问:
Chromatic polynomial and edge-chromatic number of certain graphs
I have a few questions regarding the chromatic polynomial and edge-chromatic number of certain graphs.

For an empty graph, is the edge-chromatic number $0, 1$ or not well-defined? I've come up with reasons for each ($0$ since there aren't any edges to colour; $1$ because there's one way of colouring $0$ edges; not defined because there is no edge colouring of an empty graph) but I can't decide which argument is valid - is it more of a convention type thing?

There's a chromatic polynomial for a tree on $n$ vertices, so I was wondering if there was a chromatic polynomial for a bipartite graph on $n$ vertices (the chromatic number is $2$ since you can colour one part one colour, and the other part another, so I know $P_{G}(k)$ must have a factor of $(k-1)$)?

Any help would be much appreciated, thanks in advance!

答:
I suppose the answer here is it depends. Let's view edge coloring a graph as vertex coloring its line graph (they are equivalent.) If $G$ is the empty graph, or graph with an empty edge set, then $L(G)$ is what some may call the null graph. The null graph is quite interesting in that it gives rise to puzzling questions such as yours, as well as paradoxical ones (is the null graph connected?) According to the linked Wikipedia page, the chromatic number of the null graph is $0$, and hence the chromatic index of the empty graph is $0$. I highly recommend you give $``$Is the null-graph a pointless concept?$"$ by Harary and Read a read if you can access it. It includes discussion on the chromatic polynomial of the null graph, which they determine is the constant function $1$, on the basis that $``$... no matter what the number $\lambda$ of colors, there is only one way to color the points of $K_0$ [the null graph], namely, do nothing!$"$ They also discuss the aforementioned connectedness paradox.

I don't quite understand the question here. Trees are bipartite, so are you asking if $k(k-1)^{n-1}$ is the chromatic polynomial of all bipartite graphs? Certainly not, and it may not be the case that there is a $``$standard form$"$ for chromatic polynomials of bipartite graphs. For example, even cycles have chromatic polynomial $P_{C_{2n}}(k) = (k-1)^{2n} + (k-1)$ and are chromatically unique. You may want to search for questions regarding the chromatic polynomial of the complete bipartite graph $K_{n,m}$, of which there are several.

To address your side note, any graph with an edge will have $k-1$ as a factor of its chromatic polynomial, as there is no way to properly color a graph with at least one edge with $1$ color.


问:
how can we prove that the number of nodes in a minimum node-cover C* of G is greater than the number of edges in a maximum matching M* of G
Define a node cover $C$ for a graph $G = (V;E)$ to be a set of nodes $C$ such that, for every edge
$(u, v)$ that belongs to $E$, $C$ contains at least one of u and v. Define a minimum node-cover of $G$ to be a
node-cover that minimizes the number of node in the node cover.

Let $|C*|$ denote the number of nodes in a minimum node-cover $C*$ of $G$, and let $|M*|$ denote
the number of edges in a maximum matching $M*$ of $G$. Prove that $|M*|  \leq |C*|$.

答:
Notice that each edge $(u,v) \in M*$ is incident to a vertex $w_{(u,v)} \in C*$ by the definition of $C*$. Moreover if you consider two different edges $(u_{1},v_{1}), (u_{2},v_{2}) \in M*$, then $w_{(u_{1},v_{1})} \neq w_{(u_{2},v_{2})}$. Otherwise both edges in the matching are incident in $w_{(u_{1},v_{1})} = w_{(u_{2},v_{2})}$, and it is a contradiction with the fact that $M*$ is a matching.

This shows that for each edge in $M*$ one can find a different vertex in $C*$, in other words, we have found an injection from $M*$ to $C*$. It shows that $|M*| \leq |C*|$.

ContentLicense: CC BY-SA 4.0


问:
Setting Up Boundaries of Integration
Consider the region in space bounded by the plane $z=0$ and above by $z=\frac{1}{4} -x^{2} - y^{2}$. We wish to set up the limit of integration using $\text{d}\rho\;\text{d}\phi\;\text{d}\theta$. The problem I am facing is that I can set it up using polar coordinates but to convert it to spherical of this order is quite challenging for me. Any help is much appreciated.

答:
Think of it from a geometrical perspective before you think of the conversion. You have $\rho$ is the distance from the origin to any arbitrary point $\gamma$ on the surface of the paraboloid that is given by $(P) : z=\displaystyle\frac{1}{4}-x^{2}-y^{2}$. You should see that unlike the sphere, the distance between the origin and $\gamma$ is not fixed $\forall\gamma\in P$. Thus, to translate this as a mathematical argument, the lower bound of $\rho$ is clearly $0$ but the upper bound is the surface of the paraboloid which again is not fixed. Therefore, convert the equation of the paraboloid into spherical coordinates, we get that :
$$
z=\frac{1}{4}-x^{2}-y^{2}\implies\rho\cos(\phi)=\frac{1}{4}-\rho^{2}\sin^{2}(\phi)
$$
If it is not obvious to you that this is a quadratic equation (when treating $\phi$ as a constant), then rewrite it as follow :
$$
\rho^{2}\sin^{2}(\phi)+\rho\cos(\phi)-\frac{1}{4}=0
$$
This will yield to two distinct solutions :
$$
\rho_{1,2}=\frac{-\cos(\phi)\pm\sqrt{\cos^{2}(\phi)+\sin^{2}(\phi)}}{2\sin^{2}(\phi)}
$$
However, you are only allowed to have $\rho>0$ thus :
$$
\rho=\frac{-\cos(\phi)+\sqrt{\cos^{2}(\phi)+\sin^{2}(\phi)}}{2\sin^{2}(\phi)}=\boxed{\frac{1-\cos(\phi)}{2\sin^{2}(\phi)}}
$$
To find $\phi$, it is not bad at all since by setting $x=0$, we are able to project our surface onto the $zy$-plane and by sketching the region we will get $\boxed{\displaystyle 0\leq\phi\leq\frac{\pi}{2}}$. Hence, we obtain :
$$
\boxed{\displaystyle\int_{\displaystyle0}^{\displaystyle 2\pi}\displaystyle\int_{\displaystyle0}^{\displaystyle\frac{\pi}{2}}\int_{\displaystyle0}^{\displaystyle\frac{1-\cos(\phi)}{2\sin^{2}(\phi)}}\rho^{2}\sin(\phi)\text{d}\rho\;\text{d}\phi\;\text{d}\theta}
$$

Title: Setting Up Boundaries of Integration


问:
What exactly are axioms?
The title says it all. What exactly are axioms? I mean, I know that is a statement which do not need to be proven. But what are the requirements to be an axiom? For instance, may I state something that seems to be  false as an axiom? May I state an axiom that is already a conclusion of another theorem or conclusion?

Are there any books on how to build axiomatic systems and explains this "philosophic" part of mathematics? Is Russell's a nice option for this?

Thanks

EDIT

Thanks for such many responses. I am asking this because recently, I have read an "article" in which the author tries to use the non-aggression principle as an axiom, like mathematicians and scientists do, to prove some things like mathematicians do: Establish the axioms and prove conclusions. But this principle, in my view, seems not be true, since aggression does exist.

I guess this question should be redirected to some exchange on Philosophy...

答:
It's fine to use some non-aggression principle as an axiom and prove all kinds of things from that. However, to what extent the proven theorems apply to our reality depends on the extent to which the axioms apply to our world.

This is how it always is with mathematics. Whatever we prove in mathematics will be true for any world to which the axioms apply. That may include our world, but it need not be. We can prove all kinds of things using the axioms of Euclidean geometry, for example, but if our world turns out to be non-Euclidean, then the theorems need not reflect the state of affairs in our world.

Note, though, that even if our axioms don't perfectly describe our world, they may still be a pretty good approximation. Hence, any theorems we derive from them may still be 'close enough' to our reality to learn something from.

Title: What exactly are axioms?


问:
When to stop generating cyclic subgroups?
Consider the symmetric group $S_6$. When asked to find the cyclic subgroup of $S_6$ generated by $\sigma$=(15)(263)(4), I understand $<\sigma>$ = {$\sigma$,$\sigma^{2}$,$\sigma^{3}$...,$\sigma^{6}$} = {(15)(263),(236),... (1)}, but when expanding this to other groups, how do we know when to stop. In this case we stop when the exponent is equal to six because it is in $S_6$.

答:
The order of that cyclic group is $\rm{lcm}(1,2,3)=6$.  Note that we could have, for instance, $(12)(34567)$ which has order $10$ in $S_7$.  Or, $(123)(4567)$ which has order $12$.

In general, the order of an element will divide the order of the group, by a theorem due to Lagrange.  As far as I know that's about all that can be said.


问:
Does concavity imply that $f'(a)<\frac {f(a)}a$?
Does concavity imply that $f'(a)<\frac {f(a)}a$ ? I have read this in the proof of a paper by Straub, Mian and Sufi (https://scholar.harvard.edu/files/straub/files/mss_indebteddemand.pdf), appendix B.1. and I did not know this before. Does anybody know the proof of this or why this holds?

答:
See equation (3) page 7, $\eta(a)=a\nu'(a)$. Thus if $\nu$ is concave, $\nu''<0$ so,
$$\eta'(a)=\nu'(a)+a\nu''(a)<\nu'(a)=\frac1a(a\nu'(a))=\frac{\eta(a)}{a}.$$


问:
Proof explanation of There are Infinitely many primes of the form $4n+3$.
There are infinitely many primes of the form $4n+3$.
Proof: Define $q$ by $q=2^2.3.5...p-1$. Then $q$ is of the form $4n+3$, and is not divisible by any of the primes up to $p$. It cannot be product of primes $4n+1$ only, since the product of two numbers of this form is of same form; and therefore it is divisible by a prime $4n+3$, greater than $p$.

I don't understand how "It cannot be product of primes $4n+1$ only, since the product of two numbers of this form is of same form"have conclusion "therefore it is divisible by prime greater than $p$" ? Also how $4n+1$ came out from thin air? Mine version of explanation will exclude $4n+1$ since you can literally say $4n+3$ is prime the nothing to prove else there are prime not included in that weird $q$ so using fundamental theorem of arithmetic there is one more prime number. Also how $q$ is derived? My may is you first have multiple of prime then you need to force this to be of form $4n+3$ by $4(d+q)+3$ where $q$ is $-1$ and $d$ is multiple of prime number finite amount I think it should be prime number of form $4n+3$ in $d$ which is missing from textbook definition. I hope I get absolute correct and detailed explanation. Thanks for attention!

答:
Summarizing the comments, note that all odd numbers and therefore all primes greater than $2$ can be expressed as $4n+1$ or $4n+3$.  A number of the form $4n+5=4(n+1)+1$ is of the first form.

Second, the product of numbers of the form $4n+1$ is again of the form $4n+1$.  We can see that by writing $(4n+1)(4m+1)=16nm+4(n+m)+1$

Third, assume that there are a finite number of primes of the form $4n+3$.  Among them are $3,7,11,19 \ldots$.  Let the greatest of them be $p$.  Form the number $N=2^2\cdot 3 \cdot 5 \cdot 7 \ldots p-1$ where the factors are all the primes up to $p$.  $N$ is of the form $4n+3$ and is not divisible by any of the primes up to $p$.  By our hypothesis it must be a product of primes of the form $4n+1$ because all the primes of form $4n+3$ are less than or equal to $p$.  Since the product of numbers of the form $4n+1$ is again of the form $4n+1$ we have a contradiction.

Therefore there are infinitely many primes of the form $4n+3$.


问:
prove linear partial differential operator is continuous for $C^\infty_c(\Omega)$
Let $C^\infty_c(\Omega)$ be the test function space (which has LF topology on it.) where $\Omega\subset \Bbb{R}^n$.

Prove partial differential operator defined in classical sense is contiuous (under the LF space):

$$\partial_i:C^\infty_c(\Omega) \to C^\infty_c(\Omega)$$

I don't know how to check it,since the RHS of the above map is also a LF space,this is the key obstacle,for LHS we can convert it to $C^\infty_c(K)$ which is a space generated by seminorm,much easy to deal with.Can we use sequential continuous cireterion here?

答:
We know for test function space,any linear map from test function space to an locally convex TVS is continuous if and only if it's sequential continuous.

Now if $\phi_i\to 0$ in $C_c^\infty(\Omega)$ iff:

$\text{supp}(\phi_i)\subset K$ for some compact $K$.and uniformly convergence of any derivative of any order on this compact set.

Hence $\text{supp}(\partial_j\phi_i) \subset \text{supp}(\phi_i)\subset K$.with derivative of $\partial_j\phi_i$ uniformly convergence on K ,hence the differential operator  $\partial_j$ is continuous.


问:
Number of fish one can catch in a given timeframe
The number of fish a fisher can catch follows a poisson distribution. During 300 consecutive 1-hour periods there were 45 periods where no fish are caught.

Approximately how many fish are caught in 300 hours and what is the median time it will take for the fisher to catch a fish?

For the number of fish caught I thought it would be as simple as 255 fish. However, this is assuming that only 1 fish is caught in each period.

For the median I did Ln(2)/-Ln(45/300) = 0.3654

Does this logic make sense? I am much less confident in the first part.

答: You are expected to use the fact that no fish were caught $0.15$ of the time to find the average number of fish caught per hour.  This is the $\lambda$ of a Poisson distribution that has $0.15$ chance of no events.  Then multiply the average number of fish per hour by $300$ to get the expected number of fish caught in total.  The median time to catch a fish will be the time interval that has no fish caught $0.50$ of the time.


问:
Inequality holds for all the domain
How can I show that the following inequality holds for any $0\leq y<x \leq 1$:
$$3xy^2 - y^3 - 3y + 2-x >0.$$

It seems to hold for the defined domain, but could not manipulate it accordingly. Any ideas?

Title: Inequality holds for all the domain

Body:
This is does not true, but I found
$$3xy^2 - y^3 - 3y + 2-x \geqslant \boxed{2 - \frac{8\sqrt{6}}{9}} = -0.1773$$
Indeed, we write the inequality as
$$f(x) = (3y^2-1)x-y^3-3y+\frac{8\sqrt{6}}{9} \geqslant 0.$$
If $3y^2 \geqslant 1,$ then
$$f(x) > f(y) = 2y^3-4y+\frac{8\sqrt{6}}{9} = \frac{2(3y+2\sqrt{6})(3y-\sqrt{6})^2}{27} \geqslant 0.$$
If $3y^2 < 1,$ then
$$f(x) \geqslant f(1) = 3y^2-1-y^3-3y+\frac{8\sqrt{6}}{9} = (1-y)^3+\frac{8\sqrt{6}}{9}-2>0.$$


问:
Reference for Hypergraph Expander Mixing Lemma Generalization
Wikipedia lists a hypergraph generalization of the expander mixing lemma as follows.

Let $H$ be a $k$-uniform hypergraph, i.e. a hypergraph in which every "edge" is a tuple of $k$ vertices. For any choice of subsets $V_1,\ldots,V_k$ of vertices, $$\left||e(V_1,\ldots,V_k)| - \frac{k!|E(H)|}{n^k}|V_1|\cdots|V_k|\right| \le \lambda_2(H)\sqrt{|V_1|\cdots|V_k|}.$$

However, the reference listed for it by Friedman and Widgerson doesn't appear to (at least explicitly) state such a theorem. Is there a different reference someone could suggest, or explanation of how Friedman's paper does show such a theorem? Thanks!

答:
If you want some background on the mixing lemma in hypergraph settings, you should have a look at Inverse Expander Mixing for Hypergraphs, section 3, by Cohen et al. It includes some interesting points (e.g. the eigenvalue $\lambda_2$ does not correspond to any matrix).

Then, Eigenvalues and linear quasirandom hypergraphs includes a proof of the statement.


问:
$n^{n+1} \geq (n+1)^{n} , n \in \mathbb{N}, n > 2$ Problematic induction proof,
I need to prove that:

$$n^{n+1} \geq (n+1)^{n} , n \in \mathbb{N}, n > 2$$

I mean it is pretty obvious that it is true, but I can not prove that with induction.

for $n + 1$ I get that:
$$(n+1)^{n+2} \geq (n+2)^{n+1}$$
$$(n+1)^{n}*(n+1)^2 \geq (n+2)^{n}*(n+2)$$

And I don't see how could I plug my thesis into the inequality. I could do something like this, but it doesn't get me anywhere.

$$(n+1)^{n}*(n+1)^2 \geq n^{n}*(n+1)^2 \geq  (n+2)^{n}*(n+2)$$

The problem here is $(n+2)^{n}$, I can make it only bigger to keep my inequality valid, even if I use Bernoulis inequality.

答:
Spoiler: this is not a hint, it is a solution.

The key is in the hint by @peanut . We can rewrite $n^{n+1} \ge (n+1)^n$ as:
$$ \frac{n^{n+1}}{n^n} \ge \frac{(n+1)^n}{n^n} $$
Or
$$ n \ge (1+ \frac{1}{n})^n $$
The base case is true: $3 > (1+\frac{1}{3})^3$ $\ $ (it is actually easier to see that $3^4 > 4^3$).
Now, if the inequality holds for up to $n$, we show that it also holds for $n+1$ .

Noting that $ \frac{1}{n} > \frac{1}{n+1} $ we can write:
$$ n \ge (1+ \frac{1}{n+1})^n $$
And
$$ n(1+ \frac{1}{n+1}) \ge (1+ \frac{1}{n+1})^{n+1} $$
Finally, because $ \frac {n}{n+1} < 1$ we have
$$ n + 1 \ge (1+ \frac{1}{n+1})^{n+1} $$


问:
How to calculate the distance between two vectors of 6 elements?
I have two vectors, V1 and V2 where:

```<code> V1 = [0.0084    0.1871    0.1033    0.3332    0.3156    0.0524];
 V2 = [0.0176    0.2079    0.1245    0.2954    0.2510    0.1036];
</code>```

How do I calculate the distance geometry between the two vectors ?

答:
Consider $x=(x_1,x_2, \cdots x_n)$ and $y=(y_1,y_2, \cdots y_n)$ points in $\mathbb{R^n}$ The distance between the points is given by $d(x,y)=\sqrt{(y_1-x_1)^2+(y_2-x_2)^2+ \cdots + (y_n-x_n)^2}$
in your case
$v_1=x$ and $v_2=y$ in $\mathbb{R^6}$.

Just substitute in $d(x,y)$ you should get a square root of a sum of six squares.


问:
Stopping the Coronavirus puzzle

A square region $2020 \times 2020 \text{ km}^2$ divided into $2020^2$ cells. Some cells are contaminated by covid-19. Every week the virus spread to those cell which have at least $2$ side in common with contaminated cells. Find the maximum number of contaminated cells such that no matter where they are located the covid-19 pandemic will not spread to the entire region.

My school friend gave me this problem( better to say a puzzle) may be during the lockdown period(July-August) but I forgot it and yesterday he asked me if I have been able to solve the problem or not? And then the answer was obviously not, although I put a sufficient effort behind the problem that time and after the meet yesterday and also today I gave a lots of time but unable to figure it out. Thanks for your attention!

答:
Claim: On an $n$ by $n$ grid, if there are fewer than $n$ squares initially infected, then the infection will not spread to the entire region.

Define a edge of a square to be a frontier edge if one side of the edge is infected but the other side is uninfected. (The region outside the entire $n$ by $n$ grid is considered to always be uninfected.)

Key lemma: As the infection propagates, the number of frontier edges can never increase.

Proof of key lemma: Whenever the infection spreads to a new square, then at least two of its neighbors was already infected, hence you lose at least two frontier edges and gain at most two. End of proof.

Proof of claim: Suppose the infection spreads to the entire region. At that time, the number of frontier edges is $4n$ (the entire outer edge of the board). By the key lemma, the number of initial frontier edges must be at least $4n$. Therefore, there must have been at least $n$ initial squares infected.  Put another way, if there were fewer than $n$ squares initially infected, then the infection will not propagate to the entire region.

(By the way, there are many initial configurations of size $n$ that lead to the whole board becoming infected, not just the diagonals.)

Title: Stopping the Coronavirus puzzle


问:
If $p(x)$ and $q(x)$ are polynomials in a field such that $p(a) = q(a)$ for every $a$ in that field, can I say that they are equal?
Let $p(x)$ and $q(x)$ be two polynomials in a field $A$. I know they are equal if $deg(p(x)) = deg(q(x))$ and all their coefficients are equal.

But, if $p(a) = q(a)$ for every $a\in A$, are they also equal?

答: This is false in general, for instance in $\mathbb{Z}/p\mathbb{Z}$, the polynomials $X^p -X$ and $0$ take the same values but they are not equal.


问:
Let $G$ be a finite group and $f: G \to \Bbb Z$ be a homomorphism of groups. Prove that for all $x \in G$, $f(x)=0$
I took elements in $G$ as {$a_1,a_2,...,a_n$} considering order of $G$ to be as $n$. I am not able to define a function to prove $f(x)=0$ but I am not able to prove it. Please help.

答:
Note that $f(G)$ is a finite subgroup of $\mathbb{Z}$.

But since the only finite subgroup of $\mathbb{Z}$ is the trivial group $\{0\}$, we must have $$f(G)=\{0\}.$$
That is, $f(x)=0$ for all $x\in G$.


问:
Is a proof required for the Division Algorithm for polynomials?
I understand that the Division Algorithm can be applied to polynomials. Namely, for polynomials, for any polynomials $f,g$, there exist polynomials $q,r$ such that $f = gq + r$, with $\deg(f) > \deg (g).$ What does a proof of this look like? Is it an assumption we can make that you can always divide two polynomials with a quotient and remainder? I remember we used a proof using the Well-Ordering principle to prove the Division  Algorithm for two integers, but how can this be done for polynomials? Is it a valid assumption to say that the quotient and remainder will exist?

For the record, I don't mean this with regards to $\gcd (f,g)$.

答:
The Euclidean algorithm can be proven to work in vast generality. The key part here is that you can use the fact that naturals are well ordered by looking at the degree of your remainder. The remainder's degree always strictly decreases, and so your process must terminate after finitely many steps, since each term you get a remainder with smaller and smaller degree and so eventually the remainder is constant and you're done.

Generally, Euclid's algorithm works as long as you can associate a positive integer called the norm to each nonzero element of your number system and perform some 'division' that reduces this norm.


问:
Limit of Thomae's function as x goes to 1
I am working on the following problem:

Problem: Let $t(x)$ be Thomae's function:
$$t(x)=
  \begin{cases}
    1 & \text{if }x=0, \\
    1/n & \text{if }x=m/n \in \mathbf Q\setminus\{0\}\text{ is in lowest terms with }n>0,\\\
    0 & \text{if }x\notin\mathbf Q.
  \end{cases}$$

Show that $$\lim_{x\to1}t(x) =0$$

By showing that given $\epsilon>0$ that all the points in the set $T=\{ x \in \mathbb{R} : t(x) \geq \epsilon \}$ are isolated.

Progress I understand this intuitively and I think I have proven it using the usual definition of functional convergence by picking  $\delta= |m-n|\epsilon$. However I did not use the set T. Moreover I do no understand why we would need to show that first.

Another Post about the ( maybe the same) problem: How to prove the limit of Thomae function?

I have read the solution and I am lost for the same reasons as the OP of the post. The problem is that I still do not understand the hint that was given on that post, and I have been stuck on this for a long time.

 Any help would be appreciated, thank you!!

答:
You don't need to use the set $T$ to prove continuity. It provides one way of going about it (and a handy way of thinking about limits of $t$), but it is by no means necessary. Let's do a direct approach, which works in much the same way as the hint in the linked question.

First, consider the case $\varepsilon = \frac{1}{n}$ for some $n$ (yes, $\varepsilon > 0$ should be completely arbitrary, but we'll get to that in a second). Following the hint in the linked question, we just need to show that
$$\frac{p}{q} \in \left(1 - \frac{1}{n}, 1 + \frac{1}{n}\right) \setminus \{1\} \implies |q| \ge n + 1 > n.$$
Let's say we have such a $p/q$. Then
$$\left|\frac{p}{q} - 1\right| < \frac{1}{n} \implies\frac{|p - q|}{|q|} < \frac{1}{n} \implies |q| > |p - q|n.$$
Since $p/q \neq 1$, we have $p \neq q$, hence $|p - q| \ge 1$, hence
$$|q| > |p - q|n \ge n \implies |q| \ge n + 1.$$
The benefit here is that it means
$$t\left(\frac{p}{q}\right) \le \frac{1}{n + 1} < \frac{1}{n}.$$
That is, we can simply take $\delta = \frac{1}{n}$ as well, since
$$0 < |x - 1| < \delta \implies \begin{cases}t(x) = 0 < \dfrac{1}{n} & \text{if } x \in \Bbb{R} \setminus \Bbb{Q} \\ 0 \le t(x) < \dfrac{1}{n} & \text{if } x \in \Bbb{Q}\end{cases} \implies |t(x) - 0| < \frac{1}{n},$$
proving the limit for $\varepsilon = \frac{1}{n}$.

How can we show this for more general $\varepsilon > 0$? Well, since $\frac{1}{n} \to 0$ as $n \to \infty$, we can always choose a sufficiently large $n$ such that $0 < \frac{1}{n} \le \varepsilon$. We can then use $\delta = \frac{1}{n}$ as above.

If you wanted to go about this question using the hint, let's prove the isolation of $T$. You can prove isolation directly, but here's an indirect proof.

Suppose, for the sake of contradiction $(x_n)$ is an injective sequence in $T$ such that $x_n \to x_0 \in T$. Then $x$ and $x_n$ must be rational numbers; let $x_0 = p_0/q_0$ and $x_n = p_n/q_n$, each in lowest terms, each with positive denominator. We also must have $t(x_i) = 1 / q_i < \varepsilon$, i.e. $0 < q_i < 1/\varepsilon$.

Note that there are only finitely many integers in $(0, 1/\varepsilon)$, so there must be one such integer $q$ which is used infinitely often in the sequence. That is, there must be a subsequence $x_{n_k}$ such that $q_{n_k} = q$ for all $k$. But then
$$\frac{p_{n_k}}{q_{n_k}} = \frac{p_{n_k}}{q} \to x_0 \implies p_{n_k} \to x_0 q,$$
as $k \to \infty$. This implies $p_{n_k}$ is Cauchy, and so eventually the terms must become closer than $1/2$ to each other. But since this sequence is composed of integers, this implies that $p_{n_k}$ is eventually constant, contradicting injectivity. Thus, each point in $T$ is isolated.

How does this help us prove the limit? Well, for $0 < \varepsilon \le 1$, we have $1 \in T$, as $t(1) = 1 \ge \varepsilon$. Therefore, as proven above, it is isolated. That is, there must exist some $\delta > 0$ such that $(1 - \delta, 1 + \delta) \cap T = \{1\}$. That is,
\begin{align*}
0 < |x - 1| < \delta &\implies x \in (1 - \delta, 1 + \delta) \text{ and } x \neq 1 \\
&\implies x \in (1 - \delta, 1 + \delta) \text{ and } x \notin T \cap (1 - \delta, 1 + \delta) \\
&\implies x \notin T \\
&\implies t(x) < \varepsilon.
\end{align*}


问:
$\int_0^\infty(\log x)^2(\mathrm{sech}\,x)^2\mathrm dx$
Is there any closed-form representation for the following integral?
$$\int_0^\infty(\log x)^2(\mathrm{sech}\,x)^2\mathrm dx,$$
where $\mathrm{sech}\,x$ is the hyperbolic secant, $\mathrm{sech}\,x=\dfrac{2}{e^x+e^{-x}}$.</p>

答:
Related technique: (I), (II), (III). Here is a closed form

$$ \int_0^\infty(\log x)^2(\mathrm{sech}\,x)^2\mathrm dx = \int_0^\infty(\log x)^2\left(\frac{2}{e^{x}+e^{-x}}\right)^2\mathrm dx $$

$$= \left( \ln  \left( \pi  \right)  \right) ^{2}- \left( 4\,\ln \left( 2 \right) +2\,\gamma \right) \ln  \left( \pi  \right) +2\,
 \left( \ln  \left( 2 \right) \right) ^{2}+4\,\ln  \left( 2 \right)
\gamma-2\,\gamma \left( 1 \right) +\frac{1}{4}\,{\pi }^{2}
$$

$$ \sim 1.989349759. $$

Note: $\gamma(n)$ are known as Stieltjes $\gamma$-constants

$$ \gamma(n)= \lim_{m\to \infty}\left(\sum_{k=1}^m \frac{(\ln k)^n}{k}-\frac{(\ln m)^{n+1}}{n+1}\right).$$

ContentLicense: CC BY-SA 3.0


问:
Probability with Birthdays in a School in Certain Scenarios
There are 1000 students at a school. Assume there are 365 birthdays in a year.

a. What is the probability that the birthday of a random student is January 1? What is the probability it is not January 1?

b. If you asked two random students their birthdays, what is the probability that neither of will say January 1? If you ask all 1000 students this question, what is the probability that none will say January 1?

c. What is the probability that at least person in the school will say January 1?

d. What is the probability that no one in the school has a birthday tomorrow?

Body:
Assume all dates equally probable.

(a)$\frac{1}{365}$,

(b)  $(\frac{364}{365})^2$, $(\frac{364}{365})^{1000}$

(c) $1-(\frac{364}{365})^{1000}$

(d) $(\frac{364}{365})^{1000}$


问:
IVP in differential equations
I stuck in this question. I don't know what should I do when the question wants me to use initial value.

Solve the following IVP:

$$\left(y+\sqrt{x^2+y^2}\right)dx~-~x~dy~=0~,~~~~~~~~y(1)=0$$

答:
$$\left(y+\sqrt{x^2+y^2}\right)dx~-~x~dy~=0$$
$$\implies y~dx+\sqrt{x^2+y^2}~dx~-~x~dy~=0$$
$$\implies y~dx-~x~dy~+\sqrt{x^2+y^2}~dx~=0$$
$$\implies \dfrac{y~dx-~x~dy}{x^2}~+\sqrt{1+y^2/x^2}~dx~=0$$
$$\implies d\left(\frac{y}{x}\right)~+\frac 1x\sqrt{1+y^2/x^2}~dx~=0$$
integrating we have,
$$\ln\left(y/x+\sqrt{1+y^2/x^2}\right)+\ln x=\ln c$$
$$\implies \ln\left(y+\sqrt{x^2+y^2}\right)=\ln c$$
$$\implies y+\sqrt{x^2+y^2}= c$$where $c$ is an arbitrary constant.

Given that $~y(1)=0~,$ so $~0+1=c\implies c=1~$ and hence the particular solution of the given differential equation is $$ y+\sqrt{x^2+y^2}=1~.$$

Title: IVP in differential equations


问:
Pareto allocation problem when the utility is decreasing in one good
I have to characterize the Pareto optimal allocation of the following problem:

Consider two-agents-two-goods economy. The preferences of the agents are given by the following
utility functions:

$\upsilon_1(x_{1,1},x_{2,1})=x_{1,1}-0.25x^2_{2,1}$,

$\upsilon_2(x_{1,2},x_{2,2})=x_{1,2}+x^{0.5}_{2,2}$

where $x_{j,i}$ denotes the consumption of good j by agent i and agents have the following endowements $e_1=(0,8)$ and $e_2=(16,8)$.

I know that since the second good for the first consumer is a "bad good" in a Pareto optimal allocation $x_{2,1}$ will be equal to zero.

My question is if there is a way to obtain this results mathematically?

答: With positive consumption, the marginal utility of $x_2$ is negative for the first consumer, and positive for the second consumer, $$\frac{\partial u_{1}}{\partial x_2}<0 \text{ and } \frac{\partial u_{2}}{\partial x_2}>0$$. The Pareto optimal allocations $(a,0),(b,16)$, where $a+b=16$.


Title: $n^{n+1} \geq (n+1)^{n} , n \in \mathbb{N}, n > 2$ Problematic induction proof,

Body:
I need to prove that:

$$n^{n+1} \geq (n+1)^{n} , n \in \mathbb{N}, n > 2$$

I mean it is pretty obvious that it is true, but I can not prove that with induction.

for $n + 1$ I get that:
$$(n+1)^{n+2} \geq (n+2)^{n+1}$$
$$(n+1)^{n}*(n+1)^2 \geq (n+2)^{n}*(n+2)$$

And I don't see how could I plug my thesis into the inequality. I could do something like this, but it doesn't get me anywhere.

$$(n+1)^{n}*(n+1)^2 \geq n^{n}*(n+1)^2 \geq  (n+2)^{n}*(n+2)$$

The problem here is $(n+2)^{n}$, I can make it only bigger to keep my inequality valid, even if I use Bernoulis inequality.

Body:
Consider the ratio
$r(n)
=\dfrac{(n+1)^n}{n^{n+1}}
$.

If we can show that
$r(n) \gt r(n+1)$
for all $n$ and
$r(n) < 1$
for some $n=n_0$,
then
$r(n) \lt 1$ for
$n \ge n_0$
so
$(n+1)^n < n^{n+1}
$
for $n \ge n_0$.

$r(3)
=\dfrac{4^3}{3^4}
=\dfrac{64}{81}
\lt 1
$
so we can take
$n_0 = 3$.

$\begin{array}\\
\dfrac{r(n+1)}{r(n)}
&=\dfrac{\dfrac{(n+2)^{n+1}}{(n+1)^{n+2}}}{\dfrac{(n+1)^n}{n^{n+1}}}\\
&=\dfrac{n^{n+1}(n+2)^{n+1}}{(n+1)^{n+2}(n+1)^n}\\
&=\dfrac{(n(n+2))^{n+1}}{(n+1)^{2n+2}}\\
&=\left(\dfrac{n(n+2)}{(n+1)^2}\right)^{n+1}\\
&=\left(\dfrac{n^2+2n}{n^2+2n+1}\right)^{n+1}\\
&\lt 1\\
\end{array}
$

and we are done.


问:
What's the probability of getting 2 friends in the same classroom at school?
At school, each person gets put in a class at the start of the year, so I was just thinking how likely you were to get with a certain number of your friends in every period. I formulated this question:

You are at school with 6 periods and each period has 5 teachers teaching that one period in separate classes and each class has 30 students. You have 11 friends. What is probability that you have at least 2 friends in your class per period?

Now my original approach was to look at each friend as having a choice of which class to choose from. Since there are 5 teachers to chose from per period, you could think of it that way and then sort out the rest of the students' spots. I'm not really sure where to go from here.

答:
One possible way to approach this problem is to first note the symmetry of the problem, in that the probabilities of this happening is the same for each period. It is then apparent that we just have to find the probability of getting at least 2 friends in your class for one period, then raise that probability to the power of 6 to account for each period.

The next thing that comes in mind is complementary counting. Notice the term "at least $2$". If we consider using complementary counting, that will become "$0$ or $1$", which(in my opinion) is way easier to count.

We are then to split it into two cases:

Case 1: You share a class with exactly $0$ friends

Consider it from "your" perspective. You have $5$ teachers to choose from. Then note that the placement of your friends doesn't matter at this point, as long as none of them are placed in your class. We then have that there are $\binom{150-12}{30-1}=\binom{138}{29}$ ways to choose the rest of the students in "your" class, noting that none of your friends can be placed in your class. It then follows that the number of ways to choose the rest of the students in the class is $\prod_{n=0}^3\binom{120-30n}{30}$. This comes from the fact that after every successive class, there are $30$ less students to choose from.

So the total for case 1 is $5\cdot\binom{138}{29}\cdot\prod_{n=0}^3\binom{120-30n}{30}$

Case 2: You share a class with exactly $1$ friend

Like the last case, you have $5$ teachers to choose from. You then have $11$ friends that you can choose from that can be in your class. You then have $\binom{150-12}{30-2}=\binom{138}{28}$ ways to choose the rest of "your" class. Like the previous case, there are $\prod_{n=0}^3\binom{120-30n}{30}$ ways to choose the rest of the $4$ classes.

So the total for case 2 is $5\cdot11\cdot\binom{138}{28}\cdot\prod_{n=0}^3\binom{120-30n}{30}$

We then have that the total amount of ways is the sum of those two cases. We now have to find the total ways that the 5 classes can be chosen. This is easy to calculate: $\prod_{n=0}^4\binom{150-30n}{30}$

So we have that the probability after considering complementary counting and the other 5 periods is:$$\left(1-\frac{5\cdot\binom{138}{29}\cdot\prod_{n=0}^3\binom{120-30n}{30}+5\cdot11\cdot\binom{138}{28}\cdot\prod_{n=0}^3\binom{120-30n}{30}}{\prod_{n=0}^4\binom{150-30n}{30}}\right)^6\approx\boxed{9.915\%}$$

If there are any errors, please let me know! I'm kind of a noob at math so I most likely made some error somewhere.


问:
Prove that $\sum_{k=1}^\infty \sqrt{k+1}-\sqrt{k}$ diverges.
Prove that $\sum_{k=1}^\infty \sqrt{k+1}-\sqrt{k}$ diverges.

I know that this diverges but I don't know how to prove it rigorously. Should I use the comparison theorem? I know that $\sqrt{k+1}-\sqrt{k}>-\sqrt{k}$ so I just need to show that $\sum_{k=1}^\infty -\sqrt{k}$ diverges. But that is where I am stuck.

答:
HINT:  For each positive integer $N$:

$$\sum_{k=1}^N (\sqrt{k+1}-\sqrt{k}) = \sqrt{N+1}-\sqrt{1}.$$

Why is this so, and can you finish from here.


问:
Probability that we choose a two headed coin
We have a $501$ coins on the table, and assume that they have all been flipped onto that table (i.e., there is a mix of heads and tails). This also includes a two-headed coing.

Now if we pick up $1$ coin and its heads, what is the probability it is also the two-headed coin?

I've seen questions similar to this, but not the exact same. I did this out but want to hear your thoughts, before I explain how I arrived at the probability.

答:
This seems to be too simple to be rigorous, but it gives the right answer!

There are 502 "heads" on the table.  Two of them are on the two-headed coin.  You've  picked a coin with a head, so the probability that the head you see comes from the two-headed coin is $\frac{2}{502}$ or $\frac{1}{251}$

Apparently, the fact that the coins are spread out on the table in some random heads-tails pattern is immaterial. You could do the experiment by putting the coins in a bag, and then pick one randomly, (with replacement, if the one side you look at is tails) until you see a head...


答:
If you take logs of both sides, what you need to show is
$${\ln( n + 1) \over n + 1 } \leq {\ln n \over n}$$

This follows from the fact that ${\displaystyle {\ln x \over x}}$ is decreasing for $x > e$.


答: Hint: Two relevant facts for you to prove: $f(x^n) = n f(x)$ and $f(1) = 0$

Title: Let $G$ be a finite group and $f: G \to \Bbb Z$ be a homomorphism of groups. Prove that for all $x \in G$, $f(x)=0$

Body: I took elements in $G$ as {$a_1,a_2,...,a_n$} considering order of $G$ to be as $n$. I am not able to define a function to prove $f(x)=0$ but I am not able to prove it. Please help.


问:
An interesting infinite series
What is
$$
\sum_{i=1}^{\infty} \frac{(-1)^{i+1}}{i^i}
$$

I thought this series would be pretty easy to crack with a power series, but it has proved intransigent. Is there a closed form?

答:
The series is called Sophomore's dream. There's no closed form for it, but the following equalities holds:
$$
\int _{0}^{1}x^{-x}\,dx=\sum_{i=1}^{\infty} \frac{1}{i^i}
$$
And
$$
\int _{0}^{1}x^{x}\,dx=\sum_{i=1}^{\infty} \frac{(-1)^{i+1}}{i^i}
$$

Title: An interesting infinite series


问:
Are simplicial sets equal if their non-degenerate simplices are the same?
Let $\Delta$ be the simplex category and $\Delta^{inj}\subset \Delta$ be the non-full subcategory of injective morphisms.
Then, a simplicial set $X:\Delta^{op}\to\text{Set}$ can be precomposed with the inclusion of $\Delta^{inj}$ to get a functor $X^{inj}:(\Delta^{inj})^{op}\to\textbf{Set}$ whose image are the non-degenerate morphisms of $X$.

Claim: Let $X,X'$ be simplicial sets such that $X^{inj} = X'^{inj}$. Then $X = X'$.

Proof: by Proposition 1.1.3.4 in Kerodon, any $n$-simplex $x:\Delta^n\to X$ can be factored as $\Delta^n\xrightarrow{f}\Delta^m\xrightarrow{y}X$, where $y$ is non-degenerate. Then $y\in X'$, since $X^{inj} = X'^{inj}$, hence $yf= x\in X'$. Thus $X\subset X'$. Analogously, $X'\subset X$ hence $X = X'$.

Is this correct? It would explain why the very frequent depictions of simplicial sets by their non-degenerate morphisms are mostly harmless.

答:
It's not quite true that $X^\text{inj}$ (as you defined it) is the semismplicial set of non-degenerate simplices of $X$. In fact, it has exactly the same elements as $X$. On the other hand the collection of non-degenerate simplices does not always form a semisimplicial set: this is because there may be some non-degenerate simplex of $X$ that has a face that is degenerate.

There are some statements which are true. For example:

Let $X$ and $Y$ be simplicial sets, and let $f, g : X \to Y$ be morphisms of simplicial sets. If $f (x) = g (x)$ for all non-degenerate simplices $x$ of $X$, then $f = g$.

Another example:

Let $X$ and $Y$ be simplicial sets and let $f : X \to Y$ be a morphism of simplicial sets. If $f$ restricts to a bijection between non-degenerate simplices of $X$ and $Y$, then $f$ is an isomorphism of simplicial sets.


问:
Finding $i^{i^i}$: Review My Work
Please let me know if my work is correct:

We know that: $i^i = e^{-\pi/2}e^{-2 \pi m}, m=0,\pm1,\pm2,...$
Thus,
$$
i^{i^i} = i^{e^{-\pi/2}\space\space\space\space e^{-2 \pi m}}\\
=e^{e^{-\pi/2}\space\space\space\space e^{-2 \pi m} \space\space\space\space\ln i}\\
=e^{e^{-\pi/2}\space\space\space\space e^{-2 \pi m} \space\space\space\space(\ln |i| + i \arg i)}\\
=e^{e^{-\pi/2}\space\space\space\space e^{-2 \pi m} \space\space\space\space(i (\pi/2)+2\pi ik)}\\
=e^{e^{-\pi/2} \space\space\space\space e^{i(-\pi^2m -4 \pi^2mk)}}
$$
for $m,k = 0,\pm1,\pm2,...$

Title: Finding $i^{i^i}$: Review My Work

Body: A better way.  $a=i^i=e^{-\frac{pi}{2}}$.  $x=i^a=e^{aln(i)}$, but $i=e^\frac{\pi i}{2}$, so $ln(i)=\frac{\pi i}{2}$  Net $x=e^{\frac{\pi i}{2}e^{-\frac{\pi}{2}}}$


问:
Helicoid with fixed parameters each time
Let

$A(r, \theta):=<r\cos \theta, r \sin\theta, \theta>$, $r \in[-1,1], \theta \in [0,2\pi]$ be the parameterization of an area. So this is obviously a helicoid.

(*) Let r be fixed and $\theta$ varying in the interval, then let $\theta$ be fixed and let r vary. what kind of curves would be obtain through this kind of variation? at which angle would they intersect?

So unfortunately I do not have any idea.

I tried to plot it here, https://www.desmos.com/calculator/usy8hfrtbx, for $\theta$ fixed I only got a straight line and for $r$ fixed I got kind of a helix?  Can this be right? how could I calculate the angle at which they intersect?

I would be thankful for inputs.

答: Sometimes a picture is worth 1000 words:


问:
Semi-Complicated recurrence relation to be solved via generating functions
If $(a_n)_{n\ge 1}$ is a sequence such that $a_0=3$ and for all $n \ge 1$,
\begin{equation}
\sum_{k=1}^{\lceil \frac{n}{2} \rceil}a_{n-2k+1}=na_n
\end{equation}
, use generating functions to find an explicit formula for $a_n$.

So far I've been swimming around in the algebra a bit, and I've gotten

$$A'(x)=\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=1}^\infty\left(\sum_{k=1}^{\lceil n/2 \rceil} a_{n-2k+1}\right)x^{n-1}$$

But I'm not sure how to get a nice expression in terms of $A(x)$ for the RHS of this equation.

答:
The recurrence relation hints that the derivative of $A(x)$ may be involved:
$$A'(x)=\sum_{n=0}^\infty na_nx^{n-1}=\sum_{n=1}^\infty\left(\sum_{k=1}^{\lceil n/2 \rceil} a_{n-2k+1}\right)x^{n-1}=A(x)+x^2A(x)+x^4A(x)+
\ldots,$$
and summing the geometric series one obtains the differential equation
$$A'(x)=\frac{A(x)}{1-x^2}.$$
The solution can be obtained by separation of variables:
$$A(x)=C\frac{\sqrt{x+1}}{\sqrt{x-1}}.$$
Since everything is happening for $x$ near $0$, no need to worry about convergence issues.


问:
Applying Argument Principle to find zeroes in a quadrant when there is a real root
I was trying to find the number of roots of the polynomial $p(x)=z^9+2z^5-2z^4+z+3$ that lie in the second quadrant. I know that $p(-1)<0$ and $p(0)>0$, so there is some real root on the negative real axis between $-1$ and $0$. This means that I cannot use either Rouche's Theorem or the Argument Principle to find number of roots in the second quadrant by checking the strict inequality/change in argument around the boundary of the quarter disc in the second quadrant, because there is a zero on $\delta D$. I tried applying argument principle on the quarter disc where instead of the negative real axis, I took it along $re^{i(\pi - \varepsilon)}$, but still got nowhere. Any help on how to go about this would be appreciated!

答:
We can use a few other tools quickly to form a sneaky strategy:

It's not hard to check by hand that $p(z)$ has no zeros on the imaginary axis (we can determine the zeros of its real and imaginary parts there).

Since $p(z)$ has real coefficients, its complex zeros come in conjugate pairs.

$p(z)$ has exactly one negative zero, as you've found.

Therefore the number of zeros in the left half-plane will be $2N+1$, where $N$ is the number of zeros in the second quadrant.


问:
Is Hartshorne's Remark III.2.9.1 actually a valid argument?
In Hartshorne's book Algebraic Geometry, following Proposition III.2.9 which states that on a Noetherian topological space $X$ that cohomology of abelian sheaves commutes with direct limits he follows with Remark III.2.9.1 which says

"As a special case we see that cohomology commutes with infinite direct sums."

Now I don't doubt that infinite directs sums commute with cohomology, by all means they may. I don't see how this is a special case of the above proposition however, a direct sum is not a direct limit unless you are taking the direct sum of a single object, it's defined as a colimit over a discrete category which could not be directed if there's more than a single object. Is there something that I'm missing, or is this a completely bogus proof?

答: An infinite direct sum is naturally the direct limit of its finite subsums (that is, $\bigoplus_{i\in I}A_i$ is the direct limit of $\bigoplus_{i\in F}A_i$ where $F$ ranges over the directed set of finite subsets of $I$).  Since cohomology preserves finite direct sums and direct limits, it follows that it preserves infinite direct sums.


答: The key is to get more inventive with your diagram. Instead of taking the colimit all at once, build it up in pieces: if $I$ is an infinite set, then we can consider the diagram made up of finite subsets $J\subset I$ where the arrows are inclusions. Then $\bigoplus_{i\in I} A_i$ is the colimit of $\bigoplus_{j\in J} A_j$ as $J$ varies over that diagram, and as cohomology commutes with finite direct sums and direct limits, we have a proof.

Title: Is Hartshorne's Remark III.2.9.1 actually a valid argument?

Body:
In Hartshorne's book Algebraic Geometry, following Proposition III.2.9 which states that on a Noetherian topological space $X$ that cohomology of abelian sheaves commutes with direct limits he follows with Remark III.2.9.1 which says

"As a special case we see that cohomology commutes with infinite direct sums."

Now I don't doubt that infinite directs sums commute with cohomology, by all means they may. I don't see how this is a special case of the above proposition however, a direct sum is not a direct limit unless you are taking the direct sum of a single object, it's defined as a colimit over a discrete category which could not be directed if there's more than a single object. Is there something that I'm missing, or is this a completely bogus proof?


问:
$4^{n-1} {{3n}\choose{n}} \sqrt{n} \geq 3^{3n-2}, n \in \mathbb{N} \setminus \{0\}$ Induction proof, need a hint for more efective proof.
I am looking for a smarter way of proving that inequality by induction.

$$4^{n-1} {{3n}\choose{n}} \sqrt{n} \geq 3^{3n-2}, n \in \mathbb{N} \setminus \{0\}$$

For $n + 1$:
$$4^{n} {{3n + 3}\choose{n+1}} \sqrt{n+1} \geq 3^{3n+1}$$
$$4*4^{n-1} \frac{(3n)!(3n + 1)(3n + 2)(3n + 3)}{(n)!(n+1)(2n)!(2n+1)(2n+2)} \sqrt{n+1} \geq 3^{3n+1}$$

Then I can plug my thesis into the inequality for $n + 1$:
$$4^{n-1} {{3n}\choose{n}} \sqrt{n} \geq 3^{3n-2} \implies 4^{n-1} \frac{(3n)!}{n!(2n)!} \geq \frac{3^{3n-2}}{\sqrt{n}}$$

So I get:

$$4* \frac{3^{3n-2}}{\sqrt{n}} * \frac{(3n + 1)(3n + 2)(3n + 3)}{(n+1)(2n+1)(2n+2)} \sqrt{n+1} \geq 3^{3n+1}$$

$$4*(3n + 1)(3n + 2)(3n + 3) \sqrt{n+1} \geq 27\sqrt{n} (n+1)(2n+1)(2n+2) $$

And then, if I sqare both hand sides of that ineqality (both are positive) I get 7th powers of $n$ and there is probably some smarter / easier way to prove that inequality.

答:
This isn't induction.
It uses Stirling
and it is more general.

I will show that
$0.850
\lt \dfrac{\binom{3n}{n}}{\sqrt{\dfrac{3}{4\pi n}}
\left(\dfrac{27}{4}\right)^n}
\lt 1.085
$.

Since
$n! \approx \sqrt{2\pi n}(n/e)^n$,

$\begin{array}\\
\binom{an}{bn}
&=\dfrac{(an)!}{(bn)!((a-b)n!}\\
&\sim \dfrac{\sqrt{2\pi an}(an/e)^{an}}
{(\sqrt{2\pi bn}(bn/e)^{bn})(\sqrt{2\pi n(a-b)}(((a-b)n)/e)^{(a-b)n})}\\
&= \sqrt{\dfrac{2\pi an}{2\pi bn2\pi n(a-b)}}\left(\dfrac{(an)^ae^be^{a-b}}{e^a(bn)^b((a-b)n)^{a-b}}\right)^n\\
&= \sqrt{\dfrac{ a}{2\pi bn(a-b)}}\left(\dfrac{a^an^a}{b^b(a-b)^{a-b}n^a}\right)^n\\
&= \sqrt{\dfrac{ a}{2\pi bn(a-b)}}\left(\dfrac{a^a}{b^b(a-b)^{a-b}}\right)^n\\
&=r(n, a, b)\\
\end{array}
$

If $a=3, b=1$,
$\dfrac{a}{b(a-b)}
=\dfrac{3}{2}
$
and
$\dfrac{a^a}{b^b(a-b)^{a-b}}
=\dfrac{3^3}{2^2}
=\dfrac{27}{4}
=6\frac34
$
so
$\binom{3n}{n}
\sim \sqrt{\dfrac{3}{4\pi n}}
\left(\dfrac{27}{4}\right)^n
$.

Actually
(https://en.wikipedia.org/wiki/Stirling%27s_approximation),
if
$f(n)
 =\dfrac{n!}{\sqrt{2\pi n}(n/e)^n}$,
then the following inequality holds:

$$1
 \lt f(n)
 \lt  e^{1/(12n)}
 \lt \frac{e}{\sqrt{2\pi}}
 = 1.0844...
 $$

So,
if $u < f(n) < v$,
then
$\frac{u}{v^2}
\lt \dfrac{\binom{an}{bn}}{r(n, a, b)}
\lt \frac{v}{u^2}
$.

From the simple bounds above,
we can take
$u = 1$
and
$v = \frac{e}{\sqrt{2\pi}} $,
so the bounds are
$\dfrac{2\pi}{e^2}
=0.850...
$
and
$\dfrac{e}{\sqrt{2\pi}}
=1.0844...
$.

Therefore
$0.850
\lt \dfrac{\binom{an}{bn}}{\sqrt{\dfrac{ a}{2\pi bn(a-b)}}\left(\dfrac{a^a}{b^b(a-b)^{a-b}}\right)^n}
\lt 1.085
$.

If $a=3, b=1$ then
$0.850
\lt \dfrac{\binom{3n}{n}}{\sqrt{\dfrac{3}{4\pi n}}
\left(\dfrac{27}{4}\right)^n}
\lt 1.085
$.


问:
Why does the differential equation $y' = y + 1$ have solution $y(x) = Ce^x - 1$?
I was watching a video on differential equations for a class that I'm taking. I took calculus so long ago that I can't seem to figure why the differential equation  $y' = y + 1$ has solution $y(x) = Ce^x - 1$.</p>

答:
The easy way is to plug in $y(x) = ce^x-1$ back into the differential equation and see if the differential equation is satisfied. To actually, derive the solution, note that
$$y' = y + 1 \implies (y+1)' = (y+1) \implies y+1 = ce^x \implies y = ce^x-1$$</p>


问:
Is a proof required for the Division Algorithm for polynomials?
I understand that the Division Algorithm can be applied to polynomials. Namely, for polynomials, for any polynomials $f,g$, there exist polynomials $q,r$ such that $f = gq + r$, with $\deg(f) > \deg (g).$ What does a proof of this look like? Is it an assumption we can make that you can always divide two polynomials with a quotient and remainder? I remember we used a proof using the Well-Ordering principle to prove the Division  Algorithm for two integers, but how can this be done for polynomials? Is it a valid assumption to say that the quotient and remainder will exist?

For the record, I don't mean this with regards to $\gcd (f,g)$.

答:
Basically, what you need is to be able to divide any coefficient by the leading coefficient of the divisor.

Explicitly, let $f(x) = a_nx^n+\cdots + a_1x + a_0$ be a polynomial with coefficients in some (commutative) ring, and assume that $a_n$ is a unit; that essentially means that you can “divide by $a_n$”. For example, for polynomials with integer coefficients, you would need the leading coefficient to be either $1$ or $-1$.

We prove that for every polynomial $g(x)$ (with coefficients in the same ring) there exist polynomials $q(x)$ and $r(x)$ such that $g(x) = q(x)f(x) + r(x)$, with $r(x)=0$ or $\deg(r)\lt \deg(f)$. The polynomial $r(x)$ is the “remainder”, and the polynomial $q(x)$ is the “quotient”. We prove this by induction on $\deg(g)$.

If $g=0$, then we can take $q(x)=r(x) = 0$. Assume the result holds for any polynomial of degree strictly smaller than $k$, and that $\deg(g)=k$. Write
$$g(x) = b_kx^k+\cdots + b_1x + b_0.$$
Think long division. If $k\lt n=\deg(f)$, we let $q(x)=0$ and $r(x)=g(x)$.

If $k\geq n$, then think Long Division. Look at $q_1(x)=\frac{b_k}{a_n}x^{k-n}$. This makes sense, because we said we can divide by $a_n$, and because $k-n\geq 0$. Note that the leading term of $q_1(x)f(x)$ is exactly $b_kx^k$. Therefore, $g(x) - q_1(x)f(x)$ is either $0$ or of degree strictly smaller than $k$. Either way, by induction we know that there exist a polynomial $q_2(x)$ and a remainder $r(x)$ with $r(x)=0$ or $\deg(r)\lt\deg(f)$, and
$$g(x)-q_1(x)f(x) = q_2(x)f(x) + r(x).$$
But then we have
$$g(x) = q_1(x)f(x)+q_2(x)f(x) + r(x) = (q_1(x)+q_2(x))f(x)+r(x),$$
with $r(x)=0$ or $\deg(r)\lt\deg(f)$. Letting $q(x)=q_1(x)+q_2(x)$ we are done.

Note that the Well-ordering principle for the nonnegative integers is essentially equivalent to induction.

Once you have this it is straightforward to go further and prove uniqueness, again under the assumption that you can always divide by $a_n$.

When the coefficients are in a field (like $\mathbb{Q}$, or $\mathbb{R}$, or $\mathbb{C}$) this always holds, which means you can always do division with remainder for one polynomial divided by a nonzero polynomial.


问:
Proving $(x^{p}-x)^{a} \equiv 0\pmod{p^{a}}$ as a step to another problem?
I have a problem where I have to use that $a^n\equiv a \pmod p$ to show that there exist non-zero polynomials modulo $n$ that are the zero function. I have tested some examples and figured that for the examples I tested, it seems that for $m\equiv p_1^{a_1}p_2^{a_2}\dots p_n^{a_n}$, the polynomial:

$$(x^{p_1}-x)^{a_1}(x^{p_2}-x)^{a_2}\dots(x^{p_n}-x)^{a_n} \tag{$\star$}$$

Is the zero function but I can't figure out why, I also tested some variations such as:

$$(x^{p_1^{a_1}}-x)^{a_1}(x^{p_2^{a_2}}-x)^{a_2}\dots(x^{p_n^{a_n}}-x)^{a_n}$$

But noticed they fail. I thought about writing $(\star)$ as:

$$\begin{eqnarray*}
  {(x^{p_1}-x)^{a_1}(x^{p_2}-x)^{a_2}\dots(x^{p_n}-x)^{a_n} }&\equiv &{0\pmod{p_1^{a_1}} } \\
  {(x^{p_1}-x)^{a_1}(x^{p_2}-x)^{a_2}\dots(x^{p_n}-x)^{a_n} }&\equiv &{0\pmod{p_2^{a_2}}} \\
  {\vdots\hspace{3cm}}&&{} \\
  {(x^{p_1}-x)^{a_1}(x^{p_2}-x)^{a_2}\dots(x^{p_n}-x)^{a_n} }&\equiv &{0\pmod{p_n^{a_n}}}
\end{eqnarray*}$$

And realized that if that is actually true, we must have:

$$\begin{eqnarray*}
  {(x^{p_1}-x)^{a_1} }&\equiv &{0\pmod{p_1^{a_1}} } \\
  {(x^{p_2}-x)^{a_2} }&\equiv &{0\pmod{p_2^{a_2}}} \\
  {\vdots\hspace{4mm}}&&{} \\
  {(x^{p_n}-x)^{a_n} }&\equiv &{0\pmod{p_n^{a_n}}}
\end{eqnarray*}$$

So it all boils down to proving that $(x^{p}-x)^{a} \equiv  0\pmod{p^{a}}$ for any $p$ prime and any positive integer $a$. But I have no clue on how to do that. Could you give me a hint?

答:
Hint If $b \in \mathbb Z$ you know that
$$b^p-b \equiv 0 \pmod{p}$$

This means that
$$p|b^p-b$$

Deduce from here that
$$p^a|(b^p-p)^a$$

P.S. You can also show that there exist non-zero polynomials modulo $n$ that are the zero function by a very simple counting: there are finitely many functions $f: \mathbb Z/n \mathbb Z \to \mathbb Z/n \mathbb Z$ but infinitely many polynomials. Therefore, there are two polynomials $P \neq Q$ with the same function. Then $R:=P-Q$ is your polynomial. You can even restrict the degree to $\deg(R)\leq n+1$


问:
Clifton-Pohl Torus and $\Gamma$ acting properly discontinuous
Problem: Let $M = \mathbb{R}^2 \setminus \left\{ (0,0 \right\}$ be the pseudo-Riemannian manifold with metric $$ ds^2 = \frac{2 du dv}{ u^2 + v^2}. $$ Let $\mu(u, v) = (2u, 2v)$. This is an isometry (trivially). I wish to show that the group $ \Gamma = \left\{ \mu^n \right\}$ generated by $\mu$ acts properly discontinuous on $M$. Then $T = M/\Gamma$ is a Lorentz surface, called the Clifton-Pohl Torus.

So by definition I need to prove that:  every $p \in M$ has a neighborhood $U$ such that $\phi(U) \cap U = \emptyset$ for all $\phi \in \Gamma$ with $\phi \neq Id$. (do Carmo, Riemannian Geometry, p. 165).

The way I wished to show this is as follows. If $p = (u_0, v_0) \in M$, then let $\epsilon = d((u_0, v_0), (2u_0, 2v_0))$ where $d$ denotes the distance between $(u_0, v_0)$ and $(2u_0, 2 v_0)$ for the metric $ds^2$ on $M$. Then if I take as a neighborhood $U = B_{\epsilon/2} (p)$ , i.e. the 'ball' centered at $p$ with radial distance $\epsilon/2$, then any action of a $\phi \in \Gamma$ will move all of the points of $U$ out of $U$, i.e. $\phi(U) \cap U = \emptyset$.

The problem that I have is that I'm not sure if $\epsilon > 0$, since in a semi-Riemannian manifold the distance between two distinct points can be zero.

Moreover, does one define distance in Lorentzian geometry the same way as in Riemannian geometry? In Riemannian geometry the Riemannian distance $d(p,q)$ between two points $p$ and $q$ is defined as the greatest lower bound of $\left\{ L(\alpha): \alpha \in \Omega(p, q) \right\}$, where $\Omega(p,q)$ is the set of all piecewise smooth curve segments in $M$ from $p$ to $q$ and $L(\alpha)$ denotes the arc length of $\alpha$.

Does this definition extend to Lorentzian geometry?

答:
The distance used in the definition of a properly discontinuous group action should be compatible with the topology on of the manifold. The Lorentzian distance is the wrong thing to use here. Since the group acts on $\mathbb{R}\setminus\{(0,0)\}$ you should use the euclidean distance, $\rho$.

Then given any point $(u,v)$ we can take $U=\{(x,y): \rho((u,v),(x,y))<1\}$. In your answer you seem to know what to do at this point to check the definition.

The issue in your understanding is that the Lorentzian distance is not actually a distance. It does not satisfy the definition of a distance. It is called a distance, but it is not a distance. Some authors call it the time separation function to avoid this awkwardness.

Unlike a Riemannian manifold there is no canonical choice of Lorentzian distance. There are lots of distances on the manifold that are compatible with the topology of the manifold but there is no "given" one. You just need to pick one and work with it.

In the particular case of the Clifton-Pohl torus life is easy as the construction is explicit.


问:
When are the sum of the digits of $2^n$ and $2^{n+1}$ equal and for what value of n? Stuck on this problem!
For $a \in Z$, let $p(a)$ be the sum of the digits of $2^a$. For instance, $2^5 =32$ and $p(5)=3+2$. For what values of $a$ is $p(a)=p(a+1)$.

How do I go about doing this?

I know that the last digits of $2^a$ end in 2,4,6, or 8. I tried writing equations to represent the digits but nothing is working so far.

答: Hint:  digit sums should make you think of working $\bmod 3$ or $\bmod 9$


问:
Let $G$ be a finite group and $f: G \to \Bbb Z$ be a homomorphism of groups. Prove that for all $x \in G$, $f(x)=0$
I took elements in $G$ as {$a_1,a_2,...,a_n$} considering order of $G$ to be as $n$. I am not able to define a function to prove $f(x)=0$ but I am not able to prove it. Please help.

答:
$e$ being the identity of $G$, we have

$f(e) = 0, \tag 1$

since group homomorphisms map identities to identities; if

$\exists x \in G, \; f(x) \ne 0, \tag 2$

then

$x \ne e \tag 3$

and

$\exists \Bbb Z \ni m \ge 2 , \; x^m = e; \tag 4$

thus we have

$0 \ne mf(x) = f(x^m) = f(e) = 0, \tag 5$

and this contradiction implies there is no $x$ as in (2); it follows that

$f(x) = 0, \; \forall x \in G. \tag 6$


问:
prove the following linear functional is not induced by measure
Let the linear functional $I$ below defined on $C^1_c(\mathbb{R})$. as follows:

$$I(f) = \frac{df}{dx}(0)$$

Prove it's not induced by a measure,or even signed measure.that is no such measure makes $I(f) = \int fd\mu = \frac{df}{dx}(0)$.

I try to use sequence of function $f_n$ such that makes $\frac{df_n}{dx}(0) \to \infty$,but the total mass of $\int f_nd\mu \to 0$ ?If exist such measure it can not be the Lebesgue measure ,since we can construct such function,can it holds for some other measures?

答:
Here is an argument that works for signed measures.  Suppose by contradiction that $\mu$ exists and let us analyse $\mu$ on $\mathbb R\setminus \{0\}$.

If $f$ is any function in $C_c^1(\mathbb R\setminus \{0\})$ then we may extend $f$ to the whole real line by setting $f(0)=0$, and it will follow that $f'(0)=0$, so
$$\int f d\mu=0.$$  This implies that the restriction of $\mu$ to $\mathbb R\setminus \{0\}$ vanishes so $\mu$ must be a multiple of the Dirac measure at $0$, but this clearly doesn't work either, hence $\mu$ doesn't exist.


问:
Domain of convergence for a Laurent series
The Laurent series I have is given as $$\sum_{n=-\infty}^{-1} \frac{z^n}{|n-2|^3} + \sum_{n=0}^{\infty} \frac{z^n}{(2n)!}$$ I was trying to figure out the interval of convergence for the second half, I know how to do the first half, but on the second half, I tried the ratio test and the limit ends up going to $0$ which means the test is inconclusive, what can I do with the second half to find the radius of convergence?

答:
For the second sum I get $r=\limsup_{n\to\infty}\dfrac1{\sqrt[n]{\dfrac1{2n!}}}=\infty$.

Since the first part converges for $|z|\gt1$, that will also be where the whole thing does.


问:
Proof of an Inequality with exponents
I have been given a function $$f(a, b, c) = (a^{b+c-2a})(b^{a+c-2b})(c^{a+b-2c})$$ and I need to prove that $$f(a, b, c) \le 1$$ $\mathbf{Case 1:}$ If $a=b=c$, then it is trivial that $f(a,b,c)=1$. But now, I am stuck with how to prove for $f(a,b,c)<1$? Is the further proof related to AM or GM in some way? $(a,b,c > 0)$

EDIT 1: $$log((a^{b+c-2a})(b^{a+c-2b})(c^{a+b-2c})) = ((b+c-2a)log(a))+((a+c-2b)log(b))+((a+b-2c)log(c))$$

$$a(log(bc))+b(log(ac))+c(log(ab))-2(log(a^ab^bc^c)) \le 1$$

EDIT 2: The original problem in the book I was practicing from  was to prove that $a^{(b+c)}b^{(a+c)}c^{(a+b)}\le(a^ab^bc^c)^2$. I arrived at $f(a, b, c)$ through simplification.

EDIT 3: Assuming $a \ge b \ge c$ without loss of generality, this implies that $k^{b+c} \le k^{2a}$ and similarly for other terms (if $k \in$ {$a,b,c$}, k>1 since a, b, c are positive reals). Thanks to Calvin Lin for suggesting Chebyshev.

答: Hint: Show that $ a^b b^c c^a \leq a^a b^b c^c$.

Title: Proof of an Inequality with exponents


问:
Prove the following and use it to evaluate the integral:
I want to prove that:$$\int_{-\infty}^\infty f(x)dx=\int_{-\infty}^\infty f\left(x-\frac1x\right)dx$$
And use the result of this proof to evaluate:$$\int_{-\infty}^\infty\frac{x^2}{x^4+1}dx$$

答:
Here is how to apply it to
$$\int_{-\infty}^\infty\frac{x^2}{x^4+1}dx
= \int_{-\infty}^\infty\frac{1}{(x-\frac1x)^2+2}dx
= \int_{-\infty}^\infty\frac{1}{x^2+2}dx=\frac\pi{\sqrt2}
$$


问:
How to determine effective angular range in an area of a sphere
I have populated a sphere using the Lambert's azimuthal projection method to obtain a uniform grid/area - see link for reference to the paper (https://www.aanda.org/articles/aa/full_html/2010/12/aa15278-10/aa15278-10.html).

The generated sphere is shown in the image attached 1, and I would like to calculate the angular range that each area occupies. I have identified 4 points on the sphere and their respective azimuthal and elevation angles. So my question is - how do I calculate the effective angular range that each area on the sphere contains?

Hope that is clear. Thank you.

*Edit: Added another image of the sphere from a different angle 2.

答:
We have a set of "pixels" or points on the sphere arranged in rings,
that is, the pixels can be partitioned into subsets so that the pixels in each subset lie along one line of latitude at equal intervals.
I will call one such subset a ring of pixels.

I will suppose that the sphere is divided into zones between lines of latitude so that each zone contains one ring of pixels.
Each zone is then divided into bounding boxes by a set of lines of longitude so that each bounding box contains exactly one pixel.

We require that every bounding box in every zone have the same area.
We assume the pixels are placed in such a way that this is possible.
Within these constraints, we arrange the bounding boxes so that each pixel is as close to the center of its box as possible.

Let there be $n$ rings, with $p_k$ pixels in the $k$th ring, $1\leq k \leq n,$
numbered in sequence of latitude from one pole to the other.
The total number of pixels is
$$ P = \sum_{i=1}^n p_i = p_1 + p_2 + \cdots + p_n. $$

For simplicity, we suppose the radius of the sphere is $1.$
Then the area of the sphere is $4\pi$ and the area of the bounding box of each pixel is $4\pi/P.$

The combined area of all zones up to and including the one occupied by the $k$th ring is
$$ \frac{4\pi}{P} \sum_{i=1}^k p_i = \frac{4\pi}{P}(p_1 + p_2 + \cdots + p_k). $$
These zones occupy a spherical cap;
the area of a spherical cap is proportional to its height,
with the entire sphere considered as a cap of height equal to the sphere's diameter,
so the height of this spherical cap is
$$
\frac{\frac{4\pi}{P} \sum_{i=1}^k p_i}{4\pi} \cdot 2
= \frac 2P \sum_{i=1}^k p_i.
$$

This implies that the boundary of the cap (the line of latitude between ring number $k$ and ring number $k+1$) is at colatitude
$$ \arccos \left(1 - \frac 2P \sum_{i=1}^k p_i\right).$$

(Colatitude is like latitude, but measured starting from zero the pole instead of at the equator.)

The other boundary of the zone occupied by the $k$th ring is the boundary of the spherical cap occupied by the first $k-1$ rings, which is at colatitude
$$ \arccos \left(1 - \frac 2P \sum_{i=1}^{k-1} p_i\right).$$

So that is the range of colatitudes of the bounding box of a pixel in the $k$th ring; subtract from $\frac\pi2$ radians to get the range of latitudes.

Since there are $p_k$ equal-sized bounding boxes in a ring which is a total of $2\pi$ radians around, the angular "width" of each box (measured in radians) is $2\pi/p_k.$
If the longitude of the pixes is $\lambda$, then the range of longitudes of the bounding box is
$$\left[\lambda - \frac{2\pi}{p_k}, \lambda + \frac{2\pi}{p_k} \right]. $$


问:
Showing that, for any point set in $\Bbb{R}^d$, it is possible to order them so that every consecutive three determine an acute angle
I am trying to solve this question.

Let $F$ be a point set in $\mathbb R^d$ of size $n$. Prove that it is possible to denote the points of $F$ as $X_1,\dots, X_n$ in such way that, for any $i\in \{2,\dots, n-1\}$, the angle $\angle X_{i-1}X_iX_{i+1}$ is acute.

I think I have to use permutation for this but I am unable to relate to it.

Thanks in advance.

答:
Construct a polyline $x_1x_2\ldots x_n$ using following algorithm:

Among all pairs $u, v \in F$, there are pairs which maximize their
distance. Pick any pair and call them $x_1, x_2$, we have:

$$|x_1 - x_2 | = \max\{ |u - v| : u, v \in F \}$$

For any $2 < k \le n$, let $F_k = F \setminus \{ x_1, \ldots, x_{k-1} \}$. Among all vertices in $F_k$, there are some which maximize the distance to $x_{k-1}$. Pick one of them and call it $x_k$, we have:
$$|x_k - x_{k-1}| = \max\{ |u - x_{k-1} | : u \in F_k \}$$

For any $2 \le k \le n - 1$, let
$$a = |x_{k-1} - x_k|, b = |x_k - x_{k+1}|\;\text{ and }\;c = |x_{k-1} - x_{k+1}|$$
For $k = 2$, since $x_1x_2$ is a diameter of $F$, we have $a \ge c$.

For $k > 2$, since $x_{k+1} \in F_k$, by definition of $x_k$, $a \ge c$ again.

By cosine law, the angle $\theta_k \stackrel{def}{=} \angle x_{k-1}x_kx_{k+1}$ satisfies:

$$\cos\theta_k = \frac{a^2+b^2 - c^2}{2ab} = \frac{(a^2-c^2) + b^2}{2ab} \ge \frac{b}{2a} > 0$$
This implies $\theta_k < \frac{\pi}{2}$ and the polyline has an acute angle at $x_k$.

As a side note, since $x_1x_2$ is a diameter of $F$, if we prepend $x_0 \stackrel{def}{=} x_n$ to the polyline, the angle $\theta_1 = \angle x_0x_1x_2 = \angle x_nx_1x_2$ is also acute. This means as a closed polygonal chain, $x_1x_2\ldots x_n$ has at least $n-1$ acute angles.


Title: Why does the differential equation $y' = y + 1$ have solution $y(x) = Ce^x - 1$?

Body: I was watching a video on differential equations for a class that I'm taking. I took calculus so long ago that I can't seem to figure why the differential equation  $y' = y + 1$ has solution $y(x) = Ce^x - 1$.</p>

Body:
$$\frac{dy}{dx}=y+1$$
$$\frac{dy}{y+1}=dx$$
$$\ln(y+1)=x+a$$
$$y+1=e^{x+a}=ce^x$$


问:
Positive integers $n$ that cannot be written as a sum of $r$ consecutive positive integers, with $r > 1$, are of the form n = $2^l$ for some $l ≥ 0$.
Prove that the positive integers $n$ that cannot be written as a sum of $r$ consecutive positive integers, with $r > 1$, are of the form n = $2^l$ for some $l ≥ 0$.

I wrote just  k+(k+1)+.....(k+r-1) as $2^l$ and then by applying AP formula, it gets me to something like $r/2 ( 2k + (r-1))$ = $2^l$ which leads to nowhere.. .I did to get a contradiction.
I have shown my approach and this is not a homework problem.I need some help to move from here or a new way to solve  the problem in a less cumbersome manner.
Thank you.

答: Write an expression for the sum from $k$ to $k+r-1$ using the formula for the sum of the numbers from $1$ to $n$.  Your sum is the difference of two of these.  Factor that expression and show that one of the factors is odd.  $2^l$ does not have any odd factors, but all other numbers do.


问:
Dropping the "Lowest Grade" Problem
Let $A=\{a_1, a_2, ... , a_n\}$ be a set of non-negative real numbers and $B=\{b_1, b_2, ..., b_n\}$ be sets of positive real numbers.

Let $s = \dfrac{ \sum_A a}{\sum_B b} =  \dfrac{a_1+a_2+\dots+a_n}{b_1+b_2+\dots+b_n}$.

What I want to do find $k$ so that

$s_1 = \dfrac{\sum_{A-\{a_k\}} a }{\sum_{B-\{b_k\}} b}$ is the largest.

And more generally, find

$s_p = \dfrac{\sum_{A-\{a_{k_1},a_{k_2},\dots,a_{k_p}\}} a }{\sum_{B-\{b_{k_1},b_{k_2},\dots,b_{k_p}\}} b}$

I'm not exactly sure what this is called, and I tried doing a google search but I wasn't able to find what I am looking for.

And of course, I can do this by exhaustion, but I feel that there should be a smart way to figure this out...

答:
Here is a short exposition on the problem, including some pretty good examples of why greedy algorithms or dynamic programming algorithms probably won't work.  The key insight that the authors have is that we can transform this from a "find the maximum" problem to a "find the zero of a function" problem, and that function is, fortunately, pretty easy to evaluate at arbitrary points.  Here's the transformation:

We want a subset $S$ of assignments so that:

$$ \frac{\sum_{i \in S} m_i}{\sum_{i \in S} n_i} = q  \tag 1$$

is maximized, where $m_i$ is the student's score on assignment $i$ and $n_i$ is the maximum possible score.  Rearranging this equation, we get:

$$ \sum_{i \in S} \left(m_i - qn_i\right) = 0 \tag 2$$

The LHS is a linear function of $q$ for any given $S$.  We'll just call that $f(q)$ and set about finding the zeros.  Note that at the value of $q_{opt}$ for which equation $(1)$ is maximized, any $S$ but the optimal one give the LHS $< q_{opt}$.  This translates to the LHS $< 0$ in equation $(2)$.  Let's create a new function:

$$F(q) = \text{max}_S \left(\sum_{i \in S} \left(m_i - qn_i\right), |S|=n-k\right)$$

$F$ is piecewise linear, continuous, and, since all of the pieces have negative slope ($-\sum n_i$), then it is monotonic.  This means it has exactly one zero.  We know from before that that zero is $q_{opt}$, and it occurs during the "piece" where we're using the optimal subset.  The magic of $F$ is that it is easy to evaluate at any given $q$ (just choose the largest $|S|$ $m_i - qn_i$ terms).  So, we can use arbitrary zero-finding mechanisms to find $q_{opt}$ and, as a corollary, $S$.

The paper I reference above suggests the bisection algorithm and Newton's method.  In my experiments, and also, according to them, in their experiments, Newton's method works pretty well.  However, it seems to me that $F$ may have up to (but generally has fewer than) $n\choose k$ "pieces".  This means that we might well end up doing $n\choose k$ iterations of Newton's method in the worst case.  The bisection algorithm is actually potentially worse, since we can presumably make the $q_{opt}$ "piece" arbitrarily small, and therefore arbitrarily difficult to find with the bisection algorithm.  So, this method is arguably not better than the brute-force algorithm in the worst case.  I do not have any examples of what that worst case may be.

TL; DR

The Newton's method solution seems to be worst-case $O ({n \choose k})$.  Possibly, it has a better average case.  Certainly, in the situations I've used it, it is way faster than the naïve brute force solution.

Here's python code implementing the Newton's Method solution:

```<code>import heapq
def assignment_subset_zero(q, k_drop, scores, points_possible):
    f = scores - q*points_possible
    S = heapq.nlargest(f.shape[0] - k_drop, range(f.shape[0]), key = f.take)
    return np.sum(scores[S])/np.sum(points_possible[S]), S

# find the subset of assignments to optimize the weighted average grade
# scores and points_possible should be equal-length numpy arrays
def optimal_assignment_subset(k_drop, scores, points_possible):
    previous_q = -1
    this_q = 0
    while previous_q != this_q:
        #print(previous_q, this_q)
        previous_q = this_q
        this_q, S = assignment_subset_zero(previous_q, k_drop, scores, points_possible)
    return S
</code>```

And some code to run the more interesting examples included in the paper:

```<code>def dales_quiz_test_set(c, b):
    return (
        np.array([20+c] + [20 + i - b for i in range(1,11)]),
        np.array([40 + i*2 for i in range(11)])
    )

def set_equal(s1, s2):
    return (len(s1) == len(s2)) and (len(s1 & s2) == len(s1))

assert set_equal(set(optimal_assignment_subset(5, *dales_quiz_test_set(4,1))), {0,10,9,8,7,6})
assert set_equal(set(optimal_assignment_subset(5, *dales_quiz_test_set(6,1))), {0,1,2,3,4,5})
</code>```

Title: Dropping the "Lowest Grade" Problem

Body:
Here is a short exposition on the problem, including some pretty good examples of why greedy algorithms or dynamic programming algorithms probably won't work.  The key insight that the authors have is that we can transform this from a "find the maximum" problem to a "find the zero of a function" problem, and that function is, fortunately, pretty easy to evaluate at arbitrary points.  Here's the transformation:

We want a subset $S$ of assignments so that:

$$ \frac{\sum_{i \in S} m_i}{\sum_{i \in S} n_i} = q  \tag 1$$

is maximized, where $m_i$ is the student's score on assignment $i$ and $n_i$ is the maximum possible score.  Rearranging this equation, we get:

$$ \sum_{i \in S} \left(m_i - qn_i\right) = 0 \tag 2$$

The LHS is a linear function of $q$ for any given $S$.  We'll just call that $f(q)$ and set about finding the zeros.  Note that at the value of $q_{opt}$ for which equation $(1)$ is maximized, any $S$ but the optimal one give the LHS $< q_{opt}$.  This translates to the LHS $< 0$ in equation $(2)$.  Let's create a new function:

$$F(q) = \text{max}_S \left(\sum_{i \in S} \left(m_i - qn_i\right), |S|=n-k\right)$$

$F$ is piecewise linear, continuous, and, since all of the pieces have negative slope ($-\sum n_i$), then it is monotonic.  This means it has exactly one zero.  We know from before that that zero is $q_{opt}$, and it occurs during the "piece" where we're using the optimal subset.  The magic of $F$ is that it is easy to evaluate at any given $q$ (just choose the largest $|S|$ $m_i - qn_i$ terms).  So, we can use arbitrary zero-finding mechanisms to find $q_{opt}$ and, as a corollary, $S$.

The paper I reference above suggests the bisection algorithm and Newton's method.  In my experiments, and also, according to them, in their experiments, Newton's method works pretty well.  However, it seems to me that $F$ may have up to (but generally has fewer than) $n\choose k$ "pieces".  This means that we might well end up doing $n\choose k$ iterations of Newton's method in the worst case.  The bisection algorithm is actually potentially worse, since we can presumably make the $q_{opt}$ "piece" arbitrarily small, and therefore arbitrarily difficult to find with the bisection algorithm.  So, this method is arguably not better than the brute-force algorithm in the worst case.  I do not have any examples of what that worst case may be.

TL; DR

The Newton's method solution seems to be worst-case $O ({n \choose k})$.  Possibly, it has a better average case.  Certainly, in the situations I've used it, it is way faster than the naïve brute force solution.

Here's python code implementing the Newton's Method solution:

import heapq
def assignment_subset_zero(q, k_drop, scores, points_possible):
    f = scores - q*points_possible
    S = heapq.nlargest(f.shape[0] - k_drop, range(f.shape[0]), key = f.take)
    return np.sum(scores[S])/np.sum(points_possible[S]), S

# find the subset of assignments to optimize the weighted average grade
# scores and points_possible should be equal-length numpy arrays
def optimal_assignment_subset(k_drop, scores, points_possible):
    previous_q = -1
    this_q = 0
    while previous_q != this_q:
        #print(previous_q, this_q)
        previous_q = this_q
        this_q, S = assignment_subset_zero(previous_q, k_drop, scores, points_possible)
    return S

And some code to run the more interesting examples included in the paper:

def dales_quiz_test_set(c, b):
    return (
        np.array([20+c] + [20 + i - b for i in range(1,11)]),
        np.array([40 + i*2 for i in range(11)])
    )

def set_equal(s1, s2):
    return (len(s1) == len(s2)) and (len(s1 & s2) == len(s1))

assert set_equal(set(optimal_assignment_subset(5, *dales_quiz_test_set(4,1))), {0,10,9,8,7,6})
assert set_equal(set(optimal_assignment_subset(5, *dales_quiz_test_set(6,1))), {0,1,2,3,4,5})


问:
If every path crosses a subset, then is the subset connected?

Problem

Let $X$ be a path-connected topological space and suppose that given two points $a,b$ the image of every path connecting $a$ to $b$ intersects some subset $S$. Is $S$ connected?

Context / Digression

In my Analysis class there was a problem that looked like the following.

Let $D \subset \mathbb{R}^2$ be the closed unit disk and $a = (-1,0), b = (1,0)$ and suppose $f \colon D \to \mathbb{R}$ is a continuous function where $f(a) = -10$ and $f(b) = 10$.

The two parts was that we had to show that $f$ was equal to 0 somewhere on $D$ then we were challenged to find out how many zeros there were.

We know there is a zero because the image will be a closed (because $D$ is compact) interval (because $D$ is connected) where the containing $0$ (because it contains both $-10$ and $10$). We know there are uncountably many zeros as there are uncountably many paths from $a$ to $b$ (there is at least one path intersecting each point of the $y$-axis) and each path must contain a zero as the map $[0,1] \to \mathbb{R}$ will similarly be a closed interval containing $0$.

In class I felt like there was more to say about the set of points where $f$ was $0$. I actually felt like the zero set would contain a subset $S$ such that $S$ is connected and every function would have to pass through $S$. I felt this way intuitively because if there was a path connecting $a$ to $b$ without intersecting the set then there would be a disconnection $U,V$ where $U$ is the set of points with $y$ values greater than the $y$ value of the path and $V$ contains all of the lower points. So this construction does not work because every such path contains a zero. But this is not rigorous because it can be the case that such a disconnection exists without using my $U$ and $V$.

There are more questions this problem too. For example, would the space $X - S$ be disconnected?
Do we need the path-connected hypothesis to say $S$ is connected (if the answer to the original question is yes). But that's getting ahead of myself. Any progress on this problem is greatly appreciated.

Let me know if anything is unclear.

答:
The answer is no. Let $X$ be the unit circle in $\Bbb R^2$, $a=(1,0)$, $b=(-1,0)$ and $S=\{(0,1),(0,-1)\}$.

Note also that $X\setminus S$ can be connected, for example for $X$ the Warsaw Circle and $S$ a point on the nice part of the arc.


问:
Difference between zeta sum and Euler product?
The fact that $$\sum_{n=1}^\infty \frac{1}{n^s} = \prod_{p}\frac{1}{1-p^{-s}}$$ is a consequence of unique factorization of primes.

We could form a similar sum and a similar product of irreducibles in a number field which does not have unique factorization. Since the unique factorization does not hold the vales would be probably different.

Are the always different?

Can we get any information about the class number from their difference?

答:
There are a few issues with what you are proposing: what it means for an element of $\mathcal{O}_K$ to be "positive" is unclear (so which elements are we summing over on the left?), which value of $\alpha^s$ to choose when $\alpha$ is not a positive real number is unclear, and that because not all prime ideals of $\mathcal{O}_K$ are principal, taking a product over irreducible elements (assuming we already picked which ones are "positive") will not even be capturing all the information about the number field, much less the failure of unique factorization. The fix (or at least the one I know of, perhaps there are others) is to consider ideals in $\mathcal{O}_K$ instead of positive integers, and prime ideals instead of irreducible elements, and take norms so that we only ever take positive integers to complex powers. Then because there is unique factorization of ideals into prime ideals, we have the Euler product for the Dedekind zeta function
$$\sum_{I\subseteq\mathcal{O}_K}\frac{1}{N_{\mathbb{Q}}^K(I)^s}=\prod_{P\subseteq\mathcal{O}_K}\left(1-\frac{1}{N_{\mathbb{Q}}^K(P)^s}\right)^{-1}.$$</p>


问:
Indication of Gaussian integers solutions for $x^2+y^2=z^3$.
I know the existence of infinitely many integer solutions for $x^2+y^2=z^3$. But my concern is the existence of Gaussian Integers solution of this diophantine equation. I try to search for relevant papers but most discuss in terms of integer solutions only. Thank you.

答:
$$z^3=uv,\qquad z,u,v \in \Bbb{Z}[i]$$ then $$z^3=x^2+y^2, \qquad x = \frac{u+v}{2}, y = \frac{u-v}{2i}\in \frac12 \Bbb{Z}[i]$$

$x,y$ are in $\Bbb{Z}[i]$ iff $u-v\in 2\Bbb{Z}[i]$ which implies that $u=v=z\bmod 2 \Bbb{Z}[i]$.

Every solution is of this form.

In particular there is no solution with $z=1+i$

and there is one iff $z\in 2\Bbb{Z}[i]$ or $1+2\Bbb{Z}[i]$.


问:
If the coefficient of $x$ in the penultimate term of the remainder is 1280, what is the value of $n$
$$\frac{(x+n)^n+(x+2)^n+(n+2)^n}{x^2-(n+2)x+2n}$$

From: Lumbreras Editors

I tried using the remainder theorem without success

答:
(Fill in the gaps as needed. If you're stuck, show your work and explain what you've done.)

The numerator is $ f_n(x) = (x+n)^n  + (x+2)^n + (n+2)^n$.

The denominator is $ g_n(x) = (x-n)(x-2)$.

Let $f_n(x) = g_n(x) p(x) + ax + b $, then we're after the value of $a$.

$f_n(n) = (2n)^n + (n+2)^n + (n+2)^n = na + b$.

$f_n(2) = (2+n)^n + 4^n + (n+2)^n = 2a+b$.

Then $ a = \dfrac{ f_n(n) - f_n(2) } { n - 2 }  = \dfrac{(2n)^n-4^n}{n-2}=1280$

Hence $ n = \ldots $

Note: I'm interpreting "penultimate term of the remainder" as "the coefficient of $x$ in $ax+b$", as opposed to "the coefficient of $x$ in $p(x)$".


问:
Proof of a theorem with upper/lower limits.

Theorem: If $s_n \le t_n$ for all $n$ greater than a fixed integer $N$, then $$\lim_{n \to \infty} \inf s_n \le \lim_{n \to \infty} \inf t_n$$

I would like to prove this and it would be nice if someone could check my work.

Proof: Letting $$\lim_{n \to \infty} \inf s_n = s_*$$ and $$\lim_{n \to \infty} \inf t_n = t_*$$ assume that $s_* > t_* $ while there exists $N$ such that for all $n$ >$N$, $s_n \le t_n$.
  Then there must be an integer $M$(particularly larger than $N$) such that if $m > M$, $t_m < s_*$. This contradicts the fact that for all $n > N$, $s_n \le t_n$.

答:
It is right. Another nice proof is this:

For any $N \le n$, $s_n\le t_n$, and hence $\inf \{s_n: N\le n\} \le \inf \{t_n: N\le n\}$. Then we can conclude that $$\lim_{n \to \infty} \inf s_n \le \lim_{n \to \infty} \inf t_n.$$ This complete the proof.


问:
Show $\int _{\mathbb{R}}[F(x+c) - F(x)] dx =0 $ is incorrect Proof
$F$ is CDF  (probability distribution function)

$\int _{\mathbb{R}}[F(x+c) - F(x)]  dx $

$= \int _{\mathbb{R}}F(x+c)  dx -\int _{\mathbb{R}} F(x)  dx$  ( (by linearity of the integral)

$=\int _{\mathbb{R}}F(x)  dx -\int _{\mathbb{R}}F(x)  dx $  ( By change of variables theorem )

$=0$

For change of variables we used the known theorem (see below) with $\Omega = \Omega' = \mathbb{R} $ , $T(x) = x + c $ and $ \mu= \lambda$ ( Lebesgue measure)

Question: Why is the above Proof incorrect? How can we show it is not correct analytically?

My thought:

I am not sure how to show it, maybe
because the density does not exist? ( Even assumption of continuity would not imply the existence of the density) ( e.g. Cantor function is a continuous CDF which does not have a density with respect to Lebesgue measure)

```<code>Below is the known Theorem that we used
</code>```

Why is the above proof wrong ?

In other words why is this step $=\int _{\mathbb{R}}F(x)  dx -\int _{\mathbb{R}}F(x)
> dx $  ( By change of variables theorem )  Incorrect ?

We use the change of variable theorem (see above) with $\Omega =
 \Omega' = \mathbb{R} $ , $T(x) = x + c $ and $ \mu= \lambda$ (
Lebesgue measure)

答:
For any CDF $F$ we have $\int_{\mathbb R} F(x)dx=\infty$. This is because $F(x) \to 1$ as $ x \to \infty$. If we choose $a$ such that $F(x)>\frac 1  2$ for all $x >a$ then we get $\int_{\mathbb R} F(x)dx \geq \int_a^{\infty} \frac 1 2  dx=\infty$. So splitting the integral into two terms gives you $\infty-\infty$.

For an example where the integral is not $0$ consider $exp(1)$ distribution and take $c=1$ You can easily compute the integral in this case and the answer is not $0$.

Actually the value of the integral is $c$. Let $\mu$ the probability measure corresponding to $F$. Then (by Fubini/Tonelli's Theorem)  $\int_{\mathbb R} [F(x+c)-F(x)]dx=\int_{\mathbb R} \int_{(x,x+c]} d\mu(y)dx=\int \int_{y-c}^{c}dx d\mu (y)=\int c d\mu(y)=c$.


问:
Perpendicular Vectors on Unit Sphere induce bound on components
Working through an old preliminary exam.

Let $x=(x_1,...,x_n)^{T}$ and $y=(y_1,...,y_n)^{T}$ be column vectors in $\mathbb{R}^n$ and suppose $n\geq2$. Suppose also that $\vert\vert x\vert\vert=\vert\vert y\vert\vert=1$ and that $x\perp y$. Prove that

$x_1^2+y_1^2\leq 1$

If I remember correctly, on the exam I stated that because the norm and orthogonality are both inherited from the inner product just take an isometry that maps one of the vectors to $e_1$. The desired inequality follows easily.

However, I'm trying to now actually work through this without such appeals. I've deduced that the subscript 1 is in fact arbitrary, and that

$x_i^2+y_i^2\leq 1$

must hold for all $n$. I can't seem to get a componentwise estimate. If there is such thing as a hint without giving a full answer, I'd appreciate that first. Thank you.

Body:
If $x_1 = 0$, the claim is straightforward. So assume $x_1 \ne 0$. We have

$$x \perp y \iff x_1y_1 + \cdots + x_ny_n = 0 \iff  y_1 = \frac{-(x_2y_2+ \cdots + x_ny_n)}{x_1} \tag 1$$

and

$$(x_2y_2 + \cdots + x_ny_n)^2 \le (y_2^2 + \cdots + y_n^2) (x_2^2 + \cdots + x_n^2) \tag 2$$

by the Schwarz inequality,  and

$$y_1^2 + \cdots +y_n^2 = x_1^2 + \cdots + x_n^2 \tag 3$$

since $\|x\| = \|y\| = 1$.

Substituting $(1)$ into $(3)$ and then using $(2)$ yields

$$ \frac{(y_2^2 + \cdots + y_n^2) (x_2^2 + \cdots + x_n^2)}{x_1^2} + y_2^2 + \cdots + y_n^2 \ge  x_1^2 + \cdots + x_n^2 $$

$$\iff (y_2^2 + \cdots + y_n^2) \left(\frac{x_2^2 + \cdots + x_n^2}{x_1^2} + 1 \right)\ge  x_1^2 + \cdots + x_n^2$$

$$\iff (y_2^2 + \cdots + y_n^2) \ge  \frac{1}{\frac{x_2^2 + \cdots + x_n^2}{x_1^2} + 1}(x_1^2 + \cdots + x_n^2)$$

$$\iff (y_2^2 + \cdots + y_n^2) \ge  \frac{x_1^2}{x_1^2 + \cdots + x_n^2}(x_1^2 + \cdots + x_n^2)$$

$$\iff (y_2^2 + \cdots + y_n^2) \ge  x_1^2$$

This gives the desired inequality:

$$y_1^2 + x_1^2 \le y_1^2 + (y_2^2 + \cdots + y_n^2) = 1$$

$\square$


问:
continuous linear extension from Fréchet space $C^\infty(\Omega) $ to $C(\Omega)$
Let $\Omega\subset \mathbb{R}^n$   be an open subset, can we always extend a continuous linear functional on Fréchet space $C^\infty(\Omega)$ to a continuous linear functional on Fréchet space $C(\Omega)$?

I know if the extension exists, it must be the unique one, hence the map $(C(\Omega))' \to (C^\infty(\Omega))'$ is an injection. If the above result holds, it will an linear isomorphism, it seems not need to hold?

答:
The usual Frèchet topology for $C^\infty(\Omega)$ is stronger than the norm topology in $C(\Omega)$. So, even though you have an inclusion as vector spaces, you don't have one as Frèchet spaces.

For a concrete example, consider the continuous linear functional $f\longmapsto f'(1)$ on $C^\infty[0,1]$. On $\mathbb R[x]\subset C[0,1]$ this map is discontinuous.


答:
Since $F$ is a CDF, $\displaystyle\lim_{x\rightarrow\infty}F(x)=1$. Therefore, the integral $\int_{\mathbb{R}}F(x)\,\mathrm{d}x$ diverges. When you get the expression
$$\int _{\mathbb{R}}F(x+c)\, \mathrm{d}x -\int _{\mathbb{R}} F(x)\, \mathrm{d}x$$
this becomes $\infty-\infty$, which is an indeterminate form, so it cannot be simplified to $0$.

With regard to how one would correctly evaluate this integral, there is a simple way to do it using Fubini's Theorem.

Title: Show $\int _{\mathbb{R}}[F(x+c) - F(x)] dx =0 $ is incorrect Proof

Body:
$F$ is CDF  (probability distribution function)

$\int _{\mathbb{R}}[F(x+c) - F(x)]  dx $

$= \int _{\mathbb{R}}F(x+c)  dx -\int _{\mathbb{R}} F(x)  dx$  ( (by linearity of the integral)

$=\int _{\mathbb{R}}F(x)  dx -\int _{\mathbb{R}}F(x)  dx $  ( By change of variables theorem )

$=0$

For change of variables we used the known theorem (see below) with $\Omega = \Omega' = \mathbb{R} $ , $T(x) = x + c $ and $ \mu= \lambda$ ( Lebesgue measure)

Question: Why is the above Proof incorrect? How can we show it is not correct analytically?

My thought:

I am not sure how to show it, maybe
because the density does not exist? ( Even assumption of continuity would not imply the existence of the density) ( e.g. Cantor function is a continuous CDF which does not have a density with respect to Lebesgue measure)

```<code>Below is the known Theorem that we used
</code>```

Why is the above proof wrong ?

In other words why is this step $=\int _{\mathbb{R}}F(x)  dx -\int _{\mathbb{R}}F(x)
> dx $  ( By change of variables theorem )  Incorrect ?

We use the change of variable theorem (see above) with $\Omega =
 \Omega' = \mathbb{R} $ , $T(x) = x + c $ and $ \mu= \lambda$ (
Lebesgue measure)


问:
Strong characterization of $\mathbb C$ with respect to $\mathbb R$

$\mathbb R^2$ is not a field, but $2$-tuple arithmetic rules like $(a,b)(c,d)=(ac-bd,ad+bc)$ coupled with $\mathbb R^2$ make it a field, but are there other rules than $(a,b)(c,d)=(ac-db,ad+bc)$ combined with $\mathbb R^2$ that make it a field?

Is $\mathbb C$ the only field that contains $\mathbb R$ as a subfield or are there other fields that contain $\mathbb R$ as subfields?

Is $\mathbb C$ the only that can be made by coupling $\mathbb R^2$ and rules like $(a,b)(c,d)$ $=$ $(ac-db,ad+bc)$? Are there other rules that can be coupled with $\mathbb R^2$ to make it a field?

PS: Just trying to pin down some properties that make $\mathbb C$ special/unique from other fields that can/may be constructed using some specific algebraic structure of 2 tuples in $\mathbb R$ e.g. $(\sqrt 2,\pi)$. At this point this is just a wide net as I lack the terminology to make this question specific. But any hints on how to make this question specific to listing properties that make $\mathbb C$ unique or gives strong characterisation of relationship between  $\mathbb C$ and $\mathbb R$ are welcome.

答:
For any field $F$, the field $F(t)$ of rational functions in the variable $t$ is a field that contains $F$. The degree $[F(t):F]$ is infinite, and $F(t)$ is a transcendental (i.e., non-algebraic) extension of $F$. Either of these observations suffices to check that $\mathbb{R}(t)\not\cong\mathbb{C}$ (as extensions of $\mathbb{R}$), because $[\mathbb{C}:\mathbb{R}]=2$ and $\mathbb{C}/\mathbb{R}$ is algebraic. Moreover, there is not even an abstract isomorphism between $\mathbb{R}(t)$ and $\mathbb{C}$ (i.e., one which does not necessarily respect the way $\mathbb{R}$ sits inside each of them) because $\mathbb{R}(t)$, and indeed $F(t)$ for any field $F$, is not algebraically closed, whereas $\mathbb{C}$ is.

Also, this is up to isomorphism; technically, if you've chosen what you mean by "$\mathbb{C}$" and I take your $\mathbb{C}$ and paint the elements of $\mathbb{C}\setminus\mathbb{R}$ red, that is a field different from $\mathbb{C}$ that contains $\mathbb{R}$. Similarly, $\mathbb{C}$ and $\mathbb{R}^2$ are different things; yes, $\mathbb{C}$ is a two-dimensional real vector space, and so we often think of it as a plane, but this is an identification we are making - they are not literally equal.

Yes, if $F$ is a field that contains $\mathbb{R}$ and $[F:\mathbb{R}]=2$ (i.e., $F\cong \mathbb{R}^2$ as real vector spaces), then $F\cong \mathbb{C}$. This is because $[F:\mathbb{R}]$ finite $\implies$ $F$ is an algebraic extension of $\mathbb{R}$, so $\overline{F}$ (the algebraic closure of $F$) is an algebraically closed field containing $\mathbb{R}$ and which is an algebraic extension of $\mathbb{R}$, so that $\overline{F}$ is also an algebraic closure of $\mathbb{R}$, and therefore $\overline{F}\cong\mathbb{C}$.

The most straightforward characterization of $\mathbb{C}$ is just that it is an algebraic closure of $\mathbb{R}$. It is also true that $\mathbb{C}$ is (up to isomorphism) the unique algebraically closed field of characteristic $0$ and cardinality $\mathfrak{c}$.


问:
2 dimensional representation of $\operatorname{SL}_2(\mathbb{Z})$
I am trying to work out $2$-dimensional representation of $\operatorname{SL}_2(\mathbb{Z})$. I know that $\operatorname{SL}_2(\mathbb{Z})$ is generated by $S = \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}, R = \begin{pmatrix} 0 & -1\\ 1 & 1\end{pmatrix}$ and $T= \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}$.

My lecture notes say that for $\rho\colon\operatorname{SL}_2(\mathbb{Z}) \rightarrow \operatorname{GL}_2(\mathbb{C})$, I should diagonalize $\rho(S)$ and see what choices I have for $\rho(R)$. But I am not sure how to work these calculations out. Can someone help?
Thank you in advance!

答: $\mathrm{SL}_2(\mathbb Z)$ is generated by $S$ and $R$ and we have $S^4 = \mathrm{id}$ and $S^2 = R^3 = -\mathrm{id}$. By diagonalizing $\rho$, we can assume, that $\rho(S) = \begin{pmatrix} \alpha & 0 \\ 0 & \beta\end{pmatrix}$ for $\alpha, \beta \in \{\pm 1, \pm i\}$. We have $R^6 =\mathrm{id}$, so $\rho(R)$ is diagonalizable with eigenvalues being $\gamma,\delta \in \{1,\zeta, \dots, \zeta^5\}$, where $\zeta = e^{\frac{2\pi i}{6}}$. Hence $\rho(R) = A\begin{pmatrix} \gamma & 0 \\ 0 & \delta\end{pmatrix}A^{-1}$ for some $A \in \mathrm{GL}_2(\mathbb C)$.


问:
Determining elements in a group of order 8
Suppose $G$ is non-abelian group of order 8, and it has an element $g \in G$ with order 4, and another element $f \in G$ that is not in $<g>$. How do I know that the elements are $e, g, g^2, g^3, f, fg, fg^2, fg^3$. Namely, how to prove that $f^2$ does not exist? (i.e. is just $e$) Do I just do it by trial and error?

答:
Since $|G|/|\langle g \rangle| = 2$, there are two cosets of $\langle g \rangle$ in $G$. These are $\langle g \rangle$ and $f\langle g \rangle$.

Now you can conclude that $f^{2}$ has to be in one of these two cosets. It can be seen that $f^{2} \not\in f\langle g \rangle = \{f, fg, fg^2, fg^3 \}$, so we must have $f^{2} \in \langle g \rangle = \{e, g, g^2, g^3 \}$.

I will let you continue from here.


问:
Why if a real function has a derivative with a positive lower bound, it must be surjective

Let $f$ be differentiable on $\Bbb R$ and suppose that $f'(x)\ge c >0$ for all $x$. Show that for each $y\in \Bbb R$, there is an unique $x$ such that $f(x)=y$.

By assumption and mean value theorem, $f$ is strictly increasing on $\Bbb R$. If I can prove $\lim \limits_{x\to \infty}f(x)=\infty$ and $\lim \limits_{x\to -\infty}f(x)=-\infty$. Then by definition, Given $y>0$, there exist $t_0$ and $t_1$ such that $f(t_0)>y$ and $f(t_1)<-y<y$. By intermediate value theorem, there exists $z$ between $t_0$ and $t_1$ such that $f(z)=y$, and since $f$ is strictly increasing, this $z$ is unique. We can do it for any $y\in \Bbb R$, so we are done.

I want to ask how to prove $\lim \limits_{x\to \infty}f(x)=\infty$ and $\lim \limits_{x\to -\infty}f(x)=-\infty$?

答:
For $x >0$ we have $f(x)-f(0)=xf'(t)$ for some $t$ between $0$ and $x$.Hence $f(x) \geq f(0)+cx \to \infty$ as $x \to \infty$.

Similarly. $f(0)-f(-x)=xf'(s)$ for some $s$ between $0$ and $-x$ and this gives $f(-x) \to -\infty$ as $ x \to \infty$.


问:
Laurent series centered at $z_0=1$ with $1<|z-1|<\infty$
There is a function $f(z)=\dfrac{1}{z(z-1)}$, and it can be expanded as $\dfrac{1}{z-1}=\dfrac{1}{z}\cdot \dfrac{1}{1-\frac{1}{z}}$. Furthermore, $\dfrac{1}{1-\frac{1}{z}}=1+\dfrac{1}{z}+\dfrac{1}{z^2}...$ I substituted $w=z-1$ and expanded to get my Laurent series: $\dfrac{1}{(z-1)^2}+\dfrac{1}{(z-1)^3}+\cdots$. Did I approach this correctly?

答:
It is not clear for what you are using $w = z-1$  and the link between $f(z)$ and the remainder of your argument is unclear.  However, you have nearly got the right answer.

Instead, you could justify the result by writing,
\begin{align}
f(z) &= \frac{1}{z(z-1)} \\
&= \frac{1}{(z-1)^2(1+\frac{1}{z-1})}
\end{align}
and when $\lvert z-1 \rvert > 1$ the second parentheses in the denominator can be expanded as a convergent infinite series, so that,
\begin{align}
f(z) &= \frac{1}{(z-1)^2} \cdot \Big(1 - \frac{1}{z-1}+\frac{1}{(z-1)^2}-\cdots\Big) \\
&=\frac{1}{(z-1)^2} - \frac{1}{(z-1)^3}+\frac{1}{(z-1)^4} -\cdots
\end{align}
which is nearly the same as your result, derived directly from the original expression. Notice the alternating sign in the sum.


问:
Column entries of a matrix sum to zero, so what are the properties?
What kind of properties does a matrix whose column entries sum to zero have?

$$ \begin{pmatrix} a_{11} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{m1} & \cdots & a_{mn} \end{pmatrix}$$

Where $a_{11}+\cdots+a_{m1}=0$ and so on.

答:
You have
$$a_{mk} = - \sum_{j=1}^{m-1} a_{jk}$$ for all $k$. This means the last row vector is a linear combination of the remaining row vectors. Hence, the rank of the matrix is at most $\min(m,n)-1$.</p>