问:
A linearly independent matrix doesn't have integer solution for some vector x
The whole question was too big for the title:

If a linearly independent matrix doesn't have an integer solution for some vector $x$, then can adding a linearly dependent vector to the matrix produce an integer solution?

This question arose from the thread. Answer $2$ reduced the problem to two linearly linearly independent vectors, where as original problem consisted of $4$ vectors, this seems to imply that you can just "discard" linearly dependent vectors (as they wouldn't produce integer solutions), but I'm not seeing why? Apologies if the question is very trivial.

For example: $$i\cdot\begin{pmatrix}2\\-1\end{pmatrix}+j\cdot \begin{pmatrix}1\\3\end{pmatrix} = \begin{pmatrix}1\\1\end{pmatrix}$$
This equation has no solutions in $\mathbb Z$.
How do I know for sure that $$i\cdot\begin{pmatrix}2\\-1\end{pmatrix}+j\cdot \begin{pmatrix}1\\3\end{pmatrix} + k\cdot\begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}1\\1\end{pmatrix}$$
for some $a,b \in \mathbb Z$  won't have a solution in $\mathbb Z$ as well?

答: It does have a solution.  $i=j=k=1$ and $${a \choose b} = {-2 \choose -1}$$.


问:
Understanding vital part of the proof that a closed interval is connected.
Part of the proof that a closed interval in $ℝ$ is connected usually relies on the below fact.

Given that $W \neq \varnothing$ is closed and open in a closed interval.

There exists such $x_0$ and $\epsilon > 0$ that $(x_0-\epsilon, x_0 + \epsilon) \subset W$. We then define $r$ as:
$$
r = \sup \{ p: [x_0, x_0 + p) \subset W \}
$$

Then it is stated that since $W$ is closed, then it follows that $x_0 + r \in W$.

I can't understand why the latter holds. I know it has to do with the fact that there is a sequence from $W$ and since it's closed then the limit of that sequence, which is $sup$ will belong to it.

I am confused, since the sequence is supposed to consist of $p$'s from above definition but then it is stated that $x_0 + r$ belongs to $W$ and not just $r$.

Would appreciate any help on that. Thanks.

答:
More generally:

If $A \subseteq \mathbb R$ is non-empty and bounded above, then $\sup A \in \overline A$.

Indeed, let $\alpha = \sup A$. Then, for each $n \in \mathbb N$, there is $a_n \in A$ such that $b_n=\alpha - \frac1n < a_n \le \alpha$. Then $b_n \to \alpha$ implies $a_n \to \alpha$. Thus, $\alpha \in \overline A$.

ContentLicense: CC BY-SA 4.0


问:
If $X_n$ is an $L_1$-bounded martingale, show that $\sum_n(X_n-X_{n-1})^2< \infty$ a.s.
I had asked this question previously in the following post but there were no replies.

Recently, I found a two page article(free download) with a possible alternate line of proof to the one suggested in the first link

Note: $X_n$ is an $L_1$-bounded martingale means that $\sup_nE[|X_n|]<\infty$

However in the link, the author shows that $\sum_n(X_n-X_{n-1})^2< \infty$ a.s. on a set $E$ which is defined as follows:$$E_n:=\bigcap_{k\leq n}\{\omega||X_n(\omega)|\leq M\}\text{ and }E=\bigcap_n E_n.$$ He then states that for large $M$ the measure of $E$ is near 1, since by the martingale convergence theorem, $\sup_n|X_n|<\infty$ which completes the proof.

However, I have two issues.

Why can the measure of $E$ be made close to 1?

Shouldn't we require a set $E$ with measure exactly 1 to show a.s convergence. If so how do we find a set E of measure 1?

Can someone please explain how to rectify these issues?

答:
$(E_n)_n$ is a nonincreasing sequence (for the inclusion order) of events, hence $\mathbb P(E_n)\underset{n\to+\infty}{\to}\mathbb P(E)$. Let $\varepsilon>0$. Then there exists $N$ such that $\mathbb P(E)\ge\mathbb P(E_N)-\varepsilon$. Since $\mathbb P(\sup_n\vert X_n\vert<+\infty)=1$, it is clear that $\mathbb P(E_n)\underset{M\to+\infty}{\to}1$. You can see the latter as a direct consequence of the dominated convergence theorem, for example. Indeed,
$$
\mathbb P(E_n)=\mathbb E[\prod_{k\le n}1_{\vert X_n\vert\le M}]\underset{M\to+\infty}{\longrightarrow}1.
$$

So for $M$ large enough, $\mathbb P(E_N)\ge1-\varepsilon$, hence $\mathbb P(E)\ge1-2\varepsilon$. It is not written, but the latter event $E$ depends on $M$. Let us then write $E^M$. We just showed that
$$
\forall\varepsilon>0,\quad\exists M>0,\quad\mathbb P(E^M)\ge1-2\varepsilon.
$$

Take any sequence $(\varepsilon_n)_n$ converging to $0$. Then there exists $(M_n)_n$ such that for all $n$, $\mathbb P(E^{M_n})\ge1-2\varepsilon_n$. Let $E=\bigcup_{n}E^{M_n}$. Then $\mathbb P(E)=1$ (since $\mathbb P(E)\ge\mathbb P(E^{M_n})\ge1-2\varepsilon_n$ for all $n$) and $\sum_n(X_n-X_{n-1})^2<+\infty$ on $E$ (since it is finite on all $E^{M_n}$ for all $n$).


问:
Calculate the error bound for the rounding errors of an expression
I need to calculate the error bounds for the following expression when computed under the IEEE 64-bit standard for $x\in[0,\pi/2] $:

$$
f(x)=2*\cos(x)^2-1-\cos(2x)
$$

I understand that I need to decompose the equation into several such that each rounding operation is captured independently. Therefore:

$$
u_1=(1+\delta_1)*x
$$

$$
u_2=cos(u_1)(1+\delta_2)
$$

$$
u_3=(u_2)^2 (1+\delta_3)
$$

$$
u_4=(2u_3-1)(1+\delta_4)
$$

$$
u_5=(u_4-cos(2u_1))(1+\delta_5)
$$

Is this correct? How do I proceed from here onwards?

答:
In binary based formats, multiplication with $2$ is exact. One can also assume that the input is exactly as given. Then the evaluation in floating point gives a result
$$
(\{2[\cos(x)(1+δ_1)]^2-1\}(1+δ_2)-\{\cos(2x)(1+δ_3)\})((1+δ_4)
$$
Retaining only the terms linear in $δ_j$ gives
$$
=4δ_1·\cos^2x+δ_2·(2\cos^2x-1)-δ_3·\cos(2x)+δ_4·0+...
$$
Using $|δ_1|\le\bar δ$, this gives for the leading error terms a bound of
$$
(4\cos^2x+2|\cos(2x)|)\barδ=(2+4\max(0,\cos(2x))\barδ\le 6\barδ
$$


问:
Show that if $x\geq 0$ and $n$ is a positive integer, then $\sum_{k=0}^{n-1}\left\lfloor {x+\frac{k}{n}}\right\rfloor=\lfloor {nx}\rfloor$
I need help with this question, where $\lfloor x\rfloor$ means the floor function of $x$.

Show that if $x\geq 0$ and $n$ is a postive integer, then $$\sum_{k=0}^{n-1}\left\lfloor x+\frac{k}{n} \right\rfloor =\lfloor nx\rfloor$$

答:
Here is a solution I saw many many years ago and really love:

Let $f(x)=\sum_{k=0}^{n-1}\lfloor x+\frac{k}{n}\rfloor -\lfloor nx\rfloor$.

Then
$$f(x+\frac{1}{n})= \sum_{k=0}^{n-1}\lfloor x+\frac{k}{n}+\frac{1}{n}\rfloor -\lfloor nx+1\rfloor= \sum_{k=1}^{n}\lfloor x+\frac{k}{n}\rfloor -\lfloor nx\rfloor -1  $$
$$=\sum_{k=1}^{n-1}\lfloor x+\frac{k}{n}\rfloor +\lfloor x+1 \rfloor-\lfloor nx\rfloor -1 =\sum_{k=1}^{n-1}\lfloor x+\frac{k}{n}\rfloor +\lfloor x \rfloor-\lfloor nx\rfloor =f(x) $$

Thus $f$ is periodic with period $\frac{1}{n}$.

Moreover, if $x \in [0, \frac{1}{n})$ then

$$f(x)=\sum_{k=1}^{n-1}0-0=0 \,.$$

Thus $f$ is periodic with period $T=\frac{1}{n}$ and zero on $[0,T)$, hence $f$ is the zero function.

ContentLicense: CC BY-SA 3.0


问:
Inequality $2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2) +30 abc \ge 3$ for $a+b+c=1$
Let $a,b,c$ be positive numbers with $a+b+c=1$. Show that

$$
2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2) +30 abc  \ge 3
$$

Equality holds for $a=b=c=\frac13$.  One could try homogenizing:

$$
2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2)(a+b+c)^2 +30 abc (a+b+c) - 3 (a+b+c)^4 \ge 0
$$
Expanding gives
$$
2 \sum_{cyc} a^4  -  \sum_{cyc} a^3(b+c) - 2  \sum_{cyc} a^2 b^2 + 2 a b c \sum_{cyc} a \ge 0
$$

but I didn't manage to advance further.

答:
Another idea to deal with this type of inequality.

Assume $c=\min\{a,b,c\} \to 0< c \le \dfrac{1}{3}.$

Let $$f(a,b,c)=2 \left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{2}+5\left(a^2+b^2+c^2\right)+30abc$$
$$g(a,b,c)=2 \left[ \frac{1}{2} \left( a+b \right) ^{2}+c^2 \right] ^2+\frac{5}{2}
 \left( a+b \right) ^{2}+5c^2+\dfrac{15}{2} \left( a+b \right) ^2c$$
We will so that
$$f(a,b,c)\ge g(a,b,c)\Leftrightarrow \left( a-b \right) ^{2} \left\{ 3a^2+2ab+3b^2+4{c}^{2}
+15\left(\frac{1}{3}-c\right) \right\}\ge 0\, \forall\,0<c\le \frac{1}{3}
$$

We only need to prove $$g(a,b,c)\ge 3\Leftrightarrow \frac{1}{2} c(c+1)(3c-1)^2\ge 0.$$

Done.


答:
Since $$e^{i\theta}=1 \Rightarrow \theta=0,2\pi $$
$$\oint _{ D(0,1) }f(z)dz =\int_{[0,2\pi]-  \{0,2\pi\} } ie^{i\theta}f(e^{i\theta})d\theta $$
Since $\{0,2\pi\} $ is a set of measure 0
$$\oint _{ D(0,1) }f(z)dz =\int_{0}^{2\pi} ie^{i\theta}f(e^{i\theta})d\theta $$
$$\oint _{ D(0,1) }f(z)dz =\oint _{ C(0,1) }f(z)dz$$

Title: Prove that $\oint _{C(0,1)}f(z)dz= \oint _{ D(0,1) }f(z)dz $

Body:
Define $C(0,1)= \{z: \mid z\mid=1 \}$
$D(0,1)= \{z: \mid z\mid=1, z\neq 1+0i \}$

Prove that,
$$\oint _{C(0,1)}f(z)dz= \oint _{ D(0,1)
  }f(z)dz  $$
My try

$z=1+0i$ is of measure zero. So, $$\oint _{C(0,1)}f(z)dz= \oint _{ D(0,1)
  }f(z)dz  $$


问:
Sum of products of elements of nonempty subsets of A set
Let $A_1, A_2, \ldots , A_{63}$ be the 63 nonempty subsets of $\{ 1,2,3,4,5,6 \}$. For each of these sets $A_i$, let $\pi(A_i)$ denote the product of all the elements in $A_i$. Then what is the value of $\pi(A_1)+\pi(A_2)+\cdots+\pi(A_{63})$?

Here is the solution

For size 1: sum of the elements, which is 21
For size 2: $ 1 \cdot (2 + 3 + 4 + 5 + 6) = 20 $, $ 2 \cdot (3 + 4 + 5 + 6) = 36 $, $ 3 \cdot (4 + 5 + 6) = 45 $, $ 4 \cdot (5 + 6) = 44 $, $ 5 \cdot 6 = 30 $. Sum is 175.
For size 3: Those with least element 1:  $ 6, 8, 10, 12, 12, 15, 18, 20, 24, 30 = 155 $. Those with least element 2: $ 24, 30, 36, 40, 48, 60 = 238 $. Those with least element 3: $ 60 + 72 + 90 = 222 $. Those with least element 4: only one possible subset, which is $ \{4, 5, 6\} $, the $ \pi $ of which is 120. The total sum here is 735.
For size 4: Least element 1: $ 24 + 30 + 36 + 40 + 48 + 60 + 60 + 72 + 90 + 120 = 580 $; least element 2: $ 120 + 144 + 180 + 240 + 360 = 1044 $; least element 3: only one, which is $ 3 \cdot 4 \cdot 5 \cdot 6 = 360 $. The total sum here is 1984.
For size 5: Exclude each one individually to get $ 720 + 360 + 240 + 180 + 144 + 120 = 1764 $
For size 6: $ 6! = 720 $

The final answer is $ 21 + 175 + 735 + 1984 + 1764 + 720 = \boxed{5399} $

Is there any shorter way for doing this ?

Thank a lot

答:
The answer is 5039 just because, if you add 1 (let 1 be the product of the elements of the empty set) you must get $7!$.

This is a general property: if $\pi(A)$ is the product of the elements of $A$ (with the convention $\pi(\emptyset)=1$), and $[n] = \{1,2,\dots, n\}$, then for every natural $n$
$$
\sum_{A\subseteq [n]} \pi(A) = (n+1)!\,.
$$
It is an equivalent fact that
$$
\sum_{A\subseteq [n]} \frac 1 {\pi(A)} = n+1\,,
$$
as we see by dividing each term by $n!$ (every term $\pi(A)$ becomes $1/\pi(A^c)$).

Both of them are very easily proved by induction, but an alternative proof and a nice probabilistic interpretation of the second one is as follows:

We are at the $(n+1)$-th floor, and we take a lift to get down to the ground floor. But every time we press the button the lift will stop at a random level on the way, with uniform probability.

Let $A$ be the set of levels (floors) where the lift stops before we are downstairs. If we get there in the first step then $A=\emptyset$. When we are at the $k$-th floor, we have $k$ different possible stops in the next stage. Hence the probability of any $A$ as the set of intermediate points is $1/\big((n+1)\,\pi(A)\big)$, and summing up for all $A$ we get
$$
1 = \frac 1 {n+1} \,\sum_{A\subseteq [n]} \frac 1 {\pi(A)}\,.
$$


问:
Compact $L^\infty$ spaces
I try to understand when $L^\infty(X;Y)$ is compact. Is the following statement true?

If $X\subset\mathbb{R}$ and $Y\subset\mathbb{R}$ are both compact, then $L^\infty(X;Y)$ is compact.

答:
Measurable subsets of $\sigma$-finite non-atomic measure spaces have $\sigma$-finite non-atomic induced measure, and therefore either $X$ is a null set or $X$ may be partitioned into a countably infinite union of disjoint measurable subsets of positive measure.

If $X$ is a null set or $\# Y\le 1$, then $\#L^\infty(X;Y)\le 1$ and it's therefore compact.

If $\lambda(X)>0$, $X=\bigcup_{i=1}^\infty E_i$ with the $E_i$ as above and $u,v\in Y$ with $u\ne v$, then consider $$f_n(x)=\begin{cases}u&\text{if }x\in E_n\\ v&\text{if }x\notin E_n\end{cases}$$

And notice that $d_\infty(f_n,f_m)=\lvert u-v\rvert$ for all $m\ne n$. Therefore $L^\infty(X;Y)$ isn't a compact subset of $L^\infty(X)$.

Title: Compact $L^\infty$ spaces


问:
Probability of getting a particular value or higher on at least two of the dice when rolling three six-sided dice?
I've calculated the probability for rolling a particular value or higher on both dice when rolling 2d6, as follows.

1 or higher: 100%
2 or higher: 69.44%
3 or higher: 44.44%
4 or higher: 25%
5 or higher: 11.11%
6 or higher: 2.778%

If I add a third dice to this, how does it affect the probability if I still only need the particular value or higher on two or more of the dice?

答:
This can be treated as a binomial: each die either rolls the value $v$ or higher, or it does not; the dice roll independently; there is a fixed number $n$ of trials (dice); and the probability is the same for all dice.

The general binomial formula is $$P(X=k) = {n \choose k}\cdot p^k \cdot (1-p)^{n-k}$$ where $p$ is the probability of a single die rolling at or above value $v$, which is $\frac{7-v}{6}$, $n=3$, and $k = 2$.


问:
For any primitive element $a$ in a free group of rank two we have $a^k ba^l=b$ only if $(k,l)=(0,0)$ provided $b\not\in \langle a\rangle$

Problem 1: Let $F$ be a free group of rank at least two, and $a, b\in F$ be two
non-trivial elements with $b\not \in \langle a\rangle$. Suppose,  $a\neq x^n$ for any $x\in F$ and any
$n\geq 2$, i.e. $a$ is primitive. Then, $a^kb a^l=b\iff (k,l)=(0,0)$.

Note that $a$ is primitive is necessary, otherwise, $a$ could be $b^m$ for some $m\in \Bbb Z$, and then $a^kba^l=b^{mk}bb^{ml}=b^{m(k+l)+1}$ would equal to $b$ for each $k=-l$.

Let $F$ be free on a set $\{g_i:i\in \Lambda\}$ of generators. Now, write $a$ and $b$ as reduced words. If there exist two distnict elements $g_a, g_b\in \{g_i:i\in \Lambda\}$ having contributions in the reduced representations of $a,b$, respectively, then $a^k ba^l\neq b$ unless $(k,l)=(0,0)$, and we do not need the assumption $a$ is primitive in this case.

So, we need to consider the case when both $a$ and $b$ have reduced representations obtained from the same set.

Problem 2: Suppose I replace $F$ by this non-free group $\big\langle x_1,x_2,y_1,y_2\big|[x_1,y_1]=[x_2,y_2]\big\rangle$
keeping fixed all other hypothesis on $a,b$. Is it again true that
$a^k ba^l=b$ only if $(k,l)=(0,0)$?

答:
Here is a quick proof, assuming that you know how to solve the conjugacy problem for free groups:

Note that if elements $x, y\in F$ commute then $\langle x, y\rangle$ is cyclic, which can be proven using the solution to the conjugacy problem for $F$ (or by applying the nuclear bomb which says that "subgroups of free groups are free").

Now, assume that $a^kba^l=b$ with $k,l\neq0$, where $a$ and $b$ are as in the question. The identity $a^kba^l=b$ implies that $a^k$ and $a^{-l}$ are conjugate. By applying the solution to the conjugacy problem, we see that $k={-l}$, and hence $a^k$ and $b$ commute. As subgroups of free groups are free, we have that $\langle a^k, b\rangle$ is cyclic, and in fact $\langle a\rangle=\langle a^k, b\rangle$. Hence, $b\in\langle a\rangle$, a contradiction.

Problem 2 also holds true. There are a number of ways of seeing this, all depending on what you know. I cannot see an easy, no-prior-knowledge proof though. The "slickest" way is perhaps to note that this is a surface group, and subgroups of surface groups are either surface groups of higher genus or free, and so $\langle a, b\rangle$ is free. Another way is to note that this result is true for all torsion-free hyperbolic groups (the standard phrase is "maximal cyclic subgroups are malnormal"), and the group here is a torsion-free hyperbolic group.


问:
What is the distribution of primes modulo $n$?
Let $n\geq 2$ and let $k$ be "considerably larger" than $n$ (like some large multiple of $n$). Then for each $i$ such that $0<i<n$ and $\gcd(i,n)=1$ let's define
$$c_i=\left|\{p_j\;|\; p_j\equiv i \mod n,\;\mbox{where $p_j$ is the $j$-th prime, $1\leq j\leq k$}\}\right|$$
so $c_i$ represents how many of the first $k$ primes are congruent to $i$ modulo $n$.

What can we say about $c_i$s, that is about the distribution of the first $k$ primes modulo $n$?

I thought the distribution will be seemingly random, and that is mostly true - for example $c_i$s are always very close together. But there are observable non-random patterns. For example for $n=3$, for various $k$s I've tried  (up to $10^6$) I always got $c_2>c_1$. If this were just some kind of random discrepancy due to distribution of small primes, it would eventually vanish for large $k$s, which I don't observe.

答:
I once tried to answer the particular case of primes $4n\pm 1$ to myself.
(I think it will strongly help to read the answers by Raymond, Raymond  and Greg. At the end of the last answer there's also a link to the chat, where we continued the discussion.)

Here is how far I got with an explicit formula for the number of primes of the form $4n+3$ below $x$, $\pi^*(x;4,3)$, expressed in terms of (sums of) sums of Riemann's $R$ functions over roots of Riemann's $\zeta$ resp. Dirichlet $\beta$ function:

\begin{align*}
\Pi^*(x;4,3)
&= \pi^*(x;4,3) + \tfrac12 \sum_{\substack{b\pmod 4 \\ b^2\equiv
3\pmod 4}} \pi^*(x^{1/2};4,b) + \tfrac13 \sum_{\substack{c\pmod q \\
c^3\equiv 3\pmod 4}} \pi^*(x^{1/3};4,c) + \cdots \\
\end{align*}
Then I try to complete things by adding several up

\begin{align*}
\Pi^*(x;4,3) &= \tfrac11\pi^*(x;4,3) + \tfrac13 \pi^*(x^{1/3};4,3) + \cdots \\
\tfrac12\Pi^*(x^{1/2};4,3) &= \tfrac12\pi^*(x^{1/2};4,3) + \tfrac16
\pi^*(x^{1/6};4,3) + \cdots \\
\tfrac14\Pi^*(x^{1/4};4,3) &= \tfrac14\pi^*(x^{1/4};4,3) +
\tfrac1{12} \pi^*(x^{1/12};4,3) + \cdots \\
&\vdots&\\
\hline\\
\tag{1}\sum_{k=0}^\infty
2^{-k}\Pi^*(x^{2^{-k}};4,3)&=\sum_{m=1}^\infty \tfrac1m
\pi^*(x^{1/m};4,3)
\end{align*}
Using Möbuis inversion I'll get

\begin{align*}
\pi^*(x;4,3)&=\sum_{m=1}^\infty \tfrac{\mu(m)}m\sum_{k=0}^\infty
2^{-k}\Pi^*(x^{2^{-k}/m};4,3)\\
\tag{2}&=\sum_{k=0}^\infty 2^{-k}\sum_{m=0}^\infty
\tfrac{\mu(m)}m\Pi^*(x^{2^{-k}/m};4,3)
\end{align*}
Now I use

\begin{align*}
\Pi^*(x^{2^{-k}};4,3)&=\frac1{\phi(4)} \sum_{\chi\pmod 4}
\overline{\chi(3)}\Pi^*(x^{2^{-k}},\chi)\\
\tag{3}&=\frac12 \left( \Pi^*(x^{2^{-k}},\chi_1)-
\Pi^*(x^{2^{-k}},\chi_2) \right)
\end{align*}
and then

\begin{align*}
\tag{$4_1$}\Pi^*(x^{2^{-k}},\chi_k)&=\operatorname{li}(x^{1/2^{k}})-\sum_{\rho_\zeta}
\operatorname{li}(x^{\rho_\zeta/2^k})\text{ if $k=1$}\\
\tag{$4_2$}&=\phantom{\operatorname{li}(x^{1/2^{k}})}-\sum_{\rho_\beta}
\operatorname{li}(x^{\rho_\beta/2^k})\text{ if $k=2$}\\
\end{align*}
which gives

\begin{align*}
\tag{3'}\Pi^*(x^{2^{-k}};4,3)&=\frac12 \left(
\operatorname{li}(x^{1/2^{k}})-\sum_{\rho_\zeta}
\operatorname{li}(x^{\rho_\zeta/2^k}) +\sum_{\rho_\beta}
\operatorname{li}(x^{\rho_\beta/2^k}) \right)
\end{align*}
so finally

\begin{align*}
\pi^*(x;4,3)&=\sum_{k=0}^\infty 2^{-k}\sum_{m=0}^\infty
\tfrac{\mu(m)}m\frac12 \left(
\operatorname{li}(x^{1/2^{k}})-\sum_{\rho_\zeta}
\operatorname{li}(x^{\rho_\zeta/2^k}) +\sum_{\rho_\beta}
\operatorname{li}(x^{\rho_\beta/2^k}) \right)\\
\tag{5}&=\sum_{k=0}^\infty 2^{-k-1}\left(
\operatorname{R}(x^{1/2^{k}})-\sum_{\rho_\zeta}
\operatorname{R}(x^{\rho_\zeta/2^k}) +\sum_{\rho_\beta}
\operatorname{R}(x^{\rho_\beta/2^k}) \right)
\end{align*}

I would be very, very glad to read your opinion...


问:
Prove a certain quadrilateral is actually a Saccheri Quadrilateral
Question:

"Show that a quadrilateral ABCD, which has angle C = angle D = right angle; and angle A is congruence to angle B, is a Saccheri Quadrilateral."

My attempt:

As by definition, a Saccheri Quadrilateral's (i.e. ABCD) base angles(i.e. Angle C and D) are right angles and also sides AD should be congruence to side BC.

By question, we already have the base angles C and D are equal to right angles. So we just need to show that sides AD and BC are congruence.

I had clue to prove AD // BC by alternate interior angle theorem, but I don't have clue in proving AD congruence to BC.

Please give me some help, thank you!

答:
Let $M$ be the midpoint of $CD$

Let $l$ be the line perpendicular to $ CD$ at $M$

Let $N$ be the intersection point of $l$ and $AB$

Since <$NMC$~<$NMD$, $CM$~$DM$, $\Delta$$NMC$~$\Delta$$NMD$ by SAS

Hence <$NCM$~<$NDM$

Also <$NCB$~<$NDA$ since <$MCB$~<$MDA$ and from 5

Since $NC$~$ND$ from 4, by SAA, $\Delta$$BCN$~$\Delta$$ADN$

$BC$~$AD$

Here each ~ means "congruent" for segments, angles or triangles.


问:
Closure of the set $\{f\in C[0,1] : f(x) >0 \ \forall x\in[0,1]\}$

Let $C[0,1]$ be defined with the max-norm and $A=\{f\in C[0,1] : f(x) >0 \ \forall x\in[0,1]\}$. Show that the closure of $A$ is the set $\{f \in C[0,1] : f(x) \geqslant 0 \ \forall x\in[0,1] \}$.

The closure of $A$ would be $A \cup \partial A$, but I'm not sure how to find the boundary here? It also seems that there would be alternative ways to show this?

答:
If $f(x) \geq 0$ for all $x$ then $f(x)+\frac  1n \to f(x)$ in the sup norm and $f(x)+\frac  1n >0$ for all $x$.

If $f_n(x) >0$ for all $x$ for all $n$ and  $f_n(x) \to f(x)$ in the sup norm then  $f_n(x) \to f(x)$ for each $x$,  so $f(x) \geq 0$ for all $x$.

Can you finish now?

[I am using the fact that a point belongs to the closure of  a set iff there is a sequence from the set converging to that point. Let $B=\{f \in C[0,1]: f(x) \geq 0 \, \forall x\}$. The first paragraph above shows that $B \subseteq \overline A$ and the second paragraph shows that $\overline A \subseteq B$].


问:
Surface with non-zero mean curvature means orientable
Let $M$ be a surface in $\Bbb R^3$ with non-zero mean curvature for every point. How could I  show that this implies that $M$ is orientable? By our definition, orientable means that an unitary, normal vector can be defined continuously for every point in the surface.

This is a homework question, so just hints would be appreciated :)

答:
Eventually, I found a way to prove it. I leave a hint here in case anybody finds it useful:

If $N$ is the normal vector, $N$ is not canonical (it can be "+" or "-" $N$, identifying its sign with the orientation of the base $\{D_1f, D_2f, N\}$). In the same way, $H$ is not canonical either (it can be positive or negative for each parametrization). Which operation on both would be canonical, i.e., would be uniquely determined with independence of the map?


问:
Solving a problem involving primitive and partially recursive functions.
We have that for a partial function $f : \mathbb{N}^{k+1} \to \mathbb{N}$, the $\mu-$ operator is defined as the following:

$$\mu (f)(x_1, ..., x_k)=\begin{cases}
      n, & f(n,x_1,...,x_k)=0  \land f(n',x_1,...,x_k) \neq 0 \land \text{is defined for all} \ n' < n, \\
      \bot, & \text{else}
   \end{cases}$$

We are then given the following primitive recursive functions:

$$\text{modsub}: \mathbb{N}^2 \to \mathbb{N}, \text{modsub}(x,y) = \text{max}\{0,x-y\},$$
$$\text{exp}: \mathbb{N} \to \mathbb{N}, \text{exp}(x)=2^x$$

Finally, we have that $f: \mathbb{N}^2 \to \mathbb{N}$ is defined as $f(x,y):= \text{modsub}(y, \text{exp}(x))$.

How can I show that $\mu(f)(4) = 2$ and $\mu(f)(9) =4$?

答:
$(\mu f)(4) =$ smallest $n$ such that $f(n,4) = \max\{0,4-2^n\}$ is zero, which is $n=2$.

$(\mu f)(9) =$ smallest $n$ such that $f(n,9) = \max\{0,9-2^n\}$ is zero, which is $n=4$. Here $n=3$ would lead to $\max\{0,9-2^3\}=1$.


答:
If you require $2$ or more success from $3$ dice, then visualize the 'failures'.

For throwing a $2$ or higher at least twice, we need to avoid combinations such as $116$, $151$ or $411$, and obviously $111$. There are $3$ 'legs' of dimensions $1\times1\times5$, and a cube dimension $1$.

We can repeat this pattern for the other cases, and this leads to a general formula:

$$3k^2(6-k)+k^3$$
$$=2k^2(9-k)$$

There are $216$ possible throws in total, so:

$$\text{P}(\text{two dice greater than k})=1-\frac{k^2(9-k)}{108}$$.

Hand-written cases:

```
1.1.5.3+1=16
2.2.4.3+8=56
3.3.3.3+27=108
4.4.2.3+64=160
5.5.1.3+125=200
```

Title: Probability of getting a particular value or higher on at least two of the dice when rolling three six-sided dice?

Body:
I've calculated the probability for rolling a particular value or higher on both dice when rolling 2d6, as follows.

1 or higher: 100%
2 or higher: 69.44%
3 or higher: 44.44%
4 or higher: 25%
5 or higher: 11.11%
6 or higher: 2.778%

If I add a third dice to this, how does it affect the probability if I still only need the particular value or higher on two or more of the dice?

Body:
If you require $2$ or more success from $3$ dice, then visualize the 'failures'.

For throwing a $2$ or higher at least twice, we need to avoid combinations such as $116$, $151$ or $411$, and obviously $111$. There are $3$ 'legs' of dimensions $1\times1\times5$, and a cube dimension $1$.

We can repeat this pattern for the other cases, and this leads to a general formula:

$$3k^2(6-k)+k^3$$
$$=2k^2(9-k)$$

There are $216$ possible throws in total, so:

$$\text{P}(\text{two dice greater than k})=1-\frac{k^2(9-k)}{108}$$.

Hand-written cases:

1.1.5.3+1=16
2.2.4.3+8=56
3.3.3.3+27=108
4.4.2.3+64=160
5.5.1.3+125=200


问:
Let $A_{i}$ be a a collection of connected spaces, and assume $\bigcap A_i \neq \varnothing$
Show that $\bigcup_{i \epsilon A} A_i$. is connected

I'm not sure how to even start going about this. If anyone had any advice on lemma's to look at or theorem's i'd really appreciate it.

答:
Suppose it's not connected. Then there are two non empty disjoint open sets $A,B$ such that $A \cup B = \cup A_i$.

Notice that for every $A_i$, we must have either $A_i \subseteq A$ or $A_i \subseteq B$. Otherwise, $A_i \cap A$, $A_i \cap B$ form two non empty disjoint open sets that union up to $A_i$. So pick any $A_k$, and suppose WLOG that $A_k \subseteq A$. Now given any other $A_i$, either $A_i \subseteq A$ or $A_i \subseteq B$. But notice if $A_i \subseteq B$ we have $A_i \cap A_k = \emptyset$. Thus, $A_i \subseteq A$ as well. Thus, all of the $A_i$ are in $A$ which means $\cup A_i \subseteq A$, which is a contradiction as we assumed $B$ is non empty and $A \cup B = \cup A_i$.


问:
Question regarding a proof of a inequality
Given $a,b,c, \in (0,\infty)$, then the following inequality holds
$$\sqrt{5a^2+12ab+7b^2}+\sqrt{5b^2+12bc+7c^2}+\sqrt{5c^2+12ca+7a^2} \leq 2 \sqrt6 (a+b+c)$$

What I've tried:

First, I noticed that we can factor
$$5a^2+12ab+7b^2 = (a+b)(5a+7b)$$
1.) Therefore we can use AM-GM with $a+b$as the first term and $(5a+7b)$ as the second term
$$\sqrt{(a+b)(5a+7b)} \leq \frac{(a+b)+(5a+7b)}{2}$$
So
$$\sqrt{5a^2+12ab+7b^2} \leq 3a+4b$$

Analogous we get
$$\sqrt{5b^2+12bc+7c^2} \leq 3b+4c$$
$$\sqrt{5c^2+12ca+7a^2} \leq 3c+4a$$

Summing all three inequalities,
$$\sqrt{5a^2+12ab+7b^2}+\sqrt{5b^2+12bc+7c^2}+\sqrt{5c^2+12ca+7a^2} \leq 7(a+b+c)$$
However, since $7>2\sqrt6$, this approach doesn't yield the wanted result.

2.) Another way was to see what constraints are on $a,b$ are if we have that
$$\sqrt{{5a^2+12ab+7b^2}} \leq \sqrt6 (a+b)$$
and analogue for $b,c$ and $c,a$.
Squaring both sides, and knowing that $a,b$ are positive real numbers, we can only consider the positive branch of the square-root,
$(a+b)(5a+7b) \leq 6(a+b)^2$ then $5a+7b \leq 6a+6b$, thus $b \leq a$.

Analogous,
$c \leq b \leq a$, but also $a \leq c$ which would lose generality because by the transitive property, $c \leq a$ and $a \leq c$ which is true if and only if $a=c$. Therefore this approach is also not correct.

3.) I don't think we can apply Minkowski's Inequality, I can't find a way to write $7b^2$ in a way to write our expression as a sum of two squares of binomials.

How can this inequality be proven? Can we use any of these ideas?

答:
Your inequality become the equality when $a=b=c.$ But for $a=b,$ then
$$\sqrt{(a+b)(5a+7b)} \ne \frac{(a+b)+(5a+7b)}{2}.$$
By the AM-GM inequality, we have
$$\sqrt{(a+b)(5a+7b)} = \frac{1}{\sqrt6} \cdot \sqrt{6(a+b) \cdot (5a+7b)} \leqslant \frac{6(a+b)+(5a+7b)}{2\sqrt6} = \frac{11a+13b}{2\sqrt6}.$$
Equality hold for $a=b.$

Therefore
$$\sum \sqrt{(a+b)(5a+7b)} \leqslant \sum \frac{11a+13b}{2\sqrt6} = 2\sqrt 6 (a+b+c).$$


问:
n-Queens problem possible solutions by logical equivalences
I'm studying Discrete maths recently, mainly through MIT 6042J and Rosen's Discrete Math and its applications.

In the later, I found the following problem I can't figure out how to proceed:

The context is the well-known n-Queens problem and on the textbook, the following compound preposition is given:

Let $p(i,j)$ be a proposition that is $True$ iff there's a queen in the $i$th row and $j$th column, where $i = 1...n$ and $j = 1...n$.

to check all row contains at least one queen: $Q_1 = \land_{i=1}^n\lor_{j=1}^np(i,j)$

to check at most one queen per row: $Q_2 = \land_{i=1}^n\land_{j=1}^{n-1}\land_{k=j+1}^n(\lnot p(i,j)\lor\lnot p(k,j))$

Here comes my first question. I believe it's wrong and should be $Q_2 = \land_{i=1}^n\land_{j=1}^{n-1}\land_{k=j+1}^n(\lnot p(i,j)\lor\lnot p(\textbf{i,k}))$ but I couldn't find any public errata. Does it make sense?

to check at most one queen per column: $Q_3 = \land_{j=1}^n\land_{i=1}^{n-1}\land_{k=i+1}^n(\lnot p(i,j)\lor\lnot p(k,j))$

to assert at most one queen on the diagonals:

$Q_4 = \land_{i=2}^n\land_{j=1}^{n-1}\land_{k=1}^{min(i-1,n-j)}(\lnot p(i,j)\lor\lnot p(i-k,k+j))$

$Q_5 = \land_{i=1}^{n-1}\land_{j=1}^{n-1}\land_{k=1}^{min(n-i,n-j)}(\lnot p(i,j)\lor\lnot p(i+k,j+k))$

So, to find valid results we need: $Q = Q_1 \land Q_2 \land Q_3 \land Q_4 \land Q_5$

I understand all of the proposed compound propositions and how they work. I could even easily convert them into an algorithm.

But the proposed exercises ask us to use them to find all the possible solutions for $n=4$. There are 65536 combinations for that, so, my second and final question is: is it possible to reduce these compounds with logic equivalences and, I'm not seeing it, or is it more probable that the book expects a computational solution for that?

Thanks in advance.

答:
Instead of testing all $4^{4\times 4}$ possibilities, you can do some backtracking/lazy evaluation.

Start by placing a queen in the first column, up to symetries there is only $2$ possible positions, $p(1,1)$ and $p(1,2)$, draw the two and start to explore the $p(1,1)$ branch.

This first queen induces a lot of constraints, we draw them in green and see that we have 2 possibilities for the placement of the second queen: $p(2,3)$ and $p(2,4)$. We draw these two configuration and carry on exploring the $p(1,1), p(2,3)$ branch.

Now, adding in red the constraints induced by the second queen and previously unconstrained, we see that there is no way to place a non-attacking queen in the third column, so we stop our exploration and go back to the next unexplored branch, $p(1,1),p(2,4)$.

Here we can put a non-attacking queen at $p(3,2)$ and carry on.

I won't describe the whole process, I think my drawing is way clearer.

We end up with one suitable configuration $p(1,2),p(2,4),p(3,1), p(4,3)$, and we have to keep in mind that we used symmetry to simplify one half of the computations so $p(1,3),p(2,1),p(3,4),p(4,2)$ works as well.

Hence we have $2$ possibles configurations, this is coherent with http://oeis.org/A000170

I described the process manually, but it is also very suitable for computer programming. It also works in your logical framework, by doing the lazy evaluation of your $\wedge$ (which is $False$ as soon as one proposition is).

https://en.wikipedia.org/wiki/Backtracking


问:
Showing uniform continuity
I've been trying to show that a function $f$ on the real interval $[a,b]$ which satisfies

$$
f(x)=f(a)+\int_a^xf'(s)\,ds\qquad\text{($f'$ defined almost everywhere)}
$$

must be uniformly continuous on $[a,b]$.

Since the condition above is equivalent to absolute continuity I know that I could show what I need by means of the proof that absolute continuity - from its fundamental definition - implies uniform continuity: I have seen a proof of that. However, I would like to show the above without involving another form of continuity in the process.

I know that - since what I have stated is also equivalent to there existing any integrable function in place of $f'$ -the proof should not involve the properties of the derivative. However, in establishing a bound I get only as far as

$$
\left|f(y)-f(x)\right|=\left|\int_x^yf'(s)ds\right|\leq\int_x^y\left|f'(s)\right|\,ds
$$

Is it possible to show that an integrable first derivative (or indeed any integrable function) is bounded in sup norm?

Thank you.

Marko

答:
As you mentioned, $f$ is absolutely continuous, and showing this isn't really harder than showing uniform continuity directly.

If $g$ is integrable and $\varepsilon>0$ is given, there is a $\delta>0$ such that $m(A)<\delta$ implies $\int_A|g|<\varepsilon$.  To see this, you could for instance first take $h$ bounded by $M>0$ such that $\int_a^b|g-h|<\frac{\varepsilon}{2}$, and then take $\delta = \frac{\varepsilon}{2M}$.

Once you have this, you have $|\int_x^y g|<\varepsilon$ whenever $|x-y|<\delta$.  And as mentioned, this extends to showing absolute continuity.  Boundedness of $f'$ would imply the stronger condition of Lipschitz continuity.

Title: Showing uniform continuity


问:
Show $\frac{x_1}{x_n} + \frac{x_2}{x_{n-1}} + \frac{x_3}{x_{n-2}} + \dots + \frac{x_n}{x_1} \geq n$
I was recently asked this question which stumped me.

How can you show $\dfrac{x_1}{x_n} + \dfrac{x_2}{x_{n-1}} + \dfrac{x_3}{x_{n-2}} + \dots + \dfrac{x_n}{x_1} \geq n$ for any positive reals $x_1, x_2, \dots, x_n$?

答:
HINT: Rearrangement inequality (http://en.wikipedia.org/wiki/Rearrangement_inequality).

For some permutation $(i_1,i_2,\cdots,i_n)$ of $(1,2,\cdots,n)$, the following holds:
$$x_{i_1}\ge x_{i_2}\ge \dots\ge x_{i_{n-1}}\ge x_{i_n}(>0),$$
$$\frac1{x_{i_n}}\ge \frac1{x_{i_{n-1}}}\ge \dots \ge \frac1{x_{i_2}}\ge \frac1{x_{i_1}}.$$

It follows from the rearrangement inequality that
$$x_1\cdot \frac1{x_n}+x_2\cdot\frac1{x_{n-1}}+\dots+x_n\cdot\frac1{x_1}
\ge x_{i_1}\cdot \frac1{x_{i_1}}+x_{i_2}\cdot\frac1{x_{i_2}}+\dots+x_{i_n}\cdot\frac1{x_{i_n}}=n.$$


问:
Sobolev space of disjoint union of open sets
Let $\Omega_{1}$, $\Omega_{2}$ be two disjoint bounded open sets in $\mathbb{R}^{N}$, i.e. $\Omega_{1}\cap \Omega_{2} = \emptyset$

Is it true that one can express $H_{0}^{1}(\Omega_{1} \cup \Omega_{2})$ in the form $\displaystyle H_{0}^{1}(\Omega_{1} \cup \Omega_{2})\equiv \lbrace u+v \mid u \in H_{0}^{1}(\Omega_{1}),v\in H_{0}^{1}(\Omega_{2})\rbrace$?

I have kind of convinced myself that the R.H.S. is a subspace of the L.H.S but I'm not sure about the other way, or even if it should be true.

答:
This looks correct to me. Let me give you a hint: By density it is enough to show
$$C^\infty_0(\Omega_1\cup\Omega_2)=\{u+v|\ u\in C^\infty_0(\Omega_1), v\in C^\infty_0(\Omega_2)\}.$$
Now take an $f\in C^\infty_0(\Omega_1\cup\Omega_2)$ and try to show that the support of $f$ can be splitted up into two parts, which lie in $\Omega_1$ and $\Omega_2$ respectively. What kind of regularity does $f\cdot \chi_{\Omega_1}$ now possess and where is the support of it? Here $\chi_A(x)=1$ if $x\in A$ and $0$ otherwise.


问:
How to generate a 'Discretized' beta distribution with mean and variance matching a 'Pure' beta distribution
Apologies for the rather vague title, I'm missing some terminology here. Feel free to correct me.

I would like to sample random variables from two distributions, $Z$, a Beta distribution with shape parameters $\hat{a}$ and $\hat{b}$, and $XY$ a product distribution generated by multiplying random variables sampled from a (different) Beta distribution ($X$, shape parameters $a$ and $b$) with independently sampled discrete random variables $Y$ equal to 1 with a probability $c$ and zero otherwise. The tricky part is I need the mean and variance of $Z$ and $XY$ to be identical.

Formally then (I'm not a mathematician so expect notation errors),

$Z\sim B(\hat{a}, \hat{b})$

$X\sim B(a, b)$

$P(Y=1)=c,\ P(Y=0)=1-c$

The following are the analytic expectation and variance of $Z$ and $XY$, the latter I get to using the expressions for E/var of a production distribution here:

$E(Z)=\frac{1}{1+\hat{b}/\hat{a}}$

$Var(Z)=\frac{1}{4(2\hat{b}+1)}$

$E(XY)=\frac{c}{1+b/a}$

$Var(XY)=c\left(\frac{1}{4(2b+1)} + \frac{1-c}{(1+b/a)^2}\right)$

Then setting $E(Z)=E(XY)$ and $Var(Z)=Var(XY)$ I aim to get expressions for $a$ and $b$ for known $\hat{a}$ and $\hat{b}$ and $c$. I arrive at the following:

$a=\left(\frac{1}{4}\left(\frac{1}{4c(2\hat{b}+1)} - \frac{(1-c)}{(1-\hat{\beta})^2} \right)^{-1}-1\right)\frac{1}{2\hat{\beta}}$

$b=a\hat{\beta}$

where

$\hat{\beta}=(1+\hat{b}/\hat{a})c-1$

Not super pretty in know.

The trouble is that when I compute $a$ and $b$ as above, the analytic expectation and variances of $Z$ and $XY$ don't match so clearly I have made a mistake somewhere in the analysis.

Can anyone suggest a correction or alternative approach to getting what I want?

Cheers

答: By the light of day I think this is actually correct. The issue is that for beta distributions with strong skew, the sample variance can deviate fairly substantially from the analytic variance giving the impression that the result does not hold.</p>


问:
Legendre Polynomial Identity
I was looking for a method to do the following integral:
$\int^1_{-1}(1-x^2)\frac{dP_m(x)}{dx}P_n(x)~dx$

I know there should be an explicit representation for the result but I am struggling to work it out.

答:
By Rodrigues' formula and the Legendre DE, you essentially want to find
$$
\int_{-1}^1 [(x^2-1)^m]^{(m-1)}[(x^2-1)^n]^{(n)}\,\mathrm{d}x
$$
If $n<m$, integrate by parts $n$ times gives
$$
\pm\int_{-1}^1 [(x^2-1)^m]^{(m-n-1)}\underbrace{[(x^2-1)^n]^{(2n)}}_{=(2n)!}\,\mathrm{d}x
$$
which is $0$ unless $m=n+1$, in which case you want to calculate
$$
\int_{-1}^1(x^2-1)^m\,\mathrm{d}x
$$
which with $x=\sin\theta$ and $\int_0^{\pi/2}\cos^{2m+1}\theta\,\mathrm{d}\theta=\frac{(2m)!!}{(2m+1)!!}$ nails it.  Similar for the case $n\geq m$.

Title: Legendre Polynomial Identity


问:
Probability of stepping on n-th field in game where you move using dice

We play a game where in each round we roll a six-sided dice and move the appropriate number of squares. How likely are we to step on the $n$-th field during the game, if we started with the number $0$?

The answer is sufficient in the form of a recurrent formula, which refers to a constant number of values for the lower one. Can you use only $2$ values for lower?

I am kinda lost here. I did bit know how even approach this.

Thanks to anyone who can help me guide to right direction

I made this:

```<code>| 1 | 1                                             |
| 2 | 1+1,2                                         |
| 3 | 1+1+1, 2+1, 3                                 |
| 4 | 1+1+1+1, 1+1+2, 1+3, 2+2, 4                   |
| 5 | 1+1+1+1+1, 1+1+1+2, 1+2+2, 1+1+3, 1+4, 2+3, 5 |
</code>```

But I can not see any pattern or how I can calculate probability from it. Does order of numbers matter ?

答:
For large $n$:

For $n$ that is large enough, each field have the same probability
to be landed.

The expected value for a 6-sided dice is $3.5$

Thus, the probability to land on $n$ in the limit is

$$\lim_{n \to \infty} p_n=\frac{1}{3.5}= 0.285$$

I am not an expert, so take my solution with a grain of salt.


问:
Injectivity of a map between manifolds
I'm learning the concepts of immersions at the moment. However I'm a bit confused when they define an immersion as a function $f: X\rightarrow Y$ where $X$ and $Y$ are manifolds with dim$X <$ dim$Y$ such that $df_x: T_x(X)\rightarrow T_y(Y)$ is injective.

I was wondering why don't we let $f$ be injective and say that's the best case we can get for the condition dim$X <$ dim$Y$(since under this condition we can't apply the inverse function theorem)?

Also does injectivity of $df_x$ inply the injectivity of $f$ (it seems that I can't prove it)?

How should we picture immersion as (something like the tangent space of $X$ always "immerses" into the tangent space of $Y$)?

Thanks for everyone's help!

答: Think of a particle moving around a figure 8 with nowhere zero speed. This parametric curve gives you an immersion $f\colon\mathbb R \to\mathbb R^2$ that is not injective. If you restrict the domain to make it a bijection (which you can do), the image is not a submanifold but is called an immersed submanifold.

Title: Injectivity of a map between manifolds


问:
How can this integral be evaluated?
I am trying to find the definite integral of this function. When entered into wolfram alpha the result (shown below) is given. However I do not understand what the E (elliptic integral of the second kind) is, and how it fits into the equation. So what is the E term? Can it be expanded into the rest of the definite integral?
$$ \int \sqrt{1+\dfrac{p^2 x^2}{q^2(q^2-x^2)}}dx  $$

答:
$$I=\int \sqrt{1+\frac{q^2x^2}{p^2(q^2-x^2)}}dx$$
Let $x=q \sin t$, then
$$I=\frac{1}{p}\int \sqrt{p^2\cos^2 t+q^2 \sin^2 t} dt =\int \sqrt{p^2+(q^2-p^2) \sin^2 t} dt$$
It leads to an elliptic Integral

Title: How can this integral be evaluated?


问:
Most general case of "the area of a triangle is proportional to its angular defect"
In the case of both spherical and hyperbolic geometry, the area of a triangle is proportional to the angular defect -- the difference of the sum of its angles and $\pi$.

In Euclidean (flat) geometry, one might argue that this is still true: $\frac{\pi-\pi}{A}$ is still a constant, it just so happens that that constant is $0$.

Is this true in the general case? Is some similar statement (proportional to a power of its area, proportional to the area times the curvature of the space) generally true? I ran across the Gauss-Bonnet theorem while doing preliminary Google checking, but I'm not enough of a topologist to work out its relevance -- I'm sure it's applicable but it's a tool I don't know how to use.

答:
Basically Gauss-Bonnet theorem tells you that given some distorted triangle, if you integrate the "curvature of the plane the triangle lies in(hyperbolic, spherical or any)" on the triangle, plus the total "turning" you make along the rim of the triangle gives you the $2\pi$ times Euler characteristic of triangle, which is $2\pi$
(Euler characteristic of polygon is defined by v-e+f, in this case 3-3+1=1. Euler characteristic of polygon or polyhedra is easy to follow so I'm not gonna explain it too much. see here https://en.wikipedia.org/wiki/Euler_characteristic#Polyhedra)

Now in the case of constant curvature plane with triangle with straight edges, the first integral term becomes simply $KA$, curvature times the area of the triangle. Since you do not make any turning along the edges since the edges are straight, only turning you make along the rim of the triangle is the exterior angles at each vertices. Now the formula mentioned above becomes :

$KA+\epsilon_1+\epsilon_2+\epsilon_3 = 2\pi$

where $\epsilon_i$ denotes exterior angles at three vertices of the triangle.

Since interior angle $\theta_i$ at each vertex is the $\pi - \epsilon_i$ of that vertex, we can write down above formula as this :

$KA=\theta_1+\theta_2+\theta_3 - \pi$

The right hand side is the defect of the triangle, and this tells you that the area of a triangle is proportional to the angular defect and the proportional constant is the curvature of the plane you're dealing with. So yes! in general your initial claim is correct and it is a direct consequence of the famous Gauss-Bonnet theorem.

P.s  If you want to study more about this theorem, I suggest you to look up differential geometry text book, not topology book.


答:
To find the number of ways of reaching $n$ observe the last roll that can happen it can be either 1,2,3 ... 6 thus $$ H_n = H_{n-1} + H_{n-2} + \cdots + H_{n-6}\\
H_{n+1} = 2H_{n}-H_{n-6} \ \forall n \ge6  $$

Title: Probability of stepping on n-th field in game where you move using dice

Body:
We play a game where in each round we roll a six-sided dice and move the appropriate number of squares. How likely are we to step on the $n$-th field during the game, if we started with the number $0$?

The answer is sufficient in the form of a recurrent formula, which refers to a constant number of values for the lower one. Can you use only $2$ values for lower?

I am kinda lost here. I did bit know how even approach this.

Thanks to anyone who can help me guide to right direction

I made this:

```<code>| 1 | 1                                             |
| 2 | 1+1,2                                         |
| 3 | 1+1+1, 2+1, 3                                 |
| 4 | 1+1+1+1, 1+1+2, 1+3, 2+2, 4                   |
| 5 | 1+1+1+1+1, 1+1+1+2, 1+2+2, 1+1+3, 1+4, 2+3, 5 |
</code>```

But I can not see any pattern or how I can calculate probability from it. Does order of numbers matter ?


问:
Localization and a particular exact sequence
Look at this proposition:

Let $R$ be a commutative ring with unity, and let $f_1,\ldots,f_n\in R$
  generate the unit ideal in $R$. Then the following sequence is exact:

$$0\longrightarrow R\xrightarrow{\alpha} \bigoplus_{i=1}^n
R_{f_i}\xrightarrow{\beta} \bigoplus_{i,j=1}^n R_{f_if_j}$$

where $\alpha(x)=(\frac{x}{1},\ldots,\frac{x_1}{1})$ and
  $\beta(\frac{x_1}{x_n},\ldots\frac{x_n}{f_n})=\left(\frac{x_i}{f_i}-\frac{x_j}{f_j}\;\textrm{in
 $R_{f_if_j}$}\right)$

Now, from this, should follow that for a fixed $f\in R$, also the following sequence should be exact (in this case the functions are the obvious ones arising from universal properties of localizations and moreover we suppose that $f_1\ldots,f_n$ generate the unit ideal in $R_f$ ):

$$0\longrightarrow R_f\xrightarrow{\alpha} \bigoplus_{i=1}^n
R_{ff_i}\xrightarrow{\beta} \bigoplus_{i,j=1}^n R_{ff_if_j}$$

but i don't understand why.
In general ${(R_f)}_{g}\ncong R_{fg}$ so why is the second sequence  exact?

答:
Please double check my work, but I believe that isomorphism actually does hold.

Here is why I think so. The following appears in Atiyah-MacDonald (Introduction to Commutative Algebra, p. 43, exercise 3):

Let $R$ be a ring (commutative with identity), let $S$ and $T$ be two multiplicatively closed subsets of $R$, and let $U$ be the image of $T$ in $S^{-1}R$. Show that the rings $(ST)^{-1}R$ and $U^{-1}(S^{-1}R)$ are isomorphic.

This is proved easily by showing $U^{-1} (S^{-1}R)$ satisfies the universal property of localization of $R$ at $ST$.

Now, if we take $S=\{1,f,f^2,\ldots\}$ and $T=\{1,g,g^2,\ldots\}$, then we conclude $(R_f)_{g/1} \cong R_{fg}$. One might be a little worried that $R_{fg}$ and $(ST)^{-1}R$ actually are the same ring; but...

EDIT: One can show that $(ST)^{-1}R$ satisfies the universal property of localization of $R$ at powers of $fg$.

MORE EXPLANATION: From the exercise we know $(R_f)_{g/1}$ and $(ST)^{-1}R$ are isomorphic. We only want to show $(ST)^{-1}R$ and $R_{fg}$ are the same ring. A-M give a nice criteria for showing a ring is a localization based on the universal property of localizations (corollary 3.2):

Let $S$ be a multiplicatively closed subset of a ring $R$, and suppose $\phi: R \to S^{-1}R$ is the canonical map. If $\psi: R \to Q$ is a ring homomorphism such that i) $s\in S \Rightarrow \psi(s)$ is a unit in $Q$; ii) $\psi(r)=0 \Rightarrow sr=0$ for some $s\in S$; iii) every element of $Q$ is of the form $\psi(r)\psi(s)^{-1}$; then there is a unique isomorphism $h:S^{-1}R \to Q$ such that $\psi = h\phi$.

Usually when I'm trying to show a ring is a localization, I don't build a homomorphism and then try to construct its inverse, I use this proposition.

Now, take $\phi: R \to R_{fg}$ and $\psi: R \to (ST)^{-1}R$. We verify i)-iii):

i) $\psi$ clearly maps any $(fg)^n$ to a unit in $(ST)^{-1}R$.

ii) If $\psi(r) = r/1=0$ in $(ST)^{-1}R$, then we have $f^ng^m r=0$ for some nonnegative integers $n,m$. If $n=m$, then we're done. If not, assume $m<n$ and mulitply both sides of this equality by $g^{n-m}$ (and similarly if $m>n$).

iii) Take $r/f^ng^m \in (ST)^{-1}R$. Then

$$\frac{r}{f^ng^m} = \frac{rf^mg^n}{1} \cdot \frac{1}{(fg)^n(fg)^m} = \psi(rf^mg^n) \psi((fg)^n(fg)^m)^{-1}.$$

Now invoke the above mentioned universal property.


问:
Linear algebra vs homological algebra
In linear algebra my book show the five lemma.
In wikipedia I noticed connection with the 9-lemma (=? 3x3 lemma), the snake lemma a.s.o.

In the wikipedia page this lemma are described as being part of ??homological algebra??.
Are all this lemmas also valid for linear algebra? (that is, could I take them as exercises?)

答: Yes, algebra is algebra. It's often generalised to more general settings (modules etc.) but it's certainly valid in vector spaces and so in a linear algebra setting.  It's all part of modern algebra, basically.</p>

Title: Linear algebra vs homological algebra


问:
How to find range of values for where line cuts circle?
I am doing a question on circles. The question is:

The circle  has equation $x^2 + y^2 - 6x + 10y + 9 = 0
$
The line with equation $y= kx$ , where  is a constant, cuts  at two distinct points.
Find the range of values for k

My working so far:

Found centre: (3, -5) radius = 5

Now what do I do after this?

答:
We have two equations:
\begin{align}
(x-3)^2+(y+5)^2 &= 5^2 \tag{1}\label{1} \\[4pt]
y &= kx \, . \tag{2}\label{2}
\end{align}
The points of intersection occur when $x$ and $y$ satisfy both equations $\eqref{1}$ and $\eqref{2}$. We can find these points of intersection by substituting $y=kx$ into $\eqref{1}$:
$$
(x-3)^2+(kx+5)^2=5^2 \, .
$$
This will give us a quadratic equation. Since the line cuts the circle, the quadratic should have two solutions (corresponding to two intersections). So use the discriminant to help you.


答:
The key to the problem is to forgo any consideration of a circle or of tangent lines.

Instead, you want to identify the two distinct values of $k$, such that each individual value of $k$ will result in only one value of $(x,y)$ that satisfies the following two constraints:

$x^2 + y^2 - 6x + 10y + 9 = 0$.

$y = kx$.

The above two constraints result in :

$$x^2 + k^2x^2 - 6x + 10kx + 9 = 0 \implies $$

$$x^2(1 + k^2) + x(10k - 6) + 9 = 0.\tag1$$

This is the critical moment in intuiting the analysis.

In order for equation (1) above to have exactly one solution, its discriminant must equal $(0)$.

Edit

Here is where some fun comes in.  What is the relationship between a specific value of $k$ generating only one solution to equation (1) above, and a specific value of $k$ yielding exactly one point $(x,y)$ that satisfies the two (original) constraints?

Re-stating the question, as the slope of the line $y=kx$ (in effect) goes from $0^\circ$ to $360^\circ$ is it possible that a specific slope would intersect the locus of satisfying points $(x,y)$ in more than one location, where each location had the same value of $(x)$.

Answer

The answer is no because: if a specific value of $k$ only generates one satisfying value of $x$, then there can only be one corresponding satisfying value of $y$, namely $y = kx.$

Thus, you are solving for $k$ such that

$$(10k - 6)^2 - 36(1 + k^2) = 0 \implies $$

$$100k^2 - 120k + 36 - 36 - 36k^2 = 0 \implies $$

$$0 = 64k^2 - 120k = k(64k - 120).\tag2$$

Thus, the desired values for $k$ such that $y = kx$ intersects the circle in exactly one point are

$$\left\{0, \frac{120}{64}\right\}.$$

This is the point where, absent a graph, it is easy to get confused.  Focusing on equation (2) above, you want to identify all values of $k$ such that the discriminant will be positive.  This is because when the discriminant is positive, equation (1) above will have two solutions instead of only one solution.  The pertinent values of $k$ are clearly $k < 0$ or $k > \frac{120}{64}.$

Here is the confusing part.  $k = 0$ corresponds to a horizontal line.  As the slope starts at $0$, moves into a negative slope, and then approaches a vertical slope, $k$ is actually approaching $-\infty.$  Then, when the slope crosses over from a vertical line to a positive slope, $k$ goes from $+\infty$ back to $\frac{120}{64}.$

Title: How to find range of values for where line cuts circle?

Body:
I am doing a question on circles. The question is:

The circle  has equation $x^2 + y^2 - 6x + 10y + 9 = 0
$
The line with equation $y= kx$ , where  is a constant, cuts  at two distinct points.
Find the range of values for k

My working so far:

Found centre: (3, -5) radius = 5

Now what do I do after this?


问:
Prove that $\lim_{n\to\infty} \left(\frac{\sqrt{n}(1 + 8\sqrt{n})}{4n - 1}\right) =2$

Prove that $$\left(\displaystyle\frac{\sqrt{n}(1 + 8\sqrt{n})}{4n - 1}\right)_{n \in \mathbb{N}}\to 2$$

My idea was to find some $N$ to do the definition job of convergence.

\begin{align}\left |\dfrac{\sqrt{n}(1 + 8\sqrt{n})}{4n - 1}-2 \right |&=\left |\dfrac{\sqrt{n}(1 + 8\sqrt{n})-8n+2}{4n - 1} \right |\\
&=\left |\dfrac{\sqrt{n}+8n-8n+2}{4n - 1} \right |\\
&=\left |\dfrac{\sqrt{n}+2}{4n - 1} \right |\\
&\leq \left |\dfrac{\sqrt{n}+2}{n}\right |\\
&\leq \left |\dfrac{\sqrt{n}}{n}\right |+ \left |\dfrac{2}{n}\right |\\
&\leq \dfrac{1}{\sqrt{n}}+2\\
&\leq 3
\end{align}

so I concluded that the sequence is bounded, so my idea was trying to show that this sequence is monotonous, so I could conclude that it converges. Taking $n = 4k$ I get a subsequence that converges to $2$, so if I knew that the sequence is convergent, the problem would be over.

答:
Don't be afraid to simplify even more, eliminate all additions if possible.

Get also rid of fractions by bounding them with an integer (here $\frac 23<1$).

For $n$ large then $2<\sqrt{n}$ (i.e. $n>4$) and $-1>-n$

$$\left|\frac{\sqrt{n}+2}{4n-1}\right|<\left|\frac{\sqrt{n}+\sqrt{n}}{4n-n}\right|=\left|\frac{2}{3\sqrt{n}}\right|<\frac 1{\sqrt{n}}\to 0$$ much easier to pick $N,\epsilon$ now.


问:
z complex $p, q \in \mathbb{N}, p < q$ such that $|z ^ p + \frac {1}{z ^ p}| \ge |z ^ q + \frac {1}{z ^ q}|$, so $|z + \frac{1}{z}| < 2$
I have participated yesterday in a contest. One of the problems was this one:
If $x \in \mathbb{C} \setminus \mathbb{R}$ and there exists $p, q \in \mathbb{N}, p < q$ such that $|z ^ p + \frac {1}{z ^ p}| \ge |z ^ q + \frac {1}{z ^ q}|$, we need to prove $|z + \frac{1}{z}| \lt 2$. My solution during the contest was the following: Using the triangular inequality, $|z|^p + \frac{1}{|z| ^ p} \ge |z^p + \frac{1}{z^p}| \ge |z^q + \frac{1}{z^q}|\ge |z|^q - \frac{1}{|z| ^ q}$. So $|z|^p - |z| ^ q + \frac{1}{|z| ^ p} + \frac{1}{|z| ^ q} \ge 0$. Now, here comes the mistake: I assumed that $|z| ^ p - |z| ^ q \ge 0$ which, in my mind, came form the last inequality. Of course this is wrong, but in the  heat of competition I thought it is right. I will stop here since the rest of the solution contains a really embarassing mistake and the proof is invalid once the mistake is made. So, how could I have solved it?

答:
Here I give a weaker result $\color{red}{\left|z+\dfrac1z\right|\leq 2}$ than $\left|z+\dfrac{1}{z}\right|<2$.

Let $$x_{n} = z^{n} + 1/z^{n}, n=0,1,2, \cdots.$$
Suppose
that $|x_1|=\left|z+\frac1z\right|>2$.
We have that
$$x_1x_{n} = x_{n-1} + x_{n+1},$$
therefore
$$x_1= \frac{x_{n-1}}{x_{n}}+\frac{x_{n+1}}{x_{n}}.$$
Furthermore, let
$$ y_{n} = \frac{x_{n}}{x_{n-1}},n=1,2,\cdots,$$
then we have
$$y_{n+1}=x_1-\frac{1}{y_{n}} .$$
We know that $ |y_{1}|=|x_1/2|>1 $.
By induction we can show that $ |y_{n}| > 1 $.
Indeed, if $
|y_{n}|>1 $, then
$$ |y_{n+1}|=\left|x_1-\frac{1}{y_{n}} \right|\geqslant |x_1|-\frac{1}{|y_{n}|}>2-1=1.$$
Therefore
$|y_{n}|>1 $ for every $ n=1,2,\cdots $ , so $|x_{n}|>|x_{n-1}|$ for
every $ n=1,2,\cdots$. But there exists $p, q \in \mathbb{N}, p < q$ such that $$|x_p| \ge |x_q|.$$


问:
Gram determinant
How to prove that

$$\sqrt{\Gamma(\vec{a},\vec{b},\vec{c})}=|(\vec{a},\vec{b},\vec{c})|,$$

where $\Gamma(\vec{a},\vec{b},\vec{c})= \left | \begin{array} {ccc} \vec{a} \cdot \vec{a} & \vec{b} \cdot \vec{a} & \vec{c} \cdot \vec{a} \\
\vec{a} \cdot \vec{b} & \vec{b} \cdot \vec{b} & \vec{c} \cdot \vec{b} \\
\vec{a} \cdot \vec{c} & \vec{b} \cdot \vec{c} & \vec{c} \cdot \vec{c}
\end{array} \right | $.

Thanks!

答:
Let the matrix $X$ have rows $\vec{a},\vec{b},\vec c$.

Then $XX^T$ is the matrix within the vertical bars in your definition of the gram determinant.

Then you have $\Gamma(\vec{a},\vec{b},\vec{c})=\det(XX^T)=\det(X)\det(X^T)=\dots$

Can you take it from here?

If in fact you do want to work with complex vectors, then I need to alter the above slightly. To be consistent with the order of terms in the dot products, you would have ot use column vectors instead of row vectors, and the Hermitian conjugate instead of the transpose. You'd also need the fact that the determinant of $X^H$ is the complex conjugate of the determinant of $X$, and multiplying them would give you the square modulus of the determinant of $X$.

Title: Gram determinant


问:
Solution to matrix equation with exponentials
Assuming
$$
A, M \in \mathcal{R}^{n \times n }
$$
with A, M being invertible. Could someone tell me if it is possible to find a closed form solution for the matrix $M$ in the following equation and if so, provide me with some hints on how to obtain it
$$
A = e^M + e^{M^T}.
$$

I have tried a few things but can´t seem to make it work.

Thanks.

答:
Given that $A$ is a real symmetric matrix and $e^{M^T} = (e^M)^T$, one find a solution for $M$ given one additional assumption: the diagonal of $A$ only contains non-zero elements.

Set $e^M$ equal to the lower triangular matrix of $A$ minus half the diagonal of $A$. So if $$A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}, ~~ \text{then} ~~ e^M = \begin{bmatrix} 1 & 0 \\ 2 & 3 \end{bmatrix} - \begin{bmatrix} 0.5 & 0 \\ 0 & 1.5 \end{bmatrix}$$
Given this structure of $e^M$, $A = e^M + (e^M)^T$, so the equation still holds. In order to solve $M$ for $e^M$, one has to take the matrix logarithm. Depending on the resulting properties of $e^M$, the methods could vary.


问:
Biprefix code and word factorization
Let $Y_1$ a biprefix code over a free monoid $A^{*}$. Let $u= x_{i}y_{j}$ and $v= x_{i}’y_{j}’$, with $x_i, x_i’ \in A^*$ and $y_j, y_j’ \in Y_1$.  If $u=v$, while does this imply that $y_{i} = y_{i}’$?

答:
A biprefix code is in particular a suffix code. Now $u = v$ implies that either $y_j$ is a suffix of $y'_j$ or that $y_j$ is a suffix of $y'_j$. Since your code is suffix, this implies $y_j = y'_j$.

EDIT (Answer to your comment). Would this picture help you?

Title: Biprefix code and word factorization


问:
Find all solutions: $x^2 + 2y^2 = z^2$
I'm use to finding the solutions of linear Diophantine equations, but what are you suppose to do when you have quadratic terms? For example consider the following problem:

Find all solutions in positive integers to the following Diophantine equation

$x^2 + 2y^2 = z^2$

I'd usually start by finding the gcd and use some other tricks, but I'm not sure how to approach this type of problem

答:
We can approach this in the same manner as Pythagorean Triples.

Let's only look for primitive solutions, $\gcd(x,y,z)=1$. Since
$$
z^2-x^2=2y^2
$$
$z$ and $x$ must have the same parity. That means both $z-x$ and $z+x$ are even so $y$ must be even. Therefore, for the triple to be primitive, $x$ and $z$ must be odd. Let $y=2ab$ where $(a,b)=1$ and $b$ is odd. Then, since one of $z+x$ or $z-x$ must be $2\bmod4$ and the other must be $0\bmod4$,
$$
\overbrace{2y^2}^{8a^2b^2}=\overbrace{(z+x)}^{4a^2}\overbrace{(z-x)}^{2b^2}\qquad\text{smaller factor is }2\bmod4
$$
or
$$
\overbrace{2y^2}^{8a^2b^2}=\overbrace{(z+x)}^{2b^2}\overbrace{(z-x)}^{4a^2}\qquad\text{larger factor is }2\bmod4
$$
Solving for $x$ and $z$ gives $x=2a^2-b^2$ if the smaller factor is $2\bmod4$, or $x=b^2-2a^2$ if the larger factor is $2\bmod4$. That is,
$$
x=\left|\,2a^2-b^2\right|,y=2ab,z=2a^2+b^2
$$
where $(a,b)=1$ and $b$ is odd.

For example,
$$
\begin{array}{c|cc}
a\backslash b\!\!&1&3&5&7\\\hline
1&(1,2,3)&(7,6,11)&(23,10,27)&(47,14,51)\\
2&(7,4,9)&(1,12,17)&(17,20,33)&(41,28,57)\\
3&(17,6,19)&\text{n/a}&(7,30,43)&(31,42,67)\\
4&(31,8,33)&(23,24,41)&(7,40,57)&(17,56,81)\\
5&(49,10,51)&(41,30,59)&\text{n/a}&(1,70,99)\\
6&(71,12,73)&\text{n/a}&(47,60,97)&(23,84,121)\\
7&(97,14,99)&(89,42,107)&(73,70,123)&\text{n/a}
\end{array}
$$

Title: Find all solutions: $x^2 + 2y^2 = z^2$


问:
What are the elements of $\mathbb{Q}(\sqrt[3]{2}+\sqrt{3})$?
I would like to calculate the elements of $\mathbb{Q}(\sqrt[3]{2}+\sqrt{3})$.
I know that the elements of $\mathbb{Q}(\sqrt[3]{2})$ have the form of ${a+b\sqrt[3]{2}+c\sqrt[3]{4}}$, where a,b,c $\in \mathbb{Q}$ and the elements of $\mathbb{Q}(\sqrt{3})$ have the form of $a+b\sqrt{3}$, where a,b $\in \mathbb{Q}$.
I also calculated the minimal polynomial of $(\sqrt[3]{2}+\sqrt{3})$ over $\mathbb{Q}$ , which is: $x^6−9x^4−4x^3+27x^2−36x−23$.

Can you help me to calculate the form of the elements of $\mathbb{Q}(\sqrt[3]{2}+\sqrt{3})$? I have to find the elements of the linear combinations which form the higher powers of $(\sqrt[3]{2}+\sqrt{3})$ which are also powers of $(\sqrt[3]{2}+\sqrt{3})$?
Could you give me a proper method to find the solution? Also, could you write down your calculation in your answer? Thank you for helping me!

答:
Given that $\theta=\sqrt[3]{2}+\sqrt{3}$ has degree $6$ over $\mathbb Q$, $\mathbb Q(\theta)$ is the set of polynomial expressions in $\theta$ of degree at most $5$.

Another take: $Q(\theta)=\mathbb{Q}(\sqrt[3]{2},\sqrt{3})$.


问:
Good bound for $\sum_{p\mid n}\frac{1}{p}$
I am trying to get a good asymptotic bound for $$\sum_{\substack{p\mid n\\ \text{$p$ prime}}}\frac{1}{p}.$$ One way I can think of is $$\sum_{p\mid n}\frac1p\leqslant \sum_{p\leqslant n}\frac{1}{p} \ll \log\log n,$$ by  Mertens' theorem.
Can we do better?

As a follow-up question, if you see a sum like this, is there a way to intuitively guess how large it could be?  Thanks!

答:
$$\sum_{p|n} 1/p \le \sum_{p| \prod_{q\le k_n} q}1/p= \sum_{p\le O(\log n)} 1/p= O(\log\log\log n)$$

where $\prod_{q\le k_n} q$ is the least primorial $\ge n$:

$\prod_{q\le k_n} q\ge \exp(k_n/10)$ gives $k_n\le 10\log(n)$.


问:
Is $\max(X_1,X_2,X_3)$ an unbiased estimator
Suppose that $X_1, X_2, X_3$ is a sample drawn from some population, if we use $\max(X_1, X_2,X_3)$ as an estimator of the population's mean, then can we say that its unbiased:

My approach was to let $X_{ i}$ be the max, $i=1,2,3$ and then deduce that $E(X_{ i})=\mu $, so did I miss anything?

Edit:

It appears due to the comments that $\max(X_1, X_2,X_3)$ follows a different distribution, given that the population is exponential how should I go around this sort of problem.

Thank you.

答:
given that the population is exponential

just a detail...

As known, the distribution of the max is the product of the single CDF's, thus in your case, setting

$$Z=\max(X_1,X_2,X_3)$$

you get

$$F_Z(z)=\left(1-e^{-\theta z}\right)^3$$

derivating you get the density

$$f_Z(z)=3\theta e^{-\theta z}\left(1-e^{-\theta z}\right)^2$$

with expectation

$$\mathbb{E}[Z]=\int_0^{+\infty}3z\theta e^{-\theta z}\left(1-e^{-\theta z}\right)^2dz=\frac{11}{6\theta}$$

Concluding: the estimator is biased for the mean of the population, $\mu=\frac{1}{\theta}$, but with a correctable bias


答:
Plugging the line equation in that of the circle,

$$x^2 + k^2x^2 - 6x + 10kx + 9 = 0.$$

This is a quadratic equation that has two distinct roots when

$$(5k-3)^2-9(k^2+1)>0$$

or

$$k\left(k-\frac{15}8\right)>0.$$

$$k<0\text{ or }k>\frac{15}8.$$

Alternatively, the distance of the center to the line must not exceed the radius. For convenience, we use the squared distance, and write

$$\frac{(3k+5)^2}{k^2+1}<25$$

or

$$k\left(k-\frac{15}8\right)>0.$$


问:
How to Evaluate $\int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4}dx$?
How to evaluate

$$ \int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4} \,dx \approx -0.162858 \tag{1}$$

The integral arises in the computation of

$$\left( \sum_{n=1}^{\infty} \frac{(-1)^n}{n}\right)\left(\sum_{n=1}^{\infty} \frac{(-1)^n}{4n-1}\right)$$
as

$$ \scriptsize{\frac{\pi \ln(2)}{4\sqrt{2}} + \frac{\ln(2) \ln(3-2\sqrt{2})}{4\sqrt{2}} -3\ln(2) + \frac{\pi}{2}= \int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4} \,dx + \int_{0}^{1} \frac{x^{1/4}}{2(1+x)}\left(\tan^{-1}(x^{1/4}) - \tanh^{-1}(x^{1/4}) \right)}dx $$

A similar Integral

$$ \int_{0}^{1}\left( \frac{x^2 \ln(2)}{x^4-1} - \frac{x^2 \ln(1+x^4)}{x^4-1}\right)dx = C-\frac{\pi^2}{16}+\frac{\ln^2(\sqrt{2}-1)}{4}+\frac{\pi \ln (\sqrt{2}-1)}{4} \tag{2} $$

Where $ C $ = Catalan Constant

Unfortunately the same techniques I used to evaluate $(2)$ have not worked for $(1)$.
I know only for integration - By parts, U-Sub, and  using Taylor Series as well as Mathematica.

Q = Is there a closed form for Integral $(1)$?

EDIT

$$ (1) = \frac{-\pi^2}{8\sqrt{2}} + \frac{\pi \ln(8)}{4\sqrt{2}} +\frac{\ln(8) \ln(3-2\sqrt{2})}{4\sqrt{2}} -\frac{\pi \ln(3-2\sqrt{2})}{8\sqrt{2}} + 4\sum_{k=1}^{\infty} \frac{(-1)^k}{4k-1}\sum_{n=1}^{k} \frac{1}{4n-1} $$

$$\sum_{k=1}^{\infty} \frac{(-1)^k}{4k-1}\sum_{n=1}^{k} \frac{1}{4n-1} = \frac{1}{64}\left(\psi^{(1)}\left(\frac{7}{8}\right)-\psi^{(1)}\left(\frac{3}{8}\right)\right) + W $$

Where $W$ is some value.

答:
I think (at least for the time being) tha we need to combine two things
$$\frac {x^2}{x^4+1}=\frac{1}{2 \left(x^2+i\right)}+\frac{1}{2 \left(x^2-i\right)},$$
$$\log(1-x^4)=\log(1+x^2)+\log(1-x^2)$$ Each of the four integrals
$$\int_0^1\frac{\log \left(1\pm x^2\right)}{2 \left(x^2\pm i\right)}\,dx$$ shows a complex expression involving a bunch of polylogarithms of complex arguments and, for sure, quite many terms invoving the roots of unity.

I obtained the final expression but I cannot simplify it. For sure, the numerical value is
$$I=-0.16285829160764898119151607224709931337723009033493\cdots$$

I hope and wish that this could help other users.

Edit

Consider
$$I(a)=\int_{0}^{1} x^2\frac{ \log \left(1-a^4 x^4\right)}{x^4+1}\,dx$$
$$I'(a)=-\int_{0}^{1}\frac{4 a^3 x^6}{\left(x^4+1\right) \left(1-a^4 x^4\right)}\,dx$$
$$I'(a)=\frac{a^3 \left(\pi +\log \left(3-2 \sqrt{2}\right)\right)}{\sqrt{2}
   \left(a^4+1\right)}+2\frac{\tan ^{-1}(a)-\tanh ^{-1}(a)}{a^4+1}$$ Integrating, we have
$$\frac{\log (2) \left(\pi +\log \left(3-2 \sqrt{2}\right)\right)}{4 \sqrt{2}}+2\int_0^1 \frac{\tan ^{-1}(a)-\tanh ^{-1}(a)}{a^4+1}\,da$$ Now
$$\tan ^{-1}(a)-\tanh ^{-1}(a)=-2\sum_{n=0}^\infty \frac{a^{4 n+3}}{4 n+3}$$
$$J=\int_0^1 \frac{\tan ^{-1}(a)-\tanh ^{-1}(a)}{a^4+1}\,da=-2\sum_{n=0}^\infty \frac 1{4 n+3}\int_0^1  \frac{a^{4 n+3}}{a^4+1}\,da$$ All of that leads for the integral
$$I(1)=\frac{\log (2) \left(\pi +\log \left(3-2 \sqrt{2}\right)\right)}{4 \sqrt{2}}-\frac 12\sum_{n=0}^\infty \frac{H_{\frac{n}{2}}-H_{\frac{n-1}{2}}}{4 n+3}$$
I have not been able to make the result of the summation explicit. For its evaluation, we can make
$$\sum_{n=0}^\infty \frac{H_{\frac{n}{2}}-H_{\frac{n-1}{2}}}{4 n+3}=\sum_{n=0}^p \frac{H_{\frac{n}{2}}-H_{\frac{n-1}{2}}}{4 n+3}+\sum_{n=p+1}^\infty \frac{H_{\frac{n}{2}}-H_{\frac{n-1}{2}}}{4 n+3}$$ and approximate the last summnand by
$$\frac{H_{\frac{n}{2}}-H_{\frac{n-1}{2}}}{4 n+3}=\sum_{k=2}^? \frac {c_k}{4^{k-1}\, n^k}$$ where the first $c_k$'s are
$$\{1,-5,15,-29,87,-773,2319,27859,-83577,-3812501,11437503\}$$
Using only what is given here, some results for the summation as a function of $p$
$$\left(
\begin{array}{cc}
p & \text{summation} \\
 5  & -0.16285829044608650070 \\
 10 & -0.16285829160689532584 \\
 15 & -0.16285829160764112453 \\
 20 & -0.16285829160764869429 \\
 25 & -0.16285829160764895979 \\
 30 & -0.16285829160764897866 \\
 35 & -0.16285829160764898078 \\
 40 & -0.16285829160764898111 \\
 45 & -0.16285829160764898117 \\
 50 & -0.16285829160764898119
\end{array}
\right)$$


问:
Calculate half life in days - A tumor is injected with 0.5 grams of Iodine-125, which has a decay rate of 1.15% per day.

A tumor is injected with 0.5 grams of Iodine-125, which has a decay
rate of 1.15% per day. To the nearest day, how long will it take for
half of the Iodine-125 to decay?

I get 99 days, the solution provided by my textbook says about 60 days.

Here's my working:

(Formula for the amount of Iodine remaining after $t$ days (where k is 1-1.15%))
$$A=A_oe^{\frac{ln0.5}{98.85}t}$$

The patient was injected with 0.5 grams so I want to know how long it takes for half of that, 1/4 grams to remain:

$$\frac{1}{4}=\frac{1}{2}e^{\frac{ln0.5}{98.85}t}$$
$$1=2e^{\frac{ln0.5}{98.85}t}$$
$$\frac{1}{2}=e^{\frac{ln0.5}{98.85}t}$$
$$ln\frac{1}{2}=\frac{ln0.5}{98.85}t$$
$$t=\frac{ln\frac{1}{2}}{\frac{ln\frac{1}{2}}{98.85}}=98.85$$

Where did I go wrong and how can I arrive at 60 days?

答:
Your equation $A = A_0 e^{\frac{\ln 0.5}{98.85}t}$ is incorrect.

If we start with the general model, $A(t)=A_0e^{kt}$, and use the fact that after one day, $.9885$ of the initial amount remains, we get
$$A_0e^k=.9885A_0$$
So $k=\ln(.9885)$

The model becomes $A(t)=A_0e^{\ln(.9885)\cdot t}$

From there you should be able to get the half-life.


问:
MATLAB code to find distance and eccentricity in graphs
I was trying to find the distances between vertices in graphs. But as the number of vertices are increasing up to 25 vertices or more, its becoming a tedious job for me to calculate $distance$ and $eccentricity$.

Can I draw graphs in MATLAB and then calculate the $dist$ and $ecc$?

For ex. What is the code if I want to draw $P_n$ and then the distances between vertices.

I am very new to MATLAB. Started few days back only. Can anyone help me here.

Edit : I want the code to draw the graph which are Undirected and non-weighted graphs. And also distance between every two vertices. Can I make the graph if I am taking adjacency matrix. Is it possible to make the graph when entries are like ordered pairs of adjacent vertices .

I would be very thankful. thanks

Coding for any small graph like $P_3$ or $K_4$ will be helpful for me.

答:
To work out graph distance use Dijkstra's algorithm which is available for MATLAB here

```<code>% K4 does not have edge weights in its definition
%  Make them all 1

K4 = ones(4) - eye(4) % Matrix of ones minus identity

% Find distance between nodes 1 and 2
[cost, route] = dijkstra(K4, 1, 2)

% Find the eccentricity using algorithm below

ecc = eccent(K4)
</code>```

You could download and install MATLABS graph theory toolbox (can't give a link on SE), which has the functions grEccentricity
and grShortPath
but it requires the optimization toolbox, which is propriety. So assuming you don't have the money to pay for a toolbox, here is an eccentricity function. There are surely faster algorithms than this one ($O(E^2)\times O(\text{dijkstra})$), but it is really simple - check every pair (assumes undirected edges).

```<code>function e = eccent(g)
    e = 0;

    for i=1:length(g)
        for j=1:(i-1) % only i < j

            % Find distance between nodes
            [new_e, path] = dijkstra(g, i, j);

            % If this distance is bigger than saved distance,
            %  make e the new distance

            if new_e > e
                e = new_e;
            end
        end
    end
end
</code>```

should work, but I don't own MATLAB to test it, you will need to put in a file called "eccent.m" in your working directory, because, well, it's MATLAB and that's how it works.

A function for $K_n$ is:

```<code>function g = Kn(n)
    % Puts a distance of one for all edges
    g = ones(n) - eye(n);
end
</code>```

and for $P_n$ is

```<code>function g = Pn(n)
    % Pn is a set of nodes like 1--2--3--4
    %  the edges are length one for neighbouring nodes
    %  so the matrix is tridiagonal with entries 1 on the -1, and 1 diagonals

    g = diag(ones(n-1,1),1) + diag(ones(n-1,1),-1);
end
</code>```

and the cyclic graphs just add a one to the antidiagonal corners

```<code>function g = Cn(n)
    g = Pn(n);
    g(n,1) = 1;
    g(1,n) = 1;
end
</code>```

These would need to be in files "Kn.m", "Pn.m" and "Cn.m" respectively.

To make a directed graph for a list of pairs of the form:

```<code>[[a,b];[c,d]...]
</code>```

then

```<code>function g = directed(pairs, n)
    % make graph with no edges
    g = zeros(n);

    % add each edge
    for p=1:length(pairs)
       i = pairs(p,1);
       j = pairs(p,2);

       g(i,j) = 1;

    end
end
</code>```

and for an undirected graph

```<code>function g = undirected(pairs)

    % make graph with no edges
    g = zeros(n);

    % add each edge
    for p=1:length(pairs)
       i = pairs(p,1);
       j = pairs(p,2);

       g(i,j) = 1;
       g(j,i) = 1;

    end
end
</code>```

Body:
To work out graph distance use Dijkstra's algorithm which is available for MATLAB here

% K4 does not have edge weights in its definition
%  Make them all 1

K4 = ones(4) - eye(4) % Matrix of ones minus identity

% Find distance between nodes 1 and 2
[cost, route] = dijkstra(K4, 1, 2)

% Find the eccentricity using algorithm below

ecc = eccent(K4)

You could download and install MATLABS graph theory toolbox (can't give a link on SE), which has the functions grEccentricity
and grShortPath
but it requires the optimization toolbox, which is propriety. So assuming you don't have the money to pay for a toolbox, here is an eccentricity function. There are surely faster algorithms than this one ($O(E^2)\times O(\text{dijkstra})$), but it is really simple - check every pair (assumes undirected edges).

function e = eccent(g)
    e = 0;

    for i=1:length(g)
        for j=1:(i-1) % only i < j

            % Find distance between nodes
            [new_e, path] = dijkstra(g, i, j);

            % If this distance is bigger than saved distance,
            %  make e the new distance

            if new_e > e
                e = new_e;
            end
        end
    end
end

should work, but I don't own MATLAB to test it, you will need to put in a file called "eccent.m" in your working directory, because, well, it's MATLAB and that's how it works.

A function for $K_n$ is:

function g = Kn(n)
    % Puts a distance of one for all edges
    g = ones(n) - eye(n);
end

and for $P_n$ is

function g = Pn(n)
    % Pn is a set of nodes like 1--2--3--4
    %  the edges are length one for neighbouring nodes
    %  so the matrix is tridiagonal with entries 1 on the -1, and 1 diagonals

    g = diag(ones(n-1,1),1) + diag(ones(n-1,1),-1);
end

and the cyclic graphs just add a one to the antidiagonal corners

function g = Cn(n)
    g = Pn(n);
    g(n,1) = 1;
    g(1,n) = 1;
end

These would need to be in files "Kn.m", "Pn.m" and "Cn.m" respectively.

To make a directed graph for a list of pairs of the form:

[[a,b];[c,d]...]

then

function g = directed(pairs, n)
    % make graph with no edges
    g = zeros(n);

    % add each edge
    for p=1:length(pairs)
       i = pairs(p,1);
       j = pairs(p,2);

       g(i,j) = 1;

    end
end

and for an undirected graph

function g = undirected(pairs)

    % make graph with no edges
    g = zeros(n);

    % add each edge
    for p=1:length(pairs)
       i = pairs(p,1);
       j = pairs(p,2);

       g(i,j) = 1;
       g(j,i) = 1;

    end
end


问:
How to use Permutations to create a set order of items
So for an Assignment, we are doing maximum amount of combinations with pizzas out of certain ingredients, bases, cheeses and sauces. The normal solution would be to just use: nCr * bases * cheeses * sauces. But this does not specify the order of things on the pizza, it would imply that it can be done in any order, with cheese on bottom or ingredients on top. So can permutations be used to array the items in order of Base|Sauce|Amount of ways to order the chosen 6 Ingredients (for each scenario)|Cheeses|. If so, or some other method exists, wondering if it is known?

For information amounts are here:

Bases amount = 4

Sauces amount = 3

Cheese amount = 2 * 2! (Option of extra cheese or no extra cheese)

Ingredients amount = 14 (6 ingredients allowed per pizza)

答:
The question is a bit ambiguous, so let me first write the hypothesis that I will assume:

Problem A)

Count the total number of pizza´s recipes that can be made when using:

$1$ out of $4$ bases.

$1$ out of $3$ sauces.

$1$ out of $2$ cheeses.

$1$ (or none) out of $2$ cheeses (this is the optional extra cheese).

Up to $6$ out of $14$ ingredients (when mixing these ingredients the order does not matter).

I will also assume that the 'mixing' order does NOT matter (the arrangement 'base -> sauce -> cheese -> ingredients' is equivalent to 'base -> cheese -> sauce -> ingredients').

The total ammount of recipes will be $$\dbinom{4}{1}\cdot\dbinom{3}{1}\cdot\dbinom{2}{1}\cdot\underbrace{\left[ \dbinom{2}{0}+\dbinom{2}{1}\right]}_{Extra\text{ } cheese}\cdot\underbrace{\left[ \dbinom{14}{0}+\dbinom{14}{1}+\dbinom{14}{2}+\dbinom{14}{3}+\dbinom{14}{4}+\dbinom{14}{5}+\dbinom{14}{6}\right]}_{Ingredients}$$

Problem B)

Count the total number of pizza´s recipes that can be made when using:

$1$ out of $4$ bases.

$1$ out of $3$ sauces.

$1$ out of $2$ cheeses.

$1$ (or none) out of $2$ cheeses (this is the optional extra cheese).

Up to $6$ out of $14$ ingredients (when mixing these ingredients the order does not matter).

I will also assume that the 'mixing' order DOES matter (the arrangement 'base -> sauce -> cheese -> ingredients' is NOT equivalent to 'base -> cheese -> sauce -> ingredients').

The total ammount of recipes will be $$\dbinom{4}{1}\cdot\dbinom{3}{1}\cdot\dbinom{2}{1}\cdot\underbrace{\left[ \dbinom{2}{0}+\dbinom{2}{1}\right]}_{Extra\text{ } cheese}\cdot\underbrace{\left[ \dbinom{14}{0}+\dbinom{14}{1}+\dbinom{14}{2}+\dbinom{14}{3}+\dbinom{14}{4}+\dbinom{14}{5}+\dbinom{14}{6}\right]}_{Ingredients}\cdot 5!$$
since every base-sauce-chese-extra_cheese-ingredient setting can be rearranged into $5!$ different settings.


问:
Help with Elementary number theory please
Use the second principle of finite induction to establish that for all $n\geq1$ :
$$a^n-1=(a-1)\left(a^{n-1}+a^{n-2}+a^{n-3}+\cdots+a+1\right) $$
Step by step explanation please! I'm confused how the second principle of finite induction is different from the first. </p>

答:
Second principle of finite induction establishes if the given statement is true for $n$ from $1,2,3... n$. Then it is true for $n+1$.

$$a^1 -1= a-1 \\
a^2 -1= (a-1)(a+1)$$
say $n \ge 3$
assume
$$a^{n} -1 =(a-1)(a^{n-1} +a^{n-2} +a^{n-3}.....+a+1)$$
we have to prove this for $n+1$.
$a^{n+1} -1$ can be written as $(a+1)(a^{n-1})-a(a^{n-1} -1)$
we already know $a^n -1$ is $(a-1)(a^{n-1} +a^{n-2} +a^{n-3}.....+a+1)$

Let $S=a^{n-1} +a^{n-2} +a^{n-3}.....+a+1$
so I can write $a^{n+1} -1$ as $(a+1)(a-1)S -a(S-a^{n-1})(a-1)$
which on simplifying gives
the required statement to prove

Hope it helps.


问:
Find the eigenvectors and eigenvalues of $\mathbf{C}= \left[\begin{array}{r r r} 4&6&2\\6&0&3\\2&3&-1 \end{array}\right]$

Find the eigenvectors and eigenvalues of

$$\mathbf{C}= \left[\begin{array}{r r r} 4&6&2\\6&0&3\\2&3&-1
\end{array}\right]$$

I found the the eigenvalues and are equal to $8.33, -6.63$ and $1.30.$ When I solve $(\lambda \mathbf{I} - \mathbf{C})\mathbf{E}=0$ I get the trivial solution as an answer while I shouldn't. What am I doing wrong?

答:
We are given the matrix

$$\mathbf{C}= \left[\begin{array}{r r r} 4&6&2\\6&0&3\\2&3&-1
\end{array}\right]$$

Find the characteristic polynomial using $|\lambda I - C|=0$ or $|C - \lambda I|=0$, just be consistent

$$|\lambda I - C| = \left|\begin{array}{r r r} \lambda-4&-6&-2\\-6&\lambda&-3\\-2&-3&\lambda+1\end{array}\right|$$

Use the Laplace expansion to write

$$(\lambda - 4)~ \begin{array}{|r r |} \lambda &-3\\ -3&\lambda +1\end{array}+ 6~\begin{array}{|r r |} -6&-3\\-2&\lambda +1\end{array} -2~\begin{array}{|r r |} -6&\lambda\\-2&-3\end{array}=\lambda^3 - 3 \lambda^2-53 \lambda - 72$$

The roots are

$$\lambda_{1,2,3} = 9.430018403873884, -1.5715091919701965, -4.858509211903689$$

For the eigenvectors, we solve $[\lambda I-C]v_i = 0$ using RREF, so for $9.430018403873884$, we have

$$\left(
\begin{array}{ccc}
 1 & 0 & -2.424177326987695 \\
 0 & 1 & -1.860554583299498 \\
 0 & 0 & 0 \\
\end{array}
\right)v_1 = 0$$

If you want to see steps, you can use this RREF Calculator

So, we have a free choice of $c$ and will choose $c = 1$, so $$v_1 = (2.424177326987695, 1.860554583299498, 1)$$

Repeat this process for the other two eigenvalues and find

$$v_2 = (-0.545323960202366, 0.17304624281151228, 1)\\v_3 = (1.176702188770226, -2.070637863148047, 1)$$


问:
Mathematics for cryptography.
Besides number theory, what other areas should I study for crypto. I did my undergrad in Comp Sci so the crypto course didnt have much mathematical topics. But for a grad specilisation in crypto I need to get acquainted with the required mathematical skills.

Any sort of help is appreciated.

Regards
Rahul

答:
Besides number theory, you should also know some abstract algebra. A lot of modern cryptography is based on for example elliptic curves.

You could also check out quantum cryptography. This would require that you know some quantum mechanics.

Hope this helps.

Title: Mathematics for cryptography.


问:
Delta function and $\sum_{t}\exp\{ i k t\} $
I am interested in evaluating the following, which appears in an integrand,
$$
\sum_{t=0}^{\infty}e^{ikt}
$$
where $k$ is real. Using the following relation
$$
\sum_{t=-\infty}^{\infty}e^{ikt} = 2\pi\delta(k),
$$
the real part of it is

\begin{eqnarray}
\Re\left[\sum_{t=0}^{\infty}e^{ikt}\right]&=&\frac{1}{2}+\frac{1}{2}\Re\left[\sum_{t=-\infty}^{\infty}e^{ik}\right]\\&=&\frac{1}{2}+\pi\delta(k).
\end{eqnarray}

Note the addition of $\frac{1}{2}$ term! This $\frac{1}{2}$ term does not agree with the following derivation using $\epsilon$ trick!

\begin{eqnarray}
\lim_{\epsilon\rightarrow0_{+}}\lim_{T\rightarrow\infty}\sum_{t=0}^{T}e^{ikt}e^{-\epsilon t}&=&\lim_{\epsilon\rightarrow0_{+}}\lim_{T\rightarrow\infty}\frac{e^{\epsilon}-e^{ik\left(T+1\right)}e^{-\epsilon T}}{e^{\epsilon}-e^{ik}}\\
&=&\lim_{\epsilon\rightarrow0_{+}}\frac{1}{e^{\epsilon}-e^{ik}}
\end{eqnarray}

which becomes, when $k \rightarrow 0$,

\begin{eqnarray}
\lim_{k\rightarrow0,}\lim_{\epsilon\rightarrow0_{+}}\frac{1}{e^{\epsilon}-e^{ik}}&=&\lim_{k\rightarrow0,}\lim_{\epsilon\rightarrow0_{+}}\frac{1}{\epsilon-ik}\\
&=&\lim_{k\rightarrow0}\lim_{\epsilon\rightarrow0_{+}}\frac{i}{k+i\epsilon}\\&=&\lim_{k\rightarrow0}\left[\pi\delta\left(k\right)+i\mathcal{P}\frac{1}{k}\right]
\end{eqnarray}

where Sokhotsky's formula is used.

Now when $k\nrightarrow0$,
$$
\lim_{\epsilon\rightarrow0_{+}}\frac{1}{e^{\epsilon}-e^{ik}}=\frac{1}{1-e^{ik}}
$$

Therefore, for all $k$,
$$
\sum_{t=0}^{\infty}e^{ikt}=\pi\delta\left(k\right)+\mathcal{P}\frac{1}{1-e^{ik}}
$$

Note that the real part of it does not have $\frac{1}{2}$ addition term, instead has $\Re\mathcal{P}\frac{1}{1-e^{ik}}$.

What am I doing wrong?

Building on Svyatoslav's answer which in turn supported by vitamin d, the conclusion is the following:
$$
\sum_{t=0}^{\infty}e^{ikt} = \frac{1}{2} + \pi \delta(k) + \frac{i}{2} \mathcal{P} \cot (k/2)
$$

答:
If we do formal calculations (without the proof of validity) we get

$S(k)=\sum_{t=0}^{\infty}e^{ikt}=1+\sum_{t=1}^{\infty}e^{ikt}+\sum_{t=-1}^{-\infty}e^{ikt}-\sum_{t=-1}^{-\infty}e^{ikt}=\sum_{t=-\infty}^{+\infty}e^{ikt}+1-\sum_{t=0}^{\infty}e^{-ikt}$

$\Rightarrow$ $S(k)+S^*(k)=1+2\pi\delta(k)$

On the other hand, if we use $\epsilon$-regularization

$$S(k,\epsilon)=\sum_{t=0}^{\infty}e^{ikt-\epsilon{t}}=\frac{e^{\epsilon}}{e^{\epsilon}-e^{ik}}=\frac{e^{\epsilon}(e^{\epsilon}-e^{-ik})}{1+e^{2\epsilon}-e^{\epsilon}(e^{ik}+e^{-ik})}=\frac{e^{\epsilon}-\cos{k}+i\sin{k}}{e^{\epsilon}+e^{-\epsilon}-(e^{ik}+e^{-ik})}$$

$$S(k,\epsilon)=\frac{1}{2}+\frac{\sinh\epsilon+i\sin{k}}{2(\cosh\epsilon-\cos{k})}=\frac{1}{2}+\frac{\sinh\epsilon}{2(\cosh\epsilon-\cos{k})}+\frac{i\sin{k}}{2(\cosh\epsilon-\cos{k})}$$

$S(k,\epsilon)\to\frac{1}{2}+\frac{\epsilon}{\epsilon^2+k^2}=\frac{1}{2}+\pi\delta(k)$,  $k<<\epsilon$, $\epsilon\to0$

$S(k,\epsilon)\to\frac{1}{2}+\frac{i}{2}\frac{\sin{k}}{1-\cos{k}}=\frac{1}{2}+\frac{i}{2}\cot\frac{k}{2}$, $k>>\epsilon$, $\epsilon\to0$

$S(k,\epsilon)+S^*(k,\epsilon)\to1+2\pi\delta(k)$ for all $k$


问:
Equivalence of full and sequential minimization
I would like to show that two problems are equivalent, so that by solving one of them I get the optimal solution to the second. Formally, I'm trying to prove that:
$$
\min_{\mathbf{x,y}\in\mathbf{X\times Y}}\; f(\mathbf{x,y})\; =\; \min_{\mathbf{x}\in\mathbf{X}}\; \min_{\mathbf{y}\in\mathbf{Y}} f(\mathbf{x,y})
$$
where $f$ is a closed, proper, convex function and $\mathbf{X,Y}$ are closed convex sets. Is this generally possible? So far I have the "easy" direction:
$$
 \min_{\mathbf{x}\in\mathbf{X}}\; \min_{\mathbf{y}\in\mathbf{Y}} f(\mathbf{x,y})=\min_{\mathbf{x}\in\mathbf{X}}\; \min_{\mathbf{x,y}\in\mathbb{R}^n\times\mathbf{Y}} f(\mathbf{x,y})\leq  \min_{\mathbf{x}\in\mathbf{X}}\; \min_{\mathbf{x,y}\in\mathbf{X\times Y}} f(\mathbf{x,y}) = \min_{\mathbf{x,y}\in\mathbf{X\times Y}} f(\mathbf{x,y})
$$
where the first inequality is due to the fact that $\big(\mathbf{X\times Y}\big)\subseteq\mathbb{R}^n\times\mathbf{Y}$ and the second equality is since we already minimized w.r.t. $\mathbf{x}$, hence the inner minimization produced a solution which is already optimal in $\mathbf{x}$.

However I'm not sure how to get the opposite direction. Does anyone have any ideas? Maybe a different technique should be used?

答:
Since you have minimums, they are reached at optimal points $\mathbb x^*$ and $\mathbb y^*$ ; and the value of the $\min$s is the value of $f$ at these points.

Note that if your sets and functions aren't convex and instead of minimums you have infimums (which aren't necessarily reached), this results still holds ; but instead of using points $\mathbb x^*, \mathbb y^*$ you have to use sequences that converge towards the right value, and be careful when taking the limit of limits...


问:
Write $AS$ as a multiple of $AD$ and $CS$ as a multiple of $CE$.
In a triangle $ABC$ the point $D$ is on $BC$ such that $\overline{BD}=\frac{2}{3}\cdot \overline{BC}$ and the point  $E$ is on $AB$ such that $\overline{AE}=\frac{3}{5}\cdot \overline{AB}$.

The lines $AD$ and $CE$ intersect in $S$.

Write $AS$ as a multiple of $AD$ and $CS$ as a multiple of $CE$.

We have a triangle as follows, right?

But how can we get an information for $AS$ and $CS$ ? Could you give me a hint? Do we maybe use vector operations?

$$$$

EDIT :

Using vectors we have $$\vec{AS}+\vec{SD}=\vec{AD} \Rightarrow \vec{AS}=\vec{AD}-\vec{SD} \Rightarrow \vec{AS}=\vec{AD}+\vec{DS}$$ Does this help us ? How can we continue?

答:
Here is another solution. I solve the problem for $AS$ and $AD$. You can do the same for $CS$ and $CE$.

As shown in the figure below, from $A$ , $B$ and $D$ draw $AH$ , $BK$ and $DL$ perpendicular to $CE$ .

We have:
$$\frac{AH}{BK} = \frac{AE}{BE} = \frac 32$$
$$\frac{BK}{DL} = \frac{BC}{DC} = \frac 31$$
$$\therefore \frac{AH}{DL} = \frac 32 \times \frac 31 = \frac 92 $$
Now we can calculate $\frac{AS}{AD}$:
$$\frac{AS}{SD} = \frac{AH}{DL} = \frac 92 $$
$$\therefore \frac{AS}{AD} = \frac{AS}{AS+SD} = \frac{9}{9+2} = \frac{9}{11} $$


问:
Prime counting function in large-ish intervals--two questions/conjectures
Consider two positive integers $2 < a < x$. The following two statements would seem to follow intuitively from the PNT, but I'm wondering if either has been proven or discussed. Here, $\pi(n)$ is the prime-counting function.

Conjecture 1: For any $x$, there exists some $a_0$ such that if $a \geq a_0$, the following inequality is always true:

$$\tag{1} \pi(x+a) - \pi(x) \leq \pi(x) - \pi(x-a)$$

In plainer English, consider two intervals of length $a$ on either side of $x$. This conjecture suggests that the earlier interval (which contains smaller numbers) contains more prime numbers than the later interval (which contains larger numbers).

Shown in comments to be trivially true but left for context: For every $a$ there is some large enough $x$ such that the following inequality is always true:

$$\tag{2} \frac{\pi(x) - \pi(a)}{x-a} \leq \frac{\pi(a)}{a}$$

Here, the question is about the density of primes on the two intervals. Are the primes more dense below $a$ than above $a$?

(The inequality can be written as $\frac{\pi(x)}{x} < \frac{\pi(a)}{a}$, which is true for any sufficiently different $a$ and $x$ with $a < x$ via the PNT).

We can consider an extreme example that applies to both statements. If we set $x = a$ in conjecture (1), or $x = 2a$ in conjecture (2), both conjectures simply ask whether there are more (or equal) primes between $0$ and $n$ than between $n$ and $2n$. We know this is true for $n > 2$. Outside of the extreme case, however, the PNT would suggest that equal-length intervals should have more primes if they begin at a lower value, and that the density of primes decreases as $n$ increases. I also suspect that if $a$ is very small, or in conjecture (2) if $a$ is close to $x$, that these would fail.

It seems these conjectures must have been floated at some point, but I'm wondering if they've been proven. I don't think they're just special cases of the second Hardy-Littlewood conjecture, but I'm also very much an amateur. Does anyone know of proofs/disproofs/discussions of these?

Edit: Restatement of conjecture 1, Note on conjecture 2 as it is even more trivial than I thought it even could be.

答:
For any $a \geq 2$ Let $ x = (a+2)! +a +2$ and since there is no prime in the interval $ [(a+2)!+2,(a+2)!+a+2]$ we have that $ \pi(x)  = \pi(x-a)$ and so $ \pi(x)-\pi(x-a)=0$

Now since $\pi(y)$ is non-decreasing function we have that $\pi(x+a)-\pi(x) \geq 0$, if we have $ \pi(x+a)-\pi(x)>0$ we are done otherwise $ \pi(x+a)-\pi(x) = 0$, now let $x_{new} = x_{old}+a$ and so $ \pi(x_{new})-\pi(x_{new}-a) = \pi(x_{old}+a)-\pi(x_{old}+a-a)  = \pi(x+a)-\pi(x) = 0$ (Our Assumption) and so $ \pi(x_{new})-\pi(x_{new}-a)=0$ and we repeat until $\pi(x_{new}+a)-\pi(x_{new})>0$ (Guaranteed because the infinitude of the primes).

Which implies that $ \pi(x+a)-\pi(x) > \pi(x)-\pi(x-a)$ holds infinitely many times.

With a little tweaking of the above argument we get that $ \pi(x+a)-\pi(x) < \pi(x)-\pi(x-a)$ holds infinitely many times.

So the first conjecture for any $a\geq 2$ holds infinitely many times and violated for infinitely many times for different sequences of $x$ as $ x \to \infty$.


问:
Gram determinant
How to prove that

$$\sqrt{\Gamma(\vec{a},\vec{b},\vec{c})}=|(\vec{a},\vec{b},\vec{c})|,$$

where $\Gamma(\vec{a},\vec{b},\vec{c})= \left | \begin{array} {ccc} \vec{a} \cdot \vec{a} & \vec{b} \cdot \vec{a} & \vec{c} \cdot \vec{a} \\
\vec{a} \cdot \vec{b} & \vec{b} \cdot \vec{b} & \vec{c} \cdot \vec{b} \\
\vec{a} \cdot \vec{c} & \vec{b} \cdot \vec{c} & \vec{c} \cdot \vec{c}
\end{array} \right | $.

Thanks!

答:
Presumably $\vec{a}$ and the like are column vectors. Your $\Gamma$ is just the matrix product $A^\ast A$, where $A=(\vec{a},\vec{b},\vec{c})$ and $A^\ast$ denotes the conjugate transpose of $A$ (or simply the transpose of $A$ if $\vec{a},\vec{b},\vec{c}$ are real). Using the properties that $\det(XY)=\det(X)\det(Y)$ and $\det(X^\ast)=\overline{\det(X)}$, we see that $\det(\Gamma)=\overline{\det(A)}\det(A)=|\det(A)|^2$ and hence the correct statement should be:
$$\sqrt{\det(\Gamma)}=|\det(A)|,$$
where the pair of strokes $|\cdot|$ here denotes the modulus of a complex number (or absolute value of a real number). Without taking the modulus, your statement is wrong. A counterexample is given by $\vec{a}=(-1,0,0)^\top,\ \vec{b}=(0,-1,0)^\top$ and $\vec{c}=(0,0,-1)^\top$.</p>

Title: Gram determinant


问:
Proving that the $\chi^{2}$ Distribution is a PDF
The $\chi^{2}$-distribution is given by $$f(x; k)\equiv \frac{1}{2^{\frac{k}{2}} \Gamma\left(\frac{k}{2}\right)}x^{\frac{k}{2}- 1}e^{-\frac{x}{2}}.$$

If the $\chi^{2}$-distribution is a probability-density function (PDF), then it must hold that $$
\int_{-\infty}^{+\infty} f(x; k) \, dx = 1. $$ However, when trying to prove this, I am stuck:

$$\int_{-\infty}^{+\infty} f(x;k) \, dx = \int_{-\infty}^{+\infty}\frac{1}{2^{\frac{k}{2}}\Gamma\left(\frac{k}{2}\right)}x^{\frac{k}{2}- 1}e^{-\frac{x}{2}} \, dx = \frac{1}{2^{\frac{k}{2}}\Gamma\left(\frac{k}{2}\right)}\int_{-\infty}^{+\infty}x^{\frac{k}{2}-1}e^{-\frac{x}{2}} \, dx$$

Looking only at the last integral, we get: $$\int_{-\infty}^{+\infty}x^{\frac{k}{2}-1}e^{-\frac{x}{2}}dx = \int_{-\infty}^{0} x^{\frac{k}{2}-1}e^{-\frac{x}{2}}dx + \underbrace{\int_{0}^{\infty} x^{\frac{k}{2}-1}e^{-\frac{x}{2}}dx}_{= \Gamma\left(\frac{k}{2}\right)}$$

However, I am not yet sure on what to do with the first term. Any feedback would be appreciated.

答:
The integral is

$$\frac{1}{\Gamma\left(\frac{k}{2}\right)}\int_0^{\infty}\left(\frac{1}{2}\right)^{k/2}x^{k/2-1}e^{-x/2}dx$$

Substituting

$$\frac{x}{2}=y$$

you get

$$\frac{1}{\Gamma\left(\frac{k}{2}\right)}\underbrace{\int_0^{\infty}y^{k/2-1}e^{-y}dy}_{=\Gamma\left(\frac{k}{2}\right)}=1$$


问:
Series expansion at branch cut
Basically, I have obtained the function $\rho (r)$ below as a result of integrating

$$\rho(r)=\int_{b}^{r}\frac{dx}{\sqrt{1-(b_{0}/x)^{1-q}}}$$
which results to
$$\rho=\frac{2b}{1-q}\sqrt{1-\left(\frac br\right)^{1-q}}\,_2F_1\left(\frac{1}{2},\frac{q-2}{q-1},\frac{3}{2},1-\left(\frac br\right)^{1-q}\right)$$
where $b$ is just some positive constant while $-\infty<q<1$.

What I need is $r(\rho)$ which is the inverse of the function above (which I can implement with numerics). I am also interested in the series expansion of $r(\rho)$ at $\infty$ but unknowingly, the hypergeometric function above has  a branch cut at $1-\left(\frac br\right)^{1-q}=1$ (i.e. $r\rightarrow \infty$). My question would be, am I allowed to expand  $r(\rho)$ at $\infty$ given that there is a branch cut at infinity? What are the possible complications? What can I possibly do to safely obtain an asymptotic expansion of $r(\rho)$ at $\infty$? I am also thinking of converting the integral above into a differential equation for $r(\rho)$ and expand it at infinity to bypass the hypergeometric function branch cut problem but I am not really sure of this. Thanks for the help

Title: Series expansion at branch cut

Body:
$$\rho(r)=r+a+\sum_{n\ge 1} {-1/2\choose n} \frac{b_0^{n(1-q)}}{1+n(q-1)} r^{1+n(q-1)}\in r \ \Bbb{R}[[r^{-1},r^q]]^\times$$

If $q <-1$ then
$$\rho(r)^{-1}=r^{-1} u, \qquad u\in \Bbb{R}[[r^{-1},r^q]]^\times$$
Note that $u^{-q}=\sum_{k\ge 0} {-q\choose k} (u-1)^k\in \Bbb{R}[[r^{-1},r^q]]$.

Thus $$\phi(f(r^{-1},r^q))=f(\rho(r)^{-1},\rho(r)^q)$$ is an automorphism of $\Bbb{R}[[r^{-1},r^{q-1}]]$,
there is some $u\in \Bbb{R}[[r^{-1},r^q]]$ such that $\phi(u)=r^{-1}$.

$u^{-1}$ is your asymptotic expansion of $\rho^{-1}(r)$.


答:
As pointed out in the comments, you don't integrate in $(\infty,0]$ since the chi squared distribution isn't defined there. You also have written

$$\underbrace{\int_{0}^{\infty} x^{\frac{k}{2}-1}\color{red}{e^{-\frac{x}{2}}}\mathrm{d}x}_{= \Gamma\left(\frac{k}{2}\right)}$$

Which is wrong. The integral definition of the Gamma function for $\operatorname{Re}(z)>0$ is
$$\Gamma(z)=\int_0^\infty x^{z-1}\color{red}{e^{-x}}\mathrm{d}x$$
The exponent should be $-x$, not $-x/2$. With a change of variable $t=x/2$ we can write
$$\int_0^\infty x^{k/2-1}e^{-x/2}\mathrm{d}x=2^{k/2}\int_0^\infty t^{k/2-1} e^{-t}\mathrm{d}t=2^{k/2}\Gamma(k/2)$$

Title: Proving that the $\chi^{2}$ Distribution is a PDF

Body:
The $\chi^{2}$-distribution is given by $$f(x; k)\equiv \frac{1}{2^{\frac{k}{2}} \Gamma\left(\frac{k}{2}\right)}x^{\frac{k}{2}- 1}e^{-\frac{x}{2}}.$$

If the $\chi^{2}$-distribution is a probability-density function (PDF), then it must hold that $$
\int_{-\infty}^{+\infty} f(x; k) \, dx = 1. $$ However, when trying to prove this, I am stuck:

$$\int_{-\infty}^{+\infty} f(x;k) \, dx = \int_{-\infty}^{+\infty}\frac{1}{2^{\frac{k}{2}}\Gamma\left(\frac{k}{2}\right)}x^{\frac{k}{2}- 1}e^{-\frac{x}{2}} \, dx = \frac{1}{2^{\frac{k}{2}}\Gamma\left(\frac{k}{2}\right)}\int_{-\infty}^{+\infty}x^{\frac{k}{2}-1}e^{-\frac{x}{2}} \, dx$$

Looking only at the last integral, we get: $$\int_{-\infty}^{+\infty}x^{\frac{k}{2}-1}e^{-\frac{x}{2}}dx = \int_{-\infty}^{0} x^{\frac{k}{2}-1}e^{-\frac{x}{2}}dx + \underbrace{\int_{0}^{\infty} x^{\frac{k}{2}-1}e^{-\frac{x}{2}}dx}_{= \Gamma\left(\frac{k}{2}\right)}$$

However, I am not yet sure on what to do with the first term. Any feedback would be appreciated.


问:
A property of fields of characteristic $p$ and their extensions
I am currently stuck on the following problem, taken from Herstein's Topics in Algebra, 2nd ed.. It reads:

If $F$ is of characteristic $p \neq 0$ and if $K$ is a finite extension of $F$, prove that given $a \in K$ either $a^{p^n} \in F$ for some $n$ or we can find an integer $m$ such that $a^{p^m} \notin F$ and is separable over $F$

Herstein defines an element to be separable over $F$ if, and only if, it is the root of a polynomial $p(x) \in F[x]$ with no multiple roots.

I am trying to do this by showing that, if we are not in the first case, we must be in the second case. However, I have no idea where to use that $K$ is a finite extension, and I just can't see why the second case must hold...

A couple of things I think could help are two former exercises from the same chapter. One says that the set $\{a \in K \mid a^{p^n} \in F\}$ is a subfield of $K$ and the other says that the set of separable elements of $K$ over $F$ are a subfield – but note that none of them suppose $K$ to be a finite extension...

I also thought about the fact that the polynomial $x^{p^n} - x$ has no multiple roots, but I don't know how to convert that to this problem.

In short, I am very much stuck and any help is appreciated!

Thanks in advance!

答:
Some ideas for you: Let $\;a\in K\;$ and let $\;p_1(x)\in F[x]\;$ be its minimal polynomial over $\;F\;$ . If

$\;a\in F\;$  we're done  since then any power of $\;a\;$ is again in $\;F\;$ . So assume $\;a\notin F\;$ . If $\;a\;$ is

separable then we're done, so assume $\;a\;$ isn't separable $\;\implies p_1(x)=g_1(x^p)\;$ (corollay 1), and

of course $\;\deg p_1>\deg g_1\;$.

The above means that $\;a^p\;$ is a root of $\;g_1(x)\;$ , and let $\;p_2(x)\in F[x]\;$ be the minimal polynomial

of $\;a^p\;$ over $\;F\;$ . If $\;a^p\in F\;$ we're thru, so assume $\;a^p\notin F\;$ . If $\;a^p\;$ is separable we're done, so

again assume $\;a^p\;$ is not separable $\;\implies p_2(x)=g_2(x^p)\;$ , and $\;\deg p_1>\deg g_1\ge\deg p_2>\deg g_2\;$

is a root of $\;g_2(x)\;$  and $\;(a^p)^p=a^{p^2}\;$...and continue on.

Since $\;\deg p_1\;$ is finite, the above must stop at some finite step...finish now the argument.