问:
The set of irreducible representations (over C) determines a finite group
I'm trying to understand group representations, and I think that this statement is correct (where determines means up to isomorphism), although I couldn't find a proof online (without references to more complicated things like "Tannaka duality"), so I tried to prove it, but I may have overlooked something:

Let $G = \{x_1, ..., x_n\}$, $H = \{y_1, ..., y_n\}$ be two finite groups of same cardinality.

Suppose G and H have the same set of irreducible representations in the following sense :
there is a finite set of morphisms $g_i$ (resp. $h_i$), each one from G (resp. H) to
some subgroup of $GL(n_i, \Bbb{C})$ and we can write $g_i(x_j) = P_i h_i(y_j) P_i^{-1}$ for some invertible matrix $P_i$ and all $j$.

Then if we consider the regular representation of G, we can write it as a sum of the $g_i$ representations
(with $n_i$ giving the multiplicity) and do the same for H.

That shows that the two regular representations are isomorphic as we can build a new base using a block-diagonal transition matrix using $(P_i)$.
As they are also faithful (isomorphic to the groups themselves), then G and H are isomorphic.

Is that correct ?

EDIT: I updated the "same set of irreducible representations" following the comments (that also requires the additional assumption of same cardinality)

答:
I think the "correct" way to deduce that $G$ is isomorphic to $H$, without requiring bijection between the two group is the following.

Define a straightening system $\mathcal{S}$ as a finite collection of vector spaces  $V_0, \ldots, V_n$ over $\mathbb{C}$, equipped with an hermitian product, of dimensions $d_0=1, d_1, \ldots, d_n$, plus the following data:

Numbers $n_{ij}^k$ such that
$$ d_i d_j = \sum_k n_{ij}^k d_k$$

A permutation $\sigma: \in S_{n+1}$ such that $\sigma(0) = 0$;

Unitary isomorphisms $T_{ij} : V_i \otimes V_j  \to \oplus_k (V_k) ^{n_{ij}^k} $

Unitary isomorphisms $\phi_i : V_i \to V_{\sigma(i) }$;

For $W$ a vector space equipped with an hermitian product, let $U(W) $ be the space of unitary automorphisms. Also, for $f: V \to W$ an unitary isomorphism, define
$$E(f) : U(V) \to U(W) $$
Be the conjugation by $f$.
Define the little tannaka group of the straightening system as
$$\tau(\mathcal{S}) := \{ (u_i) \in \prod_i U(V_i) : E(T_{ij}) (u_i \otimes u_j) = \oplus_k n_{ij}^k u_k, E(\phi_i) (u_i) = u_{\sigma(i)} $$

The group structure is given by componentwise multiplication. We have the following

Proposition.  *Let $G$ be a group, and let $\mathcal{S}(G) $ be the straightening system defined in this way:

Vector spaces are the irreducible representations and $V_0$ is the trivial representation, and the hermitian product is the only positive definite invariant by the $G$ action;

$n_{ij}^k = \langle \chi_i \chi_j, \chi_k \rangle $, where $\chi_s$ is the character of the representation $V_s$;

The permutation is such that $V_i^* \simeq V_{\sigma(i) }$;

Isomorphisms $T_{ij}$ are given by the representation isomorphism
$$ V_i \otimes V_j \simeq_G \oplus_k (V_k) ^{n_{ij}^k}$$

Isomorphism $\phi_i$ is given by the repn isomorphism
$$ V_i^* \simeq V_{\sigma(i) }$$

Then the map

$$ G \to \tau(\mathcal{S}(G) ) $$
given by $g \mapsto (\rho_i(g) ) $ is an isomorphism of groups*.

This is not easy to show, but it is true. Let me remark two facts:

The data of spaces and numbers $n_{ij}^k$ alone does not determine the group structure; one can show that these data are equivalent to give the character table of $G$, which is notoriously not equivalent to give the group;

The additional data of the isomorphisms are enough to determine the group, and two groups will be isomorphic one to each other iff the two straightening systems are isomorphic. Morally the flexibility you have in the straightening system is the choice of the isomorphism $T_{ij}$; you can multiply this by a constant on each summand $V_k$ of $V_i \otimes V_j$ and it will still be an isomorphism of representations. What the proposition says is that this is the only flexibility you have.


问:
I found this $\binom{2p^2}{k}\equiv 2\binom{p^2}{k}\pmod {p^2},1\le k<p^2$ ,But How prove it?
today,when I deal this problem,I found this result:

let $p>3$ be prime number, show that
$$\binom{2p^2}{k}\equiv 2\binom{p^2}{k}\pmod {p^2},1\le k<p^2$$

I found this when $k=1,2,3$ it is easy to prove it,because

$k=1$
$$\binom{2p^2}{k}=\binom{2p^2}{1}=2p^2\equiv 0=2\binom{p^2}{1}\pmod {p^2}$$
$k=2$
$$\binom{2p^2}{k}=\binom{2p^2}{2}=p^2(2p^2-1)\equiv 0=2\binom{p^2}{2}\pmod {p^2}$$
$k=3$
$$\binom{2p^2}{k}=\binom{2p^2}{3}=\dfrac{2p^2(2p^2-1)(2p^2-2)}{6}\equiv 0=2\binom{p^2}{3}\pmod {p^2}$$
but for all postive integer $k$,I think is also right.But I can't prove it

答:
Let $n$ be a positive integer and $1\le k<n$. By a straightforward counting argument we have the identity
$$\binom{2n}{k}=\sum_{a,b\ge 0, a+b=k}\binom{n}a\binom{n}b.$$
Setting $n=p^2$ and rewriting the equation gives
$$\binom{2p^2}k=2\binom{p^2}{k}+\sum_{a,b\ge 1, a+b=k}\binom{p^2}a\binom{p^2}b.$$
It therefore suffices to show that $p\mid\binom{p^2}a$ for all $1\le a<p^2$ and this follows immediately by expanding $x^{p^2}+1=(x+1)^{p^2}$ modulo $p$ using the binomial theorem.

ContentLicense: CC BY-SA 4.0


问:
Prime Numbers...
Are there any functions whose output is a prime number?

For eg. The function $x^2$ spits out squares of $x$; is there a similar function for Prime Numbers as well(which gives output as Prime Numbers)?

答: Conway's prime algortihm (see also) finds all primes using a "simple" rule if I am not mistaken.</p>

Title: Prime Numbers...


问:
Why is it that $\int^x_{-\infty} \frac{e^t}{t}dt=\gamma+\ln x+\sum^\infty_{n=1}\frac{x^n}{n\cdot n!}$?
According to MathWorld, the exponential integral $\operatorname{Ei}(x)=\int^x_{-\infty} \frac{e^t}{t}dt$ can be written as follows:

$$\int^x_{-\infty} \frac{e^t}tdt=\gamma+\ln x+\sum^\infty_{n=1}\frac{x^n}{n\cdot n!}$$

Where $\gamma$ is the Euler-Mascheroni  constant.

However, I am curious to know how exactly this was derived.

I was able to show that the derivative of both sides w.r.t. $x$ are the same:

$$\frac d{dx}\int^x_{-\infty} \frac{e^t}{t}dt=\frac d{dx}\left[\gamma+\ln x+\sum^\infty_{n=1}\frac{x^n}{n\cdot n!}\right]$$

$$\frac{e^x}x=\frac d{dx}[\gamma]+\frac d{dx}[\ln x]+\sum^\infty_{n=1}\frac d{dx}\left[\frac{x^n}{n\cdot n!}\right]$$

$$\frac{e^x}x=\frac1x+\sum^\infty_{n=1}\frac{nx^{n-1}}{n\cdot n!}$$

$$\frac{e^x}x=\frac1x+\sum^\infty_{n=1}\frac{x^{n-1}}{n!}$$

$$e^x=1+\sum^\infty_{n=1}\frac{x^n}{n!}$$

$$e^x=\sum^\infty_{n=0}\frac{x^n}{n!}$$

Which we know is true by the Maclaurin expansion of $e^x$. However, this only shows that $\int^x_{-\infty} \frac{e^t}tdt$ and $\gamma+\ln x+\sum^\infty_{n=1}\frac{x^n}{n\cdot n!}$ differ by a constant.

答:
Lets define:
$$F(x)=\int_{-\infty}^x\frac{e^t}tdt\tag{1}$$
like you said we can say:
$$\frac{e^t}t=\frac1t\sum_{n=0}^\infty\frac{t^n}{n!}=\frac1t+\sum_{n=1}^\infty\frac{t^{n-1}}{n!}$$
and so:
$$\int \frac{e^t}tdt=\ln(t)+\sum_{n=1}^\infty\frac{t^n}{n.n!}+C$$

We would like to find a point, $x$, so that we can find the exact solution ourselves and so work out this integration constant. Obviously the problem is the function around $t=0$. If we call our antiderivative $g(t)$ then we know that:
$$g(t)=\ln(t)+\sum_{n=1}^\infty\frac{t^n}{n.n!}$$
$$F(x)=g(x)-\lim_{t\to-\infty}g(t)$$

we will have a look at this limit:
$$L=\lim_{t\to-\infty}\ln(t)+\sum_{n=1}^\infty\frac{t^n}{n.n!}$$

Now remember that the definition of the Euler-Mascheroni constant is:
$$\gamma=\lim_{n\to\infty}\sum_{k=1}^n\frac1k-\ln(n)$$

can you show that $L=-\gamma$?


问:
Elementary combinatorics
You are given a string of $10$ characters, that can either be 'A' or 'B'. How many of them do not contain the string 'ABA'?

What I thought was to adopt a recursive approach. Supposing we know the number $n_7$ of strings of $7$ characters that don't contain 'ABA', then we have a choice of 8 places to put that string in, so there will be $2^{10}-(8 \cdot n_7)$ strings that satisfy the request. Similarly, we can find that $n_7$ is equal to $2^7-(5 \cdot n_4)$, and that $n_4=2^4-(2 \cdot n_1)$.

Since $n_1$ is obviously equal to $2$, we can subsitute that in and get

$n_4=2^4-4=12$

$n_7=2^7-60=68$

$n_{10}=2^{10}-544=480$

Is my reasoning correct?

答:
For nonnegative integers $n,k$ with $k\le 2$, let $f(n,k)$ be the number of strings $s$ of length $n+k$ containing at least one substring matching 'ABA' and such that the first $k$ characters of $s$ match the first $k$ characters of 'ABA'.

Then $f(n,k)$ can be computed recursively as in the following Maple code

Explanation:

If $n < 3-k$, there's not enough space for an occurrence of 'ABA', so $f(n,k)=0$.

Next assume $n\ge 3-k$.

If $k=0$, the number of qualifying strings is

$f(n-1,0)$ if the next character is 'B'.$\\[4pt]$

$f(n-1,1)$ if the next character is 'A'.

If $k=1$, the number of qualifying strings is

$f(n-1,1)$ if the next character is 'A'.$\\[4pt]$

$f(n-1,2)$ if the next character is 'B'.

If $k=2$, the number of qualifying strings is

$f(n-1,0)$ if the next character is 'B'.$\\[4pt]$

$2^{n-1}$ if the next character is 'A'.

Applying the recursion, we get

\begin{array}{|c|c|c|c|}
\hline
n&f(n,0)&f(n,1)&f(n,2)\\
\hline
0&0&0&0\\
\hline
1&0&0&1\\
\hline
2&0&1&2\\
\hline
3&1&3&4\\
\hline
4&4&7&9\\
\hline
5&11&16&20\\
\hline
6&27&36&43\\
\hline
7&63&79&91\\
\hline
8&142&170&191\\
\hline
9&312&361&398\\
\hline
10&673&759&824\\
\hline
\end{array}

Thus, since $f(10,0)=673$, the number of strings of length $10$ which contain no substring matching 'ABA' is $2^{10}-673=351$.

Title: Elementary combinatorics


问:
Derivative of a function at a constant point
Let $f(x)=\left\{\begin{array}{c}x^2\sin(1/x), x\ne 0\\0, x=0\end{array}\right.$

Why is it possible to calculate the derivative of the function when $x\neq 0$ "regularly" using the chain rule, and Lebiniz rule, but for the case when $x=0$ you have to calculate it by the definition of the derivative instead, and not just say that the derivative of a constant function is $0$?

答: Because this is NOT a "constant function"!  The fact that $f(0)= 0$ does not make it a "constant function".  Every function has some specific value at $x= 0$  (or at any $x$).  The fact that the value happens to be $0$ does not matter.


问:
Integral of $1/x^2$ without power rule
I was wondering if it was possible to evaluate the following integral without using the power rule for negative exponents

\begin{equation*}
  \int \frac{1}{x^2} \; dx
\end{equation*}

When using integration by parts, you end up with the same integral in the rhs so it seems out of luck

\begin{equation*}
  \int \frac{1}{x^2} \; dx = \frac{\ln x}{x} + \int \frac{\ln x}{x^2} \; dx
\end{equation*}

This question is inspired by this blackpenredpen's video

Using integration by parts with $u = \frac{1}{x^2}$ and $v = x$ is NOT accepted. If you write

\begin{equation*}
  \int \frac{1}{x^2} \; dx = \frac{1}{x} + 2 \int \frac{1}{x^2} \; dx
\end{equation*}

you are still implicitly using the power rule to compute the derivative of $\frac{1}{x^2}$ so it is not correct per the rules.

The same rules apply for $u$-substitution which implicitly use the power rule. See the following with $u = \frac{1}{x}$ such that

\begin{equation*}
    \int \frac{1}{\left(\frac{1}{x}\right)^2} \left(\frac{1}{x}\right)' \; dx = \int \frac{1}{u^2} \; du
\end{equation*}

and here the power rule is also considered used to compute the derivative of $\frac{1}{x}$ although it can be subject to discussion (geometric proof, limit definition of derivative, etc...)

答:
Let $f(x)=-1/x\;(x>0)$. Then
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h=\lim_{h\to0}\frac1h\left(\frac1x-\frac1{x+h}\right)=\lim_{h\to0}\frac1{x(x+h)}=\frac1{x^2}.$$Hence, for some constant $c$,
$$\int\frac1{x^2}\,\mathrm dx= -\frac1x+c.$$

Title: Integral of $1/x^2$ without power rule


问:
Orthonormal basis in $\mathbb{R^3}$ with a parameter
Let $v_1=\frac{1}{\sqrt{2}}(1,1,0)^{T},v_2=\frac{1}{\sqrt{2}}(1,-1,0)^{T}, v_3=(t+1,t+1,t)^{T}$ three vectors in $\mathbb{R^3}$ and $t\in \mathbb{R}$. For which values of $t$ is there an orthogonal application which maps canonical basis in $(v_1,v_2,v_3)$? I've been trying to solve and a necessary and sufficient condition is that $(v_1,v_2,v_3)$ is an orthonormal basis (?).

Then what remains is a matter of calculations...

答:
We have that $v_1 \perp v_2$ and that $v_2 \perp v_3$.

Furthermore

$$v_1 \perp v_3 \iff 2t+2=0 \iff t=-1 \iff v_3=(0,0,-1).$$

Consequence: for $t=-1$ we have that $(v_1,v_2,v_3)$ is an orthonormal basis.


答:
Note that the limit of your function is defined as
\begin{equation}
f^{\prime}\left(0\right) = \lim_{x\to 0}\frac{f\left(x\right)-f\left(0\right)}{x} = \lim_{x\to 0}\frac{x^{2}\sin{1/x}}{x} = \lim_{x\to 0}\frac{f\left(x\right)-f\left(0\right)}{x} = \lim_{x\to 0}{x\sin{1/x}}.
\end{equation}
Besides, a constant typically means a constant over a neighborhood. It is meaningless to say "constant over a single point". Your function only says that it is $0$ at $0$. It is not a "constant" in any neighborhood around $0$.

Title: Derivative of a function at a constant point

Body:
Let $f(x)=\left\{\begin{array}{c}x^2\sin(1/x), x\ne 0\\0, x=0\end{array}\right.$

Why is it possible to calculate the derivative of the function when $x\neq 0$ "regularly" using the chain rule, and Lebiniz rule, but for the case when $x=0$ you have to calculate it by the definition of the derivative instead, and not just say that the derivative of a constant function is $0$?


问:
Is the language $L=\{<M>|M \text{ has 24 states}\}$ recursive?
I know that I cannot use the rice theorem because there are two turing machines that calculate the same function but have different ammount of states. What techniqu can I use to prove that this language is not recursive.</p>

答:
I will assume that by $<M>$ you mean "an encoding of the machine $M$ as an integer".

It depends on your encoding of Turing Machines.
We generally suppose that this encoding is at least computable, so that we can actually use Universal Turing Machines, and so on. Hence, you should be able to retrieve the number of states of $M$ given its encoding. Therefore, it is clear that the language $L$ is recursive, and you recognize it using the machine that "decodes" its input $<M>$ and checks whether the number of states of $M$ is equal to 24.


问:
If $W_1 \subseteq V' \subseteq W_1 \oplus W_2 = V$, then $V' = W_1 \oplus (V' \cap W_2)$

Question: Let $W_1$ and $W_2$ be subspaces of a vector space $V$ such that $V = W_1 \oplus W_2$.  Prove that for every subspace $V'$ of $V$ containing $W_1$ one has $V' = W_1 \oplus (V' \cap W_2)$.

Attempt:

Let $W_1$ be a subset of $V'$ spanned by the basis vectors $u_1,\dots,u_r$. Then $W_1$ can similarly be completed to a basis of $V'$, say $u_1,\dots,u_r,u_{r+1},\dots,u_s$. $W_2$ is such that it is completed by the basis vectors $u_{r+1},\dots,u_n$, so a vector $x\in V'\cap W_2$ can therefore be written as a linear combination of the basis vectors $u_{r+1},\dots,u_s$, which are linearly independent. Hence,  $u_{r+1},\dots,u_s$ spans $V'\cap W_2$, so $V'=W_1+(V'\cap W_2)$, and clearly $W_1\cap (V'\cap W_2)=\{0\}$, so $V'=W_1\oplus (V'\cap W_2)$.

答:
Baisis of $W_1: \ \ u_1,...,u_r:=A$
Basis of $W_2: \ \ v_1,..,v_s:=B$

Since $W_1$ is completly in $V'$ , $A$ can be extendet to a Basis of $V'$. But Now again $V'$ is a subspace of $V:=W_1\oplus W_2$. This means that the basis of $V'$ , call it $C$ can be written as a lin. combination of $A$ and $B$. Hence the only way to extend $A$ to $C$ is by taking the remaing vectors from $B$ that complete $C$ , or equivalently : $C=A+B\cap C$ and since all vectors in  $A+B\cap C$ are lin. independet we can write $V'=W_1\oplus (V'\cap W_2)$

ContentLicense: CC BY-SA 3.0


答:
"...and not just say that the derivative of a constant function is $0$".

There is no interval $(-\delta,\delta)$ in which the function is equal to the constant $0$ for all $x$ in this interval.


问:
The one point compactification is actually a topology
I'm practicing for my topology exam, for this I want to show that the one point compactification is in fact a topology. The definition I am working with:

Let $X$ be a topological space. The $\textit{one point compactification}$ of $X$ is the topological space $\hat{X} = X \sqcup \{*\}$, where a subset $U$ is open if either $U$ is an open subset of $X$ or $* \in U$ and $X\setminus U$ is compact and closed.

I have a solution, which seems long and untidy, so I'm looking for advice how to streamline the argument. Also, if you find any mistakes, let me know.

$\textit{Proof}$

Denote the topology of $\hat{X}$ by $\mathcal{T}$. We check the axioms:

$\emptyset \subseteq X$, $\emptyset$ open in $X$ $\Rightarrow \emptyset \in \mathcal{T}$. Further, $X \setminus \hat{X} = \emptyset$, which is closed and compact, therefore $\hat{X} \in \mathcal{T}$.

Let $A_i \in \mathcal{T}$ for $i \in I$. If $A_i \subseteq X$ for all $i$, then $\bigcup_{i}A_i \subseteq X$ is open as a union of open sets, and so $\bigcup_{i}A_i \in \mathcal{T}$. If $* \in \bigcup_{i}A_i$, we need to show that $\left(\bigcup_{i}A_i\right)^c = \bigcap_{i}A_i^c$ is closed and compact in $X$. It is closed in $X$ as an (arbitrary) intersection of sets that are closed in $X$. If $* \notin A_j$, then $A_j$ is open in X, so $A_j^c$ is closed. If $* \in A_j$, then $A_j^c$ is closed in $X$ by the definition of $\mathcal{T}$. To see that $\bigcap_{i}A_i^c$ is compact, let $k$ be such that $* \in A_k$, so $A_k^c$ is closed and compact. Now $\bigcap_{i}A_i^c \subseteq A_k$, i.e. a closed subset of a compact set, therefore also compact.

Let $A_1, A_2 \in \mathcal{T}$. If $* \notin A_1$ and $* \notin A_2$, then $* \notin A_1 \cap A_2$ and $A_1 \cap A_2$ is open in $X$ as a finite intersection of open sets. If $* \in A_1$ and $* \in A_2$, then $* \in A_1 \cap A_2$ so we need to check that $\left(A_1 \cap A_2\right)^c = A_1^c \cup A_2^c$ is closed and compact in $X$. It is closed as the finite union of closed sets. For compactness, let $U_i, i \in I$ be an open cover of $A_1^c \cup A_2^c$. It's also an open cover of $A_1^c$ and $A_2^c$. Since both sets are compact, they admit finite subcovers, say $A_1^c \subseteq \bigcup_{j = 1}^{n}U_j =: O_1$ and $A_2^c \subseteq \bigcup_{k = 1}^{m}U_k =: O_2$. Then  $O_1 \cup O_2$ is a finite subcover of $A_1^c \cup A_2^c$ which is therefore compact. For the case where $*$ is in only one of the sets, w.l.o.g. $* \in A_1$. Then, $* \notin A_1\cap A_2$, so we need to show that $A_1 \cap A_2$ is open in X. Let $\tilde{A_1}:= A_1 \setminus \{*\} = A_1 \cap X$, then $A_1 \cap A_2 = \tilde{A_1} \cap A_2$. Further, $A_2$ is open by definition and $\tilde{A_1}$ is open in X since $\tilde{A_1}^c = \left(A_1 \cap X \right)^c = A_1^c \cup \emptyset = A_1^c$, which is closed. So $A_1 \cap A_2$ is open as a finite intersection of open sets.

Thanks a lot for checking!

答:
You're on the right track, you just need to clean up a bit. You were right thinking it was a bit long, but it wasn't wrong, so there's at least that. I'm re-writing how I think it should be written, so I may repeat some of your steps with a different notation.

Open subsets of $X$ are also open subsets of $\hat X$, so $\varnothing$ is open in $\hat X$. But your definition is a tiny bit unclean because normally we only take the complement of a subset $A \subseteq X$ and then write $X \setminus A$ in those cases only, but you take $U \subseteq \hat X$ and consider $X \setminus U$. To make things clearer, I would write $X \cap (\hat X \setminus U)$. (Some may allow this abuse of notation, it's a matter of taste.) It is clear with this notation that $X \cap (\hat X \setminus \hat X) = X \cap \varnothing = \varnothing$ is closed and compact, thus $\hat X$ is open in $\hat X$.

Given a family $\{A_i\}_{i \in I}$ of open subsets of $\hat X$, if $\bigcup_{i \in I} A_i$ is contained in $X$, then the same is true of each $A_i$, and therefore being open in $\hat X$ is equivalent to being open in $X$, so their union is open in $X$, and therefore also in $\hat X$. If this is not the case, let $j \in I$ be such that $* \in A_j$. Since that means $* \in A_j \subseteq \bigcup_{i \in I} A_i$, we have to check that $\hat X \setminus \bigcup_{i \in I} A_i = \bigcap_{i \in I} (\hat X \setminus A_i)$ is closed and compact. But $\bigcap_{i \in I} (\hat X \setminus A_i) \subseteq \hat X \setminus A_j$, and since the latter is closed and compact, and the former is an intersection of closed sets (and therefore closed), the intersection is a closed subset of a compact subset, and therefore compact (this is true in any topological space).

We can use a little trick here: if $A$ is open in $\hat X$, then $A \cap X$ is open in $X$. That's because it's trivial if $A \subseteq X$ by definition, and if $* \in A$, then $\hat X \setminus A = X \setminus (A \cap X)$ is closed in $X$, and in particular $A \cap X$ is open in $X$. So if $A_1, A_2$ are open in $\hat X$ but $*$ is not in $A_2$, then $A_1 \cap A_2 = (A_1 \cap X) \cap A_2$ is an intersection of two open subsets of $X$, and therefore is open in $X$. To finish the proof that open sets in $\hat X$ are stable under taking finite intersections, we may assume $* \in A_1 \cap A_2$, which means we can assume $\hat X \setminus A_1$ and $\hat X \setminus A_2$ are both compact and closed. Now we have to show that $\hat X \setminus (A_1 \cap A_2) = (\hat X \setminus A_1) \cup (\hat X \setminus A_2)$ is compact and closed. Both of these are subsets of $X$, so the question is equivalent to asking if in a topological space $X$, the union of two compact and closed subsets is compact and closed. The 'closed' part is trivial by the axioms of a topology, and the 'compact' part is trivial because if we have an open cover of the union of two compact sets $F_1$ and $F_2$, then a finite open subcover covers $F_1$, and another finite open subcover covers $F_2$, so both together cover the union.

This is, however, very lengthy. You have to pass to complements every time. An easier way to define this topology is to work only with closed sets. So an equivalent way to define this topology is to say that $F \in \hat X$ is closed if and only if $* \in F$ and $F \cap X$ is closed in $X$, or $* \notin F$ and $F$ is closed and compact. Another way to phrase this is: $F$ is closed in $\hat X$ if $F \cap X$ is closed in $X$, and if $* \notin F$, we require that $F$ is also compact.

$* \notin \varnothing$ and $\varnothing$ is closed in $X$ and compact, so $\varnothing$ is closed in $\hat X$. $* \in \hat X$ and $X \cap \hat X = X$ is closed in $X$, so $\hat X$ is closed in $\hat X$.

If $\{F_i\}_{i \in I}$ is a family of closed sets in $\hat X$, then
$$
X \cap \bigcap_{i \in I} F_i = \bigcap_{i \in I} (F_i \cap X)
$$
is closed in $X$ because each $F_i$ is closed in $X$. If $* \notin F_j$ for a certain $j$, then the intersection doesn't contain $*$ and it is contained in $F_j \cap X$ which is compact, and therefore our intersection is a closed subset of a compact set, thus compact as well.

If $F_1,F_2$ are closed in $\hat X$, then $(F_1 \cup F_2) \cap X = (F_1 \cap X) \cup (F_2 \cap X)$ is closed in $X$. If $*$ is in neither of $F_1$ or $F_2$, then both are compact, and therefore their union is also compact (by the same argument as above).

I find the second argument shorter because you avoid passing to complements. What do you think?

Hope that helps,


问:
Extremization of a Quartic Trigonometric Polynomial
Given positive integers $m,n$, define
$$P_{m,n}(\theta):= \left( \sum_{i=1}^m a_i^2 \sin^2( \theta-\alpha_i) \right) \left( \sum_{j=1}^n b_j^2 \sin^2( \theta-\beta_j) \right),$$

where $\{a_i\}, \{\alpha_i\}, \{b_j\} ,\{\beta_j\}$ are real numbers. I'm interested in the critical points of this function.

My attempt:

I managed to solve the $m=n=1$ case in closed form.
The minima occur at
$$ \theta \equiv \alpha_1 \pmod{\pi}, \\
   \theta \equiv \beta_1 \pmod{\pi} .$$

The maxima occur at
$$\theta \equiv \frac{\alpha_1+\beta_1}{2} \pmod{\frac{\pi}{2} }. $$

However, I couldn't solve in closed form any other general case (i.e. $m>1$ or $n>1$).

The derivative is in general
$$P_{m,n}'(\theta)=\left( \sum_{i=1}^m a_i^2 \sin( 2(\theta-\alpha_i)) \right) \left( \sum_{j=1}^n b_j^2 \sin^2( \theta-\beta_j) \right)+\left( \sum_{i=1}^m a_i^2 \sin^2( \theta-\alpha_i) \right) \left( \sum_{j=1}^n b_j^2 \sin( 2(\theta-\beta_j)) \right),$$

but I can't see how to get its roots.

My questions:

Can the critical points of $P_{m,n}$ be found in closed form?

If several minima/maxima exist, do they have the same minimum/maximum value?

Thanks.

答:
First of all rewrite the objective function as

$$P_{m,n}(\theta)=A_0+A_2 \cos 2\theta + A_4 \cos 4\theta +B_2 \sin 2\theta+B_4 \sin 4\theta ,$$
where the coefficients are given as bilinear forms
\begin{align}A_0 &= \phantom{+}\frac{1}{8} \sum_{i=1}^m \sum_{j=1}^n a_i^2 b_j^2 \left(2+\cos (2\alpha_i -2\beta_j) \right),\\
A_2&= -\frac{1}{4}\sum_{i=1}^m \sum_{j=1}^n a_i^2 b_j^2 \left( \cos (2\alpha_i) + \cos (2\beta_j) \right),\\
A_4&= \phantom{+}\frac{1}{8}\sum_{i=1}^m \sum_{j=1}^n a_i^2 b_j^2 \left( \cos \left(2\alpha_i + 2\beta_j \right) \right), \\
B_2&= -\frac{1}{4}\sum_{i=1}^m \sum_{j=1}^n a_i^2 b_j^2 \left( \sin (2\alpha_i) + \sin( 2\beta_j) \right),\\
B_4&= \phantom{+}\frac{1}{8}\sum_{i=1}^m \sum_{j=1}^n a_i^2 b_j^2 \left( \sin \left(2\alpha_i + 2\beta_j \right) \right).
\end{align}

Upon differentiation, we get

$$P_{m,n}'(\theta)=2 B_2 \cos 2 \theta +4 B_4 \cos 4 \theta - 2A_2 \sin 2 \theta - 4A_4 \sin 4\theta. $$

Use the substitution
$$z=\mathrm{e}^{i \theta} \leftrightarrow \theta = \arg z $$ to characterize critical points using roots of

$$2B_2 \frac{z^2+z^{-2}}{2}+4B_4 \frac{z^4+z^{-4}}{2}-2A_2 \frac{z^2-z^{-2}}{2i}-4A_4 \frac{z^4-z^{-4}}{2i}. $$

The equation is equivalent to the octic polynomial
$$2 (-i A_4 + B_4) + (-i A_2 + B_2) z^2 + (i A_2 + B_2) z^6 +
 2 (i A_4 + B_4) z^8=0. $$

Note that the substitution $w=z^2$ allows for a solution involving the quartic formula.


问:
Continuity in metric space
Let $F: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ be defined by

$$ F(x,y) = \left( x^3 y,\ \ln(x^2 + y^2 + 1),\ \cos(x - y^2) \right) $$

When trying to show why $F$ is continuous where should I start?

答:
To give the answer of your question let me define first the vector valued functions.

A vector-valued function is a function from $\mathbb{R}$, or a subset of $\mathbb{R}$, to a
vector space $\mathbb{R}^n$. It comprises $n$ scalar functions, one for each of the coordinates. For instance, given scalar functions $f_1$, $f_2$...$f_n$, we can construct a vector-valued function $f = (f_1, f_2, \ldots f_n) $defined by
by $t  \rightarrow (f_1(t), f_2(t), \ldots f_n(t))$.

Now come to your question:

The vector valued function is continuous at  point $a$
 if and only if all its components are continuous at $a$. Moreover,
a vector valued function is continuous on the set $S$ if and only if
all its components are continuous on the set S.

Title: Continuity in metric space


问:
Doubt in a proof on projective varieties from Hartshorne
In Algebraic Geometry by Hartshorne in the proof of theorem 3.4 in Chapter 1 he gives an isomorphism of $k[y_1,...,y_n]$ with $k[x_0,...,x_n]_{(x_i)}$ by sending $f(y_1,...,y_n)$ to $f(x_0/x_i,...,x_n/x_i)$ leaving out $x_i/x_i$.

Then he says that this isomorphism sends $I(Y_i)$ to $I(Y)S_{(x_i)}$. I don't understand this last bit.

Can some one help?

Thanks

答:
First of all, the notation Hartshorne uses is a bit confusing: by $I(Y)S(Y)_{(x_i)}$ Hartshorne means the ideal $(I(Y) \cdot S(Y)_{x_i}) \cap S(Y)_{(x_i)}$ (see here). If $g \in S(Y)$ is homogeneous and $e$ its degree, the fraction $g/x_i^e \in S(Y)_{(x_i)}$ is contained in this ideal if and only if $x_i^rg \in I(Y)$ for some $r>0$.

As in the proof of Proposition 2.2, let $\beta: k[y_1,\ldots,y_n] \rightarrow k[x_0,\ldots,x_n]$ be the "homogenisation map". It sends a polynomial $f \in k[y_1,\ldots,y_n]$ of degree, say, $d$, to $\beta(f)=x_i^{d}f(x_0/x_i,\ldots,x_n/x_i)$, omitting $x_i/x_i$. Thus, the isomorphism maps $f$ to $\beta(f)/x_i^d$.

We have to show that $f \in I(Y_i)$ if and only if $\beta(f)/x_i^d \in I(Y)S(Y)_{(x_i)}$. In view of the first paragraph, latter condition is equivalent to $x_i^r\beta(f) \in I(Y)$ for some $r>0$. This means that $Y \subseteq Z(x_i^r\beta(f))=Z(\beta(f))\cup Z(x_i)$, i.e. $Y \cap U_i = Y \setminus Z(x_i) \subseteq Z(\beta(f))$. As in Proposition 2.2, this occurs if and only if $Y_i \subseteq Z(f)$, i.e., $f \in I(Y_i)$.


问:
Number of ways to group students with certain conditions
So i have this specific question and i really can't seem to wrap my head around it. The question is as below;

There are 4 girls in a class of 16 students where 2 of the girls are twins. If we want to split them into two equal groups how many different ways can we do so for

a) when there is no restriction,

In this case i think its just 16C8

b) all girls to be in the same group

for this i think its 4C4 x 12C4

c) not wanting the twins in the same group,

This one im not too sure how to go about it.

I've always had trouble with understanding permutations and combinations and this one really got me thinking, i would appreciate any explanation i could get the answer doesn't really concern me but if i can finally understand what's happening here i would be more than grateful.

答:
Your a) is nearly correct, $\binom {16}{8}$ chooses the children to go in one group but the remaining children can also be selected in the same way; so  you can get the same division twice. You can adjust for this by dividing by two (correct), $\frac 12 \binom {16}{8}$, or by picking an arbitrary child as being on one team and choosing the other $7$ chlldren to join her, $\binom {15}7$

Your b) is correct but a little more complicated than it needs to be: $\binom 44 = 1$ ("the girls are all on one team"), so the answer calculates as expected to be just $\binom {12}4$ ("choose the $4$ boys that are on the same team with the girls"). Here the group with the girls is thus labelled and there is no ambiguity about the choice as previously.

Solution to c) can start by allocating the twins one to each group (which are now therefore distinct eg. "Brenda's team" and "Glenda's team") and picking the other $7$ children to be on one of these.


问:
Show that $\left\lfloor \sqrt{x} \right\rfloor$ is $n$
Let $n^2$ be the largest perfect square that is less than or equal to $x$, then $x$ can be defined as follows,

$$ x = n^2 + m + \epsilon, 0\lt\epsilon\lt 1, m\in\Bbb Z$$

Show that $\left\lfloor \sqrt{x} \right\rfloor$ is $n$ and show that $\left\lfloor \sqrt{\left\lfloor x \right\rfloor} \right\rfloor$ is also $n$.

Proof: let us just consider proofing why $\left\lfloor \sqrt{x} \right\rfloor$ is $n$.

Start with $\left\lfloor \sqrt{x} \right\rfloor$,

$$\sqrt{x} = \sqrt{n^2 + m + \epsilon}$$

The proof claims that since $n^2$ is the perfect largest square that is less than or equal to $x$, $\sqrt{x}$ has to be between $n$ and $n+1$,

$$\left\lfloor \sqrt{x} \right\rfloor = n$$

Could you please explain as why $\left\lfloor \sqrt{x} \right\rfloor = n$?

答:
In your specification, you overlooked something:

Since $n^2 \leq x < (n+1)^2$,

$m \leq (2n)$.

This follows, since if $m \geq (2n+1)$, then $n^2 + m + \epsilon$ would be $\geq (n+1)^2$, which is a contradiction.

Then, you have that $\sqrt{x} = \sqrt{n^2 + m + \epsilon} < \sqrt{n^2 + (2n+1)} = \sqrt{(n+1)^2} = (n+1)\implies$

$n = \sqrt{n^2} \leq \sqrt{x} < (n+1) \implies$

$\left\lfloor \sqrt{x}\right\rfloor = n.$


答:
There are $4$ girls in a class of $16$ students where $2$ of the girls are twins. If we want to split them into two equal groups, how many different ways can we do so when there is no restriction?

You have counted each group twice, once when you select the group and once when you select its complement since choosing $8$ of the $16$ students to be in one group also determines which $8$ students are in the other group.  The answer should be
$$\frac{1}{2}\binom{16}{8}$$

Another way to see this is to suppose James is one of the members.  We must choose which $7$ of the other $15$ students must be in the same group as James.  Then the other eight students must be in the other group.  Hence, there are
$$\binom{15}{7}$$
ways to divide the students into two groups of equal size.

As you can verify,
$$\frac{1}{2}\binom{16}{8} = \binom{15}{7}$$

There are $4$ girls in a class of $16$ students where $2$ of the girls are twins. If we want to split them into two equal groups, how many different ways can we do so when all the girls are placed in the same group?

Your answer is correct.

There are $4$ girls in a class of $16$ students where $2$ of the girls are twins. If we want to split them into two equal groups, how many different ways can we do so when the twins are not placed in the same group.

Suppose Katharine is one of the twins.  Since her twin cannot be in the same group, we must choose $7$ of the other $14$ students to be in the same group as Katharine, which can be done in
$$\binom{14}{7}$$
ways. The remaining students must in the group with Katharine's twin.

Title: Number of ways to group students with certain conditions

Body:
So i have this specific question and i really can't seem to wrap my head around it. The question is as below;

There are 4 girls in a class of 16 students where 2 of the girls are twins. If we want to split them into two equal groups how many different ways can we do so for

a) when there is no restriction,

In this case i think its just 16C8

b) all girls to be in the same group

for this i think its 4C4 x 12C4

c) not wanting the twins in the same group,

This one im not too sure how to go about it.

I've always had trouble with understanding permutations and combinations and this one really got me thinking, i would appreciate any explanation i could get the answer doesn't really concern me but if i can finally understand what's happening here i would be more than grateful.


问:
$f\geqslant 0$, $f''\leqslant0$ and $f$ is differentiable on $ ]1,\infty[$. $\lim_{x\to\infty} \frac{f(x)}{x}$ exist.
Let $f\geqslant 0$, $f''\leqslant0$ and $f$ is differentiable on $ ]1,\infty[$. Show that $\lim_{x\to\infty} \frac{f(x)}{x}$ exist.

I already asked quite similar question, but this time i'd like to know how to proceed on interval $ ]1,\infty[$. I was trying to consider $1<a<x<b$ and apply Mean value theorem on intervals $]a,x[$ and $]x,b[$ and then try to ''play'' with inequalities as we know that $f'$ is decreasing. But, i don't really know if it could lead me somewehere.. Thanks in advance

We could probably resolve this problem with Bernouilli-Hospital, but i'm interested in Mean value proof if it is possible.

答:
For $1 < a < x < y$ there are $c_1 \in(a, c)$ and $c_2 \in (x, y)$ such that
$$
 \frac{f(y)-f(x)}{y-x} = f'(c_2) \le f'(c_1) = \frac{f(x)-f(a)}{x-a} \le \frac{f(x)}{x-a} \, .
$$
This inequality can be rearranged to
$$
\frac{f(x)}{x-a} \ge \frac{f(y)}{y-a}
$$
which shows that $\frac{f(x)}{x-a}$ is decreasing. It is also bounded below by zero, so that $\lim_{x \to \infty} \frac{f(x)}{x-a}$ exists. But then
$$
\frac{f(x)}{x} = \frac{x-a}{x} \cdot \frac{f(x)}{x-a}
$$
also has a limit for $x \to \infty$.


问:
Having such integral, how to optimize it in maple?
So we have :

```<code> (1/3)*sig0*h^3*(int(int(sin((1/3)*arctan(y, x)), x = 0 .. r), y = 0 .. 2*Pi))
</code>```

Is it possible to optimise it? (in maple or any other way...)

How I got here:

```<code>> R := 1;
H := h;
sigma[0] := sig0;
sigma := sigma[0]*z^2*sin((1/3)*`ϕ`);
toCar := eval(sigma, {r = sqrt(x^2+y^2), z = z, `ϕ` = arctan(y, x)});

> Q := int(int(int(toCar, x = 0 .. r), y = 0 .. 2*Pi), z = 0 .. H);
</code>```

答: By optimize, I assume you are meaning simplify.  I don't see a parameter to optimize over.  The limits on the integrals look like the differentials should be $\rho \; d\rho \; d\theta$, not $dx \; dy$.  If so, you have $\int_0^{2\pi}\int_0^r\sin(\frac{\theta}{3})\rho \; d\rho \; d\theta=-3\frac{r^2}{2}\cos(\frac{\theta}{3})\mid_0^{2\pi}=\frac{9r^2}{4}$</p>


问:
Is $e^z\sum_{k=0}^\infty\frac{k^3}{3^k}z^k$ analytic inside $|z|=3$?
Am I correct that the following function is analytic at least inside $|z|=3$? (I used the ratio test.) The solutions manual says that the function is analytic on and inside |z|=1, so I wonder if I'm having a conceptual misunderstanding. $$e^z\sum_{k=0}^\infty\frac{k^3}{3^k}z^k$$

Also, is it possible to see/guess from a glance the radius of convergence of this particular function?

答:
Well, $e^z$ is an entire function, which never vanishes. So analyticity of your function is equivalent to the analyticity of the power series $\sum k^3/3^kz^k.$
This series converges for $|z|<3,$ as you suggested.</p>


问:
sum of covariant derivatives implies constant Ricci curvature

Let $R_{ij}$ denote the Ricci tensor . If
$R_{ij,k}+R_{jk,i}+R_{ki,j}=0$ , prove that the scalar curvature $R$
is constant .

A straightforward calculation gives $$R_{ij,k}=\frac{\partial R_{ij}}{\partial x^k}-\Gamma^l_{ik}R_{lj}-\Gamma^l_{jk}R_{il} \\ R_{jk,i}=\frac{\partial R_{jk}}{\partial x^i}-\Gamma^l_{ji}R_{lk}-\Gamma^l_{ki}R_{jl} \\ R_{ki,j}=\frac{\partial R_{ki}}{\partial x^j}-\Gamma^l_{kj}R_{li}-\Gamma^l_{ij}R_{kl}$$ Since $R_{ij}$ is symmetric $$R_{ij,k}+R_{jk,i}+R_{ki,j}=0 \\ \implies \frac{\partial R_{ij}}{\partial x^k}+\frac{\partial R_{jk}}{\partial x^i}+\frac{\partial R_{ki}}{\partial x^j}=2\Gamma^l_{ik}R_{lj}+2\Gamma^l_{jk}R_{il}+2\Gamma^l_{ij}R_{kl}$$ I know that $R=g^{ij}R_{ij}$ , but multiplying above expression with any one of $g^{lj},g^{il},g^{kl}$ won't help .

How to approach this one ? Any help is appreciated .

答:
The Contracted Bianchi Identity says:

$$R_{ij;}^j=\frac{1}{2}R_{;i}.\tag{1}$$

By other hand, tracing the identity
$$R_{ij;k}+R_{jk;i}+R_{ki;j}=0$$
we get
$$R_{ij;}^j+R_{;i}+R_{;j}=0.\tag{2}$$
Therefore, applying (1) in (2) we obtain
$$\frac{5}{2}R_{;i}=0,\quad\forall i.$$
(In previous equality we choose $j=i$, since (1) is valid for all $i,j,k$) Finally, we conclude
$$R\equiv C.$$
(Is necessary suppose a connected manifold, otherwise, the scalar curvature can assume a different constant value in each connected component.)


答:
Let $g(x)=\left\{\begin{array}{c}99x+x^2\sin(1/x), x\ne 0\\0, x=0\end{array}\right.$

By your logic, the derivative of $g$ at $0$ is $0$. But I hope you can see that the derivative is actually $99$.


问:
Constructing positive definite inner products
I am presently studying Rosenberg book: "The Laplacian on a Riemannian Manifold" and I came across the following exercise:

Show that the space of all positive definite inner products on $\mathbb{R}^{n}$ is in one-to-one correspondence with the set $M = GL(n; \mathbb{R}) / O(n)$. Hint: Such an inner product is determined by a basis of $\mathbb{R}^{n}$ which is orthonormal with respect to the inner product. However, two bases which differ by an orthogonal transformation (with respect to this inner product) determine the same inner product.

It also says that because $O(n)$ is a closed subgroup of the Lie group $GL(n; \mathbb{R})$, $M$ is a manifold. I understand how $O(n)$ can be a closed subgroup by construction a sequence converging to a point in $GL(n; \mathbb{R})$ and this point must necessarily be in $O(n)$ but it is not clear how this makes $M$ a manifold.

Thank you for clearing these for me.

答:
To see that $GL(n)/O(n)$ is a manifold, you can show the general statement: if $G$ is a Lie group and $H$ is a Lie subgroup of $G$, then $G/H$ is a manifold. Hint: show the action of $H$ is smooth, free and proper.

To see why this quotient is in bijection with the set of all inner-products: let $I$ be the set of all inner-products of $\mathbb{R}^n$. Let $f : I \to GL(n)/O(n)$ be the map associating to an inner-product $p$ the set of all orthonormal bases.

Show $f$ is well defined. To do this, indentify an (ordered) orthonormal basis with the matrix where the rows are the vectors of the basis, and show that if $B$ and $B'$ are two orthonormal bases for $p$, then there exists $O\in O(n)$ with $B=OB'$.

Show it is injective, that is if two inner products have some common orthonormal basis then they are equal

Show it is surjective, that is if you have an orbit of basis under $O(n)$ then you can construct an inner-product on $\mathbb{R}^n$ for which everything works

This is a difficult and abstract point of view, but it is a common thing to visualise the set of all possible structures on a set as a quotient space. For example, the set of all complex structures on $\mathbb{R}^{2n}$ is $GL(2n,\mathbb{R})/GL(n,\mathbb{C})$.


问:
Why is it that $\int^x_{-\infty} \frac{e^t}{t}dt=\gamma+\ln x+\sum^\infty_{n=1}\frac{x^n}{n\cdot n!}$?
According to MathWorld, the exponential integral $\operatorname{Ei}(x)=\int^x_{-\infty} \frac{e^t}{t}dt$ can be written as follows:

$$\int^x_{-\infty} \frac{e^t}tdt=\gamma+\ln x+\sum^\infty_{n=1}\frac{x^n}{n\cdot n!}$$

Where $\gamma$ is the Euler-Mascheroni  constant.

However, I am curious to know how exactly this was derived.

I was able to show that the derivative of both sides w.r.t. $x$ are the same:

$$\frac d{dx}\int^x_{-\infty} \frac{e^t}{t}dt=\frac d{dx}\left[\gamma+\ln x+\sum^\infty_{n=1}\frac{x^n}{n\cdot n!}\right]$$

$$\frac{e^x}x=\frac d{dx}[\gamma]+\frac d{dx}[\ln x]+\sum^\infty_{n=1}\frac d{dx}\left[\frac{x^n}{n\cdot n!}\right]$$

$$\frac{e^x}x=\frac1x+\sum^\infty_{n=1}\frac{nx^{n-1}}{n\cdot n!}$$

$$\frac{e^x}x=\frac1x+\sum^\infty_{n=1}\frac{x^{n-1}}{n!}$$

$$e^x=1+\sum^\infty_{n=1}\frac{x^n}{n!}$$

$$e^x=\sum^\infty_{n=0}\frac{x^n}{n!}$$

Which we know is true by the Maclaurin expansion of $e^x$. However, this only shows that $\int^x_{-\infty} \frac{e^t}tdt$ and $\gamma+\ln x+\sum^\infty_{n=1}\frac{x^n}{n\cdot n!}$ differ by a constant.

答:
First of all, this needs a precise formulation (if $x$ is real, then it must be negative for $\int_{-\infty}^x$ to exist, but then $\ln x$ is not defined; if we allow complex values of $x$, then the question of branching arises).

This is why $E_1(z)=\int_z^\infty\frac{e^{-t}}{t}\,dt$ is often considered instead of $\operatorname{Ei}z$ (in case of complex $z$, we choose the branch cut of this function along the negative real axis, as we do for the principal value of $\log z$; this corresponds to the path of integration chosen to stay within lower/upper halfplane according to $z$ itself). Note that for $$f(z)=\int_0^z\frac{1-e^{-t}}{t}\,dt-\int_z^\infty\frac{e^{-t}}{t}\,dt$$ we have $f(z)-f(1)=\log z$, and the equality $f(1)=\gamma$ is the subject of this question. Hence, $$E_1(z)=-\gamma-\log z+\int_0^z\frac{1-e^{-t}}{t}\,dt=-\gamma-\log z-\sum_{n=1}^\infty\frac{(-z)^n}{n\cdot n!}$$ (after termwise integration of the power series for $(1-e^{-t})/t$).


问:
Does this proof of the sequential completeness of a reflexive space require the Eberlein–Šmulian theorem?
I'm self-teaching functional analysis through Conway's book at the moment, and can't seem to get one of the steps in this proof:

The statement that $\{ x\in X : ||x||<M\}$ implies the sequence $\{x_n\}$ has a cluster point is throwing me off. I can't seem to find a way to justify this without invoking the Eberlein–Šmulian theorem - which Conway doesn't mention in his book until a few chapters later. My question is - am I missing something? The definition Conway gives for a Cauchy sequence in this situation is given below:

Is there a way to see thus statement without using such heavy machinery? Or is this just a small oversight in his book?

Edit: the problem is that weak compactness implies weak sequential compactness here. I should’ve specified that fact that ballX is weakly compact isn’t the problem

答:
I haven't checked Conway to see what has been proven up until this point, but this assertion can be proved using the following elementary argument:

Let $\mathscr Y = \overline{\operatorname{span}}\{x_n\}$ be the closed span of $\{x_n\}$ in $\mathscr{X},$ which is a separable reflexive Banach space with the inherited norm. Then I claim $\{ y \in \mathscr{Y} : \lVert y \rVert \leq M\}$ is weakly sequentially compact; this can either be proven directly by means of a diagonal argument, or by noting the set is weakly metrisable. Hence we infer there is $x \in \mathscr{Y} \subset \mathscr{X}$ such that $x_n \rightharpoonup x$ weakly in $\mathscr{Y},$ and it is easy to see this also implies weak convergence in $\mathscr{X}.$


问:
elementary mathematical analysis problem books recommendation
I'm self study rudin pma and find that exercises in the book are more concerned with proof and lack of calculation, so I want to find a problem books but not so hard,thanks a lot.</p>

答:
The book by Tomasz Radozycki ˙Solving Problems in Mathematical Analysis, Part I' might be interesting for you. If not, there are also Part II and Part III.

In these books a large number of individual problems are solved in a most detail manner.

Hope they work for you.


问:
Question on integration
Let $(p,q)\in \mathbb{R}^2$ , and $H: \mathbb{R}\times \mathbb{R} \rightarrow \mathbb{R}$

if $q'=\frac{\partial H}{\partial p} (p,q) ~~\text{and}~~ p'=-\frac{\partial H}{\partial q} (p,q)$

How to prove that $H(p(t),q(t))=1$

Please , help me

thank you

答:
Under the condition that $H\bigl(p(0), q(0)\bigr)= 1$, we can argue as follows: Let $h(t) = H\bigl(p(t), q(t)\bigr)$ for all $t$. Differentiation using the chain rule gives
\begin{align*}
  h'(t) &= \partial_p H\bigl(p(t), q(t)\bigr)p'(t) + \partial_q H\bigl(p(t), q(t)\bigr)q'(t)\\
        &= -\partial_p H\bigl(p(t), q(t)\bigr)\partial_q H\bigl(p(t), q(t)\bigr) + \partial_q H\bigl(p(t), q(t)\bigr)\partial_p H\bigl(p(t), q(t)\bigr)\\
        &= 0.
\end{align*}
So $h$ is constant, giving $h(t) = h(0) = H\bigl(p(0), q(0)\bigr) = 1$ for all $t$.</p>

Title: Question on integration


问:
Probability that three independent uniform $(0,1)$ random variables can form a triangle
I am preparing for a probability exam and while practicing stuck on this question. Do not even know how to begin.

Let $X_1$, $X_2$, $X_3$ be independent uniform $(0,1)$ random variables. What is the probability that we can form a triangle with three sticks of length $X_1$, $X_2$, $X_3$?

I am thinking of using $X_1 + X_2 > X_3$ for this to happen and there are three such combinations. But how to proceed next ?

答:
Let $x,y,z$ be the three numbers. The probability that $z > x + y$ is given by

$$
\int_0^1 dz \int_0^z dy \int_0^{z-y} dx = 1/6.
$$

Any of the three variables could be the largest, so multiply by $3$ and we obtain $1/2$.

Here's a simpler geometric argument. Suppose the largest of the three is $a$. Then the other two must lie in the square with area $a^2$. In that square, the northwest-to-southeast diagonal gives the line where the other two sum to $a$. Below that line, the other two sum to less than $a$. Clearly that line cuts the square in half. Hence, for any $a$, the probability is $1/2$.


问:
$\lim_{|z| \to 1 } |f(z)| = \infty$ for $f(z) = \sum_{k=1}^\infty z^{k!}$, a strange series manipulation
I'm dealing with a rather complicated series:

$$f(z) = \sum_{k=1}^\infty z^{k!}.$$

I know that the radius of the convergence of it is 1 but here is another problem:

$$\lim_{|z| \to 1 } |f(z)| = \infty.$$

Here is my try:

Put $z=Re^{i\theta}$ then $|z| = R$ and the limit reduces to $\lim_{R \to 1 } |f(Re^{i\theta})| $:

$$|f(Re^{i\theta})| = |\lim_{n \to \infty} \sum_{k=1}^n (Re^{i\theta})^{k!}|$$

I'm not sure that I can exchange absolute with the limit but I proceed

$$\lim_{n \to \infty}| \sum_{k=1}^n (Re^{i\theta})^{k!}|\leq \lim_{n \to \infty}\sum_{k=1}^n| R^{k!}|$$

And at this point, I have no idea how to continue.

Can anyone help me withhow to tackle it down? Maybe that's a bad approach and there is an elegant one.

答:
Your trouble may be that you think when $f\in H(\mathbb{D}^\circ)$ it should be able to continuously extend to $F\in C(\overline{\mathbb{D}})\cap H(\mathbb{D}^\circ)$, however this is generally not the case, where your trouble here is indeed reasonable.
In fact, you have given an example of the Hadamard Gap Theorem, which states that when $\liminf_{n\to\infty} c_{n+1}/c_{n}>1$, the function $f$ defined by $f(z)=\sum_{n=1}^\infty z^{c_n}$ is holomorphic on the unit disk, but it cannot be analytically extended to ANY open set properly containing $\mathbb{D}^\circ$


问:
Prove that $(K,\delta)$ is a compact metric space.
Let $(X, d)$ be a compact metric space. For $x ∈ X $ and $\epsilon > 0$, define

{$B_{\epsilon}(x) := {y ∈
X | d(x, y) < \epsilon}$}.

For $C ⊆ X$ and $\epsilon > 0$, define $B_{\epsilon}(C) := ∪_{x∈C}B_{\epsilon}(x)$.

Let K be the
set of non-empty compact subsets of X. For $C, C_0 ∈ K$, define $δ(C, C_0) = $inf{$\epsilon| C ⊆ B_{\epsilon}(C_0)$ and $C_0 ⊆ B_{\epsilon}(C)$}.

Assuming that it forms a metric space show that $(K, δ) $ is a compact metric space.

My attempt:

The metric space $(X,d)$ is compact so it is totally bounded.
Let $B=${$C_k$ } be a sequence of compact metric spaces in $K$. We need to show that the infinite set has a limit point.

$(X, d)$ is totally bounded so let $A_1$={$x_{1,1},x_{2, 1},....,x_{k_1,1}$} such that ${\cup {B_d(x_i,1) }} $ covers X for $i= (1,1),..,(k_1,1)$ and ${\cup {B_d(x_i,\frac{1}{2}) }} $ covers X for $i= (1,2),..,(k_2,2)$ ,so $A_2= x_{1,2},x_{2,2},...,x_{k_2,2}$

So we can generate a new sequence of compact metric space $A_1,A_2,...,A_n$ as mentioned above.Now,let {$\cup {(A_i)}$}=$A$ .We take the closure $cl(A)$ which is the closed subset of $(X,d)$.Hence it is compact so it is in $(K,\delta)$.

$(B_{\delta}(cl(A) \cap B)  /cl(A))\ne \phi$ for all $\epsilon>0$ is what we have to show to prove that $cl(A)$ is the limit point of the sequence {${C_k}$} to show that the sequence has bolzano weistrass property.

We pick an $\epsilon >0$ and proceed as ,

Now, $C_i \subset B_{\frac{1}{N_1}}(cl(A))$ where $\frac{1}{N_1} < \frac{\epsilon}{2}$ and $C_i \subset B_{\epsilon}(cl(A))$ where $C_i$ is any compact subset in the sequence {$C_k$} and $C_i$ can be chosen to be different from $cl(A)$.

Now, let $x \in Cl(A)$
then $x \in X$ and $x \in \cup B_d(x_i,\frac{1}{N_1})$ where $i=(1,N_1),(2,N_1),(3,N_1),...,(k_{N_1},N_1)$
then $d(x,x_i) < \frac{1}{N_1}$ and let $x_n \in C_i$ then $d(x_i,x_n) < \frac{1}{N_1}$ so $d(x,x_n) < \epsilon$ and the other case should also follow when $ x = x_i $ where $i=(1,N_1),(2,N_1),(3,N_1),...,(k_{N_1},N_1)$.
So $cl(A) \subset B_{\epsilon}(C_i)$.

Then $\delta(C_i,cl(A)) < \epsilon$ which allows me to conclude that $cl(A)$ is the limit point of the above mentioned seqeunce (as $\epsilon$ can be chosen arbitrarily).So by bolzano property I can claim the compactness.

Is my attempt ok? I have tried hard so that the notations are understandable. Thanks in advance.

答:
I will write here my idea in a more formal way, and see if it works. I will try to do as little change in your argument as possible.

Idea: Lets start with something easier. Prove that for any convergent sequence $\{ C_i\}_{i=0}^\infty$ the limit is in $K$. THat is, $K$ is closed.

First, we would like to know how the limit of a sequence of compact spaces would look like. We look at some special cases for intuition.

In the case that $C_{i+1}\subseteq C_i$ it makes sense that the limit is $\bigcap C_i$.

Next, we look at the interval [0,1] and the sequence $[0,\frac{1}{2}], [\frac{1}{2},\frac{1}{2}+\frac{1}{4}], [\frac{1}{2}+\frac{1}{4},\frac{1}{2}+\frac{1}{4}+\frac{1}{8}]... $ and so on. What is the limit here? it seems that it should be $\{1\}$ (can you prove it?) But no matter what it is, the points of the $C_i$ has to get "closer and closer" to the points of the limit set.

Take $A_i$ as a set of points (finite) such that $\bigcup\limits_{j=1}^{k_i} B_d(x_{j,i},\frac{1}{i})$ covers $C_i$. With this sets, we can construct a sequence by first taking the elements of $A_1$ (ordered by the first index), then the elements of $A_2$ and so on. This is a sequence of elements of $X$, and as $X$ is a compact metric space, it will have (at least one) converging subsequences. So it makes sense to look at the set $C$ of the limit points of those subsequences.

Why this set is a good candidate for the limit? First if you look at the examples above, it coincides with the limit, or at least our idea of what the limit should be. Second, the set $C$ does not change if you change a finite set of the $C_i$'s, and most important (if  I am not wrong) it does not change if you choose any other set of points as your $A_i$'s (if the $C_i$'s are fixed).

Now, we have to prove that $C$ is compact. It is enough to prove that is closed. Try to show it, I will complete it later if you are not able to do it.

If all of this works. What we have left is to prove that $K$ is totally bounded. I don't see an easy proof of that right now but maybe you can do it.


问:
Is there a name for function $f(x)=\max(a, \min(x, b))$, where $a \leq b$?
Is there a name for function $f(x)=\max(a, \min(x, b))$, where $a \leq b$?

What actually this function does: it keeps value in bounds of $[a, b]$.

答:
I don't think there is any official "name" for this function. It is just a fancy way to express a piecewise linear function. It is very likely that people write it in such a compact way for a purpose of programming.

When $a=1$ and $b=4$, the graph of the function:


问:
Proving $abcd+3\geq a+b+c+d$

If $a,b,c,d$ are non negatives and $a^2+b^2+c^2+d^2=3$ prove that  $$abcd+3\ge a+b+c+d$$

The inequality is not as simple as it looks.The interesting part is that the equality occurs when $a=0,b=c=d=1$ upto permutation .(I don't know if there are further equality cases.)

I tried rewriting the inequality as $$a^2+a(bcd-1)+b^2+c^2+d^2-b-c-d\ge 0$$ As its a quadratic in $a$ it suffices to show  the discriminant$$\Delta_a={(bcd-1)}^2-4(b^2+c^2+d^2-b-c-d)\le 0$$ which unfortunately is wrong (when $a=\sqrt{3},b=c=d=0$)

P.S ;I am not aware of using Lagranges multipliers

Its from here

Update :An answer has been posted in the link above (AOPS) similar to Dr Mathva's answer

答:
Since $a^2+b^2+c^2+d^2 = 3$ defines a compact set in $\mathbb{R}^4$, $f(a,b,c,d)=abcd+3-a-b-c-d$ will have a global minimum and maximum over that set, that will occur (can be easily justified) in critical points of the Lagrangian
$$
L(a,b,c,d,\lambda) = abcd+3-a-b-c-d -\lambda(a^2+b^2+c^2+d^2-3)
$$

So, just compute the critical points, compute the value of $f$ over each of these points and get the global minimum. If the global minimum is $\ge 0$, you are done.

When computing the critical points, you will get some complex solutions, some solutions with negative components but, in the end, the relevant solutions are the ones you already mentioned and also $a=b=c=d=\frac{\sqrt{3}}{2}$. The minimum value of $f$ (even admitting negative values for $a,b,c,d$) is zero (and the maximum is $\frac{57}{16}+2\sqrt{3}$).

Edit: Missed one critical point, but the result stands.

Title: Proving $abcd+3\geq a+b+c+d$


问:
Is this matrix exponential strictly positive?
Is it true that $e^{Nt}$ is strictly positive element-wise, where $t>0$ and $N$ is a real matrix with $N_{ij}\geq0$ if $i\neq j$ and $N_{ij}\leq 0$ if $i=j$?

答:
No this is not true

Take for a counterexample

$$N = \begin{pmatrix}
0 & 1\\
0 & 0
\end{pmatrix}$$


问:
Integral of $1/x^2$ without power rule
I was wondering if it was possible to evaluate the following integral without using the power rule for negative exponents

\begin{equation*}
  \int \frac{1}{x^2} \; dx
\end{equation*}

When using integration by parts, you end up with the same integral in the rhs so it seems out of luck

\begin{equation*}
  \int \frac{1}{x^2} \; dx = \frac{\ln x}{x} + \int \frac{\ln x}{x^2} \; dx
\end{equation*}

This question is inspired by this blackpenredpen's video

Using integration by parts with $u = \frac{1}{x^2}$ and $v = x$ is NOT accepted. If you write

\begin{equation*}
  \int \frac{1}{x^2} \; dx = \frac{1}{x} + 2 \int \frac{1}{x^2} \; dx
\end{equation*}

you are still implicitly using the power rule to compute the derivative of $\frac{1}{x^2}$ so it is not correct per the rules.

The same rules apply for $u$-substitution which implicitly use the power rule. See the following with $u = \frac{1}{x}$ such that

\begin{equation*}
    \int \frac{1}{\left(\frac{1}{x}\right)^2} \left(\frac{1}{x}\right)' \; dx = \int \frac{1}{u^2} \; du
\end{equation*}

and here the power rule is also considered used to compute the derivative of $\frac{1}{x}$ although it can be subject to discussion (geometric proof, limit definition of derivative, etc...)

答:
$$I=\int \frac 1{x^2} dx$$

Let
$v=x\implies dv=dx$

Let $u.v'=1\implies u=1 \implies u'=0$

$$I=\int\frac{uv'}{v^2}=\int\frac{u'}{v}-\frac{u}{v}=0-\frac{1}{x}=-\frac{1}{x}$$

Title: Integral of $1/x^2$ without power rule


问:
Show that the maximum error in a linear interpolation is given by $\frac{(x_1−x_0)^2}{8}M_2 : \{M_2= \max\{|f''(x)|, x \in[x_0,x_1]\}$
$$\frac{(x_1−x_0)^2}{8}M_2 : \{M_2= \max\{|f''(x)|, x \in[x_0,x_1]\}$$

I know that that part of the formula that goes $(x-x_0)(x-x_1)...$ is a 2nd degree polynomial, and I know that the maximum point is given by $\frac{x_1-x_0}{2}$, no troubles there. I'm just having trouble working out the rest. I did:

$$(\frac{x_1-x_0}{2}-x_0)(\frac{x_1-x_0}{2}-x_1) =\\ \frac{1}{4}(x_1-3x_0)(-x_1-x_0) = \\\frac{1}{4}(-x_1^2-x_0x_1+3x_0x_1+3x_0^2) = \\
-\frac{1}{4}(x_1^2-2x_1x_2-3x_0^2)$$

That $-3x_0^2$ isn't right. It should be $+x_0^2$. I don't see where I made a mistake. Help?

Body:
For linear interpolation

$$f(x) - P_1(x) = \dfrac{(x-x_0)(x-x_1)}{2} f''(c_x)$$

The error bound is

$$E_1 = \tag 1 |f(x) - P_1(x)| \le \max_{x_0 \le x \le x_1} \left| \dfrac{(x-x_0)(x-x_1)}{2}\right| \max_{x_0 \le x \le x_1}|f''(x)|$$

Finding $\displaystyle \max_{x_0 \le x \le x_1}  \frac{(x-x_0)(x-x_1)}{2}$

$$\dfrac{\partial}{\partial{x}} \left[(x-x_0) (x- x_1)\right] = 2x - x_0-x_1 = 0 \implies x = \dfrac{x_0+x_1}{2}$$

Substituting this max $x$

$$\left| \dfrac{(x-x_0)(x-x_1)}{2}\right| = \left|\left(\dfrac{x_0+x_1}{2} - x_0\right)\left(\dfrac{x_0+x_1}{2} - x_1\right)\right| = \left|-\frac{1}{4} (x_1 - x_0)^2\right|$$

Note

$$\left(\dfrac{x_0+x_1}{2} - x_0\right)\left(\dfrac{x_0+x_1}{2} - x_1\right) = -\frac{x_0^2}{4}+\frac{x_0x_1}{2}-\frac{x_1^2}{4} = -\frac{1}{4} (x_1-x_0)^2$$

Substituting back into $(1)$

$$E_1 = |f(x) - P_1(x)| \le \dfrac{(x_1-x_0)^2}{8} \max_{x_0 \le x \le x_1}|f''(x)|$$


答:
Is is called “clamp” in the C++ standard library, and also in the OpenGL graphics library.

References:

https://en.cppreference.com/w/cpp/algorithm/clamp

https://www.khronos.org/registry/OpenGL-Refpages/gl4/html/clamp.xhtml

https://en.wikipedia.org/wiki/Clamping_(graphics)

Body:
Is there a name for function $f(x)=\max(a, \min(x, b))$, where $a \leq b$?

What actually this function does: it keeps value in bounds of $[a, b]$.


问:
Generating student-friendly random orthgonal matrices
I am working on randomly generated exercises for a course in linear algebra. While generating random invertible (or unimodular) matrices using repeated row and column operations is a rather easy task, generating random orthogonal or unitary matrices seems to be more involved.

The first approach that comes to mind is using products of random Householder matrices. Another approach would be starting from randomly generated unimodular matrices and applying a Gram–Schmidt orthonormalization. However, in both approaches the entries of these become very ugly algebraic expressions involving sums of roots in roots in fractions quickly.

Are there any approaches to randomly generating student friendly orthogonal or unitary matrices for exercises?

I would also be interested in just static lists of exercise-suitable orthogonal matrices up to dimension $5$ to pick from, so far I couldn't find any.

答:
$\def\C{{\rm Cay}}\def\S{{\rm Skew}}$Apply the Cayley transformation
to a random skew-symmetric matrix
$$\eqalign{
\S(X) &= \tfrac 12(X+X^T) \\
\C(X) &= (I+X)^{-1}(I-X) \quad&\implies\quad&X=\C\big(\C(X)\big) \\
Q &= \C(\S(X)) \quad&\implies\quad&I=Q^TQ \\
}$$
Julia's rational arithmetic is convenient for creating the desired
student friendly matrices.

```<code>#!/usr/bin/env  julia
using LinearAlgebra

function cayley(X)
  return (I-X) / (I+X)
end

function skew(X)
  return (X-X') / 2
end

function ortho(n)
  X = convert.(Rational{Int}, rand(0:20,n,n))
  return cayley(skew(X))
end

# usage
Q2 = ortho(2)
 -21//29  -20//29
  20//29  -21//29

Q3 = ortho(3)
 -51//71  -18//71  -46//71
  26//71  -66//71   -3//71
 -42//71  -19//71   54//71
</code>```

Body:
$\def\C{{\rm Cay}}\def\S{{\rm Skew}}$Apply the Cayley transformation
to a random skew-symmetric matrix
$$\eqalign{
\S(X) &= \tfrac 12(X+X^T) \\
\C(X) &= (I+X)^{-1}(I-X) \quad&\implies\quad&X=\C\big(\C(X)\big) \\
Q &= \C(\S(X)) \quad&\implies\quad&I=Q^TQ \\
}$$
Julia's rational arithmetic is convenient for creating the desired
student friendly matrices.

#!/usr/bin/env  julia
using LinearAlgebra

function cayley(X)
  return (I-X) / (I+X)
end

function skew(X)
  return (X-X') / 2
end

function ortho(n)
  X = convert.(Rational{Int}, rand(0:20,n,n))
  return cayley(skew(X))
end

# usage
Q2 = ortho(2)
 -21//29  -20//29
  20//29  -21//29

Q3 = ortho(3)
 -51//71  -18//71  -46//71
  26//71  -66//71   -3//71
 -42//71  -19//71   54//71


问:
Integrate $\int _0^1\frac{1}{\left(9+\:x^2\right)^{0.5}}\:dx$ using integration by parts and substitution.

Integrate:
$$\int _0^1\frac{1}{(9+x^2)^{0.5}}\:dx$$

Here's my attempt, using integration by parts in the first step:

\begin{align}
\int _0^1\frac{1}{(9+x^2)^{0.5}}\:dx &= \left[\frac{x}{\left(9+x^2\right)^{\frac{1}{2}}}-\int \:-\frac{x^2}{\left(9+x^2\right)^{1.5}}dx\right]^1_0 \\
&= \left[\frac{x}{\left(9+x^2\right)^{\frac{1}{2}}}-\left(\frac{x}{\sqrt{9+x^2}}-\ln \left(\frac{1}{3} \left|x+\sqrt{9+x^2}\right|\right)\right)\right]^1_0 \\
&= \left[\ln \left(\frac{1}{3}\left|x+\sqrt{x^2+9}\right|\right)\right]^1_0 \\
&= \ln \left(\frac{1+\sqrt{10}}{3}\right)
\end{align}

I tried to solve this using substitution and I keep getting stuck at the same place. For the sake of understanding, I turned the definite integral into an indefinite one by remove the upper and lower bounds, since that's quite easy to do. I'm more concerned with the final result without the actual values. Here's my attempt:

\begin{align}
\int _0^1\frac{1}{(9+x^2)^{0.5}}\:dx &\stackrel{{u = \frac{x}{3}}, dx = 3\,du} = \int\frac{3}{\sqrt{9u^2+9}}\,\mathrm{d}u \\
&= \int \dfrac{1}{\sqrt{u^2+1}}\,\mathrm{d}u \\
&\stackrel{{u=\tan(v)} \to \mathrm{d}u=\sec^2(v)\,\mathrm{d}v} = \int \frac{\sec^2(v)}{\sqrt{\tan^2(v)+1}}\,\mathrm{d}v \\
&= \int \sec(v)\,\mathrm{d}v \\
&= \int \frac{\sec(v) (\tan(v)+\sec(v))}{\tan(v)+\sec(v)}\,\mathrm{d}v \\
&= \int \frac{\sec(v) \tan(v)+\sec^2(v)}{\tan(v)+\sec(v)}\,\mathrm{d}v \\
&= \cdots
\end{align}

Here is where I am stuck. I think that there should be another substitution here in order to get the final integral that can be solved, but I'm not sure which term to choose. I think it would have to be $w = w=\tan\left(v\right)+\sec\left(v\right)$, but I'm not $100 \%$ sure.

This really caught me off by surprise because it looks rather simple but boy is it tricky to do.

Body:
\begin{align}
\int\frac{1}{\sqrt{x^2+9}}\,dx
&=\int\sec(u)\,du\quad &(x=3\tan u)\\
&=\int \frac{\sec^2(u)+\tan(u)\sec(u)}{\tan(u)+\sec(u)}\,du\\
&=\int\frac{1}{s}\,ds&(s=\tan(u)+\sec(u))\\
&=\log(s)+C\\
&=\log(\tan(u)+\sec(u))+C\\
&=\log\bigg(\frac13(\sqrt{x^2+9}+x)\bigg)+C
\end{align}


问:
Convergence or divergence of improper integral
I want to check if the following integral converge or diverge.

$\displaystyle{\int_1^{+\infty}\frac{(\ln t)^{\beta}}{t^{\alpha}}\, dt, \alpha ,\beta\in \mathbb{R}}$

$$$$

Do we have to take cases for $\alpha$ and $\beta$ ?

If $0<\beta\leq \alpha-2$ we get $$\ln t\leq t \Rightarrow \left (\ln t \right )^{\beta}\leq t^{\beta} \leq t^{\alpha-2} \Rightarrow \frac{\left (\ln t\right )^{\beta}}{t^{\alpha}}\leq \frac{1}{t^2}$$ and since $\int_1^{\infty}\frac{1}{t^2}\, dt=\left [-\frac{1}{t}\right ]_1^{+\infty}=0+1=1<\infty$ the integral $\displaystyle{\int_1^{+\infty}\frac{(\ln t)^{\beta}}{t^{\alpha}}\, dt}$ converges also, right?

Could you give me a hint for the other cases?

答:
$$\int_1^\infty\frac{\log^\beta(t)}{t^\alpha}dt; \alpha,\beta\in\mathbb{R}$$

Let $t\rightarrow e^x:$

$$\int_0^\infty x^\beta e^{(1-\alpha)x}dx$$

For $\alpha \leq 1$, the integral will be divergent regardless of the value of $\beta$:
$$\int_0^\infty x^\beta e^{(1-\alpha)x}dx\geq \int_0^\infty x^\beta dx\rightarrow +\infty
 $$

For $\alpha > 1$,the integral can be seem like a Laplace Transform, in wich $s=\alpha-1$:
$$\displaystyle\lim_{s\rightarrow\alpha-1}\int_0^\infty x^\beta e^{-sx}dx=\displaystyle\lim_{s\rightarrow\alpha-1}\mathcal{L}[x^\beta]=\displaystyle\lim_{s\rightarrow\alpha-1}\frac{\Gamma(\beta+1)}{s^{\beta+1}}=\frac{\Gamma(\beta+1)}{\left(\alpha-1\right)^{\beta+1}}; \beta>-1$$

Therefore, the integral will converge for $\alpha>1$ and $\beta>-1$. Otherwise, it will diverge.

Edit: to prove this without resorting to the Laplace Transform, one may show that:

$$\underbrace{\int_0^\infty x^\beta e^{-sx}dx}_{sx\rightarrow z}=\frac{1}{s^{\beta+1}}\int_0^1 z^\beta e^{-z}dz+\frac{1}{s^{\beta+1}}\int_1^\infty z^\beta e^{-z}dz$$

For the first integral:
$$\int_0^1 z^\beta e^{-z}dz\leq\int_0^1 z^\beta dz=\left[\frac{z^{\beta+1}}{{\beta+1}}\right]^1_0=\frac{1}{\beta+1}$$
Hence, $\beta>-1$ for the integral converge.

For the second integral, let $\beta<N; N\in\mathbb{N}$:
$$\int_1^\infty z^\beta e^{-z}dz\leq\int_1^\infty z^N e^{-z}dz\overbrace{=}^{IBP}1+N\int_1^\infty z^{N-1} e^{-z}dz$$

Which can be further proven by Induction that will also converge.


答:
Yes. Substitute $w = \tan v + \sec v $ then $dw = \sec v \tan v + \sec ^2 v dv$.
So your integral transforms into $$ \int \frac{ dw}{w} = \log |w| = \log ( |  \tan v + \sec v|)$$ and then it is easy to get back into desired expression.

Title: Integrate $\int _0^1\frac{1}{\left(9+\:x^2\right)^{0.5}}\:dx$ using integration by parts and substitution.

Body:
Integrate:
$$\int _0^1\frac{1}{(9+x^2)^{0.5}}\:dx$$

Here's my attempt, using integration by parts in the first step:

\begin{align}
\int _0^1\frac{1}{(9+x^2)^{0.5}}\:dx &= \left[\frac{x}{\left(9+x^2\right)^{\frac{1}{2}}}-\int \:-\frac{x^2}{\left(9+x^2\right)^{1.5}}dx\right]^1_0 \\
&= \left[\frac{x}{\left(9+x^2\right)^{\frac{1}{2}}}-\left(\frac{x}{\sqrt{9+x^2}}-\ln \left(\frac{1}{3} \left|x+\sqrt{9+x^2}\right|\right)\right)\right]^1_0 \\
&= \left[\ln \left(\frac{1}{3}\left|x+\sqrt{x^2+9}\right|\right)\right]^1_0 \\
&= \ln \left(\frac{1+\sqrt{10}}{3}\right)
\end{align}

I tried to solve this using substitution and I keep getting stuck at the same place. For the sake of understanding, I turned the definite integral into an indefinite one by remove the upper and lower bounds, since that's quite easy to do. I'm more concerned with the final result without the actual values. Here's my attempt:

\begin{align}
\int _0^1\frac{1}{(9+x^2)^{0.5}}\:dx &\stackrel{{u = \frac{x}{3}}, dx = 3\,du} = \int\frac{3}{\sqrt{9u^2+9}}\,\mathrm{d}u \\
&= \int \dfrac{1}{\sqrt{u^2+1}}\,\mathrm{d}u \\
&\stackrel{{u=\tan(v)} \to \mathrm{d}u=\sec^2(v)\,\mathrm{d}v} = \int \frac{\sec^2(v)}{\sqrt{\tan^2(v)+1}}\,\mathrm{d}v \\
&= \int \sec(v)\,\mathrm{d}v \\
&= \int \frac{\sec(v) (\tan(v)+\sec(v))}{\tan(v)+\sec(v)}\,\mathrm{d}v \\
&= \int \frac{\sec(v) \tan(v)+\sec^2(v)}{\tan(v)+\sec(v)}\,\mathrm{d}v \\
&= \cdots
\end{align}

Here is where I am stuck. I think that there should be another substitution here in order to get the final integral that can be solved, but I'm not sure which term to choose. I think it would have to be $w = w=\tan\left(v\right)+\sec\left(v\right)$, but I'm not $100 \%$ sure.

This really caught me off by surprise because it looks rather simple but boy is it tricky to do.


问:
Automorphism of $\mathbb Q (\sqrt[5]{2}, \zeta_5)$

Show that if $\sigma \in \mathrm{Aut}(\mathbb Q (\sqrt[5]{2}, \zeta_5)/\mathbb Q)$, then $\sigma (\zeta_5) = \zeta^i_5$ for some $i=1,2,3,4$ and $\sigma (\sqrt[5]{2}) = \zeta^j_5 \sqrt[5]{2}$ for some $j=1,2,3,4,5$.

Help please.

答:
Hint with different approach:

Note I am using this field extension definition: an extension of a field K is a triple (i, K, L), where L is another field, and i is a monomorphism of K into L. In other words, K is embedded in L.

This becomes routine with the extension of monomorphisms theorem. I'm leaving out some gritty details. Perhaps you should fill these in.

Let i be the inclusion map of $\mathbb Q $ in $\mathbb Q (\sqrt[5]{2})(\zeta_5)$, that is $i:$ $\mathbb Q $ $\hookrightarrow$ $\mathbb Q (\sqrt[5]{2})(\zeta_5)$

Then the minimal polynomial of $\sqrt[5]{2}$ over $\mathbb Q $ is $x^5 - 2$. So i($x^5 - 2$) = $x^5 - 2$ and $x^5 - 2$ has roots $(-1)^{2/5}$$\sqrt[5]{2}$, $-(-1)^{3/5}$$\sqrt[5]{2}$, $(-1)^{4/5}$$\sqrt[5]{2}$ in $\mathbb Q (\sqrt[5]{2})(\zeta_5)$.

Hence there is a monomorphism $j_k$ extending i such that $j_k$: $\mathbb Q (\sqrt[5]{2})$ $\hookrightarrow$ $\mathbb Q (\sqrt[5]{2})(\zeta_5)$

Given by:

$j_k$: $\sqrt[5]{2}$ $\mapsto$ $\zeta_5^k$$\sqrt[5]{2}$ for k = 0, 1, 2, 3, 4.

The minimal polynomial of $\zeta_5$ over $\mathbb Q (\sqrt[5]{2})$ is $x^4 + x^3 + x^2 + x + 1$. So $j_k$($x^4 + x^3 + x^2 + x + 1$) = $x^4 + x^3 + x^2 + x + 1$ and has roots $-(-1)^{1/5}$, $(-1)^{2/5}$, $-(-1)^{3/5}$, $(-1)^{4/5}$.

Hence there is a monomorphism (which is actually an automorphism) $\sigma_{k,l}$ extending $j_k$ such that:
$\sigma_{k,l}$: $\mathbb Q (\sqrt[5]{2})(\zeta_5)$ $\hookrightarrow$ $\mathbb Q (\sqrt[5]{2})(\zeta_5)$.

Given by:

$\sigma_{k,l}$: $\sqrt[5]{2}$ $\mapsto$ $\zeta_5^k$$\sqrt[5]{2}$ for k = 0, 1, 2, 3, 4.

$\sigma_{k,l}$: $\zeta_5$ $\mapsto$ $\zeta_5^l$ for l = 1, 2, 3, 4.


问:
Improper integral doubt
I have to study the convergence of the following improper integral
$$\int_0^{+\infty} \frac{x^\alpha \sinh(\beta x)}{(\sinh(x))^\gamma}dx$$

So I split it
$$\int_0^a \frac{x^\alpha \sinh(\beta x)}{(\sinh(x))^\gamma}dx + \int_a^{+\infty} \frac{x^\alpha \sinh(\beta x)}{(\sinh(x))^\gamma}dx$$
the first is asymptotic to
$$\frac{\beta x^{\alpha+1}}{x^\gamma}=\frac{\beta}{x^{\gamma - \alpha -1}} $$
so it for $\beta=0$ converges, for $\gamma \ge \alpha +2$ diverges and for $\gamma< \alpha +2$ converges.

The second is asymptotic to
$$x^\alpha\frac{e^{x(\beta - \gamma)}}{2^{1-\gamma}} $$
Is that correct?

答:
The second integral is equal to $0$ if $\beta = 0$, otherwise it is asymptotic to
\begin{equation}
\text{sgn}(\beta)\frac{x^\alpha e^{(|\beta| -\gamma) x}}{2^{1-\gamma}}
\end{equation}
Hence it converges if $|\beta| -\gamma < 0$ or if $|\beta|-\gamma=0$ and $\alpha < -1$.

Title: Improper integral doubt


Title: Integral of $1/x^2$ without power rule

Body:
I was wondering if it was possible to evaluate the following integral without using the power rule for negative exponents

\begin{equation*}
  \int \frac{1}{x^2} \; dx
\end{equation*}

When using integration by parts, you end up with the same integral in the rhs so it seems out of luck

\begin{equation*}
  \int \frac{1}{x^2} \; dx = \frac{\ln x}{x} + \int \frac{\ln x}{x^2} \; dx
\end{equation*}

This question is inspired by this blackpenredpen's video

Using integration by parts with $u = \frac{1}{x^2}$ and $v = x$ is NOT accepted. If you write

\begin{equation*}
  \int \frac{1}{x^2} \; dx = \frac{1}{x} + 2 \int \frac{1}{x^2} \; dx
\end{equation*}

you are still implicitly using the power rule to compute the derivative of $\frac{1}{x^2}$ so it is not correct per the rules.

The same rules apply for $u$-substitution which implicitly use the power rule. See the following with $u = \frac{1}{x}$ such that

\begin{equation*}
    \int \frac{1}{\left(\frac{1}{x}\right)^2} \left(\frac{1}{x}\right)' \; dx = \int \frac{1}{u^2} \; du
\end{equation*}

and here the power rule is also considered used to compute the derivative of $\frac{1}{x}$ although it can be subject to discussion (geometric proof, limit definition of derivative, etc...)

Body:
$$I=\int \frac{1}{x^2} \, dx$$

Let $x=\cot(y)$, with $\frac{dx}{dy}=-\csc^2(y)$

$$I=-\int \frac{\csc^2(y)}{\cot^2(y)} \, dy=-\int \sec ^2(y) \, dy=-\tan(y)+C=-\frac{1}{x}+C$$

or

Let $x=\sec(y)$, with $\frac{dx}{dy}=\sec(y)\tan(y)$

$$I=\int \frac{\sec(y)\tan(y)}{\sec^2(y)} \, dy=\int \sin(y) \, dy=-\cos(y)+C=-\frac{1}{x}+C$$


答:
Firstly, a fairly standard integral that is indeed useful in your case:

$$\int\frac{1}{\sqrt{a^2+x^2}}dx=\text{arsinh}~\frac{x}{a}+C$$
Secondly, in your final integral we can factorise the integrand and reduce it to something very easy (if you know the integral of $\sec x$):

$${\displaystyle\int}\dfrac{\sec(v)\tan(v)+\sec^2(v)}{\tan(v)+\sec(v)}\,\mathrm{d}v={\displaystyle\int}\dfrac{\sec(v)(\tan(v)+\sec(v))}{(\tan(v)+\sec(v))}\,\mathrm{d}v=\int\sec(v)\mathrm{d}v$$
which is just
$$\ln\lvert\sec(v)+\tan(v)\rvert +C$$
I hope that helps.


问:
Use the one to one property of logs to solve $\ln(x^2-10)+\ln(9)=\ln(10)$
Use the one to one property of logs to solve $\ln(x^2-10)+\ln(9)=\ln(10)$.

I get $x=\sqrt{11}$ or $x=10$ whereas my textbook says it's $x=\pm\frac{10}{3}$.

My working - initial attempt:
$$x^2-10+9=10$$
$$x^2-1=10$$
$$x^2=11$$
$$x=\sqrt{11}$$

My working - another attempt:
$$\ln(\frac{x^2-10}{9})=\ln(10)$$
$$\frac{x^2-10}{9}=10$$
$$x^2-10=90$$
$$x^2=100$$
$$x=10$$

Where am I going wrong? How can I arrive at $x=\pm\frac{10}{3}$? Don't both my solutions make sense? Why are they incorrect?

答:
Note that\begin{align*}\ln(x^2-10)+\ln(9)=\ln(10)&\iff\ln\bigl(9(x^2-10)\bigr)=\ln(10)\\&\iff9x^2-90=10\\&\iff x^2=\frac{100}9\\&\iff x=\pm\frac{10}3.\end{align*}

The error from your first attempt lies in assuming that$$\ln(x^2-10)+\ln(9)=\ln(10)\iff x^2-10+9=10,$$whereas the error in your second attempt lies in assuming that$$\ln(x^2-10)+\ln(9)=\ln\left(\frac{x^2-10}9\right).$$


答:
Your first attempt simply uses the strategy "let's ignore $\ln$": but it does not work.

The second attempt is better, but you made a mistake. Actually
$$\ln (x^2-10) + \ln 9$$
becomes
$$\ln ((x^2-10) \cdot 9)$$
then you have
$$(x^2-10) \cdot 9=10$$ which is easily solved.

Body:
Use the one to one property of logs to solve $\ln(x^2-10)+\ln(9)=\ln(10)$.

I get $x=\sqrt{11}$ or $x=10$ whereas my textbook says it's $x=\pm\frac{10}{3}$.

My working - initial attempt:
$$x^2-10+9=10$$
$$x^2-1=10$$
$$x^2=11$$
$$x=\sqrt{11}$$

My working - another attempt:
$$\ln(\frac{x^2-10}{9})=\ln(10)$$
$$\frac{x^2-10}{9}=10$$
$$x^2-10=90$$
$$x^2=100$$
$$x=10$$

Where am I going wrong? How can I arrive at $x=\pm\frac{10}{3}$? Don't both my solutions make sense? Why are they incorrect?


问:
I need to find a solution for $2x+\arcsin x=1$
$$2x+\arcsin x=1$$
I didn't find way to solve this problems.
Is there a difference in solving equations between $\sin$ and $\arcsin$?
For some reason I did not find the special solutions of arc.

Is there a connection to Maclaurin or Taylor?

Thanks for your help.

Body:
Consider
$$y=2x+\sin^{-1}(x)=3x+\sum_{n=1}^\infty \frac{ (2 n)! }{4^n(2 n+1) (n!)^2}x^{2 n+1}$$ Truncate the summation and use series reversion to obtain
$$x = \sum_{n=0}^p a_n\,t^{2n+1} \qquad \text{where} \qquad t=\frac y 3$$ where the first $a_n$'s make the sequence
$$\left\{1,-\frac{1}{18},-\frac{17}{1080},-\frac{793}{136080},-\frac{72161}{29393280}
   ,-\frac{10829449}{9699782400},-\frac{2424554513}{4539498163200}\right\}$$

Using $y=1$, so $t=\frac 13$, this would give as an estimate
$$x \sim \frac{2397094567444716499}{7237426330047513600} \sim 0.331208147500$$ while the solution obtained by Newton method is
$0.331208147481$


问:
A question about the (un)derivability of Cut Rule in Sequent Calculus
This question is motivated by a previous discussion regarding How to show that a valid rule is not derivable in Intuitionistic propositional calculus.

The reference is to Gentzen (1934-35)'s Sequent Calculi $\mathsf {LJ}$ and $\mathsf {LK}$ for Intuitionistic and Classical logics respectively, or some equivalent presentation, like those in Sara Negri and Jan von Plato, Structural Proof Theory (Cambridge UP, 2008).

The basic result of structural Proof Theory is the Cut-Elimination Theorem (or Gentzen's Hauptsatz), showing that $\text {Cut Rule}$ is admissible, were admissibility of a rule in logic means that

the set of theorems of the system does not change when that rule is added to the existing rules of the system. In other words, every formula that can be derived using that rule is already derivable without that rule, so, in a sense, it is redundant.

More formally, a rule having the form:

$$\dfrac {\Sigma}{\alpha \vdash A}$$

is said to be derivable (in e.g. $\mathsf {LK}$) iff there exists a proof (in $\mathsf {LK}$) of $\alpha \vdash A$ from the hypotheses $\Sigma$.

I'll try with a silly example of derived rule in $\mathsf {LJ}$: Contraposition.

We start from $\Gamma, \varphi \vdash \psi$ and derive $\Gamma, \varphi , \lnot \psi \vdash$ followed by $\Gamma, \lnot \psi \vdash \lnot \varphi$.

This derivation amounts to the "new rule":

$\dfrac{\Gamma, \varphi \vdash \psi}{\Gamma, \lnot \psi \vdash \lnot \varphi}.$

Classical propositional calculus is structurally complete, meaning that every admissible rule is derivable, where a rule is derived if its conclusion is deducible from the premises by the rules of the calculus.

Intuitionistic propositional calculus is not.

1st Question: when we say that "Classical propositional calculus is structurally complete", does it mean CPC with $\text {Cut}$?

2nd Question: in order to show that $\text {Cut}$ is not derivable is it enough to observe that $\text {Cut}$ is the only rule that deletes a formula?

答:
Short answer. No, cut rule is not derivable in $\mathsf{LK} \smallsetminus \{\text{cut}\}$ (classical sequent calculus) or $\mathsf{LJ} \smallsetminus \{\text{cut}\}$ (intuitionistic sequent calculus).

More precisely, to answer your questions.

When someone says that classical propositional logic is structurally complete, it refers to a family of particular proof systems, which does not include sequent calculus. Actually, sequent calculus for classical propositional logic is not structurally complete because cut rule is not derivable.

Essentially, yes! See my long answer below.

Long answer.
First, let us fix some definitions. Note that Wikipedia's page about admissibility and derivability of inference rules refers to rules that act on formulas (like the one of natural deduction), not on sequents (like the one of sequent calculus). Similarly, the result about structural completeness you mentioned refers to proof systems acting on formulas, not on sequents.

So, what does it mean that cut rule is admissible in a sequent calculus? There are two possible definitions.

Definition 1 (closer to the one for proof systems acting on formulas): cut rule is admissible in $\mathsf{LK} \smallsetminus \{\text{cut}\}$ if $\Gamma, \Sigma \vdash \Delta, \Theta$ is provable in $\mathsf{LK} \smallsetminus \{\text{cut}\}$ whenever both $\Gamma, A \vdash \Delta$  and $\ \Sigma \vdash A, \Theta$  are $\mathsf{LK} \smallsetminus \{\text{cut}\}$. Similarly for admissibility in $\mathsf{LJ} \smallsetminus \{\text{cut}\}$.

Definition 2 (closer to the intended meaning, from the quote in the OP): cut rule is admissible in $\mathsf{LK} \smallsetminus \{\text{cut}\}$ if $\Gamma \vdash \Delta$ is provable in $\mathsf{LK} \smallsetminus \{\text{cut}\}$ whenever $\Gamma \vdash \Delta$ is provable in $\mathsf{LK}$. Similarly for admissibility in $\mathsf{LJ} \smallsetminus \{\text{cut}\}$.

It is easy to show to the two definitions are equivalent.

What does it mean that cut rule is derivable in a sequent calculus? In my opinion, the only definition that makes sense is the one proposed in the OP. I spell it out in the case of cut rule, for the sake of completeness.

Definition. Cut rule is derivable in $\mathsf{LK} \smallsetminus \{\text{cut}\}$ if $\Gamma, \Sigma \vdash \Delta, \Theta$ is provable in $\mathsf{LK} \smallsetminus \{\text{cut}\}$ from the hypotheses $\Gamma, A \vdash \Delta$ and $\Sigma \vdash A, \Theta$ (i.e. the leaves of the derivation-tree with conclusion $\Gamma, \Sigma \vdash \Delta, \Theta$ can be either axioms of the form $B \vdash B$ or the hypotheses $\Gamma, A \vdash \Delta$ and $\Sigma \vdash A, \Theta$), for any formula $A$. Similarly for derivability in $\mathsf{LJ} \smallsetminus \{\text{cut}\}$.

Proposition. Cut rule is not derivable either in $\mathsf{LK} \smallsetminus \{\text{cut}\}$ or in $\mathsf{LJ} \smallsetminus \{\text{cut}\}$.

Proof. Let $A$ and $B$ be distinct formulas such that $A$ is not a subformula of $B$. From the hypotheses $A \vdash B$ and $\vdash A$ it is impossible to derive $\vdash B$ in $\mathsf{LK} \smallsetminus \{\text{cut}\}$ or in $\mathsf{LJ} \smallsetminus \{\text{cut}\}$, because in $\mathsf{LK} \smallsetminus \{\text{cut}\}$ and in $\mathsf{LJ} \smallsetminus \{\text{cut}\}$ there is no inference rule to eliminate the formula $A$, hence the conclusion of any derivation from the hypotheses $A \vdash B$ and $\vdash A$ in $\mathsf{LK} \smallsetminus \{\text{cut}\}$ or in $\mathsf{LJ} \smallsetminus \{\text{cut}\}$ must contain $A$ as a subformula. $ \ \square$

The point of the proof is that cut rule is the only inference rule in a sequent calculus that allows you to remove something from the premises.

Apart from the proof, there is a maybe even more convincing argument. Cut-elimination theorem (i.e. admissibility of cut rule) is a result whose proof to prove in sequent calculi such as $\mathsf{LK}$ and $\mathsf{LJ}$. It requires advanced techniques and notions. It cut rule were derivable, the proof of cut-elimination theorem would be much easier! Quoting from a paper by Ono (Proof-Theoretic Methods for Nonclassical Logic, p. 217):

the cut rule is not a derivable rule in the system obtained from LJ by
deleting the cut rule. That is, it is impossible to replace the cut rule in a uniform way by repeated applications of other rules, as each application of the cut rule will play a different role in a given proof. Therefore, we have to eliminate the cut rule depending on how it is applied.

An aside comment. In my opinion, the fact that structural completeness holds for a proof system for classical propositional logic acting on formulas but does not hold for a proof system for classical propositional logic acting on sequents, shows that the notion of derivability for inference rules is not robust, because it depends on the proof system to which it refers. I would say that such a notion is too syntactical to be meaningful.


问:
Regular rational functions over real projective lines and projective spaces.
In Algebraic Geometry one usually says that "working with algebraically closed fields makes life easier". Today I stumbled over one instance of this: suppose $X$ is an irreducible projective variety over a field $K$. Over algebraically closed fields it is known that there are no non-constant (ie, $\not\in\,K$) rational functions regular on $X$. In symbols $$\Gamma(X,\mathcal{O}_X)=K.$$

It turns out that this seems to be false if $K$ is not algebraically closed: take $K=\mathbb{R}$ and look at the projective line $X=\mathbb{P}^1_{\mathbb{R}}$ and the rational function $f=\dfrac{x_0^2}{x_0^2+x_1^2}$ where $(x_0:x_1)$ are projective coordinates on $X$. It is non-constant and regular everywhere.

On the other hand, looking at Hartshorne's book on AG, I discovered Theorem III, 5.2 from where one deduces that under same assumptions on $X$ we have $$\Gamma(X,\mathcal{O}_X)=K^r,$$ for some $r\geqslant 1$ because, after all $K\subseteq \Gamma(X,\mathcal{O}_X)$ is still valid. What I was not able to find is how to compute this integer $r$.

Question: is $r=\dim_K\,\Gamma(X,\mathcal{O}_X)$ computable in terms of known invariants of $X$ for (at least) projective, irreducible curves over an arbitrary field $K$?

A generalization of course would be (in case of curves) how to compute $\dim_K\,\Gamma(X,\mathcal{O}_X(nP))$ where $P$ is a rational point of X. If relevant, I'm interested in $K=k(t)$ where $k$ is algebraically closed.

Thanks!

答:
You may define a locally ringed space $(X,\mathcal{O}_X)$ with $X:=\mathbb{P}^1_{\mathbb{R}}$ the real projective line in the sense of "real algebraic geometry", and the topological space $X$ is by definition the space of $\mathbb{R}$-rational points of the projective scheme $Proj(\mathbb{R}[x_0,x_1,x_2])$ with the induced topology. The structure sheaf $\mathcal{O}_X$ is discussed below. You may construct a closed embedding

$$ i:\mathbb{P}^1_{\mathbb{R}} \rightarrow \mathbb{A}^d_{\mathbb{R}}$$

where $\mathbb{R}^d_{\mathbb{R}}$ is affine $d$-space in the sense of "real algebraic geometry". The embedding $i$ exists because of regular functions of the type

$$\frac{1}{x^2+y^2+1}.$$

Hence the "real projective line" (and more generally the "real projective space") is affine and algebraic.

Remark: "It turns out that this seems to be false if $K$ is not algebraically closed: take $K=R$ and look at the projective line $P^1_R$ and the rational function $f=\frac{x^2_0}{x^2_0+x^2_1}$ where $(x_0:x_1)$ are projective coordinates on $X$. It is non-constant and regular everywhere."

Answer: You may use functions like $f$ to construct an explicit "algebraic" embedding as follows:

$$i: \mathbb{P}^1_{\mathbb{R}} \rightarrow \mathbb{A}^3_{\mathbb{R}}$$

defined by

$$i(x:y):=(\frac{x^2}{x^2+y^2}, \frac{xy}{x^2+y^2}, \frac{y^2}{x^2+y^2}).$$

You may check that $i$ is well defined and a "closed embedding". You are propably aware that $Proj(\mathbb{R}[x_0,x_1,x_2])$ cannot be realized as an affine variety/scheme.

Note: There is nothing "naive" about this defintition as is being claimed: "real algebraic geometry" is an independent field of research.

At this link you find s discussion of the ringed space $(X,\mathcal{O}_X)$:

https://math.stackexchange.com/posts/3984082/edit

If you define $\mathcal{O}(\mathbb{A}^1_{\mathbb{R}})$ as the set of functions with the following property (P1):

P1. $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ with the property that for any point $p\in \mathbb{R}^2$ there is an open subset $U(p)\subseteq \mathbb{R}^2$ and two polynomials $p(x),q(x)\in \mathbb{R}[x]$ such that the restriction $f_{U(p)}$ of $f$ to $U(p)$ satisfies $f_{U(p)}=\frac{p(x)}{q(x)}$,

then it follows (as you noted) the function $\frac{1}{x^2+1}$ is a regular function on $\mathbb{R}^2$. Hence $\mathcal{O}(\mathbb{A}^1_{\mathbb{R}})\neq \mathbb{R}[x]$.  If $z=a+ib \in \mathbb{C}$ with $b\neq 0$, it follows
the polynomial $p_z(x):=(x-z)(x-\overline{z})=x^2-2ax+a^2+b^2\in \mathbb{R}[x]$ is an irreducible polynomial - it has no real roots.
Hence if you use definition P1 to define the ring of regular functions, it follows the function $f(x):=\frac{1}{p_z(x)}$ is a regular function. If $S\subseteq A:=\mathbb{R}[x]$ is the set of polynomials on the form $p(x):=\prod_{z_i\in \mathbb{C}-\mathbb{R}} (x-z_i)(x-\overline{z_i})$ it follows $S$ is a multiplicatively closed subset, and the ring of fractions $S^{-1}A$ is a sub-ring of $\mathcal{O}(\mathbb{A}^1_{\mathbb{R}})$. Hence with definition P1 it follows the ring $\mathcal{O}(\mathbb{A}^1_{\mathbb{R}})$ is much larger than $A$.

Question: "The regular functions on $\mathbb{A}^n$ are exactly the polynomials in n variables [...], which I only agree with for K
algebraically closed and not in the case I've shown above."

Answer: You want (for many reasons) to recover the polynomial ring $\mathbb{R}[x]$ as the ring of global sections of the structure sheaf. This problem is "solved" by the affine spectrum $Y:=Spec(\mathbb{R}[x])$. What you define above is a "sheaf of rings" on the real algebraic variety $\mathbb{A}^1_{\mathbb{R}}$ in the sense of "real algebraic geometry", and this sheaf differ from the sheaf defined using the affine spectrum in the sense of Hartshornes book "Algebraic geometry". The topological space of the variety $X:=\mathbb{A}^1_K$ you consider above is the set of real numbers $r\in \mathbb{R}$. The topological space of $Y$ consists of all real numbers plus the set of all maximal ideals in $A:=\mathbb{R}[x]$ with residue field the complex numbers. Hence the two topological spaces differ. Your space $X$ is by definition the set of rational points $Y(K)$ with the induced topology.

Note: There is nothing "wrong" with the ringed space $(X, \mathcal{O}_X)$ - it has properties different from $Y$. In real algebraic geometry it follows projective space $\mathbb{P}^n_K$ may be realised as a real affine algebraic variety - there is a closed embedding into real affine space. Moreover any real projective variety is affine. The reason for this is that the structure sheaf has more local sections, as you have observed in your comment above. Hence there are "more maps". This does not happen with the Hartshorne definition: You cannot construct a closed embedding of $Proj(k[x_0,..,x_n])$ into
affine space $Spec(k[y_1,..,y_d])$.

As an example: The global section $s(x):=\frac{1}{p_z(x)}$ for $z\in\mathbb{C}-\mathbb{R}$ is a unit - the polynomial $p_z(x)$ is non-zero on $\mathbb{R}$, and with your definition of "regular map" it follows $s(x)$ is a global section - it lives in $\mathcal{O}_{\mathbb{A}^1_K}(\mathbb{A}^1_K)$. The function $s(x)$ is not a global section of $\mathcal{O}_Y(Y)\cong \mathbb{R}[x]$.

Real algebraic varieties are relevant for people studying complex algebraic varieties and their topology. You may for any complex projective manifold $X \subseteq \mathbb{P}^n_{\mathbb{C}}$ construct the Weil restriction $W_{\mathbb{C}/\mathbb{R}}(X)$. If $dim(X)=d$ it follows  $dim(W_{\mathbb{C}/\mathbb{R}}(X))=2d$ and $W_{\mathbb{C}/\mathbb{R}}(X)$ is a smooth scheme over $\mathbb{R}$ - the "underlying" real algebraic variety of $X$. The variety
$W_{\mathbb{C}/\mathbb{R}}(X)$ has the same topology as $X$ and you may realize $W_{\mathbb{C}/\mathbb{R}}(X)$ as a real affine algebraic variety. Hence the "underlying" real algebraic variety of $X$ is affine. Moreover: Any real affine algebraic variety may (topologically) be realized as a hypersurface. Hence if you study the topology of $X$ it follows there is in some cases a real affine hypersurface $Z(f)$ with the same topology as $X$ (here I am extremely "vague"): If $X\subseteq \mathbb{P}^N_{\mathbb{C}}$ is a complex projective variety of dimension $d$
there is a polynomial $f\in \mathbb{R}[y_1,..,y_{2d+1}]$ with the property that $X$ and $Y:=Z(f)\subseteq \mathbb{A}^{2d+1}_{\mathbb{R}}$ have the "same topology".

Exampl. If we consider the complex projective line $X:=\mathbb{P}^1_{\mathbb{C}}$ and it's "underlying" real algebraic variety $X(\mathbb{R})$,
there is an isomorphism

$$X(\mathbb{R}) \cong Z(f)\subseteq \mathbb{R}^3$$

where $f:=x^2+y^2+z^2-1\in \mathbb{R}[x,y,z]$. Hence the complex projective line, viewed as a real algebraic variety, is isomorphic to the real 2-sphere and in this case we may give an explicit realization of $X(\mathbb{R})$ as a hypersurface in $\mathbb{R}^3$. To construct such an explicit isomorphism we need functions on the form $f(t):=\frac{1}{t^2+1}$ and $g(x,y):=\frac{u(x,y)}{x^2+y^2}$.

You find this discussed here:

https://math.stackexchange.com/posts/3957121/edit

and here:

Different definitions of regular map (between affine varieties)?


问:
Greatest common divisor on a factorial domain proof (Integers)
Hello fellow mathematicians or students. I need some conceptual help on a proof, and I think it is not that hard, but I am just starting to study Ideals and Rings (Abstract Algebra in general to be accurate), that must be my difficulty.

Sometimes, I dare say almost always, the most simple things can be the most difficult ones to prove. And some of them are so simple that they are impossible to prove (axioms). With no further ado I'll introduce the problem.

Consider $m = q_{1}^{a_1} \cdots q_{t}^{a_t}$ and $n = q_{1}^{b_1} \cdots q_{t}^{b_t}$, where $q_i$ are prime integers and $a_i,b_i$ are non-negative integers, with $i \in \{1,\dots, t\}$.

Prove that $\gcd\{m,n\} = q_{1}^{c_1} \cdots q_{t}^{c_t}$ where $c_i = \min \{a_i,b_i\}$.

Okay, so I started with:

Let's call $d = \gcd\{m,n\}$. By definition follows that $d \mid m$. Indeed
$$
\begin{array}{ccc}
   m &=& q_{1}^{a_1} \cdots q_{t}^{a_t} \\[0.3cm]
   &=& q_{1}^{a_1 + c_1 - c_1} \cdots q_{t}^{a_t + c_t - c_t}\\[0.3cm]
   &=& q_{1}^{c_1 + a_1 - c_1} \cdots q_{t}^{c_t + a_t - c_t} \\[0.3cm]
   &=& q_{1}^{c_1}q_{1}^{a_1 - c_1} \cdots q_{t}^{c_t}q_{t}^{a_t - c_t}\\[0.3cm]
   &=& (q_{1}^{c_1} \cdots q_{t}^{c_t}) \cdot (q_{1}^{a_1 - c_1} \cdots q_{t}^{a_t - c_t}) \\[0.3cm]
   &=& d \cdot r,
\end{array}
$$
where $r = q_{1}^{a_1 - c_1} \cdots q_{t}^{a_t - c_t} \in \mathbb{Z}$. Also, $d \mid n$ can be shown analogously.

Now is where my problems begin. I need to show that

If $d' \in \mathbb{Z}$ is such that $d' \mid m$ and $d' \mid n$, then $d' \mid d$.

I'm just getting $m = d' \cdot r' = d \cdot r$ and $n = d' \cdot s' = d \cdot s$ and I don't know what to do next.

So, my path is correct? Is there another way to show this result?

Thanks in advance. Have a great day!

答:
I finally did find a way to prove this statement, and it goes as it follows:

Let there be a $d' \in \mathbb{Z}$ such that $d' \mid m$ and $d' \mid m$. It's factorization is given by $$d' = {q_1}^{e_1} \cdots {q_t}^{e_t}$$ and it is possible to conclude that there is no other prime $q_{t+1}$ in it's factorization, otherwise $d' \nmid m$ and $d' \nmid n$.

Suppose by absurd that there exists $j \in \{1, \dots, t\}$ such that $e_j > c_j$, where $c_j = \min\{a_j,b_j\}$. If $c_j = a_j$, then $d' \nmid m$. If $c_j = b_j$, then $d' \nmid n$. Which are both contradictions, ergo $e_i \leq c_i$ for all $j \in \{1, \dots, t\}$, id est, $d' \nmid d$.

Therefore, $\gcd\{m,n\} = {q_1}^{c_1} \cdots {q_t}^{c_t}$ where $c_i = \min\{a_i,b_i\}$. cqd

I deeply appreciate everyone that tried to help me, thank you! Have a great day!


问:
Definition and Doubt: Universal $\mathrm{C}^*$-Algebras given by generators and relations
Let $A:=\mathrm{C}^*(G,R)$ be the (defined) universal $\mathrm{C}^*$-algebra generated by generators $G$ and relations $R$. As far as I can discern, there are two ways of defining such objects --- either using category theory or non-commutative polynomials and ideals.

The big ticket is that given any $\mathrm{C}^*$ algebra $A$ with $|G|$ generators that satisfy $R$ there is a unique surjective $*$-homomorphism $\mathrm{C}^*(G,R)\rightarrow A$. This is the universal property.

I have an informal way of thinking about universal $\mathrm{C}^*$-algebras (that will shortly come into conflict with my second question).

My informal intuition for universal $\mathrm{C}^*$-algebras is that we more or less form a large direct sum of all $\mathrm{C}^*$-algebras that have $|G|$ generators which satisfy $R$ and that is where the universal property comes from (or sums of) projections $\bigoplus_\alpha A_\alpha\rightarrow A_\beta$.

Question 1: How bad is this intuition? Is there something big that it misses (forget if it is formally incorrect)?

That is more or less the question on definition. Now for my doubt. We can consider a particular universal $\mathrm{C}^*$-algebra $B$ which is generated by a magic unitary $u=(u_{ij})_{i,j=1}^4$. That $u$ is a magic unitary is to say that each entry is a projection $u_{ij}=u_{ij}^2=u_{ij}^*$ and that each row and column is a partition of unity:

$$\sum_{k=1}^4u_{ik}=1=\sum_{k}u_{kj}.$$

In the compact quantum group community (say p.8) this universal $\mathrm{C}^*$-algebra is the setting for the algebra of continuous functions on the compact quantum group of permutations. The definition (of compact quantum group) requires a $*$-homomorphism $\Delta :B\rightarrow B\underset{\min}{\otimes} B$ and how this is constructed in this example is to define

$$\Delta(u_{ij})=\sum_{k=1}^4u_{ik}\otimes u_{kj},$$

and the way it is shown that this is a $*$-homomorphism is to show that the matrix:

$$\left[\Delta(u_{ij})\right]_{i,j=1}^4\in M_4(B\otimes B)$$

is a magic unitary, and so $\Delta:B\rightarrow B\otimes B$ is a *-homomorphism. Then the confusion comes in because in my informal intuition from a above $B$ is the largest object with magic unitaries... but here we have $B\otimes B$ which in my understanding is LARGER again than $B$ is also a sub-object of it...

Question 2: Is there a way to resolve this confusion with my informal intuition of what a universal $\mathrm{C}^*$-algebra (perhaps by arguing that the $\mathrm{C}^*$-algebra generated by $\left[\Delta(u_{ij})\right]_{i,j=1}^4$ while living in a large space $B\otimes B$, is in fact smaller than $B$... or is my informal intuition messing up my understanding here?

答:
Question 1:  Your intuition is not only correct, but it also underlies the proof of the existence of the universal algebra.

Question 2: The universal property is only relevant in the domain.  All that is asked about the co-domain is that the relations be satisfied.


问:
Cauchy sequence of positive terms, find it's limit.
Consider the sequence of positive terms, where $x_n$ it's difined by
\begin{equation}
x_1=3\ and\ x_{n+1}=\frac{3(1+x_n)}{3+x_n}.
\end{equation}
Find   $a=\underset{n\rightarrow \infty}\lim x_n$.

All that I have found is that exists $c=0.2$ such that
\begin{equation}
|x_{n+2}-x_{n+1}| \leq 0.2|x_{n+1}-x_n|.
\end{equation}

I also tried to prove that
\begin{equation}
x_{k+1} \leq x_k,
\end{equation}
but I failed.

答:
Note that $x_{n+1}=f(x_n)$ where
$$ f(x)=\frac{3(1+x)}{3+x}=3-\frac6{x+3}.$$
We have $f(x)>3-\frac63>0$ for all $x\ne -3$.

Now if $x>y>0$ then $x+3>y+3$, then $\frac1{x+3}<\frac1{y+3}$, then $-\frac1{x+3}>-\frac1{y+3}$, and ultimately $f(x)>f(y)>0$. Therefore, $x_1=3>2=x_2$ implies by induction that $x_n>x_{n+1}>0$ for all $n$, i..e, the sequence is strictly decreasing and bounded from below, hence convergent to some $a$. For this $a$, we must have $a=f(a)$.


答:
Clearly $x_n\ge0$ for all $n\in N$. Let $y_n=x_n-\sqrt3$ and then $y_n+\sqrt3\ge0$. Note
$$ |y_{n+1}|=\bigg|\frac{(3-\sqrt3)y_n}{3+\sqrt3+y_n}\bigg|\le\frac{3-\sqrt3}{3}|y_n| $$
and hence
$$ |y_n|\le\bigg(\frac{3-\sqrt3}{3}\bigg)^{n-1}|y_1|. $$
So
$$ \lim_{n\to\infty}y_n=0 $$
or
$$ \lim_{n\to\infty}x_n=\sqrt3. $$

Title: Cauchy sequence of positive terms, find it's limit.

Body:
Consider the sequence of positive terms, where $x_n$ it's difined by
\begin{equation}
x_1=3\ and\ x_{n+1}=\frac{3(1+x_n)}{3+x_n}.
\end{equation}
Find   $a=\underset{n\rightarrow \infty}\lim x_n$.

All that I have found is that exists $c=0.2$ such that
\begin{equation}
|x_{n+2}-x_{n+1}| \leq 0.2|x_{n+1}-x_n|.
\end{equation}

I also tried to prove that
\begin{equation}
x_{k+1} \leq x_k,
\end{equation}
but I failed.