问:
Separation axioms of topology
Let $X=Z_+×Z_2, Y=Z_2×Z_+$ with the dictionary order topologies, are $X$ and $Y$,$T_1$,$T_2$ and $T_3$?

$Z_+=\{1,2,3,...\}$

$Z_2 =\{0,1\}$

The open set on the dictionary ordering will be
$I=\{x×y∈\mathbb{R×R}:a×b<x×y<c×d\}$

Now, how to deal with separation axioms $T_1$,$T_2$ and $T_3$?

I am a fresh masters student in pure mathematics. After several days, I have an exam in the Topology course while I was trying to solve of many questions I faced this question and I could not deal with it well, especially since the teaching of this semester is electronic. I want your help.

答:
Every space whose topology is determined in this way by a linear order is $T_1$ and hereditarily normal, so it is $T_1$, $T_2$, and $T_3$. The first two are very easy to prove.

Let $\langle X,\le\rangle$ be a linearly ordered set with the order topology: this is the topology generated by the subbase of open rays, the sets of the form $(\leftarrow,x)$ and $(x,\to)$. If it has no first or last element, it has a base consisting of the open intervals $(x,y)$ for $x,y\in X$ with $x<y$. If it has a first element $a$, one has to add the intervals $[a,x)$ for $a<x$ to the base, and if it has a last element $b$, one has to add the intervals $(x,b]$ for $x<b$ to the base.

Now let $x$ and $y$ be distinct points of $X$. Without loss of generality we may assume that $x<y$. Note that if we prove that $X$ is $T_2$, we can automatically conclude that $X$ is $T_1$. There are two cases; I’ll point them out but leave some of the work to you.

If there is no point $z$ such that $x<z<y$, i.e., if $(x,y)=\varnothing$, let $U=(\leftarrow,y)$ and $V=(x,\to)$. Verify that $U$ and $V$ are disjoint open nbhds of $x$ and $y$, respectively.

If there is a point $z\in(x,y)$, so that $x<z<y$, use $z$ to define disjoint open rays containing $x$ and $y$.

This is overkill for your problem, but it’s easy and worth knowing. A proof that spaces with linear order topologies are always $T_3$ is significantly harder, so for that part of the problem you should concentrate on your specific spaces. Show that one of them has the discrete topology, and the other has only one non-isolated point (i.e., one limit point). Once you’ve done that, you should have little trouble verifying that they are $T_3$, since you’ll know exactly what the closed sets are.

Title: Separation axioms of topology


问:
A question regarding Hahn Banach theorem

Let $Y=\{(a_n)\in l^2 : a_2=2a_1 , a_4=4a_3\}$.
Is there exist a functional $f \neq 0 \in X^*$ such that $f(Y)={0}$.

I know that it can be done using the Hahn-Banach corollary, which says, if $X$ is a normed space, $Y\subset X$ a closed sub-space, and $x \in X\setminus Y$, then there is a functional $f \in X^*$ such that $\|f\|=1$ ,  $f$ restricted to $Y$ equals $0$ ,and $f(x)=d=\inf_{y\in Y} \|x-y\|$.

Clearly $l_2$ is a normed space. And we can find a sequence $a_n\in l^2\setminus Y$ with $\|a_n\|_2=1$.
In order to show that $Y$ is closed in $X$ I have to take a sequence $b_n\in Y$ such that $\lim_{n\to \infty} b_n =x$ and show that $x\in X$ but is not that obvious according to $Y$'s definition?

Any support would be kindly welcomed.

答:
Since norm-convergence in $\ell^2$ implies pointwise convergence (each entry of the sequence converges), it follows that $Y$ is a closed subspace of $\ell^2$. Moreover, it is easy to see that $\ell^2 \neq Y.$ Then invoke the following corollary of the Hahn-Banach (separation) theorem:

Let $V$ be a normed space and $Y$ be a proper closed subspace of $V$. Then there exists a non-zero continuous functional $\tau \in V^*$  with $\tau(Y) = \{0\}$.

ContentLicense: CC BY-SA 4.0


问:
find a third vector to form a set that spans $\mathbb R^{3}$
i have two vectors that i need to combine with a third unknown vector to create a spanning set, but i can't figure out how to do this. i have written the two vectors i have below. this is from a singular value decomposition problem.

$u_1 = (\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0)$

$u_2 = (\frac{-1}{3\sqrt{2}}, \frac{-1}{3\sqrt{2}}, \frac{-4}{3\sqrt{2}})$

i am trying to solve a problem based on an example in my e-text, but they kind of skipped over this part and i can't figure out how to get this similar result for the vectors i'm working with. i have $u_1$ and $u_2$, but i can't figure out how to get $u_3$.

any advice? thank you in advance!

答:
For any set of $n - k$ linearly independent vectors in an $n$-dimensional vector space, we can extend to a basis by appending some $k$ of the standard basis vectors $e_1, \ldots, e_n$ (proof: add all of them in and sift).

In this case, it's easy to see that $(0,0,1)$ is linearly independent of your set of vectors, so that one will do.


问:
Assume $f:[0,1] \rightarrow \mathbb{R}$ is continuous and that $f(x) \geq 2 \quad \text{ if } 0 \leq x \lt 1$. Show that $f(1) \geq 2$
This question comes from Advanced Calculus, Fitzpatrick. Section 3.1 exercise 7

Suppose that the function $f:[0,1] \rightarrow \mathbb{R}$ is continuous and that $$f(x) \geq 2 \quad \text{ if } 0 \leq x \lt 1$$
Show that $f(1) \geq 2$

Fitzpatrick uses the sequence definition of continuity, there is a similar question [here] (Suppose that the function $f:[0,1]\rightarrow \mathbb{R}$ is continuous and that $f\left(x\right)>2$)

I wanted to do this using epsilon-delta method. Is the following valid? Is there a simpler way to write this proof?

My attempt

First, assume $f(1) < 2$. Then there exists some $\varepsilon > 0$, call it $\varepsilon_0$, such that $2 - f(1) > \varepsilon_0$. Show this creates a contradiction.

Let $x_n = 1 - 1/n$ for all $\mathbb{N}$. Then $x_n$ is a sequence in $[0,1]$ that converges to 1. By definition, this is

$$\forall \delta_1 > 0 \; \exists N \in \mathbb{N}, \forall n \geq N \quad \vert x_n - 1 \vert < \delta_1$$

Since $f$ is continuous on $[0,1]$ it is continuous at $1$, therefore we have

$$\forall \varepsilon > 0 \; \exists \delta > 0 \; \forall x \in [0,1]\quad \vert x - 1 \vert < \delta \rightarrow \vert f(x) - f(1) \vert < \varepsilon$$

If we let $\varepsilon = \varepsilon_0$ and $\delta = \delta_1$. Then we know $\vert x_n - 1 \vert < \delta_1$, therefore we can conclude $\vert f(x_n) - f(1) \vert < \varepsilon_0$.

Furthermore, we know $\{f(x_n)\} >= 2$ for all $n$, so

$$\vert 2 - f(1) \vert \leq \vert f(x_n) - f(1) \vert < \varepsilon_0$$
$$\vert 2 - f(1) \vert < \varepsilon_0$$
$$-\varepsilon_0 < 2 - f(1) < \varepsilon_0$$

However this contradicts our assumption that $2 - f(1) > \varepsilon_0$

Therefore $f(1) >= 2$

答:
You haven't exactly done this using the $\epsilon-\delta$ definition, though. Someone else has already commented on your proof so I'll just present my own argument.

Suppose that $f(1) < 2$. Since $f$ is continuous on $[0,1]$:

$$\lim_{x \to 1^-} f(x) = f(1) < 2$$

So, for each $\epsilon > 0$, there is a $\delta > 0$ such that:

$$1-x < \delta \implies |f(x)-f(1)| < \epsilon$$

Let $\epsilon = 2-f(1)$. Then, for a sufficiently small $\delta > 0$, we have:

$$1-x < \delta \implies f(1)-\epsilon < f(x) < f(1)+\epsilon$$

which implies that $f(x) < 2$ whenever $x \in (1-\delta,1]$. That's a contradiction. $\Box$


答: Yout approach is correct, but you don't need to use sequences. Suppose that $f(1)<2$. If $f(1)<2$, then there is some $\delta>0$ such that$$|x-1|<\delta\implies\bigl|f(x)-f(1)\bigr|<2-f(1).$$But\begin{align}\bigl|f(x)-f(1)\bigr|<2-f(1)&\iff f(1)-2<f(x)-f(1)<2-f(1)\\&\implies f(x)<2.\end{align}

Title: Assume $f:[0,1] \rightarrow \mathbb{R}$ is continuous and that $f(x) \geq 2 \quad \text{ if } 0 \leq x \lt 1$. Show that $f(1) \geq 2$

Body:
This question comes from Advanced Calculus, Fitzpatrick. Section 3.1 exercise 7

Suppose that the function $f:[0,1] \rightarrow \mathbb{R}$ is continuous and that $$f(x) \geq 2 \quad \text{ if } 0 \leq x \lt 1$$
Show that $f(1) \geq 2$

Fitzpatrick uses the sequence definition of continuity, there is a similar question [here] (Suppose that the function $f:[0,1]\rightarrow \mathbb{R}$ is continuous and that $f\left(x\right)>2$)

I wanted to do this using epsilon-delta method. Is the following valid? Is there a simpler way to write this proof?

My attempt

First, assume $f(1) < 2$. Then there exists some $\varepsilon > 0$, call it $\varepsilon_0$, such that $2 - f(1) > \varepsilon_0$. Show this creates a contradiction.

Let $x_n = 1 - 1/n$ for all $\mathbb{N}$. Then $x_n$ is a sequence in $[0,1]$ that converges to 1. By definition, this is

$$\forall \delta_1 > 0 \; \exists N \in \mathbb{N}, \forall n \geq N \quad \vert x_n - 1 \vert < \delta_1$$

Since $f$ is continuous on $[0,1]$ it is continuous at $1$, therefore we have

$$\forall \varepsilon > 0 \; \exists \delta > 0 \; \forall x \in [0,1]\quad \vert x - 1 \vert < \delta \rightarrow \vert f(x) - f(1) \vert < \varepsilon$$

If we let $\varepsilon = \varepsilon_0$ and $\delta = \delta_1$. Then we know $\vert x_n - 1 \vert < \delta_1$, therefore we can conclude $\vert f(x_n) - f(1) \vert < \varepsilon_0$.

Furthermore, we know $\{f(x_n)\} >= 2$ for all $n$, so

$$\vert 2 - f(1) \vert \leq \vert f(x_n) - f(1) \vert < \varepsilon_0$$
$$\vert 2 - f(1) \vert < \varepsilon_0$$
$$-\varepsilon_0 < 2 - f(1) < \varepsilon_0$$

However this contradicts our assumption that $2 - f(1) > \varepsilon_0$

Therefore $f(1) >= 2$


问:
What's the difference between an initial value problem and a boundary value problem?
I don't really see the difference, because in both case we need to determine y and the values of the constants. The only difference is that we give the value of y and y' in the former and the value of either 2 y or 2 y' in the latter.

I solve both problems the same way. I don't really understand the theory, I guess.

答:
For a simple example (second order ODE), an initial value problem would say $y(a)=p$, $y'(a)=q$.

A boundary value problem would specify $y(a)=p$, $y(b)=q$.

ContentLicense: CC BY-SA 3.0


问:
Find the value of $k$ from the following Linear Transformation
Hello beautiful community, how are you? I have a problem that I would not be understanding. Given $B = \{(1, −1,0), (0,1, −1), (0,0,1)\}$ a base of $\mathbb R^3$ and let $f:\mathbb R^3\to\mathbb R^3$ a linear transformation such that

$$M_{EB}(f)=\begin{pmatrix}1&2&4\\0&5&5\\2&4&8\end{pmatrix}$$

If $Im (f) = ⟨(k^2,6, k); (3, −6, k^2)⟩$, then $k$ is equal to:

(and they give me 4 options)

a) $\ 2$

b) $-2$

d) $\ 3$

c) $-3$

The first thing I don't understand is what you mean by $EB$ as a sub index in the matrix. Can I directly extract the Image of the columns of that matrix?
If I take the images directly from the columns of the matrix, it doesn't give me any of the options, so I must be doing something wrong.

答:
$M_{EB}$ is the representative matrix of $f$. Let me call it $\mathcal M_{\mathcal E}^{\mathcal B}(f)$. I'll show you a general method to do this kind of exercises.

The columns of the matrix are the ordered coefficients of the linear combination of the vectors of $\mathcal B$ which express the images of the canonical vectors of $\mathcal E=\{e_1,e_2,e_3\}$.

So, we have that $$f(e_1)=b_1+2b_3=\begin{pmatrix}1\\-1\\0\end{pmatrix}+\begin{pmatrix}0\\0\\2\end{pmatrix}=\begin{pmatrix}1\\-1\\2\end{pmatrix}$$
$$f(e_2)=2b_1+5b_2+4b_3=\begin{pmatrix}2\\-2\\0\end{pmatrix}+\begin{pmatrix}0\\5\\-5\end{pmatrix}+\begin{pmatrix}0\\0\\4\end{pmatrix}=\begin{pmatrix}2\\3\\-1\end{pmatrix}$$
$$f(e_3)=4b_1+5b_2+8b_3=\begin{pmatrix}4\\-4\\0\end{pmatrix}+\begin{pmatrix}0\\5\\-5\end{pmatrix}+\begin{pmatrix}0\\0\\8\end{pmatrix}=\begin{pmatrix}4\\1\\3\end{pmatrix}$$
We can observe that $f(e_3)=2f(e_1)+f(e_2)$ and it's easy to see that $f(e_1)$ and $f(e_2)$ are linearly independent $\implies\text{dim(Im}(f))=2$ and $\text{Im}(f)=\langle f(e_1),f(e_2)\rangle$.

If you want to find $k$ you have to take a particular linear combination of the vectors $f(e_i)$ in order to find vectors that can be express as you wrote in the exercise (for example the subspace generated by $f(e_2)$ is the same of $2f(e_2)=(4,6,-2)=((-2)^2,6,-2)$ and you can do something similar to find the other vector).


问:
Straightedge and compass construction of an inscribed equilateral triangle when the circle has no center
Given a circle, but not its center, construct an inscribed equilateral triangle in as few steps as possible.

I have a construction that works, but I am really having trouble understanding why the construction works. I will attach a photo of my construction below.

答:
Here is an alternative method, which requires identifying a diameter but not the center.

Select any point $A$ on the circle.  Center the compasses there and draw an arc through two point $B,C$ on the circle.

Bisect $\angle BAC$, identifying point $D$ as the angle-interior point where the bisector intersects the circle.  Center the compasses on each endpoint of $AD$ and draw an arc through the other endpoint, the two arcs intersecting at point $E$ (either of two choices).

Draw $AE$, which intersects the circle at point $F$ such that chord $DF$ measures one side of the triangle, and copy the chord around the circle accordingly.

Also $AF$ measures one side of an inscribed hexagon, so this polygon is obtainable too.


问:
What does monodromy theorem tells about log z and $z^{1/2}$
This question was asked in my complex analysis assignment and I am confused on how to attempt it.

What does monodromy theorem tells about log z and $z^{1/2}$?

I have to choose a Domain D and then define it on a subdomain$D_0$ and then think if it can be analytically continued on D , but defining it on subdomain for log z it equals ${z-1} - (z-1)^2 /2 + (z-1)^3/3-...$ defined for 0<x<1 but it can't be extended for x=0. So, I think there doesn't exists a single valued function f(z) which equalslog z on 0<x<1 and is analytic on whole $\mathbb{C}$.

For $(z)^{1/2}$  it is not analytic at z=0 ( using it's taylor series expansion ) so I think the same can be said about it.

Am I right? If wrong can you please tell what are the right observations about these functions?

答: Both have a branch point at $0$, they are not the same when continued analytically along the unit circle, for $f(z)=\log z, |z-1|<1/2$ the analytic continuation sends $f(z)$ to $f(z)+2i\pi$, for $g(z)=z^{1/2}$ it is $g(z)\to -g(z)$. Thus clearly neither of them can be extended to an analytic function on $0<|z|<2$.


问:
Show compact operators on Hilbert spaces self-adjoint
Trying to solve the following exercise.

Let H be a separable Hilbert space and $T\in K(H)$ (hence T is compact)
with infinite spectrum $\sigma(T) = \{0\}\cup\{\lambda_i|i \in \mathbb{N}\}$. Suppose that $V_{\lambda_i}=E_{\lambda_i}$
for all $i\in\mathbb{N}$ and that for all $x\in H$ we have $lim_{m\to\infty} y_m = 0$ in every
decomposition $x = y_m + z_m$ with respect to the topological direct sum $H = \cap_{i=1}^m F_{\lambda_i} \oplus \bigoplus^m_{i=1} E_{\lambda_i}$ :
Construct a sequence $(e_K)_{k\in \mathbb{N}}$ of linearly independent eigenvectors of T
for which $< e_k | k\in \mathbb{N} >$ = H.

I think I would be able to construct if I knew that T was a self-adjoint operator but I'm sure how to realize this?

答:
If $H=\cap_{i=1}^{m}F_{\lambda_i}\oplus \bigoplus_{i=1}^{m}E_{\lambda_i}$ means $$\tag{1}H=E_m^\perp \oplus E_m$$ where $E_m=\bigoplus_{i=1}^{m}E_{\lambda_i}$ ( i guess you mean that $F_{\lambda_i}=E_{\lambda_i}^\perp$ ). Then, you can procceed as follows:

Since $T$ is compact every $E_{\lambda_i}$ is finite dimensional. For every $i$ let $(e_{ik})_{1\leq k \leq N_i}$ be an orthonomal basis of $E_{\lambda_i}$. Let $$E=\overline{\text{span}\biggl(\bigcup_{i=1}^{\infty}E_{\lambda_i}\biggr)}$$
Then, the set $\{e_{ik}:\, i\in \mathbb{N},\ 1\leq k \leq N_1\}$ is an orthonomal basis of $E$ constitutes of eigenvectors of $T$. We claim that $H=E$, this shows also that the kernel of $T$ is trivial. Meaning that $0$ is not an eigenvalue of $T$. For the proof of our claim write $H=E\oplus E^\perp$. Suppose that $E^\perp\neq \emptyset$. Pick $x\in E^\perp$. Then, from $(1)$ for every $m$ there exists $y_m\in E_m^\perp,\, z_m\in E_m$ such that $x=y_m+z_m$. By our hypothesis,
$$z_m=x-y_m\to x$$
(since $y_m\to 0$). But, $z_m \in E_m\subseteq E$. Since, $E$ is closed, it follows that $x\in E$. But, since $x\in E^\perp$ we conclude that $x=0$. Hence, $E^\perp=\{0\}$ and our claim is justified. Meaning, that $H=E$. In other words,
$$H=\overline{\text{span}\{e_{ik}:\, i\in \mathbb{N},\ 1\leq k \leq N_1\}}$$
and every $e_{ik}$ is an eigenvector of $T$ corresponding to the non zero eigenvalue $\lambda_i$.


问:
"Nearly" Harmonic Series
It's well known that
$$
\sum_{n=1}^{\infty} \frac{1}{n^{1+\varepsilon}} < \infty, \ \forall \varepsilon >0.
$$

What happens if we replace $\varepsilon$ with $\varepsilon_n \downarrow 0$?

WolframAlpha says
$$
\sum_{n=1}^{\infty} \frac{1}{n^{1+\varepsilon_n}}
$$

converges for
$$
\varepsilon_n = \frac{1}{\sqrt{\log(n+1)}}
$$

and diverges for
$$
\varepsilon_n = \frac{1}{\log(n+1)}.
$$

So the question is: can we find a "borderline" decrease rate such that the series converge for $\varepsilon_n$ approaching zero slower than it and diverges if $\varepsilon_n$ decreases faster?

答:
$$\int \frac{1}{x^{1+1/\ln(x)}}dx=\int\frac{dx}{x}(1/x)^{(1/\ln x)}$$

Set $u=\ln x$, then $x=e^u$ and $1/x=e^{-u}$ so the integral becomes $$\int du (e^{-u})^{1/u}=\int e^{-1}du$$

which is divergent.  Replacing $1/\ln x$ with $(1/\ln x)^s$ will be divergent (resp. convergent) if $s\ge 1$ (resp. $s<1$).

Title: "Nearly" Harmonic Series


问:
Sequential criterion of limit
Consider a function $f: \mathbb{R} \rightarrow \mathbb{R}$. A genric variable for $\mathbb{R}$ is $x$ and a generic variable for $\mathbb{N}$ is $n$.

We know that $\lim_{x \rightarrow \infty} f(x) = L$ implies $\lim_{n \rightarrow \infty} f(n) = L$. I sense that the converse is not true, but I am not able to find/construct an appropriate counter example. Do you have any?

答: Hint: Take $f(x)=\sin(\pi x)$.

Title: Sequential criterion of limit

Body: Hint: Take $f(x)=\sin(\pi x)$.


问:
How to show that a list of sets with the parameters (4,2,1) is A VALID BLOCK DESIGN?
I have to determine whether  or not the list of sets with the parameters v=4, b=6, r=3, k=2 and $\lambda$ = 1 is a valid block design. Now my first thought was to ensure that these parameters satisfy the conditions:

b $\geq$ v

bk = vr

$\lambda$ (v-1) = r (k-1).

And they did satisfy the conditions. However this is not sufficient according to my lecturer to show its a valid block design. So my next step was to identify the design as an affine plane of order n=2 and since an affine plane is a symmetry block design then the list of sets do form a valid block design. However, I am not  sure if this is enough either. Can someone help me out please?

答:
It is not sufficient to check that these parameters satisfy the equations you've listed. These equations are necessary conditions: if the parameters violate them, then there can be no block design. But they do not guarantee that there do exist block designs with a set of parameters.

It is sufficient to invoke the affine plane of order two. To be precise, what you are saying here is "Yes, a block design with these parameters does exist. The affine plane of order $2$ is an example."

The best answer would often be to say this, and also list what the points and blocks of the affine plane of order $2$ are. These parameters are very small, and it can be easily checked by hand that the set system with $v=4$ points $\{1, 2, 3, 4\}$ and $b=6$ blocks $$\{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}$$ is a block design with the parameters we wanted. Additionally, you don't have to think of this block design as an affine plane; it is also the block design obtained by taking all $2$-element subsets of the set of points.

(That's a less interesting but also valid family of block designs, giving a $b = \binom v2$, $r=v-1$, $k=2$, $\lambda=1$ block design for any $v$.)


问:
Is the set of submonoids of $(\Bbb N,+)$ countable?
Having the monoid $(\Bbb N,+)$, I wonder if there are countable many submonoids. There are obviously infinitely many since $S_n = \{kn \mid k \in \Bbb N\}$ is a submonoid for any $n \in \Bbb N$.

My conjecture is that the set of all submonoids is countable, because I think the following statements (which I failed to prove so far) hold for any submonoid $S$ of $(\Bbb N,+)$:

there is an odd element in $S$ $\Rightarrow \exists e \forall f: (f \ge e \rightarrow f \in S)$ $\Rightarrow \Bbb N \setminus S$ is finite

all elements of $S$ are even $\Rightarrow \exists e \forall f: (2f \ge e \rightarrow 2f \in S)$ $\Rightarrow \Bbb N \setminus (S \cup \{1,3,5,...\})$ is finite

In both cases we can identify the submonoid by a finite set of numbers which are not elements of the submonoid. Therefore we have only countable many possibilities.

Can you complete this approach or provide a better one?

答: All numerical monoids have finitely many generators (called the embedding dimension), and the set  of finite subsets of $\mathbb{N}$ is countable.</p>


问:
An expression for the multilinear joint XOR polynomial over $\{1,-1\}^n$
Preliminary: If $g:\{0,1\}^n\rightarrow \{0,1\}$ then there is a unique multilinear polynomial $f:\{1,-1\}^n\rightarrow \{1,-1\}$ such that $f((-1)^{a_1},\ldots,(-1)^{a_n})=(-1)^{g(a)}$ of all $a\in \{0,1\}^n$. Indeed, one only needs to consider the indicator polynomial $$f(x_1,\ldots,x_n)=\sum_{a\in \{0,1\}^n}(-1)^{g(a)}\prod_{i=1}^n\left(\frac{1+(-1)^{a_i}x_i}{2}\right)$$ and recall that the monomials $x^S:=\prod_{j\in S}x_j$ for $S\subseteq [n]$ form a basis for the vector space of such functions. Hence, we can express $$f(x_1,\ldots,x_n)=\sum_{S\subset [n]} \hat{f_S} x^S,$$ where $\hat{f_S}$ are the Fourier coefficients (basis coefficients for $f$).

Question: Let $g$ be the XOR function i.e. $g_n(a)=1$ if $a\in \{0,1\}^n$ contains a single $1$ otherwise $g_n(a)=0$. Can we find a nice expression for the corresponding $f(x_1,\ldots,x_n)$ of the joint XOR function $g_n$?

eg. If $n=2$ we have $g_2(a_1,a_2)=a_1\oplus a_2$ and we can see that $f_2(x_1,x_2)=x_1x_2$.

So far I have tried to obtain a simplified formula by starting directly with the indicator polynomial for the XOR function but I haven't had any luck... even a formula for $n=3$ would be appreciated.

答: Observe the function $$f_n(x_1,\ldots,x_n)=1-2\sum_{k=1}^n \frac{(1-x_k)}{2}\prod_{\ell\neq k}^n\frac{(1+x_\ell)}{2}$$ attains the value $-1$ whenever the input string of $x_i$'s contains only one $x_k=-1$ (resp. $x_\ell=1$ for all $\ell\neq k$) and attains the value $1$ whenever there is a pair $x_k=x_j=-1$ ($j\neq k$) in the input.


问:
Is this function of class $L^{1}(\mathbb{R})$, $L^{2}(\mathbb{R})$, both or none?
I have the following function:

$$\frac{\sin(x)}{x^{3/2}}$$

To prove it is $L^{1}(\mathbb{R})$, one has to prove this integral converges:

$$\int_{-\infty}^{+\infty} \frac{|\sin(x)|}{|x|^{3/2}}\, dx$$

and to prove it is $L^{2}(\mathbb{R})$, one has to prove this integral converges:

$$\int_{-\infty}^{+\infty} \frac{\sin^{2}(x)}{x^{3}}\, dx$$

Could someone please help me solve these integrals?

答:
Change the domain of integration to just $[0, \infty)$ for the first integral, which we can do since the integrand is even. Write $\int_0^\infty = \int_0^1+ \int_1^\infty$. On $[0, 1]$, $|\sin(x)|\le x \implies |\sin(x)|/x^{3/2}\le 1/\sqrt{x}$, whose integral converges on this domain. For $[1,\infty)$, $|\sin(x)|/x^{3/2}\le 1/x^{3/2}$, whose integral also converges on this different set.

For the second one, the first part of the same argument fails, though the second one does work.


问:
Is it possible to specify the general solution $(\ldots,u_{-1},u_0,u_1,\ldots)$ of $\ddot{u}_n=u_{n-1}+u_{n+1}-2u_n$?
In more detail: Let $I\subset\mathbb R$ be on open interval and let $D$ be the set of all twice differentiable functions from $I$ to $\mathbb C$:
\begin{equation}
D:=\left\{f\colon I\to\mathbb C:f\text{ is twice differentiable}\right\}
\end{equation}
My goal is to determine the following set:
\begin{align}
\left\{u\colon\mathbb Z\to D:\ddot{u}_n=u_{n-1}+u_{n+1}-2u_n\text{ for all }n\in\mathbb Z\right\}
\end{align}
Observations:
If $\ddot{u}_n=u_{n-1}+u_{n+1}-2u_n$ for all $n\in\mathbb Z$, then $u_n\in C^{\infty}$ for all $n\in\mathbb Z$. This means that it suffices to determine the set
\begin{align}
X:=\left\{u\colon\mathbb Z\to C^{\infty}:\ddot{u}_n=u_{n-1}+u_{n+1}-2u_n\text{ for all }n\in\mathbb Z\right\}
\end{align}
$X$ is vector space.

Motivation: That is the differential equation one needs to solve in the case of a infinite chain of masses, where the acceleration of one specific mass at the time $t$ depends only on the distance to its neighbours at that time. More precisely, one needs to solve $m\ddot{u}_n=C(u_{n-1}-u_n)+C(u_{n+1}-u_n)$.

答:
If $v,w\in C^{\infty}$, there is exactly one $u\in X$ with $u_0=v$ and $u_1=w$.
(Simply set $u_{n+1}=\ddot{u}_n+2u_n-u_{n-1}$ for $1\leq n$ and $u_{n-1}=\ddot{u}_n+2u_n-u_{n+1}$ for $n\leq0$.)

Let $u(v,w)$ be the element of $X$ described above, then $X=\left\{u(v,w):v,w\in C^{\infty}\right\}$.


问:
Classical number theoretic applications of the $p$-adic numbers
I am sure we can all agree that the $p$-adic numbers are highly fascinating objects in their own right - just as the closely related theory of valuations.

Having independently read up on the $p$-adic numbers for a few weeks now, I have so far only seen one application of them to what I would call classical number theory - namely the proof given in Serre's Cours d'arithmétique that a natural number is expressible as the sum of $\leq 3$ squares if and only if it is not of the form $4^a(8b-1)$ for some $a,b \in \mathbb{N}$.

Since I have a tendency to appreciate the value of the higher theories of mathematics in proportion to their applications to elementary number theory, I immediately found myself wondering if there are any other applications.

So my question to the community is: What are the most delightful applications of the $p$-adic numbers and the theory of valuations to elementary number theory?

Many thanks.

P.s.: I am aware that there are already several posts on the forum about the applications of the $p$-adic numbers, but none that refers to elementary number theory specifically.

Edit: I agree that I have been too vague in what I mean by "elementary number theory", so I will try to be a little more specific: By a classical "elementary" number theoretic proposition, I mean a number theoretic proposition that Fermat might have come up with. Thus, the above proposition about the sum of three squares is an elementary number theoretic proposition, as is e.g. Fermat's Last Theorem and the Twin Prime Conjecture, while e.g. the BSD Conjecture or the Class Number Problem are not.

Edit 2: Thank you for all the answers below - they are all excellent! In case anyone should come up with another one, I should like to say that bonus-points are given for results that have so far only been proven using the theory of $p$-adic numbers, or whose proof using $p$-adic numbers is far more conceptual and insightful than the original / more elementary one.

答:
I am not sure if the Gauss (Legendre) result qualifies as "the most delightful application of the $p$-adic numbers", but it gives that
$$
n=a^2+b^2+c^2
$$
is the sum of three squares if and only if
$$
-n \text{ is not a square in } \Bbb Q_2.
$$
Of course, this says that $n$ is not of the form $4^l(8k+7)$.

Edit: I realised that you already know this application. So I looked out for other applications. This MO-post refers to elementary results specifically. Some of them are in elementary number theory.


问:
Relationship between arcsec and arccos
$$\operatorname{arcsec}(x/2) = \arccos(2/x)$$

Is this correct? If so, does this apply to other inverse trigonometric ratios and their reciprocal inverses as well?

答:
Yes indeed. It can be proven in the following way. If $y=\arccos(2/x)$, then $\cos(y)=2/x$ (with $0 \leq y \leq \pi$). Taking reciprocals, we find that
$$
\sec(y)=\frac{x}{2} \, ,
$$
which is equivalent to
$$
y=\operatorname{arcsec}(x/2) \, .
$$
Note that $\sec$, like other trigonometric functions, doesn't actually have an inverse. There are many values of $y$ such that $\sec(y)=x/2$. However, because we know that $0 \leq y \leq \pi$ (and the range of $\text{arcsec}$ is precisely that), we can draw the above conclusion. This should be somewhat intuitive, since $\sec$ is simply the reciprocal function of $\cos$.

Title: Relationship between arcsec and arccos


问:
A question regarding Hahn Banach theorem

Let $Y=\{(a_n)\in l^2 : a_2=2a_1 , a_4=4a_3\}$.
Is there exist a functional $f \neq 0 \in X^*$ such that $f(Y)={0}$.

I know that it can be done using the Hahn-Banach corollary, which says, if $X$ is a normed space, $Y\subset X$ a closed sub-space, and $x \in X\setminus Y$, then there is a functional $f \in X^*$ such that $\|f\|=1$ ,  $f$ restricted to $Y$ equals $0$ ,and $f(x)=d=\inf_{y\in Y} \|x-y\|$.

Clearly $l_2$ is a normed space. And we can find a sequence $a_n\in l^2\setminus Y$ with $\|a_n\|_2=1$.
In order to show that $Y$ is closed in $X$ I have to take a sequence $b_n\in Y$ such that $\lim_{n\to \infty} b_n =x$ and show that $x\in X$ but is not that obvious according to $Y$'s definition?

Any support would be kindly welcomed.

答:
Rather than resort to using the Hahn-Banach theorem to solve this problem, perhaps consider a constructive approach:

Define $f:l_2\to\mathbb C$ by $f(a_n)=-2a_1+a_2$.  This map is certainly linear, $f\neq 0$, $f(Y)=0$, and by Hölder's inequality, we have
$$|f(a_n)|=|-2a_1+a_2|\leq((-2)^2+1^2)^{1/2}\|(a_n)\|_2=\sqrt 5\ \|(a_n)\|_2,$$
So $f\in(l_2)^*$.


问:
Is $\pi$ more transcendent than $e$?
I’ve often wondered about this, and I conjecture the affirmative, based mainly on that it is so much easier to prove the transcendence of $e$ than that of $\pi$.

I would be surprised if, just as numbers form a linearly ordered classes ranging from algebraic of degree $n$ and culminating in the class of transcendental numbers, the transcendental numbers themselves are not further divided into a linearly ordered classes, with $\pi$ belonging to the top class, and $e$ belonging to some class below it, so this is really a reference request. Could someone please cite chapter and verse where this is done?

答:
There are different kinds of irrationality measures:

http://mathworld.wolfram.com/IrrationalityMeasure.html

As you can see in the table above $e$ is not more irrational than algebraic numbers, but it is not clear if $\pi$ is more irrational, but both are not very irrational compared to Liouville numbers.

Title: Is $\pi$ more transcendent than $e$?


答:
Use limit comparison test with harmonic series. We have
$$\dfrac{1/n}{1/n^{1+{1/\log(1+n)}}} = n^{1/\log(1+n)} \to e$$</p>

Title: "Nearly" Harmonic Series

Body:
It's well known that
$$
\sum_{n=1}^{\infty} \frac{1}{n^{1+\varepsilon}} < \infty, \ \forall \varepsilon >0.
$$

What happens if we replace $\varepsilon$ with $\varepsilon_n \downarrow 0$?

WolframAlpha says
$$
\sum_{n=1}^{\infty} \frac{1}{n^{1+\varepsilon_n}}
$$

converges for
$$
\varepsilon_n = \frac{1}{\sqrt{\log(n+1)}}
$$

and diverges for
$$
\varepsilon_n = \frac{1}{\log(n+1)}.
$$

So the question is: can we find a "borderline" decrease rate such that the series converge for $\varepsilon_n$ approaching zero slower than it and diverges if $\varepsilon_n$ decreases faster?


问:
The link between the infinity norm of the difference between matrices and the infinity norm of the difference between these matrices inverted.
I am trying to develop an inequality and I have a bit of difficulty... The problem is the following one,

$\underline{\text{notations}}$

If $A$ is a $n \times n$ matrix with entries $a_{i,j}$, then $|A|_{\infty}$ = $\text{max}_{i,j} |a_{i,j}|$

If $A$ is a $n \times n$ matrix, then $A^{-1}$ is the inverse of $A$.

$\underline{\text{problem}}$

We know that:

$A$ and $B$ are two positive definite matrices

$c$ is a strictly positive constant

$|A-B|_{\infty}$ $\leq$ $c$

From this information, is it possible to say something about the upper bound of $|A^{-1}-B^{-1}|_{\infty}$?

答: No. Suppose that $A_n=\frac1n\operatorname{Id}$ and that $B=\operatorname{Id}$. Then $(\forall n\in\Bbb N):\|A_n-B\|_\infty\leqslant1$. However,$$\left\|A^{-1}-B^{-1}\right\|_\infty=\|n\operatorname{Id}-\operatorname{Id}\|_\infty,$$which can be arbitrarily large.


问:
What is $\frac{d}{dt}\int_{t_0}^t f(t,s)g(s)ds$?
I am trying to follow through this proof involving solutions to $\dot{x}=A(t)x+g(t)$. There is a part in the proof where you have to write $\frac{d}{dt}\int_{t_0}^t f(t,s)g(s)ds$ as a function. This is a bit tricky since I cannot exactly use the fundamental theorem of calculus. I think the proof states that $\frac{d}{dt}\int_{t_0}^t f(t,s)g(s)ds=f(t,t)g(t)+\int_{t_0}^t\frac{\partial f}{\partial t}(t,s)g(s)ds$, but I am not sure, nor am I convinced since I do not have the proof. Any help is appreciated.

I tried getting dirty with the limit definition of the derivative. If I set $H(t):=\int_{t_0}^tf(t,s)g(s)ds$, then I can manipulate $\frac{dH}{dt}(t)=\lim_{\epsilon \rightarrow 0}\frac{H(t+\epsilon)-H(t)}{\epsilon}$ into $\int_{t_0}^t\frac{\partial f}{\partial t}(t,s)g(s)ds+\lim_{\epsilon \rightarrow 0}\frac{1}{\epsilon}\int_t^{t+\epsilon}f(t+\epsilon,s)g(s)dx$. I do not know where to go from here. I guess my intuition says $\lim_{\epsilon \rightarrow 0}\frac{1}{\epsilon}\int_t^{t+\epsilon}f(t+\epsilon,s)g(s)dx=f(t,t)g(t)$, but how do I show this?

答:
You are in the right track since
$$ \lim_{\epsilon\to}\frac{1}{\epsilon}\int_t^{t+\epsilon}f(t+\epsilon,s)g(s)dx=f(t,t)g(t).$$


问:
Solving a text exercise with De Moivre-Laplace
I'm working on the following exercise:

Suppose that 18% of people who reserved a seat on a plane don't actually take the flight. Suppose that the passengers take the flight independently and that the plane has 220 available seats. There are more reservations than seats, so that less seats remain unused.

How many reservations can be made at most, so that the probability of every passenger getting a seat is $ \geq 99 $%?

We have to solve this with the De Moivre-Laplace theorem. I defined $ X_n $ as the number of boarding passengers and $ n $ as the wanted number of reservations.
I wanted to approximate $ P(0\leq X_n\leq 220)\geq 0.99 $ with the standard normal distribution and solve for $ n $.
I got $ P(0\leq X_n\leq 220)\approx\frac{1}{\sqrt{2\pi}}\int_{x_2}^{x_1}e^{-\frac{t^2}{2}}dt $, where $ x_1=\frac{220-0.82n}{\sqrt{n\cdot0.82\cdot0.18}} $ and $ x_2=\frac{-0.82n}{\sqrt{n\cdot0.82\cdot0.18}} $. I'm not quite sure about that, though.

But my question would be: How to solve for $ n $? I got the hint that I can invert the distribution function by searching the standard distribution table for a value near 0.99. But I'm lost on how exactly to use this.

Thanks in advance for any help!

答:
I´ve seen that you have made some progress. Nevertheless I post my thoughts. You can use the function for the cumulative distribution of the standard normal, which is $\Phi(z)$. See here the table. But first of all you have $P(X\leq 220)=\Phi\left(\frac{220-0.82n}{\sqrt{n\cdot0.82\cdot0.18}}  \right)\geq 0.99$

Now you have to use the inverse function.

$\frac{220-0.82n}{\sqrt{n\cdot0.82\cdot0.18}}  \geq \Phi^{-1}\left(0.99\right)$

You can now use the table at look at which value of z you have a probability of 0.99 at the table. This is approximately at $z=2.33$. I don´t know where you have read off $z=2.3$.

Or you make an equation with the integral: $\int\limits_{-\infty}^z \frac1{\sqrt{2\cdot \pi}}\cdot e^{\frac12x^2} \, dx=0.99$. With wolfram alpha I get $z=2.32634$.

So the inequality is $\frac{220-0.82n}{\sqrt{n\cdot0.82\cdot0.18}}  \geq 2.32634$

The result is

 n<=251.024


问:
What's the difference between an initial value problem and a boundary value problem?
I don't really see the difference, because in both case we need to determine y and the values of the constants. The only difference is that we give the value of y and y' in the former and the value of either 2 y or 2 y' in the latter.

I solve both problems the same way. I don't really understand the theory, I guess.

答:
An initial value problem is how to aim my gun.  A boundary value problem is how to aim my gun so that the bullet hits the target.

Qualitatively the methods of solution are sometimes different, because Taylor series approximate a function at a single point, i.e.  at 0.


问:
Mods with same prime factorization as base give out same mod as base
I had a question proposed to me being "Compute $2853^{1429} $(mod $ 2039183)$". Now the answer would be $2853$ by solving a pair of congruences. But that also happens to be the base of the exponent. Now why is this? The prime factorization of $2039183$ is $1429 \cdot 1427$ which is the same as the exponent but is that why?

答:
$2853^{1429}\equiv2853\bmod1429$ by Fermat's Little Theorem.

$2853\equiv-1\bmod1427$, so $2853^{1429}\equiv(-1)^{1429}=-1\equiv2853\bmod1427$.

Therefore, by the constant case of the Chinese remainder theorem,

$2853^{1429}\equiv2853\bmod2039183$.


问:
Does the set $\left\{ \left(\begin{array}{c} x\\ y \end{array}\right)\in\mathbb{R}^{2}|x\leq y\right\} $ span all of $\mathbb{R}^2$
Does the set$\left\{ \left(\begin{array}{c}
x\\
y
\end{array}\right)\in\mathbb{R}^{2} \ s.t\ x\leq y\right\} $ span all of $\mathbb{R}^2$. I don't know because $y\in\mathbb{R}$ and then $\left(\begin{array}{c}
x\\
y
\end{array}\right)\in\mathbb{R}^{2}$ is just a general vector in $\mathbb{R}^2$ because $x\leq y$.
and does it equals to $\mathbb{R}^2$?

答: Let $V$ be the span of the specified group. Fix an arbitrary $(x,y) \in \mathbb{R}^2$. Ix $x\leq y$ then clearly $(x,y) \in V$. If $x>y$ then $-x<-y$, so $(-x,-y) = -1(x,y) \in V$. So yes, it spans.


问:
Show that $f(x)=x+\frac{1}{x^2+1}$ is strictly increasing.
Show that $f(x)=x+\frac{1}{x^2+1}$ is always increasing.

I noticed that $f'(x)=1-\frac{2x}{(x^2+1)^2}$, by looking at this function graph, I noticed it must be always positive, however I can't find a way to prove that. My attempt:

If $x\le0$, then clearly $f'(x)>0$. If $x\ge1$, it is also quite evident that $\frac{2x}{(x^2+1)^2}\le1$. Then, when $0<x<1$, I'm a bit lost. I can't seem to find an algebraic trick to show that $2x<(x^2+1)^2$, I'd love some help.

答:
$\theta = \arctan (x), g(\theta): = f(x) = \tan(\theta)+ \cos^2 (\theta)$

$g'(\theta) = \sec^2 \theta - \sin (2\theta) \ge 1-1=0$. (Can't have equality because if $\sec^2\theta=1$ then $\cos\theta=\pm1, \sin2\theta=0$.)

And $x$ increases with $\theta$.


问:
Find area of a crescent
Find the area of the shaded region that is enclosed by a full circle and a quarter circle inscribed in a square.

My math teacher gave me this question as a challenge. But I can't seem to solve it. Can you guys help? Please tell me the steps to solve it.

Thanks.

答:
Assume unit square. The area is $A_{square}-A_{purple}-A_{green}-2A_{grey}$, i.e.

$$Area
=\frac34 - \frac{3\pi}{16} -2A_{grey} \tag1
$$

where $A_{grey}$ can be obtained by evaluating the areas of the circular sectors ABP, OPE, and the triangles AFP and FPO; then subtracting them from the half square, as done below.

Let the vertex A being the origin, the polar equation of the two circles are $r=1; \space \space r(\sin\theta + \cos\theta)=r^2+\frac{1}{4}$. Eliminate $r$ to obtain $\sin\beta = \frac{5+\sqrt{7}}{8}$ and then
$\sin\alpha =\frac{\sqrt{7}-1}{4}$. The area of two triangles AFP and FPO is

$$A_q = \frac{1}{4}\sin\beta +  \frac{1}{4}\left(\frac{1}{2}-\cos\beta \right)=\frac{2+\sqrt{7}}{16}
$$

and the area of the two circular sectors is

$$A_c=\frac{1}{2}\left( \frac{\pi}{2}-\beta\right) + \frac{\alpha}{8}
=\frac{\pi}{8}-\frac{1}{8}\sin^{-1} \left(\frac{23\sqrt{7}+67}{128}\right). $$

As a result,

$$A_{grey}=\frac{1}{2}-A_c-A_q=\frac{6-\sqrt{7}-2\pi}{16}+\frac{1}{8}\sin^{-1} \left(\frac{23\sqrt{7}+67}{128}\right) $$
Substitute into (1) to obtain

$$Area = \frac{\pi+2\sqrt7}{16} - \frac{1}{4}\sin^{-1} \left(\frac{23\sqrt{7}+67}{128}\right)$$

Title: Find area of a crescent


答: I don't understand your argument, but $\dbinom 11$ and $\dbinom 01$ are elements of $\bigg\{ \dbinom uv \in \mathbb R^2 : u \leq v \bigg\}$ such that $$\binom xy = x \dbinom 11 + (y-x) \dbinom 01$$ for every $\dbinom xy \in \mathbb R^2$.

Title: Does the set $\left\{ \left(\begin{array}{c} x\\ y \end{array}\right)\in\mathbb{R}^{2}|x\leq y\right\} $ span all of $\mathbb{R}^2$

Body:
Does the set$\left\{ \left(\begin{array}{c}
x\\
y
\end{array}\right)\in\mathbb{R}^{2} \ s.t\ x\leq y\right\} $ span all of $\mathbb{R}^2$. I don't know because $y\in\mathbb{R}$ and then $\left(\begin{array}{c}
x\\
y
\end{array}\right)\in\mathbb{R}^{2}$ is just a general vector in $\mathbb{R}^2$ because $x\leq y$.
and does it equals to $\mathbb{R}^2$?


答:
The usage of "group" word is misleading here, especially with the question tagged group theory as well.

So let consider the vector space only question. $\mathbb R^2$ needs only two independent vectors to be spanned.

It is clear that $(0,1)$ and $(1,2)$ are independent and both belong to your initial set, so these two vectors just by themselves span $\mathbb R^2$. A fortiori the whole set also does.

I could have taken $(1,1)$ as in azif00's answer, but this is just to show that even with $x<y$, the question remains mostly unaffected.


问:
randomized algorithm and one solved example in probability
Example: a randomized algorithm was given and role of it, is to determines whether an input $a$ is a prime number?

This algorithm works in the following manner for one step:

I- If $a$ is prime, then our algorithm outputs YES

II- If $a$ is not prime, then our algorithm outputs NO with the probability $\frac{3}{4}$.

if we want to guarantee that achieve an output NO with the probability at least $1- \frac{1}{k}$ on step II, we should run this algorithm at least $O(\log k)$ times.

Note: first NO answer implies that a given input $a$ is not prime.

How this example solved? i.e: how $O(\log k)$ is determined?

Update:

we know the probability to tell yes in $x$ repeats of the algorithm on step II is $(\frac{1}{4})^x = 4^{-x} = 2^{-2x}$.

答: We know that the probability that it incorrectly answers "yes" on some composite in $n$ trials of the algorithm is $\left(\frac14\right)^n$ so the probability that it answers "no" at least once in $n$ runs is $1-\left(\frac14\right)^n$.  Thus, we want to find $n$ such that $$1-\left(\frac14\right)^n\geq1-\frac1k$$ or $$\frac1k\geq\left(\frac14\right)^n$$  Taking logarithms, $$-\log k\geq -n\log4$$ or $$n\geq \frac{\log k}{\log4}=O(\log k)$$


答:
Here is a rigorous proof. I assume that $f$ and $g$ are continuous. (I am not even sure if this is necessary since one can approximate integrable functions by $C^\infty$-functions)

Notice that
$$
f(t, t)g(t) = \frac{1}{\varepsilon}  \int^{t+\varepsilon}_t f(t, t) g(t)~\mathrm{d}s.
$$
Hence:
$$
\left \lvert \frac{1}{\varepsilon} \int^{t+\varepsilon}_t f(t + \varepsilon, s)g(s)~\mathrm{d}s - f(t, t)g(t)\right \rvert = \frac{1}{\varepsilon} \left \lvert \int^{t+\varepsilon}_t f(t + \varepsilon, s)g(s) -f(t, t)g(t)~\mathrm{d}s \right \rvert \leq \\
\frac{1}{\varepsilon} \int^{t+\varepsilon}_t \left \lvert f(t + \varepsilon, s)g(s) -f(t, t)g(t)~\mathrm{d}s \right \rvert \leq \frac{1}{\varepsilon} \int^{t+\varepsilon}_t \sup_{x \in [t, t + \varepsilon]} \left \lvert f(t + \varepsilon, x)g(x) -f(t, t)g(t) \right \rvert~\mathrm{d}s  = \\
\sup_{x \in [t, t + \varepsilon]} \left \lvert f(t + \varepsilon, x)g(x) -f(t, t)g(t) \right \rvert
$$
Since $[t, t+ \varepsilon]$ is compact and our functions are continuous, this $\sup$ is attained at some $x_\varepsilon \in [t, t + \varepsilon]$. Clearly,  $x_\varepsilon \rightarrow t$ for $\varepsilon \rightarrow 0$. So continuity gives us:
$$
\sup_{x \in [t, t + \varepsilon]} \left \lvert f(t + \varepsilon, x)g(x) -f(t, t)g(t) \right \rvert = \left \lvert f(t + \varepsilon, x_\varepsilon)g(x_\varepsilon) -f(t, t)g(t) \right \rvert \rightarrow 0
$$
as $\varepsilon \rightarrow 0$.

Title: What is $\frac{d}{dt}\int_{t_0}^t f(t,s)g(s)ds$?

Body:
I am trying to follow through this proof involving solutions to $\dot{x}=A(t)x+g(t)$. There is a part in the proof where you have to write $\frac{d}{dt}\int_{t_0}^t f(t,s)g(s)ds$ as a function. This is a bit tricky since I cannot exactly use the fundamental theorem of calculus. I think the proof states that $\frac{d}{dt}\int_{t_0}^t f(t,s)g(s)ds=f(t,t)g(t)+\int_{t_0}^t\frac{\partial f}{\partial t}(t,s)g(s)ds$, but I am not sure, nor am I convinced since I do not have the proof. Any help is appreciated.

I tried getting dirty with the limit definition of the derivative. If I set $H(t):=\int_{t_0}^tf(t,s)g(s)ds$, then I can manipulate $\frac{dH}{dt}(t)=\lim_{\epsilon \rightarrow 0}\frac{H(t+\epsilon)-H(t)}{\epsilon}$ into $\int_{t_0}^t\frac{\partial f}{\partial t}(t,s)g(s)ds+\lim_{\epsilon \rightarrow 0}\frac{1}{\epsilon}\int_t^{t+\epsilon}f(t+\epsilon,s)g(s)dx$. I do not know where to go from here. I guess my intuition says $\lim_{\epsilon \rightarrow 0}\frac{1}{\epsilon}\int_t^{t+\epsilon}f(t+\epsilon,s)g(s)dx=f(t,t)g(t)$, but how do I show this?


问:
Submultiplicativity of Lipschitz seminorms
I have a textbook (Nonlinear Spectral Theory, Appell et al 2004) that gives the following definitions and properties.

Let $X$ and $Y$ be Banach spaces, and $f:X\rightarrow Y$ a continuous (possibly nonlinear) operator. Define the Lipschitz seminorms as
$$[f]_{Lip} = \sup_{x\neq y} \frac{||f(x)-f(y)||}{||x-y||}$$
$$[f]_{lip} = \inf_{x\neq y} \frac{||f(x)-f(y)||}{||x-y||}$$

The book then gives the following property for two continuous operators $f$ and $g$:
$$[f]_{lip}[g]_{lip}\leq [fg]_{Lip}\leq[f]_{lip}[g]_{Lip}$$
Why is the second inequality true?  I can see why we would have submultiplicativity with $ [fg]_{Lip}\leq[f]_{Lip}[g]_{Lip}$ but don't understand the stronger inequality that is given.  The book says that the inequalities "are straightforward consequences of calculation rules for
suprema and infima" but I'm afraid it's not straightforward for me.

Thanks in advance for your help!  It would be very useful if I could apply this result, but I don't want to use something that I don't understand.

答: There may be a typo. Take $X$ and $Y$ to be two dimensional and $f:X\to Y$ linear with matrix $\begin{pmatrix}1&0 \\ 0&0.5 \end{pmatrix}$ and $g:Y\to X$ also linear with same matrix. Then $[f]_{lip}=0.5$, $[g]_{Lip}=1$ and $[fg]_{Lip}=1$ so we would have $1\leq 0.5$.  So this is not true.


问:
Sketch the region of integration for the integral $\int_{-2}^2 \int_0^{2y} f(x,y)dx\,dy$
I am slightly confused when sketching the region for the double integral $$\int_{-2}^2 \int_0^{2y} f(x,y)dx\,dy$$
When I sketch the region I get a triangle in the top right quadrant, however my textbook says the region is two triangles, one in the top right quadrant, one in the bottom left quadrant. I don't understand how this can be if the inequalities for the regions are $0<x<2y$ and $-2<y<2$.

答:
Note that, as mentioned here,
$$
\int_a^bf(x)\,\mathrm{d}x=-\int_b^af(x)\,\mathrm{d}x
$$
That is, $x$ is between $0$ and $2y$. This means that when $y\lt0$, the range for  $x$ is $2y\lt x\lt0$. Thus,
$$
\int_{-2}^2\int_0^{2y}f(x,y)\,\mathrm{d}x\,\mathrm{d}y
=\underbrace{\int_0^2\int_0^{2y}f(x,y)\,\mathrm{d}x\,\mathrm{d}y}_\text{triangle above the $x$-axis}-\underbrace{\int_{-2}^0\int_{2y}^0f(x,y)\,\mathrm{d}x\,\mathrm{d}y}_\text{triangle below the $x$-axis}
$$


答: $$\int\limits_{-2}^2 \int\limits_0^{2y} f(x,y)dx\,dy=\underbrace{\int\limits_{-4}^0 \int\limits_{-2}^{\frac{x}{2}} f(x,y)dy\,dx}_\text{left triangle}+\underbrace{\int\limits_{0}^4 \int\limits_{\frac{x}{2}}^{2} f(x,y)dy\,dx}_\text{right triangle}$$

Title: Sketch the region of integration for the integral $\int_{-2}^2 \int_0^{2y} f(x,y)dx\,dy$

Body:
I am slightly confused when sketching the region for the double integral $$\int_{-2}^2 \int_0^{2y} f(x,y)dx\,dy$$
When I sketch the region I get a triangle in the top right quadrant, however my textbook says the region is two triangles, one in the top right quadrant, one in the bottom left quadrant. I don't understand how this can be if the inequalities for the regions are $0<x<2y$ and $-2<y<2$.


问:
Markov Chains Probability Distributions Question
I have a question that I didn't understand the solution for, that I'd love some help with!

The question:

A hard-working GSI is holding her office hours. A random number of
students enter and leave her office during her OH. Let us break up
time into 1-minute segments, and assume that there is either 0 or 1
student in each time segment, and that this discrete arrival/departure
process is well modeled by a 2-state Markov Chain. For each time
segment, the transition probabilities are 0.8 for going from 0
students to 1 student in the OH, and 0.4 for going from 1 student to 0
students in the OH.

I represented this with the transition matrix:
$P = \begin{pmatrix}
0.2 & 0.8\\
0.4 & 0.6
\end{pmatrix}$

The first subpart asked

If the GSI starts off her office hours at t = 0 with 1 student, what
is the probability that she has 0 student at time t = 2?

I did $\begin{pmatrix}
1 & 0\end{pmatrix} P^2$ and got that at time=2, the state space distribution is $\begin{pmatrix}
0.36 & 0.64\end{pmatrix}$, so my final answer is 0.36. but the solution says 0.32, and they give their solution as: $0.4×0.2+0.6×0.4 = 0.32.$

For the second part of the question they ask:

What does the probability go to, as t gets large, that there is 1 student?
and give the solution:
Answer: 0.8p0 +0.6p1 = p1 p0 = 1− p1
This gives as 0.8(1− p1) +0.6p1 = p1 which gives p1 = 0.8/1.2 = 2/3.

Which I don't get either

Could I get clarification of where I went wrong and more explanation on where they got their solutions from? I'm also really shaky with Markov Chains in general, so conceptual explanations would help a lot too!

答:
For the first part:  what are the possible paths?  Given that we start in state $1$ and end in state $0$ there are only two paths, namely $$1\mapsto 0\mapsto 0\quad \&\quad 1\mapsto 1\mapsto 0$$

The first has probability $.4\times .2=.08$ and the second has probability $.6\times .4=.24$ and the sum is $\boxed {.32}$.

For the second, we let $p_0,p_1$ be the equilibrium probabilities and note that we must have $$p_0=.2p_0+.4p_1\quad \&\quad p_1=.8p_0+.6p_1\quad \&\quad p_0+p_1=1$$ which implies $$\boxed {p_0=\frac 13\quad p_1=\frac 23}$$


问:
Peano Arithmetic - Decidability
I need to show that if Peano Arithmetic does not decide a sentence $\varphi$ then the standard model
of Peano Arithmetic satisfies the negation of $\varphi$.
I know this partly has to do with Godel's Incompleteness Theorem, but I'm really lost on where to start. Any help would be greatly appreciated.

EDIT TO ADD:  "Let $\varphi$ be a $\Sigma_1$ sentence in the language of Peano Arithmetic" should be the first part of the problem, I left it out accidentally ^

答:
The only fact you need is that PA proves every true (i.e. true in the standard model) $\Sigma_1$ sentence. So if a $\Sigma_1$ sentence is undecidable in PA, it must be false.

This fact is somewhat tedious to prove, but I'm sure you have been given the hard part in some form. For instance, perhaps you've been given the fact that PA (or even the weak subtheory Robinson arithmetic) can correctly decide any instance of a computable relation. Then from there, you can use the fact that by definition, a $\Sigma_1$ sentence can be written $\exists x \varphi(x)$ for some $\varphi$ that is $\Delta_0$, hence computable. So if that's true, $\varphi(\mathbf n)$ is true for some $n,$ and since that's a $\Delta_0$ sentence, it can be proven in PA, and then PA can use existential instantiation to prove $\exists x\varphi(x)$ from there.

Title: Peano Arithmetic - Decidability


问:
Problem on Stokes' Theorem
I'm really struggling to understand Stokes' Theorem. I tried this exercise:

Let D be the portion of $z=1-x^2-y^2$ above the xy-plane, oriented up, and let $\vec{F}=\langle xy^2,-x^2y,xyz\rangle$. Compute $$\iint_{D}^{}(\nabla\times \vec{F})\cdot \hat{n}dS$$
Here is my work:
$$\nabla\times \vec{F}=\langle xz,-yz,-4xy\rangle$$

$$\vec{f}(r,\theta) = \bigl\langle r\cos\theta ,r\sin\theta ,1-r^2 \bigr\rangle$$

$$\frac{\partial\vec{f} }{\partial r}= \langle\cos\theta,\sin\theta,-2r\rangle$$

$$\frac{\partial \vec{f}}{\partial\theta }= \langle -r\sin\theta ,r\cos\theta ,0 \rangle$$

$$\frac{\partial\vec{f} }{\partial r}\times \frac{\partial \vec{f}}{\partial \theta}=\left \langle 2r^2\cos\theta ,2r^2\sin\theta ,r\right \rangle$$

$$\left \|\frac{\partial\vec{f} }{\partial r}\times \frac{\partial \vec{f}}{\partial \theta } \right \|=r\sqrt{3}$$
$$\widehat{n}= \biggl\langle \frac{2r\cos\theta}{\sqrt{3}} ,\frac{2r\sin\theta }{\sqrt{3}},\frac{1}{\sqrt{3}}\biggr\rangle$$

Integrating, I have
$$\frac{1}{\sqrt{3}}\int_{0}^{2\pi }\int_{0}^{1}2r^2\cos^2\theta (1-r^2)-2r^2\sin^2\theta (1-r\cos\theta )-4r\sin\theta\cos\theta\,dr\,d\theta $$

$$=\frac{1}{\sqrt{3}}\int_{0}^{2\pi }\int_{0}^{1}-2r^2\sin^2\theta (1-r\cos\theta )+2r^2\cos\theta\, \theta (1-r^2)-2r\sin(2\theta )\,dr\,d\theta$$
After splitting the integral into three integrals, I have
$$\frac{1}{\sqrt{3}}\int_{0}^{2\pi }\int_{0}^{1}-2r^2\sin^2\theta (1-r\cos\theta )\,dr\,d\theta=-\frac{2\pi }{3\sqrt{3}}$$

$$\frac{1}{\sqrt{3}}\int_{0}^{2\pi }\int_{0}^{1}2r^2\cos\theta (1-r^2)\,dr\,d\theta =0$$

$$\frac{1}{\sqrt{3}}\int_{0}^{2\pi }\int_{0}^{1}-2r\sin(2\theta )\,dr\,d\theta =0$$

$$=-\frac{2\pi }{3\sqrt{3}}+0+0$$

But the answer is zero. What am I doing wrong?

答:
If you don't apply Stokes' theorem (that is, you insist on computing the integral of the curl over the surface $D$), then you would have

$$\begin{align}
\iint_D(\nabla\times\vec F)\cdot\mathrm d\vec S&=\iint_D\langle xz,-yz,-4xy\rangle\cdot\vec n\,\mathrm dS\\[1ex]
&=\iint_D \langle r(1-r^2)\cos\theta,-r(1-r^2)\sin\theta,-4r^2\cos\theta\sin\theta\rangle\cdot\vec n\,\mathrm dS
\end{align}$$

where all I've done here is compute the curl of $\vec F$ and composed it with $\vec f$ to replace $x\to r\cos\theta$, $y\to r\sin\theta$, and $z\to1-r^2$. The normal vector is

$$\vec n=\frac{\partial\vec f}{\partial r}\times\frac{\partial\vec f}{\partial\theta}=\langle2r^2\cos\theta,2r^2\sin\theta,r\rangle$$

So the surface integral reduces to

$$2\int_0^{2\pi}\int_0^1 \left(r^3(\cos(2\theta)-\sin(2\theta))-r^5\cos(2\theta)\right)\,\mathrm dr\,\mathrm d\theta$$

Note that you do more work than necessary, since

$$\vec n\,\mathrm dS=\left\|\frac{\partial\vec f}{\partial r}\times\frac{\partial\vec f}{\partial\theta}\right\|\frac{\left(\frac{\partial\vec f}{\partial r}\times\frac{\partial\vec f}{\partial\theta}\right)}{\left\|\frac{\partial\vec f}{\partial r}\times\frac{\partial\vec f}{\partial\theta}\right\|}\,\mathrm dr\,\mathrm d\theta=\left(\frac{\partial\vec f}{\partial r}\times\frac{\partial\vec f}{\partial\theta}\right)\,\mathrm dr\,\mathrm d\theta$$

so you don't strictly need to normalize the normal vector.

If you do wish to apply Stokes' theorem, the integral is trivial by comparison (I've omitted the details):

$$\int_C \vec F\cdot\mathrm d\vec r=-\int_0^{2\pi}\left(\cos^3\theta\sin\theta-\cos\theta\sin^3\theta\right)\,\mathrm d\theta$$

and both do indeed have the same value.

Title: Problem on Stokes' Theorem


问:
Find the antiderivatives
Find the antiderivatives:

$\int\!\left(2x^2 +3\right)^{1/3} x\,dx $

I have hit this in my book and the way I do it I get

$3/4\left(2x^2 + 3\right)^{4/3} x^2 +c $

But my book tells me it should be $3/16(2x^2 + 3)^{4/3}$

Anyone know how they get to that?

答: Use the substitution $u=2x^2+3, du=4x$.    It is unfortunately not true that $\int f(x) g(x)dx=\int f(x)dx \int g(x)dx$; you cannot integrate the $x$ part separately.</p>

Title: Find the antiderivatives


问:
Ratio of Volume of Standard Simplex to its Dual
In the two-dimensional case, the ratio of the area of an equilateral triangle whose vertices are the mid-points of the edges of another equilateral triangle is 1/4th.  For a tetrahedron and its dual, the ratio is known to be 1/27.

In the general case, is it $\frac{1}{n^{n}}$?  By general case, I mean the standard simplex $S$ in $\mathbb{R}^n$ and the simplex whose vertices are the Euclidean barycenters of the faces of $S$.

答:
I think the key is to take the $n$th power of the ratio of the side lengths.

In what follows, I'll view the $n$-simplex as living in $\mathbb{R}^{n + 1}$ with vertices the standard basis vectors.

First, the outer simplex always has side length $\sqrt{2}$.

To get the side length of the inner simplex, we can compute the distance between the midpoints of two different faces. Since these midpoints occur at the average of $n - 1$ of the vertices, they have one coordinate 0 and the other $n$ equal to $1 / n$. Therefore the distance between two of them is $\sqrt{\frac{1}{n^2} + \frac{1}{n^2}} = \frac{\sqrt{2}}{n}$.

So, the ratio of the side lengths is
\begin{align*}
\frac{\frac{\sqrt{2}}{n}}{\sqrt{2}} = \frac 1 n ,
\end{align*}
and the $n$-volumes have ratio $1 / n^n$ as you claimed.


问:
How to derive formulas for sequences that converge to a specific value using simple functions
If I have some value $L\in \mathbb{R}$ such that a sequence $a_n$
$$\lim_{n\rightarrow\infty}{a_n}=L$$
If I state the properties of the sequence such as monotonically increasing or decreasing is it possible to find the terms in the sequence in the form
$a_{n+1}=A(a_n)$ where $A$ is a function that requires only simple functions, where simple means exactly calculable over a finite proccess. We can express this as a guess
$$|L-x_n|<\epsilon$$
And we want to rearrange it to be
$$|L-A(x_n)|<B(\epsilon)$$
And $A(x_n)$ is a better approximation if $B(\epsilon) < \epsilon$. However, we also want $A(x_n) > x_n$ since we want it to be monotonically increasing.
For example, if I want to find
$$\log_2(b)$$
We can first express it as
$$|\log_2(b)-x|<\epsilon$$
But how can I express it as $$|L-A(x_n)|<B(\epsilon)$$ where B is composed of only simple functions (i.e. exactly calcuable) and using the property that $A(x_n) > x_n$

答:
The archetypal method to evaluate "specific values" defined by transcendental functions is the Taylor series.

The approximations are polynomials with terms computed as $\dfrac{f^{(n)}(x_0)}{n!}(x-x_0)^n$, where $x$ is the value of the argument for which you want to evaluate the function, and $x_0$ a value at which the computation of these derivatives is easy. When $x$ lies in the radius of convergence of the series, you do obtain better and better approximations (except maybe for the first terms).

E.g. to obtain $\ln\dfrac32$, you use $x_0=1$, and the successive approximations are

$$\frac12,\frac12-\frac1{2\cdot2^2},\frac12-\frac1{2\cdot2^2}+\frac1{3\cdot2^3},\frac12-\frac1{2\cdot2^2}+\frac1{3\cdot2^3}-\frac1{4\cdot2^4},\cdots$$

The errors are $0.094534892,-0.030465108,0.011201559,-0.004423441,\cdots$ They are roughly halved with every new term. Notice that the computation only involves fractions.

I am not aware of a general theory to construct similar sequences using other polynomials than Taylor's, or other expressions than polynomials, except for the continuous fraction approach.


问:
Volume and triple integral
I've been studying some triple integrals and their relation to volumes in 3 dimensions and got a little confused by the following problem:

Let $$U=\{(x,y,z) \in \mathbb{R}^3; \; x^2+y^2+z^2\leq4 \; ; 0\leq z \leq5 \; ; x \leq y \leq2x\;; x\geq0 \}$$ find the integral that represents the volume of the set $U$

The solution is provided and is shown to be:

$$\int^{\frac{2}{\sqrt{5}}}_{0}\int^{2x}_{x}\int^{\sqrt{4-x^2-y^2}}_{0}\,dx\,dy\,dz+ \int^{\sqrt{2}}_{\frac{2}{\sqrt{5}}}\int^{\sqrt{4-x^2}}_{x}\int^{\sqrt{4-x^2-y^2}}_{0}\,dx\,dy\,dz$$

I just don't understand why he splits the integral into two parts and the limits of the $x$ and $y$ variables.

答: As @Arthur states:  "It's rarely a bad idea to make a drawing."  Does this help?

Title: Volume and triple integral


问:
Show that the multiplicative group of integers mod 35 is not isomorphic to the additive group of integers mod 24
I got this question on an abstract algebra test back when I was still in undergrad. I was going over some old notes from undergrad (just for fun) and I realized I had a completely wrong answer to this question so I set about trying to solve it. I am really struggling with it. To be honest, I think I just have a hard time disproving two groups are isomorphic unless they have different cardinality.

Any suggestions?

答:
Hint:

Can you show, $\forall n\in(\mathbb Z/35\mathbb Z)^\times$, $n^{12}\equiv1\bmod35$?


问:
How to find a Lyapunov function
I have the following system: $$\dot{x_1} = x_2(x_3-2) \\ \dot{x_2} = x_1(x_3-2) \\ \dot{x_3}=-x_3^3 $$ and I want to determine its equilibrium points together with their stability. To find the equilibrium points, I wrote $$x_2=0 \text{ or } x_3=2 \\ x_1=0 \text{ or } x_3=2  \\ x_3 =0$$
Thus, only equilibrium point $x^*$ for the system is the origin $0$. I found $\det{Df(0)} = 0$ so I expect that the origin is a non-hyperbolic equilibrium point, probably asymptotically stable. To determine its type, I want to find a Liapunov function. Now, I assumed I can take $V(x)=c_1x_1^2 +c_2x_2^2+c_3x_3^3$ for some positive constants $c_i$. We have $V(0)=0$ and $V(x)>0$ for any $x \neq 0$. Then, $$\dot{V}(x)=\nabla V\cdot f(x) = 2c_1x_1x_2(x_3-2)+2c_2x_1x_2(x_3-2)-2c_3x_3^4.$$

$(\star )$ Now, let $c_1=c_2=1$ and $c_3 =2$ so that we get $$\dot{V}(x)=4x_1x_2(x_3-2)-4x_3^4.$$

However, I do not know how to determine the sign of this expression to conclude the result. I want to ask

Is it enough to find a Lyapunov function which is always positive or negative to say that $x^*$ is stable/unstable/asymptotically stable?

Similar to the first question but am I allowed to set arbitary values for $c_i$'s? (The $\star$ part)

How can I deduce the result?

答:
If the system is stable, then there exists a ball centered at the origin such that all trajectories starting in this ball are bounded. Let us consider the initial condition $$x(0) = \begin{bmatrix}x_{1}(0) \\ x_{2}(0) \\ 0\end{bmatrix}.$$
From the system's dynamics, it follows that for these initial conditions $x_3(t)\equiv 0$ for all $t\ge0$. Then the dynamics of $x_1(t)$, $x_2(t)$ is
$$\begin{bmatrix}\dot{x}_{1}(t) \\ \dot{x}_{2}(t) \end{bmatrix} = \begin{bmatrix}0 & -2 \\ -2 & 0\end{bmatrix}\begin{bmatrix}{x}_{1}(t) \\ {x}_{2}(t) \end{bmatrix}.$$
This linear system is unstable, and thus all trajectories with nonzero initial conditions such that $x_3(0)=0$ are unbounded. The system is unstable.

Title: How to find a Lyapunov function


答:
Make the substitution $u=x^2+3$. Then $du=4x\,dx$ so $x\,dx=\frac{1}{4}\,du$.
Thus
$$\int (2x^2+3)^{1/3} x\,dx=\int \frac{1}{4}u^{1/3}\,du.$$

Remark: The derivative of $2x^2+3$ is "almost" part of your expression. sure, we have an $x$ instead of $4x$, but that's no problem, since $x=\frac{1}{4}(4x)$. In this kind of situation, substitution is often useful. You will meet variants often, like $\int xe^{x^2}\,dx$, or $\int (1+\sin x)^8\cos x\,dx$.

Title: Find the antiderivatives

Body:
Find the antiderivatives:

$\int\!\left(2x^2 +3\right)^{1/3} x\,dx $

I have hit this in my book and the way I do it I get

$3/4\left(2x^2 + 3\right)^{4/3} x^2 +c $

But my book tells me it should be $3/16(2x^2 + 3)^{4/3}$

Anyone know how they get to that?


问:
Gradient estimate for harmonic functions - Isn't a power missing on $N$?
In class we have seen that if $u$ is harmonic, $\Omega \subset \mathbb R^N$ and $\Omega' \subset \subset \Omega$ is a subdomain, then
$$
\|Du\| _{L^\infty(\Omega')} \leq \frac{N}{d(\Omega', \partial \Omega)}\|u\|_{L^\infty(\Omega)},
$$
which is Theorem 2.10 in Gilbarg and Trudinger for the case $|\alpha| = 1$. The proof I can see follows from the bound
$$
\left| \frac{\partial u}{\partial x_i}(x_0) \right| \leq \frac Nr \|u\|_{L^\infty(B_r(x_0))} \quad \forall x_0 \in \Omega, \forall 0 < r < d(x_0, \partial \Omega).
$$
(which in turn follows from $u_{x_i}$ being harmonic and the Mean Value Property).

Now, as I see it:
$$
\|Du\|_{L^\infty(\Omega')} = \sup_{\Omega'} |Du|
\leq \sqrt{N \sup_{i, \Omega'}\left| \frac{\partial u}{\partial x_i}(x_0) \right|^2} \leq \sqrt{N \frac{N^2}{d(\Omega', \partial \Omega)^2}\|u\|_{L^\infty(\Omega)}^2} \\ = \frac{N^{3/2}}{d(\Omega', \partial \Omega)}\|u\|_{L^\infty(\Omega)}.
$$

Why is this exponent $3/2$ appearing? What am I missing? Maybe I should have considered the maximum norm in $\mathbb R^N$ instead of the cannonical one, but wouldn't this be a problem?

Thanks in advance and kind regards.

答:
No, there is no power missing. Using the Mean-Value Property and the Gauss-Green-Theorem (see Proof of the Gauss-Green Theorem) for each component of the gradient, we have

\begin{align}
 Du(x_0)=\frac{1}{\omega_Nr^N}\int_{B_{r}(x_0)} Du(x)dx=\frac{1}{\omega_Nr^N}\int_{\partial B_r(x_0)}u(x)\nu(x)dS(x),
\end{align}

where $\omega_N$ represents the volume of the unit ball and $\nu$ the outward unit normal.
If we denote the standard inner product in $\mathbb{R}^N$ by a dot  "$\cdot $", then we have

\begin{align}
\big|Du(x_0)\big|^2&=|Du(x_0)\cdot Du(x_0)|\\
&=\frac{1}{\omega_Nr^N}\bigg|\int_{\partial B_r(x_0)}u(x)\nu(x)\cdot Du(x_0)dS(x)\bigg|\\
&\leq \frac{1}{\omega_Nr^N}\int_{\partial B_r(x_0)}|u(x)|\big|\nu(x)\cdot Du(x_0)\big|dS(x)\\
&\leq \frac{1}{\omega_Nr^N}\int_{\partial B_r(x_0)}|u(x)|\big|\nu(x)\big|\big|Du(x_0)\big|dS(x)\\
&\leq \frac{N\omega_Nr^{N-1}}{\omega_Nr^N}||u||_{L^{\infty}(B_r(x_0))}\big|Du(x_0)\big|,
\end{align}
where we used the triangle-inequality in the third line, Cauchy-Schwartz in the fourth line and the fact that the outward unit normal has euclidean norm one in the last line. If you divide by the Euclidean norm of the gradient, then you get your desired estimate without the power.


答:
That the solution wrote $dxdydz$ instead of $dzdydx$is unfortunate, but hopefully you worked out which variable's integration range each $\int$ denoted. Let's flesh out @sudeep5221's explanation of the middle $\int$ in each piece. That's the hard one, the one that explains why we have to split into two integrals in the first place.

The main thing we need to understand is why $x$'s range is $[0,\,\sqrt{2}]$, and worth splitting at $x=\tfrac{2}{\sqrt{5}}$. Since $0\le x\le y$, $2x^2\le x^2+y^2\le4-z^2\le4$, so that answers the first part. But the case $y=2x$ is consistent with $x^2+y^2\le4$ iff $x^2\le\tfrac{4}{1+2^2}=\left(\tfrac{2}{\sqrt{5}}\right)^2$, so in the first integral $y$ is capped at $2x$, whereas in the second it's capped at $\sqrt{4-x^2}<2x$.

There's also a bit of a trick question here: clearly $z^2\le4$, so $z$'s true cap is $2$ (well, $\sqrt{4-x^2-y^2}$ given how the triple integrals are nested), not the more generous-sounding $5$.

Personally, I'd rather compute $U$'s volume in cylindrical coordinates as$$\int_0^2dz\int_0^2\rho d\rho\int_{\pi/4}^{\arctan2}d\varphi=2\cdot2\cdot(\arctan2-\pi/4)=4\arctan\tfrac13.$$

Title: Volume and triple integral

Body:
I've been studying some triple integrals and their relation to volumes in 3 dimensions and got a little confused by the following problem:

Let $$U=\{(x,y,z) \in \mathbb{R}^3; \; x^2+y^2+z^2\leq4 \; ; 0\leq z \leq5 \; ; x \leq y \leq2x\;; x\geq0 \}$$ find the integral that represents the volume of the set $U$

The solution is provided and is shown to be:

$$\int^{\frac{2}{\sqrt{5}}}_{0}\int^{2x}_{x}\int^{\sqrt{4-x^2-y^2}}_{0}\,dx\,dy\,dz+ \int^{\sqrt{2}}_{\frac{2}{\sqrt{5}}}\int^{\sqrt{4-x^2}}_{x}\int^{\sqrt{4-x^2-y^2}}_{0}\,dx\,dy\,dz$$

I just don't understand why he splits the integral into two parts and the limits of the $x$ and $y$ variables.


问:
Question regarding arithmetic: When am I to simplify?
Perhaps this one is a silly question but how do I know I am to simplify an expression? For instance: Suppose $a=5+3$, why on earth I am to perform the addition and make $a=8$. Are there rules to know when not to simplify an expression: sums, fractions, etc...?

答: There is not really any standard practice for when or how much to simplify and expression, and it is not always clear what 'simplify' means. For example, the expression $(x+2)(x+1)$ can be written as $x^2+3x+2$ which is in a sense simpler, since the multiplication has been done out, but it also obscures the factorization, making it less simple. It just depends on context.


问:
What is $\frac{d}{dt}\int_{t_0}^t f(t,s)g(s)ds$?
I am trying to follow through this proof involving solutions to $\dot{x}=A(t)x+g(t)$. There is a part in the proof where you have to write $\frac{d}{dt}\int_{t_0}^t f(t,s)g(s)ds$ as a function. This is a bit tricky since I cannot exactly use the fundamental theorem of calculus. I think the proof states that $\frac{d}{dt}\int_{t_0}^t f(t,s)g(s)ds=f(t,t)g(t)+\int_{t_0}^t\frac{\partial f}{\partial t}(t,s)g(s)ds$, but I am not sure, nor am I convinced since I do not have the proof. Any help is appreciated.

I tried getting dirty with the limit definition of the derivative. If I set $H(t):=\int_{t_0}^tf(t,s)g(s)ds$, then I can manipulate $\frac{dH}{dt}(t)=\lim_{\epsilon \rightarrow 0}\frac{H(t+\epsilon)-H(t)}{\epsilon}$ into $\int_{t_0}^t\frac{\partial f}{\partial t}(t,s)g(s)ds+\lim_{\epsilon \rightarrow 0}\frac{1}{\epsilon}\int_t^{t+\epsilon}f(t+\epsilon,s)g(s)dx$. I do not know where to go from here. I guess my intuition says $\lim_{\epsilon \rightarrow 0}\frac{1}{\epsilon}\int_t^{t+\epsilon}f(t+\epsilon,s)g(s)dx=f(t,t)g(t)$, but how do I show this?

答:
Here is a complete proof that can be easily extended for functions defined on Banach spaces. Assume that $(t,s)\mapsto(\partial _tf)(t,s)g(s)$ is integrable over $[T_1,T_2]_t\times[T_1,T_2]_s$ and $s\mapsto f(t,s)g(s)$ is continuous where $[T_1,T_2]$ contains $t_0$ and consider $t\in(T_1,T_2)$. Set
\begin{align*}
 F(t)&:=\int_{t_0}^{t}f(t,s)g(s)\,\mathrm{d}s.
 \end{align*}
Pick $h\in\mathbb{R}$ sufficiently small so that $t+h\in(T_1,T_2)$ then write:
\begin{align*}
 F(t+h)-F(t)&=\int_{t_0}^{t+h}f(t+h,s)g(s)\,\mathrm{d}s-\int_{t_0}^{t}f(t,s)g(s)\,\mathrm{d}s.\tag{$\star$}
 \end{align*}
By definition of the partial derivative, there exists a neighborhood $\mathcal{V}\subset\mathbb{R}$ of $0$ as well as a function $\psi:\mathcal{V}\times(T_1,T_2)\to\mathbb{R}$ such that $\psi(h,s)\to0$ as $h\to0$ and
\begin{align*}
f(t+h,s)&=f(t,s)+h\partial_tf(t,s)+|h|\psi(h,s)
\end{align*}
for any $s\in(T_1,T_2)$. This relation implies that $\psi(h,s)$ is continuous and integrable in $s$ as are all the other terms. In particular, $\max\{|\psi(h,s)|\,\vert\,s\in[T_1,T_2]\}\to0$ as $h\to0$ by uniform continuity over a compact set. Hence $(\star)$ becomes:
\begin{align*}
F(t+h)-F(t)&=\int_{t_0}^{t+h}\big(f(t,s)+h\partial_tf(t,s)+|h|\psi(h)\big)g(s)\,\mathrm{d}s-\int_{t_0}^{t}f(t,s)g(s)\,\mathrm{d}s\\
%
&=\int_{t_0}^{t+h}f(t,s)g(s)\,\mathrm{d}s+h\int_{t_0}^{t+h}\partial_tf(t,s)g(s)\,\mathrm{d}s+|h|\int_{t_0}^{t+h}\psi(h,s)g(s)\,\mathrm{d}s\nonumber\\&\quad-\int_{t_0}^{t}f(t,s)g(s)\,\mathrm{d}s\\
%
&=\int_{t}^{t+h}f(t,s)g(s)\,\mathrm{d}s+h\int_{t_0}^{t+h}\partial_tf(t,s)g(s)\,\mathrm{d}s+|h|\int_{t_0}^{t+h}\psi(h,s)g(s)\,\mathrm{d}s.\tag{$\star\star$}
\end{align*}
Now $fg$ is continuous in $s$ so that there exists a neighborhood $\mathcal{U}\subset\mathbb{R}$ of $0$ as well as a function $\xi:\mathcal{U}\to\mathbb{R}$ (modulus of continuity) such that $\xi(h)\to0$ as $h\to0$ and
\begin{align*}
 f(t,s)g(s)&=f(t,t)g(t)+\xi(t-s)
 \end{align*}
when $|t-s|$ is small enough to belong to $\mathcal{U}$. It follows:
\begin{align*}
 \int_{t}^{t+h}f(t,s)g(s)\,\mathrm{d}s&=\int_{t}^{t+h}\big(f(t,t)g(t)+\xi(t-s)\big)\,\mathrm{d}s\\
 %
 &=hf(t,t)g(t)+\int_{t}^{t+h}\xi(t-s)\,\mathrm{d}s.
 \end{align*}
Thus, for all $\varepsilon>0$, there exists $h_0>0$ such that, for all $h\in(-h_0,h_0)$, we have $|\xi(h)|\leq\varepsilon$ and then
\begin{align*}
 \left|\int_{t}^{t+h}f(t,s)g(s)\,\mathrm{d}s-hf(t,t)g(t)\right|&\leq\left|\int_{t}^{t+h}\xi(t-s)\,\mathrm{d}s\right|\leq \varepsilon|h|.
 \end{align*}
Back into $(\star\star)$, we finally find for $h_0$ sufficiently small so that $\sup\{|\psi(h,s)|\,\vert\,s\in[T_1,T_2]\}<\varepsilon$:
\begin{align*}
 \left|\frac{F(t+h)-F(t)}{h}-f(t,t)g(t)-\int_{t_0}^{t}\partial_tf(t,s)g(s)\,\mathrm{d}s\right|&\leq\varepsilon+\int_{t}^{t+h}|\partial_tf(t,s)||g(s)|\,\mathrm{d}s+\int_{t_0}^{t+h}|\psi(h,s)||g(s)|\,\mathrm{d}s\\
 %
 &\leq\varepsilon+\int_{t}^{t+h}|\partial_tf(t,s)||g(s)|\,\mathrm{d}s+\varepsilon\int_{t_0}^{t+h}|g(s)|\,\mathrm{d}s\\
 %
 &\leq\varepsilon+\int_{t}^{t+h}|\partial_tf(t,s)||g(s)|\,\mathrm{d}s+\varepsilon\int_{t_0}^{T_1}|g(s)|\,\mathrm{d}s.
 \end{align*}
The right-hand side above is easily shown to converges to $\varepsilon\left(1+\int_{t_0}^{T_1}|g(s)|\,\mathrm{d}s\right)$ as $h\to0$, proving the formula.

More generally, under continuity or boundedness assumptions on functions $a,b:[T_1,T_2]\to\mathbb{R}$, we can show the more general formula:
\begin{align*}
 \frac{\mathrm{d}}{\mathrm{d}t}\int_{a(t)}^{b(t)}f(t,s)g(s)\,\mathrm{d}s&=f(t,b(t))g(b(t))b'(t)-f(t,a(t))g(a(t))a'(t)+\int_{a(t)}^{b(t)}\partial_tf(t,s)g(s)\,\mathrm{d}s
 \end{align*}


问:
Initial Value Problem Differential Equation Question
A question states
$y = y (x)$ which means that the variable $y$ depends on $x$. Now an initial condition is given lets say $y(0) = 10$,

Now when asking to solve the differential equation
$dy/dx + y \sin(x) = \sin^3(x)$ , how does integrating the middle term work?

Here $y$ cannot be taken as constant because question stated $y = y(x)$. I just want to know how to solve it and when to follow which rule.

答:
"How does integrating the middle term work?"

The short answer is that it doesn't, at least not right away. The usual strategy for a linear ODE such as this is to introduce what's appropriately called the "integrating factor" such that the left side can be condensed into the derivative of a product, then integrate.

In this case, we want to choose $\mu(x)$ such that

$$\mu\frac{\mathrm dy}{\mathrm dx}+\mu\sin(x)y=\frac{\mathrm d(\mu y)}{\mathrm dx}=\mu\sin^3(x)$$

By the product rule, this means

$$\frac{\mathrm d\mu}{\mathrm dx}=\mu \sin(x)$$

which is separable and easy to solve. Can you take it from here?


答: As the other users have noted, it depends on context. Integers are almost always preferable to expressions which would evaluate to make an integer, but carrying out that evaluation could obscure some key property of your expression.</p>

Title: Question regarding arithmetic: When am I to simplify?

Body: Perhaps this one is a silly question but how do I know I am to simplify an expression? For instance: Suppose $a=5+3$, why on earth I am to perform the addition and make $a=8$. Are there rules to know when not to simplify an expression: sums, fractions, etc...?


问:
$\sigma$-algebra generated by the function
Suppose function $f:\Omega \rightarrow (\Omega', Y)$ measurable. Define $$X = \sigma(\{f^{-1}(K) : K \in Y\})$$ Prove that for any $A \in X$ and any $w_1, w_2\in \Omega$ with $f(w_1) = f(w_2)$, either $w_1, w_2 \in A$ or $w_1, w_2 \in A^{\complement}$.

I was trying to establish that there exists some $D \in Y$ such that $A = f^{-1}(D)$, then the result is trivial. Is it the case that there is such a set $D$?

答:
You have to prove that
$$X=\{f^{-1}(K):K\in Y\}.$$
But this is not difficult, since
1.
$$\Omega =f^{-1}(\Omega’)\in X$$
2. $$f^{-1}(A^c) = (f^{-1}(A))^c\in X$$
3. Let be $I\neq\emptyset$ countable and $A_i = f^{-1}(B_i)$ for every $i\in I$ pairwise disjoint, then
$$
\bigcup_{i\in I} A_i = \bigcup_{i\in I} f^{-1}(B_i) =  f^{-1}(\bigcup_{i\in I} B_i)\in X
$$
Thus $\{f^{-1}(K):K\in Y\}$ is a $\sigma$-algebra (yielding $X=\{f^{-1}(K):K\in Y\}$, as $X$ is the smallest $\sigma$-algebra containing $\{f^{-1}(K):K\in Y\}$) and every $A\in X$ is of this form. With that follows also the claim.


答: The multiplicative group $(\mathbb{Z}_{35})^{\times}$ is isomorphic to $(\mathbb{Z}_{5})^{\times} \times (\mathbb{Z}_{7})^{\times}$, which is isomorphic to the additive group $\mathbb{Z}_{4} \times \mathbb{Z}_{6}$. Those three groups are not isomorphic to $\mathbb{Z}_{24}$, because $4$ and $6$ are not relatively prime.

Title: Show that the multiplicative group of integers mod 35 is not isomorphic to the additive group of integers mod 24

Body:
I got this question on an abstract algebra test back when I was still in undergrad. I was going over some old notes from undergrad (just for fun) and I realized I had a completely wrong answer to this question so I set about trying to solve it. I am really struggling with it. To be honest, I think I just have a hard time disproving two groups are isomorphic unless they have different cardinality.

Any suggestions?


问:
When does a space admit a "multiplication-defining" metric?
Define a Thales space to be a topological space $\mathcal{X}=(X,\tau)$ such that there is some $d:X^2\rightarrow\mathbb{R}$ such that the following hold:

The map $d$ is a metric and the topology it induces is $\tau$.

The map $\mathbb{R}^2\rightarrow\mathbb{R}: (x,y)\mapsto xy$ is (first-order, with parameters) definable in the two-sorted structure $$(X\sqcup \mathbb{R}; +,<,d).$$

For example, the usual metric witnesses that $\mathbb{R}^n$ is a Thales space for each $n>1$ as a consequence of Thales' intercept theorem (hence the name). [Stupid claims removed.]

Ultimately I'd love an exact characterization of Thales spaces, but I think that's rather ambitious. An easier question would be something along the lines of, "Is there a non-pathological Thales space which doesn't lean on the intersection theorem?" I'll make that precise in the following way:

Is there a Thales space which is separable and into which $\mathbb{R}^2$ does not continuously embed?

(As Eric Wofsey's answer demonstrate separability doesn't actually rule out pathological examples, but it's what I asked.)

I'd love to add "connected" but that might make the question too hard.

答:
Thales spaces have a very simple characterization: they are exactly the noncompact metrizable spaces.

First, using continuity of multiplication, to define multiplication, it suffices to be able to define multiplication on a dense subset of $\mathbb{R}$.  That is, suppose there is a formula $P(x,y,z)$ such that $P(x,y,z)$ implies $xy=z$ and for all $x$ and $y$ in some dense subset of $\mathbb{R}$, there exists a $z$ such that $P(x,y,z)$.  Then we can take a formula $Q(x,y,z)$ that says for all $\epsilon>0$, there exist $x',y'$, and $z'$ within $\epsilon$ of $x,y,$ and $z$ such that $P(x',y',z')$, and this will define multiplication.

Note moreover that for each $x,y\in\mathbb{R}^2$, there is a finite set $F_{x,y}$ of points in $\mathbb{R}^2$ which is enough to "witness" the value of the product $xy$: you just need the finitely many points involved in the geometric construction of the product, plus a few more finitely many points to witness that the two parallel lines really are parallel (for instance, the vertices of a rhombus with sides on the parallel lines).  Thus if $H$ is any inner product space and $X$ is any subspace of $H$ that contains a subset isometric to $F_{x,y}$ for each $x,y\in\mathbb{Q}$, then $X$ can define multiplication on $\mathbb{Q}$ and thus is a Thales space by the previous paragraph.

Now suppose $X$ is a noncompact metrizable space.  Then $X$ embeds in a Banach space, and every Banach space is homeomorphic to a Hilbert space, so we may assume $X$ is a subspace of a Hilbert space $H$ of dimension at least $2$. Take a sequence $(x_n)$ of distinct points in $X$ with no accumulation point.  Then there is a continuous function $f:X\to\mathbb{R}$ such that $f(x_n)\to\infty$.  Replacing $H$ with $H\times\mathbb{R}$ and using $f$ on the second coordinate, we may assume that $\|x_n\|\to\infty$ in $H$.  Choosing $f$ to grow fast enough on $(x_n)$, we may moreover assume that the norms $\|x_n\|$ are nonzero and strictly increasing with $n$.

Now take a sequence of isometric copies of $F_{x,y}$ in $H$ for each $x,y\in\mathbb{Q}$ which are eventually outside of any given ball.  We may moreover arrange that all the points in these copies have distinct norms.  Thus, we can get a sequence $(y_n)$ in $H$ with the norms $\|y_n\|$ nonzero and strictly increasing to $\infty$ such that any subspace of $H$ which contains every $y_n$ is a Thales space.

Finally, I claim there is a homeomorphism $H\to H$ that maps $x_n$ to $y_n$ for each $n$, so that $X$ is homeomorphic to a subspace of $H$ that contains each $y_n$ and thus is a Thales space.  You can construct such a homeomorphism, for instance, using polar coordinates.  On the radial coordinate, take a homeomorphism $[0,\infty)\to [0,\infty)$ which maps $\|x_n\|$ to $\|y_n\|$.  On the sphere of radius $\|x_n\|$, use a rotation of the unit sphere in $H$ that maps $x_n/\|x_n\|$ to $y_n/\|y_n\|$, and interpolate between these rotations on the other spheres.

Thus, every noncompact metrizable space is a Thales space.  Conversely, I claim multiplication can never be definable from a bounded metric, and so a compact space is never a Thales space.  Indeed, suppose $d$ is a metric on $X$ and $N$ is an integer such that $d(x,y)<N$ for all $x,y\in X$.  Note that by splitting each real number as the sum of an element of $N\mathbb{Z}$ and and element of $[0,N)$, the structure $(X\sqcup \mathbb{R}; +,<,d)$ is interpretable in the three-sorted structure $$(X\sqcup [0,N)\sqcup N\mathbb{Z};+_{[0,N)},<_{[0,N)},+_{N\mathbb{Z}},<_{N\mathbb{Z}},d_{[0,N)})$$ where $+_{[0,N)}$ is addition mod $N$ on $[0,N)$, $<_{[0,N)},+_{N\mathbb{Z}},$ and $<_{N\mathbb{Z}}$ are the obvious restrictions, and $d_{[0,N)}$ is the metric considered as a map $X^2\to [0,N)$.  But in this structure, the sort $N\mathbb{Z}$ does not interact at all with the other two sorts.  As a result, if multiplication were definable in $(X\sqcup \mathbb{R}; +,<,d)$, then multiplication would be definable in $(N\mathbb{Z};+,<)$.  But multiplication is not definable in $(N\mathbb{Z};+,<)$ (for instance, if it were, it would follow easily that multiplication is definable in $(\mathbb{N},+)$).


问:
Find a monomorphism function $U_7\rightarrow S_5$
Find a monomorphism $f$ function $U_7\rightarrow S_5$.

I know that:

$\phi(7)=6$

$|S_5|=120$

I tried to assign permutations but I'm not sure how:

$f(1)=id$

$f(2)=(1 2)$

$f(3)=(3 4)$

$f(4)=f(2)f(2)$

$f(5)=f(3)^-1$

$f(6)=f(2)f(3)$

Thank you

答:
Any cyclic subgroup of order $6$ of $S_5$ is a suitable image of $U(\Bbb Z/7\Bbb Z)\cong C_6=\{e,a,a^2,a^3,a^4,a^5\}$ under such an injective homomorphism. Such subgroups are all and only those of the form $\langle\sigma\rangle$, where $\sigma=(i_1,i_2)(i_3,i_4,i_5)$ with $\{i_1,\dots,i_5\}=\{1,\dots,5\}$ (as sets). As monomorphisms preserve elements order, we may have any of the following: $e\mapsto ()$, $a\mapsto \sigma$, $a^2\mapsto (i_3,i_5,i_4)$, $a^3\mapsto (i_1,i_2)$, $a^4\mapsto (i_3,i_4,i_5)$, $a^5\mapsto (i_1,i_2)(i_3,i_5,i_4)$.

Accordingly, there are $10$ such monomorphisms.

Yet another approach based on group action. Such a monomorphism, $\varphi$, is equivalent to a faithful action of $C_6$ (the cyclic group of order $6$) on the set $X:=\{1,\dots,5\}$. The stabilizers are in particular subgroups of $C_6$, which are: $\{e\}$, $\{e,a^3\}\cong C_2$, $\{e,a^2,a^4\}\cong C_3$, $C_6$. Now:

none of the $5$ stabilizers can be $\{e\}$, because the respective orbit would have size $6$ $(>5$) (orbit equation: $5=6+\dots$);

none of the $5$ stabilizers can be $C_6$, because the intersection of all the stabilizers wouldn't be trivial and hence the action wouldn't be faithful (orbit equations: $5=1+1+1+1+1\Rightarrow \operatorname{ker}\varphi=C_6$, $5=1+1+1+2\Rightarrow \operatorname{ker}\varphi\cong C_3$, $5=1+1+3\Rightarrow \operatorname{ker}\varphi\cong C_2$, $5=1+2+2\Rightarrow \operatorname{ker}\varphi\cong C_3$).

Therefore, the only option left is to have $3$ stabilizers equal to $\{e,a^3\}$ and $2$ stabilizers equal to $\{e,a^2,a^4\}$ (orbit equation $5=2+3\Rightarrow \operatorname{ker}\varphi=\{e\}$). Choose any $3$ elements in $X$, say $i_3, i_4, i_5$; then, $\operatorname{Stab}(i_k)=\{e,a^3\}$, $k=3,4,5$, and hence $\varphi(a^3)$ is a permutation of order $2$ of $S_5$ which fixes $\{i_3,i_4,i_5\}$, namely $\varphi(a^3)$ is any $2$-cycle of $S_5$. Similarly, $\varphi(a^2)$ and $\varphi(a^4)$ are any distinct $3$-cycles of $S_5$ which fix $X\setminus\{i_3,i_4,i_5\}=\{i_1,i_2\}$.


Title: Initial Value Problem Differential Equation Question

Body:
A question states
$y = y (x)$ which means that the variable $y$ depends on $x$. Now an initial condition is given lets say $y(0) = 10$,

Now when asking to solve the differential equation
$dy/dx + y \sin(x) = \sin^3(x)$ , how does integrating the middle term work?

Here $y$ cannot be taken as constant because question stated $y = y(x)$. I just want to know how to solve it and when to follow which rule.

Body:
$$ y' + sin(x) \ y = sin^3(x) $$
$$ \mu(x) = exp \left( \int sin(x)dx  \right)  = e^{-cos(x)} $$
$$ e^{-cos(x)} \ y' + sin(x) \ e^{-cos(x)} \ y \ = e^{-cos(x)} \ sin^3(x)   $$
$$ D_x( e^{-cos(x)} \ y) = e^{-cos(x)} \ sin^3(x) $$
$$ \int D_x( e^{-cos(x)} \ y) \ dx = \int e^{-cos(x)} \ sin^3(x) \ dx$$
$$  e^{-cos(x)} \ y = - e^{-cos(x)} \ (cos(x)+1)^2 + C  $$
$$ y(x) = -(cos(x)+1)^2 + C \ e^{cos(x)} $$


问:
Why is $[\mathbb{Q}(e^{2\pi i/p}):\mathbb{Q}] = p-1 $?
I know that $e^{2\pi i/p}$ is a root of $x^{p}-1$ and we can write:

$ x^{p}-1=\left(x-1\right)\left(\sum_{i=0}^{p-1}x^{i}\right) $

So $e^{2\pi i/p}$ is a root of $\sum_{i=0}^{p-1}x^{i}$, which means $[\mathbb{Q}(e^{2\pi i/p}):\mathbb{Q}] \leq p-1 $.

Can someone show the other direction?

答:
You can get both directions from the fact that $e^{\frac{2 \pi i}{p}}$ is a root of $P(X)=X^{p-1}+X^{p-1}+..+X+1$, and that $P(X)$ is irreducible over $\mathbb Q[X]$.

You can prove the irreducibility by observing that $P(X)=\frac{X^p-1}{X-1}$ for $x\neq 1$, thus, for $X \neq 0$ we have

$$P(X+1)=\frac{(X+1)^p-1}{X}=\sum_{k=1}^p \binom{p}{k}X^{k-1}$$

use Eisenstein Criteria.


问:
What is the inverse laplace transform of $\frac{1}{s^2+s+1}$?
So I'm trying to find the inverse laplace transform of $$\frac{1}{s^2+s+1}.$$

The first step I have taken is to change the equation to $$\frac{1}{(s+\frac{1}{2})^2+\frac{3}{4}}$$
by completing the square.

Then I use the formula: $e^{-at} = F(s+a)$ to get $$e^{-\frac{1}{2}t}\sin\left(\frac{\sqrt{3}}{2}\right).$$

This is where I am apparently taking the wrong step because the answer should be $$\frac{\sqrt{3}}{2}e^{-\frac{1}{2}t}\sin\left(\frac{\sqrt{3}}{2}\right).$$

So my question is, how do you get $\frac{\sqrt{3}}{2}$ in front of the equation?

答:
$$F(s)=\frac{1}{s^2+s+1}.$$
$$F(s)=\frac{1}{(s+\frac 12)^2+\frac 34}.$$
$$F(s)=\dfrac {2}{\sqrt 3}\frac{\frac {\sqrt 3}{2}}{(s+\frac 12)^2+\frac 34}.$$
Apply inverse Laplace transform:
$$f(t)=\dfrac {2}{\sqrt 3}e^{-t/2}\sin \left ( \dfrac {\sqrt 3 t}{2} \right)$$
Since $$ \mathcal {L^{-1}} \left (\dfrac {a}{s^2+a^2} \right) =\sin (at)$$


问:
Couple problems and classic wolf/goat/cabbage and abstraction
I was reviewing Dijkstra's approach on the problem/puzzle of how 2 married couple can cross a river with 1 boat that can carry 2 people. The original problem's restriction is that the wife can't be in the presence of another husband without hers being present as well.

His approach was to change the original restriction to be that the husbands have to stay together and the result of this abstraction is that there is no longer a need to distinguish between husband/wife so that the problem is simplified.

Although what he says is clear, I don't understand why the restriction is required in the first place.

The problem seems identical to the goat/cabbage/wolf where the cabbage and wolf had a symmetry with the goat (neither could be left alone with the goat) and hence replaced with an abstract concept "a".

Isn't this also applicable in the case of the couples? Is the approach of having an extra restriction redundant or not?

答:
For completeness, here is a solution that does separate the two husbands. Let's say the couples are Mr. and Mrs. Smith and Mr. and Mrs. Jones:

The Smiths row across together.

Mr. Smith rows back.

Mr. Smith and Mr. Jones row across together.

Mrs. Smith rows back.

Mrs. Smith and Mrs. Jones row across together.

Dijkstra would very likely have hated it, but the most effective means we have for analysing protocols like this is informed trial and error (a.k.a. model checking). Model checking does often benefit from an intelligent choice of abstraction.


问:
Order of integration change $\int_0^\infty \int_0^{x/\sqrt{t}}e^{-y^2/2}\,dy\,dt$
If we have the double integral

$\displaystyle \int_0^\infty \int_0^{x/\sqrt{t}}\exp\bigg (-\frac{y^2}{2}\bigg)dydt$, where $x > 0$

We get inequalities:

$0 \le t < \infty$ and $0 \le y \le \frac{x}{\sqrt{t}} \implies 0 \le t < \infty $ and $0 \le t \le \frac{x^2}{y^2}$

The inner integral will then be $\int_0^{x^2/y^2}\exp(-y^2/2)dt$.

The answer to the change will be:

$\displaystyle \int_0^\infty \int_0^{x/\sqrt{t}}\exp\bigg (-\frac{y^2}{2}\bigg)dydt = \int_0^\infty \int_0^{x^2/y^2}\exp\bigg (-\frac{y^2}{2}\bigg)dtdy $

My question is why does the limit of integration for $y$ become $0$ to $\infty$?

答:
Using the comment by @ancientmathematician :

WLOG let $x = 1$ as $x > 0$. Using the bounding curves $t$-axis, $y-$axis and $ty^2 = x^2$. And conditions $0 \le t < \infty, 0 \le y \le \frac{x}{\sqrt{t}}$

To get the limits of integration for $y$, let $t$ be fixed. Then

$y^2=\frac{1}{t}$ and letting $t \to \infty$ we get $y^2 \to 0$ and letting $t \to 0$ we get $y^2 \to \infty$

And we know $y \ge 0$ from the conditions above thus $0 \le y < \infty$ in the change-of-order.


问:
Is there something called the pochhammer expansion?
I'm just playing around with the Pochhammer notation where
$$n^{\underline{k}} \equiv n(n-1)\ldots(n-k+1).$$
I've established the formulas
$$\begin{split}
n &= n^{\underline{1}}\\
n^2 &= n^{\underline{2}} + n^{\underline{1}}\\
n^3 &= n^{\underline{3}} + 3n^{\underline{2}} + n^{\underline{1}}\\
n^4 &= n^{\underline{4}} + 6n^{\underline{3}} + 7n^{\underline{2}} + n^{\underline{1}}\\
n^5 &= n^{\underline{5}} + 10n^{\underline{4}} + 25n^{\underline{3}} + 15n^{\underline{2}} + n^{\underline{1}}.
\end{split}
$$
Does it exist a general formula of the form
$$
n^k = \sum_{i=k}^1 c(i,k)n^{\underline{i}},
$$
for such expansions? What are in that case the closed form of the coefficients $c(i,k)$? Does this expansion have a name?

答:
We have that

$$n^k = \sum_{q=0}^k {k\brace q} n^\underline{q}.$$

This is because we have for the RHS

$$\sum_{q=0}^k {k\brace q} q! {n\choose q}
= k! [z^k] \sum_{q=0}^k (\exp(z)-1)^q {n\choose q}.$$

Now $\exp(z)-1= z+\cdots$ so that $(\exp(z)-1)^q = z^q+\cdots$ and the
coefficient extractor enforces the upper limit of the sum and we find

$$k! [z^k] \sum_{q\ge 0} (\exp(z)-1)^q {n\choose q}
= k! [z^k] \exp(nz) = n^k.$$

We can also prove this combinatorially. The left side counts all
$k$-tuples with component values ranging from $1$ to $n.$ The right is a
classification by the number $q$ of different values that appear. We
choose these in ${n\choose q}$ ways, then we partition the slots of  the
tuple into $q$ non-empty sets in ${k\brace q}$ ways. To conclude  note
that we can assign the $q$ chosen values to the sets of slots in $q!$
ways. This is essentially the comment by @QuiaochuYuan.


问:
Factoring third degree polynomials using long division
I am sure there is a better strategy that someone smarter than me would use, but I am not that person.

I am trying to factor

$$x^2 - 3x^2 - 4x + 12$$

I don't know how so I attempt to guess with long division. I cheat and look at the answer so find out one of the factors to save myself time, so I try x - 2

I am not sure how to type out long division but I get

$$ x - 2  |  x^3 - 3x^2 - 4x + 12$$

so I know I can have an $x^2$ for how many times x goes into the leading term.

Subract it all and I am left with

$$ x - 2  |  -x^2 - 4x + 12$$

I know that the leading term goes into the inner leading term $-x$ times or however you say that.

$$ x - 2  |   - 2x + 12$$

Now -2

$$ x - 2  |   8$$

Now I don't know what to do, how did this go so wrong? I have $x^2 - x -2$ ontop and I have a remainder of 8. This can't be right, I cheated so I know that this should be a factor.

答: When you subtracted $(-x)(x-2)$ you should have gotten $-6x+12$.  You dropped a sign there.</p>


问:
Clarification of the Definition of Permutations
In set theory, when defining the permutations of a given string, does the original string count as a permutation of itself?

I.e., would a set of the permutations of "stop" include:

{ stop, stpo, sotp, sopt, spto, spot, ... }

Furthermore, if any string is a permutation of itself, does that mean the relation 'is a permutation of' is reflexive?

答: I would say "Yes" to both questions: the original string is normally regarded as a permutation of itself, and hence the relation "is a permutation  of" is reflexive.</p>


问:
Another version of recursion theorem
Let $h_1:X\times Y\to X$ and $h_2:X\times Y\to Y$ be functions, and $x\in X, y\in Y$.

Then there exist unique functions $f _1:\mathbb{N}\to X$ and $f_2:\mathbb{N}\to Y$ such that:

$f_1(0)=x$

$f_2(0)=y$

$f_i(n+1)=h_i (f_1(n),f_2(n))$ for each $i=1,2$.

I can use the following recursion theorem:

Let $A$ be a set, $a\in A$ and $g:A\times \mathbb{N}\to A$ a function. Then there exists a unique function $f:\mathbb{N}\to A$ such that $f(0)=a$ and $f(n+1)=g(f(n),n)$

Any hint?

答:
Define a function $F : \mathbb N \to X \times Y$ so that $F(0) = ( x,y )$ and

$$F(n+1) = ( h_1(F(n)) , h_2(F(n)) )$$

using the recursion theorem you have given.

$f_1$ and $f_2$ are simply the first and second components, respectively, of $F$.

Title: Another version of recursion theorem


问:
Preimage under a continuous positive definite function
Consider a continuous positive definite function:
$$
f:D \rightarrow \mathbb{R}_+\\
f(0) = 0\\
\forall x \in D \setminus \{0\}: f(x) > 0
$$
where $0 \in D \subseteq \mathbb{R}^n$ and $D$ is unbounded.

I want to show that
$$
\exists \epsilon > 0: \exists r> 0: \forall x \in D:(f(x)=\epsilon \Rightarrow \Vert x \Vert \leq r)
$$
which says that there exists $\epsilon > 0$ such that the preimage $f^{-1}[\epsilon]$ is bounded. This is a seemingly true statement but I don't know how to show that. The continuity gives that $f^{-1}[\epsilon]$ is closed in $D$ for all $\epsilon > 0$ but can we say that it is also bounded under the above conditions?

答:
The statement is not true. Let us give a counterexample in ${\mathbb R}^2$ by defining a function in polar coordinates
\begin{equation}
f(x, y) = f(r \cos\theta, r\sin\theta) = \lambda(r) a(\theta) + \mu(r)
\end{equation}
Let us choose $a(\theta) = 2 + \sin \theta$, so that $a(\theta)$ takes any value in $[1, 3]$ when $\theta$ varies in $(-\pi, +\pi]$. Let us choose $\lambda(r)$ and $\mu(r)$ such that
\begin{equation}
\begin{array}\cr
\lambda(r) + \mu(r) = \displaystyle\frac{r}{1 +r^2}\cr
3 \lambda(r) +\mu(r) = r(1 + r^2)
\end{array}
\end{equation}
On the circle $C(0, r)$, $f$ takes all the values in $[\frac{r}{1+r^2}, r(1+r^2)]$, hence if $r$ is large enough, any given value $\epsilon > 0$ is reached on that circle.


答:
The general expression for $n\ge 0$ is

$$x^n=\sum_{k=0}^n{n\brace k}x^{\underline k}\,,$$

where $n\brace k$ is the Stirling number of the second kind that counts the partitions of $[n]$ into $k$ non-empty subsets. This is easily proved by induction on $n$ once you have the identity

$$x\cdot x^{\underline k}=x^{\underline{k+1}}+kx^{\underline k}$$

and the recurrence

$${n\brace k}=k{{n-1}\brace k}+{{n-1}\brace{k-1}}\,.$$

The induction step is:

$$\begin{align*}
x^n&=x\sum_k{{n-1}\brace k}x^{\underline k}\\
&=\sum_k{{n-1}\brace k}x^{\underline{k+1}}+\sum_k{{n-1}\brace k}kx^{\underline k}\\
&=\sum_k{{n-1}\brace{k-1}}x^{\underline k}+\sum_k{{n-1}\brace k}kx^{\underline k}\\
&=\sum_k\left(k{{n-1}\brace k}+{{n-1}\brace{k-1}}\right)x^{\underline k}\\
&=\sum_k{n\brace k}x^{\underline k}
\end{align*}$$

Title: Is there something called the pochhammer expansion?

Body:
I'm just playing around with the Pochhammer notation where
$$n^{\underline{k}} \equiv n(n-1)\ldots(n-k+1).$$
I've established the formulas
$$\begin{split}
n &= n^{\underline{1}}\\
n^2 &= n^{\underline{2}} + n^{\underline{1}}\\
n^3 &= n^{\underline{3}} + 3n^{\underline{2}} + n^{\underline{1}}\\
n^4 &= n^{\underline{4}} + 6n^{\underline{3}} + 7n^{\underline{2}} + n^{\underline{1}}\\
n^5 &= n^{\underline{5}} + 10n^{\underline{4}} + 25n^{\underline{3}} + 15n^{\underline{2}} + n^{\underline{1}}.
\end{split}
$$
Does it exist a general formula of the form
$$
n^k = \sum_{i=k}^1 c(i,k)n^{\underline{i}},
$$
for such expansions? What are in that case the closed form of the coefficients $c(i,k)$? Does this expansion have a name?


问:
Behavior of Legendre polynomials $P_\ell(\cos\theta)$ under $\theta\to\pi-\theta$
I'm studying some notes on the hydrogen atom which define the Legendre polynomials via Rodrigues's formula, $$P_\ell(z)=\frac{1}{2^\ell \ell!}\frac{d^\ell}{dz^\ell}(z^2-1)^\ell,\quad z=\cos\theta.$$ I'm trying to figure out what happens when $\theta\to\pi-\theta$ or equivalently, $z\to-z$. By computing some of the lowest order Legendre polynomials I believe I've convinced myself that $P_\ell(-z)=(-1)^\ell P_\ell(z)$, but I don't see how I can prove this given the above formula, i.e. I don't see what exactly will happen to the $\ell$-th order derivative. Is it mathematically meaningful to write $$P_\ell(-z)=\frac{1}{2^\ell \ell!}\frac{d^\ell}{d(-z)^\ell}[(-z)^2-1]^\ell?$$

答: If $w = -z$, then by the chain rule $$\frac{d}{dw} = \frac{dz}{dw}\frac{d}{dz} = -\frac{d}{dz}$$ Now $\ell$-application of the above yields $$\frac{d^\ell}{dw^\ell} = (-1)^\ell \frac{d^\ell}{dz^{\ell}}$$ Since also $(w^2 - 1)^\ell = (z^2 - 1)^\ell$, then $$P_\ell(w) = \frac{1}{2^\ell \ell!}\frac{d^\ell}{dw^{\ell}} (w^2 - 1)^\ell = \frac{1}{2^\ell \ell!} (-1)^\ell \frac{d^\ell}{dz^\ell}(z^2 - 1)^\ell = (-1)^\ell P_\ell(z)$$


问:
Do all groups have initial and terminal objects?
Could someone explain this to me in basic terms and give examples explaining why?
Also, am I correct in that the initial object in a group is also the terminal object b/c all morphisms are reversible?</p>

答:
If the group is seen as a category on one object $*$ (and the group elements are considered as the arrows $*\to *$), then, as there's no more object, if it contains an initial object, it must be $*$ but that would imply that there's only one arrow $*\to *$, i.e. the group is trivial.

If a groupoid (i.e. a category where every arrow is invertible) has an initial object $0$ then it's in particular connected, there's exactly one arrow $0_x:0\to x$ for any object $x$, and thus, for any objects $x,y$ we get $0_y\circ 0_x^{-1}$ as the only arrow $x\to y$.

 If $f:x\to y$ then $f\circ 0_x:0\to y$ so by $0$ being initial, we get $f\circ 0_x=0_y$. Then use that $0_x$ is invertible.

So, a groupoid has initial object iff there's exactly one arrow between any two objects iff every object is initial iff every object is terminal iff it has terminal object iff it is equivalent to the terminal category (the discrete category on one object).

On the other hand, the category of groups and homomorphisms does have a zero object (which is initial and terminal at once), namely the trivial group $\{1\}$.

Assigning $1\mapsto 1_G$ is the unique homomorphism $\{1\}\to G$ to any group, and the constant $1$ map is the unique homomorphism $G\to\{1\}$ from any group.