问:
perfect subsets of metric space

Let $(X,d)$ be a metric space. Let $B\subseteq X.$ An accumulation point $y$ of $B$ has the property that for all $\epsilon > 0,$  $(B(y,\epsilon)\backslash \{y\}) \cap A\neq \emptyset.$ A subset $P\subseteq X$ is perfect if $P' = P$ (i.e. $P$ is the set of accumulation points of itself). Determine with proof whether the following are perfect:

$[a,b]$ in $(\mathbb{R}, |\cdot|) (a < b \in \mathbb{R})$

$\{q\in \mathbb{Q} : 0\leq q \wedge q^2 < 2\}$ in $(\mathbb{Q}, |\cdot |)$

The Cantor set $C$ in $(\mathbb{R}, |\cdot|),$ as described on Wikipedia.

$\prod_{k=1}^\infty \{0,\frac{1}{2^k}\}$ in $(\ell_1, \lVert \cdot \rVert),$ where $\ell_1$ is the subset of $\mathbb{R}^{\mathbb{N}}$ consisting of elements whose $1$-norms are finite (i.e. elements $x=(x_n)$ so that $\sum_{i} |x_i| < \infty$).

I know that if a subset is closed then it contains all its accumulation points.
I think $1$ is perfect. By the above remark it suffices to show that every element is an accumulation point. One can let $c\in [a,b]$ and consider separately the cases where $c=a$ and $a < c$ and then let $\epsilon > 0$. For instance, when $a < c,$ let $N\in \mathbb{N}$ be so that $\frac{1}{N} < \epsilon$ so that $c-\min(c-a, \frac{1}{N}) \in (B(c, \epsilon)\backslash \{c\})\cap [a,b]$ (as can be verified step by step; for example $c-\min(c-a, \frac{1}{N})\neq c$ and is within $\epsilon$ of $c$). If $c=a, c +\min\{b-c, \frac{1}N) \in (B(c, \epsilon)\backslash \{c\})\cap [a,b]$ (as can be verified similarly).

The set in 2 is clearly not closed. Also, it seems that $\sqrt{2}$ is an accumulation point of this set as $\sqrt{2}$ is a real number and $\overline{Q} = \mathbb{R}$ so it is the limit of a sequence in $\mathbb{Q}$ and this sequence can approach $\sqrt{2}$ from below (though explicitly stating what exactly this sequence is would be more convincing). So I think one can show $\sqrt{2}$ is an accumulation point but it's clearly not in the set.

As for $3$ I think it's actually perfect but I'm not really sure how to show every point is a limit point. I think it shouldn't be too hard to show the Cantor set is closed.

For $4$ I'm not sure whether it's perfect. I think it's a countable union of nowhere dense sets (point sets) so that might be useful in determining whether it's closed.

答:
You are correct about 1 and 2. For a proof of 3 see here.

For 4, to show that every point of the set is a limit point, for any sequence $(a_n)$ in it, can you find a sequence of sequence (lol)
$\{(b_n)_k\}$ in the set such that
$$
|(a_n) - (b_n)_k| < \frac{1}{2^{k-1}} ?
$$

Title: perfect subsets of metric space


问:
How would you plot co-ordinates on a grid made from equilateral triangles?
The standard square grid has two axis: x and y, how would a grid made from tesselating equilateral triangles work? (An isometric grid) Triangles don't have 2 sets of parallel sides, like squares so as far as I can tell 3 axis are needed:

But then there is no easy way of interpreting it as each axis must have its respective grid lines coming of it from either a 30' angle or a 60' angle. And you have the problem where not all coordinate combinations can actually lie on the grid, eg. (x:1, y:2, z:3).

Is there some pre-existing method/standard of plotting "isometric coordinates" or is there no other way?

答: If you use just two of the axes on your diagram then you have an unambiguous set of coordinates for your system.</p>

ContentLicense: CC BY-SA 4.0


问:
Number of solutions of a non elementary equation
I have this equation:
$$ln(x+1)=2x$$ clearly for $x>-1$.
How can I prove that it has just two solutions?
Since we can't solve it in an elementary way, I tried  graphically, thus I have plotted $ln(x+1)$ and $2x$, which is quite easy and I can see that there are the solutions $x_1=0$ and $x_2\in(-1,0)$. But now, how can I be perfectly sure that there isn't another solution $x_3>0$ in a simple way?

I tried to think about $ln(x+1)=2x$ as $g(x):=\frac{ln(x+1)}{2x}=1$ for $x>0$ but when it comes to study the sign of the derivative of $g$ I get stuck.

Thanks for the help

答: Hint: Between any two zeros there is a point where the derivative is $0$. But the derivative of $2x-\ln (1+x)$ has only one zero.


答:
Write $f(x)=\log(x+1)-2x$ and find the extrema to establish a table of variations.

We have

$$f'(x)=\frac1{x+1}-2$$ so it is an easy matter to see that $f$ is increasing in $\left(-1,-\dfrac12\right]$ and increasing in $\left[-\dfrac12,\infty\right)$. This is enough to prove that there are at most two roots.

As $f(-1^+)<0$, $f\left(-\dfrac12\right)>0$ and $f(\infty)<0$, and the function is continuous, we can affirm that there is a root in both intervals. (By inspection $x=0$ is one of them.)

Title: Number of solutions of a non elementary equation

Body:
I have this equation:
$$ln(x+1)=2x$$ clearly for $x>-1$.
How can I prove that it has just two solutions?
Since we can't solve it in an elementary way, I tried  graphically, thus I have plotted $ln(x+1)$ and $2x$, which is quite easy and I can see that there are the solutions $x_1=0$ and $x_2\in(-1,0)$. But now, how can I be perfectly sure that there isn't another solution $x_3>0$ in a simple way?

I tried to think about $ln(x+1)=2x$ as $g(x):=\frac{ln(x+1)}{2x}=1$ for $x>0$ but when it comes to study the sign of the derivative of $g$ I get stuck.

Thanks for the help


问:
Will all the models of this FOL theory represent only bipartite graphs ? And viceversa?
Will all the models of the following formula be bipartite graphs, with edges being represented by $R$ and $A$ and $\neg A$ being two disjoint groups of vertices ?

$$\forall x \forall y R(x,y) \rightarrow (\neg A(x) \leftrightarrow A(y))$$

Is this encoding right ?

答:
Literally speaking, I think YES: all models (i.e. value assignments of $R$ and $A$) of your FOL formula are bipartite graphs (together with their bipartitions). In other words, each $R$ and $A$ satisfying your formula is indeed a bipartite graph $G(V,R)$ and $A$ is a bipartition of this graph (there may exist more than one bipartitions if there are isolated vertices in the graph). Of course, to be more precise we should assume a vertex set $V$ and your formula should be written as $\forall x \in V, y \in V. R(x,y) \rightarrow (\neg A(x) \leftrightarrow A(y))$.

But this does not mean that bipartition (or in general k-colorability) is expressible in FOL. In fact, it's only expressible in MSO (or more expressible logics), where your $A$ is a set variable as existential quantifiers: $G(V,R)$ is bipartite iff $\exists A \subset V. \forall x \in V, y \in V. R(x,y) \rightarrow (\neg A(x) \leftrightarrow A(y))$.

Or maybe we can see it in this way: given any $R$ and $A$, your FOL formula can be used to test if $R$ is a bipartite graph with $A$ as its bipartition.  If the test succeeds (e.g. by loop over all edges in $R$), for sure $R$ is a bipartite graph. But if the test failed, it does not mean that $R$ is not a bipartite graph, because it could be that the $A$ part is wrong - it's not a bipartition. For instance, if we use a bipartition graph $R$ and the set of all vertices $V$ to test your formula, the formula will obviously fail, and thus this pair ($R$ and $V$) is not a model of the formula. But (by assumption) still $R$ is a bipartite graph.

In summary, just because your FOL formula doesn't give a one-one correspondence of any single input $R$ and the bipartite property of it, this FOL formula cannot be used to express (or define) bipartite graphs. Yet your original question indeed has a YES answer, just it cannot conclude that your graph property (bipartition here) is FOL-expressible.


答: $$f[x)=\ln(1+x)-2x \implies \frac{1}{1+x}-2=\frac{-1-2x}{1+x}=0, x=-1/2,f''(-1/2)<0$$ As $f(-1+h)<0, f(\infty)=-\infty$ (same sign), there are as even number $0,2,4,...$ of real roots of $f(x)=0$ in $(-1,\infty)$. Since $f(x)$ has only one max, $f_{max}=f(-1/2)=-\ln 2+1>0$ which is positive. Hence, this Eq. $\ln(1+x)=2x$ will have only two real roots.


问:
Solutions of the following Ode
I am a master student in mathematical physics. I study soliton and traveling wave solutions of the differential equations.

Let's consider the following ODE:
$$Q^{\prime}(\xi)=ln(A)(\alpha+\beta Q(\xi)+\sigma Q^2(\xi))$$
where
$A \neq 0,1.$

In a book, the solutions of the ODE are given as follows: (But, I don't understand how to derive it.)

There are twelve solution cases w.r.t coefficients of ODE.
Any help would be appreciated.

CASE I) When $\beta^{2}-4 \alpha \sigma<0$ and $\sigma \neq 0$, then
$$
\begin{array}{l}
Q_{1}(\xi)=-\frac{\beta}{2 \sigma}+\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2 \sigma} \tan _{A}\left(\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2} \xi\right) \\
Q_{2}(\xi)=-\frac{\beta}{2 \sigma}-\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2 \sigma} \cot _{A}\left(\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2} \xi\right) \\
Q_{3}(\xi)=-\frac{\beta}{2 \sigma}+\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2 \sigma}\left(\tan _{A}\left(\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)} \xi\right) \pm \sqrt{p q} \sec _{A}\left(\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right) \xi}\right)\right) \\
Q_{4}(\xi)=-\frac{\beta}{2 \sigma}-\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2 \sigma}\left(\cot _{A}\left(\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)} \xi\right) \pm \sqrt{p q} \csc _{A}\left(\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)} \xi\right)\right) \\
Q_{5}(\xi)=-\frac{\beta}{2 \sigma}+\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{4 \sigma}\left(\tan _{A}\left(\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{4} \xi\right)-\cot _{A}\left(\frac{\sqrt{-\left(\beta^{2}-4 \alpha\right) \sigma}}{4} \xi\right)\right)
\end{array}
$$

CASE II:

$\vdots$

CASE XII:
When $\beta=\lambda, \sigma=m \lambda(m \neq 0)$ and $\alpha=0,$ then
$$
Q_{37}(\xi)=\frac{p A^{\lambda \xi}}{q-m p A^{\lambda \xi}}
$$
where triangular functions are defined as
\begin{array}{l}
\sin _{A}(\xi)=\frac{p A^{i \xi}-q A^{-i \xi}}{2 i}, \quad \cos _{A}(\xi)=\frac{p A^{i \xi}+q A^{-i \xi}}{2} \\
\tan _{A}(\xi)=-i \frac{p A^{i \xi}-q A^{-i \xi}}{p A^{i \xi}+q A^{-i \xi}}, \quad \cot _{A}(\xi)=i \frac{p A^{i \xi}+q A^{-i \xi}}{p A^{i \xi}-q A^{-i \xi}} \\
\sec _{A}(\xi)=\frac{2}{p A^{i \xi}+q A^{-i \xi}}, \quad \csc _{A}(\xi)=\frac{2 i}{p A^{i \xi}-q A^{-i \xi}}
\end{array}
where $\xi$ is an independent variable, $p$ and $q$ are constants greater than zero and called deformation parameters.

答:
May be (I hope) a way to explain the differents cases

The equation is
$$y'=a+b y+c y^2$$ Switch variables to make
$$x'=\frac 1{a+b y+c y^2}=\frac 1{c(y-r)(y-s)}$$ whre $r$ and $s$  are the roots of the quadratic.
Use partial fraction decomposition
$$x'=\frac{1}{c (r-s)}\left(\frac{1}{y-r}-\frac{1}{y-s} \right)$$
$$c (r-s)x+k=\log \left(\frac{|y-r|}{|y-s|}\right)$$ Now, depending on the sign of $\Delta=b^2-4ac$, you have quite many possibilities?

Title: Solutions of the following Ode


问:
Finite groups with exactly one maximal subgroup
I was recently reading a proof in which the following property is used (and left as an exercise that I could not prove so far). Here is exactly how it is stated.

Let $G$ be a finite group. Suppose it has exactly one maximal subgroup. Then $G$ is cyclic.

Now, does this mean exactly one conjugacy class of maximal subgroups? If it really means exactly one maximal subgroup, then what can we say about a finite group that possesses exactly one class of (at least two) maximal subgroups that are all conjugate to each other?

答:
Definition: A proper subgroup $M$ of a group $G$ is maximal iff whenever $M \leq H \leq G$, we have $H = M$ or $H = G$.

Alternative Definition: A subgroup $M$ of a group $G$ is maximal iff $M \neq G$ and there is no subgroup $H$ of $G$ such that $M \subset H \subset G$.

Proposition:

Let $G$ be a finite group. Suppose it has exactly one maximal subgroup. Then $G$ is cyclic.

Proof:

Suppose $M < G$ is the maximal subgroup of finite group $G$. Take $a \in G\setminus M$.

If $H = \langle a\rangle < G$, then there is a maximal subgroup $K$ such that $H\subset K$. Clearly $K \neq M$, which leads to a contradiction to the uniqueness of the maximal subgroup. So $G = H$ must be cyclic. Furthermore, $G$ must be a $p$-group, since otherwise, $G$ would have more than one maximal subgroup.

ContentLicense: CC BY-SA 3.0


问:
Prime zeta function and squarefree divisors
Let $x_n$ and $y_n$ be two random numbers, drawn uniformly and independently from $\{1,\ldots,n\}$. Write $d_{\text{sf}}(x_n,y_n)$ for the number of squarefree divisors of $\gcd(x_n,y_n)$. We are interested in the expected value of $d_{\text{sf}}(x_n,y_n)$ as a function of $n$, as $n$ increases.

Claim. With the notation as above, the expected value of $d_{\text{sf}}(x_n,y_n)$ relates to the prime zeta function in that
$$\lim_{n \to \infty} E\big(d_{\text{sf}}(x_n,y_n)\big) = \sum_{p\text{ prime}} \frac{1}{p^2}.$$

Does anyone know of a proof of this?

答: $$\lim_{n\to \infty}\frac1{n^2}\sum_{a\le n,b\le n}\omega(\gcd(a,b))=\lim_{n\to \infty}\sum_{p\le n} \sum_{a\le n,b\le n, p| a,p| b}1$$ $$=\lim_{n\to \infty}\frac1{n^2}\sum_{p\le n}  \lfloor n/p\rfloor^2=\lim_{n\to \infty}\frac1{n^2}\sum_{p\le n}  (( n/p)^2+O(n/p))$$ $$= \lim_{n\to \infty}\sum_{p\le n} p^{-2}+O(\frac1n \sum_{k\le n} k^{-1})=\sum_p p^{-2}$$


问:
Why is $\mathbb{Z}[x]/(2, (x^3+1))\cong \mathbb{F}_2[x] /(x^3+1) $?
I am trying to understand Yuchen's answer to my other question here. The first line is

$$\mathbb{Z}[x]/(2, (x^3+1))\cong \mathbb{F}_2[x] /(x^3+1)$$

I would like to understand please two things:

$i)$ Can I tell that this is true just by looking at it? It seems weird to me that this is true, because even though the polynomial $x^3+1$ appears on both sides, polynomials seem to behave very differently in different rings.

$ii)$ What is the formal justification for this?

I have tried to prove this isomorphism by finding an explicit map; if we let $I=(2,(x^3+1))$ in $\mathbb{Z}[x]$ and we let $J=((x^3+1))$ in $\mathbb{F}_2[x]$, my best guess was to map $p(x)+I$ to $\overline{p(x)}+J$, where $\overline {p(x)}$ means reducing the coefficients $\pmod2$. However, I think this doesn't work, because our map doesn't depend at all on $(x^3+1)$ being on the left or right sides; I could have different polynomials replacing $x^3+1$ on the left and right sides. So I think the map is most likely not an isomoprhism.

I guess your exercises must come in at some point because I don't think I've used them yet.

Edit for Hurkyl:

I completed your exercises, and I'm still having trouble with your last line. I start like this: $\dfrac{\mathbb{Z}[x]}{(2, x^3+1)} = \dfrac {\mathbb{Z}[x]}{(2)+(x^3+1)} \cong \dfrac {\mathbb{Z}[x]/(2)}{((2)+(x^3+1))/(2)}$.

I know that $\mathbb{Z}[x]/(2) \cong \mathbb{F}_2[x]$. So I must show that $((2)+(x^3+1))/(2) \cong (x^3+1)$, in $\mathbb{F}_2[x]$ and then I think we are done; however, I'm not sure how to do it. Please let me know if I'm right/wrong on this, but I think that if $I \subseteq J$, then $(I+J)/I = J/I$. However, when $I \not \subseteq J$, I think $J/I$ does not make sense under the standard definition.

答:
It is very clear, you do not have to use so many isomorphism theorems.
First, you need to know if $A \cong A'$ under map $f$ and  $B \cong B'$ also under map $f$.
$B$ is an ideal of $A$, then we have $A/B \cong A'/B'$.
Then considering a natuanl map $f$ mapping from $\mathbb{Z}$ to $\mathbb{F}_2$. Then $(2,x^3+1)$ reduces to $(x^3+1)$. Using the conclusion we just said, you get
$\mathbb{Z}[x]/(2, (x^3+1))\cong \mathbb{F}_2[x] /(x^3+1)$


问:
Transforming definite integral with finite bounds using u-subs into improper integral is possible?
Can i transforming this integral:
$$\int_0^1 e^{-xt}\ln\left(\frac{1-x}{x}\right) \Bbb dx\quad t>0$$

By setting :
$$x=-\frac{1}{u}\rightarrow u=-\frac 1x$$
$$x=0\rightarrow u=-\infty$$
$$x=1\rightarrow u=-1$$

My new integral becomes an improper integral:
$$\int_{-\infty}^{-1} \frac{e^{t/u}\ln(-(1+u))}{u^2} \Bbb du$$

Tried to check with Wolfram Alpha when $t=1$, $t=2$, and $t=3$ gave me the same result. But is this allowed in math since we treat it's an undefined expression. Can you explain please?

答:
$$I=\int_0^1 e^{-tx} \log \left(\frac{1-x}{x}\right) \, dx$$
Substitute $\frac{1-x}{x}=u\to x=\frac{1}{u+1},\;dx=-\frac{1}{(u+1)^2}$

$$I=\int_0^{\infty } \frac{e^{-\frac{t}{u+1}} \log u}{(u+1)^2} \, du$$
The $u$ integral is almost as hard as the original, though.


答:
For 1. you could also note that $a=\lim_{y\to \infty} a+(b-a)/y$ and if $a<x\le b$ then $x=\lim_{y\to \infty } x-(x-a)/y.$

For 2. About 100 AD Hero (Heron) of Alexandria wrote of a method of approximating square roots. (Isaac Newton made a major generalization for approximate solutions for many other formulas). If $A>0,$ take $x_1>\sqrt A$  and let $x_{n+1}=\frac {1}{2}(x_n+\frac {A}{x_n}).$ By induction on $n$ & some elementary algebra we have  $\sqrt A< x_{n+1}< x_n$ for all $n\in \Bbb N.$ So $L=\lim_{n\to \infty }x_n$  exists and $L\ge\sqrt A >0.$ So $$0<L=\lim_{n\to \infty }x_{n+1}=\lim_{n\to \infty }\frac {1}{2}(x_n+\frac {A}{x_n})=\frac {1}{2}(L+\frac {A}{L})$$ implying $L=\sqrt A.$

Now let $A=2=x_1$ and consider the sequence $(\frac {2}{x_n})_{n\in \Bbb N}.$

For 4. Let $S= \prod_{k=1}^\infty \{0,\frac{1}{2^k}\}$. If $x=(x_n)_n\in S$ and $\epsilon >0,$ take $k_0\in \Bbb N$ with $1/2^{k_0}<\epsilon$ and let

$y=(y_n)_n$ where

$y_n=x_n$ if $n\ne k_0,$ and

$ y_{k_0}=|x_{k_0}-1/2^{k_0}|.$

So $x$ is an accumulation point of $S$.

To show $S$ is closed: Suppose $z=(z_n)_n\in l_1 \setminus S$. Take some $m\in \Bbb N$ with $z_m\not \in \{0,1/2^m\}$ and let $r=\min (|z_m|,|z_m-1/2^m|).$ The open ball $B(z,r),$ centered at $z$ with radius $r,$ is disjoint from $S$ because if $y=(y_n)_n\in B(z,r)$ then $|y_m-z_m|\le \|y-z\|<r$ so $y_m\not \in \{0,1/2^m\}.$ Therefore $z\not\in\overline S.$

Title: perfect subsets of metric space

Body:
Let $(X,d)$ be a metric space. Let $B\subseteq X.$ An accumulation point $y$ of $B$ has the property that for all $\epsilon > 0,$  $(B(y,\epsilon)\backslash \{y\}) \cap A\neq \emptyset.$ A subset $P\subseteq X$ is perfect if $P' = P$ (i.e. $P$ is the set of accumulation points of itself). Determine with proof whether the following are perfect:

$[a,b]$ in $(\mathbb{R}, |\cdot|) (a < b \in \mathbb{R})$

$\{q\in \mathbb{Q} : 0\leq q \wedge q^2 < 2\}$ in $(\mathbb{Q}, |\cdot |)$

The Cantor set $C$ in $(\mathbb{R}, |\cdot|),$ as described on Wikipedia.

$\prod_{k=1}^\infty \{0,\frac{1}{2^k}\}$ in $(\ell_1, \lVert \cdot \rVert),$ where $\ell_1$ is the subset of $\mathbb{R}^{\mathbb{N}}$ consisting of elements whose $1$-norms are finite (i.e. elements $x=(x_n)$ so that $\sum_{i} |x_i| < \infty$).

I know that if a subset is closed then it contains all its accumulation points.
I think $1$ is perfect. By the above remark it suffices to show that every element is an accumulation point. One can let $c\in [a,b]$ and consider separately the cases where $c=a$ and $a < c$ and then let $\epsilon > 0$. For instance, when $a < c,$ let $N\in \mathbb{N}$ be so that $\frac{1}{N} < \epsilon$ so that $c-\min(c-a, \frac{1}{N}) \in (B(c, \epsilon)\backslash \{c\})\cap [a,b]$ (as can be verified step by step; for example $c-\min(c-a, \frac{1}{N})\neq c$ and is within $\epsilon$ of $c$). If $c=a, c +\min\{b-c, \frac{1}N) \in (B(c, \epsilon)\backslash \{c\})\cap [a,b]$ (as can be verified similarly).

The set in 2 is clearly not closed. Also, it seems that $\sqrt{2}$ is an accumulation point of this set as $\sqrt{2}$ is a real number and $\overline{Q} = \mathbb{R}$ so it is the limit of a sequence in $\mathbb{Q}$ and this sequence can approach $\sqrt{2}$ from below (though explicitly stating what exactly this sequence is would be more convincing). So I think one can show $\sqrt{2}$ is an accumulation point but it's clearly not in the set.

As for $3$ I think it's actually perfect but I'm not really sure how to show every point is a limit point. I think it shouldn't be too hard to show the Cantor set is closed.

For $4$ I'm not sure whether it's perfect. I think it's a countable union of nowhere dense sets (point sets) so that might be useful in determining whether it's closed.


问:
How to prove that $A\cap B\subseteq C$ and $A^c\cap B\subseteq C$ imply that $B\subseteq C$?
How do you solve this problem??

Suppose that $A\cap B\subseteq C$, and $A^c\cap B\subseteq C$. Prove that $B$ is a subset of $C$.

I don't know where even to begin

Can anyone help?

Thank you

答:
Venn says:

The upper case is for $A \subseteq C$ while the lower case is for $A \nsubseteq C $.

It is evident that the region $B$ is enclosed in $C$ $(B\subseteq C)$ if and only if both the yellow and the green regions are enclosed in $C$.

We have

\begin{align}
 (A \cap B \subseteq C) \land (A^c\cap B\subseteq C) &\implies ((A \cap B) \cup (A^c\cap B))\subseteq C \\&\implies ((A \cup A^c)\cap B)\subseteq C \\&\implies (U \cap B) \subseteq C \\&\implies B \subseteq C.
\end{align}

Where $U$ is the universal set.


问:
Is the diagonal locally compact in $\mathbb{R}^{\omega}$
Is the space $\Delta = \{\mathbf{x}\in\mathbb{R}^{\omega}: x_{n} = x_{m}\text{ for all }n,m\in\mathbb{Z}^{+}\}$ locally compact in the the product space $\mathbb{R}^{\omega}$?

I believe that it is true. Here is my reasoning:

Let $\pi_{i}$ denote the projection mapping associated with the index $i$. Consider $\mathbf{0}\in\mathbb{R}^{\omega}$ and let $U = \pi_{1}^{-1}((-1, 1))\cap\Delta$ be an open neighborhood. Then since all coordinates of a point in $U$ must be equal we have $U = (-1, 1)^{\omega}\cap\Delta$. Assuming the Tychonoff theorem $U$ is contained in the compact set $[-1, 1]^{\omega}$. Since $\mathbb{R}$ is Hausdorff we can show that $\Delta$ is closed in $\mathbb{R}^{\omega}$. Since $[-1, 1]^{\omega}\cap\Delta$ is a closed subspace of a compact space it is compact (in particular, a compact subspace of $\Delta$).

Is the above reasoning correct or is the statement false?

答: $\Delta$ is homeomorphic to $\Bbb R$, so..


问:
Does the splitting principle define chern classes for vector bundles if they are known for line bundles?
Suppose we have definied chern classes $c_i$ for line bundles for all $i\geq 0$.

Let $E\to B$ be a rank $r$ vector bundle. By the splitting principle there exists a map $f:Y\to B$ such that $f^*E$ is a sum of line bundles $L_1\oplus\ldots\oplus L_r$ and $f^*:H^*(B)\to H^*(Y)$ is injective.

Does the splitting principle define chern classes for vector bundles if they are known for line bundles?

The splitting principle says only, that if the chern classes are defined in two ways (natural with respect to continuous maps), one can check on line bundles if the two ways end up to be the same. I don't see how the splitting principle could be used to define the classes because I fear that the decomposition into line bundles (or, the space $Y$ such that $f^*E$ splits) is not unique.

答:
To answer the question in the title: yes. You are really asking if the Chern classes defined via a splitting map are independent of the choice of splitting map $f$. First note that there exists a splitting map $f$ making the cohomology $H^*(Y)$ a free module over the cohomology of the base $H^*(B)$---I believe this follows from the usual construction of a splitting map via flags of subbundles.

Given another splitting map $f': Y' \rightarrow B$, form the fiber product $Y' \times_B Y$, with the canonical map to $B$ (induced by $f'$ or $f$). The map to $B$ from this fiber product corresponds to the map $R \rightarrow S' \otimes_R S$ on cohomology, where $R=H^*(B)$, $S'=H^*(Y')$ and $S=H^*(Y)$, and is hence still injective (here we use the Eilenberg-Moore spectral sequence, which actually degenerates immediately by the freeness of $H^*(Y)$ over $H^*(B)$). Inside $H^*(Y' \times_B Y)=S' \otimes_R S$ the Chern classes become equal, proving independence.


问:
Is $(3^p-1)/2$ always squarefree?
I have little conjecture. Maybe it's stupid i don't know.

Let $p>5$ be a prime number. Then $(3^p-1)/2$ is always squarefree?

It's true for $p<192$.(I used Mathematica.)

答:
This is true if and only if there exist a prime $q$ such that $p \mid q-1$ and $3$ is a $q(q-1) /p$ residue modulo $q^2$. There are $p$ of them, uniformly distributed among the $q(q-1) $ coprimes with $q$. For a fixed $q$, this happens with probability

$$A(q) := \sum_{p \mid q-1} \frac{p}{q(q-1) } $$

And one can see that the probability of finding a counterexample up to $x$ is

$$ P_x := 1- \exp \left (- \sum_{q \le x } A(q) \right )  $$

If the sum of contributions  $\sum_q A(q) $ diverges, then $P_x \to 1$ as $x$ goes to infinity and we are almost sure that a counterexample exists. However, it seems to me that $A(q) $ can be bounded from above on average by something like $\log(q) ^4 /q(q-1) $ which converges, and this would be very bad :( in practice, it would mean that if you substitute $3$ with say $5$, the probability is the same and we don't know in advance for which bases we hit such a counterexample.

I suggest you try to compute the sum $\sum_q A(q) $ and see if this converges or not; little is known about the sum of primes dividing a number, and it maybe the case that for numbers of the form $q-1$ this is bigger than on average.

Edit. Note also that the characterization at the beginning could be used to approach the problem in a direct way. There exist a version of quadratic reciprocity law for higher residues using the Hilbert Symbol, and since it is easy to understand if $q^2$ is a $q(q-1) /p$ residue modulo 3, one has some hopes to understand if $3 $ is a residue modulo $q^2$. Unluckily, I am not an expert, and I don't know how difficult it is to compute the other term (the generalization of the sign $(-1) ^{\frac{(p-1)(q-1)}{4}}$ to higher powers).

Title: Is $(3^p-1)/2$ always squarefree?


问:
Does $L^2$ convergence imply convergence of integrals?
Let $\Omega\subseteq\mathbb{R}^n$ be open such that $\mathcal{L}^n(\Omega)<\infty$ and $f_1,f_2,\ldots \in L^2(\Omega)$ a Cauchy sequence. Does there always exist an $f\in L^1(\Omega)$ such that

$\hspace{1cm}\lim_{j\to\infty}\int_{\Omega}f_jd\mathcal{L^n}=\int_{\Omega}fd\mathcal{L}^n$

holds?

This seems like it is correct by using the fact that $L^2$ is complete, hence $\lim_{j\to\infty}f_j$ exists and one should probably use that as the function $f$. Furthermore, since $\mathcal{L}^n(\Omega)<\infty,\, L^2(\Omega)\subseteq L^1(\Omega)$ holds and so the integrals are actually well-defined. However, why should $L^2$ convergence imply convergence of integrals? I also haven't been able to come up with a counterexample since canonical things like $\Omega=(0,2],f=0,\,f_j=\frac{1}{x^{i_j}}$ with $i_j$ such that $f_j$ is square integrable but not integrable don't work since such $f_j$ don't exist.

答: $L^{2}(\Omega)$ is complete. So there exists $f \in L^{2}(\Omega)$ such that $\int |f_n-f|^{2} \to 0$. By Holder's inequality $\int|f_n-f| \leq \sqrt {\mathcal L^{n} (\Omega)} \sqrt {\int |f_n-f|^{2} }\to 0$.


问:
Proving that profinite spaces form a site
I am trying to prove that the category of profinite spaces with families of finite jointly surjective maps as covers is a site (in an attempt to understand condensed sets).

The category would be all profinite spaces (Hausdorff, maximally disconnected and compact) and on a profinite space $X$, the covers on it would be any family of $(f_i : S_i \to X)_{i \in I}$ such that $\cup _{i \in J} f_i (S_i) = X$ for some finite $J \subseteq I$.

Clearly, ${\rm Hom}(-, X)$ would be a cover. I am struggling with the two other conditions to ensure it is a site.

We need that if $S$ is a covering sieve on $X$, pullback by any function $Y \to X$ is also a covering sieve. It seems to me like you must somehow use the finite set of spaces that surject onto $X$ to get morphisms that surject onto $Y$. I tried using the fact that maps between compact Hausdorff spaces are quotient maps but I am stuck.

We need that if $T$ is a sieve on $X$, $S$ is a covering sieve on $X$ and the pullback of $T$ by any function in $S$ is a covering sieve then $T$ is a covering sieve. I also have no idea how to prove this part.

答:
To avoid thinking about the huge sieves, it suffices to check that the class $\mathcal{T}'$ of finite covering families which generate them form a stable class; see Definition 2.2 in my recent preprint. Incidentally, that paper might have some handy facts about the resulting compactly generated topos that you can use.

The singleton families consisting of the identity maps are certainly in $\mathcal{T}'$. Since your covering families are defined to be jointly epimorphic, they are closed under multicomposition (finite covers of finitely many spaces covering $X$ assemble into a finite family covering $X$). If I'm not mistaken, a pullback of profinite spaces is profinite, so that given a covering family $\{f_i:Y_i \to X\}_{i \in I}$ on $X$ and a map $g:X' \to X$, we can simply pull back the $f_i$ along $g$ to obtain a covering family on $X'$ whose multicomposition with $g$ is contained in the original family. That's everything! $\mathcal{T}'$ is a stable class.

Incidentally, note that by pulling back the pairs of morphisms in a finite covering family on $X$, we produce a diagram of profinite spaces of which $X$ is the colimit, and this diagram is final in the covering sieve generated by this family, so all of the covering sieves are effective epimorphic. Moreover, since any jointly epimorphic family over a compact space has a finite covering subfamily, we've just indirectly shown that this is all of the effective epimorphic sieves, and that they are stable under pullback. So this topology coincides with the canonical topology on this category, and the canonical functor $\mathbf{Prof} \to \mathrm{Sh}(\mathbf{Prof},J_c)$ is full and faithful.

PS. I wrote this answer quickly. It relies on some facts which you should check, such as:

Profinite spaces are stable under pullback

Epimorphisms in the category of profinite spaces coincide with epimorphisms of topological spaces, which is to say surjective maps.

That diagram has $X$ as a colimit (hint: the colimit in the category of topological spaces certainly exists; check that it's $X$!)

Since these issues are purely topological in nature, they might be more accessible to you. Good luck, and let me know if it turns out I've slipped up in there somewhere.


问:
Norm of eigen vector matrix
Given a matrix $A$ of order n with distinct eigenvalues, can we bound the norm of the eigenvector matrix ($B=[v_1|\dots|v_n]: v_i$ is the eigenvector corresponding to the i-th eigen value of $A$) of $A$ using the norm of $A$ (or the coefficient of $A$)?

答:
You can scale the eigenvectors, i.e., if $v$ is an eigenvector then $\lambda v$ is as well for all real $\lambda \ne 0$. Thus, you can scale $\|B\|$ as you wish.
PS: $B$ is similar to the diagonalization of $A$, i.e. $B^{-1} A B=D$, a diagonal matrix.

Title: Norm of eigen vector matrix


问:
If $2x^4\dot{x}\ddot{x}+4x^3\dot{x}^3=-x\dot{x}$, show that $x^4\dot{x}^2=\frac{1}{2}(x_0^2-x^2)$, where $x(0)=x_0$ and $\dot{x}(0)=0$
I know that $\dot{x}\ddot{x}=\frac{1}{2}\frac{d(\dot{x}^2)}{dt}$, $x\dot{x}=\frac{1}{2}\frac{d(x^2)}{dt}$ and $x^3\dot{x}^3=\big(\frac{1}{2}\frac{d(x^2)}{dt}\big)^3$.

So this gives $x^4\frac{d(\dot{x}^2)}{dt}+\frac{1}{2}\big(\frac{d(x^2)}{dt}\big)^3=-\frac{1}{2}\frac{d(x^2)}{dt}$.

I feel like I need to integrate w.r.t $t$, but I'm not sure how to deal with the 2nd term on the LHS?

答:
To help your intuition along in such cases of second order autonomous differential equations, it is often useful to make the substitution $\dot x=u(x)$ along segments of a hypothetical solution where it is strictly monotonous. Then $\ddot x=u'(x)u(x)$ and the equation becomes
$$
2x^4u^2u'+4x^3u^3=-xu \implies u=0~\lor ~ 2x^4uu'+4x^3u^2=-x
$$
This now can be identified as a linear equation in $u(x)^2$. Then continue as usual.


问:
Effect on expected value of conditioning on inequality between random variables (do we have E[X | X>S] ≥ E[X | X>S, Y>S]?)
I've been trying to prove the following inequality:

$$
\mathbb{E}[X \mid X > S] \geq \mathbb{E}[X  \mid  X>S, Y>S]
$$

where $X$, $Y$, $S$ are mutually independent real-valued random variables (let's make them absolutely continuous to simplify). The intuition why it might be true is simple enough: if you condition on $Y>S$, the posterior on $S$ is shifted “downwards”, and then the conditioning on $X>S$ is not as informative and raises the expected value of $X$ less.

To make it simple, I have not succeeded proving this and I'm starting to wonder if it's in fact true, so I would like to know if anybody has an idea.

For completeness, my most promising attempt used the correlation inequality. First we write:

$$
\mathbb{E}[X \mid X>S, Y>S] = \frac{1}{P(X>S, Y>S)}\int_{-\infty}^{+\infty}p_S(s)P(Y>s)\int_s^{+\infty} p_X(x) x \,dx \,ds
$$

whence:

$$
\mathbb{E}[X \mid X>S, Y>S] = \frac{1}{P(X>S, Y>S)}\int_{-\infty}^{+\infty}p_S(s)P(Y>s, X>s)\mathbb{E}[X \mid X>s] \,ds
$$

Take $f(s) = \mathbb{P}(X > s, Y > s)$ and $g(s) = \mathbb{E}[X  \mid  X > s]$. Because $f$ is decreasing and $g$ is increasing, we get:

$$
\mathbb{E}[X \mid X>S, Y>S] \leq \frac{1}{\dots} \left(\int_{-\infty}^{+\infty} p_S(s) P(Y>s, X>s) \,ds\right) \left(\int_{-\infty}^{+\infty} p_S(s) \mathbb{E}[X \mid X>s] \,ds \right)
$$

i.e.:

$$
\mathbb{E}[X \mid X>S, Y>S] \leq \int_{-\infty}^{+\infty} \frac{p_S(s)}{\mathbb{P}(X > s)} \int_s^{+\infty} p_X(x) x \,dx \,ds
$$

This is frustratingly close but not quite it; if you apply the correlation inequality again you get something that is too lax (by Jensen's inequality), and I'm worrying what I got here is too lax already, or that I made a mistake.

(Why do I care? It happens to be relevant to a Bayesian model of certain linguistic phenomena. Basically, if this is true, following a certain model, you predict that upon hearing “Alice and Bridget are tall”, people will not imagine Alice to be as tall as if they hear “Alice is tall”.)

答:
Ok, so a small variation on my initial attempt works.

Let's rewrite it as:
$$
\mathbb{E}[X|X > S, Y > S] = \frac{1}{\mathbb{P}(Y>S|X>S)} \int p_S(s) \frac{\mathbb{P}(X>s)}{\mathbb{P}(X>S)} \mathbb{P}(Y>s) \mathbb{E}[X|X>s]\,ds
$$

We apply the correlation inequality, crucially taking $p_S(s) \frac{\mathbb{P}(X>s)}{\mathbb{P}(X>S)}$ to be the density (this is the density of $S$ conditional on $X > S$). We get:

$$
\mathbb{E}[X|X > S, Y > S] \leq \frac{1}{\dots} \left(\int p_S(s) \frac{\mathbb{P}(X>s)}{\mathbb{P}(X>S)} \mathbb{P}(Y>s) \,ds\right) \left( \int p_S(s) \frac{\mathbb{P}(X>s)}{\mathbb{P}(X>S)} \mathbb{E}[X|X>s]\,ds \right)
$$

The first integral is $\mathbb{P}(Y>S|X>S)$. The second one is $\mathbb{E}[X|X>S]$. Yay!


问:
A question about dominated convergence theorem
I have this question and I don't know how to proceed:

Suppose that $(f_n)$ is a sequence of measurable functions on $[0,1]$ such that $\displaystyle \lim_{n \to \infty} \int_0^1 |f_n|\, =0$
  and that there is an integrable function $g$ on [0,1] such that $|f_n|^2\leq g$ for all $n$. Show that $\displaystyle \lim_{n \to \infty} \int_0^1 |f_n|^2\, =0$.

I think that I must show $\displaystyle \lim_{n \to \infty} \int_0^1 |f_n|\, = \displaystyle  \int_0^1 \lim_{n \to \infty}|f_n|$ , in other words I must show that we can take limit inside the integral.  What does $\displaystyle \lim_{n \to \infty} \int_0^1 |f_n|\, =0$ mean? Can you help me for this question? Thanks.

答:
Start with a fact:

Let $(a_n)$ is a real sequence. $(a_n) \to L$ for some $L \in \mathbb{R}$ if and only if any subsequence $(a_{n_k})$ of $(a_n)$ has a further subsequence $(a_{n_{k_l}}) \to L$

Now take any subsequence $\displaystyle \int_0^1 |f_{n_k}|\to 0$. There exists a further subsequence  $f_{n_{k_l}}\to 0$ almost everywhere. Then as given in the question,  $|f_{n_{k_l}}|^2\leq g$ and $|f_{n_{k_l}}|^2 \to 0$. So by Dominated Convergence Theorem,

$\displaystyle \int_0^1|f_{n_{k_l}}|^2\to0$. And finally, by the fact above, $\displaystyle \int_0^1|f_n|^2\to0$.


问:
Find the probability that $3$ distinct numbers chosen from $\{1,2,\ldots,n\}$ form a geometric progression.

Find the probability that $3$ distinct numbers chosen from $\{1,2,\ldots,50\}$ form a geometric progression.

Choose numbers $a,ar,ar^2$ such that $1\le a,ar,ar^2\le 50$.

I want to know if there's a straightforward (less casework) way to go about the case where the common ratio is not an integer.

Here's my work for your reference:

Choose $a, ar,ar^2$. Make cases for $r$, and deduce the permissible range for $a$. Note that $r\le 7$. When $r\in\mathbb N$:
$$\frac{12 + 5 + 3 + 2 + 1 + 1}{{50\choose 3}}  = \frac{24}{{50\choose 3}}$$
When $r\in\mathbb{Q}\setminus\mathbb N$, choose $r = m/n$ where $m,n$ are co-prime and $m>n>1$. Note that $1 \le a\le ar \le ar^2 \le 50$ so $$1 \le a\le \left\lfloor \frac{50n^2}{m^2}\right\rfloor$$
where $n^2|a$. The probability in this case is:
$$\frac{20}{{50\choose 3}}$$
So the probability is $\displaystyle \frac{44}{{50\choose 3}}$.

Edit:
So there's a nice closed form in the comments, how is it derived?

答:
As ACheca points out, there is a closed-form formula for the number 3-term geometric progressions in $\{1,\ldots,n\}$:

$$\sum_{1<m^2\leq n}\phi(m)\lfloor n/m^2\rfloor.$$

Here is an explanation. The number of 3-term GPs is equal to the number of solutions to $ab=c^2$ with $a,b,c\in \{1,\ldots,n\}$ and $a<b$. (These give exactly 3-term GPs, with $b=ax^2$ and $c=ax$.)

If we let $d=(a,b)$, then both $b/d$ and $a/d$ must be squares, say $m^2$ and $k^2$ respectively. Note that since $a<b$ we have $1\leq k<m$.

Thus the number of 3-term GPs is equal to the number of $1\leq d\leq n$ and $1< m^2\leq n$ and $1\leq k<m$ such that $(k,m)=1$ and $1\leq m^2d\leq n$. (Since any such choice uniquely determines a valid $a,b$ and hence $c$, and vice versa.)

Thus the total count is equal to

$$\sum_{1\leq d\leq n}\sum_{1<m^2\leq n}\left(\sum_{1\leq k<m}1_{(k,m)=1}\right)1_{1\leq m^2d\leq n} = \sum_{1\leq d\leq n}\sum_{1<m^2\leq n}\phi(m)1_{1\leq m^2d\leq n}=\sum_{1<m^2\leq n}\phi(m)\lfloor n/m^2\rfloor $$

as required.


问:
How do I know when a form represents an integral cohomology class?
Suppose $M$ is an $n$-dimensional manifold, and $\omega \in \Omega^p(M)$ is a closed $p$-form. Moreover, assume that $d\omega = 0$, so that it represents a de Rham cohomology class.

I would like to understand the meaning of the following sentence that I read in a book about geometric quantisation: $\omega$ represents an integral cohomology class.

Here's what I thought: we have the Poincaré isomorphism $\Phi : H_{dR}^p(M) \rightarrow {H_p}(M, \mathbb{R})^*$, where:

$$H_{dR}^p(M) = \text{$p$th de Rham cohomology group;}$$
$${H_p}(M, \mathbb{R}) = \text{$p$th differentiable singular homology group;}$$

and the star denotes the dual. This isomorphism is given by integration of the given form over the corresponding $p$-cycle.

Now, since we have a natural inclusion ${H_p}(M, \mathbb{Z}) \subseteq {H_p}(M, \mathbb{R})$, I could say that $\omega \in \Omega^p(M)$ represents an integral cohomology class when $\Phi(\omega) \in {H_p}(M, \mathbb{Z})$. But this is basically saying that the integral of $\omega$ over any differentiable $p$-cycle is an integer. Surely this cannot hold, except for trivial cases. So what does it mean to represent an integral cohomology class?

I would also like to know: when does the volume form of a closed manifold represents an integral cohomology class, and why?

Thank you.

答:
With regard to your last question, you're on a good path for examples. Following @Qiaochu's first example, we have the closed $2$-form $\omega$ that generates $H^2(\mathbb R^3-\{0\}) \cong H^2(S^2)$:

$$\omega = \frac{x\,dy\wedge dz + y\, dz\wedge dx + z\, dx\wedge dy}{(x^2+y^2+z^2)^{3/2}}\,.$$

This should look familiar if you interpret it as the flux $2$-form of the gravitational or electric force with a particle at the origin and think of Gauss's Law. As it stands, this $2$-form is not integral, as the integral over any closed surface containing the origin, say, the unit sphere, is $4\pi$. But we can normalize and get an integral class: $\frac1{4\pi}\omega$ is now the generator of $H^2(S^2,\mathbb Z)$.

You can make this work analogously by normalizing "the" volume form for any compact, orientable $n$-manifold.

However, one of @Qiaochu's statements is wrong. If we take $M=S^2(1)\times S^2(\pi)$ (where by these numbers I mean radii), then the element of $H^2(M)$ that corresponds to the sum of the respective area forms (officially, pulled back by the canonical projections) is not an integral class, and no scalar multiple of it is (because $H^2(M)\cong \mathbb R\oplus\mathbb R$).


问:
Show that function is almost surely 0 if expectation is almost surely 0
Assume that $h(S)$ is a realvalued function of a random variable with distribution $Erlang(n,\theta)$ (h(S) not depending on $\theta$). Then conclude from $\mathbb{E}(h(S))=0$ almost surely that $h(S)=0$ a.s..

Clearly it would suffice to show the conclusion from $\int h(S) s^{n-1}e^{-\theta s}ds = 0$. However I don't see how to prove this.

My idea was to use two different arbitrary values of the parameter $\theta$ to get a contradiction for $h(S)\not = 0$ since $h(S)$ cannot depend on $\theta$. However I wasn't successful  so far with this approach. Does anybody have hint?

答:
This is essentially the completeness property of the Erlang/Gamma family of distributions under consideration, which follows from the fact that this distribution is a member of a regular (full rank) exponential family with parameter $\theta$.

Or you can directly say that $\int_0^\infty h(s) s^{n-1}e^{-\theta s}\,ds$ is the Mellin transform of $h(s)e^{-\theta s}$, so that by uniqueness of integral transforms

$$\int_0^\infty h(s) s^{n-1}e^{-\theta s}\,ds =0 \quad,\forall\,\theta>0\implies h(s)e^{-\theta s}=0 \,,\,\text{a.e.}\quad\forall\,\theta$$

And hence $$h(s)=0\,,\,\text{a.e.}$$


问:
Constant Acceleration between A and B and B and C
A train is timed between successive posts $A, B$ and $C$, each $2000$ m apart. It takes $100$ seconds to travel from $A$ to $B$ and $150$ seconds to travel from $B$ to $C$. The acceleration throughout the journey is uniform.

What is the acceleration?

I get $-2/75 ms^{-2}$

This, according to the website which marks it, is incorrect.

My method is:

[(Disp. $A$ to $B$/Time $A$ to $B$)-(Disp. $B$ to $C$/Time $B$ to $C$)]/total time

= [(Velocity $A$ to $B$)-(Velocity $B$ to $C$)]/total time

= Change in velocity/time

= acceleration

= $[(2000/150)-(2000/100)]/150+100 = -2/75ms^{-2}$

Can someone please explain why I'm wrong? Many thanks for your help

答:
Answer above has the right idea but is wrong.

Put it into a simultaneous equation and you get

20 = u + 50a

16 = u + 125a

answer should be -4/75 ms^-2


问:
Calculate the Riemann Stieltjes integral
This is not a homework question.  It is a past exam question and I would appreciate some step by step help, as I never understood this concept in class.

Let $\alpha(t) = n^2$ for $t\in[n,n+1).$  Calculate the Riemann-Stieltjes integral $$\int_0^4x^2d\alpha(x)$$ from the definition.

答:
The Riemann-Stieltjes integral $\int_a^b f(x) dg(x)$ is the limit as the step size tends to zero of $\sum_{i=0}^{n-1}f(c_i)(g(x_{i+1})-g(x_i))$ for $a=x_0<x_1<...<x_n=b$ where $c_i$ is between $x_i$ and $x_{i+1}$.

$\alpha(x_{i+1})-\alpha(x_i)$ is $0$ unless $x_i$ and $x_{i+1}$ straddle the boundary between integers, which in the limit happens at $1,2,3$.

$\int_0^4 x^2 \mathrm{d}\alpha(x)=1^2(1^2-0^2)+2^2(2^2-1^2)+3^2(3^2-2^2)=58$


问:
Converting Range with a Function Similar to Log(x)
I want to convert a number from range [x1, x2] to [y1, y2]. Except that I want the first few numbers to have more effect. Similar to the log(x) function.

For example if I want to convert [0, 100] to [0, 1], I want to have results similar to these:

```<code>0    ->   0
10   ->   0.5
50   ->   0.7
90   ->   0.96
99   ->   0.999
100  ->   1
</code>```

What's the best solution?

答:
Seems you want a function of the form $f(x) = k \cdot x^\alpha$

where

$0 \lt \alpha \lt 1$

Just pick whatever $k$ and $\alpha$ constants suit you best.

The requirement that $f(0) = 0$ is satisfied.

So try for example $\alpha=1/2$, $\alpha=2/3$, $\alpha=3/4$, etc.

and then pick the appropriate values for $k$ to satisfy $f(100)=1$.

Plot these functions and you will see what happens.


问:
Can a sequence of probability denisity functions (pdf's) converge to a function which isn't a pdf?
Let $\{\mu_n\}$ be a sequence of Borel probability measures on  $\mathcal{X} \subset \mathbb{R}^d$, $d \in \mathbb{N}\setminus\{0\}$, which are absolutely continuous with respect to the Lebesgue measure, with densities $\{f_n\}$. Similarly, let $\mu$ be an absolutely continuous probability measure, with density $f$.

A well known fact in probability theory is that convergence in distribution (weak convergence) of $\mu_n$ to $\mu$ does not entail that $f_n$ converges to $f$ pointwise almost everywhere. Another well known result (Scheffé's lemma) is that if $f_n$ converges pointwise everywhere to a probability density, then $\mu_n$ converges in total variation to the pertaining probability measure. In particular:
$$
f_n(x)\to f(x), \, a.e. \, on \, \mathcal{X} \implies \sup_{B \in \mathcal{B}(\mathcal{X})}|\mu_n(B)-\mu(B)|=0
$$
where $\mathcal{B}(\mathcal{X})$ is the Borel $\sigma$-algebra of $\mathcal{X}$.

QUESTION 1: Is it possible that $f_n$ converges almost everywhere to some nondegenerate function (i.e. nonnull on a set with positive Lebesgue measure) which is not a probability density function but still $\mu_n$ weakly converges to some probability measure?

I was starting from a simple example: let's say that $\mathcal{X}=[0,1]$. We might have that $\mu_n$ assigns no mass to the boundary $\{0,1\}$ but converges weakly to a probabity measure $\mu_*$ whose restriction to $(0,1)$ is absolutely continuous, with density $f_*$ (not a probability density) and such that $\mu_*(\{0\})>0$, $\mu_*(\{1\})>0$. Let's firstly assume that $\{f_n\}$ is dominated by an integrable function $g$ and converges to some function $f_{**}$ a.e. on $(0,1)$: in this case, for every Borel $B \subset (0,1)$, we must have that
$$
\mu_n(B)=\int_B f_n(x)dx \to \int_{B} f_{**}(x)dx =\mu_*(B)
$$

But then (by the properties of Radon-Nikodym derivatives) we must have $f_{*}=f_{**}$ almost everywhere and $f_n \to f_*$ almost everywhere.

QUESTION 2 Removing the assumption of dominance by an integrable function, can $f_n$ converge to $f_{**} \neq f_*$ on a set of non-null Lebeasgue measure?

答:
Let
$$
f_n(x)=\cases{n^2x&if $\ 0\le x\le\frac{1}{2n}$\\
              n^2\left(\frac{1}{n}-x\right)&if $\ \frac{1}{2n}\le x\le\frac{1}{n}$\\
     \frac{n}{2(n-2)}&if $\ \frac{1}{n}\le x\le1-\frac{1}{n}$\\
    n^2\left(x-1+\frac{1}{n}\right)&if $\ 1-\frac{1}{n}\le x\le1-\frac{1}{2n}$\\
    n^2\left(1-x\right)&if $\ 1-\frac{1}{2n}\le x\le1$\\
    0&otherwise.}
$$
Then $\ \lim_\limits{n\rightarrow\infty}f_n(x)=f_*(x)\ $ for all $\ x\ $, where
$$
f_*(x)=\cases{\frac{1}{2}&if $\ 0<x<1$\\
0&otherwise.}
$$
If $\ \mu_n(B)=\displaystyle\int_B f_n(x)dx\ $, then $\ \mu_n\ $ are probability measures which converge weakly to the probability measure $\ \mu_*\ $ whose distribution function $\ F_{\mu_*}\ $ is given by
$$
F_{\mu_*}(x)=\cases{0&if $\ x<0$\\
                 \frac{2x+1}{4}&for $\ 0\le x<1$\\
                 1&for $\ 1\le x\ $.}
$$
That is, $\ \mu_*\ $ has atoms of weight $\ \frac{1}{4}\ $ at $\ x=0\ $ and $\ x=1\ $, and the remaining half of its weight is uniformly distributed over the interval $\ (0,1)\ $.


问:
Expected value for a continuous random variable
Let's fix the generic probability space $(\Omega,\Sigma,P)$, and consider the generic random variable $X\colon \Omega \to \mathbb{R}$.

The expected value of $X$ (supposed to exist) is then defined as $E(X):=\int_{\Omega} X\,\text{d}P \in [-\infty,+\infty]$, namely using the Lebesgue integral of $X$ with respect to the probability measure $P\colon \Sigma \to [0,1]$.

When $X$ is a continuous variable, namely when there exists $f_X\colon \mathbb{R} \to [0,+\infty)$ such that $\forall \,x \in \mathbb{R} \quad F_X(x):=P(X \le x)=\int_{-\infty}^x f_X(t)\,\text{d}t$, we have also that $E(X)=\int_{-\infty}^{+\infty} xf_X(x)\,\text{d}x$.

My question is: how can I show that the general definition of $E(X)$ reduces to the last one for continuous random variables? Can you suggest me a textbook or a source where I can find some help?

P.S.

I have found here Wikipedia, pushforward measure that $\int_{\Omega} X\,\text{d}P=\int_{\mathbb{R}} \text{d}\mu_X$, where $\mu_X\colon \mathcal{B}(\mathbb{R}) \to [0,1] \mid \mu_X(A)=P(X \in A)$ is the pushforward measure of $X$, with $\mathcal{B}(\mathbb{R})$ the Borel $\sigma$-algebra over $\mathbb{R}$.

I know also that $F_X'=f_X$ and that $F_X(x)=\mu_X((-\infty,x])$, so maybe $\text{d}\mu_X=\text{d}F_X=f_X\,\text{d}x$, but I seriously think this is totally wrong.

Thank you for your time and help!

答:
how can I show that the general definition of $E(X)$ reduces to the last one for continuous random variables? Can you suggest me a textbook or a source where I can find some help?

This is the general definition of Expectation taken from Mood Graybill Boes


问:
Bijection between homomorphisms of rings
I am trying to show there exists a bijection of sets,

$$\hom \left( \prod_{i\in I} R_i, S\right) \to \coprod_{i\in I} \hom(R_i, S)$$

where $R_i$ are commutative rings, $S$ is a ring with no non-trivial idempotents, and $I$ is a finite set. My initial idea was the obvious but mistaken, $f \mapsto f_i = f \circ \alpha_i$ where $\alpha_i:R_i \to \prod_{j\in I} R_j$ is the naive inclusion, but this fails since $\alpha_i$ is not a homomorphism.

I am unsure of the relevance of $S$ having no idempotents. I know that I have idempotents in $\prod_{i\in I} R_i$ which correspond to $e_j$ which is $(0,\dots,1,\dots,0)$ where $1$ is at the $j$th position, and that these satisfy $e_i e_j =0$ for $i\neq j$ and $\sum_j e_j=1$.

I also know idempotents map to idempotents, so this implies for any $f \in \hom(\prod_{i\in I}R_i, S)$ that $f(e_j)$ must be $0$ or $1$. Moreover we have that, $\sum_j f(e_j) = 1$. I believe $f$ can be $1$ for only one $e_j$ and zero for all others, but I cannot prove this without assuming $S$ has characteristic zero. So there is not much I can say about $\hom(\prod_{i\in I}R_i, S)$.

I also had the idea to try formulating the bijection first the other way around, so for $f_i : R_i \to S$, to try $f_i \mapsto f$, where here $f(r_1,\dots,r_n) = f_1(r_1) + \cdots + f_n(r_n)$ (let's say $I = \{1,\dots,n\}$).

Any help would be appreciated, I am guessing the key must be something about using the idempotents of the product ring. I also believe it can be written as $\prod_{i\in I} R_i = e_1(\prod_{i\in I} R_i) + \cdots + e_n(\prod_{i\in I} R_i)$ but I do not see if this is helpful either.

答:
As a general remark, when working with a product, you do not want to use functions $R_i\to \prod_i R_i$ (which as you noted will not give anything useful), but rather the projections $\pi_i: \prod_i R_i\to R_i$. Using those, we can at least associate to every $f: R_i\to S$ some $f':\prod_i R_i\to S$ by $f'=f\circ \pi_i$. (It is similar to what you were first trying to do, but in the correct direction.)

This defines a natural function $\coprod_i \operatorname{hom}(R_i,S)\to \operatorname{hom}(\prod_i R_i,S)$, and we want to check that it is a bijection. I'll let you check that injectivity is easy, and we just need to prove surjectivity. In other words, any morphism $\prod_i R_i\to S$ is obtained by first applying a projection to some $R_i$. This amounts as in your analysis to the fact that exactly one $e_i$ is sent to $1$ and the others are sent to $0$.

First let us make sure that the hypothesis on $S$ is not useless. If $R=\mathbb{Z}$ and $S=\mathbb{Z}\times \mathbb{Z}$, then the identity morphism $R\times R\to S$ is not in the image of the above function, since it cannot be defined by first applying a projection. And indeed $S$ has nontrivial idempotents.

Now assume $S$ has no nontrivial idempotent. As you already noted, $f(e_i)$ must be $1$ or $0$, and we want to show that it is $0$ except for one $i$. If $f(e_i)=f(e_j)=1$ for $i\neq j$, then $f(e_i+e_j)=2\in S$, but $e_i+e_j$ is an idempotent because $e_i$ and $e_j$ are orthogonal, so $f(e_i+e_j)$ must be $0$ or $1$. If it is $1$, it means that $S$ is the zero ring. The question of whether your result is true for the zero ring depends on what is meant by "having no nontrivial idempotent". I would argue that the zero ring has no nontrivial idempotent, and is a counterexample since $\operatorname{hom}(\prod_i R_i,S)$ has $1$ element while $\coprod_i \operatorname{hom}(R_i,S)$ has as many elements as $I$. So I would reformulate the hypothesis to "$S$ has exactly two idempotents".

Let us assume that $S$ is not the zero ring, so $f(e_i+e_j)=0$. This means that $S$ must have characteristic $2$. But then $f(e_i)\dot f(e_i+e_j)=1\cdot 0=0$, while $f(e_i(e_i+e_j))=f(e_i^2=e_ie_j)=f(e_i)=1$: this is a contradiction.

In the end there is exactly one $i$ such that $f(e_i)=1$.

As a bonus, a geometric interpretation (if you don't know any algebraic geometry you can safely ignore this): a ring having no nontrivial idempotent amounts to saying it (or at least its spectrum) is connected. So basically your question can be rephrased as "if $Y=\coprod_i Y_i$ and $X$ is connected, there is a bijection between maps $X\to Y$ and the disjoint union of maps $X\to Y_i$", and this is somewhat clear since the image of $X$ has to be connected. Also, this becomes false when $X$ is empty, which corresponds to the counterexample of the zero ring.


问:
Why is $\dfrac{z(i-z)}{e^{2\pi z}-1}$ regular at $z=i$?
$f(z):=\dfrac{z(i-z)}{e^{2\pi z}-1} (-1<\text{Im} z<2).$

If denominator is $0$,
$e^{2\pi z}=1$.

Let $z=x+iy$ and I get $e^{2\pi x}e^{2\pi y i}=e^0,$ so $z=0,i$.

Therefore, $z=0,i $ are isolated singularities.

But my mathematics book says that $f(z)$ is regular at $z=0,i$.

I understood $f(z)$ is regular at $z=0$ because $f(z)=\dfrac{z(i-z)}{\sum_{k=0}^{\infty} \frac{z^k}{k !}-1}
=\dfrac{z(i-z)}{\sum_{k=1}^{\infty} \frac{z^{k}}{k !}}=\dfrac{i-z}{\sum_{k=0}^{\infty} \frac{z^k}{(k+1)!}}$ so $z=0$ is not isolated singularity.

But I cannot understand why $f(z)$ is regular at $z=i$.

I would like you to give me some ideas.

答: Note that\begin{align}\lim_{z\to i}\frac{e^{2\pi z}-1}{z-i}&=2\pi e^{2\pi i}\\&=-2\pi\end{align}and that therefore\begin{align}\lim_{z\to i}\frac{z(i-z)}{e^{2\pi z}-1}&=-\lim_{z\to i}\frac z{\frac{e^{2\pi z}-1}{z-i}}\\&=\frac i{2\pi}.\end{align}Therefore, yes, $i$ is a regular point (or removable singularity) of $f$.


问:
How can I solve this Bessel Integration?

prove that $\int_0^1\left(\frac{tJ_o\left(xt\right)}{\sqrt{\left(1+t^2\right)}}\right)dt=\frac{\sin x}{x}$

How can I solve this equation? Can't find an example like this.Thanks in advanced to all!!

答:
It is false.

In the picture below I plotted $y=\frac{\sin x}{x}$ in blue and $f(x)=\int_0^1 \frac{t J_0(x t)}{\sqrt{t^2+1}} \, dt$ in orange.


答:
Taylor series for the numerator at $z=i$ is
$$
-i(z-i) - (z-i)^2
$$
Taylor series for the denominator at $z=i$ is
$$
2\pi(z-i)+\dots
$$
So the Laurent series for the quotient is
$$
\frac{-i}{2\pi} + \dots
$$
Therefore the function is regular at the point.

Title: Why is $\dfrac{z(i-z)}{e^{2\pi z}-1}$ regular at $z=i$?

Body:
$f(z):=\dfrac{z(i-z)}{e^{2\pi z}-1} (-1<\text{Im} z<2).$

If denominator is $0$,
$e^{2\pi z}=1$.

Let $z=x+iy$ and I get $e^{2\pi x}e^{2\pi y i}=e^0,$ so $z=0,i$.

Therefore, $z=0,i $ are isolated singularities.

But my mathematics book says that $f(z)$ is regular at $z=0,i$.

I understood $f(z)$ is regular at $z=0$ because $f(z)=\dfrac{z(i-z)}{\sum_{k=0}^{\infty} \frac{z^k}{k !}-1}
=\dfrac{z(i-z)}{\sum_{k=1}^{\infty} \frac{z^{k}}{k !}}=\dfrac{i-z}{\sum_{k=0}^{\infty} \frac{z^k}{(k+1)!}}$ so $z=0$ is not isolated singularity.

But I cannot understand why $f(z)$ is regular at $z=i$.

I would like you to give me some ideas.


答:
HINT-For a quotient function $f(z)=\frac{p(z)}{(q(z)}$, the zeros $\alpha_i$ of the denominator $q(z)$ are the poles of $f(z)$ iff $p(\alpha_i)\ne 0$ and analytic  at $\alpha_i$. Your $p(z)$ doesn't meet the criteria of being non-zero at both $\alpha_i$. Infact you observe that

$\lim_{z\rightarrow i}f(z)$ exists so it is a removable singularity of $f(z)$.


问:
Function $\mathbb{R^2} \rightarrow \mathbb{R^3}$ which is $C^2$ but not $C^3$
I'd like to find an example of a function $f:\mathbb{R^2} \rightarrow \mathbb{R^3}$ which is $C^2$ but not $C^3$. I am not really sure how to find such an example.

 I have $f(x,y)=(f_1(x,y),f_2(x,y),f_3(x,y))$ and I need the second derivative not to be continuous, for instance.

 The first derivative w.r.t $x$ is $$\frac{\partial f}{\partial x}(x,y)=\bigg( \frac{\partial f_1}{\partial x}(x,y),\frac{\partial f_2}{\partial x}(x,y),\frac{\partial f_3}{\partial x}(x,y) \bigg)$$
and the second is given by
$$\frac{\partial f}{\partial x}(x,y)=\bigg( \frac{\partial^2 f_1}{\partial x^2}(x,y),\frac{\partial^2 f_2}{\partial^2 x}(x,y),\frac{\partial^2 f_3}{\partial^2 x}(x,y) \bigg).$$

答: Consider the function$$\begin{array}{rccc}g\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}x^2\sin\left(\frac1x\right)&\text{ if }x\ne0\\0&\text{ otherwise.}\end{cases}\end{array}$$The function $g$ is continuous (actually, it is differentiable, although not a $C^1$ function). Take $h\colon\Bbb R\longrightarrow\Bbb R$ such that $h''=g$. Finally, define$$\begin{array}{rccc}f\colon&\Bbb R^2&\longrightarrow&\Bbb R^3\\&(x,y)&\mapsto&\bigl(h(x),0,0\bigr).\end{array}$$Then $f$ is a $C^2$ function, but it's not $C^3$.


问:
general definition of curl
I am studying about 2-dimensional Euler equation's fluid vorticity, and I want to know how to calculate it.

$\omega = ∇\times u$  if $\omega$ is a fluid vorticity and u is the velocity vector of the fluid.

I tried to calculate it using polar coordinates and since $∇=(\frac{\partial}{\partial r},\frac{1}{r}\frac{\partial}{\partial \theta})$, it becomes $∇\times u=(\frac{\partial}{\partial r},\frac{1}{r}\frac{\partial}{\partial \theta},\frac{\partial}{\partial z})\times(u_r,u_\theta,0)=(0,0,\partial_r u_\theta - \frac{1}{r}\partial_\theta u_r)$.

But my book says it should be $\omega = \frac{1}{r}\partial_r (ru_\theta)-\frac{1}{r}\partial_\theta u_r$.

I think this difference is from the general definition of curl.
When I studied divergence in polar coordinate, I saw that the general definition of dot product is $\textrm{div(F)}=\frac{1}{\rho}\frac{\partial(\rho F^i)}{\partial x^i},$ where $\rho=\sqrt{\det(g)}$  and g is an metric tensor.

So similarly, I want to know what is the general definition of curl, and how to evaluate exact fluid vorticity in that case.
Does anyone help me?

答:
The dot and cross product notation only works easily with cartesian coordinates. Otherwise, you need some work to get the right expressions, because $\nabla$ is a certain differential operator, it is NOT a vector by itself. Since you're asking about the curl, we have to restrict ourselves to $3$-dimensional space. First let us fix some notation,

Let $g=dx\otimes dx+dy\otimes dy+dz\otimes dz$ denote the standard metric tensor on $\Bbb{R}^3$.

Let us suppose that we have orthogonal coordinates $(x^1,x^2,x^3)$ (not necessarily cartesian), which means $g_{ij}=0$ if $i\neq j$ (see next bullet point).

Let $g_{ij}$ be the $i,j$ component of the Euclidean metric tensor relative to these coordinates: $g_{ij}:=g\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right)$, and let $h_i:=\sqrt{g_{ii}}$. The $h_i$'s are usually what's called the scale factors (of the metric tensor relative to the coordinates $(x^1,x^2,x^3)$).

Let $\mathbf{e}_i$ denote the $i^{th}$ unit vector (in differential geometry notation, we have $\frac{\partial}{\partial x^i}= \sqrt{g_{ii}}\mathbf{e}_i = h_i\mathbf{e}_i$).

Finally, let $F^1,F^2,F^3$ denote the components of the vector field $\mathbf{F}$ relative to the basis $\{\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\}$, i.e $\mathbf{F}=F^1\mathbf{e}_1 + F^2\mathbf{e}_2 + F^3\mathbf{e}_3$.

Then, we have:
\begin{align}
\nabla \times \mathbf{F} &=\frac{1}{h_1h_2h_3}\det
\begin{pmatrix}
h_1\mathbf{e}_1 & h_2\mathbf{e}_2 & h_3\mathbf{e}_3\\
\frac{\partial}{\partial x^1}&\frac{\partial}{\partial x^2}&\frac{\partial}{\partial x^3}\\
h_1F^1& h_2F^2 & h_3F^3
\end{pmatrix}\\\\
&=\frac{1}{h_2h_3}\left[\frac{\partial(h_3F^3)}{\partial x^2}-\frac{\partial(h_2F^2)}{\partial x^3}\right]\mathbf{e}_1\\
&+
\frac{1}{h_1h_3}\left[\frac{\partial(h_1F^1)}{\partial x^3}-\frac{\partial(h_3F^3)}{\partial x^1}\right]\mathbf{e}_2\\
&+
\frac{1}{h_1h_2}\left[\frac{\partial(h_2F^2)}{\partial x^1}-\frac{\partial(h_1F^1)}{\partial x^2}\right]\mathbf{e}_3
\end{align}
The way I calculated this is using the exterior calculus: $\nabla \times \mathbf{F}:= g^{\sharp}\star d [g^{\flat}(\mathbf{F})]$.

If you wish, here are a set of lecture notes I found online which deals with this question (albeit with a bit of handwaving... which can be made more rigorous if you really want). If I remember correctly, Griffiths' electrodynamics book has an appendix which addresses these questions, and the presentation there is similar to the notes I linked to.

Title: general definition of curl


问:
If $\left| f'(x) \right| \leq A |f(x)|^\beta $ then f is a constant function
Problem Let $f(x)$ be a differentiable function on $[a,b]$ satisfying $f(a)=0$. If there exist $A \ge 0$ and $\beta \ge 1$ such that the inequality

$$\left| f'(x) \right| \leq A \left| f(x) \right|^\beta $$

holds for all $x$ in $[a,b]$, then $f(x) = 0$ for any x in that interval.

My attempted solution

We shall prove by contradiction. Suppose that for some $x$ in $[a,b]$ we have $f(x) \neq 0$.

Put $S:= \left\lbrace x \right. $ such that $a \le x \le b $ and $f(x) \neq 0 \left. \right\rbrace$ and $c:=\inf S$ (the infimum value of set $S$). If $c=b$ then $ f(x) = 0$   $\forall  x < b$ and the continuity of $f$ implies that $f(b) = 0$, too (contradiction). Therefore $c \neq b$. If $c=a$, then $f(c) = 0$ by the hypothesis. Otherwise, when $a < c < b$ we have $f(x) = 0$ $\forall x < c$ and again, by continuity, $f(c) = 0$. Thus in all cases considered, we have $f(c) = 0$ and $a \leq c c$ and close enough to $c$ such that $|f(x)| \le 1$ and $$\tag{$\star$} A(x-c) \le \frac12\quad\text{for all }x\text{ with }c \le x \le d.$$ Since $f$ is continuous on $[c,d]$, it has maximum and minimum values on this closed interval, leading to the existence of $c \le t \le d$ such that $|f(t)| = \max_{c \le x \le d} |f(x)|$. As already noted, $f(c) = 0$ and $c$ is the infimum of the set of numbers whose image under $f$ is nonzero. Therefore, for any $\epsilon>0$, there is some $c<m<c+\epsilon$ satisfying $f(m) \neq 0$. This observation, together with the definition of $t$, gives us $$\tag{$\star\!\star$}f(t) \neq 0,$$ $t \neq c$ and so $c<t$. Applying the Lagrange theorem gives:

$$ |f(t)| = |f(t) - f(c)| = |t-c||f'(u)| \leq|t-c|A|f(u)|^\beta $$ where $c<u<t\le d$. By $(\star)$:
$$ |t-c|A|f(u)|^\beta \le \frac12 |f(u)|^\beta \leq \frac12 |f(u)| $$

Combining the two inequalities and noting that $|f(t)| \ge |f(u)|$ (because of the definition of $t$), we have $f(u) = f(t) = 0$ and arrived at a contradiction with $(\star\star)$.

Question

I hope someone can verify if my solution is correct. Of course, other ideas, comments or solutions are welcome.

I like to post problems and my solutions to the forum because I think it's beneficial to the community, and for learners like me. First, I can hardly know if there's flaw in my own argument. Second, I may get new insights/solutions for my problem.

Thank you.

答:
( Edit.  The ODE analysis also explains why $\beta \geq 1$ is necessary. The condition for $|f'(x)| \leq A|H(f(x))|$ to have the property that $f$ does not change sign, when $H$ is a function such that $H(x)=0$ only at $x=0$, is that $\frac{1}{H(x)}$ an has a non-integrable singularity at $0$. Knowing that this is the answer, there is no loss in considering singularities that are powers of $f$, which is what most cases would look like and how they would be recognized.)

The differential equation $f'(x) = A(x) f(x)^\beta$ can be solved explicitly, by separation of variables.  The calculation shows that $f(a)=0$ is necessary, or we could write down nowhere zero solutions, by solving the equation.  This bears further analysis, because superficially the problem looks like a Lipschitz condition with $|f(x) - f(y)| \leq |x-y|^\beta$ which (as is very well known and often solved as an exercise) implies constant $f$ when $\beta > 1$.

The DE for constant $\beta > 1$, writing $y$ for $f(x)$, is
$$ d(\frac{1}{(\beta - 1)y^{\beta - 1}}) = A(x) dx$$

The coordinates that trivialize the equation are therefore $Y = \frac{1}{(\beta - 1)y^{\beta - 1}} $ and $X = \int A(x) dx$, in which the solutions are $Y = X + c$ for constant $c$.  This brings out the idea that $A$, which appears at first to be a harmless constant removable by a linear change of variable, has a meaningful part in the problem, as velocity of the independent variable.

Under this coordinate change, $f(a)=0$ becomes $f(a)=\infty$.  On a maximal interval where $f$ is nonzero, there cannot be any solutions that hit infinity at the boundary in finite "time" (here $X$ is time, so we mean a finite change in $\int A(x) dx$) while satisfying $Y - X = $ constant.

So the meaning of the problem seems to be: no singularity can be reached in finite time for this ODE.


问:
Maximum likelihood estimator of $\lambda$ and verifying if the estimator is unbiased
$(X_1,\ldots,X_n)$ is a random sample extracted from an exponential law of parameter $\lambda$

Calculate the  likelihood estimator $\nu$ of $\lambda$.

Then, if $n=2$: establish if $\nu$ is a unbiased estimator

$$L(\lambda: X_1,\ldots,X_n)=\prod_{i=1}^n \lambda \ e^{-\lambda \  x_i} \ \ 1_{(0,+\infty)} \  (x_i)=$$

$$=\prod_{i=1}^n (1_{(0,+\infty)} \  (x_i)) \ \ \lambda^n \ e^{-\lambda \sum_{i=1}^n  x_i} \ \ $$

$$\frac{\partial}{\partial \lambda} L(\lambda: X_1,...,X_n)=\prod_{i=1}^n (1_{(0,+\infty)} \  (x_i)) \ \ n \lambda^{n-1} \ e^{-\lambda \sum_{i=1}^n  x_i}-\sum_{1=1}^n x_i \ \lambda^n \ \ e^{-\lambda \sum_{i=1}^n  x_i}=  \ \ $$

$$=\prod_{i=1}^n (1_{(0,+\infty)} \  (x_i)) \ \ \lambda^{n-1} \ e^{-\lambda \sum_{i=1}^n x_i} \ \ (n- \lambda \sum_{i=1}^n x_i) $$

$$\frac{\partial}{\partial \lambda} L(\lambda: X_1,\ldots,X_n) \ge 0 \Longleftrightarrow \lambda \le \frac{1}{\overline{X}}$$

Maximum likelihood estimator of $\lambda$ is $\nu=\frac{1}{\overline{X}}$

If $n=2$, I think that:

$$\nu=\frac{2}{\sum_{i=1}^2 X_i}$$

and

$$\sum_{i=1}^2 X_i \sim \Gamma(2, \lambda)$$

How can I establish if $\nu$ is a unbiased estimator?

Thanks!

答:
\begin{align}
& \operatorname E\left( \frac n {X_1+\cdots+X_n} \right) \\[8pt]
= {} & \int_0^{+\infty} \frac n x \cdot f_{X_1+\cdots+X_n}(x)\, dx \\[8pt]
= {} & \int_0^{+\infty} \frac n x \cdot \frac 1 {\Gamma(n)} \cdot (\lambda x)^{n-1} e^{-\lambda x} (\lambda\, dx) \\[8pt]
= {} & \frac{n\lambda}{\Gamma(n)} \int_0^{+\infty} (\lambda x)^{n-2} e^{-\lambda x} (\lambda\,dx) = 2\lambda. \\[8pt]
= {} & \frac{n\lambda \Gamma(n-1)}{\Gamma(n)} = \frac{n\lambda}{n-1}.
\end{align}


问:
Confusion on subsets
I recently started learning a bit about set theory and thus is really confusing me Given a set $X=\{a,b,c\}$
And another set $Y=\{\{a,b\},\{a,b,c\}\}$ is $Y\subset X$
and is $X \subset Y$.

Thanks in advance.

答:
It's neither of these. In fact the sets $X$ and $Y$

don't share a common element so $X \cap Y = \emptyset $.

The set $X$ has elements $a,b,c$.

The set $Y$ has the sets {a,b} and {a,b,c} as its elements.

So obviously $X \in Y$ is valid.

But neither of $X \subset Y$ and $Y \subset X$ are valid statements.

Title: Confusion on subsets


答:
Neither $Y\subset X$ nor $X\subset Y$:

The set $Y$ is a set of subsets of $X$. Every element of $Y$ is a subset of $X$.

The set $X$ is an element of $Y$. The set $Y$ is equal to $\{\{a,b\},X\}$.

Title: Confusion on subsets

Body:
I recently started learning a bit about set theory and thus is really confusing me Given a set $X=\{a,b,c\}$
And another set $Y=\{\{a,b\},\{a,b,c\}\}$ is $Y\subset X$
and is $X \subset Y$.

Thanks in advance.


问:
Given a sequence $\{u_{n}\}$ so that $\left\{\begin{matrix} u_0=a>1\\u_{n+1}=\sqrt{1+\frac{1}{1+u_n}}\end{matrix}\right..$ Find $\lim u_n.$

Given a sequence $\left \{ u_{n} \right \}$ so that $\left\{\begin{matrix} u_{0}= a> 1\\ u_{n+ 1}= \sqrt{1+ \frac{1}{1+ u_{n}}} \end{matrix}\right..$ Find $\lim u_{n}.$

If there exists a $\lim u_{n},$ I think with $n\rightarrow \infty\Rightarrow u_{n+ 1}= u_{n}= \sqrt{1+ \frac{1}{1+ u_{n}}}\Rightarrow 1< u_{n}< 2.$ What does that help for my problem here ? I need your thought on it ? Thank you.

答:
When $n \to \infty$

We have (if the limit does exist, we need to prove it is bounded from above/below and monotinicly increasing/decreasing):

$$ u_{n+1} \sim u_{n}$$

Thus we need to solve this equation:

$$ u_n = \sqrt{ 1 + \frac{1}{1+ u_n}} \\ \implies u_n \approx 1.206 $$


问:
Calculating triple integral for 3d body

We have a 3D body $G=\left \{ (x,y,z):x^2+y^2+z^2\leq a^2,0\leq z\leq\sqrt{x^2+y^2}\right \}$

Defining $I_0=\iiint_G  \,dx\,dy\,dz$, $I_1=\iiint_G  z\,dx\,dy\,dz$

For which $a\in\mathbb{N}$ we get that $\frac{I_1}{I_0}=\frac{3\sqrt2}{4}$?

My first step was to convert into Spherical coordinate system, thus we get that $x^2+y^2+z^2=r^2$
then $0\leq r\leq a$

and $0 \leq r\cos(\varphi) \ \leq \sqrt{(r\sin(\varphi )\cos(\theta))^2+(rsin(\varphi )sin(\theta))^2}$ from converting $0\leq z\leq\sqrt{x^2+y^2}$,

now I think I need to solve the equation for $\cos(\varphi)^2 \ = (\sin(\varphi )\cos(\theta))^2+(\sin(\varphi )sin(\theta))^2$?

$$\cos(\varphi)^2 \ = (\sin(\varphi )\cos(\theta))^2+(\sin(\varphi )\sin(\theta))^2$$
$$\cos(\varphi)^2 \ = (\cos^2(\theta)+\sin^2(\theta))\ \sin^2(\varphi)$$
$$\cos(\varphi)^2 =\sin^2(\varphi)$$
$$\tan(\varphi)=1$$
$$-\frac{\pi}{4} \leq \varphi \leq \frac{\pi}{4} $$
and for $\theta$ it's $0 \leq \theta \leq 2\pi$

So the end we have

$0\leq r\leq a$

$0 \leq \theta \leq 2\pi$

$-\frac{\pi}{4} \leq \varphi \leq \frac{\pi}{4} $

Am I correct from here?

答:
In spherical coordinates, $x = \rho \cos\theta \sin\phi, y = \rho \sin\theta \sin\phi, z = \rho \cos\phi$

Here we have $0 \leq \rho \leq a, 0 \leq \theta \leq 2\pi$. For limits of $\phi$,

Please note the lower bound of $z$ in the given condition, $0 \leq z \leq \sqrt{x^2+y^2}$

As you calculated, $2a^2\cos^2\phi = a^2 \implies \cos \phi = \pm \frac{1}{\sqrt2}, \phi = \frac{\pi}{4}, \frac{3\pi}{4}$

But as $z \geq 0, \phi \leq \frac{\pi}{2}$

So the limits of $\phi \ $ is  $ \ \frac{\pi}{4} \leq \phi \leq \frac{\pi}{2}$

$I_0 = \displaystyle \int_0^{2\pi}\int_{\pi/4}^{\pi/2}\int_0^a \rho^2 \sin\phi \ d\rho \ d\phi \ d\theta = \frac{\sqrt2  \pi  a^3}{3}$

$I_1 = \displaystyle \int_0^{2\pi}\int_{\pi/4}^{\pi/2}\int_0^a \rho^3 \sin\phi \cos \phi \ d\rho \ d\phi \ d\theta = \frac{\pi a^4}{8}$

Title: Calculating triple integral for 3d body


问:
How do I get that $11^{35} \equiv 6 \hspace{0.1cm} \text{mod} (13)$ from $11^{36} \equiv 1 \hspace{0.1cm} \text{mod} (13)$
I have no clue how to do this, I manage to get I get that $11^{36} \equiv 1  \hspace{0.1cm} \text{mod} (13)$ but I can't get anywhere from there.

答:
If $11^{36}=1$ in $\Bbb Z/13$, then $11^{35}=11^{-1}$. But since $6\cdot 11=66=1$ in $\Bbb Z/13$, we have $11^{-1}=6$.

Here I just write $a=b$ in the ring $\Bbb Z/n$ for $a\equiv b\bmod n$, so that we see it a bit easier.
For $n=13$, $\Bbb Z/n$ is a field and all nonzero elements have an inverse.


问:
Is $T : \mathbb{C} \rightarrow \mathbb{C}: a + bi \mapsto \overline{a + bi} = a - bi$ a linear image?
So a question in my handbook for linear algebra is the following; Is the linear map T
$$T :\mathbb{C} \rightarrow \mathbb{C}: a + bi \mapsto \overline{a + bi} = a - bi$$

an linear image if:

$\mathbb{C}$ is interpreted as a real vector space

$\mathbb{C}$ is interpreted as a complex vector space

I know how I would solve the question itself (I would check the axiom of linear image: if $L: V \rightarrow W$ is a linear map then $L(\lambda_1v_1 + \lambda_2v_2) = \lambda_1L(v_1) + \lambda_2L(v_2)$) but I'm stuck at what they mean with the interpretation.

答:
Hint.

$T$ is a $\mathbb{R}$-linear map.

$T$ is not a $\mathbb{C}$-linear map.

Notes.

$\mathbb{C}$ is one-dimensional vector space over the filed $\mathbb{C}$   where $\alpha=(1)$ is a basis.

$\mathbb{C}$ is two-dimensional vector space over the filed
$\mathbb{R}$ where $\beta=(1,i)$ is an (ordered) basis.

When you talk about "linear maps", you need to know the underlying vector spaces (over what field). The field determines what scalar $\lambda$ you need to check for
$$
T(\lambda_1v_1+\lambda_2v_2)=\lambda_1T(v_1)+\lambda_2T(v_2)\tag{1}
$$

If $T$ is $\mathbb{C}$-linear, it means that (1) is true
for all complex numbers $\lambda_1$ and $\lambda_2$.

If $T$ is $\mathbb{R}$-linear, it means that (1) is true
for all real numbers $\lambda_1$ and $\lambda_2$.

In your example, try to see what is $T(iz)$.


答: What were your efforts ? However, if you proved that $11^{36} \equiv 1 \hspace{0.1cm} (13)$ then you have that $11^{35}\cdot 11 \equiv 1 \hspace{0.1cm}(13)$. In this line it is written that $11^{35}$ is the inverse of $11$ by definition, by uniqueness you only have to prove that $6$ does the same job, i.e $6 \cdot 11 \overset{?}{\equiv} 1 \hspace{0.1cm}(13)$, which is true, and you have the desired result.

Title: How do I get that $11^{35} \equiv 6 \hspace{0.1cm} \text{mod} (13)$ from $11^{36} \equiv 1 \hspace{0.1cm} \text{mod} (13)$

Body: I have no clue how to do this, I manage to get I get that $11^{36} \equiv 1  \hspace{0.1cm} \text{mod} (13)$ but I can't get anywhere from there.


答:
Assuming that you meant “linear map”, $T$ is a linear map if you see $\Bbb C$ as a real vector space, since, if $z,w\in\Bbb C$ and $\alpha,beta\in\Bbb R$, we have\begin{align}T(\alpha z+\beta w)&=\overline{\alpha z+\beta w}\\&=\alpha\overline z+\beta\overline w\text{ (because $\alpha,\beta\in\Bbb R$)}\\&=\alpha T(z)+\beta T(w).\end{align}

But $T$ is not a linear map if you see $\Bbb C$ as a complex vector space, because for instance, $T(i\times i)\ne i\times T(i).$

Title: Is $T : \mathbb{C} \rightarrow \mathbb{C}: a + bi \mapsto \overline{a + bi} = a - bi$ a linear image?

Body:
So a question in my handbook for linear algebra is the following; Is the linear map T
$$T :\mathbb{C} \rightarrow \mathbb{C}: a + bi \mapsto \overline{a + bi} = a - bi$$

an linear image if:

$\mathbb{C}$ is interpreted as a real vector space

$\mathbb{C}$ is interpreted as a complex vector space

I know how I would solve the question itself (I would check the axiom of linear image: if $L: V \rightarrow W$ is a linear map then $L(\lambda_1v_1 + \lambda_2v_2) = \lambda_1L(v_1) + \lambda_2L(v_2)$) but I'm stuck at what they mean with the interpretation.


问:
What could be the matrix of Transformation that has no effect on vectors? The vectors are $T(x)=(x)$.
We call this matrix Identity and denote it by $I$. But I am not clear with how to assume one matrix to proceed with its solution. The Form of Transformation here is $\mathbb R^2 \to \mathbb R^2$ subspaces. I am quoting the exact question here:

All of the transformations here have the form $T: \mathbb R^2 \to \mathbb R^2$.
Find the matrix of the transformation that has no effect on vectors, that is $T(x)= x$. We call this matrix the identity matrix and denote it by $I$.

答:
$T(x)=x\implies (T-I)x=0\forall x\in\mathbb R^2$

The minimal polynomial must be $m(t)=t-1$ which suggests $T$ is diagonizable. $m(t)$ also divides the characteristic polynomial so that is $c(t)=(t-1)^2$. Looking for the possible Jordan canonical forms for the matrix representation of $T$,  you have $T=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ ($\color{red}{Why?}$)


答:
If $A$ and $B$ are two sets then $A \subset B$ means the following statement is true: $$ x \in A \implies x \in B.$$

In your example, $X \not\subset Y$ because $a \in X$ but $a \notin Y.$

$Y \not\subset X$ because $\{a,b\} \in Y$ but $\{a,b\} \notin X.$

Title: Confusion on subsets


问:
If $f(x)=g(a^Tx)$, where $g:\mathbb{R}\to\mathbb{R}$ and $a, x\in\mathbb{R}^n$, why is the Hessian equal to $g''(a^Tx)aa^T$?
I am trying to understand the solution to below problem:

Let $f(x)=g(a^Tx)$, where $g:\mathbb{R}\mapsto\mathbb{R}$ is continuously differentiable and $a\in\mathbb{R}^n$ is a vector. What are $\nabla f(x)$ and $\nabla^2f(x)$?

The answer is

$\nabla f(x)=g'(a^Tx)a$

$\nabla^2 f(x)=g''(a^Tx)aa^T$

I understand the answer for the gradient, but why does the answer for the hessian contain $a^T$ at the end?

答:
Each successive application of the $\nabla$ operator increases
the order of the derivative of $g(\lambda)$ by one.
It also increases the tensorial order of the result by one
by liberating another $a$-vector from the function argument $\,\lambda=a^Tx,\;$ i.e.
$$\eqalign{
\nabla f(x) &= g^{\prime}(\lambda)\;&a \\
\nabla^2f(x) &= g^{\prime\prime}(\lambda)\;&a^2 \\
\nabla^3f(x) &= g^{\prime\prime\prime}(\lambda)\;&a^3 \\
\vdots\quad &= \qquad\vdots &\;\vdots \\
\nabla^nf(x) &= g^{(n)}(\lambda)\;&a^n \\
}$$
where $a^n$ and $\nabla^n$ denote $n^{th}$order tensor (aka dyadic) products. The presence of the term $aa^T$ in your formula is merely the way that $a^2$ is expressed in matrix notation.


答:
All you need to know for this is that, since $13$ is a prime number, $\mathbb{Z} / 13 \mathbb{Z}$ is a field.

In particular, every non zero element has a unique inverse for multiplication. Constating that
$$11 \times 6 = 66 \equiv 1 \hspace{0.1cm} \text{mod} (13)$$
it turns out that the multiplicative inverse of $11$ is $6$ : $\boxed{11^{-1} \equiv 6}$.

Now, since you know that $11^{36} \equiv 1 \hspace{0.1cm} \text{mod} (13)$, it turns out that :

$$11^{35} \equiv 11^{36} \times 11^{-1} \equiv 1 \times 6 \equiv 6 \hspace{0.1cm} \text{mod} (13)$$


问:
Find the relation between the dimension of the nullspace of $A$ and $A^t$
Let $A$ be a $n \times n$ matrix, what is the relation between the dimension of the nullspace of the homogeneous system of $A$ and the one of $A^t$?</p>

答: Hint: For a given matrix $A$, the dimension of the left null space is equal to the dimension of the right null space.


问:
Is it possible to draw all regular polygons on a double grid formed by a rational and an irrational grid, superimposed?
Is it possible to draw any regular polygon in such a way that its vertices be on the double grid, made of a rational and an irrational grid, superimposed, like in the image below?

If yes, what would be the step of the irrational grid?

(It is, evidently, possible to find such a double grid for drawing squares and equilateral triangles, as can be seen in the image below.)

Example of a double grid, rational and irrational (the step of the blue grid was taken $\sqrt 3$).

答:
I suppose with rational and irrational grid you mean that there are a rational number $q$ and an irrational number $i$ such that for every point of the poligon, one of its coordinates is an integer multiple of either $q$ or $i$.

In this case, the answer is negative. Pick for example a square which has opposite vertices $(\sqrt{2},\sqrt{2}), (\sqrt{3},\sqrt{3})$.

Then, if this square followed what you say, $\sqrt{2}$ and $\sqrt{3}$ should be integer multiples of some irrational $i$, that is, $\sqrt{2}=ni$ and $\sqrt{3}=mi$, for some irrational $i$. However, this is impossible, because then we would have $\sqrt{3}=\frac{m}{n}\sqrt{2}$ for some integers $m,n$, which is false.


Title: Given a sequence $\{u_{n}\}$ so that $\left\{\begin{matrix} u_0=a>1\\u_{n+1}=\sqrt{1+\frac{1}{1+u_n}}\end{matrix}\right..$ Find $\lim u_n.$

Body:
Given a sequence $\left \{ u_{n} \right \}$ so that $\left\{\begin{matrix} u_{0}= a> 1\\ u_{n+ 1}= \sqrt{1+ \frac{1}{1+ u_{n}}} \end{matrix}\right..$ Find $\lim u_{n}.$

If there exists a $\lim u_{n},$ I think with $n\rightarrow \infty\Rightarrow u_{n+ 1}= u_{n}= \sqrt{1+ \frac{1}{1+ u_{n}}}\Rightarrow 1< u_{n}< 2.$ What does that help for my problem here ? I need your thought on it ? Thank you.

Body:
Hint: Let $f(x) = \sqrt {1 + \frac{1}{{1 + x}}}$ for $x>0$. Then, for all $x>0$, $y>0$,
\begin{align*}
&\left| {f(x) - f(y)} \right| = \left| {\sqrt {1 + \frac{1}{{1 + x}}}  - \sqrt {1 + \frac{1}{{1 + y}}} } \right| \\ &= \frac{{\left| {x - y} \right|}}{{(1 + x)(1 + y)\left( {\sqrt {1 + \frac{1}{{1 + x}}}  + \sqrt {1 + \frac{1}{{1 + y}}} } \right)}} \le \frac{1}{2}\left| {x - y} \right|,
\end{align*}
i.e., $f$ is a contraction.


问:
Equation of tangents from external point to a circle
I have point $(p,q)$ and circle $ x^2 + y^2 + 2gx + 2fy + c = 0$.

I'm aware you could do substitute in $y = mx+c$ and solve a quadratic or you could use $ y-(-f) = m(x-(-g)) \sqrt{1+m^2}$ where m is the slope of the line and $(-g, - f)$ is the coordinates of the centre. But are there other ways? perhaps using calculus/vectors/complex numbers

maybe take point $(4,-5)$ and circle $ x^2 + y^2 -6x + -4y + 4 = 0$. as example

Body:
By Joachimsthal $$s_1^2=s \cdot s_{11}$$ defines the tangents from $(p,q)$ as a line pair with  $$s=x^2 + y^2 + 2gx + 2fy + c = 0,$$ $$s_{11}=p^2 + q^2 + 2gp + 2fq + c,$$ $$s_1=xp + yq + g(x+p) + f(y+q) + c.$$

I.e.

$(xp + yq + g(x+p) + f(y+q) + c)^2-(x^2 + y^2 + 2gx + 2fy + c)(p^2 + q^2 + 2gp + 2fq + c)=0$

or

${\small (-g^2+2fq+q^2+c)(x-p)^2+(-2fg-2fp-2gq-2pq)(x-p)(y-q)+(-f^2+2gp+p^2+c)(y-q)^2=0}$

which factorises as:

$${\Tiny \frac{((-g^2+2fq+q^2+c)(x-p) -(y-q) (pq+gq+fp+fg-\sqrt{D}))
((-g^2+2fq+q^2+c)(x-p)-(y-q)(pq+gq+fp+fg+\sqrt{D}))}{-g^2+2fq+q^2+c}},$$ where $$D=(g^2+f^2-c)(p^2+q^2+2gp+2fq+c)=(g^2+f^2-c)s_{11},$$

or

$${\Tiny \frac{((p^2+2gp-f^2+c)(y-q) -(x-p) (pq+gq+fp+fg-\sqrt{D}))
((p^2+2gp-f^2+c)(y-q)-(x-p)(pq+gq+fp+fg+\sqrt{D}))}{p^2+2gp-f^2+c}},$$

making the actual lines

$$(y-q)=(x-p)\frac{pq+gq+fp+fg\pm \sqrt{D}}{p^2+2gp-f^2+c}.$$


问:
In this Probability Density Function, am I doing it right?
Consider a model consisting of random variables $X,Y$ and $Z: Y=Xw+Z$, where $Z∼U[−1,1]$.

答:
Your question is right, other than talking about their independence ( which they are not  having the equality $Y=\omega X + Z$ means they cannot be ! ).

For writing it down you can do it this way (although your intuition is fine, it may not be for more complicated problems) :

Let $F_{Y|X=x}$ be the conditional CDF, and $f_{Y|X=x}$ the conditional PDF.

Recall $\frac{d}{da}F_{Y|X=x}(a)=f_{Y|X=x}(a)$ (fundamental theorem of calculus + definitions of CDF and PDF). Now calculate

For $a \in [\omega x-\frac{1}{2},\omega x+\frac{1}{2}]$

\begin{align}
F_{Y|X=x}(a) =& \mathbb{P}(Z \leq a-\omega x)
\\
=& a-\omega x - \frac{1}{2}
\end{align}

hence $\frac{d}{da}F_{Y|X=x}(a)=1=f_{Y|X=x}(a)$ for $a \in [\omega x-\frac{1}{2},\omega x+\frac{1}{2}]$.


问:
Number of homomorphisms between $\mathbb{Z}/n\mathbb{Z}$ and $\mathbb{C}^{\times}$

Problem: Let $\mathbb{C}^{\times}$ denote the nonzero complex numbers under multiplication. Show
that there are $5$ homomorphisms from $\mathbb{Z}/5\mathbb{Z}\to \mathbb{C}^{\times}$

I tried to do the following:

Let's say $\phi :\mathbb{Z}/5\mathbb{Z} \to \mathbb{C}^{\times} $. We know that $\mid \text{ker}(\phi)\mid$ divides $\mathbb{Z}/5\mathbb{Z}$. So the kernel is either full $\mathbb{Z}/5\mathbb{Z}$ or $1_G$ only. I am not sure what to do next and how to proceed. Can someone help me out?
Any hint is highly appreciated.

Thank you!

答:
Hint: Let's write $G=\mathbb{Z}/5\mathbb{Z}$ and $g=1+5\mathbb{Z}$. Note the following:

If $\varphi$ is such a homomorphism then we have $\varphi(g)^5=1$, and so $\varphi(g)$ is a fifth root of unity.

If two such homomorphisms $\varphi,\psi$ satisfy $\varphi(g)=\psi(g)$ then $\varphi=\psi$.

The first part should help you find $5$ homomorphisms. The second part should help you prove there aren't any other homomorphisms.


答:
$\newcommand{\Z}{\mathbb Z}$
$\newcommand{\C}{\mathbb C}$

I will denote by $\overline a$ the elements of $(\Z/n\Z, +)$.

You know that $\overline 1$ generates this group.
Hence, knowing the image of $\overline 1$ will leave no choice for the images of the other elements of the group !

Moreover, you know that $\overline 1$ is of order $5$, i.e. $5 \cdot \overline 1 = \overline 0$, so it is necessary that $\phi(1)^5 = 1$ in $\C$. There are exactly $5$ such numbers, namely $e^{\frac{2i\pi} 5}$ for $0 \leq i < 5$.

What you need to prove (a bit more precisely than in my semi-informal reasoning) is that this procedure correctly defines a morphism, that there are no other morphisms $\Z/n\Z \to \C^\ast$, and that they are all different.

Title: Number of homomorphisms between $\mathbb{Z}/n\mathbb{Z}$ and $\mathbb{C}^{\times}$

Body:
Problem: Let $\mathbb{C}^{\times}$ denote the nonzero complex numbers under multiplication. Show
that there are $5$ homomorphisms from $\mathbb{Z}/5\mathbb{Z}\to \mathbb{C}^{\times}$

I tried to do the following:

Let's say $\phi :\mathbb{Z}/5\mathbb{Z} \to \mathbb{C}^{\times} $. We know that $\mid \text{ker}(\phi)\mid$ divides $\mathbb{Z}/5\mathbb{Z}$. So the kernel is either full $\mathbb{Z}/5\mathbb{Z}$ or $1_G$ only. I am not sure what to do next and how to proceed. Can someone help me out?
Any hint is highly appreciated.

Thank you!


问:
Why $f$ must be monotone-decreasing function on integral test?
From proof of integral test, used this :
$$f(n)\leq f(x) \leq f(n+1)$$
and integrate all of them and sum it.when $$n<x<n+1$$

But I think monotone-decrease doesn't need if function $f$ satisfies :
$$0<f(n)\leq f(x) \leq f(n+1) \quad \forall n \in\mathbb{N}$$

So, is there any counterexample for function $f$ which satisfies
$$0<f(n)\leq f(x) \leq f(n+1) \quad \forall n \in\mathbb{N}$$
but
$$\sum f(n) <\infty \nleftrightarrow \int f(x)dx<\infty$$
?

答:
I think that you wanted to write $$f\left(n\right)\color{red}{\geq}f\left(x\right) \color{red}{\geq} f\left(n+1\right).$$

You are right, this inequality is enough for proving integral test.


答:
Writing $g(t) = g(t_0) + (t-t_0)g'(t_0) + g''(t_0)(t-t_0)^2/2 + o(|t-t_0|^2)$
with $t = a^T(x+h)$ and $t_0=a^Tx$ gives
$$f(x+h) = f(x) + h^T [a g'(t_0)] + h^T[a a^T g''(t_0)] h /2 + o(\|h\|^2).$$
The gradient $\nabla f(x)$ and Hessian $\nabla^2 f(x)$ are uniquely defined by
$$f(x+h) = f(x) + h^T \nabla f(x) + h^T[\nabla^2 f(x) ] h /2 + o(\|h\|)$$
so it provides the required answer.

Title: If $f(x)=g(a^Tx)$, where $g:\mathbb{R}\to\mathbb{R}$ and $a, x\in\mathbb{R}^n$, why is the Hessian equal to $g''(a^Tx)aa^T$?

Body:
I am trying to understand the solution to below problem:

Let $f(x)=g(a^Tx)$, where $g:\mathbb{R}\mapsto\mathbb{R}$ is continuously differentiable and $a\in\mathbb{R}^n$ is a vector. What are $\nabla f(x)$ and $\nabla^2f(x)$?

The answer is

$\nabla f(x)=g'(a^Tx)a$

$\nabla^2 f(x)=g''(a^Tx)aa^T$

I understand the answer for the gradient, but why does the answer for the hessian contain $a^T$ at the end?


问:
infinite sum of harmonic number powers $\sum_{n=0}^{\infty}x^{H_n}$
I am looking for a simplified form of the infinite sum of harmonic exponentials

$$
f(x) = \sum_{n=0}^{\infty}x^{H_n}
= 1 + x + x^{3/2} + x^{11/6} + x^{25/12} + x^{137/60} + x^{49/20} + \ldots\\
=1+x\left(
1+x^{1/2}\left(
1+x^{1/3}\left(
1+x^{1/4}\left(
1+x^{1/5}\left(
1+x^{1/6}\left(
1+\ldots
\right)\right)\right)\right)\right)\right)
$$
for $x\in[0,1/e)$.

Can it be reduced to a standard special function?

Update: a stable way of calculating $f(x)$ numerically

Inspired by Claude Leibovici's answer, we write
$$
f(x) = 1 + \sum_{n=1}^{\infty}x^{\gamma+\log(n)} - \underbrace{\sum_{n=1}^{\infty}\left[x^{\gamma+\log(n)}-x^{H_n}\right]}_{g(x)}\\
= 1+x^{\gamma}\zeta[-\log(x)]-g(x)
$$
with $g(x)$ converging rapidly on the entire interval $x\in[0,1/e]$ (and beyond) without singularities:

Specifically,
$$
g(1/e) = \sum_{n=1}^{\infty}\left[\frac{e^{-\gamma}}{n}-e^{-H_n}\right]
\approx 0.35370447459102146589\ldots\\
g'(1/e) \approx 0.38530115152397563963\ldots\\
g''(1/e) \approx -0.82751608018475931083\ldots
$$
give the approximate series expansion of $f(x)$ around $x=1/e$,
$$
f(x) \approx
\frac{e^{-\gamma-1}}{1/e-x}
+ 0.36556578362553594918\ldots
+ (1/e-x)\cdot0.11500038261356357678\ldots
+ (1/e-x)^2\cdot0.11786065053337882776\ldots
+O[(1/e-x)^3]
$$

Unfortunately, neither WolframAlpha nor AskConstants recognize any of these numbers.

Body:
As @Raffaele commented, if you use the approximation
$$H_n\sim \log( n)+\gamma$$

Using what you wrote (notice that $H_7=2.59$ while $\log(7)+\gamma=2.52$)
$$f(x)\sim1 + x + x^{3/2} + x^{11/6} + x^{25/12} + x^{137/60} + x^{49/20} +
x^\gamma\sum_{n=7}^{\infty}n^{\log(x)}$$
$$\sum_{n=7}^{\infty}n^{\log(x)}=\zeta (-\log (x))-\sum_{k=1}^6 k^{\log(x)}$$ which converges for $x\lt \frac 1e$.

Let us compare for a few values of $x$
$$\left(
\begin{array}{ccc}
 x & \text{approximation} & \text{exact} \\
 0.00 & 1.00000 & 1.00000 \\
 0.02 & 1.02424 & 1.02421 \\
 0.04 & 1.05408 & 1.05391 \\
 0.06 & 1.08957 & 1.08911 \\
 0.08 & 1.13148 & 1.13055 \\
 0.10 & 1.18100 & 1.17937 \\
 0.12 & 1.23976 & 1.23718 \\
 0.14 & 1.30998 & 1.30621 \\
 0.16 & 1.39482 & 1.38955 \\
 0.18 & 1.49874 & 1.49168 \\
 0.20 & 1.62838 & 1.61920 \\
 0.22 & 1.79396 & 1.78233 \\
 0.24 & 2.01210 & 1.99766 \\
 0.26 & 2.31175 & 2.29413 \\
 0.28 & 2.74822 & 2.72704 \\
 0.30 & 3.44199 & 3.41684 \\
 0.32 & 4.71483 & 4.68530 \\
 0.34 & 7.81187 & 7.77753 \\
 0.36 & 26.6198 & 26.5802
\end{array}
\right)$$ which seems to be quite decent.

So, you can write
$$\sum_{n=0}^{\infty}x^{H_n}=\sum_{n=0}^{m}x^{H_n}+x^\gamma\Big[\zeta (-\log (x))-\sum_{k=1}^{m+1} k^{\log(x)}\Big]\tag 1$$

As @Roman commented, if you take a large value of $m$, the asymptotics is
$$f(x)\sim  \sum_{n=0}^{m}x^{H_n}-\frac{m\, x^{\gamma+\log (m)}}{1+\log(x)}$$

Edit

Equation $(1)$ can be simplified ; in terms of the  Hurwitz zeta function, it is
$$f(x)=\sum_{n=0}^{\infty}x^{H_n}=\sum_{n=0}^{m}x^{H_n}+x^\gamma \,\zeta (-\log (x),m+1)$$ and expanded as series
$$f(x)=\sum_{n=0}^{m}x^{H_n}-x^{\gamma+\log(m)} g(t)\qquad \text{where} \qquad t=\log(x)$$
$$g(t)=\frac{m}{t+1}+\frac{1}{2}+\frac{t}{12 m}-\frac{(t-2) (t-1) t}{720 m^3}+\frac{(t-4)
   (t-3) (t-2) (t-1) t}{30240 m^5}-\frac{(t-6) (t-5) (t-4) (t-3) (t-2) (t-1)
   t}{1209600 m^7}+O\left(\frac{1}{m^9}\right)$$


问:
Is expanding via Taylor series true only for values less than 1 ??
How do I prove this result??

prove $1+x/2-x^3/8<(1+x)^{1/2}$    , for $x>0$

And while we're at it, I had a deeper doubt, Is expanding via Taylor series true only for ,
$x<1$ ?

Because, we easily write the error term as the smallest power of x that is not in the expansion i.e if we expand upto $n^\text{th}$ power, we take error term as  $(x-a)^{n+1}*f^{(n+1)thderivative}(a)$$/(n+1)!$   , which would only be true if $x<1$ otherwise the most significant term should be the last term in the series which would tend to infinity for $x>1$.

This doubt arises because I have used Maclaurin series several times and there we always take $x$ tending to zero

thanks!

答:
Yes, you are right in this case. The value of a Taylor series (or power series in general) may not be defined at some region since the value will go wild. A concept called Radius of convergence is there to determine how far can a power series converge.

See for example, the solution of this to find out how to calculate the radius of convergence.


答:
For the first result, the formula in the question appears to be in error.
I think you are supposed to use the Taylor expansion of $(1+x)^{1/2},$
which is
$$ (1+x)^{1/2} = 1 + \frac12 x - \frac18 x^2 + - \cdots $$
(continuing with alternating $+$ and $-$ signs). Note that it's $\frac18 x^2$ rather than $\frac18 x^3,$ suggesting that there's a typo in the question.
(Did you check carefully after copying the formulas?)

For $0 < x \leq 1$ this is a convergent alternating series and you can apply the Alternating Series Estimation Theorem.
(For a theorem statement and example, see Alternating Series Estimation Theorem and this series.)

For $x < 0$ the result you're supposed to prove is false.

For $x > 1$ the result is true but the Taylor series around $0$ does not converge.
There are other methods you could apply, such as squaring the quantities on both sides when $1 + \frac12 x - \frac18 x^2$ is positive.

It depends on exactly what you were supposed to prove--that is, for which $x$ you were supposed to prove the inequality.

As for Taylor series for $x > 1,$ consider a polynomial such as
$p(x) = 1 + x + 3x^2 - 4x^3.$
The formula for this polynomial is literally its own Taylor series,
with zero coefficients for the $x^4$ term and all higher terms
(because the $4$th derivative of this polynomial is zero).
The Taylor series is good for any real number $x.$

Now consider the Taylor series around $0$ for
$$f(x) = \frac1{1+x} = 1 - x + x^2 - x^3 +- \cdots .$$
This converges for $\lvert x\rvert < 1$ and
diverges for $\lvert x\rvert > 1$.

But now look at
$$g(x) = \frac1{1+2x} = 1 - 2x + 4x^2 - 8x^3 +- \cdots .$$
This diverges for $\lvert x\rvert > \frac12$.

On the other hand,
$$g(x) = \frac1{1+\frac12x} = 1 - \frac12x + \frac14x^2 - \frac18x^3 +- \cdots $$
converges for $\lvert x\rvert < 2$.

So in the case of a Taylor series around $0,$
the limit on $\lvert x\rvert$ does not have to be $1.$
It could be any number.

For any particular Taylor series centered at $a$, if there is a number $r$ for which the series converges whenever $\lvert x - a\rvert < r$
and diverges whenever $\lvert x - a\rvert > r$,
we say that $r$ is the radius of convergence of the series.
If the series converges for all real $x$, we say the radius of convergence is infinite.

Title: Is expanding via Taylor series true only for values less than 1 ??

Body:
How do I prove this result??

prove $1+x/2-x^3/8<(1+x)^{1/2}$    , for $x>0$

And while we're at it, I had a deeper doubt, Is expanding via Taylor series true only for ,
$x<1$ ?

Because, we easily write the error term as the smallest power of x that is not in the expansion i.e if we expand upto $n^\text{th}$ power, we take error term as  $(x-a)^{n+1}*f^{(n+1)thderivative}(a)$$/(n+1)!$   , which would only be true if $x<1$ otherwise the most significant term should be the last term in the series which would tend to infinity for $x>1$.

This doubt arises because I have used Maclaurin series several times and there we always take $x$ tending to zero

thanks!


问:
Fourier representation for $\tan(x)$
Q: Which Fourier representation is suitable for $f(x) = \tan(x)$: Fourier trigonometric series, Fourier half-range expansion, or Fourier integral and why?

Well I searched and found that:
$\tan(x)$ cannot be expanded as a Fourier series since $\tan(x)$ not satisfies Dirichlet’s.

Isn't there any Fourier representation for $\tan(x)$?

Thanks ^_^

答:
$\tan(x)$ is odd with period $\pi$, so the half-range fourier coefficients would be given by
$\frac{4}{\pi}\int_0^{\frac{\pi}{2}} \tan(x)\sin(\frac{nx}{2}) \mathrm{d}x$. But the integral diverges. However, if $f(x)=\tan(x)$ for $a<x<b$, where $a>-\frac{\pi}{2}$ and $b<\frac{\pi}{2}$, the periodic etension of $f(x)$ has a fourier transform.</p>

Title: Fourier representation for $\tan(x)$


问:
Nonlinear Binary Programming
I am trying to understand what is the best approach to solve this binary programming problem

$$ \max_{X\in \left\{0,1 \right\}^{N}} \left(\sum_{i=1}^{N} a_{i}X_{i}\right)^{\beta} - \sum_{i=1}^{N}c_{i}X_{i} $$

with $X\in \left\{0,1 \right\}^{N}, a_{i}>0, c_{i} >0 $ and $\beta>0$. Would linearization techniques be useful in this case? I think relaxation will not work given that the function is not concave.  Can it be formulated as a more specific well-known problem? Are there any optimized algorithms known for it?

Thanks so much in advance

答:
The title says polynomial, so I will assume that $\beta$ is an integer.  Yes, you can linearize by expanding the power and introducing a new binary variable for each resulting product of $X_i$s.  For example, if $\beta=3$, one of the products is $X_1 X_2 X_3$ and you would introduce binary variable $Y_{1,2,3}$, together with constraints
\begin{align}
Y_{1,2,3} &\le X_1 \tag1\\
Y_{1,2,3} &\le X_2 \tag2\\
Y_{1,2,3} &\le X_3 \tag3\\
X_1 + X_2 + X_3 - 2 &\le Y_{1,2,3} \tag4\\
\end{align}
Constraints $(1)$ through $(3)$ enforce $Y_{1,2,3} \le X_1 X_2 X_3$.
Constraint $(4)$ enforces $Y_{1,2,3} \ge X_1 X_2 X_3$.

You can optionally relax $Y$ to be nonnegative because it will automatically take values $\{0,1\}$ when $X$ is binary.

Here's an alternative linearization for the case that $\beta$ is not necessarily integer.  Suppose you know that $\sum_i a_i X_i \in S$ for some finite set $S$ that is easy to generate.  For example, if the $a_i$ are positive integers, you can take $S=\{0,1,\dots,\sum_i a_i\}$.  Now for each $s \in S$, let binary variable $Z_s$ indicate whether $\sum_i a_i X_i = s$.  The problem is to maximize
$$\sum_{s \in S} s^\beta Z_s - \sum_{i=1}^N c_i X_i$$
subject to
\begin{align}
\sum_{s \in S} Z_s &= 1 \\
\sum_{s \in S} s Z_s &= \sum_{i=1}^N a_i X_i \tag1\\
\end{align}

If you cannot easily generate $S$ such that $\sum_i a_i X_i \in S$, you can approximate by letting $\Delta=(\sum_i a_i)/m$, taking $S=\{0,\Delta,2\Delta,\dots,\sum_i a_i\}$,
and relaxing equality constraint $(1)$ to an inequality:
$$-\frac{\Delta}{2} \le \sum_{s \in S} s Z_s - \sum_{i=1}^N a_i X_i \le \frac{\Delta}{2}$$

Title: Nonlinear Binary Programming


问:
Find all positive integers $x$ such that $x(x+2021)$ is a perfect square
I completed the square as to get $(x+1010.5)^2-102110.25 = k^2$ but I don't know where to go from here.
Please help, thank you

I then got $(2x+2021)^2-4084441=4k^2$ then $(2k-2x-2021)(2k+2x+2021)=43^2*47^2$

Body:
$$\begin{align}x^2+2021x-a^2&=0, a\in\mathbb Z^{+}&\end{align}$$

$$\begin{align} &\Delta =2021^2+(2a)^2=T^2, T\in\mathbb Z^{+} \\
&\implies (T-2a)(T+2a)=2021^2\\
&\implies \begin {cases} T+2a =\dfrac {2021^2}{m} \\ T-2a=m \end {cases} \\
&\implies a=\dfrac {2021^2-m^2}{4m}\end{align}$$

Then, note that $2021^2=43^2×47^2$ , $a \in\mathbb {Z^{+}}$ ($a \in\mathbb {Z^{+}}$ implies $T \in\mathbb {Z^{+}}$) and  $m=1,43,47,47×43, 43×47^2, 47×43^2$ .

You can check all cases.