问:
how to get this equation of the expectation of absolute square of a random variable
I'm reading a paper which uses a lemma saying
$$
\mathbb{E}\lbrace |X|^2 \rbrace = \frac{1}{2} \int_0^\infty \Pr(|X| \ge \frac{h}{2})h d h
$$
but I have checked a lot of books, nobody telling this equation, anybody know how to derive it?

答: Hint: $E|X|^{2}=\int_0^{\infty} P(|X|^{2} \geq t) dt$. Make the substitution $h=2\sqrt t$.


问:
Abstract Algebra proof for quadratic extensions

Let $n$ be a square free integer and alpha belong to $\mathbb{Z}[\sqrt{n}]$. If $N(\alpha)$ is a prime integer, then $\alpha$ is irreducible in $\mathbb{Z}[\sqrt{n}]$.

Here $N(\alpha)$ denotes the norm of $\alpha$.

How can I show this? I'll get back in the morning because it's super late right now!

答:
The key idea is that multiplicative maps preserve multiplicative properties (a fact which is brought to the fore in divisor theory). Here we wish to pullback the property of being an atom (irreducible) along the multiplicative norm map $\rm\:N\:$ (recall that prime $\iff$ atom in a UFD, here $\rm\:Na\in \mathbb Z\:$).

Recall, by definiton, an atom (irreducible) is a nonunit which can't be split into nontrivial factors, i.e. $\rm\ x = y\ z\ \Rightarrow\ y|1\:\ or\:\ z|1\:,\:$ i.e. $\rm\:y\:$ or $\rm\:z\:$ is a unit (or, equivalently, $\rm\:x\:|\:y\:\ or\:\ x\:|\:z).$ Now consider

HINT $\ $ atom $\rm Na,\:\ a = bc\ \Rightarrow\ Na = Nb\ Nc\ \Rightarrow Nb|1\ or\ Nc|1\ \Rightarrow\ b|1\ or\ c|1\ \Rightarrow\:$ atom $\rm\:a$

In fact much of the multiplicative structure of a number ring is reflected in its monoid of norms. For example, in many favorable contexts (e.g. Galois) a number ring enjoys unique factorization iff its monoid of norms does. For references (Bumby and Dade, Lettl, Coykendall) see my sci.math post on 19 Dec 2007.

ContentLicense: CC BY-SA 3.0


问:
Show if $x^2$ is divisible by $5$ then $x$ is divisible by $5$ as well

Show if $x^2$ is divisible by $5$ then $x$ is divisible by $5$ as well

I tried to do this with contrapositive but I feel like it doesn't work. I start with if $x$ is not divisible by $5$ then $x^2$ is not divisible by $5$.

答: If $x^2$ is divisible by $5$ then it ends by $0$ or $5$ in decimal notation. Therefore, $x$ has to end by $0$ or $5$ (easy check) and, therefore, is divisible by $5$. QED </p>


问:
prove weak topology is locally convex
Prove Banach space X endowed with weak topology is locally convex.

To prove it has local convex base is easy since we can write down the neighborhood explicitly

To question is do we need to prove the space X with weak topology is topological vector space.I don't know how to prove this since the topology is no longer norm topology?

I try to prove it by definition that is prove addition map $+:X\times X \to X$ and scalar are continuous.(it seems not very hard by definition correct?)

答:
In my opinion yes, since you know have a new topology, you should prove again that the new topology
leads to a topological vector space.

Your attempt in the comments sounds correct to me so far.
Of course, one should also show that the multiplication by a scalar is continuous.

ContentLicense: CC BY-SA 4.0

Title: prove weak topology is locally convex


问:
A Householder matrix as a product of Givens matrices
I am working on the following problem:

Let $x \in \mathbb{R}^n$ and let $H$ be a Householder matrix such that $Hx = \|x\|_2e_1$. Let $G_{1,2}, \ldots, G_{1,n}$ be Givens rotation matrices such that $Gx = G_{1,n} \cdots G_{1,2}x = \|x\|_2e_1$. Is $G$ equal to $H$?

My idea: Since $H$ is a Householder matrix, it is unitary, so $H^TH = I$. If we multiply both sides of $Hx = Gx$ by $H^T$ then we obtain $H^TGx = x$. Based on this, I guess $H$ does not have to be equal to $G$; the only thing we need is that $1$ is an eigenvalue of $H^TG$. Can you guys give me a hint? I am having troubles to come up with a counterexample.

答:
If you want a counterexample, here is probably the simplest one. Let $\phi\in\mathbb{R}$ and consider $c:=\cos\phi$ and $s:=\sin\phi$. Set
$$
H:=\begin{bmatrix}c&s\\s&-c\end{bmatrix}, \quad G:=\begin{bmatrix}c&s\\-s&c\end{bmatrix}.
$$
The matrix $H$ is a reflection and $G$ is a rotation.
Clearly $H\neq G$, even though
$$
H\begin{bmatrix}c\\s\end{bmatrix}
=G\begin{bmatrix}c\\s\end{bmatrix}
=\begin{bmatrix}1\\0\end{bmatrix}.
$$

The rotation matrix $G$ is clearly a Givens rotation. We just need to show that the reflection $H$ is a Householder reflection, that is, it is of the form $H=I-2vv^T$ with $v:=[v_1,v_2]^T$, such that $\|v\|_2^2=v_1^2+v_2^2=1$.

This can be verified by comparing the equation $I-2vv^T$ with the defining expression for $H$:
$$
1-2v_1^2=c,\quad 1-2v_2^2=c, \quad s=-2v_1v_2.
$$
You can guess the solution from the first two equations and then adjust the signs of $v_1$ and $v_2$ in order to satisfy the last one.
For example, one solution is
$$
v_1=-\mathrm{sgn}(s)\sqrt{\frac{1-c}{2}}, \quad
v_2=\sqrt{\frac{1+c}{2}}.
$$


问:
Conditions that Roots of a Polynomial be Less than Unity
Is is the case that Samuelson's result is a more specific result of Rouche's Theorem, or the Routh–Hurwitz stability criterion? Is it not the goal for a polynomial to be stable that all of its roots be less than unity?

Is there a fundamental result (however primitive) that will connect all of these results? Furthermore the Jury Test or Schur's Criterion-- why do we have all of these 'tests'? Is one the most general, and understanding it will make the rest more clear?

答:
I can't answer your question well but I can point you toward a good reference.

Chapter 7 of Marden's Geometry of Polynomials (1949) is dedicated to bounding the zeros of a polynomial by functions of the coefficients of the polynomial.  A central tool in the proofs is indeed Rouché's theorem. In chapter 9 he also considers the problem of finding the number of zeros which lie in a half-plane.

As an aside, I get a lot of use out of the Eneström-Kakeya theorem, a good reference for which is this pdf.  See also this question on M.SE.  One statement of the theorem goes like

If $0 \leq a_0 \leq a_1 \leq \cdots \leq a_n$ then all zeros of the polynomial
  $$a_0 + a_1 z + \cdots + a_n z^n$$
  lie in the closed unit disk.

In the Samuelson paper you linked he considers the polynomial

$$
p(x) = -.027 - .013x + .220x^2 - .398 x^3 + x^4.
$$

By the K-E theorem the related polynomial

$$
\begin{align}
q(x) &= p(-x) + .027 \\
&= .013x + .220 x^2 + .398x^3 + x^4
\end{align}
$$

has all its roots in $|x| \leq 1$, and thus so does $p(x) + .027$.  Now on the circle $|x| = 1$ we have

$$
|p(x) + .027| \geq \left|x^4\right| - \left(|-.398|\left|x^3\right| + |.220|\left|x^2\right| + |-.013||x|\right) = .369 > |-.027|
$$

by the triangle inequality, so by Rouché's theorem we conclude that $p(x)$ has all its roots in $|x| < 1$.


问:
"Divisors" vs "factors"
Is there a difference between divisors and factors? Are they the same or are there differences?

Example: Find the factors and divisors of 18.

What I have found is answers that contradict each other:

Onlinemath4all:
Factors don't include one and the number itself. So for the example that would be 2,3,6 and 9
Divisors do include them. So for the example that would be: 1,2,3,6,9 and 18

Quora:
There is no difference between factors and divisors. So for the example 1,2,3,6,9 and 18 would all be divisors and factors.

Someone also replies there that a divisor can be any number (even with a non zero remainder). In that case 5 could also be a divisor of 18 (with a remainder).

答:
The answer is in the comment of Lonza Leggiera and can be found in the following wolfram entries:

Factor

Divisor

Proper divisor

Proper factor

Title: "Divisors" vs "factors"


问:
Symbol for multiply divide, equivalent of plus minus?
Is there a symbol for multiply or divide, which acts like the $\pm$ symbol?

答: I guess you can go with $a\cdot b^{\pm1}$, though I would generally recommend emancipation from $\pm$.


问:
Normal generation of the kernel of a surjection of free groups
Suppose that $\phi : F_n \to F_m$ is a surjective homomorphism between free groups of rank $n$ and $m$, respectively.  Is the kernel necessarily finitely normally generated?  If so, can we say anything about the number of normal generators that are needed in terms of $n$ and $m$?

答:
The answer is yes, and $n$ is an upper bound on the required number of normal generators. I don't know whether there is a better bound (such as $n-m$ perhaps)?

Let $x_i$ ($1 \le i \le n$) and $y_j$  $(1 \le j \le m)$ be free generators of $F_n$ and $F_m$, and choose elements $g_j \in F_n$ with $f(g_j) = y_j$ for $1 \le j \le m$.

Now for each $x_i$, we can write $f(x_i)$ as a word $w_i(y_j)$ in $y_j^{\pm 1}$, and $z_i := x_i^{-1} w_i(g_j) \in \ker f$ for $1 \le i \le n$.

I claim that $\ker f$ is normally generated by  $\{ z_i : 1 \le i \le n \}$. Let $N = \langle z_i \rangle^{F_n}$, so $N \le K$, and we want to prove equality.

Now, since any word in the $x_i$ can be rewritten modulo $N$ as a word in the $w_i(g_j)$, we see that  $F/N$ is generated by the elements $w_i(g_j)$, and hence also by $\{ g_j : 1 \le j \le m \}$. So $F/N$ is isomorphic to a quotient of $F_m$ and, since free groups of finite rank are known to be Hopfian (see here for example), we have $K=N$ as claimed.


问:
Phase portrait for time-optimal control problem (controllable region??)
I need to find the time-optimal control to the origin for the system:
$$\dot x_1=3x_1+x_2$$
$$\dot x_2=4x_1+3x_2+u$$ where $|u|\le1$. I have used Pontryagin's Minimum Principle and linear algebra to find the family of curves solution to be:
$$
    \begin{pmatrix}
    x_1 \\
    x_2 \\
    \end{pmatrix}
=A\begin{pmatrix}
    -1 \\
    2 \\
    \end{pmatrix}e^t+B\begin{pmatrix}
    1 \\
    2 \\
    \end{pmatrix}e^{5t}\pm\begin{pmatrix}
    \frac{1}{5} \\
    \frac{-3}{5} \\
    \end{pmatrix}$$

Since the two distinct real eigenvalues are both positive, I know $\pm\begin{pmatrix}
    \frac{1}{5} \\
    \frac{-3}{5} \\
    \end{pmatrix}$ are unstable nodes. Also, the parabolas in the phase portrait will hug the $\begin{pmatrix}
    1 \\
    2 \\
    \end{pmatrix}$ vector, and we can switch from a $u=1$ to $u=-1$ solution at most once. I have drawn a phase portrait here (not 100% sure it is correct), with the unique switching curve intersecting the origin in red (POQ). I'm not sure how to describe the controllable region, and it seems if you're to the right of the $u=1$ fixed point or to the left of the $u=-1$ fixed point, you'd be in an uncontrollable region that can never reach the origin. So I suspect the only controllable region is the rectangular one created by the eigenvectors (dotted lines in my diagram). Can anyone further elaborate? Thanks

答:
Given the general system

$$
\dot{x} = A\,x + B\,u, \tag{1}
$$

which you want to drive to $x(T)=0$ while minimizing $T$ and satisfying $|u| \leq u_\max = 1$. Minimizing time for controllable LTI systems I believe always yields a bang-bang controller, with for a two-dimensional systems at most one switch (though I do not have a source for this). All initial conditions for which no switching is required can be obtained by solving the dynamics backwards in time

$$
\begin{bmatrix}
x(0) \\ \pm u_\max
\end{bmatrix} =
e^{-\begin{bmatrix}A & B \\ 0 & 0\end{bmatrix} T}
\begin{bmatrix}x(T) \\ \pm u_\max\end{bmatrix},\ \forall\ T\in(0,\infty]. \tag{2}
$$

If $A$ is non-singular $(2)$ can also be written as

$$
x(0) = e^{-A\,T} x(T) \pm A^{-1} (e^{-A\,T} - I) B\,u_\max. \tag{3}
$$

It can be noted that if all eigenvalues of $A$ have a positive real part then $e^{-A\,T}$ vanishes as $T\to\infty$, which implies that

$$
\lim_{T\to\infty} x(0) = \mp A^{-1} B\,u_\max. \tag{4}
$$

All initial conditions for which one switch in control input is required can be obtained by again solving the dynamics backwards in time and ending at $(2)$

$$
\begin{bmatrix}
x(0) \\ \mp u_\max
\end{bmatrix} =
e^{-\begin{bmatrix}A & B \\ 0 & 0\end{bmatrix} T_1}
\underbrace{\begin{bmatrix}I & 0 \\ 0 & -1\end{bmatrix}}_\text{invert $u$}
\underbrace{e^{-\begin{bmatrix}A & B \\ 0 & 0\end{bmatrix} T_2}
\begin{bmatrix}x(T_1+T_2) \\ \pm u_\max\end{bmatrix}}_{\begin{bmatrix}x(T_1) \\ \pm u_\max\end{bmatrix}},\ \forall\ T_1,T_2\in(0,\infty], \tag{5}
$$

with $T = T_1 + T_2$.

In your case all eigenvalues of $A$ have a positive real part, which means that $x(0)$ from $(5)$ remains bounded for all $T_1,T_2\in(0,\infty]$. The border of this controllable region can be obtained by using $\lim_{T_2\to\infty}$, which by using $(3)$ and $(4)$ can be simplified to

$$
x(0) = \mp A^{-1} (2\,e^{-A\,T_1} - I) B\,u_\max,\ \forall\ T_1\in(0,\infty]. \tag{6}
$$

Plotting this together with the no switching curves yields for your system the following plot. So your guess of a rectangular region was close, but also includes a couple of initial conditions which can't be driven to the origin.

It can be noted that for $A$ with all eigenvalues having a positive real part and some non-zero imaginary parts the minimum time trajectory is not directly obvious, since in that case $(2)$ and $(6)$ can cross each other infinitely many times.


问:
Homeomorphism between topological spaces 2
Since in my previous post I couldn't find the correct solution to the following problem, I repost it, and since I'm not an expert in topology I kindly ask if somebody can give me the correct answer.

Let $X=[0,1] \times [0,1]$ and let $\sim$ be a relation defined by $(x,0) \sim (x,1)$. Show that $X/ \sim$ is homeomorphic to the cylinder $S^1 \times [0,1]$.

My goal is to get a solution without using compactness, writing a continuos and bijective function $f: X \rightarrow Y$, where $Y$ is the cylinder, constant on the equivalente classes of $X/\sim$. This implies that the function $[f]: X/\sim \rightarrow Y$ is continuos and bijective too. Since the composition $f= [f] \circ \pi$ is commutative (where $\pi$ is the natural projection) I'll have to show that also $f^{-1}$ is continuos, and then $[f]^{-1}$ is continuos too. Hence $[f]$ is an homeomorphism.

Solution:

Consider the function $f: X \rightarrow S^1 \times [0,1]$ defined as $f(\theta, t)=(\cos(2 \pi \theta), \sin (2 \pi \theta), t(t-1))$. Since $f$ is continuous, and $f(0,0)=(1,0,0)=f(0,1)$ we have that $[f]:X/ \sim \longrightarrow S^1 \times [0,1]$ is continuos too.

The problem is that $f$ is not injective, then is not invertible. Does it exist a suitable bijective function or it is impossible to claim that the two spaces are homeomorphic without using the compactness?

答:
You can construct an explicit bijection. Your attempt is close, but $f(t, \theta)$ should read $f(\theta, t)$ and $t(t-1)$ should read $t$ (it's the trig functions that identify the case where $\theta = 0$ and the case where $\theta = 1$).

Geometrically, think of the cylinder $S^1 \times [0, 1]$ being obtained from the square $[0, 1]\times[0, 1]$ by rolling it up so the top edge meets the bottom edge. (The way the question is formulated also swaps the coordinates around: as if you'd stood the cylinder on its end.)


问:
Inclusion-Exclusion Counting
I'm not sure how to use inclusion-exclusion to calculate the number of bit strings of length 9 that either begin with two 0s, have eight consecutive 0s, or end with a 1 bit.

Here is what I have done which helps me see the number of possibilities:

Condition 1: Beginning with two 0's leaves me with the rest of the position being equal to $2^{7}$ = 128

Condition 2: Eight consecutive 0's leaves me with these 3 possible combinations:

$0 0 0 0 0 0 0 0 0$

$1 0 0 0 0 0 0 0 0$

$0 0 0 0 0 0 0 0 1$

Condition 3: Ending with one bit; Since the positions can have either 1 or 0, and there are 128 possible ways that the 1's and 0's can be placed to fit the length of 9, I divided 128/2 and got that there are 64 possibilities for the last bit to be 1.

I know that the formula for inclusion-exclusion is: |A ∪ B ∪ C| = (|A| + |B| + |C|) - (|A ∩ B| + |A ∩ C| + |B ∩ C|) + |A ∩ B ∩ C|

I'm just not sure how to involve the logic I used to get the bolded values above with this formula. Thank you in advance!!

答:
$A$ is number of ways of two starting bits as zeroes.

$B$ is number of ways for at least $8$ consecutive zeroes.

$C$ is number of ways of ending bit as $1$.

$A = 2^7 = 128$ as you identified.

$B = 3$ as you identified.

$C = 2^8 = 256$ (not $64$, if I understand this point correctly)

Now,

$A \cap B = 2$ (out of $3$ cases of $B$, one of them does not have first two bits as zeroes, two of them have).

$B \cap C = 1$ (out of 3 cases of $B$, one of them has last bit as $1$)

$A \cap C = 2^6 = 64$ (First two bits are fixed and so is the last bit. There are choices only for $6$ bits in between).

$A \cap B \cap C = 1$

So your answer should become $128 + 3 + 256 - 2 - 1 - 64 + 1 = 321$.

Title: Inclusion-Exclusion Counting


问:
Norm cone is a proper cone
For a finite vector space $H$ define the norm cone $K = \{ (x, \lambda) \in H \oplus \mathbb{R} : \lVert x \rVert \le \lambda \}$ where $\lVert x \rVert$ is some norm.  There are endless lecture notes pointing out that this is a convex cone (as the pre-image of a convex set under the perspective function).  In fact, I believe it is a proper cone.  Since the proof of that is somewhat tedious, is there some reference I can cite for this fact?

答:
Obviously $K$ is convex and closed as it is defined by the inequality $g(x,\lambda):=\|x\| - \lambda\le0$ with $g$ continuous and convex.
Also it has non-empty interior. Take $x_0\ne0$, $\lambda_0>\|x_0\|$. Then $\|x\| < \lambda$ in a neighborhood of $(x_0,\lambda_0)$.
If $(x,\lambda),(-x,-\lambda) \in K$, then $0\le \|x\| \le \min(\lambda,-\lambda)$ implying $\lambda=0$ and so $x=0$. And $K$ is pointed.

Title: Norm cone is a proper cone


问:
Proof that $\sum_{k=0}^m \binom{m}{k}\frac{1}{k+1} = \frac{2^{m+1}-1}{m+1}$
Recently I needed to compute $E[\frac{1}{X+1}]$ where $X\sim Bin(m, \frac 1 2)$.
While expanding, I came across the sum $\sum_{k=0}^m \binom{m}{k}\frac{1}{k+1}$, which I was unable to solve. Plugging into Mathematica gives $\frac{2^{m+1}-1}{m+1}$, but I don't see how to derive this result.

Can anyone give a proof of this identity or an alternate way to compute $E[\frac{1}{X+1}]$?

Body:
It’s easier if you multiply through by $m+1$:

$$(m+1)\sum_{k=0}^m\binom{m}k\frac1{k+1}=\sum_{k=0}^m\binom{m}k\frac{m+1}{k+1}=\sum_{k=0}^m\binom{m+1}{k+1}=\sum_{k=1}^{m+1}\binom{m+1}k=2^{m+1}-1\;.$$


问:
Every abelian projection in a von Neumann algebra is finite
I saw a note on von Neumann algebras, it mentions that every abelian projection $p$ in a von Neumann algebra is finite.

According to the definitions, we know that $pMp$ is abelian, we need to prove that for any projection $q\in M$ such that $q\leq p$ and $q\sim p$, we have $p=q$.

It is easy to see that $pq=qp=q$, how do I use the condition $p\sim q$ to conclude that $p=q$?

答:
If $p=u^*u$ and $q=uu^*$, then
$$u=uu^*u=qu=pqu=pu=pup\in pMp.$$
Since $p$ is abelian, it now follows that
$$p=u^*u=uu^*=q.$$


问:
How to find the coefficient of $x^{l}$ in the expansion of $(1+x+x^{2}+...+x^{n})^{m}$ for some given $l$?
Suppose I'd like to find the coefficient of $x^{l}$ in the expansion of  $(1+x+x^{2}+...+x^{n})^{m}$, where $n$ and $m$ are given positive integers, for some given integer $l$ such that $n < l < mn$. Is there a straightforward formula (in terms of some multinomial coefficient, or similar)?

答:
Using the ideas of a linked post such as $\left(\sum_0^n x^s\right)^m=(1-x^{n+1})^m (1-x)^{(-m)} $ found a much more concise answer.
Again using $L$ in place of $l\;$ the coeff of $x^L$ is
$\sum_{k=0}^{min(m,\frac L{(n+1)})}(-1)^k\binom{m}{k}\binom{m+L-k(n+1)-1}{L-k(n+1)}$


问:
Let G be an abelian group. Suppose $a, b$ are elements of orders $m$ and $n.$ Let $d =\operatorname{lcm}(m, n).$ Show that $(ab)^d = 1$
Here are my thoughts so far:
G is abelian so $ab = ba.$

For $d = mn, (ab)^{mn} = abab..ab = aaaa..bbb$  as abelian $= a^{mn}b^{mn} = e$ as $a^m = e$  and $b^n = e.$

For $d< mn,$ let $d = qm$ and $d = pn.$ Then
$$\begin{align}(ab)^{qm} &= a^{qm}b^{qm} \\& = eb^{qm} \\&= b^{qm}\\
  (ab)^{pn} &= a^{pn}b^{pn} \\& = a^{pn}e \\&= a^{pn}\end{align}$$
so $b^{qm} = a^{pn}$ so $a^d = b^d.$

I'm not really sure how to continue from here/ if this is even the logically path I should be following

答: So when $G$ is abelian, we have that $(ab)^d=a^db^d=a^{qm}b^{pn}=(a^m)^q(b^n)^p=e^qe^p=e\cdot e=e$.


问:
Q: Find all values of $z such that cos(z) is purely imaginary
I haven't got a memo for the answer of this question. So I just wanted to find out if my approach and answer is correct and sufficient.
Thank you.

Let $b\in \mathbb{R}\backslash\{0\}$

$$\begin{align}cos(z)=ib\\
\frac{1}{2}(e^{iz}+e^{-iz})=ib\\
e^{iz}+e^{-iz}=2ib
\end{align}$$

Let $w=e^{iz}$, then

$$\begin{align}
w+w^{-1}=2ib\\
w^2 -2ibw + 1=0
\end{align}$$

$w=\frac{2ib \pm \sqrt{(-2ib)^2 -4(1)(1)}}{2(1)}=ib \pm\frac{\sqrt{-4b^2-4}}{2}=ib \pm\frac{2i\sqrt{b^2+1}}{2}=i(b\pm\sqrt{b^2+1})$

For $w=e^{iz}=i(b+\sqrt{b^2+1})$

$$\begin{align}
e^{iz}=i(b+\sqrt{b^2+1})\\
e^{ix}e^{-y}=e^{i\frac{\pi}{2}}(b+\sqrt{b^2+1})\\
\end{align}$$
Thus, $x=\frac{\pi}{2}+2n\pi$, $\forall n\in \mathbb{Z}$, and $y=-ln(b+\sqrt{b^2+1})$

For $w=e^{iz}=i(b-\sqrt{b^2+1})$

$$\begin{align}
e^{iz}=-i(-b+\sqrt{b^2+1})\\
e^{ix}e^{-y}=e^{-i\frac{\pi}{2}}(-b+\sqrt{b^2+1})\\
\end{align}$$
Thus, $x=-\frac{\pi}{2}+2n\pi$, $\forall n\in \mathbb{Z}$, and $y=-ln(-b+\sqrt{b^2+1})$

Therefore $cos(z)$ is purely imaginary when $\\$
$z=\frac{\pi}{2}+2n\pi+i(-ln(b+\sqrt{b^2+1}))$
or, when
$z=-\frac{\pi}{2}+2n\pi+i(-ln(-b+\sqrt{b^2+1}))$ $\forall n\in \mathbb{Z},b\in \mathbb{R}\backslash\{0\}$

答:
By using addition formula and $\begin{cases}\cos(ia)=\cosh(a)\\\sin(ia)=i\sinh(a)\end{cases}\quad$ you get for $(x,y)\in\mathbb R^2$:

$$\cos(z)=\cos(x+iy)=\cos(x)\cosh(y)-i\sin(x)\sinh(y)$$

Since all of $\cos(x),\sin(x),\cosh(y),\sinh(y)$ are real numbers, this quantity is purely imaginary when either $\cos(x)$ or $\cosh(y)$ is zero.

But $\cosh(y)\ge 1$, so the only possibility is $x=\frac{\pi}2+k\pi,\ k\in\mathbb Z$ and $$\cos(z)=-i\,(-1)^k\sinh(y)$$

Note that $\sinh^{-1}(b)=\ln(b+\sqrt{b^2+1})$ (see https://en.wikipedia.org/wiki/Hyperbolic_functions) appear in your solution but you don't really need this calculation, as $\sinh$ is a bijective function of the reals, we just can have anything for $y=\Im(z)$.

Also you have $2n\pi$ while you should have $n\pi$, check where you made the mistake.


问:
Why is linear regression considered an ill-posed problem?
Could someone please explain to me why exactly linear regression is considered ill-posed?</p>

答:
According to Wikipedia, a problem is well-posed if:

a solution exists,

the solution is unique,

the solution's behaviour changes continuously with the initial conditions.

Now suppose that you have a dataset $(X, y)$ consisting of samples $(x_i, y_i)_{i = 1}^m$, with $x_i \in \mathbb{R}^n$, $y_i \in \mathbb{R}$. I suppose that you think of linear regression as of the following problem for weights $w$ of the regression: $X w = y$ (or something similar depending on how do you stack samples in the matrix).

This problem is ill-posed because of the following reasons:

the solution is not unique if you have too few samples, namely $m < n$ (for example, there are infinitely many straight lines going through a given point in $\mathbb{R}^2$);

the solution may not exist if you have too many samples, namely $m > n$ (for example, there are not straight lines going through three given points in $\mathbb{R}^2$);

finally, there is multicollinearity issue which contradicts the last necessary condition of weight estimates being continuous in observations.

However, the problem can be adapted to fix these properties. Modifying the problem to $\|Xw - y\|_2^2 \to \min$ allows obtaining a pseudo-solution when there is no exact solution (the problem is still ill-posed because there may be many solutions).

You can also add Tikhonov regularization (a.k.a. ridge in the setting of regression) to get $\|Xw-y\|_2^2 + \lambda \|w\|_2^2 \to \min$ which allows selecting one solution if there are many solutions of the original problem.


问:
Doubts in $\Bbb Z_2 \times \Bbb Z_2$
My question asks me to prove that whether $\Bbb Z_2 \times\Bbb Z_2$ a subgroup of $S_4$ or not.

I really cant get a clear picture but the only thing that I know is that if we can let $\Bbb Z_2 $ isomorphic to $H_1=\langle(a b)\rangle$ and $\Bbb Z_2$ isomorphic to $H_2=\langle(c d)\rangle$
.Then maybe I can show some isomorphism with some subgroup of $S_4$.[I am having a problem to explicitly write this]

There is also one thing that I know is that if I take $\Bbb Z_2$ as isomorphic to a the group $H_1=\langle(12)(34)\rangle$ and $H_2=\langle(23)(14)\rangle$ Then maybe some sort of isomorphism is possible. Can someone help out please by writing the proof or giving the general structure of the proof.

答:
Yes, consider $\{e,(12)(34),(13)(24),(14)(23)\}$.  It is isomorphic to the Klein four group, as all three nonidentity elements have order two.

Actually, there are four Klein four subgroups in $S_4$.  We have $\{e,(12),(34),(12)(34)\},\{e,(13),(24),(13)(24)\}$ and $\{e,(14),(23),(14)(23)\}$.  But only the first one above is normal.

As far as an isomorphism, you can map any of the nonidentity elements to any of the nonidentity elements of $V_4$, bijectively.  So there are six.

Title: Doubts in $\Bbb Z_2 \times \Bbb Z_2$


问:
Can an arc be greater than 360 degrees or less than 0 degrees.
Can an arc be greater than 360 degrees or less than 0 degrees? Is the (angular) arc length a signed scalar?

I'm creating a Graphics and basic geometry software library. I believe an arc can be defined by 5 scalars, the x and y components of the start point, the x and y components of the centre point and the signed value of the rotation. However I'm concerned that this data definition could lead to unnecessary rounding errors on the end point of the arc, the precise value of which is generally of greater concern than the precise value of the arc's centre point. So I'm leaning towards defining arcs by the 6 scalar definition of the start point, the end point, and the mid point of the circumference of the arc. This data definition has the added advantage that you don't have to worry about reflections, but would exclude arcs of greater than 360 degrees. Is this mathematically sound?

A successful software library will get used in ways unimagined by the library authors. So this where I place a strong priority or weighting on following mathematical definitions, because those more rigorous,robust and tried and tested definitions can hopefully pre-empt problems further down the road.

答:
Heading

Circular arc case: You can define the arc using the $5$-tuple $\langle x_c, y_c, r, \theta_1, \theta_2\rangle$ where $(x_c,y_c)$ is the cartesian coordinate of the arc center, $r$ is the radius of the arc, $\theta_1$ is the start angle and $\theta_2$ is the end angle. Angles are measured in either degrees or radians and we follow the convention of counterclockwise angles as positive and clockwise angles as negative.

One optional normalization that you can perform in your routines, based on use cases of your library is to reduce the angles to the range $[0,2\pi)$ or $[-\pi, +\pi]$ radians.

If you want to avoid mixing cartesian and polar coordinates in the tuple representation, you can convert $(x_c,y_c)$ also into $(r_c,\theta_c)$ polar form so that you have $\langle r_c, \theta_c, r, \theta_1, \theta_2\rangle$.

Elliptical Arc case: The above works for circular arcs. If you need elliptical arc also to be covered, you need to specify the focal point (instead of center of circle), $r_s, r_e$ that measure the length of the line segments from the focus to the end points of the arc and the angles of the line segments relative to an axis.

If you prefer cartesian coordinates, then the $6$-tuple is $\langle x_f, y_f, x_s, y_s, x_e, y_e \rangle$ where $(x_f,y_f)$ is the focus, $(x_s,y_s), (x_e,y_e)$ are the start and end point of the arc respectively. You only need one other $x$ coordinate $x_m$ to differentiate between the reflections. So, the $7$-tuple becomes $\langle x_f, y_f, x_s, y_s, x_e, y_e, x_m \rangle$. You can calculate $y_m$ from the equation of the ellipse. If you want to avoid the performance penalty of calculation, you can store a redundant $y_m$ as well making an $8$-tuple $\langle x_f, y_f, x_s, y_s, x_e, y_e, x_m, y_m \rangle$ for the general elliptical arc.

Whether you want to treat circles and ellipses separately depends on other considerations such as the primary use of your library. Separate representations help with compact data representations for different use cases while complicating handling of the data. This decision is purely dependent on what makes sense for the primary uses of your library.

A good library should provide helper methods to convert from one representation to another for commonly used representations.


问:
Why natural numbers don't have supremum?
As I know the definition of supremum for set $ S$ is the lowest number that is greater or equal than all the members of $S$.
This means : $\forall \: m\in R \:\:m<\sup S,\exists\: s\in S \rightarrow s> x$.
Based on this definition we have supremum for all finite subset of $\mathbb N$.
For example the $\sup S$ for $S = \{1,2,\dots, k\}$ is $k$.

Is this right?

答:
Let's review several senses in which $\Bbb R$ is complete:

Each set with an upper bound has a least upper bound This works in $\Bbb N$ because $\Bbb N\subseteq\Bbb R$, so if $S\subseteq\Bbb N$ then $S\subseteq\Bbb R$. Narrowly interpreted, your question boils down to overlooking the bold part; broadly interpreted viz. the comments, other respects in which $\Bbb R$ is complete are worth comparing with $\Bbb N$.

Each Dedekind cut is generated by a real number Of course, they're not in general generated by natural numbers. (You could, I suppose, think of naturals as something similar to Dedekind cuts on themselves, but you'd gain nothing from it.)

Each Cauchy sequence converges to a value in $\Bbb R$ Any Cauchy sequence in $\Bbb N$ is eventually constant, converging to some natural number.

Intersection of interval nested, in the way the nested intervals theorem specifies, is a $1$-element set Since naturals differ by at least $1$ (crucial also in the below discussion of IVT), you eventually can't nest proper-subset intervals further.

Monotone convergence In analogy with the least-upper-bound point above, this follows because every sequence in $\Bbb N$ is also in $\Bbb R$. Bolzano-Weierstrass Ditto.

Intermediate value theorem This fails for $\Bbb N$ because if $f(a)<0<f(b)$ then there might not be values between $a,\,b$.


答:
A related problem. We start with the identity
$$ \sum_{k=0}^m \binom{m}{k}x^k = (1+x)^m. $$

Integrating both sides of the above equation with respect to $x$ form $0$ t0 $1$, the desired result follows

$$ \sum_{k=0}^m \binom{m}{k}\frac{1}{k+1} = \int_{0}^{1}(1+x)^m dx = \frac{(1+x)^{m+1}}{m+1}\Big|_{x=0}^{x=1}=\frac{2^{m+1}-1}{m+1}.$$

Title: Proof that $\sum_{k=0}^m \binom{m}{k}\frac{1}{k+1} = \frac{2^{m+1}-1}{m+1}$

Body:
Recently I needed to compute $E[\frac{1}{X+1}]$ where $X\sim Bin(m, \frac 1 2)$.
While expanding, I came across the sum $\sum_{k=0}^m \binom{m}{k}\frac{1}{k+1}$, which I was unable to solve. Plugging into Mathematica gives $\frac{2^{m+1}-1}{m+1}$, but I don't see how to derive this result.

Can anyone give a proof of this identity or an alternate way to compute $E[\frac{1}{X+1}]$?


问:
Reccursive formula as a non-recursive
The recursive formula is give as follows:

$$ s_n = n s_{n-1} + (n-1)!$$ and $s_2 = 1$. How to construct a non-recursive formula?

答:
The relation is equivalent to
$$s_{n+1}=(n+1)s_n+n!,~~s_1=1$$
Let
$$s_n=u_n\times n!$$
$$\implies u_{n+1}\times (n+1)!=(n+1)\times u_n\times n!+n!, ~~~~u_1=1$$
$$\implies (n+1)u_{n+1}=(n+1)u_n+1,~~~u_1=1$$
$$u_{n+1}=u_n+\frac{1}{n+1}$$
This is essentially the relation for the harmonic numbers, which don't have any simple closed form.

It'd be SO much easier to solve if we had
$$\implies (n+1)u_{n+1}=(n+2)u_{n+1}+1$$
instead (see Solve $(n+1)t_{n+1}=(n+2)t_n+1$, $t_1=1$):(
Are you sure it's not that, ie are you certain your relation shouldn't be
$$s_{n+1}=(n+2)s_n+n!,~~s_1=1$$

Title: Reccursive formula as a non-recursive


问:
Name and layperson's explanation for an E8 group diagram.
I am taking a risk here, but hoping it will not ignite wrath in the reader. In trying to get an intuition of Lie theory this diagram is all but impossible to ignore:

Unfortunately, there are many youtube videos on the group E8 and its applications to subatomic particles and such. Yet, the actual diagram is not clearly addressed.

So, knowing fully well that the math behind it is at this point beyond my level, I would like to ask for:

What is the specific name of this diagram? Is it a Dynkin diagram? Does it have any other names?

What is the name and/or layperson's idea of what the color coding, nodes and edges represent?

答:
The explanation by the maker of the picture, John Stembridge  (credit to Sabino Di Trani for linking there) is a great starting (and perhaps even end) point. I quote the beginning:

The Lie group E8 has a "root system" associated to it that consists of 240 points in 8-dimensional space. Similarly, the Lie group E7 has a root system of 126 points in 7-dimensional space.

These 240 points are tightly packed together in a highly symmetric way. In fact, this configuration has a total of 696,729,600 symmetries. Contrast this with what happens if you take the 8 points at the corners of a 3-dimensional cube. These 8 points have "only" 48 rotational and reflective symmetries.

Of course we can't really visualize any objects in 8 dimensions, but we can draw 2-dimensional projections of them. For example, if you imagine shining a flashlight on a cube, the shadow it casts would (depending on how you orient the cube) look like a hexagon. And if you orient the cube just right, the shadow it casts would look like a regular hexagon--a six sided figure with all sides of equal length and all angles of equal measure. Taking it one step further, if you imagine the cube as a wire frame -- 8 points together with links that connect along edges of the cube, then the projection would look like 6 dots at the corners of a hexagon, plus another dot in the center, plus lines connecting nearby dots.

What I've done with the root system of E8 is completely analogous. I picked the "just right" direction to shine a light on these 240 points so that the 2-dimensional shadow they cast is as symmetric as possible.

So the primary question this leaves open is: "what is a root system?", and related: why would we be drawing a root system when we are talking about a group?

There is a more general issue here. When people talk about $E_8$ they sometimes refer to a real Lie group, sometimes to a complex Lie group, sometimes to a complex Lie algebra, sometimes (but rarely) to a real Lie algebra, sometimes to a Coxeter group, sometimes to a lattice, sometimes to a root system and in exceptional cases to something else (e.g. singularities in two-dimensional algebraic geometry).

Often people are sloppy about to which thing they are referring and this is somewhat justified by how these things are all related and the fact that you can create the one from the other.

But lots of things are related in math and they do not all get the same name, so it is good to spend a few seconds of thought on why it happens in this case.

The short answer is that all these types of mathematical objects follow (roughly) the same classification, the ADE classification. In every class of object mentioned we have one (or a few closely related ones) called $A_n$ for every value of $n$, one (with the same caveat) called $D_n$ for every value of $n$, one (...) called $E_6$, one (...) called $E_7$ and one (...) called $E_8$. Also for some classes of objects there are others that get different letter-number combinations, but not many.

The same correspondences between groups, algebras, lattices etc we get for $E_8$ we get for every $A_n$, $D_n$ and $E_n$.

Now to the question what I recommend you to do:

If you want to get some overview of how many things follow this magical ADE classification, read John Baez on this topic. I will try and find a link later and edit it in.

If you really want to follow the full story from Lie group to Lie algebra to Coxeter group to root system (and hence this picture) I recommend doing this not for the hardest case ($E_8$) but for the second-simplest case $A_2$ (the simplest case $A_1$ is so simple that you might miss out on some of the beauty). A great resource for this is the book 'Representation Theory, a first course' by Fulton and Harris.

If you don't want the full story but only the part that is easy to understand and visualize (in as far as things beyond 3 dimensions are visualizable) I recommend you skip the Lie groups and Lie algebras and start with Coxeter groups. For this Wikipedia is a very rich source.

If you just want an informal answer here to the question what is a root system, I recommend you start with lattices.

Concretely: a lattice is just a very regular way to place points in space. In 2 dimensions you have the square lattice (the crossings in ordinary grid paper) but you also have the $A_2$ lattice, the crossings in triangular grid paper, as well as a few others. You can see that the $A_2$-lattice is really really nice (and hence deserving of its own letter) by putting circles (coins) around the points of the lattice and see how they all touch each other with no room to wiggle. This type of niceness does not exist in every dimension (the best resource for this stuff is the book by Conway and Sloane: 'Sphere packings, Lattices and [something]') but in eight dimension everything falls into place nicely again and you get a really nice lattice, the $E_8$ lattice. I like this (really informal) description because it gives me the feeling that I can almost picture it although in reality I can't of course picture anything.

Now if you have some feeling about what lattices are and why they are nice objects, the root system is pretty much just the 'inner shell' of the lattice: the points closest to the origin (together with info of their distances and angles between them) which generate the full lattice simply by adding them together as vectors.

EDITED IN LATER: let me try and answer the concrete questions from the original post.

One take-away from my last paragraph above is that lattices and hence root systems are really rigid objects. It is old school geometry: distances, angles, location in space all still have their standard meaning here. Nothing is abstracted away into topology or linear algebra or defined over an abstract field, it is just the good old way of thinking about geometry you had in primary school, except of course for the fact that you don't restrict yourself to only 3 dimensions. As a result of this it makes perfect sense to pick an arbitrary (or not so arbitrary) two-dimensional plane in the space where the lattice and root system live and consider the orthogonal projection of the root system onto that plane.

That is what this is.

So to answer the first question:

What do the vertices represent?

The vertices are the actual locations of the projections ('shadows') of the elements of the root system (which themselves are by their nature points in space) onto a cleverly chosen plane.

The 'cleverly chosen' part is important here. You can imagine that if you would have picked a random 2D-plane to project on, the configuration of vertices would be a lot less symmetric. This brings us to the second question:

Does this diagram have a name?

As far as I know: no. But the plane onto which the root system is projected, which yields the incredibly nice rotational symmetry, does. It is the Coxeter plane. So one name for the diagram is The projection of the root system [of $E_8$] onto the Coxeter plane. I put the 'of $E_8$' into block brackets because you could make similar pictures for $A_2$, $D_4$ etc.

Now:

What do the edges represent?

Edges connect each vertex to the vertices that in the actual root system are its nearest neighbors. The fact that some of the edges run between vertices that are no longer nearest neighbors in the projection suggests that the original root system was projected down from a space with more than 2 dimensions. In fact it came from a space with 8 dimensions. We sort of already knew that, of course; the subscript 8 in $E_8$ was a bit of give-away.

Finally:

What does the color code represent?

As far as I can tell from the link in the other answer: not much. One thing that is so nice and special about the Coxeter plane is that the roots (that is what elements of a root system are called) are projected into a relatively small number of rings (instead of all over the place). The colors reflect that. It seems John Stembridge assigned one color to each ring and then colored an edge in the color of the outermost ring it has a vertex in. As far as I can tell this functions mostly for making the picture more transparent. In as far as the colors have a deeper meaning which could be described in the original 8-dimensional setting without bringing in the Coxeter plane, they thus represent a property of the vertices rather than of the edges.


问:
dominated convergence theorem
I am studying the proof of a theorem and in a part of the proof I have the following situation:

Let $u : \Omega \rightarrow R$ a nonnegative measurable function, with $\Omega$ open and bounded. Consider $c>0$ a constant. Write $v_{n} = log( u + 1/n )$. Supose that  $log (u + 1/n) \in H^{1,p}_{loc} (\Omega)$.

I need to show that $lim_{n \rightarrow \infty} \int_{B} e^{-c_1 v_n} = \int_{B} u^{-c_1}$,
where B is a open ball with $2B \subset \Omega$.

I think the dominated convergence theorem can help.. Someone can give a hint to prove what i said?

Thankyou!

答: I don't think we need anything about Sobolev spaces here. The function $u^{-c_1}$ takes values in $[0,+\infty]$, so $\int_B u^{-c_1}$ is defined as a number in $[0,+\infty]$. The functions $f_n:=\exp(-c_1v_n)$ are nonnegative and form an increasing sequence: $f_{n+1}\ge f_n$. Also, $f_n\to u^{-c_1}$ pointwise. By the Lebesgue monotone convergence theorem we have $\lim \int_B f_n=\int \lim f_n$.</p>

Title: dominated convergence theorem


问:
Probability in two batches
I'm learning statistics and trying to solve the following problem:

Concentration of a chemical in water is given by a continuous random
variable with probability density:

$f(x)= \left\{\begin{matrix} \frac{3}{8}x^2, & 0\leq x\leq 2 \\  0, &
 else \end{matrix}\right.$

We assume that concentrations $X$ and $Y$ of the chemical in two
different batches are independent random variables with same same
probability distribution.

Calculate the probability, that concentration of chemical in the first batch will be higher than the concentration in the second batch.

Could you please help me on how I would solve the problem? I'm not sure how to begin.

Thanks

答:
As per independence, the joint density is the following

$$f_{XY}(x,y)=\frac{3}{8}x^2\frac{3}{8}y^2$$

in the square $[0;2]\times[0;2]$

To calculate the requested probability just solve

$$\mathbb{P}[Y>X]=\int_0^2f_X(x)\Bigg[\int_{x}^2f_Y(y)dy\Bigg]dx=\dots=\frac{1}{2}$$

Title: Probability in two batches


问:
Does $C^0$ convergence imply "eventually homotopic"?
Consider a sequence of continuous mappings $f_n:M\to N$ for $n \in \mathbb{N}$, where $M$ and $N$ are compact smooth manifolds, or in general, compact and Hausdorff metric spaces.

Suppose that $f_n \to f$ in the $C^0$ topology, which in this case reduces simply to uniform convergence. Is it true that $f_n$ is homotopic to $f$ for every $n$ sufficiently large?

I believe that this is true because of the following: in our case, the property of two mappings $f$ and $g$ being homotopic is equivalent to being in the same path-connected component of the function space $C(M,N)$ in the $C^0$ topology, i.e., just the uniform convergence topology. And, if $f_n\to f$ in this topology, I believe it should be contained in the path connected component of $f$ for sufficiently large $n$ (but I don't know how to prove it).

My goal is to use this fact to argue that $deg(f_n) = deg(f)$ for all sufficiently large $n$.

答:
For manifolds (or more generally absolute neighborhood retracts of euclidean spaces) consider embedding $N\subseteq \mathbb{R}^n$. Next pick a tubular neighborhood $U$ of $N$ in $\mathbb{R}^n$. There is a smooth retraction $r:U\rightarrow N$ and $U$ is an open subset of $\mathbb{R}^n$. For each $n\in \mathbb{N}$ consider the homotopy $h_n:[0,1]\times M\rightarrow  \mathbb{R}^n$ given by formula
$$h_n(t,m) = tf(m) + (1 - t)f_n(m)$$
Since $\{f_n\}_{n\in \mathbb{N}}$ converges uniformly to $f$, we derive that
$$h_n([0,1]\times M)\subseteq U$$
for sufficiently large $n$. Then $H = r\cdot h_n:[0,1]\times M\rightarrow N$ is a homotopy between $f_n$ and $f$.

In the case of general compact metric spaces the result does not hold. For this pick
$$M = pt,\,N = \{0\}\cup \bigg\{\frac{1}{n}\,|\,n\in \mathbb{N}\bigg\}$$
and consider continuous maps $f_n(pt) = \frac{1}{n}$ and $f(pt) = 0$. Then $f_n\rightarrow f$ uniformly, but $f_n$ are $f$ are never homotopic.


问:
The relationship between matrix rank and its generalized inverse rank..
Let's say that $r=rank(A)$ where $A$ is a rectangular matrix, and $r'=rank(A^+)$ where $A^+$ is the generalized inverse of $A$.

[$A^+ = VS^+U^T$]

Is there any relationship between $r$ and $r'$? Like a specific equality or inequality? Or maybe a way to calculate $r'$ by knowing the value of $r$?

My guess was that they're equal but then based on the definition of $A^+$ we might have some zeroes after doing the multiplication.

答:
Yes. Because the $U$ and $V$ in the singular value decomposition are unitary ( hence invertible), we have
$rank(A) = rank(S) = rank(S^+) = rank(A^+)$.

Multiplying by invertible matrices on either side does not influence rank.


问:
Propositional logic problem (getting wrong answer)
Q3. Sharky, a leader of the underworld, was killed by one of his own band of four henchmen. Detective Sharp interviewed the men and determined that all were lying except for one.
He deduced who killed Sharky on the basis of the following statements:

a. Socko: Lefty killed Sharky.

b. Fats: Muscles didn't kill Sharky.

c. Lefty: Muscles were shooting craps with Socko when Sharky was knocked off.

d. Muscles: Lefty didn't kill Sharky.

Who did kill Sharky?

Solution:

P= Socko: Lefty killed Sharky

Q= Fats: Muscles kill Sharky.

R= Lefty: Muscles were shooting craps with Socko when Sharky was knocked off.

S= Muscles: Lefty killed Sharky.

p

~q

conclusion: p and ~q (conjunction rule )

r

~s

conclusion: r and ~s (conjunction rule )

What and where I am doing wrong?

答:
If we can use the fact that all but one of the henchmen are lying, then it is possible to solve this problem in the following way. First we translate it to more mathematical terms. Let:

$A$ be the proposition that Socko did it

$B$ be the proposition that Fats did it

$C$ be the proposition that Lefty did it

$D$ be the proposition that Muscles did it

Then the four propositions of the four henchmen state:

$C$

$\neg D$

$\neg A\wedge \neg D$

$\neg C$

Now we do casework on which of the four propositions is true (which means we negate the other three):

$C,D,A\vee D,C$

$\neg C,\neg D,A\vee D,C$

$\neg C,D,\neg A\wedge\neg D,C$

$\neg C,D,A\vee D,\neg C$

The second and third possibilities have the contradictions $C$ and $\neg C$ so they are not possible. The first list is impossible because it means $C$ and $D$ both did it. So it must be the last possibility, where we find that $D$ (Muscles) did it.

This is a rather contrived problem though because in real life we would not know that all but one are lying without knowing which one is lying, and we would not know that several of the henchmen were not involved instead of just one.


问:
Find the derivative of the polar curve $r = 4\theta + \sin\theta$
I have been trying to solve this equation: Find the derivative $\frac{dy}{dx}$ of the polar curve $r = 4\theta + \sin\theta$ and I'm not sure where to go with this. Anyone mind giving a hand?

答:
HINT:

$x = r \cos \theta = (4 \theta + \sin \theta) \cos \theta$

$y = r \sin \theta = (4 \theta + \sin \theta) \sin \theta$

$\displaystyle \frac{dy}{dx} = \frac{dy}{d\theta} / \frac{dx}{d\theta}$

Can you take it from here?


问:
Mathematical problem of finding more concise representation of some function?
This question is the generalization of my question on ComputerScience.SE: How to translate automaton (Turing machine) into the program of high level programming language?

Turing machine can be considered as special, algorithmically represented function that maps binary string on the input to the binary string in the output. Essentially, Turing machine is defined by the tabular rules. My question is

Can such function be reexpressed using more traditional and conscise functions and constructions - be it imperative, functional or logic program?

Software quality metrics for the source code can provide guidance. Does mathematics have some branch to express low level function with high level, compisitional and even higher order (that takes as arguments other functions not just their return values) functions?

Of course, I know about Fouries series, about Kolmogorov-Arnold theorem, but they don't apply there, because they are somehow noncompositional and to uniform and even with infinities.

Have mathematicians seen or thought about such issues, maybe in other contexts?

答:
Reformulating your question, you look for a general approach for a Turing-complete languages to "extract" knowledge from a low-level data.
Having similar questions on the concept of emergence in mathematics, I stumbled upon a number of things that may be relevant to you. In logic: representability, definability, expressivity, particularly Beth-definability theorem, of course Godel's incompleteness theorem may be relevant (and, more generally, hierarchy of arithmetical definability). In software verification (my field): invariants and invariant representations (one article of particular interest is Padon, O., Immerman, N., Shoham, S., Karbyshev, A., & Sagiv, M. (2016). Decidability of inferring inductive invariants. ACM SIGPLAN Notices, 51(1), 217-231.), synchronization/relational invariants and program equivalence/program products (see, e.g. Mordvinov, D., & Fedyukovich, G. (2019). Property Directed Inference of Relational Invariants. 2019 Formal Methods in Computer Aided Design (FMCAD), 152-160.).</p>


答:
By the Chain Rule, we know that:

$$\frac{dy}{dx}\frac{dx}{d\theta} = \frac{dy}{d\theta}$$

Thus:

$$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$$

However, we know that $x = r\cos(\theta)$ and $y = r\sin(\theta)$. Thus, we have:

$$\frac{dy}{dx} = \frac{\frac{d}{d\theta}(r\sin(\theta))}{\frac{d}{d\theta}(r\cos(\theta))}$$

Doing some straightforward product and chain rules (remember that $r$ is also a function of $\theta$):

$$\frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin(\theta) + r\cos(\theta)}{\frac{dr}{d\theta}\cos(\theta) - r\sin(\theta)}$$

Can you take it from here? If you're still stuck:

 Taking derivatives, we find:

$$\frac{dr}{d\theta} = 4 + \cos(\theta)$$

 Thus, we have:

 $$\frac{dy}{dx} = \frac{(4 + \cos(\theta))\sin(\theta) + (4\theta + \sin(\theta))\cos(\theta)}{(4 + \cos(\theta))\cos(\theta) - (4\theta + \sin(\theta))\sin(\theta)}$$
$$\frac{dy}{dx} = \frac{4\sin(\theta) + 4\sin(\theta)\cos(\theta) + 4\theta\cos(\theta) + \sin(\theta)\cos(\theta)}{4\cos(\theta) + \cos^{2}(\theta) - 4\theta\sin(\theta) - \sin^{2}(\theta)}$$
 $$\frac{dy}{dx} = \boxed{\frac{8(\sin(\theta) + \theta\cos(\theta)) + 5\sin(2\theta)}{8(\cos(\theta) - \theta\sin(\theta)) + 2\cos(2\theta)}}$$

Title: Find the derivative of the polar curve $r = 4\theta + \sin\theta$

Body: I have been trying to solve this equation: Find the derivative $\frac{dy}{dx}$ of the polar curve $r = 4\theta + \sin\theta$ and I'm not sure where to go with this. Anyone mind giving a hand?


问:
Derivative of Multivariate Function: Chain Rule and Product Rule
How do I compute the following gradient?
$$
\nabla_{\phi} \,\,\,T(y, x(\phi, y), \phi)
$$
Should I use the product rule or the chain rule? Is it as follows?

$$
\frac{\partial}{\partial x}T(y, x, \phi) \frac{\partial x(\phi, y)}{\partial \phi} + \frac{\partial}{\partial \phi} T(y, x(\phi, y), \phi)
$$

答:
My primary interepretation would be that $\nabla_{\phi}$ refers to the partial derivative with respect to the 3rd argument, hence $$\nabla_{\phi} \; T(y,x,\phi) = \frac{\partial}{\partial \phi} T(y,x,\phi),$$ which is the second term in your answer. That is, $\nabla_{\phi}$ refers to the partial derivative of $T(y,x,\phi)$ with respect to argument $\phi$, not the total derivative of the variable $\phi$.

However, if you let $H(\phi, y) := T(y, x(\phi, y), \phi)$, then

$$\nabla_{\phi} \,\,\,H(\phi, y) = \frac{d}{d\phi} T(y,x,\phi) = \frac{d}{d\phi} T(y,x(\phi, y),\phi) = \frac{\partial T(y,x,\phi)}{\partial \phi} + \frac{\partial T(y,x,\phi)}{\partial x} \frac{\partial x}{\partial \phi},$$ which is what you wrote. That is, the total derivative with respect to $\phi$.

So I would say it is unclear what the notation refers to, although I would think it refers to the former method. To my experience, when the notation is $\nabla_{\boldsymbol v} \; T(y,x,\phi)$, with bold subscript, $\boldsymbol v$ refers to a vector and the expression refers to the directional derivative.


问:
How to conclude $\frac{d}{dt}E(f)=-\int _M (\Delta f)\dot{f} d\mu$?
Can anyone explain to me how I can conclude $\frac{d}{dt}E(f)=-\int _M (\Delta f)\dot{f} d\mu$ by using integration by parts and $\langle f_1 ,f_2 \rangle_\mu:=\int_M f_1 f_2 d\mu$?

Where $M$ is a compact Manifold, $E(f)=\frac{1}{2}\int _M |\nabla f|^2d\mu$ and $\dot{f}=\frac{d}{dt}f$.

Thanks.

答:
Well, you must have either $M $ being closed (i.e. $\partial M = \phi $) or some Neumann conditions like $ \partial f/\partial \nu $ vanishes on boundary to have the following relation from Green's formula
$$ \int_M \dot{f}\Delta f \ d\mu + \int_M \langle \nabla f,\nabla\dot{f} \rangle d\mu = \int_{\partial M}\dot{f}\frac{\partial f}{\partial \nu}dS = 0 $$
This immidiately gives
$$ \frac{d}{dt}E(f) = \frac{1}{2} \int_M \frac{d}{dt}\Bigl(|\nabla f|^2\Bigr)d\mu = \int_M \langle \nabla f, \nabla\dot{f}\rangle d\mu = -\int_M \dot{f}\Delta f\ d\mu  $$</p>


问:
Help me prove this inequality with positive real numbers
Prove that $$p(qr+s+1)(rs+p+1)(sp+q+1)+q(pq+r+1)(rs+p+1)(sp+q+1)+r(pq+r+1)(qr+s+1)(sp+q+1)+s(pq+r+1)(qr+s+1)(rs+p+1)>(pq+r+1)(qr+s+1)(rs+p+1)(sp+q+1)$$ provided that $p,q,r,s$ are positive real numbers for which $p,r>1$ and $q,s<1$. $$ $$ So far I realised that I can divide both sides of the inequality by $(pq+r+1)(qr+s+1)(rs+p+1)(sp+q+1)$ and I got $$\frac{p}{pq+r+1}+\frac{q}{qr+s+1}+\frac{r}{rs+p+1}+\frac{s}{sp+q+1}>1$$ $$$$ I'm not sure how to proceed now or what may lead me to the solution. I will be very grateful for any help on how to solve it.

Body:
$$
\sum_{cyc} \frac{p}{pq+r+1} = 1+ \sum_{cyc}\left( \frac{p}{pq+r+1} - \frac{p}{p+q+r+s} \right)\\
= 1+ \sum_{cyc}\frac{p(s+p+q-pq-1)}{(pq+r+1)(p+q+r+s)} \\
= 1+ \sum_{cyc}\frac{p(s+(p-1)(1-q))}{(pq+r+1)(p+q+r+s)} > 1.\blacksquare
$$


问:
Power Series Differential Equation evaluation
I'm trying to evaluate the solution to the following differential equation using power series:$$y''+y'+x^2y=0$$
and I got the following values to substitute:
$$y=\sum_{n=0}^\infty C_n x^n$$
$$y'=\sum_{n=1}^\infty nC_n x^{n-1}=\sum_{n=0}^\infty (n+1)C_{n+1} x^n$$
$$y''=\sum_{n=2}^\infty n(n-1)C_n x^{n-2}=\sum_{n=0}^\infty (n+1)(n+2)C_{n+2} x^n$$

and after I substitute I get

$$\sum_{n=0}^\infty [(n+1)(n+2)C_{n+2}+(n+1)C_{n+1} ] x^n + \sum_{n=0}^\infty C_nx^{n+2}=0$$

I got an expression for $C_{n+2}=\frac {-C_1}{(n+2)!}$
But $C_n$ does not seem to equate anything. Does this imply $C_n =0$?Or am I going wrong with this?

答:
it means that :

$$(2c_2+c_1)x^0+(6c_3+2c_2)x^1+\sum_{n=2}^{\infty}\Big[(n+1)(n+2)c_{n+2}+(n+1)c_{n+1}+c_{n-2}\Big]x^n =0, \forall x $$

And then you can identify the terms


问:
Describing a two person game with a pay-off matrix where i have to change the order to of utlities to find a different action profile?
I understand what weakly dominated strategies are, I understand action profiles, 3 player payoff matrixes and utilities, but I cannot seem to figure out what exactly is asked.

The utilities should be such that, by applying in two different orders the iterated
elimination of weakly dominated strategies, two different and unique action profiles are returned.

Can someone clarify this for me?

答: I can't seem to delete the weakly dominated strategies further in attempt number 2. Other ones I've tried have ended up 1 being unique, the other one appearing. I have no clue how to best approach this.


问:
What does $\langle x,3\rangle$ mean if I'm talking about $\mathbb Z[x]$
I know this sounds like a basic question but I'm really confused. What does the notation $\langle x,3\rangle$ refer to for $\mathbb Z[x]$? Can someone write out what this is?</p>

答: $\rm \langle x,3\rangle\:$ could mean a few things. Here are some possibilities, listed most-likely first. It could denote the ideal $\rm\: x\ \mathbb Z[x] + 3\ \mathbb Z[x]\:.\:$ Or it could denote the additive subgroup $\rm\:x\ \mathbb Z + 3\ \mathbb Z\:.\:$ Finally it could denote $\rm\:gcd(x,3)\:,\:$ though the gcd is more commonly denoted as $\rm\:(x,3)\:$ or $\rm\:[x,3]\:.$</p>


问:
eigenvalues for particular matrices
Let $x \in  \mathbb{C^n}\setminus \{ \vec{0} \}$ be fixed. Find all the eigenvalues and eigenvectors of the matrices $A_1=xx^*$ and
  $$A_2\ =\ \left(\begin{array}{cc}
0 & x^* \\
x & 0 \\
\end{array}\right)$$</p>

答:
Simply stick to the definition of eigenvector. Let $w=\pmatrix{z\\ v}$ be an eigenvector of $A_2$ corresponding to an eigenvalue $\lambda$. Then $A_2w=\lambda w$ and hence
\begin{align*}
x^\ast v &= \lambda z,\tag{1}\\
zx &= \lambda v.\tag{2}
\end{align*}
There are two possibilities:

$\lambda = 0$. It follows that $x^\ast v=z=0$ and $v\neq0,\lambda\neq0$. The equation $x^\ast v=0$ also shows that the eigenspace of $\lambda=0$ is $(n-1)$-dimensional.

$\lambda\ne0$. Equation $(2)$ implies that $z\neq0$ and $v\neq0$ and $v$ is a scalar multiple of $x$. Choose $v = x$, then $\lambda=z$ and you can determine $\lambda$ from $(1)$ (there are two solutions).

Eigenvalues and eigenvectors of $A_1$ can be found in a similar manner.

Title: eigenvalues for particular matrices


问:
Let B be a square matrix such that rank(B)=rank(B^2). Show that (the nullspace of B) ∩ (the column space of B) = {0}
Let B be a square matrix such that rank(B)=rank(B^2). Show that (the nullspace of B) ∩ (the column space of B) = {0}
I know that their nullspace is equal since their rank is equal. Any clues on how to proceed from here?</p>

答:
Denote by $\mathrm{Im}(B)$ and $\mathrm{Im}(B^2)$ the images (i.e. the column spaces) of $B$ and $B^2$ respectively. One has always $\mathrm{Ker}(B) \subset \mathrm{Ker}(B^2)$. But because $\mathrm{Im}(B)$ and $\mathrm{Im}(B^2)$ are supposed to have the same dimension, then by rank theorem, $\mathrm{Ker}(B)$ and $\mathrm{Ker}(B^2)$ also share the same dimension. You deduce that $\mathrm{Ker}(B^2)=\mathrm{Ker}(B)$.

Now, if $x \in \mathrm{Ker}(B) \cap \mathrm{Im}(B)$, then $x=B(y)$ for some element $y$, and $B(x)=0$, so $B^2(y)=0$, so $y \in \mathrm{Ker}(B^2)=\mathrm{Ker}(B)$, so $B(y)=0$, so $x=0$.


问:
A question regarding reason behind assumptions in 2 proofs of Cyclotomic extensions
I have self studied section Cyclotomic extensions( page 298 ) from Algebra by Thomas Hungerford and I have 1  question

Consider these 2 theorems :

Question: why does in both the theorems there is an assumption that char K doesn't divides n?

I have read both the proofs but couldn't find the reasoning behind it.

Can you please tell why it happens?

答:
For separability reasons. If the characteristic of $K$ is $p$ which divides $n$, then $$X^n-1=X^{kp}-1=(X^k-1)^p,$$
for some $k$, so the $n$-th roots of unity are actually $k$-th roots of unity. In particular if $k=1$, then there are no primitive roots of unity.


问:
Compute the limit $\lim\limits_{t \to + \infty} \int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin tx} $
I have been working on this limit for days, but I am not getting it. The question is

Compute the limit $$\lim_{t \to + \infty} \int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (tx)}$$

Note that the integral is well defined and convergent for every $t >0$. Indeed the integrand function is a positive function for every $t >0$ since
$$e^x + \sin tx > e^x-1 > x>0$$
And as $x \to + \infty$ the integrand function behaves like $e^{-x}$.

WHAT I TRIED:

I consider $t=2n \pi$ a multiple of $2 \pi$, and see what happens:
$$\int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)} = \sum_{k=0}^\infty \int_{k /n}^{(k+1) /n} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)}$$
Making the change of variables $u = 2n \pi x$ I get
\begin{align}\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{u/2n \pi}+ \sin (u)} &\ge
\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(2k+2) \pi/2n \pi}+ \sin (u)} \\&=
\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(k+1)/n}+ \sin (u)}\end{align} where I write the lower bound with the minimum of the function at $u=(2k+2) \pi$. Now I use the fact that the integrand function does is integrated over a period of $2 \pi$, and using the result for $C>1$
$$\int_0^{2 \pi} \frac{ \mathrm d u}{C+ \sin (u)} = \frac{2 \pi}{\sqrt{C^2-1}}$$
I get the estimate
\begin{align}\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(k+1)/n}+ \sin (u)} &= \sum_{k=0}^\infty \frac{1}{2n \pi} \frac{2 \pi}{\sqrt{e^{2(k+1)/n} -1 }} \\&= \frac{1}{n} \sum_{k=0}^\infty \frac{1}{\sqrt{e^{2(k+1)/n} -1 }}\end{align}
Summing all up, I got that
$$\int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)} \ge \frac{1}{n} \sum_{k=0}^\infty \frac{1}{\sqrt{e^{2(k+1)/n} -1 }}$$
As $n \to \infty$ the series converges to the Riemann integral
$$\int_0^{+ \infty} \frac{\mathrm d y}{\sqrt{e^{2y}-1}} = \frac{\pi}{2}$$

Hence the limit should be a number larger than $\pi/2$, or $+ \infty$.

Using WA I got for large values of $t$ that the integral is between $1$ and $2$, thus $\pi/2$ could be the actual limit.

答:
For each $t > 0$, we have
\begin{align}
\int_0^\infty \frac{1}{\mathrm{e}^x + \sin t x}\mathrm{d} x
&= \int_0^\infty \sum_{k=0}^\infty (-1)^k \mathrm{e}^{-(k+1)x} (\sin t x)^k\mathrm{d} x\\
&= \sum_{k=0}^\infty (-1)^k \int_0^\infty \mathrm{e}^{-(k+1)x} (\sin t x)^k\mathrm{d} x\\
&= \sum_{k=0}^\infty (-1)^k\frac{1}{k+1}
\int_0^\infty \mathrm{e}^{-y} (\sin \tfrac{t y}{k+1})^k\mathrm{d} y.
\end{align}
(Hint: Use Fubini's theorem for the interchange of integration and summation.)

Denote $I_k = \int_0^\infty \mathrm{e}^{-y} (\sin \tfrac{t y}{k+1})^k\mathrm{d} y, \ k=0, 1, 2, \cdots$. Using integration by parts, we have
$$I_k = \frac{(k-1)kt^2}{(k+1)^2 + k^2t^2}I_{k-2}, \ k=2, 3, 4, \cdots.$$
Also, $I_0 = 1$ and $I_1 = \frac{2t}{t^2+4}$.
Thus we have, for $k = 0, 1, 2, \cdots$,
$$\lim_{t\to\infty} I_{2k+1} = 0,$$
and
$$\lim_{t\to\infty} I_{2k} = \frac{2k-1}{2k}\cdot \frac{2k-3}{2k-2}\cdots \frac{1}{2}I_0 = \frac{(2k)!}{4^k (k!)^2}.$$
By Tannery's theorem (https://en.wikipedia.org/wiki/Tannery%27s_theorem),
we have
$$\lim_{t\to \infty} \int_0^\infty \frac{1}{\mathrm{e}^x + \sin t x}\mathrm{d} x
= \sum_{k=0}^\infty \frac{1}{2k+1} \frac{(2k)!}{4^k (k!)^2}
= \frac{\pi}{2}.$$


问:
Integration of the function $f\left( x \right) = \frac{{\sin \left( {\frac{{9x}}{2}} \right)}}{{\sin \left( {\frac{x}{2}} \right)}}$
Let f be the function on $[-\pi,\pi]$ given by $f(0)=9$ and $f\left( x \right) = \frac{{\sin \left( {\frac{{9x}}{2}} \right)}}{{\sin \left( {\frac{x}{2}} \right)}}$ for $x\ne 0$. Find the value of $\frac{2}{\pi }\int\limits_{ - \pi }^\pi  {f\left( x \right)dx} $

My approach is as follow. The function is continuous at $x=0$. Hence it is contionuos for $[-\pi,\pi]$

$f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\frac{{9x}}{2}} \right)}}{{\left( {\frac{{9x}}{2}} \right)}} \times 9 \times \frac{{\left( {\frac{x}{2}} \right)}}{{\sin \left( {\frac{x}{2}} \right)}} = 9$

$\frac{x}{2}=\theta$

$\frac{2}{\pi }\int\limits_{ - \pi }^\pi  {f\left( x \right)dx} $

$\int\limits_{ - \pi }^\pi  {\frac{{\sin \left( {\frac{{9x}}{2}} \right)}}{{\left( {\frac{{9x}}{2}} \right)}}dx}  \Rightarrow 2\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\sin \left( {9\theta } \right)}}{{\sin \left( \theta  \right)}}d\theta }  \Rightarrow 4\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin \left( {9\theta } \right)}}{{\sin \left( \theta  \right)}}d\theta }  $

$ \Rightarrow 4\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin \left( {8\theta  + \theta } \right)}}{{\sin \left( \theta  \right)}}d\theta }  \Rightarrow 4\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin \left( {8\theta } \right)\cos \left( \theta  \right) + \cos \left( {8\theta } \right)\sin \left( \theta  \right)}}{{\sin \left( \theta  \right)}}d\theta } $

$ \Rightarrow 4\int\limits_0^{\frac{\pi }{2}} {\left( {\frac{{\sin \left( {8\theta } \right)\cos \left( \theta  \right)}}{{\sin \left( \theta  \right)}} + \cos \left( {8\theta } \right)} \right)d\theta } $

$\int\limits_0^{\frac{\pi }{2}} {\cos \left( {8\theta } \right)d\theta }  = 0$

How so I proceed form here as expansion of $sin(8\theta)$ is cumbersome.

Is their any formula of converting $sin(n\theta)$ in terms of $sin(\theta)$

答:
Via $$\cos t={e^{it}+e^{-it}\over2},\qquad\sin t={e^{it}-e^{-it}\over2i}$$
you can easily prove that
$${\sin(9t)\over\sin t}=1+2\bigl(\cos(2t)+\cos(4t)+\cos(6t)+\cos(8t)\bigr)\ .$$


问:
Why Zorn's Lemma "fails" on nested boxes problem?

Backgound

My problem comes from the nested boxes problem. Consider a list of boxes. Each box has a length, width, and height. Since the boxes can be rotated those terms are interchangeable. The boxes have rectangular surfaces and can be nested inside each other. A box can nest inside another box if all its dimensions are less than the corresponding dimensions of the other. You may only nest a box such that the corresponding surfaces are parallel to each other. A box may not be nested along the diagonal. You cannot also put two or more boxes side by side inside another box.

Therefore, a box can be represented by a sorted (it's not hard to show that sorting the dimensions in ascending order won't affect the result of whether a box can be nested inside another) tuple of dimensions.  The set of boxes and the relation of being able to nested in another actually form a partially ordered set. And here comes my problem:

Formalization

Let's use an example. Let $S=\{(1,1,1),\,(3,4,4),\,(2,5,5)\}$. We define relation $\le$ to be
$$
(x_1,y_1,z_1)\le(x_2,y_2,z_2) \text{ if and only if } x_1\le x_2,\,y_1\le y_2,\,\text{and }z_1\le z_2
$$
We see that $(S,\le)$ is a partially ordered set.

Now, all the totally ordered subset of $S$ are:
$$
S_1=\{(1,1,1),\,(3,4,4)\}\quad S_2=\{(1,1,1),(2,5,5)\}
$$
where $(3,4,4)$ is the upper bound of $S_1$, $(2,5,5)$ is the upper bound of $S_2$.

However, Zorn's Lemma says:

Suppose a partially ordered set $P$ has the property that every chain (i.e. totally ordered subset) has an upper bound in $P$. Then the set $P$ contains at least one maximal element.

Therefore, $S$ must have a maximal element. But it's obvious that $S$ doesn't have one.

Did I miss something here? Thanks.

答:
Both $\langle 3,4,4\rangle$ and $\langle 2,5,5\rangle$ are maximal elements of $S$: a maximal element is simply one that is not strictly less than any other element. Zorn’s lemma does not guarantee that $S$ has a maximum element, which I suspect is what you have in mind.

(By the way, $S$ has three other non-empty chains: they just aren’t maximal chains. Specifically, the other non-empty chains are the three singleton subsets.)


问:
Ant Velocity antiderivative word problem
I am trying to solve the following question by getting my variables, etc mixed up.

F for velosity of an ant walking along a shelf is:
$$
v(t)=\left\{\begin{array}{ll}
5 t, & 0 \leq t<1 \\
6 \sqrt{t}, & 1 \leq t \leq 2
\end{array}\right.
$$

$t=sec, v=cm/s$.  Estimate time for ant to get to 4cm from Start Pos.

So what I have is
$$
t=0,\ v(0)=5t, \ s(0)=0\\
t=n,\ v(n)=6\sqrt{t} , \ s(n)=4
$$
For S pos,

$$
v=5t\\
s=5/2t^2 ( c=0 \ at \ s(0)=0)\\
s(1)=5/2(1)^2 = 2.5cm \ @ \ t=1sec
$$

At E pos,
$$
v(n)=6 \sqrt{t}\\
s(n)=4t^{3/2} \\
s(2)=4(2)^{3/2}+c=4cm \rightarrow c\approx-7.3137
$$

So, for 1.5cm

$$
s(n)=4t^{3/2}-7.3137=1.5\\
s(n)=4t^{3/2}=1.5+7.3137\\
s(n)=t^{3/2}=8.8137/4\ \rightarrow t\approx3.27
$$
So time to travel 4 cm = 1+3.27 = 4.27 seconds.

I know the answer is wrong but where have I made my mistakes?
Thanks.

Body:
$s(n)=4(t)^{3/2}+c, \,$ for $(1 \le t \le 2)$

At $t = 1$, ant is at $2.5$ cm. So, $c = 2.5 - 4 = -1.5$

$4 = 4(t)^{3/2} - 1.5 \implies t \approx 1.2365$


问:
Win-loss ratios, and averages of them
There are many games where player performance is ranked by ratio of wins to losses.

I mistakenly had it in my head that the average such ratio must be 1. This is not the case, as can be found from a simple example:

```<code>W   L   W:L
-----------
7   2   3.5
3   3   1
3   6   0.5
4   2   2
2   2   1
1   5   0.2
-----------
sum sum avg
20  20  1.37
</code>```

My question is this: assuming that the numbers come from a closed system, and so the total number of wins is always equal to the total number of losses, is there some way to get a result as "obvious" as I was incorrectly expecting? Perhaps there's some way to scale such ratios, whereby by averaging them you'll always get a constant?

Or to get such a result do you always need to know the absolute number of wins and losses per person? For example, I could award +1 point for a win and -1 for a loss, and then the average score is always going to be 0.

答:
I think I may have figured this out.

If I map the ratios to the range 0 to 2 like this:

For ratios 0 through 1, $f(x) = x$

For ratios 1 through infinity, $f(x) = 2 - (1/x)$

I think the average of these mapped ratios is always 1. But I don't know how to verify this mathematically.

Title: Win-loss ratios, and averages of them


问:
Inequality with integers
I am trying to find an analytical way to prove that for integers $n\ge 6$,

$$\sqrt[n]{n+1}\le \frac{1}{n}+\frac{1}{\sqrt[n]{n+1}}+\frac12$$

答:
By the binomial expansion we obtain:
$$\left(1+\frac{\sqrt{2n-1}-1}{n-1}\right)^n\geq1+\frac{n(\sqrt{2n-1}-1}{n-1}+\frac{n(n-1)}{2}\cdot\left(\frac{\sqrt{2n-1}-1}{n-1}\right)^2=n+1.$$
Id est, $$\sqrt[n]{n+1}\leq1+\frac{\sqrt{2n-1}-1}{n-1}.$$
Thus, it's enough to prove that:
$$1+\frac{\sqrt{2n-1}-1}{n-1}\leq\frac{1}{n}+\frac{1}{1+\frac{\sqrt{2n-1}-1}{n-1}}+\frac{1}{2},$$
which is true for any $n\geq14$.

Can you end it now?

Title: Inequality with integers


问:
Matching pennies : bias possible ? And setting domain of parameters.
Suppose the matching pennies game :

$$\begin{array}[ccc]1 & q & 1-q\\p&(1,-1)&(-1,1)\\1-p&(-1,1)&(1,-1)\end{array}$$

Suppose we were player 2 and choose our strategy in function of 1s choices : $q=f(p)$

The gain is $$G2(p)=(1-2p)(2f(p)-1)\\ G2'(p)=-2(2f(p)-1)+(1-2p)2f'(p)=0\\ \Rightarrow \frac{f'}{2f-1}=\frac{1}{1-2p}\Rightarrow f(p)=\frac{1}{2}\left(1+\frac{A}{1-2p}\right)$$

$$\Rightarrow G2max=A, p\neq 1/2$$

So player 2 could expect a gain of A , if A were chosen small but non zero, then player 2 could do a statistic on 1s choices, put in the formula and get the probability to make his own choices, and this could induce a bias so that player 2 could have an average gain different than zero on the long term ?

Edit : in fact the question more precisely is  : $f(p)$ has to remain in [0,1]. What is more close to reality, setting $f(p)=\cos(g(p))^2$ from the onset, giving $q=1-p$  and $G2=(1-2p)^2$ or bounding it in the above calculation ?

(Anyhow both calculations lead to unrealistic results.)

答:
Suppose we know both $p$ and $q$. Then the expected gain of the first player is $(pq + (1-p)(1-q)) - (p(1 - q) + q(1 - p)) = 2pq - p - q + 1 + 2pq - p - q = 1 - 2p - 2q + 4pq = (1 - 2p)(1 - 2q)$. And for the second player it is $-(1 - 2p)(1 - 2q)$, because the game is a zero sum one. For a given $p$ the expected gain of the second player reaches maximum, when

$$q = \begin{cases} 0 & \quad p > \frac{1}{2} \\ 1 & \quad p < \frac{1}{2} \end{cases}$$

If $p = \frac{1}{2}$, then the expectation does not depend on $q$ at all.

That means, that the largest possible expected gain of the second player is $|1 - 2p|$. Because it is a zero sum game, the first player will do their best to minimise it.

Thus, the only mixed Nash equilibrium here is when $p = q = \frac{1}{2}$. In this case everyones expected gain will be $0$.

From that we can conclude, that the game is unbiased.

And if you want to find a game that is similar, but biased, try applying the same reasoning to the following variant of matching pennies (known as "prison poker"):

$$\begin{array}[ccc]1 & q & 1-q\\p&(2,-2)&(-3,3)\\1-p&(-3,3)&(4,-4)\end{array}$$


问:
Spectrum of operator $T: \ell^{2}\supset\text{dom}(T) \rightarrow \ell^{2}$ defined by $(x_{n})_{n\in \mathbb{N}} \mapsto (nx_{n})_{n\in \mathbb {N}}$
What is the point, continuous and residual spectrum of the following operator?

$T: \ell^{2}\supset\text{dom}(T) \rightarrow \ell^{2}$, $(x_{n})_{n\in \mathbb{N}} \mapsto (nx_{n})_{n\in \mathbb {N}}$, where $\text{dom}(T) = \left\{(x_{n})_{n\in \mathbb{N}} \in \ell^{2} |
 (nx_{n})_{n\in \mathbb {N}} \in \ell^{2} \right\}$

My thoughts so far: Since $T$ is self-adjoint it must be that $\sigma_{r}(T) = \emptyset$. Now considering the point spectrum, I tried to find the eigenvalues of $T$: $Tx = \lambda x \iff (n-\lambda) x_{n} = 0$. But then I would have $x_{n}=1 $ for $n = \lambda$ and $x_{n}=0$ otherwise, so the corresponding eigenvectors are not in $\text{dom}(T)$, right? So $\sigma_{p}(T) = \emptyset$? Can anyone provide hints for $\sigma_r(T)$ or $\sigma_c(T)$, please?

答:
Let $e_1,e_2,e_3,....$ be the usual orthonormal basis of $\ell^2.$ Then $e_1,e_2,e_3,....\in dom(T)$ and $Te_k=ke_k$ for all $k \in \mathbb N.$ This gives

$$\mathbb N \subseteq \sigma_p(T).$$

If $ \lambda \in \sigma_p(T)$, then there is $x=(x_1,x_2,...) \in dom(T)$ such that $Tx= \lambda x$ and $x \ne 0.$ It follows that

$$kx_k = \lambda x_k$$

for all $k \in \mathbb N.$ This shows that $\lambda= m$ for some $m \in \mathbb N $, $x_m \ne 0$ and $x_k=0$ for $k \ne m$, hence $x=x_me_m.$

Conclusion:
$$\mathbb N =\sigma_p(T).$$


问:
On odd perfect numbers and a GCD - Part III
(Note:  This post is an offshoot of this earlier MSE question.)

In what follows, we let $\sigma(x)$ denote the sum of divisors of the positive integer $x$.  We also let $D(x)=2x-\sigma(x)$ denote the deficiency of $x$.

Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

From this paper in NNTDM, we have the equation
$$i(q) := \frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd\left(n^2,\sigma(n^2)\right). \tag{1}\label{eq1}$$

In particular, we know that the index $i(q)$ is an integer greater than $5$ by a result of Dris and Luca.

We now attempt to compute an expression for $\gcd\left(\sigma(q^k),\sigma(n^2)\right)$ in terms of $i(q)$.

First, since we have
$$\sigma(q^k)\sigma(n^2) = \sigma({q^k}{n^2}) = \sigma(N) = 2N = 2{q^k}{n^2}$$
we obtain
$$\sigma(q^k) = \frac{2 q^k n^2}{\sigma(n^2)} = \frac{2n^2}{\sigma(n^2)/q^k} = \frac{2n^2}{i(q)}$$
and
$$\sigma(n^2) = \frac{2 q^k n^2}{\sigma(q^k)} = {q^k}\cdot\bigg(\frac{2n^2}{\sigma(q^k)}\bigg) = {q^k}{i(q)},$$
so that we get
$$\gcd\left(\sigma(q^k),\sigma(n^2)\right) = \gcd\bigg(\frac{2n^2}{i(q)}, {q^k}{i(q)}\bigg).$$

Now, since $\gcd(q, n) = \gcd(q^k, 2n^2) =  1$ and $i(q)$ is odd, we get
$$\gcd\bigg(\frac{2n^2}{i(q)}, {q^k}{i(q)}\bigg) = \gcd\bigg(\frac{n^2}{i(q)}, i(q)\bigg).$$

The following is copied verbatim from this MathOverflow answer to a closely related question:

Here is a conditional proof that
$$G = \gcd(\sigma(q^k),\sigma(n^2)) = i(q) = \gcd(n^2, \sigma(n^2)).$$

As derived in the OP, we have
$$G = \gcd\bigg(\frac{n^2}{i(q)}, i(q)\bigg).$$

This is equivalent to
$$G = \frac{1}{i(q)}\cdot\gcd\bigg(n^2, (i(q))^2\bigg) = \frac{1}{i(q)}\cdot\bigg(\gcd(n, i(q))\bigg)^2.$$

But we also have
$$\gcd(n, i(q)) = \gcd\bigg(n, \gcd(n^2, \sigma(n^2))\bigg) = \gcd\bigg(\sigma(n^2), \gcd(n, n^2)\bigg) = \gcd(n, \sigma(n^2)).$$

Consequently, we obtain
$$G = \frac{1}{i(q)}\cdot\bigg(\gcd(n, \sigma(n^2))\bigg)^2 = \frac{\bigg(\gcd(n, \sigma(n^2))\bigg)^2}{\gcd(n^2, \sigma(n^2))}.$$

In particular, we get
$$\gcd(\sigma(q^k), \sigma(n^2)) = i(q) = \gcd(n^2, \sigma(n^2))$$
if and only if $\gcd(n, \sigma(n^2)) = \gcd(n^2, \sigma(n^2))$.

Here is an attempt at proving the following conjecture:

CONJECTURE: If $q^k n^2$ is an odd perfect number with special prime $q$, then the inequation
$$\gcd(\sigma(q^k), \sigma(n^2)) \neq \gcd(n^2, \sigma(n^2))$$
holds.

MY ATTEMPT AT A PROOF

Let $N = q^k n^2$ be an odd perfect number with special prime $q$.

Assume to the contrary that $$\gcd(\sigma(q^k), \sigma(n^2)) = \gcd(n^2, \sigma(n^2)).$$

By the considerations above, we obtain
$$\gcd(n^2, \sigma(n^2)) = \gcd(n, \sigma(n^2)).$$

But note that we also have
$$\frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2} = \gcd(n^2, \sigma(n^2)).$$

This implies that
$$\frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2} = \gcd(n^2, \sigma(n^2)) = \gcd(n, \sigma(n^2)),$$
from which we get the divisibility constraint
$$\frac{n^2}{\sigma(q^k)/2} \mid n$$
which gives
$$\frac{n}{\sigma(q^k)/2} \mid 1$$
and then
$$n = \frac{\sigma(q^k)}{2}.$$

We obtain
$$\frac{\sigma(n^2)}{q^k} = \frac{2n^2}{\sigma(q^k)} = \frac{D(n^2)}{\sigma(q^{k-1})} = \gcd(n^2, \sigma(n^2)) = i(q) = n = \frac{\sigma(q^k)}{2}.$$

It follows that
$$\sigma(n^2) = q^k n = \frac{q^k \sigma(q^k)}{2}.$$

In particular, note that we obtain the following equations from equation $\eqref{eq1}$, after multiplying through by $(q^k \sigma(q^k))/2$:
$$\frac{q^k \sigma(q^k)}{2}\cdot\frac{D(n^2)}{\sigma(q^{k-1})} = \frac{q^k \sigma(q^k)}{2}\cdot\gcd(n^2, \sigma(n^2)) = \frac{q^k \sigma(q^k)}{2}\cdot\gcd(n, \sigma(n^2)) = q^k n^2 = N = n\sigma(n^2).$$

Alas, this is where I get stuck.  I was hoping somebody out there might have some bright ideas on how to prove either $k=1$ or $q^k < n$ (in order to arrive at a contradiction), under the assumption that
$$\gcd(\sigma(q^k),\sigma(n^2))=\gcd(n^2,\sigma(n^2)),$$
where $q^k n^2$ is an odd perfect number with special prime $q$?

POSTSCRIPT: Note that to prove $k=1$, it suffices to show that
$$\frac{\sigma(n^2)}{q} \mid n^2.$$
Compare with what we have:
$$\frac{\sigma(n^2)}{q^k} = n \mid n^2.$$

UPDATE (October 30, 2020 - 1:53 PM Manila time) I have just posted what appears to be a complete proof of the conjecture.

答:
solution-verification

$$\frac{n^2}{\sigma(q^k)/2} \mid n$$
which gives
$$\frac{n}{\sigma(q^k)/2} \mid 1$$
and then
$$n = \frac{\sigma(q^k)}{2}.$$

Writing $\dfrac{n}{\sigma(q^k)/2} \mid 1$, you are implicitly using that $\dfrac{n}{\sigma(q^k)/2}$ is an integer, but I don't see why $\dfrac{n}{\sigma(q^k)/2}$ is an integer.

After assuming that
$$\gcd(\sigma(q^k),\sigma(n^2))=\gcd(n^2,\sigma(n^2))$$
we have
$$i(q)= \frac{\sigma(n^2)}{q^k} = \frac{2n^2}{\sigma(q^k)} = \frac{D(n^2)}{\sigma(q^{k-1})} = \gcd(n^2,\sigma(n^2))=\gcd(n,\sigma(n^2))$$
So, we can have
$$0\lt\frac{n}{\sigma(q^k)/2} = \frac{\gcd(n,\sigma(n^2))}{n}\le\frac nn=1$$
from which we can at least say that
$$\text{$\frac{n}{\sigma(q^k)/2}$ is an integer}\iff \gcd(n,\sigma(n^2))=n\iff n=\frac{\sigma(q^k)}{2}$$


问:
If $\lim f(x) = \lim g(x)$ everywhere, does it imply $f = g$ everywhere?
Prove or disprove the statement:

If $f, g ∶\mathbb R \to \mathbb R$ are functions such that
$$\lim_{x\to a}f(x) = \lim_{x\to a}g(x)$$
for any $a\in \mathbb R$, then $f = g$.

I think the statement is false, since it is not given that $f$ and $g$ are continuous.

On the other hand, it states that it works for any $a\in\mathbb R$, making it hard to find a counterexample.

Can anyone enlighten me on this issue?

答:
Disproof.
Let $f(x) = 0$ everywhere.
Let $g(x) = 0$ except for $x=0$, where $g(0)=1$.

$$\forall a\quad \lim_{x\to a}f(x) = \lim_{x\to a}g(x) = 0$$
But $f(0) \ne g(0)$.


答: I suspect you are confused as to the definition of maximal element. A maximal element of a partially ordered set $(S,\leq)$ is an element $s\in S$ such that if $t\in S$ and $s\leq t$, then $s=t$. It should be clear that the maximal elements of $S$ in your example are $(3,4,4)$ and $(2,5,5)$.</p>

Title: Why Zorn's Lemma "fails" on nested boxes problem?

Body:
Backgound

My problem comes from the nested boxes problem. Consider a list of boxes. Each box has a length, width, and height. Since the boxes can be rotated those terms are interchangeable. The boxes have rectangular surfaces and can be nested inside each other. A box can nest inside another box if all its dimensions are less than the corresponding dimensions of the other. You may only nest a box such that the corresponding surfaces are parallel to each other. A box may not be nested along the diagonal. You cannot also put two or more boxes side by side inside another box.

Therefore, a box can be represented by a sorted (it's not hard to show that sorting the dimensions in ascending order won't affect the result of whether a box can be nested inside another) tuple of dimensions.  The set of boxes and the relation of being able to nested in another actually form a partially ordered set. And here comes my problem:

Formalization

Let's use an example. Let $S=\{(1,1,1),\,(3,4,4),\,(2,5,5)\}$. We define relation $\le$ to be
$$
(x_1,y_1,z_1)\le(x_2,y_2,z_2) \text{ if and only if } x_1\le x_2,\,y_1\le y_2,\,\text{and }z_1\le z_2
$$
We see that $(S,\le)$ is a partially ordered set.

Now, all the totally ordered subset of $S$ are:
$$
S_1=\{(1,1,1),\,(3,4,4)\}\quad S_2=\{(1,1,1),(2,5,5)\}
$$
where $(3,4,4)$ is the upper bound of $S_1$, $(2,5,5)$ is the upper bound of $S_2$.

However, Zorn's Lemma says:

Suppose a partially ordered set $P$ has the property that every chain (i.e. totally ordered subset) has an upper bound in $P$. Then the set $P$ contains at least one maximal element.

Therefore, $S$ must have a maximal element. But it's obvious that $S$ doesn't have one.

Did I miss something here? Thanks.


问:
mean value property and the poisson kernel
d
How do I proceed? I have the cauchy integral formula as $f(z_0) = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{z-z_0}dz$. But I am not sure how to proceed.I would appreciate some headway. Thank you for the hints below.

For part (iii), I am using the hint given in the problem, and considering,
$\frac{1}{2\pi i} \int_{\gamma, 0,1} f(w) \frac{1}{2} (\frac{1+s/w}{1-s/w}) dw + \int_{\gamma,0,1} f(w) \frac{1}{2}\frac{1+sw}{1-sw}dw$. How do I compute this integral in two different ways?

答:
Let $ \gamma(t)=z_0+re^{it}$ for $t \in [0, 2 \pi].$

Then

$$f(z_0) = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{z-z_0}dz= \frac{1}{2 \pi i}\int_0^{2 \pi}\frac{f(z_0+re^{it})}{re^{it}}ire^{it} dt= \frac{1}{2 \pi }\int_0^{2 \pi}f(z_0+re^{it}) dt.$$

Can you proceed ?


问:
Two envelopes problem: is exchanging envelopes always a Bayes-Nash equilibrium?
Consider the following problem:

Suppose $f: [0; \infty) \to [0; \infty)$ is a continuous bijection. Suppose two people are given two envelopes with $x \$ $ and $f(x) \$ $ respectively. They know the value of $f$ and the amount of money in their envelope (but do not know, whether this is $x$ or $f(x)$). They consider getting $x \$ $ and $f(x) \$ $ to be equally likely. Then they are allowed to request exchange of envelopes (and if they both do, the envelopes are exchanged between them). Is the situation, where both of them request the exchange, always a Bayes-Nash equilibrium?

First, note that the game is symmetric.

Then, suppose, a player got $y$ in their envelope. Then, they consider the amount of money in another players envelope to be either $f(y)$ (in case, when $y = x$) or $f^{-1}(y)$ in case ($y = f(x)$) with equal probability. Thus the expected gain in case when both request exchange is $\frac{f(y) + f^{-1}(y)}{2}$ and $y$, if at least some of them refuses. Thus our question is equivalent to the following problem:

Suppose $f: [0; \infty) \to [0; \infty)$ is a continuous bijection and $y \in [0; \infty)$. Is it always true, that $f(y) + f^{-1}(y) \geq 2y$?

I managed to solve this problem (with positive answer) only for the case when $f$ is linear. Indeed, if $f(x) \equiv cx$ for $c \in (0 ; \infty)$, then $f(y) + f^{-1}(y) - 2y = cy + c^{-1}y - 2y = c^{-1}y(c - 1)^2 \geq 0$. However, I do not know, whether this method can be generalised to an arbitrary $f$...

答: It's not always true. If $f(y)=\sqrt y$, then $f(.5)+f^{-1}(.5)-2\times .5=\sqrt{\frac 12}+\frac 1 4 - 1<0.$


问:
I have a question about dealing polynomials using matrix
For example, let's say we find a constant function that is closest to $y=x^4$ on the interval $0\leq x\leq1$. Then we solve matrix

$ \begin{bmatrix}(1,1) & (1,x)\end{bmatrix}$$\begin{bmatrix}C \\ 0 \end{bmatrix}= $  $\begin{bmatrix}(1,x^4) \end{bmatrix}$

and integrate $(1,1)=\int _{0}^{1}dx=1$ and $(1,x^4)=\int _{0}^{1}x^{4}dx=\dfrac{1}{5}.$ Thus we get $C=\dfrac{1}{5}$

That's how I was instructed to do routinely without understanding.
I want to know what the ordered pairs $(1,1)$ $(1,x)$ actually mean in that matrix and why we are integrating on the interval and why $(1,1)$ equals to $\int _{0}^{1}dx$ and $(1,x^4)$ to $\int _{0}^{1}x^{4}dx$.

Not only that, it seems to me they use the matrix $\begin{bmatrix} (1,1) & (1,x)&\dots \\(x,1)&(x,x)&\dots\\(x^2,1)&(x^2,x)&\dots\\ \vdots & \vdots \end{bmatrix}$ when they deal with such polynomials problems. Where did that matrix come from and how? Why did they construct that matrix with such ordered pairs?

答:
A comment: you don't need a $1\times 2$ matrix, $1\times 1$ will do:
$$
[\langle 1,1\rangle][C]=[\langle 1,x^4\rangle]
$$
has all the information you need.

The setup. To start with we have to know what "closest" means, that is we need to say what we mean by the distance between two functions $f,g$ both defined on $[0,1]$. We have an inner product defined on the space of functions, namely $\langle f,g\rangle:=\int_{0}^{1} f(x)g(x)dx.$ The square of the distance between $f$ and $g$ is defined as $\langle f-g,f-g\rangle$, which simplifies to $\langle f,f\rangle -2\langle f,g\rangle+\langle g,g\rangle$. (This is just like the cosine rule when we are dealing with ordinary vectors.)

Constant Function. We want to find a constant function $x\mapsto C$ which is closest to the function  $x\mapsto x^4$.

The square of the distance  between these functions is $C^2 \langle 1,1\rangle -2C \langle 1,x^4\rangle +\langle x^4,x^4\rangle$. For a minimum we must find the derivative and equate it to $0$, that is we need to solve $2C\langle 1,1\rangle -2C \langle 1,x^4\rangle=0$. This gives $C=\frac{\langle 1,x^4\rangle}{\langle 1,1\rangle}=\frac{1}{5}$ as you know.

Linear Function. Now suppose we wanted to find the linear function $x\mapsto A+Bx$ closest to $x^4$.

The distance squared is $\langle A+Bx, A+Bx\rangle -2\langle A+Bx,x^4\rangle -\langle x^4,x^4\rangle.$ If we crunch this out we see it is equal to
$$
A^2\langle 1,1\rangle
+2AB\langle 1,x\rangle
+B^2 \langle x,x\rangle
-2A\langle 1,x^4\rangle
-2B\langle x,x^4\rangle
+\langle x^4,x^4\rangle.
$$

This time we need to differentiate with respect to $A$ and to $B$ and set both equal to $0$. This gives us two equations, and (after we do the calculation) we can write it in matrix terms
$$
\begin{bmatrix}\langle 1,1\rangle & \langle x,1\rangle\\
\langle 1,x\rangle & \langle x,x\rangle\end{bmatrix}
\begin{bmatrix}A \\ B \end{bmatrix}=
\begin{bmatrix}\langle 1,x^4\rangle \\ \langle x,x^5\rangle\end{bmatrix}
$$

The general case.  All this can be generalised but I am going to leave that for you to explore.


问:
How to denote a range of elements in a row array?
Let's say I have a row vector $\vec{x}_i \in \mathbb{R}^{1\times T}$. You can consider it as the $i$-th time-series from a set of time-series.
I want to index a subsequence from $\vec{x}_i$ from $t1$ till $t2$, but I do not know how to denote it properly without causing confusion.

Can I simply denote it as $\vec{x}_i(t1:t2)$ or $\vec{x}_i |_{t1}^{t2}$?

答:
Avoid vector arrows in print, when you are not in a geometric setting.

In elementary analysis we denote a sequence $(a_1,a_2,a_3,\ldots)$ by $\bigl(a_k\bigr)_{k\geq1}$. Similarly, when $I$ is an arbitrary index set used to parametrize a family of sets $A_\iota\,$, we denote this family by $\bigl(A_\iota\bigr)_{\iota\in I}$. Note that we are talking here about arrays, not about sets of objects. When $a_1=2$, $a_2=1$, $a_3=2$ then
$$\bigl\{a_i\bigm|1\leq i\leq 3\bigr\}=\{1,2\},\qquad \bigl(a_i\bigm|1\leq i\leq 3\bigr)=(2,1,2)\ .$$
For your time series I then propose  notations like
$$\bigl(x_i\bigr)_{t_1\leq i\leq t_2}\ ,\qquad\bigl(x_i\bigr)_{i\in[t_1..t_2]}\ ,\qquad\bigl(x_i\bigm|t_1\leq i\leq t_2\bigr)\ .$$


问:
Two homotopy equivalent closed $n$-manifolds are homeomorphic
I heard somewhere that there is a wrong conjecture that states:

Two homotopy equivalent closed $n$-manifolds ($n\geq 3$) are homeomorphic.

and the lens spaces are the first counterexample to this conjecture. I want to know the author of this wrong conjecture, and a reference for it. I am also curious about other counterexamples? (Here is a related question)

答:
I mentioned in the comments one can find families of counterexamples parametrized by a K-theoretic invariant that is normally nontrivial for a nontrivial fundamental group. Computing these invariants is generally difficult. Here is something on the opposite end of the spectrum; something rather easy to produce nontrivial elements of that only works for simply-connected manifolds.

Let $M$ be a simply-connected manifold of dimension $2k-1$ and $2n$ be such that $2k+2n-1>4$. If $Top$ is the colimit of the $Top(n)$, the homeomorphism group of $\mathbb{R}^n$, and $G$ is the colimit of the $G(n)$, the monoid of self homotopy equivalences of $S^{n-1}$, then it is a result of surgery theory (dependent on our conditions) that $M \times S^{2n}$ has homotopy equivalences $N \rightarrow M \times S^{2n}$, $N$ a manifold, in bijective correspondence to $[M \times S^{2n}, G/Top]$.

Another very deep result of surgery theory is that $\pi_{4n}(Top/O)=\mathbb{Z}$, $\pi_{4n+2}(Top/O)=\mathbb{Z}/2$ and trivial otherwise. So if we take the composition $M \times S^{2n} \rightarrow S^{2n} \rightarrow Top/O$, with the latter a nontrivial map as above, then this gives a nontrivial element of $[M \times S^{2n}, Top/O]$.

So we conclude that at least in every odd dimension greater than four, we have many counterexamples.


问:
Prove that $f(A \backslash S) = B \backslash f(S)$
If $f:A \rightarrow B$ is a bijection and $S \subseteq A$, prove that $f(A \backslash S) = B \backslash f(S)$

答: Let $x \in B$. One has $$x \in f(A \setminus S) \Longleftrightarrow f^{-1}(x) \in A \setminus S \Longleftrightarrow f^{-1}(x) \notin S \Longleftrightarrow x \notin f(S) \Longleftrightarrow x \in B \setminus f(S)$$


问:
Is $D^n$ defined when $n = 0$?
Define $D^n = \{x \in \mathbb{R}^n : |x| \le 1\}$.  Is $D^n$ defined when $n = 0$?  I would say no, since I don't think $\mathbb{R}^0$ is defined.</p>

答:
Using a definition of $\mathbb{R}^n$ like
$$\{(a_1,\ldots,a_n)\mid a_i\in\mathbb{R}\},$$
there would appear to be no definition of $\mathbb{R}^0$ (or it may even seem like $\mathbb{R}^0$ should be the empty set; but this would be terrible, since the rule $\mathbb{R}^{n+m}\cong\mathbb{R}^n\times\mathbb{R}^n$ would no longer be true).

For the purposes of generalization, the above definition is not ideal. A good definition to use is that $\mathbb{R}^n$ is $\color{red}{\textbf{the}}$ product (in an appropriate category) of $n$ copies of $\mathbb{R}$. $\color{red}{\textbf{The}}$ product of an empty collection of objects, when it exists, is $\color{red}{\textbf{the}}$ terminal object of the category; thus, thinking of $\mathbb{R}$ as a topological space, we would say that $\mathbb{R}^0$ is $\color{red}{\textbf{the}}$ topological space with a single point, while thinking of $\mathbb{R}$ as an abelian group, we would say that $\mathbb{R}^0$ is $\color{red}{\textbf{the}}$ trivial group.

You are thinking of $\mathbb{R}^n$ as a normed real vector space (which is a special case of both of the above examples), and as a result, we would say that $\mathbb{R}^0$ is $\color{red}{\textbf{the}}$ zero-dimensional real vector space (with its only possible norm), because such an object is terminal in the category of normed real vector spaces. Because the only element of $\mathbb{R}^0$ is its zero vector, we would therefore have that
$$D^0=\{x\in\mathbb{R}^0:|x|\leq 1\}=\mathbb{R}^0.$$
Thus, $D^0$ is $\color{red}{\textbf{the}}$ one-point topological space.

Why have I marked my uses of the word $\color{red}{\textbf{the}}$ this way? Because it really doesn't mean "the", in the sense you may be expecting. Really, any isomorphic object will do equally well, and there is no single thing to point to as being "the" product. For example, the one-point topological space
$$\{a\} \quad\text{with the topology}\quad\mathcal{T}=\{\varnothing,\{a\}\}$$ is no better or worse than the one-point topological space
$$\{b\} \quad\text{with the topology}\quad\mathcal{T}=\{\varnothing,\{b\}\}$$
so which one should we say is $\mathbb{R}^0$? The correct action is not to choose.
See this article on the nLab.

Title: Is $D^n$ defined when $n = 0$?


答:
Let $T$ be the time for the ant to travel 4cm. Then, integrate over the velocity to get the distance equation below

$$4= \int_0^T v(t)dt= \int_0^1 5t dt +\int_1^T 6\sqrt tdt = \frac52 + 4(T^{3/2}-1)
$$
Solve to obtain $T= \frac14 \sqrt[3]{121}$.

Title: Ant Velocity antiderivative word problem

Body:
I am trying to solve the following question by getting my variables, etc mixed up.

F for velosity of an ant walking along a shelf is:
$$
v(t)=\left\{\begin{array}{ll}
5 t, & 0 \leq t<1 \\
6 \sqrt{t}, & 1 \leq t \leq 2
\end{array}\right.
$$

$t=sec, v=cm/s$.  Estimate time for ant to get to 4cm from Start Pos.

So what I have is
$$
t=0,\ v(0)=5t, \ s(0)=0\\
t=n,\ v(n)=6\sqrt{t} , \ s(n)=4
$$
For S pos,

$$
v=5t\\
s=5/2t^2 ( c=0 \ at \ s(0)=0)\\
s(1)=5/2(1)^2 = 2.5cm \ @ \ t=1sec
$$

At E pos,
$$
v(n)=6 \sqrt{t}\\
s(n)=4t^{3/2} \\
s(2)=4(2)^{3/2}+c=4cm \rightarrow c\approx-7.3137
$$

So, for 1.5cm

$$
s(n)=4t^{3/2}-7.3137=1.5\\
s(n)=4t^{3/2}=1.5+7.3137\\
s(n)=t^{3/2}=8.8137/4\ \rightarrow t\approx3.27
$$
So time to travel 4 cm = 1+3.27 = 4.27 seconds.

I know the answer is wrong but where have I made my mistakes?
Thanks.


问:
How many people at a party if everyone knows 22 other people and any two who don't know each other have exactly 6 mutual friends?

You are in a strange party where everyone knows exactly 22 other person and any two who don’t know each other have exactly 6 mutual friends. How many are there in the party?

Source 40- Strange Party

My research:

I know that the adjacency matrix $M$ is symmetrical and if $M[i,j] = 0$ then i-th and j-th rows have 6 times 1 at the same index position. I am thinking about encoding problem into SAT formulae and trying to brute force SAT solver with different number of literals (elements of upper triangle matrix of adjacency matrix).

I have read some articles about Moore graphs and it seems that the upper limit for the vertices is $22^2 + 1 = 485$.

Do you have any hints to the solution?

答:
This is not a solution. This just demonstrate that $ n \leq 100$.

Take the typical graph theoretic representation.

Let friend be represented by a red edge, stranger be represented by a blue edge.

We have $d_r (v_i) = 22, d_b (v_i) = n - 23$, so total number of red edges is $11n$ and total number of blue edges is $ \frac{n(n-23)}{2}$.

We count the number of triples of red-red-blue triangles.

For each blue edge, the condition gives us exactly 6 red-red-blue triangles.

For each red-red emanating from a vertex, it could give us a red-red-red or red-red-blue triangle.

Hence, $ 6 \times \frac{ n ( n-23) } { 2 } \leq n \times { 22 \choose 2 } \Rightarrow n \leq 100$.


问:
Exponentially decaying distribution function
I don't really know what to call this problem. I need this is for one of my projects. It will be programmed in javascript so edge cases can be handled. I'll try to frame this like a textbook question.

If this is not possible, anything close to this behaviour will work too.

Problem:

I have a fixed number of some item, apples for example, that cannot exist in a broken state (i.e. must be integers). I need to giveaway these apples to villagers in an exponentially decaying fashion at certain intervals (fixed integer) before a fixed end time. The number of apples given on the first distribution will be pre-defined. There should be no apples left after the end time.

Given (user defined) parameters:

$X =$ Total number of apples

$A =$ Number of apples on first distribution

$I =$ Time interval (integer)

$T =$ Total time for distribution (integer)

The end time will be a multiple of time interval, so $T = iI$ of which, either $I$ and $i$ are given or $I$ and $T$.

I need a function $f(x)$ that returns an integer $f(nI)$ (where $n$ is an integer).
The domain of time is $(0 \le t \le T)$

so the results should be

$y(0) = A$

$y(T+) = 0$

$y(0I) + y(1I) + y(2I) + ... + y(iI) = X$

$OR$

$A + y(1I) + y(2I) + ... + y(T) = X$

Graph should look something like this: desmos, image

(note: this is not a perfect example as the values are not integers and don't add up to $X$)

My progress so far:

I have tried using the typical half- life based exponential decay functions but couldn't get the desired output

答:
After trying @vvgiri's and @md2perpe's suggestions I still couldn't figure out how to do this as intended. So I just removed the requirement for user to define $T$ and calculate it based on other parameters. And the way I've done it allows no way for all of the parameters to be user defined simultaneously.

My current solution: $f(x)$ returns the numbers of apples to distribute at a given time.

$$f(x) = \lfloor A-\frac{xA}{T} \rfloor$$

$$X = \sum_{n=0}^{\frac{T}{I}} f(nI)$$

for more context, you can refer to the actual project, which is now public: maximousblk/astatine

desmos: https://www.desmos.com/calculator/05ofbspddf

same method for exponential decay: https://www.desmos.com/calculator/mvcfqcvtwp

I'm leaving this question unanswered in hope for a better answer where all four of the parameters can be user defined.