问:
Proving $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$
Let $A$, $B$ and $C$ be sets. Prove that $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

I am not sure how to formally prove and in special write this intuitive simple question about sets.

Thank you

答:
You could say A n (B u C) is either true or false.
lets take the case when its false, well that implies omega not in A or omega not in (B u C)
for some omega in the set.
if omega not in A then the right side is always false which means false = false
if omega not in (B u C) then its not in either and therefore the right hand side is always false.
You can do the same for true. if the left hand side is true then omega is in A and it is in either B or C and for the right hand side to be true you only need it to be in two sets.
Write this out in a truth table and thats a sufficient proof.
sorry im not very good with latex.</p>


问:
Can a generator of the ring of integers of local fields can be chosen so that it is also a uniformizer at the same time?
Let $L/K$ be an extension of local fields. We can find $\alpha$ such that $\mathcal{O}_L=\mathcal{O}_K[\alpha]$. What do we know about this generating element? I think that this $\alpha$ can be selected in such a way that, in addition to the above property, it is also a uniformizing parameter at the same time. But I can not prove it.

答:
$O_L=O_K[\pi_L]$ iff $f(L/K)=1$ and $O_L=O_K[\zeta_{q_L-1}]$ iff $e(L/K)=1$. In general it is $$O_L=O_K[\zeta_{q_L-1}+\pi_L]$$

Hensel lemma is needed to construct the root unity, then (in the non-trivial case $e\ne 1,f\ne 1$) we need a closed-ness and density argument, investigating $\zeta_{q_L-1}+\pi_L-(\zeta_{q_L-1}+\pi_L)^q$

ContentLicense: CC BY-SA 4.0


问:
Linear and logistic regression
Let a sample $(x,y) \in \mathbb{R}^{2n}$ be given, where $y$ only attains the values $0$ and $1$. We can try to model this data set by either linear regression
$$y_i = \alpha_0 + \beta_0 x_i$$
with the coefficients determined by the method of least squares or by logistic regression
$$\pi_i = \frac{\exp(\alpha_1 + \beta_1 x_i)}{1+\exp(\alpha_1 + \beta_1 x_i)},$$
where $\pi_i$ denotes the probability that $y_i = 1$ under the given value $x_i$ and the coefficients are determined by the Maximum-Likelihood method. My question is whether the following statement holds true.

Claim: If $\beta_0 > 0$ ($\beta_0 < 0$), then $\beta_1 > 0$ ($\beta_1 < 0$).

I figure this could be due to the sign of the correlation coefficient. However, I could not connect this to the logistic regression.

答:
If $\beta_0>0$, i.e., $\frac{\partial}{\partial x}E[Y|X=x] = \beta_0 > 0$, is suggests that increase in $x$ will increase the probability that $Y=1$, since $E[Y|X=x] = P(Y=1|X=x) = p$. Now, the logistic model is equivalent to
$$
\ln\left( \frac{p}{1-p} \right) = \alpha_1 + \beta_1x.
$$
The right hand side can be viewed as linear approximation of $\ln(p/(1-p))$. Given that the original linear model is OK, and $\beta_0 > 0 $, it translates into $\partial / \partial x \ln(p/(1-p)) = (p(1-p))^{-1}\beta_0 > 0$ for the $\ln$ odds model, namely, $\beta_1$ must be positive as well (as the slope of the linear approximation).

Title: Linear and logistic regression


问:
Write ‘There is exactly 1 person…’ without the uniqueness quantifier
During a lecture today the prof. posed the question of how we could write "There is exactly one person whom everybody loves." without using the uniqueness quantifier.

The first part we wrote as a logical expression was "There is one person whom everybody loves.", ignoring the 'exactly one' part of the question initially. From this he wrote

$L(x,y): x$ loves $y$; domain for $x$ and $y$: $\{\text{people}\}$

$\exists x\forall y: L(y,x)$

Which I understand to mean 'There is a person $x$ such that for all $y$, $x$ is loved by $y$' AKA 'There is a person who is loved by everyone'. I get that part.

The part I don't get is how the expression of 'exactly one'.

$\forall z(\forall y(L(y,z))\to x =z)$

which then creates the joint expression

$\exists x\forall y(L(y,x))\land \forall z(\forall y(L(y,z))\to x=z)$

I just can't seem to understand how $\forall z(\forall y(L(y,z))\to x =z)$ means exactly one here.  I suppose you can take $\forall z$ here to mean 'for any given person', which means the $\forall z$ is considering every person in the world.  This would translate the second expression block to something like, "For any given person $z$, if everybody loves $z$ then $z$ is the same person as $x$".

To me though $\forall z$ generally means for every element in the domain which I see as meaning every person in the world simultaneously, as it seems to for $y$.  Is that just plain wrong?  How can I tell when $\forall$ means 'all' and when it means 'for any (one)'?  In the previous English translation the only reason I was able to translate it (if it's even right) is because I already knew what the statement was suppose to mean.

Is it just that $\forall z$ means that this statement could be true for any element, and if so what's the difference between $\forall z$ and $\exists z$?  Someone told me that $\exists z$ would be redundant because the expression says $x=z$ but how do I know that $x$ and $z$ are automatically the same person if $\exists$ is used for both?

Sorry if this is a bit long with too many questions.  I just wanted to try to make the cause of my confusion as clear as possible so you can help me figure this out.

答:
The simplest definition for $\langle \exists! x :: P(x) \rangle$ I know, which I've probably learned from Dijkstra et al.'s works, is $$\langle \exists y :: \langle \forall x :: P(x) \equiv x = y \rangle \rangle$$
which does only use $P(x)$ once.

Note how we have the following nice symmetry:
$$
\begin{array} \\
\langle \exists! x :: P(x) \rangle & \;\equiv\; & \langle \exists y :: \langle \forall x :: P(x) & \equiv & x = y \rangle \rangle \\
\langle \exists\phantom! x :: P(x) \rangle & \;\equiv\; & \langle \exists y :: \langle \forall x :: P(x) & \Leftarrow & x = y \rangle \rangle \\
\langle \phantom\exists! x :: P(x) \rangle & \;\equiv\; & \langle \exists y :: \langle \forall x :: P(x) & \Rightarrow & x = y \rangle \rangle \\
\end{array}
$$
where $\langle ! x :: P(x) \rangle$ means "there is at most one $x$ such that $P(x)$".

ContentLicense: CC BY-SA 3.0


问:
Proof of $\text{dim}(E)\leq\text{dim}(E^*)$ without using a basis
Let $E$ be a vector space over a (not necessarily commutative) field $K$. How can we show that $\text{dim}(E)\leq\text{dim}(E^*)$ without using a basis?

My proof uses a basis. Let $(b_i)_{i\in I}$ be a basis of $E$. We have a family $(b^*_i)_{i\in I}$ of elements of $E^*$ characterized by $\langle b_i,b^*_j\rangle=0$ for $i\ne j$ and $\langle b_i,b^*_i\rangle=0$. Clearly this family is free. Let $H$ be the subspace of $E^*$ generated by $(b^*_i)_{i\in I}$. Then
$$\text{dim}(E)=|I|=\dim(H)\leq\dim(E^*).$$

Is there a way to prove this without using a basis?

答:
It's a result due to Blass that "every vector space has a basis" is equivalent to the axiom of choice. Without AC this result is not necessarily true; in fact it's consistent with ZF that there exist infinite-dimensional vector spaces $E$ such that $E^{\ast} = 0$. See this MO question for some details, giving the example of $E = \mathbb{R}$ as a vector space over $\mathbb{Q}$.

Of course without bases / AC we have the further problem that it's no longer even clear what $\dim E$ means. We could take it to mean either the smallest size of a generating subset or the largest size of a linearly independent subset and it's no longer clear that these coincide. But in any case it's clear that with either of these definitions $\dim_{\mathbb{Q}} \mathbb{R}$ is infinite (and with a little effort uncountable; this, requires explicitly constructing an uncountable linearly independent subset which is done here), whereas it's consistent that $\text{Hom}_{\mathbb{Q}}(\mathbb{R}, \mathbb{Q}) = 0$ which certainly has dimension $0$.


问:
Elementary equivalence $\mathbb{Z}$ and $\mathbb{Q}$ in language of addition only.
I am stuck on a seemingly easy question. When considering the language $\mathcal{L}=\{f\}$ where $f$ is a 2-ary function that we will interpret as $+$ in models $(\mathbb{Z},+)$ and $(\mathbb{Q},+)$.

The question is whether these models are elementary equivalent. I think they are since we are given no constant symbols like $1$ and we have no order $<$ in our language, I can't see how there could be a statement about addition that is only true in $\mathbb{Q}$.

Since they are not isomorphic, I thought the only option was direct proof but I couldn't work that out either.

Is my intuition correct and is there a way out of a direct proof?

thanks in advance!

答:
I'm not sure because my knowledge on logic models is poor, but I believe this argument proves they're not equivalent. It only uses addition.

For every $x\in \mathbb Q$ there exists $y \in \mathbb Q$ such that $y+y=x$. The same is not true for $\mathbb Z$.

Edit: Sniped by TonyK


问:
How many permutations of numbers $(1,2,3, .., n)$ are there in which at least $2$ elements are in their original place?
How many permutations of numbers $$(1,2,3, .., n)$$ are there in which at least $2$ elements are in their original place?

My idea for solving:
$n! - |\text{no element stands in its original place}| - |\text{exactly one element is in its original place}|$

Is it correct?

答:
We first compute the number of permutations with zero or one elements in
their original place. Using combinatorial classes we have the following
class $\mathcal{P}$  of permutations with fixed points marked:

$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{P} = \textsc{SET}(
\mathcal{U} \times \textsc{CYC}_{=1}(\mathcal{Z}) +
\textsc{CYC}_{=2}(\mathcal{Z})  +
\textsc{CYC}_{=3}(\mathcal{Z})  + \cdots).$$

This gives the EGF

$$G(z, u) = \exp\left(uz+\frac{z^2}{2}
+ \frac{z^3}{3}
+ \frac{z^4}{4}
+ \cdots \right)
\\ = \exp\left(\log\frac{1}{1-z} + (u-1)z\right)
= \frac{1}{1-z} \exp\left((u-1)z\right)
\\ = \frac{1}{1-z} \exp\left(-z\right)
\exp\left(uz\right).$$

We have for the desired statistic

$$n! [z^n] ([u^0] + [u^1]) G(z, u)
= n! [z^n] \frac{1}{1-z} \exp(-z) (1 + z).$$

Extracting the coefficients we find

$$n! \times
\left( \sum_{k=0}^n \frac{(-1)^k}{k!}
+ \sum_{k=0}^{n-1} \frac{(-1)^k}{k!} \right).$$

This is

$$2 n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}
- (-1)^n.$$

The desired answer is thus

$$\bbox[5px,border:2px solid #00A000]{
n! + (-1)^n - 2n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}.}$$

which is

$$\bbox[5px,border:2px solid #00A000]{
n! + (-1)^n - 2 \times !n.}$$

This gives the sequence

$$0, 1, 1, 7, 31, 191, 1331, 10655, 95887, 958879, \ldots$$

which is OEIS A155521, where these data are
confirmed.


问:
If $G=⨝_{i\in I} H_i$ and $K\unlhd G$ is perfect, then $K=⨝_{i\in I}(H_i\cap K)$.
This is Exercise 5.11(b) of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to Approach0, it is new to MSE.

The Details:

On page 72 . . .

Definition: If $\mathcal{F}=\{H_i\mid i\in I\}$ is a nonempty family of normal subgroups of a group $G$, we write $$H_{(i)}:=\bigvee \{H_j\mid j\in I, i\neq j\}$$ for the join of all members of $\mathcal{F}$ except $H_i$.

[. . .]

We say $\mathcal{F}$ is strongly disjoint if $$H_i\cap H_{(i)}=\{1\}$$ for all $i\in I$.

I use $⨝$ instead of $\times$ for a reason. It means something specific in Roman's book; namely, on page 153,

Definition: A group $G$ is the (internal) direct sum or (internal) direct product of a family $\mathcal{F}=\{H_i\mid i\in I\}$ of normal subgroups if $\mathcal{F}$ is strongly disjoint and $G=\bigvee\mathcal{F}$. We denote the internal direct product of $\mathcal{F}$ by $$⨝H_i\quad \text{ or }\quad ⨝\mathcal{F}.$$

Since definitions vary, here is a

Definition: A subgroup $H$ of $G$ is normal in $G$, written $H\unlhd G$, if $$aH=Ha$$ for all $a\in G$.

This is on page 65.

Finally,

Definition: A group $G$ is perfect if $G=[G, G].$

This is on page 84.

The Question:

Let $G=⨝_{i\in I}H_i$. Suppose $K\unlhd G$ is perfect. Show that $$K=⨝_{i\in I}(H_i\cap K).$$

Thoughts:

The question concerns an equality, not an isomorphism, so the isomorphism theorems don't help here. (My first attempt, essentially, was to apply the second isomorphism theorem, since intersections are at play.)

One thing I'm not sure of is the definition of $\bigvee \mathcal{S}$ (if it is not just the least upper bound of the family $\mathcal{S}$ with respect to the subgroup relation).

I don't see how to use the fact that $K$ is perfect. Do I need to start with

$$⨝_{i\in I}(H_i\cap [K,K])$$

then manipulate this using properties of the derived subgroup of $K$?

I'm not sure where normality comes in either. I know that $$\mathcal{H}\unlhd \mathcal{G}\quad\iff\quad [\mathcal{H},\mathcal{G}]\le\mathcal{H}.$$

Please help :)

答:
Since $G$ is the internal direct product of the subgroups $H_i$, each element $g \in G$ can be written as $\prod_{i \in I} g_i$ with $g_i \in H_i$, and all but finitely many of the $g_i$ trivial.

Then the commutator of two elements $x = \prod x_i$ and $y = \prod y_i$ of $K$ is $[x,y] = \prod [x_i,y_i]$.

Since $K \unlhd G$, we have $[x,y_i] = [x_i,y_i] \in K$ for each $i \in I$.

Now let $k = \prod k_i \in K$. Since $K$ is perfect, the element $k$  is a product $\prod_{j=1}^n [x(j),y(j)]$ of commutators $[x(j),y(j)]$ of elements $x(j),y(j) \in K$,
and then $k_i = \prod _{j=1}^n [x(j)_i,y(j)_i]$ for each $i \in I$.

But we saw above that each commutator $[x(j)_i,y(j)_i] \in K$, and so $k_i \in K$ for all $i \in I$, and the result follows.


问:
Are there infinitely many Mersenne primes?
known facts :

$1.$ There are infinitely many Mersenne numbers : $M_p=2^p-1$

$2.$ Every Mersenne number greater than $7$ is of the form : $6k\cdot p +1$ , where $k$ is an odd number

$3.$ There are infinitely many prime numbers of the form $6n+1$ , where $n$ is an odd number

$4.$ If $p$ is prime number of the form $4k+3$ and if $2p+1$ is prime number then $M_p$ is composite

What else one can include in this list above in order to prove (or disprove) that there are infinitely many Mersenne primes ?

答:
A Mersenne number is a string of binary digits of the form ab, where the lengths of a and b are n and n >= 1. The length of ab is 2n. There are two types of Mersenne number. The first type is a periodic binary number where a=b and all the digits of a and b are 1's eg 11111111. The second type is similar to the first, except that the first bit of a is 0, so a<>b, for example 01111111.

Using the prime number theorem, the total number of Mersenne primes is therefore approximately:

Since the ratio between the successive terms of the series is < 1, D'Alembert's criterion, the series is absolutely convergent. Thus the number of Mersenne primes is finite.

The series converges very slowly, so slowly in fact that for n=512 (the MS Excel floating point limit) only 10 Mersenne primes are approximated, where 14 already exist. This series has not yet been further enumerated.

This work is available in more detail in the appendix of my recent paper

The supplementary materials
contain example MS Excel spreadsheets illustrating my approach.


问:
Prove that exist a matrix $A_{m\times n}$ such that $T(X)=AX$ for any $X\in F^{n}$
I have to prove this:

Let be $F$ a field and $T$ a linear operator from $F^{n}$ to $F^{m}$.
Prove that there exists a matrix $A_{m\times n}(F)$ such that $T(X)=AX$, for
any $X\in F^{n}$

My attempt:

Let be $\mathfrak{B}_{n}=\left \{ e_{1}, \cdots , e_{n} \right \}$ an ordered basis of $F^{n}$, and let be $\mathfrak{B}_{m}=\left \{ e_{1}, \cdots , e_{m} \right \}$  an ordered basis of $F^{m}$.

Any $X\in F^{n}$, can be written as a linear combinations of the vectos of the basis $\mathfrak{B}_{n}$. That is $X=\sum_{i=1}^{n}\lambda_{i}e_{i}$ for some $\lambda_{i}\in F$.

Any $Y\in F^{m}$, can be written as a linear combinations of the vectos of the basis $\mathfrak{B}_{m}$. That is $Y=\sum_{i=1}^{m}\gamma_{i}e_{i}$ for some $\gamma_{i}\in F$.

So, we have this:

\begin{align}
T(X)&=T\left ( \sum_{i=1}^{n} \lambda_{i}e_{i} \right )\\&=\sum_{i=1}^{n} \lambda_{i}T(e_{i})\\&=\sum_{i=1}^{n} \lambda_{i}\sum_{j=1}^{m} \gamma_{j}e_{j}\\&=\sum_{i=1}^{n} \sum_{j=1}^{m}\lambda_{i} \gamma_{j}e_{j}
\end{align}

But I'm stocked here, I'm not sure how to continue. What I want is to show that in that last double sum, is hidden a matrix $A_{m\times n}$.

I would really appreciate your help.

答:
Let $(e_1,...,e_n)$ a basis of $F^n$ and $(f_1,...,f_m)$ a basis of $F^m$. Write $T(e_i)=\sum a_{ij}f_j$.

Let $x\in F^n, x=\sum_i x_ie_i$ implies that $T(x)=T(\sum_i x_ie_i)=\sum x_iT(e_i)=\sum_i x_i\sum_j a_{ij}f_j =\sum_j\sum_i a_{ij}x_if_j$. This implies that the coordinates of $T(x)$ in the basis $(f_1,...,f_m)$ are the line of $AX$ where $A$ where $A$ is the matrix whose $i$-column is $a_{i1},...,a_{im}$ and $X$ the coordinates of $x$.


问:
Infinite order of a quotient of functions' groups
Let $G=\{f:\mathbb N\to \mathbb Z\}$ be a commutative group with the following operation for $f,g\in G$ and $n\in\mathbb N$:
$$(f+g)(n):=f(n)+g(n).$$
I've already showed that the set $H=\{h:\in G:\forall n\in\mathbb N,h(n)\in2\mathbb Z\}$ is a subgroup of $G$.

The second request of the exercise asks me to show that $|G:H|$ is infinite.

This part of the problem seemed to be quite easy but I can't find a consistent way to prove this.

I've showed that $H\trianglelefteq G$, since $\forall n\in N$ the conjugated of a generic $h\in H$ with an element of $f\in G$ is such that
$$(\bar f+h+f)(n)=(-f+h+f)(n)=-f(n)+h(n)+f(n)\in H.$$
I was trying to find a surjective homomorphism $\varphi: G\to X$ such that $H$ is the Kernel of this homomorphism in order to apply the first isomorphism theorem with $G/H\cong\varphi (G)\cong X$, where X is an infinite group, but I can't find it. Using the corespondence theorem on $G/H$,supposing that $|G:H|$ is finite, also seems to be inconclusive.

I really thank you for the support.

答: Following your idea, we can consider the map $\varphi \colon G \rightarrow (\mathbb{Z}/2 \mathbb{Z})^{\mathbb{N}}$ defined by $$\varphi(f) = \left( f(n) \bmod{2} \right)_{n \in \mathbb{N}} \, \text{.}$$ The map $\varphi$ is a surjective group homomorphism whose kernel is $H$, so it induces an isomorphism from $G/H$ to the group $(\mathbb{Z}/2 \mathbb{Z})^{\mathbb{N}}$, which is infinite.


问:
What is a more rigorous method of solving modulus function inequalities?
I have the inequality

$$\vert x-3\vert < 3x-4$$

To solve this I first break the modulus containing function into the two cases using the critical value

$$\vert x-3 \vert  \left\{\begin{array}{cc}
x-3 & x > 3 \\
3-x & x<3
\end{array}\right.  $$

For the case where $x>3$:

$$x-3<3x-4$$
$$x>\frac{1}{2}$$

For the case where $x<3$:

$$3-x<3x-4$$
$$x>\frac{7}{4}$$
However it is apparent that these two statements are contradictory and by simply looking at the graph of the two functions it is clear that the value $x>\frac{1}{2}$ does not hold.

The value does however seem to hold for the function $f(x)=x-3$

My question is this, why has my method given a false answer and what improvements can be made to this method to make it such that these false answers no longer appear?

答:
Don't forget about the words 'either', 'or', and 'and',

$$\vert x-3 \vert  =\left\{\begin{array}{cc}
x-3 & x > 3 \\
3-x & x<3
\end{array}\right.$$

So we have that

Either

$$ x>3  \text{ and } x-3>3x-4$$

or

$$ x<3 \text{ and } -x+3>3x-4 $$

The first option resolves into

$$ x>3 \text{ and } x<\frac{1}{2} $$

which is impossible. But the second option is

$$ x<3 \text{ and } x<\frac{7}{4} $$

which simplifies to the correct solution: $x<\frac{7}{4}$.


问:
Prove König's theorem using Dilworth's theorem
I am trying to derive König's theorem from Dilworth's theorem, but it seems like I'm stuck. I know that I have to define some kind of binary relation on the set of a bipartite graph's vertices, then use Dilworth's theorem. However, every relation I've come up with so far either did not satisfy some properties needed for Dilworth's theorem or seemed to be completely useless. Can anyone point me in the right direction?

Thanks.

答:
First, let us recall the following facts:

In any graph, a subset of vertices is a vertex cover iff its complement is an independent set.

In any graph, the size of a matching is a lower bound for the minimal size of a vertex cover.

Now, let $G = (V, E)$ be a bipartite graph, i.e., we have a disjoint union $V = V_1 \cup V_2$ such $V_1$ and $V_2$ are independent sets. We want to show that the maximum size of a matching in $G$ equals the minimum size of a vertex cover in $G$ using Dilworth's theorem.

We define a partial order on $V$ by setting $u \le v$ iff $u \in V_1$ and $\{u, v\} \in E$ (which implies $v_2 \in V_2$). By construction, as $G$ is bipartite, any chain has length at most two, i.e., it is either a single edge or a single vertex.
Hence,
\begin{align*}
 |V| & = \mbox{# of single element chains} + 2 \cdot \mbox{# of two element chains} \\
     & = \mbox{size of the chain partition} + \mbox{# of two element chains}.
\end{align*}
If we only take those chains containing precisely two vertices, i.e., the edges, they form a matching. So, with the previous equation,
$$
 |V| - \mbox{size of the chain partition} \le \mbox{maximum size matching}.
$$
By the first fact mentioned above
and as an antichain is precisely an independent set in $G$,
$$
 \mbox{minimum size vertex cover} \le |V| - \mbox{maximum size of an antichain}.
$$
By Dilworth's theorem,
$$
 |V| - \mbox{maximum size of an antichain} = |V| - \mbox{size of the chain partition}.
$$
So, we find,
$$
 \mbox{minimum size vertex cover} \le \mbox{maximum matching}.
$$
By the above mentioned second fact, we have equality. QED


问:
Counter example in topology
We known that every subspace of second countable space is second countable.  I was think if A is subspace of X and A is second countable space need to be the whole space is second countable space?  My statements is it true? Given an example plz</p>

答:
Suppose that whenever $A$ is a second countable subspace of a space $X$, then $X$ is also second countable. This proposition is logically equivalent to its contrapositive: if a space $X$ is not second countable, and $A\subseteq X$, then $A$ is not second countable. But this statement is clearly false, because every finite space is second countable, and there are spaces that are not second countable.

For instance, if $X$ is an uncountable space with the discrete topology, then $X$ is not second countable, but every countable subset of $X$ is second countable.

Title: Counter example in topology


问:
Need help with substitution for integration
I'm trying to go through the solution for a SIS epidemic model, but I'm stuck figuring out the following steps.

So we have

$i′=r(k−i)i−αi$

And to solve this, they did a substitution $y=i^{−1}$. They integrated and came up with

$ye^{(rk−α)t}=∫re^{(rk−α)t}dt+C$

I have no idea how they ended up with this equation. So taking substitution $y = i^{-1}$, we get $\frac{dy}{dt} = -\frac{1}{i^2}$, but i don't know how to use that for the first equation.

Did they use separation of variables to solve this after the substitution? Anyone who could show me how they got to the second equation would be appreciated.

Thank you!

答:
From $
-\mathrm{y}^{\prime}=(\mathrm{rk}-\alpha) \mathrm{y}-\mathrm{r}
$, we get $\mathrm{y}^{\prime}=-(\mathrm{rk}-\alpha) \mathrm{y}+\mathrm{r}$.

Thus we obtain the following linear first - order differential equation
$$ \mathrm{y}^{\prime}+(\mathrm{rk}-\alpha) \mathrm{y}=\mathrm{r}.$$
We get that the integrating factor is
$$
\mu(\mathrm{t})=\mathrm{e}^{\int(\mathrm{rk}-\alpha) \mathrm{dt}}=\mathrm{e}^{(\mathrm{rk}-\alpha) \mathrm{t}}.
$$

Multiplying the differential equation by the integrating factor $\mu(\mathrm{t})$, we obtain
$$\mathrm{e}^{(\mathrm{rk}-\alpha) \mathrm{t}} \mathrm{y}^{\prime}+\mathrm{e}^{(\mathrm{rk}-\alpha) \mathrm{t}}(\mathrm{r} \mathrm{k}-\alpha) \mathrm{y}=\mathrm{re}^{(\mathrm{rk}-\alpha) \mathrm{t}},$$

which we can write as $$\left(e^{(\mathrm{rk}-\alpha) \mathrm{t}} \mathrm{y}(\mathrm{t})\right)^{\prime}=\mathrm{re}^{(\mathrm{rk}-\alpha) \mathrm{t}}.$$

Integrating both sides with respect to $t$,

$$
y e^{(r k-\alpha) t} = \int r e^{(r k-\alpha) t} d t+C
 $$


问:
Simple Derivative $\frac{x}{\ln x}$
Okay,
I have find the derivative of $$\frac{x}{\ln x}$$ so I used the quotient rule and got: $$\frac{\ln x(1)-x(\frac{1}{x})}{(\ln x)^2}$$

I then simplified this to: $$\frac{-x}{\ln x}.$$

Now here come the problem, when I put $e$ into $f'(x)$ I am supposed to get $-1.5271$ and I do not and I cannot figure out why.  Could someone point out what I am doing wrong?

Thanks

答:
The answer you were given is wrong.

$f'(e) = 0$.

Plot it and see.

Title: Simple Derivative $\frac{x}{\ln x}$


答:
When in doubt, element chase.  $M = N$ will be true if $x \in M \iff x \in N$.

If $x \in A\cap (B\cup C)$ then $x\in A$ and $x \in B\cup C$.  So $x \in B$ or $x \in C$.  So we have two possibilities i) either $x \in A$ and $x \in B$ so $x \in A\cap B$.  or ii) $x \in A$ and $x \in C$ so $x \in A\cap C$.

So either $x \in A\cap B$ or $x \in A\cap C$.  So $x \in (A\cap B)\cup (A\cap C)$.

So $x \in A\cap (B\cup C) \implies x \in (A\cap B)\cup (A\cap C)$.

Now if $x \in (A\cap B) \cup (A\cap C)$ then we have to options i) $x \in A\cap B$ so $x \in A$.  Or ii) $x \in A\cap C$ so $x \in A$. In either case we have $x \in  A$ so $x\in A$ is a certainty.

Now we either have i)$x\in A\cap B$ so $x \in B$ or we have ii) $x \in A\cap C$ so $x \in C$.  So we have either $x \in B$ or $x \in C$.  So $x \in B\cup C$.

So we have $x \in A$ and $x \in B\cup C$ so $x \in A\cap (B\cup C)$.

So $x\in (A\cap B)\cup (A\cap C) \implies  x \in A\cap (B\cup C)$

so $x \in A\cap (B\cup C) \iff x \in (A\cap B)\cup (A\cap C)$

So $A\cap (B\cup C)= (A\cap B)\cup (A\cap C)$.

Title: Proving $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

Body:
Let $A$, $B$ and $C$ be sets. Prove that $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

I am not sure how to formally prove and in special write this intuitive simple question about sets.

Thank you


问:
Quick Volume with Vertical Lines Question
So I was given the following prompt with the graph:

"Let $M$ be the region in the first quadrant bounded by the graphs of $y=\arctan(x)$ and $y=\frac{\pi}{4}x$, as shown in the figure below. What is the volume of the solid generated when region $M$ is revolved around the vertical line $x=2$?"

So I guess I'm just a bit confused about what my volume equation might look like here. I understand that the equations here would both have to be expressed in terms of $y$, but I'm a bit lost about where to go after that. Any help would be appreciated!

答:
You have to find volume of the area bound between curves $C1$ and $C2$ when rotated around $x = 2$

$C1: y=\frac{\pi}{4} x$ and $C2: y=\arctan x$ $(0 \leq x \leq 1, 0 \leq y \leq \frac{\pi}{4}$)

When you rotate it around $x = 2$, please note the radius of the rotation of both curves at any given $y$.

For $C1$, $r = 2 - \frac{4y}{\pi}$

For $C2$, $r = 2 - \tan y$

So your triple integral to find volume should be $\displaystyle \int_0^{2 \pi} \int_0^{\pi/4} \int_{2 - \frac{4y}{\pi}}^{2 - \tan y} r \ dr \ dy \ d\theta$


问:
Definable types over stably embedded sets
Let $T$ be a complete $\mathcal{L}$-theory and $\mathfrak{U}$ a monster model, with $\theta(v)$ an $\mathcal{L}$-formula and $M\preccurlyeq \mathfrak{U}$ a (not necessarily small) model. We say that $\theta(M)$ is stably embedded in $M$ if, for every $\mathcal{L}_M$-formula $\varphi(v_1,\dots,v_n,\overline{m})$, the formula $\varphi(v_1,\dots,v_n,\overline{m})\wedge\bigwedge_{i=1}^n\theta(v_i)$ is equivalent to some $\mathcal{L}_{\theta(M)}$ formula. The following exercise is from Tent and Ziegler:

Exercise 8.3.4: Show that $\theta(M)$ is stably embedded in $M$ if and only if, for all $n$, every $p\in S_n(\theta(M))$ which contains $\bigwedge_{i=1}^n\theta(v_i)$ has a unique extension $\overline{p}\in S_n(M)$. If $\theta$ is absolutely stably embedded and $p$ is definable, show that $\overline{p}$ is definable over $\theta(M)$.

I've done the first part of the problem, for which one direction is clear; indeed, if $p_1,p_2\in S_n(M)$ are distinct extensions of $p$, let $\varphi(\overline{v},\overline{m})$ be a separating formula, with $\varphi\in p_1$ and $\neg\varphi\in p_2$. Since $\bigwedge_{i=1}^n\theta(v_i)\in p_1,p_2$, letting $\varphi'=\varphi(\overline{v},\overline{m})\wedge\bigwedge_{i=1}^n\theta(v_i)$ still gives $\varphi'\in p_1$ and $\neg\varphi'\in p_2$. If $\theta(M)$ were stably embedded, then $\varphi'$ would be equivalent to an $\mathcal{L}_{\theta(M)}$-formula, and then either that formula or its negation would lie in $p$, which would make one of $p_1$ or $p_2$ inconsistent.

The other direction seems to require a bit more work; suppose any type in $S_n(\theta(M))$ containing $\bigwedge_{i=1}^n\theta(v_i)$ has a unique extension to $S_n(M)$, and let $\varphi(v_1,\dots,v_n,\overline{m})$ be any $\mathcal{L}_M$-formula such that $\varphi\to\bigwedge_{i=1}^n\theta(v_i)$. We wish to show that $\varphi$ is equivalent to some $\mathcal{L}_{\theta(M)}$ formula. To see this, enumerate the stone space $S_n(\theta(M))$ as $(p_\alpha)_{\alpha\in\lambda}$ for some cardinal $\lambda$, and define
\begin{align}
A&=\{\alpha\in\lambda:p_\alpha\cup\{\varphi(\overline{v},\overline{m})\}\text{ is consistent}\} \\
B&=\{\beta\in\lambda:p_\beta\cup\{\neg\varphi(\overline{v},\overline{m})\}\text{ is consistent}\}.
\end{align}
We claim $S_n(\theta(M))=A\sqcup B$. That $S_n(\theta(M))=A\cup B$ is clear, and disjointness follows from the hypothesis on types. Indeed, if $p_\alpha$ is consistent with $\varphi(\overline{v},\overline{m})$, then it must contain $\bigwedge_{i=1}^n\theta(v_i)$, and so by the hypothesis has a unique extension to $S_n(M)$. This extension must contain $\varphi(\overline{v},\overline{m})$, and so $p_\alpha$ is inconsistent with $\neg\varphi(\overline{v},\overline{m})$, whence $\alpha\notin B$, as desired. Thus, for any $\alpha\in A$ and $\beta\in B$, we can choose $\psi_{\alpha\beta}$ with $\psi_{\alpha\beta}\in p_\alpha$ and $\neg\psi_{\alpha\beta}\in p_\beta$; note in particular that $\psi_{\alpha\beta}$ is an $\mathcal{L}_{\theta(M)}$-formula.

Now, fix $\alpha\in A$. Since any type over $M$ containing $\neg\varphi$ must contain $p_\beta$ for some $\beta\in B$, we have $\{\psi_{\alpha\beta}\}_{\beta\in B}\models \varphi$, and thus there are $\beta_1,\dots,\beta_k$ such that $\{\psi_{\alpha\beta_i}\}_{i=1}^k\models\varphi$. Let $\eta_\alpha=\bigwedge_{i=1}^k\psi_{\alpha\beta_i}$, so that $\eta_\alpha\to\varphi$ and $\eta_\alpha\in p_\alpha$. Since every type over $M$ containing $\varphi$ must be $p_\alpha$ for some $\alpha\in A$, this latter condition means $\{\neg\eta_\alpha\}_{\alpha\in A}\models \neg\varphi$, and thus there are $\alpha_1,\dots,\alpha_k$ with $\{\neg\eta_{\alpha_i}\}_{i=1}^k\models\neg\varphi$. Then, letting $\eta=\bigvee_{i=1}^k\eta_{\alpha_i}$, we have $\varphi\leftrightarrow\eta$, and by construction $\eta$ is an $\mathcal{L}_{\theta(M)}$-formula, as desired.

So this wraps up the first part. Unfortunately, I'm struggling a bit with the second part. For one thing, I'm not actually sure what T&Z mean by "absolutely" stably embedded; this isn't a term they've defined. I assume this result doesn't hold in the general case, or else they wouldn't add that qualifier, but for now I've only been trying to approach it in general. Perhaps it will be clearer once I know the right hypotheses?

Anyway, let $\varphi(\overline{v},\overline{w})$ be an $\mathcal{L}$-formula; we may assume $\varphi(\overline{v},\overline{w})\to\bigwedge_{i=1}^n\theta(v_i)$. For any $\overline{m}\in M$, we then know that $\varphi(\overline{v},\overline{m})$ is equivalent to some $\theta(M)$-formula $\psi_\overline{m}(\overline{v},\overline{a}_\overline{m})$, and so $\varphi(\overline{v},\overline{m})\in\overline{p}$ if and only if $\psi_\overline{m}(\overline{v},\overline{a}_\overline{m})\in p$. Thus, if there were some way to choose all the $\psi_\overline{m}(\overline{v},\overline{u})$ to be equal, we would be done, since we would have $$\varphi(\overline{v},\overline{m})\in \overline{p}\iff\mathfrak{U}\models\exists \overline{u}\left[\bigwedge_{i=1}^k \theta(u_i)\wedge\left[\forall\overline{v}\varphi(\overline{v},\overline{m})\leftrightarrow\psi(\overline{v},\overline{u})\right]\wedge d_p\overline{v}\psi(\overline{v},\overline{u})\right],$$ where $\psi=\psi_\overline{m}$ for all $\overline{m}$ and $d_p\overline{v}$ denotes the definition of $p$ over $\theta(M)$. However, I'm not so sure how we might go about removing the dependence on $\overline{m}$, and presumably we can't in general; does the "absolute" hypothesis somehow make this possible? Is this even a good approach, or does it not work?

答:
You've correctly identified the sticking point in your suggested proof: Every formula $\varphi(v,m)$ with $m\in M$ is equivalent to some formula $\psi_m(v,n)$ with $n\in \theta(M)$, but which formula $\psi_m$ works apparently depends on $\varphi$ and $m$, and we would like it to depend just on $\varphi$.

One of the most important tricks in the model theory toolbox is using the compactness theorem to get "uniformity for free". We often know something of the form "In every model $M$, for every definable set $X$, there exists a definable set $Y$ such that..." Then we can use the compactness theorem to show that when we restrict to those $X$ which are defined by instances of a formula $\varphi$, there must be finitely many formulas $\psi_1,\dots,\psi_k$ such that the witnessing $Y$ is defined by an instance of one of the $\psi_i$. And finally, we can code the finitely many $\psi_i$ into a single formula and conclude "For every formula $\varphi$ there is a formula $\psi$ such that for every instance $X$ of $\varphi$ there is an instance $Y$ of $\psi$ such that..."

At this point, you might like to use the above as a hint and try to solve the exercise yourself. But if you get stuck, here are the details.

Suppose $\theta(M)$ is stably embedded in $M$ for every $M\models T$. To simplify notation, I'll write all tuples of variables as if they were singletons. Fix a formula $\varphi(v,w)$ which implies $\theta(v)$. Consider the following partial type in the variables $w$: $$T\cup \{\lnot \exists z\,(\theta(z)\land \forall v\, (\varphi(v,w)\leftrightarrow \psi(v,z)))\mid \psi(v,z)\in \mathcal{L}\}.$$
This partial type is inconsistent (this is where we need to use stable embeddedness in every small model, or equivalently in the monster model). So by compactness there are finitely many formulas $\psi_1(v,z_1),\dots,\psi_k(v,z_k)\in \mathcal{L}$ such that $$T\models \forall w\, \bigvee_{i=1}^k \exists z_i\,(\theta(z_i)\land \forall v\, (\varphi(v,w)\leftrightarrow \psi_i(v,z_i))).$$
Now by the standard trick of coding finitely many formulas by a single formula (which I described, for example, here), we may assume there is a single formula $\psi(v,z)$ such that for every model $M$ and every instance of $\varphi(v,w)$ with parameters from $M$, there is an equivalent instance of $\psi(v,z)$ with parameters from $\theta(M)$.

[Technical note: To do this coding, we need to assume that $\theta(M)$ has as least two elements. But if $\theta(M)$ has at most one element, then any complete type which implies $\theta(v)$ is realized in $M$, and hence definable.]

Finally, we can conclude as you did, writing down an explicit definition over $\theta(M)$ for any type containing $\theta(v)$ which is definable over $M$.


问:
How is this property equivalent to the Reiter Property?
We have the Reiter Property $(R_2)$ for an action of a group G on a set X:
For any $\epsilon>0$, any finite subset $S$ of G, there exists $\phi\in{\ell^2(X)}$ such that $\|s\phi-\phi\|_{\ell^2}<\epsilon{\|\phi\|_{\ell^2}}$ for all $s\in{S}$.
I am trying to show this is equivalent to the alternative property $(R_2)'$: for any $\epsilon>0$, any finite subset $S$ of G, there exists $\phi\in{\ell^2(X)}$ such that $$\left\|\frac{1}{|S|}\sum_{s\in{S}}{s\phi}\right\|_{\ell^2}>(1-\epsilon)\|\phi\|_{\ell^2}$$
but I am completely stuck.
I have tried using some uniform convexity since $\ell^2$ has an inner product, but can only get anything out of it when $|S|=2$, I have heard from someone else that this is related to adjoint operators, so I have tried defining $T:\ell^2(X)\rightarrow\ell^2(X)$ by $T(\phi)=\frac{1}{|S|}\sum_{s\in{S}}{s\phi}$ and can deduce that it has norm 1 and, if we extend S to also contain the inverses of all its elements, is self adjoint, but I can't see how this could be helpful to solve the problem.

Many thanks.

答:
It is enough to prove the equivalence for finite sets $S$ such that $S = S^{-1}$ and $1 \in S$. Given a such a set, let $T := \frac{1}{|S|} \sum_{s \in S} s \phi$. As you point out, this is a self-adjoint operator of norm at most $1$, and $(R_2)'$ is equivalent to: $\| T \| = 1$ for all $S$.

Suppose $(R_2)$. Then for all $S$ there exists a sequence $\phi_i$ such that $\| \phi_i \| = 1$ and $\| s \phi_i - \phi_i \| \to 0$ for all $s \in S$. It follows that $\| T \phi_i - \phi_i \| \to 0$. Therefore $\| T \phi_i \| \geq \| \phi_i \| - \| T \phi_i - \phi_i \| \to 1$, and $(R_2)'$ follows.

Suppose $(R_2)'$. Since $T$ is self-adjoint and has norm $1$, we have $1 = \| T \| = \sup_{\| \phi \| = 1} | \langle T \phi, \phi \rangle |$. So there exists a sequence $\phi_i$ with $\| \phi_i \| = 1$ and $|\langle T \phi_i, \phi_i \rangle | \to 1$. Developing and using the Cauchy-Schwarz inequality:

$$|\langle T \phi_i, \phi_i \rangle| \leq \frac{1}{|S|}\sum_{s \in S} |\langle s \phi_i, \phi_i \rangle| \leq \frac{1}{|S|} \sum_{s \in S} \| s \phi_i \| \cdot \|\phi_i \| = 1.$$

Since the left-hand-side converges to $1$, this implies that $|\langle s \phi_i, \phi_i \rangle| \to 1$ for all $s \in S$.

The idea now is to use an approximate version of the equality case from the Cauchy-Schwarz inequality. More precisely, (see e.g. Proof 2 of Cauchy-Schwarz from wikipedia), if $u$ and $v$ have norm $1$, then we have: $|\langle u, v \rangle^2 - 1| = \| u - \langle u, v \rangle v \|^2$. So in our case taking $u = s \phi_i$ and $v = \phi_i$, and letting $\lambda_s = \lim_{i \to \infty} \langle s \phi_i, \phi_i \rangle = \pm 1$ we obtain
$$\lim_{i \to \infty} \| s \phi_i - \lambda_s \phi_i \| = \lim_{i \to \infty} \| s \phi_i - \langle s \phi_i, \phi_i \rangle \phi_i \| = 0.$$
We are left to take care of the sign. First, note that $\lim_{i \to \infty} \langle T \phi_i, \phi_i \rangle = \frac{1}{|S|} \lambda_s$, and since the left-hand side is $\pm 1$ this shows that all of the $\lambda_s$ are equal. But clearly $\lambda_1 = 1$ (recall that we are assuming $1 \in S$) and so $\lambda_s = 1$ for all $s \in S$.


问:
Convergence in cdf and pdf
This question will probably be trivial to most. However I failed to find a clear answer on the internet.

When one says that n random variables converge "in distribution", does it mean that this random variables converge to the same cumulative distributive function (cdf) ? What does it say about their convergence in probability density function (pdf) ? Can two random variables have the same cdf but different pdf or the other way around ?

答:
Given a sequence of random variables, say $(X_n)_{n \in \mathbb{N}}$. We will say this sequence converges in distribution to the random variable $X$ if we take the sequence $(F_n(x))_{n \in \mathbb{N}}$ where $F_i(x)$ is the CDF of $X_i$ and have the following result

$\displaystyle \lim_{n \to \infty } F_n(x) = F(x)$ where $F(x) $ is the CDF of $X$.

Note: If we only have $m$ many random variables then we can form a sequence by taking $X_1,X_2, \ldots X_m$ and for each successive term after $X_m$, set $X_{m+j} = X_i$ where $j \in \mathbb{N}$ and $X_i \in \{ X_1, \ldots X_m \}$ and such a sequence trivially converges in distribution to $X_i$

If we have that $f_i(x)$ is the density of $X_i, \forall i$. That is if we have that the density exists for each random variable in our sequence then from Scheffé’s theorem we get that the convergence of the densities will imply convergence in distribution.

For the last question, say we have two continuous random variables of which both have a density function. Namely, say we have $X \sim F_X(z) $ and $Y \sim F_Y(z)$  then the denisties will be, respectively, $F'_X(z) = f_X(z)$ and $F'_Y(z) = f_Y(z)$.

Hence if $F_X(z) = F_Y(z)$ then $f_X(z) = f_Y(z)$. We can see the converse will hold in this case as well.

Title: Convergence in cdf and pdf


问:
Expansion of a modified Struve function in terms of modified Bessel functions
The Wolfram functions site includes (without reference) the identity (source)

$$L_0(z)=\frac{4}{\pi}\sum_{k=0}^\infty \frac{I_{2k+1}(z)}{2k+1}$$
where $L_0(z)$ is the modified Struve function of order zero and $I_n(z)$ is the $n$th modified Bessel function of the second kind. But from my work in another question, I think this is incorrect as written and should include $(-1)^k$, i.e., it should be alternating. Can someone provide a reference to either result or a direct proof of such?

答:
By http://dlmf.nist.gov/11.2.E2, http://dlmf.nist.gov/11.4.E21 and http://dlmf.nist.gov/10.27.E6, we have
$$
\mathbf{L}_0 (z) =  - \mathrm{i}\mathbf{H}_0 (\mathrm{i}z) =  - \mathrm{i}\frac{4}{\pi }\sum\limits_{k = 0}^\infty  {\frac{{J_{2k + 1} (\mathrm{i}z)}}{{2k + 1}}}  = \frac{4}{\pi }\sum\limits_{k = 0}^\infty  {( - 1)^k \frac{{I_{2k + 1} (z)}}{{2k + 1}}} .
$$


问:
Finding the gradient of $r = \sqrt{x^2+y^2+z^2}$ .
In David J. Griffiths' Introduction to Electrodynamics - Example $1.3$, the author solves the problem of

Find the gradient of $r = \sqrt{x^2+y^2+z^2}$ . (As in the magnitude of the position vector).

He does
$$
\nabla r = \frac{\delta r}{\delta x} \hat{x}+\frac{\delta r}{\delta y} \hat{y}+\frac{\delta r}{\delta z} \hat{z}
\tag{1}
$$
nothing wrong there, but then he does the following
$$
\nabla r = \frac{1}{2}\frac{2x}{\sqrt{x^2+y^2+z^2}} \hat{x}+\frac{1}{2}\frac{2y}{\sqrt{x^2+y^2+z^2}} \hat{y}+\frac{1}{2}\frac{2z}{\sqrt{x^2+y^2+z^2}} \hat{z}
\tag{2}
$$
which I am confused as to how he got to, I can't seem to understand/find if he just manipulated any of the values: I don't understand why he makes $\delta x$ into $2{\sqrt{x^2+y^2+z^2}}$ for all dimension variables. Can someone elaborate? Thank you.

答:
To elaborate what was already spoke in @leoli's comment, if $g(x) = x^{1/2}$ then $g'(x) = x^{-1/2}/2$, so if $h(x) = \sqrt{f(x)} = g(f(x))$, then by chain rule,
$$
h'(x)
 = g'(f(x))f'(x)
 = \frac{f'(x)}{2\sqrt{f(x)}},
$$
as you are seeing in all your examples.


问:
Differential or Derivative of vector and vector transpose product
I have tried to get $\frac{\partial (ab^T)}{\partial a}$ and $\frac{\partial (ab^T)}{\partial b}$ or $d(ab^T)$, where $a,b \in \mathbb{R}^{n \times 1}$.

But all I searched on the Internet are $\frac{\partial (a^Tb)}{\partial a}$ and $\frac{\partial (a^Tb)}{\partial b}$.

So how to solve $\frac{\partial (ab^T)}{\partial a}$ and $\frac{\partial (ab^T)}{\partial b}$ ?

Thanks.

答:
$\def\R#1{\mathbb R^{#1}}\def\p#1#2{\frac{\partial #1}{\partial #2}}$The gradient of a vector (or transpose vector) with respect to its own components is
$$\p{a}{a_k}=e_k\qquad\quad\p{b^T}{b_k}=e_k^T$$
where $e_k$ is the $k^{th}$ cartesian basis vector for $\R{n}$

Use the above to compute component-wise gradients
$$\eqalign{
\p{(ab^T)}{a_k} &= e_kb^T\qquad\quad
\p{(ab^T)}{b_k} &= ae_k^T
}$$
Note that both $e_kb^T{\;\rm and\;}ae_k^T\;$ are themselves $\,\R{n\times n}$ matrices, hence the full gradients are third-order tensors which do not lend themselves to standard matrix/vector notation.


问:
What's wrong with this equal probability solution for Monty Hall Problem?
I'm confused about why we should change door in the Monty Hall Problem, when thinking from a different perspective  gives me equal probability.

Think about this first: if we have two doors, and one car behind one of them, then we have a 50/50 chance of choosing the right door.

Back to Monty Hall: after we pick a door, one door is opened and shows a goat, and the other door remains closed. Let's call the door we picked A and the  other closed door B. Now since 1 door has already been opened, our knowledge has changed such that the car can only be behind A or B. Therefore, the problem is equivalent to: given two closed doors (A and B) and one car, which door should be chosen (we know it's a 50/50 thing)?

Then, not switching door = choosing A, and switching door = choosing B. Therefore, it seems that switching should be equally likely, instead of more likely.

Another way to think: no matter which door we choose from the three, we know BEFOREHAND that we can definitely open a door with a goat in the remaining two. Therefore, showing an open door with a goat reveals nothing new about which door has the car.

What's wrong with this thinking process? (Note that I know the argument why switching gives advantage, and I know experiments have been done to prove that. My question is why the above thinking, which seems legit, is actually wrong.) Thanks.

答:
You can better see it imagining you had a box with $100$ balls, from which $99$ are black and only $1$ is white, which is what you want. You grab one randomly and keep it in your hand without seeing it. Do you agree that you are $99/100$ likely to have picked a black ball, so in $99$ out of $100$ attempts (on average) the white ball would still be in the box?

In case you agree, let's keep that ball in your hand and now suppose that another person deliberately pulls out $98$ black balls from the box. With "deliberately" I mean that that person sees what he is pulling; there is no risk that he removes the white ball by accident. In this way, there are only two balls remaining, one in your hand and one in the box, and one of them is necessarily the white.

What do you think is the probability that the white ball is which is in the box? If you say $50$%, what happened with the $99$ out of $100$ attempts in which it was still in the box? The revelation of the $98$ black ones didn't move it from the box to your hand.

Before the revelation of the $98$ black balls, the cases are:

```<code>                                        Hand       ||          Box
                                     =============================================
1) In 99 out of 100 attempts ->        1 black     ||    98 black ones and 1 white
2) In 1 out of 100 attempts ->         1 white     ||        99 black ones
</code>```

So, when the other person removes the $98$ black balls from the box:

```<code>                                        Hand       ||          Box
                                     =============================================
1) In 99 out of 100 attempts ->        1 black     ||         1 white
2) In 1 out of 100 attempts ->         1 white     ||         1 black
</code>```

So, it is true that you always end with two balls, one white and one black, but the important thing is that they are in two different positions (hand or box), and those two positions depend on the first selection. Moreover, that first selection determines that the white ball will end more frequently in the "box" position than in the "hand" position.

The way you are thinking the Monty Hall problem is like since you are always going to end with two balls, it would be the same if you started with both in the box and you had to grab one. But it is not the same. One thing is the probability to get the correct one when you randomly pick from two, and another different thing is the probability that the correct is already set in one position or in the other.

Note that if you randomly decide if you will pick the ball in the box or the ball in your hand, like flipping a coin, then you will get the white $50$% of the time. But that does not mean that it is $50$% of the time in the hand and $50$% in the box. It is because the extra times that you guess right picking the one from the box are compensated with the extra times you guess wrong picking the one from the hand. The $50$% $= 1/2$ is the average of the two cases:

$$1/2 * 99/100 + 1/2 * 1/100$$
$$= 1/2 * (99/100 + 1/100)$$
$$= 1/2$$

But if you always pick the ball that is in the box, your chances are:

$$1 * 99/100 + 0 * 1/100$$
$$= 1 * 99/100$$
$$= 99/100$$

In Monty Hall it occurs the same. Since there are two incorrect doors and a correct one, it is like if the $3$ doors were $3$ balls in the box, $2$ blacks and $1$ white. The initial selection is like when you start grabbing one ball randomly, and after the revelation the switching door is like the other ball that was left in the box. In $2$ out of $3$ attempts you pick a wrong door (like you would pick a black ball $2$ out of $3$ times) so in $2$ out of $3$ attempts the correct one will be the other the host leaves closed.

Body:
You can better see it imagining you had a box with $100$ balls, from which $99$ are black and only $1$ is white, which is what you want. You grab one randomly and keep it in your hand without seeing it. Do you agree that you are $99/100$ likely to have picked a black ball, so in $99$ out of $100$ attempts (on average) the white ball would still be in the box?

In case you agree, let's keep that ball in your hand and now suppose that another person deliberately pulls out $98$ black balls from the box. With "deliberately" I mean that that person sees what he is pulling; there is no risk that he removes the white ball by accident. In this way, there are only two balls remaining, one in your hand and one in the box, and one of them is necessarily the white.

What do you think is the probability that the white ball is which is in the box? If you say $50$%, what happened with the $99$ out of $100$ attempts in which it was still in the box? The revelation of the $98$ black ones didn't move it from the box to your hand.

Before the revelation of the $98$ black balls, the cases are:

                                        Hand       ||          Box
                                     =============================================
1) In 99 out of 100 attempts ->        1 black     ||    98 black ones and 1 white
2) In 1 out of 100 attempts ->         1 white     ||        99 black ones

So, when the other person removes the $98$ black balls from the box:

                                        Hand       ||          Box
                                     =============================================
1) In 99 out of 100 attempts ->        1 black     ||         1 white
2) In 1 out of 100 attempts ->         1 white     ||         1 black

So, it is true that you always end with two balls, one white and one black, but the important thing is that they are in two different positions (hand or box), and those two positions depend on the first selection. Moreover, that first selection determines that the white ball will end more frequently in the "box" position than in the "hand" position.

The way you are thinking the Monty Hall problem is like since you are always going to end with two balls, it would be the same if you started with both in the box and you had to grab one. But it is not the same. One thing is the probability to get the correct one when you randomly pick from two, and another different thing is the probability that the correct is already set in one position or in the other.

Note that if you randomly decide if you will pick the ball in the box or the ball in your hand, like flipping a coin, then you will get the white $50$% of the time. But that does not mean that it is $50$% of the time in the hand and $50$% in the box. It is because the extra times that you guess right picking the one from the box are compensated with the extra times you guess wrong picking the one from the hand. The $50$% $= 1/2$ is the average of the two cases:

$$1/2 * 99/100 + 1/2 * 1/100$$
$$= 1/2 * (99/100 + 1/100)$$
$$= 1/2$$

But if you always pick the ball that is in the box, your chances are:

$$1 * 99/100 + 0 * 1/100$$
$$= 1 * 99/100$$
$$= 99/100$$

In Monty Hall it occurs the same. Since there are two incorrect doors and a correct one, it is like if the $3$ doors were $3$ balls in the box, $2$ blacks and $1$ white. The initial selection is like when you start grabbing one ball randomly, and after the revelation the switching door is like the other ball that was left in the box. In $2$ out of $3$ attempts you pick a wrong door (like you would pick a black ball $2$ out of $3$ times) so in $2$ out of $3$ attempts the correct one will be the other the host leaves closed.


问:
Can a generator of the ring of integers of local fields can be chosen so that it is also a uniformizer at the same time?
Let $L/K$ be an extension of local fields. We can find $\alpha$ such that $\mathcal{O}_L=\mathcal{O}_K[\alpha]$. What do we know about this generating element? I think that this $\alpha$ can be selected in such a way that, in addition to the above property, it is also a uniformizing parameter at the same time. But I can not prove it.

答:
Hint: If $\mathcal{O}_L = \mathcal{O}_K[\alpha]$ with $\alpha$ a uniformiser of $\mathcal{O}_L$, then the canonical inclusion of their residue fields $\mathcal{O}_K/(\pi_K) \hookrightarrow \mathcal{O}_L/(\alpha)$  is surjective.

Conversely, if the extension is totally ramified (which is what the above says), then ...

reuns' answer says the same and more.


问:
How to show that $\mathbb{R}$ in the language of rings does not admit quantifier elimination?
I am studying for an exam in an introductory mathematical logic course. Suppose we are working in the language of rings $\mathcal{L}=(+,-,\cdot,0,1)$. Then $\mathbb{R}$ does not admit quantifier elimination.

First of all I am a bit unsure of what it means for a structure to admit quantifier elimination, but I assume that $\mathbb{R}$ is shorthand for $\operatorname{Th}(\mathbb{R})=\{\phi \text{ an } \mathcal{L}\text{-sentence}\mid \mathbb{R}\models \phi\}$, since in this case it would make sense (quantifier elimination is defined for theories). Please correct me if this is a false assumption.

So now I want to show that $\mathbb{R}$ does not admit quantifier elimination. It seems difficult to choose a specific formula which has no equivalent quantifier-free formula and then showing this is the case. I do however have the following theorem at my disposal:

Let $T$ be an $\mathcal{L}$-theory, $n\geq 1$ a natural number and $\phi(x_1,\ldots,x_n)$ an $\mathcal{L}$-formula. Then the following are equivalent:

There is a quantifier-free $\mathcal{L}$ formula $\psi(x_1,\ldots,x_n)$ such that $\phi$ and $\psi$ are equivalent in $T$.

Let $\mathcal{M},\mathcal{N}$ be models of $T$ and let $\mathcal{A}$ be a common substructure. Then for any $\overline{a}\in A^n$ one has $\mathcal{M}\models \phi[\overline a]\iff \mathcal{N}\models \phi[\overline a]$.

If we use this, it suffices to find some elementary equivalent sub- or superstructure in which there is a non-true formula which becomes true or vice versa.

For example, one can take $\mathbb{R}_\text{alg}= \{r\in\mathbb{R}\mid \exists 0\neq p(X)\in \mathbb{Q}[X] \text{ such that } p(r)=0\}$. I tried looking at the formula $\exists x x^2+y=z$ but I believe this to be equivalent in both models.

So can anyone help me find a formula which is not equivalent in both models (or in other cases where we do not choose $\mathbb{R}_\text{alg}$). Or is there perhaps a different method which is better suited for this problem? I would prefer a method which is applicable in similar situations as I believe the one I describe is, but any proof is welcome.

答:
I agree with Andreas Blass that the argument suggested in his answer is the most straightforward way to see that the theory of the real field does not have QE. But here's a way to see this using the criterion for QE which you quoted in your question.

First of all, the precise strategy you tried cannot work. The theory $\text{RCF}$ of the real field is model complete, which means that if $M\subseteq N$ are both models of $\text{RCF}$, then $M$ is an elementary substructure of $N$. So you cannot "change the truth value" of a formula by moving down to a smaller model or up to a larger model. The syntactic consequence of this is that every formula is equivalent to both an existential and a universal formula.

But we can still use the criterion for QE to show that the formula $\varphi(x)$: $\exists y\, (y^2 = x)$ is not equivalent to a quantifier-free formula. It suffices to find $M\models \text{RCF}$ and $N\models \text{RCF}$ with substructures (not themselves models of $\text{RCF}$) $A\subseteq M$ and $B\subseteq N$ and an isomorphism $f\colon A\cong B$ such that $M\models \varphi(a)$ and $N\not\models \varphi(f(a))$ for some $a\in A$.

Take $M = N = \mathbb{R}$ and $A = B = \mathbb{Q}[\sqrt{2}]$. Now $\mathbb{Q}[\sqrt{2}]$ has an automorphism $f$ mapping $\sqrt{2}\mapsto-\sqrt{2}$. And we have $\mathbb{R}\models \varphi(\sqrt{2})$ but $\mathbb{R}\not\models \varphi(f(\sqrt{2}))$.

One can boil this down to a very concrete argument by repeating the proof of (the easy direction of) the criterion. Suppose for contradiction that $\varphi(x):\exists y\, (y^2 = x)$ is equivalent in $\mathrm{RCF}$ to a quantifier-free formula $\psi(x)$. Since $\sqrt{2}$ is a square in $\mathbb{R}$, $\mathbb{R}\models \varphi(\sqrt{2})$. Since  $\mathbb{R}\models \mathrm{RCF}$, $\mathbb{R}\models \psi(\sqrt{2})$. Since $\psi$ is quantifier-free and $\sqrt{2}\in \mathbb{Q}[\sqrt{2}]$, $\mathbb{Q}[\sqrt{2}]\models \psi(\sqrt{2})$. Since we can map $\sqrt{2}$ to $-\sqrt{2}$ by an automorphism of $\mathbb{Q}[\sqrt{2}]$, $\mathbb{Q}[\sqrt{2}]\models \psi(-\sqrt{2})$. Since $\psi$ is quantifier-free, $\mathbb{R}\models \psi(-\sqrt{2})$. Since $\mathbb{R}\models \mathrm{RCF}$, $\mathbb{R}\models \varphi(-\sqrt{2})$. But this is a contradiction, since $-\sqrt{2}$ is not a square in $\mathbb{R}$.


问:
Proving that $\gcd(a,ak+c)=\gcd(a,c)$
I'm having some trouble proving the following proposition:

Let $a,c \in \mathbb Z\setminus\{0\} $ and $k \in \mathbb Z$, then:$$\gcd(a,ak+c)=\gcd(a,c)$$

If $D_a$ and $D_c$ are the set of all numbers that devide $a$ and $c$ respectively I was able to prove that, if $\lambda := \gcd(a,ak+c)$ then $\lambda \in D_a \cap D_c$, but now, I can't arrive at the conclustion that $\max(D_a \cap D_c) = \lambda$. How can I conclude this proof?

答:
HINT

Sufficient to prove $\gcd(a,c) = \gcd(a,a+c)$.

This is the basis for the Euclidean algorithm, which finds $\gcd(a,b)$ with $a>b$ as $\gcd(a,b) = \gcd(a-b,b)$.

Title: Proving that $\gcd(a,ak+c)=\gcd(a,c)$


答:
Let $d = \gcd(a,ak+c)$ and $d_{1}=\gcd(a,c)$.

We have that $d|a \implies d|ak$ so $d|(ak+c)-ak=c \implies d|d_{1}$.

$d_{1}|a \implies d_{1}|ak$. Since $d_{1}|c, d_{1}|ak+c$.

Hence $d|d_{1}$ and $d_{1}|d$ with $d,d_{1} > 0.$

Therefore $d = d_{1}.$

You may also use $d \leq d_{1}$ and $d_{1} \leq d \implies d=d_{1}$ instead.

Title: Proving that $\gcd(a,ak+c)=\gcd(a,c)$

Body:
I'm having some trouble proving the following proposition:

Let $a,c \in \mathbb Z\setminus\{0\} $ and $k \in \mathbb Z$, then:$$\gcd(a,ak+c)=\gcd(a,c)$$

If $D_a$ and $D_c$ are the set of all numbers that devide $a$ and $c$ respectively I was able to prove that, if $\lambda := \gcd(a,ak+c)$ then $\lambda \in D_a \cap D_c$, but now, I can't arrive at the conclustion that $\max(D_a \cap D_c) = \lambda$. How can I conclude this proof?


问:
$f:\mathbb{R}\to \mathbb{R}$, $f^n(x) = x^n$
I have been wondering, could there be any function $f: \mathbb{R} \to \mathbb{R}$, such that $f^n(x) = x^n$. I have tried to make it easier by restricting it to $(0, \infty)$, but I couldn't find it...What do you think?

A variation of this problem was given to 10th grade students at a regional contest in Romania.

答:
Fix $n\ge 2$.

$f^{[2n]}(x)=x^{2n}$

$f^{[2n]}(x)=f^{[2n-1]}(f(x))=f(x)^{2n-1}$.

So $f(x)=x^{\frac{2n}{2n-1}}$. Since the expression of $f(x)$ depends on $n$, such $f$ cannot exists.


问:
People say that |x| is not differentiable, but isn't the derivative of |x|, |x|/x?
I have heard that $|x|$ is not differentiable, but I found a derivative that meets all the requirements necessary.
Here is my proof that the derivative of $|x|$ is $|x|/x$. First, note that $|x|$ has a slope of $1$ when x is positive. $|x|/x$ also equals $1$ when $x$ is positive. $|x|$ has a slope of $-1$ when $x$ is negative. $|x|/x$ equals $-1$ because $|x|$ is positive and $x$ is negative. $|x|$ has an undefined slope at $x=0$ and |x|/x is also undefined at $x=0$. Therefore, $|x|$'s derivative is $|x|/x$.

Anything wrong with this?

答: If $A\subset\Bbb R$ and $f$ is a function from $A$ into $\Bbb R$, we say that $f$ is differentiable if $f$ is differentiable at every $a\in A$. So, the absolute value function is not differentiable, since it is not differentiable at $0$. Otherwise, you are right, but I would simply say that if we differentiate it at a point $a\ne0$, then we get $1$ if $a>0$ and $-1$ if $a<0$.


问:
Find the interval in which $a$ lies such that the roots of the equation $(a+1)x^2 - 3ax + 4a$, is greater than 1.
Question:
Find the interval in which $a$ lies such that the roots of the given function, $f(x)$, is greater than 1.
$$f(x) = (a+1)x^2 - 3ax + 4a$$

My Attempt:

Using the quadratic formula,
$$x = \frac{3a±\sqrt{9(a)^2-16(a)^2-16a}}{2(a+1)}>1$$

First of all, the radicand must be greater than or equal to $0$. From that, we can restrict $a$ to
$$a \in [-\frac{16}{7}, 0]$$

Here, we can make two cases:

Case I:
$a+1>0$

$$±\sqrt{9(a)^2-16(a)^2-16a}>2 - a$$

Now squaring both sides and simplifying it further,

$$-7(a)^2-16a >a^2 + 4 -4a$$
$$2(a)^2+3a+1 <0$$
Hence, $a \in (-1, -\frac{1}{2})$

Intersecting this with the restricted set, we get $$a \in (-1, -\frac{1}{2})$$

Case II:
$a+1 < 0$

Here, we will end up getting,
$$a \in [-\frac{16}{7}, -1)$$

But this is not the required answer. Using desmos, it can be clearly seen that for $a\in(-1, -\frac{1}{2})$ there is only one root greater than $1$ and not both.

I suspect that I messed it up while squaring the radical. By squaring, I was trying to find the interval for $a$ where at least one root has a value greater than $1$. Right?

How am I supposed to get around it?
Also, is there a better approach for this particular problem?

答:
To find the values of $a$ such that real roots of the equation
$$ (a+1)x^2 - 3ax + 4a=0$$
are greater than $1$.

First the discriminant must be positive
$$(3a)^2-16a(a+1)\ge 0\to -\frac{16}{7}\leq a\leq 0\tag{1}$$

Second, if we want both roots greater than $1$, then the sum must be greater than $2$

The sum of the roots is $x_1+x_2=\frac{3a}{a+1}$

$$x_1+x_2>2\to \frac{3a}{a+1}>2\to a<-1\lor a>2\tag{2}$$

Putting together $(1)$ and $(2)$ we have that the roots of the given equation are greater than $1$ if $-\frac{16}{7}\leq a<-1$


问:
What is the relationship between "recursive" or "recursively enumerable" sets and the concept of recursion?
I understand that "recursive" sets are those that can be completely decided by an algorithm, while "recursively enumerable" sets can be listed by an algorithm (but not necessarily decided).  I am curious why the word "recursive" appears in their name.  What does the concept of decidability/recognizability have to do with functions that call themselves?</p>

答:
Today in the context of the branch of mathematics called recursion theory or computability theory, the term recursive and computable mean the same thing. The term decidable also mean the same thing but more often used in the context of logical theories or in textbook more focused on computer science. All these terms means there is an effective procedure to determine membership in some sets. Effective formally means accepted by a Turing Machine, $\mu$-recursive function, unlimited register machine, $\lambda$ calculus, etc.

The terms computable enumerable, recursively enumerable, recognizable all mean the same thing. They are the domain of the algorithms, the range of a total algorithm, the $\Sigma_1^0$ definable subsets of $\omega$, etc.

As I mentioned above, there are several formal model of computation that define the concept of being computable (decidable, recursive). I suspect the name "recursion" comes from one of the earlier method of computation. I believe Kleene proved many of the basic theorems of computability theory using primitive recursive functions and the $\mu$-recursive functions. Possibly this is where the term recursion comes from.

Soare argues in the 1990s that the name of the field called Recursion Theory should be called Computability Theory. The following papers discusses the history of the name "recursion" and "computability" and why he thinks computability is more appropriate.

http://www.people.cs.uchicago.edu/~soare/History/compute.pdf
http://www.people.cs.uchicago.edu/~soare/History/siena.pdf


问:
Solving quadratic inequalities algebraically without graphing/common-sensing the equality signs
I've just been presented an inequality at GCSE (high school) level that, solving algebraically (the method used for linear inequalities), doesn't work correctly which has caught me by surprise. Research always leads to graphed or common-sensed solutions. Can anyone explain why this doesn't work and if there's an algebraic manipulation that leads to the right answers?

$$x^2 - 49 > 0$$
Using the quadratic identity
$$(x + 7)(x - 7) > 0$$
Divide both sides by (x+7)...
$$x - 7 > 0$$
$$x > 7$$
Divide both sides by (x-7)...
$$x + 7 > 0$$
$$x > -7$$

Alternatively,
$$x^2 - 49 > 0$$
$$x^2 > 49$$
$$x > \pm7$$

The negative solution has to be $$x < -7$$ but I don't know how that's arrived at through algebra.

答:
The problem arises when you divide by $x+7$ since you can't be certain at this point that it is positive. Essentially you disregard the situation where $x+7$ and $x-7$ are both negative.

The alternative does not work as raising both sides to a certain exponent does not work with inequalities as you do not know that you are doing the same thing to both sides. This is more explained here but with squaring, but the same logic applies to this case also.

There are several ways of doing higher degree inequalities, but I will show what I was taught in high school (please correct me if this is not an algebraic way of doing this).

$$x^2-49>0$$
Let's look at the corresponding equation.
$$x^2-49=0$$
$$x=\pm 7$$
Now, we know that these won't work as the original one isn't $x^2-49\geq 0$. If you don't want to do this common-sense as in your title, you can just plug these results back into the inequality.

Now, knowing that these are the only zeroes of the function $f(x)=x^2-49$, we can check the values on the either "sides" of these zeroes and know it will never be zero again. Thus it can't change the sign. This is provable, of course, but this might not be the way you wanted to hear and in that case I will remove this answer.

So, $x=0$ doesn't fulfill the inequality, so $x\notin [-7, 7]$.

$x=100$ and $x=-100$ fulfill the inequality, so $x<-7\lor x>7$


答:
The issue is that $f(x) = |x|$ is not differentiable at $x = 0$.  Looking at the definition of the derivative at $x = 0$:
$$\lim_{h\to 0}\frac{f(h) - f(0)}{h} = \lim_{h\to 0}\frac{|h|}{h}.$$ Now, since $$|h| = \begin{cases}
-h, & h\le 0\\
h, & h>0
\end{cases}$$ we have that $$\lim_{h\to 0^{-}}\frac{|h|}{h} = \lim_{h\to 0^{-}}\frac{-h}{h} = -1$$ whereas $$\lim_{h\to 0^{+}}\frac{|h|}{h} = \lim_{h\to 0^{+}}\frac{h}{h} = 1,$$ which means that $f'(0)$ does not exist.  If we consider $x\ne 0$ then we can say that $f'(x) = -1$ for $x < 0$ and $f'(x) = 1$ for $x>0$, i.e.
$$f'(x) = \begin{cases}
-1, &x<0\\
\text{DNE}, &x=0\\
1, &x>0
\end{cases} = \begin{cases}
\frac{|x|}{x}, &x\ne 0\\
\text{DNE}, &x = 0
\end{cases}
$$
So, we can't say that $f$ is differentiable on $\mathbb{R}$, but we can say it is differentiable on $\mathbb{R}{\setminus\{0\}}$.

Title: People say that |x| is not differentiable, but isn't the derivative of |x|, |x|/x?

Body:
I have heard that $|x|$ is not differentiable, but I found a derivative that meets all the requirements necessary.
Here is my proof that the derivative of $|x|$ is $|x|/x$. First, note that $|x|$ has a slope of $1$ when x is positive. $|x|/x$ also equals $1$ when $x$ is positive. $|x|$ has a slope of $-1$ when $x$ is negative. $|x|/x$ equals $-1$ because $|x|$ is positive and $x$ is negative. $|x|$ has an undefined slope at $x=0$ and |x|/x is also undefined at $x=0$. Therefore, $|x|$'s derivative is $|x|/x$.

Anything wrong with this?


答: I think what the OP is trying to say is that there isn't generally a single, unified formula given for the derivative of $|x|$, and he has provided one.  I think the formula looks correct, and has the same domain as the piecewise derivative as well.


答:
Suppose this holds
for a particular value of $n$
(not all $n$).

Let's try
$f(x) = x^a$.
Then
$f^{[2]}(x)
=f(f(x))
=(x^a)^a = x^{a^2}
$,
$f^{[3]}(x)
=f^{[2]}(f(x))
=(x^a)^{a^2}
= x^{a^3}
$,
and,
in general,
$f^{[n]}(x)
= x^{a^n}
$.

Therefore
$n = a^n$
so
$a = n^{1/n}
$.

If we compose the other way
so that
$f^{[n+1]}(x)
=f(f^{[n]}(x))
$
then,
assuming that
$f(x) = x^a$,
$f^{[2]}(x)
=f(f^{[1]}(x))
=(x^a)^a
=x^{a^2}
$,
$f^{[3]}(x)
=f(f^{[2]}(x))
=(x^{a^2})^a
=x^{a^3}
$,
and we get
the same result as before.

Title: $f:\mathbb{R}\to \mathbb{R}$, $f^n(x) = x^n$

Body:
I have been wondering, could there be any function $f: \mathbb{R} \to \mathbb{R}$, such that $f^n(x) = x^n$. I have tried to make it easier by restricting it to $(0, \infty)$, but I couldn't find it...What do you think?

A variation of this problem was given to 10th grade students at a regional contest in Romania.


问:
Finding joker numbers
Let $n$ be an integer, $q(n)$ be the smallest prime number which divides $n$ and $r(n)$ be the biggest prime number less than or equal to $\sqrt{n}$. We say that $n$ is joker if $q(n)=r(n)$.

Except $8$, joker numbers  are of the form $p_n\cdot p_{n+k}$, with $k\geq 0$.
The converse is obviously false.
If $(p_n)$ is the prime numbers sequence, it is easy to prove that a number of the form $p_n\cdot p_{n+k}$ is joker if and only if $p_{n+1}^2 > p_n\cdot p_{n+k}$. In particular, if a number is not joker for $k$, it is not joker for $k+1$ (with the same $n$).
So, all the numbers of the form $p_n^2$ and $p_n\cdot p_{n+1}$ are jokers.
Using a spreadsheet, we can see that for $n\leq 1000$, the percentage of numbers of the form $p_n\cdot p_{n+k}$ which are jokers, with $k=2, 3, 4, 5, 6$ and $7$ is respectively $52 \%, 19.2 \%, 7.8 \%, 2 \%, 0.5 \%$ and $0.1 \%$.

Question : for any $k$, does there exist $n$ such that the number $p_n\cdot p_{n+k}$ is joker ?

答:
NOT AN ANSWER. An extended discussion to restate the question being asked, which won't fit as a comment

OP uses $n$ both as the name for joker numbers and as the primary index of primes. For clarity, I will avoid using $n$ as the primary index of primes.

A number $n$ with $3$ or more prime factors can have at most one prime factor larger than $\sqrt{n}$; if it had more than one such prime factor, their product alone would be greater than $n$. Thus if $n$ has $3$ or more prime factors, at least two of them must be smaller than $\sqrt{n}$. The only way that such a number might be a joker number is if the two prime factors $p_a$ smaller than $\sqrt{n}$ are equal: $n=(p_a)^2\cdot c$. Then $\sqrt{n}=p_a\sqrt{c}$ where $c$ has prime factors $\ge p_a$.

Specific cases: $p_a=c=2$ affords the noted exception $n=8$. $p_a=2,\ c=3 \Rightarrow n=12$, and $12$ is not a joker number by examination. $p_a\ge 3 \Rightarrow n\ge 27$, and since $27>25=p_{a+1}^2$, this affords no joker numbers. $c\ge 4 \Rightarrow p_a\sqrt{c}\ge 2p_a$ and by Bertrand's postulate, $p_a$ is not the smallest prime $\le \sqrt{n}$. Since $n$ has at least $2$, and (excepting $8$) cannot have $3$ or more prime factors, $n$ is a semiprime.

For any particular $p_a$, the joker numbers containing it as a prime factor lie within the interval $(p_a)^2$ to $(p_{a+1})^2$. OP's question can be stated: Let $k$ be the count of numbers in that interval that are products $p_ap_b$ such that $p_a<p_b<\frac{(p_{a+1})^2}{p_a}$; does there always exist $p_a$ such that $k$ can be arbitrarily large?

Let $p_{a+1}=p_a+g_a$, where $g_a$ is the gap between $p_a$ and $p_{a+1}$. Then acceptable $p_b$ to be counted are defined as $p_a<p_b<\frac{(p_{a}+g_a)^2}{p_a}=p_a+2g_a+\frac{g_a^2}{p_a}$. For sufficiently large $p_a$ we can consider $\frac{g_a^2}{p_a}$ to be an error term. We also know that there is only one prime to be counted within the interval from $p_a$ to $p_{a+1}$. So the question becomes: In the interval from $p_{a+1}$ to $p_{a+1}+g_a$, can there be arbitrarily many prime numbers?

At this time, I don't have an answer to that question.

Title: Finding joker numbers


问:
Cohomological dimension of a group via infinite direct product
Let $R$ be a ring and $G$ an infinite group. The $R$-cohomological dimension of $G$ can be defined as $cd_R(G) = \min \{ n \geq 0 : H^{n+1}(G, R[G]) = 0\}$, which implies that $H^i(G, M) = 0$ for all $i > cd_R(G)$ and any $R[G]$-module $M$.

Here $R[G]$ denotes the group ring, which as a module is simply the direct sum of copies of $R$ indexed by $G$, with the action of $G$ permuting the coordinates. Denote instead by $R(G)$ the direct product of copies of $R$ indexed by $G$ with the same action (is there a standard notation?). This is not a ring anymore but it is still an $R[G]$-module, and so we can ask when does $H^n(G, R(G)) = 0$.

My question is,can we use this module to compute the cohomological dimension instead? That is, does $cd_R(G) = \min \{ n \geq 0 : H^{n+1}(G, R[G]) = 0\}$? Or what are some examples in which this does not hold? Does it hold under some finiteness hypothesis on the group (say, finitely generated, finitely presented...)

In case it helps I am interested in particular in the case in which $R$ is finite, or even a finite field, and $n = 1$.

答:
$H^n(G,R(G))$ is always zero for $n>0$.

As Qiaochu Yuan pointed out in comments, $R(G)=\operatorname{Hom}_R(R[G],R)$. So $$\operatorname{Hom}_{R[G]}(?, R(G))\!=\operatorname{Hom}_{R[G]}(?,\operatorname{Hom}_R(R[G],R))=\operatorname{Hom}_R(?\otimes_{R[G]}R[G],R)=
\operatorname{Hom}_R(?,R).$$

So this shows that, calculating $H^n(G,R(G))=\operatorname{Ext}^n_{R[G]}(R,R(G))$ by applying $\operatorname{Hom}_{R[G]}(?, R(G))$ to a projective $R[G]$-module resolution of $R$, it is isomorphic as an $R$-module to $\operatorname{Ext}^n_R(R,R)$, which is zero for $n>0$.


问:
Solving $\cos^2(x)+\sin^2(x)-1=0$ for $x\in[0,2\pi]$
Which of these is a possible solution for
$$\cos^2(x)+\sin^2(x)-1=0$$
in the interval $x\in[0,2\pi]$

a. $x=\frac{\pi}{4}$

b. $x=\pi$

c. $x=\frac{5\pi}{3}$

d. all of the above

答:
All of the above.

The equation you wrote is basically a n identity and is true for all $x\in[0,2\pi]$


答:
You're essentially correct, except that what you've stated isn't exactly "differentiability"; if we let $f(x)=|x|$ then it's true that:

Wherever $|x|/x$ is defined, the derivative $f'(x)$ exists and equals $|x|/x$.

The domain on which $|x|/x$ is defined is the same as the domain on which $f'(x)$ exists.

That said, this is a bit of an unusual combination of facts - generally, mathematicians like functions to be defined on some domain fixed ahead of time rather than inferred later and aren't so concerned about trying to get the domains of things to match up by cleverly dividing something by zero at various places. If you defined derivatives and algebra using partial functions, then there would be some literal equality between the partial functions $f'(x)$ and $\frac{|x|}x$ - but this would be a non-standard definition, much as it may get informally used in calculus classes.

This doesn't mean that $|x|$ is differentiable - that means that the derivative exists everywhere and, clearly, $f'(0)$ doesn't exist since $\frac{f(x)}x$ is $1$ if $x>0$ and $-1$ if $x<0$ - so, in particular, that quotient has no limit as $x$ goes to $0$. It's also worth noting that, if $x\neq 0$, it is more common to write this using the notation on the right hand side:
$$\frac{|x|}x=\begin{cases}1 & \text{if }x >0 \\-1&\text{if }x<0 \end{cases}$$
where the reader can be immediately informed that this is just two constant functions pasted together, rather than having to struggle to identify what $\frac{|x|}x$ is.


答:
like aryan said it is an identity.
$\sin^2(x)+\cos^2(x)=1$ for all $x$ therefore with a simple substitution you have $1-1=0$
so it is true for all $x$

Title: Solving $\cos^2(x)+\sin^2(x)-1=0$ for $x\in[0,2\pi]$

Body:
Which of these is a possible solution for
$$\cos^2(x)+\sin^2(x)-1=0$$
in the interval $x\in[0,2\pi]$

a. $x=\frac{\pi}{4}$

b. $x=\pi$

c. $x=\frac{5\pi}{3}$

d. all of the above


答: By Bezout $g=au+cv=a(u-kv)+(ak+c)v$ and since $\begin{cases}g\mid a\\g\mid c\end{cases}\implies g\mid ak+c$ then $g=\gcd(a,ak+c)$

Title: Proving that $\gcd(a,ak+c)=\gcd(a,c)$


问:
Proof Check: $4x^2+3x+17$ is continuous by using a $\epsilon$, $\delta$ Argument
Can anyone double check my following epsilon delta proof.

I want to prove that the following function is continuous with an Epsilon Delta Argument.
$$ f: x \in \mathbb R \mapsto (4x^2+3x+17) \in \mathbb R $$

So I started with
$$\left\lvert f(x)-f(y) \right\rvert =$$
$$= \left\lvert (4x^2+3x+17) - (4y^2+3y+17) \right\rvert$$
$$= \left\lvert (4x^2+3x)-(4y^2+3y) \right\rvert$$
$$= 3\left\lvert (4x^2+x)-(4y^2+y) \right\rvert$$
$$= 3\left\lvert (\frac {4}{3}x^2+x)-(\frac {4}{3}4y^2+y) \right\rvert $$
$$= 3\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 + x-y \right\rvert$$
$$= 3\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 + x-y \right\rvert$$
(Triangel inequality)

$$\le 3 \left(\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 \right\rvert + \left\lvert x-y \right\rvert \right)$$

Note that $\delta \le 1$

$$\le 3 \left(\frac {4}{3}\left\lvert x^2 - y^2 \right\rvert + \delta \right)$$

$$= 3 \left(\frac {4}{3}\left\lvert (x - y) \right\rvert\left\lvert (x +y) \right\rvert + \delta \right)$$

$\delta \le 1$
$$\le 3 \left(\frac {4}{3}\left\lvert x + y \right\rvert \delta + \delta \right)$$
Adding zero in form of y - y
$$= 3 \left(\frac {4}{3}\left\lvert x + y - y + y \right\rvert \delta + \delta \right)$$
$$= 3 \left(\frac {4}{3}\left\lvert x - y + 2y \right\rvert \delta + \delta \right)$$
Triangel inequality
$$\le 3 \left(\frac {4}{3}(\left\lvert x - y \right\rvert +\left\lvert 2y \right\rvert) \delta + \delta \right)$$
Note that $\delta \le 1$

$$\le 3 \left(\frac {4}{3}(\delta +\left\lvert 2y \right\rvert) \delta + \delta \right)$$
$$= 4\delta^2+4\left\lvert 2y \right\rvert \delta + 3\delta $$
$$= \delta(4\delta+4\left\lvert 2y \right\rvert + 3) $$
Note that $\delta \le 1$
$$\le \delta(4 +4\left\lvert 2y \right\rvert + 3) $$
$$\le \delta(\left\lvert 8y \right\rvert + 7) = \epsilon$$
Therefore $$\delta = \frac {\epsilon}{(\left\lvert 8y \right\rvert + 7)} \;\; with \; \delta \le1$$

Thanks in advance for the help, I really appreciate it. :)

答:
To prove that $f(x)=4 x^2 + 3 x + 17$ is continuous $\forall a\in\mathbb{R}$

$$\underset{x\to a}{\text{lim}}\left(4 x^2+3 x+17\right)=f(a)$$
Equivalently
$$\underset{h\to 0}{\text{lim}}\left(4 (a+h)^2+3 (a+h)+17\right)=4 a^2 + 3 a + 17$$
$\forall \varepsilon>0$ we must find a $\delta>0$ such that if $|h|<\delta$ then
$$|4 (a+h)^2+3 (a+h)+17-(4 a^2 + 3 a + 17)|<\varepsilon$$
that is
$$|8 a h+4 h^2+3 h|<\varepsilon$$

$$0<\delta<\left|\frac{1}{8} \sqrt{64 a^2+48 a+16 e+9}+\frac38-a\right|$$
if $|h|<\delta$ then $|f(a+h)-f(a)|<\varepsilon$


问:
Proof of the uniqueness of maximal ideal
Let $R$ be a commutative ring with $1$. Let $M$ be a maximal ideal of $R$ such that $M^2 = 0$. Prove that $M$ is the only maximal ideal of $R$.</p>

答: The first thing I thought of was the fact that if $M'$ is a maximal ideal of $R$ distinct from $M$, then $M$ and $M'$ are comaximal, i.e., $M + M' = R$. But then we can write $1=am+bm'$, where $m$ and $m'$ belong to $M$ and $M'$, respectively, and $a,b$ belong to $R$. Now, let $x$ be any element of $M$. Then $x=x\cdot1=axm+bxm'=bxm'\in M'$, since $xm\in M^2$ and $M^2=0$. Since $x$ is arbitrary, $M\subseteq M'$, which is impossible unless $M'=M$.</p>


Body:
The derivative is easily found by the classical definition when $x\ne0$.

Let $f(x)=|x|$

\begin{eqnarray}
f^\prime(x)&=\lim_{h\to0}\frac{|x+h|-|x|}{h}\\
&=\lim_{h\to0}\frac{|x+h|-|x|}{h}\cdot\frac{|x+h|+|x|}{|x+h|+|x|}\\
&=\lim_{h\to0}\frac{|x+h|^2-|x|^2}{h(|x+h|+|x|)}\\
&=\lim_{h\to0}\frac{(x+h)^2-x^2}{h(|x+h|+|x|)}\\
&=\lim_{h\to0}\frac{h(2x+h)}{h(|x+h|+|x|)}\\
&=\lim_{h\to0}\frac{2x+h}{|x+h|+|x|}\\
&=\frac{x}{|x|}
\end{eqnarray}

Note also that $$\frac{x}{|x|}=\frac{x}{|x|}\cdot\frac{|x|}{|x|}=\frac{x|x|}{|x|^2}=\frac{x|x|}{x^2}=\frac{|x|}{x}$$


问:
Proof that subspace $U$ of a finite dim vector space $V$ is also finite dim and $\dim(U)\leq\dim(V)$
In my class lecture there was a longer proof of this than what I thought, so I want to check with you guys what flaws my argument has:

Since $V$ is a finite-dimensional VS, by definition, it has a finite unordered basis $S=\{s_1,\ldots,s_m\}$ such that $V=\text{span}(S)$. Since $U$ is a subspace of $V$, any $u\in U$ also lies in $V=\text{span}(S)$. So $u=\sum_i a^is_i$.

Thus any arbitrary vector in $U$ can be expressed as a linear combination of vectors in $S$. More generally it may be that coefficients corresponding to a subset of vectors in $S$ are zero in expressions for all vectors in $U$ (e.g. $a^1$ is always zero, so that subset would be $\{s_1\}$). In that case a subset of $S$ would span $U$.

So either $S$ or a subset of it (both independent) span $U$, are finite and therefore form a finite unordered basis of $U$.

This completes the argument, but is there any flaw in it?

答:
Sorry, but it is incorrect. Only some subspaces are generated by subsets of a basis, precisely $2^n$ of them.

For instance, assuming $n\ge2$, $\operatorname{span}\{v_1+v_2\}$ is not generated by a subset of $S$: indeed, it has dimension $1$ and no vector of $S$ belongs in it.

The proof is done by recursively constructing new vectors in $U$, ensuring to get a linearly independent set.

If $U=\{0\}$, then there is nothing to prove. Otherwise there exists $u_1\in U$, $u_1\notin U_0=\{0\}$.

Suppose we have built the linearly independent subset $\{u_1,\dots,u_k\}$ of $U$ and that $U_k=\operatorname{span}\{u_1,\dots,u_k\}\ne U$. Then there is $u_{k+1}\in U$, $u_{k+1}\notin U_k$ and the set
$$
\{u_1,\dots,u_k,u_{k+1}\}
$$
is easily seen to be linearly independent. Since the process cannot go beyond $n=\dim V$, because no linearly independent set in $V$ can have more than $n$ elements, we see that at some stage $k$ we must have $U_k=U$. Then $k=\dim U\le \dim V$.


问:
Would $ \lim_{\sigma\rightarrow0^+}\int_{-\sigma}^{\sigma}\frac{1}{\sigma}f(x)dx$ be equal to zero?
Suppose a limit of the form

$$ \lim_{\sigma\rightarrow0^+}\int_{-\sigma}^{\sigma}\frac{1}{\sigma}f(x)dx$$

where $f(x)$ is an integrable function.

I think that this limit would be zero, since we would be taking an integral at a single point, even if the argument was divergent. Is this correct?

答:
Not necessarily.

Consider the case when $f(x)=1$ for $x\in[-1,1]$, and $f(x)=0$ outside the interval.

In general, you may want to take a look at the Lebesgue differentiation theorem.


问:
Is there a function of bounded variation on $[0,1]$, say $f$, such that $f(x)-2x \sqrt{V^x_0(f)}= \frac{5}{4}x^2$ for all $x \in [0,1]$?
Is there a function of bounded variation on $[0,1]$, say $f$, such that $f(x)-2x \sqrt{V^x_0(f)}= \frac{5}{4}x^2$ for all $x \in [0,1]$?
Here, $V^b_a(f) = \sup_{\mathcal{P}}\{ \sum_{j=1}^{N}|f(x_j) - f(x_{j-1})|: \mathcal{P}=\{a=x_0<x_1<...<x_N =b\} \}$

And how do we find such a function $f$? Would it be unique?

I am working on the above question for my qual prep but I am stuck. As a hint, I was told that I would start with showing if $f$ is monotonic but I have no clue how it helps. Any helps would be highly appreciated!

答:
Observing that $f$ is monotonic is indeed the key to this problem. We will now prove the existence and uniqueness of such $f$.

We can rewrite the main equation as follows
$$
\tag{1} f(x) = 2x \sqrt{V_0^x (f)} + \frac 54 x^2.
$$
The function $V_0^x(f)$ is non-decreasing in $x$ for any $f$ of bounded variation, hence the right-hand side of $(1)$, as a sum of two monotonic functions is itself monotonic. Thus $f$ is increasing on $[0,1]$. Observe also, that substituting $x=0$ in $(1)$ implies that $f(0) = 0$.
Now, from the definition of variation and the fact that $f$ is increasing we have $V_0^x(f) = f(x) - f(0) = f(x)$ as $f(0) = 0$. Substituting this in $(1)$ we arrive at
$$
f(x) = 2x \sqrt{f(x)} + \frac 54 x^2.
$$
Completing the square in the last equation implies
$$
\left(\sqrt{f(x)} - x\right)^2 = \frac 94 x^2.
$$

From the last expression we have $\sqrt{f(x)} -x = \pm \frac 32 x$ for any $x\in [0,1]$ and since $f$ is increasing with $f(0) = 0$ we get that $\sqrt{f(x)} - x = \frac 32 x$ for all $x \in [0,1]$, arriving thus at
$$
f(x) = \frac {25}{4} x^2,
$$
which is the unique solution to the given functional equation.


答:
Since the $1/\sigma$ is just a constant and can move outside the integral this is essentially just half the average value of the function on small intervals centered at the origin.

So it can be any value you like.

Title: Would $ \lim_{\sigma\rightarrow0^+}\int_{-\sigma}^{\sigma}\frac{1}{\sigma}f(x)dx$ be equal to zero?

Body:
Suppose a limit of the form

$$ \lim_{\sigma\rightarrow0^+}\int_{-\sigma}^{\sigma}\frac{1}{\sigma}f(x)dx$$

where $f(x)$ is an integrable function.

I think that this limit would be zero, since we would be taking an integral at a single point, even if the argument was divergent. Is this correct?


问:
Inverses in $\mathbb{Z}[\zeta_p]$
I'm trying to solve the following exercise from my algebraic number theory class:

Let $\zeta_p$ be a p-th root of unity. We want to show that for $1\leq k,j\leq p-1$, $$\frac{1-\zeta_p^k}{1-\zeta_p^j}$$ is invertible in $\mathbb{Z}[\zeta_p]$.

I tried using the cyclotomic polynomial identity to factor this differently, but I ended up with $\frac{1+\zeta_p+\cdots+\zeta_p^k}{1+\zeta_p+\cdots+\zeta_p^j}$ and it's not clear to me how to divide this cleanly to pop back in $\mathbb{Z}[\zeta_p]$. I feel like I'm missing something obvious but I don't see it.

答: Hint (assuming $p$ is prime): $\zeta_p^j$ is primitive $p$-th root of unity, hence there is some $l\in \Bbb Z$ with $(\zeta_p^j)^l=\zeta_p^k$. Now use the geometric sum formula to show $$\frac{1-(\zeta_p^j)^l}{1-\zeta_p^j}=\frac{1-\zeta_p^k}{1-\zeta_p^j}\in\Bbb Z[\zeta_p]$$. Similarly show $\frac{1-\zeta_p^j}{1-\zeta_p^k}\in\Bbb Z[\zeta_p]$

Title: Inverses in $\mathbb{Z}[\zeta_p]$


答: Your proposition is wrong. This is a counterexample: Consider any integrable function which is constantly 1 on some symmetric non-empty interval around 0. For small enough values of sigma, the integral is exactly 2. Therefore the limit is also 2.</p>


问:
Geometric Interpretation of Intersection Multiplicities
I'm taking a course on Algebraic curves, and this is excerpted from a portion of the lecture notes.

Theorem: Let $p$ be a point of $\mathbb{P}_{\mathbb{C}}^{2}$. There is a unique way to associate with each couple $(P, Q)$ of non-zero homogeneous polynomials in $\mathbb{C}\left[x_{0}, x_{1}, x_{2}\right]$ an invariant $\mathbf{I}(p, P, Q)$ in $\mathbb{N} \cup\{\infty\}$ that satisfies the following six axioms.

Symmetry: $\mathbf{I}(p, P, Q)=\mathbf{I}(p, Q, P)$

Detects intersection points: $\mathbf{I}(p, P, Q) \neq 0$ if and only if $P(p)=Q(p)=0$.

Detects common components: $\mathbf{I}(p, P, Q)=\infty$ if and only if $P$ and $Q$ have a common irreducible factor that vanishes at $p$.

Transversality: suppose that $P$ and $Q$ both have degree one and that $Q$ is not of the form $\lambda P$ for some $\lambda \in \mathbb{C}^{*} .$ If $p$ is the unique point in $\mathbb{P}_{\mathbb{C}}^{2}$ such that $P(p)=Q(p)=0,$ then $\mathbf{I}(p, P, Q)=1$

Additivity: if $R$ is another non-zero homogeneous polynomial in $\mathbb{C}\left[x_{0}, x_{1}, x_{2}\right],$ then $\mathbf{I}(p, P, Q R)=$ $\mathbf{I}(p, P, Q)+\mathbf{I}(p, P, R)$

Deformation: assume that $\operatorname{deg}(Q) \geq \operatorname{deg}(P),$ and let $R$ be a homogeneous polynomial in $\mathbb{C}\left[x_{0}, x_{1}, x_{2}\right]$ of degree $\operatorname{deg}(Q)-\operatorname{deg}(P)$ such that $Q+P R \neq 0 .$ Then $\mathbf{I}(p, P, Q)=\mathbf{I}(p, P, Q+P R)$
Moreover, if $P$ and $Q$ are non-constant and have no repeated factors, and we denote by $C$ and $D$ their zero sets in $\mathbb{P}_{\mathbb{C}}^{2},$ then $\mathbf{I}(p, P, Q)=\mathbf{I}(p, C, D)$

The question is about axiom 5. The geometric translation of this axiom reads,

If $E$ is another projective plane curve over $\mathbb{C}$ that does not have a common component with $D,$ then $\mathbf{I}(p, C, D \cup E)=\mathbf{I}(p, C, D)+\mathbf{I}(p, C, E)$.

The question is, why do we require $E$ to have no common component with $D$? Under the convention $\infty + \infty = \infty$, isn't whether or not $E$ has a common component with $D$ irrelevant?

答: It's possible for $D,E$ to have a common component without having a common component with $C$, and this makes the additivity condition behave quite poorly. In the silliest case where $D=E$, we would have $I(p,C,D\cup E)=I(p,C,D)=I(p,C,E)$ for any $p$, which would imply that $I(p,C,D)=0$ for any $p,C,D$ where the invariant is not $\infty$. So intersection multiplicity could only ever give $0$ or $\infty$ if we didn't enforce the "no common component" requirement you're asking about.


答:
There's a history here. In thumbnail version, in the 1930s various attempts where made to formally characterize the intuitively computable numerical functions, and relatedly the effectively decidable sets of numbers (i.e. those whose membership can be decided by a computable function). Thus we encounter as various formal accounts of computability Church's idea of $\lambda$-computability, Turing computability, Herbrand-Gödel computability, and Gödel and Kleene's $\mu$-recursiveness (and more!). Of these, it is in the latter formal definition of computability where the notion of recursion in the sense of a function calling itself centrally features.

Now as a matter of technical fact, the $\lambda$-computable function, the Turing-computable functions, the  Herbrand-Gödel computable functions, and the ($\mu$)-recursive functions turn out to be the same class of functions. And for various reasons the preferred term for this class of functions became "recursive".

The technical fact that all these attempts (and other later ones) to characterize the intuitively computable functions converge on the recursive functions leads to the Church-Turing thesis that the computable functions in the intuitive sense just are these recursive functions (and the decidable sets of numbers are the recursive sets, i.e. those with a recursive characteristic function).

I wouldn't myself say that "computable" means "recursive" (nor would I recommend a linguistic reform to this effect). Rather I'd put it like this: it is a discovery that the intuitive notion of an algorithmically computable function (dressed up a bit) picks out the class of recursive functions.

Title: What is the relationship between "recursive" or "recursively enumerable" sets and the concept of recursion?

Body: I understand that "recursive" sets are those that can be completely decided by an algorithm, while "recursively enumerable" sets can be listed by an algorithm (but not necessarily decided).  I am curious why the word "recursive" appears in their name.  What does the concept of decidability/recognizability have to do with functions that call themselves?</p>


问:
Triple volume integral of divergence
S is given by $z = \sqrt{x^2+ y^2} \ $ for $1 ≤ z ≤ 4$, then calculate $\nabla \cdot \vec{F} \ $ for the vector field $\vec{F} = (xz, −zy, x^3)$.

Evaluate $I = \iiint \ (\nabla \cdot \vec{F}) \ dV$ ,
where $V$ is the volume enclosed by the surface S and the two discs
{(x, y, z) : x^2+y^2 ≤ 1 , z = 2} and {(x, y, z) : x^2+y^2 ≤ 9 , z = 3}$.

I got a negative answer it thet right?

答:
There is no reason why the surface integral cannot be negative. Also the divergence of a vector field can be negative or negative in certain regions and positive in others.

As far as your integral, you have not completely written it so difficult to say where you went wrong but if the first limit mentioned is of $r$, it is not correct.

In your case, $\vec{F} = (z, -zy, x^2)$

The surface is cone $z = \sqrt{x^2+y^2} \ $ along with two discs $x^2+y^2 \leq 1$ at $z = 1$ and $x^2+y^2 \leq 9$ at $z = 3$. So we have a closed surface and applying divergence theorem,

$div{\vec{F}} = \nabla \cdot \vec{F} = -z$ as you mentioned. So your integral is,

$\displaystyle \int_0^{2\pi} \int_1^3 \int_0^z \ (-z) \ r \ dr \ dz \ d\theta = -20 \pi$

Title: Triple volume integral of divergence


答:
It is much fun and pretty enlightening to do this kind of exercises with a ping-pong between algebra and geometry. Anyway, if you want to avoid that , then we have the following algebraic facts which, funny enough, are pretty easy to prove precisely by means of geometry...but not only, as they follow from the definition of absolute value:

$$\text{For}\;0a\iff  x<-a\;\;\text{or}\;\;x>a\end{cases}$$

Thus, in your case, we could do as follows (all algebra, not geometry):

$$x^2-49>0\implies x^2>49\stackrel{\text{taking square roots}}\implies |x|>7\iff x<-7\;\;\text{or}\;\;x>7$$

and in intervals notation, the solution is $\;x\in(-\infty,-7)\cup(7,\infty) \;$

Title: Solving quadratic inequalities algebraically without graphing/common-sensing the equality signs

Body:
I've just been presented an inequality at GCSE (high school) level that, solving algebraically (the method used for linear inequalities), doesn't work correctly which has caught me by surprise. Research always leads to graphed or common-sensed solutions. Can anyone explain why this doesn't work and if there's an algebraic manipulation that leads to the right answers?

$$x^2 - 49 > 0$$
Using the quadratic identity
$$(x + 7)(x - 7) > 0$$
Divide both sides by (x+7)...
$$x - 7 > 0$$
$$x > 7$$
Divide both sides by (x-7)...
$$x + 7 > 0$$
$$x > -7$$

Alternatively,
$$x^2 - 49 > 0$$
$$x^2 > 49$$
$$x > \pm7$$

The negative solution has to be $$x < -7$$ but I don't know how that's arrived at through algebra.

Body:
It is much fun and pretty enlightening to do this kind of exercises with a ping-pong between algebra and geometry. Anyway, if you want to avoid that , then we have the following algebraic facts which, funny enough, are pretty easy to prove precisely by means of geometry...but not only, as they follow from the definition of absolute value:

$$\text{For}\;0<a\in\Bbb R\;,\;\;\begin{cases}|x|\le a\iff -a\le x\le a\\{}\\
|x|>a\iff  x<-a\;\;\text{or}\;\;x>a\end{cases}$$

Thus, in your case, we could do as follows (all algebra, not geometry):

$$x^2-49>0\implies x^2>49\stackrel{\text{taking square roots}}\implies |x|>7\iff x<-7\;\;\text{or}\;\;x>7$$

and in intervals notation, the solution is $\;x\in(-\infty,-7)\cup(7,\infty) \;$


问:
Prove Hölder continuity of power function
How can I prove that $f(x)=x^s$ for $0< s ≤1$ is Hölder-continuous with constant s?

答: Hint: The result is true for $s=1,$ so assume $0<s<1.$ Suppose $0\le x.$ We'll be done if we show $y^s-x^s \le (y-x)^s$ for $y\ge x.$ Define $f(y) = y^s-x^s - (y-x)^s.$ Show $f'(y) < 0$ on $(x,\infty)$ to see $f$ decreases on $[x,\infty).$


问:
What is a more rigorous method of solving modulus function inequalities?
I have the inequality

$$\vert x-3\vert < 3x-4$$

To solve this I first break the modulus containing function into the two cases using the critical value

$$\vert x-3 \vert  \left\{\begin{array}{cc}
x-3 & x > 3 \\
3-x & x<3
\end{array}\right.  $$

For the case where $x>3$:

$$x-3<3x-4$$
$$x>\frac{1}{2}$$

For the case where $x<3$:

$$3-x<3x-4$$
$$x>\frac{7}{4}$$
However it is apparent that these two statements are contradictory and by simply looking at the graph of the two functions it is clear that the value $x>\frac{1}{2}$ does not hold.

The value does however seem to hold for the function $f(x)=x-3$

My question is this, why has my method given a false answer and what improvements can be made to this method to make it such that these false answers no longer appear?

答:
recommend putting everything on one side together and graphing that; a sum of absolute values of linear functions with another linear function gives a continuous piecewise linear function. The changes in slope occur exactly where each absolute value becomes zero...

It is possible to have horizontal segments if the coefficients work out just right. For your problem, that does not happen.

So, graph $|x-3| + 4 - 3x,$  if possible by hand on actual graph paper.

Note how there are corners, changes of slope, at $x=-5$ and $x=3$ in the graph below for $|x-3| + 2|5+x| - 3x-5$


问:
Does the "$\neq$" in, for instance, "Prove that $(x+y)^2\neq x^2+y^2$" mean "never equal"? "not equivalent"?
I have the problem:

Prove that $(x + y)^2 \not= x^2 + y^2$

I have answered it with something like this:

$$(x+y)^2 = x^2 + 2xy + y^2$$

$$(x^2 + 2xy + y^2) - (x^2 + y^2) = 2xy$$

$$\therefore (x + y)^2 = x^2 + y^2 \Rightarrow 2xy = 0$$

$$\therefore (x+y)^2 = x^2 + y^2 \Rightarrow x=0 \lor y=0$$

(Apologies if I'm using symbols incorrectly here, I'm very new to this sort of thing, and as I'm teaching myself I have no one to point these sorts of things out to me)

My problem is that, as far as I can tell, this proves that $(x + y)^2 = x^2 + y^2$ for at least some combination of values for $x$ and $y$. So I suppose my question is, for this kind of proof question, am I being asked to prove that the two statements are not equivalent, or that they are never equal?

Here's a picture of the problem:

答:
The statement $(x+y)^2\ne x^2+y^2$ contains "free variables" and whether it is true or not depends on the values of $x$ and $y$.

If $x=y=0$, then this statement is clearly false.

It is correct to say that "there exist real numbers $x$ and $y$ such that $(x+y)^2\ne x^2+y^2$".


问:
Prove $A = \{x ∈ X : f(x) ≥ a\}$ has finite measure
Let $(X, \mathcal{A}, \mu)$ be a measurable space. I want to prove that is $f: X \longrightarrow \mathbb{R}$ is measurable, non-negative and its integral on $X$ is finite, then $\forall a>0$ the set
$$A = \{x ∈ X : f(x) ≥ a\}$$
is $\mu-$finite (that is, $\mu (A) < \infty$). I think I could use this lemma to prove this statement but I don't really know how to do it. Could someone help me?

答:
Hint

$$\int_X f\geq \int_A f\geq a\mu(A).$$