问:
How to prove $ \prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{(-1)^{n+1}n} \,= \frac{\pi}{2e}$
How can I prove that

$$ \prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{\large{(-1)^{n+1}n}} \,= \frac{\pi}{2e}$$

The result is given here (result 48).

The source: Prudnikov et al. 1986, p. 757 is given, however I have been unable to find the book online.

Some of my attempts include:

Multiplying known infinite products for $\frac{\pi}{2}$  and
$\frac{1}{e}$

$$ \frac{\pi}{2} = \frac{2\cdot2\cdot4\cdot4\cdot6\cdot6\cdots}{1\cdot3\cdot3\cdot5\cdot5\cdot7\cdots} $$
$$ \frac{1}{e} =  \frac{1}{2} \left(\frac{3}{4}\right)^{\large{\frac{1}{4}}}\left(\frac{5\cdot7}{6\cdot8}\right)^{\large{\frac{1}{4}}} \cdots$$

as well as others given in https://arxiv.org/pdf/1005.2712.pdf

Trying to find the partial products from ${n=1} $ to $k $, Mathematica Gives:

$$\scriptsize{ \frac{\exp\left(2\left(-\zeta(-1,k+\frac{1}{2})^{(1,0)}+\zeta(-1,k+\frac{3}{2})^{(1,0)}+\zeta(-1,k+1)^{(1,0)}-\zeta(-1,k+2)^{(1,0)}\right)\right)\pi\, \Gamma(k+2)^2}{2\,\Gamma(k+\frac{3}{2})^2}} $$

However, I have been unsuccessful producing partial products on my own.

Taking the $\ln$ of the product to try to find the partial sums;

Trying to transform the following integral into infinite product.

$$\int_{0}^{\infty} \frac{\cos(x)}{1+x^2} = \frac{\pi}{2e} $$

From what I have seen in some papers, the partial products are found, then the Stirling's Approximation is used to find the limit.

Question: How can I prove the value of the given Infinite Product? Proofs or hints are both welcome.

Thank you kindly for your help and time.

答:
Maybe try splitting it into two products to get rid of the alternating sign:
$$\prod_{n=1}^\infty \left(1+\frac2n\right)^{(-1)^{n+1}n}=\prod \left(1+\frac2{2n}\right)^{-2n}\prod\left(1+\frac2{2n+1}\right)^{2n+1}$$
then look at these individually, the first one:
$$P_1=\prod_{n=1}^\infty\left(1+\frac1n\right)^{-2n}=\left[\prod_{n=1}^\infty\left(1+\frac1n\right)^n\right]^{-2}$$
although this diverges you may be able to get it to cancel with something from the other expression?


问:
Prerequisite to Stevin's decimal expansion construction - it's a complete totally ordered set.
Update: The theory here has been checked by Brian M. Scott and with the New Year in mind I am changing the solution-verification tag to recreational-mathematics and adding this question,

Given any integer $p \le 0$, define a natural (using $\text{base-}10$ representations) injective mapping

$\quad \Gamma_p: \Bbb N^{\gt 0} \to \Bbb M$.

Of course this won't be possible unless you've invested some thought and have at least a minimal 'buy-in'. But, to my way of thinking, there is only one answer. I'll be using this 'puzzle piece' in defining addition (and multiplication?!?).

I want to check if the following arguments are valid.

Let $D = \{0,1,2,3,4,5,6,7,8,9\}$, a totally ordered set.

The set all functions $a \in D^{\,\Bbb Z}$,

$\quad \large a =(a_k)_{\,k \in \Bbb Z}$

satisfying the following two conditions

$\tag 1 (\exists\, A \in \Bbb Z) \; [ k \gt A \implies a_k = 0]$

$\tag 2 (\forall \, B \in \Bbb Z) \;(\exists \, k \in \Bbb Z) \; [ k \lt B \land a_k \ne 0]$

will be denoted by the symbol $\Bbb M$.

For $a,b \in \Bbb M$ with $a \ne b$ we can associate the subscript $\delta(a,b) \in \Bbb Z$ where

$\tag 3 \delta(a,b) = \text{max}\bigr(\{k \in \Bbb Z \mid a_k \ne b_k\}\bigr)$

Define a relation $\le$ on $\Bbb M$ by

$\quad a \le b \quad \text{ if } \; \; [a = b] \; \lor \; \large[a \ne b \, \land \, a_{\delta(a,b)} \lt    b_{\delta(a,b)}]$

Proposition 1: The relation $\le$ is a total ordering on $\Bbb M$.

Proof

Antisymmetry/Connexity: If $a \ne b$ then upon comparing $a_{\delta(a,b)}$ with $b_{\delta(a,b)}$ one can write

$\quad a \lt b \; \text{XOR} \; b \lt a$

Transitivity: Let $\{a,b,c\} \subset \Bbb M$ satisfy $a \lt b$ and $b \lt c$ and $a \ne c$.

If $\delta(a,b) \ge \delta(b,c)$ then clearly $\delta(a,c) = \delta(a,b)$  and

$\quad a_{\delta(a,c)} = a_{\delta(a,b)} \lt b_{\delta(a,b)} \le  c_{\delta(a,b)}= c_{\delta(a,c)}$

If $\delta(a,b) \lt \delta(b,c)$ then clearly $\delta(a,c) = \delta(b,c)$  and

$\quad a_{\delta(a,c)} = a_{\delta(b,c)} = b_{\delta(b,c)} \lt c_{\delta(b,c)}= c_{\delta(a,c)}$
$\quad \blacksquare$

Theorem 2: The totally ordered set $(\Bbb M, \le)$ is complete.

Proof Sketch

Let $M$ be a nonempty subset of $\Bbb M$ bounded above. There is therefore an integer $A$ and an upper bound $b$ of the form

$\quad b_k = 0 \text{ for } k \gt A \;  \land \; b_k = 9 \text{ for } k \le A$

We can certainly improve $b$ by defining $b_A$ to be the maximum digit found in the elements of $M$ at the $A$ subscript.

Using recursion, we can continue this technique, reducing $M$ while constructing the 'best' possible upper bound $\hat b$ that will also be the least upper bound for $M$

(verified using induction, recursion's 'twin sister').
$ \quad \blacksquare$

My work

I'm interested in answering the question

$\quad$ How to define operations on Stevin's construction of real numbers?

by, constructing, up to isomorphism, the set

$\quad (0,+\infty) = \{x \in \Bbb R \mid x \gt 0\}$

together with its accompanying binary operation of addition, $+$.

I have not put all the pieces together, but the thinking is that the construction will not be awkward and can be understood with very little set theory (no equivalence relations) or real analysis (no limits/infinite sums).

Note: With this perspective there is no need to worry about $1 = .9999\dots$; we have the element $\hat 1 \in \Bbb M$ defined by

$\quad k \mapsto 9 \; \text{ for } k \lt 0, \text{ else }  k \mapsto 0$

This model is abstract, which leads to perhaps an oxymoron,

$\quad$ Stevin's abstract construction

答:
The OP will no doubt need the following 'puzzle pieces' (notation/definitions) to define addition on $\Bbb M$.

Piece #1:

We define the function $\\K : \Bbb N_1 \to \Bbb M$ as follows:

If $n \ge 1$ then set $a = n - 1$ and write $a$ in $\text{base-}10$ format,

$\tag 1 \displaystyle \large a = \sum_{k = 0}^{+\infty} a_k \, \beta^{k} \quad \text{where } 0 \le a_k \le 9$

Defining $a_k = 9$ for $k \lt 0$ completes the specification,

$\tag 2 \LARGE  \\K(n) = (a_k)_{\,k \in \Bbb Z}$

Piece #2:

For $a \in \Bbb M$ with $a = (a_k)$ define a function $\\Q: \Bbb M \to \Bbb N$ by writing

$\tag 3 \displaystyle \large \\Q(a) = \displaystyle \sum_{k = 0}^{+\infty} a_k \, \beta^{k}$

Piece #3:

We'll need a function to translate/shift the subscripts.

If $p \in \Bbb Z$ then $\large \Psi_{p}: \Bbb M \to \Bbb M$ is specified as follows.

If $a = (a_k)$ define the bilateral sequence $\large ({a'}_k)_{\,k \in \Bbb Z}$ by writing

$\tag 4 \displaystyle \large {a'}_k = a_{k+p}$

and then

$\tag 5 \displaystyle \large \Psi_{p}(a) =  ({a'}_k) \quad \text{(denoted also by } {\hat a}^p \text{)}$

ContentLicense: CC BY-SA 4.0


问:
Multiplication properties in rings of matrices
Let $R$ be an arbitrary ring and let $M_n(R)~~(n>1)~$ be the ring of all $n ~ х ~ n$ matrices with elements from $R$ with usual matrix addition and multiplication.

1) Is it right that there are zero divisors in $M_n(R)$ iff $R$ is non-trivial?

2) Which necessary and sufficient conditions on $R$ do we need for $M_n(R)$ to be non-commutative?

3) Suppose $R$ has no unity. Is it right that this condition is sufficient for each element of $M_n(R)$ to be a zero divisor? If not, how can we construct an example of matrix that isn't zero divisor?

Thanks in advance.

EDIT

By zero-divisors i mean NON-ZERO elements.

答:
Perhaps the meaning of (1) is that there are non-trivial zero divisors...etc.? But

$$\begin{pmatrix}0&0&\ldots&0&r\\0&0&&\ldots&0\\\ldots&\ldots&\ldots&\ldots&\ldots\\0&0&0&\ldots&0\end{pmatrix}$$

is a nilpotent matrix, and thus it is a zero divisor (which is non zero if $\,0\ne r\in R\,$)

Also

$$\begin{pmatrix}0&r\\0&0\end{pmatrix}\begin{pmatrix}s&0\\0&0\end{pmatrix}\neq \begin{pmatrix}s&0\\0&0\end{pmatrix}\begin{pmatrix}0&r\\0&0\end{pmatrix}$$

whenever $\;sr\neq 0\;$ , so assuming the ring $\,R\,$ is non trivial and with non-trivial multiplication (like the one we can define on any abelian group and defining every product to be zero), if $\,n>1\,$ the product won't be commutative.

The last part: $\,2\Bbb Z\,$ is a non-unitary ring...

ContentLicense: CC BY-SA 3.0


问:
Is this solution for an homomorphism prove correct?
given $f$ : $a+bi$ $; a,b \in  \mathbb{Z}$ ; $ i^2=-1$; is a function over itself defined as $f(a+bi)=a-bi$ ; prove if $f$ is an homomorphism.

in some examples I had seen, it is only to write the functions like this: $(f\cdot g)(x)=f(x) \cdot g(x)$ but I am given a definition of $i^2 = -1$ and I don't know what to do with it

this is my solution, but I am not sure about it:

$f \rightarrow f \rightarrow  \mathbb{Z} $

for  $\forall c,d \in \mathbb{Z}$

a) sum

$(f+f)(a+bi)=(a-bi)+(c-di)$

$(f+f)(a+bi)=f(a-bi)+f(c-di)$

b) product

$(f \cdot f)(a+bi)=(a-bi) \cdot (c-di)$

$(f \cdot f)(a+bi)=f(a-bi) \cdot f(c-di)$

then  $f$ is an homomorphism.

Can anyone with more knowledge in this topic tell me if this is correct or if not, show how can I solve this problem.

答:
Let $X = \{a+bi : a, b \in \mathbb{Z}, i^2 =-1\}$ (although you can just take $\mathbb{C}$ as well with no problem). What you want to show is that $f : X \rightarrow X$ is a homomorphism on the addition and/or multiplication structure. This just means that $f$ preserves the addition and/or multiplication structure (respectively). In other words, given $z, w \in X$ (say $z = a+bi, w = c+di$) you want that

$f(z+w) = f(z) + f(w)$ (preserving the addition structure)

and/or

$f(z\cdot w) = f(z) \cdot f(w)$ (preserving the multiplication structure)

$f(z+w) = f(a+c + (b+d)i) = a+c - (b+d)i = a-bi + c-di = f(z) + f(w)$ so $f$ is an addition preserving homomorphism.

$f(z \cdot w) = f(ac-bd+(ad+bc)i) = ac - bd - (ad + bc)i = (a-bi)(c-di) = f(z) \cdot f(w)$ so $f$ preserves the multiplication structure as well. So $f$ is indeed a homomorphism (in this case a ring homomorphism since it preserves both addition and multiplication).

In general, given a group $(G, +)$, a group homomorphism only needs to preserve the addition structure. Given a ring $(R, + , \cdot)$ a ring homomorphism needs to preserve both the addition and multiplication structures, and so on.


问:
Triple integral: cylinder inside a sphere
I'm trying to solve the following exercise about triple integrals, but I need a check.

Let $V \subset \mathbb{R}^3$ which is made by one sphere of radius $2$ centered at $(0,0,0)$ and a cylinder with base at $(0,0,0)$ and radius $1$. Formally, this is

$$V=\{x^2+y^2+z^4 \leq 4 \} \cup \{(x,y,z): x^2 + y^2 \leq 1, z \in [0,4] \}$$

Goal: Compute the Volume of $V$

My attempt:

$|V|=|\text{Sphere}| + |\text{Cylinder}| - |\text{Portion of Sphere inside Cylinder}|$

The first two terms are straightforward: $|\text{Sphere}|=\frac{16}{3} \pi $, $|\text{Cylinder}|=4 \pi$

The tricky part for me is $|\text{Portion of Sphere inside Cylinder}|$

To  do that, I consider the $Oxz$ plane: here I can see that cylinder intersects the sphere at $(1,\sqrt{3})$ and the angle $\varphi = \frac{\pi}{6}$ ($\varphi$ is intended with this convenction)

Now I want to use spherical coordinates to compute the volume of that region: to this aim, I let the $\rho$ of the spherical coordinates range in $(0,1)$, in order to be inside the cylinder. All in all, the triple integral do to is the following:

$$\int_{0}^{2 \pi} \int_0^{\frac{\pi}{6}} \int_{0}^{1} \rho^2 \sin(\varphi) d \rho d \varphi d \theta$$

which can easily be solved.

答: Please, read what you wrote. It is not clear what does the region look like. Draw it and make sure it has finite volume. Than compute the integral in convenient coordinates system, in this case, cylindrical or spherical.</p>


问:
How to solve this analytical geometry problem?-parable inscribed within a square
This problem appeared on the network, and although it looks simple I am not sure of the result.

The polygon $ABCD$ is a square with side $4$ cm and the curve inscribed inside the square is a parabola, determine the length of the segment $AP$, if $QC$ is $3$ cm.

problem

答:
The quadratic

Passes through the coordinates $(0,0), (4, 3)$.

Has a maximum at $X < 4$

Has the form $ y = - A (x-X)^2+4$.

Can you take it from here? If not, explain what you've tried and where you're stuck.

 $y = -\frac{9}{16}(x-\frac{8}{3})^2 + 4$.

 Hence $AP = \frac{8}{3}$.


问:
Existence of a multigraph
I'm an undergraduate student majoring in mathematics. I'm solving my home assignment and got stuck at this problem.

Suppose $d_1\geq d_2\geq\dots \geq d_n\geq 1$ and $\sum_{i=1}^nd_i$ is
  even. Prove that there is a multigraph with degree sequence $d_1, d_2,
> \dots, d_n$ if and only if $d_1 \leq \sum_{i=2}^nd_i$.

So, please help me.

答:
An alternative construction (that doesn't use induction):

First observe that if $d_1 = \sum_{i=2}^n d_i$, we can form a multigraph with degree sequence $d_1, \dots, d_n$ by simply adding $d_i$ edges between $v_1$ and $v_i$ for each $i \in \{2,\dots, n\}$.

In general, of course, $d_1$ may be less than $\sum_{i=2}^n d_i$, but we can always reduce to the special case above if we can somehow add $k \equiv \frac{1}{2}(\sum_{i=2}^n d_i - d_1)$ edges between vertices in $\{v_2,\dots, v_n\}$ (to make up for the difference between $d_1$ and $\sum_{i=2}^n d_i$. Note that $\sum_{i=2}^n d_i - d_1 = \sum_{i=1}^n d_i - 2d_1$ is even, since $\sum_{i=1}^n d_i$ is even by assumption. Note also that $k \leq \frac{1}{2}(\sum_{i=2}^n d_i - d_2) = \frac{1}{2}\sum_{i=3}d_i$ since $d_1 \geq d_2$.

To add the $k$ edges between vertices in $\{v_2,\dots, v_n\}$, consider adding edges between $v_i$ and $v_{i-1}$, starting with $i = n$, until $d_i$ has been "depleted." That is, add $d_n$ edges between $v_n$ and $v_{n-1}$, $d_{n-1} - d_n$ edges between $v_{n-1}$ and $v_{n-2}$, $d_{n-2} - (d_{n-1} - d_n)$ edges between $v_{n-2}$ and $v_{n-3}$, and so on, finally adding edges between $v_3$ and $v_2$ until no more edges can be added to $v_3$. This is possible because $v_2 \geq v_3 \geq \dots \geq v_n$, so we can clearly add $\leq d_i$ edges between $v_i$ and $v_{i-1}$. Moreover, at least $\frac{1}{2}\sum_{i=3}d_i$ edges are added in total, since we've depleted degrees $d_3,\dots, d_n$. Therefore, from the $k \leq \sum_{i=3}^n d_i$ bound above, we see that we can add the $k$ edges this way.

Title: Existence of a multigraph


答: This image may help.  (Where's the "cone"??)

Title: Triple integral: cylinder inside a sphere

Body:
I'm trying to solve the following exercise about triple integrals, but I need a check.

Let $V \subset \mathbb{R}^3$ which is made by one sphere of radius $2$ centered at $(0,0,0)$ and a cylinder with base at $(0,0,0)$ and radius $1$. Formally, this is

$$V=\{x^2+y^2+z^4 \leq 4 \} \cup \{(x,y,z): x^2 + y^2 \leq 1, z \in [0,4] \}$$

Goal: Compute the Volume of $V$

My attempt:

$|V|=|\text{Sphere}| + |\text{Cylinder}| - |\text{Portion of Sphere inside Cylinder}|$

The first two terms are straightforward: $|\text{Sphere}|=\frac{16}{3} \pi $, $|\text{Cylinder}|=4 \pi$

The tricky part for me is $|\text{Portion of Sphere inside Cylinder}|$

To  do that, I consider the $Oxz$ plane: here I can see that cylinder intersects the sphere at $(1,\sqrt{3})$ and the angle $\varphi = \frac{\pi}{6}$ ($\varphi$ is intended with this convenction)

Now I want to use spherical coordinates to compute the volume of that region: to this aim, I let the $\rho$ of the spherical coordinates range in $(0,1)$, in order to be inside the cylinder. All in all, the triple integral do to is the following:

$$\int_{0}^{2 \pi} \int_0^{\frac{\pi}{6}} \int_{0}^{1} \rho^2 \sin(\varphi) d \rho d \varphi d \theta$$

which can easily be solved.


问:
Understanding implications
I have a set of questions where to task is to add the correct implication arrow.

I was able to make sense of most of them, but for two problems, I can't figure out why I was wrong.

Here is what I put:

Statement $S1$
Statement $S2$
I answered
The answer in the book is

$\frac{x}{x+1} = 0$
$x=0$
$S1\iff S2$
$S1 \Rightarrow S2$

$(a,b)$ is a point on the line $y=2x-1$
$b=2a-1$
$S1 \iff S2$
$S1 \Rightarrow S2$

For the first one, I can't see how when $x=0$, $\frac{x}{x+1}$ can equal anything other than $0$, so surely the implication works both ways? I presume I'm missing something obvious.

For the second, the point $(a, b)$ is on line $y = 2x -1$, so $b$ must equal $2a - 1$. Again I presume I've missing something here.

Really would appreciate any help here, I've run out of ideas.

答:
The symbols represent the direction of the implication. It means you can read back and/or forwards substituting by "then". The one with two arrows is "similar" to the equivalence symbol.

In the first, x = 0 does not imply only that x/(x+1)=0, but much other expressions. The same for the line.

Strictly speaking, the statement has a direction which must be rigidly read.

Title: Understanding implications


问:
A proof from Grillet's Abstract Algebra: an extension containing roots of finitely many polynomials
This is Lemma 4.3 on page 167 of Grillet's book Abstract Algebra.

Every field $K$ has an algebraic extension that contains a root of every nonconstant polynomial with coefficients in $K$.

I don't understand the first step in the proof. Grillet claims that for a finite number of polynomials $f_1,....,f_n$, we can divide them into monic irreducible factors $q_{1i}, ..., q_{ni}$ and repeatedly use the following theorem:

If $q \in K[x]$ is monic irreducible, then $E = K[x]/(q)$ is a field extension of $K$ such that $\alpha = x + (q)$ is a root of $q$ in $E$.

However, I don't see how we can repeatedly use it. Suppose, for instance that $f_1 = a_1 q_{11}\dots q_{1m}$. where $q_{1i}$ are monic irreducible. We have used constructed the field extension $E = F[x]/(q_{11})$ in which $q_{11}$ has a root. To proceed, we need to know that $q_{21}$ is also irreducible over $E$, but we only know that it is irreducible over $K$.

答:
For simplicity, you can start out with only one polynomial $f:=f_1\cdot f_n$, and look for only the first irreducible factor $q_1$, so $f=q_1^kg$.

Then you obtain $E=K[x]/(q_1)$ and factor $g$ to irreducibles over $E$, but use only the first factor.

And so on..


问:
Given certain conditions, prove a function G(x) is always equal to 4.
Theres a question I've been having trouble with:

$G(x)$ is the function $|x+2|+|x-2|$. Show that if $-2<x<2$ then $G(x) = 4$.

Any help would be greatly appreciated. No calculus please :D

答:
Draw a number line. In general, $|x-a|$ is the distance from $x$ to $a$. So $|x-2|$ is the distance from $x$ to $2$. And $|x+2|=|x-(-2)|$ is the distance from $x$ to $-2$.

Thus $|x+2|+|x+2|$ is the sum of the distances from $x$ to $-2$ and to $2$.  If $x$ is anywhere in the interval $=2\le x\le 2$, the sum of its distances from $-2$ and $2$ is always equal to the distance between $-2$ and $2$, which is $4$.

A more informal way of seeing it is to suppose that you are standing on the number line at $-2$, and start walking slowly to the right. At the beginning, the sum of your distances from $-2$ and $2$ is $4$. For every tiny step $s$ that you take, your distance from $-2$ increases by $s$. But your distance from $2$ decreases by $s$, so the sum of your distances from $-2$ and $2$ does not change. This continues until you reach $2$, at which point the sum of your distances from $-2$ and $2$ starts to increase.


问:
Question on Weierstrass $\wp$ function

I omitted the proof of Lemma 1.5

In remark, how do we conclude from lemma 1.5 that the given series converges uniformly? Don't we only know the point-wise convergence (with fixed $\tau$) of the series?

The below argument seems to show how to use Lemma 1.5 to conclude that Weierstrass $\wp$ function has double poles at lattices, but I'm really lost what's happening here. Could you explain what Stein is doing here, or suggest any relevant concept/theorem to understand this? Thanks!

答:
$$\frac1{z^2}+\sum_{n,m\ne (0,0)} \frac1{(z+n+m\tau)^2}-\frac1{(n+m\tau)^2}=\frac1{z^2}+\sum_{n,m\ne (0,0)} \frac{-z^2-2z(n+m\tau)}{(n+m\tau)^2(z+n+m\tau)^2}$$ converges locally uniformly on $\Bbb{C-(Z+Z\tau)}$ thus it is analytic there.

Moreover removing one term of the series gives that the series for $\wp(z)-\frac1{(z-a-b\tau)^2}$ converges uniformly and is analytic on a disk around $a+b\tau$.

Title: Question on Weierstrass $\wp$ function


问:
Finding a transformation matrix $\hat{T}$ such that $\hat{T}A \hat{T}^{-1}$ is diagonal?
Given an example matrix $A$ as follows:
$$A = \begin{bmatrix}
3 & 1 & -1 \\
-1 & 1 & 1 \\
2 & 2 & 0
\end{bmatrix} $$

I know how to find the matrix $T$ such that $T^{-1}AT$ is diagonal. ie: You simply find the eigenvectors and you are good to go.

Question:

What if I wanted to find the matrix $\hat{T}$ such that $\hat{T}A \space\hat{T}^{-1}$ is diagonal? Is this related to $T$ in any way? That is, can I still use the eigenvectors of $A$ in some way to create this matrix? I feel like I'm missing something obvious as this doesn't seem like a difficult task.

Thanks!

答: If $T^{-1}AT$ is diagonal and you want $\hat{T}A\hat{T}^{-1}$ to be diagonal, then just take $\hat{T}=T^{-1}$.


问:
How can I solve this ODE using a predictor-corrector method?
I'm looking to solve an ODE system for $y$ over time using the predictor-corrector method below. I'm not sure how predictor-corrector for ODEs work. I know other numerical methods for ODEs, like RK and forward Euler.

$\ y = 0$

Timestep (h) is 0.001s.

$\ \frac{dy}{dt}  = PI-QB \cos\beta\, $

Predictor:

$\ y(n+1) = y(n)+(\frac{h}{2}) [3dy/dt(n)-dy/dt(n-1)]$ is used to calculate $\ dy/dt(n+1)$

Corrector:

$y(n+1) = \frac{1}{3} [4y(n)-y(n-1)+2h dy/dt(n+1)]$

For the corrector, do I just substitute in the predictor instead of the $\ dy/dt(n+1)]$ on the right hand side of the equation?

Then $\ y(n+1)$ is $\ y(n)$  added to the corrector.
Is that right?

Link to my spreadsheet,  with my Predictor corrector formulas on the right hand side, that I think are wrong. - Google Sheet

答: Predictor-corrector are integrators which compute an estimation and then utilizes it to update the current value to the states of the ode. If you do this way, you will do it correctly.</p>


问:
Grid Combinations with Positive and Negative Numbers
If there is an $n$ by $n$ matrix where each element is either 1 or -1, how many unique matrices are there such that each row and each column multiplies to 1?

I solved for the trivial case of $n = 2$, which is 2. However, for larger $n$, I'm not sure how to find a systematic way to count. I have tried using the fact that you can transform an existing correct matrix into another correct matrix by selecting 4 points that form a rectangle to all become negative, but I am not sure how I could count these in an organized way.

答:
Consider any $(n-1) \times (n-1) $ minor.

This could be filled with any option of $ \pm 1$.

Claim: The unfilled cells are uniquely determined by the column / row that they are in, and there is no conflict.

Proof: Do it yourself.

Corollary: There are $ 2^{(n-1)^2 } $ ways to fill up the matrix.


问:
Hyperelliptic iff there exist points $P,Q$ such that $\dim L(P+Q)=2$?
I want to prove that a curve $C$ of genus $\geq3$ is hyperelliptic iff there exist points $P,Q\in C$ such that $\dim L(P+Q)=2$. My definition of hyperelliptic curve is that it has genus $\geq2$ and it has a degree $2$ morphism into $\mathbb{P}^1$.

One direction is obvious, if there exist points $P,Q\in C$ such that $\dim L(P+Q)=2$, then any non-constant $f\in L(P+Q)$ gives a degree $2$ morphism into $\mathbb{P}^1$.

In the other direction, if $C$ is hyperelliptic, then it has a degree $2$ morphism into $\mathbb{P}^1$, I want to choose the preimage of $\infty$ as $P,Q$, but I can't see why $\dim L(P+Q)=2$. By using the general bound of $L(D)\leq1+\deg D$ for effective divisors, I know $\dim L(P+Q)\leq3$, but why $\dim L(P+Q)\neq3$?

Could anyone help me?

答:
$\newcommand\P{\mathbb P}\newcommand\msL{\mathscr L}\newcommand\pull[1]{#1^*}\newcommand\msO{\mathscr O}\newcommand\msF{\mathscr F}$
Say $C$ is hyperelliptic with morphism $f:C\xrightarrow2\P^1$, and let $\msL=\pull f\msO(1)$. In other words $H^0(\msL)=L(P+Q)$ where $P,Q$ are the preimages of $\infty\in\P^1$, as you suggest. Our strategy will be to use $f$ to related cohomology groups on $C$ to those on $\P^1$, and then stare at a diagram until the answer pops out.

We have a short exact sequence $0\to\msO_{\P^1}\to\msO_{\P^1}(1)\to\msO_\infty\to0$. Further, it is a general fact that the morphism $f$ induces a (unique) map $H^\bullet(\P^1,-)\to H^\bullet(C,\pull f(-))$ of $\delta$-functors extending the natural map in degree $0$ (i.e. the map $H^0(\P^1,\msF)=\msF(\P^1)\to (\pull f\msF)(C)=H^0(C,\pull f\msF)$). The upshot is that our short exact sequence induces a homomorphism of long exact sequences
$$\require{AMScd}\begin{CD}
0 @>>> H^0(\msO_{\P^1}) @>>> H^0(\msO_{\P^1}(1)) @>>> H^0(\msO_\infty) @>>> H^1(\msO_{\P^1}) @>>> H^1(\msO_{\P^1}(1)) @>>> 0\\
& @VVV @VVV @VVV @VVV @VVV\\
0 @>>> H^0(\pull f\msO_{\P^1}) @>>> H^0(\pull f\msO_{\P^1}(1)) @>>> H^0(\pull f\msO_\infty) @>>> H^1(\pull f\msO_{\P^1}) @>>> H^1(\pull f\msO_{\P^1}(1)) @>>> 0
\end{CD}$$

This is a bit of a mouthful, so let's rewrite this a little. We know that $\pull f\msO_{\P^1}=\msO_C$, that $\pull f\msO_{\P^1}(1)=\msL$, and that $\pull f\msO_\infty=\msO_{P+Q}$. We also know $h^1(\msO_{\P^1})=g(\P^1)=0$ and $h^1(\msO_{\P^1}(1))=h^0(\msO_{\P^1}(-3))=0$ by Serre duality. Letting $k$ be our ground field, we have

$$\begin{CD}
0 @>>> k @>>> k^{\oplus2} @>>> k @>>> 0 @>>> 0 @>>> 0\\
& @VVV @VVV @VVV @VVV @VVV\\
0 @>>> k @>>> H^0(\msL) @>>> k^{\oplus2} @>>> k^{\oplus g} @>>> H^1(\msL) @>>> 0
\end{CD}$$
which now has a shorter scroll bar. Now, we're in luck. The map
$$k=H^0(\msO_\infty)\to H^0(\msO_{P+Q})=k^{\oplus2}$$
above is injective; it sends a function on the point $\infty\in\P^1$ to the induced function on the subscheme $P+Q\subset C$. Thus, commutativity of the square (with top row $k\to0$) shows that
$$\dim\mathrm{im}(H^0(\msL)\to k^{\oplus2})=\dim\ker(k^{\oplus2}\to k^{\oplus g})\ge1$$
and
$$\dim\ker(k^{\oplus g}\to H^1(\msL))=\dim\mathrm{im}(k^{\oplus2}\to k^{\oplus g})\le1.$$

Since $k^{\oplus g}\twoheadrightarrow H^1(\msL)$ is surjective, we see that $h^1(\msL)\le g-1$. Hence,
$$h^0(\msL)=\chi(\msL)+h^1(\msL)=(3-g)+h^1(\msL)\le2.$$
At the same time, exactness of the bottom row of our diagram + the bound $\dim\mathrm{im}(H^0(\msL)\to k^{\oplus2})\ge1$ shows that $h^0(\msL)\ge2$

Edit: To anyone looking at this answer, user10354138 has a comment on the question giving a much simpler proof that $h^0(\msL)\le2$.


答:
Consider the games, which happen to be numbers, $G = \{ 0\,|\,\}$ and $H = \{ 0.5\,|\,\}$.

$G = 1$, directly from the simplicity rules. You know that $H = 1$, but that is not a direct implication. What you can do is show that $G \leq H \leq G$.

$H \leq G$, because $0.5$ is the only Left option of $H$, there are no Right options in $G$ and $0.5 < G = 1$.

$G \leq H$, because $0$ is the only Left option of $G$, there are no Right options in $H$ and $0 < H$. The last inequality, $0 < H$, is true because in the game $H$, Left wins no matter who starts, and this means the game is positive.

Now it is clear that $\{ 0\,|\,\} = \{ 0.5\,|\,\} =  1$. I hope that it helps you understanding why Polichinelle's answer is correct.

Title: Real equivalent of the surreal number {0.5|}

Body:
I've been reading up on Surreal numbers, but have some questions.

Some equivalent real and surreal numbers.

```<code>2.5 =
    {2|3} =
    {{{0|}|}|{{{0|}|}|}} =
    {{{{{|}}|{}}|{}}|{{{{{|}}|{}}|{}}|{}}}

0 =
    {-1|1} = {-2|1} = {-2,-1|1} =
    {{|0}|{0|}} = {{|{|0}},{|0}|{0|}} =
    {{{}|{{|}}}|{{{|}}|{}}}

-3/8 = {-0.5|-0.25} = {{-1|0}|{{-1|0}|0}}
    {{{{}|{{|}}}|{{{}|{{|}}}|{{|}}}}|{{{}|{{|}}}|{{|}}}}
</code>```

What about the real number for {0.5|}?


问:
Counting the number of good functions
Hy, happy new year everyone.

I have been stuck on the following problem for a while now, so I am posting it here in other to discuss it.

A function $g: [[1,n]] \to [[0,n]]$ is called good if :

$$\forall j \in[[1,n]] , \exists i~ \text{integer} \geq 0 , g^{i} (j)=0 $$

where $g^{i}=g\circ \dots \circ g ~~(i ~~\text{times})$

How many  such good functions  there is?

答:
Let $A_n,B_m$ be two disjoint sets of size $n,m\in \mathbb{N}$.

Take $a(m,n)$ as the number of functions $f:A_n\rightarrow A_n \cup B_m$ satisfying the following condition: $$\forall a\in A_n \text{  } \exists i\geq 0 \text{ such that } f^i(a)\in B_m$$ You're looking to compute $a(1,n)$. Evidently $a(m,1)=m$ and $$a(m,n)=m^n+\sum_{k=1}^{n-1}m^k {n \choose k}a(k,n-k)$$ It's easily shown that $$a(n,m)=m(m+n)^{n-1}$$ solves this relation. Taking $m=1$ gives $a(1,n)=(n+1)^{n-1}$ as required.

Title: Counting the number of good functions


问:
eig function MATLAB on complex symmetric matrix
Suppose I have a complex symmetric matrix D (not Hermatian) which after diagonalization becomes E. Let, D = F + iG and E = H + iJ. E, H and J are diagonal matrices. I am using 'eig' function in MATLAB to diagonalize D, i.e., [V,E] = eig(D). I always find that if I apply 'eig' on matrices F and G, I always get H and J respectively. It is easy to understand if the matrix formed by V is always real, i.e., $\text{V}^{-1}\text{DV}=\text{V}^{-1}\text{FV}+\text{i}\text{V}^{-1}\text{GV}=\text{H}+\text{iJ}$. Why does it happen? Is there a theorem behind it? Or, it is just a shortcoming of the algorithm used in 'eig' in MATLAB?

答: $\mathbb{C}$ is an isomorphism to $R^2$.


问:
Continuous bijection between compacts of $\mathbb{R}^n$
Is it possible to find continuous bijection between $[0;1]$ and any compact $E \subset \mathbb{R}$?
And more generally between $[0;1]^n$ and any compact $E \subset \mathbb{R}^n$?

$\mathbb{R}^n$ has a finite dimension so the compact set are the closed and bounded set of $\mathbb{R}^n,$ can we deduce from this the above result?

答: No. Consider a set, which consists of one point.</p>


问:
Finding limits from cdf and expectation

So I tried to find how each segment was formed in order to try and get an expression for expectation and solve from there, for example from c to 1, the probability is 0.2. From 1 to 2 it is also 0.2 and so on, except this seem to fall apart once I react the upper limit and thus can't get an expression, assuming this is even the correct approach, although I assume there is possibly a more simple one I'm missing?

答:
What you seem to be missing is

not that "from $c$ to $1$, the probability is $0.2$"

but that when $c \le x \lt 1$ you have the probability that $X$ is less than or equal to $x$ is $0.2$ as this is a cumulative distribution function

Combine this for example with knowing that when $1 \le x \lt 2$ you have the probability that  $X$ is less than or equal to $x$ is $0.4$, and it tell you that the probability that $X=1$  must be $0.4-0.2=0.2$

So you can calculate the probability that $X=c$, $1$, $2$, $3$ and so find $\mathbb E[X]$ in terms of $c$.  Then solve to find the $c$ which makes $\mathbb E[X]=1.5$

Title: Finding limits from cdf and expectation


问:
Solving the cryptarithm $\mathrm{TWO}+\mathrm{TWO}=\mathrm{FOUR}$ in base $10$ and base $7$

Find the numerical value of each of the letters in the following expression $$\mathrm{TWO}+\mathrm{TWO}=\mathrm{FOUR}$$
in (a) base $10$ and (b) base $7$.

I don't even know how I would approach this problem. Any help would be appreciated.

答:
For (a), base 10:

The easiest place to start with is F (in four) and that must be 1, as

TWO + TWO

when the letters are at the highest possible value (987+987) = 1974, so

F=1

Now we move to O.

$O$ cannot be 1,
so can be $2,3,4,5,6,7,8,9,0$

If $O$=2, then $R$ is 4. $W$ cannot be assigned a number, as if $W$ is 5 and more, we will have a carry over into $T$ and 11 (12-1) cannot be divisible by 2, so then no number will be assigned to $T$, so 5 or more cannot be correct.

$W$ cannot be 1 because of $F$. $W$ cannot be 2 because of $O$. If $W$ was 3, then $U$=6 but $T$=6 (12 divided by 2), so it cannot be 3. $W$ cannot be 4 because $R$ is 4.

So, $O$ isn't 2.

If $O$ was 3, then $R$=6. If $R$= 6 then $T$ cannot be assigned a number, since if there is a carry over of 1, 12 divided by 2 is 6, but $R$ is 6.

SO, $O$ isn't 3.

If $O$ is 4, then $R$=8.

Now we have
$$\begin{array}{ccccccc}
&&&&T&W&O\\
+&&&&T&W&O\\
\hline
&&&F&O&U&R\\
\end{array}$$

=
$$\begin{array}{ccccccc}
&&&&?&?&4\\
+&&&&?&?&4\\
\hline
&&&1&4&?&8\\
\end{array}$$

If there was a carry over into the $T$$T$$O$ column, then 13 divided by 2 is not a whole number. So $T$ has to be 7.

Now,

$$\begin{array}{ccccccc}
&&&&7&?&4\\
+&&&&7&?&4\\
\hline
&&&1&4&?&8\\
\end{array}$$

Now, we find $W$ and $U$.

$W$ cannot be 0, 1 and 4 from just looking, so we have $2,3,5,6,7,8,9$

If $W$ was 2, then $U$= 4 but $O$ is 4.

If $W$ was 3, then $U$= 6 and that is the answer.

$$\begin{array}{ccccccc}
&&&&7&3&4\\
+&&&&7&3&4\\
\hline
&&&1&4&6&8\\
\end{array}$$

This is one solution. Other solutions are:
765+765==1530

836+836==1672

846+846==1692

867+867==1734

928+928==1856

938+938==1876

For b)

I used brute force for this, and found an unique solution:

$$\begin{array}{ccccccc}
&&&&5&2&3\\
+&&&&5&2&3\\
\hline
&&&1&3&4&6\\
\end{array}$$


问:
Probably of Fukushima like event
I was reading an article where an author was discussing the relative safety of power plants. In the article there was statement that a tsunami with a wave of 15 meters happens once every 5000 years.

One of the commenter said that if we assume that the lifetime of an reactor is 50 years the probability of similar event is 50/5000 = 0.01 or 1%. And knowing that the are 23 nuclear power plants that operate in a high risk tsunami areas, the probability that it happens in one of them is 1 - (1-0.01)^23 = 20%

It seems to be too high to be true. What is wrong and what would be more accurate estimation?

答: Your math is right, but there's one catch - the $1\%$ probability, at least to my understanding, applies to a tsunami in Japan as a whole rather than just near power plants. The real probability is then far less, because the chances that a tsunami hits a plant is far less than $1\%.$

Title: Probably of Fukushima like event


问:
$f$ is continuous on $\mathbb{R}$ and $f(x) \cdot f(−x)\leq 0$ for all $x$. Is $f$ a onto function and does $f$ have a zero?
Question is:
Given that $f$ is continuous on $\mathbb{R}$ and $f(x) \cdot f(−x)\leq 0$ for all $x$.

Is $f$ always an onto function ?

Does $f$ has zero in all examples?

My thoughts:

Firstly I think $f$ is an odd function so its onto. Can you give an anti example for this? I couldn't find.

Frankly, I don't fully understand this part. What is mean that $f$ has zero? Could you explain?

答:
Even simpler, $f(x) = 0$ satisfies $f(x) \cdot f(-x) \leq 0$ for all $x$, but is not onto.

By the hypothesis we must have $0 \leq f(0)^{2} \leq 0 \implies f(0) = 0$ and $f$ has a zero.

The following is a longer argument based on the intermediate value theorem:

If $f(x) < 0$ for all $x$, then $f(x) \cdot f(-x) > 0$ and similarly for the case where $f(x) > 0$ for all $x$. This means the codomain must contain at least one nonnegative number and at least one nonpositive number.

Hence there must be points $x_{1}$ and $x_{2}$, such that $f(x_{1}) \leq 0$ and $f(x_{2}) \geq 0$ with $x_{1} \leq x_{2}$. If either of the points are zeroes, we are done. Otherwise, $f(x_{1}) < 0$ and $f(x_{2}) > 0$, and  $x_{1} < x_{2}$ since $f$ is a function.

Since $f$ is continuous by the hypothesis, one may use the intermediate value theorem to see that there exists a $c \in (x_{1},x_{2})$ such that $f(c) = 0$, which means $f$ has a zero.


问:
Any book for understanding the concept of Random measures
Could anyone please give me some references for understanding the concept of random measure? I'm beginner in this field.</p>

答: Kallenberg, Olav. Random measures, theory and applications.  Springer International Publishing, 2017.</p>


问:
Geometric question?
First of all, is it Geometric?

Image of the drafted:

I need help solving this question, and I am completely lost on how can I solve this.
Could anyone explain the way of solving this geometric question?

Here is a drafting of the two lines I and II.
With 3 formulas (a), (b), (c).

(a) $y = 2x + 8$,

(b) $y = - 2x + 8$

(c) $y = x + 2$.

A. Every of the straights I and II, please find the correct formula out of the formulas (a) , (b) , (c), and explain your answer.

b. Find the points of intersection of the straight rates.

c. Find the straight formula that's going through the way of the point (5, 2) and parallels to the straight II.

How do I do this?

This is a translated version.

答:
Hints: for the straight line $\,y=ax+b\,$:

$\bullet\;\;$ The line above is ascending iff $\,a>0\,$ , and it is constant iff $\,a=0\,$ ;

$\bullet\;\;$ the $\,y$-intersection of the above line is the point $\,(0,b)\,$ ;

$\bullet\;\;$ The line above intercepts the line $\,y=mx +n\,$ at the point $\,(x_0,y_0)\,$ which is the solution of the linear system

$$\begin{cases}y=ax+b\\{}\\y=mx+n\end{cases}$$

How to solve: compare $\,y$-s:

$$ax+b=mx+n\implies(a-m)x=n-b\implies \,\text{if}\;a\neq b\;,\;x=\frac{n-b}{a-m}$$

Iff $\,a=b\,$ the lines are parallel and thus they are the very same line or else they have no common point.

Title: Geometric question?


问:
Centroid and "Centre of mass"
I thought that the centroid was also the point at which the area in all directions was net 0. Aka area above the cenrroid was the same area as below.

But for a triangle everything I can see seems to suggest that the area above the centroid of a triangle is about 55% of the area, I just dont see how this fits the fact of it being the "centre of mass"

Even integrating it or just simple equations (rearranging the area of a traingle formula) and even when I checked in Geogebra seemed to show that the point at which half the area is reached vertically is ${\sqrt2}/2$ from the apex for a unit triangle.

I am so sure I'm wrong but I cant see how I could be

答:
The centroid is the point at which the figure will balance when you prop it up on a pin.

As you note, that does not mean that every line through the centroid bisects the area. Balancing takes into account the distances from the balance point to the points of the figure.

If you think about this a bit you will see that very few figures have a point such that every line through that point bisects the area. (Probably only circles do - I don't know a proof offhand.)

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

Title: Centroid and "Centre of mass"


问:
Does Gradient flow of a distance function preserve geodesics (distance)?
Let $(M,g)$ be a smooth Riemannian manifold, and for some $p,q\in M$, let $\gamma_{p,q}:[0,1]\to M$ be a constant speed minimizing geodesic with $\gamma(0)=p,\gamma(1)=q$. Consider the function $d_p(x):M\to \mathbb{R}^+$ defined by $d_p(x)=d_g(x,p)$ where $d_g$ is the distance induced by the Riemannian metric $g$, and consider the gradient flow of $-\nabla d_p$ denoted by $\phi_s$.

Now let $t\in(0,1)$ and fix $r>0$ small so that $d_p$ is smooth in $B_r(\gamma_{p,q}(t))$, if it is true that for $x,y\in B_r(\gamma_{p,q}(t))$ we have $\gamma_{\phi_s(x),\phi_s(y)}=\phi_s(\gamma_{x,y})$ (or if $d_g(\phi_s(x),\phi_s(y))=d_g(x,y)$?)for at least small $s>0$?

There are several conditions one could add to the manifold, like completeness, but I don't know if it is needed, so please add suitable conditions if needed. Thanks in advance.

答: There is a paper from Paul Ehrenfest which extends the manifold and its metric by one dimension in case some defines a function in it. The curve from A to B by action of the gradient flow is not geodesic preserving, but in this extended dimension, it is.</p>


答:
f(x) = arctan(x) disproves part 1 since it satisfies the hypothesis but fails to be onto.

The IVT guarantees there is a 0 in all examples.

Title: $f$ is continuous on $\mathbb{R}$ and $f(x) \cdot f(−x)\leq 0$ for all $x$. Is $f$ a onto function and does $f$ have a zero?

Body:
Question is:
Given that $f$ is continuous on $\mathbb{R}$ and $f(x) \cdot f(−x)\leq 0$ for all $x$.

Is $f$ always an onto function ?

Does $f$ has zero in all examples?

My thoughts:

Firstly I think $f$ is an odd function so its onto. Can you give an anti example for this? I couldn't find.

Frankly, I don't fully understand this part. What is mean that $f$ has zero? Could you explain?


答:
John Conway and others discussed this is 2001 and I produced a brief web page illustrating this with a no-longer functional Java applet.  You are correct that apart from the three medians (blue below), none of the area bisectors (green below) pass through the centroid.

But if the triangle is solid then any line through the centroid will balance the triangle, in the sense that the first moments on each side will be equal and opposite, even if the two areas are different;  regions of the same area but further away will have a greater moment.

Title: Centroid and "Centre of mass"

Body:
I thought that the centroid was also the point at which the area in all directions was net 0. Aka area above the cenrroid was the same area as below.

But for a triangle everything I can see seems to suggest that the area above the centroid of a triangle is about 55% of the area, I just dont see how this fits the fact of it being the "centre of mass"

Even integrating it or just simple equations (rearranging the area of a traingle formula) and even when I checked in Geogebra seemed to show that the point at which half the area is reached vertically is ${\sqrt2}/2$ from the apex for a unit triangle.

I am so sure I'm wrong but I cant see how I could be


问:
Checking if a matrix is positive semidefinite

Determine whether the following $2 \times 2$ matrix is positive semidefinite (PSD)

$$\begin{bmatrix}\frac{2}{x} & \frac{-2y}{x^2} \\\frac{-2y}{x^2} & \frac{2y^2}{x^3}\end{bmatrix}$$

where $x > 0$ and $y \in \mathbb R$.

A matrix is PSD if $v^T A v \geq 0$. So, do I just multiply by a vector $v = (v_1, v_2)$ and check if it is $\geq 0$? Thanks for any help.

答:
I would like to add that you can also check if a symmetric matrix is positive semi-definite by checking that all of its eigenvalues are non-negative.

If they are also all positive then the matrix is actually positive definite.

Reference: https://en.wikipedia.org/wiki/Definite_symmetric_matrix#Eigenvalues


问:
having trouble intuiting analyticity
My textbook seems to suggest that the analytic functions are precisely the functions that can be written in terms of $z$ alone (no $x$ or $y$ or conjugate-$z$).

Am I inferring correctly?

Does this mean that $\sin (z+x)$ is not analytic?   [where $z=x+iy$]

答:
I think that $f$ "can be written in terms of $z$" alone means $\partial f/\partial \bar{z}=0$ here. It is easy to check that $\partial f/\partial \bar{z}$ is "equivalent" to the Cauchy-Riemann equation.

Note that you can obviously express a complex function in terms of $x$ and $y$, i.e. the real part and the imaginary part, respectively, and that, more importantly, you can also express it in terms of $z$ and $\bar{z}$ since both $x$ and $y$ can be expressed in terms of $z$ and $\bar{z}$, and that you can calculate $\partial f/\partial z$ and $\partial f/\partial \bar{z}$.

Title: having trouble intuiting analyticity


问:
Cardinality of almost disjoint families of sets of reals
Let $\Delta$ be a set of infinite subsets of reals, suppose for any elements $X\neq Y$ of $\Delta$, $X\cap Y$ is finite. Does that imply anything about the cardinality of $\Delta$? Notice that if we are looking at a set of infinite subsets of naturals, then the set can be as small as $\omega$ and as large as $2^\omega$.

I am interested in knowing an answer assuming the axiom of choice, because I imagine without the axiom of choice things are pretty wild. For instance if $\mathbb{R}$ is a countable union of pairwise disjoint countable sets, then we could have $\Delta$ being countable. Or if we partition $\mathbb{R}$ into more equivalence classes than $|\mathbb{R}|$ itself, then we could have $\Delta$ being "larger" than $|\mathbb{R}|$.

答:
We get the same range of possibilities as for almost disjoint families in $\wp(\omega)$.

Let $\mathscr{D}$ be an almost disjoint family of infinite subsets of $\Bbb R$. For each $D\in\mathscr{D}$ choose a countably infinite $C_D\subseteq\Bbb R$ such that $C_D\subseteq D$, and let $\mathscr{C}=\{C_D:D\in\mathscr{D}\}$; clearly $\mathscr{C}$ is almost disjoint. Moreover, if $D,E\in\mathscr{D}$, and $D\ne E$, then $D\cap E$ is finite, so $C_D\cap C_E$ is finite, and $C_D\ne C_E$. Thus, $|\mathscr{D}|=|\mathscr{C}|\le 2^\omega$.


问:
Gaussian distribution and independence
Suppose you have Z=$(Z_1,...,Z_n)^T$, which has the standard Gaussian distribution on $\Bbb R^n$, and the $m$ x $n$ matrix A. Let W=$(W_1,...,W_m)^T$ be defined by W=AZ. Since Z is standard Gaussian, it means that each $Z_i$ are independent. Since W is a function of Z, W has a multivariate normal distribution. However, will the $W_i$, i=1,2,...,m be independent i.e will the variance covariance matrix of W, which is $AA^T$, be a diagonal matrix?

答:
No. $AA^\intercal$ has no reason to be diagonal e.g. $A=\begin{pmatrix}1&1\\1&0\end{pmatrix}$, then $AA^\intercal=\begin{pmatrix}2&1\\1&1\end{pmatrix}$.

You should know that any Gaussian vector is of the form $\mu+\sigma Z$, where $Z$ is the standard Gaussian vector. It is certainly false in general that the components of a Gaussian vector must be independent.

Title: Gaussian distribution and independence


问:
Prove: for $x$ large enough, there are more primes in the interval $(1,x]$ than in $(x,2x]$
I want to show that there are more primes in the interval $(1,x]$ than in $(x,2x]$, when $x$ is large enough. Let $\pi$ be the prime counting function, ie. $\pi(x)=\# \{ p \leq x | p \text{ is a prime} \}$. Then the above statement is equivalent to showing:

$\pi (2x)-2\pi(x)=-\log(2)\frac{x}{\log ^2x}+O\left(\frac{x}{\log ^3 x}\right)$

I have a hint saying to use the prime number theorem. I'm trying to use that $\pi(x)= \frac{x}{\log x} + \frac{x}{\log ^2 x} + O \left( \frac{x}{\log ^3 x} \right)$, but can't seem to rewrite it fitting. Maybe I need to use other aspects of the prime number theorem.

Added: It's useful to see alternative ways of solving the problem, but I'm still specifically seeking to show the equation above using PNT.

答:
From Dusart we know that $\frac{x}{\ln x}\left(1+\frac{1}{\ln x} \right) < \pi(x )<\frac{x}{\ln x}\left(1+\frac{1.3}{\ln x} \right) $ for all $x \geq 600$ and so you need to prove that $\pi(2x) -2 \pi(x) \leq \frac{2x}{\ln 2x} \left(1+\frac{1.3}{\ln 2x} \right)-2\frac{x}{\ln x}\left(1+\frac{1}{\ln x} \right) << 0  $

Or Equivalently $\frac{1}{\ln 2x} \left(1+\frac{1.3}{\ln 2x} \right) << \frac{1}{\ln x}\left(1+\frac{1}{\ln x} \right) => \ln x (1+\frac{1.3}{\ln 2x}) < (\ln x +\ln 2)(1+\frac{1}{\ln x}) => \ln x +\frac{1.3 \ln x}{\ln 2x} < ln x +1.3< \ln x+ 1 + \ln 2 < \ln x+1+\ln 2 + \frac{\ln 2}{\ln x} $

Checking for smaller cases the inequality is true for all $x\geq 11$


答:
There is no need for error estimates.  Use the PNT in the form $\pi(x) \sim x/\log x$. That implies $\pi(2x)/\pi(x) \sim (2x/\log(2x))/(x/\log x) = 2(\log x)/\log(2x)$.  Now convince yourself that this tends to $2$ as $x\to \infty$, so the difference
$$
\pi(2x) - \pi(x) = \pi(x)\left(\frac{\pi(2x)}{\pi(x)} - 1\right)
$$
is large for large $x$.  Similar reasoning shows for each $c > 1$ that $\pi(cx) - \pi(x)$ is large for large $x$.

Title: Prove: for $x$ large enough, there are more primes in the interval $(1,x]$ than in $(x,2x]$

Body:
I want to show that there are more primes in the interval $(1,x]$ than in $(x,2x]$, when $x$ is large enough. Let $\pi$ be the prime counting function, ie. $\pi(x)=\# \{ p \leq x | p \text{ is a prime} \}$. Then the above statement is equivalent to showing:

$\pi (2x)-2\pi(x)=-\log(2)\frac{x}{\log ^2x}+O\left(\frac{x}{\log ^3 x}\right)$

I have a hint saying to use the prime number theorem. I'm trying to use that $\pi(x)= \frac{x}{\log x} + \frac{x}{\log ^2 x} + O \left( \frac{x}{\log ^3 x} \right)$, but can't seem to rewrite it fitting. Maybe I need to use other aspects of the prime number theorem.

Added: It's useful to see alternative ways of solving the problem, but I'm still specifically seeking to show the equation above using PNT.


问:
The limit $\lim_{x \to ∞} (3x^4-5x^2+1)(1-e^{1/x^2})^2$
$\lim_{x \to ∞} (3x^4-5x^2+1)(1-e^{1/x^2})^2$

I found +∞ but the limit is 3. How to compute this limit ?

Body:
Since $e^{x} = \sum_{k = 0}^{\infty} x^{k}/k!$, we have

$$\lim_{x\to \infty} (3x^4 - 5x^2 + 1) (1 - e^{1/x^2})^{2} = \lim_{x \to \infty} (3x^{4} - 5x^{2} + 1)\left(x^{-4} + \mathcal{O}(x^{-6})\right) = \boxed{3} $$


问:
Inverses of Surjective and Injective Functions
Can you explain if the inverse of a bijective function is always a bijection, and the same for the inverses of a surjection and injection (i.e. is the inverse of a surjective function always surjective)

Additionally, can you explain why compositions of bijections are always bijections (and same for surjection/injection)

This isn't for homework or anything, my textbook notes are just vague and I would appreciate some more background on this topic!

答:
$
\newcommand{\range}{\mathrm{range}}
\newcommand{\i}[1]{{#1}^{-1}}
$
In short:

The inverse of a bijective function is itself a bijection. If $f$ has inverse $f^{-1}$, then $f$ is the inverse function to $f^{-1}$ specifically. (Of course, you can prove this.)

The inverse of an injective function $f : X \to Y$ need not exist (unless it is a bijection); however, it can have a left inverse $f_L : Y \to X$ such that $(f_L \circ f)(x)=x$ for every $x \in X$. However, $f_L$ need not be a function itself. Consider the function $f : \{1,2\} \to \{1,2,3\}$ given by $f(x) = x$. Then $f_L(x)=x$ as well, but since $f_L(3)$ isn't defined, $f_L$ isn't actually a function, but rather a partial function.

Alternatively, we may say $f_L$ is a function on $\range(f)$, or, rather, $f : X \to \range(f) \subseteq Y$ has a left inverse $f_L : \range(f) \to X$. However, in this sense, since every function is surjective on its range, $f$ is a bijection anyways.

Meanwhile, to every surjective $f : X \to Y$, we can associate a right inverse $f_R : Y \to X$ such that $(f \circ f_R)(y) = y$ for each $y \in Y$. In this case, it becomes very clear that a right inverse need not be unique, even: let $f : \Bbb Z \to \{1\}$ be given by $f(x) = 1$. Any right inverse can be given by $f_R(1) = n$ for any $n \in \Bbb Z$, so we have infinitely many of them. Intuitively, if $f_R$ is not injective, we have problems, because then there would exist a point in the domain of $f$ which maps to two different points in $f$'s codomain -- so $f_R$ not being injective implies that $f$ is a multi-valued function, a case we typically exclude in our study at this level. Of course, typically $f_R$ isn't a surjection, so there's not much more we can say, I think.

You can easily prove that the composition of bijections is a bijection. Note that if $f : A \to B,g : B \to C$ are bijections, then there are inverses $f^{-1} : B \to A$ and $g^{-1} : C \to B$. You can show that $f \circ g$ is a bijection as well, with inverse $g^{-1} \circ f^{-1}$. The details are as simple as showing that $f \circ g$ composed on either side of this results in the identity. For instance,

\begin{align*}
&\Big( (g^{-1} \circ f^{-1}) \circ (f \circ g) \Big)(x)\\
&= \Big( \i g \circ \underbrace{\i f \circ f}_{=\mathrm{id}_A} \circ g \Big)(x) \tag{function association} \\
&= \Big( \i g \circ g \Big)(x) \tag{inverses} \\
&= \mathrm{id}_B(x) \tag{inverses}\\
&= x \tag{def. of identity}
\end{align*}

as desired. The other side is very similar.

Similarly, if $f : A \to B,g : B \to C$ are surjections, then they have right inverses $f_R : B \to A$ and $g_R : C \to B$. Of course, you can show that the right inverse to $f\circ g$ is $g_R \circ f_B$.

And, of course, if $f : A \to B$ and $g : B \to C$ are injections, you have left inverses $f_L : B \to A$ and $g_L : C \to B$, and $f \circ g$ has left inverse $g_L \circ f_L$.

I'll leave verifying these details to you.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.


Title: The limit $\lim_{x \to ∞} (3x^4-5x^2+1)(1-e^{1/x^2})^2$

Body:
$\lim_{x \to ∞} (3x^4-5x^2+1)(1-e^{1/x^2})^2$

I found +∞ but the limit is 3. How to compute this limit ?

Body:
$\displaystyle \lim_{ x \to \infty} (3x^4-5x^2+1)(1-e^{1/x^2})^2$
$=\displaystyle \lim_{x \to \infty} \left(\frac{e^{1/x^2}-1}{\frac{1}{x^2}}\right)^2\left(\frac{3x^4-5x^2+1}{x^4}\right)$
$=1\times 3$

You know that: $$\lim_{u \to 0^+} \frac{e^u-1}{u}=1$$
Putting $u=1/x^2$ we get:$$\lim_{x \to \infty} \frac{e^{\frac{1}{x^2}}-1}{\frac{1}{x^2}}=1$$


问:
Multiplicities of Polynomials
I'm trying to understand the following explanation. This comes up as a first step into defining intersection multiplicities for Bezout's theorem.

Let $P$ be a non-zero polynomial in $\mathbb{C}\left[x_{0}, \ldots, x_{n}\right]$ and let $a=\left(a_{0}, \ldots, a_{n}\right)$ be a point in $\mathbb{C}^{n+1}$. The multiplicity $\operatorname{mult}_{a} P$ of $P$ at $a$ is the smallest non-negative integer $m$ such that there exists $\alpha \in \mathbb{N}^{n+1}$ with $|\alpha|=m$ and $\partial^{\alpha} P(a) \neq 0$.
Such a non-negative integer $m$ exists, because, up to some non-zero factors, the partial derivatives $\partial^{\alpha} P(a)$ are the coefficients in the Taylor expansion of $P$ around the point $a .$ Thus if $P$ is non-zero, then at least one of these partial derivatives is different from zero. This argument also shows that mult $_{a} P \leq \operatorname{deg}(P),$ and that $\operatorname{mult}_{a}(P Q)=\operatorname{mult}_{a} P+\operatorname{mult}_{a} Q$ for every non-zero polynomial $Q$ in $\mathbb{C}\left[x_{0}, \ldots, x_{n}\right],$ since the Taylor expansion of $P Q$ around $a$ is the product of the Taylor expansions of $P$ and $Q$.

Specifically, I'm trying to comprehend the last line. If I recall correctly, a taylor expansion of a multivariate polynomial at point $a$ is given $$ P(x) = \sum_\alpha \frac{\partial^\alpha P(a)}{\alpha !} (x-a)^\alpha.$$
According to the above explanation, $$\frac{\partial^\alpha P(a)}{\alpha !} (x-a)^\alpha \frac{\partial^\beta Q(a)}{\beta !} (x-a)^\beta = \frac{\partial^{\alpha+\beta} PQ(a)}{(\alpha+\beta) !} (x-a)^{\alpha+\beta} $$
but I don't get why this is. We have, simply using definitions, that $ (x-a)^{\alpha} (x-a)^{\beta} = (x-a)^{\alpha+\beta}$. But I still don't get why the above expression holds.

Edit: We do NOT have that $\alpha ! \beta != (\alpha+\beta)!$, upon closer reflection.

答:
You have to combine like terms. So when you multiply $P(x)$ and $Q(x)$ you get

$$P(x)Q(x) = \sum_{\gamma} \left[ \sum_{\alpha + \beta = \gamma} \frac{\partial^\alpha P(a)}{\alpha!} \frac{\partial^\beta Q(a)}{\beta!} \right] (x - a)^\gamma. \tag{1} $$

Now comparing coefficients with the Taylor expansion of $PQ$:

$$ (PQ)(x) = \sum_\gamma \frac{\partial^\gamma (PQ)(a)}{\gamma!} (x - a)^\gamma \tag{2} $$

we get this identity:

$$ \frac{\partial^\gamma (PQ)(a)}{\gamma!} = \sum_{\alpha + \beta = \gamma} \frac{\partial^\alpha P(a)}{\alpha!} \frac{\partial^\beta Q(a)}{\beta!}. \tag{3} $$

But I like to think about this a different way. The set $\{ (x - a)^\alpha : \alpha \in \mathbb{N}^{n+1}\}$ is a basis for $\mathbb{C}[x_0, \dots, x_n]$. That means there is a unique way of writing any polynomial $P(x)$ as $P(x) = \sum c_\alpha (x - a)^\alpha$. Now, of course we can take derivatives to get the identity

$$ c_\alpha = \frac{\partial^\alpha P(a)}{\alpha!}. \tag{4} $$

And just so everyone's on the same page, this is short for

$$\frac{\partial_{x_0}^{\alpha_0} \cdots \partial_{x_n}^{\alpha_n} P(a_0,\dots,a_n)}{\alpha_0! \cdots \alpha_n!} (x_0 - a_0)^{\alpha_0} \cdots (x_n - a_n)^{\alpha_n}.$$

But then it necessarily follows, that however we write $(PQ)(x)$ in this basis, those must be the coefficients. And remember: these are not power series, so there are no questions of convergence that need to be addressed.

So because $(1)$ is true by multiplication and $(2)$ is true by fact $(4)$, it must follow that $(3)$ is true since we are working with a basis. Notice that this is something you can't do (like this anyways) with power series since $\{ (x - a)^\alpha : \alpha \in \mathbb{N}^{n+1}\}$ is not a basis for the set of power series.

Title: Multiplicities of Polynomials


答:
Replacing  $x^2$ by $t,$ your limit
becomes $$\lim_{t \to \infty} 3t^2(1-e^{1/t})^2.$$
Now let $s=\frac{1}{t}$ so that the limit becomes
$$\lim_{s \to 0^+} 3(\frac{1-e^s}{s})^2=3(-1)^2=3.$$


问:
Evaluating a convergent improper triple integral over the unit sphere
In Exercise 5 (f) of Angus Taylor's Advanced calculus (p. 659) one is asked to find the value of the following integral if convergent:

$$I:=\underset{R}{\iiint}\dfrac{x^2 y^2 z^2}{r^{17/2}}\mathrm dV$$

where $R$ is the unit sphere $x^2+y^2+z^2\leq 1$ and $r^2=x^2+y^2+z^2$.

Observing that $\dfrac{x^2 y^2 z^2}{r^{17/2}}\leq \dfrac{r^6}{r^{17/2}}=r^{-5/2}$ I proved that $I$ is convergent.

Using spherical co-ordinates $r$, $\theta $, $\phi $ i.e.

$$\begin{align*}x&=r\sin \phi \cos \theta\\y&=r\sin \phi \sin \theta\\z&=r\cos \theta\end{align*}$$

I transformed the integral $I$ into

$$I=\int\nolimits_0^{2\pi }\left(\int_0^{\pi }\left(\lim_{\delta \to 0}\int_{\delta }^1\left(r^2 \sin \phi\right)\dfrac{x^2 y^2 z^2}{r^{17/2}}\;\mathrm dr\right)\;\mathrm d\phi \right)\;\mathrm d\theta$$

$$=\lim_{\delta \to 0}\left( \int_{\delta }^1 r^{-1/2}\mathrm dr\right)\int_0^{2\pi }\cos^4 \theta \sin^2 \theta \mathrm d\theta\int_0^{\pi }\sin^5 \mathrm d\phi $$

$$=2\cdot \dfrac18 \pi \cdot \dfrac{16}{15}=\dfrac4{15}\pi $$

In the solutions the answer is $\dfrac8{105}\pi$. Since sometimes there
are a few book typos (in the exercises) to prevent undue copying, I ask the
following

Question: What is the correct solution, $\dfrac4{15}\pi $ or $\dfrac8{105}\pi $?

UPDATE (Correction): instead of $z=r\cos \theta $ it is

$z=r\cos \phi $

See a comment from whuber.

The integral $I$ is transformed into

$$I=\int_0^{2\pi }\left(\int_0^{\pi }\left(\lim_{\delta \to 0}\int_{\delta }^1\left(r^2 \sin \phi\right)\dfrac{x^2 y^2 z^2}{r^{17/2}}\;\mathrm dr\right)\;\mathrm d\phi \right)\;\mathrm d\theta$$

Since

$$(r^2 \sin \phi )\dfrac{x^2 y^2 z^2}{r^{17/2}}=(r^2\sin \phi )\dfrac1{r^{17/2}}\left( r\sin \phi \cos \theta \right)
^{2}\left( r\sin \phi \sin \theta \right) ^{2}\left( r\cos \phi \right) ^{2}$$

$=r^{-1/2}\cos ^{2}\theta \cdot\sin ^{2}\theta \cdot\cos ^{2}\phi \cdot\sin ^{5}\phi $,

the transformed integral becomes (if I am right):

$$I=\left(\lim_{\delta \to 0} \int_{\delta
}^1 r^{-1/2}\mathrm dr\right)\int_0^{2\pi }\cos^2\theta\cdot\sin^2\theta \;\mathrm d\theta
\int_0^{\pi }\cos^2 \phi \cdot\sin^5 \phi \;\mathrm d\phi$$

$$=2\cdot \dfrac14 \pi \cdot \dfrac{16}{105}=\dfrac8{105}\pi$$

The correct solution will be $\dfrac8{105}\pi $ as in the book.

答:
As a double check, let's perform the integral in a completely different manner (so that my mistakes are unlikely to overlap yours!).

It is handy to use a condensed notation.  I will write $[i,j,k;a]$ for the value of $x^{2 i}y^{2 j}z^{2 k} \rho^{2 a}$ integrated over the unit sphere, where $i,j,k$ are integral and $a$ is real.  These relations are easy to establish:

$[i,j,k;a]$ is invariant under permutations of $(i,j,k)$.

$6 [1,1,1;a] = [0,0,0;a+3] - 3[3,0,0;a] - 18[2,1,0;a]$ is a consequence of expanding $\rho^6 = (x^2+y^2+z^2)^3$ as a multinomial, using the first property to collect equal terms, and isolating $[1,1,1;a]$ on the lhs.

$[i,j,k;a+1] = [i+1,j,k;a] + [i,j+1,k;a] + [i,j,k+1;a]$ follows from $\rho^2 = x^2 + y^2 + z^2$.  Use this to compute $[2,1,0;a]$ in terms of $[3,0,0;a]$ and $[2,0,0;a+1]$.

$[n,0,0;a] = \frac {4 \pi} {(2 n + 1) (2 n + 3 + 2 a)}$ can be obtained via integration by parts in cylindrical coordinates or by directly performing the integration of $ z^{2 n}  \rho ^ {2 a}$ over the sphere, which is very easy to do (because the angular part only involves $\phi$ and the integrand is exact).  (The numerator is the area of the unit sphere and the factors in the denominator come from integrating a $2 n$ power (in the angular part of $z$, which separates from the $\rho$ integral) and a $2 n + 2 a + 2$ power due to the $\rho^{2 n} \rho^{2 a} \rho^2 d \rho$ term in the integrand.)  Even when $a \lt 0$, this is justified whenever $2 n + 3 + 2 a \gt 0$, because the integral still converges at the origin.

From these we algebraically obtain

$$\eqalign{
[1,1,1;a] = &\frac{1}{6} \left( [0,0,0;a+3] - 3 [3,0,0;a] - 18 [2,1,0;a] \right) \cr

= &\frac{4 \pi} {6} \left( \frac{1}{2 a+9} - \frac{3}{7 (2 a+9)} - \frac{18}{35(2 a+9)} \right) \cr

= &\frac{4 \pi}{105 (2 a+9)}.
}$$

Setting $2 a = -17/2$ gives the textbook answer $\frac{8 \pi}{105}$.

ContentLicense: CC BY-SA 2.5


问:
How do I show that the degree $n$ Taylor polynomials of $f$ about two points are equal?
Question

Suppose that $f(x)$ is a polynomial of degree $d$, and that $n \ge d$. Let $x_0 \neq x_1$. Show that the degree $n$ Taylor polynomials of $f$ about $x_0$ and $x_1$ are equal.

Attempt

Let the polynomial be $f(x) = \sum_{k=0}^{d}a_kx^k$. Then the Taylor polynomial of $f$ about $x_0$ is $$P_n(x) = \sum_{k=0}^{n} \dfrac{f^{(k)}(x_0)(x-x_0)^k}{k!}$$

and the Taylor polynomial of $f$ about $x_1$ is $$Q_n(x) = \sum_{k=0}^{n} \dfrac{f^{(k)}(x_1)(x-x_1)^k}{k!}.$$

For them to be equal, I think it would have to be the case that $P_n(x) - Q_n(x)= 0$

I tried with a simple example. Let $d=1$ and $n=3$. Then

$$f(x) = \sum_{k=0}^{1}a_kx^k = a_0 + a_1x ,$$

$$P_3(x) = \sum_{k=0}^{3} \dfrac{f^{(k)}(x_0)(x-x_0)^k}{k!} = a_0 + a_1(x-x_0) ,$$

$$Q_3(x) = \sum_{k=0}^{3} \dfrac{f^{(k)}(x_1)(x-x_1)^k}{k!} = a_0 + a_1(x-x_1) .$$

However, $P_n(x) - Q_n(x)= 0 \Rightarrow x_0 = x_1$ or $a_1 = 0$. Where did I go wrong?

Thank you for your time.

答: The $k=0$ term in $P_3(x)$ should be $f(x_0)$, which is $a_0+a_1x_0$, not $a_0$. Similar remarks apply to $Q_3(x)$. </p>


问:
Does adding two vectors from different subspaces have meaning?
Given two vectors $\langle1,1\rangle$ and $\langle2,2,2\rangle$ in $\mathbb{R}^2$ and $\mathbb{R}^3$ respectively, is it possible to add them together?

If we add them in order, increasing the resulting dimension to $\mathbb{R}^5$, we get $
\langle1,1,2,2,2\rangle$. However, this is not a commutative result as $\langle1,1\rangle + \langle2,2,2\rangle = \langle1,1,2,2,2\rangle$ would not be equal to $\langle2,2,2\rangle+ \langle1,1\rangle =\langle2,2,2,1,1\rangle$.

Additionally, adding the vectors component wise seems illogical as $\langle1,1\rangle + \langle2,2,2\rangle = \langle1+2,1+2, \text{unknown}+ 2\rangle$. And assuming the unknown to be $0$ seems arbitrary.

I have a gut reaction, that this operation shouldn't be undefined, but I am new to linear algebra and don't have a strong conceptualization of it yet. (This is my first post. Yay!!)

答:
You've proposed two kinds of operations here. The first corresponds to what is commonly called 'lifting,' while the second is certainly illegal (techincally, the first is called concatenation, but that is literally just conjoining ordered values, and doesn't have the kind of linear algebraic interest as lifting).

To explain why the first is called, lifting, we turn to $\mathbb{R}^2$ and $\mathbb{R}^3$ for intuition.

Consider a vector $v \in \mathbb{R}^2$. It is natural to consider this vector on the $z = 0$ plane in $\mathbb{R}^3$, so when we form the vector:

$$\begin{bmatrix} v \\ 1 \end{bmatrix}$$

we are 'lifting' the plane up one unit. It's certainly not like adding, and has none of the nice properties that addition has. That said, it is a useful tool used to derive identities that show certain things to be true; the idea is that if we take some point arrangement and want to prove something about it (often in convex geometry), it might be useful to lift up the arrangement which keeps its properties, but forces information on the values which allows us to retain some of that information when we 'bring it back down' to its original dimension.

The second is ill-defined, unless you stipulate what value fills the missing spots. It is true that such a thing is arbitrary, but it may be useful to embed the lower-dimensional vector in the higher dimensional space, in which case you can add them in a way that is analogous to the second way.


问:
Drawing level curves on x-y axes
Suppose I sketched the domain of a function $f(x, y) = \sqrt{(x-1)^2 + (y-2)^2 - 1}$ on the x-y plane. There's no problem with that. But suppose on that same set of x-y axes I wanted to add (by hand) some level curves of the function for $z = \sqrt{3}, \sqrt{7}$ and $\sqrt{12}$ (Feel free to use any $\sqrt{}$ values).

If we allow $ z = k = (x-1)^2 + (y-2)^2 ≥ 1$ I could maybe manoeuvre $x = y-1$ which yields $2(y-2)^2 ≥ 1$

If this is correct, what steps are required next to input and graph the z-values?

答:
Why not use ubiquitous plotting programs rather than "by hand"?

Such a 3D plot shows the structure extremely well.

But if you're required to lose the information (such as the relative heights of contours), you can easily simplify such graphs:

All the above took no more than one minute in software.  I wonder how long the OP will work on doing such plots "by hand" (and not making a mistake) and whether all that hand work really helps.

Title: Drawing level curves on x-y axes


问:
Examining properties of the relation $R = \{(S_1,S_2) \mid |S_1| < |S_2|, S_1,S_2 \subseteq S\}$ where $S$ is finite

The Problem: Given a finite set $S$, let the relation $$R = \{(S_1, S_2) \mid |S_1| < |S_2|, S_1, S_2 ⊆ S\}$$ Show
whether or not $R$ is reflexive, symmetric, antisymmetric or transitive.

I'm shaky on how to approach this problem. Any help would be greatly appreciated.

I think that it's antisymmetric only.

Suppose $S = \{1, 2, 3, 4, 5\}$. Let $S_1 = \{1, 2\}$ and $S_2 = \{3, 4, 5\}$. $S_1$ and $S_2$ are subsets of $S$, and $|S_1| < |S_2|$.

Then $R = \{(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)\}.$

$R$ is not reflexive because $S_1$ will never equal $S_2$ and a reflexive relation must act on a set, not two different sets.

$R$ is clearly not symmetric, nor is it transitive.

答:
Some comments on your work:

You're correct on reflexivity: $(S_1,S_1) \in R$ iff $|S_1| < |S_1|$, which is nonsense. For instance, if $S = \{1,2,3\}$ and you have $A = \{1\}$, $|A| = 1 \not < 1 = |A|$, so $(A,A) \not \in R$.

You're correct on symmetry. For instance, take $S = \{1,2,3\}$, then it holds that $(A,B) =(\{1\},\{1,2\}) \in R$ since $|A| = 1 < 2 = |B|$. However, the reverse isn't true, since $|B| = 2 \not < 1 = |A|$, so $(B,A) \not \in R$, and thus no symmetry.

You're correct on anti-symmetry, but probably for the wrong reasons since it's a bit nuanced and subtle. If $(A,B),(B,A) \in R$, then $|A| < |B|$ and $|B|<|A|$. Thus there never exists such a double pair in $R$, so we have anti-symmetry vacuously.

You're incorrect on transitivity. If $(A,B),(B,C) \in R$, then $|A|<|B|$ and $|B|<|C|$. Since $<$ is transitive on cardinals, it holds that $|A|<|C|$ and thus $(A,C) \in R$.

Note, however, the elements of $R$ are ordered pairs whose components are sets, not numbers. Your example of $R$ has ordered pairs of the form $(|A|,|B|)$ as opposed to $(A,B)$.

In general, for work like this, it might be best to avoid claiming things "clearly" hold, when they tie directly into the things you want to prove. For adjacent results, it might be okay, but for situations like these or posts on MSE, it obfuscates what you're actually thinking and what you know, so it makes it harder to help you.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.


问:
Solving the cryptarithm $\mathrm{TWO}+\mathrm{TWO}=\mathrm{FOUR}$ in base $10$ and base $7$

Find the numerical value of each of the letters in the following expression $$\mathrm{TWO}+\mathrm{TWO}=\mathrm{FOUR}$$
in (a) base $10$ and (b) base $7$.

I don't even know how I would approach this problem. Any help would be appreciated.

答:
I do not know what 102152111's "brute force" solution is for base 7, but we can use a relatively small number of trials by focusing on $O$ as our first target.  Since $F=1$, the distinctness criterion rules out that value for $O$, so we try the six others in turn.  Each such value of $O$ corresponds to a unique value for $T$ and a unique carry digit from the $7^1$ column to the $7^2$ column, thus simplifying the analysis.

$O=0$ fails quickly, since it would give $R=0$ in the units place.

$O=2$ corresponds to $T=4$ and a carry digit of $1$ ($4×4+1=12$ base $7$), but then it leads to $R=4$ in the units place which fails because $T=4$.

$O=3$ corresponds to $T=5$ and a carry digit of $0$, and leads to $R=6$ in the units place.  We are good so far on distinctness.  To get the right carry digit into the $7^2$ place for this case we then need $W<4$.  $W=1$ and $W=3$ fail distinctness from previously determined letters, $W=0$ gives $U=0$ because $2O<7$, but $W=2$ makes it through.  This is the solution $523+523=1346$ given by 102152111.

$O=4$ gives $R=1$ in the units place, which gets an $F$, so to speak.

$O=5$ corresponds to $T=6$, a carry digit of $0$ into the $7^2$ place, and $R=3$.  This looks good so far, but  no values of $W$ work.  $W=3$ is too large because $35+35>100$ base $7$; $W=1$ is not distinct from $F$; $W=0$ and $W=2$ fail to hold up distinctness when we calculate $U$.  So $O=5$ falls just short of making it through.

$O=6$ requires $T=6$ and a carry digit of $1$, the former of which already fails distinctness.

So $523+523=1346$ derived from the $O=3$ case is the unique solution.


问:
Properties of a hyperbola slice of a cone
I'm attempting to locate some points (in a 3D coordinate system) on the surface of a cone by slicing the cone with a plane, and using the resultant ellipse, parabola or hyperbola to calculate the points.  I've had success with the ellipse & parabola solutions, but am seeking help with the hyperbola case.

Known data:  The slope of the cone, the slope of the plane, and distance from the vertex of the cone to the intersection of the plane and cone's longitudinal axis.  Given these knowns, I'm looking for "a" & "b".

From reading, I have the formulas x²/a² - y²/b² = 1 & c² = a² + b².  I understand distance "a" to be half the distance between the two vertices.  Question 1 is, is "a" always half the distance between the two vertices, even when the plane is sloped?  If not always half, how is "a" calculated when the plane is sloped?  Question 2 is how to determine "b"?  Question 3 is how to determine the angle of the asymptotes? ...is that the angle of the cone, or if the plane is sloped, some other angle?

I'm concerned only with the upper nappe of the cone.

As I massage the above formula, attempting to solve for an "x" & "y" & "b", as the plane angle gets shallower, "a" gets longer.  And when "a" gets longer than an "x" value that I am using, my solutions fail due to an attempt to take the square root of a negative number.  So, I know I'm messing up somewhere.  Any help is appreciated.

This is my first post here so I don't know my way around.  I'm going to attempt to post a sketch.

Thanks in advance.

答:
Here's the plan of attack.

Give equations that describe the cone and plane in the standard basis

Define "plane coordinates"

Find the coordinate transform matrix between the two coordinate systems

Find the equation of the conic section in plane coordinates

Rotate the coordinate system so our conic is oriented along the coordinate axes and thus find the major and minor axis lengths

Once we have the equation for the conic, there will still be a lot of work before we can get the semi major and semi minor axes, but hopefully someone else can chime in on this matter. First a preface on notation - scalars will be denoted with normal italic font, e.g $s$. Vectors will be denoted with upright (Roman) font, e.g $\mathrm{v}$. Matrices will be denoted with bold, e.g $\mathbf{M}$. With that out of the way, let's get started!

1: Cartesian equations of the plane and cone

First, the plane. Any plane in $\Bbb{R}^3$ is a level surface of a linear polynomial in $x,y,z$. What I mean by this is given $a,b,c\in\Bbb{R}$ we can represent the corresponding plane as
$$\mathcal{P}=\{(x,y,z)\in\Bbb{R}^3:ax+by+cz=d\}$$
Yes, apologies on using $a$ and $b$ - I will probably use them again later.

Now, the cone. The upward facing right angled cone is the surface corresponding to $\phi=\pi/4$, in spherical coordinates. The inverse transformation tells us that this means
$$\arccos\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)=\frac{\pi}{4}$$
Taking the cosine of both sides we see that
$$\frac{z^2}{x^2+y^2+z^2}=\frac{1}{2}$$
So we get
$$\mathcal{C}=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2-z^2=0\}$$
Now while we initially only wanted the upward part of the cone, we can see that rewriting this as
$$z^2=\pm\sqrt{x^2+y^2}$$
Means that our equation gives the lower part of the cone as well. So as long as our plane isn't vertical, i.e $c=0$, the system of equations
$$\begin{bmatrix}
ax+by+cz\\
x^{2} +y^{2} -z^{2}
\end{bmatrix} =\begin{bmatrix}
d\\
0
\end{bmatrix}$$
Which will always have at least one solution.

2: Plane coordinates

Our goal here is to create a new "basis" (this is technically an abuse of terminology) for $\mathbb{R}^3$ that we can write a vector as an affine transformation
$$x\hat{\mathrm{i}}+y\hat{\mathrm{j}}+z\hat{\mathrm{k}}=(x'\hat{\mathrm{i}}'+y'\hat{\mathrm{j}}'+z'\hat{\mathrm{k}}')+\mathrm{p}$$
Where $\mathrm{p}$ is some reference point on the plane, and the primed and unprimed basis vectors are related through via a linear transformation
$$\mathbf{M}\begin{bmatrix} \hat{\mathrm{i}}\\ \hat{\mathrm{j}}\\ \hat{\mathrm{k}} \end{bmatrix} =\begin{bmatrix} \hat{\mathrm{i}}'\\ \hat{\mathrm{j}}'\\ \hat{\mathrm{k}}' \end{bmatrix}$$
The inverse transformation between coordinate systems is
$$\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\mathbf{M}^{-1}\begin{bmatrix} x'\\ y'\\ z' \end{bmatrix}+\mathrm{p}$$
We can multiply both sides by $\mathbf{M}$ to get the forward transformation:
$$\mathbf{M}\left(\begin{bmatrix} x\\ y\\ z \end{bmatrix}-\mathrm{p}\right)=\begin{bmatrix} x'\\ y'\\ z' \end{bmatrix}$$
First, we find the point $\mathrm{p}$. Though any point will do, it makes sense to use the one closest to the origin (of our original coordinate system). It turns out that this point is
$$\mathrm{p}=\frac{d}{a^2+b^2+c^2}(a,b,c)$$
This can be obtained either with simple coordinate geometry or, as I did it, with Lagrange multipliers. I can elaborate on how I did this if you wish. We let our third new basis vector be a unit vector parallel to this vector, that is
$$\hat{\mathrm{k}}'=\frac{(a,b,c)}{\sqrt{a^2+b^2+c^2}}$$
We want our new basis to be orthonormal, that is we want the other two basis vectors to satisfy the equations
$$\hat{\mathrm{i}}'\boldsymbol{\cdotp}\hat{\mathrm{k}}'=\hat{\mathrm{j}}'\boldsymbol{\cdotp}\hat{\mathrm{k}}'=\hat{\mathrm{i}}'\boldsymbol{\cdotp}\hat{\mathrm{j}}'=0$$

3: Finding the precise transformation matrix between the unit vectors

As of now, we have three equations for six (three components for both i and j) unknowns, so we need three extra conditions so we can uniquely determine our coordinate system. We'll only impose two extra conditions, namely $i'_1=j'_2$ and $j'_1=i'_2$. Letting $i'_1=1$ for now (this will be normalized later) we get the solutions
$$\hat{\mathrm{i}}'\propto \left(1,\frac{-a^2-b^2-2c^2+\sqrt{(a^2+b^2+2c^2)^2-4a^2b^2}}{2ab},\frac{-a^2+b^2+2c^2-\sqrt{(a^2+b^2+2c^2)^2-4a^2b^2}}{2ac}\right)$$
$$\hat{\mathrm{j}}'\propto \left(\frac{-a^2-b^2-2c^2+\sqrt{(a^2+b^2+2c^2)^2-4a^2b^2}}{2ab},1,\frac{a^2-b^2+2c^2+\sqrt{(a^2+b^2+2c^2)^2-4a^2b^2}}{2bc}\right)$$
For first one I'll use the abbreviations $\mu,\nu_1$ (in that order) and for the second one I'll use the abbreviations $\mu,\nu_2$ in that order. I'll also let $\kappa_1=\sqrt{1+\mu^2+\nu_1^2}$ and $\kappa_2=\sqrt{1+\mu^2+\nu_2^2}$ so we can normalize the above two expressions and write
$$\hat{\mathrm{i}}'=\frac{1}{\kappa_1}(1,\mu,\nu_1)~~;~~\hat{\mathrm{j}}'=\frac{1}{\kappa_2}(\mu,1,\nu_2)$$
The gammas can be found exactly in terms of $a,b,c$, but the expressions are rather lengthy so I won't bother posting them here. In practice you would do this numerically, not symbolically, anyway. Regardless, from the above expressions it's clear that
$$\hat{\mathrm{i}}'=\frac{1}{\kappa_1}(\hat{\mathrm{i}}+\mu\hat{\mathrm{j}}+\nu_1\hat{\mathrm{k}})~~;~~\hat{\mathrm{j}}'=\frac{1}{\kappa_2}(\mu\hat{\mathrm{i}}+\hat{\mathrm{j}}+\nu_2\hat{\mathrm{k}})$$
And of course
$$\hat{\mathrm{k}}'=\frac{a}{\sqrt{a^2+b^2+c^2}}\hat{\mathrm{i}}+\frac{b}{\sqrt{a^2+b^2+c^2}}\hat{\mathrm{j}}+\frac{c}{\sqrt{a^2+b^2+c^2}}\hat{\mathrm{k}}$$
Hence we can write the inverse transformation as
$$\begin{bmatrix} \hat{\mathrm{i}}'\\ \hat{\mathrm{j}}'\\ \hat{\mathrm{k}}' \end{bmatrix} =\underbrace{\begin{bmatrix} \frac{1}{\kappa _{1}} & \frac{\mu }{\kappa _{1}} & \frac{\nu _{1}}{\kappa _{1}}\\ \frac{\mu }{\kappa _{2}} & \frac{1}{\kappa _{2}} & \frac{\nu _{2}}{\kappa _{2}}\\ \frac{a}{\sqrt{a^{2} +b^{2} +c^{2}}} & \frac{b}{\sqrt{a^{2} +b^{2} +c^{2}}} & \frac{c}{\sqrt{a^{2} +b^{2} +c^{2}}} \end{bmatrix}}_{\mathbf{M}}\begin{bmatrix} \hat{\mathrm{i}}\\ \hat{\mathrm{j}}\\ \hat{\mathrm{k}} \end{bmatrix}$$
We can, of course, invert this matrix to get the forward transformation, but the expressions become extremely long, so I'll once again omit them. Recalling the inverse transformation, points $(x,y,z)$ sitting on the plane will have a zero $z'$ coordinate, precisely, they will be expressible as
$$\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\mathbf{M}^{-1}\begin{bmatrix} x'\\ y'\\ 0 \end{bmatrix}+\mathrm{p}$$

4: Equation of the conic section

Recalling that the equation of the cone is $x^2+y^2-z^2=0$, we can plug in the RHS from the previous section into the equation for the cone to get the intersection surface between our plane and cone. We can represent our conic section as the quadric plane curve in our plane $\mathcal{P}$ using our primed plane coordinates
$$\begin{array}{l} f(x',y')=\\ +\left( (\mathbf{M}^{-1} )_{1,1} \ x'+(\mathbf{M}^{-1} )_{1,2} \ y'+\frac{da}{a^{2} +b^{2} +c^{2}}\right)^{2}\\ +\left( (\mathbf{M}^{-1} )_{2,1} \ x'+(\mathbf{M}^{-1} )_{2,2} \ y'+\frac{db}{a^{2} +b^{2} +c^{2}}\right)^{2}\\ -\left( (\mathbf{M}^{-1} )_{3,1} \ x'+(\mathbf{M}^{-1} )_{3,2} \ y'+\frac{dc}{a^{2} +b^{2} +c^{2}}\right)^{2}\\ =0 \end{array}$$
So it seems computing the third row of the inverse matrix is actually not needed. However, numeric matrix inversion is very fast in software such as Mathematica, so I won't worry about it too much.

5: Reorienting our coordinate system

Let's drop the primes to make life easier for ourselves.
We need to convert the above expression for $f$ into something of the form
$$f(x,y)=A{x}^2+Bxy+C{y}^2+D=0$$
Which will be more useful for further analysis.  We can avoid a pile of algebra (useful for numerical calculations) by simply noting
$$\begin{array}{l} D=f( 0,0)\\ A=f( 1,0) -D\\ C=f( 0,1) -D\\ B=f( 1,1) -( A+C+D) \end{array}$$
Recall the standard (no tilting) equation for a conic is
$$ux^2+vy^2-1=0$$
So the goal here is to find some change of coordinates $(x,y)\to(X,Y)$ that will eliminate the mixed $XY$ terms. We want to find some polynomial function $F$ such that
$$f(x,y)=F(X,Y)\propto u X^2+vY^2-1$$
It turns out that a linear, not just affine, transformation will do - we don't need to move the origin. And we don't need to stretch the coordinate grid either, since our plane coordinates are already orthonormal. So we can simply use a rotation. So we are looking for a transformation of the form
$$\begin{bmatrix} X\\ Y \end{bmatrix} =\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix}$$
The inverse transformation is of course
$$\begin{bmatrix} x\\ y \end{bmatrix} =\begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}\begin{bmatrix} X\\ Y \end{bmatrix}$$
Which means that
$$\begin{array}{l} F(X,Y)=\\ A( X\cos \theta +Y\sin \theta )^{2}\\ +B( X\cos \theta +Y\sin \theta )( -X\sin \theta +Y\cos \theta )\\ +C( -X\sin \theta +Y\cos \theta )^{2}\\ +D \end{array}$$
So we expand out and set the mixed term equal to zero:
$$\text{mixed coefficient}=2A\cos\theta\sin\theta+B(\cos^2\theta-\sin^2\theta)-2C\sin\theta\cos\theta=0$$
Using some double angle identities we get
$$(A-C)\sin(2\theta)+B\cos(2\theta)=0$$
Obviously if $B=0$ this whole process is unnecessary and our original coordinates are already correctly aligned. Apart from that however, we break the solution of this equation into two cases.

Case 1 - $A=C$

When $A=C$, our equation reduces to

$$B\cos(2\theta)=0$$
So our first solution is $\theta=\pi/4$. Now since $\sin(\pi/4)=\cos(\pi/4)$ we can factor them out from our expression for $F$:
$$F(X,Y)=\frac{1}{2}\left(C(X+Y)^2+B(Y+X)(Y-X)+C(X-Y)^2+2D\right)$$
Doing some algebra, we get
$$F(X,Y)=\left(C-\frac{B}{2}\right)X^2+\left(C+\frac{B}{2}\right)Y^2+D$$
So
$$F(X,Y)=-D\left(\underbrace{\frac{B-2C}{2D}}_{\equiv u}X^2+\underbrace{\frac{-B-2C}{2D}}_{\equiv v}Y^2-1\right)$$
If either $u=0$ or $v=0$ then we have a parabola. If not, then if the signs of $u$ and $v$ are different, we have a hyperbola. If both are nonzero and the same sign, then we have an ellipse or circle, and
$$"b"=\text{semi minor axis}=\beta=\sqrt{\frac{1}{\min(|u|,|v|)}}$$
$$"a"=\text{semi major axis}=\alpha=\sqrt{\frac{1}{\max(|u|,v|)}}$$
Meaning if $|u|>|v|$ our conic can be written in the form
$$\pm \frac{x^2}{\alpha^2}\pm \frac{y^2}{\beta^2}=1$$
Which perhaps is more familiar. The alpha and beta go in opposite places if $|u|<|v|$.

Case 2: $A\neq C$

In this case the equation for $\theta$ is not as simple. Recall that we want
$$(A-C)\sin(2\theta)+B\cos(2\theta)=0$$
We can restate this as
$$\tan(2\theta)=\frac{B}{C-A}$$
So
$$\theta=\frac{1}{2}\arctan\left(\frac{B}{C-A}\right)$$
Our mixed term is zero, by construction. The other coefficients are
$$X\text{ squared  coefficient}=A\cos^2\theta-B\cos\theta\sin\theta+C\sin^2\theta$$
$$Y\text{ squared coefficient}=C\cos^2\theta+B\cos\theta\sin\theta+A\sin^2\theta$$
So then
$$u=\frac{A\cos^2\theta-B\cos\theta\sin\theta+C\sin^2\theta}{-D}$$
And
$$v=\frac{C\cos^2\theta+B\cos\theta\sin\theta+A\sin^2\theta}{-D}$$
Once again if either is zero, then the figure we have is a parabola. If the signs are different, we have a hyperbola. If both are nonzero with the same sign, we have an ellipse or hyperbola, and the same equations hold:
$$"b"=\text{semi minor axis}=\beta=\sqrt{\frac{1}{\min(|u|,|v|)}}$$
$$"a"=\text{semi major axis}=\alpha=\sqrt{\frac{1}{\max(|u|,|v|)}}$$

In response to comments - some examples with plots

Shown below is a graphic of the intersection between the plane $ax+by+cz=d$ with the cone $x^2+y^2-z^2=0$, using $a=0.7,b=4.1,c=1,d=-3.3$:

And now the plot of the conic of intersection $Ax^2+Bxy+Cy^2+D=0$, which we achieved through "plane coordinates" - in this case the coefficients are $A=0.934531,B=-0.291272,C=-1.57866,D=0.530046$:

Finally, we rotate our coordinate system by an amount $\theta=0.0576914$ to get the conic in the form $ux^2+vy^2-1=0$, in this case $u=-1.71572,v=2.99421$. The signs are different, hence we have a hyperbola -

This is just a rotated version of the previous plot.

A quick recap

We started with the equations for the plane and the cone, $ax+by+cz=d$ and $x^2+y^2-z^2=0$. Now we have an equation for their intersection, which I have as
$$ux^2+vy^2-1=0$$
Which perhaps you're more used to seeing as
$$\pm\frac{x^2}{"a"^2}\pm\frac{y^2}{"b"^2}=1$$
The quotation marks are because I used a and b earlier and I don't want to cause any confusion.

Slope and angle

Let's assume now that $u$ and $v$ have different signs, i.e, we have a hyperbola. How can we find the equations of the asymptotes and the angle between them? First, the equations. We can solve the equation for $y$ to get
$$y=\pm\sqrt{\frac{1-ux^2}{v}}$$
Which, since $u$ and $v$ are opposite signs, is
$$y=\pm\sqrt{\frac{1}{v}+\frac{|u|}{|v|}x^2}$$
So for large $x$, the asymptotes are the lines
$$y=\pm\sqrt{\frac{|u|}{|v|}}x$$
The line with the positive slope makes an angle of $\arctan(\sqrt{|u|/|v|})$ with the positive $x$ axis whereas the other makes an angle of $\pi-\arctan(\sqrt{|u|/|v|})$ with the positive $x$ axis meaning that the angle between them is $\phi_1=\pi-2\arctan(\sqrt{|u|/|v|})$. The other angle between them is $\phi_2=\frac{2\pi-2\phi_1}{2}$.


问:
Identifying properties of relations
I was studying binary relations and, while solving some exercises, I got stuck in a question.

I am sharing the question and my thoughts on solving it, and I am looking for some advice and comments about my attempt (what is wrong or what should I do to improve it).

Question. Let $A = \mathbb{N} \setminus \{1\}$ and consider the following binary relations on $A.$
$$R_1 = \{(x,X) : X \in \mathcal{P}(A) \wedge x \in X\}, \quad R_2 = \{(x,y) \in A^2 : x < y\}, \quad \quad R_3 = \{(x,y) \in A^2 : y > x^2\}.$$

Determine, justifying, if each of the above relations are reflexive, symmetric, transitive or antisymmetric.

Sketch. Of course, to solve this problem, one must understand what does it mean for a binary relation to be reflexive, symmetric, transitive or antisymmetric. So, let’s, first, recall the definition of each concept.

Let $A$ be a set $R \subseteq A^2$ a binary relation on $A.$ The binary relation $R$ is

$1.\quad$ reflexive, if $\quad \forall a \in A, aRa$;

$2.\quad$ symmetric, if $ \quad \forall a,b \in A, aRb \implies bRa$;

$3.\quad$ transitive, if $ \quad \forall a, b, c \in A, aRb \wedge bRc \implies aRc$;

$4.\quad$ antisymmetric, if $\quad \forall a,b \in A, aRb \wedge bRa \implies a = b.$

Hence, we must check if these conditions are satisfied for each of the above relations.

Attempt.

$\qquad 1. \quad R_1.$

This relation was include in this exercise, but I don’t agree with this. Because, $R_1 \subseteq \mathcal{P}(A) \times A,$ and the question states that the relations that we are working on are relation on $A.$

But forgetting this for a moment, those properties were only defined for binary relations on a set $A$ and not for a binary relation from $A$ to $B.$ Therefore, it makes no sense in talking about those properties in this example.

$\qquad 2. \quad R_2.$

Let $n \in A.$ The proposition $n < n$ is false, hence $(n,n) \notin R_2.$ Therefore, $R_2$ is not reflexive.

Let $m, n \in A.$ Suppose that $(m,n) \in R_2.$ Then, by definition of $R_2$ we have that $m < n.$ Then, it is not true that $n < m.$ So, $(n,m) \notin R_2.$ Therefore, $R_2$ is not symmetric.

Let $m, n, p \in A.$ Suppose that $m R_2 n$ and $n R_2 p.$ Then, $m < n$ and $n < p.$ Since $<$ is transitive, then $m < p$ and so $m R_2 p.$ Therefore, $R_2$ is transitive.

Let $m, n \in A.$ Suppose that $m R_2 n$ and $n R_2 m.$ Hence, we have that $m < n$ and $n < m$ which is a contradiction and so
$$\forall a,b \in A, aRb \wedge bRa \implies a = b$$
is vacuously true, Therefore, $R_2$ is antisymmetric.

$\qquad 3. \quad R_3.$

Let $n \in A.$ Since $n \geq 2,$ then $n^2 > n.$ So, it is not true, that $n > n^2.$ Hence, $(n,n) \notin R_3.$ Therefore, $R_3$ is not reflexive.

Let’s $m, n \in A.$ Suppose that $m R_3 n.$ Then, $n > m^2.$ It follows that $n^2 > m^4$ and $m^4 > m.$ Hence, $n^2 > m.$ Therefore, $R_3$ is symmetric.

Let’s $m, n, p\in A.$ Suppose that $mR_3n$ and $nR_3p.$ Then, $n > m^2$ and $p > n^2.$ Because $n^2 > n,$ then $p > m^2.$ Therefore, $R_3$ is transitive.

Let’s $m,n \in A.$ Suppose that $mR_3n$ and $nR_3m.$ Then $n > m^2$ and $m > n^2.$ Since, $m^2 > m$ then $n > m.$ So $n \neq m.$ Therefore, $R_3$ is not antisymmetric.

My biggest doubt is definitely on $R_3.$ I don’t know why, but that looks a bit suspicious to me. Although I have no clue of what is wrong. Can you help me?

Thank you in advance for your attention.

答:
You’re right about $R_1$, except that it’s a subset of $X\times\wp(A)$, not of $\wp(A)\times A)$.

Your analysis of $R_2$ is correct.

Note that $R_3$ would not be reflexive even if $1$ were in $A$: as long as there is at least one $a\in A$ such that $\langle a,a\rangle\notin R_3$, $R_3$ is not reflexive.

$R_3$ is not symmetric: if $\langle n,m\rangle,\langle m,n\rangle\in R_3$, then $m>n^2$ and $n>m^2$, so

$$m>n^2>n>m^2\,,$$

and hence $m>m^2$, which is false for every $m\in A$. Thus, not only is $R_3$ not symmetric, it is asymmetric: if $m\mathrel{R_3}n$, then $n\not\mathrel{R_3}m$. It is true that if $n>m^2$, then $n^2>m^4>m$, so $n^2>m$, but that actually implies that $n\not\mathrel{R_3}m$: $n\mathrel{R_3}m$ means that $m>n^2$.

Your argument for transitivity of $R_3$ is correct. But as I showed above, $R_3$ is asymmetric, so it, like $R_2$, is vacuously antisymmetric.

Title: Identifying properties of relations


答:
From the proof of the prime number theorem by the de la Vallée Poussin we have
$$\frac{x}{\log -(1-\epsilon)} < \pi(x) < \frac{x}{\log x -(1+\epsilon)}$$
for $x\ge x_{\epsilon}$. In other words, if we write $\pi(x) = \frac{x}{\log x - B(x)}$, then $B(x) \to 1$ ( see Legendre's constant).  In fact, we do have an explicit estimate of Dusart, but the above is enough.

We get
$\pi(2 x) < \frac{2x}{\log(2x) -(1+\epsilon)}$
and $\frac{2x}{\log x- (1-\epsilon) } < 2\pi(x)$
so we should have
$$\log (2x) -(1+\epsilon) > \log x - (1-\epsilon)$$
or $\log 2 > 2 \epsilon$. So we should take $\epsilon < \frac{\log 2}{2}$, and then get the inequality for $x \ge x_{\epsilon}$.

Note that in general for every $M>1$, we can get $\pi(Mx) < M \pi(x)$ for $x> \bar x_{M}$.

$\bf{Added:}$  Let us use the explicit Dusart bounds
$$\frac{x}{\log x -1} < \pi(x)$$
for $x\ge 5393$, and
$$\pi(x) < \frac{x}{\log x -1.1}$$
for $x \ge 60184$.  Therefore, for $x\ge 30 092$ we get
$$\pi(2x)< \frac{2x}{\log (2x) - 1.1} < \frac{2x}{\log x -1}< 2\pi(x)$$

Consider for  integer $4\le x \le 60183$,  the values of $\frac{\pi(x)}{ 2\pi(x/2)}$. It reaches $1$ at $4$, $7$, $8$, $9$, $13$, $19$, $20$, $21$, and values $< 1$ otherwise.


问:
Looking for help with reasoning about a function defined on congruence relations
Let:

$T_{p,r}(n,x)$ be the count of integer $i$ such that:

$p$ is an odd prime

$n,r<p,x$ are integers

$n - x \le i < n$

$i \equiv r \pmod {p}$

for all prime $q > 2$ where $i \equiv r \pmod {q}$, it follows that $p \le q$

Does it now follow that for any odd prime $p$, any integers $r, n, x$:

$$T_{p,1}(0,x) \ge T_{p,r}(n,x) \ge T_{p,0}(0,x)$$

I am finding it difficult to prove that this is the case in light of the 5th bullet point which is my attempt to generalize the concept of least prime factor to modular arithmetic.

Am I wrong in my assumption?

Edit:  Attempting an argument for $T_{p,1}(0,x) \ge T_{p,r}(n,x)$ where $r=1$:

It occurs to me that this special case of my assumption is straight forward to show:

(1)  Assume it is not true so that there exists $p,n,x$ such that:

$$T_{p,1}(0,x) < T_{p,1}(n,x)$$

(2)  Let:

$S_{0,x,q}$ be the set of integers $i$ where $-x \le i < 0$ and there exists a minimal prime $q$ such that $q < p$ and $i \equiv 1 \pmod q$ and $i \equiv 1 \pmod p$

$S_{n,x,q}$ be the set of integers $j$ where $n-x \le j < n$ and there exists a prime $q$ such that $q < p$ and $j \equiv 1 \pmod q$ and $j \equiv 1 \pmod p$

$|S_{0,x,q}|$ be the count of elements in $S_{0,x,q}$ and $|S_{n,x,q}|$ be the count of elements in $S_{n,x,q}$

(3) There must exist at least one $q$ where $m=|S_{0,x,q}| > |S_{n,x,q}|$

(4) But then, $x \ge mpq+1$  Otherwise, it would be impossible for $m$ such elements to exist.

(5) It follows $n-x, n-x+1, n-x+2, \dots, n-x+mpq$ forms a complete residue system modulo $n-x+mpq+1$

(6) But then $|S_{n,x,q}| \ge m$ which contradicts step(3).

(7) So, we can reject our assumption at step(1).

Edit 2: Attempting an argument for $T_{p,1}(0,x) \ge T_{p,r}(n,x)$ where $r>1$:

(1)  Assume it is not true so that there exists $p,n,x$ such that:

$$T_{p,1}(0,x) < T_{p,1}(n,x)$$

(2)  From the argument above, we can assume that $x \ge pqm+1$ and $x < pqm+r$.  Otherwise, the same argument applies.

(3)  We can assume that:

$$\sum_{\text{all }q}|S_{0,x,q}| = \sum_{\text{all }q}|S_{n,x,q}|+1$$

Assume that $\sum\limits_{\text{all }q}|S_{0,x,q}| > \sum\limits_{\text{all }q}|S_{n,x,q}|+1$

Then, there exists primes $q_1, q_2$ such that $q_1p +1 < q_1p + p +1 \le x < q_1p + r$

But this is impossible since by assumption $r < p$

(4) Let $U_{p,r}(n,x)$ be the count of integers $i$ where $n-x \le i < n$ and $i \equiv r \pmod p$ so that:

$$T_{p,r}(n,x) = U_{p,r}(n,x) - \sum_{\text{all }q}|S_{n,x,q}|$$

(5) It follows from step(1) that $U_{p,1}(0,x) = mpq$ while $U_{p,r}(n,x) = mpq-1$

(6) Combining step(5) and step(3):

$$T_{p,1}(0,x) = U_{p,r}(n,x)+1 - \sum\limits_{\text{all }q}|S_{n,x,q}|-1 = T_{p,r}(n,x)$$

(7) This contradicts step (1) so we can reject our assumption in step(1).

答:
Note with your definition of $T_{p,r}(n,x)$, your fourth & fifth bullet points combined basically state you're looking for where the least odd prime factor (I'm calling call it $\operatorname{lopf}$ for simplicity) of $r + (-i)$ is $p$.

Your conjecture is equivalent to all of the following $3$ inequalities being true:

$$T_{p,1}(0,x) \ge T_{p,r}(n,x) \tag{1}\label{eq1A}$$

$$T_{p,r}(n,x) \ge T_{p,0}(0,x) \tag{2}\label{eq2A}$$

$$T_{p,1}(0,x) \ge T_{p,0}(0,x) \tag{3}\label{eq3A}$$

Of course, if the first $2$ are true, then so is the third one. However, a counter-example to \eqref{eq1A} is $T_{3,1}(0,1) = 0$, but $T_{3,1}(5,1) = 1$. Also, a counter-example to \eqref{eq2A} is $T_{5,0}(16,5) = 0$ (since with $q = 3$, then $q \mid 15$ and $q \lt p$, so $15$ is excluded), but $T_{5,0}(0,5) = 1$. Nonetheless, \eqref{eq3A} is always true, as I show later.

Regarding your Edit attempt to prove $T_{p,1}(0,x) \ge T_{p,r}(n,x)$ where $r = 1$, this is a special case of \eqref{eq1A}, and I already gave a counter-example with $r = 1$, i.e., with $p = 3$, $x = 1$ and $n = 5$. In your attempted proof by contradiction, with your $(3)$, you seem to be assuming that if the number of integers where the $\operatorname{lopf}$ is $p$ in one range is greater than in a second range, the second range must have at least one prime $q \lt p$ (you don't state your $q$ must be odd, but it only makes sense since your $T_{p,r}(n,x)$ definition only worries about odd prime factors) where the number of values with $\operatorname{lopf}$ being $q$ is greater than that in the first range. However, as my example shows, this is not always the case. For one thing, if $p = 3$, there is not even any such $q \lt p$. However, even if $p \gt 3$, then it's still not necessarily true, e.g., with $T_{5,1}(0,1) = 0$ and $T_{5,1}(7,1) = 1$.

Your $(4)$ states $x \ge mpq+1$. However, with some $q \lt p$, have $i = -x = -pq + 1$. Then $i \equiv 1 \pmod{p}$ and $i \equiv 1 \pmod{q}$, so $m = 1$. However, $x = pq - 1$. The correct statement to make is actually $x \ge mpq - 1$.

This, of course, affects your $(5)$. Also, you state it forms a complete residue system modulo $n - x + mpq + 1$. However, for example if $n = 0$ and $x = mpq + 1$, this modulo expression is $0$. I believe you meant to state it is for modulo $mpq + 1$ instead. Nonetheless, a minor point is you actually want to instead state the range of integers contains a complete residue system modulo $mpq$, to then try to conclude your $(6)$. However, even then, there's a potential complication if any of those values which have a factor of $pq$ also have an odd prime factor smaller than $q$.

With your Edit 2 attempt to prove $T_{p,1}(0,x) \ge T_{p,r}(n,x)$ where $r \gt 1$, a specific counter-example is $T_{3,1}(0,1) = 0$ and $T_{3,2}(6,1) = 1$. Since the rest of your argument is based on the faulty logic and results of the first edit proof, this proof likewise is not reliable.

To prove \eqref{eq3A}, first here is a definition of the $\operatorname{lopf}$ function, where $m \in \mathbb{Z}$ and $\mathbb{P}$ is the set of all primes:

$$\operatorname{lopf}(m) = \begin{cases}
1 & m = \pm 2^j, j \in \mathbb{Z}, j \ge 0 \\
\min\{q : q \mid m, \, q \in \mathbb{P}, \, q \gt 2 \} & \text{otherwise}
\end{cases} \tag{4}\label{eq4A}$$

Thus, your $T_{p,r}(n,x)$ can be defined as

$$T_{p,r}(n,x) = \left|\{i : \operatorname{lopf}(r - i) = p, \, n - x \le i \lt n \}\right| \tag{5}\label{eq5A}$$

As $x$ increases starting from $1$, both sides of \eqref{eq3A} stay at $0$ until $x$ gets to the first value where $\operatorname{lopf}(x + 1) = \operatorname{lopf}(-x - 1) = p$. In that case, since $-x \equiv 1 \pmod{p}$, then $T_{p,1}(0,x) = 1$, but $T_{p,0}(0,x) = 0$ still. However, when $x$ increases by $1$ more, then $\operatorname{lopf}(x) = p$, so $T_{p,0}(0,x) = 1$, which means both sides are equal. This continues until the next larger $x$ where $\operatorname{lopf}(x + 1) = p$, with the pattern repeating, i.e., the left side becomes $1$ greater then, but becomes equal with the next larger $x$. In general, whenever $\operatorname{lopf}(x + 1) = p$, then $T_{p,1}(0,x) = T_{p,0}(0,x) + 1$ and, otherwise, $T_{p,1}(0,x) = T_{p,0}(0,x)$.


问:
Solving $(x-1)y'' -xy' +y = 0$ using Frobenius Series
I wish to solve $(x-1)y'' -xy' +y = 0$ using Frobenius series. The issue I am running into is the $x$ coefficient of the first derivative as I wish to find a solution centered at $x=1$. How would one take care of this? Can you express $x$ as a finite power series of $x = (x-1) + (x-1)^0$?

答:
Let's see... we start with $z=x-1$. Then
$$\frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}=\frac{dy}{dz}$$
And so
$$\frac{d^2y}{dx^2}=\frac{d^2y}{dz^2}$$
So that
$$z\frac{d^2y}{dz^2}-(z+1)\frac{dy}{dz}+y=0$$
We let
$$y=\sum_{k=0}^{\infty}a_kz^{k+r}$$
Which leads to
$$\sum_{k=0}^{\infty}(k+r)(k+r-1)a_kz^{k+r-1}-\sum_{k=0}^{\infty}(k+r)a_kz^{k+r}-\sum_{k=0}^{\infty}(k+r)a_kz^{k+r-1}+\sum_{k=0}^{\infty}a_kz^{k+r}=0$$
The coefficient of $a_0z^{r-1}$ is $r(r-1)-r=r(r-2)=0$, so the difference between the roots is an integer. But we catch a break: for $k\ge1$ matching coefficients of like powers of $z$ we get
$$\left[(k+r+1)(k+r)-(k+r+1)\right]a_{k+1}=(k+r+1)(k+r-1)a_{k+1}=(k+r-1)a_k$$
Thus if $r=0$ and $k=1$ we find that $a_2$ is unconstrained so there is no logarithmic solution. We could start with $r=0$ so that $a_1=a_0$ and then let the unconstrained $a_2=0$. This yields the solution $y=a_0(z+1)=a_0x$. Or we could let the indeterminate
$$\frac{k+r-1}{k+r-1}=1$$
even when $k=1$ and $r=0$. This leads to
$$(k+1)a_{k+1}=\frac{(k+1)!}{k!}a_{k+1}=a_k$$
so
$$(k+1)!a_{k+1}=k!a_k=0!a_0=a_0$$
and we have
$$a_k=\frac{a_0}{k!}$$
which results in
$$y=a_0\sum_{k=0}^{\infty}\frac{z^k}{k!}=a_0e^z=\frac{a_0}ee^x$$
Our overall solution is thus
$$y=c_1x+c_2e^x$$
confirmed by differentiation. Now, can you edit your post to explain where you encountered a problem with all of this?