问:
Complete connected manifold implies nonextendible
If $M$ is a connected riemannian manifold, we can define it as extendible if there is another connected riemannian manifold $M'$ and there exists a proper open $A \subset M'$ ($A \neq M'$) such that $M$ is isometric to $A$.

Well, let $M$ be a connected riemannian manifold, I have to see that if $M$ is complete, then $M$ is nonextendible. I cannot see how to show it, however I have proved a characterization of completeness, which could be helpful: a riemannian manifold $M$ is complete iff each divergent curve in $M$ has no finite length (a curve $\gamma : [0 , 1) \to M$ is divergent if for each compact $K \subset M$ there exists $t \in [0 , 1)$ such that $\gamma([0 , 1)) \cap K = \emptyset$.

The beginning probably is taking a connected extendible manifold $M$, but how can I take a Cauchy and not convergent sequence $\{x_n\}$ in $M$?

答:
For the proof it suffices to show that if $A \subset M'$ is a proper, nonempty, open subset of a connected Riemannian manifold then $A$ is not complete.

For the proof, pick a point $p \in A$, and a point $q \in M'-A$, and a smooth path $\gamma : [0,1] \to M'$ such that $\gamma(0)=p$ and $\gamma(1)=q$ (the existence of this path follows from a theorem of topology saying that every connected, locally path connected topological space is path connected).

Let ${\cal T}  = \gamma^{-1}(M'-A) = \{t \in [0,1] \mid \gamma(t) \in M'-A\}$. This set ${\cal T}$ is a closed subset of $[0,1]$, because $A$ is open in $M'$ hence $M'-A$ is closed in $M'$. Also ${\cal T}$ is nonempty, because $1 \in \cal T$. Also $0 \not \in T$.

Let $T = \min {\cal T}$, and so $T > 0$. Choose a sequence $t_n \in [0,T)$ that converges to $T$. It follows that $x_n = \gamma(t_n)$ is a sequence in $A$ that converges to $\gamma(T) \in M'-A$. Thus, $x_n$ is a Cauchy sequence in $A$ that has no limit in $A$, so $A$ is not complete.


问:
From Ideals, Varieties and Algorithms: Monomial Ideals
From Cox, Little and O'Shea's book Ideals, Varieties and Algorithms. I really don't understand their proof on the following lemma about monomial ideals.

Let $I=\langle x^{\alpha}|\alpha \in A\rangle$ be a monomial ideal. Then a monomial $x^{\beta}$ lies in $I$ iff $x^{\beta}$ is divisible by $x^{\alpha}$ for some $\alpha \in A$.

Here's the proof.

If $x^{\beta}$ is a multiple of $x^{\alpha}$ for some $\alpha \in A$ then $x^{\beta} \in I$ by definition of an ideal. Conversely, $x^{\beta} \in I$ then $x^{\beta} =\sum_{i=1}^s h_ix^{\alpha (i)}$ where $h_i \in k[x_1,...,x_n]$ and $\alpha (i) \in A$. If we expand each $h_i$ as a linear combination of monomials we see that every term on the right side of the equation is divisible by some $x^{\alpha (i)}$. Hence the left side must have the same property.

Firstly, $x^{\beta}$ is a "monomial" right? If I understood monomials correctly, it should be $x_1^{\alpha _1}x_2^{\alpha _2}...$, so basically, it's a multiple of exponents of variables. However, $x^{\beta} =\sum_{i=1}^s h_ix^{\alpha (i)}$ seems to imply, since it's a "sum" notation, that the monomial $x^{\beta}$ is somehow a "linear combination" of monomials (or polynomials, but in the end, I guess it is essentially the same as linear combinations of monomials).

How? I can only think of expressing a monomial as a linear combination of monomials only with every single coefficient being $0$ except for the term where the exponent $n$-tuple ($\beta$ in this case) coincide. Then, why bother with the "sigma" notation? Isn't it redundant?

Second, what does it mean by "expand $h_i$ as a linear combination of monomials"? Does it literally want me to express each coefficient $h_i$ in terms of a liner combination of monomials? Even if so, how does this immediately tell me that "every term on the right side of the equation is divisible by some $x^{\alpha (i)}$?

I mean, so this proof is saying $x^{\beta}= h_1x^{\alpha (1)}+h_2x^{\alpha (2)}+...+h_sx^{\alpha (s)}$ and further, if I expand each $h_i$, namely,
$$x^{\beta}= (h_{1,0}x_1^{\gamma_{1,0}}+h_{0,1}x_2^{\gamma_{1,1}}+...+h_{1,n}x_n^{\gamma {0,n}})x^{\alpha (1)}+(h_{2,0}x_1^{\gamma_{2,0}}+h_{2,1}x_2^{\gamma_{2,1}}+...+h_{2,n}x_n^{\gamma {2,n}})x^{\alpha (2)}+...+(h_{s,0}x_1^{\gamma_{s,0}}+h_{s,1}x_2^{\gamma_{s,1}}+...+h_{s,n}x_n^{\gamma {s,n}})x^{\alpha (s)}$$ where $h_i=h_{i,0}x_1^{\gamma_{i,0}}+h_{i,1}x_2^{\gamma_{i,1}}+...+h_{i,n}x_n^{\gamma {i,n}} \in k[x_1,...,x_n]$.
I mean, this expanding business, to me, seems like it's just made it super complicated. How is this supposed to make me "see" that it is divisible by some $x^{\alpha (i)}$? I really can't see anything even close to it.

What's going on with this proof? Can somebody please please help me? Thanks...

答:
For $\alpha, \beta \in \Bbb Z^n_{\ge 0}$, we write $\beta \ge \alpha$ to mean that $\beta_i \le \alpha_i$ for all $1 \le i \le n$, where $\alpha = (\alpha_1, \ldots, \alpha_n)$ and $\beta_i$ defined similarly.

Under this notation, note that $x^\alpha \mid x^\beta \iff \alpha \le \beta \iff \beta - \alpha \in \Bbb Z^n_{\ge 0}$.

We have $x^\beta = \displaystyle\sum_{i = 1}^s h_i x^{\alpha(i)}$. We can expand the $h_i$ in terms of monomials as well to get

$$x^\beta = \sum_{i = 1}^s\left(\sum_{j = 1}^{n_i}a_{i, j}x^{\beta(i, j)}\right)x^{\alpha(i)}$$

where $a_{i, j} \in \Bbbk$ and $\beta(i, j) \in \Bbb Z^n_{\ge 0}$ for all valid $i, j$.

Now, recalling that the monomials form a $\Bbbk$-basis for $\Bbb k[x_1, \ldots, x_n]$, there must be some $i \in \{1, \ldots, s\}$ and some $j \in \{1, \ldots, n_i\}$ such that $\beta(i, j) + \alpha(i) = \beta$.

(Why? Otherwise, $x^\beta$ would be a linear combination of monomials distinct from $x^\beta$, contradicting linear independence.)

In turn, $\alpha(i) \le \beta$ and thus, $x^{\alpha(i)} \mid x^\beta$, as desired.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
Upper bound for - $\sum_{i=1}^n i {2n \choose n+i} $.
I am tryin to calculate an upper bound for the following sum:
$\sum_{i=1}^n i {2n \choose n+i} $.
I know that the limit should be $\Theta (4^n\sqrt n)$ but I am not sure how to proceed. I already tried to bound by the biggest element, but the bound was too high.

答:
Another proof, neater but from a probabilistic point of view. If you are comfortable with discrete probability theory (random variables, expectations etc.) you might find this one more to your liking.

Note that the quantity you are trying to analyze is equal to

$$ \frac{1}{2}\cdot \ \ \ \  \underbrace{\sum_{i=-n}^n |i| \binom{2n}{n+i}}_S $$.

I'll ignore the $1/2$ because the question is about asymptotics and consider only the sum $S$ above.

Consider the following probablistic scenario. Let $X_1,\ldots, X_{2n}$ be $2n$ independent random variables each uniformly distributed over $\{\pm 1\}.$ Let $X$ denote $X_1+X_2+\cdots+X_{2n}.$ The sum $S$ is exactly $4^n \cdot \mathbb{E}[|X|]$ (i.e. the expectation of $|X|$).

By the Cauchy-Schwarz inequality, we have $\mathbb{E}[|X|]\leq \sqrt{\mathbb{E}[X^2]}$. An easy analysis allows us to show that $\mathbb{E}[X^2]=n.$


问:
A question related to cyclic extension and simple extension
The following question was asked in mid term exam of field theory ( Exam has concluded and I was unable to solve it. I tried it today again and failed to solve it. So, I need help with this.

If F is a cyclic extension of K of degree $p^n$ (p prime) and L is an intermediate
field such that F = L(u) and L is cyclic over K of degree $p^{n-1}$, then F= K(u).

It can be easily deduced that [L(u): L] =p.  I have read some results in cyclic extension from Hungerford's Algebra but I am not able move foreward in this case and need help.

Thank you !

答: Let $Gal(F/K)=<\sigma>$ and $Gal(F/L)=<\sigma^{p^{n-1}}>$. It equals to prove $Gal(F/K(u))=\{id\}$, which we can supposed to be $\{\sigma^{p^i}\}$ with $i<n$, then u is fixed by $\sigma^{p^i}$, and it’s fixed by $\sigma^{p^{n-1}}$, causing a contradiction.


问:
Prove the following theorem without Hensel's lemma
Let $P(X)\in\mathbb{Z}[X]$. Suppose that $P(X)\equiv 0$ has a root in $\mathbb Z/p^r\mathbb Z$ for every $r\geqslant 1$ and a fixed prime $p$. Prove that there exists a sequence of integers $(x_r)$ such that $P(x_r)\equiv 0\pmod{p^r}$ and $x_r\equiv x_{r-1}\pmod{p^{r-1}}$.

Can one prove this without any notion of Taylor's expansion or Hensel's lemma?

I tried to consider $P(x_{r-1}+tp^{r-1})$ for $0\leqslant t<p$. However I cannot connect the roots of $P(X)$ modulo $p^r$ to the roots of $P(X)$ modulo $p^{r-1}$. I cannot see the correct place to apply the condition that $P(X)\equiv 0$ has a solution in $\mathbb{Z}/p^r\mathbb{Z}$ for all $r$. I am not sure if I should do this forward or backward. From Hensel's lemma, it seems that one can obtain this sequence of integers "backward" since the integers $x_i$ are unique modulo $p^i$. However, I cannot prove this fact from the given conditions.

Thanks in advance; any help will be appreciated.

答:
It is the same as proving that $\Bbb{Z}_p$ (the ring of $p$-adic integers) is compact.

For each $r$ take one integer $y_r$ such that $P(y_r)\equiv 0\bmod p^r$.

Let $A_r = \{ a\in \Bbb{Z/p^r Z}, y_n \equiv a \bmod p^r$ for infinitely many $n\}$.

Since $\Bbb{Z/p^rZ}$ is finite it is clear that $A_r$ is non-empty.

Moreover $A_r=A_{r+1}\bmod p^r$.

Let $x_0=0$ then by induction on $r$, given $x_r$ take some $x_{r+1}\in A_{r+1}$ such that $x_{r+1}\equiv x_r\bmod p^r$.


问:
Does $\int^\infty_1 \frac {e^x}{x^{x^2}}$ coverge or diverge? (comparison test).
I am trying to decide if this integral converges or diverges, so I'm trying to find a function that behaves like the integrand at infinity to use the limit comparison test since the integrand is positive.

I couldn't find a function to compare my integral with, I know that at infinity the highest power is what the function behaves like ($\frac {1}{x^2}$ and $\frac {1}{1+x^2}$ behave the same at infinity for example), and I have been solving a couple of improper integral questions with comparison test with a little list I made about what functions to compare with, but none was close to this integrand. So here's what I tried:

$\int^\infty_1 \frac {e^x}{x^{x^2}}=\int_1^\infty\frac {e^x}{e^{x^2ln(x)}}$, I know that $\int^\infty_2\frac {1}{xln^\alpha(x)}$ converges if and only if $\alpha>1$.

But here I got stuck and I don't know how to continue or if the function I'm trying to compare with isn't the right one.

I would appreciate any help in how to find a function that works for comparison test for this integrand, also if anyone knows where I can find a list of known improper integrals that converges or diverges that would help me alot.
Thanks in advance

答:
It converges.

HINT: For $x>e^2$ note that $x^{x^2} \ge e^{2x^2} \ge e^{2x}$.


问:
Solving differential equation of the form $u_x=u^n$
This is my second week of learning PDE's.
I have two equations,

$u_t=u^2$ and $y_t=3y^{2/3}$.

I know how to solve $u_x=x$ but I can't figure out what to to with those.

If you could give me a helping hand on how to solve the following PDE's and give a method of solving similar to them.

I'm still a bit confused because it is a new subject, please be respectful.

答:
I am not going to give you a complete answer. I just want to give you the common idea to solve equations like yours (and I leave it to you to fix the details). Let $f$ be a differentiable function to be determined later. You can write your first equation as
$$
f'(u)u'=f'(u) u^2.
$$
That is (by chain rule)
$$
[f(u(t))]'=f'(u) u^2.
$$
If you are able to find a function $f$ such that $f'(t)=1/t^2$, then your original equation becomes
$$
[f(u(t))]'=1.
$$
Now you can compute $f(u)$ by integration. If your $f$ is injective then you are done.
This is the case of your first equation. You can take $f(t)=-(1/t)$. The equation you have to solve is then
$$
(1/u(t)))'=-1
$$
which gives you
$(1/u(t)))=-t+const$. Hence $u(t)=1/(const-t)$.


问:
eigenvalues of k-regular bipartite graph adjacency matrix.
I need some help with this proof:

Let G be a k-regular graph. prove that:
a) If G is bipartite then -k is an eigenvalue of G's adjacency matrix.
b) If -k is an eigenvalue of G's adjacency matrix then G is bipartite.

Got stuck in figuring out what the general form of such an adjacency matrix.
Thanks!

答:
For (a) you first prove that $k$ is an eigenvalue of $G$'s adjacency matrix $A$. This is simple and is already explained in Hidalgo's answer: $A-kI$ is not invertible.

Now I will show (a) in a different way from Hidalgo. This is taken from Bartlett's lecture notes: write
$$\mathbf A=\begin{bmatrix}\mathbf0&\mathbf B\\\mathbf B^T&\mathbf0\end{bmatrix}\qquad\mathbf v=\begin{bmatrix}\mathbf a\\\mathbf b\end{bmatrix}\qquad \mathbf{Av}=k\mathbf v$$
where $\mathbf B$ has dimensions $m×n$, $\mathbf a$ has length $m$ and $\mathbf b$ has length $n$. It is then easy to check that $(\mathbf a,-\mathbf b)^T$ is an eigenvector of $A$ associated to $-k$.

To prove (b) without counting walks, assuming $G$ is connected (otherwise, as Hidalgo's example shows, this is false in general) normalise the eigenvector $\mathbf v$ associated to $-k$ such that some component (say it lies at index $i$; these indices also label $G$'s vertices) is $1$ and all other components have absolute value at most $1$. Then, looking at the $i$th component of $\mathbf{Av}=-k\mathbf v$:
$$\sum_{j=1}^na_{ij}v_j=\sum_{j\text{ adjacent to }i}v_j=-kv_i$$
With the magnitude restrictions, $G$'s $k$-regularity and $v_i=1$, this means $v_j=-1$ for all neighbours $j$ of $i$. Negate $\mathbf v$ and repeat the argument for all the neighbours $j$; iterate until the whole graph is covered, by connectivity. This shows that all elements of $\mathbf v$ are $+1$ or $-1$ with signs differing if the corresponding vertices are adjacent in $G$.

Now assume $G$ is not bipartite, then it contains an odd cycle. However, the sign assignment above will run into a contradiction on this odd cycle. Hence $G$ is bipartite.

Combining the (b) argument with the proof in Bartlett's notes that $G$ having $\Delta(G)$ as an eigenvalue implies it is regular gives a proof that if a connected $G$ has $-\Delta(G)$ as an eigenvalue it is regular and bipartite.


问:
REVISTED$^1$ - Order: Modular Arithmetic

Relevant Literature:

Question:

Observe that $2^{10}=1024≡−1 \pmod{25}$.Find the order of $2$ modulo $25$.

Thoughts:

Direct answers are OK, but I'd like to know if I'm right that what I'm really looking for is this:
$$\inf\left\{\frac{2^x-2}{25}~:~x\in \mathbb{N}\setminus \{0\}\right\}$$

EDIT$^1$:

What I meant was, I'm trying to find the smallest such $x$ so that $2^{10}+25k+2=2^x$, where $k\in \mathbb{Z}$. How do I write that in $\inf$ terms?

EDIT$^2$:

Am I right in thinking that I could think of it this way too:

$$ \inf{\{\log_2(2^{10}+25x+2)~:~x\in \mathbb{Z}\}} $$

答: Hint: The order is $\le 20$, since $2^{20}\equiv (-1)^2\pmod{25}$.  And the order divides $20$, but does not divide $10$. That does not leave many possibilities to rule out.


问:
What's the difference between a "computational algorithm" and a "numerical algorithm"?
I'm struggling to understand what the difference is between a computational and a numerical algorithm.

On Wikipedia, Monte Carlo methods fall under the rubric of computational algorithms; however the Newton-Raphson method is described as a numerical algorithm.

Both algorithms can be (i) run on a computer, (ii) involve calculations, (iii) and are iterative. The only broad distinction is that MC requires the simulation of random numbers (which we often designate to computers).

Can someone demystify this distinction in definitions?

N.B. I'm not sure what the correct tags for this question should be - I typed in "nomenclature" but it wasnt listed. Editors feel free to edit the tags.

References :

https://en.m.wikipedia.org/wiki/Monte_Carlo_method

https://en.m.wikipedia.org/wiki/Newton's_method

答: I wouldn't say there's a difference. Neither "numerical algorithm" nor "computational algorithm" is a very well-defined term anyway.</p>


问:
Some algebraic manupilation of Fermat's last theorem

Meta

Some elementary manipulation of Fermat's theorem is tried and an approach to prove it in very few lines is suggested. The question is to pinpoint errors in the analysis, as usually these stuff is junk. Anyhow algebra is fun and I hope you'll enjoy the faulty ride. Can you see the mistake? I can't. I verified the last explicit formula for $n=3$ and $a=7$, $b=13$, $c = (a^3+b^3)^{1/3}$

Here is the approach:

Consider the following equation over integers with $a \in \mathbb N, b \in \mathbb N, a < b, c \in
\mathbb N, n \geqslant 2, \text{odd}, \gcd(a,b)=1, \gcd(a,c)=1,\gcd(b,c)=1$,
$$ a^n + b^n = c^n. $$
Put $y = b / c$, $r = a / b$ and we get
$$ y^k = \frac{1}{1 + r^k} = \frac{1}{(1 + r) (1 - r + r^2 -r^3+ \cdots + r^{k -
   1})} = \frac{1}{(1 + r) u_{k - 1} (r)}. $$
Now we can rewrite this as
$$ y^k u_{k - 1} (r) = \frac 1{1+r} = \frac {1+r^k - r^k}{1+r} = u_{k - 1} (r) - r^k \frac{1}{1 +
   r} = u_{k - 1} (r) - r^k b / (b + a). $$
Multiply by $c^k b^{k - 1}(b+a)$ yields,
$$ (b + a) b^k b^{k - 1} u_{k - 1} (r) = (b + a) c^k b^{k - 1} u_{k - 1} (r) -
   c^k a^k. $$
Now note that,
$$ b^{k - 1} u_{k - 1} = b^{k - 1} - b^{k - 2} a^1 + \ldots \dotplus a^{k - 1}
   = b v (a, b) + a^{k - 1}, $$
And using this we conclude,
$$ (b + a) b^k (b v + a^{k - 1}) = (b + a) c^k b v + (b + a) c^k a^{k - 1} -
   c^k a^{k} = (b + a) c^k b v + c^k b a^{k - 1}. $$
Divide by $b$,
$$ (b + a) b^{k - 1} (b v + a^{k - 1}) = ((b + a) v + a^{k - 1}) c^k $$
Now, if $p \backslash (a + b)$ Then either $p \backslash c$ or $p \backslash
a$. If the latter then $p \backslash b$ which contradicts $\gcd (a, b) = 1$.

Hence we can write
$$ c = \alpha (a + b) $$
But $c < a + b$ hence a contradiction. This shows that Fermat's last theorem
holds for odd $n$ larger then 2. Then it holds also for every $n$ that has an odd prime as
a factor and the remaining cases follows from the special case $n = 4$ and Fermat's last theorem is proven.

答:
I am not sure if this is the crux of the problem, but in the final step the implication

$$(p \mid (a+b) \implies p\mid c)\implies c = \alpha(a+b)$$

is false in general if $\alpha$ is expected to be an integer.

For example, take $a+b = 27$ and $c=15$. (If you insist, take $a = 13, b=14$) then every prime that divides $a+b$ also divides $c$, but $c$ is not a multiple of $a+b$; moreover $a+b>c$ which does not lead to a contradiction.


问:
Characterization of proper maps using filters
A continuous function $f \colon X \to Y$ is called proper if it is closed and if $f^{-1}(\{y\})$ is compact for every $y \in Y$. I want to prove that a continuous function $f \colon X \to Y$ is proper if and only if for every ultrafilter $\mathcal U$ on $X$ and every point $y \in Y$ such that $f[\mathcal U]$ converges to $y$ (here, $f[\mathcal U]$ is the filter generated by sets of the form $f(U)$, $U \in \mathcal U$), there is an $x \in X$ such that $\mathcal U$ converges to $x$ and $f(x) = y$.

Here is what I have been able to argue so far: Suppose $f$ is proper. Pick an ultrafilter $\mathcal U$ on $X$ and a $y \in Y$ such that $f[\mathcal U]$ converges to $y$. Since $f$ is proper, $A := f^{-1}(\{y\})$ is compact. Since $\mathcal U$ is an ultrafilter, it either contains $A$ or it doesn't. In the form case, $\mathcal U$ converges to some $x \in A$ and we're done. In the latter case, $\mathcal U$ contains some set $B$ such that $A \cap B$ is empty. I do not know how to proceed from here. Any tips? I have not attempted the reverse implication.

Edit: In the argument above, I meant to say that either $A \in \mathcal U$ or $A' := X \setminus A$ is in $\mathcal U$. In this latter case, $f(A')$ is in $f[\mathcal U]$. Since $f[\mathcal U]$ converges to $y$, the closure of every set in $f[\mathcal U]$ contains $y$, which means $y \in \overline{f(A')}$. Since $f$ is closed, $\overline{f(A')}=f(\overline{A'})$. Thus, there is an $x$ in the closure of $A'$ such that $f(x) = y$. Now I just need to argue that $\mathcal U$ converges to $x$. Suppose it does not. Then $\mathcal U$ contains the compliment $U'$ of some open set $U$ containing $x$, which means that $f(U')$ is contained in $f[\mathcal U]$, and since $f(U)$ is in $f[\mathcal U]$, $f(U) \cap f(U') \in f[\mathcal U]$. I'm not sure where to go from here...

答:
$\Rightarrow)$ Suppose $y\in Y$ is arbitrary and $C=f^{-1}(\{y\})$.
Let $\mathcal U$ be an ultrafilter on $X$ with $f(\mathcal U)\to y$. then
$$y\in \bigcap_{U\in \mathcal U }\overline{f(U)}\subseteq \bigcap_{U\in \mathcal U }\overline{f(\overline U)}\subseteq \bigcap_{U\in \mathcal U }{f(\overline U)}$$
So
$$(\forall U\in \mathcal U)(\overline{U}\cap C\ne \emptyset) $$
Let $\mathcal F$ be the filter generated by $\{\overline{U}\cap C \mid U\in \mathcal U\}$ on $C$. Since $C$ is compact $\mathcal F$ has some limit(=cluster) point $c\in C$; that is
$$c\in \bigcap_{F\in \mathcal F}\overline{F}\cap C\subseteq \bigcap_{U\in \mathcal U}\overline{\overline{U}\cap C}\subseteq \bigcap_{U\in \mathcal U}\overline{U}$$
and so $c$ is a limit point of $\mathcal U$. Since $\mathcal U$ is an ultrafilter:
$$\mathcal U\to c$$
and so
$$f(\mathcal U)\to f(c)=y$$

$\Leftarrow)$ Suppose $y\in Y$ is arbitrary and $C=f^{-1}(\{y\})$. Suppose $\mathcal F$ is a filter on $C$. It is enough to prove $\mathcal F$ has limit point in space $C$. $\mathcal F $ is a base for a filter $\mathcal G$ on $X$. $\mathcal G$ is contained in some ultrafilter $\mathcal U$. We have

$$f(\mathcal F) \to y$$

So

$$f(\mathcal U) \to y$$
By assumption there's some $c\in C$ with
$$\mathcal U\to c$$
which implies $c$ is a limit point of $\mathcal F$ in space $X$ and so in space $C$.

closedness of $f$ is not proved in $\Leftarrow$. If anything is wrong let me know.


问:
Let $R$ be an equivalence relation on $X$ and $S$ an equivalence relation on $X/R$.....

Let $R$ be an equivalence relation on $X$ and $S$ an equivalence relation on $X/R$. Find an equivalence relation $T$ on $X$ such that $(X/R)/S$ is in one-to-one correspondence with $X/T$ under the mapping $[[x]R]S\mapsto[x]T$.

Not much to add. I found this question in Elements of Abstract Algebra by Allen Clark. Any help would be appreciated. Cheers!

答:
Hint

Define

$$(u,v) \in T \iff (\exists \mathcal S \in (X/R)/S)(u \in \mathcal S \wedge v \in \mathcal S)$$


问:
Meaning cone, ray, fan for polytopes
I'm trying to understand some mathematical operations and definitions for my project. Could you explain the meanings? $P_b =\{x∈\mathbb{R}^d : Ax≤b\}$
is a polytope. Let's have a 10x3 matrix $Ax≤b$

$$
\begin{bmatrix}
0 & 1 & 1 \\
1 & 0 &1 \\
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -1 \\
0 &  0& 1\\
0 &  -1& 0\\
0 &  -1& 1\\
-1 &  0& 0\\
-1 &  0& 1
\end{bmatrix}
\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix}
\leq
\begin{bmatrix}
4\\
4\\
3\\
3\\
0\\
2\\
0\\
1\\
0\\
1
\end{bmatrix}
$$

Let us denote the set of such right hand side vectors $b$ by
$U(A)=\{b∈\mathbb{R}^m : P_b \neq ∅\}.$

Switching from the H-representation to its V- representation, the cone U (A) is generated by 9 rays and 3 lines in $Z^{10}$.
The normal cone of a face F of a polytope P in $R^d$ is the set
$N(F;P)=\{v∈\mathbb{R}^dd : v^⊤x=h(P,v)~for~all~ x∈F\}.$
The dimension of the normal cone of a $k$-dimensional face is $(d − k)$. The normal fan $N(P)$ of $P$, which is the collection of the normal cones of all faces of $P$ , is a complete fan in $\mathbb{R}^d$.

I have difficulty in understanding of the terms.
So, what is the exact meaning of $U(A)$ here? What are the $b_{1,2,...}$ letters and how obtained? How is the 9 rays and 3 lines expression written? Can you explain also meanings of normal fan,cone and complete fan?

Article can be found from here. I want to grasp all of the article also I have to, but at least up to there is enough.

@Edit:
I think it's a kinda Millenium Problem. If not, isn't there any helper mathematician?

答:
The article investigates the set $U(A) = \{b : \exists x\; Ax \leq b \}$. In words, that is the set of vectors $b$ for which $Ax \leq b$ has a solution, that is, for which the set $\{x : Ax \leq b\}$ is not empty. Via Farkas Lemma, they express the set as $U(A) = \{ b : y^Tb \geq 0 \; \forall y \in S\}$ with $S = \{ y \geq 0 : A^Ty=0 \}$.

Without going into how the set $S$ is derived, we can already make a few remarks. Since $A \in \mathbb{R}^{10 \times 3}$ in the example, $S \subset \mathbb{R}_+^{10}$. Moreover, if $y \in S$, then also $\alpha y \in S$ for any $\alpha \geq 0$. Therefore, $S$ is a cone. The set $S$ is also convex, so it is a convex cone. The set $\{ \alpha y : \alpha \geq 0 \}$ is called a ray, which is just a half-line with one endpoint at the origin. A ray of $S$ that cannot be expressed as a convex combination of other rays of $S$ is called an extreme ray of $S$. Figure 1 of this article has an illustration of the extreme rays of a cone. If $-y\in S$ for some $y\in S$, $S$ is said to contain a line. A cone that does not contain any lines is a pointed cone. A cone is defined (or generated) by the set of its (possibly infinite number of) lines and rays.

To verify if a given vector $b$ is an element of $U(A) = \{ b : y^Tb \geq 0 \; \forall y \in S\}$ one does not need to check all elements in $S$, but it suffices to check whether $y^Tb \geq 0$ for one nonzero element of each extreme ray of $S$ and $y^Tb=0$ for one nonzero element of each extreme line of $S$ (it then automatically holds for all elements in $S$). The remainder of the article is about deriving the extreme rays of $S$. As a spoiler they already provide the final result (those 9 inequalities involving $b_i$), but you cannot understand why they are correct without reading the rest of the paper. At this point you can however check that indeed $A^Ty=0$ for each vector $y$.

The set $S$ is a polytope, since it is the finite intersection of closed halfspaces. The face of a polytope is visually one of the sides, edges or corners of the polytope. The mathematical definition of a face is $F = S \cap\{y:c^Ty=d\}$ where $c$ and $d$ are such that $c^Ty \leq d$ for all $y \in S$ ($c^Ty \leq d$ is a valid inequality). Each face $F$ has an associated normal cone, $N(F,S) = \{ v : v^Ty \geq v^Tz \; \forall y \in F \; \forall z \in S\}$. The normal cone (which is a convex cone) is the set of all vectors that are orthogonal to the face. An illustration can be found in this question (left panel). The set of all normal cones is called the normal fan. An illutration is found here, and indeed it covers every angle which means it is a complete fan.

Title: Meaning cone, ray, fan for polytopes


问:
Translating a English Sentence to Predicate Logic
Let $C(a,b)$ be "$a$ and $b$ have communicated over the web," and the domain of discourse is all students in the class. What is the translation of "Everyone has communicated over the web with the same student" Note: the student hasn't communicated with themselves.

Very stumped on this one. I am initially stating $\forall a \exists b ((a \neq b) \implies C(a,b)).$ Is this the correct way to show that someone is not themselves in this statement?

答:
You're very close, but currently your statement says "for all people in the class, there exists someone they've communicated with" but these $b$s needn't be the same. You're looking to say that "there exists someone in the class such that everyone else (bar themselves) has communicated with them". In practice this is the difference between:

$$
\forall a \exists b ((a \neq b) \implies C(a,b)).
$$

and

$$
\exists b \forall a ((a \neq b) \implies C(a,b)).
$$

The latter of which being the correct answer as @amWhy explains better, in the comments here.


问:
$\int_0^1 \sqrt {\tan^{-1}x}\space dx=$?
How to evaluate $$\int_0^1 \sqrt {\tan^{-1}x}\space dx\qquad ?$$ Is there an elementary expression or value for it? (though I know that there is no elementary expression for $\int \sqrt {\tan^{-1}x}\space dx$)  . </p>

答: By using the formula http://upload.wikimedia.org/wikipedia/en/math/7/2/a/72a1058ad2087aec467af24bddcf9479.png, we have $\int_0^1\sqrt{\tan^{-1}x}~dx=\sum\limits_{n=1}^\infty\sum\limits_{m=1}^{2^n-1}\dfrac{(-1)^{m+1}}{2^n}\sqrt{\tan^{-1}\dfrac{m}{2^n}}$</p>


问:
Recurrence relation $a_{n+1} = 2 a_n -1$ with $a_0=3$. What is $a_n$ expressed as a function of $n$?
I'm having a hard time finding an explanation how to solve the following problem.

Given the recurrence relation $a_{n+1} = 2 a_n -1$ with $a_0=3$.   What is $a_n$ when expressed as an explicit function
of $n$?

Supposedly the correct answer is $a_n = 2^{n+1}+1$. But what is the step by step approach to arrive at this solution? All of the resources I've found begin with a recurrence relation in the form an = (some expression) but how to handle this type of problem in the form an+1 = (some expression) ?

Is it a matter of moving the variable to the other side and then solving iteratively? Or is there some clever/magic approach we should use here? Super appreciate if anyone can detail the steps to take to arrive at the correct solution.

Thanks in advance!

答: First of all note that assuming the form of recurrence, the answer is linear. Assume $a_n=na+b$ and assert it into the condition: $na+a+b=2na+2b-1$ and also $a_0=b=3$ so $b=3$ and $a=2$.


答:
First way

Verify by induction that the provided solution is the right one...

Second way

If you don't know the solution. Take $b_n = a_n +c$ in order to get rid of the term $-1$ in the recurrence relation. That gives

$$b_{n+1}- c = 2 (b_n-c) -1$$ and therefore $b_{n+1} = 2 b_n$ if $c=-1$. The relation $b_{n+1} = 2b_n$ is easy to solve in $b_n = 2^n b_0$ and finally $a_n = 2^n (a_0-1) + 1 =  2^{n+1} + 1$ knowing that $a_0=3$.

This solution avoids guessing!

Title: Recurrence relation $a_{n+1} = 2 a_n -1$ with $a_0=3$. What is $a_n$ expressed as a function of $n$?

Body:
I'm having a hard time finding an explanation how to solve the following problem.

Given the recurrence relation $a_{n+1} = 2 a_n -1$ with $a_0=3$.   What is $a_n$ when expressed as an explicit function
of $n$?

Supposedly the correct answer is $a_n = 2^{n+1}+1$. But what is the step by step approach to arrive at this solution? All of the resources I've found begin with a recurrence relation in the form an = (some expression) but how to handle this type of problem in the form an+1 = (some expression) ?

Is it a matter of moving the variable to the other side and then solving iteratively? Or is there some clever/magic approach we should use here? Super appreciate if anyone can detail the steps to take to arrive at the correct solution.

Thanks in advance!


问:
Find the area of an equilateral triangle
Point $M$ lies on the side $AB$ of an equilateral triangle $ABC$ such that $AM:BM=1:2$. If $CM=m$, find the area of the triangle $ABC$.

 The answer given in my book is $\dfrac{9m^2\sqrt3}{28}$.

So we have $$S_{ABC}=S_{AMC}+S_{BMC}\iff\\\iff\dfrac{AB\cdot CH}{2}=\dfrac{AM\cdot CH}{2}+\dfrac{BM\cdot CH}{2}\\AB\cdot CH=AM\cdot CH+BM\cdot CH.$$
This doesn't seem to help very much. Any help would be appreciated.

答: The cosine rule on triangle $MBC$ gives $$m^2=\dfrac49a^2+a^2-2\cdot\dfrac23 a\cdot a\cdot\cos60^\circ\\\Rightarrow m=\dfrac{a\sqrt7}{3}.$$ Now using $S=\dfrac{1}{2}a\cdot a\cdot\sin60^\circ=\dfrac{a^2\sqrt3}{4}...$


答:
You can compute the first terms:

$a_0=3$;

$a_1=5$;

$a_2=9$;

$a_3=17$

$\vdots$

These numbers are $2+1$, $2^2+1$, $2^3+1$, $2^4+1$, and so on, which suggests that $a_n=2^{n+1}+1$. And now you can prove by induction that this is indeed correct.


问:
Special case of graph-theory cycle
In graph theory, is there a name for a (simple) cycle with no path between any of its non-adjacent vertices? When the graph is planar, this could be called a face, but how about non-planar graphs?</p>

答:
I think you're talking about a chordless cycle, or sometimes an induced cycle.

A face in a planar graph doesn't necessarily conform to this so there is some conflict in your description.

Title: Special case of graph-theory cycle


问:
Intuition for Brownian motion time-inversion formula
Background: I am studying a course in Stochastic Financial Models, and have been introduced to Brownian motions (BMs) and Gaussian processes.

The BM time-inversion formula states that if $W_t$ is a BM, then $$B_t := tW_{1/t}$$ is also a BM.

I know how to prove the result, through Gaussian processes, similar to the answer here.

However, the proof is unenlightening and does not offer any intuition. In fact, my course lecturer does not offer any intuition either.

Question: Please provide some intuition why the formula “should be” true.

Example of my intuitions for another formula:

“Scaling” formula: if $W_t$ is a BM, then $cW_{t/c^2}$ is also a BM.

Intuition: picture time on the $x$ axis and the motion on the $y$ axis. BMs “don’t go that far up or down”, i.e. their standard deviation is “only” sd$(W_t)=\sqrt t$, so to compensate for the factor of $c$ stretch in the $y$ axis, we have to stretch by a factor of $c^2$ in the $x$ axis.

答: This property and all similar ones stem from a simple fact: BM is the sum of uncorrelated normal random variables. It does not make any difference if the time indices are reversed, inverted, or shuffled, as long as the sigma algebra is not compromised. The correlation depends on the intervals that overlap, the corresponding time index of which could show up as minimum, maximum, or some complicated combination of indices.


问:
Show that for every $n$ there are integers $a,b$ such that $X_n = a^2 + 2b^2$
We have a sequence $X_0, X_1,...$ which is determined by $X_0=1, X_1=3$ and $X_n = 6X_{n-1} - X_{n-2}$

Show that for every $n$ there are integers $a,b$ such that
$X_n = a^2 + 2b^2$

I thought of changing $a^2 + 2b^2 = (a + \sqrt 2 i)(a - \sqrt 2 i)$, since I thought it might lead to something good, but I'm not getting anything.
Any hints or solutions are appreciated!

Body:
Let $(a_n,b_n)$ be solutions to Pell's equation $a_n^2-2b_n^2=\pm1$.

The fundamental solution is $1^2-2\times1=(1+\sqrt2)(1-\sqrt2)=-1$,

so solutions are $a_n^2-2b_n^2=(a_n+\sqrt2b)(a_n-\sqrt2b)=(1+\sqrt2)^n(1-\sqrt2)^n=(-1)^n$, where

$a_n+\sqrt2b_n=(1+\sqrt2)^n=(a_{n-1}+\sqrt2b_{n-1})(1+\sqrt2)=(a_{n-1}+2b_{n-1})+(a_{n-1}+b_{n-1})\sqrt2$,

so $$a_n=a_{n-1}+2b_{n-1}\tag1$$ and $$b_n=a_{n-1}+b_{n-1}\tag2$$.

Solutions to $X_n^2-2Y_n^2=1$ are given by $X_n+\sqrt2Y_n=(1+\sqrt2)^{2n}=a_{2n}+\sqrt2b_{2n}$

$=[(1+2\sqrt2)^n]^2=(a_n+\sqrt2b_n)^2=a_n^2+2b_n^2+2a_nb_n\sqrt2$,

so $X_n=a_n^2+2b_n^2=a_{2n}$ (and $Y_n=2a_nb_n=b_{2n}$).

From (1)-(2) we have $b_{n-1}=a_n-b_n=b_n-a_{n-1}$, so $a_n+a_{n-1}=2b_{n}$,

so $a_n=a_{n-1}+2b_{n-1}=a_{n-1}+(a_{n-1}+a_{n-2})=2a_{n-1}+a_{n-2}$.  Therefore,

$a_n=2(2a_{n-2}+a_{n-3})+a_{n-2}=5a_{n-2}+2a_{n-3}=5a_{n-2}+(a_{n-2}-a_{n-4})=6a_{n-2}-a_{n-4}$,

so $a_{2n}=6a_{2(n-1)}-a_{2(n-2)}$, so $X_n=6X_{n-1}-X_{n-2}$, and $X_n$ matches your sequence.


问:
Prove or disprove: if f(x) is continuous at $1$, and $\lim_{x\to 1}\frac {f(x)}{x-1}=1$, then $f(1)=0$.
I'm having a little trouble of where to start in this question.

Intuitively, this seems true, just based on the limit $\frac {sinx}{x}$ when $x\to 0$, but intuition isn't enough, and I'm not getting how to start proving this statement (if it's true anyway).
 I appreciate any help or hints, thanks in advance.

Body: Since $\lim_{x\to1}\frac{f(x)}{x-1}=1$,$$\lim_{x\to1}f(x)=\lim_{x\to1}(x-1)\frac{f(x)}{x-1}=0\times1=0.$$But, since $f$ is continuous at $1$, $\lim_{x\to1}f(x)=f(1)$. Therefore, $f(1)=0$.


问:
Jarnik's theorem in Falconer's book: what number?
Simple reference question:

What is the theorem number/page number of Jarnik's theorem in Kenneth Falconer's Fractal Geometry (third ed.)?

Full reference:

Falconer, Kenneth, 2014.

Fractal geometry: Mathematical foundations and applications.
Third edition. John Wiley & Sons, Ltd., Chichester.

答: Reference for 3rd Edition: Page 173, Theorem 10.3.</p>


答: HINT: $\lim_{x\to1} f(x)/(x-1) = 1$ implies that $f(x)$ near $1$ should be of the form $f(x) = (x-1)^\alpha g(x)$ where $g$ is continuous at $x = 1$ (since $f$ is continuous at $x = 1$).

Title: Prove or disprove: if f(x) is continuous at $1$, and $\lim_{x\to 1}\frac {f(x)}{x-1}=1$, then $f(1)=0$.

Body:
I'm having a little trouble of where to start in this question.

Intuitively, this seems true, just based on the limit $\frac {sinx}{x}$ when $x\to 0$, but intuition isn't enough, and I'm not getting how to start proving this statement (if it's true anyway).
 I appreciate any help or hints, thanks in advance.


答: Hint. Let $AM=x.$ Then $MB=2x.$ Apply Heron to the equilateral triangle to find its area $$\sqrt{s(s-3x)^3},$$ where $s=9x/2.$ Then calculate the areas of the component triangles to find the area of the equilateral triangle in another form, to get $$\sqrt{t(t-x)(t-m)(t-3x)}+\sqrt{u(u-2x)(u-3x)(u-m)},$$ where $t=\frac{x+m+3x}{2}$ and $u=\frac{2x+3x+m}{2}.$ Then you can find the area of the equilateral triangle in terms of $m$ alone.

Title: Find the area of an equilateral triangle

Body:
Point $M$ lies on the side $AB$ of an equilateral triangle $ABC$ such that $AM:BM=1:2$. If $CM=m$, find the area of the triangle $ABC$.

 The answer given in my book is $\dfrac{9m^2\sqrt3}{28}$.

So we have $$S_{ABC}=S_{AMC}+S_{BMC}\iff\\\iff\dfrac{AB\cdot CH}{2}=\dfrac{AM\cdot CH}{2}+\dfrac{BM\cdot CH}{2}\\AB\cdot CH=AM\cdot CH+BM\cdot CH.$$
This doesn't seem to help very much. Any help would be appreciated.


答:
In general, if $\lim_{x\to a}g(x)=L$ and $\lim_{x\to a}h(x)=M$, you have
$$
\lim_{x\to a}g(x)h(x)=LM.
$$

For your problem, pick $g(x)=x-1$ and $h(x)=\frac{f(x)}{x-1}$, $a=1$. Then use the continuity of $f$ at $x=1$.


问:
Proving that the outer measure of a closed interval $[a,b]$ is $b-a$
In Sheldon Axler's book, Measure Integration, and Real Analysis, he defines outer measure of a set as $|A| = \inf\big\{\sum_{k=1}^\infty \ell(I_k): I_1, I_2, \dots \text{are open intervals such that} A\subset \bigcup_{k=1}^\infty I_k\big\}$, where $\ell(I)$ for an open interval $(a,b)$ is just $b-a$. He later proves that outer measure preserves order, i.e. $A\subset B \Rightarrow |A| \le |B|$.

Later, we are trying to prove that the outer measure of the closed interval $[a,b]$ is $b-a$. We bound it from above by saying for $\varepsilon > 0$, $(a-\varepsilon, b+\varepsilon), \varnothing, \varnothing,\dots$ is a sequence of open intervals whose union contains $[a,b]$, so $|[a,b]|\le b-a+2\varepsilon$ which with the definition of outer measure implies $|[a,b]| \le b-a$. The next section is confusing to me:

Is the inequality in the other direction obviously true to you? If so,
think again, because a proof of the inequality in the other direction
requires that the completeness of $\mathbf{R}$ is used in some form...Thus
something deeper than you might suspect is going on with the
ingredients needed to prove that $|[a, b]| ≥ b − a$.

He then goes onto prove it using the Heine-Borel theorem. However, because outer measure preserves order and $(a,b)$ is a subset of $[a,b]$, couldn't we easily bound it from below with that? Is the open interval not thought of as a subset? I don't quite understand the reasoning and feel I'm missing something obvious. Any help would be appreciated.

答:
You are arguing that
$$
|[a, b]| \ge |(a, b)| \ge b-a \, ,
$$
but the right inequality needs to be justified. We know that $\ell((a, b)) = b-a$, but it not obvious from the definition that $|(a, b)| = \ell((a, b))$.

It is in fact easier to prove
$$
|[a, b]| \ge b-a
$$
first, because any open covering $\bigcup_{k=1}^\infty I_k$ of the compact interval $[a, b]$ contains a finite sub-covering, that is where the Heine-Borel theorem comes into play.


问:
Polynomials/ remainder theorem
What must be added to p(x)=x^4+2x^3-2x^2+x-1 to make it exactly divisible by g(x)=x^2+2x-3?

The book says that the remainder will be a linear polynomial and so the expression that must be added to p(x) will be of the form ax+b; the book essentially says we add the remainder. The divisor has roots 1 and -3. We can solve by putting p(1) and p(-3) equal to zero and finding the values of a and b. I know how to solve this but I am confused as to why we say that the expression to be added is of the form ax+b. How do we know this. Essentially what the book seems to be saying is add the remainder. But if we take the example7/3, the remainder is 1 but adding the remainder will not make 7 exactly divisible by 3. Here we subtract 1 from 3 and add the result to make 7 exactly divisible by 3. I believe the same principle must hold in the case of polynomials and the explanation in the book that we add the remainder to the dividend must be wrong. Please tell me the reason why we add ax+b. Please explain.

Also, can it be solved through polynomial long division? If so, how?

Are there any other easier ways to solve this? Feel free to ignore the last question and even the second last question if it will make the answer too long.

答: I think that polynomial long division is the way to go. Since we have$$x^4+2x^3-2x^2+x-1=(x^2+2x-3)(x^2+1)-x+2,$$then we have$$(x^4+2x^3-2x^2+x-1)+(x-2)=(x^2+2x-3)(x^2+1).$$

Title: Polynomials/ remainder theorem


答:
The long division of $p$ by $q$ is

$$p(x) = q(x) (x^2 + 1)+  2 - x$$ and therefore

$$p(x) + x-2 = q(x) (x^2 + 1)$$

So adding $x-2$ to $p$ would make it divisible by $q$.

Title: Polynomials/ remainder theorem

Body:
What must be added to p(x)=x^4+2x^3-2x^2+x-1 to make it exactly divisible by g(x)=x^2+2x-3?

The book says that the remainder will be a linear polynomial and so the expression that must be added to p(x) will be of the form ax+b; the book essentially says we add the remainder. The divisor has roots 1 and -3. We can solve by putting p(1) and p(-3) equal to zero and finding the values of a and b. I know how to solve this but I am confused as to why we say that the expression to be added is of the form ax+b. How do we know this. Essentially what the book seems to be saying is add the remainder. But if we take the example7/3, the remainder is 1 but adding the remainder will not make 7 exactly divisible by 3. Here we subtract 1 from 3 and add the result to make 7 exactly divisible by 3. I believe the same principle must hold in the case of polynomials and the explanation in the book that we add the remainder to the dividend must be wrong. Please tell me the reason why we add ax+b. Please explain.

Also, can it be solved through polynomial long division? If so, how?

Are there any other easier ways to solve this? Feel free to ignore the last question and even the second last question if it will make the answer too long.


问:
help needed to find the 5th term on the number series
Can someone help me calculate the 5th term of these two series of numbers? They are somewhat difficult because they have no common difference or ration, which means they are neither arithmetic nor geometric.
7,1,2,3,___ the 5th term seems to be 5, but I don't know how to get it.
6,7,7,12___ The 5th term is 7, also I don't know how to get it.

Many thanks in advance.

答:
It is obviously 42, as Douglas Adams brilliantly observed in "The Hichhiker's Guide to the Galaxy" as the answer to the ultimate question.

Seriously, any sequence (and thus any following values) are possible. Without any further information on the sequence, my guess is as good as any other.


问:
There are $2^{\omega(a)}$ pairs $(b,c)$ such that $a=bc$ and $\gcd(b,c)=1$
It looks as there are $2^{\omega(a)}$ ordered pairs $(b,c)$ such that $a=bc$ and $\gcd(b,c)=1$, where $\omega(a)$ is the number of different prime factors of $a$.

Proof?

It's also true that there are  $n^{\omega(a)}$ n-tipples $(a_1,\dots ,a_n)$ such that $a=a_1\cdots a_n$.

答:
Bear with me.  I'm going to introduce components for the sake of lining up my ducks in a row.  And then I will shoot.

(If this whole thing is too long you can skip to the very last paragraph:

LAST PARAGRAPH:

Likewise for each prime factor of $a$ we can either have $p|b$ or $p|c$ so there are two choices for each prime factor of $a$ and so there are $2^{\text{number of prime factors}}$ choices of pairs.

If that is too terse read on for a duck guide:)

let $\omega(a) = n$.  Let $U = \{p_1,p_2,....,p_n\}$ be the distinct prime factors of $a$.  Then we can write the unique prime factorization of $a$ as $a = \prod_{p_i\in U} p_i^{m_i}$ where the $m_i$ are natural numbers (the powers to which the prime factors divide $a$)

Suppose $b, c$ are natural numbers.  And let $B=\{$ the distinct prime factors of $b\}$ and $C = \{$ the distinct prime factors of $c\}$ and $b,c$ have  unique prime factorizations of $b= \prod_{q_i\in B} q_i^{w_i}$ and $c= \prod_{r_i\in B} r_i^{u_i}$

Case 1: If $b|a$ and if $c|a$ then $B\subset U$ and $b=
    \prod\limits_{p_i\in B}p_i^{w_i}$ where $w_i \le m_i$.  And $C
    \subset U$ and $c= \prod\limits_{p_i\in C}p_i^{u_i}$ where $u_i \le
    m_i$.

Case 2: If $\gcd(b,c) = 1$ then $B\cap C = \emptyset$

Case 3: If $a = bc$ (then $b|a$ and $c|a$ so Case 1 holds)  then
$B\cup C = U$ and $bc = \prod\limits_{p_i\in
B}p_i^{w_i}\prod\limits_{p_i\in C}p_i^{u_i} = a = \prod_{p_i\in U}
p_i^{m_i}$.  And if $p_i \in B\cap C$ then $w_i + u_i = m_i$.  And
if $p_j\in B\setminus C$ then $w_i = m_i$ and if $p_k \in C\setminus
B$ then $u_i = m_i$.

So finally Case 4:  If $a = bc$ and $\gcd(b,c) = 1$ then $B\cup C =
U$ and $B\cap C = \emptyset$ so if we consider $U$ as our universal set we have $C = B^c$ and we have $b = \prod_{p_i\in B}p_i^{m_i}$ and $c = \prod_{p_j\in B^c} p_j^{m_j}$.

Now..... bang.

To determine a pair of $b,c$ so that $a = bc$ and $\gcd(b,c) = 1$, I must consider each prime factor, $p_i$ of $a$.

Either $p_i\mid b$ or $p_i \not \mid b$.  If $p_i\mid b$ then $p_i^{m_i}\mid b$.

So $b$ is uniquely determined by which subset $B\subseteq U$ we use. And $c$ is uniquely determined by ... what the complement of $B$ is.

And....

$|\{(b,c): a = bc; \gcd(b,c)=1\}|=$

$|\{b:b\mid a, c=\frac ab; \gcd(b,c) =1\}|=$

$|\{\prod_{p_i\in B} p_i^{m_i}: B\subseteq U\}|=$

$|\{B: B\subseteq U\}|=$

$|\mathcal P(U)|=$

$2^{|U|} = 2^{\omega(a)}$.

Or if you are not that comfortable with set theory:

Remember.  $\mathcal P(U)$ is the set of all subsets of $U$, and the number of subsets of $U$ are determined by considering each element of $x\in U$ and either $x$ may be in the subset or $x$ may not be in the subset.  That is two choices for each element of $U$ so there are $2^{\text{number of elements of U}}$ possible subsets.

Likewise for each prime factor of $a$ we can either have $p|b$ or $p|c$ so there are two choices for each prime factor of $a$ and so there are $2^{\text{number of prime factors}}$ choices of pairs.


问:
Expectation value of an operator
I am new to quantum mechanics and have some questions regarding expectation values of operators. I have the non-degenerate eigenstates $|\psi_n \rangle $ of an operator $A$. I have the normalized superposition of the eigenstates given by:

$$|\psi\rangle=\alpha|\psi_1 \rangle +\beta|\psi_2\rangle+\gamma|\psi_3 \rangle$$

Where $\alpha,\beta,\gamma$ are numbers. What the expectation value of $A$ is equal to?

答:
Assume $A$ is a self-adjoint operator.

Since $|\psi_n\rangle$ are eigenstates of $A$ then $A|\psi_n\rangle = \lambda_n|\psi_n\rangle$. In particular, we have that
\begin{align}
\langle \psi \mid A\ \mid \psi\rangle =&\ \langle\alpha\psi_1 + \beta\psi_2 + \gamma \psi_3 \mid A\mid \alpha\psi_1 + \beta\psi_2 + \gamma \psi_3\rangle\\
=&\ \langle\alpha\psi_1 + \beta\psi_2 + \gamma \psi_3 \mid \alpha\lambda_1\psi_1 + \beta\lambda_2\psi_2 + \gamma\lambda_3 \psi_3\rangle\\
=&\ \lambda_1|\alpha|^2\langle \psi_1\mid \psi_1\rangle +\lambda_2|\beta|^2\langle \psi_2\mid \psi_2\rangle +\lambda_3|\gamma|^2\langle \psi_3\mid \psi_3\rangle\\
=&\ \lambda_1|\alpha|^2+\lambda_2|\beta|^2+\lambda_3|\gamma|^2
\end{align}

Title: Expectation value of an operator


问:
Determine value of flow
For a flow $\phi^t \in \mathbb{R}^m$ we know that $\phi^{-1}(A)=B,\phi^2(B)=C,\phi^{-2}(B)=A$. Determine the value of $\phi^1(C)$ or show that this flow cannot exist.

I know that flow has a group property, so $\phi^t\circ\phi^s=\phi^{t+s}$. So I know that $\phi^1(A)=C$ and $\phi^{-3}(B)=B$ but it hasn't helped me much.

Title: Determine value of flow

Body:
Okay, so I solved it, it goes like this
$$\phi^1(C)=\phi^1(\phi^2(B))=\phi^3(B)=\phi^3(\phi^{-1}(A))=\phi^2(A)=\phi^2(\phi^{-2}(B))=\phi^0(B)=B$$


答:
The additional $+1$ is annoying, but can be gotten rid of. Let $b_n=a_n+c$ for a constant $c$ (that we do not know yet). Equivalently, $a_n=b_n-c$, so the recursion formula becomes
$$ b_{n+1}-c = 2(b_n-c)-1$$
or:
$$ b_{n+1}=2b_n-(c+1).$$
Now if we let $c=-1$, the recursion simplifies to
$$ b_{n+1}=2b_n$$
from which we easily guess (and verify) that
$$ b_n=2^nb_0,$$
where $b_0=a_0+c=3-1=2$ so that explcitly
$$b_n=2^n\cdot 2=2^{n+1} $$
and
$$ a_n=b_n-1=2^{n+1}+1.$$


问:
Norm of functional on $C[-1,1]$ defined as $P(f) = f(1)+f(-1)-2f(0)$
I am studying for my final exam and kinda struggling with the following:

Calculate the norm of functional on $C[-1,1]$ defined as $P(f) = f(1)+f(-1)-2f(0)$

This seems like one-liner but I am clueless anyway. I appreciate your time.

Body:
Clearly $\|P\| \le 4$. Let $f(x) = 2x^2-1$. Then
$$f(1) = f(-1) = 1, \quad f(0) = -1$$
so $P(f) = 4$. On the other hand, we have
$$\|f\|_\infty = \sup_{x\in[-1,1]}|f(x)| = \sup_{x\in[0,1]}(2x^2-1) =1$$
so we conclude $\|P\|=4$.


问:
Sum of a number of terms of $\left\lceil \frac{n}{2} \right\rceil $
The sum of integers from $1$ to $n$ is the following if we ignore the first half of the terms on the left side below:

$ 1+2+...+n\ge \left\lceil \frac{n}{2} \right \rceil + (\left\lceil \frac{n}{2} \right \rceil +1) + ... +n$

$\left\lceil \frac{n}{2} \right \rceil + (\left\lceil \frac{n}{2} \right \rceil +1) + ... +n \ge \left\lceil \frac{n}{2} \right \rceil +...+\left\lceil \frac{n}{2} \right \rceil$

$\left\lceil \frac{n}{2} \right \rceil +...+\left\lceil \frac{n}{2} \right \rceil = (n-\left\lceil \frac{n}{2} \right \rceil+1)*\left\lceil \frac{n}{2} \right \rceil$

$(n-\left\lceil \frac{n}{2} \right \rceil+1)\left\lceil \frac{n}{2} \right \rceil \ge (\frac{n}{2})(\frac{n}{2})$

Could you please explain how we get:

$\left\lceil \frac{n}{2} \right \rceil +...+\left\lceil \frac{n}{2} \right \rceil = (n-\left\lceil \frac{n}{2} \right \rceil+1)*\left\lceil \frac{n}{2} \right \rceil$

$(n-\left\lceil \frac{n}{2} \right \rceil+1)\left\lceil \frac{n}{2} \right \rceil \ge (\frac{n}{2})(\frac{n}{2})$

答:
The number of terms in $a+(a+1)+\dots +b$ is $b-a+1$.

Check the two cases $n=2k$ and $n=2k+1$.


问:
For what values of $l$ and $j$ (simultaneously) is the following congruence solvable?($a \leq b$)
Here is the congruence I want to know for what values of $l$ and $j$ (simultaneously) is it solvable?

$$ p^{b-a}l (xp^s)^2 \equiv xp^{s}j  + yp^t\pmod{p^a} $$ where $x,y$ are not divisible by $p.$ and $s,a,t, b$ are any non-negative integers and $a \leq b.$

We have dealt here For what values of $l$ and $j$ (simultaneously) is the following congruence solvable? with a very similar case but with $a \geq b.$
Now this is my trial for the solution:

We will use the same theorem 1 used in the above mentioned question (together with its proof):

Theorem 1

Let $M=\min\{A,B,C,D\}$. Then the equation
$$Xp^A+Yp^B+Zp^C \equiv 0\pmod {p^D}$$

can be solved with $XYZ \not\equiv 0\pmod p$ if and only if either

$\bullet \hspace{1cm}M=D$

$\bullet \hspace{1cm}M$ is equal to precisely two of $\{A,B,C\}$

$\bullet \hspace{1cm}M$ is equal to all three of $\{A,B,C\}$ and $p\ne 2$.

Proof

CASE 1. If $M=D$

Then each of $Xp^A,Yp^B$ and $Zp^C$ is divisible by ${p^D}$ whatever the values of $X,Y$ or $Z$.

CASE 2. If $M=A=B,M<C$

The equation is solved by, for example, $Y=Z=1,X=-1-p^{C-M}$. By symmetry, we have similar solutions when $M=A=C$ and $M=B=C.$

CASE 3. If $M=A=B=C,M<D$

If $p\ne 2$ then the equation is solved by, for example, $Y=Z=1,X=p^{D-M}-2$. However, if $p=2$ then $X+Y+Z$ has to be even and so at least one of $X,Y,Z$ is even.

CASE 4. If $B,C$ and $D$ are all greater than $M=A$

Dividing the equation through by $p^M$ gives
$$X+Yp^{B-M}+Zp^{C-M} \equiv 0\pmod {p^{D-M}}$$ and then we have the contradiction that $p$ divides $X$.

By symmetry, we have similar contradictions when $M=B$ and $M=C.$

The transformation

Let $l=Lp^U$ and $j=Jp^V$ where $LJ \not\equiv 0\pmod p$. Then **(Note here is a difference between the 2 questions)**let $$A = s+V, B=t,C=2s + b -a + U ,D = a.$$

Theorem 2

For non-negative integers $a,b,s$ and $t$ with $a \leq b$, the equation
$$p^{b-a} l (xp^s)^2 \equiv xp^{s}j  + yp^t\pmod{p^a}$$
can be solved with $xy \not\equiv 0\pmod p$ if and only if the equation
$$Xp^A+Yp^B+Zp^C \equiv 0\pmod {p^D}$$
can be solved with $XYZ \not\equiv 0\pmod p$.

Proof

If the $x,y$ equation can be solved then simply take $$X=xJ,Y=y,Z=-Lx^2.$$

Conversely, suppose we have a solution of the $X,Y,Z$ equation. Since $X$ and $L$ are coprime to $p$ we can get the following congruence equations:

$$x \equiv -\frac {JZ}{LX}\pmod {p^D},y \equiv -\frac {J^2YZ}{LX^2}\pmod {p^D}.$$

Details of their calculations:

Details of solution for $x \equiv -\frac {JZ}{LX}\pmod {p^D}$

$LX$ and $p^D$ are coprime. Therefore there are integers $\alpha$ and $\beta$ such that $\alpha LX+\beta p^D=1$.\

Then  $\alpha LX\equiv 1 \pmod {p^D}$  and so  $-\alpha JZ LX\equiv -JZ \pmod {p^D}$. Then take $x=-\alpha JZ$.\

Details of solution for $y \equiv -\frac {J^2YZ}{LX^2}\pmod {p^D}$.

$LX^2$ and $p^D$ are coprime. Therefore there are integers $\alpha' $ and $\beta '$ such that $\alpha' LX^2 + \beta' p^D=1$.\

Then  $\alpha' LX^2 \equiv 1 \pmod {p^D}$  and so multiplying by $-J^2YZ$ we will get  $-\alpha'LX^2 J^2YZ\equiv -J^2YZ \pmod {p^D}$. Then take $y = -\alpha' J^2 YZ$.

Then $$  xp^{s}j + yp^t - p^{b-a}l(xp^s)^2=xJp^A+yp^B-Lx^2p^C$$

and by substituting the previous congruences of $x$ and $y,$ we get

$$ -\frac {J^2Z}{LX^2}(Xp^A+Yp^B+Zp^C )\equiv 0\pmod {p^D}.$$ Because we know that $ Xp^A+Yp^B+Zp^C \equiv 0\pmod {p^D}.$

Conclusion

We can now see that the original equation can be solved if and only if the quantities $$A = s + V,B=t,C= 2s + b -a + U, D = a$$ satisfy the conditions of Theorem 1. Let $v_p(n)$ represent the highest power of $p$ which divides the number $n$. You will notice that the precise values of $j$ and $l$ do not affect the existence of solutions; all that matters is $v_p(j)$ and $v_p(l)$ .

CASE 1. $t\ge a$,  $v_p(l)\ge 2a - 2s -b$ and  $v_p(j)\ge a-s$.

Where the first inequality because $B \geq D,$ the second inequality because $C \geq D$ and the third inequality because $A \geq D.$

CASE 2. There are three possibilities.

$v_p(l)> t-2s -b +a$ and  $v_p(j)= t-s$.

Where the inequality because $B < C$ and the equality because $A = B.$

$v_p(j)< t - s$ and  $v_p(l) = -b +a -s + v_p(j)$.

Where the inequality because $A \leq B$ and the equality because $A = C.$

$v_p(l)= t-2s - b + a$ and  $v_p(j) > t-s$.

Where the equality because $B = C$ and the inequality because $B < A.$

CASE 3. If $p\ne 2$, then $t< a$,  $v_p(j)= t-s$ and  $v_p(l)= a - b + t - 2s$.

Where the inequality because $B < D,$ the first equality because $B = A$ and the second equality because $B = C.$

Is this a correct solution or I need to adjust anything?

答:
Well done. All the calculations are correct.

In the statement of the problem it might be best to say $a$ is positive so as not to have an equation modulo $1$.

Also, there is a special possibility in CASE 3 if $p=2$. (Unless this case is of interest to you it might be simpler to say $p$ is odd from the outset.)

In the case that $p=2$ then, for a solution, we require at least one of $X=xJ,Y=y,Z=-Lx^2$ to be even. However, these quantities cannot be multiples of $p$ and so are all odd. So there is no solution for CASE 3 when $p=2$.


问:
Check if $X$ is completely metrizable

Let: $$D_t = \{(x,y,t) \in \mathbb{R}^3 : x^2 +y^2 \leq 1\},$$ $$S=\{(p,q,0)\in \mathbb{R}^3: p^2 +q^2<1, p\in \mathbb{Q}, q\in \mathbb{Q} \}$$ $$I(a,b) \text{ - open line segment between points } a, b$$
Now let's define set $X$: $$X=D_0 \cup \Biggl( \bigcup_{(p,q,0)\in S} I\big((p,q,0),(p,q,p)\big) \Biggr) \subset \mathbb{R}^3$$ Check if $X$ is completely metrizable.

I have an answer of this task but I don't understand one statement:

Let: $$A=X\cap \{(x,y,z): z=\frac 12\}=\{(p,q,\frac 12): p\ge \frac 12; \; p^2+q^2<1; \; p,q\in \mathbb Q\}=$$  $$=\bigcup\limits_{p,q\in \mathbb Q} \{(p,q,\frac 12): p\ge \frac 12; \; p^2+q^2<1\}$$ So $A$ is the countable sum of points i.e. closed and boundary set.
 Assume that $X$ is completely metrizable. Then $A$ as closed subset $X$ is also completely metrizable. However $A$ isn't a boundary set so Baire's condition is not satisfied. So we got a contradiction - $X$ is not completely metrizable.

Can someone explain to me why $ A $ is not a boundary set?

EDIT: $X$ is a boundary set when Int$(X)=\varnothing$

答: I don´t know how they define boundary set. However, it is easy to check that $A$ is not a Baire space. For if you pick an enumeration $a_n$ of the points of $A$, and you pick the open sets $A-\{a_n\}$, they are all dense open sets in $A$, and their intersection is empty. Thus $A$ is not a Baire space.

Title: Check if $X$ is completely metrizable


问:
How do you find the area of a Venn diagram?
If we have a Venn diagram (only two circles)

And the larger circles have a radius of 2...

If the middle circle has a radius of 1, how can we find the area of the intersection part?

I understand that side is the middle of the larger circles, but I'm still confused on what to do to find the area of the middle intersection part.

Can anyone help? Thanks.

答:
I think you want to compute twice the area of the circular segment delimited by the line connecting the two intersection points: https://en.wikipedia.org/wiki/Circular_segment

What you were drawing in red is a bit confusing (the line you drew to indicate the "radius" of the big circles doesn't seem to pass through their centre, as it should be given the numbers you were posting). However if the two large circles have a radius equal to 2, and the radius of the inner circle tangent to the two bigger ones is equal to 1, then the distance between the centres of the big circles will be $d_c = 2R - 2r = 2$, i.e. each big circle pass through the centre of the other one. Thus, the distance between each centre and the line connecting the intersection points is $d = d_c/2 = 1$. Now, following the formula for the area of the circular segment, the whole area of the intersection is $$A = 2R\biggl(R\arccos(d/R)-d\,\sqrt{1-(d/R)^2} \biggr) = 4\biggl(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\biggr) = \frac{8\pi-6\sqrt{3}}{3}$$


问:
Is learning (theoretical) physics useful/important for a mathematician?
I'm starting to read The Princeton Companion to Mathematics, at the beginning it says:

A proper appreciation of
  pure mathematics requires some knowledge of applied
  mathematics and theoretical physics.

Some of my professors have told me that modern Mathematics require some knowledge about Quantum Mechanics and theoretical Physics.

I attended the second and third day of the José Adem Memorial Lecture Series by Matilde Marcolli at the beginning of this year, they were about Number theory, Quantum statistical Mechanics and Quantum Field theory. I did not understand a single word but somehow the few things I understood had a big impression on me.

Please give me some examples of pure Mathematics that require/use Physics.

答:
It is certainly possible to study all kinds of topics in pure mathematics without any knowledge of physics, because you will always find literature/researchers who are used to explain the key concepts to fellow mathematicians without any knowledge in physics.

But here are some examples of useful interconnections, in arbitrary order:

The most prominent example of recent history is probably the work of Edward Witten on low dimensional topology that earned him the Fields medal. Wittens work is inspired by thinking about quantum field theory and extensions, using heuristic tools like path integrals. He was very successful, so that other mathematicians have tried to learn tools from theoretical physics, too, which led to seminars which led, e.g., to the two volume book "Quantum Fields and Strings: A Course for Mathematicians". His methods resulted in reducing the length of some proofs of Donaldson from 100+ pages to < 10 pages.

Classical mechanics is the study of symplectic manifolds (this is a simplification, of course). Much of the work of Poincaré on ordinary differential equations was inspired by the study of classical mechanics.

Principle bundles is about the study of both classical and quantum gauge theories, which encompasses electromagnetics, for example. The close ties have led Dale Husemöller to (co-) author "Basic Bundle Theory and K-Cohomology Invariants (Lecture Notes in Physics)", which is an extension of his classic textbook "Fibre bundles".

Quantum mechanics is about linear operators on Hilbert spaces. Most of linear functional analysis has been developed in order to understand quantum mechanics, espacially everything that von Neumann did.

Differential geometry on Lorentzian manifolds is the study of the theory of general relativity = gravitation.

Operator algebras (C* and von Neumann algebras) is about axiomatic quantum statistical mechanics, see Robinson, Bratteli: Operator Algebras and Quantum Statistical Mechanics 1 and two. And it is about axiomatic quantum field theory, see Haag: "Local Quantum Physics". The structure theory of von Neumann algebras finds deep applications here, for example the classification of factors is applied to local algebras, modular theory is connected to representations of the Poincaré group etc. You already mentioned the work of Connes et. alt. about noncommutative geometry and their application to quantum field theory and the standard model. Noncommutative geometry is also applied in order to construct quantum versions of spacetimes (i.e. quantum geometry).

Fluid dynamics is about solutions to the Navier-Stokes equations. The flow of ideal fluids forms infinite dimensional manifolds, see "V.I. Arnold, ; B.A. Khesin: Topological methods in hydrodynamics."

Someone else than me should explain the applications of complex and algebraic geometry to string theory. (And n-categories.)

All of these topics can be studied without mentioning theoretical physics, although I think it would be a shame to do that.


问:
Prove if the set is closed? Bounded? Compact?
Consider the metric space $(\mathbb{Q},d)$ where $\mathbb{Q}$ denotes the rational numbers and $d(x,y)=|x-y|$.  Let $$E:=\{x \in\mathbb{Q}:x>0, 2<x^2<3\}$$

Is $E$ closed and bounded in $\mathbb{Q}?$  Is it compact?  Justify your answers.

答: It is a metric space.  A compact metric space is necessarily complete.  This will tell you about compactness.</p>


问:
Is there a short symbology to express that $f$ is a morphism in $C$?
For the simple purpose of taking notes without writing too much, I'm wandering two things.

Given the category $C$

is it ok to write $c \in C$ to express the idea that the object $c$ is in the category $C$?

Is there a symbol to express that $f: a \to b$ is a morphisms between objects of $C$? I guess even if the answer to the previous question was yes, $f \in C$ would be ambiguous because the reader wouldn't know whether $f$ is a morphism or an object.

As far as I know, the symbol $\in$ is a thing in set theory at least, but I'm not sure if it's appropriate for category theory, hence I'm asking this question.

答:
A common notation seems to be
$$c\in\operatorname{Ob}(C)$$
and $$f\in\operatorname{Mor}(C)$$
(or if you want to be more  specific: $f\in \operatorname{Mor}_C(a,b)$).

Using $\in$ in this context is okay because $\operatorname{Ob}(C)$ and $\operatorname{Mor}(C)$ are at least classes, though typically not sets.


问:
Characterisation of complete riemannian manifolds using divergent curves.
Let $M$ be a riemannian manifold. Then $M$ is complete iff each divergent curve in $M$ has no finite length (a curve $\gamma : [0 , 1) \to M$ is divergent if for each compact $K \subset M$ there exists $t \in [0 , 1)$ such that $\gamma([0 , 1)) \cap K = \emptyset$.

I have shown that if $M$ is complete then each divergent curve in $M$ has no finite length. But how can I prove the converse?

答: For the converse, assume that $M$ is not complete, and let's find a divergent curve with finite length. By incompleteness, there is an inextendible geodesic $\alpha\colon [0,b) \to M$, with $b<+\infty$. We claim that $\alpha$ is already divergent: indeed, if there is a compact $K\subseteq M$ with $\alpha(t) \in K$ for all $t \in [0,b)$, we may take an increasing sequence $(t_n) \subseteq [0,b)$ with $t_n \to b$ and pass to a subsequence to assume that $\alpha(t_n) \to p$ and $\alpha'(t_n) \to v$ for some $p \in M$ and $v \in T_pM$. With this in hands, we may pick $\epsilon > 0$ small enough and extend $\alpha$ to a geodesic $\widetilde{\alpha}\colon [0,b+\epsilon) \to M$ by $$  \widetilde{\alpha}(t) = \begin{cases} \alpha(t), &\mbox{ if }t\in[0,b) \\ \exp_p((t-b)v), &\mbox{ if }t \in [b,b+\epsilon)\end{cases},   $$contradicting the inextendability of $\alpha$. And if $\alpha$ has constant speed $c$, clearly $L[\alpha] = cb<+\infty$, concluding the proof. Note that we may reparametrize this $\alpha$ to something whose domain is $[0,+\infty)$, but the length and the property of being divergent remain both unchanged.


答:
$\triangle CAB$ is equilateral, i.e. $\text{ar}\triangle CAB=\frac{\sqrt3}4(2)^2=\sqrt3$. Area of sector $CAB$ is $\frac{60}{360}\pi(2)^2=2\pi/3$.

Thus the area of the shaded region is $2\pi/3-\sqrt3$ and the area of the top-half of the intersection region is the sum of the areas of sector $CAB$ and the shaded region, i.e. $4\pi/3-\sqrt3$, giving the required area as its double $8\pi/3-2\sqrt3$.

Title: How do you find the area of a Venn diagram?

Body:
If we have a Venn diagram (only two circles)

And the larger circles have a radius of 2...

If the middle circle has a radius of 1, how can we find the area of the intersection part?

I understand that side is the middle of the larger circles, but I'm still confused on what to do to find the area of the middle intersection part.

Can anyone help? Thanks.


问:
Finding a strong enough solution to a specific PDE problem.
Let $U\subset \mathbb{R}^n$ with smooth boundary $\partial U$. And consider the expression

$$\Delta u = f.$$

$$\text{+"convenient boundary conditions"}$$

In my specific case $f\in H^2_0$. Under what (boundary) conditions is it possible to conclude a weak solution which is also in $H^2_0$ of this problem?

The problem is mainly concerned with the subscript zero in $H^2_0$. For instance if I introduce the boundary condition

$$u=0 \text{ on } \partial U\ \text{ (trace-wise),}$$

then for any $f\in L^2$, there exists a weak solution in $H^1_0$. I can regularize the solution to be also in $H^2$, hence it is in $H^1_0\cap H^2$. (I use Evans for referrences concerning to Regularity theorems.) But as far as I know $H^1_0\cap H^2$ is a strictly bigger class of functions then $H^2_0$.

Help is very much appreciated.

答:
Under what (boundary) conditions is it possible to conclude a weak solution to $\Delta u = f$ which is also in $H^2_0$ of this problem?

The answer is that: no boundary conditions can do this.

Rule of thumb: If the imposed boundary condition leads to a well-posed weak formulation (solution exists and is unique), then no matter what this boundary condition is, we can not get a weak solution $u\in H^2_0$ in general.

First notice $H^2_0$ means:
$$
H^2_0(U) := \{u\in H^2: \; u\big|_{\partial U} = 0, \; \nabla u\big|_{\partial U} = 0\}.
$$
Also note if $u$ is a constant on boundary, then $u$ does not change tangentially along any curve on boundary, so the directional derivative of $u$ along any tangent vector of $\partial U$ is 0. The only thing we wanna deal with is the normal derivative:
$$
u\big|_{\partial U} = 0 \text{ and } \nabla u\big|_{\partial U} = 0 \implies u\big|_{\partial U} = 0\text{ and } \frac{\partial u}{\partial n}\Big|_{\partial U} = 0
$$

Now if $u$ is a weak solution of the Poisson equation, and $u\in H^2_0$, this means
$$\left\{
\begin{aligned}
-\Delta u &= f \quad \text{ in } U,
\\
u &= 0  \quad \text{ on } \partial U,
\\
\partial u/\partial n &= 0  \quad \text{ on } \partial U,
\end{aligned}\right.\tag{$\star$}
$$
for some $f$.

However, $(\star)$ is not a well-posed problem. Looking for a function $u\in H^2_0$  is like imposing Dirichlet and Neumann boundary conditions at the same time (a Cauchy boundary problem), which can be proved ill-posed in several ways.

A simple argument can be: if neglecting Neumann boundary at first, Poisson equation together with the Dirichlet boundary has a unique solution for any $f\in L^2(U)$. This means $\nabla u$ can be evaluated a.e. in $U$, and if $\partial u/\partial n|_{\partial U}$ is defined as a limit from the interior to the boundary in the direction of the normal , then $\partial u/\partial n$ is already determined in this case.
In general we can't expect $\partial u/\partial n = 0$ coincides with what you obtained from the solution of Dirichlet boundary problem.

A counterexample would be: suppose we have a classical solution which coincides with a strong and weak solution to the Dirichlet boundary problem except a measure zero set, and $f\leq 0$. By maximum principle the maximum is achieved on boundary, $u|_{\partial U}=0> u(x)$ for $x\in U$, then $\partial u/\partial n$ must be strictly greater than 0 pointwisely (see Gilbarg and Trudinger Page 32).

Another argument would be if $\partial u/\partial n|_{\partial U} = 0$, by divergence theorem:
$$
\int_U f\,dx = \int_U \mathrm{div}(\nabla u)\,dx = \int_{\partial U} \frac{\partial u}{\partial n} \,dS = 0,
$$
which means $f$ has average 0 in $U$. Hence this says that, if $f$ does not have zero average, the $(\star)$ does not have a solution.

Some comment about your terminology, "regularization" doesn't mean what you have written in your question, what you applied is just elliptic regularity.


问:
Another form of the divergence theorem
The task is to calculate an integral of the form
$$\iint_s x dydz + ydxdz +zdxdy. $$
The text I'm studying says this is just another form of the famed divergence theorem,
$$\iint_s F.n \,ds = \iiint_v \operatorname{div} F \,dV,$$
where $F$ is simply $(x, y, z)$, so that the value of the integral is
$$\iiint (1+1+1) dV = 3V.$$

The question is why is the original integral equal to the divergence integral?
More generally, why is it true that
$$\iint_s  Pdzdy+ Qdxdz +Rdxdy =
\iint_s (P, Q, R)\cdot n \,ds?$$

答:
It is indeed true that
$$(F\cdot \hat{n})\,dA = F_x dy\wedge dz + F_y dz\wedge dx + F_z dx\wedge dy.$$
(Be careful about the orientation of each term!)

This follows from the fact that
$$\hat{n}_x dA = dy\wedge dz$$
on the surface of $S$, and similarly for the other coordinates; you can prove this by direct calculation on a coordinate chart, but probably the easiest way is to notice that the Hodge stars of both term,
$$\hat{n}_x (\hat{n}_x dx + \hat{n}_y dy + \hat{n}_z dz), \qquad dx$$
agree on vectors $\alpha \hat{n}$ orthogonal to $S$.

Title: Another form of the divergence theorem


问:
Does we have this equality $\lim_{n\to\infty}\int_{-a}^a K_n(s)f(s)\,ds=\frac{\lim_{x\nearrow 0}f(x)+\lim_{x\searrow 0}f(x)}{2}$?
I claim that we do have the following equality. And I give my proof. Am I correct?

Let $I=(-a, a)$ be an interval (finite or infinite). Suppose that $\{K_n\}_{n=1}^\infty$ is a positive summation kernel, i.e. it is a sequence of real-valued, Riemann-integrable functions defined on $I$, with the following properties:
\begin{align}&(a)K_n(t)\geq 0,\forall n\in\mathbb N,t\in\mathbb R\notag\\
&(b)\int_{-a}^aK_n(t)\,dt=1\notag\\
&(c)\text{If }\delta>0, \lim_{n\to\infty}\int_{a>|t|>\delta}K_n(t)\,dt=0.
 \end{align}  If $f$ is integrable and bounded on $I$ and the left and right limits exist, say $$\lim_{x\nearrow 0}f(x)<\infty\text{ and }\lim_{x\searrow 0}f(x)<\infty.$$ Then $$\lim_{n\to\infty}\int_{-a}^a K_n(s)f(s)\,ds=\frac{\lim_{x\nearrow 0}f(x)+\lim_{x\searrow 0}f(x)}{2}.$$

Here is my proof:

Let $\lim_{x\nearrow 0}f(x)=L\text{ and }\lim_{x\searrow 0}f(x)=R.$ $\exists \delta_1>0$ such that if $x\in (-\delta_1,0),$ we get  $$|f(x)-L|<\epsilon ,\forall \epsilon>0.$$ $\exists \delta_2>0$ such that if $x\in(0,\delta_2)$, then $$|f(x)-R|<\epsilon ,\forall \epsilon>0.$$ Let $\delta=\min\{\delta_1,\delta_2\},$ then if $x\in(-\delta,\delta)\setminus\{0\},$ we have $$\left|f(x)-\frac{L+R}{2}\right|=\frac{1}{2}|2f(x)-L-R|\leq \frac{1}{2}|f(x)-L|+\frac{1}{2}|f(x)-R|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$ Then $$
 \Delta=\int_{-a}^aK_n(s)f(s)\,ds-\frac{L+R}{2}=\int_{-a}^aK_n(s)f(s)\,ds-\frac{L+R}{2}\int_{-a}^aK_n(s)\,ds=\int_{-a}^aK_n\left[f(s)-\frac{L+R}{2}\right]\,ds.$$ We want to show that $\Delta\to 0$ as $n\to\infty.$ Since $f(x)$ is bounded on $I,$ then $\exists M'\in \mathbb R$ such that $$|f(x)-\frac{L+R}{2}|\leq 2M,$$ where $M=\max\{M',(L+R)/2\}.$ Then
\begin{align}
  |\Delta|&=\left|\int_{-a}^aK_n(s)\left[f(s)-\frac{L+R}{2}\right]\,ds\right|\leq \int_{-a}^aK_n(s)\left|f(s)-\frac{L+R}{2}\right|\,ds\notag\\
  &=\int_{(-\delta,\delta)}K_n(s)\left|f(s)-\frac{L+R}{2}\right|\,ds+\int_{(-a,a)\setminus (-\delta,\delta)}K_n(s)\left|f(s)-\frac{L+R}{2}\right|\,ds\notag \\
  &=\lim_{\alpha \nearrow 0}\int_{(-\delta,\alpha )}K_n(s)\left|f(s)-\frac{L+R}{2}\right|\,ds+\lim_{\beta \searrow 0}\int_{(\beta,\delta)}K_n(s)\left|f(s)-\frac{L+R}{2}\right|\,ds\notag  \\
  &\ \ \ \ +\int_{(-a,a)\setminus (-\delta,\delta)}K_n(s)\left|f(s)-\frac{L+R}{2}\right|\,ds\notag \text{ ------- }(1) \\
  &\leq \epsilon\lim_{\alpha \nearrow 0}\int_{(-\delta,\alpha )}K_n(s)\,ds+\epsilon\lim_{\beta \searrow 0}\int_{(\beta,\delta)}K_n(s)\,ds+2M\int_{(-a,a)\setminus (-\delta,\delta)}K_n(s)\,ds\notag\\
  &= \epsilon\int_{(-\delta,\delta)}K_n(s)\,ds+2M\int_{(-a,a)\setminus (-\delta,\delta)}K_n(s)\,ds\notag \text{ ------- }(2)\\
  &\leq\epsilon+2M\int_{(-a,a)\setminus (-\delta,\delta)}K_n(s)\,ds\notag.
 \end{align} $\exists N\in\mathbb N$ such that if $n\geq N,$ then $$\int_{(-a,a)\setminus (-\delta,\delta)}K_n(s)\,ds\leq \frac{\epsilon}{2M}$$ $\Longrightarrow$ $$|\Delta|<2\epsilon.$$
Did I make mistakes in (1) and (2) above?

答: As stated, the result is false. Take $a=1.$ Let $K_n(x) = (n+1)(1-x)^n, x\in [0,1],$ $K_n=0$ elsewhere. Define $f=1$ on $[0,1],$ $f=0$ on $[-1,0).$ We then have $\int K_n f \to 1,$ not $(1+0)/2.$


问:
Equivalent condition for a collection of subsets to be a base for a topology
I wish to prove the following claim.

Claim. Let $(X,\tau)$ be a topological space. Then a collection of subsets of $X$, $\sigma$ is a base of $\tau$ if and only if:

Every $x\in X$ is contained in at least one set in $\sigma$. (That is, $\sigma$ covers $X$).

For all $B_1,B_2\in \sigma$, we have $B_1\cap B_2$ can be written as a union of elements of $\sigma$.

Certainly it is obvious that $\sigma$ being a base implies the conditions (1) and (2). However, I have been having more trouble with the other way round.

I can see that for any set $U\in \tau$, we will have $U\subset\bigcup_{\alpha \in A}B_\alpha,$ for some indexing set $A$. This follows from (1), but I am unsure how (2) helps us prove that $U$ can be written as a union of base elements.

Any guidance would be useful.

Note: I had recently asked a similar question to this, but realised I made a fundamental typo which invalidated comments and answers, so I have posted this as a new question with the correct information.

答:
The question is false as stated, because you start with a fixed $(X,\tau)$.

In order to get a base for that topology we at least need the base elements to be open themselves so $\sigma \subseteq \tau$, or else any $B$ that is not open will be in the topology generated by $\sigma$ but not in $\tau$...

But even that extra requirement is not enough: let $(X,\tau)$ be the Sorgenfrey line, so $X=\Bbb R$ in particular and we could choose $\sigma$ the collection of all open intervals (which are indeed Sorgenfrey open), and obey the two axioms. But they only generate the smaller Euclidean topology, not $\tau$.

You can salvage it by omitting $\tau$ altogether and start with a set $X$ and a collection $\sigma$. Then these two conditions are necessary and sufficient for the set of unions of subcollections of $\sigma$ to form some topology on $X$. Try to prove that fact instead.


问:
Problem on real-valued harmonic function.
Problem: Let $g$ be harmonic on $U$ where $U \subseteq \mathbb{R}^2$ is open and connected. Show that if $g$ is non-constant on $U$, then $g(U)$ is open in $\mathbb{R}$.

I noticed that we could show that $g$ does not have a minimum in $U$ by taking the contrapositive of the Minimum Principle:

Let g be harmonic on $U$ where $U$ is open and connected. If $g$ has a minimum in $U$, then $g$ is constant.

I'm currently stuck here and not too sure how to proceed. Any suggestion?

答:
$f(U)$ will be an interval (or a ray, or $\mathbb{R}$), because $U$ is connected and $f$ is continuous.

If $x\in U$, $f(U)$ must contain a nhood of $f(x)$. Because if not, $x$ would be an extreme point of the interval, that is, there would be a maximum of a minimum at $x$, which can´t happen. Thus $f(U)$ is open.


问:
Cantor's Diagonal Argument - Rational
I have read other answers about this same subject, however, no answers seem to formally prove, why the Cantor Diagonal Argument applied on a list of all the rational does not create another rational as Cantor Diagonal Argument applied on the Reals create another real.

Thank you

答: Because if you have a sequence $(d_n)_{n\in\Bbb N}$ of digits, there is some real number whose decimal expansion is $0.d_1d_1d_3\ldots$ So, if you apply the Cantor diagonal argument to the numbers of $[0,1]$, you will get a number from $[0,1]$. But if you apply it only to the rationals form $[0,1]$, it may well happen that the number that you created is not rational, since most sequences $(d_n)_{n\in\Bbb N}$ of digits lead you to a number $0.d_1d_1d_3\ldots$ which is irrational.

Title: Cantor's Diagonal Argument - Rational


答:
HINT: You should have no trouble deciding whether $E$ is bounded in $\Bbb Q$. To decide whether it’s closed in $\Bbb Q$, ask yourself whether $\Bbb Q\setminus E$ is open in $\Bbb Q$: can you find an open set $U$ in $\Bbb R$ such that $U\cap\Bbb Q=\Bbb Q\setminus E$? Remember, $\sqrt2$ and $\sqrt3$ are irrational.

For compactness, does the sequence $\langle 1.7,1.73,1.732,1.7320,1.73205,\dots\rangle$ in $E$ have a convergent subsequence?

Title: Prove if the set is closed? Bounded? Compact?

Body:
Consider the metric space $(\mathbb{Q},d)$ where $\mathbb{Q}$ denotes the rational numbers and $d(x,y)=|x-y|$.  Let $$E:=\{x \in\mathbb{Q}:x>0, 2<x^2<3\}$$

Is $E$ closed and bounded in $\mathbb{Q}?$  Is it compact?  Justify your answers.


答:
Since $U$ is connected and $g$ is continuous, the image $g(U)$ must be connected. The connected subsets of $\Bbb R$ are precisely the intervals. (Possibly unbounded on one or both ends.)

Now, let $y \in g(U)$. Then, $y = g(x)$ for some $x \in U$. As you noted: $g$ cannot have a minimum (and the same reasoning applied to $-g$ shows that it cannot have a maximum either).

Thus, there exist $x_1, x_2 \in U$ such that $g(x_1) < g(x) < g(x_2)$. Since $g(U)$ is an interval, we see that $$g(x) \in (g(x_1), g(x_2)) \subset g(U),$$ as desired.

Title: Problem on real-valued harmonic function.

Body:
Problem: Let $g$ be harmonic on $U$ where $U \subseteq \mathbb{R}^2$ is open and connected. Show that if $g$ is non-constant on $U$, then $g(U)$ is open in $\mathbb{R}$.

I noticed that we could show that $g$ does not have a minimum in $U$ by taking the contrapositive of the Minimum Principle:

Let g be harmonic on $U$ where $U$ is open and connected. If $g$ has a minimum in $U$, then $g$ is constant.

I'm currently stuck here and not too sure how to proceed. Any suggestion?


问:
Polynomials/ remainder theorem
What must be added to p(x)=x^4+2x^3-2x^2+x-1 to make it exactly divisible by g(x)=x^2+2x-3?

The book says that the remainder will be a linear polynomial and so the expression that must be added to p(x) will be of the form ax+b; the book essentially says we add the remainder. The divisor has roots 1 and -3. We can solve by putting p(1) and p(-3) equal to zero and finding the values of a and b. I know how to solve this but I am confused as to why we say that the expression to be added is of the form ax+b. How do we know this. Essentially what the book seems to be saying is add the remainder. But if we take the example7/3, the remainder is 1 but adding the remainder will not make 7 exactly divisible by 3. Here we subtract 1 from 3 and add the result to make 7 exactly divisible by 3. I believe the same principle must hold in the case of polynomials and the explanation in the book that we add the remainder to the dividend must be wrong. Please tell me the reason why we add ax+b. Please explain.

Also, can it be solved through polynomial long division? If so, how?

Are there any other easier ways to solve this? Feel free to ignore the last question and even the second last question if it will make the answer too long.

答:
The division algorithm says that if $p(x)$ is a polynomial of degree $n$ and $g(x)$ is a polynomial of degree $m\le 0$ then there is a unique quotient polynomial $q(x)$ of degree $n-m$ and a unique remainder polynomial of degree $< m$ so that:  $p(x) = q(x)g(x) + r(x)$.

So if $p(x) = x^4+2x^3-2x^2+x-1$ is of degree $4$ and $g(x) = x^2+2x-3$ is of degree $2$ then there is a unique polynomial of degree $4-2 = 2$ called $q(x)$ and a unique polynomial $r(x)$ of degree less than $2$ so or degree of at most $1$ so that

$p(x) = q(x)g(x) + r(x)$.

Now as $r(x)$ is at most of degree $1$ it is of the form $ax + b$ (where $a$ could be $0$).

And

$x^4+2x^3-2x^2+x-1 = (x^2+2x-3)(\alpha x^2 + \beta x + \gamma) + (ax+b)$

so to make $p(x)$ exactly divisible by $g(x)$ we must add the NEGATIVE of the remainder (I don't understand why the book says to add rather than subtract the remainder) to get:

$x^4 + 2x^3 - 2x^2 + x-1 +(-ax -b) = (x^2+2x-3)(\alpha x^2 + \beta x + \gamma)$

And the most straightforwar way to do this is long division.

But typing long division on a keyboard is too hard for me so I'll talk us through:

$(x^2+2x-3)(\alpha x^2 + \beta x + \gamma)= \alpha x^4 + (2\alpha + \beta)x^3 +(\gamma+2\beta -3\alpha) x^2 + (2\gamma -3\beta)x- 3\gamma$.

We want that to be as close to $x^4 + 2x^3 -2x^2 + x-1$ as possible.

So $\alpha = 1$ and $2\alpha + \beta = 2$ and $\gamma +2\beta -3\alpha = -2$.  And that's as fair as we will be able to go.

$\alpha = 1$ and $\beta = 0$ and $\gamma = 1$.

And so we have

$x^4+2x^3-2x^2+x-1 = (x^2+2x-3)(x^2 +1) + (ax+b)$

$x^4+2x^3-2x^2+x-1 = x^4 +2x^3 - 2x^2 +2x -3 +(ax+b)$.

And so the remainder is $ax + b = -x +2$

so what we must add is $-(-x+2) = x -2$.

That is to say

$(x^4+2x^3-2x^2+x-1) + (x-2) = (x^2+2x-3)(x^2 +1)$

.......

But that is the most straightforward way.

The easiest way is to note:

If $p(x) + (mx+n) = (x^4 + 2x^3 - 2x^2 + x-1) + (mx + n) = (x^2 + 2x -3)q(x)=g(x)q(x)$.

And the roots of $g(x) = x^2 + 2x -3)$ are $x = 1$ and $x=-3$ then

$p(1) + (m*1 + n) = g(1)q(1) = 0*q(1) = 0$ so

$(1^4 + 2*1^3 -2*1^2 + 1-1) + (m+n) = 0$

$1+m+n = 0$

And $p(-3) + (-3m + n) = 0$ and

$(3^4 - 2*3^3 - 2*3^2 - 3-1) -3m +n =0$

$81 - 54 -18 -4-3m+n = 0$

$5- 3m + n=0$

So $5-3m + n = 1+m+n$

$5-3m = 1+m$

$4=4m $ and $m =1$ and $1+m+n = 0$ so $1+1+n =0$ so $n=-2$.

Title: Polynomials/ remainder theorem


问:
Solve Diophantine Equation
Show that $ x^3 - 6 = 25y^2  + 35y $ doesn't have any non-zero integer solution.

What I have tried

$$ 25y^2 +  35y - x^3 + 6 = 0 $$
solving for y we get
$$ y = \frac{-35 \pm \sqrt { 625 + 100 x^3} } {50} $$
so  I need to show that
$$ 25 + 4x^3= z^2 $$
doesn't have a solution
or if it does I need to show that
the fraction is not equal to an integer
but I can't proceed.

答:
$x^3-6=25y^2+35y$ can be transformed to $v^2=u^3+400$ with $x=u/4$ and $y=(v-28)/40$.

The $(u,v)$ equation is an elliptic curve which has $2$ finite torsion points at $(0, \pm 20)$, giving $x=0,y=-1/5$ and $x=0,y=-6/5$ as solutions.

The rank of the curve is $0$ (from standard software) so there are NO other rational points on the curve.

Thus, there are no integral solutions.

Title: Solve Diophantine Equation


答: If you mean the version of the diagonal argument that deals with decimal digits, or with binary digits (etc.), then the answer is that we know how to create an enumeration of the rationals, i.e. a sequence in which for every rational there is some finite number of steps after which we will reach that rational in the list, and the diagonal argument shows that the real number that emerges differs from every number in the list.</p>

Title: Cantor's Diagonal Argument - Rational

Body:
I have read other answers about this same subject, however, no answers seem to formally prove, why the Cantor Diagonal Argument applied on a list of all the rational does not create another rational as Cantor Diagonal Argument applied on the Reals create another real.

Thank you


问:
Linear diophantine equation with 3 unknowns
I have the following linear diophantine equation with $3$ unknowns:

$$10x+3y+{z \over 2}=100$\qquad\text{ with }\qquad x,y,z \ge 1.$$

I only know how to solve with 2 unknowns, how do I approach with 3 unknowns?

答:
If we want positive integer solutions, we can see
$$10x+3y+{z \over 2}=100\implies 200-z = 20 x + 6 y\\
\implies
13 \le (10x+3y) \le 99\\
\quad\implies\quad x\le\bigg\lfloor\frac{99-3}{10}\bigg\rfloor=9\\
\quad\implies\quad y\le\bigg\lfloor\frac{99-10}{3}\bigg\rfloor=29
\\ \land\quad
200-(20+6)=174\implies z\le 174$$

Now cycling through two nested loops of $x,y$ with the limits above, integer values of $z$ reveal all the solutions.

$$z = 200 -20 x - 6 y$$

Heuristic shows, there are solutions whenever
$$x\in\{1,2,3,4,5,6,7,8,9\}\quad\land\quad y\in\big\{1,2,3,\cdots,
\big\lfloor(10-x)\log_2 10\big\rfloor\big\}$$

where $\quad\big\lfloor(10-x)\log_2 10\big\rfloor\quad$ yeilds $\qquad Set_{y_{max}}=\{29,26,23,19,16,13,9,6,3\}$


问:
The splitting field of polynomial over $Z/2Z$
I am trying to figure out the splitting field of $x^4 ＋ x^3 + 1$ over $\mathbb{F}_2$ .

I know $x^4 ＋ x^3 + 1$ over $\mathbb{F}_２$ is irreducible.

Let $α$ be a root of $x^4 ＋ x^3 + 1$, then the spitting field will be the field $\mathbb{F}_2(\alpha)$. So, the spitting field is isomorphic to finite field of $16$ elements.

My question is:

How to formally prove that the spitting field is the field
$\mathbb{F}_2(\alpha)$

Let the other roots of $x^4＋x^3＋1$ be $\beta,\gamma,\delta$, I want to show $\mathbb{F}_2(\alpha,\beta,\gamma,\delta)=$$\mathbb{F}_2(\alpha)$.

Why are $α^2, α^4,α^8$ also roots of $x^4＋x^3＋1$? Thank you for your help.

After reading a lot of good comments,
could you someone write in answer form, the proof with normality and not using normality? Thank you.

答:
For $a\in \Bbb{F}_{p^k}$ (some field with $p^k$ elements) let $$\prod_{m=0}^{k-1} (x-a^{p^m})=\sum_{n=0}^k c_m x^k$$
Because $b\to b^p$ is an automorphism and $a^{p^k}=a$ we get that $$\sum_{n=0}^k c_m^p x^k=\prod_{m=0}^{k-1} (x-(a^{p^m})^p)=\prod_{m=0}^{k-1} (x-a^{p^m})=\sum_{n=0}^k c_m x^k$$
Whence $c_m=c_m^p$ which implies that $c_m\in \Bbb{F}_p$ so that the minimal polynomial $f$ of $a$ divides $\sum_{n=0}^k c_m x^k$ which means that $f$ splits completely in $\Bbb{F}_p(a)$.


问:
Inequalities involving polynomials
Let $A_k=\{x \mid  2^k < x^k +x^{k+1} < 2^{k+1}\},\, k=1, 2, 3, \ldots$.

Find the largest positive integer $n$ such that
$$
A_1\cap A_2\cap\cdots \cap A_n \ne \emptyset.
$$

Title: Inequalities involving polynomials

Body:
$A_1\cap A_2\cap A_3\neq \emptyset$, but $A_1\cap A_4=\emptyset$.  I have no analytic proof, just a calculation on wolframalpha:

$A_4\approx(1.58,1.83)$, while $A_1\approx (-2.56,-2)\cup (1,1.56)$

Also $A_2\approx (1.31, 1.72)$ and $A_3\approx(-2.31,-2)\cup(1.48,1.79)$, so $A_1\cap A_2\cap A_3\approx(1.48,1.56)$

Proof that $A_1\cap A_2\cap A_3\neq \emptyset$:  Take $x=1.5$. $1.5+1.5^2=3.75$ and $2<3.75<4$.  $1.5^2+1.5^3=5.625$ and $4<5.625<8$.  $1.5^3+1.5^4=8.4375$ and $8<8.4375<16$.

Proof that $A_1\cap A_4=\emptyset$: Take $x=1.57$.
 On the interval $(0,+\infty)$, $f(x)=x+x^2$ is increasing and $f(1.57)=4.0349>4$.  But also $g(x)=x^4+x^5$ is increasing on the same interval, and $g(1.57)=15.6146312657<16$.  Hence $1.57$ is strictly above $A_1$ and strictly below $A_4$.


问:
Is it possible to evaluate the integral $I=\int_0^\infty \frac{\sqrt{x}\arctan(x)}{1+x^2}dx$ with residue theorem?
let's consider the following integral

$$I=\int_{0}^{+\infty}{\frac{\sqrt{x}\arctan{x}}{1+x^{2}}\,\mathrm{d}x}$$
I know exactly how to evaluate it with sneaky methods and obtained $I=\frac{\pi\sqrt{2}}{8}(\pi+2\ln{\left(2\right)})$, but would it be possible to evaluate it with contour integration (Residue theorem)? Of course I have done some simple contour integrations before for example a half circle to evaluate $J=\int_{0}^{+\infty}{\frac{\mathrm{d}x}{1+x^{2}}}$, but for some reason I cannot come up with ideas for this.

Thanks in advance.

In case it is possible, I would like to get hints.

答:
Notice that : \begin{aligned}I=2\int_{0}^{+\infty}{\frac{x\arctan{x}}{2\sqrt{x}\left(1+x^{2}\right)}\,\mathrm{d}x}&=2\int_{0}^{+\infty}{\frac{x^{2}\arctan{\left(x^{2}\right)}}{1+x^{4}}\,\mathrm{d}x}\\ &=\int_{-\infty}^{+\infty}{\frac{x^{2}\arctan{\left(x^{2}\right)}}{1+x^{4}}\,\mathrm{d}x}\\ I&=\int_{0}^{1}{\int_{-\infty}^{+\infty}{\frac{x^{4}}{\left(1+x^{4}\right)\left(1+x^{4}y^{2}\right)}\,\mathrm{d}x}\,\mathrm{d}y}\end{aligned}

Now we can apply the residue theorem to $ f_{y}:z\mapsto\frac{z^{4}}{\left(1+z^{4}\right)\left(1+z^{4}y^{2}\right)} $, for $y\in\left(0,1\right] $, on the countour $ \mathscr{C}_{R}=\left[-R,R\right]\cup\Gamma_{R} $, where $ R> \frac{1}{y} $ :

We know : \begin{aligned} \int_{\mathscr{C}_{R}}{f_{y}\left(z\right)\mathrm{d}z}&=2\pi\,\mathrm{i}\left(\mathrm{Res}\left(f_{y},\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}\right)+\mathrm{Res}\left(f_{y},-\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}\right)+\mathrm{Res}\left(f_{y},\frac{\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}}{\sqrt{y}}\right)+\mathrm{Res}\left(f_{y},-\frac{\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}}{\sqrt{y}}\right)\right)\\ &=2\pi\,\mathrm{i}\left(\frac{\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}}{4\left(1-y^{2}\right)}-\frac{\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}}{4\left(1-y^{2}\right)}-\frac{\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}}{4\sqrt{y}\left(1-y^{2}\right)}+\frac{\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}}{4\sqrt{y}\left(1-y^{2}\right)}\right)\\ \int_{-R}^{R}{f_{y}\left(x\right)\mathrm{d}x}+\int_{\Gamma_{R}}{f\left(z\right)\mathrm{d}z}&=\frac{\pi\sqrt{2}}{2\sqrt{y}\left(1-y^{2}\right)}-\frac{\pi\sqrt{2}}{2\left(1-y^{2}\right)} \end{aligned}

We have : $$ \left|\int_{\Gamma_{R}}{f_{y}\left(z\right)\mathrm{d}z}\right|\leq\int_{\Gamma_{R}}{\left|f\left(z\right)\right|\left|\mathrm{d}z\right|}\leq\frac{R^{2}}{\left(R^{4}-1\right)\left(R^{4}y^{2}-1\right)}\int_{\Gamma_{R}}{\left|\mathrm{d}z\right|}=\frac{R^{4}\pi}{\left(R^{4}-1\right)\left(R^{4}y^{2}-1\right)}\underset{R\to +\infty}{\longrightarrow}0 $$

Thus, taking $ R $ to $ +\infty $, we get : $$ \int_{-\infty}^{+\infty}{f_{y}\left(x\right)\mathrm{d}x}=\frac{\pi\sqrt{2}}{2\sqrt{y}\left(1+\sqrt{y}\right)\left(1+y\right)} $$

That means : \begin{aligned} I=\pi\sqrt{2}\int_{0}^{1}{\frac{\mathrm{d}y}{2\sqrt{y}\left(1+\sqrt{y}\right)\left(1+y\right)}}&=\pi\sqrt{2}\int_{0}^{1}{\frac{\mathrm{d}y}{\left(1+y\right)\left(1+y^{2}\right)}}\\ &=\frac{\pi\sqrt{2}}{2}\int_{0}^{1}{\left(\frac{1}{1+y}+\frac{1-y}{1+y^{2}}\right)\mathrm{d}y}\\ &=\frac{\pi\sqrt{2}}{2}\left[\ln{\left(1+y\right)}+\arctan{y}-\frac{\ln{\left(1+y^{2}\right)}}{2}\right]_{0}^{1}\\ I&=\frac{\pi\sqrt{2}}{8}\left(\pi+\ln{4}\right) \end{aligned}


问:
What is $\int_0^1 \frac{(1-x^2)xm^2}{m^2x^2+\lambda^2(1-x)} \, dx$?
What is $$\int_0^1 \frac{(1-x^2)xm^2}{m^2x^2+\lambda^2(1-x)} \, dx \text{?}$$ I tried to compute it using Mathematica but it couldn't solve the integral. In fact, I am only interested in the part that diverges for $\lambda \rightarrow 0$, i.e., in terms up to $\mathcal{O(\lambda^0)}$.

答:
Let $\lambda = \gamma m,$ in order to remove the $m^2$s. $$ I(\gamma) := \int_0^1 \frac{(1-x^2) x}{x^2 + \gamma^2(1-x)}\,\mathrm{d}x \\ = \int_0^1 \frac{x -x^3}{(x - \gamma^2/2)^2 + (\gamma^2 - \gamma^4/4)}\,\mathrm{d}x.$$

Since we're interested in the behaviour for small $\gamma$, we may assume that $\gamma^2 > \gamma^4/4$. Let $\gamma^2 - \gamma^4/4 = \alpha^2$, and $x - \gamma^2/2 = u$. Also, let $\gamma^2/2 = \beta$ to ease notation. We have

$$\int_{-\beta}^{1-\beta} \frac{u + \beta - (u+\beta)^3}{u^2 + \alpha^2}\\ = \int_{-\beta}^{1-\beta} -u - 2\beta + \frac{(1-3\beta^2 - \alpha^2) u + \beta(1-3\alpha - \beta^2)}{u^2 + \alpha^2}$$

It is possible but tedious to exactly calculate the above. But we can see the behaviour as $\gamma \to 0$ more directly - notice that $\alpha = \gamma(1 + O(\gamma))$ and $\beta = \gamma^2/2$. Thus $\beta/\alpha = \gamma/2(1 + O(\gamma))$

Now, the first two integrals are easily seen to be $\int_{-\beta}^{1-\beta} - u = -\frac12 + O(\gamma)$, $\int - 2\beta = -2\beta = O(\gamma)$. Next note that $$ \int_{-\beta}^{1-\beta} \frac{\beta}{u^2 + \alpha^2} = \frac{\beta}{\alpha} \left(\arctan \frac{1- \beta}{\alpha} + \arctan \frac{\beta}{\alpha}\right).$$ Since $\arctan$ is bounded, and $\beta/\alpha = O(\gamma)$, this whole thing is $O(\gamma)$. This leaves us with $$ I(\gamma) = -\frac12 + O(\gamma) + \frac{(1-3\beta^2 - \alpha^2)}{2} \log \frac{(1-\beta)^2 + \alpha^2}{\beta^2 + \alpha^2}.$$ It's a matter of computation that $(1-\beta)^2 + \alpha^2 = 1$ and $\beta^2 + \alpha^2 = \gamma^2,$ giving us $$ I(\gamma) = \log(1/\gamma) - \frac{1}{2} + O(\gamma),$$ or, setting $\gamma = \lambda/m,$ giving us that the original integral behaves as $\log(1/\lambda) + \log(m) - \frac12 + O(\lambda)$.


问:
Why does $\int_0^1 \frac {dx}{x^{0.5}(1-x)^2}$ diverge?
I solved this question and was sure about my answer that it converges but the answer was it diverges, with no explanation,
 I still don't get what's my mistake, here's what I did:

I wanted to use the limit comparison test, $\displaystyle \frac {1}{x^{0.5}(1-x)^2}=\frac {1}{x^{0.5}(1-2x+x^2)}=\frac {1}{x^{0.5}-2x^{1.5}+x^{2.5}}$, at $x\to 0$ , this function behaves like $\displaystyle \frac {1}{x^{0.5}}$.
 $\displaystyle \lim_{x\to0} \frac {x^{0.5}}{x^{0.5}-2x^{1.5}+x^{2.5}}$ then I divided by $x^{0.5}$ and got $\displaystyle \lim_{x\to0} \frac {1}{1-2x^{}+x^2}=1$.

And the integral $\displaystyle \int_0^1 \frac{dx}{x^{0.5}}$ converges, then my integral should converge too.

Now I'm not sure where is my mistake, I would appreciate any help.

Thanks in advance.

答:
You are right about $x=0$. It's what happens at $x=1$ that makes your integral diverge.

To be more precise: your integral converges if and only if both integrals$$\int_0^{1/2}\frac1{\sqrt x(1-x)^2}\,\mathrm dx\quad\text{and}\quad\int_{1/2}^1\frac1{\sqrt x(1-x)^2}\,\mathrm dx$$converge. Your computations show indeed that the first one converges. But the second one diverges.


问:
Confusing question on $\sigma$ algebra
If $A \subset \sigma(B)$, does it mean $A \subset B$ ? Here $\sigma(B)$ implies the minimal $\sigma$ field generated by $B$.

答: Take $B = \{\emptyset\}$ and $A=\{X\}$. Now $\sigma(B) = \{\emptyset, X\}$.

Title: Confusing question on $\sigma$ algebra


答:
Cantor's diagonal argument on a given countable list of reals does produce a new real (which might be rational) that is not on that list.

The point of Cantor's diagonal argument, when used to prove that $\mathbb{R}$ is uncountable, is to choose the input list to be all the rationals. Then, since we know Cantor produces a new real that is not on that input list, that real can't be rational.

Title: Cantor's Diagonal Argument - Rational


问:
Prove if the set is closed? Bounded? Compact?
Consider the metric space $(\mathbb{Q},d)$ where $\mathbb{Q}$ denotes the rational numbers and $d(x,y)=|x-y|$.  Let $$E:=\{x \in\mathbb{Q}:x>0, 2<x^2<3\}$$

Is $E$ closed and bounded in $\mathbb{Q}?$  Is it compact?  Justify your answers.

答: It is not closed. You can always find a sequence of $x \in E$ such that converge to a real number.</p>


问:
Let $z_1$, $z_2$ and $z_3$ be complex vertices of an equilateral triangle. Show $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$.
Prompt:

Let $z_1$, $z_2$ and $z_3$ represent vertices of an equilateral triangle in the complex plane. Show $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$.

Question:

I hope the following question isn't too soft. I know a solution to the above question by taking the sides of the triangle and equating them by rotations in the complex plane of $e^{i\pi/3}$.

What, if anything, is the significance of the result other than showing some clever algebraic manipulation? Does it tell us something unique about equilateral triangles defined by vertices in the complex plane?

Is there any intuition to be gained here?

答:
@Milo Brandt is completely correct.

Now, there aren't any things left to know about the equilateral triangles, but there are a lot of amazing things with the complex numbers.

I am talking about the relationships of complex numbers and the coordinate geometry...

With the help of complex numbers, it becomes very easy for us to calculate or prove things like this. Imagine proving the same thing with the coordinates/coordinate geometry. It makes our work easier.

Now, since you want to know about the equilateral triangles, these properties,or I should say, these results may help you:

let the circum centre of the triangle be $z_0$ , then, $3z^2_0 = z^2_1 + z^2_2 + z^2_3$

The angle between the points must be $60°$ or $π/3^c$(most obvious). Thus, $arg((z_3 - z_1)/(z_2 - z_1)) = π/3^c$. And similar for all other angles.

Best ever property for equilateral triangle:

The triangle with vertices as $z_1,z_2$ and $z_3$ can only be equilateral if and only if:
$1/(z_1 - z_2) + 1/(z_2 - z_3) + 1/( z_3 - z_1) =0$.

I am going to prove the last point for your convenience...

Let,

$a = z_1 - z_2$

$ b = z_2 - z_3$

$ c = z_3 - z_1$

Thus,

$a + b + c = 0$

Also,

$1/a + 1/b + 1/c = 0$. [ GIVEN IN QUESTION]

Thus,

$1/a + (b+c)/bc = 0$

$1/a - a/bc = 0$ [ as $a + b + c =0$]

Thus,

$a^2 = bc$

$|a^3| = |abc|$

Similarly,

$|b^3| = |abc|$ and $|c^3| = |abc|$

Now, add these equations

$|a^3| + |b^3| + |c^3| = 3|abc|$

Which is only possible when $|a| + |b| + |c| = 0$ or $ |a| = |b| = |c| $

But since $|a|,|b|,|c|>0$
,

Thus $|a| = |b| = |c|$.

Thus the triangle is an equilateral triangle!

Hence, Proved!