问:
Highest Twin primes such that the number in between twins is the $\text{lcm}$ of first $N$ numbers
$\text{lcm}(1,2,3,4,5,6,7) = 420$ and this number is placed between two Twin primes: $419,\ 421$.
This happens again for $\text{lcm}(1,2,3...19) = 232792560$ and
$\text{lcm}(1,2,3,4,...,47)=442720643463713815200$ that is  placed between the 2 Twin primes:
$$442720643463713815199 \\
442720643463713815201$$
However it does not happen again up to where I could reach about $\text{lcm}(1,2,3..., 44983).$

Is $N=47$ the largest number such that $\text{lcm}(1,2,3,...,N)$ is placed between two Twin primes?

I find this huge gap between 47 and, at least, 44983, highly counter-intuitive. There could be a good reason for this.

Complete list :
$$\text{lcm}(1,2,3) = 6 \text{  Twin primes: 5 7}$$
$$\text{lcm}(1,2,3,4) = 12 \text{  Twin primes: 11 13}$$
$$\text{lcm}(1,2,3,4,5) = 60 \text{  Twin primes: 59 61}$$
$$\text{lcm}(1,2,3,4,5,6) = 60 \text{  Twin primes: 59 61}$$
$$\text{lcm}(1,2,3,4,5,6,7) = 420 \text{  Twin primes: 419 421}$$
$$\text{lcm}(1,2,3,...,19) = 232792560 \text{  Twin primes: 232792559 232792561}$$

(then 47 already mentioned above)

I also add a very simple code in python, using the sympy library, so everyone can verify this.

```<code>    from sympy import *

    lcmi=2
    for i in range(3,100000):
        print "Evaluating lcm of first", i, "numbers"
        ni= lcm(lcmi,i)
        if ni != lcmi:
         if isprime(ni-1) and isprime(ni+1):
            print i,"lcm:", ni,"Twin primes:", ni-1, ni+1
         lcmi=ni
</code>```

答:
Write $a_n := \operatorname{lcm}(1, 2, \ldots, n)$.

When you're looking at $n=7$ you're considering the pair $420\pm 1$.  The form of these numbers excludes the prime factors 2, 3, 5, and 7, and $\sqrt{420} \approx 20.5$, so the only potential remaining prime factors are 11, 13, 17, and 19.

On the other hand, when you're looking at, say, $n = 9$, you're considering $2520\pm 1$, and $\sqrt{2520} \approx 50.2$.  The form of these numbers still excludes only 2, 3, 5, and 7 as prime factors, but now in addition to 11, 13, 17, and 19 you have to worry about 23, 27, 29, 31, 37, 41, 43, and 47.

Now $a_n = e^{n + o(n)}$ (see Chebyshev's function), so the number of primes less than $\sqrt{a_n}$ grows faster than linear in $n$, while the number of primes less than $n$ (i.e. the number of primes excluded as factors of $a_n \pm 1$) grows slower than linearly.

As a result, going only by this we'd expect exactly the sort of behavior you see here: true for all sufficiently small $n$, and then sporadically true for a few larger numbers if the pieces fall into place.

Of course that doesn't prove anything, but it does tell us that the numerical evidence here isn't pointing towards anything unusual, and therefore likely gives us no help in trying to prove or disprove that there are infinitely many $n$ for which $a_n \pm 1$ are relatively prime.


问:
What is a good book to train drawing graphs of function?
I am about undergraduate here , and I just want to ask what is the best book for drawing graph of functions as I am very weak in drawing graphs</p>

答: Drawing graphs correctly is a matter of practice. And also from the elementary point of view the more you know about a function the better you can draw it's graph analytically. For starting, you may use some graphing calculators or programs, show some inquisitiveness about the graph of any function you encounter and you will gradually learn how to draw the related ones. And a vast knowledge about different graphs also come handy. I also faced that problem, so keep patience, if you follow this advice you will surely do good. So once again $-$play with graphs.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
Bolzano and Cauchy's proof of Intermediate value theorem
I am not sure if it is the right place to ask, but I have been searching for the proof of Intermediate value theorem made by Bolzano and by Cauchy and I am a bit confused because as I understood it the theorem was first proven by Bolzano and then Cauchy provided a proof of the theorem few years later and it was not the same proof. But where can I find the proofs of the theorem?</p>

答:
Proof by Bolzano is in Steve Russ - The mathematical works of Bernard Bolzano-Oxford University Press (2004), page 250.

Proof by Cauchy is in Robert E. Bradley, C. Edward Sandifer (auth.) - Cauchy’s Cours d’analyse_ An Annotated Translation-Springer-Verlag New York, (2009) page 32.


问:
Distributing $n$ different things among $r$ distinct groups such that all of them must get atleast $1$
In how many ways can we arrange $7$ different things to $3$ people, such that all of them must get at least one?

We know that if we have $n$ identical items which will be distributed in $r$ distinct groups where each must get at least one then the number of way  is $\binom {n-1}{r-1}$ (i.e in this case $\binom62$. But in the question it is said that $7$ DIFFERENT things then what will be the approach?

MY TRY ::

At first selecting $1$ by $1$ for each $3$ people so that each get at least one : $\binom71 \times \binom 61 \times \binom 51$

Now each of $4$ can go to any of $3$ so :$3^4$ ways, so $\binom71\times\binom61\times\binom51\times (3^4)$

MY TRY #2 ::

total possibilities - {any one get $0$  $(0,6,1)(0,5,2)(0,4,3)$} - {any two get $0$ $ (0,0,7)(0,7,0)(7,0,0)$}
so

$3^7$ - {($\binom76 + \binom 75 + \binom 74$  )$\times$ 3!} - 3

答:
In any distribution let there be groups $z_1, z_2, z_2,...,z_r$ and consider the distribution where blanks are allowed.

The total number of these are $ r^n$.
The number in which $z_1$ is blank is
$$(r-1)^n$$
$\therefore$ the number in which $z_1$ is not blank is ,
$$r^n-(r-1)^n$$
of these last,the umber in which $z_2$ is blank, is
$$(r-1)^n-(r-2)^n$$
$\therefore$, the number in which $z_1$ and $z_2$ are not blank,is
$$r^n-2(r-1)^n+(r-2)^n$$
If we keep doing this for $z^r$ groups, we will get this expression,
$$r^n-{x\choose1}(r-1)^n+{x\choose2}(r-2)^n-...+(-1)^x(r-x)^n$$
The number of distribution in which no one of x assigned groups is blank, is
when x=r, then
$$r^n-{r\choose1}(r-1)^n+{r\choose2}(r-2)^n-...+(-1)^(r-1){r\choose{r-1}}$$
$$or$$
$$\sum^{r}_{p=0} (-1)^p{{r}\choose{p}}(r-p)^n$$
$$or$$
$$Coefficient\,\,of\,\,x^n\,in\,\,n!(e^x-1)^r$$

So finally answer to you question,In how many ways can we arrange 7 different things to 3 people, such that all of them must get at least one?

--> $3^7-{3\choose1}2^7+{3\choose2}1^7-{3\choose3}0^7$
=1806

Bibliography:
Goyal, S., Dr. (n.d.). Chapter-5, Session-6, Arrangement in groups,Multinomial Theorem, Multiplying Syntetically. In Skill in Mathematics - Algebra for JEE Main and Advanced 2020. Arihant.


问:
What is the meaning of ⊊?
I have encountered this when referencing subsets and vector subspaces. For example, T ⊊ span(S) should mean that T is smaller than span(S)--at least from what I've gathered.

Is ⊊ a sort of ≤ or < but for subsets? What would that little crossing line mean? I have tried looking it up on the net but could not find an answer.

Thank you.

答:
Short answer:

$A\subsetneq B$ means that $A$ is a subset of $B$ and $A$ is not equal to $B$.

Long answer:

There is some confusion on mathematical textbooks when it comes to the symbols indicating one set is a subset of another.

It's relatively clear what the symbol "$\subseteq$" means. This symbol is more or less universally understood as the following:

$$A\subseteq B \iff \forall a: (a\in A: A\in B)$$

that is, $A$ is a subset of $B$.

The problem comes when we want to talk about the relation of "proper subset". That is, the relation described by the property
$$A\subseteq B \land A\neq B$$

Some textbooks use the sybol $\subset$ for this relation. That is, they define "$\subset$" as

$$A\subset B \iff A\subseteq B \land A\neq B.$$

The problem arises when some authors and mathematitians use the symbol "$\subset$" when they are talking about "any subset" instead of "proper subset". In other words, some authors use "$\subset$" in places where other authors would use $\subseteq$. This has caused the symbol "$\subset$" to become ambiguous, and authors tend to want to avoid it. To describe the relation "proper subset", they instead use the symbol "$\subsetneq$", which is less ambiguous.

TLDR: It's a mess, and because we don't want to use $\subset$, we use $\subsetneq$ to avoid ambiguity.

Title: What is the meaning of ⊊?


答: It means the space $T$ is a propper subspace of $span(S)$

Title: What is the meaning of ⊊?

Body:
I have encountered this when referencing subsets and vector subspaces. For example, T ⊊ span(S) should mean that T is smaller than span(S)--at least from what I've gathered.

Is ⊊ a sort of ≤ or < but for subsets? What would that little crossing line mean? I have tried looking it up on the net but could not find an answer.

Thank you.


问:
Expected number of tests in batch testing
Suppose we have $n=k^L$ people and we want to test who has a disease. Suppose for each person, the probability of having a disease is $p$. We do the test in batches, first divide these people in $k$ groups each with $k^{L-1}$ people. If the pooled test of a group is positive, we recurse in this group by dividing that group further into $k$ subgroups and continue the process.

What is the expected number of tests needed to pin down the infected individuals? I come up with a recurrence relation but do not know how to solve and simplify. Let $a_n$ be the expected total number of tests. Then we have
$$a_n=k(1-(1-p)^{n/k})a_{n/k}+k$$
How to solve this and simplify the result? How does $a_n$ depend on $k$ and $p$?

答:
Fix an integer $k \ge 2$ and a real number $p\in [0,1]$.

For each positive integer $l$ let
$$
\left\lbrace
\begin{align*}
n(l)&=k^l\\[4pt]
Q(l)&=(1-p)^{n(l)}\\[4pt]
P(l)&=1-Q(l)\\[4pt]
\end{align*}
\right.
$$
and let $e(l)$ be the expected number of tests, assuming

A population of size $n(l)$.$\\[4pt]$

Each member of the population has probability $p$ of having the disease.

I don't think that a closed form for $e(l)$ can be had.

You proposed a recursion which can be stated as
$$
e(l)
=
\begin{cases}
k+kP(l-1)e(l-1)&&\text{if}\;l > 1\\[4pt]
k&&\text{if}\;l=1\\[4pt]
\end{cases}
$$
but it yields incorrect results for $l > 2$.

A correct recursion can be stated as
$$
e(l)
=
\begin{cases}
{\displaystyle{
k
+
\sum_{h=1}^k
h
\binom{k}{h}
P(l-1)^h
Q(l-1)^{k-h}
e_1(l-1)
}}
&&\text{if}\;l > 1\\[4pt]
k&&\text{if}\;l=1\\[4pt]
\end{cases}
$$
where $e_1$ is defined recursively by
$$
e_1(l)
=
\begin{cases}
{\displaystyle{
k
+
\frac{1}{P(l)}\sum_{h=1}^k
h
\binom{k}{h}
P(l-1)^h
Q(l-1)^{k-h}
e_1(l-1)
}}
&&\text{if}\;l > 1\\[4pt]
k&&\text{if}\;l=1\\[4pt]
\end{cases}
$$
For example, using $k=2$ and applying the above, we get
\begin{align*}
e(1)&=2\\[4pt]
e(2)&=-4p^2+8p+2\\[4pt]
e(3)&=-4p^4+16p^3-32p^2+32p+2\\[4pt]
\end{align*}
As a numerical example, using $k=2$ and $p={\large{\frac{1}{10}}}$, we get
$$
e(3)=\frac{12239}{2500}\approx 4.90
$$
which agrees with results obtained from a simulation.


问:
If the derivative tends to $0$ at infinity, then so does the function divided by $x$
Suppose that $$f\colon (a,\infty) \to \mathbb{R}$$
is differentiable, and that $$\lim_{x\to\infty} f'(x) =0. $$

Show that $$\lim_{x\to\infty} \frac{f(x)}{x} = 0.$$

This can be done easily with L'Hôpital's rule, but it appears as a textbook exercise, in a chapter discussing Rolle's theorem, Mean Value Theorem, etc., before L'Hôpital's rule is introduced.

答: Let $x >T$. $\frac {f(x)} x=\frac {f(x)-f(T)} x +\frac {f(T)} x=\frac {x-T} x \frac {f(x)-f(T)} {x-T} +\frac {f(T)} x$By MVT this gives $\frac {f(x)} x=\frac {x-T} x f'(s) +\frac {f(T)} x$ for some $s \in (T,x)$. Can you finish?


答:
If you look into some random books dealing with set theory (maybe as introductory to other subjects), you'll find that in some of them
$$
A\subset B
$$
means “for all $x$, if $x\in A$ then $x\in B$” (that is, “$A$ is a subset of $B$”).

In other books, you'll find the same relation denoted by $A\subseteq B$, whereas $A\subset B$ would mean that $A$ is a proper subset of $B$.

However, due to the notational conflict, somebody prefers to be as clear as possible and will write
$$
A\subsetneq B
$$
to mean that $A$ is a proper subset of $B$ (that is $A$ is a subset of $B$, but $A\ne B$). People who write $A\subset B$ for “subset, equality possibly happening” will probably use this notation for proper subsets.

Different traditions. When you see $A\subset B$, look in the initial pages to see what it is bound to mean. If there's no hint, try and infer from the usage.

Title: What is the meaning of ⊊?


问:
Existence of subgroup of Abel group
$G$ is a Abel group, the group of order is $n$. If there is an integer $m$ that is a factor of $n$, then there exists a subgroup $H$ of $G$ which order is $m$ .How to prove that?

答:
Let G be our group of order $n$.
First you can use the Fundamental Theorem of Arithmetic to say that $n$ is a unique product of distinct primes $p_i$: $$n = p^{k_1}_{1} \cdot p^{k_2}_{2} \cdot ... \cdot p^{k_{s}}_{s}.$$
Since $m$ is a divisor of $n$, we can write $m$ as a product of the same primes where the exponents of the primes may be $0$.
$$m =  p^{l_1}_{1} \cdot p^{l_2}_{2} \cdot ... \cdot p^{l_{s}}_{s}, \text{where } 0 \leq l_i \leq k_i. $$
The First Sylow Theorem gurantees the existence of groups of order equal to any power of a prime dividing $n$, we will choose the groups whose orders are $p^{l_i}_{i}$ and call them $H_{i} < G$. We know that these groups are all normal, because $G$ is abelian.
The last step will be the construction of the group $ H_1  H_2  ... H_s = H$ which we will show is isomorphic to $ H_1 \times  H_2 \times  ... \times H_s$ a group with order $m$.

Take any two $H_{i}, H_{j}$ where $i \neq j$ and suppose $a \in H_{i} \cap H_{j}$, then $o(a) \mid p_{i}$ and $o(a) \mid p_j$, which can only happen if $o(a) = 1$, since the primes $p_i, p_j$ are distinct. Then the intersection of any two such groups is trivial. (This step is needed to ensure unique representation of any element from $H$).

It's obvious elements from any two such groups commute since they lie in $G$ which is abelian, then we can show that $H$ is a subgroup of $G$ with the required order.

Consider the map $\phi: H \rightarrow H_1 \times  H_2 \times  ... \times H_s, \phi(h_1h_2...h_s) = (h_1, h_2, h_3, ..., h_s)$.

$\phi$ is a group homomorphism: Obvious since $G$ is abelian.

$\phi$ is injective: ker$\phi = \{h_1h_2h_3...h_s \in H \mid h_1 = h_2 = ... = h_s = id \} = \{id\}$

$\phi$ is surjective: Obvious by construction.

Then we have shown $\phi$ is a group isomorphism between $H < G$ and $H_1 \times  H_2 \times  ... \times H_s$. Then $H$ is the subgroup of $G$ with order $m$.

Title: Existence of subgroup of Abel group


问:
If $a<b$ why is $a \le b$
I recently showed a problem to someone and they stated the following:

If $a<b$ then $a\le b$.

I find this very confusing how can if we stated that $a$ is strictly less than $b$ how can we then say that $a$ is less than $b$ and could be equal to $b$?
For example if I wanted to solve an equation where a strict inequality was stated putting $x\le b$ would be incorrect right?

答: $a\le b$ is an abbreviation for $a<b\lor a=b$ ($\lor$ denoting the inclusive or, which satisfies $p\implies(p\lor q)$).

Title: If $a<b$ why is $a \le b$


答:
Define $\leq$ first say on the set of integers:

$a\leq b :\Leftrightarrow \exists c\in\Bbb Z_{\geq 0}:\; a+ c =b$.

Then define $<$ as follows

$a<b :\Leftrightarrow a\leq b \wedge a\ne b.$

Title: If $a<b$ why is $a \le b$

Body:
I recently showed a problem to someone and they stated the following:

If $a<b$ then $a\le b$.

I find this very confusing how can if we stated that $a$ is strictly less than $b$ how can we then say that $a$ is less than $b$ and could be equal to $b$?
For example if I wanted to solve an equation where a strict inequality was stated putting $x\le b$ would be incorrect right?


问:
Lifting maps of quotient modules
Today I tried to check this, but couldn't see how to do it. I think it is probably a standard result, but a brief check of Atiyah-Macdonald didn't yield anything, and I don't know what to google for. A reference is also appreciate.

Consider a associative, unital, $K$-algebra $R$ and a nontrivial ideal $\bar{R}$. Now consider two $R$ modules $B$ and $B'$. Also consider the quotients $\bar{B}=B/\bar{R}B$ and $\bar{B'}=B'/\bar{R}B'$.

If we have a map $\varphi:\bar{B}\to\bar{B}'$, is there some condition that we can lift to a map of $B\to B'$?

If the quotient map on $B'$ splits, we have this lift, but this is not iff.

Thanks in advance!

答:
Here is one way to look at this problem.  I will write $I$ for $\bar{R}$.

Firstly, $B/IB = (R/I) \otimes_R B$, that is, $B/IB$ is the image of $B$ under the induction functor (change of rings) $R$-mod $\to R/I$-mod.  This functor will be written $\uparrow$.  Frobenius reciprocity tells us

$$ \hom_{R/I}( B\uparrow, B'\uparrow) \cong \hom_R (B, B'\uparrow\downarrow) $$
where $\downarrow$ is restriction $R/I$-mod $\to R$-mod.  We'll make this an identification, so we consider $\phi$ as an element of the right hand side.  The question is then: "is $\phi$ in the image of $\hom_R(B,B') \to \hom_R(B,B'\uparrow \downarrow)$ ?" (the map on homs arises from $B' \to B\uparrow \downarrow$, the morphism given by the universal property of induction).

This stuff fits into the long exact sequence obtained by applying $\hom_R(B, -)$ to $IB' \to B' \to B'\uparrow\downarrow$.  We get

$$0\to \hom_R(B,  IB') \to \hom_R(B,  B') \to \hom_R(B,  B'\uparrow\downarrow) \stackrel{\omega}{\to} \operatorname{Ext}^1 _R(B, IB') \to \cdots $$

$\phi$ being in the image is equivalent to its being in the kernel of the connecting homomorphism $\omega$, which is multiplication by the short exact sequence above.  So one answer to your question is:  $\phi$ lifts if and only if it is killed by multiplication by the short exact sequence.

Really though, this is just a restatement of the problem.  Clearly $B$ being projective is enough to guarantee the lift exists, as is any assumption that makes that ext group $\operatorname{Ext}^1 _R(B, IB')$ vanish.  But for arbitrary $\phi$ I don't know general conditions other than the one above.  It's sufficient for the induction functor to be full.  $R/I$ being flat over $R$ (so induction is exact) certainly isn't enough.

Title: Lifting maps of quotient modules


问:
partials of a PDF with no closed form solution
I need to estimate partial derivatives for all N parameters denoted $\theta_{N}$ of a probability density function(PDF) $\mathcal{f}$.

This PDF $\mathcal{f}$ has no closed form solution and is instead expressed in terms of a characteristic function $\phi$ and an FFT $\mathcal{F}^{-1}$.

$\mathcal{f}(x) = \mathcal{F}^{-1} [\phi(x)]$

Is it possible to estimate the partial derivative with respect to $\theta_{N}$, $\frac {\partial \mathcal{f}} {\theta_{N}}$  for all N parameters?

答:
$$
\frac{\partial f}{\partial\theta_{N}}=\mathcal F^{-1}\left(\frac{\partial \phi}{\partial\theta_{N}}\right)
$$</p>


问:
Integration of weak $L^p$ norm
I want an accurate understanding of weak norm.

The definition of weak $L^p$:
$$ \left\lVert f \right\rVert_{p,\infty}=\sup_{a>0} a \mu \{|f|>a\}^{1/p}.  $$

The question is:

Let $v$: velocity, $x$: location in $\mathbb{R}^3$, $t$: time variable, $E(t, x) : \mathbb{R}^4 \rightarrow \mathbb{R}^3$.

Show that
$$\left\lVert E(t,x-s\cdot) \right\rVert_{\frac{3}{2},\infty}=s^{-2} \left\lVert E(t,\cdot) \right\rVert_{\frac{3}{2},\infty} $$
(In this problem, how to change of variable without integration? Can I use Jacobian?)

答:
$$ \{y : sy-x \in A \} = \tfrac1s (A+x) $$ and $$\mu\left(\tfrac1s (A+x)\right) = \tfrac1{s^3} \mu(A) .$$
Now let $$A = \{|E(t,\cdot)| > a\}.$$

Title: Integration of weak $L^p$ norm


问:
Does every subgroup of an abelian group have to be abelian?
My original problem is to show that E/L is an abelian extension over L and L/F is an abelian extension over F, given that E/F is an abelian extension over F and that L is a normal extension of F such that $F\subseteq L \subseteq E$.

So far I have proved that E is a normal extension of F, E is a normal extension of L, and L is a normal extension of F. I know that to prove abelian extension I must also prove that Gal(E/L) is an abelian group. I have shown that Gal(E/L) $\subseteq$ Gal (E/F). In my mind it makes sense that I cannot lose commutativity therefore my subgroup must be Abelian too. How do I show this in a proof? Is it enough to show two elements in the subgroup must also exist in the larger group and that they must be commutative in the larger group? I feel like I know what needs to be done, just not how to phrase it.

答: Just use proof by contradiction. Suppose $H$ is not abelian and thus contains two non-commuting members $x$ and $y$. Then $xy \neq yx$. But $x$ and $y$ are also in $G$, and thus $G$ is not abelian. Contradiction. Therefore $H$ is abelian.


问:
Does the series $\sum_{n=2}^{\infty}{\frac{\sin(n^3)}{\ln(n)}}$ converge?
Do you have any idea if the series $\sum_{n=2}^{\infty}{\frac{\sin(n^3)}{\ln(n)}}$ converges? I am totally lost!</p>

答:
The series is divergent.

We adopt the result and extend the idea of this posting by Terry Tao: https://mathoverflow.net/questions/201250/is-sum-k-1n-sink2-bounded-by-a-constant-m?sfb=2

The result of the posting is $\sum_{n\leq x} \sin n^3$ is unbounded.

By the unboundedness of $\sum_{n\leq x} \sin n^3$, the abscissa of convergence  $\sigma_c$
of $\sum (\sin n^3)/n^s$ satisfies $\sigma_c\geq 0$. If we assume that the series is convergent, then we have the convergence of the Dirichlet series $\sum (\sin n^3)/n^s$ for $\Re(s)>0$. Thus, $\sigma_c=0$ and we have
$$
\limsup_{x\rightarrow\infty} \frac{\log |\sum_{n\leq x} \sin n^3|}{\log x} =0.
$$
Hence, if the series is convergent, then for any $\epsilon>0$ and $x\geq 1$, there is $C_{\epsilon}>0$ such that
$$
|\sum_{n\leq x} \sin n^3|\leq C_{\epsilon} x^{\epsilon}. \ \ (1)
$$
We will obtain a contradiction to this statement.

By this posting by David Speyer (based on Weil's inequality): Convergence of $\sum \limits_{n=1}^{\infty}\sin(n^k)/n$
we have an absolute constant $\delta>0$ such that the following hold.

Define $X_1=\sin((k+1)^3)$, $\ldots$, $X_h=\sin((k+h)^3)$ be random variables (as empirical distributions with corresponding outputs for $k=1,\ldots n$ with probability $1/n$ each). We have
$$
EX_i=O(n^{-\delta}), \ \text{Var}(X_i)=\frac12 + O(n^{-\delta}),
$$
and
$$
\text{Var}\sum_{i\leq h} X_i = \frac h2 + O(hn^{-\delta})+2\sum_{1\leq i<j\leq h} \text{Cov}(X_i,X_j)$$
$$
=\frac h2+O(hn^{-\delta})+O(h^2 n^{-\delta}).$$
Let $h=\lceil n^{\delta/2}\rceil$. Assuming all length $h$ partial sums are $O(h^{1/4})$, we have by comparing two estimates of $\text{Var}\sum_i X_i$,
$$
\frac h2 + O(h^2 n^{-\delta}) = O(h^{1/2}).
$$
These estimates do not agree as $n\rightarrow\infty$. Thus, there is an absolute constant $c>0$ and some length $h$ partial sum of $\sin k^3$ which is $\geq c h^{1/4}$. That is, there is $k\leq n$ such that
$$
|\sum_{i\leq h} \sin (k+i)^3| \geq c h^{1/4}\geq c n^{\delta/8}.
$$
Since $\sum_{i\leq h} \sin (k+i)^3 = \sum_{j\leq k+h} \sin j^3 - \sum_{j\leq k} \sin j^3$, one of the two partial sums' absolute value on the RHS must satisfy $\geq (c/2) n^{\delta/8}$.

If we take $\epsilon=\delta/9$ in (1), then we have a contradiction.

Therefore, the series is divergent.

Remark

We may explicitly write $\delta$ in terms of the irrationality measure of $\pi$.


问:
prove or disprove on bounded variation
Prove or disprove :

If a function $f$ is bounded on $[\alpha,\beta]$ and Riemann integrable on $[\alpha,\beta]$ then it is integral function $F(x)= \int_{\alpha}^{x} f(t) dt$
is of bounded variation on $[\alpha,\beta]$.

My attempt:

I tried to find a counter example on a function $F(x)$ which is bounded on $[\alpha,\beta]$ and continuous but it's not bounded variation on $[\alpha,\beta]$
and it's derivative is bounded on $[\alpha,\beta]$ and Riemann integrable
But almost I didn't find any example satisfying the above conditions.

答:
$F(x)$ must be of bounded variation thats why you couldnt find any counterexample. To see this, observe that if $|f(t)|\leq M$ for every $t\in [\alpha,\beta]$ then, for every $x,y$
\begin{align}
|F(x)-F(y)|&=\biggl|\int_{\alpha}^x f(t)\,dt-\int_{\alpha}^y f(t)\,dt\biggr|\\
&=\biggl|\int_{x}^y f(t)\,dt\biggr|\\
&\leq \int_{x}^y |f(t)|\,dt\\
&\leq M|x-y|
\end{align}
This shows that $F$ is Lipschitz continuous with constant $M$. Now using the definition of bounded variation try to prove that every Lipshcitz continuous function on $[\alpha,\beta]$ is of bounded variation.

Title: prove or disprove on bounded variation


答:
Since $f$ is bounded on $[a,b],$ there exists $M > 0$ such that
$$ \vert f(t) \vert < M \qquad \forall t \in [a,b]$$
Therefore $F$ has bounded variation since
\begin{align*}
\sum_{i = 0}^{n-1} \vert F(x_{i+1}) -F(x_i)\vert  & = \sum_{i=0}^{n-1}\left \vert \int_{x_i}^{x_{i+1}} f(t)dt\right\vert \\
&\leq \sum_{i = 0}^{n-1} \int_{x_i}^{x_{i+1}} \vert f(t)\vert dt \\
&\leq \sum_{i = 0}^{n-1} \int_{x_i}^{x_{i+1}} M dt \\
&= M \int_a^b dt \\
&= M(b-a)
\end{align*}
for any partition $P = \{x_0,...,x_n \}$ of $[a,b]$.

Title: prove or disprove on bounded variation

Body:
Prove or disprove :

If a function $f$ is bounded on $[\alpha,\beta]$ and Riemann integrable on $[\alpha,\beta]$ then it is integral function $F(x)= \int_{\alpha}^{x} f(t) dt$
is of bounded variation on $[\alpha,\beta]$.

My attempt:

I tried to find a counter example on a function $F(x)$ which is bounded on $[\alpha,\beta]$ and continuous but it's not bounded variation on $[\alpha,\beta]$
and it's derivative is bounded on $[\alpha,\beta]$ and Riemann integrable
But almost I didn't find any example satisfying the above conditions.


问:
How to prove this function is increasing $\frac{e^x - x -1}{x^2}$
How to prove that the function $\displaystyle h(x) = \frac{e^x - x - 1}{x^2} = \sum_{k=2}^{\infty} \frac{x^{k-2}}{k!}$ is increasing?

Edit for context: I was asking because this function appears in Wainwright's high dimensional statistics textbook, as part of proving a Bernstein type probability inequality for random variables that are bounded on one side.

答:
For $x\not =0$ we have :

$$\int_0^1 v\exp(uvx)du=\frac{\exp(vx)-1}{x}$$ and
$$\int_0^1 \frac{\exp(vx)-1}{x}dv =\frac{1}{x}(\frac{\exp(vx)}{x}-v)_0^1=\frac{\exp(x)-1-x}{x^2}$$
Hence with $D=[0,1]\times [0,1]$
$$\int \int_Dv\exp(uvx)dudv= \frac{\exp(x)-1-x}{x^2}$$ and this is also true for $x=0$. Now use that the function $x\to v\exp(uvx)$ for fixed $u,v\in [0,1]$is increasing.


问:
Let $G$ be a group and $H,K \le G$,then - If $H \le N_{G}(K)$ then $HK \le G$ - If $H \le K$ then $ C_{G}(K) \le C_{G}(H)$
Let $G$  be a group and $H,K \le G$,then show that:

If $H \le N_{G}(K)$ then $HK \le G$

If $H \le K$ then $ C_{G}(K) \le C_{G}(H)$

Where $C_{G}(H)$ is the centralizer of $H$ defined as:

$${\displaystyle \mathrm {C} _{G}(H)=\{g\in G\mid gh=hg{\text{ for all }}h\in H\}.}$$

And $N_{G}(K)$ is the normalizer  of $K$ defined as:

$${\displaystyle \mathrm {N} _{G}(K)=\{g\in G\mid gK=Kg\}.}$$

For the first part I have that:

$N_G(K)$ is the normalizer of $K \le G$.  It's the set of all $g$ such that $gKg^{-1} = K$.

Since $H \subset N_G(K)$, then for all  $h \in H$ we know that they commute  with $K$ like the $g$. Hence for all $h$ in $H$ we have that $hK = Kh, $  which means $HK = KH$.

On the other hand from this link we see that $HK \le G$,as desired.

The second part is the part that I don't know how to prove,I know that for $H \le K$ we have that $C_G(H)<N_G(H)$,but does that help?

答:
Take an arbitrary element $c \in C_G(K)$. We want to show that $c \in C_G(H)$.

Note that $c \in C_G(K) \implies ck = kc, \forall k \in K$. But since $H \le K$, we have $ch = hc, \forall h \in H \implies c \in C_G(H)$.


问:
Nonlinear second order differential equation involving sine
I need to solve this ODE:

$A +B\frac{dy}{dt}+C\sin(y)+\frac{d^2y}{dt^2} = 0$

where A, B, and C are constants, and  $y$  is  $y(t)$.

I manage to reduce the order of the ODE by using dimension analysis to:

$A +B\frac{dy}{dt}+\sin(y) = 0$

where A and B are constant, and $y$ is $y(t)$

I would like to know the analytic solution for the 1st order ode (and if it's a possibility for the 2nd order as well).

I'm working on a numerical solution just in case there is no solution for that problem, in the meanwhile.

Thanks in advance!

答:
$$y''(x)+By'+c \sin y +A=0~~~~(1)$$
Let $y'=p$, then
$$\frac{dp}{dx}+Bp+c \sin y+A=0~~~~(2)$$
$\frac{dp}{dx}=p\frac{dp}{dy}$, then
$$p\frac{dp}{dy}+Bp+C \sin y+A=0$$
$$pdp+(Bp+C \sin y+A)dy=0~~~(3)$$
Multiply (3) by $e^{By}$
$$p~e^{By} dp+e^{By}(Bp+C \sin y+A)dy=0~~~(3)$$
(3) becomes an Exact Eq.
Whose solution is
$$\int pe^{By} dp \text{(treat $y$ as const)}+\int e^{By}(C \sin y+A) dy=G$$
$$\implies e^{By}\frac{p^2}{2}+e^{By}\left( \frac{B^2C \sin y- BC \cos y+A+AB^2}{B+B^3}\right)$$
$$\implies p=\frac{dy}{dx}=\sqrt{\sqrt{2}G+\sqrt{2}\frac{Bc\sin y-B^2C \sin y-A-AB^2}{(B+B^3)}}$$
Given the values of A,B,C, this integral can be solved to get $y(x)$ with two constants $G$ and one more $H$.


问:
Minimize difference of difference of binomial coefficients
Let
$$ f_n(i) = \binom{n}{i} - \binom{n}{i-1}$$
Is there an expression in terms of $n$ for
$$\min_{1 \leq i \leq n/2} f_n(i)$$
I know $f_n(1)=n-1$ but this does not equal the minimum value for all $n$, for example $f_4(2) = 2$, but I'm hoping I can find something in terms of $n$.
Maybe
$$f_n(i) = \Big( 1 - \frac{2i}{n+1} \Big) \binom{n+1}{i}$$
is useful?

答:
Just as @K.defaoite, I thought that the problem was to maximize
$$f_n(k) = \binom{n}{k} - \binom{n}{k-1} \qquad \text{for} \qquad 1 \leq k \leq \frac n 2$$

So, as @K.defaoite did, I shall try to work the maximization.

Using the gamma function and differentiating, we have to find the solution of
$$\frac{k^2+(k-n-1)^2}{k (k-n-1) (2 k-n-1)}=\psi (-k+n+1)-\psi(k) $$ Writing the second order Taylor expansion of the above expression around $k=\frac n2$ we need to solve for $k$ the quadratic equation
$$a \left(k-\frac{n}{2}\right)^2+b \left(k-\frac{n}{2}\right)+c=0$$ where
$$a=-\frac{\left(n^2+2 n-2\right) \left(n^2 \psi ^{(1)}\left(\frac{n}{2}\right)-4\right)}{n^2}$$
$$b=n-\frac{4}{n}+\frac{1}{2} (n+2) n \psi ^{(1)}\left(\frac{n}{2}\right)-1\qquad \text{and} \qquad c=\frac {n(n+1)}{2}$$ Retain the smallest solution (this is only an estimate)
$$k\sim \frac {n} 2+\frac{b-\sqrt{b^2-4 a c}}{2 a}$$

For illustration, let $n=10^p$ and compare
$$\left(
\begin{array}{ccc}
p & \text{approximation} & \text{solution} \\
 1 & 3.8284441 &  3.7898465 \\
 2 & 45.474367 &  45.458418 \\
 3 & 484.68068 &  484.67544 \\
 4 & 4950.4975 &  4950.4958 \\
 5 & 49842.385 &  49842.385
\end{array}
\right)$$

For very large values of $n$, from the quadratic equation, the asymptotics would be
$$k\sim \frac {n+1} 2-\frac 12 \sqrt n \left(1+\frac{1}{2n}+\frac{37}{24 n^2}+O\left (\frac {1} {n^{5/2}} \right) \right)$$

Using this result as base point for a new expansion of the equation to solve gives
$$k\sim \frac {n+1} 2-\frac 12 \sqrt n \left(1+\frac{5}{6 n}-\frac{41}{360 n^2}-\frac{1721}{3024 n^3}+O\left(\frac{1}{n^4}\right)\right)$$

Used for $n=10$, this gives $k=3.7898002$

Edit

In view of the previous result, we can even do better : let $k=\frac{n+1-x\sqrt{n} }{2} $ which makes the equation to solve
$$2\frac{ n \left(n+2+x^2\right)+1}{x\sqrt{n}  \left((n+1)^2-n x^2\right)}=H_{\frac{1}{2} \left(n-1+x \sqrt{n}\right)}-H_{\frac{1}{2} \left(n-1-x \sqrt{n}\right)}$$

Expanded as series around $x=0$, we have to solve for $x$
$$2=\sum_{p=1}^\infty a_p x^{2p}$$ where the  coefficients are
$$a_p=4n^p\Bigg[\frac { \psi ^{(2 p-1)}\left(\frac{n+1}{2}\right)} {4^p\, (2 p-1)! }-\frac {1}{(n+1)^{2 p} }\Bigg]$$

Now, using series reversion
$$k\sim \frac {n+1} 2-\frac   {\sqrt n}2 \left(1+\sum_{k=1}^6 \frac{a_k}{n^k}+O\left(\frac{1}{n^7}\right)\right)$$ where the coefficients $a_k$ form the sequence
$$\left\{\frac{5}{6},-\frac{61}{360},-\frac{227}{5040},-\frac{23483}{1814400},\frac{44933951
   }{119750400},-\frac{3444044587}{130767436800}\right\}$$

Trying for $n=2^p$, the results
$$\left(
\begin{array}{ccc}
 1 & 0.524771687500380499473261 &  0.529176960274205650239913\\
 2 & 1.302651240758724990905955 &  1.302745705191306338073585 \\
 3 & 2.942329575624503533168965 &  2.942330978833762361174524 \\
 4 & 6.397178792509027687945137 &  6.397178809463434940801070 \\
 5 & 13.59838783848710969755951 &  13.59838783867800346779702 \\
 6 & 28.44808282869740112407188 &  28.44808282869951434549459 \\
 7 & 58.80637589759390516640941 &  58.80637589759392849270466 \\
 8 & 120.4739790389675902908572 &  120.4739790389675905483369 \\
 9 & 245.1678845786669264194795 &  245.1678845786669264223226 \\
 10 & 496.4869817528550288338029 &  496.4869817528550288338343 \\
 11 & 1001.863376796720752704879 &  1001.863376796720752704879 \\
 12 & 2016.493489906543937479655 &  2016.493489906543937479655 \\
 13 & 4051.240562558556105524810 &  4051.240562558556105524810 \\
 14 & 8128.496744832066025772882 &  8128.496744832066025772882 \\
 15 & 16293.98803024251845644220 &  16293.98803024251845644220 \\
 16 & 32640.49837240088319177524 &  32640.49837240088319177524 \\
 17 & 65355.47951312808588722713 &  65355.47951312808588722713 \\
 18 & 130816.4991861985478970517 &  130816.4991861985478970517 \\
 19 & 261782.4607525877391722967 &  261782.4607525877391722967 \\
 20 & 523776.4995930990372370715 &  523776.4995930990372370715
\end{array}
\right)$$

Final point

Concerning the maximum value, using the above, we have
$$\frac{2 }{\sqrt{ne} }\binom{n}{\frac{n}{2}}\left(1-\frac{7}{12 n}+\frac{239}{480 n^2}-\frac{184189}{362880
   n^3}+O\left(\frac{1}{n^4}\right) \right)$$

Used fo $n=100$, this gives a maximum value of $1.216794929\times 10^{28}$ while a rigorous optimization leads to
$1.216794935\times 10^{28}$.


问:
Is there a equation that can make the answer negative when it is odd, and positive when it is even?
One of the exercises I got asked how it would be possible to build a set such that $B = \{−1,\frac{1}{2},\frac{−1}{3},\frac{1}{4},...\}$. Is there an equation that can make the answer negative when it is odd, and positive when it is even?

答:
A simple, well known example of this is
$$
f(n) = (-1)^{n+1}
$$


问:
Understanding a particular proof that a Hermitian matrix has an orthonormal basis whose diagonal contains its eigenvalues
From pg. 36 of Shankar's Principles of Quantum Mechanics:

Here in red (which confuses me) Shankar seems to be appealing to this proof:

Lemma: If an n-dimensional square operator has an eigenpair, it necessarily has (n) (not necessarily distinct) eigenvalues.

Lemma proof sketch:

Suppose $Ω |V\rangle = ω |V\rangle$ (that is assume $Ω$ has at least one eigenpair $|V\rangle$ and $ω$).

Then $\det(Ω - ω I) = 0$, else $Ω - ω I)$ would be invertible, leading to a contradiction.

But $\det(Ω - ω I)$ can be expanded to a characteristic polynomial, which when set to 0 contains $n$ roots (a fundamental fact from complex analysis).

Hence these $n$ (not necessarily distinct) roots must be eigenvalues via (2).

NOTE: The full details of this proof, as actually presented in the Shankar's book, can be found here.  I suspect there's a chance I've misunderstood it, though I don't see how?

Problem 1: I am having trouble understanding this argument, since it seems to me this might merely imply that $Ω$ doesn't have any eigenpairs. That is, (4) may simply be false, since this would only imply that (1) is false. Or put differently: the above proof merely seems to me to suggest the claim that "If (1) is true, then (4) must be true", but (1) might not be true!

The proof continues with more confusion for me:

Problem 2: Why must the first column of the basis be $[ω_1, 0, …, 0]^T$? This seems equivalent to saying that

$$
\Omega | \omega_1 \rangle = \omega_1 |1\rangle
$$

but I don't see why this must be so.  Perhaps it has something to do with the following two facts, but I don't quite see the connection?

$\Omega | ω_1 ⟩ = ω_1 | ω_1 ⟩$

$‖ω_1 | ω_1 ⟩‖ = |ω_1| ‖ |ω_1 ⟩‖ = |ω_1| 1 = |ω_1|$

答:
Problem 1: What is left implicit here is that the condition $\det(\Omega - \omega I) = 0$ is in fact equivalent to $\omega$ being an eigenvalue of $\Omega$. Indeed, we have
\begin{aligned}
\det(\Omega - \omega I) = 0 &\iff \Omega - \omega I\text{ is not an isomorphism} \\
&\iff \Omega - \omega I \text{ is not injective} \\
&\iff \text{ there is a vector $V$ with }(\Omega - \omega I)V = 0 \\
&\iff \text{ there is a vector $V$ with } \Omega V = \omega V
\end{aligned}
So it is really true that a complex number $\omega$ is an eigenvalue of $\Omega$ if and only if it is a zero of the characteristic polynomial.

Problem 2: The first column being $[\omega_1, 0, \dots, 0]^T$ just means that $\Omega \vert\omega_1\rangle = \omega_1 \vert\omega_1\rangle$. Recall: what does it mean to write a linear transformation as a matrix? It means that for every element $e_i$ of some chosen basis $\{e_1, \dots, e_n\}$, you write each $\Omega e_i$ as
\begin{aligned}
\Omega e_i = \sum_{j=1}^n \Omega_{ij} e_j,
\end{aligned}
and the coefficients $\Omega_{ij}$ will be the matrix coefficients appearing in the $i$-the column. In this case, we choose the basis $\{\lvert \omega_1 \rangle, V^1, \dots, V^{n-1}\}$, and we have
\begin{aligned}
\Omega \vert\omega_1\rangle = \omega_1 \vert\omega_1\rangle = \omega_1 \vert\omega_1\rangle + 0 V^1 + \dots + 0 V^{n-1},
\end{aligned}
and thus the matrix coefficients on the first column are $[\omega_1, 0, \dots, 0]$.

Note that there is no mention of a vector $\vert 1 \rangle$.


问:
Proving a $Z- $transform
I am having trouble demonstrating the $Z$ transform of $a^{n-1}u(n-1)$ is $\frac{1}{(z-a)}$, as it says in this table.

I try using the definition of the z transform, but it comes out different than what the table says:

$$\sum_{k=-\infty}^{\infty}a^{k-1}u(k-1)z^{-k}=\sum_{k=1}^{\infty}a^{k-1}z^{-k}=\frac{1}{a}\sum_{k=1}^{\infty}(\frac{a}{z})^k$$

This is a geometric series, so the result should be:

$$\frac{1}{a}\frac{1}{1-\frac{a}{z}}=\frac{z}{a(z-a)}$$

Why is my answer coming out different than what it says in the table?

答:
The geometric series $\sum_{k\geq k_0} r^k$ converges if and only if $|r|<1$. You have a confusion regarding its actual sum, which depends where the summation starts. In particular
$$
\sum_{k\geq 1}r^k=\frac{r}{1-r}\qquad\mbox{and}\qquad\sum_{k\geq 0}r^k=\frac{1}{1-r}.
$$
More generally,
$$
\sum_{k\geq k_0}r^k=\frac{r^{k_0}}{1-r}.
$$
To see where this all comes from, you just need to observe that the partial sums satisfy, for $K\geq k_0$:
$$
\sum_{k=k_0}^Kr^k=r^{k_0}(1+r+\ldots+r^{K-k_0})=r^{k_0}\left(\frac{1-r^{K-k_0+1}}{1-r}\right)=\frac{r^{k_0}-r^{K+1}}{1-r}.
$$

In your case, you do get
$$
\frac{1}{a}\sum_{k\geq 1}\left(\frac{a}{z}\right)^k=\frac{1}{a}\cdot \frac{\frac{a}{z}}{1-\frac{a}{z}}=\frac{1}{z-a}.
$$

Title: Proving a $Z- $transform


答:
The function should be extended by continuity at $0$ with $h(0)=1/2$ and it becomes differentiable at $0$ with $h'(0)=1/6$.

For $x\ne0$, it holds
$$
h'(x)=\dfrac{(x-2)e^x+x+2}{x^3}
$$
and we can consider $g(x)=(x-2)e^x+x+2$. Then $g'(x)=(x-1)e^x+1$ and $g''(x)=xe^x$. Thus we infer that $g'$ has a minimum at $x=0$, with $g'(0)=0$. Therefore we have $g'(x)>0$ for $x\ne0$, which proves that $g$ is strictly increasing.

Since $g(0)=0$, we conclude that $g(x)<0$ for $x<0$ and $g(x)>0$ for $x>0$.

Therefore $h'(x)>0$ for every $x$.

Title: How to prove this function is increasing $\frac{e^x - x -1}{x^2}$

Body:
How to prove that the function $\displaystyle h(x) = \frac{e^x - x - 1}{x^2} = \sum_{k=2}^{\infty} \frac{x^{k-2}}{k!}$ is increasing?

Edit for context: I was asking because this function appears in Wainwright's high dimensional statistics textbook, as part of proving a Bernstein type probability inequality for random variables that are bounded on one side.


答:
$$y''+by'+c\sin y+a=0$$
Multiply it by $2y'$ to get
$$2y'y''+2by'^2+2c\sin y y'+2ay'=0$$
$$\implies \frac{dy'^2}{dx}+2by'^2=-(2c\sin y+2a) \frac{dy}{dx}.$$
$$\frac{d (e^{2by}y'^2)}{dx}=-2(c\sin y+2) \frac{dy}{dx} e^{2by}$$
Integrate both sides w.r. $x$, to get
$$y'^2e^{2by}=-2\int e^{2by}(c\sin y+a) dy$$
$$y'^2=-2e^{-2by}\int e^{2by}(c\sin y+a) dy +ge^{-2by}$$
From here one can carry on.

Title: Nonlinear second order differential equation involving sine

Body:
I need to solve this ODE:

$A +B\frac{dy}{dt}+C\sin(y)+\frac{d^2y}{dt^2} = 0$

where A, B, and C are constants, and  $y$  is  $y(t)$.

I manage to reduce the order of the ODE by using dimension analysis to:

$A +B\frac{dy}{dt}+\sin(y) = 0$

where A and B are constant, and $y$ is $y(t)$

I would like to know the analytic solution for the 1st order ode (and if it's a possibility for the 2nd order as well).

I'm working on a numerical solution just in case there is no solution for that problem, in the meanwhile.

Thanks in advance!


问:
How to simplify $\operatorname{P} ( { C |B,A} )P( { B |A} )P( A ) + P( { {\bar B} |A} )P( A )$?
How to simplify the following probability

$\operatorname{P} ( { C |B,A} )P( { B |A} )P( A ) + P( { {\bar B} |A} )P( A )$

$ = P\left( {A,B,C} \right) + P\left( {A,\bar B} \right)$

Can $P\left( {A,B,C} \right) + P\left( {A,\bar B} \right)$ be further simplify ?

Also, are there any systematic way to do the simplification instead of following heuristic. I see that there are some link between Boolean algebra and the simplification of these problem.

Particularly, my thinking is that $ = P\left( {A,B,C} \right) + P\left( {A,\bar B} \right)$ may be somehow related to
$F\left( {A,B,C} \right) = \left( {ABC} \right)\,OR\,\left( {A\bar B} \right)$ but I do not know if my guts feeling is correct or not.

Please help me clear out these doubt, thank you for your enthusiasm !

Body: Hint: $ P(C|B,A)P(B|A)P(A)=P(A,B,C)$.


问:
Binomial distribution (i think)
This may seem like a simple question but I honestly don't really know how I would go about solving it.
So essentially the question is as follows:

2 players play a match whereby the winner is the first person to win n games.

the probability of player 1 winning a single game is 0.835
the probability of player 2 winning a single game is 0.165

what is the smallest value for n such that the probability that player 1 is the overall winner of the match is at least 0.9

I feel like i need to use binomial distribution to solve the question but i wasn't able to do so.
Any help will be greatly appreciated.

答:
Bram28's hints point in the right direction. Still, a more general formulation would be helpful for posterity.

Say that player 1 wins the match (i.e., they are the first to win $n$ games), and player 2 wins $k < n$ games. The probability that this happens (given that player 1 wins a game with probability $p$) is

$$\text{Prob}_{n, k}(p) = p\binom{n+k-1}{n-1} p^{n-1} q^{k}$$

We use $\binom{n+k-1}{n-1}$ rather than $\binom{n+k}{n}$ because if player 1 wins the match, they must win the final game and thus there are only the remaining $n-1$ game wins to be distributed amongst the $n+k-1$ possible game slots. The initial factor of $p$ is the probability of player 1 wining the final game; the $p^{n-1}q^{k}$ factor is the probability of player 1 wining $n-1$ and losing $k$ prior games.

The probability that player 1 is the first to win $n$ games is the sum of the above quantity over the possible values of $k$, that is

$$\text{Prob of player 1 winning match} =   p^n\sum_{k=0}^{n-1}\binom{n+k-1}{n-1}  q^{k}. $$

We could find a closed form expression for the summation by differentiating the identity for a finite geometric series, but here it's easier to just plug in values. For $n=1$, we find 0.835 which isn't sufficient. For $n=2$, we find

\begin{align}
\text{Prob of player 1 winning match} &= (0.835)^2\left[\binom{2-1}{2-1} (0.165)^0 + \binom{2+1-1}{2-1}(0.165)^1\right]\\
& = 0.9273,
\end{align}

which is.

Title: Binomial distribution (i think)


问:
Proving an inequality: $|1-e^{i\theta}|\le|\theta|$
We have been using this result without proof in my class, but I don't know how to prove it. Could someone point me in the right direction?

$$|1-e^{i\theta}|\le|\theta|$$

I believe this is true for all $\theta\in\mathbb{R}$. It is easy to show that the left side is bounded by 2 (triangle inequality), but I'm stuck otherwise.

答: Geometically, $|1-e^{i\theta}|$ can be viewed as the distance between $1$ and $e^{i\theta}$, which is smaller than the arc from $1$ to $e^{i\theta}$(portion of the unit circle).</p>


问:
What is heavy tailed distribution?
I want to understand the heavy tailed distribution. I took help of Internet but I'm not able to comprehend mathematical definition of heavy tailed distribution. Can anyone explain me the Heavy-tailed distribution in layman terms with some example.</p>

答: Say we have a continuous distribution. Its tails are the values below $\underline x$ or above $\bar x$. If a distribution is heavy tailed, the probabilities $P(x<\underline x)$ or $P(x>\bar x)$ are large, i.e., larger than those of the exponential distribution

Title: What is heavy tailed distribution?


问:
Surjectivity of the degree homomorphism
Regarding the following question and discussion:

Is the degree homomorphism $\text{deg}: \text{Pic}(X)\to \mathbb{Z}$ surjective?

We agree that if $X$ is a curve over an algebraically closed field the answer is yes, and not allways otherwise.
What is the answer in the case that $X$ is a curve over a separably closed field?

Is this a sufficient condition for the degree homomorphism to be surjective?

答:
The answer is yes, assuming you take "curve" to mean "geometrically reduced". Every geometrically reduced $k$-scheme of finite type has a $k^s$ point by proposition 3.2.20 of Liu's Algebraic Geometry and Arithmetic Curves (the idea of the proof is that geometrically reduced means $k(X)/k$ is separable, and the locus of such behavior is open for schemes of finite type over a field). This proves your claim, assuming you take geometrically reduced as part of your definition of a curve: pick a regular $k^s$ point.

You may also consult Stacks 04QM for details on the statement above.

Title: Surjectivity of the degree homomorphism


问:
How can I find $6N^2 + 3NS + S^2 - 21N -61$, without finding N and S itself?
Let $N$ and $S$ be the roots of $A(D) = 5D^2 - 21D - 61$. How can I find $6N^2 + 3NS + S^2 - 21N -61$, without finding N and S itself?
I know how to get $N^2 + S^2$ from $(N+S)^2 - 2(NS)$ via Vieta's formulas but that's about it, I'm not sure how to proceed from here.

答:
Here is a systematic way.

Consider $P=6N^2 + 3NS + S^2 - 21N -61$ and $A(N)$ as polynomials in $N$. Let $R$ be the remainder of $P$ mod $A(N)$. Now consider $R$ and $A(S)$ as polynomials in $S$. Compute the remainder of $R$ mod $A(S)$. The result is
$$
3 NS + 21(N+S)/5 + 122/5
$$
and you can now use Vieta's formulas.


问:
There exist $\beta \in$ Aut$(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z})$ such that $\phi'(f)=\beta \circ \phi(f) \circ \beta^{-1}$
Let $\phi, \phi' : \mathbb{Z}/3\mathbb{Z} \rightarrow$ Aut$(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z})$ 2 non-trivial homomorphisms. Show that there exist $\beta \in$ Aut$(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z})$ such that $\phi'(f)=\beta \circ \phi(f) \circ \beta^{-1}$ $\forall f \in \mathbb{Z}/3\mathbb{Z}$. We know that Aut$(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}) \cong GL(2,\mathbb{F}_2) \cong S_3$ and that we must have $\phi([1]_3) = \sigma \in S_3$ such that $\sigma$ is a 3-cycle (if not we have a trivial homomorphism). Then i have problem to show that there exist $\beta \in$ Aut$(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z})$ such that $\phi'(f)=\beta \circ \phi(f) \circ \beta^{-1}$ $\forall f \in \mathbb{Z}/3\mathbb{Z}$. I don't really see what happens when we want to ''jump'' from $S_3$ to Aut$(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z})$. Thanks in advance

答:
It was proven in that question that $\mathrm{Aut}(\Bbb Z / 2 \Bbb Z \times \Bbb Z / 2 \Bbb Z)$ is isomorphic to $\mathcal S_3$. Therefore, it is enough to prove the result for $\mathcal S_3$ instead of $\mathrm{Aut}(\Bbb Z / 2 \Bbb Z \times \Bbb Z / 2 \Bbb Z)$.

A non-trivial homomorphism $\phi \in \mathrm{Hom}(\Bbb Z / 3 \Bbb Z, \mathcal S_3)$ is fully defined by $\phi(1)$ as $\Bbb Z / 3 \Bbb Z$ is cyclic. And $\phi(1)$ is a 3-cycle $\sigma_\phi$as the 3-cycles are the only elements of order 3 in $\mathcal S_3$. In $\mathcal S_3$, there are only two 3-cycles. Namely $\sigma_1 = (1 \ 2 \ 3)$ and $\sigma_2 = (1 \ 3 \ 2)$. Therefore $\mathrm{Hom}(\Bbb Z / 3 \Bbb Z, \mathcal S_3)$ also has only two non-trivial elements $\phi_1,\phi_2$ with $\phi_1(1) = \sigma_1$ and $\phi_2(1) = \sigma_2$.

Now in $\mathcal S_n$, any two cycles of the same lengths are conjugate. Applying the result to our specific case, it exists $\beta \in \mathcal S_3$ such that $\sigma_2 = \beta \circ \sigma_1 \circ \beta^{-1}$. By the way $\beta = (2 \ 3)$ is such an element.

This conjugacy relation means that

$$\phi_2(1) = \beta \circ \phi_1(1) \circ \beta^{-1}$$ and implies

$$\phi_2(2) = \phi_2(1+1)=(\beta \circ \phi_1(1) \circ \beta^{-1})^2=\beta \circ (\phi_1(1))^2 \circ \beta^{-1} = \beta \circ \phi_1(2) \circ \beta^{-1}$$

As $$\phi_2(0) = \mathrm{Id}= \beta \circ \phi_1(0) \circ \beta^{-1}$$

we have proven the desired result.


问:
Central limit theorem - std dev away from mean
I was reading about the CLT and found something that I think people use interchangeably. On one hand I found that 68% of the means are 1 standard deviations from away and 95% are 2 std dev. On the other hand, if I take a look at the Z-table I found that 68% is approx 1 std dev whereas 95% is 1.96 std dev away.

Can someone clarify this please? Are these values of "std devs away" interchangeable? Or better yet, what am I confusing?

答: 1.96 is equal to 2 for many practical purposes.</p>


问:
How many ways to select $4$ people at a round table such that no two of the selected people are adjacent?

There are $100$ people sited in a circular table with $100$ chairs. How
many ways to select $4$ such that each chosen person is not adjacent to
any other chosen person.

The problem is equivalent to the following

Let there be $100$ empty chairs in a circular table. How many ways to choose $4$ chairs such that each chosen chair is not adjacent to any other chair

When it's a straight table, we can remove a chair between each chosen one yielding $\binom{97}{4}$ by bijection argument. It can maybe be applied when the table is circular.

I found a solution using PIE

$$\binom{100}{4}-\frac{100\binom{98}{2} \cdot 4-100 \cdot 9 7 \cdot 6+4 \cdot 100}{4}= 3460375$$

答:
I will answer the part which you say is "equivalent to" the question you want to ask because the original question says $100$ people seated in $100$ chairs. Also when "people" implies distinct entities whereas in the "equivalent" question they are faceless.

If we cut the circle to straighten it out $X ..................Y$, two cases arise: (i) we use neither $X$ nor $Y\;$ (ii) we use one of $X$ or $Y$

in either case, we have $95$ slots available,

thus the number of ways $= \dbinom{95}4 + 2\dbinom{95}3 = 3460375 $

Extra Answer

You may find this answer even simpler.

$1.$ Remove $4$ chairs and place them in the $96$ interstices.

$2.$ The chairs have only $96$ "starting" places instead of the actual $100$, so multiply by $\frac{100}{96}$

Thus $\dbinom{96}4\dfrac{100}{96} = 3460375$


答:
$$\begin{align}&~~~~~~\mathsf P(C\mid B,A)\,\mathsf P(B\mid A)\,\mathsf P(A)+\mathsf P(\overline B\mid A)\,\mathsf P(A)\\&=\big(\mathsf P(B,C\mid A)+\mathsf P(\overline B\mid A)\big)\,\mathsf P(A)&&\small\text{by definition of conditional probability}\\&=\mathsf P\big((B\cap C)\cup\overline B\mid A\big)\,\mathsf P(A)&&\small\text{by sigma additivity of disjoint events}\\&=\mathsf P(C\cup\overline B\mid A)\,\mathsf P(A)&&\small\text{by absorption}\\&=\mathsf P(A\cap(C\cup\overline B)
)\end{align}$$

Title: How to simplify $\operatorname{P} ( { C |B,A} )P( { B |A} )P( A ) + P( { {\bar B} |A} )P( A )$?

Body:
How to simplify the following probability

$\operatorname{P} ( { C |B,A} )P( { B |A} )P( A ) + P( { {\bar B} |A} )P( A )$

$ = P\left( {A,B,C} \right) + P\left( {A,\bar B} \right)$

Can $P\left( {A,B,C} \right) + P\left( {A,\bar B} \right)$ be further simplify ?

Also, are there any systematic way to do the simplification instead of following heuristic. I see that there are some link between Boolean algebra and the simplification of these problem.

Particularly, my thinking is that $ = P\left( {A,B,C} \right) + P\left( {A,\bar B} \right)$ may be somehow related to
$F\left( {A,B,C} \right) = \left( {ABC} \right)\,OR\,\left( {A\bar B} \right)$ but I do not know if my guts feeling is correct or not.

Please help me clear out these doubt, thank you for your enthusiasm !


问:
Solve THIS ODE $ (2x-y\sin(2x))dx = (\sin^2x-2y)dy $
The following problem is in Mathematical Methods in physical science CH8 miscellaneous problems
$$
(2x-y\sin(2x))dx = (\sin^2x-2y)dy
$$
it isnt an exact equation the only difference is $-$ sign
because it not an exact equation I tried to rearrange it to produce linear first order but it did not work
I dont know what to do
I assume there an integration factor for such problem to make it exact or methods

Also I tried to use Mathematica DSolve function to find a solution

```<code>DSolve[{y'[x] == (2 x - y[x] sin (2 x))/(sin^2 x - 2 y[x])}, y[x], x]
</code>```

This what it produced

```<code>DSolve::deqn: Equation or list of equations expected instead of True in the first argument {True}.
DSolve[{True}, y[x], x]
</code>```

thank you

答: Since $d[x^2+y^2-y\sin^2x]=(2x-y\sin2x)dx+(2y-\sin^2x)dy=0$, the solution is $x^2+y^2-y\sin^2x=c$.


答:
Recall we say a differential equation of type
$$N(x,y)dx+M(x,y)dy=0$$ is exact if $$\frac{\partial N}{\partial y}=\frac{\partial M}{\partial x}$$

So re-writing the above differential equation as
$$
(2x-y\sin(2x))dx - (\sin^2x-2y)dy
=0$$

we can see that $$\frac{\partial N}{\partial y}=-\sin(2x)=\frac{\partial M}{\partial x}$$
so it is in fact exact.

Title: Solve THIS ODE $ (2x-y\sin(2x))dx = (\sin^2x-2y)dy $

Body:
The following problem is in Mathematical Methods in physical science CH8 miscellaneous problems
$$
(2x-y\sin(2x))dx = (\sin^2x-2y)dy
$$
it isnt an exact equation the only difference is $-$ sign
because it not an exact equation I tried to rearrange it to produce linear first order but it did not work
I dont know what to do
I assume there an integration factor for such problem to make it exact or methods

Also I tried to use Mathematica DSolve function to find a solution

```<code>DSolve[{y'[x] == (2 x - y[x] sin (2 x))/(sin^2 x - 2 y[x])}, y[x], x]
</code>```

This what it produced

```<code>DSolve::deqn: Equation or list of equations expected instead of True in the first argument {True}.
DSolve[{True}, y[x], x]
</code>```

thank you


答:
Considering$$A +B\,y'+C\sin(y)+y'' = 0$$ start switching variables to make
$$A+\frac B {t'}+C\sin(y)-\frac{t''}{[t']^3}=0$$ which makes the reduction of order quite obvious. So
$$A+\frac B {p}+C\sin(y)-\frac{p'}{p^3}=0$$ If I am not mistaken, we should arrive to something like
$$\frac 1 p=-\frac A B+\frac{C (\cos (y)-B \sin (y))}{B^2+1}+k\, e^{-B y}$$

Doing the same with

$$A +B\,y'+\sin(y) = 0\implies t'=-\frac{B}{A+\sin (y)}$$ which is quite simple using the tangent half-angle subsitution.

Making the story short, we should get
$$t=k-\frac{2 B }{\sqrt{A^2-1}}\tan ^{-1}\left(\frac{A \tan
   \left(\frac{y}{2}\right)+1}{\sqrt{A^2-1}}\right)$$ which can be inversed.


问:
Confusing limit problem within proof of the chain rule
Let $f:G\to G'$, $G\subset \Bbb{R^n}$ and $G'\subset \Bbb{R^m}$ be differentiable at $x_0 \in G$ and let $g:G' \to \Bbb{R^p}$ be differentiable at point $y_0 = f(x_0)$. Then the mapping $g\circ f$ is differentiable at $x_0$ and $D_{g\circ f}(x_0)=D_g(y_o)\cdot D_f(x_0)$

Proof.

$g(f(x_0 + u)=g(f(x_0)+D_f(x_0)u+\vert u\vert\epsilon_1(u))$ , let $h:=D_f(x_0)u+\vert u\vert\epsilon_1(u)$

$=g(y_0)+D_g(y_0)(D_f(x_0)u+\vert u \vert\epsilon_1(u))+\vert h \vert\epsilon_2(h)$

$=g(y_0)+D_g(y_0)\cdot D_f(x_0)u+D_g(y_0)\vert u\vert\epsilon_1(u)+\vert h\vert\epsilon_2(h)$

At this point, I substract $g(y_0)+D_g(y_0)\cdot D_f(x_0)u$ and divide by $\vert h \vert$.
The proof is done, if I could solve the limit $$ \lim_{h\to 0} \frac{D_g(y_0)\vert u\vert\epsilon_1(u)+\vert h \vert\epsilon_2(h)}{\vert h \vert}$$

How can I do this?

答:
You have things a little mixed up. You want the error term to be $o(|u|)$, not $o(|h|)$, since $u$ was the pertubation term you used for the composition $g\circ f$. With that in mind we have
\begin{equation}\tag{1}
\frac{D_g(y_0)\vert u\vert\epsilon_1(u)+\vert h \vert\epsilon_2(h)}{\vert u \vert} = D_g(y_0)\epsilon_1(u)+\frac{|h|}{|u|}\epsilon_2(h).\end{equation}
On the other hand,
\begin{equation}\tag{2}
|h| = |D_f(x_0)u+|u|\epsilon_1(u)|\leq \big(\|D_f(x_0)\|+|\epsilon_1(u)|\big)|u|.
\end{equation}
Thus, in norm, (1) becomes
$$\frac{|D_g(y_0)\vert u\vert\epsilon_1(u)+\vert h \vert\epsilon_2(h)|}{\vert u \vert}\leq \|D_g(y_0)\||\epsilon_1(u)|+\big(\|D_f(x_0)\|+|\epsilon_1(u)|\big)|\epsilon_2(h)|.$$
Since $u\to 0$ implies $h\to 0$ as well by (2), the right-hand side of the above tends to $0$ as $u\to 0$. This completes the proof.

One last note: I've kept the same notation as you used in your statement of the problem, but $D_f(x)$ is really nonstandard. $D_{x}f$, $(Df)(x)$, and $df_x$ are all much more common.


问:
finding all possible. numbers
When a number is divided by $11$,the obtained number equals the sum of the squares of the digits of the starting number. If the starting number has three digits, find all such numbers.

Detailed solution please.

答:
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight.

I wrote and ran some Mathematica-code:

```<code>In[1]:=Clear["Global`*"];
ParallelTable[
  If[TrueQ[IntegerQ[n/11] &&
     n/11 == (Last@IntegerDigits[n])^2 + (Part[IntegerDigits[n],
          1 + Floor[Log10[n]] - 1])^2 + (Part[IntegerDigits[n],
          1 + Floor[Log10[n]] - 2])^2], n, Nothing], {n, 100,
   999}] //. {} -> Nothing
</code>```

Running the code gives:

```<code>Out[1]={550, 803}
</code>```

This is true, because:

$$\frac{550}{11}=50=5^2+5^2+0^2\tag1$$

$$\frac{803}{11}=73=8^2+0^2+3^2\tag2$$

Title: finding all possible. numbers

Body:
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight.

I wrote and ran some Mathematica-code:

In[1]:=Clear["Global`*"];
ParallelTable[
  If[TrueQ[IntegerQ[n/11] &&
     n/11 == (Last@IntegerDigits[n])^2 + (Part[IntegerDigits[n],
          1 + Floor[Log10[n]] - 1])^2 + (Part[IntegerDigits[n],
          1 + Floor[Log10[n]] - 2])^2], n, Nothing], {n, 100,
   999}] //. {} -> Nothing

Running the code gives:

Out[1]={550, 803}

This is true, because:

$$\frac{550}{11}=50=5^2+5^2+0^2\tag1$$

$$\frac{803}{11}=73=8^2+0^2+3^2\tag2$$


Body:
$$(2x-y\sin(2x))dx = (\sin^2x-2y)dy$$
$$(2x-y\sin(2x))= (\sin^2x-2y)y'$$
$$2x+2yy'=y'\sin^2 x +y \sin (2x)$$
$$2x +(y^2)'=(y \sin^2 x)'$$
Integrate.


问:
Find all solutions for $x \cdot y = 5$ in $\mathbb{Z}[i] $
Let $\mathbb{Z}[i] = \{a+ib \in \mathbb{C}, a, b \in \mathbb{Z} \}. $ Find all $x, y \in \mathbb{Z} [i] \backslash \{i, - i, 1,-1\} $ such that $$x \cdot y = 5.$$
I've denoted $x=a+ib, y=c+id $ and substituted that into the equation. By the equality of complex numbers it follows that
$$ac-bd=5$$
$$ad+bc=0.$$
I'm not sure how to use the assumption that all numbers are integers and that the solution is not from $i, - i, 1,-1.$

Any hints would be appreciated!

答: If $xy=5$, then $|xy|^2=25$. In other words, $|x|^2|y|^2=25$. But the numbers $|x|^2$ and $|y|^2$ are natural numbers. And you can express $25$ as the product of two natural numbers in two ways and only two ways: as $1\times25$ or as $5\times5$. Can you take it from here?


答:
Hint: The equation $ad-bc=0$ corresponds to the zero-determinant of the real-valued matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$.
In this case, the columns of the matrix must be collinar.

Title: Find all solutions for $x \cdot y = 5$ in $\mathbb{Z}[i] $

Body:
Let $\mathbb{Z}[i] = \{a+ib \in \mathbb{C}, a, b \in \mathbb{Z} \}. $ Find all $x, y \in \mathbb{Z} [i] \backslash \{i, - i, 1,-1\} $ such that $$x \cdot y = 5.$$
I've denoted $x=a+ib, y=c+id $ and substituted that into the equation. By the equality of complex numbers it follows that
$$ac-bd=5$$
$$ad+bc=0.$$
I'm not sure how to use the assumption that all numbers are integers and that the solution is not from $i, - i, 1,-1.$

Any hints would be appreciated!


问:
Help with truth statements "For all $x \in \varnothing, p(x)$" and "$\exists x \in \varnothing, p(x)$ "
Why is the statement "For all $x \in \varnothing, p(x)$" always true, and why is the statement "$\exists x \in \varnothing, p(x)$ always false for a statement $p(x)$ that depends on $x$?

答:
Consider this statement : I am older than my father & I have a car. Is this true?

Well, you do not know whether I have a car ( in fact I have one) but you can surely know that I'm not older than my father, since it is impossible. Being given the whole " and" statement contains an impossible conjunct, this whole statement must be false.

In the same way : it is impossible that $\exists (x)\space [ x\in \emptyset]$

so, a fortiori, it is impossible that $\exists (x) \space [ x\in \emptyset \space \& \space \phi(x)] $   (whatever the property  $\phi$ may be).

And, since what is impossibly true just isn't true : it is false that $\exists (x) \space [ x\in \emptyset \space \&\space  \phi(x)\space] $

The statement "   $\forall (x)_{\in \emptyset}, \phi(x)$ " is a conditional, namely, the conditional : " $\forall(x) [ x\in\emptyset \rightarrow\phi(x) ] $".

Let's drop the quantifier and only consider the " if... then" statement inside, that is :

$x\in\emptyset \rightarrow\phi(x)$

and let's try to find an object $x$ that would make the statement false . So consider whatever object you want: number 5, The Moon, The Eiffel Tower, whatever, and let this object play the role of $x$ in the conditional statement.

What do we notice? That in each case, the antecedent of the " if ... then " is false ( for, of course, it is always false to say that an object belongs to the empty set, since it is impossible ).

Now, look at the truth table of the " if ... then " operator.

You can see that, whatever object you choose to play the role of " x" , you are on line 3 or on line 4 of the truth table. And on these lines, the " if ... then " is always true .

Note that here the truth value of the consequent  "$\phi(x)$ has no impact on the truth value of the whole conditional.

Conclusion : since what we discovered holds for any $x$, it also holds for all $x$, hence the statement

$\forall(x) [ x\in\emptyset \rightarrow\phi(x) ] $


问:
$f_n \rightarrow f$ in $L^1$ if $\mu(X) < \infty$, $f_n \rightarrow f$ almost uniformly and $\int_X |f_n|^2 d \mu \leq 1$
Suppose $(f_n)$ is a sequence of measurable functions on measure space $(X, \mathcal{A}, \mu)$. Suppose that $\mu(X) < \infty$, $f_n$ converges to $f$ almost everywhere and $\int_X |f_n|^2 d \mu \leq 1$ for all $n \in \mathbb{N}$. Prove that $f_n$ converges to $f$ in $L^1$.

Attempt:

By Egoroff's theorem, $f_n$ converges to $f$ almost uniformly (i.e. for all $\epsilon >0$, there exists $E \subseteq X$ such that $\mu(E) < \epsilon$ and $f_n$ converges to $f$ uniformly on $E^c$.)

\begin{align*}
\lim_{n \rightarrow \infty} \|f_n-f\|_1
&= \lim_{n \rightarrow \infty} \int_X |f_n-f| d \mu \\
&= \lim_{n \rightarrow \infty} \left( \int_E |f_n-f| d \mu + \int_{E^c} |f_n-f| d \mu \right) \\
&= \lim_{n \rightarrow \infty} \int_E |f_n-f| d \mu + \lim_{n \rightarrow \infty}  \int_{E^c} |f_n-f| d \mu \\
\end{align*}

Notice
\begin{align*}
\lim_{n \rightarrow \infty} \int_{E^c} |f_n-f| d \mu
 &\leq \lim_{n \rightarrow \infty}  \int_{E^c} \sup_{E^c} |f_n-f| d \mu \\
 &=  \lim_{n \rightarrow \infty}  \sup_{E^c} |f_n-f| \int_{E^c} d \mu \\
&=  \left( \lim_{n \rightarrow \infty}  \sup_{E^c} |f_n-f| \right) \mu(E^c) \\
&= 0
\end{align*}
by uniform convergence of $f_n$ to $f$ on $E^c$.

My professor said to use Egoroff's theorem to do this proof but I am not sure how to proceed.

答:
So because of $f_n \in L^2$ and $\mu (E ) < \infty$ we have the existence of some $C>0$ such that $\int \vert f_n \vert ~ \mathrm d \mu \leq C$ (so it is still uniformly bounded, in other words). Therefore $$\int_E \vert f_n - f \vert ~ \mathrm d \mu \leq \int_E \vert f_n \vert ~ \mathrm d \mu + \int_E \vert f \vert ~ \mathrm d \mu \leq \mu (E) \cdot \big(C + 1\big) \quad\text{for any } n \in \mathbb N .$$
With $\mu (E) \to 0$ we are done.

Note that you run into problems when handling the other integral like that, below 'Notice'. Because although $f_n \to f$ almost everywhere it could still be, that $\sup \vert f_n - f \vert = \infty$ for every $n$. So I would just use, that by uniform convergence, given an $\varepsilon >0$ we have $\vert f_n - f \vert < \frac \varepsilon{\mu (E)}$ on $E^\mathrm c$ and therefore $$\int_{E^\mathrm c} \vert f_n - f \vert ~ \mathrm d \mu < \varepsilon$$ for large enough $n$.


问:
regarding derivatives
Does the derivative (the slope value) only give us the instantaneous rate change at a point and nothing else? Is this the only significance of that value?

For example I took the function $f(x)=x^2$,lets consider at $x=3$. $f(x)=x^2=9$ and $f'(x)=2x=6$. Does the value $6$ only tells us rate of change at point $x=3$?

Lets consider average rate change, e.g. suppose the average rate of change in y with respect to x over some interval is $7$; that is, for every single unit by which x changes, "y" on average changes by $7$ units. Here the average value "$7$" is related to the y value of the function means y on average changes by $7$ units.

Some of my friends said that the value $f'(x)=6$ in the above example only gives the rate of change at that point, i.e it is just the slope of the tangent at that point and nothing else, it has no effect on the y value ($x^2=9$) of the function. But if it doesn't have to do anything with 'y' value, then why is the slope at point called "the instantaneous rate of change of y with respect to x"?can anyone explain to me this, I really need help.

答:
Does the derivative (the slope value) only give us the instantaneous rate change at a point and nothing else? Is this the only significance of that value?

The expressions "only" and "nothing else" in mathematical statements generally require a considerate effort in finding the proof, which is often nontrivial.

I'll attempt to show the inverse: derivatives can be used in an unexpected way to solve problems.  For example, to prove Young's inequality
$$\forall\,a,b > 0, p \ge 1, q:= \frac{p}{p-1}, ab \le \frac{a^p}{p} + \frac{b^q}{q},$$
one might consider the function $f(x) = \dfrac1p x^p − x +\dfrac1q$ and its derivatives.

Apart from giving "instantaneous rate change" at the points, it actually provides a way for identifying extrema, which help solving lots of optimization problems.

Derivatives help in exploring repeated roots of a polynomial defined on (structures defined over) a field with characteristic zero.  If you loosen the meaning of "derivatives" in your question context, you might find that "algebraic derivatives" (defined explicitly for $x^n$ with respect to $x$) have that functionality as well on (structures defined over) a field with nonzero characteristic, where one might not talk about the order '$<$' that the definition of limits replies on.

If you further allow me to move even further away from the topic, the application of $q$-derivatives in quantum calculus in the proof of Ramanujan's summation of $_1\Psi_1$ might amaze you.

References:

Kac V., Cheung P. (2002) Ramanujan Product Formula. In: Quantum Calculus.  Universitext. Springer, New York, NY. https://doi.org/10.1007/978-1-4613-0071-7_15

Andrews, G. E., & Askey, R. (1976). A simple proof of Ramanujan’s summation of the $_1\Psi_1$.  University of Wisconsin–Madison Mathematical Research Center Technical Summary Report #1669

Title: regarding derivatives


答: Do you know all prime elements in the UFD $\mathbb{Z}[i]$? Moreover $5=1^2 + 2^2$. Now it's your turn.


答:
Both are approximations, though the second is a very good one. The actual percentage of the area under the normal curve that is within two standard deviations of the mean is about $95.44\%$; the cutoff within which you get $95\%$ of the area is very close to $1.96$ standard deviations on each side of the mean.

The rule of thumb that says that about $68\%$ of the area is within one standard deviation, about $95\%$ is within two, and about $99.7\%$ is within three is just that: a rule of thumb. These are approximate numbers that are easy to remember.

Title: Central limit theorem - std dev away from mean

Body:
I was reading about the CLT and found something that I think people use interchangeably. On one hand I found that 68% of the means are 1 standard deviations from away and 95% are 2 std dev. On the other hand, if I take a look at the Z-table I found that 68% is approx 1 std dev whereas 95% is 1.96 std dev away.

Can someone clarify this please? Are these values of "std devs away" interchangeable? Or better yet, what am I confusing?


问:
Condensed notation for specifying an interval for several variables
What is the most correct notation to specify an interval for several variables at the same time? E.g. if both $\alpha$ and $\beta$ are in the interval $(0,\pi/4)$;

$$0<\alpha,\beta<\pi/4$$

$$0<\{\alpha,\beta\}<\pi/4$$

And perhaps even a third and better alternative?

TIA.

答:
Regarding your proposal, I prefer the first one, I think most people would understand it.

You can use $(\alpha,\beta)\in(0,\frac{\pi}4)^2$

Generally for an n-uplet $(x_1,x_2,\cdots,x_n)\in\mathbb R^n$ or $I^n$ for some interval $I$.

Since $I^n$ is a notation for set product $I\times I\cdots\times I$, you can also specify different intervals for different variables.

$\, z=re^{i\theta}$ with $(r,\theta)\in[0,3)\times[-\frac{\pi}2,\frac{\pi}2]$

$x\in\mathbb Q\iff x=\frac pq$ with $(p,q)\in\mathbb Z\times\mathbb N^*$


问:
What general techniques are there to show that two finitely-generated groups are bi-Lipschitz equivalent?
We define bi-Lipschitz equivalency as follows.

We say two metric spaces $(X,d)$, $(Y,e)$ are bi-Lipschitz
equivalent if there exists a bijection $f : X \to Y$ and some $ L
 \geq 1$ such that  $$ \frac 1 L \ d(x,y) \leq e(f(x),f(y)) \leq L \
 d(x,y), $$ for all $x,y \in X$.

Viewing finitely-generated (f.g.) groups as metric spaces under their word metric, we have that any two word metrics on a f.g. group are bi-Lipschitz equivalent. Thus it makes sense to discuss whether two f.g. groups $G$, $H$ are bi-Lipschitz equivalent, without fixing any generating sets.

I understand that to show two groups are not bi-Lipschitz,  we may compare their growth functions, for example. However I am not aware of any general tools for showing two groups are bi-Lipschitz equivalent, outside of just exhibiting such a map $f$ as above (and even that seems incredibly difficult).

Any references to literature which discusses this topic or contains examples would be appreciated. Thanks!

答:
I'd say:

For non-amenable groups, the theorem by Whyte mentioned by Moishe without comment is that every quasi-isometry is at bounded distance from a bilipschitz map. Hence the question boils down to the same with quasi-isometry, which is broad and widely studied. Among others, all cocompact lattices in the same locally compact group are quasi-isometric.

For amenable groups, this is trivially false (finite groups of distinct cardinal are QI but not bilipschitz), Much harder, Dymarz exhibited (solvable) infinite examples. So I guess there are few general methods. It is enough to have bilipschitz subgroups of the same finite index. For instance, $\mathbf{Z}$ and $D_\infty$ are bilipschitz since they both have an index 2 copy of $\mathbf{Z}$. Also, some constructions pass to bilipschitz maps (rather than QI): if $A,B$ are bilipschitz, then $A\wr\Gamma$ and $B\wr\Gamma$ are bilipschitz (the analogous statement with QI being false), the proof being the obvious naive one.

There are many open questions, for instance it's not known whether there exists two QI but not bilipschitz infinite f.g. nilpotent groups. It's not known if there exists a f.g. nilpotent group that is  not bilipschitz to one of its finite index subgroups.


问:
Solving $2\log_2 (\log_2 x) + \log_{1/2} (\log_2 x) = 1$
just looking for some help with the following:

$$2\log_2 (\log_2 x) + \log_{1/2} (\log_2 x) = 1$$

I know how to solve a log equation with a single nested log that's $= 1$ or $0$ or some number, but I'm unsure of what to do when summing/ subtracting nested logs like these.

If possible, a hint would be much more appreciated than the flat out answer.

Thanks in advance to anyone who stops to help.

答:
I would first perform a change of basis in the second term.

Recall: $\log_{(1/2)}a = \frac{\log_2(a)}{\log_2(1/2)}$

(Of course, then see that $1/2 = 2^{-1}$)


问:
The domain of convergence
Help me to find the domain of the function sequence where it convergence uniformly.
$$
x_{n}(t) =  \frac{1}{t^2-nt+1}
$$
For $n > 2$ the function would have points where it is undefined and this confuses me

答:
For $n \gt 2$ $x_n$ is defined on $D_n= (-\infty, \frac{n -\sqrt{n^2 - 4}}{2}) \cup (\frac{n +\sqrt{n^2 - 4}}{2}, \infty)$.

Analyzing $D_n$, you will see that $\bigcap\limits_{n \in \mathbb N} D_n = D=(-\infty , 0]$. Therefore, studying the convergence of the sequence of functions $\{x_n\}$ only makes sense on $D$.

As $x_n(0)=1$ for all $n \in \mathbb N$, $x_n(0)$ converges to $1$. And for $t \lt 0$, $x_n(t)$ converges to $0$. Therefore $\{x_n\}$ converges pointwise on $D$ to the always vanishing map $x(t)$ except at zero where $x(0)=1$.

As $x$ is discontinuous, $\{x_n\}$ which is a sequence of continuous maps doesn't converge uniformly to $x$ on $D$.

Title: The domain of convergence


问:
Evaluate: $\int_{0}^{\infty}\sqrt{x}e^{-x^2}\ln^2(x)\left[\frac{1}{x\Gamma\left(\frac{1}{4}\right)}-\frac{1}{\Gamma\left(\frac{3}{4}\right)}\right]dx$
According to wolfram integrator:
$$\int_{0}^{\infty}\sqrt{x}e^{-x^2}\ln^2(x)\left[\frac{1}{x\Gamma\left(\frac{1}{4}\right)}-\frac{1}{\Gamma\left(\frac{3}{4}\right)}\right]\mathrm dx=2C+\frac{\pi}{4}\large[\gamma+3\ln(2)]\tag1$$

Does anyone know how to prove $(1)?$

We may try this substitution, $u=\ln\sqrt{x}$, $\mathrm dx=2x\mathrm du=2e^{2u}\mathrm du$

$$\int_{-\infty}^{\infty}8u^2e^{u-e^{4u}}\left[\frac{1}{\Gamma\left(\frac{1}{4}\right)}\right]\mathrm du-\int_{-\infty}^{\infty}8u^2e^{3u-e^{4u}}\left[\frac{1}{\Gamma\left(\frac{3}{4}\right)}\right]\mathrm du=I+J$$

Applying IBP to $I$:

$$I=u^2(1-4e^{4u})e^{u-e^{4u}}-2\int u(1-4e^{4u})e^{u-e^{4u}}\mathrm du$$

getting more difficult to deal with.

答: Since $$\int_{0}^{+\infty}x^{s-1}e^{-x}dx=\Gamma(s),\,\mathrm{Re}(s)>0$$ we have $$\int_{0}^{+\infty}x^{s-1}e^{-x}\log^{2}(x)dx=\Gamma(s)(\psi(s)^{2}+\psi_{1}(s))$$ where $\psi(x),\,\psi_{1}(x)$ are the Digamma and Trigamma function, respectively. I think you can solve the problem from here.


问:
Solutions to the functional equation $ ( x - 2 ) f ( y ) + f \big( y + 2 f ( x ) \big) = f \big( x + y f ( x ) \big) $
This is one of my practice problems, which may be solved with a substitution.

Find all functions $ f : \mathbb R \to \mathbb R $ which verify the relation
$$ ( x - 2 ) f ( y ) + f \big( y + 2 f ( x ) \big) = f \big( x + y f ( x ) \big) $$
for all $ x , y \in \mathbb R $.

I have tried to find whether such function is surjective, or injective, but unsuccessfully.

答:
You can show that the only functions $ f : \mathbb R \to \mathbb R $ satisfying
$$ ( x - 2 ) f ( y ) + f \big( y + 2 f ( x ) \big) = f \big( x + y f ( x ) \big) \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $ are those of the form $ f ( x ) = 0 $ and $ f ( x ) = x - 1 $. It's straightforward to verify that these are indeed solutions. To see the converse, let $ a = f ( 0 ) $ and $ b = f ( 2 ) $. Setting $ y = 0 $ in \eqref{0} gives
$$ a ( x - 2 ) = f ( x ) - f \big( 2 f ( x ) \big) \text , \tag 1 \label 1 $$
while putting $ x = 2 $ in \eqref{0} shows that
$$ f ( y + 2 b ) = f ( b y + 2 ) \text . \tag 2 \label 2 $$
By respectively setting $ x = y + 2 b $ and $ x = b y + 2 $ in \eqref{1} and using \eqref{2} you get $ a ( y + 2 b - 2 ) = a b y $, which for $ y = 3 $ can be simplified to $ a ( b - 1 ) = 0 $.

Now, assume that for some $ x _ 0 \ne 1 $ we have $ f ( x _ 0 ) = 0 $. Setting $ x = x _ 0 $ in \eqref{0} you can see that $ f $ will be the constant zero function, which is one of the mentioned solutions. This, in particular, happens if we take $ x _ 0 = 0 $, which settles the case $ a = 0 $. So, from now on, assume that we have $ b = 1 $. This, in particular, shows that you can't have $ a = 0 $, as then $ f $ must be constantly zero, which contradicts $ b = 1 $.

Consider any $ x _ 1 $ with $ f ( x _ 1 ) = 1 $. By the facts that $ b = 1 $ and $ a $ cannot be equal to zero, you can put $ x = x _ 1 $ in \eqref{1} to see that you must have $ x _ 1 = 2 $. This means that for any $ x \ne 2 $ you have $ f ( x ) \ne 1 $, and you can let $ y = \frac { 2 f ( x ) - x } { f ( x ) - 1 } $ in \eqref{0}. In that case, you'll have $ y + 2 f ( x ) = x + y f ( x ) = \frac { 2 f ( x ) ^ 2 - x } { f ( x ) - 1 } $, and since $ x \ne 2 $, \eqref{0} gives you $ f \left( \frac { 2 f ( x ) - x } { f ( x ) - 1 } \right) = 0 $. As $ f $ is not constantly zero, by the above argument we must have $ \frac { 2 f ( x ) - x } { f ( x ) - 1 } = 1 $, which can be simplified to $ f ( x ) = x - 1 $. We also have $ b = 2 - 1 $, and hence we can conclude that $ f ( x ) = x - 1 $ for all $ x \in \mathbb R $, which gives the other mentioned solution, and we're done.


问:
Matrix of shape operator of the sphere
Problem: Let $M$ be a sphere of radius $a$ in $\mathbb{R}^3$, defined by $$ x^2 + y^2 + z^2 = a^2. $$ Parametrize the sphere using spherical coordinates $$x = a \sin \phi \cos \theta \\ y = a \sin \phi \sin \theta \\ z = a \cos \phi $$ where $0 \leq \phi < \pi, 0 \leq \theta < 2 \pi$.

Then for each $p \in M, e_1 = \partial/\partial\phi$ and $e_2 = \partial/ \partial \theta$ is a basis for the tangent space $T_p M$ for $p \in M$. Let $N_p$ be the unit outward normal vector at $p$ on the sphere.

Find the matrix of the shape operator of the sphere with respect to the basis $e_1, e_2$.

Attempt: I need to calculate $L(e_j) = - D_{e_j} N$ and then find the matrix $[a^{i}_{j}]$ such that $L(e_j) = \sum a^{i}_{j} e_i. $ Here $L$ denotes the shape operator.

I know that in cartesian coordinates, a normal unit vector is $N = \frac{1}{a} (x, y, z)$. But do I have to write this first in the $(\theta, \phi)$ basis?

I would calculate e.g. $$ D_{\partial_{\phi}} \frac{1}{a} (a \sin \phi \cos \theta, a \sin \phi \sin \theta, a \cos \phi) \\ = ( \cos \phi \cos \theta, \cos \phi \sin \theta, - \sin \phi). $$ But I need to write this as a linear combination of the basis vectors $\partial_{\phi}$ and $\partial_{\theta}$ to find the matrix representation?

答:
I guess that this exercise is from the book Differential Geometry of Loring Tu. It is misleading that in the body of the exercise appear the map of the change from polar to Cartesian coordinates, or at least I didn't use this map to solve the exercise.

This is what I did: define the map $f(x,y,z):=\sqrt{x^2+y^2+z^2}-a$, then $f^{-1}(0)=a\mathbb{S}^2$, and as zero is a regular value of $f$ then we find that $\nabla f(x,y,z)=\tfrac1{a}(x,y,z)$ is the outward pointing unit normal vector to every $(x,y,z)\in a\mathbb{S}^2$, therefore
$$
N=\frac1{a}\left(x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}+z\frac{\partial}{\partial z}\right)
$$
and from the definition of the shape operator you get
$$
L(e_1)=-\frac1{a}\left(\frac{\partial x}{\partial \phi }\frac{\partial}{\partial x}+\frac{\partial y}{\partial \phi }\frac{\partial}{\partial y}+\frac{\partial z}{\partial \phi }\frac{\partial}{\partial z}\right)=-\frac1{a}\frac{\partial}{\partial \phi }
$$
where the last identity follow easily applying the coordinate functions $x,y,z$ to the right of both expressions for the vector field. Therefore in basis $e_1,e_2$ we find that $L=-\tfrac1a\left[\begin{smallmatrix}1&0\\0& 1\end{smallmatrix}\right]$.∎

Title: Matrix of shape operator of the sphere


问:
Does category theory provide a rigorous definition of the cartesian product?
Let $X$ and $Y$ be sets. As far as I know, the defining property of the cartesian product
\begin{equation}
X\times Y=\left\{(x,y):x\in X\text{ and }y\in Y\right\}
\end{equation}
is the "universal property"
\begin{equation}
(x,y)=(x',y')\Leftrightarrow x=x'\text{ and }y=y'
\end{equation}
for all $x,x'\in X$ and $y,y'\in Y$. It is known that the definition $(x,y):=\{\{x\},\{x,y\}\}$ has the above property. But I guess this is only one of many possible definitions.

I have no background in category theory, but I got the impression that it is always needed to get the full picture when dealing with something that has a universal property (e.g. the tensor product), that's why I thought that it might help me get a more precise definition of the cartesian product.

If your answer contains terms specific to category theory, it would be much appreciated if you also gave some reference, so that I have a starting point for a rigorous introduction to the topic.

答:
Here's a different formulation of the universal property. It should be equivalent to the other answer in the sense that sets that have the following property also have the universal property described in the other answer and vice versa.

Consider a set $I$ and let $X_i$ be a set for all $i\in I$.
\begin{equation*}
    X:=\bigcup_{i\in I}X_i
\end{equation*}
Now, let $M$ be the set of all maps from $I$ to $X$ and
\begin{equation*}
    \mathcal{M}=\{x\in M:x_i\in X_i\text{ for all }i\in I\}.
\end{equation*}

A Cartesian product of the family $\{X_i:i\in I\}$ consists of a set $P$ and a family of maps $\{\pi_i\colon P\to X_i:i\in I\}$ such that the map $f\colon P\to\mathcal{M}$ that assigns to each $x\in P$ the map
\begin{align*}
    f(x)\colon I&\to X\\
    i&\mapsto\pi_i(x)
\end{align*}
is bijective. We identify $x\in P$ with $f(x)$.

Let $P$ to $P'$ be Cartesian products of $\{X_i:i\in I\}$. Then there is exactly one natural bijection from $P$ to $P'$:

According to the definition of Cartesian products, we can construct bijective maps $f\colon P\to\mathcal{M}$ and $f'\colon P'\to\mathcal{M}$. The map $f^{-1}\circ f'\colon P'\to P$ is the natural bijection that allows us to identify elements of $P$ and $P'$ that correspond the same element of $\mathcal{M}$: If $x'\in P'$ and $x=F(x')$, $f(x)=f'(x')\in\mathcal{M}$.

Comment. I think this is a very natural way to describe the universal property: It often happens that the Cartesian product is defined as $\mathcal{M}$, just like the tensor product is sometimes defined as the set of multilinear maps. In both cases, the universal property can be described making use of the "naive" definitions.


问:
Given a natural number $n< 10^{9},$ find the maximum number of the multiple of $3,$ which differs by exactly one digit from the given one

I have an algorithm to solve this problem  * given a natural number $n< 10^{9},$ find the maximum number of a multiple of $3,$ which differs by exactly one digit from the given one." My algorithm is finding $m= n\mod 3.$ Then I will check from the left from the first digit, i.e. $i,$ if that at least $9- m,$ I must continue with the next digit, so on.. Else if $i< 9- m,$ substracting $i:=i- m,$ then adding $i:=i+ 3,$ not until $i> 9.$ But it seems not good.

And my algorithm is wrong by tested, how can I fix it up, I need to the help, thanks for all your comments !

答:
Your algorithm takes the whole of $n$ modulo $3$ into account. You need to take "$n$-each digit $\bmod 3$"  into account. See my definition of $t_c$ below.

Let $\displaystyle n = \sum_{k=0}^{s-1}10^kd_k$ be an $s$-digit number when written in base $10$.

Let $\displaystyle t_c = \large (\sum_\stackrel{k=0}{k \neq c}^{s-1}d_k \large ) \bmod 3$ be the sum, modulo $3$, of all the digits except the $c$'th one.

The algorithm goes:

Step $1$

Set $c=s-1$.

Step $2$

If digit $d_c \lt 9-t_c$, replace digit $d_c$ with $9-t_c$ and exit.

Step $3$

Set $c=c-1$.

If $c \ge 0$ return to Step $2$.

If $c < 0$ exit: there is no solution for this $n$.

A few examples follow.

$n=523, t_2=2, t_1=2, t_0=0$. Step $2$ ($c=2$) replaces the $5$ with $9-2=7$ and exits with $723$.

$n=826, t_2=2, t_1=2, t_0=1$. Step $2$ ($c=2$) decides that the $8$ is too big to be replaced. Step $2$ ($c=1$) replaces the $2$ with $9-2=7$ and exits with $876$.

$n=879, t_2=1, t_1=2, t_0=0$. Step $2$ decides that all the digits are too big to be replaced. ($8 \ge 8, 7 \ge 7, 9 \ge 9$ ). There is no solution for $879$.


问:
What is the perimeter of a sector?
I don't understand this.

So we have:

\begin{align}
r &= 12 \color{gray}{\text{ (radius of circle)}} \\
d &= 24 \text{ (r}\times2) \color{gray}{\text{ (diameter of circle)}} \\
c &= 24\pi \text{ (}\pi\times d) \color{gray}{\text{ (circumference of circle)}} \\
a &= 144\pi \text{ (}\pi\times r^2) \color{gray}{\text{ (area of circle)}}
\end{align}

And we have:

\begin{align}
ca &= 60^\circ \color{gray}{\text{ (Central Angle of sector)}} \\
ratio &= \frac{60}{360} = \frac{1}{6} \color{gray}{\text{ (ratio of ca to circle angle which is 360 degrees)}}
\end{align}

So now we can calculate:

\begin{align}
al = \frac{1}{6} \times 24\pi &= 4\pi \color{gray}{\text{ (arc length of SECTOR = ratio X circumference of circle)}}
sa &= \frac{1}{6} \times 144per = 24\pi \color{gray}{\text{ (sector area = ratio X area of circle)}}
\end{align}

So my question is: What is meant by the perimeter of a Sector. Is it the arch length or the are of a Sector? And what is $24 + 4\pi$?

答:
The perimeter of the sector includes the length of the radius $\times 2$, as well as the arc length. So the perimeter is the length "around" the entire sector, the length "around" a slice of pizza, which includes it's edges and its curved arc.

The arc length is just the curved portion of the circumference, the sector permimeter is the length of line $\overline{AC} = r$ plus the length of line $\overline{BC} = r$, plus the length of the arc ${AOC}$.

The circumference of the circle is the total arc length of the circle.

Length is one-dimensional, the length of a line wrapped around the circle. Area is two dimensional; All of what's inside the circle.

Title: What is the perimeter of a sector?


问:
If the trace of the product of two matrices is zero,is the product of the matrices zero?
Does $\:\operatorname{ Tr}(XY) = 0 \Rightarrow XY=O $, for all $X,Y \succeq O$ ?

答: Hint: Let $A$ and $B$ be the positive semi-definite square roots of $X$ and $Y$ respectively, i.e. $X = A^2$ and $Y = B^2$ where $A,B \succeq O$. Try to prove $\text{tr}(XY) = \|AB\|_F^2$ (the cyclic property of the trace might help). After doing this, if $\text{tr}(XY) = 0$, what can you say about $AB$, and consequently $XY$?


问:
Asymptotic formula/closed form for $\sum_{n=1}^{x}\frac{1}{\log n}$
I was trying to get an asymptotic formula for the sum
$$\sum_{n=2}^{x}\frac{1}{\log n}$$
this looks a bit tricky, but I still tried it. We have

\begin{align}
\sum_{n=2}^{x}\frac{1}{\log n}&=-\sum_{n=2}^{x}\frac{1}{\log(nt)}\Bigg{|}_{t=1}^{t=\infty}\\
&=\sum_{n=2}^{x}\int_{1}^{\infty}\frac1{t\log^2(nt)}dt
\end{align}
Now I don't know what to do further. Any help would be appreciated. Any asymptotic formula or closed form would help.

Note: I will probably update this question to post more of my work.

Update: I proved that this sum is $O(x/\log x)$, and this was also mentioned in the comments. But I want to improve this. How can one improve the error term?

答:
By Riemann-Stieltjes integration we have

$$
\begin{aligned}
\sum_{2\le n\le x}{1\over\log n}
&=\sum_{2<n\le x}{1\over\log n}+{1\over\log2} \\
&=\int_2^x{\mathrm d(t-\{t\})\over\log t}+{1\over\log2} \\
&=\int_2^x{\mathrm dt\over\log t}+{1\over\log2}-\int_2^x{\mathrm d\{t\}\over\log t} \\
&=\operatorname{Li}(x)+{1\over\log2}-{\{x\}\over\log x}-\int_2^x{\{t\}\over t\log^2t}\mathrm dt \\
&=\operatorname{Li}(x)+A+\mathcal O\left(1\over\log x\right)
\end{aligned}
$$

in which

$$
A={1\over\log2}-\int_2^\infty{\{t\}\over t\log^2t}\mathrm dt
$$

and $\operatorname{Li}(x)=\operatorname{li}(x)-\operatorname{li}(2)$ is the logarithmic integral function satisfying

$$
\operatorname{Li}(x)\sim{x\over\log x}
$$


问:
Flux Homomorphism has right inverse
I'm reading about the Flux homomorphism in Symplectic Topology and I'm trying to show that it is surjective.

I know that if $\psi_{t}$ is the flow of a symplectic vector field $X$, then Flux({$\psi_{t}$}) = [$i(X) \omega$] (here $\omega$ is the symplectic form, so it must be nondegenerate) I'm trying to use this to construct a right inverse, but I don't really know how to proceed.

答:
Fix a class $[\alpha]\in H^{1}(M)$. Since $\omega$ is non-degenerate, there exists a unique vector field $X\in\mathfrak{X}(M)$ such that
$$
\iota_{X}\omega=\alpha.
$$
Since $\alpha$ is closed, the vector field $X$ is symplectic:
$$
L_{X}\omega=d\iota_{X}\omega+\iota_{X}d\omega=d\alpha=0.
$$
Let $\psi_{t}$ denote the flow of $X$, and denote by $\{\psi_t\}$ its homotopy class. Then
$$
Flux\{\psi_{t}\}=[\iota_{X}\omega]=[\alpha].
$$

Title: Flux Homomorphism has right inverse


问:
Embedding of $W^{1,n}(U)\hookrightarrow L^q(U)$
Let $U$ be a $C^1$ bounded region in $\Bbb{R}^n$,we know the Sobolev inequality given as follows:

Asssume $1\le p <n$ for any $u \in W^{1,p}(U)$ we have $u\in L^{p^\star}(U)$ where $p^\star = 1/(\frac{1}{p} - \frac{1}{n})$.moreover the inequality holds.

$$\|u\|_{L^{q^\star}(U)} \le C \|u\|_{W^{1,p}(U)}$$

The question is how to deduced the embedding relationship for $p = n$ that is :
$$W^{1,n}(U)\hookrightarrow L^q(U)$$

For any $1\le q <\infty$

答: Since $L^n(U)\subset L^p(U)$ for $p<n$, $W^{1,n}(U)\subset W^{1,p}(U)\subset L^{p^*}(U)$. Since $p^*\to \infty$ as $p\to n$, the result follows


问:
Help with: $p$ and $q$ are primes, then a group of order $p^2q$ cannot be simple
I am trying to understand the proof of the following statement:

If $p$ and $q$ are primes, than a group of order $p^2q$ cannot be simple.

The proof is given as follow:

Assume that G is simple and $ p \neq q$. The number of $p$-Sylow subgroups must be $q$ and hence $q \equiv 1 \mod p$. Hence, $q > p$. The number of $q$-Sylow subgroups is either $p$ or $p^
2$. It cannot be $p$, because this would imply that $p \equiv 1 \mod q$ implying $p > q$.

Hence, the number of $q$-Sylow subgroups must be $p^2$. But any pair of these subgroups can have only the identity as a common element.
Here I got confused. Why is it true? Suppose we have that $p^2$ $q$-Sylow subgroups $Q_1,\dots, Q_{p^2}$. Why we have that $$Q_j \neq Q_i \;\Rightarrow Q_i \cap Q_j = \{e\}$$

Furthermore why we get than that $\vert Q_1 \cup \dots \cup Q_{p^2} \vert=1 +p^2*(q-1)$?

EDIT: In my textbook it uses also the following statement: Let $P_1, \dots, P_q$ be the $p$-Sylow subgroups of $G$. Then $\vert P_1 \cup \dots \cup P_q \vert \geq \vert P_1 \cup P_2 \vert \geq  p^2+(p^2-p)$. How we get there the last inequality $\vert P_1 \cup P_2 \vert \geq  p^2+(p^2-p)$?

Many thanks for your help!

答:
If $H$ and $K$ are different subgroups of $G$ of order $q$ a prime, then $H \cap K$ is a subgroup of $H$ and by Lagrange $|H \cap K|$ divides $|H|=q$. Since $q$ is prime, $H \cap K=H$, impossible since $H$ and $K$ are different, or $H \cap K=1$.

To finish the argument: hence there are $p^2(q-1)$ elements of order $q$. That leaves $|G|-p^2(q-1)=p^2$ elements of $p$-power order, so these consitute the Sylow $p$-subgroup of order $p^2$ which must be normal, whence $G$ is not simple.


问:
How to Understand the Domain of a Function
Typically, a function $f: D \mapsto R$ is described as
$$\forall x \in D, \exists ! y \in R, \left(x,y\right) \in f \wedge P\left(x,y\right),$$
where $P$ is a predicate that specifies the relation between an input and the output, for instance, $y = 3x$ if $D \subseteq \mathbb{R}$ and $R \subseteq \mathbb{R}$. However, this definition does not specify what happens outside $D$. Should we explicitly specify that no element outside $D$ corresponds to an output under $f$? That is, should we use the following proposition
$$\left(\forall x \in D, \exists ! y \in R, \left(x,y\right) \in f \wedge P\left(x,y\right) \right) \wedge \left(\forall x \not\in {D}, \lnot \exists y, \left(x,y\right)\in f\right)$$
to describe $f$?

My own understanding is as follows. In mathematical proofs, people are more concerned with the existence of such a function. That is, what happens outside $D$ does not matter. What matters is, such a function $f$ exists, so that the proof can move on. Specifically, when people talk about a function, they are saying:
$$\exists f, \forall x \in D, \exists ! y \in R, \left(x,y\right) \in f \wedge P\left(x,y\right).$$
And they can proceed with a particular example of such a function. In such a circumstance, the information of existing $f$s outside $D$ is not of interest.

答:
Consider this relation $f$  from $ D= \{a, b,c\}$ to $T=\{0,1\}$ :

$f = \{(a,1), (b,1), (c,2)\}$

Is it true that

for all $x$, [ if $x\in D$ then, there is a unique $y$ in $T$ such that
$(x,y)\in f ]$ ?

If $x = a$ or $x = b$ or $x=c$ , the antecedent is true and the consequent is true, so the whole condiitonal is true.

If $x$ is neither $a$ nor $b$ , nor $c$ ( that is if $x\notin D$) the antecedent is false, and ( by the truth table of the " if ... then" operator) the whole conditional is true.

So, the conditional is true for all $x$ ... and relation $f$ is a function.


问:
Proof using trigonometry that circle circumference is $2 \pi R$
Using trigonometry, I would like to prove that the circumference of a circle is $2\pi$ times its radius. Can someone help please?</p>

答:
The question as it stands is not well posed as you are asking the wrong question.  In order to prove that the circumference formula $C = 2\pi r$ holds for all circle of radius $r$, we first have to understand what $\pi$ is.

First, let's discuss what $\pi$ is not.  In grade school, $\pi$ is typically defined as a number which is about $3.14159\dots$  In reality, $\pi$ is more than just a number.  It is defined as the proportion of a circle's circumference to its diameter.  First, we need to ask ourselves: Why is that no matter what diameter of a circle we have, the ratio of the circumference to the diameter are always the same? The answer to this question can be found here, here, or at any of these links via a google search.  Once we know that the circumference of a circle and its diameter are always proportional, then we can ask: What is the proportionality constant?  This amounts to precisely evaluating digits of pi, which is a computation problem.

With these two question behind us, your original question now answers itself.  We have $C = 2 \pi r$, since $\pi = \frac{C}{d} = \frac{C}{2r}$ by definition.


问:
Usage of generating functions in the given enumeration problem
There are $48$ flags in a ship that should be sent to the other ships,the flags are white,blue,black,red and for each one of the colors there are $12$ flags.
We choose $12$ of the flags.

How any messages can be sent assuming an odd number of black flags and an even number of blue flags have been used?

Consider the following exponential generating function:

$$\sum_{k=0}^{n}P(n,k)\frac{x^k}{k!}\tag{I}$$

Where $P(n,k)=k!\binom{n}{k}=k!C(n,k)$ is the number of ways to distribute $k$ distinct objects into $n$
distinct bins such that no bin can contains more than one object.

And so $$[x^k/k!]\sum_{k=0}^{n}P(n,k)\frac{x^k}{k!}=P(n,k)$$

I want to use the same approach to answer the question:
We can consider the coefficient of $x^{12}/(12)!$ in $f(x)$ where:

$$f(x)=\sum_{k=0}^{48}P(n,k)\frac{x^{2k}}{(2k)!}\sum_{k=0}^{48}P(n,k)\frac{x^{2k+1}}{(2k+1)!}\left[\sum_{k=0}^{48}P(n,k)\frac{x^{k}}{k!}\right]^2$$

And I "guess" since we only interested in the coefficients so the convergence is not what we should be worry about,and so I think the answer is given by :

$$[x^{12}/(12)!]\sum_{k \ge 0}^{}P(n,k)\frac{x^{2k}}{(2k)!}\sum_{k \ge 0}^{}P(n,k)\frac{x^{2k+1}}{(2k+1)!}\left[\sum_{k \ge 0}^{}P(n,k)\frac{x^{k}}{k!}\right]^2$$

Since $P(n,k)=k!C(n,k)[0 \le k \le n]$,where $[]$ is Iverson bracket and so the terms with $k > n$ will be equal to $0$ (ignoring the convergence ).

But the book I'm working with says that we should be looking for the coefficient of $x^{12}/{(12)!}$ in :
$$ \sum_{k \ge 0}^{}\frac{x^{2k}}{(2k)!}\sum_{k \ge 0}\frac{x^{2k+1}}{(2k+1)!}\left[\sum_{k \ge 0}\frac{x^{k}}{k!}\right]^2$$

But I don't understand why that is,because it looks to me a conflict with $\text{(I)}$

答:
If we'd multiply

$$\sum_{k \ge 0}{x^{2k}}\sum_{k \ge 0}x^{2k+1}\left[\sum_{k \ge 0}x^k \right]^2$$

out, then each summand of the form $x^n$ would correspond to a composition of $n$ into 4 parts (i.e. $x_1+...+x_4=n$) with the first part ($x_1$) being an even number, and the second part ($x_2$) being an odd number.

The number of words we can create from (the letters corresponding to) a particular composition $2x_1+2x_2+1+x_3+x_4=n$ is
$$
\binom{n}{2x_1,2x_2+1,x_3,x_4}
$$

And the number of words corresponding to some composition of the form $2x_1+2x_2+1+x_3+x_4=n$ is:

$$
[x^n]\sum_{k_1 \ge 0}\sum_{k_2 \ge 0}\sum_{k_3 \ge 0}\sum_{k_4 \ge 0}\binom{n}{2k_1,2k_2+1,k_3,k_4}x^{(2k_1)+(2k_2+1)+(k_3)+(k_4)}
$$

Factoring as much as possible, we obtain:
$$
[x^n] n!\cdot \sum_{k \ge 0}^{}\frac{x^{2k}}{(2k)!}\sum_{k \ge 0}\frac{x^{2k+1}}{(2k+1)!}\left[\sum_{k \ge 0}\frac{x^{k}}{k!}\right]^2
$$
as the quantity we're looking for.

That the bounds don't matter comes from the problem description - there is no summand $c\cdot x^{12}$ that uses more flags of a type than there are.

The question remaining is: What does your generating function do?

To each particular composition $2x_1+2x_2+1+x_3+x_4=n$ it creates a summand
$$
\frac{P(n,x_1)P(n,x_2)P(n,x_3)P(n,x_4)}{(2x_1)!(2x_2+1)!(x_3)!(x_4)!}
$$

I don't see any concrete thing this could count, especially because of the factor $\frac{P(n,x_2)}{(2x_2+1)!}$.

If your generating function were
$$
f(x)=\sum_{k=0}^{48}P(n,2k)\frac{x^{2k}}{(2k)!}\sum_{k=0}^{48}P(n,2k+1)\frac{x^{2k+1}}{(2k+1)!}\left[\sum_{k=0}^{48}P(n,k)\frac{x^{k}}{k!}\right]^2$$

then to each particular composition $2x_1+2x_2+1+x_3+x_4=n$ it'd create a summand
$$
\frac{P(n,2x_1)P(n,2x_2+1)P(n,x_3)P(n,x_4)}{(2x_1)!(2x_2+1)!(x_3)!(x_4)!}
$$

This would count the number of messages of the following scenario:

Each message has 4 parts. Each part has 12 slots. You fill the first part with an even number of blue flags, the second part with an odd number of black flags, the third part with white flags and the fourth with red flags, so that all 4 parts together contain exactly 12 flags.

So the problem of your generating function is that it would create 4 distinct parts, one for each of the factors
$\frac{P(n,2x_1)}{(2x_1)!}
,\frac{P(n,2x_2+1)}{(2x_2+1)!}
,\frac{P(n,x_3)}{(x_3)!}
,\frac{P(n,x_4)}{(x_4)!}
$.