问:
On a step of Marcus' "Number Fields", Theorem 22 Chapter 3
I've been reading Marcus' Number Fields and there is a step in the proof of Theorem 22 that I can't understand. Specifically, it's about the proof of 22(b).

Theorem 22. Let $K$ and $L$ be number fields, $K\subset L$, let $R=\mathbb{A}\cap K$ (the set of algebraic integers in $K$) and $S=\mathbb{A}\cap L$, and $n=[L:K]$.

(a) For ideals $I$ and $J$ in $R$, $\lVert IJ\rVert=\lVert I\rVert\,\lVert J\rVert$.

(b) Let $I$ be an ideal in $R$. For the $S$-ideal $IS$, $\lVert IS\rVert = \lVert I\rVert^n$.

(c) Let $\alpha\in R$, $\alpha\neq 0$. For the principal ideal $(\alpha)$, $\lVert(\alpha)\rVert = |N_{\mathbb{Q}}^K(\alpha)|$.

Inside the proof of 22(b), Marcus states and proves (well, sort of) a Lemma.

Lemma: Let $A$ and $B$ be nonzero ideals in a Dedekind domain $R$, with $B\subset A$ and $A\neq B$. Then there exists $\gamma\in K$ such that $\gamma B\subset R$, $\gamma B\not\subset A$.

Then, he applies the Lemma with $A=P$ and $B=(b_1,\ldots,b_{n+1})$. Of course, we want a contradiction and so we assumed that $B$ is a subset of $A$. This is why we can apply the Lemma. But I am having trouble filling in the details. I can't get a contradiction.
Thanks a lot!

答:
I have a copy of Marcus with me, so I can fill in some of the details. However, unless you wanted the audience of your post to be limited to those who (i) have access to a copy of the book; and (ii) take the trouble to pull it off the shelf, look up the theorem, and backtrack enough to see what all the symbols mean; then you should try, in the future, to provide enough context in your post so that even those without access may have a shot at understanding the problem and your query about it, and potentially offer help.

Marcus will prove (b) in the special case of $I=P$ a prime ideal, relying on (a) to deduce the general case. Now, $S/PS$ is a vector space over $R/P$, and we want to show it has dimension exactly $n$. First, we show it has dimension at most $n$, by showing that any collection of $n+1$ elements is necessarily linearly dependent. To that end, let $\alpha_1,\ldots,\alpha_{n+1} \in S$, and we want to show that their images in $S/PS$ are linearly dependent over $R/P$. We know the original elements are linearly dependent in $L$ over $K$, since $[L:K]=n$. And we know that we can multiply the linear dependence equation by some integer so that all coefficients lie in $R$, rather than in $K$ (this was proven in an exercise in the previous chapter: if $\alpha\in K$, then there exists $m\in\mathbb{Z}$ such that $m\alpha\in R$). This gives us an equation of the form
$$\beta_1\alpha_1+\cdots+\beta_{n+1}\alpha_{n+1} = 0,$$
where $\beta_i\in R$. We want to reduce this modulo $P$, but to prove that we don't have a trivial linear combination after reduction, we need to ensure that not all $\beta_j$ lie in $P$. This is where the Lemma comes in.

If at least one $\beta_j\notin P$, we are done. Assume, however, that we are unlucky enough to have all $\beta_i\in P$. This could happen: for example, maybe all of our original $\alpha_i$ are in $P$, and so we pick  all $\beta_j$ in $P$. So we aren't looking for a contradiction. Instead, we want to show we can tweak the linear dependence relation so that the resulting one does  not have all coefficients in $P$.

We apply the Lemma with

$A=P$, $B=(\beta_1,\ldots,\beta_{n+1})$, ideals in the Dedekind domain $R$, we have $B\subseteq A$, $A\neq R$ (since it is prime). Let $\gamma\in K$ be guaranteed by the Lemma, so that $\gamma B\subset R$ and $\gamma B\not\subset P$. Now take the original linear dependent equation and multiply through by $\gamma$:
$$0 = \gamma(\beta_1\alpha_1+\cdots+\beta_{n+1}\alpha_{n+1}) = (\gamma\beta_1)\alpha_1+\cdots+(\gamma\beta_{n+1})\alpha_{n+1}.$$
Now notice that since $\gamma B$ is generated by $\gamma\beta_1,\ldots,\gamma\beta_{n+1}$, it cannot be the case that all the new coefficients lie in $P$ (since $\gamma B\not\subset P$). Thus, we are now in the situation where not all coefficients lie in $P$, and so reducing modulo $P$ we obtain a nontrivial linear dependence relation between $\overline{\alpha_1},\ldots,\overline{\alpha_{n+1}}$, which is what we wanted to show. This proves that $S/PS$ has dimension at most $n$, as claimed.


问:
Solving limit which can't be (maybe I'm wrong) solved by L'Hospital
I'm unable to solve the following limit:
$$\lim_{x \to 0+}\frac{\sin( x \sqrt x )-x \sqrt
x}{x\sqrt{x+1}}$$

My attempt:
Using L'Hospital rule:
$$\frac{ \cos( x \sqrt x ) \frac{3}{2} \sqrt x - \frac{3}{2} \sqrt x }  { \sqrt{x+1} + x \frac{1}{2\sqrt{x+1}} }$$
$$\frac{\sqrt{x+1} (\cos( x \sqrt x ) \frac{3}{2} \sqrt x - \frac{3}{2} \sqrt x)}{2x+1}$$
Since $(\sqrt x)'=\frac{1}{2\sqrt x}$  we can see that we will get $\frac{0}{0}$ when $x \to 0$ if my calculations are right.

答:
Here is how it can be solved without L'hopital or series expansion :

Observe that for any $ x\in\mathbb{R}^{*} $, we have : $$ \frac{x-\sin{x}}{x^{2}}=\int_{0}^{1}{\left(1-y\right)\sin{\left(xy\right)}\,\mathrm{d}y} $$

Thus, if $ x $ is a real different from $ 0 $, we have : \begin{aligned}\left|\frac{x-\sin{x}}{x^{2}}\right|&=\left|\int_{0}^{1}{\left(1-y\right)\sin{\left(xy\right)}\,\mathrm{d}y}\right|\\ &\leq\int_{0}^{1}{\left(1-y\right)\left|\sin{\left(xy\right)}\right|\mathrm{d}y}\\ &\leq\int_{0}^{1}{\left(1-y\right)\left|x\right|y\,\mathrm{d}y}=\frac{\left|x\right|}{6}\end{aligned}

That means : $$ \lim_{x\to 0}{\frac{x-\sin{x}}{x^{2}}}=0 $$

Thus : \begin{aligned}\lim_{x\to 0^{+}}{\frac{\sin{\left(x\sqrt{x}\right)}-x\sqrt{x}}{x\sqrt{x+1}}}&=\lim_{x\to 0^{+}}{\left(-\frac{x^{2}}{\sqrt{x+1}}\times\frac{x\sqrt{x}-\sin{\left(x\sqrt{x}\right)}}{\left(x\sqrt{x}\right)^{2}}\right)}\\ &=-0\times 0\\ &=0\end{aligned}

ContentLicense: CC BY-SA 4.0


问:
Does an integrable function vanish at infinity in a suitable sense?
I'm searching for a generalization of the following observation: If $f\in L^2(\mathbb R)$, then \begin{equation}\begin{split}\int_{[-n,\:n+1]}f(x)\:{\rm d}x&=\left\langle1_{[-n,\:n+1]},1_{\mathbb R\setminus[-n,\:n)}g\right\rangle_{L^2(\mathbb R)}\\&\le\left\|f-1_{[-n,\:n)}f\right\|_{L^2(\mathbb R)}\xrightarrow{n\to\infty}0\end{split}\tag1\end{equation} by the Cauchy-Schwarz inequality and Lebesgue's dominated convergence theorem.

$(1)$ captures the idea that an integrable function should "vanish at infinity" in a suitable sense.

Can we generalize this result to $f\in L^p(\mu)$, where $(E,\mathcal E,\mu)$ is an arbitrary measure space and $p\ge1$? Maybe assuming that $\mu(E)=\infty$ and/or that $E$ is a metric/normed space with $\mathcal E=\mathcal B(E)$?

In the case of a metric/normed space, the intuition tells me that an integrable function should "essentially be supported in a compact set".

What's obviously true is that $$\forall f\in L^p(\mu):\not\exists B\in\mathcal E:\mu(B)=\infty\text{ and }\inf_{x\in B}|f(x)|>0\tag2.$$

答:
I suppose there might be lots of different generalizations, and the most appropriate will depend on the finer points one wishes to preserve.  Here is a very simple generalization:

If $F_n$ is an increasing family of measurable sets covering $E$, and if $f$ is in $L^1(E,\mu)$ then
$$
\lim_{n\to\infty} \int_{F_{n+1}\setminus F_n} f(x)\,d\mu(x) =0.
$$


问:
Prove or disprove that $x_1 = x_2= \cdots = x_n$, if $x_1^{x_2} = x_2^{x_3} = \cdots = x_{n-1}^{x_n} = x_n^{x_1}$, $x_i > 1, \forall 1 \le i \le n$.
The case $n=3$ is from here. It's straightforward to prove it's true:

First we notice that if any two of $x_1, x_2, x_3$ are equal then all must be equal.

Denote $a=x_1^{x_2}=x_2^{x_3}=x_3^{x_1}$ then
$$x_1 = a^{1/x_2}, x_2=a^{1/x_3}, x_3=a^{1/x_1}$$
WLOG assume $x_1 \ge x_2, x_3$. There are two cases:

$x_1 \ge x_2 \ge x_3 \implies \frac{1}{x_2} \ge \frac{1}{x_3} \ge \frac{1}{x_1} \implies x_1 \ge x_3 \ge x_2 \implies x_2=x_3 \implies x_1=x_2=x_3$.

$x_1 \ge x_3 \ge x_2 \implies \frac{1}{x_2} \ge \frac{1}{x_1} \ge \frac{1}{x_3} \implies x_3 \ge x_1 \ge x_2 \implies x_1=x_3 \implies x_1=x_2=x_3$.

If $n$ is even, then $x_{2i-1} = 2, x_{2i}=4$ is a counterexample.

When $n=5$ the above method will need to examine $4!=24$ cases. Basically we map $x_i$ to $x_{(i \mod 5) +1}$ and reverse the order. In many of cases we can deduce $4$ or all $5$ of the $x_i$'s are equal. For example, if $$x_1 \ge x_3 \ge x_2 \ge x_5 \ge x_4 \tag 1$$ then
$$
x_5 \ge x_1 \ge x_3 \ge x_4 \ge x_2 \tag 2
$$
Since the order of $x_2$ and $x_5$ reversed from $(1)$ to $(2)$, they must be equal and so are everything in between them from both $(1)$ and $(2)$ hence all $x_i$'s are equal.

There are other cases that are different.

Example 2: $$x_1 \ge x_3 \ge x_4 \ge x_2 \ge x_5 \tag 3$$ then
$$x_1 \ge x_3 \ge x_5 \ge x_4 \ge x_2 \implies x_1 \ge x_3 \ge x_2=x_4=x_5 \tag 4$$

Example 3: $$x_1 \ge x_3 \ge x_4 \ge x_5 \ge x_2 \tag 5$$ then
$$x_3 \ge x_1 \ge x_5 \ge x_4 \ge x_2 \implies x_1=x_3 \ge x_2=x_4=x_5 \tag 6$$

But they all lead to the conclusion that $x_1=x_2=x_3=x_4=x_5$.

Now my questions:

Question #1: Is it always true if $n>1$ and $n$ is odd?

Question #2: What if we allow $x_i>0$ instead of $x_i>1$?

答:
It can be proved to be false for n = 2.

Let x1^x2 = x2^x1. Then we raise both sides with power 1/(x1*x2).

This means x1^((x2/x2)*x1) = x2^((x1/x1)*x2). This means x1^(1/x1) = x2^(1/x2).

This means we have to check if the function f(x) = x^(1/x) is one-one or not. It turns out it is not.

A function whose derivative changes sign is not one-one.

f(x) = x^(1/x) has a maxima at x=e.

So this proves the existence of distinct values of x1,x2 for n=2.


问:
Which is true $A$ is subset of $B$ or $B$ is subset of $A$.

Consider the sets dened by the real solutions of the inequalities $$A=\{(x,y):x^2+y^4\le 1\}$$ and $$B=\{(x,y):x^4+y^6\le 1\}$$Then which is true $A$ is subset of $B$ or $B$ is subset of $A$.

This is Question No. $10.$ of this paper. Please help me.Thanks in advance.

答: Hint: Try the following points: $$(0.5,0.5),~~(0.8,0.8),~~\{(0.7,0.9),(0.7,0.8)\}$$ I know this is an strange way based on trial and error method. I did it just because your question is a multiple choices.</p>

ContentLicense: CC BY-SA 3.0


问:
Counterexample for approximation by convolution
I'm currently reading these lecture notes. In the remark to theorem 3.7, it is explicitly stated that the convergence doesn't hold for $p= \infty$. I'm not really sure about that, as I am not able to find a counterexample.

Is there an easy counterexample?

答:
Let $f=1$ on $[0, 1]$ and $f=0$ elsewhere, let $\varphi=1$ on $(-\frac 12, \frac 12)$ and $\varphi=0$ elsewhere. Thus $\varphi$ has integral $1$. We see for all $\epsilon<\frac 12$, $\varphi_\epsilon* f$ is continuous: for example near $0$ we have $\varphi_\epsilon*f=0$ for $x\leq -\frac{\epsilon}{2}$, and then increases linearly to $1$ at $x=\frac{\epsilon}{2}$. Thus for all $\epsilon<\frac 12$ we have
$$
\|\varphi_\epsilon*f-f\|_\infty=\frac 12.
$$.


问:
Prove that $ o( g) =m$ iff is a multiplication of $ \frac{n}{m}$ disjoint cycles.
The problem:

Given a group $G$ or order $n$, and a Cayley embedding $\phi \ :\ G\to S_{n}$. Prove that some $g\in G$ is of order $m$ iff $\phi ( g)$ is a multiplication of $\frac{n}{m}$  disjoint cycles of length $m$.

I was able to prove that if $\phi ( g) \ $ is s a multiplication of $\displaystyle \frac{n}{m}$ disjoint cycles of order $m$, then the order of $g$ is $m$. This was fairly straightforward given the fact that:

The order of a multiplication of disjoint cycles is the $lcm$ of the length of the cycles.

In our case the $lcm$ is obviously $m$, so it was easy to prove from here that the order of $g$ is $m$.

The other direction, meaning if the order of $g$ is $m$, then $\phi ( g)$ is a multiplication of $\frac{n}{m}$ disjoint cycles of length $m$, I'm struggling to prove it.

I also want to admit that my understanding of Cayley embeddings, and Cayley's theorem in general, is very poor. What I know is simply that it's a homomorphism and also injective, not much more.

Any help?

答:
Let's consider the action by left multiplication of $\langle g\rangle$ on $G$. For $x\in G$, the stabilizer is given by:

\begin{alignat}{1}
\operatorname{Stab}(x) &= \{g^k\in\langle g\rangle\mid g^kx=x\} \\
&= \{g^k\in\langle g\rangle\mid g^k=e\} \\
&= \{e\} \\
\tag 1
\end{alignat}

Therefore, $G$ is partitioned into $r:=\frac{n}{m}$ orbits each of size $m$ (orbit-stabilizer theorem). This means that there are $r$ elements $\tilde g_i\in G$, $i=1,\dots,r$, such that $G$ reads:

$$G=\bigsqcup_{i=1}^rO(\tilde g_i) \tag 2$$

where:

$$O(\tilde g_i):=\{g^k\tilde g_i, k=1,\dots,m\} \tag 3$$

Let's now take the permutation by left multiplication by $g$, $\phi_g\in S_G$. By $(2)$ and $(3)$, $\forall x\in G, \exists i\in \{1,\dots,r\}, k\in \{1,\dots,m\}$ such that $x=g^k\tilde g_i$. Therefore, $\forall l\in \{1,\dots,m\}$:

\begin{alignat}{1}
\phi_g^l(x) &\stackrel{(*)}{=}g^lx \\
&=g^{l+k \pmod m}\tilde g_i \\
\end{alignat}

[$(*)$ induction on $l$], whence, $\forall x\in G$:

$$x\in O(\tilde g_i) \iff \phi_g^l(x)\in O(\tilde g_i), \space\forall l\in\{1,\dots,m\} \tag 4$$

Now, let's define $\alpha_i$ as the extension by the identity map of the restriction of $\phi_g$ to the orbit $O(\tilde g_i)$, namely:

\begin{alignat}{2}
&x \in O(\tilde g_i) &&\Longrightarrow \alpha_i(x) :=\phi_g(x) \\
&x \in O(\tilde g_{j\ne i}) &&\Longrightarrow \alpha_i(x) :=x \\
\tag 5
\end{alignat}

Firstly, $\alpha_i \in S_G$ for every $i=1,\dots,r$, because ${\phi_g}_{|O(\tilde g_i)}$ is a bijection on $O(\tilde g_i)$. Then, by $(4)$ and $(5)$:

\begin{alignat}{2}
& x \in O(\tilde g_i) &&\Longrightarrow \alpha_i^m(x)=\phi_g^m(x)=x\\
& x \in O(\tilde g_{j\ne i}) &&\Longrightarrow \alpha_i^m(x)=x\\
\tag 6
\end{alignat}

and finally, for every $i=1,\dots,r$:

$$\alpha_i^m=Id_G \tag 7$$

so all the $\alpha_i$'s are $m$-cycles. Moreover, again by $(4)$ and $(5)$, for $j\ne i$ we get:

\begin{alignat}{2}
&x \in O(\tilde g_i) &&\Longrightarrow (\alpha_i\alpha_j)(x)=\alpha_i(\alpha_j(x))=\alpha_i(x)=\phi_g(x) \\
&x \in O(\tilde g_j) &&\Longrightarrow (\alpha_i\alpha_j)(x)=\alpha_i(\alpha_j(x))=\alpha_i(\phi_g(x))=\phi_g(x) \\
&x \in O(\tilde g_{l\ne i,j}) &&\Longrightarrow (\alpha_i\alpha_j)(x)=\alpha_i(\alpha_j(x))=\alpha_i(x)=x \\
\tag 8
\end{alignat}

or, equivalently:

\begin{alignat}{2}
&x \in O(\tilde g_i)\sqcup O(\tilde g_j) &&\Longrightarrow (\alpha_i\alpha_j)(x)=\phi_g(x) \\
&x \in O(\tilde g_{l\ne i,j}) &&\Longrightarrow (\alpha_i\alpha_j)(x)=x \\
\tag 9
\end{alignat}

By induction on $(9)$:

\begin{alignat}{2}
&x \in O(\tilde g_1)\sqcup\dots\sqcup O(\tilde g_r)=G &&\Longrightarrow (\alpha_1\dots\alpha_r)(x)=\phi_g(x) \\
\tag {10}
\end{alignat}

namely:

$$\phi_g=\alpha_1\dots\alpha_r \tag {11}$$

Note that $\alpha_i(x)\ne x \iff x \in O(\tilde g_i)$, and hence $\alpha_i$ and $\alpha_j$ have disjoint supports as soon as $i\ne j$. Therefore, $(11)$ (along with $(7)$) reads: $\phi_g$ is the product (composition) of $\frac{n}{m}$ disjoint cycles of length $m$.


问:
Showing the infinite product space of $[0, 1]^\omega$ is not compact with respect to the box topology using the notion of open covers
I wish to show that the space $[0,1]^\omega$ in the box topology is not compact using the notion of open covers. The box topology has been a topic where I have struggled in my studies of topology, and I am having trouble trying to come up with an open cover that has no finite sub-cover for this space (I understand that there are other ways to prove that this space is not compact, but my goal is to use the notion of open covers and finite sub-covers in the proof). Any help would be greatly appreciated

答:
HINT: For $n\in\omega$ let $x_n$ be the point

$$x_n(k)=\begin{cases}
1,&\text{if }k=n\\
0,&\text{otherwise}
\end{cases}$$

in $X=\square[0,1]^\omega$. For $n\in\omega$ let $F_n=\{x_k:k\ge n\}$, and let $U_n=X\setminus F_n$.

Show that each $F_n$ is closed. You may find it helpful to begin by shoing that $F_0$ is a closed, discrete subset of $X$.

Let $\mathscr{U}=\{U_n:n\in\omega\}$; show that $\mathscr{U}$ is an open cover of $X$ with no finite subcover.


问:
Open Zariski subsets of $\mathbb{P}^1$
Are open subsets of $\mathbb{P^1}$ connected? I believe yes because they are complement of finite sets. Is this right?

答: The projective line is irreducible. An open subset of an irreducible space is irreducible, which is a much stronger property than connectedness.</p>

Title: Open Zariski subsets of $\mathbb{P}^1$


问:
Notation for different sizes of infinity?
i realize that there are multiple sizes of infinity so one can be larger than another, but how do you show that one infinity is larger. I'm not looking for proofs or anything but I just want the notation that would be used in a problem or anything. I read something about alphas and omegas to distinguish the sizes of infinity but I can't find any explanations.</p>

答:
Let $X$ and $Y$ be sets; we say that $X$ and $Y$ are equipollent if and only if there is a bijection $f\colon X\to Y$, and denote it by writing $|X|=|Y|$. The relation of equipollence is an equivalence relation, as is easy to prove.

Though the class of all sets is not a set, we can still think of this equivalence relation as partitioning the sets into equivalence classes.

We say that $X$ "has cardinality less than or equal to $Y$" if there is an injection from $X$ into $Y$. We write this as $|X|\leq |Y|$. We say $|X|\lt |Y|$ if and only if $|X|\lt |Y|$ and $|X|\neq|Y|$.

It is equivalent to the Axiom of Choice that given any two sets $X$ and $Y$, either $|X|\lt |Y|$, $|X|=|Y|$, or $|Y|\lt|X|$ (and exactly one holds); that is, the relation is trichotomic.

If we assume the Axiom of Choice, then there is also a special collection of representatives for the equivalence classes under equipollence, called the cardinal numbers. The classes corresponding to finite sets are represented by the natural numbers, $0$, $1$, $2$, etc; we write things like $|X|=7$ to mean that $X$ is equipollent with $7=\{0,1,2,3,4,5,6\}$, i.e., has seven elements.

For infinite sets, the equipollence classes are represented by the infinite cardinals, which are denoted by the aleph numbers (this may be what you dimly recall as "alphas"). The smallest such infinite cardinal is $\aleph_0$ (called "aleph nought" or "aleph zero", or "aleph null"); followed by $\aleph_1$ ("aleph one"), then $\aleph_2$, etc. The aleph numbers are indexed by ordinals, which are another special kind of sets (which correspond to "well orderings" in much the same way that cardinals correspond to "sizes"). The first infinite ordinal is $\omega$ (omega), which corresponds to the well-ordered set of natural numbers. There is an arithmetic of alephs, so that they can be added, multiplied, and other operations (such as exponentiation or infinite products).

And now, to put all of that together: to denote that one infinite set $X$ is (strictly) smaller than another set $Y$, we write $|X|\lt |Y|$. When working with alephs or more generally cardinal numbers,  we dispense with the "absolute value bars" and simply write the expressions involving alephs, e.g., $\aleph_1\leq 2^{\aleph_0}$, or $\kappa\lambda = \kappa+\lambda$ if at least one of the cardinals $\kappa$ and $\lambda$ are infinite, etc.

I'm pretty sure you can find most of this in Halmos's Naive Set Theory, but I'm away from the office right now; I'll check tomorrow, and if not, give another reference.

Added. Halmos uses $X\sim Y$ for the case where $X$ and $Y$ are bijectable, and $X\stackrel{\lt}{\sim} Y$ for the case where $X$ is bijectable with a subset of $Y$ (that is, when there is an injection from $X$ to $Y$. (Section 22, on the Schroeder-Bernstein Theorem). He then uses $\prec$ for strict inequality of cardinals. At first he uses $\mathrm{card}\;a$ for the cardinality of the set $a$, then uses roman letters for cardinal numbers, and switches to the more common $\lt$ in Section 24 (Cardinal Arithmetic); he introduces the Alephs in the penultimate page of his book.


问:
Build graph with exactly n automorphisms
Construct graph with exactly distinct n automorphisms. For n $\geq$ 2.

I wonder if we can just take an asymmetric graph, such as this one as building block.

答:
Frucht's Theorem says that every finite group can be realized as the group of automorphisms of a graph, and the Wikipedia essay gives some idea of the proof, and some links. </p>

<p>Of course, for every $n$ there is a group of order $n$. </p>

Body:
Frucht's Theorem says that every finite group can be realized as the group of automorphisms of a graph, and the Wikipedia essay gives some idea of the proof, and some links.

Of course, for every $n$ there is a group of order $n$.


问:
Intuition of integration
While trying to derive the volume of a spherical cap, I noticed that my method of integrating small discs($\int \pi r^2 \, dr$) was wrong, since it resulted in the final formula $\frac{\pi r^3}{3}$, where r is the radius of the cap face.

Edit: this's what I did; $$\int \pi r^2 \, dr= \pi \frac{(r^3-0)}{3}=\frac{\pi R^3}{3}$$

If this was used to find the full volume of a sphere, it'd come up short, at $\frac{2\pi R^3}{3}$. This seems to be because integration seems to take a linear sum of values. For example, if you calculate the lower hemisphere volume using my incorrect method, the result will essentially only be that of a cone with a vertice at the bottom, since the hemisphere radius increases at a faster rate than the linear rate of the cone radius.

What is the intuitive reason for integration being a linear sum in this way?

答:
In your method, you are finding the area of a parabola $y=\pi r^2$ from $r=0$ to $R$. [I am assuming that you meant the radius $R$ of the sphere instead of $r$ in your answer.] Note, that if this were valid, it would be the volume of the sphere, not the volume of the cap. You are using $r$ as a variable whereas, evidently it is supposed to be the fixed radius of the cap face. To find the volume of the cap face, you should be 'summing' the varying areas of the cross-sections parallel to the face of the cap.

If $r$ is the fixed radius the face of a cap of a sphere of radius $R$ and you want to compute the volume of the cap by integrating the cross-sections parallel to the fixed face, then you need to integrate with respect to some variable $x$ measured along the central axis of the cap. For each $x$ you will have some $y$ corresponding to the radius of the cross section of the cap parallel to the face.
Then your integral will be expressed as

$$ \int \pi y^2\,dx $$

You have enough information to find the limits of integration in terms of $R$ and $r$ and $y$ in terms of $x$ and $R$.

Title: Intuition of integration


问:
A question about Banach spaces and closed graph theory

Let $X, Y$ be Banach spaces on the same field.
Let $T \in B(X,Y)$ a bijection (surjective and injective). Where $B(X,Y)$ denotes the collection of the linear, bounded operators.
Prove that: $\inf_{\|x\|=1} \|Tx\| > 0$.

How to use the closed graph theorem to show this?

答:
Since $T:X\to Y$ is a bijection and $X,Y$ Banach spaces it follows that $T^{-1}:Y\to X$ it is also continuous. Indeed, we can show that $T^{-1}$ is continuous by showing that $T^{-1}$ has closed graph. So, suppose $T^{-1}y_n\to x$ and $y_n\to y$. We wish to show that $T^{-1}y=x$. By the continuity of $T$, $y_n\to Tx$. But, $y_n\to y$ also. Hence, $y=Tx$ and it follows that $T^{-1}y=x$. Hence, $T^{-1}$ is continuous. It follows that
$$||T^{-1}y||\leq ||T^{-1}||\cdot ||y||$$
Plugging $y=Tx$ we obtain
$$||x||\leq ||T^{-1}||\cdot ||Tx||$$
for every $x$. Now, $T^{-1}\neq 0$ implies that $||T^{-1}||>0$ (otherwise $X=Y=\{0\})$ Hence,
$$||Tx||\geq \frac{||x||}{||T^{-1}||}$$
for every $x$. In particular,
$$\inf_{||x||=1}||Tx||\geq \frac{1}{||T^{-1}||}>0$$


答:
The data given does not immediately imply that $T$ is a homeomorphism: the inverse of a continuous bijection may be discontinuous, so what you write is not quite right. In any case, what you write does not really lead you in the right direction (it would only give you the opposite inequality, which is more or less given).

Here's an outline of a correct proof:

$T$ is continuous (because it is bounded).

The graph of $T$ is closed, as you say (because $Y$ is a Hausdorff space).

This implies that the graph of $T^{-1}$ is also closed.

By the closed graph theorem, you can conclude that $T^{-1}$ is continuous.

This implies that $T^{-1}$ is bounded, which can be used to obtain what you need.

Title: A question about Banach spaces and closed graph theory

Body:
Let $X, Y$ be Banach spaces on the same field.
Let $T \in B(X,Y)$ a bijection (surjective and injective). Where $B(X,Y)$ denotes the collection of the linear, bounded operators.
Prove that: $\inf_{\|x\|=1} \|Tx\| > 0$.

How to use the closed graph theorem to show this?


问:
Prove that $7\mid n^7+7n^3-n$ for every positive integer n.
Let $P(n)=7\mid n^7+7n^3-n$ for integers $n\ge 1$. Then $P(1)$ is true because $1+7-1=7$ is multiple of $7$.

Assume that $P(n)$ is true and $n\in \mathbb{Z}$ and $n\ge 1$. Then $P(n+1)$ is true because;

$n^7+7n^3-n=7m$ for some natural number $m$

So, $$(n+1)^7+7(n+1)^3-(n+1)=...$$

I'm stuck here, and I can't continue the proof.

Can you help?

答: HINT: Instead of induction here, we can use Fermat's Little Theorem as it says $n^7 \equiv n \mod 7$ for every positive integer $n$.


答:
Note that $7  \vert 7n^3$ so the problem reduces to proving $n^7 \equiv n \pmod 7$.  This is obviously true for $n=1$.

Assume $k^7 \equiv k \pmod 7$.  Then $(k+1)^7 = \sum_{i=0}^7 \binom  7i k^i \equiv k^7+1 \pmod 7$ because $7 \vert \binom 7i$ for $ 1 \leq i \leq 6$.  By our inductive hypothesis, $k^7 \equiv k \pmod 7$, so $(k+1)^7 \equiv k^7+1 \equiv k+1 \pmod 7$ and we are done.

Title: Prove that $7\mid n^7+7n^3-n$ for every positive integer n.

Body:
Let $P(n)=7\mid n^7+7n^3-n$ for integers $n\ge 1$. Then $P(1)$ is true because $1+7-1=7$ is multiple of $7$.

Assume that $P(n)$ is true and $n\in \mathbb{Z}$ and $n\ge 1$. Then $P(n+1)$ is true because;

$n^7+7n^3-n=7m$ for some natural number $m$

So, $$(n+1)^7+7(n+1)^3-(n+1)=...$$

I'm stuck here, and I can't continue the proof.

Can you help?


问:
Integrate $\frac{1}{x\sqrt{a - bx^2}}$
I've been toiling for hours trying to answer this one:

Find $\int \frac{1}{x\sqrt{a - bx^2}}\,dx$

It's from Silvanus Thompson's 'Calculus Made Easy' (apparently not easy enough for me!).

The answer given is $\frac {1}{\sqrt{a}}\log\left(\frac{\sqrt{a}-\sqrt{a-bx^2}}{x\sqrt{a}}\right)$

I've attempted the question by letting $x=\sqrt{\frac ab}\sin(u)$, which I think is on the right track as I end up with the $\frac1{\sqrt a}$ term in my answer, but nothing else falls into place in terms of the given answer.

Thank you in advance for any help!

Andrew

答:
Substitute $x=\sqrt{\frac ab }\frac1t$ to get

$$\int \frac{dx}{x\sqrt{a - bx^2}}=-\frac1{\sqrt a}\int \frac{dt}{\sqrt{t^2-1}}= - \frac1{\sqrt a}\cosh^{-1}t=- \frac1{\sqrt a}\ln\left(t+\sqrt{t^2-1}\right)
$$

which is equal to $\frac {1}{\sqrt{a}}\ln\left(\frac{\sqrt{a}-\sqrt{a-bx^2}}{x\sqrt{a}}\right)$.

Title: Integrate $\frac{1}{x\sqrt{a - bx^2}}$


问:
Two problems about Structure Theorem for finitely generated modules over PIDs
1) Let $A$ be a real 4 by 4 matrix. Supose $i,-i$ are the eigenvalues of $A$. Show that there exists an invertible matrix $P$ such that $PAP^{-1}$ is either

$$\begin{pmatrix}
0&-1&0&0\\
1&0&0&0\\
0&1&0&-1\\
0&0&1&0
\end{pmatrix},
\begin{pmatrix}
0&-1&0&0\\
1&0&0&0\\
0&0&0&-1\\
0&0&1&0
\end{pmatrix}.$$

2) Let
$A=
\begin{pmatrix}
1&2&3\\
4&5&6\\
7&8&9
\end{pmatrix}
$. Find an integer $r>0$ and positive integers $1<d_1\mid d_2\mid\cdots\mid d_s$ such that $\mathbb{Z}^3/A\mathbb{Z}^3$ is isomorphic to $\mathbb{Z}/d_1\mathbb{Z} \times \cdots \times \mathbb{Z}/d_s\mathbb{Z} \times \mathbb{Z}^r$.

I think both problems have something to do with the Structure Theorem for finitely generated modules over PIDs, but I am not familiar with that, would someone please help. Actually I don't even understand what is meant by $\mathbb{Z}^3/A\mathbb{Z}^3$...

答:
1) The matrix $A$ has the characteristic polynomial $(X^2+1)^2$. Then its minimal polynomial can be $X^2+1$ or $(X^2+1)^2$.

If the minimal polynomial of $A$ is $X^2+1$, then its characteristic matrix is equivalent to the canonical diagonal matrix having $1,1,X^2+1,X^2+1$ on the main diagonal. This corresponds to the decomposition of the $\mathbb R[X]$-module $\mathbb R^4$ given by $$\mathbb R[X]/(X^2+1)\oplus\mathbb R[X]/(X^2+1).$$
In this case your matrix is similar to
\begin{pmatrix}
0&-1&0&0\\
1&0&0&0\\
0&0&0&-1\\
0&0&1&0
\end{pmatrix}
which is a matrix made by two (identical) companion matrices associated to the polynomial $X^2+1.$

On the other side, if the minimal polynomial of $A$ is $(X^2+1)^2$, then its characteristic matrix is equivalent to the canonical diagonal matrix having $1,1,1,(X^2+1)^2$ on the main diagonal. This corresponds to the decomposition of the $\mathbb R[X]$-module $\mathbb R^4$ given by $\mathbb R[X]/((X^2+1)^2)$. In this case your matrix is similar to
\begin{pmatrix}
0&0&0&-1\\
1&0&0&0\\
0&1&0&-2\\
0&0&1&0
\end{pmatrix}
which is the companion matrix associated to the polynomial $(X^2+1)^2$.

As you can see this is not the given matrix, so you need to work more. What you have to do now is to show that this matrix and the given one are similar and to prove this use the following result: two matrices are similar iff their characteristic matrices are equivalent. This means that you have to consider the characteristic matrix of the given matrix and to show that it has the same canonical diagonal form as the matrix above.

2) I don't know if the first question has something to do with the structure theorem of modules over a PID, but the second definitely has via the Smith Normal Form. Here $A\mathbb Z^3$ denotes the $\mathbb Z$-submodule of $\mathbb Z^3$ generated by the rows (or columns) of $A$ and coincides to the $\mathbb Z$-submodule of $\mathbb Z^3$ generated by the rows (or columns) of its SNF. Since the SNF of $A$ is
\begin{pmatrix}
1&0&0\\
0&3&0\\
0&0&0
\end{pmatrix}
we get that $\mathbb{Z}^3/A\mathbb{Z}^3$ is isomorphic to $\mathbb Z/3\mathbb Z\oplus\mathbb Z$.


问:
Mathematical analysis book with a specific result of continuity.
Let $I^{n}\overset{_\mathrm{def}}{=}\left\{x=\left(x_{1}, \ldots, x_{n}\right) \in \mathbb{R}^{n}|\;\;a_i\leq x_{i} \leq b_{i}\,;\; i=1, \ldots, n\right\}$ be an $n$-dimensional closed interval and $I$ a closed interval $[a, b] \subset \mathbb{R}$.

Proposition.  Let $f \in C^0(I^n\times I,\mathbb{R})$. If for all $(x_1,\ldots,x_n)\in I^n$ there is a $t^\ast\in [a,b]$ such that $f(x_1,\ldots,x_n,t^\ast)=0$ then the function
$$
   m(x_1,\ldots,x_{n-1},x_{n})=\inf\{t\in [a,b] : f(x_1,\ldots,x_{n-1},x_{n},t)=0\}
$$
is also continuous on the  $I^n$.

Can anyone suggest a mathematical analysis textbook that contains proof of this Proposition?

I have already sought proof of this result in classic textbooks of mathematical analysis such as

Real Mathematical Analysis by Charles Chapman Pugh.

Principles of Mathematical Analysis by Walter Rudin

Mathematical Analysis I by V. A. Zorich

but without success.

答:
An example:
Let $g(t)=1$ for $t\leq 1$ and $g(t)=2-t$ for $t>1$.
Let $[a, b]=[-3, 3]$. Let
$$
f(x_1, ...,x_n, t)=(x_1^2+x_2^2+...+x_n^2+t^2)g(t),
$$
Thus if $(x_1, ...,x_n)\neq 0$ we have
$m(x_1, ...,x_n)=2$, while $m(0, ... ,0)=0$.

In general one can only show $m$ is lower semicontinuous.

If one assumes $t^*$ is unique, one can prove the conclusion, but then the $\inf$ in the definition of $m$ looks funny.


问:
Closest point on a̶ ̶s̶u̶r̶f̶a̶c̶e̶ 3d bell curve to a point
I am developing a game using SDFs and ray marching. High school starts next year so I don't really know any calculus (not even pre calc) or linear algebra.

Here's the problem:

A surface is defined by this function (removed constants, not sure if it changes anything):
$$S(x,y) = e^{-(x^2+y^2)}$$
graph of S

A vector A is on the surface if it satisfies $S(A_x,A_y) = A_z$ .This is the distance function (with constants removed):
$$\mathrm{dist}(a,b) = \sqrt{(a_x - b_x)^2 + (a_y - b_y)^2 + (a_z - b_z)^2}$$

Given any vector A, how can I find a vector B on the surface, such that $\mathrm{dist}(A, B)$ is minimum, or at least approximate it?

答:
The distance between a point $A(x_a,y_a,z_a)$ and the surface is
$$d(x,y)=\sqrt{(x-x_a)^2+(y-y_a)^2+\left(e^{-x^2-y^2}-z_a\right)^2}$$
to minimize this function we should solve the system
$$\begin{cases}
 (x-x_a)-2 x e^{-x^2-y^2} \left(e^{-x^2-y^2}-z_a\right)=0\\
 (y-y_a)-2 y e^{-x^2-y^2} \left(e^{-x^2-y^2}-z_a\right)=0\\
\end{cases}
$$
which is not solvable with elementary formulas.

One should implement a subprogram that solves the system above numerically.

For instance I tried to solve approximately when $A(2,5,3)$ and got

$x = 0.36, \;y= 0.86$

The real problem is finding the values where to start the approximation.

Linearizing the problem I found that a good start is

$$x_0= \frac{x_a}{2 z_a-1},y_0= \frac{y_a}{2 z_a-1}$$


问:
Proving integral with finite number of values greater than any small number is $0$
I was given this question and think it is mostly wrong,there were also more mistakes than what I am going to write, but I would like to know how wrong it is.So would there be any validity to this approach, or is it completely incorrect?

Question Suppose that $f:[0,1]\rightarrow [0,1]$ satisfies the following properties:given any $\epsilon>0,f(x)<\epsilon$ except at finitely many points, prove $\int_{0}^{1}f=0$

My attempt:

Let $T=\{y_1,...,y_m\}$ be the set of points such that $f(x) \geq \epsilon$ let $P=\{x_0,...,x_n\}$ be a partition of $[0,1]$.

Let $A=\{k|[x_{k-1},x_k] \cap T\neq \emptyset\}$

Let $B=\{k|[x_{k-1},x_k]\cap T=\emptyset\}$

Then there are at most $2m$ intervals having endpoints with indices in $A$ containing points in $T$. Let $P*$ be a partition containing these intervals such that

$\Delta x_k\leq\frac{\epsilon}{4m}$

Then $U(f,P*)=\sum_{k \in A}\text{sup}\{f(x)|x \in [x_{k-1},x_k]\}\Delta x_k<\frac{\epsilon}{2}$

similarly $L(f,P*)=\sum_{k \in A}\text{inf}\{f(x)|x \in [x_{k-1},x_k]\}\Delta x_k<\frac{\epsilon}{2}$

Now consider a partition $Q$ of intervals with endpoints with indices in $B$.

Let $\Delta x_k<\frac{1}{2(n-2m)}$ Then since $f(x)<\epsilon$ for $f$ on intervals having endpoints with indices in $B$

$U(f,Q)<\frac{\epsilon}{2}$ and $L(f,Q)<\frac{\epsilon}{2}$

Then $L(f,Q)+L(f,P*)<\epsilon$ and $U(f,Q)+U(f,P*)<\epsilon$

My question is was this on the right track of a correct approach?Is there a different way to approach this? When I saw the question I felt a similarity between this question and the proof that thomae's function is integrable, but I do not think I studied/understood that proof well enough, to successfully come up for a proof for this. And I realized in that proof, that didn't come up with separate partitions for intervals having endpoints with indices in $A,B$ like I did.

答:
For each $n$ define $s_n=(x,f(x)\ge \frac{1}{n})$.  Hypothesis $s_n$ is finite.  Therefore $S=\cup_ns_n$ consisting of all points where $f(x) \gt 0 $ is countable.  Countable sets have measure zero, so the measure of the set where $f(x)=0$ is one and the integral $=0$.

Without measure theory: Since $S$ is countable, order the list $(x_k)$ and for each $k$ bracket the point by an interval $(x_k-\frac{\epsilon}{2^k},x_k+\frac{\epsilon}{2^k})$  Union of covering intervals has length $\lt 2\epsilon$. Let $\epsilon \to 0$, so integral over non-zero points $=0$.


问:
Sequence $(a_n)$ s.t $\sum\sqrt{a_na_{n+1}}<\infty$ but $\sum a_n=\infty$
I am looking for a positive sequence  $(a_n)_{n=1}^{\infty}$ such that  $\sum_{n=1}^{\infty}\sqrt{a_na_{n+1}}<\infty$ but  $\sum_{n=1}^{\infty} a_n>\infty$.

Thank you very much.

答: Let $a_n=1+(-1)^n+\epsilon^{2n}$ for $0<\epsilon<1$.  Then $\sum_{n=1}^{+\infty}a_n=+\infty$ and  $\sum_{n=1}^{\infty}\sqrt{a_na_{n+1}}<\infty$.</p>


问:
Why is it injective?
Let $G$ be group and $M$ be a set. Furthermore let $G_m$ be a the stabilizer of $m\in M$ and $Gm$ be the orbit of $m$ under the action.  Now my problem:

Let $\psi: G /G_m \to Gm$, $gG_m \mapsto g(m)$. I wanna show that $\psi$ is injective. Then the argument is given as follows:
$$\begin{align}
\psi(gG_m)=\psi(hG_m)& \Rightarrow g(m)=h(m)\\
& \Rightarrow (g^{-1}h)(m)=m\\
& \Rightarrow g^{-1}h \in G_m\\
& \Rightarrow gG_m =hG_m.
\end{align}$$
I just don't understand the last part, why we get $g^{-1}h \in G_m \Rightarrow gG_m =hG_m$?

Many thanks for your help!

答:
Here is the fact that is being used:

If $H$ is a subgroup of a group $G$, then for $x,y\in G$ we have $xH=yH\iff x^{-1}y\in H$.

Here is a proof:

Suppose $xH=yH$. Then $y=y{\cdot}1\in yH=xH$ so we can write $y=xh$ for some $h\in H$. Then $x^{-1}y=h\in H$.

On the other hand suppose $x^{-1}y\in H$, say $x^{-1}y=h$. Notice then that $y=xh$ so given an element $yh'\in yH$ we have $yh'=(xh)h'=x(hh')\in xH$, which shows $yH\subseteq xH$. For the other inclusion notice $x=yh^{-1}$ so if $xh'\in xH$ then $xh'=(yh^{-1})h'=y(h^{-1}h')\in yH$.

Title: Why is it injective?


问:
Expected number of coin tosses until $HH$ in a row
I was asked the following problem:

Given a coin with $\displaystyle P( H) =p,\ P( T) =1-p$, what is the expected number of tosses until we receive two $\displaystyle H$ in a row?

I may use the fact that the expected number of tosses until the first $\displaystyle H$ is $\displaystyle \frac{1}{p}$.

My solution:

Let's denote $\displaystyle X_{2} =number\ of\ tosses\ until\ HH\ ( including\ the\ HH)$.

We want to find $\displaystyle E[ X_{2}]$.

At the same time, let's denote $\displaystyle X_{1} =number\ of\ tosses\ until\ H\ ( including\ H)$.

From the given we know that $\displaystyle E[ X_{1}] =\frac{1}{p}$.

Let's now try to simplify the expression $\displaystyle E[ X_{2}]$ with the Law of Total Probability.

Let's denote two events:

$\displaystyle A=after\ the\ 1^{st} \ H\ appears,\ the\ next\ toss\ is\ an\ H$.

$\displaystyle A^{c} =after\ the\ 1^{st} \ H\ appears,\ the\ next\ toss\ is\ a\ T$.

Let's also assume that the number of tosses until the first $\displaystyle H$ is some number $\displaystyle r$.

Since every toss is independent from each other, we know that $\displaystyle P( A) =P( H) =p$.

At the same, from the same reason, we will get $\displaystyle P\left( A^{c}\right) =P( T) =1-p$.

From the \ Law of Total Probability then, we can define:

$\displaystyle E[ X_{2}] =P( A) \cdotp E[ X|A] +P\left( A^{c}\right) \cdotp E\left[ X|A^{c}\right]$.

$\displaystyle =p\cdotp E[ X|A] +( 1-p) \cdotp E\left[ X|A^{c}\right]$.

But,

• $\displaystyle E[ X|A]$ = the expected number of tosses until the first HH, if we know that we had exactly $\displaystyle r$ tosses until the first $\displaystyle H$, and after that we had an $\displaystyle H$.

Basically, this means we received $\displaystyle HH$ consecutively for the first time! So we will stop tossing here.

The number of tosses, in this case, will be is $\displaystyle r+1$.

We know the expected value of $\displaystyle r$ is $\displaystyle E[ X_{1}] =\frac{1}{p}$.

So: $\displaystyle E[ X|A] =\frac{1}{p} +1$.

• $\displaystyle E\left[ X|A^{c}\right]$ = the expected number of tosses, if we know that we had exaclty $\displaystyle r$ tosses until the first $\displaystyle H$, and after that we had a $\displaystyle T$.

Also, in this case, we tossed the coin $\displaystyle r+1$ times, but we won't be stopping, since we didn't achieve the $\displaystyle HH$ consecutively. The opposite, we are left facing the same problem recursively. Each toss is independent from the previous one, so after the $\displaystyle r+1$, we have to "start over", and have again $\displaystyle E[ X_{2}]$ expected tosses.

So: $\displaystyle E\left[ X|A^{c}\right] =\frac{1}{p} +1+E[ X_{2}]$.

Overall plugging into our initial equation we get:

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
E[ X_{2}] =p\cdotp \left(\frac{1}{p} +1\right) +( 1-p) \cdotp \left(\frac{1}{p} +1+E[ X_{2}]\right)\\
=1+p+\frac{1}{p} +1+E[ X_{2}] -1-p\cdotp E[ X_{2}]\\
\Rightarrow \\
p\cdotp E[ X_{2}] =\frac{1}{p} +1\\
\Rightarrow \\
E[ X_{2}] =\frac{1+p}{p^{2}} .
\end{array}$

I'm confident this solution is correct, by looking at this other question: Expected value of number of trials to get k SUCCESSIVE successes.

I basically plugged in my numbers into their solution.

Despite this, they used a different method it seems in the first passage, which I don't really understand where it's derived. At the same time, I feel like my answers use a lot of words especially for the cases of $\displaystyle E[ X|A] ,\ E\left[ X|A^{c}\right]$. I hoped to have some formula to use instead of having to explain the solution "logically".

Any advice?

答:
Best to solve this sort of thing via states.  There are $3$ states we care about, labeled by the portion of $HH$ we have going.  The states are $\emptyset, H, HH$.  We start in $\emptyset$.  For a state $S$ let $E_S$ denote the expected number of tries it will take to get to $HH$. We want $E_{\emptyset}$ Of course $E_{HH}=0$.

We remark that state $\emptyset$ either goes to $H$ (prob. $p$) or stays in $\emptyset$ (prob. $1-p$).  Thus $$E_{\emptyset}=p\times E_H+(1-p)\times E_{\emptyset}+1$$

Similarly $$E_H=p\times 0 + (1-p)\times E_{\emptyset}+1$$

which quickly implies that $$E_{\emptyset}=\frac {1+p}{p^2}$$

as desired.

As a variant:  Consider the possible initial sequences.  Only three starts are possible:  Either you get $T$ initially, in which case the game restarts after $1$, or you get $HT$ in which case the game restarts after $2$, or you get $HH$ in which case the game is over in $2$.  Thus $$E=(1-p)\times (E+1)+p(1-p)\times (E+2)+p^2\times 2\implies E=\frac {1+p}{p^2}$$

Note:  as the question of finiteness was raised, it is not terribly difficult to resolve it.  Consider the "coarser" problem in which you toss the con twice at a time and require that you get the $HH$ in a single turn.  That has an expectaion of $\frac 1{p^2}$ so the expected value of the current game must be $≤\frac {2}{p^2}$.  (Of course, if $p=0$ none of these quantities are finite).


问:
"When p is a prime number, $\mathbb{Z_p}$ behaves a lot like $\mathbb{R}$"
I came across this quote while studying Linear systems over $\mathbb{Z_p}$ where the author of the book I am reading casually states

When p is a prime number, ℤp behaves a lot like ℝ in particular, we can add, subtract, multiply, and divide (by nonzero numbers)

I am very confused through this statement, why would that be the case, I've tried myself and I can't find a difference between behaviour in $\mathbb{Z_3}$ and $\mathbb{Z_4}$. Do you know what he is referring to?

Thank you.

答:
The key difference is that $\mathbb{Z}_p$ forms what's called a field, like $\mathbb{R}$, but unlike e.g. $\mathbb{Z}_4$. For instance, we can't define $1/2$ in $\mathbb{Z}_4$ because there's no number $b$ such that $2\cdot b=1$, but in $\mathbb{Z}_5$, say, we have $1/2=3$ since $3\cdot 2\equiv 1\pmod 5$. The essential property is that there are no zero divisors in $\mathbb{Z}_p$, so division is well-defined for all (non-zero) members of it.

Incidentally, there are other finite fields, including one with four elements; more specifically, for any prime $p$ and any $n$ there's a field $F_{p^n}$ with $p^n$ elements. But they're a bit more complicated to think about, as addition and multiplication in the field don't quite map to addition and multiplication as integers. For instance, in $F_4$ we can write the elements as $0$, $1$, $\alpha$, and $1+\alpha$; we have $a+a=0$ for all $a\in F_4$ so each element is its own additive inverse, and $\alpha\cdot(1+\alpha)=1$, so these two elements  are multiplicative inverses of each other. (See if you can figure out what $\alpha^2$ and $(1+\alpha)^2$ are, knowing these facts; there are multiple ways of doing this).
The fields with $2^n$ elements see quite a bit of use in the mathematics of encryption, since it's handy for calculation purposes to be able to do all the relevant arithmetic operations on $n$-bit 'numbers'.


答:
The author most likely refers to the fact that $\mathbb{Z}_p$ is a field if and only if $p$ is a prime number. A field is an ordered triple $(S,+,\cdot)$ where $S$ is a set and $+$, $\cdot$ are binary operations defined on $S$ which satisfy the field axioms. (Note that the "addition" and "multiplication" operations here are not necessarily the usual addition/multiplication operations you had previously encountered on real numbers).

These axioms are chosen in a way to allow us to meaningfully define addition, subtraction, multiplication and division in a similar way that we would define them on $\mathbb{R}$:

Axiom 1 Addition is commutative: $\forall a,b \in S$, $a+b=b+a$

Axiom 2 Addition is associative: $\forall a,b,c \in S$, $a+(b+c)=(a+b)+c$

Axiom 3 Addition has an identity element: $\exists 0_S \in S$  $\forall a \in S$, $a+0_S=a$ (This is referred to as the "Zero element" of the field, but keep in mind that this is not necessarily the number $0 \in R$)

Axiom 4 Additive inverses exist: $\forall a \in S$ $\exists -a \in S$, $a + (-a) = 0_S$
(Note that $-a$ is only a symbol here, the property of "Minus $a$" being the additive inverse of $a$ is only meaningful when we have this axiom. In other words, we state that there exists a "Minus $a$" for every $a$.)

Axiom 5 Multiplication is commutative: $\forall a,b \in S$, $a \cdot b = b \cdot a$

Axiom 6 Multiplication is associative: $\forall a,b,c\in S$, $a \cdot (b \cdot c) = (a \cdot b) \cdot c$

Axiom 7 Multiplication has an identity element: $\exists 1_S \in S$  $\forall a \in S$, $a \cdot 1_S = a$
(This is referred to as the "One element" of the field, but keep in mind that this is not necessarily the number $1 \in R$)

Axiom 8 Multiplicative inverses exist: $\forall a \in S \backslash \{0_S\}$ $\exists a^{-1} \in S$, $a \cdot a^{-1} = 1_S$
(Similar to Axiom 4, we state that there exists a "reciprocal of $a$" for every $a$, but this time except $0_S$.)

Axiom 9 Multiplication is distributive over addition: $\forall a,b,c\in S$, $a \cdot (b+c) = a \cdot b + a \cdot c$

All of these might seem obvious in a sense, but that's only because you're used to the algebraic manipulation in real numbers that involves these properties of real numbers. Indeed, all of these axioms hold for the set of real numbers and usual addition/multiplication. Turns out there are many other structures that satisfy all of these axioms, and thus behave like real numbers. (You might want to pick up an abstract algebra textbook for interesting examples).

As to why $\mathbb{Z}_n$ only behaves like $\mathbb{R}$ when $n$ is a prime, consider the following example:

Assume that you don't know what rational numbers are, and the only number system that you're familiar with is the set of integers $\mathbb{Z}$ (With the usual addition and multiplication, of course). Now, does the expression $5^{-1}$ make sense in this context?

The answer is obviously no. $5^{-1}$ refers to the "reciprocal of 5", or rather the multiplicative inverse of 5, which does not exist in $\mathbb{Z}$. We can safely say that $(\mathbb{Z},+,\cdot)$ does not satisfy axiom 8, and thus is not a field.

In a very similar way, $\mathbb{Z}_n$ satisfies axiom 8 only when $n$ is a prime. When $n$ is composite, there exist (possibly multiple) elements $a\in \mathbb{Z}_n$ with no multiplicative inverse. (Again, you might want to take a look at an introductory abstract algebra textbook for a proof of this)

Sorry for the wall of text, I hope I could make everything clear.

Title: "When p is a prime number, $\mathbb{Z_p}$ behaves a lot like $\mathbb{R}$"

Body:
I came across this quote while studying Linear systems over $\mathbb{Z_p}$ where the author of the book I am reading casually states

When p is a prime number, ℤp behaves a lot like ℝ in particular, we can add, subtract, multiply, and divide (by nonzero numbers)

I am very confused through this statement, why would that be the case, I've tried myself and I can't find a difference between behaviour in $\mathbb{Z_3}$ and $\mathbb{Z_4}$. Do you know what he is referring to?

Thank you.


问:
differential geometric notation

I have solved the following question. I don't understand the choice of notation in the question. Can someone explain to me how the author assigns the superscript and lower script ?

For completeness here is my solution:

$L^{\wedge}(\bar{\alpha}^i)(e_j) = \Sigma_{k} b_k^i \alpha^k(e_j) = \Sigma_k b_k^i \delta_j^k = b_j^i$

On the other hand:
$L^{\wedge}(\bar{\alpha}^i)(e_j) = \bar{\alpha}^i(L(e_j)) = \bar{\alpha}^i(\Sigma_k a_j^k \bar{e}_k) = \Sigma_k a_j^k \bar{\alpha}^i(e_k) = \Sigma_k a_j^k \delta_k^i(e_k) = a_j^i$.

答:
You are correct, except for the extra $(e_k)$ at the last step.

There are two points with the upper and lower indices. One is that formulas should be compatible with the Einstein summation convention — so that you sum over an index that appears once up and once down. The other is that the position of the index keeps track of whether you're dealing with a contravariant object (vector or multi-vector) or covariant object (dual vector, etc.). Of course, the basis vectors of a vector will have a lower index ($e_j$) and the basis vectors of a dual vector will have an upper index ($\alpha^j$), so the corresponding coordinates are the reverse — the coordinates of a vector have upper indices and the coordinates of a dual vector have lower indices.

Note that a linear map maps vectors to vectors, so it can be written as a tensor of type $(1,1)$. The linear map $T$ with matrix $(a^i_j)$ mapping $V$ to $V$, using basis $\{e_i\}$ on both sides, can be represented as $T = \sum\limits_{i,j} a^i_j e_i\otimes \alpha^j$ (where $\{\alpha^j\}$ is the dual basis for $V^\vee$). Notice that this is equivalent to writing $T(e_j) = \sum\limits_i a^i_j e_i$.

Title: differential geometric notation


问:
Prove that we have at least one set with at least $4$ red dots.
We have a set of points, some of which are red.  We form sets $A_{1},A_{2},...,A_{68}$ so that each set has exactly $5$ points, of which at least one is red. Moreover, any triplet in the set of points is in exactly one set. Prove that we have at least one set with at least $4$ red dots.

The hint in the book is counting in two ways.

答:
Let $N$ be the number of points. Counting triplets in two ways we get ${5\choose 3}\cdot 68 = {N \choose 3}\implies N = 17$.

Let $R$ be the number of red points. Then, counting triplets with no red points we have ${17-R\choose 3}\leq 68\cdot {4\choose 3} \implies R\geq 5$.

At this point, I think that some Pigeonhole Principle argument will work.


答: The author of your book is probably refering to the fact that both $\mathbb{Z_p}$ and $\mathbb{R}$ are fields. The main difference between $\mathbb{Z_3}$ and $\mathbb{Z_4}$ is that the former is a field and the latter is only a ring, which is a structure with less requisites (all fields are rings but not all rings are fields). Observe that 2 doesn't have an inverse in $\mathbb{Z_4}$ because $2\cdot 0=0$, $\space$ $2\cdot 1=2$, $\space$ $2\cdot 2=4=0$ and $2\cdot 3=6=2$. We cannot find an element of $\mathbb{Z_4}$  that multiplied by 2 gives the identity 1. Another important thing is that we have zero divisors, that is, a non zero element such that multiplied by another non zero element gives 0. In this case 2 is a zero divisor since $2\cdot 2=0$. $\mathbb{Z_n}$ is a field only when n is prime.


问:
Why the direct product does not form a ring?

Prove the direct product of rings never form a ring.

And be the direct product $A\times B$:

$+$: $(x,y) + (z,t) = (x+z,y+t)$

$*$: $(x,y) * (z,t) = (x*z, y*t)$

$x,z\in A$.

If we define $A$ and $B$ both commutative rings, with units and inverses itself (So $A$ and $B$ are fields).

Now, $A\times B$ is commutative, it is a ring, there is a unity element.

And since $A$ and $B$ have inverses, the inverse of each element in $A\times B$ $(x,y)$ is just $(x^{-1},y^{-1})$, being $x^{-1}$ the inverse of $x$ in $A$, and so to $y$ in $B$.

So
$$(x,y) * (x^{-1},y^{-1}) = (e_{a},e_{b})$$

Defining the unity of $A\times B$ as $(e_{a},e_{b})$

I can't see the error in my writings.

答:
The direct product of fields is never a field; because we have zero divisors, $(1,0)$ and $(0,1)$, for instance, it is not even an integral domain.

But the direct product of rings is a ring.


问:
Why is the Math-tea argument wrong?
The Math-tea argument says there must be some real numbers that cannot be specified because we only have countably many definitions.

Why is that wrong?

Is there a simple explanation?

I don't understand this blog.

https://alexanderpruss.blogspot.com/2020/11/the-math-tea-argument.html

I don't understand what does this mean:

For instance, there is no guarantee that there is a set of all the definable real numbers.

Why does it matter that there's no set that contains the enumeration?

There is no set of all sets, yet we can enumerate through every set of sets that has a finite description by considering all combinations of the following characters from least length to greatest length: { },,

```<code>{}
{{}}
{{{}}}
{{{{}}}}
{{},{{}}}
{{{{{}}}}}
{{{},{{}}}}
{{{{{{}}}}}}
{{{{},{{}}}}}
{{},{{},{{}}}}
</code>```

etc.

The axioms of set theory tell us that for any predicate F in the language of set theory there is a set of all the numbers that satisfy F.

I think this makes sense - we can specify a predicate that corresponds to a subset of real numbers that satisfy the predicate

I don't follow the logic here:

Elegant as this argument is, it has crucial set-theoretic flaws. For instance, there is no guarantee that there is a set of all the definable real numbers. The axioms of set theory tell us that for any predicate F in the language of set theory there is a set of all the numbers that satisfy F. But the predicate "is definable" is in English, not in set theory.

答:
I would describe the situation as follows: it's not that the math tea argument is wrong, so much as that it is easy to misapply it. Specifically, here's a theorem which on the face of it may seem to contradict the math tea argument:

$(@)$ Assuming $\mathsf{ZFC}$ is consistent, there is a model $M$ of $\mathsf{ZFC}$ such that every element of $M$ is parameter-freely definable in $M$. In particular, every real number in $M$ is parameter-freely definable in $M$.

(Really it's more convenient to talk about $\omega$-models here, but I'll ignore this since it makes things more technical and isn't the key point.)

The issue is that we have to be very careful about what system the math tea argument takes place in. There's a version of the math tea argument which we can run inside $M$, but it is substantially weaker than one might expect.

Specifically, for $n\in\mathbb{N}$ consider the sentence

$(*)_n$: "There is a real which is not parameter-freely $\Sigma_n$-definable over $M$.

Each $(*)_n$ is outright provable in $\mathsf{ZFC}$ via essentially the math tea argument, and so holds in $M$ in particular. However, the following sentence

$(\dagger)$: "There is a real which is not definable over the universe"

is not even expressible in the language of set theory - this is a consequence of Tarski's undefinability theorem - and so we can't talk about whether or not $M$ thinks it's true.

We can (express and) prove in $\mathsf{ZFC}$ the statement

$(\dagger)_{set}$: "For every model $A$ of $\mathsf{ZFC}$, there is a real $r$ which is not definable over $A$,"

again using the math tea argument. This is because we can talk about truth within a set, rather than over the universe. However, note that the real $r$ is not guaranteed to live in $A$. So e.g. the fact that $(@)$ is a $\mathsf{ZFC}$-theorem doesn't contradict the fact that $\mathsf{ZFC}$ proves each $(*)_n$ and $(\dagger)_{set}$.

Title: Why is the Math-tea argument wrong?


问:
How to do calculus with split-complex (hyperbolic) numbers?
TL;DR: how do I define a "split-holomorphic" function?

As far as I've heard, there is a notion of split-complex numbers: $z = x+uy$, with $u\not\in \Bbb R $ and $u^2 = 1$. One defines the conjugate as $\overline{z} = x - uy$ and the modulus as $|z| = \sqrt{|z\overline{z}|}$, where the absolute value inside the root is the one for real numbers, as usual.
Apparently this can be used to study the pseudo-Riemannian geometry of $\Bbb L^2 = \Bbb R^2_1$ in the same way thay $\Bbb C$ is identified with $\Bbb R^2$.

But these split-complex numbers form a ring only, and not a field: elements in the diagonals don't have multiplicative inverses. Incidentally, these diagonals correspond to lightlike directions in $\Bbb L^2$ (and such directions cause every sort of problems).

I don't know if there is a standard notation for this set of split-complex numbers, but given a function of said set to itself, how do I define things like "being holomorphic"? I can't make sense of the limit $$\lim_{h\to 0} \frac{f(z_0+h)-f(z_0)}{h}$$ since $h$ could approach zero from lightlike directions. Do I just ignore these directions? Most likely I'm overthinking this.

We could define continuity of $f$ by continuity of its components, and work formally with the above limit to obtain a revised version of the CR equations, and say that $f$ is split-holomorphic if the revised CR equations hold and if the partial derivatives of the components are continuous (copying the Looman-Menchoff theorem), but I am speculating too much and I am unsure about these things.

What is the correct way to do calculus with split-complex numbers, if this is possible? More references are also welcome.

P.s.: there is a nice book called "Geometry of Minkowski Space-Time", by Zampetti, et al, but they do a more algebraic approach and do not talk about the things I'm asking.

答:
You've probably found an answer in the past 4 years, but just in case anyone else is curious split-complex analysis is typically referred to using the term "motor variable", and a notion of holomorphic functions can be found here:
https://en.wikipedia.org/wiki/Motor_variable#D-holomorphic_functions</p>


问:
When is $\sqrt{2}$ in $\mathbb{Q}(\zeta_n)$?
The following question cropped up in a physics context when studying some properties of conformal field theory. I know a little Galois Theory, but not enough to be able to answer it, though I suspect it is elementary.

When $n$ is a multiple of $8$, we can see that $\sqrt{2}$ lies in the cyclotomic field $\mathbb{Q}(\zeta_n)$ thanks to the identity

$$ \sqrt{2} = \zeta_n^{n/8} + \zeta_n^{-n/8} $$

Experiments with Sage seem to suggest that these are also the only values of $n$ for which this is true. Is this correct, and if so how does one prove it?

Moreover, in the interests of others viewing this question, is there a straightforward generalisation to other quadratic irrationals?

答:
Since the intersection $\mathbb Q(\zeta_m)\cap \mathbb Q(\zeta_n)=\mathbb Q(\zeta_{\gcd(m,n)})$ (for example here), it suffices to show that $\sqrt2$ is not contained in any cyclotomic field strictly inside $\mathbb Q(\zeta_8)$. The largest such field is $\mathbb Q(\zeta_4)=\mathbb Q(i)$, and it's clear that $\sqrt2$ is not in this field (since the only reals in $\mathbb Q(i)$ are the rationals), so the only $n$ for which $\sqrt2\in\mathbb Q(\zeta_n)$ are multiples of $8$.

For the more general case of $\sqrt{m}$ with $m=\pm p_1\cdots p_s$, we can use that, for an odd prime $p$,
$$\sqrt{(-1)^{(p-1)/2}p}\in\mathbb Q(\zeta_p).$$
(This can be proven, for example, by using Gauss sums.) From this, we definitely get that
$$\sqrt m\in\mathbb Q(\zeta_{4|m|}),$$
since $\sqrt{\pm p_i}$ is in this field for all $i$ (since if $2|m$, $8|4m$, this is true for $p=2$ as well). If $m$ is odd and of correct sign, i.e.
$$m=\prod_{i=1}^s (-1)^{\frac{p_i-1}{2}}p_i,$$
then we can reduce $4|m|$ to $|m|$.

We can prove that this in fact optimal by strong induction on $s$. For the base case, we need only worry about factors of $p$ or $4p$, and so the cases aren't that bad since $\sqrt{\pm p}$ isn't in $\mathbb Q(i)$ regardless of $p$ and $\mathbb Q(\zeta_n)=\mathbb Q(\zeta_{2n})$ for all odd $n$. For the inductive step, using any factor of $|m|$ or $4|m|$ either reduces the numbers of factors of $2$ (in which case arguments similar to the base case work) or removes a subset of the prime factors, in which case our strong inductive hypothesis can be used directly.


问:
Intersection of contractible spaces is contractible?
Given that the intersection is nonempty, is it true?

Can I argue that if $1_U$ and $1_V$ are null-homotopic, then $1_{U\cap V}$ is null homotopic as well? Also, I'm assuming that $U$ and $V$ are path connected(though I don't know if they matter)

答: No. Let $U = \{ (x,y) \in \mathbb R^2 \mid x^2 + y^2 = 1, y \ge 0 \}$ and $V = \{ (x,y) \in \mathbb R^2 \mid x^2 + y^2 = 1, y \le 0 \}$. These sets describe the upper and lower half of the unit circle in the plane. Both contain $\{(1,0)\}$ as a strong deformation  retract, but $U  \cap V = \{(1,0), (-1,0)\}$ is not contractible. Note that also $U \cup V$ is not contractible.


问:
Trying to show that $r = \frac{1}{C}\left( \frac{1}{1 + e\cos{(\theta + \omega)} } \right)$ is an ellipse
I am trying to prove that planets move in ellipses,

I watched this video: https://www.youtube.com/watch?v=DurLVHPc1Iw

and read this: https://arxiv.org/pdf/1009.1738.pdf .

But both sources end up with this as an equation for the paths that planets follow :
$$
 r = \frac{1}{C}\left( \frac{1}{1 + e\cos{(\theta + \omega)} } \right),
 $$
where $C=G\frac{m_{1}+m_{2}}{h^{2}}$ which is just a constant, as is $\omega$, and $e$ is the eccentricity, so this is an ellipse with the focus at the origin as long as $e<1$ and when I put the equation in graphing software it gives me ellipse.

But I don't understand why it is an ellipse? Can anyone show me how it is an ellipse? Either from the normal polar coordinates of a ellipse centred at the origin and shifting it (I couldn't get the algebra to work out when I added ae to the x coordinate) or just by explaining it conceptionally?

答:
Let's define $p=1/C$ (the semi-latus rectum) and set WLOG $\omega=0$. Let's construct then a line (the directrix) perpendicular to the $\theta=0$ axis and lying at a distance $FH=p/e$ from the the origin $F$.

An ellipse can be defined as the locus of points $P$ having a fixed ratio $e$ between their distance $PF$ from the focus and their distance $PN$ from a fixed lined (the directrix).

Let's take then a point $P$ whose distance from the focus is given by your equation:

$$
PF=r={p\over1+e\cos\theta}.
$$

The distance of $P$ from the directrix is given by:

$$
PN=FH-FK={p\over e}-r\cos\theta=
{p\over e}\left(1-{e\cos\theta\over1+e\cos\theta}\right)={r\over e}.
$$

Hence $P$ lies on the ellipse having $F$ as focus and $HN$ as directrix.


问:
prove that $\sum_{n=0}^{\infty} {\frac{n^2 2^n}{n^2 + 1}x^n} $ does not converge uniformly in its' convergence radius
In calculus:

Given $\displaystyle \sum_{n=0}^{\infty} {\frac{n^2 2^n}{n^2 + 1}x^n} $, prove that it converges for $-\frac{1}{2} < x < \frac{1}{2} $, and that it does not converge uniformly in the area of convergence.

So I said:

$R$ = The radius of convergence $\displaystyle= \lim_{n \to \infty} \left|\frac{C_n}{C_{n+1}} \right| = \frac{1}{2}$ so the series converges $\forall x$ : $-\frac{1}{2} < x < \frac{1}{2}$.

But how do I exactly prove that it does not converge uniformly there? I read in a paper of University of Kansas that states: "Theorem: A power series converges uniformly to its limit in the interval
of convergence." and they proved it.

I'll be happy to get a direction.

答:
According to Hagen von Eitzen theorem, you know that it will uniformly (and even absolutly) converge for intervals included in ]-1/2;1/2[. So the problem must be there.

The definition of uniform convergence is :
$ \forall \epsilon>0,\exists N_ \varepsilon \in N,\forall n \in N,\quad [ n \ge N_ \varepsilon \Rightarrow \forall x \in A,d(f_n(x),f(x)) \le \varepsilon] $

So you just have to prove that there is an $\varepsilon$ (1/3 for instance or any other) where for any $N_ \varepsilon$ there will always be an $x$ (look one close from 1/2) where $d(f_n(x),f(x)) \ge \varepsilon $ for an $n \ge N_ \varepsilon$


问:
injective map between the semi-unit circle and a parabola
How can we injectively map the semi-unit circle to a parabola bounded between two points?

I earnestly think this can be done, as the curves are homeomorphic (one can be "bent" to the other), but the explicit map is out of my imagination. Any hints? Thanks beforehand.

答: Thanks to the hint provided by @AdamRubinson, supposing for the sake of representation, we wish to find a map between the upper unit semicircle and the parabolic segment $x=y^2$ bounded between $(0,0)$ and $(0,1)$. Then, the map $f(x,y)=\left(\frac{(x+1)^2}{4},\frac{x+1}{2}\right)$ does the job. For other cases, appropriate scaling can be done.


问:
Determine time spent by orbits near an equilibrium
Consider the system of ODEs
\begin{align*}
\dot{x}_1 = x_1(1-x_1-\alpha x_2-\beta x_3)\\
\dot{x}_2 = x_2(1-\beta x_1 - x_2 - \alpha x_3)\\
\dot{x}_3 = x_3(1-\alpha x_1 - \beta x_2 - x_3)
\end{align*} for $0 < \beta < 1 < \alpha$ and $\alpha + \beta > 2$. I am trying to show that the time intervals which a solution spends near the rest points $(1,0,0), (0,1,0), (0,0,1), (1,0,0),...$ increase geometrically with factor $(\alpha - 1)/(1-\beta) > 1$. I'm not sure where to start on this problem. I first tried finding an expression for $dt$ so that I could integrate over a region of time where a solution is near any of the aforementioned points, but the dimensionality of the problem makes this impossible (this method works great for one-dimensional problems).

So then I tried "reducing" the system, e.g. in a neighborhood of $(1,0,0)$, $x_2 \approx 0$ and $x_3 \approx 0$, so that the corresponding derivatives are negligible, leaving $\dot{x}_1 \approx x_1(1-x_1)$, implying the dynamics near the equilibrium are well-approximated by a 1-D system. Employing the aforementioned technique for this new system I still came up short, and feel as though this isn't the intended approach. Any help is greatly appreciated.

答:
Disclaimer: this is not a solution. It provides

a graphical representation and

exhibits an exceptional solution which does not enter into the general scheme.

A general remark: This system of differential equations is a kind of generalization of the one-dimensional logistic differential equation (see here).

Let us assume that the initial values are such that

$$x_1(0) > 0, \ \ x_2(0)>0, \ \ x_3(0)>0$$

(Otherwise there is divergence).

I have written a Matlab program (using a Runge-Kutta method) for the numerical computation of a certain number of trajectories for a certain common time range $[0,T]$. They all remain into the first octant ; their terminal positions (for $t=T$) is materialized by a small red circle, evidencing the "attracting power" of points $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$.

The trajectory singled out in red follows the line $x=y=z$ ; it corresponds to an exceptional solution.

Indeed, let us look for a common equation

$$\varphi(t):=x(t)=y(t)=z(t) \tag{1}$$

with

$$\varphi(t)>0 \ \text{and} \ \varphi'(t)<0$$

If we plug (1) in any of the 3 equations, we get the same differential equation:

$$\varphi'(t)=\varphi(t)(1- \gamma \varphi(t)) \ \ \text{where} \ \gamma:=1+\alpha+\beta \tag{2}$$

which is a classical logistic differential equation with solution:

$$\varphi(t)=\dfrac{1}{\gamma +c e^{-t}}$$

where $c$ is an arbitrary constant.

We can remark that

$$\lim _{t \to \infty}\varphi(t)=\dfrac{1}{\gamma}$$

The limit point of this trajectory $(\varphi(t), \varphi(t), \varphi(t))$ is therefore $(\dfrac{1}{\gamma},\dfrac{1}{\gamma},\dfrac{1}{\gamma})$.

Final remark: I am wondering if looking for solutions of the form
$$x(t)=X(t)+\varphi(t), y(t)=Y(t)+\varphi(t), z(t)=Z(t)+\varphi(t)$$ wouldn't bring a simplification...


问:
What's the probability the cumulative average of multiple gaussian variables exceed a certain value?
Consider a number $n$ of independent, normally distributed variables $Y_1$, $Y_2$, ..., $Y_n$.
Consider also the $n$ variables defined by the average of the last $h$ terms, $X_h = \frac{\sum^{n}_{k = n -h + 1} Y_k}{h}$ for $h \in \{1,.., n\}$. The variables $\sqrt{h}X_h$ are normally distributed.

I'd like to understand what is the probability of having $P\{\sqrt{h}X_h > T \text{ or } \sqrt{h-1}X_{h-1} > T \text{ or .. or } X_1>T \}$, in words having one or more of the $\sqrt{h}X_h$ to output above a threshold value.

Do you have any reference or suggestions about this problem?
I can work out the $n = 2$ case through geometry working on the expression $P\{Y_1 + Y_2 > \sqrt{2}T \, |\, Y_1<T \}$. However I soon get lost into integral hell when dimension increases.

Thank you

EDIT:

I add my solution when $n = 2$. As remarked by @ConMan, the probability we are looking for is equal to one minus the probability for all $\sqrt{h}X_h$ to output under threshold $T$, so:

$$ P = 1 - p(Y_1 + Y_2 < \sqrt{2}T\,|\,Y_1< T)p(Y_1<T) $$

We can calculate the term $ p(Y_1 + Y_2 < \sqrt{2}T\,|\,Y_1< T) $ geometrically (see picture), through:

$$ p(Y_1 + Y_2 < \sqrt{2}T\,|\,Y_1< T)  = \int_{-\infty}^{\sqrt{2}T -T} \text{d}x N(x) + \int^{\infty}_{\sqrt{2}T -T}\text{d}x N(x)\Big[ 1 - \int^{T}_{\sqrt{2}T -x}\text{d}y N(y)\Big]=\\
 = 1 - \int^{\infty}_{\sqrt{2}T -T}\text{d}x N(x)\Big( F(T) - F(\sqrt{2} T - x) \Big)=\\
 = 1 - \Phi(T)\big(1 - \Phi(\sqrt{2}T - T)\big) + \int^{\infty}_{\sqrt{2}T -T}\text{d}x N(x) \Phi(\sqrt{2} T - x) $$

Naming $H(T)$ the last integral term, the probability we are looking for become:

$$P = 1 - \Phi(T)\Big( 1 -\Phi(T)\big(1-\Phi(\sqrt{2}T - T)\big) + H(T) \Big)$$

If $T = 3$, we find a probability $P \sim 2.5\%$ which implies a value larger than the probability $1 - \Phi(T)$ by a factor $\frac{P(T)}{1-\Phi(T)} \sim 1.823$.

Where we noted with $\Phi$ and $N$ the normal partition and probability density function respectively.
This seems confirmed by simulations, in python:

```<code>import scipy.stats as sts
import numpy as np

N = 1000000
M = 2
host_matrix = np.zeros((N,M))

for n in range(N):
    arr = sts.norm(0).rvs(M)
    wind = np.cumsum(arr)/np.sqrt(np.arange(M) + 1)
    host_matrix[n] = wind
</code>```

Here a routine for calculating the value of the formula above:

```<code>def fp_prb(thr):
    import scipy.integrate as integrate
    def hard_term_f(x, T):
        return sts.norm.pdf(x)*sts.norm.cdf(sqrt(2)*T - x)

    hard_term = integrate.quad(lambda x: hard_term_f(x,thr),thr*(sqrt(2) - 1),+np.inf)[0]
    F_t = F(thr)
    F_sqrt = F(thr*(sqrt(2)-1))

    prb = 1 - F_t*( 1 - F_t*(1 - F_sqrt) + hard_term )
    return  prb
</code>```

The observed frequency and probability ratio can be computed through:

```<code>thr = 3

no_fp = len(np.argwhere((host_matrix[:,0] < thr ) & (host_matrix[:,1] < thr )))
fp_freq =  1 - no_fp/N
norm_freq = 1 - sts.norm.cdf(thr)
print('FP frequency: {:.4f}'.format(fp_freq))
print('Normal detections frequency: {:.4f}'.format(norm_freq))
print('Ratio: {:.3f}'.format(fp_freq/norm_freq))
</code>```

An example output:

```<code>FP frequency: 0.0025
Normal detections frequency: 0.0013
Ratio: 1.841
</code>```

答:
I'm going to do all the calculations assuming you're taking the average of the first $h$ variables rather than the last, because it makes the algebra a little neater without really changing the results. Also, I'm going to assume you always look at the full sequence of $n$ variables, since you can always just drop off the ones you're not working with and set $n = h$ to get the same effect.

The first step is to flip that probability around. Instead of looking at "the probability that at least one is greater", instead look at "the probability that all of them are less", i.e. $P(X_1 \leq T \land X_2 \leq T \land \ldots \land X_h \leq T)$, because these two probabilities are complements and hence will add to 1, but this one is easier to work with.

Second, let's write everything in vectors.

$\mathbf{X} = \begin{bmatrix} X_1 \\ X_2 \\ \vdots \\ X_n \end{bmatrix} =
\begin{bmatrix} 1 & 0 & 0 & \ldots & 0 \\
\frac{1}{2} & \frac{1}{2} & 0 & \ldots & 0 \\
& & &\ddots \\
\frac{1}{n} & \frac{1}{n} & \frac{1}{n} & \ldots & \frac{1}{n} \end{bmatrix} \begin{bmatrix} Y_1 \\ Y_2 \\ \vdots \\ Y_n \end{bmatrix} = \mathbf{MY} $

So our probability becomes $P(\mathbf{X} \leq T\mathbf{1}_n) = P(\mathbf{MY} \leq T \mathbf{1}_n) = P(\mathbf{Y} \leq T\mathbf{M}^{-1}\mathbf{1}_n)$.

So now we just have to invert that matrix. I am going to propose, and suggest that you prove for yourself, that:

$\mathbf{M}^{-1} = \begin{bmatrix} 1 & 0 & 0 & \ldots & 0 & 0 \\
-1 & 2 & 0 & \ldots & 0 & 0 \\
0 & -2 & 3 & \ldots & 0 & 0 \\
 & & & \ddots \\
0 & 0 & 0 & \ldots & -(n-1) & n \end{bmatrix}$

Which, neatly, means that $\mathbf{M}^{-1} \mathbf{1}_n = \mathbf{1}_n$, and so our probability is literally just $P(\mathbf{X} \leq T\mathbf{1}_n) = P(\mathbf{Y} \leq T\mathbf{1}_n) = P(Y_1 \leq T \land Y_2 \leq T \land \ldots \land Y_n \leq T) = P(Y_1 \leq T)P(Y_2 \leq T) \ldots P(Y_n \leq T) = \Phi(T)^n$ because the $Y_i$ are all independent.

So, the probability that all of the averages are less than the threshold is $\Phi(T)^n$, meaning that the probability that at least one crosses the threshold is $1 - \Phi(T)^n$.


问:
How to integrate $\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$
The question given is to calculate
$$\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$$

My attempt

I managed to figure out that the denominator is given out as a perfect square of $$\{1+2\sin(x + \frac{\pi}{3})\}$$ and broke the $\sin(x+\frac{\pi}{3})$ so it looks like
$$\int \frac{\cos(x) + \sqrt 3}{(1+ \sin(x) +\sqrt 3 \cos(x))^2}{\rm d}x$$
I can't figure out how to approach further. Please guide me through this question.

答:
Hint

Let $y=\dfrac{a\cos x+b\sin x +c}{1+\sin x+\sqrt3\cos x}$

Find $\dfrac{dy}{dx}$

Compare
$-(a\cos x+b\sin x+c)(\cos x-\sqrt3\sin x)+(1+\sin x+\sqrt3\cos x)(-a\sin x+b\cos x)$

with $\cos x+\sqrt3$

to find $a,b,c$

For example, by comparing the coefficient of $\cos x\sin x$ we get $b=\sqrt3a$

By $\sin x,a=-\sqrt3c$

So we reach at the expression below with $c$ as constant multiple
$$(-\sqrt3\cos x+3\sin x+1)(\cos x-\sqrt3\sin x)+(1+\sin x+\sqrt3\cos x)(\sqrt3\cos x+3\sin x)$$

Can you take it from here?


问:
A sum with binomial coefficients
Show that $$\sum_{k=0}^{n}(-1)^k\binom{n}{k}(n-2k)^{n+2}=\frac{2^{n}n(n+2)!}{6}.$$</p>

Title: A sum with binomial coefficients

Body:
Using these three identities for $n^{\text{th}}$ differences:
$$
\begin{align}
\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}k^n&=n!\\
\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}k^{n+1}&=n!\binom{n+1}{2}\\
\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}k^{n+2}&=n!\left(3\binom{n+2}{4}+\binom{n+2}{3}\right)
\end{align}
$$
and the fact that the $n^{\text{th}}$ difference of a polynomial of degree less than $n$ is $0$, we get
$$
\begin{align}
&\sum_{k=0}^n(-1)^k\binom{n}{k}(n-2k)^{n+2}\\
&=2^{n+2}\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}(k-n/2)^{n+2}\\
&=2^{n+2}\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}\left(k^{n+2}-\frac{(n+2)n}{2}k^{n+1}+\frac{(n+2)(n+1)n^2}{8}k^n+\dots\right)\\
&=2^{n+2}n!\left(\left(3\binom{n+2}{4}+\binom{n+2}{3}\right)-\frac{(n+2)n}{2}\binom{n+1}{2}+\frac{(n+2)(n+1)n^2}{8}\right)\\
&=2^nn!\binom{n+2}{3}\\[6pt]
&=\frac{2^nn(n+2)!}{6}
\end{align}
$$


问:
Prove or disprove that $x_1 = x_2= \cdots = x_n$, if $x_1^{x_2} = x_2^{x_3} = \cdots = x_{n-1}^{x_n} = x_n^{x_1}$, $x_i > 1, \forall 1 \le i \le n$.
The case $n=3$ is from here. It's straightforward to prove it's true:

First we notice that if any two of $x_1, x_2, x_3$ are equal then all must be equal.

Denote $a=x_1^{x_2}=x_2^{x_3}=x_3^{x_1}$ then
$$x_1 = a^{1/x_2}, x_2=a^{1/x_3}, x_3=a^{1/x_1}$$
WLOG assume $x_1 \ge x_2, x_3$. There are two cases:

$x_1 \ge x_2 \ge x_3 \implies \frac{1}{x_2} \ge \frac{1}{x_3} \ge \frac{1}{x_1} \implies x_1 \ge x_3 \ge x_2 \implies x_2=x_3 \implies x_1=x_2=x_3$.

$x_1 \ge x_3 \ge x_2 \implies \frac{1}{x_2} \ge \frac{1}{x_1} \ge \frac{1}{x_3} \implies x_3 \ge x_1 \ge x_2 \implies x_1=x_3 \implies x_1=x_2=x_3$.

If $n$ is even, then $x_{2i-1} = 2, x_{2i}=4$ is a counterexample.

When $n=5$ the above method will need to examine $4!=24$ cases. Basically we map $x_i$ to $x_{(i \mod 5) +1}$ and reverse the order. In many of cases we can deduce $4$ or all $5$ of the $x_i$'s are equal. For example, if $$x_1 \ge x_3 \ge x_2 \ge x_5 \ge x_4 \tag 1$$ then
$$
x_5 \ge x_1 \ge x_3 \ge x_4 \ge x_2 \tag 2
$$
Since the order of $x_2$ and $x_5$ reversed from $(1)$ to $(2)$, they must be equal and so are everything in between them from both $(1)$ and $(2)$ hence all $x_i$'s are equal.

There are other cases that are different.

Example 2: $$x_1 \ge x_3 \ge x_4 \ge x_2 \ge x_5 \tag 3$$ then
$$x_1 \ge x_3 \ge x_5 \ge x_4 \ge x_2 \implies x_1 \ge x_3 \ge x_2=x_4=x_5 \tag 4$$

Example 3: $$x_1 \ge x_3 \ge x_4 \ge x_5 \ge x_2 \tag 5$$ then
$$x_3 \ge x_1 \ge x_5 \ge x_4 \ge x_2 \implies x_1=x_3 \ge x_2=x_4=x_5 \tag 6$$

But they all lead to the conclusion that $x_1=x_2=x_3=x_4=x_5$.

Now my questions:

Question #1: Is it always true if $n>1$ and $n$ is odd?

Question #2: What if we allow $x_i>0$ instead of $x_i>1$?

答:
After reading this answer to another question, as well as Andreas's answer I decide to write up the following solution. I believe John Omielan had already figured out how to do it, though he chose to leave that to the one who asked the case $n=3$ question last year. Then Andreas's answer made everything clear to me.

Assume $n=2m+1$.
First notice
$$x_2 \ln x_1 = x_3 \ln x_2 = \cdots = x_n \ln x_{n-1} = x_1 \ln x_n$$
Then there are only three possibilities: 1) $x_i = 1, \forall i$; 2) $x_i > 1, \forall i$;
3) $x_i < 1, \forall i$.

For case 2), if not all $x_i$'s are equal, WLOG we assume $x_1 > x_2$.

Then $x_2 \ln x_1 = x_3 \ln x_2 \implies x_2 < x_3$.

Then $x_3 \ln x_2 = x_4 \ln x_3 \implies x_3 > x_4$.

...

$x_{2m+1} \ln x_{2m} = x_1 \ln x_{2m+1} \implies x_{2m+1}>x_1$

$x_1 \ln x_{2m+1} = x_2 \ln x_1 \implies x_1 < x_2$, a contradiction. Therefore all $x_i$'s must be equal.

For case 3), since $\ln x_i < 0$ we'd have $x_i > x_{i+1}$, and $x_{2m+1}>x_1$, also a contradiction (in fact this also works when $n$ is even).


问:
Absolute continuity on $[a,b]$ implies mapping of sets of measure zero to sets of measure zero
I am reviewing some materials for real analysis and measure theory.

I want to show that if we have an absolutely continuous function on a closed interval, the function maps sets of measure zero to sets of measure zero.

I found the following that I have a few questions about, found at http://www.ms.uky.edu/~perry/676-s19/_assets/final-answers.pdf on the last page :

Where the author writes that $m(O) < \delta$, is this because $m(O) < m(E) + \delta = \delta$ ?

How does the author assert that $f$ is continuous on each interval $[a_k,b_k]$ ? Since $Z$ is contained in $O$, can't it be that $[a_k,b_k]$ is not necessarily contained in $[a,b]$ ?

Why can we conclude from $m(f(O)) = 0$ that $m(f(Z)) = 0$ ?

Thank you ! This is my first time here.

答:
Where the author writes that $m(O) < \delta$, is this because $m(O) < m(E) + \delta = \delta$ ?

$\mathcal O$ is just chosen to be an open set containing $Z$ of measure less than $\delta$.  There exist open sets of arbitrary small measure containing $Z$.

How does the author assert that $f$ is continuous on each interval $[a_k,b_k]$ ? Since $Z$ is contained in $O$, can't it be that $[a_k,b_k]$ is not necessarily contained in $[a,b]$ ?

If $S \subseteq T \subseteq \mathbb R$, and $f$ is a continuous function on $T$, its restriction to $S$ is always continuous.

Why can we conclude from $m(f(O)) = 0$ that $m(f(Z)) = 0$ ?

They proved that given any $\epsilon > 0$, there exists an open set $\mathcal O \subset [a,b]$ such that  $m(f(Z)) \leq m(f(\mathcal O)) < \epsilon$.  So for any $\epsilon >0$, we have $m(f(Z)) < \epsilon$.  This can only happen if $m(f(Z)) = 0$.


问:
Uniform distribution of points on a sphere: only Platonic solids?
I'm quite sure the only way to uniformly distribute $n$ points on the sphere $S^2$ is by inscribing one of the 5 Platonic solids, thus there only exists a solution for $n=4,6,8,12,20$.

But am I right? And how can this be proven?

Addendum

My intuitiv definition of a uniform distribution is: Any 2 adjacent points have the same (spherical) distance; the points adjacent to $p$ are those that have minimal distance from $p$.

This should describe the uniformity of the vertices of a Platonic solid distributed on its circumscribed sphere.

答:
I can think of at least two more uniform distributions that fit your current description, one with 14 points, and one with 32 points.

You obtain them by combining the vertices of a pair of dual regular polytopes:

six vertices of the octahedron + eight vertices of the cube.
The adjacent points form a regular triangle or square.

twelve vertices of the icosahedron + 20 vertices of the dodecahedron.
The adjacent points form a regular triangle or regular pentagon.


问:
How do regions of integration combine when performing a product of integrals?
I'm a physicist attempting to symbolically evaluate a path integral. In order to do that, I need to know the following: I think that the double integral, $I:=\int_{E_1}d\mu\int_{E_2}d\mu f$ is writable as $I=\int_{g(E_1,E_2)}d^2\mu f$, however I'm not sure as to the form of $g$? So, my question is: how do the sets $E_1$ and $E_2$ generally combine into another set, $g(E_1,E_2)$ when they are being integrated over? (The specific integral I'm working with is integrated over a series of infinitely many shells, if that helps...)

答: I think you mean $\int_{E_1}d\mu_1\int_{E_2}d\mu_2f=\int_{E_1\times E_2}d^2\mu f$, with $E_1\times E_2$ the Cartesian product. If $E_2$ is $\mu_1$-dependent, the integration range is more generally $\{(x,\,y)|x\in E_1,\,y\in E_2(x)\}$. Note the product of two contours is a surface.


问:
simple function $f(x)$ for which $x^{1/x} -1 \sim f(x)$ as $x \rightarrow \infty$.
$f(x) \sim g(x)$ iff $\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = 1$. Find a simple function $f(x)$ for which $x^{1/x} -1 \sim f(x)$ as $x \rightarrow \infty$.

I know that $\lim_{x \rightarrow \infty} x^{1/x} = 1$ so I was thinking of using a function that tends towards $0$. However, I didn't really get very far past that.

答:
$$x^{\frac{1}{x}}-1=e^{\frac{\log x}{x}}-1\underset{x\rightarrow +\infty}{\sim}\frac{\log x}{x}$$
because $\lim\limits_{x\rightarrow +\infty}\frac{\log x}{x}=0$ and $e^t-1\underset{t\rightarrow 0}{\sim}t$.


问:
Why presheaves are generalized objects?
While self studying category theory (Yoneda lemma), I came across the statement that for any category $\mathsf{C}$ the functor category $\mathsf{Fun}(\mathsf{C}^{op}, \mathsf{Set})$ represents generalized objects of $\mathsf{C}.$

Here generalized means bunch of objects of $\mathsf{C}$ glued together.

Because of the Yoneda embedding $$Y:\mathsf{C}\to\mathsf{Fun}(\mathsf{C}^{op}, \mathsf{Set}),$$ I can imagine that $\mathsf{C}$ lives inside of $\mathsf{Fun}(\mathsf{C}^{op}, \mathsf{Set})$ as $Y(\mathsf{C}),$ however I can not see why other objects in this category acts like generalized objects of $\mathsf{C}.$

Can anybody explain me why this philosophy works, possibly with some example.

答:
This may not be technical as other answers. But I haven't seen this very intuitive explanation written anywhere, and therefore wanted to share with others. Suppose we have a category $\mathsf{C}$ that we want to somehow (sorry, this is too vague) generalize without adding more morphisms, but adding more objects. Lets say this new (unknown generalized) category is $\tilde{\mathsf{C}},$ which comes with a fully faithful functor $\iota: \mathsf{C}\hookrightarrow\tilde{\mathsf{C}}$. For any generalized object $A\in\tilde{\mathsf{C}},$ we have a functor $$Y_A:\mathsf{C}^{op}\to\mathsf{Set}$$ given by $Y_A(X)=\mathsf{mor}_{\tilde{\mathsf{C}}}(\iota X, A).$ This functor is contravariant (or covariant according to my notation) because any morphism $X\xrightarrow{f} X'$ in $\mathsf{C}$ induces a function $Y_A(X')\xrightarrow{(\_)\circ\iota f}Y_A(X)$ by precomposition. Now we can implement this construction to another functor $$Y: \tilde{\mathsf{C}}\to\mathsf{Fun}(\mathsf{C}^{op}, \mathsf{Set})$$ such that $Y(A)=Y_A.$ For any morpsism $A\xrightarrow{g}A'$ in $\tilde{\mathsf{C}}$ we have a natural transformation $Y(A)\Rightarrow Y(A')$ whose components are given in the commutative square
$\require{AMScd}$
\begin{CD}
\mathsf{mor}_{\tilde{\mathsf{C}}}(\iota X', A) @>(\_)\circ f>> \mathsf{mor}_{\tilde{\mathsf{C}}}(\iota X, A)\\
@V g\circ(\_) V V @VV g\circ(\_) V\\
\mathsf{mor}_{\tilde{\mathsf{C}}}(\iota X', A') @>>(\_)\circ f> \mathsf{mor}_{\tilde{\mathsf{C}}}(\iota X, A')
\end{CD}

By the Yoneda lemma $Y\vert_{\mathsf{C}}$ is fully faithful, therefore contains all the information of $\mathsf{C}$ (and possibly more interesting objects in the restricted target). The most important part of this seemingly trivial construction is to realize that $Y\vert_{\mathsf{C}}$ has no reference to its generalization $\tilde{\mathsf{C}}.$ So, we declare that unknown category $\tilde{\mathsf{C}}$ to be the category of preshaves $\mathsf{Fun}(\mathsf{C}^{op}, \mathsf{Set}).$ In other words, $Y$ is an isomorphism of categories and $Y\vert_{\mathsf{C}}$ is the Yoneda embedding.

Next, the co-Yoneda lemma says that we haven't added any wild objects while generalizing $\mathsf{C}$ in this manner. In technical terms it say every presheaf is a colimit of objects in the image of $Y\vert_{\mathsf{C}}$ (this image is called the representable presheaves). Also, the category of preshaves is cocomplete. So, we think of it as the  free cocompletion of $\mathsf{C}.$

Title: Why presheaves are generalized objects?


问:
Show that P the set of all primes is closed in the Euclidean topology
Theorem: Prove the set P of all prime numbers
is a closed set of R

I am using TOPOLOGY WITHOUT TEARS by Sydney Morris

My thoughts on it

It would suffice to show that R$\setminus$P
is opened

The set (n,n+1) is opened

So R$\setminus$P=$\bigcup_{n\in P}$(n,n+1) is
opened so P is closed..

Any help would be appreciated.

答:
Your idea is right. We can think of the complement of the set of prime numbers as a countable union of open intervals, so open.

Another way to see this is to use the sequence convergence characterization of closed sets. If $n_i$ is a sequence in $P$ with limit $n,$ then it could be checked (easily) that $n=n_i$ for some $i,$ so $n\in P$ also. Thus $P$ is closed since any limit of sequences in $P$ still belongs to $P.$


问:
General solution to wave equation using traveling waves
In Stein and Shakarachi's Fourier Analysis, they prove the general solution to the $1$D wave equation. There is one step I cannot follow.

Let $u(x,t)$ be a solution to $\frac{\partial^2 u}{\partial t^2} = \frac{\partial^2 u}{\partial x^2}$. Then define $\xi = x+t$ and $\eta = x-t$ and set $v(\xi, \eta) = u(x,t)$. They then claim that this implies $\frac{\partial^2}{\partial \xi \partial \eta} v(\xi, \eta) = 0$. I can sort of see how this works out, since there are two derivatives in $x$ and in $t$, with a minus sign so it turns into the wave equation for $u$, but I don't see the precise steps. Particularly, I'm trying to plug in $v(\xi, \eta)$ into the wave equation for $u(x,t)$ and am having trouble dealing with $\frac{\partial}{\partial x} v(x+t, x-t)$ and the other similar terms. How does one evaluate this? Is there is nicer was to see the relation for $v$?

答:
Notice that $\frac{\eta + \xi}{2} = x$ and $\frac{\xi- \eta}{2} = t$. Then by the chain rule,
\begin{align*}
\frac{\partial v}{\partial{\eta}} &= \frac{\partial u}{\partial x}\frac{\partial x}{\partial \eta} + \frac{\partial u}{\partial t}\frac{\partial t}{\partial \eta}\\
&= \frac{1}{2}(u_x - u_t).
\end{align*}
Taking one more derivative, we have
\begin{align*}
\frac{\partial v}{\partial \xi \partial\eta} &= \frac{\partial u_x}{\partial x}\frac{\partial x}{\partial \xi} +  \frac{\partial u_x}{\partial t}\frac{\partial t}{\partial \xi} -\frac{\partial u_t}{\partial x}\frac{\partial x}{\partial \xi} -  \frac{\partial u_{t}}{\partial t}\frac{\partial t}{\partial \xi} \\
&= \frac{1}{4}(u_{xx} + u_{xt} - u_{tx} - u_{tt})
\end{align*}
The mixed partials are equal, so $u_{xt} = u_{tx}$. Also, $u_{xx} = u_{tt}$ since we assumed that $u$ solves the wave equation. Thus, it follows that
$$\frac{\partial v}{\partial \xi \partial\eta} = 0.$$


问:
I'm having a hard time finding the $c_1$ and $c_2$ coordinates of a system after finding the Eigenvalues and Eigenvectors.
My task is to solve a system given the initial value.

Solve System : $\frac{{d\vec{x}}}{dt} = \left | \matrix{-4 & 3 \\ 2 & -3 } \right |$ with init value $\vec{x_0} = \left | \matrix{1 \\ 0}\right | $

What I have done so far:

So I have a matrix :

$\frac{{d\vec{x}}}{dt} = $$\left | \matrix{-4 & 3 \\ 2 & -3 } \right | $

I found the Eigenvalues, $\lambda_1 = -6$ , $\lambda_2 = -1$
I found the Eigenvectors by solving the system of the Eigenvalues

Which are $\vec{v1} = span(\left | \matrix{3 \\ -2} \right |)$

and $\vec{v2} = span(\left | \matrix{1 \\ 1} \right |)$

So now, I have to find a c(t) = $S^{-1}\vec{x(t)}$

What I did, then was applied my eigenvectors to a new matrix
$S =\left | \matrix{3 && 1 \\-2 && 1} \right |$

and get the $S^{-1}=\det(\left | \matrix{3 && 1 \\-2 && 1} \right | )$

which is $\frac{1}{5}$

The point where I am confused:

So, according to the theorem in my book, to find the scalars $c_1 , c_2,...,c_n$ according to the basis
$\vec{v_1}, \space \vec{v_2}, \space ,...\vec{v_n}$ I have to write the matrix form as.

$S \lambda^{t}_t S^{-1}\vec{x_0} = S =\left | \matrix{3 && 1 \\-2 && 1} \right |\left | \matrix{e^{\lambda_1t}_t & 0 \\ 0 & e^{\lambda_2t}_t} \right| \frac{1}{5}\left | \matrix{3 && 1 \\-2 && 1} \right |\left|\matrix{1 \\ 0}\right| $

with $\lambda^{t}_t$ as a stand in for

$\left | \matrix{e^{\lambda_1t}_t & 0 \\ 0 & e^{\lambda_2t}_t} \right|$

So now, I would do the math and end up with :

$\left| \matrix{\frac{9}{5}e^{-6t} & \frac{-2}{5}e^{-t} \\ \frac{-6}{5}e^{-6t} & \frac{-2}{5}e^{-t}}\right|$

But I look in the back of the book, aaaand the coordinates $c_1$ && $c_2$ with respect to $\vec{v1}$ &&  $ \vec{v2}$ are actually $c_1 = \frac{-1}{5}$ and $c_2 = \frac{2}{5}$

I'm clearly doing something wrong here, but I do not know what exactly.

答:
To find this answer, all we need to do is abide by the initial value point.

$\vec{x_0} = \left |\matrix{1 \\ 0} \right |$

and therefore,

$x(t)=c_1e^{-6t}\left | \matrix{3 \\ -2} \right | + c_2e^{-1t} \left| \matrix{1\\1} \right|$

so,

$x(0)=c_1e^{-6(0)}\left | \matrix{3 \\ -2} \right | + c_2e^{-1(0)} \left| \matrix{1\\1} \right|$

Which reduces our equation to :

$x(0) = c_1\left | \matrix{3 \\ -2} \right | + c_2\left | \matrix{1\\1} \right|$

Now recall since we have an initial value $x(0) = \left | \matrix{1\\1} \right|$

$ \left | \matrix{1\\0} \right| = c_1\left | \matrix{3 \\ -2} \right | + c_2\left | \matrix{1\\1} \right|$

So now apply c1 and c2 to the matrices and we have a system that can be represented by an augmented matrix.

$\left | \matrix{3c1 & 1c2 & | 1 \\ -2c1 & 1c2 & | 0} \right|$

Now solve using row reduction and we get $\left| \matrix{1 & 0 & | \frac{1}{5}\\ 0 & 1 & |\frac{2}{5}}\right|$

Plug these values into the equation $3c_1 + c_2 = 1$
and $-2c_1 + c_2 =0$ and you get

$\frac{3}{5} + \frac{2}{5} = 1$

and

$\frac{-2}{5} + \frac{2}{5} = 0$

So our finished general equation is

$\frac{1}{5}e^{-6t}\left | \matrix{3 \\ -2} \right | + \frac{2}{5}e^{-t} \left |\matrix{1 \\ 1} \right|$

and our coordinates $c_1 = \frac{1}{5} \space  c_2 = \frac{2}{5}$


问:
Does every ordinal have cardinality no greater than $\aleph_\mathbb{0}$?
My notes say that the ordinals $\omega + 1, \omega + 2, ... , 2 \omega, ... , 3 \omega, ... \omega^2, ... $ are all countable, and hence have cardinality equal to $\omega = \aleph_\mathbb{0}$. So I was wondering if it's fair to say that every ordinal has cardinality no greater than $\aleph_\mathbb{0}$?

Alternatively, I guess it's possible that that the sequence of infinite ordinals listed above, does not include some of the ordinals with a cardinality greater than the naturals.. but I wasn't sure.

答: (In ZFC) by Cantor's theorem, $2^{\aleph_{0}}=\vert \mathcal{P}(\omega)\vert> \vert \omega\vert=\aleph_{0}$, so that $\vert P(\omega)\vert (\approx \mathbb{R})$ is an uncountable cardinal and therefore an uncountable (limit) ordinal (here $2^{\aleph_{0}}$ is cardinal exponentiation). The claim that $2^{\aleph_{0}}=\aleph_{1}$, i.e that the least cardinal strictly greater than $\aleph_{0}$ is $2^{\aleph_{0}}$, is known as Continuum Hypothesis.


问:
Matrix for a linear map
Consider $V= \left \{ v= \begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix} : x_1-x_2+2x_3=0\right \}\subset K^3$ and the linear map $f:V \rightarrow K^2$ defined by

$f\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=\begin{pmatrix}
x_1-x_3\\
x_2-3x_3
\end{pmatrix}$

Now, what is the matrix $M$ that represents $f:V \rightarrow K^2$ with respect to the basis of $V$ and the standard basis $(e_1,e_2)$ for $K^2$.

So far

I found the basis of V of dimension 2 to be the vectors  ${(1, 1, 0), (-2, 0, 1)}$. But then I do not know how to continue. Maybe coordinate shift?

答:
In the general (finite dimensional) case, given a linear transformation between two vector spaces $V$ and $W$ as well as basis $\mathcal{B}_{V} = \{v_{1},v_{2},\ldots,n\}$ and $\mathcal{B}_{W} = \{w_{1},w_{2},\ldots,w_{m}\}$ where $n = \dim V$ and $m = \dim W$, the matricial representation of $T:V\to W$ is given by
\begin{align*}
[T]_{\mathcal{B}_{V}}^{\mathcal{B}_{W}} = [[T(v_{1})]_{\mathcal{B}_{W}} [T(v_{2})]_{\mathcal{B}_{W}} \ldots [T(v_{n})]_{\mathcal{B}_{W}}]
\end{align*}

At your case, we are given $V$ (which is a plane) and $W$ (which coincides with $\mathbb{K}^{2}$, where $\mathbb{K}$ is the field of scalars, I assume).

Given that $B_{V} = \{(1,1,0),(-2,0,1)\}$ and $\mathcal{B}_{W} = \{(1,0),(0,1)\}$, we have that
\begin{align*}
\begin{cases}
f(1,1,0) = (1,1) = (1,0) + (0,1)\\\\
f(-2,0,1) = (-3,-3) = -3(1,0) -3(0,1)
\end{cases}
\end{align*}

Consequently, one can conclude that
\begin{align*}
[f]_{\mathcal{B}_{V}}^{\mathcal{B}_{W}} =
\begin{bmatrix}
1 & -3\\
1 & -3
\end{bmatrix}
\end{align*}

Hopefully this helps!

Title: Matrix for a linear map


答:
There are multiple errors/unclear parts in your question. I'm not very sure what the set $K$ is, so I will be assuming it's the field $\mathbb{R}$.

<s>Also, the codomain of $f$ seems to be $K^2$ rather than $K^3$, so I assume $f:V \to K^3$ was a typographic error on your part.</s>

Now, a "matrix representation of a linear transformation $g:A \to B$" only makes sense when we're given specific ordered bases of $A$ and $B$. You gave us $B_{K_2}=(e_1,e_2)$ for $K^2$ but ordered the basis of $V$ with respect to which we will find the matrix representation is not very clear. I will assume it is the ordered basis $B_{V}=((1,1,0),(-2,0,1))$.

I will assume that you are familiar with the theory behind matrix representations of a linear transformation and I will simply find the representation. If anything within this process is unclear, just let me know.

First, we will find the images of elements of $B_V$ under $f$, expressed as linear combinations of elements of $B_{K^2}$.

$$f(1,1,0)=(1-0,1-3 \cdot 0)=(1,1)=1 \cdot (1,0) + 1\cdot (0,1) = \textbf{1} \cdot e_1 + \textbf{1} \cdot e_2$$
$$f(-2,0,1)=(-2-1,0-3\cdot 1)=(-3,-3)=(-3) \cdot(1,0) + (-3) \cdot (0,1) = \textbf{(-3)} \cdot e_1 + \textbf{(-3)} \cdot e_2$$

Now, we will simply take the coefficients written in bold and place them in our representation matrix. Remember, the i-th column represents the coefficients of the linear combination of the i-th element of $B_V$ in terms of the elements of $B_{K^2}$.

$$[f]^{B_V}_{B_{K^2}}= \begin{bmatrix}
1 & -3 \\
1 & -3
\end{bmatrix}$$

Title: Matrix for a linear map

Body:
Consider $V= \left \{ v= \begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix} : x_1-x_2+2x_3=0\right \}\subset K^3$ and the linear map $f:V \rightarrow K^2$ defined by

$f\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=\begin{pmatrix}
x_1-x_3\\
x_2-3x_3
\end{pmatrix}$

Now, what is the matrix $M$ that represents $f:V \rightarrow K^2$ with respect to the basis of $V$ and the standard basis $(e_1,e_2)$ for $K^2$.

So far

I found the basis of V of dimension 2 to be the vectors  ${(1, 1, 0), (-2, 0, 1)}$. But then I do not know how to continue. Maybe coordinate shift?


问:
Is there generally understood set theory term for $A \cap B \setminus C$
I'm interested in formalising/documenting a program plotting Venn charts (well UpSet plots actually). Is there a term that can unambiguously refer to the coloured partition of the visualisation below (obtained from Wolfram alpha from $A \cap B \setminus C$) in relation to $A$ and $B$? It was previously referred to as distinct intersection of $A$ and $B$ (because it does not include the intersection of $A$ and $B$ with any other set) - is there a formal term for this concept?

Edit: in general I am interested in a case of $n$ sets. I am looking for a generic name for a region of an intersection of $k$ sets ($k \leq n$) such that it is exclusive to the sets forming the intersection this is it excludes overlap with any other set that do not for this $k$ intersection.

答: No, there is not any commonly used term for this.  Simply writing it as $(A\cap B)\setminus C$ or $A\cap B\cap C^c$ (with ${C}^c$ referring to the complement of $C$) would be a reasonable choice.  If you want to refer to it in words, you could say something like "the region that is in $A$ and $B$ but not $C$", or "the intersection of $A$ and $B$ and the complement of $C$".  You could also say something like "the region that is only in $A$ and $B$".  I would strongly recommend against "distinct intersection"; that's not the way "distinct" is normally used in mathematics.