问:
Criteria for being a true martingale
Could you kindly list here all the criteria you know which guarantee that a continuous local martingale is in fact a true martingale? Which of these are valid for a general local martingale (non necessarily continuous)? Possible references to the listed results would be appreciated.

答:
I found by myself other criteria that I think it is worth adding to this list.

5) $M$ is a local martingale of class DL iff $M$ is a martingale

6) If $M$ is a bounded local martingale, then it is a martingale.

7) If $M$ is a local martingale and $E(\sup_{s \in [0,t]} |M_s|) < \infty \, \forall t \geq 0$, then $M$ is a martingale.

8) Let $M$ be a local martingale and $(T_n)$ a reducing sequence for it. If $E(\sup_{n} |M_{t \wedge T_n}|) < \infty \, \forall t \geq 0$, then $M$ is a martingale.

9) Suppose we have a process $(M_t)_{t\geq 0}$ of the form $M_t=f(t,W_t)$. Then $M$ is a local martingale iff $(\frac{\partial}{\partial t}+\frac{1}{2}\frac{\partial^2}{\partial x^2})f(t,x)=0$.
If moreover $\forall \, \varepsilon >0$ $\exists C(\varepsilon,t)$ such that $|f(s,x)|\leq C e^{\epsilon x^2} \, \forall s \geq 0$, then $M$ is a martingale.

Title: Criteria for being a true martingale


问:
Existence of a graph G?
The Q i am asked is as follows.

Q:Let G=(V,E) be a graph with nine vertices  such that each vertex has a degree 5 or 6. Show that G has at least 5 vertices of degree 6, or at least 6 vertices of degree 5.

first lets bust the middle 5 of6 = 9-5 thus 4of 5 and  6 of 5 = 9-6 hence 3 of 6.

so they are claiming that 4 vertex of degree 6 and 5 of degree 5 cannot be a graph. and anything else is fine.
clearly $4*6 + 5*5 = 24+25 = 49$  thus $E= 24.5$ which i would like to see you show me half an edge?

Now i could draw pictures of all the other graphs ( me being lazy i want to find a better way)  or i could ask is the existence of an even sum here sufficient to prove that there exists a simple connected graph? i know that having half an edge is sufficient to disprove the existence  but for a set number of vertices if there exists an even number of edges does a graph need to exist?

EDIT: ( to downvote and comment on duplicate)

My question is for a set number of vertices if there exists an even number of edges does a graph need to exist?

答:
Let $k_5$ denote the number of degree $5$ vertices and $k_6$ the number of degree $6$ vertices. Since every vertex is of degree $5$ or $6$ we must have
$$k_5 + k_6 = |V| = 9$$
From the handshaking lemma,
$$5k_5 + 6k_6 = 45 + k_6 = 2|E|$$
This shows that $k_6$ is odd and $k_5$ is even. Therefore, we have the following possibilities
$$(k_5,\ k_6) \in \{(2,7),\ (4,5),\ (6,3),\ (8,1)\}$$
Therefore for all possibilities, we have either at least $5$ vertices of degree $6$ or at least $6$ vertices of degree $5$.</p>

ContentLicense: CC BY-SA 3.0

Title: Existence of a graph G?


问:
Finite sum of combinations
Suppose $a$ and $b$ are two positive integers with $a \leq b$. Do we have the following?
$$\sum_{k=0}^{a}\binom{a}{k}\binom{b}{b-k} = \binom{a+b}{a}$$

This is very intuitive to me: we want to choose $a$ items from a set of $a+b$ items. We can choose $k \leq a$ items from the first $a$ items and $b-k$ items from the last $b$ items.

答:
Note that
$$ (1+x)^{a+b}=(1+x)^a(1+x)^b. $$
By using
$$ (1+x)^n=\sum_{k=0}^n\binom{n}{k}x^k $$
one has
$$ (1+x)^{a+b}=\sum_{k=0}^n\binom{a+b}{k}x^k \tag1$$
and
$$ (1+x)^a(1+x)^b=\sum_{k=0}^a\binom{a}{k}x^k\sum_{k=0}^b\binom{b}{k}x^k=\sum_{j=0}^a\sum_{k=0}^b\binom{a}{j}\binom{b}{k}x^{j+k}. \tag2 $$
From (1), the coefficient of $x^a$ is
$$\binom{a+b}{a} $$
and from (2), the coefficient of $x^a$ is
$$ \sum_{j=0}^a\binom{a}{j}\binom{b}{a-j}. $$
So
$$ \sum_{j=0}^a\binom{a}{j}\binom{b}{a-j}=\binom{a+b}{a}. $$

ContentLicense: CC BY-SA 4.0

Title: Finite sum of combinations


问:
If independent sequences weakly converge, their product weakly converges.
This is a duplicate of weak convergence of independent sequence , but I don't think that answer actually answers the question.

Let $(X_n), (Y_n)$ be independent sequences, weakly converging to $X$ and $Y$ respectively, who are also independent. I want to show that $\langle X_n, Y_n \rangle \xrightarrow{w} \langle X, Y \rangle$.

For $(X_n)$ to weakly converge to $X$, means by definition that $\int f d \mathbb{P}^{X_n}$ converges to $\int f d \mathbb{P}^X$ for any continuous bounded $f$, or equivalently that $\int f \circ X_n d \mathbb{P}$ converges to $\int f \circ X d \mathbb{P}$.

For showing that $\int h \circ \langle X_n, Y_n \rangle d \mathbb{P}$ converges to $\int h \circ \langle X, Y \rangle d \mathbb{P}$, I think the idea is to look at $"h \circ X"$ and $"h \circ Y"$, where the "" are because $h$ takes its arguments in $\mathbb{R}^2$ and there is no canonical injection $\mathbb{R} \to \mathbb{R}^2$.

To get an idea on how to solve this, I thought we could first have a look at $h$ an indicator function $1_{A \times B}$. Then we get that

$\int 1_{A \times B} \circ \langle X_n, Y_n \rangle d \mathbb{P} = \mathbb{P}(X_n^{-1}(A) \cap Y_n^{-1}(B)) = \mathbb{P}(X_n^{-1}(A)) \cdot \mathbb{P}(Y_n^{-1}(B)) =$

$\int 1_{A} \circ X d \mathbb{P} \cdot \int 1_{B} \cdot Y d \mathbb{P}$ by independence, which looks promising as it at least uses the independence and splits up the integral, but I don't know how to continue from here.

答:
Your approach is on the right track.  What you're getting at is the fact that any function of the form $f(x,y) = g(x)h(y)$ satisfies
$$
\int f \circ (X_n,Y_n) \,d\mathbb{P} = \int (g\circ X_n) (h \circ Y_n) \,d\mathbb{P} = \int g \circ X_n \,d\mathbb{P} \cdot \int h \circ Y_n \,d\mathbb{P} \to \int g \circ X \,d\mathbb{P} \cdot \int h \circ Y \,d\mathbb{P}
$$
as $n \to \infty$.  Then you can use the Stone-Weierstrass theorem to prove that the set of functions of the form $(x,y) \mapsto g(x)h(y)$ is dense in the set of all continuous functions.


问:
Use jacobian matrix to solve equation.
our teacher told us that Jacobian matrices can be used to solve a system of nonlinear equations, and I am wondering exactly how this work (He never actually showed us an example). if I have the following system to solve ($u,v$ are $C^1$) :

(local inversion of $f(x,y)$ next to the point $(x,y,u,v)= (1,1,-2,2)$ , and the origin? )
$$ u =x^3-3xy^2 $$
$$v=3x^2y-y^3$$

Here's my attemps :

Let ,  $f=x^3-3xy^2 $ and we have : $g = 3x^2y-y^3$

So I found $f_x = 3x^2-3y^2 = g_y$ and also $f_y = 6xy = g_x$ ,

I used the jacobian to find the  matrix determinant:

$\frac{\partial(f,g) }{\partial (x,y)}=9x^4-54x^2y^2+9y^4$ .

Then I used the given point$(1,1)$ to find it equal to $-36 \neq 0 $ , But I don't know what are the next steps , since the results is not $0$ we can now use the inverse function theorem ?

I'm really not too sure about my steps

And for the point at the origin should we take $(0,0,-2,2) $ or $(0,0,0,0)$ ?

Thanks in advance , any help would be a lot appreciated.

答:
Your notation is clumsy so let's clean up the problem and see what we are doing.

You have a function $F:\mathbb{R}^2\times \mathbb{R}^2\to \mathbb{R}^2$ given by $F(x,y,u,v)=(f_1(x,y,u,v),f_2(x,y,u,v))^T$, where
\begin{align}
f_1(x,y,u,v)&=x^3-3xy^2-u \\
f_2(x,y,u,v)&=3x^2y-y^3-v
\end{align}
(The order of the $x,y$ terms and the $u,v$ terms may be flipped).
Then we have $F(1, 1, -2, 2)=(0,0)^T$ and $F(0,0,0,0)=(0,0)^T$.

By the implicit function theorem, we may write $F(x(u,v),y(u,v),u,v)=(0,0)^T$ locally around the points (i.e. there exists a neighborhood such that this is valid) of the Jacobian with respect to $x$ and $y$ has nonzero determinate.

Alternatively, if you wanted to write, $F(x,y,u(x,y),v(x,y))=(0,0)^T$, you would have to show that the Jacobian with respect to $u$ and $v$ has nonzero determinate. In this problem, this is really easy because you can solve for $u$ and $v$ explicitly.

Title: Use jacobian matrix to solve equation.


问:
inequality $\frac{2(x + y)^2}{2x^2 + y^2} \leq 3$
I'm looking for a "nice" way to show that the following inequality holds, i.e. without differentiating and determining the maximum:

$$
\frac{2(x + y)^2}{2x^2+ y^2} \leq 3 \quad \forall x,y \in \mathbb{R}
$$

It's rather easy to show

$$
\frac{4xy}{x^2 + y^2} \le 2
$$

and obviously

$$
\frac{2x^2 + 2y^2}{2x^2+y^2} \le 2
$$

but combining these two does not give me a sufficient low bound.

Body:
How about this method?
$$\frac{2(x + y)²}{2x² + y²}-3=\frac{2x^2+4xy+2y^2-(6x^2+3y^2)}{2x^2+y^2}=-\frac{4x^2+4xy+y^2}{2x^2+y^2}=-\frac{(2x-y)^2}{2x^2+y^2}\le0..$$


问:
Finding $\frac{\sum_{r=1}^8 \tan^2(r\pi/17)}{\prod_{r=1}^8 \tan^2(r\pi/17)}$
I have tried to wrap my head around this for some time now, and quite frankly I am stuck.

Given is that :
$$a=\sum_{r=1}^8 \tan^2\left(\frac{r\pi}{17}\right) \qquad\qquad b=\prod_{r=1}^8 \tan^2\left(\frac{r\pi}{17}\right)$$

Then what is the value of $a/b$?

I tried evaluating the quantities on DESMOS to realise that: a=136 and b=17 which gives 8 as the answer. Any helpful insight about how to reach at these values !?

答:
Here is a solution using the observation in Cauchy's proof of Basel problem.

Lemma. We have
$$ \prod_{k=1}^{n} \left( t - \tan^2\left(\frac{k\pi}{2n+1}\right)\right) = \sum_{k=0}^{n} (-1)^{n-k}\binom{2n+1}{2k+1} t^k. \tag{*} $$

Using this, we immediately know that

$$ a = \sum_{k=1}^{n} \tan^2\left(\frac{k\pi}{2n+1}\right) = \binom{2n+1}{2} = n(2n+1) $$

and

$$ b = \prod_{k=1}^{n} \tan^2\left(\frac{k\pi}{2n+1}\right) = \binom{2n+1}{1} = 2n+1. $$

So the ratio $a/b$ is exactly $n$. In OP's case, $n = 8$ and hence the answer is $8$.

Proof of Lemma. Write $m = 2n+1$. Then by the de Moivre's formula,

\begin{align*}
\frac{\cos(mx) + i\sin(mx)}{\cos^m x} = \left( \frac{\cos x + i\sin x}{\cos x} \right)^m = (1 + i\tan x)^m = \sum_{k=0}^{m} \binom{m}{k} (i\tan x)^k.
\end{align*}

Taking the imaginary parts only and plugging $m = 2n+1$ back,

\begin{align*}
\frac{\sin((2n+1)x)}{\cos^{2n+1} x} = \sum_{k=0}^{n} (-1)^k \binom{m}{2k+1} \tan^{2k+1} x,
\end{align*}

and so,

\begin{align*}
(-1)^n \frac{\sin((2n+1)x)}{\sin x \cos^{2n} x} = \sum_{k=0}^{n} (-1)^{n-k} \binom{m}{2k+1} \tan^{2k} x.
\end{align*}

However, the left-hand side becomes zero if $x_k = \frac{k\pi}{2n+1}$ for $k = 1, \dots, n$, and also note that $\tan^2(x_k)$ for $k = 1, \dots, n$ are all different. So, denoting the right-hand side of $\text{(*)}$ by $P(t)$, i.e., writing

$$ P(t) = \sum_{k=0}^{n} (-1)^{n-k}\binom{2n+1}{2k+1} t^k, $$

then this implies that $P(\tan^2(x_k)) = 0$ for $k = 1, \dots, n$, hence $\tan^2(x_k)$ for $k = 1, \dots, n$ are $n$ distinct zeros of $P(t)$. Since the leading coefficient of $P(t)$ is $1$, this proves $\text{(*)}$.


问:
Appearance of sentence parameters in a theorem
Is it true that if $A$ is a formula in a Hilbert system $H$, then if $B_1,B_2,\ldots,B_n$ is a proof of $A$ in $H$, any sentence parameter not appearing in $A$ doesn't appear in $B_1,\ldots,B_n$? If it's true, how to show it?</p>

答:
What counts as a Hilbert system?

Do you mean a theory whose deductive apparatus is Hilbertian? Well, suppose $T$ is a theory with a classical Hilbertian deductive system, and the non-logical axiom  $P \land Q$. Then $T$ will prove $Q$. But  $Q$ isn't provable from the logic alone, so any proof of $Q$ will have to involve the premiss $P \land Q$. So there is a propositional letter earlier in proof which isn't in the conclusion.

Or do you mean a pure Hilbertian system of logic? Well note that a propositional logic with particular sentences as axioms (rather than with axiom schemata) and a substitution rule  and modus ponens as the sole rules of inference would often (usually?) be counted as Hilbertian. And obviously in such systems where e.g. $P \to P$ is an axiom and $(Q \land R) \to (Q \land R)$ is derived in one step by substitution, then again there is a propositional letter earlier in proof which isn't in the conclusion.


问:
Prove $f:[a,b) \longrightarrow (a,b)$ cannot be bijective and continuous (verification)
I would just like to make sure my proof for this is correct. If I made any wrong assumptions, please do let me know so I can correct them or even try a different approach. Thank you!

Let $f(a)=c$, by the bijectivity of $f$. Now, consider the case where we remove $c$ from $(a,b)$ and what we are left with is a disconnected set which is the union of two connected sets, namely $(a,c) \cup (c,b)$. We now impose a restriction on $f$ defined by $$g: (a,b) \longrightarrow (a,c) \cup (c,b), \ g(x) = f(x)$$This restriction is continuous (by Theorem proved in class). The image of $g$ is the union of two connected components. Since $(a,b)$ is connected, its image must also be connected, which means it must lie entirely within one of the two components. But then this would imply that the original function, $f$, is a proper subset of $(a,b)$ which contradicts the bijectivity of it. As such, it cannot be bijective and continuous.

The intervals are regarded with the subset topology induced by the standard topology of $\mathbb R$

答:
The proof has a clean idea, but it's a bit confusing the way you've written it - the assumptions are a bit scattered into the proof. I would start the proof as follows, where you lay out clearly what your goal is:

We will show that, if $f:[a,b)\rightarrow (a,b)$ is continuous and injective, it cannot be surjective.

This expresses, for the reader, exactly what your proof is going to do ahead of time, rather than haphazardly bringing in assumptions during it. This helps the reader since they can verify that your proof strategy actually works without needing to parse through the details. This also reflects the structure of your proof better than a straight proof by contradiction would, since your proof is very much "we prove this property of continuous injective functions then note it's not compatible with surjectivity" rather than "we prove a property of bijective functions, then realize it's a contradiction."

Then, you can proceed:

Let $f(a)=c$.

There is no need to mention anything about $f$ here - a mere evaluation of a function does not require bijectivity, as your proof suggests. Then we can bring in the first condition we promised to assume:

By injectivity of $f$, observe that the image of $(a,b)$ under $f$ must not include $c$, hence must be a subset of $(a,c)\cup (c,b)$.

The rest of your proof is clear as is, so I won't repeat or rewrite it - you restrict the domain of $f$ to $(a,b)$, use the assumed condition of continuity to show something about connectivity, and then show that this implies that $f$ isn't surjective. Mostly, the proof you have just needs a bit of clarity at the start and a couple of improvements to be more explicit about assumptions, but is otherwise good as is.


问:
Topology of Mobius Ring
I'm having some difficulty with the finer points of the construction of the topology of the Mobius ring from Tu's manifolds. Given the square $ R = \{(x,y) \in \mathbb{R}^2 | \, 0 \leq x \leq 1, -1 < y < 1\}$, $ R /\sim $ is formed with the relation $ (0,y) \sim (1,-y) $. I understand that the interior points of $ R $ are only related only to themselves (and are mapped into their own equivalence classes by the projection $ \pi: R \rightarrow R / \sim $), and the edges are glued together into equivalence classes of two elements, $ \pi\big( (0,y) \big) = \pi\big( (1,-y) \big) = \{ (0,y),(1,-y) \}, \, \forall y \in \, ]-1,1[ $.

What I don't understand are the open sets of $ R / \sim $. I know that a set $ U \in R/\sim $ is open if $ \pi^{-1}(U) $ is open in $ R $, but I'm trying to consider an open set in $ R/\sim $ that contains the "glued-together" region. This hypothetical region $ V $ contains some of the points in the glued-together equivalence classes as well as points on both "sides" of the interior. When mapped back to $ R $, this region contains parts on both sides of the square, as well as points on both boundary regions, so how can such a region be open in $ R $? I know that the overlap region of $ R/\sim $ shouldn't be seen any differently from any other region $ R/\sim $, but such regions seem to map back to non-open sets in $ R $. Isn't it true that a set containing boundary points of $ R $ (i.e. at $ x = 0 $ and $ x = 1 $) isn't open in $ R $?

答:
The topology of R is the one inherited in R as a subset of $\mathbb{R}^2$. This means that an open set in R can contain parts of the edge of the square, as long as this open set is the intersection of a regular open ball in $\mathbb{R}^2$ with R.

For clarity, picture a ball of small radius and center one of the vertices.

Title: Topology of Mobius Ring


问:
Truth, Proof and Axiomatic Systems
I still struggle mighty with basic conceptions of truth and proof.

For example: The Continuum Hypothesis (CH) is either true or false, i.e. either CH or ~CH holds. Now, Goedel and Cohen proved that CH/~CH are independent from ZFC, so ZFC + CH and ZFC + ~CH are consistent (in case ZFC is consistent but mathematicians assume that anyway). But since we know that CH or ~CH must be false, how can that be? One of those axiom systems must be inconsistent since it has no models (because one of its axioms is false).

Another example is the parallel axiom (P) in euclidean geometry. P is true or false, i.e. P or ~P. That would mean that either the euclidean or non-euclidean geometry system has to be inconsistent (= no model).

Can somebody explain where I make a mistake?

答:
As in this other question of yours, the issue comes from conflating truth/falsity (which are relevant to structures) with provability/disprovability (which are relevant to theories).

More specifically, you're trying to apply bivalence - "everything is either correct or incorrect" - in a way it can't be applied. The satisfaction relation for individual structures is indeed bivalent: if $\mathcal{A}$ is a structure and $\varphi$ is a sentence (in the same language), then either $\mathcal{A}\models\varphi$ xor $\mathcal{A}\models\neg\varphi$. However, bivalence does not hold for provability: e.g. there is no sense in which "$\mathsf{CH}$ is either true in $\mathsf{ZFC}$ or false in $\mathsf{ZFC}$."

So when you invoke bivalence to claim

The Continuum Hypothesis (CH) is either true or false, i.e. either CH or ~CH holds,

that's correct iff we're tacitly referring to some specific structure $\mathcal{S}$. However, your later claim

One of those axiom systems must be inconsistent since it has no models (because one of its axioms is false)

is then mixing up provability and truth. We can see this clearly if we make explicit the tacit assumption above:

One of those axiom systems must be inconsistent since it has no models (because one of its axioms is false $\color{red}{\mbox{in the structure $\mathcal{S}$}}$).

It should be clear that this is unwarranted: just because a theory is not satisfied in some particular structure doesn't mean it's not satisfied in any structure.

To avoid exactly this sort of confusion, we should be careful - at least until the fundamentals are mastered - to only use true/false/etc. when talking about structures, and use provable/disprovable/etc. when talking about theories. This way we won't be tempted to improperly use our intuitions about truth/falsity when talking about provability/disprovability (or at least, we'll be less tempted).

Title: Truth, Proof and Axiomatic Systems


问:
Bernoulli trial and asymptotic analysis
I want to find such function $p(n)$ in a way that obtaining one succes will be asymptotically-equivalent to obtaining two failures.
I basically want to have something like this:
$$ \lim_{n \to \infty}{\frac{{n \choose 1}p(n)(1-p(n))^{n-1}}{{n \choose 2}(1-p(n))^2p(n)^{n-2}}}= $$
$$= \lim_{n \to \infty}{\frac{np(n)(1-p(n))^{n-1}}{\frac{n!}{2(n-2)!}(1-p(n))^2p(n)^{n-2}}}=...=$$
$$= \lim_{n \to \infty} {\frac{2(1-p(n))^{n-3}}{(n-1)p(n)^{n-3}} } = 1$$
And now I'm kind of stuck... what do do with $n-1$ in denominator? I could use some hints.

答:
We don't even need asymptotic here. Just write
$$\frac{2(1 - p(n))^{n - 3}}{(n - 1)p(n)^{n - 3}} = 1$$
and solve w.r.t. $p(n)$:
$$\sqrt[n - 3]{2} (1 - p(n)) = \sqrt[n - 3]{n - 1} p(n)$$
$$p(n) = \frac{\sqrt[n - 3]{2}}{\sqrt[n - 3]{n - 1} - \sqrt[n - 3]{2}}$$
and if we replace $1$ in the first equation with $1 + o(1)$, we get general solution
$$p(n) = \frac{\sqrt[n - 3]{2}}{\sqrt[n - 3]{n - 1} - \sqrt[n - 3]{2} + o(1)}$$

Title: Bernoulli trial and asymptotic analysis


问:
What is $\frac{(an)!}{n!}$?
How can $\frac{(an)!}{n!}$ be expressed in terms of $a!$, $n!$, $a$, $n$ (and maybe Pochhammers)?</p>

答:
For $n>m$,
$$
\frac{n!}{m!} = P(n,n-m) = \text{the number of permutations of $n-m$ in a set of $n$}.
$$
This is one context in which it doesn't make sense to define "permutation" as a bijection from a set to itself, but rather one should define it as an ordered list of length $n-m$ of members of a set of size $n$.</p>

Title: What is $\frac{(an)!}{n!}$?


问:
Does This Proof Make Sense?
I've been working on understanding the following proof for some time: Link

Can somebody please tell me if my solution makes sense (and, in some places, help me patch up the proof)?

We want to prove that if we have a group, $G$, and endomorphism, $\phi$ such that the subset of elements, $x \in G$ satisfying $\phi(x) = x^2$, makes up more than $\frac{3}{4}$ of the elements of $G$, then $G$ is abelian and $\phi(x) = x^2$ for all $x\in G$.

To begin, let $S$ be the subset of $G$ where any element, $x$, in $S$ satisfies $\phi(x) = x^2$

Now, let $x, y, xy$ be elements that are all in $S$. We claim that then, $xy = yx$. This follows naturally from the definition of homomorphism. We must have $\phi(xy) = \phi(x)\phi(y).$ Since $x, y, xy \in S$, it means $$(xy)^2 = x^2 y^2.$$

I don't understand how to prove from here that xy = yx.

Then, denote $x^{-1}S$ as the set of elements: $\{x^{-1}s, s \in S \}.$ Each element in $x^{-1}S$ corresponds to the inverse of some element in $S.$ That means they both have the same number of elements. Furthermore, by the Principles of Inclusion and Exclusion, we have the following statement: $$|S \cup x^{-1}S| + |S \cap x^{-1}S| = |S| + |x^{-1}S|.$$

Because we know that $|S| = |x^{-1}S| > \frac{3}{4} |G|$, $$|S \cup x^{-1}S| + |S \cap x^{-1}S| > 1.5 |G|.$$

Additionally, $|S \cup x^{-1}S| \le |G|.$ The elements in $x^{-1}S$ correspond to the inverses of the elements in $S$. By the definition of groups, $G$ must contain all of the inverses of its elements. Hence, it contains the inverses of $S$. This means that all elements in $|S \cup x^{-1}S|$ are contained in $G \rightarrow |S \cup x^{-1}S| < |G|.$

Taking that into account, we then get $$|S \cap x^{-1}S| > 0.5 |G|.$$

Now, let $y \in S \cap x^{-1}S.$ Notice that $y \in S, S \cap x^{-1}S.$

Because $y$ is an element in $x^{-1}S$, there must exist an $s \in S$ such that $y = x^{-1}s.$

Rearrangement gives $xy = s$ which means all three of $x, y,$ and $xy$ are all in $S$. We proved above that when this happens, $xy = yx.$ That means that $y \in C_G(x)$ where $C_G(x)$ is the centralizer of the element $x$ in group $G$.

Because $y$ is just some generic element in $S \cap x^{-1}S$ and we showed that it is in the centralizer of $x$, it follows that $S \cap x^{-1}S \subseteq C_G(x).$

However, earlier we showed that $|S \cap x^{-1}S| > 0.5 |G|.$ This means that $$|C_G(x)| > 0.5 |G|.$$

The centralizer is a subgroup of $S$. Since it is at least half as big as $G$, it must be $G$. Thus, $C_G(x) = G.$

Because this statement holds for all $x \in S$, it follows that $S \subseteq Z(G).$

But, we know that $|S| > \frac{3}{4} |G|.$ Because $S \subseteq Z(G),$ $$|Z(G)| > \frac{3}{4} |G|.$$

$Z(G)$ is also a subgroup of $S$ and because it contains more than half of the elements in $G$, it also must equal $G$. Thus, $Z(G) = G.$

By the definition of abelian groups, a group is abelian iff $G = Z(G)$. We just showed that which means $G$ is abelian.

Now, consider the subgroup $\langle S \rangle$ made by the elements of $S.$ Because $|S| > \frac{1}{2} G,$ $| \langle S \rangle | > \frac{1}{2} |G| \rightarrow \langle S \rangle = G.$

By definition, $S$ must be a generating set of $G$.

Because $G$ is abelian, there must exist an endomorphism, $\tau$ that maps each element to its square. Realize that $\tau$ and $\phi$ behave the same over the elements in $S$. Since we just showed that $S$ is a generating set of $G$, and an endomorphism's behavior over a generating set uniquely defines it, $\tau$ and $\phi$ must be the same. This implies that $S$ maps every element to its square, as desired.

Does my solution make sense?

Can somebody help me patch up the part about $xy = yx$?

Did I understand/use the $x^{-1}S$ thing correctly?

答:
Once you have $(xy)^2=x^2y^2$, multiply on the left by $x^{-1}$ and on the right by $y^{-1}$ to get $yx=xy$.

I don’t understand what you mean by the statement that each element of $x^{-1}S$ corresponds to the inverse of some element of $S$. $|x^{-1}S|=|S|$ because the map $S\to x^{-1}S:s\mapsto x^{-1}s$ is a bijection.

I don’t understand the paragraph that begins Additionally, and it is in any case unnecessary. You know that

$$|S\cup x^{-1}S|+|S\cap x^{-1}S|>\frac32|G|\tag{1}$$

and

$$|S\cup x^{-1}S|\le|G|\,,\tag{2}$$

and subtracting $(2)$ from $(1)$ immediately gives you

$$|S\cap x^{-1}S|>\frac12|G|\,.$$

For the next step in the argument you should explicitly state that $x\in S$, since you need that assumption to draw the conclusion that $xy=yx$. The rest appears to be okay.

Title: Does This Proof Make Sense?


问:
Group homomorphism between $\mathbb Z/n\mathbb Z$ and $S_n$.
How does one work out which elements map from, say, $(\mathbb{Z}/n\mathbb{Z},+)$ to $S_n$, i.e. if $\theta(1)$ is some permutation in $S_n$, for example?

I understand that ($\mathbb Z/n\mathbb Z,+$) is additive and $S_n$ is multiplicative, and thus
$$\theta(u+v) = \theta(u)*\theta(v).$$
My problem is understanding what $u$ and $v$ are in $S_n$.

For an example with numbers: $G$ is the $group ($\mathbb Z/18\mathbb Z,+$), $H$ is the group $S_6$ and $\theta(1) = (13)(56)$.

Why does $[1]$ map to $(13)(56)$ and how could I use this to work out the kernel? I've worked out that $[1]$ has order $18$ in $G$ and $(13)(56)$ has order $2$ in $H$.

答:
Here we are in the lucky case where "order of the group $=$ degree of symmetric group", and Cayley theorem gives us the chance to show at least a particular homomorphism among all those sought for, namely the embedding by left multiplication $\Bbb Z/n\Bbb Z\stackrel{\epsilon}{\hookrightarrow} S_{\Bbb Z/n\Bbb Z}\to S_{n}$ (for a given bijection $f\colon \Bbb Z/n\Bbb Z\to\{1,\dots,n\}$):

\begin{alignat}{1}
(f\epsilon(m)f^{-1})(i) &= f(\epsilon(m)(f^{-1}(i))) \\
&= f(m+ f^{-1}(i)) \\
\tag 1
\end{alignat}

For example, take $n=3$ and the bijection $f(0)=1, f(1)=2, f(2)=3$. Then:

\begin{alignat}{1}
&(f\epsilon(0)f^{-1})(1)=f(0+f^{-1}(1))=f(f^{-1}(1))=1 \\
&(f\epsilon(0)f^{-1})(2)=f(0+f^{-1}(2))=f(f^{-1}(2))=2 \\
&(f\epsilon(0)f^{-1})(3)=f(0+f^{-1}(3))=f(f^{-1}(3))=3 \\
\end{alignat}

whence the homomorphic image of $0\in\Bbb Z/3\Bbb Z$ in $S_3$ is given by $()$ (the identity of $S_3$). Similarly:

\begin{alignat}{1}
&(f\epsilon(1)f^{-1})(1)=f(1+f^{-1}(1))=f(1+0)=f(1)=2 \\
&(f\epsilon(1)f^{-1})(2)=f(1+f^{-1}(2))=f(1+1)=f(2)=3 \\
&(f\epsilon(1)f^{-1})(3)=f(1+f^{-1}(3))=f(1+2)=f(0)=1 \\
\end{alignat}

whence $1\mapsto (123)$, and finally:

\begin{alignat}{1}
&(f\epsilon(2)f^{-1})(1)=f(2+f^{-1}(1))=f(2+0)=f(2)=3 \\
&(f\epsilon(2)f^{-1})(2)=f(2+f^{-1}(2))=f(2+1)=f(0)=1 \\
&(f\epsilon(2)f^{-1})(3)=f(2+f^{-1}(3))=f(2+2)=f(1)=2 \\
\end{alignat}

whence $2\mapsto (132)$. To sum up, this particular (injective) homomorphism sends $\Bbb Z/3\Bbb Z$ to $\langle (123)\rangle\le S_3$.


问:
Analytic continuation of the integral $\int_{-1}^{1} \mathrm{d} x \frac{\sqrt{1 - x^{2}}}{x - \omega}$
I am interested in computing the complex integral
$$
f (\omega) = \int_{-1}^{1} \!\! \mathrm{d} x \, \frac{\sqrt{1 - x^{2}}}{x - \omega} ,
$$
where ${ \omega \!=\! \omega_{0} + \mathrm{i} \eta }$ is a complex number.
This integral is very similar to integrals that appear in the plasma dispersion function,
up to two changes: (i) the range of integration is finite, ${ -1 \!\leq\! x \!\leq\! 1 }$; (ii) the integrand function ${ x \!\mapsto\! \sqrt{1 - x^{2}} }$ suffers from branch cuts when extended to the whole complex plane.

As is usually done in plasma physics, I want to compute this integral using Landau's prescription.
So, I first start with the case ${ \mathrm{Im} [\omega] > 0 }$. In that case, the integral can be done without any particular care. Using Mathematica, I got
$$
f (\omega) = -\pi (\omega_{0} + \mathrm{i} \eta) \bigg( 1 - \sqrt{1 + \frac{1}{(\eta - \mathrm{i} \omega_{0})^{2}}} \bigg) \quad \text{ for } \quad \mathrm{Im} [\omega] > 0
$$

Then, I compute the integral for ${ \mathrm{Im} [\omega] = 0 }$. In that case, I need to be careful with the pole that lies along the real line, noting that this pole only contributes if ${ |\omega_{0}| < 1 }$. And following Plemelj's formula, it only contributes ''half'' a residue, hence the factor $\mathrm{i} \pi$. Using once again Mathematica, I obtained
$$
f (\omega) =
\begin{cases}
\displaystyle - \pi \omega_{0} + \mathrm{Sign} [\omega_{0}] \, \pi \sqrt{\omega_{0}^{2} - 1} & \text{if} \quad |\omega_{0}| > 1 ,
\\
\displaystyle - \pi \omega_{0} + \mathrm{i} \pi \sqrt{1 - \omega_{0}^{2}} & \text{if} \quad |\omega_{0}| < 1 ,
\end{cases}
\quad \text{for} \quad \mathrm{Im}[\omega] = 0 .
$$

Finally, I tried to compute the same integral, this time for ${ \mathrm{Im} [\omega] < 0 }$. Naively, I thought that I could re-use the expression from the case ${ \mathrm{Im}[\omega] > 0 }$,
and simply add the contribution from ${ 2 \mathrm{i} \pi }$ times the residue of the pole in $\omega$. As such, I wrote
$$
f(\omega) =
\begin{cases}
\displaystyle -\pi (\omega_{0} + \mathrm{i} \eta) \bigg( 1 - \sqrt{1 + \frac{1}{(\eta - \mathrm{i} \omega_{0})^{2}}} \bigg) & \text{if} \quad |\omega_{0}| > 1
\\
\displaystyle -\pi (\omega_{0} + \mathrm{i} \eta) \bigg( 1 - \sqrt{1 + \frac{1}{(\eta - \mathrm{i} \omega_{0})^{2}}} \bigg) + 2 \mathrm{i} \pi \sqrt{1 - (\omega_{0} + \mathrm{i} \eta)^{2}} & \text{if} \quad |\omega_{0}| < 1
\end{cases}
\quad \text{for} \quad \mathrm{Im}[\omega] < 0 .
$$

Unfortunately, when I look at the behaviour of the function ${ (\omega_{0} , \eta) \!\mapsto\! f (\omega_{0} + \mathrm{i} \eta)}$, I notice that my function is not continuous. Indeed, it suffers from a discontinuity along the lines ${ \omega_{0} \!=\! \pm 1 }$ and ${ \eta < 0 }$. I believe that the issue comes from missing contributions associated the branch cuts of the function ${ (\omega_{0} , \eta) \!\mapsto\! \sqrt{1 - (\omega_{0} + \mathrm{i} \eta)^{2}} }$ that is discontinuous along the lines ${ |\omega_{0}| > 1 }$ and ${ \eta = 0 }$. But, it is unclear to me how those should be accounted for.

How should one proceed to compute explicitly (and correctly!) the present integral?

PS: In practice, I must also compute a second integral, where one makes the replacement ${ \sqrt{1 - x^{2}} \!\to\! x \sqrt{1 - x^{2}}}$ in the integrand. It is likely that computing this second integral requires solving the exact same difficulties.

答:
1. Focusing only on the integral

$$ f(\omega) = \int_{-1}^{1} \frac{\sqrt{1-x^2}}{x-\omega} \, \mathrm{d}x \tag{*} $$

mathematically, OP's first formula works for all $\omega \in \mathbb{C}\setminus[-1,1]$, i.e.,

$$ f(\omega) = \pi \omega \left( \sqrt{1 - \frac{1}{\omega^2}} - 1 \right), \tag{1} $$

where $\sqrt{\,\cdot\,}$ is the principal square root. Indeed, it is not hard to check that $\text{(1)}$ holds for $\omega \in (1, \infty)$. Since both sides of $\text{(1)}$ are analytic on the open connected set $\mathbb{C}\setminus[-1,1]$, the identity extends to all of $\mathbb{C}\setminus[-1,1]$ by the principle of analytic continuation. (Note that the branch cut of $\sqrt{1-\omega^{-2}}$ is precisely $[-1, 1]$.) Also by the continuity, $\text{(1)}$ continues to hold at $\omega = \pm 1$.

Figure 1. The following is a domain coloring of the function $f(\omega)$ in $\text{(1)}$. (Here, hue represents the argument of $f(z)$ and brightness represents the magnitude of $f(z)$. The discontinuities in brightness is introduced at every level of the form $2^n$ to better visualize the growth. The white line represents the branch cut of $f$.)

2. However, thinking that this question concerns the Landau prescription, I guess OP is actually interested in the integral

$$ f(\omega) = \int_{\mathcal{C}} \frac{\sqrt{1-\xi^2}}{\xi - \omega} \, \mathrm{d}\xi $$

where $\mathcal{C}$ is a Landau contour from $-1$ to $1$. Saying differently, $f(\omega)$ is the analytic continuation of the function $\text{(*)}$, initially defined on $\operatorname{Im}(\omega) > 0 $ and then extended across and beyond the slit $[-1, 1]$. The resulting function is

$$ f(\omega) = \begin{cases}
- \pi \omega + \pi \omega \sqrt{1 - \omega^{-2}}, & \operatorname{Im}(\omega) > 0, \\
- \pi \omega + \pi i \sqrt{1 - \omega^2}, & \omega \in [-1, 1], \\
- \pi \omega + \pi \omega \sqrt{1 - \omega^{-2}} + 2\pi i \sqrt{1 - \omega^2}, & \operatorname{Im}(\omega) < 0.
\end{cases} \tag{2} $$

This function is analytic outside $(-\infty,-1]\cup[1,\infty)$.

We also note that it is impossible to find an analytic continuation of $f$ on all of $\mathbb{C}$, since $\pm 1$ are branch points of $f$.

Figure 2. The following is a domain coloring of the function $f(\omega)$ in $\text{(2)}$. Again, the white line represents the branch cut of $f$.

3. Finally, we have

\begin{align*}
\int_{-1}^{1} \frac{x\sqrt{1-x^2}}{x-\omega} \, \mathrm{d}x
&= \int_{-1}^{1} \frac{(x - \omega + \omega)\sqrt{1-x^2}}{x-\omega} \, \mathrm{d}x \\
&= \int_{-1}^{1} \sqrt{1-x^2} \, \mathrm{d}x + \omega \int_{-1}^{1} \frac{\sqrt{1-x^2}}{x-\omega} \, \mathrm{d}x \\
&= \frac{\pi}{2} + \omega f(\omega).
\end{align*}

This relation can be used to find the expression for $g$.


问:
Induction proof $1 + \frac{1}{\sqrt[3]{2}} + ... + \frac{1}{\sqrt[3]{n}} < \frac{3}{2}\sqrt[3]{n^2}$
I need to prove that:
$$1 + \frac{1}{\sqrt[3]{2}} + ... + \frac{1}{\sqrt[3]{n}} < \frac{3}{2}\sqrt[3]{n^2}$$

The problem is that I can't find a way to separate that "$1$" on the right side ($\frac{3}{2}\sqrt[3]{(n+1)^2}$) to get a sum on the let AND right side. I thought about Bernoulli's inequality but it doesn't seem to help much either.

Edit: I've got some bomb idea, I will post it as a comment

Body: Since $k^{-1/3}=\int_{k-1}^kk^{-1/3}dx<\int_{k-1}^kx^{-1/3}dx$,$$\sum_{k=1}^nk^{-1/3}<\sum_{k=1}^n\int_{k-1}^kx^{-1/3}dx=\int_0^nx^{-1/3}dx=\tfrac32n^{2/3}.$$


问:
Let $A$ be a matrix sized $p\times p$, where $2\le p$. Using recurrence relations, describe $A^k$.
Let $A$ be a matrix sized $p\times p$, Where $2\le p$.
The matrix values in the main diagonal are $0$ and the rest are $1$'s.

Example for $A$ where $p=5$:
$$\begin{bmatrix} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \\ \end{bmatrix}
$$

Using reccurences relations, describe $A^k$.

答:
HINT: Calculating a few powers of such matrices quickly suggests that $A^k$ always has just two distinct entries, one appearing on the diagonal, and the other appearing off the diagonal. You might try proving this by induction: the induction step, if it works, is likely to tell you what the recurrences actually are.

Let the $(i,j)$-entry of $A^k$ be $a_{ij}^k$, and suppose that $a_{ii}^k=b_k$ for $i=1,\dots,n$ and $a_{ij}^k=c_k$ when $i\ne j$. Then

$$a_{ii}^{k+1}=\sum_{j=1}^{i-1}c_k+\sum_{j=i+1}^nc_k=(n-1)c_k\;;$$

why? This is independent of $i$, so all of the diagonal entries of $A^{k+1}$ are equal to $b_{k+1}(n-1)c_k$. See if you can finish it by dealing with the off-diagonal elements.


问:
Reference Request: Algorithms for computation on curves (more generally varieties)
I'm looking for a book explaining how algorithms allowing the computations of

poles and zeros,

singular points,

divisors,

genera,

residues,

etc,

on algebraic curves work.
I expect most of those actually can work in higher dimensions: if so, an appropriate reference would be welcome.
Probably this starts with something like Gröbner bases, but I don't know.

Thanks!

答: Probably you want Using Algebraic Geometry by O'Shea, Little, and Cox.


问:
Show that $\int_{\mathbb{R}^3}\nabla u\cdot\nabla a\, dx=-\int_{\mathbb{R}^3}u\Delta a\, dx$.
Let $u\in H^1(\mathbb{R}^3)$ and $a\in C^\infty(\mathbb{R}^3)$ such that $|\nabla a|,\Delta a \in L^\infty(\mathbb{R}^3)$, where the norm in the Sobolev space $H^1(\mathbb{R}^3)$ is given by $\|u\|_{H^1}=\|u\|_{L^2}+\|\nabla u\|_{L^2}$. I would like to show that $$\int_{\mathbb{R}^3}\nabla u\cdot\nabla a\, dx=-\int_{\mathbb{R}^3}u\Delta a\, dx.$$ My thoughts are to use integration by parts and show that the integral over the boundary is zero, which intuitively makes sense because $|\nabla a |$ is bounded and $u$ decays. In the reference given in  this answer the authors states, in particular, that if $B_r\subset \mathbb{R}^3$ is an open ball with radius $r$ and $f,g\in C^1(\mathbb{R}^3)$, then $$\int_{B_r} D_j f(x)g(x)dx=\int_{\partial {B_r}}(fg)(y)\nu_j(y)d_{2}y-\int_{B_r} f(x)D_jg(x)dx, $$ where $d_{2}y$ is the euclidean $2$-dimensional density on $\partial B_r$ and $\nu (y)$ the outer normal at $y\in\partial B_r$, which in this case i believe is $y/r$. Since $u$ is not necessarily $C^1$, i thought that i would need to approximate $u$ by smooth functions and take the limit $r\rightarrow \infty$, does this work? Also, i currently don't know how to evaluate the integral $$\int_{\partial {B_r}}(uD_ja)(y)\nu_j(y)d_{2}y.$$

Hints or references will be appreciated.

As
$S(\mathbb{R}^3)\subset L^1(\mathbb{R}^3)$, taking $u_n\rightarrow u$ a.e., $u_n\in S(\mathbb{R}^3)$ Schwartz functions, for all $n\in\mathbb{N}$, we have
$$
\left|\int_{\partial {B_r}}(u_nD_ja)(y)\frac{y_j}{r}dy\right|\leq \|{\nabla a}\|_{\infty}\int_{\partial {B_r}}|u_n|dy\stackrel{r\rightarrow\infty}{\longrightarrow} 0.
$$
it follows that
$$\int_{\mathbb{R}^3} D_j u_n(x)D_ja(x)dx=-\int_{\mathbb{R}^3} u_n(x)D^2_ja(x)dx.$$

but now i don't know what ensures that i can take the limit $n\rightarrow\infty$ inside the integrals. The dominated convergence theorem seems to require $u\in L^1(\mathbb{R}^3)$, i cannot think of any $L^1$ function to bound the sequence.

On the other hand, i could take $a_n \rightarrow a$ with $a_n$ smooth and compactly supported. As $u\in H^1(\mathbb{R}^3)$, integration by parts holds by the definition of weak derivate, but i think i cannot take $n\rightarrow \infty$.

答:
The idea is indeed to approximate $u$ by smooth functions, and to let $r\to\infty$. To deal with the integral over $\partial B_r$, you do not need to evaluate the integral - it suffices to show it tends to $0$ as $r\to\infty$.

The basic reason is that you can choose an approximating sequence $(u_n)$ to $u$ with each $u_n\in C^\infty\cap H^1$ such that $u_n(x)\to 0$ as $|x|\to\infty$. In particular, the space of Schwartz functions $\mathcal{S}$ is a dense subspace of $C^\infty\cap H^1$ (with respect to the $H^1$ norm) and every Schwartz function satisfies this property. You could also work with the space of smooth compactly supported functions if you want - these are dense in $C^\infty\cap H^1$.

If you choose an approximating sequence $(u_n)$ of functions in $C^\infty\cap H^1$ in this way, then you can show that for fixed $n$,
$$
\int_{B_r} u_n(x)D_jg(x)dx\to 0
$$
as $r\to\infty$, since $u_n$ restricted to $\partial B_r$ tends to $0$; the claim then basically follows from dominated convergence.
The proof then boils down to choosing your approximating sequence $(u_n)$ and the order in which to take the double limits $r\to\infty,n\to\infty$ carefully.


Title: Induction proof $1 + \frac{1}{\sqrt[3]{2}} + ... + \frac{1}{\sqrt[3]{n}} < \frac{3}{2}\sqrt[3]{n^2}$

Body:
I need to prove that:
$$1 + \frac{1}{\sqrt[3]{2}} + ... + \frac{1}{\sqrt[3]{n}} < \frac{3}{2}\sqrt[3]{n^2}$$

The problem is that I can't find a way to separate that "$1$" on the right side ($\frac{3}{2}\sqrt[3]{(n+1)^2}$) to get a sum on the let AND right side. I thought about Bernoulli's inequality but it doesn't seem to help much either.

Edit: I've got some bomb idea, I will post it as a comment

Body:
You seek to verify the inequality $\frac{3}{2} \sqrt[3]{n^2} + \frac{1}{\sqrt[3]{n+1}} \overset{?}{<} \frac{3}{2} \sqrt[3]{(n+1)^2}.$
Cubing both sides yields
$$n^2 + 3 (2/3) n^{4/3}(n+1)^{-1/3} + 3(2/3)^2 n^{2/3}(n+1)^{-2/3} + (2/3)^3 (n+1)^{-1/3} \overset{?}{<}  (n+1)^2$$
$$3 (2/3) n^{4/3}(n+1)^{-1/3} + 3(2/3)^2 n^{2/3}(n+1)^{-2/3} + (2/3)^3 (n+1)^{-1/3}\overset{?}{<} 2n + 1$$

Note that the first term is $< n$, the second term is $<4/3$, and the last term is $<(2/3)^3 2^{-1/3}< 1/3$ for $n \ge 2$. So the left-hand side is bounded by $ < n+\frac{5}{3} < 2n+1$ for $n \ge 2$.


问:
Correct reasoning? About an iid random variable series
The statement I want to prove is that, given a sequence $\lbrace X_n \rbrace_n$ of independent identically distributed random variables, non degenerated in $0$, that is, none of them are almost sure $0$, then

$P(\sum_{i=1}^n X_i$ converges$)$ = $0$

I have tried the following:

Let ($\Omega, \sigma$, $P$) be the measure space where the random variables are defined.

$\{\sum_{i=1}^n X_i$ converges $\}$ is an asymptotic event related to the independent random variables $\lbrace X_n \rbrace_n$, then, by the $0-1$ Kolmogorov's law we have $P(\sum_{i=1}^n X_i$ converges$)$ $\in \{0,1\}$.

Notice that $(\{\sum_{i=1}^n X_i$ does not converge $\}$ = $\{\sum_{i=1}^n X_i$ converges $\}^c$.

As the variables are identically distributed we have for all $\omega \in \Omega$ that $X_i(\omega)$ = $X_1 (\omega)$. Thus we have $\sum_{i=1}^n X_i(\omega) = \sum_{i=1}^n X_1(\omega) = n\cdot X_1(\omega)$. (I don't know if this is correct)

Now, as the random variables are non-degenerated in $0$, we have that there should exist an element $\omega_0 \in \Omega$ such as $P(\omega_0) > 0$, with $X_1(\omega_0) = k \neq 0$, $k \in \mathbb{R} $.

Therefore we have that $\sum_{i=1}^n X_i(\omega_0) = \sum_{i=1}^n X_1(\omega_0) = n\cdot X_1(\omega_0) \longrightarrow \infty$ as $n \longrightarrow \infty$.

Then $\omega_0 \subset \{\sum_{i=1}^n X_i$ does not converge$\}$ and $P(\{\sum_{i=1}^n X_i$ does not converge $\}) \geq P(\omega_0) > 0$, but by $0-1$ Kolmogorov's law that must be $1$, and then $P(\sum_{i=1}^n X_i$ converges$)$ = $0$.

答:
No, variables been identically distributed doesn't mean $X_i(\omega) = X_1(\omega)$.

However, we don't need any complex tools to show that the series diverges a.s.

As $X_i \neq 0$, we have $P(|X_i| > \epsilon) > \epsilon$ for some $\epsilon$.

Recall that if the series converges, all terms after $n$-th for some $n$ is less than $\epsilon$ (necessary condition of series convergence).
So $$P(\text{series converge}) \leq P(\exists n \forall i > n: |X_i| < \epsilon) \leq \sum_n P(\forall i > n: |X_i| < \varepsilon)$$

$$P(\forall i > n: |X_i| < \epsilon) \leq\\ P(|X_{n + 1}| < \epsilon, P(|X_{n + 2}| < \epsilon, \ldots, P(|X_{n + N}| < \epsilon) =\\ P(|X_1| < \epsilon)^N <\\ (1 - \epsilon)^N \xrightarrow{N \to \infty} 0$$

So $P(\forall i > n: |X_i| < \epsilon) = 0$. By sigma-additivity, $\sum_n P(\forall i > n: |X_i| < \varepsilon) = 0$too. And so probability of series convergence is zero too.


问:
Where do the powers go?
I have this weird identity between power series. It's kind of like the relation between a geometric series and $\frac{1}{1-x}$. I was wondering if there was some theory developed along these lines I could read up on.

Consider
$$F(x) = (1+x^{1/2})(1+x^{1/4})\cdots$$
then
$$F(x^2) = (1+x)F(x)$$
so $$\frac{F(x^2)}{1+x} = F(x)$$

and finally,
$$F(x) = \frac{1}{1+x}\frac{1}{1+x^2}\frac{1}{1+x^4}\cdots = 1-x.$$

The main thing I'm having trouble wrapping my head around is where all the fractional powers "go"? In an expression like:
$$(1+x)(1+x^2)\cdots = 1+ x + x^2 + \ldots = \frac{1}{1-x}$$
everything still makes sense to me, all the powers still appear. But in the above expression you're summing $x$ to the power of all dyadic rationals and somehow getting a $-x$ term.

Edit: I think I have a formulation where this makes sense. As a Hahn-series where formal infinite products are defined analogously. The support will be well-ordered as we are only taking finite sums. But this would mean that Hahn series are not unique, the same Hahn series can have two different representations. Either I'm still making some sort of mistake or the above is true.

https://en.wikipedia.org/wiki/Hahn_series

答:
The last line of your argument is invalid; you've hidden an argument in the use of "$\dots$" that, if you expand it out, does not actually work at all. You've written

$$F(x) = \frac{1}{1 + x} \frac{1}{1 + x^2} \frac{1}{1 + x^4} \dots$$

when what you mean is

$$F(x) = \frac{1}{1 + x} F(x^2) = \frac{1}{1 + x} \frac{1}{1 + x^2} F(x^4) = \dots.$$

This does not imply that $F(x) = 1 - x$, because all of the power series $F(x^{2^n})$, if expanded out formally, continue to have nonzero terms with arbitrarily small exponents; the sequence of formal power series $F(x^{2^n})$ simply does not in any reasonable sense converge to $1$, as it would if $F$ were an ordinary formal power series with constant term $1$.

It is in fact possible to talk about formal power series with non-negative real exponents; this is a valuation ring. For multiplication to be well-defined we need to only consider formal power series $\sum_{r \in \mathbb{R}_{\ge 0}} c_r x^r$ where the support $\{ r : c_r \neq 0 \}$ is well-ordered. If we expand out the infinite product $F(x)$ formally (ignoring the question of whether or not it makes sense in this valuation ring) we get that $c_r \ge 1$ for all $r \in [0, 1]$ (each such $r$ has at least one infinite binary expansion, and some of them have two) so the support contains $[0, 1]$ and therefore can't be well-ordered. So this infinite product can't exist in the valuation ring. (Concretely, even if it might sort of make sense to expand $F(x)$ out formally, there's no corresponding formal expansion for $F(x)^2$; computing the coefficients of this would involve adding up uncountably many terms which are bounded below!)

I don't know what the "standard" choice of topology on a non-discrete valuation ring is, but this probably means that the infinite product doesn't converge.

Title: Where do the powers go?


问:
Induction to prove Fibonacci's sequence grows exponentially fast
How to solve this using induction?

Use induction to prove that

$F_n\ge2^{\frac n2}\;,\;$ for $n\ge6\;.$

$F_0=0\;,\;F_1=1\;,\;F_n=F_{n-1}+F_{n-2}\;.$

答:
There are three steps to any induction problem.

Step 1, prove your base cases. In this case, first show that $F(n)\geq 2^\frac n2$ when $n=6$ (i.e. $F(6)\geq 2^3=8$). Working out $F(6)$ suffices here. Since the recursive definition is based on two prior cases ($F(n)=F(n-1)+F(n-2)$, there are two terms on the RHS), you also need a second base case, the next one along, here that is $n=7$, so show that $F(7)\geq 2^\frac 72$ (Hint: square this number to find its approximate numerical value)

Step 2, Assume that the property holds true for a given $k,k+1$. That is, assume $F(k)\geq 2^\frac k2, F(k+1)\geq 2^{\frac{k+1}{2}}\tag1$

Step 3: Use this to prove the statement holds for the next case, i.e. that $(1)$ implies that $$F(k+2)=F(k+1)+F(k)\geq 2^{\frac{k+2}{2}}$$


问:
Is this a correct way to show that $\sum_{n \geq 0} \frac{n^3}{n!}=5e$
Is this a correct way to show that $\sum_{n \geq 0} \frac{n^3}{n!}=5e$ ?

$$S_3 = \sum_{n \geq 0} \frac{n^3}{n!}=\sum_{n \geq 1} \frac{n^2}{(n-1)!} \implies$$

$$S'_3=S_3-e=\sum_{n \geq 1} \frac{n^2-1^2}{(n-1)!}=\sum_{n \geq 2} \frac{n+1}{(n-2)!} \implies$$

$$S'_3-3e=\sum_{n \geq 2} \frac{(n+1)-3}{(n-2)!}=\sum_{n \geq 3} \frac{1}{(n-3)!}=e\implies$$

$$S'_3-3e=e\iff S_3=5e$$

Exploring $ \sum_{n=0}^\infty \frac{n^p}{n!} = B_pe$, particularly $p = 2$. : One of the answers shows how $\sum_{n \geq 0} \frac{n^2}{n!}=2e$ and asserts that $\sum_{n \geq 0} \frac{n^3}{n!}=5e$ can be showed in the same manner of reasoning. Is my "proof" correct?

答:
Seems good to me. An alternate way - which is slightly neater - would be to first write $n^3$ in a convenient way:
\begin{align} n^3 &= n^2(n-1) + n^2 \\
&= n(n-1)(n-2) + 2n(n-1) + n^2 \\
&= n(n-1)(n-2) + 2n(n-1) + n(n-1) + n.\end{align}

Then, since all the sums involved converge, \begin{align} \sum_{n \ge 0} \frac{n^3}{n!} &= \sum_{n \ge 0} \frac{n(n-1)(n-2)}{n!} + \frac{3n(n-1)}{n!} + \frac{n}{n!} \\ &= \sum_{n\ge 3} \frac{1}{(n-3)!} + \sum_{n \ge 2} \frac{3}{(n-2)!} + \sum_{n \ge 1} \frac{1}{(n-1)!} = 5e.\end{align}

Expressions such as $n(n-1)(n-2)$ are called falling factorials:
$(n)_k = n (n-1) \dots(n-k+1)$. These are sometimes also denoted $n^{\underline{k}}$. There is a standard way to write powers as falling factorials: $$ n^k = \sum_{i = 0}^k \begin{Bmatrix} n \\ i\end{Bmatrix} (n)_i, $$ where $\begin{Bmatrix} n \\ i\end{Bmatrix}$ are the Stirling numbers of the second kind. Using this, and the observation that $\frac{(n)_i}{n!} = \frac{1}{(n-i)!},$ you can generalise the above decomposition to any $k$ instead of $3$, and derive a similar result, that $$ \sum_{n \ge 0} \frac{n^k}{n!} = \left( \sum_{i = 0}^k \begin{Bmatrix} n \\ i\end{Bmatrix}\right) e = B_k e,$$ where $B_k$ are the Bell numbers, that are simply the sum in the parenthesis above.


问:
Having trouble understanding nilpotent and supersolvable group in terms of short exact sequence

$G$ is supersolvable if $G_{i+1}/G_i$ is cyclic for every $i\geq 0$
where $G_i$ is a normal sequence of $G=G_n$.

According to the lecture note I read, it says that
in the following short exact sequence,

$1\to G_{i+1}/G_i\to G/G_i\to G/G_{i+1}\to 1$

supersolvable group implies that we have successive cyclic extensions of $\{1\}$ for all $i$. From my understanding, this means that for every mapping we have above, kernel of it is cyclic. So following this new definition, we have to have that $1$, $G_{i+1}/G_i$, $G/G_i$ should be cyclic (since short exact sequence means that the image of mapping in the former is equal to the kernel of the latter). The original definition of supersolvable demands that $G_{i+1}/G_i$  is cyclic, but not necessarily $G/G_i$. Where did I have a mistake?

Also, I have another question.

If $G$ is supersolvable, any central
extension of $G$ is also super solvable. Hence nilpotent group is
supersolvable group.

How do I see this? Is there any good reference to it? I would greatly appreciate any help.

答:
Every finitely generated nilpotent group is supersolvable.  It is not true that any nilpotent group is supersolvable.

In fact, all supersolvable groups are finitely generated.  So take the free nilpotent group on an infinite set as a counterexample.  Or any uncountable nilpotent group, which are known to exist, if we assume the generalized continuum hypothesis (CH), for instance.


问:
Given a complex number set, how do you prove which set is it in the complex plane?
Let $a,b \in \mathbb{C}$ is it true that:
$|a| < |b|$ implies $|a|^{2} < |b|^{2}$
I want to use this to build sets of points in the complex plane since I have things like:
$$|z-1| < |z|$$
So I want to apply power to both sides and then change $z = x+iy$ to discover what the set is in terms of the real coordinates.

Is it useful?

Which other strategies do you recommend in order to discover what the set is in terms of the real coordinates?

答: Geometrically, $|z-1| < |z|$ means that the distance from $z$ to the point $(1,0) $ is less than its distance to $(0,0)$, which covers the half plane to the right of $x=\frac12$.


问:
On the proof that $x^2+y^2-z^2=1$ is a topological manifold
Let $S=\{(x,y,z)\in\mathbb{R}^3\ |\ x^2+y^2-z^2=1\}$ and, for every $\theta \in \mathbb{R}$: $$L_\theta = \{(x,y,z)\in \mathbb{R}^3\ |\ (x-z)\cos\theta=(1-y)\sin\theta,\ (x+z)\sin\theta=(1+y)\cos\theta\}$$
One can easily prove that $L_\theta=\{(\sin2\theta,-\cos2\theta,0)+t(\cos2\theta,\sin2\theta,1)\ |\ t,\theta\in \mathbb{R}\}$ for all $\theta \in \mathbb{R}$ and that $\bigcup_{\theta \in \mathbb{R}} L_\theta=S.$

This proves that the map $\sigma:\mathbb{R}^2\rightarrow S, \ (\theta,t)\mapsto (\sin2\theta+t\cos 2\theta, -\cos 2\theta+t\sin2\theta, t)$ is well defined. I'm convinced of the fact that one can find a family $\{U_i\}_{i \in I}$ of open sets of $\mathbb{R}^2$ such that the restriction of $\sigma$ to $U_i$ (which I'll indicate with $\sigma_i:U_i\rightarrow \sigma(U_i)$) is a homeomorphism for all $i \in I$ and such that $\bigcup_{i\in I}\sigma(U_i)$, but I'm having trouble coming up with such an atlas. Does this lead to some viable solution? How would one define the charts?

答: If you don't care about constructing explicit chart maps, you could say that $S = F^{-1}(0)$ is the zero set of the polynomial equation $F(x,y,z) = x^2 + y^2 -z^2 -1$. Since $F$ is smooth, and the Jacobian $\begin{pmatrix} 2x & 2y & -2z\end{pmatrix}$ has full rank on $S$, you can locally write it as the graph of a function using the Implicit Function Theorem. That gives you a local homeomorphism to $\mathbb{R}^2$.


问:
Existence of a graph G?
The Q i am asked is as follows.

Q:Let G=(V,E) be a graph with nine vertices  such that each vertex has a degree 5 or 6. Show that G has at least 5 vertices of degree 6, or at least 6 vertices of degree 5.

first lets bust the middle 5 of6 = 9-5 thus 4of 5 and  6 of 5 = 9-6 hence 3 of 6.

so they are claiming that 4 vertex of degree 6 and 5 of degree 5 cannot be a graph. and anything else is fine.
clearly $4*6 + 5*5 = 24+25 = 49$  thus $E= 24.5$ which i would like to see you show me half an edge?

Now i could draw pictures of all the other graphs ( me being lazy i want to find a better way)  or i could ask is the existence of an even sum here sufficient to prove that there exists a simple connected graph? i know that having half an edge is sufficient to disprove the existence  but for a set number of vertices if there exists an even number of edges does a graph need to exist?

EDIT: ( to downvote and comment on duplicate)

My question is for a set number of vertices if there exists an even number of edges does a graph need to exist?

答:
For simple graphs, i.e., those without loops or multiple edges, the complete answer to your final question is given by the Erdős-Gallai theorem: it gives a straightforward necessary and sufficient condition for a finite sequence of positive integers to be the degree sequence of a simple graph. Specifically, if the numbers are $d_1\ge d_2\ge\ldots\ge d_n$, a simple graph with vertices of these degrees exists if and only if

$$\sum_{i=1}^kd_i\le k(k-1)+\sum_{i=k+1}^n\min\{d_i,i\}$$

for $k=1,\dots,n$.

Title: Existence of a graph G?


问:
Finding parametric expression for epicycloid
This is a question that comes from Shifrin's Multivariable Mathematics [Edited after being put in place by the author :)]

A circle of radius $b$ rolls without slipping outside a circle of radius $a > b$. Give the parametric equations of a point $P$ on the circumference of the rolling circle (in terms of the angle $\theta$ of the line joining the centers of the two circles). Attached is a figure of the question:

In the text a method to approach this question is by connecting vectors and the idea of associating polar coordinates to the usual $x,y$ values if a cartesian system was used. Using these ideas along with
right triangles I was able to get the first part of the idea being captured:

$$\begin{bmatrix}
(a+b)\cos\theta \\
(a+b)\sin\theta \\
\end{bmatrix}
$$

Fortunately there is a solution to the question. After looking at it I'm stuck trying to understand how the full solution cam about which is the following:

$$\begin{bmatrix}
(a+b)\cos\theta - b\cos((a+b)\theta/b) \\
(a+b)\sin\theta - b\sin((a+b)\theta/b) \\
\end{bmatrix}
$$

So how do we reason the subtraction portions of these expressions? From what I understand I know there is the notion of using the arc length and equating it to a distance of one of the sides, but I couldn't see what side that may be, the other thing that is troubling me is the trig expression $\frac{((a+b)\theta)}{b}$. What is the trig relationship between the triangles I'm missing to use in this situation?

答:
Here's a picture with some more annotations:

I've highlighted, in blue, an arc $c$ of the larger circle and, in red, an arc $d$ of the smaller circle, as well as drawn, in green, an arrow from the center of the smaller circle $B$ to $P$. The part of the expression you understand is essentially the position of $B$, so let's look at how to determine the vector from $B$ to $P$.

The first thing to observe is that the arcs $c$ and $d$ must have the same length, since they are rolling on one another. Let $\alpha$ be the angle $\angle ABP$. The length of $c$ is $\theta a$ and the length of $d$ is $\alpha b$, since arc length is just angle times radius. Thus $\alpha = \frac{a}b\cdot \theta$.

To figure out the vector for the green arrow, however, we really want the angle between $BP$ and a horizontal reference line - the same way that we figured out the vector from $A$ to $B$. Suppose that the angle of $BP$ counterclockwise from horizontal is $\kappa$, the same way that $\theta$ is the angle of $AB$ counterclockwise from horizontal. We can determine $\kappa$ by extending the horizontal line through $A$ and also extending the line $BP$. These extended lines intersect at some point $I$. The triangle $ABI$ has angles $\theta$ and $\alpha$ and $-\kappa$ - the negative sign being since the angle measured at $\angle BIA$ is actually how far clockwise of horizontal the vector $BP$ is from horizontal where we really wanted to know a counterclockwise angle for consistency.  Thus, since the sum of the angles of a triangle is $\pi$, we get $$\kappa = -\pi + \alpha + \theta = -\pi + \frac{a}b \theta + \theta = -\pi + \frac{a+b}b \theta$$
This gives that the displacement of the vector $BP$ is
$$\begin{bmatrix}b\cos\left(-\pi+\frac{a+b}b\theta\right)\\b\sin\left(-\pi+\frac{a+b}b\theta\right)\end{bmatrix}$$
which, since $\sin(x+\pi)=-\sin(x)$ and $\cos(x+\pi)=-\cos(x)$,
$$\begin{bmatrix}-b\cos\left(\frac{a+b}b\theta\right)\\-b\sin\left(\frac{a+b}b\theta\right)\end{bmatrix}.$$
Summing this to the center of the smaller circle gives the answer.

You could also get this by imagining that instead of the smaller circle rolling on the larger one, they are both keeping their centers constant and spinning without slipping (like gears). Then, the calculation of $\alpha$ would tell you that the smaller circle is spinning at a ratio of $\frac{a}b$ times the speed of the larger one. In this view, a point on the smaller circle is just tracing the circumference of the smaller circle at a fixed rate - which is easy enough to describe in cartesian coordinates.

If you then imagined that you were an observer standing on and rotating with the larger circle, you would realize that, from your point of view, the smaller circle was revolving around you and moving without slipping - as in the original problem. You could thus also get the answer by starting with the "both circles spin in place" case and shifting reference frames appropriately - essentially by rotating all of space over time. This is definitely not easier algebraically, but it might be helpful conceptually. (Or might not - it's at least a fun way to look at it)


Title: Induction to prove Fibonacci's sequence grows exponentially fast

Body:
How to solve this using induction?

Use induction to prove that

$F_n\ge2^{\frac n2}\;,\;$ for $n\ge6\;.$

$F_0=0\;,\;F_1=1\;,\;F_n=F_{n-1}+F_{n-2}\;.$

Body:
$F_6=8\ge2^{\frac62}\;.$

$F_7=13\ge\sqrt{2^7}=2^{\frac72}\;.$

Now, for $\;n\ge8\;,$ we suppose that $\;F_{n-2}\ge2^\frac{n-2}{2},\;F_{n-1}\ge2^\frac{n-1}{2}$ and prove that $\;F_n\ge2^\frac n2.$

$F_n=F_{n-1}+F_{n-2}\ge2^{\frac{n-1}{2}}+2^{\frac{n-2}{2}}=2^{\frac{n-2}{2}}\big(\sqrt 2+1\big)\ge$

$\ge2^{\frac n2-1}\cdot 2=2^{\frac n2}.$

Hence, by induction on $n$, it follows that

$F_n\ge2^\frac n2\quad$ for all $\;n\ge6\;.$


问:
A integral computation problem
For $a,b \geq 0$, integrate
$$\int_0^\infty \dfrac{x^\frac{1}{a}}{b}e^{-x} dx. $$

I came across this integral and I have no idea where to begin.

答:
Notice that if we let $\alpha=\frac1a$ your integral is:
$$I=\frac1b\int_0^\infty x^\alpha e^{-x}dx$$
and the Gamma function is defined as:
$$\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}dx$$
and so:
$$I=\frac1b\Gamma(\alpha+1)=\frac1b\Gamma\left(1+\frac1a\right)$$

Title: A integral computation problem


答:
If you start with $F(x)=1-x= (1-x^{1/2^k}) \prod_{n=1}^k (1+x^{1/2^n})$ then yes everything is fine in $\bigcup_{n\ge 1}\Bbb{C}[[x^{1/2^n}]]$, but you can't let $k\to \infty$ as you did, ie. $\lim_{k\to \infty} \prod_{n=1}^k (1+x^{1/2^n})$ doesn't converge in any topological integral domain where $x\ne 0$

(except in $\overline{\Bbb{F}}_2$ with the discrete topology and $x=1$  which gives $F=0=1-x$ by the way)

Title: Where do the powers go?

Body:
I have this weird identity between power series. It's kind of like the relation between a geometric series and $\frac{1}{1-x}$. I was wondering if there was some theory developed along these lines I could read up on.

Consider
$$F(x) = (1+x^{1/2})(1+x^{1/4})\cdots$$
then
$$F(x^2) = (1+x)F(x)$$
so $$\frac{F(x^2)}{1+x} = F(x)$$

and finally,
$$F(x) = \frac{1}{1+x}\frac{1}{1+x^2}\frac{1}{1+x^4}\cdots = 1-x.$$

The main thing I'm having trouble wrapping my head around is where all the fractional powers "go"? In an expression like:
$$(1+x)(1+x^2)\cdots = 1+ x + x^2 + \ldots = \frac{1}{1-x}$$
everything still makes sense to me, all the powers still appear. But in the above expression you're summing $x$ to the power of all dyadic rationals and somehow getting a $-x$ term.

Edit: I think I have a formulation where this makes sense. As a Hahn-series where formal infinite products are defined analogously. The support will be well-ordered as we are only taking finite sums. But this would mean that Hahn series are not unique, the same Hahn series can have two different representations. Either I'm still making some sort of mistake or the above is true.

https://en.wikipedia.org/wiki/Hahn_series


问:
Euclidean division in quotients of polynomial rings
I know that given a field $\mathbb{K}$, the one variable polynomial ring $\mathbb{K}[x]$ is an euclidean domain. This helps to figure out how the quotient $\dfrac{\mathbb{K}[x]}{(f(x))}$ (where $f(x) \in \mathbb{K}[x]$) is made: its elements are the polynomials $h(x)$ such that $\text{deg}(h(x))<\text{deg}(f(x))$, because of the euclidean division. If we have a generic $A[x]$ where $A$ is a commutative ring with $1$, what can we say about $\dfrac{A[x]}{(f(x))}$ with $f(x) \in A[x]$?

答:
Let $c \in A$ be the leading coefficient of $f$. If $c$ is a unit, then every element of $A[x]/(f)$ can still be represented as a polynomial in $x$ of degree less than $\deg(f)$, by exactly the same argument. However, if $c$ isn't a unit, the situation is a bit more complicated.

For example, consider $\newcommand{\ZZ}{\mathbb{Z}}\ZZ[x]/(2x - 1)$. If $g \in \ZZ[x]$ has odd leading coefficient, then for any $h \in \ZZ[x]$, we have $\deg(g + (2x - 1)h) \geq \deg(g)$. In particular, any system of representatives of $\ZZ[x]/(2x - 1)$ requires polynomials of arbitrarily high degree. (One can also see this by observing that $\ZZ[x]/(2x - 1) \cong \ZZ[\frac{1}{2}]$, and one can have arbitrarily high powers of $2$ in the denominator of elements of $\ZZ[\frac{1}{2}]$.)


问:
If $\operatorname{sp}(A) \cup \operatorname{sp}(B)=\operatorname{sp}(A\cup B) \Rightarrow A\cup B$ is linearly dependent
$\newcommand{\sp}{\operatorname{sp}}$

Let $V$ be a vector space over $F$ field, and let $A,B$ be two different, disjoint, non empty sets of vectors from $V$.

If $\sp(A) \cup \sp(B)=\sp(A\cup B) \Rightarrow A\cup B$ is linearly dependent

I've started by saying that if $\sp(A) \cup \sp(B)=\sp(A\cup B)$
then $\sp(A) \subseteq \sp(B)$ or $\sp(B)\subseteq \sp(A)$.

Thus if we assume, WLOG, that $\sp(A) \subseteq \sp(B)$ and we take $v_1 \in A$ and multiply it by scalars $\alpha_1,\ldots,\alpha_k$ we'll get a vector from $B$.

Thus $ A\cup B$ is linearly dependent.

I feel like this proof is not good enough, where is it failing?

答:
Your idea is good, but not expressed in the best way.

You can assume, as you did, $\def\sp{\mathrm{sp}}\sp(A)\subseteq\sp(B)$. If the only vector in $A$ is $0$, you have nothing to prove, because any set containing the zero vector is linearly dependent. Otherwise, take $v\in A$, $v\ne0$. Then you can write
$$
v=\alpha_1w_1+\alpha_2+\dots+\alpha_nw_n
$$
for suitable $w_1,w_2,\dots,w_n\in B$ and so $A\cup B$ is not linearly independent.


答:
Alternatively, let $E=\sum \frac{z^n}{n!}=e^z$. Then

$A=zE'=\sum \frac{nz^n}{n!}=ze^z$

$B=zA'=\sum \frac{n^2z^n}{n!}=(z^2+z)e^z$

$C=zB'=\sum \frac{n^3z^n}{n!}=(z^3+3z^2+z)e^z$

Therefore, $C(1)=5e$.

Body:
Is this a correct way to show that $\sum_{n \geq 0} \frac{n^3}{n!}=5e$ ?

$$S_3 = \sum_{n \geq 0} \frac{n^3}{n!}=\sum_{n \geq 1} \frac{n^2}{(n-1)!} \implies$$

$$S'_3=S_3-e=\sum_{n \geq 1} \frac{n^2-1^2}{(n-1)!}=\sum_{n \geq 2} \frac{n+1}{(n-2)!} \implies$$

$$S'_3-3e=\sum_{n \geq 2} \frac{(n+1)-3}{(n-2)!}=\sum_{n \geq 3} \frac{1}{(n-3)!}=e\implies$$

$$S'_3-3e=e\iff S_3=5e$$

Exploring $ \sum_{n=0}^\infty \frac{n^p}{n!} = B_pe$, particularly $p = 2$. : One of the answers shows how $\sum_{n \geq 0} \frac{n^2}{n!}=2e$ and asserts that $\sum_{n \geq 0} \frac{n^3}{n!}=5e$ can be showed in the same manner of reasoning. Is my "proof" correct?


问:
Assigning tasks to minimize cost
I have a set of jobs that must be done on a given day in sequence. $J_1, J_2, J_3$, with deadlines $D_1, D_2, D_3$. There are three workers $W_1, W_2, W_3$ that can execute the tasks each one with varying effects of proficiency $V_1, V_2, V_3$, so a highly proficient worker will do the task in less time, but will also incur a higher cost $C_1, C_2, C_3$. I want to figure out the constraints for a linear program that will execute all tasks within their respective deadline, while minimizing my costs.

I have:
$x_{ij}=
\begin{cases}
1 &\text{if job $i$ executed by worker $j$},\\
0 &\text{otherwise}
\end{cases}$

and one of the constraints I believe should be the fact that I need to choose one, and only, one worker to do the job, so maybe:

$$\sum_{j=1}^3 X_{ij}=1 \quad \text{for $i\in J$}$$

The next constraints that we should consider is obviously the deadline; time to execute previous jobs plus time for current job, since they must be done in sequence (production line), so maybe:

$$\sum_{i=1}^3 \sum_{j=1}^3 T_{ij}X_{ij}\le D_i$$

$$ T_{ij}, i \in J, j \in W $$ is the time that worker $j$ needs to execute job $i$.

Are these constraints correct, or do they make sense in the first place?
If not what should I consider?

答:
The first constraint should be for $i\in J$ instead of $i\in W$.

The second constraint is missing $i\in J$, and you should not mix $\sum_i$ and $i\in J$.  It might help to write this as three separate constraints, one per job.  Also, you have not used $V_i$ anywhere.

Title: Assigning tasks to minimize cost


问:
Using properties of Cauchy sequence to prove uniform convergence
So I have a partial solution to the problem below but I reached a dead end that I'm hoping someone could assist me with.

Let $V$ be a complete normed vector space.

We will use $\|\cdot\|$ to indicate the norm on $V$.

Part i) Let $(f_n)$ be a Cauchy sequence in $C([a,b],V)$. Show that for each $x \in [a,b], (f_n(x))$ is Cauchy, and so define the pointwise limit $f(x) = \lim\limits_{n \to \infty} f_n(x)$.

Solution: We know $\forall x \in [a,b], \forall \epsilon > 0, \exists N$ such that if $m,n \geq N$, then $\|f_n(x) - f_m(x)\| < \epsilon$.

So let $x \in [a,b]$. Hence, $\forall \epsilon > 0, \exists N$ such that if $n \geq N$, then $\|f_n(x) - f_m(x)\| < \epsilon$.

This shows that $(f_n(x))$ is Cauchy in $V$. Since $V$ is complete, $(f_n(x))$ is convergent. Hence, we can define $f(x) = \lim\limits_{n \to \infty} f_n(x)$.

I'm confident that I did this portion of the problem correctly, but I am not completely certain.

Part ii) Prove that $(f_n)$ converges uniformly. HINT: Use the Cauchy criterion to obtain an estimate for $\|f_n(x) - f(x)\|$ that is independent of the point $x$.

So I'm having a bit of trouble with this part.

So I know $\|f_n(x) - f(x)\| = \lim\limits_{m \to \infty}\|f_n(x) - f_m(x)\|$ but I don't know how to bound this by $\epsilon$.

Any help or corrections to the work I have done will be appreciated.

答:
If this is word-for-word what your question says, then if I have to make an educated guess, I would say we put the sup norm on $C([a,b],V)$. The way the question is phrased does indeed suggest that the objective is to prove that $C([a,b],V)$ with sup norm is a Banach space (and usually in my experience, if nothing else is said, this is the norm we consider on the space of continuous functions).

For any function $g:[a,b]\to V$, I shall write $\lVert g \rVert_{\infty}:= \sup_{x\in [a,b]}\lVert g(x)\rVert$, where $\lVert\cdot \rVert$ is the norm on $V$. Now, by definition of $(f_n)$ being Cauchy in $C([a,b],V)$ (with sup norm), given $\epsilon>0$, there exists $N\in\Bbb{N}$ such that for all $m,n\geq N$, we have
\begin{align}
\lVert f_n-f_m\rVert_{\infty}\leq \epsilon.
\end{align}
Now, let $x\in [a,b]$ be arbitrary, and let $n\geq N$. Then,
\begin{align}
\lVert f_n(x)-f(x)\rVert&= \lVert f_n(x)-\lim_{m\to \infty}f_m(x)\rVert\\
&=\lim_{m\to \infty}\lVert f_n(x)-f_m(x)\rVert \tag{$*$}\\
&\leq \limsup_{m\to \infty}\lVert f_n-f_m \rVert_{\infty}
\end{align}
(by the way I put a $\limsup$ in the end because for a given $n$, $\lim_{m\to \infty}\lVert f_n-f_m\rVert_{\infty}$ need not exist, but the limit superior of a sequence of real numbers always exists).
Try to justify why you can pull the limit out in $(*)$. Now, can you finish up the proof that $f_n\to f$ uniformly (i.e $\lVert f_n-f \rVert_{\infty}\to 0$)?


问:
Calculating the matrix exponential of rotation matrix
I have the matrix:

$$A=\begin {pmatrix}
0 &-1\\
1 & 0\\
\end{pmatrix}$$
And I want to try to calculate its exponential using this formula $$\ e^{M} = \sum_{k=0}^\infty \frac{1}{k!}\ M^{k}\\$$

I have worked out that $$A^{2} = -I$$ $$A^{3}=-A$$ and $$A^{4}=I$$ where $I$ is the identity matrix. I have then tried to use the fact that the sum will cycle through these matrices to separate the sum and then recombine it into one matrix. However what I get out cannot easily be expressed as a sum. Any help would be greatly appreciated!

答:
$${\rm A}^2 = - {\rm I}_2$$

Hence, matrix ${\rm B} := \frac{{\rm A}}{i}$ is involutory, i.e., ${\rm B}^2 = {\rm I}_2$. Using Euler's formula,

$$\begin{aligned} \exp({\rm A}) = \exp \left( i {\rm B} \right) &= \cos \left( {\rm B} \right) + i \sin \left( {\rm B} \right)\\ &= \cos \left( 1 \right) \, {\rm I}_2 + i \sin \left( 1 \right) \, {\rm B}\\ &= \color{blue}{\cos \left( 1 \right) \, {\rm I}_2 + \sin \left( 1 \right) \, {\rm A}}\end{aligned}$$

where $\cos \left( {\rm B} \right) = \cos \left( 1 \right) \, {\rm I}_2$ and $\sin \left( {\rm B} \right)=  \sin \left(1 \right) \, {\rm B}$ because ${\rm B}$ is involutory.


问:
What can we say about this conditional expectation of a conditional expectation?
I have just started on probability theory and I was thinking about this object.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. Let $\mathcal{F_1},\mathcal{F_2}$ be sub $\sigma$-algebras. If we just have any random variable on the original probability space, can we say anything about $\mathbb{E}[\mathbb{E}[X|\mathcal{F_1}]|\mathcal{F_2}]?$

Can we simplify this in some way? I know there is some sort of tower property given the inclusion, however I am explicitly not assuming it here. In particular I am conjecturing $\mathbb{E}[\mathbb{E}[X|\mathcal{F_1}]|\mathcal{F_2}]=\mathbb{E}[X|\sigma(\mathcal{F_1},\mathcal{F_2})]$ but I am not sure if that is true?

Many thanks in advance!

答:
No, you are wrong.  If $\mathcal F_1 \subseteq \mathcal F_2$, then
$$
\mathbb{E}[\mathbb{E}[X|\mathcal{F_1}]|\mathcal{F_2}] =
\mathbb{E}[\mathbb{E}[X|\mathcal{F_2}]|\mathcal{F_1}] =
\mathbb{E}[X|\mathcal{F_1}] .
$$
We used to call this rule "coarse wins".  But in this case we have
$\sigma(\mathcal{F_1},\mathcal{F_2}) = \mathcal F_2$, so your conjecture makes the wrong choice.

If neither of $\mathcal F_1 , \mathcal F_2$ contains the other, then the composition $\mathbb{E}[\mathbb{E}[X|\mathcal{F_1}]|\mathcal{F_2}]$ in general will not be of the form $\mathbb E[X | \mathcal G]$ for some sigma-algebra $\mathcal G$.


问:
Find the formula of the sequence using generating functions
I have the following sequence given:

$$\sum_{k=1}^{n} (-1)^{k}k^{2}$$

How can I do this? The sequence goes like this:

$$-1 + 4 - 9 + 16 - 25 + 36 - ...$$

So it doesn't have any variables inside. It looks like a geometric sequence for me, so the simple known formula would do the whole thing, but I actually doubt it's that simple.

So there are some generating functions that are quite similar, like:

$$\sum (-1)^{n}x^{n} = x - x^{2} + x^{3} - x^{4} + ...$$

But, uhm, well...what's next? The lack of x variable seems a bit strange to me.

答:
Start with:
$$
\sum_{0 \le k \le n} (-1)^k z^k = \frac{1 - (-z)^{n + 1}}{1 - (-z)}
$$
Then note that:
$$
z \frac{d}{dz} \sum_{k} a_k z^k = \sum_k k a_k z^k
$$
This hints at differentiating the above sum twice and evaluating at $z = 1$:
$$
z \frac{d}{dz} \left(z \frac{d}{dz} \frac{1 - (-z)^{n + 1}}{1 + z} \right) \big\lvert_{z = 1}
  = \frac{(-1)^n n (n + 1)}{2}
$$

Maxima help with the mess is gratefully acknowledged.


问:
How many pillars do we need to surround a triangular area?
Suppose, I have a triangular garden with sides of $25$ feet, $40$ feet and $55$ feet. If I want to surround my garden with pillars set $5$ feet apart, how many pillars will I need?

I calculated the triangle's perimeter and divided it by $5$. I got 24.

Since it is a enclosed area, I will need one less pillar. So I will need a total of $23$ pillars.

But someone suggested me that I actually need $22$ pillars, but he didn't explain why. So am I miscalculating something?

答:
Since it is a enclosed area, I will need one less pillar.

If you mean that it's a closed curve, it is true that closed curves need one fewer than open curves. However, you are mistaken in thinking that the number for open curves is p/d, and the number for closed curves is one less than that. Rather, the number for closed curves is p/d, and the number for open curves is one more than that.

The person who suggested $22$ seems to also be aware of this "off by one", but doesn't realize that you've already subtracted one.

One tactic when you're trying to remember how a rule works is to take a simple case. If you have a perimeter of 3 feet, and want a post every foot, how many posts you need?


问:
How do I calculate the number of permutations for this list [1, 2, 3, 1, 2, 3, 1, 2, 3]?
How do I calculate the number of permutations for the list,
$$[1, 2, 3, 1, 2, 3, 1, 2, 3]?$$
I thought the number of permutations is given by
$$\frac{9!}{3\times3!}.$$
However, I created a python code to test this and it suggests that the number of permutations is much less than this. Can you see where I have gone wrong? Is my formula incorrect or have I made a mistake in the code?

Here is the python code:

```<code>from itertools import permutations
from math import factorial

perm = permutations([1, 2, 3, 1, 2, 3, 1, 2, 3])

# Generate all permuations based on index number
mylist = list(perm)
print('The total number of permuations based on cell index =', len(mylist))
print('Note that 9! =', factorial(9))

# Remove all repeated entries based on cell value
mylist = list(dict.fromkeys(mylist))
print('The total number of permuations based on cell value is ', len(mylist))
print('Note that 9! / (3 * 3!) = ', int(factorial(9) / (3 * factorial(3))))
</code>```

Here is the output:

```<code>The total number of permuations based on cell index = 362880
Note that 9! = 362880
The total number of permuations based on cell value is  1680
Note that 9! / (3 * 3!) =  20160
</code>```

答: The denominator should be $(3!)^3$ and not $3\cdot 3!$. This will give you the right value $1680$.


问:
Bounded functions and inclusion on the $L^{p} $ spaces
Let $f \in L^p(X) $ be a bounded function, show that $\forall q \geq p, \; f \in L^q(X)$

I'm aware that this result holds for the case where $\mu(X)<\infty$ but I can't see how to prove this for when the measure of the whole space is not necessarily finite. I've tried using Holder's inequality but that didn't seem to help much.

答:
Since $f$ is bounded there is some $K >0$ such that
$$
|f| \leq K.
$$
Hence
$$
\int |f|^q d\mu= \int |f|^p|f|^{q-p} d\mu \leq \int K^{q-p}|f|^p d\mu=K^{q-p} \int |f|^p d\mu.
$$
Since $f \in L_p$, $\int |f|^p d\mu$ is finite. Hence $\int |f|^q d\mu$ is finite and $f \in L_q$.


答:
The denominator should not be $3\times 3!$ but $(3!)^3$.

Namely, you start with a situation where you distinguish all the $1$'s, all the $2$'s and all the $3$'s, which gives you $9!$ permutations, but then need to divide by $3!$ three times (to account for the fact that you don't distinguish them), which is effectively dividing by $(3!)^3$.

Title: How do I calculate the number of permutations for this list [1, 2, 3, 1, 2, 3, 1, 2, 3]?

Body:
How do I calculate the number of permutations for the list,
$$[1, 2, 3, 1, 2, 3, 1, 2, 3]?$$
I thought the number of permutations is given by
$$\frac{9!}{3\times3!}.$$
However, I created a python code to test this and it suggests that the number of permutations is much less than this. Can you see where I have gone wrong? Is my formula incorrect or have I made a mistake in the code?

Here is the python code:

```<code>from itertools import permutations
from math import factorial

perm = permutations([1, 2, 3, 1, 2, 3, 1, 2, 3])

# Generate all permuations based on index number
mylist = list(perm)
print('The total number of permuations based on cell index =', len(mylist))
print('Note that 9! =', factorial(9))

# Remove all repeated entries based on cell value
mylist = list(dict.fromkeys(mylist))
print('The total number of permuations based on cell value is ', len(mylist))
print('Note that 9! / (3 * 3!) = ', int(factorial(9) / (3 * factorial(3))))
</code>```

Here is the output:

```<code>The total number of permuations based on cell index = 362880
Note that 9! = 362880
The total number of permuations based on cell value is  1680
Note that 9! / (3 * 3!) =  20160
</code>```


问:
Which is the correct application of Chebyshev's inequality in the below case?
Starting from a probability space $\left(\Omega,\mathcal{F},\mathbb{P}\right)$, a standard normal random variable $Z\sim\mathsf{N}\left(0,1\right)$ and a positive constant quantity $\varepsilon$, if I have to say something about the following probability:
$$\mathbb{P}\left(Z<-\dfrac{\sqrt{\varepsilon}}{\Delta t}\left(1+\left(1-\dfrac{1}{\varepsilon}\right)\Delta t\right)\right)\tag{1}$$
By applying Chebyshev's inequality, doesn't one have that (since $Z$ is a standard normal random variable, its second moment equals $1$):
$$\mathbb{P}\left(Z<-\dfrac{\sqrt{\varepsilon}}{\Delta t}\left(1+\left(1-\dfrac{1}{\varepsilon}\right)\Delta t\right)\right)\leq 1-\dfrac{1}{\dfrac{\varepsilon}{\Delta t^2}\left(1+\Delta t-\dfrac{\Delta t}{\varepsilon}\right)^2}\tag{2}$$
?

If so, how can $(2)$ be furtherly expanded/simplified? My goal is to be able to say how $\mathbb{P}\left(Z<-\dfrac{\sqrt{\varepsilon}}{\Delta t}\left(1+\left(1-\dfrac{1}{\varepsilon}\right)\Delta t\right)\right)$ behaves as $\varepsilon$ increases/decreases and as $\Delta t$ increases/decreases.

If my reasoning is wrong, which would be the correct application of Chebyshev's inequality to $(1)$?

答:
Let $M := M(\epsilon,\Delta t) = \frac{\sqrt{\epsilon}}{\Delta t}\left(1 + \left(1 - \frac{1}{\epsilon}\right)\Delta t\right)$. Recall that for any random variable $X$ with a finite second moment and non-negative number $x$, Chebyshev's inequality states:

$$\mathbb{P}(|X| > x) \leq \frac{\mathbb{E}[X^2]}{x^2}.$$

I'm guessing your application of the inequality roughly followed these steps:

$$\mathbb{P}(Z < -M) = 1 - \mathbb{P}(Z > -M) \leq 1 - \frac{1}{M^2}.$$

If I got that right, then the inequality is wrong for two reasons. First, Chebyshev's inequality should be applied to $\mathbb{P}(|Z| > M)$ instead of $\mathbb{P}(Z > -M)$. Second, the direction of the inequality is wrong as $a \leq b$ implies that $1 - a \geq 1 - b$. Instead, note that by symmetry of the standard Gaussian, $\mathbb{P}(Z < -M) = \frac{1}{2}\mathbb{P}(|Z| > M)$. By Chebyshev,

$$\mathbb{P}(Z < -M) = \frac{1}{2}\mathbb{P}(|Z| > M) \leq \frac{1}{2M^2} = \frac{(\Delta t)^2}{2\epsilon\left(1 + \left(1 + \frac{1}{\epsilon}\right)\Delta t\right)^2}.$$

From here you can use standard calculus to analyze the expression as $\epsilon$ and $\Delta t$ increase, decrease and converge to $0$ or whatever else you want to study.


答: A simple counter-example is obatined by taking $\mathcal F_1=\{\emptyset, \Omega\}$ and $\mathcal F_1=\mathcal F$. In this case LHS is $EX$ and RHS is $X$.

Title: What can we say about this conditional expectation of a conditional expectation?

Body:
I have just started on probability theory and I was thinking about this object.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. Let $\mathcal{F_1},\mathcal{F_2}$ be sub $\sigma$-algebras. If we just have any random variable on the original probability space, can we say anything about $\mathbb{E}[\mathbb{E}[X|\mathcal{F_1}]|\mathcal{F_2}]?$

Can we simplify this in some way? I know there is some sort of tower property given the inclusion, however I am explicitly not assuming it here. In particular I am conjecturing $\mathbb{E}[\mathbb{E}[X|\mathcal{F_1}]|\mathcal{F_2}]=\mathbb{E}[X|\sigma(\mathcal{F_1},\mathcal{F_2})]$ but I am not sure if that is true?

Many thanks in advance!


问:
$x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$
I'm doing some exercise to prepare for my multivariable analysis exam. I didn't understand the second part of this question.

Given the function

$$f(x,y)=(x^2+y^2+1)^2 - 2(x^2+y^2) +4\cos(xy)$$

Prove that the taylor polynomial of degree $4$ of $f$ is equal to
  $5+x^4+y^4$.

First, $4\cos(xy) = 4 - 2(xy)^2 + 4R_3 $

$(x^2+y^2+1)^2=x^4+2 x^2 y^2+2 x^2+y^4+2 y^2+1$

Therefore: $(x^2+y^2+1)^2 - 2(x^2+y^2)=x^4+2 x^2 y^2+y^4+1$

Therefore: $f(x,y)=x^4+y^4+5+4R_3$

I don't know exactly why I can now conclude that Taylor Polynomial of degree 4 must be $5+x^4+y^4$, but I don't know exactly why.

Now the second question is: $x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$

New edit

I understand this now thanks to hint of Hagen von Eitzen, thanks !

The third question is:

Determine what kind of stationary point you have in $(0,0)$.

Title: $x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$

Body:
Write $x^4=2x^4-x^4$ and similarly $y^4=2y^4-y^4$

$$(x^2-y^2)^2 \ge 0$$

$$x^4+y^4-2x^2y^2 \ge 0$$

$$2x^4-x^4+2y^4-y^4-2x^2y^2 \ge0$$

$$2x^4+2y^4 \ge x^4+y^4+2x^2y^2$$

$$x^4+y^4 \ge \dfrac{(x^2+y^2)^2}{2}$$


问:
Let $f(x,y)=\frac{x^3-y^3}{x^2+y^2}$. Is f differentiable in $(0,0)$?
Let
$$f(x,y)=\frac{x^3-y^3}{x^2+y^2}$$

My solution manual says that this function is not diffb. in $(0,0)$ because it is not linear. Well my problem is that I don't see why this function is linear, and I also don't see why that would imply that $f$ is not differentiable.

答:
We find the directional derivatives at $(0,0)$. Assume $\Vert v\Vert =1$. Then

$$f'((0,0);v)=\lim_{h\to 0}\frac{f((0,0)+h\cdot v)-f(0,0)}{h}$$

I take $f(0,0)=0$ to ensure continuity.

$$f'((0,0);v)=\lim_{h\to 0}\frac{f(hv_1,hv_2)}{h}$$

$$f'((0,0);v)=\lim_{h\to 0}\frac{h^3v_1^3-h^3v_2^3}{h^3}\frac{1}{\Vert v\Vert}$$

$$f'((0,0);v)=\lim_{h\to 0}\frac{h^3v_1^3-h^3v_2^3}{h^3} $$

$$f'((0,0);v)=v_1^3-v_2^3$$

Since the directional derivatives do not follow a linear dependence on $(v_1,v_2)$, we conclude $f$ cannot be differentiable at $(0,0)$.


问:
For a hyper cube of dimension N what number of its vertices can be covered by intersection with a hyperplane
If you look at intersecting a binary cube (the set contained by [0,1]^n) with a plane in $\mathbb{R}^3$. Then the plane can potentially intersect with the corners of the cube. Depending on the choice of plane you might intersect with 0,1,2,3 or 4 corners, but it’s impossible to intersect with any other number of corners.

If we shift our attention to $\mathbb{R}^4$ it gets a little trickier, all the $\mathbb{R}^3$ numbers must be valid but we also know that hyperplanes such as $x_0+x_1+x_2+x_3=2$ let you intersect with 6 corners.

This leads to our problem:

Given a dimension $n$, what does the list of possible intersection counts of a hyperplane with vertices of the Boolean cube look like? Even for $n=4$ at best we can show the list includes (but as of right now I haven’t exhausted) 0,1,2,3,4,6,8.

Powers of 2 and Pascal’s triangle numbers are all present but there definitely are more exotic coverings out there especially for $n=5$ onwards

答:
This is not a complete answer, but was too long to be a comment.

For $n=4$ your list is complete.

First, note that given a hyperplane, we can split the $n$-cube $C_n$ into two disjoint $(n-1)$-cubes along an axis so that the intersection of the plane with each of the $(n-1)$-cubes is a hyperplane in $\mathbb{R}^{n-1}$. In fact, only an axis normal to the chosen hyperplane can fail to work here. So every solution in dimension $n$ is a sum of two solutions in dimension $n-1$. (Note that in this decomposition, the two $(n-2)$-dimensional hyperplanes are parallel within $C_{n-1}$ if projected into the same $(n-1)$-dimensional space by collapsing our axis of the cube.)

This shows that $2^n$ is always an upper bound for the number of intersections $k$ in $C_n$, so it remains to rule out the cases of $k=5$ and $k=7$.

Suppose a hyperplane intersects $C_4$ in $7$ points. Then it must be formed from two parallel planes in $C_3$, one of which intersects in $4$ points and one of which intersects in $3$ points. But all planes passing through $3$ points in $C_3$ have the same orientation, up to rotation, and their translates all pass through $0,1,$ or $3$ points.

Suppose a hyperplane intersects $C_4$ in $5$ points. Then it is formed either from a $3$-point intersection and a $2$-point intersection, or a $4$-point intersection and a $1$-point intersection. The former case is ruled out by the same reasoning as the above paragraph, while the latter case is dealt with by noting that there are two kinds of $4$-point planes in $C_3$: one containing all the points in face, and one "diagonal" cut whose intersection with $C_3$ consists of the vertices of a $1$ by $\sqrt{2}$ rectangle. Translates of these planes intersect $C_3$ in $\{0,4\}$ or $\{0,2,4\}$ points respectively, which rules out the second case.

For $n=5$, the possible intersections (by the sum decomposition proven above) are a subset of $\{{\color{green} 0}, {\color{green} 1}, {\color{green} 2}, {\color{green} 3}, {\color{green} 4}, {\color{green} 5}, {\color{green} 6}, {\color{green} 7}, {\color{green} 8}, \color{red} 9, \color{green}{10}, \color{red}{11}, \color{green}{12}, \color{red}{14},  {\color{green}{16}}\}$, where I have colored green the cases which I can find examples of.

The $7$-point case here is $x_1+x_2+x_3+2x_4+2x_5=3$; everything else is naturally given by a $4$-simplex, an element of Pascal's triangle, or a doubling of the $n=4$ case. I found no other examples after searching through $1,000,000$ random planes with integer coefficients from $0$ to $10$ for each coordinate.

For $n=6$, I have found examples for $k=0,1,2,3,4,5,6,7,8,9,10,11,12,14,15,16,20,24,32$.

If I am correct that the $12$ values of $k$ given above are complete, then the sequence of the number of solutions for $C_n$ which starts $2,3,5,7,12,(19-28),\ldots$ does not seem to have any good matches in the OEIS.


问:
Induction proof $1 + \frac{1}{\sqrt[3]{2}} + ... + \frac{1}{\sqrt[3]{n}} < \frac{3}{2}\sqrt[3]{n^2}$
I need to prove that:
$$1 + \frac{1}{\sqrt[3]{2}} + ... + \frac{1}{\sqrt[3]{n}} < \frac{3}{2}\sqrt[3]{n^2}$$

The problem is that I can't find a way to separate that "$1$" on the right side ($\frac{3}{2}\sqrt[3]{(n+1)^2}$) to get a sum on the let AND right side. I thought about Bernoulli's inequality but it doesn't seem to help much either.

Edit: I've got some bomb idea, I will post it as a comment

答:
So I got stuck at that point with my induction:

$$1 + \frac{1}{\sqrt[3]{2}} + ... + \frac{1}{\sqrt[3]{n}} + \frac{1}{\sqrt[3]{n+1}} < \frac{3}{2}\sqrt[3]{(n+1)^2}$$

Then, trying to separate elements on the right hand side I got:

$$1 + \frac{1}{\sqrt[3]{2}} + ... + \frac{1}{\sqrt[3]{n}} + \frac{1}{\sqrt[3]{n+1}} < \frac{3}{2}\sqrt[3]{n^2(1 + \frac{2}{n} + \frac{1}{n^2})}$$
$$1 + \frac{1}{\sqrt[3]{2}} + ... + \frac{1}{\sqrt[3]{n}} + \frac{1}{\sqrt[3]{n+1}} < \frac{3}{2}\sqrt[3]{n^2}\sqrt[3]{(1 + \frac{2}{n} + \frac{1}{n^2})}$$
And now I see that the growth of values on the right hand side associated with changes of that ($\sqrt[3]{(1 + \frac{2}{n} + \frac{1}{n^2})}$) part is the smallest when that ($\frac{1}{\sqrt[3]{n}}$) part is the smallest and it is allways bigger than 1. So I know that:
$$1 + \frac{1}{\sqrt[3]{2}} + ... + \frac{1}{\sqrt[3]{n}} + \frac{1}{\sqrt[3]{n+1}} < \sqrt[3]{(1 + \frac{2}{n} + \frac{1}{n^2})}<\frac{3}{2}\sqrt[3]{n^2}\sqrt[3]{(1 + \frac{2}{n} + \frac{1}{n^2})}$$

Now I see that:

$ \frac{1}{\sqrt[3]{n+1}} $ is falling for all $ n \in \mathbb{N}$

$\sqrt[3]{(1 + \frac{2}{n} + \frac{1}{n^2})}$ is growing for all $ n \in \mathbb{N}$

So I just check both valuse for n = 1 and I get accordingly:

$ \frac{1}{\sqrt[3]{2}} < 1$

$\sqrt[3]{4} > 1$

So, that is it. To make sure I can calculate:

$\lim_{x \to +\infty} \frac{1}{\sqrt[3]{n+1}} = 0$

$\lim_{x \to +\infty}\sqrt[3]{(1 + \frac{2}{n} + \frac{1}{n^2})} = \infty $


问:
Looking for nomenclature! What is it called to remove all parts of a cycle that have no area?
Given this directed graph:

And the cycle A->B->C->D->E->F->E->D->G->B->A (so simply going around the graph clockwise):

What is it the operation called to eliminate all parts of this cycle that have no area, so that the result would be B->C->D->G->B?

Another problem description maybe would be: To find the minimal cycle with the same area.

I read on wikipedia that A->B->A for example would be a digon (cycle of length 2), so am I "eliminating digons" from the cycle, or how would I describe this, so that other people familiar with graph theory understand what I'm trying to do here?

I do this exact thing in a programm of mine (to prevent 0-area polygons to be formed), and am really struggling with appropriate naming.

答: I don't think there's a standard term for this: you should introduce one of your own and clearly define it when it's introduced. (This happens pretty frequently, there are a lot of simple constructions out there which don't have specific names.) "Removing digons" makes sense, and another one which appeals to me is "cancel edges" (thinking of an edge $AB$ as the inverse of an edge $BA$ - this has the advantage of making sense even when we allow multiple arrows with the same source and target and we care about multiplicity).


问:
Average number of head streaks in $N$ coin tosses
You throw $N$ times a fair coin. What is the expected number of subsequences of $k$ consecutive heads?

For instance, with $N=6$ and $k=2$, for the outcome HTHHHH, we would get $3$ such subsequences.

I know the solutions to two related problems, but did not find this one being asked:

what is the expected number $N$ needed to get $k$ consecutive heads when throwing coins (see here)

what is the probability of having at least $k$ consecutive heads for $N$ throws (see here)

答:
For $j = 1,\ldots,n$, let $X_j = 0$ if the $j$-th coin flip is tails and $X_j = 1$ if the $j$-th coin flip is heads. The number of subsequences of $k$ consecutive heads is $$N = \sum_{j = 0}^{n-k}X_{j+1}X_{j+2}\cdots X_{j+k}.$$ Do you see why this is true? Now, using linearity of expectation followed by the fact that $X_1,\ldots,X_n$ are independent, we have
\begin{align*}
E[N] &= E\left[\sum_{j = 0}^{n-k}X_{j+1}X_{j+2}\cdots X_{j+k}\right]
\\
&= \sum_{j = 0}^{n-k}E[X_{j+1}X_{j+2}\cdots X_{j+k}]
\\
&= \sum_{j = 0}^{n-k}E[X_{j+1}]E[X_{j+2}] \cdots E[X_{j+k}].
\end{align*}

Can you finish the problem from here?


问:
Limit of the ratio of two non-Riemann sums.
Let $\left[ {a,b} \right] \subset \mathbb{R}$  and $f,g:\left[ {a,b} \right] \to \mathbb{R}$  be two Riemann-integrable functions.

Let $a = {x_0} < {x_1} < {x_2}... < {x_n} = b$  be a partition of $\left[ {a,b} \right]$  and let $\Delta x = \mathop {\max }\limits_{i = 0}^{n - 1} \left( {{x_{i + 1}} - {x_i}} \right)$.

Let ${t_i} \in \left[ {{x_i},{x_{i + 1}}} \right],\;i = 0,n - 1$  and let  $k \in {\mathbb{N}^*}$.

I’d like to prove that

$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sum\limits_{i = 0}^{n - 1} {{{\left( {{x_{i + 1}} - {x_i}} \right)}^k}f\left( {{t_i}} \right)} }}{{\sum\limits_{i = 0}^{n - 1} {{{\left( {{x_{i + 1}} - {x_i}} \right)}^k}g\left( {{t_i}} \right)} }} = \frac{{\int\limits_a^b {f\left( x \right){\text{d}}x} }}{{\int\limits_a^b {g\left( x \right){\text{d}}x} }}$

It is obvious for equally spaced partitions  ${x_{i + 1}} - {x_i} \equiv \Delta x$

$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sum\limits_{i = 0}^{n - 1} {\Delta {x^k}f\left( {{t_i}} \right)} }}{{\sum\limits_{i = 0}^{n - 1} {\Delta {x^k}g\left( {{t_i}} \right)} }} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sum\limits_{i = 0}^{n - 1} {\Delta xf\left( {{t_i}} \right)} }}{{\sum\limits_{i = 0}^{n - 1} {\Delta xg\left( {{t_i}} \right)} }} = \frac{{\int\limits_a^b {f\left( x \right){\text{d}}x} }}{{\int\limits_a^b {g\left( x \right){\text{d}}x} }}$

But I don’t see how to do it in the general case?

答:
This result can fail to hold with non-uniform partitions. For a counterexample, we look for Riemann integrable functions $f$ and $g$ and  a sequence of partitions

$$P_n: a = x_0^{(n)}<x_1^{(n)} < \ldots < x_{n-1}^{(n)} < x_n^{(n)} = b$$

along with a choice of tags $t_j^{n} \in [x_{j},x_{j+1}]$ where

$$\tag{*}\Delta x := \|P_n\| = \underset{0 \leqslant j \leqslant n-1} \max \left(x_{j+1}^{(n)}-x_j^{(n)}\right) \underset{n \to \infty}\longrightarrow 0$$ and, such that for some $k > 1$,

$$\lim_{\Delta x \to 0, \,n \to \infty}\frac{\sum_{j=0}^{n-1} f(t_j^{(n)})(x_{j+1}^{(n)}- x_j^{(n)})^k}{\sum_{j=0}^{n-1} g(t_j^{(n)})(x_{j+1}^{(n)}- x_j^{(n)})^k} \neq \frac{\int_a^bf(x) \, dx}{\int_a^b g(x) \, dx}$$

Note that condition (*) is an essential requirement here as it ensures convergence of Riemann sums (with $k=1$) to the respective integrals.

Take $[a,b] = [1,e]$, $f(x) = 1$, $g(x) = x$, $k = 2$, partition points $x_j^{(n)} = e^{j/n}$ and tags $t_j^{(n)} = e^{j/n}$ for $j=0,1,\ldots,n$.

In this case, the $n$th partition is $P_n : 1 < e^{1/n} < e^{2/n} < \ldots <  e^{(n-1)/n}< e$, and we have

$$\Delta x = \|P_n\| = \max_{0 \leqslant j  \leqslant n-1}(e^{(j+1)/n}-e^{j/n}) = \max_{0 \leqslant j  \leqslant n-1}e^{j/n}(e^{1/n}-1) = e^{(n-1)/n}(e^{1/n}-1), $$

where $\Delta x = e^{(n-1)/n}(e^{1/n}-1)\to e\cdot 0 = 0$ as $n \to \infty$.

Note that, with $k=1$,

$$\sum_{j=0}^{n-1} f(t_j^{(n)})(x_{j+1}^{(n)}- x_j^{(n)})= \sum_{j=0}^{n-1} 1 \cdot (e^{(j+1)/n}- e^{j/n)})= (e^{1/n} - 1)\sum_{j=0}^{n-1}e^{j/n} =  (e^{1/n} - 1)\frac{e-1}{e^{1/n} -1}\\ \sum_{j=0}^{n-1} g(t_j^{(n)})(x_{j+1}^{(n)}- x_j^{(n)})= \sum_{j=0}^{n-1} e^{j/n} \cdot (e^{(j+1)/n}- e^{j/n)})= (e^{1/n} - 1)\sum_{j=0}^{n-1}e^{(2j)/n} =  (e^{1/n} - 1)\frac{e^2-1}{e^{2/n} -1} ,$$

and, as we expect for Riemann sums,

$$\lim_{n \to \infty}\sum_{j=0}^{n-1} f(t_j^{(n)})(x_{j+1}^{(n)}- x_j^{(n)})= e-1 = \int_1^e f(x) \, dx\\ \lim_{n \to \infty}\sum_{j=0}^{n-1} f(t_j^{(n)})(x_{j+1}^{(n)}- x_j^{(n)})= \frac{e^2-1}{2} = \int_1^e g(x) \, dx$$

However, for $k=2$,

$$\sum_{j=0}^{n-1} f(t_j^{(n)})(x_{j+1}^{(n)}- x_j^{(n)})^k= \sum_{j=0}^{n-1} 1 \cdot (e^{(j+1)/n}- e^{j/n)})^2= (e^{1/n} - 1)^2\sum_{j=0}^{n-1}e^{(2j)/n} =  (e^{1/n} - 1)^2\frac{e^2-1}{e^{2/n} -1}\\ \sum_{j=0}^{n-1} g(t_j^{(n)})(x_{j+1}^{(n)}- x_j^{(n)})^k= \sum_{j=0}^{n-1} e^{j/n} \cdot (e^{(j+1)/n}- e^{j/n)})^2= (e^{1/n} - 1)^2\sum_{j=0}^{n-1}e^{(3j)/n} =  (e^{1/n} - 1)^2\frac{e^3-1}{e^{3/n} -1} ,$$

and,

$$\lim_{\Delta x \to 0, \,n \to \infty}\frac{\sum_{j=0}^{n-1} f(t_j^{(n)})(x_{j+1}^{(n)}- x_j^{(n)})^k}{\sum_{j=0}^{n-1} g(t_j^{(n)})(x_{j+1}^{(n)}- x_j^{(n))})^k} = \lim_{n \to \infty}\frac{e^2-1}{e^3-1}\frac{e^{3/n}-1}{e^{2/n}-1} = \frac{3}{2}\frac{e^2-1}{e^3-1} \\\neq \frac{2}{e+1}  = \frac{e-1}{\frac{e^2-1}{2}}= \frac{\int_a^bf(x) \, dx}{\int_a^b g(x) \, dx}$$