问:
Demonstrate $\mathrm{Spin}(3, 3) = \mathrm{SL}(4, R)$
It seems that there is an accidental isomorphism
$$
\mathrm{Spin}(3, 3) = \mathrm{SL}(4, R).
$$

Some facts I am aware is that

$\mathrm{SL}(4, R)$ has the Lie algebra of 15 generators.

the homotopy group of $\pi_1(\mathrm{SL}(4, R))=\pi_1(\mathrm{SO}(4))=\mathbf{Z}/2$.

How to show or convince the accidental isomorphism
$
\mathrm{Spin}(3, 3) = \mathrm{SL}(4, R)?$

答:
$SL_4(\mathbb{R})$ has a defining $4$-dimensional representation on $V = \mathbb{R}^4$. Its exterior square is a $6$-dimensional representation $\wedge^2(V) = \wedge^2(\mathbb{R}^4)$, and the action of $SL_4(\mathbb{R})$ on this representation preserves the wedge product $\wedge^2(V) \otimes \wedge^2(V) \to \wedge^4(V) \cong \mathbb{R}$ (the last isomorphism is an isomorphism of $SL_4(\mathbb{R})$-representations), which is a nondegenerate symmetric bilinear form in this degree.

Now we need to compute the signature of the wedge product. If $e_1, \dots e_4$ is the standard basis of $\mathbb{R}^4$ then $e_i \wedge e_j, i < j$ is a basis of $\wedge^2(\mathbb{R}^4)$. The wedge product acting on this basis has the property that $(e_i \wedge e_j) \wedge (e_k \wedge e_{\ell}) = 0$ unless $i, j, k, \ell$ is a permutation of $1, 2, 3, 4$, in which case it is the sign of this permutation times the volume form $e_1 \wedge e_2 \wedge e_3 \wedge e_4$. It follows that as a symmetric bilinear form the wedge product is the direct sum of $3$ copies of the "hyperbolic" form, which has signature $(1, 1)$, so altogether it has signature $(3, 3)$. This gives a map

$$SL_4(\mathbb{R}) \to SO(3, 3).$$

The action of $SL_4(\mathbb{R})$ is faithful so the induced map on Lie algbras is injective, and since the two have the same dimension ($15$) the induced map on Lie algebras is an isomorphism, so this map is a covering map. We have $\pi_1(SO(3, 3)) \cong \mathbb{Z}_2 \times \mathbb{Z}_2$, and in fact the block-diagonal inclusion of the connected component of the maximal compact $SO(3) \times SO(3)$ is an isomorphism on $\pi_1$. Now it suffices to show that the image of $\pi_1(SL_4(\mathbb{R}))$ in $\pi_1(SO(3, 3))$ is the diagonal copy of $\mathbb{Z}_2$, and it will then follow that the map above lifts to $\text{Spin}(3, 3)$. The inclusion of $SO(4)$ is an isomorphism on $\pi_1$, so in turn it suffices to show that the induced map

$$SO(4) \to SO(3) \times SO(3)$$

exhibits $\pi_1(SO(4)) \cong \mathbb{Z}_2$ as the diagonal copy of $\mathbb{Z}_2$ in $\pi_1(SO(3) \times SO(3))$. I can't actually think of a really slick way to do this off the top of my head but it ought to be doable by a direct if unenlightening computation, or by a consideration of the other two double covers of $SO(3) \times SO(3)$, which are $SU(2) \times SO(3)$ and $SO(3) \times SU(2)$. Maybe someone else has a clever idea here.


问:
Show that exists $x \in \mathbb{R} $ such that $x^3+x=3$
Show that exists $x \in \mathbb{R} $ such that $x^3+x=3$

I did:

I take $S=\{x\in \mathbb{R} \mid x^3+x<3\}$

So, this set is bounded and by completeness of the real numbers exists $a=\sup(S)$

Then I want to prove that $a^3+a=3$

And then in order to get a contradiction I suppose that $a^3+a>3$

and how $a $ is the supremum then $\forall\epsilon\in\mathbb{R} $ exists $b \in S$ such that
$a-\epsilon<b$ then $(a-\epsilon)^3+b<b^3+b<3$

But I'm stuck from here. I'm not able to use derivatives, or continuity so I can't use IVT

答:
$S = \{x| x^3 + x < 3\}$ is one set but $T= \{x|x^3 + x < 3\}$ is another.

If $s = \sup S$ and $t = \inf T$ it's easy to prove $\sup S \le \sup T$.

Now if you can prove $s^3 + s \ge 3$ then you can probably use a similar argument to prove $t^3 + t \le 3$.

That means $t^3 + t \le s^3 + s$.  It's easy to prove $x^3 + x$ is increasing so $t^3 + t \le s^3 + s\iff t\le s$.

so you have $t \le s$ and $s \le t$ and $t^3 +t \le 3 \le s^3 +s$ which altogether shows $s=t=3$.

so what argument can we use to prove that $s^3 + s \ge 3$?

Consider that if $u > s$ then $u \not \in S$ so $u^3 + u \ge 3$. If we assume $s^3 +s < 3$, the would mean for all $\epsilon > 0$ then $(s+\epsilon)^3 + (s+\epsilon) \ge 3$

If we can show there is an $\epsilon > 0$ so that $(s+\epsilon)^3 + (s + \epsilon) < 3$ that would be a contradiction and that would prove $s =\sup S \ge 3$.

Let's find such an epsilon we need

$(s+\epsilon)^3 + (s + \epsilon) < 3$

$s^3 + 3s^2\epsilon + 3s\epsilon^2 + \epsilon^3+s + \epsilon <3$.

$(3s^2+1)\epsilon + 3s\epsilon^2 + \epsilon^3 < 3 - (s^3 + s)$

Now we know if $s \ge 2$ then $s^3 + s \ge 10$ so $s< 2$ and if $\epsilon < 1$ we have:

$(3s^2+1)\epsilon + 3s\epsilon^2 + \epsilon^3 <13\epsilon + 6\epsilon + \epsilon =20 \epsilon$

So if we can find and $\epsilon < 1$ so that $20\epsilon < 3-(s^3 + s)$ we'd be done.

So as we are assuming $3-(x^3 + s) > 0$ we can just let $\epsilon < \min(1, \frac {3-(x^3 + s)}{20}$.

And we are done.  That's a a contradiction so $s^3 + s \ge 3$.

ContentLicense: CC BY-SA 4.0


问:
To find $R\circ R^{-1}$ in Discrete mathematics
Today I came across a question in DMS which says:

If $R$ is the relation “Less Than” from $A = \{1, 2, 3, 4\}$ to $B = \{1,3,5\}$ then find $R\circ R^{-1}$.

Now what is $R\circ R^{-1}$?

I know how to find $R\circ R$,
like in this question firstly we will find $R$ where

$$R= \{(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)\}$$

and then $R\circ R$ would be:

$$\{(1,5),(2,5),(3,5),(4,5)\}.$$

Please correct me if I'm wrong.
And also please explain how to find $R\circ  R^{- 1}$?
Thanks in advance :-)

答:
I wanna introduce the domain and the range of $R$, which are defined in Elements of Set Theory by Herbert B. Enderton:

\begin{align}
x \in \mathrm{dom} R &\Leftrightarrow \exists y :\: \left( x,\, y \right) \in R,\\
x \in \mathrm{ran} R &\Leftrightarrow \exists t :\: \left( t,\, y \right) \in R.
\end{align}

It'll help you think.

Because $ R = \left\{ \left(1,\,3\right),\;\left(1,\,5\right),\;\left(2,\,3\right),\;\left(2,\,5\right),\;\left(3,\,5\right),\;\left(4,\,5\right) \right\} $.

We can get $\mathrm{dom} R = A = \left\{1,\,2,\,3,\,4\right\}$, and $\mathrm{ran} R = \left\{3,\,5\right\}$.

To get $R \circ R$, evaluate $\mathrm{ran}R \cap \mathrm{dom}R = \left\{ 3,\, 5 \right\} \cap \left\{ 1,\, 2,\, 3,\, 4 \right\} = \left\{ 3 \right\}$, use the elements of it in the second $R$. It's obvious that the only possible routes are $1 \rightarrow 3 \rightarrow 5$ and $2 \rightarrow 3 \rightarrow 5$, i.e. $R \circ R = \left\{ \left(1,5 \right),\, \left( 2,\, 5 \right) \right\}$.

By definition, $R^{-1} = \left\{ (y,\,x) \mid xRy \right\}$. So $R^{-1} = \left\{ \left(3,\,1\right),\;\left(5,\,1\right),\;\left(3,\,2\right),\;\left(5,\,2\right),\;\left(5,\,3\right),\;\left(5,\,4\right) \right\} $.
$\mathrm{dom}R^{-1} = \mathrm{ran}R = \left\{ 3,\,5 \right\}$.

Similarly, to get $R \circ R^{-1}$, evaluate $\mathrm{ran}R \cap \mathrm{dom}R^{-1} = \left\{ 3,\,5 \right\}$. Therefore we can get $R \circ R^{-1} = \left\{ \left(1,\,1\right),\;\left(2,\,2\right),\;\left(3,\,3\right),\;\left(4,\,4\right),\;\left(1,\,2\right),\;\left(2,\,1\right),\;\left(2,\,3\right),\;\left(3,\,2\right),\;\left(3,\,4\right),\;\left(4,\,3\right),\;\left(1,\,3\right),\;\left(3,\,1\right),\;\left(2,\,4\right),\;\left(4,\,2\right),\;\left(1,\,4\right),\;\left(4,\,1\right) \right\}$.

ContentLicense: CC BY-SA 3.0


问:
How to simplyfy the expression ${(m+k-1 \text{ mod } k) + 1}$?
$m$ takes values from 1 to an even number $M$ and $k = M/2$. This expression ${(m+k-1 \text{ mod } k) + 1}$ gives the repeated sequence of numbers from 1 to k. How can I further simply the expression to give the same repeated sequence?

For example M = 24 the expression gives the sequence

1     2     3     4     5     6     7     8     9    10    11    12     1     2     3     4     5     6     7     8     9    10    11    12

答: Obviously $k\equiv0\bmod k$, so in the end we get $(m-1)\bmod k+1$ (assuming the computing definition of $\bmod$).


问:
Maximum modulus of a complex rational function $r(z)$ on a unit circle
If $r(z) =\frac{p(z)}{\prod_{j=1}^n(z-z_j)}$ where all coefficients of the polynomial $p(z)$ are non negative and real and all $z_j$ are real and $z_j> 1$ for all $j$. Then why is the maximum modulus of $r(z)$ on the unit circle obtained at $z=1$?

答:
The numerator achieves its maximum at $1$ since its coefficients are non-negative. In fact, for any $z$ on the circle, by the triangle inequality, if $p(z)=a_nz^n+...+a_0,$
$$|p(z)|\le |a_n||z|^n+...+|a_0|=a_n+...+a_0=p(1).$$

The norm of the denominator is the product of $|z-z_j|$, which achieves minimum at $1$ (shortest distance). Thus the norm of the whole function achieves its maximum at $1.$


问:
Show that $8 \mid (a^2-b^2)$ for $a$ and $b$ both odd

If $a,b \in \mathbb{Z}$ and odd, show $8 \mid (a^2-b^2)$.

Let $a=2k+1$ and $b=2j+1$. I tried to get $8\mid (a^2-b^2)$ in to some equivalent form involving congruences and I started with
$$a^2\equiv b^2 \mod{8} \Rightarrow 4k^2+4k \equiv 4j^2+4j \mod{8}$$
$$\Rightarrow k^2+k-j^2-j=2m$$
for some $m \in \mathbb{Z}$ but I am not sure this is heading anywhere that I can tell.

Second attempt: Use Euler's Theorem and as $\gcd(a,8)=\gcd(b,8)=1$ and $\phi(8)=4$, $a^4 \equiv b^4 \equiv 1 \mod 8$ so $a^4-b^4\equiv 0 \mod{8}$.

I haven't gotten too much further are there any hints?

答:
HINT:

$k^2-j^2+k-j=(k-j)(k+j+1)$

As $(k+j+1)-(k-j)=2j+1$ which is odd, they must be of opposite parity, exactly one of them must be divisible by $2$

Method 2:

If $a,b$ are odd, observe that one of $(a-b),(a+b)$ is divisible by $4,$ the other by $2$

Method 3:

$(2a+1)^2=4a^2+4a+1=8\frac{a(a+1)}2+1\equiv 1\pmod 8$


问:
Gradient of a function with matrix variables $V(Q) = \|Q-Q_N\|^2_F\big[-\ln(-\frac{\langle D-dI,Q \rangle}{2})\big]$
I followed the following discussion:  Gradient and Hessian of a function with Matrix Variables

$V(Q) = \|Q-Q_N\|^2_F\big[-\ln(-\frac{\langle D-dI,Q \rangle}{2})\big]$

I want to calculate $\nabla_Q V(Q)$, where $Q_N$, $D$ are constant square matrix. $Q$ is a square matrix variable.

So I did the following

Find $$\nabla_Q \|Q-Q_N\|_F^2=\nabla_Q\text{tr}[(Q-Q_N)^T(Q-Q_N)]=\nabla_Q\langle Q,Q \rangle-2\nabla_Q\langle Q,Q_N \rangle+\nabla_Q\langle Q_N,Q_N \rangle=2(Q-Q_N)$$

Find $$\nabla_Q -\ln\bigg(-\frac{\langle D-dI,Q \rangle}{2}\bigg)=-\frac{D-dI}{\langle D-dI,Q\rangle}$$

So $$\nabla_Q V(Q)=2(Q-Q_N)\bigg[-\ln\bigg(-\frac{\langle D-dI,Q \rangle}{2}\bigg)\bigg]+ \|Q-Q_N\|_F^2\bigg[ -\frac{D-dI}{\langle D-dI,Q\rangle}\bigg]$$

Am I correct?  Please let me know your suggestion, thanks!

答:
Looks right to me. The result actually holds in any real Hilbert space; if we set

$$ V_1(v):= \|v\|^2, \quad V_2(v) := v-v_0,\quad  V_2(v):=-\ln\langle c,v\rangle$$
then
\begin{align}
dV_1(v)h&=2\langle v,h\rangle
\\
dV_2(v)h &= h
\\
dV_3(v)h &= \frac{-1}{ \langle c,v\rangle}\langle c,h\rangle
\end{align}
And by chain and product rule, the derivative of $V(v) := (V_1\circ V_2)(v)V_3(v)=-\|v-v_0\|^2\ln \langle c,v\rangle $ is
$$ dV(v)h = -2\langle v-v_0,h\rangle\ln \langle c,v\rangle  - \|v-v_0\|^2\frac{\langle c,h\rangle}{\langle c,v\rangle} $$
So the gradient vector (i.e. the vector $u=\nabla V(v)$ such that $\langle u,h\rangle = dV(v) h)$  is
$$\nabla V(v) = -2\ln\langle c,v\rangle (v-v_0) -  \|v-v_0\|^2\frac{ c}{\langle c,v\rangle} $$


问:
Show that $G$ has an element of order $p$.

Let $G$ be a group of order $p^n$, for some prime $p$ and positive integer $n$. Show that $G$ has an element of order $p$.

I know this question has been asked before, but I have not found a complete answer and I am quite confused.

I have tried to solve it and so far this is what I have achieved:

If $|G|=p^n$ and $e\neq a\in G$, we have that order of $a$ divides $p^n$, then $|a|\in \{p^m:1\leq m \leq n\}$. Also, if order of $G$ is $p^n$, for all $a\in G$, $a^{p^{n}}=e$. Then write $p^n=p^{m}p^s$ , $1\leq s\leq n$, we have that $(a^{p^{m-1}p^s})^p=e$. Then we can conclude that $b=a^{p^{m-1}p^s}\in G$ is an element of orden $p$.

Is what I did correct?

答:
According to the_fox hint.

If $G$ is cyclic then it must have an element of order $p$.

So let $G$ be not cyclic and $a\in G$. Let $H=\left \langle a  \right \rangle$ then $|H|$ divides $|G|$ and so that it is of the form $p^m$ where $m<n$.Then, as $H$ is cyclic hence it must have an element of order $p$ and this element is then required element of order $p$ in $G$


问:
Why is spatial tensor called spatial?
I am reading Blackadar's book on operator algebras, and came upon the notion of spatial tensor product of $C^*$-algebras. It turns out to be the same as the minimal tensor product. To me, the term ''minimal'' makes sense. But why is it called ''spatial'' in the first place? What is the ''space'' being alluded to?

答: When you represent your C*-algebras faithfully on Hilbert spaces, say $A \subseteq B(H), B \subseteq B(K)$, the spatial/minimal tensor product is just the norm closure of algebraic tensor product $A \odot B \subseteq B(H \otimes K)$. In my mind, this is why its called spatial - it is defined as some algebra of operators on the underlying Hilbert space $H \otimes K$. It just so happens that the resulting C*-algebra is independent of the choices of faithful representations and so you have "the" spatial tensor product or "the" minimal tensor product.

Title: Why is spatial tensor called spatial?


问:
Question regarding inward pointing vector fields help
Let $M$ be a n manifold with boundary and $p\in \partial{M}$.

$v\in T_pM\backslash T_p\partial{M}$ is inward pointing if there exists some $\epsilon>0$ and a smooth curve $c:[0,\epsilon)\rightarrow M$ such that $c(0)=p$ and $c((0,\epsilon))\subseteq intM$ and $c'(0)=v$.

Problem:

Let $M$ be a smooth n manifold and $p\in \partial{M}$. Let $(U,x^1,.....,x^n)$ be a chart about $p$. $v=\sum_{i=1}^nv^i\frac{\partial}{\partial{x^i}} \in T_pM$ is inward pointing if and only if $v^n>0$.

My attempt:

Suppose $v\in T_pM$ is inward pointing. This means $v\notin T_p\partial{M}$ and there exists some $\epsilon>0$ and a smooth curve $c:[0,\epsilon)\rightarrow M$ such that $c(0)=p,$ $c((0,\epsilon))\subseteq intM$ and $c'(0)=v$. By definition of $\mathbb{H}^n$, we have that on $U$, $x^n\geq 0$. Also, it is easy to show that $v^n=\frac{d}{dt}|_{t=0}(x^n\circ c)$. Also note that we may assume $x^n>0$; for otherwise, $v\in T_p\partial{M}$.

However, I'm not sure what else to say. If $(U,\phi)$ was centered at $p$, then I can prove the result, but in this case, the chart may not necessarily be centered at p.,

答: Note, $(U,\phi)=(U,x^1,....,x^n)$ is a chart in $M$. Thus $\phi:U\rightarrow \phi(U)$ is a diffeomorphism. Hence $\phi(p)=\phi(c(0))\in \partial{\mathbb{H}}^n$. So $x^n(c(0))=0$. Thus $v^n=\frac{d}{dt}|_{t=0}(x^n\circ c)=\lim_{t\rightarrow 0^{+}}\frac{x^n(c(t))}{t}\geq 0$.  We cannot have the latter expression equal to $0$, for as you said, this would imply that $v\in T_p(\partial{M})$. Therefore we must have $v^n>0$. Conversely, if $v^n>0$, define for sufficiently small $\epsilon>0$, $\gamma: (-\epsilon,\epsilon)\rightarrow M$ by $\gamma(t)=\phi^{-1}(v^1t,.....,v^nt)$. We must show that $\gamma((0,\epsilon))\subseteq intM$. Indeed, for $t'\in (0,\epsilon)$, $\gamma(t')\in U$. Thus, $(v^1t,.......,v^nt)\in \phi(U)$ , where $v^nt>0$, and so $\gamma(t')\in intM$. $\gamma$ is the curve we seek. $\square$


问:
Ultrafilters on naturals determine p-adic numbers
While preparing for a short lecture on ultrafilters for undergraduates, I realized some interesting things I have never read about.

Though I'm asking now a specific question, any reference about this idea is welcome.

Let $\beta(X)$ denote the set of ultrafilters on $X$.

By the principal filters we obtain an embedding $X\hookrightarrow \beta(X)$, so ultrafilters can be regarded as 'generalized elements' joined to our set $X$, as this is often done in topology.

The main observation is, that unary predicates can be perfectly evaluated to ultrafilters themselves (without referring to any ultraproduct or such), as the interpretation of unary predicates are just the subsets of $X$.

For example, every nonprincipal ultrafilter $U$ on $\Bbb N$ satisfies $x>n$ for every fixed $n\in\Bbb N$, by which I simply mean $$\{x\in\Bbb N:x>n\}\ \in\  U\,.$$

Specifically, every ultrafilter $U\in\beta(\Bbb N)$ has a last digit in any number system $n$, in the sense that exactly one of the below sets will belong to $U$: $\ n\Bbb N,\ n\Bbb N+1,\ \dots,\ n\Bbb N+(n-1)$, since their disjoint union is the full set $\Bbb N$.

But this gives rise to a well defined function $\Phi:\beta(\Bbb N)\to\hat{\Bbb Z}=\prod_p\Bbb Z_p$ where $\Bbb Z_p$ denotes the set of $p$-adic integers.

Question: Is $\Phi$ surjective? In particular, does there exist an ultrafilter for every $p$-adic integer?

Attempt. Assume $a\in\Bbb Z_p$ is given. I'd like to define a filter and then extend it to ultrafilter.

$$\mathcal F:=\big\{A\subseteq\Bbb N \mid \forall n\,\exists x\in A: x\equiv a\!\!\!\pmod{p^n}\big\}$$
but the problem is that it's not a filter, though nonempty and increasing system, it's not closed under intersection, because if $a_n$ denotes the last $n$ digits of $a$ modulo $p$, then we can take $A=\{a_n\}\in\mathcal F$ and the disjoint $B=\{p^n+a_n\}\in\mathcal F$.

答:
Suppose that $a=\langle a_n:n\in\Bbb Z^+\rangle\in\prod_{n\in\Bbb Z^+}\Bbb Z/p^n\Bbb Z$ is in $\Bbb Z_p$, i.e., is such that $a_m\equiv a_n\pmod{p^m}$ whenever $m\le n$. For $n\in\Bbb Z^+$ let

$$A_n=\left\{k\in\Bbb Z^+:k\equiv a_n\pmod{p^n}\right\}\,;$$

if $m\le n$, then $A_n\subseteq A_m$, so there is a $\mathscr{U}\in\beta\Bbb N$ such that $A_n\in\mathscr{U}$ for each $n\in\Bbb Z^+$, and clearly $\big(\Phi(\mathscr{U})\big)(p)=a$. Thus, for each prime $p$ and each $p$-adic integer there is an ultrafilter that picks out that $p$-adic integer. (I don’t, however, know whether $\Phi$ is surjective.)


问:
Express roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta, $

If roots of the equation $ax^2+bx+c=0$ are $\alpha, \beta, $  find roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta$

Here's what I have tried so far,

I know that $\alpha+ \beta=\frac{-b}{a} $ and $\alpha \beta=\frac{c}{a} $

So I can express $b=-a(\alpha+\beta)$

$c=a.\alpha\beta$

Once I substitute for b and c in the equation I can get, $$\alpha\beta x^2+(\alpha+\beta)(\alpha\beta+1)x+(\alpha\beta+1)^2=0$$

I want to know whether there is any different approach other than this method?

Any hint is higly valued. thank you!

答:
As given by me in comment:

let $g(x)=acx^2-b(c+a)x+{(c+a)}^2=0$ and $f(x)=ax^2+bx+c$

we see that $$g(x)={(c+a)}^2\cdot \frac{1}{c}\cdot f(\frac{-cx}{c+a})$$ thus $g(x)=0$ implies $$\frac{-cx}{c+a}=\alpha ,\beta$$ now use vieta and rearrange .....


问:
Counter example relating the ramification
I want to prove the following:

If $K/F$ is unramified, then $LK/LF$ is unramified.

Let $f, K, L$ be number fields. Let $\mathfrak{p}$ be a prime in $F$, and $\mathfrak{P}$ a prime above $p$ in $LF$. If the following statement is true, then it is easy to conclude the above proposition.

If $\mathfrak{p}$ is unramified in $K/F$, then $\mathfrak{P}$ is unramified in $LK/LF$.

Is it true? Or do we need a more suitable assumption on $F$ or $K$?

答: The local version of this claim is Proposition 7.2 in Chapter II of Neukirch. To deduce the global version from the local version, just complete $ F, K, LF, LK $ at appropriate primes lying over $ \mathfrak p $ and verify the claim for each possible combination where the extension lattice is compatible with the lying over relations between the primes.


问:
How to prove an element is idempotent
In a commutative ring $R$, if $x$ is a non unit element outside Jacobson radical and $Rx +Ry =R$ for all non units $y$ outside Jacobson radical other than x, then how to prove $x$ is idempotent in $R$?

答:
Let $J=\bigcap_{M\in\operatorname{Max}(R)} M$ denote the Jacobson radical of $R$, and suppose $x$ is not idempotent. Then we claim we must have $x^2\in J$; indeed, if $x^2\notin J$, then $x^2$ is a non-unit of $R$ distinct from $x$ and lying outside the Jacobson radical, so we have $R=Rx+Rx^2=Rx$, contradicting that $x$ is not a unit. Thus $x^2\in J$, so $x^2$ lies in every maximal ideal of $R$, so $x$ lies in every maximal ideal of $R$, so $x\in J$, a contradiction. Thus $x$ must indeed be idempotent.

To answer your question below, the hypotheses does indeed force $J=\{0\}$. Indeed, suppose we had $j\in J\setminus\{0\}$. Then certainly $x+j\notin J$, since $x\notin J$, and $x+j\neq x$. Thus, by hypothesis, we have $Rx+R(x+j)=R$, so there are some $a,b\in R$ with $ax+b(x+j)=1$. But this means that $$(a+b)x=1-bj,$$ and we claim $1-bj$ is a unit; indeed, if it were not, then it would be contained in some maximal ideal $M<R$. But we have $j\in J\leqslant M$, so this would mean $1=(1-bj)+bj\in M$, a contradiction. Thus $1-bj$ is a unit, so $x$ is a unit, contrary to the hypothesis, and thus we must have $J=\{0\}$.

(Exercise: Show that, for any commutative ring $R$ and any element $x\in R$, $x$ lies in the Jacobson radical of $R$ if and only if $1-rx$ is a unit for all $r\in R$. One direction of the proof can be argued as above.)

Also, to show that this discussion is not vacuous – ie to show that there actually are rings satisfying the desired conditions – consider the ring $R=\mathbb{Z}\big/6\mathbb{Z}$. The Jacobson radical of $R$ is $\{\overline{0}\}$, and the non-zero non-units of $R$ are $\{\overline{2},\overline{3},\overline{4}\}$, so $\overline{3}$ satisfies the conditions of your problem, since it is coprime with both $\overline{2}$ and $\overline{4}$. For another very tractable example, consider the ring $$R=\mathbb{F}_2[x]\big/\langle x(x+1)\rangle,$$ the non-zero non-units of which are $\overline{x}$ and $\overline{x+1}$, which are of course coprime with each other.

Title: How to prove an element is idempotent


答:
The set $\beta(\mathbb{N})$ has a natural topology that makes it the Stone-Cech compactification of $\mathbb{N}$, and your map $\Phi$ is just the unique continuous extension of the inclusion map $\mathbb{N}\to\hat{\mathbb{Z}}$.  In particular, since $\Phi$ is continuous, its image is closed in $\hat{\mathbb{Z}}$.  Since $\mathbb{N}$ is dense in $\hat{\mathbb{Z}}$, this implies $\Phi$ is surjective.

Concretely, given $a\in\hat{\mathbb{Z}}$ and nonzero $n\in\mathbb{Z}$, let $X_{a,n}$ be the set of integers that have the same mod $n$ residue as $a$.  For fixed $a$, these sets $X_{a,n}$ generate a proper filter (since $X_{a,n}\cap X_{a,m}=X_{a,\operatorname{lcm}(m,n)}$) which can then be extended to an ultrafilter, and this ultrafilter maps to $a$ under $\Phi$.

Title: Ultrafilters on naturals determine p-adic numbers

Body:
While preparing for a short lecture on ultrafilters for undergraduates, I realized some interesting things I have never read about.

Though I'm asking now a specific question, any reference about this idea is welcome.

Let $\beta(X)$ denote the set of ultrafilters on $X$.

By the principal filters we obtain an embedding $X\hookrightarrow \beta(X)$, so ultrafilters can be regarded as 'generalized elements' joined to our set $X$, as this is often done in topology.

The main observation is, that unary predicates can be perfectly evaluated to ultrafilters themselves (without referring to any ultraproduct or such), as the interpretation of unary predicates are just the subsets of $X$.

For example, every nonprincipal ultrafilter $U$ on $\Bbb N$ satisfies $x>n$ for every fixed $n\in\Bbb N$, by which I simply mean $$\{x\in\Bbb N:x>n\}\ \in\  U\,.$$

Specifically, every ultrafilter $U\in\beta(\Bbb N)$ has a last digit in any number system $n$, in the sense that exactly one of the below sets will belong to $U$: $\ n\Bbb N,\ n\Bbb N+1,\ \dots,\ n\Bbb N+(n-1)$, since their disjoint union is the full set $\Bbb N$.

But this gives rise to a well defined function $\Phi:\beta(\Bbb N)\to\hat{\Bbb Z}=\prod_p\Bbb Z_p$ where $\Bbb Z_p$ denotes the set of $p$-adic integers.

Question: Is $\Phi$ surjective? In particular, does there exist an ultrafilter for every $p$-adic integer?

Attempt. Assume $a\in\Bbb Z_p$ is given. I'd like to define a filter and then extend it to ultrafilter.

$$\mathcal F:=\big\{A\subseteq\Bbb N \mid \forall n\,\exists x\in A: x\equiv a\!\!\!\pmod{p^n}\big\}$$
but the problem is that it's not a filter, though nonempty and increasing system, it's not closed under intersection, because if $a_n$ denotes the last $n$ digits of $a$ modulo $p$, then we can take $A=\{a_n\}\in\mathcal F$ and the disjoint $B=\{p^n+a_n\}\in\mathcal F$.


问:
R is symmetric and transitive relation, if and only if $R = R^{-1}\circ R$
How do I prove that the following:

Let $R$ be a (binary) relation on a set. Prove that $R$ is symmetric and transitive relation, if and only if $R = R^{-1}\circ R$.

Here, $R^{-1} \circ R$ is defined as the relation $\{(u,v) | \exists x, uR^{-1}x \land xRv\}$, where $xRu$ means $\left(x,u\right) \in R$.

I only have a problem with assuming the 2nd, and deriving the first, any tips on the proof technique?

答:
Preparations:

$\newcommand{\op}[2]{ \left\langle #1 ,\, #2 \right\rangle }$
$\newcommand{\rel}[1]{ \mathcal{#1} }$
$\newcommand{\dom}[1]{ \mathrm{dom} \ {#1} }$
$\newcommand{\ran}[1]{ \mathrm{ran} \ {#1} }$

Define ordered pair:

$$\op{x}{y} = \left\{\left\{x\right\},\,\left\{x,\,y\right\}\right\}$$

Define domain and range of relation:

\begin{align}
\dom{R} &= \left\{ x \in \cup\cup\rel{R} \mid \exists y :\: \op{x}{y} \in \rel{R} \right\}, \\
\ran{R} &= \left\{ x \in \cup\cup\rel{R} \mid \exists t :\: \op{x}{y} \in \rel{R} \right\}.
\end{align}

If $X$ and $Y$ are sets, and $R$ is a relation between $X$ and $Y$,
i.e. $\rel{R} \subseteq X \times Y$.

Let $\rel{T}$ is a composite relation on $Y$,
i.e. $\rel{T} \subseteq Y \times Y$,
which satisfies $\rel{T} = \rel{R}^{-1}\circ\rel{R}$.

Assume $\op{x}{y} \in \rel{R},\; x \in \dom{R} \subseteq X,\; y \in \ran{R} \subseteq Y$.

\begin{align}
\op{x}{y} \in \rel{R}
&\implies \op{y}{x} \in \rel{R}^{-1} \\
&\implies \op{y}{y} \in \rel{R}^{-1}\circ\rel{R} \\
&\implies \op{x}{x} \in \rel{T}
\end{align}

Therefore, the composite relation $\rel{T}$ is reflexive.

Assume $\op{a}{b} \in \rel{T},\; a \in \dom{T} \subseteq Y,\; b \in \ran{T} \subseteq Y$.

\begin{align}
\op{a}{b} \in \rel{T}
&\implies \op{a}{b} \in \rel{R}^{-1}\circ\rel{R} \\
&\implies \exists t \in X :\: \op{a}{t} \in \rel{R}^{-1} \;\land\; \op{t}{b} \in \rel{R} \\
&\implies \exists t \in X :\: \op{b}{t} \in \rel{R}^{-1} \;\land\; \op{t}{a} \in \rel{R} \\
&\implies \op{b}{a} \in \rel{R}^{-1}\circ\rel{R} \\
&\implies \op{b}{a} \in \rel{T}
\end{align}

Therefore, the composite relation $\rel{T}$ is symmetric.

Sufficiency:

Given $\rel{R} = \rel{R}^{-1}\circ\rel{R}$, because $\rel{R}^{-1}\circ\rel{R}$ is both reflexive and symmetric, $\rel{R}$ has the same properties.

Suppose $\ran{R} \cap \dom{R} \neq \varnothing$, hence $\exists b :\: \op{a}{b} \in \rel{R}$, $\op{b}{c} \in \rel{R}$,
\begin{align}
\op{a}{b} \in \rel{R} \land \op{b}{c} \in \rel{R}
&\implies
\op{a}{b} \in \rel{R}^{-1} \land \op{b}{c} \in \rel{R} \\
&\implies
\op{a}{c} \in \rel{R}^{-1}\circ\rel{R} \\
&\implies
\op{a}{c} \in \rel{R}
\end{align}

Therefore, $\rel{R}$ is transitive as well.

Necessity:

If $\rel{R}$ is both symmetric and transitive, assume $\op{x}{y} \in \rel{R},\;\op{y}{z} \in \rel{R},\;\op{x}{z} \in \rel{R}$,
\begin{align}
\left\{ \begin{array}{l}
\op{x}{z} \in \rel{R} \\
\op{y}{z} \in \rel{R}
\end{array} \right.
&\implies
\left\{ \begin{array}{l}
\op{z}{x} \in \rel{R} \\
\op{z}{y} \in \rel{R}
\end{array} \right. \\
&\implies
\left\{ \begin{array}{l}
\op{x}{z} \in \rel{R}^{-1} \\
\op{z}{y} \in \rel{R}
\end{array} \right. \\
&\implies
\op{x}{y} \in \rel{R}^{-1}\circ\rel{R} \\
\left\{ \begin{array}{l}
\op{x}{y} \in \rel{R}^{-1}\circ\rel{R} \\
\op{x}{y} \in \rel{R}
\end{array} \right.
&\implies
\rel{R} = \rel{R}^{-1}\circ\rel{R}
\end{align}

All in all, $\rel{R}$ is symmetric and transitive if and only if $\rel{R} = \rel{R}^{-1}\circ\rel{R}$.


问:
Find the limit $ \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n$
For $a,b,c>0$, Find $$ \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n$$

how can I find the limit of sequence above?
Provide me a hint or full solution.
thanks ^^

答:
This answer is inspired by @Aryabhata's answer here.

Suppose $x_k>0$.
By $\mathrm{AM}\ge\mathrm{GM}$ we have
$$\sqrt[m]{x_1\cdots x_m} \le \frac{x_1+\ldots +x_m}{m}.$$
By weighted $\mathrm{AM}\ge\mathrm{GM}$ we have
$$\begin{eqnarray*}
\frac{m}{x_1+\ldots +x_m}
&=& \frac{x_1\cdot 1/x_1+\ldots +x_m\cdot 1/x_m}{x_1+\ldots +x_m} \\
%&\ge& \left(\frac{1}{x_1^{x_1}\cdots x_m^{x_m}}\right)^{1/(x_1+\ldots +x_m)} \\
%&\ge& \left(\frac{1}{x_1^{x_1}\cdots x_m^{x_m}}\right)^{\frac{1}{x_1+\ldots +x_m}} \\
&\ge& \sqrt[x_1+\ldots +x_m]{\frac{1}{x_1^{x_1}\cdots x_m^{x_m}}}.
\end{eqnarray*}$$
Therefore,
$$\sqrt[m]{x_1\cdots x_m}
\le \frac{x_1+\ldots +x_m}{m}
\le \sqrt[x_1+\ldots +x_m]{x_1^{x_1}\cdots x_m^{x_m}}.$$

Letting $x_k = a_k^{1/n}$ and taking the $n$th power we find by the squeeze theorem that
$$\lim_{n \to \infty}\left(\frac{a_1^{1/n}+\ldots +a_m^{1/n}}{m}\right)^n
= \sqrt[m]{a_1\cdots a_m}.$$
In particular,
$$\lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n
= \sqrt[3]{a b c}.$$


问:
Components of double covariant derivative.
My question is simple: Could you please explain (unfortunately, it means perform an step-by-step calculation) why the components of double covariant is given by $(1)$?

$$ (\nabla_{u}\nabla_{v}w)^{a} = u^{c}v^{b}\nabla_{c}\nabla_{b}w^{a}+(u^{c}\nabla_{c}v^{b})\nabla_{b}w^{a} \tag{1}$$

答: It's a direct computation using the Leibniz rule: $$\nabla_u\nabla_vw = u^b\nabla_b(v^c\nabla_cw) \stackrel{(\ast)}{=} u^bv^c\nabla_b\nabla_cw + (u^b\nabla_b v^c)\nabla_cw.$$


问:
Diferential equation with $\sinh$
I'm trying to solve the following differential equation for $z\in [0,\infty]$:

\begin{equation}
\frac{\partial^{2}\phi}{\partial z^{2}}=A\sinh\left[B\phi\left(z\right)\right]
\end{equation}
with boundary condition $\phi(z=0)=W$ and $\frac{\partial\phi(z)}{\partial z}|_{z\rightarrow \infty} = 0$.
I started solving this by multiplying both sides by $\frac{\partial\phi(z)}{\partial z}$ and integrating in z. This gave me

\begin{equation}
\frac{1}{2}\left(\frac{\partial\phi}{\partial z}\right)^{2}=\frac{1}{B}\cosh\left[B\phi\left(z\right)\right]+C
\end{equation}

However, I have no idea how to proceed from here. I initially thought about squaring everything, but it seems only more complicate to integrate $\sqrt{\cosh[] + C}$. Any ideas?

Thanks!

I have a book that says there is a closed analytical formula for that

答:
Starting with the equation
$$\frac{d^{2}\phi}{dz^{2}}=A\sinh\left(B\phi\right)$$ switch variables to get
$$-\frac{z''}{[z']^3}=A\sinh\left(B\phi\right)$$ So, reduction of order $(p=z')$
$$-\frac{p'}{p^3}=A\sinh\left[B\phi\right]\implies p=\pm\frac{\sqrt{B}}{\sqrt{2A \cosh (B \phi)+k_1}}$$ The problem is that now we shall face elliptic integrals.

Before any simplification (still ignoring the signs of $A$, $B$ and $k_1$
$$z+c_2=\pm\frac{2 i \sqrt{\frac{2 A \cosh (B \phi)+k_1}{2 A+k_1}}}{\sqrt{B} \sqrt{2 A \cosh (B
   \phi)+k_1}}\, F\left(i\frac{ B \phi}{2}|\frac{4 A}{2 A+k_1}\right)$$

Inversing the monster would not be very pleasant.

Title: Diferential equation with $\sinh$


问:
For a graph, is its degree sequence a function defined on its vertex set?
In West's Introduction to Graph Theory

1.3. Vertex Degrees and Counting

1.3 1. Definition. The degree of vertex $v$ in a graph $G$, written $d_G(v)$ or $d(v)$, IS the number of edges incident to $v$, except
that each loop at $v$ counts twice.

GRAPHIC SEQUENCES

Next we consider all the vertex degrees together.

1.3.27. Definition. The degree sequence of a graph is the list of vertex degrees .

Every graph has a degree sequence, but which sequences occur? That is,
given nonnegative integers $d_l , . . . , d_n$, is there a graph with
these as the vertex degrees? The degree-sum formula implies that
$\sum_i d_i$ must be even.  When we allow loops and multiple edges,
the obvious necessary condition is also sufficient:

1.3.28. Proposition. The nonnegative integers $d_l , . . . , d_n$ are the vertex degrees of some graph if and only if $\sum_i d_i$ is even.

In 1.3.1 definition, is $d_G$ a function defined on $V(G)$?

Are $d_G$ and $d_l , . . . , d_n$  in 1.3.27 definition  equivalent to each other?

Why do we need two representations of the same thing? (Why do we need the concept of degree sequence of a graph, instead of just its $d_G$?)

Thanks.

答:
In 1.3.1 definition, is $d_G$ a function defined on $V(G)$?

Yes.

Are $d_G$ and $d_l,\ldots,d_n$ in 1.3.27 definition equivalent to each other?

No. The degree sequence forgets which value corresponds to which vertex. So it's not a function anymore, because it's not a relation between inputs and outputs — it's only a list of numbers. We should think of it as a new object, a finite nondecreasing sequence of nonnegative integers, built from the given graph as a whole.

Why do we need two representations of the same thing? (Why do we need the concept of degree sequence of a graph, instead of just its $d_G$?)

For example, in some applications where we'd want to model things with graphs, we may come up with a desired degree sequence first, and then build a graph from it.


答: What you’ve written $d_l$ should be $d_1$. The numbers $d_1,\ldots,d_n$ are simply non-negative integers; they may or may not be the degrees of the vertices of some graph, and even if they are, the sequence, unlike the function $d_G$, does not actually associate each of them with a vertex. The question is which such sequences actually are the values of the function $d_G$ on the vertex set $V(G)$ of some graph $G$.

Title: For a graph, is its degree sequence a function defined on its vertex set?

Body:
In West's Introduction to Graph Theory

1.3. Vertex Degrees and Counting

1.3 1. Definition. The degree of vertex $v$ in a graph $G$, written $d_G(v)$ or $d(v)$, IS the number of edges incident to $v$, except
that each loop at $v$ counts twice.

GRAPHIC SEQUENCES

Next we consider all the vertex degrees together.

1.3.27. Definition. The degree sequence of a graph is the list of vertex degrees .

Every graph has a degree sequence, but which sequences occur? That is,
given nonnegative integers $d_l , . . . , d_n$, is there a graph with
these as the vertex degrees? The degree-sum formula implies that
$\sum_i d_i$ must be even.  When we allow loops and multiple edges,
the obvious necessary condition is also sufficient:

1.3.28. Proposition. The nonnegative integers $d_l , . . . , d_n$ are the vertex degrees of some graph if and only if $\sum_i d_i$ is even.

In 1.3.1 definition, is $d_G$ a function defined on $V(G)$?

Are $d_G$ and $d_l , . . . , d_n$  in 1.3.27 definition  equivalent to each other?

Why do we need two representations of the same thing? (Why do we need the concept of degree sequence of a graph, instead of just its $d_G$?)

Thanks.


答:
No, the degree sequence is not a function on $V(G)$, although it involves a function on $V(G)$ indirectly in its definition, namely the function $d : V(G) \to \{0,1,2,3,...\}$ which assigns to each $v \in V(G)$ its degree $d(v)$.

Amongst all enumerations of the vertex set $V_G$, choose an enumeration $v_1,v_2,...,v_n$ of $V_G$ having the property that
$$d(v_1) \ge d(v_2) \ge d(v_3) \ge ... \ge d(v_n)
$$
Then the degree sequence $d_1,d_2,d_3,...,d_n$ is defined by $d_i = d(v_i)$. So in the degree sequence, you have recorded the degrees in order, but you have lost the information of what degree goes with what vertex. So no, the function $d$ and the degree sequence $d_1,d_2,...,d_n$ are not equivalent to each other.

To address your question of why we need two representations of the same thing, well, because sometimes we want to ask a question or formulate a statement about the degrees themselves, such as Proposition 1.3.28, so that's what the degree sequence is used for. Other times we want to know the actual degrees of each vertex, that's what the degree function $d(v)$ is for.


问:
Simultaneous Equations with partial derivatives
The function $x^3 - 3x^2 - 6xy + 2y^2+ 8y + 8$ yields

$f_x = 3x^2 - 6x - 6y$

$f_y = 4y + 8 - 6x$

When we equate both of these to $0, y = \frac{x^2}{2} - x$ and $y = \frac{3}{2}x - 2$

Plotting these on a graph we can see the points of interaction are $(1, -0.5)$ and $(4, 4)$ and although that is the answer, I was wondering how to get those results using simultaneous equations, as the $x^2$ means that Gaussian elimination is inadequate, but I can't seem to figure it out using simultaneous equations.

Also, is the equation of a tangent plane the same formula as linear approximation? The tangent plane equation I found when at $(1, -2)$ is 13 + 9x - 6y

答:
Just equate them. We need $y=\frac{x^2}2-x=\frac{3x}2-2$ giving $x^2-5x+4=0\iff x=1,4$ and corresponding $y=-1/2,4$.

$13+9x-6y$ is not an equation. For the functional equation $f(x,y,z)=0$, the equation of the tangent plane at $(x_0,y_0,z_0)$ is $[(x,y,z)-(x_0,y_0,z_0)]\cdot\vec\nabla f(x_0,y_0,z_0)=0$ where $\vec\nabla f=(f_x,f_y,f_z)$ represents the gradient or normal vector to the surface at a given point. Simplifying, you get$$(x-x_0)f_x(x_0,y_0,z_0)+(y-y_0)f_y(x_0,y_0,z_0)+(z-z_0)f_z(x_0,y_0,z_0)=0$$

We also have$$f(x-x_0,y-y_0,z-z_0)=\underbrace{\require{cancel}\cancelto{0}{f(x_0,y_0,z_0)}}_{(x_0,y_0,z_0)\text{ lies on the surface }f(x,y,z)=0}\\+(x-x_0)f_x(x_0,y_0,z_0)+(y-y_0)f_y(x_0,y_0,z_0)+(z-z_0)f_z(x_0,y_0,z_0)+\text{higher order terms}$$so indeed the equation of the tangent plane of $f(x,y,z)=0$ resembles the linear approximation of $f(x,y,z)$.


问:
Area of a circle smaller than one?
If a circle has radius less than one, does it mean that the circumference of the circle is bigger than the area? How can that even be possible?

For example, if a circle has radius $0.5$, then the circumference is $2 \pi \cdot 0.5 = \pi$, and the area will be $\pi \cdot 0.5^2 = 0.25 \pi$. How can this be the case?

答:
As mentioned above, the units can have an affect on the ratios

For circumference of a circle $C=2\pi r\quad$ for area $A=\pi r^2\quad$ for ratio
$$R=C/A=\frac{2\pi r}{\pi r^2}=\frac{2}{\pi r}\quad \land\quad r=1\implies R=\frac{2}{3.14\cdots}\approx 0.64$$

Management says the measure was in yards but should have been in feet

$$r=3\implies  R=\frac{2}{3.14(3)\cdots}\approx 0.21$$

Management now says it WAS feet and should have been in yards $\quad (1/3$ that of feet)

$$r=1/3\implies  R=\frac{2}{3.14(1/3)\cdots}\approx 1.91$$

So yes. If your radius is less than one, circumference can be larger than area.

Title: Area of a circle smaller than one?


问:
Evaluate the Integral using Contour Integration (Theorem of Residues)
$$
J(a,b)=\int_{0}^{\infty }\frac{\sin(b x)}{\sinh(a x)} dx
$$

This integral is difficult because contour integrals normally cannot be solved with a sin(x) term in the numerator because of singularity issues between the 1st quadrant to the 2nd quadrant in the upper half circle of the contour.

I've assumed:
$$
\sin(bx)=e^{ibz}
$$

since it effectively IS the sine term in the upper half circle.

I've also used the following substitutions:
$$
x=z  ;  z=re^{i\theta}   ;  z=\cos(\theta)+i\sin(\theta)
$$

From what I understand, the x range has to include all values somehow.
When I plug in the substitutions and the Euler forms of sin and sinh, the integral becomes:
$$
2\int_{-\infty }^{\infty }\frac{e^{ibx}}{e^{ax}-e^{ax}}dx
$$

Have I screwed up the subs?  Did I reduce incorrectly?  Not sure what to go from here. If anyone could help shed some light it would be very much appreciated.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\on{J}\pars{a,b} & \equiv
\bbox[5px,#ffd]{\int_{0}^{\infty }{\sin\pars{bx} \over \sinh\pars{ax}}\,\dd x} =
\on{sgn}\pars{a}\int_{0}^{\infty }{\sin\pars{bx} \over \sinh\pars{\verts{a}x}}\,\dd x
\\[5mm] & \stackrel{x\ =\ t/\pars{2\verts{a}}}{=}\,\,\,
\left.{1 \over 2a}\int_{0}^{\infty }{\sin\pars{\beta t/2} \over \sinh\pars{t/2}}\,\dd t
\,\right\vert_{\,\ds{\beta\ \equiv\ b/\verts{a}}}
\\[5mm] & =
-\,{\ic \over 2a}\int_{0}^{\infty }
{\expo{-\pars{1 - \beta\,\ic}t/2}\,\,\, -\,
\expo{-\pars{1 + \beta\,\ic}t/2} \over 1 - \expo{-t}}\,\dd t
\\[5mm] & =
-\,{\ic \over 2a}\left[%
\int_{0}^{\infty }
{1 - \expo{-\pars{1 + \beta\,\ic}t/2} \over 1 - \expo{-t}}
\,\dd t\right.
\\[2mm] & \phantom{-\,{\ic \over 2a}\,\,\,\,\,}\left.
-\int_{0}^{\infty }
{1 - \expo{-\pars{1 - \beta\,\ic}t/2} \over 1 - \expo{-t}}
\,\dd t\right]\label{1}\tag{1}
\\[5mm] & =
-\,{\ic \over 2a}\bracks{%
\Psi\pars{{1 \over 2} + {\beta\,\ic \over 2}} -
\Psi\pars{{1 \over 2} - {\beta\,\ic \over 2}}}
\label{2}\tag{2}
\\[5mm] & =
-\,{\ic \over 2a}\braces{%
\pi\cot\pars{\pi\bracks{%
{1 \over 2} - {\beta\,\ic \over 2}}}} =
-\,{\pi\ic \over 2a}\tan\pars{\beta\,\ic \over 2}
\label{3}\tag{3}
\\[5mm] & =
\bbx{{\pi \over 2a}\tanh\pars{b \over 2\verts{a}}} \\ &
\end{align}

$\ds{\Psi}$: Digamma Function.

(\ref{2}): See $\ds{\color{black}{\bf 6.3.22}}$ in A & S Table.

(\ref{3}): Euler Reflection Formula. See $\ds{\color{black}{\bf 6.3.7}}$ in A & S Table.


问:
Divergence of $\zeta(z)$ tamed or not tamed by any analytic continuation
We know the conjecture about the Riemann hypothesis is about the nontrivial zeros are on
$$(1/2 + r i)$$
for some $r \in \mathbb{R}$ of Riemann zeta function.

My question is how many divergences of $\zeta(z)$ can be tamed by analytic continuations and how many divergences of $\zeta(z)$ canNOT be tamed by analytic continuations?

For example,
$$\zeta(1)=1+2+3+4+\dots=\infty$$
seem not to be tamed by any analytic continuation. How about other $\zeta(z)$? Are there others cannot be done by analytic continuation as a finite value?

答: The only pole of the Riemann zeta function is at $ z = 1 $, so that's the only divergence that's not "tamed" by analytic continuation.


问:
Confusion over joint probability and joint expectation
This question is likely mainly a notational issue, but I would greatly appreciate some clarification.

Consider two (discrete) random variables X and Y. It makes sense to think about the joint probability distribution between the two rv’s, which can be written as $f(x,y)=P(X=x,Y=y)$. Note the commas. Does it make sense to also think about a joint expectation, denoted as $E(X,Y)$? Is there a requirement that the expectation operator only takes one “argument”? Why?

I often see $E(XY)$ but I don’t think this is the same thing, as this is the expectation of a new random variable $Z=XY$ which the product of X and Y.

答:
I think the confusion is due to the fact that joint probability density functions do not have expected values, but instead, random variable have expected values.

Informally, the expected value is an operator that takes a random variable as input and returns the most likely value of the random variable. Thus, the notation $\text{E}[X,Y]$ is not valid and doesn't make much sense. However, the notation $\text{E}[XY]$ is valid and is equivalent to $\text{E}[Z]$ where $Z = XY$. This is valid because $Z$ is a random variable. You might be interested in the law of the unconscious statistician, which is related to computing the expectation of a function of random variables.

Formally, the expected value of a random variable $X$ with probability density function $f(x)$ is given by

$$ \text{E}[X] = \int_{\mathbb{R}} x f(x) dx $$

You can find a more general definition here with further explanation in this answer, but given the question, I don't think the general definition is necessary.


答:
Let $\bar\alpha$, $\bar\beta$ be the roots of $acx^2-b(c+a)x+(c+a)^2=0$, which can be written as

$$\left(1+\frac ac\right)^2\frac c{x^2}
-\left(1+\frac ac\right)\frac b{x}+a=0
$$
Compare with $ax^2+bx+c=0$ written as
$$\frac c{x^2} +\frac bx +a=0$$

to establish
$$\left(1+\frac ac\right)\frac1{\bar\alpha}=-\frac1{\alpha},\>\>\>\>\>\left(1+\frac ac\right)\frac1{\bar\beta}=-\frac1{\beta}
$$
Then, with $\frac ca = \alpha \beta$
$$\bar\alpha=- \alpha -\frac1{\beta}
,\>\>\>\>\>
\bar\beta= -\beta -\frac1{\alpha}
$$

Title: Express roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta, $

Body:
If roots of the equation $ax^2+bx+c=0$ are $\alpha, \beta, $  find roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta$

Here's what I have tried so far,

I know that $\alpha+ \beta=\frac{-b}{a} $ and $\alpha \beta=\frac{c}{a} $

So I can express $b=-a(\alpha+\beta)$

$c=a.\alpha\beta$

Once I substitute for b and c in the equation I can get, $$\alpha\beta x^2+(\alpha+\beta)(\alpha\beta+1)x+(\alpha\beta+1)^2=0$$

I want to know whether there is any different approach other than this method?

Any hint is higly valued. thank you!


问:
Inequality regarding the Brownian motion stopping time
I try to solve a problem related with my previous post. I believe if I can show following inequality then I am done.

Let $D$ be $a$ bounded domain in $\mathbb{R}^{d}$, let $B_{t}$ be a standard Brownian motion starting at $x \in D,$ and let
$$
T=\inf \left\{t: B_{t} \notin D\right\}
$$
Let $q_{t}=\sup _{x \in D} \mathbb{P}^{x}\{T>t\}$, where superscript means that "starting at  $x$".

By strong Markov property, it is easy to see that $q_{s+t} \leq q_{s} q_{t}$. I try to show there exists constant $c$ such that $q_{s+t} \geq c q_{s} q_{t}$, still by strong Markov property, if I can show: for any $x \in D$, there exist a constant $c_1$ such that

$$\mathbb{P}^{x}\{T>t\}\geq c_1 q_t$$
then I am done. But I have no idea how to show this inequality. Can anyone help? Really appreciate your help!

答:
Unfortunately the inequality you want doesn't hold. The reason being that it would mean that for fixed $t$, $\mathbb{P}^{x}(T>t)$ is bounded below by some positive constant, independent of $x$. However, intuitively, it is clear that as $x$ tends towards the boundary of our domain $D$, $\mathbb{P}^{x}(T>t)$ tends to $0$.

More precisely, if $x^{*}\in{\partial{D}}$ is regular, then $\mathbb{P}^{x}(T>t)\rightarrow{0}$ as $x\rightarrow{x^{*}}$.


问:
Show that $G$ has an element of order $p$.

Let $G$ be a group of order $p^n$, for some prime $p$ and positive integer $n$. Show that $G$ has an element of order $p$.

I know this question has been asked before, but I have not found a complete answer and I am quite confused.

I have tried to solve it and so far this is what I have achieved:

If $|G|=p^n$ and $e\neq a\in G$, we have that order of $a$ divides $p^n$, then $|a|\in \{p^m:1\leq m \leq n\}$. Also, if order of $G$ is $p^n$, for all $a\in G$, $a^{p^{n}}=e$. Then write $p^n=p^{m}p^s$ , $1\leq s\leq n$, we have that $(a^{p^{m-1}p^s})^p=e$. Then we can conclude that $b=a^{p^{m-1}p^s}\in G$ is an element of orden $p$.

Is what I did correct?

答:
This follows from Cauchy's theorem.

The proof of Cauchy that I remember is that by Sylow we have, for any prime $p$ dividing the order of the group, a subgroup of order $p^n$, where $p^{n+1}$ does not divide the order of the group.  Then, as in the other answer, either that Sylow subgroup is cyclic, in which case it has a subgroup of every order dividing $p^n$, including one of order $p$, or take the cyclic subgroup generated by any element, and apply Lagrange.  Then we get a cyclic subgroup of order $p^m$ for some $m\lt n$, and we are done.


问:
Find real solution $x$ in order to $x + \sqrt{2020}$ and $\dfrac{5}{x} -\sqrt{2020}$ are integers
$x + \sqrt{2020}$ and $\dfrac{5}{x} - \sqrt{2020}$ are integers

$\Rightarrow x + \dfrac{5}{x}$ is an integer

$\Rightarrow \dfrac{x^2 + 5}{x}$ is an integer

$\Rightarrow x^2 + 5\ \vdots\ x$

$\Rightarrow x^2 + 5 - x^2\ \vdots\ x$

$\Rightarrow 5\ \vdots\ x$

Is it the right way?

Any idea? Or any better way to solve this problem?

Please help !!!

答:
As Edward H's question comment hint suggests, let $a$ be the integer where

$$a = x + \sqrt{2020} \implies x = a - \sqrt{2020} \tag{1}\label{eq1A}$$

Then let $b$ be the other integer, and use \eqref{eq1A}, to get

$$\begin{equation}\begin{aligned}
b & = \frac{5}{a - \sqrt{2020}} - \sqrt{2020} \\
b + \sqrt{2020} & = \frac{5}{a - \sqrt{2020}} \\
(b + \sqrt{2020})(a - \sqrt{2020}) & = 5 \\
ab - b\sqrt{2020} + a\sqrt{2020} - 2020 & = 5 \\
ab + \sqrt{2020}(-b + a) & = 2025 \\
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Since $ab$ and $2025$ are integers, this means $\sqrt{2020}(-b + a)$ must also be an integer. However, due to $\sqrt{2020}$ being irrational and $-b + a$ being an integer, this is only possible if it's $0$, i.e.,

$$b = a \tag{3}\label{eq3A}$$

Using this in \eqref{eq2A} gives

$$a^2 = 2025 \implies a = \pm 45 \tag{4}\label{eq4A}$$

Finally, \eqref{eq1A} gives that

$$x = a - \sqrt{2020} \implies x = \pm 45 - \sqrt{2020} \tag{5}\label{eq5A}$$


问:
Continuous functions equal except on a nowhere dense set
Consider two continuous functions $f,g : \mathbb{R} \supset I \to \mathbb{R}^k$ for which every subinterval $J \subset I$ contains a subinterval $K \subset J$ such that $\forall t \in K, f(t) = g(t)$.

Is it then true that $\forall t \in I \setminus E, f(t) = g(t)$ where $E$ is a nowhere dense subset of $I$? If so, can you point me to a solid reference?

Thanks for the help!

答:
I don't have a reference for this, but it's a relatively straightforward application of the definition of a nowhere dense set. That's just to say that I wouldn't necessarily expect to find this recorded in a textbook somewhere, but I'll freely admit I haven't checked.

Here's the proof. To begin, let $\mathcal{J}$ be the collection of all open intervals $J \subseteq I$ where $f \equiv g$ on $J$. Letting $U = \bigcup_{J \in \mathcal{J}} J$, we have that $U$ is open, and $f \equiv g$ on $U$. Now let $F = I \setminus U$, so that $F$ is closed in $I$. Evidently $F$ contains the set $E = \{ t \in I : f(t) \neq g(t) \}$, so the closure of $E$ in $I$ is contained in $F$. As a result, to show that $E$ is nowhere dense it suffices to show that $F$ contains no (non-degenerate) interval.

This follows almost at once from the hypotheses: if $J'$ were an interval contained in $F$, we know there's a sub-interval $K \subseteq J'$ where $f \equiv g$. So then the interior of $K$ is actually in $U$, contradicting it being a part of $F$.


问:
Show that $\lim_{x \to 1} \sum_{n=1}^{\infty}\frac{n x^2}{n^3 + x^2}= \sum_{n=1}^{\infty} \frac{n}{n^3 +1}$.
What I understand from the question is that the author (N. L. Carothers) what me to show that the series $\sum_{n=1}^{\infty}\frac{n x^2}{n^3 + x^2}$ converges uniformly. Hence, the exchange between the limit and the summation is possible. However, It's been a week that I have been trying to find a summable upper bound so that the M-test will be applicable with no improvements. I tried AM-GM but did not get any results. Honestly, I do not publish the post here unless I try everything I know.

答:
It is not necessary to prove uniform convergence. The series on both sides of the equation are convergent so we can consider the differenece and show that it goes to $0$. The differnce is $(x^{2}-1)\sum \frac {n^{4}} {(n^{3}+x^{2}) (n^{3}+1)}$ which tends to $0$ since  $0 \leq\sum \frac {n^{4}} {(n^{3}+x^{2}) (n^{3}+1)} \leq \sum\frac  1{n^{2}}$.

The series does not converge uniformly on the real line since the general term does not tend to $0$ uniformly. [ Choose $x$ such that $\frac {nx^{2}} {n^{3}+x^{2}} =1$ to see this]. However, the series converges uniformly on $[-L,L]$ for any $L$: Check that  $\frac {nx^{2}} {n^{3}+x^{2}} \leq \frac {L^{2}} {n^{2}}$ for $|x| \leq L$. Of course,  this is good enough for proving the result.


问:
Joint Continuity and Separate Continuity in Topological rings
I have been stuck with the following problem for quite some time, as I will next describe, by first introducing the definitions at hand. Here I have two definitions of joint continuity and two definitions of separate continuity. It is supposed to be easy to prove, but I am having trouble proving claim 1 and claim 2. I would be grateful for any suggestions and happy to clarify any irregularities concerning the post

Joint Continuity

J1 A binary operation $\beta: R \times R \rightarrow R$ defined for all $x,y \in R$ by $\beta(x,y)=x * y$ is said to be jointly continuous at $(x,y) \in R \times R$ if and only if for each neighbourhood $W$ of $x * y$, there exists a product of open sets $U \times V \subseteq R \times R$ containing $(x,y)$ such that $U*V \subseteq W$. If $\beta$ is jointly continuous at each point in $R \times R$ then $\beta$ is called (globally) jointly continuous.

J2 Suppose that $X,Y$ and $Z$ are topological spaces. Let $f: X \times Y \rightarrow Z$ be a mapping. Then $f$ is called jointly continuous on $X \times Y$ if and only if the map $f$ is continuous from $X \times Y$, equipped with the product topology $\tau^{\times}$.

Claim 1 In the event that $X=Y=Z$, we have equivalence for J1 and J2.

Separate Continuity

S1 A binary operation $\beta: R \times R \rightarrow R$ defined for all $x,y \in R$ by $\beta(x,y)=x * y$ is said to be separately continuous if and only if for all $a \in R$ and for all $U \in \mathcal{N}_0$ there exists $V \in \mathcal{N}_0$ such that $a*V \subseteq U; V*a \subseteq U$

S2 Suppose $X,Y$ and $Z$ are topological spaces. Let $f: X \times Y \rightarrow Z$ be a mapping. Then $f$ is separately continuous on $X \times Y$ if and only if for all $(x_0,y_0) \in X \times Y$, both functions $x \mapsto f(x,y_0)$ and $y \mapsto f(x_0,y)$ are continuous on $X$ and $Y$ respectively.

Claim 2 In the event that $X=Y=Z$, we have equivalence for S1 and S2.

答:
Claim 1 In the event that $X=Y=Z$, we have equivalence for J1 and J2.

Implication $(J1 \Rightarrow J2)$ holds since $U\times V\in\tau^\times$ for each open subsets $U$ and $V$ of $R$. On the other hand, suppose that J2 holds. Let $(R,\tau)$ be a topological space, $(x,y) \in R \times R$ be any point, and $W$ be any neighbourhood of $x*y$. By the definition of a neighborhood, there exists an open set $W’\ni (x,y)$ such that $W’\subseteq W$. J2 implies that a set $\beta^{-1}(W’)$ is an open neighborhood of $(x,y)$. Since a family $\{ U \times V :U,V\in\tau \}$ is a base of the topology $\tau^\times$, there exist  $U,V\in\tau$ such that $(x,y)\in U \times V\subseteq  \beta^{-1}(W’)$. Then
$$U*V=\beta(U\times V) \subseteq \beta(\beta^{-1}(W’))\subseteq W’\subseteq W.$$


问:
How to evaluate $\sum_{k=1}^{n} \frac{f^3_k}{3(f^2_k-f_k)+1}, ~~f_k=k/n$
Very often the summations at pre-degree level are done by using differencing-telescoping and by some  symmetry properties. The following summation is fabricated keeping one thing in mind, I may tell it later. $$\sum_{k=1}^{n} \frac{f^3_k}{3(f^2_k-f_k)+1},~~ f_k=k/n.$$

The question is: How will you evaluate this sum?

Body: let $$S= \sum_{k=0}^{n} \frac{f_k^3}{3(f_k^2-f_k)+1}=\sum_{k=1}^{n} \frac{f_k^3}{3(f_k^2-f_k)+1}$$ writing sum in reverse order $$S=\sum_{k=0}^n\frac{f_{n-k}^3}{3(f_{n-k}^2-f_{n-k})+1}$$ now using the beatiful property $f_k=1-f_{n-k}$ $$S=\sum_{k=0}^n \frac{{(1-f_k)}^3}{3(f_k^2-f_k)+1}$$ Adding $$2S=\sum_{k=0}^n 1=n$$ $$S=(n+1)/2$$


问:
About the YouTube video "The Riemann Hypothesis, Explained"
About the video The Riemann Hypothesis, Explained at 658 secs and on:

What are the meanings of the curves (and their analytic forms) that create the spiral or heart shapes and that keep hitting the zeta zeros several times?

What do these mean to have the curves keep hitting the zeta zeros multiple say N times?
What is the meaning of this N? if hitting N times? (or can the N be bounded finite or be infinite?)

答:
I recognize the spiral as being the curve parametrized by $\zeta(\frac{1}{2}+it)$ for $t$ in some finite interval $[0,T]$; the spiral is growing larger by making $T$ larger. The value $\zeta(\frac{1}{2})=1.46035...$ is the leftmost point on the spiral in the video, where it starts. If you look closely, the "critical line" $\mathrm{Re}(s)=\frac{1}{2}$ (the vertical orange line halfway between $0$ and $1$) is the one that turns into the spiral.

In general, the straight lines of the grid moments before are a "before" picture, and the collection of curves moments later are the "after" picture. That is, the curves are the images of $\zeta$ applied to the original straight lines. The dark teal curves are those that can be obtained using the original series definition of $\zeta$, while the orange ones are those that can only be obtained by using analytic continuation to extend the domain of $\zeta$.

(I am unsure but very curious how they smoothly transitioned the lines into curves.)

Every time the spiral hits the origin is a zero of the zeta function. So the multiplicity with which the spiral intersects the origin is the number of zeros $\zeta$ has on the critical line with imaginary part restricted to the interval $[0,T]$. As $T$ grows larger, this number will keep growing, forever.


问:
Efficient Proof for all cases of triangle inequality
The question asks Find an efficient proof for all the cases at once by first demonstrating

$$
(a+b)^2 \leq (|a|+|b|)^2
$$

My attempt at the proof:

for $a,b\in\mathbb{R}$

$$
\begin{align*}
|a+b|^2 =&(a+b)^2 \text{ (Since $\forall x\in\mathbb{R}$, $x^2=|x|^2$)}\\
 =& a^2+2ab+b^2\\
 \leq&|a|^2+2|a||b|+|b|^2 \text{( Since $\forall x\in\mathbb{R}, x\leq |x|$)}\\
=&(|a| +|b|)^2
\end{align*}
$$

But this is where I get stuck, I can arrive at the inequality but I do not know how to continue you from here. The question states that this should be efficient proof for all the cases, but jumping from that step to $|a+b|\leq |a|+|b|$ seems like a big jump with some steps missing. Any push in the right direction would be appreciated!

答:
All you need to say is that if $0<r^2<s^2$, then $r<s$. This follows from the fact that $r\geq s>0$ implies $r^2\geq s^2$.

Also, you are missing a square on $|a|$, and your last $\leq$ should be $=$.


问:
Problem of relations between the coefficients of a quadratic equation.how to proceed?
I have tried several ways to solve this problem of using the root properties of a second degree equation but I can't come up with anything favorable.

If the roots of $ax^{2}+bx+c=0$  are of the form  $\dfrac{\lambda }{\lambda -1}$ and  $\dfrac{\lambda+1 }{\lambda}$, then the value $(a+b+c)^{2}$ in terms of $a$, $b$, $c$ is?

Options

$(a) \quad b^{2}-4ac$

$(b) \quad b^{2}-2ac$

$(c) \quad 2b^{2}-ac$

$(d) \quad 4b^{2}-2ac$

$(e) \quad b^{2}-4abc$

Body:
\begin{align}
  \frac{c}{a} &= \frac{\lambda}{\lambda-1} \times \frac{\lambda+1}{\lambda} \\
  &= \frac{\lambda+1}{\lambda-1} \\
  \lambda &= \frac{c+a}{c-a} \\
  x_1 &= \frac{\lambda}{\lambda-1} \\
  &= \frac{c+a}{2a} \\
  x_2 &= \frac{\lambda+1}{\lambda} \\
  &= \frac{2c}{c+a} \\
  -\frac{b}{a} &= x_1+x_2 \\
  &= \frac{(c+a)^2+4ac}{2a(c+a)} \\
  -2b(c+a) &= (c+a)^2+4ac \\
  0 &= (c+a)^2+2b(c+a)+4ac \\
  \color{red}{b^2}-4ac &= (c+a)^2+2b(c+a) \color{red}{+b^2} \\
  (a+b+c)^2 &= b^2-4ac
\end{align}


问:
What functions have $f (t + 1)=f (t)+ 2$?
The functions have to be linear. Do I have to find the functions that satisfy the condition or is there a specific name for functions that satisfy these conditions. This question is the $26$th question in Strang's calculus at section $1.1$, If it is to find functions that satisfy these conditions, how do I do it?

Thinking without any math procedures, I find that if you take $f(t) = 2t$, then $f(t+1)$ will be $2t+2$ which will be same as $f(t) + 2 = 2t + 2$.

Are there any more functions that will satisfy these? Is there any procedure to find these or we can find this only through trial and error?

答:
Since $f$ is specified as linear, let $f(t)=at+b$. Then
$$a(t+1)+b=at+b+2$$
$$at+a+b=at+b+2$$
$$a=2$$
and $b$ is unrestricted. Thus the possible functions $f$ are $f(t)=2t+b$ where $b\in\mathbb R$.


问:
Show $K(\alpha)$ is a splitting field of $\text {Irr}(\alpha,K)$ over $K$ $\iff$ $K \subset K(\alpha)$ is a normal extension.

Show $K(\alpha)$ is a splitting field of $\text {Irr}(\alpha,K)$ over $K$ $\iff$ $K \subset K(\alpha)$ is a normal extension.

I see that if $K \subset K(\alpha)$ is a normal extension, then $\text {Irr}(\alpha,K)$ splits over $K(\alpha)$ by definition and $K(\alpha)$ is by definition a splitting field of $\text {Irr}(\alpha,K)$ over $K$.

However, I cannot prove the other way: $K(\alpha)$ is a splitting field of $\text {Irr}(\alpha,K)$ over $K$ $\Rightarrow$ $K(\alpha)$ is a normal extension.

I know the equivalent definition of a normal extension $K \subset K(\alpha)$ as: Each irreducible polynomial $p \in K[X]$ that has at least one root in $K(\alpha)$ splits over  $K(\alpha)$ into linear factors.

Can someone help me out ?

答:
Let $f(x)=\text {Irr}(\alpha,K)(x). $ Given $\sigma:K(\alpha) \hookrightarrow \overline K$ fixing $K$, since  $f(\sigma(\alpha))=\sigma(f(\alpha))=0, \sigma(K(\alpha))\subseteq K(\alpha)$. For any $\beta\in K(\alpha)$, let $T=\{\gamma\in K(\alpha)\mid \text{Irr}(\gamma,K)(\beta)=0\}$. Then $K(T)\subseteq K(\alpha)$, and for any $\gamma\in T, \text{Irr}(\sigma(\gamma),K)(\beta)=0$, so $\sigma(\gamma)\in T $ as well, i.e. $\sigma(K(T))\subseteq K(T)$. Since $\sigma$ is a non-trivial field homomorphism, its kernel is an ideal of $K(\alpha)$ viewed as a ring, which must be trivial. By the first isomorphism theorem of rings, we get $\sigma(K(T))\cong K(T)$ and hence $\sigma(K(T))=K(T)$. This means that $\beta\in \text{Im}(\sigma)$, and since $\beta$ was an arbitrary element in $K(\alpha)$, we get $\sigma(K(\alpha))= K(\alpha)$ and thus $\sigma$ is in fact an automorphism.

Lemma: Let $F(\alpha)/F$ be algebraic and $\sigma: F\rightarrow L$. Then $\sigma$ has an extension into $F(\alpha)$ if and only if $\exists \beta\in L$ such that it is a root of $\text{Irr}(\alpha,F^{\sigma})$.

Let $p(x)$ be an irreducible polynomial in $K[x]$ with a root $\xi\in K(\alpha)$. We may assume it is monic and therefore the minimal polynomial of $\xi$ over $K$. If $\eta$ is another root of $p$, then by the previous lemma, $\tau: K\rightarrow K(\eta)$ has an extension into $K(\xi)\subseteq K(\alpha)$. This extension may be viewed as a monomorphism from $K(\alpha)$ into $\overline K$, so it is in fact an automorphism, which means that $\eta\in K(\alpha)$. Therefore, every root of $p$ lies in $K(\alpha)$ and hence $p(x)$ splits in $K(\alpha)$.


问:
Drawing dynamic circles based on input value
Is there a formula that will allow me to calculate the radius of a circle based on an input value?  The input value could be as small as zero or as large as $10^7$, or larger.  The circle is restricted to a minimum radius of $10$ and a maximum radius of $100$.

Does anyone know how to calculate something like this?

UPDATE

The input values correspond to state/country population.  I want to calculate the radius (how big the circle should be) of the circle based on the input value.

答:
I'll expand on what I understood from your comment.

Let $\{x_i\}$ be the set of inputs.

Then the range of inputs is given by $\max\{x_i\}-\min\{x_i\}$.

The radius for the $i$th input is then given by

$$R=10+90\left(\frac{x_i-\min\{x_i\}}{\max\{x_i\}-\min\{x_i\}}\right)$$


答:
Note that for positive numbers $a$ and $b$, $$a^2 \le b^2 \implies a\le b $$
Thus $$|a+b|^2 \le (|a|+|b|)^2 \implies |a+b|\le|a|+|b|$$

Title: Efficient Proof for all cases of triangle inequality

Body:
The question asks Find an efficient proof for all the cases at once by first demonstrating

$$
(a+b)^2 \leq (|a|+|b|)^2
$$

My attempt at the proof:

for $a,b\in\mathbb{R}$

$$
\begin{align*}
|a+b|^2 =&(a+b)^2 \text{ (Since $\forall x\in\mathbb{R}$, $x^2=|x|^2$)}\\
 =& a^2+2ab+b^2\\
 \leq&|a|^2+2|a||b|+|b|^2 \text{( Since $\forall x\in\mathbb{R}, x\leq |x|$)}\\
=&(|a| +|b|)^2
\end{align*}
$$

But this is where I get stuck, I can arrive at the inequality but I do not know how to continue you from here. The question states that this should be efficient proof for all the cases, but jumping from that step to $|a+b|\leq |a|+|b|$ seems like a big jump with some steps missing. Any push in the right direction would be appreciated!


问:
why $f_n$ is decreasing function?
Given  $f_n : X \to [0,\infty)$ define by

$f_n =\begin{cases} 1 \ \text{if }\ x\in  [n,\infty) \\ 0 \ \text{if} \ x \in (-\infty,n) \end{cases}$

Here It's mention that $f_n$ is decreasing function

My question is that    why $f_n$ is decreasing function ?

My thinking : $f_n$ is neither decreasing nor increasing .$f_n$ take only value $0$ and $1$ .Therefore we can said that $f_n$ is oscillating

答:
I think you misunderstood the difference between an increasing/decreasing function and an increasing/decreasing sequence of functions. In the link you have attached, they say that the sequence of functions $(f_n)$ is decreasing. This is true, since for every $x\in \mathbb{R}$ one has
$$f_n(x)\geq f_{n+1}(x).$$
However, the definition of an increasing/decreasing function $f$ is different. Namely for all $x<y$ one has for an increasing function $f(x)\leq f(y)$ and for a decreasing function $f(x)\geq f(y)$. For fixed $n$, your function $f_n$ is clearly increasing.

Title: why $f_n$ is decreasing function?


问:
Change of variable with g(u,v) = (u-v+1,u+v)
I want compute the integral of f(x,y) = 1 over $R = \{ (x,y) : 0\leq x \leq 1 \land 0\leq y \leq x^2 \} $ with the transformation $g(u,v) = (u-v+1,u+v)$ and check that is the same that compute without change of variable. i.e. $ \int_{0}^{1} \int_{0}^{x^2} dy dx = \frac{1}{3} $

I have problems for determinate the new region, I know that if I get the inverse function of $g(u,v)$, then I can map the vertices as follow

$$ g^{-1}(x,y) = \left( \frac{x}{2} + \frac{y}{2} - \frac{1}{2} , \frac{-x}{2} + \frac{y}{2} + \frac{1}{2} \right) $$

$$ g^{-1}(0,0) = \left(\frac{-1}{2}, \frac{1}{2} \right), \ g^{-1}(1,1) = \left(\frac{1}{2}, \frac{1}{2} \right), \ g^{-1}(1,0) = (0,0) $$

But, how I determine the ranges of $u$ and $v$? for the integral $\int\int J_g(u,v) du dv$

答:
$0 \leq x \leq 1, 0 \leq y \leq x^2$

Transformation $x = u - v + 1, y = u + v$

$J = \begin{vmatrix} 1 & -1 \\ 1 & 1\end {vmatrix} = \, 2$

For the upper bound of $v$,

$y = x^2 \implies u + v = (u-v+1)^2$ (which is equation of another parabola)

$v^2 - (2u+3)v + (u^2+u+1) = 0$ which is a quadratic in $v$.

$(v - \frac{1}{2}\big(2u+3)\big)^2 = \big(\frac{2u+3}{2}\big)^2 - (u^2+u+1)$

$v = \frac{1}{2} (2u+3 \, \pm \sqrt{8u+5})$. As we have to find area below the parabola, we take the lower part of the curve.

So your integral becomes,

$\displaystyle \int_{-1/2}^0 \int_{-u}^{\frac{1}{2} (2u+3 - \sqrt{8u+5})} 2 \,dv \, du + \int_0^{1/2} \int_u^{\frac{1}{2} (2u+3 - \sqrt{8u+5})} 2 \,dv \, du = \frac{1}{3}$


答:
I'll suggest you hints how to do it in a way-

Product of roots is c/a. Use componendo-dividendo to express lambda in terms of c and a.

Sum of roots is -b/a. Substitute lambda in this from above to get an equation in a, b and c.

Simplify this equation in the form $(a+b+c)^2=..... $

Title: Problem of relations between the coefficients of a quadratic equation.how to proceed?

Body:
I have tried several ways to solve this problem of using the root properties of a second degree equation but I can't come up with anything favorable.

If the roots of $ax^{2}+bx+c=0$  are of the form  $\dfrac{\lambda }{\lambda -1}$ and  $\dfrac{\lambda+1 }{\lambda}$, then the value $(a+b+c)^{2}$ in terms of $a$, $b$, $c$ is?

Options

$(a) \quad b^{2}-4ac$

$(b) \quad b^{2}-2ac$

$(c) \quad 2b^{2}-ac$

$(d) \quad 4b^{2}-2ac$

$(e) \quad b^{2}-4abc$


问:
Do we have to use Bernoulli polynomials in the Euler-Maclaurin summation formula?
It may be that I have not picked up the proof, but I cannot see where the third condition of Bernoulli polynomials, given below, is used in the derivation of the Euler-Maclaurin summation formula.

The Bernoulli polynomials are defined inductively by
$$
b_0(x) = 1,
$$
$$
b_n'(x) = nb_{n-1}(x) \:\text{ and }\: \int_0^1 b_n(x) = 0 \:\text{  for  }\: n \geq 1.
$$

Suppose instead that we defined the alternative Bernoulli polynomials as follows,
$$
a_0(x) = 1,
$$
$$
a_n'(x) = na_{n-1}(x) \:\text{ and }\: a_n(0) = 0 \:\text{  for  }\: n \geq 1.
$$
so that the new third condition essentially gives $a_n(x) = x^n$ for all $n$, thus simplifying matters. Then the periodic Bernoulli functions used in the proof would be $A_n(x) = a_n(\{x\})$, where $\{x\}$ denotes the fractional part of $x$.

Using these functions, can anyone show me where the proof of Euler-Maclaurin breaks down? (Assume we are using the elementary proof by induction that is explained in detail on Wikipedia, for example.)

答: The problem is that your functions don't have $a_n(0)=a_n(1)$ for $n>1$. The proof you linked to uses the fact that $B_1(0)=-B_1(1)$ and that $B_n(0)=B_n(1)$ for $n>1$; this is why the sum of the function values appears in the integration by parts for $n=1$ but only boundary terms appear in the further integrations by parts for $n>1$.</p>


问:
How can we show $\left\|g\ast f\right\|_{L^p}\le\left\|g\right\|_{L^1}\left\|f\right\|_{L^p}$?
Let $p\ge1$, $f\in L^p(\mathbb R^d)$ and $g\in L^1(\mathbb R^d)$. I've read that the inequality $$\left\|g\ast f\right\|_{L^p}\le\left\|g\right\|_{L^1}\left\|f\right\|_{L^p}\tag1$$ would follow from applying the Minkowski inequality to $$F(x,y):=g(y)f(x-y)$$ and using that $$\left\|\lambda_yf\right\|_{L^p}=\left\|f\right\|_{L^p}\tag2,$$ where $(\lambda_yf)(x):=f(x-y)$.

However, I absolutely don't get why $(1)$ does hold. How can we show it

The "usual" version of Young's inequality, as stated on Wikipdia, is clear to me.

答: $\|\int f(x-y)g(y)dy\|_p \leq \int \|f(x-y)g(y)\|_p dy$ where the norm inside the integral is w.r.t. the variable $x$. Since $g(y)$ is a constant we get $\|\int f(x-y)g(y)dy\|_p \leq  \int |g(y)| \|f(x-y)\|_p dy=\|f\|_p \int |g(y)|dy=\|f\|_p \|g\|_1$.


问:
Divergence of $\zeta(z)$ tamed or not tamed by any analytic continuation
We know the conjecture about the Riemann hypothesis is about the nontrivial zeros are on
$$(1/2 + r i)$$
for some $r \in \mathbb{R}$ of Riemann zeta function.

My question is how many divergences of $\zeta(z)$ can be tamed by analytic continuations and how many divergences of $\zeta(z)$ canNOT be tamed by analytic continuations?

For example,
$$\zeta(1)=1+2+3+4+\dots=\infty$$
seem not to be tamed by any analytic continuation. How about other $\zeta(z)$? Are there others cannot be done by analytic continuation as a finite value?

答:
The series $\sum_{n\ge 1}n^{-s}$ diverges whenever $\Re(s)\le 1$,

Iff $s=1$ or $\Re(s)<1$ then $\lim_{N\to \infty}|\sum_{n\ge N}n^{-s}|= \infty$.

For $\Re(s)\le 1$,$s\not\in \Bbb{R}$ then $\frac{\sum_{n\ge N}n^{-s}}{|\sum_{n\ge N}n^{-s}|}$ oscillates.

For $\Re(s)\le 1,s\ne 1$
$$\lim_{N\to \infty} \frac{\sum_{n\ge N}n^{-s}}{\int_1^N x^{-s}dx}=\lim_{N\to \infty} \frac{1-s}{N^{1-s}}\sum_{n\ge N}n^{-s}=1$$

It is mostly similar for $\eta(s)$ with $\Re(s)=0$ instead of $\Re(s)=1$.

For the analytic continuation: you really have to compare with $\frac1{1-s}$ which is the analytic continuation of $\sum_{n\ge 0} s^n,|s|<1$ to $\Bbb{C}-\{1\}$, the setting is exactly the same for $\zeta(s)$.

Next, consider the functional equation saying that $\pi^{-(1/2+s)/2}\Gamma((1/2+s)/2)\zeta(1/2+s)$ is an even function which is real-valued for $s$ on the imaginary axis, thus having a zero at each sign change (the non-trivial zeros, under the RH).


答:
I believe the following works:

Here is another way:

$f(1)-f(0)=2$   $+$

$f(2)-f(1)=2$    $+$

$\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot$

$f(n)-f(n-1)=2$
$\implies f(n)-f(0) = 2n \implies f(n)=2n+f(0)$ for all $n \in \mathbb{N}$ where $n$ is extended to $\mathbb{Z}$ using a similar argument where one starts at $f(0)-f(-1)=2$ and so on.

Since $f$ is assumed to be linear we must connect each $[x,x+1]$ interval from endpoint to endpoint with a line (which can be done in one unique way), and this extends the domain of $f$ to $\mathbb{R}$.

Title: What functions have $f (t + 1)=f (t)+ 2$?

Body:
The functions have to be linear. Do I have to find the functions that satisfy the condition or is there a specific name for functions that satisfy these conditions. This question is the $26$th question in Strang's calculus at section $1.1$, If it is to find functions that satisfy these conditions, how do I do it?

Thinking without any math procedures, I find that if you take $f(t) = 2t$, then $f(t+1)$ will be $2t+2$ which will be same as $f(t) + 2 = 2t + 2$.

Are there any more functions that will satisfy these? Is there any procedure to find these or we can find this only through trial and error?


问:
A Negative binomial problem $P(X \ge 5)$ equals?

Consider a sequence of independent Bernoulli trials with the
probability of success in each trial being $\dfrac{1}{3}$. Let $X$
denote the number of trials required to get the second success. Then
$P(X \ge 5)$ equals.

$A=\dfrac{3}{7}$

$B=\dfrac{16}{27}$

$C=\dfrac{2}{3}$

$D=\dfrac{9}{13}$

This is a problem of negative binomial $r=2$

$P(X=x)= {x+r-1 \choose r-1}p^rq^x ;\ \ x=0,1,2..$

$P(X\ge5)=1-P(X < 5) \implies 1-(P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4))$

This is very time consuming how do I save time on this problem.

How do I utilize CDF of negative binomial distribution in this problem?

${\displaystyle k\mapsto 1-I_{p}(k+1,\,r),}$ the regularized incomplete beta function

答:
Another way to think about the question is to observe that if $X \ge 5$, that means that at least $5$ trials are needed to observe the second success.  This is equivalent to saying that no more than $1$ success occurs in the first four trials.

So there are two cases:  $0$ successes in four trials, or exactly $1$ success in $4$ trials.

In the first case, the probability is $(1-p)^4$.  In the second, the probability is $\binom{4}{1} p (1-p)^3$.  So the total probability is $$(1-p)^4 + 4p(1-p)^3$$ where $p = 1/3$.


问:
How to solve multiple inequalities?
$$100!=K(2!)^x(5!)^y(7!)^z$$
Find the maximum value of $x+y+z$.

I tried to open all the brackets and I got $$100!=K(2)^{x+3y+4z}(3)^{y+2z}(5)^{y+z}(7)^{z}$$

By using greatest integer function, I got the following inequalities:
$$x+3y+4z\le97$$ $$y+2z\le48$$ $$y+z\le24$$ $$z\le16$$
By adding the last two, I got $$y+2z\le40$$
Now I can't figure out how to proceed from here.

Edit: $x, y$ and $z$ are non-negative integers.

答:
If $x,y,z$ are nonnegative integers, as I suppose, then the answer is $97$, achieved when $x=97, y=z=0$.

Just look at the first inequality.  If we decrease $z$ by $1$, we can increase $x$ by $4$, giving a net increase of $3$ to $x+y+z$.  Similarly, if we decrease $y$ by $1$, we can increase $x+y+z$ by $2$.

Title: How to solve multiple inequalities?


问:
Inspection method doubt
how do I solve (y^2 + y + x) dy - ydx = 0 using inspection method?
I reached this point but I'm not sure how to proceed because when I integrate, my answer isn't correct</p>

Title: Inspection method doubt

Body:
$$(y^2 + y + x) dy - ydx = 0 \implies (y^2+y)dy = (ydx-xdy)$$

$$ (1+1/y)dy = (ydx-xdy)/y^2$$

$$y+\ln y = x/y + C$$


问:
Properties, bounds and limits about difference of two inverse standard normal CDF variables and extreme value distribution
I'm interested in the variable: $$\sigma_n=\Phi^{-1}\left(1-{1\over n}e^{-1}\right)-\Phi^{-1}\left(1-{1\over n}\right),$$ where $\Phi(\cdot)$ is the CDF of standard normal distribution. I want to prove $\sigma_n$ is monotonically decreasing for $n\ge 2$, which is shown by my simulation, but I do not know how to prove it. If this property holds, then it implies that $\sigma_n$ also has a non-negative limit as $\sigma_n>0$ for all $n\ge 2$.

Or you can just tell me $\sigma_n$ is bounded or has a finite limit.

More info:

$\sigma_n$ comes from the distribution $GEV(x;\mu_n,\sigma_n,0)$, which is the distribution of the max of $n$ i.i.d. standard normal distribution (Extreme value Type I distribution). See https://en.wikipedia.org/wiki/Generalized_extreme_value_distribution.

My simulation shows:

when $n$ increases from 2 to 2000, $\sigma_n$ decreases from 0.9 to 0.27

$n=10, 10^2, ..., 10^7$, $\sigma_n=0.90,0.51,0.35,0.29,0.25,0.22,0.20,0.18$, respectively

Body:
We have to start
$$\Phi^{-1}\left(1-\frac{x}{n}\right) = \sqrt{2}\ \text{erf}^{-1}\left(1-\frac{2x}{n}\right)$$ which makes
$$\sigma_n=\sqrt{2}\Bigg[\text{erf}^{-1}\left(1-\frac{1}{en}\right)-\text{erf}^{-1}\left(1-\frac{1}{n}\right) \Bigg]$$

For small $x$ we have
$$\text{erf}^{-1}\left(1-x\right)=\sqrt{\frac{1}{2} \left(\log \left(\frac{2}{\pi  x^2}\right)-\log \left(\log
   \left(\frac{2}{\pi  x^2}\right)\right)\right)}$$ (have a look here). It is very good for $0\leq x \leq 0.1$.

Using  it, we have
$$\sigma_n=\sqrt{\log \left(\frac{2 e^2 n^2}{\pi }\right)-\log \left(\log \left(\frac{2 e^2 n^2}{\pi
   }\right)\right)}-$$ $$\sqrt{\log \left(\frac{2 n^2}{\pi }\right)-\log \left(\log
   \left(\frac{2 n^2}{\pi }\right)\right)}$$ and the limit is $0$.

Making $n=10^k$, the table contains the approximate and exact values of $\sigma_n$
$$\left(
\begin{array}{ccc}
k & \text{approximation} & \text{exact} \\
 1 & 0.430284 & 0.443256 \\
 2 & 0.328501 & 0.328637 \\
 3 & 0.272163 & 0.271588 \\
 4 & 0.236707 & 0.236187 \\
 5 & 0.211987 & 0.211577 \\
 6 & 0.193551 & 0.193231 \\
 7 & 0.179145 & 0.178892 \\
 8 & 0.167499 & 0.167296 \\
 9 & 0.157836 & 0.157671 \\
 10 & 0.149656 & 0.149518 \\
 11 & 0.142615 & 0.142498 \\
 12 & 0.136474 & 0.136354 \\
 13 & 0.131056 & 0.131152 \\
 14 & 0.126231 & 0.126569 \\
 15 & 0.121898 & 0.133751
\end{array}
\right)$$


问:
Is a torus a subset of $\mathbb{R}^3$ or $\mathbb{R}^4$?
Defining a torus $T$ as $S^1 \times S^1$, it should follow that $T \subseteq \mathbb{R}^4$.   But you can also think of a torus as a bagel, which means it's a subset of $\mathbb{R}^3$.  Can anyone clarify this point?</p>

答:
Neither!  A torus is an abstract topological space defined, as you say, to be the product $S^1 \times S^1$.  Since, as it happens, $S^1 \subset \mathbb{R}^2$, we get a natural embedding of "the" torus into $\mathbb{R}^4$.  However, there are of course other ways of representing this space; one can change coordinates, apply functions to it, and so on.  And among these alterations are some that "project" the torus down into $\mathbb{R}^3$, where it resembles a bagel.

This is a little bit of an aberration, since not every two-dimensional manifold can be embedded in $\mathbb{R}^3$, the Klein bottle being a counterexample (also, this is related to your question in that it is more or less a twisted torus).  And more generally, an $n$-dimensional manifold can be embedded in $\mathbb{R}^{2n}$ but not necessarily lower.  So if you insist on thinking of a manifold as "being" a subset of some $\mathbb{R}^N$, that's the one to choose.


问:
Show that $f(x) = \sum_{n=1}^{\infty} c_n I(x-x_n)$ defines a bounded function on $\mathbb{R}$ that is continuous except, possibly, at the $x_n$
Define $I(x)=0$ $x\le 0$ and $I(x)=1$ $x>1$. Given sequences $(x_n)$ and $(c_n)$ in $\mathbb{R}$, with $\sum_{n=1}^{\infty} |c_n| < \infty$, show that $f(x) = \sum_{n=1}^{\infty} c_n I(x-x_n)$ defines a bounded function on $\mathbb{R}$ that is continuous except, possibly, at the $x_n$

I has been trying to prove that the function is continuous. The problem is when I take the term $|f(x) - f(y)|$ to make it less the $\epsilon$. Could you please show me how to prove this problem.

答:
Let $x\not= x_n,\,\,\, \forall n \in \mathbb{N}$. So we need to show that $f(x) = \sum_{n=1}^{\infty} c_n I(x-x_n)$ is continuous for every $x$ that located between any two points of the sequence $x_n$. Since the step function is continuous on every point $x \not x_n$, because;

For each $x \in (x_{n-1},x_{n})$ or $x \in (x_n, x_{n+1})$, where $n \in \mathbb{N}$, Let $\epsilon >0$, choose $\delta_0 = x_n -x_{n-1} >0$ and $\delta_1 = x_{n+1}- x_n >0$ such that

if $|x-y| < \delta_0$ then $|I(x-x_n) - I(y-x_n) =|0-0|=0 <\epsilon$.
and
if $|x-y| < \delta_1$ then $|I(x-x_n) - I(y-x_n) =|1-1| <\epsilon$.

Since the function $I(x-x_n)$ is continuous for any $x \not=x_n$, then $ c_nI(x-x_n)$ is continuous too (means there is $\delta_n$ for every $n$).

Now, doing the steps, Mr. @GEdgar showed in his comments.

Choose  $1<N \in \mathbb{N}$, so large such that $\sum_{n=N+1}^{\infty} |c_n| \le \frac{\epsilon}{2}.$ Since (As we showed) $c_n I(x -x_n)$ is continuous on any $x \not= x_n$ for all $n \in \mathbb{N}$. consequently, it is continuous on all $x \not = x_n,\,\,\, n\in \{ 1,...,N\}$. Choose $\delta_n$ so small such that

if $|x-y|< \delta_n$ then $|c_n I(x -x_n) - c_n I (y-x_n)| < \frac{\epsilon}{2 N}$.

let $\delta_n = \min \{ \delta_1,...,\delta_N \}$, so

if $|x-y| < \delta$ then

$|f(x)-f(y)|= |\sum_{n=1}^{\infty} c_n I(x-x_n) - \sum_{n=1}^{\infty} c_n I(y-x_n)| $
$\le \sum_{n=1}^{\infty} |c_n I(x-x_n) c_n I(y-x_n)| $

$\le \sum_{n=1}^{N} \frac{\epsilon}{2 N} + \sum_{n=N+1}^{\infty} |c_n|$
$< \frac{\epsilon}{2} + \frac{\epsilon}{2}= \epsilon.$


答:
Easy is $B:\frac{16}{27}$

$$P(X\geq 5)=1-P(X\leq 4)=1-\binom{2-1}{2-1}\Bigg(\frac{1}{3}\Bigg)^2\cdot\Bigg(\frac{2}{3}\Bigg)^0-\binom{3-1}{2-1}\Bigg(\frac{1}{3}\Bigg)^2\cdot\Bigg(\frac{2}{3}\Bigg)^1-\binom{4-1}{2-1}\Bigg(\frac{1}{3}\Bigg)^2\cdot\Bigg(\frac{2}{3}\Bigg)^2=$$

$$=1-\frac{3}{27}-\frac{4}{27}-\frac{4}{27}=\frac{16}{27}$$

Note that there are at least 2 parametrizations of NBinomial. I used the one counting the trials before s successes whose pmf is

$$P(X=n)=\binom{n-1}{s-1}\theta^s(1-\theta)^{n-s}$$

Where the support is $n=s,s+1,s+2,...$

Title: A Negative binomial problem $P(X \ge 5)$ equals?

Body:
Consider a sequence of independent Bernoulli trials with the
probability of success in each trial being $\dfrac{1}{3}$. Let $X$
denote the number of trials required to get the second success. Then
$P(X \ge 5)$ equals.

$A=\dfrac{3}{7}$

$B=\dfrac{16}{27}$

$C=\dfrac{2}{3}$

$D=\dfrac{9}{13}$

This is a problem of negative binomial $r=2$

$P(X=x)= {x+r-1 \choose r-1}p^rq^x ;\ \ x=0,1,2..$

$P(X\ge5)=1-P(X < 5) \implies 1-(P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4))$

This is very time consuming how do I save time on this problem.

How do I utilize CDF of negative binomial distribution in this problem?

${\displaystyle k\mapsto 1-I_{p}(k+1,\,r),}$ the regularized incomplete beta function


问:
How to solve decics like $x^{10}+100x^2+160x+64=0$ having Galois group 10T33?
Using the approach described in Smart way to solve octics like $x^8+5992704x-304129728=0$ (the method DecomPoly available in GAP) the decic quadrinomial from this question can be decomposed into two quintics:
$$x^{10}+100x^2+160x+64=(x^5+10ix+8i)(x^5-10ix-8i)$$
where $i$ is the imaginary unit. But here I got stuck. How to get explicit formula for roots of those solvable quintic trinomials?

I have tried to use the RadiRoot's method RootsOfPolynomialAsRadicals but without success. It seems Galois group 10T33 of the original quadrinomial is too big, and the method does not accept (quintic) polynomials with complex coefficients.

I am afraid the answer will be too complicated for me, but still have hope there is some tool / built-in method that can run calculations without too deep digging into Galois theory...

答:
Since both quintic factors are in Bring–Jerrard normal form, Malfatti's method (described here) may be applied. Consider the following mpmath program (so all roots are principal unless otherwise stated):

```<code>#!/usr/bin/env python3
from mpmath import *

i = root(-1, 2, s)
a = sqrt(1+2*i)
l1 = [1, 4*(-2 + i + a - i*a)/5, 4*(-7*a + i*a)/3125]
z1 = root(polyroots(l1)[0], 5, n1)
z2 = root(polyroots(l1)[1], 5, n2)
l2 = [1, 4*(-2 + i - a + i*a)/5, 4*(+7*a - i*a)/3125]
z3 = root(polyroots(l2)[0], 5, n3)
z4 = root(polyroots(l2)[1], 5, n4)
r = z1 + z2 + z3 + z4
print(polyval([1, 0, 0, 0, 10*i, 8*i], r))
</code>```

Then for $s=0,1$ the $n_i$ must be set to these values so that $r$ will end up being a root of the quintic beneath it:
$$s=0:\begin{array}{cccc}
n_1&n_2&n_3&n_4\\
0&4&1&0\\
1&3&4&2\\
2&2&2&4\\
3&1&0&1\\
4&0&3&3\end{array}\qquad s=1:\begin{array}{cccc}n_1&n_2&n_3&n_4\\
0&1&4&0\\
1&0&2&2\\
2&4&0&4\\
3&3&3&1\\
4&2&1&3\end{array}$$
Since it is easy to solve a quadratic, this leads to explicit reproducible formulas for all the initial decic's roots.

Body:
Since both quintic factors are in Bring–Jerrard normal form, Malfatti's method (described here) may be applied. Consider the following mpmath program (so all roots are principal unless otherwise stated):

#!/usr/bin/env python3
from mpmath import *

i = root(-1, 2, s)
a = sqrt(1+2*i)
l1 = [1, 4*(-2 + i + a - i*a)/5, 4*(-7*a + i*a)/3125]
z1 = root(polyroots(l1)[0], 5, n1)
z2 = root(polyroots(l1)[1], 5, n2)
l2 = [1, 4*(-2 + i - a + i*a)/5, 4*(+7*a - i*a)/3125]
z3 = root(polyroots(l2)[0], 5, n3)
z4 = root(polyroots(l2)[1], 5, n4)
r = z1 + z2 + z3 + z4
print(polyval([1, 0, 0, 0, 10*i, 8*i], r))

Then for $s=0,1$ the $n_i$ must be set to these values so that $r$ will end up being a root of the quintic beneath it:
$$s=0:\begin{array}{cccc}
n_1&n_2&n_3&n_4\\
0&4&1&0\\
1&3&4&2\\
2&2&2&4\\
3&1&0&1\\
4&0&3&3\end{array}\qquad s=1:\begin{array}{cccc}n_1&n_2&n_3&n_4\\
0&1&4&0\\
1&0&2&2\\
2&4&0&4\\
3&3&3&1\\
4&2&1&3\end{array}$$
Since it is easy to solve a quadratic, this leads to explicit reproducible formulas for all the initial decic's roots.


问:
How comes that integrating factor = $1/(x\,y)$
I'm just curious how the integrating factor m(x,y) = $\frac{1}{x\,y}$ of $\frac{\mathrm{d}}{\mathrm{d}x}f(x,y) = x+y-\frac{x^2}{y}\,y'(x)$ was determined. It was just given. When I was introduced to this topic this week, my fist impression was that integrating factor just can depend on x or y. Because determinations like $m(x) = \exp(\int \alpha(x)\mathrm{d}x)$ whereas $\alpha(x) = \frac{\partial_y\partial_x(f(x,y))-\partial_x\partial_y(f(x,y))}{\partial_xf(x,y)}$ or such just seem to work this way. So loosely speaking: Are there other methods?

答:
To add to what Oscar said, you don't necessarily have to guess the form of your integrating factor, but if you guess it right it will expedite your process. I'll lay out what I do when I'm trying to find an integrating factor.
To find the integrating factor for the expression
\begin{align}
f(x,y)+g(x,y)\frac{\mathrm{d}y}{\mathrm{d}x},
\end{align}
multiply by $m(x,y)$,
\begin{align}
m(x,y)f+m(x,y)g\frac{\mathrm{d}y}{\mathrm{d}x}
\end{align}
and assume that our equation is now a perfect differential of some function $\lambda$,
\begin{align}
mf+mg\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}\lambda(x,y).
\end{align}
Writing out the chain rule for $\lambda$ gives
\begin{align}
mf+mg\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\partial\lambda}{\partial x}+\frac{\partial\lambda}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}x}.
\end{align}
Giving two equations,
\begin{align}
mf=\frac{\partial\lambda}{\partial x},\\
mg=\frac{\partial\lambda}{\partial y}.
\end{align}
Assuming symmetry of second derivatives,
\begin{align}
\frac{\partial}{\partial y}mf&=\frac{\partial}{\partial x}mg\\
\frac{\partial m}{\partial y}f+m\frac{\partial f}{\partial y}&=\frac{\partial m}{\partial x}g+m\frac{\partial g}{\partial x}.
\end{align}
Rearranging gives the linear PDE
\begin{align}
f(x,y)\frac{\partial m}{\partial y}-g(x,y)\frac{\partial m}{\partial x}+\left(\frac{\partial f}{\partial y}-\frac{\partial g}{\partial x}\right)m(x,y)=0.
\end{align}
This PDE can be solved for the desired integrating factor, which in this specific case can be done via separation of variables.


问:
Arrangement of singular values in the SVD of A^{-1}
I know that the SVD of the inverse of A may be expressed as:
$$
A^{-1} = (U\Sigma V^T)^{-1} = (V^T)^{-1}\Sigma^{-1}U^{-1},
$$
where the inverse of a diagonal matrix $\Sigma$ is just the reciprocal of each diagonal element. But if this is the case, then the diagonal elements of $\Sigma^{-1}$ (the singular values) are in ascending order. I thought they should be arranged in descending order for the SVD. Is this still considered a valid SVD? Or is it only valid with the correct arrangement of singular values?

P.S. For motivation, most textbooks speak of the $\sigma_1$ as the max singular value of the SVD, whereas of course here $\sigma_1$ is the minimum value. I haven't found any other questions on SE relating to the specific arrangement of singular values in the SVD of the inverse of A.

答:
In my view, a singular value decomposition just requires the diagonal entries of $\Sigma$ to be nonnegative, so $A^{-1} = V \Sigma^{-1} U^\top$ does indeed constitute a singular value decomposition. This is also the view Wikipedia takes, though some authors may adopt a different definition. If this is something you might see on a true/false question on an exam, I would certainly consult the reference being used in your particular course to see what their official definition of an SVD is.

The ordering of the singular values in descending order is a useful convention, which is worth enforcing if one does not have a good reason not to. Wanting to order the singular values in ascending order for the inverse of a matrix so the SVD of a matrix and its inverse is certainly a good reason to loosen this convention. In any kind of mathematical exposition, it is helpful to make explicit note of any of these convention issues to avoid possible confusion with the reader.

Certainly by reversing the order of the columns in $U$ and $V$ and the diagonal entries of $\Sigma$ to matrices $\tilde{U}$, $\tilde{V}$, and $\tilde{\Sigma}$, one easily converts the (reverse-ordered) SVD of $A^{-1} = V \Sigma^{-1} U^\top$ to an equivalent SVD $A^{-1} = \tilde{V} \tilde{\Sigma}^{-1} \tilde{U}^\top$ with singular values in descending order.


问:
Can an algebraic structure have indistinguishable elements?
Sometimes, a topological space has indistinguishable points - we call those spaces non-$T_0$. But given such a space, we can always identify indistinguishable points, thereby yielding a $T_0$ space. (Technically, we've taken the Kolomogorov quotient).

Does this sort of thing ever happen in abstract algebra?

Here's two more examples.

A preordered set can have comparable, distinct points - in other words it can fail to be antisymmetric. But that's cool, we can identify comparable points to obtain a partially ordered set.

Sometimes a pseudometric space has distinct points that are zero distance apart. But that's okay, we can identify zero-distance points to obtain a metric space.

Edit: It would be nice to see a definition of 'indistinguishable' for the elements of arbitrary structures. It would then be a consequence of this more general definition that for an arbitrary preordered set $X$ (order relation $\leq$) it holds that $x,y \in X$ are indistinguishable iff $x \leq y$ and $y \leq x$.

Here's an example. Consider the function $f : \mathbb{N} \rightarrow \mathbb{N}$, with $f(n)=0$ for all $n \in \mathbb{N}$. The associated notion of indistinguishability for the structure $(\mathbb{N},f)$ should probably be the relation $\sim$ such that $a \sim b$ iff both $a$ and $b$ equal $0$, or both $a$ and $b$ are distinct from $0$.

Edit2: On the other hand, perhaps it does not make sense to speak of 'the natural notion of indistinguishability in a structure $X$' without first situating that structure in a category. After all, if we're going to quotient out by the indistinguishability relation, epimorphisms will probably show up at some point.

答:
Yes.

add to a ring some null elements $\eta_i$ with $\eta_i x = 0$ for all $x$.

add extra coordinates $X_i$ and equations $X_i = 0$ to the presentation of an algebraic variety by equations.

for an algebraic structure that has a notion of representation, consider elements that act the same in all representations as indistinguishable.

The quotient implied in the last example is to prevent stupidity like taking a finite ring, adjoining uncountably many null elements, and thinking of that as a "big" structure.


问:
What are the positive/negative intertial exponent of the quadratic form $q(A)=tr(A^2)$ for $A\in \Bbb R^{n\times n}$.
What are the positive/negative intertial exponent of the quadratic form $q(A)=tr(A^2)$ for $A\in \Bbb R^{n\times n}$.

Choosing the canonical base $E_{ij}$, then ...

答: Let $\mathcal H$ and $\mathcal K$ be respectively the subspace of all symmetric matrices and the subspace of all skew symmetric matrices. Then $q$ is positive on $\mathcal H\setminus0$ and negative on $\mathcal K\setminus0$. Also, $\mathbb R^{n\times n}=\mathcal H\oplus\mathcal K$. Therefore $n_+(q)=\dim\mathcal H$ and $n_-(q)=\dim\mathcal K$.


问:
How to evaluate $\sum_{k=1}^{n} \frac{f^3_k}{3(f^2_k-f_k)+1}, ~~f_k=k/n$
Very often the summations at pre-degree level are done by using differencing-telescoping and by some  symmetry properties. The following summation is fabricated keeping one thing in mind, I may tell it later. $$\sum_{k=1}^{n} \frac{f^3_k}{3(f^2_k-f_k)+1},~~ f_k=k/n.$$

The question is: How will you evaluate this sum?

Body:
There a many ways to do it, I presume.
$$S_n=\sum_{k=1}^n \frac{k^3}{3 k^2 n-3  n^2k+n^3}=\sum_{k=1}^n \frac{k^3}{3(k-a)(k-b) }$$ where
$$a=\frac{n}{6} \left(3-i \sqrt{3}\right)\qquad \text{and} \qquad b=\frac{n}{6} \left(3+i \sqrt{3}\right)$$
$$S_n=\sum_{k=1}^n \Bigg[\frac{a^3}{3 (a-b) (k-a)}-\frac{b^3}{3 (a-b) (k-b)}+\frac{a+b}{3}+\frac{k}{3}\Bigg]$$
$$S_n=\frac{2 a^3 \left(H_{n-a}-H_{-a}\right)+n (a-b) (2 a+2 b+n+1)+2 b^3
   \left(H_{-b}-H_{n-b}\right)}{6 (a-b)}$$
$$S_n=\frac{1}{18} \left(9 n+9+\frac{4 \pi  n \sin (\pi  n)}{\cos (\pi  n)-\cosh
   \left(\frac{\pi  n}{\sqrt{3}}\right)}\right)=\frac{n+1}2$$


问:
If $A$ and $B$ are sets, do $\mathcal{P}(A - B)$ and $\mathcal{P}(A) - \mathcal{P}(B)$ equal?
Question:

If $A$ and $B$ are sets, do $\mathcal{P}(A - B)$ and $\mathcal{P}(A) - \mathcal{P}(B)$ equal?

My Attempt:

My instinct tells me that $\mathcal{P}(A - B) \neq \mathcal{P}(A) - \mathcal{P}(B)$ always holds.

So I made some examples to help myself think.

(1) If $A = B \neq \varnothing$, then $A - B = \varnothing$, hence $\mathcal{P}(A - B) = \mathcal{P}(\varnothing) = \{ \varnothing \}$. Meanwhile, $\mathcal{P}(A) - \mathcal{P}(B) = \mathcal{P}(A) - \mathcal{P}(A) = \varnothing$. Obviously, $\varnothing \neq \{ \varnothing \}$.

(2) If $A \neq \varnothing$ but $B = \varnothing$, then $A - B = A$, $\mathcal{P}(A - B) = \mathcal{P}(A) = \{ \varnothing,\, A,\, \ldots{} \}$, while $\mathcal{P}(A) - \mathcal{P}(B) = \mathcal{P}(A) - \mathcal{P}(\varnothing) = \{ \varnothing,\, A,\, \ldots{} \} - \{ \varnothing \} = \{ A,\, \ldots{} \}$.

I don't know what to do next. Please help me proceed. Thx in advance. :)

答:
It seems like you have everything you need to finish the problem, but I want to make it more clear:

$P(A-B)$ represents the set of all subsets of $A-B$ which implies that $\emptyset \in P(A-B)$ for all sets $A,B$.

But $P(A)-P(B) = \{\emptyset,...,A,...\}-\{\emptyset,...,B,...\} \implies \emptyset \notin P(A)-P(B).$

Therefore $P(A-B) \neq P(A)-P(B)$ for all sets $A,B$.


答: These are two different embeddings of the same topological space, i.e. the surface of the bagel in $\mathbb R^3$ and the product of two circles in $\mathbb R^4$ are homeomorphic. Note that they are not isometric though; the torus embeded in $\mathbb R^4$ as a product of circles is flat, while the surface of the bagel has regions of positive curvature and of negative curvature.

Title: Is a torus a subset of $\mathbb{R}^3$ or $\mathbb{R}^4$?

Body: Defining a torus $T$ as $S^1 \times S^1$, it should follow that $T \subseteq \mathbb{R}^4$.   But you can also think of a torus as a bagel, which means it's a subset of $\mathbb{R}^3$.  Can anyone clarify this point?</p>


问:
Is $a$ bigger than $0$ or not?
Consider Goldbach's original conjecture (no need worry, because we don't talk about "Goldbach" itself):

every integer $n> 2$ could be detached as a sum of three primes. (In Goldbach's age, $1$ is regarded as a prime.)

For any $n>2$, we denote the statement above as $p(n)$. Then define

$$a_n:=\begin{cases}
0, &\text{if $p(n)$ is true},\\
1, &\text{otherwise}.
\end{cases}$$

Note that $a(n)$ is definite, since every $p(n)$ could be verified, at least by machine, as long $n$ is given. However, consider the decimal number $$a:=0.00a_3a_4a_5\cdots.$$

What's $a$? Is it a real number?

If $a$ is a real number, then we can compare the magnitude of $a$ and
$0$, since the real number field is ordered.

But we can't in fact. If $a=0$, then "Goldbach" is proven, and if not, then "Goldbach" is overturned. However, it's in suspense hitherto whether Goldbach is true or not.

Probably, we may say $a$ is indefinite, that's to say, we do not know what $a$ is yet. But a new question is coming about:

Why we know any digit but do not know the number itself ? In order
to know the number, what else we need ?

Further, let's compare $a$ with another classic number
$$\pi=3.1415926\cdots.$$

Indeed, we do not know all the digits of $\pi$ untill we compute every one. But we can safely say it's $\pi$. What's the exact difference between $a$ and $\pi$ on earth? Why we think we know $\pi$ but do not know $a$? In another word, what does "knowing a number" mean exactly? And in what sense can we say we know a number?

答:
What you have defined is a real number. Indeed, from the definition given it is also a computable number. Does $a=0$? Well, this is something we simply do not know. Of course, there is nothing wrong with not knowing a number from a definition. For example, the Ramsey number $R(6,6)$ is bounded between $102$ and $165$. However, we have no idea what this number is. Another example, define the number $x$ to be $0$ if the Riemann hypothesis is true and $1$ if it is false. This is a perfect definition, but I have no idea what $x$ is and if you can tell me you will get a million dollars.

To reiterate: One can write a perfectly sound mathematical definition and not know what that number actually is. This is not a paradox, but simply something that can arise naturally.

Title: Is $a$ bigger than $0$ or not?