答: When mean $\lambda$ follows geometric distribution itself,$$\begin{align*}P(N(1)<5)&=\sum_{k=0}^\infty P(N(1)<5\wedge\lambda=k)\\&=\sum_{k=0}^\infty \color{red}{P(N(1)<5|\lambda=k)}\color{blue}{P(\lambda=k)}\\&=\sum_{k=0}^\infty\left[\color{red}{\sum\limits_{x=0}^4 e^{-k}\frac{k^x}{x!}}\right]\color{blue}{(1/2)^{k+1}}\end{align*}$$

Title: Poisson process, mean is random variable

Body:
What is the probability that group of $5$ items will last in warehouse for more than $1$ day, if sale of this item can be described by the Poisson process:

a) with mean of $5$ items per day

b) with mean being geometrically distributed random variable with parameter $\frac{1}{2}$

a) $P(N(1)<5)=\sum\limits_{k=0}^4 P(N(1)=k)=\sum\limits_{k=0}^4 e^{-4} \frac{4^k}{k!}$

Is a) correct? How to do b)?


问:
Analogies between Prime Ideals and Knots
While reading this question posted at this link:

Subjects studied in number theory

i interestingly landed up on this Wikipedia page, and was quite amazed to see the variety of branches opening up. That's how I came to know about the subject Arithmetic Topology. Wikipedia describes this subject as:

Arithmetic topology is an area of mathematics that is a combination of algebraic number theory and topology. In the 1960s topological interpretations of class field theory were given by John Tate based on Galois cohomology, and also by Michael Artin and Jean-Louis Verdier based on Étale cohomology. In subsequent years, Barry Mazur and Yuri Manin pointed out a series of analogies between prime ideals and knots.

Can anyone explain me as to how Prime Ideals and Knot's are analogous (elementary explanation would be helpful) and why are we interested in studying these analogies.

答:
This is briefly described in Baez's week257 as well as in a post on lieven le bruyn's blog. I do not think there is a particularly elementary explanation here, but as I understand it there are three basic ideas:

From the point of view of étale cohomology, $\text{Spec } \mathbb{Z}$ is like a $3$-manifold. I confess I don't really understand what this means, although there is a nice heuristic explanation in the last bit of week257. Apparently it means there is an analogue of Poincaré duality in this setting with $n = 3$.

In fact, it is like a $3$-manifold with trivial fundamental group, so is like the $3$-sphere $S^3$.

The category of finite extensions of a finite field $\mathbb{F}_q$ looks exactly like the category of finite covers of the circle $S^1$; there's exactly one extension / cover for every positive integer $n$ with automorphism group $\mathbb{Z}/n\mathbb{Z}$. So the map $\text{Spec } \mathbb{F}_p \to \text{Spec } \mathbb{Z}$ is supposed to look like an inclusion of $S^1$ into $S^3$, which is a knot.

As for why we are interested in studying these analogies... I don't know about you, but I find this connection inherently interesting. Analogies of this kind have traditionally been very fruitful for both of the fields they connect in mathematics; if nothing else, it's nice to have a different source of intuition and motivation to make conjectures from.

ContentLicense: CC BY-SA 3.0


问:
coordinate transformation - reparameterization map: polar coordinates
I'm trying to find the reparameterization map $\phi$ (bijective and smooth) for the coordinate transformation of a parameterized curve $\sigma(u,v)$ which is reparameterized from $\sigma(r,\theta)$. Where $\phi :(u,v) \rightarrow  (r,\theta)$ and $\phi^{-1}: (r, \theta) \rightarrow (u,v) $ and where $u = rcos\theta$ and $v = rsin\theta$.

In order to find the jacobian of $\phi$.

Does any one have any suggestions for this map $\phi$?

答:
$$ \phi: (r, \theta) \mapsto (r \sin \theta, r\cos \theta), $$
$$ \phi^{-1}: (u,v) \mapsto ( \sqrt{u^2 + v^2 }, \arctan \frac{u}{v}) $$</p>


问:
Is a closure-preserving union of closed paracompact subspaces paracompact?
Let $X$ be a space. Suppose that $X$ has a closure-preserving closed covering $\mathcal{A}$ by paracompact subspaces. Then $X$ is regular.

Does it follow that $X$ itself is paracompact?

The answer is affirmative under the stronger assumption that $\mathcal{A}$ is hereditarily closure-preserving, and in particular under the assumption that $\mathcal{A}$ is locally-finite.

Note that any point-finite closure-preserving closed family is necessarily locally-finite, so perhaps I'm just not looking hard enough for a counterexample.

If the previous question is answered in the negative, then is there a counterexample where $\mathcal{A}$ is point-countable?

Edit: Thanks to Brian Scott for pointing out that my own attempt at a counterexample was flawed.

答:
This is a minor modification of the nice example due to H.B. Potoczny in the paper noted by Ulli in the comments.

Let $X=\omega_1\times\omega_1$, and let $D=\{\langle\alpha,\alpha\rangle:\alpha\in\omega_1\}$ be the diagonal. For convenience let $p_\alpha=\langle\alpha,\alpha\rangle$ for each $\alpha\in\omega_1$. Points of $X\setminus D$ are isolated. For each $\alpha\in\omega_1$ let

$$U_\alpha=\big(\{\alpha\}\times\omega_1\big)\cup\big(\omega_1\times\{\alpha\}\big)\,,$$

and for each finite $F\subseteq U_\alpha\setminus\{p_\alpha\}$ let $U_\alpha(F)=U_\alpha\setminus F$; these sets $U_\alpha(F)$ are the basic open nbhds of $p_\alpha$. It’s straightforward to verify that this defines a clopen base $\mathscr{B}$ for a Hausdorff topology, so $X$ is a Tikhonov space.

However, $X$ is not normal, so it is not paracompact. Let $H=\{p_n:n\in\omega\}$ and $K=D\setminus H$; $H$ and $K$ are disjoint closed sets. Let $V$ be any open nbhd of $H$; for each $n\in\omega$ there is a $U_n(F_n)\in\mathscr{B}$ such that $p_n\in U_n(F_n)\subseteq V$. Let $C=\bigcup_{n\in\omega}F_n$; $C$ is a countable subset of $X$, so there is an $\alpha\in\omega_1$ such that $\alpha\ge\omega$ and $\xi,\eta<\alpha$ for each $\langle\xi,\eta\rangle\in C$. Then $p_\alpha\in K$, since $\alpha\ge\omega$, and I claim that $p_\alpha\in\operatorname{cl}V$, so that $K$ cannot have an open nbhd disjoint from $V$. To see this, let $U_\alpha(F)\in\mathscr{B}$ be any basic open nbhd of $p_\alpha$; then

$$\langle n,\alpha\rangle,\langle\alpha,n\rangle\in U_\alpha(F)\cap U_n(F_n)\subseteq U_\alpha(F)\cap V$$

for all but finitely many $n\in\omega$, since $F$ is finite.

Finally, if $\alpha<\beta<\omega_1$ let

$$K(\alpha,\beta)=\{p_\alpha,p_\beta,\langle\alpha,\beta\rangle,\langle\beta,\alpha\rangle\}\,,$$

and let $\mathscr{K}=\{K(\alpha,\beta):\alpha<\beta<\omega_1\}$; this is clearly a cover of $X$ by compact sets and hence by paracompact closed sets, and I claim that it is closure-preserving. Let $\mathscr{A}\subseteq\mathscr{K}$, and let $A=\bigcup\mathscr{A}$. If $p_\gamma\in\operatorname{cl}A$ for some $\gamma\in\omega_1$, then there is a $K(\alpha,\beta)\in\mathscr{A}$ such that $U_\gamma\cap K(\alpha,\beta)\ne\varnothing$, but then $\gamma\in\{\alpha,\beta\}$, and $p_\gamma\in K(\alpha,\beta)\subseteq A$. Clearly the only isolated points in $\operatorname{cl}A$ are those already in $A$, so $A$ is closed, and $\mathscr{K}$ is closure-preserving.

The family $\mathscr{K}$ is not, however, point-countable: for each $\alpha<\omega_1$, $p_\alpha\in K(\alpha,\beta)$ for each $\beta$ such that $\alpha<\beta<\omega_1$.

ContentLicense: CC BY-SA 4.0


问:
Terence Tao Analysis I on defining subtraction operation problem
I have been reading Tao’s Analysis I, Chapter 4 which he builds up the definition of integers and in particular, the definition of subtraction using a placeholder “ $ —$” such that $a —b$ implicitly implies $a-b$.
He goes on to define several definitions of the subtraction placeholder and its properties using the previous chapters before coming to the conclusion that the placeholder and the minus sign of subtraction (as we know it) are equivalent to one another.

I am a little sceptical with this way of looking at subtraction from Tao’s foundational standpoint. It seems that the logic is circular in the fact that subtraction is already assumed before defining it, thus rendering all the sidelines of placeholders unnecessary. One could merely take subtraction as it is common known and show its properties. Why did Tao them require the placeholder for subtraction in the first place?

I also do know that he at Pg. 74 mentioned that Tao had admitted integers, defined as what you can get by subtracting two natural numbers to be circular. This I guess is pointing towards the same line of argument as I am making?

Also, proposition 2.2.6, defining the cancellation property (i.e given $a+b = a+c$ thereby $b=c$ ) already a definition of subtraction by itself? Why did he still take it as a “virtual subtraction” when by itself it could be well defined as a core of all subtraction cases?

What am I missing out here?
Is there a better way to look at the idea of subtraction from a less circular point of view?

答:
As stressed by the author, the place-holder is not subtraction (still to be defined): it is written as: $(a−−b)$.

Usually (see 2nd line page 75) we write $(a,b)$ and we mean exactly that the new defined entity is a pair of naturals.

We define rules for "manipulating" pair and we check that they give us the expected properties of integers.

The first step is to verify equality: $(a,b)=(c,d)$ is defined as $a+d=c+b$. As you can see, the RHS of the definition does not use the "$-$" symbol at all.

Thus, no circularity.

Proposition 2.2.6 (Cancellation law) is about natural numbers:

Let $a, b, c$ be natural numbers such that $a + b = a + c$. Then we have $b = c$.

It is not a "full" subtraction: from it we cannot retrieve: $(a+b)-b=a$, simply because subtraction is not defined in $\mathbb N$.

The next steps are:

(i) defining sum for the new entities: $(a,b)+(c,d)=((a+c),(b+d))$ (and verifying that the definition is consistent).

(ii) defining the "inverse" of new entities: $-(a,b)=(b,a)$.

(iii) "identify" the natural $n$ with the new entity $(n,0)$, verifying (see page 77) that $(n,0)=(m,0)$ if and only if $n = m$ (using property above: $n+0=m+0$).

(iv) prove that $(n,n)=(0,0)$ for every $n$ (because: $n+0=n+0$).

Now we can start checking that the new entities have the expected properties of integers: is is true that $(a,b)+ [-(a,b)] = (0,0)$ ?

Let check it: $(a,b)+ [-(a,b)] = (a,b) + (b,a)$, by (ii), $=((a+b),(b+a))$, by (i), $=((a+b),(a+b))$, by property of naturals, $=(0,0)$, by (iv).


问:
How do you "regularize" infinite integrals?
This question was inspired by the post:
" Is there a solid reason why some people assume the fundamental theorem of calculus should still hold for divergent integrals with improper bounds? " (and the follow-up discussion). Long story short, the OP was using some technique to "regularize" infinite integrals (and claims, for example, that $\int_2^{\infty}\frac{dx}{x}$ has a "regularized value" of $-\log 2$) - and then complaining that whatever regularization they were using did not play well with the rules for change of variable in an integral.

So... What is the "regularization" of an infinite integral? Are there multiple ways to do it, or do all of them boil down to the same thing whenever they "work"? Is there a preferred way of doing it? What generalizations of this notion have been used/studied/are practical? Is it a topic of active research? What are known unsolved problems related to it?

(Of course, I don't expect all those questions to be answered here, I am more after some sort of reference that would possibly be understandable with basic knowledge of real and complex analysis and Lebesgue integration, or perhaps slightly more than that. Ideally an article of the type "Everything you've always wanted to know about regularization but were afraid to ask...")

答:
Basic answer :There is no such thing as regularization of an integral. Either it is convergent, or it makes no sense.
The examples in the cited post are just another example of manipulations involving infinity that lead to absurd result (hence this should not be done).

Longer answer :
-In courses on complex analysis you have integrals with (complex) parameters that define holomorphic functions and can be extended on a larger set than the set where the integral is convergent. See for example the holomorphic extension of the Gamma function, this is in any book on complex analysis.

The next answers are well beyond undergrad level.

-Physicists often use the notion of renormalization, which consists in removing the divergent part of the integral of some physical quantity, unfortunately I never really understood what part of it was rigorous, and what part was just cooking.

-To some extent, the work of Martin Hairer (Fields medalist 2014) is also centered on renormalization issues for stochastic partial differential equations. I do not know much about it, but I can tell that it requires graduate level knowledge in probabilities to understand the issues.
Survey paper https://arxiv.org/pdf/1803.03044.pdf


问:
Prove that a function is differentiable everywhere
Let $c \in \mathbb{R}$. Let $g$ be a function with domain $\mathbb{R}$. We know the following:

$g(0) = 0$

$g$ is differentiable at $0$ and $g' (0) = c$

for every $x, y \in \mathbb{R}$, g(x+y) = g(x) + 2xy + g(y).

I have to prove that $g$ is differentiable everywhere and find a formula for $g '$. I have tried using the formal definition of derivative by limit but I can't figure out where to begin with this. Could you help me out?

Body:
Hint: from $ g(x+h) = g(x) + 2xh + g(h)$ we get

$$\frac{g(x+h)-g(x)}{h-0}=2x+\frac{g(h)-g(0)}{ h-0}.$$


问:
How to prove : $|A_{1} \Delta \cdots \Delta A_{n}| = \sum_{i} |A_{i}| - 2 \sum_{i<j}|A_{i} \cap A_{j}| + \cdots$?
The Problem is under the topic 'Inclusion-Exclusion' of Combinatorics.

The 'Indicator function' $\chi_{A}(x)$ is defined as :

$\chi_{A}(x)= 1, if x \in A$ otherwise $ 0, if x \notin A$.

And has the properties :

(1) $\chi_{A \cap B}(x) = \chi_{A}(x) \chi_{B}(x)$,

(2) $\chi_{A \cup B}(x) = \chi_{A}(x) + \chi_{B}(x) - \chi_{A}(x) \chi_{B}(x)$,

(3) $\overline {\chi_{A}(x)} = 1- \chi_{A}(x)$, and analogiclly so on.

With the help of this above defined function we've to prove the following equality :

$|A_{1} \Delta \cdots \Delta A_{n}| = \sum_{i} |A_{i}| - 2 \sum_{i<j}|A_{i} \cap A_{j}| +4 \sum_{i<j<k}|A_{i} \cap A_{j} \cap A_{k}| - \cdots$

I started by using this formula :

$|A \Delta B| = |A| + |B| - 2 |A \cap B|$.

Edit (my Second attempt) :

$A_{1} \Delta A_{2} = (A_{1} - A_{2}) \cup (A_{2} - A_{1}) - (1)$

By the theorem of Inclusion- Exclusion principle, we know that :

$|A_{1} \cup \cdots \cup A_{n}| = \sum_{i} |A_{i}| - \sum_{i<j}|A_{i} \cap A_{j}| + \sum_{i<j<k}|A_{i} \cap A_{j} \cap A_{k}| - \cdots + (-1)^{n+1} |A_{1} \cap \cdots \cup A_{n}| -(2)$.

I've been trying to combine these two equations (1) and (2) to get the desired result but no success so far.

Any suggestion for this method will be very much helpful.

But, I don't know how to proceed next? Please help me!

答:
HINT:

Use the formula
$$\mathcal{X}_{A\Delta B} = \mathcal{X}_A + \mathcal{X}_B- 2\mathcal{X}_A \cdot \mathcal{X}_B$$

Then use induction on $n$ to show that
$$\mathcal{X}_{A_1\Delta A_2\Delta \ldots \Delta A_n}= \sum \mathcal{X}_{A_i}- 2 \sum \mathcal{X}_{A_i} \mathcal{X}_{A_j} + 4 \sum \mathcal{X}_{A_i} \mathcal{X}_{A_j} \mathcal{X}_{A_k} -\cdots$$

$\bf{Added:}$

Note that
$$1- 2\mathcal{X}_{A\Delta B} = (1-2 \mathcal{X}_A)(1-2 \mathcal{X}_B)$$

so
$$1-2\mathcal{X}_{A_1\Delta A_2\Delta \ldots \Delta A_n} = \prod_{i=1}^n (1- 2 \mathcal{X}_{A_i})$$


问:
Measure Not Preserving Homeomorphism
Are there homeomorphisms that do not preserve measure (Let us consider Lebesgue Measure)? I ask this because, I currently cannot think one but while googling, I get Measure Preserving Homeomorphisms. So this specification is made perhaps because there are homeomorphisms that do not preserve measure?

P.S. I am fairly beginner in this topic.

答: $x\mapsto x^2$ is a homeomorphism from $[0,1]$ onto $[0,1]$ that does not preserve measure. Moreover, every continuous, strictly increasing function $f:\mathbb{R}\to\mathbb{R}$ that is not bounded below or above is a homeomorphism. Clearly, "most" such functions do not preserve measure.  </p>

Title: Measure Not Preserving Homeomorphism


问:
Minimum degree for an irreducible polynomial
Is it true that for every non-closed field $\mathbb F$ exists a polynomial of degree 2 that is irreducible over $\mathbb F$?
This holds for $\mathbb Q, \mathbb R$ but I can't understand if it is always true.

答:
No. See for instance the field of straightedge-and-compass (complex) constructible numbers.

Another example is like this: consider $\overline {\Bbb F}_p$, inside which you have exaclty one copy of each $\Bbb F_{p^n}$. Then, $\bigcup_{m\in\Bbb N}\Bbb F_{p^{2^{m}}}$ is a subfield and it's as you mention.


问:
Comparing proof of $R[[x]]^*$ consists of $\sum_{n \geq 0}a_{n}x^n (a_{n} \in R) ,a_{0} \in R^* ,R$ an integral domain to $R$ a ring.
I want to answer letter $(b)$ in this question:

Let $R$ be an integral domain and let $R[[x]]$ be the corresponding ring of formal power series.

$(a)$ Show that $R[[x]]$ is an integral domain.

$(b)$Show that $R[[x]]^*$ consists of the series $\sum_{n \geq 0}a_{n}x^n (a_{n} \in R)$ such that $a_{0} \in R^*.$

My question is:

I found a solution here If $a_0\in R$ is a unit, then  $\sum_{k=0}^{\infty}a_k x^k$ is a unit in $R[[x]]$ in case of $R$ a ring, I am wondering how will my solution differ as my $R$ is an integral domain?

答:
Is it clear to you that for $R$ any unital ring (it may be non-commutative in which case $x$ commutes with the elements of $R$) and $f=\sum_{n\ge 0}a_n x^n\in R[[x]]$,

If $a_0\in R^\times$ then $$g=\sum_{k\ge 0} (1-a_0^{-1}f)^k\in R[[x]]$$ and (using that $a_0^{-1}f$ commutes with $1-a_0^{-1}f$) $$a_0^{-1}fg= 1$$ thus $ga_0^{-1} = f^{-1}$ ?

Conversely if $(\sum_{n\ge 0}a_n x^n)(\sum_{n\ge 0}b_n x^n)=1$ then $a_0b_0=1$ ie. $a_0\in R^\times$.


问:
$\ker p(T) = \ker p_1(T)\oplus...\oplus\ker p_k(T)$ where $T\in \mathcal{L}(V)$

Let $p(x)$ be a polynomial, and assume that $p(x) = p_1(x)...p_k(x)$ where $p_1,...,p_k$ are relatively prime. Let $T:V\to V$ be a linear operator on $V$. Show that $\ker p(T) = \ker p_1(T)\oplus\dots\oplus\ker p_k(T)$.

My plan:

To show that $\ker p_i(T) \cap\ker p_j(T) = \{0\}$ for all $i,j$

To show that $\ker p_1(T) + \dots + \ker p_k(T) \subseteq \ker p(T)$

To show that $\ker p(T) \subseteq \ker p_1(T) + \dots + \ker p_k(T)$

Will this be sufficient? I'm unable to show the third condition.
For the first two, here's what I did:

Let $v\in \ker p_i(T) \cap \ker p_j(T)$, so that $p_i(T)v=0$ and $p_j(T)v=0$. Since $p_i,p_j$ are relatively prime, there exist polynomials $q_i,q_j$ such that $q_i(T)p_i(T) + q_j(T)p_j(T) = I$. Multiplying LHS and RHS by $v$, we get $0+0 = v$, so $v=0$. This proves (1).

By definition, $p(x) = p_1(x)\dots p_k(x)$, so $p(T) = p_1(T)\dots p_k(T)$. Clearly, $\ker p_i(T) \subseteq \ker p(T)$ for all $i$, hence $\sum_{i=1}^k \ker p_i(T) \subseteq p(T)$. This proves (2)

For (3), I started with $v \in \ker p(T)$ so that $p(T)v = 0$ and consequently $p_1(T)...p_k(T)v = 0$. I also know that there exist polynomials $q_1,\dots ,q_k$ such that $p_1(T)q_1(T) + \dots + p_k(T)q_k(T) = I$. What do I do next?

答:
Let $v \in\ker p(T)$, i.e. $p(T)v = 0$. Consider polynomials $p_i'(T) = p(T)/p_i(T)$, and observe that all such $p_i'(T)$ are also relatively prime. So, there exist polynomials $f_i(T)$ such that $\sum_{i=1}^k f_i(T)p_i'(T) = I$.

$v = Iv = \sum_{i=1}^k f_i(T)p_i'(T)v = \sum_{i=1}^k v_i$ where $v_i = f_i(T)p_i'(T)v$. We show that $v_i \in \ker p_i(T)$.

$p_i(T)v = p_i(T)p_i'(T)f_i(T)v = f_i(T)p(T)v = 0$ for all $i=1,2,...,k$ which shows that $v \in \ker p(T) \implies v \in \sum_{i=1}^k \ker p_i(T)$ and finishes the proof.


问:
Are rational points dense on every circle in the coordinate plane?
Are rational points dense on every circle in the coordinate plane?

First thing first I know that rational points are dense on the unit circle.
However, I am not so sure how to show that rational points are not dense on every circle.

How would one come about answering this.
Any hits are appreciate it.

答:
First of all, I want to thank @Hidaw for having answered such an interesting question, which was swirling around in my head some days ago. After reading all the answers to this post with great pleasure, I discovered that Humke and Krajewski completely solved the problem in a very simple and beautiful article on the American Mathematical Monthly in 1979 (a few years before I was born!) called A Characterization of Circles Which Contain Rational Points.

In particular they establish the very relevant result that if the radius $r$ is irrational, but $r^2=p/q$, with $p, q \in \mathbb{Q}$, then the circle $C$ of equation $x^2+y^2=r^2$ contains no rational point if $pq$ is not the sum of two square integers, while $\mathbb{Q}$ is dense on $C$ if $pq$ is the sum of two square integers.


问:
What is the name of the answer to exponentiation?
What is the name of the answer to exponentiation? Adding two numbers produces a sum. Multiplying two numbers produces a product, but I cannot think of or find the name for the solution to exponentiation. </p>

答:
I agree that calling both the exponent and answer power is a bad idea. We need a more specific name. Already we can call any result from any operation " a result, answer, number, sum, output, yield, yielding answer/result...etc.".

The result of powering is called the power. The confusion lies in the ambiguity. When I say "the power" am I referring to the exponent or the answer? Product seems safest because it is already well defined in terms of multiplication and all we are doing is repeated multiplication of a base number.

We have the base (or radix), the exponent (or index or power) and the power (or product). The fear is whatever unique name we try to call the answer will also became a name for the exponent. We should choose a name that is unique to the answer and forbidden from being used as a name for the exponent or else it defeats the point.

A power is a square or cube or higher dimensional entity. We should call the exponent the dimension and the answer the "entity". Entity is hands down the best choice when used in this context as it means "thing" and thing is defined as " I can not give a name to this but would like to".

The radix is the Latin root forming "radical". The we could take index and turn it into "indical" as it is referring to the exponent but not the exponent itself hence saying it is the "power" of the number represented as an answer but not "power" as an exponent.

Word play using synonyms for "object", "shape" or "number" do not produce as meaningful new words. Using a funny sounding make up word like "indical" you can see the connection to the index and hence the power but it sounds too funny to use for anything thing else than the answer. As such it would only be used when you need to specifically refer to the answer. It sounds legit enough to gain acceptance while being funny enough that most people will just say "product" to avoid confusion with the exponent.

Your best choices are entity, indical, power, product, in that order.
Your worst choices are power or a name stemming from shape. I tried to use nary or n-ary or some change of n-tuple. Exponent-tuple = e-tuple? It is almost there but not simple enough and that is what the poster is asking for. I tried using modified versions of rank or ordinal but they were too confusing, defined elsewhere, and/or they can be used for the exponent itself. While this is not a bad thing it is something that lacks specificity and I wished to avoid.

We can "square or cube" a number so a generic form for any higher dimensional shape could work but the simplest and easiest to accept is "entity" which means thing or dimension. I vote for entity, product, or indical.

Edit: It should be called the proponent. Exponet is latin,  "ex" means outside and "ponent" means position. We can take this to mean that the exponent is outside of the base number, and proponent is after the base number being on the other side of the equal sign. The word component was considered but lost to entity since a component is defined as being part of an entity. Component is also used in vector sums and wished to avoid dual meaning if possible. Proponent means before or after a position. When we are taking about base we are talking about a positional system of where the digits are. 3^2 = 9 has 3 as the base, 2 outside of the base and 9 after the base.


问:
Let $Z_m \to c$ in probability. When can we say $EZ_m \to c, Var[Z_m] \to 0?$ Could you please see if my proof is correct in a special case?
Let $Z_m \to c$ in probability. Then when can we say $EZ_m \to c$ and $Var[Z_m] \to 0?$

I think a sufficient condition is that: $|Z_m|$ is uniformly bounded, i.e. $|Z_m| \le K \forall m.$

QUESTION 1: Here's my proof of the above claim: could you please check if it's correct?

Fix $\epsilon > 0.$ Then:  $Z_m \to c$ in probability implies $\exists N \in \mathbb{N}$ so that $\forall m \ge N, P[|Z_m - c| \ge \epsilon]  \le \epsilon .....(1)$

$|EZ_m - c| \le E|Z_m - c| = \int_{|Z_m - c| \ge \epsilon}  |Z_m - c|dP + \int_{|Z_m - c| < \epsilon}  |Z_m - c| dP  \le \int_{|Z_m - c| \ge \epsilon}  (|Z_m| + c)dP +  \int_{|Z_m - c| < \epsilon}  |Z_m - c| dP \le (K+ |c) \epsilon  + \epsilon,$ where we used (1) above to get the left summand on the right side of the last inequality. This shows that $|EZ_m - c| \le (K + |c| + 1) \epsilon,$ proving what we wanted: is the proof correct?

The proof of $Var[Z_m] \to 0$ is very similar as above.

QUESTION 2: When can we weaken the condition i.e. $|Z_m| \le K \forall m$ and still have $EZ_m \to c$ and $Var[Z_m] \to 0$ when $Z_m \to c$ in probability?

答:
Your claim is true. In fact, it is a special case of the result found under "Relation to convergence of random variables" here: https://en.wikipedia.org/wiki/Uniform_integrability. Note that $(|Z_m|)_m$ being uniformly bounded implies that also $(|Z_m|^2)_m$ is uniformly bounded (by $K^2$), which implies that both are uniformly integrable and hence $L^1$ convergent, which gives the result you want. This also gives a weaker condition: it is sufficient that $(Z_m)_m$ are uniformly integrable to obtain that $EZ_m\to c$. For the variance, it is sufficient that $(Z_m^2)_m$ are uniformly integrable.

Your proof looks fine, only minor detail is that you write $\forall n\geq N$, but that should be $m$ rather than $n$.


问:
Is a group isomorphic to the internal product of its Sylow p-subgroups?
Let $G$ be a group such that its order is a product of distinct primes $p_1, \dots, p_n$ and let $P_i$ denote each Sylow $p_i$-subgroup. Is $P_1 \dots P_n$ (the internal or Frobenius product) equal to $G$, that is, $G = P_1 \dots P_n$?

答:
For $G$ finite, one of the many theorems of P. Hall is that your condition holds whenever $G$ is solvable (regardless of order).  A note of Rowley and Holt discusses the general problem (and includes a reference for said result of Hall), and provides a few non-solvable examples.
They also show that such a product does not exist for the finite simple group $G=U_3(3)$.  So in full generality the answer is "no".  However, if $|G|$ is square-free, as you assume, then $G$ is solvable (see also these notes for more details), so the answer will be "yes" in this case.

OwnerDisplayName: Pedro Minicz

OwnerDisplayName: zibadawa timmy


答:
There is a more direct proof than quoting the fairly deep Theorem of P. Hall, but you do need to know a little transfer theory. The argument that follows is well-known and may be found in many group theory texts. We proceed by induction, there being nothing to prove when $n = 1$. Suppose then that $n > 1$ and that the result is true for smaller values of $n$.
If $|G| = p_{1}p_{2} \ldots p_{n}$ where $p_{1} < p_{2} < p_{3} < \ldots < p_{n}$ are primes, and if we let $P_{i}$ be a Sylow $p_{i}$-subgroup of $G$ for each $i$, then we note that the order of $N_{G}(P_{1})/C_{G}(P_{1})$ divides $p_{1}-1.$ But since $p_{1}$ is the smallest prime divisor of $|G|$, we see that $N_{G}(P_{1}) = C_{G}(P_{1})$.

By Burnside's transfer theorem, $G$ has a normal $p_{1}$-complement, which means that $G$ has a normal subgroup $H_{1}$ of order $p_{2}p_{3} \ldots p_{n}.$ Then $H_{1}$ contains all elements of $G$ of order coprime to $p_{1}$, and we have $G = H_{1}P_{1} = P_{1}H_{1}$, since $H_{1} \lhd G$.

By induction, we have $H_{1} = P_{2} P_{3} \ldots P_{n}$, so that $G = P_{1}H_{1}
= P_{1}P_{2} \ldots P_{n}.$

Title: Is a group isomorphic to the internal product of its Sylow p-subgroups?

Body: Let $G$ be a group such that its order is a product of distinct primes $p_1, \dots, p_n$ and let $P_i$ denote each Sylow $p_i$-subgroup. Is $P_1 \dots P_n$ (the internal or Frobenius product) equal to $G$, that is, $G = P_1 \dots P_n$?

OwnerDisplayName: Geoff Robinson


答: Following @Fred 's answer, I would just add that once we establish that $g'(x)= 2x + c$, we can even obtain the exact expression $g(x) = x^2 + cx$. This is not required as part of the answer but it is still nice that we can fully characterise such functions.

Title: Prove that a function is differentiable everywhere

Body:
Let $c \in \mathbb{R}$. Let $g$ be a function with domain $\mathbb{R}$. We know the following:

$g(0) = 0$

$g$ is differentiable at $0$ and $g' (0) = c$

for every $x, y \in \mathbb{R}$, g(x+y) = g(x) + 2xy + g(y).

I have to prove that $g$ is differentiable everywhere and find a formula for $g '$. I have tried using the formal definition of derivative by limit but I can't figure out where to begin with this. Could you help me out?


问:
construct a sequence Sn that does not converge but its arithmetic mean converges to 0
I ran into a problem which requires me to construct a sequence call it Sn which does not converge, but its arithmetic mean call it An, converges to 0 as n-->infinity. im having a hard time solving this so i looked up the solution on a website called slader but im not convinced with the solution it's probably wrong because the Sn used is 1/n+1which converges to one. slader solution. i left the image of the solution if anyone wants to see it. help would be appreciated thanks original question. ive also left a picture of the original question. (it's part B im stuck on)

Body:
$1,-1,1,-1,-1,...$ diverges, but
$\frac{1+(-1)+1+(-1)+...}{n}$ converges to $0$.


问:
image of the st. line $y=mx+c$

What is the slope of the image of the st. line $y=mx+c$ under the mapping $f(z)=az+b,a,b\in\mathbb C?$

I think I don't need the constant $b$ for translation don't rotates any shape. Letting $A$ is the angle the line makes with the postive real axis I have found that the image line makes the angle $A+\arg a$ with the positive $x$-axis as follows:

Thus the slope of the new line is $\tan (A+\arg a)=\tan(\tan^{-1}m+\arg a).$ Where did I go wrong?

答:
When we talk about the image of a complex valued function, we are talking about the image in $u,v$ plane.

$$f(z) = a(x+iy)+b = a(x+i(mx+c))+b = (ax+b)+a(mx+c)i = u(x)+i\cdot v(x)$$
$$u(x) = ax+b,\;\;v(x) = a(mx+c)$$
So the image of $f$ will be
$$\{(ax+b,a(mx+c))\;|\;x\in\mathbb{R}\}$$

Notice while $x$ takes all value in $\mathbb{R}$, so will $u$. We have $v = a(mx+c) = mu+(ac-mb)$.

The image then can be rewritten as
$$\{(u,mu+(ac-mb)\;|\;u\in\mathbb{R}\}$$

That is, in the $uv$ plane, it looks like the graph of linear affine $v = mu+(ac-mb)$, a line with slope $m$ and intercept $(ac-mb)$. Therefore the slope is still $m$.

Title: image of the st. line $y=mx+c$


问:
How do I solve the inequality $2(x-3)^2 < 32$?

How do I solve this inequality?
$$2(x-3)^2 < 32$$

Currently, I solve such inequalities by sketching. This just seems weird and I think there is much more to it than I know. What insights have you gained from thinking about such inequalities?

答:
We notice that $2$ is a factor common to both the LHS and RHS. So we essentially reduced the problem to finding the solutions to the inequality:
$$(x-3)^2 < 16$$
Noting that $16 = 4^2$, we bring it to the LHS. We then get,
$$(x-3)^2 - 4^2 < 0$$
Applying the algebraic identity $a^2-b^2 = (a-b)(a+b)$
$$\implies (x-3-4)(x-3+4) < 0$$
$$\implies (x-7)(x+1) < 0$$
Since the product of these $2$ factors is negative, we must conclude that one of them must be negative while the other is positive. So, two cases are possible:

Case 1:    $x-7 > 0 \implies x>7$ and $x+1<0 \implies x < -1$  (Not possible)

Case 2:    $x-7 < 0 \implies x<7$ and $x+1>0 \implies x > -1$

Hence, via case 2, we conclude that the solution to the inequality is $x \in \ (-1,7) $


问:
Proof that $\frac{3-x^{4}}{8x^{2}}$ is the only solution to the functional equation $f(x) + 3f\left(\frac{1}{x}\right) = x^{2}$
How can we prove that $f(x) + 3f\left(\frac{1}{x}\right) = x^{2}$ has only one solution and no other solution and that solution is $\frac{3-x^{4}}{8x^{2}}$.

I know that this is a linear function equation, but can't find any sources which say that all linear functional equation will have only one solution.

答:
We always have $x \ne 0.$

From

$$(1) \quad f(x) + 3f(\frac{1}{x}) = x^{2}$$

we get

$$(2) \quad f( \frac{1}{x})+3f(x)=\frac{1}{x^2}.$$

If we add $(1)$ and $(2)$ we derive

$$(3) \quad f(x)+f(\frac{1}{x})= \frac{1}{4}(x^2+\frac{1}{x^2}).$$

$(3) - (1)$ shows that

$$(4) \quad -2f(\frac{1}{x})=\frac{1}{4}(x^2+\frac{1}{x^2})-x^2.$$

Now substitute $\frac{1}{x} \to x$ in $(4)$. Easy computations give now the desired result.


答:
Try this
$$
2(x-3)^2 < 32
\\
(x-3)^2 < 16
\\
|x-3| < 4
\\
-4 < x-3 < 4
\\
-1 < x < 7.
$$

The important lesson is: When you have $(x-3)^2 < 16$ and you "take square root of both sides", you do not get $x-3 < 4$.

Title: How do I solve the inequality $2(x-3)^2 < 32$?

Body:
How do I solve this inequality?
$$2(x-3)^2 < 32$$

Currently, I solve such inequalities by sketching. This just seems weird and I think there is much more to it than I know. What insights have you gained from thinking about such inequalities?


问:
Number of solutions of $x^m-y^m=a \pmod p$
I'm interested in the total number of solutions of $x^m-y^m\equiv a \pmod p$ where $p$ is prime.

Is there any way to estimate the total number of solutions ?

I found https://www.degruyter.com/view/journals/crll/1935/172/article-p151.xml?language=en is related, but I cannot read German.

答:
I am interpreting the question to have fixed $p,m,a$ and $x,y$ unknown.

We are obviously only interested in pairwise non-congruent solutions. Or rather, we let $x,y$ range over $\Bbb{Z}_p$, and want to estimate the number of pairs $(x,y)\in\Bbb{Z}_p^2$ satisfying $x^m-y^m=a$. I am replacing congruences modulo $p$ with equations in $\Bbb{Z}_p$ in all that follows. Denote the number of such pairs by $N(m,p,a)$.

Let's first record the standard fact:
$$
N(m,p,a)=N(d,p,a),
$$
where $d=\gcd(m,p-1)$. This follows from the basic properties of cyclic groups. Clearly $x^m=0$ iff $x=0$. The multiplicative group $\Bbb{Z}_p^*$ is known to be a cyclic group of order $p-1$. Therefore the mapping $f:x\mapsto x^m$ is a homomorphism of groups. Its kernel $N$ has size $d$ and its image $M$ thus has size $(p-1)/d$, each element of $M$ the image of a coset of $N$. The exact same thing happens with the mapping $\tilde{f}:x\mapsto x^d$, and the claim follows from this. Ask, if you need something fleshed out (most likely that's done in some other thread on the site already).

The point of that observation was to make $m$ smaller as we can replace it with $d$, and assume that $m\mid p-1$. On with the main business.

The case $a=0$, i.e. the equation $x^m=y^m$, is a special case. With the assumption $m\mid p-1$ in place we can give the exact number of solutions. We have the trivial solution $(x,y)=(0,0)$. If $x$ is non-zero, then $x^m=y^m$ is equivalent to $1=(y/x)^m$
which holds if and only if $y/x\in N$. Therefore to each non-zero choice of $x$ there are exactly $m$ solutions $y$. Hence we can conclude that
$$
N(m,p,0)=1+m(p-1).
$$

On with the main case $a\neq0$, when we bring out the big guns of algebraic geometry.
The equation $x^m=y^m+a$ defines a plane curve. Its projective version, call it $C$, is defined by
the homogenized polynomial equation
$$
F(X,Y,Z):=X^m-Y^m-a Z^m=0.
$$
Here the gradient
$$\nabla F(X,Y,Z)=(mX^{m-1},-mY^{m-1},-amZ^{m-1})$$
vanishes only at $(X,Y,Z)=(0,0,0)$ so $C$ is non-singular. Therefore
the genus-degree formula says that $C$ has genus
$$
g=\frac12(m-1)(m-2).
$$
Next we count the number of points of $C$ on the line at infinity, $Z=0$. We are back in (the projective version of) the case $a=0$, and can conclude that there are exactly $m$ points on the line at infinity, namely $[X:Y:Z]=[1:\zeta:0]$ with $\zeta\in N$ and $m$th root of unity.

It remains to call upon the results of André Weil giving an upper bound on the difference $\#C(\Bbb{Z}_p)-(p+1)$
of the number of $\Bbb{Z}_p$-rational points of $C$ and $p+1$ (=what we would get if $C$ were a line).

$$|N(m,p,a)+m-(p+1)|\le 2g\sqrt p=(m-1)(m-2)\sqrt p.$$

Basically it says that the number of solutions is $\approx p$, if $m$ is very small in comparison to $p$, and gives an upper bound on the deviation from that expected number otherwise.

Weil's bound becomes in some sense the best possible, if we also consider solutions in the extension field $\Bbb{F}_q, q=p^n$ of $\Bbb{F}_p=\Bbb{Z}_p$. A lot is also known about some special values of $m$. The case $m=3$ being a particularly interesting case as then $g=1$, so we have an elliptic curve.


问:
Probabilistic inequalities involving random variables on both sides?
Can I break down $P(h \geq (A + B)$, given all $ A,B,h$ are all random variables. Will the following rule works?

$$P[h \geq (A + B)] = P(h\geq A) +  P(h\geq B)$$

Actually, in one of my mathematical analysis, I end up with a complex expression which can be simplified to $P[h \geq (A + B)]$. I believe I can move forward if I can break it down somehow. Further explanations of variables are as below.

$h \sim \exp(\lambda')$ and $g \sim \exp(\lambda'')$

$A = a(1 + e^{sh})$, $B = bg(1+e^{sh})$

$a,b,s, \lambda', \lambda''$ are constants.

答: As has already been stated, your approach does not work. Here is an idea: Since $h,g$ are independent (this has not directly been stated in the question but I assume it's true), the following should hold: \begin{align*}\textbf{P}(h\geq A+B)&=\textbf{P}(h\geq(1+e^{sh})(a+bg))=\textbf{P}(h\geq(1+e^{sh})(a+bg))\\&=\textbf{P}(h/(1+e^{sh})\geq(a+bg))=\textbf{E}[\textbf{P}(h/(1+e^{sh})\geq(a+bg)|h)]\\&=\textbf{E}[F_g((h/(1-e^{sh})-a)/b)]\end{align*} where $F_g$ is the CDF of $g$. This is the case if $b>0$. Cases $b=0$ and $b<0$ follow analogously.


问:
Convergence in distribution of fraction of Brownian increments
I would like to prove that the following random variables converge in distribution:
$X_h(t)=\frac{1}{h}(B(t+h)-B(t))$ where $(B(t):t\geq0)$ is standard Brownian Motion.

In my attempts at this I used the definition that:
$\lim_{h \to +\infty}\mathop{\mathbb{E}}(f(X_h))=\mathop{\mathbb{E}}(f(X))$ for all continuous bounded functions $f$.

My thoughts so far are that the process $X$ will be equal to 0 (though I haven't proved pointwise convergence either) so $\mathop{\mathbb{E}}(f(X))$ will just be a constant value.

I can also see that $X_h(t)\stackrel{}{\sim}N(0,\frac{1}{h})$. Would it then suffice to show that $F_h(t)\rightarrow F(t)$ where $F_h(t)$ is the cumulative distribution function for $X_h$ and $F(t)=0$? If so, how is this done?

EDIT: Corrected the definition of the process I am looking at, previously was $X_h(t)=\frac{1}{h}(B(t+h)-B(h))$

答: $EX^{2}_h(t)=\frac 1 {h^{2}} var (B_{t+h}-B_h)=\frac t {h^{2}} \to 0$ as $ h \to \infty$. So  $X_h(t) \to 0$ in mean square. This is much stronger than convergence in distribution. (It implies convergence in probability).


答: Any homeomorphism that isn't equal to the identity or the map $x\mapsto 1-x$ on the space $[0,1]$ with the Lebesgue measure will not preserve measure I think.</p>

Title: Measure Not Preserving Homeomorphism

Body:
Are there homeomorphisms that do not preserve measure (Let us consider Lebesgue Measure)? I ask this because, I currently cannot think one but while googling, I get Measure Preserving Homeomorphisms. So this specification is made perhaps because there are homeomorphisms that do not preserve measure?

P.S. I am fairly beginner in this topic.


答:
If in addition $g(x) + 3g(1/x) = x^2$ then $$f(x) - g(x) + 3f(1/x) - 3g(1/x) = 0.$$ Define $h = f-g$ so that
$$h(x) = -3 h(1/x).$$ Now swap $x \mapsto 1/x$ to find  $$h(1/x) = -3h(x)$$ so that $$h(x) = 9 h(x).$$ This forces $h = 0$ and thus $f=g$.

Title: Proof that $\frac{3-x^{4}}{8x^{2}}$ is the only solution to the functional equation $f(x) + 3f\left(\frac{1}{x}\right) = x^{2}$

Body:
How can we prove that $f(x) + 3f\left(\frac{1}{x}\right) = x^{2}$ has only one solution and no other solution and that solution is $\frac{3-x^{4}}{8x^{2}}$.

I know that this is a linear function equation, but can't find any sources which say that all linear functional equation will have only one solution.


问:
Range of adoint operator
We consider infinite dimension.

$X,Y$: Banach Spaces
$T:X→Y$ is a bounded linear operator.

I want to prove
$(\ker\, T)^\bot = \overline {R(T^*)}$.

$(\ker\, T)^\bot = \{f\in X^*|f(x)=0\ (x\in \ker\,T) \}$

But I can't.
If you have an answer, please tell me hints.
I'm sorry my broken English,I'm Japanese.

答:
Counterexample. Let $\lambda\in c_0$ with non zero entries. Consider diagonal operators
$$
T:\ell_1\to\ell_1:x\mapsto\lambda x\\
S:\ell_\infty\to\ell_\infty:x\mapsto\lambda x
$$
This is straightforward to check that $S=T^*$.
Obviously $\operatorname{Ker}(T)^\perp=\{0\}^\perp=\ell_1^*=\ell_\infty$. Since $\lambda\in c_0$ then $\operatorname{Im}(S)\subset c_0$. So
$$
\overline{\operatorname{Im}(T^*)}=\overline{\operatorname{Im}(S)}\subset \overline{c_0}=c_0\subsetneq \ell_\infty=\operatorname{Ker}(T)^\perp
$$

Fact.
$$
\overline{\operatorname{Im}(T^*)}=\operatorname{Ker}(T)^\perp
\Longleftrightarrow \operatorname{Im}(T)\text{ is closed}
$$

Title: Range of adoint operator


问:
$f(x) \in F[x]$ is irreducible over $F$ if $f(x) = g(x)h(x)$ implies that $g(x) \in F$ or $h(x) \in F$ where $F$ is a field.

$f(x) \in F[x]$ is irreducible over $F$ if $f(x) = g(x)h(x)$ implies that $g(x) \in F$ or $h(x) \in F$ where $F$ is a field.

If $g(x) \in F$ or $h(x) \in F$, then doesn't that make $f(x)$ reducible since it has factors in $F$?

答: No, because being reducible (if $f(x)\ne0$) means that you can factor it as a product of non-invertible elements of $F[x]$.


问:
Order the basis of a Lie algebra so that multiplication is increasing?
Let $\frak{g}$ be an arbitrary Lie algebra over a field $K$ with a fixed basis $e_1,\ldots,e_m$. Is it possible to impose a total order $e_1<\ldots<e_m$, so that
$$\forall e_i\!<\!e_j\: \forall e_k\!\in\![e_i,e_j]: e_i<e_j\leq e_k?\tag{1}$$
In other words, is it always possible to permute the basis elements of $\frak{g}$ so that whenever $[e_i,e_j]=\alpha_1e_{k_1}\!+\!\ldots\!+\!\alpha_re_{k_r}$, where $\alpha_1,\ldots,\alpha_1\in K$ are the structure constants, there holds $e_j\leq e_{k_1},\ldots,e_{k_r}$?

If not, I'd appreciate a counterexample. If yes, a procedure to find this order is desired.

Remarks

The condition I wish to satisfy reminds me of $\mathbb{Z}$-graded Lie algebras: $\mathfrak{g}=\bigoplus_{i\in\mathbb{Z}}\mathfrak{g}_i$ such that $[\mathfrak{g}_i,\mathfrak{g}_j]\subseteq \mathfrak{g}_{i+j}$. Indeed, if $\mathfrak{g}$ admits a $\mathbb{N}$-grading and $[e_i,e_j]=0$ for all $e_i\neq e_j\in\mathfrak{g}_0$, then if we impose any order on each $\mathfrak{g}_i$ and let $\mathfrak{g}_0<\mathfrak{g}_1<\mathfrak{g}_2<\ldots$, the property (1) is satisfied. Thus the Lie algebras $\mathfrak{sol}_n=\langle e_{uv}; 1\!\leq\!u\!\leq\!v\!\leq\!n\rangle$ and $\mathfrak{nil}_n=\langle e_{uv}; 1\!\leq\!u\!<\!v\!\leq\!n\rangle$ of all (strictly) upper triangular matrices, with grading $\deg e_{uv}=v-u$, admit $(1)$.

答:
The Lie algebra $\mathfrak{gl}_2=\langle e_{11},e_{12},e_{21},e_{22}\rangle$ is a counterexample. If $e_{12}<e_{21}$, then by multiplication, $e_{12}<e_{21}<e_{11},e_{22}$, but then $[e_{12},e_{22}]=e_{22}$ does not satisfy (1). If $e_{21}<e_{12}$, then by multiplication, $e_{21}<e_{12}<e_{11},e_{22}$, but then $[e_{21},e_{11}]=e_{21}$ does not satisfy (1).

Using @darijgrinberg's comments, I've found on Wiki, that $\frak{g}$ is solvable iff there is a finite sequence of subalgebras $\mathfrak{a}_{i}$ of $\mathfrak{g}$, such that:
$$\mathfrak{g}=\mathfrak{a}_0\supset \mathfrak{a}_1\supset ...\supset \mathfrak{a}_m=0,\quad \dim \mathfrak{a}_i/\mathfrak{a}_{i+1}=1,\;\; \mathfrak{a}_{i+1}\text{ is an ideal in }\mathfrak {a}_{i},\;\;\forall i.$$
This corresponds precisely to my condition (1): $e_i$ is the basis element of $\mathfrak{g}_i/\mathfrak{g}_{i+1}$.


问:
$100$ people have $100$ one-dollar bills. Some give bills to others until all have different amounts. What's the least number of people to give money?

In a group of $100$ people, each of them has a wallet with $100$ one-dollar bills. Some of them gave one or more dollars to one or more of the others and eventually everyone ended up with a different amount $>0$. What is the least number of people who gave some of their money?

I have started by trying to make $10,000$ as a sum of different integers. One way is $4+6+7+\cdots+141$, which means that $95$ people have given money to others but I don’t think this is the correct approach.

Also tried taking one person's amount and sharing it to others, so for example, $4+5+\cdots+14 = 99$ but again this will take for ever and we won't know if it's the optimal.

Can anybody help me out? Thank you!

答:
For everyone to have a different amount of money, everyone (except perhaps one person) has to either give or receive some money. Without loss of generality we may assume that nobody is both receiving and giving money; if person $A$ receives money from person $B$ and gives money to person $C$, then we get the same result if person $B$ gives money directly to person $C$.

To minimize the number of people giving money, we must maximize the number of people not giving money. Every pair of people not giving must receive a different amount of money. To minimize the amount of money given, the first receiver receives $\$0$, the next receiver receives $\$1$, etcetera, until everyone not giving has received money (possibly $\$0$). The total amount received is then
$$0+1+2+\ldots+(n-1),$$
where $n$ is the number of people not giving money. To minimize the number of people giving this amount of money, the first giver gives $\$99$, the next giver gives $\$98$, etcetera, until there is sufficient money to give to everyone.

So now you want to find the maximal positive integer $n$ such that
$$0+1+2+3+\ldots+(n-1)\leq 99+98+97+\ldots+n.$$
The number of people giving money is then $100-n$.


问:
Two aspects of randomness
Consider a random sequence of integers

1, 4, 3, 8, 2, 5, 3, 8 ...

The only sufficient condition for the sequence to be random is its unpredictability ie. probability of any number coming next must be equal to $\frac{1}{10}$.

Now consider that we are getting only numbers less than 5 in the sequence, it then implies that for the sequence to be random the probability of getting numbers greater than 5 is now more, this does not follow the randomness criteria as numbers are now in some form more predictable.

Do the two aspects of randomness contradict with each other?

Or am I wrong somewhere in this deductive thinking?

答:
The only sufficient condition for the sequence to be random is its
unpredictability

"Randomness" in the context of probability models†, unlike in ordinary speech, has in a sense a contrary meaning to "haphazard and unpredictable".

After all, a stochastic process (which describes a random phenomenon) evolving in space or time does have an underlying structure, and so can be used for forecasting: while individual trials are unpredictable, by the law of large numbers, the process's underlying pattern (the relative frequencies of different outcomes) emerges over numerous trials.

probability of any number coming next must be equal to $\frac1{10}.$

This process, as in classical probability, involves equally-likely outcomes: as such, the next digit is a uniformly—not haphazardly—distributed random variable.

consider that we are getting only numbers less than 5 in the sequence,
it then implies that for the sequence to be random the probability of
getting numbers greater than 5 is now more

This is the Gambler's Fallacy—based on the false implicit assumption that small samples are representative of the larger population—a misapplication of the Law of Large Numbers.

† To what extent probability models correspond to the actual workings of reality or actual randomness—and to what degree this question is simplistic—is a different, philosophical discussion

Title: Two aspects of randomness


问:
Solve $(3x^2y^4 +2xy)dx + (2x^3y^3 - x^2)dy=0$

Solve $(3x^2y^4 +2xy)dx + (2x^3y^3 - x^2)dy$

This is not one of the standard forms and neither is it an exact form. How do I go about doing this question?

答:
$$
(3x^2y^4 + 2xy)dx + (2x^3y^3 - x^2)dy = 0
$$

$$
\implies 3x^2y^4dx + 2x^3y^3dy + 2xydx - x^2dy = 0 \tag{$1$}
$$

We know that:

$$
d\left(\frac{x^2}{y}\right) = \frac{2xydx - x^2dy}{y^2}
$$

Substituting this back into $(1)$:

$$
\implies 3x^2y^4dx + 2x^3y^3dy + d\left(\frac{x^2}{y}\right)y^2 = 0
$$

Assuming $y \not= 0$:

$$
\implies 3x^2y^2dx + 2x^3ydy + d\left(\frac{x^2}{y}\right) = 0 \tag{$2$}
$$

Also:

$$
d(x^3y^2) = 3x^2y^2dx + 2x^3ydy
$$

Substituting this into $(2)$:

$$
\implies d(x^3y^2) + d\left(\frac{x^2}{y}\right) = 0\\
\implies x^3y^2 + \frac{x^2}{y} = c
$$

I don't know if this is formally legal in mathematics (I only have high school mathematics under my belt), but this is how I would have done it.


问:
Finding roots $x_1$, $x_2$, $x_3$ of $x^3+ax+a=0$ (for real $a\neq 0$) if $\sum_{cyc}\frac{x_1^2}{x_2}=-8$
This question is taken from AMTI 1994:

The solutions $x_1$, $x_2$, $x_3$ of the equation $x^3+ax+a=0$, where $a\ne0$ is real, satisfy
$$\frac{x_1^2}{x_2}+\frac{x_2^2}{x_3}+\frac{x_3^2}{x_1}=-8$$
Find $x_1$, $x_2$, $x_3$.

I tried to solve this by using Vieta's relations and tried to factorize the the equation they gave, but wasn't able to something solid out of that factorization.

Body:
Hint: If $x$ is a root, $x^3=-a(x+1)$
$$\implies -8a = \sum_{cyc} x_1^3x_3 = -a\sum_{cyc} (x_1+1)x_3 = -a\left(\sum_{cyc} x_1x_3 + \sum_{cyc}x_3 \right)=a^2$$


问:
Having trouble understanding (intuitively) a combinatorics problem
I was given the following puzzle:

You manage a bank and you have 5 employees. With an infinite number of locks (and keys), you need to create a system that enables each triplet of workers open the vault of the bank, but a pair of employees should never be able to open together the vault.

Intuitively I immediately asked myself how many possible triplets of employees are there, and I know the answer from the combinations formula ($n!/r!(n-r)!$), which I also understand intuitively.

So we can represent all of our cases with this table:

https://upload.wikimedia.org/wikipedia/commons/6/65/Combinations_without_repetition%3B_5_choose_3.svg

We can say each case (or triplet) is represented by a row, and each worker is represented by a column. Obviously, when I see the table I immediately understand that the rows can also represent keys, but I can’t explain to myself in my head, logically, why this isomorphism (if I use the term correctly) is possible here. Perhaps I would have been able to get to this table myself, but it is not clear to me how (or why) the transformation from a table of triplets or cases, to a table of keys, occur. I try to imagine it somehow in my head but can’t. We can create 10 (unique) triplets (without repetition) from a set of 5, but why this necessarily implies that in order for each set of (only) three employees to be able to open a vault, we need 10 locks, and to give each employees 6 keys? Can you explain the isomorphism between the possible triplets and the number of locks, and the number of keys per employee?

I hope my question is making sense.

答:
Actually, the question should have been to find the minimum number of locks and keys, otherwise you can go on adding locks to which everyone holds a key !

So each group of two must be encountering one lock that they alone can't open, thus minimum number of locks needed = $^5C_2 = 10$

Also, each group must hold the missing key for each other group, and thus hold $4C_2$ = $6$ keys, totalling $5\cdot6$ = $30$ keys


问:
Calculate the convolution of $xe^{-x^2}$
$h=xe^{-x^2}$. Calculate the convolution $(h\ast h)(x)$

$\int^{\infty}_{-\infty}ye^{-y^2}(x-y)e^{-(x-y)^2}\,dy$

The answer is $\frac{1}{4}\sqrt{\frac{\pi}{2}}(x^2-1)e^{-x^2/2}$

Could someone give me some hints?

I know how to solve $\int^{\infty}_{-\infty}e^{-y^2}e^{-(x-y)^2}\,dy=\int^{\infty}_{-\infty}e^{-2(y-\frac{1}{2}x)^2-X^2/2}\,dy=e^{-X^2/2}\frac{1}{\sqrt{2}}\int^{\infty}_{-\infty}e^{-u^2}\,du=\sqrt{\frac{\pi}{2}}e^{-x^2/2}$

Above is another question of my homework.

答:
You have:

$\int^{\infty}_{-\infty}ye^{-y^2}(x-y)e^{-(x-y)^2}\,dy \underset{\text{adding and removing $\frac{x^2}{2}$ to the exponent of $e$}}{=} e^{-\frac{1}{2}x^2}\int^{\infty}_{-\infty}(yx-y^2)e^{-(\sqrt 2 y - \frac {x}{ \sqrt {2}})^2}dy \underset{\text{defining $z=\sqrt 2 y - \frac{x}{\sqrt 2}$}}{=} $

$=\sqrt 2 e^{-\frac{1}{2}x^2}\int^{\infty}_{-\infty}((\frac {z}{\sqrt 2}+\frac x 2)x-(\frac {z}{\sqrt 2}+\frac x 2)^2)e^{-z^2}dy=\sqrt 2 e^{-\frac{1}{2}x^2}[\frac {x^2}{ 4} \int^{\infty}_{-\infty}e^{-z^2}dz-\frac 1 2 \int^{\infty}_{-\infty} z^2 e^{-z^2}dz]$

that brings you to the result thanks to the fact that:

(i) $\int^{\infty}_{-\infty}e^{-z^2}dz=\sqrt{\pi}$

(ii) $\int^{\infty}_{-\infty}z^2e^{-z^2}dz=\sqrt{\frac{\pi}{2}}$


问:
If A1, A2, …, A10 are to speak and “either of A1, A2 should speak before A3, A4, A5”, how many orders are possible?
The constraint on the order is:

Either of A1, A2 should speak before A3, A4, A5.

A1, A2 > A3, A4, A5

I had a vague idea of doing it by selecting 5 members from 10. Out of 5, I selected 3 and permuted them and then selected 2 and permuted them. Then I permuted the remaining 5 members, which gives
$$
 {10 \choose 5} {5 \choose 3}\cdot 3! \cdot {2 \choose 2}\cdot2!\cdot5!
$$
But the answer isn’t this; rather, it is missing $5 \choose3$ and $2 \choose 2$.

But I believe that we are choosing specifically 2 and 3 people so these terms should be there.

答:
Considering either of $A1$ or $A2$ speaks before $A3, A4, A5$ (which means, it can be both as well).

Selection of $5$ speaking positions out of $10 = {10 \choose 5}$

Permuting $A6$ to $A10$ in remaining $5$ positions $= 5!$

Now of the chosen $5$ positions, the first position cannot go to $A3, A4, A5$. So that leads to $2$ possibilities in the first place - $A1$ or $A2$.

Remaining $4$ places can be permuted in $4!$ ways.

So total number of ways $=  {10 \choose 5} \times 5! \times 4! \times 2$.

I think this is the right interpretation of the question but in cases -

i) If only one of them has to be speaking before $A3, A4, A5$ then

Number of ways $=  {10 \choose 5} \times 5! \times 3! \times 6$. This is because the second position has to be out of $A3, A4, A5$ ($3$ ways). So $2$ possibilities for first position and $3$ for the second. Rest $3$ can be permuted in $3!$ ways.

ii) If both of them have to be speaking before $A3, A4, A5$ then

Number of ways $=  {10 \choose 5} \times 5! \times 3! \times 2!$


问:
Riemann-Stieltjes Integral
Define:

Partition $P = \{x_0,\dots,x_n\}$

$L(P) = \sum_{i=1}^n (\alpha(x_i)-\alpha(x_{i-1}) \inf (f(x): x \in [x_{i-1}, x_i])$

$U(P) = \sum_{i=1}^n (\alpha(x_i)-\alpha(x_{i-1}) \sup (f(x): x \in [x_{i-1}, x_i])$

$L(P) \le I \le U(P)$. I is the Riemann-Stieltjes Integral.

Using this definition, and assume f(x) is a continues function then:
1)  $\int_0^1 f(x)2^{-[x]} d(2^{[x]}) = 0.5f(1)$
2)  $\int_1^2 f(x) d(x[x+3]) = 4 \int_1^2 f(x) dx + 2f(2)$

My attempt  to 1)

$\sum_{i=1}^n  2^{[x_i]} – 2^{[x_{i-1}]} = 2^1 – 2^0 = 1$

$L(P) = 1 \cdot \inf (f(x)2^{[-x]}) = 1 \cdot 2^0f(0) = f(0)$

$U(P) = 1 \cdot \sup (f(x)2^{[-x]}) = 1 * 2^{-1}f(1) = 0.5f(1)$

Clearly my L(P) is wrong but I’m not sure how/why? And I’m completely stuck on the second question. Can someone please help me? I may be using the definition incorrect.
NB: I know nothing about the extreme value theorem etc. Basically no other knowledge apart from this definition! Lol

Question 2: If $f’(x)$ is continuous show
$2 \int_a^b f(x)d(f(x)[x]) = [f^2(x)[x]] (a,b) + \int_a^b f^2(x)d[x]$

My attempted solution:
Using integration by parts:
2(fx)f(x)[x] between (a,b) – int (a,b) f(x)[x]df(x)
= 2(fx)f(x)[x] between (a,b) – int (a,b) f(x)[x]f’(x)dx
Not sure how to get that second term!

Question 3: How do you start to calculate these:
$\int_1^n x^3 d([x]/x^2)$

$\int_1^\infty  d([\log x]/x^2)$

$\int_1^n x(x-1) d(1/[x])$

答:
I suppose that by $[x]$ you mean the floor function.

$\Delta(P)$ denotes the norm of partition, i.e., the maximal length of an interval of the partition $P$.

1) Using the fact that $2^{-[x]}$ is either 1 (for $x<1$) or $1/2$ (for $x=1$) you can see that for any partition such that $\Delta(P)\le\varepsilon$ you get $U(P)\le \frac12 \sup\{f(x); x\in [1-\varepsilon,1]\}$ and $L(P)\ge \frac12 \inf\{f(x); x\in (1-\varepsilon,1)\}$. Both quantities tend to  $\frac{f(1)}2$ as $\varepsilon\to 0^+$. (Using continuity of $f$.)

2) By $x[x+3]=4x+x[x-1]$ (which holds for $x\in[1,2]$) you can express the function $x[x+3]$ as a sum of 2 non-decreasing functions.

Now  $\int_1^2 f(x) d(x[x+3]) = \int_1^2 f(x) d(4x) + \int_1^2 f(x) d(x[x-1])= 4\int_1^2 f(x) dx + 2f(2)$.

In the first integral we used $(4x)'=4$, the second one is similar to the first part.

In both cases we used R-S integral for a function having single jump. You can read more about it at this blog: Integrating Across a Jump. We have also used additivity in integrator, see another entry at the same blog.

Hint for question 2: Use the additivity in integrator again, in this way you can get two integrals, in one case the integrator is differentiable, in the second one it is a function having single jump. I guess question 3 can be handled in the similar way, although you'll have more jumps, not a single one.

Title: Riemann-Stieltjes Integral


问:
One 2 Class Example and multiple choice question answer different?
I select one example from my notes (this is multiple choice) as follows:

Question and Parameters attached to my solution is present as following image:

but when I plot these as the interval of $x$ is differ from above calculation, why?

and the answer option of multiple choice for this question is quite differ:

答:
You started working with the assumption that $ 0 \le x \le 1$.

Notice that $\frac{8-\sqrt{76}}2 < 0$ and $\frac{8 + \sqrt{76}}2 > 1$.

$$\left[ \frac{8-\sqrt{76}}{2}, \frac{8+\sqrt{76}}{2} \right] \cap [0,1]=[0,1]$$


问:
Equivalence of $\square_\kappa\Leftrightarrow\square'_\kappa$
In Devlin - Constructibility on page 158f. he defines $\square_\kappa$ and $\square_\kappa'$ and then proceeds to show that these two are equivalent.

$\square_\kappa$ holds if there is a sequence $(C_\alpha\mid\alpha<\kappa^+\wedge\alpha\in\text{Lim})$ s.t.

$C_\alpha$ is a club of $\alpha$,

$\operatorname{cf}(\alpha)<\kappa\implies|C_\alpha|<\kappa$ and

if $\beta$ is a limit point of $C_\alpha$, then $C_\beta=\beta\cap C_\alpha$.

$\square_\kappa'$ holds if there is a sequence $(B_\alpha\mid\alpha<\kappa^+\wedge\alpha\in\text{Lim})$ s.t.

$B_\alpha$ is a closed subset of $\text{Lim}\cap\alpha$,

$\operatorname{cf}(\alpha)>\omega\implies B_\alpha$ is unbounded in $\alpha$,

$\operatorname{otp}(B_\alpha)\leq\kappa$ and

$\beta\in B_\alpha\implies B_\beta=\beta\cap B_\alpha$

Assuming that $(B_\alpha\mid\alpha<\kappa^+\wedge\alpha\in\text{Lim})$ is a $\square_\kappa'$-sequence, he defines inductively

$C_\alpha=\bigcup\lbrace C_\gamma\mid \gamma\in B_\alpha\rbrace$, if $\sup B_\alpha=\alpha$

$C_\alpha=\bigcup\lbrace C_\gamma\mid \gamma\in B_\alpha\rbrace\cup\lbrace\theta_n^\alpha\mid n<\omega\rbrace$ else, where the $\theta_n^\alpha$ are strictly increasing, bigger than $\sup B_\alpha$, and cofinal in $\alpha$.

Then he says that if $\operatorname{otp}(C_\alpha)>\kappa$ then $\operatorname{otp}(B_\alpha)>\kappa$ just because $B_\alpha$ is the set of all limit points of $C_\alpha$ below $\alpha$ and because $\kappa$ is uncountable. Why is $\operatorname{otp}(C_\alpha)=\kappa+\omega$ and $\operatorname{otp}(B_\alpha)=\kappa$ not possible? Maybe i'm just being dumb here.

Maybe its useful to state the rest of the proof since it's also not that long anymore.

If $\kappa$ is regular we need that $\operatorname{otp}(B_\alpha)\leq\kappa$, because we then can assume that it's equal and have a contradiction, since then, if $\operatorname{cf}(\alpha)<\kappa$ then $\operatorname{cf}(\kappa)\leq\operatorname{cf}(\alpha)<\kappa$.
(So maybe in this case we use

3'. $\operatorname{otp}(B_\alpha)\leq\kappa$ and if $\operatorname{cf}
(\alpha)=\omega$ then $\operatorname{otp}(B_\alpha)<\kappa$

instead of 3. in the first definition.)

In the singular case, we define a sequence $(C_\alpha'\mid\alpha<\kappa^+)$ as follows.

Let $(\theta_\nu\mid\nu<\operatorname{cf}(\kappa))$ be a strictly increasing, continuous sequence sequence of limit ordinals, which is cofinal in $\kappa$.
Set $\theta_{\operatorname{cf}(\kappa)}=\kappa$. (And i think one could set $\theta_{\operatorname{cf}(\kappa)+n}=\kappa+n$ for all $n<\omega$ and $\theta_{\operatorname{cf}(\kappa)+\omega}=\kappa+\omega$)
Now let

$C_\alpha'=\lbrace\gamma\in C_\alpha\mid\operatorname{otp}(C_\alpha\cap\gamma)\geq\theta_\nu\rbrace$,

if there is a $\nu<\operatorname{cf}(\kappa)$ s.t. $\theta_\nu<\operatorname{otp}(C_\alpha)\leq\theta_{\nu+1}$. Otherwise $\operatorname{otp(C_\alpha)}=\theta_\nu$ for a limit $\nu$, then we set

$C_\alpha'=\lbrace\gamma\in C_\alpha\mid(\exists\tau<\nu)\operatorname{otp}(C_\alpha\cap\gamma)=\theta_\tau\rbrace$

Then $(C_\alpha'\mid\alpha<\kappa^+)$ is a $\square_\kappa'$-sequence (and i think this is still true even with $\theta_{\operatorname{cf}(\kappa)+n}$ and $\theta_{\operatorname{cf}(\kappa)+\omega}$ defined).

Since Devlin only uses $\square_\kappa'$ to show that $\square_\kappa$ holds in $L$, i think that it would be okay to change 3. from the definition of $\square_\kappa'$ to 3' in the regular case, because in that proof we get clubs $B_\alpha$ which have order type $\leq\kappa$ and if $\kappa$ is regular, then we should get 3'.

My question is, is everything stated here correct?

答: If $\operatorname{otp}(C_\alpha)=\kappa+\omega$, then $\operatorname{otp}(\operatorname{acc}(C_\alpha))=\kappa+1$ and $\operatorname{acc}(C_\alpha)=B_\alpha$. So i think what Devlin stated is true.


问:
Computation of a quotient group in the direct sum of three copies of Z
My question is simple: How to compute the quotient group
$$\frac{\langle(1,1,0),(0,0,1)\rangle}{\langle(1,1,-1),(1,1,1)\rangle}$$
of subgroups of $\mathbb Z \oplus \mathbb Z \oplus \mathbb Z$? The answer would be graet if each of the steps are explained.

This is my try: Letting $v=(1,1,0)$ and $w=(0,0,1)$, then the problem is to compute
$$\frac{\langle v,w\rangle}{\langle v+w, v-w \rangle}.$$
I wish this is isomorphic to the quotient
$$\frac{\mathbb Z \oplus \mathbb Z}{\langle (1,1),(1,-1) \rangle} \simeq \mathbb Z_2.$$

How to verify this argument? Or is there any other good argument? I want to know as many ways as possible. The answer may contain any undergraduate level mathematics.

Thank you.

答:
By definition of quotient:

\begin{alignat}{1}
\frac{\langle(1,1,0),(0,0,1)\rangle}{\langle(1,1,-1),(1,1,1)\rangle} &= \{(i,i,j)+\langle(1,1,-1),(1,1,1)\rangle\mid i,j\in\Bbb Z\} \\
&= \{(i,i,j)+\{(k+l,k+l,-k+l)\mid k,l\in \Bbb Z\}\mid i,j\in\Bbb Z\} \\
&= \{\{(i+k+l,i+k+l,j-k+l)\mid i,j,k,l\in\Bbb Z\}\} \\
&= \{\{(m,m,j-m+2l+i)\mid i,j,l,m\in\Bbb Z\}\} \\
&= \{(m,m,j-m)+\{(0,0,2l+i)\mid i,l\in \Bbb Z\}\mid j,m\in\Bbb Z\} \\
&= \{(m,m,j-m)+\{(0,0,r)\mid r\in \Bbb Z\}\mid j,m\in\Bbb Z\} \\
&= \{(m,m,j-m)+\langle(0,0,1)\rangle\mid j,m\in\Bbb Z\} \\
&= \frac{\langle(1,1,-1),(0,0,1)\rangle}{\langle(0,0,1)\rangle}
\end{alignat}

Now, $\varphi\colon \langle(1,1,-1),(0,0,1)\rangle\to \langle(1,1,0)\rangle$ defined by $(m,m,j-m)\mapsto (m,m,0)$ is a surjective homomorphism with kernel $\operatorname{ker}\varphi=\langle(0,0,1)\rangle$, whence (First Homomorphism Theorem):

$$\frac{\langle(1,1,-1),(0,0,1)\rangle}{\langle(0,0,1)\rangle}\cong \langle(1,1,0)\rangle$$

Finally, $\langle(1,1,0)\rangle\cong \Bbb Z_2$; in fact, consider the mapping $(m,m,0)\mapsto 0$ if $m$ is even and $(m,m,0)\mapsto 1$ if $m$ is odd.

(Remark. I wasn't smart enough to use the FHT directly on the first quotient, but it's certainly possible.)


问:
Prove intersection is ${0_v}$
I asking this question because I'm not sure how to prove it exactly.

Let $B$ be a basis of vector space $V$ , $B=B_1∪B_2$. so $U_1=span{B_1}$ and $U_2=span{B_2}$.

I need to show that $U_1∩U_2 = 0_v$.
I assume there is a vector $v∈U_1∩U_2$ such that $v≠0_v$.
because $v∈U_1∩U_2$ then $v∈U_1 ∧ v∈U_2$.
$\implies$ $v$ can be written as linear equation of $U_1$ and $U_2$. $\implies$ $v=c_1v_1+...+c_nv_n = c_1v_1+...+c_mv_m$
when $U_1=span(v_1,...,v_n) , U_2=span({v_1,...,v_m})$
which means we have a common vector between $U_1$ and $U_2$ so $B$ is lineary dependent which contradicts the fact that it isn't (We can take the scalars to be equal to $v$ and $-v$ and then not all scalars are 0 $\implies$ not independent).

答: That is not correct. You wrote that $U_1=U_2=\operatorname{span}(v_1,\ldots,v_n)$, which is not true. The set $U_1$ is spanned by some vectors $v_{n_1},v_{n_2}\ldots,v_{n_k}$, with $n_1,\ldots,n_k\in\{1,2,\ldots,n\}$ and $n_1<\cdots<n_k$. And the set $U_2$ is spanned by $v_{m_1},\ldots,v_{m_l}$, with$$\{m_1,\ldots,m_l\}=\{1,2,\ldots,n\}\setminus\{{n_1},{n_2}\ldots,{n_k}\}.$$So, if$$v=\alpha_{n_1}v_{n_1}+\cdots+\alpha_{n_k}v_{n_k}=\alpha_{m_1}v_{m_1}+\cdots+\alpha_{m_k}v_{m_k}$$and $v\ne0$, then not all $\alpha$'s are $0$, but$$0=v-v=\alpha_{n_1}v_{n_1}+\cdots+\alpha_{n_k}v_{n_k}-\alpha_{m_1}v_{m_1}-\cdots-\alpha_{m_k}v_{m_k},$$which is impossible, since $\{v_1,\ldots,v_n\}$ is linearly independent.

Title: Prove intersection is ${0_v}$


问:
Solving mixed system of equalities and inequalities
I have a mixed system of equalities and inequalities, for example:

$x + 2y +3z = 10\\
2x + 4y + 10z = 20\\
4x + y + z < 10$

where unknown variables $x, y, z$ are all real-valued.

How can get a solution (or the range of all feasible solutions) for this system? I understand this problem can be converted into a linear constrained optimisation problem, but I do not understand how.

Please help

答:
Multiply the first equation with $2$, the eq.1 and eq. 2 read:

$2x + 4y +6z = 20\\
2x + 4y + 10z = 20\\$

Subtraction gives: $4z=0$ hence $z=0.$

From the first eq. we now derive $x=10-2y.$ Thus the Third eq. becomes: $4(10-2y)+y<1=$ or $y > \frac{30}{7}.$

Consequence: the set of solutions is given by:

$$ \{(10-2y, y,0): y > \frac{30}{7}\}.$$


问:
Abelian groups related to $(\mathbb{Z}/p\mathbb{Z})^*$, with order close to $p$ -- analogous to elliptic curve groups, but simpler
I'm looking for a construction of a group $H$ that is "a sister" to $(\mathbb{Z}/p\mathbb{Z})^*$, in the following rough sense:

Each element of $H$ can be represented by one or a few elements of $\mathbb{Z}/p\mathbb{Z}$ (possibly together with some auxiliary information), and $H$'s group operations (multiplication, inverse, pick a random element of $H$) can be computed using a few field operations from $\mathbb{Z}/p\mathbb{Z}^*$.  Computing the group operations in $H$ should not require knowledge of $p$ directly, merely the ability to work in $\mathbb{Z}/p\mathbb{Z}$.

$H$ is an abelian group whose order is close to $p$, or not too much larger than $P$.

$H$ is elementary and can be described relatively simply.  (It should not need fancy machinery or a huge buildup.)

Here $p$ is a prime.

Maybe this is easier to explain with a few examples:

$(\mathbb{Z}/p\mathbb{Z})^*$ satisfies all these requirements.  Obviously, you can compute group operations directly.  Its order is $p-1$, and it is simple to explain.

An elliptic curve $H=\mathbb{E}(\mathbb{Z}/p\mathbb{Z})$ defined by some equation $y^2 = x^3 + ax + b$ (all modulo $p$) is another example.  Each element of $H$ can be represented as a pair of elements of $\mathbb{Z}/p\mathbb{Z}$.  The group law can be computed using a few operations on $\mathbb{Z}/p\mathbb{Z}$.  The group order is a number near $p$. Unfortunately, this is not elementary: it requires a bunch of setup to learn about elliptic curves.

I think there might be another group $H$ based upon Lucas sequences, defined by $V_0=2$, $V_1=A$, $V_j=AV_{j-1} - V{j-2}$ (all taken modulo $p$) and where $A$ is chosen so that the Jacobi symbol $(A^2-4/p)$ is $-1$.  An element of $H$ is a value in the Lucas sequence.  We have the relation $V_{2n} = V_n^2 - 2$, so then I think it might be possible to define the group operation of $H$ so that $h^2$ can be computed from $h$ in the way that $V_{2n}$ can be computed from $V_n$ (the one thing I don't know is how to define the multiplication operation).  I have a suspicion this might lead to a group of order $p+1$, but I'm not sure.  This is not super-elementary, though, and I'm not sure whether it is possible to fill in the details to make this work -- maybe you can fix this idea up.

Does anyone know of any other examples for sister groups $H$? (Preferably ones where the group order is something other than $p-1$, as I already know an examples for that.)

[Motivation: each group $H$ has a chance of leading to a factoring method.  For instance, the first example leads to the $p-1$ factoring algorithm, while the second example leads to the elliptic curve (ECM) factoring algorithm.  I was hoping to find some simpler, more elementary examples than elliptic curves as a way of introducing the conceptual idea behind the ECM factoring algorithm, without requiring to explain all of the machinery behind the theory of elliptic curves.  Thus my request for a group $H$ that requires less mathematical background to understand.]

答:
There is a rather general theorem saying that affine algebraic groups over the finite field with $p$ elements are all built from groups of orders $p-1$, $p$ or $p+1$. The case of order $p-1$ is the classical residue class group, that of order $p$ the additive group, and that of order $p+1$ a group constructed from finite fields with $p^2$ elements, Lucase sequences, or conics. These lecture notes, for example, explain how to factor integers with the unit circle.

If you want groups different from that, you have to consider projective algebraic groups, that is, abelian varieties, the simplest case of which are given by elliptic curves.


问:
Let $x $ be a solution of $(E) $. Show that $x $ is Lipschitz continuous?
We consider differential equation
$$
(E):\qquad x'(t)=\cos(2\pi(x(t)-t))
$$

Let $x $ be  a solution of $(E) $. Show that $x $ is Lipschitz continuous.

An idea please.

答: A ($C^1$-derivable) function is $k$-Lipschitz continuous iff the absolute value of the derivative is majored by $k$ (ie, its normal derivative is bounded by $-k$ and $k$). Here, $x'(t)$ is a $cos$ of something, so its values are necessarily between $-1$ and $1$. This means $x(t)$ is $k$-Lipschitz continuous for $k = 1$ and above. There might be a better bound $k$ to find, though.


问:
Prove by induction that $3^{4n+2}+1$ is divisible by $5$ when $n \ge 0.$

Prove by induction that $3^{4n+2}+1$ is divisible by $5$ when $n \ge 0.$

(1) When $n=0$ we have that $3^2+1 = 10$ which is divisible by $5$ clearly.

(2) Assuming that the condition hold for $n=k.$

(3) Proving that it holds for $n=k+1$

$$3^{4(k+1) + 2} + 1 = 3^{4k + 6} + 1 = 3^4 \cdot 3^{4k+2} + 1$$

Since we assumed that $5 \mid 3^{4k+2} + 1$ we have that

$$3^4 \cdot 3^{4k+2} + 1 = 3^4 \cdot 5t, \text{ where $t \in \Bbb Z$}.$$

Thus $5 \mid 3^{4(k+1) + 2} + 1$.

Is this a valid proof? I'm not entirely sure I'm correct with this...

答:
An alternative strategy for this kind of problem is to consider the difference between consecutive terms.

Let $f(n)=3^{4n+2}+1$. Then $f(n+1)-f(n)=80 \cdot 3^{4 n + 2}$ is a multiple of $5$.

The claim follows by induction since $f(0)=10$ is a multiple of $5$.

(Actually, this proves that  $f(n)$ is always a multiple of $10$.)


问:
Simple PDE classification question
Benjamin-Bona-Mahony equation:
$$\displaystyle u_t+u_x+uu_x-u_{xxt}=0$$
both the paper I was reading and wikipedia claimed that it is nonlinear. It has been some time since I studied classification so I wasn't sure it is nonlinear because of $uu_x$ or $u_{xxt}$ or both.

So I looked it up and found that
"... semilinear equations are ones in which the coeﬃcients of
the terms involving the highest-order derivatives of u depend only on x, not on u or its
derivatives."
And claims that $$\displaystyle u_t+u_{xxx}+u u_{x}=0$$
is semiliner.

So it seems to me that if BBM is indeed nonlinear, it is not because of $uu_{x}$, but because of $u_{xxt}$, but that feels unnatural for me.

Please shed some light on me.

答:
Consider first order PDE depending on two independent variables. It is

linear, if it has the form
$$
a(x,y)\partial_x u+b(x,y)\partial_yu=c(x,y)u+f(x,y),
$$
Example:
$$
\partial_xu+\partial_yu=0.
$$

semi-linear if it has the form
$$
a(x,y)\partial_x u+b(x,y)\partial_yu=c(x,y,u),
$$
Example:
$$
\partial_xu+\partial_yu=u^2
$$

quasi-linear if it has the form
$$
a(x,y,u)\partial_x u+b(x,y,u)\partial_yu=c(x,y,u),
$$
Example:
$$
\partial_xu+u\partial_yu=0
$$

fully non-linear if it has the form
$$
F(x,y,u,\partial_xu,\partial_yu)=0.
$$
Example:
$$
(\partial_xu)^2+(\partial_yu)^2=V(x,y).
$$

Note that semi-linear, quasi-linear, and fully nonlinear equations are nonlinear.

Can you generalize this classification to equations of any order?

Title: Simple PDE classification question


问:
Skolem hulls in arbitrary models of some fragments of ZFC
So I am trying to learn a little bit about iteration trees, and have decided to read a note by Steel, found here. On page 2, just after exercise 3, he uses the Hull "operator" in a way which I can't make sense of. To give some context:

Let $M\models \mathsf{ZFC}^-$, $\mathsf{ZFC}^-$ is $\mathsf{ZFC}$-$\mathsf{Powerset}$, $j:M\rightarrow N$ be elementary with $\kappa = \operatorname{crit}(j)$ and $\kappa \in \operatorname{wfp}(N)$(the well-founded part of $N$) and let $U_j$ be the usual ultrafilter derived from $j$ and let $k:\operatorname{Ult}(M, U_j)\rightarrow N$ be the factor map.

Now it is claimed that $\operatorname{Ult}(M, U_j)$ is isomorphic to $\operatorname{Hull}^N(\operatorname{ran}(j) \cup \{\kappa\})$ via $k$. What I don't understand is how do we construct the hull for general $N$ and what does it mean? In the case that we have a definable well-ordering of $N$, we can define something like "canonical" Skolem terms, just as we do for $L$. But in general, I don't see how this can be done.

And also another problem is a canonicity problem. Say we have managed to define the hull for general $N$ using some Skolem functions. But then it is claimed that the isomorphism is witnessed via $k$, and since different Skolem functions may give different hulls, we need a canonical way of finding such Skolem terms/functions. Which complicates the matter further for me.

答:
You do not need much to recover the full ultrapower. In fact, the $\Sigma_1$-weak Skolem hull should suffice, where the latter is defined by using not all Skolem functions but only those for $\Sigma_1$-formulas, and not even that, but only those functions defined as follows: given a $\Sigma_1$ formula $\varphi(t,y_1,\dots,y_n)$, let $f_\varphi:{}^nN\to N$ be the map given by $f_\varphi(a_1,\dots,a_n)=\emptyset$, unless there is a unique $b\in N$ such that $$N\models``\mbox{$b$ is the unique $t$ such that }\varphi(t,\vec a)\mbox{"},$$ in which case $f_\varphi(a_1,\dots,a_n)=b$. (Note that, even if $N$ is a proper class, we only need a partial satisfaction predicate to define this hull, and our theory should be enough to verify that $N$ has such a predicate.)

The point is that there is a very simple way of recovering every element of the ultrapower from the range of the embedding and its critical point.


问:
For which value of K does this process have zero drift?
The question is as follows:

For what value of $k \in R $, does the process
$$ Y_t = (W_t + t)e^{-W_t+kt}, t\geq0$$
have zero drift?

I've tried solving this using Ito's Lemma 1 and substituting coefficient of $dt$ (which is the drift) to be equal to zero but that just makes the right side $ln(0)$. Any help or suggestion would be appreciated. Thanks!

Computation:

According to Ito’s Lemma, we know

$$\ dF = \frac{\partial F}{\partial W_t}dW_t
      + \frac{1}{2}\frac{\partial^2 F}{\partial W_t^2}dt$$

Here, the coefficient of dt is the drift so as stated in the question $\frac{1}{2}\frac{\partial^2 F}{\partial W_t^2} = 0$.

$\frac{\partial^2 F}{\partial W_t^2} = (W_t+t-2)e^{kt-W_t}$

When I equate $  \frac{1}{2}(W_t+t-2)e^{kt-W_t} = 0$, there is no way to derive the equation in terms of k as $ln(0)$ is undefined.

答:
The formula which you stated holds for functions $F$ of the form $F(W_t)$. Here, however, we are dealing with a function of the form $F(t,W_t)$, and then Itô's lemma looks slightly different:

$$dF = \frac{\partial F}{\partial x} dW_t + \left(\frac{1}{2} \frac{\partial^2 F}{\partial x^2}  + \frac{\partial}{\partial t} F \right) \, dt. \tag{1}$$

As $$F(t,x) = (t+x) e^{-x+kt}$$ implies that \begin{align*} \partial_x F(t,x) &= (1-t-x) e^{-x+kt} \\ \partial_x^2 F(t,x) &= (-2+t+x) e^{-x+kt} \\ \partial_t F(t,x) &= e^{-x+kt} (1+k(t+x)) \end{align*} we see that the drift term in (1) equals zero if $$\frac{1}{2} e^{-W_t+kt} (-2+t+W_t) + e^{-W_t+kt} (1+kt+kW_t)=0, $$ i.e. if $$ \frac{t}{2} + kt + \frac{W_t}{2} + kW_t=0.$$ Consequently, the proper choice for $k$ is $k=-\frac{1}{2}$.


问:
example of a particular homogeneous topological space
I encountered a problem a few days ago: what is an example of a homogeneous $T_0$ space which is not $T_1$? I tried to solve this using some symmetrical properties of separation axioms which I saw before, but I couldn't. I also find this on web, but I can't prove that the so called space in example 4.4 is homogeneous (the article says that it is an example of what I'm searching for.) because I have some problems taking the greater element to the smaller, applying order topology on an ordered set.

Any help would be appreciated!

答:
Let $X$ be the set $(0,\infty)$ of non-negative real numbers and give it the left-order topology. Thus it has a base of open sets given by $\{(0,x)\mid x\in(0,\infty)\}$.

Then $X$ is $T_0$, since if $x<y\in X$, then $x\in (0,z)\not \in y$ whenever $x<z<y$. On the other hand $X$ is not $T_1$, since for any $x\in X$ we have $\overline{\{x\}}][x,\infty)$.

If $\lambda>0$ is any real number, then the map $f_\lambda:X\rightarrow X$, $x\mapsto \lambda\cdot x$, is continuous. It has a continuous inverse given by $f_{1/\lambda}$. To see that $X$ is homogeneous consider $x\neq y\in X$ and take $f_\lambda$ with $\lambda=x/y$.

Edit: In fact a slightly stronger conclusion is true: the multiplication $X\times X\rightarrow X$, $(x,y)\mapsto x\cdot y$, is a continuous group operation. The inversion, however, fails to be continuous. In any case $X$ is an example of a first-countable $T_0$ paratopological group which fails to be $T_1$.


问:
length of composition of module homomorphisms

Let $R$ be a commutative ring. Let $M_1,M_2,M_3$ be indecomposable $R$ - modules such that $l(M_1),l(M_2), l(M_3) \geq s$, where $l$ denotes the length of a module.
Let $f_1: M_1 \rightarrow M_2$, $f_2: M_2 \rightarrow M_3$ be either injective oder surjective module homomorphisms.

Then $l($im$(f_2f_1))\geq s$.

My ideas:

Since $f_1,f_2$ are either injective or surjective (but not bijective), it follows from the additivity from $l$, that it must hold $l(M_1) \neq l(M_2)$ and $l(M_2) \neq l(M_3)$.

If for $i \in \{ 1,2\}$ $f_i$ is injective $\Rightarrow l($im$(f_i))=l(M_i) \geq s$.
If for $i \in \{ 1,2\}$ $f_i$ is surjective $\Rightarrow l($im$(f_i))=l(M_{i+1}) \geq s$

Assuming that $f_1$ is surjective:
im$(f_2f_1)=f_2(f_1(M_1))=f_2(M_2)=$im$(f_2)$ and therefore im$(f_2f_1) \geq s$

I  don't know how to proceed in the case that $f_1$ is injective. Can someone help?

答:
If you allow $f_1$ to be injective and $f_2$ to be surjective, then the result is false. For example, take $R=k[t]/(t^2)$ with $k$ a field, $M_1=k=M_3$ and $M_2=R$. Then $s=1$ and we have a short exact sequence
$$ 0 \to k \to R \to k \to 0 $$
so the composition $f_2f_1:k\to k$ is zero.

If both are injective, then $s\leq l(M_1)\leq l(f_2f_1(M_1))$, and if both are surjective, then $l(f_2f_1(M_1))\geq l(M_3)\geq s$.


问:
Convergence Newton's Method for this system
I'd like to prove that the following non-linear system

$$
 F(x) =
 \begin{pmatrix}
  x_1^3 + x_2^3 - 4 \\
  x_1^3 - x_2^3 \\
 \end{pmatrix}
=
 \begin{pmatrix}
 0 \\
 0 \\
 \end{pmatrix}
$$
will converge when using the Newton's Method for the start vectors $x^{(0)} = (x_1^{(0)}, x_2^{(0)})^T \in [1,2] \times [1, 2] \subset \mathbb R^2$.

I'm trying to use this lemma I learned on class

Let $f: D \subset \mathbb R^n \rightarrow \mathbb R^n$ be continuously
  differentiable and let $D$ be convex as well open, $f'$ is the jacobian matrix. If $$q = \sup_{x\in
 D}||f'(x)|| < 1$$ then $f$ is a contraction mapping using the Norm $||
 \cdot ||$ with a contraction rate $q$.

I have calculated the Jacobian matrix, and it is as follows
$$
\begin{pmatrix}
  3x_1^2 & 3x_2^2 \\
  3x_1^2 & - 3x_2^2 \\
 \end{pmatrix}
$$
Using the maximum norm (sum of lines) for matrixes, I get:
$||F' (x_1, x_2)||_\infty = \max \{3x_1^2 + 3x_2^2, 3x_1^2 - 3x_2^2\} = 3x_1^2 + 3x_2^2$ which is for sure not $< 1$ in the desired interval. Where have I gone wrong? Thanks a lot guys!

答:
Suppose that $x$ is one vector root of $F(x)=0$, then
$$F(x_k)-F(x)=F'(x)(x_k-x)+(O(x_k-x)^2)$$, here $F'(x)$ is a Jocobian matrix.
$x_{k+1}=x_k-F(x_k)/F'(x_k)$

as for this problem $F'(x)$ is
$$
\begin{pmatrix}
  3x_1^2 & 3x_2^2 \\
  3x_1^2 & - 3x_2^2 \\
 \end{pmatrix}
$$

Using Newton's method: $x_{k+1}=x_k-F(x_k)/F'(x_k)$, then
$$x_{k+1}-x=x_k-x-(F(x_k)-F(x))/F'(x_k)=F^{-1}(x_k)(O(x_k-x)^2)$$. If $|F^{-1}(x_k)|\le M$, then newton's method will converge.

where you also need to think which root the formula will converge to, because sometime you can find several roots, at different initial condition will have different root's convergence.


问:
How to check whether a function $f : \mathbb{R}^2 \to \mathbb{R}$ is differentiable or not
It is not clear for me how to check whether a function $f : \mathbb{R}^2 \to \mathbb{R}$ is differentiable or not. For example, $f : \mathbb{R}^2 \setminus\{(0, 0)\} \to \mathbb{R}$ defined by $$f(x, y) = \frac{x}{\sqrt{x^2 + y^2}}.$$
Can someone help me, please?

答: The partial derivatives $f_x$ and $f_y$ of $f(x, y) = \frac{x}{\sqrt{x^2 + y^2}}$ exist and are continuous on $\mathbb{R}^2 \setminus\{(0, 0)\}.$ Thus $f$ is differentiable on $\mathbb{R}^2 \setminus\{(0, 0)\}.$


问:
Find the smallest subspace of a vector space
I am given the vector space of $\mathbb{R}^2$ and one of its element which is $(5,7)$. How can I determine the "smallest" subspace $A$ of $\mathbb{R}^2$, that contains $(5,7)$?

答:
The smallest subspace which contains $(5,7)$ is one-dimensional:

$$\{t(5,7): t \in \mathbb R\}.$$


问:
How to prove $ y_1, \dots , y_n$ generate module M？
Let $M$ be a finitely generated module over PID $R$, and $M$ is generated by $x_1, x_2, \dots, x_n$. Let $y_1=a_1\cdot x_1+\dots +a_n\cdot x_n$, and $\gcd(a_1, \dots , a_n)=1.$ Prove that there exist $y_2, \dots, y_n$ such that $y_1, \dots, y_n$ generate $M$.

Since $R$ is PID, there exist $u_1, \dots , u_n \in R$ such that $u_1a_1+ \dots +u_na_n=1$. I want to use this equation to prove it, but I have trouble constructing $y_2, \dots, y_n$. Any hint would be helpful！

答:
For convenience, I'll reword the problem as follows . . .

Let $M$ be an $R$-module where $R$ is a PID, let $x_1,...,x_n\in M$ with $n\ge 2$, and let $y_1=a_1x_1+\cdots+a_nx_n$ where $a_1,...,a_n\in R$ are such that $\text{gcd}(a_1,...,a_n)=1$.

Claim:$\;$There exist $y_2,...,y_n\in \langle{x_1,...,x_n}\rangle$ such that $\langle{y_1,...,y_n}\rangle=\langle{x_1,...,x_n}\rangle$.

Proof:

First suppose $n = 2$.

Let $a_1,a_2\in R$ be such that $\text{gcd}(a_1,a_2)=1$, and let $y_1=a_1x_1+a_2x_2$.

Since $\text{gcd}(a_1,a_2)=1$,  there exist  $u_1,u_2\in R$ such that $u_1a_1+u_2a_2=1$.

Let $y_2=-u_2x_1+u_1x_2$.

It's clear that $y_1,y_2\in \langle{x_1,x_2}\rangle$.

But identically we have
$$
\left\lbrace
\begin{align*}
u_1y_1-a_2y_2
&=
u_1(a_1x_1+a_2x_2)-a_2(-u_2x_1+u_1x_2)
\\[4pt]
&=
(u_1a_1+u_2a_2)x_1+(u_1a_2-u_1a_2)x_2
\\[4pt]
&=
(1)(x_1)+(0)(x_2)
\\[4pt]
&=
x_1
\\[10pt]
u_2y_1+a_1y_2
&=
u_2(a_1x_1+a_2x_2)+a_1(-u_2x_1+u_1x_2)
\\[4pt]
&=
(u_2a_1-u_2a_1)x_1+(u_1a_1+u_2a_2)x_2
\\[4pt]
&=
(0)(x_1)+(1)(x_2)
\\[4pt]
&=
x_2
\\[4pt]
\end{align*}
\right.
$$
hence $x_1,x_2\in \langle{y_1,y_2}\rangle$.

Thus $\langle{y_1,y_2}\rangle=\langle{x_1,x_2}\rangle$, so the claim holds for $n=2$.

Before continuing, we prove a simple lemma . . .

Lemma:

If $R$ is a UFD and $r_1,r_2\in R$, there exist $s_1,s_2\in R$ with $\text{gcd}(s_1,s_2)=1$ such that $r_1s_1+r_2s_2=0$.

Proof of the lemma:

If $r_1=0$ then letting $s_1=1$ and $s_2=0$ we have $\text{gcd}(s_1,s_2)=1$ and
$$
r_1s_1+r_2s_2
=
(0)(1)+(r_2)(0)
=
0+0
=
0
$$
Next suppose $r_1\ne 0$ and let
$$
\left\lbrace
\begin{align*}
s_1&=\frac{r_2}{d}\\[4pt]
s_2&=-\frac{r_1}{d}\\[4pt]
\end{align*}
\right.
$$
where $d=\text{gcd}(r_1,r_2)$.

Then we have $\text{gcd}(s_1,s_2)=1$ and
$$
r_1s_1+r_2s_2
=
(r_1)\left(\frac{r_2}{d}\right)+(r_2)\left(-\frac{r_1}{d}\right)
=
\frac{r_1r_2-r_1r_2}{d}
=
\frac{0}{d}
=
0
$$
which completes the proof of the lemma.

Returning to the proof of the main claim . . .

Proceeding by induction on $n$, let $n\ge 3$ and assume the claim holds for the case $n-1$.

Let $a_1,...,a_n\in R$ be such that $\text{gcd}(a_1,...,a_n)=1$, and let $y_1=\sum_{k=1}^na_kx_k$.

Since $\text{gcd}(a_1,...,a_n)=1$, at least one of $a_1,...,a_n$ must be nonzero.

Relabeling if necessary, assume $a_1\ne 0$.

Let $g=\text{gcd}(a_1,...,a_{n-1})$.

Let $r_1,...,r_{n-1}\in R$ be such that $\sum_{k=1}^{n-1}r_ka_k=g$ and  $\text{gcd}(r_1,...,r_{n-1})=1$.

Applying the lemma, let $s_1,s_2\in R$ be such that $\text{gcd}(s_1,s_2)=1$ and $r_1s_1+r_2s_2=0$.

Then letting $s_k=0$ for $3\le k\le n-1$, it follows that $\text{gcd}(s_1,...,s_{n-1})=1$ and $\sum_{k=1}^{n-1}r_ks_k=0$.

Let $y_n=x_n+\sum_{k=1}^{n-1}s_kx_k$ and let $y_1'=y_1-a_ny_n$.

Then $y_1'=\sum_{k=1}^{n-1}b_kx_k$ where $b_k=a_k-s_ka_n$.

Suppose $f\in R$ is a common divisor of $b_1,...,b_{n-1}$.

Identically we have
\begin{align*}
\sum_{k=1}^{n-1}r_kb_k
&=
\sum_{k=1}^{n-1}r_k(a_k-s_ka_n)
\\[4pt]
&=
\sum_{k=1}^{n-1}r_ka_k-a_n\sum_{k=1}^{n-1}r_ks_k
\\[4pt]
&=
g-(a_n)(0)
\\[4pt]
&=
g
\\[4pt]
\end{align*}
hence $f{\,\mid\,}g$.

Then $f{\,\mid\,}a_k$ for all $k\in\{1,...,n-1\}$ and $\text{gcd}(f,a_n)=1$.

Then for all $k\in\{1,...,n-1\}$
\begin{align*}
&
f{\,\mid\,}b_k
\\[4pt]
\implies &
f{\,\mid\,}a_k-s_ka_n
\\[4pt]
\implies &
f{\,\mid\,}s_ka_n
\\[4pt]
\implies &
f{\,\mid\,}s_k
\\[4pt]
\end{align*}
hence, since $\text{gcd}(s_1,...,s_{n-1})=1$, it follows that $f$ is a unit of $R$.

Thus we have $\text{gcd}(b_1,...,b_{n-1})=1$.

Applying the inductive hypothesis, there exist $y_2,...,y_{n-1}\in \langle{x_1,...,x_{n-1}}\rangle$ such that $\langle{y_1',y_2,...,y_{n-1}}\rangle=\langle{x_1,...,x_{n-1}}\rangle$.

Now consider $\langle{y_1,...,y_n}\rangle$.

It's clear that $y_1,...,y_n\in\langle{x_1,...,x_n}\rangle$.

From $y_1'=y_1-a_ny_n$, we get $y_1'\in \langle{y_1,...,y_n}\rangle$.

Hence from $\langle{y_1',y_2,...,y_{n-1}}\rangle=\langle{x_1,...,x_{n-1}}\rangle$, we get $x_1,...,x_{n-1}\in \langle{y_1,...,y_n}\rangle$.

And then from $y_n=x_n-\sum_{k=1}^{n-1}s_kx_k$, we get $x_n\in \langle{y_1,...,y_n}\rangle$.

Thus $x_1,...x_n\in \langle{y_1,...,y_n}\rangle$, hence $\langle{y_1,...,y_n}\rangle=\langle{x_1,...,x_n}\rangle$, which completes the induction, and thus completes the proof.


问:
Name for $K' = \{y \in \mathbb R^m: y^T A \ge 0\}$ when $K = \{x \in \mathbb R^n: Ax \le 0\}$?
The following question came up in the context of an optimization problem.

Let $A \in \mathbb R^{m \times n}$ be a matrix and let
$$K = \{x \in \mathbb R^n: Ax \le 0\}$$
be the polyhedral cone defined by the set of vectors in $\mathbb R^n$ that make a nonpositive angle with each of the rows of $A$.

Now consider the following polyhedral cone:
$$K' = \{y \in \mathbb R^m: y^T A \ge 0\}.$$
$K'$ is the set of vectors in $\mathbb R^m$ whose angle with each of the columns of $A$ is nonnegative.

Question: Is there a name for $K'$? If not, what can we say about the (geometric) relation between the sets $K$ and $K'$?

答:
I'm not aware of a term for the relationship between $K$ and $K'$. However, we can characterize $K'$ as the dual cone to the convex cone generated by the columns of $A$. That is, $K = C^*$, where $C$ is the convex cone defined by $C = \{Ax : x \in \Bbb R^n, x \geq 0\}$.

Similarly, $K$ is the polar cone to the convex cone generated by the rows of $A$.


问:
Graphing $f(2-x)$
Sorry for this very trivial question, but I've become slightly confused by this question. Consider a graph
$y=f(x)$. How would I draw the graph $y=f(2-x)$?

It seems to me that as this is obviously equal to $y=f(-(x-2))$ this should represent the graph being translated $2$ units in the positve $x$ direction and then reflected in the $y$ axis.

Is that true? It doesn't seem to be from the graphs I have plotted using Desmos. If not, please explain why it is incorrect.

Thanks for your help.

EDIT:
I have now slept over my problem and I believe that it lies in the following statement I have been led to believe in class:

The graph of $f(\text{Blah}+a)$ is ALWAYS  a translation of $a$ units of the graph $f(\text{Blah})$ in the negative direction.

More specifically, I thought that the as graph of $f(x+a)$ is a translation of $a$ units of the graph $f(x)$ in the negative direction, then the graph of $f(-x+a)$ is a translation of $a$ units of the graph $f(-x)$ in the negative direction as well. After thinking it over logically however, I now think this is wrong.

This is my reasoning:

Consider $y=f(x+a)$. For a given $y$ value on the $y=f(x+a)$ graph, the $x$ value needed for it must be $a$ smaller than the $x$ value needed if it was just the function $y=f(x)$; hence the graph $y=f(x+a)$ must be the graph of $y=f(x)$ but shifted $a$ units to the negative $x$ direction.

But, if we consider $y=f(-x+a)$: For a given $y$ value on the $y=f(-x+a)$ graph, the $x$ value needed for it must be $a$ bigger than the $x$ value needed if it was just the function $y=f(-x)$; hence the graph $y=f(-x+a)$ must be the graph of $y=f(-x)$ but shifted $a$ units to the positive $x$ direction.

Is my reasoning correct now?  Thanks again for your help.

答: Label the point $x=0$ as $a$ and $x=2$ as $b$. Now exchange $a,b$.

Title: Graphing $f(2-x)$


问:
Is cartesian product of the Cantor set and real numbers ($C\times\Bbb R$) has measure zero in $\Bbb R^2$
I'm trying to show that the cartesian product of real numbers and the Cantor set ($C$) has (Lebesgue) measure zero in $\Bbb R^2$, I was successful at showing (I think) that $\Bbb R$ has measure zero in $\Bbb R^2$ by covering it with many "rectangles" with Volume(area) zero. Now my plan was to show that $C\times\Bbb R$ is a subset of the union of many "rectangles" and some set that contains $C$, but I can't seem to find that second set.
Now, I know how to show that $C$ has measure zero in $\Bbb R$, but I can't seem to translate this into $\Bbb R^2$.

Could anybody help me? I think that it would be useful to find how to cantor set looks in R$^2$ because co-student told me that it would be the same as in R but with infinite strips going up from the small intervals, but this doesn't fit right with me.

答: Hint: First show that $S_n = (C \times \Bbb R) \cap ([-n,n] \times [-n,n])$ has measure $0$ for any $n \in \Bbb N$. Using the fact that $C \times \Bbb R = \bigcup_{n \in \Bbb N} S_n$, conclude that $C \times \Bbb R$ has measure $0$.


问:
dimension of a module and simple modules
I am trying to wrap my head around these new concepts. Suppose I have a scenario where $V$ is a free module and also $V$ is a finite dimensional module, if V is simple, must $V$ be dimension $1$?

Def:

A module is said to be simple if any submodule of such module is either $\{0\}$ or itself.

My thought is that if we have a $A$-module $V$ which happens to be also finite dimensional then say $V=\langle v_1,\cdots,v_n\rangle$, with the scalars come from $A$, then surely if we take any proper subset of such basis vectors, their span would be a proper submodule of $V$? Hence $V$ is simple only if $n=1$? (of course the converse is not true, if $n=1$ then it wouln't necessarily be simple)

I feel like there is something wrong with my argument, I must have confused something from modules with Linear Algebra.

答:
if $V$ is simple, must $$ be dimension 1?

Nope.  Simplicity of a module of an algebra over a field does not say much about its dimension.

Consider the matrix ring $R=M_2(F)$ for a field $F$.

Then $F\times F$, viewed as a right $R$ module using matrix multiplication, is a simple module that is $2$ dimensional over $F$.

Another example with the same idea: let $V$ be an infinite dimensional vector space, and $E$ be its ring of linear transformations. Then $V$ is in fact a simple $E$ module, and yet $V$ has infinite $F$ dimension.

This all assumes you meant "dimension" in the way it is most commonly used: as an invariant of a vector space. If in fact you meant some other type of dimension, you'll have to specify. There are several you could be talking about.  For example, if by dimension you meant "composition length", then yes, a simple module has composition length $1$.


问:
Help with proof concerning ordered fields and smallest elements
I have been trying to prove that every ordered field has no smallest positive element for a while now, and I think I have it worked out. However, I feel like something is missing in my proof. Here is what I have so far:

Let $F$ be an ordered field

Let $x$ exist in $F$ such that $0$ is less than $x$

Then $x/2$ is less than $x$

Therefore, $F$ can have no smallest element

This does seem to prove that the field has no smallest element, but it seems like there are some holes. Am I correct in thinking this way? I saw a similar proof involving the positive real numbers, but $x/2$ exists in the positive reals.  How do I know that $x/2$ exists in $F$?  I am just concerned that this is not enough.  Any helpful pushes would be most appreciated. Thank you!

答:
The proof you showed is really a proof by contradition.

Suppose there is a smallest positive number in $F$, call it $x$.

1) Since it is positive by assumption, $0<x$.

2) Now we show $0<x/2<x$, that is $x/2$ is a positive number that is smaller than $x$.

3) Thus $x$ cannot be the smallest positive number, since we found some even smaller positive number, a contradiction.


问:
A relation between Lebesgue and Riemann integrable functions
Is it true that a bounded function $ f:[a,b]\to \mathbb{R} $ is Lebesgue integrable if and only if there exists a function $f_1:[a,b]\to \mathbb{R}$ that is Riemann integrable and
$$  m^*\lbrace x: (f-f_1)(x)\ne 0\rbrace =0 $$

One side of the above assumption is true and obvious.

答: No. Take $f=1_A$, where $A$ is a measurable set with the property that both $A$ and $A^c$ have positive measure intersection with any interval (See for example, Construction of a Borel set with positive but not full measure in each interval). It is not too hard to see that modification on a null set $N$ cannot make such an $1_A$ continuous a.e. (because $A\setminus N$ and $A^c\setminus N$ must accumulate at every point).


问:
Likelihood test ratio for random sample with p.d.f. $f(x\mid \theta) = e^{-(x-\theta)}$
I don't know how to find the max region of the denominator since doing the usual gives me a constant as the derivate of the likelihood function of theta.
The problem is as follows

Let $X_1 , \ldots , X_n$ be a random sample with
$$f(x\mid \theta) = e^{-(x-\theta)}$$
where $x \in [\theta,\infty)$

find $\lambda(x)$ where $H_0: \theta = \theta_0$ and $H_1: \theta > \theta_0$

so I know that the numerator since it's just a point, is equal to the maximum likelihood function on $\theta_0$

but if I try to maximize the denominator since the Likelihood function is
$$ \prod^n_{i=1} e^{-(x_i-\theta)} = e^{n\theta-n \bar{x} } $$
but I can't maximize the likelihood from its derivative in the region $[\theta_0 , \infty]$

答:
The main error you did is that the likelihood you stated is WRONG. You did not take into consideration that the suppport is depending on the parameter.

Let's see how to proceed in this case:

$$f(x|\theta)=e^{\theta}e^{-x}$$

$x\geq\theta$

$\theta \in \mathbb{R}$

This implies that $min(\mathbf{X})=X_{(1)}\geq\theta$

The likelihood is the following

$$L(\theta)=e^{n \theta}e^{-\Sigma_i X_i}\cdot\mathbb{1}_{[\theta;\infty)}(x_{(1)})$$

Now if we take two points $\theta'<\theta''$ we get the following likelihood ratio

$$\frac{L(\theta';\mathbf{x})}{L(\theta'';\mathbf{x})}=\frac{e^{n\theta'}\mathbb{1}_{[\theta';\infty)}(x_{(1)})}{e^{n\theta''}\mathbb{1}_{[\theta'';\infty)}(x_{(1)})}=\begin{cases}
+\infty,  & \text{if $x_{(1)}<\theta''$ } \\
e^{n(\theta'-\theta'')}, & \text{if $x_{(1)}\geq\theta''$ }
\end{cases}$$

Thus in this case we have a "Monotone Likelihood Ratio" that is a "Non increasing LR" with respect to the Statistic $T=X_{(1)}$ and so we can apply the following theorem