^2}{k}=\binom{50}{5}=\dfrac{50\cdot 49\cdot 48\cdot 47\cdot 46}{5\cdot 4\cdot 3\cdot 2\cdot 1}\ne 0\pmod {25}$$,so I think $(1)$ is not right?


答:
Double angle theorem once:

$\cos 4x = \cos^2 2x + \sin^2 2x$.

Double angle theorem twice:

$= (\cos^2 x -\sin^2 x)^2 + (2\sin x\cos x)^2$.

Expand chicken soup and rice:

$= \cos^4 x- 2\cos^2 x\sin^2 x + \sin^4 x +4\sin^2 x\cos^2 x$

Can you finish?

To get it entirely in terms of $\sin^k x$ replace $\cos^2 x$ with $1-\sin^2 x$ (keeping in mind $\cos^4 x = (\cos^2 x)^2$)

Can you finish?

$= (1 - \sin^2 x)^2  - 2(1-\sin^2 x) \sin^2 x + \sin^4 x +4\sin^2 x(1-\sin^2 x) $

Title: Find constants $a$, $b$, $c$, $d$ and $e$ such that $\cos4x=a\sin^4x+b\sin^3x+c\sin^2x+d\sin x+e$ for all angles $x$

Body:
Basically, write $\cos4x$ as a polynomial in $\sin x$.

I've tried the double angles theorem and $\cos2x=\cos^2x-\sin^2x$. I'm still having trouble right now though.

Please help! Thanks!


问:
Why are singletons of Real Numbers an interval according to this definition?
"Definition: A subset S of R is said to be an interval if it has the following property: if x∈S, z∈S, and y∈R are such that x < y < z, then y∈S"

According to the book, singletons are intervals from this definition, but since singletons only have one element, why would that be the case?

答:
In order to prove that $S=\{p\}$ were not an interval, logically you have to give $x \in S$, $z \in S$ and some $y \in \Bbb R$ such that $x < y < z$ but nevertheless $y \notin S$:

(in first order logic the definition translates (quantifier domain is $\Bbb R$) as

$$\forall x,y,z: ((x \in S) \land (z \in S) \land (x < y) \land (y < z)) \to (y \in S)\tag{1}$$

which negates via standard rules (an implication is only falsified by having the antecedent true and the consequent false etc.)

$$\exists x,y,z: ((x \in S) \land (z \in S) \land (x < y) \land (y < z)) \land (y \notin S)\tag{2}$$

But as $x$ and $z$ are forced to be equal to $p$ but then $(p < y)$ and $y < p$ forces $p < p$ contradiction.

So $(2)$ (= not $(1)$) cannot hold for $S=\{p\}$ so because we have two-valued logic, $(1)$ does hold for $\{p\}$.

This is just an expansion of the vacuous truth notion that was already mentioned, in the hope it adds to the understanding of the why of it. You cannot falsify it so it's true.


问:
Picard group of a Parabolic subgroup subgroups of a simple or reductive algebraic group.
Let $G$ be a reductive algebraic group for eg. $G=GL(n,\mathbb{C}).$ Let $P$ be a Parabolic subgroup of $G.$   What is the Picard group of a Parabolic subgroup? Is it the same as that of its Levi factor?

答:
If your field is perfect, the unipotent part $U_P$ is isomorphic to an affine space and hence ${\rm{Pic}}(U_P) = 0$. Since as variety $P \cong L \times U_P$ the claim follows.

If not, the question looks more complicated, see this article for more details. The same article gives right away an example of a unipotent subgroup with non-trivial Picard group.


问:
What is an alternative definition of local homeomorphism?
The definition I was given is for local homeomorphism:

" Let $(X,\tau)$ and $(Y,\tau_1)$ be topological spaces. A map $f:X \to Y$ is said to be a local homeomorphism if each point $x \in X$ has an open neighbourhood $U$ such that the restriction of $f$ to $U$ maps $U$ homeomorphically onto an open subspace $V$ of $(Y,\tau_1)$; that is, if the topology induced on $U$ by $\tau$ is $\tau_2$ and the topology induced on $V=f(U)$ by $\tau_1$ is $\tau_3$, then $f$ is a homeomorphism of $ (U,\tau_2)$ onto $(V,\tau_3)$. The topological space $(X,\tau)$ is said to be locally homeomorphic to $(Y,\tau_1)$ if there exists a local homeomorphism of $(X,\tau)$ into $(Y,\tau_1)$."

Are there any alternative definitions that are better or gives more insight into what exactly local homeomorphisms are? Or are there better definitions out there that makes it easier when doing proofs.

答: This definition is fine of doing proofs. You could (should?) check that a circle in the plane is locally homeomorphic to $\Bbb R$ and the $\{x \in \Bbb R^3\mid \|x\|=1\}$, a sphere surface, is locally homeomorphic to $\Bbb R^2$. The notion mostly comes up when studying manifolds. It's not an important notion outside of that area, really.


问:
What is the smallest integer $n$>1 such that $n^{5000}+n^{2013}+1$ is prime?
Which is the smallest integer $n>1$, such that $$n^{5000}+n^{2013}+1$$ is prime ?
 Since $x^{5000}+x^{2013}+1$ is irreducible over $\mathbb{Q}$ and has value $1$ for $x=0$,
 there should be infinitely many such $n$, if Bunyakovsky's conjecture is
 true.</p>

答:
$n=23205$ produces the smallest prime value of the polynomial (aside from the trivial $n=1$). Interestingly, $23205=3\times 5\times 7\times 13\times 17$.

$n\in\{44579, 55754, 78120, 78515, 94154, 99045\}$ produce all the remaining primes with $n<10^5$. In all cases, OpenPFGW has been used to find the primes and prove primality using Brillhart-Lehmer-Selfridge method.


问:
Is $\mathbb{R}^J$ metrizable for $J$ uncountable?
Example 2 section 21 Munkres' Topology says that an uncountable product of $\mathbb{R}$ with itself is not metrizable. Same book Theorem 43.5 says that if a space $Y$ is complete in the metric $d$ then $Y^J$ (J any indexed set) is complete in the uniform metric corresponding to $d$.

My question is that if we can define a metric (uniform metric in here) on $\mathbb{R}^J$ so how it still is not metrizable?!

答:
The product $\Bbb R^J$ by default has the product topology and in that topology the space is not metrisable when $J$ is uncountable. The uniform metric on that set only coincides with the product topology when $J$ is finite (even for countable $J$ they're different: one is separable and the other is not, but then both are metrisable topologies, no problem as a set can have many metrics and thus many different metrisable topologies)

$\Bbb R^J$ is called metrisable iff the product topology coincides with the topology induced by some metric and this holds iff $J$ is at most countable (the uniform metric only works for the finite case; Munkres also gives the correct metric for the countable case in another theorem).


问:
Is it possible to produce a field with *more* than three operations?
This question is a sequel to my question Is it possible to produce a field with three operations?.

Using the same definition of/notation for fields, is it possible to have $(F, +, \cdot, *, \dagger)$ such that all of $(F, +, \cdot)$, $(F \setminus \{0\},\cdot, *)$ and $(F \setminus \{0,1\},*, \dagger)$ are fields?

What about if we wanted to chain more operations on?

答:
The answer to the first question, as helpfully pointed out by @JasonDeVito in a comment to my original question, is yes.

To preserve and expand upon it: The field with four elements, $\mathbb{F}_4 = \{0, 1, \alpha, 1+\alpha\}$, can support the structure of $(F, +, \cdot, *, \dagger)$, with $+$ and $\cdot$ defined in the usual way for $\mathbb{F}_4$; with $*$ as the operation defined by @lonzaleggiera, with $\alpha$ defined as the $*$-identity; and with $\dagger$ as the operation defined by @DavidPement, with the somewhat dubious convention that $0 < 1 < \alpha < 1+\alpha$.

It is also the only such field (up to isomorphism). Why? Because both $(F, +, \cdot, *)$ and $(F \setminus \{0\}, \cdot, *, \dagger)$ must be fields with three operations. By the classification collectively provided by the answers to my previous question, this means that, either both $|F|$ and $|F \setminus \{0\}|$ are Mersenne powers of $2$ (that is, powers of $2$ one greater than a prime), or one of the two sizes is $3$ and the other is a Mersenne power of $2$. Since $|F| = |F \setminus \{0\}| +1$, and there are no consecutive Mersenne powers of $2$, the only option is $|F| = 4$ and $|F \setminus \{0\}| = 3$, which, by the classification collectively provided, also forces the operations to be as defined (although $\alpha$ and $1+\alpha$ could be switched, provided it is done for both $*$ and $\dagger$, but this is isomorphic).

As for the second question, the answer is no, and the proof is straightforward:

Suppose $(F, +, \cdot, *, \dagger, \#, ...)$ is a field with at least five operations.

This implies $(F, +, \cdot, *, \dagger)$ and $(F \setminus \{0\}, \cdot, *, \dagger, \#)$ must both be fields with four operations, which implies $|F| = |F \setminus \{0\}| = 4$. But $|F| = |F \setminus \{0\}|+1 = 4+1 = 5$, a contradiction.

Thus, there is no field with five or more operations, QED.


问:
Proof Verification: Nested Interval Theorem $\implies$ Completeness of $\mathbb{R}$
My proof is similar in spirit to this one, but I'm not able to figure out if it's correct or not.

Consider $S\subset \mathbb{R}$ that is bounded above. Consider an upper bound of $S$, say $u$. If $u\in S$, we are done, so let $u\notin S$. We will construct $I_n = [a_n,b_n]$ to apply the nested-interval theorem. Define $b_i = u$, for all $i$.

For $a_1$, pick some $s\in S$. Now define $a_n$ recursively (is it okay to make recursive definitions like this one):

If there exists $s'\in S$, such that $s' > a_i$ put $s_{i+1} = s'$.

Otherwise, $s_{i+1} = s_i$.

We can see that $$I_1\supseteq I_2 \supseteq ...$$
By the nested interval theorem $$\bigcap_{n\ge 1} I_n = [a,b] \neq \varnothing$$
We can now show that $a$ is the least upper bound of $S$.

Is this proof alright?

答: I think your definition of $a_n$ is a problem. Suppose $S = [0,2]$. This is clearly bounded. Pick $u=3$ and the we can pick $a_n=1-\frac{1}{n}$. We then get a nested series of closed intervals but their intersection is $[1,3]$. $1$ is clearly not an upper bound for $S$.


问:
Could we evaluate these two indefinite integrals?
I have tried much with the following two integrals, even with Mathematica and Matlab but in vain

$$\int \frac{x^{4}-3x^{3}+x^{2}+2x-2}{\sqrt{x^{2}+2x+3}}dx$$
and
$$\int \frac{x^{\frac{2}{5}}}{\sqrt[3]{1+4x^{\frac{5}{2}}}}dx$$

Any help with that is very appreciated.

答:
Mathematica 12.0 gives (up to the integration constant) the first integral as
$$\frac1{12}\left((3x^3-19x^2+40x+18)\sqrt{x^2+2x+3}-162\sinh^{-1}\frac{x+1}{\sqrt2}\right)$$
and the second as
$$\frac57x^{7/5}{_2F_1}\left(\frac13,\frac{14}{25};\frac{39}{25};-4x^{5/2}\right)$$


问:
How is that possible to be $X$ discrete random variable?

Suppose that the cumulative distribution function of discreate random variable $X$ is given by,
$$F(x) =
\begin{cases}
0  & \text{$x$ < 0 } \\[1.5ex]
\dfrac{x}{4} & \text{$0 \leq x<1$}\\[1.5ex]
\dfrac{1}{2}+\dfrac{x-1}{4} & \text{$1 \leq x<2$}\\[1.5ex]
\dfrac{11}{12}   &\text{$2 \leq x<3$}\\[1.5ex]
1 &\text{$3 \leq x$}\\
\end{cases}$$
Find $P(X=i),i=1,2,3$.

My problem is, how is that possible to be $X$ discrete random variable as I think it is mixed random variable(both discrete and continuous) since there are $x$ in $F(X)$. Can anyone help me to understand this?

答:
It is neither discrete nor continuous.  $P(X=1)=\frac 12  -\frac1  4=\frac  1 4$ (the jump at $x=1$) etc.

For any random variable $X$ we have $P(X=x)=F_X(x)-F_X(x-)$ where $F_X(x-)$ is the left hand limit of $F_X$ at $x$. In this case the right hand limit of $F$ at $1$ is $\frac 1  2$ and the left hand limit is $\frac  1 4$ so the difference is $\frac 1 4$.


问:
Norm/Mesh of partitions in multiple integrals

Here, $||P||$ is the norm of the partition which is defined by $max${$\Delta x,\Delta y$}, i.e., the maximum of either side of each sub-rectangle. Why can we not define $||P||$ as $max${$\Delta x \Delta y$}? Can someone provide a counter-example of a function explaining why the following assumption as the norm fails in certain cases? I was told that the Darboux Integral failed to hold for certain integrable functions for the above assumed norm. Also, can we not define $||P||$ as the length of the largest diagonal of all such sub-rectangles? Would there be any problem in that case?

答:
Consider $f(x,y) =x$ on the rectangle $[0,1]^2$, where

$$\int_{[0,1]^2} f = \frac{1}{2}$$

Let the partition $P_n$ have subrectangles $[0,1]\times [(j-1)/n,j/n]$ for $j=1,\ldots,n$. The upper Darboux sum $U(P_n,f)$ remains constant with value $1$ even as $n \to \infty$ and does not converge to the integral.  However, the norm of the partition defined as the maximum area of subrectangles is $1/n$ and tends to $0$.


答:
The following rephrasing may be easier to think about at first, although ultimately you should be comfortable with the given one:

A map $f: X\rightarrow Y$ is a local homeomorphism iff there is some open cover $\mathcal{C}$ of $X$ such that for each $U\in \mathcal{C}$, $f[U]$ is open as a subset of $Y$ and $f\upharpoonright U$ is a homeomorphism between the subspaces $U$ of $X$ and $f[U]$ of $Y$.

(It's a good exercise to prove that this is in fact equivalent to the definition given by the text.)

This rephrasing captures the idea that $f$ is a local homeomorphism iff we can "break $X$ into pieces" such that $f$ looks like a homeomorphism from the perspective of each piece. We have to be careful, though: "break into pieces" suggests disjoint pieces, and that's definitely not wanted here. Hence the more technical language of open covers: $\mathcal{C}$ is just supposed to be some collection of open sets which together account for all the points in $X$, they don't have to be disjoint.

For what it's worth, I'd say that when starting out the original definition is meaningfully easier to use in proofs, while the rephrasing above is slightly easier to grasp conceptually. But that's just my own experience with a small number of students.

Title: What is an alternative definition of local homeomorphism?

Body:
The definition I was given is for local homeomorphism:

" Let $(X,\tau)$ and $(Y,\tau_1)$ be topological spaces. A map $f:X \to Y$ is said to be a local homeomorphism if each point $x \in X$ has an open neighbourhood $U$ such that the restriction of $f$ to $U$ maps $U$ homeomorphically onto an open subspace $V$ of $(Y,\tau_1)$; that is, if the topology induced on $U$ by $\tau$ is $\tau_2$ and the topology induced on $V=f(U)$ by $\tau_1$ is $\tau_3$, then $f$ is a homeomorphism of $ (U,\tau_2)$ onto $(V,\tau_3)$. The topological space $(X,\tau)$ is said to be locally homeomorphic to $(Y,\tau_1)$ if there exists a local homeomorphism of $(X,\tau)$ into $(Y,\tau_1)$."

Are there any alternative definitions that are better or gives more insight into what exactly local homeomorphisms are? Or are there better definitions out there that makes it easier when doing proofs.


问:
How to solve a system of a single equation?
I’m trying to solve an apparently simple problem but couldn’t find a solution not involving advanced math.

Let $10ax + (b-3a) = x - 3$

How do we solve equations of the given form? The solution of the above quation is $10a=1$ and $b-3a=-3$, so they basically turned a single equation into two different equations, but there is no explanation of how and I couldn’t  find a explanation online.
Can somebody help me?

答: The equation implies that the equality holds for any real $x$. Setting two different values of $x$ (such as $0$ and $1$) will give you two different equations with the above solutions. I those two equalities hold, then $ 10ax + (b-3a) = x - 3$ for all real $x$.


问:
Show that the number of faces in a planar connected graph on $n$ vertices is bounded from above by $2n-4$

Let $G$ be a planar (and connected) graph on $n \geq 3$ vertices. Show that the number of faces of $G$ is bounded from above by $2n-4$.

Can someone help me with an idea on this problem?

答:
It is well known that the number of edges is at most $3n-6$ in a planar graph on $n$ vertices. This can be easily proven using the following idea: Take a planar graph $G$ and take an arbitrary face which is not a triangle. We quickly realize that we can always add diagonal edges in each non-triangular face without destroying planarity until we have contructed a plane triangulation. Using Euler's formula, given by $n-m+f=2$, whereby $n$ is the number of vertices, $m$ is the number of edges and $f$ is the number of faces, the above bound is quickly found.

Corollary: A planar graph $G$ with $n \geq 3$ vertices has at most $3n-6$ edges.

Proof: Assume that $G$ is a maximal planar graph, that is a planar graph for which we cannot add any more edges without destroying planarity. By Euler's formula: $n-m+f=2$. Each face must be a triangle by maximality of $G$ (or we could add diagonals). Since every edge bounds two faces, $2m=3f \implies n-m+ \frac{2}{3}m=n-\frac{1}{3}m=2 \iff m=3n-6$.

Now, use Euler's formula again and plug in the above results.

$n - m +f \leq 2\\ n-3n+6 +f \leq 2 \\ f \leq 2n -4$

Hence the number of faces in a planar graph is at most $2n -4$.


问:
Show reflexive normed vector space is a Banach space
$X$ is a normed vector space. Assume $X$ is reflexive, then $X$ must be a Banach space.

I guess we only need to show any Cauchy sequence is convergent in $X$.

答:
Hint: (1) If $X$ is reflexive, $X$ is isomorphic to $X^{**}$. (2) Dual spaces are allways complete.

Regarding (2), we will prove, that $L(X,Y)$ the space of bounded linear operators from $X$ to $Y$ is complete in the operator norm if $Y$ is complete. Then (2) follows, as $X^* = L(X, \mathbb K)$ and $\mathbb K$ is complete. So let $(T_n)$ be an operator norm Cauchy sequence, then $(T_n x)$ is Cauchy for each $x$, as $\def\norm#1{\left\|#1\right\|}$
$$ \norm{T_nx-T_mx} \le \norm{T_n - T_m}\norm x $$
As $Y$ is complete, we may define $T\colon X \to Y$ by $Tx := \lim_n T_n x$. $T$ is linear, as the $T_n$ and the limit is, an bounded since
$$ \norm{Tx} \le \sup_n\norm{T_n x} \le \sup_n\norm{T_n}\cdot \norm x $$
and Cauchy sequences are bounded. Now given $\epsilon > 0$, we can find a $N$, such that
$$ \norm{T_n - T_m} < \epsilon, \text{  all $n,m \ge N$} $$
giving
$$ \norm{T_n x - T_m x} < \epsilon, \text{   all $\norm x \le 1$, $n,m \ge N$} $$
for $m \to \infty$
$$ \norm{T_n x - T x} \le \epsilon, \text{   all $\norm x \le 1$, $n\ge N$} $$
that is $\norm{T_n - T} \le \epsilon$, $n \ge N$. So $T_n \to T$.


问:
Comparing two large polynomials
How would one show that $g(n) > h(n)$ for $n \in \mathbb{Z} : n \geq 2$

$g(n) = 2048n^{15} + 45568n^{13} +112768n^{11} +70624n^9 +11968n^7+674n^4+68n^2+1$
$h(n) = 14336n^{14} + 87296n^{12} +104064n^{10} +35184n^8 +1796n^6+764n^5+268n^3+11n$

I was originally going to say
since all the coefficients are positive, $\deg(g(n)) > \deg(h(n))$ and there exist $p \in \mathbb{R} : g(p) = h(p)$ then any $n>p$ will satisfy $g(n) > h(n)$

However I can't understand/grasp why any number greater than $p$ will satisfy $g(n) > h(n)$.

答:
Note that if $\ n\ge8\ $ then every term on the right side of the following equation is obviously positive:
\begin{align}
g(n)-h(n)&=n^{14}(2048n-14336)+n^{12}(45568n-77296)\\
&\hspace{2em}+n^{10}(112768n-104064)+n^8(70624n-35184)\\
&\hspace{2em}+n^5(11968n^2-1796n-764)+n^3(674n-268)\\
&\hspace{2em}+n(68n-11)+1\ .
\end{align}
For $\ n=7\ $ the first term in the sum on the right side is zero, but all the other terms are positive.
Therefore, to show that the expression is positive for all integers $\ n\ge 2\ $, you simply have to check that it's positive for the $5$ integers $\ n=2,3,4,5,6\ $.

Title: Comparing two large polynomials


问:
Find $ \lim_{n \to \infty}n \log\left(1+ \left(\frac{f(x)}{n}\right)^p\right)$
Find $$\lim_{n \to \infty}n \log\left (1+ \left(\frac{f(x)}{n}\right)^p\right)$$  where $0<p<1$

my attempt :By using L-Hospital Rule i got

$$\lim_{n\to \infty} \dfrac  {\dfrac{1}{1+\left( \dfrac{f(x)}{n}\right)^p}
\cdot\dfrac {f(x)^p}{-n^{2p}} }{\dfrac{-1}{n^2}}$$

After that I’m   not able to   proceed further

答:
Let $y = f(x)$. Consider the essential case $y \neq 0$.
\begin{align*}
n\log\left(1 + \left(\frac{y}{n}\right)^p\right) &
= y^pn^{1-p}\log \left(\left(  1 + \left( \frac{1}{\left(\frac{n}{y} \right)^p}\right)\right)^{\left(\frac{n}{y}\right)^p}\right).
\end{align*}
Since $\left\{\left(\frac{n}{y}\right)^p \right\}_{n \in \mathbb{N}}$ is monotonic,
\begin{align*}
\log \left(\left(  1 + \left( \frac{1}{\left(\frac{n}{y} \right)^p}\right)\right)^{\left(\frac{n}{y}\right)^p}\right) \to 1.
\end{align*}
But
\begin{align*}
y^pn^{1-p} \to +\infty \ or \ -\infty.
\end{align*}

Thus the limit is $+\infty$ or $-\infty$.


问:
Is it possible to rewrite $\frac{m\cdot \ln2}{\frac{n(1-(1-p)^r)}{p}}$ as $\frac{p\cdot \ln2}{(1-(1-p)^r)\frac{n}{m}}$?
Is this rewriting possible? If so, how?

$$\frac{m\cdot \ln2}{\dfrac{n(1-(1-p)^r)}{p}} = \frac{p\cdot \ln2}{(1-(1-p)^r)\dfrac{n}{m}}$$

答: If you are multiplying $n/m $ by $(1-(1-p)^r)$, then its the same thing. It is possible.


答: Yes, indeed it is. To understand this, send terms of both m and p to the numerator, on both sides, and the equation will then look exactly the same.</p>

Title: Is it possible to rewrite $\frac{m\cdot \ln2}{\frac{n(1-(1-p)^r)}{p}}$ as $\frac{p\cdot \ln2}{(1-(1-p)^r)\frac{n}{m}}$?

Body:
Is this rewriting possible? If so, how?

$$\frac{m\cdot \ln2}{\dfrac{n(1-(1-p)^r)}{p}} = \frac{p\cdot \ln2}{(1-(1-p)^r)\dfrac{n}{m}}$$