问:
Question concerning Galois Fields( GF(p) notation)
I've just been starting to learn about Fields and encountered Galois Fields today: Here, the question arises to me, whether there exists only one field with 15 elements, meaning that $GF(16)^{*}$ denotes only one field. How can one understand the GF() notation in general, does it denote a set of fields or something like that? (I know that my question must sound very stupid)

答:
There is no field with 15 elements, in contrary to 16.

$GF(p^k)$, the field with $p^k$ elements, is the splitting field of $x^{p^k}-x\in \Bbb{F}_p[x]$, it is unique up to isomorphism, concretely it is $\Bbb{F}_p[x]/(f(x))$ for any polynomial $f\in \Bbb{F}_p[x]$ irreducible of degree $k$.

$\Bbb{F}_p[x]/(f(x))$ is the quotient of a polynomial ring by an ideal, similar to that $\Bbb{C}=\Bbb{R}[i]=\Bbb{R}[x]/(x^2+1)$.


问:
Function of $f : \mathbb{R} → \mathbb{R}$ satisfies $f(x + y) = f(x) + f(y)$ for each $x, y ∈ \mathbb{R}.$

Suppose that $f : \mathbb{R} → \mathbb{R}$ satisfies $f(x + y) = f(x) + f(y)$ for each $x, y ∈ \mathbb{R}.$

$(a).$ Show that $f(nx) = nf(x)$ for all $x ∈ \mathbb{R}$ and $n ∈ \mathbb{Z}.$

$(b).$ Prove that $f(mx) = mf(x)$ for all $x ∈ \mathbb{R}$ and $m ∈ \mathbb{Q}.$

$(c).$ Prove that $f$ is continuous at $0$ if and only if $f$ is continuous on $\mathbb{R}$.

$(d).$ Prove that if $f$ is continuous at $0$, then there is an $m ∈ \mathbb{R}$ such that $f(x) = mx$ for all $x ∈ R$.

$\underline{Attempt}$

When $x=y=0$, we can get $f(0)=0$ and also when $x=-y$, $f(x)=-f(-x)$

$(a).$ For this part I considered  $3$ cases when $n> 0,$ $n<0$ and $n=0$

When $n=0$ it is obviously true.

When $n> 0$, $$f(nx)=f(\underbrace{x+x+x...+x}_{n})=\underbrace{f(x)+f(x)...+f(x)}_{n}=nf(x)$$

When $n<0$, $$f(nx)=f(\underbrace{(-x)+(-x)+(-x)...+(-x)}_{|n|})=\underbrace{f(-x)+f(-x)...+f(-x)}_{|n|}=nf(x)$$

$(b)$. Take $m \in \mathbb{Q}$ such that $m=\dfrac{p}{q}$
$$f(mx)=f \left(\dfrac{px}{q} \right)=f\left(p \left(\dfrac{x}{q}\right)\right)=pf\left(\frac{x}{q}\right)$$ using $y=\dfrac{x}{q}$ substitution $f(yq)=f(x)=qf(y)$ hence
$$f(mx)=f \left(\dfrac{px}{q} \right)=\frac{p}{q}f(x)$$

$(c)$.  If $f$ is continuous at $0$,

$\forall \epsilon>0$ $ \exists \delta>0$  such that $|x-0|<\delta$ and $x \in \mathbb{R}$ imply $|f(x)-f(0)|=|f(x)|< \dfrac{\epsilon}{(1+|p|)}$

let consider  $a$ and $p$ $\in\mathbb{R}$ such that $a=px$, from part $(a)$ and $(b)$

$$|f(x)-f(a)| \leq |f(x)|+|f(a)|=|f(x)|(1+|p|)<\epsilon \text{ whenever }  |x-a|<\delta_1$$

If $f$ is continuous on $\mathbb{R}$ it also continuous at $0$, So we can conclude
$f$ is continuous at $0$ if and only if $f$ is continuous on $\mathbb{R}$.

$(d).$ Define a function $f : \mathbb{R} → \mathbb{R}$ satisfies $f(x + y) = f(x) + f(y)$ for each $x, y ∈ \mathbb{R}$ such that $f(x)=mx$

given that $f$ is continuous at $0$ so from part $(c)$ $f$ is continuous on $\mathbb{R}$ and $m=f(1)$

This is what I tried. If there Is anything wrong correct me and not sure part $(c)$ and $(d)$

答:
$ a) $ : First of all $ f\left(0\right)=0 $. And for any $ x\in\mathbb{R} $, $ 0=f\left(x-x\right)=f\left(x\right)+f\left(-x\right)\iff f\left(-x\right)=-f\left(x\right)$.

$\ \ \ \ \ \ \ $Let $ x\in\mathbb{R} $ and $ n\in\mathbb{N}^{*} $, we have :
\begin{aligned} f\left(nx\right)=\sum_{k=0}^{n-1}{\left(f\left(\left(k+1\right)x\right)-f\left(kx\right)\right)}&=\sum_{k=0}^{n-1}{\left(f\left(\left(k+1\right)x\right)+f\left(-kx\right)\right)}\\&=\sum_{k=0}^{n-1}{f\left(\left(k+1\right)x-kx\right)}\\&=\sum_{k=0}^{n-1}{f\left(x\right)}\\&=nf\left(x\right) \end{aligned}

$\ \ \ \ \ \ \ $If $ n\in\mathbb{Z}_{-}=\mathbb{Z}\setminus\mathbb{N}^{*} $, then : $$ f\left(nx\right)=-f\left(\left(-n\right)x\right)=-\left(-n\right)f\left(x\right)=nf\left(x\right) $$

$\ \ \ \ \ \ \ $Thus $ \left(\forall \left(n,x\right)\in\mathbb{Z}\times\mathbb{R}\right),\ f\left(nx\right)=nf\left(x\right) $.

$ b) $ : Let $ p\in\mathbb{Z} $, and $ q\in\mathbb{N}^{*} $. $ m=\frac{p}{q} $. We have : $$ pf\left(x\right) = f\left(q\times\frac{p}{q}x\right)=q f\left(\frac{p}{q}x\right)\iff f\left(mx\right)=mf\left(x\right)$$

$\ \ \ \ \ \ \ $Thus $ \left(\forall m\in\mathbb{Q}\right),\ f\left(mx\right)=m f\left(x\right) $.

$ c) $: $ \Longleftarrow )$ If $ f $ is continuous on $ \mathbb{R} $ then it is continuous at $ 0 $.

$\ \ \ \ \Longrightarrow )$ Suppose $ f $ is continuous at $ 0 $. Let $ a\in\mathbb{R} $ :

$$ \lim_{x\to a}{f\left(x\right)}=\lim_{x\to 0}{f\left(x+a\right)}=\lim_{x\to 0}{\left(f\left(x\right)+f\left(a\right)\right)}=f\left(0\right)+f\left(a\right)=f\left(a\right) $$

$\ \ \ \ \ \ \ \ \ \ \ \ \ $Since $ a $ is an arbitrary real. then $ f $ is continuous on $ \mathbb{R} $.

$ d): $ Suppose $ f $ is continuous at $ 0 $.

$\ \ \ \ \ \ \ $ Let $ x\in\mathbb{R} $, and $ \left(x_{n}\right)_{n}\in\mathbb{Q}^{\mathbb{N}} $ such that $ x_{n}\underset{n\to +\infty}{\longrightarrow}x $.

$\ \ \ \ \ \ \ $We have for all $ n\in\mathbb{N} $ : $$ f\left(x_{n}\right)=x_{n}f\left(1\right) $$

$\ \ \ \ \ \ \ $Since $ f $ is continuous, we have $ \lim\limits_{n\to +\infty}{f\left(x_{n}\right)}=f\left(x\right) $. That means if we took $ n $ to $ +\infty $ in the previous equality, we obtain : $$ f\left(x\right)=x f\left(1\right) $$

$\ \ \ \ \ \ \ $ $ f\left(1\right) $ is a constant, it can be anything, let's denote it $ m $. We've just proven that, for any $ x\in\mathbb{R} $, we have : $$ f\left(x\right)=mx $$

ContentLicense: CC BY-SA 4.0


问:
Show that $f(x)$ cannot be zero
Let $f: \mathbb{R} \to \mathbb{R}$ be differentiable everywhere, $f’(x) = f(x)$ for all $x$ and $f(0) = 1$ then show that $f(x)$ cannot be zero for any $x$.

I can see that this will not be the case for at least $x$ strictly greater than zero. I can also see that if $f$ is zero for some $x <0$ then we can’t have $f(0) = 1$ but I don’t know how to formally prove the proposition. Please help me out.

答: Let $g(t)=f(t)f(-t)$. Then $g'(t)=f(t)f(-t)-f(t)f(-t)=0$, and so, since $g(0)=1$, $(\forall t\in\Bbb R):f(t)f(-t)=1$. So, $f$ has no zeros.

Title: Show that $f(x)$ cannot be zero


问:
Does convex and radially open imply open?
I want to show that a convex set $A$ is radially open iff $A\cap W$ is open in W, for every finite dimensional linear subspace.

Here the 'openness' we are talking about is from any normed space.

Help would be much appreciated.

Definition : We call a set $A$ of a vector space $V$ $\underline{\text{radially open}}$ if $\forall v \in V, x \in A$ $\exists \lambda \in \mathbb{R}$ such that $x + \alpha v \in A$ for any $0 < \alpha < \lambda$.

答:
It seems the following.

Let $A$ be a radially open convex subset of a vector space $V$ (that is,
 if $\forall v \in V, \forall w \in A$ $\exists \lambda>0$ such that $w + \alpha v \in A$ for any $0< \alpha < \lambda$).  Let $W$ be a finite dimensional linear subspace of $V$. Let $\{a_1,\dots,a_n\}\subset W$ be the  vertices of a simplex $\Delta$ such that $0\in\operatorname{int}_W\Delta$. Then $0\in\operatorname{int}_W\lambda\Delta$ for each $\lambda >0$. Now let $v\in A\cap W$ be an arbitrary point. Since the set $A$ is radially open, there exists a number $\lambda>0$ such that $v+\lambda a_i\in A$ for each $i$. Since the set $A$ is convex and $\Delta=\operatorname{conv} \{a_1,\dots,a_n\}$, we see that $v+\lambda\Delta\subset A$. Moreover, $v, a_1,\dots a_n\in W$ implies $v+\lambda\Delta\subset W$. Since $0 \in\operatorname{int}_W\lambda\Delta$,  $v\in v+\operatorname{int}_W\lambda\Delta\subset \operatorname{int}_W (A \cap W)$.

ContentLicense: CC BY-SA 3.0


问:
Strong clustering and degree distributions
I didn't find an anywhere good explanation of what "typical
properties of complex networks such as heterogeneous degree distributions and strong clustering" is?

In "Hyperbolic Geometry of Complex Networks" (2010) there is a correspondence between:

The hierarchical organization and heterogeneous degree distribution.

The metric structure and strong clustering.

But does strong clustering mean like some special metric structure?

Thanks in advance for your reply!

答:
I do not think there is any consensus on a precise definition for heterogeneous degrees and strong clustering, but there is a consensus on the following weakly defined criteria:

heterogeneous degree distribution means that degrees span several orders of magnitude, with a very high variance that makes the average degree of little interest; this distribution if sometimes well (but often poorly) fitted by a power law with an exponent typically around $2$.

strong clustering is generally measured as the average node clustering coefficient, which is the fraction of a node neighbours linked by an edge, aka the density of the sub-graph induced by their neighbours; if this average is orders of magnitude larger than the one in a comparable random graph, clustering is considered high, or strong; considered comparable graphs are generally the ones with the same number of vertices and edges, obtained with the ER model, or (more rarely) the ones with the same degree sequence, obtained with the CM model.

Hierarchical structure and presence of an underlying metric are interpretations for these empirical observations: various models that rely on such underlying principle lead to graphs with these features, which makes them good candidates for being their causes. Other interpretations do exist, though, like preferential attachment or node similarities, for instance.


问:
Specific permutation
Let $p : A → A$ be a permutation where $A = \{1,...., 2n\}$. Probability that all the cycles in $p$ are of length less than or equal to $n$?

答:
You can have it by remarking that $P(A) = 1- P(\overline{A})$ whith $A$ the event "all the cycles in p are of length less than or equal to n".

And $\overline{A}$ is "at least one cycle in p is of length higher or equal to n+1".

To count all such cycles, we can divide them according to the lenght of the longest cycle, so if we note $C$ the set of all those permuations, and $C_k$ the set of all the permutations for which the longest cycle is of length k, then :
$$ |C| = \sum\limits_{k=n+1}^{2n} |C_k|$$

And we notice that if $p$ a permutation has a cycle of length $k\geqslant n+1$, then all the other cycles are of lengths $k\leqslant n$.

So this gives that for $n+1 \leqslant k\leqslant 2n$, $|C_k| = \binom{2n}{k} (k-1)! (2n-k)!$.

($\binom{2n}{k}$ to choose the $k$ elements, $(k-1)!$ to permute the $k$ elements, and $(2n-k)!$ is the number of permutations with the other elements)

So we get that $|C|= \sum\limits_{k=n+1}^{2n} \binom{2n}{k} (k-1)! (2n-k)! = \sum\limits_{k=n+1}^{2n} \frac{(2n)!}{k}$.

Then $$P(A)= 1- P(\overline{A}) = 1 - \frac{|C|}{(2n)!} = 1 - \sum\limits_{k=n+1}^{2n} \frac{1}{k}$$

Title: Specific permutation


Title: Show that $f(x)$ cannot be zero

Body:
Let $f: \mathbb{R} \to \mathbb{R}$ be differentiable everywhere, $f’(x) = f(x)$ for all $x$ and $f(0) = 1$ then show that $f(x)$ cannot be zero for any $x$.

I can see that this will not be the case for at least $x$ strictly greater than zero. I can also see that if $f$ is zero for some $x <0$ then we can’t have $f(0) = 1$ but I don’t know how to formally prove the proposition. Please help me out.

Body: $f'(x)=f(x)$ implies $f'(x)f(x)=f^2(x)$, so $$f^2(x)=f^2(0)+\int^x_0 2f'(t)f(t)\,dt=f^2(0)+2\int^x_0 f^2(t)\,dt\ge f^2(0)=1$$ for $x\ge0$.


问:
How the element $x$ is different from $\{x\}$?
In Discrete Math, it's trivial to say that $x \ne \{x\}$, why is that?

答:
Because $x$ is an element (such as a real number $x\in \mathbb{R}$, or an integer, ...) while $\{x\}$ is a set where the only element is $x$ : it is called a singleton.

So $x\in \{x\}$ but these are not the same types of objects.

And if you take another element y, then $x\in \{x,y\}$ and $y\in \{x, y\}$.

I'm not sure the analogy will clarify it, but it's the same thing as a socket, and a socket in a drawer.
The socket is the only thing in the drawer now, but it's not the same as the drawer containing the socket. And you can add other sockets in your drawer...

But be careful : $x$ can also be a set itself, for instance $x=\{2, 3\}$, then $\{x\} = \{\{2, 3\}\}$ so $\{x\}$ is a set of set...


问:
What's happening when you have more variables than equations?
This is probably a dumb question but I am confused about why you can simply choose any number for the free variables in a system of equations where you have more variables than equations. I can't wrap my head around it.</p>

答:
If $x_k$ are the variables, then best what you can achieve from a single equation is $x_k=\ldots$ where the variable $x_k$ doesn't occur on the right side. Then you can substitute all occurrences of $x_k$ by that result and obtain a system with one variable and one equation less. Repeating the process will leave you with a remaining set of variables and no more equations. Such variables are free and can be set to any value without changing the system at all.

However, the result $x_k=\ldots$ cannot always be obtained. E.g. if we have an equation $x_k^2=x_1+2x_2^3$ then we would have to take a square root, which isn't unambiguous anymore. Should we take the positive or the negative solution? So only in case our equations are linear we can do the simple addition "variables with equations plus free variables = all variables".


问:
Exercise 2.11 of Waterhouse about Hopf algebra
I asked part (c) of this exercise already in this link Categorization of Group Scheme of rank 2. But I still have some difficulty solving next parts of this exercise. So I hope anyone can help me to clear it out. Thank you

Let $A$ be a Hopf algebra over $k$ (a base ring) which is free of rank 2. I proved that $I = Ker(\epsilon) = kx$ for some $x$, and $\Delta(x) = x \otimes 1 + 1 \otimes x + b x\otimes x$ for some $b$ in k. Also, $x^2 + ax = 0$ for some $a$, and $A = k[X]/(X^2 + aX)$.

(d) Use $\Delta(x^2) = (\Delta x)^2$ to show $(2 - ab)^2=2-ab$.

(e) Show $Sx = cx$ with $c^2 = 1$. Then use (e) and $0 = (S, id)\Delta x$ to show $c = 1$ and $ab = 2$.

For part (d), after replacing the expression of $\Delta x$ into $\Delta(x^2) = (\Delta x)^2$, I got $ax(2-ab)^2 = ax(2-ab)$. Because $1, x$ is a basis of $A$, so we must have $a(2-ab)^2 = a(2-ab)$. But because $k$ may be not an integral domain, so $a$ can be a zerodivisor. In this case, I can't conclude that $(2-ab)^2 = (2-ab)$.

For part (e), the fact that $Sx = cx$ for some $c^2=1$ seems to imply that $S^2 = id$, which is not stated in the book of Waterhouse. But in the book Algebraic Groups of Milne, Exercise 3.3 (a), it states that $S^2=id$. The difference between the two definitions is, the definition of Waterhouse assumes $k$ is a general ring, while the definition of Milne assumes $k$ is a field. So, is this relation always true for all Hopf algebra? And if not, is it always true in case $k$ is a field? Thank a lot for helping me.

答:
We have
$$\begin{align*}
\Delta(x^2) &= \Delta(-ax) = -a \Delta(x) \\
&= -a (x \otimes 1 + 1 \otimes x + b x \otimes x), \\
\Delta(x)^2 &= (x \otimes 1 + 1 \otimes x + b x \otimes x)^2 \\
&= x^2 \otimes 1 + 1 \otimes x^2 + b^2 x^2 \otimes x^2 + 2 x \otimes x + 2b x^2 \otimes x + 2b x \otimes x^2 \\
&= -ax \otimes 1 - 1 \otimes ax + a^2 b^2 x \otimes x + 2 x \otimes x - 2ab x \otimes x - 2ab x \otimes x \\
&= -a(x \otimes 1 + 1 \otimes x) + (2 - 4ab + a^2 b^2)(x \otimes x).
\end{align*}$$

Equating the coefficients of $x \otimes x$, we have $$-ab = 2 - 4ab + a^2 b^2.$$  Simplifying, we have
$$\begin{align*}
0 &= (ab)^2 - 3ab + 2 \\
&= (2 - ab)(1 - ab) \\
&= (2 - ab)((2 - ab) - 1 \\
&= (2-ab)^2 - (2-ab),
\end{align*}$$
so that $(2-ab)^2 = 2 - ab$ as desired.

For the rest of the question, a Hopf algebra for which $S^2 = \mathrm{id}$ is called involutive. Not all Hopf algebras are involutive (even over a field), but there are general conditions that would imply that a Hopf algebra is involutive.  Googling this word will probably turn up more references.


问:
Find the highest natural number which is divisible by $30$ and has exactly $30$ different positive divisors.

Find the highest natural number which is divisible by $30$ and have exactly $30$ different positive divisors.

What I Tried: I am not sure about any specific approach to this problem. Of course as the number is divisible by $30$ , some of the factors of the number would be $1,2,3,5,6,10,20,30$ , but that only makes $8$ divisors and there will be $12$ more which I have no idea of .

Can anyone help me?

答:
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did only a quick search with the following bound: $1\le\text{n}\le10^8$.

I wrote and ran some Mathematica-code:

```<code>In[1]:=Clear["Global`*"];
\[Alpha] = 10^8;
ParallelTable[
  If[TrueQ[Length[Divisors[n]] == 30 && IntegerQ[n/30]], n,
   Nothing], {n, 1, \[Alpha]}] //. {} -> Nothing
</code>```

Running the code gives:

```<code>Out[1]={720, 1200, 1620, 4050, 7500, 11250}
</code>```

We can see that for $1\le\text{n}\le10^8$ the number $\text{n}=11250$ is the biggest possible with the desired properties:

```<code>In[2]:=11250/30

Out[2]=375

In[3]:=Divisors[11250]

Out[3]={1, 2, 3, 5, 6, 9, 10, 15, 18, 25, 30, 45, 50, 75, 90, 125, 150, 225,
250, 375, 450, 625, 750, 1125, 1250, 1875, 2250, 3750, 5625, 11250}
</code>```

Body:
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did only a quick search with the following bound: $1\le\text{n}\le10^8$.

I wrote and ran some Mathematica-code:

In[1]:=Clear["Global`*"];
\[Alpha] = 10^8;
ParallelTable[
  If[TrueQ[Length[Divisors[n]] == 30 && IntegerQ[n/30]], n,
   Nothing], {n, 1, \[Alpha]}] //. {} -> Nothing

Running the code gives:

Out[1]={720, 1200, 1620, 4050, 7500, 11250}

We can see that for $1\le\text{n}\le10^8$ the number $\text{n}=11250$ is the biggest possible with the desired properties:

In[2]:=11250/30

Out[2]=375

In[3]:=Divisors[11250]

Out[3]={1, 2, 3, 5, 6, 9, 10, 15, 18, 25, 30, 45, 50, 75, 90, 125, 150, 225,
250, 375, 450, 625, 750, 1125, 1250, 1875, 2250, 3750, 5625, 11250}


答:
(I'm assuming you know the number theoretic way to count the number of divisors of an integer $ n = 2^a \times 3^b \times 5^c \ldots $, which is $ \sigma_0 (n) = (a+1)(b+1)(c+1) \ldots$. If not, go read up on it.)

Let $n$ be such a number.

Hint: Since $ 30 = 2 \times 3 \times 5 $, show that $ n = 2^a \times 3^b \times 5^c$.

In particular, conclude that $n$ cannot have any other prime factors.

 Note that every distinct prime factor of $n$ contributes at least one prime factor (not necessarily distinct) of $ \sigma_0 (n)$.

 Since $30 \mid n$, so $n$ has at least 3 prime factors (2, 3, 5), which contribute at least 3 (not necessarily distinct) prime factors to $ \sigma_0 (n) = 30$. Since this has 3 prime factors, we conclude that $n$ can have no other prime factors.

 Hence, $n = 2^a \times 3^b \times 5^c$.

Hence, the largest natural number is $ 5^4 \times 3^2 \times 2^1$.

Generlization: If $N = \prod p_i$ is the product of distinct primes, then the highest natural number which is divisible by $N$ and has exactly $N$ different positive divisors is $\prod p_i ^ { p_i - 1 } $.

Title: Find the highest natural number which is divisible by $30$ and has exactly $30$ different positive divisors.

Body:
Find the highest natural number which is divisible by $30$ and have exactly $30$ different positive divisors.

What I Tried: I am not sure about any specific approach to this problem. Of course as the number is divisible by $30$ , some of the factors of the number would be $1,2,3,5,6,10,20,30$ , but that only makes $8$ divisors and there will be $12$ more which I have no idea of .

Can anyone help me?


问:
Clustering analysis of a weighted graph
My data consists of a large weighted undirected graph of $n$ nodes. I need to group the nodes into $m$ clusters ($m < n$), such that nodes in a cluster are connected with heavy weights. What algorithms are available for this problem?</p>

答:
Walktrap and the Louvain algorithms work very well with weighted undirected graphs.

You may find them in most modern graph libraries.

Title: Clustering analysis of a weighted graph


问:
Is it possible to "hide" auxiliary statements introduced by universal and existential instantiations in Hilbert's system?
Let's consider a simple derivation in Hilbert's style. We start from two statements that we treat as "axioms":

$\forall x : [P(x) \implies Q(x)] $

$\forall x : P(x)$

We want to prove the following "theorem" (which is an obvious consequence of the two "axioms"):

$\forall x : Q(x)$

First, we apply the rule of "universal instantiation" to the first axiom and get the following statement:

$P(y) \implies Q(y)$

Second, we apply the rule of "universal instantiation" the second axiom and get:

$P(y)$

Now we combine the last two statements using "Modus ponens" and get the following statement:

$Q(y)$

Now we apply "generalization" the last statement to finally get what we wanted to prove:

$\forall x : Q(x)$

My problem with this proof is that between the two axioms and the final statement in the proof we have some "intermediate" statements that do not look to me like "valid" statements because they have some auxiliary variable $y$.

In other words, my expectation was that by applying derivation rules to "valid" statements (axioms of proven theorems) we always get some other "valid" statements. But it looks like it is not always the case.

So, my question is if the Hilbert's style derivation system can be "modified" a bit such that we "hide" those statements such that a proof is a directed acyclic graph with axiom being "input nodes" and the proven theorem is "output node" and, what is important, all the intermediate nodes correspond to some "valid" expressions that make sense (meaning) out of the context of the proof (by themselves).

ADDED:

The proof is taken from here.

答:
In your previous question, you seem to have been seeking a proof system where each proof consists of a list of sentences of first-order classical logic, such that every element of the list is in fact validated by the usual semantics of first-order classical logic (i.e. it holds under all assignments). Here you seek something very similar, except you wish to allow additional mathematical axioms as well. I'll focus on your previous question, since you can freely pass between the two perspectives via a deduction theorem in any Hilbert-style proof system.

The usual Hilbert system H fails your requirement only because of two perceived issues:

each H-proof consists of a list of formulae, as opposed to sentences, and

you cannot interpret what it means for a formula to be semantically valid, due to the presence of free variables.

The second is not an issue in practice. We call a sentence semantically valid if it holds under every assignment. As Eric Towers noted in the comments, the very same definition applies to formulae. Each assignment assigns values to free variables, so the formula $P(y)$ with only $y$ free in $P$ holds under every assignment precisely if its universal closure, the sentence $\forall y. P(y)$ holds under every assignment. This fact is often summarized (badly) as "free variables are implicitly universally quantified".

So if you dispense with your first requirement (that each proof consist of sentences only), the usual Hilbert system H satisfies your second one.

This leaves your first requirement only, which can now be reformulated as follows: is there a Hilbert-style proof system that generates only sentences? The answer is positive, e.g. the book Introduction to Mathematical Logic by Elliott Mendelson constructs such a Hilbert system in the section "Properties of First-order Theories" of Chapter 2 (in my edition the axioms are given in the statement of Exercise 2.28, but each edition is different).


问:
basis of k-vector space V depending on characteristic of K
Let $K$ be a field and $V=K^3$
$$B = \lbrace f_1 = (1,2,3),f_2=(-1,2,4),f_3=(2,1,5)\rbrace$$
for what characteristic of K is $B$ a basis of $V$.

I am not really sure how to approach this question besides trying different cases for the characteristic of K but that does not seem to be the right way to approach this question for me.

答:
Hint: The determinant has to be non-zero in characteristic k, for it to be a basis.

Can you find the determinant of the matrix spanned by those vectors?


问:
Card game, choose composition to minimize opponent's expected value
There are infinite cards numbered 1-5. You construct a deck. Someone guesses a value and draws a card. If they guessed correctly, they get the value. What composition of the deck that minimizes their expected value?

One way to proceed in this question is to aim to make the opponent's expected value from choosing any of the numbers 1-5 the same. If you set these expected values to be equal, you get that the proportion of each card in the deck that you construct should be inversely proportional to its value. E.g the number of cards with the number 4 should be proportional to 1/4 etc. However, I'm not clear on how we get this condition. How do we know that the composition that minimizes the opponent's expected value is also one where the opponent is indifferent between all the cards?

答:
$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5} =\frac{137}{60}$

The equal expectation approach makes each outcome $x$ happen with probability $\frac{60}{137 x}$, so if the opponent choses $x$ their expected value is a constant $\frac{60}{137}$; this makes the opponent indifferent between the card numbers.

If you choose any other distribution, then for at least one $x$ the probability is greater than $\frac{60}{137 x}$ and for another $x$ the probability is less than $\frac{60}{137 x}$; the opponent will prefer the former over the latter as it gives a higher expected value.


问:
Does newton's method work for non-roots?
So if you have $f(x)$ and the associated tangent line $y=f'(x_0)(x-x_0)+f(x_0)$, then standard newton's method is $x_{n+1} = x_n + \frac{f(x_n)}{f'(x_n)}$ because you are looking for points $(x_{root}, 0)$.

But what if you are looking for any $y$ in the range of $f$?

Can we simply use the tangent line and set $y=c$ to get $x_{n+1} = x_n + \frac{c - f(x_n)}{f'(x_n)}$?

Does it hold up in the multi-variable case $\vec{x_{n+1}}=\vec{x_n} + [Df(\vec{x_n)}]^{-1}(\vec{c}-f{\vec{x_n}})$

答:
So we must iterate

$$x_{n+1} = x_n - \dfrac{f(x_n)-c}{f'(x_n) }$$


问:
summation of this series as $ x \to \infty $ ??
given the series for the Mangoldt function $ \Lambda (n) $

$$ f(x)= \sum_{n=1}^{\infty}\frac{\Lambda (n)}{\sqrt{n}}\cos(\sqrt{x} \log n+\pi /4) $$

if we truncate the series, can we say that

$$\lim_{x\to\infty} \frac{f(x)}{x^{1/4}}=0$$

Body:
Hint:

$$ \lim_{x\to \infty} \frac{\cos( a\sqrt{x}+b )}{x^{1/4}}=0, $$

since

$$ \Big|\frac{\cos( a\sqrt{x}+b )}{x^{1/4}}\Big| \leq \frac{2}{x^{1/4}}. $$


问:
Showing explicitly by using the valuation that DVRs are Dedekind domains.
We know that DVRs are Dedekind domains by the characterization of Dedekind domains in terms of localizations at maximal ideals. There are many nice abstract ways to prove that DVRs are Dedekind domains. This is not what this question is about. I would like to see the following:

Can we see explicitly how the existence of a valuation map into $\mathbb Z$ ensures our ring is a Dedekind domain?

Here are the definitions I am using:

$A$ is a  dedekind domain if every non zero ideal $\mathfrak{a}$ has a factorization into primary ideals, $\mathfrak{a}=\prod q_i$ where the $q_i$.

$A$ is DVR if it is a valuation ring and the valuation map has image in $\mathbb Z$.

How can we see that every non zero ideal has a factorization into primary ideals using just that we have a valuation map in $\mathbb Z$.

答:
The ideals of a DVR are all of the form $(\pi^n)$ where $\pi$ is any element with valuation $1$ (let's assume that the image of $v$ is $\mathbb{Z}$ and not $k\mathbb{Z}, k >1$).

To see this, suppose $I$ is an ideal and pick $a \in I$ with minimal valuation $n$. Then $v(a/\pi^n) = 0$ so $a/\pi^n$ is a unit, so $\pi^n \in I$. This means that $(\pi^n) \subseteq I$.

Conversely, if $b \in I, b \neq 0$ then $v(b) \ge v(a)$ so $v(b/\pi^n) \ge 0$. Therefore $b/\pi^n \in R$ and therefore $b = (b/\pi^n) \cdot \pi^n \in (\pi^n)$.

So given that the ideals of $A$ are all of the form $(\pi^n)$, we immediately get unique factorization of (fractional) ideals because $(\pi^n) = (\pi)^n$.

The ideal $(\pi)$ is maximal since $x \in (\pi) \iff v(x) \ge 1$ so $A - (\pi) = \{x : v(x) = 0\} = A^\times$.


问:
Why is the total derivative of a diffeomorphism invertible?
I'm trying to brush up on some differential geometry, but there's a subtle point I don't understand. Suppose $h$ is a diffeomorphism. Then the lecture notes here suggest that it's derivative $df_x$ is an invertible linear map. Why precisely does the invertibility of $df_x$ follow from that of $f$?

Apologies if this is a trivial question - I'm a little out of practise with total derivatives!

答:
Let me give an explanation based on intuition (not a rigorous proof).

For every tangential vector $v$ at $x$, the tangential vector $df_x \cdot v$ at $f(x)$ has the following significance: if you stand at $x$ and move in the direction $v$, then your image at $f(x)$ moves in direction $df_x \cdot v$.

Suppose that $df_x$ is not invertible. Then there exists $v$ such that $df_x \cdot v = 0$. In other words, moving in direction $v$ does not change your image $f(x)$, which intuitively contradicts the idea that $f$ is a diffeomorphism.


问:
Has this been solved: $\int_{-\infty}^{\infty} \frac{dx}{x^{2n}+1}$?
It gets complicated pretty quickly.

For $n = 1$
$$ \int_{-\infty}^{\infty} \frac{1}{x^2+1}dx = \arctan(x)_{-\infty}^{\infty}$$
For $n=2$
$$\int_{-\infty}^{\infty} \frac{1}{x^4+1}dx = \left[\frac{1}{4\sqrt{2}}\log{\frac{x^2+\sqrt2x+1}{x^2-\sqrt2x+1}} + \frac{1}{2\sqrt2}\arctan{\frac{\sqrt2x}{1-x^2}}\right]_{-\infty}^{\infty}$$
Is there a formula for all $n$ and if so where can I find it.

答:
Use the known result

$$\int_0^\infty \frac {x^{a-1}}{x^m +1} = \frac\pi m \csc\frac{\pi a}m $$


问:
Probability of at least N events occuring
I have a series of N events, each with its own probability of occurring. How would I calculate the probability that at least M of the N events actually do occur?

I think this is conditional, in that getting at least M occurrences depends on getting at least M-1 occurrences. Past that I'm getting stuck.

答:
Let us consider that we have four independent events A,B,C,D each with probability Pa,Pb,Pc,Pd associated with it.
Now we have to find the probability of at least two events occurring.

First we find the probability of exactly two events occurring.

These two events can be AB, AC, AD, BC, BD, CD.

event space = {AB, AC, AD, BC, BD, CD, }

Since all these events are mutually exclusive.

Probability of any one them occurring is Sum of all probabilities in the events space.

P(AB) = Pa * Pb * (1-Pc) * (1-Pd)

P(AC) = Pa * (1-Pb) * Pc * (1-Pd)

P(AD) = Pa * (1-Pb) * (1-Pc) * Pd

P(BC) = (1-Pa) * Pb * Pc * (1-Pd)

P(BD) = (1-Pa) * Pb * (1-Pc) * Pd

P(CD) = (1-Pa) * (1-Pb) * Pc * Pd

After that we can sum all these to get the probability of Two events occurring. In a similar way we can calculate probability of Three and Four events occurring.

Final probability = Prob(Two_events) + Prob(Three_events) + Prob(Four_events)

However I am unable to generalize for at least M events out of N.

I hope it will shed more light on your problem.


答:
Well, we have the following integral:

$$\mathscr{S}_\text{n}:=\int_\mathbb{R}\frac{1}{1+x^{2\text{n}}}\space\text{d}x\tag1$$

Where $\text{n}\in\mathbb{N}$.

Let's first observe that the integrand is an even function, this implies:

$$\mathscr{S}_\text{n}=2\int_0^\infty\frac{1}{1+x^{2\text{n}}}\space\text{d}x\tag2$$

Now, we can split the integral:

$$\int_0^\infty\frac{1}{1+x^{2\text{n}}}\space\text{d}x=\int_0^1\frac{1}{1+x^{2n}}\space\text{d}x+\int_0^1\frac{x^{2\text{n}-2}}{1+x^{2\text{n}}}\space\text{d}x\tag3$$

And it is not very hard to see that we can use an infinite series to rewrite the integrands:

$$\int_0^\infty\frac{1}{1+x^{2\text{n}}}\space\text{d}x=\int_0^1\left(1-x^{2\text{n}}+x^{4\text{n}}-x^{6\text{n}}+\dots\right)\space\text{d}x+$$
$$\int_0^1\left(x^{2\text{n}-2}-x^{4\text{n}-2}+x^{6\text{n}-2}+\dots\right)\space\text{d}x\tag4$$

I will let you prove the fact:

$$\int_0^\infty\frac{1}{1+x^{2\text{n}}}\space\text{d}x=\frac{1}{2\text{n}}\sum_{\text{k}\space\in\space\mathbb{Z}}\frac{(-1)^\text{k}}{\text{k}+\frac{1}{2\text{n}}}=\frac{\pi}{2\text{n}}\csc\left(\frac{\pi}{2\text{n}}\right)\tag5$$

The last step uses the result from "An Infinite Alternating Harmonic Series".

So, in the end we end up with:

$$\mathscr{S}_\text{n}:=\int_\mathbb{R}\frac{1}{1+x^{2\text{n}}}\space\text{d}x=\frac{2\pi}{2\text{n}}\csc\left(\frac{\pi}{2\text{n}}\right)=\frac{\pi}{\text{n}}\csc\left(\frac{\pi}{2\text{n}}\right)$$

Title: Has this been solved: $\int_{-\infty}^{\infty} \frac{dx}{x^{2n}+1}$?

Body:
It gets complicated pretty quickly.

For $n = 1$
$$ \int_{-\infty}^{\infty} \frac{1}{x^2+1}dx = \arctan(x)_{-\infty}^{\infty}$$
For $n=2$
$$\int_{-\infty}^{\infty} \frac{1}{x^4+1}dx = \left[\frac{1}{4\sqrt{2}}\log{\frac{x^2+\sqrt2x+1}{x^2-\sqrt2x+1}} + \frac{1}{2\sqrt2}\arctan{\frac{\sqrt2x}{1-x^2}}\right]_{-\infty}^{\infty}$$
Is there a formula for all $n$ and if so where can I find it.


问:
Monotone Involutory Set Functions and Unions/Intersections of Families
I am currently attempting to solve the following problem in elementary set theory/introductory real analysis:

Let $A$ be a nonempty set and let $f:2^{A} \to 2^{A}$ be a function satisfying:

$X \subset Y \implies f(Y) \subset f(X)$ for all $X, Y \in 2^{A}$, i.e., it is monotone decreasing (in a certain sense, but do correct me if I'm wrong);

$f(f(X)) = X$ for all $X \in 2^{A}$, i.e., $f$ is an involution.

With these properties at hand, I'm asked to show that for any family $(X_\lambda)_{\lambda \in L}$ of subsets of $A$, we have:

$$f\left(\bigcup_{\lambda \in L} X_\lambda \right) = \bigcap_{\lambda \in L} f(X_\lambda),$$

and

$$f\left(\bigcap_{\lambda \in L} X_\lambda \right) = \bigcup_{\lambda \in L} f(X_\lambda).$$

I have been able to show that

$$f\left(\bigcup_{\lambda \in L} X_\lambda \right) \subset \bigcap_{\lambda \in L} f(X_\lambda)$$

in two steps: first, note that for all fixed $\lambda_0 \in L$, we have: $X_{\lambda_0} \subset \bigcup X_\lambda$, and therefore property 1 yields:

$$f\left(\bigcup_{\lambda \in L} X_\lambda \right) \subset f(X_{\lambda_0}),$$

and since $\lambda_0$ was arbitrarily chosen, this holds for all $X_\lambda$ in the family. It therefore follows that

$$f\left(\bigcup_{\lambda \in L} X_\lambda \right) \subset \bigcap_{\lambda \in L} f(X_\lambda),$$

seeing as the intersection of the $f(X_\lambda)$ is the largest set that is contained in each member of the family $(X_\lambda)_{\lambda \in L}$ (I have already proved this statement in a previous exercise). By means of a similar argument one can show the analogous set inclusion for the second statement.

I'm stumped as to how I would go about showing the opposite set inclusion. It would obviously require that we invoke property 2, otherwise it wouldn't be there in the first place, but even so, any naïve application of it leads me to dead ends. Any hints or solutions would be much appreciated.

P.S.: This question has already been posed here, but a proper answer to it was never given. See William Elliot's comment on it.

答:
As suggested by William Elliot's comment in the post you linked, one way of thinking about the problem is to pretend $f$ is set complementation, in which case the results you're trying to show are De Morgan's laws. So if you write down a proof of De Morgan's laws and then replace complementation with $f$, this should produce a proof of your result. Go ahead and do this on your own if you want, but I've included a proof of the missing portion of your argument below for the sake of completeness.

 We wish to show $$\bigcap_\lambda f(X_\lambda)\subset f\Big(\bigcup_\lambda X_\lambda\Big).$$ Take $x$ to be an element of the left-hand side. Then $x\in f(X_\lambda)$ for each $\lambda\in L$. Equivalently, $\{x\}\subset f(X_\lambda)$ for each $\lambda\in L$, so applying $f$ and using assumptions (1) and (2) we have $f(\{x\})\supset f(f(X_\lambda)) = X_\lambda$ for each $\lambda\in L$. Thus $$f(\{x\})\supset\bigcup_{\lambda\in L}X_\lambda.$$ Applying $f$ to the above and using (1) and (2) again, we have $$\{x\} = f(f(\{x\}))\subset f\Big(\bigcup_{\lambda\in L} X_\lambda\Big).$$ Thus $x$ is an element of the right-hand side, and the desired conclusion follows. Since $$\bigcap_\lambda f(X_\lambda) = f\Big(\bigcup_\lambda X_\lambda\Big),$$ we can prove the other relation by replacing $X_\lambda$ in the above with $f(X_\lambda)$ and applying involutivity again.


问:
Compact property
I want to prove that if $\Omega$ is complete metric space, $\mathcal{F}$ is Borel $\sigma$-algebra $\Omega$, then for any positive $\varepsilon$ there is a compact set $\mathit{K}$, that $\mathbb{P}(\mathit{K})>1-\varepsilon$. I don't know where to start. I would be very grateful for help

答:
The answer is here:

Billingsley - Convergence of probability measures (1999, Wiley)

See p.8, Theorem 1.3 and Problem 1.12, p.13

Title: Compact property


问:
A function that is continuous at $c \in \mathbb Z$ and discontinuous at $c \in \mathbb R \setminus \mathbb Z$
I'm looking for a function that is continuous on $\mathbb Z$  and discontinuous on $\mathbb R \setminus \mathbb Z$.
I know the function floor of $x$, denoted $\lfloor x \rfloor$, that is continuous at all $c \in \mathbb R \setminus \mathbb Z$ and discontinuous at all $c \in \mathbb Z$.
But I'm looking for the opposite one.
Thank you.

答:
I think the function $f$ such that :

if $x\in \mathbb{R}-\mathbb{Q}, f(x) = 0$

if $x\in \mathbb{Q}, f(x) = \sin \pi x$

will work.


问:
How is n = 4q + r here?

If it says to divide quotient by 4 , then it should q/4 = r .

答:
When n is divided by 4 , the quotient is q.

Now the value n will be equal to 4q + r.It is easily understandable from the photo now I guess.

Example is :

$i^{135}$ = Here n is divided by 4 .

We are left with 3 as remainder.

Notice , 4q + 3 = 135 where q = 33

I hope it it clear now

Title: How is n = 4q + r here?


问:
Proof of Rank–nullity theorem
I'm trying to understand the proof of Rank–nullity theorem,but there are parts that I don't understand:

Steinitz exchange lemma

If ${\displaystyle U=\{u_{1},\dots ,u_{m}\}}$ is a set of ${\displaystyle m}$ linearly independent vectors in a vector space ${\displaystyle V}$, and ${\displaystyle W=\{w_{1},\dots ,w_{n}\}}$ span ${\displaystyle V}$, then:

${\displaystyle m\leq n}$;

There is a set ${\displaystyle W'\subseteq W}$ with ${\displaystyle |W'|=n-m}$ such that ${\displaystyle U\cup W'}$ spans ${\displaystyle V}$.

Rank–nullity theorem

Let ${\displaystyle V}, {\displaystyle W}$ be vector spaces, where ${\displaystyle V}$ is finite dimensional. Let ${\displaystyle T\colon V\to W}$ be a linear transformation. Then

$${\displaystyle \operatorname {Rank} (T)+\operatorname {Nullity} (T)=\dim V}$$

Proof

Let ${\displaystyle V,W}$ be vector spaces over some field ${\displaystyle \mathbb {F} }$  and ${\displaystyle T}$ defined as in the statement of the theorem with ${\displaystyle \dim V=n}$.

As ${\displaystyle \operatorname {Ker} T\subset V}$ is a subspace, there exists a basis for it. Suppose ${\displaystyle \dim \operatorname {Ker} T=k}$

I know that $\text{ker T}$ is a subset of a finite set $V$,and hence is finite,but how does this imply that $\text{ker T}$ does have a basis?

and let ${\displaystyle {\mathcal {K}}:=\{v_{1},\ldots ,v_{k}\}\subset \operatorname {Ker} (T)}$
be such a basis.

We may now, by the Steinitz exchange lemma, extend ${\displaystyle {\mathcal {K}}}$ with ${\displaystyle n-k}$ linearly independent vectors ${\displaystyle w_{1},\ldots ,w_{n-k}}$ to form a full basis of ${\displaystyle V}$.

Let

${\displaystyle {\mathcal {S}}:=\{w_{1},\ldots ,w_{n-k}\}\subset V\setminus \operatorname {Ker} (T)}$
such that

${\displaystyle {\mathcal {B}}:={\mathcal {K}}\cup {\mathcal {S}}=\{v_{1},\ldots ,v_{k},w_{1},\ldots ,w_{n-k}\}\subset V}$
is a basis for ${\displaystyle V}$.

What is $V,W,W',U$ here? Since the proof uses Steinitz exchange lemma,however I can't recognize  $V,W,W',U$.

From this, we know that ${\displaystyle \operatorname {Im} T=\operatorname {Span} T({\mathcal {B}})=\operatorname {Span} \{T(v_{1}),\ldots ,T(v_{k}),T(w_{1}),\ldots ,T(w_{n-k})\}=\operatorname {Span} \{T(w_{1}),\ldots ,T(w_{n-k})\}=\operatorname {Span} T({\mathcal {S}})}$

Why $\text{Im}\; T=\text{span} \;T (\mathcal B)$?

And why  $\operatorname {Span} \{T(v_{1}),\ldots ,T(v_{k}),T(w_{1}),\ldots ,T(w_{n-k})\}=\operatorname {Span} \{T(w_{1}),\ldots ,T(w_{n-k})\}$?

Thanks for your help.

答:
let $v_1,v_2,...,v_n$ be a basis for $V$ then for all $v \in V$

$v=c_1v_1+...+c_nv_n$ $\Rightarrow$  $T(v)=T(C_1v_1+...+c_nv_n)=c_1T(v_1)+...+c_nT(v_n) \in \text{span} \;T (\mathcal B)$

similarly, $w \in \text{span} \;T (\mathcal B) \Rightarrow w=T(c_1v_1+...+c_nv_n) \in
Im(T)$

that is why $Im(T)=\text{span} \;T (\mathcal B)$

Title: Proof of Rank–nullity theorem


答:
$\ker T$ is a subspace, hence it is a vector space, so it has a basis.

$V$ is $V$, $W$ is any fixed basis of $V$, $U=\mathcal K$ and $W'=\mathcal S$.

It's because $\mathcal B$ is a basis of $V$, thus in particular every vector is a linear combination of them, and $T$ is linear.

$T(v_i)=0$ since $v_i\in\ker T$.

Title: Proof of Rank–nullity theorem

Body:
I'm trying to understand the proof of Rank–nullity theorem,but there are parts that I don't understand:

Steinitz exchange lemma

If ${\displaystyle U=\{u_{1},\dots ,u_{m}\}}$ is a set of ${\displaystyle m}$ linearly independent vectors in a vector space ${\displaystyle V}$, and ${\displaystyle W=\{w_{1},\dots ,w_{n}\}}$ span ${\displaystyle V}$, then:

${\displaystyle m\leq n}$;

There is a set ${\displaystyle W'\subseteq W}$ with ${\displaystyle |W'|=n-m}$ such that ${\displaystyle U\cup W'}$ spans ${\displaystyle V}$.

Rank–nullity theorem

Let ${\displaystyle V}, {\displaystyle W}$ be vector spaces, where ${\displaystyle V}$ is finite dimensional. Let ${\displaystyle T\colon V\to W}$ be a linear transformation. Then

$${\displaystyle \operatorname {Rank} (T)+\operatorname {Nullity} (T)=\dim V}$$

Proof

Let ${\displaystyle V,W}$ be vector spaces over some field ${\displaystyle \mathbb {F} }$  and ${\displaystyle T}$ defined as in the statement of the theorem with ${\displaystyle \dim V=n}$.

As ${\displaystyle \operatorname {Ker} T\subset V}$ is a subspace, there exists a basis for it. Suppose ${\displaystyle \dim \operatorname {Ker} T=k}$

I know that $\text{ker T}$ is a subset of a finite set $V$,and hence is finite,but how does this imply that $\text{ker T}$ does have a basis?

and let ${\displaystyle {\mathcal {K}}:=\{v_{1},\ldots ,v_{k}\}\subset \operatorname {Ker} (T)}$
be such a basis.

We may now, by the Steinitz exchange lemma, extend ${\displaystyle {\mathcal {K}}}$ with ${\displaystyle n-k}$ linearly independent vectors ${\displaystyle w_{1},\ldots ,w_{n-k}}$ to form a full basis of ${\displaystyle V}$.

Let

${\displaystyle {\mathcal {S}}:=\{w_{1},\ldots ,w_{n-k}\}\subset V\setminus \operatorname {Ker} (T)}$
such that

${\displaystyle {\mathcal {B}}:={\mathcal {K}}\cup {\mathcal {S}}=\{v_{1},\ldots ,v_{k},w_{1},\ldots ,w_{n-k}\}\subset V}$
is a basis for ${\displaystyle V}$.

What is $V,W,W',U$ here? Since the proof uses Steinitz exchange lemma,however I can't recognize  $V,W,W',U$.

From this, we know that ${\displaystyle \operatorname {Im} T=\operatorname {Span} T({\mathcal {B}})=\operatorname {Span} \{T(v_{1}),\ldots ,T(v_{k}),T(w_{1}),\ldots ,T(w_{n-k})\}=\operatorname {Span} \{T(w_{1}),\ldots ,T(w_{n-k})\}=\operatorname {Span} T({\mathcal {S}})}$

Why $\text{Im}\; T=\text{span} \;T (\mathcal B)$?

And why  $\operatorname {Span} \{T(v_{1}),\ldots ,T(v_{k}),T(w_{1}),\ldots ,T(w_{n-k})\}=\operatorname {Span} \{T(w_{1}),\ldots ,T(w_{n-k})\}$?

Thanks for your help.


问:
Given triangle ABC and its inscribed circle O, AO = 3, BO = 4, CO = 5, find the perimeter of ABC
As title,

I could only figure out that

$$a^2 + r^2 = 9$$
$$b^2 + r^2 = 16$$
$$c^2 + r^2 = 25$$
$$r^2 = \frac{abc}{a+b+c}$$

But couldn't get how to derive $2(a+b+c)$.

答:
In a triangle $ABC$ with sides $a,b,c$ opposite to $A,B,C$, respectively, and with incenter $I$, one has $$AI^2 = bc\cdot \frac{b+c-a}{a+b+c}, \quad BI^2 = ca\cdot \frac{c+a-b}{a+b+c}, \quad CI^2 = ab\cdot \frac{a+b-c}{a+b+c}.$$
The conditions $AI=3$, $BI=4$ and $CI=5$ can be rewritten as
$$\lambda := \frac{2abc}{a+b+c} = bc-9 = ca-16 = ab - 25.$$
One gets $$a^2=\frac{ab\cdot ac}{bc} = \frac{(\lambda+16)(\lambda+25)}{\lambda+9}$$ and similarly $$b^2=\frac{(\lambda+9)(\lambda+25)}{\lambda+16}, \quad c^2=\frac{(\lambda+9)(\lambda+16)}{\lambda+25}.$$
Substituting this into $\lambda = \dfrac{2abc}{a+b+c}$ one obtains
$$\lambda = \frac{2(\lambda+9)(\lambda+16)(\lambda+25)}{(\lambda+9)(\lambda+16)+(\lambda+9)(\lambda+25)+(\lambda+16)(\lambda+25)}$$
which, according to wolframalpha, is equivalent to $$\lambda^3-769\lambda-7200=0,$$
hence the answer is
$$\sqrt{\frac{(\lambda+16)(\lambda+25)}{\lambda+9}} + \sqrt{\frac{(\lambda+9)(\lambda+25)}{\lambda+16}} + \sqrt{\frac{(\lambda+9)(\lambda+16)}{\lambda+25}}$$ where $$\lambda = \frac{\sqrt[3]{32400 + i \sqrt{314509827}}}{3^{2/3}} + \frac{769}{\sqrt[3]{3(32400 + i \sqrt{314509827})}} \approx 31.5756803606316$$
is the positive root of $\lambda^3-769\lambda-7200=0$.


答:
I don't really like calling it "division". Instead, I would say that every $n\in \Bbb N$ can be represented as one (and only one) of the following: $$\begin{cases} 4q \\ 4q+1 \\ 4q+2 \\ 4q+3\end{cases}$$
that is, there is a $q\in \Bbb N$ and an $r\in\{0,1,2,3\}$ such that $n=4q+r$ for every $n\in\Bbb N$. That is what the question is driving at.

So, instead of calculating $i^n$, it calculates $i^{4q+r}=i^r$ because $i^4=1$, and so because $i^0=1, i^1=i, i^2=-1, i^3=-i$, we can construct that: $$i^n=\begin{cases} 1 &n= 4q \\ i &n=4q+1 \\ -1 &n=4q+2 \\-i & n= 4q+3\end{cases}$$ which is the aim of this question, to generalise $i^n$

Title: How is n = 4q + r here?

Body: If it says to divide quotient by 4 , then it should q/4 = r .


问:
Has anything useful come from Ackermann's Function?
Here is the function:

```<code>if (m == 0)
    return n + 1;
else if (n == 0)
    return A(m-1, 1);
else
    return A(m-1, A(m, n-1));
</code>```

This seems like an interesting function, especially since its values grow quite rapidly (my computer crashes if I try to run A(4,2) or A(3,10)).

From the wikipedia page it seems like it was only invented to show that a total computable function does not have to be primitive recursive. Has anything practical come from this function, other than homework problems in a computer science class?

答: We used the inverse of Ackermann's function in section 5 of our paper “Untangling planar graphs from a specified vertex position - Hard cases” devoted to straight line drawings of planar graphs. </p>


问:
Order of a subgroup H in a multiplicative group.

Let $G$ be the group of all $2 \times 2$ matrices $\left(\begin{array}{c c}a & b \\ c & d\end{array}\right)$ where $ad-bc \ne 0$ and $a,b,c,d$ are integers modulo 3, relative to matrix multiplication. Show that $o(G) = 48$. Now, If we modify the example of G by insisting that determinant = 1; then what is |G|?

I read a solution that says since this subgroup H of determinant 1 has index 2 (*), the order of this subgroup is 24 . But i am a little confused

(*) Why does this subgroup H has index 2? Because 1H forms a coset belonging to G?
But any other subgroup forms it too, no?

答:
An invertible matrix with integer entries has a determinant that is an invertible element of $\mathbb Z$, i.e. belongs to $L=\{-1,1\}$. Now, the determinant is a group homomorphism $\varphi$ between $G$ and the group $(L, \cdot)$. According to the first group homomorphism theorem, $\varphi$ induces an isomorphism between $G/H$ and $L$ where $H$ is the kernel of $\varphi$.

Therefore $H$ is a subgroup of index $2$ in $G$ as the order of $L$ is equal to two.


问:
Geometric construction (problem from Swedish 12th grade ‘Student Exam’ from 1932)
The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.

Inscribe in a given circle a quadrangle in which de two diagonals have given lengths and, in addition, where the ratio between two adjacent sides are equal to a given ratio.

I have no clue on how to solve problems of this kind, possibly since the topic was dropped in the Swedish schools in the mid 1900s and will probably never be taken up again, and I find them very difficult. (Usually the solutions are very elegant and ‘easy’ once you see them but never easy to initially spot for the untrained eye.) The problem should be solved with only a pen, a straightedge and a compass. Problems of this kind were often the final question on the exams, and therefore the hardest one. All help and hints on how to solve this problem are much appreciated.

Initial ideas

Consider the image below where the given circle has been drawn. To the left are the given ratio $r$ and the given two diagonals.

We can place the first diagonal in the circle vertically as we can always rotate the problem around $O$.

We can (I think) remove the quadangle from the problem and rewrite it as: “Set the end points $P$ or $P'$ of the second diagonal so that they are $d_2$ apart and that the distances $PA$ and $P'A$ have the ratio $r$.” There is only need to set one of the end points of the second diagonal as the other end point is easy to draw when the first end point is placed.

The image below shows three such second-diagonal lines. For the 3rd I have drawn the adjacent sides in red. As $P_i$ “slides” on the circle (and $P_i'$ tags along) we ge a ratio from $0$ to $\infty$ (or $0$ to $1$, depending on how you look at the ratio). The question is how to place the second diagonal so it results in the specified ratio. Currently I have no idea.

The original problem

答:
Comment

Not clear in what manner the adjacent sides bear a ratio $p$. Assuming equal sides are placed at opposite positions as

$( a, ap,a, ap),$ then by Ptolemy theorem

$$ a^2(1+p^2) = d_1 d_2 $$

where $ (d_1,d_2) $ are the diagonals...

Using Ruler/Compass a rectangle  can be constructed as a special (trivial?) case inside a cyclic quadrilateral.


答:
\begin{aligned}
f(x)=\frac{1}{2}\left(l(x)+r(x)-|l(x)-r(x)| \right),
\end{aligned}

where

\begin{aligned}
l(x)=\begin{cases}\lceil x \rceil-x & \text{if }x\in\mathbb{Q}\\
0 & \text{if }x\notin\mathbb{Q}\end{cases}
\end{aligned}

and

\begin{aligned}
r(x)=\begin{cases}x-\lfloor x \rfloor & \text{if }x\in\mathbb{Q}\\
0 & \text{if }x\notin\mathbb{Q}\end{cases}
\end{aligned}

Title: A function that is continuous at $c \in \mathbb Z$ and discontinuous at $c \in \mathbb R \setminus \mathbb Z$

Body:
I'm looking for a function that is continuous on $\mathbb Z$  and discontinuous on $\mathbb R \setminus \mathbb Z$.
I know the function floor of $x$, denoted $\lfloor x \rfloor$, that is continuous at all $c \in \mathbb R \setminus \mathbb Z$ and discontinuous at all $c \in \mathbb Z$.
But I'm looking for the opposite one.
Thank you.


问:
Let $A\in \mathbb{R}^{3\times 3} $ such that: ...
$A \begin{bmatrix}1&1&a\\2&1&b\\3&1&c\end{bmatrix}=\begin{bmatrix}1&1&1\\1&1&2\\1&1&3\end{bmatrix}$

and let $Y=\begin{bmatrix}1\\1\\1\end{bmatrix}$.

Then $AX=Y$ has infinite solutions for all $a,b,c$.

Why is this correct? I can't seem to understand how I am supposed to reach that conclusion (it was a multiple choice question and this was the correct answer).

Any help or explanation is appreciated.

答:
Since $A \begin{bmatrix}1&1&a\\2&1&b\\3&1&c\end{bmatrix}=\begin{bmatrix}1&1&1\\1&1&2\\1&1&3\end{bmatrix}$, you can easily see that :

$A \begin{bmatrix}1\\2\\3\end{bmatrix} = \begin{bmatrix}1\\1\\1\end{bmatrix}$

and $A \begin{bmatrix}1\\1\\1\end{bmatrix} = \begin{bmatrix}1\\1\\1\end{bmatrix}$

So for every $\lambda \in \mathbb{R}$, $A (\lambda \begin{bmatrix}1\\1\\1\end{bmatrix} + (1-\lambda) \begin{bmatrix}1\\2\\3\end{bmatrix}) = \begin{bmatrix}1\\1\\1\end{bmatrix}$.

So there are infinite solutions, and this does not depend on the values of $a,b,c$.


问:
A Hilbert class field that is not contained in the Hilbert class field of any imaginary quadratic field.
I realize that this question may be very wild and unusual.

Is every Hilbert class field, contained in a Hilbert class field of an imaginary quadratic field?

I think the answer is probably no. If the answer is no, can anyone give an explicit counter-example?

Give a counter-example for the above statement: Give a field whose Hilbert class field is not contained in the Hilbert class field of any imaginary quadratic field.

答:
No. Let $K$ be an imaginary quadratic field, and let $H$ be its Hilbert class field. Then $\mathrm{Gal}(H/K)$ is abelian, since it is isomorphic to the class group of $K$.

Since $K/\mathbb Q$ is Galois, $H$ is a Galois extension of $\mathbb Q$ whose Galois group sits inside a short exact sequence
$$1\to A=\mathrm{Gal}(H/K)\to \mathrm{Gal}(H/\mathbb Q)\to B = \mathrm{Gal}(K/\mathbb Q)\to 1,$$
where $A$ is an abelian group and $B\cong C_2$.

There are plenty of fields that cannot be a subfield of $H$ and, if $L$ is not a subfield of $H$, neither is its Hilbert class field.

An explicit example is the splitting field of $f(x) = x^5 + x +1$ over $\mathbb Q$, whose Galois group $S_5$ is not solvable.


问:
Prove that $\sum_{n=0}^\infty c_nx^n$ and $\sum_{n=0}^\infty n\log n c_nx^{n+3}$ have same radius of convergence.
Prove that $\sum_{n=0}^\infty c_nx^n$ and $\sum_{n=0}^\infty n\log n c_nx^{n+3}$ have same radius of convergence.

Now for $\sum_{n=0}^\infty c_nx^n$ the radius of convergence is $\frac{1}{\limsup\left|\frac {c_{n+1}}{c_{n}}\right|}$. Again, for the series $\sum_{n=0}^\infty n\log n c_nx^{n+3}$ we have
$$\limsup\frac{(n+1)\log (n+1) c_{n+1}|x|^{n+4}}{n\log n c_n|x|^{n+3}}<1$$
this will imply that $$|x|<\frac{1}{\limsup\left|\frac {c_{n+1}}{c_{n}}\right|\limsup\left|\frac {{n+1}}{{n}}\right|\limsup\left|\frac {\log {n+1}}{\log {n}}\right|}.$$

Now $\limsup\left|\frac {{n+1}}{{n}}\right|=1$, what should I do with $\limsup\left|\frac {\log {n+1}}{\log {n}}\right|$?

Please help me from here. Also, let me know if there is any mistake in this attempt and also if you know any other methods.

答:
Supposing the limit exists. Be $R$ the radius of convergence, this is:

$1/R = \limsup\limits_{n\to\infty} \sqrt[n]{|c_n|}$

By according to Cauchy, Hadamard's Theorem. But it's evident that:

$  \limsup\limits_{n\to\infty} \sqrt[n]{|c_n|} \leq \limsup\limits_{n\to\infty} \sqrt[n]{\frac{|c_{n+1}|}{c_n}}$

Then, we have that

$ \limsup\limits_{n\to\infty} \sqrt[n]{|c_n|} \leq \limsup\limits_{n\to\infty} \sqrt[n]{\frac{|c_{n+1}|}{c_n}}$

It's clear that,

$ \limsup\limits_{n\to\infty} \sqrt[n]{\frac{|c_{n+1}|}{c_n}} = \lim\limits_{n\to\infty} \frac{(n+1) \log (n+1)}{n \log n} = 1$,

we got this result applying L'Hopital's Rule.

this happens because whenever a limit does exist, the lim sup and lim inf also.

There is also something we should notice is that:

Suppose we have a series $\sum_{i = 0}^{\infty} c_n x^{ap + n}$ converges to the same radius of convergence r of $x^{ap} \sum_{i = 0}^{\infty} c_n x^{n}$. And this is important because of our $x^3$ in the problem. This way it doesn't make difference at all!


问:
Interesting question regarding generating function on words
Let $W_n$ be the set of all words of length n, on alephbet {a,b,c}.
Let $L$ be the maximal length of consecutive $a$ letters in a word.

A. Find the generating function of the number of words in $W_n$ such as $L<3$.

B. Find the generating function of the number of words in $W_n$ such as $L<k$.

C. Find the expectation of $L$, i.e $W(q,x)=\sum_n \sum_\pi x^n q^{L(\pi)}$. Then find its asymptotic.

My solution:

A. Every word in $W_n$= "empty word" or "a ( )" or "b(every word)" or "c(every word)".
Let $x$ count a lenght of a single number.

So, $W(x)=1+A_1(x) + 2xW(x)$

"a()" ="a" or "aa( )" or "ab(every word)" or "ac(every word)".

So, $A_1(x)=x+A_2(x)+2x^2W(x)$

"aa( )= "aa" or "aab(every word)" or "aac(every word)".

So, $A_2(x)=x^2+2x^3W(x)$.

Affer substituting we get that the gf is:

$W(x)=\frac{1+x+x^2}{1-2x-2x^2-2x^3}=\frac{1-x^3}{1-3x+2x^4}$.

B. As we did in part A:
Now it is the generalized problem given in A.

$W(x)=1+A_1(x)+2xW(x)$

$A_1(x)=x+A_2(x)+2x^2W(x)$

$A_2(x)=x^2+A_3(x)+2x^3W(x)$

$A_3(x)=x^3+A_4(x)+2x^4W(x)$
.
.

$A_{k-1}(x)=x^{k-1}+2x^kW(x)$

After solving this, we get:

$W_k(x)=\frac{1-x^k}{1-3x+2x^{k+1}}$

C.Let $x$ count the length of a single letter, $q$ counts a letter $a$.
We can do the same approach as part B, and see that:

$W_k(q,x)=\frac{1-(qx)^k}{1-(q+2)x+2q^kx^{k+1}}$.

Then, we need to find the coefficients of $W(1,x)=W(x)$
And,
$[d/dq W_k(q,x)]_{q=1}$
Then, their proportion will be the expectation.
Another way, is to look at $W^{L+1}-W_L$ then $E=\sum_L L*(W_{L+1}-W_L$).
But both ways are complicated for me, with these not simple generatinh functions.

How can the coefficients in the two generating functions be calculated (and analyzed asymptotically)?
Or in other words how can the expectation be computed and analyzed asymptoticaly?

答:
The following answer is based upon the Goulden-Jackson Cluster Method which provides a convenient technique to solve problems of this kind. We consider the set of words of length $n\geq 0$ built from an alphabet $\mathcal{V}=\{a,b,c\}$ and the set $B=\{a^k\}$ containing a bad word of length $k$, which is not allowed to be part of the words we are looking for.

Generating function $W_k(x)$: We derive a generating function $W_k(x)$ with the coefficient of $x^n$ being  the number of valid words of length $n$.

According to the paper (p.7) the generating function $W_k(x)$  is
\begin{align*}
W_k(x)=\frac{1}{1-dx-\text{w}(\mathcal{C})}\tag{1}
\end{align*}
with $d=|\mathcal{V}|$ the size of the alphabet and $\mathcal{C}$ the weight-numerator of bad words with
\begin{align*}
\text{w}(\mathcal{C})=\text{w}(\mathcal{C}[a^k])\tag{2}
\end{align*}

We calculate according to the paper:
\begin{align*}
\text{w}(\mathcal{C}[a^k])&=-x^{k}-x\text{w}(\mathcal{C}[a^k])-x^2\text{w}(\mathcal{C}[a^k])-\cdots-x^{k-1}\text{w}(\mathcal{C}[a^k])\tag{3}\\
\end{align*}
The term $x^k$ at the right-hand side represents a $k$-run of $a$. Since there might occur more than one bad words, i.e. more than one $k$-runs of $a$ we have to take care of overlappings. This is done in the expressions $x^j\text{w}(\mathcal{C}[a^k])$ which account for overlappings of $k-j$ $a$'s of two $k$-runs of $a$. Some more details of this technique is given in this answer.

We obtain from (3):
\begin{align*}
\text{w}(\mathcal{C}[a^k])\left(1+x+\cdots+x^{k-1}\right)&=-x^k\\
\text{w}(\mathcal{C}[a^k])&=-\frac{x^k(1-x)}{1-x^k}
\end{align*}
It    follows from (1) - (3):
\begin{align*}
\color{blue}{W_k(x)}&=\frac{1}{1-dx-\text{weight}(\mathcal{C})}\\
&=\frac{1}{1-3x+\frac{x^k(1-x)}{1-x^k}}\\
&\,\,\color{blue}{=\frac{1-x^k}{1-3x+2x^{k+1}}}\tag{4}
\end{align*}
and conclude the number of valid words of length $n\geq 0$ is the coefficient of $x^n$ in (4) in accordance with OPs results (a) and (b).

In the following we use the coefficient of operator $[x^n]$ to denote the coefficient of a series.

Coefficient extraction:  We obtain from (4) for $n\geq 1$:
\begin{align*}
\color{blue}{[x^n]}\color{blue}{W_k(x)}&=[x^n]\frac{1-x^k}{1-x\left(3-2x^k\right)}\\
&=[x^n]\sum_{j=0}^{\infty}x^j\left(3-2x^k\right)^j\left(1-x^k\right)\tag{4.1}\\
&=\sum_{j=0}^{n}[x^{n-j}]\left(3-2x^k\right)^j\left(1-x^k\right)\tag{4.2}\\
&=\sum_{j=0}^{n}[x^{j}]\left(3-2x^k\right)^{n-j}\left(1-x^k\right)\tag{4.3}\\
&=\sum_{j=0}^{\left\lfloor\frac{n}{k}\right\rfloor}
[x^{kj}]\sum_{l=0}^{n-kj}\binom{n-kj}{l}(-2)^l3^{n-kj-l}x^{kl}\\
&\qquad-\sum_{j=0}^{\left\lfloor\frac{n}{k}\right\rfloor}
[x^{kj}]\sum_{l=0}^{n-kj}\binom{n-kj}{l}(-2)^l3^{n-kj-l}x^{k(l+1)}\tag{4.4}\\
&\color{blue}{=\sum_{j=0}^{\left\lfloor\frac{n}{k}\right\rfloor}\binom{n-kj}{j}(-2)^j3^{n-(k+1)j}}\\
&\quad\,\,\color{blue}{-\sum_{j=1}^{\left\lfloor\frac{n}{k}\right\rfloor}
\binom{n-kj}{j-1}(-2)^{j-1}3^{n+1-(k+1)j}[[n\geq k]]}\tag{4.5}\\
\end{align*}

Comment:

In (4.1) we use the geometric series expansion.

In (4.2) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$. We also set the upper limit of the series to $n$, since other values of $j$ do not contribute.

In (4.3) we change the order of summation $j\to n-j$.

In (4.4) we use the binomial theorem and note that powers of $x$ are multiples of $k$, so we take only corresponding summands $k$: $j\to kj$.

In (4.5) we select the coefficient of $x^{kj}$ and use Iverson brackets.

Number of occurrences of $a$ (part I):

Next we want to calculate the number of occurrences of the letter $a$ in $W_k(x)$. In order to do so, we mark in (4)  the letter $a$ with $q$ and obtain the generating function
\begin{align*}
\color{blue}{W_k(x;q)}&=\frac{1}{1-(q+2)x+\frac{(qx)^k(1-qx)}{1-(qx)^k}}\\
&\,\,\color{blue}{=\frac{1-(qx)^k}{1-(q+2)x+2q^kx^{k+1}}}\tag{5}
\end{align*}
where $q$ in $(q+2)$ marks the letter $a$ and the summand $2$ respresents the occurrence of the other letters $b$ and $c$.

Before we go on it's time for a check.

Plausibility check $W_3(x;q)$:

We take as small plausibility check $k=3$, consider the bad word $\mathcal{V}=\{aaa\}$ and obtain with some help of Wolfram Alpha:
\begin{align*}
W_3(x;q)&=\frac{1-(qx)^3}{1-(q+2)x+2q^3x^4}\\
&=1+(q+2)x+(q^2+4q+4)x^2\\
&\qquad+(6q^2+\color{blue}{12}q+8)x^3\\
&\qquad+(4q^3+24q^2+32q+16)x^4+\cdots\tag{6}
\end{align*}
Here we have for instance as coefficient of $x^3$ the expression $$\left.\left(6q^2+\color{blue}{12}q+8\right)\right|_{q=1}=26$$ which gives all valid words of length $3$, namely all $3^3=27$ three-letter words from $\mathcal{V}=\{a,b,c\}$ minus the one bad word $aaa$. The blue marked term $\color{blue}{12}$ indicates we have $12$ valid words containing exactly one $a$, which are
\begin{align*}
&abb\quad abc\quad acb\quad acc\\
&bab\quad bac\quad cab\quad cac\\
&bba\quad bca\quad cba\quad cca\\
\end{align*}

Number of occurrences of $a$ (part II):

The number of occurrences of $a$ in valid words of length $n$ can be calculated as already indicated by OP as the coefficient of $x^n$ of
\begin{align*}
\left.\frac{\partial}{\partial q}W_k(x;q)\right|_{q=1}
&=\left.\frac{\partial}{\partial q}\left(\frac{1-(qx)^k}{1-(q+2)x+2q^kx^{k+1}}\right)\right|_{q=1}\\
&=\left(-\frac{kx(qx)^{k-1}}{1-(q+2)x+2q^kx^{k+1}}\right.\\
&\qquad\quad\left.\left.-\frac{(2kq^{k-1}x^{k+1}-x)\left(1-(qx)^k\right)}{\left(1-(q+2)x+2q^kx^{k+1}+\right)^2}\right)\right|_{q=1}\\
&=\frac{x-kx^k+(k-1)x^{k+1}}{\left(1-3x+2x^{k+1}\right)^2}\tag{7}
\end{align*}

Another plausibility check: We take $k=3$ in (7) and get with some help of WA:
\begin{align*}
\frac{x-3x^3+2x^{4}}{\left(1-3x+2x^4\right)^2}
=x+6x^2+24x^3+\color{blue}{92}x^4+332x^5+\cdots
\end{align*}
We look at the blue marked value $\color{blue}{92}$ and we see, the coefficient $(4q^3+24q^2+32q+16)$ of $x^4$ in (6) gives
\begin{align*}
3\cdot 4+2\cdot 24+1\cdot 32 = 92
\end{align*}
as expected.

Coefficient extraction:  We obtain from (7) similarly as we did in (4) for $n\geq 1$:
\begin{align*}
\color{blue}{[x^n]}&\color{blue}{\left.\frac{\partial}{\partial q}W_k(x;q)\right|_{q=1}}\\
&=[x^n]\frac{x-kx^k+(k-1)x^{k+1}}{\left(1-x\left(3-2x^k\right)\right)^2}\\
&=\left([x^{n-1}]\left(1+(k-1)x^k\right)-k[x^n]x^k\right)\sum_{j=0}^{\infty}\binom{-2}{j}(-x)^j\left(3-2x^k\right)^j\\
&=\sum_{j=0}^{n-1}(j+1)[x^{n-1-j}]\left(3-2x^k\right)^j\left(1+(k-1)x^k\right)\\
&\qquad-k\sum_{j=0}^n(j+1)[x^{n-j}]\left(3-2x^k\right)^jx^k\\
&=\sum_{j=0}^{n-1}(n-j)[x^{j}]\left(3-2x^k\right)^{n-j-1}\left(1+(k-1)x^k\right)\\
&\qquad-k\sum_{j=0}^n(n-j+1)[x^{j}]\left(3-2x^k\right)^{n-j}x^k\\
&=\sum_{j=0}^{\left\lfloor\frac{n-1}{k}\right\rfloor}(n-kj)
[x^{kj}]\sum_{l=0}^{n-1-kj}\binom{n-1-kj}{l}(-2)^l3^{n-1-kj-l}x^{kl}\\
&\quad+(k-1)\sum_{j=0}^{\left\lfloor\frac{n-1}{k}\right\rfloor}(n-kj)\\
&\qquad\quad\cdot[x^{kj}]\sum_{l=0}^{n-1-kj}\binom{n-1-kj}{l}(-2)^l3^{n-1-kj-l}x^{k(l+1)}\\
&\quad-k\sum_{j=0}^{\left\lfloor\frac{n}{k}\right\rfloor}(n-kj+1)
[x^{kj}]\sum_{l=0}^{n-kj}\binom{n-kj}{l}(-2)^l3^{n-kj-l}x^{k(l+1)}\\
&\color{blue}{=\sum_{j=0}^{\left\lfloor\frac{n-1}{k}\right\rfloor}(n-kj)\binom{n-1-kj}{j}(-2)^j3^{n-1-(k+1)j}}\\
&\quad\color{blue}{+(k-1)\sum_{j=1}^{\left\lfloor\frac{n-1}{k}\right\rfloor}(n-kj)
\binom{n-1-kj}{j-1}(-2)^{j-1}3^{n-(k+1)j}[[n\geq k]]}\\
&\quad\,\,\color{blue}{-k\sum_{j=1}^{\left\lfloor\frac{n}{k}\right\rfloor}(n-kj+1)
\binom{n-kj}{j-1}(-2)^{j-1}3^{n+1-(k+1)j}[[n\geq k+1]]}\\
\end{align*}


问:
A $2021$ problem: $20\sim 21$ and $43\times 47$
Notice that $2021$ is a concatenation of consecutive integers:  $20\sim 21$

Also $2021$ is a product of consecutive primes: $43\times 47$.

What is the next number with both of these properties?

$24502451$ is close, $4943\times 4957$  but $4951$ is in between.

$2484224843$ is close, $49831\times 49853$ but $49843$ is in between.

$715353612\sim 715353607$ isn't quite there but is $845785793\times 845785799$

Are there any other numbers with the $2021$ property?

Body:
I've found two more solutions, bringing the total to 4 solutions.
I've posted code here.

$794018604377235322848433897872605582\sim794018604377235322848433897872605583 = 891077215721081784886888257701070827\times891077215721081784886888257701070829$

$2518711810848159770018909254809359591672377471484881441744436703324716\sim2518711810848159770018909254809359591672377471484881441744436703324717 = 5018676928083894672666012088036109843105301546773725102790665815794437\times5018676928083894672666012088036109843105301546773725102790665815794441$

$353879205744237011544616255111782082608671961515039134082358165448687146\sim353879205744237011544616255111782082608671961515039134082358165448687147 = 594877471202462845078583461328011525336167267541426222873827376039101347\times594877471202462845078583461328011525336167267541426222873827376039101401$


问:
Prove that there is a sequence $j_1<j_2<j_3<\cdots$ such that for every $i$ we have $(x_{i,j_n})_{n=1}^\infty$ is convergent.
$x_{i,j}$, $i,j\in \Bbb N$ be real numbers bounded by $1$ i.e., $|x_{i,j}|<1$ for all $i,j \in \Bbb N$. Prove that there is a sequence $j_1<j_2<j_3<\cdots$ such that for every $i$ we have that the sequence $(x_{i,j_n})_{n=1}^\infty$ is convergent.

The way I am thinking it is that for $i=1$ we have $(x_{1,j})$ has a convergent subsequence $(x_{1,j_1})$ then we will consider $(x_{2,j_1})$ which is bounded and hence have a convergent subsequence $(x_{2,j_2})$ then we will consider $(x_{3,j_2})$ and so on inductively we will choose for $i=n$ we get convergent sequence $(x_{n,j_n})$ and then if we take sequences $(x_{1,j_n}),(x_{2,j_n}), \cdots, (x_{n,j_n})$ all of them are convergent but I can't generalise it to for all $i$. Please help here.

答:
Following my comment here is my awnser using diagonlisation. If you've never seen this before it can be a notational nightmare and I'm sorry about that. A tip would be to write out the first few terms of sequences I define below stacked on top of one another and see where the diagonliastion comes from.

We will need Bolzano-Weierstass, which sates any bounded sequence in $\mathbb{R}$ has a convergent subsequence.

As you did in the first step we extract a sequence $(j_n^{(1)})_n$ such that $(x_{1,j_n^{(1)}})_n$ converges. Next notice that $(x_{2,j_n^{(1)}})_n$ is bounded and hence we can again extract a subsequence of $(j_n^{(1)})_n$ call it $(j_n^{(2)})_n$ such that $(x_{2,j_n^{(2)}})_n$ is convergent. But also note that $(x_{1,j_n^{(2)}})_n$ is convergent and converges to the same limit as $(x_{1,j_n^{(1)}})_n$  as it is a subsequence. Repeat this process indefinitely so that we a get an infinite collection of nested sequences. i.e. we get a sequence for each $k\in \mathbb{N}$,  $(j_n^{(k)})_n$ such that $(j_n^{(k+1)})_n \subset (j_n^{(k)})_n$ for all $k$. Then we can extract the "diagonal" sequence (hence the name diagonalisation) which will be precisely the sequence $(j_m^{(m)})_m$ which has tail subsequence of $(j_n^{(k)})_n$ for all $k$. By the construction of our sequences $(j_n^{(k)})_n$ it must follow that $(x_{i,j_m^{(m)}})_m$ must be convergent for all $i$.

Hope this helps!


问:
Is a certain open subset of a connected space also connected?
If $S$ is a connected space, what additional properties are required for $S$ to make the following a theorem: Suppose $p\in S$ is a cut point of $S$, and thus, there exist open $V\subset S$ and open $W\subset S$ satisfying $S-\{p\}=V\cup W$ and $V\cap W=\emptyset$. If $A\subset S$ is open and $p\in A$, then $A\cap V\not=\emptyset$ and $A\cap W\not=\emptyset$.

If this is not true in Euclidean space in particular, can you supply a simple counter example?

答:
This should hold in general, as the converse would violate the connectedness:

Theorem Let $S$ be any topological space and $p\in S$ a cut point decomposing $S\setminus \{p\}$ into disjoint nonempty open sets $V, W$. Then any open set $A$ containing $p$ must intersect both $V$ and $W$.

Proof If $A$ would not intersect one of these sets, say w.l.o.g. $V$, then $A\cup W$ and $V$ would be a nontrivial partition of $S$ into open (and hence also closed) sets: Indeed, if both $A$ and $W$ are disoint from $V$, so is their union. Furthermore, These two sets contain all of $S$ since $S = \{p\} \cup W \cup V \subseteq A \cup W \cup V$.
Also, they are clearly open.


答:
The general idea for these is to partition $\mathbb R$ into two distinct but dense partitions.  The traditional way is $\mathbb Q$ and $\mathbb R \setminus \mathbb Q$ but any other such ways.  Then set $f(x)$ to be one thing for $x$ in one partition and $f(x)$ to be something else significantly far away for $x$ in the other partition.

For example if $f(x) = 1$ if $x\in \mathbb Q$ and $f(x)= 0$ if $x \not \in \mathbb Q$.  Such a function is clearly continuous nowhere as in any interval $(x-\delta, x + \delta)$ there will be $x_0\in \mathbb Q$ and $x_1\in \mathbb R\setminus \mathbb Q$ so $|f(x)- f(x_1)|+ |f(x)-f(x_0)| \ge |f(x_0) - f(x_1)|=1$ so either $|f(x)-f(x_1)| \ge 12$ or $|f(x) - f(x_1)| \ge \frac 12$ so we can conclude $|f(x) -f(y)| < \epsilon < \frac 12$ for all $y; |x-y| < \delta$ can not be true.

To make this continuous at some points we need $\lim_{x\to a;x \in \mathbb Q} f(x) = \lim_{x\to a; x\not \in \mathbb Q} f(x)$ for the points where we want an $a$ to be a point of continuity.

So for instance $f(x) = x$ if $x\in \mathbb Q$ and $f(x) =0$ if $x \not \in \mathbb Q$ (or vice versa) is continuous at $x = 0$ and discontinuous every else.

To make it continuous at all integers but nowhere else we can do:

$f(x) = 0$ if $x$ is irrational.  But $f(x)=0$ and $x \in \mathbb Z$ and $f(x)\ne 0$ for $x \in \mathbb Q\setminus \mathbb Z$ but $\lim_{x\to n; x\in \mathbb Q} f(x)= 0$ for all $n\in \mathbb Z$ that will be continuous at all integers.

Such an $f(x)_{x\in \mathbb Q}$ could be $f(x)= $ distance to nearest integer $= \min(x-\lfloor x\rfloor x, \lceil x\rceil - x)$.

So that will do it:

$f(x) = \begin{cases}0& x\not \in \mathbb Q\\\min(x-\lfloor x\rfloor x, \lceil x\rceil - x)&x\in \mathbb Q\end{cases}$.

It's continuous at $n\in\mathbb Z$ as $\lim_{x\to n} f(x) = \lim_{x\to n;x\in \mathbb Q} f(x)=\lim_{x\to n;x\not \in \mathbb Q} f(x)=f(n) = 0$.

But for any $a\not \in \mathbb Z$ we have $\lim_{x\to a;x\in \mathbb Q} \min(a-\lfloor a\rfloor a, \lceil a\rceil - a)\ne 0$ but $\lim_{x\to a;x\not\in \mathbb Q}f(x) = 0$.

=== Old function that turns out to be continuous on irrationals ====

$f(x)=\begin{cases} 0 & x\not \in \mathbb Q\\ 0& x\in \mathbb Z\\\frac 1b& x\in \mathbb Q\setminus \mathbb Z;x = \frac ab, a\in \mathbb Z; b\in \mathbb N; \gcd(a,b) = 1\end{cases}$


问:
Projecting a point from a diameter to the circumference of a circle.
What I want to know is if I have a unit circle centered at the origin with a diameter drawn along the y axis and I know the distance that a point on the diameter has from the center what would be the x coordinates of the points on the circumference with the same y coordinates. further more how would you repeat this process on bigger or smaller circles.</p>

答:
Alternatively, you can solve this using trigonometry.

Let's say the point along the diameter (diameter being twice the radius, $r$) above the origin is $y$, $(0,y)$.
The angle the swept radius makes from the horizontal axis so that the $y$ component of the triangle is,$$y=r\cdot \sin(\theta)$$The points of this triangle being: $(0,0), (x,0),$ and $(x,y)$.

We can solve for $\theta$ using $$\theta = \sin^{-1}\left(\frac{y}{r}\right)$$
Then, we can solve for the x-component using$$x=r\cdot \cos(\theta)$$
or,$$x=\frac{y}{\tan(\theta)}$$
Since you want both sides of the x-axis, the xs you are looking for are $\pm x$ of the above.


问:
Integrate $x^3\sqrt{x^2-9}\,dx$ with trig substitution
$\newcommand{\arcsec}{\operatorname{arcsec}}$This expression can be integrated without trig substitution, but I wanted to try using it anyways and got a different answer.

Starting with finding the indefinite integral $$\int x^3\sqrt{x^2-9} \, dx$$
I then substituted $x = 3\sec\theta$, $\theta = \arcsec(\frac{x}{3})$, and $d\theta = \frac{3}{\sqrt{x^2-9} |x|} \, dx$.
Plugging these values into the original integral,
$$\int 27\sec^3\theta\sqrt{9\sec^2\theta - 9}\frac{3}{\sqrt{9\sec^2\theta-9} |3\sec\theta|} \, d\theta$$
$$\int \frac{81\sec^3\theta3\tan\theta}{3\tan\theta |3\sec\theta|} \, d\theta$$
$$\int \frac{81\sec^3\theta}{|3\sec\theta|} \, d\theta$$
$$\int 27\sec^2\theta \, d\theta$$
$$27\tan\theta + C$$
Now substituting $x$ back in and simplifying:
$$27\tan(\arcsec(\frac{x}{3})) + C$$
$$9\operatorname{sgn}(x)\sqrt{x^2-9} + C$$

This does not seem close at all to the solution I found by $u$-substitution,
$$\frac{(x^2-9)^\frac{3}{2}(x^2+6)}{5} + C$$

I am relatively new to integration so I think I made a mistake. What did I do wrong? Any help appreciated.

答:
The mistake is in the differential $d\theta$. You accidentally switched $d\theta$ for $dx$. If you fix that it should be OK. Alternatively, differentiate $x=3\sec{\theta}$ on both sides, so you obtain:
$$dx=3\sec{\theta}\tan{\theta}\,d\theta$$

Try to do the rest from here


问:
Reference requested for homotopy theory theorem
I came across this post:
Homotopy groups of compact topological manifold
which states exactly the result I need for a theorem I'm working on.  However, I would need a reference, since the audience need not be very well-versed in homotopy theory.

Could someone suggest where I can find the result:

Theorem: Every closed, connected smooth $d$-manifold $M$ has a continuous and not nullhomotopic map $f: S^{d'} \rightarrow M$ for some sphere $S^{d'}$ with $1 \leq d' \leq \dim(M)$.

In other words, if $M$ is a closed and connected smooth manifold then there is a non-trivial $\pi_{d'}(M)$ for some $d'\leq \dim(M)$.

答:
This is not a reference but a short proof:

if not, then with $d'=1$ we see that $M$ would have to be simply-connected.

In particular, if its homology groups all vanish, then $M$ is contractible. But the homology groups in dimension $> \dim(M)$ always vanish, and the hypothesis implies (by Hurewicz) that the homology groups in dimension $\leq \dim(M)$ vanish too.

This implies that $M$ is contractible, which is impossible by Poincaré duality (either mod $2$, or integrally because $M$ is simply-connected)

More simply put: $M$ is mod $2$-orientable, so it must have nontrivial mod $2$-cohomology, this must be in dimension $\leq \dim(M)$, but the hypothesis implies it doesn't, by the Hurewciz theorem.


问:
$\int \frac{\cos x}{x} dx $ using contour integration
I saw a solution of how to use complex analysis to integrate
$\int_{-\infty}^{\infty} \frac{\cos x}{x} dx$ which uses the contour that traverses an upper semi circle of radius $R$ in the positive direction followed by the line segment from $-R$ to $-\epsilon$ then the upper semi circle of radius $\epsilon$ in the negative direction followed by the line segment from $\epsilon$ to $R$. Everything makes sense to me except computing $\lim_{\epsilon \rightarrow 0} \int_{C_\epsilon} \frac{e^{iz}}{z}dz$. Everywhere I have looked, the limit is computed by taking the limit inside the integral. Why can this be done? I think it has to do with dominated convergence but what is the dominating integrable function?

Thanks in advance.

答:
This is a shameless adaptation from my own set of notes, but it does help (me) tie a few things together mentioned in the relevant PDF linked below the question.

Consider the principal value integral \begin{equation}\label{principalval}I = \text{P} \int_{-\infty}^\infty \frac{f\left(x\right) \: dx}{x-x_0} \:,\end{equation} where by shifting $x \rightarrow z$, we assume that $f\left(z\right)$ is analytic except for a finite number of poles, and that $\left|f\right| \rightarrow 0$ on the upper (or lower) infinite semicircle in the complex plane.

Since the pole $x_0$ lies on the real axis, the integration contour cuts directly through $x_0$. This is handled by regularization of the denominator, which entails introducing a small factor $\delta>0$ as $$I = \text{P} \int_{-\infty}^\infty \frac{f\left(x\right) \: dx}{x-x_0} = \lim_{\delta\rightarrow 0} \int_{-\infty}^\infty \frac{\left(x-x_0\right)f\left(x\right) \: dx}{\left(x-x_0\right)^2 + \delta^2} = \lim_{\delta\rightarrow 0} \oint_C \frac{\left(z-x_0\right)f\left(z\right) \: dz}{\left(z-x_0\right)^2 + \delta^2} \:.$$ After a little complex algebra, find $$I = \lim_{\delta\rightarrow 0} \oint_C \frac{f\left(z\right) \: dz}{z-x_0+i\delta} + \lim_{\delta\rightarrow 0} \oint_C i\delta \frac{f\left(z\right) \: dz}{\left(z-x_0-i\delta\right)\left(z-x_0+i\delta\right)} \:,$$ which indicates one simple pole $z_0 = x_0 + i\delta$ inside the upper-half plane. The first integral in fact excludes the pole, so $x_0$ is skipped in subsequent residue calculations. (Use the $\delta$-term as a reminder to skip $x_0$.) The second integral is solved by standard residue calculus, i.e., let $g\left(z\right)=f\left(z\right)/\left(z-x_0+i\delta\right)$, resulting in $\pi i f\left(x_0\right)$.

Pulling the results together, we write \begin{equation}\label{principalvaltwoterms}I^+ = \pi i f\left(x_0\right) + \lim_{\delta\rightarrow 0} \oint_C \frac{f\left(z\right) \: dz}{z-x_0+i\delta} \:,\end{equation} where if we started with $\delta<0$ instead, the integration contour would flip to the lower-half plane, resulting in $$I^- = -\pi i f\left(x_0\right) + \lim_{\delta\rightarrow 0} \oint_C \frac{f\left(z\right) \: dz}{z-x_0-i\delta} \:.$$ In tighter notation (regardless of path or the sign of $\delta$), one may write \begin{equation}\label{principalvalcomplex}I = \text{P} \int_{-\infty}^\infty \frac{f\left(x\right) \: dx}{x-x_0} = \text{P} \oint_C \frac{f\left(z\right) \: dz}{z-x_0} \:,\end{equation} reminding us to include $x_0$ inside integration contour, but take the residue with a factor of $1/2$.

So for instance, to evaluate $$I = \int_{-\infty}^\infty \frac{\sin x}{x} \: dx \:,$$ the answer is almost too trivial to show off the method, but here it is:

$$I = \text{Im} \left( \pi i \: e^{i \cdot 0} + \lim_{\delta\rightarrow 0} \oint_C \frac{e^{iz} \: dz}{z+i\delta} \right) = \text{Im} \left( \pi i \: e^{i \cdot 0} + 0 \right) = \pi$$

By the same apparatus, we can show the cosine-version to be zero.


问:
Proving a statement about a logarithmic function
I'm trying to prove the function I've contracted is giving me a negative result in some equation but I'm having trouble doing it either with a direct way or by contradiction as I'm new to the log functions..

My function is $f:(0,\infty)->R$ such that $f\left(x\right)\ =\ \ln (x)-0.5$

I need to show that
$$f\left(x\right)\cdot f\left(\frac{1}{x}\right)<0$$
$$\left(\ln\left(x\right)-0.5\right)\cdot\left(\ln\left(\frac{1}{x}\right)-0.5\right)<0$$

After I'm done with proving that (is it even correct?) I need to assume f(x) is continous at 1 and that f(1) does not equal zero..

答:
This is not true since we have

$$f(1) \cdot f\left(\frac11\right) = f(1)^2 \ge 0$$

To make things more fun, let's study when does $f(x) < 0$.

For $f(x) < 0$, we need $\ln (x) < 0.5$ and hence $x < e^\frac12$.

Hence $f\left(\frac1x\right) < 0$ we need $x > e^{-\frac12}$.

$$f(x) f\left(\frac1x \right) < 0 \iff x  < e^{-\frac12} \lor x > e^{\frac12}$$


问:
Clever method to integrate $ \int \frac{\sqrt{x}}{1+x^{2}}\,dx $
I'm curious in knowing if there is any other method to solve this indefinite integral
$$
\int \frac{\sqrt{x}}{1+x^{2}}\,dx
$$

besides set $ \sqrt{x} = t $ and obtain
$$
\int \frac{2t^{2}}{1+t^{4}}\,dt
$$
where I reduce the problem to
$$
\int \frac{t^{2}}{1+t^{4}}\,dt
$$
and factorize $1+t^{4}=(t^{2}+\sqrt{2}t+1)(t^{2}-\sqrt{2}t+1)$
to reduce the integral to

$$
\int \frac{t^{2}}{(t^{2}+\sqrt{2}t+1)(t^{2}-\sqrt{2}t+1)}\,dt
$$
This is not impossible with partial fraction but rather tedious.

答:
Here is a shortcut

\begin{align}
\int \frac{\sqrt{x}}{1+x^{2}}\,dx
&= \int \frac{\frac1{\sqrt{x}}}{x+\frac1x}\,dx
=\int \frac{d(\sqrt x- \frac1{\sqrt{x}}) + d(\sqrt x+\frac1{\sqrt{x}}) }{x+\frac1x}\\
&= \int \frac{d(\sqrt x- \frac1{\sqrt{x}}) }{(\sqrt x-\frac1{\sqrt x})^2+2} +\int \frac{d(\sqrt x+\frac1{\sqrt{x}}) }{(\sqrt x+\frac1{\sqrt x})^2-2}\\
&=\frac1{\sqrt2}\tan^{-1}\frac{ \sqrt x-\frac1{\sqrt x}}{\sqrt2}
-\frac1{\sqrt2}\coth^{-1}\frac{ \sqrt x +\frac1{\sqrt x}}{\sqrt2}+C
\end{align}


答:
$\ln \frac {1}{x} = -\ln x$

$(\ln x - \frac 12)(-\ln x - \frac 12)<0\\
-(\ln x)^2 + \frac 14<0\\
\frac 14< (\ln x)^2\\
\frac 12< |\ln x|$

Your proposition is only true when x is outside the interval $[e^{-\frac 12}, e^{\frac 12}]$

Title: Proving a statement about a logarithmic function

Body:
I'm trying to prove the function I've contracted is giving me a negative result in some equation but I'm having trouble doing it either with a direct way or by contradiction as I'm new to the log functions..

My function is $f:(0,\infty)->R$ such that $f\left(x\right)\ =\ \ln (x)-0.5$

I need to show that
$$f\left(x\right)\cdot f\left(\frac{1}{x}\right)<0$$
$$\left(\ln\left(x\right)-0.5\right)\cdot\left(\ln\left(\frac{1}{x}\right)-0.5\right)<0$$

After I'm done with proving that (is it even correct?) I need to assume f(x) is continous at 1 and that f(1) does not equal zero..


答:
HINT

Make the change of variable $x = 3\cosh(u)$. Thus we get that
\begin{align*}
\int x^{3}\sqrt{x^{2} - 9}\mathrm{d}x & = 243\int\cosh^{3}(u)\sinh(u)\sqrt{\cosh^{2}(u) - 1}\mathrm{d}u\\\\
& = 243\int\cosh^{3}(u)\sinh^{2}(u)\mathrm{d}u\\\\
& = 243\int\cosh(u)\cosh^{2}(u)\sinh^{2}(u)\mathrm{d}u\\\\
& = 243\int\cosh(u)(1 + \sinh^{2}(u))\sinh^{2}(u)\mathrm{d}u\\\\
& = 243\int[\cosh(u)\sinh^{2}(u) + \cosh(u)\sinh^{4}(u)]\mathrm{d}u\\\\
& = 81\sinh^{3}(u) + \frac{243\sinh^{5}(u)}{5} + C
\end{align*}

Can you take it from here?

Title: Integrate $x^3\sqrt{x^2-9}\,dx$ with trig substitution

Body:
$\newcommand{\arcsec}{\operatorname{arcsec}}$This expression can be integrated without trig substitution, but I wanted to try using it anyways and got a different answer.

Starting with finding the indefinite integral $$\int x^3\sqrt{x^2-9} \, dx$$
I then substituted $x = 3\sec\theta$, $\theta = \arcsec(\frac{x}{3})$, and $d\theta = \frac{3}{\sqrt{x^2-9} |x|} \, dx$.
Plugging these values into the original integral,
$$\int 27\sec^3\theta\sqrt{9\sec^2\theta - 9}\frac{3}{\sqrt{9\sec^2\theta-9} |3\sec\theta|} \, d\theta$$
$$\int \frac{81\sec^3\theta3\tan\theta}{3\tan\theta |3\sec\theta|} \, d\theta$$
$$\int \frac{81\sec^3\theta}{|3\sec\theta|} \, d\theta$$
$$\int 27\sec^2\theta \, d\theta$$
$$27\tan\theta + C$$
Now substituting $x$ back in and simplifying:
$$27\tan(\arcsec(\frac{x}{3})) + C$$
$$9\operatorname{sgn}(x)\sqrt{x^2-9} + C$$

This does not seem close at all to the solution I found by $u$-substitution,
$$\frac{(x^2-9)^\frac{3}{2}(x^2+6)}{5} + C$$

I am relatively new to integration so I think I made a mistake. What did I do wrong? Any help appreciated.


问:
When can we interchange the order of limits of the expected value of products of random variables?
Suppose that the sequence of random variables $X_n$ converges to a limiting random variable $X$, and let $Y$ be an arbitrary random variable.

Is there any specific mode of convergence of the sequence $X_n$ that guarantees that
$$\lim_{n \to \infty} E[X_n Y] = E[X Y]?
$$
I'm pretty sure that convergence in distribution is not good enough, but I'm not sure about (e.g.) convergence in probability or convergence a.s.

What about if there are multiple sequences (i.e. $E[X_n Y_n]\stackrel{?}{\to} E[X Y]$), or if $X$ or $Y$ are constant or otherwise constrained? (I know that if $Y$ is constant, then it suffices for $X_n$ to converge in mean.)

答:
As it was mentioned by Dasherman, we can get some conditions, using Hölder's inequality.

Here is the result:

Statement 1: if $p, q \ge 1$ where $\frac1{p} + \frac1{q} = 1$,$X_n \to X$ in $L_p$ and $Y$ in $L_q$ then
$$ |E(X_n - X)Y| \le ||(X_n-X) Y ||_1 \le ||X_n -X||_p || Y||_q \to 0.$$

What may we say if conditions of statement 1 don't hold true?

Notice that $X_n$, $X$ and $Y$ should be defined in common probability space. Otherwise $EX_nY$ or $EXY$ are not defined.

Suppose that $X_n \to X$ in probability. Hence $(X_n,Y) \to (X,Y)$ in probability and $X_n Y \to X Y $  in probability.

Statement 2a:

If  $X_n \to X$ in probability and $|X_n Y| \le \xi$ for some $\xi$ such that $E\xi < \infty$ then $EX_n Y \to EXY$. Moreover, $E|X_n Y - XY| \to 0$.

Proof: as $X_n Y \to X Y $  in probab. then Statement 2 follows from Dominated convergence theorem.

With the help of notion of uniform integrability assumptions may be weakened.

Statement 2b: If  $X_n \to X$ in probability and if $\{X_n Y\}_{n\ge 1}$ is uniformly integrable class of r.v. then $EX_n Y \to EXY$. Moreover, $E|X_n Y - XY| \to 0$.

Statement 2a follows from statement 2b.

What may we say if we have only weak convergence?

Statement 3a: If  $X_nY \to XY$ in distribution and if $\{X_n Y\}_{n\ge 1}$ is uniformly integrable class of r.v. then $EX_n Y \to EXY$.

Proof. By Skorokhod's representation theorem there are $Z_n$ and $Z$ such that $Z_n \to Z$ a.s., distribution of $Z_n$ coinsides with distribution of $X_n Y$ and distribution of $Z$ coinsides with distribution of $X Y$. Obviously, $Z_n$ is uniformly integrable class of r.v. Thus $EX_n Y = EZ_n \to  EZ = EXY$.

First corollary 3a. If  $X_nY \to XY$ in distribution and  $\sup_n E G(|X_n Y|) < \infty$ for some non-negative increasing convex function $G(t)$ such that $\frac{G(t)}t \to \infty$ as $t \to \infty$ then $EX_n Y \to EXY$.

Proof: Corollary 3a follows immediately form de la Vallée-Poussin theorem and Statement 3a.

Second corollary 3a. If  $X_nY \to XY$ in distribution and  $\sup_n E |X_n Y|^{1 + \varepsilon} < \infty$ for some $\epsilon > 0$ then $EX_n Y \to EXY$.

Proof: follows immediately form First corollary 3a.

Statement 3b: If  $X_n \to X$ in distribution, $P(Y = const) =1 $  and if $\{X_n \}_{n\ge 1}$ is uniformly integrable class of r.v. then $EX_n Y \to EXY$.

Proof. As $X_n \to X$ in distribution and $Y = const$ a.s. then $X_n Y \to X Y$ and statement 3b follows from statement 3a.

First corollary 3b. If  $X_n \to X$ in distribution, $P(Y = const) =1 $  and  $\sup_n E G(|X_n|)< \infty$ for some non-negative increasing convex function $G(t)$ such that $\frac{G(t)}t \to \infty$ as $t \to \infty$ then $EX_n Y \to EXY$.

Proof: similarly to proof of First Corollary 3a.

Second corollary of 3b. If   $X_n \to X$ in distribution, $P(Y = const) =1 $  and  $\sup_n E |X_n|^{1 + \epsilon} < \infty$ for some $\epsilon > 0$ then $EX_n Y \to EXY$.

Proof: similarly to proof of Second Corollary 3  a.

Conclusion.
If $X_n \to X$ in $L_p$  then statement 1 works.
If $X_n \to X$ in probability (or, moreover, a.s.) then statements 2a and 2b work.
If $X_n \to X$ in distribution, last statements work.


问:
Prove that $a ( n ) = b( n + 2)$

let $a(n)$ denotes the number of ways of expressing the positive integer $n$ as an ordered sum of 1's and 2's. Let $b(n)$ denote the number of ways of expressing n as an ordered sum of integers greater than 1. prove that  $a(n) = b(n+2)$. for $n=1,2,3...$

My approach:
$a(1) = 1$ (only) , $b(3) = 3$

$a(2) = 2$, because $2=1+1=2$ ,and  $b(4)=2$, because $4=2+2=4$

$a(3) = 3$, because $3=1+1+1=1+2=2+1$ and, $b(5) = 3$ , because $5=2+3=3+2=5$

$a(4)= 5$, because $4=1+1+1+1=2+2=1+1+2=1+2+1=2+1+1$ and, $b(6)=5$, because $6=3+3=2+4=4+2=2+2+2=6$

$a(5)=8$, because $5=1+1+1+1+1=2+1+1+1=1+2+1+1=1+1+2+1=1+1+1+2=2+2+1=2+1+2=1+2+2$ and, $b(7)=8$, because
$7=3+2+2=2+3+2=2+2+3=3+4=4+3=2+5=5+2=7$

By this way i am able to show that  $a(n)=b(n+2)$. but is there any general method for this problem. I mean any recursion relation which  i can  understand.

Background:-This problem is from pathfinder for Olympiad mathematics.

答:
For compositions into ones and twos we have the OGF

$$A(z) = \sum_{q\ge 1} (z+z^2)^q =
\frac{z+z^2}{1-z-z^2}
= -1 + \frac{1}{1-z-z^2}.$$

Compositions into parts at least two have OGF

$$B(z) = \sum_{q\ge 1} (z^2+z^3+\cdots)^q
= \frac{z^2/(1-z)}{1-z^2/(1-z)}
= \frac{z^2}{1-z-z^2}.$$

We then have for $n\ge 1$

$$[z^n] A(z) = [z^n] \left(-1 + \frac{1}{1-z-z^2}\right)
= [z^n] \frac{1}{1-z-z^2}
\\ = [z^{n+2}] \frac{z^2}{1-z-z^2} = [z^{n+2}] B(z)$$

as claimed.

Title: Prove that $a ( n ) = b( n + 2)$


问:
The positive solution of $a x e^{2x} + b x e^x - c = 0$ where $a,c > 0$
I've stumbled upon the equation
$$f(x) := a x e^{2x} + b x e^x - c = 0$$
where $a,c > 0$ while doing my research. I'm only interested in the positive solution, i.e. $x>0$. It is easy to show a unique positive solution exists since $f(0)=-c < 0$ and if $f(x_0) > 0$ then $f(x)>0, \forall x\geq x_0$ (obviously such $x_0$ exists). This shows the existence of the unique positive solution because of continuity.

I'm interested in the analytical expression of the solution (if such an expression exists). I tried to put this equation in one of the forms in https://arxiv.org/pdf/1902.08910.pdf without any luck.

Is there any known solution to this kind of equation or am I stuck with numerical solutions?

答:
Since $a>0$, WLOG we can fix it as $a=1$ and

consider zeros of the function
with two parameters instead of three,
\begin{align}
f(x) &= x \exp(2x) + \alpha x \exp(x) - \beta = 0
\tag{1}
\end{align}

where
$\alpha=\tfrac{b}{a}$, $\beta=\tfrac{c}{a}$.

Unfortunately, I don't think there is a closed form solution for (1),
but the numeric can be just fine.
As a heuristic for initial guess of the positive root of (1)
the root $x_0$ of a closely related equation

\begin{align}
f_2(x) &= x^2 \exp(2x) + \alpha x \exp(x) - \beta = 0
,
\tag{2}
\end{align}

which has a positive root,
\begin{align}
x_0&=
\operatorname{W}\Big(-\tfrac12\,\alpha+\tfrac12\,\sqrt{\alpha^2+4\beta}\Big)
,
\end{align}
can be used.

Here is an example python code, which uses
Halley's method
to iteratively find the approximation of the root of (1)
for $\alpha=1.2$, $\beta=1.6$,
which is approximately $x\approx 0.39903258116278510968$:

```<code>import decimal
decimal.getcontext().prec = 20
exp=decimal.getcontext().exp

from scipy.special import lambertw as W
from numpy.lib.scimath import sqrt as sqrt

a=1.2; b=1.6
alpha = decimal.Decimal(a)
beta = decimal.Decimal(b)

def f(x):
  return x*exp(2*x) + alpha*x*exp(x) - beta

def df(x):
  return exp(x)*((1+2*x)*exp(x)+alpha*(1+x))

def ddf(x):
  return exp(x)*(4*exp(x)*(1+x)+alpha*(2+x))

def F(x):
  fx=f(x)
  dfx=df(x)
  ddfx=ddf(x)
  return x-2*fx*dfx/(2*dfx**2-fx*ddfx)

def fx0():
  return W(1/2*(-a+sqrt(a*a+4*b))).real

x=decimal.Decimal(fx0()); print(x)
# 0.49006785880157982537

x=F(x); print(x)
x=F(x); print(x)
x=F(x); print(x)
x=F(x); print(x)

# 0.39948996502974891810
# 0.39903258122349558881
# 0.39903258116278510968
# 0.39903258116278510968
</code>```

And this is a contour plot of solutions of (1)
for $\alpha\in[-3,3]$, $\beta\in[0,6]$, $x=0.1,\dots,1.2$:

Body:
Since $a>0$, WLOG we can fix it as $a=1$ and

consider zeros of the function
with two parameters instead of three,
\begin{align}
f(x) &= x \exp(2x) + \alpha x \exp(x) - \beta = 0
\tag{1}
\end{align}

where
$\alpha=\tfrac{b}{a}$, $\beta=\tfrac{c}{a}$.

Unfortunately, I don't think there is a closed form solution for (1),
but the numeric can be just fine.
As a heuristic for initial guess of the positive root of (1)
the root $x_0$ of a closely related equation

\begin{align}
f_2(x) &= x^2 \exp(2x) + \alpha x \exp(x) - \beta = 0
,
\tag{2}
\end{align}

which has a positive root,
\begin{align}
x_0&=
\operatorname{W}\Big(-\tfrac12\,\alpha+\tfrac12\,\sqrt{\alpha^2+4\beta}\Big)
,
\end{align}
can be used.

Here is an example python code, which uses
Halley's method
to iteratively find the approximation of the root of (1)
for $\alpha=1.2$, $\beta=1.6$,
which is approximately $x\approx 0.39903258116278510968$:

import decimal
decimal.getcontext().prec = 20
exp=decimal.getcontext().exp

from scipy.special import lambertw as W
from numpy.lib.scimath import sqrt as sqrt

a=1.2; b=1.6
alpha = decimal.Decimal(a)
beta = decimal.Decimal(b)

def f(x):
  return x*exp(2*x) + alpha*x*exp(x) - beta

def df(x):
  return exp(x)*((1+2*x)*exp(x)+alpha*(1+x))

def ddf(x):
  return exp(x)*(4*exp(x)*(1+x)+alpha*(2+x))

def F(x):
  fx=f(x)
  dfx=df(x)
  ddfx=ddf(x)
  return x-2*fx*dfx/(2*dfx**2-fx*ddfx)

def fx0():
  return W(1/2*(-a+sqrt(a*a+4*b))).real

x=decimal.Decimal(fx0()); print(x)
# 0.49006785880157982537

x=F(x); print(x)
x=F(x); print(x)
x=F(x); print(x)
x=F(x); print(x)

# 0.39948996502974891810
# 0.39903258122349558881
# 0.39903258116278510968
# 0.39903258116278510968

And this is a contour plot of solutions of (1)
for $\alpha\in[-3,3]$, $\beta\in[0,6]$, $x=0.1,\dots,1.2$:


问:
Linear algebra - a unit, U, vector that is perpendicular to u. How many possibilities for U?
Find a unit vector u in the direction of v = (3, 4). Find a unit vector U that is
perpendicular to u. How many possibilities for U?

u should be (3/5,4/5) And because of perpendicular to U. So, u dot U = 0.

The solution is the only perpendicular unit vector would be -U = (4/5, -3/5)

My question is why it can't be (-4/5, 3/5)

答:
Below we can see a plot of the vectors $U=(3,4)$, $u=(\frac{3}{5},\frac{4}{5})$, $\omega =(-\frac{4}{5}, \frac{3}{5})$ and $w$, where $\omega=-w$. Here we can clearly see, graphically, that:

$$\omega \perp u$$ and

$$w \perp u$$

Computing dot products we can also see that:

$$\omega \cdot u=0$$ and

$$w \cdot u=0$$

So we arrive at the same conclusion. There are two vectors that are perpendicular to $u$ ($\omega$ and $w$), as is always the case with nonzero vectors in $\mathbb{R}^2$, and such vectors are always the negatives of each other.


问:
Homomorphisms between $\mathbb{Q}(x)$ and $\mathbb{C}$
I want to describe all homomorphisms between the ring of rational $\Bbb{Q}(x)$ functions and the ring of complex numbers $\Bbb{C}$.

my idea: Let $\phi:\Bbb{Q}(x)\longrightarrow\Bbb{C}$ be a homomorphism such that $\phi(1)=1$. Then it is easy to show that $\phi|_{\Bbb{Q}}=\rm id_{\Bbb{Q}}$. Then for any $\frac{p(x)}{q(x)}\in\Bbb{Q}(x)$, we have $\phi(p(x)/q(x))=\frac{p(\phi(x))}{q(\phi(x))}$. Then $\phi$ is determined by the image og $x$. therefore $x$ must go to some transcendental number in $\Bbb{C}$, otherwise $q(\phi(x))$ maybe $0$. Therefore there exists an unique homomorphism $\phi_{\alpha}$ for any transcendental number $\alpha\in\Bbb{C}$. Am I right about this? Is this description perfect?

答: Yes, your description is correct, and moreover it uses no special features of $\mathbb{C}$: this is a description of homomorphisms from $\mathbb{Q}(x)$ to any field, which exhibits the universal property of $\mathbb{Q}(x)$ in the category of fields, namely, it is the free field on a transcendental element.


问:
transformation of trigonometric graphs
Thomas calculus for the following function:

$$
  f(x) = A\ Sin(\frac{2π}B (x-C))+D
$$

says that |A| is the amplitude , |B| is the period, C is the horizontal shift, and D is the vertical shift.

which general formula,

horizontal shifts have the form:  y=f(x+h)

vertical shifts have the form:    y=f(x)+h

period is supposed to be the positive number p such that f(x+p)=f(x)

I can see how A is the amplitude and D is the vertical shift, but I don't really get how |B| is the period and C is the horizontal shift

I could simplify the function to this:

$$
  f(x) = A\ Sin(\frac{1}B (x-C))+D
$$

what else can we do here? How do we simplify this to the above mentioned general form or something to make the value of the horizontal shift and period more popping out clearly in the equation? How does the period become B and the horizontal shift C?

答:
To illustrate how $|B|$ is the period, simply evaluate $f(x+\lvert B \rvert)$,
\begin{align}
f(x+\lvert B \rvert) &= A \sin\Big(\frac{2\pi}{B}(x+|B|-C)\Big) +D \\
&=A\sin\Big(\frac{2\pi}{B}(x-C) \pm 2\pi \Big) + D \\
&=f(x)
\end{align}
so that $|B|$ is a period.  The $\pm$ enters the equation because $B$ and $|B|$ may have different signs and the period by convention is usually taken to be positive.  Strictly you need to show $|B|$ is the least such number, which you can try do do, possibly using the addition formula for the sine function.

That $C$ is the $x$-shift should be obvious by your own definition since it applies a fixed subtraction directly and only to the $x$ variable.

Title: transformation of trigonometric graphs


问:
Graph of a single point continuous function.
We often come across functions like f(x)= x, x belongs to rational and -x, when x belongs to irrational, which are continuous at just one (or more isolated) point, in this example at x=0. I googled to look at the graph of such a function, with no luck. Even desmos wouldn't let me type such a function. So, does anyone know how to find graph of such a function, can anybody share it here?</p>

答: One can imagine what the graph would look like: since the rationals and irrationals are both dense, the graph would appear to be an X shape: the union of both the lines $f(x)=x$ and $f(x)=-x$. Of course, if one could "zoom in infinitely", one would see that the lines are not really lines, but points that are infinitesimally close.