答:
Several comments.

tl;dr You are overthinking this.

When you write $\mathbb{Z}_5$ as $\{0, 1, 2, 3, 4\}$ those elements are not integers. They are just the names you have chosen for the five elements of that field. Of course those symbols also name integers, and the coincidence is not an accident. The coincidence together with the idea of "modulo $5$" tells you how to do the arithmetic in that field.

The wikipedia definition tells you that a standard way to name the inverse of the element $a$ is to write $-a$. Then in  $\mathbb{Z}_5$ you write $-1 = 4$ since $1 + 4 = 0$. So "$-1$" is just another name for the element names "$4$". It's not some other element of the field you forgot to write down.

Every field has a multiplicative identity which is always called $1$ although it is not the integer 1. Then by repeated addition, for every positive integer every field contains an element you could reasonably call $n$. If those elements are all different from one another then the field contains a copy of the integers (even a copy of the rationals) and is said to have characteristic $0$. If there are repeats in the sequence of field elements named by the positive integers then ...

Title: Notation for fields and the elements of the sets used to define them.

Body:
Context:

While trying to figure out how to prove that the characteristic of a field $\mathcal{F}$ is either $0$ or prime, I realized that I was conflating the elements of the field with integers, because of the notation.

Question:

Because articulating this question abstractly is beyond my ability, I'll illustrate my confusion with an example:

If $\mathbb{Z}_5$ is the set $\{0,1,2,3,4\}$, along with modular addition and multiplication (i.e. in this case, after performing the standard addition or multiplication, the least remainder after ordinary division by 5; e.g. $4\cdot4 = 1$ and $3+4 = 2$) defines a field. Here,
$$
4 + 1 = 0,
$$
i.e. $1$ is the additive inverse of $4$.

Now, all the classical definitions I've seen usually define the additive inverse as follows (this example is taken from Wikipedia):

for every a in F, there exists an element in F, denoted −a, called the additive inverse of a, such that a + (−a) = 0.

Does the phrase "denoted -a" imply that, in the example provided above, we should denote "$1 = -4$"? Or are such definitions simply stating what they mean when they use the symbol (i.e. it is simply saying we should think of $1$ as the additive-inverse-of-4)? In other words, is "$-4$" an element of this field, even if it is not an element of the set used to define the field (does it even make sense to talk of "an element of a field" in this way)?

Just to be clear, if we were to instead denote the additive inverse of a by na, is it then appropriate to write something like $1 = n4$, for the above field? While I can understand what is meant by this, what bothers me is that $n4$ is not an element of any set described or defined, elsewhere.

Final Considerations

Up until here, I'd be tempted to more or less ignore the whole issue; however, while trying to work out a proof that a field's characteristic is either zero or prime, I found myself wanting to use integers to describe the operations, regardless of whether or not the elements of the set were, themselves, numbers. For example, given the field $\mathbb{F}$ defined by:
$$
\left\{
  \begin{array}{lll}
    \mathbb{S} =  \{a, b\},
    &
     \begin{array}{c|c|c}
       \pmb{+} & \pmb{a} & \pmb{b} \\ \hline
       \pmb{a} & a       & b       \\ \hline
       \pmb{b} & b       & a
     \end{array},
    &
     \begin{array}{c|c|c}
       \pmb{\cdot} & \pmb{a} & \pmb{b} \\ \hline
       \pmb{a} & a       & a       \\ \hline
       \pmb{b} & a       & b
     \end{array},
  \end{array}
\right.
$$

it seems natural and useful to define an expression such as $a + a + a$ as $3a$; however, I'm not sure how to think about the relationship of the symbols introduced by this notation to the field itself (this becomes even more confusing if I use "1" to denote the multiplicative identity). This seems -- at least to my confused mind -- somehow related to the confusion expressed in the first example, above.

Is the use of 3 in "3a" and -4 in "1 = -4" nothing more than a convenient notation to describe the elements of the field? While I feel like that is all they are, I also feel that they carry with them some sort of implied structure (perhaps that of an abelian group?), that isn't accounted for, formally / explicitly.


问:
Fréchet Derivative of $\frac{1}{\|x\|}$
In a Hilbert Space $V$, for function $f:V\to\mathbb{R}$, if $f$ is Fréchet differentiable at $x_0$, the Fréchet Derivative $\nabla f(x_0)$ is  $v$ such that $$ \lim_{x\to x_0} \frac{|f(x) - f(x_0) - \langle v, x-x_0 \rangle|}{\|x-x_0\|} = 0$$

As an example, for $f(x) = \|x\|$, the Fréchet derivative is $\nabla f(x) = \frac{x}{\|x\|}$, as shown in this Wikipedia proof.

Now my question is, what is the Fréchet derivative of $f(x) = \frac{1}{\|x\|}$. My gut says something like $-\frac{x}{\|x\|^3}$ following normal derivative rules, but I can't seem to be able to solve it.

ContentLicense: CC BY-SA 4.0

Title: Fréchet Derivative of $\frac{1}{\|x\|}$

Body:
You can apply the product rule on the relation
$$1 = \|\cdot\| \cdot \frac1{\|\cdot\|}$$

to obtain

$$D1(x_0) = \frac1{\|x_0\|}D(\|\cdot\|)(x_0) + \|x_0\|D\left(\frac1{\|\cdot\|}\right)(x_0)$$
$$0 = \frac{x_0}{\|x_0\|^2} + \|x\|D\left(\frac1{\|\cdot\|}\right)(x_0) \implies D\left(\frac1{\|\cdot\|}\right)(x_0) = -\frac{x_0}{\|x_0\|^3}.$$


问:
Find factor of $x^{11} + x + 1$ of degree 2 in $\mathbb{Z}[x]$

Find factor of $x^{11} + x + 1$ of degree 2 in $\mathbb{Z}[x]$

To be honest, it is the first time I see problem like this. I was trying to find divisor by just taking different polinomials of degree 2 and trying to divide this one by them. That was disgusting, I was mixing up in calculations and couldn't finally get answer.

So, Is there any better way to find such a polinomial (divisor of this one with degree 2)? It would be awesome if proposed way would not require super advanced knowledge. Thanks in advance!

答:
standard trick. We have $11 \equiv 2 \pmod 3$ from which follows that the nontrivial cube roots of unity are roots,
$  \omega = \frac{-1+i \sqrt3}{2} , \bar{\omega} = \omega^2 =  \frac{-1-i \sqrt3}{2}$
This means the original polynomial is divisible by
$$ (x-\omega) ( x - \bar{\omega} ) = \; \;  x^2 + x +1 $$

$$  \left(   x^{11}  +  x  + 1 \right)  =  \left(   x^{2}  +  x  + 1 \right)  \cdot \color{magenta}{  \left(   x^{9}  -  x^{8}  +  x^{6}  -  x^{5}  +  x^{3}  -  x^{2}  + 1 \right) }  $$


问:
Proof $-(x+y) = (-x) + (-y) = -x + -y$
This was left as an exercise in Apostol's Calculus Vol II and I'm not very sure how to proceed and have my doubts on whether the following would be sufficient due to step $(1)$:
$$-(x+y) \stackrel{(1)}{=} -1(x+y) \stackrel{(2)}{=} -1x + -1y \stackrel{(3)}{=} (-x) + (-y) = -x-y$$

Where $(2)$ is allowed by the distributive multiplication property for addition in V and $(3)$ is allowed because $(-a)x = -(ax)$, which is proven by:

Let $z = (-a)x$
$$ z + ax = (-a)x + ax \stackrel{(a)}{=} (-a + a)x = 0x \stackrel{(b)}{=} O $$
Where $(a)$ is allowed by the distributive multiplication property for scalars addition and $(b)$ is true because $0x = O$ (I'm not providing this proof here, but it can be assumed already proven). We find that $z$ is the negative of $ax$ and so $z = -(ax)$.

Would this be a sufficient proof?

答:
Notice that $(x + y) - (x + y) = 0$ but we also have $x + y - x - y = 0$. Hence we have $(x + y) - (x + y) = x + y - x - y = (x + y) - x - y$. This can be reduced to $(x + y) - (x+y) = (x + y) -x -y$. Then simply cancel and we reach $-(x + y) = -x - y$.

I'm assuming that you've proven the uniqueness of inverses in a vector space and the additive cancellation rule.

Title: Proof $-(x+y) = (-x) + (-y) = -x + -y$


答:
Let $f:H\to \mathbb R: f(x)=\|x\|.$ Then, if $x\neq 0,\ f$ is differentiable at $x$. Note that $f(x)\neq 0$ whenever $x\neq 0.$ Let $g:\mathbb R^+\to \mathbb R^+:g(x)=x^{-1}.$ If $x\neq 0,$ then $g$ is differentiable at $x$. By the chain rule, we have

$$(g\circ f)'(x)h=g'(f(x))\circ f'(x)h$$  Calculating, we get

$$(g\circ f)'(x)h=g'(f(x)(\langle\frac{x}{\|x\|},h\rangle)=\langle\frac{x}{\|x\|},h\rangle g'(f(x))=\langle\frac{x}{\|x\|},h\rangle\left(\frac{-1}{\|x\|^2}\right)=\langle\frac{-x}{\|x\|^3},h\rangle$$

Title: Fréchet Derivative of $\frac{1}{\|x\|}$

Body:
In a Hilbert Space $V$, for function $f:V\to\mathbb{R}$, if $f$ is Fréchet differentiable at $x_0$, the Fréchet Derivative $\nabla f(x_0)$ is  $v$ such that $$ \lim_{x\to x_0} \frac{|f(x) - f(x_0) - \langle v, x-x_0 \rangle|}{\|x-x_0\|} = 0$$

As an example, for $f(x) = \|x\|$, the Fréchet derivative is $\nabla f(x) = \frac{x}{\|x\|}$, as shown in this Wikipedia proof.

Now my question is, what is the Fréchet derivative of $f(x) = \frac{1}{\|x\|}$. My gut says something like $-\frac{x}{\|x\|^3}$ following normal derivative rules, but I can't seem to be able to solve it.


问:
An excercise about omitting types theorem
I am studying the omitting types theorem, which I know in the following form (where we implicitly work inside a monster model of a complete theory without finite models)

Theorem (Omitting types) Assume $L(A)$ is countable. Then for every consistent type
$p(x) ⊆ L(A)$ the following are equivalent:

All models containing $A$ realize $p(x)$ $\quad\quad\quad$

$A$ isolates $p(x)$

Now, I am asked to proove the following:

Let $p(x) ⊆ L(B)$ and $p_n (x) ⊆ L(A)$, for $n < ω$, be consistent types such
that:
$$p(x) →\lor_{n<\omega}\ p_n(x)$$
Prove that there is an $n<ω$ and a formula $φ(x) ∈ L (A)$ consistent with $p(x)$ such
that
$$p(x) ∧ φ(x) → p_n (x) $$

Now, I know that it must hold that $p(x)\rightarrow p_m(x)$ for some $m$. And intuitively I would like to find some $\phi(x)\in L(A)$ which isolates something like $p(x)\rightarrow p_m(x)$, but I don't know how to go further, since I cannot translate this "implication" between types into a type itself, which moreover should be something over $A\cup B$

I am also asked to prove the following:

Let $A\subset B$ and $p(x)\subset L(A)$ suppose that for any solution $a$ of $p(x)$ the complete type over $B$, $tp(a/B)$ is isolated. Show that $p(x)$ is isolated.

Now, applying the theorem (with $B$) instead of $A$, I am able to conclude that any model containing $B$, contains also, for each solution $a$ of $p(x)$ an element which is in the same orbit (under automorphisms fixing $B$) of $a$. Then this will also be a solution of $p(x)$, since it is the image of $a$ through an automorphism which fixes $B$ and hence also $A$. But how to extend to models which contain $A$ but maybe not $B$ in order to apply the theorem again proving the claim?

Thanks in advance for any help

答:
First question:

Let's prove the contrapositive. Assume that for all $n$ and all $L(A)$-formulas $\varphi(x)$, $p(x)\cup \{\varphi(x)\}\not\models p_n(x)$. We want to prove that $p(x)\not\models \bigvee_{n\in\omega} p_n(x)$, i.e. we want to realize $p(x)$ by an element which does not satisfy any of the $p_n(x)$.

One way to go is to introduce a new constant symbol $c$, consider the theory $T\cup p(c)$, and use the omitting types theorem to omit all the $p_n(x)$. But this is a little messy, what with the changing of languages, and it's also overkill: you don't really have to omit all the types $p_n(x)$, just make sure that $c$ doesn't realize them.

So I would prefer to prove this result by redoing a small part of the proof of the omitting types theorem. Note that we don't have to assume the language is countable in this proof, which is a big advantage over using omitting types!

We build a sequence of $L(A)$-formulas $(\varphi_n(x))_{n\in \omega}$ by induction, ensuring as we go that for all $n$, $p(x)\cup \{\varphi_k(x)\mid k\leq n\}$ is consistent. To pick $\varphi_n(x)$, we let $\psi_n(x) = \bigwedge_{k<n}\varphi_k(x)$. By induction, $\psi_n(x)$ is consistent with $p(x)$. So by our assumption, $p(x)\cup \{\psi_n(x)\}\not\models p_n(x)$. Thus we can find some element $b$ satisfying $p(x)\cup \{\psi_n(x)\}$ but not satisfying $p_n(x)$. Let $\varphi_n(x)$ be some formula which is true of $b$, but whose negation is in $p_n(x)$. Then $p(x)\cup \{\varphi_k(x)\mid k\leq n\}$ is consistent, since it is realized by $b$.

Now let $q(x) = p(x)\cup \{\varphi_n(x)\mid n\in \omega\}$. By compactness, $q(x)$ is consistent. Let $c$ be some realization. Then $c$ realizes $p$, but $c$ does not realize any $p_n(x)$, since for any $n$, $c$ satisfies $\varphi_n(x)$, and $\lnot\varphi_n(x)\in p_n(x)$.

Second question:

To me, this question seems totally unrelated to the first question and the omitting types theorem... I would prove it like this:

Let $Q$ be the set of complete types $q(x)$ over $B$ with $p(x)\subseteq q(x)$. For each $q(x)\in Q$, $q(x)$ is isolated by an $L(B)$-formula $\varphi_q(x)$. So $p(x)\cup \{\lnot \varphi_q(x)\mid q\in Q\}$ is inconsistent. By compactness, there are finitely many types $q_1,\dots,q_n$ such that $p(x)\models \bigvee_{i=1}^n \varphi_{q_i}(x)$. Let's call this disjunction $\psi(x)$. Note that also $\psi(x)\models p(x)$, since if $\psi(a)$, then $a$ realizes $q_i(x)$ for some $i$, and  $p(x)\subseteq q_i(x)$. Thus we have shown that $p(x)$ is isolated by a formula over $B$.

The question as stated is a bit ambiguous, but I suppose the goal is to show that $p(x)$ is isolated by a formula over $A$.  But this follows from the above, since if a set definable over $B$ is invariant over $A\subseteq B$ (i.e. fixed by all automorphisms of the monster model which fix $A$ pointwise), then the set is actually definable over $A$. I'll leave that as an exercise for you.


问:
What well-orders are definable over $V$?
Let $V$ be a (the) universe of sets, and $On^V$ denote the ordinals of $V$.

It is well known that there are formulae that seem to define orderings `longer than' $On^V$. For example:

$\alpha < \beta$ iff either:

(a) $\alpha$ is a limit ordinal and $\beta$ is a successor ordinal.

(b) $\alpha$ and $\beta$ are both limits and $\alpha < \beta$.

(c) $\alpha$ and $\beta$ are both successors and $\alpha < \beta$.

Prima facie such a formula defines a (non-set-like) well-order of twice the `length' of $On^{V}$. There's more interesting definable well-orderings of this kind (the mouse ordering is a good example).

My question: Putting aside foundational worries (we can work over a $V_\kappa$ or $H_\kappa$ if people feel queasy), what is the limit on the `length' of the well-orders that can be defined this way?

My guess is that it's going to be the least non-recursive ordinal above $On^V$. So, for example, in the case of $V_\omega$, the upper bound of ordinals definable in $ZFC -$ Infinity will be $\omega_1^{ck}$. The reason for this being that one can't take the supremum of objects one doesn't have -- we're only looking at what orderings are definable over a fixed model. However, it might be lower, or I might have missed something, I'm not sure.

答:
This is perhaps surprisingly subtle. The short version is that this ordinal is at least sometimes, and I suspect always, much smaller than the relevant analogue of $\omega_1^{CK}$.

First, suppose $A$ is a countable structure in a finite language (e.g. a countable well-founded model of $\mathsf{ZFC}$). There are various ordinals associated to $A$ (sadly there's no standard notation here, so I'm making it up as I go along):

$\omega_1^{CK}(A)$ is the least ordinal not Muchnik reducible to $A$; that is, the least ordinal such that there is some copy of $A$ not computing any copy of that ordinal.

$\omega_1^{def}(A)$ is the least ordinal which is not interpretable in $A$ with parameters; that is, the least $\alpha$ for which there are no formulas $\varphi,\psi$ with parameters from $A$ such that $\varphi$ defines an equivalence relation on $A$ and $\psi$ defines a well-ordering of the $\varphi$-equivalence classes of ordertype $\alpha$. If we take $A$ to be a countable well-founded model of $\mathsf{ZFC}$, this is the ordinal you're asking about.

$\omega_1^{ad}(A)$ is the smallest ordinal $\alpha$ such that there is a transitive set $X$ of height $\alpha$ such that $X\models\mathsf{KP+Inf}$ and there is some structure $y\in X$ which is isomorphic to $A$. (The "ad" is short for "admissible.")

These are each reasonable notions of "$A$'s version of $\omega_1^{CK}$;" the first two are pretty obviously motivated, while the third is motivated by a technical theorem of Sacks. It turns out that we always have $\omega_1^{CK}(A)=\omega_1^{ad}(A)$, basically as a consequence of Sacks' theorem, but in general $\omega_1^{def}(A)$ could be surprisingly small. In particular:

$(*)\quad$ Any "sufficiently closed" countable well-founded model $A$ of $\mathsf{ZFC}$ will have $\omega_1^{CK}(A)>\omega_1^{def}(A)$.

Indeed, I suspect that no countable well-founded model $A$ of $\mathsf{ZFC}$ has $\omega_1^{CK}(A)=\omega_1^{def}(A)$; however, I haven't been able to prove this. The stumbling block is that I haven't been able to find any plausible proxy for "sufficiently closed;" in particular the theory of $A$ itself (does $A=L^A$? does $A$ have lots of large cardinals?) doesn't seem to be relevant.

What about uncountable models of $\mathsf{ZFC}$?

Well, the definitions of $\omega_1^{def}(A)$ and $\omega_1^{ad}(A)$ lift immediately to uncountable structures, and in fact there's a snappier and easier-to-prove analogue of $(*)$ in this context:

$(**)\quad$ Suppose $A$ is a well-founded model of $\mathsf{ZFC}$ and $A^\omega\subseteq A$. Then $\omega_1^{def}(A)<\omega_1^{ad}(A)$.

See this MO answer of mine for the proof. The original statement $(*)$ can then be (stated precisely and) proved by thinking a bit more carefully about the proof of $(**)$.

And we can go further: it turns out that we can also lift the definition of $\omega_1^{CK}(A)$ to uncountable structures $A$. To do this we replace Muchnik reducibility with generic Muchnik reducibility. Given two structures $A,B$ of arbitrary cardinality, we roughly say that $A$ is generically Muchnik reducible to $B$ if $A$ is Muchnik reducible to $B$ in any generic extension where $\vert A\vert+\vert B\vert\le\aleph_0$. By Shoenfield absoluteness this is a pretty well-behaved notion; in particular, if we let $\omega_1^{CK*}(-)$ be the "generic Muchnik" version of $\omega_1^{CK}(-)$, we get that $\omega_1^{CK*}(A)=\omega_1^{ad}(A)$ for every structure $A$ whatsoever. So even for uncountable models we have:

For at least some, and I suspect for all, transitive models $A$ of $\mathsf{ZFC}$ we have $\omega_1^{def}(A)<\omega_1^{CK*}(A)$.

(Meanwhile, if you're familiar with some higher recursion theory note as a caveat that the $\alpha$-recursion analogue of $\omega_1^{CK}(-)$ is actually surprisingly small in this context - see here.)

ContentLicense: CC BY-SA 3.0


问:
Question regarding the natural isomorphism given by a triangulated functor
If we have two triangulated categories, let's say $\mathcal{T}_{1}$ and $\mathcal{T}_{2}$, with translation functors $\Sigma_{1}$ and $\Sigma_{2}$, respectively. A exact functor or triangulated functor between these two categories is defined by an additive functor $F:\mathcal{T}_{1} \to \mathcal{T}_{2}$ such that there is a natural isomorphism

$$\eta:= \lbrace \eta_{X}:F\Sigma_{1}(X) \to \Sigma_{2}F(X) \rbrace_{X \in \operatorname{Obj}(\mathcal{T_{1}})}.$$

In a way that for every distinguished triangle $X \xrightarrow{u} Y \xrightarrow{v} Z \xrightarrow{w}  \Sigma_{1}X$ in $\mathcal{T_{1}}$ then $FX \xrightarrow{F(u)} FY \xrightarrow{F(v)} FZ \xrightarrow{\eta_{X} \circ F(w)}  \Sigma_{2}FX$ is a distinguished triangle  in $\mathcal{T_{2}}$. My question is that if we have that this natural transformation $\eta$ induced by the triangulated functor $F:\mathcal{T}_{1} \to \mathcal{T}_{2}$ we also have a natural isomorphism $\Sigma_{2}^{-i}F \cong F \Sigma_{1}^{-i}$ for every $i \in \mathbb{Z}$?? I guess this is true by checking out on proof of "the adjoint functor of a triangulated functor is also triangulated". I tried to prove this simple case $\Sigma_{2}^{-1}F \cong F \Sigma_{1}^{-1}$.

I was thinking maybe  I didn't understand the definition of a triangulated functor correctly and there is a natural isomorphism $\eta$ for each distinguished triangle in $\mathcal{T}_{1}$?? Can anyone tell me if this is true? Otherwise I don't have any idea how to obtain this natural isomorphism :/

I came with this question since I was following for this proof  Daniel Huybrechts' book "Fourier-Mukai Transforms in Algebraic Geometry" page 15 Proposition 1.41. As there is one step "using my notation" where he uses an isomorphism:

$$\operatorname{Hom}_{\mathcal{T}_{2}}(\Sigma_{-2}F(X),Y) \cong \operatorname{Hom}_{\mathcal{T}_{2}}(F \Sigma^{-1}(X),Y).$$

This question has been stuck in my head for so many day so I will really appreciate your help. Thanks!

答:
$\Sigma_2$ is an equivalence, so to construct such an isomorphism it suffices to construct an isomorphism $F\cong\Sigma_2^i F\Sigma_1^{-i}$ (to which you can then apply $\Sigma_2^{-i}$)

Similarly, by using precomposition instead of postcomposition, it suffices to construct an isomorphism $F\Sigma_1^i\cong \Sigma_2^iF$. But this one is given to you by iterating $\eta$, so we are done.


问:
Is the kernel of a homomorphism from a Boolean ring to $\mathbb{Z}_2$ always a maximal ideal?
Let $(B, +, \cdot)$ be a ring (not necessarily unital!) with the property that every $x \in B$ satisfies $x \cdot x = x$.

How does one show that the kernel of any non-zero homomorphism of rings $h:B\rightarrow \mathbb{Z}_2$ is a maximal ideal of $(B, +, \cdot)$?

I'm looking for an elementary proof, not requiring anything more than the definitions of a ring, of an ideal, and of $(B, +, \cdot)$, and, if necessary, the easily shown facts that $x + x = 0$ and $x\cdot y = y\cdot x,\,\forall\, x,y \in B$.

答:
Let $f:B\to\mathbb{Z}_2$ be a non-zero homomorphism, and let $I=\ker(f)$. Suppose $I\subsetneq J\subseteq B$. There exists an element $x\in J\setminus I$ (because $I\subsetneq J$), and we must have $f(x)=1$ (because there are only two elements of $\mathbb{Z}_2$ to go to), so that $$f(x-1_B)=f(x)-f(1_B)=1-1=0,$$ and hence $x-1_B\in \ker(f)=I\subset J$. Thus
$$1_B=x-(x-1_B)\in J,$$
and therefore $J=B$. Thus $I$ is maximal.

The standard proof:

Any non-zero homomorphism of rings to $\mathbb{Z}_2$ is surjective; apply the first isomorphism theorem. Then use that an ideal $I$ of a ring $R$ is maximal $\iff$ $R/I$ is a field.

(None of this depended on any properties of $B$.)


问:
Evaluate $\int _0^{\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx$
The function

$$f\left(z\right)=\frac{z^6}{\left(z^4+a^4\right)^2}$$

Has the following poles of order 2:

$$ z(k)=a \exp\left( \frac{\left(2k+1\right)}4 i\pi \right)$$

$f$ is even, therefore: $$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx =\frac{1}{2}\int _{-\infty }^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx$$

$$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx=i\pi \sum _k\:Res\left(f,\:z\left(k\right)\right)$$

$$Res\left(f,\:z\left(k\right)\right)=\lim _{z\to z\left(k\right)}\left(\frac{1}{\left(2-1\right)!}\left(\frac{d}{dz}\right)^{2-1}\frac{z^6\left(z-z\left(k\right)\right)^2}{\left(z^4+a^4\right)^2}\right)$$

$$z^4+a^4=z^4-z_k^4\implies\dfrac{z^6(z-z_k)^2}{(z^4+a^4)^2}=\dfrac{z^6}{(z^3+z_k z^2+z_k^2 z+z_k^3)^2}$$

$$Res\left(f,\:z_k\right)=\lim _{z\to \:z_k}\left(\frac{d}{dz}\left(\frac{z^6}{\left(z^3+z_kz^2+z_k^2z+z_k^3\right)^2}\right)\right)$$

$$Res\left(f,\:z_k\right)=\frac{2z_kz^5\left(z^2+2z_kz+3z_k^2\right)}{\left(z^3+z_kz^2+z_k^2z+z_k^3\right)^3}=\frac{2z_k^6\cdot 6z_k^2}{\left(4z_k^3\right)^3}$$

$$Res\left(f,\:z_k\right)=\frac{12z_k^8}{64z_k^9}=\frac{3}{16z_k}$$

$$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi }{16a}\sum _{k=0}^n\:e^{-\frac{\left(2k+1\right)}{4}i\pi }$$

We consider only the residues within the upper half plane, that is to say those corresponding to $k=0$ and $k=1$.

$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi \:}{16a}\left(e^{-\frac{i\pi }{4}\:\:}+e^{-\frac{3i\pi \:}{4}\:\:}\right)$$

$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi \:}{16a}\left(\frac{\sqrt{2}}{2}\:-i\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right)$$

$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3\pi \sqrt{2}\:}{16a}$$

答: Consider $x=a \tan^{1/2}u$ to obtain$$\frac{1}{2|a|}\int_0^{\pi/2}\cos^{-1/2}u\sin^{5/2}udu.$$ Now recall the Beta fucntion.


问:
Roll $1$ die and square or roll $2$ dice and multiply; which has higher mean?
I am wondering if there is a fast way to tell which expected value is higher: "Roll 1 die and take the square of the number that comes up or roll 2 dice and multiply them". Thank you!</p>

答:
Let $\xi$ and $\eta$ be values of the two die rolls. The expected value of squaring the the first roll is $\operatorname{E}(\xi^2)$, and the expected value of multiplying the two values is $\operatorname{E}(\xi\eta)$. Since we can assume the two die rolls are iid, we have $$\operatorname{E}(\xi\eta)=\operatorname{E}(\xi)\operatorname{E}(\eta)=\operatorname{E}(\xi)^2$$

Since the difference between these two expected values is equal to the variance,
$$\operatorname{Var}(\xi)=\operatorname{E}(\xi^2)-\operatorname{E}(\xi)^2$$
it must be nonnegative. Therefore, we have
$$\operatorname{E}(\xi^2)\ge\operatorname{E}(\xi)^2$$
so the expected value of squaring the result of one die roll is higher.


答:
Not sure whether you consider this faster.  The first probability is
$$\frac{1}{n}\sum_{k=1}^n k^2 = \frac{1}{n}\cdot\frac{n(n+1)(2n+1)}{6}=\frac{(n+1)(2n+1)}{6}.$$
The second probability is
$$\left(\frac{1}{n}\sum_{k=1}^n k\right)^2 = \left(\frac{1}{n}\cdot\frac{n(n+1)}{2}\right)^2 = \frac{(n+1)^2}{4}.$$
The first one is larger when $(2n+1)/6 > (n+1)/4$, that is when $n > 1$.

Title: Roll $1$ die and square or roll $2$ dice and multiply; which has higher mean?

Body: I am wondering if there is a fast way to tell which expected value is higher: "Roll 1 die and take the square of the number that comes up or roll 2 dice and multiply them". Thank you!</p>


问:
Prove convergence in probability
$\{X_n\}$ and $\{X\}$ are random variables defined on $(S,σ,P)$ with $E(X_n^2)<\infty$, for all $n=1,2,3,\dots$, and $E(X)<\infty$. If
$E\left((X-X_n)^2\right)<\infty$ as $n\to\infty$, then I'd like to show that the sequence $\{X_n\}$ converges to $\{X\}$ in probability as $n\to\infty$.

My teacher asked me to use convergence in moments however I am not able to make any progress in this problem.

答:
There is a typo in your post.

If

$$\lim\limits_{n\to\infty}\mathbb{E}[X_n-X]^2=0$$

That is

$$X_n\xrightarrow{L^2}X$$

than it is also

$$X_n\xrightarrow{\mathcal{P}}X$$

Proof:

Using Chebishev inequality we get

$$\lim\limits_{n\to\infty}\{|X_n-X|\leq \epsilon\}\geq 1-\frac{\mathbb{E}[X_n-X]^2}{\epsilon^2}$$

Title: Prove convergence in probability


问:
Does convergence imply continuity in topology when a topology is first countable?
I am reading Simmons 'An Introduction to topology' and came across this question that I haven't been able to solve.

If $f:X\rightarrow Y$ is a mapping of one topological space into another, and if $x_n \rightarrow x$ in $X$ implies $f(x_n)\rightarrow f(x)$ in $Y$ then $f$ is continuous if $X$ is first countable.

Here, $x_n \rightarrow x$ means for the sequence $\{x_n\}$ in $X$ each neighbourhood $G$ of $x$ has a positive integer $n_0$ such that $x_n$ is in $G$ for all $n \geq n_0.$

I am not very familiar with how to prove a function is continuous, all I know is $f$ is continuous if $f^{-1}(G)$ is open when $G$ is open,but I was not able to prove that here.

答:
Welcome to MSE ^_^

Hint:

Instead of checking that $f^{-1}(\text{open}) = \text{open}$, you can check that
$f^{-1}(\text{closed}) = \text{closed}$. If this isn't obvious, it's also a nice exercise.

But one of the most important properties about first countable spaces is that $x \in \overline{A}$ if and only if $x$ is the limit of some sequence $a_n \in A$.
(If you're looking for googleable buzzwords, this says that every first countable space is Frechet-Urysohn)

So now take a closed set $F \subseteq Y$, and look at $f^{-1}(F)$ in $X$. We want to show that $f^{-1}(F)$ is closed, so let's take some limit point and show that it's already in the set. By the previous paragraph, limit points look like sequences. But we have a hypothesis on how $f$ interacts with sequences... Can you finish it from here?

I hope this helps ^_^


问:
The Unit Circle over $\mathbb R$ is not a UFD
This is exercise 14.2 N in Vakil, self-study. Similar questions have been asked a few times on this site, but none of the answers use a method that I believe Vakil intended: here, here, and here.

We are to show $\mathbb R[x, y]/(x^2 + y^2 -1)$ is not a UFD, but over $\mathbb C$, it is, using exercise 14.2 L, which says, among other things, that $\mathbb P^n - Y$ is not the spectrum of a UFD if $Y$ is a hypersurface of degree $d > 1$.

The issue is that I don't see how to write the unit circle as the complement of a degree $d > 1$ hypersurface in a projective space, if indeed that is what we are supposed to do, nor do I see how, supposing we have done this, the result will change over $\mathbb C$, since presumably 14.2 L still applies.

答:
Attempt at an answer:

Over $\mathbb C$, Euler's formula and parameterization of the unit circle give us that our ring is isomorphic to $\mathbb C[x]_x$, a UFD.

Over $\mathbb R$, the projectivized unit circle is isomorphic to $\mathbb P^1_{\mathbb R}$, and the image of the affine unit circle under this isomorphism is the complement of $x^2 + y^2 = 0$ (thinking of points on this curve as points in $\mathbb C$), so 14.2 L concludes.


问:
Distributions Theory $ {u}\,{\sin((\pi)x)}=1.$
I am using Theory of distributions by Friedlander and Joshi. I want to find all $u$ in $D'(\mathbb R)$ such :

$$ {u}\cdot{\sin((\pi)x)}=1.$$

I see that any integer is a pole. Are there any simplifications?

答:
$$f = pv(\frac{1}{\sin(\pi x)})$$ is a solution.

Then any solution is of the form $$u=f+w, \qquad w\sin(\pi x)=0$$ Given such a $w$, fix some $\psi \in C^\infty_c(-1,1)$, $\psi(0)=1$.

For all $\phi \in C^\infty_c(-1,1)$ we have
$$\langle w,\phi-\phi(0)\psi\rangle = \langle w\sin(\pi x),\frac{\phi-\phi(0)\psi}{\sin(\pi x)}\rangle=0$$
ie. $$\langle w,\phi\rangle= \langle C\delta,\phi\rangle,\qquad C = \langle w,\psi\rangle$$
Can you finish ?


问:
Show that for any A, there exists { dom R | R in A }
I want to show that

(a) $ \forall \alpha \; \exists \beta$ such that $\beta = \{ dom R \; | \; R \in \alpha \}$

I'm having difficulty proving this axiomatically. I'm not even certain that it can be done. I'm trying to solve this as part of the larger problem of showing that

(b) $\forall  \alpha \; dom\bigcup\alpha = \bigcup\{dom R \; | \; R \in \alpha\}$

My approach to solving (a), so far, is the following (supposed to be a fitch-style proof - I couldn't figure out how to do fitch in TeX):

START OF PROOF

(1) Take $A$

(2) Take $s$

(3) Assume $s \in dom \bigcup A$

      (4) [Somehow show that $\{dom R \; | \; R \in \alpha\}$ exists for any $\alpha$]

      (5) Therefore, there exists $B$ = $\{dom R \; | \; R \in A\}$

      (6) From (3), $\exists \psi \;(<s,\psi> \in \bigcup A)$

      (7) Fix $y$ : $<s,y> \in \bigcup A$

      (8) From (7), $\exists \mu \; (<s,y> \in \mu \; \wedge \; \mu \in A)$

      (9) Fix $R$ : $<s,y> \in R \; \wedge \; R \in A$

      (10) $dom R \in B$    [Remember (5): $B$ = ${dom R \; | \; R \in A}$]

      (11) $s \in \bigcup B$    [Because $s \in dom R \; \wedge \; dom R \in B$]

[This gets me the first side of the bi-conditional]

(12) Assume $s \in \bigcup\{domR \; | \; R \in A\}$

      (13) [Show that $s \in dom \bigcup A$]

[This gets me the second side of the bi-conditional]

    (14) From (3-13) : $s \in dom \bigcup A \; \leftrightarrow \; s \in \bigcup \{ dom R \; | \; R \in A\}$

(15) $\forall \varphi \; ( \varphi \in dom \bigcup A \; \leftrightarrow \; \varphi \in \bigcup \{ dom R \; | \; R \in A \} )$

$\forall \alpha \forall \varphi \; ( \varphi \in dom \bigcup \alpha \; \leftrightarrow \; \varphi \in \bigcup \{ dom R \; | \; R \in \alpha \} )$

END OF PROOF

My problem is figuring out step (4).

(This is not homework - I'm teaching myself set theory from Enderton's in my spare time, and am a little stumped on this one.)

Thanks!

Max

答:
First let $X = \bigcup \bigcup \bigcup \alpha$. Then let $Y$ be the powerset of $X$. Now for any $R \in \alpha$, we have $\operatorname{dom} R \in Y$. This is because an ordered pair $(a,b)$ is coded in the Kuratowski style as $\{\{a\}, \{a,b\}\}$, so if you unwrap three layers of $\{ \}$ from $\alpha$ you will get to $a$. If you use a different pairing operation, you may need a different number of unions in the definition of $X$.

Now, using $Y$, we can use the axiom of separation to form
$$
\{ \operatorname{dom}(R) : R \in A\} = \{ D \in Y : \exists R ( R \in A \land D = \operatorname{dom}(R))\}.
$$
To turn this into a formal proof, you just need to use the axioms of whatever set theory you have to follow the three steps.

This is the general pattern you use to show that various sets exist in formal set theory: first, you move to some set like $Y$ of high enough rank that all the elements of your target set are elements of $Y$. Then you use separation to pull out just the elements of $Y$ that you want to be in your target set, assuming that the property that you want to select for can be expressed in the language of set theory.


问:
Laplace transform $f(t) = f(t + \frac{2\pi }{a}) = \frac{\sin at}{|\sin at|}$
I had done the laplace transform with the period fucntion (T): $F(s) = \frac{1}{1 - e^{ - sT}}\int_o^T {e^{ - st}}f(t) \, dt $
Howerver, in this fuction:

$$f(t) = f \left(t + \frac{2\pi}{a}\right) = \frac{\sin at}{|\sin at|}$$

What is the period? And You can give me detail of solution

Body:
Note that $T = 2 \pi/a$, and

$$\frac{\sin{a t}}{|\sin{a t}|} = \begin{cases} \\ +1 & a\, t \in (0,\pi) \\ -1 & a \,t \in (\pi,2 \pi)\end{cases}$$

Then

$$\begin{align}\int_0^{2 \pi/a} dt\, \frac{\sin{a t}}{|\sin{a t}|} \, e^{-s t} &= \int_0^{\pi/a} dt \, e^{-s t} - \int_{\pi/a}^{2 \pi/a} dt \, e^{-s t}\\ &= \frac{1-e^{-\pi s/a}}{s} -\frac{e^{-\pi s/a}-e^{-2 \pi s/a}}{s} \\ &= \frac{(1-e^{-\pi s/a})^2}{s} \end{align}$$

The full Laplace transform is then

$$\int_0^{\infty} dt\, \frac{\sin{a t}}{|\sin{a t}|} \, e^{-s t} = \frac{(1-e^{-\pi s/a})^2}{s(1-e^{-2\pi s/a})} = \frac{1}{s}\tanh{\frac{\pi s}{2 a}}$$


问:
Are Lie algebra representations $\rho$ and $\rho\circ F$ isomorphic?
Let $F: \mathfrak g \to \mathfrak g$ be a Lie algebra automorphism, and let $\rho: \mathfrak g \to GL(V)$ be a representation.

Question: Are $\rho$ and $\rho \circ F$ isomorphic?

Motivation: This question is motivated from the physics literature. Physicists often deal with 2nd quantization formalism (where operators can be thought of as elements of a Lie algebra), and they sometimes employ transformations on these operators that preserve the commutation relations. Then, they (maybe naively) use the transformed operators. When I carefully examine these arguments, there is a gap which is precisely stated by my question.

答:
In general, these representations are not equivalent.

Example: Let $\mathfrak g := \mathfrak{sl}_3(\mathbb C)$ and $F: X \mapsto -X^{tr}$. Let $\rho: \mathfrak{sl}_3(\mathbb C) \subset \mathfrak{gl}_3(\mathbb C)$ the standard representation. Then it is an exercise to check that there is no $A \in M_3(\mathbb C)$ such that $F(X) = AXA^{-1}$ for all $X\in \mathfrak g$, i.e. the representations $\rho$ and $\rho \circ F$ are not equivalent.

However, note the following which might redeem the approach: An isomorphism of Lie algebras $F: \mathfrak g_1 \stackrel{\simeq}\rightarrow \mathfrak g_2$ does induce an equivalence of categories between (say, finite-dimensional complex) representations of $\mathfrak g_1$ and (finite-dimensional complex) representations of $\mathfrak g_2$ via

$$ Rep (\mathfrak g_2) \rightarrow Rep(\mathfrak g_1)$$

$$ (\rho, V) \mapsto (\rho \circ F, V).$$

In particular, any automorphism $F: \mathfrak g \rightarrow \mathfrak g$ induces an auto-equivalence of its category of representations, however as the above example shows, this auto-equivalence does not need to be isomorphic to the identity. However, if our interest is to investigate the general structure of that category, i.e. all (fin.-dim., complex) representations, this is kind of harmless. In a way, we only "shuffle them around".

[Actually, in the above example, which generalises to all $\mathfrak{sl}_{n \ge 3}$, the involution $F$, being "the" outer automorphism of the Lie algebra, on first sight seems to induce (up to ...) the duality functor on the category, i.e.: $$\rho \circ F \simeq \rho^* $$
However it is not actually the duality functor, because that one also flips arrows around (is contravariant), whereas the one defined as above via $F$ is covariant: They only do the same on objects. Now this estranged me at first, but I think is redeemed by the category also being semisimple (and duals commuting with direct sums), i.e. basically every arrow can be reversed.]


问:
Relationship between the graphs of $nCk$ and continuous functions
I was curious about functions that involve $nCk$ namely "n choose k" so I just thought of this function:

$$ f(x) = \binom{x+2}{x}^{-1}$$

This function looks like this:

And I thought what is the integral of this on the interval $[0, \infty)$

$$ \int_0^{\infty} \binom{x+2}{x}^{-1} dx$$

It looks like a lot of rectangular areas together. However if we use the fact that:
$$ nCk = \binom{n}{k} =\frac{n!}{k!(n-k)!} \\ \binom{x+2}{x}^{-1} = \frac{x!(x+2-x)!}{(x+2)!} = \frac{2x!}{x!(x+1)(x+2)} = \frac{2}{(x+1)(x+2)}$$

Which is a continuous function that looks like:

And together on the same grid:

Question: Is there a relationship between these? It does not seem like the integral

$$ \int_0^{\infty} \frac{2}{(x+1)(x+2)} dx$$

Is the same as the prior integral, but the functions are the same aren't they?
I would like to hear your ideas! Thank you :-)

答:
You are right, it is because, a priori, the binomial number is defined for integer valued numbers. What is probably happening in your CAS is that they are rounding to the nearest integer the input number (denoted by $\lfloor x\rceil =\arg \min _{n}\{|n-x|\},$ for example $\lfloor 7.3\rceil=7,\lfloor 7.6\rceil=8$). So the orange is actually:
$$\frac{2}{(\lfloor x \rceil+1)(\lfloor x \rceil+2)}.$$ which is not the same function.

If you want the area for the orange, notice that is the sum of the areas of those rectangles of width $1$ and height $\frac{2}{(n+1)(n+2)}=2\left (\frac{1}{n+1}-\frac{1}{n+2}\right ).$ You can use the telescopic sum to find your integral.

If, insted, you want the area of the blue one, a similar decomposition shows the presence of a logarithm.

It is worth noticing that the area of the orange is an approximation of the area of the blue one, by construction of Riemann integral.


问:
Uniqueness of automorphisms on root
The following is from A. Białynicki-Birula, "Algebra" (the translation is mine).

Example 3. Let $p$ be a prime number. Let us consider the splitting field $L$ of the polynomial $\frac{x^p - 1}{x - 1}$ over $Q$. The polynomial $f$ is (see p. 214) irreducible in $Q[x]$. The roots of this polynomial are primitive roots of $p$-th degree. Let $\zeta \in L$ be a fixed primitive root of $p$-th degree. Then $L_{f_p} = \{\zeta, \zeta^2, ..., \zeta^{p-1}\}$. For every $i \in N_+$, $1 \le i \le p-1$ there is exactly one automorphism $\alpha \in Aut(L \backslash K)$ such that $\alpha(\zeta) = \zeta^i$. Indeed, the existence of such an automorphism is implied by the theorem 11.3. Besides that, if $\alpha(\zeta) = \beta(\zeta)$ for $\alpha, \beta \in Aut(L \backslash K)$, then for every natural number $j, 1 \leq j \leq p-1$ we have
$$\alpha(\zeta^j) = (\alpha(\zeta))^j = (\beta(\zeta))^j = \beta(\zeta^j),$$
hence
$$\alpha|L_{f_p} = \beta|L_{f_p}.$$
This - by the final remark of the \$5, example 7, implies that $\alpha = \beta$.

(...)

Example 4. Let us assume that $n$ is a natural number and that the field $K$ contains $n$ different $n$-th degree roots of unity (thus it contains - by theorem VI.7.9 - a primitive $n$-th degree root). Let $a \in K$; let us consider the polynomial $f=x^n-a \in K[x]$ and let $L$ be the extension of splitting field of this polynomial over $K$. Let us denote by $\sqrt[n]a$ whichever root of polynomial $f$ in the field $L$. Then
$$K(\sqrt[n]a) = L \text{ and } L_f = \{\sqrt[n]a, \zeta \sqrt[n]a, \zeta ^2 \sqrt[n]a, ..., \zeta^{n-1} \sqrt[n]a\}$$
where $\zeta$ is any fixed $n$-th degree primitive root. Applying the reasoning similar to that in the example 3 one can easily show that for every integer number $i     $ there exists at most one such an automorphism $\alpha \in Aut(L \backslash K)$ that $\alpha(\sqrt[n]a) = \zeta^i \sqrt[n]a$.

The final remark of the \$5, example 7 says (a bit roughly speaking, but I think that for these needs it is sufficient) that if the restrictions of $\alpha$ and $\beta$ to the set of roots of the polynomial are the same, then $\alpha = \beta$.

Now, how in the case of example 4 the fact that $\alpha (\sqrt[n]a) = \beta (\sqrt[n]a)$ would imply that $\alpha(\zeta^j \sqrt[n]a) = \beta(\zeta^j \sqrt[n]a)$? In the case of example 3, the situation was relatively easy because we had $\alpha(\zeta) = \beta(\zeta)$. Here it seems that $\alpha(\zeta) = \beta(\zeta)$ would also be immediately useful - then we would have $$\alpha(\zeta^j \sqrt[n]a) = \alpha(\zeta)^j \alpha (\sqrt[n]a) = \beta(\zeta)^j \beta (\sqrt[n]a) = \beta(\zeta^j \sqrt[n]a).$$
I tried some other things like raising the whole expression in the parentheses to the $j$-th power, but all these attempts were in vain.

答:
The key is that both $x_1:=\root n\of a$ as well as $x_2:=\zeta\root n\of a$ are roots of the polynomial $x^n-a$. So if the restrictions of $\alpha$ and $\beta$ to the set of roots are equal, then we have, in addition to $\alpha(x_1)=\beta(x_1)$ also
$$
\alpha(\zeta)=\alpha(\frac{x_2}{x_1})=\frac{\alpha(x_2)}{\alpha(x_1)}=
\frac{\beta(x_2)}{\beta(x_1)}=\beta(\frac{x_2}{x_1})=\beta(\zeta).
$$
All because we were also given that $\alpha(x_2)=\beta(x_2)$.

In other words, we can write $\zeta$ in terms of $x_1$ and $x_2$. So if we know where both $x_1$ and $x_2$ are mapped, the image of $\zeta$ is also determined.

Title: Uniqueness of automorphisms on root


问:
If the sum of positive integers $a$ and $b$ is a prime, their gcd is $1$. Proof?
I feel this is an intuitive result.
If, for example, I was working with the prime number $11$, I could split it in the following ways: $\{1, 10\}$, $\{2, 9\}$, $\{3, 8\}$, $\{4, 7\}$, $\{5, 6\}$.

Then clearly there is no way that the $2$ numbers can have a $\gcd$ of anything other than $1$. However, I am sort of lost on how to start a formal proof for this. Any pointers would be much appreciated.

答:
Here is a direct proof that is independent of the others:

Let $g = \gcd(a,b)$. Then we can see that $$g\mathbb{Z} = a\mathbb{Z} + b\mathbb{Z} = \{n\in \mathbb{Z} | ra + sb = n, \text{for } r,s \in \mathbb{Z} \}.$$

In other words, $g$ generates the set of all integer combinations of $a$ and $b$. (If you are not familiar with this, it is most likely presented as a theorem in your textbook.) Since we are $a+b=p$ (given) and $a+b$ is some integer combination of $a$ and $b$, then $p$ must be in $g\mathbb{Z}$.

If $p \in g\mathbb{Z}$, then $g$ must be equal to $1$ or $p$ (since $p$ is prime). But by the fact that $a$ and $b$ are both positive and less than $p$ (the sum of the two equals $p$), then $p$ can't possibly be the greatest common divisor, let alone a divisor of any kind. Therefore $g = 1$.


问:
Is it necessary to include "for all" when using set builder notation to define the image set?
Image Set

Let F be a function from A to B, that is F is a subset from AxB. And let S be a subset of A.

Then the image set of S under F is:
0⃣ F[S] := {y in B: there exist x in S such that xRy}

Is the above definition equivalent to:

1⃣ F[S] := {all y in B such that there exist x in S such that xRy}

I claim definition 0⃣ is not equivalent to 1⃣ to show that let S={7,9} and B={2,4,8} and F={(7,2),(9,8)}

Then be 0⃣ F[S] could be {2} or {8} or {2,8} but 1⃣ must be {2,8}. This is because the ambiguity for “y in B” it could be every y in B in this case it is equivalent to 1⃣ or some y in B which is not equivalent to 0⃣ the core idea is ”y in B” could mean any y or every y in B.

Or we can say F[S]={y: there exist x in A such that xFy} without saying “y in B” or equivalently we can say F[S]=ran(F|S) where F|S means the restriction of F to S.

My whole idea is that “y in B” is ambiguous, and I don’t like that. Am I right?

答:
Based on common conventions for set builder notation, I believe the "for all" or "any" is implied.

For example, to refer to the set $A$ of all real numbers from 0 to 5, it is common to write something like:

$$
A = \{ x : x \in \mathbb{R} \text{ and } 0 \leq x \leq 5 \},
$$

where it's clear that the set is meant to refer to the entire collection of number $[0,5]$, and not just one.

Personally, if I felt there was some risk of misunderstanding, due to context (e.g. conflating the image of a single element Vs. image of a set), I would include a "for all" or "any"; however, it is usually clear from context.

[edit]

I got curious enough to check Wikipedia, according to their definition, the notation always implies inclusion of all elements that meet the criteria (emphasis added):

Set-builder notation can be used to describe sets that are defined by a predicate, rather than explicitly enumerated. In this form, set-builder notation has three parts: a variable, a colon or vertical bar separator, and a logical predicate. Thus there is a variable on the left of the separator, and a rule on the right of it. These three parts are contained in curly brackets:
$$\{\ x \, | \, \Phi(x) \ \} \quad \text{ or  }\quad
\{ \ x : \Phi(x) \ \}$$
The vertical bar (or colon) is a separator that can be read as "such that", "for which", or "with the property that". The formula $\Phi(x)$ is said to be the rule or the predicate. All values of $x$ for which the predicate holds (is true) belong to the set being defined.

Of course, this is a convention, so it may vary from one source to another. FWIW I believe this is the only convention I've encountered.

Edit 2: to clarify that the convention holds on both sides of the statement.

In the comments, you basically asked if the "for all" applies to both the variable and the logical predicate statement, and I would say yes.

Again, pointing to Wikipedia's entry regarding more complex expressions, an expression such as:

$\{ 2n \mid n \in \mathbb{N} \}$ where $\mathbb{N}$ is the set of all natural numbers, is the set of all even natural numbers.

Clearly here the assumption is that the set is the collection of all 2n.

While I completely empathize with your confusion, (mathematical notation can feel very unintuitive when first encountered) I think if you stake a step back, you'll see that this is the obvious convention to adopt: if there is only one element in the set, there's no need for all this elaborate notation. The whole point of set builder notation is to efficiently and unambiguously refer to the members of a set that would be impractical (or impossible) to list, explicitly. If there were only one member, it would be much simpler (& unambiguous) to refer to that single member.

One of my best math teachers once told me that mathematical notation is founded on two principles:

mathematicians are pedantic (i.e. they want to be precise and unambiguous), and - the element that makes it tricky for new-comers -

they are lazy (i.e. they want to write the absolute minimum necessary).

This is a caricaturization, no doubt, but it can be very helpful, in situations like this :)


问:
Prove $C([0,1]),||.||)$ is not complete with $||x||= \text{sup}_{t \in [0,1]}|tx(t)|$
Prove $C([0,1]),||.||)$ is not complete where $||x||= \text{sup}_{t \in [0,1]}|tx(t)|$ is a norm.

The solution given in my lectures:

We take a Cauchy sequence defined as follows:
$$ f_n(t)= \begin{cases} \frac{1}{\sqrt t} && ,\frac{1}{n} \leq t \leq 1 \\ \sqrt n &&, 0 \leq t \leq \frac{1}{n}  \end{cases}   $$

To prove it is Cauchy we take n>m

$$||f_n(t)-f_m(t)||=\text{sup}_{t \in [0,1]}|t(f_n(t)-f_m(t))|=\text{max}\{\text{sup}_{t \in [0,\frac{1}{n}]}|t(\sqrt n -\sqrt m)| ,\text{sup}_{t \in [\frac{1}{n},\frac{1}{m}]}|t(\frac{1}{\sqrt t}-\sqrt m)|,0\} $$
Since in $[\frac{1}{n},\frac{1}{m}], \frac{1}{\sqrt t}-\sqrt m \leq \sqrt n-\sqrt m$
$$||f_n(t)-f_m(t)||= \text{max}\{\frac{1}{n}(\sqrt n -\sqrt m) ,\frac{1}{m}(\sqrt n -\sqrt m) \} ....(\alpha)$$
$$||f_n(t)-f_m(t)||  \to 0 , \text{ when } n,m\to \infty....(\beta)$$

If $\exists f \in C([0,1]) \text{ such that } ||f_n-f|| \to 0$, then

$ \text{sup}_{t \in [0,1]}|t(f_n-f)|\to 0 $,

$\forall t_0 \in [0,1], |f_n(t_0)-f(t_0)| \to 0$,

$\forall t_0 \in (0,1] ,f(t_0)=\lim_{n\to \infty} f_n(t_0)=\frac{1}{\sqrt t_0} ....(\gamma)$

, but no continuous function in $[0,1]$ exists , that coincides with $\frac{1}{\sqrt t}$ in $(0,1] $

Can someone clarify these points:

-In $(\alpha)$ how is that they found that second sup? It doesn't seem right that $ \text{sup}_{t \in [\frac{1}{n},\frac{1}{m}]}|t(\frac{1}{\sqrt t}-\sqrt m)||= \frac{1}{m}(\sqrt n -\sqrt m)$, that would be like requiring $t$ to be $1/m$ and $1/n$ at the same time

-In $(\beta)$ why are they taking a double limit when $m, n \to \infty$ is this legal?, I 've only seen that in Cauchy sequences proofs we have to find the expresion that is less that an $\varepsilon$ for some $n$ and $m$,whya aren't they doing that?

-In $(\gamma)$ they are just rewriting the previous line using $lim$, so why are they leaving out the $0$?

答:
($\alpha$) It doesn't matter. All you need is an inequality, and $\frac{1}{m}(\sqrt n -\sqrt m)$ is an upper bound.

($\beta$) It's the same. Being Cauchy is $\lim_{m,n\to\infty}\|f_n-f_m\|=0$, which is the same as the common phrasing of "Cauchy" that you use.

($\gamma)$  Because it doesn't fit the expression $1/\sqrt{t_0}$, and it doesn't hurt the argument.


问:
Show the fields $\mathbb{Q[\sqrt[3]{3}]}\cong\frac{\mathbb{Q[x]}}{\langle x^3-3\rangle}$ .
show the fields
$$
\mathbb{Q[\sqrt[3]{3}]}\cong\frac{\mathbb{Q[x]}}{\langle x^3-3\rangle}
$$
where $\mathbb{Q[\sqrt[3]{3}]}=\{a+b\sqrt[3]{3}+c\sqrt[3]{9}:a,b,c\in\mathbb{Q}\}$. I am having trouble to show the homomorphism between $\mathbb{Q[\sqrt[3]{3}]}$ and $\mathbb{Q[x]}$, and $\langle x^3-3\rangle$ is the kernel and the maximal ideal of $\mathbb{Q[x]}$. Additionally, I am also confused by what the form of $\mathbb{Q[x]}$ should be. As I only know it means polynomial over rational coefficients but I am thinking in this case it should look like this $\mathbb{Q[x]}=\{a+bx+cx^2:a,b,c\in\mathbb{Q}\}$(which may be wrong.). I am a beginner to Abstract Algebra so a hint or explanations that are simple to understand is important to me. Thanks.

答:
You have the right idea! Here are some hints to push you in the right direction.

First, $\mathbb{Q}[x]$ is the collection of all polynomials with rational coefficients. We allow polynomials of arbitrarily high degree.

Now let's try to use the first isomorphism theorem to attack this problem. We want to show

$$\left . \mathbb{Q}[x] \right / \langle x^3 - 3 \rangle \cong \mathbb{Q}[\sqrt[3]{3}]$$

So it suffices to find a surjective homomorphism $f : \mathbb{Q}[x] \to \mathbb{Q}[\sqrt[3]{3}]$ whose kernel is $\langle x^3 -3 \rangle$ (Do you see why? Here is where you'll use the first isomorphism theorem).

But when we write it like that, there's a natural choice of $f$ to try!

Consider the function $f$ which sends the rational numbers to the rational numbers, and with $f(x) = \sqrt[3]{3}$. Be careful! In this case $x$ is a member of $\mathbb{Q}[x]$, it does not stand for an arbitrary element like it normally does when you see $f(x)$.

Do you see how to extend $f$ to a homomorphism on all of $\mathbb{Q}[x]$? You know what it does to elements of $\mathbb{Q}$, and you know what it does to the element $x$. But you also know that $f(ab) = f(a) f(b)$ and that $f(a+b) = f(a) + f(b)$. Since all polynomials can be built out of $x$ and $\mathbb{Q}$ using multiplication and addition, the rest of the homomorphism is determined by where we send $x$. This is another worthwhile exercise.

Finally, we have a map $f : \mathbb{Q}[x] \to \mathbb{Q}[\sqrt[3]{3}]$ which is obviously surjective (why?). What is the kernel? Can you find some polynomial in $x$ which becomes $0$ in $\mathbb{Q}[\sqrt[3]{3}]$? What do you want the kernel to be?

I hope this helps ^_^


问:
show that differentiation of Fourier sine transform is its own Fourier transform.
I'm trying to solve this problem:

By differentiating the definition of the Fourier sine transform $\tilde f_s(w)$ of the function $f(t) = t^{−1/2}$ with respect to ω, and then integrating the resulting expression by parts, find an elementary differential equation satisfied by $\tilde f_s(w)$. Hence show that this function is its own Fourier sine transform, i.e. $\tilde f_s(w)$  =  Af(w), where A is a constant. Assume that the limit as x → ∞ of $x^{1/2} sin (\alpha x)$ can be taken as zero.

My attempt:

Fourier sine transform of f is given by

$$\sqrt{\frac{2}{\pi}} \int^{\infty}_0 t^{-1/2}sin(wt)\, dt $$

Differentiating this w.r.t w gives:

$$\sqrt{\frac{2}{\pi}} \int^{\infty}_0 t^{1/2}cos(wt)\, dt $$

integrating by parts:

$$\sqrt{\frac{2}{\pi}} [\frac{t^{1/2}sin(wt)}{w} |^{\infty}_0 - \int^{\infty}_0 \frac{sin(wt)}{2wt^{1/2}}\, dt] $$

by assumption as x → ∞ of $x^{1/2} sin (\alpha x)$ can be taken as zero:

$$-\sqrt{\frac{2}{\pi}}  \int^{\infty}_0 \frac{sin(wt)}{2wt^{1/2}}\, dt $$

My question is how do I take this integral?

Thanks!

答:
Let $$F(w) = \int_0^\infty t^{-1/2} \sin (w t)\, dt$$

$$F(w) = w^{-1/2}  \int_0^{\infty} w^{-1/2}t^{-1/2}  \sin wt \, w dt$$

Let $q=wt$.

$$F(w) = w^{-1/2} \int_0^\infty q^{-1/2} \sin q \, dq  = \text{const.}\cdot w^{-1/2}.$$

It is known that $\int_0^\infty q^{-1/2} \sin q \, dq = \sqrt{\frac{\pi}{2}}$, but we don't need to evaluate this integral to show that $t^{-1/2}$ is its own Fourier transform (up-to-a-constant).

UPDATE:

One way to compute $\int_0^\infty q^{-1/2} \sin q\, dq = 2\int_0^\infty \sin p^2 \, dp, $ is to consider the contour integral
$$ \oint_C e^{i z^2} \, dz$$ with C the segment $0$ to $R$ along the $x$-axis, a quarter circle of radius $R$ from $R$ to $z^\star=Re^{i\pi/4}$ and a segment from $z^\star$ to the origin (a slice (1/8) of a pie of radius $R$).

Then $$0=\oint_C e^{i z^2} \, dz = \int_0^\infty e^{ip^2} \, dp - e^{i\pi/4} \int_0^\infty e^{-y^2} \, dy + \int_{\Gamma_R} \exp (Re^{i2\theta})R i e^{i\theta}\,d\theta,$$ and

$$ \begin{aligned} \left| \int_{\Gamma_R} \exp (Re^{i2\theta})R i e^{i\theta}\,d\theta \right| &\le  \int_{\Gamma_R} \left| \exp (R^2e^{i2\theta})R i e^{i\theta}\right| \,d\theta \\ &= \int_0^{\pi/4} Re^{-R^2\sin 2\theta} \,d\theta \\&<\int_0^{\pi/4} Re^{- \frac{4}{\pi} R^2\theta} \, d\theta \\&=\left(1-e^{-R^2} \right)\frac{\pi}{4R} \to 0.\end{aligned} $$

So

$$ \int_0^\infty \cos p^2 \, dp = \frac{1}{2} \sqrt{\frac{\pi}{2}},$$

$$ \int_0^\infty \sin p^2 \, dp = \frac{1}{2} \sqrt{\frac{\pi}{2}},$$

and the Fourier sine transform of $t^{-1/2}$ is $w^{-1/2}$.

$$t^{-1/2} \iff w^{-1/2}.$$


问:
Notation for fields and the elements of the sets used to define them.
Context:

While trying to figure out how to prove that the characteristic of a field $\mathcal{F}$ is either $0$ or prime, I realized that I was conflating the elements of the field with integers, because of the notation.

Question:

Because articulating this question abstractly is beyond my ability, I'll illustrate my confusion with an example:

If $\mathbb{Z}_5$ is the set $\{0,1,2,3,4\}$, along with modular addition and multiplication (i.e. in this case, after performing the standard addition or multiplication, the least remainder after ordinary division by 5; e.g. $4\cdot4 = 1$ and $3+4 = 2$) defines a field. Here,
$$
4 + 1 = 0,
$$
i.e. $1$ is the additive inverse of $4$.

Now, all the classical definitions I've seen usually define the additive inverse as follows (this example is taken from Wikipedia):

for every a in F, there exists an element in F, denoted −a, called the additive inverse of a, such that a + (−a) = 0.

Does the phrase "denoted -a" imply that, in the example provided above, we should denote "$1 = -4$"? Or are such definitions simply stating what they mean when they use the symbol (i.e. it is simply saying we should think of $1$ as the additive-inverse-of-4)? In other words, is "$-4$" an element of this field, even if it is not an element of the set used to define the field (does it even make sense to talk of "an element of a field" in this way)?

Just to be clear, if we were to instead denote the additive inverse of a by na, is it then appropriate to write something like $1 = n4$, for the above field? While I can understand what is meant by this, what bothers me is that $n4$ is not an element of any set described or defined, elsewhere.

Final Considerations

Up until here, I'd be tempted to more or less ignore the whole issue; however, while trying to work out a proof that a field's characteristic is either zero or prime, I found myself wanting to use integers to describe the operations, regardless of whether or not the elements of the set were, themselves, numbers. For example, given the field $\mathbb{F}$ defined by:
$$
\left\{
  \begin{array}{lll}
    \mathbb{S} =  \{a, b\},
    &
     \begin{array}{c|c|c}
       \pmb{+} & \pmb{a} & \pmb{b} \\ \hline
       \pmb{a} & a       & b       \\ \hline
       \pmb{b} & b       & a
     \end{array},
    &
     \begin{array}{c|c|c}
       \pmb{\cdot} & \pmb{a} & \pmb{b} \\ \hline
       \pmb{a} & a       & a       \\ \hline
       \pmb{b} & a       & b
     \end{array},
  \end{array}
\right.
$$

it seems natural and useful to define an expression such as $a + a + a$ as $3a$; however, I'm not sure how to think about the relationship of the symbols introduced by this notation to the field itself (this becomes even more confusing if I use "1" to denote the multiplicative identity). This seems -- at least to my confused mind -- somehow related to the confusion expressed in the first example, above.

Is the use of 3 in "3a" and -4 in "1 = -4" nothing more than a convenient notation to describe the elements of the field? While I feel like that is all they are, I also feel that they carry with them some sort of implied structure (perhaps that of an abelian group?), that isn't accounted for, formally / explicitly.

答:
For every a in F, there exists an element in F, denoted −a, called the additive inverse of a, such that a + (−a) = 0.

This actually means: For every $a \in F$ there exists a unique element $b \in F$ such that $a+b=0$. Since the element $b$ is unique for a given $a,$ the set $\{ (a,b) \in F \times F \mid a+b=0 \}$ is a function (I hope that you know the set-theoretic definition of function) which will be denoted with a prefix minus sign, i.e. $-a$ denotes the unique element in $F$ such that $a+(-a)=0.$

For example, in $\mathbb{Z}_5 = \{ 0,1,2,3,4 \}$ (with operations modulo 5) the notation $-3$ denotes the element $2$ since $3+2=0.$ It does not really denote the negative integer $-3$ which is not in $\{0,1,2,3,4\}.$

Is the use of 3 in "3a" and -4 in "1 = -4" nothing more than a convenient notation to describe the elements of the field?

That is correct.

We define a mapping $\mathbb{N}\times F \to F,$ given by
$(n,x) \mapsto \underbrace{x+\cdots+x}_{n\text{ copies of }x}.$
For simplicity, and since there's no risk of confusion, as KCd points out, we denote this with $nx$.

For example, in $\mathbb{Z}_5,$ we have $7x = x + x + x + x + x + x + x$. It can be shown that this equals $2x = x+x,$ which in turn equals $2x$ as the product of the elements $2$ and $x$ in $\mathbb{Z}_5.$ (Thus, there is no risk of confusion.)

This notation can be extended to negative $n \in \mathbb{Z}$ by $nx=-(-n)x,$ where the inner minus sign is ordinary negative on the integers, and the outer/front minus sign is negative in $F$.

For example, in $\mathbb{Z}_5,$ we have $(-7)x=-(7x)$ by definition, and therefore can skip the parentheses: $-7x$ is unambiguous.


问:
Why are addition and multiplication commutative, but not exponentiation?
We know that the addition and multiplication operators are both commutative, and the exponentiation operator is not.  My question is why.

As background there are plenty of mathematical schemes that can be used to define these operators. One of these is hyperoperation
where

$H_0(a,b) = b+1$   (successor op)

$H_1(a,b) = a+b$   (addition op)

$H_2(a,b) = ab $   (multiplication op)

$H_3(a,b) = a^b$   (exponentiation op)

$H_4(a,b) = a\uparrow \uparrow b$   (tetration op: $a^{(a^{(...a)})}$ nested $b$ times )

etc.

Here it is not obvious to me why $H_1(a,b)=H_1(b,a)$ and $H_2(a,b)=H_2(b,a)$ but not $H_3(a,b)=H_3(b,a)$

Can anyone explain why this symmetry breaks, in a reasonably intuitive fashion?

Thanks.

答:
I think I can dress Qiaochu's answer up in geometric terms. Commutativity of addition and multiplication (in $\bf N$) can be seen as an observation of symmetry of counting blocks that are arranged in certain ways. I recall in elementary school seeing big numbers depicted concretely in our textbooks as piles of blocks.

(Their system involved $a_3$-many $10\times10\times10$ cubes of blocks, $a_2$-many  $10\times10$ squares of blocks, $a_1$-many height-$10$ columns of blocks, and $a_0$-many single blocks in order to depict the number with decimal representation $a_3a_2a_1a_0$.)

Let's represent whole numbers by themselves as columns of blocks. Then $a+b$ involves a column with $a$ blocks underneath a column with $b$ blocks, and $b+a$ vice-versa. Both $a+b$ and $b+a$ are represented by the addition operation as a single column. Dissimilarly, we represent $a\times b$ as a rectangle of blocks of dimensions $a$ by $b$.

The symmetry in both of these cases are clear: flip the column $a+b$ upside-down to get $b+a$, and flip the sheet or plane in which the rectangle of $a\times b$ resides over to obtain that of $b\times a$.

Exponentiation, or repeated multiplication, clearly does not manifest the advantage of yielding the same dimension of object after application. That is, $a^b$ should be represented by a $b$-dimensional hypercube of sidelength $a$, whereas $b^a$ a $a$-dimensional hypercube of sidelength $b$. Thus it becomes obvious that no flipping can give us $a^b$ from $b^a$, and the source of symmetry has been taken away.

In conclusion, commutativity of operations on counting numbers which can be represented via these visual constructions can be seen as a flipping-symmetry which is a property requiring a constant dimension. Thus, dimensional-breaking is the cause of symmetry-breaking.

One may of course fight back with, "well, there is no flipping symmetry, but why isn't there any type of symmetry at all that yields commutativity?" In a sense, "most" binary operations are noncommutative, so believing a "random" binary operation is noncommutative is the natural tentative position to take until evidence to the contrary arises. The evidence to the contrary (flipping symmetry) is strikingly obvious in the cases of addition and multiplication, but not exponentiation.


问:
Find factor of $x^{11} + x + 1$ of degree 2 in $\mathbb{Z}[x]$

Find factor of $x^{11} + x + 1$ of degree 2 in $\mathbb{Z}[x]$

To be honest, it is the first time I see problem like this. I was trying to find divisor by just taking different polinomials of degree 2 and trying to divide this one by them. That was disgusting, I was mixing up in calculations and couldn't finally get answer.

So, Is there any better way to find such a polinomial (divisor of this one with degree 2)? It would be awesome if proposed way would not require super advanced knowledge. Thanks in advance!

Body:
Factorize as follows

\begin{align}x^{11} + x + 1
&= \frac{ x^{11}(x-1) + (x+1)(x-1)}{x-1}\\
&= \frac{x^{12}-x^{11}+x^2-1}{ x-1}
= \frac{(x^{12}-1 )-x^2(x^{9}-1)}{ x-1}\\
&= \frac{(x^{3}-1)(x^9+x^6+x^3+1)-x^2(x^{3}-1)(x^6+x^3+1)}{ x-1}\\
&= \frac{x^{3}-1}{x-1}(x^9+x^6+x^3+1-x^{8}-x^5-x^2)\\
 &= (x^2+x+1)(x^9-x^8+x^6-x^{5}+x^3-x^2+1)
\end{align}
To generalize, follow the same steps above
\begin{align}
x^5+x+1 &= (x^2+x+1)(x^3-x^2+1)\\
x^{17}+x+1 &= (x^2+x+1)(x^{15}-...-x^2+1)\\
...&= \> ...\\
x^{6n-1}+x+1 &= (x^2+x+1)(x^{6n-3}-...-x^2+1)\\
\end{align}


问:
Proving the number of sides of a graph
"A graph of 10 nodes, each of degree 6, has e = 30 sides " , how can i show that ?</p>

答:
Each vertex corresponds to $6$ edges, and so $10 \cdot 6 = 60$.  But of course this overcounts the edges by a factor of $2$ because each edge links two vertices.

QED.

Title: Proving the number of sides of a graph


问:
Show $\sum_{k=1}^n \frac{(n-1)!}{k!(n-k)!(10n)^{k-1}} \leq 1.06$
I was reading Lecture 7 of Afternotes on Numerical Analysis by G. W. Stewart, and he mentions without proof the following result:

If $n \epsilon_M \leq 0.1$ and $\epsilon_i \leq \epsilon_M (i = 1,2,...,n)$, then

$(1+\epsilon_1)(1+\epsilon_2)\cdots(1+\epsilon_n) = 1 + \eta$

where

$\eta \leq 1.06n \epsilon_M$.

I wanted to figure out where this comes from, and, from pretty standard manipulations, I found that it suffices to show that

$\sum_{k=1}^n \frac{(n-1)!}{k!(n-k)!(10n)^{k-1}} \leq 1.06$. I graphed this on Desmos, and it does appear that this inequality is true. However, I have no idea how to go about showing this or where to even really start with this. Any insight would be appreciated!

Body: $\sum \limits_{k=1}^{n} \frac{(n-1)!}{k! (n-k)! (10 n)^{k-1}} = \frac{1}{n} \sum \limits_{k=1}^{n} \frac{(n)!}{ k! (n-k)! } (\frac{1}{10n})^{k-1} = 10\sum \limits_{k=1}^{n} \binom{n}{k}(\frac{1}{10n})^{k} = 10((1+\frac{1}{10n})^n-1) \to 10 (e^{\frac{1}{10}}-1) $ as $n \to \infty$ and $ 10 (e^{\frac{1}{10}}-1) \approx 1.05171$


问:
Product of adjacency matrices involving paths.
Let $A=\textbf{A}_{P_n}$, where $\textbf{A}_{P_n}$ is the adjacency matrix of a path. When we do a graph representation of $A^2-I_n\odot A^2$, we get a graph whose vertex set is defined as $V=\{v_1,v_2,..,v_n\}$ and edge set as $E=\{v_iv_j\;|\;i-j=2\;,\;i>j \}$. I have not been able to prove why this is true.

Edit: ($\odot$ is the Hadamard Product)

答:
The $(i,j)$ entry of $A^k$ counts the number of length-$k$ walks from $v_i$ to $v_j$.

Therefore $A^2$ counts the length-$2$ walks. For each $i$, there is one length-$2$ walk from $v_i$ to $v_{i \pm 2}$ (whenever that makes sense), and one or two length-$2$ walks from $v_i$ back to itself.

Subtracting $A^2 \odot I$ clears out the diagonal, getting rid of the second type of walk and leaving only the first.


问:
Linearizing the product of a binary and a continuous variable
I have an MIP optimization problem that has a constraint $p\geq xy$, where $x$ is a binary variable, $p$ and $y$ are non-negative continuous variables.

I tried the Big-M method. However, the upper bound for $y$ is quite large ($M \geq 10^{15}$). All the MIP solver I tried have numerical problems and cannot solve the problem correctly. I tried to set the solver parameters and the numerical problem still exists.

Since the value range of $y$ is from $10^{-5}$ to $10^{14}$, it seems the values cannot be scaled directly in the problem.

Is there any other way to linearize the product $xy$ which can avoid numerical problems?

答:
As Rob Pratt suggested, if you have one binary variable, you can just solve two problems. If you have a few (say $n$) binary variables, and $2^n$ is small, you can just solve a problem for each combination.

If that's impractical, you might look at combinatorial Benders decomposition [1]. You put $x$ and any other integer variables in the master problem; $p$, $y$ and any other continuous variables go in the subproblem. Given a feasible solution $\hat{x}$ to the master problem, you include the constraint $p\ge y$ if $\hat{x}=1$ and omit the constraint if $\hat{x}=0$. If the subproblem is feasible and its optimal value matches the claimed objective component in the master, accept the master solution. Otherwise, generate either a Benders optimality cut or a Benders feasibility cut. The cuts are similar but not identical to the cuts you get in conventional Benders decomposition. The advantage of the combinatorial Benders approach is that you do not need a constant $M$. The disadvantage is that, as an outer approximation approach, Benders may be slower than solving a single MIP model ... but that apparently is not an issue, since your minimal valid $M$ triggers numerical problems.

[1]
Codato, G. & Fischetti, M. Combinatorial Benders' Cuts for Mixed-Integer Linear Programming. Operations Research, 2006, 54, 756-766.


问:
Why we need to differentiate between the deg. of the polynomial being stationary and the leading coefficient being stationary?
Here is the question I was reading the answer to it by Xam:

Prove that if $R$ is an integral domain and has ACCP, then $R[X]$ has ACCP

Here is the answer by Xam:

This answer is for item ii) given in the question. The other answer is fine, with the corrections given in the comments, but there is not necessary at all to use the fact that $F[X]$ satisfies the ACCP.

As usual, let $$(P_1)\subseteq (P_2)\subseteq \ldots$$ be a chain of principal ideals of $R[X]$. Then $P_{i+1}\mid P_i$ for all $i\ge 1$, and this implies that $\deg(P_1)\ge \deg(P_2)\ge \ldots$ is a decreasing sequence of natural numbers, then it must stabilize and thus there is some $n\in \Bbb{N}$ such that $\deg(P_n)=\deg(P_{n+i})$ for all $i\ge 0$. As $P_{n+i+1}\mid P_{n+i}$ it follows that $P_{n+i}=r_iP_{n+i+1}$ for some $r_i\in R$.

Now, let's denote $a_i$ the leading coefficient of $P_{n+i}$, therefore $a_i=r_ia_{i+1}$ for all $i\ge 0$ and then $a_{i+1}\mid a_{i}$, which lead us to the following chain of principal ideals of $R$: $$(a_0)\subseteq (a_1)\subseteq \ldots$$ Since $R$ satisfies the ACCP, then there exists $k\in \Bbb{N}$ such that $(a_k)=(a_{k+j})$ for all $j\ge 0$. Let's set $t=n+k$, we claim that $(P_t)=(P_{t+j})$ for all $j\ge 0$. Indeed, since $P_{t+j}\mid P_t$, it's enough to prove that $P_t\mid P_{t+j}$. Let's write $P_t=dP_{t+j}$, with $d\in R$, then $da_{k+j}=a_k=ua_{k+j}$, with $u\in R^{\times}$ because $a_k\sim a_{k+j}$. As $a_{k+j}\neq 0$, we deduce that $d=u$. Therefore $d^{-1}P_t=(d^{-1}d)P_{t+j}=P_{t+j}$,
so $P_t\mid P_{t+j}$. Hence, $R[X]$ satisfies the ACCP.

I am asking about the last paragraph:

Why we need to differentiate between the deg. of the polynomial being stationary and the leading coefficient being stationary? could anyone explain that to me please?

NOTE: Below is part of a hint from @Bill Dubuque in my previous question about this step but still I am not understanding:

Finally by R a domain: if $0≠P,Q∈R[x]$ have equal degree and $P∣Q$ then they are associate ⟺ their lead-coefs are associate (and divisibility minimal elements are associate since they must divide each other).

答:
You do not need stationary degrees to conclude abot the leading coefficients.
Ignoring anything about degrees, let $a_k$ be the leading coefficient of $P_k$.
As $P_k\in(P_{k+1})$, we know that $P_k=QP_{k+1}$ for some $Q$ (with $\deg Q=\deg P_k-\deg P_{k+1}$) and by looking at the leading coefficients we see that $a_k\in(a_{k+1})$. So we have our ascending chain
$$ (a_1)\subseteq (a_2)\subseteq (a_3)\subseteq \ldots$$
even without considering degrees. As this gets stationary, we find $n$ such that $(a_k)=(a_n)$ for all $k\ge n$. In particular $a_k=r_ka_n$ for some $r_k\in R^\times$. Now it follows for example for all $k\ge n$ that
$X^{\deg P_n-\deg P_k}P_k-r_kP_n$ is $\in (P_k)$ and due to cancellation of leading terms is of degree $<\deg P_n$.

However, this does not help us at all because having degree $<\deg P_n$ is nothing unusual when all we can say about the general non-zero element of $(P_k)$ is that it has degree $\ge \deg P_k$; after all, there is nothing that prevents $\deg P_k$ from being smaller than $\deg P_n$. Or is there?

Cue the non-increasing and hence eventually stationary degrees. That is, by perhaps switching to some larger $n$, we can ensure that $\deg P_k=\deg P_n$ for all $k\ge n$. So now we find from $P_k-r_kP_n\in(P_k)$ and $\deg (P_k-r_kP_n)<\deg P_n=\deg P_k$ that $P_k-r_kP_n=0$, i.e., $(P_k)=(r_kP_n)=(P_n)$.


问:
show that $R^n$ and $C^n$ are not compact
i need to prove it using the definition of compactness by successions "A metric space X is said to be
compact if every sequence in X has a convergent subsequence".

my attempt:

a)in the case of $R^n$:

suppose that $R^n$ is compact and take the sequence $x_k = (k+1,...,k+n)$ this sequence don't have a convergent subsequence, it's a contradiction, hence $R^n$ is not compact.

For the case of $ C ^ n $ I would try something similar but I don't know how to take a proper sequence.

答:
First of all, by Heine Borel Theorem, a set $X$ is compact iff it's closed and bounded. Now, neither $\mathbb{R}^n$ not $\mathbb{C}^n$ is bounded, so you can directly state that they are not compact.

However, if you want to prove using the sequential criterion, what you've written for $\mathbb{R}^n$ is true. You can take a simpler example anyway, e.g. $x_k := (k, 0, 0, \cdots , 0)$ . This sequence doesn't have any convergent subsequence. So, $\mathbb{R}^n$ is not compact.

Now, remember that $\mathbb{C}^n$ is a superset of $\mathbb{R}^n$ . So you can simply consider the same sequence $\{x_k\}_{k=1}^\infty ~\subset \mathbb{R}^n \subset \mathbb{C}^n$ which certainly shows that $\mathbb{C}^n$ is not compact.


答:
The $i,j$ entry of $A^2$ is given by
$$
[A^2]_{ij} = \sum_{k = 1}^n a_{ik}a_{kj}.
$$
We therefore see that $v_i v_j \in E$ iff there is a $k$ for which $v_iv_k$ and $v_kv_j$ are edges in the original (path) graph. In other words, we have a non-zero entry iff $|i-j| = |j-k| = 1$. This occurs when $i = j$ and when $|i-j| = 2$. Note that the diagonal entries of $A^2$ are "zeroed out" by the subtraction.

Title: Product of adjacency matrices involving paths.

Body:
Let $A=\textbf{A}_{P_n}$, where $\textbf{A}_{P_n}$ is the adjacency matrix of a path. When we do a graph representation of $A^2-I_n\odot A^2$, we get a graph whose vertex set is defined as $V=\{v_1,v_2,..,v_n\}$ and edge set as $E=\{v_iv_j\;|\;i-j=2\;,\;i>j \}$. I have not been able to prove why this is true.

Edit: ($\odot$ is the Hadamard Product)


问:
Rudin Real and Complex Analysis - Baire's Category Theorem
I have been trying to understand Rudin's proof of Baire's Category theorem and I think I understood most of it except eqn (2), where he mentions $0 \leq r_1 \leq 1$, and later generalizes this to $0 \leq r_n \leq 1/n$. My understanding of open set $X$ is that for every $x \in X$, there exists $r > 0$ such that $B(x, r) \subset X$. So how can we know that $0 \leq r_n \leq 1/n$? This inequality is important since this helps us to determine that the sequence $\{x_n\}$ is Cauchy. I know I'm missing something and I would really appreciate much clarity on this concept.

答: If $\varepsilon_n$ is such that$$\overline{S(x_n,\varepsilon_n)}\subset V_n\cap S(x_{n-1},\varepsilon_{n-1}),$$let $r_n=\left\{\varepsilon,\frac1n\right\}$. Then $r_n\leqslant\frac1n$ and$$\overline{S(x_n,r_n)}\subset V_n\cap S(x_{n-1},r_{n-1}).$$


问:
Some questions about a proposition that leads to definition of affine morphism
The question is about a proposition that leads to definition of affine morphism, in section 2.3 of "Geometry I" by Marcel Berger et al. To be searchable, I type the proposition and related text as follows:

Let $(X,\vec X, \Theta)$ and $(X',\vec {X'}, \Theta')$ be two affine spaces (over the same field), and $f:X\to X'$ a map (in set-theoretic sense). The following conditions are equivalent:

i)  $f\in L(X_a; X'_{f(a)})$ for some $a\in X$;

ii) $f\in L(X_a; X'_{f(a)})$ for all $a\in X$;

There are more to come in the proposition, but I wish to prove i)$\Rightarrow$ii) first, with the hope that this proof can help me deal with the remaining. If the notations are not familiar to you, in affine space $(X,\vec X, \Theta)$, $X$ is the affine space, $\vec X$ is the underlying vector space, $\Theta$ is the map from $X\times X$ to $\vec X$ decided by simple transitivity. $L(E;F)$ is the set of all linear transformations from vector space $E$ to $F$. $X_a$ is the vectorialization of $X$ at $a$ (cf 2.1.9 of this text).

The first question is that I don't know if my understanding of vectorialization is correct. When writing $X_a$, does it mean the vector space structure is given, though I don't know what the concrete vector addition and scale multiplication is? If so, does the condition i) mean that, for some $a\in X$, $X$ is endowed with a vector space structure which is isomorphic with $\vec X$ under the map $\Theta_a$, while at the same time, $X'$ is endowed with a vector space structure which is isomorphic with $\vec{X'}$ under the map $\Theta'_{f(a)}$?

The second question is about the proof. To establish ii) under arbitrary $b\in X$, I think I have to construct a vector space structure on $X$ so that $X$ is vectorializable at $b$, as well as the same thing for $X'$ at $f(b)$. But how to do it? I just have no idea what first step to take from the given vectorialization. After that, I need to prove $f$ is a linear transformation between $X_b$ and $X'_{f(b)}$. But before I can endow the two affine spaces with a vector space structure, I don't know how to proceed.

I hope I have formulated my question clearly. Please let me know if there is any clarification needed. Thank you in advance for you help with the proof.

PS, perhaps a little off-topic, is there a text that have the same coverage, depth and rigor as that of Marcel Berger's book, but more accessible to nonmathematicians? Thanks for your recommendation.

答:
I solved the problem myself, so I post the answer here. For the first question, let me assume my guess is correct. That is, when we write $X_a$, the vector space structure is given and therefore we can assume the vector space (and its operations) can be denoted $(X_a, +_a, \cdot_a)$. This vector space is isomorphic with $\vec X$ under $\Theta_a$. So, we are given two vector spaces $(X_a, +_a, \cdot_a)$ and $\bigl(X'_a, +'_{f(a)}, \cdot'_{f(a)}\bigr)$ in i).

At first, I had thought the vector space structure remains the same when $b\in X$ changes. This incorrect assumption led to an additional vector $\vec{ab}$ which made the additive linear relation in the isomorphism unbalanced. As a result, I later decided to fix this unbalance by adding a term in $X$ to the addition to make up an equality. Specifically, I define the vector addition corresponding to $b$ to be $$x+_by=x+_ay+_a\Theta_a^{-1}(-\vec{ab}).$$ Then I tried to define scale multiplication for $X_b$ by changing the scale, but this attempt seemed not working because there was always an additive term that I could not remove. So I went back to fix it by adding a term, and it turns out that the scale multiplication in $X_b$ should be $$c\cdot_bx=c\cdot_ax+_a\Theta_a^{-1}(-c\vec{ab}).$$ Note that the above two operation formulas are for vector space structures in $X$. For vector space structures in $X'$, we need to change them accordingly, say, by adding primes.

With the relation $\Theta_b(x)=\Theta_a(x)-\vec{ab}$ (the Chasles'relation), it can be verified that for $\forall b\in X$ ($a$ is fixed), the above formulas indeed define a vector space structure $X_b$ and make it isomorphic to $\vec{X}$ under $\Theta_b$. Similarly, $(X'_{f(b)},+'_{f(b)},\cdot'_{f(b)})$ is the desired vector space structure for $X'$ at $f(b)$ with the isomorphism being $\Theta'_{f(b)}$.

Next, with the given condition i) which can be translated into $f(x+_ay)=f(x)+'_{f(a)}f(y)$ and $f(c\cdot_ax)=c\cdot'_{f(a)}f(x)$, we can verify that $f$ is indeed a linear transformation between two vector spaces $(X_b, +_b, \cdot_b)$ and $(X'_{f(b)},+'_{f(b)},\cdot'_{f(b)})$ defined in previous paragraph. The only possible obstacle in the verification might be the equality $f(\Theta_a^{-1}\bigl(-c\vec{ab})\bigr)={\Theta'}_{f(a)}^{-1}\bigl(-c\overrightarrow{f(a)f(b)}\bigr)$. To show this, notice that since $\Theta_a$ is an isomorphism, $\Theta_a^{-1}$ is an isomorphism too. So, we have $\Theta_a^{-1}(-c\vec{ab})=-c\cdot_a\Theta_a^{-1}(\vec{ab})=-c\cdot_ab$, and hence $f\bigl(\Theta_a^{-1}(-c\vec{ab})\bigr)=f(-c\cdot_ab)=-c\cdot'_{f(a)}f(b)$. For the same reason, ${\Theta'}_{f(a)}^{-1}$ is an isomorphism so ${\Theta'}_{f(a)}^{-1}\bigl(-c\overrightarrow{f(a)f(b)}\bigr)=-c\cdot'_{f(a)}{\Theta'}_{f(a)}^{-1}\bigl(\overrightarrow{f(a)f(b)}\bigr)$ which is just $-c\cdot'_{f(a)}f(b)$.

========== Edit ==========

I have just understood vectorialization defined in 2.1.9 of the text. The vector space structure assigned to $X_a$ is not something given but unknown, but can be constructed (uniquely) based on the affine space $(X, \vec X, \Theta)$. Specifically, for any $a\in X$, define $x+y=\Theta_a^{-1}(\vec{ax}+\vec{ay})$ and $cx=\Theta_a^{-1}(c\vec{ax})$. It can be verified that it is indeed a vector space and is isomorphic with $\vec X$ under $\Theta_a$. Note that in general the vector space structure changes as $a$ changes. Fortunately, it can proved that the operations defined in my original answer are consistent with the vector space structure just described, so it is still valid.


问:
How do I find the limits to this double integral and convert into polar?
With $M = \{(x, y) ∈ R^2: 1 ≤ x^2 − y^2 ≤ 9, 2 ≤ xy ≤ 4,     x, y ≥ 0\}$, calculate
$$\int_M (x^2+y^2) \mathrm dV$$

What I´ve managed so far is to convert $x$ and $y$ into polar coordinates with $x=r\cos(\theta)$ and $y=r\sin(\theta)$ and got
$$\iint \left[(r\cos(\theta))^2+(r\sin(\theta))^2\right] r\mathrm dr\mathrm d\theta$$

is it right to use polar coordinates? if so, how do I set the limits of this double integral? Thx

答:
Our region (highlighted in yellow) is bound by curves $x^2 - y^2 = 1, x^2 - y^2 = 9, xy = 2$ and $xy = 4$ in the first octant as ($x \geq 0, y \geq 0$).

Equating $x^2 - y^2 = 1, xy = 2$,

$(\frac{2}{y})^2 - y^2 = 1 \implies y^4 + y^2 = 4$. Consider this as a quadratic in $y^2$ and solve for $y^2$. We find $y \approx 1.25$ and so $x = \frac{2}{y} = \approx 1.6$.

Similarly find intersection with $xy = 4$.

Then find intersection of $x^2-y^2 = 9$ with $xy = 2$ and $xy = 4$.

We obtain intersection points as $(1.6, 1.25), (2.13, 1.88), (3.08, 0.65)$ and $(3.25, 1.23)$.

Keeping things simple, I will set up the integral in $3$ parts (please see sketch) -

i) $1.6 \leq x \leq 2.13$ between curves $xy \geq 2$ and $x^2-y^2 \geq 1$

ii) $2.13 \leq x \leq 3.08$ and $2 \leq xy \leq 4$

iii) $3.08 \leq x \leq 3.25$ between $x^2-y^2 \leq 9$ and $xy \leq 4$.

$\displaystyle I_1 = \int_{1.6}^{2.13}\int_{2/x}^{\sqrt{x^2-1}} (x^2+y^2) dy dx \approx 1.48$

$\displaystyle I_2 = \int_{2.13}^{3.08}\int_{2/x}^{4/x} (x^2+y^2) dy dx \approx 6.02$

$\displaystyle I_3 = \int_{3.08}^{3.25}\int_{\sqrt{x^2-9}}^{4/x} (x^2+y^2) dy dx \approx 0.5$

$I = I_1 + I_2 + I_3 \approx \fbox{8}$


问:
bipartite graph, no odd cycles, one question about the proof
I found a proof of the fact that if a graph G is bipartite(1), then it cannot have any odd cycles(2). I have a question about $(2) \Rightarrow (1)$. Why is it sufficient to assume that  G is connected?</p>

答:
It's a consequence of the following:

A graph is bipartite if and only if each of its components are bipartite.

So, for example, this graph is bipartite, since each of its components are bipartite:

This generalises to:

A graph is $k$-colourable if and only if each of its components are $k$-colourable.

This generalises to:

The chromatic polynomial of a graph is the product of the chromatic polynomials of its components.

This generalises to:

The Tutte polynomial of a graph is the product of the Tutte polynomials of its components.


问:
Gram-Schmidt process for polynomial space
I'm looking for a orthonormal basis to 2 polynomials $f,g ∈ \Bbb R_2[x]$

we will define inner product space:

$<f,g>=f\left(0\right)\cdot g\left(0\right)+f\left(1\right)\cdot \:g\left(1\right)+f\left(2\right)\cdot \:\:g\left(2\right)$

I know how to do Gram-Schmidt process to vectors

But how do you make it for polynomial?

**
For example in vectors it is easy to see vectors like $\vec v=(0,1),\vec u=(1,0)$

And see that all vectors in $\Bbb R_2$ can appear as linear combination of $\vec v,\vec u$
**

But how does it work with polynomials?

Thank you.

答:
Definition: A vectorial space $V$ over some field $F$ is a vectorial space with inner product if there exists and function $<.>: V \times V \longrightarrow F$ su h that for any
$u,v,w\in V$ and $\alpha \in F$:

$<v,v> \, >0$

$<v,v>=0$ iff $v=0$

$=\overline{}$

$=+$

$=+<w,v>$

$<\alpha u,v>=\alpha $

Definition: Let $V$ a vector space with inner product over some field $F$,given $v\in V$ we define the norm of $v$ by  $||v||=\sqrt{<v,v>}$

${\displaystyle \mathrm {proj} _{\mathbf {u} }\,(\mathbf {v} )={\langle \mathbf {u} ,\mathbf {v} \rangle  \over \langle \mathbf {u} ,\mathbf {u} \rangle }{\mathbf {u} },}$

Finally given a finite vector space $V$ over $F$, with inner product, then $V$ have a ortonormal basis.

Construction

Let $S= \lbrace v_1,v_2, \cdots v_n \rbrace $ basis of $V$.Pick $v_1\in S$.
$u_1=v_1$

$u_2=v_2-Proj_{u_1}(v_2)$

$u_3=v_3-Proj_{u_1}(v_3)-Proy_{u_2}(v_3)$

and notice

$u_n=v_n-\sum_{j=1}^{n}{Proy_{u_j}(v_n)}$

And consider $e_i=\frac{u_i}{||U_i||}$

Then the set $U=\lbrace e_1,e_2, \cdots, e_n \rbrace$ is a ortonormal basis of $V$.

In your case you had that the polinomials are a vector space with inner product given by
$
=\sum_{i=0}^{n}{a_ib_i}$ where $a_i,b_i$  are the coeficients of your polinomial(you can comprove it).

Now you only had to follow the process.

I hope that my comment helps you

Body:
Definition: A vectorial space $V$ over some field $F$ is a vectorial space with inner product if there exists and function $<.>: V \times V \longrightarrow F$ su h that for any
$u,v,w\in V$ and $\alpha \in F$:

$<v,v> \, >0$

$<v,v>=0$ iff $v=0$

$<u,v>=\overline{<u,v>}$

$<u,v+w>=<u,v>+<u,w>$

$<u+w,v>=<u,v>+<w,v>$

$<\alpha u,v>=\alpha <u,v>$

Definition: Let $V$ a vector space with inner product over some field $F$,given $v\in V$ we define the norm of $v$ by  $||v||=\sqrt{<v,v>}$

${\displaystyle \mathrm {proj} _{\mathbf {u} }\,(\mathbf {v} )={\langle \mathbf {u} ,\mathbf {v} \rangle  \over \langle \mathbf {u} ,\mathbf {u} \rangle }{\mathbf {u} },}$

Finally given a finite vector space $V$ over $F$, with inner product, then $V$ have a ortonormal basis.

Construction

Let $S= \lbrace v_1,v_2, \cdots v_n \rbrace $ basis of $V$.Pick $v_1\in S$.
$u_1=v_1$

$u_2=v_2-Proj_{u_1}(v_2)$

$u_3=v_3-Proj_{u_1}(v_3)-Proy_{u_2}(v_3)$

and notice

$u_n=v_n-\sum_{j=1}^{n}{Proy_{u_j}(v_n)}$

And consider $e_i=\frac{u_i}{||U_i||}$

Then the set $U=\lbrace e_1,e_2, \cdots, e_n \rbrace$ is a ortonormal basis of $V$.

In your case you had that the polinomials are a vector space with inner product given by
$<p,q>=\sum_{i=0}^{n}{a_ib_i}$ where $a_i,b_i$  are the coeficients of your polinomial(you can comprove it).

Now you only had to follow the process.

I hope that my comment helps you


问:
Homomorphisms between $ \mathbb{Z} $ modules.

Calculate $\newcommand\Hom{\operatorname{Hom}}\Hom(\mathbb Z \oplus \mathbb Z_{p^\infty},\mathbb Z \oplus \mathbb Z_{p^\infty})$. Where $ \mathbb Z_{p^\infty}= \bigcup_{k=1}^{\infty}\bar{\langle\frac1{p^k}\rangle}$ for $p$ prime.

As $\mathbb Z$ and  $\mathbb Z_{p^\infty}$ are $\mathbb Z$-modules then applying the theorem $\Hom_A ( \bigoplus_{i \in I} M_i, \prod_{j \in J} N_j ) \cong\prod_{(i,j) \in I \times J} \Hom(M_i,N_j)$ because for finite indices have $\bigoplus_{i=1}^{n}M_i = \prod_{i=1}^n M_1$. Then the problem reduces to find those $\Hom$ where

$$ \Hom(\mathbb{Z}, \mathbb{Z}) \cong \mathbb{Z} $$
$$ \Hom(\mathbb{Z},\mathbb{Z}_{p^\infty}) \cong \mathbb{Z}_{p^\infty}$$
but  $\Hom(\mathbb{Z}_{p^\infty},\mathbb{Z})$ and  $\Hom(\mathbb{Z}_{p^\infty},\mathbb{Z}_{p^\infty})$ not how to calculate them.

答:
To understand $\newcommand\Hom{\operatorname{Hom}}\Hom$, think about generators.  Since any generator of $\mathbb{Z}_{p^\infty}$ is torsion, it has to map to $0$ in $\mathbb{Z}$.  So,
$$
\Hom(\mathbb{Z}_{p^\infty}, \mathbb{Z}) = 0.
$$

In order to understand $\Hom(\mathbb{Z}_{p^\infty}, \mathbb{Z}_{p^\infty})$, consider the definition of $\mathbb{Z}_{p^\infty}$ as a direct limit.
$$
\mathbb{Z}_p \overset{p}{\to} \mathbb{Z}_{p^2} \overset{p}{\to} \mathbb{Z}_{p^3} \overset{p}{\to} \cdots \to \mathbb{Z}_{p^\infty}
$$
Any map from a direct limit is characterized by the commuting diagram of maps from each of the groups in the directed system.  So, if $x_n$ is a generator of $\mathbb{Z}_{p^n}$, then its image in $\mathbb{Z}_{p^{n+1}}$ is $px_{n+1}$ for some generator of $\mathbb{Z}_{p^{n+1}}$.  These are identified in the direct limit $\mathbb{Z}_{p^\infty}$, so for any well-defined $f \in \Hom(\mathbb{Z}_{p^\infty}, G)$ for any group $G$,
$$
f(x_n) = p f(x_{n+1}) = p^2 f(x_{n+2}) = \cdots
$$
Thus, $f(x_n)$ must be infinitely divisible:  $f(x_n) = p^k f(x_{n+k})$ for any $k \ge 0$.  In fact, one can show that
$$
\Hom(\lim_{\to} A_n, G) \cong \lim_{\leftarrow}\Hom(A_n, G),
$$
for any abelian groups.  In your case,
$$
\begin{align}
\Hom(\mathbb{Z}_{p^\infty}, \mathbb{Z}_{p^\infty}) &\cong \Hom(\lim_{\to} \mathbb{Z}_{p^n}, \mathbb{Z}_{p^\infty}) \\
&\cong \lim_{\leftarrow}\Hom(\mathbb{Z}_{p^n}, \mathbb{Z}_{p^\infty}) \\
&\cong \lim_{\leftarrow}\mathbb{Z}_{p^n}.
\end{align}
$$
The last group is an uncountable abelian group called the $p$-adic integers.


问:
Dimension of the dual space of an infinite dimensional vector space using Hahn-Banach
I have to prove that if $\dim(X)=\infty$ then $\dim(X^*)=\infty$. But this question is from the section of excercise of Hahn-Banach extension theorem of my professor. I really don't know how to start this, please if you can give me a couple of tips it would be very helpfull. Thanks.

答:
Lemma 1:
If $F \in X^{*}$ is nonzero and $\mathcal{N}=\{x \in X: F(x)=0\},$ then there exists a one-dimensional subspace $\mathcal{M}$ of $X$ such that
$$
X=\mathcal{N}\oplus\mathcal{M}=\{x+y: \quad x \in \mathcal{N} \quad \text { and } \quad y \in \mathcal{M}\} \quad \text { and } \quad \mathcal{N} \cap \mathcal{M}=\{0\}.
$$

Lemma 2:
Given $x\neq 0$, then there is an $F \in X^*$ such that
$$||F||=1 \qquad \text{ and } \qquad F(x)=||x||_X $$

Both lemmas are direct consequences from Hahn-Banach Theorem and can be easily verified.

Now, let's suppose (by contradiction) that $\dim (X^*)=n\leq \infty$, then we can consider a base $\mathcal{B}=\{F_1,\dots,F_n \}$ of $X^*$ and apply the lemma 1 for each $F_i$, this gaves the one dimentional subspaces $\mathcal{M}_i$ and the $\mathcal{N}_i$ with the above mentioned properties. Note that by the definition of the $\mathcal{M}_i$, $X=\mathcal{M}_1\oplus \mathcal{M}_2\oplus \dots \oplus \mathcal{M}_n \oplus X_0$, where $X_0$ is the infinitely dimensional space given by $\bigoplus \mathcal{N}_i/\bigoplus M_i$.

So, consider $x$ an arbitrary element of $X_0$, and $F\in X^*$, note that: $$ F(x)=F_1(x)+F_2(x)+\dots+F_n(x)=0,$$ where the first equality is due to the fact that $F_i$ is a base for $X^*$, and the second come from the fact that $X_0\cap M_i = \{0 \}$, note that is for any linear functional and any $x\in X_0$. In particular given an $x\in X_0$ we can consider the $F$ given by lemma 2, as $F(x)=0=||x||$, we can conclude that $||x||=0$ for all $x\in X_0$, therefore $x=0$ and hence $X_0=\{0\}$ which is a contradiction to the fact that $X_0$ is infinitely dimensional.

With this we have that $\dim X^* = \infty$.


问:
Counting subsets (Permutations) when certain numbers must exist
Suppose I have a list of $n$ numbers $\{a1,a2...an\}$, and I need to create subsets of this list with $k$ numbers, REPETITIONS ALLOWED!, and the order does matter. The number of such subsets is simply $n^k$.

But, I also need the subset to include $m$ specific numbers from the list at least once. So if, say, $m=3$, then the first 3 numbers $(a1,a2,a3)$ must exist in the subset at least once.

I have tried to subtract like this: $n^k-m*(n-1)^k$ but got wrong results.

So, what is the correct formula?

答:
Call $$A_i = \text{# of lists that do not contain element }i.$$
You want, by the principle of Inclusion-Exclusion, $$\text{All}-\left | \bigcup _{i=1}^mA_i\right |=\sum _{i=0}^{m}(-1)^i\binom{m}{i}(n-i)^k.$$
The problem with your formula is that you are taking too many elements out that you want to include back. For example, if $k=5,m=3,n=5$ and your list is $(a_4,a_4,a_4,a_4,a_4)$ you were taking out this element $3$ times.

It is worth noticing the connection with the problem of finding surjective functions i.e., Stirling numbers of second kind


问:
How to maximize $n!\sum^n_{k = 0}\frac{a^k(1+(-1)^{n-k})}{k!(n-k)!} \pmod{a^2}$ for a given $a$?
Short Version of the Question:

How do I maximize the value of $n!\sum^n_{k = 0}\frac{a^k(1+(-1)^{n-k})}{k!(n-k)!} \pmod{a^2}$ for a given $a$?

Long Version of the Question:

I'm currently attempting a Project Euler question using as much mathematics as possible before resorting to coding.

The problem itself is as follows:

```<code>Let r be the remainder when (a-1)^n + (a+1)^n is divided by a^2.

For example, if a = 7 and n = 3, then r = 42: 63^3 + 83^3 = 728  42 mod 49. And as n varies, so too will r, but for a = 7 it turns out that rmax = 42.

For 3 < a < 1000, find the sum of all rmax.
</code>```

My first attempt at this problem was to change $(a-1)^n + (a + 1)^n$ to make it easier for coding purposes. And so I used the binomial theorem to come up with the following:

$(a-1)^n + (a + 1)^n$ = $n!\sum^n_{k = 0}\frac{a^k(1+(-1)^{n-k})}{k!(n-k)!}$

Obviously what I want to do now, is to maximize this new equation with the modulo. However, I cannot come up with how to do this. It may very well be that the original statement would be easier to simplify!

Nonetheless, I've come to ask: how can I maximize the value of $n!\sum^n_{k = 0}\frac{a^k(1+(-1)^{n-k})}{k!(n-k)!} \pmod{a^2}$ for a given $a$?

On this point, I would rather a hint, as opposed to a full answer. That would ruin the fun!

Note: I have a limited knowledge of calculus, and have yet to find the derivative of a modulo statement, or a summation. Would this need to be learned?

答: Hint: If you expand $(a-1)^n$, you will find that every term but two are multiples of $a^2$.  Similarly for $(a+1)^n$.  Hence you can have a much simpler expression,  with no $\Sigma$,  $\mod{a^2}$.</p>


问:
How can a vector space $V$ over a field have $V$ as a basis?
I think my question is related to Free vector space over a set, Non-numerical vector space examples and Formal (series/sum/derivative…).

I'm reading a paper, which starts with the following sentence (the original source is "Flag Algebras" by Razborov)

Let $\mathbb{R}F$ be the real vector space with basis comprised of all elements of $F$; in other words, the space of all formal, finite $\mathbb{R}$-linear combinations of elements of $F$.

Let me summarize my understandings first. We can define the set $\mathbb{R}F$ as a formal sum as $\mathbb{R}F=\{\sum_{f \in F} \lambda_f f | \lambda_f \in \mathbb{R}\}$ and construct the structure of a vector space as described in Free vector space over a set.

Now, here is my trouble: If $\mathbb{R}F$ is supposed to be a vector space, then, due to the vector space axioms, the following must be satisfied: Let $a, b \in F$, then $a+b=c \in F, \forall a,b \in F$, without defining the actual operation "+" of our vector space (since we speak of "formal" sum here). But if $F$ is the basis of our vector space, then all elements of $F$ must be linear independent, thus there cannot exist a $c \in F$ for which $a+b=c$ with $a,b \in F$. (my definition for a vector space is taken from "Advanced Linear Algebra" by Steven Roman").

I would suppose that we define the vector space as the set of all formal sums of elements of $F$ with real coefficients and take the axioms for granted. This works fine for associativity, the inverse element, etc., since we do not define "+" and "juxtaposition" anyway, but I don't see how the basis can be the whole set $F$.

My question is, how can the vector space axioms be satisfied? Can someone give me some examples, or clarifications?

答: You did not apply the vector space axioms correctly.  If $\mathbb{R} F$ is a vector space, then for all $a,b \in \mathbb{R} F$ (not $F$, as you stated), we have some $c \in \mathbb{R} F$ (not $F$) with $c = a + b$.


问:
Prove that inclusion is a "universal" partial order
I wanted to prove a theorem in my textbook which states that inclusion can be viewed as a "universal" partial ordering (how cool is that!). The theorem goes like this: Let $(X, \leq)$ be a partially ordered set. Then, there exists a family of sets $A \subseteq P(X)$ such that partially ordered sets $(X, \leq)$ and $(A, \subseteq |_A)$ are isomorphic. Note: $P(A)$ means power set of set $A$ and $\subseteq|_A$ means that the inclusion was restricted to $A$.

Proof. Define $f(x)=\{a \in X:a \leq x\}$ for every $x \in X$. First let's show that $f:X \rightarrow P(X)$ is injective. Assume that $f(x)=f(y)$, where $x,y \in X$. We see that $x \in f(x)$ because $\leq$ is reflexive by definition. Since $f(x)=f(y)$, $x \in f(y)$, so $x \leq y$. Conversely, $y \in f(y)$, but $f(x)=f(y)$, so $y \leq x$. By antisimmetric property of $\leq$, we get that $x \leq y$ and $y \leq x \iff x=y$. Hence, $f$ is indeed injective. Great. Now we would like to show that $f(x) \subseteq f(y) \iff x \leq y$. Let's define $A=f(X)$. Note that $f(x)$ is a set, so $f(X)$ is a set of sets a.k.a. a family of sets. Now we should probably use the definition of subset.
$$f(x) \subseteq f(y) \iff \forall_\alpha (\alpha \in f(x) \Rightarrow \alpha \in f(y)).$$
We see that $\alpha \in f(x) \iff \alpha \leq x$ and $\alpha \in f(y) \iff \alpha \leq y$, thus
$$\forall_\alpha (\alpha \in f(x) \Rightarrow \alpha \in f(y)) \iff \forall_\alpha(\alpha \leq x \Rightarrow \alpha \leq y) $$
I would like to show that this is equivalent to $x \leq y$, but I have no idea how to proceed from this point onward. It makes sense for the "standard" less or equal to relation, but I don't know how to prove it rigorously. I would really like to prove this think cause it's pretty awesome in my opinion.

EDIT: In case anyone was curious how to finish the proof (basing on the kindly provided answers below).
First, show that $$\forall_\alpha (\alpha \leq x \Rightarrow \alpha \leq y) \Rightarrow x \leq  y.$$ Assume $\forall_\alpha (\alpha \leq x \Rightarrow \alpha \leq y)$ and choose $\alpha = x$. By the associative property, $x \leq x$, so by assumption $x \leq y$.

Second, prove that
$$x \leq  y \Rightarrow \forall_\alpha (\alpha \leq x \Rightarrow \alpha \leq y).$$
Suppose that $x \leq y$. Definition of transitivity tells us that $\forall_\alpha (\alpha \leq x \wedge x\leq y \Rightarrow \alpha \leq y)$. Which, by assumption, reduces to $\forall_\alpha (\alpha \leq x \Rightarrow \alpha \leq y)$. QED

答: If $x\le y$ and $\alpha\le x,$ then $\alpha\le y$ by transitivity. If for all $\alpha\le x$, $\alpha\le y,$ then, in particular, $x\le x$, so $x\le y.$