问:
On Dummit's paper
I have been recently reading Dummit's paper on solvability of quintic polynomials. At certain point, he mentiones that all results are valid "over any field $K$ of characteristic different from 2 and 5."

I would like to know if the real field satisfies that. Also what does it mean in order words?

Also, any elementary references on Abstract Algebra would be appreciated.

Thanks

答: One way of thinking about the characteristic of a field is it's the smallest number of times you add $1$ to itself to get back to $0$. For example, in $\mathbb{Z}/2\mathbb{Z}$, $1 + 1 = 0$, so the characteristic is $2$. For fields like $\mathbb{R}$ and $\mathbb{Q}$ where you will never get to $0$ by adding strings of $1$, we say the characteristic is $0$.

Title: On Dummit's paper


问:
How to generate REAL random numbers with some random and pseudo random
I'm doing (with Java) a very simple simulator (Queueing Systems..) that needs many random numbers (more than $10^5$).

I know that Java Random class would give me all the random numbers I need, and they pass all the test I'm doing, but.. they are pseudo-random! I don't like it.

I know that random number web site gives real random numbers from atmospheric noise (or something like that) but it limits the download. In particular you cannot download more than 10000 numbers with one HTTP request.

So the question is: is there a way to generate more than $X$ random numbers with $X$ random numbers and infinite (...) pseudo-random?

答:
There's absolutely no reason to use "real" random numbers. Furthermore, those "real" random numbers may be worse that pseudorandom ones. For example, it could happen that for some physical reasons, the "real" random bits are slightly skewed or have slight correlations. That can be corrected using a "randomness extractor" but then, why not just use the same techniques without all the hassle?

There are surely some reasons not to settle for pseudorandomness, but none of them apply in your case. For example, you don't mind that your numbers are deterministic. You don't care if they're cryptographically secure. And you'd want all your experiments to be reproducible (though that's still possible under your "real" model).

As to your concrete final question, yes, it's possible to use outside "real" random bits to influence a cryptographic model generating random numbers. You can probably find some methods in the literature (don't try to invent anything yourself, though). Just keep it mind that instead of focusing on the simulation itself, this will change your focus to something which, in my mind, is completely bogus.

ContentLicense: CC BY-SA 3.0


问:
How to prove Character space of $M_n$(C) is empty by following question?
Prove that Mn(C) is spanned by {AB−BA : A, B ∈ Mn(C)}. Deduce that Ω
(
Mn(C)
) is empty, ie character space/Spectrum is empty.</p>

答:
For the second part of your question (independent from the first part): the characters here are $\star$-homomorphisms from $M_n(\mathbb{C})$ to $\mathbb{C}$. The kernel of this homomorphism must be a two-sided ideal. $M_n(\mathbb{C})$ has no nontrivial two-sided ideals, so the kernel can only be either the full algebra or zero. However, the range being at most one-dimensional, by rank nullity, kernel must have dimension at least $n^2-1$. So, the only choice is the zero homomorphism, which is not a character, which means, the character space is empty.

Regarding obtaining the second part from the first part: this follows directly from the definition of characters ($\star$-homomorphisms), since the range $\mathbb{C}$ is commutative.

ContentLicense: CC BY-SA 4.0


问:
arrange k identical robots in 15 chairs, with limitations
I ran into this question, want to challenge you.

in how many ways can you arrange a number of identical robots on $15$ chairs?

limitations:
1) $2$ robots cannot sit next to each other.
2) each empty chair has at least one neighbour with a robot.

what i got so far: i know that it has something to do with $\sum_{5}^{8}$ because the minimum amount of robots is $5$ and maximum is $8$.

i'm not sure how to proceed from here though.

Thank you very much in advance,

Yaron

答:
Hint:  Think about a recurrence.  Let $R(n)$ be then number of acceptable strings of length $n$ ending in $R, E(n)$ be the number of acceptable strings of length $n$ ending in $RE$, and $EE(n)$ the number of acceptable strings of length $n$ ending in $EE$.  We have $R(1)=1, E(1)=0, EE(0)=0, R(n)=EE(n-1)+E(n-1), E(n)=R(n-1), EE(n)=E(n-1)$  Can you see why?

For small $n$, a spreadsheet makes this easy.  For larger $n$, making a matrix and diagonalizing will solve it.


问:
Exercises about cardinality of sets
Cardinality of both following sets are $3^n$, but I can not understand how:

$$ {\{(S,T) \mid S \in \mathcal{P}(A),\, T \in \mathcal{P}(A),\, S \cap T = \emptyset\}}$$
$$ {\{(S,T) \mid S \in \mathcal{P}(A),\, T \in \mathcal{P}(A),\, S \cup T = A\}}$$

$\mathcal{P}$ indicates power set and $A = \{1,2,3, \dots ,n\}$.
I really appreciate if anyone can answer my question. Thanks a lot.

答:
For the first set,
\begin{align}
{\{(S,T) | S \in P(A), T \in P(A), S \cap T = \oslash\}} &= {\{(S,T) | S \in P(A), T \subset S^c\}} \\
&= \bigcup_{S \subset A} \bigcup_{T \subset S^c} {\{ (S,T) \}}
\end{align}
where the two unions are disjoint so the cardinal of this set is
$$ \sum_{S \subset A}\sum_{T \subset S^c}1= \sum_{S \subset A} 2^{n-card(A)}=2^n\sum_{k=0}^n\binom{n}{k} \left( \frac{1}{2} \right)^k = 3^n.$$
For the second set it is the same idea :
\begin{align}
{\{(S,T) | S \in P(A), T \in P(A), S \cup T = A\}} &= {\{(S,T) | S \in P(A), A\setminus S \subset T\}} \\
&= \bigcup_{S \subset A} ~~ \bigcup_{T:A \setminus S \subset T}~~ {\{(S,T)\}}
\end{align}
and for the cardinal,
$$ \sum_{S \subset A} ~~\sum_{T: A\setminus S \subset T}~~1= \sum_{S \subset A} 2^{card(S)} = \sum_{k=0}^n \binom{n}{k}2^k=3^n.$$

Title: Exercises about cardinality of sets


问:
Find the trigonometric limit
Find the limit $$\lim_{x \to 0} \frac{\tan x-\sin x}{x^4} $$

Here is my solution:

$$\lim_{x \to 0} \frac{\tan x-\sin x}{x^4} = \lim_{x \to 0}\frac{\sin x(\frac{1}{\cos x}-1)}{x^4}=\lim_{x \to 0}\frac{\sin x}{x^4}*(\frac{1}{\cos x}-1)=\lim_{x \to 0}\frac{\sin x}{x^4}*(0)$$

$$ \lim_{x \to 0}\frac{\sin x}{x^4}$$

$$ \to \lim_{x \to 0+0}\frac{\sin(0+0)}{(0+0)^4}=\frac{0}{0} \space AND \space \lim_{x \to 0-0}\frac{\sin(0-0)}{(0-0)^4}=\lim_{x \to 0-0}\frac{-\sin(0)}{(0)^4}=-\frac{0}{0}, Hence \space the \space limit \space DNE $$

So, is that an appropriate way? I have troubles with the limits that have no solution (do not exist). I can easily find the limit with L'Hôpital's Rule, but then it turns out that the limit one-sided and do not exist. Is there any universal method for that type of limits?

答:
You were on the right track with the second equality, all that was left was to multiply numerator and denominator by a "conjugate"

$$\frac{\sin x(\sec x - 1)}{x^4}\cdot\frac{\sec x+1}{\sec x + 1} = \frac{\sin x}{x}\cdot \frac{\tan^2x}{x^2} \cdot\frac{1}{x} \cdot\frac{1}{\sec x + 1}$$

The limit to $0$ of $xf(x)$ is

$$ 1\cdot 1^2 \cdot \frac{1}{1+1} = \frac{1}{2}$$

Thus the limit of $f$ to $0$ does not exist.

Title: Find the trigonometric limit


问:
Definable order types without infinity axiom.
Denote by $ZF^\times$ the theory of $ZF$ without the axiom of infinity. We know that $V_\omega$, the set of all hereditarily finite sets in a model of $ZF$, is a model of $ZF^\times$.

We further know that $\newcommand{Ord}{\operatorname{Ord}}\Ord^{V_\omega}=\omega$. Consider the following formula: $$\varphi(x,y)=(x \text{ is an ordinal})\land (y\text{ is an ordinal})\land \Big((x\neq0\land x\in y)\lor y=0\Big)$$

It is not hard to see that the class $\{\langle x,y\rangle\mid \varphi(x,y)\}$ is a well-ordering of order type $\omega+1$. Similarly we can define $\omega+2$, even $\omega+\omega$. We can go even further, much further.

However, we can obviously go so far, there are only countably many formulas and countably many parameters so there can only be countably many order types definable.

Questions:

What is the least ordinal not definable in $V_\omega$ from $ZF^\times$?

We can push this question into $NBG^\times$, defined as $ZF^\times$. Now we can quantify over classes, surely we can push this even higher?

Vaguely related:

Surreal numbers without the axiom of infinity

答:
To move this off the unanswered queue (actually user 喻良 already posted the correct answer but it was deleted as a link-only answer; also, it only addressed the first question):

The answer is indeed $\omega_1^{CK}$, the least noncomputable ordinal. To see this it's convenient to switch from $V_\omega$ to $\mathbb{N}$; they are bi-interpretable (via e.g. the Ackermann interpretation), so we can basically always switch from one to the other.

By definition, $\omega_1^{CK}$ is the smallest ordinal with no computable copy - that is, the least ordinal $\alpha$ such that there is no computable relation $R$ on $\omega$ such that $R$ defines a well-ordering of $\omega$ of type $\alpha$. Since computable = $\Delta^0_1$, every ordinal with a computable copy in fact has a definable copy. So your ordinal is $\ge\omega_1^{CK}$. But in fact $\omega_1^{CK}$ is gigantic: it is also the smallest ordinal with no hyperarithmetical copy. This is a theorem due to Spector - see Corollary $5.6$ of chapter $1$ of Sacks' book. Since $\mathbb{N}$-definable = arithmetical $\subsetneq$ hyperarithmetical, this shows that your ordinal is $\le\omega_1^{CK}$.

What about bringing classes into the picture? Well, by implementing the $L$-construction in the setting of second-order arithmetic (see the last section of Simpson's reverse math book) this gets us all the reals in $L_\alpha$ for $\alpha$ the smallest "gap ordinal," that is, the smallest $\alpha$ such that $L_{\alpha+1}\cap\mathbb{R}=L_\alpha\cap\mathbb{R}$. This ordinal is absolutely gigantic, certainly vastly bigger than $\omega_1^{CK}$. So yes, things increase at that point.

That's such a big jump that it feels a bit rude. A gentler hike up the countable ordinals goes through looking at the smallest ordinal with no copy definable over $L_\alpha$ for appropriate $\alpha$. This is always at most the next admissibile above $\alpha$, and can be strictly smaller - indeed it usually is, in the sense that this inequality is strict on a club. This is proved similarly to here. One important issue to keep in mind is that, contra the $\omega$-setting, "definable over definable $\not=$ definable" for ordinals over general admissible sets. The robustness of $\omega_1^{CK}$ is rather special.

Incidentally, there's a good "atlas" of countable ordinals here.


答:
For $x\approx0$, we know that $\sin x\approx x-\frac{x^3}{6}$ and $\tan x \approx x+\frac{x^3}{3}$.* Hence,
\begin{align}
\frac{\tan x-\sin x}{x^4} &\approx \frac{x+\frac{x^3}{3}-x+\frac{x^3}{6}}{x^4} \\[5pt]
&= \frac{\frac{x^3}{2}}{x^4} \\[5pt]
&= \frac{1}{2x} \, .
\end{align}
We thus conclude that the limit from the right is $\infty$, and the limit from the left is $-\infty$.

*(The higher-order terms of the Taylor series of $\sin x$ and $\tan x$ can safely be ignored because they contain powers of $x$ which are larger than $4$.)

Title: Find the trigonometric limit

Body:
Find the limit $$\lim_{x \to 0} \frac{\tan x-\sin x}{x^4} $$

Here is my solution:

$$\lim_{x \to 0} \frac{\tan x-\sin x}{x^4} = \lim_{x \to 0}\frac{\sin x(\frac{1}{\cos x}-1)}{x^4}=\lim_{x \to 0}\frac{\sin x}{x^4}*(\frac{1}{\cos x}-1)=\lim_{x \to 0}\frac{\sin x}{x^4}*(0)$$

$$ \lim_{x \to 0}\frac{\sin x}{x^4}$$

$$ \to \lim_{x \to 0+0}\frac{\sin(0+0)}{(0+0)^4}=\frac{0}{0} \space AND \space \lim_{x \to 0-0}\frac{\sin(0-0)}{(0-0)^4}=\lim_{x \to 0-0}\frac{-\sin(0)}{(0)^4}=-\frac{0}{0}, Hence \space the \space limit \space DNE $$

So, is that an appropriate way? I have troubles with the limits that have no solution (do not exist). I can easily find the limit with L'Hôpital's Rule, but then it turns out that the limit one-sided and do not exist. Is there any universal method for that type of limits?


问:
Why doesn't the circle retract to a point?
OK, this appears to me like perhaps a dumb question. I am reading Allen Hatcher's Algebraic Topology. I've seen bits and pieces of further material here and there before, now I'm restarting from the beginning.

OK, visually I can see why, say, the 2-sphere $S^2$ can not deformation retract onto it's equator. Intuitively, we can't do this without tearing a hole (or rather two holes). Even with visualizing the 'cylinder' $S^2$ x I (well as a volume) for the function $F$ to $S^1$,

$F:S^2 \times I \rightarrow S^1$,

$F(x,t) = f_t(x)$ the family of homotopic maps. You can see that this mapping cylinder won't work in trying to identify points $F(x,1)$ to points in the circle.

The mapping cylinder for a map and spaces $f:X \rightarrow Y$, is the quotient space

$[(X \times I) \amalg Y]/((x,1) \sim f(x))$.

Or even $S^1$ to its equator, $S^0$ the union of two separate points: not "deformation-retractable". Again, looking at $S^1 \times I$ doesn't appear to have a cont. function to $\{x\} \cup \{y\}$ for distinct points $x,y$ in $S^1$. It doesn't seem to have any mapping cylinder as well.

Disclaimer: OBVIOUSLY a circle is NOT homotopic to a point. Just in case anyone gets any wrong ideas as to where I'm going with this. It's just a question on my part :) If anyone could guide me to the a better intuition or visualization to correct the error of what I'm seeing.

However, if we just pick one point $x_0$ from $S^1$, then the mapping cylinder looks just like a cone, with $F(x,0) = x$, $F(x_0, t) = x_0$ (which is a line going down the cone from the circle to the bottom apex), and $F(x,1) = x_0$. And it looks like the mapping cylinder shows how to continuously deform the circle to the point, even if the point is an element of the circle.

I am thinking this "homotopic picture" of what merely looks like, the circle def. retracting to a point, is somehow misleading me, in the sense that I am just missing something or looking at it wrong or etc...

Can anyone elucidate?

答:
If I understand correctly, you are asking why a circle does not deformation retract onto a point. Recall that a deformation retract of $S^1$ onto a point $x_0\in S$ is a function
$f:S^1\times I\to S^1$ such that $f(x,0)=x$ for all $x\in S^1$ and $f(x,1)=x_0$ for all $x\in S^1$. You propose drawing a cone with one end (corresponding to $0$) equal to $S^1$ and the other (corresponding to $1$) equal to $x_0$, and defining $f(x,t)$ to be the point obtained by travelling a distance $t$ along the line segment connecting $x$ at one end to $x_0$ at the other. The issue is that for $t\ne 0,1$ and $x\ne x_0$, we will have $f(x,t)\notin S^1$, so this map does not actually have range in $S^1$. Essentially, the problem with your cone visualization is that part of the cone lies outside $S^1$.</p>


问:
Prove the following bound $\sum_{n=1}^\infty n^{-1/2}|f_\epsilon(n)|\lesssim \frac{1}{\sqrt{\epsilon}}$
Let $f_\epsilon(x) $ be a smooth function and suppose for all integer $k>0$ there exists some $C_k\in\mathbb{R}$ such that $$|f_\epsilon(x)| \le C_k (1+|\epsilon x|)^{-k}$$

Prove the following bound $$\sum_{n=1}^\infty n^{-1/2}|f_\epsilon(n)|\lesssim \frac{1}{\sqrt{\epsilon}}$$

Here is the alternative solution after receiving the solution:

This proof is motivated by Gary's comment,As we know for monotone sequence we have the following fact(you may see it from Rudin's first textbook)  $$\sum a_n \sim \sum 2^ka_{2^k}$$

So $$\sum_{n=1}^\infty n^{-1/2}|f_\epsilon(n)|\lesssim \sum_{k}(2^k)^{-1/2}(1+\epsilon2^k)^{-N} =  \sum_{k}(2^k)^{-1/2}(\epsilon2^k)^{-N}(1+(\epsilon2^k)^{-1})^{-N}$$ for some sufficient large $N$. RHS is almost a geometric series. We can see if $k>-\log \epsilon$ then $\epsilon2^k$ will dominate 1:

$$\sum_{k\le\log_2\epsilon}(2^k)^{-1/2}(1+\epsilon2^k)^{-N}\lesssim\sum_{k\le -\log_2\epsilon}2^{-k/2}\sim 2^{-1/2\ log_2\epsilon} = \frac{1}{\sqrt{\epsilon}}$$

To analyse the tail:

$$\sum_{k\ge -\log_2\epsilon} (2^k)^{-1/2}(\epsilon2^k)^{-N}(1+(\epsilon2^k)^{-1})^{-N}\lesssim\sum_{k\ge -\log_2\epsilon} (2^k)^{-1/2}(\epsilon2^k)^{-N} \sim \epsilon^{-1/2 - 2N}$$

Since N can be arbitary small as the assumption,this tail as function of $\epsilon$ vanish.

答:
Take $k=1$ and note that
\begin{align*}
\sum\limits_{n = 1}^\infty  {\frac{{\left| {f_\varepsilon  (n)} \right|}}{{\sqrt n }}}  & \le C_1 \sum\limits_{n = 1}^\infty  {\frac{1}{{\sqrt n }}\frac{1}{{1 + \varepsilon n}}}  \le C_1 \sum\limits_{n = 1}^\infty  {\int_{n - 1}^n {\frac{1}{{\sqrt x }}\frac{1}{{1 + \varepsilon x}}dx} }
\\ & = C_1 \int_0^{ + \infty } {\frac{1}{{\sqrt x }}\frac{1}{{1 + \varepsilon x}}dx}
\mathop  = \limits^{t = \varepsilon x}  \frac{{C_1 }}{{\sqrt \varepsilon  }}\int_0^{ + \infty } {\frac{1}{{\sqrt t }}\frac{1}{{1 + t}}dt}  = \frac{{C_1 \pi }}{{\sqrt \varepsilon  }}.
\end{align*}


问:
Need help with mathematical notation in defining a line
What I'm trying to do is describe a generalized form of the following piecewise function using proper notation.

What I have so far is (excuse my non-existent latex skills):

f(x) is defined between [0, 1]

f(x; A, B, C) -> the function is parameterized by A, B, C (three arbitrarily named variables representing, respectively, the y-value of the flat portion of the function, the x-value at which the function takes on that sloping function, and C is the y-value at f(1.0)

```<code>f(x) = {
   A; if X < B
   [the linear function]; if X >= B
}
</code>```

That is, the two components are f(x < B) = A and f(x >= B) = the line containing the points f(B) = 1, f(1) = C. The graph above is not correct because there is a discontinuity at f(B), so ignore that.

Thanks!

答:
This function is

$$f(x)=\begin{cases}
a,&0\leq x<b\\
\frac{c-1}{1-b}(x-b)+1,&b\leq x\leq 1
\end{cases}$$

Given two points $(x_1,y_1)$ and $(x_2,y_2)$, the line passing through them is $y-y_1 = \frac{y_2-y_1}{x_2-x_1}(x-x_1)$; this is where the linear function for $x \in [b,1]$ comes from.


问:
Let $A$ be a ring, and $I$ be an ideal of $A$, then $ AI＝I$
Let $A$ be a ring, and $I$ be an ideal of $A$, then is $ AI＝I$ true?

Here, $AI$ means the product as ideals.

Does this holds when we change product as sets, that is, $A・I＝I$?

Sorry to ask very basic question, thank you in advance.

答:
Notice that $AI$ is equal to the additive subgroup of $A$ generated by $A{\cdot}I=\{ax\mid a\in A,x\in I\}$ (let me know if I am misreading your notation).

By definition of an ideal we have $A{\cdot}I\subseteq I$, and then because $I$ is an additive subgroup you also get $AI\subseteq I$.

If $A$ is a unital ring then equality always holds, because for $x\in I$ you have $x=1_A{\cdot}x\in A{\cdot}I$.

On the other hand equality may not hold if the ring is not unital: for instance if you take $A=2\mathbb Z$ and $I=4\mathbb Z$ then $A{\cdot}I=AI=8\mathbb Z$.


问:
Threshold probability of the 3-uniform hypertriangle
I am stuck on a random hypergraph problem (I am encountering random hypergraphs for the very first time).

Let $G_{3}(n, p)$ be a binomial 3-uniform hypergraph. Find a threshold probability for containing a 3-uniform hypertriangle.

What I have tried: I saw a similar problem for the existence of triangles in $G(n, d/n)$. I am trying the same technique here for the hypergraphs. So according to my calculations I am getting $E[x]=\binom{n}{6}p^{3}$, where $x$ is the number of hypergraphs. Then for the variance part, I get here 5 cases: when the number of common edges are :

(i) up to 2, in which case its contribution is $E^{2}[x]$,

(ii) 3, the contribution is $\binom{n}{9}p^{6}$,

(iii) 4, the contribution is $\binom{n}{8}p^{5}$,

(iv) 5, the contribution is $\binom{n}{7}p^{4}$ and

(v) 6, the contribution is $\binom{n}{6}p^{3}$.

After this, I do not know how to proceed and somehow, I feel that all this is completely wrong and something entirely different needs to be done for hypergraphs. Any suggestions would be helpful. Thanks in advance.

$\textbf{Edit}$:

A 3-uniform hypergraph is a pair $(V, E)$, where $V$ is a set of its vertices, and $E\subset\binom{V}{3}$ is a set of its edges.

And a 3-uniform hypertriangle is a 3-uniform hypergraph with $|V | = 6$
with three edges such that every pair of edges share 1 vertex, and all three edges
do not have a common vertex.

$G_{3}(n, p)$ is a binomial 3-uniform hypergraph such that $V = \{1, . . . , n\}$ and every hyperedge is chosen independently with probability $p$.

And I got my ideas from a similar calculation for graphs on page 6 of this link : https://www.cs.cmu.edu/~avrim/598/chap4only.pdf

答:
The same technique works. If we let $\mathbf X$ be the number of hypertriangles in $G_3(n,p)$, then $\mathbb E[\mathbf X] = O(n^6 p^3)$. It is not $\binom n6 p^3$ because not all $6$ vertices play the same role: if we were being careful, it would be $\frac{6!}{3!} \binom n6 p^3$. But even without being that careful, we see that for $p \ll \frac1{n^2}$, $\mathbb E[\mathbf X] \to 0$, and with high probability there are no hypertriangles.

To show that with high probability, hypertriangles are present when $p \gg \frac1{n^2}$, we must first to compute $\text{Var}[\mathbf X]$, which requires computing $\mathbb E[\mathbf X^2]$: the expected number of ordered pairs of hypertriangles in $G_3(n,p)$. It will be convenient to consider $p$ in some small range, say $\frac1{n^2} \ll p \ll \frac1{n^{1.99}}$. For larger $p$, the claim follows by monotonicity.

However, I'm not sure your division into cases for computing $\mathbb E[\mathbf X^2]$ makes sense. A hypertriangle has only $3$ edges, so how can two hypertriangles share $6$ edges?

Instead, consider:

We have pairs of hypertriangles that share no edges. An upper bound on the number of pairs like this is $\mathbb E[\mathbf X]^2$. Here, we must be slightly careful, because a bound of $O(n^{12}p^6)$ is not precise enough. However, there is some $f(n) = O(n^6)$ which tells us the number of potential hypertriangles. We have $\mathbb E[\mathbf X] = f(n)p^3$. Meanwhile, $f(n)^2$ is an upper bound on the number of potential pairs that share no edges - it is the total number of potential pairs. Each of the ones that share no edges has probability $p^6$ of appearing, so $f(n)^2 p^6 = \mathbb E[\mathbf X]^2$ is our upper bound.

We have pairs of hypertriangles that share one edge. An upper bound on the number of pairs like this is $O(n^9 p^5)$.

We have pairs of hypertriangles that share two edges. An upper bound on the number of pairs like this is $O(n^7 p^4)$.

We have pairs of hypertriangles that share all three edges. There are exactly $\mathbb E[\mathbf X]$ of these.

Remember how I decided to look at $p$ such that $\frac1{n^2} \ll p \ll \frac1{n^{1.99}}$? That's so that the second and third cases are $o(1)$ terms. More precisely, we see that this holds as long as $p \ll \frac1{n^{1.8}} = n^{-9/5}$.

Anyway, putting together all three cases, we get $\mathbb E[\mathbf X^2] \le \mathbb E[\mathbf X]^2 + \mathbb E[\mathbf X] + o(1)$, so $\text{Var}[\mathbf X] \le \mathbb E[\mathbf X] + o(1)$, so $\frac{\text{Var}[\mathbf X]}{\mathbb E[\mathbf X]^2} \le \frac{\mathbb E[\mathbf X] + o(1)}{\mathbb E[\mathbf X]^2}$. For $p \gg \frac1{n^2}$, $\mathbb E[\mathbf X] \to \infty$, so $\frac{\text{Var}[\mathbf X]}{\mathbb E[\mathbf X]^2}  \to 0$.

By a standard application of Chebyshev's inequality, $\mathbf X > 0$ with high probability (and actually, $\mathbf X = (1+o(1))\mathbb E[\mathbf X]$ with high probability). So we get the threshold we wanted.


问:
An easier evaluation of $\det\limits_{1\leqslant i,j\leqslant n}\left\{\frac{x_i-x_j}{x_i+x_j}\right\}$
I'm looking for an easier proof of the identity (attributed to K. F. W. Rohn)
$$R_n(\bar{x}):=\det_{1\leqslant i,j\leqslant n}\left\{\frac{x_i-x_j}{x_i+x_j}\right\}=\prod_{i<j}\left(\frac{x_i-x_j}{x_i+x_j}\right)^2,$$
where $\bar{x}:=(x_1,\ldots,x_n)$ and $n$ is even (if $n$ is odd then clearly $R_n(\bar{x})=0$).

This identity is used in the end of this answer of mine.

The proof I have (see my own answer below) obtains it from the formula for
$$D_n(\bar{x},\bar{y},\bar{z}):=\det_{1\leqslant i,j\leqslant n}\left\{\frac{x_i+z_j}{x_i+y_j}\right\}$$
found by H. J. A. Duparc; this time, $n$ is not necessarily even.

But it looks convoluted. I don't have access to Rohn's works - did he have simpler arguments?..

答:
One can prove the identity with induction on the dimension $n$, using elementary transformations on the determinant.

The identity holds for $n=0$, with the empty product and the empty determinant both being equal to $1$.

For the induction step $n \to n+2$ we compute
$$
R_{n+2}(x_1, \ldots, x_n, a, b) =
\begin{vmatrix}
q(x_1, x_1) & \cdots & q(x_1, x_n) & q(x_1, a) & q(x_1, b) \\
\vdots & \ddots & \vdots &\vdots &\vdots \\
q(x_n, x_1) & \cdots & q(x_n, x_n) & q(x_n, a) & q(x_n, b) \\
q(a, x_1) & \cdots & q(a, x_n) & 0 & q(a, b) \\
q(b, x_1) & \cdots & q(b, x_n) & q(b, a) & 0
\end{vmatrix}
$$
where I have used
$$
 q(x, y) = \frac{x-y}{x+y}
$$
to shorten the expressions a bit. We add multiples of the last two rows to the first $n$ rows to eliminate  the last two entries in those rows. This gives
$$
R_{n+2}(x_1, \ldots, x_n, a, b) =
\begin{vmatrix}
d_{1,1} & \cdots & d_{1,n} & 0 & 0 \\
\vdots & \ddots & \vdots &\vdots &\vdots \\
d_{n,1} & \cdots & d_{n,n} & 0 & 0 \\
* & \cdots & * & 0 & q(a, b) \\
* & \cdots & * & q(b, a) & 0
\end{vmatrix}
$$
with
$$
d_{i,j} = q(x_i, x_j) - \frac{q(x_i, a)q(b,x_j)}{q(b, a)} - \frac{q(x_i, b)q(a,x_j)}{q(a, b)} \, .
$$
Now the magic (?) happens: This expression simplifies to
$$
 d_{i,j} = q(x_i, a)q(x_i, b)q(x_j, a)q(x_j, b)q(x_i, x_j) \, .
$$

The  common factors $q(x_i, a)$ and $q(x_i, b)$ can be extracted from each row and each column in $ \det \{ d_{i,j} \}$, and it follows that
$$
R_{n+2}(x_1, \ldots, x_n, a, b) = \det \{ d_{i,j} \} q(a, b)^2 \\
 = \det \{ q(x_i, x_j) \}\left(\prod_{i=1}^n q(x_i, a)^2 q(x_i, b)^2\right)  q(a, b)^2 \\
= R_{n}(x_1, \ldots, x_n)\left(\prod_{i=1}^n q(x_i, a)^2 q(x_i, b)^2\right)  q(a, b)^2 \, ,
$$
which is exactly the induction step.


问:
Using Cauchy Integral, solve: $\oint_C \frac{1}{(2z+1)^2}dz$ without use residue theorem
C is a unitary circle.

If I simplify, I get:  $\oint_C \frac{1/2}{(z+1/2)(z+1/2)}$
How can i solve? If I use Cauchy Integral Theorem, $z_0 = -1/2$ and f(z) = 2, so answer is $2\pi i * (-1/2)$?

Idk how to solve with 1/(z-a)^n

答: Unlike the logarithm, $-\tfrac{1/2}{2z+1}$ doesn't have branches, so an integral of $\tfrac{1}{(2z+1)^2}$ around a closed loop is $0$.


问:
Find the winding number of $f$ with $f(z)=f(e^{2\pi i /m} z)$
I have the following problem:

Let $m\geq 1$ be a natural number and $f:S^1\rightarrow S^1$ a
continuous map such that $f(z)=f(e^{2\pi i /m} z)$ $\forall z \in S^1$. Prove that the degree (or winding number) of $f$ is a multiple
of $m$.

I tried to, without explicitly finding an elevation, use the fact that $f(z)=f(e^{2\pi i /m} z)$ without success.

Any help would be appreciated!

答:
Consider the covering projection $p : \mathbb R \to S^1, p(t) = e^{2\pi i t }$.  The winding number $n(f)$ of $f$ is given as follows.

For any closed path $u : [a,b] \to S^1$ (i.e. we have $u(a) = u(b)$) take any lift $l_u : [a,b] \to \mathbb R$. Then $w(u) = l_u(b) - l_u(a)$ is an integer which does not depend on the choice the  lift $l_u$.

Consider the closed path $f' = f \circ p \mid_{[0,1]} : [0,1] \to S^1$. Then take $n(f) = w(f')$.

We have $f(p(t)) = f(e^{2 \pi i/m}p(t)) = f(e^{2 \pi i/m}e^{2\pi i t}) = f(e^{2\pi i (t + 1/m)})  = f(p(t+1/m))$ for all $t$. That is, $f \circ p$ has period $1/m$. In particular $f'_1 = f \circ p \mid_{[0,1/m]} : [0,1/m] \to S^1$ is a closed path. Lift it to $l_1 : [0,1/m] \to \mathbb R$. Let $d = l_1(1/m) - l_1(0) \in \mathbb Z$. Define
$l : [0,1] \to \mathbb R, l(t)  = l_k(t) = kd + l_1(t -k/m)$ for $t \in [k/m, (k+1)/m]$. This is a well-defined continuous map because for $t = (k+1)/m$ we have
$$l_k((k+1)/m) =  kd + l_1((k+1)/m -k/m) = kd + l_1(1/m) = kd + l_1(1/m) -l_1(0) + l_1(0) \\= (k+1)d + l_1((k+1)/m - (k+1)/m) = l_{k+1}((k+1)/m) .$$
$l$ is a lift of $f'$ because for $t \in [k/m, (k+1)/m]$ we have
$$p(l(t)) = p(kd + l_1(t -k/m)) = p(kd)\cdot p(l_1(t -k/m)) = 1 \cdot f'_1(t -k/m) = f'(t) .$$
Therefore
$$n(f) = w(f') = l(1) - l(0) = l_{m-1}(1) - l_1(0) = (m-1)d + l_1(1- (m-1)/m) - l_1(0) \\ = (m-1)d + l_1(1/m) - l_1(0) = md. $$


答:
The natural density of a set $S\subseteq\mathbb{N}$ of naturals is defined as

$$ \mu(S)=\lim_{N\to\infty} \frac{|S\cap[1,B]|}{N}. $$

Not all subsets of $S$ have a density - indeed we may construct $S$ so that $|S\cap[1,N]|$ oscillates indefinitely if we want. This density is also not countably additive the way probability measure would be, so for instance every singleton $S=\{n\}$ has density $0$ but their union has density $\mu(\mathbb{N})=1$. It's also worth pointing out it is difficult to prove things about the natural density of certain sets, however it is upgraded by something called the Dirichlet density, defined by a limit with Dirichlet series at $s=1$: whenever the natural density exists, so does the Dirichlet density and they are equal, but the Dirichlet density may exist even when the natural density doesn't and in practice it is easier to work with the Dirichlet density for sets encountered and under scrutiny within analytic number theory.

It should be clear how to extend the definition of natural density to all of $\mathbb{Z}$ if we want a more "symmetric" treatment of cosets. The natural density of a coset is simply $\mu(a+m\mathbb{Z})=\tfrac{1}{m}$. You are interested in the relative density $\mu(A\cap B)/\mu(B)$ for two cosets $A=a+7\mathbb{Z}$ and $B=2+13\mathbb{Z}$.

Fortunately, the intersection of two cosets is itself a coset (or empty). The elements of $A\cap B$ where $A=a+m\mathbb{Z}$ and $B=b+n\mathbb{Z}$ are solutions to the system of congruences

$$ \begin{cases} x \equiv a \mod m \\ x \equiv b \,\mod n \end{cases} $$

The Chinese Remainder Theorem says this is empty if $a\not\equiv b\mod \gcd(m,n)$, or else it is a residue mod $\mathrm{lcm}(m,n)$ (which can be constructed explicitly using a version of Lagrange interpolation and modular inverses, which I can explain if you want). Therefore we have

$$ \mu\big((a+m\mathbb{Z})\cap(b+n\mathbb{Z})\big) =
\begin{cases}
\frac{1}{\mathrm{lcm}(m,n)} & \textrm{ if } a\equiv b\mod\gcd(m,n) \\[5pt]
~~~~~0 & \textrm{ if } a\not\equiv b \mod \gcd(m,n)
 \end{cases} $$

Title: The "covering rate" of cosets $7 \Bbb{Z} + a$ by some coset $b \Bbb{Z} +c$.

Body:
The covering rate of the cosets $7\Bbb{Z} + a$ by the coset $13\Bbb{Z} + 2$:

$$
7(0) -4 = 13(-1) + 2 \\
7(1) -5 = 13(0) + 2 \\
7(2) + 1 = 13(1) + 2 \\
7(4) + 0 = 13(2) + 2 \\
7(6)  -1 = 13(3) + 2 \\
7(8)  -2 = 13(4) + 2 \\
7(10)  -3 = 13(5) + 2 \\
7(12) - 4 = 13(6) + 2 \\
7(14) -5 = 13(7) + 2 \\
$$

It takes minimally an offset of $14$, i.e. $7(x + 14) + a = 13y + 2$ before the cycle starts over again (look at the residue cycling: $-4,-5,1,0,-1,-2,-3,-4,-5, \dots$, the offset separating the two $-5$'s is $14$.

Is $\dfrac{1}{14}$ or for all fixed $x, a \in \Bbb{Z}$, the limit

$$\lim\limits_{y \to \infty} \dfrac{\#(\{7x + a, 7(x+1) + a, \dots, 7(x + y) + a\} \cap (13 \Bbb{Z} + 2))}{y - x} = \dfrac{1}{14}$$.

The covering rate of $7\Bbb{Z} + a$ by $19 \Bbb{Z} + 3$:

$$
7(3) + 1 = 19(1) + 3\\
7(5) + 6 = 19(2) + 3 \\
7(8) + 4 = 19(3) + 3 \\
7(11) + 2 = 19(4) + 3 \\
7(14) + 0 = 19(5) + 3 \\
7(16) + 5 = 19(6) + 3 \\
7(19) + 3 = 19(7) + 3 \\
7(22) + 1 = 19(8) + 3 \\
7(24) + 6 = 19(9) + 3 \\
$$

is therefore $\dfrac{1}{19}$.  How can we compute this rate without going through the trouble of enumerating?


答: I think your confusing homotopy equivalent and homeomorphic. The fact that you can wrap a disc into a point just means that they are homotopy equivalent, but not that they are homeomorphic, for that you would need a bijection. Since the map you describe is clearly not bijective, a point and $\mathbb{S}^1$ are not homeomorphic.</p>

Title: Why doesn't the circle retract to a point?

Body:
OK, this appears to me like perhaps a dumb question. I am reading Allen Hatcher's Algebraic Topology. I've seen bits and pieces of further material here and there before, now I'm restarting from the beginning.

OK, visually I can see why, say, the 2-sphere $S^2$ can not deformation retract onto it's equator. Intuitively, we can't do this without tearing a hole (or rather two holes). Even with visualizing the 'cylinder' $S^2$ x I (well as a volume) for the function $F$ to $S^1$,

$F:S^2 \times I \rightarrow S^1$,

$F(x,t) = f_t(x)$ the family of homotopic maps. You can see that this mapping cylinder won't work in trying to identify points $F(x,1)$ to points in the circle.

The mapping cylinder for a map and spaces $f:X \rightarrow Y$, is the quotient space

$[(X \times I) \amalg Y]/((x,1) \sim f(x))$.

Or even $S^1$ to its equator, $S^0$ the union of two separate points: not "deformation-retractable". Again, looking at $S^1 \times I$ doesn't appear to have a cont. function to $\{x\} \cup \{y\}$ for distinct points $x,y$ in $S^1$. It doesn't seem to have any mapping cylinder as well.

Disclaimer: OBVIOUSLY a circle is NOT homotopic to a point. Just in case anyone gets any wrong ideas as to where I'm going with this. It's just a question on my part :) If anyone could guide me to the a better intuition or visualization to correct the error of what I'm seeing.

However, if we just pick one point $x_0$ from $S^1$, then the mapping cylinder looks just like a cone, with $F(x,0) = x$, $F(x_0, t) = x_0$ (which is a line going down the cone from the circle to the bottom apex), and $F(x,1) = x_0$. And it looks like the mapping cylinder shows how to continuously deform the circle to the point, even if the point is an element of the circle.

I am thinking this "homotopic picture" of what merely looks like, the circle def. retracting to a point, is somehow misleading me, in the sense that I am just missing something or looking at it wrong or etc...

Can anyone elucidate?


Title: Algebraic manipulation of equations and determinant

Body:
$$a_{11}^2 + a_{12}^2 + a_{13}^2 = 1$$
$$\dfrac{a_{11}}{a_{33}a_{22}-a_{23}a_{32}} = \dfrac{a_{12}}{a_{21}a_{33}-a_{31}a_{23}} = \dfrac{a_{13}}{a_{22}a_{31}-a_{21}a_{32}} = k$$

$$k(a_{11}(a_{33}a_{22}-a_{23}a_{32})+a_{12}(a_{21}a_{33}-a_{31}a_{23})+a_{13}(a_{22}a_{31}-a_{21}a_{32})) = k \det
\begin{bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{bmatrix} = 1. $$

Does someone know how to obtain the last equation combining the first and second ones?

Body:
The trick is to express $a_{1i}^2$ as $a_{1i}.k(...)$

$\dfrac{a_{11}}{a_{33}a_{22}-a_{23}a_{32}} = \dfrac{a_{12}}{a_{21}a_{33}-a_{31}a_{23}} = \dfrac{a_{13}}{a_{22}a_{31}-a_{21}a_{32}} = k$

$a_{11}^2=a_{11}k({a_{33}a_{22}-a_{23}a_{32}})$

$a_{12}^2=a_{12}k({a_{21}a_{33}-a_{31}a_{23}})$

$a_{13^2}=a_{13}k({a_{22}a_{31}-a_{21}a_{32}})$

$a_{11}^2 + a_{12}^2 + a_{13}^2 = 1$

$a_{11}^2 + a_{12}^2 + a_{13}^2$

$ = k(a_{11}({a_{33}a_{22}-a_{23}a_{32}})$$+a_{12}({a_{21}a_{33}-a_{31}a_{23}})+a_{13}({a_{22}a_{31}-a_{21}a_{32}}))$

=k $\det\begin{bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{bmatrix}$=1


问:
Finding $f^{(n)}(0)$ if $f(x)= \frac{x^5}{ 1 + x^2}$
In an exercise I'm asked the following:

Let $f: \Bbb R \to \Bbb R$ such that: $$f(x)= \frac{x^5}{ 1 + x^2}$$
Determine the value of: $$f^{(n)}(0),\ \ \forall n \in \mathbb N$$

Because we have a fraction, then $f^{(n)}(0)$ will take of the following form, because of the quotient rule for derivatives:

$$f^{(n)}(x)=\frac{P_n(x)}{(1-x^2)^{2n}}$$

So if follows that: $$f^{(n)}(0) = P_n(0)$$

So to solve this we basically need to find out what the numerator of that quotient will be for any $n \in \mathbb N$. That is precisely the part that I'm struggling with. How can I solve this?

答:
Hint :

prove that the numerator is $x \cdot Q_n(x)$ for every $n$


Title: Finding $f^{(n)}(0)$ if $f(x)= \frac{x^5}{ 1 + x^2}$

Body:
In an exercise I'm asked the following:

Let $f: \Bbb R \to \Bbb R$ such that: $$f(x)= \frac{x^5}{ 1 + x^2}$$
Determine the value of: $$f^{(n)}(0),\ \ \forall n \in \mathbb N$$

Because we have a fraction, then $f^{(n)}(0)$ will take of the following form, because of the quotient rule for derivatives:

$$f^{(n)}(x)=\frac{P_n(x)}{(1-x^2)^{2n}}$$

So if follows that: $$f^{(n)}(0) = P_n(0)$$

So to solve this we basically need to find out what the numerator of that quotient will be for any $n \in \mathbb N$. That is precisely the part that I'm struggling with. How can I solve this?

Body:
You have that $$\frac{1}{1+x^2}=\sum_{k=0}^\infty (-1)^nx^{2n}.$$
Therefore $$\frac{x^5}{1+x^2}=\sum_{k=0}^\infty (-1)^nx^{2n+5}=\sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!}x^{k}.$$
I let you find $f^{(n)}(0)$ for all $n$.


问:
why $F(g(a)) \subset F(a)?$
Let $f(x)$ a irreducible polynomial of degree $n$ over a field $F$. Let $g(x)$ be a polynomial in $F[x]$. Prove that every irreducible factor of the composition $f(g(x))$ has a degree which is divisible by $n$.

My attempt : I got the answer  here  .But  I have some confusion

my confusion is given below  marked in red box

My confusion:  why  $F(g(a)) \subset F(a)?$

Here   $a  \subset g(a)   $ so i think $F(a) \subset F(g(a))$?

答:
What you wrote, "$a \subset g(a)$," does not make sense.  $a$ and $g(a)$ are elements, not sets.

We have $g(a) = c_0 + c_1a + \cdots + c_na^n$ for some $c_i \in F$.  Since $a$ is an element of the field $F(a)$, and $F(a)$ is closed under addition and multiplication, this implies that $g(a)$ is also an element of $F(a)$.

Therefore every element of the form $cg(a)^n$ for $c \in F$ and $n \geq 0$ is in $F(a)$.  Therefore, the field $F(g(a))$ is contained in $F(a)$.

Title: why $F(g(a)) \subset F(a)?$


问:
Adriaan van Roomen's 45th degree equation in 1593
Adriaan van Roomen proposed a 45th degree equation in 1593(see this book, picture reference as follows):

$$
\begin{gathered}
f(x) = x^{45} - 45x^{43} + 945x^{41} - 12300x^{39} + 111150x^{37} - \color{red}{740459}x^{35} + 3764565x^{33} \\- 14945040x^{31} + 46955700x^{29} - 117679100x^{27} + 236030652x^{25} - 378658800x^{23} \\+ 483841800x^{21} - 488494125x^{19} + 384942375x^{17} - 232676280x^{15} + 105306075x^{13}
\\ - \color{red}{34512074}x^{11} + 7811375x^9 - 1138500x^7 + 95634x^5 - 3795x^3 +45x \\ = \sqrt{\frac{7}{4}-\frac{\sqrt{5}}{4}- \sqrt{\frac{15-3\sqrt{5}}{8}}} \approx 0.4158234.
\end{gathered}\tag{1}
$$
Another source said the right side is: $\sqrt{\frac{7}{4}-\frac{\sqrt{5}}{4}-\frac{5\sqrt{3}}{8}} $. ( EDIT: According to the answer, the coefficients in red are wrong.)

Another mathematician Viète noticed that left side is nothing but the expansion of sine using
$$
\sin3\theta = 3\sin\theta - 4\sin^3\theta,
$$
recursively, and the equation (1) reduced to:
$$
\sin(\alpha) = c, \quad \text{ for } \alpha \in (0,\frac{\pi}{2}) \tag{2}
$$
where $x = \sin(\alpha/45)$
(EDIT: According to the answer, this is not correct, should be  $x = 2\sin(\alpha/45)$).

Now $2k\pi+\alpha$ satisfies (2) too, hence the solutions are:
$$
x = \sin\left(\frac{\alpha + 2k\pi}{45}\right), \quad \text{ for } k = 0,1,2,\ldots,44.
$$
Totally 45 roots in $(-1,1)$.

However, when drawing the graph of $f(x)-c$, something does not look quite right:

There are two other roots near roughly 3 and -3? Am I missing something here?

Body:
Typos! Scroll to the end to see which coefficients are typos.

$$\begin{align}
\sin(45t)&=\Im\left(e^{45it}\right)\\
&=\Im\left(\left(e^{it}\right)^{45}\right)\\
&=\Im\left(\left(\cos(t)+i\sin(t)\right)^{45}\right)\\
&=\Im\left(\sum_{k=0}^{45}\binom{45}{k}\left(i\sin(t)\right)^k\cos^{45-k}(t)\right)\\
&=\frac{1}{i}\left(\sum_{k=0}^{22}\binom{45}{2k+1}\left(i\sin(t)\right)^{2k+1}\cos^{44-2k}(t)\right)\\
&=\sum_{k=0}^{22}\binom{45}{2k+1}(-1)^k\sin^{2k+1}(t)\cos^{44-2k}(t)\\
&=\sum_{k=0}^{22}\binom{45}{2k+1}(-1)^k\sin^{2k+1}(t)\left(1-\sin^2(t)\right)^{22-k}\\
&=\sum_{k=0}^{22}\binom{45}{2k+1}(-1)^k\sin^{2k+1}(t)\sum_{j=0}^{22-k}\binom{22-k}{j}(-1)^j\sin^{2j}(t)\\
&=\sum_{k=0}^{22}\sum_{j=0}^{22-k}\binom{45}{2k+1}\binom{22-k}{j}(-1)^{j+k}\sin^{2k+2j+1}(t)\\
&=\sum_{k=0}^{22}\sum_{j=k}^{22}\binom{45}{2k+1}\binom{22-k}{j-k}(-1)^{j}\sin^{2j+1}(t)\\
&=\sum_{j=0}^{22}\sum_{k=0}^{j}\binom{45}{2k+1}\binom{22-k}{j-k}(-1)^{j}\sin^{2j+1}(t)\\
&=\sum_{j=0}^{22}(-1)^{j}\left[\sum_{k=0}^{j}\binom{45}{2k+1}\binom{22-k}{j-k}\right]\sin^{2j+1}(t)\\
&=\sum_{j=0}^{22}(-1)^{j}\left[\sum_{k=0}^{j}\binom{45}{2k+1}\binom{22-k}{j-k}\right]x^{2j+1}\\
&=45x+\cdots+17592186044415x^{45}
\end{align}$$

But now if we replace $x$ by $\frac{x}{2}$ and multiply the whole thing by $2$, we have

$$
\sum_{j=0}^{22}\left(\frac{-1}{4}\right)^{j}\left[\sum_{k=0}^{j}\binom{45}{2k+1}\binom{22-k}{j-k}\right]x^{2j+1}
$$
which expands as $$
\begin{gathered}
f(x) = x^{45} - 45x^{43} + 945x^{41} - 12300x^{39} + 111150x^{37} - 740\mathbf{2}59x^{35} + 3764565x^{33} \\- 14945040x^{31} + 46955700x^{29} - 117679100x^{27} + 236030652x^{25} - 378658800x^{23} \\+ 483841800x^{21} - 488494125x^{19} + 384942375x^{17} - 232676280x^{15} + 105306075x^{13}
\\ - 3451207\mathbf{5}x^{11} + 7811375x^9 - 1138500x^7 + 95634x^5 - 3795x^3 +45x \\
\end{gathered}
$$
and the coefficients of $x^{35}$ and of $x^{11}$ are typos.

Note that $x=2\sin(\alpha/45)$ is the substitution, not $x=\sin(\alpha/45)$, and that we multiplied by $2$, so that doubles the value of $c$.


问:
Most ambiguous and inconsistent phrases and notations in maths
What are some examples of notations and words in maths which have been overused or abused to the point of them being almost completely ambiguous when presented in new contexts?

For instance, a function $f$:

$f^{-1}(x)$ can be an inverse and a preimage and sometimes even $\frac{1}{f(x)}$.

$f^2(x)$ can be $(f\circ f)(x)$ and $(f(x))^2$.

$f^{(2)}(x)$ on the other hand, is the second derivative, even though adding parentheses to a number usually does nothing.

And for some functions the parentheses for the argument are omitted: $f\:x = f(x)$.

So how should $f^{(2-3)}(x)$ be interpreted? $f^{(-1)}$, an integral of $f$? or a composition, $\left(f^{(2)}\circ f^{(-3)}\right)(x)$? Or just $f^{2-3}(x) = f^{-1}(x)$?

Another example is mathematicians notorious use of the word normal to describe... normal things?

Using similar symbols and expressions for different things is unavoidable, but it can create some ambiguity when first introduced to their other uses.

答:
$L/K$ for field extensions.

First, this is not a quotient, and second, this is not even some other kind of operation, just the statement that $K \subseteq L$ as fields.


问:
Integrating around a circle
I have a function $$f(x,y)=1/2x^2y-2x+1/6y^3-5/2 y$$ and I need to find $\int_{d1}\nabla f\cdot dr$ with $d_1$ being orientated from the second quadrant towards the fourth quadrant.

$d_1$ will be the straight edge of D which is $$d_1 = \{(x,y):  x^2 +y^2\le 9 \mathrm{\ and\ } x+y=0\}$$

Ok so D is just the whole circle which means the line will intersect at $3\pi/4$ and $-\pi/4$ and I know we are going to be integrating from top left to bottom right (quadrant 2 - 4) would this mean we are going counter clockwise around the circle??

Furthermore I did manage to calculate the gradient which came out to be $∇f=<xy-2, 1/2$ $x^2-y^2/2$ $-5/2>$ so how exactly would I go about setting up this integral and solving it with polar coordinates and am I correct in assuming it will be going counter clockwise?

答:
The question says $d_1$ is the straight edge of $D$ which is $x+y = 0$ and $x^2 + y^2 \leq 9$. That means straight line segment from $A \ (-\frac{3}{\sqrt2}, \frac{3}{\sqrt2})$ to $B \ (\frac{3}{\sqrt2}, -\frac{3}{\sqrt2})$.

So parametrize the line segment as $ \ \vec {r}(t) = (t, -t), -\frac{3}{\sqrt2} \leq t \leq \frac{3}{\sqrt2}$.

Now please note that you are finding line integral of $\vec F$ which is a conservative vector field as $\vec F = \nabla f$.

So your line integral will simply be $f(B) - f(A)$

Title: Integrating around a circle


问:
Winding number in complex plane homework
If $\gamma_1,\gamma_2$ are closed paths s.t. $|\gamma_1(t)|>|\gamma_2(t)|$ for all $t$, and $\Gamma(t) := \gamma_1(t) + \gamma_2(t)$, show that $n(\Gamma,0) = n(\gamma_1,0)$, where $n$ is the winding number defined by $n(\Gamma,0) :=\frac{1}{2\pi i} \int_\Gamma \frac{dz}{z}$.

I have tried looking at $n(\Gamma,0) - n(\gamma_1,0)$ but am still stuck. Please can somebody give me a hint??

答:
HINT:

Consider the homotopy $H(t,s): \mathbb{C}\times[0,1] \to \mathbb{C}$ with $H(t,0) = \gamma_1(t)$, $H(t, 1) = \Gamma(t)$, and for $0 < s < 1$ assign $H(t,s)$ to be the corresponding point on the line segment connecting $\gamma_1(t)$ and $\Gamma(t)$. The realization is that since $|\gamma_1(t)| > |\gamma_2(t)|$, this smooth deformation of $\gamma_1$ into $\Gamma$ never passes through the origin. And therefore the winding numbers are the same.

Let's restate the same idea without any algebraic topology speak. First, draw a picture. Start with $\gamma_1(t)$, some closed path around the origin. We're going to deform $\gamma_1$ into $\Gamma$ without passing through the origin, so that the winding numbers are the same. To do this, for each $t$, imagine sending $\gamma_1(t)$ to $\Gamma(t)$ by having it move down the line segment connecting the points $\gamma_1(t)$ to $\Gamma(t)$. Notice that since $|\gamma_1(t)| > |\gamma_2(t)|$, these line segments do not pass through the origin and therefore we have deformed $\gamma_1$ into $\Gamma$ without passing through the origin.

Ok - this is less of a hint and more of the entire idea behind the solution. But there are some details left to be filled in, I think.


问:
Fixed points of differential equation
Suppose we know that $\frac{dA}{dt}=h-A+kA^2$ for constants h,k. If we wish to find the fixed points for all h,k, surely we can just use the quadratic formula to get $A=\frac{1\pm\sqrt{1-4hk}}{2k}$ provided $hk<0.25$ and notice there are no fixed points for $hk>0.25$?

答:
Yes, that is correct.  The fixed points of

$\dfrac{dA}{dt} = h - A + kA^2 \tag 1$

occur where

$\dfrac{dA}{dt} = 0; \tag 2$

combining (1) and (2) yields

$kA^2 - A + h = h - A + kA^2 = 0; \tag 3$

at this juncture we invoke the quadratic formula, and write

$A = \dfrac{1 \pm \sqrt{1 - 4hk}}{2k}; \tag 4$

we see the solutions are real provided that

$4hk \le 1, \tag 5$

and that there are no real solutions when

$4hk > 1. \tag 6$

as

$hk \to 1 \tag 7$

from below, the two distinct values of $A$ approach one another, until the meet and become a single zero when $hk = 1$, at which point single zero of (1) disappears leaving no critical point of (1) whilst (6) binds.

Title: Fixed points of differential equation


问:
If $c_1,\dots, c_h$ are the orders of the centralisers of elements of distinct conjugacy classes of a finite group, then $1/c_1+\dots+1/c_h=1$
This is part of Exercise 1.6.8 of Robinson's "A Course in the Theory of Groups (Second Edition)".

The Details:

The centraliser of an element $x$ in a group $G$ is defined as

$$C_G(x)=\{g\in G\mid xg=gx\}.$$

The conjugacy class of $x$ is defined as

$$Cl(x)=\{gxg^{-1}\mid g\in G\}.$$

The previous part of the exercise in question is as follows.

Lemma: Let $G$ be a finite group. The elements of the same conjugacy class have conjugate centralisers.

Proof: Let $x,y\in Cl(a)$ for some $a\in G$. Then there exist $g,h\in G$ such that $x=gag^{-1}$ and $y=hah^{-1}$. Note that $gh=f$ for $f\in G$, solved for $g$, has the unique solution $g=fh^{-1}$. We have

$$\begin{align}
C_G(x)&=\{k\in G\mid kx=xk\}\\
&=\{k\in G\mid kgag^{-1}=gag^{-1}k\}\\
&=\{k\in G\mid (kg)a(kg)^{-1}=gag^{-1}\}\\
&=\{k\in G\mid (kfh^{-1})a(kfh^{-1})^{-1}=(fh^{-1})a(fh^{-1})^{-1}\}\\
&=\{k\in G\mid (f^{-1}kf)h^{-1}ah(f^{-1}kf)^{-1}=h^{-1}ah\}\\
&=\{b=f^{-1}kf\in G\mid byb^{-1}=y\}\\
&=f\{k\in G\mid (fkf^{-1})y=y(fkf^{-1})\}f^{-1}\\
&=f\{K\in G\mid Ky=yK\}f^{-1}\\
&=fC_G(y)f^{-1},
\end{align}$$

because $fkf^{-1}$ runs through $G$ just as an arbitrary $K\in G$ does.$\square$

(I'm sorry about the garbled proof. I'm trying to piece together my understanding without recourse to external sources.)

The Question

If $c_1,\dots, c_h$ are the orders of the centralisers of elements of distinct conjugacy classes of a finite group, then $1/c_1+\dots+1/c_h=1$.

Thoughts:

This might require a combinatorial approach (in the numerical sense), since it involves counting, but of reciprocals; that's where I'm struggling: I can't use anything beyond the number of conjugacy classes being

$$\frac{1}{|G|}\sum_{\pi\in G}|{\rm Fix}(\pi)|,$$

by Exercise 1.6.2, where ${\rm Fix}(\pi)$ is the number of fixed points of the action of conjugacy on $\pi$. (I'm a little unsure of this. Have I got it right?)

I know that conjugation partitions the underlying set of $G$. This is mentioned earlier in the book.

Please help :)

答:
Hint : The class equation is
$$|G|=\sum_{g \in G}|Cl(g)|=\sum_{g \in G}|G:C_G(g)|=\sum_{g \in G}{|G|\over |C_G(g)|}$$
where we're taking only one $g$ from each conjugacy class.


问:
Branch cuts of the Hypergeometric function with one complex parameter
I am trying to understand if the hypergeometric function with the following choice of parameters has any branch cut in the complex plane
$$
\ _{2}F_{1}\left(a,b - i \rho ,c ,1-z^{2}\right)
$$

here $a,b,c \in \mathbb{N}$ and $\rho \in \mathbb{R}$ and $\rho > 0$.

I did some numerical explorations with Mathematica and I was not able to find any branch cuts on the real line. Is this property true?

I was not able to find any reference on the branch cuts of the hypergeometric function with complex parameters.

答:
Complex parameters have nothing to do with the choice of branch cuts. The conventional way to fix the branch cut on the principal sheet of the Riemann surface of $_2F_1(a,b;c;x)$ is to draw it from $1$ to $+\infty$ along the real axis in the complex plane of $x$. For example, this is how $_2F_1$ is implemented in Mathematica.

In terms of your $z$-variable, this branch cut corresponds to two halves of the imaginary axis. Specifically, $z\mapsto 1-z^2$ maps each of the half planes $\Re z\gtrless0$ to $\mathbb C\backslash[1,+\infty)$. So the signature of the branch cut will be the difference  between the values of $_2F_1(1-z^2)$, calculated for $z=\pm i\alpha+0$ (or $z=i\alpha\pm 0$) for $\alpha\in\mathbb R$.


问:
Let $A$ be a ring, and $I$ and $J$ are distinct maximal ideals of $A$. Then, $I$ never contains $J^n$ for every positive integer $n$
Let $A$ be a ring, and $I$ and $J$ are distinct maximal ideals of $A$.
Then, $I$ never contains $J^n$ for every positive integer $n$?
I want to figure out contradiction using $J$'s maximality.
But I cannot say $J^n$ contains $J$, so I'm having trouble.

Thank you in advance.

答: Assume that $J^n \subseteq I$. Taking radicals, we get $r(J^n) \subseteq r(I)$. Then we have $J \subseteq r(I)$. We also have $I \subseteq r(I)$, which is just a property of the radical. But, since $I$ and $J$ are maximal, this implies that $I = r(I) = J$, which is a contradiction.


问:
Proving $\{a\}$ is a set
I recently came across the following post Proving $\{a\}$ is a also set given that $a$ is a set. Introduction to Set Theory..
While I understand the answer given I am still unsure whether or not the proof given in the OP is correct, i.e. if it is legitimate to use the axiom of extensionality. It seems to me that the answer given only suggests that there is no need to use the axiom of extensionality, not that this is not permitted.
Any help would be appreciated. Many thanks!</p>

答:
A set can be defined in three ways:

1- By extension

2- By comprehension

3- By Venn diagram.

In your case $ \{a\} $ is a set defined by extension. it contains only one element which is $ a$.

In set Theory, same objects can be seen as sets or as elements.

$$\{a\} \text{ is an element of the set } \{\{a\}\}$$

Title: Proving $\{a\}$ is a set


问:
Does $\mathbb{E}(X^2+Y^2)=\mathbb{E}(X^2)+\mathbb{E}(Y^2)$?
For random variables X and Y, does $\mathbb{E}(X^2+Y^2)=\mathbb{E}(X^2)+\mathbb{E}(Y^2)$? I know that the expectation of $X+Y$ is $\mathbb{E}(X)+\mathbb{E}(Y)$, so does it work for squares, too?

答:
The short answer is: Yes, it works for squares too.

But it is important to see why this follows directly from $\mathbb{E}(X + Y) = \mathbb{E}(X) + \mathbb{E}(Y)$. This equality holds for any random variables, and since $X^2$ and $Y^2$ are random variables in their own right, it also holds for them.

If you want, we can set $U=X^2$ and $V=Y^2$. Then since $U$ and $V$ are random variables, we have $\mathbb{E}(U+V) = \mathbb{E}(U) + \mathbb{E}(V)$, or, by substituting the $X^2$ and $Y^2$ back in, that
$$\mathbb{E}(X^2+Y^2) = \mathbb{E}(X^2) + \mathbb{E}(Y^2).$$


问:
Is symmetric group on natural numbers countable?
I guess it is too difficult a question to ask about the cardinality of $S_{\mathbb{N}}$ so I would like to ask whether it is countable or not.

I tried to prove it is uncountable somewhat mimicking the Cantor's diagonal argument but failed.

答:
You can prove this easily via diagonal argument. Let's start with supposing that $\mathrm{Aut}(\mathbb{N})$ is countable. Then we can list such automorphisms: $\forall j: \pi_{j} \in \mathrm{Aut}(\mathbb{N})$. In table: $$\begin{array}{ccc}
\pi_{1}(1) & \pi_{1}(2)&\dots\\
\pi_{2}(1) & \pi_{2}(2)&\dots \\
\vdots &\vdots&\ddots&
\end{array}$$
Now consider automorphism $\pi_{j}$ such that $\forall k\in\mathbb{N} \Rightarrow \pi_{j}(k)\neq k$ (i.e. without fixed points). It is pretty easy to construct though: divide numbers into groups of, say, four numbers (it must be an even number!) and reverse the order (you get $4,3,2,1,8,7,6,5,\dots$) Now consider permutation $\delta=\pi_{1}(1)\pi_{2}(2)\dots$ (consisting of diagonal elements). Now it is easy to see that permutations $\delta$ and $\delta\circ \pi_{j}$ are different ($\forall i \in\mathbb{N}\Rightarrow \pi_{j}(i)\neq (\delta\circ\pi_{j})(i)$). Indeed, permutations $a$ and $a\circ b$ are equal at some point iff $b$ has fixed point. Here is contradiction: permutation $\delta\circ \pi_{j}$ coincides with $\delta$ on a diagonal.


问:
Is there a non-measurable function $f$ where $\sqrt{f}$ is measurable?
The title is basically the question:

Does there exist a non-measurable function $f$ where $\sqrt{f}$ is measurable?

I'm honestly not sure. My book does have a proposition that if $f$ is measurable, then $f^n$ where $n \in \mathbb{N}$ are measurable. However, it states this for naturals, not rationals. While no example is given, I'd assume the reason the statement is given this way is that $f^{q}$ where $q \in \mathbb{Q}$ need not be measurable. If it was true, I think they'd state the more general statement.

I know I am asking a different question than that proposition, but being that I don't believe the proposition would be true for a $q$ power, I'm not sure what to make of the question I'm being asked. I can't seem to think of a counterexample, but I'm also not convinced it's true. Can anyone shed some light on this?

答:
Part 1: If you don't assume $f > 0$,  then $f^q$ may not uniquely defined for $q\in \Bbb Q \setminus \Bbb N$. For instance, if $f(x)=0$ and $q=-1$, $f^q(x)$ is not defined.  If $f(x)=-1$ and $q=1/2$, $f^q(x)$ can be $i$ or $-i$.

So, the real issue is what you meant by $\sqrt{f}$, when $f$ has negative values. You need to choose one branch of the complex square root and that leads to a more sophisticate environment. That is why your book (and other books) stay with $q\in \Bbb N$.

Part 2:"Does there exist a non-measurable function $f$ where $\sqrt{f}$ is measurable?"
As we have seen in Part 1, the real issue is to assure that  $\sqrt{f}$ is well-defined as function. If it is well-defined and it is a measurable function, then $f$ is measurable.

Proof: Let $g= \sqrt{f}$. Then $g$ is measurabe, So $f= g^2$ is measurable. $\square$

So, there exist not a non-measurable function $f$ where $\sqrt{f}$ is measurable.


问:
Identities for Euclidean norms
Maybe I am looking too closely and am failing to utilise some known identities but I wish to prove elegantly that for any $X, Y \in \mathbb{R}^n$

$$\frac{a}{2}\|Y\|^2-aX\cdot Y = \frac{a}{2}\left(\|Y-X\|^2-\|X\|^2\right),$$
where $X\cdot Y = X^TY $.

The identity is easy to show when manually unwinding the Euclidean norms into their sums however I cannot help be feel there is a much easier way to show this!

If anything is unclear about my question please let me know!

答:
$$\begin{aligned}
\Vert y- x \Vert^2 - \Vert x \Vert^2 &= (y-x-x) \cdot(y-x+x)\\
&=(y-2x)\cdot y\\
&=\Vert y \Vert^2- 2x \cdot y
\end{aligned}$$

Title: Identities for Euclidean norms


问:
Representing vector as a column vector with respect to two bases
I have never done anything like this and the only examples I can find online basically tell me how to do the change of basis for "change-of-coordinates matrix from $B$ to $C$". If anyone is able to help me solve this, I would be very grateful. I really don't know where to start.

This is the problem:

Represent the vector as a column vector with respect to each of the two bases.
$$..\vec{v}=\begin{pmatrix}\begin{array}{r}
3\\
-1
\end{array}\end{pmatrix}\quad B_{1}=\left<\begin{pmatrix}\begin{array}{r}
1\\
-1
\end{array}\end{pmatrix},\begin{pmatrix}\begin{array}{r}
1\\
1
\end{array}\end{pmatrix}\right>,\;\quad B=\left<\begin{pmatrix}\begin{array}{r}
1\\
2
\end{array}\end{pmatrix},\begin{pmatrix}\begin{array}{r}
1\\
3
\end{array}\end{pmatrix}\right>$$..

答:
You seem to be confused.  When you talk about the matrix, I take you to mean a matrix whose columns are the basis vectors.  In the case of $B_1$ that would be $$M :=\begin{pmatrix}1&1\\-1&1\end{pmatrix}$$  If we multiply a vector $\begin{pmatrix}a\\b\end{pmatrix}$ by $M$, we get
$$\begin{pmatrix}1&1\\-1&1\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}
=\begin{pmatrix}a+b\\-a+b\end{pmatrix}=a\begin{pmatrix}1\\-1\end{pmatrix}+b\begin{pmatrix}1\\1\end{pmatrix}$$

Expressing $\begin{pmatrix}1\\-1\end{pmatrix}$ as a linear combination of the basis vectors means to find $a$ and $b$ such that $$\begin{pmatrix}3\\-1\end{pmatrix}=a\begin{pmatrix}1\\-1\end{pmatrix}+b\begin{pmatrix}1\\1\end{pmatrix}\tag1$$
That is, you don't want to multiply the given vector by the matrix; you want to find a matrix which, when multiplied by the vector, produces the given vector.  In other words, you need to solve $(1)$ for $a$ and $b$.  This simply means solving a system of two linear equations in two unknowns.


问:
Questions about SVD, Singular Value Decomposition
I am not a mathematician, so I need to understand what SVD does and WHY more than how it works exactly from the math perspective. (I understand at least what is the decomposition though).

This guy on youtube gave the only human explanation of SVD saying, that the U matrix maps "user to concept correlation" Sigma matrix defines the strength of each concept, and V maps "movie to concept correlation" given that initial matrix M has users in the rows, and movie (ratings) in the columns.

He also mentioned two concept specifically "sci fi" and "romance" movies. See the picture below.

My questions are:

How SVD knows the number of concepts. He as human mentioned two - sci fi, and romance, but in reality in resulting matrices are 3 concepts. (for example matrix U - that one with blue titles - has 3 columns not 2).

How SVD knows what is the concept after all. I mean, what If i shuffle the columns randomly how SVD then knows what is sci fi, what is romance. I mean, I suppose there is no rule, group the concepts together in the column order. What if scifi movie is the first and last one? and not first 3 columns in the initial matrix M?

What is the practical usage of either U, Sigma or V matrices? (Except that you can multiply them to get the initial matrix M)

Is there also any other possible human explanation of SVD than the guy up provided, or it is the only one possible function? Matrices of correlations.

答:
I've made the following code to illustrate what I mean about the line of best fit. I hope you find it useful

```<code>import numpy as np
import scipy.linalg as la
import matplotlib.pyplot as plt

# Adjustable parameters
N = 20           # Number of data points
var1 = 3         # Horizontal stretch factor
var2 = 5         # Vertical stretch factor
mean1 = 0        # Horizontal mean (distribution)
mean2 = 1        # Vertical mean
line_scale = 5   # determines length of best fit line

#Generate data
R = np.random.rand(2,N)              #Random matrix to generate data
# R = np.random.normal(size = (2,N))
A = (R-.5) * np.asarray([[var1],[var2]]) + np.asarray([[mean1],[mean2]])

# Plot data
plt.scatter(A[0,:],A[1,:])                 # Plot data points
mu = np.average(A,axis=1)
plt.scatter(mu[0],mu[1],marker='*',s=250)  # Plot data mean (as orange star)

# Find best-fit line using SVD
U = la.svd(A)[0]
v = np.reshape(U[:,0],(-1,1))
B = np.hstack((-v,v))*line_scale
plt.plot(B[0,:],B[1,:])

# Uncomment to set x and y scales equal
# plt.xlim((-5,5))
# plt.ylim((-5,5))
</code>```

Here's an example output. The trendline inferred by the SVD slopes upward, whereas the true trendline should be more horizontal. This can be seen as the effect of the mean (the orange star).

Update: this version includes the true best fit line (which requires zeroing the column-mean)

```<code>import numpy as np
import scipy.linalg as la
import matplotlib.pyplot as plt

# Adjustable parameters
N = 20           # Number of data points
var1 = 3         # Horizontal stretch factor
var2 = 5         # Vertical stretch factor
mean1 = 0        # Horizontal mean (distribution)
mean2 = 1        # Vertical mean
line_scale = 3   # determines length of best fit line

#Generate data
R = np.random.rand(2,N)              #Random matrix to generate data
# R = np.random.normal(size = (2,N))
A = (R-.5) * np.asarray([[var1],[var2]]) + np.asarray([[mean1],[mean2]])

# Plot data
plt.scatter(A[0,:],A[1,:])                 # Plot data points
mu = np.average(A,axis=1)
plt.scatter(mu[0],mu[1],marker='*',s=250)  # Plot data mean (as orange star)
mu = np.reshape(mu,(-1,1))

# Find best-fit 1-D subspace using SVD
U = la.svd(A)[0]
v = np.reshape(U[:,0],(-1,1))
B = np.hstack((-v,v))*line_scale
plt.plot(B[0,:],B[1,:])

# Find best-fit line (1-D affine subspace) using PCA
M = A - mu
U = la.svd(M)[0]
w = np.reshape(U[:,0],(-1,1))
C = mu + np.hstack((-w,w))*line_scale
plt.plot(C[0,:],C[1,:])

# Uncomment to set x and y scales equal
# plt.xlim((-5,5))
# plt.ylim((-5,5))
</code>```

Sample output:

Body:
I've made the following code to illustrate what I mean about the line of best fit. I hope you find it useful

import numpy as np
import scipy.linalg as la
import matplotlib.pyplot as plt

# Adjustable parameters
N = 20           # Number of data points
var1 = 3         # Horizontal stretch factor
var2 = 5         # Vertical stretch factor
mean1 = 0        # Horizontal mean (distribution)
mean2 = 1        # Vertical mean
line_scale = 5   # determines length of best fit line

#Generate data
R = np.random.rand(2,N)              #Random matrix to generate data
# R = np.random.normal(size = (2,N))
A = (R-.5) * np.asarray([[var1],[var2]]) + np.asarray([[mean1],[mean2]])

# Plot data
plt.scatter(A[0,:],A[1,:])                 # Plot data points
mu = np.average(A,axis=1)
plt.scatter(mu[0],mu[1],marker='*',s=250)  # Plot data mean (as orange star)

# Find best-fit line using SVD
U = la.svd(A)[0]
v = np.reshape(U[:,0],(-1,1))
B = np.hstack((-v,v))*line_scale
plt.plot(B[0,:],B[1,:])

# Uncomment to set x and y scales equal
# plt.xlim((-5,5))
# plt.ylim((-5,5))

Here's an example output. The trendline inferred by the SVD slopes upward, whereas the true trendline should be more horizontal. This can be seen as the effect of the mean (the orange star).

Update: this version includes the true best fit line (which requires zeroing the column-mean)

import numpy as np
import scipy.linalg as la
import matplotlib.pyplot as plt

# Adjustable parameters
N = 20           # Number of data points
var1 = 3         # Horizontal stretch factor
var2 = 5         # Vertical stretch factor
mean1 = 0        # Horizontal mean (distribution)
mean2 = 1        # Vertical mean
line_scale = 3   # determines length of best fit line

#Generate data
R = np.random.rand(2,N)              #Random matrix to generate data
# R = np.random.normal(size = (2,N))
A = (R-.5) * np.asarray([[var1],[var2]]) + np.asarray([[mean1],[mean2]])

# Plot data
plt.scatter(A[0,:],A[1,:])                 # Plot data points
mu = np.average(A,axis=1)
plt.scatter(mu[0],mu[1],marker='*',s=250)  # Plot data mean (as orange star)
mu = np.reshape(mu,(-1,1))

# Find best-fit 1-D subspace using SVD
U = la.svd(A)[0]
v = np.reshape(U[:,0],(-1,1))
B = np.hstack((-v,v))*line_scale
plt.plot(B[0,:],B[1,:])

# Find best-fit line (1-D affine subspace) using PCA
M = A - mu
U = la.svd(M)[0]
w = np.reshape(U[:,0],(-1,1))
C = mu + np.hstack((-w,w))*line_scale
plt.plot(C[0,:],C[1,:])

# Uncomment to set x and y scales equal
# plt.xlim((-5,5))
# plt.ylim((-5,5))

Sample output:


问:
pointwise convergence and convergence in $\ell_{1}$ norm
Assume that we have a sequence $\{\mathbf{x}_{n}\}_{n=1}^{\infty}$ of non-negative vectors in $\ell_{1}$, i.e. $x_{n,i}\geq 0$ for all $n = 1,2 \dots $ and $i= 1,2 \dots $. Next, assume that for any $i$ we have
$$
x_{n,i} \to a_{i}
$$
and $||\textbf{a}||_{1} = b >0$.

The statement: $\mathbf{x}_{n} \to \textbf{a}$ in $\ell_{1}$.

I have the following proof, but I am note 100 % sure:

Let's fix some $\delta>0$. Then there exists $M$ such that
$$
\sum_{i\leq M}a_{i} \geq b - \frac{\delta}{2}.
$$

Since $\mathbf{x}_{n}$ converges to $\mathbf{a}$ pointwise, then there exists $n_{1}$ such that for all $n\geq n_{1}$ we have
$$
\sup_{i\leq M}|x_{n,i} - a_{i}| \leq \frac{\delta}{2M}.
$$
From those two inequalities above, it follows that $\sum_{i\leq M}x_{n,i} \geq b - \delta$, for all $n\geq n_{1}$.

Therefore, using inequalities above, we have
$$
\sum_{i}|x_{n,i} - a_{i}| = \sum_{i\leq M}|x_{n,i} - a_{i}| + \sum_{i>M}|x_{n,i} - a_{i}| \leq \sum_{i\leq M}|x_{n,i} - a_{i}| + \sum_{i>M} x_{n,i} + \sum_{i>M}a_{i} \leq \frac{\delta}{2} + \delta + \frac{\delta}{2} = 2 \delta
$$
holds for all $n\geq n_{1}$, which proves the oroginal statement.

edit:
I see the problem now. In order to have the last inequality, we must also have that $||\textbf{x}_{n}||_{1} = b$, for all $n$.

hypothesis: I think, the statement is also true when we just have $\sup_{n}||\textbf{x}_{n}||_{1} < \infty$

答:
This is not true in general. Suppose $a_1=1$ and $x_{1,n}=1$ for all $n\in\mathbb N$. Let $a_i=0$ and $x_{i,n}=\frac{1}{n}$ for all $n\geq 1$ and $2\leq i\leq n$. For $i\geq n$ let $x_{i,n}=0$. Clearly $x_{i,n}\to a_i=0$, and $\|a\|_1=1>0$. However, for any $n\in \mathbb N$
$$\|x_n-a\|_1=\frac{1}{n}\sum_{i=2}^n1=\frac{n-1}{n}\to 1.$$


答:
As you mentioned in your edit, the proof will be correct if $\|\mathbf{x}_n\|_1 = b$ for all sufficiently large $n$. However, there is a more general condition under which it will hold:

Suppose that for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that

$$\sup_n \sum_{k=N}^\infty |\mathbf{x}_{n,k}| < \epsilon.$$

(I believe this is the $\ell_1$ analogue of uniform integrability). Then your proof works if we also assume $M$ is large enough that the above equation holds for $\epsilon = \delta$ and $N = M$.

Edit: This is actually necessary and sufficient. Suppose $\{\mathbb{x}_n\}$ does not satisfy the condition above. Then there must exist some $\epsilon > 0$ and sequences $n_i,N_i \to \infty$ such that $\sum_{k=N_i}^\infty |\mathbf{x}_{n_i}| > \epsilon$ for all $i$.

Then $x_{n_i}$ cannot converge to $a$. To see why, let $M$ be large enough that $\sum_{k=M}^\infty |a_k| < \frac{\epsilon}{2}$. Then,

$$\sum_{k=1}^\infty |\mathbf{x}_{n_i,k} - a_k| \geq \sum_{k=M}^\infty |\mathbf{x}_{n_i,k} - a_k| \geq \sum_{k=M}^\infty |\mathbf{x}_{n_i,k}| - \sum_{k=M}^\infty|a_k| > \frac{\epsilon}{2}.$$

Since this holds for infinitely many $i$ (all $i$ such that $N_i > M$), $\mathbf{x}_n$ does not converge to $a$ in $\ell^1$.

Title: pointwise convergence and convergence in $\ell_{1}$ norm

Body:
Assume that we have a sequence $\{\mathbf{x}_{n}\}_{n=1}^{\infty}$ of non-negative vectors in $\ell_{1}$, i.e. $x_{n,i}\geq 0$ for all $n = 1,2 \dots $ and $i= 1,2 \dots $. Next, assume that for any $i$ we have
$$
x_{n,i} \to a_{i}
$$
and $||\textbf{a}||_{1} = b >0$.

The statement: $\mathbf{x}_{n} \to \textbf{a}$ in $\ell_{1}$.

I have the following proof, but I am note 100 % sure:

Let's fix some $\delta>0$. Then there exists $M$ such that
$$
\sum_{i\leq M}a_{i} \geq b - \frac{\delta}{2}.
$$

Since $\mathbf{x}_{n}$ converges to $\mathbf{a}$ pointwise, then there exists $n_{1}$ such that for all $n\geq n_{1}$ we have
$$
\sup_{i\leq M}|x_{n,i} - a_{i}| \leq \frac{\delta}{2M}.
$$
From those two inequalities above, it follows that $\sum_{i\leq M}x_{n,i} \geq b - \delta$, for all $n\geq n_{1}$.

Therefore, using inequalities above, we have
$$
\sum_{i}|x_{n,i} - a_{i}| = \sum_{i\leq M}|x_{n,i} - a_{i}| + \sum_{i>M}|x_{n,i} - a_{i}| \leq \sum_{i\leq M}|x_{n,i} - a_{i}| + \sum_{i>M} x_{n,i} + \sum_{i>M}a_{i} \leq \frac{\delta}{2} + \delta + \frac{\delta}{2} = 2 \delta
$$
holds for all $n\geq n_{1}$, which proves the oroginal statement.

edit:
I see the problem now. In order to have the last inequality, we must also have that $||\textbf{x}_{n}||_{1} = b$, for all $n$.

hypothesis: I think, the statement is also true when we just have $\sup_{n}||\textbf{x}_{n}||_{1} < \infty$


答: In identities such as $\mathrm{E}(X+Y)=\mathrm{E}(X)+\mathrm{E}(Y)$, $X$ and $Y$ are both simply placeholders for any two random variables. $X^2$ and $Y^2$ are also random variables, meaning that the identity still applies. It's for the same reason that $\sin^2x+\cos^2x=1$ implies $\sin^2(x^2)+\cos^2(x^2)=1$.

Title: Does $\mathbb{E}(X^2+Y^2)=\mathbb{E}(X^2)+\mathbb{E}(Y^2)$?

Body: For random variables X and Y, does $\mathbb{E}(X^2+Y^2)=\mathbb{E}(X^2)+\mathbb{E}(Y^2)$? I know that the expectation of $X+Y$ is $\mathbb{E}(X)+\mathbb{E}(Y)$, so does it work for squares, too?


问:
Arranging numbers in an array (Swedish Math Olympiad 1986)
Consider an $m\times n$ array of real numbers.  Let $d>0$.  Suppose that the difference between the maximum number and the minimum number in each row is at most $d$.  We then sort the numbers in each column so that the maximum element is in the first row and the minimum element is in the  last row (so each column is arranged in decreasing order).  Show that the difference between the maximum number and the minimum number in each row (after the change) is still $\le d$.

Source : Swedish Math Olympiad 1986, also Principles and Techniques in Combinatorics

Here is my attempt.  Let $M_k$ and $m_k$ be the original largest and smallest elements of the $k$th row.  Let $M'_k$ and $m'_k$ be the new largest and smallest elements of the $k$th row.  We have $M'_1\ge M'_2 \ge ... \ge M'_m$ and $m_1'\ge m_2' \ge ... \ge m'_m$.  Let's try induction on $k$.

For $k=1$, suppose that $M'_1$ was originally in row $i$.  Then $M'_1\le M_i$.  We know that $m'_1$ is in the same column as some element originally from row $i$.  Let that element be $a_i$.  Therefore $m'_1\ge a_i\ge m_i$ so $M'_1-m'_1\le M_i-m_i\le d$.

Let $1<k\le m$.  Suppose that $M'_k$ was originally in row $i$ so $M'_k\le M_i$.  We know that $m'_k$ is in the same column as some element $a_i$ originally from row $i$.  Suppose that $a_i$ is in row $j$ after the change.  If $j\ge k$, then $m'_k\ge a_i\ge m_i$ so $M'_k-m'_k\le M_i-m_i\le d$.  We now assume $j<k$.  What to do here?

BTW in the book "Principles and Techniques in Combinatorics" this problem is in chapter 3 which is about pigeonhole principle.  So if you have a solution with pigeonhole that'd be great.

答:
Here's an approach, but it doesn't use PP.

Claim: Given $n$ intervals $[a_i, a_i + d]$ on the real line.

In each interval, one real number $b_i$ is chosen.

Sort the $a_i, b_i$ in increasing order to get the sequences $a'_i, b'_i$

Then, $ b'_k \in [ a'_k , a'_k + d]$.

Proof: Do this yourself. If stuck, show what you've tried.

Show that $b'_k \geq a'_k$. Prove by contradiction.

Show that $b'_k \leq a'_k+d$. Prove by contradiction.

Corollary: The problem holds.

Let $ a_i = \min_k m_{i, k}$ be the minimum in each row.

Add an additional $n+1$st column, with entries  $m_{i,n+1} =d +  a_i$.

Fix a column $k$.

By definition, each element $m_{i, k} \in [a_i, a_i + d]$.

Now, apply the claim to conclude that $m'_{i, k}\in [a'_k, a'_k + d] \quad \forall i$.

Fix a row $i$.

We have  $m'_{i, k } \in [a'_k, a'_k + d] \quad \forall k$,

Hence, the difference between the max and min in each row (for $n+1$ columns) is still $ \leq d$.

Hence, the difference between the max and min in each row (for $n$ columns) is still $ \leq d$.

Note: In the claim, I used intervals of exactly length $d$ to ensure that we could sort the intervals cleanly. Otherwise, we could have nested intervals, and then it might not be clear how to sort the rows in increasing order.

As it turns out, this was unnecessary. We have a stronger claim:

Given $n$ intervals $[a_i, b_i]$, with elements $c_i \in [a_i, b_i]$, and we sort each individual sequence $a_i, b_i, c_i$ in increasing order, then $ c'_i \in [a'_i, b'_i]$.

Note: If we sorted the intervals just by the first coordinate, then the claim need not be true. IE $c'_i \not \in [a_{i'}, b_{i'}]$.

However, I can't apply the stronger claim to the problem. Any help here is greatly appreciated.  The trouble that I run into is that ultimately, we still have to show that $ b'_i - a'_i \leq d$. While this is recognizable as the 2 column case, it is essentially the entire argument (IE the induction on the number of columns is trivial).

My solution avoids this because $b'_i - a'_i = d$ by construction. Doing so felt a bit forced / unnatural, though I know why I thought of doing it.


问:
Given a circle of radius $3\rm{cm}$ inscribed in an equilateral triangle $\triangle ABC$ and $EZDU$ is a square inscribed in the circle.
Given a circle of radius $3\rm{cm}$ inscribed in an equilateral triangle $\triangle ABC$ and $EZDU$ is a square inscribed in the circle. Find the divsion of the area of the square divided by the area of the triangle. (Romania 1961)

The area of the square is obvious from the radios of the circle. We have that $2ZE^2=36\implies ZE^2=18$ hence the area is $18$. This is where I got stuck. I don't know how to calculate the area of the triangle. Could you please explain to me how to solve this question?

答:
$O$ is the centroid of $\triangle ABC$
$\implies OC=2OD$.

Can you take it from here?


问:
Convergent sequence that satisfy $y_{n+1}=3y_n^2-3y_n+1$
Let be the sequence $y_n$ with $0<y_0<1$ such that
$$y_{n+1}=3y_n^2-3y_n+1$$
and
$$0<y_n<1$$
for every $n$ positive integer.
Prove that $y_n$ is convergent with the limit $\frac{1}{3}$.

This is what I dis
$$y_{n+1}-\frac{1}{3}=\left(y_n-\frac{1}{3}\right)\left(3y_n-2\right)$$
Therefore
$$y_{n+1}-\frac{1}{3}=\left(y_0-\frac{1}{3}\right)\left(3y_n-2\right)\left(3y_{n-1}-2\right)\left(3y_{n-2}-2\right)...\left(3y_0-2\right)$$
then
$$\left|y_{n+1}-\frac{1}{3}\right|=\left|\left(y_0-\frac{1}{3}\right)\right|\left(\left|3y_n-2\right|\right)\left(\left|3y_{n-1}-2\right|\right)...\left(\left|3y_0-2\right|\right)$$
if we denote $$z_n=\left|3y_n-2\right|$$
then
$$0<z_n<1$$
And we have to prove that
$$\lim _{n\to \infty \:\:}\left(z_1z_2...z_n\right)=0$$
and we are done.
I suppose that the right hand side is convergent to $0$, but how should I prove that? Please help!!!

答: $l=\frac{1}{3}$ is surely not a limit of the sequence as the map $f(x)=3x^2-x+1$ is continuous and $l$ is not one of its fixed point.


问:
Atiyah Macdonald 2.19 is a tautology?

The image reads:

If $T_n$ is exact, that is to say, tensoring with $N$ transforms all exact sequences into exact sequences, then $N$ is said to be a flat $A$-module.

Proposition 2.19: The following are equivalent for any $A$ module $N$:
(i) $N$ is flat
(ii) If $0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$ is any exact sequence of $A$ modules, the tensored sequence $0 \rightarrow M' \otimes N \rightarrow M \otimes N \rightarrow M'' \otimes N \rightarrow 0$ is exact

As far as I understand, (i) and (ii) are identical: (i) says $N$ is flat, (ii) spills out the definition of flat. What is there to prove? The book says:

(i) is implied and implied by (ii) by taking a long exact sequence and splitting into short exact sequences

I don't understand what long exact sequence I need to consider, because to me, the proposition looks like a tautology.

答:
Atiyah-Macdonald says "exact sequence" to mean "long exact sequence". The definition of flatness says that tensoring preserves long exact sequences. Characterisation (ii) tells us that it if a module preserves short exact sequences on tensoring, then it must preserve long exact sequences on tensoring.

So if we are given an LES, we can break it up into SESes, apply flatness-on-SES to prove flatness-on-LES.

I was confused, since I implicitly assume "exact sequence = SES" as I am following multiple books at the same time.

Title: Atiyah Macdonald 2.19 is a tautology?


问:
Best linear spline - inequaltiy.
For a ﬁxed set of knot points $\omega^N$, let $L$ be the operator that maps the continuous function $f$ to its linear spline interpolant $l ∈ \mathcal{L}$.

Let $l=L(f)$ prove that for every $l^* \in \mathcal{L}$: $||f-l||_\infty \le ||f-l^*||_\infty$.

答: I don't think that's true. For example, consider $f(x)=x^2$ on the interval $[-1,1]$ with knot points $-1,1$. The linear interpolant is $l(x)\equiv 1$, with $\|f-l\|_\infty=1$. But $l^*\equiv 1/2$ has $\|f-l^*\|_\infty=1/2$.</p>

Title: Best linear spline - inequaltiy.


问:
You are making cookies and add N chips to dough randomly, and split it into 100 equal cookies, again at random. How many chips should go into dough?
Question: You are making chocolate chip cookies. You add N chips randomly to the dough and you randomly split the dough into 100 equal cookies. How many chips should go into the dough to give a probability of at least 90% that every cookie has at least one chip?

I tried to attempt to solve this using IID random variables. I am not sure how to set the problem up. I know that there should at least be 100 chocolate chips or else the cookies will not meet the "at least 1 chip per cookie" requirement and that there is 10% chance that the cookies do not have a chip.

答:
The general approach is as follows. This is only an approximation but it's an okay one as we see at the end.

Let's define $P_c$ as the probability that every cookie has at least 1 chip. Then we have the complement of this $P_{nc} = 1-P_c$ which is the probability that one or more cookies don't have a chip. We can now represent this event as a sum of simple events. Define $C_i$ as the event that the i-th cookie has no chips. We can now write

$$\begin{align}
P_{nc} & =  P(C_0 \cup C_1 \cup \cdots \cup C_{100}) \\
& = P \left( \bigcup_{i=1}^{100} C_i \right) \leq \sum_{i=1}^{100} P(C_i)
\end{align}$$

and we use Boole's inequality to place an upper bound on this. So now what's left is to determine $P(C_i)$ probability of a single cookie of having no chips. Here's one way to think about it:

each chip arrives independently and there is a $\frac{1}{100}$ chance it will land in this cookie $C_i$

there is a $\frac{99}{100}$ chance it will not land in cookie $C_i$

for a total of N chips this becomes $\frac{99}{100}^N$

Thus $P(C_i) = \frac{99}{100}^N$ is the probability of a single cookie having no chips when N chips are randomly distributed amongst 100 cookies. Continuing, we have

$$\begin{align}
P_{nc} & =  \sum_{i=1}^{100} P(C_i) \\
& = \sum_{i=1}^{100} \frac{99}{100}^N = 100 \frac{99}{100}^N
\end{align}$$

Consequently $P_c \ge 1 - 100 \frac{99}{100}^N$, which we solve by taking the equality and obtain

$$\begin{align}
1 - 100 \frac{99}{100}^N & = 0.9 \\
\frac{99}{100}^N &=  1000^{-1} \\
N \ln{\frac{99}{100}}  & = -\ln{1000} \\
N & = -\frac{\ln{1000}}{\ln{\frac{99}{100}}} \\
N & \approx 687.31
\end{align}$$

Which is the upper bound solution to the problem.

Note: that by using Boole's Inequality we are assuming $C_i$ events do not overlap, i.e. we will not have more than 1 chipless cookie. These higher-order terms include probabilities of more than 1 cookie having no chips. These terms range from i=1 (1 chipless cookie, which we considered here) to 99, which represents a situation where all the chips would be in a the 100th cookie, and the remainder 99 would be chipless. It can be shown that such higher-order terms contribute little to the sum above, but I am not mathematical enough to do it.

However, numerical simulations yield a value of $N = 682-683$, so we can see it's a great approximation.


问:
Show that complex Integral of 1/z on an ellipse is equal to the integral on a circle
I am trying to show that if I have two curves:

Where $a,b > 0$ and $a \neq b$. Then,

There is also a hint that suggests splitting curves into two parts such as lower and upper half-planes and using the Cauchy Integral Theorem to calculate both integrals. However, I could not calculate it. How can I show that both integrals are equal?

Edit: On the textbook "Complex Analysis 2nd ed. Freitag, Busam" there is a solution, but the explanation is not clear to me:

One can split the curve α, and correspondingly β, into the pieces in the upper and lower half-plane, α = α+ ⊕ α−, α+ = α|[0, 1/2], α− = α|[1/2, 1]. The Cauchy Integral Theorem can be used to show that the integrals along the two pieces are correspondingly equal. For this, slit the plane along the positive, and respectively the negative imaginary axis

答:
Suppose $0<a<b.$ Let $\alpha_1$ be the top of $\alpha,$ and $\beta_1$ be the top of $\beta.$ Then by Cauchy

$$\int_{\alpha_1}\frac{1}{z}\,dz - \int_{\beta_1}\frac{1}{z}\,dz =0.$$

Similarly, let $\alpha_2$ be the bottom of $\alpha,$ and $\beta_2$ be the bottom of $\beta.$ Then

$$\int_{\alpha_2}\frac{1}{z}\,dz - \int_{\beta_2}\frac{1}{z}\,dz =0.$$

Therefore

$$\int_{\alpha_1}\frac{1}{z}\,dz+\int_{\alpha_2}\frac{1}{z}\,dz = \int_{\beta_1}\frac{1}{z}\,dz +\int_{\beta_2}\frac{1}{z}\,dz.$$

The left side of this is $\int_{\alpha}\dfrac{1}{z}\,dz.$ The right side is $\int_{\beta}\dfrac{1}{z}\,dz.$


Body:
Given a circle of radius $3\rm{cm}$ inscribed in an equilateral triangle $\triangle ABC$ and $EZDU$ is a square inscribed in the circle. Find the divsion of the area of the square divided by the area of the triangle. (Romania 1961)

The area of the square is obvious from the radios of the circle. We have that $2ZE^2=36\implies ZE^2=18$ hence the area is $18$. This is where I got stuck. I don't know how to calculate the area of the triangle. Could you please explain to me how to solve this question?

Body:
HINT:

$\frac{S_{EZDU}}{S_{\triangle ABC}}$= $\frac{(3\sqrt 2)^2}{\frac{\sqrt 3}{4}{(6\sqrt 3)}^2}$=$\frac{2\sqrt 3}{9}$


问:
Good book for Riemann Surfaces
I am considering reading one of 'Algebraic curves and Riemann Surfaces' by Rick Miranda or 'Lectures on Riemann Surfaces' by Otto Forster. Which one of these is more advanced and comprehensive ? What are the differences in the approaches of these two books ?</p>

答:
Ted Shifrin notes above:

I know Forster's book quite well, having taught out of a good portion of it a few times. It is extremely well-written, but definitely more analytic in flavor. In particular, it includes pretty much all the analysis to prove finite-dimensionality of sheaf cohomology on a compact Riemann surface. It also deals quite a bit with non-compact Riemann surfaces, but does include standard material on Abel's Theorem, the Abel-Jacobi map, etc. I would also recommend Griffiths's Introduction to Algebraic Curves — a beautiful text based on lectures.

Title: Good book for Riemann Surfaces


问:
$\int_{[-a,a]} h(x) = \int_{[-a,a]} h(-x) ?$
Suppose $h$ is a $2a$-periodic function ($a > 0)$ that is integrable on $[-a,a]$, is it true that $\int_{[-a,a]} h(x) = \int_{[-a,a]} h(-x) ?$

I'm currently stuck and not sure what to do. Is there any suggestion?

答:
Sub in the last integral:
$$\int_{[-a,a]} h(-x)=-\int_{[a,-a]} h(x)=\int_{[-a,a]} h(x)$$


问:
If $c_1,\dots, c_h$ are the orders of the centralisers of elements of distinct conjugacy classes of a finite group, then $1/c_1+\dots+1/c_h=1$
This is part of Exercise 1.6.8 of Robinson's "A Course in the Theory of Groups (Second Edition)".

The Details:

The centraliser of an element $x$ in a group $G$ is defined as

$$C_G(x)=\{g\in G\mid xg=gx\}.$$

The conjugacy class of $x$ is defined as

$$Cl(x)=\{gxg^{-1}\mid g\in G\}.$$

The previous part of the exercise in question is as follows.

Lemma: Let $G$ be a finite group. The elements of the same conjugacy class have conjugate centralisers.

Proof: Let $x,y\in Cl(a)$ for some $a\in G$. Then there exist $g,h\in G$ such that $x=gag^{-1}$ and $y=hah^{-1}$. Note that $gh=f$ for $f\in G$, solved for $g$, has the unique solution $g=fh^{-1}$. We have

$$\begin{align}
C_G(x)&=\{k\in G\mid kx=xk\}\\
&=\{k\in G\mid kgag^{-1}=gag^{-1}k\}\\
&=\{k\in G\mid (kg)a(kg)^{-1}=gag^{-1}\}\\
&=\{k\in G\mid (kfh^{-1})a(kfh^{-1})^{-1}=(fh^{-1})a(fh^{-1})^{-1}\}\\
&=\{k\in G\mid (f^{-1}kf)h^{-1}ah(f^{-1}kf)^{-1}=h^{-1}ah\}\\
&=\{b=f^{-1}kf\in G\mid byb^{-1}=y\}\\
&=f\{k\in G\mid (fkf^{-1})y=y(fkf^{-1})\}f^{-1}\\
&=f\{K\in G\mid Ky=yK\}f^{-1}\\
&=fC_G(y)f^{-1},
\end{align}$$

because $fkf^{-1}$ runs through $G$ just as an arbitrary $K\in G$ does.$\square$

(I'm sorry about the garbled proof. I'm trying to piece together my understanding without recourse to external sources.)

The Question

If $c_1,\dots, c_h$ are the orders of the centralisers of elements of distinct conjugacy classes of a finite group, then $1/c_1+\dots+1/c_h=1$.

Thoughts:

This might require a combinatorial approach (in the numerical sense), since it involves counting, but of reciprocals; that's where I'm struggling: I can't use anything beyond the number of conjugacy classes being

$$\frac{1}{|G|}\sum_{\pi\in G}|{\rm Fix}(\pi)|,$$

by Exercise 1.6.2, where ${\rm Fix}(\pi)$ is the number of fixed points of the action of conjugacy on $\pi$. (I'm a little unsure of this. Have I got it right?)

I know that conjugation partitions the underlying set of $G$. This is mentioned earlier in the book.

Please help :)

答:
If we consider the action of conjugacy, the stabilizer of an element is exactly its centralizer. Let $\gamma_i$ be the cardinality of the conjugacy class of the i-th conjugacy class. Since the product of the orbit of an element and its stabilizer is $|G|$, we have that $\gamma_i c_i = |G| $. Hence

$$ \sum_{i=1}^h 1/c_i = \sum_{i=1}^h \gamma_i / |G| = \frac{1}{|G|} \sum_{i=1}^h \gamma_i  = 1$$

Since the sum of the cardinalities of all the conjugacy classes sum up to $|G|$ (the conjugacy classes are a partition of $G$).


问:
very slow convergence of Picard method for solving nonlinear system of equations
I have a nonlinear system of equations as
$$
\left(\mathbf{K}_{\mathbf{L}}+\mathbf{K}_{\mathbf{N L}}(\mathbf{X})\right) \mathbf{X}=\mathbf{F}
$$
in which $\mathbf{K}_{\mathbf{N L}}(\mathbf{X})$ represents the nonlinear stiffness matrix which is dependent to $\mathbf{X}$.
I'm solving with Picard iteration like this:

first ignore the nonlinear stiffness matrix and solve the linear
matrix for $\mathbf{X}$.

put the resulting $\mathbf{X}$ in the nonlinear stiffness matrix and solve the full equation for $\mathbf{X}$.

check convergence and repeat 2 if the convergence is not satisfied.

the problem i have here is when the force vector($\mathbf{F}$) is small the nonlinear equation solves very fast but when i increase the force beyond some threshold it gets ages to converge.
i have tried to solve it using Matlab fsolve function with algorithms like 'trust-region' and 'levenberg-marquardt' but the same thing happens with large force vectors.

is there any way i can improve the convergence speed ?

p.s.
heres a gif of the result vector $\mathbf{X}$ inside the convergence loop with a force vector slighly over the threshold.

edit(more details):
so my problem is bending of a nonlinear timoshenko beam that has three governing equations as below:
$$
-\frac{d}{d x}\left\{A_{x x}\left[\frac{d u}{d x}+\frac{1}{2}\left(\frac{d w}{d x}\right)^{2}\right]+B_{x x} \frac{d \phi_{x}}{d x}\right\}=0
$$
$$
-\frac{d}{d x}\left\{A_{x x} \frac{d w}{d x}\left[\frac{d u}{d x}+\frac{1}{2}\left(\frac{d w}{d x}\right)^{2}\right]+B_{x x} \frac{d w}{d x} \frac{d \phi_{x}}{d x}\right\}-\frac{d}{d x}\left[S_{x x}\left(\frac{d w}{d x}+\phi_{x}\right)\right]=q
$$
$$
-\frac{d}{d x}\left\{D_{x x} \frac{d \phi_{x}}{d x}+B_{x x}\left[\frac{d u}{d x}+\frac{1}{2}\left(\frac{d w}{d x}\right)^{2}\right]\right\}+S_{x x}\left(\frac{d w}{d x}+\phi_{x}\right)=0
$$
along with the proper boundary conditions and using finite difference, when assembled they form:
$$
\left(\mathbf{K}_{\mathbf{L}}+\mathbf{K}_{\mathbf{N L}}(\mathbf{X})\right) \mathbf{X}=\mathbf{F}
$$

答: You can try some of the convergence acceleration algorithms which can work very well for fixed-point (Picard-type) iterations.  If you are using R, there is a package called SQUAREM which implements a reliable, convergence acceleration scheme.  It is based on the paper (Varadhan and Roland, Simple and Globally Convergent Methods for Accelerating the Convergence of Any EM Algorithm, Scandinavian Journal of Statistics, 2008).  EM algorithms are essentially Picard-like algorithms - they are contraction mappings which converge slowly.</p>


问:
Why do $\prod_{i=1}^{n} c_{i} !$ out of the $n!$ different ways to label a tree give rise to the same unordered labelled tree?
My question is about a sentence in the proof of the following proposition:

Proposition 2.3. The uniform random tree $\mathbb{T}_{n}$ has the same distribution as a tree generated
as follows:

Take a Galton-Watson tree with Poisson(1) offspring distribution;

Condition it to have total progeny precisely (n);

Assign the vertices random labels chosen from [n] and forget the original ordering.

Proof. Recall the standard labelling of the vertices of a Galton-Watson tree $\mathbf{T}$, and let $\mathbf{t}$ be a particular tree with $\#(\mathbf{t})=n$ and numbers of offspring $c\left(v_{1}\right)=c_{1}, \ldots, c\left(v_{n}\right)=c_{n}$. Then
$$
\mathbb{P}(\mathbf{T}=\mathbf{t})=\prod_{i=1}^{n} \frac{e^{-1}}{c_{i} !}=e^{-n} \prod_{i=1}^{n} \frac{1}{c_{i} !}
$$
Now observe that $\mathbb{P}(\#(\mathbf{T})=n)$ is a function only of $n$. Hence,
$$
\mathbb{P}(\mathbf{T}=\mathbf{t} \mid \#(\mathbf{T})=n)=f(n) \prod_{i=1}^{n} \frac{1}{c_{i} !}
$$
for some function $f$. Now consider labelling the vertices of $\mathbf{T}$ with $[n]:=\{1,2, \ldots, n\}$. There are $n!$ different ways to do this, of which $\prod_{i=1}^{n} c_{i} !$ give rise to the same unordered labelled tree, once we forget the ordering. Hence, the probability of obtaining a particular labelled unordered tree $t$ is $f(n) / n !$. Since this depends only on $n$, and not on any other feature of the tree, it must be the case that
the tree is uniformly distributed on $\mathbb{T}_{n}$.

(From An introduction to random trees by Christina Goldschmidt )

There is one part I do not understand, why do $\prod_{i=1}^{n} c_{i} !$ out of the $n!$ different ways give rise to the same unordered labelled tree?

答:
Permuting the offspring of any subset of the vertices does not change the unlabelled tree, because it preserves all of the parent-child relationships that define the tree. For each $k\in[n]$ there are $c_k!$ ways to permute the offspring of $v_k$, so there are altogether $\prod_{k=1}^nc_k!$ ways to permute the offspring of each vertex: you can combine any of the $c_1!$ permutations of the offspring of $v_1$ with any of the $c_2!$ permutations of the offspring of $v_2$ and so on without changing the underlying unlabelled tree.

Or you can turn it around. Suppose that $T$ is an unlabelled (rooted) tree on $n$ vertices. In the standard labelling the root must be labelled $\varnothing$. If it has $c_1$ offspring, they can then be labelled $1,2,\ldots,c_1$ in any of $c_1!$ orders. If the vertex now labelled $1$ has $c_2$ offspring, these can be labelled $11,12,\ldots,1c_2$ in any of $c_2!$ different orders. Continuing in this fashion, it the numbers of offspring of the $n$ vertices are $c_1,\ldots,c_k$, we have $\prod_{k=1}^nc_k!$ different ways to label the vertices using the standard labelling, depending on how we order the offspring of each vertex.


问:
Why is $|x-y|^2 \geq \frac 12 |y|^2$ when $|x|$ is sufficiently small?
Why is $|x-y|^2 \geq \frac 12 |y|^2$ when $|x|$ is sufficiently small. Here $x,y \in \Bbb R^n$.

Logically it seems obvious because as $|x|$ is small $|x-y|$ is eventually very close to $|y|$ but I can't prove it using Triangle inequality. Maybe I am missing something. The furthest I went;

$$||x|-|y||\leq |x-y| \Rightarrow -|x-y| \leq |x|-|y|\leq |x-y|$$
If we take $|x|\leq \frac{|y|}{2}$ then
$$\Rightarrow  \frac{|y|}{2} -|y|\leq |x|-|y|\leq |x-y|$$
$$\Rightarrow  -\frac{|y|}{2}\leq |x-y|$$
but still I can't conclude the result as $-\frac{|y|}{2}<0$.

答:
If $|x|<\varepsilon|y|,$ then:
$$|y-x|^2\geq ||y|-|x||^2=(|y|-|x|)^2>|y|^2(1-\varepsilon)^2$$
for a small $\varepsilon.$ So you can even quantify what exactly is sufficiently small in terms of this $\varepsilon.$


答:
I think this post is almost a duplicate. See the following:

Perspectives on Riemann Surfaces

I do recommend the recent published book by Donaldson on this subject. It is really interesting to read.

Title: Good book for Riemann Surfaces

Body: I am considering reading one of 'Algebraic curves and Riemann Surfaces' by Rick Miranda or 'Lectures on Riemann Surfaces' by Otto Forster. Which one of these is more advanced and comprehensive ? What are the differences in the approaches of these two books ?</p>


问:
Fundamental group of $(A \sqcup B)/\{A \ni (z,0) \sim (z^p,z^q) \in B\}$ where $A = B = T^2$ and $p,q$ coprime
Let $A,B$ be two copies of the torus $T^2:=\mathbb{S^1}\times\mathbb{S^1}$. Fix coprime integers $p, q \in \mathbb{Z}$. Compute the fundamental group of
$$C := A \sqcup B/ (A\ni (z, 0) \sim (z^p, z^q)\in B).$$
Clearly, we want to compute this using van Kampen. However, the glueing of the spaces confuses me.
As I have seen in another question, we could remove something simple from $C$ twice, which makes the fundamental group structure of both $U$ and $V$ easier to compute.
What would be a good choice of path-connected opens $U, V$ such that we can use their fundamental groups to compute the fundamental group of $C$?

答:
Let's take a moment to visualize the objects in question, at least in 2D.

We have a torus (with an embedded copy of $S^1$, shown squiggled):

which we're gluing to another torus along a more complicated copy of $S^1$ (also shown squiggled):

you can see that this copy of $S^1$ wraps around our square like a candy cane. In fact, before it meets itself, it wraps around the square $p$ many times "vertically",and $q$ many times "horizontally". See also here:

Let's call the first torus the "straight" torus, and the second torus the "candy cane" torus.

So what happens when we glue our toruses (tori?) together along the squiggly lines?

Well, we can do the standard Van Kampen trick: let $U$ be the straight torus, living as a subset of the final figure $C$. This isn't open, so we puff out just a little along the squiggly line into the candy cane torus. There's always an $\epsilon$ small enough to guarantee we keep a nice happy non-self-intersecting tube around the candy-cane squiggle, so this is an open set.

Similarly, we let $V$ be the candy cane torus (again, as a subset of $C$). This isn't open in the whole figure, so we puff out a little bit into the straight torus.

Now we find ourselves in the following situation:

$U$ deformation retracts onto the straight torus

$V$ deformation retracts onto the candy cane torus

$U \cup V$ is the entire figure

$U \cap V$ deformation retracts onto the squiggly $S^1$

We're now perfectly poised to use the Van Kampen theorem! This buys us a formula for the fundamental group of our whole figure:

$$\pi_1 C \cong \pi_1 U \  \underset{\pi_1 U \cap V}{\ast} \ \pi_1 V$$

Of course, we know the fundamental groups of torices and circles:

$$\pi_1 C \cong \mathbb Z^2 \underset{\mathbb{Z}}{\ast} \mathbb Z^2$$

This might be enough, but let's keep going! How might we fully compute this?

Well, we need to understand how $\mathbb{Z}$ sits inside of $\pi_1 U$ and inside $\pi_1 V$. That is, what loop does the squiggle represent?

For the straight torus, this is fairly easy: If our generators are $a$ and $b$ (say), then the squiggle corresponds to $a$.

For the candy cane torus, this takes a bit of thought. Call our generators $c$ and $d$. We want to find a homotopy from the squiggly line to some combination of these generating loops... A useful idea that shows up all the time in problems like this is to think outside the box:

From this image it becomes clear that the squiggly line is homotopic to $c^pd^q$. Just push it down towards the bottom right.

And now, when we amalgamate, we want to make the copy of $\mathbb{Z}$ generated by the straight squiggle equal to the copy of $\mathbb{Z}$ generated by the candy cane squiggle (since they're really the same squiggle!). How do we do that?

$\pi_1 U = \langle a, b \mid [a,b] \rangle$

$\pi_1 V = \langle c, d \mid [c,d] \rangle$

$\pi_1 U \cap V = \langle s \rangle$ (for squiggle)

$\pi_1 U \cap V \hookrightarrow \pi_1 U$ by $s \mapsto a$

$\pi_1 U \cap V \hookrightarrow \pi_1 V$ by $s \mapsto c^pd^q$

This looks like a lot of information, but really we're just naming the things we already knew about so that we can write down the explicit inclusion maps for the fundamental group of the squiggle.

But now we see the light at the end of the tunnel: We amalgamate to see

$$
\begin{aligned}
\pi_1 C
&= \pi_1 U \underset{\pi_1 U \cap V}{\ast} \pi_1 V \\
&= \langle a, b \mid [a,b] \rangle
\underset{\langle s \rangle}{\ast}
\langle c, d \mid [c,d] \rangle \\
&= \langle a, b, c, d \mid [a,b], [c,d], a = c^pd^q \rangle
\end{aligned}
$$

It's possible this can be simplified more, but I'll leave that to a better combinatorial group theorist than me.

I hope this helps ^_^