问:
Solve the following equation if x, y and z are prime numbers
$x$, $y$ and $z$ are prime numbers. Solve the following equation:
$$105 (x + y + z) = xyz$$

Knowing that
$$105 = 3 × 5 × 7$$
it seems that there's no solution for this equation considering that $xyz$ has to be divisible by 3, 5 and 7 and the only prime numbers divisible by 3, 5 and 7 are 3, 5 and 7.

Thanks in advance!

答: For $x\in\mathbb{Z}$, let $\Omega(x)$ be the number of divisors of $x$ that can be written as $p^i$ where $p$ is a prime number and $i\geqslant 1$. Then if there is a solution for your equation, we have $\Omega(xyz)=\Omega(105(x+y+z))$, that is to say $3=3+\Omega(x+y+z)$ and thus $\Omega(x+y+z)=0$ which leads to the contradiction $x+y+z=1$. You can also, with your argument $105|xyz$, say that $\{x,y,z\}=\{3,5,7\}$ and thus $x+y+z=1$.


问:
Is the vector field $F(x,y)=(x+y,x^2)$ conservative?
let's consider the vector field $F(x,y)=(x+y,x^2)$. It's defined over the whole $\mathbb{R}^2$. We can see immediately that $$\partial_x F_2 = \partial_y F_1$$

iff $x = \frac{1}{2}$

Since the domain (the plane) is simply connected, by Poincarè's lemma we know that if the form is closed, then it admits a potential. But in this case, it seems that it can be conservative only for $x=\frac{1}{2}$ and $y \in \mathbb{R}$. What does this mean? Actually, by integration of

$$\partial_x U(x,y)=x+y$$
$$\partial_y U(x,y)= x^2$$
I wasn't able to find a potential $U(x,y)$ as I found something whose gradient is not the vector field itself.

So I'm suspecting that this vector field is not conservative: but how can I deduce it from the fact that the differential form is closed only on the line $(\frac{1}{2},y)$?

答:
You have to get $$2x = 2x$$.

Since you got: $$2x = 1$$ it is not conservative.

Here is a graph.  Note that if you go up via the left or right and come back down the middle you wouldn't get a zero vector for a closed loop.
desmos link

ContentLicense: CC BY-SA 4.0


问:
A closed and convex subset of an Hilbert space on which the norm does not attain its maximum
I want to prove that the norm $\lvert \lvert \cdot \rvert \rvert _2$ has no maximum on  $X=\{x\in \ell^2(k):\sum_{n \in \mathbb{N}} 2^{1/n}\lvert x_n \rvert^2 \le 1\}$, where $k=\mathbb{R}$ or $\mathbb{C}$.

I have managed to see that this is close and bounded, I can't figure out if this is also convex and I don't know if this would be helpful.

I've tried to do some contradiction argument, I have observed is that since $2^{1/n}>1$, if there is $x \in X$ which attains the maximum for $\lvert \lvert \cdot \rvert \rvert _2$, then this maximum, say $M$, must be strictly smaller than $1$, otherwise $x$ would be outside of $X$. Now my idea would be that in this case we could add a small $\epsilon$ to some term $x_n$ of $x$ in such a way that $\sum_{n \in \mathbb{N}} 2^{1/n}\lvert x_n \rvert^2$ stays below one, but I can't carry out the computations, even because if $\lvert \lvert x \rvert \rvert _2<1,$ then it does not follow that $\sum_{n \in \mathbb{N}} 2^{1/n}\lvert x_n \rvert^2<1$, as we can see taking the element $(2^{-1/4}, 0, 0, \ldots)$.

答:
TOO LONG FOR A COMMENT - NOT AN ANSWER.

The set $X$ is indeed convex, and this is interesting; see the remark at the bottom of this post.

Proof of convexity: Let $$
Lx=(2^{\frac1{2n}}x_n)_{n\in \mathbb N}, $$ and consider $x_1, x_2\in X$, which by definition means that $\langle Lx_j|x_j\rangle \le 1$ for $j=1, 2$.

For $\alpha, \beta>0,\alpha+\beta=1$, let $$x' = \alpha x_1+\beta x_2.$$
Then  $$\langle Lx'|x'\rangle\le \alpha^2+\beta^2+2\alpha\beta \langle Lx_1|x_2\rangle.$$
Now since $L$ is a positive definite operator, we have the Cauchy-Schwarz inequality
$$
\lvert \langle Lx_1|x_2\rangle \rvert\le \sqrt{\langle Lx_1|x_1\rangle\langle Lx_2|x_2\rangle}\le 1.$$
We conclude that
$$\langle Lx'|x'\rangle\le (\alpha+\beta)^2=1,$$
that is, $x'\in X$.

Remark. As I mentioned in a comment to deb's answer, the fact that $X$ is convex is interesting. Indeed, it is a standard theorem that the norm attains its minimum on closed convex sets. Here we see that the norm can fail to attain its maximum.


问:
Variance and expected value of $2^X$
I have 2 similar problems:

given $X \sim Po(3)$, find the variance of $2^X$

given $X \sim B\left(200, \frac{1}{9}\right)$, find the expected value of $2^{X+3}$

I'm not really certain what I'm expected to calculate should it be $Var(2^X)$ and $E[2^{X+3}]$ respectively or should it be $2^{Var(X)}$ and $2^{E[X+3]}$. If it is the second one then: for (1) $$X \sim Po(3)\Longrightarrow\lambda=3\Longrightarrow Var(X)=\lambda = 3 \Longrightarrow 2^{Var(X)}=2^3$$  and for (2) $$X \sim B\left(200, \frac{1}{9}\right)\Longrightarrow E[X+3]=\frac{200}{9} + 3\Longrightarrow 2^{E[X+3]}=2^{\frac{200}{9} + 3}$$

But if it's the first one I don't know how I can find the variance and expected value with the given information

after reading the answers I did for (1): $$ E[2^X]=\sum^\infty_{i=0}2^i\cdot e^{-\lambda}\frac{\lambda ^i}{i!} = e^{-\lambda}\sum^\infty_{i=0}\frac{(2\lambda)^i}{i!}=e^{-\lambda}\cdot e^{2\lambda} = e^\lambda=e^3$$
and same way
$$ E[(2^X)^2]=E[(4^X)]=\sum^\infty_{i=0}4^i\cdot e^{-\lambda}\frac{\lambda ^i}{i!} = e^{-\lambda}\sum^\infty_{i=0}\frac{(4\lambda)^i}{i!}=e^{-\lambda}\cdot e^{4\lambda} = e^{3\lambda}=e^9 $$
hence
$$ Var(X) = E[(2^X)^2] - (E[2^X])^2 = e^9 - (e^3)^2 = e^9 - e^6$$

as for (2): $$ E[2^X]=\sum^{200}_{i=0}2^i\binom{200}{i}(\frac{1}{9})^i(1-\frac{1}{9})^{200-i} = \sum^{200} _{i=0}\binom{200}{i}(\frac{2}{9})^i (\frac{8}{9})^{200-i} =(\frac{2}{9} + \frac{8}{9})^{200} = (\frac{10}{9})^{200} $$
so $$ E[2^{X+3}]=E[8\cdot 2^X]=8\cdot E[2^X]=8(\frac{10}{9})^{200} $$

答:
Find variance of $2^X$ for X ~ Poi(3).

First, what is $E(2^X)$? Intuitively, this is just $2^3$. Let’s check:

$$\sum_{k=0}^\infty2^k\frac{e^{-3}3^k}{k!} = \frac{e^{-3}}{e^{-6}}\sum_{k=0}^\infty\frac{e^{-6}6^k}{k!} = e^3.$$

Well, that was surprising. In fact, using the same calculations, $E(b^X)$ for X ~ Poi(a) is $\exp(ab-a)=\exp(a(b-1))$.

I invite you to use similar calculations to find $E((2^X)^2)=E(2^{2X})$ and finish the calculation. Feel free to post your solution as I’d be interested in the result.

Title: Variance and expected value of $2^X$


问:
Is it possible to find a null vector of matrix $A$ of size $m\times n$ when $m >> n$?
Is it possible to find a null vector (or null space) of matrix $A$ of size $m\times n$ when $m >> n$? or is it the case that in such a case, the null space does not exist?

Thank you,

答:
An $m\times n$ matrix is a linear transformation form ${\mathbb R}^n$ to ${\mathbb R}^m$:

the columns span the image; and

the null space is the orthogonal complement of the row space.

If the row space has dimension less than $n$,  then the null-space is nontrivial.


问:
Computing $\iiint_\mathbb{R^3} e^{-x^2-y^2-z^2}dxdydz$ using substitution
Consider this integral:
$$\iiint_\mathbb{R^3} e^{-x^2-y^2-z^2}dxdydz$$
How would you compute it?

I already solved this problem this way:
$$\iiint_\mathbb{R^3} e^{-x^2-y^2-z^2}dxdydz = \left( \int_\infty^\infty e^{-x^2} \right)^3 = \pi^{3/2}$$

But I wanted to find it using substitution (spherical coordinates) but this is all I could do:
$$\iiint_\mathbb{R^3} e^{-x^2-y^2-z^2}dxdydz = \lim_{j\,\rightarrow\infty}\int_0^jdu\int_0^\pi dv\int_0^{2\pi} e^{-u^2}u^2\sin(v) dw=$$

$$=2\pi\lim_{j\,\rightarrow\infty}\int_0^jdu\int_0^\pi e^{-u^2}u^2\sin(v)dv=4\pi\lim_{j\,\rightarrow\infty}\int_0^je^{-u^2}u^2du$$

But it doesn't get me anywhere. Help would be very appreciated, thanks in advance.

答:
To finish off your problem, you only need

$$\int_{0}^{\infty}e^{-x^2}x^2dx=\frac{\sqrt{\pi}}{4}$$

which can be shown by integrating

$$\int_{0}^{\infty}e^{-x^2}dx=\frac{\sqrt{\pi}}{2}$$

by parts.  That is, we have:

$$\frac{\sqrt{\pi}}{2}=\int_{0}^{\infty}e^{-x^2}dx=xe^{-x^2}\vert_{0}^{\infty}+2\int_{0}^{\infty}e^{-x^2}x^2dx=2\int_{0}^{\infty}e^{-x^2}x^2dx$$

ContentLicense: CC BY-SA 3.0


问:
Number Theory: Prove for all $1\leq n \in \mathbb{N}$ $F_n \mid 2^{F_n}-2$ with $F_n$ is Fermat numbers
Number Theory: Prove for all $1\leq n \in \mathbb{N}$ $F_n \mid 2^{F_n}-2$ with $F_n$ is Fermat numbers

Follow the following sections They will guide you in proof

Use polynomial division to prove that $2^{2^{n+1}}-1\mid2^{F_n-1}-1$.

Use polynomial division to prove that $F_n\mid2^{2^{n+1}}-1$.

Draw the conclusion from these two sections.

Attempt:

if $2^{2^{n+1}}-1$ divide $2^{F_n-1}-1$ and $F_n$ divide $2^{2^{n+1}}-1$ so $F_n$ divide $2^{F_n-1}-1$

and i have no idea how to make it happen.

Body:
Hints:

$2^{2^{2^n}}-1=(2^{2^{n+1}}-1)(2^{2^{2^n}-2^{n+1}}+2^{2^{2^n}-2\cdot2^{n+1}}+2^{2^{2^n}-3\cdot2^{n+1}}+\cdots+2^{{2^{n+1}}}+1)$

$2^{2^{n+1}}-1$ is the difference of two squares

if $a|b$ and $b|c$ then $a|2c$


问:
Modular automorphism group associated with a normal state on a von Neumann algebra
Let $\tau_1$ and $\tau_2$ be two normal states on a von Neumann algebra $M$. Denote $\sigma_t^{\tau_i}(i=1, 2)$ by the modular automorphism group associated with $\tau_i$. If $\sigma_t^{\tau_1}(x)=\sigma_t^{\tau_2}(x)$ for any $x\in M$, can we conclude that $\tau_1=\tau_2$?

What about the case when $M$ is a factor?

答:
If $M$ is finite but it is not a factor, you can construct two different traces. As the modular group of a trace is trivial, you get two distinct normal states with the same modular group.

If I remember correctly, the condition $\sigma^\varphi_t=\sigma^\psi_t$ is equivalent to the existence of invertible positive $h\eta  M_\psi\cap M_\varphi$ with $\psi=h\,\varphi$. What I don't remember is whether the centralizer can be non-trivial in a factor.


问:
Question about the definition of an algebra homomorphism
https://en.wikipedia.org/wiki/Algebra_over_a_field

I understand the operational definiton of an algebra homomorphism ($f$ is a K-linear ring homomorphism).
But I don't see how this definition is implied by the the fact that $ f \circ \eta_A=\eta_B$.

答:
The point of the map $\eta_A$ is that it defines the scalar multiplication on $A$ as a $K$-vector space. In particular, for $\lambda \in K$ and $a \in A$, we define scalar multiplication by
$$
\lambda\cdot a := \eta_A(\lambda)a,
$$
where the right hand side is just a product in the ring $A$. Now, for a map $f:A \to B$, the condition that $f \circ \eta_A = \eta_B$ means that for $\lambda\in K$ and $a \in A$, we have
$$
f(\lambda\cdot a) = f(\eta_A(\lambda)a) = f(\eta_A(\lambda))f(a) = \eta_B(\lambda)f(a) = \lambda\cdot f(a),
$$
so $f\circ \eta_A = \eta_B$ makes $f$ a $K$-linear transformation (the other properties of $K$-linear multiplication come from the fact that $f$ is a ring homomorphism).


问:
Understanding nonstandard Peano arithmetic
I've had the idea of nonstandard Peano arithmetic introduced to me in the comments of this question. The concept that we could write down the axioms which produce the natural numbers and also produce something else from them is totally crazy to me! I did not expect this.

I've been doing some reading here, to try and understand what these non-standard models of the Peano axioms are. It's difficult for me to follow the details of this paper, but I've understood that the most simple non-standard model of the Peano axioms is
$$
M = \{\mathbb{N}, +, ×, ≤, 0, 1, c\}
$$

I understand this to be the natural numbers along with an additional constant $c$, is that correct?

So from this understanding, it makes sense to me to ask the questions,

What is $n+c$ for $n$ a natural number?

What is $n\cdot c$ for $n$ a natural number?

What is $c+c$?

What is $c\cdot c$?

Is $c≤n$  for $n$ a natural number?

Are these sensible questions? If not, is there some similar concrete way to understand this model?

答:
Delving into the proof of completeness theorem, as Noah suggests, is one good way to gain intuition about nonstandard models of PA. Here's another approach.

First, we should see why there are nonstandard models of PA. It's not hard to find proofs that "Peano's Axioms" have a unique model, up to isomorphism. (Peano's predecessor Dedekind may have given the first such proof, in his paper "Was sind und was sollen die Zahlen" ("The nature and meaning of numbers")). This doesn't contradict the existence of nonstandard models of PA, because "PA" nowadays doesn't mean Peano's Axioms, but the first-order theory of arithmetic. Peano's induction axiom says that an arbitrary subset of $\mathbb{N}$ is $\mathbb{N}$ if it contains 0 and is closed under succession. This is a second-order axiom, meaning it talks about arbitrary sets of natural numbers.

The induction axioms of PA talk only about subsets of the system that can be defined by a first-order formula: a formula built up with +, $\times$, etc., and the basic constructs of logic ($\forall$, $\neg$, $\wedge$, etc.)  A nonstandard model, call it $N$, will have subset that is isomorphic to $\mathbb{N}$; we might as well call it $\mathbb{N}$. This subset violates the second-order version of induction in $N$, since it's a subset closed under succession and containing 0 but not all of $N$. But it's "invisible" to first-order logic: it can't be defined by a first-order formula.

There is a weaker set of first-order axioms known as PA$^-$. (See here for a list of its axioms.) Basically this includes all the "algebraic" and "ordering" properties of PA, but omits all the induction axioms. So for example we have $x+y=y+x$ and $x<y\rightarrow x+z<y+z$ as axioms. Here it's much easier to construct a nonstandard model. Start with $\mathbb{N}$, and add a constant $c$; you may think of $c$ as a kind of "infinite number". Because you have $c$, you also must have $c\pm n$ for all $n\in\mathbb{N}$, $c^2$, $2c^2$, etc. In fact, it's not hard to see that the model must contain elements corresponding to all polynomials of the form
$$a_nc^n+\cdots+a_0,\qquad a_n,\cdots,a_0\in\mathbb{Z},\;a_n>0$$
And that's enough! The set of such polynomials, together with 0, with the natural definitions for $+,\times,<$, is a model of PA$^-$ (call it $M$).

The induction axioms do not hold in $M$. For example, you can prove by induction that $$\forall x\exists y(y+y=x\vee y+y+1=x)$$
in other words, every number is even or odd. But there is no such $y$ in $M$ when $x=c$.

What to do, if we want a nonstandard model of PA? One idea: extend $M$ to a model $N$ of PA. Say we add a new constant $d$ and stipulate that $d+d+1=c$. (Or $d+d=c$, but not both!) Just as adjoining $c$ forced us to add lots of other elements, adjoining $d$ will force us to add a lot more. And that's just for one instance of one induction axiom! Still, the basic idea is to keep adjoining new elements whenever PA proves that something should exist, and it doesn't yet.

It is kind of remarkable that this approach can be made to work. With the model $M$, each polynomial defined a unique element, and different polynomials defined different elements. Also it was not hard to decide on the meaning of $+,\times,<$ for the elements of $M$. This becomes much more problematic with the wholesale introduction of constants designed to satisfy the induction axioms.

That it does work is the beauty of Henkin's construction, that Noah sketched. But we have to sacrifice something. The model $M$ of PA$^-$ is recursive, i.e., you could write a computer program to implement it. Tennenbaum's theorem says that no recursive nonstandard model of PA exists. In fact, you can't even make a model where the + operation is computable, not to say $\times$. (And vice versa.)

Plug: I've been blogging about nonstandard models of PA with a couple of other people (John Baez and Bruce Smith).


问:
How is this read correctly in maths , and meaning (about: sets)?

$J \subseteq \left\{1,2,..,n\right\}$ is an index set ....   Fix $J
\subseteq \left\{1,2,..,n\right\}$ and let $S=\left\{x_j : j \in
J\right\}$

The problem I'm having here is "Fix" , what does that mean ? I think "fix" refers to some constant value, or in this case SOME set we are chosing?

My other question is how is the set $S$ read correctly? I would read it like that: The set $S$ contains elements $x_j$ where each index $j$ is from index set $J$"

I hope someone can help :)

答:
The "fix" means "consider some particular $J$" or "for each $J$ define $S$ to be ..." . Your description of $S$ in terms of $J$ is correct.

No need to overthink this.


问:
Number of elements in an orbit and proving isomorphism.
I'm asked to find the number of elements in the orbit of the permutation $\sigma=(123)(45)(67)$ in $S_{7}$, and prove that its centralizer is isomorphic to $\mathbb{Z}_3\times D_4$.

I was able to prove that its orbit contains $420$ elements - all the elements in $G({\sigma})$ (its orbit) are in the form of $(x_{1}x_{2}x_{3})(x_{4}x_5)(x_6x_7)$ so using combinatorics and the fact that for example $(x_{1}x_{2}x_{3})=(x_{3}x_{1}x_{2})=(x_{2}x_{3}x_{1})$ I get that the number of elements is 420 ($\frac{7\times 6\times 5}{3}$ options for $(x_{1}x_{2}x_{3})$, $\frac{5\times 4}{2}$ for $(x_{4}x_5)$ and $\frac{2}{2}$ for $(x_6x_7)$) .

Now I tried finding its centralizer. I was able to find an element with order $4$ - $(6475)$ and an element of order $2$ - $(45)$ and an element of order $3$ - $(123)$. Moreover, I've proven that $\langle(123)\rangle, \langle(6475),(45)\rangle $ are subgroups of $C_{s_{7}}(\sigma)$ (that is the centralizer of $\sigma$ in $S_7$). So if I prove that these are all the elements in $C_{s_{7}}(\sigma)$, I'll get that $\mathbb{Z}_3\times D_4$ is isomorphic to $C_{s_{7}}(\sigma)$. But, using the orbit - stabilizer theorem I get that $|C_{s_{7}}(\sigma)|$=$\frac{|S_7|}{|G(\sigma)|}=\frac{7!}{420}=12\neq|\mathbb{Z_3}\times D_4|=24$.

Where is my mistake?

答:
I think your mistake is that there are only $210$ conjugates of $\sigma$ in $S_7$;

you counted each twice, once as  $(x_1x_2x_3)(x_4x_5)(x_6x_7)$ and once as
$(x_1x_2x_3)(x_6x_7)(x_4x_5)$.

Here is a convenient on-line reference about $S_7$.


问:
proof verification on separable metric space
let $(X,d)$ and $(Y,p)$ be metric spaces and $f:X\rightarrow Y$ continuous, if $A\subseteq X$  and $(A,d|_{A})$ is separable show that $$(f(A),p|_{f(A)})$$ is also separable.

Since $f$ is continuous and $A$ is countable then $f(A)$ is countable.
because $\overline{A}= X\Rightarrow B(x,\delta  )\cap X\neq \varnothing (x\in A)$

$\Rightarrow  f(B(x,\delta  ))\cap f(X)\neq \varnothing $ (can i do that?)

$f$ is continuous so $B(f(x),\varepsilon  )\cap Y\neq \varnothing $

答:
$f|_A:A\to f[A]$ is a continuous surjection.

There are many properties that are equivalent to continuity, and some are better tools than others for some problems.

$f:U\to V$ is continuous iff $f(p)\in Cl_V (f[D])$ whenever $D\subseteq U$ and $p\in Cl_U(D).$

So in your Q, let $D$ be a countable dense subset of $A.$ For any $b\in f[A]$ we have $b=f(a)$ for some $a\in A,$ and $a\in Cl_A(D)$. So $b=f(a)\in Cl_{f[A]}(f[D]).$ Therefore $f[D]$ is dense in $f[A].$

Alternatively:

$f:U\to V$ is continuous iff $f[Cl_U(D)]\subseteq Cl_V(f[D])$ for every $D\subseteq U.$

So in your Q, let $D$ be a countable dense subset of $A.$ Then $$f[A]=f[Cl_A(D)]\subseteq Cl_{f[A]}(f[D])\subseteq f[A].$$ So $f[D]$ is dense in $f[A]$ because $Cl_{f[A]}(f[D])=f[A].$


问:
Request to check a proof for existence of prime decomposition of integers
I wrote a proof by contradiction for the existence of a prime decomposition for all positive integers, I came up with it while I was trying solidify my understanding of the theorem itself. (Just existence, not uniqueness). If you are reading this and are interested, would you check for any errors? Thank you for reading!

-Proof-

Take $z\geq2\in\mathbb{Z}$ where z is not prime.

Then $$ \exists\;a_{01},a_{02}\geq2\;s.t.\;a_{01}a_{02}=z$$

If both factors are prime, than we are done.

If not, then $$\exists\;a_{11},a_{12}\geq2\;s.t. a_{11}a_{12}=a_{0i_{0}}$$ where $i_{0}=1$ or $2$

If $a_{11},a_{12}$ prime, we are done.

If one is not, then
$$ \exists a_{21}a_{22}\geq2\; s.t.\; a_{21}a_{22}=a_{1i_{1}}\; where\; i_{1}=1 \;or\; 2  $$

The process continues ad infinitum $\iff \nexists $ a factorization into primes for z

Suppose this process would continue ad infinitum.

Where $ j_{n}=1 \;or\; 2 \;and\; j_{n}\neq i_{n} $

$$z>a_{0j_{0}}\cdot a_{1j_{1}}\cdot a_{2j_{2}}\cdot....=\prod_{n\geq0}^{\infty}a_{nj_{n}} $$

But $$a_{nj_{n}}\geq2\; \forall\;n$$

Thus $$ z>\infty \;and\; z=\infty $$

But we took z finite thus this process must stop at some point, and thus  $\exists$ prime factorization for all positive integers

答:
You implicitly imply induction (that there is no infinite descending chain of naturals). This must be made explicit for the proof to be rigorous. Below is a  way to do so that highlights innate structure.

${\small {\bf Lemma}}\,\ \Bbb N_{>1} =$ only set $S$ of naturals $>\! 1$ containing  $\rm\color{#c00}{all\ primes}$ & $\rm\color{#0a0}{closed\,  under\, multiplication}$

Proof $\,\ $  If $\,n\,$ is  $\rm\color{#c00}{ prime}$ then by hypothesis $\,n\in S.\,$ Else $\,n\,$ is composite, $ $ so $\ n = j k\ $ for $\, 1 < j,k < n.\,$ By strong induction the smaller $\ j,k\in S,\,$ $\rm\color{#0a0}{hence}$ $\, n = jk\in S.$

Corollary $ $ Every natural $>1$ has a prime factorization.

Proof $ $ The set $S$ of naturals $>1$ having prime factorizations contains $\rm\color{#c00}{all\ primes}$ (each is a singleton product) and is $\rm\color{#0a0}{closed\ under\ multiplication},\,$ so the lemma yields $\,S = \Bbb N_{> 1}$.


问:
How can I find which function corresponds to a set of data points?
Suppose I have a set of data points like this:

```<code>1;1
2;4
3;9
4;16
5;25
6;36
...
</code>```

The first column is the input of the function and the second one is the result. I can tell if I look at it that the function here is y = f(x^2). The problem is that the data points are not always this obvious and I want to write a program which can tell me the definition of the function.

How can I do so?

Some context: I'm trying to write a program which can be used to analyze asymptotic time complexity of algorithms. So in fact I'm only interested in a predefined set of functions:

y = f(n)

y = f(log(n))

y = f(n * log(n))

y = f(n ^ m)

y = f(2 ^ n)

答:
Here is a Python Jupyter script for that purpose.

If you can restrict yourself to polynomials, NumPy (a Python math library) will offer an easy solution:

```<code>import numpy as np
from numpy.polynomial import Polynomial as P
import matplotlib.pyplot as plt

x = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27]
y = [1,2,4,7,6,3,2,3,4,5,9,12,11,9,7,3,1,2,3,5,8,12,14,16,12,8,3]

plt.plot(x,y)  # plot the original dataset

polynomial = P.fit(x, y, 14) # 14 is the degree of the polynomial
fx, fy = p.linspace(100)  # generate 100 sample points on this graph
plt.plot(fx, fy)  # plot the calculated polynomial
</code>```

I am not sure whether this applies to your usecase, but restricting oneself to polynomials might not be as bad as it seems since most functions can be approximated with a Taylor series and compared to other functions, polynomials are usually easier to work with.

The provided code sample produces the following output (the blue line being the original dataset and the orange line being the calculated function):

Body:
Here is a Python Jupyter script for that purpose.

If you can restrict yourself to polynomials, NumPy (a Python math library) will offer an easy solution:

import numpy as np
from numpy.polynomial import Polynomial as P
import matplotlib.pyplot as plt

x = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27]
y = [1,2,4,7,6,3,2,3,4,5,9,12,11,9,7,3,1,2,3,5,8,12,14,16,12,8,3]

plt.plot(x,y)  # plot the original dataset

polynomial = P.fit(x, y, 14) # 14 is the degree of the polynomial
fx, fy = p.linspace(100)  # generate 100 sample points on this graph
plt.plot(fx, fy)  # plot the calculated polynomial

I am not sure whether this applies to your usecase, but restricting oneself to polynomials might not be as bad as it seems since most functions can be approximated with a Taylor series and compared to other functions, polynomials are usually easier to work with.

The provided code sample produces the following output (the blue line being the original dataset and the orange line being the calculated function):


问:
Definition on multiplication in rings
Let's say I have a ring $R=(\mathbb{R}, +, *)$. Is

$a+a=2*a$

($a \in \mathbb{R}$)

always true for any ring¹? (So that when this equation is not true then I can definitively infer that $R$ is not a ring.)

1 ) With "any" I mean any other ring which is also using $\mathbb{R}$ as underlying set, thank you for your comment nick.

答:
If your ring has a unit, i.e. a multiplicative identity, (and the definition that almost everybody uses these days requires that a ring does have an identity), then yes.

As the commenters point out, $2$ is defined* to be $1 +1$, where $1$ is the multiplicative identity, and so it follows from the distributive law and the fact that $1$ is the multiplicative identity.

The only thing to be careful about is that it is possible that $ 2 = 0$ (e.g. in $\mathbb Z_2$), or perhaps $2 = -1$ (e.g. in $\mathbb Z_3$), so these "integers" inside your ring might not behave the way you expect integers to behave.

BTW, if you are dealing with an algebraic structure that doesn't have a $1$, people will often define an "action" of $\mathbb Z$ on your elements, and use multiplication to denote it, where

$$ n \cdot a = a + a + .... + a \text{  (n times)}$$

Edit: Okay, you added "With 'any' I mean any other ring which is also using $\mathbb{R}$ as underlying set", and this needs to be addressed: You can take the underlying set $\mathbb R$, and define a wacky new addition and multiplication on it. The simplest is $a \oplus b = a + b -1$ and $a \otimes b = ab - a -b + 2$.

Let's use the symbol $S$ to denote this new ring $\langle \mathbb R, \oplus, \otimes\rangle$. Then the number 1 in $\mathbb R$ (which I'm going to write as $1_{\mathbb R}$) is not the multiplicative identity for the ring $S$. $1_S$, which is the standard notation for the multiplicative identity in a ring named $S$, is, in fact $2$, by which I mean good old 2 in good old $\mathbb R$, which we might want to write as $2_{\mathbb R}$, and yes $2_{\mathbb R} = 1_{\mathbb R} + 1_{\mathbb R}$.

But what your question asks is still true in $S$, i.e. $a \oplus a =2_{S} \otimes a$; however notice that you have to be sure to use the ring operations of $S$, and remind yourself that you're using $2_{S}$, which is defined to be $1_{S} \oplus 1_{S}$. (And corresponds to the underlying real number $3_{\mathbb R}$!)

The ring $S$ is of course extremely confusing to work with, and I've never seen it used seriously, only to break the brains of undergraduate math majors, to show them how we can define groups, rings, fields, etc. that behave very differently from what they are used to. I.e. $\langle \mathbb R, \oplus, \otimes\rangle$ is a cautionary tale, not a commonly used mathematical tool, but the only requirement you put was that $\mathbb R$ was the underlying set, and so you left it open to me to define really weird addition and multiplication. I wouldn't spend too much time agonizing over it, but it can be a fun example to contemplate and sharpen your wits.

*If someone uses the symbol "$2$" and says it is not equal to $1+1$, you get to look at them funny, ask just what the heck they think they are doing, and demand that they explain why they are using that symbol.

Title: Definition on multiplication in rings


问:
Why is the spectrum of $L^\infty[0,1]$ inseparable?
In the Gelfand correspondance any measurable subset of $[0,1]$ corresponds to a clopen subset of $\text{Spec}(L^\infty[0,1])$, so clearly the space contains a bunch of clopens. I'm not sure if this is relevant and remain stuck on the problem. The text I read said that it was essentially due to the Borel measure on $[0,1]$ being atomless, so this might be a more general result.

答:
Let $X=\operatorname{Spec}(L^\infty[0,1])$.  A point of $X$ corresponds to an ultrafilter on the Boolean algebra of measurable subsets of $[0,1]$ modulo null sets.  In particular, any such ultrafilter must contain sets of arbitrarily small positive measure (this follows from atomlessness: we can keep splitting elements of the ultrafilter into two subsets of half the measure, and one  of the halves must be in the ultrafilter).

Now suppose $\{x_n:n\in\mathbb{N}\}$ is a countable subset of $X$.  For each $n$, we can pick a set $A_n$ in the ultrafilter corresponding to $x_n$ such that $\mu(A_n)<\epsilon_n$, for any $\epsilon_n>0$ that we choose.  In particular, we can choose a sequence $(\epsilon_n)$ such that $\sum_n\epsilon_n<1$.  Now consider the measurable set $A=[0,1]\setminus\bigcup_n A_n$.  By our choice of $(\epsilon_n)$, $A$ has positive measure.  However, $A$ is not in the ultrafilters corresponding to any of our $x_n$.  Thus the set of points of $X$ whose corresponding ultrafilters contain $A$ is a nonempty open subset of $X$ which does not contain any $x_n$.  That is, $\{x_n\}$ is not dense in $X$.


问:
Oddity: Determinant of Skew-Symmetric $n\times n$ Matrices
Let $A \in M_{n×n}(\mathbb{R})$ be a skew-symmetric matrix, i.e., $A^t = −A$. Prove that if $n$ is odd, then $\det{A} = 0$.</p>

答:
Amire's answer is certainly the most straightforward way. </p>

<p>Another way to see it is that the eigenvalues of a skew-symmetric matrix are necessarily imaginary. Being a real matrix, they have to appear in conjugate pairs. If the dimension is odd, then one eigenvalue is also real, and so it has to be zero. As the determinant is the product of the eigenvalues, the result follows. </p>

Body:
Amire's answer is certainly the most straightforward way.

Another way to see it is that the eigenvalues of a skew-symmetric matrix are necessarily imaginary. Being a real matrix, they have to appear in conjugate pairs. If the dimension is odd, then one eigenvalue is also real, and so it has to be zero. As the determinant is the product of the eigenvalues, the result follows.


问:
From monoids to groups
I was looking at the case when you go from the monoid of natural numbers to the group of integers by means of a suitable equivalence relation. The key here was to find inverses for each natural and I was wondering how it could be generalized. I mean:

Given any monoid, when can I find inverses for your elements?

In other words, can I always go from a monoid to a group?
If not, when is it? When do not?

If you could tell me where I can find information about this (a book or paper) it would be very helpful. Thank you.

答:
If $M$ is a semigroup, the group presentation $$\langle (x_m)_{m\in M}\mid (x_mx_n=x_{mn})_{m,n\in M}\rangle$$
defines a group $\hat{M}$ and the canonical map $M\to\hat{M}$, $m\mapsto x_m$ is a semigroup homomorphism (monoid homomorphism if $M$ is a monoid), and this satisfies the universal property. As mentioned by Arturo Magidin it is not injective in general (by the universal property, it's injective iff $M$ embeds into a group; a necessary condition is that $M$ is left and right cancelative).

While this construction is theoretically obvious, it is practically not very useful and not "practically explicit" (since it's a huge set modulo a huge equivalence relation).
If $M$ is commutative, a much more practical construction is provided in Ottavio's answer; in particular if $M$ is commutative and cancelative then $M\to\hat{M}$ is injective.

Title: From monoids to groups


问:
Reflection Over Planes in 3D
Reflecting a point over a flat plane in 3D is not super simple, but relatively simple and I have seen on stackexchange how matrices are used to figure out where the reflected point is projected.

However, throughout this I thought how could this be applied to curved planes like cones, spheres, parabola and more.

For example, say you have a generic line L1 [never intersects the plane with equation ax+by+cz+d=0] and L1 is reflected over said plane, where would L1 be projected to?

Also what would happen if this plane is now curved like a parabola in 3D? Is it possible to derive a single equation for this?

答:
Let (P) be the plane with equation

$$ax+by+cz=0$$

(I have dropped the $d$ but it can easily been included).

$N(a,b,c)$ be a normal vector to (P); we will assume WLOG that $\|N\|^2=a^2+b^2+c^2=1$.

Suppose one wants to reflect a surface with equation

$$f(x,y,z)=0\tag{1}$$

for example the sphere with center $(1,0,0)$ and unit radius

$$(x-1)^2+y^2+z^2-1=0\tag{2}$$

The symmetry $X \leftrightarrow X'$ with respect to plane (P) is obtained through the equation (see explanation below):

$$X'=(I_3-2 NN^T)X \ \ \iff  \ \ X=(I_3-2 NN^T)X' \tag{3}$$

(because symmetry is involutive ; N is assumed a column vector)

i.e., detailed in terms of coordinates:

$$\begin{pmatrix}x\\
y\\
z \end{pmatrix}=\begin{pmatrix}(1-2a^2)& -2ab&-2ac\\
-2ab&(1-2b^2)&-2bc \\
-2ac&-2bc&(1-2c^2)& \end{pmatrix}\begin{pmatrix}x'\\
y'\\
z'\end{pmatrix}\tag{4}$$

It is enough now to replace $x,y,z$ in (1) by their expressions issued from (4) to get the equations of the symmetric surface.

For example, if you reflect the sphere with equation (2) into vertical plane with equation $x=0$ (meaning that $a=1$, $b=c=0$), you will get $x=-x',y=y',z=z'$, giving:

$$(x'+1)^2+y'^2+z'^2=1$$

Remark: If you want to symmetrize a straight line, consider it classically as the intersection of two planes, and symmetrize separately the equations of two planes using the method just seen.

Explanation of formula (3):

the image of $X=N$ is $(I_3-2NN^T)N=N-2N\underbrace{(N^TN)}_{\|N\|^2=1}=-N$ as expected.

if $M \perp N$, $(I_3-2NN^T)M=M-2N\underbrace{(N^TM)}_0=M$.

Edit: (following a remark of @mr_e_man):

in fact, one can "mirror" any geometrical object with respect to a circle in 2D or a sphere in 3D using a transformation called "inversion" which is described here. The mirroring property can be seen on the example of the fish "reflected" with respect to his bowl. A circle is in general reflected into a circle. For extended information and references, see there.

Title: Reflection Over Planes in 3D


问:
Annihilator of a direct sum and product of modules
Let be $R$ a ring and  $\{M_i\}_{i \in I} $ a family of $R$-modules. I am trying to proof that $\text{Ann}_R(\prod_{i \in I} M_i)
 = \text{Ann}_R(\oplus_{i \in I}M_i)  $

Applying definitions :

$\prod_{i \in I} M_i = \{(x_i)_{i\in I}\hspace{2mm}|\hspace{2mm}x_i \in M_i \hspace{2mm} \forall i \in I\}$

$\text{Ann}_R(M)=\{r \in R \hspace{2mm}|\hspace{2mm} rx=0 \ \forall x \in M\}$

so:

$\text{Ann}_R(\prod_{i \in I} M_i)=\{r \in R \hspace{2mm}|\hspace{2mm} r x_i=0 \forall x_i \in M_i,i\in I\}$

and that is : $\cap_{i \in I}\text{Ann}_R(M_i)$

I arrive to the exact same set when applying the definition for direct sum. It almost seems trivial, so I want to know if there is any way to do a more "formal " proof.

答:
Your approach is fine. In general though, if you want to show that two sets $A$ and $B$ are equal, I would suggest showing that $x \in A \implies x \in B$ and then that $x \in B \implies x \in A$. This leads to cleaner arguments, and reduces the chance of error. I'll show you what I mean in this case.

Let $r \in \operatorname{Ann}_R(\prod_{i\in I} M_i)$. Since $\bigoplus_{i\in I}M_i\subseteq \prod_{i\in I}M_i$, certainly $r$ kills every element of $\bigoplus_{i\in I}M_i$, so $r \in \operatorname{Ann}_R(\bigoplus_{i \in I}M_i)$.

Now let $r \in \operatorname{Ann}_R(\bigoplus_{i\in I}M_i)$. Let $\alpha \in \prod_{i\in I}M_i$. Then we may write $\alpha$ as $(\alpha_i)_{i\in I}$ where $\alpha_i \in M_i$ is the component of $\alpha$ in $M_i$. For each $j$, there is an element of $\bigoplus_{j\in I}M_j$ whose $M_i$-component is $\alpha_i$, and whose other components are $0$. This must be killed by $r$, so $r\cdot \alpha_i = 0$ for all $i$. Therefore $r\cdot \alpha = 0$. Since $\alpha$ was arbitrary, $r \in \operatorname{Ann}_R(\prod_{i\in I}M_i)$.

The reason I prefer this is that it argues in a linear fashion; each statement follows the previous ones. When - as you did in your proof - you rearrange definitions of sets to make them equal, it requires reading back and forth a few times, which I think makes for a less pleasing argument.


问:
Trouble proving a tricky inequality
I'm working on proving something that basically boils down to proving a particular inequality that I can't seem to make any progress on.  Given $$G(x) = \frac{x(1+x)}{1+x+x^2}$$ I need to show that for all $a\in (0,1)$ and $x > 0$ we have $$G'\left(\frac{1}{ax}\right) + xG'(ax) > G'\left(\frac{1}{x}\right) + xG'(x).$$  Graphically this seems to be true, and it basically boils down to showing $$\frac{a^{3}x^{2}\left(2+ax\right)+1+2ax}{\left(1+ax+\left(ax\right)^{2}\right)^{2}} > \frac{x^{2}\left(2+x\right)+1+2x}{\left(1+x+x^{2}\right)^{2}}$$ but I haven't gotten further than that.  I had thought that maybe the racetrack principle would be applicable, but it's not.

答:
If you cross multiply and factorize, you WTS

$$ ( a - 1)^2 x^2 (x^2+x+1)(a^2 x + 2ax + 2a + 1 ) > 0. $$

This is obvious.

Title: Trouble proving a tricky inequality


问:
Suppose $C_1,C_2,...,C_n$ are compact sets on $\mathbb{R}$. Prove that $\underset{i = 1}{\bigcup}^n C_i$ is compact
This was one of the problems on my last exam. The proof I provided was mostly correct, but my professor took off a couple points for a mistake I made. I'll number the lines of the proof for reference.

Proof

1 Let $\mathcal{U}$ be an open cover of $C_n$

2 Then, $\mathcal{U}\cup\{\mathbb{R} \backslash C_n\}$ is an open cover of $C_1,C_2,...,C_{n-1}$

3 Since $C_n$ is compact, there is a finite collection of open intervals $U_1,U_2,...,U_k \in \mathcal{U}$ such that $\underset{i = 1}{\bigcup}^k U_i \supseteq C_n$

4 So $\underset{i = 1}{\bigcup}^k U_i \cup\{\mathbb{R} \backslash C_n\} \supseteq C_i$ for $1 \leq i \leq n$

5 Then $\underset{i = 1}{\bigcup}^k U_i \cup \{\mathbb{R} \backslash C_n\} \supseteq \underset{i = 1}{\bigcup}^n C_i$

6 Thus, $\underset{i = 1}{\bigcup}^n C_i$ is compact

So my professor thought the proof was mostly correct except for Line 2. On Line 2, when I said that

"$\mathcal{U}\cup\{\mathbb{R} \backslash C_n\}$ is an open cover of $C_1,C_2,...,C_{n-1}$," he said, "Not necessarily."

So why isn't $\mathcal{U}\cup\{\mathbb{R} \backslash C_n\}$ an open cover of $C_1,C_2,...,C_{n-1}$ ?

We know that $\{\mathbb{R} \backslash C_n \}$ covers all of $C_1, C_2, ..., C_{n-1}$ if all of the $C_i$ are disjoint.

If the $C_i$'s are not disjoint (meaning at least one $C_i$ has $C_i \cap C_n \neq \emptyset$), then $\mathcal{U} \cup \{\mathbb{R} \backslash C_n\}$ covers all of them. What am I missing here?

答:
$\mathcal U\cup\{\Bbb R\setminus C_n\}$ covers all the $C_i$-s because it covers $\Bbb R$.

Your mistake lies else where. Other than the small imprecision of you saying that $\mathcal U$ is an open cover, but then claiming to take "intervals" from $\mathcal U$ (in fact, $\mathcal U$ may not contain intervals at all), there's the catastrophic issue that you have only proved that all covers coming from a construction such as yours admit a finite subcover: effectively, just the covers $\mathcal V$ of $\bigcup_{i=1}^n C_i$ such that $\Bbb R\setminus C_n\in \mathcal V$ (or, in a broader sense, the ones such that there is some $i$ such that $\Bbb R\setminus C_i\in\mathcal V$). But you have to prove existence of a finite subcover for all open covers of $\bigcup_{i=1}^n C_i$.


问:
Variance and expected value of $2^X$
I have 2 similar problems:

given $X \sim Po(3)$, find the variance of $2^X$

given $X \sim B\left(200, \frac{1}{9}\right)$, find the expected value of $2^{X+3}$

I'm not really certain what I'm expected to calculate should it be $Var(2^X)$ and $E[2^{X+3}]$ respectively or should it be $2^{Var(X)}$ and $2^{E[X+3]}$. If it is the second one then: for (1) $$X \sim Po(3)\Longrightarrow\lambda=3\Longrightarrow Var(X)=\lambda = 3 \Longrightarrow 2^{Var(X)}=2^3$$  and for (2) $$X \sim B\left(200, \frac{1}{9}\right)\Longrightarrow E[X+3]=\frac{200}{9} + 3\Longrightarrow 2^{E[X+3]}=2^{\frac{200}{9} + 3}$$

But if it's the first one I don't know how I can find the variance and expected value with the given information

after reading the answers I did for (1): $$ E[2^X]=\sum^\infty_{i=0}2^i\cdot e^{-\lambda}\frac{\lambda ^i}{i!} = e^{-\lambda}\sum^\infty_{i=0}\frac{(2\lambda)^i}{i!}=e^{-\lambda}\cdot e^{2\lambda} = e^\lambda=e^3$$
and same way
$$ E[(2^X)^2]=E[(4^X)]=\sum^\infty_{i=0}4^i\cdot e^{-\lambda}\frac{\lambda ^i}{i!} = e^{-\lambda}\sum^\infty_{i=0}\frac{(4\lambda)^i}{i!}=e^{-\lambda}\cdot e^{4\lambda} = e^{3\lambda}=e^9 $$
hence
$$ Var(X) = E[(2^X)^2] - (E[2^X])^2 = e^9 - (e^3)^2 = e^9 - e^6$$

as for (2): $$ E[2^X]=\sum^{200}_{i=0}2^i\binom{200}{i}(\frac{1}{9})^i(1-\frac{1}{9})^{200-i} = \sum^{200} _{i=0}\binom{200}{i}(\frac{2}{9})^i (\frac{8}{9})^{200-i} =(\frac{2}{9} + \frac{8}{9})^{200} = (\frac{10}{9})^{200} $$
so $$ E[2^{X+3}]=E[8\cdot 2^X]=8\cdot E[2^X]=8(\frac{10}{9})^{200} $$

答:
The probability-generating function (PGF) of a random variable $X$ is given by $$\mathcal P_X(t) = \operatorname{E}[t^X]$$ whenever this expectation is well-defined.  For $X \sim \operatorname{Poisson}(\lambda)$, we have
$$\begin{align}
\mathcal P_X(t) &= \sum_{x=0}^\infty t^x \Pr[X = x] \\
&= \sum_{x=0}^\infty t^x e^{-\lambda} \frac{\lambda^x}{x!} \\
&= e^{-\lambda} \sum_{x=0}^\infty \frac{(\lambda t)^x}{x!} \\
&= e^{-\lambda} e^{\lambda t} \sum_{x=0}^\infty e^{-\lambda t} \frac{(\lambda t)^x}{x!} \\
&= e^{\lambda (t-1)},
\end{align}$$
where in the last step, we recognize that the summand is the probability mass function for a Poisson random variable with parameter $\lambda t$, thus the sum equals $1$.

Similarly, I leave it as an exercise to show that if $X \sim \operatorname{Binomial}(n,p)$, then
$$\mathcal P_X(t) = ((1-p) + pt)^n = (1 + p(t-1))^n.$$
Hint:  use the binomial theorem $$(a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k.$$

With these in hand, we can now easily evaluate the desired moments.  We have
$$\operatorname{Var}[2^X] = \operatorname{E}[(2^X)^2] - \operatorname{E}[2^X]^2 = \operatorname{E}[4^X] - \operatorname{E}[2^X]^2 = \mathcal P_X(4) - (\mathcal P_X(2))^2,$$ and $$\operatorname{E}[2^{X+3}] = \operatorname{E}[8 \cdot 2^X] = 8 \operatorname{E}[2^X] = 8 \mathcal P_X(2).$$  I leave it to you to evaluate these for their respective distributions.

Title: Variance and expected value of $2^X$


答:
There is nothing wrong with Line $2$. $C_n\subseteq\bigcup\mathscr{U}$, so $(\Bbb R\setminus C_n)\cup\bigcup\mathscr{U}=\Bbb R$, and therefore $\mathscr{U}\cup\{\Bbb R\setminus C_n\}$ is in fact an open cover of $\Bbb R$. However, I would have given very little credit for your argument: you have not shown that an arbitrary open cover of $\bigcup_{k=1}^nC_k$ has a finite subcover.

Start with an arbitrary open cover $\mathscr{U}$ of $\bigcup_{k=1}^nC_k$. $\mathscr{U}$ is an open cover of $C_k$ for each $k\in\{1,\ldots,n\}$, so for each $k\in\{1,\ldots,n\}$ there is a finite subset $\mathscr{U}_k$ of $\mathscr{U}$ that covers $C_k$. Let $\mathscr{V}=\bigcup_{k=1}^n\mathscr{U}_k$; clearly $\mathscr{V}$ is a finite subset of $\mathscr{U}$ that covers $\bigcup_{k=1}^nC_k$, which is therefore compact.

Title: Suppose $C_1,C_2,...,C_n$ are compact sets on $\mathbb{R}$. Prove that $\underset{i = 1}{\bigcup}^n C_i$ is compact

Body:
This was one of the problems on my last exam. The proof I provided was mostly correct, but my professor took off a couple points for a mistake I made. I'll number the lines of the proof for reference.

Proof

1 Let $\mathcal{U}$ be an open cover of $C_n$

2 Then, $\mathcal{U}\cup\{\mathbb{R} \backslash C_n\}$ is an open cover of $C_1,C_2,...,C_{n-1}$

3 Since $C_n$ is compact, there is a finite collection of open intervals $U_1,U_2,...,U_k \in \mathcal{U}$ such that $\underset{i = 1}{\bigcup}^k U_i \supseteq C_n$

4 So $\underset{i = 1}{\bigcup}^k U_i \cup\{\mathbb{R} \backslash C_n\} \supseteq C_i$ for $1 \leq i \leq n$

5 Then $\underset{i = 1}{\bigcup}^k U_i \cup \{\mathbb{R} \backslash C_n\} \supseteq \underset{i = 1}{\bigcup}^n C_i$

6 Thus, $\underset{i = 1}{\bigcup}^n C_i$ is compact

So my professor thought the proof was mostly correct except for Line 2. On Line 2, when I said that

"$\mathcal{U}\cup\{\mathbb{R} \backslash C_n\}$ is an open cover of $C_1,C_2,...,C_{n-1}$," he said, "Not necessarily."

So why isn't $\mathcal{U}\cup\{\mathbb{R} \backslash C_n\}$ an open cover of $C_1,C_2,...,C_{n-1}$ ?

We know that $\{\mathbb{R} \backslash C_n \}$ covers all of $C_1, C_2, ..., C_{n-1}$ if all of the $C_i$ are disjoint.

If the $C_i$'s are not disjoint (meaning at least one $C_i$ has $C_i \cap C_n \neq \emptyset$), then $\mathcal{U} \cup \{\mathbb{R} \backslash C_n\}$ covers all of them. What am I missing here?


答:
Further to my comment, this is what a proof would look like:

Let $\{U_{j}\}_{j}$ be an open cover of $\bigcup_{i=1}^{n} C_{n}$. Then $\{U_{j}\}_{j}$ covers each $C_{i}$. Since each $C_{i}$ is compact, you can find $U^{i}_{1}, \dots, U^{i}_{k_{i}}$ that cover $C_{i}$. Now consider the cover
$$
\bigcup_{i=1}^{n} (U^{i}_{1} \cup \dots \cup U^{i}_{k_{i}})
$$
This has only finitely many sets from your original cover and covers all $C_{i}$, so it covers $\bigcup_{i=1}^{n} C_{i}$, and we're done.

The conceptual error you made in your answer is that you started with an open cover of $C_{n}$ (just one of the sets). When using this definition (open cover, finite subcover) to show that something is compact, the proof will (pretty much) always go this way:

Pick an arbitrary open cover of the set that you want to prove is compact (!!!)

Use the assumptions given in the question to try to find a subset that covers the set


答:
$$A_2 = \begin{bmatrix}1 & 1\\ 1 & 2 \end{bmatrix} = \begin{bmatrix}1 & 0\\ 1 & 1\end{bmatrix}\begin{bmatrix}1 & 1\\ 0 & 1\end{bmatrix}$$
$$A_3 = \begin{bmatrix} 1 & 1 & 1\\ 1 & 2 & 2\\1 & 2 & 3 \end{bmatrix} = \begin{bmatrix}1 & 0 & 0\\ 1 & 1 & 0\\ 1 & 1 & 1\end{bmatrix}\begin{bmatrix}1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1\end{bmatrix}$$
$$A_4 = \begin{bmatrix} 1 & 1 & 1 & 1\\ 1 & 2 & 2 & 2\\1 & 2 & 3 & 3\\ 1 & 2 & 3 & 4\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 & 0\\ 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 0\\ 1 & 1 & 1 & 1\end{bmatrix}\begin{bmatrix}1 & 1 & 1 & 1\\ 0 & 1 & 1 & 1\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 1\end{bmatrix}$$
Can you see the pattern here? Prove this is the case in general.

Use this, along with the fact that $\det(XY) = \det(X) \cdot \det(Y)$ and determinant of a triangular matrix is the product of the diagonal entries to conclude the answer.

EDIT

Note that $$A_{ij} = \min(i,j) = \sum_{k=1}^{\min(i,j)} 1 = \sum_{k=1}^{n} \mathbb{I}_{k \leq i}\mathbb{I}_{k \leq j} = \sum_{k=1}^n B_{ik} C_{kj}$$
where $B_{ik} = \begin{cases} 1 & k \leq i\\ 0 & \text{else}\end{cases}$ and $C_{kj} = \begin{cases} 1 & k \leq j\\ 0 & \text{else}\end{cases}$.

This should enable you to get the decomposition.

Title: Evaluation of a specific determinant.

Body:
Evaluate $\det{A}$, where $A$ is the $n \times n$ matrix defined by $a_{ij} = \min\{i, j\}$, for all $i,j\in \{1, \ldots, n\}$.

$$A_2
\begin{pmatrix}  1&  1\\
  1&  2
\end{pmatrix};
A_3 = \begin{pmatrix}  1&  1&  1\\
  1&  2&  2\\
  1&  2&  3
\end{pmatrix};
A_4 = \begin{pmatrix}  1&  1&  1&  1\\
  1&  2&  2&  2\\
  1&  2&  3&  3\\
  1&  2&  3&  4
\end{pmatrix};
A_5 = \begin{pmatrix}  1&  1&  1&  1&  1\\
  1&  2&  2&  2&  2\\
  1&  2&  3&  3&  3\\
  1&  2&  3&  4&  4\\
  1&  2&  3&  4&  5
\end{pmatrix}$$

$$A_6 = \begin{pmatrix}  1&  1&  1&  1&  1&  1\\
  1&  2&  2&  2&  2&  2\\
  1&  2&  3&  3&  3&  3\\
  1&  2&  3&  4&  4&  4\\
  1&  2&  3&  4&  5&  5\\
  1&  2&  3&  4&  5&  6
\end{pmatrix};
A_7 = \begin{pmatrix}  1&  1&  1&  1&  1&  1&  1\\
  1&  2&  2&  2&  2&  2&  2\\
  1&  2&  3&  3&  3&  3&  3\\
  1&  2&  3&  4&  4&  4&  4\\
  1&  2&  3&  4&  5&  5&  5\\
  1&  2&  3&  4&  5&  6&  6\\
  1&  2&  3&  4&  5&  6&  7
\end{pmatrix}
$$


问:
Product of elements in a list in GAP
I just started learning GAP and I'm trying to do simple exercises.
The following problem is something I have not been able to figure out.

So I found the character degrees of $S_4$ using the following code:

CharacterDegrees(CharacterTable($"S_4"$));

Now this gives the output [ [ 1, 2 ], [ 2, 1 ], [ 3, 2 ] ]
, which means that there are two irreducible character of degree 1, one irreducible character of degree 2 and two irreducible characters of degree 3. I want to calculate the product of irreducible character degrees from this list. i.e., I want to calculate 1.1.2.3.3 directly from this list. Is there a command or a function that can be used in a situation like this?

Thak you!

答:
I don't know if there's any inbuilt function to do this, but you could easily make one as follows:

```<code>myFunction := function( L )
    return Product( List( L, x -> x[1]^x[2] ) );
end;
</code>```

You can then use it like this:

```<code>gap> L := CharacterDegrees( CharacterTable( "S4" ) );
[ [ 1, 2 ], [ 2, 1 ], [ 3, 2 ] ]
gap> myFunction( L );
18
</code>```

Title: Product of elements in a list in GAP

Body:
I don't know if there's any inbuilt function to do this, but you could easily make one as follows:

myFunction := function( L )
    return Product( List( L, x -> x[1]^x[2] ) );
end;

You can then use it like this:

gap> L := CharacterDegrees( CharacterTable( "S4" ) );
[ [ 1, 2 ], [ 2, 1 ], [ 3, 2 ] ]
gap> myFunction( L );
18


问:
Prove that $\frac{\partial f}{\partial z} =\overline{ \frac{\partial \overline{f}}{\partial{\overline{z}}}}$
Prove that if $f :\Omega \to \mathbb{C}$ has all its first partial derivatives defined at any
point in the open set $\Omega\subset\mathbb{C}$, then at that point  $$\frac{\partial f}{\partial z} =\overline{ \frac{\partial \overline{f}}{\partial{\overline{z}}}}$$

Let $f(z) = u(x,y) + i v(x,y)$

Then, $\overline{f(z)} = u(x,y) - i v(x,y)$

I know that $ \frac{\partial f}{\partial z} = \frac{1}{2}(\frac{\partial f}{\partial x} - i\frac{\partial f}{\partial y}) $ and $ \frac{\partial f}{\partial \overline{z}} = \frac{1}{2}(\frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y}) $

Also, $ \frac{\partial f}{\partial z} = u_x + i v_x = v_y - iu_y$

How can I use this to prove the equation above?

答:
Write
$$
f=u+iv
$$
then
$$
\overline f=u-iv\;.
$$
Since
$$
\frac{\partial}{\partial \overline z}
=\frac12\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)
$$
you get
\begin{align*}
\frac{\partial \overline f}{\partial\overline z}
&=\frac12\left(u_x-iv_x+iu_y+v_y\right)\\
&=\frac12\left(\overline f_x+i \overline f_y\right)\\
&=\overline{\frac12\left(f_x- if_y\right)}\\
&=\overline{\frac{\partial f}{\partial z}}
\end{align*}
as wanted.


问:
A closed and convex subset of an Hilbert space on which the norm does not attain its maximum
I want to prove that the norm $\lvert \lvert \cdot \rvert \rvert _2$ has no maximum on  $X=\{x\in \ell^2(k):\sum_{n \in \mathbb{N}} 2^{1/n}\lvert x_n \rvert^2 \le 1\}$, where $k=\mathbb{R}$ or $\mathbb{C}$.

I have managed to see that this is close and bounded, I can't figure out if this is also convex and I don't know if this would be helpful.

I've tried to do some contradiction argument, I have observed is that since $2^{1/n}>1$, if there is $x \in X$ which attains the maximum for $\lvert \lvert \cdot \rvert \rvert _2$, then this maximum, say $M$, must be strictly smaller than $1$, otherwise $x$ would be outside of $X$. Now my idea would be that in this case we could add a small $\epsilon$ to some term $x_n$ of $x$ in such a way that $\sum_{n \in \mathbb{N}} 2^{1/n}\lvert x_n \rvert^2$ stays below one, but I can't carry out the computations, even because if $\lvert \lvert x \rvert \rvert _2<1,$ then it does not follow that $\sum_{n \in \mathbb{N}} 2^{1/n}\lvert x_n \rvert^2<1$, as we can see taking the element $(2^{-1/4}, 0, 0, \ldots)$.

答:
Take $x_n=(y_k)_k$, where $y_k=0$, if $k\neq n$ and $y_n=2^{-1/2n}$. Then $x_n\in X$ and $\|x_n\|_2 \to 1$. So the sup of the norm over the set $X$ is one.

Now, if $x\in X$ and $\|x\|_2=1$, let $k$ be the smallest number such that $\sum_{i=1}^k|x_k|^2\geq 1/2$. Write $a:=\sum_{i=1}^k|x_k|^2$ and $b:=\sum_{i>k}|x_k|^2$. So $$\sum_{n\in\mathbb{N}}2^{1/n}|x_k|^2\geq 2^{1/k}a+b>a+b=1,$$ which is a contradiction. So the sup of the norm is unattained.


问:
ordering of a group
An ordering of a group $G$ is a linear ordering $<$ on (the underlying set of) $G$
that satisfies, in addition,
$a < b \implies ac < bc ∧ ca < cb$,
for all $a,b, c \in G$.

Show that a group $G$ can be ordered if each finitely generated subgroup of $G$
can be ordered.

答:
If you know the compactness theorem for propositional logic, then this is an easy consequence.  Imagine a language in which the propositional variables (also called atomic sentences) are the inequalities $a<b$ for all elements $a,b$ of your group $G$.  You can write the requirements for an ordering of $G$ as a set $S$ of formulas built from these inequalities by means of propositional connectives.  That is, you arrange $S$ so that a truth assignment $v$ satisfies $S$ if and only if the binary relation $\{(a,b): v\text{ gives value true to }a<b\}$ is an ordering of $G$.  Then check that orderings of finitely generated subgroups of $G$ give you satisfying truth assignments for arbitrary finite subsets of $S$, so the compactness theorem gives you the result.

If you don't know the compactness theorem, there are different-looking but basically equivalent approaches, one using Tychonoff's theorem, another using ultrafilters, another (for the case of countable $G$) using König's infinity lemma.  Of course, there are also proofs that start from scratch and essentially re-prove the relevant special case of one of the tools I've mentioned here.

Title: ordering of a group


问:
Interpretation of probability density greater than one
Good morning to everyone,

I'm trying to figure out how to interpret a probability density greater than one, but I didn't find any explanation that I considered satisfactory... Let me formalize my question.

Suppose that you have a random variable $p$ uniformly distributed over the support $[0,\frac 1 2]$. Obviously, it the probability density of this random variable is $f(p)=2$. Does this mean that if I marginally increase the realization of my random variable, the amount of probability "clustered around that new realization" (I couldn't find a better term to express the concept, I'm thinking about the graph of the cumulative, which has slope $2$ in this case) doubles?.

Sorry if I'm not being clear, I couldn't find better words to express my doubts.

答:
The density function of a continuous random variable is not an uncountably infinite 'list' of probabilities.
A continuous random variable has no probability at any one point.

A continuous random variable has positive probabilities only for intervals. (Intervals can be very short, but they cannot shrink of length $0.)$
The density function of a random variable $X$ provides
a way to find probabilities such as $P(0 < X < 0.1).$ By convention, one
writes $P(X = 0.0300)=0,$ and similarly for any other individual value.

Example 1. Let $X \sim \mathsf{Unif}(-.2, .2),$ with density function $f_X(t) = 2.5,$ for $-.2 \le x \le 2$ and $0$ elsewhere.

The total area under the density curve is $1.$
In order for that to be true, notice that the height of the density function must exceed $1$ for some values of $t.$ (See plot below.)

Then
$$P(0 < X < 0.1) = \int_0^{0.1} f_X(t)\,dt
= \int_0^{0.1} 2.5\, dt = 0.25.$$

Consider the plot below:

```<code>curve(dunif(x,-.2,.2), -.5,.5, col="blue",
      lwd=2, n=10001, ylab="PDF", xlab="t",
      main="Density of UNIF(-.2, .2)")
 abline(h=0, col="green2")
 abline(v = c(0, .1), col="red")
</code>```

The desired probability is the area beneath the
(blue) density curve between the red vertical lines.

Example 2.  Suppose that $Y \sim \mathsf{Norm}(\mu = 10, \sigma=0.1),$ with density function $f_Y(t).$ Then
$P(Y \le 9.8) = \int_{-\infty}^{9.8} f_Y(t)\, dt = 0.02275.$

However, the this integral cannot be evaluated
using the ordinary methods of calculus. One must use numerical integration or printed tables (obtained by numerical integration). In R,
one can evaluate this integral using a normal
cumulative distribution function (CDF) pnorm, (also obtained by numerical means).

```<code>pnorm(9.8, 10, .1)
[1] 0.02275013
</code>```

In the figure below, the desired probability is the area under the density curve to the left of
the vertical red line.

```<code>curve(dnorm(x, 10,.1), 9.5,10.5, col="blue",
      ylab="PDF", xlab="t",
      main="Density of NORM(10, .1)")
 abline(h=0, col="green2")
 abline(v = 9.8, col="red")
</code>```

Body:
The density function of a continuous random variable is not an uncountably infinite 'list' of probabilities.
A continuous random variable has no probability at any one point.

A continuous random variable has positive probabilities only for intervals. (Intervals can be very short, but they cannot shrink of length $0.)$
The density function of a random variable $X$ provides
a way to find probabilities such as $P(0 < X < 0.1).$ By convention, one
writes $P(X = 0.0300)=0,$ and similarly for any other individual value.

Example 1. Let $X \sim \mathsf{Unif}(-.2, .2),$ with density function $f_X(t) = 2.5,$ for $-.2 \le x \le 2$ and $0$ elsewhere.

The total area under the density curve is $1.$
In order for that to be true, notice that the height of the density function must exceed $1$ for some values of $t.$ (See plot below.)

Then
$$P(0 < X < 0.1) = \int_0^{0.1} f_X(t)\,dt
= \int_0^{0.1} 2.5\, dt = 0.25.$$

Consider the plot below:

curve(dunif(x,-.2,.2), -.5,.5, col="blue",
      lwd=2, n=10001, ylab="PDF", xlab="t",
      main="Density of UNIF(-.2, .2)")
 abline(h=0, col="green2")
 abline(v = c(0, .1), col="red")

The desired probability is the area beneath the
(blue) density curve between the red vertical lines.

Example 2.  Suppose that $Y \sim \mathsf{Norm}(\mu = 10, \sigma=0.1),$ with density function $f_Y(t).$ Then
$P(Y \le 9.8) = \int_{-\infty}^{9.8} f_Y(t)\, dt = 0.02275.$

However, the this integral cannot be evaluated
using the ordinary methods of calculus. One must use numerical integration or printed tables (obtained by numerical integration). In R,
one can evaluate this integral using a normal
cumulative distribution function (CDF) pnorm, (also obtained by numerical means).

pnorm(9.8, 10, .1)
[1] 0.02275013

In the figure below, the desired probability is the area under the density curve to the left of
the vertical red line.

curve(dnorm(x, 10,.1), 9.5,10.5, col="blue",
      ylab="PDF", xlab="t",
      main="Density of NORM(10, .1)")
 abline(h=0, col="green2")
 abline(v = 9.8, col="red")


问:
Given two functions $f$ and $g$, prove that $f=g$ iff dom$f=$ dom $g$ and for every $x\in$ dom $f$, $f(x)=g(x)$
Given two functions $f$ and $g$, prove that $f=g$ iff dom$f=$ dom $g$ and for every $x\in$ dom $f$, $f(x)=g(x)$

Proof:

Assume $f=g$. Then we have

$$(a_0,b_0)\in f\Rightarrow (a_0,b_0)\in g\wedge(a_1,b_1)\in g\Rightarrow (a_1,b_1)\in f$$
so
$$a_0\in\mbox{dom }f\Rightarrow a_0\in\mbox{dom }g\wedge a_1\in\mbox{dom }g\Rightarrow a_1\in\mbox{dom }f$$
and
$$\mbox{dom }f\subseteq\mbox{dom }g\wedge\mbox{dom }g\subseteq\mbox{dom }f$$
therefore,
$$\mbox{dom }f=\mbox{dom }g$$

Now let $x\in\mbox{dom }f$. Then $\exists y=f(x)\in\mbox{rng }f$ such that $(x,y)\in f$. But if $f=g$, then $(x,y)\in g$.

$$\therefore\forall x\in\mbox{dom }g, f(x)=g(x)$$

Now to show the converse, let $\mbox{dom }f=\mbox{dom }g$ and $\forall x\in\mbox{dom }f$, $f(x)=g(x)$.

Then we have $\forall a_0\in\mbox{dom }f$, $\exists b_0\in\mbox{rng }f$ such that $(a_0,b_0)\in f$. But since $\forall x\in\mbox{dom }f$, $f(x)=g(x)$, we have

$$(a_0,b_0)\in f\Rightarrow(a_0,b_0)\in g$$

Likewise, $\forall a_1\in\mbox{dom }g$, $\exists b_1\in\mbox{rng }g$ such that $(a_1,b_1)\in g$. But again, since $\forall x\in\mbox{dom }f$, $f(x)=g(x)$, we have

$$(a_1,b_1)\in g\Rightarrow(a_1,b_1)\in f$$

Therefore, we have $f\subseteq g$ and $g\subseteq f$, and thus $f=g$.

$\blacksquare$

I'm just looking for proof validation here. Is my argument sound? Thanks in advance.

I'm using the usual set theoretic definition of a function:

Let $A$ and $B$ be sets. A function from $A$ to $B$ is a nonempty relation $f\subseteq A\times B$ that satisfies the following two conditions:

Existence - $\forall a\in A$, $\exists b\in B$ such that $(a,b)\in f$.

Uniqueness - If $(a,b)\in f$ and $(a,c)\in f$, then $b=c$.

答: Your argument is fine, but you are doing more than you need to do in the first part: you are doing the right thing in the second part showing that $f = g$ as sets of pairs if $f$ and $g$ are functions with the same domain that agree on any element of that domain. For the first part, there isn't really anything to prove: if $f = g$, then any property that holds of $f$ also holds of $g$ (because they are the same thing).


问:
Question about Notation of Function
we represent function by letters f,g,h....etc, and the variable of the function are represented by x,y,.....etc.
Thus, the complete representation of function is done by f(x)= RULE . Now The letter y represents the output of the function.
Here, f(x) represents our function, we read f(x) as --> "function of the variable x" meaning we have the Rule as our function and x is the variable in it. thus, this is our complete representation of function with its output.

Now, if this is our complete representation of Function along with its output. Then why many people say f(x) and y both represents the output of the function if we represent the [output of the function with y] and [our function with f(x)] then why people mix both the representations ?
if we have seprate representation (y) for the output of the function and and seprate representation f(x) for our function. then why people say f(x) and y both represents the output of the function.

This is making me very much confused please help me.

Title: Question about Notation of Function

Body: $f = \{ (x, y) : x \in \mathbb{R}, y = 2x\}$


答:
This is basically true by definition, although there are some things you should be aware of.

Some people require that any ring $(R,+_R,\cdot_R)$ contains a multiplicative identity $1_R,$ and that ring homomorphisms $f : (R,+_R,\cdot_R)\to (S,+_S,\cdot_S)$ satisfy $f(1_R) = 1_S.$ If you require this condition, then for any ring $(R,+_R,\cdot_R)$ there is a unique ring homomorphism $i_R : (\Bbb{Z},+,\cdot)\to(R,+_R,\cdot_R).$ In this case, even if the set $R$ does not literally contain $2,$ you may think of $i_R(2)\in R$ as being $2$ (you might even write $i_R(2) = 2_R$). It is true then that for any $r\in R,$
$$
2_R\cdot_R r = i_R(2)\cdot_R r = r +_R r,
$$
because
$$
\begin{align*}
i_R(2)\cdot_R r &= i_R(1 + 1)\cdot_R r\\
&= (i_R(1) + i_R(1))\cdot_R r\\
&= (1_R + 1_R)\cdot_R r\\
&= r + r.
\end{align*}
$$
As JonathanZ supports MonicaC notes, it might be the case that $i_R(2)$ behaves differently than you'd expect, or looks different than you'd expect. It could be that $i_R(2) = -1_R$ or even $i_R(2) = 0_R$! See the last paragraph for a particularly outrageous example of this.

If you do not require that your rings have multiplicative identitites and/or that ring homomorphisms need not send multiplicative identities to multiplicative identities, then this is still true to some extent, although we should be careful about what we mean.

Let $(R,+_R,\cdot_R)$ be our possibly non-unital ring. In this case, we can't use the unique homomorphism $i_R :(\Bbb{Z},+,\cdot)\to(R,+_R,\cdot_R)$ from before -- there might be more than one ring homomorphism now! Additionally, the set $R$ might not contain $2.$

So, what do we do? Well, remember that any ring has an underlying abelian group $(R,+_R).$ An abelian group is the same as a $\Bbb{Z}$-module (see here for the definition of a module over a ring if you're not familiar). This means explicitly that we have an action of $\Bbb{Z}$ on $R$ which interacts nicely with addition. We define this action by setting
$$
n\cdot r :=\begin{cases}
\underbrace{r + \dots + r}_{n\textrm{ times}},&n > 0,\\
0,&n=0,\\
\underbrace{-r + \dots + -r}_{-n\textrm{ times}}, &n <0.
\end{cases}
$$
Notice that I'm not writing $n\cdot_R r$ -- that's because there's not necessarily an element $n\in R$ which behaves like $n.$ However, it's still sensible to think of adding the element $r$ to itself $n$ times, which is what $n\cdot r$ means by definition. The $\cdot$ refers to the action of $\Bbb{Z}$ on the underlying abelian group of $(R,+_R,\cdot_R),$ not multiplication in the ring itself. In this sense, the equality
$$
2\cdot r = r+r
$$
always holds, and this is basically by definition!

One last remark. You asked if this is true of any ring which has $\Bbb{R}$ as its underlying set. You should be slightly careful here. Consider the following ring structure on $\Bbb{R}$:
$$
\begin{align*}
+' : \Bbb{R}\times\Bbb{R}&\to\Bbb{R}\\
(r,s)&\mapsto r+'s:=\sqrt[3]{r^3 + s^3},\\
\cdot' : \Bbb{R}\times\Bbb{R}&\to\Bbb{R}\\
(r,s)&\mapsto r\cdot's := rs.
\end{align*}
$$
This is not the standard ring structure on $\Bbb{R}$ -- the multiplication is the same, but the addition is "twisted." In this case, $2\in \Bbb{R}$, but it is not true that $2\cdot' r = r +' r.$ Suppose $r = 2.$ Then:
$$
\begin{align*}
2 +' 2 &= \sqrt[3]{2^3 + 2^3}\\
&= \sqrt[3]{16}\\
&= 2\sqrt[3]{2}.
\end{align*}
$$
On the other hand,
$$
2\cdot'2 = 4.
$$
What happened? I'll let you think about this for yourself before revealing the answer below!

 What happened here is that $2\in\Bbb{R}$ is no longer playing the same role it was before. Our ring $(\Bbb{R},+',\cdot')$ still has a multiplicative identity, but our ring homomorphism $i_{(\Bbb{R},+',\cdot')} : (\Bbb{Z},+,\cdot)\to(\Bbb{R},+',\cdot')$ now sends $$i_{(\Bbb{R},+',\cdot')}(2) = i_{(\Bbb{R},+',\cdot')}(1) +' i_{(\Bbb{R},+',\cdot')}(1) = 1 +' 1  = \sqrt[3]{2}.$$So there is an element of $(\Bbb{R},+',\cdot')$ which behaves like $2$ should -- it's $\sqrt[3]{2}$. We thus have$$\sqrt[3]{2}\cdot' r = r +' r$$for any $r\in\Bbb{R}.$ This is very confusing, because we already have $2\in\Bbb{R}$! In this case, it would be very important to distinguish between $2\cdot r$ (which is $2\in\Bbb{Z}$ acting on $r,$ giving $r +'r$) and $2\cdot' r$ (which as we computed, is not $r +' r$ in general). In the notation of the first paragraph, $2_{(\Bbb{R},+',\cdot')} = \sqrt[3]{2}$ and $2\neq 2_{(\Bbb{R},+',\cdot')}$.

 To be even more explicit about what happened, given any set $X,$ any ring $(R,+_R,\cdot_R),$ and any bijection $f : X\to R,$ we can give $X$ the structure of a ring by defining addition on $X$ by $x +_X y := f^{-1}(f(x)+_R f(y))$ and $x\cdot_X y := f^{-1}(f(x)\cdot_R f(y)).$ We're taking the ring structure on $R$ and transporting it to $X$ via the bijection $f$: first, take your elements $x$ and $y$ in $X,$ send them over to $R$ where you add or multiply them, and then bring them back to $X.$ In my example above, I'm using the bijection $\Bbb{R}\to\Bbb{R}$ which sends $x$ to $x^3.$

Title: Definition on multiplication in rings

Body:
Let's say I have a ring $R=(\mathbb{R}, +, *)$. Is

$a+a=2*a$

($a \in \mathbb{R}$)

always true for any ring¹? (So that when this equation is not true then I can definitively infer that $R$ is not a ring.)

1 ) With "any" I mean any other ring which is also using $\mathbb{R}$ as underlying set, thank you for your comment nick.


问:
$\nabla \frac{1}{|x|}$ is Lipschitz continuous
Let $x, y \in \mathbb{R}^3.$
How to prove $\nabla \frac{1}{|x|}$ is Lipschitz continuous for $|x|\geq1$?

I tried that WLOG suppose $|y|\geq |x|$, then

$$ \left| \frac{x}{|x|^3} - \frac{y}{|y|^3} \right| =  \frac{1}{|x|^3} \left| x - \frac{|x|^3}{|y|^3}y \right| \leq C(?)|x-y|$$

I don't know how to construct $C$.

答: Hint: $$\frac{x|y|^3-y|x|^3}{|x|^3|y|^3}=\frac{(x-y)|y|^3}{|x|^3|y|^3}+\frac{y(|y|^3-|x|^3)}{|x|^3|y|^3}.$$Now factor the numerator in the second fraction on the right and note that $\left||y|-|x|\right|\le|x-y|.$


答:
You can just try to be a bit more explicit and compute the square norm:

$$\left|\dfrac{x}{|x|^3} - \dfrac{y}{|y|^3}\right|^2 = \dfrac{(x^T|y|^3 - y^T|x|^3)(x|y|^3 - y|x|^3)}{|x|^6|y|^6} = \dfrac{|x|^4+|y|^4 - 2|x||y|x^Ty}{|x|^4|y|^4}.$$

Now use Cauchy-Schwarz as:
$$x^Ty\leq |x||y|$$
then you are basically done.

Body:
Let $x, y \in \mathbb{R}^3.$
How to prove $\nabla \frac{1}{|x|}$ is Lipschitz continuous for $|x|\geq1$?

I tried that WLOG suppose $|y|\geq |x|$, then

$$ \left| \frac{x}{|x|^3} - \frac{y}{|y|^3} \right| =  \frac{1}{|x|^3} \left| x - \frac{|x|^3}{|y|^3}y \right| \leq C(?)|x-y|$$

I don't know how to construct $C$.


问:
Computing $\iiint_\mathbb{R^3} e^{-x^2-y^2-z^2}dxdydz$ using substitution
Consider this integral:
$$\iiint_\mathbb{R^3} e^{-x^2-y^2-z^2}dxdydz$$
How would you compute it?

I already solved this problem this way:
$$\iiint_\mathbb{R^3} e^{-x^2-y^2-z^2}dxdydz = \left( \int_\infty^\infty e^{-x^2} \right)^3 = \pi^{3/2}$$

But I wanted to find it using substitution (spherical coordinates) but this is all I could do:
$$\iiint_\mathbb{R^3} e^{-x^2-y^2-z^2}dxdydz = \lim_{j\,\rightarrow\infty}\int_0^jdu\int_0^\pi dv\int_0^{2\pi} e^{-u^2}u^2\sin(v) dw=$$

$$=2\pi\lim_{j\,\rightarrow\infty}\int_0^jdu\int_0^\pi e^{-u^2}u^2\sin(v)dv=4\pi\lim_{j\,\rightarrow\infty}\int_0^je^{-u^2}u^2du$$

But it doesn't get me anywhere. Help would be very appreciated, thanks in advance.

Body: $$\int_0^{\infty}e^{-x^2}x^2dx = \frac 1 2 \int_0^{\infty } e^{-x^2} x \;  2x dx = \frac 1 2 \int_0^{\infty } e^{-u} \sqrt u du = \frac{\Gamma (3/2)}{2} = \frac 1 4\sqrt{\pi}$$


问:
$M\otimes A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}\neq0$ if and only if $\text{Ann}(M)\subseteq \mathfrak{p}$?
Let $A$ be a Noetherian ring, $M$ be a finitely generated $A$-module, and $\mathfrak{p}$ be a prime ideal in $A$.

I want to ask how to prove that $M\otimes A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}\neq0$ if and only if $\text{Ann}(M)\subseteq\mathfrak{p}$.

Thanks.

答:
(We don't need to assume that $A$ is Noetherian.)

$M_{\mathfrak{p}} \cong M \otimes_A A_{\mathfrak{p}}$ is finitely generated over $A_{\mathfrak{p}}$, so by Nakayama, we get $0 \neq M_{\mathfrak{p}}/\mathfrak{p}M_{\mathfrak{p}} \cong M \otimes_A A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$ if and only if $M_{\mathfrak{p}} \neq 0$ which is by definition the case if and only if $\mathfrak{p} \in \operatorname{Supp}(M)$, so your question reduces to the relation $V(\operatorname{Ann}(M))=\operatorname{Supp}(M)$ which is shown for example here.


问:
What is the probability that some people are chosen in different groups?
Let's say that I have $n$ people and $m \le n$ of this people are malicious. Then I run a process that chooses 6 people randomly and assigns each of them to one of two different groups $A$ and $B$ such that each group in the end has 3 people. What is the probability that there will be at least a malicious person in every group?</p>

答:
I think our approach to this problem should ultimately depends on whether or not group $A$ and group $B$ are distinguishable, and they certainly could be. For example, if the individuals in group $A$ are assigned to control a company's finances while the individuals in group $B$ are assigned to control a company's internal affairs, then group $A$ and group $B$ are distinguishable.

First suppose $A$ and $B$ are distinguishable. How many ways can we construct two distinguishable groups of three people so that at least one group contains no malicious people? Well, there are $${n-m \choose 3}{n-m-3 \choose 3}+2{n-m \choose 3}\sum_{k=1}^3{m \choose k}{n-m-3 \choose 3-k}$$ ways to do this. Since there are ${n \choose 3}{n-3 \choose 3}$ different ways to select the two distinguishable groups of three people, the probability that both distinguishable groups contain at least one malicious individual equals $$1-\frac{{n-m \choose 3}{n-m-3 \choose 3}+2{n-m \choose 3}\sum_{k=1}^3{m \choose k}{n-m-3 \choose 3-k}}{{n \choose 3}{n-3 \choose 3}}$$

Now suppose group $A$ and group $B$ are indistinguishable. Let's count the number of ways we can choose the two indistinguishable groups of three people so that at least one group contains no malicious people:
$$\frac{1}{2!}{n-m \choose 3}{n-m-3 \choose 3}+{n-m \choose 3}\sum_{k=1}^3{m \choose k}{n-m-3 \choose 3-k}$$ Because there are $\frac{1}{2!}{n \choose 3}{n-3 \choose 3}$ ways to choose the two indistinguishable groups, the desired probability in this situation equals $$1-\frac{\frac{1}{2!}{n-m \choose 3}{n-m-3 \choose 3}+{n-m \choose 3}\sum_{k=1}^3{m \choose k}{n-m-3 \choose 3-k}}{\frac{1}{2!}{n \choose 3}{n-3 \choose 3}}$$


问:
Why $\int{\frac{dr}{r\sqrt{c^{2}r^4 -1}}} = -\frac{1}{2} \arcsin\left(\frac{1}{cr^2}\right)$?
I must solve the following integral, where $c$ is a constant $$\int{\frac{dr}{r\sqrt{c^{2}r^4 -1}}}$$ When I compute and do the calculations, I obtain that $$\int{\frac{dr}{r\sqrt{c^{2}r^4 -1}}} = \frac{1}{2} \arctan \left(\sqrt{c^{2}r^4 -1} \right)$$ I have rectified my calculations many times and I always get the same result, but in the book I'm studying I have that $$\int{\frac{dr}{r\sqrt{c^{2}r^4 -1}}} = -\frac{1}{2} \arcsin\left(\frac{1}{cr^2}\right)$$  Why? Is it some mistake in the book or is it my own mistake? The book I mention is: The calculus of variations by Van Brunt page 47, there the equation arises through an example on the invariance of the Eule-Lagrange equation.

Attached images:

Body:
In fact
$$ \tan\bigg(\cos^{-1}\bigg(\frac{1}{cr^2}\bigg)\bigg)=\frac{\sin\bigg(\cos^{-1}\bigg(\frac{1}{cr^2}\bigg)\bigg)}{\cos\bigg(\cos^{-1}\bigg(\frac{1}{cr^2}\bigg)\bigg)}=\frac{\sqrt{1-\frac1{c^2r^4}}}{\frac{1}{cr^2}}=\sqrt{c^2r^4-1}. $$
Since $\frac{1}{cr^2}>0$, $\cos^{-1}\bigg(\frac{1}{cr^2}\bigg)\in(0,\pi/2)$ and hence
$$ \cos^{-1}\bigg(\frac{1}{cr^2}\bigg)\bigg)=\tan^{-1}\sqrt{c^2r^4-1}. $$
Using
$$ \sin^{-1} x+\cos^{-1} x=\frac{\pi}{2} $$
and hence
$$ -\sin^{-1}x=\cos^{-1}x-\frac{\pi}{2}. $$
one has
$$ -\sin^{-1}\bigg(\frac{1}{cr^2}\bigg)=\cos^{-1}\bigg(\frac{1}{cr^2}\bigg)-\frac{\pi}{2}=\tan^{-1}\bigg(\sqrt{c^2r^4-1}\bigg)-\frac{\pi}{2}. $$


问:
Fourier transform through graph
I have to find the Fourier transform of this function:

I wrote that :

$x(t) = \sum_{k= -\infty}^{\infty}\sin(\frac{2\pi k}{3})\delta(t-k)$ where $k \in Z$

It looks like a convolution but the $t$ should be $\in Z$
Can I do this (?)

$X(\omega) = $
$\int_{-\infty}^{\infty}\sum_{k= -\infty}^{\infty}\sin(\frac{2\pi k}{3})\delta(t-k)e^{j\omega t}dt$   $=\sum_{k= -\infty}^{\infty}\int_{-\infty}^{\infty}\sin(\frac{2\pi k}{3})\delta(t-k)e^{j\omega t}dt = \sum_{k= -\infty}^{\infty}sin(\frac{2\pi k}{3})\int_{-\infty}^{\infty}e^{j\omega t}\delta(t-k)dt$

And the Fourier tranform of :  $\delta(t-k) \rightarrow e^{j\omega k}$

\
So : $X(\omega) = \sum_{k= -\infty}^{\infty}\sin(\frac{2\pi k}{3})e^{j\omega k}$

And : $\sin(\frac{2\pi k}{3}) =\frac{1}{2j}(e^{j\frac{2\pi k}{3}}  - e^{-j\frac{2\pi k}{3}})$

$X(\omega) = \sum_{k= -\infty}^{\infty}\frac{1}{2}( e^{k(j\frac{2\pi}{3} + j\omega)}  - e^{k(-j\frac{2\pi k}{3}+ j\omega)})$  but how do I proceeed?

update**
I thought another way :
this signal is periodic so I may can express it with fourier - analysis :
$ c_n = \begin{cases}
      \delta(t-n) & n = 3k + 1 \\
      -\delta(t-n) & n = 3k +2 \\
      0 & n = 3k + 3
   \end{cases}$ where $k \in Z$

Then we can express $x(t)$ as :

$x(t) = \begin{cases}
      \sum_{n=-\infty}^{\infty}\delta(t-n)& n =3k+1 \\
       \sum_{n=-\infty}^{\infty}-\delta(t-n)& n=3k+2 \\
      0 & n = 3k + 3
   \end{cases} \rightarrow  x(t) = \sum_{n=-\infty}^{\infty}\delta(t-nT -1)  -\delta(t-nT -2)$

In my table with fourier transforms (provided by my uni) there this property:

$\sum_{n=-\infty}^{\infty}\delta(t-nT) \leftrightarrow \frac{2\pi}{T}\sum_{n=-\infty}^{\infty}\delta(\omega - n\omega_0)$

$\\ \\$

So :
$X(\omega) =\sum_{n=-\infty}^{\infty}e^{-j\omega}\delta(\omega -\omega_0)-e^{-j\omega 2}\delta(\omega -\omega_0) = \sum_{n=-\infty}^{\infty}e^{-j\omega}\delta(\omega -\omega_0)e^{-j\omega}(1 - e^{-j\omega}) $

And $e^{-j\omega}(1 - e^{-j\omega}) = e^{-j\omega}e^{\frac{-j\omega}{2}}(e^{\frac{j\omega}{2}} - e^{\frac{-j\omega}{2}} ) = e^{\frac{-3j\omega}{2}}(2jsin(\frac{\omega}{2}))$

And

$X(\omega) =
     \frac{2\pi}{T} e^{\frac{-3j\omega}{2}}(2jsin(\frac{\omega}{2}))\sum_{n=-\infty}^{\infty}\delta(\omega -\omega_0)
$

答:
First of all, due to the rule $f(x)\delta(x-a)=f(a)\delta(x-a)$:

$$x(t) = \sum_{k= -\infty}^{\infty}\sin(\frac{2\pi k}{3})\delta(t-k)= \sum_{k= -\infty}^{\infty}\sin(\frac{2\pi t}{3})\delta(t-k)$$

Hereafter, I advise you to use the formalism of Dirac comb(s):

$$x(t)=\sin(\frac{2\pi t}{3})\underbrace{\sum_{k= -\infty}^{\infty}\delta(t-k)}_{\text{unit Dirac comb }\text{Ш}}\tag{1}$$

where, more generally:

$$\displaystyle \operatorname {\text{Ш}} _{T}(t)\ \triangleq \ \sum _{k=-\infty }^{\infty }\delta (t-kT)={\frac {1}{T}}\operatorname {\text{Ш}} \left({\frac {t}{T}}\right)$$

with the property that:

$$\displaystyle \operatorname {\text{Ш}} _{T}(t){ \ \ \stackrel {\mathcal {F}} {\longleftrightarrow }} \ \ {\frac {\sqrt {2\pi }}{T}}\operatorname {\text{Ш}} _{\frac {2\pi }{T}}(\omega ) \ \ \text{or } \ {\frac {1}{\sqrt {2\pi }}}\sum _{n=-\infty }^{\infty }e^{-i\omega nT}\tag{2}$$

(the last expression making the link with Fourier series, but we haven't to use it here).

Therefore, using (2), the Fourier Transform of product (1) is convolution product

$$(\mathcal{F}x)(\omega)=-i \pi (\delta(\omega-\frac{2\pi}{3})-\delta(\omega+\frac{2\pi}{3})) \color{red}{\star}
 \sqrt {2\pi } \sum _{k=-\infty }^{\infty }\delta (\omega-2 \pi k)$$

It remains to take all the convolution products, an easy task using the shifting property of delta distribution:

$$\delta(\omega-a) \star \delta(\omega-b)=\delta(\omega-(a+b)).$$

Remark 1: Formula (2) is rather natural: a Dirac comb with quite spaced "teeth" will be transformed into a Dirac comb with quite clenched "teeth".

Remark 2: I see in your computation that you come back to the definition of Fourier Transform (as an integral). Why not, but in virtually all practical cases, we only need tables and formulas.

Title: Fourier transform through graph


问:
Order of permutation $a \in S_n$ if $a^k$ is a cycle of length n

Let $a$ be an element of $S_n$ ,the permutation group of order n. $a^k$ is a cycle of length n. Then what is the order of $a$?

If $n$ is prime then a should be a cycle of length $n$.But if $n$ is not prime then how to find the order of $a$? Let $a$  can be expressed as the product of some disjoint cycles. Then the order of $a$ is lcm of the length of the disjoint cycles and $n$ should divide the order of $a$. After that, I'm not getting anything.

答: Let $l:=|a|$ be the order of $a$. Hence $n=|a^k|=\frac{l}{(l,k)}$. Therefore, the order of $a$ is a solution $l$ of the integral equation $$l=n(l,k),$$ $(l,k)=\textrm{gcd}\{l,k\}$.


问:
What are the irreducible elements of the polynomial ring $\mathbb Q[x]$?

Let $R$ be the ring of $f(x) \in \mathbb{Q}[x]$ such that $f(0) \in \mathbb{Z}$. Find the irreducible elements of $R$.

I believe I need to prove three things. First, that elements of the form $\pm p$ for $p$ prime and functions $f(x)$ where $f(0) = \pm 1$ are irreducible. Second, I need to show that non-constant polynomials where $f(0) \neq \pm 1$ are not irreducible. Third, I need to show that constant polynomials that are not prime and not $\pm 1$ are reducible.

The last one is straightforward. If $f(x) = c$, then $f(x) = ab$ where $a,b$ are neither $c$ nor $1$ or $-1$. Since $\pm 1$ are the only units in $R$, neither $a$ nor $b$ are units, so $f(x)$ is reducible.

I can't figure out how to prove the first two parts or if I'm even on the right track.

答:
First, if $f(0)=0$, then $f/2\in R$ hence $f=2\frac f2$ is reducible.

Similarly if $|f(0)|>1$, then $f/f(0)\in R$ and thus $f$ is reducible, unless it's constant, when it needs to be prime in order to be irreducible.

Now prove that if $f(0)=\pm1$ then $f$ is irreducible in $R$ iff it's irreducible in $\Bbb Q[x]$ (which is not necessarily straightforward to decide, by the way).

 If $f=gh$ in $R$ then $g$ and $h$ are not constant (unless $\pm1$) since $f(0)=\pm1$, so it shows reducibility in $\Bbb Q[x]$.

 For the converse, if $f=gh$ with nonconstant $g,h\in\Bbb Q[x]$ and $f\in R,\,f(0)=\pm1$, then we have $f(0)=g(0)h(0)=\pm1$ and thus $f=f(0)\cdot g/g(0)\cdot h/h(0)$ is a decomposition of $f$ in $R$.


问:
Inequality in proving the Isoperimetric Inequality
The question came from reading the following post: A proof of the Isoperimetric Inequality - how does it work?

I almost can follow the whole proof, but I am stuck at one point. Why does $(x^2 +\bar{y}^{2})⋅({y_s}^2 +{x_s}^2)=x^2+\bar{y}^{2}$?

I actually came across this proof in Do Carmo's "Differential Geometry" and had the same question. I'm still new to this site so if this is the wrong place to post this, please let me know.

Thanks!

答: The curve is parametrized by arc length, i.e. the ''speed'', $x_s'^2 + y_s'^2$, equals $1$.</p>


问:
Proving a necessary condition for an egyptian fraction with odd denominators
In the question posted in  Proving an equality involving cyclic sums, I realized that all the possible solutions to the following egyptian fraction with denominators from a set of odd positive integers $S=\{x_1,x_2,x_3,x_4\}$ $$1 = \frac{2}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}$$ have the following property:

Property

There is one $x_j\in S$ such that for every $x_i\in S$, $$\gcd(x_j,x_i)=x_i$$

I copy from the question mentioned the possible solutions of the above egyptian fraction, which have motivated my  present question, for you to check:

$$[3, 5, 9, 45]
[3, 5, 15, 15]
[3, 7, 7, 21]
[3, 9, 9, 9]
[3, 5, 5, 15]
[5, 5, 5, 5]$$

However, I am stuck when looking for a proof of this property as a necessary condition for every solution, other than the fact that the solutions listed are all the possible solutions. I am able to prove that every $x_k$ divides $\prod _{i\neq k} x_i$, but nothing more. Any help on this proof would be welcomed!

答:
I think I have a proof of the property for most cases, I post it for your revision and looking for ideas for the case left.

Proof

We have that $$1 = \frac{2}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}$$

This is only possible if $1<\min\{x_1,x_2,x_3,x_4\}\leq5$. Therefore, the only two possibilities left are that $\min\{x_1,x_2,x_3,x_4\}=3$ or $\min\{x_1,x_2,x_3,x_4\}=5$. An interesting fact here is that both are prime numbers.

If $\min\{x_1,x_2,x_3,x_4\}=5$, then necessarily the property holds, as necessarily $x_1=x_2=x_3=x_4$ is the only possible solution.

If $\min\{x_1,x_2,x_3,x_4\}=3$, all the inmmediate greater odd numbers are 5, 7, and 9. If $x_1=3$, then at least one of the others denominators must be equal or lesser than 9, because $1 = \frac{2}{3}+\frac{3}{9}$ (in which case the property holds, as all the denominators are different combinations of 3 and 9). Therefore, any other solution other than the ones mentioned implies that some other denominator is either 5 or 7, which are prime numbers.

As I commented when posting the question, it is easy to show that every $x_k$ divides $\prod _{i\neq k} x_i$. For the proof, the numerators of the fractions are irrelevant, so we will assume without loss of generality that $\min\{x_1,x_2,x_3,x_4\}=x_1$. Therefore, $x_1\mid x_2x_3x_4$, where $\mid$ means "divide".

As $x_1$ is prime, then some other denominator must be equal to $x_1$ or some multiple of $x_1$; in any of those cases, the property holds for both denominators. As the numerators of the fractions are irrelevant, we will assume without loss of generality that $x_1\mid x_2$.

Let us assume, also without loss of generality, that $x_3$ is prime (as stated, either 5 or 7). We have that $x_3\mid x_1x_2x_4$. It could happen:

(i)That $x_3\mid x_2$, in which case the property will continue holding for $x_1$, $x_2$ and $x_3$,

(ii) That $x_3\mid x_4$. If this were the case, when considering that $x_4\mid x_1x_2x_3$, we would have the possibility of $x_4\mid x_1x_3$, in which case the property would hold for $x_1$, $x_3$ and $x_4$, and otherwise that $x_4\mid x_3$, which would imply that $x_3=x_4$.

In all the cases exposed where the property holds for three of the four denominators, we can show that necessarily it must hold for all of them. For instance, let us assume that $\gcd(x_1,x_2)=x_1$ and $\gcd(x_3,x_2)=x_3$. Then, multiplying by $x_2$, we would have that $$x_2 = \frac{2x_2}{x_1}+\frac{x_2}{x_2}+\frac{x_2}{x_3}+\frac{x_2}{x_4}$$ As all the fractions are integer numbers, then necessarily we get that $x_4\mid x_2$, and otherwise the equation would not be possible. The same process can be applied for all the mentioned cases where the property holds for three of the four denominators.

The only case left for consideration is the one where we had that $x_1\mid x_2$ and $x_3=x_4$. Substituting $x_4$ for $x_3$, we have this two possible cases, depending on the numerator being in a fraction with $x_1$ or $x_2$ as denominator, or $x_3$ as denominator:

(i) $$x_2 = \frac{2x_2}{x_1}+\frac{x_2}{x_2}+\frac{x_2}{x_3}+\frac{x_2}{x_3}$$ In this case, the equation has solution only if $\frac{2x_2}{x_3}$ is integer, which can only be possible if $x_3\mid x_2$, and in which case the property would hold;

(ii) $$x_2 = \frac{2x_2}{x_3}+\frac{x_2}{x_2}+\frac{x_2}{x_1}+\frac{x_2}{x_3}$$ In this case, the equation has solution only if $\frac{3x_2}{x_3}$ is integer, which can only be possible if $x_3\mid x_2$, and in which case the property would hold.

However, it is still not proved the case $x_4\mid x_2x_3$ when $x_4>x_3$, and $x_2\mid x_1x_4$, when $x_2>x_1$.


答:
A complementary of Lukas's excellent answer.

Proposition. If $M$ is a finitely generated $A$-module, then $\operatorname{Supp}(M)=V(\operatorname{Ann}(M))$.

Proof. If $\mathfrak{p}\notin\operatorname{Supp}(M)$, then $M_\mathfrak{p}=0$. Then for any $m\in M$ and $a\in A-\mathfrak{p}$, there exists $s\in A-\mathfrak{p}$ such that $sm=0$. Thus for generators $m_1,...,m_n$, we can find $s_1,...,s_n\in A-\mathfrak{p}$ such that $s_1\cdots s_nm=0$ for any $m\in M$, thus $\mathfrak{p}$ doesn't contain $\operatorname{Ann}(M)$.

In the other side, if $\mathfrak{p}$ doesn't contain $\operatorname{Ann}(M)$, then there exists an element $t\in A-\mathfrak{p}$ such that $tm=0$ for any $m$. Therefore $M_\mathfrak{p}=0$.

Title: $M\otimes A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}\neq0$ if and only if $\text{Ann}(M)\subseteq \mathfrak{p}$?

Body:
Let $A$ be a Noetherian ring, $M$ be a finitely generated $A$-module, and $\mathfrak{p}$ be a prime ideal in $A$.

I want to ask how to prove that $M\otimes A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}\neq0$ if and only if $\text{Ann}(M)\subseteq\mathfrak{p}$.

Thanks.


问:
Does conjugation preserve trace?
If I were to conjugate by a matrix, what kind of matrix would it have to be in order to preserve trace? Does the trace of a matrix have anything to do with its spectrum? Any insights are appreciated, I'm just a bit confused.</p>

答:
For any square matrix $M$ whatsoever and any invertible matrix $Q$ of the same size, we have that the trace of $M$ equals the trace of $QMQ^{-1}$. No conditions on $Q$ or $M$. This follows immediately once you have the fact that $\text{tr}(AB) = \text{tr}(BA)$.

You asked for the relationship to the spectrum -- the trace is the sum of the eigenvalues (with appropriate multiplicities).

Title: Does conjugation preserve trace?


问:
Different definitions of Kodaira dimension
Let X be a smooth projective variety with canonical class K.

Let a be defined to be the maximum dimension of the image of X under the rational map induced by the linear system |nK| as n ranges over all positive integers.

Let b be defined to be the smallest positive integer c such that the sequence (dim(H^0 (X, nK))/n^c) is bounded (as n gets large).

The second definition I guess doesnt include the case with kodaira dimension smaller than 0 (ie -1 or -infinity depending on how you like to define it), so we're only talking about non-negative kodaira dimension.

Will someone please tell me why a=b?

答:
This is by no means a trivial result!

It is usually attributed to Iitaka, and is more or less contained in the so-called Iitaka fibration theorem.
If you are looking for references, there is a book from Ueno, "Classification Theory of Algebraic Varieties and Compact Complex Spaces", and of course the book from Lazarsfeld "Positivity in Algebraic Geometry, I".

Two side remarks:

this is not only suited for the canonical bundle, but holds for any line bundle on a smooth projective variety.

we define $\kappa(L)=- \infty$ when $|nL|=\emptyset$ for all $n$.

OwnerDisplayName: Robert Garbary

ContentLicense: CC BY-SA 2.5


问:
Is the winding number of a map $\Omega: \; S^n \mapsto S^n$ dependent on the radius of both spheres?
I have 2 questions regarding homotopy groups.

My first questions comes from the fact that I've read different definitions of homotopy groups. The book by Manton and Sutcliffe defines them as:

''The set of homotopy classes of based maps $\Psi: S^n \mapsto Y$'', where $S^n$ is a n-dimensional unit sphere. But most other sources that I've read about about this topic do not define $S^n$ necessary as a unit sphere, they don't define the radius of the sphere at all, therefore I'm not sure what is right or wrong.

My second question is for the situation where we consider a mapping from $\Psi: S^n \mapsto S^n$, i.e. $\pi_n(S^n)=\mathbb{Z} \; \forall \; n$. Now I've read in a lot of sources (for instance in Magnetic Monopoles by John Preskill) that say the winding number is a conserved quantity. I understand that the winding number is "preserved by continuous deformations''.
However, I wonder if the winding number is also conserved if the radius of either n-spheres is changing with time. For instance, there exist a situation where $\pi_1(S^1)=2$, but there is surely also a situation where $\pi_1(S^1)=3$. I.e. they have different winding numbers, in my mind (I'm most likely wrong) this is because the radius of either spheres have changed. I don't, intuitively, understand how else this is possible.

Just in case my background is important: I'm physics student, so have no mathematical skills regarding topology. I'm not very good in mathematical proofs, but I would really like to intuitively understand this topic.

Thanks in advance!

答: The radius of the sphere doesn't matter. Moreover, if you take a map from a particular $n$-sphere to $\mathbb R^{n+1}-\{0\}$, then you really see it as a winding number around the origin. As long as you never hit the origin by mistake as $t$ varies, this number will stay constant.</p>


问:
Prove that a function is continuous (square integrability)
I need help for the following proof of continuity:

Let $E=L_2([t_0,t_1],\mathbb R)$ be a Hilbert space of
  square-integrable real-valued functions on $[t_0,t_1]$. Let
  $K(t,\tau):[t_0,t_1]\times[t_0,t_1]\to\mathbb R$ be such that:

$$ \int_{t_0}^{t_1}\int_{t_0}^{t_1} |K(t,\tau)|^2dtd\tau<\infty $$

Define $A:E\to E$ by:

$$ (A(u))(t)=\int_{t_0}^{t_1} K(t,\tau)u(\tau)d\tau\quad\forall t\in [t_0,t_1] $$

Show that $A$ is continuous.

Attempt at solution

Let $u_1,u_2\in E$ such that $\|u-u'\|_{L_2}<\delta$, in other words:

$$
\begin{array}{rcl}
\left\langle u-u',u-u' \right\rangle & = & \int_{t_0}^{t_1} \overline{(u-u')(t)}^T(u-u')(t)dt<\delta^2 \\
& = & \int_{t_0}^{t_1} (u(t)-u'(t))^2dt<\delta^2
\end{array}
$$

since the functions maps to real numbers, we can get rid of the complex conjugate and the transpose. Now consider:

$$
\begin{array}{rcl}
\left\langle A(u-u'),A(u-u') \right\rangle & = &
\int_{t_0}^{t_1} (A(u(t)-u'(t)))^2 dt \\
& = &
\int_{t_0}^{t_1} \left( \int_{t_0}^{t_1} K(t,\tau)(u(\tau)-u'(\tau))d\tau \right)^2 dt \\
& \le &
\int_{t_0}^{t_1} \int_{t_0}^{t_1} K(t,\tau)^2(u(\tau)-u'(\tau))^2d\tau dt
\end{array}
$$

But at this point I get stuck. I feel like I'm on the right track because I can see this leading to an integral that I can bound by some $M\in\mathbb R$ due to the property of $K(t,\tau)$ in the question and by $\delta$ coming from the assumption $\|u-u'\|_{L_2}<\delta$. However, I don't know how to get this.

Thanks for helping!

答:
Firstly, we verify that $A$ is well-defined. Let $u\in E$. By Cauchy-Schwarz
inequality, we have that
\begin{eqnarray*}
|A(u)(t)|^{2} & = & \left(\int_{t_{0}}^{t_{1}}K(t,\tau)u(\tau)d\tau\right)^{2}\\
 & \leq & \left(\int_{t_{0}}^{t_{1}}K^{2}(t,\tau)d\tau\right)\left(\int_{t_{0}}^{t_{1}}u^{2}(\tau)d\tau\right).
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
\int_{t_{0}}^{t_{1}}|A(u)(t)|^{2}dt & \leq & \left(\int_{t_{0}}^{t_{1}}\int_{t_{0}}^{t_{1}}K^{2}(t,\tau)d\tau dt\right)\left(\int_{t_{0}}^{t_{1}}u^{2}(\tau)d\tau\right)\\
 & < & \infty.
\end{eqnarray*}
This shows that $A(u)\in E$. It is routine to verify that $A$ is
linear, i.e., $A(\alpha u+v)=\alpha Au+Av$. Finally,
\begin{eqnarray*}
||A(u)||^{2} & = & \int_{t_{0}}^{t_{1}}|A(u)(t)|^{2}dt\\
 & \leq & \left(\int_{t_{0}}^{t_{1}}\int_{t_{0}}^{t_{1}}K^{2}(t,\tau)d\tau dt\right)\left(\int_{t_{0}}^{t_{1}}u^{2}(\tau)d\tau\right)\\
 & = & \left(\int_{t_{0}}^{t_{1}}\int_{t_{0}}^{t_{1}}K^{2}(t,\tau)d\tau dt\right)||u||^{2}.
\end{eqnarray*}
Hence, $A$ is a bounded linear operator on the Hilbert space $E$
and therefore $A$ is continuous (with repsect to the norm topology).


问:
Showing that $\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha)$ with $\alpha=\sqrt{(2+\sqrt{2})(3+\sqrt{3})}$ is a Galois extension.
Let $L=\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha)$  with $\alpha=\sqrt{(2+\sqrt{2})(3+\sqrt{3})}$

Show that:

i) $\alpha$ is a primitive element

ii) $L/\mathbb{Q}$ is a Galois extension.

iii) $L/\mathbb{Q}\simeq Q_8$

iv) There is no irreducible polynomial of the form  $p(x)=x^4-ax^2+b\in\mathbb{Q}[x]$, which has a root in $L$

Next I put my solution omitting some calculation details and topics such as proving that the extensions obtained by each root are in effect automorphisms.

Let $K:=\mathbb{Q}(\sqrt{2},\sqrt{3})$

i)

Affirmation 0 . $\alpha$ is a primitive element of $L/\mathbb{Q}$

Proof: I did something similar to what is in this topic
$\mathbb{Q}(\sqrt2,\sqrt3,\sqrt{(2+\sqrt{2})(3+\sqrt{3})})$ is Galois over $\mathbb{Q}$

ii)

Affirmation 1. $\alpha^2$ is not a square in $K$.

Proof. If $\alpha^2=c^2$ some $c\in K$. Let $\varphi\in \text{Aut}(K/\mathbb{Q})$ given by $\varphi(\sqrt{2})=\sqrt{2}$ and $\varphi(\sqrt{3})=-\sqrt{3}$ then $\alpha^2\varphi(\alpha^2)=6(2+\sqrt{2})^2$
therefore $6=\left(\frac{c\varphi{c}}{2+\sqrt{2}}\right)^2$
Because $c\in K$ then $c=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$ then $c\varphi(c)=(a+b\sqrt{2})^2-3(c+d\sqrt{2})^2\in\mathbb{Q}(\sqrt{2})$
Therefore $c\varphi(c)\in\mathbb{Q}(\sqrt{2})$
Therefore $6=\left(\frac{c\varphi{c}}{2+\sqrt{2}}\right)^2$ with $\frac{c\varphi{c}}{2+\sqrt{2}}\in \mathbb{Q}(\sqrt{2})$
Therefore $\sqrt{6}\in\mathbb{Q}(\sqrt{2})$ a contradiction.

Therefore $\alpha\not\in K$

Afirmation 2. $[L:\mathbb{Q}]=8$

Indeed, $[L:K][K:\mathbb{Q}]=[L:K][\mathbb{Q}(\sqrt{3})(\sqrt{2}):\mathbb{Q}(\sqrt{3})][\mathbb{Q}(\sqrt{3}):\mathbb{Q}]$ and $[L:K]=2$ with $\text{Irr}_{K,\alpha}(x)=x^2-\alpha^2$, $[\mathbb{Q}(\sqrt{3})(\sqrt{2}):\mathbb{Q}(\sqrt{3})]=2$ with $\text{Irr}_{\mathbb{Q}(\sqrt{3}),\sqrt{2}}(x)=x^2-2$ and $[\mathbb{Q}(\sqrt{3}):\mathbb{Q}]=2$ with $\text{Irr}_{\mathbb{Q},\sqrt{3}}(x)=x^2-3$

Affirmation 3. $|\text{Aut}(L/\mathbb{Q})|=8$.

Indeed, by afirmation 2, $\text{Irr}_{K,\alpha}(x)=0$ implies $x\in \left\{\pm \alpha\right\}$ (in $L$)
therefore exists two extension $\alpha\mapsto \pm \alpha$
Analogously, $\text{Irr}_{\mathbb{Q}(\sqrt{3}),\sqrt{2}}(x)=0$ implies $x\in\left\{\pm \sqrt{2}\right\}$ therefore exists two extension   $\sqrt{2}\mapsto \pm \sqrt{2}$

$\text{Irr}_{\mathbb{Q},\sqrt{3}}(x)=0$ implies $x\in\left\{\pm \sqrt{3}\right\}$ therefore exists two extension $\sqrt{3}\mapsto \pm\sqrt{3}$

Therefore exists 8 $\mathbb{Q}$-automorphisms, this are:
$\alpha\mapsto \pm \sqrt{ (2+\pm\sqrt{2})(3\pm\sqrt{3})}$

Therefore $|\text{Aut}(L/\mathbb{Q})|=8=[L:\mathbb{Q}]$ therefore $L/\mathbb{Q}$ is a Galois extension.

Why $\pm\alpha=\pm\sqrt{ (2+\sqrt{2})(3+\sqrt{3})},\, \pm \beta=\pm\sqrt{ (2-\sqrt{2})(3+\sqrt{3})}$,
$ \pm\gamma=\pm\sqrt{ (2+\sqrt{2})(3-\sqrt{3})},\, \pm\delta=\pm\sqrt{ (2-\sqrt{2})(3-\sqrt{3})}$ (conjugates of $\alpha$) are in $L$?

This is because

$\beta=\frac{\alpha \beta}{\alpha}$ and $\alpha\beta=\sqrt{2}(3+\sqrt{3})\in \mathbb{Q}(\sqrt{2},\sqrt{3})\subset L$ then
$\alpha\beta\in L$ therefore $\beta\in L$ etc.

iii)

Affirmation 4. $\text{Gal}(L/\mathbb{Q})\simeq Q_8$.

Indeed, let $\varphi_{\beta},\varphi_{\gamma}\in \text{Gal}(L/\mathbb{Q})$ given by $\varphi_{\beta}(\alpha)=\beta$ and $\varphi_{\gamma}(\alpha)=\gamma$ with $\beta:=\sqrt{ (2-\sqrt{2})(3+\sqrt{3})},\gamma:=\sqrt{(2+\sqrt{2})(3-\sqrt{3})}$ then $\varphi_{\beta}\circ \varphi_{\gamma}\neq \varphi_{\gamma}\circ \varphi_{\beta}$ therefore $\text{Gal}{L/\mathbb{Q}}$ is non abelian, this implies is not cyclic.

Moreover,  $\varphi_{-\alpha}(\alpha):=-\alpha$ is the unique $\mathbb{Q}$-automorphism with order 2.

Because any group not cyclic and with a unique element of order $2$ is isomorphism to $Q_8$, then $\text{Gal}(L/\mathbb{Q})\simeq Q_8$.

To proves that iv) I am somewhat lost. I put below what I know about the minimal polynomial of $\alpha$.

Because $Irr_{L,\alpha}(x)=x-\alpha$ then

$Irr_{\mathbb{Q},\alpha}$
$=\prod_{\sigma\in \text{Gal}(L/\mathbb{Q})}(x-\sigma(\alpha))$
$=(x-\alpha)(x+\alpha)(x-\beta)(x+\beta)(x-\gamma)(x+\gamma)(x-\delta)(x+\delta)$

with $\pm\alpha=\pm\sqrt{ (2+\sqrt{2})(3+\sqrt{3})},\, \pm \beta=\pm\sqrt{ (2-\sqrt{2})(3+\sqrt{3})}$,
$ \pm\gamma=\pm\sqrt{ (2+\sqrt{2})(3-\sqrt{3})},\, \pm\delta=\pm\sqrt{ (2-\sqrt{2})(3-\sqrt{3})}$ (conjugates of $\alpha$) (this is a knows results)

By  contradiction, if $p(x)=x^4-ax^2+b\in\mathbb{Q}[x]$ has a root in $L$ then I should get a contradiction but I can't think of how.

How can I demonstrate this last part iv)?
thank you

Actualization 1. I fix part iv)

iv) If $p(x):=x^4-b$ (irreducible add) with $b\in\mathbb{Q}$ is not a square show that $p(x)$ has not root in $L$.

答:
answer to the original question: $\sqrt 2+\sqrt 3 \in L$ has minimal polynomial $x^4-10x^2+1$, so iv) is simply not true.

answer to the revised question: There are two cases: either $x^4-b$ is irreducible over $\Bbb Q$ or $b=-4c^4$ for some $c \in \Bbb Q$, see here.
Let us treat the second case first, so assume that $b=-4c^4$, then $x^4-b=x^4+4c^4=(x^2+2cx+2c^2)(x^2-2cx+2c^2)$. This has the four roots $x=(\pm1\pm i)c$. So if $x^4-b$ has root in $L$, then $\Bbb Q(i) \subset L$. But we also have $\Bbb Q(\sqrt{2}) \subset L$ and $\Bbb Q(\sqrt{3}) \subset L$ and $\Bbb Q(\sqrt{6}) \subset L$. By the Galois correspondence, having these four degree 2 subextensions implies that the Galois group $\operatorname{Gal}(L/\Bbb Q)$ has four subgroups of order $4$. But $Q_8$ only has three subgroups of order $4$.

For the second case, we may assume that $x^4-b$ is irreducible. Then because $L$ is a normal extension of $\Bbb Q$, if there is one root of $x^4-b$ in $L$, then all roots are in $L$. This implies that $L$ contains $\frac{i\sqrt[4]{b}i
}{\sqrt[4]{b}}=i$, as both $i\sqrt[4]{b}$ and $\sqrt[4]{b}$ are roots of $x^4-b$. This implies that $\Bbb Q(i) \subset L$ which is impossible, as we have already seen.