问:
An $R_0$ space can't have Sierpinski space as its subspace. Proof Verification.
I'm not sure if my proof is right and if it is not could you please point me to the mistake.

Proof by contradiction.

Defining Sierpinski Space as: $\underline S = (\{0,1\}, \{\{0, 1\}, \{1\}, \emptyset\})$

Since $\underline S$ is a subspace of some space $\underline Y$, which is $R_0$, then such an initial map exists:

$m: \underline S \rightarrow \underline Y$

We denote:

$m(0) = y_0, m(1) = y_1$

Due to Y being $R_0$ and $m$ being continuous, we have:

$0 \in cl_S(\{1\}) \Rightarrow y_0 \in cl_Y(\{y_1\}) \Rightarrow y_1 \in cl_Y(\{y_0\})$

Since $\underline S$ is an initial topology and $\{ 0 \}$ is closed in $\underline S$, then:

$\exists B \subset Y, B \ \text - closed: m^{-1}[B] = \{ 0 \}, \ \text{also note that} \ y_0 \in B$

Then:

$cl_Y\{y_0\} \subset B$

And so:

$y_1 \in cl_Y\{y_0\} \subset B$, so $y_1 \in B$

But then:

$1 = m^{-1}(y_1) \in m^{-1}[B] = \{0\}$

From the last follows that $1 \in \{0\}$, which is a contradiction.

Therefore, no such initial map $m$ can exist, so $\underline S$ is not a subspace of $\underline Y$. $\blacksquare$

答:
The proof is correct, but it’s far more complicated than necessary. Suppose that $\langle Y,\tau\rangle$ is $R_0$ and has a subspace homeomorphic to the Sierpiński space; then there are distinct points $y_0,y_1\in Y$ such that

$$U\cap\{y_0,y_1\}\in\big\{\varnothing,\{y_1\},\{y_0,y_1\}\big\}$$

for each $U\in\tau$, and there is at least one $U_1\in\tau$ such that $U_1\cap\{y_0,y_1\}=\{y_1\}$. $Y$ is $R_0$, and $U_1$ is an open nbhd of $y_1$ that is not an open nbhd of $y_0$, so $y_0$ and $y_1$ are separated in $Y$, i.e., each has an open nbhd that does not contain the other. But then there is a $U_0\in\tau$ such that $U_0\cap\{y_0,y_1\}=\{y_0\}$, which is impossible. Thus, $Y$ has no such subspace.


问:
Weyl's inequality for largest eigenvalues of two real symmetric matrices
I am trying to show that

$| \lambda_n(A) - \lambda_n(B)| \leq \lVert{\mathbf{A} - \mathbf{B}} \rVert_F $

where $A, B \in \mathbb{R}_{sym}^{nxn}$ and $\lambda_n(A)$ denotes the largest eigenvalue of $A$.

The largest eigenvalue for $A$ is defined as $\max_{v \in \mathbb{R}^n,   \lVert v \rVert_2 = 1}  v^TAv$, and for B, is similar.

I know this is related to Weyl's Inequality in some sense but cannot construct proof to show this.

答:
With an additional assumption, you don't even need to argue with $\operatorname{max}$.

Suppose that A and B are positively semidefinite. We have [1, 2], $A=M^2, \ B=N^2$ for some $M,N$ symmetric. $\Vert{A-B}\Vert_F \\ = \Vert M^2-N^2\Vert_F \\ \geq \Vert M^2-N^2\Vert_2 \\ \geq \left| \Vert M^2 \Vert_2 - \Vert N^2\Vert_2 \right| \\
= \left| \Vert M \Vert_2^2-\Vert N \Vert_2^2\right| \\
= \left| \sqrt{\lambda_{max}(M^2)}^2 - \sqrt{\lambda_{max}(N^2)}^2\right| \\
= \left |\lambda_{max}(A) - \lambda_{max}(B)\right|$.

ContentLicense: CC BY-SA 4.0


问:
Precursor to Serre's criterion in Eisenbud
Theorem $11.2$ in Eisenbud states the following:

A Noetherian domain $R$ is normal iff for every prime $P$ of $R$ associated to a principal ideal, $P_P$ is principal.

Since $R$ is an integral domain, then for any principal ideal $Q=(r)$, the associated primes must all be zero since an integral domain has no zero divisors. Furthermore, localizing $0$ at $0$ is just zero...

What is going on here??

答: In the paragraph right after the definition of associated primes, Eisenbud writes that the associated primes of an ideal $I$ will be understood to mean the associated primes of $R/I$.

ContentLicense: CC BY-SA 3.0


问:
Find $a\in \mathbb R$ so that $\{(a, 1, 1), (1,a,1), (1,1,a)\}$ is a base of $\mathbb R^3$
Find $a\in \mathbb R$ so that the list $v=[v_1, v_2, v_3]^t$ is a base of $\mathbb R^3$, where $$v_1 = (a, 1, 1),\ v_2= (1,a,1),\ v_3 = (1,1,a),$$
where $^t$ means transpose.

Any help is highly appreciated, because I don't really know how I could solve this.

答:
$n$ vectors in $ \mathbb {R} ^{n}$ are linearly independent if and only if the determinant of the matrix formed by taking the vectors as its columns is non-zero.

$\begin{vmatrix}a&1&1\\1&a&1\\1&1&a\end{vmatrix}=\begin{vmatrix}a+2&1&1\\a+2&a&1\\a+2&1&a\end{vmatrix}=(a+2)\begin{vmatrix}1&1&1\\1&a&1\\1&1&a\end{vmatrix}=(a+2)\begin{vmatrix}1&1&0\\1&a&0\\1&1&a-1\end{vmatrix}$

$=(a+2)(a-1)(a-1) .$

Can you take it from here?


问:
What exactly are Cayley-Klein geometries?
Everywhere I see Cayley-Klein geometries discussed, they're presented in a manner that seems wholly disconnected from the rest of geometry.  Just what sorts of mathematical object are Cayley-Klein geometries/spaces?  What structure is considered to be part of the abstraction and what is considered accidental?  And how do these fit in with the rest of geometry?

Like, the cases of elliptic angle -- elliptic/projective geometry, Euclidean geometry, hyperbolic geometry -- these are all familiar as Riemannian manifolds.  And the cases of hyperbolic angle -- de Sitter / co-hyperbolic geometry, Minkowski geometry, anti-de-Sitter / doubly hyperbolic geometry -- these are all familiar as Lorentzian manifolds.  But what about the cases of parabolic angle (co-Euclidean, Galilean, and co-Minkowski geometries)?  Do we formalize these as manifolds where tangent spaces are equipped with a degenerate metric tensor?  Does that even result in something usable, and how would one get the appropriate notion of angle from that?

Of course, ideally I'd like an answer that doesn't just work separately for individual Cayley-Klein geometries but provides a more unified answer.  And again I'm asking above -- just what exactly is a Cayley-Klein geometry?  They have points, lines, incidence, distance (sort of) and angle between lines (sort of), but is there more included in the abstraction, or is it really just incidence geometry but with distance and angle information?

Thank you all!  I'm pretty confused by this...

Edit: At the suggestion of brainjam, I'm updating this question to explain why e.g. I don't find the Wikipedia article a sufficient answer.

So, this article talks a field of study called Cayley-Klein geometry, which involves looking at symmetries of Cayley-Klein metrics.  But I'm asking about a mathematical object called a Cayley-Klein geometry; I don't want just the symmetries, I want the thing we're looking at the symmetries of!  So the Cayley-Klein metric is what I want, then, right?  Well, not so fast...

The problem is that this Cayley-Klein metric doesn't seem to explain any of what I think I know about Cayley-Klein geometries.  What is a line, and how do we get the expected incidence properties (specifically, the appropriate variants of the parallel postulate and the dual parallel postulate)?  Why does "angle" seem to mean something different from usual?  (Especially in the case of parabolic angle.)  And this article talks about a Cayley-Klein metric, but "distance" in Cayley-Klein spaces as I know them is not necessarily a metric -- it's not even defined between all pairs of points! -- and is only a metric in the cases where angle is elliptical.  These are the sorts of questions I want answers to... hoping that clears things up!

Edit again: OK, it seems that I still haven't expressed the question clearly enough.  Let me try again.

Very frequently in mathematics, we define an area of study by defining a set of features we want to study and what conditions those features are required to satisfy.  E.g., groups -- the feature is a binary operation, the requirements are the group axioms.  Or metric spaces -- the feature is a two-input function to nonnegative reals, the requirements are the metric space axioms.  Etc.  Obviously there may be multiple features; perhaps there are multiple operations, an operation and a topology, etc, in which case typically the requirements demand that those features be somehow compatible with one another.  Some times some of the features have no requirements, as with e.g. graphs or incidence geometries.  But you get what I'm talking about.

In geometry -- not of the algebraic sort -- there isn't one idea of what makes a geometry, but rather there are several of these abstractions, depending on what features we are interested in capturing.  Incidence geometries.  Metric spaces and length spaces.  Riemannian manifolds and pseudo-Riemannian manifolds.  Etc.

I want to know if it is possible to fit Cayley-Klein geometries into one of these (beyond incidence geometries, which are kind of, well, disappointing).  As I said above -- the elliptic angle cases can be realized as Riemannian manifolds; the hyperbolic angle cases can be realized as Lorentzian manifolds.  What I want to know is, is there some notion of a "geometry" which:

Is broad enough that it can include all the Cayley-Klein geometries;

Is also broader than that, and can serve as a useful general notion of a geometry, such that all sorts of other familiar spaces can be fit into it, rather than only these nine;

Ideally, incorporates the Cayley-Klein geometries in a natural way, such that the relevant notions of angle and distance fall out naturally rather than needing definitions with explicit case distinctions.

So, e.g., just saying, well, here's the definition of a Cayley-Klein geometry, is not what I'm looking for, because it fails condition (2).  I don't want just Cayley-Klein geometries, I want an abstraction that covers both those and other geometries.  Plus, the definition of Cayley-Klein geometries is more of a construction than a set of requirements.  I mean technically it can be phrased as a set of requirements, but morally it's more of a construction.

So e.g., if the only Cayley-Klein geometries were the ones with elliptic angle measure, then "Riemannian manifolds" would be a suitable answer.  Since that's not the case, though, it isn't.  If the only Cayley-Klein geometries were the ones with elliptic or hyperbolic angle measure, then "pseudo-Riemannian manifolds" would be a somewhat acceptable answer; it wouldn't be ideal though because it would fail condition (3), e.g. with having to define angle differently depending on the signature (and not having angle at all in most signatures).

If the above cannot be done, then that is what I would call being disconnected from the rest of geometry.

I am hoping that this will finally make it clear what I am asking.

答:
Richter-Gebert's Perspectives on Projective Geometry - A Guided Tour Through Real and Complex Geometry devotes Chapters 20-23 to working out many details and examples of C-K (Cayley-Klein) geometries.  The preceding two chapters build up Euclidean geometry from projective geometry, giving a foretaste of the general C-K methods and concepts. (There's an online draft here.) In the following, I'll assume the planar case.

To answer your question, a C-K geometry is the projective plane along with a fundamental conic (also called the absolute conic).  There are dual recipes for deriving the distance between points and the angle between lines using the absolute conic.  The allowable transformations are the projective transformations that preserve the reference conic.

section 20.5 discusses distance, which can be real, imaginary, or complex.  As OP points out, if one is thinking of a real distance, then that may not be defined between all pairs of points.

parts of the space may be removed for various purposes.  Often, for example, the fundamental conic is removed because it is at infinite distance from the rest of the points.  If the fundamental conic has an interior (the hyperbolic geometry case), the non-interior can be removed, and the remaining line segments can be regarded as the lines of the hyperbolic plane (they are infinite according to the induced metric).  In the Euclidean case removal of the fundamental conic (the circular $I$ and $J$ points) and the line they are on amounts to removing the line at infinity from the projective plane, thus allowing the existence of parallel lines that have no intersection in the geometry.  I haven't thought about it much, but maybe these removals could also be implemented by removing all points that don't have a finite real distance from some chosen point.

section 2.6 is a census of the planar C-K geometries, and includes many (but not all) listed in the OP.  There are frameworks other than C-K that include more geometries.

it's worth remembering that the C-K idea is a way of constructing models of various geometries.  Perhaps this is why OP sees them as disconnected from the rest of geometry - because they are perhaps not the best way to reason within a given geometry.  As for whether the various parallel postulates are part of a C-K geometry, I'd say it's more a matter of verifying which postulate applies to a given geometry.

the paper On Klein’s So-called Non-Euclidean geometry is worth a look.  It discusses Klein's C-K paper in light of the ideas in the air at the time, and earlier mathematical ideas and mathematicians that led up it.

As a side note, it's interesting how projective geometry, which is famously devoid of measurements such as distance and angle, can generate them once a reference conic is given.  It's reminiscent of the Poncelet–Steiner theorem, which states that whatever can be constructed by straightedge and compass together can be constructed by straightedge alone, provided that a single circle and its centre are given.


问:
Picard-Lindelöf theorem variations
There are so many slight variations of this theorem around. I wrote the following in my own words from different sources online in a way that makes sense to me. Is what I have written accurate? I want to make sure I have understood the theorem correctly.

Let $U$ be an open subset of $\mathbb{R}\times \mathbb{R}^m$ containing $(t_0,y_0)$ and let $f:U\to \mathbb{R}^m$. If $f$ is continuous in $t$ and satisfies for some $K\geqslant 0$, $|f(t,x)-f(t,y)|\leqslant K\,|x-y|$ for every $x,y \in U$. Then for some $\tau >0$ there exists a unique solution $y(t)$ on the interval $[t_0 - \tau, t_0 +\tau]$ for the initial value problem
$$y'(t)=f(t,y(t)), \,\,\,\,\, y(t_0)=y_0$$

答:
The most general statements are:

Peano: $f$ is continuous on $U$, then a local solution to the IVP exists.

Picard-Lindelöf - local variant: $f$ is continuous on $U$ and is locally Lipschitz in $y$ direction (for instance if $f$ is continuously differentiable in that direction), then a local solution to the IVP exists and is unique.

Where the idea of a local solution is as you wrote. Then there is

Picard-Lindelöf - global variant: If $U=(a,b)\times \Bbb R^n$ and $f$ is continuous with a global Lipschitz condition in $y$ direction, then a global solution $y:(a,b)\to\Bbb R^n$ exists and is unique.

As you can see, I avoided the naming of constants in the theorem, as they are really only needed in the proof. This would change if the statement of the theorem is to serve at the same time as definition of the constants and properties. For instance to make the theorem pseudo-constructive.

Your formulation is in the middle of all of this,

it is a local theorem with a global Lipschitz condition, and thus slightly more restrictive than the more general formulation, especially if $U$ is not bounded,

and it is naming the constant in the Lipschitz condition, but not with the full construction of the interval radius $τ$.

Remember that the proof of the general local theorem first selects a cylindrical domain and 1.) determines the maximum of $|f|$ over it, then restricts this to a smaller cylinder where all solutions would have to exit through the sides, then 2.) on that determines the local $y$-Lipschitz constant and then 3.) further makes the cylinder narrower so that the Picard iteration is contractive. In your formulation the first two steps are exchanged, resp. the second is left out as there is a global $y$-Lipschitz constant.

Title: Picard-Lindelöf theorem variations


问:
If $u,v$ are orthogonal vectors in $\mathbb{R}^3$ and $||u|| = 3||v||$, then the set $\{u+3v,u-3v\}$ is also orthogonal
I can come up with examples where this is true, but I am having trouble generating a formal proof for this. Tried to prove this false with a counter-example but soon realized that wasn't gonna work.</p>

Body:
Consider the inner product $<u + 3v, u - 3v>$. Then,

$$<u + 3v, u - 3v> = <u, u> - <u, 3v> + <3v, u> - <3v, 3v>$$

However, $<u, u> = <3v, 3v>$ since :

$$||u|| = 3||v|| \implies ||u||^2 = 9||v||^2 \implies <u, u> = <3v, 3v>$$

Also, $<u, 3v> = 3<u, v> = 3<\overline{v, u}> = 3<v, u> = <3v, u>$.

Thus, $<u + 3v, u - 3v> = 0$


问:
Extension of real functions to Riemann surface
Let $f:\mathbb{R}^*_+\to \mathbb{R}$ be a function that is locally the restriction of an holomorphic function.

Notating $R$ the Riemann surface of the complex logarithm, with coordinates $(r,\theta)$ with $r>0$ and $\theta\in\mathbb{R}$, we find that the Cauchy-Riemann conditions in this system of coordinates is given by:
$$
\frac{\partial f}{\partial r} = \frac{1}{ir}\frac{\partial f}{\partial \theta}.
$$

My question is: the Cauchy-Riemann are a system of partial differential equations, if we use the function $f$ as an initial condition for $\theta=0$, are there conditions for the existence of a solution that is everywhere holomorphic on $R$? If yes, is it possible to know its form?

Thank you!

答:
If a solution exists, then its real part is the harmonic function $u$ with the given initial values $u=f$, $u_\theta=0$ (and the imaginary part is the conjugate function of $u$).

The problem is that the Cauchy problem for an elliptic equation (such as $u_{xx}+u_{yy}$) is ill-posed. The solution exists if and only $f$ is given by a power series that extends analytically to the entire region. But in this case, it's the form of the power series that tells you that the solution exists, not the PDE. There is nothing in the PDE that says that the initial value problem will have a solution; rather the opposite, the PDE does its best to kill the solution.


问:
Probability question on random pick

Looking at the table, if we pull a business unit's annual profitability at random, and find that they made more than £8,000, what is the probability that it was from 2019?

Would the correct answer be 1/4 if we consider only the TOTAL?

Or how would you calculate it considering the single units, so that we have like 2/4 in 2016, 0 in 2017, 1/4 in 2018, and 3/4 in 2019?

Thanks in advance!

答: There are $6$ cells in which the value is more than $8000$. Of these, $3$ of them are in the $2019$ row. Thus, the probability of selecting a cell from $2019$ given its value is more than $8000$ is $\displaystyle\frac{3}{6}=\boxed{\frac{1}{2}}$

Title: Probability question on random pick


问:
Easy Question about P-names
Notation(from Kunnen): $M$ is countable transitive model of $ZFC$, $P$ is a partial order that belongs to $M$, $G$ is a $P$-generic filter over $M$, $\hat{b}$ will be a $P$-name for $b\in M$, and $M[G]$ is the smallest model of $ZFC$ containing $G$ and $M$.

The exercise i'm trying to do is the following: If $f$ is a function with range contained in $M$ and $f$ belongs to $M[G]$ then there is a set $B$ that belongs to $M$ containing the range of $f$. The hint given is to consider

$B=\left \{ b:\exists p  \in P(p\Vdash \hat{b} \in \operatorname{Rng}(\tau ) ) \right \} $ where $\operatorname{val} (\tau,G) = f$

I understand that this describes the desired $B$ as a class in $M$ but i can't find a way to show that this is actually a set in $M$.

Thanks for the help and sorry for the trivial question.

答:
Here is a quick proof, which doesn't use the hint.

If $f$ is a function in $M[G]$ with $\text{ran}(f)\subset M$, then since $\text{ran}(f)$ is a set in $M[G]$, it has some von Neumann rank $\alpha$ there. Thus, every element of $\text{ran}(f)$ has rank less than $\alpha$. Since rank is absolute and $\alpha$ is an ordinal in $M$, it follows that $\text{ran}(f)\subset (V_\alpha)^M$, and this is a set in $M$.

(This proof requires you already to know that $M$ and $M[G]$ have the same ordinals and that $M[G]$ satisfies ZFC.)

Title: Easy Question about P-names


问:
An explanation on variances
I am having trouble understand what the following represent.

答:
Let $\mathbb X^{p\times 1}$ denotes a column vector of random components $X_i,i=1(1)p.$ Then the expectation of $\mathbb x$ is defined as $$E(\mathbb X)=E\begin{pmatrix}X_1 \\ \vdots\\X_p\\\end{pmatrix}=\begin{pmatrix}EX_1 \\ \vdots\\EX_p\\\end{pmatrix}=\begin{pmatrix}\mu_1 \\ \vdots\\\mu_p\\\end{pmatrix}=\mathbb{\mu}
$$
Define $\text{Cov}(X_i,X_j)=E[(X_i-EX_i)(X_j-EX_j)^T]=\sigma_{ij}$. We extend the variance notation to the p-dimentional random vector $\mathbb X$ by the following matrix:
$$E[(\mathbb X-E \mathbb X)(\mathbb X-E \mathbb X)^T] \\=E[(\mathbb X- \mathbb \mu)(\mathbb X-\mathbb \mu)^T] \\=E\left[\begin{pmatrix}X_1-\mu_1 \\ \vdots\\X_p-\mu_p\\\end{pmatrix} \begin{pmatrix}X_1-\mu_1& \cdots &X_p-\mu_p\\\end{pmatrix}\right]\\=\begin{pmatrix}\sigma_{11} &\sigma_{12} &\cdots &\sigma_{1p}\\ \vdots & \vdots &\ddots &\vdots \\\sigma_{p1} &\sigma_{p2}&\cdots& \sigma_{pp}\\\end{pmatrix}=\Sigma$$ In similarly way co-variance between $X$ and $Y$ is defined.

That's all from me.

Title: An explanation on variances


问:
Gre Algebra question, the least possible value of $3^x + 3^{-2}$
if $x$ is an integer, what's the least possible value of $3^x + 3^{-2}?$

$A.0$

$B.2$

$C. 2\frac{2}{3}$

$D.3$

E.$3\frac{1}{3}$

The way I did was I know both $3^x$ and $3^{-2}$ are positive so there's no way they can add up to $0$, which I can guess that the least possible value is greater than $0$ but it's the least value in the answer choices, which is $2$, and indeed the answer is $2$ as well.

I wonder is there any other safer ways to do this question except guessing.

答:
This question makes no sense.  If $x$ is an integer, we can surely choose $x = 0$ and get $3^0 + 3^{-2} = 1 + \frac{1}{9} < 2$.  That eliminates choices (B) through (E).  But (A) is invalid because we know $3^x > 0$ for all real $x$ and $3^{-2} = 1/9 > 0$.

I have no idea what this question is trying to say.  If $x$ is allowed to be any integer, we can choose $x$ to be an arbitrarily small negative number such that $3^x \to 0$ consequently $3^x + 3^{-2}$ is bounded below by $1/9$ but can never attain this value; i.e., $$\inf_{x \in \mathbb Z} 3^x + 3^{-2} = \frac{1}{9}.$$  If $x$ is required to be a positive integer, then the answer is $3 + 1/9 = 28/9$, which is not any of the provided answer choices.  So I cannot find any interpretation of the question that would admit an answer that is one of the possible choices.


问:
Probability of picking bad egg on given day?

I have a basket with n eggs. Each day, I eat an egg at random from the basket and then I put a fresh on. After a time T spent in the basket an egg goes bad. What is the probability I pick a bad egg a given day? What is the probability I pick bad eggs two days in a row?

My attempts: Any egg that has been in the basket for T days goes bad, at least that is my interpretation. My initial approach was to pick a couple of values for T and n, and work out what happens, but it gets unruly pretty quickly. I'm wondering if there is a better method using maybe first step analysis, or even markov chains that I'm missing that could be used to solve this...

答:
Same result as @joriki by a different approach.  Let's write:
$$P( {\rm egg_1}>T\cap{\rm egg_2}>T)=P( {\rm egg_2}>T|{\rm egg_1}>T)P({\rm egg_1}>T)$$ so that we can recycle @joriki's first result, $P({\rm egg_1}>T)=(1-1/n)^T$. Then, we just need to find $P( {\rm egg_2}>T|{\rm egg_1}>T)$.  Hopefully the notation is clear: $P( {\rm egg}_i>T)$ is the probability that the $i$-th egg we choose is greater than $T$ days old.

Assume $T\ge 1$. Then, in order to both choose a second bad egg and satisfy the ${\rm egg_1}>T$ conditioning criteria, we have to do the following:

The $T-1$ draws preceding ${\rm egg}_1$ can only be taken from the $n-1$ distinct slots not corresponding to ${\rm egg}_1$, by the conditioning criteria.  From these, the draws must also not choose the ${\rm egg}_2$ slot, so the probability for this process is $\left(\frac{n-2}{n-1}\right)^{T-1}$.

The final ${\rm egg}_2$ draw must correspond to a slot other than the ${\rm egg}_1$ slot, for which we have probability $\frac{n-1}{n}$.

Combining, this gives $$P( {\rm egg_2}>T|{\rm egg_1}>T)=\frac{n-1}{n}\times \left(\frac{n-2}{n-1}\right)^{T-1}=(1-1/n)^{2-T}\times (1-2/n)^{T-1}$$
or, all together:

$$P( {\rm egg_1}>T\cap{\rm egg_2}>T)=(1-2/n)^{T-1}(1-1/n)^{2}$$


问:
If $\zeta(1+it_0)=0$ then $|\zeta (\sigma)^3 \zeta (\sigma +it_0)^4 \zeta(\sigma + 2it_0)| \longrightarrow 0$ as $\sigma \longrightarrow 1$
I need to prove that if $\zeta(1+it_0)=0$ then $|\zeta (\sigma)^3 \zeta (\sigma +it_0)^4 \zeta(\sigma + 2it_0)| \longrightarrow 0$ as $\sigma \longrightarrow 1$, where $\zeta$ is the Riemann Zeta function. I guess I need to show that $\zeta(\sigma+it_0)^4$ approaches $0$ faster than $\zeta (\sigma)^3\zeta (\sigma + 2it)$ approaches $\infty$ when $\sigma \longrightarrow 1$. But don't know how to get around that.

Edit: The purpose of this computation is to show, by contradiction, that $\zeta$ has no zeros at the line $1+it$.

答:
$\zeta(1+it_0)=0$, thus if $n$ denotes the multiplicity of $1+it_0$ as a zero of $\zeta$, there exists $g$ holomorphic such that $\zeta(\sigma+it_0)=(\sigma-1)^ng(\sigma)$ and $g(1)\neq 0$ around $\sigma=1$. We thus have
$$ |\zeta(\sigma)^3\zeta(\sigma+it_0)^4\zeta(\sigma+2it_0)|=(\sigma-1)^3\zeta(\sigma)^3|\cdot(\sigma-1)^{4n-3}|g(\sigma)|\cdot|\zeta(\sigma+2it_0)|\underset{\sigma\rightarrow 1^+}{\longrightarrow}0 $$
because $\lim\limits_{\sigma\rightarrow 1^+}(\sigma-1)^3\zeta(\sigma)^3=1$ and $4n-3\geqslant 1$. (Which btw leads to a contradiction since $|\zeta(\sigma)^3\zeta(\sigma+it_0)^4\zeta(\sigma+2it_0)|\geqslant 1$ for all $\sigma>1$)


问:
show that the ways to pick 2 subsets A and B of [n] when A is subset of B equals to n^3
I know that the total of subsets of a set is 2^n.
I am assuming I have to use combinations but I cannot think of something to start.
Anybody can help
Thanks in advance</p>

答:
For each element of $[n]$ you have $3$ choices:

Put it in $A$ (so necessarily also in $B$).

Put it in $B$ but not in $A$.

Don't put it in $A$ nor $B$.


问:
Taylor series with function composition
Pretty simple, but I want to take the first order taylor series expansion of the following:

$f(g(x,y+Δy))$

Would the following be correct?

$f(g(x,y+Δy)) = f(g(x,y) + \frac{\partial}{\partial y}g(x,y)*Δy)$

$= f(g(x,y)) + \frac{\partial}{\partial g}f(g(x,y))*\frac{\partial}{\partial y}g(x,y)*Δy$

答: Yes, this is correct. In fact, since $x$ never changes, we don't have to recognize it as a variable at all. Just consider $g$ as a function of $y$ only. The chain rule says that $(f\circ g)'=(f'\circ g)g'$, and rewriting this in terms of differentials gives the result.</p>

Title: Taylor series with function composition


答:
This is a standard computation. By Euler's product identity, for all complex numbers with $\operatorname{Re}(s):=\sigma>1$
\begin{align*}
  \log(\zeta(s))=\sum_{n,m}\frac{1}{m}\frac{1}{p^{ms}_n}
\end{align*}
where $\{p_n:n\in\mathbb{N}\}$ is the enumeration of the prime numbers listed according to the usual order in $\mathbb{N}$. Then, for any real $\tau\neq0$ and $\sigma>1$
\begin{align*}
  \log\Big(\big|\zeta^3(\sigma)\zeta^4(\sigma+i\tau)\zeta(\sigma+i2\tau)\big|\Big) =  \sum_{n,m}\frac{1}{m}\frac{1}{p^{\sigma m}_n} \operatorname{Re}\Big(3+\frac{4}{p^{4m\tau i}_n}+\frac{1}{p^{2m\tau i}_n}\Big)\geq0
\end{align*}
since $\operatorname{Re}(3+4e^{i\theta}+e^{2i\theta})=2(1+\cos\theta)^2\geq0$ for all $\theta\in\mathbb{R}$. It follows that
$$\begin{align}
  |(\sigma-1)\zeta(\sigma)|^3\Big|\frac{\zeta(\sigma+i\tau)}{\sigma-1}\Big|^4|\zeta(\sigma+i2\tau)| \geq\frac{1}{\sigma-1} \tag{1}\label{riemann-at-1}
\end{align}
$$
If $\zeta(1+it_0)=0$ for some $t_0\in\mathbb{R}$, the left--hand--side of~\eqref{riemann-at-1} converges to  $|\zeta'(1+i\tau)|^4|\zeta(1+i2\tau)|$ as $\sigma\searrow1$, where $\tau=t_0$. The right--hand--side of \eqref{riemann-at-1} however, tends to $\infty$ and so, we get a contradiction. Therefore, $\zeta$ has no zeroes along the line $1+i\tau$, $\tau\in\mathbb{R}$

Title: If $\zeta(1+it_0)=0$ then $|\zeta (\sigma)^3 \zeta (\sigma +it_0)^4 \zeta(\sigma + 2it_0)| \longrightarrow 0$ as $\sigma \longrightarrow 1$

Body:
I need to prove that if $\zeta(1+it_0)=0$ then $|\zeta (\sigma)^3 \zeta (\sigma +it_0)^4 \zeta(\sigma + 2it_0)| \longrightarrow 0$ as $\sigma \longrightarrow 1$, where $\zeta$ is the Riemann Zeta function. I guess I need to show that $\zeta(\sigma+it_0)^4$ approaches $0$ faster than $\zeta (\sigma)^3\zeta (\sigma + 2it)$ approaches $\infty$ when $\sigma \longrightarrow 1$. But don't know how to get around that.

Edit: The purpose of this computation is to show, by contradiction, that $\zeta$ has no zeros at the line $1+it$.


问:
Prove $(2.n+1)!!=\frac{(2.n+1)!}{2^n.n!}$ using Gamma Function
Prove $(2.n+1)!!=\frac{(2.n+1)!}{2^n.n!}$ using Gamma Function.

$\Gamma(1+z).\Gamma(z+1/2)=2^{-2.z}.\sqrt{\pi}.\Gamma(2.z+1)$

$\Gamma(z+1/2)=\sqrt{\pi}.2^{-z}.(2.z-1)!!$

and

$\Gamma(1+z)=(2.z)!!.2^{-n}$

I think that at least one of these expressions is useful for this problem but I don't know which.

答:
You can do this verbally without any equations.

The double factorial is the product of the odd numbers from one through 2n+1. What is omitted are the even numbers from 2 to 2n. There are n of these, so their product is $2^nn!$. And this is what you want.

I don't see how the gamma function comes into this at all unless you define the double factorial in terms of it.


问:
Why is the derivative Operator defined over $C^2([-1,1])$ not closed?
The question is: Let A be the Operator defined by $A:C^2([-1,1])\subset C([-1,1])\to C(-1,1]),\; x\mapsto x'$. Show that the operator is not closed.

I suppose an example where a derivative sequence converge to a point that is not in the graph of A would suffice, but isn't a property of a sequence of differentiable functions $f_n$, that if the sequence of derivatives $f'_n$ converges uniformly to a function $g$, then the sequence of functions converges to a function $f$ with $f'=g$?

Thank you in advance.

答:
Comment: "Not closed" would be shown by an example where
$$
f_n \to f,\qquad f_n' \to g,\qquad f' \ne g
\tag1$$

Of course the derivative operator is closed in the $C^2$ metric.  But perhaps the way of stating it $$C^2([-1,1])\subset C([-1,1])\to C(-1,1])$$ is meant to mean in the $C$ topology, that is: uniform convergence.

So find an example of $(1)$ where $f_n, f \in C^2$, $f_n',g \in C$, and $\to$ means uniform convergence.


问:
One dimensional subspaces of a 3 dimensional vector space over $\mathbb Z/3\mathbb Z$, multiple choice
Let $V$ be a 3-dimentional vector space over the field $F_3=\Bbb Z/3 \Bbb Z$  of $3$ elements.the number of distinct 1 dimentional subspaces of $V$ is

$13$

$26$

$9$

$15$

答:
Hint:

If $V_n(q)$ be a vector space of dimension $n$ on a field $F=GF(q)$, the number of all one-dimensional subspace of $V$ will be: $$\frac{q^n-1}{q-1}$$ In fact, all non zero scalar products of $\langle e\rangle\subset V$ generate the same subspaces of dimension one.


问:
How does this difference equation (discrete function map) behave?
Consider the iterated map

$x_{n+1} = rx_n(1 - x_n^2)$

As this is non-linear, I'm a bit confused on how to approach it's analysis. I'd like to find out/justify the following things I found out with plotting and numeric testing:

Is it true that $[x_n \in (0,1) \implies x_{n+1} \in (0,1)], \  \forall r \in (0,\frac{3\sqrt{3}}{2})$. Why?

This iterated map has two fixed points, which are $x_0=0$ and $x_0=\sqrt{1-\frac{1}{r}}$. Why does this second fixed point only exist when $r \gt 1$? How do I determine it's stability, which depends on r?

How do I show that $x_0=0$ is an attractive fixed point?

What happens when $r \gt 2$?

How must I procede? Either hints or solutions are welcome.

答:
Let's draw the functions
$$
\left\{ \matrix{
  f(x) = x\left( {1 - x^2 } \right) \hfill \cr
  g(x) = x/r \hfill \cr}  \right.
$$
both of which are odd, and the crossing point $C$ on the first quadrant
$$
x/r = x\left( {1 - x^2 } \right)\quad  \Rightarrow \quad
 C = \left( {\sqrt {{{r - 1} \over r}} ,\,{1 \over r}\sqrt {{{r - 1} \over r}} } \right)
$$
the other being the origin and the symmetric of $C$,   then the maximum $M$ of $f(x)$
$$
M = \left( {{{\sqrt 3 } \over 3},\;2{{\sqrt 3 } \over 9}} \right)
$$
and the point $L$ at which $x/r = M_y$
$$
L = \left( {2r{{\sqrt 3 } \over 9},\;2{{\sqrt 3 } \over 9}} \right)
$$
and finally the point $S$ at which $f(x) = - M_y$ for positive $x$
$$
S = \left( {2{{\sqrt 3 } \over 3},\; - 2{{\sqrt 3 } \over 9}} \right)
$$

The recurrence is represented by the path shown in blue.

Then it is easy to understand what happens when, at varying $r$ the point $L$:

lies before the point $I$ , i.e. $r \le 1$, the stable point is the origin;

$L \in (I,M]$, i.e. $1 < r \le 3/2$, the stable point is $M$;

$L \in (M,S]$, i.e. $3/2< r \le 2$, the stable point is $C$;

$L \in (S, \infty )$, i.e. $2< r $, the sequence diverges, the last terms oscillating in sign;


问:
Finding an inverse using the square and multiply algorithm

Given $7^{45}\equiv32 \bmod 101$, find $32^{-1}$ using square and multiply.

What I did was since $(7^{45})^{-1}$ is equal to $(32)^{-1}$, and $-45\bmod 101 = 56$ then $32^{-1} = 7^{56}$. I then used square and multiply but I get $16$ for $7^{56}$ but I know that the inverse of $32$ modulo $101$ is $60$.

Am I missing a property of square and multiply or of modular arithmetic for this not to work?

答:
People have pointed you to the right direction. I'll just do it here for my own sake. Everything below is $\pmod{101}$.

$$
32^{-1}\equiv 7^{55}\equiv 108^{55}\equiv (2^2 \cdot 3^3)^{55}\equiv 2^{110} \cdot 3^{165} \equiv 2^{10} \cdot 3^{65} \equiv 2^{10} \cdot 3 \cdot (81)^{16} \\
\equiv 2^{10} \cdot 3 \cdot (-20)^{16} \equiv 2^{10} \cdot 3 \cdot 2^{16} \cdot 10^{16} \equiv 2^{26} \cdot 3 \cdot 100^8 \equiv \frac{(2^9)^3 \cdot 3}{2}\\
\equiv \frac{512^3}{2}\cdot 3 \equiv \frac{7^3}{2} \cdot 3 \equiv \frac{343}{2}3 \equiv \frac{40}{2}\cdot 3 \equiv 60.
$$


问:
Conceptual Question about Memoryless Property an Exponential Distribution
Let's say I am given $X\sim \exp$ with mean=3, and then I am also asked to find $E[X-8|X>8]$. By memoryless property of the exponential distribution  $E[X-8|X>8]=E[X]=3$. However, I am trying trouble understanding how it is mathematically possible in such a situation for the expected value to be = 3 when 3 < 8 seeing as we have the condition that X > 8?

答:
Remember that $E(X + Y) = E(X) + E(Y)$, because expectation is linear. And in the case where $Y$ is a constant $y$, $E(Y) = E(y) = y$.

So $E(X - 8 | X > 8) = E(X | X > 8) + E(-8 | X > 8) = E(X | X > 8) - 8$.

Without knowing the distribution of $X$, the only thing you can say about $E(X | X > 8)$ (assuming it even exists) is that it is going to be at least 8. So we're going to find something that's at least 8, and subtract 8 from it, which means we're going to get a value that's greater than 0, but that's all we can know for sure.

Imagine flipping a coin 20 times, and calling $H$ the number of heads. What's $E(H - 15)$? We know that $0 \leq H \leq 20$ always holds, so $-15 \leq H - 15 \leq 5$ and hence $-5 \leq E(H - 15) \leq 5$. If our coin is fair, then $E(H - 15) = -5$, which is a negative number despite the number of heads being positive.


问:
Reversible and irreversible operations in elementary algebra
I am in high school algebra, solving typical equations such as rational, irrational, quadratic, etc., and I have come across the idea of extraneous solutions. My textbook does not touch upon the idea of extraneous solutions and how they relate to reversible and irreversible operations, and I can't find much online. Could I get an explanation of what exactly reversible and irreversible operations mean?

I know that squaring is irreversble, as is multiplying both sides of an equation by $0,$ but why is multiplying both sides by $x$ (where $x$ is the variable in the equation) irreversible?

Is it also irreversible if I start with $2x^2 = x,$ and divide by $x$ to get $2x = 1,$ in which case $0$ is no longer a solution?

Why is it irreversible if I have $\displaystyle \frac{x(2x + 1)}{x} = 0$ and cancel out $x$ to get $2x + 1 = 0$?

Finally, why is it irreversible if I have $2x + 1 = 0$ and multiply both sides to get $\displaystyle \frac{x(2x + 1)}{x} = 0$?

These are the four cases I am most interested in. I want to understand them specifically, because I don't want my maths to be ambiguous.

答:
If you are given Equation B and the operation that resulted in Equation B (pretend this is all you know) and you can determine the exact equation in the previous line (call it Equation A), then this is a reversible step. For example, if I told you that I added 3 to both sides of Equation A to get x=5 (x=5 would be Equation B in my explanation), then you could reverse this operation by subtracting 3 from both sides to find that Equation A is x-3=2.

If you can't be sure what the line before is, then this is an irreversible step.  For example, if I told you that I squared both sides of Equation A to get x^2=4, there's no way of knowing exactly what Equation A is.  It could be either x=-2 or x=2.  Or if I told you that I multiplied both sides of an equation by zero to get 0=0, then you have no idea what the previous line is because it could be anything. These steps are irreversible.


答:
Let $$Y = X - 8 \mid X > 8.$$  What does this mean?  Let's consider an example.  Say you are waiting at a train platform and $$X \sim \operatorname{Exponential}(\mu = 3)$$ represents the random amount of time (in minutes) it takes for the next train to arrive.  So on average, your waiting time is $3$ minutes.

Now, say you have been waiting and $8$ minutes have passed and the train has not yet arrived.  The random variable $Y$ represents the amount of additional time that you will wait for the next train to arrive.  The total time you wait is still $X$, and you know it has been more than $8$ minutes, but you won't know exactly what $X$ is until the train does arrive.

Because $X$ is exponentially distributed, and because of the memoryless property, it actually turns out that $Y$ is also exponentially distributed with mean $3$ minutes.  That is to say, given you have already waited more than $8$ minutes with no train, the average time to wait until the next train is still $3$ minutes!  This seems rather counterintuitive (and is one of criticisms of exponentially distributed event time models).  After all, if the average wait time was originally $3$ minutes and you already waited $8$ minutes, wouldn't it be more likely that a train will arrive soon?  Perhaps in a real-world scenario, yes.  But as we said earlier, this is just a hypothetical example to illustrate how the exponential distribution behaves.  Memorylessness is a property that, given how much time you have already waited, the amount of additional time to wait does not depend on how much time has passed.  A discrete analogy to this phenomenon that does make sense is coin flipping:  the outcome of a coin toss does not depend on the outcomes that were observed before.  The familiar "gambler's fallacy" is that if you toss a fair coin many times and see, say, $10$ heads in a row, you might be inclined to think that the next toss will be more likely to be heads than tails.  But if the coin is by definition fair, this is not true--the probability of heads remains $1/2$ even if the previous tosses were all heads.  That's why it's called "memoryless"--the coin does not remember what it did in the past.

Title: Conceptual Question about Memoryless Property an Exponential Distribution

Body: Let's say I am given $X\sim \exp$ with mean=3, and then I am also asked to find $E[X-8|X>8]$. By memoryless property of the exponential distribution  $E[X-8|X>8]=E[X]=3$. However, I am trying trouble understanding how it is mathematically possible in such a situation for the expected value to be = 3 when 3 < 8 seeing as we have the condition that X > 8?


问:
What is the largest possible number of moves that can be taken to color the whole grid?
Consider a $10\times 10$ grid. On every move, we color $4$ unit squares that lie in the intersection of some two rows and two columns. A move is allowed if at least one of the $4$ squares is previously uncolored. What is the largest possible number of moves that can be taken to color the whole grid?

Attempt:
It is easy to get $81$ moves: By always choosing the first line, the first
column and a square of the remaining $9\times 9$ grid as the lower
right square, the whole grid can be colored in $81$ moves. But is this a maximum number of moves?

答:
We can think of this table as a matrix $M$ filled with nonegative numbers. If the cell is colored it has number $>0$ and if it is not it has $0$. So, at the start we have $M= 0$. Now in any moment of process, if we want to color $4$ entries marked with $*$ say $$M=
\begin{bmatrix}
   & \vdots &  & \vdots &   \\
  \cdots & * & \cdots & * & \cdots  \\
   & \vdots &  & \vdots &   \\
  \cdots & * & \cdots & * &  \cdots\\
   &\vdots  &  &\vdots &   \\
\end{bmatrix} $$
then at least one $*$ is $0$, so we can color those cells or equivalently we can add to $M$ matrix: $$E=
\begin{bmatrix}
   & \vdots &  & \vdots &   \\
  \cdots & 1 & \cdots & 1 & \cdots  \\
   & \vdots &  & \vdots &   \\
  \cdots & 1 & \cdots & 1 &  \cdots\\
   &\vdots  &  &\vdots &   \\
\end{bmatrix} $$
which has exactly four $1$ and all other entries $0$. The process ends when there is no $0$ entry.

Now observe $81$ special matrix of form as $E$ is, which has first $1$ in down left corner, second $1$ in first colum in $i$-th row ($i>1$), third $1$ in first row in $j$-th column $j>1$ and fourth $1$ in $i$-th row and $j$-th column. Mark this special matrix with $S_{i,j}$.

Clearly, in whole process we can add each matrix of type $E$ at most once and in particulary any special matrix at most once.

Also, if we add to $M$ any matrix of form as $E$ is (but it is not special), with wich we want to color $k$ uncolored cells in green area, (so $1\leq k\leq 4$) we can substitute this $E$ with $k$ special matrices. So we can consider this process as adding only special matrices to $M$, but this we can not do more than $81$ times and we are done.


问:
Sequence uniform convergence but the derivatives are not.

Give a sequence $(f_n)_{n\in \mathbb{N}}$ of differentiable functions which uniformly converge to $0$, but for which the seqeunce $(f_n')_{n\in \mathbb{N}}$ of the derivatives isn't even pointwise convergent.

I found this one in my textbook marked as "Fun things to solve" and although it may be fun, it's kind of hard to do.

To be honest I already failed at the first hurdle. I couldn't even find a sequence of functions which uniformly converges to $0$.

Is there a certain way of dealing with this kind of problem? Because it seems kind of hard for me to come up with sequences without a mathematical of way of doing so (or maybe there is a mathematical way I just don't know yet).

答:
Consider:

$f_n:(0,1]\to\mathbb{R}$, where $f_n(x)=\frac{1}{\sqrt{n}}x^n$. Clearly, $f_n$ is differentiable for each $n$ and $f_n\to\hat{0}$ (uniformly). As $f_n'(x)=\sqrt{n}x^{n-1}$, we immediately see that the sequence is unbounded for $x=1$ and hence not pointwise convergent.


问:
Concretely, what is the horn $\Lambda_k^n$?
Recall that we define the simplex category $\mathbf{\Delta}$ and define the category of simplicial sets $\mathbf{\Delta\mbox{-}Set}$ to have as objects, functors $\mathbf{\Delta}^\mathrm{op}\to\mathbf{Set}$, and as morphisms, natural transformations between these functors.
We then define, for $n\in\mathbb{N}$, the simplicial n-simplex $\Delta^n$ s.t. $\Delta^n(i)=\mathrm{Hom}_\mathbf{\Delta}([i],[n])$ with the induced face and degeneracy maps.

Then we define the $(n,k)$-horn to be the minimal simplicial subset $\Lambda_k^n$ of $\Delta^n$ such that $$\Big(\delta^i:[n-1]\to[n]\Big)\in\Lambda_k^n(n-1)$$ for all $i\neq k$.

However, I'm having trouble making any concrete statements about $\Lambda_k^n$. How do we know that $\Lambda_k^n\ne\Delta^n$? How can it be shown that the non-degenerate simplices of $\Lambda_k^n$ do in fact form a horn, as the name suggests?

答:
The sharpest description I found is the one given at Exercise 1.1.2.14 of Kerodon (it can be found elsewhere).

A horn $\Lambda^n_k\to X$ is a sequence $(\sigma_0,\dots,\sigma_{k-1},\bullet,\sigma_{k+1},\dots,\sigma_n)$ of $(n-1)$-simplices such that $d_i(\sigma_j) = d_{j-1}(\sigma_i)$ whenever $i<j$ and $i\neq k\neq j$.

When you picture $\Lambda^n_k$ as the $n$-simplex $\Delta^n$ with its $k$-th face missing, the $\sigma_i$ represent the faces (notice the $k$-th input is missing), and $d_i(\sigma_j) = d_{j-1}(\sigma_i)$ is saying that, as expected, "the $k$-th face of $\sigma_j$ is the $(j-1)$-face of $\sigma_i$".

For instance, a horn $\Lambda^2_0\to X$ is a sequence $(\bullet,f,h)$, where $f,h\in X_1$ are such that $d_1(f) = d_1(h)$.
This horn (in fact, its non-degenerate simplices) is depicted by

Here, $x,y,z\in X_0$ are the $0$-th simplices defined by $x = d_1(f) = d_1(h)$, $y = d_0f$ and $z = d_0(h)$.

In particular, the horns themselves are simplicial subsets $\Lambda^n_k\hookrightarrow \Delta^n$, characterized by the sequence $(\delta^0,\dots,\delta^{k-1},\bullet,\delta^{k+1},\dots,\delta^n)$, where $\delta^i\in\Delta^n_{n-1}$ are the cofaces.


问:
Convergence of the sequence $\frac{(\ln{k})^s}{k^t}$ where $t, s \in \bf{R}^{+}$ and $k \in \{2, 3, ....\}$
I am trying to figure out under what conditions the sequence $\frac{(\ln{k})^s}{k^t}$ converges, where $s, t$ are positive real numbers. I believe that this should converge $\forall s,t > 0$, but I am not sure if this is the case. This would be based on the growth rate of a polynomial versus a power of a logarithmic function.

Similary, if
$$\frac{k^p}{(e^{k})^q} = \frac{k^p}{e^{qk}}$$
this would still hold for any $p, q > 0$, $p, q \in \bf{R}$. This is because exponentials would overpower any polynomials. Is this thought process correct in the intuition for the sequence convergence?

答:
Use this
$$\lim_{x \to \infty} {\log(x)^s\over x^t}
= \lim_{x \to \infty} {\log(e^x)^s\over (e^x)^t}$$


问:
Find the minimum of $|a+\frac 2 {a-1}|$ where $|a|\leq2$.

Find the minimum of $|a+\frac 2 {a-1}|$ where $|a|\leq2$.

I tried using differentiation, but the absolute makes things troublesome...

Please help. Thank you.

答: Let $f(x)=x+\frac{2}{x-1}$.   Find where the derivative is $0$ in our interval, and evaluate $|f|$ at these and at the endpoints $\pm 2$. We will end up wanting to evaluate $|f|$ at $-2$, $1-\sqrt{2}$ and $2$. </p>


问:
multiplying a convergent series by a bounded sequence
If $\sum{r_j}$ is an absolutely convergent series bounded by R, and the sequence $\{s_{j,k}\}$ is bounded by S and $\lim_{k \to \infty} s_{j,k} = 0$, then how do we show that $\lim_{n \to \infty} \sum^{\infty}_{j=0}r_j s_{j,k}$ go to $0$? By the answer here Under what condition we can interchange order of a limit and a summation? I can see that we can interchange the limit and the sum, but I was wondering what you need to be able to show this a bit more intuitively.

答: The Dominated Convergence Theorem will justify this rigorously. If $|s_{j,k}|\le S$ always, then each series is dominated by $\sum S|r_j|$ which converges by hypothesis; therefore the limit can be taken inside the infinite series.


问:
Arithmetic mean is smaller or equal than geometric mean to the power of two
We consider $\displaystyle{n \in \mathbb{N}}$ and $\displaystyle{a_{1}, a_{2}, . . . , a_{n} \in \big( 1, n \big]}$ then prove or disprove that
$$ \frac{a_{1} + a_{2} + \cdots + a_{n}}{n} \leqslant \Big( \sqrt[n]{a_{1} \cdot a_{2} \cdot . . . \cdot a_{n}} \Big)^{2} $$

Let $f \big( a_{1}, a_{2}, . . . , a_{n} \big) = \frac{a_{1} + a_{2} + \cdots + a_{n}}{n} - \Big( \sqrt[n]{a_{1} \cdot a_{2} \cdot . . . \cdot a_{n}} \Big)^{2} \leqslant 0 , \forall a_{1}, a_{2}, . . . , a_{n} \in \big( 1, n \big]$ which is a continous function and we find that $ b_{n} = f \big( 1, 1, . . . , n \big) = \frac{2 \cdot n-1+}{n} - \big( \sqrt[n]{n} \big)^{2}$
so the limit $ \lim\limits_{n \to + \infty} b_{n} = \lim\limits_{n \to + \infty} f \big( 1, 1, . . . , n \big) = 1 \Leftrightarrow \forall \epsilon > 0 : \exists n_{0} \in \mathbb{N} : \big| b_{n} - 1 \big| < \epsilon , \forall n \geqslant n_{0} $, so for $ \epsilon = \frac{1}{2} > 0 : \exists n_{0} \in  \mathbb{N} : \frac{1}{2} < b_{n} < \frac{3}{2} , \forall n \geqslant n_{0} $, so then for $ n \geqslant n_{0} $ : $ \frac{1}{2} < b_{n} = f \big( 1, 1, . . . , n \big) = \lim\limits_{\big( a_{1}, a_{2}, . . . , a_{n} \big) \to \big( 1^{+}, 1^{+}, . . . ,n \big)} f \big( a_{1}, a_{2} , . . . , a_{n} \big) \leqslant 0 $, is contradict.

答:
Update We can prove if $n\ge 6$, we always have $$ \frac{(n-1)\cdot 1+n}{n}>n^{\frac 2n}$$

For $n=6$ we have shown it at the end of the proof below. If it's true for $n$, the above inequality is equivalent to $$\left( 2 - \frac 1n\right)^n > n^2$$

Now,$$\left( 2 - \frac{1}{n+1} \right)^{n+1} - (n+1)^2 > \left( 2 - \frac{1}{n} \right)^{n+1} - (n+1)^2 \\> n^2 \left( 2-\frac 1n\right) -(n+1)^2 = n(n-3)-1 > 0$$

Conclusion: the inequality holds if and only if $n\le 5$.

If $n=1$ it's trivial. If $n\ge2$, consider the function $$f(x)=\frac{x+S}{n}-(xP)^{\frac{2}{n} }$$
where $S>0,P>0$.

$$f'(x)=\frac 1n - \frac 2n P^\frac 2n x^{\frac 2n-1} \implies f''(x)=\frac 2n \frac{n-2}{n}P^\frac 2n x^{\frac 2n -2}\ge 0$$

Therefore $f(x)$ obtains its maximum at either $x=1$ or $x=n$.

Hence we only need to show that

$$\frac{k\cdot n + (n-k) \cdot 1}{n} \le \left(n^k \cdot 1^{n-k}\right)^{\frac 2n}, \forall k \in \{0, 1, \ldots, n \} $$

For $n\le 5$ we can prove the above is true by manual calculation.

If $n=6$ there is a counter example: $a_1 = a_2 = \cdots = a_5=1, a_6=6$ (if $a_i=1$ is not allowed, just make it arbitrarily close to $1$): AM $\approx 1.83333$, but GM$^2 \approx 1.8171$.


问:
Prove T is a linear operator
Let $\mathcal{H}$ be a Hilbert space. Given that $T: \mathcal{H} \to \mathcal{H}$ $(Tx,Ty) = (x,y)$, how should we prove that T is an linear operator?

It is not hard to show that
$$(T(x+y), Tz) = (Tx + Ty, Tz),(T(ax), Tz) = (aTx, Tz)$$

But, how may I show that $Tz$ can be any element of $\mathcal{H}$?

It seems that all theorems relies on linear operator...

答:
Use the inner product relation to calculate $\|T(x+y)-(Tx+Ty)\|^2$ and likewise $\|aTx-T(ax)\|^2$. You'll find that they are both zero. I'll do the second one and leave the first for you to check:

\begin{align*}
\|aTx-T(ax)\|^2 & = (aTx-T(ax), aTx-T(ax)) \\
& = (aTx,aTx)-2(aTx,T(ax)+(T(ax),T(ax)) \\
& = a^2(Tx,Tx)-2a(Tx,T(ax))+(T(ax),T(ax))\\
& = a^2(x,x)-2a(x,ax)+(ax,ax) \\
& = a^2\|x\|^2-2a^2\|x\|^2+a^2\|x\|^2 = 0.
\end{align*}

Note that I've implicitly assumed the Hilbert space is real. The calculation works basically the same way in the complex case though.

After you check for yourself that $\|T(x+y)-(Tx+Ty)\|^2=0$, just appeal to the fact that $\|v\|=0$ if and only if $v=0$. Thus $T(x+y) = Tx+Ty$ and $T(ax)=aTx$.

Title: Prove T is a linear operator


问:
Calculate $\vec{A}$ $\cdot$ $(\vec{B} \times\vec{C})$ = $\vec{A}$ $\cdot$ $\vec{B}$ $\cdot$ $\vec{C}$
Let $\vec{A} = (1,2,3), \vec{B} = (3,-1,4), \vec{C}= (0,3,0) $

Calculate $\vec{A}$ $\cdot$ $(\vec{B} \times\vec{C})$ = $[\vec{A}$ $\cdot$ $\vec{B}$ $\cdot$ $\vec{C}]$

I first calculate $\vec{A}$ $\cdot$ $(\vec{B} \times\vec{C})$ = $1 \times\begin{bmatrix}-1 & 4\\3 & 0 \end{bmatrix}$-2 $\begin{bmatrix}3 & 4\\0 & 0\end{bmatrix}$ + 3  $\begin{bmatrix}3 & -1\\0 & 3\end{bmatrix}$

$=1(-12)-2(0) +3(9)$ = $15$

I then try to calculate $\vec{A}$ $\cdot$ $\vec{B}$ $\cdot$ $\vec{C}$, I know that $\vec{A}$ $\cdot$ $\vec{B}$ can be calculated as $a_xb_x+a_yb_y+a_zb_z$ I then try to similarly calculate $\vec{A}$ $\cdot$ $\vec{B}$ $\cdot$ $\vec{C}$ in the same way as $a_xb_xc_x+a_yb_yc_y+a_zb_zc_z$ = $(1\times3 \times0) + (2 \times -1 \times 3) + (3 \times 4 \times 0) = 0 +-6+0 = -6$

$-6\neq 15$,

but $\vec{A}$ $\cdot$ $(\vec{B} \times \vec{C})$ = $\vec{A}$ $\cdot$ $\vec{B}$ $\cdot$ $\vec{C}$

So I'm assuming that I'm calculating the dot product wrong... Any solutions hints or tips??

答: Here: The equality you are trying to prove is just a definition. The point is that there is no triple dot product as you intended in your question.</p>


问:
Constant function formal definition
A generic function $F : X \rightarrow Y$ can be defined as $\forall x \in X \exists !y \in Y (x, y) \in F$. How about the formal definition of a constant function? The formula $\exists !y \in Y \forall x \in X (x, y) \in F$ does not seem right to me, since it does not prevent that the same element in the set $X$ is mapped to multiple different elements in the set $Y$. Instead, it only prevents that two different constant lines can be mapped from the same set $X$.

答:
A generic function $F$ mapping set $X$ to set $Y$ can also be defined as follows:

$~~~~~~\forall a\in X: F(a)\in Y~~~~~$ or $~~~~~\forall x: [x\in X \implies F(x)\in Y]$

Then a constant function would be defined as:

$~~~~~~\forall a\in X:F(a)=y_0~~~~~$ or $~~~~~\forall a:[a\in X \implies F(x) = y_0]~~~ $for some $y_0 \in Y$

Title: Constant function formal definition


问:
What single estimator of a 5000 series of 20 elements for a non-normal distribution should be used?
The data are traffic counts of one minute (20 different days) of 5000 streets. The problem is that the mean of the 20x5000 observations does not explain much as it is not a normal distribution.  Also, there is high variability in the street itself,  due to the issues such as traffic jams, traffic lights, accidents, etc., there are big variations, even a lot of zeros.  Probably a smooth maximum? I lack of deeper knowledge, but sure that there is a mathematical approach that provides a (one!) metric that can better represent the “intensity” or “flow” of that city.
Thank you!</p>

答:
Just because the data is non-normal, doesn't mean that the mean isn't potentially a useful value. If you data are from a simple random sample of the full data (or of the underlying model), then the sample mean is still an unbiased estimator of the population (or model) mean.

However, it sounds like your problem is more trying to understand what property you even want to estimate, which is a conceptual one. Traffic is a flow, so static counts might not even be the best thing to measure - this is a problem in real-life statistics all the time, where you have to determine whether the data you have actually provides any information about the thing you're actually looking to describe. Sometimes, the answer is no, and you can only answer another, related question.

My first suggestion is to get in contact with someone whose expertise is related to measuring traffic - e.g. local road authorities - and see if they have any information on what they tend to measure, or design roads and signals based on. From the looks of it, traffic flow is a pretty dense theoretical field and there are a lot of factors other than just "number of cars at an intersection", so you may have to heavily cut down your expectations.

My second suggestion is to consider something relating to how long traffic stays at its peak capacity, or how long it takes to recover from being at peak. There might even be a nice maximum-likelihood estimator somewhat in the vein of the German tank problem (although you can't apply that one directly since you don't have unique sequential identifiers for the vehicles).


问:
What is the mean and variance of $1/X$ if $X$ is skew normal distributed?
The density function of skew normally distributed random variable $X$ is given as

$$f(x)=\frac{2}{\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)\Phi\left(\alpha \left(\frac{x-\mu}{\sigma}\right)\right)$$
where $\phi$ is standard Gaussian density, $\Phi$ is standard Gaussian cumulative distribution function with location $\mu$, scale $\sigma$ and shape $\alpha$ parameters.

What is the mean and variance of the random variable $Y=1/X$?

答: If $X$ is skew normal, then $Y = 1/X$ does not possess a first moment. Thus mean and variance are not defined.


问:
Proving that $\sim$ defines an Equivalence relation on $S_3$
I am studying for my final in Abstract Algebra, and I need some help.
Here is the question:

Define a relation on the group $S_3$ by $\sigma \sim \tau$ iff there exists a $\rho \in S_3$ such that $\tau=\rho \sigma \rho^{-1}$

Prove that $\sim$ defines an Equivalence relation on $S_3$.
I know I need to show $\sim$ is reflexive, symmetric, and transitive.

答:
It's a matter of writing up the definitions, there's nothing deep here:

Reflexive: $\sigma \sim \sigma$ because $\sigma = e\sigma e^{-1}$, where $e\in S_3$ is the identity.

Symmetric: if $\sigma\sim \tau$, there is $\rho \in S_3$ such that $\tau = \rho \sigma \rho^{-1}$, and this implies that $\sigma = \rho^{-1}\tau \rho = (\rho^{-1})\tau (\rho^{-1})^{-1}$ --- this means that $\tau \sim \sigma$.

Transitive: if $\sigma \sim \tau$ and $\tau \sim \theta$, there are $\rho, \upsilon\in S_3$ such that $\tau = \rho\sigma\rho^{-1}$ and $\theta = \upsilon \tau \upsilon^{-1}$ --- this implies that $\theta = \upsilon \rho\sigma\rho^{-1}\upsilon^{-1}=(\upsilon\rho)\sigma(\upsilon\rho)^{-1}$, meaning that $\sigma\sim\theta$.

The fact that you're dealing with permutations and $S_3$ is irrelevant and this is a particular instance of a general phenomenon: if $G$ is a group acting on a set $X$ and we define $\sim$ on $X$ via $x\sim x'$ if there is $g\in G$ with $x'=g\cdot x$, then $\sim$ is an equivalence relation (and $X/_{\sim}$ is denoted by $X/G$).


问:
How to find a matrix $A$ given two vectors which are solutions to $Ax=0$ and two vectors which are not
How do I find a $3\times 4$ matrix $A$ given two $4\times 1$ vectors which are solutions to $Ax=0$ and two $4\times 1$ vectors which cannot be solutions?

I have this question and so far the only method I could think of is working out the value of each number in the matrix using simultaneous equations from the two linear systems which do actually give $Ax=0$. I found that I can't seem to work out how to do this simultaneously and I can't actually find the value of any of the numbers in $A$ without them being in terms of the other numbers in the matrix.

答:
So you're given vectors $x$ and $y$ which are solutions. And you're also given $u$ and $w$ which are not solutions.

Let us assume that $x$, $y$ are linearly independent. (Otherwise one of them is redundant.) Let us also assume that $z,w\notin\operatorname{span}(x,y)$. (Otherwise no solution is possible.)

There is a standard way to find system of two linear equations such that the solution space is exactly $\operatorname{span}(x,y)$. (We simply solve the conditions on the coefficients implied by the fact that $x$ and $y$ is solution.)

Then we simply add the third equation $0=0$. (Or some linear combination of the two equations we got.)

Since the solution space of this system is exactly $\operatorname{span}(x,y)$, the vectors $z$ and $w$ are not solutions.

Notice that it was relevant that if we have two linearly equations with four unknowns, then the dimension of the solution space is equal to $2$. In general, the dimension of the solution space is  $\dim(S)=n-\operatorname{rank}(A)$, where $n$ is the number of variables. This is Rank–nullity theorem.


问:
I'm having trouble with a definition of the upper and lower limits, and a theorem that follows it.
The following is the definition.

Let $\{s_n\}$ be a sequence of real numbers. Let $E$ be the set of numbers $x$ such that $s_{n_{k}}\rightarrow{x}$. This set $E$ contains all subsequential limits, plus possibly the numbers $+\infty$, $-\infty$. Now, putting $s^* = \sup (E)$ and $s_{*} = \inf (E)$. The numbers $s^{*}$ and $s_{*}$ are called upper and lower limits of ${s_n}$. $${\lim_{n \to \infty}} \sup s_n = s^{*}$$ $${\lim_{n \to \infty}}\inf s_n = s_{*}$$

I have no idea what this means.

From my understanding, if a sequence has a limit the limit is unique. Why is this definition implying that a sequence has multiple limits?

Or is it implying that a subsequence of a sequence can have a different limit ?

My book lacks in examples and I cannot figure out what's going on at all ...

Also, I learned that $\infty$ is not a number. Why is this definition treating it as if it is one ?

Theorem. Let ${s_n}$ be a sequence of real numbers. Then $s^*$ has the following properties.

  a) $s^* \in E$

  b) If $x>s^*$, $\exists N \in \Bbb Z$ such that $\forall n \ge N$, $s_n < x$.

Moreover, $s^*$ is unique.

I was able to understand the proof of a), and b), but I couldn't really understand the proof of the uniqueness. The book says,

Suppose $p<q$ where both $p$ and $q$ are upper limits. Choose $x$ such that $p < x < q$. Since $p$ satisfies b), we have $s_n<x$ for $n \ge N$. But then $q$ cannot satisfy a).

1) Why is it guaranteed that such $x$ between $p$ and $q$ exist?

2) Why will $q$ not satisfy a) ?

答:
Why is this definition implying that a sequence has multiple limits?

Andres has given a good answer to this in the comments. In short, even if a sequence has no unique limit, subsequences of the sequence can have a unique limit. The quoted definition is talking about limits of subsequences of a sequence.

Why is it guaranteed that such x between p and q exist?

We assumed that $p<q$, this means that $q-p=\varepsilon$ for some $\varepsilon>0$. You can take $x$ to be $p+\frac{\varepsilon}{2}$.

Why will q not satisfy a) ?

Since $s_n<x<q$ for $n \ge N$,

$${\lim_{n \to \infty}} \sup s_n \le x<q.$$

So $q$ can't be the supremum of $s_n$ as $n\rightarrow \infty$ because $x$ will always be a tighter bound.


问:
Inverse of matrix - definition
Usually the inverse of a square $n \times  n$ matrix $A$ is defined as a matrix $A'$ such that:

$A \cdot A' = A' \cdot A = E$

where $E$ is the identity matrix.

From this definition they prove uniqueness but using significantly the fact that $A'$ is both right and left inverse.

But what if... we define right and left inverse matrices separately. Can we then prove that:

(1) the right inverse is unique (when it exists)

(2) the left inverse is unique (when it exists)

(3) the right inverse equals the left one

I mean the usual definition seems too strong to me. Why is the inverse introduced this way? Is it because if the inverse is introduced the way I mention, these three statements cannot be proven?

答:
The proposed, accepted answer, has a bug. I mean, the conclusion $(AB=I)\Rightarrow (BA=I)$ is true for $n\times n$ matrices, but the proof is wrong.

Suppose that $AB=I$.

IF there exists $X$ so that $BX=I$ -- which you don't know --

THEN you can conclude that $ABX=AI$, whence $X=A$.

But may be $X$ does not exist. In fact the proposed proof never used the fact that the matrices are square $n\times n$, which is crucial.

Example. Suppose $A=\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}$ and $B=A^t=\begin{pmatrix}1&0\\ 0&1\\ 0&0\end{pmatrix}$

Then $AB=I$, more precisely $AB=I_2$ the identity $2\times 2$.

IF it would exist $X$ so that $BX=I$, more precisely $BX=I_3$ the identidy $3\times 3$, then one could conclude that $X=A$. BUT in fact $BA=\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&0\end{pmatrix}$

It is still true, for $n\times n$ matrices, that $AB=I$ if and only if $BA=I$, but this fact has not a two-lines proof. (See for instance this question).

Note: the above example can be adapted to "square" matrices of infinite size. That is to say, endomorphisms of infinite dimensional vector spaces that have a left inverse need not to have a right inverse.

Finally, to give a complete answer to the original question, observe that any matrix $C=\begin{pmatrix}1&0\\ 0&1\\ x&y\end{pmatrix}$ is a right-inverse of $A$. (And so $C^t$ is a left inverse of $B$). Which shows that right and left inverses are not unique outside the realm of $n\times n$ matrices.

So yes, the fact that for $n \times n$ matrices a right inverese is also a left inverse is crucial.

Title: Inverse of matrix - definition


问:
ILP constraint on sum of k variables out n
Say I have $n$ variables, denoted $v_n$ and I want to constrain them in the following way
$\sum_{j=1}^{j=k}v_j\le1$ $\forall$ groups of $v$ with cardinality $k$. Obviously there are $n \choose k$ groups, which would lead to a naive implementations with $n \choose k$ constraints. This does not scale and I was wondering if there is a better way of encoding this in ILP or SAT formulation. Thanks.

答: The single constraint $$\sum_{j=1}^n v_j \le 1$$ suffices.  If more than one $v_j=1$, you would have a $k$-subset that violates one of your required constraints.


问:
What about the middle terms in this computation?
Let $A$ be a real symmetric matrix and $u,v$ be real ortoghonal vectors, $a, b\in\mathbb{R}$.
On my notes I wrote:
$$(au-bv)^T A(au-bv) = a^2 (u^T Au) + b^2 (v^T A v).$$
What about the middle terms?
It seems that here the middle terms
$$-ab (u^T Av)-ab(v^T Au)=0.$$
Could anyone please help me to understand why?

Thank you in advance!

答:
The statement that you have written in your notes is not generally correct. For instance, take $u = (1,0)^T, v = (0,1)^T$, and
$$
A = \pmatrix{2&1\\1&3}.
$$
Verify that
$$
(au - bv)^TA(au - bv) = 2a^2 + 3b^2 - 2ab.
$$

For the specific case that $a = u_n$ and $b = v_n$, consider
$$
u = (1,-2)^T, \quad v = (2,1)^T
$$
with the same matrix $A$. If we take $a = -4$ and $b = 3$, we get
$$
(au^T - bv)^TA(au - bv) = 9a^2 + 15b^2 - 10ab \neq
9a^2 + 15b^2.
$$


问:
Does there exist a smooth function which is nowhere analytic?
Smooth means has derivatives of all order, and analytic means can be given as a convergence of power series.</p>

答:
A usual example is
$$
f(x)=\sum_{k\in A} e^{-\sqrt{k}}\cos kx
$$
where $A=\{2^n:n\in\mathbb{N}\}$.</p>


问:
Determine linear independence when....
let , $ys=-(x_1t_1+...+x_nt_n)$

If $y\not= 0$, then

$s=-(1/y)(x_1t_1+...+x_nt_n)$

$s=(-x_1/y)t_1+...+(-x_n/y)t_n$

Thus, $s \in \operatorname{span}(x_1,...,x_n)$

Is my proof correct here?

答: No, it is not. Actually, $w\notin\operatorname{span}(v_1,\ldots,v_k)$. Otherwise, there would be scalars $\alpha_1,\ldots,\alpha_k$ such that $w=\alpha_1v_1+\cdots\alpha_kv_k$. But then$$1\times w-\alpha_1v_1-\cdots-\alpha_kv_k,$$and therefore (since $1\ne0$) the set $\{w,v_1,\ldots,v_k\}$ would be linearly dependent.

Title: Determine linear independence when....


答:
Your proof is not correct.  If $a_1v_1 + \ldots + a_kv_k + bw=0$ and $\{ v_1, \ldots, v_k, w \}$ is a linearly independent set, then you know each and every coefficient (including $b$) is equal to $0$.  That's the definition of linear independence.

So what you want to do is assume that $w \in \operatorname{span} (v_1, \ldots, v_k)$ and arrive at a contradiction (by finding a non-trivial linear combination of a linearly independent set that's equal to $0$).  The answer above shows you how to do that.

Title: Determine linear independence when....

Body:
let , $ys=-(x_1t_1+...+x_nt_n)$

If $y\not= 0$, then

$s=-(1/y)(x_1t_1+...+x_nt_n)$

$s=(-x_1/y)t_1+...+(-x_n/y)t_n$

Thus, $s \in \operatorname{span}(x_1,...,x_n)$

Is my proof correct here?


问:
Can we find two subgroups $A,B \leqslant D_8$ such that $AB \not\leqslant D_8$?

Give an example of a group $G$ and two subgroups $A,B \leqslant G$ such that $AB$ is not a subgroup of $G$.

For reference, the presentation of $D_8$ is
$$D_{8} = \langle r,s: r^4 = s^2 = 1, rs = sr^{-1} \rangle.$$

I know that we can find an example that suffices if we inspect $S_3$ already so I am trying to see if I can do the same for $D_8$. Let $A = \langle r \rangle$ and $B = \langle s \rangle$. Then $AB = \langle rs \rangle$, but I suspect that $\langle rs\rangle$ doesn't form a subgroup in $D_8$.

I think that the set $AB$ would not be closed with under inverses. However I haven't been able to convince myself of the fact. Here is my idea so far:

Using the relation $rs = sr^{-1}$, we have
\begin{align}
(rs)(sr^{-1})^{-1} &= e.
\end{align}
Then $(sr^{-1})^{-1} = (s^{-1}r^{-2})$, but that element is not in $AB$. I'm not sure if I am allowed to distribute the inverse though. If I am not, I don't think my argument makes any sense then. Any clarification if I'm taking the right approach or not would be very helpful, thanks.

答:
There is a very large number of elementary errors in what you write.

First, if $A$ and $B$ are subgroups, then they are not elements. You should not have "$A,B,\in G$", you should have $A,B\subseteq G$ or $A,B\leq G$.

Second, if $A$ and $B$ are subsets/subgroups of a group, then $AB=\{ab\mid a\in A, b\in B\}$. If you take $A=\langle r\rangle = \{1, r, r^2, r^3\}$ and $B=\langle s\rangle = \{1, s\}$, then
$$AB = \{1, r, r^2, r^3, s, rs, r^2s, r^3s\} = D_8,$$
and not, as you claim $\langle rs\rangle$.

Third: by definition, the notation "$\langle X\rangle$" means "the smallest subgroup that contains $X$". So $\langle rs\rangle$ is, by definition, the smallest subgroup of $D_8$ that contains $rs$; it would be $\{e,rs\}$, and it is of course a subgroup, by definition. You seem to think that $AB=\{rs\}$. This is not true, since $A\neq\{r\}$ (if it were, then it wouldn't be a subgroup either), and $B\neq\{s\}$ (same reason).

Three such elementary errors are way too many; one would be bad, two is catastrophic, three is... a very large number.

As it happens, you will not be able to get any examples if $A$ (or $B$) is contained in $\langle r\rangle = \{1,r,r^2,r^3\}$; the reason is that every subgroup of $\langle r\rangle$ is normal in $D_8$, and for a normal subgroup $A$, $AB$ is always a subgroup whenever $B$ is a subgroup (though you may not know this yet). So you should look for examples in which neither of the two subgroups is contained in $\langle r\rangle$. And you should really be much more careful with what the objects you are working with are. Almost nothing


问:
Show that $\pi^{-1} (\pi(U))=U$ for $\pi:X \to X/A$ canonical projection and $U$ open in $X$ such that $A \subset U$ or $A \cap U = \emptyset$,
Let $(X, \tau )$ be a topological space and let $A \subset X$ be a closed subset. Let $X/A$ be equipped with the quotient topology, and let $\pi : X  \to  X/A$ be the canonical projection (by collapsing $A$ to a point in $X/A$).

If $U$ is an open subset of $X$ such that either $A \subset U$ or $A \cap U = \emptyset$, show that $$ \pi^{-1} (\pi(U))=U.$$

We wish prove that  $ \pi^{-1} (\pi(U))=U$ via mutual inclusion. Suppose first that $x\in U$, we then see that $\pi(x) \in \pi(U)$. We look at what the inverse image means for $S=\pi(U)\subset X/A$:
$$\pi^{-1}(S) = \{ z \in X | \pi(z) \in S \}.$$
$$\pi^{-1}(\pi(U)) = \{ z \in X | \pi(z) \in \pi(U) \}.$$
We see that $x\in U\subset X$  and indeed  $\pi(x) \in \pi(U)$, so we have that $ x \in \pi^{-1}(\pi(U))$.  We have our first inclusion: $$U\subset  \pi^{-1} (\pi(U)).$$  This was the easy direction I think. This is just a set-theoretic notion and this inclusion always holds, we did not use any properties.

Now we would easily get the other direction via injectivity of $\pi$ (it is always surjective, but not injective since it collapses all of $A$ to a single point), but somehow the conditions imply that   $U\supset  \pi^{-1} (\pi(U)).$  It seem to, in a sense, be "injective on $U$" if I am interpreting this correctly, but I do not see how to get the inclusion out of the assumptions.

答:
If $A\cap U=\varnothing$, you can easily check that $\pi\upharpoonright U$ is injective and hence that $U=\pi^{-1}\big[\pi[U]\big]$.

Now suppose that $A\subseteq U$, and let $\pi[A]=\{p\}$, so that $\pi(x)=p$ iff $x\in A$. Then $\pi\upharpoonright U\setminus A$ maps $U\setminus A$ injectively to $\pi[U]\setminus\{p\}$, so

$$\begin{align*}
\pi^{-1}\big[\pi[U]\big]&=\pi^{-1}\big[\pi[U\setminus A]\big]\cup\pi^{-1}\big[\pi[A]\big]\\
&=(U\setminus A)\cup\pi^{-1}[\{p\}]\\
&=(U\setminus A)\cup A\\
&=U\,.
\end{align*}$$


问:
Number of partitions of $n$ in which parts can occur at most 7 times
I have got the following answer without detailing the whole answer as I am looking precisely for the notation used here as I am new to graph theory, so please pardon me. After a series of operations, I got the following

$$A(x)=\sum\limits_{n=1}^{\infty} p(n)q^n = ... = \frac{1}{(9^6\times9^6)_{\infty}(9^6\times9^6)_{\infty}
(9^6\times9^2)_{\infty}(9^4\times9^6)_{\infty}}$$

What does $\infty$ notation here referring to?

答:
It looks like a bad OCR scan.  I suspect that some of the $9$s are supposed to be $q$s, and you have mixed up $x$ and $q$.

If $p(n)$ is the number of such partitions, the correct generating function is
$$A(x) = \sum_{n=0}^\infty p(n) x^n = \prod_{k=1}^\infty \sum_{i=0}^7 x^{ki} = \prod_{k=1}^\infty \frac{1-x^{8k}}{1-x^k}$$


答: Take $B=\langle s\rangle$ and $A=\langle rs\rangle$.  Then $AB=\{e,s,rs,r\}$, is not a subgroup of $D_8$.

Title: Can we find two subgroups $A,B \leqslant D_8$ such that $AB \not\leqslant D_8$?

Body:
Give an example of a group $G$ and two subgroups $A,B \leqslant G$ such that $AB$ is not a subgroup of $G$.

For reference, the presentation of $D_8$ is
$$D_{8} = \langle r,s: r^4 = s^2 = 1, rs = sr^{-1} \rangle.$$

I know that we can find an example that suffices if we inspect $S_3$ already so I am trying to see if I can do the same for $D_8$. Let $A = \langle r \rangle$ and $B = \langle s \rangle$. Then $AB = \langle rs \rangle$, but I suspect that $\langle rs\rangle$ doesn't form a subgroup in $D_8$.

I think that the set $AB$ would not be closed with under inverses. However I haven't been able to convince myself of the fact. Here is my idea so far:

Using the relation $rs = sr^{-1}$, we have
\begin{align}
(rs)(sr^{-1})^{-1} &= e.
\end{align}
Then $(sr^{-1})^{-1} = (s^{-1}r^{-2})$, but that element is not in $AB$. I'm not sure if I am allowed to distribute the inverse though. If I am not, I don't think my argument makes any sense then. Any clarification if I'm taking the right approach or not would be very helpful, thanks.


问:
Linearizing constraint ( multiplication of binary variables)
I am trying to think of contraint(s) that can linearize constraint below.

$\sum_{T} \sum_{TR} Z_{(T,D)}* Y_{(TR,T)}  \leq CAP_{(D)} \forall D$

Both Z nad Y are Binary Variables and CAP is capacity parameter.

I tried to apply big M but without luck.

答:
Although the usual linearization of a product of two binary variables requires three inequalities, you can get by with only one here.  You can enforce
$(Z \land Y) \implies X$ via linear constraint $Z + Y - 1 \le X$.


问:
Find $a\in \mathbb R$ so that $\{(a, 1, 1), (1,a,1), (1,1,a)\}$ is a base of $\mathbb R^3$
Find $a\in \mathbb R$ so that the list $v=[v_1, v_2, v_3]^t$ is a base of $\mathbb R^3$, where $$v_1 = (a, 1, 1),\ v_2= (1,a,1),\ v_3 = (1,1,a),$$
where $^t$ means transpose.

Any help is highly appreciated, because I don't really know how I could solve this.

答:
I have never liked to use matrices or determinants to answer questions about vector spaces.  A "basis" for a vector space of dimension n is a set of vectors that has three properties:

The vectors span the space.

The vectors are independent.

There are n vectors in the set.

Further any two of them are sufficient to prove the third.

This is a set of three vectors so to prove it is a basis for $R^3$ it is sufficient to prove the vectors are independent.  That is, we want to show that if numbers p, q, and r are such that $pv_1+ qv_2+ rv_3= p(a, 1, 1)+ q(1, a, 1)+ r(1, 1, a)= (pa+ q+ r, p+ qa+ r, p+ q+ ra)= (0. 0, 0)$ then p= q= r= 0.

We have pa+ q+ r= 0, p+ qa+ r= 0, and p+ q+ ra= 0..  Subtract the first equation from the second to get (1- a)p+ (a- 1)q= 0.  Divide by 1- a to get p- q= 0 so p= q.  The first two equations become (1+ a)p+ r= 0 so r= -(1+ a)p and the third equation becomes p+ q+ ra= 2p- (1+ a)p= (3+ a)p= 0.  As long as $a\ne -3$ the vectors are independent so a basis for $R^3$.


问:
Prove or refute: If $\pi_1(X,x_0)\simeq 0$ for all $x_0 \in X$, then $X$ is path connected

Prove or refute: If $\pi_1(X,x_0)\simeq 0$ for all $x_0 \in X$, then $X$ is path connected

Let $x_0,y_0\in X \Rightarrow \pi_1(X,x_0)\simeq \pi_1(X,y_0)$, but I don't know more, or if it's true.

答:
The statement is not true. Let $X=\{p,q\}$ with $p\neq q$ be equipped with the discrete topology. The fundamental group depends only on the path-connected component the basepoint is in, so one has that $\pi_1(X,p)\cong 0$ and $\pi_1(X,q)\cong 0$ (each path-connected component is trivially contractible, as they consist of only one point), but $X$ is not path-connected.

In fact, this argument generalizes to any finite set with the discrete topology.


问:
Normal Distribution, Probability a point is inside of a circle in $\mathbb{R^2}$
We have $X$~$Normal(0,1)$,$Y$~$Normal(0,3)$. What is the Probability that $M \in [(x,y) \in \mathbb{R^2} : 1 \leq 3x^2 + y^2 \leq 3]$?
\n

What I did, I set $\tilde{Y} := \frac{Y}{\sqrt{3}},\tilde{Y}$~$Normal(0,1)$. Now we have $\frac{1}{3} \leq x^2 + \tilde{y}^2 \leq 1$, a donut. I'm lost here, my professor hinted polar co-ordinates but I don't really see how, because if I set

$$
X:= \cos\theta, Y:= \sin\theta,
$$
$\theta$ would have 2 different distribution function.

答: The probability density for $\theta$ is uniform on $[0, 2\pi)$ because $(X,\tilde Y)$ is radially symmetric.  So you need to calculate the density of $R = \sqrt{X^2 + \tilde Y^2}$ and compute $\Pr[1 \le R \le 3]$ or equivalently, $\Pr[1 \le R^2 \le 9]$.  Hint:  $R^2$ is a chi-squared random variable with $2$ degrees of freedom.


问:
A question concerning polynomial solvable by radicals
We know from Galois Theory that a polynomial is solvable by radicals if and only if its Galois group is solvable. On the other hand solvable by radicals for example means that the equation $X^n-1=0$ is always solvable by radicals (its Galois group is abelian), but this only means that we can find a solution by saying it is $1^{1/n}$, which is a radical. As for $n\le 6$ we can find its solution in the form $a+ib$ where $a,b$ are representable by real radicals. I guess this is not always possible for larger $n$. (I can see how it could be up to $n\le10$, but any higher?) Is there a theory concerning this kind of problem (whether a polynomial can be solved by "real radicals")?</p>

答:
This article "Solution of Polynomials by Real Radicals" might interest you

http://www.jstor.org/stable/2323164

It closes with the observation:

"We close with the observation that solvable polynomials with real roots but which are not solvable by real radicals seem to abound. For example, for any prime p, the polynomial
$$f(x) = x^3 - 2px + p$$
has this property... It is amusing to solve this polynomial by Cardan's method to see where nonreal numbers come in."


问:
Prob question on dice rolling and expected value
So let's say I roll 3 dice.

If they show all the same number, I will EARN 10£

If all numbers are different, I will LOSE 2£

If they show two numbers equal and one different, I will EARN 5£

What would be my expected return per roll?

If we calculate the probabilities (if I am not wrong)

6/216 --> earn 10£

90/216 --> earn 5£

120/216 --> lose 2£

How would I proceed to calculate the expected value per roll?

EDIT

Would the EV be: (6/216)x10 + (90/216)x5 - (120/216)x2 = 1.25?

答:
By definition for discrete random variable, the expected value is the sum of the product of possible values of the random variable with it's probability. Say, X can be $10$ with probability $\frac{1}{2}$, X can be $\left(-100\right)$ with probability $\frac{1}{4}$, and X can be $0$ with probability $\frac{1}{4}$. Then the expected value is $10 \cdot\frac{1}{2} + \left(-100\right)\cdot\frac{1}{4} + 0\cdot\frac{1}{4} = \left(-20\right)$

So, the expected value is $10 \cdot\frac{6}{216} + 5\cdot\frac{90}{216} + \left(-2\right)\cdot\frac{120}{216} = \frac{270}{216}$


答:
So... square matrices over a field are special: one can prove that if a square matrix has a left inverse, then it also has a right inverse. And once you know that, it is easy to show that the left and right inverse must be equal. This holds in general for groups/rings: if $x$ is an element, and there exist $a,b$ such that $ax=1$ and $xb=1$, then $a=b$:
$$a = a1 = a(xb) = (ax)b = 1b = b.$$

(Use capital letters and the identity for matrices).

Now, if $A$ is a square matrix over a field, and there exists a matrix $B$ such that $AB=I$, then interpret $A$ and $B$ as linear transformations. Then $AB$ is bijective, hence $A$ is surjective; but that means that $A$ is full rank, hence by the Dimension Theorem has trivial nullity, hence $A$ is also injective. Since $A$ is therefore a bijection, it has a left inverse $C$ (which is also linear, and thus corresponds to a matrix), and thus $A$ has a two-sided inverse.

Similarly, if $A$ has a left inverse $C$ with $CA=I$, then $A$ is one-to-one, hence full rank and so surjective, so it has a right inverse and the argument proceeds as before.

In arbitrary rings you can have elements that have a left inverse but no right inverse; in such cases, you will have multiple left inverses but no right inverse. For if $x$ is an element such that there exists $a$ with $ax=1$, but $xb\neq 1$ for all $b$, then $xa\neq 1$, so $xa-1\neq 0$. Then $(xa-1)x = xax-x = x-x = 0$, so then $(a+xa-1)x = ax+(xa-1)x = 1$, but $a+xa-1\neq a$. Thus, $x$ has at least two left inverses. A symmetric argument shows that if $x$ has a right inverse but no left inverse, then it has at least two right inverses.

And yes, it is possible to have rings in which some elements have left inverses but no right inverses. Consider the vector space $\mathbb{R}[x]$ of all polynomials with coefficients in $\mathbb{R}$, and the ring of all linear transformations from $\mathbb{R}$ to itself. The linear transformation $T(p(x)) = xp(x)$ is one-to-one but not onto, so it has left inverse but no right inverse. In fact, it has infinitely many left inverses. Similarly, the linear transformation $T(p(x)) = p'(x)$ (the derivative) is onto, but not one-to-one, so it has a right inverse (in fact, infinitely many), but no left inverse.

Title: Inverse of matrix - definition

Body:
Usually the inverse of a square $n \times  n$ matrix $A$ is defined as a matrix $A'$ such that:

$A \cdot A' = A' \cdot A = E$

where $E$ is the identity matrix.

From this definition they prove uniqueness but using significantly the fact that $A'$ is both right and left inverse.

But what if... we define right and left inverse matrices separately. Can we then prove that:

(1) the right inverse is unique (when it exists)

(2) the left inverse is unique (when it exists)

(3) the right inverse equals the left one

I mean the usual definition seems too strong to me. Why is the inverse introduced this way? Is it because if the inverse is introduced the way I mention, these three statements cannot be proven?


问:
Additional criteria to establish graph isomorphism apart from moving vertice and renaming?
The task is to check whether the two graphs are isomorphic:

The solutions is given as:

The instructions say we can rearrange the vertices and rename them. However looking at the solution I could also rearrange the vertices V5 and V2 in as such that their position switches. (Both are connected to V1 and V3). How can I tell which solution is correct?

答:
They are both correct. In general, we say that two objects, graphs in this case, are isomorphic if there exists an isomorphism between them. In this context that means that there is a way to relabel/rearrange the vertices to transform one graph into another. However, that does not preclude the possibility that there is more than one isomorphism, or in your case that there is more than one way to rearrange and relabel the vertices to make the two graphs the same. So both solutions are correct.

A nice example of this in the graph theoretic context is given by the $C_n$, the cycle graph on $n$ vertices. Obviously $C_n \cong C_n$, but there are actually $n$ possible isomorphism, corresponding to the $n$ relabeling of the vertices that leave the vertices in the same order.


问:
The proof that $\kappa$ is the only measurable in L[U]
Let $\kappa$ be measurable and $U$ be its normal measure. I didn't understand a step in the proof that $\kappa$ is the only measurable cardinal in $L[U]$.

So suppose there is another measurable $\lambda < \kappa$ such that $\lambda$ is measurable in  $L[U]$. Then we consider the elementary embedding $j: L[U] \rightarrow M$ given by $\lambda$. We now wish to show that $j(U) = U\cap M$. As $\kappa$ is a strong limit with $cf(\kappa) > \lambda$, $j(\kappa)=\kappa$. Then the textbook (e.g., Kanamori p 262) says that $U\cap M \subseteq j(U)$ since $j(U)$ is an ultrafilter on $\kappa$ in $M$. I don't see how this follows. What am I missing here?

Thanks for the help!

答: Since $j(U)$ is an ultrafilter on $\kappa$ in $M$, for every $A\subseteq\kappa$ that is in $M$, either $A\in j(U)$ or $\kappa\setminus A\in j(U)$.  This means we cannot have a strict inclusion $j(U)\subset U\cap M$, since the only way $U\cap M$ could be strictly larger is to contain both a set and its complement, which is impossible since $U$ is a proper filter.  So once you show $j(U)\subseteq U\cap M$, you also get the reverse inclusion for free.


问:
Joint probability of uniform random variables and absolute difference
Let $X,Y$ be independent standard uniform random variables. How do we compute the joint density
$$P(a\leq X\leq 1-a, a \leq Y \leq 1-a, |X-Y|\geq a) $$

for some positive value $a$. I can split the absolute value into two separate conditions $X-Y\geq a$ and  $-(X-Y) \geq a$. Now, I have way too many conditions. Any leads is appreciated.

Many thanks.

答:
By symmetry the required probability is $2P(a\leq X \leq 1-a,a\leq Y \leq 1-a, X \geq Y+a)$. This gives $\int_a^{1-2a} \int_{y+a}^{1-a} dxdy$ provided $a<1-2a$ or $a<\frac1 3$. [The inside integral requires $y+a <1-a$ or $y<1-2a$. That is why the outside integral is from $a$ to $1-2a$].

I will leave the case $a\geq \frac 1 3$ to you.


问:
Apply the principle of mathematical induction or its variants to show that for all positive integers $m$ and $n$ such that $0\le m<n$:
As the title says

Apply the principle of mathematical induction or its variants to show that for all positive integers $m$ and $n$ such that $0\le m < n$,

$$\sum_{k=0}^m(-1)^k\binom{n}k=(-1)^m\binom{n-1}m\,.$$

Not sure, I tried using traditional combinational proofs, but I'm a little lost on where to go.

答:
Do the induction on $n$. For your induction hypothesis suppose that

$$\sum_{k=0}^m(-1)^k\binom{n}k=(-1)^m\binom{n-1}m$$

for all $m$ such that $0\le m<n$; then you want to show that

$$\sum_{k=0}^m(-1)^k\binom{n+1}k=(-1)^m\binom{n}m$$

whenever $0\le m<n+1$. For $0\le m<n$ you can follow the usual pattern for such proofs, starting with this:

$$\begin{align*}
\sum_{k=0}^m(-1)^k\binom{n+1}k&=\sum_{k=0}^m(-1)^k\left(\binom{n}k+\binom{n}{k-1}\right)\\
&=\sum_{k=0}^m(-1)^k\binom{n}k+\sum_{k=0}^m(-1)^k\binom{n}{k-1}\\
&=(-1)^m\binom{n-1}m+\sum_{k=1}^m(-1)^k\binom{n}{k-1}\\
&=(-1)^m\binom{n-1}m+\sum_{k=0}^{m-1}(-1)^{k+1}\binom{n}k\\
&=(-1)^m\binom{n-1}m-\sum_{k=0}^{m-1}(-1)^k\binom{n}k\,;
\end{align*}$$

see if you can finish it from there.

The remaining case, $m=n$, has to be handled separately; for now I’ll leave that to you, merely noting that it follows from an application of the binomial theorem.


问:
Why $a^{\frac{q(q-1)}{2}}= -1$ over the finite field $GF(q)$ when q is odd and $a$ is primitive of the field?
Why $a^{\frac{q(q-1)}{2}}= -1$ over the finite field $GF(q)$ when q is odd and $a$ is primitive of the field?

It's true for q that is even since the characteristic of the in this case is 2 and therefore 1=-1, and $$(a^{(q-1)})^{q/2}=1^{q/2}=1=-1$$
But for q that is odd, I'm missing something.

Body:
Hint:
$(a^{\frac{q-1}{2}}-1)(a^{\frac{q-1}{2}}+1)=a^{q-1}-1=0$


问:
Equivalency of forms of Bolzano-Weierstrass theorem
Apparently the result "every bounded infinite set in $\mathbb{R}^n$ has a limit point" is equivalent to the Bolzano-Weierstrass theorem (every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence): see e.g. Proving every bounded infinite set has a limit point without using Bolzano-Weierstrass or https://uccs.edu/goman/sites/goman/files/inline-files/A%20short%20proof%20of%20the%20Bolzano-Weierstrass%20Theorem.pdf. I am not sure how this can be, given that the Bolzano-Weierstrass theorem gives you a convergent sequence in the bounded infinite set, but there is no guarantee that its limit would be a limit point (as here: Is the limit of a convergent sequence always a limit point of the sequence or the range of the sequence?). So can someone explain please? Thanks very much.

答:
As pointed out in the comments, the difference has to do with sequences which are eventually constant. Let $(x_n)$ be a bounded sequence. If we consider the set $A = \{x\in\mathbb{R}^d: x=x_n\text{ for some $n$}\}$, we cannot immediately apply the limit-point formulation of Bolzano-Weierstrass. Indeed, this set $A$ is bounded since $(x_n)$ is a bounded sequence, but $A$ may only have finitely many points and thus not satisfy the "infinite set" part of the hypotheses.

I claim that if $A$ is a finite set, then any convergent subsequence $(x_{n_k})$ of $(x_n)$ is eventually constant. To do this, just pick $\epsilon>0$ so small that $B(a,\epsilon)\cap B(a',\epsilon)=\emptyset$ for any $a,a'\in A$ with $a\neq a'$ (which is possible since $A$ is finite). If $x_{n_k}\to x$ as $k\to\infty$, then $x\in A$ since $A$ is closed (it is a finite union of closed singletons). With $\epsilon$ as above, there should be $K$ such that $k\geq K$ implies $x_{n_k}\in B(x,\epsilon)$. Since $x_{n_k}\in A$ though, our choice of $\epsilon$ forces $x_{n_k} = x$. Thus $x_{n_k}=x$ for $k\geq K$, i.e., the subsequence was eventually constant.

If the set $A$ happens to be infinite, then we infer there is a limit point $x$, and by definition of limit point we may construct a (sub)sequence of points from $A$ which converges to $x$.


问:
Taylor second order approximation for an implicit curve
I'm given the curve:

$$f(x,y)=e^{x^2+y^2}+3y-4x^2-1=0$$

and I want to find the second order Taylor approximation for y(x) at the point (0,0):

$$y(x)\approx y(0)+ y'(0)x+\frac{1}{2!}y''(0)x^2+...$$

I thought about finding the derivatives with the chain rule like:

$$y'(x)=-\frac{F_x}{F_y}$$

But I don't have $y(0)$ and I don't know how to get it. Is my approach incorrect?

I appreciate any tip!

答:
You may perform implicit differentiation:

$$\frac{dy}{dx} = \frac{2(4-e^{x^2+y^2})x}{3+2y e^{x^2+y^2}},$$ hence at $(0,0)$ we have $y'(0) = 0$.  Repeating the process yields

$$\frac{d^2 y}{dx^2} = \frac{-8 e^{3
   \left(x^2+y^2\right)} \left(x^2+y^2\right)+8 e^{2
   \left(x^2+y^2\right)} \left(8 x^2+y (4 y-3)\right)-2
   e^{x^2+y^2} \left(16 y \left(8 x^2 y+6 x^2-3\right)+82
   x^2+9\right)+72}{\left(2 y
   e^{x^2+y^2}+3\right)^3}$$

and again at $(0,0)$ we obtain $$y''(0) = 2.$$

This kind of thing is best left to a computer.  It is not strictly necessary to express the second derivative in terms of $x$ an $y$ only; if you know the lower order derivatives, the computation can be simplified.

The plot below shows the curve $f(x,y)$ in blue and the second-order approximation $y = x^2$ in orange:


问:
Describing Domain of Integration (Triple Integral)
I'm really struggling to go about starting the following problem:

This question concerns the integral,

$\int_{0}^{2}\int_0^{\sqrt{4-y^2}}\int_{\sqrt{x^2+y^2}}^{\sqrt{8-x^2-y^2}}\!z\ \mathrm{d}z\ \mathrm{d}x\ \mathrm{d}y$.

Sketch or describe in words the domain of integration.

I've tried sketching the projections of this integral's domain onto the $xy$-plane, the $yz$-plane and the $xz$-plane. I established that the projection onto the $xy$-plane is the portion of the circle $x^2+y^2=4$ on $0\le x\le2,\ y\ge0$; that the projection onto the $yz$-plane is the portion of the circle $y^2+z^2=8$ above the line $z=y$, on $0\le y\le2$; and that the projection onto the $xz$-plane is the portion of the circle $x^2+z^2=8$ above the line $z=x$, on $0\le x \le2$.

I'm not completely confident that these findings are useful in any way, and if they are, I am struggling to see the link which will allow me to solve the problem.

Do you care to get me started?

Thank you very much.

答:
The bounds on $z$ correspond to equations:

$$z^2=x^2+y^2\\x^2+y^2+z^2=8$$

The first is a cone, the second is a sphere.  They intersect above the $xy$-plane in a circle of radius $2$.  The bounds on $x$ and $y$ show that we are considering the region in between the cone and the sphere with $x>0$ and $y>0$.  Can you see it?

Let's first just consider the $xy$-plane.  As you've said, the given bounds correspond to the quarter circle of $x^2+y^2=4$ that lies in the first quadrant.  So whatever region we consider, it will lie completely above this quarter circle.  Now $z$ lies in between the cone and the sphere as I mentioned above.  You can think of this region as an ice cream cone.  However, we're only considering part of the ice cream cone above our quarter circle, so its like a quarter of an ice cream cone.

Hopefully this is helpful, along with Babak's picture.


问:
How do you calculate arrival rates and service rates?
If I have to observe a queue at an electronics shop and I need to perform at least 10 observations, how can I determine what the mean rate of arrival is? or the mean rate of service?

Like lets say that I choose to observe 10 people between 9 AM and 12 PM

```<code>Person 1 entered at 9:10   -- Left at 9:15
Person 2 entered at 9:15   -- Left at 9:20
Person 3 entered at 9:40   -- Left at 9:50
Person 4 entered at 9:59   -- Left at 10:20
Person 5 entered at 10:30  -- Left at 10:45
Person 6 entered at 10:45  -- Left at 11:00
Person 7 entered at 11:10  -- Left at 11:25
Person 8 entered at 11:20  -- Left at 11:30
Person 9 entered at 11:30  -- Left at 11:45
Person 10 entered at 11:35 -- Left at 11:50
</code>```

How can I determine from those observations what the mean arrival rate is and the mean service rate? Lets say its for hours as well as for every 3 hours.

答:
Mean arrival rate can be estimated as total number of arrivals divided by total time spent observing the system, if you assume the arrival rate is constant. If you think the arrival rate varies during the day (for instance, busy at lunch and dinner, slow between meals), then you need to estimate the arrival rate over multiple time intervals (for instance, count arrivals in each 30 minute window and double to get a per-hour rate).

Service rate is trickier. The difference between when a customer enters and when they leave combines waiting time (if there is a line) and service time. One way to estimate mean service rate would be to count the number of customers exiting (after service) per unit time while there is a continuous line (always someone waiting). Another way would be to record when each customer starts and ends service, take the difference to get their time in service, and average that. If you assume Markovian service times, the mean service rate is the reciprocal of the mean service time.


答:
You do know $y(0)$... Since $(0,0)$ belongs to the curve, $y(0)=0$. You can get $y'(0)$ and $y''(0)$ by differentiating the equality. Assuming $y=y(x)$,
$$
e^{x^2+y^2} + 3y-4x^2-1 = 0 \Rightarrow (2x + 2 y' y)e^{x^2+y^2}+3 y'-8x =0.
$$

substituting $x=0$, you get $y'(0)=0$. Differentiating again,
$$
(2+2y'' y +2y'y')e^{x^2+y^2}+(2x+2y'y)^2e^{x^2+y^2}+3y''-8 = 0$$

and substituting $x=0$ you get $y''(0) =2$.

So, finally
$$
y(x) \approx y(0) + y'(0) x  + \frac 12 y''(0) x^2 = x^2.
$$

Title: Taylor second order approximation for an implicit curve

Body:
I'm given the curve:

$$f(x,y)=e^{x^2+y^2}+3y-4x^2-1=0$$

and I want to find the second order Taylor approximation for y(x) at the point (0,0):

$$y(x)\approx y(0)+ y'(0)x+\frac{1}{2!}y''(0)x^2+...$$

I thought about finding the derivatives with the chain rule like:

$$y'(x)=-\frac{F_x}{F_y}$$

But I don't have $y(0)$ and I don't know how to get it. Is my approach incorrect?

I appreciate any tip!


答: Hint: Write $a^{\frac{q(q-1)}{2}}= a^0 a^1 \cdots a^{q-1}$ and use Wilson's theorem for abelian groups.

Body:
Why $a^{\frac{q(q-1)}{2}}= -1$ over the finite field $GF(q)$ when q is odd and $a$ is primitive of the field?

It's true for q that is even since the characteristic of the in this case is 2 and therefore 1=-1, and $$(a^{(q-1)})^{q/2}=1^{q/2}=1=-1$$
But for q that is odd, I'm missing something.


问:
Confusion about a simple proof of prime ideal implies irreducibility in UFD.
Let $R$ be a UFD if $f\neq 0$ and $\langle f\rangle$ is a prime ideal then $f$  is irreducible. Here is a short and easy proof for this claim but I do not understand the last bit.

Assume for a contradiction that $f$ is not irreducible then $f=uv$ where $u$ and $v$ are not units. By $\langle f\rangle$ a prime ideal we have either $f|u$ or $f|v.$ WLOG, assume former, then $uv=f|u$, this contradicts irreducibility of $u$.

My question:

Where in this proof have we assumed $u$ is irreducible?

{I can see that we can alter this proof a bit by using UFD property to say $f=r_1...r_k$  say where $r_i$ are irreducible then the same can follow. Alternatively $uv|u$ also contradicts $v$ is not a unit. But my question is where on this proof did we assume $u$ is irreducible?}

Many thanks in advance!

答:
This only contradicts the assumption that $v$ is not a unit, not the irreducibility of $u$.

Assume $f\mid u$. Then $u=af$ for some $a\in R$. Consequently, $$u=af=auv=av\cdot u.$$Since $R$ is a UFD (also an integral domain), we must have $av=1$. That is, $v$ is a unit, contradicting the assumption.

In fact, it has nothing to do with the irreducibity of $u$. You can see the proof goes through as long as $R$ is an integral domain.