问:
Proving the concurrence of three lines.
Let $p_{1}, p_{2} \text{ and } p_{3}$ be three planes which intersect in a straight line (and not a point, which is generally the case).

Let a fourth plane $p_{4}$ cut these planes (not at the line of intersection of the three planes). Evidently, the intersections of the planes with $p_{4}$ would form three distinct straight lines.

Introduction to Higher Geometry by Graustein says that the three lines are concurrent. Any help regarding the proof would be great.

Thanks in advance.

答:
As you mentioned, generally three planes intersect in a point. You supposed that the common line $l$ of $p_1$, $p_2$, $p_3$ is not in $p_4$ . So, for example, $p_1$, $p_2$ and $p_4$ meet in a point $P$. This point must be on the lines $p_1\cap p_4$ and $p_2\cap p_4$. $P$ must also be on $l$, because it is a common point of $p_1$ and $p_2$. Thus $P=l\cap p_4$.
Similarly, consider the planes $p_1$, $p_3$ and $p_4$, and let their common point be $Q$. $Q$ is on $p_1\cap p_4$ and $p_3\cap p_4$, and $Q$ is also on $l$. So $Q=l\cap p_4$, thus $P=Q$.
Summarizing: $P$ is on $p_1\cap p_4$, $p_2\cap p_4$ and $p_3\cap p_4$ as we had to prove.</p>

Title: Proving the concurrence of three lines.


问:
When reducing $A$ to $I$ via elementary matrices $E_k...E_1A=I$, why is it true also that $AE_k...E_1=I$
Say we have a $3 \times 3$ matrix $A$ and we perform the row operation $Z_1:(R_1 \iff R_2)$ of switching the first and second row to get $A'$.

Now, if we perform the same row operation $Z_1:(R_1 \iff R_2)$ on the identity matrix $I_3$ we get the elementary matrix $E_1$ and it is a common fact of linear algebra that $E_1A = A'$. Note that it is important that we multiply $E_1$ on the left of $A$.

Now, say further that $A$ is invertible and that we want to find $A^{-1}$. Then what we do is set up the augmented matrix $[A | I_3]$ and perform row operations $Z_1,...,Z_k$ until we we get an equivalent augmented matrix of the form $[I_3|B]$. It is a common fact of linear algebra that $B = A^{-1}$.

Now, the row operations $Z_1,...,Z_k$ correspond to the elementary matrcies $E_1,...,E_k$ and we can thus see that $A^{-1}=E_k...E_1$.

This is great and makes perfect sense! We reduced $A$ to $I_3$ via row operations corresponding to elementary matrices, so it is clear that $A^{-1}A = E_k,...E_1A=I_3$.

However, it is also true that $AA^{-1}=I_3$ and thus we have that $AE_k,...E_1 = I_3$. In the case when applying a string of elementary matrices to a matrix $A$ such that the string reduces $A$ to the identity, it is no longer important whether we apply the multiplication of the matrices on the right or left.

The bold font was meant to emphasize the phenomonon that I was hoping somebody could clarify. Understand that I am not asking why $A^{-1}A = AA^{-1}$ but rather seeking to clarify the more subtle observation of how the elementary matrices that embody the row operations can be applied to either side of $A$ in the case when they reduce $A$ to $I$.

Thank you.

答:
I'm going to work with a $2\times 2$ example so that it is explicit and clear in each step what is happening. Let $A = \begin{pmatrix} 4 & 5 \\ -2 & 3 \end{pmatrix}$, then

$$ \begin{pmatrix} \frac{1}{4} & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & -5 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{11} \end{pmatrix} \begin{pmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{pmatrix} \begin{pmatrix} 4 & 5 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \tag{1}$$

Multiplying on the left

Step by step (so we can compare with multiplying on the right shortly):

$$ \begin{pmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{pmatrix} \begin{pmatrix} 4 & 5 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} 4 & 5 \\ 0 & \frac{11}{2} \end{pmatrix} \tag{2}$$

Rescaling,

$$ \begin{pmatrix} 1 & 0 \\ 0 & \frac{2}{11} \end{pmatrix} \begin{pmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{pmatrix} \begin{pmatrix} 4 & 5 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} 4 & 5 \\ 0 & 1 \end{pmatrix} \tag{3}$$

Then

$$ \begin{pmatrix} 1 & -5 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & \frac{2}{11} \end{pmatrix} \begin{pmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{pmatrix} \begin{pmatrix} 4 & 5 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} 4 & 0 \\ 0 & 1 \end{pmatrix} \tag{4}$$

and finally

$$ \begin{pmatrix} \frac{1}{4} & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & -5 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & \frac{2}{11} \end{pmatrix} \begin{pmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{pmatrix} \begin{pmatrix} 4 & 5 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \tag{5}$$

Let's then work this the opposite way to see what the row operations above are doing when we multiply on the right instead.

Multiplying on the right

$$ \begin{pmatrix} 4 & 5 \\ -2 & 3 \end{pmatrix} \begin{pmatrix} \frac{1}{4} & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 5 \\ -\frac{1}{2} & 3 \end{pmatrix} \tag{6}$$

So in the first part, this gave us a unit pivot in the upper left and it does the same here (which makes sense). Note that it actually became a column operation on the first column though as opposed to a row operation on the first row when multiplied on the left. Of course that distinction doesn't matter when multiplying diagonal matrices as in $(5)$..

$$ \begin{pmatrix} 4 & 5 \\ -2 & 3 \end{pmatrix} \begin{pmatrix} \frac{1}{4} & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & -5 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ -\frac{1}{2} & \frac{11}{2} \end{pmatrix} \tag{7}$$

So our matrix $E_3$ again eliminated the upper right element.

$$ \begin{pmatrix} 4 & 5 \\ -2 & 3 \end{pmatrix} \begin{pmatrix} \frac{1}{4} & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & -5 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & \frac{2}{11} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ -\frac{1}{2} & 1 \end{pmatrix} \tag{8}$$

This one again was a column operation and instead of scaling the bottom row, it scaled the rightmost column. Finally,

$$ \begin{pmatrix} 4 & 5 \\ -2 & 3 \end{pmatrix} \begin{pmatrix} \frac{1}{4} & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & -5 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & \frac{2}{11} \end{pmatrix} \begin{pmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \tag{9}$$

So again $E_1$ eliminated the bottom left element.

As we saw above, our rescaling elementary matrices keep that behavior, it's just a matter of whether it's a row or a column rescaling depending on if it is multiplied on the left or on the right. And you can see easily that if you had to switch rows, the same logic would apply. So the question then is: what are the elimination elementary matrices doing?

Let's look at equations $(4)$ and $(7)$ more closely. In $(3)$, our previous operations had made our matrix into upper triangular form and we were starting to eliminate upwards and so we had a unit pivot in the bottom right before eliminating upwards. In step $(6)$, we rescaled the upper left element. However in step $(7)$, we didn't get a $0$ below, but rather to the right. So what that means is that the elimination matrices do not eliminate upwards when multiplied on the right but rather eliminate to the right. Similarly, eliminating downwards is actually the same as eliminating to the left when multiplying on the right which you can see when you compare equation $(2)$ to equation $(9)$. The pivot that was created in the bottom right was then used to eliminate the element to the left of it. (To know that it's right/left, pay close attention to what the matrix multiplication is doing and which elements are at play.)

So in short: row rescalings get turned into column rescalings, row swaps get turned into column swaps, and eliminating downward (upward) gets turned into eliminating leftward (rightward).

ContentLicense: CC BY-SA 4.0


问:
Prove on cardinality of a countable subset of real numbers
I'm asked to prove that, for a countable subset of real numbers, either (a) the cardinality of such subset with non-negative real numbers equals the cardinality of natural numbers (i.e. that resulting intersection is countable) or (b) the cardinality of such subset with negative real numbers equals the cardinality of natural numbers (i.e. that resulting intersection is countable), this is:

Let B be countable subset of $\mathbb{R}$.

Prove that either $card(B\cap[0,\infty))=card(\mathbb{N})$ or $card(B\cap(-\infty,0))=card(\mathbb{N})$.

Any insights on suitable approaches to prove this? I'm struggling to put this into formal and general terms.

Many thanks.

答:
Put it into simpler words and concepts.

They are asking you to show that if you have a countably infinite set, and you partition it into two subsets, then at least one of the two subsets (or maybe both) must be countably infinite.

Note that a subset of a countably infinite set is either countable or finite (that should be a proposition you have proven already).  So you have four options.

Both subsets are finite

One subset is finite and the other is countable infinite

One subset is countably infinite and the other is finite

Both are countably infinite.

You have to prove 1) is impossible.

Note that the union of two finite sets is finite (that should be a proposition you have proven already).  So if 1) is true than the original set is the union of two finite sets and therefore the original set was finite.

That contradicts that you were told the original set was countably infinite.


问:
A more logical approach to a word problem
I was taught in high school algebra to translate word problems into algebraic expressions. So when I encountered this problem I tried to reason out an algebra formula for it

For every dollar Ben spent on bagels, David spent 25 cents less. Ben
paid $12.50 more than David. How much did they spend in the bagel store
together?

To solve this I imagined a series of comparisons when Ben spends $x$, David spends $.75x$. Loop this relationship until $x - .75x \approx 12.50$. Good. Done. $x = 50$, then add David's for the answer. Coming from computers, I would have set this up in code where a loop (recursion) would increase $x$ until the condition $x - .75x = 12.50$ was met, then the "loop counter/accumulator" would be how much Ben spent, i.e., $50$, etc.

I'm a beginner with math, but it seems like there should be a better approach, something with series and sequences or even calculus derivatives, something better than my brute-force computer algorithm. Can someone enlighten? The "answer" given at the site (see link) is its own brute-force and hardly satisfying. I'm thinking there should be something more formal -- at least for the first part that derives $50$.

Update

I think everyone so far has missed my point. Many of you simply re-did the problem again. I'm wondering if there is a more formal way to do this other than just "figuring it out" (FIO). The whole FIO routine is murky. It looks like a limit problem; it looks like a system of equations, but I'm not experienced enough to know exactly. If there isn't, then let's call it a day....

答: You can use linear algebra, by defining something as x, say the amount paid by Ben. Then Dave's will be 0.75x. Then we can get an easily solvable equation.</p>


问:
How to calculate fixed points and plot bifurcation diagram for non-linear ODE system
I am trying to understand how to analyse a system of coupled, non-linear ODEs taken from this  paper. I want to perform a fixed point analysis and plot a bifurcation diagram to show how fixed points and their stability depend on the parameters.The setup is we have for example two cells, each characterized by its level of $n$ and $d$.

-$n_{i}$ represses the activity of its own $d_{i}$

-$d_{i}$ increases the activity of $d$ of its neighbour $d_{j}$

This is the non-dimensionalized form of the equations

$$\dot n_{i}=f(d_{j}) - n_{i}$$
$$\dot d_{i}=\nu (g(n_{i}) - d_{i})$$

where, $f=\frac{x^n}{a-x^n}$ and is monotonically increasing, $g=\frac{1}{1+bx^n}$ and is monotonically decreasing and $\nu$ represents a ratio of time constants $\frac{\gamma_d}{\gamma_{n}} $.

I honestly have no idea how to approach finding the fixed points, performing stability analysis and doing the bifurcation diagram, I care about dependence on $\nu,a,b$ and $n$.

If I take the case where $\nu >> 1$, I can simplify by assuming $d_{i}$ is in quasi steady state and hence $g(n_{i})=d_{i}$, therefore
$$\dot n_i = f(g(n_j)) - n_i$$
$$\dot n_j = f(g(n_i)) - n_j$$

which can be simplified further by substituting

$$\dot n_i = f(g(f(g(n_i))) - n_i$$
$$\dot n_j = f(g(f(g(n_j))) - n_j$$

and this is where i'm stuck. Usually in the 1d case you would solve for the fixed point and then check stability dependent on derivative and parameter values. I have no idea where to start from here.

Any pointers/help will be vastly appreciated. Also if this post is too long I will edit it down, I couldn't really decide what background is necessary so to understand things..

答:
I ended up brute-forcing it by simplifying fgfg using WolframAlpha, formulating the equation fgfg(x) - x = 0  and sovling it with a newton-rhapson root finder.

Producing the bifurcation diagram from there was easy.

I guess I was looking for a closed form way, where I get formulas for the fixed points dependent on parameters but I don't think that is possible or atleast I couldn't do it.


问:
Do fractals necessarily have to be self-similar? What is the definition of a fractal? Is this figure a fractal?
I am doing a dissertation in Materials Engineering. I obtained the following images regarding crystal growth:

In the literature they usually call these structures "fractals" and calculate their fractal dimension using the box-counting method. From what I read on the internet, a fractal has to have self-similarity.

However, these structures appear to be so irregular that they do not appear to have any kind of repetition. The fractals according to the DLA (diffusion limited aggregation) have a fractal dimension of approximately 1.70, which is close to that of these structures. Why do the authors consider these structures fractal if there is no self-similarity? What are the characteristics necessary for a figure to be considered a fractal?

答:
This is more a comment than an answer, but I don't have the reputation.

Just a guess: A way that fractals arise is by iteration of a map. Perhaps what the authors have in mind is that the growth of these structures comes about by iteration of some simple rules, after translation or some simple function perhaps? The reason I'm saying this is that as far as I know, this sense of fractal (really just an approximate fractal) is what I would expect to see in nature, more than bona fide fractals.


问:
A question about the use of successions in the quantifier clause of the definition of satisfaction
In first order logic, why do we say that:
$\exists$ $xi$ $\phi$ is satisfied by succession si iff there is a different succession from s that satisfied $\phi$?

For instance, maybe the only succession that satisfies $\phi$ is actually si, and there is no other succession different from si that satisfies $\phi$. Why there must be a different succession?

答: The definition is not that there is a different succession. The definition is that there is some succession. This may be different or identical to the original one.


问:
Fourier transform and Limit of complex functions multiplied trigo functions
I would like to calculate the Fourier Transform for
$$
\frac{1}{1+x^2}.
$$

By definition
\begin{align}\DeclareMathOperator{atan}{atan}
\int_{-\infty}^{\infty}\frac{1}{1+x^2}e^{-i\omega x}dx
&= e^{-i\omega x}\atan(x) + i\omega\int_{-\infty}^\infty e^{-i\omega x}\atan(x)dx \\
&= e^{-i\omega x}\atan(x) + i\omega\frac{\atan(x)e^{-i\omega x}}{-i\omega} - i\omega\int_{-\infty}^{\infty}\frac{1}{1+x^2}e^{-i\omega x}.
\end{align}

Thus
$$
\int_{-\infty}^{\infty}\frac{1}{1+x^2}e^{-i\omega x}dx = \left.\frac{e^{-i\omega x}\atan(x) + i\omega\frac{atan(x)e^{-i\omega x}}{-i\omega}}{1+i\omega}\right |_{-\infty}^{\infty}
$$

and now I'm stuck with how I should carry on.

Original question :
How can I calculate the following limit
$$
\lim_{x\rightarrow\infty}\atan(x)e^{-i\omega x}
$$

where $\omega\in \mathbb{R}$.

答:
Except for $\omega=0$ the limit doesn't exist, since $\lim_{x\to\infty}\arctan(x)=\pi/2$ and $$\lim_{x\to\infty}e^{-i\omega x}\overset{\omega\neq0}{=}\lim_{x\to\infty}e^{\pm ix}$$ ($\pm$ dependent on the sign of $\omega$) which doesn't exist as $$e^{\pm ix}=\cos(x)\pm i\sin(x)$$ is a oscillating function.

Regarding your Edit:

I think you are following a wrong path. Because of the reasons above, you can't do integration by parts. Instead you could try to use
$$\frac{1}{1+x^2}=\frac{1}{x+i}\cdot\frac{1}{x-i}$$
and the formular for the Fourier transform of products. With the substitution $u:=x\pm i$ you just have to calculate the Fourier transform of the function $f(u)=1/u$, which has been described here.


答: $\arctan(x)\to \frac{\pi}{2}+n\pi$ and $e^{-i\omega x}$ has no limit, so the product has no limit.

Title: Fourier transform and Limit of complex functions multiplied trigo functions

Body:
I would like to calculate the Fourier Transform for
$$
\frac{1}{1+x^2}.
$$

By definition
\begin{align}\DeclareMathOperator{atan}{atan}
\int_{-\infty}^{\infty}\frac{1}{1+x^2}e^{-i\omega x}dx
&= e^{-i\omega x}\atan(x) + i\omega\int_{-\infty}^\infty e^{-i\omega x}\atan(x)dx \\
&= e^{-i\omega x}\atan(x) + i\omega\frac{\atan(x)e^{-i\omega x}}{-i\omega} - i\omega\int_{-\infty}^{\infty}\frac{1}{1+x^2}e^{-i\omega x}.
\end{align}

Thus
$$
\int_{-\infty}^{\infty}\frac{1}{1+x^2}e^{-i\omega x}dx = \left.\frac{e^{-i\omega x}\atan(x) + i\omega\frac{atan(x)e^{-i\omega x}}{-i\omega}}{1+i\omega}\right |_{-\infty}^{\infty}
$$

and now I'm stuck with how I should carry on.

Original question :
How can I calculate the following limit
$$
\lim_{x\rightarrow\infty}\atan(x)e^{-i\omega x}
$$

where $\omega\in \mathbb{R}$.


答:
I have to admit I'm having a hard time seeing how you "imagine a series of comparisons " and how you "loop" them.

As a mathematician, I simply view this as Ben did a single transaction where in one go he spent $x$ dollars on bagels.

Therefore David spend $0.75x$ on bagels (in one go).

Set one equation (no series of loops) to get $x - 0.75x = 12.50$ and solve:

$x-0.75x = 12.50$

$0.25x = 12.50$

$\frac {0.25}{0.25} x = \frac {12.50}{0.25}$

$x = 50$

So Ben spent $50$ dollars on bagels.

Now add Davids answer.  David spent $0.75x$ and $x = 50$ so David spent $0.75 \times 50 = 37.50$.  Adding that they added $50+37.50 = 87.50$.

Or.... Ben spent $x$ and David spent $0.75x$ so together they spent $x + 0.75x = 1.75x$ and as $x = 50$ together they spent $1.75\times 50 = 87.50$.

Title: A more logical approach to a word problem

Body:
I was taught in high school algebra to translate word problems into algebraic expressions. So when I encountered this problem I tried to reason out an algebra formula for it

For every dollar Ben spent on bagels, David spent 25 cents less. Ben
paid $12.50 more than David. How much did they spend in the bagel store
together?

To solve this I imagined a series of comparisons when Ben spends $x$, David spends $.75x$. Loop this relationship until $x - .75x \approx 12.50$. Good. Done. $x = 50$, then add David's for the answer. Coming from computers, I would have set this up in code where a loop (recursion) would increase $x$ until the condition $x - .75x = 12.50$ was met, then the "loop counter/accumulator" would be how much Ben spent, i.e., $50$, etc.

I'm a beginner with math, but it seems like there should be a better approach, something with series and sequences or even calculus derivatives, something better than my brute-force computer algorithm. Can someone enlighten? The "answer" given at the site (see link) is its own brute-force and hardly satisfying. I'm thinking there should be something more formal -- at least for the first part that derives $50$.

Update

I think everyone so far has missed my point. Many of you simply re-did the problem again. I'm wondering if there is a more formal way to do this other than just "figuring it out" (FIO). The whole FIO routine is murky. It looks like a limit problem; it looks like a system of equations, but I'm not experienced enough to know exactly. If there isn't, then let's call it a day....


问:
Relationships between three subspaces
Let, $V$ be a finite dimensional vector space over $\mathbb R$ and $ W_1,W_2,W_3$ be subspaces of $V$. Which of the following is true?

If $ W_1+W_2+W_3=V$ then
$$\mathrm{span}(W_1\cup W_2) \cup \mathrm{span}(W_2\cup W_3) \cup \mathrm{span}(W_3\cup W_1) = V.$$

If $W_1 \cap W_2 =\{0\}$ and $W_1 \cap W_3 =\{0\}$ then $W_1 \cap (W_2+W_3)=\{0\}$.

If $W_1+W_2=W_1+W_3$ then $W_2=W_3$

If $V\neq W_1$ then $\mathrm{span}(V-W_1)=V$.

My attempt: Take, $V=\mathbb R^3$ and $W_1=\{(x,y,z):\ y=0,z=0\}$,  $W_2=\{(x,y,z):\ x=0,z=0\}$ and $W_3=\{(x,y,z):\ x=0,y=0\}$. Then option 1 is discarded.

Take $V=\mathbb R^2$ and $W_1=\{(x,y):\ y=x\}$, $W_2=\{(x,y):\ x=0\}$ and $W_3=\{(x,y):\ y=0\}$. Then option 2 is discarded.

Take $V=\mathbb R^2$ and $W_1=V$, $W_2=\{(x,y):\ x=0\}$ and $W_3=\{(x,y):\ y=0\}$. Clearly option 3 is discarded.

So option 4 is right. But I cannot prove it. Please help to prove this

答:
Here is a proof for option 4: if $V \neq W_1$, then $V - W_1 \neq \emptyset$. Let $u \in V - W_1$.

We wish to show that $v \in V \implies v \in \operatorname{span}(V - W_1)$. Note that, if $v \in V - W_1$, then $v \in \operatorname{span}(V - W_1)$ trivially. So, we really only need to worry about $v \in W_1$.

Suppose $v \in W_1$, and let us suppose for the sake of contradiction that $v + u \in W_1$. Then $u = (v + u) - v$, the difference of two vectors in $W_1$. Since $W_1$ is a subspace, this makes $u \in W_1$, against assumption. Thus, $v + u \in V - W_1$.

Similarly, we have $v - u \in V - W_1$, for otherwise $u = v - (v - u) \in W_1$ against assumption. Now we can see that
$$v = \frac{1}{2}(v + u) + \frac{1}{2}(v - u) \in \operatorname{span}(V - W_1),$$
completing the proof.

Title: Relationships between three subspaces


问:
Two limits problems (statement and l'hopital)

Given: $\lim \limits_{x \to \infty }\frac{x^2+p(x)}{1-x^2}=5$ $p(x)$ is a polynomial.
What can we conclude about $p(x)$ power?

What I did was just solve an equation, $\frac{x^2+y}{1-x^2}=5$ then we get $y=5-6x^2$ then I applied it to the limit ($p(x)$ instead of y)
$\lim \limits_{x \to \infty }\frac{x^2+(5-6x^2)}{1-x^2}=5$ after calculating the limit I got $5=5$ . So the power is $2$.

"A student claims that there is a one and only polynomial that fulfills the questions answer"

I think that the answer is right because if the power is greater than $2$ we will get $+\infty$ and I belive that the Coefficient also has role but I am not sure , can anyone explain this part better?.

The second question is:

given $f(x)=\frac{sin(\pi x)}{x-x^2}$ and also given $
g(x)=
\begin{cases}
 0&\text {if  } \, x= 1\\
 0.5&\text{if }\, x≠1\\
     \end{cases}$
also given the limits of the functions: $\lim \limits_{x \to 1 }f(x)=A$ and $\lim \limits_{x \to 1 }f(g(x))=B$
find the values of A and B (the answer can be " the limit does not exist")

so for A it wasn't much of a problem as I used l'hopital.$\lim \limits_{x \to 1 }\frac{sin(\pi x)}{x-x^2}$ after doing l'hopital rule we get $\lim \limits_{x \to 1 }\frac{\pi cos(\pi x)}{1-2x}$ $=1=A$
for the second part of the question I got confused because I didn't know if when we do ${x \to 1 }$ I should get $0$ of $g(x)$ or get $0.5$ of $g(x)$ as when we do a limit we only get close to the number according to what I know , not exactly $1$.

if we look at it as in $x=1$ we get $\lim \limits_{x \to 1 }f(0)$ = $\lim \limits_{x \to 1 }\frac{sin(0)}{0-0}$ and then even l'hopital won't help

and if we look at it as in $x≠1$ we get $\lim \limits_{x \to 1 }f(0.5)$ = $\lim \limits_{x \to 1 }\frac{sin(\frac {\pi}2)}{0.5-0.25}$ = 4

My question is , is my way correct? and from the second part how should I look at the function when the limit is to $1$? should it be $x=1$ or  $x≠1$? and of course about the first part.

Appreciate all the help , and sorry for the English mistakes.

答:
In part two, the limit is different:

$$\lim_{x\to1}\frac{\sin\pi x}{x-x^2}\stackrel{L'H}=\lim_{x\to1}\frac{\pi\cos\pi x}{1-2x}=\frac{-\pi}{-1}=\pi$$

The rest of part two and part one looks fine: the function $\;f(g(x))=4\;$ is a constant for any $\;x\neq1\;$ , thus that it is the limit when $\;x\to1\;$ .


问:
simple concave two-variable function
I'm looking for a simple continuous concave two-variable function $f(x,y)$, where $0\leq x,y \leq 1$, that satisfies all of the following:
$$\frac{\partial f(x,y)}{\partial x}>0$$ $$\frac{\partial f(x,y)}{\partial y}<0$$ $$\frac{\partial^2 f(x,y)}{\partial x^2}<0$$ $$\frac{\partial^2 f(x,y)}{\partial y^2}<0$$

I was able to find one: $1-e^{-(x\cdot(1-y))}$. However, I would like to find another one without an exponential function. Could anyone give other suggestions or hints please? Thank you!

EDIT: Forgot to mention one another condition. I need $0\leq f(x,y)\leq 1, \forall 0\leq x\leq 1, 0\leq y\leq 1$

答:
Not sure why you want to avoid using an exponential function, but the following function satisfies your constraints:

$f(x, y) = -(y+1)^2 - \frac{1}{x + 1}$

Title: simple concave two-variable function


问:
Is there an analogue of the moduli space of the torus in semi-Riemannian signature?
I'm starting to study Riemann surfaces and already met the fact that Riemann surfaces have both a complex structure and a conformal structure that are, in fact, very closely related.

If we consider a Riemann surface one can classify the different conformal structures and if I correctly understand the space whose points label these different conformal structures is the so-called Riemann moduli space.

For the torus the moduli space is $${\cal M}={\cal H}/{\rm PSL(2,\mathbb{Z}})$$

where ${\cal H}$ is the upper half plane in $\mathbb{C}$ and ${\rm PSL}(2,\mathbb{Z})={\rm SL}(2,\mathbb{Z})/\mathbb{Z}_2$. So the distinct conformal structures of the torus are identified by a complex number $\tau$ with ${\rm Re}(\tau)\leq 1/2$, ${\rm Im}(\tau)>0$ and $|\tau|>1$.

Now, all of this is in the context of Riemann surfaces which are equipped with a conformal structure and hence with an equivalence class of Riemannian metrics.

But I would say nothing stops us from picking the manifold $\mathbb{T}^2=S^1\times S^1$ and endowing it with a Lorentzian metric, like $$g=-d\phi\otimes d\phi+d\psi\otimes d\psi,$$

where $\phi$ and $\psi$ are angle coordinate functions on $S^1$ and $(\phi,\psi)$ is the product chart.

Question: Is there an analogue of the moduli space and of the modular parameter for such a Lorentzian torus?

My intuition says that there should be because we can still consider two Lorentzian metrics conformally equivalent if they are a Weyl rescaling of one another, and we can still talk about a conformal structure as an equivalence class of such metrics.

答:
This is a corrected version of my earlier answer.

Many things are different in the pseudo-Riemannian setting. First of all, while every pseudo-Riemannian metric on a surface locally conformally-flat, there are  pseudo-Riemannian metrics on 2-dimensional tori which are not globally conformally-flat (unlike in the Riemannian case):

Sánchez, Miguel, Structure of Lorentzian tori with a Killing vector field, Trans. Am. Math. Soc. 349, No. 3, 1063-1080 (1997). ZBL0873.53046.

I am unaware of any classification of pseudo-conformal structures on tori in full generality. Hence, I will consider only conformal classes of flat pseudo-Riemannian tori: Quotients $T_{\Lambda}$ of ${\mathbb R}^{1,1}$ by lattices, i.e. subgroups $\Lambda$ of ${\mathbb R}^2$ generated by two translations $T_u, T_v$ along two linearly independent vectors $u, v$.
The choice of the generators $T_u, T_v$ of the lattice means that we are actually working with "marked tori;" the space of marked globally conformally flat Lorentzian tori is the Lorentzian analogue ${\mathcal T}_{1,1}$ of the classical Teichmuller space ${\mathcal T}$ of marked conformal structures on 2-dimensional tori. (The space  ${\mathcal T}$ is identified with the upper half-plane $U$, which is a model of the hyperbolic plane; the modular group $SL(2, {\mathbb Z})$ acts on $U$ via linear-fractional transformations and the moduli space ${\mathcal M}$ of conformal structures on the torus is the quotient of $U$ by the action of $SL(2, {\mathbb Z})$. Below, I will explain, to which extent the same holds for
globally conformally flat Lorentzian tori. First of all, I will consider oriented tori and I will assume that conformal maps are orientation-preserving (as it is done in the classical setting).

This orientation will translate into the assumption that the basis $\{u, v\}$ in ${\mathbb R}^2$ has the standard orientation.

One can show that two marked pseudo-Riemannian tori $T_{\Lambda}$,  $T_{\Lambda'}$ (with the given generators $u, v$ for $\Lambda$ and $u', v'$ for $\Lambda'$)
are pseudo-conformally equivalent  if and only if there exists a pseudo-conformal transformation $A$ of ${\mathbb R}^{1,1}$ such that
$$
Au= u', Av=v'$$
where $A$ is an (invertible) linear transformation preserving the given Lorentzian inner product up to a scale factor. It will be convenient for my answer to assume that this inner product is given by the Gram matrix
$$
\left[\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right],
$$
i.e. is $dx dy$ in terms of differentials. Then the matrix $A$ has to be diagonal with the diagonal entries of the same sign. If I were to write $u, v$ as column vectors in a matrix $B$, then the above condition reads:
$$
AB=B'.
$$
I can also consider the effect of this multiplication on the rows of $B$ and will find that each row gets rescaled. The matrix $B$, of course, contains all the information about the vectors $u, v$. The fact that $u, v$ form a positively oriented basis means simply that $\det(B)>0$. It is convenient (at first) to allow more matrices $B$ than that and assume only that both rows $b_1, b_2$ of $B$ are nonzero. Thus, ${\mathcal T}_{1,1}$ will be contained in a space $X$ obtained from pairs of nonzero vectors $b_1, b_2\in {\mathbb R}^2$
we would be identifying pairs if they differ by rescaling $b_1$ by a positive factor and $b_2$ by a (possibly different) positive factor. This rescaling is done independently for $b_1$ and $b_2$.  The next observation is that the space of nonzero vectors $b\in {\mathbb R}^2$ up to such rescaling is simply the unit circle, since you can just normalize the vector $b$ (divide it by its Euclidean length) to make it a unit vector. Thus, our space $X$ is nothing but the 2-dimensional torus $T^2=S^1\times S^2$ (with points $(z_1, z_2)$, where $z_1, z_2$ are complex numbers of unit absolute value). Now, the torus $T^2$ contains the subset $D$ which corresponds to
pairs of vectors $u, v$ which do not form a basis because they are linearly dependent; in other words, $z_1=\pm z_2$, hence, $D$ consists of two disjoint circles. Removing these two circles, results in two open annuli, one, $A_+$, for the positively oriented bases and the other one for negatively oriented ones. (In fact, we need
to do one more identification, since we can multiply both $u, v$ by  $-1$; this sends $(z_1, z_2)$ to $(-z_1, -z_2)$. This quotient of  $T^2$ is still the 2-dimensional torus.)

Thus, the annulus $A_+$ is our Teichmuller space  ${\mathcal T}_{1,1}$. The trouble comes from an attempt to form the moduli space ${\mathcal M}_{1,1}$ which is the quotient of
${\mathcal T}_{1,1}$ by the action of $SL(2, {\mathbb Z})$. This amounts to "forgetting the basis." Unlike the classical case, this action of $SL(2, {\mathbb Z})$, however, is quite bad: Almost every orbit is dense. Hence, taking the naive quotient we get a highly non-Hausdorff space: almost every point is "infinitesimally close" to any other point.

Algebraic geometers teach us how to work around this problem (the keyword is "stacks"), but I will stop here. The bottom line: Yes, there is a moduli space, just it does not look like a surface (or, for this matter, anything you have seen before in your math or physics classes).

Two more references:

The most comprehensive treatment of pseudo-Riemannian surfaces:

Weinstein, Tilla, An introduction to Lorentz surfaces, De Gruyter Expositions in Mathematics. 22. Berlin: de Gruyter. xiv, 213 p. (1996). ZBL0881.53001.

This one should be highly relevant, but I do not have access to it:

Smyth, Robert; Weinstein, Tilla, How many Lorentz surfaces are there?, Gindikin, Simon (ed.), Topics in geometry. In memory of Joseph D’Atri. Boston, MA: Birkhäuser. Prog. Nonlinear Differ. Equ. Appl. 20, 315-330 (1996). ZBL0860.53039.


问:
Impossible integral?

$$\int_{0}^{x^2-1} f(t) dt= x^6+x^4+3x^2$$

I saw this problem in a calculus exam. $f$ is assumed to be continuous. Using the Fundamental Theorem of Calculus, I calculated $f(t)= 3t^2+8t+8 $. But when I integrate $f$ it gives me $\int_{0}^{x^2-1} f(t) dt= x^6+x^4+3x^2-5$:

$$\frac {d}{dx}\int_{0}^{x^2-1} f(t) dt= \frac {d}{dx}[x^6+x^4+3x^2]$$

$$f(x^2-1)2x = 6x^5+4x^3+6x$$
$$f(x^2-1) = 3x^4+2x^2+3$$
$$f(x^2-1) = 3(x^2-1)^2+8(x^2-1)+8$$
$$f(t) = 3t^2+8t+8$$
$$\int_{0}^{x^2-1} 3t^2+8t+8 dt= x^6+x^4+3x^2-5$$

This means that there is no continuous function that satisfies the equation of the problem?

答:
Suppose $f(t)=at^2+bt+c$

we have
$$\int_0^{x^2-1} \left(a t^2+b t+c\right) \, dt=$$
$$=x^6+x^4+3 x^2\equiv \frac{a x^6}{3}+x^4 \left(\frac{b}{2}-a\right)+x^2 (a-b+c)-\frac{a}{3}+\frac{b}{2}-c$$
and
$$
\begin{cases}
\frac{a }{3}=1\\
\frac{b}{2}-a=1\\
a-b+c=3\\
-\frac{a}{3}+\frac{b}{2}-c=0\\
\end{cases}
$$
the system has no solution, therefore there is no polynomial $f(t)=at^2+bt+c$ which satisfies the equation.

Title: Impossible integral?


问:
Sum $\sum_{x}\sum_{y}xy\left(\frac{1}{2}\right)^{x-1}\cdot\left(\frac{1}{3}\right)^y$?
$P(X=x,Y=y)=\left(\dfrac{1}{2}\right)^{x-1}\cdot\left(\dfrac{1}{3}\right)^y;x,y\ge1$. Find $E(XY)$ .

$$E(XY)=\sum_{x}\sum_{y}xy\left(\frac{1}{2}\right)^{x-1}\cdot\left(\frac{1}{3}\right)^y$$

$\sum_{y}y\left(\dfrac{1}{3}\right)^y\sum_{x}x\left(\dfrac{1}{2}\right)^{x-1}$

here $x,y\ge1$

I used this formula which is the sum of Arithmetic-Geometric progression.

$$a+(a+d)r+(a+2d)r^2\cdots = \frac{a}{1-r}+\frac{dr}{(1-r)^2}$$

$r$= common ratio(GP)

$d$= common difference(AP)

$a$= first term

$\bigg(1+2\cdot\dfrac{1}{2}+3\cdot\dfrac{1}{2^2}....\bigg)\cdot\bigg(1\cdot\dfrac{1}{3}+2\cdot\dfrac{1}{3^2}+3\cdot\dfrac{1}{3^3}....\bigg)=\bigg(\dfrac{1}{1-\frac{1}{2}}-\dfrac{1\cdot\frac{1}{2}}{(1-\frac{1}{2})^2}\bigg)\cdot\dfrac{1}{3}\bigg(\dfrac{1}{1-\frac{1}{3}}-\dfrac{1\cdot\frac{1}{3}}{(1-\frac{1}{3})^2}\bigg)=\bigg(4\cdot\dfrac{1}{3}\cdot\dfrac{3}{4}\bigg)=1$

But this is wrong.  Answer lies between $2.50$ to $3.50$

答:
You have made a miscalculation.  The formula has a plus sign where you have a minus sign:  $$\frac{a}{1-r} \color{red}{+} \frac{dr}{(1-r)^2}.$$

The sum over $x$ is $$\frac{1}{1-1/2} + \frac{1/2}{(1-1/2)^2} = 4.$$  The sum over $y$ is $$\frac{1}{3}\left(\frac{1}{1-1/3} + \frac{1/3}{(1-1/3)^2}\right) = \frac{3}{4}.$$ Their product is $3$.  You have an extra $1/3$ in your calculation because when you calculated the sum over $y$, you already included the $1/3$, but forgot that you did that so you multiplied again by $1/3$.


问:
correlation of product with its normally distributed factors
If x and y are normally dist. with standard deviation of 10%, and they are independent, then their product X.Y is 71% correlated with Y (or X).

I can show this empirically, but how to I prove it in closed form?

答:
Given: Random variables $X$ ~ $N(\mu_1,\sigma_1^2)$ and $Y$ ~ $N(\mu_2,\sigma_2^2)$ are independent with joint pdf $f(x,y)$:

(source: tri.org.au)

You seek:

$$correlation(X Y, Y) = \frac{\text{Cov}(X Y,\text{}Y)}{ \sqrt{\text{Var} (X Y)} \sqrt{\text{Var} Y}}$$

The solution, obtained here using mathStatica, is simply:

(source: tri.org.au)

Given some numerical values, for instance:

(source: tri.org.au)

... the answer would be 2/3 for those particular parameter values. It is easy to do a quick Monte Carlo check:

```<code>  xdata = RandomReal[NormalDistribution[2, 1], 1000000];
  ydata = RandomReal[NormalDistribution[2, 1], 1000000];
  Correlation[xdata * ydata, ydata]
</code>```

0.666552

Looks fine. Your empirical case of 0.71 would presumably fit the parameter values you have been using ...

More generally ...

One can obtain an exact symbolic solution to the correlation you seek, without assuming a specific parametric form (Normality), and without even needing to assume independence.  Here, again derived by mathStatica, is the general solution for any 2 independent random variables (whose first 2 moments exist):

$$\text{correlation}(X Y, Y) = \frac{\text{Var}(Y) \mathbb{E}[X]}{\sqrt{\text{Var}(X) (\mathbb{E}[Y])^2+\text{Var}(Y) \left(\text{Var}(X)+(\mathbb{E}[X])^2\right)}\sqrt{\text{Var}(Y)} }$$

The dependent case is just a bit more messy.

ContentLicense: CC BY-SA 3.0

Body:
Given: Random variables $X$ ~ $N(\mu_1,\sigma_1^2)$ and $Y$ ~ $N(\mu_2,\sigma_2^2)$ are independent with joint pdf $f(x,y)$:

(source: tri.org.au)

You seek:

$$correlation(X Y, Y) = \frac{\text{Cov}(X Y,\text{}Y)}{ \sqrt{\text{Var} (X Y)} \sqrt{\text{Var} Y}}$$

The solution, obtained here using mathStatica, is simply:

(source: tri.org.au)

Given some numerical values, for instance:

(source: tri.org.au)

... the answer would be 2/3 for those particular parameter values. It is easy to do a quick Monte Carlo check:

  xdata = RandomReal[NormalDistribution[2, 1], 1000000];
  ydata = RandomReal[NormalDistribution[2, 1], 1000000];
  Correlation[xdata * ydata, ydata]

0.666552

Looks fine. Your empirical case of 0.71 would presumably fit the parameter values you have been using ...

More generally ...

One can obtain an exact symbolic solution to the correlation you seek, without assuming a specific parametric form (Normality), and without even needing to assume independence.  Here, again derived by mathStatica, is the general solution for any 2 independent random variables (whose first 2 moments exist):

$$\text{correlation}(X Y, Y) = \frac{\text{Var}(Y) \mathbb{E}[X]}{\sqrt{\text{Var}(X) (\mathbb{E}[Y])^2+\text{Var}(Y) \left(\text{Var}(X)+(\mathbb{E}[X])^2\right)}\sqrt{\text{Var}(Y)} }$$

The dependent case is just a bit more messy.


问:
If $f(x)\in\mathbb Q[x]$ and $f(f(x)),f(f(f(x)))\in\mathbb Z[x]$ prove that $f(x)\in\mathbb Z[x]$

If a polynomial $f(x)\in\mathbb Q[x]$ and $$f(f(x)),f(f(f(x)))\in\mathbb Z[x]$$
prove that $f(x)\in\mathbb Z[x]$.

This problem is from 2019 Japan Mathematical Olympiad: see problem 9 (the third day last problem) https://www.imojp.org/archive/mo2019/selection_camp/problems/2019.pdf

Here is what I tried:: Let $f(x)=\dfrac{P(x)}{d}$ where $P(x)\in\mathbb Z[x],d$ is the common denominator of the coefficients of $f(x)$. If let $P(x)=\sum_{i=0}^{n}a_{i}x^i,a_{i}\in\mathbb Z$, then we have
$$f(f(x))=\dfrac{1}{d}\sum_{i=0}^{n}\dfrac{a_{i}}{d^i}\left(\sum_{i=0}^{n}a_{i}x^i\right)^i\in\mathbb Z[x]$$
and $f(f(f(x)))$ is very ugly, so how to solve this problem?

答:
$\newcommand{\d}[1]{\mathtt{\underline{\large{#1}}}}$ This answer can be found in the Math Overflow post by David Speyer here. I give him due credit for it in the comments. I wish to bring to the notice of everybody the arguments used here, so they may attempt to use it in the future. I request feedback from people who wish for me to elaborate on further things.

The Newton Polygon of a $p$-adic polynomial

The Newton polygon of a polynomial has many uses : probably the most relatable use is to prove that the polynomial is irreducible if one can locate the roots of the polynomial. Please read the answer by Captain Lama here for more details on the Newton polygon and its uses.

Nevertheless, we must extend the notion provided in the linked post. For this purpose, I will require knowledge of the following notions.

The $p$-adic numbers $\mathbb Q_p$. Details can be found on the Wikipedia page here.

The $p$-adic integers $\mathbb Z_p$, which is the ring of integers of $\mathbb Q_p$, and to be fair is not a stretch away from $\mathbb Q_p$ in terms of the definition. See the Wolfram Mathworld post here.

The $p$-adic valuation $\nu$ , which may have a couple of different definitions, but in the case of this answer is defined as the biggest integer $e$ such that $p^e$ divides that number. Basically, if we express the number as a traditional $p$-adic sum, then the smallest non-zero index of the sum is the valuation. There is some confusion regarding this since an alternate definition is also known : you can see here for some details. I will use the given definition.

The notion of an algebraic closure of a field, which $\mathbb Q_p$ is. One can see here and here for details. We call the (unique up to isomorphism) algebraic closure as $\mathbb Q_p^{alg}$.

The fact that the $p$-adic valuation extends to a valuation on $\mathbb Q_p^{alg}$. This is explained by David and many others here.

Before going into things , some motivation. The reason we wish to go into $\mathbb Q_p$ is because, by a coefficient-to-coefficient mapping ,it is easy to figure out a map from $\mathbb Z[x]$ to $\mathbb Z_p[x]$. Furthermore, note that if a polynomial $g(x) \in \mathbb Z_p[x]$ for all primes $p$ then $g(x) \in \mathbb Z[x]$, since one can see the contrapositive by taking the LCM of denominators or any prime dividing the denominator.

Definition : Let $f = f_0 + f_1x + ... + f_dx^d$  be a polynomial with coefficients in $\mathbb Q_p$. The Newton polygon of $f$ is the lower convex hull of the points $(0,\nu(f_0)), ..., (d,\nu(f_d))$. That is, it is the unique piecewise straight line path , such that each of these points either lies above the path or on the path, the points corresponding to $0$ and $d$ must lie on the path, and the path can be non-linear only at the given points. Examples can be found on the Wiki page here.

Now, the Newton polygon has slopes associated to it, which are the slopes of the lines forming it. Suppose that the polygon passes through $(j,N_j)$ for $0 \leq j \leq d$. The quantities $s_j = N_j - N_{j-1}$ are called the slopes of the polygon, and by convexity, $s_1 \leq s_2 \leq ... \leq s_d$. (Note : If one wishes to try , please calculate the slope sequence of $P(x) = 1+5x + \frac 15 x^2 + 35x^3$ with respect to the $5$-adic valuation : the Newton polygon is on the Wiki page. I will hide the answer below.)

Note that a slope sequence , along with the location of the left and rightmost points, uniquely determine the polygon.

Before moving , one notation : $\mathbb Q_p[x]$ is the set of all polynomials having coefficients in $\mathbb Q_p$, and likewise $\mathbb Z_p[x]$.

Two results concerning the Newton Polygon

What's beautiful about the Newton polygon is the fact that you can figure out the Newton polygon of a product of polynomials, from the Newton polygon of the individual polynomials. The reason why this is true is that the coefficients of the product can be easily described in terms of the coefficients of the individual polynomials, and then one just needs to do a little bit of hard work figuring out the details. The details may be found in the paper linked here.

Ostrowski's lemma (OL) : Let $f,f'$ be $p$-adic polynomials with slope sequences $s_1,...,s_n$ and $t_1,...,t_m$ respectively. Then the slope sequence of $ff'$ is the sequence $s_1,...,s_n,t_1,...,t_m$ sorted in ascending order.

The next fact links the roots of $f$, which will lie in the algebraic closure $\mathbb Q_p^{alg}$. Recall that these roots have a valuation. The next result is a genius result. There is a proof in Matthew Baker's notes, and an otherwise excellent explanation of another result, in his blog post here.

The fundamental theorem of the Newton polygon (FNTP) : Let $f$ be a $p$-adic polynomial and $\theta_1,...,\theta_d$ be the roots which lie in $\mathbb Q_p^{alg}$. Then , up to a reordering of the roots, we have $\nu(\theta_j) = -s_j$.

Truly astonishing : this links the $p$-adic values of the coefficients to the $p$-adic values of the roots of the polynomial, via the slopes of the Newton polygon. You can see why this result is so amazing : to link coefficients to roots , all you need is the Newton polygon!

All this serves to illustrate the really powerful machinery we are working with. Indeed, if I were to provide a proof sketch now of the theorem stated by the OP, then I would say this : using the Newton polygon of $f$, we can compute part of the Newton polygon of $f \circ f$. The answer then uses the link between the roots of $f$ and $f \circ f$ , and then from the FTNP transfers to the link between the coefficients of these polynomials, which is what we want to study. So a condition on the coefficients of $f$ becomes a condition on the coefficients of $f \circ f$ and so on.

The main lemma

The main lemma proved by David is the following :

Let $f \in \mathbb Q_p[x]$ not have all its coefficients in $\mathbb Z_p$ , but have the constant term $f_0 \in \mathbb Z_p$. Then $f \circ f$ cannot have all its coefficients in $\mathbb Z_p[x]$.

Proof : Note that when we write $f = f_0 +... + f_dx^d$, we get $f \circ f = f_0 + ... + f_d^{d+1}x^{d^2}$. So the leading term is $f_d^{d+1}$. Clearly if $f_d \notin \mathbb Z_p$ then $f_d^{d+1} \notin \mathbb Z_p$, and we are done. Therefore we assume that $f_d \in \mathbb Z_p$.

What can we infer about the Newton polynomial of $f$? Well, $\nu(f_0) , \nu(f_d) \geq 0$, but by assumption one of $f$'s coefficients is not an integer, so $\nu(f_j) < 0$ for some $j$. Thus, the Newton polygon of $f$ strictly dips below zero to accommodate $(j,\nu(f_j))$, and then recovers. This means that the polygon slopes are first negative and then positive : we can order them as $s_1 \leq s_2 \leq ... \leq s_k < 0 < s_{k+1} \leq ... \leq s_d$. Then $(k,N_k)$ is the lowermost point of the Newton polygon. Note that $N_k <0$  since it is the lowermost point and we know that the polygon comes below zero. We keep track of this fact.

Using FTNC, let us get the roots of $f$, labelled so that $\nu(\theta_j) = -s_j$. Make now a very key observation, which links back to the proof sketch I gave :
$$
f(x) = f_d \prod_{j} (x-\theta_j) \implies f\circ f(x) = f_d \prod_{j} (f(x)-\theta_j)
$$

and therefore $f \circ f(x)$ has been expressed as a product of polynomials, whose Newton polygons can be ascertained. We are firmly in OL territory. However, we must break into two cases, for $j$ being on the left and right hand side of $k$ on the polygon.

Let me first explain what we are trying to achieve in the next few lines : essentially, $f \circ f \notin \mathbb Z_p[x]$ means that the Newton polygon of $f \circ f$ dips below the $x$-axis somewhere, since wherever it dips, the coefficient can't be a $p$-adic integer. So we need to study height changes among the slopes of the polygons of $f(x)-\theta_j$ and transfer them to that of $f \circ f$ using OL.

If $1 \leq j \leq k$, then $\nu(\theta_j) \geq 0$. The constant term of $f - \theta_j$ is $f_0 - \theta_j$ , which by the ultrametric nature of the valuation will have a non-negative valuation. The other coefficients don't change. In particular, the increasing part of the slope sequence of $f$ is retained by $f - \theta_j$, and remains $(s_{k+1},...,s_d)$. Thus, the height change of the Newton polygon of $f-\theta_j$ between  $(k,N_k)$ and $(d,N_d)$ is $s_{k+1} + ... + s_d = N_d - N_k$.

If $k+1 \leq j \leq d$ then $\nu(\theta_j) < 0$. The constant term of $f-\theta_j$ is $\nu(f - \theta_j) = \nu(\theta_j)  = -s_j< 0$ by the ultrametric property of the valuation. The right hand end point is $(d,N_d)$. Thus, the height change of the Newton polygon of $f-\theta_j$ between $(k,N_k)$ and $(d,N_d)$ is at least $N_d + s_j$, since the polygon will dip at $(0,-s_j)$ and then go up. It turns out the estimate is sufficient.

Now, we are trying to find the dip of $f \circ f$ from the right endpoint, and if this dip is more than the right endpoint's y coordinate, we are done. By OL, we know that all the slopes of $f(x)-\theta_j$ are to be sorted into increasing order, and for finding the lowest point of the Newton polygon of $f \circ f$ we need to find the non-negative slopes and add them together to get the total drop.

What are those non-negative slopes? Well, there's those from $1 \leq j \leq k$, giving a total drop by $N_d - N_k$, and there are $k$ of those, giving a combined drop of $k(N_d - N_k)$ from these cases. For $k+1 \leq j \leq d$, we have a drop of at least $N_d + s_j$ in each case. Finally the total drop is at least $k(N_d - N_k) + \sum_{j=k+1}^d (N_d + s_j)$.

Recall that the leading term of $f \circ f$ is $f_d^{d+1}$. Using the multiplicative property of the valuation , we get that the right endpoint of the Newton polygon of $f \circ f$ is $(d^2, \nu(f_d^{d+1})) = (d^2,N_d(d+1))$.

So, we are left to ask ourselves : is $k(N_d - N_k) + \sum_{j=k+1}^d (N_d + s_j) > N_d(d+1)$?

Exercise you can give your children : Provide them the context, and let them prove that $$k(N_d - N_k) + \sum_{j=k+1}^d (N_d + s_j) - N_d(d+1) = -(k+1)N_k$$

which is a positive quantity since $N_k < 0$. Therefore, the dip is more than the right endpoint! So the Newton polygon of $f_j$ goes below the $x$ axis at some point : and exhale.

The main theorem

The main theorem can be stated and proved easily now.

Let $g$ have coefficients in $\mathbb Q_p$ and let $g \circ g , g \circ g \circ g $ have coefficients in $\mathbb Z_p$. Then $g$ has coefficients in $\mathbb Z_p$.

Proof : Note that $g(g(0)), g(g(g(0)))  \in \mathbb Z_p$, these are the constant terms of the respective polynomials. Let $f(x) = g(x + g(g(0))) - g(g(0))$. Now, note that $$
f \circ f(x) = g(g(x + g(g(0)))) - g(g(0))  = g \circ g( x + g \circ g(0)) - g \circ g(0)
$$

is a difference of polynomials in $\mathbb Z_p[x]$ hence belongs in $\mathbb Z_p[x]$. Finally, $f(0) = g \circ g \circ g(0) - g \circ g(0)$, hence $f(0) \in \mathbb Z_p$. By the contrapositive of the main lemma , we get that $f$ has coefficients in $Z_p$ (if not, then the main lemma would be contradicted).

But then $g$ has coefficients in $\mathbb Z_p$ too, from the following logic : if $f$ has coefficients in $\mathbb Z_p$,then so does $f - g(g(0))$. But then this is just a $g(g(0))$ shift of the polynomial $g(x)$, and any $\mathbb Z_p$ integer shift retains coefficient membership in $\mathbb Z_p$ from a simple binomial theorem type argument and the ultrametric nature of the valuation.

Thus, we are done. But we can go one step further. Define $h^{(r)}$ to be the $r$-fold composition of $h$, with $h^{(1)} = h$.

Let $h \in \mathbb Q_p[x]$ and $k_1,k_2$ be relatively prime such that $h^{k_1},h^{k_2} \in \mathbb Z_p[x]$ for relatively prime $k_1,k_2$. Then $h \in \mathbb Z_p[x]$.

Proof : Compositions retain membership of coefficients in $\mathbb Z_p$, so from Bezout's lemma with $k_1$ and $k_2$, we conclude that if $S = \{r : h^r \notin Z_p[x]\}$ then $S$ is bounded above. Let $r  = \max S$. But then from the previous theorem, $2r,3r \notin S$ so $r \notin S$ as well, a contradiction. Thus, $S$ must be empty i.e. $1 \notin S$, as desired.

Back to our problem

We haven't addressed the OP's problem yet! We will do that easily.

Let $f(f(x)) , f(f(f(x))) \in \mathbb Z[x]$. Then $f \in \mathbb Z[x]$.

Proof : Well, $f(f(x)) , f(f(f(x))) \in \mathbb Z_p[x]$ for all primes $p$, so $f(x) \in \mathbb Z_p[x]$ for all primes $p$, and hence $f(x) \in \mathbb Z[x]$.

Exercise you can give your children : if for co-prime $k_1,k_2$ we have $f^{(k_1)},f^{(k_2)} \in \mathbb Z[x]$ then $f \in \mathbb Z[x]$.

Well , that was worth it. There's another extension we can make as well.

Let $k_1,...,k_n$ be positive numbers such that $\gcd(k_1,...,k_n) = 1$. Then if $h^{(k_i)} \in \mathbb Z[x]$ for all $x$ then $h \in \mathbb Z[x]$.

Proof : Let $S = \{k : h^{(k)} \in \mathbb Z[x]\}$. Since each $k_i \in S$ we know that $k_i$ is closed under positive linear combinations. Thus two consecutive numbers will be contained in $S$ by a generalization of Bezout's lemma. Then we apply the child case, and we are done!

This result helps , for example, if none of the $k_i$ are pairwise coprime : for example, if $h^{(6)}, h^{(15)} , h^{(21)} \in \mathbb Z[x]$ then $h \in \mathbb Z[x]$.

Conclusions

The theory of Newton polygons is useful, since the fundamental lemma helps link the coefficients and roots of a polynomial together, and then OL helps in putting together the Newton polynomial for product polynomials. Aside from what we've done here, the theory can be used to prove a large number of irreducibility results and can give information about the nature of the Galois groups of the extension formed by the splitting field of polynomials over fields, see Baker's blog for the same.

But somewhere in Japan, I am sure they have a five-line solution waiting for me.


问:
Telescope of CW-complexes
Let $X_n$ the n-th term of an ascending sequence of pointed CW-complexes in their homotopy category. Let $X’= \bigcup_{n\geq 1} [n-1, n]^{+} \wedge X_n$ and let $A_1 = \bigcup_{k\geq 1, k odd} [k-1, k]^{+} \wedge X_k$ and $A_2 = \bigcup_{k\geq 0, k even} [k-1, k]^{+} \wedge X_k$.

Why is it true that $A_1 \cap A_2 = \bigvee_k X_k$ and $A_1$ is homotopically equivalent to $\bigvee_{k,odd} X_k$?

This is my very first approach to smash products and wedge sums, so I might be failing to understand some fundamental aspects of their definition. Any hints?

答:
We have an ascending sequence of pointed CW-complexes $X _1\subset X_2 \subset X_3 \subset \ldots$ with a common basepoint $* \in X_1$. Noting that $X'_n = [n-1,n]^+ \wedge X_n = ([n-1,n] \times X_n) / ([n-1,n] \times *)$, we see that $X'_n$ deformation retracts to the subspace $\bar X_n = \{n\} \times X_n' \subset X'$ which is a homeomorphic copy of $X_n$.

The subspace $\bigcup_n \bar X_n \subset X'$ is nothing else than the disjoint sum of the $X_n$ with all basepoints $* \in X_n$ identified to a single point, i.e. $\bigcup_n \bar X_n \approx \bigvee_n X_n$.

Similarly the subspace $A_ 1 = \bigcup_{n \text{ odd}} X'_n \subset X'$ is nothing else than the disjoint sum of the $X'_n$, $n$ odd,  with all basepoints $* \in X'_n$ identified to a single point, i.e. $\bigcup_{n \text{ odd}} X'_n \approx \bigvee_{n \text{ odd}} X'_n$.

By construction $A_1 \cap A_2 = \bigcup_n \bar X_n$, i.e. $A_1 \cap A_2 \approx \bigvee_n X_n$.

Since each $X'_n$ deformation retracts to $\bar X_n$, we see that $A_1$ deformation retracts to $\bigcup_{n \text{ odd}} \bar X_n \approx \bigvee_{n \text{ odd}} X_n$.

Title: Telescope of CW-complexes


问:
sequence limits - proof by definition
I would like to ask how should I show from the definition of sequence limits that if $\lim a_n = 0$ then $\lim \sqrt{a_n} = 0$. I know how to show it for a specific example, but I don't know how to show it "resulting".

答:
The definition gives
$$\forall\varepsilon>0,~\exists N \in \mathbb N,~ \forall n > N: \quad a_n < \varepsilon^2$$
Then, for these $n$'s we have $\sqrt{a_n} < \varepsilon$. What can we conclude now?

As @DonAntonio has mentioned, I assumed $a_n \ge 0$.

Title: sequence limits - proof by definition


问:
System of differential equations with parameter

Let $x_1$, $x_2$ create the fundamental system of solutions of
$$
\begin{cases}
x_1'=tx_1+x_2\sin(t+1)\\
x_2'=x_1+t^2x_2
\end{cases} \tag{1} \label{1}
$$
Check for which values of parameter $a\in \mathbb{R}$ functions $y_1=2x_1-ax_2$ and $y_2=ax_1-8x_2$ create the fundamental system of solutions of \eqref{1} too.

In order for $y_1$, $y_2$ to create the fundamental system of solutions, they must:

a) be a solution to \eqref{1}

b) be linearly independent
I checked b) from definion of linear independence and got that they are independent $\iff a\in \mathbb{R}\setminus \{-4;4\}$, but I don't know how to check a).

Also, checking my result for b) would be appreciated too.

答:
Hint.

The system

$$
\cases{
x_1'=tx_1+x_2\sin(t+1)\\
x_2'=x_1+t^2x_2
}
$$

can be represented as $\mathcal{D}[x_1,x_2]=0$ and $\mathcal{D}[2x_1-ax_2,ax_1-8x_2]=\left(\begin{array}{cc}2 & -a\\ a&-8\end{array}\right)\mathcal{D}[x_1,x_2]=0$. This occurs uniquely when $a^2\ne 16$


问:
Last 3 digits of $3^{999}$
I know that it's $3^{999} \mod 1000$ and since $\varphi(1000) = 400$ and $3^{400}\equiv1 \mod1000$ it will be equivalent to $3^{199} \mod 1000$ but what should I do from then? Or am I wrong about this from the start?</p>

答:
Giving some structure to the answer by @fruitbat that is effectively using exponentiation by squaring, this answer ignores the advantages that are possible by the consideration of modular order (cycling of values) to focus on an example of what is possible if using a larger or less convenient modulus:

So, here's a tabular form of exponentiation by squaring:

starting with the exponent, work down the left-hand column, subtracting $1$ from odd numbers and halving even numbers. then

starting with the base, work up the right-hand column. multiplying by the base, $3$, or squaring as appropriate, taking the modulus each time.

$\newcommand{oddnote}{{x \mathit{\text{ odd, }{\times}\text{base}}}}
\newcommand{evennote}{{x \mathit{\text{ even, square}}}}$
\begin{array}{c|c}
x & 3^x \bmod 1000 & \mathit{\text{notes}}\\\hline
\bbox[yellow]{\;999\;} & \fbox{667} & \oddnote \\
998 & 889 & \evennote \\
499 & 667 & \oddnote \\
498 & 889 & \evennote  \\
249 & 83 & \oddnote \\
248 & 361 & \evennote  \\
124 & 481 & \evennote  \\
62 & 809 & \evennote  \\
31 & 947 & \oddnote \\
30 & 649 & \evennote  \\
15 & 907 & \oddnote \\
14 & 969 & \evennote  \\
7 & 187 & \oddnote \\
6 & 729 & \evennote  \\
3 & 27 & \oddnote \\
2 & 9 & \evennote \\
1 & \bbox[yellow]{\;3\;} & \mathit{\text{base}} \\
\phantom{\large\Rightarrow}\Downarrow\Downarrow\underset{\large\Rightarrow}{}
&\underset{\large\Rightarrow}{}\Uparrow\Uparrow\phantom{\large\Rightarrow} \\
\end{array}

Title: Last 3 digits of $3^{999}$


问:
Are initial submanifolds a thing?

Question

While studying Lie groups, I came across the notion of an "initial submanifold" mentioned in Definition 2.14. here. Seeing a technical definition like this (see details below) I consulted the great wisdom of the internet (looking at you, Wikipedia and nLab); futilely. In fact, I couldn't even find any other independent source defining the notion of an initial manifold.

On the other hand, a quick google scholar search revealed that this seems to be a well-known concept in the sense that people use it, but no one bothers to define it. Some of these papers are even older than the book I got the definition from (e.g. this one).

So I started to wonder (excuse me if that's not a well-defined question):

How established is the concept of an initial submanifold?

  Why are there so few sources containing an actual definition of it?

And a question that rather addresses math community as a whole:

Can a concept be seen as folklore even though it is so hard to find an actual source defining it?

Background

Recall the "usual" definition of a submanifold:

Definition (embedded submanifold): Let $N$ be an $n$-dimensional manifold.

A subset $M \subset N$ is called an embedded submanifold if for each $p \in M$ there exists a chart $(U,u)$ around $p$ such that $u(U \cap M)) = u(U) \cap \left(\mathbb{R}^m \times \mathbf{0} \right) \subset \mathbb{n}$.

We can weaken this definition by requiring the essential condition only on path-connected components:

Definition (initial submanifold): Let $N$ be a an $n$-dimensional manifold.

1) For any point $p \in N$ and any subset $A \subset N$ we write $C_p(A)$ for the path-connected component of $p$ in $A$.

2) A subset $M \subset N$ is called an initial submanifold if for all $p \in M$ there exists a chart $(U,u)$ around $p$ such that $u(C_p(U \cap M))) = u(U) \cap \left(\mathbb{R}^m \times \mathbf{0} \right) \subset \mathbb{R}^n$.

The name initial submanifold already hints towards a certain universal property:

Definition (universal property): Let $i:M \rightarrow N$ be an injective immersion. We say that $i$ satisfies the universal property if for any manifold $Z$ a mapping $f: Z \rightarrow M$ is smooth if and only if $i \circ f: Z \rightarrow N$ is smooth.

Indeed we have

Lemmas:

1) Let $f: M \rightarrow N$ be an injective immersion between manifolds satisfying the universal property. Then $f(M)$ is an initial submanifold of $N$.

2) Let $M$ be an initial submanifold of a manifold $N$. Then there is a unique $C^{\infty}$-manifold structure on $M$ such that the injection $i: M \rightarrow N$ is an injective immersion.

3) Any initial submanifold $M$ of a manifold $N$ with injective immersion $i: M \rightarrow N$ has the universal property.

And hence (the authors don't mention that one explicitly, so I hope I understood this right):

Theorem: An immersed submanifold is initial if and only if it has the universal property.

答:
I don't know whether initial submanifolds are a 'thing'; however, they are also known as weakly embedded submanifolds and as such have been asked about on Math.SE here.

They're important as:

The inverse image of an intial submanifold by a transversal map is again initial and with the same codimension as the original submanifold has.

Integral submanifolds of an distribution are always initial submanifolds

The leaves of a singular foliation are also initial submanifolds.

Lie subgroups are always initial submanifolds

These examples are all found in the book by Michor, Kolar & Slovak, Natural Operations in Differential Geometry. In fact, they give a short proof that a Lie subalgebra integrates to a Lie subgroup by using the initial properties of a distribution defined by the subalgebra.

Title: Are initial submanifolds a thing?


问:
How to prove that the system of ODE's won't have any negative solutions?
I have a system of ODE's:

\begin{equation}
\begin{aligned}
\dot{S}=-\alpha S(t)I(t),\\
\dot{I}=\alpha S(t)I(t)-\beta I.
\end{aligned}
\end{equation}
I have already drew the graphs for both of the functions. I can see that the functions will never be less than 0, wich is also true because functions represent the number of population.

I know that the $\dot{S}$ function for $S$ in $t=0$ equals 0, which means that the function is constant.

My question is how do I prove that the solutions to the system of ODE's will never be less than 0.

答:
There are in fact negative solutions: if your initial condition has a negative value, it will stay negative (at least for a while).  The correct statement is that if the initial values $S(0)$ and $I(0)$ are positive, the solutions will stay positive for all future time.

One way to see this is that if a solution ever becomes negative, it must pass through $0$.  But the solution with $I(t_0) = 0$, $S(t_0) = s_0$ is constant $I(t) = 0$, $S(t) = s_0$, while the solution with $I(t_0) = i_0$, $S(t_0) = 0$ is $I(t) = i_0 e^{-\beta (t-t_0)}$, $S(t) = 0$.  Thus if $I(t)$ is ever $0$, it must always be $0$, and if $S(t)$ is ever $0$, it must always be $0$
(and have always been $0$).


答: You are correct that this does mean there are no continuous functions satisfying the equation. (There's a quicker way to see this: set $x=1$, then $LHS=0$, but $RHS=5$.)

Title: Impossible integral?

Body:
$$\int_{0}^{x^2-1} f(t) dt= x^6+x^4+3x^2$$

I saw this problem in a calculus exam. $f$ is assumed to be continuous. Using the Fundamental Theorem of Calculus, I calculated $f(t)= 3t^2+8t+8 $. But when I integrate $f$ it gives me $\int_{0}^{x^2-1} f(t) dt= x^6+x^4+3x^2-5$:

$$\frac {d}{dx}\int_{0}^{x^2-1} f(t) dt= \frac {d}{dx}[x^6+x^4+3x^2]$$

$$f(x^2-1)2x = 6x^5+4x^3+6x$$
$$f(x^2-1) = 3x^4+2x^2+3$$
$$f(x^2-1) = 3(x^2-1)^2+8(x^2-1)+8$$
$$f(t) = 3t^2+8t+8$$
$$\int_{0}^{x^2-1} 3t^2+8t+8 dt= x^6+x^4+3x^2-5$$

This means that there is no continuous function that satisfies the equation of the problem?


问:
Injectivity of map $f:\mathbb{Z}\oplus\mathbb{Z}/(2)\rightarrow B$
I have oneissues with this problem

"Let $f:\mathbb{Z}\oplus(\mathbb{Z}/(2))\rightarrow B$ a map of abelian groups, then $f$ is injective if and only if $f(e_{1})$ is free torsion and $f(e_{2})\not=0$ where $e_{1}=(1,0)$ and $e_{2}=(0,1)$"

For the part $(\Rightarrow)$ I proved it by proving the contrapositive and I only needed check two cases when  $f(e_{2})=0$  and when  $f(e_{1})$ is not free torsion. In both cases I can give two different elements whose images are equal and thus the function is not injective.

My question is about the direction $(\Leftarrow)$ I can not see how to check that argument. I tried by contradiction but but I have cases where I am not sure how to do it.

Any suggestion would be great!

Thanks you!

答:
Take an element $\;x=m(1,0)+n(0,1)=(m,n)\in\Bbb Z\oplus\Bbb Z/2\Bbb Z\;$ , then

$$x\in\ker f\implies 0=fx=mf(1,0)+nf(0,1)\;,\;\;\text{for some}\;\;m,n\in\Bbb Z$$

But $\;f(1,0)\;$ is a torsion free element in $\;B\;$, so it must be $\,m=0\;$ (can you see why?), and thus since also $\;f(0,1)\neq0\;$ , it thus must be $\;n=0\;$ , so in fact

$$x=(0,0)\implies \ker f=\{(0,0)\}\;$$

and $\;f\;$ is injective


问:
Intersection of perpendicular chords: is this true for a sphere?
It is true that in a circle with radius $R$, if the intersection of any two perpendicular chords divides one chord into lengths $a$ and $b$ and divides the other chord into lengths $c$ and $d$, then
$$a^2 + b^2 + c^2 + d^2 = 4R^2.$$

Does this equation hold for a sphere with radius $R$, and three mutually perpendicular chords:
$$a^2 + b^2 + c^2 + d^2 + e^2 + f^2 = 6R^2?$$

答:
For circles:

From the circle center draw perpendicular lines to the chords. Let the intersection of the lines with the chords be $X$ and $Y$, respectively. Let $|OX|=x$, $|OY|=y$.

We have (see figure):

$$\begin{align}
a&=\sqrt{R^2-x^2}+y,\\
b&=\sqrt{R^2-x^2}-y,\\
c&=\sqrt{R^2-y^2}-x,\\
d&=\sqrt{R^2-y^2}+x,
\end{align}$$
so that
$$
a^2+b^2+c^2+d^2=4R^2,
$$
as claimed.

Try to apply the same construction for the sphere case.

 Drawing from the center of the sphere the planes perpendicular to the chords and denoting the intersection points $X,Y,Z$ you will obtain: $$a^2+b^2+c^2+d^2+e^2+f^2=6R^2-2(x^2+y^2+z^2),$$ so that the claim does not hold for a sphere.


问:
Checking if a solution exists across two inequalities
If I have

$ay < x < by$

$cx < y < dx$

With $a,b,c,d$ as known (they are real-valued, can be positive or negative or 0) and $x,y$ unknown, is there a methodical way to see if a solution exists for $x$ and $y$?

答:
If solution points $(x; y)$ exist, linearly scaled points $(kx; ky)$ must also be solutions, as no inequality is violated by multiplying both sides with the same factor.

Therefore, we can assert an arbitrary non-zero value for $x$ and calculate the corresponding solution interval for $y$. If the interval is empty, no solution exists.

Example:

$x = 1$

$ay < 1 < by$

$c < y < d$

With known values for $a, b, c, d$ we can immediately see if there exists a feasible value interval for $y$. It cannot be more than one interval as the intersection of two intervals is either empty or one uninterrupted interval.


问:
Square root of a block matrix
Given the $2n \times 2n $ matrix

$$A=\begin{pmatrix}
\bar{A} & A^* \\
A^* & \bar{A}
\end{pmatrix}$$

let $X$ the matrix such that $XX=A$, that is $X=\sqrt{A}.$

So $$X=\begin{pmatrix}
X_{1,1}&X_{1,2}\\
X_{2,1}&X_{2,2}
\end{pmatrix}$$
How can i show that $X_{1,2} X_{1,1} =  X_{1,1}X_{1,2}$, that is that the matrices commutes?

答:
$\def\m#1{\left[\begin{array}{c}#1\end{array}\right]}$Multiplying
two block-wise centrosymmetric matrices yields
$$\eqalign{
\m{A&B\\B&A}\cdot\m{X&Y\\Y&X} &= \m{(AX+BY)&(AY+BX)\\(BX+AY)&(BY+AX)} \\
}$$
Since matrix addition commutes
$\big({\rm i.e.}\;(BX+AY) = (AY+BX)\big)$
the product is also block-wise centrosymmetric. Therefore any integer power of such a matrix maintains the symmetry. And any function of such a matrix (assuming it can written as a convergent power series) exhibits the same symmetry.

Title: Square root of a block matrix


问:
hypothesis testing result problem
For a normal distribution of $N(30,10)$ and sagnificence of 5% (α=0.05),

the sample mean is 32.5 and the population size is 64.

H0: $μ=30$

H1: $μ>30$

A. What is the probability for a second type error?(β)

What I did was, $=P( accept H_0 |   H_0  false)=P(X ̅<30 | μ=32.5)=Φ((30-32.5)/(10/√64))=1-Φ(2)=0.0227$

B. What is the sample size needed for a $90$% discovery chance of the second type error?

what I did was,

$n≥((Z_(1-α)+Z_(1-β) )^2 σ^2)/(μ_1-μ_0 )^2 =((1.65+1.285)^2 ×10^2)/(32.5-30)^2 =137.83$

My problem is,

In part B, the probability for a second type error is $0.1$, and the sample size I get is 138, BUT in part A the probability for a second type error I get is 0.027 and the sample size is 64.

How is that possible that I get a bigger sample size for a higher probability for an error?

The full excercise,

答:
In D they are asking you to calculate $n$ with a power of $90\%$ thus here $\beta=10\%$ is fixed.

This results in

$$P(\overline{X}_n>32.06|\mu=32.5)\geq 90$$

That is

$$\frac{32.5-32.06}{10}\sqrt{n}>1.28$$

$$n=847$$

I did the calculations with paper z-table thus the result can be better approximate

Title: hypothesis testing result problem


问:
Which property of convex functions was used in this step?
I am reading a larger proof about convex functions and came across this step, which is not at all obvious to me.
Take $f$ convex and points $x_1, x_2, x_3$ such that $x_2$ is a convex combination of $x_1$ and $x_3$. Set
$$\lambda=\frac{\lVert x_2-x_1\rVert}{\lVert x_3-x_1\rVert}\in(0,1),$$
then by the convexity of $f$ we have
$$f(x_2)-f(x_1)\le \lambda(f(x_3)-f(x_1)) = \lVert x_2-x_1\rVert\frac{f(x_3)-f(x_1)}{\lVert x_3-x_1\rVert}.$$
This doesn't look like any transformation for convex functions I've ever seen. Can someone please tell me why the inequality is obvious?

答:
It is assumed that $x_2$ lies on the line segment between $x_1$ and $x_3$. That is, there exists some $\lambda \in [0, 1]$ such that
$$x_2 = \lambda x_3 + (1 - \lambda)x_1.$$
Intuitively, you can think of $\lambda$ and $1 - \lambda$ as weights for a weighted average. When $\lambda = 0$, we get back $x_1$. When $\lambda = 1$, we get back $x_3$. When $\lambda = 1/2$, we get a perfect balance between the two: the midpoint. If you want $x_2$ to be twice as far from $x_3$ as from $x_1$, then you want $\lambda = 1/3$.

So, intuitively (to me at least) it makes sense that $\lambda$ could be expressed as the proportion of the distance between $x_2$ and one of the endpoints (in this case, $x_3$), and the length of the whole interval from $x_1$ to $x_3$.

It's also super easy to prove rigorously:
\begin{align*}
x_2 = \lambda x_3 + (1 - \lambda)x_1 &\implies x_2 - x_1 = \lambda x_3 - \lambda x_1 \\
&\implies \|x_2 - x_1\| = \lambda\|x_3 - x_1\| \\
&\implies \lambda = \frac{\|x_2 - x_1\|}{\|x_3 - x_1\|}.
\end{align*}
(Note: $\lambda \ge 0$, so we don't need to put absolute values around $\lambda$.)

What they didn't make clear was the significance of $\lambda$: that it satisfies $x_2 = \lambda x_3 + (1 - \lambda)x_1$. From this, using the definition of convexity in functions,
\begin{align*}
f(x_2) - f(x_1) &= f(\lambda x_3 + (1 - \lambda)x_1) - f(x_1) \\
&\le \lambda f(x_3) + (1 - \lambda)f(x_1) - f(x_1) \\
&= \lambda( f(x_2) - f(x_1)) \\
&= \frac{\|x_2 - x_1\|}{\|x_3 - x_1\|}(f(x_2) - f(x_1)).
\end{align*}


问:
Prove that if $ h \circ f = g $ then $ h $ is an $A$-algebra homomorphism.
Let $f:A\rightarrow B,\ g:A\rightarrow C$ be ring homomorphisms. An $A$-algebra homomorphism $h:B\rightarrow C$ is a ring homomorphism which is also an $A$-module homomorphism.
Please prove that if $h \circ f = g$ then $h$ is an $A$-algebra homomorphism.</p>

答:
We have $h(ax)=h(f(a)x)=h(f(a))h(x)=g(a)h(x)=ah(x)$ for all $a\in A$ and $x\in B$. (Note that $B$ is an $A$-module via $f$, that is, $ax:=f(a)x$ for all $a\in A$ and $x\in B$, while $C$ is an $A$-module via $g$.)

Edit. The question was unclear and I assumed the $h$ is a ring homomorphism, otherwise the claim being (almost) obviously wrong.


问:
Proof that $\;\dfrac{x}{1+x^2}\;$ is a bounded function.
I don't quite understand how $\dfrac1 x$ is unbounded while $\dfrac{x}{1+x^2}$ is not.

答: Look at where $\dfrac 1 x$ goes to infinity,  what causes that unbounded behavior?  Notice that that is avoided in the second function.


问:
Notation for iterating functions
Is there any well established notation for iterating a function a certain amount of times? What I mean is that, let's say you have a function $f(x)$ that you are iterating a certain amount of times:

$$\underbrace{f(f(f(f(\cdots(x)\cdots))))}_\text{$n$ times}$$

Is there some similar notation to e.g. the $\sum$ symbol if we would be talking about sums?

答:
As both the $f^n \circ x$ and the $f^{(n)} \circ x$ notations have their own alternate meanings ($f^n \circ x = \displaystyle\prod_{i = 1}^{i = n} (f\circ x)$, $f^{(n)} \circ x = {d^n f\over dx^n}$), that are distinct from function composition / recomposition.

One proposed operator is the recomposition operator, the reuse of the composition ($\circ$) operator.

This way, your case would become $f\overset{n} \circ x$, where $$f\overset{m + 1} \circ = f\overset{m} \circ f = f\circ f\overset{m}\circ$$

This is also, for bijective functions, extensible from $n\in\mathbb N^*$ to $n\in\mathbb Z$ (or even $n\in\mathbb R$).

Let's say we have $f\circ x = f(x) = 2x$: $f\overset 2\circ x = f(f(x)) = 2(2x) = 4x$, which is easily extensible to $\mathbb R_+$: $f\overset r\circ x = 2^r x$.

For the same function $f$, $f\overset{- 1}\circ x = {x\over 2}$, the reverse¹ of $2x$.

And even $f\overset 0\circ x = x$, the identity function.

Other pro of the recomposition proposal is killing the $\sin^{- 1}, \cos^{- 1}, \tan^{- 1}$ problems.

Ex: $\sin^{- 1} = {1\over\sin} = \csc$, while $\sin\overset{- 1}\circ x = \arcsin\circ x \overset 2= x\circ \sin$

Notes:

Inverse of $f$: ${1\over f}$. The reverse of $f\circ x$: $f\overset{- 1}\circ x$.

Informally / Not so formally, $x\circ f$ ($x$ of $f$) may appear as the reverse of $f\circ x$ ($f$ of $x$).

I did not work all the domains and images / codomains problems / restrictions. It's always good remembering of doing so, mainly with reversions.

LaTeX for the operator: \overset n \circ, or even \overset{n}\circ.

Title: Notation for iterating functions


问:
Uniform distribution on the unit sphere rotated by a random orthogonal matrix
Question 1. Let $u\in \mathbb{R}^n $ be a random vector uniformly distributed on $\mathbb{S}^{n-1}$, and $T\in \mathbb{R}^{n\times n}$ be a random orthogonal matrix. If $u$ and $T$ are independent, is $Tu$ uniformly distributed on $S^{n-1}$ and statistically independent of $T$?

I get this question when thinking about the following one.

Question 2. Let $u, v\in\mathbb{R}^n$ be two independent random vectors, and $u$ be uniformly distributed on $\mathbb{S}^{n-1}$. Consider the inner product $u^\top v$. Take a constant vector $v_0\in \mathbb{S}^{n-1}$. It seems that

$u^\top v$ has the same distribution as $u^\top v_0$;

$u^\top v$ and $v$ are statistically  independent.

Is this true?

Towards a positive answer to Question 2, we let $T$ be the Housholder matrix such that $Tv = v_0$. Note that $T$ is independent of $u$. Then
$$
u^\top v = u^\top (T^\top v_0) = (Tu)^\top v_0.
$$
If the answer to Question 1 is yes, then $Tu$ and $u$ are identically distributed, and $Tu$ is independent of $v$, and hence the answer to Question 2 is yes. (BTW, are Question 1 and Question 2 equivalent?)

Any comments or criticism will be appreciated. Thank you.

(A different but related question: $X$ is independent of $\mathcal{G}$, $f(X ,Y)$ is independent of $Y$, $Y$ is $\mathcal{G}$-measurable, then $f(X,Y)$ is independent of $\mathcal{G}$?)

答:
Let $X$ be a standard Gaussian in $\mathbb{R}^n$. In spherical coordinates, this may be expressed as $(R,u)$ where $R \in [0,\infty)$ and $u$ are independent, and $u$ is uniform on $\mathbb{S}^{n-1}.$

Let $\mathbf T$ is a random orthogonal matrix independent of $X$, with law $\mu$. Notice that $|\det(T)| = 1$ for any orthogonal matrix. So, using Fubini's theorem a couple of times, and a change of coordinates $$P(\mathbf{T}X \in A) = \int_T\int_{x \in T^{-1}(A)} e^{-\|Tx\|^2/2}\mathrm{d}x\,\mu(\mathrm{d}T) = \int_T\int_{y \in A} e^{-\|y\|^2/2}\,\mathrm{d}y\,\mu(\mathrm{d}T)\\ = \int_{x \in A} e^{-\|x\|^2/2}\mathrm{d}x = P(X \in A),$$ and so $\mathbf{T}X$ is also a standard Gaussian. But $\mathbf{T} (R,u) = (R, \mathbf{T} u),$ since $\mathbf{T}$ is norm preserving, so it follows that $\mathbf{T}u$ has the same law as $u$.

Finally, notice that the above calculation was completely agnostic to the law of $T$. This implies that if I replaced the law of $T$ with something else - for instance a conditional law given that $T$ lies in some set, I'll get the same calculation out. Of course this means that $$ P(\mathbf{T} X \in A| \mathbf{T} \in \tau) = P(X \in A) = P(\mathbf{T}X \in A).$$ It follows that $\mathbf{T}X,$ and thus $\mathbf{T}u,$ are indepdendent of $\mathbf{T}$.


问:
A more logical approach to a word problem
I was taught in high school algebra to translate word problems into algebraic expressions. So when I encountered this problem I tried to reason out an algebra formula for it

For every dollar Ben spent on bagels, David spent 25 cents less. Ben
paid $12.50 more than David. How much did they spend in the bagel store
together?

To solve this I imagined a series of comparisons when Ben spends $x$, David spends $.75x$. Loop this relationship until $x - .75x \approx 12.50$. Good. Done. $x = 50$, then add David's for the answer. Coming from computers, I would have set this up in code where a loop (recursion) would increase $x$ until the condition $x - .75x = 12.50$ was met, then the "loop counter/accumulator" would be how much Ben spent, i.e., $50$, etc.

I'm a beginner with math, but it seems like there should be a better approach, something with series and sequences or even calculus derivatives, something better than my brute-force computer algorithm. Can someone enlighten? The "answer" given at the site (see link) is its own brute-force and hardly satisfying. I'm thinking there should be something more formal -- at least for the first part that derives $50$.

Update

I think everyone so far has missed my point. Many of you simply re-did the problem again. I'm wondering if there is a more formal way to do this other than just "figuring it out" (FIO). The whole FIO routine is murky. It looks like a limit problem; it looks like a system of equations, but I'm not experienced enough to know exactly. If there isn't, then let's call it a day....

答:
Another way, with no $x$'s needed:

The first condition says the difference between their spending is $1/4$ of what Ben spends.  That difference is $\$12.50$, so Ben's amount is $4 \times \$ 12.50 = \$50$.


问:
We got $x \cdot f\left (\frac {x} {2} \right) - f\left (\frac {2} {x} \right) = 1$? Can you find all $f : \mathbb {R} _0 \to \mathbb{R} $
Hello We got $x \cdot f\left (\frac {x} {2} \right) - f\left (\frac {2} {x} \right) = 1$? Can you find all $f : \mathbb {R} _0 \to \mathbb{R} $?

答:
Let $x \in \mathbb R \setminus \{0\}$. Then, by plugging in $x$ and $\frac4x$ into the functional equation, we have

$$1 = x f\left(\frac{x}2\right) - f\left(\frac2x\right) = \frac4x f\left(\frac2x\right) - f\left(\frac{x}2\right).$$

In particular, by isolating the "$f\left(\frac{x}2\right)$ terms" on one side and the "$f\left(\frac2x\right)$ terms" on the other side, we obtain

$$(x+1)\cdot f\left(\frac{x}2\right) = \frac{x+4}x \cdot f\left(\frac2x\right).$$

For $x = -4$, it follows that $f(-2) = 0$.

For $x \neq -4$, we have

$$f\left(\frac2x\right) = \frac{x(x+1)}{x+4} \cdot f\left(\frac{x}2\right).$$

Substituting back into the original functional equation, we obtain that

$$1 = \frac{3x}{x+4} \cdot f\left(\frac{x}2\right) \ \ \ \ \ \text{ for each } x \neq 0, -4.$$

Hence, after replacing $x$ by $2x$,

$$f(x) = \frac{x+2}{3x}.$$

Note that this solution is also defined at $-2$ and gives the desired value $f(-2) = 0$.


问:
What do we call the domain and codomain of a functor? Do they have special names?
Do the domain and codomain categories of a functor have special names, or are they simply called the domain and codomain? It seems to me that it might not be appropriate to call them as such, since we usually do this when we're talking about functions, whose domains have to be sets. But the "domain" of a functor need not be a set.</p>

答: The usual terminology for morphisms (so in particular functors) in category theory is that of source and target. So if $C,D$ are categories and $F\colon C\rightarrow D$ is a functor, you can refer to $C$ as the source (or source category, if there is possibility of ambiguity) of $F$ and $D$ as the target (or target category) of $F$.


答:
Regarding $x/(1+x^2)$, it may help to analyze the cases $|x| \leq 1$ and $|x| > 1$ separately.

In the former case you have a continuous function everywhere defined on a closed interval (here it's important that there's squaring in the denominator; nothing is getting close to zero down there), so it's going to be bounded. You can also do algebra with the hypothesis $|x| \leq 1$ and derive an explicit bound, or be more sophisticated and verify that a particular number is the least upper bound for the function on that interval.

For $|x| > 1$ one can leverage the fact that in this instance $|x| < |x|^2$ and hence $|x/(1 + x^2)| = |x|/(1 + |x|^2) \leq |x|^2/(1 + |x|^2)$ and this last fraction is less than $1$ because it has the form $a/b$ with $a$ and $b$ positive numbers and $a < b$.

Title: Proof that $\;\dfrac{x}{1+x^2}\;$ is a bounded function.

Body: I don't quite understand how $\dfrac1 x$ is unbounded while $\dfrac{x}{1+x^2}$ is not.


问:
If $\int_0^\infty f(x)\,dx$ exists, then there exists some function $g$ such that $f=o(g)$ and $\int_0^\infty g(x)\mathrm\,dx$ exists
If $f$ is a positive function such that $\int_0^\infty f(x)\,\mathrm dx$ converges, show that there is another function $g$ such that $\lim_{x \to \infty} \frac {g(x)} {f(x)} = \infty$ and $\int_0^\infty g(x)\,\mathrm dx$ also converges.

答:
For every $x\geqslant0$, let $F(x)=\int_x^\infty f(t)\,\mathrm dt$ and $g(x)=\frac{f(x)}{\sqrt{F(x)}}$, then:

$\int_0^\infty g(t)\,\mathrm dt=2\sqrt{F(0)}$ is finite,

$f(x)=o(g(x))$ when $x\to\infty$.

These two assertions hold for the same reason, which is that $F(x)\to0$ when $x\to\infty$.

Exercise: Adapt this example to show that for every nonnegative sequence $(a_n)$ such that $\sum\limits_na_n$ converges there exists some nonnegative sequence $(b_n)$ such that $a_n=o(b_n)$ and $\sum\limits_nb_n$ converges.


问:
Probability of lines in a pentagon intersecting internally or at vertices.
I'm studying maths as a hobby.

The points A,B,C,D & E are the vertices of a regular pentagon. All possible lines joining the pairs of these points are drawn. If two of these lines are chosen at random, what is the probability that their point of intersection is (a) inside the pentagon, (b) one of the points A,B,C,D,E?

I start out by saying the number of ways of choosing 2 lines from 10 is $\binom{10}{2} = 45$

Then I started getting confused. I can see there are 5 internal points of intersection, which I mark in red, so that would give me $\frac{5}{45} = \frac{1}{9}$ probability of choosing lines which meet internally. And this is indeed the answer my text book gives. But I'm not sure if this is just coincidence.

As for (b), I'm not sure how to proceed. The book gives the answer as $\frac{2}{3}$

答:
Your approach for (a) is correct. I would clarify that the intersection points in the inside of the pentagon are formed by unique pairs of diagonals, which means that there would be $5$ pairs of lines that would intersect inside the pentagon. If, for example, two distinct pairs of diagonals intersected at a single intersection point inside the pentagon(which is impossible), then it would not be $5$ possible pairs.

For (b), we would first calculate the number of possible pairs, which is $45$ as you stated. Now assume that the intersection point is $A$ for now. There are $4$ lines extending out from $A$, and choosing any two of those $4$ lines will result in a pair of lines with an intersection point at $A$. Therefore there is $\binom42=6$ pairs that would intersect at $A$. By symmetry, the other vertices will have the same amount of pairs that would intersect at that vertex, so we can multiply by $5$, the number of vertices in a pentagon, to obtain $5\cdot6=30$ total pairs.

Thus, the final probability is $\frac{30}{45}=\frac23$ as stated in your question.


答:
a) There are 5 intersections and each of the intersection is an intersection of two line segments. So you need to choose both of them to make them itersect at that point. So you have one possibility for each intersection. That is, 5 possibilities out of 45. The probability is $\frac19$.

b) There are 5 intersections and each of them is an intersection of four line segments. To make the choosen line segments intersect in a given intersection, you have to choose 2 out of 4 line segments, i.e. there are $\binom42=6$ possibilities. And there are 5 intersections, so you have 30 possibilities out of 45 in total. That means, the probability is $\frac23$.

Title: Probability of lines in a pentagon intersecting internally or at vertices.

Body:
I'm studying maths as a hobby.

The points A,B,C,D & E are the vertices of a regular pentagon. All possible lines joining the pairs of these points are drawn. If two of these lines are chosen at random, what is the probability that their point of intersection is (a) inside the pentagon, (b) one of the points A,B,C,D,E?

I start out by saying the number of ways of choosing 2 lines from 10 is $\binom{10}{2} = 45$

Then I started getting confused. I can see there are 5 internal points of intersection, which I mark in red, so that would give me $\frac{5}{45} = \frac{1}{9}$ probability of choosing lines which meet internally. And this is indeed the answer my text book gives. But I'm not sure if this is just coincidence.

As for (b), I'm not sure how to proceed. The book gives the answer as $\frac{2}{3}$


问:
The initial value problem $|\frac{dy}{dx}|+2|y|=0$, $\,$ $y(0)=1$ has no solution.
The initial value problem $|\frac{dy}{dx}|+2|y|=0$, $\,$ $y(0)=1$  has no solution.

How to do this?

Should I find the find the solutions of  $\frac{dy}{dx}+2y=0$ $\,$ and$\,$  $\frac{dy}{dx}-2y=0$ $\,$ these O.D.E's?

Any hint will also help me.

答: At $x=0$, we must have $\big\lvert\frac{dy}{dx}\big\rvert_{x=0}=-2|y(0)|=-2$, which is obviously impossible. (Basically, because you want the sum of two absolute values to be $0$, it follows that they both must always be $0$.)


答:
We need to organize the situation properly. There are three types of intersections:- side-side ($SS$), side-diagonal ($SD$), diagonal-diagonal ($DD$).

Now a very important observation : every point of intersection counts a pair of lines.

The inside five points result from $DD$. These are $5$ pairs. Hence probability $5/45=1/9$.

The vertices result from one $SS$ pair, one $DD$ pair and four $SD$ pairs. For example $A$ is intersection point of $SS$ - $(AE,AB)$; $DD$ - $(AD,AC)$; and $SD$ - $(AE,AD)$, $(AE,AC)$, $(AB,AD)$, $(AB,AC)$. Hence probability of

$$\frac{5+5+4\cdot 5}{45}=\frac{30}{45}=\frac{2}{3}$$


答:
Notice that there is exactly one pair of lines for each of the red dots, so there are in fact only five pairs that intersect inside the pentagon.

As for the vertices, there are four lines for each one, so for each vertex there are six ways to choose a pair of lines that meet at that vertex. Now, if two lines meet at one vertex they will not meet at any other vertex, so the total number of lines that meet at a vertex is 6*5 = 30 and the probability is then 30/45 = 2/3.


问:
Finding solutions to $[x]^2 + 2[x]-3x=0$
Related question, I was attempting the question in an alternative way I wrote $ \{ x \} = x-[x]$, then after subbing that into the equation I got the following quadratic :

$$ [x]^2 + 2[x]-3x=0$$
I rearrange this as:

$$ [x]^2 + 2 [x]  = 3x$$

Adding one:

$$ ( [x] +1)^2 = 3x+1$$

Or,

$$ [x] + 1 = \pm \sqrt{3x+1}$$

Now, I use property to take in $1$ to the box function:

$$ [1+x] = \pm \sqrt{3x+1}$$

Now , I substitute $ x+1 = u$, this leads to:

$$ [u] = \pm \sqrt{3u-2}$$

The bound on $u$ is given as $ u \geq \frac23$, it is clear that the $ [u]= - \sqrt{3u-2}$ has no solutions because left side is always positive for $u \geq \frac23$ , hence can't be equal to negative number of RHS. This leads to a single inequality:

$$ [u] = \sqrt{3u-2}$$

This leads for checking when $3u-2$ is perfect square:

$$ 3u-2 = 0$$

$$ 3u-2=1 \implies u=1$$
$$ 3u -2 = 4 \implies u=2$$
$$ 3u-2 =9 \implies u= \frac{11}{2}$$

Seems like there are no solutions for $ u >2$ , so how do I proof that this is the case?

答:
$[u] = \sqrt{3u - 2}$ has no solution if $\sqrt{3u-2} < u-1$ since $[u] > u-1$.

This is equivalent to $3u-2 < u^2-2u+1$, or $u^2-5u+3 = (u-2.5)^2 - 3.25 > 0$

Hence there is no solution for $u$ if $u > \sqrt{3.25} + 2.5 \approx 4.30278$.


问:
Transitive groups
Someone told me the only transitive subgroup of $A_6$ that contains a 3-cycle and a 5-cycle is $A_6$ itself.

(1) What does it mean to be a "transitive subgroup?"  I know that a transitive group action is one where if you have a group $G$ and a set $X$, you can get from any element in $X$ to any other element in $X$ by multiplying it by an element in $G$.  Is a transitive subgroup just any group that acts transitively on a set?  And if so, does its transitiveness depend on the set it's acting on?

(2) Why is $A_6$ the only transitive subgroup of $A_6$ that contains a 3-cycle and a 5-cycle?

Thank you for your help :)

答:
Let $H\leq A_6$ be transitive and generated by a 3-cycle and a 5-cycle.

Let if possible, $H\neq A_6$, and let us compute $|H|$.

$|H|$ is divisible by 15, and divides 360$=|A_6|$, so it is one of $\{15,30,45,60,90,120,180\}$.

$|H|$ can not be $\{90,120,180\}$, otherwise we get a subgroup of $A_6$ of index less than $6$.

$|H|$ can not be 15, since then $A_6$ will have an element of order 15, which is not possible,

$|H|$ can not be 45, since a group of order 45 is abelian and so it contains an element of order 15.

$|H|$ can not be 30, since a group of order 30 has normal Sylow-5 subgroup, and so it will contain a subgroup of order 15, hence an element of order 15.

Hence  $|H|$ should be $60$.

Now in this subgroup of order 60, Sylow-5 subgroup can not be normal, since if it is normal, then it will also be normalized by an element of order 3, giving a subgroup of order 15, hence an element of order 15.

So $H$ is a group of order $60$, which has no normal Sylow-5 subgroup; $H$ must be isomorphic to $A_5$. There are 6 Sylow-5 subgroups of $H\cong A_5$, hence 24 elements of order 5; they will be 5 cycles, hence fixing an element in $\{1,2,...,6\}$. Let $(12345)\in H$.

As $H$ is transitive subgroup, there will be an element $\sigma \in H$ such that $\sigma(6)=1$, so $\sigma (12345)\sigma^{-1}\in H$, will be a 5-cycle fixing 1; in this way all Sylow-5 subgroups of $A_6$, and hence all element of order 5 of $A_6$ will be in $H$, exceeding the size of $H$.

Hence we must have $H=A_6$.

Title: Transitive groups


问:
Norm of the operator $T(x_1,x_2,\cdots , x_n,\cdots )=(x_2-x_1, x_3-x_2,\cdots, x_{n+1}-x_n, \cdots).$
Let $T$ be a linear operator from  $l_1$ space  to itself defined as  $$T(x_1,x_2,\cdots , x_n,\cdots )=(x_2-x_1, x_3-x_2,\cdots, x_{n+1}-x_n, \cdots).$$ Then which of the following statements are true ?

$1.$ $\|T\|=1$.

$2.$ $\|T\|>2$.

$3.$ $1<\|T\|\leq 2.$

$4.$ none .

I tired it as $$\|T(x)\|=\|T(x_1,x_2,\cdots , x_n,\cdots )\|=\|(x_2-x_1, x_3-x_2,\cdots, x_{n+1}-x_n, \cdots)\|=\sum |x_{i+1}-x_i|\leq 2\|x\|.$$ So $\|T\|\leq 2$. Now how to process further ? If I choose $x=(1,0,0,\cdots)$ then $$\|T(x)\| =\|x\|$$ Now I am confused. Thank you .

Body:
You can choose $x = (0,1,0,\ldots)$ and have
$$\|Tx\| = \|(1,-1,0,\ldots)\| = 2 = 2 \|x\|.$$
Thus, $\|T\| \ge 2$.


问:
simple concave two-variable function
I'm looking for a simple continuous concave two-variable function $f(x,y)$, where $0\leq x,y \leq 1$, that satisfies all of the following:
$$\frac{\partial f(x,y)}{\partial x}>0$$ $$\frac{\partial f(x,y)}{\partial y}<0$$ $$\frac{\partial^2 f(x,y)}{\partial x^2}<0$$ $$\frac{\partial^2 f(x,y)}{\partial y^2}<0$$

I was able to find one: $1-e^{-(x\cdot(1-y))}$. However, I would like to find another one without an exponential function. Could anyone give other suggestions or hints please? Thank you!

EDIT: Forgot to mention one another condition. I need $0\leq f(x,y)\leq 1, \forall 0\leq x\leq 1, 0\leq y\leq 1$

答: $p(x, y) = -x^2 - xy - y^2 + 4x - y$ is an example. The eigenvalues of the Hessian are $-3$ and $-1$.

Title: simple concave two-variable function


问:
An onto function $f: \mathbb{R}^2 \rightarrow \mathbb{R}^4$
Does there exist an onto function $f: \mathbb{R}^2 \rightarrow \mathbb{R}^4$? If $f$ is required to be linear then the answer is no, this follows from basic ideas of linear algebra.

In the case where we allow for $f$ to be an arbitray function, I believe the answer is that yes, such a function $f$ does exist, simply because the domain and codomain both have the same cardinality (uncountably infinite) and thus there is a bijection between the two.

How far does this generalize? If it is true then it must also be true that for each $n$ there exists a function $f$ such that $f: \mathbb{R} \rightarrow \mathbb{R}^n$ is onto.

What about $f: \mathbb{R} \rightarrow \mathbb{R}^{\mathbb{N}}$?

What about $f: \mathbb{R} \rightarrow \mathbb{R}^I$ where $I$ is an uncountable infinite set?

Thank you.

答:
Finite case

Yes, it is true that for each $n \in \Bbb N$, there exists a function $f_n : \Bbb R \to \Bbb R^n$ which is onto.

This is simply because $\Bbb R$ and $\Bbb R^2$ have the same cardinality and thus, by induction, so does $\Bbb R^n$. This means that there exists a bijection from $\Bbb R$ to $\Bbb R^n$.

Countable case

This is again true. As earlier, $\Bbb R^\Bbb N$ has the same cardinality as $\Bbb R$.

Here's a sketch of the proof: Show that $\{0, 1\}^\Bbb N \cong \Bbb R$ ($\cong$ denotes bijection of sets) and hence note

$$\Bbb R^\Bbb N \cong \{0, 1\}^{\Bbb N \times \Bbb N} \cong \{0, 1\}^\Bbb N \cong \Bbb R.$$

Uncountable case

As noted in the comments, simply saying that $I$ is uncountable does not tell us the cardinality of $I$. For example, $|I|$ could be $|\Bbb R|$ or $|\cal P(\Bbb R)|$ (or something else altogether).

I suspect that you wished to say $I \cong \Bbb R$. In this case, note that

$$|\Bbb R^\Bbb R| \ge |\{0, 1\}^\Bbb R| = |\cal P(\Bbb R)|.$$
It is a standard (and easy) theorem that there is no surjection from any set onto its power set. Thus, we see that there's no onto function from $\Bbb R$ onto $\Bbb R^\Bbb R$.


答:
Here it is an alternative approach for the sake of curiosity.

\begin{align*}
|x^{2} + 1| = x^{2} + 1 \geq 2\sqrt{x^{2}} = 2|x| \Longleftrightarrow \left|\frac{x}{x^{2}+1}\right| \leq \frac{1}{2} \Longleftrightarrow -\frac{1}{2} \leq \frac{x}{x^{2}+1}\leq\frac{1}{2}
\end{align*}

and we are done.

Hopefully this helps!


问:
Solving $e^{-x} + x e^{-x} + \frac12x^2e^{-x} + \frac16x^3 e^{-x}=0.99$
Equation:

$$e^{-x} + x e^{-x} + \frac12 x^2 e^{-x} + \frac16x^3e^{-x}=0.99$$

I tried simplifying it and got it down to

$$e^{-x}\left(6+6x+3x^2+x^3\right)=5.94$$

although I am not sure how I would solve the $(6+6x+3x^2+x^3)$ portion.

Thanks to Wolfram Alpha I know the roots (solution for $x$), so I am really looking for guidance on how you would approach this equation and how far you can simplify it before having to use a calculator.

答:
The roots of the cubic polynomial are distinct, so this requires a generalization of Lambert's W function to a cubic polynomial. You can find an exposition on this here:

Numerics of the Generalized Lambert W Function - Scott, Fee:
https://www.researchgate.net/publication/275483330_Numerics_of_the_Generalized_Lambert_W_function

I'm not sure if there's an easy closed-form solution for every possible root, but you can follow the above guide to do a simple iteration to converge on the roots.


问:
Tips for proving: $\Gamma(n-a)/\Gamma(-a)=(-1)^na(a-1)(a-2)...(a-n+1),~$ for $~n = 1, 2, ...$?
I am trying to prove how $\Gamma(n-a)/\Gamma(-a)=(-1)^na(a-1)(a-2)...(a-n+1),~$ for $~n = 1, 2, ...$? So I believe that the numerator is $\Gamma(n-a)=(n-a-1)!=((n-a)!)/(a-n)$, and I proved this by using an analogy to the recurrence relation, and checked it with Wolfram Alpha. However, I do not know what to do with the denominator. The lecture notes state, $1/\Gamma(-n)=0$, where n = 0, 1, 2... However, the domain of "a" is not specified in the homework (the problem is given just as I have stated it in the title), so I am assuming that "a" can be any number, complex, negative, noninteger, etc.. So now I don't know where to start.

One route I was thinking of is using integration by parts and using a pattern that may come up to solve the integral (since simply integration by parts will go on endless without recognizing a pattern). In class today, he implied that the answer should be relatively simple to get, so it makes me think that integration by parts is not what he intended. Another route I was wondering about is if there is a relevant identity/formula for the Gamma function that I can use that may help me by analogy? Any starting point would be great, and especially in generic terms/concepts so that I can apply those tips to any similar problem (rather than an answer to the question I have).

This is my first time posting so hopefully I included all relevant information.

答: The right-hand side is $\prod_{i=0}^{n-1}(i-a)$. Write $i-a$ in the form $\frac{\Gamma(z+1)}{\Gamma(z)}$, then telescope.