问:
If $\sum a_n b_n <\infty$ for all $(b_n)\in \ell^2$ then $(a_n) \in \ell^2$
I'm trying to prove the following:

If $(a_n)$ is a sequence of positive numbers such that $\sum_{n=1}^\infty a_n b_n<\infty$ for all sequences of positive numbers $(b_n)$ such that $\sum_{n=1}^\infty b_n^2<\infty$, then $\sum_{n=1}^\infty a_n^2 <\infty$.

The context here is functional analysis homework, in the subject of Hilbert spaces.

Here's what I've thought:

Let $f=(a_n)>0$. Then the problem reads: if $\int f\overline{g}<\infty$ for all $g>0,g\in \ell^2$, then $f\in \ell^2$. This brings the problem into the realm of $\ell^p$ spaces.

I know the inner product is defined only in $\ell^2$, but it's sort of like saying: if $\langle f,g\rangle <\infty$ for all $g>0,g\in \ell^2$ then $f\in \ell^2$.

I read this as: "to check a positive sequence is in $\ell^2$, just check its inner product with any positive sequence in $\ell^2$ is finite, then you're done", which I find nice, but I can't prove it :P

From there, I don't know what else to do. I thought of Hölder's inequality which in this context states:
$$\sum_{n=1}^\infty a_nb_n \leq \left( \sum_{n=1}^\infty a_n^2 \right)^{1/2} \left( \sum_{n=1}^\infty b_n^2 \right)^{1/2}$$

but it's not useful here.

答: It can be proven using the Hölder inequality — see lemma 1.8 (page 8) of these notes. (There may be mistakes; I typeset these from lecture notes.) There's no need to invoke heavy machinery like the uniform boundedness principle or even the Riesz representation theorem; it's a special case of a general result for $\ell^p$ spaces.</p>


问:
Intersection of collection of sets not in the collection of sets?
Hello I was doing a practice problem and I came across something that kind of tripped me up. The question says:

Suppose the Collection $F$ is given by $F=\{ [1, 1+ 1/n] : n\in \mathbb{N}\}$

Find the union of this collection of sets and find the intersection of these sets (I'm struggling to type this).

I looked at the solution and I don't understand why the intersection is $\{1\}$. This element is not actually contained in any of the sets since all the elements of the set are intervals. Can anyone please explain why this is? Intuitively on some level it makes sense since $1$ is the only common "thing" for lack a of a better, word but isn't the intersection of a collection of sets be the elements that are common to all the sets?

答: First note that $1$ is an element of each of the intervals of the form $[1,1+\frac{1}{n}]$, and thus $1$ is in the intersection. Now, assume that $x$ is in the intersection. Then $x$ is in all the mentioned intervals, thus $1\le x\le x+\frac{1}{n}$ holds for all $n\ge 1$. That is only possible if $x=1$, this shows thus that the only element in the intersection is $1$. In other words: the intersection is the set $\{1\}$.</p>

ContentLicense: CC BY-SA 3.0


问:
Is calculating a mean of sample means a standard practice?
In order to learn about the distribution of a statistic in the long-run, a common practice is to construct the sampling distribution of that statistic across a number of samples, each of a certain sample size $n$. One example of this is constructing a sampling distribution of sample means by constructing a density histogram of the sample means considered across a number of random samples of size $n$.

What if, after one collects the sample mean from each of a number of random samples of size $n$, one takes the mean of these sample means ? Is this a common practice to investigate the overall population mean ? If not, why not ?

Thanks!

答:
The mean of sample means would simply be the mean of all of those samples. Sure, the random variable given by the mean of means would have lower variance than just one sample mean, but you could have gotten the same result just by increasing the size of your sample in the first place.

For example, if we took the average of 3 samples each of size k:

$\frac{(x_1+x_2+...+x_k)/k + (x_{k+1}+...+x_{2k})/k + (x_{2k+1}+...+x_{3k})/k}{3}$

= $\frac{(x_1+x_2+...+x_k) + (x_{k+1}+...+x_{2k}) + (x_{2k+1}+...+x_{3k})}{3k}$

= $\frac{(x_1+x_2+...+x_{3k})}{3k}$

it would have been the same as taking a sample of size 3k in the first place.

ContentLicense: CC BY-SA 4.0


问:
Homotopy equivalence: three touching circles vs 3-bouquet
Are these two spaces homotopic equivalent? This question is already posed here but the answer is missing.

答: Hint: for the right hand picture, give each $S^1$ a CW-structure of two $0$-cells on the equator, and the two $1$-cells connecting them. Take any spanning tree of the complex and contract it to a point. One idea is to take the union of bottom hemispheres of each $S^1$, and pinch them to a point.


问:
Smallest group for finite groups.
Does it make sense to talk about a smallest group imaginable in terms of this group being unique? I know that given the prerequisites for a group (closure, identity, inverse, associativity) and if we don't necessarily require that the subgroup be a generator for the set I could construct a group for example: $\langle \mathbb{Z}_n, +\rangle$ for the smallest possible n, but would there not be other groups containing only 1 element? Also if we are dealing with non-finite groups what would be the possible smallest group imaginable?

(Sorry beforehand if this is poorly formulated as I have only recently started to learn about group theory.)

答: Note that for induction purposes (in case of finite groups) and classification, one is interested in deriving structure and properties of so-called minimal non-$\mathfrak{X}$ groups, where $\mathfrak{X}$ is a property of groups. This means that all proper subgroups of $G$ possess the property $\mathfrak{X}$, but $G$ itself not. For $\mathfrak{X}$ being abelian, Miller and Moreno classified these groups already in 1903, while later on, O.J. Schmidt did the same for $\mathfrak{X}$ being nilpotent. Minimal non-$M$ groups (monomial groups, a class between nilpotent and solvable) were treated late 1970ties by R.W. van der Waall (see for example here).

Title: Smallest group for finite groups.


问:
Additive closure of a category
Given a category $\mathcal{C}$, there is a (I believe) well-known way to obtain an additive category from that, called the additive closure of $\mathcal{C}$ (see eg Bar-Natan's Khovanov’s homology for tangles and cobordisms, Definition 3.2): first one turns $\mathcal{C}$ into an Ab-enriched category by replacing the Hom sets by their $\mathbb{Z}$-linearisations and later one considers formal direct sums (coproducts) of elements of $\mathcal{C}$.

As everything in category theory, it is better to define things by their universal properties. The following is an attempt to give the universal property of this additive closure.

Claim: Let $\mathcal{C}$ be a category. There exists a unique (up to equivalence) additive category $\overline{\mathcal{C}}$ together with a fully faithful functor $i: \mathcal{C} \hookrightarrow  \overline{\mathcal{C}}$ such that every object of $\overline{\mathcal{C}}$ is isomorphic in a unique way to the coproduct of a finite family of objects in the image of $i$.

Is the previous statement the correct characterisation of the additive closure of $\mathcal{C}$? If not, what is its correct universal property?

答:
Your description works well enough, in that it clearly specifies $\overline{\mathcal C}$ essentially uniquely as the category of tuples $(c_1,\cdots,c_n)$ of objects from $\mathcal C$ in which morphisms are given by viewing such a tuple as the biproduct of its objects, taken in the $\mathbb Z$-linearization of $\mathcal C$. Note that we must be a little careful about the meaning of "unique", here, since for instance biproducts are commutative and associative up to isomorphism: it's a bit clearer to say that each object comes equipped with the structure maps making it a biproduct of some objects of $\mathcal C$'s linearization.

Another of describing this category is as follows:
$\mathcal C$ is an additive category admitting a fully faithful embedding of $\mathcal C$ such that, if $F:\mathcal C\to \mathcal D$ is any functor from $\mathcal C$ into (the underlying ordinary category of) an additive category $\mathcal D$, there is an essentially unique additive functor $\overline F:\overline{\mathcal C}\to \mathcal D$ supplied with a natural isomorphism $\overline F\circ i\cong F$.

This kind of characterization works much better for constructions like "the free completion of $\mathcal C$ under filtered colimits", in which objects do not have a unique-up-to-isomorphism decomposition as a filtered colimit of objects of $\mathcal C$. Such constructions arise often; for instance the category of abelian groups is the free cocompletion under filtered colimits of the category of finitely presentable abelian groups. An abelian group $A$ does have a canonical description as a filtered colimit of finitely presentable groups (namely $A$ is the colimit of the diagram of all maps to it from such groups) but this description is non-unique. For instance $A$ is also the filtered colimit of all its finitely-generated subgroups.

Title: Additive closure of a category


问:
Interesting binomial coefficients sum: $\sum\limits_{i = 0}^b(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}$
Any idea on whether or not $$\sum_{i = 0}^b(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}$$
has a closed formula on $a$ and $b$ (and on what it is, in case it does)?

It is supposed that $b \le a$.

答:
Here's a combinatorial proof that the expression is $0$.  Both sides of the identity count the number of $(b-1)$-subsets of $\{1,\dots,a+b-1\}$ that include $\{1,\dots,b\}$.  Because $b > b-1$, this count is obviously $0$, establishing the RHS.  For the LHS, apply inclusion-exclusion, where the $b$ properties to be avoided are that $j$ does not appear for $j \in \{1,\dots,b\}$.  More generally, this argument shows that
$$\sum_{i=0}^b (-1)^i \binom{b}{i} \binom{a+b-1-i}{k} = 0$$
for $k < b$, and it does not require $b \le a$.


问:
$\Delta ABC$ is right angled triangle. $AP$ and $AQ$ meet $BC$ and $BC$ produced in $P$ and $Q$ and are equally inclined to $AB$.
$\Delta ABC$ is right angled triangle at $A$. $AP$ and $AQ$ meet $BC$ and $BC$ produced respectively in $P$ and $Q$ and are equally inclined to $AB$ ($\angle BAP=\angle BAQ$). Show that $\frac{BP}{BQ}=\frac{CP}{CQ}$.

I used angle bisector theorem and found that $\frac{BP}{BQ}=\frac{AP}{AQ}$. But I don't know how to proceed further. Can this be solved without trigonometry? Thanks in advance.

答:
Hint: Finish off with the Sine Law.

 From angle bisector theorem, $$\frac {CP}{CQ} = \frac {AP}{AQ}$$ The notations are different since I extended $BC$ towards $C$.
 By Sine Law, $$\frac {AP}{\sin B} = \frac{BP}{\sin \angle PAB} = \frac {BP}{\sin (90^\circ - \angle PAC)}$$ $$\frac {AQ}{\sin B} = \frac{BQ}{\sin \angle QAB} = \frac {BQ}{\sin (90^\circ + \angle QAC)} = \frac {BQ}{\sin (90^\circ - \angle PAC)}$$
 Hence $$\frac {AQ}{BQ} = \frac {AP}{BP}$$$$\leadsto\frac {AP}{AQ} = \frac {BP}{BQ} = \frac {CP}{CQ}$$


答:
If $\cal A$ is a collection of sets then we define the two operations:

The union of $\cal A$ which is $\bigcup{\cal A}=\{x\mid\exists A\in{\cal A}: x\in A\}$. So the elements of the union are all those which are elements of elements of the collection $\cal A$.

The intersection of $\cal A$ which is $\bigcap{\cal A}=\{x\mid\forall A\in{\cal A}:x\in A\}$. Similarly, the elements of the intersection are those which are elements in all the elements of $\cal A$.

So now we look for $\bigcup F$ and $\bigcap F$. What real numbers are in all the intervals? How can you describe that set nicely? What is the collection of real numbers which appear in some of the sets?

Let me hint you for the first answer, $1$ is clearly in all the intervals and no number smaller than $1$ is in any interval to begin with. For every number $x>1$ we can find an interval $[1,1+\frac1n]$ such that $x$ is not in that interval. So which numbers appear in all the intervals?

To your final paragraph, let me add that $\{1\}$ is in fact the interval $[1,1]$. And it goes to show you that the intersection of non-degenerate intervals doesn't have to be a non-degenerate interval.

Title: Intersection of collection of sets not in the collection of sets?

Body:
Hello I was doing a practice problem and I came across something that kind of tripped me up. The question says:

Suppose the Collection $F$ is given by $F=\{ [1, 1+ 1/n] : n\in \mathbb{N}\}$

Find the union of this collection of sets and find the intersection of these sets (I'm struggling to type this).

I looked at the solution and I don't understand why the intersection is $\{1\}$. This element is not actually contained in any of the sets since all the elements of the set are intervals. Can anyone please explain why this is? Intuitively on some level it makes sense since $1$ is the only common "thing" for lack a of a better, word but isn't the intersection of a collection of sets be the elements that are common to all the sets?


问:
Relationship between smooth sections and group actions on manifold
This question may be a bit naive, but I am relatively new to the study of differential topology. I am currently reading through John Lee's Smooth Manifolds. He devotes a section to the theory of integral curves of vector fields on manifolds, and the associated flow on the manifold. In the case of a complete vector field, the flow corresponds to a smooth $\mathbb{R}$-action on $M$.
In later sections, vector fields on $M$ get fit into the more general context of smooth sections of bundles over $M$ (vector bundles, tensor bundles, etc.) In this language, certain smooth sections of the tangent bundle (ie vector fields) induce smooth $\mathbb{R}$ actions on $M$ (in the form of the flow).
I am wondering if this phenomenon in anyway generalizes to smooth sections of other types of bundles. More explicitly, are there other contexts in which a smooth section of a bundle over $M$ induce some smooth Lie group action on $M$? Or is this phenomenon specific to vector fields?
Thank you in advance!

答:
Among bundles over $M$, $TM$ is somewhat special in that vector fields correspond to differential equations on $M$, while there is no such correspondence for sections of other bundles.

There is, however, a generalization of the notion of a generator of a Lie group action. The space of vector fields $\mathfrak{X}M$ is a Lie algebra under Lie brackets, and a Lie algebra homomorphism $\varphi:\mathfrak{g}\to\mathfrak{X}M$ can, under certain conditions, give rise to a Lie group action $\theta:G\times M\to M$ where $G$ is a Lie group whose lie algebra is $\mathfrak{g}$. In this way, Lie group actions correspond not with sections of $TM$, but with Lie subalgebras of $\mathfrak{X}M$. In the case of an $\mathbb{R}$-action generated by a vector field $V$, the relevant lie algebra morphism is $t\mapsto tV$.


问:
Solution for $y(x^2+y^2)dx + x(3x^2-5y^2)dy = 0$.
I am teaching assistant of an elementary course of ODE and a student asked me to solve the following equation:

$$y(x^2+y^2)dx + x(3x^2-5y^2)dy = 0$$

The standard procedure to solve it consists in multiplying the equation by some function $\mu = \mu(x,y)$. Once we do this we must solve
$$-8\mu(x,y)\cdot (x^2-y^2) +\partial_y\mu(x,y)\cdot \left(yx^2 + y^3\right) - \partial_x\mu(x,y)\left(3x^3-5xy^2\right) = 0.$$
In general one assumes that $\mu$ is constant in one of each variable, but this can not be the case here.

I tried lots of manipulations, but could not handle it. I do appreciate any hints.

Body:
$$y(x^2+y^2)dx + x(3x^2-5y^2)dy = 0$$
Multiply by $y^2$:
$$y^3x^2dx+y^5dx + x^3dy^3-xdy^5 = 0$$
$$(y^3x^2dx + x^3dy^3)+(y^5dx-xdy^5) = 0$$
Divide by $x^2$:
$$y^3dx + xdy^3-d\left (\dfrac {y^5}x \right) = 0$$
$$d(xy^3)-d\left (\dfrac {y^5}x \right) = 0$$
Integrate:
$$xy^3-\dfrac {y^5}x = K$$


问:
If $\lambda = \sum k_i \alpha_i$ and $P_\lambda \subseteq \cup_{\alpha \in \Phi} P_\alpha \implies \lambda = c\alpha$ for some $\alpha \in \Phi$.
This is an exercise 10.10 in Humphreys book on Lie algebras.

Let $\Phi$ be a root system lying in the euclidean space $E$ and let $\Delta = \{\alpha_1,\cdots,\alpha_\ell\}$ be a basis for $\Phi$.  Let $\lambda = \sum_i k_i \alpha_i$ with all $k_i\geq 0$ or all $k_i\leq 0, k_i \in \mathbb Z.$ Prove that either $\lambda$ is a multiple (possibly 0) of a root, or else there exists $\sigma \in \mathscr W$ (Weyl group) such that $\sigma \lambda = \sum_i k_i'\alpha_i$ with some $k_i'>0$ and some $k_i'<0$.

He gives the following tip: If $\lambda$ is not a multiple of any root, then the hyperplane $P_\lambda$ orthogonal to $\lambda$ is not included in $\bigcup_{\alpha \in \Phi} P_\alpha$. Take $\mu \in P_\lambda \setminus \bigcup P_\alpha$ and then find $\sigma \in \mathscr W$ for which all $(\alpha_i,\sigma\mu)>0$.

I couldn't prove that $P_\lambda \not \subseteq \bigcup P_\alpha$, altough I managed to finish the exercise as follows. Taking any such a $\mu$, since every point in $E$ is $\mathscr W$-conugate to a point in the fundamental Weyl chamber, there exists $\sigma \in \mathscr W$ satisfying $(\sigma\mu, \alpha_i)>0$ as claimed. In particular, each $\sigma \alpha_i \in \Phi$, so we may write $\sigma\lambda = \sum k_i' \alpha_i$ for some (possibly new) integers $k_i'$.  Now, $\mu \in P_\lambda$, so

$$ 0 = (\mu,\lambda ) = (\sigma\mu, \sigma \lambda) = \sum k_i'(\sigma\mu,\alpha_i)$$
implies that some $k_i'>0$ and some $k_i'<0$, as the terms $(\sigma\mu ,\alpha_i)$ are all positive.

The question then is: how to prove that $P_\lambda \not\subseteq \bigcup P_\alpha$? All the computations I did so far were useless, stuff like $0 = (\lambda,x) = \sum k_i (\alpha_i,x)$ cannot imply anything. I also tried to begin simple with  $P_\lambda \subset P_\alpha \implies \lambda = c\alpha$ by supposig $\lambda - c\alpha\neq 0$ and $P_\lambda \subseteq P_\alpha$, but that only yelds $P_\lambda \subseteq P_{\lambda - c\alpha}$.

Any help? Thank you.

答:
Lemma: If $H, H_1, ... H_r$ are hyperplanes (i.e. $(n-1)$-dimensional subspaces) in some $n$-dimensional space over an infinite field, and $H \subseteq \bigcup_{i=1}^n H_i$, then $H = H_j$ for some $1 \le j \le r$.

Proof: By assumption we have

$$H = H \cap \left( \bigcup_{i=1}^r H_i \right) = \bigcup_{i=1}^r (H \cap H_i).$$

Now the intersection of any two hyperplanes has dimension $n-2$ unless the two hyperplanes are equal. But if all the spaces in the union on the RHS are $(n-2)$- dimensional, their union cannot be the $(n-1)$-dimensional space on the LHS. QED.

To apply this to your problem: If $P_\lambda \subseteq \bigcup P_\alpha$, then by the lemma there is a root $\alpha$ such that $P_\lambda = P_\alpha$, consequently $\langle \lambda \rangle = P_\lambda^\perp = P_\alpha^\perp = \langle \alpha\rangle$, i.e. $\lambda$ is a scalar multiple of $\alpha$.


问:
What does it mean when a theorem can be proved without induction?
Recently I learned about interpolation with Newton polynomials and divided differences. Specifically we learned how $f[x_0,...,x_k] = \frac{f[x_1,...,x_k] - f[x_0,...,x_{k-1}]}{x_k - x_0}$, where $f[x_0,...,x_i]$ is the coefficient of $x^i$ of the unique polynomial of degree at most $i$ that interpolates $f$ at distinct $x_0,...x_i$. We were able to prove this by choosing an arbitrary $k$ and showing this theorem holds, without induction.

I was wondering what is implied about a theorem that can be proved without induction vs a theorem that requires induction to be proved. Does this have any relation to complexity classes maybe?

答:
Practically every interesting family of facts depending on a natural number parameter $n \in \mathbb{N}$ has to be proven "with induction" in the sense that if you didn't assume induction in your foundations you wouldn't be able to prove it. As a simple example, every time you assume the commutativity or associativity of addition in the integers, the rationals, the reals, or the complex numbers, that fact ultimately boils down to commutativity or associativity of addition of natural numbers, which must be proven by induction; in fact you need induction even to define addition.

On the other hand, there's a different more colloquial notion of what it means to prove something "with induction" or "without induction," which is whether you explicitly perform an induction in the proof. When you don't do this you are basically shoving the induction into another fundamental fact you've proven with induction at some point (such as the commutativity or associativity of addition).

For example, consider the classic induction exercise

$$1 + 2 + \dots + n = \frac{n(n+1)}{2}.$$

Induction is required even to define the LHS so in that sense it's unavoidable. On the other hand, we can give a classic proof that does not require explicitly performing an induction, by instead considering

$$2(1 + 2 + \dots + n) = (1 + n) + (2 + (n-1)) + \dots + (n + 1) = n(n+1).$$

This proof implicitly uses the commutativity and associativity of addition many times, which as mentioned above ultimately depends on induction (Edit: in two ways! Each individual use of commutativity or associativity requires induction, and using them repeatedly also requires induction!) But it doesn't perform any new inductions; all the induction necessary is subsumed into commutativity and associativity, which are usually invoked without explicitly even naming them, let alone noting that they require induction to prove.


问:
Calculating matrix exponential

Given matrix $$M = \begin{pmatrix} 7i& -6-2i\\6-2i&-7i\end{pmatrix}$$ how do I calculate matrix exponential $e^M$?

I know I can use that $e^A=Pe^DP^{-1}$ where $D=P^{-1}AP$. I computed the characteristic polynomial of the above matrix as

$$P(\lambda)=\lambda^2+89$$

Is there an easier way to do this than trying to compute the diagonalized matrix?

Title: Calculating matrix exponential

Body:
Via Cayley-Hamilton, ${\rm M}^2 + 89 \, {\rm I}_2 = {\rm O}_2$. Hence,

$$\begin{aligned} {\rm M}^2 &= - 89 \, {\rm I}_2\\ {\rm M}^3 &= - 89 \, {\rm M}\\ {\rm M}^4 &= 89^2 {\rm I}_2\\ {\rm M}^5 &= 89^2 {\rm M}\\ &\vdots\\ {\rm M}^{2k} &= (-1)^k 89^k {\rm I}_2\\ {\rm M}^{2k+1} &= (-1)^k 89^k {\rm M} \end{aligned}$$

and

$$\exp({\rm M}) = \sum_{k=0}^{\infty} \frac{{\rm M}^k}{k!} = \cdots = \color{blue}{\cos( \sqrt{89} ) \, {\rm I}_2 +\frac{\sin( \sqrt{89} )}{\sqrt{89}} {\rm M}}$$


问:
Proof of a proposition involving universal quantifiers and a cover
I started to learn topology recently. Solving some examples in a textbook, I found the solutions for some problems make use of a cover in a way that can be described as follows (if I correctly understood):

Consider the topological space $X$, the set $S$, and the family of sets $C$ such that $\forall A \in C: A\subseteq X$, $S\subseteq X$, and $S \subseteq \cup C$, that is, $C$ covers $S$. $P$ is a predicate. Then, the following proposition seems true.
$$
(\forall A \in C: (x \in A \cap S \implies P(x))) \iff (x\in S \implies P(x))
$$

But I could not figure out how to prove that. I am not familiar with logic and don't know how to correctly handle universal quantifiers.

The last question is, are $x\in S \implies P(x)$ and $\forall x \in S : P(x)$ equivalent in this case?
Any advice will be appreciated.

答:
Let $P=\{x\in X:P(x)\}$. Then $\forall A\in C\,\big(x\in A\cap S\to P(x)\big)$ clearly just says that $A\cap S\subseteq P$ for each $A\in C$, and $x\in S\to P(x)$ says that $S\subseteq P$. Thus, your question boils down to asking whether $\forall A\in C\,(A\cap S\subseteq P)$ is equivalent to $S\subseteq P$.

Suppose that $A\cap S\subseteq P$ for each $A\in C$, and let $x\in S$. $C$ covers $S$ so there is an $A\in C$ such that $x\in A$. Clearly $x\in A\cap S\subseteq P$, so $x\in P$, and since $x$ was an arbitrary element of $S$, we’ve shown that $S\subseteq P$.

Now suppose that $S\subseteq P$, and let $A\in C$. Then $A\cap S\subseteq S\subseteq P$, so $A\cap S\subseteq P$. Thus, $A\cap S\subseteq P$ for all $A\in C$ if and only if $S\subseteq P$.

The answer to your last question is yes.


问:
Plotting Voronoi decomposition in Matlab
Let $x_0,x_1,...,x_k\in R^n$. Consider a set $V_i:=\{x\in R^n:\|x-x_i\|_p\leq\|x-x_j\|_p \mbox{, } \forall i\neq j\}$

For $p=2$, show that $V_i$ is a polyhedral set.

Generate $10$ points randomly in $R^2$, which are well spread from each other. Take $p=2$ and plot $V_i$ for each $i=1,...,10$.

Construct an algorithm to find $A\in R^{m\times n}$ and $b\in R^n$ such that $V_i = \{x\in R^n: Ax\leq b\}$. Justify each step of the algorithm rigorously.

For the first part, I tried this:
$$\|x-x_i\|_2\leq\|x-x_j\|_2 \implies (x-x_i)^T(x-x_i) \leq (x-x_j)^T(x-x_j)$$
$$\implies x^Tx-2x_i^Tx+x_i^Tx_i \leq x^Tx-2x_j^Tx+x_j^Tx_j$$
$$\implies 2(x_j-x_i)^Tx \leq x_j^Tx_j-x_i^Tx_i$$

which is enough to define a half space and hence, a polyhedron I think. Is this proof correct? My main issue starts with the next parts of this question. I have been asked to plot these sets in Matlab and I have no idea how to do it. I know about plotting Voronoi diagrams in Matlab, but this involves the decompositions of the Voronoi sets and I am not sure how to proceed. Secondly, can someone give an insight of what this set represents? Maybe that will help me a bit. Finally, I have no idea on how to go about the third part. All hints are appreciated.

Thanks

答:
can someone give an insight of what this set represents?

The set $V_i$ consists of the points closer to $x_i$ than to any other $x_j$.

I know about plotting Voronoi diagrams in Matlab, but this involves the decompositions of the Voronoi sets and I am not sure how to proceed

To plot $V_i$, for each $j$ you can plot the set of points that is separated as far from $x_i$ as from $x_j$ (equidistant). That set is a line. After doing that for each $j$, some of those lines will be the boundary of $V_i$. Let $x_i, x_j \in \mathbb{R}^2$. Let's find two points that are equidistant. The midpoint $(x_i+x_j)/2$ is equidistant. Another point can be found using the observation that the line is perpendicular to the line through $x_i$ and $x_j$; so is $(x_i+x_j)/2 + v$ where $v$ is a nontrivial solution to $v^T(x_i-x_j) = 0$. After you find such $v$, you can draw the line with Matlab's line function (and then repeat for the other $j$ to finish $V_i$).

I have no idea on how to go about the third part

You have shown that:
$$V_i = \left\{ x \in \mathbb{R}^n : 2(x_j-x_i)^Tx \leq x_j^Tx_j-x_i^Tx_i  \forall i \neq j\right\}$$
So for part 3, you can create the matrix $A \in \mathbb{R}^{(n-1)\times n}$ with $2(x_j-x_i)^T$ as rows (one row for each $j$), and $b \in \mathbb{R}^{n-1}$ with elements $x_j^Tx_j-x_i^Tx_i$ (one element for each $j$).


问:
Binomial type sum
Does the following reduce to something simpler

$\sum_j {k \choose 2j} x^{2j}$

I have perused the combinatorial identities by could but I did not find anything that fits my case.

Title: Binomial type sum

Body:
\begin{align}
\sum_{j \ge 0} \binom{k}{2j} x^{2j}
&= \sum_{j \ge 0} \frac{1+(-1)^j}{2}\binom{k}{j} x^j \\
&= \frac{1}{2}\sum_{j \ge 0} \binom{k}{j} x^j + \frac{1}{2}\sum_{j \ge 0} \binom{k}{j} (-x)^j \\
&= \frac{(1+x)^k + (1-x)^k}{2}
\end{align}


答:
$\forall x \in S : P(x)$ is shorthand for $\forall x, (x \in S) \wedge P(x)$, so the second answer is "absolutely".

For the first, let's use that second one:
\begin{align}
(\forall A& \in C : (x \in A \cap S \implies P(x))) \iff (x\in S \implies P(x)) \\
(\forall A  &: ((A \in C ) \wedge (x \in A \cap S \implies P(x)))  \iff (x\in S \implies P(x)) & \text{applying rule from above}\\
\end{align}
Now let's look at that last thing. Suppose $x \in S$. Then because $C$ is a cover, there's some element $A$ of $C$ such that $x \in A$. That means that $x \in A$ and $x \in S$, so $x \in A \cap S$, so $P(x)$ holds. That proves the left-to-right implication.

Now suppose that the right-hand side is known. Let $A$ be any element of $C$. Then $x \in A \cap S$ tells use $x \in S$, hence $P(x)$ (by our assumption). So that proves the right-to-left implication.

Title: Proof of a proposition involving universal quantifiers and a cover

Body:
I started to learn topology recently. Solving some examples in a textbook, I found the solutions for some problems make use of a cover in a way that can be described as follows (if I correctly understood):

Consider the topological space $X$, the set $S$, and the family of sets $C$ such that $\forall A \in C: A\subseteq X$, $S\subseteq X$, and $S \subseteq \cup C$, that is, $C$ covers $S$. $P$ is a predicate. Then, the following proposition seems true.
$$
(\forall A \in C: (x \in A \cap S \implies P(x))) \iff (x\in S \implies P(x))
$$

But I could not figure out how to prove that. I am not familiar with logic and don't know how to correctly handle universal quantifiers.

The last question is, are $x\in S \implies P(x)$ and $\forall x \in S : P(x)$ equivalent in this case?
Any advice will be appreciated.


问:
Pre Calculus Math Equation With Logarithms
Please Help me with this I think i figured out question 1... but I get no solution...
please help me start number 2 or if you can show full solution that be sick thanks.

$\log_{3x}(81)=2$

$\log_3(\log_x(\log_4 16))=-1$

my attempt for number 1 ! :)

$$\frac{1}{81}\log_{3x}(81)=\frac{2}{81}$$
$$\log_{3x}=2/81$$
$$10^{\large{\log_{3x}}}=10(2/81)$$
$$3x=10 (2/81)$$
$$x=10 (2/81) /3 $$

答:
I make multiple use of the rule: $\log_a x=b\implies a^b=x$

The first problem is already addressed in one of the other answers, but when you see $\log_{3x}(81)=2$, you should think $(3x)^2=9x^2=81\implies x=3$

As for the second problem, since $4^2=16$, we know that $\log_4 16=2$ and we can make the following simplification:

$$\log_3(\log_x(\log_4 16))=-1\\
\log_3(\log_x(2))=-1$$

Now, if we think of $\log_x(2)$ as a variable, call it $y$, we can rewrite the above equality like so:
$$\log_3 y=-1\\
y=3^{-1}=\frac{1}{3}$$
Now that we have a value for $y$, lets put it back in context:
$$y=\log_x(2)=\frac{1}{3}\\
x^{\frac{1}{3}}=2\\
x=8
$$


问:
Determining what relation means
If we have to consider the relation R from {1,2,3,4} to {3,4,6,7,9} given by aRb ↔ b = a + 3 then would the relation given be {1,2,3,4,6,7,9}? I'm a little confused on what is meant when I'm being told to consider a relation R from one relation to another.</p>

答:
A relation on sets $A$ and $B$, $a R b$, $a \in A, b \in B$ is a subset of $A \times B.$ If $a R b,$ then $(a,b) \in R.$ So if $b=a+3,$ then $(a,b) \in R.$

$$R= \{ (1,4), (3,6), (4,7)\}.$$

Title: Determining what relation means


问:
$\int_{\mathbb{R}^n} f \, d\lambda=0$ if and only if $f(x)=0$ almost everywhere
let's say $f: \mathbb{R}^n \to [0,+\infty]$ is a positive measurable function. I want to prove that $\int_{\mathbb{R^{n}}} f \, d\lambda=0$ if and only if $f(x)=0$ almost everywhere.

Let's say $f(x)=0$ almost everywhere. Name $g$ the zero function. Now it follows that $g$ and $f$ are almost everywhere the same. This means that the intergal of $f$ equals the integral of $g$ (which exists because we know that the integral of the zero function existst). This means that $\int_{\mathbb{R}^n} f \, d\lambda=\int_{\mathbb{R}^n} g \, d\lambda=0$

Now the other side. Let's say $\int_{\mathbb{R}^n} f \, d\lambda=0$ .
I know I need to do something with $f$ is positive because when it's not this implication wouldn't work. I just don't know how to start with this side. How do you start to prove this or what are thet tools?

答: Hint: suppose that $\lambda(\{x\mid f(x)>0\})>0$. Then argue that there exists $n$ so that $\lambda(\{x\mid f(x)>0)>1/n\})>0$. Now recall that the Lebesgue integral of $f$ is the supremum of the integral of all simple functions less than $f$. Can you find a simple function less than $f$ with positive integral to get a contradiction?


答:
$$
\{ x : f(x)>0 \} = \{x: f(x)\ge 1\} \cup \bigcup_{n=1}^\infty \left\{x:\tfrac 1 {n+1} \le f(x) < \tfrac 1 n \right\}
$$
If the set on the left has positive measure then one of the terms in this union has positive measure, and that measure times the corresponding lower bound on values of $f(x)$ in that set gives you a positive lower bound on the integral.

Title: $\int_{\mathbb{R}^n} f \, d\lambda=0$ if and only if $f(x)=0$ almost everywhere

Body:
let's say $f: \mathbb{R}^n \to [0,+\infty]$ is a positive measurable function. I want to prove that $\int_{\mathbb{R^{n}}} f \, d\lambda=0$ if and only if $f(x)=0$ almost everywhere.

Let's say $f(x)=0$ almost everywhere. Name $g$ the zero function. Now it follows that $g$ and $f$ are almost everywhere the same. This means that the intergal of $f$ equals the integral of $g$ (which exists because we know that the integral of the zero function existst). This means that $\int_{\mathbb{R}^n} f \, d\lambda=\int_{\mathbb{R}^n} g \, d\lambda=0$

Now the other side. Let's say $\int_{\mathbb{R}^n} f \, d\lambda=0$ .
I know I need to do something with $f$ is positive because when it's not this implication wouldn't work. I just don't know how to start with this side. How do you start to prove this or what are thet tools?


问:
Inverse and derivative of a function
Find an example of an inverse function f(x) such that its derivative is the same as its inverse.

I tried many different functions but non of them worked.

答:
If I understood your question correctly and assuming that you are asking about functions over $R,$ there is no such function.
Indeed, if $f'(x)=g(x)$ and $f(g(x))=x=g(f(x))$ then $f'(f(x))=x.$ But $f$ has to be a bijection so it is monotone and derivative cannot change the sign.</p>

Title: Inverse and derivative of a function


问:
Proper Terminology for Difference via Subtraction vs. Division
When comparing normalized data sets, I'm struggling to find the correct terminology for $X_1-X_2$ vs $ X_1\div X_2$. Often I see the word "Difference" used interchangeably. Is there a most correct terminology that differentiates the two?

If it helps I'm working in an engineering field, not a pure math environment.

答:
Sets aren't ordered, so it doesn't make sense to subtract componentwise. Normally $A-B$ where $A$ and $B$ are sets is the set $C$ which consists only of elements of $A$ which are not in $B$.
$$A-B = \{a \in A : a \not \in B\}$$


问:
Proving the existence of the minimum of a function
I need to prove that, given a real continuous function $f(x)$ defined in $(a,\infty)$ with $a$ finite and such that $\lim_{x \to a} f(x) = l$,  $\lim_{x \to\infty} f(x) = \infty$, if there is a point $x_0>a$ where $f(x_0)<l$ then the function has (at least) a minimum. I was thinking to use Weierstrass theorem, but the theorem applies to continuous functions defined on compact intervals... Any hint?

答: as $\lim_{x\rightarrow \infty}f(x)=\infty$, there is a $\delta>0$ such that $f(x)>l$ if $x>\delta$ cleary that $\delta>x_0$, now as $\lim_{x\rightarrow a}f(x)=l$ there is a $\epsilon>0$ such that $f(x)>f(x_0)$ if $a<x<a+\epsilon$, now use Weiertrass theorem on $[a+\epsilon,\delta]$.


答: Rolles theorem should do this one! Rolle's theorem essentially states that any real-valued differentiable function that attains equal values at two distinct points must have at least one stationary point somewhere between them</p>

Title: Proving the existence of the minimum of a function

Body: I need to prove that, given a real continuous function $f(x)$ defined in $(a,\infty)$ with $a$ finite and such that $\lim_{x \to a} f(x) = l$,  $\lim_{x \to\infty} f(x) = \infty$, if there is a point $x_0>a$ where $f(x_0)<l$ then the function has (at least) a minimum. I was thinking to use Weierstrass theorem, but the theorem applies to continuous functions defined on compact intervals... Any hint?


问:
Moore-Penrose Inverse Formula
Let $A^+$ denote the Moore-Penrose inverse of a matrix $A$.

Problem. Let $A$ and $B$ be two compatible matrices where $B$ has full row rank. Show that $$AB(AB)^+=AA^+.$$

The verification is easy: Both sides are symmetric and $AA^+AB=AB=AB(AB)^+AB$. Of course we cannot cancel $AB$ on both sides, but I can't figure out how to start and what the condition $B$ has full row rank is used for. Hope anyone has good suggestions.

Update. Here are some of my recent findings but they are not sufficient to solve this problem.

Since $B$ has full row rank, we have $B^+=B^T(BB^T)^{-1}$ and thus $BB^+=I$. Thus, we have
$$AA^+=AIA^+=ABB^+A^+.$$
If we can show that $(AB)^+=B^+A^+$. Then we are done. Nonetheless, this does not generally hold. We can see here that $B^+A^+$ is merely a $(1,2,3)$-inverse of $AB$:

$ABB^+A^+AB=AA^+AB=AB$.

$B^+A^+ABB^+A^+=B^+A^+AA^+=B^+A^+$.

$ABB^+A^+=AA^+$ is symmetric.

The last condition $B^+A^+AB$ is symmetric cannot be verified (at least by me). Of course, if $A$ has full column rank, then it would be easier because $A^+A=I$ provided $A$ has full column rank.

答:
Here is an almost one-line proof obtained just now from my instructor.

Note that $AA^+$ is a projection matrix onto the space $R(A)$.

Since $B$ has full row rank, we indeed have $R(AB)=R(A)$. Thus we must have $(AB)(AB)^+=AA^+$.

Title: Moore-Penrose Inverse Formula


问:
Find the matrix of the linear transformation $T : \Bbb R^3 \to \Bbb R^3$ that projects a vector onto the plane $x − y + 2z = 0$

Find the matrix of the linear transformation $T : \Bbb R^3 \to \Bbb R^3$ that projects a vector onto the plane $x − y + 2z = 0$.

I know how to find matrix of linear transformation when projecting onto the xy/xz/yz planes but not sure what to do when then plane is an equation. Any help would be greatly appreciated.

答:
Hint:

An orthonormal basis for your plane $x-y+2z=0$ is given by
$$v_1 = \frac1{\sqrt2}(1,1,0), \quad v_2 = \frac1{\sqrt3}(1,-1,1)$$
so your $T$ is given by $Tx = \langle x,v_1\rangle v_1 + \langle x,v_2\rangle v_2.$


问:
Showing $\alpha(\beta+1)\leq\frac{5}{3}\alpha^2+\frac{1}{3}\beta^2$ for nonnegative integers $\alpha$ and $\beta$.
I found this lemma in a paper I was reading and it was not proved. There doesn't seem to be any obvious factorized form so how would one go about proving the inequality holds?

$$\alpha(\beta+1)\leq\frac{5}{3}\alpha^2+\frac{1}{3}\beta^2$$

$$\frac{5}{3}\alpha^2+\frac{1}{3}\beta^2-\alpha(\beta+1) \geq 0$$

$$5\alpha^2+\beta^2-3\alpha\beta-3\alpha \geq 0$$

答:
Alternative proof inspired by Michael Hardy:

$$5\alpha^2+\beta^2-3\alpha\beta-3\alpha \geqslant 0 \iff 20\alpha^2+4\beta^2-12\alpha
\beta-12\alpha \geqslant 0$$

$$20\alpha^2+4\beta^2-12\alpha \beta-12\alpha
= (2\beta-3\alpha )^2+11\alpha^2-12\alpha \\
=\begin{cases}
(2\beta-3\alpha )^2+\alpha (11\alpha-12)>0,  & \text{if $\alpha>1$} \\
(2\beta-3)^2-1 \geqslant 0, & \text{if $\alpha=1$} \\
4\beta^2 \geqslant 0 & \text{if $\alpha=0$}
\end{cases}
$$

$$
"=" \iff \alpha=1, 2\beta-3=\pm 1 \text{ or } \alpha=\beta=0\\ \iff (\alpha, \beta)\in \{(0,0), (1,1), (1,2)\}
$$


问:
Number of diagonals that cut the center of a regular polygon
An $n$-sided regular polygon ($n \geq 3$) has, as we know, $\dfrac{n(n-3)}{2}$ diagonals. If $n$ is odd, then none of those diagonals pass through the center of it. If $n$ is even, then $\dfrac{n}{2}$ diagonal passes through its center.

That's the thing: how to prove it?

All I've found was: drawing a diagonal that passes through the center of a $n$-sided polygon, the other points are separated in two equal groups (because of its symmetry) of $p$ points. So the total of points is $p + p + 2=2p+2=2(p+1)$, which is even. But this doesn't really feel like proof to me, since it uses arguments based on nothing.

答:
Welcome to MSE.  I think you have a proof, of half of it at least, though you might want to state it a bit more carefully.

Consider the line joining a vertex to the center.  If it passes, through a second vertex, then every vertex lies on a diagonal passing through the center, by symmetry.  Since there are two vertices on each such diagonal, there must be an even number of vertices.

What we haven't shown is that, in the case that $n$ is even, these $\frac n2$ diagonals actually exist.  The only way I can see to do that is by construction, but that's easy.


答:
Here is an outline of an approach slightly ddifferent from mechanoidroid's.

First, note that the vector $v=(1,-1,2)$. is orthogonal to the plane $x-y+2z=0$ (why?).  Next construct a map $P$ that projects onto the line spanned by $v$.  Finally, note that if $T$ is the projection onto the plane, $P+T=I$ (why?), similar to how projection onto the z axis and projection onto the xy-plane add to $I$.

Body:
Find the matrix of the linear transformation $T : \Bbb R^3 \to \Bbb R^3$ that projects a vector onto the plane $x − y + 2z = 0$.

I know how to find matrix of linear transformation when projecting onto the xy/xz/yz planes but not sure what to do when then plane is an equation. Any help would be greatly appreciated.


答:
We find a function defined for all positive $x$, with the right properties.

There may be no general theory, so let's fool around and look for a function of the shape $f(x)= cx^a$, for constants $a$ and $c$.

Then the derivative of $f$ is given by $f'(x)=acx^{a-1}$. The inverse function is given by $f^{-1}(x)=\frac{x^{1/a}}{c^{1/a}}$. So we need
$$acx^{a-1}=\frac{x^{1/a}}{c^{1/a}}.$$
For the the identity above to hold, we want $a-1=\frac{1}{a}$. Rewrite as $a^2-a-1=0$, which has $a=\frac{1+\sqrt{5}}{2}$ as a root. Nice number!

We also want $ca=\frac{1}{c^{1/a}}$. There is such a $c$. It is given by
$$c=\left(\frac{\sqrt{5}-1}{2}\right)^{1/\sqrt{5}}.$$

Title: Inverse and derivative of a function

Body:
Find an example of an inverse function f(x) such that its derivative is the same as its inverse.

I tried many different functions but non of them worked.


问:
Comparison convergence test $\sum_{n=1}^\infty (2n+1)\beta^{-\sqrt{2n+1}}$
Wolfram Alpha tells me that the series $$\sum_{n=1}^\infty (2n+1)\beta^{-\sqrt{2n+1}}$$ converges by comparison test for $\beta > 1$, but I'm struggling to find a function to compare it with. Any ideas?

Edit 1 (Motivation): I got to this series because in the paper I link below there is a proposition (2.7) which uses the Dominated Convergence Theorem in the proof, but they don't specify the dominating function. I am using that proposition in my final degree Thesis, so I have to formally proove it, and this is the natural dominance function. This part of my thesis is about the Hermite-Fourier expansions and the heat and Hermite semigroups.

https://www.sciencedirect.com/science/article/pii/S0022123603000831

Edit 2 (Current progress): This actually converges by the integral test, but I would like to find a solution using convergence test.

答:
If $b > 1$
then,
since
$\dfrac{\log(n)}{\sqrt{n}}
\to 0$,
$\sqrt{2n+1}
\gt 2\log(n)/\log(b)
$
for all large enough $n$,
so

$\begin{array}\\
 b^{-\sqrt{2n+1}}
&=\dfrac1{b^{\sqrt{2n+1}}}\\
&\lt\dfrac1{b^{2\log(n)/\log(b)}}\\
&=\dfrac1{e^{2\log(b)\log(n)/\log(b)}}\\
&=\dfrac1{e^{2\log(n)}}\\
&=\dfrac1{n^2}\\
\end{array}
$

and the sum of these converges.


问:
Why do the triangular numbers initially form long cycles mod $2^k$?
As discussed at Triangular numbers ($\text{mod } 2^n$) as a permutation of $\{0,1,2,\dots,2^n-1\}$ and What is the set of triangular numbers mod $n$?, mapping the integer $n$  for $0\le n\lt2^k$ to the residue of the corresponding triangular number $\frac12n(n+1)$ modulo $2^k$ yields a permutation. For example, for $k=3$:

$$
01234567\\
01362754
$$

I noticed that up to $k=5$, all elements except for $0$ and $1$ (which are always mapped to themselves) form a single cycle of length $2^k-2$. The probability for a uniformly random permutation of length $n$ to consist of a single cycle is $\frac1n$, so if these permutations (excluding $0$ and $1$) could be considered uniformly random, the probability for this to happen would be only $\frac12\cdot\frac16\cdot\frac1{14}\cdot\frac1{30}=\frac1{5040}$. That was reason enough to check whether the pattern continues for greater $k$.

It turns out that it doesn't, as for $k=6$ there is a $3$-cycle: $(4,10,55)$. Nevertheless, at first unusually large cycle lengths persist: For all $k$ from $2$ to $12$, except for $k=7$, the largest cycle contains more than half the elements, whereas the probability for this to happen in a random permutation is roughly $\ln 2$. In fact, in $9$ of these $11$ cases (all except $k=6$ and $k=7$), the largest cycle contains more than $\frac45$ of the elements; the probability for that is roughly $\ln\frac54\approx0.223$ per case, so the probability for it to happen at least $9$ times out of $11$ is only $\sum_{k=9}^{11}\binom{11}k\left(\ln\frac54\right)^k\left(1-\ln\frac54\right)^{11-k}\approx5\cdot10^{-5}$.

However, this pattern, too, doesn't continue: For $k$ from $2$ to $30$, there are $21$ cases with cycles of more than half the elements, which is about the expected number $29\ln2\approx20.1$; and for $k$ from $13$ to $30$ there are only $4$ cases with cycles of more than $\frac45$ of the elements, which is almost exactly the expected number $18\ln\frac54\approx4.0$.

My question is: Is there an explanation for this initial tendency to form long cycles? Or should we put it down to coincidence?

For your convenience, here's the code I used to find the cycle lengths, and here are the results up to $k=30$:

```<code>4 : 2
8 : 6
16 : 14
32 : 30
64 : 40, 19, 3
128 : 55, 48, 14, 6, 3
256 : 247, 4, 3
512 : 488, 7, 6, 6, 3
1024 : 818, 157, 47
2048 : 1652, 371, 23
4096 : 4060, 25, 9
8192 : 3754, 3609, 412, 321, 79, 12, 3
16384 : 15748, 292, 190, 71, 24, 22, 13, 13, 9
32768 : 20161, 11349, 333, 305, 281, 218, 72, 44, 3
65536 : 20128, 17231, 16759, 8072, 2377, 579, 295, 60, 33
131072 : 85861, 26603, 9389, 3887, 3365, 682, 594, 488, 118, 49, 23, 6, 5
262144 : 159827, 89991, 5749, 5465, 592, 231, 118, 100, 42, 24, 3
524288 : 211265, 176243, 59029, 35639, 28496, 6122, 4245, 1239, 713, 632, 244, 146, 133, 59, 39, 36, 6
1048576 : 620076, 216520, 131454, 68118, 7535, 2111, 1235, 1028, 225, 213, 36, 20, 3
2097152 : 993084, 583840, 394263, 87941, 31835, 3389, 1648, 459, 306, 273, 45, 35, 14, 10, 8
4194304 : 1487646, 1119526, 942359, 429054, 118037, 64446, 28806, 3238, 323, 291, 186, 126, 118, 102, 12, 11, 10, 7, 4
8388608 : 2542051, 2462220, 2040680, 1138236, 93072, 45880, 19664, 16473, 14243, 6319, 2917, 2598, 2160, 1414, 514, 118, 23, 19, 5
16777216 : 12137774, 4004239, 271250, 253890, 43860, 33597, 25495, 4132, 2575, 157, 116, 67, 35, 9, 8, 6, 4
33554432 : 28169497, 2552414, 1401622, 1019221, 356682, 21006, 14735, 10242, 8223, 566, 135, 45, 21, 15, 6
67108864 : 32223531, 29360424, 3530597, 932310, 809707, 99109, 83093, 67418, 1612, 364, 248, 248, 166, 21, 14
134217728 : 87591110, 34361487, 3360928, 3343185, 3291274, 1345478, 353498, 323522, 158252, 47767, 17776, 11159, 5927, 2681, 2343, 530, 235, 208, 162, 84, 59, 31, 30
268435456 : 232647749, 24918738, 5559122, 3742461, 525140, 384941, 278834, 197080, 62977, 48736, 21684, 16632, 13525, 8993, 3073, 2721, 1625, 1262, 153, 5, 3
536870912 : 379598603, 129063661, 26279056, 665648, 483286, 222289, 137686, 106713, 94323, 80276, 59199, 41767, 15498, 10615, 5066, 2816, 2699, 1579, 113, 10, 7
1073741824 : 877039442, 181409872, 7571387, 6549459, 921247, 240525, 3924, 3416, 1602, 894, 54
</code>```

答:
You say that "the largest cycle contains more than half the elements, whereas the probability for this to happen in a random permutation is roughly $\ln(2)$". But this doesn't account for a thing or two. (1) we can prove there are no 1-cycles, so only derangements on $2^k-2$ need be considered. And (2) maybe we can prove there are no 2-cycles after the $k=2$ case, so we would only need consider permutations lacking a transposition.

Let's just eliminate $1$-cycles for starters. The number of derangements on $2^k-2$ elements is within an integer of $\frac{\left(2^k-2\right)!}{e}$ while the number of $(2^k-2)$-cycles is $(2^k-3)!$. So the probability of a random derangement being a full cycle is about $\frac{e}{2^k-2}$. The exact values for $k=2,3,4,5$ are $1,\frac{120}{265},\frac{479001600}{2467007773},\frac{304888344611713860501504000000}{3364864615063302680426807870189}$. Their product is about $0.00796$, relatively a good deal greater than $\frac{1}{5040}$.

Now if we can prove that there are no $2$-cycles beyond the $k=2$ case then these numbers become a lot larger, roughly $\frac{e^{1.5}}{2^k-2}$ according to OEIS. Then for $k=3,4,5$ we have $\frac{3}{4},\frac{1247400}{3896663},0.149\ldots$, and the product is about $0.035$.

[So far I'm only able to prove there is no $2$-cycle in certain cases, so it's not yet clear that $2$-cycles can be tossed out.]

This has only looked at the probabilities when asking for a full cycle permutation. If we ask for the permutations with a cycle of at least half length, the numbers rise dramatically. For example with $k=3$, once you exclude 1-cycles and 2-cycles, all that is left is 6-cycles and pairs of disjoint 3-cycles. So there is a 100% chance of a random transposition-free derangement having a cycle of at least half-length. For $k=4$, I worked out there is about an 84% chance of a random transposition-free derangement having a cycle of at least half-length.

This effect of excluding singletons (and possibly transpositions) will taper off quickly as $k$ rises. But I think it is a significant factor in raising the odds for this observation to be happening at low $k$.


答:
Another way to see this is to connect all the vertices to the center. The regular polygon would be divided into $n$ equal parts, and form an angle of $\dfrac {2\pi}n$ at the center.

For a diagonal to pass through the center, we need to show that the vertices involved, when connected to the center, forms one straight line. The angle formed at the center would then be $\pi$. This must also be equal to $k\left(\dfrac {2\pi}n\right)$ for some integer $k$.

Now it is clear that for odd $n$, $\dfrac {2k\pi}n \ne\pi$, or $2k \ne n$ for any $k$, so no diagonals satisfy this property.

For even $n$, the unique $k$ that satisfy this is $k=\dfrac n2$. Hence for each vertex, there is one unique "opposite" vertex that forms a diagonal passing through the center. Since we have $n$ vertices, we can draw $n$ diagonals. However we have double-counted, since each diagonal is formed by $2$ vertices. Hence the number of diagonals passing through the center in this case is precisely $\dfrac n2$.

Title: Number of diagonals that cut the center of a regular polygon

Body:
An $n$-sided regular polygon ($n \geq 3$) has, as we know, $\dfrac{n(n-3)}{2}$ diagonals. If $n$ is odd, then none of those diagonals pass through the center of it. If $n$ is even, then $\dfrac{n}{2}$ diagonal passes through its center.

That's the thing: how to prove it?

All I've found was: drawing a diagonal that passes through the center of a $n$-sided polygon, the other points are separated in two equal groups (because of its symmetry) of $p$ points. So the total of points is $p + p + 2=2p+2=2(p+1)$, which is even. But this doesn't really feel like proof to me, since it uses arguments based on nothing.


问:
What implications does $\sum_{n=1}^\infty |a_n|^\beta<\infty, \ 0<\beta<1$ have for $\sum_{n=1}^\infty|a_n|$?
It is rather obvious that $\sum_{n=1}^\infty |a_n|^\beta<\infty$ for some $0<\beta<1$ implies that $|a_n|^\beta<1$ for almost all $n$.

Hence, $|a_n|^\beta \geq |a_n|$ for almost all $n$. Therefore $\sum_{n=1}^\infty |a_n|^\beta<\infty$ implies $\sum_{n=1}^\infty |a_n|<\infty$.

My question is whether we can get any quantitative results? Intuitively, the convergence of the first series implies "faster" convergence of the second series. Can we formalize this somehow? Is the rate of convergence affected?

I am wondering about this because I read a book on harmonic analysis in which the convergence of $\sum_{n=1}^\infty |f_n|^\beta<\infty$ is studied where $f_n$ are the Fourier coefficients of some function $f$. For $\beta=1$ this is clearly interesting because it implies uniform convergence of the Fourier series but the book also gives results for $\beta<1$ and I am unsure how this is relevant to the actual approximation theory? Does the convergence of this series imply any "better" qualities for the Fourier approximation? If not, why are these results even mentioned?

For refrence the book is "Commutative Harmonic Analysis IV" by V.P. Khavin and N.K. Nikol'skiǐ.

If this is a common question studied within harmonic analysis I would be more then thankful for any book or paper refrences, my search so far has only come up with convergence results for above series under certain conditions. I havn't found any explanations of what that convergence actually means.

答:
I'll assume throughout that $a_n\ge0$.
Define $E_\beta(x) = \sum_{n>x} a_n^\beta$, so that the convergence of $\sum_{n=1}^\infty a_n$ is equivalent to $E_\beta(x) \to 0$ as $x\to\infty$. I interpret your question to be: what can we say about $E_1(x)$ in terms of $E_\beta(x)$ alone? A slight elaboration of your argument shows that $E_1(x)/E_\beta(x) \to 0$; can we say anything stronger?

In general, I believe the answer is no. (I'm writing this quickly, so double-checking will be welcome.) In other words, given any function $d(x)$ decaying to $0$, no matter how slowly, we should be able to find an example of a sequence $\{a_n\}$ such that $E_1(x)/E_\beta(x) > d(x)$ for all sufficiently large $x$. Indeed, I think an example can be found using a fixed convergent series like $1/n^2$, but spreading the positive terms out by inserting longer and longer strings of $0$s between the positive terms.

So we would need additional information to get better bounds. A natural quantity to consider is $M(x) = \max\{a_n\colon n>x\}$, which must tend to $0$ since $\sum a_n$ converges. (If $a_n$ is decreasing then $M(x)$ is simply $a_{\lfloor x\rfloor+1}$.) It's clear that
$$
E_1(x) = \sum_{n>x} a_n = \sum_{n<x} a_n^{1-\beta} a_n^\beta \le M(x)^{1-\beta} \sum_{n<x} a_n^\beta = M(x)^{1-\beta} E_\beta(x),
$$
which is a quantitative improvement over the mere $E_1(x)/E_\beta(x) \to 0$.
I suspect that nothing better than $E_1(x) \le M(x)^{1-\beta} E_\beta(x)$ (other than perhaps a multiplicative constant) is true in general: it is already the correct order of magnitude when $a_n$ is polynomially decaying and when $a_n$ is exponentially decaying.


问:
Evaluate $\int_{-\infty}^{\infty}\sin(ax)\operatorname{sech}^4(b(x-c))dx$, for $a,b,c>0$
$\newcommand{\sech}{\operatorname{sech}}$
I am trying to compute the following integral:

$$\int_{-\infty}^{\infty}\sin(ax)\sech^4(b(x-c))dx; a,b,c>0$$

I cannot seem to tackle this integral. I have tried integration by parts, but this leads nowhere since we need to evaluate $\sin(ax)$ at the infinities. How should I be approaching this integral?

答:
$\newcommand{\sech}{\operatorname{sech}}$
$\newcommand{\res}{\operatorname{Res}}$

Put $x=z/b+c$ and use angle-addition:
$$
\int_{-\infty}^{\infty} \sin(ax)\sech ^4(b(x-c))\,dx
$$
$$
=\frac{1}{b}\int_{-\infty}^{\infty} \sin(a(z/b+c))\sech ^4(z)\,dz
$$
$$
=\frac{\sin(ac)}{b}\int_{-\infty}^{\infty} \cos(\frac{a}{b}z)\sech ^4(z)\,dz+\frac{\cos(ac)}{b}\int_{-\infty}^{\infty} \sin(\frac{a}{b}z)\sech ^4(z)\,dz
$$Now $\sech^4$ is improperly integrable (in fact, rapidly decaying) and even, so the second integral vanishes by symmetry. Note that
$$
\int_{-\infty}^{\infty} \cos(\frac{a}{b}z)\sech ^4(z)\,dz=\Re\left(\int_{-\infty}^{\infty} \exp(\frac{a}{b} i z)\sech ^4(z)\,dz\right)
$$ We will use the Residue Theorem. Let $f(z)$ denote the integrand. A pole of order four occurs at $z=\pi i/2$, so it makes sense to take our contour as the rectangle $\gamma$ (with sub-contours $\gamma_1,\gamma_2,\gamma_3,\gamma_4$) whose path is $$\gamma:-R\longrightarrow R \longrightarrow R+\pi i \longrightarrow -R+\pi i \longrightarrow -R.$$

The details are left to the reader (or to Mathematica or some such), but we have:
$$
\int_{\gamma_3}f(z)\,dz = -\exp(-\frac{a}{b}\pi) \int _{\gamma_1}f(z)\,dz;\\
\lim_{R\to \infty} \int_{\gamma_2}f(z)\,dz=\lim_{R\to \infty} \int_{\gamma_4}f(z)\,dz=0
$$

Computing the residue is tedious, but the end result is:
$$
\res(f(z),\pi i /2)
 = \frac{1}{6}  \lim_{z\to \pi i/2} \frac{d^3}{dz^3}\frac{\exp(\frac{a}{b} iz)(z-\pi i/2)^4}{\sech^4(z)}
$$
$$
= -\frac{i a \left(a^2+4 b^2\right) \exp(-\frac{a}{b}\pi/2)}{6 b^3}
$$ Then in the limit, we have
$$
\lim_{R\to\infty}\int _{\gamma_1}f(z)\,dz = \Re\left((1-\exp(-\frac{a}{b}\pi))^{-1}\cdot 2\pi i \cdot \left(\frac{-i a \left(a^2+4 b^2\right) \exp(-\frac{a}{b}\pi/2)}{6 b^3}\right)\right)
$$
$$=\frac{\pi  a \left(a^2+4 b^2\right) \text{csch}\left(\frac{\pi  a}{2 b}\right)}{6 b^3}
$$

So in summary,
$$
\int_{-\infty}^{\infty} \sin(ax)\sech ^4(b(x-c))\,dx =  \frac{\pi  a \left(a^2+4 b^2\right) \text{csch}\left(\frac{\pi  a}{2
   b}\right) \sin (a c)}{6 b^4}
$$

Note: originally, I took the contour to be the semicircle enclosing the upper half plane, with poles at $t=\pi i(k+1/2)$, $k\in\mathbb{N}^0$. Their sum is a geometric series but showing the integral vanished along the circular arc was slightly tricky because the integrand is large, of size $ \cosh(\frac{a}{b}R)\sec^4(R)$, near $\theta=\pi/2$. If one were careful with the approximation by bounding the size of this region (the integrand rapidly decreases elsewhere), you could do the problem the same way.


问:
Uniform Convergence $f_n(x)=\sin\left(\frac{x}{n}\right)$ on $[-R, R]$
I wish to show uniform convergence of $f_n(x)=\sin\left(\frac{x}{n}\right)$ to $g(x)=0$ for all $x \in [-R, R]$, where $R$ is a positive real number.

To do this, we need:

$\forall  \epsilon>0, \forall x\in [-R, R], \exists N\in \mathbb N$ such that $n>N$ means $|f_n(x)-g(x)| < \epsilon$.

So we are interested in estimating:

$$|\sin(x/n)|$$

The familiar problem, how do we pick $N$? For $n$ sufficiently large we see that since $x$ is bounded by $R$ and finite we know $x/n$ can be made as small as we like, at which point $\sin(x/n)$ behaves like $x/n$ which tends to $0$. How do I make this more precise?

答:
A classical estimate is $|\sin t|\le |t|$ for all $t\in\Bbb R$. So
$$|f_n(x)|=|\sin(x/n)|\le |x/n|\le R/n$$ for all $x\in[-R,R]$.
So Given $\varepsilon>0$, let $N\in\Bbb N$ such that $\frac{R}{N}<\varepsilon$. So for $n\ge N$ and $x\in[-R,R]$
$$|f_n(x)|\le \frac{R}{n}<\varepsilon$$ and you're done.


答:
hint

By MVT, For $ X>0 $, $$|\sin(X)-\sin(0)|=|X\cos(c)|\le |X|$$
For $ X<0 $,
$$|\sin(-X)|\le |-X|$$

So,

$$(\forall X\in \Bbb R)\;\; |\sin(X)|\le |X|$$

to get

$$(\forall x\in[-R,R])\; \left|\sin\left(\frac xn\right)\right|\le \left|\frac xn\right|\le \frac Rn$$

and
$$\sup_{[-R,R]}|f_n-g|\le \frac Rn$$

with $$\lim_{n\to+\infty}\frac Rn=0$$

Title: Uniform Convergence $f_n(x)=\sin\left(\frac{x}{n}\right)$ on $[-R, R]$

Body:
I wish to show uniform convergence of $f_n(x)=\sin\left(\frac{x}{n}\right)$ to $g(x)=0$ for all $x \in [-R, R]$, where $R$ is a positive real number.

To do this, we need:

$\forall  \epsilon>0, \forall x\in [-R, R], \exists N\in \mathbb N$ such that $n>N$ means $|f_n(x)-g(x)| < \epsilon$.

So we are interested in estimating:

$$|\sin(x/n)|$$

The familiar problem, how do we pick $N$? For $n$ sufficiently large we see that since $x$ is bounded by $R$ and finite we know $x/n$ can be made as small as we like, at which point $\sin(x/n)$ behaves like $x/n$ which tends to $0$. How do I make this more precise?


问:
Are product measures the only $T$-invariant measures if $T$ is the left shift on $\{0,1,\ldots,k-1\}^\mathbb Z$?
Consider $X=\{0,1,\ldots,k-1\}^\mathbb Z$ equipped with the $\sigma$-algebra $\mathcal C$ generated by the cylinder sets $$\{x\in X:x_m=i_m,\ldots,x_n=i_n\}\tag{$m\leq n$}.$$ It is well known that the $p$-Bernoulli measure $\mu$, defined by letting $$\mu\{x\in X:x_m=i_m,\ldots,x_n=i_n\}=p_{i_m}\cdots p_{i_n}$$ is $T$-invariant (and ergodic) for the left shift $T:X\to X:x\mapsto y$, where $y_n=x_{n+1}$. However, I was wondering if there are any other $T$-invariant (not necessarily ergodic) measures. If I understand it correctly, this post seems to imply that no other $T$-invariant measure exists, but I don't see why such another measure is impossible.

答:
There are many, many other shift-invariant measures.  One very easy example is something like $\mu = \frac12 \delta_{x} + \frac12 \delta_{Tx}$, where $x$ is the periodic sequence $010101\cdots$.

Another natural example is the Markov measure generated by transition probabilities $(P_{ij})_{0 \leq i,j \leq k-1}$ and the corresponding stationary distribution $(\pi(0), \dots, \pi(k-1))$.  The Markov measure $\mu$ is defined on cylinders by
$$
\mu\{x \in X : x_m \cdots x_{n} = i_{m} \dots i_n\} = \pi(i_m) P_{i_m i_{m+1}} \cdots P_{i_{n-1} i_n}.
$$
This example is called a Markov measure because it's just the dynamical systems way of talking about the joint distribution for sample paths of a finite state Markov chain.

Also, you can check that the first example is ergodic, and it's a classical theorem that the Markov measure is ergodic if the Markov chain is irreducible.

There are plenty of other examples besides these.  For shift systems, the set of invariant measures and the set of ergodic invariant measures are both huge.


问:
Orthogonal with respect to killing form
Let $k$ denote the killing form on the Lie algebra $\mathfrak{gl}_n(\mathbb{C})$ and let $\mathfrak{h},\mathfrak{n}_+, \mathfrak{n}_-$ denote the subspaces of diagonal matrices, strictly upper triangular matrices, and strictly lower triangular matrices respectively. How can I show that $\mathfrak{h}$ and $\mathfrak{n}_+\oplus \mathfrak{n}_-$ are orthogonal with respect to the killing form $k$?

答: It should not be necessary to compute $ad(x)\circ ad(y)$ explicitly. The point is that you can use a basis of $\mathfrak{gl}_n(\mathbb C)$ that consits of joint eigenvectors for $ad(x)$ for all elements $x\in\mathfrak h$. By bilinearity, it suffices to consider the case that $y$ is one of these basis elements cotained in $\mathfrak n_+$ or $\mathfrak n_-$. Then the action of $ad(y)$ maps any eigenvector for $ad(\mathfrak{h})$ to an eigenvector with different eigenvalue, while $ad(x)$ preserves each of the eigenspaces. This shows that in a matrix representation with respect to the above basis, the map $ad(x)\circ ad(y)$ has all entries on the main diagonal equal to zero, so in particular, the trace vanishes.

Title: Orthogonal with respect to killing form


问:
If $\kappa_i, i\in I,$ are distinct cardinal numbers, then $|I|\leq\sup\{\kappa_i\mid i\in I\}$
It's used in the book I'm reading without even being mentioned explicitly. It seems intuitive, but I cannot prove it.</p>

答:
Presumably $I$ or at least some $\kappa_i$ are infinite.

Let $\lambda=\sup\{\kappa_i\mid i\in I\}$, then $\lambda$ is an infinite ordinal. So $I\mapsto\lambda+1$, the ordinal, by mapping $i\mapsto\kappa_i$. Since for an infinite ordinal, $|\lambda+1|=|\lambda|$, we are done.

If you want to be a bit more specific about it, you can note that since the $\kappa_i$ are cardinals, if the above happens, then there is some "gaps" between them (i.e. not all the cardinals are finite, or $\{\kappa_i\mid i\in I\}=\omega+1$ in which case we can manually move everything around) so we can put the maximum cardinal, if it exists, into one of these gaps.

Finally, note that if $I=\{0\}$, and $\kappa_i=0$, then $\sup\{\kappa_i\mid i\in I\}=0$, but $|I|=1$. So some assumption of infinitude is necessary here.


问:
Monty Hall Problem - what is the probability space for the problem, and a question on $\sigma$-algebras generated by events?
So we have the following problem where we have 3 doors, 1 has a car behind it and 2 don't:

i) We first choose door 1

ii) Monty opens one of the other doors which he knows for certain doesn't have a car behind it

iii) We wither keep door 1 or choose the remaining unopened door

So in my mind the sample space is $\Omega=\{\{1,2,1\},\{1,2,3\},\{1,3,1\},\{1,3,2\}\}$ (1st element is our pick, 2nd is Monty's pick, 3rd is our final pick). Is this correct?

And so $\mathcal{F}=\mathcal{P}(\Omega)$.

I'm now unsure exactly how we would define $\mathbb{P}$? Of course, the probability a  car is behind a given door is 1/3, and if door 1 does have the car behind it then Monty can pick either door 2 or 3 with probability 1/2 each. Or if door 1 doesn't have a car behind it then Monty picks either door 2 or 3 with certainty. Is there a neater way in which to define this probability measure?

Now my final issue is how we would define:

(a) The event that door 2 has a no car behind it (call A)

(b) The $\sigma$-algebra generated by A, $\sigma(A)$

For (a) we know that the outcomes $\{1,2,1\},\{1,2,3\}$ both indicate that door 2 has no car behind it. So is event A just the event we pick either one of these outcomes? (however door 1 or 3 may still not have a car behind it, so am I missing outcomes here? If so, this may help me determine $\sigma(A)$?)

答:
Really interesting question! Tricky and fiddly stuff indeed! Hopefully I can help!

How about alternatively we define our sample space, $\Omega := \{D_{1,2}, D_{1,3}, D_{2,3} , D_{3,2} \} $ Where each $D_{i,j} $ is the event that the car is behind Door i and monty opens j. Notice we are always guessing door 1. You should not include the third argument, our choice of final selection, as this is non random.

One could define $P(D_{i,j})$ =
\begin{array}{ll}
      \frac{1}{6} & i =1 \\
      \frac{1}{3} & i \not = 1\
\end{array}

Onto your second part: The event A that there is no car behind door 2. We have 2 ways to write this. Firstly: A = $\Omega \setminus D_{2,3} $ = $\Omega \cap D_{2,3}^c$.

Alternatively:  A = $D_{1,2} \cup  D_{1,3} \cup D_{3,2} $

Now onto the sigma algebra $\sigma(A)$

Firstly: $\Omega , \emptyset \in \sigma(A) $ By definition of a sigma algebra. A is in our sigma algebra obviously, so we have $\Omega, \emptyset $ and A = $D_{1,2} \cup  D_{1,3} \cup D_{3,2} $  are in $\sigma(A)$. Now we can take any interesection, compliment or union to find more sets in our sigma algebra. Infact the only one left is A$^c$. So $\sigma(A) = \{\emptyset , \Omega , A , A^c \}$

Hope this helped!


问:
What exactly is $\mathbb{Z_n}$?
My impression has been that $\mathbb{Z_n}$ is the set $\{0,1,...,n-1\}$ under binary operation addition modulo $n$. However I'm also coming across this notion that $\mathbb{Z_n}$ is actually a set of equivalence classes of equivalence relation $x\sim y \iff x \equiv y$ mod $n$ and the addition here is actually addition of equivalence classes rather than simply addition of integers modulo $n$. Is this correct? So would it be correct to say $\mathbb{Z_n} = \{n,n+1,...,2n-1\}$? if we are considering these elements as being equivalence classes?

答: To my knowledge, possibly the main plus of the definition of $\Bbb Z_n$ as the quotient set (group under addition modulo $n$, actually) $\Bbb Z/\sim$, where $x\sim y \stackrel{(def.)}{\iff} x-y \in n\Bbb Z$, is that it leads to the generalization to any group $G$ (in place of $\Bbb Z$) and any subgroup $H$ (in place of $n\Bbb Z$), and finally to the powerful notion of coset.

Title: What exactly is $\mathbb{Z_n}$?


问:
Calculating the Surface Area of an Egg Using 29 Sample Points on Outline
I am calculating the surface area of an egg. To do so, I have taken a picture of the egg and have plotted 29 points on the outline of the egg based on the image. The egg was calibrated to a scale of 1cm=1 unit, and a axis was placed on the middle of the egg, with (0,0) being the centre of the egg. With my 29 points plotted on the half portion of the egg above the x axis, how do I use rules such as the Riemann Sum or the Trapezium Rule to calculate the surface area of the egg? Each of my points are 0.25units apart on the x axis, and have varying y values based on the height of the egg. My teacher told me that it is possible to find the surface area of the egg using the Riemann Sum, but based on my research, the Riemann Sum approximates the area underneath the curve. I do not understand how this would be helpful for me in terms of calculating the surface area, because by revolving this area under the curve around the y axis, I would only be getting the volume of the egg which is not the aim of my investigation.</p>

Body:
Suppose your data are $(x_k,y_k)$ with $k=1,\ldots,n$

The surface of the solid of revolution around $x$ axis generated by a curve $f(x)$ in the interval $[a,b]$ is
$$A_x=2\pi \int _{a}^{b}f(x){\sqrt {1+{\big (}f'(x){\big )}^{2}}}\,dx$$
With $n$ discrete data we can get the surface area of revolution around $x$-axis with the formula
$$A_x=2\pi\sum_{k=1}^n \left[y_k\sqrt{1+\left(\Delta_k y/\Delta_k x\right)^2}\right]$$
If you need more help, send the actual data

Suppose you have a set of data $Y=(y_k),1\le k \le 21$ and $X=(x_k)=k$
$$Y=\{0,4.359,6.000,7.141,8.000,8.660,9.165,9.539,9.798,9.950,10.00,9.950,9.798,9.539,9.165,8.660,8.000,7.141,6.000,4.359$$
Differences are
$$\Delta y_k=\{4.359,1.64,1.14,0.86,0.66,0.50,0.37,0.26,0.15,0.05,-0.05,-0.15,-0.26,-0.37,-0.50,-0.66,-0.86,-1.14,-1.64,-4.359\}$$
and in this case, as $\Delta x_k =1$ for all $k$ we have
$$\left(\frac{\Delta y_k}{\Delta x_k}\right)^2=\{19.00,2.69,1.30,0.74,0.44,0.25,0.14,0.07,0.023,0.003,0.003,0.023,0.07,0.14,0.25,0.44,0.74,1.30,2.69,19.00\}$$
add $1$ and take square root
$$\sqrt{1+\left(\frac{\Delta y_k}{\Delta x_k}\right)^2}=\{4.472,1.922,1.518,1.318,1.198,1.120,1.068,1.033,1.011,1.0013,1.0013,1.011,1.033,1.068,1.120,1.198,1.318,1.518,1.922,4.472\}$$
Now take $Y'=(y_k),1\le k \le 21$ and $X=(x_k)=k$
$$Y'=\{4.359,6.000,7.141,8.000,8.660,9.165,9.539,9.798,9.950,10.00,9.950,9.798,9.539,9.165,8.660,8.000,7.141,6.000,4.359\}$$
which is $Y$ less the first component. This is necessary because differences are $20$, while data are $21$

Now perform the dot product $$Y\cdot \sqrt{1+\left(\frac{\Delta y_k}{\Delta x_k}\right)^2}=199$$

$S=2\pi\cdot 199=1251$ is the approximated surface.

The data were not random. They were the coordinates of $21$ points on a semicircle and the actual surface is the surface of a sphere with radius $10$ that is $S=4\pi\cdot 10^2\approx 1257$.


问:
Is the direction of inequality preserved when multiplying the linear matrix inequality (LMI) by a nonsingular square matrix left and right?
Let $Y$ be a square and nonsingular matrix (invertible).

Given $X \prec 0$, if we multiply $Y^*$ on the left and $Y$ on the right to the $X$ (assuming the dimension matches),

I have a claim that

$$
X \prec 0
\iff
Y^*XY \prec 0
$$

Question:
I can get an intuitive understanding with a scalar analogy but I am not sure how I can prove this.

Attempt:

I first tried transforming this into the quadratic form (making it scalar).

For instance, multiply them by the vectors $z^*$ and $z$, left and right, respectively:

$$ X \prec 0 \implies \forall z  \neq 0, \quad z^*Xz < 0. $$
By letting $z = Yw$ for non zero $w$ ($z^{*} = w^{*}Y^{*}$),
$$ \implies \forall w  \neq 0, \quad w^*Y^*XYw < 0 \implies Y^*XY \prec 0.$$

For the opposite direction, we have:

$$ \forall w  \neq 0, \quad w^*Y^*XYw < 0, $$

Let $z = Yw$ and since $Y^{-1}$ exsits by the assumption, we have $w = Y^{-1}z$.

By substituting $w$ into the above inequality:

$$ \forall z  \neq 0, \quad z^*Xz < 0 \implies X \prec 0.$$

Is my proof correct?

Some side question I have:

Is the assumption of $Y$ being square and nonsingular essential? I obviously used it in my attempt above. If I don't have it, is it possible still to get the same result?

答:
Your proof is right. For the side question, you need $Y$ to be square and invertible to show that the relation is iff.

Example for $\implies$: let $X=-1$ and $Y=[1 \ 1]$, then you have $X<0$, but $$Y^\top X Y=\begin{bmatrix} -1 & -1 \\ -1 & -1\end{bmatrix}\leq 0$$

Example for $\Longleftarrow$: let $X=\begin{bmatrix} -1 & -1 \\ -1 & -1\end{bmatrix}$ and $Y=\begin{bmatrix} 1\\ 1 \end{bmatrix}$, then $Y^\top X Y=-4<0$, but $X\leq 0$.


问:
Solve $y'+a(x)y=b(x)$ where $b(x)$ is not continuous

Find all the solutionsof the equation:
$$y'+ay=b(x),\ 0<x<\infty,\ $$
where $a$ is a constant and $b(x)=1$ for $0\le x\le \alpha$, and $b(x)=0$ for $x\gt \alpha$ and $\alpha$ here is a positive constant.

Well, I really got stuck on this problems where the right hand side of  the differential equation $b(x)$ is not continuous. Could anyone give me some hints on this, please?

答: Use the method of integrating factors on the corresponding intervals of $b(x)$.


问:
Prove that the Archimedes property doesn't occur in the Polynomials
I found this question and I can't solve it.

So by Definition of Archimedes property - Archimedes property doesn't occur if The $\mathbb{N}$ set is bounded.

Which means - this is what we need to prove - that $\mathbb{N}$ is bounded over Polynomials.

My try :

Prove by contradiction:

Lets assume that $\mathbb{N}$ is not bounded over $P(x)$ which means exist $n∈\mathbb{N}$:

such that :
$∀P(x) : P(x)<n$

$n∈\mathbb{N}$ and $\mathbb{N}⊆\mathbb{R}$ $\Rightarrow$ $n∈\mathbb{R}$.

Lets take the polynomial $P(x)=x+1$ and do $P(n)=n+1$ and by assumption $n+1<n$
which is a contradiction.

Which means $\Rightarrow$ $\mathbb{N}$ is bounded above $P(x)$ $\Rightarrow$ The polynomials doesn't "have" the archimedes axiom.

is it ok to assume that ? or I did a mistake somewhere?

答:
That problem only makes sense if we define an order relation on the polynomials. The usual one is this: $p(x)<q(x)$ if one of these conditions holds:

$\deg p(x)<\deg q(x)$

$\deg p(x)=\deg q(x)$ and, if $n$ is that common degree, then the coefficient of $x^n$ in $p(x)$ is smaller than the coefficient of $x^n$ in $q(x)$.

With this definition, it is clear that $(\forall n\in\Bbb N):n<x$. So, indeed, the Archimedes property doesn't hold in this context.


问:
Trying to understand a double summation formula
I am trying to understand a double summation mentioned in this paper: https://doi.org/10.1016/0024-3795(95)00696-6. How can we prove that this is true?

$$\sum_{l = 1}^n\;\; \sum_{k = l}^{n+l} = \sum_{k \,=\, 1}^n \;\; \sum_{l \,=\, 1}^{k} \,+\, \sum_{k \,=\, n+1}^{2n}\;\; \sum_{l \,=\, k-n}^{n}$$

Body:
$$
\begin{array}{c} n=3 \\[8pt]
\begin{array}{|c|c|ccccccc|c|}
% {}_{\large\ell} \, \diagdown \, {}^{\large k}
\hline
& k & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
\ell & (\ell,k)\\
\hline
1 & & (1,1) & (1,2) & (1,3) & (1,4) &   &   &   & \sum_{k=1}^{n+1} \\
2 & &   & (2,2) & (2,3) & (2,4) & (2,5) &   &   & \sum_{k=2}^{n+2} \\
3 & &   &   & (3,3) & (3,4) & (3,5) & (3,6) &   & \sum_{k=3}^{n+3} \\
\hline
  & & \sum_{\ell=1}^1 & \sum_{\ell=1}^2 & \sum_{\ell=1}^3 & \sum_{\ell=4-3}^3 & \sum_{\ell=4-2}^3 & \sum_{\ell=4-1}^3 \\
\hline
\end{array}
\end{array}
$$


问:
Homomorphism from $\mathbb{Z}/n\mathbb{Z}$
Does there exist a non-zero homomorphism from $\mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}$ ? If yes, state the mapping. How is this map exactly? </p>

答: Hint: Every element of $\Bbb{Z/nZ}$ has finite order.


问:
Does E[|X|] = |E[X]| always hold?
Let $X$ be some random variable, does the equality $$E\left[ \left| X \right| \right] = \left| E[X] \right|$$
always holds? Here $X$ can be real and can be complex.

答:
That is not true. Consider a random variable $X$ with $P(X=1)=1/2=P(X=-1)$.
We have $|E[X]|=0$ and $E[|X|]=1$.

Title: Does E[|X|] = |E[X]| always hold?


答: If $X$ is a nonzero random variable with $E[X]=0$, then $|E[X]| = 0$, but $E[|X|] > 0$.

Title: Does E[|X|] = |E[X]| always hold?

Body:
Let $X$ be some random variable, does the equality $$E\left[ \left| X \right| \right] = \left| E[X] \right|$$
always holds? Here $X$ can be real and can be complex.


问:
Prove the following map is an isometry.
I have a map $$T :C[0,1] \to C[a,b]$$ such that $$T(f)=f \circ s$$ where $s=a+t(b-a)$ is a homeomorphism from $[0,1] \to [a,b]$.
($C[0,1]$ and $C[a,b]$ are the spaces of continuous functions on $[0,1]$ and $[a,b]$ respectively).
Prove $$T(f)=f \circ s$$ is an isometry on the metric $$d(f,g)=\sup_{t\in[a,b]}\bigl|f(t)-g(t)\bigr|.$$
So for $f,g \in C[0,1]$ I have $$d_{C[0,1]}(T(f),T(g))=d_{C[0,1]}(f \circ s,g \circ s)$$
Now I can't proceed from there, I was thinking that the fact that $s$ is a homeomorphism gives $f \circ s=f$ since the domain and range of $s$ are similar enough but I am not sure if this is true.
Can you help me proceed?

答: If the distance in $C[a,b]$ is $d(f,g)=\sup_{t\in[a,b]}\bigl|f(t)-g(t)\bigr|$, then that's just because$$\sup_{t\in[0,1]}\bigl|f(s(t))-g(s(t))\bigr|=\sup_{t\in[a,b]}\bigl|f(t)-g(t)\bigr|.$$This follows from the fact that $s$ is a bijection.

Title: Prove the following map is an isometry.


问:
Counting with cards
Suppose you draw five cards from a deck of 52 cards. What is the probability of getting three cards (but not four cards) of one kind (all of their suits are same)?

Basically, it seems intuitively to me that $\frac{C(13,3)C(4,1)C(48,1)}{C(52,5)}$ is the correct answer, since we pick arbitrarily $5$ from $52$ cards, 3 cards of same kind from each $13$ possibilities, multiplying it with $4$ and choosing remaining card.

For the sake of completeness, 2 2 2 K 6 and 2 2 2 K K are each three cards but 2 2 2 2 K is not.

答:
"Three of a kind" in card problems usually means three of the same value (Ace, King, Queen, etc.) rather than three of the same suit.  The OP, however, has parenthetically explained that they mean three of the same suit. Given that interpretation, $C(13,3)C(4,1)$ does describe the number of ways of choosing those three cards. But it remains to choose two more cards, not just one, and they must come from one of the other three suits (since the OP explains parenthetically that there are to be exactly three, not four), so the final multiplicative factor should be $C(39,2)$, not $C(48,1)$. The probability is thus

$$C(13,3)C(4,1)C(39,2)\over C(52,5)$$

Just for completeness, if "three of kind" is given its usual meaning, the numerator for the probability is $C(13,1)C(4,3)C(48,2)$. I.e., choose the value for the three of a kind, then choose three of the four cards of that value, and finally choose two cards from the remaining cards of the other values.

Title: Counting with cards


问:
Is there a way to plot separate functions in 3D using Wolfram Alpha?
Is there a way to plot separate functions in 3D using Wolfram Alpha? Am I doing something wrong?

Similar problem with GeoGebra

答: Just to give an alternative via CalcPlot3D:


问:
Comparison convergence test $\sum_{n=1}^\infty (2n+1)\beta^{-\sqrt{2n+1}}$
Wolfram Alpha tells me that the series $$\sum_{n=1}^\infty (2n+1)\beta^{-\sqrt{2n+1}}$$ converges by comparison test for $\beta > 1$, but I'm struggling to find a function to compare it with. Any ideas?

Edit 1 (Motivation): I got to this series because in the paper I link below there is a proposition (2.7) which uses the Dominated Convergence Theorem in the proof, but they don't specify the dominating function. I am using that proposition in my final degree Thesis, so I have to formally proove it, and this is the natural dominance function. This part of my thesis is about the Hermite-Fourier expansions and the heat and Hermite semigroups.

https://www.sciencedirect.com/science/article/pii/S0022123603000831

Edit 2 (Current progress): This actually converges by the integral test, but I would like to find a solution using convergence test.

答:
$$\begin{align}
&  \sum_{n=1}^\infty (2n+1)\beta^{-\sqrt{2n+1}} \\\\
& = \sum_{3\le m < \infty,\\\text{m odd}} m\beta^{-\sqrt{m}}\\\\
& < \sum_{m=1}^\infty m\beta^{-\sqrt{m}}\\\\
& = \sum_{n=1}^\infty n\beta^{-\sqrt{n}}\end{align}$$

So proving the last series converges is sufficient.

We can use the limit comparison test with the series

$$\sum_{n=1}^{\infty}\frac{1}{n^2}$$

So we need

$$\lim_{n \to \infty} \frac{n\beta^{-\sqrt{n}}}{\frac{1}{n^2}}$$

$$=\lim_{n \to \infty} \frac{n^3}{\beta^{\sqrt{n}}}$$

For convenience, switch to functional form

$$=\lim_{x \to \infty}\frac{x^3}{\beta^\sqrt{x}}$$

Then we can do a variable substitution $y=\sqrt{x}$

$$=\lim_{y \to \infty}\frac{y^6}{\beta^y}$$

Repeated use of l'Hopital's rule will reduce the degree of the numerator, leaving the denominator the same except for multiplying by $\ln(\beta)$. So the limit goes to zero, because $\beta>1$.

So since the limit is zero and the bottom series converges, the top series also converges.


问:
finding closed formula for the recurrence relations with non-constant coefficient
We know writing closed formula for recursive relations.For example;

If $a_n=7a_{n-2}+6a_{n-3} $ with $a_0=9,a_1=10,a_2=32 $ , then closed formula will be equal to

$a_n=8(-1)^{n}+4(3)^{n}+(-3)(-2)^{n}$.  ( I did not need to write all process here.)

My question is that what would be happen if the coefficients were variables such as $(n-1) ,(n) $ instead of $6,7$.

Is there any procedure to find the closed formula of recursive relations with non-constant coefficient?

For example ; if the recursion were in the form of $a_n=(n-1)a_{n-2}+na_{n-3} $ with $a_0=9,a_1=10,a_2=32 $ , what would be the closed formula ?

答:
I'm not sure what process you used, as different processes generalize differently to this case. I'm personally partial to generating functions, so I'll discuss that method.

The key observation is that coefficients $(n-1)a_{n-2}$ look a lot like a derivative. After all, if we let

$$A = \sum a_n x^n$$

Then $$\frac{d}{dx} A = \sum n a_n x^{n-1}$$.

In the example you posed, $a_n = (n-1)a_{n-2} + na_{n-3}$, we could multiply everything in sight by $x^n$ and sum to get

$$A = \sum a_n x^n = \sum \left ( (n-1) a_{n-2} x^n + n a_{n-3} x^n \right )$$

We split up the right hand side as

$$x^2 \sum (n-1) a_{n-2} x^{n-2} + x^3 \sum n a_{n-3} x^{n-3}$$

Now by using derivative rules, you can simplify this to get a differential equation for $A$. Then by solving the differential equation, you can get a closed form for $A$. All of this can be automated in a computer algebra system like sage, though you can do it by hand with some persistence.

You can read more about this technique in Wilf's fantastic book generatingfunctionology.

I hope this helps ^_^


问:
How to prove that $|u_{n+1} - \alpha | \le \frac{1}{2} | u_n - \alpha |$
I'm preparing for a high school math exam and I came across this question in an old exam.

Let $f(x) = \dfrac{1}{2(1+x^3)}$.

$\alpha \in (0, \frac{1}{2})$ is the only real number such that $f(\alpha) = \alpha$.

$(u_n)$ is the series such that
$$
\begin{cases}
u_0 = 0 \\
u_{n+1} = f(u_n) \quad \forall n \in \mathbb{N}
\end{cases}
$$
Prove that $|u_{n+1} - \alpha| \le \frac{1}{2} |u_n - \alpha|$

The end goal is to prove that $(u_n)$ converges to $\alpha$. But we have to do it this way instead of finding that both $(u_{2n})$ and $(u_{2n+1})$ converge to $\alpha$.

It's supposed to be a very easy question. But I have been trying for the last two hours and I couldn't find it.

It's easy to prove that $u_n \in [0, \frac{1}{2}]$ by induction.

Here's what I tried:

I tried separating the question into two cases $u_n \le \alpha$ and $u_n \ge \alpha$.
I got that I just need to prove that $u_{n+1} + \frac{1}{2} u_n \ge \frac{3}{2} \alpha$. But I didn't know how to go from here. Substituting $u_{n+1}$ with $\dfrac{1}{2(1+u_n^3)}$ seems to only make the problem more complicated.

I tried squaring both sides. It only made the expression more complicated.

The problem vaguely reminds me of the epsilon-delta definition of $\lim_{x \to \alpha} f(x) = \alpha$. But that seems like it won't lead anywhere.

答:
As mentioned by others, the key is to show that

Prove that for $x,y \in \left(0,\frac12\right)$ we have
$$|f(x)-f(y)| \le \frac12 |x-y|.$$
and then apply this to $x = u_n$ and $y = \alpha$.

We can do so directly, without the need of calculus / MVT.

The LHS is

$$ | \frac{1}{2(1+x^3 ) } - \frac{1}{2 ( 1 + y^3) } | = \frac{1}{2} |(x-y)| \times | \frac{ x^2 + xy + y^2  } { (1+x^3 ) ( 1+y^3)}|.$$

It remains to show that the last term is $\leq 1$, which can be done in several ways like

$$x^2 + xy + y^2 \leq (x+y)^2 \leq 1 \leq ( 1 + x^3 ) ( 1+y^3 ). $$


问:
What is the probability of $B^2\ge 4AC$ if A, B and C are independent exponential random variables with parameter 1?
I tried the following integral, is it correct?

$$\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\frac{b^2}{4a}}e^{-b}e^{-a}e^{-c}\,dc\,da\,db$$

答: An easier approach (but please double-check my calculations): fix $A,\,C$ so$$P(B\ge2\sqrt{AC})=\exp(-2\sqrt{AC}).$$With $a=s\cos^4\theta,\,b=s\sin^4\theta$, averaging gives$$\begin{align}\int_{[0,\,\infty)^2}\exp(-(\sqrt{a}+\sqrt{c})^2)dadc&=4\int_0^{\pi/2}\cos^3\theta\sin^3\theta d\theta\int_0^\infty s\exp(-s)ds\\&=4\frac{1}{12}1\\&=\frac13\end{align}$$by Beta and Gamma functions. Python agrees with me.


答:
Suppose there is a homomorphism $f: \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}$. What could $f(1)$ be? Let's call it $a$. Then $f(2) = f(1+1) = f(1)+f(1) = a+a = 2a$, and likewise $f(3) = 3a$, $\dots,$ $f(\underbrace{1+\cdots+1}_{n\text{ times}}) = na$. But $\underbrace{1+\cdots+1}_{n\text{ times}} = 0$, and $f(0) = 0$. So what can $a$ be?

Once you've worked out which value(s) of $a$ is/are allowed, can you check that the associated map $f$ is a homomorphism? (We've already done this, really, but make sure you're sure of that.)

Title: Homomorphism from $\mathbb{Z}/n\mathbb{Z}$

Body: Does there exist a non-zero homomorphism from $\mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}$ ? If yes, state the mapping. How is this map exactly? </p>


问:
Why is the Absolute value / modulus function used?
Why is the absolute value function or modulus function $|x|$ used ? What are its uses?

For example the square of a modulus number will always be positive, but why is it used when for example the square of any number whether positive or negative is always positive ? For example,  $X^2$, will give a positive number whether negative or positive where $X$ is any number positive or negative.

答:
In computer science, it can be useful for computing vectors.

Let x be a projection of a vector into 1 dimension such that it has a direction (plus or minus), and a magnitude.

If I want to square the vector but preserve its direction (say, to find air resistance from velocity), then I can take:
$$ a = x^2 \frac{|x|}{x} $$
or simplified:
$$ x|x| $$
This gives me the same direction, but the magnitude is squared. Therefore, the vector I am calculating acts in the correct direction.


问:
Numerically solving an ODE with intermediate condition on MATLAB
I am trying to solve the following ODE numerically-

$\frac{dn}{dt} = \frac{1-n}{T_A} - \frac{|u(t)|^2}{\omega_a}n$

with the condition

$n(-T_R/2) = 0$

$T_R, T_A$ and $\omega_0$ are constants

I have usually used a standard template to solve such DEs on MATLAB when the condition defines the value of the function at zero. How do I solve this particular equation? I would also love to know if this equation can be solved analytically, and if so, how?

答:
You may introduce another integration variable, $s$ which is related to $t$ as
$s = t + \frac{T_R}{2}$.

Your problem changes to
$$
\frac{dn}{ds} = \frac{1-n(s)}{T_A} - \frac{|u(s - T_R / 2)|^2}{\omega_A} n(s)\\
n(s=0) = n\left(t = -\frac{T_R}{2}\right) = 0.
$$
which is the standard form for integrating on $s \in \left[0, T_0 + \frac{T_R}{2}\right]$ interval.

If your ODE solver cannot integrate backwards in time (for the interval $s \in \left[-T_0 + \frac{T_R}{2}, 0\right]$ you may use the same trick. Introduce another integration variable $r = -s = -t - \frac{T_R}{2}$. Then
$$
\frac{dn}{dr} = - \frac{dn}{dt} = -\frac{1 - n(r)}{T_A} + \frac{|u(r - T_R / 2)|^2}{\omega_A} n(r)\\
n(r = 0) = n\left(t = -\frac{T_R}{2}\right) = 0.
$$
For this case the integration domain will be $r \in \left[0, T_0 - \frac{T_R}{2}\right]$


问:
Determine the gcd of f(x) and f'(x)
Let $n$ be a natural number, and set $f(x)=1+\frac{x}{1!}+\frac{x^2}{2!}+...+\frac{x^n}{n!}$ in $\mathbb{Q}[x]$. Determine the gcd of $f(x)$ and $f'(x)$. Then, determine for which natural numbers $n$ the polynomal $f(x)$ has multiple roots.

I've calculated $f'(x)=1+\frac{x}{1!}+\frac{x^2}{2!}+...+\frac{x^{n-1}}{(n-1)!}$. I then tried to do polynomial long division by examining just a few of the highest degree terms and a few of the lowest degree terms, but that didn't seem to be getting me anywhere.

答:
Here $f(x)$ and $f'(x)$ has no common roots:

If $f(u)=f'(u)=0$, then we must have $u^n/n!=0$; that is, $u=0$. Nonetheless, $f(u)=f'(u)=1\neq 0$, a contradiction.

Therefore, $\gcd(f(x),f'(x))=1$.

Note that if the gcd of $f$ and $f'$ has degree higher than $1$, then $f(x)$ and $f'(x)$ must have at least one common root in $\mathbb{C}$.

Title: Determine the gcd of f(x) and f'(x)


问:
Showing that $ \sum \limits_{m=1}^{n} b_m x_{m-n}~\to~ ab$ as $n~\to~\infty$
If $x_n ~\to ~a$ as $n~ \to~ \infty$

Does:

$ \sum \limits_{m=1}^{n} b_m x_{n-m}~\to~ ab$ as $n~\to~\infty$?

$b_m ~\geq~0$ and $ b~\equiv~ \sum \limits_{m=1}^{\infty} b_m < \infty$

My attempt:

$\lim\limits_{n\to\infty}x_n~=~a$

I would like to have something like this I think

$$\inf x_n \sum_{m=1}^n b_m~\leq~\sum b_m x_{n-m}~\leq ~\sup x_n \sum_{m=1}^n b_m$$

Then taking the limits will squeeze it in to $ab$, I am having trouble getting this form, is there any easier way to do this?

i.e. $$\sum_{m=1}^n b_m\inf x_{n-m} ~\leq~\sum b_m x_{n-m}~\leq ~\sum_{m=1}^n b_m\sup x_{n-m} $$

Then $\sup x_{n-m} \leq \sup x_n$ and $\inf x_{n-m} \geq \inf x_n$

Is $b_m \geq 0$ necessary?

答: You can rewrite your sum as $$\sum_{m=0}^{n-1} b_{n-m} x_m$$ Define $$a_{n,m}=\frac{b_{n-m}}{\sum_{k=1}^n b_k}, m=0,\cdots,n$$ and $a_{n,m}=0$ for $m>n$. Then $\sum_{m=0}^\infty a_{n,m}=1$ for all $n$ and $\lim_{n\to\infty}a_{n,m}=0/b=0$. By Silverman-Toeplitz Theorem $$\lim_{n\to\infty}\sum_{m=0}^\infty a_{n,m}x_m=\lim_{n\to\infty}\frac{\sum_{m=0}^{n-1} b_{n-m} x_m}{\sum_{k=1}^n b_k}=a$$ Since $\lim_{n\to\infty}\sum_{k=1}^n b_k=b$, we get $$\lim_{n\to\infty}\sum_{m=0}^{n-1} b_{n-m} x_m=ab$$</p>


问:
Transitive Mutual Information Bernoulli Random Variables in a fork structure
Suppose we have Bernoulli random variables $X$, $Y$, and $Z$, whose dependency structure is $Y \leftarrow X \rightarrow Z$  (i.e. $Y \perp Z \mid X)$. If we know $\mathrm{I}(X:Y)$ and $\mathrm{I}(X:Z)$,
can we get an exact value or a lower bound (or a tighter upper bound) on  $\mathrm{I}(Y:Z)$?

From the data processing inequality, we know that

$\mathrm{I}(Y:Z) \leq min(\mathrm{I}(X:Z), \mathrm{I}(X:Y))$

Can we exploit the fact that all the variables can only carry up to 1 bit of information to make a stronger statement about  $\mathrm{I}(Y:Z)$?

答:
Notice the elementary equality $$I(Y;Z) = I(Y;X,Z) - I(Y;X|Z) = I(Y;X) - I(Y;X|Z),$$ where we've used the chain rule first, and then used the conditional indepdence to argue $I(Y;Z|X) = 0.$ In general, $I(Y;Z)$ can, of course, be as small as $0$, since the conditional independence allows for actual independence. This is irrespective of how large $I(X;Y)$ and $I(X;Z)$ are because we can take $X = (U,V)$ for independent $U,V,$ and $Y = U, Z = V$.

We can use the about to write down a symmetric expression, which also sheds clearer light on when $I(Y;Z)$ is positive.

Running the same expansion but with $Y$ and $Z$ interchanged, and then averaging tells us that

$$ I(Y;Z) = \frac{ I(X;Y) + I(X;Z) - I(X;Y|Z) - I(X;Z|Y)}2 \\ = H(X) + H(X|Y,Z) - H(X|Z) - H(X|Y),\\ = I(X;Y) + I(X;Z) + H(X|Y,Z) - H(X)\\ = I(X;Y) + I(X;Z) - I(X;Y,Z).$$ where in the first line we have used that $I(X;Y|Z) = H(X|Z) - H(X|YZ)$ and $I(X;Z) = H(X) - H(X|Z),$ and similarly for the other terms.

This tells us that $I(Y;Z)$ is positive iff $I(Y;X) + I(Z;X)$ exceeds $I(X;Y,Z)$. This should make sense - loosely, the only way $Y$ can tell us about $Z$ is if there is some 'common' information about $X$ between the two. This means that the sum of the marginal informations must exceed the information that they have jointly.

From the above, we can also work out the simpler lower bound $$I(Y;Z) \ge \max(0, I(X;Y) + I(X;Z) - H(X))$$ by bounding $I(X;Y,Z) \le H(X)$. The advantage of this is that it is only a function of the marginal laws $P_{XY}, P_{XZ}$ (like the upper bound in the question). Notice that this is tight, since we can take $X = (U,V,W)$ for independent $U,V,W$ and $Y = (U,W), Z = (V,W)$ to together attain any possible triple of $(H(X), I(X;Y), I(X;Z)),$ and in this setting $$I(Y;Z) = H(W) = (H(W) + H(U)) + (H(W) + H(V)) - (H(U) + H(V) + H(W)) \\ = I(X;Y) + I(X;Z) - H(X).$$

Note - the answer is incomplete, because the tightness argument it presents needs at least $X$ to be non-binary. However, note that the argument does work as long as the alphabet of $X$ has at least $8$ letters, and of $Y,Z$ at least $4$, so it isn't very far off. It's certainly possible that more can be said in the binary setting.


问:
Proof that $x^2$ is not uniformly continuous on $\mathbb R$
This is my attempt. My instructor said it was incorrect. I would like to know exactly where my mistake is.

Suppose $x^2$ is uniformly continuous on $\mathbb R$.

So, $\forall \epsilon > 0 \ \exists \delta > 0 \ \forall x,y \in \mathbb R \ s.t. \ if \ |x - y| < \delta \ \text{then} \ |f(x) - f(y)| < \epsilon$

In this case, $|x - y| < \delta \ \text{and} \ |x^2 - y^2| < \epsilon$

$|(x - y)(x + y)| < \epsilon$

$|(x - y)| |(x + y)| < \epsilon$

$\delta|x + y| < \epsilon$

Notice, $|x + y|$ is unbounded on the real line. So $x$ and $y$ can be made arbitrarily large. So there will not always be a $\delta$ that will make all values of $x$ and $y$ less than an arbitrarily small $\epsilon$.

Therefore, $x^2$ is not uniformly continuous on $\mathbb R$.

答:
The intuition is not bad, but it is incorrect. From $|(x-y)(x+y)|<\varepsilon$ you cannot go to $\delta|x+y|\le\varepsilon$. Indeed, by applying $|x-y|<\delta$, you have $|(x-y)(x+y)|<\delta|x+y|$. So you have that one expression (say $|(x-y)(x+y)|$) is bounded by two different estimates ($\delta|x+y|$ and $\varepsilon$) but you cannot conclude anything about wich one is bigger.

A correct approach is trying to find a concrete value of $\varepsilon$ for wich for any $\delta, you have the oppose inequality.

For example, take $\varepsilon=1$ and try to prove that for any $\delta>0$ you always can find $x,y\in\Bbb R$ such that $|x-y|<\delta$ and $|x^2-y^2|\ge 1$. And to do that, your intuition about the factor $(x+y)$ will helps you.


答:
Do it with sequences

Let
$$x_n=n \;\; and\;y_n=n+\frac 2n$$

we have
$$\lim_{n\to+\infty}|x_n-y_n|=0$$

But

$$\lim_{n\to+\infty}|x_n^2-y_n^2|=4\ne0$$

thus
$x\mapsto x^2 $ is not uniformly continuous at $\Bbb R$.

We proved that with $\epsilon=3 $ and any $ \eta>0 $, we take $ n $ such that $ \frac 1n<\eta $ and check that
$$|x_n-y_n|=\frac 1n <\eta$$
but
$$|x_n^2-y_n^2|=4+\frac{4}{n^2}>3$$

Title: Proof that $x^2$ is not uniformly continuous on $\mathbb R$

Body:
This is my attempt. My instructor said it was incorrect. I would like to know exactly where my mistake is.

Suppose $x^2$ is uniformly continuous on $\mathbb R$.

So, $\forall \epsilon > 0 \ \exists \delta > 0 \ \forall x,y \in \mathbb R \ s.t. \ if \ |x - y| < \delta \ \text{then} \ |f(x) - f(y)| < \epsilon$

In this case, $|x - y| < \delta \ \text{and} \ |x^2 - y^2| < \epsilon$

$|(x - y)(x + y)| < \epsilon$

$|(x - y)| |(x + y)| < \epsilon$

$\delta|x + y| < \epsilon$

Notice, $|x + y|$ is unbounded on the real line. So $x$ and $y$ can be made arbitrarily large. So there will not always be a $\delta$ that will make all values of $x$ and $y$ less than an arbitrarily small $\epsilon$.

Therefore, $x^2$ is not uniformly continuous on $\mathbb R$.


答:
There is no single mistake. Even if every step were correct, which is not the case, this doesn’t have the right logical structure to be a proof that the function $f(x)=x^2$ is not uniformly continuous. For that you need to show that there is some $\epsilon>0$ such that no matter what $\delta>0$ you choose, there are points $x_\delta,y_\delta\in\Bbb R$ such that $|x_\delta-y_\delta|<\delta$, but $|x_\delta^2-y_\delta^2|\ge\epsilon$. For example, you could let $\epsilon=1$ and for each $\delta>0$ show how to find points $x_\delta$ and $y_\delta$ such that $|x_\delta-y_\delta|<\delta$, but $|x_\delta^2-y_\delta^2|\ge 1$.

I mentioned that not every step is correct: in particular, $|x-y|\cdot|x+y|<\epsilon$ and $|x-y|<\delta$ do not imply that $\delta|x+y|<\epsilon$.

That said, I suspect that you do have a reasonable intuitive understanding of why $f(x)=x^2$ is not uniformly continuous on $\Bbb R$.