问:
Confusion of the “size n” used in the central limit theorem
First of all, let me re-state a version of the central limit theorem:-

If $X_1, X_2, …, X_n$ is a random sample of size $n$ from any infinite population with mean $\mu$ and variance $\sigma^2$, then , with $\bar{X}=\frac{1}{n}(X_1+X_2+...+X_n)$, we have $E[\bar{X}] = \mu$ and $Var [\bar{X}] = \dfrac {\sigma^2}{n}$

In which, I found only one “n”.

Now, consider a numeric example trying to verify the fact.

Let {2, 3, 6, 8, 11} be a population with size $ = 5 = n_0$, say. [This “$n_0$” will not cause any trouble in the following discussion.]

From that we get $\mu = … = 6$ and $\sigma^2 = ... = \dfrac {54}{5}$.

Next, we take all possible samples of size $2 = n_1$, say, from the population.

These samples are (2,2), (2, 3), …, (11, 11) with a total of $25 = n_2$, say.

From these order pairs, we get a random variable of means with $25 = n_2$ in total.

The corresponding probabilities are in the form of $\dfrac {j}{25}$ where j could be 1, 2, 3, or 4

Using these data, we get the mean of the means $ = \mu_{\bar X } = … = \dfrac {150}{25} = 6 = \mu $. [So far so good.]

Also, $\sigma_{\bar X}^2 = \dfrac {(2 – 6)^2 + …}{25} = … = \dfrac {27}{5}$, which can be interpreted as $\dfrac {1}{2}(\dfrac {54}{5})$ in order to agree with the formula $\sigma_{\bar X}^2 = \dfrac {\sigma^2}{n}$, if $n$ is $2$.

My questions is:-
In the theorem, there is only one “n”, but in the example, we have two (namely $n_1$, $n_2$, and not counting the $n_0$). Which is which?

According to my understanding about the theorem, $n$ is $n_2 = 25$.

However, in the very last step, the “$n$” used in the formula $\sigma_{\bar X}^2 = \dfrac {\sigma^2}{n}$ is $n_1 = 2$.

Added: Although this is a finite example, one can imagine that the population is large and after taking samples from it, we still have to distinguish the $n, n_1,n_2$ problem.

答: You have in your example, named the size of the population, the number of samples taken and the total possible number of samples. Only one of these three appears in the theorem (and only one makes sense to do calculations with infinite populations). That's the number of samples taken, your $n_1$.


问:
Proof P by premises $(\neg P\vee Q)\to P$
I have to proof next question by a formal proof that when the premise is equal to:
$(\neg P\vee Q)\to P$ has the result ,$P$, I first though that I must write:

$\neg P\vee Q\to P$, as:
$\neg(\neg P\vee Q)\vee P$ and work further from there.

The problem is that I only may use Reit ,introduction and Elimination. Can someone help me?

答:
Suuppose ( for refutation) that  P is false.

It means that the true premise  $(\neg P \lor Q) \rightarrow P$ has a false consequent.

The only way for a true conditional with a false consequent to be true is to have a false antecedent ( by the truth table of the " if ... then " operator). So $(\neg P \lor Q)$ has to be false.

By DeMorgan's , the negation of a disjunction amounts to a conjoint negation of its disjuncts. So $\neg ( \neg P \lor Q) \equiv ( \neg \neg   P \& ~Q )$

Appliy $\&$ Elim to get $\neg \neg   P$

The hypothesis $\neg P$ led you to the negation of this hypothesis, namely $\neg \neg   P$.

By Neg Intro Rule, you're entitled to conclude that the hypothesis was false, wich means that $\neg \neg   P$ is true, and ( by Double Negation ) that P is true.

ContentLicense: CC BY-SA 4.0


问:
Inequality about Euclid norm. $|y_1-x_1|+ \cdots +|y_n-x_n| \leqq \sqrt{n} ||y-x||.$
Does this inequality about Euclid norm hold?

\begin{align}
&|y_1-x_1|+ \cdots +|y_n-x_n|
\leqq \sqrt{n} ||y-x||. \\
&\text{ where }
||y-x||:=
\sqrt{(y_1-x_1)^2+ \cdots +(y_n-x_n)^2}
\end{align}

I substituted some concrete examples (when $n=3$),

$x=
\begin{pmatrix}
1 \\
0\\
2\\
\end{pmatrix},
y=
\begin{pmatrix}
4 \\
3\\
1\\
\end{pmatrix}$
or
$x=
\begin{pmatrix}
7 \\
0\\
1\\
\end{pmatrix},
y=
\begin{pmatrix}
12 \\
4\\
6\\
\end{pmatrix}$
and so on, and the inequality above held.

But I couldn't prove $|y_1-x_1|+ \cdots +|y_n-x_n|
\leqq \sqrt{n} ||y-x||.$

How can I prove this inequality?

答: Hint: Apply Cauchy-Schwarz inequality to $(|x_1-y_1|,|x_2-y_2|,\dots,|x_n-y_n|)$ and $(1,1,\dots,1)$.


问:
Counterexample to slight alteration of Theorem
One can quite easily prove that if $(A,<)$ is linearly ordered and for all $a\in A$,
$$|\{b\in A\mid b\leq a\}|<\aleph_\alpha,\quad (*)$$
then $|A|\leq \aleph_\alpha$.

The strict inequality in $(*)$ is essential for the proof. I'm wondering if there is an easy counter-example to this statement when $<$ has been replaced by $\leq$ in $(*)$?

If $A=\mathbb R, \mathbb Q,\mathbb N$, a counter-example cannot be found.

答: Let $A=\omega_1$ - the first uncountable ordinal. Then, for every $a\in A$, $a$ is countable, so the sets $\{b\in A\mid b\le a\}=\{a\}\cup a$ are countable as well, i.e. $|\{b\in A\mid b\le a\}|=\aleph_0$, however $|A|=\aleph_1$.


问:
Average path length of unconnected graphs
I'm reading graph theory and I found a concept named average path length. Basically, it is the average of all the shortest lengths between any two nodes of a graph. In case two nodes are not connected, the length between them is set to zero (i.e., $d(v_i, v_j) = 0$).

I find this definition not meaningful because a connected graph may have the average path length higher than an unconnected graph. Should the shortest distance between two unconnected nodes be set to a large value?

My questions are:

Is there another definition of the average path length, or is the above definition correct?

In determining whether a graph is a small world graph, should we take into account the average path length in case the graph is unconnected?

答:
As you say, that definition doesn't make much sense: if you want to be able to get easily between nodes then having a pair of nodes which are inaccessible from each other is a bad thing, not a good thing.

A more sensible (and I would have thought more usual) definition is that $d(v_i,v_j)=\infty$ if there is no path from $v_i$ to $v_j$. This means any disconnected graph has infinite average path length. When faced with a disconnected graph, it therefore makes more sense to view it as a collection of different connected graphs and look at the average path length of the components. After all, if there is no connection between two parts of a network, what makes you think it is all the same network to start with?

A reference for this definition is here. Finding out which definition is more prevalent is likely to be difficult, since many of the papers that look at average path length work with networks which are known to be connected (for example, preferential attachment networks), and so they probably won't bother to specify what happens with unconnected ones.


问:
Using an Infinite Universe in a Formal Langauge Structure
I'm working through Leary and Kristiansen's "a friendly introduction to mathematical logic". When I get to page 23, they have introduced the idea of a formal language, along with the concepts of atomic and composite terms, and similarly formulae and sentences, and induction on complexity. On page 23 they introduce $\mathcal L$-structures, and at this point I get confused.

They have the langauge of number theory, $\mathcal L_{NT}:=\{0,S,+,\cdot,<,E\}$,  and they introduce the associated $\mathcal L$-structure $\mathfrak{N} :=(\mathbb{N},0,S,+,\cdot,<,E).$ I have some philosophical difficulties with introducing $\mathbb{N}$ at this point.

My understanding was that we were going to go through building up to different mathematical formalism necessary to be able to construct $\mathbb{N}$. So why does it make sense to just introduce this set at this point as though it's something that we already have? We haven't yet assigned any meaning to the symbols in $\mathcal L_{NT}$, (ie. as yet no Peano axioms). My understanding is that it doesn't make sense to talk about an infinite set unless we have a definition for what $<$ means. I think I would be happy if we had put a finite set in here as the universe, but to me putting in $\mathbb{N}$ results in some kind of circular definitions.

Can anyone clear up my confusion on this? I'm not even really sure if my question is mathematical in nature or philosophical.

答:
My understanding was that we were going to go through building up to different mathematical formalism necessary to be able to construct $\mathbb N$.

It is not so; we assume the existence of a structure: a domain of mathematical objects with the relevant properties. The structure $\mathfrak{N}$ with domain $\mathbb N$ is the usual domain of natural numbers.

We set-up a formal language that we will use to "talk about" the mathematical object.

More specifically, we have a previous understanding of the relation "less than" between numbers, and we use it to interpret the binary relation symbol of the language: $<$.

See page 22:

Let us return to the language $\mathcal L_{\text{NT}}$ of number theory.
[...] We now want to discuss the possible mathematical structures in which we can interpret these symbols, and thus the formulas and sentences of $\mathcal L_{\text{NT}}$.

It is certainly the case that you know an interpretation for these symbols.
The point of this section is that there are many different possible
interpretations for these symbols, and we want to be able to specify which
of those interpretations we have in mind at any particular moment.

See page 23: "Def.1.6.1. Fix a language $\mathcal L$. An $\mathcal L$-structure $\mathfrak A$ is a nonempty set $A$, called the universe of $\mathfrak A$, together with..."

The semantical clauses of the definition assign meaning to symbols of the language using the domain of mathematical objects with their properties: this is an interpretation.

The interpretation assigns to the symbols some objects7 of the domain: constant symbols are mapped to designated elements of the domain, unary predicate symbols to subsets of the domain, binary predicate to relation on the domain, and so on.

Thus, we can re-phrase the paragraph preceding that definition as follows: "determining the structure under consideration, we decide how we wish to interpret the symbols of the language; in this way, we have a way of talking about the truth or falsity sentences, like e.g. $(\forall v_1)(v_1 < S(v_1))$."


问:
Continuous dependence of inverse functions on parameters
If we have a bijective, continuous function $f\colon \mathbb{R} \to \mathbb{R}$, then it is proven in every introductory course to analysis that the inverse function $f^{-1}\colon \mathbb{R} \to \mathbb{R}$ will be also continuous.

Recently I needed this fact for continuous families of such functions, i.e., assume that $f\colon \mathbb{R} \times U \to \mathbb{R}$ is a continuous function, where $U \subset \mathbb{R}^r$ is an open subset (the parameter space), such that for each $s \in U$ the function $f_s\colon \mathbb{R} \to \mathbb{R}, x \mapsto f(x,s)$ is bijective. Since each $f_s$ is also continuous, we get a family of continuous inverse functions $\{f_s^{-1}\}_{s \in U}$. Does this family depend continuously on the parameter $s$, i.e., do these inverse functions assemble to a single continuous function $F\colon \mathbb{R} \times U \to \mathbb{R}, (y,s) \mapsto f_s^{-1}(y)$?

The answer is 'yes', but is there any way of proving this without an $\varepsilon$-$\delta$-massacre?

答:
Let $s_i \to s$ and $y_i \to y$. We have to show that $f_{s_i}^{-1}(y_i) \to f_s^{-1}(y)$.

Let us first show that the sequence $(f_{s_i}^{-1}(y_i))_{i \in \mathbb{N}}$ is bounded. Since $y_i$ converges, there are numbers $c$ and $d$ and an $\varepsilon > 0$ such that the set $\{y_i,y\colon i \in \mathbb{N}\}$ is contained in the open interval $(c-\varepsilon, d + \varepsilon)$. Since the function $s' \mapsto f(f_s^{-1}(d),s')$ is continuous, there exists a $\delta > 0$ such that $|f_{s'}(f_s^{-1}(d)) - d| < \varepsilon$ for all $s'$ with $|s' - s| < \delta$. A similar statement holds for $c$ instead of $d$. Since each $f_{s'}^{-1}$ is monotone (since it is continuous and bijective), we conclude $\{f_{s'}^{-1}(y_i)\colon i \in \mathbb{N}\} \subset [f_s^{-1}(c),f_s^{-1}(d)]$ (or in $[f_s^{-1}(d),f_s^{-1}(c)]$ if it is decreasing) for all $s'$ with $|s'-s| < \delta$. This proves the claim.

Since the sequence $(f_{s_i}^{-1}(y_i))_{i \in \mathbb{N}}$ is bounded, it has an accumulation point $\bar x$. Let $x_{i_k} = f_{s_{i_k}}^{-1}(y_{i_k})$ be a subsequence converging to it. Then we have $$f_s(\bar x) = f(\lim_{k \to \infty}(x_{i_k},s_{i_k})) \stackrel{!}= \lim_{k \to \infty} f(x_{i_k},s_{i_k}) = \lim_{k \to \infty} f_{s_{i_k}}(f_{s_{i_k}}^{-1}(y_{i_k})) = y = f_s(f_s^{-1}(y))\,,$$
where at the marked equality we used continuity of $f$. Since $f_s$ is injective, we conclude that $\bar x = f_s^{-1}(y)$. But $\bar x$ was the limit of $f_{s_{i_k}}^{-1}(y_{i_k})$. All in all, we conclude that the sequence $(f_{s_i}^{-1}(y_i))_{i \in \mathbb{N}}$ is bounded and only has one accumulation point, namely $\bar x = f_s^{-1}(y)$. It follows that the sequence $(f_{s_i}^{-1}(y_i))_{i \in \mathbb{N}}$ converges to it, proving continuity of our family of inverse functions.


问:
Conjugate gradient method's convergence steps
I met a question that asks me to show that if the $A$ of $Ax = b$ is a symmetric positive definite matrix and all its eigenvalues are 1, 2, or 8. Then the conjugate gradient method converges in 3 or fewer steps.

I don't know where to start, did I missed some important knowledge between its convergence and $A$'s eigenvalues?

答:
The error $e_k:=x-x_k$ at step $k$ of CG satisfies
$$\tag{1}
\|e_k\|_A=\min_{p\in\Pi_k \\ p(0)=1}\|p(A)e_0\|_A,
$$
where $\Pi_k$ is the set of polynomials of degree at most $k$. If $A$ has only $k$ distinct eigenvalues, you can take $p$ in (1) to be the minimal polynomial of $A$. Since $p(A)=0$ you get that $e_k=0$.


问:
Finding matrices whose multiplication does not change the trace
Let $C$ be an $n\times n$ matrix. Then find matrices $A$ of size $n \times n$ such that $trace(AC)=trace(C)$. One obvious choice for $A$ is the Identity matrix $I$, but are there other choices?

A similar question is, if $A$ and $C$ are $m\times m$ matrices and $B$ and $D$ are $n\times n$ matrices such that $trace(C)=trace(D)$ and $trace(AC)=trace(BD)$, then what is the explicit relationship between $A$ and $B$? Here also we can choose $A$ and $B$ to be Identity matrices, but are there other choices of matrices?

答:
The expression $\mathrm{trace}(X^*C)$ can be thought of as an inner product between matrices.

So the equation $\mathrm{trace}(AC)=\mathrm{trace}(C)$ is equivalent to $\langle A^*-I,C\rangle=0$. So $A^*=I+D$ where $D$ is any matrix 'perpendicular' to $C$. One can create a basis for this space by starting with the elementary matrices $E_1,\ldots,E_{n^2}$ and modifying them: $$D_i=E_i-\frac{\langle E_i,C\rangle}{\langle C,C\rangle} C=E_i-\frac{\mathrm{trace}(E_i^*C)}{\mathrm{trace}(C^*C)}C$$

Hence the answer is $A=I+\sum_{i=1}^{n^2-1}\alpha_i\left(E_i^*-\frac{\mathrm{trace}(E_i^*C)}{\mathrm{trace}(C^*C)}C^*\right)$ for any $\alpha_i\in\mathbb{C}$.

[For $C=0$, then any matrix $A$ is valid.]


问:
What is the time-integral of motion for first order differential equations?
For a second order differential equation (many physical systems) in one variable, I know "procedures" to compute the energy. Given $$q''(t)=f(q(t),q'(t)),\ \  q(0)=q_0,\ \ q'(0)=v_0,$$
if we're lucky we can read off the related Lagrangian $L$, introduce $p=\frac{\partial L}{\partial q}$, do a Legendre transform and we got the Hamiltonian function $H(q,p)$ for which $\frac{\text d}{\text dt}H(q(t),p(t))=0$ for solutions of the differential equation.

We can be more exact and give all the conditions for Noethers theorem to hold and the result is that along the flow $X(t)=\langle q(t),p(t)\rangle$ in phase space given by a solution with initial conditions $X(0)$, the function $H(q,p)$ always takes the same values. It defines surfaces in phase space indexed by initial conditions $X(0)$.

I wonder how to view this for a priori first order systems
$$q'(t)=f(q(t)),\ \ q(0)=q_0,$$
where I think this must be even simpler.
E.g. I thing some functional
$$q(t)\mapsto q(t)-\text{an integral over}\  f(q(t))\  \text{something},$$
should exists which will be constant for solutions of the equation, i.e. only depend on $q_0$. For each $f$, the functional dependence of this "energy" on "$q_0$" will be different.

However, I can't seem to find a general relation. What's the theory behind this, is there an energy'ish time integral of motion? What is the functional dependence on the intial condition, for a suitable constant of motion for for first order systems. And what would be the interpretation, given that we speak of a situation with only one initial condition?

If it is that case that the system is too restricted so that there is no meaningful geometrical interpretation, then let's think about a system of first order differential equations. This is like the one we generated from phase space, except that it doesn't really come from a second order situation and so the intial conditions aren't really e.g. intial position and velocity. I'm pretty sure there are situation where one considers such a directly generated flow (the equation might be more complicated than exponential flow $\dot x\propto x$), but I don't recall any talk about the time-constant of motion in these systems, or how to interpret it.

答:
Assuming the first order system is Hamiltonian, for one degree of freedom, we may construct the hamiltonian by integrating the pair of ode's using the relation:

$
q'= \frac{\partial H}{\partial p}\qquad
p'= -\frac{\partial H}{\partial q}
$

Higher order systems have the same "symplectic" structure, and the integrals could have no closed forms. So we would have:

$ \frac{\partial H}{\partial p} = f(q),\quad p'=??$

Things seem to get notationally ambiguous. The point is that we must look at variables as conjugated pairs when working in the Hamiltonian formulation of mechanics. As a definitive answer to the OP, if we knew that the system was Hamiltonian, we could in principle integrate the above relations to generate the "Hamiltonian". Also, energy and time are considered to be conjugated in this framework. May I suggest the Poisson bracket as a starting point.

ContentLicense: CC BY-SA 3.0


问:
Why is it okay to take out the constant when finding asymptotes or hyperbola?
$\require{cancel}$I have been tasked to find the equation(s) of the asymptotes for the graph of the following hyperbolic equation:$$16x^2-25y^2-400=0$$My teacher explains that you can just take out the constant term to find the asymptotes. For this specific case:$$16x^2-25y^2\cancel{-400}=0\implies16x^2-25y^2=0$$I have graphed both $16x^2-25y^2-400=0$ and $16x^2-25y^2=0$ and confirmed that $16x^2-25y^2=0$ indeed does graph the asymptotes of $16x^2-25y^2-400=0$.

Can someone provide a simple explanation as to why this is the case?

答:
Recall that when you sketch the graphs of
$$\gamma : 16x^2-25y^2-400=0$$
and of
$$\delta : 16x^2-25y^2 = 0$$
you are representing in the Cartesian plane the points whose coordinates solve the above equations.

So when your teacher says that $\delta$ represents the asypmptotes of $\gamma$ he means that the the solutions to the equations $\gamma$ and $\delta$ (the latter being coordinates of the points on the straight lines $y = \pm \frac45 x$) get closer and closer, the larger are the values of $x$ (or $y$) you consider.

To convince yourself of this, you could try to analyze the situation just in the first quadrant. With this limitation, the above mentioned solutions can be expressed in these terms (can you doublecheck that?)
$$y = \frac45 x$$
and
$$y=\frac45 \sqrt{x^2-25}.$$

Now you can ask yourself: for a given $x$, what is the difference between the $y$-coordinate of the points lying on the hyperbola and on the straight line? What we are trying to determine is therefore
$$\Delta = \frac45 x - \frac45 \sqrt{x^2-25}.$$
If you multiply and divide the expression by the sum of the two terms and use the identity $(A+B)(A-B) = A^2-B^2$ (fill in the details yourself) you get
$$\Delta = \frac{20}{x^2+\sqrt{x^2-25}},$$
from which you can see that you can make this difference arbitrarily small, by just taking larger and larger values of $x$.

You could try to generalize these results, and see that the conclusions are not affected by a different choice of the original equation's coefficients.

Hope this helped you.


问:
Intuition on the quality of the approximation of the sample average in the central limit theorem
Let's say that a sequence of random variables $(X_n)_{n \in \mathbb{N}}$ is $\mathcal {o}(Y_n)$ for some other sequence of random variables $(Y_n)_{n \in \mathbb{N}}$, if $\lim_{n \to \infty}\frac{X_n}{Y_n}=0$ in distribution.

We can use this to interpret the central limit theorem for such a sequence of random variables of mean $\mu$ and variance $\sigma^2$ as an asymptotic expansion (of the distribution)
$$ \frac{1}{n}\sum_{j=1}^{n} X_j = \mu + \frac{\xi}{\sqrt{n}} + \mathcal{o}\left(\frac{1}{\sqrt{n}}\right), $$
where $\xi$ is a random variable distributed as $\mathcal{N}(0,\sigma^2)$.

Intuitively the first term comes from the law of large numbers, and the second term is a correction of order $\frac{1}{\sqrt{n}}$ coming from the central limit theorem.
Is there some intuition in why this term should be of order $\frac{1}{\sqrt{n}}$? Why isn't the expansion in some other power of $n$, e.g. $\frac{1}{n}$ as in the definition of differentiability.

Assuming that it is intuitively clear that for large $n$, the sample average should be approximately distributed as a normal distribution, why should the quality of this approximation be exactly $\frac{1}{\sqrt{n}}$ and not some other numerical value?

答: Do you have good intuition for random walks? After $n$ steps, you're within $O(\sqrt{n})$ of the origin with high probability. That's why the second term is of order $\frac{1}{\sqrt{n}}$. As to why it's exactly $\frac{\xi}{\sqrt{n}}$, I think you're gonna need a proof....


问:
How to state precisely this requirement relative to the application of natural deduction rules?

Contrary to replacement  rules, in natural deduction, rules of infrence need to be applied " to an entire line" .

For example, I cannot use Modus Ponens in the following way

Premise (1)  : $( A \rightarrow (B\rightarrow C)) $

Premise (2) :  $B$

Conclusion : $C$  ( wrongly applying MP to the consequent in premise (1) ) .

What is the ( standard) formal statement of this requirement?

答:
You cannot apply Modus Ponens that way in no proof system, because the inference is not sound.

If $B$ is true and both $A$ and $C$ are false, we have that both premises are true while the conclusion $C$ is false.

To correctly apply MP you have to assume $A$ and then (using MP twice) derive $C$.

Then conclude with $A \to C$ by CP.

The (now correct) result is the derivation:

$A → (B → C), B \vdash A \to C$.


问:
Euclid's Algorithm.

This is a Theorem I am constructing. I have to add the last equation (the one with zero remainder) right underneath the 3 dots, but for some reason I am getting very confused. Any tips?

答: I hope this helps. I solved it in assignment of Ring Theory in this semester only.

Title: Euclid's Algorithm.


问:
Find $\lim\limits_{n\rightarrow \infty} \left| \ln \frac{1}{\left| x_n-A \right|} \right|^{1/n}$
The sequence $\{x_n\}$ satisfies $x_{n+1}=x_n+\sin x_n,n=1,2,\cdots $.

Prove $\{x_n\}$ is convergent and find its limit $A$.

Let $x_1\neq A$. Find the limit $$\lim_{n\rightarrow \infty} \left| \ln \dfrac{1}{\left| x_n-A \right|} \right|^{1/n}$$

My Attempt: For the first question, let $k\in \mathbb Z$ and $f(x)=x+\sin x$. When $x=k\pi$, it is trivial that $A=k\pi$. When $x\in (k\pi,(k+1)\pi)$, since $f(x)\in (k\pi,(k+1)\pi)$ and $f(x)$ monotonously increases on $\mathbb R$, obviously $\{x_n\}$ monotonously increases and is bounded from above. Take the limits of both sides of the recurrence, and we get $A=A+\sin A$, so $A=(k+1)\pi$.

My Question is, did I correctly do the first question? And how to solve the second one? I don't have any ideas yet.

答:
There is a more elementary way to solve the second part of your question.

Since $\;x_1\ne\lim\limits_{n\to\infty}x_n\;,$ by looking at the first and second cases of my previous answer, it follows that $\;\lim\limits_{n\to\infty}x_n=A=\pi+2k\pi\;$ where $\;k\in\mathbb{Z}\;$ and $\;x_n\ne A=\pi+2k\pi\;$ for all $\;n\in\mathbb{N}\;.$

$\begin{align}
\lim\limits_{n\rightarrow\infty}&\left|\ln\dfrac{1}{\left|x_n-A\right|}\right|^{1/n}=\lim\limits_{n\rightarrow\infty}\sqrt[n]{\big|-\ln\left|x_n-A\right|\big|}=\\
&=\lim\limits_{n\rightarrow\infty}\sqrt[n]{\big|\ln\left|x_n-A\right|\big|}
\end{align}$

and, by using an application of Stolz-Cesàro Theorem (geometric mean application), it is sufficient to calculate the following limit

$\lim\limits_{n\to\infty}\dfrac{\big|\ln\left|x_{n+1}-A\right|\big|}{\big|\ln\left|x_n-A\right|\big|}=\lim\limits_{n\to\infty}\left|\dfrac{\ln\left|x_n-A+\sin x_n\right|}{\ln\left|x_n-A\right|}\right|.$

It is possible to use that application because there exists $\;n_0\in\mathbb{N}\;$ such that $\;\big|\ln|x_n-A|\big|>0\;$ for all $\;n\ge n_0\;$, indeed $\;\lim\limits_{n\to\infty}\big|\ln|x_n-A|\big|=+\infty\;.$

First of all, I am going to calculate the limit

$\begin{align}
\lim\limits_{x\to A}&\dfrac{\ln\left|x-A+\sin x\right|}{\ln\left|x-A\right|}=\lim\limits_{x\to A}\dfrac{\ln\left|x-A-\sin(x-A)\right|}{\ln\left|x-A\right|}=\\
&=\lim\limits_{t\to0}\dfrac{\ln\left|t-\sin t\right|}{\ln\left|t\right|}=\lim\limits_{t\to0}\dfrac{\ln\left|t-\sin t\right|-3\ln|t|+3\ln|t|}{\ln\left|t\right|}=\\
&=\lim\limits_{t\to0}\left[\dfrac{\ln\left|\dfrac{t-\sin t}{t^3}\right|}{\ln\left|t\right|}+3\right]=\dfrac{\ln\left|\dfrac16\right|}{-\infty}+3=3\;.
\end{align}$

Hence,

$\lim\limits_{x\to A}\left|\dfrac{\ln\left|x-A+\sin x\right|}{\ln\left|x-A\right|}\right|=3$

and, by applying the Sequential Criterion for a limit of a function, it follows that

$\lim\limits_{n\to\infty}\dfrac{\big|\ln\left|x_{n+1}-A\right|\big|}{\big|\ln\left|x_n-A\right|\big|}=\lim\limits_{n\to\infty}\left|\dfrac{\ln\left|x_n-A+\sin x_n\right|}{\ln\left|x_n-A\right|}\right|=3\;.$

Finally, by using an application of Stolz-Cesàro Theorem (geometric mean application), it follows that

$\lim\limits_{n\rightarrow\infty}\left|\ln\dfrac{1}{\left|x_n-A\right|}\right|^{1/n}=\lim\limits_{n\rightarrow\infty}\sqrt[n]{\big|\ln\left|x_n-A\right|\big|}=3\;.$


问:
Proving inequality of probabilty to derive upper bound for moment-generating functions
I'm stuck with proving following inequalities.

Let $X_i$ be independent r.v in [0,1]. Denote $S=X_1+\dots+X_n$.

Note that for any $a\in \mathbb{R}$ and $0\leq x\leq 1$, we have $$e^{ax}\leq 1-x+xe^a \tag{0}$$

How to show the following inequalities, which provide upper bounds for moment-generating functions?

For any $a\in \mathbb{R}$,

$$M_{X_i}(a) := \mathbb{E}[e^{aX_i}]\leq e^{\mathbb{E}[X_i](e^a-1)} \tag{1.1}$$ and $$M_S(a) := \mathbb{E}[e^{aS}]\leq e^{\mathbb{E}[S](e^a-1)} \tag{1.2}$$

assuming each $M_{X_i}(a) := \mathbb{E}[e^{aX_i}]$ exists (and thus $M_S(a)$ exists)

Also, how do I extend this to the following?

For any $0<t<1$,

$$\mathbb{P}(S\geq (1+t)\mathbb{E}[S])\leq(\frac{e^t}{(1+t)^{1+t}})^{\mathbb{E}[S]} \tag{2.1}$$

and

$$\mathbb{P}(S\geq (1-t)\mathbb{E}[S])\leq(\frac{e^{-t}}{(1-t)^{1-t}})^{\mathbb{E}[S]} \tag{2.2}$$

答:
(formerly partial answer. now full answer, I hope).

To prove 1.1:

There's another inequality to use: We have for any $b \in \mathbb R$ that  $$1+b \le e^b \tag{A}$$

By $(0)$, $M_X(a) \le 1+(e^a-1)E[X]$. By $(A)$ with $b=(e^a-1)E[X]$, we get $1+(e^a-1)E[X] \le e^{E[X](e^a-1)}$.

Note: I actually got the inequality $(A)$ from here: https://en.wikipedia.org/wiki/Moment-generating_function#Other_properties --> in the part with 'This follows from the simple inequality'.

To prove 1.2 from 1.1:

By independence (independence of $X_i$ implies independence of $e^{aX_i}$ because $f(x)=e^{ax}, f: \mathbb R \to \mathbb R$ is measurable or something. see here or here or my new question here...or use definition of independence in terms of joint density splitting up or something),

$$\mathbb{E}[e^{a(\sum_i X_i)} = \prod_i e^{aX_i}] = \prod_i E[e^{aX_i}]$$

Then 1.1 says $E[e^{aX_i}] \le e^{\mathbb{E}[X_i](e^a-1)}$. Hence, from non-negativity of each $E[e^{aX_i}]$,

$$\mathbb{E}[e^{a(\sum_i X_i)}] \le \prod_i e^{\mathbb{E}[X_i](e^a-1)} = e^{\sum_i \mathbb{E}[X_i](e^a-1)} = e^{E[S](e^a-1)}$$

For 2 (attempt 1):

Not sure Markov's inequality is helpful, but...

Markov's inequality, assuming applicable, gives

$$\mathbb{P}(S\geq (1 \pm t)\mathbb{E}[S])\le \frac{E[S]}{(1 \pm t)\mathbb{E}[S]} = \frac{1}{1 \pm t}$$

Here, Markov's inequality is applicable because $S$, $1 \pm t$ and $\mathbb{E}[S]$ are non-negative, BUT IF $\mathbb{E}[S]$ is nonzero. At this point, I guess let's use that non-negative random variables have zero mean if and only if they are almost surely zero.

For 2 (attempt 2):

I think Markov's inequality is helpful but not for $P(S \ge (1 \pm t)\mathbb{E}[S])$ directly but rather for $P(S \ge (1 \pm t)\mathbb{E}[S]) = P(e^{mS} \ge e^{m[(1 \pm t)\mathbb{E}[S])]})$, for any $m  > 0$. We get

$$P(S \ge (1 \pm t)\mathbb{E}[S]) \le e^{-m((1 \pm t)\mathbb{E}[S])} M_S(m) \tag{B}$$

based on (see here) that for any random variable $Z$ (doesn't have to be non-negative) and for any $z$ (doesn't have to be positive) and for any $m > 0$

$$P(Z \ge z) \le e^{-zm} M_Z(m) := e^{-zm} E[e^{mZ}]$$

I think I got it for (2.1):

In $(B)$, choose $m=\ln(1+t)$, which is indeed positive for $0 < t$ (not sure where $t < 1$ is used. maybe this is used more in (2.2)) and then use $(0)$ on $E[e^{mS}]$ with $a=m$:

$$P(S \ge (1 + t)\mathbb{E}[S]) \le e^{(e^m-1-m-mt)E[S]}$$

and then $e^{(e^m-1-m-mt)E[S]}$ is indeed $\le (\frac{e^t}{(1+t)^{1+t}})^{\mathbb{E}[S]}$, which (I hope!) is true if and only if $(e^m-1-m-mt)E[S] \le \mathbb{E}[S] \ln(\frac{e^t}{(1+t)^{1+t}})$, which is true if and only if $e^m-1-m+mt \le \ln(\frac{e^t}{(1+t)^{1+t}} = t - (1+t)\ln(1+t)$

Similarly for 2.2: (maybe related: Borel-Cantelli-related exercise: Show that $\sum_{n=1}^{\infty} p_n < 1 \implies \prod_{n=1}^{\infty} (1-p_n) \geq 1- S$.)

Do $m=ln(1-t)$. Now this uses $t < 1$ (not sure if $0 < t$ is used). Instead of having to prove '$e^m-1-m-mt \le t - (1+t)\ln(1+t)$', we have to prove...either

$e^m-1-m+mt \le - t \pm (1-t)\ln(1-t)$. You double check which of the $\pm$ but regardless I believe we have

$e^m-1-m+mt \le - t - (1-t)\ln(1-t)$. Finally, in case it's the $+$, just use that $- t - (1-t)\ln(1-t) \le - t + (1-t)\ln(1-t)$.

Remark: Remember to prove Markov's inequality, if you haven't in class!


问:
If $2n+1$ and $4n+3$ are prime, then $2n-1$ and $4n+1$ are not when $n>2$
How do you prove that, for $n>2$, if $2n+1$ and $4n+3$ are prime numbers, then $2n-1$ and $4n+1$ are composite numbers?</p>

答:
HINT:  $3$ divides $2n(2n+1)(2n-1)$ and $3$ divides $(4n+1)(4n+2)(4n+3)$

As $n>2, 4n+1>2n+1>3$

Hence, $4n+1, 2n+1$ are not divisible by $3$

If $4n+3$ is prime, $3$ does not divide $n\implies 3$ divide  $2n-1$

If $2n+1$ is prime, $3$ does not divide $2(2n+1)\implies 3$ divide $4n+1$

which can also be derived from $4n+1=2(2n-1)+3$


问:
Schur decomposition of matrix $X - \alpha \, u_1 u_1^T$
Consider the Schur decomposition $X = URU^T$ of a real matrix $X$, where $U$ is orthogonal and $R$ is upper triangular.

Is there a nice way to compute the Schur decomposition of the matrix $X - \alpha \, u_1 u_1^T$ for any $\alpha \in \mathbb{R}$, where $u_1u_1^T$ is the outer product of the first column of $U$ with itself?

I'm not sure how to use the decomposition of $X$ here.

答:
The same orthogonal matrix $U$ will give the Schur decomposition:

$$U^\top(X-\alpha u_1u_1^\top)U=U^\top XU-\alpha U^\top u_1u_1^\top U=R-\alpha e_1e_1^\top=\tilde{R}$$ $$\therefore\ X-\alpha u_1u_1^\top=U\tilde{R}U^\top$$ where $\tilde{R}$ is the same as matrix $R$ except $\tilde{R}_{11}=R_{11}-\alpha$. Note that $U^\top u_1=U^{-1}u_1=e_1$.


问:
variance of standard sample deviation in a normal distribution
I need help. I Know that using Fisher Lemma, we know that in a normal distribution
$ Var(\hat{\sigma^2})= \frac{2\sigma^4(n-1)}{n^2}$, since $\frac{n\hat{\sigma^2}}{\sigma^2}\approx \chi^2_{n-1}$.
I need now to calculate $Var(\hat{\sigma})$, but i don´t know the relation between
$var(X^2) $ and $Var(X)$ in a normal distribution.
Thank you everyone.

答: Hint: If $Y\sim N(0,1)$ the $X \stackrel {d} {=}\mu+\sigma Y$ where $\mu$ is the mean of $X$ and $\sigma^{2}$ is the variance.  Hence, $EX^{4}=E(\mu+\sigma Y)^{4}$ and $EX^{2}=E(\mu+\sigma Y)^{2}$ You can compute these by expanding the powers using Binomial Theorem.


答:
I need now to calculate $Var(\hat{\sigma})$

Let's state $\hat{\sigma}^2=S^2$

you can express

$$\mathbb{V}[S]=\frac{\sigma^2}{n}\mathbb{V}\Bigg[\sqrt{\frac{nS^2}{\sigma^2}\Bigg]}$$

the calculation is not immediate but not difficult because now you can calculate the simple moments using the distribution of $\frac{nS^2}{\sigma^2}$

In particular, set $\frac{nS^2}{\sigma^2}=Y$ you have

$$\mathbb{V}[S]=\frac{\sigma^2}{n}\Bigg\{\mathbb{E}[Y]-\mathbb{E}^2[\sqrt{Y}]\Bigg\}=\frac{\sigma^2}{n}\Bigg\{n-1-\Bigg[\int_0^{\infty}\sqrt{y}f_Y(y)dy\Bigg]^2\Bigg\}$$

Now the only hurdle to jump over is to calculate

$$\int_0^{\infty}\sqrt{y}\frac{\Big(\frac{1}{2}\Big)^{\frac{n-1}{2}}}{\Gamma\Big(\frac{n-1}{2}\Big)}y^{\frac{n-1}{2}-1}e^{-\frac{y}{2}}dy=\frac{\Gamma\Big(\frac{n}{2}\Big)\sqrt{2}}{\Gamma\Big(\frac{n-1}{2}\Big)}\underbrace{\int_0^{\infty}\frac{\Big(\frac{1}{2}\Big)^{\frac{n}{2}}}{\Gamma\Big(\frac{n}{2}\Big)}y^{\frac{n}{2}-1}e^{-\frac{y}{2}}dy}_{=1}=\frac{\Gamma\Big(\frac{n}{2}\Big)\sqrt{2}}{\Gamma\Big(\frac{n-1}{2}\Big)}$$

Usually $\frac{(n-1)S^2}{\sigma^2}\sim \chi_{(n-1)}^2$ is taken into consideration because it is used to consider the unbiased estimator of $\sigma^2$. It is not forbidden to use the statistic you showed; the procedure remains the same, just modify it accordingly

Title: variance of standard sample deviation in a normal distribution

Body:
I need help. I Know that using Fisher Lemma, we know that in a normal distribution
$ Var(\hat{\sigma^2})= \frac{2\sigma^4(n-1)}{n^2}$, since $\frac{n\hat{\sigma^2}}{\sigma^2}\approx \chi^2_{n-1}$.
I need now to calculate $Var(\hat{\sigma})$, but i don´t know the relation between
$var(X^2) $ and $Var(X)$ in a normal distribution.
Thank you everyone.


问:
Some big-theta and big-omega notation question

Let $f : \mathbb{N} \rightarrow \mathbb{R}$ be the function $f ( n ) = n ^ { 2 } + \sqrt { n }$. Determine whether the following statement is true or false, providing a proof for your answer.

$ f(n) = \Theta(n^2) $

$ \log_2[f(n)] = \Theta(\log_2n) $

$ 2^{f(n)} = \Theta(2{n^2}) $

$2 ^ { 2 ^ { f ( n ) } } = \Omega ( g )$, where $g(1) = 1$, $g(2) = 2$, and $g(n) = [g(n - 1)] ^2 + [g(n - 2)] ^2$ for $n \ge 3$.

How can I get a prove to determine the abovementioned is true or false

for a),

what I have got for the first questionnow is:

I have $ c_1, c_2 , n_0 > 0$

c_1 $ \leq  n^2 + \sqrt(n) \leq c_2 $

c_1 $ \leq  1 + \frac{1}{n^\frac{3}{2}} \leq c_2 $

So how can I continue to prove the Big theta notation? I don't understand how to choose the c

答:
Perhaps this will set you on the right track
$$c_1 n^2\le n^2+\sqrt{n}\le c_2 n^2$$
Now just find $c_1$ and $c_2$, both greater than zero, that makes the inequality true as $n$ increases past some $n_0$. Once you achieve this, then you can say
$$n^2+\sqrt{n}\in\Theta\left(n^2\right)$$
The same strategy applies to the other Big Theta problems.


问:
How to find the integral for $\int^1_{\frac{1}{1+e^\sigma}}(\frac{y}{1-y})^{1/r}\frac{1}{y-1} dy$
How can I find the solution for following integral.

$$\int^1_{\frac{1}{1+e^\sigma}}(\frac{y}{1-y})^{1/r}\frac{1}{y-1} dy$$

where $r,\sigma$ are positive constants.

Body:
$$I=\int^1_{y_0}(\frac{y}{1-y})^{1/r}\frac{1}{y-1} dy, y_0=\frac{1}{1+e^\sigma}<1$$
$$-I=\int_{0}^{1} y^{1/r}(1-y)^{-1/r-1} dy-\int_{0}^{y_0} y^{1/r}(1-y)^{-1/r-1} dy $$
$$-I=B(1+1/r,-1/r)-B(y_0, 1+1/r,-1/r)= \frac{\Gamma(1+1/r) \Gamma(-1/r)}{\Gamma(1)}-B(y_0,1+1/r,-1/r)$$
$$\implies -I=\frac{1}{r}\Gamma(1/r)\Gamma(-1/r)-B(y_0,1+1/r,-1/r) $$$$\implies I=\pi \csc(\pi/r)+B(y_0,1+1/r,-1/r).$$
$B(a,b)$ is Beta function and $B(x,a,b), 0<x<1$; is incomplete Beta function.

See:https://en.wikipedia.org/wiki/Beta_function


问:
Discrete Exponential Martingale - Properties
This question is about the discrete exponential martingale.

Let $(Y_n)_n$ be a sequence of independent and identically distributed random variables
with $m_{Y}(t) :=\mathbb{E}\left[e^{t Y_{1}}\right]<\infty, t \in \mathbb{R}$. I want to show that

\begin{align*}
M_{n} :=\frac{1}{m_{Y}(t)} \exp \left(t \sum_{i=1}^{n} Y_{i}\right)
\end{align*}
is a martingale.
But I didn't know how to proof that \begin{align*}
\begin{aligned} \mathbb{E}\left[M_{n+1} | \mathcal{F}_{n}\right] &=\frac{1}{m_{Y}(t)} \mathbb{E}\left[\exp \left(t \sum_{i=1}^{n+1} Y_{i}\right) | \mathcal{F}_{n}\right] \\ &=\frac{1}{m_{Y}(t)}\left\{\mathbb{E}\left[\exp \left(t Y_{n+1}\right) | \mathcal{F}_{n}\right]+\sum_{i=1}^{n} \mathbb{E}\left[\exp \left(t Y_{i}\right) | \mathcal{F}_{n}\right]\right\} = ... = M_n\end{aligned}
\end{align*}

答: Yeah, it sounds good. I did not realize that $\frac{\exp^{(t\sum_{i=1}^{n}Y_{i})}}{m_Y(t)^{n}}>0$ so $\mathbb{E}|\frac{\exp^{(t\sum_{i=1}^{n}Y_{i})}}{m_Y(t)^{n}}|=\mathbb{E}(\frac{\exp^{(t\sum_{i=1}^{n}Y_{i})}}{m_Y(t)^{n}})$. Next $\frac{1}{m_Y(t)^{n}}\in\mathbb{R}$ and all $\mathbb{E}e^{tY_{i}}<\infty$.


问:
Prevalence of ergodic rotations on a compact group
So I know that any compact group $G$ carries a Haar probability measure $\mu$. For any $g \in G$, let $T_g : G \to G$ be the rotation $T_g : x \mapsto gx$. In the cases where $G = \mathbb{R}^n / \mathbb{Z}^n$ and $g = (t_1, \ldots, t_n)$, the rotation $T_g$ is ergodic iff $\{ t_1, \ldots, t_n, 1 \}$ is linearly independent over $\mathbb{Q}$.

In the case of toral rotations, I notice that almost every rotation is ergodic, i.e. $T_g$ is ergodic for almost all $g$. Is this typical behavior for group rotations, i.e. that almost every element of the compact group gives rise to an ergodic rotation?

答: Let $G = \{0,1\}^\mathbb{N}$ with component-wise mod $2$ addition. Then for all $g$, the rotation $x \mapsto gx$ is not ergodic, since $g^2 = 0$.


问:
How to prove irreducibility of $P(x)=x^4-2x^3-2x^2+2x-1$ over $\mathbb{Q}$?
I have polynomial $P(x)=x^4-2x^3-2x^2+2x-1$ and I have strong suspicions it's irreducible. Any ideas how to prove that?

Thank you.

答:
The polynomial is irreducible over $\Bbb Z$ and $\Bbb Q$, because it is already irreducible modulo $3$, so over $\Bbb F_3$, which is easy to see.

Alternatively, you can use Eisenstein with a shift $x\mapsto x+1$, or just see it directly, like in this post: Show that polynomial is irreducible over $\mathbb{Q}$


问:
Definition of $I(\varphi, M)$ in Buchsbaum's and Eisenbud's "What makes a complex exact?"
In their paper What makes a complex exact?, Buchsbaum and Eisenbud define an ideal $I(\varphi, M)$, where $M$ is a module over a commutative ring, and $\varphi: F \to G$ is a homomorphism of projective $R$-modules. I'm not sure that I understand the definition correctly. First they define $I(\varphi)$ to be the image of the induced morphism
$$(\Lambda^k G)^* \otimes \Lambda^kF \to R, \tag{$*$}$$
for $k = \operatorname{rank}(\varphi)$, that is the smallest $k$ such that $0 \neq \Lambda^k \varphi: \Lambda^k F \to \Lambda^k G$.
Then they write

Similarly, if $M$ is an $R$-module, we define $I(\varphi, M)$ to be the image of the above map where $k = \operatorname{rank}(\varphi, M)$.

So is $\operatorname{rank}(\varphi, M)$ really the only property of $M$ that enters into the definition of $I(\varphi, M)$? Or do I have to tensorize something in $(*)$ by $M$? This is somehow also suggested in this question, though it's not the point of discussion there.

答: So I think that the image of $(\Lambda^k G)^* \otimes \Lambda^kF \to R$ is the same as the image of $(\Lambda^k G \otimes M)^* \otimes (\Lambda^kF \otimes M)\to R$, at least if $k = \operatorname{rank}(\varphi, M)$. The reason here is that the canonical map $M^* \otimes M \to R$ is surjective.


问:
Form Dirichlet Log Density with DCP ruleset.
I would like to formulate the log likelihood of the Dirichlet density as a disciplined convex programming (DCP) optimization problem with respect to $\alpha \in \mathbb{R}^K_{\succ 0}$. The log likelihood

$$\mathcal{L}(\textbf{p}, \alpha) = \log \frac{\Gamma \left(\sum_k \alpha_k \right)}{\prod_k \Gamma (\alpha_k)} \prod_k p_k^{\alpha_k - 1}$$
$$= \log \Gamma \left(\sum_k \alpha_k \right) - \sum_k \log \Gamma(\alpha_k) + \sum_k (\alpha_k - 1) \log p_k $$

despite being concave in $\alpha$ is not formulated as DCP because is involves the difference of $\log \Gamma \left(\sum_k \alpha_k \right)$ and $\sum_k \log \Gamma(\alpha_k)$ which each have positive curvature. Note that despite this, the function is concave because $\log \Gamma (\alpha)$ is convex on $\alpha \succ 0$ and thus

$$\log \Gamma \left(\sum_k \alpha_k \right) \leq \sum_k \log \Gamma(\alpha_k)$$

holds from Jensen's inequality, which is sufficient to show that their difference is concave. However, in order to use certain convex optimization solvers such as cvxpy, I need to formulate this function using the DCP ruleset, which does not allow the difference of two functions with the same curvature in general. Is it possible to do so, and what approaches might I take to achieve this?

答:
Let
$$ \beta(x_1,x_2) = \frac{\Gamma \left( x_1 \right) \Gamma \left( x_2 \right)}{\Gamma \left(x_1 + x_2 \right)} $$

be the beta function, and

$$ B(x) = \frac{\prod_{j=1}^n \Gamma \left( x_j \right)}{\Gamma \left(\sum_{j=1}^n x_j \right)}$$

be the multivariate beta function, and note their beautiful relation (just expand and see how everything cancels out to give you exactly what you expect):

$$ B(x) = \prod_{k=2}^{n} \beta \left(\sum_{j=1}^{k-1} x_j,x_k \right) $$

What you are asking for is a way to represent the concave expression
$$\log \left( B(x)^{-1} \prod_{j=1}^n p_j^{(x_j - 1)} \right)$$

which, since $\log(x^{-1}) = -\log(x)$, is equivalent to representing the convex expression:
$$\log \left( B(x) \prod_{j=1}^n p_j^{(1 - x_j)} \right)\\
= \sum_{k=2}^{n} \log \left( \beta \left(\sum_{j=1}^{k-1} x_j,x_k \right) \right) + \sum_{j=1}^n \log \left(p_j^{(1 - x_j)} \right)\\
= \sum_{k=2}^{n} \log \left( \beta \left(\sum_{j=1}^{k-1} x_j,x_k \right) \right) + \sum_{j=1}^n \log (p_j) (1 - x_j)\\
= \sum_{k=2}^{n} y_k + \sum_{j=1}^n \log (p_j) (1 - x_j) $$

where
$$ y_k \geq \log \left( \beta \left(\sum_{j=1}^{k-1} x_j,x_k \right) \right) \quad \mbox{ for } k=2,\ldots,n. $$

In summary, the DCP representation costs one linear expression plus $n-1$ convex inequalities representable by this atom:
$$\log \left( \beta \left( x_1,x_2 \right) \right)$$

This atom is convex by logarithmic convexity of the bivariate beta function as shown, e.g., in this paper:

D. S. Dragomir, R.P. Agarwal, and N. S. Barnett, Inequalities for Beta and Gamma functions via some classical and new integral inequalities, Journal of Inequalities and Applications,5(2000), no. 2, 103–165.

Unfortunately, this atom is neither available in cvxpy nor treated in the MOSEK cookbook, so what remains of you is to device an approximation of it. You can find inspiration, e.g., in the approximation for $\log \left( \Gamma (x) \right)$ found in the cvxpy issue tracker.


问:
Example of G-invariant ideal
Definition:
Let $G$ be a group acting on $R_n:=K[x_1,\dots,x_n]$ with
$$\begin{aligned}
G\times R_n &\rightarrow R_n\\
(g,f) &\mapsto f^g
\end{aligned}
$$
where $(f^g)$ acts on a point $p$ in the follow way:
$$(f^g)(p)=f(g\cdot p)$$
Let $I\subset R_n$ an ideal. $I$ is a G-invariant ideal if and only if for all $g \in G$ and for all $f \in I$, then $f^g \in I$.

Goals: My aim is to look for an example of such an ideal, for $R_n$ and $G$ fixed.

The most immediate thing that came to my mind is to take $G:=S_n$.
But then which ideal of $R_n$ can I take to be $G$-invariant?

答:
Consider che polynomials of the form

$h_k(x_1, \dots, x_n)= \sum_{i_1\dots i_k \in [n]} x_{i_1}\dots x_{i_k}$

It is easy to observe that these polynomials are $S_n$ invariants (the action on $h_k$ comes from the action of $S_n$ on the sets of the form $\{i_1\dots i_k\} $)

The ideal generated by the $h_k$ for $k \in [n]$ is $S_n$ invariant.

Title: Example of G-invariant ideal


问:
Zeroes of homogeneous polynomial on projective curve
Given a curve in projective $n$-space over an algebraically closed field and a homogeneous polynomial of degree $\geq 1$. Does it then always have a zero on the curve?

答:
Let $C$ be the given curve, and $f$ the homogeneous polynomial, and consider the hypersurface $V = \{f = 0\}$. The general theorem about intersections of projective varieties is Bezout's theorem. But there are weaker statements that suffice in this situation. The following can be found in Hartshorne, Algebraic geometry, Theorem I 7.2:

Theorem Let $Y, Z$ be varieties of dimensions $r,s$ in $\mathbb P^n$. If $r + s - n \geq 0$, then the intersection $Y \cap Z$ is nonempty.

In our situation, $r = \dim C = 1$ and $s = \dim V = n-1$, so $r + s - n = 0$, and hence $C \cap V$ is nonempty.


问:
how to find probability of intersection
Given that events S and T such that $P(S)=\dfrac{1}{5}$, $P(T)=\dfrac{1}{4}$, and $P[(S \cap T') \cup (S' \cap T) ]=\dfrac{1}{6}$.

i) Find $P[(S \cap T)$ and determine if the events S and T are independent.

ii) Find $P[(S \cap T') / S ]$.

Part (i), I am trying to find the probability of $P(S \cap T)$,

$P[(S \cap T') \cup (S' \cap T) ]=\dfrac{1}{6}$

$P(S \cap T')+P(S' \cap T)-P[(S \cap T') \cap (S' \cap T) ]=\dfrac{1}{6}$

$P(S)-P(S \cap T)+P(T)-P(S \cap T)-P[(S \cap T') \cap (S' \cap T) ]=\dfrac{1}{6}$

$P(S)+P(T)-2P(S \cap T)-P[(S \cap T') \cap (S' \cap T) ]=\dfrac{1}{6}$

$P(S)+P(T)-2P(S \cap T)-P[(S \cap T') \cap (S' \cap T) ]=\dfrac{1}{6}$

$\dfrac{1}{5}+\dfrac{1}{4}-2P(S \cap T)-P[(S \cap T') \cap (S' \cap T) ]=\dfrac{1}{6}$

$P(S \cap T)=\left(\dfrac{17}{60}-P[(S \cap T') \cap (S' \cap T)]\right)\dfrac{1}{2}   $

I am not sure how to get the value of $P[(S \cap T') \cap (S' \cap T)]$ and how to check if S and T are independent or not, can someone help me?

For part (ii), what is "/S" inside $P[(S \cap T') / S ]$?

Thank you in advance.

答:
Set $(S\cap T^{\complement})\cup(S^{\complement}\cap T)$ is usually denoted as $S\Delta T$.

Observe that it is a union of disjoint sets so that:
$$P(S\Delta T)=P(S\cap T^{\complement})+P(S^{\complement}\cap T)$$

Make a Venn-diagram and set:

$a:=P(S\cap T^{\complement})$

$b:=P(S^{\complement}\cap T)$

$c:=P(S\cap T)$

Then we have the equalities:

$a+c=\frac15$

$b+c=\frac14$

$a+b=\frac16$

This can easily be solved and results in:

$a=\frac7{120}$

$b=\frac{13}{120}$

$c=\frac{17}{120}$

I leave the rest to you.

(ii) Probably you are asked to find: $$P(S\cap T^{\complement}\mid S)$$denoting the probability that event $S\cap T^{\complement}$ under the condition that event $S$ occurs.

You can use the general rule:$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$$

Title: how to find probability of intersection


问:
Complex analysis: The taylor series for $\frac{z\cos z}{z-2}$ around $z=0$
I'm stucking trying to solve the following problem:

Given a function $f$ given as $f(z)=\frac{z\cdot \cos z}{z-2}$ I am asked to show that the Taylor series for $f$ around $0$ is given as
$$f(z)=\sum_{k=0}^{\infty}a_kz^k $$
where $a_0=0$ and
$$a_{2k}=a_{2_k-1}/2 $$
$$ a_{2k+1}=\frac{\left(a_{2k}+\frac{(-1)^{k+1}}{(2k)!}\right)}{2}$$

My work so far:

Given
$$f(z)=\frac{z\cdot \cos z}{z-2}=\sum_{k=0}^{\infty}a_kz^k$$
it must be the case that
\begin{alignat*}{2}
z\cdot \cos z&=\sum_{k=0}^{\infty}a_kz^k(z-2)\\
&= \sum_{k=0}^{\infty}-2a_kz^k+a_kz^{k+1} \\
&= \sum_{k=0}^{\infty}-2a_kz^k+a_{k-1}z^{k} \\
&= \sum_{k=0}^{\infty}(-2a_k+a_{k-1})z^k \\
&= \sum_{k=0}^{\infty}(-a_k+\frac{a_{k-1}}{2})z^k
\end{alignat*}

From here I'm not sure how to proceed. I know that $z\cdot \cos z=z\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}z^{2k}=\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}z^{2k+1}$, but I'm not sure how this connects with what I've done so far.

Any tips are appreciated.

Body:
Hint:
$$
\frac{1}{z-2} = \sum_{n=0}^\infty \frac{-1}{2^{n+1}}z^n \quad\text{for } |z|<2
$$


答:
Observe that the function has a simple pole at $z=2$, so the series will converge for $|z| < 2$, in which region we have

$$\frac{1}{z-2} = -\frac{1}{2} \frac{1}{1-\tfrac{z}{2}}=- \sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}}$$

We also, as you point out, have

$$z \cos(z) = z \sum_{n=0}^{\infty} (-1)^n \frac{z^{2n}}{(2n)!} = \sum_{n=0}^{\infty} (-1)^n \frac{z^{2n+1}}{(2n)!}.$$

So the Taylor series will be the product

$$f(z) = \bigg( - \sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}} \bigg) \bigg( \sum_{n=0}^{\infty} (-1)^n \frac{z^{2n+1}}{(2n)!} \bigg).$$

Now, from this point, you might be able to find an expression for the $n$-th term in this series using the Cauchy product formula, or you could just write out the first few terms is that is satisfactory.

Title: Complex analysis: The taylor series for $\frac{z\cos z}{z-2}$ around $z=0$

Body:
I'm stucking trying to solve the following problem:

Given a function $f$ given as $f(z)=\frac{z\cdot \cos z}{z-2}$ I am asked to show that the Taylor series for $f$ around $0$ is given as
$$f(z)=\sum_{k=0}^{\infty}a_kz^k $$
where $a_0=0$ and
$$a_{2k}=a_{2_k-1}/2 $$
$$ a_{2k+1}=\frac{\left(a_{2k}+\frac{(-1)^{k+1}}{(2k)!}\right)}{2}$$

My work so far:

Given
$$f(z)=\frac{z\cdot \cos z}{z-2}=\sum_{k=0}^{\infty}a_kz^k$$
it must be the case that
\begin{alignat*}{2}
z\cdot \cos z&=\sum_{k=0}^{\infty}a_kz^k(z-2)\\
&= \sum_{k=0}^{\infty}-2a_kz^k+a_kz^{k+1} \\
&= \sum_{k=0}^{\infty}-2a_kz^k+a_{k-1}z^{k} \\
&= \sum_{k=0}^{\infty}(-2a_k+a_{k-1})z^k \\
&= \sum_{k=0}^{\infty}(-a_k+\frac{a_{k-1}}{2})z^k
\end{alignat*}

From here I'm not sure how to proceed. I know that $z\cdot \cos z=z\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}z^{2k}=\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}z^{2k+1}$, but I'm not sure how this connects with what I've done so far.

Any tips are appreciated.


答:
Hint

think about the terms mod 3.

2n +1 can only be congruent to 2 mod 3. why?

now if it's congruent to 2 mod 3, what are the other two terms congruent to mod 3?

Title: If $2n+1$ and $4n+3$ are prime, then $2n-1$ and $4n+1$ are not when $n>2$

Body: How do you prove that, for $n>2$, if $2n+1$ and $4n+3$ are prime numbers, then $2n-1$ and $4n+1$ are composite numbers?</p>


问:
Using row reduction to solve for a variable in a matrix
With given column vectors x = [1,3,2], y = [3,1,2], and z = [7,a,6], I need to use row reduction to find the value of a such that the vectors {x, y, z} are linearly dependent.

I am seeking to understand how to find this variable via row reduction. I reduced the matrix entirely to RREF when I remembered that I was supposed to be solving for a. Could I get some assistance?

答: Hint: A set of vectors $v_1, ..., v_n$ is linearly independent if the $c_1v_1 + ... + c_nv_n = 0 \implies c_i = 0$ for $i = 1, ..., n$. When row-reducing a matrix, you are essentially solving a system of equations like this and trying to show that there is only one solution s.t. all the unknowns of the system are $0$. If you pay close attention, you'll notice that a system has only one solution when the coefficient matrix is the identity i.e. if the system of equations being solved has one solution that is $0$ in every entry. Thus, a matrix $A \in R^{nxn}$ has linearly independent columns if RREF($A$) = $I_{nxn}$, where $I_{nxn}$ is the identity matrix.


问:
Finding the automorphism of upper half plane given three distinct points

Suppose $(x_1,x_2,x_3)$ and $(y_1,y_2,y_3)$ are two pairs of three
distinct points on the real axis with $x_2<x_2<x_3$ and $y_1<y_2<y_3$.
Prove that there exists (a unique) automorphism $\phi$ of $\mathbb{H}$
so that $\phi(x_j)=$

I already know the uniqueness part, and want to prove the existence.

I know that $Aut(\mathbb{H})=\{\phi:\phi(z)=\frac{az+b}{cz+d}:a,b,c,d\in\mathbb(R), ad-bc>0\}$. Automorphisms of the upper half plane suggests me to "scale and conjugate" $\frac{(z-x_1)(x_2-x_3)}{(z-x_3)(x_2-x_1)}$ but I don't seem to understand how this works out. I tried $(y_2-y_1)\frac{(z-x_1)(x_2-x_3)}{(x_2-x_1)(z-x_3)}+y_1$ so that it sends $x_1$ to $y_1$ and $x_2$ to $y_2$ but I don't know why I initially send $x_3$ to $\infty$. Could someone help me with this question? Thank you.

答:
The idea is to show that

$T(z; x_1, x_2, x_3) = \dfrac{(z-x_1)(x_2-x_3)}{(z-x_3)(x_2-x_1)}$ is an automorphism of $\Bbb H$ which maps $x_1, x_2, x_3$ to $0, 1, \infty$, respectively, and

the automorphisms of $\Bbb H$ form a group.

Therefore
$$
    T(z; y_1, y_2, y_3)^{-1} \circ T(z; x_1, x_2, x_3)
$$
is an automorphism of $\Bbb H$ with the desired property.

Remark: $\dfrac{(z-x_1)(x_2-x_3)}{(z-x_3)(x_2-x_1)}$ is, up to some permutation of the arguments, the so-called “cross-ratio” or “double ratio“ of $z, x_1, x_2, x_3$.


问:
Show that an abelian group cannot have exactly 4 elements of order 2.
This question is from the book Abstract Algebra by Gregory Lee.

What I got so far is:

Abelian group is commutative.

The answer key says suppose $a$ and $b$ are elements of this group $G$ such that they are distinct (a.k.a. $a\neq b$) and $|a|=|b|=2$

Please help me to show this.

Thanks.

答:
Here is an elementary argument.

If $a$ and $b$ are two elements of order $2$, then there are at least three elements of order $2$: $a,b,ab$.

If $c$ is a fourth element of order $2$, then there are at least seven elements of order $2$: $a,b,c,ab,ac,bc,abc$.


问:
Partial Derivatives : Given $f(x) = Ax^3 + By^3 - Cx - Dy + E$
Given $f(x) = Ax^3 + By^3 - Cx - Dy + E$

Propose any value for $A, B, C, D$ and $E$ so that these will give three(3) critical points.

Please help. My choice of value only gives me two(2) critical point therefore it is not accurate.

I randomly pick value for $A, B, C, D$ and $E$, then I differentiate the equation.

I find $\frac{\partial}{\partial x} (Ax^3 + By^3 - Cx - Dy + E )$ and
$\frac{\partial}{\partial y}(Ax^3 + By^3 - Cx - Dy + E )$.

Both equation that I differentiated is then equal = $0$

Then I determine the value of $x$ and $y$ which will give me the critical points

p/s: Very sorry I did not know how to use the math format

答:
To calculate the critical points of the function
$$f(x,y) = Ax^3 + By^3 - Cx - Dy + E$$
you have to find appropriate values $A,\ldots,F$ where
$$\frac \partial {\partial x}f=0$$
$$\frac \partial {\partial y}f=0$$
, so
$$3Ax^2-C=0$$
$$3By^2-D=0$$
The solutions of these quadradic equations are independent of each other. The first quadratic equation has two real solutions if the C is not $0$ and $A$  has the same sign as $C$. If $A=C=0$ we have infinitely many solutions.
If  $x_1$ is a solution of the first euation and $y_1$ is a solution of the second one, then $(x_1,y_1)$ is a solution of both equations.

So if we choose $A=B=1$, $C=D=3$, $E=1$ then the 4 critical points are $(1,1),(1,-1),(-1,1),(-1,-1)$


答:
By the Inclusion-Exclusion Principle,
$$\Pr(S \cup T) = \Pr(S) + \Pr(T) - \Pr(S \cap T)$$
Since $S \cup T$ is the union of the disjoint sets $S \cap T'$, $S \cap T$, and $S' \cap T$,
$$\Pr(S \cup T) = \Pr(S \cap T') + \Pr(S \cap T) + \Pr(S' \cap T)$$
Since the sets $S' \cap T$ and $S \cap T'$ are disjoint,
$$\Pr[(S \cap T') \cup (S' \cap T)] = \Pr(S \cap T') + \Pr(S' \cap T)$$
Equating the two results for $\Pr(S \cup T)$ yields
$$\Pr(S) + \Pr(T) - \Pr(S \cap T) = \Pr(S \cap T') + \Pr(S \cap T) + \Pr(S' \cap T)$$
Solving for $\Pr(S \cap T)$ yields
$$\Pr(S \cap T) = \frac{1}{2}[\Pr(S) + \Pr(T) - \Pr(S \cap T)' - \Pr(S' \cap T)]$$

Since you know $\Pr(S)$, $\Pr(T)$, and $\Pr[(S \cap T') \cup (S' \cap T)]$, you can substitute for these values in equation to find $\Pr(S \cap T)$.

Events $S$ and $T$ are independent if $\Pr(S \cap T) = \Pr(S)\Pr(T)$.

Title: how to find probability of intersection

Body:
Given that events S and T such that $P(S)=\dfrac{1}{5}$, $P(T)=\dfrac{1}{4}$, and $P[(S \cap T') \cup (S' \cap T) ]=\dfrac{1}{6}$.

i) Find $P[(S \cap T)$ and determine if the events S and T are independent.

ii) Find $P[(S \cap T') / S ]$.

Part (i), I am trying to find the probability of $P(S \cap T)$,

$P[(S \cap T') \cup (S' \cap T) ]=\dfrac{1}{6}$

$P(S \cap T')+P(S' \cap T)-P[(S \cap T') \cap (S' \cap T) ]=\dfrac{1}{6}$

$P(S)-P(S \cap T)+P(T)-P(S \cap T)-P[(S \cap T') \cap (S' \cap T) ]=\dfrac{1}{6}$

$P(S)+P(T)-2P(S \cap T)-P[(S \cap T') \cap (S' \cap T) ]=\dfrac{1}{6}$

$P(S)+P(T)-2P(S \cap T)-P[(S \cap T') \cap (S' \cap T) ]=\dfrac{1}{6}$

$\dfrac{1}{5}+\dfrac{1}{4}-2P(S \cap T)-P[(S \cap T') \cap (S' \cap T) ]=\dfrac{1}{6}$

$P(S \cap T)=\left(\dfrac{17}{60}-P[(S \cap T') \cap (S' \cap T)]\right)\dfrac{1}{2}   $

I am not sure how to get the value of $P[(S \cap T') \cap (S' \cap T)]$ and how to check if S and T are independent or not, can someone help me?

For part (ii), what is "/S" inside $P[(S \cap T') / S ]$?

Thank you in advance.


问:
Necessary and sufficient condition for weak convergence of gamma distribution
Let $(X_n)_n$ be a sequence of random variable such that $f_{X_n}(x)=\frac{1}{\Gamma(\alpha_n)}\lambda_n^{\alpha_n}x^{\alpha_n-1}e^{-\lambda_nx}1_{]0,+\infty[}(x),$ where $\alpha_n>0,\lambda_n>0.$

Suppose that $\alpha_n=1,\forall n \in \mathbb{N}.$ Find a necessary and sufficient condition on $(\lambda_n)_n$ such that $(X_n)_n$ converges in distribution.

More generally, find a necessary and sufficient condition on $\alpha_n,\lambda_n$ so that $(X_n)_n$ converges in distribution.

The first part is easy, it converges in distribution if and only if $0<\liminf_n\lambda_n=\limsup_n\lambda_n$.

Concerning part 2), is it true that a condition of weak convergence is the convergence of $(\alpha_n)$ and $(\lambda_n)$?

答:
In this comment I will give a full answer to the question about convergence of $X_n$.

We will say that the sequence $(\alpha_n, \lambda_n)$ is "nice" if it has the next property.
If for some subsequence $\alpha_{n_k}$ we have $\alpha_{n_k} \to c \in [0, \infty]$ then:

if $c=0$ then $\liminf_{k} \lambda_{n_k}^{\alpha_{n_k}} \ge 1$

if $c>0$ then $\lim_{k} \frac{\alpha_{n_k}}{\lambda_{n_k}}=0$.

Theorem 3:
If $X_n \overset{d}{\to}  X$ then $X$ may be $constant \ge 0$ or $X \sim \Gamma(\lambda, \alpha)$.

$$\Bigl( X_n \overset{d}{\to} 0 \Bigr) \Longleftrightarrow  \Bigl( (\alpha_n, \lambda_n) \text{ is "nice"} \Bigr).$$

$$\Bigl(X_n \overset{d}{\to}  X = C = const>0 \Bigr) \Longleftrightarrow \Bigl( \alpha_n \to \infty \text{ and } \frac{\alpha_n}{\lambda_n} \to C \in (0, \infty)  \Bigr)$$

\begin{gather}
\Bigl(X_n \overset{d}{\to}  X  \sim \Gamma(\lambda, \alpha) \Bigr) \Longleftrightarrow
\Bigl( \lim_n \alpha_n = \alpha \in (0, \infty) \text{ and } \lim_n \alpha_n = \lambda \in (0, \infty) \Bigr)\nonumber
\end{gather}

Corollary 1.

$X_n$ is convergent iff one of the following conditions holds true:

$(\alpha_n, \lambda_n)$ is "nice"

$\Bigl( \alpha_n \to \infty \text{ and } \frac{\alpha_n}{\lambda_n} \to C \in (0, \infty)  \Bigr)$

$\lim_n \alpha_n = \alpha \in (0, \infty) \text{ and } \lim_n \alpha_n = \lambda \in (0, \infty)$.

Proof.

At first let us present results of my $2$ previous comments in a concise way.

If $\alpha_n \ge c > 0$ and $\lambda_n \to 0$ then $X_n$ diverges.

If $\alpha_n \to 0$ and $\lambda_n \ge c > 0$ then $X_n \overset{w}{\to} 0$.

If $\alpha_n \to 0$ and $\lambda_n \to 0$ then

$$X_n \overset{w}{\to} \Longleftrightarrow X_n \overset{w}{\to} 0 \Longleftrightarrow \Bigl(\lim_{n \to \infty} \alpha_n \ln \lambda_n = 0.\Bigr)$$

If for some $0<c_1 \le c_2 < \infty$ we have $\alpha_n, \lambda_n  \in [c_1, c_2]$  then

$ \Bigl( X_n \overset{w}{\to} \Bigr) \Longleftrightarrow \Bigl(\exists \lim_n \lambda_n$ and  $\exists \lim_n \alpha_n \Bigr)$. In this case $X_n \overset{w}{\to} \Gamma(\lim_n \lambda_n, \lim_n \alpha_n)$.

If $\alpha_n \le c_2$ and $\lambda_n \to \infty$ then $X_n  \overset{w}{\to} 0$.

If $\alpha_n \to \infty$ and $\lambda_n  \le c_2$ then $X_n$ diverges.

If  $\alpha_n \to \infty$ and $\lambda_n \to \infty$ then
$$X_n  \overset{w}{\to} \Longleftrightarrow \exists \lim_n \frac{\alpha_n}{\lambda_n},$$
and if $X_n$ is convergent then $X_n  \overset{w}{\to} \lim_n \frac{\alpha_n}{\lambda_n}$.

It's easy to see that if $X_n \overset{d}{\to}  X$ then $X$ may be $constant \ge 0$ or $X \sim \Gamma(\lambda, \alpha)$.

Moreover, it follows immediately that
$$\Bigl(X_n \overset{d}{\to}  X = C = const>0 \Bigr) \Longleftrightarrow \Bigl( \alpha_n \to \infty \text{ and } \frac{\alpha_n}{\lambda_n} \to C  \Bigr)$$
and
\begin{gather}
\Bigl(X_n \overset{d}{\to}  X  \sim \Gamma(\lambda, \alpha) \Bigr) \Longleftrightarrow
\Bigl( \lim_n \alpha_n = \alpha \in (0, \infty) \text{ and } \lim_n \alpha_n = \lambda \in (0, \infty) \Bigr)\nonumber
\end{gather}
These conditions were mentioned here by JGWang.

The most interesting case is case when $X_n \overset{d}{\to}  0$.

We got convergence $X_n \overset{d}{\to}  0$ only in these cases:

$\alpha_n \to 0$ and $\lambda_n \ge c > 0$,

$\alpha_n \to 0$, $\lambda_n \to 0$ and $\lim_{n \to \infty} \alpha_n \ln \lambda_n = 0$,

$\alpha_n \le c_2$, $\lambda_n \to \infty$,

$\alpha_n \to \infty$, $\lambda_n \to \infty$ and $\lim_n \frac{\alpha_n}{\lambda_n}=0$.

These cases can be summarized as follows:

$\alpha_n \to 0$ and $\Bigl( \limsup_n (-\alpha_n \ln \lambda_n) \le 0 \Longleftrightarrow \liminf_n \lambda_n^{\alpha_n} \ge 1 \Bigr)$

$0 \le c_1 \le \alpha \le c_2 < \infty$ and $\lambda_n \to \infty$

$\alpha_n \to \infty$ and $\lim_n \frac{\alpha_n}{\lambda_n}=0$.

Hence $X_n \overset{d}{\to} 0$ iff $(\alpha_n, \lambda_n)$ is "nice", q.e.d.


问:
If $f(0) = g(0)$ and f is not a constant function, show that $f(x)\neq g(x)$ for any other value of $x$
If $$f = 2\arctan(e^{x+C}),\quad g = 2\arctan(e^{({\sqrt\pi \operatorname{erf} (x)}/{2})+C)})$$

and $f(0) = g(0)$, and $f$ is not a constant function, how do I show that $f$ does not equal $g$ for any other value of $x$?

答:
Tou are considering the two functions
$$f(x)=2 \tan ^{-1}\left(e^{x+c}\right)\qquad \text{and} \qquad g(x)=2 \tan ^{-1}\left(e^{\sqrt{\pi }\, \text{erf}(x)+c}\right)$$
There are always three roots to the equation $f(x)=g(x)$ and one is trivial $x=0$.

What happens is that
$$f'(0)= \frac{2 e^c}{e^{2 c}+1} \qquad \text{and} \qquad g'(0)=\frac{4 e^c}{e^{2 c}+1}=2f'(0)$$ So, at for small values of $x$, $g(x)$ increases twice faster than $f(x)$.

But, the asymptotic value of $f(x)$ is a constant $\pi$ while it is systematically lower the $\pi$ for $g(x)$.

Comparing the arguments, in practice, we look for the zero's of function
$$h(x)=x-\sqrt{\pi }\, \text{erf}(x)$$ which does not make any trouble using Newton method, the solution being $x=\pm 1.74871$

If you do not want to use numerical methods, you could make some approximation. The function $h'(x)$ cancels at $x=\pm \sqrt{\log (2)}$. Making a second order Taylor expansion around this value $x_0=\sqrt{\log (2)}$ and solving the quadratic
$$x_1=x_0+\sqrt{-2\frac {h(x_0)}{h''(x_0)}}$$ that is to say
$$x_1=\sqrt{\log (2)}+\sqrt{\sqrt{\frac{\pi }{\log (2)}} \text{erf}\left(\sqrt{\log (2)}\right)-1} \sim 1.61999$$

We could also make the infinte series
$$\frac {h(x)}x=-1+2\sum_{n=1}^\infty (-1)^n\frac{  x^{2 n}}{(2 n+1)\, n!}$$ and, truncating it to order $O(x^{2(p+1)})$ use series reversion.

As a function of $p$, the numerical values of the estimate will be in the form
$$x_{(p)}=\frac {a_p} {b_p}\sqrt{\frac{3}{2}}$$ Numerically, the $x_{p}$ generate the sequence
$$\{1.22474,1.50031,1.61890,1.67850,1.71030,1.72769,1.73729,1.74259,1.74549\}$$

If you use Newton mathod starting with $x_1$, the iterates will be
$$\left(
\begin{array}{cc}
 n & \text{iterate} \\
 0 & 1.619987725 \\
 1 & 1.752776245 \\
 2 & 1.748711937 \\
 3 & 1.748708965
\end{array}
\right)$$

Just for the fun, the solution is close to $$\frac{3+\sqrt[3]{3}}{2+\cos (1)}\sim 1.748709026$$


问:
Orbit of a subgroup
$G$ is a group acting on $S$. $H$ is a subgroup of $G$ and let $s$ be anything in $S$. Is it true that the size of the orbit of $s$ under action of $G$ is divisible by the size of the orbit of $s$ under action of $H$?

答:
If G is the symmetric group on n letters, then the orbit of s has exactly n elements. Saying that H ≤ G means that H is a permutation group on n letters.  Does every permutation group on n letters have orbits that are divisors of n?

Perhaps check n = 3 to start.  A similar investigation n = 5 should also be instructive.

Title: Orbit of a subgroup


问:
Homology of $S^1 \times (S^1 \vee S^1)$
I'm trying to solve question 2.2.9(b) in Hatcher's Algebraic Topology.

Question: Calculate the homology groups of $X=S^1 \times (S^1 \vee S^1)$.

My attempt: I try to use the Mayer-Vietoris sequence.

Let $A=S^1 \times (S^1 \vee \text{small bit}) =S^1\times S^1$ and $B=S^1 \times ( \text{small bit} \vee S^1)=S^1 \times S^1$. (Unsure how to express this, but hopefully this is clear enough.)

Then $A\cap B=S^1\times \{\text{point}\}=S^1$ and $A\cup B=X$.

The Mayer-Vietoris sequence then gives us an exact sequence in reduced homology as follows:

$$0 \to \mathbb{Z}^2 \to \tilde{H}_2(X) \to \mathbb{Z} \to \mathbb{Z}^2 \to \tilde{H}_1(X) \to 0.$$

Is it possible from this exact sequence to determine the homology groups, in some algebraic fashion? Or do I need to determine what the maps are? How do I determine the maps?

答:
Firstly, I believe that $H_1(\mathrm{Torus})=\mathbb{Z}^2$ hence the sequence should be
$$0 \xrightarrow 0 \mathbb{Z}^2 \xrightarrow f \tilde{H}_2(X) \xrightarrow g \mathbb{Z} \xrightarrow h \mathbb{Z}^2 \oplus \mathbb{Z}^2 \to \tilde{H}_1(X) \to 0.$$
Now, you need to determine at least one map to solve this. In Mayer Veatoris sequence, f is $(t,s) \mapsto i_*t+i_*s$, g is $t\mapsto \partial t$, and h is $t \mapsto (i_*t,-i_*t)$ where all of my $i_*$'s are the homomorphims related to the appropriate canonical inclusions. (At this point you should try solving the question yourself and check the answer for feedback).

We'll start by using the exactness of the sequence: $\operatorname{im}0=0=\operatorname{ker}f$ means $f$ is injective, $\operatorname{im}f=\operatorname{ker}g$ tells us nothing yet, and $\operatorname{im}g=\operatorname{ker}h$.

It will be most natural to determine $h$, since its the only one whose both domain and range are known. So, $h$ takes a generator of the circle, embeds it in each of the tori, and the embedded circle inside each of the tori we know to be a generator of one of his $\mathbb{Z}$'s. ($H_1(T)=\mathbb{Z}^2$, a $\mathbb{Z}$ for the "horizontal" circle and a $\mathbb{Z}$ for the "vertical" circle, in our case, the generator of the original $\mathbb{Z}$ is being mapped to the horizontal circle, that is if you draw the torus "horizontally" as you normally would, and draw $X$ as two tori ("tires") stacked on top of each other).

So we can write $h(1)=((1,0),(1,0))$ and so $\operatorname{im}g=\operatorname{ker}h=0$, hence $g=0$, so $\operatorname{im}f=\operatorname{ker}0=$ entire group, meaning $f$ is surjective and hence an isomorphism. So $H_2(X)=\mathbb{Z}^2$. $H_1$ could be obtained similarly or by using $H_1=\operatorname{Ab}(\pi _1)$.

Title: Homology of $S^1 \times (S^1 \vee S^1)$


问:
Find the derivative of the following
Can someone help me solve this?. I have hard time understanding the lesson, and our teacher will give us a quiz next meeting. She leave this as our exercise. I just want to see the answer and solution. I can't comprehend of what she had been saying and of her given examples.

Find the derivative of
 $$
f(x)= x^2+x+2
$$

答:
For $f(x) = x^n$ for $n \in \mathbb{R}$ then the derivative can be written

\begin{align}
\frac{d}{dx} f(x) &= \frac{d}{dx}(x^n) \\
&=n x^{n-1}
\end{align}

Constants, like $2$ can be written as $2x^0$, and the above can still be implemented.

Title: Find the derivative of the following


Title: Using row reduction to solve for a variable in a matrix

Body:
With given column vectors x = [1,3,2], y = [3,1,2], and z = [7,a,6], I need to use row reduction to find the value of a such that the vectors {x, y, z} are linearly dependent.

I am seeking to understand how to find this variable via row reduction. I reduced the matrix entirely to RREF when I remembered that I was supposed to be solving for a. Could I get some assistance?

Body: $\begin{pmatrix} 1 & 3& 2\\3 & 1& 2\\7 & a& 6 \end{pmatrix}\rightarrow \begin{pmatrix} 1 & 3& 2\\0 & -8& -4\\0 & a-21& -8 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 3& 2\\0 & -8& -4\\0 & a-21+16& -8+4*2 \end{pmatrix} $, so $a=5$


问:
How to understand the sequential characterization of continuity
I'm studying Real Analysis and struggle a lot on how to interrupt the definitions. I'm on the sequential characterization of continuity and cannot comprehend what the statement in my textbook actually means. I am posting a photo because I'm still unfamiliar with how to write limits on math stack

Could someone offer a worded interpretation on what this definition is trying to tell me? What properties it is representing? I also need to prove the sequential characterization of continuity but first thing is understanding what the definition is trying to tell me.

答: It says: if the sequence $x_n$ gets arbitrarily close to $x$ and stays arbitrarily close to it- in other words, then $f(x_n)$ gets arbitrarily close to $f(x)$ (and stays there). Which is what the $\varepsilon-\delta$-definition says, too, really, but using distance measurements instead of convergence. It's the whole idea of continuity: if the inputs of a function get close, then so do the outputs.


问:
Subfield of cyclotomic extension of degree 2 is either $\Bbb Q(\sqrt{p})$ or $\Bbb Q(\sqrt{-p})$.
From my previous post, I conlcude that 'If $\zeta$ is a primitive $p$th root of 1 ($p\geq 3$), then there is a unique subfield $K\subset\Bbb Q(\zeta)$ such that $[K:\Bbb Q] = 2$'. I want to find the field $K$ explicitly. I hear that such $K = \Bbb Q(\sqrt{p})$ or $\Bbb Q(\sqrt{-p})$. Why is that?

I also heard that this is the special case of 'Kroneker-Weber' theorem.

答:
The cyclotomic extension $E := \mathbb{Q}(\zeta)$ of $\mathbb{Q}$ is Galois with Galois group isomorphic to $(\mathbb{Z}/p\mathbb{Z})^{\times} \cong \mathbb{Z}/(p-1)\mathbb{Z}$. Since $p \geq 3$, $p-1$ is odd, and so $\mathbb{Z}/(p-1)\mathbb{Z}$ contains a (unique) subgroup of index $2$. If we call this subgroup $H$, then the fixed field $K := \mathbb{Q}(\zeta)^{H}$ is of degree $2$ over $\mathbb{Q}$ by Galois theory. We want to show that $K = \mathbb{Q}(\sqrt{p})$ or $K = \mathbb{Q}(\sqrt{-p})$.

Suppose that $K = \mathbb{Q}(\sqrt{D})$. Recall that the discriminant of $E$ over $\mathbb{Q}$ is (up to some sign) $p^{p-1}$. Since the discriminant of $K$ divides the discriminant of $E$, we conclude that the only prime divisor of the discriminant of $K$ is $p$. But the discriminant of $K$ is known: it is $D$ if $D \equiv 1 \pmod 4$, and $4D$ otherwise. Hence, since the only divisor of the discriminant of $K$ is $p$, we must have that $D$ is a power of $p$ or $-p$, depending on which of the two is congruent to $1$ mod $4$. We can ignore any even powers of $p$ in $D$, whence we conclude that $K = \mathbb{Q}(\sqrt{p})$ or $K = \mathbb{Q}(\sqrt{-p})$.


问:
To find the position of the ant after 2020 moves is $(p, q)$,
An ant is moving on the coordinate plane. Initially it was at (6, 0). Each move
of the ant consists of a counter-clockwise rotation of 60◦ about the origin fol-
lowed by a translation of 7 units in the positive x-direction. If the position of
the ant after 2020 moves is $(p, q)$, find the value of $p^2$ + $q^2$

For this problem,I just drew some and tried to figure out how this is going to go... But I have absolutely no idea what is the trajectory of the ant after 2020 moves.
I need help.

答:
Let $e^{i\pi/3}=:\omega$. Then we have to iterate the map
$$z\mapsto T(z):=\omega z+7\ .$$
This $T$ has a fixed point $a:={7\over1-\omega}$, and is in fact a $60^\circ$  rotation of the plane around $a$. We can therefore write
$$T(z)-a=\omega(z-a)\ ,$$
and this implies
$$T^n(z)-a=\omega^n(z-a)\ .$$
As $2020=6\cdot336+4$ and $\omega^4=-\omega$ we obtain the equation
$$T^{2020}(6)-a=-\omega(6-a)\ .$$
Solving this leads to
$$\bigl|T^{2020}(6)\bigr|^2=57\ .$$


问:
Does the syntax $\sum_{n \geq 1}$ mean the same thing as $\sum_{n = 1}^{\infty}$?
Is the syntax $\sum_{n \geq 1}$ equivalent to $\sum_{n = 1}^{\infty}$?
Is it just another way of writing the same thing?

答:
The quick answer is: it depends. It might mean the same thing, but some people would see a difference, which can be important in some situations.

About a given $(u_n)_{n \in \mathbb{N}^*}$, a sequence of real numbers, one can say different things:

the sequence ("of partial sums") $\left(\sum^n_{i=1} u_i\right)_{n \in \mathbb{N}}$ is convergent; one usually says that the series (associated to the sequence) converges;

the value of the series is some real number (i.e. the limit of the above sequence of partial sums);

the sequence is summable and the value of the sum is some real number (to be defined below).

For pedagogical purposes, it might be worthwhile to forbid students to write things whose existence has not been previously shown. For example, some teachers don't like things like

$\lim_{n\to \infty} \frac{\sqrt{n+1}}{\sqrt{n}} = \lim_{n\to \infty} \sqrt{1+\frac{1}{n}} = 1$ so the sequence $(\frac{\sqrt{n+1}}{\sqrt{n}})_{n \in \mathbb{N}}$ converges

because one shouldn't write $\lim_{n\to \infty} \frac{\sqrt{n+1}}{\sqrt{n}} = \cdots$ before having proved that the sequence is convergent.

Something like

the series $\sum^{+\infty}_{n=1} u_n$ is convergent and $\sum^{+\infty}_{n=1} u_n = \frac{\pi^2}{6}$"

would also be weird, since the same notation $\sum^{+\infty}_{n=1} u_n$ is used to denote two different things in the same sentence: the series as an object in itself and its value. In other words, if we replace $\sum^{+\infty}_{n=1} u_n$ by $\frac{\pi^2}{6}$ in the above sentence, we get

the series $\frac{\pi^2}{6}$ is convergent

which is definitely weird.

For this reason, one of my teachers used to prefer the notation $\sum_{n\geq 1}$ to denote the series in itself, and $\sum^{\infty}_{n=1}$ to denote the value of this series, once it has been proved to be convergent. The $\sum^{\infty}_{n=1}$ notation for the value of a series, is, I think, quite universal; but I'm not sure about the notation $\sum_{n\geq 1}$.

The story could end here without further worries if you haven't heard of summable families or Lebesgue integration... But if you have, there is something else that can be denoted in a similar way and could produce another confusion.

Let $(u_n)_{n \in \mathbb{N}^*}$ be a sequence of, let's say, real numbers. The notation $\sum_{n \in \mathbb{N}^*} u_n$ refers to (if it exists) the only real number $S$ such that for every $\epsilon >0$, there exists $J \subset \mathbb{N}^*$ finite such that for every finite subset $J'$ of $\mathbb{N}^*$ containing $J$, we have $\vert \sum_{n \in J'} u_n - S \vert < \epsilon$.

If $\sum_{n \in \mathbb{N}^*} u_n$ exists, the sequence $(u_n)_{n \in \mathbb{N}^*}$ is said to be summable. Like S.K. mentioned it in his/her answer, this is equivalent to say that the function $ u : \mathbb{N}^* \rightarrow \mathbb{R}$ is Lebesgue-integrable, with respect to the counting measure.

One can, in fact, show that a sequence of real numbers is summable if and only the series associated to it is absolutely convergent.

So, for a convergent but not absolutely convergent series, $\sum^{+\infty}_{n=1} u_n$ exists (it is the value of the series) but $\sum_{n \in \mathbb{N}^*} u_n$ does not.

All that said, my advice would be: if you are a student, ask the teacher to tell you exactly what notation choices he/she makes and, when writing an exam, give precisions on what your notations refer to.


问:
Non singular matrices

Assume matrices $A$, $I-A$, $I-A^{-1}$ are all non-singular. Show that $$(I-A)^{-1} + (I-A^{-1})^{-1} = I$$

答:
$$\color{red}{(I-A)}\color{blue}{(I-A^{-1})}=I-A-A^{-1}+I = \color{red}{(I-A)}+\color{blue}{(I-A^{-1})}$$ Hence multiplying with the inverses:

first with $\color{red}{(I-A)^{-1}}$ on the left $$\color{blue}{I-A^{-1}}=I+\color{red}{(I-A)^{-1}}\color{blue}{(I-A^{-1})}$$
then with $\color{blue}{(I-A^{-1})^{-1}}$ on the right
$$I=\color{blue}{(I-A^{-1})^{-1}}+\color{red}{(I-A)^{-1}}$$

Title: Non singular matrices


答:
Your teacher used limits, so I’m assuming that you’re required to use first principles.

The derivative of a function $f(x)$ is defined as $$f’(x)=\lim_{t\to x} \frac{f(t)-f(x)}{t-x} $$ Now just plug in the definition of $f$: $$f’(x) = \lim_{t\to x} \frac{(t^2+t+2)-(x^2+x+2)}{t-x} =\lim_{t\to x} \frac{(t-x)(t+x) +(t-x)}{t-x}=\lim_{t\to x} (t+x+1) = x+x+1 =2x+1$$

Title: Find the derivative of the following

Body:
Can someone help me solve this?. I have hard time understanding the lesson, and our teacher will give us a quiz next meeting. She leave this as our exercise. I just want to see the answer and solution. I can't comprehend of what she had been saying and of her given examples.

Find the derivative of
 $$
f(x)= x^2+x+2
$$


问:
Help Understanding Proof of Replacement Theorem?
Sorry if this is a trivial question.

The book is Linear Algebra Done Right by Axler, page 25-26.

Theorem: In a finite-dimensional vector space, the length of every linearly independent list of vectors is less than or equal to the length of every spanning list of vectors.

Proof: Suppose that $(u_1 ,\ldots, u_m)$ is linearly independent in $V$ and
  that $(w_1,\ldots ,w_n)$ spans V. We need to prove that $m \leq n$. We do so
  through the multistep process described below; note that in each step
  we add one of the $u$'s and remove one of the $w$'s.

Step 1: The list $(w_1,\ldots, w_n)$ spans $V$, and thus adjoining any vector to it
  produces a linearly dependent list. In particular, the list $(u_1,w_1, \ldots,w_n)$ is linearly dependent.

Question:
Why is $(u_1,w_1, \ldots,w_n)$ is linearly dependent?

答:
Here is a proof of the result I mention in the comment I made above which might be helpful to someone who comes across this question.

Let $V$ be a vector space over the field $F$. If $S \subset V$ is linearly independent and $v \in V$ and $v \notin S$ then $S \cup \{v\}$ is linearly dependent if and only if $v \in$ span$(S)$.

Proof:

($\Rightarrow$) If $S \cup \{v\}$ is linearly dependent then there exists finitely many distinct vectors $u_1, \ldots, u_n \in S \cup \{v\}$ and scalars, not all zero, $a_1, \ldots, a_n \in F$ such that

$$a_1u_1 + \dots + a_nu_n = 0.$$

Now one of the $u_i$'s must be equal to $v$, as otherwise we would have a case of linear dependence for $S$, so after re-ordering let $u_1 = v$ and we have

$$a_1v + a_2u_2 + \dots + a_nu_n = 0$$

And solving for $v$:

$$v = (-a_1^{-1}a_2)u_2 + \dots + (-a_1^{-1}a_n)u_n.$$

As $u_2, \ldots, u_n \in S$ we have $v \in$ span$(S)$.

($\Leftarrow$) If $v \in$ span$(S)$ then we can write, for $u_1,\ldots,u_n \in S$ and $a_1,\ldots, a_n \in F$,

$$v = a_1u_1 + \dots + a_nu_n$$

As $v \notin S$ we have,

$$a_1u_1 + \dots + a_nu_n + (-1)v = 0.$$

Therefore, $S \cup \{v\}$ is linearly dependent. $\Box$

Hope this helps someone!


答:
$\sum_{n\geq 1} a_n$ and $\sum_{n=1}^\infty a_n$ coincide if $(a_n)$ is absolutely convergent. The first symbol means "integration w.r.t. the counting measure", i.e. is defined as $$ \sum_{n\geq 1} a_n = \sup_{S \subset \mathbb{N}, |S| < \infty } \sum_{s \in S} \max(a_s,0) - \sup_{S \subset \mathbb{N}, |S| < \infty } \sum_{s \in S} \max(-a_s,0) $$ while the second symbol means
$$ \sum_{n=1}^\infty a_n = \lim_{N\rightarrow\infty} \sum_{n=1}^N a_n. $$

The first symbol is not defined for some series, e.g. $a_n=\frac{(-1)^n}{n}$, because its positive and negative parts may both (!) be separately divergent, so we would get $\infty-\infty$. The advantage here is that the notation can be extended to arbitrary sets, even uncountable ones, e.g. $\sum_{x \in \mathbb{R}} f(x)$ makes sense for some $f$, namely those for which there are either at most countably many negative or positive terms and whose sum converges.

On the other hand, we have $\sum_{n=1}^\infty a_n = -\log(2)$. The important point is that both definitions agree for absolutely convergent sequences. In this case the summation order is immaterial. The first definition has the advantage, that one can use Fubini's theorem to justify the exchange of summation order, etc.

EDIT: The above convention for $\sum_{n\geq 1}$ is non-standard and depends on the field/context in question.

Title: Does the syntax $\sum_{n \geq 1}$ mean the same thing as $\sum_{n = 1}^{\infty}$?

Body:
Is the syntax $\sum_{n \geq 1}$ equivalent to $\sum_{n = 1}^{\infty}$?
Is it just another way of writing the same thing?


问:
Prove if $f$ and $g$ are coprime then $C(fg) \sim C(f) \oplus C(g)$
In my course we recently learned about rational canonical form. The version I learned in lectures wrote the RCF as the direct product of irreducible polynomials e.g. $C(f_1(x)^{a_1}) \oplus \cdots \oplus C(f_k(x)^{a_k})$ where each $f_i$ is irreducible. $C(f(x))$ denoting the companion matrix of $f(x)$. However I've also come across a RCF theorem that takes the product of coprime polynomials as the blocks.

As I understand it the key lemma that connects these two forms is that if $f$ and $g$ are coprime then $C(fg) \sim C(f) \oplus C(g)$ but I can't seem to come up with or find any simple proof of this.

答:
If you have seen the $k[X]$-module point of view: the result you are looking for is just saying that if $f,g$ are coprime, then we have an isomorphism of $k[X]$-module $k[X]/(fg)\simeq k[X]/(f)\times k[X]/(g)$.

If not, let $u$ be the endomorphism of your $k$-vector space $E$, whose is cyclic with characteristic (or minimal, that the same here) polynomial is $fg$.

Since $f,g$ are coprime and $\chi_u=fg$, we have $E=\ker g(u)\oplus \ker(f(u))$.

Let $v$ be a vector which is $u$-cyclic (meaning  $(v,u(v),\ldots, u^{n-1}(v))$ is a $k$-basis of $E$. Then using a Bézout relation $Uf+Vg=1$, you should be able to prove that
$\ker(g(u))=Im((Uf)(u)), \ker(f(u))=Im((Vg)(u))$, and that
$w=(Vg)(u)(v)$ is  a cyclic vector for the restriction of $u$ to $\ker(f(u))$, and $w'=(Uf)(u)(v)$ is  a cyclic vector for the restriction of $u$ to $\ker(g(u))$ (or something like that, I didn't do the computations...but it is the translation of the $k[X]$-module isomorphism above in terms of endomorphisms, so it should work that way).

Of course, you should also convince yourself along the way that the characteristic polynomials over  each subspace are $f$ and $g$ respectively.


问:
asymmetric Dual of the asymmetric Dual is Primal?
It is known that in Linear Programming the Dual of the Dual is the Primal.

In wikipedia I saw that apart from the symmetric Dual I knew, there is also the asymmetric one:
https://en.wikipedia.org/wiki/Dual_linear_program#Vector_formulations

I tried to show that the "Dual of the Dual is the Primal" would still hold in this case but I didn't manage to.

My question is: Does the "Dual of the Dual is the Primal" still holds for the asymmetric case? And if yes, could you please provide a comprehensive proof ?

答:
Okey. I managed.

The asymmetric duality says:

Primal

max ${c^Tx}$

s.t. $Ax ≤ b$

produces the folowing

Dual

min $b^Ty$

s.t. $A^Ty = c$

$y ≥ 0$

I want to prove that I can go back to Primal from the Dual applying only the knowledge of this asymmetric transformation.

First I need to bring the Dual in the 'standard' form as in Primal:

$$
max\; -b^Ty\\
s.t.\quad
\begin{align}
   A^Ty & \le c\\
   -A^Ty & \le -c\\
   -y & \le 0\\
\end{align}\\
$$
$$
\implies
\begin{align}
max\; -b^Ty\\
s.t. \qquad \begin{bmatrix}A^T\\-A^T\\-I\end{bmatrix}
\begin{bmatrix}y\\y\\y\end{bmatrix} =
\begin{bmatrix}c\\-c\\0\end{bmatrix}
\end{align}
$$
Where $I$ is the identity matrix.
Applying the assymetric Dual transformation we get:
$$
min\; \begin{bmatrix}c^T & -c^T & 0\end{bmatrix}w\\
s.t.
\begin{align}
\quad \begin{bmatrix}A & -A & -I\end{bmatrix}w & = -b\\
w & \ge 0
\end{align}\\
$$
let $w = \begin{bmatrix}u & v & k\end{bmatrix}^T$. Then:
$$
min\; c^Tu -c^Tv\\
s.t.
\begin{align}
\quad Au -Av -Ik & = -b\\
u,v,k & \ge 0
\end{align}\\
$$
reforming:
$$
max\; c^T(v - u)\\
s.t.
\begin{align}
\quad A(v - u) & = b - Ik\\
u,v,k & \ge 0
\end{align}\\
$$
let $x = v-u$
$$
max\; c^Tx\\
s.t.
\begin{align}
\quad Ax & = b - Ik\\
k & \ge 0
\end{align}\\
$$
We no longer have guarantee for the sign of $x$.

Since $k \ge 0$ the Right Hand Side will be greater if we omit the last term:
$$
max\; c^Tx\\
s.t.
\begin{align}
\quad Ax \le b \\
\end{align}\\
$$
Which is exactly the Primal Problem.


问:
Generalized Deduction Theorem
In my comment on this question, I mentioned that there was a generalized form of the deduction theorem that looks something like this

For any finite set of formulae, $\Gamma$
$$\dfrac{\Gamma\vdash\psi}{\vdash\bigwedge\Gamma\to\psi}$$

The only problem is, I can't figure out where this came from! I'm almost certain that this is a named theorem that I've seen in a textbook or article somewhere and the fact that I can't find a source is extremely frustrating.

Does anyone know what this theorem is called or have a reference for it? I'm not trying to figure out the proof (which is easy) - I just need to know that I haven't gone mad.

Note:

I think that I saw this somewhere in the context of sequent calculus; the original form might have been

$$\dfrac{\Gamma,\Delta\vdash\psi}{\Gamma\vdash\bigwedge\Delta\to\psi}$$

答:
In the literature on proof systems, given a finite set of formulas $\Delta$, there are no "official" names for theorems of the form

$$\tag{1}\dfrac{\Delta \vdash\psi}{\vdash\bigwedge\Delta\to\psi}$$
and its "generalization"

$$\tag{2}\dfrac{\Gamma,\Delta\vdash\psi}{\Gamma\vdash\bigwedge\Delta\to\psi}$$

other than deduction theorem, because $(1)$ and $(2)$ just are the deduction theorem, written in a slightly different syntactical form.

Indeed, since $\Delta$ is finite, in any reasonable proof system the fact that $\Gamma, \Delta \vdash \psi$ (i.e. $\psi$ is derivable from the sets of formulas $\Gamma$ and $\Delta$) is equivalent to say that $\Gamma, \bigwedge \Delta \vdash \psi$ (i.e. $\psi$ is derivable from the set of formulas $\Gamma$ and the conjunction of the formulas in $\Delta$). By deduction theorem (in its apparently restricted form), from $\Gamma, \bigwedge \Delta \vdash \psi$ it follows that $\Gamma \vdash \bigwedge \Delta \to \psi$.

According to Occam's razor, "entities should not be multiplied without necessity", which is always a good principle. Here, $(1)$ and $(2)$ are essentially the same as deduction theorem, so there is no need to give them another name such as "generalized deduction theorem" or whatever, just because of their slightly different syntactical notation. I would distrust textbooks, articles and notes that call $(1)$ or $(2)$ in a way different from deduction theorem, because they would just complicate something that is actually simpler.

The fact that $(1)$ and $(2)$ are nothing but the deduction theorem, is evident
in the context of sequent calculus. Here, the rules $(1)$ and $(2)$
are derivable, which roughly means that they can be simulated by means of the (more "atomic") inference rules of sequent calculus.
More precisely, the rule $(2)$ is derivable by means of the inference rules
$$\dfrac{\Gamma, \varphi_1, \varphi_2 \vdash \psi}{\Gamma, \varphi_1 \land \varphi_2 \vdash \psi}\land_L
\qquad\qquad
\dfrac{\Gamma, \varphi \vdash \psi}{\Gamma \vdash \varphi \to \psi}\to_R
$$
Indeed, we have

$$
\dfrac
{\dfrac
  {\dfrac{\Gamma,\Delta\vdash\psi}{\vdots}\land_L}
  {\Gamma, \bigwedge\Delta \vdash \psi}
\land_L}
{\Gamma\vdash\bigwedge\Delta \to \psi}\to_R
$$

The rule $\to_R$ internalizes the (apparently "restricted" version of) deduction theorem within the proof system, and $\land_L$ is an inference rule that is valid in any reasonable logic system (as a rule or a meta-rule).
Summing up, to prove that $(2)$ (and hence $(1)$, take $\Gamma = \emptyset$) is derivable in the sequent calculus, we just need a rule that is always valid and the sequent calculus version of deduction theorem. So, $(1)$ and $(2)$ do not express anything more than deduction theorem.

Title: Generalized Deduction Theorem


问:
Prevalence of ergodic rotations on a compact group
So I know that any compact group $G$ carries a Haar probability measure $\mu$. For any $g \in G$, let $T_g : G \to G$ be the rotation $T_g : x \mapsto gx$. In the cases where $G = \mathbb{R}^n / \mathbb{Z}^n$ and $g = (t_1, \ldots, t_n)$, the rotation $T_g$ is ergodic iff $\{ t_1, \ldots, t_n, 1 \}$ is linearly independent over $\mathbb{Q}$.

In the case of toral rotations, I notice that almost every rotation is ergodic, i.e. $T_g$ is ergodic for almost all $g$. Is this typical behavior for group rotations, i.e. that almost every element of the compact group gives rise to an ergodic rotation?

答:
This also fails for nonabelian compact connected nontrivial Lie groups (e.g. $G=SO(3)$): Every $g\in G$ is contained in a compact abelian subgroup (a compact torus) $T$ and the latter does not act ergodically on $G$ because $G/T$ is a manifold which is not a singleton.

Edit. I'll say that a compact Hausdorff topological group $G$ satisfies Property E if almost every (for a Haar measure on $G$) $g\in G$ acts ergodically on $G$ via left multiplication. The class of groups satisfying Property E is very restrictive:

Theorem. $G$ satisfies Property E if and only if $G$ is the inverse limit of a family of compact tori $T^{n_\beta}$, where $n_\beta\in {\mathbb N}\cup \{0\}$ for each $\beta$.

Besides tori, examples of groups satisfying E include (but are not limited to) solenoids and infinite products of circles.

Proof of the theorem (one direction). Suppose that $G$ satisfies Property E and $g\in G$ acts ergodically on $G$. Let $H$ denote the closure in $G$ of the cyclic subgroup $\langle g\rangle$ generated by $g$. Then $H$ is abelian (since it is the closure of an abelian subgroup). If $G$ is nonabelian, $G\ne H$. Let $Hx\subset G$ be a coset different from $H$.
Since $G$ is Hausdorff, there exists $H$-invariant disjoint neighborhoods $U, V$ of $H$ and $Hx$ in $G$. Thus, $H$ does not act ergodically on $G$ and neither does $g$.

We, thus, conclude that $G$ is abelian.

I next claim that $G$ is connected. Let $G_0< G$ be the identity component ; it is a closed subgroup of $G$. The quotient  $Q=G/G_0$ is a totally disconnected Hausdorff compact topological group. Then $1_Q$ has a basis of clopen neighborhoods $V_\alpha$ consisting of subgroups of $Q$. If $Q$ is nontrivial, there exists $V_\alpha\ne Q$. The preimage $U_\alpha$ of $V_\alpha$ in $G$ has positive measure (since it is open). Taking any element $g\in U_\alpha$, we see that $g^nU_\alpha=U_\alpha$ for all $n\in {\mathbb Z}$. At the same time, if $yV_\alpha\subset Q$ is a coset different from $V_\alpha$ and $x\in G$ is an element projecting to $y$, then $xU_\alpha$ has positive measure and  $g^n x U_\alpha\cap U_\alpha=\emptyset$ for all $n$. Thus, $g$ does not act ergodically on $G$, which is a contradiction.

We conclude that $G$ has to be connected. Now, it follows from the classification of (locally) compact connected Hausdorff topological groups that $G$ is isomorphic (as a topological group) to the inverse limit of a sequence of quotient groups $G_\beta$ of $G$ (obtained by dividing $G$ by "small" subgroups of $G$) such that each $G_\beta$ is a Lie group. In our case, each $G_\beta$ is compact, connected and abelian, hence a torus. Thus, $G$ is isomorphic to an inverse limit of a family of tori.  If $G$ were also 1st countable (equivalently, metrizable) we would could say that this family of tori is a sequence.

The following reference contains much more than you need regarding (locally) compact groups:

Montgomery, Deane; Zippin, Leo, Topological transformation groups, Mineola, NY: Dover Publications (ISBN 978-0-486-82449-9). xi, 289 p. (2018). ZBL1418.57024.

I will add a proof for the converse later on...


问:
Proving that $\int_a^\infty e^w \phi_{\mu,\sigma}(w)dw = e^{A+\frac{\sigma^2}{2}} \Phi(\frac{\mu+\sigma^2 - a}{\sigma})$
We have been given this equality in a class, and I would like to prove it.

$$\int_a^\infty e^w \phi_{\mu,\sigma}(w)dw = e^{\mu+\frac{\sigma^2}{2}} \Phi(\frac{\mu+\sigma^2 - a}{\sigma})$$

Where $\phi$ is the normal pdf with mean $\mu$ and variance $\sigma$ and $\Phi$ is the standard normal (0,1) CDF.

The teacher's assistant refused to prove it but said, OK it is so easy, and can be done using first principles and a change in variables. Here is the proof I came up with.

$$e^z f(z) = \frac{1}{\sigma \sqrt{2 \pi}}e^{z-\frac{1}{2}\left(\frac{z - \mu}{\sigma}\right)^2}$$ And define the indefinite integral
(used Mathematica). $$\int \frac{1}{\sigma \sqrt{2 \pi}}e^{z-\frac{1}{2}\left( \frac{z - \mu}{\sigma}\right)^2} dz =
-\frac{1}{2} e ^{(\sigma^2 / 2 + \mu)} \text{Erf} \left(\frac{\sigma^2 + \mu - z }{\sqrt{2} \sigma} \right)$$

The Erf function above is the Gauss error function, defined by
$\text{Erf}(z) = \frac{2}{\sqrt{\pi}}\int_0^z e^{-t^2} dt$. I then use
this to evaluate the definite integral, which uses the fact that the
Erf function has a value of -1 for $-z$ as $z$ tends to infinity.

$$\int_{a}^\infty \frac{1}{\sigma \sqrt{2 \pi}}e^{z-\frac{1}{2}\left(
\frac{z - \mu}{\sigma}\right)^2} dz = -\frac{1}{2} e ^{(\sigma^2 / 2 +
\mu)} (-1)+ \frac{1}{2} e ^{(\sigma^2 / 2 + \mu)} \text{Erf}
\left(\frac{\sigma^2 +u - a }{\sqrt{2} \sigma} \right) $$ $$ =
\frac{1}{2} e ^{(\sigma^2 / 2 + \mu)}\left(1+ \text{Erf}
\left(\frac{\sigma^2 +u - a }{\sqrt{2} \sigma} \right) \right)$$

The Erf is related to the standard normal cumulative distribution
function. Let $N(x)$ be the CDF for a standard normal distribution (
to match the notation used in Björk), then

$$N(x)= \frac{1}{2}\left(1+ \text{Erf}\left( \frac{x}{\sqrt{2}}\right)
\right)$$

So we surely have the following relation.

$$N \left( \frac{1}{\sigma} (\sigma^2 +u - a )
\right)= \frac{1}{2}\left(1+ \text{Erf}\left( \frac{\sigma^2 +u - a
}{\sqrt{2} \sigma}\right) \right)$$

So all together for this step we have evaluated the following definite
integral. $$\int_{a}^\infty e^z f(z) dz = \int_{a}^\infty
\frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{1}{2}\left( \frac{z -
\mu}{\sigma}\right)^2} dz = e ^{(\sigma^2 / 2 + \mu)} N \left( \frac{1}{\sigma} (\sigma^2 +u - a ) \right)$$

The T.A. was suggesting that the way I used mathematica for the indefinite integral isn't needed and this conclusion can be made in a much "simpler" manner. Perhaps with just using basic principles of the PDF, CDF, etc. I am just not seeing how I can make this conclusion without evaluating the indefinite integral as I have done above, in terms of the Erf and then converting to the CDF.

Thanks!

答:
\begin{align}
\frac{1}{\sqrt{2\pi}\sigma}\int_a^\infty e^w e^{-\frac{(w-\mu)^2}{2\sigma^2}} dw &= \frac{1}{\sqrt{2\pi}\sigma}\int_a^\infty e^{-\frac{(w-\mu)^2 - 2\sigma^2w}{2\sigma^2}} dw \\
&= \frac{1}{\sqrt{2\pi}\sigma}\int_a^\infty e^{-\frac{(w^2+\mu^2-2(\mu+\sigma^2)w}{2\sigma^2}} dw \\
&= \frac{1}{\sqrt{2\pi}\sigma}\int_a^\infty e^{-\frac{(w-(\mu+\sigma^2))^2 + \mu^2 - (\mu + \sigma^2)^2}{2\sigma^2}} dw \\
&= e^{\mu + \frac{\sigma^2}{2}}\frac{1}{\sqrt{2\pi}\sigma}\int_a^\infty e^{-\frac{(w-(\mu+\sigma^2))^2}{2\sigma^2}} dw
\end{align}

The result follows from the symmetry of Normal pdf i.e. $\Phi(-x) = 1 - \Phi(x).$


问:
Proof Verification of a Linear Algebra problem

Consider a symmetric and positive semidefinite matrix $A \in \mathbb{R}^{n \times n}$ Assume additionally that there exists no vector $x \in \mathbb{R}^n \setminus \{ 0 \}$ which satisfies $x^T A y = 0$ for all $y \in \mathbb{R}^n$. Prove that $A$ is positive definite.

My attempt:

Since $A$ is psd, for any $\bf z\in\mathbb{R}^n\setminus\{0\}$ we have $\bf z^{T}A\bf z\ge0$. Now according to the question $\not\exists\bf x\in\mathbb R^n\setminus\{0\}$ such that $\bf x^{T}A\bf y=0$ for all $\bf y\in\mathbb R^n$. In particular, if we take $\bf y=\bf z$, then $\bf z^{T}A\bf z\ne0$, but since $A$ is psd $\bf z^{T}A\bf z\ge0$. This implies $\bf z^{T}A\bf z >0$. Thus $A$ is positive definite.

Is this proof correct? If not where is the flaw?

Thank you.

答: As stated in the comments, your proof does not work. However, the result is true. Indeed, if $A$ was merely positive semidefinite and not positive definite, then there would have to be some nonzero vector $x$ such that $x^t A x = 0$. Now by the spectral theorem, there is some orthogonal matrix $B$ such that $A = BDB^t$ for $D$ diagonal. Now let $y = B^{-1}x = B^t x \neq 0$. $x^t A x = 0$ implies that $(By)^t (BDB^t) (By) = 0$. This left hand side is equal to $y^t B^tBDB^tBy = y^t D y$. Let $y^t = (y_1, \dots, y_n)$. Then $0 = y^t D y = \sum D_{ii} y_i^2$. As $A$ is positive semi-definite $D$ is as well, so all the $D_{ii} \geq 0$. Indeed, if $D_{ii} < 0$ then $e_i^t D e_i < 0$, a contradiction. Thus, each $D_{ii} y_i^2 \geq 0$ and their sum is zero, so each $D_{ii} y_i^2 = 0$. Now, $y \neq 0$ so some $y_j \neq 0$. Thus, that corresponding $D_{jj} = 0$. In other words, $e_j$ is an eigenvector of $D$ of eigenvalue $0$. $ABe_j = B D e_j = 0$, so $0 \neq Be_j$ is an eigenvector of $A$ of eigenvalue $0$. Now, let $z \in \mathbb R^n$ and we shall compute $(B e_j)^t A z$. Indeed, as $A$ is symmetric this is $(Be_j)^t A^t z = (ABe_j)^t z = 0$. But $z$ was chosen arbitrarily and as $B$ is orthogonal, $B e_j \neq 0$. We assumed that no such nonzero vector would exist, so $A$ must have been positive definite.


问:
Show a Series Converges Uniformly to a Doubly Periodic Meromophic Function
This is a problem from Gamelin's textbook that I am trying to solve:

Let $w_1$ and $w_2$ be two complex numbers that do not lie on the same line through $0$. Ket $k \geq 3$. Show that the series:
$$
\sum_{m,n = -\infty}^\infty \frac{1}{(z-(mw_1+nw_2))^k}
$$
converges uniformly on any bounded subset of the complex plane to a doubly periodic meromorphic function $f(z)$, whose periods are generated by $w_1$ and $w_2$.

I am having an issue seeing the image of the term inside the summation, which I think is essential for me to understand this question. But I could be totally wrong, too.

答:
Let $U = \{ sw_1+tw_2, s,t\in (-3/4,3/4)\}$. Then
$$z^k f(z)=\sum_{m,n = -\infty}^\infty \frac{z^k}{(z-(mw_1+nw_2))^k}$$
converges uniformly on $U$ and it is holomorphic since its derivative is $$(z^k f(z))'=\sum_{m,n} \frac{(z-(mw_1+nw_2)) k z^{k-1}-k z^k}{(z-(mw_1+nw_2))^{k+1}}$$

So $z^k f(z)$ is analytic on $U$.

Thus $f$ is meromorphic on $U$. Since it is obvious that $f(z)=f(z+w_1)=f(z+w_2)$ we get that $f$ is meromorphic on $$\bigcup_{n,m}( U+mw_1+nw_2)=\Bbb{C}$$
Do you see where are its poles ?