问:
Determine the degree of the field extension

Determine the degree of the field extension

($1$st task) $\mathbb{Q}(\sqrt[3]{5},\sqrt[4]{6},\sqrt[5]{7})/\mathbb{Q}$,

($2$nd task) $\mathbb{Q}(\sqrt{3}, \sqrt[10]{75})/\mathbb{Q}$,

($3$rd task) $N/\mathbb{Q}$, where $N$ is the splitting field  of the polynomial $X^5-3\in \mathbb{Q}[X]$.

Studying for my upcoming algebra exam, I found this exercise from an old algebra exam.

So far I have:

The first task

For the $\sqrt[3]{5}$ we have
$\alpha = \sqrt[3]{5} \rightarrow \alpha^3-5=0$ is the minimal polynomial, so we have $f(t)=t^3-5$. The Eisenstein criteria is fulfilled with $p=5$, so the degree is $\deg(f(t))=3$.

So for $\sqrt[4]{6}$ the degree should be $4$ and for $\sqrt[5]{7}$ the degree should be $5$. But I am not sure how to proceed from here, because normally you should factorize the given numbers which is not possible here.

The second task

$\sqrt[2]{3}$ so the minimal polynomial should be $\alpha^2-3=0$ and the degree is $2$. The term $\sqrt[5]{75}$ can be factorized into $\sqrt[5]{75}=\sqrt[5]{25} \cdot \sqrt[5]{3}=\sqrt[5]{5} \cdot \sqrt[5]{5} \cdot \sqrt[5]{3}$, but I am unsure how to proceed to determine degree from here.

The third task

For the third task I unfortunately have no idea how to start solving it. Maybe someone could give me hint?

Edit Thanks for the quick answer. For the third task the splitting field is $\mathbb Q(\zeta_5,\sqrt[5]3)$. So I think the degree of $\mathbb{Q}[\sqrt[5]{3}]$ over $\mathbb{Q}$ is 5 because the polynomial is irreducible in $\mathbb{Q}$. And the solutions for the polynomial are $\alpha=\sqrt[5]{3}$ times the $5$th roots of unity $\zeta_5=e^{\frac{2\pi i}{5}}$, that is $\alpha, \alpha \zeta_5, \alpha \zeta_5^2, \alpha \zeta_5^3$ and $\alpha \zeta_5^4$. Then the degree of $\mathbb{Q}[\zeta_5]$ over $\mathbb{Q}$ is 4. So the degree of the splitting field should be $[\mathbb{Q}[\alpha, \zeta_5]:\mathbb{Q}]=20$, because $4$ and $5$ have no common divisor.

Is this right?

Thanks in advance!

答:
For the first task: You already noted that \begin{align*}
[\Bbb Q(\sqrt[3]{5}):\Bbb Q]&=3\\
[\Bbb Q(\sqrt[4]{6}):\Bbb Q]&=4\\
[\Bbb Q(\sqrt[5]{7}):\Bbb Q]&=5
\end{align*}
Since these degrees are pairwise coprime the corresponding field extensions are pairwise linearly disjoint and we get as you did in task three:
\begin{align*}
[\Bbb Q(\sqrt[3]{5},\sqrt[4]{6},\sqrt[5]{7}):\Bbb Q]&=[\Bbb Q(\sqrt[3]{5}):\Bbb Q][\Bbb Q(\sqrt[4]{6}):\Bbb Q][\Bbb Q(\sqrt[5]{7}):\Bbb Q]=3\cdot 4\cdot 5 = 60
\end{align*}

For the second task: We can do this actually a bit quicker than the way I hinted in the comment. Note that $$(\sqrt[10]{75})^5=5\cdot\sqrt{3}$$
Hence $$\Bbb Q(\sqrt 3,\sqrt[10]{75})=\Bbb Q(\sqrt[10]{75})$$
And $\Bbb Q(\sqrt[10]{75})$ has degree $10$ over $\Bbb Q$ since $t^{10}-75$ is irreducible.


问:
Sequence of linear operators with bounded norm converges pointwise
Let $U$ and $V$ be Banach spaces and $L(U,V)$ the space of bounded linear operators. Suppose $T_{n}\in L(U,V), \forall n\in \mathbb{N}$ and the sequence $(||T_{n}||_{L(U,V)})_{n\in \mathbb{N}}$ is bounded. Then if $\lim_{n \to +\infty} T_{n}(u)$ exists for each $u$ in some dense subset $D\subset U$ then $T_{n}$ converges pointwise on all of $U$ to some $T\in L(U,V)$.

I tried to use the density of $D$ in $U$ to define $T_{n}(x)$ for general $x\in U$ by constructing a sequence $(u_{k})_{k\in \mathbb{N}}$ in $D$ converging to $x$ and then setting $T_{n}(x):= \lim_{k\to +\infty} T_{n}(u_{k})$. But then I wasn't sure how to then obtain a limit  $T\in L(U,V)$. I wondered if maybe I needed to take the limit as $n\to +\infty$ but I wasn't sure if this made sense while I was taking the limit from $k\to +\infty$.

答:
Let $u \in U$ and take  a sequence $(d_n) \subset D$ converging to $u$. Then $\|T_nu-T_mu\|\leq \|T_nd_k-T_md_k\|+2M\|d_k-u\|$ where $M=\sup_n \|T_n\|$. Choose $k$ such that $2M\|d_k-u\|<\epsilon /2$. Then choose $n_0$ such that $\|T_nd_k-T_md_k\| <\epsilon /2$ for $n, m \geq n_0$. We now get $\|T_nu-T_mu\|<\epsilon$ for $n, m \geq n_0$. Since $U$ is complete it follows that $Tu=\lim T_nu$ exists.
Linearity of $T$ is clear. It is also easy to see that $\|Tu\| \leq M\|u||$ for all $u$, so $T$ is bounded.

ContentLicense: CC BY-SA 4.0


问:
Can the integral value of ${\int}_{0}^{t} \frac{{x^2} exp(x)}{(exp(x)-1)^2} dx$ for t>0 be obtained analytically?
In the process of solving the exercise on the qualitative discussion
of the specific heat of a one-dimensional lattice,
I need to consider the integral of the following f(x).

$$f(x)= \frac{{x^2} \exp(x)}{(\exp(x)-1)^2}  \tag{1-1}$$

The problem I present to you is as follows;

(1) Can the following integral value be obtained analytically for $t>0$?
$$F(t):= {\int}_{0}^{t} \frac{{x^2} \exp(x)}{(\exp(x)-1)^2}  dx \tag{1-2}$$

(2) If the above is not possible, can the limit of $t\to\infty$ be found analytically? In other words, please calculate the following limit values.
$$\lim_{t \to \infty}F(t) \tag{1-3}$$

(3) If none of the above is possible, is it possible to approximate "this integral is almost independent of t when t is sufficiently large"? ($F(t)$ can be regarded as a constant)?

As for (3), I plotted a graph of f(x) and found that,
$\lim_{t\to0}f(t)=1$, and
the $f(x)$ rapidly approaches $0$ when $x$ becomes large.

Fig.A graph showing $F(t)$ from 0 to 15.

Body:
Let $x=\log(u)$ to make
$$I=\int\frac{e^x x^2}{\left(e^x-1\right)^2}dx=\int\frac{\log ^2(u)}{(u-1)^2}\,du=\int\frac{\log ^2(v+1)}{v^2}\,dv$$Back to $x$
$$I=\int\frac{e^x x^2}{\left(e^x-1\right)^2}dx=2 \text{Li}_2\left(e^x\right)+x \left(\frac{e^x x}{1-e^x}+2 \log
   \left(1-e^x\right)\right)$$
$$J=\int_0^t\frac{e^x x^2}{\left(e^x-1\right)^2}dx=2 \text{Li}_2\left(e^t\right)+t \left(2 \log \left(1-e^t\right)-\frac{e^t
   t}{e^t-1}\right)-\frac{\pi ^2}{3}$$

$J$ tends very fast to $\frac {\pi^2}{3}$. Solving for $t$ such that $J(t)=k\frac{\pi^2}{3}$
$$\left(
\begin{array}{cc}
k & t_k \\
 0.9000 & 4.59526 \\
 0.9900 & 7.77037 \\
 0.9990 & 10.6318 \\
 0.9999 & 13.3521
\end{array}
\right)$$


问:
How to solve equation with multiple variables with each of them having different "weight"?
I need to solve equation where the goal is to buy specific number of snacks for the lowest price.

There are 7 types of snack packages, each of them contains different number of snacks and has different price, here is the list:

A: 0.0051 $ / 4 snacks

B: 0.0102 $ / 4 snacks

C: 0.0204 $ / 8 snacks

D: 0.0408 $ / 17 snacks

E: 0.0816 $ / 35 snacks

F: 0.1632 $ / 58 snacks

G: 0.3264 $ / 58 snacks

Now I need to know, how to get the best "combo of packages" for lowest price when buying 10 snacks, 20 snacks..... up to 100 snacks. (When 10 is needed, buying some extra is not problem - such as buying 12 instead of 10.)

I tried bruteforcing (such as starting from maximum number of cheapest package and then combining with other packages), but I think there mast be some smarter way to solve this. :)

答:
Maybe a too long answer for what the outcome is.

This is generally referred to as the Knapsack problem.

Mathematically formulated as:

$$minimize \sum_{i=1}^n G_i(s_i)$$

s.t.

$$ \sum_{i=1}^n s_i = V,$$
$$ s_i \ge 0$$

where $G_i(s_i)$ is the price per snack times the number of snacks and $V$ is your total number of snacks you want to buy.

You can solve this via multiple ways (e.g. dynamic programming, Lagrangian relaxation, ...)

You can also do a marginal analysis algorithm, where you first calculate $\Delta G_i(s_i)$ for all $i$ and then as long as $\sum_{i=1}^n s_i < V$ you iteratively increases the amount for the snack where $ \frac{\Delta G_i}{s_i} $ is the smallest.
So in your problem: where the price per snack is the smallest.

Thus in your problem you just need to buy A...


问:
$\mathbb{C}$ as the Lie algebra of $\mathbb{C}$
It is known that $\mathbb{C}$ is a linear algebraic group as the set of matrices $\begin{pmatrix} 0 & a \\ 0 & 0 \end{pmatrix}$ with $a \in \mathbb{C}$. I read in some notes that $Lie(\mathbb{C})$, the Lie algebra associated to $\mathbb{C}$, is isomorphic to $\mathbb{C}$, but I cannot see why. I only know that if $G$ is a linear algebraic group, $Lie(G)$ is the set of all $\phi : O(G) \to O(G)$ (where $O(G)$ is the set of regular functions on $G$) that satisfy Leibniz rule and such that $g \cdot \phi(f) = \phi(g \cdot f), \forall g \in G, \forall f \in O(G)$. It looks hard to use this definition, so I was thinking of using the fact that $Lie(\mathbb{C})$ is isomorphic to the tangent space, but I am not really sure how. Any hints?

答:
Actually, if you want to see $\Bbb C$ as a linear group, you should see it isomophic to$$\left\{\begin{bmatrix}1&a\\0&1\end{bmatrix}\,\middle|\,a\in\Bbb C\right\};$$that set that you mentioned is not a group.

Clearly, this group is $1$-dimensional. Therefore, its Lie algebra $\mathfrak g$ is a $1$-dimensional complex Lie algebra. And the only way of defining a Lie braket $[\cdot,\cdot]$ on a $1$-dimensional vector space $V$ is to define $[x,y]=0$ for every $x$ and every $y$ in $V$. So, yes, it is naural to see $\mathfrak g$ as $\Bbb C$ endowed with the trivial braket.


问:
Tiling $3k\times (3k-2)$ board with $L$- Trominoes

Consider a $3k\times (3k-2)$ board. For which values of $k$ can we cover the board with $L$- trominoes?

For this, It was clear that for $k$ even, we are done, because then we can divide the board into $2\times 3$ rectangles. And we can cover the $2\times 3,$ as shown in the image.

I then tried for odd $k$, which wasn't working out for me. So I think it's not possible. Hence I am trying to prove that for odd $k$ it's not possible.

Consider the following colouring, which I did for $9\times 7,$ but can be done for any odd $3k\times 3k-2$ (Just make sure that the corners have same colour):

I was first trying to prove for $9\times 7$ board.

Prove that in a $9\times 7$ board, we can't cover it with $L-$ trominoes.

In this coloring, note that there are $5\times 4 + 4\times 3=32$ yellow cells and $63-32=31$ white cells. Since $L-$ trominoes are of $3$ cells area, we get that a total of $21$ $L-$ trominoes were used to cover the board.

Also note that $L-$ trominoes, have $2$ yellow and $1$ white cells or $1$ white and $2$ yellow cells. Let $k$ be the number of $L-$ trominoes which have $2$ yellows cells. So we get $2\cdot k+(21-k)\cdot 1=32.$

From here, we get $k=11$. So there are $11$ $L-$ trominoes which have $2 $ yellow cells and $10$ $L-$ trominoes which have $2$ white cells.

Now consider the $10$ white dominated $L-$ Trominoes. Labelling the rows, with $1,2\dot 7$ (as shown in image ), we get that  these white dominated $L-$ trominoes, can be placed only in row $2,4,6$ rows only ( so that it's white dominated).

Hence by PHP, we that that there will exist one row with $4$ white dominated cells, centred at the yellow cells of the row. But note that there are only $4$ yellow cells.

This couldn't give me a contradiction, but this is the progress I had till now.

This question was actually a lemma I wanted to prove for another question.

The actual question was,

Consider an $n\times n$ board with the four corners removed. For which values of $n$ can you cover the board with $L-$ trominoes.

And by some coloring pattern, I reduced to prove the starting question.

Any hints?

答: Not an answer, but a diagram showing a possible $9\times 7$ tiling.


问:
Joint Distribution of Two Dependent Variables having the Marginals
From two Independent Normal Random Variables: $X \sim N(\mu_1,\sigma) $ and $Y\sim N(\mu_1,\sigma)$, I created two DEPENDENT random variables $Z$ and $W$:

$Z= X - Y$

$W= X - g(Y)$           where  $g(\cdot)$ monotonic transformation

I would like to find the joint distribution $f_{Z,W}$.

(I have read that I might need the Copula, but I believe that having the marginals PDFs and the equations of $Z$, $W$ and $g(\cdot)$ the Copula should be specified somehow and not chosen randomly)

Thanks in Advance!

答:
From two Independent Normal Random Variables: $X\sim N(\mu_1,\sigma)$ and $Y\sim N(\mu_1,\sigma)$ , I created two DEPENDENT random variables Z and W:

Be careful because for a particular choice of $g$ the random variables you created are INDEPENDENT

Counterexample:

Set $g(Y)=-Y$ thus you have

$$Z=X-Y$$

$$W=X+Y$$

and obviously

$$Cov(Z,W)=\mathbb{E}[X^2]-\mathbb{E}[Y^2]=0$$

in a gaussian model, this is equivalent as independence.


问:
Closed subschemes and sheaves
Let $X$ be a Noetherian scheme and $Y$ an integral closed subscheme of $X$. Why the structure sheaf $\mathcal{O}_Y$ is indecomposable? (Indecomposable means that it cannot be expressed as a non-trivial product of sheaves)

答:
the closed immersion $i:Y \hookrightarrow X$ corresponds to the structure sheaf exact sequence $$0 \rightarrow \mathcal{I}\rightarrow \mathcal{O}_X \rightarrow i_*\mathcal{O}_Y \rightarrow 0$$
The structure sheaf of $\mathcal{O}_Y$ is hence given by a quotient of $\mathcal{O}_X$ by the ideal sheaf $\mathcal{I}$. $Y$ being integral is equivalent to $Y$ being reduced and irreducible. $Y$ being integral means that vor every  affine open $\operatorname{Spec} A \subseteq Y$, $A$ is an integral domain. So "affine locally", this exact sequence above comes from taking a quotient by a prime ideal $\mathcal{I}$. This quotient is integral, so it cannot split into a product.

Title: Closed subschemes and sheaves


问:
Surjectivity of linear transformation over an infinite dimensional Vectorspace.
Let $W$ be the complex vectorspace of all functions from $\mathbb{R}\to\mathbb{C}$.

$T:W \to W$ defined as $f\mapsto T(f)$ with $T(f)(t):=f(t+\frac{1}{2})$ for every $t\in\mathbb{R}$.

Prove that:

$T$ is a linear transformation.

$T$ is bijective.

To show that $T$ is indeed linear we need to show that $T(f+\lambda g)=T(f)+\lambda T(g)$ for every $t$, so I wrote:  $$T(f+\lambda g)(t)=(f+\lambda g)(t+\frac{1}{2})=f(t+\frac{1}{2})+\lambda g(t+\frac{1}{2})=T(f)(t)+\lambda T(g)(t)$$
My argument for injectivity, feels a bit handwavy however.
Let $f$ be an element of the kernel of $T$: $$T(f)=f(t+\frac{1}{2})=0$$ Since $f(t+\frac{1}{2})$ has to be  $0$ for every $t$ it follows that $f$ can only be the zero function and thus the kernel is trivial. Is that a valid argument? Whats left to show is the surjectivity of $T$. Here I don't know how to proceed since $W$ is of infinite dimension. Any help would be greatly appreciated.

答: If $g$ is a function from $\mathbb R$ to $\mathbb C$ then $f(t)=g(t-\frac1 2 )$ defines another function from $\mathbb R$ to $\mathbb C$ and we have $Tf=g$. Hence, $T$ is surjective. Your argument for injectivity is valid.


问:
Peano axioms and natural numbers
Suppose I have an ordered field $F$ and $1 \in F$. Then we define the set of natural numbers recursively, here we call it $\mathbb{N}_F$ in the field $F$ via

$1 ∈ \mathbb{N}_F$, and if $\textbf{n} \in \mathbb{N}_F $ then $\textbf{n}+1 ∈ \mathbb{N}_F$, and $\textbf{n} = 1+ .... +1$, where the right side has $n$ "ones".

I want to prove this recurrence relation satisfies the Peano axioms. These kinds of questions look so basic so I don't what tools should I use to prove it/... I just know $\mathbb{N}_F$ is defined as an element $1$ and some successors to complete the set.

$1.$For proving 1 is not the successor of any $k$ ∈ $\mathbb{N}_F$. I am thinking contradiction and showing that there is some $k ∈ \mathbb{N}_F$, and I have that $k+1=1$ (since its a successor). But I don't know what to state next, I know I cannot explicitly define $0$ since it's not an element of $\mathbb{N}_F$.

$2.$ For proving if $\textbf{n},\textbf{m} ∈ \mathbb{N}_F$, and $\textbf{m}+1=\textbf{n}+1$, then $\textbf{n}=\textbf{m}$, this also seems obvious for me but I don't know how to prove it rigorously, as I don't know how to get rid of the $1$ on each side. Maybe I have to apply the fact that its an ordered field but I don't know where to start

$3.$ For $S ⊆ \mathbb{N}_F$, I want to show that $S=\mathbb{N}_F$ if i) $1 ∈ S$ and ii) if $\textbf{n} ∈ S$, then $\textbf{n}+1∈S$

Here I know that $S⊆\mathbb{N}_F$ is already assumed and I only have to prove that $\mathbb{N}_F⊆S$, but its very obvious since $S$ is defined exactly like $\mathbb{N}_F$ but how do I write it formally?

Any inputs are appreciated!

答:
First, you need to define the successor function: $s(n) = n+1$. (Then the simplest definition of $\mathbb{N}_F$ would be as the smallest subset of $F$ that contains $1$ and is closed under $s$. This gives you (3) just by definition.)

You condition (2) follows from the axioms of a field: Assuming that $m,n\in\mathbb{N}_F$ are such that $s(m)=s(n)$ translates into $m+1=n+1$ in $F$. Adding $-1$ (by definition: the additive inverse of $1\in F$) to each side of this equation gives $m=n$, as desired.

To show that $1\ne s(n)$ for all $n\in\mathbb{N}_F$, you need to use that the characteristic of $F$ is zero. Try it yourself.

Title: Peano axioms and natural numbers


问:
The digital root of factorials from 6! to infinity! is always 9.
While observing digital roots of factorials, i observed that ,

digital root of (5+n)! where 'n' is any natural number , is always 9.

The reason lies in the number 720.
It can also be written as 6! .
And digitalroot of (6!× n) where 'n' is always a natural number, is always 9.
And thus the statement holds true for factorials after 6! .
Has it been observed before?
Does it's proof exist?
I am stuck at writting it's proof using mathematical induction.
Can anybody help?

答:
The factorial numbers greater than 5! are all divisible by 9, as we have 3 twice in their expansion. The divisibility test for 9 is that the digital root is 9.

The proof for the divisibility rule for 9 is essentially the same as the proof for the divisibility rule for 3.

For any integer x written as an· · · a3a2 a1a0 we will prove that if 9|(a0 + a1+ a2+ a3 ... + an), then 9|x and vice versa.

First, we can state that

x = a0 + a1×10 + a2×102 + a3×103... + an×10n

Next if we let s be the sum of its digits then

s = a0 + a1 + a2 + a3 + ... + an .

So

x - s = (a0 - a0) + (a1 × 10 - a1) + (a2×102 - a2) + ... + (an×10n - an)

```<code>                           = a1(10 - 1) + a2(102 - 1) + ... + an(10n - 1).
</code>```

If we let bk = 10k - 1, then bk = 9...9 (9 occurs k times) and bk =9(1…1) and we can rewrite the previous equation as

x - s = a1(b1)+ a2(b2)+ ... + an (bn)

It follows that all numbers bk are divisible by 9, so the numbers ak×bk are also divisible by 9. Therefore, the sum of all the numbers ak×bk  (which is x-s) is also divisible by 9.

Since x-s is divisible by 9, if x is divisible by 9, then so is s and vice versa.

ContentLicense: CC BY-SA 3.0

Body:
The factorial numbers greater than 5! are all divisible by 9, as we have 3 twice in their expansion. The divisibility test for 9 is that the digital root is 9.

The proof for the divisibility rule for 9 is essentially the same as the proof for the divisibility rule for 3.

For any integer x written as an· · · a3a2 a1a0 we will prove that if 9|(a0 + a1+ a2+ a3 ... + an), then 9|x and vice versa.

First, we can state that

x = a0 + a1×10 + a2×102 + a3×103... + an×10n

Next if we let s be the sum of its digits then

s = a0 + a1 + a2 + a3 + ... + an .

So

x - s = (a0 - a0) + (a1 × 10 - a1) + (a2×102 - a2) + ... + (an×10n - an)

                           = a1(10 - 1) + a2(102 - 1) + ... + an(10n - 1).

If we let bk = 10k - 1, then bk = 9...9 (9 occurs k times) and bk =9(1…1) and we can rewrite the previous equation as

x - s = a1(b1)+ a2(b2)+ ... + an (bn)

It follows that all numbers bk are divisible by 9, so the numbers ak×bk are also divisible by 9. Therefore, the sum of all the numbers ak×bk  (which is x-s) is also divisible by 9.

Since x-s is divisible by 9, if x is divisible by 9, then so is s and vice versa.


问:
Calculate a point's position from 4 semi-accurate distances
I've got 4 sensors with known positions in a square. Each sensor reports the distance, but not vector, between itself and an object (point P). Unfortuanately the sensors aren't perfectly accurate.

Point P can be anywhere inside or outside the square formed by the sensors.

I'm trying to calculate the best-guess position of the point P.

Diagram of sensors and point

Just to make it a bit more real - the sides of the sensor square are 2 meters and the sensor inaccuracy is between 0 and 15cm.

Thinking about it visually - it feels like it could be the geometric center of the area intersection between donuts drawn around each sensor point.

Doing it visually with donuts

The grey area is where point P could be.

答:
Here is a less than beautiful solution using the donuts as described. It places random points and then progressively eliminates irrelevant ones with a simple check whether it's inside/outside a circle bound.

The speed and accuracy are very affected by the number of sampled points.

Creating a polygon from the intersections to get it's centroid would be preferrable, but I haven't been able to figure it out so far.

https://ht-random.s3-eu-west-1.amazonaws.com/sampled-position.html


Title: Can the integral value of ${\int}_{0}^{t} \frac{{x^2} exp(x)}{(exp(x)-1)^2} dx$ for t>0 be obtained analytically?

Body:
In the process of solving the exercise on the qualitative discussion
of the specific heat of a one-dimensional lattice,
I need to consider the integral of the following f(x).

$$f(x)= \frac{{x^2} \exp(x)}{(\exp(x)-1)^2}  \tag{1-1}$$

The problem I present to you is as follows;

(1) Can the following integral value be obtained analytically for $t>0$?
$$F(t):= {\int}_{0}^{t} \frac{{x^2} \exp(x)}{(\exp(x)-1)^2}  dx \tag{1-2}$$

(2) If the above is not possible, can the limit of $t\to\infty$ be found analytically? In other words, please calculate the following limit values.
$$\lim_{t \to \infty}F(t) \tag{1-3}$$

(3) If none of the above is possible, is it possible to approximate "this integral is almost independent of t when t is sufficiently large"? ($F(t)$ can be regarded as a constant)?

As for (3), I plotted a graph of f(x) and found that,
$\lim_{t\to0}f(t)=1$, and
the $f(x)$ rapidly approaches $0$ when $x$ becomes large.

Fig.A graph showing $F(t)$ from 0 to 15.

Body:
$$I=\int_{0}^{\infty}\frac{x^2 e^x}{(e^x-1)^2}dx = \int_{0}^{\infty}\sum_{k=0}^{\infty}  (k+1) x^2 e^{-x} e^{-kx} dx.$$
Use the series for $(1-z)^2=\sum_{k=0}^{\infty} (k+1)z^k, |z|<1.$
$$I=\sum_{k=0}^{\infty} \int_{0}^{\infty} (k+1) x^2 e^{-(k+1)x} dx$$
$$I=2\sum_{k=0}^{\infty} \frac{ k+1}{(k+1)^3}=2 \sum_{k=0}^{\infty} \frac{1}{(k+1)^2}=2 \zeta(2)=\frac{\pi^2}{3}$$


问:
Prove that $6|n(n+1)(2n+1)$, $n\in\mathbb{Z}$
I am trying to solve this question without modular arithmetic.

Let $n\in\mathbb{Z}$. Prove that $6$ divides $n(n+1)(2n+1)$.

Using induction:

$\text{For }n=1
\\P(1) = 1(1+1)(2(1)+1)=6
\\\text{For }n=k
\\P(k)=k(k+1)(2k+1)
$

$P(k)$ is divisible by $6$ so $k(k+1)(2k+1)=6h, h\in\mathbb{Z}$. Once by the induction hypothesis, $P(k)$ and $P(k+1)-P(k)$ being divisible by $6$ so $P(k+1)$ will be divisible by $6$. as well.

$P(k+1)= (k+1)(k+1+1)(2(k+1)+1) = (k+1)(k+2)(2k+3)$

So,

\begin{align}
P(k+1)-P(k) & =(k+1)(k+2)(2k+3)-k(k+1)(2k+1) \\
& = 2k^3+9k^2+13k+6 - (2k^3+3k^2+k) \\
& = 6k^2+12k+6 \\
& = 6(k^2+2k+1) \\
& = 6(k+1)^2
\end{align}
Therefore, $P(k+1)$ is true and $6$ divides $n(n+1)(2n+1)$.

But once $n\in\mathbb{Z}$ the proof is not complete. I was wondering about considering $6|(-n)(-n+1)(-2n+1)$, $n\in\mathbb{Z}$ but I am not really sure how to proceed. It doesn't seem to be the right approach. How can I keep doing that or what would be a good proof? Proofs without modular arithmetic are preferable, but if you got a nice proof it would be welcomed. I've seen similar questions, but they considered $n\in\mathbb{N}$.

答:
We can use The Division Algorithm:

$n = 6m+r$ with $0\le r <m$, So $r \in \{0,1,2,3,4,5\}$
$$n = 6m; n=6m+1; n=6m+2; n=6m+3; n=6m+4; n=6m+5 $$

Now you have to just plug these values to:
$$\frac{n(n+1)(2n+1)}{6}$$
And you will get an integer every time, which means that $6 \mid n(n+1)(2n+1)$


问:
Prove that a map is Injective
How can I prove that $f: (0, \infty) \times (0,\pi) \to \mathbb{R}^2$ where $f(x,y) = (\sinh(x)\sin(y),\cosh(x)\cos(y))$ is injective? </p>

答:
Suppose $\cosh a\cos b=\cosh c\cos d$ and $\sinh a\sin b=\sinh c\sin d$. Square the two relations:
\begin{gather}
\cosh^2 a\cos^2 b=\cosh^2 c\cos^2 d\\
\sinh^2 a\sin^2 b=\sinh^2 c\sin^2 d
\end{gather}
Sum and subtract the two:
\begin{gather}
\cosh^2 a + \sinh^2 a = \cosh^2 c + \sinh^2 c\\
\cos^2 b - \sin^2 b = \cos^2 d - \sin^2 d
\end{gather}
Therefore
$$
\cosh 2a = \cosh 2c
$$
which means that $2a=2c$ because $\cosh$ is injective in $(0,\infty)$.

Furthermore
$$
\cos 2b = \cos 2d
$$
that is, $2b=2d$ or $2b=2\pi-2d$, from which
$$
b=d\quad\text{or}\quad b=\pi-d
$$
The second possibility, together with $\cosh a\cos b=\cosh c\cos d$, gives $\cos b=\cos(\pi-b)$, that is $\cos b=0$ and, again, $b=d=\pi/2$.

Title: Prove that a map is Injective


答:
Use integration by parts with $u = x^2$ and $dv = \frac{e^x \: dx}{(e^x - 1)^2}$. Then $$\int_0^t \frac{x^2 e^x}{(e^x - 1)^2} \: dx = \frac{x^2}{1 - e^x} \biggr|_0^t + \int_0^t \frac{2x}{e^x - 1} \: dx$$
The first term goes to zero as t approaches infinity. The second denominator can be replaced by its Maclaurin series:
$$\int_0^t \frac{2x}{e^x - 1} \: dx = \int_0^t \frac{2x}{x + x^2/2 + x^3/6 + \ldots} \: dx \leq \int_0^t \frac{2}{1 + x/2 + x^2/6} \: dx$$
The antiderivative of the last integral is $$\frac{24}{\sqrt{15}} \arctan \left( \frac{2x + 3}{\sqrt{15}} \right)$$ This converges as $x$ approaches infinity.
To four decimal places: $$\lim_{t \rightarrow \infty} \int_0^t \frac{x^2 e^x}{(e^x - 1)^2} \: dx \approx 3.2899$$


问:
Find an expression for the $n$-th derivative of $f(x)=e^{x^2}$
I need to find an expression for $n$th derivative of $f(x) = e^{x^2}$. Really need help.</p>

答:
I love to see questions like this in patterns/scheme. After the already given complete answers (especially the comment of Peter Taylor contains a very nice formal statement) here one more version, but which focuses on the visibility of the pattern in the coefficients. In Pari/GP I simply called

```<code>   (exp(x^2) )        / exp(x^2)
   (exp(x^2) ')       / exp(x^2)
   (exp(x^2) '' )     / exp(x^2)
   (exp(x^2) ''' )    / exp(x^2)
</code>```

with so many apostrophes as needed for the n'th derivative. In the result

```<code>  1 + O(x^34)
  2*x + O(x^33)
  2 + 4*x^2 + O(x^32)
  12*x + 8*x^3 + O(x^31)
  12 + 48*x^2 + 16*x^4 + O(x^30)
  120*x + 160*x^3 + 32*x^5 + O(x^29)
  120 + 720*x^2 + 480*x^4 + 64*x^6 + O(x^28)
  1680*x + 3360*x^3 + 1344*x^5 + 128*x^7 + O(x^27)
  1680 + 13440*x^2 + 13440*x^4 + 3584*x^6 + 256*x^8 + O(x^26)
</code>```

I find the pattern:

```<code>                                                   1      = exp(x^2)/exp(x^2)
                                                   2*x    = exp(x^2)'/exp(x^2)
                                    2* 1* 1      + 4*x^2  = exp(x^2)''/exp(x^2)
                                    2* 2* 3*x    + 8*x^3  = ...
                     3*4* 1* 1     +2* 4* 6*x^2 + 16*x^4  = ...
                     3*4* 2* 5*x   +2* 8*10*x^3 + 32*x^5  = ...
    4*5*6* 1* 1    + 3*4* 4*15*x^2 +2*16*15*x^4 + 64*x^6  = ...
    4*5*6* 2* 7*x +  3*4* 8*35*x^3 +2*32*21*x^5 +128*x^7  = ...
</code>```

After that I find it obvious how this continues (formal description&proof are in the other answers above) and leave that schematic description just for the accidental later reader here.

Body:
I love to see questions like this in patterns/scheme. After the already given complete answers (especially the comment of Peter Taylor contains a very nice formal statement) here one more version, but which focuses on the visibility of the pattern in the coefficients. In Pari/GP I simply called

   (exp(x^2) )        / exp(x^2)
   (exp(x^2) ')       / exp(x^2)
   (exp(x^2) '' )     / exp(x^2)
   (exp(x^2) ''' )    / exp(x^2)

with so many apostrophes as needed for the n'th derivative. In the result

  1 + O(x^34)
  2*x + O(x^33)
  2 + 4*x^2 + O(x^32)
  12*x + 8*x^3 + O(x^31)
  12 + 48*x^2 + 16*x^4 + O(x^30)
  120*x + 160*x^3 + 32*x^5 + O(x^29)
  120 + 720*x^2 + 480*x^4 + 64*x^6 + O(x^28)
  1680*x + 3360*x^3 + 1344*x^5 + 128*x^7 + O(x^27)
  1680 + 13440*x^2 + 13440*x^4 + 3584*x^6 + 256*x^8 + O(x^26)

I find the pattern:

                                                   1      = exp(x^2)/exp(x^2)
                                                   2*x    = exp(x^2)'/exp(x^2)
                                    2* 1* 1      + 4*x^2  = exp(x^2)''/exp(x^2)
                                    2* 2* 3*x    + 8*x^3  = ...
                     3*4* 1* 1     +2* 4* 6*x^2 + 16*x^4  = ...
                     3*4* 2* 5*x   +2* 8*10*x^3 + 32*x^5  = ...
    4*5*6* 1* 1    + 3*4* 4*15*x^2 +2*16*15*x^4 + 64*x^6  = ...
    4*5*6* 2* 7*x +  3*4* 8*35*x^3 +2*32*21*x^5 +128*x^7  = ...

After that I find it obvious how this continues (formal description&proof are in the other answers above) and leave that schematic description just for the accidental later reader here.


问:
Help finding parameterization for a surface integral

Find the surface area above the xy plane which is defined by: $x^2+y^2+z^2=1, x^2+y^2 \leq x$

I am first trying to find a parameterization which should be simple enough to work with.

I though about using spherical coordinates:
$$ x=\sin(\theta)\cos(\phi)$$
$$ y=\sin(\theta)\sin(\phi) $$
$$z=\sqrt{1+\sin^2(\theta)}$$

Which should work but the problem here is that I'm not sure how to express where $\phi$ is defined.

Because we have $0 \leq \theta \leq \frac{\pi}{2}$, and $\sin^2(\theta) \leq \sin(\theta) \cos(\phi) \Rightarrow  \sin(\theta) \leq \cos(\phi) $.

Help appreciated.

答: Note that$$x^2+y^2\leqslant x\iff\left(x-\frac12\right)^2+y^2\leqslant\frac14.$$So, you're after the area of the surface $z=\sqrt{1-x^2-y^2}$ with $x$ and $y$ such that $\left(x-\frac12\right)^2+y^2\leqslant\frac14$. This area is equal to\begin{multline}\int_0^1\int_{-\sqrt{x-x^2}}^{\sqrt{x-x^2}}\sqrt{\frac{x^2}{1-x^2-y^2}+\frac{y^2}{1-x^2-y^2}+1}\,\mathrm dy\,\mathrm dx=\\=\int_0^1\int_{-\sqrt{x-x^2}}^{\sqrt{x-x^2}}\frac1{\sqrt{1-x^2-y^2}}\,\mathrm dy\,\mathrm dx.\end{multline}In polar coordinates, this is\begin{align}\int_{-\pi/2}^{\pi/2}\int_0^{\cos(\theta)}\frac\rho{\sqrt{1-\rho^2}}\,\mathrm d\rho\,\mathrm d\theta&=\int_{-\pi/2}^{\pi/2}\left[-\sqrt{1-\rho^2}\right]_{\rho=0}^{\rho=\cos(\theta)}\,\mathrm d\theta\\&=\int_{-\pi/2}^{\pi/2}1-\bigl|\sin(\theta)\bigr|\,\mathrm d\theta\\&=2\int_0^{\pi/2}1-\sin(\theta)\,\mathrm d\theta\\&=\pi-2.\end{align}


问:
Normal between two lines
$$l_1:x=(-1,1,2)+t(1,-3,-2)$$
$$l_2:y=(1,1,1)+s(-2,6,-4)$$
Q: Find the equation of plane passing through $l_1$ and $l_2$.

My teacher wrote answer like this:

Since they're parallel, choose vector $u=(1,-3,-2)$ and choose normal vector $n=u$x$(x_0-y_0)=(-3,-5,6)$. Can someone please explain me how did he find $n=(-3,-5,6)$.

No matter how I choose vectors $x_0,y_0$ it gives $n=(-3,3,-6)$. Is there a way to obtain $n=(-3,-5,6)$?

Thanks!

答:
The vector joining points A(-1,1,2) and B(1,1,1) is $\vec L=-2i+k$ which lies on the plane. Another vector lying in the plane is given as $\vec U=i-3j-2k$ so the normal to the plane is given by $\vec N=\vec L \times \vec U=3i-3j+6k$. So the equation of plane  through the point B can be written as
$\vec N.[\vec r-\vec B]=0 \implies (3i-3j+6k).\vec r=\vec N.\vec B \implies 3x-3y+6z=3-3+6 \implies x-y+2z=2.$

OP's $\vec N$ is correct.

Title: Normal between two lines


问:
Can we always construct a sequence $(x_k)_{k \geq 1}$ such that it is dense in $X$ and $x_k \in \text{int}(X), \forall k$ for a compact set $X$?
Can we always construct a sequence $(x_k)_{k \geq 1}$ such that it is dense in $X$ and $x_k \in \text{int}(X), \forall k$ for a compact set $X \subset \mathbb{R}^n$?

For example, if a compact set $X$ has a isolated point, then it seems impossible.
I guess that if $X$ has no isolated point, then it seems possible.
Is there any condition for this? Further, is there any way to construct such dense subset?

EDIT: suppose that $\text{int}(X)$ is nonempty.

EDIT2: Sorry you all. I think that my question is too ambiguous and not clear. I specify my question here.

答: If $C$ is the Cantor set then it has no isolated points and no point of $C$ is a limit of a sequence from its interior since its interior is empty.


问:
Expected value of the square of the sum of two uniformly distributed independent random variables
What is $E[(X + Y)^2]$, where $X,Y ~ U(0,1)$ and independent ?

edit: I have tried expanding the brackets which results in the following:
$E[X^2 + 2XY + Y^2]$, can you write this expression as $E[X^2] + 2E[XY] + E[Y^2]$ ?

答:
Assuming that $X \perp\!\!\!\!\!\!\perp Y$  you expansion is

$$\mathbb{E}[X+Y]^2=\mathbb{E}[X^2]+\mathbb{E}[Y^2]+2\mathbb{E}[X]\mathbb{E}[Y]$$

Furthermore being

$$\mathbb{E}[X^2]=\mathbb{V}[X]+\mathbb{E}^2[X]$$

...it is enough to substitute to get the result


问:
Bertrand's postulate in another point of view
I was just wondering why can't we use prime number's theorem to prove Bertrand's postulate.
We know that if we show that for all natural numbers $n>2, \pi(2n)-\pi(n)>0$ we are done.
Why can't it be proven by just showing (By using the prime number's theorem) that for every natural numbers $n>2, \frac{2n}{ln(2n)}-\frac{n}{ln(n)}>0$?

答:
The reason is Prime number theorem is much deeper than Bertrand's postulate. Once you have proven PNT, you can do away with Bertrand's postulate and prove much stronger versions of it.

For every integer $k$, there exists a natural number $n_k$ which depends only on $k$ such that for all $n \ge n_k$ there is always at least one prime between $kn$ and $(k+1)n$.

For for every small $\epsilon$, $0 < \epsilon \le 1$, there exists an inter $n_{\epsilon}$ which depends only on $\epsilon$ such that for all $n \ge n_{\epsilon}$, there is always a prime between $n$ and $(1+\epsilon)n$.


问:
Complex Map From the Unit Disc to the Right Half Plane
So I have a map from the unit disc to the upper half plane:

$$f(z) = \frac{z+i}{iz+1}$$

My logic is to now rotate this map by multiplying by $-i$ to arrive at the right half plane - though this does not seem to be a correct answer - what is the mistake I am making here?

Thanks

答:
I find it simpler to remember what the conformal mappings from a half plane to a disk are. As an example, the right half plane is the locus of points which are closer to $+1$ than to $-1$, i.e. the points $w$ for which $|w-1| < |w+1|$. It follows that
$$
 T(w) = \frac{w-1}{w+1}
$$
is the Möbius transformation which maps the right half plane onto the unit disk. Now you can solve the equation $z = \frac{w-1}{w+1}$ for $w$ to get the inverse mapping
$$
 S(z) = T^{-1}(z) = \frac{1+z}{1-z}
$$
which maps the unit disk onto the right half plane.

This is related to your result via
$$
 -i f(z) = -i \frac{z+i}{iz+1} = \frac{-iz+1}{iz+1} = S(-iz)
$$
which is therefore correct as well: $-if$ is a rotation by $-i$ which maps the unit disk onto itself, followed by $S$ which maps the unit disk onto the right half plane.


问:
How can the Cantor set be uncountable, yet its complement is the union of only countably many disjoint open intervals?
$C$ = Cantor set. $C$ is closed, so $C^c$ is the countable union of disjoint open intervals. So let $$C^c = \bigcup\limits_{\substack{i~\in~\mathbb{N} \\}} A_i $$

where $A_i$ are disjoint open intervals.

For each $i \in \mathbb{N},\ A_i\ $ differs from a closed set by just two points, namely the endpoints of the interval $A_i.$ For each $i \in \mathbb{N},\ $ let

$$ B_i = \{x_{i1}, x_{i2} \} = \{ \text{ the two endpoints of } A_i \}. $$

Then $\left|\bigcup\limits_{\substack{i~\in~\mathbb{N} \\}} B_i\right| \leq 2 \left|\mathbb{N}\right| $ is countable. However, $\bigcup\limits_{\substack{i~\in~\mathbb{N} \\}} B_i$ is surely the Cantor set, which is uncountable.

Something is incorrect in all of this, but what?

答: There are lots of points in the Cantor set other than the end points of the intervals $A_i$. Infinite sums of the type $\sum \frac {a_i} {3^{i}}$  ($a_i \in \{0,2\}$) are in $C$ and they are not end points  unless the sum reduces to a finite sum.


问:
Fixed Points in RSA
I need to prove the following:

For different primes, p and q, there exist, as a consequence of Bezout's Lemma, two natural numbers $a$ and $b$ with $ap - bq = 1$. Prove, if we encrypt $x = ap \bmod n$ (for $n = pq$), than the chiffre is equal to $x$.

How can I do this? I need to show that $D(x,(n,e)) = x^e \bmod n = (ap \bmod pq)^e = \ldots = x$

I have tried using Fermat's Little Theorem (from $x^p \equiv x \pmod p$), and also to show by contradiction: if $x = ap \bmod n$ is not a fixed point then this and that happens, but none of them seemed to help. How cand I solve this?

Thanks in advance :D

答: $x$ is the unique integer such that $0\leqslant x<n$, $x\equiv 0\pmod p$ and $x\equiv 1\pmod q$. Since $y=x^e\bmod n$ satisfies $y\equiv 0\pmod p$ and $y\equiv 1\pmod q$ as well, we must have $y=x$.

Title: Fixed Points in RSA


答:
There are three main reasons why it's usually not proved like you proposed:

It has been proved a few decades earlier than the prime number theorem.

Proving PNT is considerably harder than proving the postulate directly.

PNT talks about the asymptotic behaviour of $\pi(n)$ and might not reflect the reality for small numbers. Of course, one can strengthen the PNT to obtain explicit bounds on $n$-th prime valid even for small values of $n$. These can then be used to prove Bertrand's postulate too.

Title: Bertrand's postulate in another point of view

Body:
I was just wondering why can't we use prime number's theorem to prove Bertrand's postulate.
We know that if we show that for all natural numbers $n>2, \pi(2n)-\pi(n)>0$ we are done.
Why can't it be proven by just showing (By using the prime number's theorem) that for every natural numbers $n>2, \frac{2n}{ln(2n)}-\frac{n}{ln(n)}>0$?


问:
Show that $s − r = \sup \{ a − a^\prime \vert a, a^\prime \in A \}$ and $r − s = \inf \{a − a^\prime \vert a, a^\prime \in A \}$
So far this is what I have:
I've seen online the proofs for:

inf(A) = - sup(-A)

inf(B) = − sup(A)

sup(A+B) = sup(A) + sup(B)

Using some of the above formula I came up with:
s - r = sup(A) - inf(A)
= sup(A) - [-sup(-A)]
= sup(A) + sup(-A)
= sup(A - A)

Here is where it doesn't quite make sense. Is the way I'm trying to prove (a) wrong? Any help is very appreciated thank you.

答:
As always when dealing with questions about $\sup, \inf$ and some equations involving those quantities, it is useful to go back to the basic definition and characterizations. Indeed, you yourself say that "you've seen online the proofs of" some facts. This is not per se but a lot of equalities involing those quantities seem true, and are not, so it is good practice to actually prove things.

Now, back to our problem. We will simply use the basic definitions, and nothing more.

For a set $A \subset \mathbb R$, $\sup A$ is the smallest upper bound of $A$, and $\inf A$ its greatest lower bound - assuming $A$ is bounded.

Now, to show that $s - r = \sup\underbrace{\{a-a'\;|\; a, a' \in A\}}_{A_1}$, we need to show that it is an upper bound of this set, and that no other upper bound is smaller than $s-r$.

Let $x \in A_1$. By definition, there exist $a, a' \in A$ such that $x = a - a'$. We know that $a \leq s$ because $s = \sup A$, and $a' \geq r$ so $-a' \leq -r$.
Combining those two inequalities, we get $a-a' \leq s-r$ and so $s-r$ is indeed an upper bound of $A_1$.

Let $p$ be an upper bound of $A_1$. Suppose that $p < s-r$. This means that $p+r < s$ so $p+r$ cannot be an upper bound of $A$ because $s = \sup A$ is its least upper bound. Hence, there exists $a \in A$ with $a > p+r$. Using the same reaoning, we find $a' \in A$ such that $a' < s-p$. Now combine those inequalities and observe that $a-a' > (p+r) + (p-s)$ i.e. $a-a' > 2p + r-s > p$. This contradicts the fact that $p$ is an upper bound of $A_1$, so we do not have have $p < s-r$.

We have proved that $s-r$ is the least upper bound of $A_1$, i.e. it is its supremum.

You should be able to do the second question now !

The takeaway is that whenever you are in doubt, just go back to the basic definitions, and you won't feel as if you were invoking some kind of magic formula such as $\sup  A + sup B = sup(A+B)$ without understanding anything.


问:
The formulas of prostapheresis: memorization technique
This question is related purely for my students of an high school and indirectly for me. The formulas below are the formulas of prostapheresis,

\begin{cases}
\sin\alpha+\sin\beta=2\,\sin \dfrac {\alpha+\beta}{2}\, \cos \dfrac {\alpha-\beta}{2} \\
\sin\alpha-\sin\beta=2\sin \dfrac {\alpha-\beta}{2} \,\cos \dfrac {\alpha+\beta}{2}\\
\cos\alpha+\cos\beta=2\cos \dfrac {\alpha+\beta}{2}\,\cos \dfrac {\alpha-\beta}{2}\\
\cos\alpha-\cos\beta=-2 \,\sin \dfrac {\alpha+\beta}{2} \,\sin \dfrac {\alpha-\beta}{2}
\end{cases}

and while I am able to find them, I am not able to find a technique to memorize them.

Is there a technique to be able to memorize them?

答:
You can memorize the pattern
$$f(a)+\varepsilon f(b)=2\delta g\left(\frac{a+b}{2}\right)h\left(\frac{a-b}{2}\right)$$
where $f$, $g$ and $h$ are either $\sin$ or $\cos$, and $\varepsilon$ and $\delta$ are either $1$ or $-1$. Given $f$ and $\varepsilon$, you then need a strategy to find $g$, $h$ and $\delta$.

Step 1: Find $g$ and $h$

Specializing this for $b=a$ we get
$$f(a)+\varepsilon f(a)=2\delta g\left(a\right)h\left(0\right)$$
and for $b=-a$ we get
$$f(a)+\varepsilon f(-a)=2\delta g\left(0\right)h\left(a\right)$$

Note that if the function that is given $0$ as input is $\sin$ then the right-hand side is $0$ for all $a$, while it if is $\cos$, the right-hand side is non-zero for some $a$. Using the (a)symmetry of $f$, we can easily determine if the left-hand side is zero for all $a$, and hence what $g$ and $h$ are:

If $f(a)+\varepsilon f(a)=0$ for all $a$ then $h=\sin$ and otherwise $h=\cos$;

If $f(a)+\varepsilon f(-a)=0$ for all $a$ then $g=\sin$ and otherwise $g=\cos$.

Step 2: Find $\delta$

Taking $a=\frac{\pi}{2}$ and $b=0$ yields

$$f\left(\frac{\pi}{2}\right)+\varepsilon f(0)=2\delta g\left(\frac{\pi}{4}\right)h\left(\frac{\pi}{4}\right)$$

Since both $\sin$ and $\cos$ evaluate to $\frac{\sqrt{2}}{2}$ at $\frac{\pi}{4}$, we therefore have

$$f\left(\frac{\pi}{2}\right)+\varepsilon f(0)=\delta $$

which gives us $\delta$.

Plugging other specific values for which both sides are easy to evaluate would also work as long as both $g\left(\frac{a+b}{2}\right)$ and $h\left(\frac{a-b}{2}\right)$ are non-zero (so you do not need to remember to take $a=\frac{\pi}{2}$ and $b=0$).

Example

$$\cos(a)- \cos(b)=2\delta g\left(\frac{a+b}{2}\right)h\left(\frac{a-b}{2}\right)$$

Taking $a=b$ on the left-hand side makes it $0$ so $h=\sin$. Taking $a=-b$ also makes it zero so $g=\sin$:
$$\cos(a)- \cos(b)=2\delta \sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)$$

Specializing to $b=0$ we get
$$\cos(a)-\cos(0)=2\delta \sin\left(\frac{a}{2}\right)\sin\left(\frac{a}{2}\right)$$
so we just need to pick a value of $a$ such that $\sin\left(\frac{a}{2}\right)\not = 0$. Taking $a=\pi$ gives:
$$\cos(\pi)-\cos(0)=2\delta \sin\left(\frac{\pi}{2}\right)\sin\left(\frac{\pi}{2}\right)$$
i.e.
$$-2=2\delta $$
which allows to conclude $\delta=-1$ and hence:
$$\cos(a)- \cos(b)=-2 \sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)$$


问:
I know the integral of a function. Can I find the arc length over an interval without knowing the function?
Can I find the arc-length of a function knowing the integral only?

More specific: is there a way to find the arc-length of an ellipse without using elliptic integrals as we now the area under the first quadrant = πab/4 = integral from 0 to a?

答:
The wording of your question is awkward: ask if you can determine the arc length of a curve knowing only the area under the curve rather than "the integral" since arc length is itself described with integrals.

You ask for an arc length formula that does not use an elliptic integral. The real world is not as nice as you wish it to be: merely because the area has a tidy little formula does not mean the arc length should have a nice formula.  And it doesn't: for an ellipse $x^2/a^2 + y^2/b^2 = 1$ that is not a circle, so without loss of generality we can take $0 < b < a$, the arc length integral for the perimeter of the ellipse is (after some changes of variable)
$$
4a\int_0^1 \frac{\sqrt{1 - mt^2}}{\sqrt{1-t^2}}\,dt =
4a\int_0^1 \frac{1-mt^2}{\sqrt{(1 - t^2)(1-mt^2)}}\,dt
$$
where $m = 1 - b^2/a^2 \in (0,1)$.  Liouville proved in the 1830s that the integrand here does not have an elementary antiderivative.  That doesn't mean one particular value of the definite integral (with specific upper bounds, like $0$ and $1$) can't possibly have a simple exact formula not mentioning integrals, comparable to the case of a formula for the total area inside an ellipse. For example, Liouville showed $e^{-t^2}$ has no elementary antiderivative but we can compute $\int_{-\infty}^{\infty} e^{-t^2}\,dt = \sqrt{\pi}$ by other methods.
Still, because the integrand for the total arc length of an ellipse does not have an elementary antiderivative, you should not be surprised that nobody has ever found a simple algebraic formula for an ellipse's arc length.

In fact, even the formula $\pi ab$ for the area of the whole ellipse is in a sense not a simple formula: it involves the transcendental number $\pi$, which we are familiar with only because long ago someone gave that number a name and you've learned about in your previous math courses.


答:
You need a precise estimate of the form $$c_1<\frac{\pi(n)\ln n}{n}<c_2.$$ With that you can derive
$\pi(2n)>c_1\frac{2n}{\ln(2n)}>2c_1\frac{n}{\ln 2+\ln n}$. If you are lucky, you can continue$2c_1\frac n{\ln2+\ln n}>c_2\frac n{\ln n}>\pi(n)$.
However, for this you better have $\frac{2c_1}{c_2} >1+\frac{\ln 2}{\ln n}$. So at least if $c_2<2c_1$, your idea works for sufficiently big $n$.</p>


问:
An idempotent operator is compact if and only if it is of finite rank
Would you help me to solve this problem. Show that an idempotent operator on hilbert space is compact if and only if it has finite rank.</p>

答:
If you are working on with the Conway's A Course in Functional Analysis, the proof gets even more easier.

Your question is in fact in this book, Exercise II.4.4. The right-to-left is immediate consequence of the Exercise I.4.3. For the left-to-right proof, by II.3.2 Proposition, $ranE=ker(I-E)=ker(E-I)$. Hence by II.4.13 Proposition, the proof is complete.


问:
Can the integral value of ${\int}_{0}^{t} \frac{{x^2} exp(x)}{(exp(x)-1)^2} dx$ for t>0 be obtained analytically?
In the process of solving the exercise on the qualitative discussion
of the specific heat of a one-dimensional lattice,
I need to consider the integral of the following f(x).

$$f(x)= \frac{{x^2} \exp(x)}{(\exp(x)-1)^2}  \tag{1-1}$$

The problem I present to you is as follows;

(1) Can the following integral value be obtained analytically for $t>0$?
$$F(t):= {\int}_{0}^{t} \frac{{x^2} \exp(x)}{(\exp(x)-1)^2}  dx \tag{1-2}$$

(2) If the above is not possible, can the limit of $t\to\infty$ be found analytically? In other words, please calculate the following limit values.
$$\lim_{t \to \infty}F(t) \tag{1-3}$$

(3) If none of the above is possible, is it possible to approximate "this integral is almost independent of t when t is sufficiently large"? ($F(t)$ can be regarded as a constant)?

As for (3), I plotted a graph of f(x) and found that,
$\lim_{t\to0}f(t)=1$, and
the $f(x)$ rapidly approaches $0$ when $x$ becomes large.

Fig.A graph showing $F(t)$ from 0 to 15.

Body:
Another version using integration by parts only
$$\int_0^t \frac{x^2 e^x}{(e^x-1)^2} \, {\rm d}x = -\frac{t^2}{e^t-1} + \underbrace{\int_0^\infty \frac{2x}{e^x-1} \, {\rm d}x}_{=2\zeta(2)} - \int_t^\infty \frac{2x}{e^x-1} \, {\rm d}x \\
=-\frac{t^2}{e^t-1}  + \frac{\pi^2}{3} + 2t \log(1-e^{-t}) + 2 \underbrace{\int_t^\infty \log(1-e^{-x}) \, {\rm d}x}_{-{\rm Li}_2(e^{-t})} \\
= \frac{\pi^2}{3} - e^{-t} \left( t^2 + 2t + 2 \right) + {\cal O}\left(t^2 e^{-2t}\right)$$
where ${\rm Li}_2(x)=\sum_{n=1}^\infty \frac{x^n}{n^2}$.


问:
If $\textbf{x} \in \mathbb{R}^n$, then show that the following inequalities hold.
If $\textbf{x} \in \mathbb{R}^n$, then show that the following inequalities hold

$||\textbf{x}||_\infty \leq ||\textbf{x}||_2 \leq \sqrt{n}||\textbf{x}||_\infty$

$\frac{1}{\sqrt{n}}||\textbf{x}||_1 \leq ||\textbf{x}||_2 \leq \sqrt{n}||\textbf{x}||_1$

I have been trying to read about $L^\infty$ spaces and not sure if this translates to vector norms. I read that a vector norm is the max entry value. I would like some help on a definition of the terms shown above and how to being this problem. I have tried to think about how an infinite vector could possibly have an entry with larger value than a vector in $\mathbb{R}^2$. I am not sure if I am even on the right track in thinking about this problem.

答:
If you are looking for definitions, in the finite dimensional space $\mathbb R^n$, if $\mathbf x \in \mathbb R^n$ with coordinates $x_i$, for $1 \leqslant i \leqslant n$
\begin{align}
\lVert \mathbf x \rVert_\infty &= \max_{1 \leqslant i \leqslant n} \lvert x_i \rvert \\
\lVert \mathbf x \rVert_2 &= \sqrt {\sum_{i=1}^n |x_i|^2 }\\
\lVert \mathbf x \rVert_1 &= \sum_{i=1}^n |x_i|.
\end{align}
The inequalities you state then follow pretty directly from these definitions.

The concepts also extend to infinite dimensions.  There are three vector spaces that correspond to those above.

The first, written $\ell_\infty$ is the space of all bounded sequences $x_i, 1 \leqslant i$ in which the norm is defined as $\lVert \mathbf x \rVert_\infty = \sup_{1 \leqslant i} | x_i |$.  The supremum exists because the sequence must be bounded to be contained in the space $\ell_\infty$.

A few examples:  the sequence $(x_i) = (1,2,3,4,\cdots)$ is not in $\ell_\infty$ because it is unbounded.  The two sequences $(1,1,1,\cdots)$ and $(1,1/2,1/3,\cdots)$ are both in $\ell_\infty$ because they are bounded.  Both have $\ell_\infty$ norm of $1$.

The second space is $\ell_2$ with corresponding norm $\sqrt{\sum_{i=1}^\infty |x_i|^2}$, where members of the space are required to have finite norm.  Of the three sequences above, only the last is in $\ell_2$ with norm $\pi/\sqrt 6$,as the norms of the other two would be infinite, thereby disqualifying the sequences from inclusion in the space $\ell_2$.

The third, $\ell_1$ has norm $\sum_{i=1}^\infty |x_i|$, and none of the three examples are in the space because all there have unbounded norms.


问:
A function with period 1 and $\alpha$ (irrational)
I'm looking for a non-constant function $f(x)$, which there is a irrational number $\alpha$ such that
$$
f(x)=f(x+1)=f(x+\alpha)
$$
for all $x\in\mathbb{R}$.

Note that though the function $f(x)$ do not have a smallest period, but that doesn't mean $f(x)$ must be a constant (the Dirichlet function is a good example, but it failed in this situation).

答:
Let $S=\{a+b\alpha\mid a,b\in\mathbb Z\}$, and let $\sim$ be relation on $R$ defined by $$x_1\sim x_1\iff x_1-x_2\in S.$$
Relation $\sim$ is equivalence relation, and it has infinitely many equivalence classes (because $S$ is countable). Let $S_1$ be one equivalence class. Now set $$f\left(x\right)=\begin{cases}1, & x\in S_1 \\ 0, &x\not\in S_1\end{cases}.$$ Function $f$ is periodic with periods $1$ and $\alpha$.


答:
Fix $\alpha$ irrational. How about :
$$
f(x) = \begin{cases}
0 & \exists r,s \in \mathbb Q , x = r+s\alpha \\
1 & \mathrm{otherwise}
\end{cases}
$$

Then, if $f(x)=0$, $x = r+s\alpha$, so $x+\alpha = r+(s+1)\alpha$ and $x+1 = (r+1) + s\alpha$ ,so $f(x+1) = f(x+\alpha) = 0$.

If $f(x) =1$ then $x$ is not a rational plus a rational multiple of $\alpha$ , then $x+1$ and $x+\alpha$ also cannot be. Thus this function works.

We can see that $f$ doesn't have a smallest period , however. This is because if $f$ is a function such that $f(x) = f(x+1) = f(x+\alpha)$ for some $\alpha$ irrational, then we can easily write $f(x) = f(x +r +s\alpha)$ for any $r,s$ rational, and one can easily show that $\inf\{|r+s \alpha| : r,s  \in \mathbb Q\} = 0$.

From the theory of functional equations, all we need to define such an $f$ is that $f$ be defined on one element of every equivalence class of the relation on $\mathbb R$ given by $x \sim y$ if and only if $x-y = r+s\alpha$ for some $r,s$ rational. Since there are many such equivalence classes, one may easily define $f$ individually on each class and get a multitude of functions which work out.

Title: A function with period 1 and $\alpha$ (irrational)

Body:
I'm looking for a non-constant function $f(x)$, which there is a irrational number $\alpha$ such that
$$
f(x)=f(x+1)=f(x+\alpha)
$$
for all $x\in\mathbb{R}$.

Note that though the function $f(x)$ do not have a smallest period, but that doesn't mean $f(x)$ must be a constant (the Dirichlet function is a good example, but it failed in this situation).


问:
Is it possible for each vector bundle to find another one that their direct sum is trivial on an affine?
Given a smooth projective variety $X$ and a hyperplane section complement $U$. For any vector bundle $E$ on $X$, is it possible to find another vector bundle $F$ such that $E\oplus F$ is a trivial vector bundle on the affine scheme $U$?

答:
The bundle $E(n)$ is globally generated for $n \gg 0$. Therefore, there is an exact sequence
$$
0 \to F \to \mathcal{O}_X^{\oplus m} \to E(n) \to 0.
$$
Since $\mathcal{O}_X(1)\vert_U \cong \mathcal{O}_U$, its restriction to $U$ takes the form
$$
0 \to F\vert_U \to \mathcal{O}_U^{\oplus m} \to E\vert_U \to 0.
$$
Finally,
$$
\mathrm{Ext}^1(E\vert_U,F\vert_U) = H^1(U,E^\vee \otimes F\vert_U) = 0
$$
since $U$ is affine, hence the second exact sequence splits, hence $E\vert_U \oplus F\vert_U$ is trivial.


答:
Unit sphere: $x^2+y^2+z^2=1$, Cylinder: $x^2+y^2 \leq x, z \geq 0$. Please note the equality in equation of the sphere and inequality in the equation of the cylinder which means we need to find the surface area of the sphere that is bound by the cylinder for $z \geq 0$.

As you mentioned in spherical coordinates, $x = \rho \cos\theta \sin\phi, y = \rho \sin\theta \sin\phi, z = \rho \cos\phi$.

For sphere, the surface is $\rho = 1$

Equation of the cylinder $x^2+y^2 \leq x$ in spherical coordinates becomes

$\rho = \cos \theta \csc \phi$

At the intersection of the cylinder and the sphere,

$\rho = \cos\theta \csc \phi = 1 \implies \phi = \arcsin(\cos\theta)$

Also note the cylinder has radius of $\frac{1}{2}$ and its center is at $(\frac{1}{2}, 0, z)$. For the cylinder in $XY$ plane, $\theta$ varies from $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.

For the given sphere, $|r'_{\theta} \times r'_{\phi}| = \sin\phi$

So the integral for the surface area is

$\displaystyle \int_{-\pi/2}^{\pi/2} \int_0^{\arcsin(\cos\theta)} \sin \phi \ d\phi \ d\theta \approx 1.1416$

In cylindrical coordinates,

$x = \rho \cos\theta, y = \rho \sin\theta, z$

Sphere is given by $r^2 + z^2 = 1$. Cylinder is given by $r = \cos\theta, -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.

For the given sphere, $|r'_{\rho} \times r'_{\theta}| = \frac{\rho}{\sqrt{1-\rho^2}}$

So the surface area is

$\displaystyle \int_{-\pi/2}^{\pi/2} \int_0^{\cos\theta} \frac{\rho}{\sqrt{1-\rho^2}} \ d\rho \ d\theta \approx 1.1416$

Title: Help finding parameterization for a surface integral

Body:
Find the surface area above the xy plane which is defined by: $x^2+y^2+z^2=1, x^2+y^2 \leq x$

I am first trying to find a parameterization which should be simple enough to work with.

I though about using spherical coordinates:
$$ x=\sin(\theta)\cos(\phi)$$
$$ y=\sin(\theta)\sin(\phi) $$
$$z=\sqrt{1+\sin^2(\theta)}$$

Which should work but the problem here is that I'm not sure how to express where $\phi$ is defined.

Because we have $0 \leq \theta \leq \frac{\pi}{2}$, and $\sin^2(\theta) \leq \sin(\theta) \cos(\phi) \Rightarrow  \sin(\theta) \leq \cos(\phi) $.

Help appreciated.


问:
self-adjoint operator is bounded in Hilbert space
$X$ is a Hilbert space. $A$ is a linear and  defined everywhere on $H$. $A$ satisfies $$<x,Ay>=<Ax,y>$$ for all $x,y\in H$, then $A$ is bounded. $<,>$ denote the inner product

答:
There is another way to prove this:
For fixed $y \in H$ let $T_y: H \rightarrow \mathbb{R}$ be the operator
$$
T_y(x) := \langle Ax, y \rangle_H.
$$
Because of symmetry and Cauchy-Schwarz, it is continuous ($x \in H$ arbitrary):
$$
T_y(x) = \langle x, Ay \rangle_H \leq \lVert x \rVert_H \lVert Ay \rVert_H
$$
Now let
$$
\Gamma := \lbrace T_y \in L(H, \mathbb{R}): \lVert y \rVert_H = 1 \rbrace
$$
be a family of linear operators. Then we have for fixed $x \in H$ by Cauchy-Schwarz:
$$
\sup_{T_y \in \Gamma} \lvert T_y(x) \rvert = \sup_{\lVert y \rVert_H = 1} \lvert \langle Ax, y \rangle \rvert \leq \sup_{\lVert y \rVert_H = 1} \lVert Ax \rVert_H \lVert y \rVert_H = \lVert Ax \rVert_H \leq \infty
$$

Now the Banach-Steinhaus theorem tells us:
$$
\sup_{T_y \in \Gamma} \lVert T_y \rVert_{L(H, \mathbb{R})} = \sup_{\lVert y \rVert_H = 1} \sup_{\lVert x \rVert_H = 1} \lvert \langle Ax, y \rangle_H \rvert < \infty
$$

Hence:
$$
\sup_{\lVert x \rVert_H = 1} \lVert Ax \rVert_H = \sup_{\lVert x \rVert_H = 1} \frac{\lvert \langle Ax, Ax\rangle_H \rvert}{\lVert Ax \rVert_H} = \sup_{\lVert x \rVert_H = 1} \left \lvert \left \langle Ax, \frac{Ax}{\lVert Ax \rVert_H} \right \rangle_H \right\rvert \leq \sup_{\lVert y \rVert_H = 1} \sup_{\lVert x \rVert_H = 1} \lvert \langle Ax, y \rangle_H \rvert < \infty
$$

So $A$ is bounded.


问:
What is a Polyhedral Set?
Is the following set a Polyhedral ?

$S = \{x \in \mathbb R^2 : |x_1|+ |x_1| \leq 1 \}$

Can anyone please tell me what is the exact meaning of Polyhedral set ?

I was given to check if this set is Polyhedral or not. I do not know the term Polyhedral. But I know Polyhedron.

I have not got he meaning of the term 'Polyhedral'  in the internet as well.

答:
A polyhedral set is a set which can be expressed as the intersection of a finite set of closed half-spaces, i.e., as $\{x \in \mathbb{R}^n : Ax \leq b\}$ for some matrix $A \in \mathbb{R}^{m \times n}$ and vector $b\in\mathbb{R}^m$.

Your set is polyhedral, take
$$A = \begin{pmatrix}1 & 1 \\ 1 & -1 \\ -1 & 1 \\ -1 & -1\end{pmatrix} \quad \text{and} \quad b = \begin{pmatrix}1 \\ 1 \\ 1 \\ 1\end{pmatrix}.$$

Title: What is a Polyhedral Set?


问:
Is a topology a set or a family?
So I've just started learning topology and a lot of the definitions confuse me.
My main problem is that some of the definitions are quite inconsistent to me for example in topology without tears the author says that a topology $\tau$ is a set of subsets of $X$ and then he proceeds with the axioms.
However I've also seen a lot from other sources that a topology is a family of subsets of $X$ and then they proceeds to describe the same axioms. However they do not refer to the topology $\tau$ as being a set at all which confuses me.

I think my confusion is the definition on what a family actually is ive seen lots of confusing definitions on what a family is like it is a surjective function however I cannot seem to grasp the idea.

So why do they define them differently and what is a family?

Thanks in advance.

答:
I think the word "family" is somewhat ambiguous. For example, Wikipedia says

In set theory and related branches of mathematics, a collection $F$ of subsets of a given set $S$ is called a family of subsets of $S$, or a family of sets over $S$. More generally, a collection of any sets whatsoever is called a family of sets or a set-family or a set-system.

The term "collection" is used here because, in some contexts, a family of sets may be allowed to contain repeated copies of any given member, and in other contexts it may form a proper class rather than a set.

Here allowed to contain repeated copies of any given member means what Wikipedia denotes as an indexed family :

More formally, an indexed family is a mathematical function together with its domain $I$ and image $P$. Often the elements of the set $P$ are referred to as making up the family. In this view indexed families are interpreted as collections instead of as functions. The set $I$ is called the index (set) of the family, and $P$ is the indexed set.

The first quotation shows that a family of sets can be understood either as a set of sets or as function  $f :  I \to P$ where $P$ is a set of sets. This is indeed vague and leaves much scope for interpretation. The same vagueness of notation can be found in  many textbooks.

For a given set $X$ we may consider

sets $\tau$ of subsets of $X$, i.e. subsets of $\tau \subset \mathfrak P(X)$ = power set of $X$.

"indexed collections" of subsets of $X$, i.e. functions $\theta : I \to \mathfrak P(X)$.

Each subset $\tau \subset \mathfrak P(X)$ can be canonically identified with the inclusion function $\iota(\tau) : \tau \hookrightarrow  \mathfrak P(X)$; this produces a "self-indexed collection" of subsets of $X$. Conversely, each function $\theta : I \to \mathfrak P(X)$ determines the subset $\text{im}(\theta) = \theta(I) \subset \mathfrak P(X)$. Clearly $\text{im}(\iota(\tau)) = \tau$, but in general $\iota(\text{im}(\theta)) \ne \theta$. In fact, for a given $\tau \subset \mathfrak P(X)$ there are many $\theta : I \to \mathfrak P(X)$ such that $\text{im}(\theta) = \tau$. One can even show that the "collection" of all $\theta : I \to \mathfrak P(X)$ such that $\text{im}(\theta) = \tau$ is not even a set, but a proper class.

The standard is to define a topology on a set $X$ as a set of subsets of $X$ satisfying suitable axioms.

If you regard the word family as a synonym for set, then you do not get conflicting definitions. If you regard a family as a function, then a topology on $X$ would be some function $\theta : I \to \mathfrak P(X)$ from an index set $I$ to the power set of $X$ which is often written in the form $\{U_i\}_{i \in I}$ (indexed collection of subsets).  You can easily modify the axioms for a topology to obtain similar axioms for an indexed collection of subsets of $X$. It is fairly obvious that $\theta$ satisfies these modified axioms if and ony if $\text{im}(\theta)$ is a topology in the standard sense. The essential disadvantage of the alternative definition is that indices are completely unnessary - many formally distinct families of "indexed topologies" give the same "standard topology".

Title: Is a topology a set or a family?


问:
If set is closed under some operation, what does that tell us about the structure of the groupoid?

If set is closed under some operation,
  what does that tell us the about structure
  of the groupoid?

It's a question from one of the previous math exams which was shown to me by one of my friends.

Unfortunately, I can't find a good answer to it. As far as I know, the answer is nothing, because of the definition of groupoid we use. Here's what (among other things) my math textbook has to say about groupoids:

Ordered pair (S,*), where arbitrary
  binary operation closed in  set S
  is marked with *, is called groupoid.

答:
You settled that the original formulation was a mistake. But another meaning could have been

If a set inside a groupoid is closed under some operation, what does that tell us about the groupoid?

In that case, one answer would be:

If the inside set is also closed under the same operation as the groupoid operation, then the groupoid

(1) has a non trivial sub-structure (a subgroupoid),

(2) the groupoid can be restricted to this set and form an early pre-ring (a groupoid with two different operations on the same set).

ContentLicense: CC BY-SA 2.5


问:
Shuffle a poker deck between 4 players, with least required entropy
We are shuffling a standard poker deck of 52 cards between 4 players, each getting 13 cards. The order of cards for a particular player does not matter.

A naive algorithm is to first shuffle the whole deck, then split it into 4 parts. However, this requires $log_2(52!) \approx 226$ bits of entropy.

Combinatorics says that such a shuffle should only require $log_2(52! / 13! / 13! / 13! / 13!) \approx 96$ bits of entropy, more than half less.

What is the algorithm that would produce such an efficient shuffle, using a random generator that produces integers in a range?

I was thinking of this algorithm:

Start with a sorted deck, for each card:

generate a number between 1 and 4, to determine which player the card goes to.

once one of the players gets 13 cards, only generate numbers between 1 and 3 (and so on)

Stop when only one player remains, give them all undealt cards

Would such algorithm produce an unbiased deal, and if not, what is the correct one?

答:
Let's approach this problem backward.

So shuffle the deck, deal the cards, have each player sort their hand.
From this state, how much information can we gather?

As you already pointed out, we cannot retrieve the order of the deck after shuffling, before dealing. A shuffled deck contains more information than four hands of cards.

So how much information is there?
For simplicity, we call the cards just card $1$, card $2$, etc.
Imagine, each player sorted their hands and put the cards in front of them, face down, the lowest card on top.
Then we know, that one of the top cards must be card $1$, and the probability is $1/4$ for each one. So the question "Which hand contains card $1$?" is like rolling a four-sided fair dice.

Let us say, Player 1 had card 1. We remove the card from their stack and ask "Which hand contains card 2?". Again, it must be one of the top cards, but the probabilities are no longer equal since Player 1 has just 12 cards while the others have 13. So this question is equivalent to rolling four-sided dice, where one side has a slightly lower probability.

We can continue asking for each card in turn to retrieve as much information as we can.

Reverting this process gives us a means of distributing the cards:
Take a fair four-sided dice, label the sides Player 1 to Player 4.
Roll the dice and give card 1 to the respective player.
Then adjust the probabilities of the dice to 12/51 for the player with card 1 and 13/51 for the others.
Continue rolling and adjusting until all cards have been distributed.

To sum up, we have a means of distributing the cards and a way of looking at a dealt deck containing the same information, so the distribution method should be optimal in this sense.


问:
Is the Cantor set made of interval endpoints?
The Cantor set is closed, so its complement is open.  So the complement can be written as a countable union of disjoint open intervals.  Why can we not just enumerate all endpoints of the countably many intervals, and conclude the Cantor set is countable?</p>

答:
Just because the open intervals approach a number, doesn't mean the number is the endpoint of one of the intervals. Consider:

$$D_1= \bigcup\limits_{\substack{n~\in~\mathbb{N},~n~\text{odd}  \\}} \left(\frac{1}{n+1}, \frac{1}{n}\right). $$

This looks like:

$$$$
Next, let
$$D_2= \bigcup\limits_{\substack{n~\in~\mathbb{N},~n~\text{odd}  \\}} \left(\frac{-1}{n}, \frac{-1}{n+1}\right). $$

This looks like:

Now let $D = D_1 \cup D_2.\ $ Then $0 \in D^c\ $ and $0$ is a limit point of $D^c$. But $0$ is not the end-point of any of the intervals in $D$. The same goes for members of the Cantor set- most members of the Cantor set, it turns out.

Just because the intervals "close in" on $0\ $ doesn't make $0$ an endpoint of one of the intervals.


问:
$\sum^\infty_{n=0}\frac {x^{4n}}{(2n)!}$ is Taylor's series of which function?
$\sum^\infty_{n=0}\frac {x^{4n}}{(2n)!}$

a) $e^{x^4}$

b) $1+ x^2*\frac {\sin(x^2)}{2}$

c) $\cos^2(x^2)$

d) $(1+x^2)\ln(1+x^2)$

e) $\frac {1}{2} (e^{x^2} + e^{-x^2})$

What I tried to do: since $e^x$ Taylor series is $\sum^\infty_{n=0}\frac {x^{n}}{n!}$, I'm gonna investigate the options (a) and (e) first.

(a) $\sum^\infty_{n=0}\frac {x^{4n}}{n!}$ , if we substitute ($x^4$).

(e) if $e^x$ Taylor series is $\sum^\infty_{n=0}\frac {x^{n}}{n!}$, then $e^{-x}$ Taylor series is $\sum^\infty_{n=0}\frac {(-x)^{n}}{n!} = \sum^\infty_{n=0}\frac {(-1)^nx^{n}}{n!}$, from here I tried to do $\frac {e^x + e^{-x}}{2}$, and I realized I have no idea how to calculate these sums.

I also couldn't think of similar ways to write the series of (b)(c)(d), I would appreciate any help if there's an easy way to find these functions series without having to take derivatives, just using the know Taylor's series for known functions, (if it's not possible I would appreciate any help of how to reach a $\sum$  from just calculating derivatives and substituting in Taylor's polynomials formula).
 Thanks in advance.

答:
Hint:

$$e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$$

Replace $y$ with $x^2,-x^2$ one by one to find $e^{x^2}+e^{-x^2}$


问:
Integral of CDF Difference as Comparison Measure
Given two r.v. $X_1$ and $X_2$ taking values in $[0,2]$, and their CDFs (cumulative distribution functions) $F_1$ and $F_2$, I'm computing the following measure:

$M = \int_0^2 F_1(t) - F_2(t) dt$

From its sign and magnitude, this measure $M$ produces information that can intuitively be described as: what values are larger on average, values from $X_1$ or values from $X_2$.

I'm looking for some already known statistical notion/metric for a distribution comparison,  that is related to $M$ in some way. Does $M$ has a mathematical name? Or is there any metric that has a tight connection to $M$?

答:
Below all existing relations are described.

The Kolmogorov-Smirnov statistic:

$KS = \sup_t |F_1(t) - F_2(t)|$

and the signed Kolmogorov-Smirnov statistic:

$sKS = \sup_t F_1(t) - F_2(t)$

are popular choices in the setting of two-sample test. They both have some similarities with $M$, yet there are also major differences (e.g. integral vs $\sup$).

Likewise, the first Wasserstein distance between two densities can be shown to be equal to [Aaditya Ramdas et al.]:

$W_1 = \int_{0}^{2} |F_1(t) - F_2(t)| dt$

It is very close to $M$, but still has the absolute value.

Finally, denoting by $\mu_1$ and $\mu_2$ the means of $X_1$ and $X_2$ respectively, and using the known relation between the expected value and CDF $\mu_i = \int_{0}^{2} 1 - F_i(t) dt$, we can conclude that:

$M = \int_{0}^{2} F_1(t) - F_2(t) dt = \mu_2 - \mu_1$.

Hence, $M$ is just a difference between the corresponding expected values.


问:
Can $n!$ start with the digits 2020...
Can anyone give me any hints on how to tackle this? I assume the answer is no otherwise why ask the question. I am aware that $k \mid n!$
for all $k\leq n$, and I am aware that divisibility by $3$, $9$ and $11$ in particular have nice and easy tests involving sums of digits etc. I know that $2020 = 2^2 \cdot 5 \cdot 101$ written as a product of primes. However I am not seeing how these things help..

Thanks

Edit: Plot twist... $384!$ begins with the digits $2020$ so now the question is how do we prove this can happen (let's pretend we don't have computers)?

答:
The main idea to prove that this works for any starting sequence of numbers is that if you have a number $n$ that starts with a $1$, then many $0$'s, then a $1$ then many more zeros, the factorials of $n$, $n+1$,a.s.o will not change fast in the first however many digits you are interested (as you are effectily multipliying with $1.00\ldots00100\ldots xyz$. If you can prove that over many such factorials they do change to "wrap around" and come back to the initial value, you know they must at some time have been of the value that you are interested in.

Formally, set $n=10^{3k}+10^{2k}$ and look at

$$n!, (n+1)!, \ldots, (n+10^{2k})!,$$

then you notice that each factorial will be generated from the one before by multiplication with a number between $10^{3k}$ and $10^{3k}+2\cdot10^{2k}$ and that this will increase the number formed by the first $k-1$ digits of the factorial by at most $1$ (both multiplications by $10^{3k}$ and $10^{2k}$ are just shifts by many zeros, multiplication with $2$ may move the leading digit one to the left, and the adding means there might be a potential carry from the $k$-the digit from the start to the $k-1$-th digit.

That means if you look at the first $k-1$ digits of those factorials, they will change by at most $1$ from one factorial to the next.

OTOH, since

$$(1+\frac1{10^k})^{10^{2k}} \ge 1 + 10^{2k}\frac1{10^k}=1+10^k > 10$$

we have that the "renormalized" factorials

$$n!, \frac{(n+1)!}{10^{3k}}, \ldots, \frac{(n+i)!}{10^{3ki}},\ldots,\frac{(n+10^{2k})!}{10^{3k10^{2k}}}$$

where the quotient of one to the one before is

$$\frac{\frac{(n+i)!}{10^{3ki}}}{\frac{(n+i-1)!}{10^{3k(i-1)}}} = \frac{n+i}{10^{3k}} \ge \frac{n}{10^{3k}} = 1+\frac1{10^k}$$

actually increase such that from the first $n!$ to the last $\frac{(n+10^{2k})!}{10^{3k10^{2k}}}$ it covers a factor of at least $10$.  That means the initial $k-1$ digits from $n!$ have come around again, so the proof is done.

Title: Can $n!$ start with the digits 2020...


答:
You can do addition on this sums as you would do on finite sums, at least in this case where they converge everywhere.

Hence, for (e), you can keep pushing your initial idea:

$e^{x^2} = \sum\limits_{n=0}^{\infty} \frac{x^{2n}}{n!}$ and $e^{-x^2} = \sum\limits_{n=0}^{\infty} \frac{(-x^2)^n}{n!}$

So now $e^{x^2} + e^{-x^2} = \dots$ and you can add "term-by-term" the Taylor series and see where this leads you.

Title: $\sum^\infty_{n=0}\frac {x^{4n}}{(2n)!}$ is Taylor's series of which function?

Body:
$\sum^\infty_{n=0}\frac {x^{4n}}{(2n)!}$

a) $e^{x^4}$

b) $1+ x^2*\frac {\sin(x^2)}{2}$

c) $\cos^2(x^2)$

d) $(1+x^2)\ln(1+x^2)$

e) $\frac {1}{2} (e^{x^2} + e^{-x^2})$

What I tried to do: since $e^x$ Taylor series is $\sum^\infty_{n=0}\frac {x^{n}}{n!}$, I'm gonna investigate the options (a) and (e) first.

(a) $\sum^\infty_{n=0}\frac {x^{4n}}{n!}$ , if we substitute ($x^4$).

(e) if $e^x$ Taylor series is $\sum^\infty_{n=0}\frac {x^{n}}{n!}$, then $e^{-x}$ Taylor series is $\sum^\infty_{n=0}\frac {(-x)^{n}}{n!} = \sum^\infty_{n=0}\frac {(-1)^nx^{n}}{n!}$, from here I tried to do $\frac {e^x + e^{-x}}{2}$, and I realized I have no idea how to calculate these sums.

I also couldn't think of similar ways to write the series of (b)(c)(d), I would appreciate any help if there's an easy way to find these functions series without having to take derivatives, just using the know Taylor's series for known functions, (if it's not possible I would appreciate any help of how to reach a $\sum$  from just calculating derivatives and substituting in Taylor's polynomials formula).
 Thanks in advance.


问:
Prove the identity for all positive integers m and n
I'm reading Concrete Mathematics and stumbled on this question:

Prove the identity

$$
\sum_{0 \leq k<m}\left(\left\lfloor\frac{m+k}{n}\right\rfloor-\left\lfloor\frac{k}{n}\right\rfloor\right)=\left\lfloor\frac{m^{2}}{n}\right\rfloor-\left\lfloor\frac{\min (m \bmod n,(-m) \bmod n)^{2}}{n}\right\rfloor
$$

for all positive integers m and n.

Any help would be much appreciated!

I worked on the case where $m<=1/2n$, both LHS and RHS are 0. But I don't know how to solve other cases.

答:
It could occur that provided solutions from books may add extra confusion. At least, this is my impression regarding the solution (perhaps better call it as hint) to problem 3.37 from the book (the one depicted in the original question). Even the authors of the book Concrete Mathematics 2nd Edition warn against prematurely peeking into solutions (this is stated in the introductory section) to avoid being "locked" into specific way of thinking from the very beginning.

Setup

That being said, let us see how may we tackle this problem in a different way. It is beneficial to first perform some exploration about what is going on, and try to decipher pertinent properties of the components from the problem. We can identify three major components:

The floor function $\lfloor x \rfloor$ with a real argument x. It is a non-decreasing function with the following properties:

$\forall x, y\in R: (y > x \land \lfloor x \rfloor=\lfloor y \rfloor) \Rightarrow \forall z \in (x, y): \lfloor z \rfloor = \lfloor x \rfloor$.

$\lfloor n+x \rfloor=n+\lfloor x \rfloor$, where n is an integer.

The function $g(m,n)=min(m \bmod n, (-m) \bmod n)$.  It maps pairs of integers m and n into a closed interval $\left[0,\frac{n}{2}\right]$. We will use this fact for structuring the argument m, as this will allow a nice cancellation to happen on the right hand side (RHS).

The summation on the left hand side (LHS) with summands involving the floor function.

Structuring our Parameter m

This is the most crucial part of the solution. It is worthwhile to try applying this heuristic, i.e., to think whether breaking down quantities into smaller constituent parts may help express terms in smoother fashion. It turns out, that it is useful to define $m=pn+qr$, where p and r are non-negative integers, $r \in \left[0,\frac{n}{2}\right]$ and $q \in \{-1, 1\}$. Observe that r is taking values from the image of the function g(m,n) (see above).

Suppose than n=200 and we would like to represent m=175 in the previously mentioned format (it is always possible to do this). We would have $m=2\times100-1\times25$, with p=2, q=-1, and r=25. The number 150 could be described both as $m=1\times100+1\times50$ or $m=2\times100-1\times50$. Finally, m=125 would become $m=1\times100+1\times25$.

Handling the Right Hand Side of the Identity

On the RHS, using our definition of m, we will get:
$$\Bigl\lfloor\frac{p^2n^2+2pnqr+(qr)^2}{n}\Bigr\rfloor-\Bigl\lfloor\frac{min(pn+qr \bmod n, (-pn-qr) \bmod n)^2}{n}\Bigr\rfloor=p^2n+2pqr$$

Notice that the remainder in the second term is $min(qr,-qr)^2$, which is exactly the one from the first summand. Just make sure to properly analyze the cases when q=1 and q=-1.

Handling the Left Hand Side of the Identity

Similarly as we have structured m, let us represent $k=p'n+r'$, where $r' \in [0, n)$.

Case with q=1

Here is the sum expanded with our definitions of m and k:

$$\sum_{p'=0}^{p-1} \sum_{r'=0}^{n-1} \left( \Bigl\lfloor\frac{pn+qr+p'n+r'}{n}\Bigr\rfloor-\Bigl\lfloor\frac{p'n+r'}{n}\Bigr\rfloor\right) + \sum_{r'=0}^{qr-1} \left(\Bigl\lfloor\frac{pn+qr+pn+r'}{n}\Bigr\rfloor-\Bigl\lfloor\frac{pn+r'}{n}\Bigr\rfloor\right)$$

Observe that the factors p' vanish from the whole expression. We are left with the following:

$$\sum_{p'=0}^{p-1} \sum_{r'=0}^{n-1} \left(p+\Bigl\lfloor\frac{qr+r'}{n}\Bigr\rfloor\right) + \sum_{r'=0}^{qr-1} \left(p+\Bigl\lfloor\frac{qr+r'}{n}\Bigr\rfloor\right)$$

How many times do we encounter the term p in total? Well, exactly m times, which gives us the first important term $p(pn+qr)$.

Now we just need to count how many times do we encounter "overflows" in the first sum, i.e., when $qr+r'>n$. Observe that the second sum turns out to be zero as $2qr-1<n$. As r' goes from 0 to n-1, once it reaches n-qr we will have this situation. Therefore, for each p' we will have qr extra 1s. This amounts to the additional pqr term.

All in all, our LHS equals to $p(pn+qr)+pqr=p^2n+2pqr$, which is identical to the RHS.

Case with q=-1

When q=-1 the floor function will emit -1s instead of 1s. Furthermore, these are going to be issued while r' goes from 0 to $\vert qr\vert-1$. The easiest way to handle this case is to "reformat" $m=pn+qr$ into $m=(p-1)n+1 \times(n+qr)$ when defining p' and r'. For example, when n=200 and $m=2\times100-1\times25$ make it $m=1\times100+1\times75$. Here is how the initial LHS looks like in terms of the original p, q, and r:

$$\sum_{p'=0}^{p-2} \sum_{r'=0}^{n-1} \left(\Bigl\lfloor\frac{pn+qr+p'n+r'}{n}\Bigr\rfloor-\Bigl\lfloor\frac{p'n+r'}{n}\Bigr\rfloor\right) + \sum_{r'=0}^{n+qr-1} \left(\Bigl\lfloor\frac{pn+qr+(p-1)n+r'}{n}\Bigr\rfloor-\Bigl\lfloor\frac{(p-1)n+r'}{n}\Bigr\rfloor\right)$$

At the end, you will end up with the same expressions as in case 1.

Conclusion

This might seem like a lot of work, but once you figured out how to express m all the rest happens rather mechanically (you just need to watch out for trivial mistakes). At any rate, as the LSH is always equal to the RHS we have established the requested proof.


问:
Choose a positive number $k > 1$ so that $b^k \equiv a \pmod{91}$

Variables a and b are integers: $b ≡ a \pmod {91}$ and $\gcd(a, 91) = 1$.

(a) Choose a positive number $k > 1$ so that $b^k ≡ a \pmod {91}$.

(b) What is $a \pmod {91}$ if $b = 53$?

Here is how I tried to do it:

a)

$b^k ≡ a \mod 91$

$b^{k-1}*b ≡ a*1 \pmod {91}$

$b^{k-1} ≡ 1\pmod {91}$

Now $\Phi (91) = 72$ gives $b^{72} = 1 \pmod {91}$
so k = 73.

b)

$53^{73} = 53^{72} * 53 ≡ 53\ \pmod {91}$

I am wondering if I got it right and if I made any algebraic blunders.

Thank you

答:
Hint:

Actually we need to use Carmichael Function

$\lambda(91)=[6,12]=12$  which should be the highest exponent for any integer $a,(a,91)=1$

Now $53\equiv-3\pmod7\implies53^3\equiv(-3)^3\equiv1\pmod7$

Again, $53\equiv1\pmod{13}$

$\implies53^{[1,3]}\equiv1\pmod{[13,7]}$


问:
Removing factor $x$ from a number
I am required to implement this code function but I have no idea what it is asking for. Can someone explain this to me in simpler terms so that I can try to turn it into code?

Given $x$ and $y$, I need to return $Z$.

The description of the function:

If $x$ can be expressed as $x=qy^n$, then $z = q$ i.e. remove factor $y$ from the number $x$, such that $z$ no longer has $y$ as a factor.

Some example results:

x: 2835, y = 3, result = 35

x: 2357, y = 19, result = 2357

I just need a clearer idea of what they are actually asking for.

Thank you

答:
The solution was to simply divide the x value as much as possible by y, and return the final divisible value.

Thank you to the kind souls for helping me out! Point noted on the formatting in future.

Title: Removing factor $x$ from a number


答: Since $e^x=\sum_{n\ge0}\dfrac{x^n}{n!}$, what we have here is  $e)$, or the "odd terms" of $e^{x^2}$.


问:
lemma to the primitive divisor theorem.
I want to prove a lemma to the primitive divisor theorem. Where the theorem is given as: For each pair of coprime integers $a$ and $b$, $u_{n} = a^{n} - b^{n}$ has a primitive prime factor for $n > 6$.

and the Lemma to be proved which gives arithmatic properties of $u_{n}$ is : $\gcd(u_{n},u_{m}) = u_{\gcd(n,m)}$

I have attemted it in the following way: let $d=\gcd(n,m) \implies d\lvert n, d\lvert m\\
 a^d -b^d \lvert a^n -b^n  \space\text{   by definition of  } u_{n}\\
 = (a^d)^{\frac{n}{d}} -(b^d)^{\frac{n}{d}} = (a^d -b^d)((a^d)^{\frac{n}{d}-1} + \cdots+ (b^d)^{\frac{n}{d}-1})\\
\text{and similarly for m}\\
 a^d -b^d \lvert a^m -b^m  \space\text{   by definition of  } u_{m}\\
 = (a^d)^{\frac{m}{d}} -(b^d)^{\frac{m}{d}} = (a^d -b^d)((a^d)^{\frac{m}{d}-1} + \cdots+ (b^d)^{\frac{m}{d}-1})\\
\text{so }\\
u_d\lvert\gcd(u_m,u_n)$.

Could I please get some assistance with proving the reverse implication so that I can show the lemma to be true.

答:
Let gcd$(u_m,u_n)=G$ and let $d=mx+ny$ for integers $x$ and $y$ (Euclid).

Then $a^m\equiv b^m \pmod G$ and so $a^{mx}\equiv b^{mx} \pmod G$. Similarly $a^{ny}\equiv b^{ny} \pmod G$.

Then $a^{mx+ny}\equiv b^{mx+ny} \pmod G$ and $a^d-b^d\equiv 0 \pmod G$ as required.

Title: lemma to the primitive divisor theorem.


问:
Calculate $\mathbb{F}_2(\alpha, \beta)$ where $\alpha^3 + \alpha + 1 = 0, \beta^2 + \beta + 1 = 0$
Calculate $\mathbb{F}_2(\alpha, \beta)$ where $\alpha^3 + \alpha + 1 = 0, \beta^2 + \beta + 1 = 0$ where $\alpha, \beta$ are elements of a field extension $L$ of $\mathbb{F}_2$.

I know $\mathbb{F}_2(\alpha) \cong \frac{\mathbb{F}_2[X]}{X^3+X+1}$ and $\mathbb{F}_2(\beta) \cong \frac{\mathbb{F}_2[X]}{X^2+X+1}$.

There are 2 methods that I think could work but I am not sure which one is correct.

I know $\mathbb{F}_2(\alpha) \cong \frac{\mathbb{F}_2[X]}{X^3+X+1}$ and $\mathbb{F}_2(\beta) \cong \frac{\mathbb{F}_2[X]}{X^2+X+1}$. Since $X^3+X+1$ is irreducible, it means that $\mathbb{F}_2(\alpha) \cong \mathbb{F}_{2^3}$ since the irreducible polynomial has degree 3. Then since $\mathbb{F}_2(\alpha, \beta) = (\mathbb{F}_2(\alpha))(\beta)$, I thought of maybe doing a similar thing and then getting that it is isomorphic to $\mathbb{F}_{{2^3}^2}$ but I don't know how to check that $X^2 +X +1$ is irreducible in $\mathbb{F}_2(\alpha)$.

Since $X^2+X+1$ is in the ideal, I know that $X = -X^2 - X$, by filling this in the other polynomial we get $X^6 + X^4 + X^2 + 1$. This is reducible over $F_2$ since in $\mathbb{Z}_2$ it holds that 1 is a root. But then I'm not sure what I can do, I wanted to divide everything by $(x-1)$ to make it reducible but this doesn't really work since that would be for $\mathbb{Z}_2$.

I know that $\mathbb{Z}_2(\alpha) \cong \frac{\mathbb{Z}_2[X]}{X^3+X+1} \cong \mathbb{F}_8$ and $\mathbb{F}_4 \cong \frac{\mathbb{Z}_2[X]}{X^2+X+1} \cong \mathbb{Z}_2(\beta)$. Could I maybe use this in this exercise?

Which method is correct, or is there another method that I could use?

答:
Well, in view of dimensions

$[F_2(\alpha,\beta)] = [F_2(\alpha,\beta):F_2(\beta)] \cdot [F_2(\beta):F_2]$,

where $F_2[\alpha] = GF(8)$ and $F_2[\beta] = GF(4)$ and both extensions are have trivial intersection: $F_2[\alpha] \cap $F_2[\beta] = F_2$.

Trivial intersection since $GF(p^m)$ is a subfield of $GF(p^n)$ iff $m\mid n$.

The last property shows that

$[F_2(\alpha,\beta)] = [F_2(\alpha):F_2] \cdot [F_2(\beta):F_2]$

This should clarify the situation.


问:
Confusion about pushforwards of Lie Brackets
This is a very simple example that's bugging me, there's a basic gap in my understanding.

Let $\gamma : [0,1]^2 \rightarrow M$ be a curve, show that $[\frac{\partial \gamma}{\partial t}, \frac{\partial \gamma}{\partial s}] = 0$.

I can immediately solve it using the fact that $[\gamma_*(\frac{\partial}{\partial t}), \gamma_*(\frac{\partial}{\partial s})] = \gamma_*[\frac{\partial }{\partial s},\frac{\partial }{\partial s}]$, and the coordinate vector fields vanish in the bracket, but if I try and do things the long way round I start getting quite muddled.

If I try to do things in coordinates, I end up having to calculate stuff like $X(\frac{\partial \gamma^i}{\partial s})$ for $X = \left.\frac{\partial \gamma^j}{\partial t}\frac{\partial}{\partial x^j}\right|_{\gamma(t)}$, but $\frac{\partial \gamma^i}{\partial s}$ is a function of $t$ and $s$, so how can it make sense to differentiate it with the $x^i$ coordinates on $M$?

I suspect this has something to do with pullback bundles, indeed, if you have access to a connection then I "feel compelled" to want to write something like $X(\frac{\partial \gamma^i}{\partial s}) = \nabla_X(\frac{\partial \gamma^i}{\partial s})   = \gamma^*\nabla_\frac{\partial }{\partial t}(\frac{\partial \gamma^i}{\partial s}) = \frac{\partial^2 \gamma^i}{\partial t\partial s}$, but this is wrong (I'm not getting any $\frac{\partial \gamma^j}{\partial t}$ terms, and I shouldn't have to invoke a connection anyway).

Any explanation appreciated.

Following Andreas Cap's answer, I have wish to understand when this works. If $\gamma(s,t) = (x(s,t), y(s,t))$ is a diffeomorphism of $2$ dimensional manifolds, then it makes sense to pushforward the vector fields $\frac{\partial}{\partial s}, \frac{\partial}{\partial t}$. I can say

$\gamma_*(\frac{\partial}{\partial s}) = \left.\frac{\partial x}{\partial s}(s,t)\frac{\partial}{\partial x}\right|_{\gamma(s,t)} + \left.\frac{\partial y}{\partial s}(s,t)\frac{\partial}{\partial y}\right|_{\gamma(s,t)}$

$\gamma_*(\frac{\partial}{\partial t}) = \left.\frac{\partial x}{\partial t}(s,t)\frac{\partial}{\partial x}\right|_{\gamma(s,t)} + \left.\frac{\partial y}{\partial t}(s,t)\frac{\partial}{\partial y}\right|_{\gamma(s,t)}$

If I want to try and take the Lie bracket then I have $$\left[\frac{\partial x}{\partial s}(s,t)\frac{\partial}{\partial x}\right. + \left.\frac{\partial y}{\partial s}(s,t)\frac{\partial}{\partial y}\right., \left.\frac{\partial x}{\partial t}(s,t)\frac{\partial}{\partial x}\right. + \left.\frac{\partial y}{\partial t}(s,t)\frac{\partial}{\partial y}\right]$$

This is the same problem, $\frac{\partial x}{\partial t}(s,t)$ are functions of $s,t$ but we must differentiate them with $\frac{\partial}{\partial x}$ in this bracket.

Edit: I was being foolish, you are supposed to use $\gamma$ to write the reconcile the coordinates using the fact its a diffeomorphism, so the calculation indeed works when the pushforwards are well defined.

答:
The problem is that there is no well defined operation of push forward of vector fields along smooth maps. Indeed, the "vector fields" that you are trying to use are certainly only defined on the image of $\gamma$. Even there, they may not be well defined if $\gamma$ is not injective. So it is not obvious how the initial question makes sense, the question being what the Lie bracket should be.

The expressions $\frac{\partial\gamma}{\partial t}$ and $\frac{\partial\gamma}{\partial s}$ certainly make sense as "vector fields along $\gamma$, i.e. as maps $\xi:[0,1]^2\to TM$ such that $p\circ\xi=\gamma$. (These can be viewed as sections of the pullback bundle $\gamma^*TM$ over $[0,1]^2$ but I don't think that there is a Lie bracket in this seeting.

A way to make sense of the original question is the setting of $\gamma$-related vector fields. This means that you suppose that there are vector fields $\xi$ and $\eta$ on $M$ such that $\xi\circ \gamma=T\gamma\circ \frac{\partial}{\partial t}$,  and similarly for $\eta$ and $\frac{\partial}{\partial s}$. (This just mean that $\xi$ and $\eta$ are vector fields whose restriction to the image of $\gamma$ gives the objects you were trying to construct. Such fields will not exist in general, they certainly do if e.g. $\gamma$ is an embedding.) Then there is a general result saying that the brackets of $\gamma$-related vector fields are $\gamma$-related, which then shows that $[\xi,\eta]$ vanishes on the image of $\gamma$.


问:
Is it possible to calculate the input variable of a specific encryption function without brute-forcing it?
I didn't know where to post this question, here or SO but I feel it's more of a math problem.

Lately I have been reverse engineering a game, and figured out how it takes user input, encodes it, and verifies it against a specific value to see if the user input is a correct cheat code or not. The game has a unique id for every player and a specific key that it uses for this function. The algorithm looks like this:

```<code>userInput = ReadUserInput();
userInputEncrypted = userInput MOD encryptionKey;
for i = 0; i<16; i++{
    userInputEncrypted = userInputEncrypted^2 MOD encryptionKey;
}
userInputEncrypted = userInputEncrypted * userInput;
userInputEncrypted = userInputEncrypted MOD encryptionKey;

if (userInputEncrypted == correctValue)
    //Bingo!
</code>```

Now my problem is this:
I know what the correct values are, and I know the encryptionKey, is there a way mathematically to do this in reverse and instantly calculate what the starting userInput should be?

Or am I doomed to bruteforce it as the modulo that occurs 16 times gives me no other choice?

As you can see in the last step, the algorithm re-uses the starting userInput, so my hopes lie there.

Thanks in advance!

答:
Let $m$ be the encryption key, and $x$ be the userInput, modulo $m$. Then, as mentioned by mihaild, the encryption algorithm is essentially

$$x ^ {2^{16} + 1} \equiv a \pmod m$$

Let $E = 2^{16} + 1 = 65537$. To reverse the encryption, we need to find $D$ such that
$$x^{ED}\equiv x \pmod m$$
for all $x, 0 \le x < m$ and thus
$$a^D\equiv x \pmod m$$

The way to do that, is to solve
$$ED \equiv 1 \pmod \phi(m)$$
where $\phi()$ is Euler's Totient function.

For the given
$$m = 19474436171537127533$$
$$\phi(m) = 19474436152198143420$$

Once we know $\phi(m)$, we can find $D$ by using the extended Euclidean algorithm, since
$$ED - k\phi(m) = 1$$
for some $k$, provided that E is coprime to $\phi(m)$. That's certainly true in this case. In fact, $E$ is prime, and
$$phi(m) \equiv 35631 \pmod E$$

Below is a short Sage Python script that performs the necessary calculations. For convenience, I used Sage to find $\phi(m)$, but its easy to calculate it from the prime factors of $m$:

$p = 1065734711$, and

$q = 18273249403$,

since
$$\phi(pq) = (p-1)(q-1)$$
when $p$ and $q$ are prime.

Sage isn't necessary for the subsequent calculations, since plain Python has arbitrary precision integer arithmetic, and an efficient power modulus function.

Code

```<code>m = 19474436171537127533
#print(factor(m))
#m = 1065734711 * 18273249403
#print(euler_phi(m))
phi = 19474436152198143420
E = 65537 # 2^16 + 1
#print(xgcd(E, phi))
# g, s, t : g = s * E + t * phi
# (1, -6901944375041675347, 23227)
#D = -6901944375041675347
#print(D+phi)
D = 12572491777156468073
# D*E % phi == 1
#print(D*E % phi)

@interact
def test(n=12345):
    # Encrypt
    a = pow(n, E, m)
    print(a)
    # Decrypt
    x = pow(a, D, m)
    print(x)
</code>```

Output

```<code>10240177983085547351
12345
</code>```

And here's a live version of the script that you can play with, running on the SageMathCell server. (The script isn't stored on the server, it's encoded into the URL).

FWIW, this is essentially the process of cracking RSA, except that proper RSA uses a much larger modulus; these days it's recommended that $m$ be at least 1024 bits so $p$ & $q$ can't feasily be found.

Body:
Let $m$ be the encryption key, and $x$ be the userInput, modulo $m$. Then, as mentioned by mihaild, the encryption algorithm is essentially

$$x ^ {2^{16} + 1} \equiv a \pmod m$$

Let $E = 2^{16} + 1 = 65537$. To reverse the encryption, we need to find $D$ such that
$$x^{ED}\equiv x \pmod m$$
for all $x, 0 \le x < m$ and thus
$$a^D\equiv x \pmod m$$

The way to do that, is to solve
$$ED \equiv 1 \pmod \phi(m)$$
where $\phi()$ is Euler's Totient function.

For the given
$$m = 19474436171537127533$$
$$\phi(m) = 19474436152198143420$$

Once we know $\phi(m)$, we can find $D$ by using the extended Euclidean algorithm, since
$$ED - k\phi(m) = 1$$
for some $k$, provided that E is coprime to $\phi(m)$. That's certainly true in this case. In fact, $E$ is prime, and
$$phi(m) \equiv 35631 \pmod E$$

Below is a short Sage Python script that performs the necessary calculations. For convenience, I used Sage to find $\phi(m)$, but its easy to calculate it from the prime factors of $m$:

$p = 1065734711$, and

$q = 18273249403$,

since
$$\phi(pq) = (p-1)(q-1)$$
when $p$ and $q$ are prime.

Sage isn't necessary for the subsequent calculations, since plain Python has arbitrary precision integer arithmetic, and an efficient power modulus function.

Code

m = 19474436171537127533
#print(factor(m))
#m = 1065734711 * 18273249403
#print(euler_phi(m))
phi = 19474436152198143420
E = 65537 # 2^16 + 1
#print(xgcd(E, phi))
# g, s, t : g = s * E + t * phi
# (1, -6901944375041675347, 23227)
#D = -6901944375041675347
#print(D+phi)
D = 12572491777156468073
# D*E % phi == 1
#print(D*E % phi)

@interact
def test(n=12345):
    # Encrypt
    a = pow(n, E, m)
    print(a)
    # Decrypt
    x = pow(a, D, m)
    print(x)

Output

10240177983085547351
12345

And here's a live version of the script that you can play with, running on the SageMathCell server. (The script isn't stored on the server, it's encoded into the URL).

FWIW, this is essentially the process of cracking RSA, except that proper RSA uses a much larger modulus; these days it's recommended that $m$ be at least 1024 bits so $p$ & $q$ can't feasily be found.


问:
Is the image of a weakly convergent sequence also weakly convergent?
This question comes from my thinking on the learning of functional analysis. Suppose that $S:L^2(\mathbb{R})\to L^2(\mathbb{R})$ is an operator.Suppose that  $f_n$ is convergent weakly to $f$ in $L^2(\mathbb{R})$ and $\lVert S(f)\rVert \leq M\lVert f\rVert$ for any $f\in L^2(\mathbb{R})$.Can we show that $S(f_n)$ converges to $S(f)$ weakly in $L^2(\mathbb{R})$.I think it is right,but I don't know how to show it.If it is not true,could you offer some example?

答:
I do not know how you defined weak convergence. I assume that it means
$$
\phi(f_n) - \phi(f) \rightarrow 0
$$
for every bounded linear functional $\phi:L^2(\mathbb{R}) \rightarrow \mathbb{R}$.

Note that $\phi \circ S$ is a bounded linear functional, since both $\phi$ and $S$ are bounded and linear. It follows that
$$
\phi(S(f_n)) - \phi(S(f)) = (\phi \circ S)(f_n) - (\phi \circ S) (f) \rightarrow 0.
$$

Note that this also works for every Banach space and not only $L^2(\mathbb{R})$.


问:
Determining if $R=\left\{(f,g)\mid \exists k\in\Bbb Z,\forall x\in\Bbb Z, \ f(x)g(x)\lt k\right\}$ is an equivalence
I need help with proving whether or not the relation $R=\left\{(f,g)\mid \exists k\in\Bbb Z,\forall x\in\Bbb Z, \ f(x)g(x)\lt k\right\}$ is an equivalence relation on $\mathbb{Z}\times\mathbb{Z}$

I understand the basic key components needed in order to determine if the relation is an equivalence relation, like reflexivity, symmetry, and transitivity but i can't seem to find a way to proof reflexivity.

答:
Consider the set $A$ of all functions $f:\mathbb{Z}\to\mathbb{Z}$.  Define an equivalence relation on $A$ by:

$$R=\{(f,g)|\exists k,\forall x,f(x)g(x)<k\}\subset A\times A$$

I claim the relation is symmetric, but not reflexive or transitive.  For symmetry, it suffices to note that $f(x)g(x)=g(x)f(x)$.

As a counterexample to reflexivity, consider $f(x)=x$, and note that $(f,f)\notin R$ because $x^2$ is not bounded.

Finally, consider the functions $f(x)=h(x)=x$ and $g(x)=0$.  Then we have $(f,g),(g,h)\in R$, but $(f,h)\notin R$ so the relation is not transitive.

It follows that the relation is not an equivalence relation.


答:
Recall that the Maclaurin series (or Taylor series if you prefer) of $\cosh x$ is:
$$\cosh x=\sum_{r=0}^{\infty}\frac{x^{2r}}{(2r)!}$$
Replace $x$ by $x^2$ and we find that
$$\cosh x^2=\sum_{r=0}^{\infty}\frac{x^{4r}}{(2r)!}$$


问:
$\sum_{k=1}^n [X_k-E(X_k)]$ converge in probability
$\{X_n\}$ are independent, $\sum_{k=1}^\infty Var(X_k)<\infty$,prove
$$\sum_{k=1}^n [X_k-E(X_k)]
$$converge in probability.

I have tried chebyshev inequality, but it's too loose.

答:
Let $S_n:=\sum_{i=1}^n (X_i-\mathsf{E}X_i)$. Then $\{S_n\}$ converges in $L^2$ (and in probability) because
$$
\operatorname{Var}(S_n-S_m)\to 0\quad\text{as}\quad n,m\to\infty,
$$
and so $\{S_n\}$ is $L^2$-Cauchy convergent.


问:
Differential equation with partial fraction
How do you separate this differential equation into a partial fraction?

Solve the following differential equation:
$$\frac{dy}{dx}=\frac{2y^2-xy+x^2}{xy-x^2}$$

答:
Let $\frac{y}{x}=v$

$$\frac{{\rm d}y}{{\rm d}x} = v + x\frac{{\rm d}v}{{\rm d}x} =\frac{2v^2-v+1}{v-1} \\
x\frac{{\rm d}v}{{\rm d}x} = \frac{v^2+1}{v-1} \\
\frac{v-1}{v^2+1}dv = \frac {{\rm d}x}{x} \\
\text{Integrating  both  sides} \\
\int \frac{v-1}{v^2+1}dv =\int \frac {{\rm d}x}{x} \\
\int \frac{v}{v^2+1}dv - \int \frac{1}{v^2+1}dv= \frac {{\rm d}x}{x} \\
\frac{1}{2} \ln(v^2+1) - \tan^{-1}v = lnx + C\\
\frac{1}{2} \ln \left(\frac{y^2+x^2}{x^4}\right) = \tan^{-1}\left( \frac{y}{x}\right) + C $$

Where $C$ is the constant of integration.