问:
Why is $\langle \operatorname{grad} f, X\rangle_g$ independent of the metric on a Riemannian manifold?
Let $(M,g)$ be a Riemannian manifold and let $f \in C^{\infty}(M)$. Let $X$ be a smooth vector field on $M$. In smooth local coordinates $(x^i)$ on $M$, we can write $g = g_{ij} dx^i \otimes dx^j$ as well as $X = X^i \partial_{x^i}$. Now I have computed that $\langle\operatorname{grad} f, X \rangle_g =  X^i\frac{\partial f}{\partial x^i}$ as follows. If we let $g^{ij}$ be the inverse matrix of $g_{ij}$ We have
$$\begin{eqnarray*} \langle\operatorname{grad} f, X \rangle_g &=&  \left\langle g^{ij} \frac{\partial f}{\partial x^i}\partial_{x^j} , X^k\partial_{x^k} \right\rangle_g\\
&=& g^{ij} \frac{\partial f}{\partial x^i}X^k\left\langle \partial_{x^j},\partial_{x^k} \right\rangle_g\\
&=& g_{jk}g^{ij}  \frac{\partial f}{\partial x^i}X^k \\
&=& g_{kj}g^{ji}\frac{\partial f}{\partial x^i}X^k \hspace{1cm} (\text{by using that $g_{ij}$ is a symmetric matrix})\\
&=& X^i\frac{\partial f}{\partial x^i}. \end{eqnarray*}$$

My question is: Is there any geometric reason as to why $\langle \operatorname{grad} f, X \rangle_g$ should be independent of the metric $g$?

答:
Absolutely. The exterior differential of a function, $df$, is a one-form which is defined independently of metric: $df(X) = X(f)$. In coordinates, $df = \partial_if dx^i$.

Choice of a nondegenerate metric induces an isomorphism between vector fields and $1$-forms by pairing: $g(X)(V) = \langle X,V\rangle$. In coordinates, this pairing corresponds to multiplying by the inverse of the metric, or "raising indices." This isomorphism is also known as a musical isomorphism: if $X$ is a vector field, then the one-form $X^\sharp$ is defined by $X^\sharp(V) = \langle X,V\rangle$.

The gradient $\operatorname{grad}(f)$ is defined to be the metric dual of the one-form $df$, i.e., $\operatorname{grad}(f) = (df)^\sharp$, so if you pair $\operatorname{grad}(f)$ with an arbitrary vector field $X$, you get $df(X) = X(f)$, which does not depend at all on the metric. The gradient is constructed so that, in pairing, all mentions of the metric cancel.


问:
Prove by induction: $\forall n \in \mathbb{n}, n \geq 1 \Rightarrow 2^{2^{n}}-1$ is divisible by at least n distinct primes
I tried to play around with $2^{2^{n + 1}} - 1$, such as splitting it to $2^{2^{n} \cdot 2} - 1$, but it didn't helped much. I am kind of stuck and don't know what to do next

答:
As guided in the comments, taking $2^{2^n}-1$ as a difference of squares allows for the $n+1$ term to be written as

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

On the RHS, any prime which divides both of the factors must divide their difference, which is $2$.  But both factors are odd, so these factors are relatively prime to one another.

ContentLicense: CC BY-SA 4.0


问:
How can one justify if the graph is planar from adjacency matrix?
Given an adjacency matrix of a graph $G$, I was asked to do the following without drawing the graph:

A) Find the vertex of largest degree.

B) Does the graph have an Euler Circuit?

C) Is the graph Planar?

I did A and B. How can one justify if the graph is planar from adjacency matrix?

答:
There are some simple conditions which, if transgressed, would mean that the graph is non-planar but in general there is nothing which could easily determine whether the graph is planar.

Since you said 'without drawing the graph' can we assume that the matrix is fairly small?

For a small number of vertices (e.g. $6$) it is easy to check whether or not the graph   contains $K_{3,3}$ or a subgraph homeomorphic to $K_5$.

The reference you have already been given

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

should be useful for the simple conditions.


问:
How many prime numbers can satisfy $p_{1}|p_{2}^{2}-1,..., p_{n-1}|p_{n}^{2}-1, p_{n}|p_{1}^{2}-1$?
The problem is as follows:

Find all natural numbers $ n$  for which $ n$ primes $p_{1},p_{2},p_{3},...,p_{n}$ exist, such that $p_{1}|p_{2}^{2}-1,..., p_{n-1}|p_{n}^{2}-1, p_{n}|p_{1}^{2}-1$.

I've tried to look the the specific recursion without the prime number constraint, but I haven't been able to find anything useful to the problem. Do I try to find the specific primes or can I get around that a find $n$ directly?

Thank you for your insights.

答: Hint: Consider primes $p,q$ for which $p|q^2-1$. This means that $p|q-1$ or $p|q+1$; in particular, $p$ is at most $q+1$, and is actually much less if $q+1$ isn't prime (which it usually isn't, being even if $q$ is even). As a result, we should expect $p_i<p_{i+1}$ most of the time; see if you can turn this into a somewhat explicit description of what such sequences $(p_1,\dots,p_n)$ should look like.


问:
Find the sum: $\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n$

Find the sum:
$$\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n$$

My try:

I played a bit with the coefficient to make it look easier/familiar:

First attempt:
$$\begin{align}
\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n
&= \sum_{n=0}^\infty \frac{n!}{2^n(2n-1)!!}x^n \\
&= \sum_{n=0}^\infty \frac{n!}{(2n-1)!!}\left(\frac x2\right)^n
\end{align}$$
Second attempt:
$$\begin{align}
\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n
&= \sum_{n=0}^\infty \frac{n!\cdot n!}{(n+n)!}x^n \\
&= \sum_{n=0}^\infty \frac{1}{{2n \choose n}}x^n
\end{align}$$
However, I could not proceed with any of them. Also, I have figured out that the convergence radius is $4$.

My research:

I have also found the same sum has been discussed at AoPS, which unfortunately uses Beta function that my course has not covered yet.

Entering the sum to Wolphram Alpha, I got the following output for the partial sum:

$$\sum_{n=0}^k\frac{(n!)^2x^n}{(2n)!}=\frac{4\sqrt{x}\left(\sin^{-1}\left(\frac{\sqrt{x}}{2}\right)-\frac{2^{2k}k!(k+1)!B_\frac{x}{4}\left(k+\frac{1}{2},\frac{3}{2}\right)}{(2k)!}\right)}{(4-x)^{3/2}}+\frac{4}{4-x}.$$

My background:

As I have already mentioned, I cannot use Gamma, Beta or similar functions. I only know about the convergence theorems on functional series and operations on them. So, I'm looking for some method that uses quite elementary tricks.

Thanks in advance.

答:
Suppose that $\displaystyle f(x) = \sum_{n\geqslant 0} a_n x^n$ with $a_0=1$.  We shall assume that the series has non-zero radius of convergence.  Then,
\begin{align*}
\frac{1}{2}\frac{d}{dx} \Big( x^2 f(x^2)\Big) = \sum_{n \geqslant 0} (n+1)a_n x^{2n+1} \quad \tag{1}\label{A}
\end{align*}
and
\begin{align*}
2x \frac{d}{dx} \Bigg( \frac{f(x^2)-1}{x} \Bigg)
&= \sum_{n \geqslant 1} 2(2n-1)a_n x^{2n-1} \\
&=\sum_{n \geqslant 0}2(2n+1) a_{n+1} x^{2n+1} \tag{2}\label{B}
\end{align*}
If $f$ now satisfies the differential equation,
\begin{align*}
2x \frac{d}{dx} \Bigg( \frac{f(x^2)-1}{x} \Bigg) = \frac{1}{2} \frac{d}{dx} \Big(x^2 f(x^2)\Big) \tag{3}\label{C}
\end{align*}
we can equate coefficients in the power series \eqref{A} and \eqref{B} to derive,
\begin{align*}
a_{n+1} = \frac{n+1}{2(2n+1)} a_n, \quad a_0 = 1
\end{align*}
which means
\begin{align*}
a_n = \frac{n!}{2^n(2n-1)(2n-3) \cdots 1} = \frac{(n!)^2}{(2n)!}.
\end{align*}
and
\begin{align*}
f(x) = \sum_{n \geqslant 0} \frac{(n!)^2}{(2n)!} x^n.
\end{align*}
Introduce $\phi(x) = (f(x^2)-1)/x$ so that the differential equation \eqref{C} becomes,
\begin{align*}
(4-x^2)\phi' -3x \phi  = 2
\end{align*}
Multiply this by the integrating factor $(4-x^2)^{1/2}$ to obtain,
\begin{align*}
(4-x^2)^{3/2} \phi' - 3x (4-x^2)^{1/2}\phi=2(4-x^2)^{1/2}
\end{align*}
which is the same as
\begin{align*}
\frac{d}{dx} \Big( (4-x^2)^{3/2} \phi \Big) = 2(4-x^2)^{1/2} \tag{4}\label{D}.
\end{align*}
We  note that by construction $\phi(0)=0$.

The right hand side may be integrated using the substitution $x=2\sin\theta$ to give,
\begin{align*}
\int_0^\xi 2(4-x^2)^{1/2} dx
&= 8\int_0^{\sin^{-1}(\xi/2)} \cos^2\theta   d\theta \\
&=8 \Big[  \sin\theta\cos\theta \Big]_0^{\sin^{-1}(\xi/2)} + 8\int_0^{\sin^{-1}(\xi/2)} \sin^2\theta d\theta \\
&=8 (\xi/2) (1-\xi^2/4)^{1/2}+8\int_0^{\sin^{-1}(\xi/2)} 1-\cos^2\theta d\theta \\
&= 2 \xi (4-\xi^2)^{1/2} + 8\sin^{-1}(\xi/2) - 8\int_0^{\sin^{-1}(\xi/2)} \cos^2\theta d\theta
\end{align*}
from which we see,
\begin{align*}
\int_0^\xi 2(4-x^2)^{1/2} dx = \xi(4-\xi^2)^{1/2} + 4\sin^{-1}(\xi/2).
\end{align*}
Substituting this in \eqref{D}, and recalling $\phi(0)=0$, we get
\begin{align*}
\phi(x)(4-x^2)^{3/2} = x(4-x^2)^{1/2} + 4\sin^{-1}(x/2)
\end{align*}
yielding
\begin{align*}
f(x) &= \sqrt x \phi(\sqrt x) + 1 \\
&=  1 + \frac{x}{4-x}+\frac{4\sqrt x \sin^{-1}(\frac{\sqrt x }{2})}{(4-x)^{3/2}} \\
&=\frac{4}{4-x}\Bigg(1 + \frac{\sqrt{x} \sin^{-1}(\frac{\sqrt{x}}{2})}{\sqrt{4-x}} \Bigg)
\end{align*}


问:
Recovering random variable from its moments
The problem is: can you recover a distribution of random variable if you know all its moments?

My first guess was to use moment-generating function (MGF). It is known that if two random variables have the same MGF, then they have equal distributions. So if $M(X) = 1 + \sum_{n = 1}^\infty \mathbb{E} X^n \frac{t^n}{n!}$ converges, then it uniquely corresponds to some distribution; we can apply inverse Laplace transform and get the PDF of X, which solves the problem.
But there are some issues:

MGF may not exist.

Even if MGF exists, the approach with Laplace transform works only with continuous random variables.

Can you help me with this problem? Thank you!

答: The characteristic function may be what you need. It is connected closely to MGF but it always exists and there is one-to-one correspondence between characteristic functions and cumulative distributions. Look for example into quoted Wiki article.</p>

ContentLicense: CC BY-SA 3.0


答:
If any prime is $2$ then suppose $p_n=2$. Then $p_{n-1}|2^2-1=3$ and so $p_{n-1}=3$. Repeating this process, $p_{n-2}|3^2-1=8$ and so $p_{n-2}=2$.

So one solution is the cycle $(2,3)$ for $n=2$.

Now suppose all $p_i$ are odd. But then $$p_i|(p_{i+1}-1) \left( \frac{p_{i+1}+1}{2}\right)$$ and so  $p_i<p_{i+1}$ for all $i$ and so the sequence can never cycle round. There is only one solution.

Title: How many prime numbers can satisfy $p_{1}|p_{2}^{2}-1,..., p_{n-1}|p_{n}^{2}-1, p_{n}|p_{1}^{2}-1$?

Body:
The problem is as follows:

Find all natural numbers $ n$  for which $ n$ primes $p_{1},p_{2},p_{3},...,p_{n}$ exist, such that $p_{1}|p_{2}^{2}-1,..., p_{n-1}|p_{n}^{2}-1, p_{n}|p_{1}^{2}-1$.

I've tried to look the the specific recursion without the prime number constraint, but I haven't been able to find anything useful to the problem. Do I try to find the specific primes or can I get around that a find $n$ directly?

Thank you for your insights.


答:
If you have adjancency matrix like this one $$A = \pmatrix{0&1&1&1&1&1 &1\\1&0&1&1&1&1&1\\1&1&0&1&1&1&1 \\1&1&1&0&1&1&1 \\1&1&1&1&0&1&1\\1&1&1&1&1&0&1\\1&1&1&1&1&1&0}$$
then this graph is clearly not planar since all degrees are $6$ and in planar graph must exists vertex of degree less than $6$.

Probably you have some matrix from which is easy to deduce some numercial estimation which says that it is not planar, like $e> 3n-6$ or something like this.

Body:
Given an adjacency matrix of a graph $G$, I was asked to do the following without drawing the graph:

A) Find the vertex of largest degree.

B) Does the graph have an Euler Circuit?

C) Is the graph Planar?

I did A and B. How can one justify if the graph is planar from adjacency matrix?


问:
How to write $P(\omega_{n} \neq 1| \omega_{n-1} \neq 1)$ in terms of the stochastic matrix $P$ of a Markov chain
Suppose we have a Markov chain with finite state space $X = \{1, \cdots, r \}$ and stochastic matrix $P = (p_{ij})$ with initial distribution $\pi$ which is also stable.

How does one express the conditional probability $\textbf{P}(\omega_{n} \neq 1| \omega_{n-1} \neq 1)$ in terms of $P$?

答:
Use the total probability formula to force the condition to be $\omega_{n-1}=i$:

$$P(\omega_n \neq 1 \mid \omega_{n-1} \neq 1)=\sum_{i \neq 1} P(\omega_n \neq 1 \mid \omega_{n-1}=i) P(\omega_{n-1}=i \mid \omega_{n-1} \neq 1).$$

Now for the first factor you can just expand the event $\{ \omega_n \neq 1 \}$ into the disjoint union $\bigcup_{j \neq 1} \{ \omega_n=j \}$.

For the second factor, you can use the knowledge that the chain is in equilibrium to evaluate the prior probability that $P(\omega_{n-1}=i)$ and then update it based on knowing that $\omega_{n-1} \neq 1$. So that results in $\frac{\pi_i}{1-\pi_1}$.


问:
$g(x) = 1/(1+x^2)$ is continuous everywhere epsilon delta approach
I have a function

g: R→R given by the function $g(x) = 1/(1+x^2)$.

I want to prove that this is continuous everywhere. I was reading my real analysis textbook and it seems like a great approach would be to do the delta-epsilon method. I don't fully understand how to use this method.

I know that I can choose an ε > 0 and then I need to find a corresponding δ. How would I show that at a point $p$, $|p - x| < δ$ then $|g(p) - g(x)| < ε$.

Thanks for the help!

答:
Let a $\in \mathbb{R}$ and $\varepsilon > 0$. Set $\delta := \min\{1, \frac{\varepsilon}{ 2 |a| +1}\}$.
$$\forall x \in \mathbb{R}: |x-a|<\delta \implies |f(x)-f(a)|=|\frac{1}{1+x^2}-\frac{1}{1+a^2}|=\frac{|a^2-x^2|}{(1+x^2)(1+a^2)}<|a^2-x^2|\\=|x-a||x+a|< \delta |x-a+2a| \leq \delta (|x-a|+2|a|)\leq \delta(1+2|a|)\leq \frac{\varepsilon}{2|a|+1}(1+2|a|)=\varepsilon.$$
The last two inequalities are valid because $\delta\leq 1$, as well as $\delta \leq \frac{\varepsilon}{ 2 |a| +1}$, since $\delta$ is defined as their minimum.


问:
How to calculate the absolute extrema for this equation
Basically, I am trying to find the absolute extrema for this problem, but I am stuck after a few steps.

$$y=\frac{4}{x}+\tan\left(\frac{\pi x}{8}\right),\quad x\in[1,2].$$

I have figured out the derivative, which is

$$y'=-\frac{4}{x^2}+\sec^2\left(\frac{\pi x}{8}\right)\cdot\frac{\pi}{8}.$$

However, after reaching this step, I am stuck. I have tried setting the derivative equal to zero, but I do not get anywhere as I don't know what to do next. How can I solve this problem? Any help is appreciated. Thanks.

答:
You can use the inequality $\quad\cos(x)\ge 1-\dfrac{x^2}{2}$

Once factored $y'(x)=\dfrac{\pi x^2-32\cos(\frac{\pi x}8)^2}{8x^2\cos(\frac{\pi x}8)^2}$, and since the denominator is always positive in $[1,2]$, we can focus on the numerator.

We have $N(x)\le \pi x^2-32(1-\frac 12(\frac{\pi x}8)^2)^2=\pi x^2-32+\frac 12\pi^2 x^2-\frac 1{512}\pi^4 x^4$

This is a quadratic in $u=x^2$ and it can be solved classically (though a bit ugly...), in particular $N(x)\le 0$ between the roots $x\approx \pm 2.103$ which cover the interval of interest $[1,2]$.

So $y'<0$ on $[1,2]$ so the minimum is $y(2)$.


问:
Congruences on the lattice $N_5$.
I am studying Universal Algebra using the book of Clifford Bergman and I encounter trouble in one of the exercises.

Task (2.12 - 3.)  Let $L$ be a copy of $N_5$ with elements $\{a,b,c,u,v\}$ and $u<a<b<v$. Let $\theta$ be a congruence on $L$. Prove that $\theta\neq O_L$ implies $(a,b)\in\theta$.

Notation Here $N_5$ is the pentagon-shaped five element lattice, where $v$ serves as 1, $u$ as 0 and we have two linear sublattices $u<a<b<v$ and  $u<c<v$. Other then this, elements are uncomparable.

By $0_L$ we mean trivial congruence where each block of equivalence is a singleton.

My approach So one can understand congruences just as partitioning elements of lattice into subsets (classes of equivalence) such that if I take two elements from one class of elements then both their meet and their join still lie in a given class of equivalence.

But if I take partition $\theta$ as $\{a\}$ and $\{b,c,u,v\}$ clearly $(a,b)\notin \theta$ but it seems to me that this is indeed a congruence.

I tried taking couples of elements, I tried thinking about it but it seems to me that if I take $x,y\in \{b,c,u,v\}$ then both their meet and join cannot be $a$

So I am clearly missing something here. Can anyone point out my mistake? Why $\theta$ is not congruence?

答:
Consider the interval $[u,b]$ in $L$. Writing $x\equiv y$ for $\langle x,y\rangle\in\theta$, we have

$$u=u\land a\equiv b\land a=a\,,$$

since $u\equiv b$ and $a\equiv a$. But then $u\equiv a$, contradicting the definition of $\theta$. (The same argument shows that in general lattice congruence classes are convex.)

Title: Congruences on the lattice $N_5$.


问:
In a $C^*$ algebra why $\|a\|\not =\rho(a)$ for any $a$?
In a $C^*$ algebra why $\|a\|\not =\rho(a)$ for any $a$? Where $\rho(a)$ is the spectral radius.

It can be shown that the equality holds for self-adjoint elements. Then that can be used to show that the Gelfand transform is an isometry. Thus $\|a\|=\|\hat{a}\|=max \{|\phi(a)| \hspace{0.2cm}| \phi\in Spec(A)
\}$. First equality follows from the Gelfand transformation being an isometry. The second follows form the definition of the norm on $C(Spec(A))$. But we know that $spec(a)=\{\phi(a) \hspace{0.2cm}| \phi\in Spec(A)
\}$. thus $\rho(a)=\rho(\hat{a})$ and the result follows. My professor said this is incorrect. What I am doing wrong here?

Edit:

I am assuming the algebra is commutative

答:
I think your proof is correct and your professor is wrong.

A faster proof is using the full power of Gelfand transformation. Indeed, without loss of generality we may assume that $A= C(X)$ where $X$ is a compact Hausdorff space. Then if $f \in C(X)$, then $\operatorname{spec}(f) = f(X)$ and thus
$$\rho(f) =\max_x |f(x)|  = \|f\|_\infty.$$

For non-commutative $C^*$-algebras, the statement is false. Consider
$$A:=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}\in M_2(\mathbb{C}).$$
Then $A^2=0$ and thus $\operatorname{spec}(A) = \{0\}$ but $\|A \| \neq 0$.


问:
Differential equations and linear algebra way
I am a beginner at differential equations. Never learned this before but do know calculus.

I am reading Section 6.3 (Systems of Differential Equations) in Introduction to Linear Algebra by Gilbert Strang.

At the beginning of the section:

So, e^ut is equal e^lambdatx. I don't understand this part.

So, here are what I understand:

$$
\frac{\mathrm{d}u }{\mathrm{d} t} = u(t) = {e^{ut}}
$$

And in the linear algebra:

It turns out,

$$
u(t) = e^{\lambda t}x
$$

Can anyone help to explain or let me know the key word to find the relation?

答:
If $Ax=\lambda x$, and $u(t)=e^{\lambda t}x$, then direct substitution tells you that
$$
\frac{du}{dt}=\frac{d}{dt}e^{\lambda t}x=\lambda e^{\lambda t}x=e^{\lambda t}(\lambda x)=e^{\lambda t}(Ax)=A(e^{\lambda t}x)=Au
$$

Therefore, $u(t)=e^{\lambda t}x$ solves that differential equation
$$
\frac{du}{dt}=Au.
$$


问:
Showing that the set $\mathbb{R^2} \setminus \{(n,0) \mid n \in \mathbb{N} \}$ is an open subset of the plane.

For every $n \in \mathbb{N}$ denote $x_n=(n,0) \in \mathbb{R^2}.$ Show that the set $\mathbb{R^2} \setminus \{x_n \mid n \in \mathbb{N} \}$ is an open subset of the plane.

The set $\mathbb{R^2} \setminus \{x_n \mid n \in \mathbb{N} \}$ is the plane excluding the positive $x-$axis? It seems that I cannot use the definition of a ball here to conclude that the set would be open. Other definition that I know states that the union of open sets is open also, but it doesn't seem applicable here also. What other definitions might I use here?

答:
First of all, the most direct way to proceed in these situations is exactly what hamam_Abdallah proposed in his answer. Recall that closed sets are by definition complements of open sets, so if you manage to do that, it will immediately follow that $\mathbb{R}^2 \setminus (x_n)_{n \in \mathbb{N}}$ is open. Indeed, let $(a_k)_{k \in \mathbb{N}} \subseteq \{x_n: n \in \mathbb{N}\}$ be a converging sequence and denote its limit by $\lambda.$ Since convergence in the topology of $\mathbb{R}^2$ is the same as convergence on components, we deduce that $\lambda = (\mu, 0)$ for some $\mu \in \mathbb{R}.$ You should be familiar with the fact that a converging sequence with elements in $\mathbb{Z}$ is constant from a certain point on, but for the sake of completeness, I will provide a proof of this fact here. Note that this would complete our endeavour since it would follow that $\mu = n$ for some $n \in \mathbb{N},$ which would enable us to conclude that $(x_n)_{n \in \mathbb{N}}$ is closed (or, equivalently, that its complement is open). Thus, let $(y_n)_{n \in \mathbb{N}} \subseteq \mathbb{Z}$ be a converging sequence and let $\lambda$ be its limit. By the definition of limit, this means that there is $n_0 \in \mathbb{N}$ such that $y_n \in (\lambda - \frac{1}{2}, \lambda + \frac{1}{2})$ for any $n \geq n_0.$ But any such open interval of length $1$ contains at most one integer, so this interval contains exactly one integer $k$ and that integer must in fact be equal to $\lambda$ seeing as $y_n = k$ for all $n \geq 0.$ Thus, we proved our claim.

However, there is another more direct way to approach this problem (i.e. directly from  the definition of what being an open set means). Let $x \in \mathbb{R}^2 \setminus (x_n)_{n \in \mathbb{N}}$ and write $x = (x_1, x_2).$ Let $f := \lfloor x_1 \rfloor$ and $r := \frac{1}{2} \min \{d(x, (f, 0)), d(x, (f+1, 0))\},$ where $d$ stands for the standard euclidian distance. Then we infer that the ball $B(x, r)$ is included in $\mathbb{R}^2 \setminus (x_n)_{n \in \mathbb{N}}.$ Thus, the set $\mathbb{R}^2 \setminus (x_n)_{n \in \mathbb{N}}$ is open.


问:
longest path two nodes in common
Let $G$ be a connected graph. Prove that two longest paths in $G$ have atleast one node in common. Note that two longest paths do not neccessarily have the same length.

I began by defining two paths namely $P_1 = <v_1,...,v_k>$ and $P_2 =<u_1,...,u_m>$ where $m\neq k$ because of the note given. Now I state the definition of a conneted graph.

A graph G is connected if for each pair of vertices u; v, there is a u; v-walk
  in G.

now my first intuition is to consider the fact that if no node was shared than connecting the two paths would give a (slightly ?) longer path which would make them not the largest paths. Given the fact that they are the two largest (say $n$ and $n-1$). They should be connected via atleast one node.

HOWEVER it seems odd for me to now state that this is a proof.

答:
Let $P_1 = \langle v_0, \dots, v_m \rangle$ and $P_2 = \langle u_0, \dots, u_n \rangle$, such that $P_1$ and $P_2$ have no common vertices, be two different longest inextensible paths in a graph. Let us assume that $m \ge n$. Furthermore, let $v_k$ and $u_l$ be connected (in the original graph) be a path $\langle v_k, w_1, \dots, w_p, u_l \rangle$ such that neither of $w_i$ belong to either $P_1$ or $P_2$ (the length of such path is, obviously, $p+1$).

Without the loss of generality, we can assume that $k \ge m/2$ and $l \ge n/2$. This is just a matter of notation: we're labeling vertices in a way that $\langle v_0, \dots, v_k \rangle$ is longer than or equally long as $\langle v_k, \dots, v_m \rangle$ and that $\langle u_1, \dots, u_l \rangle$ is longer than or equally long as $\langle u_l, \dots, u_n \rangle$.

We construct a new path
$$P := \langle v_0, \dots, v_k, w_1, \dots, w_p, u_l, \dots u_0 \rangle$$
which has length
$$k + (p + 1) + l \ge m/2 + p + 1 + n/2 \ge n/2 + p + 1 + n/2 = n + p + 1 > n,   $$
so $P$ is strictly longer than $P_2$ (and obviously different from $P_1$), which is a contradiction with the assumption that $P_2$ is the second longest inextensible path in the graph.

Title: longest path two nodes in common


答:
To continue after your last step . Simplifying after setting the derivative to zero,

$$ \dfrac{x}{2}=\cos \sqrt{\dfrac{\pi x}{8}}$$

Can be solved by iterative methods numerically, but no closed form.

Title: How to calculate the absolute extrema for this equation

Body:
Basically, I am trying to find the absolute extrema for this problem, but I am stuck after a few steps.

$$y=\frac{4}{x}+\tan\left(\frac{\pi x}{8}\right),\quad x\in[1,2].$$

I have figured out the derivative, which is

$$y'=-\frac{4}{x^2}+\sec^2\left(\frac{\pi x}{8}\right)\cdot\frac{\pi}{8}.$$

However, after reaching this step, I am stuck. I have tried setting the derivative equal to zero, but I do not get anywhere as I don't know what to do next. How can I solve this problem? Any help is appreciated. Thanks.


问:
Prove $\int_a^bf(x+x_0)\cos{\alpha x}dx=\int_a^bf(x)\cos{(\alpha(x-x_0))dx}$
I want to prove that $\int_a^bf(x+x_0)\cos{\alpha x}dx=\int_a^bf(x)\cos{(\alpha(x-x_0))dx}$.

To do this, I used the substitution $u=x+x_0$, and so $x=u-x_0$, and also $\frac{du}{dx}=1$. Is this enough to prove the result? Or is it wrong?

答:
You have forgotten to change the integration bounds:

$\int_a^bf(x+x_0)\cos{\alpha x}dx=\int_{a+x_0}^{b + x_0}f(x)\cos{(\alpha(x-x_0))dx} \tag{1}$

Considering this, it is easy to give a counter-example to your equality if $x_0 \neq 0$.


问:
Continuous version of Kingman's subadditive ergodic theorem.
Let $T$ be a measure-preserving transformation on the probability space $(\Omega, \mathcal{A},\mathbb{P})$.

Kingsman's subadditive ergodic theorem states the following :

Let $(g_n)_n$ a sequence of random variables in $L^1(\Omega)$ such that

$$g_{n+m}(\omega) \leq g_n(\omega) + g_m(T^n \omega), \quad \forall n,m>0$$
for almost every $w \in \Omega$. Then there exists $g : \Omega \rightarrow \mathbb{R}$ a $T$-invariant function such that

$$\underset{n \rightarrow + \infty}{\lim} \frac{g_n(\omega) }{n} = g(\omega)$$
for almost every $w \in \Omega$. Moreover, if $T$ verify the ergodic property (every $T$-invariant function is constant almost everywhere on $\Omega$), then $g$ is a constant function.

I am looking for a continuous version of this theorem.

I would like a theorem that works on the following framework :

Let $(\Omega, \mathcal{A}, \mathbb{P})$ a probability space equipped with a group of transformation $(\tau_x)_{x \in \mathbb{R}^d}$ such that $(\Omega, \mathcal{A}, \mathbb{P},(\tau_x)_{x \in \mathbb{R}^d} )$ forms a $d$-dimension dynamic system. Let $(w \mapsto g(x,\omega))_{x \in \mathbb{R}^d}$ a family of random variables in $L^1(\Omega)$ indexed by $x \in \mathbb{R}^d$ and assume that

$$g(x+z,\omega) \leq g(x,\omega) + g(x,\tau_z \omega), \quad \forall x,z \in \mathbb{R}^d$$
and for a.e $w \in \Omega$.

Is it true that we can find $h$ a $\tau$-invariant function (i.e $h(\tau_x \omega)=\omega, \ \forall x \in \mathbb{R}^d, \ a.e \ \omega \in \Omega$) such that

$$\underset{R \rightarrow + \infty}{\lim} \frac{1}{|B_R|} \int_{B_R} g(x,\omega) \ \mathrm{d}x = h(\omega)$$
for almost $w \in \Omega$, where $B_R$ stands for the ball in $\mathbb{R}^d$ of radius $R$ and center $0$.

Moreover, if I assume the following :

the field $g : \mathbb{R}^d \times \Omega$ is stationary (i.e there exists a random variable $\tilde{g} \in L^1(\Omega)$ such that $g(x,\omega)=\tilde{g}(\tau_x \omega)$ for all $x \in \mathbb{R}^d$ and almost all $w \in \Omega$)

the system $(\Omega, \mathcal{A}, \mathbb{P},(\tau_x)_{x \in \mathbb{R}^d} )$ is ergodic

can I say that $h$ is a constant function (yes thanks to ergodicity) and do we have $h=\mathbb{E}[\tilde{g}]$ ?

答:
$d$-dimensional, continuous ergodic theorem

If $g$ is stationary and integrable, then a continuous version of the ergodic theorem implies that
\begin{equation*}
\lim_{R \to \infty} \frac{1}{|B_{R}|} \int_{B_{R}} g(x,\omega) \, dx = \tilde{g}(\omega),
\end{equation*}
where $\tilde{g}$ is a $\{\tau_{x}\}_{x \in \mathbb{R}^{d}}$-invariant random variable (i.e. $\tilde{g}(\tau_{x}\omega) = \tilde{g}(\omega)$).  When the group action is ergodic, $\tilde{g} = \mathbb{E}(g)$.  Note that none of this requires the sub-additivity property you postulated concerning $g$.

On the contrary, the way to see the above is to define a function $A(\cdot,\omega)$ on measurable sets by $A(E,\omega) = \int_{E} g(x,\omega) \, dx$.  This has the following additivity property:
\begin{equation*}
A(E,\omega) = \sum_{i = 1}^{N} A(E_{i},\omega) \quad \text{if} \, \, E = \bigcup_{i = 1}^{N} E_{i}, \, \, E_{i} \cap E_{j} = \phi.
\end{equation*}
Further, it is stationary and integrable in the sense that
\begin{equation*}
A(E + x, \omega) = A(E,\tau_{x}\omega), \quad \mathbb{E}(|A(B_{R},\omega)|) \leq C |B_{R}|.
\end{equation*}
Using these three properties, a continuous version of the ergodic theorem implies that
\begin{equation*}
\lim_{R \to \infty} \frac{1}{|B_{R}|} A(B_{R},\omega) = \tilde{g}(\omega).
\end{equation*}

$d$-dimensional, continuous sub-additive ergodic theorem

As in the case of space averages, the $d$-dimensional version of sub-additivity I will mention uses sub-additivity over sets instead of points.

Let's go back a step and simply consider the $d$-dimensional sub-additive lemma.  Here is a typical example: pick a constant $F$ and define the electrostatistic energy $E_{F}(U)$ in a domain $U \subseteq \mathbb{R}^{d}$ by
\begin{equation*}
E_{F}(U) = \inf \left\{ \int_{U} \left(\frac{1}{2} \|Du(x)\|^{2} + F u(x) \right) \, dx \, \mid \, u \in C^{\infty}_{c}(U) \right\}.
\end{equation*}
If $U = \cup_{i = 1}^{N} U_{i}$ and $\{U_{1},\dots,U_{N}\}$ is pairwise disjoint, then
\begin{equation*}
E_{F}(U) \leq \sum_{i = 1}^{N} E_{F}(U_{i}).
\end{equation*}
Working directly with cubes and mimicking the proof of Fekete's sub-additive lemma, one can show that there is a constant $\epsilon_{F} \geq 0$ such that
\begin{equation*}
\epsilon_{F} = \lim_{R \to \infty} \frac{1}{|Q_{R}|} E_{F}(Q_{R}) = \inf \left\{ \frac{E_{F}(Q_{R})}{|Q_{R}|} \, \mid \, R > 0 \right\}.
\end{equation*}
(Here $Q_{R}$ is the ball centered at $0$ with radius $R$ with respect to the $\ell^{\infty}$-norm, i.e. a cube centered at zero with sidelength $2R$.
Actually, the Euclidean norm would give the same result, which is another good exercise.)

[Actually, one can show that $E_{F}(Q_{R}) = \epsilon_{F} R^{d}$, but that's not the point!]

The same thing is true in a random setting.  Let $f : \mathbb{R}^{d} \times \Omega \to [0,\infty)$ be a stationary random field satisfying $\mathbb{E}\left(\int_{B(0,1)} |f(x)| \, dx\right) < \infty$.  Imagine $f$ captures the microscopic properties of some material, which, in this case, vary from one place to another.  This time, let's define the electrostatic energy $E(U,\omega)$ by
\begin{equation*}
E(U,\omega) = \inf \left\{ \int_{U} \left(\frac{1}{2} \|Du(x)\|^{2} - f(x) u(x) \right) \, dx \, \mid \, u \in C^{\infty}_{c}(U) \right\}.
\end{equation*}

Notice that $E(,\omega)$ has the following sub-additivity property: if $U = \sum_{i = 1}^{N} U_{i}$, then
\begin{equation*}
E(U,\omega) \leq \sum_{i = 1}^{N} E(U_{i},\omega).
\end{equation*}
Additionally, it is stationary in the sense that
\begin{equation*}
E(U + x, \omega) = E(U,\tau_{x} \omega).
\end{equation*}
One can show that there is a $\tau$-invariant random variable $\tilde{\varepsilon}$ such that
\begin{equation*}
\tilde{\varepsilon} = \lim_{R \to \infty} R^{-d} E(Q(0,R),\omega),
\end{equation*}
where $Q(0,R)$ is the ball centered at $0$ with radius $R$ with respect to the $L^{\infty}$-norm.  (One could replace $Q(0,R)$ by $B(0,R)$ with respect to any norm, in fact.)  If you want, $\tilde{\varepsilon}$ is the macroscopic electrostatic energy density of the material.

When everything is ergodic, the density is constant and is given by
\begin{equation*}
\tilde{\epsilon} = \inf \left\{ \frac{\mathbb{E}(E(Q_{R}))}{|Q_{R}|} \, \mid \, R > 0\right\}.
\end{equation*}

Reference:

The standard reference for this is Krengel's book Ergodic Theorems.


问:
Exponential distribution - Calculate the probability
Let $X$ be exponential distributed with parameter $\lambda>0$, i.e. it holds that $P[X\leq x]=1-e^{-\lambda x}$.

Calculate $P[X\cdot E[X]\geq 1]$.

$$$$

It holds that $E[X]=\frac{1}{\lambda}$, right?

So we get \begin{align*}P[X\cdot E[X]\geq 1]&=P\left [X\cdot \frac{1}{\lambda}\geq 1\right ]=P\left [X\geq \lambda \right ]\\ & =1-P[X<\lambda ]=1-P[X\leq\lambda ]+P[X=\lambda ] \\ & =1-\left (1-e^{-\lambda \cdot \frac{1}{\lambda}}\right )+\frac{\lambda^{\lambda}}{e^{-\lambda}}\cdot \lambda!\\ & =e^{-1}+\frac{\lambda^{\lambda}}{e^{-\lambda}}\cdot \lambda!\end{align*}
Is that correct?

答:
It is correct that $E[X]= \frac{1}{\lambda}$. But the equation
$$P[X\geq \lambda] = 1-P[X \leq\lambda] + P[X=\lambda]$$
is not correct since for a continuos distribution the term $P[X=\lambda]$ is not really defined (as A rural reader pointed out). Because of the definition of the distribution function with the help of an integral, namely
$$P[X \leq x] := \int_{0}^x \lambda e^{-\lambda t} dt$$
for $x \geq 0$ and $P[X \leq x] = 0$ for $x \leq 0$ one can see that
$$P[X \geq \lambda] = 1- \int_0^{\lambda} \lambda e^{-\lambda t} dt = 1- (1-e^{-\lambda^2}) = e^{-\lambda^2}$$


问:
$x^3+y^3-3xy+C=0$ is a smooth curve for $C\neq0$ or $1$
Show that $x^3+y^3-3xy+C=0$ describes a smooth curve for $C\neq0$ and $C\neq1$.

For $C=0$, the curve self-intersects at the origin, so cannot be smooth. For $C=1$, the curve contains an isolated point at $(x,y)=(1,1)$. However, I'm not sure how to show that the curve is smooth for all other values of $C$.

I guess I'd be done if I could find some smooth parametrisation for the curve (that is injective and has non-zero derivative).

答:
The singular points are characterized by the vanishing of both partial derivatives of the polynomial describing the curve. Since
$$
\frac{\partial f}{\partial x}=3x^2-3y,\qquad \frac{\partial f}{\partial y}=3y^2-3x
$$
they can both vanish only at a point where $x^4=x$.

For $x=0$, we have also $y=0$, which is not on the cubic unless $C=0$.

For $x=1$, we have $y=1$, which is not on the cubic unless $C=1$.

For $x=\omega$ (a complex cube root of $1$), we have $y=\omega^2$ and
$$
\omega^3+\omega^6-3\omega\omega^2+C=0
$$
if and only if $C=1$. Similarly for $x=\omega^2$ and $y=\omega^4=\omega$.

What about points at infinity? The homogeneous equation is $F(x,y,z)=x^3+y^3-3xyz+Cz^3$ and the partial derivatives are
$$
\frac{\partial F}{\partial x}=3x^2-3yz,\qquad
\frac{\partial F}{\partial y}=3y^2-3yz,\qquad
\frac{\partial F}{\partial z}=-3xy+3Cz^2
$$
For $z=0$, we get $x=y=0$. So the improper points are not singular for any $C$.


问:
Complex quintic equation
Given the equation $x^5=i$, I need to show by both algebraic and trigonometrical approaches that
$$\cos18^{\circ}=\frac{\sqrt{5+2\sqrt5}}{\sqrt[5]{176+80\sqrt5}}$$
$$\sin18^{\circ}=\dfrac1{\sqrt[5]{176+80\sqrt5}}$$
Trying by trigonometric approach,
$x^5$ = i     $\;\;\;\;$ -- eqn. (a)

=> x = $i\sin(\dfrac{\pi}{2} +2k\pi)$ => $i\sin(\pi\dfrac{4k + 1}{2}) $

Taking the value of k=0, for getting the principal root of 18$^{\circ}$, have x = $i\sin(\dfrac{\pi}{10}) $

Solving algebraically, the solution approach is : $(a+bi)^5$ = i $\;\;\;\;$ -- eqn. (b)

=> $a^5 + 5ia^4b -10a^3b^2 -10ia^2b^3 +5ab^4 +ib^5$

Separating the real & imaginary parts:

$a^5 -10a^3b^2 +5ab^4=0$$\;\;\;$ -- eqn. (c); $\;\;\;\;$$5a^4b -10ia^2b^3+b^5=1$$\;\;\;$ -- eqn. (d)

Solving (c), we have : $a(a^4 -10a^2b^2 +5b^4)=0$$\;\;\;$ -- eqn. (c);

Either $a$ = $0$, or $(a^4 -10a^2b^2 +5b^4)=0$$\;\;$ -- eqn. (c'),

dividing both sides by $b^4$, and having c = a/b,  $(c^4 -10c^2 +5)=0$$\;\;$ -- eqn. (c''),

having d = $c^2$, get : $(d^2 -10d +5)=0$$\;\;$ -- eqn. (c'''), with factors as : d =$5\pm 2\sqrt5$

finding value of c for the two values, get square roots of the two values for d.

//Unable to proceed any further with (c''').

Only root of significance, from eqn. (c) is $a = 0$.

Taking eqn.(d), and substituting $a = 0$, we get:$\;\;\;b^5$=1 => $b =1$

//Unable to prove any of the two values for $\sin18^{\circ}$, or $\cos18^{\circ}$

答:
This problem seems a bit wonky to me because there are simpler expressions for these functions and easier ways to prove them, but let's roll with it.

Once you render $c^2=5\pm2\sqrt5$, take its square roots to get four roots

$c=a/b=\pm\sqrt{5\pm2\sqrt5}$

with the $\pm$ signs independent.  Now from DeMoivre's Theorem we know that the complex root we are seeking is $\cos18°+i\sin18°$, so $a/b$ must be the cotangent of $18°$, therefore greater than $1$. Only by choosing both $\pm$ signs as positive can we get that result; $\sqrt{5-2\sqrt5}$ is too small with $2\sqrt5=\sqrt{20}$ lying between $4$ and $5$.

So, $a=(\sqrt{5+2\sqrt5})b$.  Plug that into the imaginary part of $(a+bi)^5$ you computed above and equate this to $1$, which gives $1/b^5=176+80\sqrt5$.  This will lead easily to the final claims.

Title: Complex quintic equation


问:
Counting problem question
"How many 3 digit integers are possible, such that none of the digits appear more than twice, and none of the digits are equal to zero?"  At first glance the problem seemed easy enough:  you have 9 possible digits and 3 slots, with no two digits appearing more than twice. I assumed $9 \times 9 \times 8 = 648$ would give the solution. To my surprise, my answer was not correct. The correct answer is 720. The way they solved it was by taking all possibilities $9 \times 9 \times 9 - 9$.  (Here $-9$ expresses the break on the constraint.)  I understand what they did, but my question is why does $9 \times 9 \times 8$ fail to account for missing possibilities? I cannot wrap my head around it, $9 \times 9$ would account for any 2 digits and allowing repetition and you would have 8 more possibilities. Whats wrong with my logic?

答:
The total number of digits from $000$ to $999$ is 1000.

Remove all the "numbers" starting with the digit $0$ e.g. $014$: there are 100 of these.

Also remove all the numbers with only $0$ as either the second of third digit. There are $9 \times 19 = $171 of these. [I suppose in this part I did use inclusion-exclusion].

Finally, remove $111,222,...,888,999.$ There are 9 of these.

These three removals are mutually exclusive (therefore we don't need to worry about inclusion-exclusion), so the answer is $1000 - 100 - 171 - 9 = $720.

Title: Counting problem question


答:
Consider the function $\mathbb R^2 \to \mathbb R^3$ given by $f(x,y)=(\sin(x \pi),x,y)$. Then $f$ is continuous.

Let $C=\{0\} \times [0,\infty) \times \{0\}$. Then $C$ is closed in $\mathbb R^3$.

Therefore, $f^{-1}(C)=\mathbb N \times \{0\}$ is closed in $\mathbb R^2$ and so its complement is open.

(I assume that $0 \in \mathbb N$. Otherwise, take $[1,\infty)$ instead of $[0,\infty)$.)

Title: Showing that the set $\mathbb{R^2} \setminus \{(n,0) \mid n \in \mathbb{N} \}$ is an open subset of the plane.

Body:
For every $n \in \mathbb{N}$ denote $x_n=(n,0) \in \mathbb{R^2}.$ Show that the set $\mathbb{R^2} \setminus \{x_n \mid n \in \mathbb{N} \}$ is an open subset of the plane.

The set $\mathbb{R^2} \setminus \{x_n \mid n \in \mathbb{N} \}$ is the plane excluding the positive $x-$axis? It seems that I cannot use the definition of a ball here to conclude that the set would be open. Other definition that I know states that the union of open sets is open also, but it doesn't seem applicable here also. What other definitions might I use here?


问:
Calculate $\int_{\mathcal{C}(0,3)} \frac{e^{2/z}}{z-2}\,dz$
$$\int_{\mathcal{C}(0,3)} \frac{e^{2/z}}{z-2}\,dz$$
Solution approach:
I thought of calculating that problem using the residue theorem, therefore I calculated the residue at $z=2$, which turned out to be equal to $e$ (Wolfram confirmed). However, I am not sure if I should calculate the residue at $z=0$, but I am having problem doing that. Thanks for help!

答: Another way is to kick the essential singularity $z=0$ out of the contour. That is, substitute $z=2/w$ (to get rid of these $2$'s). Then $dz=-2\,dw/w^2$, and the "$-$" is compensated by the opposite orientation of the resulting contour $|w|=2/3$; choosing the standard orientation instead, $$\oint_{|z|=3}\frac{e^{2/z}}{z-2}\,dz=\oint_{|w|=2/3}\frac{e^w}{w(1-w)}\,dw.$$ Now the only singularity inside is the simple pole at $w=0$, with the residue clearly equal to $1$.


问:
Under what conditions does $\log a^b=b \log a$ for complex numbers $a,b$

Under what conditions does $\log a^b=b \log a$ for complex numbers $a,b$ Use the branch of $\log$ with $-\pi< \theta <\pi$.
For solve the problem I need understand the meaning of brach, and the definition of $\log z$ for $z\in \mathbb{C}$.

First of all I understand that given $z\in \mathbb{C} \setminus \lbrace 0 \rbrace $ we define $\log z=\log(|z|)+(arg(z)+2 \pi k)i$ for $k\in \mathbb{Z}$

Too I know that if $a,b\in \mathbb{C}$ we define $a^b=e^{b \log a}$ where $a\neq 0$.

Now for give a correct answer I need know the correct interpretation of the  branch of logarithm which I don´t understand.

according to my low skills in complex variable I Imagine that,first I should take the LHS
$$\log a^b=log(e^{b \log a})=b \log a$$ which is true for any $a,b\in \mathbb{C}, a\neq 0$.

Someone can give my a explanation about my mistake with my assumption, give me a step-by step solution or much better give me a intuitive and formally explanation of the meaning of Branch of logarithm

答:
By one definition, $\log$ is an inverse function to $\exp$, i.e. $\log(z)$ should be a
number $w$ with $\exp(w) = z$.  The trouble is, there are infinitely many $w$ that work: $\exp(2 \pi i) = 1$, so if $\exp(w) = z$ then also $\exp(w + 2 \pi i k) = z$ for any integer $k$.  To be able to give a definite meaning to $\log(z)$, we have to make a choice from these infinitely many possibilities.  We can do it in such a way that $\log(z)$ is an analytic function on the complex plane with a curve starting at $0$ and going out to complex infinity removed.  Such a function is called a branch of the logarithm, and the excluded curve is called a branch cut.  If you choose the version of $\log(z)$ with imaginary part in $(-\pi, \pi)$, that's called the principal branch, and its branch cut is the negative real axis.

$a^b$ is defined, for complex numbers, as $\exp(b \log(a))$.  Again, in general that's ambiguous, so there are branches to consider.  The principal branch uses the
principal branch of logarithm.

Now $\log(a^b) = w$ where $\exp(w) = a^b = \exp(b \log(a))$, so $w = b \log(a) + 2 \pi i k$ for some integer $k$.  If you're using the principal branch of $\log$ (both for $\log(a^b)$ and $\log(a)$, the imaginary parts of both $w$ and $\log(a)$ must be in $(-\pi, \pi)$.  You want $k = 0$ according to your formula, so what you need is
that the imaginary part of $b \log(a)$ is between $-\pi$ and $\pi$.


问:
How to find the number of compound divisors of the smallest product from two unknown numbers?
The problem is as follows:

The number of panadol pills at a pharmacy is a positive whole number
that it has two prime divisors and 45 positive divisors. The number of
tylenol pills at the same pharmacy is also a positive whole number
which has 66 positive divisors. Assuming that both numbers which
represents panadol and tylenol pills have the same prime divisors. How
many compount divisors will it have the smallest value of the product
of both numbers?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{210}\\
2.&\textrm{214}\\
3.&\textrm{187}\\
4.&\textrm{207}\\
\end{array}$

The official solution according to my book is to assume the following:

It states:

$\textrm{A: The number of panadol pills}$
$\textrm{B: the number of tylenol pills}$

$A=p^a q^b$
$B=p^c q^d$

According to this in order to get the minimum product of pills then $A\times B$ must be minimum.

For this to happen:

$\textrm{p<q and a>b and c>d}$

(Comment: This part it doesn't seem much sense)

Then it states: Using the notation of $\textrm{QD = quantity of divisors}$

$\operatorname{QD}(A)=(a+1)(b+1)=45$

$\operatorname{QD}(B)=(c+1)(d+1)=66$

From this it concludes that:

$a=9$

$b=4$

Assuming that the combination can only be: $9\times 5=45$

It doesn't say why it discards other choices.

and for the other number it assumes:

$22\times 3 =66$

Therefore

$c=21$

$d=2$

Again it doesn't say why it does have to be those numbers and in that order.

Then it declares:

$A\times B = p^8 \cdot q^4 \times p^{21}\cdot q^2= p^{29} \cdot q^6$

Now it comes the most weird conclusion:

$\operatorname{QD}(A\times B)= 19\times 10=190$

Hence:

It uses the formula:

$\operatorname{QD}=\textrm{prime divisors}+1+\textrm{compount divisors}$

I believe this arises from the fact that the number of divisors is equal to the number of prime divisors plus one plus the compound divisors.

Then:

$190=2+1+\textrm{compount divisors}$

Therefore it concludes:

$\textrm{compound divisors}=187$

Hence the answer is the third option.

But this doesn't make any sense. I mean at the point I reached this part I got lost.

As it can be seen I don't feel happy with this method of solution because it lacks of the proper explanation on why it must be those peculiar divisors it could be any other combination as well. This confuses me a lot.

As I mentioned it isn't very clear why that must be the quantity of divisors and that the smallest number of pills.

Does it exist a way to solve this using those concepts but better explained?

I'm lost at this ans since I feel there are many missing steps it would help me a lot if someone could help me here with an answer which would be easier to understand and the most detailed as possible.

答:
First you need to be familiar with the fact that the divisors of $p^aq^b$ are all the numbers of the form $p^xq^y$ with $0\le x\le a,  0\le y\le b$. There are $(a+1)(b+1)$ of these.

So we know that
$$(a+1)(b+1)=45$$ $$(c+1)(d+1)=66.$$

Now $45=1\times 45=3\times 15=5\times 9=9\times 5=15\times 3=45\times 1$. The possible solutions of the first equation are $$(a,b)\in \{(2,14),(4,8),(8,4),(14,2)\}.$$

By the same method
$$(c,d)\in \{(1,32),(2,21),(5,10),(10,5),(21,2),(33,1)\}.$$

There are of course lots of ways of combining these. You should have a go at trying this out and for each possibility working out the number of divisors of the product of the two numbers, $(a+c+1)(b+d+1)$.

You can see that one possibility is $$(a,b)=(8,4)\text { and } (c,d)=(10,5)$$
This gives a number of divisors $19\times10=190$. Now, three of the divisors are $ 1,p,q$ and these are not composite and so we have the 'book' solution of $187$ composite factors.

Note

I don't think you should expect to be able to 'see' that $(a,b)=(8,4)\text{ and } (c,d)=(10,5)$ is obviously best. However if you try out some of the other possibilities you will get a feel for whether any other combination is likely to be better.

For example

$(a,b)=(2,14)\text { and } (c,d)=(2,21)$ gives $5\times 36=180$, a better answer! However the wording of the question concerning what is 'smallest' might mean that the product itself is smallest (although we don't know the primes). Do you have an image of the actual question?


问:
Finding global max./min.
my task is to figure out the critical points of $f(x,y)=e^y(x^4-x^2+y)$, $\ $$\mathbb{R}^2 \rightarrow \mathbb{R}$, and show which of them is a maximum or minimum. As far as I got, I've shown that the critical points are:

1.: $(0,-1)$ which is neither max. nor min. (char. pol. of Hessian is indefinite)

2.: $\left(\frac{1}{ \sqrt2},-\frac{3}{4}\right)$, which is a local minimum and

3.: $\left(-\frac{1}{ \sqrt2},-\frac{3}{4}\right)$ which is the second local minimum.

Moving towards my question, is there any way to easily show if any global maximum or minimum exists (in general and/or concerning this example)? I’ve used the char. poly. of the Hessian, is there any faster possibility to finding local min./max.?

答:
First find the limit as x and y approach infinity. In single variable functions, you only have to check to "ends", $-\infty$ and $\infty$. In functions of two variables, there are four, $x\to-\infty$, $x\to\infty$, $y\to-\infty$, and $y\to\infty$.

$$\lim_{x\to-\infty}f(x,y)=\infty$$

$$\lim_{x\to\infty}f(x,y)=\infty$$

$$\lim_{y\to-\infty}f(x,y)=-\lim_{y\to\infty}\frac{e^{-y}}{\frac{1}{y}}=0$$

$$\lim_{y\to\infty}f(x,y)=\infty$$

Evaluate $f(x,y)$ at the critical points you calculated and compare them to these values.

Title: Finding global max./min.


问:
Checking an "equivalence of products" in the notes on deformation quantization by Allen C. Hirshfeld and Peter Henselder
These notes are a great introduction to deformation quantization but I failed to check the validity of the statement p.9, right before (5.18).

Context: let $(\mathcal{A},+,\mu)$ be an algebra. $\mu:\mathcal{A}\times \mathcal{A} \to \mathcal{A}$ standing for multiplication. Deformation consists in considering a family (paramatrized by $\nu$ in a yet to be chosen space) of product on $\mathcal{A}[[\nu]]$ (formal power series with coefficients in $\mathcal{A}$) generically given by
$$ \forall\ f,g\in \mathcal{A},\quad  \mu_{\nu}(f,g) := \mu(f,g) + \sum_{k=1}^{+\infty} \nu^k \mu_k (f,g) \label{1}\tag{1}$$
i.e. by a family of bilinear maps $\mu_k:\mathcal{A}\times \mathcal{A} \to \mathcal{A}$ satisfying some conditions and extended to elements $F, G\in \mathcal{A}[[\nu]]$ of the form $F=\sum_{k=1}^{+\infty} \nu^k f_{k},\ f_k \in \mathcal{A}$ by $\mathbb{K}[[\nu]]$-bilinearity. (the idea behind formal power series, as far as I understand, is to ignore convergence issues but still have a structure where one can compare terms of the same degree in $\nu$").

Two of these star-product are equivalent if there exists an invertible algebra isomorphism (transition map) $T:(\mathcal{A}[[\nu]],+,\mu_{\nu}) \longrightarrow (\mathcal{A}[[\nu]],+,\rho_{\nu})$, i.e. a map such that
$$ \forall\ F,G \in \mathcal{A}[[\nu]], \quad T\big(\mu_{\nu}(F,G)\big)= \rho_{\nu}(T(F),T(G))$$

Question: Let $\mathcal{A}=\mathcal{C}^{\infty}(\mathbb{R}^2)$ and denote $(a, \overline{a})$ or $(b, \overline{b}$) the variables of the functions. I want to check that the normal product ( (5.4) p.8; with the more usual notation for products)
$$f \ast_N g := \sum_{k=0}^{+\infty} \frac{\hbar^k}{k!} \frac{\partial^k f}{\partial a^k} \frac{\partial^k g}{\partial \overline{a}^k} = f\, e^{\hbar \overleftarrow{\partial}_a \overrightarrow{\partial}_{\overline{a}}}\, g \label{2}\tag{2}$$
is equivalent to the Moyal product ((5.15) p.9, one can consider $\hbar$ as the deformation parameter... although there is usually the factor as in (\ref{5}))
$$ f \ast_M g := \sum_{k=0}^{+\infty} \left(\frac{\hbar}{2} \right)^k \frac{1}{k!}  \left. \left( \frac{\partial }{\partial a} \frac{\partial }{\partial \overline{b}} - \frac{\partial }{\partial \overline{a}} \frac{\partial }{\partial b}\right)^k f\big(a, \overline{a}\big) g\big(b, \overline{b}\big)  \right|_{\genfrac{}{}{0pt}{1}{a=b}{\overline{a}=\overline{b}}} = f\, e^{\frac{\hbar}{2}\big( \overleftarrow{\partial}_a \overrightarrow{\partial}_{\overline{a}} - \overleftarrow{\partial}_{\overline{a}} \overrightarrow{\partial}_{a} \big)}\, g \label{3}\tag{3}$$
i.e.
$$ T\big(f \ast_N g \big) = T(f) \ast_M T(g)\quad \text{with}\quad T = \genfrac{}{}{0pt}{0}{"}{}\!\!\exp\left(-\frac{\hbar}{2} \frac{\partial^2}{\partial a\, \partial \overline{a}} \right)\genfrac{}{}{0pt}{0}{"}{} \label{4}\tag{4}$$

Remarks:

In fact I already checked (\ref{4}) up to second order in $\hbar$ but it did not work at order 3 (although I'm not sure as the calculations were quite tedious...). It was not a priori clear that (\ref{4}) hold, one could have the other way round $ T\big(f \ast_M g \big) = T(f) \ast_N T(g)$ instead but this seems to fail at order 1. I only want to check the first few orders, but I would gladly take a proof for all order. I will soon write what I have done, but as I mentionned, it's tedious.

The Moyal product is first defined in the text (3.5-3.6) p.5 by
$$ f \ast_M g := \sum_{k=0}^{+\infty} \frac{\nu^k}{k!} \underbrace{\left(\frac{\partial }{\partial q_1} \frac{\partial }{\partial p_{2}} - \frac{\partial }{\partial p_{1}} \frac{\partial }{\partial q_2} \right)^k f(q_1,p_1)\, g(q_2,p_2)}_{\mu_k(f,g)}\left.\vphantom{\frac{T}{T}}\right|_{\genfrac{}{}{0pt}{1}{q_1=q_2}{p_{1}=p_{2}}}\quad \text{with}\quad \nu = \frac{i\hbar}{2} \label{5}\tag{5}$$
and it does coincide with (\ref{3}) via (these are the correct $\sqrt{2}$ factors...)
$$ \left\lbrace \begin{aligned}
 a & := \frac{1}{\sqrt{2}} \left(q  + i\hspace{.5pt} p \right) \\
 \overline{a} & := \frac{1}{\sqrt{2}} \left( q - i\hspace{.5pt} p \right)
 \end{aligned} \right. \enspace \Longrightarrow\quad
 \left\lbrace \begin{aligned}
 \frac{\partial}{\partial\hspace{.7pt} q} & = \frac{\partial\hspace{.7pt} a}{\partial\hspace{.7pt} q} \frac{\partial}{\partial\hspace{.7pt} a} + \frac{\partial\hspace{.7pt} \overline{a}}{\partial \hspace{.7pt} q} \frac{\partial}{\partial\hspace{.7pt} \overline{a}} =\frac{1}{\sqrt{2}} \left( \frac{\partial}{\partial\hspace{.7pt} a} +  \frac{\partial}{\partial\hspace{.7pt} \overline{a}} \right) \\
 \frac{\partial}{\partial\hspace{.7pt} p} & = \frac{\partial\hspace{.7pt} a}{\partial\hspace{.7pt} p} \frac{\partial}{\partial\hspace{.7pt} a} + \frac{\partial\hspace{.7pt} \overline{a}}{\partial \hspace{.7pt} p} \frac{\partial}{\partial\hspace{.7pt} \overline{a}} = \frac{i}{\sqrt{2}} \left( \frac{\partial}{\partial\hspace{.7pt} a} - i\, \frac{\partial}{\partial\hspace{.7pt} \overline{a}}\right)
 \end{aligned} \right. $$

To make (\ref{3}) (same for (\ref{5})) more explicit, let me write the $k=2$ term: (notation $\displaystyle \partial_a=\frac{\partial}{\partial a},\ \partial_{ab}= \frac{\partial^2}{\partial a \partial b}$ etc.)
$$\begin{split}
 \mu_2(f,g) &= \Big(\partial_{aa\overline{b}\overline{b}} - 2 \partial_{a\overline{a}b\overline{b}} + \partial_{\overline{a}\overline{a}bb} \Big) f(a,\overline{a})g(b,\overline{b})\left.\vphantom{\frac{T}{T}}\right|_{\genfrac{}{}{0pt}{1}{a=b}{\overline{a}=\overline{b}}}\\
 &= (\partial_{aa}f)(\partial_{\overline{a}\overline{a}}g) - 2 (\partial_{a\overline{a}}f)(\partial_{b\overline{b}}g) + (\partial_{\overline{a}\overline{a}}f) (\partial_{aa}g)
\end{split} \label{6}\tag{6}$$
One can also use the $\overleftarrow{\partial}$ or $\overrightarrow{\partial}$ notations or a tensorial notation.

答:
It is in fact possible to prove

$$\begin{split} T\big(f \ast_N g \big) = T(f) &\ast_M T(g)\quad \text{with}\quad T = \genfrac{}{}{0pt}{0}{"}{}\!\!\exp\left(-\frac{\hbar}{2} \frac{\partial^2}{\partial_{a}\, \partial_{\overline{a}} } \right)\genfrac{}{}{0pt}{0}{"}{} \\
\Longleftrightarrow \quad e^{-\frac{\hbar}{2} \partial_{ a\, \overline{a}}} \left(f\, e^{\hbar \overleftarrow{\partial}_a \overrightarrow{\partial}_{\overline{a}} }\, g \right) & = \left(e^{-\frac{\hbar}{2} \partial_{ a\, \overline{a}}} f\right)  e^{\frac{\hbar}{2}\big( \overleftarrow{\partial}_a \overrightarrow{\partial}_{\overline{a}} - \overleftarrow{\partial}_{\overline{a}} \overrightarrow{\partial}_{a} \big)} \left(e^{-\frac{\hbar}{2} \partial_{ a\, \overline{a}}} g\right)
\end{split}\tag{4}$$
by brute force. The idea is to reckognize the simplification that occurs for the exponents of exponential functions suggested by Cosmas but in terms of series:
$$\begin{split}
\text{L.h.s.} &= e^{-\frac{\hbar}{2} \partial_{ a\, \overline{a}}} \left(f\, e^{\hbar m s }\, g \right) = e^{-\frac{\hbar}{2} \partial_{ a\, \overline{a}}} \left( e^{\big(ma + n\overline{a} \big) + \hbar m k  + \big(ka + s\overline{a} \big)} \right)\\
& = e^{-\frac{\hbar}{2} (m+k)(n+s)} \left(f\, e^{\hbar m s }\, g \right)\\
&= \left( e^{-\frac{\hbar}{2} m(n+s)} f\right) e^{\hbar m s } \left(\,e^{-\frac{\hbar}{2} k(s+n)} g \right)
\end{split} \label{4L}\tag{4L}$$

$$ \begin{split}
\text{R.h.s.} &= \left(e^{-\frac{\hbar}{2} mn\ + \big(ma + n\overline{a}\big)} \right)  e^{\frac{\hbar}{2}\big( \overleftarrow{\partial}_a \overrightarrow{\partial}_{\overline{a}} - \overleftarrow{\partial}_{\overline{a}} \overrightarrow{\partial}_{a} \big)} \left(e^{-\frac{\hbar}{2} ks\ + \big(ka + s\overline{a}\big)}\right) \\
&= \left(e^{-\frac{\hbar}{2} mn\ + \big(ma + n\overline{a}\big)} \right)  e^{\frac{\hbar}{2}\big( ms - nk \big)} \left(e^{-\frac{\hbar}{2} ks\ + \big(ka + s\overline{a}\big)}\right)
\end{split} \label{4R}\tag{4R}$$
The simplification for the exponents to go from L.h.s. to R.h.s. $-\frac{\hbar}{2} ms + \hbar m s -\frac{\hbar}{2} kn = \frac{\hbar}{2} (ms-nk) $.

Remark: One could probably exploit this argument by decomposing a certain class of functions in terms of exponentials, probably with a Laplace transform in analogy with the case of complex exponentials: a tempered distribution can be written $u =\mathcal{F}^{-1}(\hat{u})=\genfrac{}{}{0pt}{0}{"}{}\!\!\frac{1}{(2\pi)^2}\int_{\mathbb{R}^2}\hat{u}(k,\overline{k})\, e^{ika +i\overline{k}\overline{a}}\, dk\, d\overline{k}\, \genfrac{}{}{0pt}{0}{"}{}$.

Notice that Leibniz' rule can be written: $\partial^j_{a}(fg) =f \left(\overleftarrow{\partial}_a +\overrightarrow{\partial}_a \right)^j g$ and similarly with two variables: $\partial^j_{a\overline{a}}(fg)=f \left(\overleftarrow{\partial}_a +\overrightarrow{\partial}_a \right)^j \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^j g$. Hence
$$\begin{split}
\text{L.h.s.} &=\sum_{j=0}^{+\infty} \frac{1}{j!} \left(-\frac{\hbar}{2} \partial_{ a\, \overline{a}}\right)^j \left(\sum_{k=0}^{+\infty} \frac{\hbar^k}{k!} \partial^k_a f\, \partial^k_{\overline{a}} g\right)\\
&= \sum_{j=0}^{+\infty} \left(-\frac{\hbar}{2}\right)^j \frac{1}{j!} f\left(\overleftarrow{\partial}_a +\overrightarrow{\partial}_a \right)^j \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^j \left(\sum_{k=0}^{+\infty} \frac{\hbar^k}{k!} \overleftarrow{\partial}^k_a \, \overrightarrow{\partial}_{\overline{a}} ^k \right)  g \\
&= f \left( \sum_{j=0}^{+\infty} \left(-\frac{\hbar}{2}\right)^j \frac{1}{j!} \left(\sum_{0\leq i\leq j} {j\choose i} \overleftarrow{\partial}_a^{j-i} \overrightarrow{\partial}_a^i \right) \left(\sum_{k=0}^{+\infty} \frac{\hbar^k}{k!} \overleftarrow{\partial}^k_a \, \overrightarrow{\partial}_{\overline{a}} ^k \right)  \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^j \right) g \\
&= f \left( \sum_{\genfrac{}{}{0pt}{1}{j,k=0}{i+l=j}}^{+\infty} \left(-\frac{\hbar}{2}\right)^i \!\! \frac{1}{i!}  \overleftarrow{\partial}_a^{i}\ \left(-\frac{\hbar}{2}\right)^l\!\! \frac{1}{l!} \overrightarrow{\partial}_a^l \left( \frac{\hbar^k}{k!} \overleftarrow{\partial}^k_a \, \overrightarrow{\partial}_{\overline{a}} ^k \right)  \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^{i+l} \right) g \\
&=f \underbrace{\left( \sum_{i=0}^{+\infty} \left(-\frac{\hbar}{2}\right)^i \!\! \frac{1}{i!}  \overleftarrow{\partial}_a^{i} \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^i \right)}_{"e^{-\frac{\hbar}{2} m(n+s)}"}  \underbrace{\left(\sum_{k=0}^{+\infty} \frac{\hbar^k}{k!} \overleftarrow{\partial}^k_a \, \overrightarrow{\partial}_{\overline{a}} ^k \right)}_{"e^{\hbar ms}"} \underbrace{\left( \sum_{l=0}^{+\infty}  \left(-\frac{\hbar}{2}\right)^l\!\! \frac{1}{l!} \overrightarrow{\partial}_a^l  \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^{l} \right)}_{"e^{-\frac{\hbar}{2} k(s+n)}"} g
\end{split}$$
Now it suffices to check the equivalent of $e^{-\frac{\hbar}{2} m(n+s)} = e^{-\frac{\hbar}{2} mn} e^{-\frac{\hbar}{2} ms},\ e^{-\frac{\hbar}{2} ms} e^{\hbar ms}= e^{\frac{\hbar}{2} ms}$ and $e^{\frac{\hbar}{2} ms} e^{-\frac{\hbar}{2} nk}= e^{\frac{\hbar}{2} (ms-nk)}$. Since they are all similar, let us just write the first one:
\begin{equation}
\begin{split}
& f \left( \sum_{i=0}^{+\infty} \left(-\frac{\hbar}{2}\right)^i \!\! \frac{1}{i!}  \overleftarrow{\partial}_a^{i} \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^i \right) g\\
&= f \left( \sum_{\genfrac{}{}{0pt}{1}{i=0}{0\leq h\leq i}}^{+\infty} \left(-\frac{\hbar}{2}\right)^{i-h+h} \!  \overleftarrow{\partial}_a^{i-h+h} \, \frac{1}{(i-h)!} \overleftarrow{\partial}^{i-h}_{\overline{a}}\, \frac{1}{h!}  \overrightarrow{\partial}_{\overline{a}}^h  \right) g\\
&= f \left( \sum_{\genfrac{}{}{0pt}{1}{i=0}{h +l =i}}^{+\infty} \left(-\frac{\hbar}{2}\right)^{l} \! \frac{1}{l!} \overleftarrow{\partial}^{l}_{a\overline{a}}\ \left(-\frac{\hbar}{2}\right)^{h} \! \frac{1}{h!}  \overleftrightarrow{\partial}^h_{\overline{a}}  \right) g\\
&= f \left( \sum_{l=0}^{+\infty} \left(-\frac{\hbar}{2}\right)^{l} \! \frac{1}{l!} \overleftarrow{\partial}^{l}_{a\overline{a}}\right) \left(\sum_{h=0}^{+\infty} \left(-\frac{\hbar}{2}\right)^{h} \! \frac{1}{h!} \overleftarrow{\partial}^h_a \overrightarrow{\partial}_{\overline{a}}^h  \right) g
\end{split}
\end{equation}

I also thought about the fact that a linear map from certain classes of functions of $\mathbb{R}^n$ (different versions, e.g. $\mathcal{C}^{\infty}_c(\mathbb{R}^n)$) which commutes with translations is a convolution operator (there is also a continuity condition, so one has to specify the topology on the space of functions, and this will constrain the convolution operator: convolution with what kind of function or distribution). Unfortunately, a quick computation shows that it would be a convolution with the Fourier transform of $e^{\frac{\hbar}{2}k^2}$ which does not make sense...
$$ \begin{split}
T(f) &= e^{-\frac{\hbar}{2} \partial_{ a\, \overline{a}}} \frac{1}{(2\pi)^2} \int_ {\mathbb{R}^2} \hat{f}(k,\overline{k})\, e^{ika +i\overline{k}\overline{a}}\, dk\, d\overline{k} = \frac{1}{(2\pi)^2} \int_ {\mathbb{R}^2} \hat{f}(k,\overline{k})\,e^{\frac{\hbar}{2} k\overline{k}}\, e^{ika +i\overline{k}\overline{a}}\, dk\, d\overline{k}
\\
&= \frac{1}{(2\pi)^2} \int_ {\mathbb{R}^2} \left(\int_ {\mathbb{R}^2} f(b,\overline{b})\,e^{-ikb -i\overline{k}\overline{b}} db\, d\overline{b}\right) e^{\frac{\hbar}{2} k\overline{k}}\, e^{ika +i\overline{k}\overline{a}}\, dk\, d\overline{k}\\
&=  \frac{1}{(2\pi)^2} \int_ {\mathbb{R}^2} f(b,\overline{b}) \left(\int_ {\mathbb{R}^2}  e^{\frac{\hbar}{2} k\overline{k}}\, e^{ik(a-b) +i\overline{k}(\overline{a}-\overline{b})}\, dk\, d\overline{k}\right) db\, d\overline{b}
\end{split}$$
Writing things differently also leads to the same result:
$$\begin{split}
T(f) &= \mathcal{F}^{-1}\circ\mathcal{F}\left(\sum_{j=0}^{+\infty} \frac{1}{j!} \left(-\frac{\hbar}{2} \partial_{ a\, \overline{a}}\right)^j f\right) \\
&=\sum_{j=0}^{+\infty} \frac{1}{j!} \left(-\frac{\hbar}{2}\right)^j (ik)^j (i\overline{k})^j f
\end{split}$$

Way out: $T$ seems to be unbounded but one can always imagine that its "inverse" (we didn't define its domain and target space...) could be convolution by a Gaussian. Another possibility would be to take a modified Fourier transform, $ \hat{f}(k,\overline{k}) := \int_ {\mathbb{R}^2} f(a,\overline{a})\, e^{- ika +i\overline{k}(\overline{a}-\overline{b})}\, da\, d\overline{a}$ or as remarked, take a Laplace transform. cf. also Inverse Weierstrass transform


问:
Transform Limit definition of $e$
The limit definition of $e$ can be written as:

$$ \lim_{x \to \infty} {\left(1 + \frac{1}{x}\right)^x} . $$

This is equal to another similar form:

$$ \lim_{x \to 0} {(1 + x)^{\frac{1}{x}}} . $$

I am wondering how to prove these 2 forms are equivalent, any help is appreciated thank you.

答: Yes!  We can start with $$\lim_{x \to \infty} {\left(1 + \frac{1}{x}\right)^x}$$ and use the substitution $$y = \frac{1}{x}$$ and note that $$x\to\infty \implies y\to 0^{+}.$$  This tells us that $$\lim_{x \to \infty} {\left(1 + \frac{1}{x}\right)^x} = \lim_{y\to 0^{+}}\left(1+y\right)^{\frac{1}{y}}.$$  Note now that it doesn't matter what the new variable is called, so we can say $$\lim_{x \to \infty} {\left(1 + \frac{1}{x}\right)^x} = \lim_{x\to 0^{+}}\left(1+x\right)^{\frac{1}{x}}.$$

Title: Transform Limit definition of $e$


问:
ode formation when constant appears with arc length.
How to form ode and solve for

$$ s( 4-s) = y^2+5 $$

where $s$ is arc length of a curve in x-y plane ? The constant 5 disappears when derivating.

答:
Your condition can be written as

$$1+y^2=-(s-2)^2$$

which is impossible.


问:
If A and B are both pathwise-connected, and the intersection of A and B is nonempty, show that A union B is pathwise-connected
My thought process here was to use a theorem in our textbook that says that a subset A of R is pathwise-connected iff A is an interval. Therefore it suffices to show that A union B is an interval. Since A and B are both intervals and their intersection is nonempty, it should follow that A union B is an interval, but I do not know if I need more for this section of the proof to show that AuB is an interval. Any help would be greatly appreciated.</p>

答: HINT: You don’t need intervals: you just need the definition of path. Let $p\in A\cap B$, and let $x,y\in A\cup B$. There is a path $f$ from $x$ to $p$ in $A\cup B$ (why?), and there is a path $g$ from $p$ to $y$ in $A\cup B$ (again, why?). Now show how to combine $f$ and $g$ to get a path from $x$ to $y$ in $A\cup B$.


问:
Collinearity in bicentric pentagon
Can you provide a proof for the following claim:

Claim. The circumcenter, the incenter, and the incenter of the pentagon formed by diagonals in a bicentric pentagon are collinear.

GeoGebra applet that demonstrates this claim can be found here.

My idea is to show that $|OJ|=|OI|+|IJ|$ . By Fuss formula we know that $$r(R-|OI|)=(R+|OI|)\sqrt{(R-r+|OI|)(R-r-|OI|)}+(R+|OI|)\sqrt{2R(R-r-|OI|)}$$

where $R$ and $r$ are circumradius and inradius of bicentric pentagon ,respectively. But how to express lengths $|OJ|$ and $|IJ|$ in terms of $R$ and $r$ ?

答:
The short answer is that the three circles are coaxal, and therefore their centers are collinear.  The circles are coaxal because of the following lemma (screen shot from Johnson, Modern Geometry aka Advanced Euclidean Geometry, pg 92):

Proof later, but you might take a look at another question and answers.

All of this is related to Poncelet's Porism, one general version of which is (excerpted from Hraskó, Poncelet's theorem, well worth reading in full):

Poncelet's General Theorem: Let $e$ be a circle of a non-intersecting pencil and let $a_1,a_2,\ldots,a_n$ be (not necessarily
different) oriented circles in the interior of $e$ that belong to the
same pencil. Starting at an arbitrary point $A_0$ of the circle $e$, the
points $A_1,A_2,\ldots,A_n$ are constructed on the same circle, such that the
lines $A_0A_1, A_1A_2, \ldots, A_{n-1}A_n$ touch the circles $a_1,a_2,\ldots, a_n$,
respectively, in the appropriate direction. It may happen that at the
end of the construction, we get back to the starting point, that is,
$A_n=A_0$. The theorem states that in that case, we will always get back
to the starting point in the $n$-th step, whichever point of $e$ we start
from. We do not even need to take care to draw the tangents to the
circles in a fixed order.

Here's a page from Poncelet's Traité des propriétés projectives des figures, 1827 that illustrates the lemma.

But we don't need the general theorem, other than to observe that a Poncelet Porism will involve (or generate) circles in a coaxal system.

By Poncelet's porism, you can move the pentagon in the OP continuously with its vertices on the circumcircle and sides touching the incircle.  Along for the ride will go triangles like $A_1A_2A_3$ with sides $A_2A_1,A_2A_3$ tangent to the incircle and line $d_2$ which must be tangent to the diagonals incircle (it's shape will vary).  By the lemma, the diagonals incircle must be coaxal with the other two circles.

The proof of the lemma is excerpted here (Johnson, pg 93):

Title: Collinearity in bicentric pentagon


问:
Constructing a point D on AB such that PD is perpendicular to AB
I need to write the steps of construction from this exercise.

Construct a line segment $AB$, and a point $P$, not on $AB$. Use the theorem on the previous page or one of its alternative statements to construct a point $D$ on $AB$ such that $PD ⊥ AB$. Write down your steps of construction, and explain how you used the theorem to do the construction.

This is the previous theorem: In an isosceles triangle, the median to the base is the perpendicular bisector (of the base) and also the angle bisector for the angle opposite to the base.

So far I have learned how to create an equilateral triangle with side $AB$ by using the intersections of circle $(A,AB)$ and circle $(B,BA)$.

I am unsure how to start this. I think I need to somehow create a isosceles triangle that would be triangle $PAB$ but not sure how to construct it?

答: Construct circle such as P is center and PA is radius. P,A, and other inserction of circle and line form isosceles triangle</p>


问:
$\bar{\partial}$ operator of a function
Let $\bar{\partial}$ be the Cauchy–Riemann operator that is for $z=x+i y$,
$$\bar{\partial} = \frac{1}{2i}\left( i\partial_x - \partial_y\right).$$
Now assume that $z=u+ve^{ia}$, where $u$ and $v$ are real variables, $a\in (0,1)$ a real constant, and we want to calculate $\bar{\partial} f(z)$ in terms of $\partial_u$ and $\partial_v$.

My question is that the following claim correct?
$$\bar{\partial} f(z)=\frac{1}{2 i \sin a} \left( e^{i a} \partial_u f-\partial_v f\right).$$
In my attempt, I used the chain rule. I found the same first term on r.h.s, but I got a different term for the second.

Edit:
Here is my calculation:

Write $z=u+ve^{ia}=(u+v\cos a)+i (v\sin a)=u'+iv'$. Then
$$\bar{\partial} f(z)=\frac{1}{2 i} \left(i \partial_{u'} f-\partial_{v'} f\right).$$
By the chain rule, we have $\partial_{u'} f=\partial_{u} f + \frac{1}{\cos a} \partial_{v} f$ and $\partial_{v'} f=-\cot a \;\partial_{u} f + \frac{1}{\sin a} \partial_{v} f$. When replacing we get
$$\bar{\partial} f(z)=\frac{1}{2 i \sin a} \left( e^{i a} \partial_u f- (i \tan a -1)\partial_v f\right).$$

答:
Your idea seems to be allright, but I think you made some mistake using the chain rule (I'm still not sure what the error was).

Remember the chain rule says

$$\begin{aligned}
\partial_x&=\frac{\partial u}{\partial x}\partial_u+\frac{\partial v}{\partial x}\partial_v \\
\partial_y&=\frac{\partial u}{\partial y}\partial_u+\frac{\partial v}{\partial y}\partial_v.
\end{aligned}$$
(There's no need to introduce $u'$ and $v'$, since these are the usual $x$ and $y$.)

The partial derivatives that appear in these equations tell you that you need $u$ and $v$ in terms of $x$ and $y$.
Solving from your equation
$$x+iy=(u+v\cos a)+i(v\sin a),$$
you get
$$\begin{aligned}
u&=x-y\cot a \\
v&=y/\sin a.
\end{aligned}$$

Can you finish?

Title: $\bar{\partial}$ operator of a function


问:
Integral of the square of the integral of a function in $L^1 \cap L^2(S^1 \times \mathbb{R})$
Suppose that I have a function $f: S^1 \times \mathbb{R} \to [0,+\infty]$, which is in $L^1(S^1 \times \mathbb{R}) \cap L^2(S^1 \times \mathbb{R})$. I am wondering about the integral $$I := \int_{S^1} \left( \int_{\mathbb{R}} f(t,x)dx \right)^2 dt.$$Can I conclude that this integral is finite, and, better yet, can I estimate it in terms of $||f||_{L^1(S^1 \times \mathbb{R})}$ and $||f||_{L^2(S^1 \times \mathbb{R})}$? I don't consider the spaces $S^1$ and $\mathbb{R}$ of vital importance - the only important thing is that $S^1$ has finite measure and $\mathbb{R}$ has infinite measure. I couldn't really make much progress on this, the best thing I could do is try to think of an example where $I$ would be infinite and come up with $f(x,t) = \frac{1}{2}e^{-x}\frac{1}{\sqrt{t}}$ (where I replaced $S^1$ by $[0,1]$, but I'm sure a similar example could be given for $S^1$ too). However, this $f$ is not in $L^2([0,1] \times \mathbb{R})$.

答:
Let's consider a slightly simpler scenario: Let $E=(0,1)\times(0,\infty)$ with area measure. For $(t,x)\in E,$ define

$$f(t,x) = \frac{1}{1+tx^2}.$$

Looks to me like $f\in (L^1\cap L^2)(E).$ What do you get for

$$\int_0^1 \left (\int_0^\infty f(t,x)\,dx\right )^2 \,dt?$$


问:
Joint Probability function in expected value of sum of 2 random variable and product of 2 random variable
I realized that same joint probability $f_{xy}(x,y)$ is used in calculation of expected value for sum $E[X+Y]$ and product of 2 random variables $E[XY]$.
How can we use same joint probability in totaly 2 different cases? What is the intuition here?

In case of calculation of sum, I was expecting to use $$f_{X+Y}(x,y)$$ and for calculation of product $$f_{XY}(x,y)$$

As you can see from the pictures, $f_{XY}(x,y)$ is used for both calculations.

Thank you

答:
In general, if you have two random variables $X$ and $Y$ with joint density $f(x,y)$, then the expectation of the random variable $g(X,Y)$ is given by
$$
E(g(X,Y))=\iint g(x,y)f(x,y)dxdy\tag{0}
$$

This is sometimes called the Law of the unconscious statistician.

Notes 1.

The "density" $f_{X+Y}(x,y)$ does not makes sense. Similar for the other one. If you write $Z=X+Y$, then this is one random variable. It expectation is given by
$$
E(Z)=\int zf_Z(z)dz
$$
if you know the density function of $Z$. The mentioned law above allows you to calculate the expectation without knowing the density $f_Z$.

Notes 2.

Given two random variables $X$ and $Y$ and a function $g:\mathbb{R}^2\to\mathbb{R}$ (with good properties), you can define another random variable $Z$ by $Z=g(X,Y)$. For instance, if $g(x,y)=x+y$, then you have $Z=X+Y$; if $g(x,y)=xy$, you have $Z=XY$.

By definition of the expectation, for any continuous random variable $W$ with density function $f_W$, you have
$$
E(W)=\int xf_W(x)dx:=\int_{-\infty}^\infty xf_W(x)dx\tag{1}
$$

So in the example of $Z=X+Y$, there are three ways to express its expectation.

you can use the linearity of expectation to get
$$
E(X+Y)=E(X)+E(Y)=\int xf_X(x)dx+\int yf_Y(y)dy
$$
where $f_X$ and $f_Y$ are the density functions for $X$ and $Y$. Note that in the second integral, $y$ is a dummy variable; you can also write it as $\int xf_Y(x)dx$.

you can use (0):
$$
E(X+Y)=\iint (x+y)f(x,y)dxdy
$$
where $f$ is the joint density function of $X$ and $Y$.

If you denote $f_Z$ as the density function for $Z=X+Y$, then
$$
E(X+Y)=\int zf_Z(z)dz
$$

In these three ways, the functions $f$, $f_X$, $f_Y$, $f_Z$ are related.


问:
Equivalent metrics in Complex numbers
In complex numbers $\Bbb{C}=\{z=x+iy:x,y\in \Bbb{R}\}$ $$d(z_1,z_2)=|z_1-z_2|$$
$$p(z_1,z_2)=\frac{2.|z_1-z_2|}{{\sqrt{1+|z_1|^2}}.{\sqrt{1+|z_2|^2}}}$$

Show that $p$ and $d$ aren't equivalent using

$$\exists M,m>0 \ \ \text{such that} \ \ m.d \le p \le M.d.$$

I showed
$p \le 2d$ but I can't find $m.d \le p$.

答: Note that there is no $m > 0$ such that $m d \leq p.$ Indeed, if this were the case, then we would have that $m \vert z_1 - z_2 \vert \leq \frac{2 \vert z_1 - z_2 \vert}{\sqrt{1 + \vert z_1 \vert^2} \sqrt{1 + \vert z_2 \vert^2}}$ for all $z_1, z_2 \in \mathbb{C}.$ The inequality is of course trivial when $z_1 = z_2,$ so assuming $z_1 \neq z_2,$ we would get that $m \leq \frac{2}{\sqrt{1 + \vert z_1 \vert^2} \sqrt{1 + \vert z_2 \vert^2}}$ for all $z_1, z_2 \in \mathbb{C}$ with $z_1 \neq z_2.$ In particular, let $z_1^n := 0,$ $z_2^n := n$ for all $n \in \mathbb{N}.$ Then $m \leq \frac{2}{\sqrt{n^2 + 1}}$ for all $n \in \mathbb{N},$ which yields the contradiction $0 < m \leq 0.$ Hence, these two metrics are not equivalent.

Title: Equivalent metrics in Complex numbers


问:
You roll a die 8 times. What is the probability of getting an odd number at least twice and an even number at least three times?
My initial idea was to do something such as $3^5 * 6^3$ for the possible cases. However, I believe this is not appropriate for this question, since simply multiplying is usually used in cases in which order matters (e.g. combination of a locker), and here order does not matter.
I'd imagine that maybe we should use something like Bernoulli trials or Combinations, but I'm not sure.

答:
The probability of getting  an odd number at least 2 times = the probability of getting an odd number exactly 2 times + probability of getting odd number 3 times

$n$=8.

Hint: Probability of getting an odd number exactly twice:

$8*(\frac{1}{2})^2$*$\frac{1}{2}$.


问:
How to optimally adjust the probabilities for the random graph-directed IFS algorithm?
In this related question it is shown that the optimal probabilities for an iterated function system of similarities (that satisfy the open set condition) with contraction ratios $r_k$ and fixed point attractor fractal similarity dimension $s$ are $p = r_k^s$.

How does this change when you have a graph-directed iterated function system, where some transitions are disallowed (a non-graph-directed IFS can be seen as a GDIFS with a complete graph)?  There will be a matrix $p_{k_{\operatorname{from}}k_{\operatorname{to}}}$ with some items exactly $0$ (determined by the graph structure) and $\sum_{k_{\operatorname{to}}} p_{k_{\operatorname{from}}k_{\operatorname{to}}} = 1$ for each $k_{\operatorname{from}}$.  The dimension of the whole fractal can be calculated from the graph structure and contraction ratios, via spectral radius of a matrix:

Hausdorff Dimension in Graph Directed Constructions;
R Daniel Mauldin, S C Williams;
Transactions of AMS, vol309 no2, October 1988, 811-829;
http://cas.unt.edu/~mauldin/papers/no67.pdf

I implemented a numerical algorithm to try to calculate optimal probabilities by adjusting weights during image generation, and it sort of worked sometimes, but I'm not even sure if there is a unique optimal set of probabilities...

答:
Let's denote the digraph by $G$, the set of vertices of the digraph by $V$, and the set of edges by $E$. Thus, any $e \in E$ is a directed edge from $u\in V$ to some $v\in V$; conversely, we could denote all the edges from $u$ to $v$ as $E_{uv}$.

Now, to transform this into a digraph IFS, we associate a similarity transformation $f_e$ with each edge $e\in E$. Then, under suitable hypotheses, there is a unique collection of compact sets $K_v$, one for each $v\in V$, such that for every $u\in V$,

$$K_u = \bigcup_{v\in V,e\in E_{uv}} f_e(K_v).$$

A major result of the paper of Mauldin and Williams is the computation of the common Hausdorff dimension of these sets, under the assumption that the digraph is strongly connected. Specifically, we define a matrix $A(s)$ whose rows and columns are indexed by $V$; the entry in row $u$ and column $v$ is
$$A_{uv}(s) = \sum_{e\in E_{uv}} r_e^s.$$
The paper shows that there is a unique value of $s$ such that $A(s)$ has spectral radius 1 and that this value is the common Hausdorff dimension of the sets.

The transition ratios that you seek are proportional to those numbers $r_e^s$. To obtain them, simply normalize them so that the sum of all the numbers coming into any edge is 1.

Here's an example that's worked out in code in this Observable notebook. Our example consists of the following digraph fractal pair:

Each blue piece is a scaled copy of the spiral on the left and each orange piece is a scaled copy of the spiral on the right. Thus, the left image consists of one copy of itself together with one copy of the other. The image on the right consists of one copy of itself together with two copies of the other. This implies the following directed graph:

Note that the edges are labeled with the corresponding contraction ratios. From here, we can read off the matrix $A(s)$:

$$
A(s) = \begin{pmatrix}
  0.96^s & 0.1^s \\
  2\times0.1^s & 0.96^s
\end{pmatrix}.
$$

As it turns out, the largest eigenvalue of $A(s)$ is quite close to 1 when $s\approx1.40449$; thus, the dimension is approximately $1.40449$.

That suggests that each $0.96$ in the transition digraph should be replaced by $0.96^{1.40449} \approx 0.94445$ and that each $0.1$ should be replaced by $0.1^{1.40449} \approx 0.0398$. We still need to normalize, though, so that the sum probability of all edges coming into each node is $1$. For example, the self loop on the left should have probability
$$\frac{0.96^{1.40449}}{0.96^{1.40449} +2\times0.1^{1.40449}} \approx0.922975.$$
The self-loop on the right should have probability
$$\frac{0.96^{1.40449}}{0.96^{1.40449} + 0.1^{1.40449}} \approx 0.959945.$$
Similar computations can be performed for the other two edges to yield the following transition probabilities:

Finally, I might mention that these spirals fit together quite nicely:

But the image looks not nearly so nice, if we use equal probabilities:


问:
How can we integrate a function for surface area of revolution?

This is the equation (with my values plugged in) for surface area of revolution which I found online and I understand the derivation, thus I wanted to use it for a math project I am currently working on. I attempted to use the quadratic function (outside the radical) and inside the radical is the first derivative squared. However, I am stuck in evaluating the integral. I tried u substitution where u is equal to -1/12x+2/3 but was unable to finish it.

Is trigonometry required and does the u substitution I attempted not work in this case? What is the Surface area (answer to the integral)?

答:
Note that
$$\begin{aligned}
S &= 2 \pi \int_0^8 \left( \frac{2}{3} x - \frac{1}{12} x^2 \right) \sqrt{1 + \left( \frac{2}{3} - \frac{1}{12} x \right)^2 } \, d x \\
~ &= \frac{2 \pi}{12} \int_0^8 \left( 8 x - x^2 \right) \sqrt{1 + \left( \frac{2}{3} - \frac{1}{12} x \right)^2 } \, d x \\
~ &= \frac{2 \pi}{12} \int_0^8 \left( 8 x - x^2 \right) \sqrt{\frac{1}{12^2} \left( x^2 - 16 x + 208 \right) } \, d x \\
~ &= \frac{2 \pi}{12^2} \int_0^8 \left( 8 x - x^2 \right) \sqrt{x^2 - 16 x + 208 } \, d x \\
~ &= \frac{\pi}{72} \int_0^8 x ( 8 - x ) \sqrt{12^2 + (8 - x)^2} \, d x \\
\end{aligned}$$
and with $y = 8 - x$, noting that $d y = - d x$, but we also need to swap the integration boundaries, so we get
$$\begin{aligned}
S &= -\frac{\pi}{72} \int_8^0 ( 8 - y ) y \sqrt{12^2 + y^2} \, d y \\
~ &= \frac{\pi}{72} \int_0^8 \left( 8 y - y^2 \right) \sqrt{12^2 + y^2} \, d y \\
\end{aligned}$$
I would continue by integration by parts, via
$$\begin{aligned}
S &= \frac{\pi}{72} \int_0^8 f(y) \, g(y) \, d y \\
f(y) &= 8 y - y^2 \\
g(y) &= \sqrt{12^2 + y^2} \\
\end{aligned}$$
The exact result is
$$\begin{aligned}
S &= 36 \pi \log\left(\frac{2}{3} + \sqrt{1 + \left(\frac{2}{3}\right)^2} \right) + \frac{\pi \, 424 \, \sqrt{13}}{27} - 64\pi \\
~ &= 36 \pi \log\left(\frac{2}{3} + \frac{\sqrt{13}}{3} \right) + \frac{\pi \, 424 \, \sqrt{13}}{27} - 64\pi \\
~ &= 36 \pi \operatorname{asinh}\left(\frac{2}{3}\right) + \frac{\pi \, 424 \, \sqrt{13}}{27} - 64\pi \\
~ &\approx 47.519 \\
\end{aligned}$$


问:
Why the $\nabla f(x)$ in the direction orthogonal to $f(x)$?
Why the $\nabla f$ in the direction orthogonal to $f$?

I can't understand it intuitively.

Is it because of the definition of gradient?

答:
The vector field $\operatorname{grad}(f)$ realizes the total derivative of $f$, in the sense that the inner product of $\operatorname{grad}(f)$ and a vector $v$ gives $df(v)$. But if $v$ is parallel to a level curve, then $df(v) = 0$, so it is perpendicular to the gradient.

Let $f:M\to \mathbb{R}$ be a smooth function with $n_0$ a regular value. Equip $M$ with a Riemannian metric. Suppose $f(m) = n_0$ and $v\in T_mM$ is parallel to the level set $f^{-1}(n_0)$. Then the trick is:

$$\langle \operatorname{grad}(f),v\rangle = df(v) = 0$$

because $v$ is annihilated by $df$.

So in a very real sense, this is a defining property of the gradient. We've constructed the gradient so that it is perpendicular to level sets.


问:
How to prove using contrapositive on $\forall a \in \mathbb{Z}^+$, if $a$ is even, then $a$ does not divide $2a^2+a+7$?
I've tried to prove it in contrapositive $\forall x \in \mathbb{Z}^+$ ($\neg$($a$ does not divide $2a^2+a+7$) $\rightarrow$ $\neg (a$ is a even number)), which basically means that if a divide the equation then a is an odd number. So:

$a$ is odd: $a = 2k+1$ for an arbitrary positive integer k.

$2(2k+1)^2+(2k+1)+1 = (2k+1)x$, since if $a$ can divide the equation, there must be another integer x such that $ax=2a^2+a+7$

But i don't know what to do next or even I'm on the right track.

答: So you want to assume that $a$ divides $2a^2+a+7 = a(2a+1)+7$. From this you should see that $a$ must divide $7$. Can you finish?


答:
HINT: It might help to phrase the contrapositive a little more understandably: if $a$ does divide $2a^2+a+7$, then $a$ is odd. Now what does it mean to say that $a$ divides $2a^2+a+7$? It means that there is an integer $n$ such that $2a^2+a+7=an$. And

$$2a^2+a+7=a(2a+1)+7\,,$$

so $a(2a+1)+7=an$. What does this tell you about $7$?

Title: How to prove using contrapositive on $\forall a \in \mathbb{Z}^+$, if $a$ is even, then $a$ does not divide $2a^2+a+7$?

Body:
I've tried to prove it in contrapositive $\forall x \in \mathbb{Z}^+$ ($\neg$($a$ does not divide $2a^2+a+7$) $\rightarrow$ $\neg (a$ is a even number)), which basically means that if a divide the equation then a is an odd number. So:

$a$ is odd: $a = 2k+1$ for an arbitrary positive integer k.

$2(2k+1)^2+(2k+1)+1 = (2k+1)x$, since if $a$ can divide the equation, there must be another integer x such that $ax=2a^2+a+7$

But i don't know what to do next or even I'm on the right track.


问:
Every path in $S^n$ is homotopic to a non-surjective path
I want to prove that the fundamental group of $S^n$ is trivial.

I found this demonstration by Nersés Aramian which can be found in this pdf.

It is essentially demonstrated that every path in $S^n$ is homotopic to a path that is not surjective.

I would like to understand this demonstration. My problem with this proof is at the end of it.

Why is the comment made about the $\sigma$ and $\tilde{\sigma}$ maps?

And the last paragraph was very confusing for me.

答:
The bit about $\sigma$ and $\tilde{\sigma}$ is claiming that any continuous function $\sigma: J \to D^n$ such that the endpoints of $J$ are not mapped to the center of $D^n$ is homotopic to a continuous function $\tilde{\sigma}: J \to D^n$ which does not go through the center at all. You can picture in your head what such a $\tilde{\sigma}$ could look like: Either a straight line between the endpoints of the path, or if that happens to go through the center, then a curved path between them. Perhaps a semicircle. Or any continuous path you care to imagine that avoids the center.

Aramian simply says that "Clearly $\sigma \simeq \tilde{\sigma}$" without giving an explicit proof, but it's actually true that all paths with the same endpoints are homotopic to each other in $D^n$. After all, following one path and then the other in reverse gives a loop, and we know that $\pi_1(D^n) = 0$, so the loop must be null homotopic, implying that the two paths are homotopic.

In the last paragraph, Aramian constructs a homotopy of $f$ to a new path $g$. To construct $g$, they consider each point in which $f$ crosses $*$ and bend $f$ locally to avoid $*$. They show that this is done a finite number of times (once within each open interval $I_n$) and use the argument from the previous paragraph to show that there is a homotopy from $f$ to $g$ within each of these $I_n$. Since $f$ and $g$ agree outside of these intervals, we get a homotopy from $f$ to $g$ in total.


问:
If a sequence is not bounded, does it mean it's divergent?
If a sequence is not bounded, does it mean it's divergent?</p>

答: You can't prove it because it's not true.  Think about how $n$ is going to infinity.  Is it possible for $a_n$ to go to infinity too, just much more slowly than $n$ does?  That may allow the quotient $\frac{a_n}{n}$ to tend to $0$ even though both numerator and denominator are going to infinity.


问:
Gradient and Hessian of a matrix
For $g(y) = f(D^{\frac{1}{2}}y)$ where $D^{\frac{1}{2}}$ is a matrix to the power half and $ x = D^{\frac{1}{2}}y$

Then $\nabla g(y) = \nabla f(D^{\frac{1}{2}}y) = D^{\frac{1}{2}} \nabla f(D^{\frac{1}{2}}y) = D^{\frac{1}{2}} \nabla f(x)$

What I struggle with is the form of the Hessian given in Amir Beck's Introduction to non-linear optimisation on page 63. Why is the $D^{\frac{1}{2}}$ mutliplied on the RHS in the Hessian?

$\nabla^2 g(y) = D^{\frac{1}{2}} \nabla^2 f(D^{\frac{1}{2}}y) D^{\frac{1}{2}} = D^{\frac{1}{2}} \nabla^2 f(x) D^{\frac{1}{2}}$

答:
Imagine that you have $A^{m\times n}, y^{n\times 1}$, and have the function $f:\mathbb R^m\rightarrow\mathbb R$. Let $g(y)=f(Ay)=f(x)$ with $x^{m\times 1}=Ay$

The gradient of g at y is given by

$$\begin{split}\underbrace{\nabla g(y)}_{\in\mathbb R^n}&=A^T\nabla f(Ay)\\
&=\underbrace{A^T}_{\in\mathbb R^{n\times m}}\underbrace{\nabla f(x)}_{\in\mathbb R^{m\times 1}}\end{split}$$

Then the hessian is given by

$$\begin{split}\underbrace{\nabla^2 g(y)}_{\in\mathbb R^{n\times n}}&=\nabla\left[A^T\nabla f(Ay)\right]\\
&=A^T\nabla^2f(Ay)A\\
&=\underbrace{A^T}_{\in\mathbb R^{n\times m}}\underbrace{\nabla^2f(x)}_{\in\mathbb R^{m\times m}}\underbrace{A}_{\in\mathbb R^{m\times n}}\end{split}$$

Thus $(D^{\frac 1 2})^T=D^{\frac 1 2}$ is multiplied on the left and $D^{\frac 1 2}$ is multiplied on the right.

Title: Gradient and Hessian of a matrix


问:
Marcus, Number Fields, Problem 19(b), Chapter 4
It would be really helpful if somebody could check my argument! Marcus suggests something else but I (think) I found another proof. So here goes:

First of all let's set the stage: Let $L/K$ be a Galois extension of number fields with Galois group $G$. Let $\mathfrak{q}$ be a prime of the ring of integers $\mathcal{O}_L$ of $L$ lying above a prime $\mathfrak{p}$ of the ring of integers $\mathcal{O}_K$ of $K$. We define the ramification groups for $m \geqslant 0$ as follows: $$ V_m = \{\sigma \in G: \sigma(\alpha) \equiv \alpha \pmod{\mathfrak{q}^{m+1}}\quad \forall \alpha \in \mathcal{O}_L \}  .$$

Fix any element $\pi \in \mathfrak{q}-\mathfrak{q}^2$ (such an element exists by unique factorization of ideals in a Dedekind domain). Let $\sigma \in V_{m-1}$, $m>0$ (this will be referred to as Assumption 1). We assume that $\sigma(\pi) \equiv \pi\pmod{\mathfrak{q}^{m+1}}$ (this will be referred to as Assumption 2).

In the first part of the problem we proved that $ \sigma(\alpha) \equiv \alpha \pmod{\mathfrak{q}^{m+1}}\quad\forall \alpha \in \pi\mathcal{O}_L $.

Now we want to prove $\sigma(\alpha) \equiv \alpha\pmod{\mathfrak{q}^{m+1}}\quad \forall \alpha \in \mathfrak{q}$. So now I'll start writing my proof.

So let $\alpha \in \mathfrak{q}$. Of course, $\pi \alpha \in \pi \mathcal{O}_L$. Applying what we proved in the first part of the problem for the element $\pi \alpha$ we get $\sigma(\pi \alpha) \equiv \pi \alpha \pmod{\mathfrak{q}^{m+1}}$. That it we get: $$ \sigma(\pi \alpha) - \pi \alpha \in \mathfrak{q}^{m+1}. \tag{1} $$
On the other hand, Assumption 1 implies that $\sigma(\pi) \equiv \pi\pmod{\mathfrak{q}^m}$ (since $\pi \in \mathcal{O}_L$). That is, $\sigma(\pi) - \pi \in  \mathfrak{q}^m$. By assumption though, $\alpha \in \mathfrak{q}$. Hence, $\alpha \cdot (\sigma(\pi) - \pi) \in \mathfrak{q}^{m+1}$. Hence,
$$ \alpha \sigma(\pi) - \alpha \pi \in \mathfrak{q}^{m+1}. \tag{2}$$
Combining relations $(1)$ and $(2)$  yields $\sigma(\pi \alpha) - \alpha \sigma(\pi) \in \mathfrak{q}^{m+1}$. That is $$ \sigma(\pi) \cdot \sigma(\alpha) - \sigma(\pi) \alpha \in \mathfrak{q}^{m+1}.\tag{3}$$
Now, we know that $\sigma^{-1}(\mathcal{O}_L)=\mathcal{O}_L$. Therefore $\sigma^{-1}(\pi) \in \mathcal{O}_L$. Multiplying relation $(3)$ with $\sigma^{-1}(\pi)$ (and taking of course advantage of the fact that $\mathfrak{q}^{m+1}$ is an ideal of $\mathcal{O}_L$ ) gives us: $\sigma^{-1}(\pi) \cdot (\sigma(\pi) \sigma(\alpha) - \sigma(\pi) \alpha) \in \mathfrak{q}^{m+1} $. That is $\sigma(\alpha) - \alpha \in \mathfrak{q}^{m+1}$. Of course, this is equivalent to $\sigma(\alpha) \equiv \alpha \pmod{\mathfrak{q}^{m+1}}$, which is what we wanted to prove.

If something doesn't seem right please let me know! Thanks !

答:
Take $z\in O_L/q^{m+1}$ such that $z$ has of order $N(q)-1$ in $(O_L/q)^\times$,  then $\zeta=z^{N(q)^{m+1}}$ has order $N(q)-1$ in $(O_L/q^{m+1})^\times$.

Let $S=\{ \zeta^n, n \ge 0\}\cup 0$.

As a set $$O_L/q^{m+1}=\{ \sum_{j=0}^m s_j \pi^j, s_j\in S\}$$

If $\sigma\in V_0$ then $\sigma(q)=q$ so that $\sigma(q^{m+1})=q^{m+1}$ and $\sigma\in Aut(O_L/q^{m+1})$. Moreover $\sigma(\zeta)=\zeta$.

Therefore $\sigma(a)=a$ for all $a\in O_L/q^{m+1}$ iff $\sigma(\pi)=\pi\in O_L/q^{m+1}$.


问:
Prove the solution of differential equation is an odd function and can't be extended on $\mathbb{R}$

Prove that the solution to Cauchy's problem
$$
\begin{cases}
y'=y^2+x^2\\
y(0)=0
\end{cases}
$$
is an odd function, and that this solution can't be extended on $\mathbb{R}$.

I tried to substitute $y(x)$ with $-y(-x)$ and check if cancels out:
$$
\begin{align}
(-y(-x))'=(-y(-x))^2+x^2\\
y'(-x)=y^2(-x)+(-x)^2 && \text{substiute } t=-x\\
y'(t)=y^2(t)+t^2
\end{align}
$$
I think that proves the first part.

As for the second part of the question:

let $y$ be a solution to the Cauchy's problem, then:
$$
\begin{align}
y'=y^2+x^2\\
y'\leq y^2\\
\int \frac{dy}{y^2}\leq \int dx\\
\frac{-1}{y}\leq x+c\\
\frac{1}{y}\geq -x+c\\
y\geq \frac{1}{-x+c} \stackrel{x\rightarrow c}{\longrightarrow}\infty
\end{align}
$$

therefore $y$ does not extend on $\mathbb{R}$. Is that correct?

答:
You almost got it correctly:

You have only proven that $-y(-x)$ is also a solution. You need to patch this result together with a uniqueness result to achieve that $-y(-x)$ is in fact equal to $y(x)$.

You got two mistakes with the inequality signs but they cancel each other out...
$$
\begin{align}
y'=y^2+x^2\\
y'\color{red}{\geq} y^2\\
\int \frac{dy}{y^2}\geq \int dx\\
\frac{-1}{y}\geq x+c\\
\frac{1}{y}\leq -x+c\\
y\color{\red}{\geq} \frac{1}{-x+c} \stackrel{x\rightarrow c}{\longrightarrow}\infty
\end{align}
$$
Additionally, to achieve the last line, you should give an explanation why $y>0$, which is straightforward for $x>0, y(0)=0$ and $y'(x)>0$.


问:
What is the logic behind the balls colours in MarkSix lottery?
I am not familiar with this kind of problem and I am trying to find what could be the general term or pattern used for the color assignment in the Mark Six lottery (This is the matrix of colors).

There are three sequences:

Red balls: [1, 2, 7, 8, 12, 13, 18, 19, 23, 24, 29, 30, 34, 35, 40, 45, 46];

```<code>Red(n) = ?
</code>```

Blue balls: [3, 4, 9, 10, 14, 15, 20, 25, 26, 31, 36, 37, 41, 42, 47, 48];

```<code>Blue(n) = ?
</code>```

Green balls: [5, 6, 11, 16, 17, 21, 22, 27, 28, 32, 33, 38, 39, 43, 44, 49];

```<code>Green(n) = ?
</code>```

How do you think would be the best approach to find the pattern followed there? What would be the general term for those number sequences?

Thanks in advance!

答:
Here's a table of the ball colors in a format which makes the pattern a little more visible:

$$\begin{array}{|rr|rr|rr|}
\hline
\color{red}{1} & \color{red}{2} & \color{blue}{3} & \color{blue}{4} & \color{green}{5} & \color{green}{6}\\
\color{red}{7} & \color{red}{8} & \color{blue}{9} & \color{blue}{10} & ** & \color{green}{11}\\
\color{red}{12} & \color{red}{13} & \color{blue}{14} & \color{blue}{15} & \color{green}{16} & \color{green}{17}\\
\color{red}{18} & \color{red}{19} & \color{blue}{20} & ** & \color{green}{21} & \color{green}{22}\\
\color{red}{23} & \color{red}{24} & \color{blue}{25} & \color{blue}{26} & \color{green}{27} & \color{green}{28}\\
\color{red}{29} & \color{red}{30} & ** & \color{blue}{31} & \color{green}{32} & \color{green}{33}\\
\color{red}{34} & \color{red}{35} & \color{blue}{36} & \color{blue}{37} & \color{green}{38} & \color{green}{39}\\
\color{red}{40} & ** & \color{blue}{41} & \color{blue}{42} & \color{green}{43} & \color{green}{44}\\
\color{red}{45} & \color{red}{46} & \color{blue}{47} & \color{blue}{48} & \color{green}{49} & \\
\hline
\end{array}$$

The pattern is easier to express mathematically if we subtract 1 from the ball number, so that the numbers start from zero.

We can calculate the color of a number $i$ using this formula:

Let $j = i - 1$

Then
$$k = floor(((j + floor(j / 10)) \mod 6) / 2)$$
$k$ will be in ${{0, 1, 2}}$, with 0 = red, 1 = blue, 2 = green.

So subtract 1 from the ball number $i$ to get $j$. Then divide $j$ by 10, rounding down, and add that to $j$. Now find the remainder of that total, mod 6. Finally, divide that remainder by 2 (rounding down) to get the color number.

Here's a short Python script which tests that formula.

```<code>red = [1, 2, 7, 8, 12, 13, 18, 19, 23, 24, 29, 30, 34, 35, 40, 45, 46]
blue = [3, 4, 9, 10, 14, 15, 20, 25, 26, 31, 36, 37, 41, 42, 47, 48]
green = [5, 6, 11, 16, 17, 21, 22, 27, 28, 32, 33, 38, 39, 43, 44, 49]

groups = [[], [], []]
for i in range(1, 50):
    j = i - 1
    u = (j + j // 10) % 6 // 2
    groups[u].append(i)

print(red == groups[0])
print(blue == groups[1])
print(green == groups[2])
</code>```

You can run it on the SageMathCell server here. The table at the start of this answer was also created in Sage / Python using the above formula. Here's the table script.

Body:
Here's a table of the ball colors in a format which makes the pattern a little more visible:

$$\begin{array}{|rr|rr|rr|}
\hline
\color{red}{1} & \color{red}{2} & \color{blue}{3} & \color{blue}{4} & \color{green}{5} & \color{green}{6}\\
\color{red}{7} & \color{red}{8} & \color{blue}{9} & \color{blue}{10} & ** & \color{green}{11}\\
\color{red}{12} & \color{red}{13} & \color{blue}{14} & \color{blue}{15} & \color{green}{16} & \color{green}{17}\\
\color{red}{18} & \color{red}{19} & \color{blue}{20} & ** & \color{green}{21} & \color{green}{22}\\
\color{red}{23} & \color{red}{24} & \color{blue}{25} & \color{blue}{26} & \color{green}{27} & \color{green}{28}\\
\color{red}{29} & \color{red}{30} & ** & \color{blue}{31} & \color{green}{32} & \color{green}{33}\\
\color{red}{34} & \color{red}{35} & \color{blue}{36} & \color{blue}{37} & \color{green}{38} & \color{green}{39}\\
\color{red}{40} & ** & \color{blue}{41} & \color{blue}{42} & \color{green}{43} & \color{green}{44}\\
\color{red}{45} & \color{red}{46} & \color{blue}{47} & \color{blue}{48} & \color{green}{49} & \\
\hline
\end{array}$$

The pattern is easier to express mathematically if we subtract 1 from the ball number, so that the numbers start from zero.

We can calculate the color of a number $i$ using this formula:

Let $j = i - 1$

Then
$$k = floor(((j + floor(j / 10)) \mod 6) / 2)$$
$k$ will be in ${{0, 1, 2}}$, with 0 = red, 1 = blue, 2 = green.

So subtract 1 from the ball number $i$ to get $j$. Then divide $j$ by 10, rounding down, and add that to $j$. Now find the remainder of that total, mod 6. Finally, divide that remainder by 2 (rounding down) to get the color number.

Here's a short Python script which tests that formula.

red = [1, 2, 7, 8, 12, 13, 18, 19, 23, 24, 29, 30, 34, 35, 40, 45, 46]
blue = [3, 4, 9, 10, 14, 15, 20, 25, 26, 31, 36, 37, 41, 42, 47, 48]
green = [5, 6, 11, 16, 17, 21, 22, 27, 28, 32, 33, 38, 39, 43, 44, 49]

groups = [[], [], []]
for i in range(1, 50):
    j = i - 1
    u = (j + j // 10) % 6 // 2
    groups[u].append(i)

print(red == groups[0])
print(blue == groups[1])
print(green == groups[2])

You can run it on the SageMathCell server here. The table at the start of this answer was also created in Sage / Python using the above formula. Here's the table script.