问:
Determine if a grid (as a list of unique tuples) is full or sparse
I need a quick way to determine if a grid is full or sparse. The grid (in my case of dimension 3) is represented as a list of unique 3-tuples $(x,y,z)$ (for context each tuple is stored as a record in a relational database). The list if of size $m$ and is known. Example (with a full grid) :

```<code>{1,1,1},
{1,1,2},
{1,2,1},
{1,2,2},
...
{3,2,1},
{3,2,2}
</code>```

What I mean by a full grid is an Cartesian product of 3 sets $X,Y,Z$ of size $i,j,k$ respectively. However $X,Y,Z$ and $i,j,k$ are not given. I only have the list of tuples.

Let $i,j,k$ be the number of unique elements in each dimension in the list (in the example above, the unique elements would be $\{1,2,3\},\{1,2\},\{1,2\}$ so $i=3$ , $j=2$ and $k=2$).

With $m$ the number of tuples in the list, intuitively, I would say that
$$m=i\times j\times k$$ is a sufficient condition to say that the grid is full.

In other words, if the length of the list of tuples is equal to the product of the counts of unique elements in each dimension, then the grid must be full (and the list of tuples is indeed the Cartesian product of the unique elements).

This seems true and I cannot find any counter examples, however I would prefer to have a solid proof.

Sorry if the problem is not set properly. Any help appreciated!

答: To justify Brian Scott's answer (yes), let $A$ be the collection of all tuples that appear and let $B$ be the collection of all tuples in the box. Then $A\subseteq B$. If $|A|=|B|$ (they have the same size), then $A=B$, and the collection is exactly the box. This works whenever the two sets are finite. You can rigorously justify this using induction.


问:
score the evenness spread of data across time
I'm trying to feature engineer a data frame with a time delta in seconds column. I would like to create a new feature that will score higher for data that spreads more evenly across time.

I've created this sample DF with 2 time lines, one is evenly distributed while the other one is not:

```<code>df = pd.DataFrame({'id':[1,1,1,1,1,2,2,2,2,2],'timestamp':[
            '2020-09-01 18:14:00',
            '2020-09-01 18:14:01',
            '2020-09-01 18:14:02',
            '2020-09-01 18:14:03',
            '2020-09-01 18:14:04',
            '2020-09-01 19:14:05',
            '2020-09-01 19:14:16',
            '2020-09-01 19:14:18',
            '2020-09-01 19:14:30',
            '2020-09-01 19:14:59'
]})

df['timestamp'] = pd.to_datetime(df.timestamp)
df['delta'] = (df['timestamp']-df['timestamp'].shift()).fillna(pd.Timedelta(seconds=0)).astype('int64') / 1000000000
</code>```

I've googled around and I'm kinda lost, is standard deviation the way to go?

答:
Well, that depends on how severely you want to punish the "non-uniformity" and whether you want some extra features from your scoring method. I would use some extremal problem. For instance, if we have $n$ points on the circle of length $1$ with pairwise distances $d_1,\dots, d_n$, then the sum is fixed (to $1$) but the sum of the squares is at least $1/n$. So, if your time interval is of length $T$ and the successive distances from each point to the next one (wrapping the interval around into a circle, so you can go from the rightmost point to the leftmost one for the last distance) are $D_1,\dots D_n$, it would be natural to put $d_j=D_j/T$ and to define $V=n\sum_{j=1}^n d_j^2-1$ and put $\rm{Score}=e^{-\lambda V}$, say, so the score of $1$ is given to a perfect equi-spaced configuration and everything else would get it between $0$ and $1$.

The choice of $\lambda$ (which can also be made depending on $n$) will determine how sensitive you want your scoring function to be. I would put it at $1$ to start with and see whether you like the result it gives or not. Probably you will not like it but then you'll understand what exactly you are unhappy about and we can modify the definition of $V$ (which is technically any reasonable function of positions with unique global minimum at the equi-spaced configurations) to eliminate some particular effects.

In general, it is always this way for a problem with a vague objective: just start with something that is not too outlandish and modify it a few times to get its properties closer and closer to those of the "ideal scoring function" you have in mind. Of course, it would be faster if you could formulate all those ideal properties clearly in advance, but very few people can do it immediately and I doubt you are one of them, so just figure out what you really want by trial and error.

ContentLicense: CC BY-SA 4.0


问:
What is Riemann sum of this integral and using it to prove this
This is Question 11.19 (e) part of Tom Apostol's Mathematical Analysis.

Let $f$ be a function of period $2\pi$  whose values on $[-\pi ,\pi]$ are $f(x)=1$ if $0<x < \pi$, $f(x)=0$ if $x =\pi$, or $0$ and $f(x)=-1$ if $-\pi <x<0$ and  $f(x)= 4/\pi \sum_{n=1}^{\infty} \frac{\sin(2n-1) x} {(2n-1)} dx$, for every $x$.
Consider the function$s_n(x)=\frac{2}{\pi} \int_{0}^{x} \frac{\sin (2nt)} {\sin (t)} dt $ . I proved that this function has local maxima at $x_1, x_3, x_5,\dots, x_{2n-1}$ where $x_m=\frac{m \pi } {2n}$.

The question is Interpret   $s_n ( \frac{\pi} {2n}) $  as a Riemann sum and prove that $\lim_{n\to \infty} s_n( \frac{\pi} {2n}) = \frac{2}{\pi} \int_{0}^{\pi} \frac{\sin t}{t} dt$ .

I am getting different answer/ answer which is not even close to this representation when I used Riemann sum method on $f(x)= 4/\pi \sum_{n=1}^{\infty} \frac{\sin(2n-1) x} {(2n-1)}\, dx$ so I request you to kindly outline a proof. No need to fully  solve it.

I have tried this a lot but couldn't solve it.

Thanks

答:
The partial sum of the Fourier series for $f$ evaluated at $x = \frac{\pi}{4n}$ is

$$s_n\left(\frac{\pi}{2n}\right) = \frac{4}{\pi}\sum_{k=1}^n\frac{\sin\frac{(2k-1)\pi}{2n} }{2k-1}$$

Consider a partition $P_n: 0 < \frac{\pi}{2n}< \frac{3\pi}{2n} < \frac{5\pi}{2n} < \ldots < \frac{(2n-1)\pi}{2n}< \pi$ of the interval $[0,\pi]$ and the function

$$g(x) = \begin{cases}\frac{\sin x }{x}, & 0 < x \leqslant \pi \\1, & x = 0 \end{cases}$$

The partition points are given by  $x_0 = 0$, $x_k = \frac{(2k-1)\pi}{2n}$ for $k = 1, \ldots, n$, and $x_{n+1} = \pi$ and the Riemann sum for $g$ using right endpoints of the subintervals as tags is

$$S(P_n,g) = \sum_{k=1}^{n+1}f(x_k) (x_k - x_{k-1}) = \\\frac{\sin \frac{\pi}{2n}}{\frac{\pi}{2n}}\left( \frac{\pi}{2n}-0\right)+\sum_{k=1}^{n}\frac{\sin\frac{(2k-1)\pi}{2n} }{\frac{(2k-1)\pi}{2n}}\left(\frac{(2k-1)\pi}{2n}- \frac{(2k-3)\pi}{2n} \right) +\frac{\sin \pi}{\pi}\left( \pi - \frac{(2n-1)\pi}{2n}\right)\\ = \sin \frac{\pi}{2n}+ \sum_{k=1}^{n}\frac{\sin\frac{(2k-1)\pi}{2n} }{\frac{(2k-1)\pi}{2n}}\frac{\pi}{n} = \sin \frac{\pi}{2n}+ 2\sum_{k=1}^{n}\frac{\sin\frac{(2k-1)\pi}{2n} }{2k-1}$$

Hence,

$$s_n\left(\frac{\pi}{2n}\right)= \frac{2}{\pi}\cdot 2\sum_{k=1}^{n}\frac{\sin\frac{(2k-1)\pi}{2n} }{2k-1} = \frac{2}{\pi}S(P_n,g) - \frac{2}{\pi}\sin \frac{\pi}{2n}$$

Since the Riemann sum $S(P_n,g)$ converges to $\int_0^\pi g(x) \, dx = \int_0^\pi \frac{\sin x}{x} \,dx$ as $n \to \infty$, we have

$$\lim_{n \to \infty}s_n\left(\frac{\pi}{2n}\right) = \lim_{n \to \infty}\frac{2}{\pi}S(P_n,g) -  \lim_{n \to \infty}\frac{2}{\pi}\sin \frac{\pi}{2n} = \frac{2}{\pi} \int_0^\pi \frac{\sin x}{x} \, dx$$

The expression for $s_n(x)$ given in the problem is equal to the partial sum used above.

Differentiating  $s_n\left(x\right) = \frac{4}{\pi}\sum_{k=1}^n\frac{\sin(2k-1)x }{2k-1}$ we get

$$s_n'\left(x\right) = \frac{4}{\pi}\sum_{k=1}^n\cos(2k-1)x,$$

which can be summed in closed form and integrated to obtain

$$s_n(x) = \frac{2}{\pi} \int_{0}^{x} \frac{\sin 2nt} {\sin t} dt$$

In this form we have

$$s_n\left(\frac{\pi}{2n}\right) = \frac{2}{\pi} \int_{0}^{\frac{\pi}{2n}} \frac{\sin 2nt} {\sin t}\, dt = \frac{2}{\pi} \int_{0}^{\pi} \frac{\sin u} {2n \sin \frac{u}{2n}}\, du =  \frac{2}{\pi} \int_{0}^{\pi} \frac{\sin u}{u} \frac{\frac{u}{2n}}{ \sin \frac{u}{2n}}\, du,$$

and by the dominated convergence theorem

$$\lim_{n \to \infty}s_n\left(\frac{\pi}{2n}\right) = \frac{2}{\pi} \int_{0}^{\pi} \frac{\sin u}{u}\, du$$


问:
Why is $\sigma(u,v)=(\cos u,\sin u,v)$ a homeomorphism on $0< u <2\pi$?
Take $S=\{(x,y,z)\in \mathbb{R}^3\ |\ x^2+y^2=1\}$ and define $\sigma:\mathbb{R}^2\rightarrow S,\ \sigma(u,v)=(\cos u, \sin u, v).$

If $U= \{ (u,v)\in \mathbb{R}^2\ |\ 0<u<2\pi\}$, then $\sigma_{|U}:U\rightarrow \sigma(U)$ is a homeomorphism. Why is that?

More generally I don't know how to prove that trigonometric parametrizations such as this have continuous inverses (on the right domain and codomain). Can someone help?

答:
Define $\phi : \mathbb R \to S^1, \phi(u) = e^{iu} = (\cos(u),\sin(u))$. Here $S^1 =\{(x,y) \in \mathbb R^2 \mid x^2 + y^2 = 1 \} \subset \mathbb R^2$ is the unit circle.
It is well-known that $\phi(u) = \phi(u')$ iff $u - u' = 2k\pi$ for some $k \in \mathbb Z$ and $\phi([0,2\pi)) = S^1$. Thus $\phi \mid_{[0,2\pi)} : [0,2\pi) \to S^1$ is a continuous bijection (but of course no homeomorphism since the domain is non-compact, but the range is compact).

We claim that $\phi \mid_{(0,2\pi)} : (0,2\pi) \to \phi((0,2\pi)) = S^1 \setminus \{(1,0)\}$ is a homeomorphism. To show this, it suffices to show that $\phi \mid_{(0,2\pi)}$ is an open map. So let $V \subset (0,2\pi)$ be open. Then $C = [0,2\pi] \setminus V$ is compact, hence $\phi(C) \subset S^1$ is compact and $W = S^1 \setminus \phi(C)$ is open in $S^1$. But $\phi(C) = \phi((0,2\pi) \setminus V)) \cup \phi(\{0,2\pi\}) = \phi((0,2\pi) \setminus V)) \cup \{(1,0)\}$, thus
$$W = (S^1 \setminus \{(1,0)\}) \setminus \phi((0,2\pi) \setminus V)) = \phi((0,2\pi))  \setminus \phi((0,2\pi) \setminus V)) \\ = \phi(0,2\pi)  \setminus ((0,2\pi) \setminus V)) = \phi(V). $$
We have $\sigma \mid_U = \phi \mid_{(0,2\pi)} \times id_{\mathbb R}$ which shows that it is a homeomorphism.


问:
Continuity in the mean: $\lim_{y \to 0}\int_{\mathbb{R}^d}|f(x+y)-f(x)|dx=0$
Let $f \in L_{\mathbb{R}^+}^1(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d),\lambda_d).$ Consider the sequence $f_k=\min(f,k)\min(1,\max(0;k+1-|x|))$ to show the continuity in the mean:
$$\lim_{y \to 0}\int_{\mathbb{R}^d}|f(x+y)-f(x)|dx=0.$$
This result remains true if we take a Radon measure on $(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d))$?

Is the following attempt correct:

Let $\phi_k(x)=\min(1,\max(0;k+1-|x|)),$ $\phi_k$ is continuous with compact support. Let $(y_n)_n$ be a sequence converging to $0,$ so there exist $n_0$ such that for all $n \geq n_0,|y_n| \leq 1$

$$\int_{\mathbb{R}^d}|f(x+y_n)-f(x)|dx \leq2 \int_{\mathbb{R}^d}|f_k(x)-f(x)|dx+\int_{\mathbb{R}^d}|f_k(x+y_n)-f_k(x)|dx$$
So, $$\int_{\mathbb{R}^d}|f(x+y_n)-f(x)|dx \leq2 \int_{\mathbb{R}^d}|f_k(x)-f(x)|dx+k\int_{\mathbb{R}^d}|\phi_k(x+y_n)-\phi_k(x)|dx$$
taking $\limsup_n,$ we have $\int_{\mathbb{R}^d}|\phi_k(x+y_n)-\phi_k(x)|dx \to 0$ using the dominated convergence theorem (since $\phi_k$ are continuous with compact support),

then taking $k \to \infty$ and again using the dominated convergence theorem, we conclude that $$\lim_n \int_{\mathbb{R}^d}|f(x+y_n)-f(x)|dx=0$$

答:
It'd help to know what level you're at to know what level of detail would be good to comment on this!

Looks good to me but here are a few suggestions. If you're allowed to its a lot quicker to just use density of continuous functions with compact support are dense in $L^1$. However you've essentially proved this using your $\phi$ functions.
Secondly I'd just make a few points a bit clearer, your first integral inequality could have one more intermediate line and when you use dominated convergence at the end its not clear which limit you are doing first are you sending $k$ to infinity first and then $n$ or the other way round. One way leads to a horrible mess the other is fine.

Anyway good job looks good to me mathematically.


问:
Given 3 events, $S_1$, $S_2$, $S_3$, how do I express $((S_1 \cup S_2) \cap (S_2 \cup S_3) \cap (S_3 \cup S_1))$ more concisely?
Suppose I have three events: $S_1$, $S_2$, $S_3$. Is there a way to express $((S_1 \cup S_2) \cap (S_2 \cup S_3) \cap (S_3 \cup S_1))$ in a simpler manner?

答:
zkutch's claim isn't quite right. If you draw the Venn diagram, you get:

Which is not the same as $A\cap B\cap C$. I guess you can also express your event as $$(S_1\cap S_2)\cup(S_2\cap S_3)\cup(S_3\cap S_1)$$
but this isn't really simpler.


问:
Write the integral $\int_0^\infty \left(\frac{1}{1 + x^2} \right)^\alpha x^\beta dx$ in terms of the Euler Beta Function?
How can I write the integral $$\int_0^\infty \left(\frac{1}{1 + x^2} \right)^\alpha x^\beta dx$$
with $\alpha, \beta >0$, in terms of the Euler Beta Function? For which $\alpha, \beta$ does the integral actually converge? Maybe some change of variable can do the trick?

答:
$$I=\int_0^\infty(x^2+1)^{-\alpha}x^{\beta}dx$$
now if we split up our regions, it is fair to say that:
$$\int_0^\infty f(x)dx=\int_0^1f(x)dx+\int_1^\infty f(x)dx$$
now let $u=\frac 1x\Rightarrow dx=-x^2du=-\frac{du}{u^2}$ so:
$$\int_0^\infty f(x)dx=\int_0^1f(x)dx-\int_1^0f(1/x)\frac{dx}{x^2}=\int_0^1\left[f(x)+\frac{f(1/x)}{x^2}\right]dx$$
in your case:
$$f(x)=(x^2+1)^{-\alpha}x^\beta$$
so:
$$f(x)+\frac1{x^2}f(1/x)=(x^2+1)^{-\alpha}x^\beta+\frac1{x^2}(1/x^2+1)^{-\alpha}(1/x)^{\beta}$$
and since:
$$\frac{1}{\left(\frac1{x^2}+1\right)^\alpha}x^{-\beta}=\frac{x^{2\alpha-\beta}}{(x^2+1)^{\alpha}}=(x^2+1)^{-\alpha}x^{2\alpha-\beta}$$
so you now have:
$$I=\int_0^1(x^2+1)^{-\alpha}\left[x^\beta+x^{2\alpha-\beta}\right]dx$$
now try using the substitution $v=x^2$ and splitting it into two integrals


问:
solving the recurrence relation using the fastest method
what is the fastest method in solving recurrence relation that we can solve the following recurrence and get the answer $O(\log^3n)$

$T(n)=T(\frac{n}{4})+O(\log^2n)=O(\log^3n)$?

答:
Welcome to MSE!

You might be interested in the Master Theorem, which gives a quick way to determine the asymptotics of many recurrences of the type you're interested in.

In particular, your question falls into case $2a$ of the linked wikipedia article with $c_\text{crit} = \log_4(1) = 0$.

Then $f = O(n^0 \log^2(n))$, so by the master theorem $T = O(\log^{2+1}(n))$, as desired.

I hope this helps ^_^


问:
Show that $\,f(x) = \sum_{n=0}^\infty a_nx^n$, for $x \in [0,1]$, is of bounded variation
Let $\{a_n\} \subset \mathbb{R}$, be such that $\sum_{n=0}^\infty \lvert a_n\rvert < \infty$. Define
$$f(x) = \sum_{n=0}^\infty a_nx^n \quad \text{for } x \in [0,1]$$
Prove that $f$ is of bounded variation.

Attempt:

Let $P = \{ x_0, \ldots , x_n \}$ be any partition of $[0,1]$. Then
$$
\sum_{i=1}^n \left| \sum_{j=0}^\infty a_j x_i^j - \sum_{j=0}^\infty a_j x_{i-1}^j \right| = \sum_{i=1}^n \left| \sum_{j=0}^\infty a_j [ x_i^j - x_{i-1}^j ] \right|.
$$
Since
$\sum_{n=0}^\infty \lvert a_n\rvert < \infty$ and $x_i \in [0,1]$ $\forall i$,
$$
\sum_{j=0}^\infty \left| a_j [ x_i^j - x_{i-1}^j ] \right|
\leq \sum_{n=0}^\infty \lvert a_n\rvert < \infty.
$$
Hence, $f(x)$ is of bounded variation since the elements of the sum are finite the index of the outer sum is also finite.

Is this a valid approach? What other ways are there to prove this?

答:
Take $g(x) = \sum_{a_{k}>=0}a_{k}*x^{k}$ and $h(x) = \sum_{a_k < 0} |a_{k}|*x^{k}$
Then both g(x) and h(x) are increasing so they are of finite variance.
Clearly f = g - h so it must be of finite variance as well (triangle inequality).

ContentLicense: CC BY-SA 3.0


问:
Let $A$ and $B$ be sets; $f: A \times B \rightarrow B \times A$ is given by $f(x,y)=(y,x)$ and $g: B \times A \rightarrow B$ is given by $g(y,x)=y$.
Let $A$ and $B$ be sets; $f: A \times B \rightarrow B \times A$ is given by $f(x,y)=(y,x)$ and $g: B \times A \rightarrow B$ is given by $g(y,x)=y$. Find $g \circ f$.

I'm not sure how to do the composition when dealing with $(y,x)$ or $(x,y)$ instead of just $x$. I know that $(y,x)=f(x,y)$ so $g(f(x,y))=g(y,x)=y$.

答:
Your solution is correct: $g\circ f:A\times B\to B$ mapping
$$(x,y)\mapsto y\,.$$


问:
Calculate the probability for the total insurance benefit -- solution verification

A family buys two policies from the same insurance company. Losses under the two policies are independent and have continuous uniform distributions on the interval from $0$ to $10$. One policy has a deductible of $1$ and the other has a deductible of $2$. The family experiences exactly one loss under each policy. Calculate the probability that the total benefit paid to the family does not exceed $5$.

This question has been discussed here and here. I've read them both but couldn't figure out why my solution is incorrect based on those discussions.

My attempt:
Let $P$ denote the loss amount under the first policy and let $Q$ denote that under the second policy. Let $X$ denote the benefits received under the first policy and let $Y$ denote that under the second policy. Then $f(p) = \frac{1}{10}$ when $0\le p\le 10$ and $f(q) = \frac{1}{10}$ when $0\le q\le 10$ and it follows that $f(p, q) = \dfrac{1}{100}$ when $0\le p, q\le 10$. Also,

$$X =
\begin{cases}
0 & 0\le P<1\\
P-1 & 1\le P\le 10
\end{cases}$$

$$Y =
\begin{cases}
0 & 0\le Q<2\\
Q-2 & 2\le Q\le 10
\end{cases}$$

We need to find $\Pr(X+Y\le 5 | X, Y>0)$. I am not sure what distributions $X$ and $Y$ follow (do they necessarily follow the uniform distribution?), so I tried to express the desired probability in terms of the variables whose joint distribution I derived.
$$\Pr(X+Y\le 5| X, Y>0)= \Pr(P+1+Q+2\le 5| P>1, Q>2) =\Pr(P+Q<8 | P>1, Q>2) $$
$$= \int_1^8 \int_2^{8-p} \frac{1}{100} \; \text{d}Q \, \text{d}P$$

This, however, did not lead to the correct response. Where have I gone wrong? Did I compute the conditional probability correctly? Should I divide $\int_1^8 \int_2^{8-p} \frac{1}{100} \; \text{d}Q \, \text{d}P$ by the product of the marginal distributions of $P$ and $Q$?

答:
You've mixed up your random variables.  $P$ and $Q$ are the ground-up loss variables, and $X$ and $Y$ are the claim payment variables.  $P, Q$ are uniform, but $X, Y$ are not.  You defined $X = P - 1$ if $P > 1$, yet when you wrote $$\Pr[X + Y \le 5 \mid X, Y > 0] = \Pr[P + 1 + Q + 2 \le 5 \mid P > 1, Q > 2] + \cdots,$$ you are now saying $X = P+1$ and $Y = Q + 2$.  This is not correct.  Your probability statement should instead read:
$$\begin{align}
\Pr[X + Y \le 5]
&= \Pr[P-1 + Q-2 \le 5 \mid P > 1, Q > 2]\Pr[P > 1, Q > 2] \\
&\quad + \Pr[P-1 \le 5 \mid P > 1, Q \le 2]\Pr[P > 1, Q \le 2]  \\
&\quad + \Pr[Q-2 \le 5 \mid P \le 1, Q > 2]\Pr[P \le 1, Q > 2] \\
&\quad + \Pr[0 \le 5 \mid P \le 1, Q \le 2]\Pr[P \le 1, Q \le 2].
\end{align}$$

This further simplifies to

$$\begin{align}
\Pr[X + Y \le 5] &= \Pr[P + Q \le 8 \mid P > 1, Q > 2]\frac{72}{100} + \Pr[P \le 6 \mid P > 1]\frac{18}{100} \\
&\quad + \Pr[Q \le 7 \mid Q > 2]\frac{8}{100} + \frac{2}{100}.
\end{align}$$  I leave it as an exercise to conclude that $$\Pr[P + Q \le 8 \mid P > 1, Q > 2] = \frac{25}{2(8)(9)} = \frac{1}{8}, \\ \Pr[P \le 6 \mid P > 1] = \frac{5}{9}, \\ \Pr[Q \le 7 \mid Q > 2] = \frac{5}{8},$$ hence $$\Pr[X + Y \le 5] = \frac{59}{200}.$$


问:
Recurrence inequality for Dirichlet's eta function.
I'm studying the following function:

$\theta(p)=\eta(p)\eta(p-2)-\frac{p-1}{p}\eta^2(p-1)$,

where $\eta$ - Dirichlet's eta function. If we build a plot for $p\in [1,150]$, it's easy to see that it's positive and decreasing for $p\in [3,150]$. I have a hypothesis that this function remains positive for all $p\ge 1$.

Are there any analytical results on this subject?

答:
You can say stronger things. Since
$1-2^{-x}<\eta(x)<1$ for real $x>0$
$$
1>\eta(p)\eta(p-2)>(1-2^{-p})(1-2^{-p+2})>1-\frac{5}{2^p} \\
1>\eta(p-1)^2>(1-2^{-p+1})^2>1-\frac{4}{2^p}
$$
So
$$
-\frac{5}{2^p} < \eta(p)\eta(p-2)-\eta^2(p-1) < \frac{4}{2^p}
$$
And for a sequence $\epsilon(p)$ taking on values in $[0,1)$
$$
\epsilon-\frac{4\epsilon+5}{2^p}<\eta(p)\eta(p-2)-(1-\epsilon)\eta^2(p-1)
< \epsilon+\frac{4}{2^p}
$$
In your case $\epsilon(p)=1/p$, so
$$
\frac{1}{p}-\frac{5+4/p}{2^p}<\theta(p)<\frac{1}{p}+\frac{4}{2^p}
$$
It follows that $\theta(p)>0$ for $p\ge 5$ and that it converges quickly to $1/p$.</p>


问:
Finding a counterexample of the following statement
This statements is false so I have to find a counterexample.

Statement: If $f:A \to B$ is injective iff $g\circ f = h\circ f$ for all function $g,h:B \to A$.

Attempt:
Let $A=\Bbb Z_{\ge 0}$ and $B = \Bbb Z$. I define $f:A \to B$ as: $f(x) = \frac{1}{4}\left(1-(-1)^x(2x+1)\right)$ and define $g,h:B \to A$ as:
\begin{equation*}
g(x) =
\begin{cases}
2x, & x \ge 0 \\
-(2x+1), & x < 0
\end{cases}
\end{equation*}
and $h(x) = |x|$.

Is it true?

答:
It is correct, but you still have to prove that it is correct by showing that $f$ is injective and that $g\circ f\ne h\circ f$. Your definition of $f$ is far more complicated than is necessary or desirable: the function is simply

$$f(n)=(-1)^{n+1}\left\lceil\frac{n}2\right\rceil=\begin{cases}
-\frac{n}2,&\text{if }n\text{ is even}\\
\frac{n+1}2,&\text{if }n\text{ is odd}\,.
\end{cases}$$

The first of these definitions is about as compact as you can get, and the second is probably the easiest to work with. Then you can check that

$$(g\circ f)(n)=\begin{cases}
n+1,&\text{if }n\text{ is odd}\\
n-1,&\text{if }n\text{ is even}
\end{cases}$$

and

$$(h\circ f)(n)=\begin{cases}
\frac{n}2,&\text{if }n\text{ is even}\\
\frac{n+1}2,&\text{if }n\text{ is odd}
\end{cases}$$

and easily demonstrate that these are different functions.

I’ll point out, however, that there are much simpler counterexamples. The simplest is this one:

Let $A=B=\{0,1\}$, $f(a)=a$ for $a\in A$, $g(b)=b$ for $b\in B$, and $h(b)=1-b$ for $b\in B$.


问:
Proving that intersection number of divisor is 0
My question arises from proposition 8.3 in the algebraic geometry text by Shigeru Iitaka:

Let $f:W\rightarrow S$ be a surjective morphism between nonsingular projective surfaces. Let $E$ be a divisor on $W$ such that $f(E) =p$ is a point. Then for any divisor $D$ on $S$, one has their intersection number on $W$: $(f^*D, E)_W=0$.

Here is the proof the author gave:

Basically we want to show that $(f^*D,E)_W =\operatorname{deg}(\mathcal{O}(f^*D)|_E)$ is zero. The author claims that $\mathcal{O}(f^*D)|_E \cong \mathcal{O}_E$ and it is proven as follows:

\begin{align*}
\mathcal{O}(f^*D)|_E & = (f^* (\mathcal{O}(D))|_E \\
& = f|_E^* (\mathcal{O}(D)|_p)  \quad \quad - \star \\
& \cong \mathcal{O}_E \quad \quad -\star\star
\end{align*}

Where $f|_E:E\rightarrow \{p\}$ denotes the restriction of $f$ to $E$. I am lost at the steps labelled ‘$\star$‘ and ‘$\star\star$’

Q1) What exactly does $\mathcal{O}(D)|_p$ mean in this context? I believe you need to restrict to a divisor on $S$. But $\{p\}$ is a point in $S$ and so restricting to just merely $’p’$ does not make any sense. Given that this restriction makes sense, then do we jump from the preceding step to $\star$?

Q2) Naturally, I am unable to understand how $\star\star$ is derived, i.e. why is the preceding line isomorphic to $\mathcal{O}_E$.

I believe if I understand how $\star$ is derived I will be able to work out the rest. Would greatly appreciate any help given! Thank you, and wishing you guys a happy new year. :)

答:
Looking at $f^*(\mathcal{O}(D))|_E$, we're pulling back $\mathcal{O}(D)$ along the two maps $E\to W\to S$, and because this composition $E\to S$ factors through $\{p\}$, it will equal the pullback of $\mathcal{O}(D)$ along $E\to \{p\} \to S$, which is what's going on in the first starred step.

In the second starred step, we're just using the fact that $\mathcal{O}(D)|_p$ is a line bundle on a point, so it's trivial, and pulling it back to $E$ again gives the trivial line bundle.


答:
If by 'more concisely' you meam 'fewer symbols', here's an equivalent expression that uses $15$. It does however, use the symmetric difference operation to achieve this concision, which you might regard as cheating!
$$((S_1 \triangle S_2) \cap S_3) \cup (S_1 \cap S_2)$$

Also, it's not unique: any permutation of the indices gives the same set.

Title: Given 3 events, $S_1$, $S_2$, $S_3$, how do I express $((S_1 \cup S_2) \cap (S_2 \cup S_3) \cap (S_3 \cup S_1))$ more concisely?

Body: Suppose I have three events: $S_1$, $S_2$, $S_3$. Is there a way to express $((S_1 \cup S_2) \cap (S_2 \cup S_3) \cap (S_3 \cup S_1))$ in a simpler manner?


问:
Number Theory - Differences of powers of 2
I'm not really familiar with number theory, or mathy stuff for that matter.
However I recently noticed that if you take the even powers of 2 ( $2^n$ where $n$ is even)
and write the difference of each two consecutive numbers ( $2^4 - 2^2$ which is 12),
and turn those differences into a sequence and repeat, like so:

```<code>1, 4, 16, 64,256
   3, 12, 48,192
       9, 36,144
          27,108
              81 (let's call this 'm')
</code>```

the final numbers is a power of 3.

Why does this happen? Is there any way to prove this? Is it a proven thing?

答:
The original sequence is $a_n = 2^{2n} = 4^n$. The first number in the second row is $4^1-4^0$, the first in the third row is $4^2-2\cdot 4^1 + 4^0$, etc.  In general the first number of the $k^{th}$ row is

$$4^k - \binom{k}{1} 4^{k-1} + \binom{k}{2} 4^{k-2} - \cdots + \binom{k}{k} 4^0 (-1)^k = (4-1)^k=3^k$$
via the binomial theorem.

Using the forward shift operator it's more clear: the $k^{th}$ difference operator is just
$$(\mathbb E -1)^k = \sum_{i=1}^k \binom{k}{i} (-1)^{k-i} \mathbb E^i $$
Therefore
$$(\mathbb E -1)^k a_0 = \sum_{i=1}^k \binom{k}{i} (-1)^{k-i} \mathbb E^i a_0 = \sum_{i=1}^k \binom{k}{i} (-1)^{k-i} a_i \\= \sum_{i=1}^k \binom{k}{i} (-1)^{k-i} 4^i = (4-1)^k$$


问:
What is wrong with this proof that x = |x|
First, take $x$. $x=x^1$. $1=\frac22$, so $x^1 = x^\frac22 = (\sqrt{x})^2 = |x|$. Therefore $x = |x|$.

I checked my proof and think that all my other steps are valid. I think my mistake has to do with
$x^\frac22 = (\sqrt{x})^2$

So, if there is anything wrong with this step, what is it?

答:
You must use the definition of $|x|$ and consider the cases $x > 0$ and $x \leq 0$.

What you have done does not work as $\sqrt{x}$ is not defined for $x < 0$ when working over the real numbers. Even then, after you square the square root of $x$ you are left with $|x| = |x|$...


问:
Condition to on $k$ to make the two sequences "disjoint"?
Suppose I have two sequences $a_n = n + \frac12 $ and $ b_m = k ( m + \frac12) $ for some constant $k>0$, and $n,m \in \mathbb{N}$

I am trying to find $k$ such that $a_n$ and $b_n$ have an empty intersection if they are viewed as sets. I think it should be that $k$ cannot be odd and if it is any other real number that would work.

For some really strange reason, I am struggling to prove this. Clearly, if we assume there is some $n$ and $m$ for which $a_n = b_m$, then $k$ must be rational, but I can't seem to go from there. I also tried to argue that $b_m - \frac12 $ must be an integer, but again I got stuck.

Appreciate any help massively, thank you!

答:
$a_n \not= b_m => (n+\frac{1}{2}) \not= k (m+\frac{1}{2}) => 2n+1 \not= k(2m+1)$

Which means that $ k \not= \frac{2n+1}{2m+1}$ and so for instances $k\not=1,\frac{3}{5} ,3,\cdots$


问:
Find a jacobian matrix of a function

Here's the problem:

Let $F:R^4 \to R^2, g:R^2\to R^2, h:R\to R$ all differentiable, such that:

$g(x_0,y_0) = (2,3), h(x_0) = 1, h(y_0) = 4$.

Suppose that $F(h(x),g(x,y),h(y)) = (0,0), \forall(x,y) \in R^2$.

The following Jacobian Matrices are given:

$JF(1,2,3,4) = \bigg[\matrix{1\\-2}\matrix{7\\0}\matrix{0\\-5}\matrix{2\\5}\bigg], Jh(x_0) = \big[\matrix{14}\big], Jh(y_0) = \big[\matrix{10}\big]$.

Calculate the jacobian matrix of $g$ in the point $(x_0,y_0)$.

What I Know

I'm really lost here... I know how to calculate a Jacobian Matrix, given a function, but I don't know how to start here or what I can use to solve it. I just need a tip to start.

答:
Rewriting some equations of your problem for convenience

$$h(x) = h_1(x), \, h(y) = h_2(y)$$

$$g(x,y) = \left[\begin{matrix}g_1(x,y)\\g_2(x,y)\end{matrix}\right]$$

$$f(h(x),g(x,y),h(y)) = f(h_1(x),g_1(x,y),g_2(x,y),h_2(y)) = \left[\begin{matrix}f_1(h_1(x),g_1(x,y),g_2(x,y),h_2(y))\\f_2(h_1(x),g_1(x,y),g_2(x,y),h_2(y))\end{matrix}\right] = \left[\begin{matrix}0\\0\end{matrix}\right]$$

$$J_f(h_1(x),g_1(x,y),g_2(x,y),h_2(y)) = \left[\matrix{\frac{\partial f_1}{\partial h_1}\\\frac{\partial f_2}{\partial h_1}}\matrix{\frac{\partial f_1}{\partial g_1}\\\frac{\partial f_2}{\partial g_1}}\matrix{\frac{\partial f_1}{\partial g_2}\\\frac{\partial f_2}{\partial g_2}}\matrix{\frac{\partial f_1}{\partial h_2}\\\frac{\partial f_2}{\partial h_2}}\right]$$

$$J_g(x,y) = \left[\matrix{\frac{\partial g_1}{\partial x}\\\frac{\partial g_2}{\partial x}}\matrix{\frac{\partial g_1}{\partial y}\\\frac{\partial g_2}{\partial y}}\right]$$

$$J_{h_1}(x) = \frac{\partial h_1}{\partial x}$$

$$J_{h_2}(y) = \frac{\partial h_2}{\partial y}$$

From the chain rule, we know that for $i={1,2}$

$$ \frac{\partial f_i}{\partial x} = 0 = \frac{\partial f_i}{\partial h_1} \frac{\partial h_1}{\partial x} + \frac{\partial f_i}{\partial g_1} \frac{\partial g_1}{\partial x} + \frac{\partial f_i}{\partial g_2} \frac{\partial g_2}{\partial x} $$

$$ \frac{\partial f_i}{\partial y} = 0 = \frac{\partial f_i}{\partial h_2} \frac{\partial h_2}{\partial y} + \frac{\partial f_i}{\partial g_1} \frac{\partial g_1}{\partial y} + \frac{\partial f_i}{\partial g_2} \frac{\partial g_2}{\partial y} $$

It is given that

$J_f(h(x_0),g_1(x_0,y_0),g_2(x_0,y_0),h(y_0)) = \bigg[\matrix{1\\-2}\matrix{7\\0}\matrix{0\\-5}\matrix{2\\5}\bigg], J_{h_1}(x_0) = \big[\matrix{14}\big], J_{h_2}(y_0) = \big[\matrix{10}\big]$

So it is possible to write the system

$$ 14 + 7 \frac{\partial g_1}{\partial x} (x_0,y_0) = 0 $$

$$ 20 + 7 \frac{\partial g_1}{\partial y} (x_0,y_0) = 0 $$

$$ -28 -5 \frac{\partial g_2}{\partial x} (x_0,y_0) = 0 $$

$$ 50 - 5 \frac{\partial g_2}{\partial y} (x_0,y_0) = 0 $$

Title: Find a jacobian matrix of a function


问:
Galois extension of $\mathbb{Q}_2$ with Galois group $(\mathbb{Z}/2\mathbb{Z})^3$
I am trying to solve this exercise:
Prove that $\mathbb{Q}_2$ has a unique Galois extension F with Galois group $(\mathbb{Z}/2\mathbb{Z})^3$. Compute it's ramification groups.

Here is what I have done so far:

Im will use the following lemma: If F has residue class degree f, then it contains $K=\mathbb{Q}_2(\zeta_{q^f-1})$ and F/K is totally ramified, while $K/\mathbb{Q}_2$ is unramified. The extension is of degree 8 and so it has residue class degree f = 1,2,4 or 8.

f=8: This can't happen since unramified extensions of $\mathbb{Q}_2$ have cyclic Galois group.

f=4: This can't happen either, since the extension has an unramified part of order 4, which is $\mathbb{Q}_2(\zeta_{15})/\mathbb{Q}_2$ and the Galois group of this extension is cyclic of order 4. (F has no element of order 4)

f=2: This works for example for $F=\mathbb{Q}_2(\zeta_3,\zeta_8)$, so this must be the only solution... Im not sure how to prove it though. By the lemma I've been using it must contain $\zeta_3$ and the other extension is an Eisenstein polynomial of order 4.

f=1: F is a totally ramified extension, so it comes from adjoining a root of an Eisensteinpolynomial. How do I get a contradiction here?

答:
I'm not sure I follow your attempted solution. However, here is how I would proceed: If $K$ is a field of characteristic not equal to $2$, then every quadratic extension of $K$ is of the form $K(\sqrt{a})$ for some $a$. (Recall that $\mathbb{Q}_2$ has characteristic zero, not two, so this is relevant.) If $\mathrm{Gal}(L/K) \cong (\mathbb{Z}/2)^3$, then $L$ must be the compositum of three disjoint quadratic extensions of $K$. So you should be trying to find $a$, $b$ and $c$ in $\mathbb{Q}_2$ such that $\mathbb{Q}_2(\sqrt{a}, \sqrt{b}, \sqrt{c})$ is a degree $8$ extension.

As a warm up, which elements in $\mathbb{Q}_2$ are squares?


问:
$ \sum_{y=1}^\infty {}_1F_1(1-y;2;-\pi\lambda c) \frac{\lambda^y}{y!} $
I am not able to solve the following sum. Can you please provide any hints ?
$$ \sum_{y=1}^\infty {}_1F_1(1-y;2;-\pi\lambda c) \frac{\lambda^y}{y!} $$

Note that the 3rd parameter of the Confluent Hypergeometric Function includes parameter λ.

Thank you for your time

答:
Integral method using the associated Laguerre polynomials with $\;x:=-\pi\lambda c$
\begin{align}
\sum_{n=1}^\infty {}_1F_1(1-n;2;x) \frac{\lambda^n}{n!}&=\sum_{n=1}^\infty \left[1+\sum_{k=1}^\infty \frac{(1-n)_k}{(2)_k}\frac{x^k}{k!}\right] \frac{\lambda^n}{n!}\\
&=\sum_{n=1}^\infty \sum_{k=0}^{n-1} \frac{(n-1)!}{(n-k-1)!(k+1)!}\frac{(-x)^k}{k!} \frac{\lambda^n}{n!}\\
&=\sum_{m=0}^\infty \sum_{k=0}^m \frac{m!}{(m-k)!(k+1)!}\frac{(-x)^k}{k!} \frac{\lambda^{m+1}}{(m+1)!}\quad\text{for}\ m:=n-1\\
&=\sum_{m=0}^\infty \frac{ L_m^1(x)}{(m+1)!}\frac{\lambda^{m+1}}{m+1}\\
&=\int \sum_{m=0}^\infty \frac{ L_m^1(x)}{(m+1)!}\lambda^m\;d\lambda\\
&=\int \frac{e^{\lambda}J_1\bigl(2\sqrt{x\lambda}\bigr)}{\sqrt{x\lambda}}\;d\lambda\\
&=\int_0^{2\sqrt{x\lambda}} \frac{e^{u^2/(4x)}}x\;J_1(u)\;du\quad\quad\text{(using}\ u:=2\sqrt{x\lambda}\,\text{)}\\
\end{align}
Since $\ \displaystyle L_m^1(x)= \sum_{k=0}^m \frac{(m+1)!}{(m-k)!\,(1+k)!\,k!}(-x)^k\ $ and $\ \displaystyle \sum_{m=0}^\infty \frac{L_m^1(x)}{(m+1)!}\lambda^m=\frac{e^{\lambda}J_1\bigl(2\sqrt{x\lambda}\bigr)}{\sqrt{x\lambda}}$

(formula $(5)$ and $(18)$ from the MathWorld link)

Initial method (returning alternative formulations...)

Let's expand the hypergeometric series using the Pochhammer symbols :
\begin{align}
\sum_{n=1}^\infty {}_1F_1(1-n;2;-\pi\lambda c) \frac{\lambda^n}{n!}&=\sum_{n=1}^\infty \left[1+\sum_{k=1}^\infty \frac{(1-n)_k}{(2)_k}\frac{(-\pi\lambda c)^k}{k!}\right] \frac{\lambda^n}{n!}\\
&=e^\lambda-1+\sum_{n=1}^\infty \sum_{k=1}^{n-1} \frac{(n-1)!}{(n-k-1)!(k+1)!}\frac{(\pi\lambda c)^k}{k!} \frac{\lambda^n}{n!}\\
&=e^\lambda-1+ \sum_{k=1}^\infty \frac{(\pi\lambda c)^k}{k!(k+1)!} \sum_{n=k+1}^\infty\frac{\lambda^n}{n\;(n-k-1)!}\\
&=e^\lambda-1+ \sum_{k=1}^\infty \frac{(\pi\lambda c)^k}{k!(k+1)!} \int \sum_{n=k+1}^\infty\frac{\lambda^{n-1}}{(n-k-1)!}d\lambda\\
&=e^\lambda-1+ \sum_{k=1}^\infty \frac{(\pi\lambda c)^k}{k!(k+1)!} \int \lambda^k\sum_{m=0}^\infty\frac{\lambda^m}{m!}d\lambda\\
&=e^\lambda-1+ \sum_{k=1}^\infty \frac{(\pi\lambda c)^k}{k!(k+1)!} \int \lambda^k e^{\lambda}\;d\lambda\\
&=e^\lambda-1-\sum_{k=1}^\infty \frac{(-\pi\lambda c)^k\;\gamma(k+1,-\lambda)}{k!(k+1)!}\\
\end{align}

(using the definition of the incomplete gamma function :
$\ \displaystyle\gamma(a,x)=\int_0^xt^{a-1}e^{-t}\;dt\ $)

\begin{align}
\sum_{n=1}^\infty {}_1F_1(1-n;2;-\pi\lambda c) \frac{\lambda^n}{n!}&=e^\lambda-1-\sum_{k=1}^\infty \frac{(-\pi\lambda c)^k\;k!\left(1-e^{\lambda}\sum_{m=0}^k\frac {(-\lambda)^m}{m!}\right)}{k!(k+1)!}\\
&=e^\lambda-1-\sum_{k=1}^\infty \frac{(-\pi\lambda c)^k\;\left(1-e^{\lambda}\sum_{m=0}^k\frac {(-\lambda)^m}{m!}\right)}{(k+1)!}\\
&=e^\lambda+\frac{e^{-\pi\lambda c}-1}{\pi\lambda c}+e^{\lambda}\sum_{k=1}^\infty \frac{(-\pi\lambda c)^k\;\sum_{m=0}^k\frac {(-\lambda)^m}{m!}}{(k+1)!}\\
\end{align}

At this point it would be interesting to know if :
$$f(x,y):=\sum_{k=0}^\infty \frac{x^k}{k!}\;\sum_{m=0}^k\frac {y^m}{m!}$$
may be simplified (I don't know...)


问:
Since dividing $x=x^6$ by $x$ gives $1=x^5$, how can I get to $x=0$ as a root?
this might sound like a stupid question,
bear with me it probably is.

I know the solutions for $x=x^6$ are 1 and 0.

Now, since $1 \cdot x = 1 \cdot x^6 $ and it follows $ 1 \cdot x = 1 \cdot x \cdot x^5$ and I divide both sides by $ 1 \cdot x$ to get $ 1=x^5 $, how can I find $x=0$?

Thank you so much in advance.

答: You cannot divide by $0$, it is not allowed. If you have $x=x^6$, it is not necessarily equal to $1=x^5$ if $x$ can be $0$. Take this example of $0x=0y$. If you divide both sides by $0$, you get $x=y$ even though $x$ doesn't have to equal $y$ for $0x=0y$ to be true. Likewise, if you let $x=0$, $x=x^6$ can be rewritten as $0\times1 = 0\times x^5$, and if you divide both sides by $0$, you get $1=x^5$. As we have shown before, $1=x^5$ isn't necessarily true because $x$ can be $0$, according to the original equation.


问:
Fake coin detection and relationship between coins and comparisons
I am reading about the problem of detecting a fake coin among a collection of $m$ coins optimally.

The first task is to define what we mean by optimally.

It starts with $m$ coins as a given and the following reasoning:

If we compare 2 coins by their weights we have 3 possible outcomes of
such a comparison which are tip to the left, tip to the right and even
balance.   So with $n$ comparisons we have $3^n$ different outcomes at
most

So far so good. Then

Suppose we have $m$ coins and at most 1 is fake. Then there are $1 + 2m$
possible possibilities which are 1 (all genuine) + $m$ (one of the m
coins is fake/lighter) + $m$ (one of the m coins is fake/heavier)

(a) At this point I am not really sure why it is $2m$ and not $\frac 2 m$ since it is $\frac 1 m$ that is fake (heavier/lighter).

Then it says:

with $n$ comparisons, the number of coins among which a fake can be
detected is at most $m$ where $1 + 2m = 3^n$.   More precisely if the
number of m coins is greater than $\frac {(3^n - 1)} 2$ then it is not possible
to guarantee we can find the fake coin with n comparisons

So optimally here means with $n$ comparisons.

(b) At this point though I don't really understand how the $n$ (as number of comparisons) showed up and determined/defined $m$ when all we had as given when we started is that we have $m$ coins.

Could someone please help me with (a) and (b)

答:
For a, let there be $m=5$ coins.  We can call them $A,B,C,D,E$.  It is asserted that there are $2 \cdot 5+1=11$ possibilities.  Either one of the coins is heavy ($5$ possibilities because it can be any coin), all the coins are correct, or one of the coins is light (again $5$ possibilities.)

For b, let us continue with $m=5$.  Each weighing has $3$ possible results.  If we weigh twice, there are $3 \cdot 3=9$ possible results.  As we can only distinguish $9$ possibilities with two weighings, it must take at least $3$ weighings to find a possible fake coin among $5$.  You determine $n$ by finding the smallest $n$ so that $3^n \ge 2m+1$ so you can distinguish enough possibilities.  This does not prove that the coin can be found in $n$ tries, it only proves that it cannot be found with certainty in $n-1$.  To prove that $n$ will work, one needs to lay out a strategy of weighings.

$m=5, n=3$ should be easy because $3^3=27$ is so much more than $11$.  We first weigh $AB$ vs $CD$.  If they balance $ABCD$ are all good and we weigh $E$ against any other coin and have the result.  If $AB$ are heavier than $CD$ we can have either of $AB$ heavy or $CD$ light.  We can then weigh $AC$ against $BD$.  If $AC$ is heavy either $A$ is heavy or $D$ is light.  We weigh $A$ against $B$ and have the answer.  The other results are similar.  We have shown that $3$ weighings suffice for $m=5$ coins.


问:
Quantifier elimination of $\exists x\bigwedge \Delta$
I'm currently studying logic and theory of computation (right now I'm concerned with quantifier elimination). Consider the successor theory generated by the following axioms
$$(S1)\,\forall\, x\neg S(x)=0;$$
$$(S2)\,\forall x\forall y\,(S(x)=S(y)\rightarrow x=y);$$
$$(S3)\,\forall x\,((\neg x=0)\rightarrow \exists y\,S(y)=x);$$
$$(S4)\,\forall x\,\neg S^k(x)=x.$$
Say there is a set $\Delta$ with elements of the form
$$S^n(x)=S^m(y),$$
where $0<n<m.$ My goal is to eliminate the quantifiers of $\exists x\bigwedge \Delta$, but I can't find a method that solves this. Any help? Thanks in advance

答: Given $0 < n < m$, $S^n(x) = S^m(y)$ is equivalent to $x = S^{m-n}(y)$ (by using $S2$ $n$ times). But then $\exists x \phi$, where $\phi$ is a conjunction including the formula $S^n(x) = S^m(y)$ is equivalent to $\phi[S^{m-n}(y)/x]$ (the resulting of substituting $S^{m-n}(y)$ for each occurrence of $x$ in $\phi$).


问:
How big can a quotient space be?
Let $X$ be a space of weight $w(X)=\kappa$. Suppose $q:X\rightarrow Y$ is a quotient map. If $q$ is open, or if $Y$ is compact, then $w(Y)\leq w(X)$. In general it is possible for $w(Y)$ to be larger than $w(X)$ (consider $\mathbb{R}/\mathbb{Z}$).

How much larger than $w(X)$ can $w(Y)$ be?

Really what I'm looking for is a machine to produce countexamples for any fixed infinite cardinal $\kappa$. I'm particularly interested in the case that $X$ is second-countable.

答:
Let $T_X$ be the topology on $X.$ Then $|T_X|\le 2^{w(X)}.$

Now $q:X\to Y$ is a continuous surjection. So if $B$ is a base for $Y$ then $A=^{def}\{q^{-1}b: b\in B\}\subset T_X. $

So $|A|\le 2^{w(X)}.$

So $|B|=|\{q[a]:a\in A\}\le |A|\le 2^{w(X)}.$

Title: How big can a quotient space be?


问:
General solution for $\cos(5x) = \sin(5x)$
I want to find the general solution of the following equation for $x∈\mathbb R$:

$\cos(5x) = \sin(5x)$

Believe me, I have no clues how to begin....

答:
You can start by letting $y=5x$.

Then $$\cos y = \sin y$$

First note that no solution to $\cos y =0$ solves this equation, so division throughout by $\cos y$ is allowed.

You get $\tan y = 1$.

At this point, you should be able to continue. Remember the periodicity of the tangent function (careful, the period is not $2\pi$) and don't forget to reverse the original substitution.


答:
$\cos(5x)=\sin(5x)\iff \tan(5x)=1\iff 5x=\dfrac\pi 4+n\pi, n\in\mathbb Z$.

Can you take it from here?

Title: General solution for $\cos(5x) = \sin(5x)$

Body:
I want to find the general solution of the following equation for $x∈\mathbb R$:

$\cos(5x) = \sin(5x)$

Believe me, I have no clues how to begin....


问:
How would I go about solving for the grayed out area?
I was asked to get the grayed out area that you see in this image.

All that I'm given to solve this is the three points of the main triangle (A, B, C) and the center of the circle (X) Also the dashed lines are parallel to the sides of the main triangle and tangent to the circle. All ways I can think of, for solving this problem seem over complicated, like: rotating a copy of the main triangle, like the dashed lines imply, and then computing the intersections between the sides of both triangles to get the vertices of the hexagon shaped area and then computing that area and subtracting it from the main triangle area.

Is there a better more intuitive way of solving this?

Thanks btw.

答:
The construction of the dotted line triangle appears to be the $180^\circ$ rotation of the original triangle about the incenter, or equivalently the construction of lines parallel to each side and shifted by twice the radius of the incircle.

Then the required residual triangles are all scaled-down copies of the original triangle.

The easiest general process seems to be:

Calculate the side lengths $|AB|,|BC|,|AC|$

Calculate the semiperimeter, $s = (|AB|+|BC|+|AC|)/2$

Calculate the area $A_T$ either using the coordinates (bounding rectangle - three triangles) or Heron's formula $A_T = \sqrt{\strut s(s-|AB|)(s-|BC|)(s-|AC|)}$

Calculate the incircle radius using $r_I = \dfrac{A_T}{s}$

Calculate the three altitudes $ h_A, h_B, h_C$ from the area and side lengths e.g $h_A = \dfrac{2A_T}{|BC|}$

Calculate the scaled-down area of each grey section, eg. $A_A = A_T\left(\dfrac{h_A-2r_I}{h_A}\right)^2$

Sum


问:
What is a direct proof, formally?
In my logic class I took in college, I sometimes would get wrong marks because I proved a statement indirectly instead of directly. Informally, a direct proof of a conditional $P \implies Q$, is one where you assume $P$ and try to deduce $Q$ "directly", without using the contrapositive of that conditional. But no one has ever given me a formal, rigorous definition of direct proof. Can someone give me one, and with one, can someone give examples of statements which can't be proven directly, only indirectly?

答:
From wikipedia at https://en.wikipedia.org/wiki/Direct_proof

"In mathematics and logic, a direct proof is a way of showing the truth or falsehood of a given statement by a straightforward combination of established facts, usually axioms, existing lemmas and theorems, without making any further assumptions.In order to directly prove a conditional statement of the form "If p, then q", it suffices to consider the situations in which the statement p is true. Logical deduction is employed to reason from assumptions to conclusion."

This sentence "In order to directly prove a conditional statement of the form "If p, then q", it suffices to consider the situations in which the statement p is true.", is particularly important because it tells us that a direct proof only examines what happens if the hypothesis is true. A true hypothesis can't yield a false conclusion, so a direct proof doesn't examine false conclusions, either. However, an indirect proof examines what would happen if the conclusion were false for its proof.

I am not sure if there is any theorem that has been proven to have indirect proof, but here's an indirect proof of the irrationality of $\sqrt2$ as an example.
https://www.math.utah.edu/~pa/math/q1.html

Title: What is a direct proof, formally?


问:
Show that $\gamma:[a,b]\rightarrow \mathbb{C}$ has bounded variation iff exists $h:[0,1]\rightarrow [a,b]$ with $\gamma \circ h$ Lipschitz.
Let $~\gamma:[a,b]\rightarrow \mathbb{C}~$ be a continuous curve: show that $~\gamma~$ has bounded variation if and only if there exists $~h:[0,1]\rightarrow [a,b]~$ continuous and surjective such that $~\gamma \circ h~$ is a Lipschitz curve.

My only idea is for the proof of the reciprocal proposition: if $$
|\gamma(h(x))-\gamma(h(y))|\leq k|x-y|~\forall x,y\in [0,1],$$
then, given a partition $P=\{a=x_0,x_1,\dots, x_n=b\}$, I have that exists $~y_i~$ with $~h(y_i)=x_i~$ for each $i$ from $1$ to $n$. Then,
$$\sum_{i=1}^n|\gamma(x_i)-\gamma(x_{i-1})|\leq \sum_{i=1}^n|\gamma(h(y_i))-\gamma(h(y_{i-1}))|\leq k\sum_{i=1}^n|y_i-y_{i-1}|$$
But I don't know how to continue, because I think that $h$ is not necessarily monotonic.

Edit: For the direct implication I found a reference that helped me a lot, here goes https://mathoverflow.net/questions/253024/from-bounded-variation-to-1-lipschitz-function

答: The intuition here: $\gamma$ has finite variation if and only if the curve $\gamma$ has finite length. Now the new parametrization is by arc-length, and it should be clear that it is Lipschitz with constant $1$. Works for maps into metric spaces $\gamma\colon [a,b] \to X$, continuous, and of finite variation.


问:
Confusion about Lemma 13.2 in Munkres' topology (property which implies that a collection is a basis for a topology)
Lemma 13.2 and its proof confuse me.

$X$ is a topological space and $C$ is a collection of open sets of $X$ satisfying a property. A specific topology is not mentioned in the lemma. In the proof, he assumes the properties of topologies are properties of $C$. Is $C$ a topology? Is this something that should be proven, or was it implicit in the language of the lemma?

He says he will often omit mentioning the topology $T$. But he defined open just in terms of the basis, and so in my mind, I don't associate an open set with any particular topology. Or in this case, a collection of open sets doesn't translate to topology. Is it supposed to?

答:
Lemma 13.2 on p.80 states: Let $X$ be a topological space. Suppose that $C$ is a collection of open sets of $X$ such that for each open set $U$ of $X$ and each $x \in U$, there is an element $K$ of $C$ such that $x \in K \subset U$. Then $C$ is a basis for the topology of $X$.

$C$ is not a topology, but a basis for a topology. A basis generates a topology by taking all unions of all elements of $C$.

I think your confusion is that "\'a priori", you don't know which sets are open. He assumes that you have a topology (so you DO know which sets are open), and then proceeds to show that the conditions  on $C$ he provides are sufficient to guarantee that $C$ is a basis and generates the same topology as the assumed one.


问:
Derivations of the algebra $K[x,y]/(y^2-x^3)$
Let $K$ be a field of characteristic zero, and let $S=K[x,y]/(y^2-x^3)$. It is easy to see that $S\cong R:=K[t^2,t^3]$ via the isomorphism induced by the ring homomorphism $K[x,y]\to R$ given by $x\mapsto t^2, y\mapsto t^2$. It is known that for any $K$-algebra $A$ and any ideal $J$,
$$
\operatorname{Der}_K(A/J) \cong \operatorname{Der}_J(A)/J\operatorname{Der}_K(A),
$$
where $\operatorname{Der}_K(A)$ is the space of $K$-linear derivations of $A$, and
$$
\operatorname{Der}_J(A) = \{D\in \operatorname{Der}_K(A) \; | \; D(J)\subseteq J\}.
$$
In Exercise 3.6 of Chapter 3 in the book A primer on Algebraic $D$-modules by Coutinho, I'm asked to show that the derivations
$$
D_1 = 2y\partial_x + 3x^2\partial_y \quad \text{and} \quad D_2 = 3y\partial_y - 2x\partial_x
$$
generate $\operatorname{Der}_K(S)$ as $S$-module. Then I have to conclude that $\operatorname{Der}_K(R)$ is generated as an $R$-module by $t\partial_t$ and $t^2\partial_t$.

I have two issues here: First, it seems to me that $D_2$ must be $3y\partial_y + 2x\partial_x$, as the application of the chain rule gives
$$
t\partial_t f = t \frac{\partial f}{\partial x} \frac{dx}{dt} + t \frac{\partial f}{\partial y} \frac{dy}{dt} = 2t^2 \frac{\partial f}{\partial x} + 3t^3 \frac{\partial f}{\partial y} = (2x\partial_x + 3y\partial_y)f
$$
and
$$
t^2\partial_t f = t^2 \frac{\partial f}{\partial x} \frac{dx}{dt} + t^2 \frac{\partial f}{\partial y} \frac{dy}{dt} = 2t^3 \frac{\partial f}{\partial x} + 3t^4 \frac{\partial f}{\partial y} = (2y\partial_x + 3x^2\partial_y)f = D_1f.
$$
Is it a typo in the book or am I misunderstanding something?

The second one is about the proof that $D_1$ and $D_2$ generates $\operatorname{Der}_K(S)$ as $S$-module. First of all, an element of $\operatorname{Der_J(K[x,y])}$, where $J=(y^2-x^3)$ has the form $d=f_1 \partial_x + f_2\partial_y$, and the condition $d(J)\subseteq J$ means that
$$
-3x^2f_1 + 2yf_1 \in J,
$$
and the condition that that $D_1$ and $D_2$ generates $\operatorname{Der}_K(S)$ as $S$-module means that there are polynomials $g,h\in K[x,y]$ such that
$$
f_1 \partial_x + f_2\partial_y - gD_1 -hD_2 \in J\operatorname{Der}_K(K[x,y]),
$$
but I can't manage to show that such $g$ and $h$ would exist.

Working a little bit this attempt, I need to find polynomials $u,v\in K[x,y]$ such that
$$
f_1 \partial_x + f_2\partial_y - gD_1 -hD_2 = (y^2-x^3)(u\partial_x + v\partial_y).
$$
Taking $D_2 = 3y\partial_y + 2x\partial_x$ with the "right" sign, we can see that the above equation is equivalent to the following equalities:
$$
f_1 - 2yg - 2xh = (y^2-x^3u) \quad \text{and} \quad f_2 - 3x^2g - 3yh = (y^2-x^3)v,
$$
and, as $-3x^2f_1 + 2yf_2 \in J$ there exists some polynomial $w\in K[x,y]$ such that $-3x^2f_1 + 2yf_2=(y_2-x^3)w$. A little manipulation gives
$$
-6h = -3x^2v + 2yu - w
$$
and I don't know if I can do more, or even if my attempt is correct, so any help will be appreciated.

答:
In terms of the modules of Kähler differentials, we have an exact sequence of $S$-modules
$$J/J^2 \overset{\delta}{\rightarrow} S \,dx \oplus S \,dy \rightarrow \Omega^1_{S/K} \to 0.$$
(Here, the middle term is $\Omega^1_{K[x,y]/K} \otimes_{K[x,y]} S$, and $J / J^2 \simeq S$; and $\delta : y^2 - x^3 \mapsto 2y\,dy - 3x^2\,dx$.)  Taking the duals, we get an exact sequence
$$0 \to \operatorname{Der}_K(S) \to S \partial_x \oplus S \partial_y \to S.$$
Therefore, $\operatorname{Der}_K(S)$ is the kernel of the map $\varphi : S \partial_x \oplus S \partial_y \to S$ given by $\partial_x \mapsto -3x^2, \partial_y \mapsto 2y$.  (In particular, I think you're right that there's a typo in the book: $3y \partial_y + 2x \partial_x \mapsto 3y (2y) + 2x (-3x^2) = 6(y^2 - x^3) = 0$; so $3y \partial_y + 2x \partial_x$ is indeed in the kernel.)

Note that the above is essentially equivalent to your calculation based on $\operatorname{Der}_K(A/J) \simeq \operatorname{Der}_J(A) / J \operatorname{Der}_K(A)$ where $\operatorname{Der}_J(A) = \{ \partial \in \operatorname{Der}_K(A) \mid \partial J \subseteq J \}$.

Now, it is easy to see that the graph of $\varphi$, as a submodule of $S \partial_x \oplus S \partial_y \oplus S$, is generated by
\begin{align*}
e_1 & = ( 1, 0, -3x^2 ) \\
e_2 & = ( 0, 1, 2y).
\end{align*}
This is equivalent to looking at the $K[x,y]$-submodule of $K[x,y] \partial_x \oplus K[x,y] \partial_y \oplus K[x,y]$ generated by
\begin{align*}
\bar e_1 & = (1, 0, -3x^2 ) \\
\bar e_2 & = (0, 1, 2y) \\
f_1 & = (y^2 - x^3, 0, 0) \\
f_2 & = (0, y^2 - x^3, 0) \\
f_3 & = (0, 0, y^2 - x^3).
\end{align*}
From here, calculate a Gröbner basis for this submodule with respect to an elimination order in which the domain coordinates are smaller than the codomain coordinates.  That will make it easy to get a generating set for the intersection of the graph of $\varphi$ with $S \partial_x \oplus S \partial_y \oplus 0$, from which it will be easy to deduce a generating set for $\operatorname{ker}(\varphi) \simeq \operatorname{Der}_K(S)$.


问:
Injection between posets $(\Phi([-1,1]^d)\setminus\{\emptyset\},\subseteq)$ and $(\{-, 0,+\}^d,\preceq)$, where $-\preceq 0$ and $+\preceq 0$.

Let $(\{-, 0,+\}, \preceq)$ be the poset defined by the relations $-\preceq 0$ and $+\preceq 0$ and let $\left(\{-, 0,+\}^d, \preceq\right)$ be its $d$ -fold product. Show that:
$$
(\Phi([-1,1]^d)\setminus\{\emptyset\},\subseteq)\cong(\{-, 0,+\}^d,\preceq),
$$
where $\Phi([-1,1]^d)$ is the face lattice of the $d$-dimensional cube.

I have an idea of what the isomorphism would be, but I cannot yet prove that it’s injective.

First, a remark is that, since the product of posets is a binary operation, the elements of $(\{-, 0,+\}^d, \preceq)$ technically have a lot of extra parentheses, which could be omitted without losing any important information. So, for example, instead of $((0,+),-)$, we’d simply have $(0,+,-)$.

Let $F \in \Phi([-1,1]^d)$ denote a face of the cube, and in particular let $v$ denotes a vertex. Then the mapping I’m thinking of is defined as:

\begin{align}
&\phi:(\Phi([-1,1]^d)\setminus\{\emptyset\},\subseteq)\to(\{-, 0,+\}^d,\preceq) \\
&\phi(F) = (\text{sgn}(\sum_{v_i \in F}v_{i1}),…,\text{sgn}(\sum_{v_i \in F}v_{id}))
\end{align}
i.e. we look at the vertices contained in $F$, summing up their entries component-wise, and then taking their signs.

Now I want to show that this map is injective. Given two faces $F_1$ and $F_2$ such that $\phi(F_1) = \phi(F_2)$, it’s sufficient to show that a vertex $v_i = (v_{i1},…,v_{id})$ in $F_1$ is also in $F_2$ (and vice versa). Consider an entry $j$ in $\phi(F_1) = \phi(F_2) \in \{-, 0,+\}^d$. If $F_1$ and $F_2$ contain just one vertex, then this entry determines exactly the sum at position $j$ of the vertices in $F_1$ and $F_2$, respectively (namely, the “sum” of just one summand, resulting in value $-1$ or $1$).

If $F_1$ and $F_2$ both contain $2^m$ vertices, then we can continue in an inductive fashion to show that the summands that result in the $j$ entry of $\phi(F_1)$ and $\phi(F_2)$ match each other one by one: in summing the $m$ entries in the $j$-th position of vertices in $F_1$ and those in $F_2$, assuming that the partial sums up to a certain position $n < m$ match, then the $j$ entry of $\phi(F_1)$ and $\phi(F_2)$ determines what the value of the next summands must be.

Of course, this only works if $F_1$ and $F_2$ have the same number of vertices. I got stuck here since I can’t show that given $\phi(F_1) = \phi(F_2)$, it cannot be the case that $F_1$ contains $m$ vertices, and $F_2$ contains $m’$ vertices, where $m \neq m’$.

Edit: I’ve found another map, which might be easier to check for the properties of an isomorphism. It’s not yet entirely rigorous, but I’m just putting it here for comments.

For $F \in \Phi([-1,1]^d), \ \phi(F) = (u_1,…u_d) := u \in (\{-, 0,+\}^d, \preceq)$, where $u$ satisfies the following:

$\cdot$ $u$ has |dim($F$)| $0$’s, at positions corresponding to the axes of $\mathbb{R}^d$ that $F$ is “parallel” to.

$\cdot$ The other positions in $u$ correspond to the remaining axes in $\mathbb{R}^d$, have values either $1$ or $-1$, depending whether $F$ lies in the positive or negative side of those axes, respectively.

答:
First, let me clarify some notation:

The poset $(\{-, 0,+\}^d, \preceq)$ to me means the set of d-tuples $(u_1,\dots,u_d)$ with each component $u_i\in \{-,0,+\}$, and with $u\preceq u'$ iff $u_i\preceq u_i'$ for all $i$.

The face lattice of the d-dimensional cube, $\Phi([-1,1]^d)$, can be expressed as follows: A face of a d-dimensional cube has the form $F=f_1 \times f_2 \times \cdots \times f_d$, where each $f_i$ is either the singleton $\{-1\}$, the singleton $\{+1\}$, or the interval $[-1,1]$. Then $\Phi([-1,1]^d)$ is the set of all such faces, together with the empty set $\emptyset$ (required to make it a lattice), and the order is usual set inclusion $\subseteq$.  In particular, $f_1\times\cdots\times f_d \subseteq f_1'\times\cdots\times f_d'$ iff $f_i \subseteq f_i'$ for all $i$.

Note that both $(\{-, 0,+\}^d, \preceq)$ and $\Phi([-1,1]^d) \setminus \emptyset$ contain $3^d$ elements, so to show they are isomorphic it suffices to find an injection which preserves the order.

The isomorphism is essentially what the OP suggested in the edit.  Define $\psi:(\{-, 0,+\}^d, \preceq) \rightarrow \Phi([-1,1]^d) \setminus \emptyset $ by $\psi(u_1,\dots,u_d) = f_1\times \cdots \times f_d$ where $f_i = \{\pm 1\}$ if $u_i=\pm$ and $f_i = [-1,1]$ if $u_i = 0$.  This is clearly a bijection, so it just remains to show that it preserves the order.

For $u,u'\in \{-, 0,+\}^d$ we have

$$
\begin{align*}
u\preceq u' & \iff \forall (i) \;\; u_i \preceq u_i' \\
& \iff \forall(i)\;\; \psi(u)_i \subseteq \psi(u')_i \\
& \iff \psi(u) \subseteq \psi(u')
\end{align*}
$$
where the second line is the only non-trivial one, and follows immediately from our definition of $\psi$ and the properties of the order $\preceq$.  Hence $\psi$ preserves the order, and we are done.


答:
Firstly, as some users have noted it should be $b_m=k(m+\frac 12)$.

If $a_n=b_m$ for some $n,m\in \mathbb N$ ,then $k$ must be positive. If you take $k$ as any even positive integer, say $2p$ then $a_n\notin \mathbb N$  but $b_m=2pm+p\in \mathbb N\,\,\forall n,m\in \mathbb N$. Thus you are done. If $k$ is any odd positive integer, say $2p+1$ then $b_m=p(2m+1)+m+\frac 12=a_n$  for
$n=p(2m+1)+m$ .

In general if $k\notin \{{2n+1\over 2m+1}:n,m\in \mathbb N\}$, then $a_n\neq b_m\,\,\forall n,m\in \mathbb N$ .

Title: Condition to on $k$ to make the two sequences "disjoint"?

Body:
Suppose I have two sequences $a_n = n + \frac12 $ and $ b_m = k ( m + \frac12) $ for some constant $k>0$, and $n,m \in \mathbb{N}$

I am trying to find $k$ such that $a_n$ and $b_n$ have an empty intersection if they are viewed as sets. I think it should be that $k$ cannot be odd and if it is any other real number that would work.

For some really strange reason, I am struggling to prove this. Clearly, if we assume there is some $n$ and $m$ for which $a_n = b_m$, then $k$ must be rational, but I can't seem to go from there. I also tried to argue that $b_m - \frac12 $ must be an integer, but again I got stuck.

Appreciate any help massively, thank you!


问:
Convergence in probability question clarification
Suppose you have a sequence of independent random variables {$ X_i, i \geq 1$} where:
$$\Bbb P(X_i = 1 )=1-\frac{1}{i}, \Bbb P(X_i =i )=\frac{1}{i}.$$
Let $Y_n=\frac{1}{n}X_n.$
How would you show that $Y_n$ converges to $0$ in probability?
i.e $\lim_{n \to \infty}\Bbb P(|Y_n| \gt \epsilon)=0$     $  \forall \epsilon \gt0.$

For a similar question, I had the distribution for $X_i$ [$\Bbb P(X_i = \sqrt{i} )= \Bbb P(X_i = -\sqrt{i} )= \frac{1}{i+1}$ and $\Bbb P(X_i=0)=1-\frac{2}{i+1}$ ] and where $Y_n=\frac{1}{n}\sum_{i=1}^{n}X_i$. I calculated $E(Y_n)=0, Var(Y_n)=\frac{2i}{n(i+1)}$. Therefore by Chebyshev's inequality, I obtained that the probability $\Bbb P(|Y_n|\gt\epsilon)\leq$Var($X$)/$\epsilon^2$= $\frac{2i}{\epsilon^2n(i+1)}$. Since Var($X$)/$\epsilon^2$ converges to $0$, the probability converges to $0$ and we say that $Y_n \to 0$ in probability.

However, that approach became particularly messy with the first question above.

答: Let $\epsilon > 0$ and $n>1+\lfloor\frac{1}{\epsilon}\rfloor$ ($\implies n\epsilon >1$). You note that $\mathbb{P}(|Y_n|>\epsilon) = \mathbb{P}(X_n >n\epsilon) = \mathbb{P}(X_n = n) = \frac{1}{n}$ for all $n>1+\lfloor\frac{1}{\epsilon}\rfloor$ and taking $n\to\infty$, you have done.


问:
What is the formulation of the Least Upper Bound propierty in First Order Logic?
I've been readining about the completeness Godel's theorems. Accordingly, the axioms of $R$ in first order logic make up one of these sets that is complete and consistent. But I've always seen the axiom of the least upper bound formulated in second order logic. What is the equivalent formulation in first order logic?</p>

答:
There is no properly equivalent formulation in first-order logic to the completeness property of $(\Bbb R,\leq)$.

One reason is that the second-order axiom allows us to quantify over all the subsets of $\Bbb R$, and there are $2^{2^{\aleph_0}}$ of those.

We can write a schema which says that every definable subset - with parameters - if it has an upper bound, then it has a least upper bound. If we don't allow parameters then there are only $\aleph_0$ subsets that we can define, and that is far from being enough; and if we do allow parameters then we can at most define $2^{\aleph_0}$ subsets, which is still far from enough to cover all the subsets.

There is, however, a small trick that we can use: We define a new language which contains another predicate $S$ and another binary symbol relation $\in$. We add the axioms which say that if $x\in y$ then $S(y)$ and $\lnot S(x)$; and that the axioms of $\leq$ are only applicable to the elements for which $\lnot S$ holds.

In short, we allow two sort of objects in our universe, sets and real numbers and we want to make sure that the things that were true in $(\Bbb R,\leq)$ are true for the things which are not sets.

Now we can write the following statements:

$$\varphi(A,x)=S(A)\land\forall y(y\in A\rightarrow y\leq x)$$
This is a formula stating that $x$ is an upper bound of the set $A$. And finally we can write:

$$\forall A(S(A)\land\exists y(\varphi(A,y)\rightarrow\exists y(\varphi(A,y)\land\forall z(\varphi(A,z)\rightarrow y\leq z))$$

We still have to slightly modify our model, now its universe is $\Bbb R\cup\cal P(\Bbb R)$ and the relations are $\leq$ and $\in$ as we normally think of them. It is not hard to verify that indeed the above statement is true.


问:
Show the Statistic is Complete
Consider a random sample $Y_1,\ldots,Y_n$ of the Uniform Distribution on the Interval $[-\phi,\phi]$

I'm wondering how I can show that the Statistic $$ T(\mathbf{Y}) = ( Y_{(1)} , Y_{(n)}) $$

is a Complete Statistic.

Thoughts so far :

the pdf of $T$ can be represented as $$ f(x,y) = n(n-1) \left(\frac{y-x}{2 \phi} \right)^{n-2} \frac{1}{4 \phi^2}  \; \; \; \; - \phi < x < y < \phi $$
The expectation of any measurable function  of $T$ can be represented as $$E[g(T)] = \int_{- \phi}^{\phi} \int_{x}^{\phi} g(x,y) f(x,y) \,dy\,dx $$

Setting this equal to $0$ however does not really let me continue very far. I cant deduce completeness

I can only guess that  I'm on the wrong track but I dont know how else to attempt this.

答:
Actually, $(Y_{(1)},Y_{(n)})$ is not a complete statistic for $U(-\phi,\phi)$.

Proof:

Method 1

If $(Y_{(1)},Y_{(n)})$ is complete statistic, since $(Y_{(1)},Y_{(n)})$ is sufficient it is also a minimal sufficient statistic by Bahadur's theorem.

We know that a minimal sufficient statistic for $U(-\phi,\phi)$ is $\max\{-Y_{(1)},Y_{(n)}\}$. Since minimal sufficient statistics are one-to-one corresponded, there should be a one-to-one map between $\max\{-Y_{(1)},Y_{(n)}\}$ and $(Y_{(1)},Y_{(n)})$ if $(Y_{(1)},Y_{(n)})$ is a minimal sufficient statistic. However, this is not possible since two different $(y_{(1)},y_{(n)})$ can have same $\max\{-y_{(1)},y_{(n)}\}$, e.g. when $(y_{(1)},y_{(n)})=(-1,0)$ or when $(y_{(1)},y_{(n)})=(0,1)$, $\max\{-y_{(1)},y_{(n)}\}$ are both $1$.

Method 2

We can write the p.d.f of $Y_{(n)}$ $$\delta_{Y_{(n)}}(y_n)=\frac{n}{(2\theta)^n}(y_n+\theta)^{n-1},y_n\in[-\phi,\phi]$$
and the p.d.f of $Y_{(1)}$ $$\delta_{Y_{(1)}}(y_1)=\frac{n}{(2\theta)^n}(\theta-y_1)^{n-1},y_1\in[-\phi,\phi].$$
Thus $$E_{\phi}[Y_{(n)}]=\frac{\phi(n-1)}{n+1}\text{ and }E_{\phi}[Y_{(1)}]=\frac{\phi(1-n)}{n+1}$$
This means if we let $g(x,z)=\frac{n+1}{n-1}z-\frac{n+1}{1-n}x$ we will get $$\mathbb{E}_{\phi}(g(Y_{(1)},Y_{(n)}))=0,\forall\phi\in\mathbb{R}\text{ but } g\not\equiv0,$$
which means $(Y_{(1)},Y_{(n)})$ is not complete.

Title: Show the Statistic is Complete


问:
Prove that there are prime numbers such as $p_{i}$ such that $A = p_{1}^{a_{1}}*p_{2}^{a_{2}} * ... * p_{k}^{a_{k}} , a_{i}> 0 $

A and B are natural numbers.

Prove that there are prime numbers such as $p_{i}$ such that   $A = p_{1}^{a_{1}}*p_{2}^{a_{2}} * ... * p_{k}^{a_{k}} , a_{i}> 0 $

Then we consider $ B = p_{1}^{b_{1}}*p_{2}^{b_{2}}*...*p_{k}^{b_{k}}$ , prove that $ (a,b) = p_{1}^{c_{1}}*...*p_{k}^{c_{k}} --- c_{i} = min (a_{i} , b_{i})$ , and $[a,b]=p_{1}^{d_{1}}*...*p_{k}^{d_{k}}--- d_{i}= max(a_{i},b_{i})$ . for all : 1<= i <= k

notice: The first part of the proof originates directly from the Fundamental Theorem of Arithmetic, but I have no idea for the second part of the prove, thanks for your help

答:
For each prime $p$, you can talk about the multiplicity with which the prime occurs in a given natural number.  For example $8$ has multiplicity $3$ at the prime $2$, and it has multiplicity $0$ at every other prime.

It's not difficult to see that a natural number $n$ is divisible by another natural number $m$ if and only if at each prime $p$, the multiplicity of $p$ in $n$ is greater than or equal to the multiplicity of $p$ in $m$.

Using this principle, it is pretty straightforward to interpret greatest common divisor and least common multiple in terms of prime factorizations.


答:
As you say, the topology is usually omitted, i.e. left implicit. The statement

Let $X$ be a topological space.

means

Let $X$ be a set, together with a chosen topology $\mathcal{T}$ on $X$.

the $\mathcal{T}$ being implicit. The collection $\mathcal{C}$ is exactly what he says in the statement of the theorem, namely,

$\mathcal{C}$ is a collection of open sets of $X$ such that, for each open set $U$ of $X$ and each $x\in U$, there is an element $C$ of $\mathcal{C}$ such that $x\in C\subset U$.

In other words,

$\mathcal{C}$ is a subset of $\mathcal{T}$ such that, for each $U\in\mathcal{T}$ and each $x\in U$, there is an element $C\in\mathcal{C}$ such that $x\in C$ and $C\subset U$.

The topology on $X$ was already chosen; he did not define the topology on $X$ in terms of any basis. The goal here is to prove that, if $\mathcal{C}$ is a collection of subsets of $X$ satisfying the above criterion, then $\mathcal{C}$ is a basis for the topology $\mathcal{T}$ on $X$.

Because elements of $\mathcal{C}$ are open subsets of $X$ (as defined by the topology $\mathcal{T}$), or in other words $\mathcal{C}\subset\mathcal{T}$, the elements of $\mathcal{C}$ will have the properties of open sets: for example, an intersection of elements of $\mathcal{C}$ will be an open set, because the intersection of any two open sets is again an open set. It need not be the case that an intersection of elements of $\mathcal{C}$ will again be an element of $\mathcal{C}$.

Title: Confusion about Lemma 13.2 in Munkres' topology (property which implies that a collection is a basis for a topology)

Body:
Lemma 13.2 and its proof confuse me.

$X$ is a topological space and $C$ is a collection of open sets of $X$ satisfying a property. A specific topology is not mentioned in the lemma. In the proof, he assumes the properties of topologies are properties of $C$. Is $C$ a topology? Is this something that should be proven, or was it implicit in the language of the lemma?

He says he will often omit mentioning the topology $T$. But he defined open just in terms of the basis, and so in my mind, I don't associate an open set with any particular topology. Or in this case, a collection of open sets doesn't translate to topology. Is it supposed to?


问:
How to show that this language is Turing recognizable
So this question has two questions and i have to use the answer from 1 to answer question 2. Assuming that my answer for 1 is good. I need help with 2. ( Correct me if wrong please.)

Question 1 :

Show that this language is decidable

The language is $L=$ { $< G,w > $ : $G$ is a grammar in normal form of Chomsky and $w$ is a word on the terminal alphabet that can be derivated in $G$ by at least 2 differents derivations trees. }

In other words, find an algorithm that when given a grammar that is in normal form of Chomsky and a word $w$, it concludes after a finite number of steps if yes or no $w$ can be derivated by two derivations trees.

Here's my attempt (correct me if this is wrong) :

Entry : < G, w >;

Try every derivations in $G$ of length $2|w| -1$.

If two derivations generate w stop and accept < G,w >. Else stop and reject < G,w >.

Question 2

Show that this language is Turing recognizable $K =$ { < G > : G is a grammar in normal form of Chomsky and G is ambiguous ( can have two derivation tree for same word).} In other words, give an algorithm that when given an grammar in normal form of Chomsky, it returns "true" if $G$ is ambiguous and return "false" or doesn't stop when $G$ is not ambiguous. ( Give an answer absed on the algorithm used in question 1)

My reflexion on question 2 :

For this part i am kinda lost, i know that we will use algorithm of question 1 to be able to return true but for the infinite loop part i am not sure on how to say this.

I am also not sure if i can use a word "w" since our entry is a grammar G.

Entry : < G >

Apply algorithm from question 1 if it returns true,  return true. ( Because question 1 basically confirm that the CNF is ambiguous).

note sure how to handle the rest for it to loop.

Thanks a lot for your help.

答:
In order to recognize a ambiguous CFG in Chomsky normal form, you only need to find a certificate string to show that it can be derived by at least two heterogeneous trees. Now that you have figured out how to verify whether a given string $w$ can be derived ambiguously, a method of simply enumerating every string over the alphabet in the string order works.

Aside:
This language is notoriously known to be undecidable. PCP can be reduced to it by a computable function (It is easy to modify the original function in a way such that it always produces unambiguous Chomsky normal form for every possible list of symbols on the top of the dominos and on the bottom alike). Hence, it is $\Sigma_1$-complete.


问:
$f(x)=\cos(x)$ if $x=1/n$ for $n\geqslant1$ and $f(x)=-1$ if not. Show that $f(x)$ is Riemann integrable on$ [0,1]$
Let $f:[0,1] \rightarrow \mathbb{R}$ such that$f(x)=\cos(x)$ if $x=1/n$ for $n\geqslant1$ and $f(x)=-1$ if not.

So, to show that $f$ is integrable, i need to proof that $\bar{S}_{\sigma}(f) \leqslant \underline{S}_{\sigma}(f) + \epsilon$ $\forall \epsilon > 0$. $\sigma$ denotes a subdivision of $[0,1]$, $\bar{S}$ superior Darboux sum and $\underline{S}$ lower Darboux sum.

I remarked that by density of irrational numbers i can write that $\underline{S}_{\sigma}(f)=-1$ for all subdivision $\sigma$. So, i have to show that

$\bar{S}_{\sigma}(f) +1 \leqslant\epsilon$.

So now i don't really see how to choose correctly the subdivision of $[0,1]$.. If someone could explain it with details i would appreciate it. Thanks in advance.

答:
Fix $\epsilon > 0$.

Let $M=m^2$ where $m > 1$ is an integer such that ${\large{\frac{6}{m}}} < \epsilon$.

Let $\sigma$ be the partition of $[0,1]$ consisting of the $M+1$ intervals $I_1,...,I_{M+1}$ with $I_k=[x_{k-1},x_k]$ where
$$
0=x_0 < x_1 < \cdots < x_{M+1}=1
\;\;\;\;\;
$$
and $x_1,...,x_M$ are defined by
\begin{align*}
x_1&=\frac{1}{m}\\[4pt]
x_k&=x_1+(k-1)d,\;\,\text{for}\;2\le k\le M\\[4pt]
d&=\frac{1-x_1}{M}=\frac{m-1}{m^3}\\[4pt]
\end{align*}
If $A,B$ are given by
\begin{align*}
A&=\left\{{\small{\frac{1}{n}}}{\;{\Large{\mid}}\;}1\le n\le m\right\}
\qquad\qquad\;\;\;\,
\\[4pt]
B&=\left\{{\small{\frac{1}{n}}}{\;{\Large{\mid}}\;}n > m\right\}
\\[4pt]
\end{align*}
then the following claims are immediate:

Each element of $B$ lies in the interior of $I_1$.$\\[4pt]$

Each element of $A$ lies in at most two of $I_2,...,I_{M+1}$.$\\[4pt]$

Now let $S$ be the Riemann sum given by
$$
S=\sum_{k=1}^{M+1}f(x_k^*)\Delta x_k
$$
where $x_k^*\in I_k$ and $\Delta x_k=x_k-x_{k-1}$.

Then we get the lower bound
\begin{align*}
S
&=
\sum_{k=1}^{M+1}f(x_k^*)\Delta x_k
\\[4pt]
&\ge
\sum_{k=1}^{M+1}(-1)\Delta x_k
\\[4pt]
&=
(-1)\sum_{k=1}^{M+1}\Delta x_k
\\[4pt]
&=
(-1)(1)
\\[4pt]
&=
-1
\end{align*}
and since

$f(x_k^*)\le 1$ for all $k$.$\\[4pt]$

For $2\le k\le M+1$, we have $f(x_k^*)=-1$ with at most $2m$ exceptions.

we get the upper bound
\begin{align*}
S
&=
\sum_{k=1}^{M+1}f(x_k^*)\Delta x_k
\\[4pt]
&=
f(x_1^*)\Delta x_1+\sum_{k=2}^{M+1}f(x_k^*)\Delta x_k
\\[4pt]
&=
f(x_1^*)x_1+d\sum_{k=2}^{M+1}f(x_k^*)
\\[4pt]
&\le
(1)(x_1)+d\bigl((1)(2m)+(-1)(M-2m)\bigr)
\\[4pt]
&=
-1+\frac{2(3m-2)}{m^2}
\\[4pt]
&<
-1+\frac{6}{m}
\\[4pt]
&<
-1+\epsilon
\\[4pt]
\end{align*}

Thus $-1\le S < -1+\epsilon$.

It follows that $f$ is Riemann integrable on $[0,1]$ and ${\displaystyle{\int_0^1 f(x)\,dx = -1}}$.


问:
The Empty Function
I refer to the book from Deiser, Einführung in die Mengenlehre, 3. Aufl. because I wanna know if I think correctly purely based on his definitions.

Deiser first defines the ordered pair for any mathematical objects a,b as (a,b) = {{a}, {a,b}}. From this definition follows that even if a = $\emptyset$ then the ordered pair is just ($\emptyset$,b) = {{$\emptyset$}, {b}}. So far so good.

Then he defines the function as a set f if for all a, b1, b2 it holds: if (a,b1) $\in$ f and (a,b2) $\in$ f then b1=b2.

Then at some point he says that the empty function (f: $\emptyset$ -> A) is indeed a function.

But IS IT according to what he wrote?

Here's a counterexample: let's assume f: $\emptyset$ -> A and A = {1,2}. So f = {($\emptyset$,1), ($\emptyset$,2)}. This violates the definition of a function and therefore it shows that the empty function is at least not always a function. But I always read that the empty function is a function and it's graph is the empty set. So where does it go wrong: does Deister give bad definitions or do I think wrongly somewhere and he's innocent?

答:
You have confused $\emptyset$ and $\{\emptyset\}$.

If the domain of $f$ is empty, then $f = \emptyset$.

Title: The Empty Function


问:
$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}\geqslant 3$
Let $a,b,c>0$ and $a^2+b^2+c^2=3$, prove

$$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}\geqslant 3$$

This inequality looks simple but I do not know how to solve it. The straightforward method is to bash the inequality with brutal computation method and assume $x = \min \{ x,y,z \}$, $y = x+a$, $z = x + b$, etc. I wonder if we can prove it using classical inequality.

答:
If you are unable to solve this equation by other means, this may help

WolframAlpha has a solution

$$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}\ge 3
\implies(a,b,c)\\= (-3,3,3), (0,1,1), (1,0,1),(1,1,0), (1,1,1), (3,-3,3), (3,3,-3) $$

WolframAlpha has a solution
$$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}= 4\implies
(a,b,c)\\
=(-5,3,7), (-4,4,4), (-3,9,5),(-2,2,0), (-2,4,2), (0,-2,2), (2,-2,4), (2,0,-1) $$

WolframAlpha has a solution
$$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}=  5\implies
(a,b,c)\\
= (-5,5,5), (5,-5,5), (5,5,5) $$

You can continue this for $6,7,8,\cdots$ and get different solutions for each. This may not be the "proper" way to solve an inequality but it may help in revealing "patterns".


问:
Can somebody explain this to me? Qual Prep.
Prepping for the qual I was given a list of questions for RA, among them was this:

Give a definition of the Borel $\sigma$-algebra $\mathcal{B}$ of $\mathbb{R}^{n}$. Let $\mathcal{I}$ be the smallest $\sigma$-algebra containing all the open cubes in $\mathbb{R}^{n}$. Show $\mathcal{I}=\mathcal{B}$.

So, the definition... straightforward...

A "borel $\sigma$-algebra of $\mathbb{R}^{n}$" is the smallest $\sigma$-algebra containing all open subests of $\mathbb{R}^{n}$.

But how am I suppose to prove the other part? He used the definition to form $\mathcal{I}$...

答:
$\mathcal B$ is the smallest sigma algebra containing all the open sets in $\mathbb R^n$, and $\mathcal I$ is the smallest sigma algebra containing all the open cubes in $\mathbb R^n$.  Clearly $\mathcal B$ contains all the open cubes in $\mathbb R^n$.  Therefore, $\mathcal I \subseteq \mathcal B$.

To show that $\mathcal B \subseteq \mathcal I$, it suffices to show that $\mathcal I$ contains all the open sets in $\mathbb R^n$.  This will follow if you can show that every open set in $\mathbb R^n$ is a countable union of open cubes.

The case $n = 1$ (every open set in $\mathbb R$ is a countable union of open intervals) is well known.  In fact, more is true: there is a countable collection of open intervals in $\mathbb R$ of which every open set in $\mathbb R$ is the union.  In particular, $\mathbb R$ is second countable.

The general case follows from an inductive argument and the definition of the product topology.


问:
Clarification regarding solving ODEs
Let consider the differential equation $y'(t)=(y(t)^2-a^2)t-b$. Is there any method to solve this ODE? I was trying to solve by variations of parameters but it is not solvable.

答: This is a Ricatti equation. You can find out how to solve it in many places, here is one. https://www.math24.net/riccati-equation/</p>

Title: Clarification regarding solving ODEs


问:
Integrate $\int\limits_0^\infty \frac{{1}}{(a^2+x^2)^7}\,dx\\$
$$\int\limits_0^\infty \frac{{1}}{(a^2+x^2)^7}\,dx\\$$
tried with $x = a\tan(y)$.

Body:
Let's consider a more general case,let $$I_n=\int_0^{+\infty} \frac{{1}}{(a^2+x^2)^n}dx$$Using partial integration, we have$$I_n=\frac{x}{(a^2+x^2)^n}\bigg|_{0}^{+\infty}+2n\int_0^{+\infty} \frac{{x^2}}{(a^2+x^2)^{n+1}}dx=2n\int_0^{+\infty} \frac{{a^2+x^2-a^2}}{(a^2+x^2)^{n+1}}dx=2nI_n-2na^2I_{n+1}$$We got it$$2na^2I_{n+1}=(2n-1)I_n,(2n-2)a^2I_n=(2n-3)I_{n-1}$$
So$$I_n=\frac{2n-3}{2n-2}\frac{2n-5}{2n-4}\cdots\frac{1}{2}\cdot I_1\cdot\left(\frac{1}{a^2}\right)^{n-1}=\frac{2n-3}{2n-2}\frac{2n-5}{2n-4}\cdots\frac{1}{2}\frac{1}{a^{2n-1}}\int_{0}^{+\infty}\frac{dx}{a^2+x^2}$$

Thus$$I_n=\frac{2n-3}{2n-2}\frac{2n-5}{2n-4}\cdots\frac{1}{2}\frac{1}{a^{2n-1}}\frac{\pi}{2}\cdot\varepsilon$$
Among$$\varepsilon=\begin{cases}
1,&a>0\\
-1,&a<0
\end{cases}$$


问:
Rings with zero divisors
Is there a ring  $~R~$ with non-trivial multiplication (i.e. $~\exists a,b\in R ~~~ ab\neq 0$) such that each non-zero element of $~R~$ is a zero-divisor? </p>

答: A (trivialish) example is $\{0, 1\}$ with $1 \cdot 1 = 0$.</p>

Title: Rings with zero divisors


问:
Computing the expectation $E[2^{X1+X2+X3}]$
Suppose that $X_1$, $X_2$ and $X_3$ are three independent random variables satisfying

$$ P(X_i=1)=P(X_i=−1)=\frac{1}{2} \qquad \text{for} \ i = 1,2,3 $$

I am asked to find $E[2^{X1+X2+X3}]$.

So far I have worked out $E[X_i]=0$ from symmetry and tried to separate using Fubini's:

$$ E[2^{X_1}]E[2^{X_2}]E[2^{X_3}] $$

however I am stuck on how to proceed or whether this is the right route.

答:
Observe that $Y=\frac{X+1}{2}$ is bernulli $B(1/2)$ thus $\Sigma_i Y_i\sim Bin(3;1/2)$ and thus $\Sigma_i X_i$ takes the values

$$\{-3;-1;1;3\}$$

with probabilities

$$\{1/8;3/8;3/8;1/8\}$$

And thus $2^{\Sigma_i X_i}$ takes the values

$$\{1/8;1/2;2;8\}$$

with the same probabilities...

Concluding:

$$\mathbb{E} [2^{\Sigma_i X_i}]=\frac{1}{64}+   \frac{3}{16}+  \frac{6}{8} +1=\frac{125}{64} $$

This matches  with your reasoning being

$$E(2^{X_1})=\frac{1}{2}\Bigg[\frac{1}{2}+2\Bigg]=\frac{5}{4}$$


问:
Show that $y[x_0,x_1,\ldots ,x_n]=0$
Show using the Newton representation of interpolation polynomial that $y[x_0,x_1,\ldots ,x_n]=0$ if $y_i=p(x_i)$, $i=0,  \ldots , n$ for a polynomial $p$ of degree $n-1$.

The Newton representation of interpolation polynomial using the divided differences is \begin{align*}p(x)&=\sum_{i=0}^ny[x_0,x_1,\ldots ,x_i](x-x_0)\cdots (x-x_{i-1})\end{align*}

We have that
$$p(x)= y[x_0] + (x - x_0)y[x_0, x_1] + \ldots + \prod_{j = 0}^{n - 1} (x - x_j) y[x_0, \ldots, x_n]$$
Do we take $x=x_i$ for $0\leq i\leq n$ to get desired result?

答:
Let $q$ be the unique polynomial of degree at most $n$ such that $q(x_i) = y(x_i)$ for $i = 0, 1, \dots, n$. Then by Newton's representation, we have
$$q(x) = \sum_{i = 0}^{n}y[x_0, \dots, x_i](x - x_0)\dots(x - x_{i - 1}).$$
By uniqueness, $q = p$. Therefore the coefficient of $x^n$ in $q(x)$ is $0$. This means exactly that
$$y[x_0, \dots, x_n] = 0.$$

Title: Show that $y[x_0,x_1,\ldots ,x_n]=0$


问:
Effective methods for this qustion.
In the beginning, the numbers $1, 2, 3, ..., 57$ were written on the board. In each process, the numbers a and b are selected and deleted from the board so that the number $a$ divides the number $b$ completely. The number $\frac{b}{a}$ is written instead of the deleted numbers (For example, in the first operation, the numbers $a = 3$ and $b = 9$ can be deleted and the number $\frac{9}{3} = 3$ can be written on the board again). At least how many numbers can be left on the board after a few actions?

It is clear that  the remaining numbers are primes $29,31,37,41,43,47,53$.
In each of my attempts, I find $2$ different numbers in addition to these prime.  But I am waiting for your answers to see more different points of view.

答:
You might try an approach like this in EXCEL because you don't have to worry about word-wrap, and you can see the entire string if the cells to the right are empty.

Beginning with $\{1,2,3,\cdots,57\}\\$

\begin{align*}
Column_A\quad & Column_B\\
19,57\rightarrow3\quad &\{1,2,3,3,4,5,6,7\cdots 56\}\\
8,56\rightarrow7\quad &\{1,2,3,3,4,5,6,7,7\cdots 55\}\\
11,55\rightarrow5\quad &\{1,2,3,3,4,5,5,6,7,7\cdots 54\}\\
27,54\rightarrow2\quad &\{1,2,2,3,3,4,5,5,6,7,7,\cdots,26,28\cdots 53\}\\
\end{align*}

Now, perhaps it is time to cleanup such as

\begin{align*}
2,2\rightarrow1\quad &\{1,1,3,3,4,5,5,6,7,7,\cdots,26,28\cdots 53\}\\
3,3\rightarrow1\quad &\{1,1,1,4,5,5,6,7,7,\cdots,26,28\cdots 53\}\\
5,5\rightarrow1\quad &\{1,1,1,4,6,7,7,\cdots,26,28\cdots 53\}\\
7,7\rightarrow1\quad &\{1,1,1,1,4,6,\cdots,26,28\cdots 53\}\\
1,1\rightarrow1\quad &\{1,1,1,4,6,\cdots,26,28\cdots 53\}\\
1,1\rightarrow1\quad &\{1,1,4,6,\cdots,26,28\cdots 53\}\\
1,1\rightarrow1\quad &\{1,4,6,\cdots,26,28\cdots 53\}\\
13,52\rightarrow4\quad &\{1,4,4,6,\cdots,12,14\cdots,26,28\cdots,51,53\}\\
\end{align*}

Number will be deleted, added, and deleted again from the list. For example, the interactions of $\quad11,55\quad 4,44 \quad 11,33 \quad 2,22\quad 11,11\quad$ will end up deleting $11$ from the list permanently. As an alternative to the above, you can, in one step, drop the numbers $\quad 11,22,33,44,55\quad$ from the list, knowing $11$ and multiples will all go. For other "factors" you will have to test them for having odd or even interactions as we did with $11x$.

Title: Effective methods for this qustion.


问:
$\int \sqrt{x^4+1}\ dx$
I have tried to find $\int \sqrt{x^4+1} dx$ but can not find a way to get the answer.

答:
I think you'll have to go to an integral table or software.  Regardless, here's the answer:

$$\frac{1}{3} \left(x \sqrt{x^4+1}-2 \sqrt[4]{-1} F\left(\left.i \sinh
   ^{-1}\left(\sqrt[4]{-1} x\right)\right|-1\right)\right) ,$$

where $F$ is an elliptic integral.

Title: $\int \sqrt{x^4+1}\ dx$


问:
Does $\sum a_i b_i $ converge, if $ \sum a_i^2 , \sum b_i^2 $ converge?
I came across a question in linear algebra,
Prove that $W = \{ \langle a_n \rangle \in V : \sum_{n=1}^\infty a_i^2$ is finite } is a vector space, where V is vector space of sequences in R.
Now, I tried to proceed as usual, but got stuck in proving that $\sum_{n=1}^\infty {ab}$ is finite. This question is not from a standard textbook, and hence I am unsure whether the statement is even true.

With some examples, it is easy to see that this statement is definitely true for series of the form $\sum \frac{1}{n^k}, k\in R$. However, whether this extends to all sequences, I am not sure.
If your answer is in the affirmitive, please provide a proof. I did try to prove the statement, but couldn't (Cauchy series doesn't help). I think Cauchy's inequality might help, but I am unsure how to use it(It has been time since I studied it in functional analysis class while in undergraduate class).

答:
Just remembered the inequality in its correct form (credit- Wikipedia). The statement itself is a one-line proof:
$|\langle u,v\rangle|^2 \leq \langle u,u\rangle\cdot\langle v,v\rangle$
I was writing it as $\sum{a_ib_i \leq (\sum a_i)^{1/2}(\sum b_i)^{1/2}}$
Credit : Qiaochu Yuan and Copper.hat in the comments section.


答:
With this antiderivative, you face elliptic integrals of the first kind.

If you are not aware of them, I prefer to just give the result
$$\int \sqrt{x^4+1}\,dx=\frac{1}{3} \left(x \sqrt{x^4+1}-(1+i) \sqrt{2} F\left(\left.i \sinh
   ^{-1}\left(\frac{(1+i) }{\sqrt{2}}x\right)\right|-1\right)\right)$$

Oherwise, use series expansions
$$\sqrt{x^4+1}=\frac 1{2\sqrt \pi}\sum_{n=0}^\infty (-1)^{n+1}\frac{ \Gamma \left(n-\frac{1}{2}\right)}{n!} x^{4n}$$ and integrate term wise.

Title: $\int \sqrt{x^4+1}\ dx$

Body: I have tried to find $\int \sqrt{x^4+1} dx$ but can not find a way to get the answer.


问:
Laurent expansion problem
Expand the function $$f(z)=\frac{z^2 -2z +5}{(z-2)(z^2+1)} $$ on the ring $$ 1 < |z| < 2 $$

I used partial fractions to get the following $$f(z)=\frac{1}{(z-2)} +\frac{-2}{(z^2+1)}  $$
then
 $$ \frac{1}{z-2} = \frac{-1}{2(1-z/2)} = \frac{-1}{2} \left[1+z/2 + (z/2)^2 + (z/2)^3 +\cdots\right]
 $$

but I'm stuck with $$ \frac{-2} {z^2+1} $$ how can I expand it?

答: Hint: Find $A,B$ such that $$\frac{-2}{z^2+1}=\frac{A}{z+i}+\frac{B}{z-i},$$ then proceed in a similar fashion to what you did with the term $\frac1{z-2}.$ Alternately, note that $1<|z|$ if and only if $1<|z|^2=|z^2|$ if and only if $|\frac1{z^2}|<1,$ and just work with $\frac{-2}{z^2+1}.$

Title: Laurent expansion problem


问:
How to prove undecidability other than Rice Theorem?
Im studying Rice Theorem and I would like to verify its consistency. If I am able to prove de undecidability in other ways, the Rice Theorem would prove to be useful after all.

Im trying to find out two different ways to prove that the following Decision problem is undecidable:

Is it true that the Turing Machine $M$ accepts exactly a finite and odd number of words?

So far:

One way is using the Rice Theorem. Let $X$ be the property of being finite and have an odd number of words. Rice Theorem tells us that for every property of semi-decidable languages, the Test of Property is undecidable only if $X$ is non trivial. Now I need to verify if $X$ is non-trivial correct? I got stuck on that part.

Talking about other way to prove, I was thinking about Reductio ad absurdum. I have an idea but I don't know how to build this reduction.

If I have an entry $(M,w)$ of HALTING that produces $r(M,w) = M'$ of ODD.

If $(M,w)$ is an affirmative instance of HALTING then $M'$ accepts only one word and nothing more.
If $(M,w)$ is an negative instance of HALTING then $M'$ accepts no words at all.

Is that clear? Sorry, english is not my mother language, so translating the way we refer to problems and theorems can be tricky.

Thanks for everything.

答:
There are indeed other ways to prove that the problem is undecidable. However, the technique can be used to prove the more generous Rice's theorem as well.

That is, by virtue of recursion theorem. It allows for a computable process to self-replicate for any Turing machines, which makes the diagonalization technique ready to use under many circumstances.


问:
Average shortest distance between some random points in a box
Suppose there is a square box with side length $m$ (measured in pixels). Let there be $n$ points in this box, distributed uniformly within the box (with integer coordinates, aligned to a pixel grid). If we take from each point the Euclidean distance to its nearest neighbor, what would be the expected value of this distance averaged over all points?

My actual problem is about a discrete pixel grid, an $m\times m$ bitmap image, but if that's easier I would be happy with a continuous solution. A more general solution e.g. for a rectangle instead of a square box is welcome, but at this point it is not necessary. I found similar questions about the continous case, but without answers. For me it wouldn't be easy to generalise the case for two points only.

答:
For the continuous version, assume that you pick $n$ points randomly from $[0, 1]^2$. You are trying to find $$E\left[ \frac{1}{n}\sum_{k=1}^n \min_{i, 1 \le i \le n, i \not=k}||X_k-X_i|| \right] \tag 1$$

where each of $X_i$ are randomly and uniformly chosen from $[0, 1]^2$.

Because the expected value of a sum of random variables is equal to the sum of the expected values, this simplifies to $$ \frac{1}{n}\sum_{k=1}^n E\left[\min_{i, 1 \le i \le n, i \not=k}||X_k-X_i|| \right] \tag 2$$

By symmetry, this is $$E\left[\min_{i, 1 \le i \le n-1}||X_n-X_i|| \right] \tag 3$$ Let $F_n(x) = P\left(\min_{i, 1 \le i \le n-1}||X_n-X_i|| > x\right)$, which would be $1$ for $x < 0$ and $0$ for $x > \sqrt{2}$. Then $(3)$ is $\int_0^{\sqrt{2}}F_n(x) dx$. $F_n(x)$ is equal to $$P\left(||X_n-X_1|| > x\right)^{n-1} \tag 4$$

The probability $P\left(||X_n-X_1|| > x\right)$ is given using formula $(6)$ from this paper as $\int_x^{\sqrt{2}}g(t)dt$ where $$g(x) = \begin{cases}
      2x^{3}-8x^{2}+2\pi x & 0 \leq x < 1 \\
      -2x^{3}-\left(2\pi+4\right)x+8x\operatorname{arccsc}\left(x\right)+8x\sqrt{x^{2}-1} & 1\leq x < \sqrt{2}
   \end{cases} \tag 5$$

Therefore, $P\left(||X_n-X_1|| > x\right)$ is given by $$\begin{cases}
-\frac{x^{4}}{2}+\frac{8x^{3}}{3}-\pi x^{2}+1 & 0 \leq x < 1 \\
\frac{2}{3}-\frac{4}{3}\left(2x^{2}+1\right)\sqrt{x^{2}-1}+\frac{1}{2}x^{4}+\left(2+\pi\right)x^{2}-4x^{2}\operatorname{arccsc}\left(x\right) & 1 \leq x < \sqrt{2}
\end{cases} \tag 6$$

The integral $\int_0^{\sqrt{2}}P\left(||X_n-X_1|| > x\right)^{n-1}dx$ is unlikely to have a closed form (or if it does, it would probably be very complicated). Instead, for an approximation, split it up as $$\int_0^{1}P\left(||X_n-X_1|| > x\right)^{n-1}dx + \int_1^{\sqrt{2}}P\left(||X_n-X_1|| > x\right)^{n-1}dx$$

The second integral has an upper bound of $\left(\frac{19}{6}-\pi\right)^{n-1}\left(\sqrt{2}-1\right)$, obtained because each value in that interval has an upper bound of $P\left(||X_n-X_1|| > 1\right)^{n-1} = \left(\frac{19}{6}-\pi\right)^{n-1}$. The first integral is given by $$\sum_{k=0}^{n-1}\sum_{m=0}^{n-k-1}\sum_{r=0}^{n-k-1}\frac{\binom{n-1}{k, m, r}}{\left(1+2k+3m+4r\right)}\left(-\pi\right)^{k}\left(\frac{8}{3}\right)^{m}\left(-\frac{1}{2}\right)^{r}$$

Numerically, this approaches $\frac{1}{2\sqrt{n-1}}$, but I'm having trouble proving it.


问:
Law of total covariance for random vectors
Suppose $Y_1, Y_2,$ and $X$ are random variables, then the iterated covariance formula is as follows:

$$Cov(Y_1, Y_2) = E(Cov(Y_1, Y_2|X)) + Cov(E(Y_1|X), E(Y_2|X))$$

Now let $Y = \begin{bmatrix}Y_1 \\ Y_2 \end{bmatrix}$. And let $X$ denote the set of random variables $X_1, \ldots, X_n$. And I'm interested in writing out the formula for $Cov(Y)$. Is the following correct?

$$Cov(Y) = E(Cov(Y|X)) + Cov(E(Y|X))$$

Or should I write it as

$$Cov(Y) = E(Cov(Y|X)) + Cov(E(Y_1|X), E(Y_2|X))$$

答:
If $\text{Cov}(Y)$ denotes the covariance matrix of $Y$, then I think your first equation is fine.
Since $E[Y \mid X]$ is a random vector, $\text{Cov}(E[Y\mid X])$ is another covariance matrix, which is fine.

You can check that it is true by checking each entry of the matrix. The equality of the off-diagonal entries is true by the law of total covariance. The equality of the diagonal entries is true by the law of total variance.


问:
Upper bound sum of entries of a vector in terms of L2 norm
Consider a vector $\mathbf{x}\in\mathbb{R}^n$ with entries $x_i$. Given that $\|\mathbf{x}\|^2_2\le n$, and no other information, what is the best possible upper bound on $\sum_{i=1}^n x_i$ that we can have?

答:
Apply the Cauchy-Schwarz inequality and check that it is tight.

 $$\sum_{i=1}^n x_i \le \sqrt{\sum_{i=1}^n x_i^2} \sqrt{\sum_{i=1}^n 1^2} = \sqrt{n}\|x\|_2$$ Equality holds when $x_1=\cdots=x_n$.


Title: Laurent expansion problem

Body:
Expand the function $$f(z)=\frac{z^2 -2z +5}{(z-2)(z^2+1)} $$ on the ring $$ 1 < |z| < 2 $$

I used partial fractions to get the following $$f(z)=\frac{1}{(z-2)} +\frac{-2}{(z^2+1)}  $$
then
 $$ \frac{1}{z-2} = \frac{-1}{2(1-z/2)} = \frac{-1}{2} \left[1+z/2 + (z/2)^2 + (z/2)^3 +\cdots\right]
 $$

but I'm stuck with $$ \frac{-2} {z^2+1} $$ how can I expand it?

Body:
So, on $|z|>1$ we have
$$\frac1{z^2+1}=\frac1{z^2}\frac1{1+\frac1{z^2}}=\frac1{z^2}\left(1-\frac1{z^2}+\frac1{z^4}-\dots\right)\,.$$


问:
Is the Banach Algebra generated by a separable subset separable?
Let $A$ be a Banach algebra and $S \subset A$ a subset. Suppose $B$ is the Banach algebra generated by $S$ and that $S$ is a separable subset of $A$.

Does this imply that $B$ is separable?

My intuition says the answer is yes, but I can't quite prove it. Here's my progress. For any  $T \subset A$, I define
$$
W(T):=\{ t_1\cdots t_m : m \in \mathbb{N}, t_j \in T\} \subset A
$$
By definition of Banach algebra generated by a set we have that
$$
B_1:=\left\{ \sum_{k=1}^n \lambda_k w_k  : n \in \mathbb{N}, \lambda_k \in \mathbb{C} , w_k \in W(S) \right\}
$$
is a dense subset of $B$. Let $D$ be a countable dense subset of $S$. I claim that
$$
B_2:=\left\{ \sum_{k=1}^n \lambda_k w_k : n \in \mathbb{N}, \lambda_k \in \mathbb{Q}+i\mathbb{Q} , w_k \in W(D) \right\}
$$
is a countable dense subset of $B$. That $B_2$ is countable is clear. I can't quite prove that is dense because I am having issues showing that an any element of $W(S)$ is arbitrarily close to an element in $W(D)$.

I would appreciate any help with this. It might as well be that my intuition is wrong or that I am trying to prove it in a complicated way.

Thanks in advance.

答:
Yes, it's true, and your proof idea works. The Banach algebra generated by $S$ is the closure of the (noncommuting) polynomials in $S$ as you say. So it suffices to show that any polynomial in $S$ (with complex coefficients) can be approximated by a sequence of polynomials in $D$ (with coefficients in $\mathbb{Q}(i)$), since then the closure of the latter contains the former and hence the closure of the former.

By linearity (and the continuity of addition and scalar multiplication), together with the density of $\mathbb{Q}(i)$ in $\mathbb{C}$, it suffices to show that a monomial $\prod_{i=1}^n s_i, s_i \in S$ can be approximated by a sequence of monomials in $D$. This follows from the continuity of multiplication: take for each $s_i$ a sequence $d_{i, j} \in D$ converging to $s_i$ and consider the sequence of monomials $j \mapsto \prod_{i=1}^n d_{i, j}$.


问:
Finding the dimension of certain tensor product with flat $A$-algebra
Assume $A$ is a noetherian local ring with $\mathfrak{m}_y$ being the unique maximal ideal and $\dim A=0$.  We have the exact sequence
$$0\to \mathfrak{m}_y\to A\to k(y)\to 0,$$
where $k(y)$ is the residue field $A_{\mathfrak{m}_y}/\mathfrak{m}_y$.

Let $B$ be a finitely generated flat $A$-algebra by $\phi:A\to B$, then
$$0\to \mathfrak{m}_y\otimes_A B\to B\to k(y)\otimes_A B\to 0$$
is an exact sequence of $A$-algebra. From the fact that $\dim A=0$, $\mathfrak{m}_y$ is nilpotent.

Let $\phi^{-1}(\mathfrak{p}_x)=\mathfrak{m}_y$ for a prime ideal $\mathfrak{p}_x$ in $B$. My question is: Why would this implies $$\dim (k(y)\otimes_A B)_{\mathfrak{p}_x}=\dim B_{\mathfrak{p}_x}?$$
An approach that I can think of is to consider the dimension of the objects in the second exact sequence above. But is it true that $\dim (\mathfrak{m}_{y}\otimes_A B)_{\mathfrak{p}_x}=0$ and does it make sense?

This question will be helpful in understanding Hartshorne III.9.5, the dimension formula of fiber of flat morphism.

答:
As mentioned in the comments you need to use the fact that modding out by a niplotent ideal does not change the dimension. The reason is that a nilpotent ideal is contained in every prime ideal, or in other words the topological space defined by taking $\operatorname{Spec}$ are isomorphic (but not as schemes!).

Now $k(y)\otimes_A B\cong B/\mathfrak{m}_yB$ and $\mathfrak{m}_yB$ is nilpotent in $B$.

Thus your result follows.


问:
in which base is the number 221 the factor of 1215
I basically tried to find a number $n$ which would be the base and then divide both the numbers $221$ and $1215$ so as to get a combo which has same remainder
But I am not able to proceed in doing so.
Help would be appreciated

答:
$$ 2  \left(   x^{3}  + 2 x^{2}  +  x  + 5 \right)  =  \left(  2 x^{2}  + 2 x  + 1 \right)  \cdot  \left(     x  + 1   \right)    -  x  + 9     $$

so try base $9$

We have
$$ 221_{nine} = 181  $$
$$ 1215_{nine} = 905   $$
As both are odd, it did no damage when I multiplied the larger number by $2$


问:
Closed form of an integral
Is there a closed form of $$\int\limits_0^1\frac{\arctan(\sqrt{x^2+2})}{(1+x^2)\cdot(\sqrt{x^2+2})}dx \quad \text{?}$$

I just know that $\int\limits_0^1\frac{\arctan(\sqrt{x^2+2})}{(1+x^2)\cdot(\sqrt{x^2+2})}dx = 0.514042...$

答:
This is called the Ahmed integral.

$$ \int_{0}^{1} \frac{\tan^{-1}\sqrt{x^2 + 2}}{(x^2 + 1)\sqrt{x^2 + 2}} \, dx = \frac{5\pi^2}{96} . $$

You can see a solution here.

Title: Closed form of an integral


问:
Do some rules of syntax permit a sentence to literally be an element of a set?
In meta-logic, it's routine to consider sets of sentences. However, are there rules of syntax that permit a sentence to be literally an element of a set?

Is there always an explicit or implicit encoding of a sentence, so that we replace each sentence with an object that can be an element of a set, and strictly speaking study sets of those non-sentence objects, rather than study sets of sentences?

答:
The most common set theory, ZFC, has no prohibition on anything being a member of a set.  It does not require them, but you can have a model with things that are not sets being members of sets.  Those things might as well be sentences.  This will not allow you to consider the structure of the sentences themselves.  Each one is an atom.  You could add some axioms to define the structure of sentences and how to relate them to each other.

There is also the technique used in the original Gödel proof of independent propositions in arithmetic.  It maps each string of characters or set of strings to a natural number.  He gave a function that would take a natural number and report whether it was a syntactically correct sentence.  Any set of properly chosen natural numbers can then be interpreted as a set of sentences.