问:
Prove that for all $n \in \mathbb{N}$, $\sum_{k=0}^{n-1}{n+k-1\choose k}\frac 1{2^{n+k}}=\frac12$
$$  \sum_{k=0}^{n-1}{n+k-1\choose k}\frac 1{2^{n+k}}=\frac12$$

To be honest, I can't really get started,

I would like to ask everyone to give me an idea of how to solve it, give me a starting push, thank you.

答:
By the way, the product being summed looks like the PMF of the negative binomial distribution, that to flip a fair coin repeatedly until having $n$ heads, there's $\frac 12$ probability that there were at most $n-1$ tails before the $n$th head.

Let $\Pr(A)$ be the probability that there were at most $n-1$ tails before the $n$th head.

Then $1-\Pr(A)$ would be the probability that the $n$th tail appears while there were at most $n-1$ heads.

For a fair coin, head and tail are symmetric, and so

$$\Pr(A) = 1-\Pr(A) = \frac 12$$


问:
Critical Nonplanar Graphs
In Alan Tucker's Applied Combinatorics, it is stated that "A graph G is critical nonplanar if G is nonplanar but any subgraph obtained by removing a vertex is planar."

The question is "Show that critical nonplanar graphs must be connected and cannot have a
vertex whose removal disconnects the graph."

I am not sure how to show this, I have an example of a case where a vertex is removed, resulting in a planar graph, but is there something in the definition of a planar graph that it has to be connected?

答: The deal is that if you have more than one component, then either one or more of the components are non-planar. To see this, suppose $G=G_1\cup G_2,$ such that the is no edge connecting any vertex of $G_1$ to $G_2$ and suppose further that $G_1$ is non-planar, then take any vertex of $G_2$ and is going still to be non-planar. What happens now if there is such vertex that disconnects the graph? It does not belong to a cycle(why?), so when you disconnect it, one of the parts is going to be non-planar. Can you show this?

ContentLicense: CC BY-SA 4.0

Title: Critical Nonplanar Graphs


问:
Expectation of a random variable conditioned on two random variables and an event
I'm trying to work out a conditioned expectation of a variable that depends on two other via the Law of Total Probability. I've got some intuition on what it should be but there is an important detail that I can't work out, and I need help with it. I've got three discrete random variables $X,Y,Z$, where $Y$ depends on $X,Z$. By the Law of Total Expectation, $E[Y] = E[E[Y|X,Z]]$. Now consider the event $A=X < Z$. My first question is: what is the expression for $E[Y|A]$: is it 1) or 2)?
$$
\begin{align}
1) \qquad
E[Y|A]
    &= E[E[Y|X,Z,A]] \\
    &= \sum_x \sum_z P(X=x,Z=z|A)E[Y|X=x,Z=z,A] \\
2) \qquad E[Y|A]
    &= E[E[Y|X,Z,A]] \\
    &= \sum_x \sum_z P(X=x,Z=z,A)E[Y|X=x,Z=z,A]
\end{align}
$$
Note that the only difference between the two expressions is the conditioning of the probability inside the double summation. Maybe it is neither of these? Am I applying the Law of Total Expectation correctly?

I would like to ask another question: am I right when I claim that 3) is correct?
$$
\begin{equation}
3) \qquad E[Y|X=x,Z=z,A] = \sum_y yP(Y=y|X=x,Z=z,A)
\end{equation}
$$

I've seen posts across this site discussing similar topics, but I haven't found anything that could help me. Any help will be very much appreciated.

答:
$E(Y|A) = E(Y1_A) = E(E(Y1_A | X, Z)) = \sum_{x,z}E(Y1_A | X=x, Z=z)P(X=x, Z=z) = \sum_{x<z} E(Y|X=x, Z=z)P(X=x, Z=z)$

As far 3, think about what the event $\{X=x, Z=z, A\}$ really means given that $A = \{X < Z\}$.


问:
Union of normal spaces
Let $(X, \mathcal{T})$ a topological space and $X= \bigcup_{i \in I} U_i$ where $U_i$ are open for all $i \in I$ and $U_i \cap U_j = \emptyset$ if $i \neq j$, if each $(U_i, \mathcal{T}_{U_i})$ is normal then $X$ is normal.

I don't know how to start the proof. I know that I have to take $A,B \subset X$ closed, but I can't continue

答:
The key point is that if $V \subseteq U_i$ is open in $U_i$ (i.e, with the topology $\tau_{U_i}$), then $V$ is open in $X$. (why?)

Suppose $A, B \subseteq X$ are closed and disjoint. By definition of the subspace topologies then, $A \cap U_i, B \cap U_i$ are closed in $U_i$ for each $i \in I$. Then since $U_i$ are normal we have disjoint $V_i, W_i$ open in $U_i$ such that $A \cap U_i \subseteq V_i$, $B \cap U_i \subseteq W_i$.

Then $V_i, W_i$ are open and disjoint in $X$ for every $i \in I$, and then notice that $V = \cup_{i \in I}V_i$ and $W = \cup_{i\in I}W_i$ does the trick of separating $A$ and $B$.

Title: Union of normal spaces


问:
How do I show $\int_{\mathbb{R}}f(x) d\mu(x)=\int_{\mathbb{R}}f(x) w(x)dx$?
i)Let $\mu$ is a measure on $\mathbb{R}$ with the density $w$ i.e $\mu(A)=\int_{A}w(x)dx$.

Prove that, using the definition of integral that for any measurable function $f: \mathbb{R} \rightarrow \mathbb{C}$

$\int_{\mathbb{R}}f(x) d\mu(x)=\int_{\mathbb{R}}f(x) w(x) dx$.

ii) Let $F: \mathbb{R} \rightarrow (0,\infty) $ be an increasing, differentiable function, and $\mu((a,b])=F(b)-F(a)$.Prove that for any measurable function $f: \mathbb{R} \rightarrow \mathbb{C}$

$\int_{\mathbb{R}}f(x) d\mu(x)=\int_{\mathbb{R}}f(x) w(x) dx$.

I proved part i), by taking the measurable function as

Indicator function

Simple function

Non-negative function

Signed function

Complex valued function.

I prove every case except #3 Non-negative function case. Can anyone suggest some hint for this case and ii) part of the question?

答: Once you have proved (2), (3) follows by Monotone Convergence Theorem. Recall that if $f:\mathbb{R}\rightarrow [0,\infty)$ is a non-negative measurable function, we can choose a sequence of simple function $(f_n)$ such that $0\leq f_1 \leq f_2 \leq \ldots \leq f$ and $f_n\rightarrow f$ pointwisely.


问:
Why is the collection of zeroes of all derivatives of an analytic function also open?
Let $f$ be analytic on $\Omega\subset\mathbb{C}$

Let $E_n = \{w\in\Omega | f^{(n)} (w) = 0\}$

Let $E = \cap_{n\geq 0} E_n$

Clearly, $E_n$ are closed because they are pre-image of the closed set $\{0\}$ and $f^{(n)}$ are continuous

As $E_n$ are closed, so is $E$

But how can I show that $E$ is also open set ?

答: If $E$ is empty then you're done, so suppose $w \in E$. Then $f$ has a convergent power series expression in some neighborhood $U$ of $w$. By the Cauchy integral formula, the coefficients of this series are $a_n = f^{(n)}(w)/n!$. But all of these derivatives vanish, so the power series is just the constant term $a_0$. This holds on all of $U$, so all of the derivatives of $f$ vanish on $U$. Hence, $U \subseteq E$. As $w \in E$ was arbitrary, $E$ is open.


问:
$\Vert f_n\Vert_p \longrightarrow \Vert f\Vert_p$ implies the convergence of $f_n$ to $f$ in $\textit{L}^p$ with the right hypotesis?
I think this might work, but how i don't know:

Let $p\in (1,+\infty)$ and $\{f_n\}$, $f\in \textit{L}^p(E)$, if we know that:

$\Vert f_n\Vert_p \longrightarrow \Vert f\Vert_p$ as $n\longrightarrow +\infty$;

$\forall$ measurable $F\subseteq E$ with $|F| < +\infty \Longrightarrow \int_Ff_n\longrightarrow \int_Ff$, as $n\rightarrow +\infty$;

is it true that $f_n$ converges to $f$ in $\textit{L}^p(E)$ in norm (strong convergence)?

Any suggestions?

答:
I'll procede as suggested by gerw.

For the first of the two hipothesys i have that
$$\forall \epsilon > 0 \mbox{, } \exists N\in \mathbb{N} \mbox{ : } \forall n \ge N \mbox{, } \bigl| ||f_n||_p-||f||_p \bigr|< \epsilon$$
so $\forall n\ge N$, $||f_n||_p <||f||_p + \epsilon$, now letting $M:=\max\{||f1||_p$, ...,$||f_{N-1}||_p$, $||f||_p+\epsilon\}$, we obtain $||F_n||_p \le M$, that is the boudedness of the sequence. With what we have we can apply:

Theorem 1.1. $(1 < p < +\infty)$ Let $\{f_n\}$ be a bounded sequence in $\textit{L}^p(U)$. Then there exists a subsequence $\{f_{n_k}\}$ which converges weakly to some $f\in \textit{L}^p(U)$. For a proof click here.

So given $\{f_{n_k}\}_{k\in \mathbb{N}}$, $\overline{f}\in \textit{L}^p(E)$ such that $f_{n_k} \rightharpoonup \overline{f}$ we have shown the first step.

Now, passing to the second one, by the definition of weak convergence:
$$\int_Egf_{n_k} \longrightarrow \int_Eg\overline{f} \mbox{ as } k\rightarrow +\infty \mbox{, } \forall g\in \textit{L}^p(E) \mbox{ (for Riesz's rep. theorem, with } \frac{1}{p}+\frac{1}{q}=1\mbox{).}$$
So for $g=\chi_F\in  \textit{L}^p(E)$ if $|F|<+\infty$, gives us
$$\int_Ff_{n_k} \longrightarrow \int_F\overline{f} \mbox { as } k\rightarrow +\infty \mbox{, } \forall \mbox{ measurable } F \mbox{ s.t. } |F| <+\infty \mbox.$$
Furthermore, by hipothesys
$$\int_Ff_n \longrightarrow \int_Ff \mbox { as } k\rightarrow +\infty \mbox{, } \forall \mbox{ measurable }F \mbox{ s.t. } |F| <+\infty$$
and since $\{\int_Ff_{n_k}\}_{k\in \mathbb{N}}\subseteq \{\int_Ff_n\}_{n\in \mathbb{N}}\subset \mathbb{R}$, also the real subsequence converges.

So
$$\forall F:|F|<+\infty \mbox{, } \int_Ff=\int_F\overline{f} \Longrightarrow f=g \mbox{ a.e. on these } F\mbox;$$
if we pick $F_k:=B(0,k)$, then $\forall k\in \mathbb{N}$, $f=\overline{f}$ on $F_k$ except on a $N_k$ such that $|N_k|=0$; so, observing that $E=\cup_{k\in \mathbb{N}}F_k$, we obtain that $f=\overline{f}$ on $E$ except on the measurable set $\cup_{k\in \mathbb{N}}N_k$, but $\bigl|\cup_{k\in \mathbb{N}}N_k\bigr|\le \sum_{k\in \mathbb{N}}|N_k|=0$, and we can deduce that $f=\overline{f}$ a.e., which means that they are the same element in $\textit{L}^p(E)$. This concludes the second point.

Now to show the third point we first suppose that $f_n \not \rightharpoonup f$, then sow it is absurd, implying it is not true, which means weak convergence is true.

So if $f_n \not \rightharpoonup f$, then
$$\exists \overline{g}\in \textit{L}^p(E) \mbox{ s.t. } \int_E\overline{g}f_n \longrightarrow \int_E\overline{g}f \mbox{ as } n\rightarrow +\infty \mbox{ is NOT true.}$$
This means that there exists a subsequence $\{f_{n_i}\}_{i\in \mathbb{N}}$ that doesn't converge weakly to $f$ and neither do all of it's subsequences, but by hipothesys $||f_n||_p \longrightarrow ||f||_p$ so that also $||f_{n_i}||_p \longrightarrow ||f||_p$ as $i\rightarrow +\infty$; also $||f_{n_i}||_p \le M$ defined as before.
So now we can apply to $\{f_{n_i}\}_{i\in \mathbb{N}}$ the first two points already proven to obtain that
$$\exists \{f_{n_{i_k}}\}_{k\in \mathbb{N}} \mbox{ subsequence, s.t. } \int_E\overline{g}f_{n_{i_k}} \longrightarrow \int_E\overline{g}f \mbox{ as } k\rightarrow +\infty \mbox,$$
which is not possible since we said the contrary before. So we deduce that $f_n \rightharpoonup f$, proving also the third point.

To finish the proof we just have to apply the Radon-Riez Theorem (Theorem 5 here)

In fact under the condition $f_n \rightharpoonup f$, the following are equivalent as $n \rightarrow +\infty$:
$$||f_n||_p \longrightarrow ||f||_p \iff f_n\longrightarrow f \mbox{ in norm }\textit{L}^p(E) \mbox{ (strong topology).}$$

Thanks to @gerw who gave me valuable hints.


问:
Why is true this part of Krull's height Theorem
I am stuck trying to prove this part of Krull's height Theorem. I checked this proof and, in page 18, it says that if a prime ideal $\mathfrak{p}$ contains $f_1, g_2,\dots,g_n$ this implies that $f_1, f_2^{m},\dots,f_n^{m}\in\mathfrak{p}$ and hence $f_1, f_2,\dots,f_n\in\mathfrak{p}$.

It was easy to proof the first implication, but I don't know why the second one holds.

That fact is used to prove that if you quotient $R$ by $(g_2,\dots,g_n)$, then $\overline{\mathfrak{p}}$ is minimal over $(\overline{f_1})$. That's the only thing I need to (finally) understand this proof.

答: If $f_j^m$ is in a prime ideal, either $f_j^{m-1}$ or $f_j$ is in the prime ideal. Repeat this enough times to conclude that $f_j$ is in the prime ideal after all.


答: If a prime ideal $\mathfrak p$ contains some power of an element $f^m$ then it contains $f$ itself. Indeed, $f^m \in \frak p$ then $f^m = 0$ in $R/\mathfrak p$. As $\mathfrak p$ is a prime ideal, this is a domain. Hence, $f = 0$ in $R/\mathfrak p$, i.e. $f \in \mathfrak p$. An ideal satisfying this property ($f^m \in I \implies f \in I$) is called radical.

Title: Why is true this part of Krull's height Theorem

Body:
I am stuck trying to prove this part of Krull's height Theorem. I checked this proof and, in page 18, it says that if a prime ideal $\mathfrak{p}$ contains $f_1, g_2,\dots,g_n$ this implies that $f_1, f_2^{m},\dots,f_n^{m}\in\mathfrak{p}$ and hence $f_1, f_2,\dots,f_n\in\mathfrak{p}$.

It was easy to proof the first implication, but I don't know why the second one holds.

That fact is used to prove that if you quotient $R$ by $(g_2,\dots,g_n)$, then $\overline{\mathfrak{p}}$ is minimal over $(\overline{f_1})$. That's the only thing I need to (finally) understand this proof.


问:
Give example of 2 standard uniform random variables with given Pearson correlation
Give example of r.v. $X$ and $Y$ with $X \sim U(0,1)$ and $Y \sim U(0,1)$ and $corr(X,Y) = .25$.

How does one approach such a problem?

答:
I think there are a few approaches to generating random variables with given marginal distributions and specified correlations, but, in general, the ones I've seen all seem to be ad hoc.

One such method is to do the following: Let $U_1$ and $U_2$ be $U(0,1)$ i.i.d. r.v. and let  $B$ be a r.v. distributed according to Bernoulli($0.25$). One way to construct $X$ and $Y$ as you've described is to set $X=U_1$ and $Y=BU_1+(1-B)U_2$. Intuitively, $Y$ 'copies' the same value as $X$ $25\%$ of the time, and the other $75\%$, they are independent uniform random variables. You can check that $X$ and $Y$ have the desired marginal distributions and correlation.


问:
Need help solving Laplace Transform Question
For context: I am just starting to learn LaPlace Transform and am stumped with this question.

Find the PS of $x''+x'-12x=sin(3t) + e^{-4t}+e^{3t}$
Where $x(0)=0, x'(0)=0$ using LaPlace Transform

My Approach:

Let X(s)=$\mathscr{L}[x(t)]$

Apply the LaPlace Transform to both sides of the equation

$\mathscr{L}[x''] + \mathscr{L}[x'] - 12\mathscr{L}[x]$ = $\mathscr{L}[sin(3t)] +\mathscr{L}[e^{-4t}] +\mathscr{L}[e^{3t}]$

to get $s^2X(s)-sx(0)-x'(0)+sX(s)-x(0)-12X(s) = \cfrac{3}{s^2+9}+\cfrac{1}{s+4}+\cfrac{1}{s-3}$

Plugging in the IC's $x(0) = 0$ and $x'(0)=0$ and then factoring $X(s)$ we get

$X(s)[s^2+s-12] = \cfrac{3}{s^2+9}+\cfrac{1}{s+4}+\cfrac{1}{s-3}$

You should then make the RHS have a common denominator so you get

$X(s)[s^2+s-12] = \cfrac{3(s+4)(s-3)}{(s^2+9)(s+4)(s-3)}+\cfrac{(s^2+9)(s-3)}{(s+4)(s^2+9)(s-3)}+\cfrac{(s+4)(s^2+9)}{(s+4)(s-3)(s^2+9)} = \cfrac{2s^3+4s^2+21s-27}{(s^2+9)(s-3)(s+4)}$

Dividing both sides by $[s^2+s-12]$ AKA $(s+4)(s-3)$ to get $X(s)$ by itself

We get

$X(s) = \cfrac{2s^3+4s^2+21s-27}{(s^2+9)(s-3)^2(s+4)^2}$

Now I Believe I'm supposed to use partial fraction decomposition here to make taking the Inverse LaPlace Transform easier so...

$\cfrac{2s^3+4s^2+21s-27}{(s^2+9)(s-3)^2(s+4)^2} = \cfrac{As+B}{s^2+9} + \cfrac{C}{s-3} + \cfrac{D}{(s-3)^2} + \cfrac{E}{(s+4)} + \cfrac{F}{(s+4)^2}$

Then multiplying both sides by the LHS denominator and simplifying we get

$2s^3+4s^2+21s-27 = (As+B)(s-3)^2(s-4)^2+C(s^2+9)(s-3)(s-4)^2+D(s^2+9)(s+4)^2+E(s+4)(s-3)^2(s^2+9)+F(s^2+9)(s-3)^2$

Here is where I get stuck. How do I solve for A,B,C,D,E,F? This seems ridiculously complex and seems like so much algebra that it makes me think that partial fraction decomposition isn't the right way. So what I've come to ask is partial fraction decomposition to right way to tackle this question? Is there an more efficient way besides partial fraction decomposition? If there's not and partial fraction decomposition is the correct way, How can I solve for A,B,C,D,E,F? Can someone help me determine the algebra, I'm not used to partial fractions being this long unless I've made an error somewhere...

Thank You For Any Help

答:
It's more simple to decompose fractions from this step:
$$X(s)[s^2+s-12] = \cfrac{3}{s^2+9}+\cfrac{1}{s+4}+\cfrac{1}{s-3}$$
$$X(s)(s+4)(s-3) = \cfrac{3}{s^2+9}+\cfrac{1}{s+4}+\cfrac{1}{s-3}$$
$$X(s)=\dfrac 1{(s+4)(s-3)}\left ( \cfrac{3}{s^2+9}+\cfrac{1}{s+4}+\cfrac{1}{s-3}\right)$$
And you have:
$$\dfrac 1{(s+4)(s-3)}=\dfrac 1 {7}\left (\dfrac 1{(s-3)}-\dfrac 1{(s+4)}\right)$$

Now what we have is this;
$$X(s)=\color {red}{\dfrac 37\left (\dfrac 1{(s-3)}-\dfrac 1{(s+4)}\right)\left ( \cfrac{1}{s^2+9}\right)}+\color {blue}{\dfrac 17\left (\dfrac 1{(s-3)^2}-\dfrac 1{(s+4)^2}\right)}$$
For the term in red you still need to decompose a little bit but for the term in blue you can already apply inverse Laplace Transform.
Note that:
$$\dfrac 1{(s-3)^2}=-\dfrac {d}{ds}\dfrac 1{(s-3)}$$
The inverse Laplace transform is therefore:
$$te^{3t}$$
And:
$$\dfrac 1{(s+4)^2}=-\dfrac {d}{ds}\dfrac 1{(s+4)}$$
The inverse Laplace transform is:
$$te^{-4t}$$

How to decompose this term ?
$$\dfrac 1{(s-3)}\cfrac{1}{(s^2+9)}$$
$$\dfrac {s+3}{(s^2-9)}\cfrac{1}{(s^2+9)}$$
$$(s+3)\dfrac {1}{(s^2-9)}\cfrac{1}{(s^2+9)}$$
$$(s+3)\left (\dfrac {A}{(s^2-9)}+\cfrac{B}{(s^2+9)}\right)$$
That's easy now ....
$$\dfrac {(s+3)}{18}\left (\dfrac {1}{(s^2-9)}-\cfrac{1}{(s^2+9)}\right)$$
$$\dfrac {1}{18}\left (\dfrac {1}{(s-3)}-\cfrac{s+3}{(s^2+9)}\right)$$


问:
Every Planar Graph is 6-colorable
Some proofs for the theorem that states that every planar graph is 6-colorable usually assume both the $G$ is simple and connected. The simple assumption is because coloring is not affected by loops or parallel edges, but what about connectivity?

I guess the reason behind connectivity is to connect each component separately because different components don't interact with each other with respect to coloring, but this justification for connectivity assumption is not clear for me, could you please rephrase or explain what does that mean?

答: If you can 6-color each connected component, then you can 6-color the whole graph, by taking the union of the 6-colorings. So you only need to prove the theorem for a connected graph, and then it extends to unconnected graphs as a trivial corollary.</p>

Title: Every Planar Graph is 6-colorable


问:
If $L^1$ is the space of lebesgue-integrable functions on $(X, \mathcal{F}, \mu)$, does it matter whether or not $\mu$ is complete?
Let $L_1(\mu)$ be the space of Lebesgue-integrable functions on $(X, \mathcal{F}, \mu)$. That is, $L^1$ is the set of equivalence classes of the relation $f \sim g \iff f = g\ \mu-$a.e., equipped with the metric $d(f, g) = \int |f - g|\ d\mu$.

Is it important whether or not $(X, \mathcal{F}, \mu)$ is a complete measure space? What are some properties of $L^1$ that hold only if $\mu$ is complete?

答:
Yes it sometimes depends.

For example, the Dominated Convergence Theorem is saying that if $f_n$ is dominated by $g$ and converges pointwise to $f$ then $f$ is integrable and
$$
\lim_n \int |f_n-f| d \mu =0
$$

When $\mu$ is complete, it is enough to assume that $f_n$ is dominated by $g$ and converges pointwise $\mu$-almost everywhere to $f$, but this is not true in general.

For a counterexample, pick a subset $E \subset X$ which is a subset of a null set but no measurable and observe that $f_n=0$ converges to $f=1_{E}$ $\mu$-almost everywhere.

[Side note I actually run into this issue once in a paper, luckly I could assume that my measure ws complete.]


答:
If you expand along the bottom row, then most of the $(n-1)\times(n-1)$ subdeterminants are $0$ because the $(n-1)\times(n-1)$ submatrices have their two right-most columns identical.

So expansion along the bottom row leaves only the last two terms: $$(-1)(n-1)\det(A_{n-1})+n\det(A_{n-1})$$ which is just $\det(A_{n-1})$. So inductively, $\det(A_{n})=\det(A_{n-1})=\cdots=\det(A_{1})=1$

ContentLicense: CC BY-SA 3.0

Title: Evaluation of a specific determinant.

Body:
Evaluate $\det{A}$, where $A$ is the $n \times n$ matrix defined by $a_{ij} = \min\{i, j\}$, for all $i,j\in \{1, \ldots, n\}$.

$$A_2
\begin{pmatrix}  1&  1\\
  1&  2
\end{pmatrix};
A_3 = \begin{pmatrix}  1&  1&  1\\
  1&  2&  2\\
  1&  2&  3
\end{pmatrix};
A_4 = \begin{pmatrix}  1&  1&  1&  1\\
  1&  2&  2&  2\\
  1&  2&  3&  3\\
  1&  2&  3&  4
\end{pmatrix};
A_5 = \begin{pmatrix}  1&  1&  1&  1&  1\\
  1&  2&  2&  2&  2\\
  1&  2&  3&  3&  3\\
  1&  2&  3&  4&  4\\
  1&  2&  3&  4&  5
\end{pmatrix}$$

$$A_6 = \begin{pmatrix}  1&  1&  1&  1&  1&  1\\
  1&  2&  2&  2&  2&  2\\
  1&  2&  3&  3&  3&  3\\
  1&  2&  3&  4&  4&  4\\
  1&  2&  3&  4&  5&  5\\
  1&  2&  3&  4&  5&  6
\end{pmatrix};
A_7 = \begin{pmatrix}  1&  1&  1&  1&  1&  1&  1\\
  1&  2&  2&  2&  2&  2&  2\\
  1&  2&  3&  3&  3&  3&  3\\
  1&  2&  3&  4&  4&  4&  4\\
  1&  2&  3&  4&  5&  5&  5\\
  1&  2&  3&  4&  5&  6&  6\\
  1&  2&  3&  4&  5&  6&  7
\end{pmatrix}
$$


问:
Show that the Niemytzki's tangent disc topology is completely regular.
I need to show that the Niemytzki's tangent disc topology is completely regular. In by attempt I could show that there is a Urysohn function for a point $b$ on the upper half plane $\{(x,y): x\in\mathbb{R},y>0\}$ and a closed set $F$ that does not contain $b$. However I am stuck in the case where $b$ is a point on the line $y=0$. Could someone please help? Thanks. </p>

答:
As it is a Hausdorff space (why?) it suffice to show that it is functionally regular,

Let $F$ be a closed non-empty subset of $(Y,\mathcal{T}_N)$ (being the upper half place with the Niemytzki's topology) and $p=(x,y)$ a point not in $F$, we want to find a continuous function which completely separates $F$ and $p$, we consider the two cases;

If $p\in Y'$ (denote by Y' the open upper half plane and D the x-axis line), recall that $Y\setminus F$ is open, so there exist an open $U$ in the canonical topology such that $p\in U \subseteq Y\setminus F$, so $Y\setminus U$ is closed in both the Moore topology and the canonical topology, following regularity of the latter there exists a Urysohn function $f$ such that $f(Y\setminus U)= 0,\ and\ f(p)= 1$, $f$ is continuous relatively to the canonical topology, constant on $D$ as $f(D)=0$ and vary only in $U\subset Y'$ hence continuous with respect to $\mathcal{T}_N$ (why?), as $F\subseteq Y\setminus U$, the same function is a Urysohn function for $F$ and $p$.

If $p\in D$, we can have an element from the basis that contains $p$ and does not intersect $F$ (since $F$ is closed), say $B_p= B((x_0,\epsilon), \epsilon)\cup \{p\}$, let us define a function $f$ that satisfy $f(p)= 0$, $f(q)= 1\ if\ q\notin B_p$, and at any point $(x,y)$ inside $B_p$\ ; $f(x,y)= \frac{1}{2 y\epsilon}\big((x-x_0)^2+ y^2\big) $.

We claim that $f$ is continuous, to see this, notice that $f^{-1}([0,\alpha))= \{p\}\cup B\big((x_0,\alpha\epsilon), \alpha\epsilon\big)$ (why?) is open, and $f^{-1}((\alpha,1])= Y\setminus \overline{B\big((x_0,\alpha\epsilon), \alpha\epsilon\big)}$ is also open,

we conclude that $f$ is a Urysohn function for $p$ and $F$.


答:
(1) If you have a single measurable function (or countably many measurable functions) then anything you define from them is again measurable.  So completeness is not needed.

(2) But if you have uncountably many, that no longer works.  For example, in probability theory, a "stochastic process" may be a collection $(f_t)_{t \ge 0}$ of random variables indexed by the nonnegative reals.  And you may want some uncountable constructions still to yield something measurable.  That is where completeness is important.  Example:
$$
T(x) = \inf\{t \; : \; |f_t(x)| \ge 1\}
$$
We may want $T$ to be measurable.  In some cases, we can provide natural conditions on the process $(f_t)$ that will insure this when the measure is complete.

(3)  If $E \subseteq \mathbb R^2$ is a Borel set, then the projection onto the $x$-axis is
$$
p(E) = \{x \in \mathbb R\;:\;\exists y, (x,y \in E\} .
$$
If $E$ is Borel in $\mathbb R^2$, then $p(E)$ need not be a Borel set in $\mathbb R$.  But in fact $p(E)$ is measurable with respect to any complete Borel measure on $\mathbb R$.

Title: If $L^1$ is the space of lebesgue-integrable functions on $(X, \mathcal{F}, \mu)$, does it matter whether or not $\mu$ is complete?

Body:
Let $L_1(\mu)$ be the space of Lebesgue-integrable functions on $(X, \mathcal{F}, \mu)$. That is, $L^1$ is the set of equivalence classes of the relation $f \sim g \iff f = g\ \mu-$a.e., equipped with the metric $d(f, g) = \int |f - g|\ d\mu$.

Is it important whether or not $(X, \mathcal{F}, \mu)$ is a complete measure space? What are some properties of $L^1$ that hold only if $\mu$ is complete?


问:
Is there a way to prove that the 2-norm of a matrix is greater than the largest element?
I would like to prove

$$\max_{ij} |A_{ij}| \leq \| A \|_2$$

Any help would be greatly appreciated!

答:
The notation $\|A\|_2$ usually refers to the Frobenius norm, also called the Hilbert-Schmidt norm, defined to be
$$
\|A\|_2 = \sqrt{\sum_{i,j}|a_{i,j}|^2}.
$$
If this is what the OP has in mind, then the question is trivial since it is obvious that
$$
|a_{k,l}|^2 \leq \sum_{i,j}|a_{i,j}|^2,
$$
for every $k$ and $l$.

Most other matrix norms share this property, so it would be nice to know exactly which matrix norm is of interest to the OP.


问:
natrual deduction is this model wrong answer?
This model entails premise but not the conclusion

But in my opinion if we use the b also in conclusion we have a proof that this premise entails conclusion or?

答:
No, the answer given is correct.

If I understand the picture correctly, the extracted countermodel is

$\mathcal{M} = \langle \{a, b\}, P \mapsto \{a\}, S \mapsto \text{False} \rangle$.

Then $\mathcal{M} \models \exists x (P(x) \to S)$ because with $x \mapsto b$, $P(x)$ is false so $P(x) \to S$ is true so $\exists x (P(x) \to S)$ is true, but $\mathcal{M} \not \models \exists x P(x) \to S$ because with $x \mapsto a$, $P(x)$  is true so $\exists x P(x)$ is true but $S$ is false so $\exists x P(x) \to S$ is false.

For $x \mapsto b$, sure, $P(x)$ is false. But there exists an $x$ such that $P(x)$ is true, so $\exists x P(x)$ is true, whereas $S$ is false, so the conclusion $\exists x P(x) \to S$ is still false in that model.

Even if we find a model that makes both the premise and the conclusion true, e.g. by interpreting $S$ as $\text{True}$, that doesn't make the premise entail the conclusion. Entailment means that in all models in which the premises are true, the conclusion must be true as well. The existence of a single model in which the conclusion is false despite the premises being true already makes the sequent invalid; that there may be other models which do satisfy the conclusion is irrelevant. The one counter example above is enough to show the invalidity and hence unprovability of the sequent.


问:
Show that if $G$ and $H$ are groups then $Z(G \times H) = Z(G) \times Z(H)$
Show that if $G$ and $H$ are groups then $Z(G \times H) = Z(G) \times Z(H)$.

I'm not too sure how to begin, I know I can find some element from $Z(G \times H)$ that is also in $Z(G) \times Z(H)$ and vice versa but I don't know how to set up the initial procedure.

答:
Let $(g,h) \in G \times H$. If $(g,h) \in Z(G \times H)$, then $(g,h)(x,y) = (x,y)(g,h)$ for every $(x,y) \in G \times H$. We have that
$$(g,h)(x,y) = (gx,hy) = (xg,yh) = (x,y)(g,h).$$
Use it to show that $Z(G \times H) \subset Z(G) \times Z(H)$. The converse can be solved in a similar way.


问:
Context-Free Grammar for {w| w contains at least as many 0's as 1's and |w| >= 2} over the alphabet {0,1}.
I have first created the following CFG for the same number of 0's and 1's:

S → S0S1S | S1S0S | ε

To create more 0's than 1's, do I need to introduce a new variable such as the following?

S → SAS1S | S1SAS | 0A

A → 0A | 0

But it's not working!

答:
The problem with the rule
$$S \to SAS1S \mid S1SAS \mid 0A,$$
is that, as soon as a $1$ is introduced, you are automatically spawning many more zeros all around the $1$. For example, you couldn't make the string $001$ with the given rules. Instead, consider creating the rule $S \to \varepsilon$ to allow each $S$ to be replaced with no $0$s (and no $1$s).

The above adjustment means that the string no longer satisfies $|w| \ge 2$, which we now need to address. What we'll need to do is replace $S$ with a new state, and have the starting state feed into it when we're comfortable that at least two symbols will be produced. Here's my suggestion:

\begin{align*}
S &\to TAT1T \mid T1TAT \mid 0A \\
T &\to TAT1T \mid T1TAT \mid \varepsilon \\
A &\to 0A \mid 0.
\end{align*}


问:
Confusion about Jointly Gaussian

From here, it says that, linear combination of two Gaussian distribution, are always Gaussians.

However, Let  be standard normal and =±1 with probability 1/2 each, independently of . Let =. Then  is also standard normal, but =+ is exactly equal to zero with probability 1/2 and is equal to 2 with probability 1/2.

But (2) contradicts with (1). Am I missing anythings?

答:
The simplest case is when you take linear combinations of independent Gaussian random variables, these will be Gaussian. However, $X$ and $Y$ are not independent. A more general case in which linear combinations of correlated Gaussian random variables are Gaussian is if $X$ and $Y$ are bivariate Gaussian (aka jointly Gaussian). You can read more about this difference between (1) $X$ and $Y$ are both Gaussian and (2) $X$ and $Y$ are bivariate (jointly) Gaussian here: https://en.wikipedia.org/wiki/Multivariate_normal_distribution

Essentially, there are many ways for variables to correlated. Being jointly Gaussian random variables is a special way for Gaussian variables to be correlated that implies this property we've discussed. In fact, jointly Gaussian variables can actually be defined by this property that all linear combinations are normally distributed.

Title: Confusion about Jointly Gaussian


问:
Deduce that there exists a prime $p$ where $p$ divides $x^2 +2$ and $p≡3$ (mod 4)
I am revising for a number theory exam and have a question that I am struggling with, any help would be greatly appreciated.

First I am asked to show that for an odd number $x$, $x^2+2 ≡3$(mod 4).

I can do this part of the question, but next I am asked to deduce that there exists a prime $p$ where $p$ divides $x^2 +2$ and $p≡3$ (mod 4)

I am struggling to see how to attempt the second part and how the first part relates.

My thoughts so far are that I want to show $x^2≡-2$(mod p) ?
 And perhaps Fermat's Little Theorem could be of use here somehow?

Not sure if I'm barking up the wrong tree though.

Thanks in advance.

答: Our number $x^2+2$ is odd, and greater than $1$, so it is a product of odd not necessarily distinct primes. If all these primes were congruent to $1$ modulo $4$, their product would be congruent to $1$ modulo $4$. But you have shown that $x^2+2\equiv 3\pmod{4}$.</p>


答: Linear combinations of jointly Gaussians (also known as multivariate Gaussians) are always Gaussian; however, X and Y are not jointly Gaussian. (One of the easiest ways to define a joint Gaussian $X$ is to say that $X = AZ + b$ where $A$ is some matrix, $b$ is some vector and $Z$ is a vector of i.i.d. Gaussians.)

Title: Confusion about Jointly Gaussian

Body:
From here, it says that, linear combination of two Gaussian distribution, are always Gaussians.

However, Let  be standard normal and =±1 with probability 1/2 each, independently of . Let =. Then  is also standard normal, but =+ is exactly equal to zero with probability 1/2 and is equal to 2 with probability 1/2.

But (2) contradicts with (1). Am I missing anythings?


问:
A computation in the field of rational functions.
In Dummit and Foote 3 ed., Chapter 14, Section 2, Exercise 30, I am asked the following:

Let $ k $ be a field, $ k(t) $ the field of rational functions in the variable $ t $.  Define the maps $ \sigma $ and $ \tau \in Aut(k(t)/k) $ by
$$
\sigma f(t) = f \left( \frac{1}{1-t} \right) \quad \tau f(t) = f \left( \frac{1}{t} \right)
$$
for $ f(t) \in k(t) $.  Prove that the fixed field of $ \langle \tau \rangle $ is $ k \left( t + \frac{1}{t} \right) $, the fixed field of $ \langle \tau \sigma^2 \rangle $ is $ k(t(1-t)) $; determine the fixed field of $ \langle \tau \sigma \rangle $ and $ \langle \sigma \rangle $.

The only part of this I am struggling with is the fixed field of $ \langle \sigma \rangle $.  Call this fixed field $ E = k(s) $, where $ s = P(t) / Q(t) \in k(t) $ is some rational function.  Note, I am making an assumption here that $ E $ is of the form $ k(s) $, and so far cannot justify this a priori.  I have shown in a previous exercise from the last chapter that $ [k(t) : k(s)] = \max \left\{ \deg P(t), \deg Q(t) \right\} $, so, since $ k(t)/k(s) $ is a Galois extension ($k(s)$ being the fixed field of a subgroup of automorphisms), I expect
$$
\max \left\{ \deg P(t), \deg Q(t) \right\} = [k(t) : k(s)] = |\langle \sigma \rangle| = 3
$$
All I have been able to accomplish at this point was brute-force equation-solving by computer, setting
$$
s = \frac{a_3 t^3 + a_2 t^2 + a_1 t + a_0}{b_3 t^3 + b_2 t^2 + b_1 t + b_0}
$$
and solving the equations resulting from $ \sigma s = s $.  I thereby found the element $ s = \frac{t^3 - 3t + 1}{t(t-1)} $.  Hence I am inclined to conclude that $ k \left( \frac{t^3 - 3t + 1}{t(t-1)} \right) $ is the fixed field of $ \langle \sigma \rangle $.  This approach feels inelegant, and I would like to know what tools I might have used to avoid an unsatisfying and opaque computer-search.

答:
For $G$ a finite subgroup of $Aut(k(t)/k)$ then the fixed subfield is $k(t)^G=k(a_0(t),\ldots,a_{|G|-1}(t))$ where $\prod_{g\in |G|} (X-g(t))=\sum_{m=0}^{|G|} a_m(t) X^m$.

Then take any non-constant coefficient $a_m(t)$, because each $g(t) = \frac{e_g t+b_g}{c_g t+d_g}$ is a Möbius transformation we get that $a_m(t)$ has at most $|G|$ poles counted with multiplicity (including the pole at $\infty$), thus $[k(t):k(a_m(t))]\le |G|=[k(t):k(t)^G]$ which implies that $$k(t)^G=k(a_m(t))$$

Edit by OP: for this problem, the technique produces the element $ a_2(t) = \frac{t^3 - 3t + 1}{t(t-1)} $, reifying the computer calculations.


问:
Proof for $\sin(x) > x - \frac{x^3}{3!}$
They are asking me to prove $$\sin(x) > x - \frac{x^3}{3!},\; \text{for} \, x \, \in \, \mathbb{R}_{+}^{*}.$$ I didn't understand how to approach this kind of problem so here is how I tried:
$\sin(x) + x -\frac{x^3}{6} > 0 \\$
then I computed the derivative of that function to determine the critical points. So:
$\left(\sin(x) + x -\frac{x^3}{6}\right)' = \cos(x) -1 + \frac{x^2}{2} \\ $
The critical points:
$\cos(x) -1 + \frac{x^2}{2} =  0 \\ $
It seems that x = 0 is a critical point.
Since $\left(\cos(x) -1 + \frac{x^2}{2}\right)' =  -\sin(x) + x \\ $
and $-\sin(0) + 0 = 0 \\$
The function has no local minima and maxima. Since the derivative of the function is positive, the function is strictly increasing so the lowest value is f(0).
Since f(0) = 0 and 0 > 0 I proved that $ \sin(x) + x -\frac{x^3}{6} > 0$. I'm not sure if this solution is right. And, in general, how do you tackle this kind of problems?

答:
Check that
$$\int_0^x (x-t)^2 \sin^2\frac{t}{2} d\,t= \sin x - (x - \frac{x^3}{6})$$

WA link


问:
Prove the following relation:
I must prove the relation $$\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^k}=2\sum_{k=0}^n\binom{n+k}{k}\frac1{2^k}.$$

I got this far before I got stuck:

$\begin{eqnarray*}
\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^k} & = & \sum_{k=0}^{n+1}\left\{\binom{n+k}{k}+\binom{n+k}{k-1}\right\}\frac1{2^k}\\
& = & \sum_{k=0}^n\binom{n+k}{k}\frac1{2^k}+\sum_{k=0}^n\binom{n+k}{k-1}\frac1{2^k}+\binom{2n+1}{n}\frac1{2^k}.
\end{eqnarray*}$

If I can combine the second and third terms and get something same as first term, I am done but I could not do that.

答:
The following is proving a scaled version of the required equation:

$$\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^{n+k+2}}=\sum_{k=0}^n\binom{n+k}{k}\frac1{2^{n+k+1}},$$

and based on my answer in another question.

The product being summed is the PMF of the negative binomial distribution, which considers flipping a fair coin repeatedly until having a fixed number of heads, and counts the number of tails that showed up.

The left hand side models flipping a fair coin repeatedly until having $n+2$ heads, then add the probabilities of having $n+1$ or fewer tails. Let $\Pr(A)$ be the sum of probabilities.

Then $1-\Pr(A)$ is the probability of having the $(n+2)$th tail while only having $n+1$ heads or fewer.

For a fair coin, head and tail are symmetric, and so

$$\Pr(A) = 1-\Pr(A) = \frac 12$$

The right hand side is similar, this time modelling flipping a fair coin repeatedly until having $n+1$ heads, and adding the probabilities of having $n$ or fewer tails.

So both the left and right hand sides are equal to $\frac12$, which is also the "base case" where $n=0$:

$$\sum_{k=0}^0 \binom {0+k}k \frac{1}{2^{0+k+1}} = \binom00\frac{1}{2} = \frac12$$

Title: Prove the following relation:


问:
Could someone help me calculate the areas in this map?
Map of my book - Please click here to see it

So, I'm writing a book and I'm trying to be really detailed. Right now I'm writing a wiki of it and I want to specify the areas of continents and cities.

But I'm having a lot of trouble doing this. I'm not good at math and the shapes are complex, so I won't find an equation to solve everything.

Is there a way to solve this? More than just the answers, I really need to understand so I can calculate this alone later.

If it helps, each one of the rectangles in the grid have 3 million km² (I built it in Illustrator with 1cm height and 2cm width).

答:
The way the map looks (equidistant latitudes), you should probably weigh pixels at latitude $\phi$ with a factor of $\cos\phi$. That is (as your image is $539\times 508$ pixels):</p>

```<code>area = 0.0
planetsurface = 0.0
for y=0 to 507
   weight = sin(pi*(y+0.5)/508)
   planetsurface = planetsurface + weight*539
   for x=0 to 538
      if PixelAt(x,y) is land then
         area = area + weight
      endif
   next
next
print area*100/planetsurface, " percent of the surface are land"
</code>```

Body:
The way the map looks (equidistant latitudes), you should probably weigh pixels at latitude $\phi$ with a factor of $\cos\phi$. That is (as your image is $539\times 508$ pixels):

area = 0.0
planetsurface = 0.0
for y=0 to 507
   weight = sin(pi*(y+0.5)/508)
   planetsurface = planetsurface + weight*539
   for x=0 to 538
      if PixelAt(x,y) is land then
         area = area + weight
      endif
   next
next
print area*100/planetsurface, " percent of the surface are land"


问:
$\lim_{x \to 4} \frac{\sqrt{5x-4-x^2}}{\lfloor x \rfloor - 3}= 0$
Is the limit $\lim_{x \to 4} \frac{\sqrt{5x-4-x^2}}{\lfloor x \rfloor  - 3}= 0$?

I was trying to solve this using the extremes.

$$\lim_{x \to 4^{+}} \frac{\sqrt{5x-4-x^2}}{\lfloor x \rfloor  - 3}$$ is equal to zero because the denominator is different from 0 but
$$\lim_{x \to 4^{-}} \frac{\sqrt{5x-4-x^2}}{\lfloor x \rfloor  - 3}$$ is $\frac{0}{0}$

so I thought in using L'Hopital but I was doubtful if i can, because the floor function is not continuous in $[n,n+1]$.

答:
The expression in the square root needs to be non-negative:

$$5x-4-x^2=\tfrac94-(x-\tfrac52)^2\geq0$$

$$(x-\tfrac52)^2\leq\tfrac94$$

$$|x-\tfrac52|\leq\tfrac32$$

$$-\tfrac32\leq x-\tfrac52\leq\tfrac32$$

$$1\leq x\leq4$$

On the other hand, the denominator needs to be non-zero:

$$\lnot\big[\,\lfloor x\rfloor-3=0\,\big]$$

$$\lnot\big[\,3\leq x<4\,\big]$$

$$x<3\quad\text{or}\quad x\geq4$$

Combining these, we find that the domain of the function is $[1,3)\cup\{4\}$; that is,

$$1\leq x<3\quad\text{or}\quad x=4$$

There is nothing in the domain around $x=4$, so the left and right limits are both undefined.


问:
Does the categorical notion of a coequalizer coincide with the usual quotient objects considered in algebra?
I heard that the category-theoretic formulation of "quotient object" is that of a coequalizer. Instead of taking the quotient of, say, a group $G$ by some normal subgroup, one takes the quotient (really, the coequalizer) by some diagram of the form $H\rightrightarrows G$ consisting of two arrows from some group $H$ to $G$.

I wonder: Can each quotient $G/N$ of $G$ by some normal subgroup $N$ be written as the coequalizer of some diagram $H\rightrightarrows G$? What about the converse: Can each coequalizer be written as the quotient by some normal subgroup?

Of course, I'm not only interested in groups, but I also want to know whether the same things are true for rings/ideals, modules/submodules, and so on.

答:
Yes. In a fairly general class of categories, we can define an internal equivalence relation or congruence on an object $X$ to be a subobject $R \subseteq X \times X$ satisfying the axioms of an equivalence relation, suitably generalized so that they can be stated categorically (see the link for details). A congruence, being a subobject of $X \times X$, is equipped with a pair of maps to $X$, and the quotient by a congruence is the coequalizer of this pair of maps.

For groups you can check that every congruence on a group $G$ has the form

$$\{ (g, h) \in G \times G : g^{-1} h \in N \}$$

for a unique normal subgroup $N$. If we write the congruence as an equivalence relation $\sim$ on $G$, then $N = \{ n \in G : n \sim e \}$. So talking about congruences is equivalent to talking about normal subgroups, and furthermore the quotient by the congruence (the coequalizer above) associated to a normal subgroup is the quotient in the usual sense.

The point of writing things this way instead of in terms of cokernels is that this recipe generalizes to settings other than groups. For example, for rings you can check that every congruence on a ring $R$ has the form

$$\{ (r, s) \in R \times R : r - s \in I \}$$

for a unique two-sided ideal $I$, and again quotienting by this ideal corresponds to taking the coequalizer. For more general objects such as monoids and semirings it's not possible to do this kind of reduction and we really just need to work with congruences.


问:
Probability problem. Coin tosses.
I have the following problem:

A coin is tossed until it gives either 10 heads or 10 tails. Player A bets on 10 heads and player B bets on 10 tails. The game is unexpectedly interrupted after 15 tossings with 8 heads and 7 tails observed. What would be the fair ratio to split the prize pool between player A and B?

I know the answer to that question and I know how to calculate the probability of 8 heads after 15 tosses. However I do not understand how to calculate the answer. Any help is highly appreciated

Update: as Lulu mentioned below those 15 toses are irrelevant. We have to try 4 toses more. If we enumerate them we will get 11 possibilities  to get 2 H and 5  possibilities to get 3 T. Thus we should divide the prize fund in 11/5 proportion.

答:
The idea is that you should divide the prize according to the probability that a player would eventually have won.  If, say, the probability that $A$ would eventually have won was $p$, then you give $p$ times the prize money to $A$.

Thus:  The question here is:  "What is the probability that you get $2$ Heads before you get $3$ Tails?"

Note that this must be decided in $4$ tosses.  It follows that it is equivalent to the question "What is the probability that you get at least $2$ Heads in $4$ tosses?"  That is easily seen to be $\frac {11}{16}$, so that is the portion of the prize you should give to $A$.

Title: Probability problem. Coin tosses.


问:
Proving convergence in probability of discrete variables
I'm working with the following sequence of independent random variables:

$$X_1 \equiv 0$$
and for $n \geq 2$:
$$
P(X_n = j) =\begin{cases}
  \frac{1}{n^3}&j = \pm1, \pm2, ..., \pm n\\
  1-\frac{2}{n^2}&j=0
\end{cases}
$$

And trying to prove the average converges in probability to $0$:
$$
S_n = \sum_{i=1}^{n} \frac{X_i}{n} \xrightarrow{P} 0
$$

My main issue is I don't exactly know how to work with $S_n$, not about the convergence itself. By definition of convergence in probability, I'm trying to prove $P(|S_n| > \varepsilon ) \xrightarrow{} 0$ for all $\varepsilon > 0$. I tried finding the pmf of $S_n$ by using the convolution operator for distributions, but has gotten me no where.

Any tips/hints on how to work with $S_n$ will be greatly appreciated. Thanks in advance.

答: Note that $|S_n| \le \frac{1}{n} \sum_j{ |X_j|} $. But $E(|X_j|) = \frac{1+j}{j^2} \le \frac{2}{j}$, so $E(|S_n|) \le \frac{1}{n} \sum_{j=1}^n{ \frac{2}{j}} = 2\frac{ln(n)}{n}(1+o) \rightarrow 0$. Apply Markov's inequality to the result.


问:
Definition: Gauss Sum - Where is the error?
In my algebraic number theory lecture we defined Gauss sums as follows. However, I am quite unsure whether this definition is correct. My intuition says "there is a mistake somewhere". I tried double-checking with Pierre Samuels book (§5.5, page 78), but Samuel uses slightly different notation.

If there is indeed a mistake in my defintion, would somebody be so kind and give me a clean (standard) defintion of Gauss sums.

答: I would think that $\zeta$ should be a primitive $q$-th (not $p$-th) root of unity.  In characteristic $p$, the only $p$-th root of unity is unity.  That's because $x^p-1=(x-1)^p$ in characteristic $p$; the cross-terms that you'd ordinarily expect have coefficients that are divisible by $p$ and thus vanish in characteristic $p$.</p>


问:
How to compute the relative cohomology groups of a simple example?
I'm starting to learn homology and cohomology, and I would like to try an example to understand more concretely what's going on. Let $T$ be the 2-torus, and $S^1$ the circle. Let $i: S^1 \to T$ be an embedding. I want to compute the relative cohomology groups $H^k(T,S^1)$ in two cases: the embedding $i$ is trivial (contractible), and $i$ maps the circle once around the handle of the torus. I'm working in $\mathbb{Z}$ coefficients.

I'm reading through Hatcher to try to solve this, and I see two ways to approach it. Using the long exact sequence of cohomology, and doing everything from scratch using cocycles and so on. I tried the long exact sequence but I couldn't see how to use the specific embeddings to get an answer. And I'm pretty much lost when it comes to doing it from scratch. All I know is that in homology, the trivial embedding of the circle is a boundary since it is the manifold-boundary of a submanifold (a disc), while the non-trivial embedding is a cycle that is not a boundary. So the homology class of the first is $0$ while that of the second is non zero. I don't understand the cohomology counterpart to these facts.

Could I get some guidance for this problem?

答:
Alright, with the help of Eric Wofsey's hints, I think I have an answer to my question.

First we have that $H^k(T,S^1) \cong \tilde{H}^k (T/S^1)$ for all $k$, where the quotient map is induced from the embedding of the circle. (I cannot find a reference for this fact.) Note that whenever a (nice enough?) space is path connected, the $0$th reduced homology is always $0$.

Now in the first case, let $S^1$ be embedded in $T$ such that it is contractible. Then $T/S^1$ is a torus wedge a $2$-sphere. Call this space $X_1$. Using the Mayer-Vietoris sequence, we get that $\tilde{H}^k (X_1) \cong \tilde{H}^k (T) \oplus \tilde{H}^k (S^2)$. The summands are widely known, so we get that

\begin{equation}
H^k(T,S^1) \cong
  \begin{cases}
  0 & \text{if}\ k=0\\
  \mathbb{Z}^2 & \text{if}\ k=1,2\\
  0 & \text{if}\ k \geq 3.
  \end{cases}
\end{equation}

Next, the second case, where $S^1$ is wrapped once around the handle of the torus, we get that $T/S^1$ is the pinched torus, and the wiki contains its cohomology groups. Changing the $0$th to be $0$, we get that

\begin{equation}
H^k(T,S^1) \cong
  \begin{cases}
  0 & \text{if}\ k=0\\
  \mathbb{Z} & \text{if}\ k=1,2\\
  0 & \text{if}\ k \geq 3.
  \end{cases}
\end{equation}

The next challenge would be to apply this to other non-trivial loops in the torus.


问:
Confusion Regarding Notation in a Proof about two Well-Ordered Sets
This partial proof is taken from A Course in Mathematical Logic for Mathematicians by Yu. I. Manin.

Lemma. Let X and Y be two well-ordered sets. Then exactly one of the following alternatives holds:

(a) X and Y are isomorphic

(b) X is isomorphic to an initial segment in Y.

(c) Y is isomorphic to an initial segment in X.

PROOF.

Let X be well-ordered, and let $f:X \to X$ be a monotonic map. Then, for all $Z \in X$ we have $f(Z) \geq Z$. Therefore, $X$ is not isomorphic to any of its initial segments, $\hat X_1$. Now, let $X$ and $Y$ be well-ordered. We set $f = \{<X_1, Y_1>| X_1 \in X, Y_1 \in Y, \text{and there exists an isomorphism of } \hat X_1 \text{ with } \hat Y_1 \}$. First of all, $f$ is a graph of a one-to-one mapping of $\text{pr}_1f$ onto $\text{pr}_2 f$. In fact, if $X_1 \neq X_2$, say $X_1 < X_2$, then by the previously shown result, $\hat X_1$ is not isomorphic to $\hat X_2$...

I omitted parts of the proof that don't seem to be pertinent to the question (it's a bit lengthy). I'm confused by the notation $\text{pr}_1f$ and $\text{pr}_2f$. What does this mean? I don't think that it's defined anywhere else in the book*.

Also, this is unrelated, but can somebody provide me a link to an alternate proof (that's around the same level as this one) of this lemma? I like to see multiple proofs to make sure I really understand what's going on.

*The proof that I'm currently reading is in an appendix to a later chapter that the author recommends reading before continuing the book, so perhaps the notation is defined somewhere else in the book I just haven't gotten to yet (but I did look and didn't see anything)

答:
Here $\operatorname{pr}$ is for projection: $\operatorname{pr}_1$ is the projection map to the first factor,

$$\operatorname{pr}_1:X\times Y:\langle x,y\rangle\mapsto x\,,$$

that takes each ordered pair in $X\times Y$ to its first component, and $\operatorname{pr}_2$ is the projection map to the second factor,

$$\operatorname{pr}_2:X\times Y:\langle x,y\rangle\mapsto y\,.$$

Note that here $f$ is a subset of $X\times Y$, so this does make sense. In more familiar terms, $\operatorname{pr}_1f$ is simply the domain of $f$, and $\operatorname{pr}_2f$ is the range of $f$.


问:
Prove that $G= \ker(f) \times{\rm im}(f)$
Let $f:G\to G$ be a homomorphism such that $f\circ f = f$. Prove that if $im(f)$ is normal in $G$, then $G = \ker(f) \times{\rm im}(f)$. Hint: If $x\in G$, then $x = [xf(x^{-1})]f(x)$.

My first try:

We know ${\rm im}(f)\in G$. Let $g\in G$ but $g\not\in {\rm im}(f)$.

Let $x\in {\rm im}(f)$. Since ${\rm im}(f)$ is normal in $G$, we know:
$$gxg^{-1} = x$$
Applying $f$ to both sides, we get:
$f(gxg^{-1}) = f(x)$ and since $f$ is a homomorphism, we know:
$$f(gxg^{-1}) = f(g)f(x)f(g^{-1})$$
So $f(g)f(x)f(g^{-1}) = f(x)$.

I need to some how prove that $f(g) = e$, so that $g\in \ker(f)$, does anyone know what to do next, or if this method will even work?

答:
The hint shows that $\rm{im}f\cdot\rm{ker}f=G$, and follows from the homomorphism property.

It remains to show $\rm{im}f\cap\rm{ker}f=\{e\}$.  Let $h\in\rm{im}f\cap\rm{ker}f$.  Then $h=f(g)=f\circ f(g)=f(h)=e$.

Since both the image and the kernel of $f$ are normal, we have the result.


答:
First: since $f\circ f= f$, then I claim that $\mathrm{Im}(f)\cap\ker(f)=\{e\}$. To see this, note that $\mathrm{Im}(f)\cap\ker(f)$ certainly contains $e$. For the converse inclusion, assume that $x\in\mathrm{Im}(f)\cap\ker(f)$. Then $f(x)=e$, and there exists $g\in G$ such that $f(g)=x$. But now, since $f\circ f=f$, applying $f$ again, we conclude... what?

Now, since $\mathrm{Im}(f)$ is normal by assumption, and $\ker(f)$ is normal always, and $\mathrm{Im}(f)\cap\mathrm{ker}(f)=\{e\}$, this tells us that $\langle \mathrm{Im}(f),\mathrm{ker}(f)\rangle=(\ker(f)) (\mathrm{Im}(f))\cong \mathrm{Im}(f)\times\mathrm{Im}(f)$.

So the only remaining thing is to show that every element of $G$ can be written as the product of something in the kernel and something in the image. That’s the point of the hint. Write
$$x = x f(x)^{-1}f(x) = (xf(x^{-1}))f(x).$$
Clearly $f(x)\in\mathrm{Im}(f)$. So, we just need to show that $xf(x^{-1})\in \ker(f)$. To do that, we just need to apply $f$ and see if we get $e$. Do we?

Title: Prove that $G= \ker(f) \times{\rm im}(f)$

Body:
Let $f:G\to G$ be a homomorphism such that $f\circ f = f$. Prove that if $im(f)$ is normal in $G$, then $G = \ker(f) \times{\rm im}(f)$. Hint: If $x\in G$, then $x = [xf(x^{-1})]f(x)$.

My first try:

We know ${\rm im}(f)\in G$. Let $g\in G$ but $g\not\in {\rm im}(f)$.

Let $x\in {\rm im}(f)$. Since ${\rm im}(f)$ is normal in $G$, we know:
$$gxg^{-1} = x$$
Applying $f$ to both sides, we get:
$f(gxg^{-1}) = f(x)$ and since $f$ is a homomorphism, we know:
$$f(gxg^{-1}) = f(g)f(x)f(g^{-1})$$
So $f(g)f(x)f(g^{-1}) = f(x)$.

I need to some how prove that $f(g) = e$, so that $g\in \ker(f)$, does anyone know what to do next, or if this method will even work?


问:
Cyclic normal subgroup of perfect group is in the center
I've been trying for a while to solve an exercise/prove a proposition, which at first seemed elementary, but now I even doubt if it's a true proposition. The proposition is:

Let $G$ be a perfect group and let $K$ be a cyclic, normal subgroup of $G$. Show that $K$ is contained in the center of $G$ (i.e. $Z(G)$).

Obviously it's enough to show that the generator of $K$ is in the center, i.e. it commutes with every element of the group, but I can't figure out why that's true.

I thought about using Grun's lemma and show that if the generator of $K$ wasn't in the center, then its $Z(G)$-coset would be in the center of $G/Z(G)$, but it turned out to be the same approach as the first one.

Then I thought about showing that the orbit (under the action of conjugation) of the generator is a singleton and I found that $\sqrt{|G|}$ is a lower bound for the size of the stabilizer $|Stab(x)|$, where $x$ is the generator, but that's just for a finite group $G$, and I couldn't actually find a higher bound (it'd be fine if I could show that x is actually stable under conjugation from any $g\in G$).

Thank you in advance for any help.

答:
You need to take advantage of the fact that the normal subgroup is cyclic somehow.

So let $H$ be cyclic and normal in $G$, where $G$ is perfect. Then by the $N/C$ Theorem, $G/C_G(H)$ is isomorphic to a subgroup of $\operatorname{Aut}(H)$. But since $H$ is cyclic, $\operatorname{Aut}(H)$ is abelian. Thus $G/C_G(H)$ is abelian which implies that $C_G(H) \geq G'=G$. Therefore, $C_G(H)=G$ and that is equivalent to $H \leq \operatorname{Z}(G)$.


问:
Compute $\int_{|z| = r} \frac{|dz|}{|z-a|^2}$ where $|a| \neq r$
Compute $\int_{|z| = r} \frac{|dz|}{|z-a|^2}$ where $|a| \neq r$

I know that $|dz| = -ir\frac{dz}z$ and that $z\overline{z} = r^2$. So I get

$\int_{|z| = r} \frac{|dz|}{|z-a|^2} = -i\int_{|z| = r} \frac{rdz}{(z-a)(\overline{z} - \overline{a})} $

But I don't know how to proceed from here

答:
The denominator of your integrand is missing a factor of $z$. The denominator should be $z(z-\alpha)(\bar{z}-\bar{\alpha})$. The idea now is to multiply this out and replace $z\bar{z}$ with $r^2$. After some multiplication and factoring, the denominator of your integrand becomes $(r^2-\bar{\alpha}z)(\alpha-z)$.

Thus, your integral takes the form:
$$\int_{|z|=r}\frac{-ir}{(r^2-\bar{\alpha}z)(\alpha-z)} = \int_{|z|=r}\frac{ir}{(r^2-\bar{\alpha}z)(z-\alpha)} = \int_{|z|=r}\frac{ir}{(r^2-\bar{\alpha}z)}\cdot\frac{1}{(z-\alpha)}$$

The answer now depends on whether $|\alpha| > r$ or $|\alpha| < r$. In the latter case, use Cauchy's integral formula.


问:
Prove that for all $n \in \mathbb{N}$, $\sum_{k=0}^{n-1}{n+k-1\choose k}\frac 1{2^{n+k}}=\frac12$
$$  \sum_{k=0}^{n-1}{n+k-1\choose k}\frac 1{2^{n+k}}=\frac12$$

To be honest, I can't really get started,

I would like to ask everyone to give me an idea of how to solve it, give me a starting push, thank you.

Body:
we rewrite the sum like this ;$$\sum_{k=0}^{k=n-1}\binom{n+k-1}{n-1}\frac{1}{2^{n+k}}$$

this is just the coefficient of $x^{n-1}$ in expansion $$\frac{{(1+x)}^{n-1}}{2^n}+\frac{{(1+x)}^n}{2^{n+1}}...+\frac{{(1+x)}^{2n-2}}{2^{2n-1}}$$

Recognise this as a GP: hence we seek the coefficient of $x^{n-1}$ in $$\frac{1}{2^{n-1}}{(1+x)}^{n-1}\frac{1-{(\frac{x+1}{2})}^n}{1-x}$$

or $$\frac{{(1+x)}^{n-1}(1+x+x^2..)-{(1+x)}^{2n-1}(1+x+x^2+x^3....)}{2^{2n-1}}$$ The coefficent is $$\frac{\left(\binom{n-1}{0}+\binom{n-1}{1}..+\binom{n-1}{n-1}\right)-\left(\binom{2n-1}{0}+\binom{2n-1}{1}+...\binom{2n-1}{n-1}\right)}{2^{2n-1}}$$
$$=\frac{2^{2n-1}-\frac{1}{2}2^{2n-1}}{2^{2n-1}}=\frac{1}{2}$$


问:
Probability: Theory and Examples Version 5a Durrett Exercises 4.8.5

4.8.5. Variance of the time of gambler's ruin. Let $\xi_{1}, \xi_{2}, \ldots$ be independent with $P\left(\xi_{i}=1\right)=p$ and
$P\left(\xi_{i}=-1\right)=q=1-p$ where $p<1 / 2 .$ Let
$S_{n}=S_{0}+\xi_{1}+\cdots+\xi_{n}$and let $V_{0}=\min \left\{n \geq
> 0: S_{n}=0\right\}$. Theorem 4.8 .9 tells us that $E_{x} V_{0}=x /(1-2
> p)$ The aim of this problem is to compute the variance of $V_{0} .$ If
we let $Y_{i}=\xi_{i}-(p-q)$ and note that $E Y_{i}=0$
and$$\operatorname{var}\left(Y_{i}\right)=\operatorname{var}\left(X_{i}\right)=E
> X_{i} u^{2}-\left(E X_{i}\right)^{2}$$then it follows that
$\left(S_{n}-(p-q) n\right)^{2}-n\left(1-(p-q)^{2}\right)$ is a
martingale. (a) Use this to conclude that when $S_{0}=x$ the variance
of $V_{0}$ is$$x \cdot \frac{1-(p-q)^{2}}{(q-p)^{3}}$$

I know that there are some answers to this question on the Internet, but what I am curious about is not his answer, but how this formula is derived:
$\left(S_{n}-(p-q) n\right)^{2}-n\left(1-(p-q)^{2}\right)$. I would be grateful if anyone could help me.

答:
Let $ \{ \zeta_i \} $ be a sequence of iid random variables with $ \mathbb{E} \zeta = 0 $ and $ \mathbb{E} \zeta^2 = \sigma^2 < \infty $. If we let $ T_n = \zeta_1 + \cdots + \zeta_n $, then $
T_n^2 - n\sigma^2 $ is a martingale. (See Example 4.2.2.) Now you should be able to show that this is actually a uniformly integrable martingale.

The formula you've mentioned can exactly be derived from this when we centralize $ \xi $'s.


问:
Constant of Integration of the Integrating Factor
I was reading my notes and at some point I write that in solving a first order linear DE using the integrating factor the final solution should have only one arvitrary constant arising from the integration of $Q(x)e^{\int P(x)dx}$ and so $e^{\int P(x)dx}$ does not contain an arbitrary constant.

Essentially what this means is that if $$\frac{dy}{dx}+P(x)\cdot y=Q(x) \Rightarrow y=e^{-\int P(x)dx}\int Q(x)e^{\int P(x)dx}dx,$$the only constant that would determine a particular solution would be the one arising from the second integral and not the one in the exponent. Why is this?

I understand that since this is a firstorder DE is should only involve one constant, but what about what I ask above?

If I define $\int P(x)dx=f(x) + C$ and substitute above, this would yield $$y=e^{-f(x)+c}\int Q(x)e^{-f(x)+c}dx=e^{-f(x)}e^{2c}\int Q(x)e^{-f(x)}dx$$and now also define $\int Q(x)e^{-f(x)}dx=g(x)+k$ we have$$y=e^{-f(x)}e^{2c}\left(g(x)+k\right)=e^{2c}\cdot e^{-f(x)}g(x)+ke^{2c}\cdot e^{-f(x)}$$
so wouldn't the constant of integration $c$ also affect the final solution?

Thanks in advance!

答:
Take a close look at your result.  Everything is divisible by $e^{2c}$, leaving $k$ as the only relevant constant.

In other words, if you do it without $c$, you can still get to your solution by just multiplying both sides by $e^2c$, which you are allowed to do anyway since it is a constant.


问:
Alternative form of Gradient Descent
We define the standard gradient descent as follows:

$$x^{(k+1)} = \arg\min_{x\in\mathbb{R}^n} f(x^{(k)}) + \langle \nabla f(x^{(k)}), x - x^{(k)}\rangle + \frac{1}{2\alpha_k} ||x - x^{(k)}||_2^2$$

Here, $x^{(k)}$ is the result of the $k$-th iteration of the gradient descent. Solving the above equation gives us $x^{(k+1)} = x^{(k)} - \alpha_k \nabla f(x^{(k)})$, which is our usual gradient descent.

How do we prove that the gradient descent problem is equivalent to the following version:

$$x^{(k+1)} = \arg\min_{x\in\mathbb{R}^n}\langle \nabla f(x^{(k)}), x - x^{(k)}\rangle $$ subject to $$||x - x^{(k)}||_2 \leq \alpha_k||\nabla f(x^{(k)})||_2$$

答:
Two more elementary solutions, as requested by OP.

We need to show that $x=x_k-\alpha g_k$ is the optimal solution to the following problem:
$$\min_x\ \langle g_k, x\rangle\quad \mbox{s.t. } \|x-x_k\| \le \alpha_k \|g_k\|.\tag{*}$$

Solution 1. Using Cauchy-Schwarz inequality we have
$$|\langle \alpha_k g_k, x-x_k\rangle| \le \alpha_k\|g_k\| \|x-x_k\| \le \alpha_k^2 \|g_k\|^2,$$
which implies $\langle g_k, x - x_k\rangle \ge -\alpha_k \|g_k\|^2$, or equivalently, $\langle g_k, x\rangle \ge \langle g_k, x_k\rangle -\alpha_k \|g_k\|^2$. It is easy to check that equality occurs iff $\alpha_k g_k = x_k - x$. We conclude that the optimal solution to $(*)$ is $x=x_k-\alpha g_k$. QED

Solution 2. Notice that
$$2\langle \alpha_k g_k, x-x_k\rangle + 2\alpha_k^2 \|g_k\|^2 \ge 2\langle \alpha_k g_k, x-x_k\rangle + \alpha_k^2 \|g_k\|^2 + \|x-x_k\|^2 = \|\alpha_k g_k + x-x_k\|^2 \ge 0,$$
with equality iff $\alpha_k g_k = x_k - x$. We conclude that the optimal solution to $(*)$ is $x=x_k-\alpha g_k$. QED

Title: Alternative form of Gradient Descent


问:
Is a linear map between two vector spaces with dimension $n$ that maps a set whose cardinality is $n$ to a basis in range bijective?
Let $f:V \to W$ be a linear map, where $V,W$ are finite-dimensional and $\dim V =\dim W = n$.

I know that if $f$ maps a basis of $V$ to a basis of $W$ then $f$ is bijective. But I wonder if $f$ maps a set $X \subseteq V$ whose cardinality is $n$ to a basis of $W$ implies $f$ is bijective.

This is my attempt:
Let $X =\{\vec{x}_1 \cdots,\vec{x}_n\} \subseteq V$. Suppose that $f(X) = \{f(\vec{x}_1) \cdots,f(\vec{x}_n)\}$ is a basis of $W$. To show $X$ is a basis, we need only to prove that $X$ is linearly independent. If it was not, then we could find two different group of scalars $a_i,b_i$ such that
$$
\sum_{i=1}^n{a_i\vec{x}_i} = \sum_{i=1}^n{b_i\vec{x}_i}
$$
, which would imply that
$$
\sum_{i=1}^n{a_if(\vec{x}_i)} = \sum_{i=1}^n{b_if(\vec{x}_i)}
$$
, contradicting to the fact that $f(X)$ is a basis.

Therefore, $f$ maps a basis to a basis, and thus it is bijective.

Did I do it right?

答:
That's correct, but you can prove it directly. Suppose $x_1,\ldots,x_n$ satisfy a linear relation
$$
a_1x_1+ \cdots a_nx_n = 0.
$$
Then
$$
a_1f(x_1) + \cdots a_nf(x_n) = 0
$$
since $f$ is linear. Since $f(x_1),\ldots,f(x_n)$ is a basis, $a_1 = \cdots = a_n = 0$.


问:
Can we define a set containing all sets within ZFC and then prove that it does not exist?
I am currently reading "Elements of Set Theory" by Herbert B. Enderton. The first theorem proven in the book states:

There is no set to which every set belongs.

But isn't it the case that phrase "set to which every set belongs" cannot even be formulated within a theory. According axiom schema of specification (also called the axiom schema of separation or of restricted comprehension) a new set can be defined only using the following formula:

$\forall z \forall w_1 \forall w_2\ldots \forall w_n \exists y \forall x [x \in y \Leftrightarrow (( x \in z )\land \phi )]$

In other words, any new set $y$ can be defined only as a subset of some other set ($z$ in the formula). So, how can we define a set containing all the other sets under this restriction?

So, my concern not about the prove that "set containing all sets" does not exist but about the fact that we prove existence (non-existence) of something that cannot be formulated in terms of the language of theory. It looks like we try to prove within ZFC that "cats" exist. ZFC does not know what is a "cat".

答:
Noah's answer suffices for ZFC. But even in a foundational system with both sets and non-sets that has axioms similar to ZFC (e.g. ZFA), your argument is still flawed. The sentence "there is no set to which every set belongs" translates obviously and directly to "¬∃S∈Set ∀T∈Set ( T∈S )". Absolutely nothing in this makes any claim about the existence of any set of all sets. Here "Set" denotes a sort, not a set. In fact, the claim is that there does not exist any such set. In simple FOL, quantifiers are unrestricted and range over the entire domain. In many-sorted FOL, each quantifier ranges over a specific sort of objects. For example, a vector space can be easily axiomatized as a 2-sorted structure with one sort for vectors and the other sort for scalars. You can translate many-sorted FOL into simple FOL by using a predicate-symbol for each sort, so the extra expressiveness does not actually matter.

As Noah pointed out, "set to which every set belongs" is a property, not a set, and can be captured in ZFC by a suitable formula with one parameter. I want to point out further that every sentence over a set theory is full of quantifiers over sets in the first place, and there is not even any need to paraphrase any of them to be in terms of properties. For instance, "for every sets $S,T$ there is a set to which an object belongs if and only if it belongs to $S$ or $T$" is a perfectly good sentence using the same kind of phrasing, and you should just translate it using quantifiers.

Also, what Noah means by "set" being a dummy phrase in ZFC is that actually none of the ZFC axioms say anything about "sets". Rather, all of their quantifiers range over the whole (intended) universe. And whatever that universe might be is not actually specified by ZFC. ZFC axiomatizes what some people believe are meaningful FOL statements when the quantifiers are interpreted to range over "sets". (There is in fact a substantial amount of philosophical assumptions underlying some axioms of ZFC.) So the ZFC axioms by themselves are technically meaningless; only after you have interpreted them to be about "sets" do they become statements about "sets". Compare with the FOL theory of groups; none of the axioms say anything about "elements", and they only make sense when you interpret their quantifiers to range over the elements of an actual group and interpret the binary function-symbol to be the actual binary operation in that group.


问:
How to check if a ray intersects a circle
Given a ray (Origin(x,y) and Direction(x,y)) and a circle (Center(x,y) and Radius) determine if the ray intersects the circle (touches one or two points in the circle).

I've followed this solution (both geometic and analityc) but I can't get it to work on my python code. Anyone can see my mistake?

The code is the following:

```<code>def dot(x1, y1, x2, y2):
    return (x1*x2) + (y1*y2)

def intersects(x, y, vx, vy, cx, cy, r):
# r is radius, x and y are the ray origin, vx and vy are the ray direction and cx and cy are the circle center

    dirx = vx-x
    diry = vy-y

    Lx = cx - x
    Ly = cy - y

    tca = dot(Lx,Ly,dirx,diry)

    if tca < 0:
        return False

    Lsq = dot(Lx,Ly,Lx,Ly)
    tcasq = tca*tca

    dsq = Lsq - tcasq

    rsq = r*r

    if dsq>rsq:
        return False

    thc = math.sqrt(rsq-dsq)

    t0 = tca - thc

    t1 = tca + thc

    if t0 < 0:
        t0 = t1
        if t0 < 0:
            return False

    return True

</code>```

答:
I think this might be your mistake

dirx = vx-x

diry = vy-y

needs to be

dirx = vx

diry = vy

Since vx,vy is already a direction vector.


问:
Difference quotients and Lipschitz
Is there any example of a function $f$ on $R$ such that
$$\left|\frac{f(x+h) + f(x-h) -2f(x)}{h}\right| \leq C$$
for all $x$ and all $h\neq 0 $ but not Lipschitz continuous(Lip$_1$) in $R$?

答: Any discontinous additive function will work because it will have a. graph dense in $\mathbb{R}^2$ and the given quotient is $0$. Read about the construction here https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation

Title: Difference quotients and Lipschitz


问:
Showing every open cover of [0,1] has a finite sub-cover with connectedness.
Question is as in the title. We are not allowed to invoke compactness.

Professor provides this argument: Suppose $F=\{O_\alpha|\alpha \in I\}$ is an open cover of $[0,1]$. Define $J\subseteq [0,1]$ such that $J$ can be covered by some finite subset of $F$. Then $J$ is open because for any point $a\in J$, $a$ is in some $O_1$, and $O_1\cap J \subseteq J$. Then $a$ is an interior point of $J$. I don't understand why it is interior. I know there is an open interval around $a$ that is in $O_1$.

Also $J$ is closed, because any cluster point $x_0$ of $J$ in $[0,1]$ is contained in some $O_\beta$. Then $x_0 \in J$. I don't see why $x_0$ has to be in $J$ at all, $O_\beta$ could have non-trivial intersection with the finite sub-cover of $J$ but not be a member of the sub-cover.

In fact I think the above argument is broken. Any ideas to make a cogent argument out of this?

答:
Are you sure that you’ve accurately reported the definition of $J$? The argument works if

$$J=\{x\in[0,1]:[0,x]\text{ can be covered by a finite }F_x\subseteq F\}\,.$$

Let $x\in J$. Then there are an $O\in F_x$ such that $x\in O$ and an $\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\subseteq O$; clearly $[0,x+\epsilon)\subseteq\bigcup F_x$, so $\left[0,x+\frac{\epsilon}2\right]\subseteq J$, and $\left(0,x+\frac{\epsilon}2\right)$ is an open nbhd of $x$ contained in $J$. This shows that $J$ is open in $[0,1]$.

Now let $x$ be any cluster point of $J$ in $[0,1]$. There are an $O\in F$ and an $\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\subseteq O$, and since $x$ is a cluster point of $J$, there is a $y\in(x-\epsilon,x+\epsilon)\cap J$. Let $F_y$ be a finite subset of $F$ covering $[0,y]$; then $F_y\cup\{O\}$ is a finite subset of $F$ covering $[0,y]\cup(x-\epsilon,x+\epsilon)$. In particular, $F_y\cup\{O\}$ covers $[0,x]$, so $x\in J$. This shows that $J$ is closed in $[0,1]$.

Clearly $0\in J$, so $J\ne\varnothing$, and $[0,1]$ is connected, so $J=[0,1]$, $1\in J$, and there is a finite $F_1\subseteq F$ covering $[0,1]$.


问:
Fourier series $\sum_{m=0}^\infty \frac{\cos (2m+1)x}{2m+1}$
Does anyone know the sum of Fourier series $$\sum_{m=0}^\infty \frac{\cos (2m+1)x}{2m+1}?$$
I tried WA; it does not return a function.</p>

答:
Rewrite the series as

$$\Re{\left [ \sum_{n=0}^{\infty} \frac{\left ( e^{i x}\right )^{2 n+1}}{2 n+1} \right]} = \Re{[\text{arctanh}{(e^{i x})}]} = \frac12 \Re{\left[\log{\left(\frac{1+e^{i x}}{1-e^{i x}}\right)}\right]}$$

With some manipulation,noting that $\log{z} = \log{|z|} + i 2 \pi \arg{z}$, we find that

$$\sum_{n=0}^{\infty} \frac{\cos{(2 n+1) x}}{2 n+1} = \frac12 \log{\left |\cot{\frac{x}{2}}\right |}$$


问:
Countable intersection of open dense subsets in a locally compact space is dense
Let $(X, \mathcal{T})$ be a locally compact space and, for each $n \in \mathbb{N}$, let $U_n$ be an open dense subset of $X$. Prove that $\bigcap_{n \in \mathbb{N}} U_n$ is dense in $X$.

My idea I have to prove that $ X= \overline{\bigcap_{n \in \mathbb{N}} U_n}$, let $x \in X$ and let $U \in \mathcal{T}$ such that $x \in U$, since $U_n$ is dense in $X$ we have $\overline{U_n}=X$, so $x \in \overline{U_n}$ and $U \cap U_n \neq \emptyset$, and we have that $X$ is locally compact so there is $V \in \mathcal{T}$ and a compact subset $K$ of $X$ such that $x \in V$ and $V \subset K$. I don't know how to use the fact that $K$ is compact

答: HINT: Choose $V_0\in\mathcal{T}$ so that $x\in V_0\subseteq\operatorname{cl}V_0\subseteq U\cap U_0$. Given $V_n$ for some $n\in\Bbb N$, there is a $V_{n+1}\in\mathcal{T}$ such that $$x\in V_{n+1}\subseteq\operatorname{cl}V_{n+1}\subseteq V_n\cap U_{n+1}\,.$$ Now consider $\bigcap_{n\ge 0}\operatorname{cl}V_n$.


问:
Questioning a proof that the Cantor set is uncountable.
The 2nd Ed. of Gamelin & Greene's Intro. to Topology contains as exercise 2.7 (c), p15, show the Cantor set is uncountable.  A solution offered, on p198 relies on exercise 2.3, which states:
"The set of isolated points of a countable complete metric space X forms a dense subset of X."   The theory of the proof that C (Cantor Set) is uncountable is, as I understand it,  is that since there is a dense subset of C and "none of these is isolated", C is uncountable.  The complete metric space here is C.  The dense subset of C are the endpoints of the middle-third intervals.  The clincher is, since these are not isolated, C has no isolated points, and "by exercise 2.3 is uncountable."

In applying exercise 2.3, C is X and the set of middle-third endpoints is dense in C.  The argument seems to be that if C were countable, its dense subset would have to have isolated points.  My reading of 2.3 is if C had a set of isolated point and C were countable, then that set would be dense in C.   I don't think that is the same as arguing that since the dense set has no isolated points, C is uncountable.

Comments please on the logic of this proof.

答: Your interpretation of the result in Exercise $2.3$ is incorrect. It’s actually a very strong result: it says that if $X$ is any countable, complete metric space, then $X$ has a dense set of isolated points. Thus, if $C$ were countable, it would be a countable, complete metric space and would therefore have a dense set of isolated points. In fact $C$ has no isolated points, so it certainly does not have a dense set of isolated points. Thus, it cannot be countable.


问:
Express an integral in terms of another integral.
Question:

Let $$I=\int_0^{\pi/6}\tan^{10}(2x)\sec(2x)\,dx.$$ Express the value of $$\int_0^{\pi/6}\tan^{12}(2x)\sec(2x)\,dx$$ in terms of $I$.

Attempts:

$$\begin{align}\int_0^{\pi/6}\tan^{12}(2x)\sec(2x)\,dx&=\int_0^{\pi/6}\tan^{10}(2x)(\sec^2(2x)-1)\sec(2x)\,dx\\&=\int_0^{\pi/6}\tan^{10}(2x)\sec^3(2x)\,dx-I\end{align}$$

From here I can’t continue. How should I express $\int_0^{\pi/6}\tan^{10}(2x)\sec^3(2x)\,dx $? Thanks a lot.

答:
You can apply Integration by Parts to the original integral $\displaystyle \int_0^{\pi/6}\tan^{12}(2x)\sec(2x)\,dx$.

If we set
$$u=\tan^{11}(2x) \quad \text{and} \quad dv=\tan(2x)\sec(2x)\,dx,$$
then
$$du=22\tan^{10}(2x)\sec^2(2x)\,dx \quad \text{and} \quad v=\frac{1}{2}\sec(2x),$$
and the integral turns into
$$\int_0^{\pi/6}\tan^{12}(2x)\sec(2x)\,dx=\left.\frac{1}{2}\tan^{11}(2x)\sec(2x)\right|_0^{\pi/6}-11\int_0^{\pi/6}\tan^{10}(2x)\sec^3(2x)\,dx.$$
And to the last integral you can apply what you did but backwards.


问:
Double integral in a triangular region
Integrate:
$\iint_D e^{-y-x} dx\ dy $  where $D$ is the triangle with vertices (0,2), (-3,4) and (4,4).  If someone could guide me through the question I'd really appreciate it.

答:
Yes the equations of the lines are correct.

Here is your region -

You can integrate over $x$ first going from $(3 - \frac{3}{2}y)$ to $(2y-4)$ between two given lines. Then integrate over $y$ from $2$ to $4$.

Title: Double integral in a triangular region


问:
Ball-counting problem (Combinatorics)
I would like some help on this problem, I just can't figure it out.

In a box there are 5 identical white balls, 7 identical green balls and 10 red balls (the red balls are numbered from 1 to 10). A man picks 12 balls from the box. How many are the possibilities, in which:

a) exactly 5 red balls are drawn --
b) a red ball isn't drawn --
c) there is a white ball, a green ball and at least 6 red balls

Thanks in advance.

答:
For part (a), since we require exactly $5$ red balls, we know that the $7$ other balls must be composed of only white and green balls.  How many ways can we choose $5$ distinct red balls from $10$?  This is exactly what the symbol $\binom{10}{5}$ measures.  It's formula is given by:

$$\binom{10}{5}=\frac{10!}{5!(10-5)!}=\frac{10\cdot9\cdot8\cdot7\cdot6}{5\cdot4\cdot3\cdot2\cdot1}=630$$

Now, we have to ask ourselves how many ways can we make a collection of $7$ balls from $5$ identical white balls and $7$ identical green balls.  Since there are only $5$ white balls, we can have anywhere from $0$ to $5$ white balls, making $6$ possibilities.  This means, altogether, there are $6\times 630=3780$ possible different collections of $12$ balls with exactly $5$ red balls.

Notice that since the red balls are distinguishable, there are many more ways to choose them than there are to choose the identical white and green balls.

As is mentioned in the comments, part (b) is relatively straightforward.  See if you can do part (c) applying some of the concepts we used for part (a) (just do part (a) for exactly $6,7,8,9,$ and $10$ red balls, and add the results).

Title: Ball-counting problem (Combinatorics)


Title: Double integral in a triangular region

Body:
Integrate:
$\iint_D e^{-y-x} dx\ dy $  where $D$ is the triangle with vertices (0,2), (-3,4) and (4,4).  If someone could guide me through the question I'd really appreciate it.

Body:
The integral is
$$\iint_D e^{-y-x} dx\ dy =\int_2^4 \int_{\frac{6-3y}2}^{2y-4} e^{-y-x} dx\ dy =\frac13e^{-8}-\frac73e^{-2}+2e^{-1}
$$


问:
What is the asymptotic expansion of $x_n$ where $x_{n+1} = x_n+1/x_n$?
Let
$x_{n+1} = x_n+1/x_n,
x_0 = a \gt 0$
and
$y_n = x_n^2$.

What is the asymptotic expansion
of $x_n$ ($y_n$ will do)?

I can show that
$y_n
=2n+\dfrac12 \ln(n) + O(1)
$.

Is there an explicit form
for the constant
implied by the $O(1)$?

What is the asymptotic form
of the terms following that constant
(e.g.,
$O(\frac{\ln(n)}{n}),
O(\frac1{n}), ...
$)?

答:
Not a complete answer. Let me first reproduce what I imagine is your argument. We have $y_{n+1} = y_n + \frac{1}{y_n} + 2$. In particular $y_{n+1} \ge y_n + 2$ which gives $y_n \ge 2n + a^2$ and hence $\frac{1}{y_n} \le \frac{1}{2n + a^2}$. This gives

$$y_{n+1} \le y_n + 2 + \frac{1}{2n + a^2}$$

which gives

$$y_n \le 2n + \sum_{i=0}^{n-1} \frac{1}{2i + a^2} + a^2 = 2n + \frac{1}{2} \log n + C$$

for some constant $C$. Write $y_n = 2n + \frac{1}{2} H_{n-1} + e_n$, where we now know that $e_n$ is bounded from above by a constant (and it's not hard to show that it's bounded from below by a constant also). Then the recurrence relation gives

$$y_{n+1} - y_n = 2 + \frac{1}{2n} + e_{n+1} - e_n = \frac{1}{2n + \frac{1}{2} H_{n-1} + e_n} + 2$$

and rearranging a bit gives

$$e_n - e_{n+1} = \frac{1}{2n} - \frac{1}{2n + \frac{1}{2} H_{n-1} + e_n} = \frac{ \frac{1}{2} H_{n-1} + e_n}{2n \left( 2n + \frac{1}{2} H_{n-1}+ e_n) \right)}.$$

Heuristically this gives something like $e_n = C + \frac{\ln n}{8n} + \dots$ but I don't know how to prove it off the top of my head, mostly because I can't think of a reasonable way to describe the constant $C$.


问:
Is there a function whose second derivative is the inverse square of itself?
Namely, is there a function $f(x)$ that satisfies

$$\frac{d^2f(x)}{dx^2} = \frac{1}{(f(x))^2}$$

I've been messing with an overly simplified physical system in which a body has a gravitational attraction towards another body much more massive (allowing me to assume that the second body doesn't move), and I got the following differential equation:

$$\frac{d^2x(t)}{dt^2} = \frac{Gm}{(x(t))^2}$$

That got me wondering if such differential equation has a known solution.

答:
Another way to solve
$$f''(x) = \frac{1}{(f(x))^2}$$ Consider instead $x(f)$ and the equation is
$$-\frac {x''(f)} {[x'(f)]^3}=\frac 1{f^2}$$ Reduction of order $p=x'(f)$ gives

$$-\frac {p'}{p^3}=\frac 1{f^2}\implies p=\frac{\sqrt{f}}{\sqrt{2} \sqrt{c_1 f-1}}$$
$$x(f)=c_2+\frac{\log \left(c_1\sqrt{f}+\sqrt{c_1} \sqrt{c_1f -1}\right)}{\sqrt{2}   c_1^{3/2}}+\frac{\sqrt{f} \sqrt{c_1f -1}}{\sqrt{2} c_1}$$

Now the problem is to inverse $x(f)$.


问:
Riemann-Hurwitz formula and degree of a mapping
I'm working over my exercise sheet in my course on Riemann surfaces and I have to verify the Riemann-Hurwitz formula $$\chi(X)=n\chi(Y)-\sum_{x\in X}(ord_x(f)-1)$$ for the mapping $f(z)=\frac{(z+1)^2}{z^4},f(\infty)=0$ between $\mathbb{P}^1\to \mathbb{P}^1$. So far only $z=0,z=-1$ contribute to the sum and give multiplicity $4$ and $2$.This would imply that the degree of the map $f$ is -1 since $\chi(\mathbb{P}^1)=2$ but I'm actually not quite convinced on how to compute the degree of this map. There's no definition in my notes, any help is welcome :)

答:
You did not consider all critical points of the mapping. We have
$$
 f'(z) = -\frac{(z+1)(z+2)}{z^5}
$$
so that the critical points are

$f(0) = \infty$ with multiplicity $4$,

$f(-1) = 0$ with multiplicity $2$,

$f(-2) = 1/16$ with multiplicity $2$,

$f(\infty) = 0$ with multiplicity $2$.

This gives $\sum_{x\in X}(ord_x(f)-1) = 3+1+1+1 = 6$ and $n=4$.


问:
Show that there exists a holomorphic function $f:C\backslash[0,1] \to{C}$ such that $\exp(f(z))=\frac{z-1}{z}$ for all $z \in \Bbb{C}\backslash[0,1]$
The hint for the problem says use the Schwartz reflection principle. Can I get some clues on how to proceed?</p>

答: Show that $z \to w = \frac{z-1}{z}$ maps the segment [0,1] to the closed half line $\{\infty\} \cup (-\infty,0] $, so the complement of [0,1] is mapped to the complex plane with a cut along the negative real numbers, and there is an analytic branch of logarithm defined in that domain.


问:
How to compute $\int_0^{\infty}\sqrt x \exp\left(-x-\frac{1}{x}\right) \, dx$?
How to compute this integral? :

$$\int_0^{\infty}\sqrt x \exp\left(-x-\frac{1}{x}\right) \, dx$$

Wolframalpha gives the answer $\dfrac{3\sqrt{\pi}}{2e^2}$, but how to compute this?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}\root{x}
\exp\pars{-x - {1 \over x}}\,\dd x}
\\[5mm] \stackrel{x\ =\ \expo{2\theta}}{=}\,\,\,&
\int_{-\infty}^{\infty}\expo{\theta}
\exp\pars{-2\cosh\pars{2\theta}}\pars{2\expo{2\theta}}\,\dd\theta
\\[5mm] = &\
4\int_{0}^{\infty}\cosh\pars{3\theta}
\exp\pars{-2\cosh\pars{2\theta}}\,\dd\theta
\\[5mm] = &\
4\int_{0}^{\infty}\
\overbrace{\cosh\pars{\theta}\bracks{4\sinh^{2}\pars{\theta} + 1}}^{\ds{\cosh\pars{3\theta}}}\
\exp\pars{-4\sinh^{2}\pars{\theta} - 2}\,\dd\theta
\\[5mm] \stackrel{t\ =\ \sinh{\theta}}{=}\,\,\,\,\,\,&
4\expo{-2}\int_{0}^{\infty}\pars{4t^{2} + 1}\expo{-4t^{2}}\,\dd t =
\bbx{3\root{\pi} \over 2\expo{2}} \approx 0.3598 \\ &
\end{align}


答:
It seems that I misunderstood the problem. I assumed that the balls are distinguishable objects, as one would do when counting probabilities of the corresponding events. If green and white balls are not supposed to be distinguishable, then the below is wrong.

Because there seems to be nothing to gain from deleting a solution, I'm leaving it with the above disclaimer.

For a) you essentially want to choose $5$ out of $10$ red balls, and then $7$ of the remaining $12$ balls. Altogether, ${10 \choose 5} \cdot {12 \choose 7}$

For b) you just want to draw $12$ non-red balls. It so happens that there are $12$ non-red balls in the box, hence the answer is just $1$.

For c), fix the number of red balls you draw, call it $k$. Then, there are ${10 \choose k}$ ways to draw $k$ red balls, and $12 \choose 12 - k$ ways to draw the rest. We want to exclude the situations in which only white or only green balls are drawn apart from the red ones. This happens in, respectively, $5 \choose k$ and $7 \choose k$ ways. Hence, for given $k$, there are:
$${10 \choose k} \cdot \left( {12 \choose 12 - k} - {5 \choose k} - {7 \choose k} \right)$$
I am afraid there is no other way than laboriously sum this for $k=6,7,8,9,10$.

Title: Ball-counting problem (Combinatorics)

Body:
I would like some help on this problem, I just can't figure it out.

In a box there are 5 identical white balls, 7 identical green balls and 10 red balls (the red balls are numbered from 1 to 10). A man picks 12 balls from the box. How many are the possibilities, in which:

a) exactly 5 red balls are drawn --
b) a red ball isn't drawn --
c) there is a white ball, a green ball and at least 6 red balls

Thanks in advance.


问:
Compute: as $\displaystyle{\lim_{n\to \infty} \int_0^\pi \sin^n x dx}$
$$\lim_{n\to \infty} \int_0^\pi \sin^n x dx$$

I have been trying to compute this integral for months and have gotten nowhere. I seem to be going in circles. Any help at all would be seriously appreciated.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\lim_{n \to \infty}
\int_{0}^{\pi}\sin^{n}\pars{x}\,\dd x} =
\lim_{n \to \infty}
\int_{-\pi/2}^{\pi/2}\cos^{n}\pars{ x}\,\dd x =
2\lim_{n \to \infty}\int_{0}^{\pi/2}\cos^{n}\pars{ x}\,\dd x
\end{align}
The main contribution to the integration comes from values of $\ds{x \gtrsim 0}$. Then,
\begin{align}
&\bbox[5px,#ffd]{\lim_{n \to \infty}
\int_{0}^{\pi}\sin^{n}\pars{x}\,\dd x} =
2\lim_{n \to \infty}
\int_{0}^{\pi/2}\expo{n\ln\pars{\cos\pars{x}}}\,\,\,\dd x
\\[5mm] = &\
2\lim_{n \to \infty}
\int_{0}^{\infty}\expo{-nx^{2}\,/2}\,\,\dd x\qquad
\pars{~Laplace\,'s\ Method~}
\\[5mm] = &\
2\lim_{n \to \infty}\pars{\root{2 \over n}{\root{\pi} \over 2}} =
\bbx{0} \\ &
\end{align}


问:
Can we give an equivalent definition for computable numbers without mentioning turing machines?
Let $\mathbb{R}$ be the set of all real numbers, as defined by dedekind cuts. Given $\mathbb{R}\subset\mathscr{P}(\mathbb{Q})$, and we have a construction of $\mathbb{Q}$ as as a set of equivalence classes on $\mathbb{Z}$ which is itself a set of equivalence classes on $\mathbb{N}$ and of course we can take $\mathbb{N}$ to be the set of finite von neumann ordinals, which means that $\mathbb{R}$ is a genuine set that exists in any model of ZF.
Lets define the set of 'provable real numbers' $\mathbb{R}_p$ in some second order meta theory, to be the set of all sets which are real numbers, and are sets in every model of ZF.
So we know that $\mathbb{R}_p$ is going to be countable, my question however is: Is $\mathbb{R}_p$ at all related to the set of all computable real numbers? Is it equivalent to, a subset of, or a super set of the set of computable real numbers? or is it none of those?

答:
Unfortunately this is one of those situations where a very natural question has a surprisingly technical answer. Right now I'm erring on the side of concision, but let me know if there's a particular point you'd like me to expand on. Some points however really can't be treated elementarily - e.g. the proof that $\mathsf{HYP}$ is an upper bound for $\mathbb{R}_p$ really requires some work.

It takes a bit of work to make your notion of $\mathbb{R}_p$ fully precise. I think the following definition captures your intuition, however:

Let $\mathbb{R}_p$ be the set of reals whose corresponding Dedekind cuts are in every $\omega$-model of $\mathsf{ZF}$.

(To make this nontrivial we assume that $\mathsf{ZF}$ does in fact have $\omega$-models.)

An $\omega$-model of $\mathsf{ZF}$ is a model which is "correct about $\omega$:" all the things that model thinks are finite are actually finite. By the compactness theorem, not all models are $\omega$-models, and in non-$\omega$-models we have "nonstandard rational numbers" which makes the definition of $\mathbb{R}_p$ no longer clear.

The key here is absoluteness: there are certain facts which all $\omega$-models are correct about, and so any real defined entirely in terms of those facts will be in $\mathbb{R}_p$. For example, all computable reals are in $\mathbb{R}_p$ since Turing machine computations are ultimately just statements about natural numbers, and all $\omega$-models agree on what the natural numbers look like. But this immediately pushes us far beyond the computable: for the same reason we get all the arithmetic reals such as the (highly non-computable) Chaitin's $\Omega$, and with a bit of thought we get the hyperarithmetic reals as well. A general argument (which is unfortunately a bit technical) also establishes the set of hyperarithmetic reals as an upper bound, so in fact we get exactly that $\mathbb{R}_p$ is the set of hyperarithmetic reals.

Incidentally, this characterization is extremely robust. For any set theory $T$ we can similarly define $\mathbb{R}_p^T$ to be the set of reals in every $\omega$-model of $T$. It turns out that basically any reasonable $T$ will yield $\mathbb{R}_p^T=\mathsf{HYP}$ as above, for the same reason: as long as $T$ has $\omega$-models in the first place, is reasonably simple (= hyperarithmetically axiomatizable), and can prove basic facts about the Turing jump operation, the argument hinted at above goes through. So for example we can add choice, or large cardinals, or drop all the way down to something like $\mathsf{KP}$, and we won't change the answer.

You mention in the comments the possibility of shifting to a more "proof-theoretic" notion, of the real numbers which "provably exist." Here we run into a serious problem:

What does it mean for $\mathsf{ZF}$ to prove that a real number $r$ exists?

The problem is that it only makes sense to ask whether $\mathsf{ZF}$ proves a sentence in the language of set theory. So to even ask whether $\mathsf{ZF}$ proves "$r$ exists," we need to already have some fixed definition of $r$ in the language of set theory. This motivates the following definition:

Let $\mathbb{R}_{named}$ be the set of reals of the form $\varphi^V$, for $\varphi$ a formula which $\mathsf{ZF}$ proves defines some real number.

Here "$\varphi^V$" is the interpretation of $\varphi$ in the set-theoretic univere $V$: that is, the thing $\varphi$ actually names.

The problem is that this is highly undetermined. For example, and shifting attention to binary sequences for simplicity, let $c:\mathbb{N}\rightarrow\{0,1\}$ be the continuum pattern: $$c(n)=1\quad\iff\quad 2^{\aleph_{2n}}=\aleph_{2n+1}.$$ Obviously $\mathsf{ZF}$ proves that $c$ exists, but Easton's theorem says that $\mathsf{ZF}$ can't prove any particular facts about $c$! As far as $\mathsf{ZF}$ knows, $c$ could be any sequence whatsoever.

(Moreover, $\mathbb{R}_{named}$ can't even be defined in set theory per Tarski, unlike $\mathbb{R}_p$. So the "models-only" definition is actually much better behaved.)

EDIT: Let me say a bit more about the proof that $\mathbb{R}_p\subseteq \mathsf{HYP}$ that I alluded to above. The real result being proved is:

Suppose $r$ is a non-hyperarithmetic real. Then there is a countable $\omega$-model of $\mathsf{ZFC}$ not containing $r$.

Note the restriction to countable models. This is a situation where strengthening the goal makes the result easier to prove: countable structures live in the realm of descriptive set theory, and we can use powerful tools there.

Basically, we look at the relations $R\subseteq\mathbb{N}^2$ such that $(\mathbb{N};R)$ is a (necessarily countable) $\omega$-model of $\mathsf{ZFC}$. The set of all binary relations on $\mathbb{N}$ whatsoever is naturally viewed as a topological space (indeed, one homeomorphic to the Cantor space), and the set $\mathfrak{X}$ of relations representing $\omega$-models of $\mathsf{ZFC}$ is $\Sigma^1_1$ in the sense of that space. A result of Gandy says that for every non-hyperarithmetic real $r$, any nonempty $\Sigma^1_1$ set has an element $s$ to which $r$ is not hyperarithmetically reducible. Now the point is this: if $R\in\mathfrak{X}$ and $r$ is a real in the model represented by $R$ in the appropriate sense, we can "read off" $r$ from the relation $R$ itself - precisely, we have that $r$ is hyperarithmetic relative to $R$ (in fact we can strengthen that to "computable relative to," but that takes a bit of thought). This means that we can apply Gandy to get the result stated above.

(Actually, Gandy's result is even stronger: see Gandy's basis theorem. But we only need the "cone-avoiding" aspect of it here.)

Now this really is a technical argument: not only have I not even remotely sketched how to prove Gandy's result, I also haven't defined "$\Sigma^1_1$." But I think the above is still worth writing down, if only to motivate the blackboxed bits by demonstrating how they can be used.


问:
Set that is not well-ordered
I want to find a set that is transitive, every non-empty subset of it has $\in$-maximum and is not $\in$-well ordered.

Any hint?

Title: Set that is not well-ordered

Body: $\{\varnothing,\{\varnothing\},\{\{\varnothing\}\}\}$


问:
Proving that the least upper bound of  is less than or equal to the greatest lower bound of B
I'm failing to understand the basics concepts of upper/lower bounds and I was hoping I could get some guidance through a simple question.

Suppose that $A$ and $B$ are two non-empty bounded sets of real numbers, such that $x \leq y$ for every $x \in A$ and $y \in B$. Show $lubA \leq glbB$

I have no idea how to even start this proof. I obviously need to somehow relate $x \leq y$ to the boundaries of their sets, that is $lubA \leq glbB$. Although I'm not even sure if my idea is correct. Any help?

答:
I'm going to use $\sup$ and $\inf$ for lub and glb, respectively, as these are more natural for me. These are exactly equivalent terms.

Pick $x \in A$. (We can do this because $A$ is nonempty.) Then $x \leq \sup A$ (by definition). But $x \leq y$ for all $y \in B$, so this $x$ is a lower bound of $B$. But any lower bound of $B$, by definition, must be less than or equal to the greatest lower bound of $B$. So $x \leq \inf B$. Since this $x$ was arbitrary, we have $x \leq \inf B$ for all $x \in A$. But then $\inf B$ is an upper bound of $A$, so that means that it cannot possibly be less than the least upper bound of $A$, so $\sup A \leq \inf B$.


答:
Hints: (assuming $A,B\neq \varnothing$)since $x\leq y$ for every $x\in A$
Therefore $y$ is an upper bound of $A$.
Thus $\text{lub}A\leq y$.
Similarly, $x\leq \text{glb}B$.

Title: Proving that the least upper bound of  is less than or equal to the greatest lower bound of B

Body:
I'm failing to understand the basics concepts of upper/lower bounds and I was hoping I could get some guidance through a simple question.

Suppose that $A$ and $B$ are two non-empty bounded sets of real numbers, such that $x \leq y$ for every $x \in A$ and $y \in B$. Show $lubA \leq glbB$

I have no idea how to even start this proof. I obviously need to somehow relate $x \leq y$ to the boundaries of their sets, that is $lubA \leq glbB$. Although I'm not even sure if my idea is correct. Any help?


问:
Compute integral $\int_{-\infty}^\infty \frac{x^4}{1+x^8} \, dx$
Compute the integral.

$$\int_{-\infty}^\infty \frac{x^4}{1+x^8} \, dx$$

The answer at the back of the book is
$$\frac{\pi}{4\sin(\frac{3\pi}{8})}$$

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{-\infty}^{\infty}{x^{4} \over 1 + x^{8}} \,\dd x} = 2\int_{0}^{\infty}{x^{4} \over 1 + x^{8}} \,\dd x
\\[5mm] \,\,\,\stackrel{x^{8}\ \mapsto\ x}{=}\,\,\, &
{1 \over 4}\int_{0}^{\infty}{x^{\color{red}{5/8} - 1} \over 1 + x} \,\dd x
\end{align}
Note that $\ds{{1 \over 1 + x} = \sum_{k = 0}^{\infty}
\color{red}{\Gamma\pars{1 + k}}{\pars{-x}^{k} \over k!}}$. Then,
\begin{align}
&\!\!\!\!\!\bbox[5px,#ffd]{\int_{-\infty}^{\infty}{x^{4} \over 1 + x^{8}} \,\dd x} = {1 \over 4}\,
\Gamma\pars{\color{red}{5 \over 8}}
\Gamma\pars{1 -\color{red}{5 \over 8}}\,,\
\pars{\substack{\ds{Ramanujan's}\\[0.5mm] \ds{Master}\\[0.65mm] \ds{Theorem}}}
\\[5mm] = &\
\bbx{\pi \over 4\sin\pars{3\pi/8}} \approx 0.8501 \\ &
\end{align}


问:
Endomorphism that maps Center to "not Center"
I have a question about group endomorphism. Suppose $G$ is a group and endomorphism $f : G \to G$. Is it true that $f(Z(G)) = Z(G)$ ?

If not, suppose I have $G= D_{8}$ dihedral group of order $8$.
Can we find an endomorphism of $D_{8}$ that maps $Z(D_{8}) \to D_{8} \setminus Z(D_{8})$ ?

Thanks in advance.

答: Neither of those claims are true. For the first one, consider the trivial homomorphism. For the second, the image of a homomorphism can't possibly not contain the identity, as a homomorphism always maps the identity to the identity; and the center always contains the identity.</p>


答:
Take the map $f:D_8 \rightarrow D_8/\langle r\rangle \cong \mathbb{Z/2Z}$.

$Z(D_8) = \langle r^2 \rangle$ but $f(Z(G))=\{0\} \neq \mathbb{Z/2Z} =Z(\mathbb{Z/2Z})$ because $\mathbb{Z/2Z}$ is abelian.

We know that the subgroup generated by $r^2$ is of order $2$. Hence we have a map $g:\mathbb{Z/2Z} \cong \langle r^2 \rangle \rightarrow D_8 $ which is the inclusion map. So $g\circ f:D_8 \rightarrow D_8$ serves as a counter-example to the first question $f(Z(G))=Z(G)$

As for the second question whether there is an endomorphism $f:D_8 \rightarrow D_8$ such that it maps $Z(D_8)$ to some element in $D_8-Z(D_8)$? This cannot happen.

The centre of $D_8$ is $\{1,r^2\}$. Any endomorphism of $D_8$ can map $r$ to  an order $4$ element or order $2$ element. Only order $4$ element in $D_8$ are $r,r^3$. Hence if $r$ is mapped to either of them we can see that $r^2$ is mapped to itself. When $r$ is mapped to any order $2$ element then $r^2$ is mapped to identity element of $D_8$ which is again in the centre. Hence in both cases $f(Z(D_8)) \subset Z(D_8)$

Title: Endomorphism that maps Center to "not Center"

Body:
I have a question about group endomorphism. Suppose $G$ is a group and endomorphism $f : G \to G$. Is it true that $f(Z(G)) = Z(G)$ ?

If not, suppose I have $G= D_{8}$ dihedral group of order $8$.
Can we find an endomorphism of $D_{8}$ that maps $Z(D_{8}) \to D_{8} \setminus Z(D_{8})$ ?

Thanks in advance.