问:
Imagine graph that has 10 vertices and 38 edges. Prove that there exist $K_4$ induced subgraph.

Imagine graph that has 10 vertices and 38 edges. Prove that there exist $K_4$ induced subgraph.

Such question already was asked here but I wanted to clarify some things.

One of people who answered claims that "There are only 7 edges missing from $K_{10}$" (what is obviously true). However is next statement is so: Each missing edge can prevent up to 28 different 4-vertex sets from inducing a $K_4$. If this is actually correct then we are done with proof. But I don't see why this is the case. Can you, please, explain this statement?

I was trying to prove it a bit differently: since we have 38 edges we have total sum of degrees equal to $38 \cdot 2 = 76$. Let us find pair of vertices with total degree equals 16 (such one definitely exists otherwise our total degree is not greater than 75 $\Rightarrow$ contradiction). There are basically two situations that we should consider. When one of vertices of this pair has degree 8 and another has degree 8 either and when one has degree 7 and another has degree 9. The second case is trivial. However, I am not managing to finish the first one.

Any help on that? (Please, do not use Turan's theorem if you want to show some alternative proof)

答:
Let $f(n)$ be the maximum possible number of edges in a $K_4$-free graph on $n$ vertices; we have to show that $f(10)\lt38$.

Let $G$ be a $K_4$-free graph with $n$ vertices ($n\ge3$) and $f(n)$ edges. Let $p$ be the number of pairs $(e,v)$ where $v\in V(G)$, $e\in E(G)$, and $e$ is not incident with $v$. Picking the edge first, we see that $p=f(n)(n-2)$. On the other hand, picking the vertex first, we see that $p\le nf(n-1)$. Thus, for $n\ge3$, we have
$$f(n)\le\left\lfloor\frac n{n-2}f(n-1)\right\rfloor.$$
Starting with $f(4)=5$ we get successively $f(5)\le8$, $f(6)\le12$, $f(7)\le16$, $f(8)\le21$, $f(9)\le27$, and finally $f(10)\le33$.


问:
Residue of Fractional Function.
$P(z), Q(z)$ : regular analytic function.

$f(z):=\dfrac{P(z)}{Q(z)}$.

$P(z_0) \neq 0, Q(z_0)=0$ and $z_0$ is $1$st order of $f$.

Then, is residue of $f$ at $z=z_0$ equal to $\dfrac{P(z_0)}{Q'(z_0)}$ ?

My darivation is following.

\begin{align*}
\text{Res}(z_0,f)
&=\lim_{z \to z_0} (z-z_0)\dfrac{P(z)}{Q(z)}\\
&=_{\text{L'Hôpital's rule}} \lim_{z \to z_0} \dfrac{P(z)+(z-z_0)P'(z)}{Q'(z)}\\
&=\dfrac{P(z_0)}{Q'(z_0)}.
\end{align*}

Is this idea correct?

答:
That is correct if $f$ has a simple pole at $z_0$, i.e. if $Q'(z_0) \ne 0$. Alternative derivations are
$$
\operatorname{Res}(z_0,f) =\lim_{z \to z_0} (z-z_0)\frac{P(z)}{Q(z)} \\
= \lim_{z \to z_0} \frac{P(z)}{\frac{Q(z)-Q(z_0)}{z-z_0}} = \frac{P(z_0)}{Q'(z_0)} \, ,
$$
using the definition of the derivative, or
$$
\frac{P(z)}{Q(z)} = \frac{P(z_0) + O(z- z_0)}{Q'(z)(z-z_0) + O(z-z_0)^2} \\
= \frac{P(z_0)}{Q'(z_0)} \frac{1}{z-z_0} + O(1) \, ,
$$
using asymptotic development at $z=z_0$.

Note that the condition $P(z_0) \ne 0$ is not needed as long as $Q$ has a simple zero at $z_0$: If $P(z_0) = 0$ then the formula gives $\operatorname{Res}(z_0,f) = 0$, which is correct because $f$ has a removable singularity at $z_0$.

ContentLicense: CC BY-SA 4.0

Title: Residue of Fractional Function.


问:
Can you prove this inequality for positive-definite matrices?
Given two positive definite matrices $A=(A_{ij})$ and $B=(B_{ij})$ prove that $A_{ij}\leq (\sum_{k,l}A_{kl}B_{kl})(B^{-1})_{ij}$ for each $i,j$.

My idea was to first observe that this is trivial if the two matrices are diagonal. Then show that diagonalization does not affect the inequality. But I could not prove the latter.

答:
The inequality does not always hold. Let $A=B$ be any matrix that is both positive definite and entrywise positive. Then $B^{-1}$ must contain some negative off-diagonal entries. Hence the LHS of the inequality is positive but the RHS is negative for some $(i,j)$. E.g. when $A=B=\pmatrix{2&1\\ 1&1}$, we have
$$
A_{12}=1>-7
=\operatorname{tr}(AB)\left(B^{-1}\right)_{12}
=\left(\sum_{k,l}A_{kl}B_{kl}\right)\left(B^{-1}\right)_{12}.
$$


问:
How do I prove that $\sum _{n=1}^{\infty }\:\ln\left(1+\frac{1}{n^2}\right)$ converges without applying the series integral test?
I am at a loss as to how I can prove that the sum $\sum_{n=1}^{\infty }\:\ln\left(1+\frac{1}{n^2}\right)$ converges. At first I tried the ratio test, but that came out inconclusive. Then I tried it by showing that the sequence of partial sums is bounded, to no avail. If anyone could give me a hint on this, it would be much appreciated!

答:
Since you have been given the answers for the comparison test ...

$$\sum _{n=1}^{\infty }\log\left(1+\frac{1}{n^2}\right)=\log\left(\prod _{n=1}^{\infty }\left(1+\frac{1}{n^2}\right) \right)=\log \left(\frac{\sinh (\pi )}{\pi }\right)$$


问:
Reference for another equivalent of Riemann hypothesis
I don't know why but i assumed that $\ln \theta(x) + E < \sum \limits_{p \leq x} \frac{\ln p}{p} $ for all real numbers $x \geq 2$ and with $E = -\gamma - \sum \limits_{P} \frac{\ln p}{p(p-1)}  \approx -1.33258$ is equivalent to RH, i saw it somewhere

When i went back to find it in the articles in my laptop I just found the one directly related to Robin's inequality $\ln \ln \theta(x) +B < \sum \limits_{p \leq x} \frac{1}{p} $ for all $x\geq 3$.

Is there a ref talking about the first inequality being equivalent to RH, or a direct proof of it from Robin,Nicolas inequality ?

Thanks.

答:
Define von Mangold function as $\Lambda(n)=\log p[n=p^k]$ then

$$
\sum_{n\le x}{\Lambda(n)\over n}<\sum_{p\le x}\log p\sum_{k\ge1}p^{-k}=\sum_{p\le x}{\log p\over p}+\sum_{p\le x}\log p\sum_{k\ge2}p^{-k}<\sum_{p\le x}{\log p\over p}+\sum_p{\log p\over p(p-1)}
$$

As a result, we can use the definition of the constant $E$ to get

$$
\sum_{n\le x}{\Lambda(n)\over n}+\gamma+E<\sum_{p\le x}{\log p\over p}
$$

If RH is equivalent to saying

$$
\log\vartheta(x)\le\sum_{n\le x}{\Lambda(n)\over n}+\gamma
$$

then your statement will be another equivalent form of RH.


问:
How to find a square in a measurable set with positive measure?
Let $E \subset \mathbb{R}^{2}$ be a measurable set with positive measure $($i.e. $m(E)>0).$ The question is to show that we can find four points in E such that the four points are the vertexes of a square.

I was trying to prove that the Lebesgue integral $\int f$
is positive, where $f = \chi_{E}$ $\chi_{E+(a,b)}$ $\chi_{E+(-b,a)}$ $\chi_{E+(a-b,a+b)}$, so that there exists a pair $(a,b)$ such that $E\cap E+(a,b) \cap E+(-b,a) \cap E+(a-b,a+b)$ is nonempty, but I'm stuck here.

It is known that if $E \subset \mathbb{R}$ is a measurable set with positive measure in $\mathbb{R}$, then $E-E$ contains an interval. Is there any relation between the fact and the question?

Any help would be appreciated.

答:
HINT:

Assume that $E$ is a measurable subset and  $S$ a square  such that $m(E\cap S) > \frac{3}{4} m(S)$. Wiggle $E$ to $E_i =E+ v_i$, $i=1,4$, ($v_i$ vertices of a small square) such that we still  have $m(E_i \cap S) > \frac{3}{4} m(S)$.  Show that the $E_i$ have a common point (in $S$). What can you conclude?


问:
Counting exercise
Three players a,b,c take turns in a game according to the following rules:

At the start A and B play (so C does not play). The winner of the first trial plays against C and so on until one of the players wins two trials in a row.
  Possible outcomes are
  aa,acc,acbb,acbaa,
  bb,bcc,bcaa,bcabb etc.

We have to prove that probability of A winning equals p(A) = 5/14, p(B) = 5/14, p(C) = 2/7

I have been stuck on this problem for a long time. The only thing I have been able to find out so far is that C can never win on even turns.

Re-edit

Players continue to play until one of them wins two times consecutively and all players are equally good at playing the game.

Apologies for leaving such crucial information.

I have finally solved this problem with a different approach

Sample space

aa ,acc ,acbb ,acbaa ,acbacc ,acbacbb ,acbabcaa ,...

bb ,bcc ,bcaa ,bcabb ,bcabcc ,bcabcaa ,bcabcabb ,...

Each point in the sample space has a probability (1/2^k) associated with it where k is the number of turns.

For example probability of point (aa) = 1/4

Now let us construct a table enumerating probabilities -

Let us consider the event that C wins overall. But C can only win if k = 3,6,9,12,...

P(C) = P(C3) + P(C6) + P(C9) ......

where P(Ck) = Probability of C winning overall after k turns.

P(C) = 1/4 + 1/32  .....

P(C) = a / (1 - r)       ( Sum of an infinite GP )

a = 1/4                  ( First Term )

r = 1/8                  ( Ratio )

P(C) = 2/7

P(A) = 5/14 = P(B)

答:
Perhaps the problem is supposed to be,

3 players A, B and C take turns at a game according to the following rules. At the start A and B play while C is out. The loser is replaced by C and at the second trial the winner plays against C while the loser is out. The game continues in this way until a player wins twice in succession, thus becoming the winner of the game. What is the probability that Player A will be the winner? What is the probability that player C will be the winner? (assume in each match, A has .6 chance to beat B, and .4 chance to beat C, B has .5 chance to beat C.)

which I found at http://www.chegg.com/homework-help/questions-and-answers/3-players-b-c-turns-game-according-following-rules-start-b-play-c--loser-replaced-c-second-q2427231

(and, yes, I know this is not an answer, but I don't think it would fit as a comment).

ContentLicense: CC BY-SA 3.0

Title: Counting exercise


问:
Looking for a scoring formula that takes into account how quickly was answered
I'm not that great at math,  so looking for your help on this one...

The goal is to end up with a formula f(q, t) where q is a value between 0 and 1 inclusive, representing the percentage of correct answers,  and t is the time it took to answer the questions, with a maximum.

The challenge I'm facing is that I want to avoid that a team can enter 1 correct question very quickly and win. the team with more correct answers should always be preferred over others.

I have the feeling this is quite basic but not knowing the right words to google for it, trying my luck here.

thanks!

答:
A team gets $q$ right out of $Q$, and takes a time $t$ out of a maximum $T$, should score
$$\frac{q+1-(t/T)}{Q+1}$$


问:
Calculate the derivatve of the differentiable function $f ( x, y, z ) = 2x^2 + 7y^2 + 2z^2$
Calculate the derivatve of the differentiable function $f ( x, y, z ) = 2x^2 + 7y^2 + 2z^2$   according to the direction in the direction of the vector $v_p$, as $v = (1, 2, 3) \in\mathbb{R}^3$,  $P= (3,2,1)\in \mathbb{R}^3$ and $v_p$ is the tangent vector.

Here is my question, I found it on my differential geometry book without a solution and I couldn't solve. Can you help? Thankss..

答:
First you have to calculate the gradient of your function:

$$f'_x=4x,f'_y=14y,f'_z=4z \implies\nabla f(x,y,z)=(4x,14y,4z).$$
If we now evaluate the gradient of $f$ in $(3,2,1)$ we find $\nabla f{(3,2,1)}=(12,28,4)$.

The vector $v=(1,2,3)$ is such that $||(1,2,3)||=\sqrt{14}$, so we have to normalize it in order to find a versor whose norm is $1$: $v'=\dfrac{(1,2,3)}{||(1,2,3)||}=\Big(\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}\Big)$. In conclusion:
$$D_{v'}f(3,2,1)=\Big\langle\nabla f(3,2,1),\Big(\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}\Big)\Big\rangle= \frac{12}{\sqrt{14}} +\frac{56}{\sqrt{14}}+\frac{12}{\sqrt{14}}=\frac{40\sqrt{14}}{7}$$


答:
$\sum\limits_{k=0}^n (k+1) \binom n k = \sum\limits_{k=0}^n \binom n k + \sum\limits_{k=0}^n k\binom n k = 2^n + n\sum\limits_{k=0}^{n-1} \binom {n-1} k $.

The last "=" is because
$k \binom n k = \frac{k*n!}{k!(n-k)!} = n\frac{(n-1)!}{(n-k)!(k-1)!}$.

Because $\sum\limits_{k=0}^{n-1} \binom {n-1} k = 2^{n-1}$, $\sum\limits_{k=0}^n (k+1) \binom n k = 2^{n-1}(n+2)$.

Done.

Title: Prove $\sum_{k=0}^n (k + 1)\binom{n}{k}= 2^{n - 1}(n + 2)$

Body:
I am looking at the following problem and can not get any further.

$$\sum_{k=0}^n (k + 1)\binom{n}{k}= 2^{n - 1}(n + 2)$$

Induction begin for $n = 0$

$$
\sum_{k = 0}^{0} (k + 1)\binom{0}{k} = 2^-1 (0 + 2) \\
1 = 1
$$

Shows $n$ holds, then $n + 1$ holds:

$$
\sum_{k = 0}^{n + 1} (k + 1)\binom{n + 1}{k} = 2^n (n + 3) \\
(n + 2)\binom{n+1}{n+1} + \sum_{k = 0}^{n} (k + 1)\binom{n + 1}{k} = 2^n (n + 3) \\
(n + 2) + \sum_{k = 0}^{n} (k + 1) (\binom{n}{k - 1} + \binom{n}{k})= 2^n (n + 3) \\
(n + 2) + \sum_{k = 0}^{n} (k + 1) \binom{n}{k - 1} + \sum_{k = 0}^{n} (k + 1) \binom{n}{k} = 2^n (n + 3) \\
(n + 2) + \sum_{k = 0}^{n} (k + 1) \binom{n}{k - 1} + 2^{n - 1}(n + 2) = 2^n (n + 3) \\
$$


问:
Proving $E\left[G(X)\right]=\int_0^\infty P(X\ge t)\,dG(t)$ where $X\ge 0$ and $G(\ge 0)$ is an increasing, right-continuous function
Suppose $G$ is an increasing, right-continuous function such that $G(x)\ge 0$ for all $x\ge 0$ and $G(0)=0$. If $X$ is an arbitrary non-negative random variable, then I am trying to show that

$$E\left[G(X)\right]=\int_0^\infty P(X\ge t)\,dG(t) \tag{1}$$

Since $G(X)\ge 0$ a.e., comparing with this popular question, I think I should have

$$E\left[G(X)\right]=\int_0^\infty P(G(X)>t)\,dt \tag{2}$$

But does $(2)$ reduce to $(1)$ from a change of variables?

Starting from scatch, if $F$ is the distribution function of $X$, then I have $$E\left[G(X)\right]=\int_0^\infty G(x)\,dF(x)=\int_0^\infty \left\{\int_0^{G(x)}\,dy\right\}dF(x)$$

Here I thought of using Fubini's theorem but I am not sure how to proceed with that.

Does $G^{-1}$ exist? If $G^{-}$ is some sort of generalized inverse of $G$, then I could perhaps say $$0< y< G(x),0<x<\infty\implies G^{-}(y)< x<\infty,0<y<G(\infty)$$

Using integration by parts I think it is true that $$\int_0^\infty G(x)\,dF(x)=\int_0^\infty P(X\ge t)\,dG(t)$$

But then I am not sure how the conditions on $G$ come into play. Any hint would be great.

答: This is a simple application  of Fubini/Tonelli's Theorem: $\int_0^{\infty} P(X \geq t)dG(t)=\int_0^{\infty}\int I_{t \leq X} dPdG(t)= \int  \int_0^{\infty} I_{t \leq X} dG(t)dP=\int G(X) dP=E[G(X)]$.


问:
any monic polynomial of degree $n$ whose root is $A$ is the characteristic polynomial of $A$?
Consider any matrix $A$ of order $n\times n.$ How to show that any monic polynomial of degree $n$ whose root is $A$ is the characteristic polynomial of $A?$

I have used the result several times without knowing the proof.

答: This result is true if and only if the minimal and characteristic polynomial of $A$ are the same. It is true for example if $A$ has distinct eigenvalues...</p>


问:
Determining whether two free groups are conjugates
Let $A= \langle a,b \rangle$, i.e. the free group on two generators. Also, let $B=\langle b^{-1}a^{-1}bab,b^{-1}a^2b, b^2, a^2, ab^2a^{-1}, abab^{-1}a^{-1} \rangle$, $C= \langle a^{-1}b^{-1}aba,a^{-1}b^2a, a^2, b^2, ba^2b^{-1}, baba^{-1}b^{-1} \rangle$.

I'm trying to determine whether $B$ and $C$ are conjugate in $A$. From a glance, they don't look like they are but I'm not entirely sure how to formally write it. I know that the abelinations of $B$ and $C$ are the same in $C$, but that doesn't tell me whether they are actually conjugate or not.

答:
To prove that $B=C$, we just have to prove that all of the generators of $B$ are in $C$ and vice versa. Many of these are clear. Here is a proof that the first generator of $C$ is in $B$ - the others are similar. (Of course this was all guided by computer calculations.)

$$a^{-1}b^{-1}aba=(a^{-2})(ab^2a^{-1})^{-1}(abab^{-1}a^{-1})(ab^2a^{-1})(a^2).$$


问:
Standard deviation (sample) / average leads to constant when all values but one are zero
I was doing some database work for a client, and I came across an oddity that I hoped someone could explain.

Take the sequence 0,0,0...,0,n (where n is any non-zero number), calculate the standard deviation (sample), and divide that by the average, and you always get the square root of the total number of samples.

e.g.(for 5 samples)

```<code>0,0,0,0,100 = 2.236067977
0,0,0,0,-0.21 = -2.236067977
</code>```

(for 12 samples)

```<code>0,0,0,0,0,0,0,0,0,0,0,22 = 3.464101615
0,0,0,0,0,0,0,0,0,0,0,-0.1 = -3.464101615
</code>```

Is there an obvious explanation for this?

答:
Assuming $ x_i = 0 $ for $ i \neq N $ as in yout example.

Mean:
$$ \bar x = \frac{1}{N}\sum_{i=1}^Nx_i = \frac{x_N}{N} $$
Variation:
$$ \sigma^2 = \frac{1}{N-1}\sum_{i=1}^N(x_i-\bar x)^2 $$
Ratio $ \sigma/\bar x $
$$ \frac{\sigma^2}{\bar x^2} = \frac{1}{N-1}\sum_{i=1}^N\left(\frac{x_i}{\bar x} - 1\right)^2 = \frac{1}{N-1}(N-1) + \frac{1}{N-1}\left(\frac{x_N}{x_N/N} - 1\right)^2 = 1 + (N-1) = N $$
Note how the sum becomes $ N-1 $ terms of $(-1)^2=1$ and the the final different term.


问:
Optimality of Khintchine's inequality, convergence in distribution and convergence of moments
Let $b_1,..b_n$ be real numbers and $\varepsilon_1,...,\varepsilon_n$  be independant Rademacher random variables. The Khintchine's inequality states that $$\mathrm{E}\left [ \left (  \sum_{i=1}^{n} b_i\varepsilon_i \right )^{2p}\right ]\leqslant \frac{\left ( 2p \right )!}{2^pp!}\left ( \sum_{i=1}^{n}b_i^2 \right )^p$$ for every integer $p \geqslant 1$.

I'm trying to prove that the constant $\frac{\left ( 2p \right )!}{2^pp!}$ is optimal, in the sense that it is impossible to obtain an inequality that holds for every Rademacher sum with a strictly smaller constant that does not depend on the dimension $n$.

Since $\frac{\left ( 2p \right )!}{2^pp!}$ is the $2p$-th moment of a standard normal variable, my idea was to approximate a well chosen Rademacher sum with a standard normal variable to obtain the optimality.

Let $b_1=...=b_n=1$. The central limit theorem ensures that $Z_n=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\varepsilon_i$ converges in distribution towards a random variable $X$ of distribution $\mathcal{N}(0,1)$.

If that implied that $$\lim_{n\rightarrow\infty}\mathrm{E}[Z_n^{2p}] = \mathrm{E}[X^{2p}]$$ then we would have $$\lim_{n\rightarrow\infty}\frac{1}{n^p}\mathrm{E}\left [  \left ( \sum_{i=1}^{n}\varepsilon_i \right )^{2p}\right ] = \frac{\left ( 2p \right )!}{2^pp!}$$ which proves the optimality.

So my question really is : is it true that $\lim_{n\rightarrow\infty}\mathrm{E}[Z_n^{2p}] = \mathrm{E}[X^{2p}]$ ? I don't think the dominated convergence theorem works here since $Z_n$ is not bounded.

The interpretation of convergence in distribution in terms of pointwise convergence of the characteristic functions yields $\forall t \in \mathbb{R}, \lim_{n\rightarrow\infty} \cos(\frac{t}{\sqrt{n}})^n=e^{-\frac{t^2}{2}}$. Could that be of any use ?

答:
Indeed, the dominated convergence theorem does not work directly, since $\sup_n \lvert Z_n\rvert$ is not integrable.

However, we can conclude the wanted convergence if we can show that for each $p\geqslant 1$,  $\left(Z_n^{2p}\right)$ is uniformly integrable, see here for the details.

If we show that $\sup_{n\geqslant 1}\mathbb E\left[Y_n^{2p}\right]$ is finite for each $p\geqslant 1$, we will get the uniform integrability. Indeed, if we want to show that $\left(Z_n^{2p_0}\right)$ is uniformly integrable for some $p_0$, we use the fact that $\left(Z_n^{2(p_0+1)}\right)$ is bounded in $\mathbb L^1$.

Now $\mathbb E\left[Y_n^{2p}\right]$ can be estimated by Khintichine inequality and can be bounded independently of $n$.

Alternatively, one can start from
$$
\left\lvert\mathbb E\left[Y_n^{2p}\right]-\mathbb E\left[X^{2p}\right]\right\rvert =2p\left\lvert \int_0^\infty t^{2p-1}\left(
\mathbb P\left(\left\lvert Y_n\right\rvert >t\right)-\mathbb P\left(\left\lvert X\right\rvert>t\right)\right)dt\right\rvert,
$$
split the integral into two parts: on $(0,A)$, which can be bounded using the uniform convergence of the c.d.f. of $Y_n$ to that of $X$; for the integral on $(A,+\infty)$, one can use Hoeffding's inequality to show that its contribution vanishes as $A$ goes to infinity uniformly on $n$.


问:
The volume between 2 planes and a octant
The problem that i tried to solve have this particular case:

Problem : In $R^3$, Let
$P_1$ be the plane
$x+y+z=l$, $l$ greater than $0$, and let $P_2$ be a generic plane with equation $a x+b y+c z=0$, with {a,b,c} not all with the same sign (otherwise, our problem would'n be well definied since the vector (a,b,c) would be in the first octant). Find the 2 volumes of the regions determined by the the 2 planes and the first octant.

The general case is a generalization of this problem to higher dimensions that i come up and can't solve it:
In $R^n$, Let
$P_1$ be the plane
$x_1+x_2+ ... + x_n=l$, $l$ greater than $0$, and let $P_2$ be a generic plane with equation $a_1 x_1+a_2 x_2+...+a_nx_n=0$. Supposing that this problem is well defined, Find the 2 volumes of the regions determined by the the 2 planes and the first "octant" (i.e., the points $(x_1,x_2,...,x_n)$ with $x_i>0, i = 1,...,n$).

答:
Let us consider the two planes with equations:

$$\begin{cases}(P_1): \ &x+y+z&=&L\\(P_2): \ &ax+by&=&z \end{cases}$$

(with an unimportant difference of notation with you for the second equation),

The intersection of plane $(P_1)$ with the first octant is an equilateral triangle (yellowish).  Here we have taken $L=3$. Plane $(P_2)$ (dark green) has here equation $z=-2x+3y.$

we obtain intersection points with:

$$\begin{cases}A(0,L/(b+1),Lb/(b+1))\\B(-bL/(a-b),aL/(a-b),0)\\C(0,L,0)\end{cases}$$

The volume of tetrahedron $OABC$ will be obtained classically as

$$\dfrac16 \det[\vec{OA},\vec{OB},\vec{OC}]=\dfrac{L^3b^2}{6(b - a)(b + 1)}$$

The other volume can be easily computed by difference.


问:
Numerical analysis , using Taylor Theorem
On interval [$-\pi ,\pi$ ] , show that $1-\frac{{x}^{2}}{2}\le \cos\left(x\right)$ by using Taylor theorem.

I thought that choose ${x}_{0}=0$ and $f(x)=\cos\left(x\right)$. Then,

$f\text{'}(x)=-\sin\left(x\right)$ , $f\text{'}\text{'}(x)=-cos\left(x\right)$ and $f\text{'}\text{'}\text{'}(x)=sin\left(x\right)$ then by Taylor theorem, expansion of function is ,

$\cos(x)=1-\frac{{x}^{2}}{2}+\sin\left(\xi (x)\right)\frac{{x}^{3}}{6}$

But then I stuck here. Any help will be appreciated.

Body:
Case 1: $x \in [0, \pi]$, then $ \xi(x) \in [0,\pi]$, hence $\sin\left(\xi (x)\right)\frac{{x}^{3}}{6} \ge 0.$ Thus $1-\frac{{x}^{2}}{2}\le \cos\left(x\right)$

Case 2: $ x \in [- \pi,0]$. Then  $ -x \in [0,\pi]$. Case 1 now shows:

$1-\frac{{x}^{2}}{2}= 1-\frac{{(-x)}^{2}}{2} \le \cos(-x) = \cos(x).$


问:
Show that $\{(X_{n},X_{n+1})\}_{n\geq 0}$ is a Markov Chain where $\{X_{n}\}_{n\geq 0}$ is a Markov Chain
Show that $\{(X_{n},X_{n+1})\}_{n\geq 0}$ is a Markov Chain where $\{X_{n}\}_{n\geq 0}$ is a Markov Chain.

Remark: We know that $\mathbb{P}(A|\emptyset)$ is undefined, I am right?
This fact is important in my attempt and is what motivates this post.

My attempt: We define $Y_{n}:=(X_{n},X_{n+1})$, then we are going to verify if we have the Markov property. On the one hand, we know that
\begin{align}
\mathbb{P}[Y_{n}=(i,j)\:|\:Y_{n-1}=(i_{n-1},j_{n-1})]&=\mathbb{P}[X_{n}=i,X_{n+1}=j\:|\:X_{n-1}=i_{n-1},X_{n}=j_{n-1}]\\
&=\left\{\begin{array}{ll}0 &\mbox{ if }i\neq j_{n-1}\\ p_{ij} &\mbox{ if } i=j_{n-1}. \end{array}\right.
\end{align}
where $P$ is transition matrix of $\{X_{n}\}_{n\geq 0}$. On the other hand, we have

\begin{align}
&\mathbb{P}[Y_{n}=(i,j)\:|\:Y_{0}=(i_{0},j_{0}),\ldots,Y_{n-2}=(i_{n-2},j_{n-2}),Y_{n-1}=(i_{n-1},j_{n-1})]\\
&=\mathbb{P}\left[X_{n}=i,X_{n+1}=j\:|\:X_{0}=i_{0},X_{1}=j_{0},\ldots,X_{n-2}=i_{n-2},X_{n-1}=j_{n-2},X_{n-1}=i_{n-1},X_{n}=j_{n-1}\right]\\
&=\left\{\begin{array}{ll}\mbox{undefined} & \mbox{if }\exists k\in\{0,1,\ldots,n-2\}\mbox{ such that }j_{k}\neq i_{k+1}\\ 0 & \mbox{if }\forall k\in\{0,1,\ldots,n-2\}\mbox{ such that }j_{k}=i_{k+1} \mbox{ and }j_{n-1}\neq i\\ p_{ij} & \mbox{if }\forall k\in\{0,1,\ldots,n-2\}\mbox{ such that }j_{k}=i_{k+1} \mbox{ and }j_{n-1}= i \end{array}\right.
\end{align}
My problem is in the undefined case, this is making me think that this chain is not a Markov chain. What do you think about this?

This exercise appears in Adventures in Stochastic Processes; Resnick, Sidney I.

答:
Usually, in the definition of a Markov chain (with state space $\mathbb N$),
we require the equality
$$
\mathbb P\left(X_n=j\mid X_{n-1}=i,X_{k}=i_k,0\leqslant k\leqslant n-2\right)=\mathbb P\left(X_n=j\mid X_{n-1}=i\right)
$$
to hold for each $i$, $i_1,\dots,i_{n-2}$ such that the event $A:=\{X_{n-1}=i,X_{k}=i_k,0\leqslant k\leqslant n-2\}$ has a positive probability. This is possible that the previously event $A$ has probability zero, for example if $i_1$ and $i_2$ are two states which do not communicate. So an undefined case may also hold for Markov chains which are not defined as a vector like $(X_n,X_{n+1})$ hence this "issue" (quote mark because it is not really one) is not specific to this problem.


问:
What is the expected present discounted value of a variable following a geometric Brownian motion?
According to Investment under Uncertainty by Dixit & Pindyck (1993, p.82), the expected present discounted value of a variable following a geometric Brownian motion is defined as: $$E\left[\int_{0}^{\infty}{F(x(t))}\exp[-rt]dt\right]$$
Suppose $$\tag 1 F(x)=x^\theta$$ and $$\tag 2 dx={α}{x}{dt}+{σ}{x}{dz}$$ where $dz$ is the increment of a Weiner process. Equation $(1)$ then follows a geometric Brownian motion. Exapanding equation $(1)$ by Itô's lemma, we can derive $$\tag 3 \theta{x}^\theta({α}{x}{dt}+{σ}{x}{dz})+\frac{1}{2}\theta(\theta-1){x}^\theta\sigma^2dt$$
Substituting equation $(1)$ for ${x}^\theta$ and rearranging gives $$\tag 4 [\theta{α} + \frac{1}{2}\theta(\theta-1)\sigma^2]Fdt + \theta\sigma{F}dz$$
If $x(0)$ is currently ${x_0}^\theta$, the expected value of $F(x)$ is given by equation $(5)$. $$\tag 5 {x_0}^\theta\exp[t(\theta{\alpha}+\frac{1}{2}\theta(\theta-1)\sigma^2)]$$
According to the textbook mentioned above, the present discounted value of equation $(5)$ is given by equation $(6)$ $$\tag 6 \frac{x_0^\theta}{r-(\theta{\alpha}+\frac{1}{2}\theta(\theta-1)\sigma^2)}$$

As far as I understand, deriving this result involves multiplying equation $(5)$ by $\exp[-rt]$ and integrating with respect to time between $t=0$ and $t=\infty$. Doing this, we arrive at the integral given by equation $(7)$ $$\tag 7 \int_{0}^{\infty}\frac{x_0^\theta}{\exp[t(r-(\theta{\alpha}+\frac{1}{2}\theta(\theta-1)\sigma^2))]}dt$$
However, I do not understand how $(5)$ follows from $(7)$, if this is indeed the correct approach. Any assistance in understanding this problem would be appreciated.

答:
The anti-derivative of (7) is $$\left[\frac{- x_o^\theta}{r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)} exp^{-t (r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)}\right]^\infty_0 $$ Now evaluate as $t \rightarrow \infty$ and $t \rightarrow 0$
$$\lim_{t \rightarrow \infty} \left[\frac{- x_o^\theta}{r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)} exp^{-t(r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)}\right] + \lim_{t \rightarrow 0} \left[\frac{x_o^\theta}{r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)} exp^{-t(r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)}\right]$$ $$0 + \frac{ x_o^\theta}{r - (\theta \alpha + \frac{1}{2} \theta (\theta - 1) \sigma^2)}$$ Which is equal to equation (6).


答:
This works when the minimal polynomial of $A$, $m_A(x)$, is $n$th order.

Why? Because given another $n$th degree polynomial $f$ with leading coefficient 1, $f-m_A$ has $A$ as a root but has degree lower than $n$. Therefore it must be 0.

Title: any monic polynomial of degree $n$ whose root is $A$ is the characteristic polynomial of $A$?

Body:
Consider any matrix $A$ of order $n\times n.$ How to show that any monic polynomial of degree $n$ whose root is $A$ is the characteristic polynomial of $A?$

I have used the result several times without knowing the proof.


问:
Every non-empty perfect set in $\mathbb{R}^{k}$ is uncountable
In the Principles of Mathematical Analysis, 3rd Ed. by Rudin, he mentions in a Theorem 2.43 on page 41 that every non-empty perfect set $P$ in $\mathbb{R}^{k}$ is uncountable.

In the proof, he assumes contrariwise that the points of $P$ are denumerable and labelled $\overline{x_{1}}, \overline{x_{2}}, \overline{x_{3}}, ...$ . He then constructs a sequence of neighborhoods $\{V_{n}\}$ by induction as follows:

Let $V_{1}$ be the neighborhood of $\overline{x_{1}}$, defined as $V_{1} = \{\overline{y} \in \mathbb{R}^{k} : |\overline{x_{1}} - \overline{y}| < r\}$, and the closure of $V_{1}$ denoted $\overline{V_{1}}$ defined $\overline{V_{1}} = \{\overline{y} \in \mathbb{R}^{k} : |\overline{x_{1}} - \overline{y}| \leq r\}$.

Now, suppose $V_{n}$ has been constructed, with $V_{n} \cap P \neq \emptyset$. Then, there exists $V_{n+1}$ such that:

$\overline{V_{n+1}} \subset V_{n}$ (the closure of $V_{n+1}$ is a proper subset of $V_{n}$)

$\overline{x_{n}} \notin \overline{V_{n+1}}$

$\overline{V_{n+1}} \cap P \neq \emptyset$.

The third condition ensures our induction hypothesis holds and allows for the construction to continue.

Now, let $K_{n} = \overline{V_{n}} \cap P$. Since $\overline{V_{n}}$ is closed and bounded, it must be compact as well (proved earlier). Since $\overline{x_{n}} \notin K_{n+1}$ by construction, no point of $P$ is in $\cap_{1}^{\infty}K_{n}$. Since $K_{n} \subset P$, this implies that $\cap_{1}^{\infty}K_{n} = \emptyset$. This fact in turn contradicts a corollary of an earlier theorem that such an intersection must be non-empty.

Where my understanding falls short is in the statement that is in bold. The contradiction relies on it. Now, since since $P$ is perfect, $P$ is closed and every point of $P$ must be a limit point of $P$. Further, since $P$ is non-empty, it must have infinitely many points (or else, it cannot have any limit points). Now, when I take a point $x_{n+1}$ inside the neighborhood $V_{n}$ of $x_{n}$, and construct a neighborhood $V_{n+1}$ in a manner that doesn't include $x_{n}$, $V_{n+1}$ still contains infinitely many points; it just doesn't contain $x_{n}$. So, when I continue this induction, doesn't each subsequent neighborhood $V_{m}$ have points in it? And, when I take an infinite intersection as suggested, it needn't contain the specific labelled points at the centre of the neighborhood; but it still does contain infinitely many points.

Where am I going wrong?

For some visual context, I am visualizing my argument as a set of circles with each successive circle (denoting $V_{n}$) drawn within the previous one but drawn so that it doesn't contain the centre of the previous one.

答: You are making things too complicated. $\overline {x_n} \notin K_{n+1}=\overline {V_{n+1}} \cap P$ by property 2). If $\overline {x_i}$ is a point of $P$ which belongs to every $K_n$ then  $\overline {x_i} \in K_{i+1} \subset \overline {V_{i+1}} $ again contradicting 2).


问:
Is $\mathbb{Z}[√13] $ a Unique factorization domain?
I think it is not so as $12 $ can be written in two ways $12=2.6=(1+\sqrt{13})(-1+\sqrt{13})$. Are these two factorization unique upto irreducibles? Please help.

答: $\mathbb{Z}[\sqrt{13}]$ is not even integrally closed (consider for instance $\displaystyle \frac{3+\sqrt{13}}{2}$), Hence it can't be a principal ideal domain. It also can't be a unique factorization domain because a Dedekind domain is a PID if and only if it is a UFD, but $\mathbb{Z}[\sqrt{13}]$ is not even integrally closed thus not Dedekind. At any rate: if you're not integrally closed then you're not a UFD.


问:
Could the integral be calculated by complex analysis?
The following integral seems to be related to a complex integration:
$$
\int_{0}^{1}e^s\left(\frac{1}{s}-\frac{1}{s^2}\right)ds.
$$
However, the upper end is $1$ instead of $+\infty$ in which case usually can be dealt with a Cauchy integral. Please give me some help or hints. Thanks a lot.

答:
The integral is equal to $-\infty$ since
$$
\int_0^1e^s({1\over s}-{1\over s^2})ds{\le \int_0^1{1\over s}-{1\over s^2}ds
\\=\ln |s|+{1\over s}\Big|_0^1
\\=-\infty
}
$$


问:
Integration of product of logarithm
For integration of
$ \int_0^1 (\log(1-u)^5)( \log(u)^5)/(u-1) du ,$
I have tried integral-by-parts, change of variable $ \log u =x $, and also expand it in series.
But I didn't get an answer. Could any one help me out?

答:
Use $\int_{0}^{a} f(x) dx=\int_{0}^{a} f(a-x) dx.$
$$I=\int_{0}^{1} \frac{\ln(1-x)^5 \ln x^5}{x-1}dx=-25\int_{0}^{1} \frac{\ln(1-x) \ln x}{x}dx.~~~~(1)$$
Let us use polylog functions defined as $$\text{Li}_s(x)=\sum_{k=1}^{\infty} \frac{x^k}{k^s}$$
See https://en.wikipedia.org/wiki/Polylogarithm
$$\int \frac{\ln(1-x)}{x} dx= -\text{Li}_2(x), \int \frac{\text{Li}_2(x)}{x} dx= \text{Li}_3(x), \lim_{|x|\to 0} \text{Li}_s(x)=x$$
Let us do integration by parts of (1) taking $\ln x$ as first function, then
$$I=-25\left(\left . -\ln x \text{Li}_2(x)\right|_{0}^{1}+\int_{0}^{1} \frac{1}{x} \text{Li}_2(x) dx\right)=-25~\text{Li}_3(x)|_{0}^{1}=-25 \sum_{k=1}^{\infty} \frac{1}{k^3}=-25 \zeta(3)$$

Title: Integration of product of logarithm


答:
The sample standard deviation divided by the average (the coefficient of variation, or abbreviated 'CV) can be denoted as:

$$CV = \frac{\sigma}{\mu}$$

or (with the sample standard deviation)

$$ CV= \frac{\sqrt{\frac{1}{n-1}\sum_{i=1}^n(x_i-\bar x)^2}}{\frac{1}{n}\sum_{i=1}^nx_i} $$

When all entries (n-entries) from the sequence are zero, except for one (let's call this entry $E$), we know that the average will be uniformly distributed over the sequence, so $$\mu= \frac{E}{n}$$

And then we can also write the $CV$ differently, as:

$$ CV= \frac{\sqrt{\frac{1}{n-1}\sum_{i=1}^{n-1}(\frac{E}{n})^2+(E-\frac{E}{n})^2}}{\frac{E}{n}} $$

or as

$$ CV= \frac{\sqrt{\frac{1}{n-1}\times[(n-1)(\frac{E}{n})^2+(E-\frac{E}{n})^2}]}{\frac{E}{n}} $$

easier to simplify, we take:

$$ CV= \frac{\sqrt{\frac{1}{n-1}\times[(n-1)(\frac{E}{n})^2+(E-\frac{E}{n})^2}]n}{E} $$

which simplifies to:

$$\frac{n\frac{E}{\sqrt{n}}}{E}$$ and leaves you $$\sqrt{n}$$

Title: Standard deviation (sample) / average leads to constant when all values but one are zero

Body:
I was doing some database work for a client, and I came across an oddity that I hoped someone could explain.

Take the sequence 0,0,0...,0,n (where n is any non-zero number), calculate the standard deviation (sample), and divide that by the average, and you always get the square root of the total number of samples.

e.g.(for 5 samples)

```<code>0,0,0,0,100 = 2.236067977
0,0,0,0,-0.21 = -2.236067977
</code>```

(for 12 samples)

```<code>0,0,0,0,0,0,0,0,0,0,0,22 = 3.464101615
0,0,0,0,0,0,0,0,0,0,0,-0.1 = -3.464101615
</code>```

Is there an obvious explanation for this?


问:
Find a polynomial $p$ of degree $3$ such that $-1,2$,and $3$ are zeroes of $p$ and $p(0)=1$ where book solution lacks rigour.
Of course, I considered $p(x) = (x+1)(x-2)(x-3)$, but it is clear that $p(0) \neq 1$. A bit stumped by this I see that the book solution says that we MUST have $p(x) = c(x+1)(x-2)(x-3)$ for some $c$. I understand why this is an idea to get to a polynomial that satisfies the conditions. What I don't understand is how it is obvious $p(x)$ MUST be of this form. If I knew this I believe I would have considered this idea on my own.

Is there a rigorous argument to see why $p(x)$ MUST be able to be reduced in this way (preferably using only precalculus knowledge)?

答: Suppose that $-1$, $2$, and $3$ are roots of $p(x)$ and that $\deg p(x)=3$. Divide $p(x)$ by $(x+1)(x-2)(x-3)$; you will get$$p(x)=(x+1)(x-2)(x-3)q(x)+r(x)$$where $\deg r(x)<3$. But then since $\deg p(x)=3$, $\deg q(x)=0$; that is, $q(x)=c$, for some constant $c$. So, you have$$p(x)=c(x+1)(x-2)(x-3)+r(x)$$and, since $r(-1)=r(2)=r(3)=0$ and $\deg r(x)<3$, $r(x)=0$.


问:
Does every smoothing operator have a unique kernel?
Consider smooth manifolds $M, N$ and the spaces of smooth functions, $C^\infty(M)$, $C^\infty(N)$ and $C^\infty(M \times N)$. A smoothing operator $T_k$ with kernel $k$ is defined as
$$
(T_k f)(x) = \int_N k(x, y) f(y) dy,
$$
where $f \in C^\infty(N)$ and $k \in C^\infty(M \times N)$ and $T_k f \in C^\infty(M)$.

In this way, every smooth kernel $k \in C^\infty(M \times N)$ induces a linear operator $T_k : C^\infty(N) \rightarrow C^\infty(M)$. My first question is whether the kernel $k$ of a smoothing operator $T_k$ is unique, or whether it is possible that $T_k = T_{k'}$ for $k \neq k'$. My second question is whether any (bounded?) linear operator $T : C^\infty(N) \rightarrow C^\infty(M)$ can be written as an integral transform with a smooth kernel $k \in C^\infty(M \times N)$. Finally, I'm interested in the same questions for the case of smooth vector valued functions and operator-valued / matrix-valued kernels (i.e. $f : N \rightarrow V$, $k : M \times N \rightarrow \operatorname{Hom}(V, W)$ and $T_k f : M \rightarrow W$).

I think that the Schwartz kernel theorem is relevant here, but I am not too familiar with the theory of distributions, and for my application I need to stay within the realm of smooth functions and kernels.

In addition to an explanation or proof, references for the theorems would be much appreciated.

答:
There is a follow-up theorem to the Schwartz-Kernel Theorem in Hörmanders Linear Partial Differential Equations 1, Thm., 5.2.6, saying that every continiuous map from $\mathcal{C}^\infty_0(N)\rightarrow \mathcal{C}^\infty(M)$ is defined by a smooth integral kernel (with no requirements to the support of $k$) and vice versa.

Jumping into distribution theory, the extention from $\mathcal{C}^\infty_0(N)\rightarrow \mathcal{C}^\infty(M)$ to $\mathcal{E}'(N)\rightarrow \mathcal{C}^\infty(M)$ gives you the uniqueness of $k$, since $\delta_q\in\mathcal{E}'(N)$  defined by $\delta_q g:=g(q)$ yields that
\begin{align*} \int_N k(x,y)\delta_q(y)dy=k(x,q) \end{align*}
and hence for any continiuous operator $T:\mathcal{C}^\infty_0(N)\rightarrow \mathcal{C}^\infty(M)$ we obtain a function $f_q\in\mathcal{C}^\infty(M)$ by $f_q:=T\delta_q$ for each $q\in N$. Define $k(\cdot,q):=f_q$ which gives you the integral kernel.
Note that the "integral" is to be understood as "applying $\delta_q$ to $k$", which turns out to be usefull in the theory of distributions. You can probably recover the same result by assuming $T$ is given by two different integral kernels $k$ and $k'$ and then evaluate these at a sequence $\tau_n$ of compactly supported simple functions with the condition that $\{q\}=\cap_n\text{supp}(\tau_n)$  and $\int_N \tau_n dy=1$ (i.e., $\tau_n\rightarrow \delta_q$ in the topology of distributions).

You can see that the compactness of $f$ is required for the integral to be finite if you allow non-compact kernels.


问:
Find density functions of $Z=X+Y$ when the joint density function is known.
Find the density function of $Z=X+Y$, $X$, $Y$ where the joint density function of $(X,Y)$ is given by $f(x,y) = \frac{1}{2} (x+y) e^{-(x+y)},\, x,y \geq 0$.

My initial idea is to calculate the distribution function of $Z$ like this:

$P(Z < z) = P(X+Y < z) = P(X < z-Y)$

$F_{z} = \frac{1}{2}\int\limits^{z}_{0}\int\limits^{z-y}_{0} (x+y) e^{-(x+y)} dx\, dy$

and then calculate its derivative $F_{z}^{'}$

Is this the way to go ?

答:
Joint density is $f(x,y) = \frac{1}{2} (x+y) e^{-(x+y)},\, x,y \geq 0$ and the distribution function is

$F_Z(z) = P(Z < z) = P (X+Y < z) = P ( X < z, Y < z - X)$

then we have for the density of Z

\begin{array}[rl]{l}
f_{Z}(z) &= \frac{\partial}{\partial z} F_z(z) = \frac{\partial}{\partial z} \int\limits_{0}^{z}\int\limits_{0}^{z-x} f(x,y)\,dy\,dx \\
         &= \frac{\partial}{\partial z} F_z(z) =  \int\limits_{0}^{z}\frac{\partial}{\partial z} \int\limits_{0}^{x-z} f(x,y)\,dy\,dx \\
         &=   \int\limits_{0}^{z} f(x,z-x)\cdot 1 - f(x,0)\cdot 0 \,dx \\
           &= \frac{1}{2} \int\limits_{0}^{z} z\, e^{-z}\, dx \\
           &= \frac{1}{2} z\cdot e^{-z} \int\limits_{0}^{z} \, dx \\
f_{Z}(z) &= \frac{1}{2} z^2\,e^{-z} \\
\end{array}


答:
Hint for the title question: We have $${\displaystyle (1+{\sqrt {13}})^{2}=2(7+{\sqrt {13}})}.$$
Since $2$ is irreducible in ${\displaystyle \mathbb {Z} [{\sqrt {13}}]}$ but does not divide ${\displaystyle 1+{\sqrt {13}}}$, we see that ${\displaystyle \mathbb {Z} [{\sqrt {13}}]}$ is not a UFD.

Title: Is $\mathbb{Z}[√13] $ a Unique factorization domain?

Body: I think it is not so as $12 $ can be written in two ways $12=2.6=(1+\sqrt{13})(-1+\sqrt{13})$. Are these two factorization unique upto irreducibles? Please help.


Title: Could the integral be calculated by complex analysis?

Body:
The following integral seems to be related to a complex integration:
$$
\int_{0}^{1}e^s\left(\frac{1}{s}-\frac{1}{s^2}\right)ds.
$$
However, the upper end is $1$ instead of $+\infty$ in which case usually can be dealt with a Cauchy integral. Please give me some help or hints. Thanks a lot.

Body:
$$\int e^s\left(\frac{1}{s}-\frac{1}{s^2}\right)\,ds=\frac{e^s}{s}$$
$$\int_{\epsilon}^{1}e^s\left(\frac{1}{s}-\frac{1}{s^2}\right)\,ds=e-\frac{e^{\epsilon }}{\epsilon }$$


问:
On the notation for indicator random variables
I'm studying a number of highly cited papers and I read the following notation:

$$\mathbb{1}_{d\in D}$$

which, from my understanding, is a shortcut for $\mathbb{1}_D(d)$ where $\mathbb{1}$ is the indicator function. In another paper, I found another notation for an indicator random variable:

$$\mathbb{1}\{X_1,\ldots,X_n\}$$

where $X_1,\dots,X_n$ are elements of a set.

I could not find any evidence about such ways to write indicator functions. In the first case, I don't think that writing $\mathbb{1}_{d\in D}$ is more readable than $\mathbb{1}_D(d)$. In the second case, a possible explanatio is that
$$\mathbb{1}\{X_1,\ldots,X_n\}$$

is more readable than

$$\mathbb{1}_{\{X_1,\ldots,X_n\}}$$

Since these are prominent papers written by well-known scientists, I'm afraid that I'm missing something. Do you know why people use these alternative notations?

Thanks!

答:
The reason that people use alternative notations is because there is no/little consensus in some areas of mathematics.
Note, that mathematics is the science that deals with logic and arrangements. Moreover, it is all around us.
Whether or not you believe that mankind had a single origin, by the time humans started writing, in particular writing down mathematics, they were well spread out over our globe. These groups of humans all had there own language and symbols. The aftermath, pun intended, can still be seen today. One can only hope the consensus in mathematical symbols increases over time. However, in particular fields, like machine learning, they often do not really care about the notational integrity.

I think you hoped for a different, more rational, reason but I do not believe there is any.


问:
Show that $4\mid f_n$ if and only if $6\mid n$.
We define the Fibonacci sequence via $f_1 = f_2 = 1$, and if $n > 2$, then $f_n = f_{n−1} + f_{n−2}$. Show that $4\mid f_n$ if and only if $6\mid n$.

I have been trying to make a copy of the following solution for the exercise "$ 3 \mid f_n $ iff $ 4 \mid n $".

But so far I have not found anything satisfactory. In this question $f_n$ is divisible by $4$ if and only if $n$ is divisible by $6$ they answer it by redefining the fibonacci sequence, but I wonder if it can be done as I am trying.
By the way, what they do here $f_n$ is divisible by $4$ if and only if $n$ is divisible by $6$  I don't understand very well.

答:
You can just go on and see that
$$
f_{n+6}=8f_{n+1}+5f_n
$$
If $f_n$ is divisible by $4$, then also $f_{n+6}$ is.

If $f_n$ is not divisible by $4$, then neither $f_{n+6}$ is.

(In simpler terms: $f_{n+6}\equiv f_n\pmod{4}$.)

Now just look at $f_0=1,f_1=1,f_2=1,f_3=2,f_4=3,f_5=5$, which is the start of the induction.


问:
When can the level of the test be exactly $\alpha ?$ in non randomized test. And how to use CLT to find the critical value.
Let $X_{1}, \ldots, X_{n}$ be a sample from the Bernoulli distribution with parameter $p$ Consider testing $H_{0}: p=p_{0}$ versus $H_{1}: p=p_{1}$ where $p_{0}<p_{1}$ are known numbers.

(a) Using the Neyman-Pearson lemma, find the most powerful test (Non-Randomized) among tests with level at most $\alpha$.   When can the level of the test be exactly $\alpha ?$

(b) Use the CLT to find a critical value such that the level of the test is approximately (asymptotically for large $n$ ) $\alpha$.

To use Neyman Peason lemma I calculate $r$ as :
$$r= \frac{f_1}{f_0} = \frac{p_1^{\Sigma x_i}(1-p_1)^{n-\Sigma x_i}}{p_0^{\Sigma x_i}(1-p_0)^{n-\Sigma x_i}}$$
For $p_1>p_0$ we have non randomized test is $\phi(x) =$ Reject $H_0$ for $\Sigma x_i \geq k$.

For given at most level $\alpha$ :
$$\alpha \geq E_{p_0} \phi(x) = P_{p_0}\{\Sigma x_i \geq k\} = \Sigma_{r=k+1}^n (n_{C_r}) p_{0}^r (1-p_{0})^{n-r}$$

(a) I don't know how to proceed after this and what to say about When can the level of the test be exactly $\alpha ?$

(b) How to use CLT to find the critical value which satisfies the given condition.

Please help me with this problem. Thankyou.

答:
I will get you started with couple of particular numerical examples that illustrate the main issues involved in a general answer.

Example 1. Let $n = 10, p_0 = .5, p_1 = .7, \alpha = 0.5.$ Then NP says to reject
when $T = \sum_{i=1}^{10}X_i \ge k.$

The issue is to
find $k$ so that $\alpha \le 0.05.$ Under $H_0,$
$T \sim \mathsf{Binom}(n=10, p=.5).$ Then using $k = 8$ would give $\alpha = 0.0547 > 0.05,$ so use $k = 9$ to
get $\alpha=0.0107 < 0.05.$  Computations in R,

```<code>qbinom(.95, 10,.5)
[1] 8
sum(dbinom(8:10, 10, .5))
[1] 0.0546875
sum(dbinom(9:10, 10, .5))
[1] 0.01074219
</code>```

Thus, there is no nonrandomized test exactly at level $\alpha = 0.05.$

Under $H_0,$ we have $E(T) = np = 5,$ $SD(T) = \sqrt{np(1-p)} = \sqrt{2.5} = 1.581.$

So $T\stackrel{aprx}{\sim}\mathsf{\mu=5, \sigma=\sqrt{2.5}}.$ So an approximate value of $k = 7.6.$
This is not a good approximation for $n$ as small as $n=10,$ but this kind of approximation is useful for large $n.$

```<code>qnorm(.95, 5, sqrt(2.5))
[1] 7.600742
</code>```

```<code>t = 0:10;  pdf = dbinom(t, 10, .5)
plot(t, pdf, type="h", lwd = 3, col="blue")
 abline(h=0, col="green2"); abline(v=0, col="green2")
 abline(v = 8.5, lwd=2, lty="dotted")
 curve(dnorm(x, 5, sqrt(2.5)), add=T, col="red")
</code>```

Example 2. Now suppose $n=100,$ keeping the other parameters the same. Then we have
$T\sim\mathsf{Binom}(100, .5)$ under $H_0$ and $k = 59$ gives a test at level $\alpha = 0.044$ and the normal
approximation says to use $k = 58.22.$

```<code>qbinom(.95, 100, .5)
[1] 58
sum(dbinom(58:100, 100, .5))
[1] 0.06660531
sum(dbinom(59:100, 100, .5))
[1] 0.04431304
qnorm(.95, 50, 5)
[1] 58.22427
</code>```

```<code>t = 0:100;  pdf = dbinom(t, 100, .5)
plot(t, pdf, type="h", col="blue")
 abline(h=0, col="green2"); abline(v=0, col="green2")
 abline(v = 58.5, lwd=2, lty="dotted")
 curve(dnorm(x, 50, 5), add=T, col="red")
</code>```

Body:
I will get you started with couple of particular numerical examples that illustrate the main issues involved in a general answer.

Example 1. Let $n = 10, p_0 = .5, p_1 = .7, \alpha = 0.5.$ Then NP says to reject
when $T = \sum_{i=1}^{10}X_i \ge k.$

The issue is to
find $k$ so that $\alpha \le 0.05.$ Under $H_0,$
$T \sim \mathsf{Binom}(n=10, p=.5).$ Then using $k = 8$ would give $\alpha = 0.0547 > 0.05,$ so use $k = 9$ to
get $\alpha=0.0107 < 0.05.$  Computations in R,

qbinom(.95, 10,.5)
[1] 8
sum(dbinom(8:10, 10, .5))
[1] 0.0546875
sum(dbinom(9:10, 10, .5))
[1] 0.01074219

Thus, there is no nonrandomized test exactly at level $\alpha = 0.05.$

Under $H_0,$ we have $E(T) = np = 5,$ $SD(T) = \sqrt{np(1-p)} = \sqrt{2.5} = 1.581.$

So $T\stackrel{aprx}{\sim}\mathsf{\mu=5, \sigma=\sqrt{2.5}}.$ So an approximate value of $k = 7.6.$
This is not a good approximation for $n$ as small as $n=10,$ but this kind of approximation is useful for large $n.$

qnorm(.95, 5, sqrt(2.5))
[1] 7.600742

t = 0:10;  pdf = dbinom(t, 10, .5)
plot(t, pdf, type="h", lwd = 3, col="blue")
 abline(h=0, col="green2"); abline(v=0, col="green2")
 abline(v = 8.5, lwd=2, lty="dotted")
 curve(dnorm(x, 5, sqrt(2.5)), add=T, col="red")

Example 2. Now suppose $n=100,$ keeping the other parameters the same. Then we have
$T\sim\mathsf{Binom}(100, .5)$ under $H_0$ and $k = 59$ gives a test at level $\alpha = 0.044$ and the normal
approximation says to use $k = 58.22.$

qbinom(.95, 100, .5)
[1] 58
sum(dbinom(58:100, 100, .5))
[1] 0.06660531
sum(dbinom(59:100, 100, .5))
[1] 0.04431304
qnorm(.95, 50, 5)
[1] 58.22427

t = 0:100;  pdf = dbinom(t, 100, .5)
plot(t, pdf, type="h", col="blue")
 abline(h=0, col="green2"); abline(v=0, col="green2")
 abline(v = 58.5, lwd=2, lty="dotted")
 curve(dnorm(x, 50, 5), add=T, col="red")


答: Use the integration formula $$\int e^x[f(x)+f'(x)] dx=e^x f(x),$$ which comes by integration by parts. Then $$\int_{0}^{1}e^s\left( \frac{1}{s}-\frac{1}{s^2} \right) ds=\frac{e^s}{s}|_{0}^{1}=-\infty$$


问:
Is a $R \otimes S$-module of the form $V \otimes W$ with $V$ simple $R$-module and $W$ simple $S$-module a simple module?

Let $R$ and $S$ be $\mathbb{C}$-algebras. Show that $X$ is a simple $R \otimes S$-module if and only if it is isomorphic to a $R \otimes S$-module of the form $V \otimes W$ with $V$ simple $R$-module and $W$ simple $S$-module.

Starting with the backwards direction, I know that if we assume the semisimplicity of $R$ and $S$, I can show that the endomorphism ring of $V \otimes W$ is a division ring (it is isomorphic to $\mathbb{C}$), since we can also show $R \otimes S$ is finite dimensional and semisimple, that would imply that $V \otimes W$ is simple.

But now I would like to show both directions $without$ assuming the semisimplicty of $R$ and $S$, and I'm lost.

答:
Let's begin with some generalities. Let $K$ be any field, $R$, $S$ two $K$-algebras, and $V$ (resp. $W$) a simple right module over $R$ (resp. $S$). As Qiaochu noted, $V\otimes_KW$ need not be simple if $V$ and $W$ are not finite dimensional, so let's restrict to that case. Set $D=\mathrm{End}_R(V)$ and $E=\mathrm{End}_S(W)$, division algebras acting on the left on $V$ and $W$ respectively.

Passing to $R/\mathrm{Ann}(V)$ we may assume that $R$ is a finite dimensional simple $K$-algebra. Applying Morita theory, we may assume that $R=D$ (no opposite needed since endomorphisms act on the left), and $V=R$ as right modules. Similarly for $S$. Thus $V\otimes_KW$ is just the regular representation for $D\otimes_KE$.

Let $Z(D)$ be the centre of $D$, and set $L:=Z(D)\otimes_KZ(E)$. Then $D\otimes_KE\cong D\otimes_{Z(D)}L\otimes_{Z(E)}E$. The properties of this are all controlled by the properties of $L$. In particular, $V\otimes_KW$ is simple if and only if $L$ is a field.

To see this, suppose $L$ is a field. Then $D$ is a central simple algebra (CSA) over $Z(D)$, so $D\otimes_{Z(D)}L$ is a CSA over $L$. Similarly for $L\otimes_{Z(E)}E$, and so their tensor product over $L$ is again a CSA over $L$.

If $L$ is a proper product of rings, then so too is $D\otimes_KE$, and hence $V\otimes_KW$ has proper summands. If $L$ has nilpotent elements, then so too does $D\otimes_KE$, and $V\otimes_KW$ has a proper submodule.

Finally, assume $K$ is algebraically closed. Then $D=K=E$, so $L=K$, and the tensor product $V\otimes_KW$ is simple.


问:
Limit of a composite function using trigonometric identities
Show that if $\lim_{x\to a} f(x) = L$, then     $\lim_{x \to a} cos(f(x))=cos(L)$.

Hint: looking up some trigonometric identities may help you

I have been using delta-epilson to prove this statement, but the hint that has been given is throwing me off. Why would using trigonometric identities help with this question?

答:
$$\cos(A)-\cos(B)=-2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$$
Apply this with $A=f(x)$ and $B=L$. Then
$$\left|\cos(f(x))-\cos(L)\right|=2\left|\sin\left(\frac{f(x)+L}{2}\right)\sin\left(\frac{f(x)-L}{2}\right)\right|.$$
Now, recall that $|\sin(t)|\le|t|$, so,
$$\left|\cos(f(x))-\cos(L)\right|\le\frac{1}{2}|f(x)+L||f(x)-L|.$$ And from here, it is easy to conclude a $\delta-\varepsilon$ type proof.

For the sake of completeness,
$$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$$
$$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$$
by substracting
$$\cos(a+b)-\cos(a-b)=-2\sin(a)\sin(b).$$
Now, calling $A=a+b$ and $B=a-b$, gives $a=\frac{A+B}{2}$ and $b=\frac{A-B}{2}$.


问:
$f(\mathbb{R})$ continuous iff $m(k)=\lbrace x\in \mathbb{R} \mid f(x)<k\rbrace$ and $M(k)=\lbrace x \in \mathbb{R} \mid f(x)>k \rbrace$ are open

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ a continuous function if and only if $m(k)=\lbrace x\in \mathbb{R} \mid f(x)<k\rbrace$ and $M(k)=\lbrace x \in \mathbb{R} \mid f(x)>k \rbrace$ are both open.

Attempt

Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ a continuous function and $k\in \mathbb{R}$ consider $\alpha \in m(k)$, then $f(\alpha)<k$.We shall prove that exists $V_{\delta}(\alpha)\subset m(k)$

Since $f$ is continuous function $\forall \varepsilon>0$ $\exists \delta>0$ such that $|x-\alpha|<\delta$ then $|f(x)-f(\alpha)|< \varepsilon$ .

Let $\varepsilon=k-f(\alpha)>0$.Then $\exists \delta>0$ such that $|x-\alpha|<\delta$ implies
$$|f(x)-f(\alpha)|<k-f(\alpha) $$
$$-k+f(\alpha)<f(x)-f(\alpha)<k-f(\alpha) $$
$$-k+2f(\alpha)<f(x)<k$$
this is $V_{\delta}(\alpha) \subset m(k)$ As desired.
Now consider $\alpha\in M(k)$.Then $f(\alpha)>k$.
Since $f$ is continuous function $\forall \varepsilon>0$ $\exists \delta>0$ such that $|x-\alpha|<\delta$ then $|f(x)-f(\alpha)|< \varepsilon$
now
let $\varepsilon=f(\alpha)-k>0$ and notice that
$$|f(x)-f(\alpha)|<f(\alpha)-k $$
$$k<f(x)<2f(\alpha)-k $$
Hence $V_{\delta} \subset M(k)$
Therfore $m(k),M(k)$ are both open sets in $\mathbb{R}$

Conversely Suppose that $m(k),M(k)$ are both open we should prove that $f:\mathbb{R} \rightarrow \mathbb{R}$ is a continuous function.

if $f$ is not a continuous function then exists an open set $U=(a,b):a<b\subset \mathbb{R}$ such that
$f^{-1}(U)$ is not open in $\mathbb{R}$.
Since $f^{-1}(U)=\lbrace x\in \mathbb{R} \mid f(x)\in U \rbrace $ that is
$f^{-1}(U)=\lbrace x\in \mathbb{R} \mid a<f(x)<b \rbrace$

Notice the fact that
$a<b$ then $f^{-1}(U)=\lbrace x\in \mathbb{R} \mid a<f(x)<b \rbrace$
then since $m(K),M(k)$ are both open for any $k\in \mathbb{R}$, set $k=a$ and $M(k)$
then $f^{-1}(U)=M(k)\cup m(k)$ wich is open.Contradiction.
Then $f$ is a continuous function.
is my proof right?

答:
Your proof is correct (with one inattentiveness in line -2: $f^{-1}(U) = M(a) \cap m(b)$). For a shorter proof see William Elliot's anwer.

It sticks out that in the first part you work with the $\varepsilon$-$\delta$-definition of continuity, whereas in the second part you use a theorem about continuity: A function $f$ is not continuous if for some open interval $U$ the set $f^{-1}(U)$ is not open. I conclude that you know that a function $f$ is continuous if and only if preimages of open sets are open, the latter being equivalent to "preimages of open intervals are open". The final equivalence is due the facts that each open subset of $\mathbb R$ is the union of open intervals and that $f^{-1}(\bigcup_\alpha M_\alpha) = \bigcup_\alpha f^{-1}(M_\alpha)$.

Using this characterization, you can shorten your proof.

We have $m(k) = f^{-1}((-\infty,k))$ and $M(k) = f^{-1}((k,\infty))$.

Let $f$ be continuos. Since $(-\infty,k), (k,\infty)$ are open, also $m(k)$ and $M(k)$ are open for all $k$.

Let $m(k)$ and $M(k)$ be open for all $k$. Since $(a,b) = (a,\infty) \cap (-\infty,b)$, we get $f^{-1}((a,b)) = m(a) \cap M(b)$ which is open. Thus $f$ is continuous.

Note that if all $m(k)$ are open, then $f$ is called upper semi-continuous and if all $M(k)$ are open, then $f$ is called lower semi-continuous.


问:
Probability of getting split pill from bottle?
I have a bottle of 100 pills.  The daily dose is 1/2 pill, so if the first pill I extract is a whole pill, I split it and put 1/2 back.

Just out of my own general curiosity, I'd like to model the probability of extracting a whole pill vs. a half pill over time, but I'm not sure how to start.

答: I have written a paper on this which will be published in the American Mathematical Monthly sometime in the next year or so.  The title is "A drug-induced random walk."  The main theorem is this:  Consider a bottle of $n$ pills.  Every day, you remove a pill from the bottle at random (with each pill equally likely to be chosen).  If it is a whole pill, you cut it in half, take half of the pill, and return the other half to the bottle.  If it is a half pill, then you take it and nothing is returned to the bottle.  At any time, let $x$ be the fraction of the original pills in the bottle that are still whole, and let $y$ be the fraction that are now half pills.  ($x+y$ may be less than $1$, since some pills may have been used up completely.)  Then the point $(x,y)$ executes a random walk in the plane, starting at the point $(1,0)$ (all pills whole) and ending at $(0,0)$ (no pills left).  The theorem says that for large $n$, the random walk will approximately follow the curve $y = -x \ln x$.  More precisely, the theorem says that for every $\epsilon > 0$, the probability that the walk stays within $\epsilon$ of the curve $y = -x \ln x$ approaches $1$ as $n$ approaches infinity.  The paper also answers the questions "What is the expected number of whole pills removed before the first half pill is removed?" and "What is the expected number of half pills removed after the last whole pill is removed?"</p>


问:
Prove double negation introduction with this axiom system.
I want to find a proof for ( $⊢A \rightarrow ¬¬A $ ) with these four axioms:

A1: $ A \rightarrow (B\rightarrow A) $

A2: $ (A \rightarrow (B\rightarrow C)) \rightarrow ((A\rightarrow B)\rightarrow(A\rightarrow C)) $

A3: $ \lnot\lnot A \rightarrow A $

A4: $ \lnot A \leftrightarrow (A\rightarrow \bot) $

These axioms are some Hilbert system so I tried to prove A3 which is:

Ax3 : $ (( \lnot B \rightarrow \lnot A)\rightarrow(( \lnot B \rightarrow A)\rightarrow B))) $

But i couldn't do that. so i thought maybe this system with these 4 Axioms cant do that, but i think with double-negation elimination rule and first two rules of Hilbert system it should be able to do such thing.

Thanks.

答:
By modus ponens, we see

$$A, (A \to \bot) \vdash \bot$$

Then applying the deduction theorem twice, we find

$$ \vdash A \to (A \to \bot) \to \bot$$

Lastly, we apply axiom $4$ twice to see

$$ \vdash A \to \lnot \lnot A $$

I hope this helps ^_^


问:
Finitely generated Dedekind domains
Let $R$ be a Dedekind domain. Suppose that $R$ is finitely generated as a ring, say $R = \mathbb{Z}[x_1,\dots,x_n]/I$ for some ideal $I \subseteq \mathbb{Z}[x_1,\dots,x_n]$.

<s>Is $R$ necessarily the ring of integers of a global field?</s>

Is the ring of fractions of $R$ necessarily a global field? What would be a good way to prove this?

答:
Here is a proof, with the help of the comments above. I tried to be as detailed as possible.

Typically, the definition of Dedekind domains includes fields. The fields that are finitely generated as rings are precisely the finite fields $\mathbb{F}_{p^n}$, and these are not global fields.

If additionally, $\mathrm{dim}(R)=1$, then there are two cases:

$\mathrm{char}(R)=p$. Then $R$ is a finitely generated $\mathbb{F}_p$-algebra, so by the Noether Normalization Lemma there is an injective ring morphism $\mathbb{F}_p[t] \to R$ that turns $R$ into a finitely generated $\mathbb{F}_p[t]$-module. So $R$ is a finite $\mathbb{F}_p[t]$-extension, in particular it is integral. This implies that the fraction field $K$ of $R$ is generated by finitely many algebraic elements, as a field extension of $\mathbb{F}_p(t)$. So it is a finite extension of $\mathbb{F}_p(t)$, see here. So it is a global field.

$\mathrm{char}(R)=0$. Because $\mathbb{Z}$ is not a field, the standard Noether Normalization Lemma does not apply, but there is a more general version of Noether normalization in the notes by Hochster here. It implies that $R[\tfrac{1}{n}]$ is finitely generated as $\mathbb{Z}[\tfrac{1}{n}]$-module, for some natural number $n$. Now the same proof applies as in characteristic $p$: since $R$ is integral and finitely generated, its fraction field will be generated as a field by finitely many algebraic elements. As a result, it is a finite extension of $\mathbb{Q}$. So it is a global field as well.

Title: Finitely generated Dedekind domains


问:
How do we define $\frac{\mathbb{Z}}{p\mathbb{Z}}$ for any $p\in \mathbb{N}$?
Can someone explain in simple terms, how $\mathbb{Z}/p\mathbb{Z}$ is defined as? Is it the same as $\mathbb{Z}_p$ or 'integers mod p'? If not what is the difference? Please keep it simple.

答:
You probably mean the group "$\mathbb{Z}/p \mathbb{Z}, +$", which is indeed the group of integers modulo $p$ (for a $p \in \mathbb{N}_0$). This notation comes from the fact that it is the quotient group of the group of integers $\mathbb{Z}$ (i.e. $\{0,1,2,3...\}, +$)  by the normal subgroup $n \mathbb{Z}$ of integer multiples of $n$ (i.e. $\{0,n,2n,3n,...\}, +$).

Since $p$ is not necessarily a prime number it is best to use $n$ instead, to avoid confusion.

I would not use $\mathbb{Z}_p$, as this notation is often used to designate the ring of p-adic integers (for a prime number $p$).


问:
Proving inequality using mean value theorem
I'm having some trouble proving the following inequality:

$$\sqrt x \leq \frac{x+1}{2}\ , \forall x \geq0$$

In this exercise, I'm supposed to use the mean value theorem. I always find myself having some trouble in these kinds of exercises. How can I prove this?

答:
It is not necessary to use MVT.
$$0\le(\sqrt x-1)^2=x-2\sqrt x+1 \iff 2\sqrt{x}\le x+1 \iff \sqrt{x}\le\frac{x+1}{2}$$
I use $x\ge0$ twice: first to be $\sqrt{x}$ well defined. Second, $(\sqrt{x})^2=|x|=x$.


Title: Proving inequality using mean value theorem

Body:
I'm having some trouble proving the following inequality:

$$\sqrt x \leq \frac{x+1}{2}\ , \forall x \geq0$$

In this exercise, I'm supposed to use the mean value theorem. I always find myself having some trouble in these kinds of exercises. How can I prove this?

Body:
For $x \ge 0$ we have:

$$\sqrt x \leq \frac{x+1}{2} \iff x \le \frac{1}{4}(x+1)^2 \iff 4x \le x^2+2x+1 \iff 0 \le x^2-2x+1 \iff 0 \le (x-1)^2.$$


答:
$\newcommand{\SES}[1]{\mathbb Z /#1 \mathbb Z }$
Well, there are many equivalent definitions, one classical definition is to take the integers over the congruence relation $a\sim b \iff \exists n\in \mathbb Z, a-b=pn$

Or, simply, $a\sim b$ if and only if their difference $a-b$ is divisible by p.

this congruence relation divides $\mathbb Z$ into $p$ equivalence classes, where an equivalence class is a set all whose elements are equivalent under $\sim$.

The equivalence classes are $\{...-3p,-2p,-p,0,p,2p,3p...\},\{...-3p+1,-2p+1,-p+1,1,p+1,2p+1,3p+1...\}...\{...-3p+p-1,-2p+p-1,-p+p-1,p-1,2p+p-1,39+p-1...\}$

Now I will use a concrete example, $p=3$:
our equivalence relation, $a\sim b$ is just if the difference of $a$ and $b$ is divisible by $3$ or simply if they are equivalent modulo $3$.
the equivalence classes formed are $\{...-6,-3,0,3,6\},\{...-5,-2,1,4,7...\},\{...-4,-1,2,5,8...\}$

The construct created is $\SES p$ where each element is an equivalence class.

It is provable that these classes respect multiplication and addition, i.e. we can interchange elements of classes and still have the result.

For example, $4+2=6$ if we think of this in terms of $\SES 2$ we get $0+0=0$, in terms of $\SES 3$, we get $1+2=0$, which seems not to make sense, but then we remember that in $\SES 3$, $3$ and $0$ are interchangeable.
Note: We denote it by $\SES p$ because $p\mathbb Z$ is the set of integers dividing $p$ and a normal subgroup(or subring) of $\mathbb Z$. and we use the symbol of division as we divide $\mathbb Z$ using(or "by") $p\mathbb Z$.

Title: How do we define $\frac{\mathbb{Z}}{p\mathbb{Z}}$ for any $p\in \mathbb{N}$?

Body: Can someone explain in simple terms, how $\mathbb{Z}/p\mathbb{Z}$ is defined as? Is it the same as $\mathbb{Z}_p$ or 'integers mod p'? If not what is the difference? Please keep it simple.


问:
Calculating volumes and surface areas (Calc 1)
I am having some trouble understanding why the integrals setup for finding surfaces consider the slant of the function, whereas while calculating volumes we don't bother with them.

Setup diagram

Consider the setup in the link.

If I were to rotate the graph about the x-axis, and then try to write the equation for infinitesimal surface area, I would end up with

$$∆Surface = 2y * \sqrt(1 + (\frac{∆y}{∆x})^2) * ∆x$$

Whereas for Volume I would have,

$$∆Volume = y^2 * ∆x$$

In case of volume why don't we consider the slant, or why isn't the formula for infinitesimal volume

$$∆Volume = y^2 * ∆x + \frac{1}{2} * ∆y * ∆x * 2y$$ where the additional part would be the volume due to the enlarged triangle in the image rotated about the x - axis.

答:
You are using the fundamental theorem of calculus,
$$
V(b)=\int_a^b V'(x)\,dx,
$$
to compute the volume. For both of your two expressions for $\Delta V$ the limit
$$
\lim_{\Delta x\to0}\frac{\Delta V}{\Delta x}=V'(x)
$$
is the same.


答:
Equality holds for $x=1$, which suggests to use apply the mean value theorem to $f(x) - f(1)$ with $f(x) = \sqrt x$.

We get
$$
 \sqrt x - \sqrt 1 = (x-1) \frac{1}{2\sqrt c}
$$
for some $c$ between $1$ and $x$. Now consider the cases $0 \le x < 1$ and $x > 1$ and show that the right-hand side is always
$$
 \le \frac{x-1}{2} \, .
$$

(Note that we have essentially used is that the derivative of $f$ is decreasing. That is equivalent to $f$ being concave so that its graph lies below its tangent line at $x=1$.)


问:
Linear programming problem neither max nor min
Heres the actual question:

television provider broadcasts two movie channels, A and B. Channel A broadcasts 1 romantic
  movie, 3 action movies and 3 comedies per month at a cost of 50 Euro. Channel B broadcasts
  3 romantic movies, 4 action movies and 1 comedy per month at a cost of 25 Euro. Suppose that
  you like to see 8 romantic movies, 19 action movies and 7 comedies at minimal cost in the coming
  months. For how many months should you request both channels?

a. Formulate this problem as an LP problem

Now my way of thinking was that I would first write it down more clearly, A and B are the television channels.

$A: 1R,3A,3C => 50 euro$

$B: 3R, 4A, 1C => 25 euro$

And I would define a new variable $x_i :=$ cost of taking a channel one month.

But then I stumble upon the problem of defining the hard limits they have set, the 8 R, 19 A, and  7 C.

In short: how do I define a constraint that is hard (e.g. not a maximum)

答:
$min \quad 50A + 25B$

Such that

$A + 3B \geq 8$

$3A + 4B \geq 19$

$3A + B \geq 7$

$A,B\geq 0$


问:
What does adding two random variables mean?
In my assignment, I came across with a question asking me that if $X+Y$ is independent of $Z$ if $X, Y, Z$ are three random variables that are pairwise independent. I solved the problem by repeating what the after-class exercise did (comparing probabilities and stuff). But I don't understand the geometric implications, if any, of adding two random variables $X+Y$ together.

In the realm of the real number, adding two numbers is simply manoeuvres over the real number line. But this idea makes no sense in the context of adding random variables.

答:
Welcome to MSE!

Remember the intuitive idea of a random variable: It simply chooses some real number $r$ according to a probability distribution.

Consider the random variable $X$ which takes values in $\{1,\ldots,6\}$ based on the roll of a dice.

Consider also the random variable $Y$ which takes values in $\{0,1\}$ based on the flip of a coin.

Then we can consider the random varaible $X+Y$, which takes values in $\{1,\ldots,7\}$ depending on both the dice roll and the coin flip.

I'm not sure if there are "geometric implications" (unless your random variables are geometric in nature). Here's an example:

For instance, you could imagine random variables $X$ and $Y$ that each select a random number in the interval $[0,1]$. Then the random variable $\frac{X + Y}{2}$ has some geometric significance: it's the midpoint of the two points you chose randomly.

I hope this helps ^_^


问:
Closure proof by a subfield
I have to show that the subset $L \subseteq \mathbb{K} $ of the field
$$ \mathbb{K} := (\mathbb{K},+,*,0,1) $$

$*$ stands for multiplication

is a subfield of it. Which means that $L$ is a field.

I know that $L$ has at least to elements and that the following property is satisfied for every $ x,y,z \in L$
$$(x-y)z^{-1} \in L$$

My attempts so far:

Splitting $L$ into two objects and proving that $L$ with multiplication and $1$ is a monoid.
$L$ with addition and $0$ is a abelian group. As a part of doing that one can conclude that the commutativity associativity and distributivity are being inherited from $ \mathbb{K}$.

The closure however is not being inherited and must be proven, the question is how?

I thought about showing through substitution that every product $xz,xy,yz \in L$ but am now sure that it is rigorous enough.

答:
Referring to your comment above, Ozk, writing it that way runs into problems since $y = 0$ and $z^{-1}=y$ here would imply $z^{-1} = 0$ and all you've done is prove $0 \in L$.

As you have already shown the other properties and are trying to prove additive closure, substitute $-y$ for $y$ and $z = 1$ so that $(x-y)z^{-1}=x+y\in L$.

For multiplicative closure, since $z^{-1} \in L$, one can substitute into the given expression with $y = 0$ and $z^{-1}$ for $z$ so that $(x-y)z^{-1} = x(z^{-1})^{-1}=xz \in L$ for every $x,z \in L$.

Title: Closure proof by a subfield


问:
Theorem for convergence of a series
I was studying Infinite Series. The book I use is Thomas and Finney - Calculus and Analytical Geometry 9th Edition. On pg 635, the following theorem 6 was provided :

Theorem 6 says that when the sum , say

S = a1 + a2 + a3 + a4 + ..... + a(n)

is continued for infinite terms and converges resulting into a unique and finite value, then the limit of n tending to infinity of a(n) must be zero. This means that as we go further and further with our sum S, we reach to a term that would become, well, very close to zero.

I have tried thinking about it, but I could not really justify this theorem in any manner. The proof of the theorem is not given in the book. Could anyone please help ?

答:
Suppose $a_n$ does $not$ converge to $0.$

Then we have some $r>0$ such that the set $S=\{n\in \Bbb N: |a_n|>r\}$ is infinite.

Now "$\sum_{n=1}^{\infty} a_n$ converges" means $b_n$ converges, where $b_n=\sum_{j=1}^na_j.$

But there are infinitely many $n>1$ for which $|b_n-b_{n-1}|>r,$ i.e. when $1<n\in S.$

So $b_n$ does not converge.

Title: Theorem for convergence of a series


问:
Determine the points where $f$ is has a local minimum/maximum. Multivariable calculus question.
This is not homework, but it is in my book and I find it hard to solve:

Determine the points where $f$ is has a local minimum/maximum.
  Determine if it strong/weak and absolute/relative and interior
  extremum or boundary extremum.

$$f(x,y)= (x^2-y^2)e^{-x^2-y^2}$$
  The domain of $f$ is $E=\{(x,y): x^2+y^2 \leq 4 \}$

A worked solution would be really appreciated. Or if someone knows some worked solutions of exercises like these (somewhere on the internet), that would be great as well. Or a step-by-step plan how to solve exercises like these. In other words, any help is really appreciated.

答:
First you have to use KKT conditions with a slack variable $s\ge 0$
$$max_{x,y}\quad (x^2-y^2)e^{-x^2-y^2}\qquad s.t.\quad x^2+y^2+s= 4$$
and the Lagrangian becomes
$$L=(x^2-y^2)e^{-x^2-y^2}+\lambda\big(x^2+y^2+s- 4\big)$$
The first order conditions are
$$\frac{\partial L}{\partial x}=2xe^{-x^2-y^2}(\lambda e^{x^2+y^2}-x^2+y^2+1)=0$$
$$\frac{\partial L}{\partial y}=2ye^{-x^2-y^2}(\lambda e^{x^2+y^2}-x^2+y^2-1)=0$$
$$\frac{\partial L}{\partial \lambda}=x^2+y^2+s- 4=0$$
$$s\cdot \frac{\partial L}{\partial s}=s\cdot \lambda=0$$
where the last one is KKT condition.

Let's check the boundary where $s=0$. The solution sets are $(0,\pm2)$ and $(\pm2,0)$

For interior where $\lambda=0$ the solution sets are $(0,\pm1)$ and $(\pm1,0)$

By second derivative test you can find that

$(\pm1,0)$ global max

$(0,\pm1)$ global min

$(\pm2,0)$ local max only on the boundary

$(0,\pm2)$ local min only on the boundary


问:
Taking exponents of two numbers to make them equal
If $3^a = 5^b = m$ and $\frac{1}{a} + \frac{1}{b} = 2$, what is the value for m?

Since $a, b \neq 0$, I assumed both a and b are fractions, so I did:

Let $a = \frac{a_1}{a_2}; b = \frac{b_1}{b_2}$

Then we subsitute them in and get $3^{a_1 b_2} = 5^{a_2 b_1} = m$

Now we're stuck into a similar situation as our original equation. How can I move further?

答:
Observe that we have $m>0.$

We get $a \ln 3 = \ln m$ and $b \ln 5= \ln m.$ This gives

$$2= \frac{\ln 3+ \ln5}{\ln m}= \frac{\ln 15}{\ln m}.$$

Can you take it from here ?


Title: Taking exponents of two numbers to make them equal

Body:
If $3^a = 5^b = m$ and $\frac{1}{a} + \frac{1}{b} = 2$, what is the value for m?

Since $a, b \neq 0$, I assumed both a and b are fractions, so I did:

Let $a = \frac{a_1}{a_2}; b = \frac{b_1}{b_2}$

Then we subsitute them in and get $3^{a_1 b_2} = 5^{a_2 b_1} = m$

Now we're stuck into a similar situation as our original equation. How can I move further?

Body: Hint: $m^{\frac 1a}=3$ and $m^{\frac 1b}=5$.


问:
Substituting $u=\ln$ when lower bound is $0$
The following substitution came up in the solution set when I was practicing Diff. Geom., and it seems wrong to me:
$$\int_0^t\cos(\ln x) dx = \int_0^{\ln(t)} \cos(u)e^u du$$
When $u$ is evaluated at the lower bound, it is not defined! Additionally, the limit $\ln s\downarrow 0 = -\infty$.

Is this just very sloppy notation in the solution set, or am I missing something else?

答:
You are right, it should read:

$$\int_0^t\cos(\ln x) dx = \int_{- \infty}^{\ln(t)} \cos(u)e^u du.$$


问:
Trying to show $|a|\geq |b|-|b-a|$ for $a,b \in \mathbb{R}$
I'm currently solving a question that boils down to showing this inequality. The only tool I've got in my belt to show this the triangle inequality essentially.

My first step is $|a|=|a-b+b|$, so that I can apply the triangle inequality. I've tried many manipulations on $|a-b+b|$ to prove my inequality but I cannot seem to land on the right step.

Any help would be greatly appreciated!

答:
|a|-|b|≥-|b-a|
since |b| always greater or equal to zero, there are only two cases:
case 1: |a|-|b| = |a-b| this happens if a ≥ b
case 2: |a|-|b| = |b-a| this happens if a ≤ b
furthermore, |a-b|≥ 0
and |b-a| ≥ 0
therefore in both cases |a|-|b| ≥ -|b-a|
rearrange again to prove |a|≥|b|-|b-a|</p>


问:
How can I tell if this serises converge or diverge?
I need to check if this series converges absolutley:$$
\sum_{n=1}^{\infty}\frac{\sin(3n)}{\sqrt{4n+1}} $$

I showed it converge without the absolute value but didn't succeded for its absolute value:

$$\sum_{n=1}^{\infty}\frac{|\sin(3n)|}{\sqrt{4n+1}} $$

I don't know how to tell if it converges or diverges,
I tried to use some comparing tests, but I didn't find it fruitful in that case.

Many thanks.

答: That series is not absolutely convergent. Note that$$|\sin(3n)|\geqslant\sin^2(3n)=\frac{1-\cos(6n)}2$$and that the series$\sum_{n=1}^\infty\frac{\cos(6n)}{2\sqrt{4n+1}}$ converges, by Dirichlet's test. However, the series $\sum_{n=1}^\infty\frac1{2\sqrt{4n+1}}$ diverges.


问:
Coefficient Extraction of Products of Generating Functions

Is there an easier way to extract coefficients from the product of two generating functions?

In Levin's book, the only way mentioned was $${a_0}{b_0} + ({a_0}{b_1}+{a_1}{b_0})x + (a_0b_2+a_1b_1+a_2b_1)x^2...$$ and so on.

It's hard when the sequence does not have a definite pattern. Is there any easier way to do this?

答:
This is known as the "cauchy product", and as far as I know it is the only way to do it. However, it's not as complicated as it looks.

The $n$th coefficient of $x^n$ in $AB$ is $\displaystyle \sum_{i+j=n} a_i b_j$. That is

$$a_0b_n + a_1b_{n-1} + \ldots + a_{n-1}b_1 + a_nb_0$$

Notice each product sums to $n$:

$a_0 b_n$ is $0+n = n$

$a_1 b_{n-1}$ is $1 + (n-1) = n$

and so on

In this way, knowing the first $n$ coefficients of $A$ and the first $n$ coefficients of $B$ is enough to know the first $n$ coefficients of $AB$.

I hope this helps ^_^


问:
Sufficient conditions for secant method to converge
Are there some sufficient conditions for secant method
$$x_{k+1}=x_k-\dfrac{x_k-x_{k-1}}{f(x_k)-f(x_{k-1})}f(x_k)$$

to converge?

答:
You can easily deduce a local result. The error satisfies
$$
e_{n+1} = -\dfrac{f''(\xi_n)}{2f'(\eta_n)} e_{n} e_{n-1}.
$$

If $f\in C^2$ and $f'(z)\ne 0$, the method will converge if $x_0, x_1$ are sufficiently close to $z$. the rate of convergence will be $\alpha = (1+\sqrt{5})/2 \approx 1.618$.


问:
Convention for Coefficient of Variation of a Poisson Process
The Coefficient of Variation (CV) according to the linked Wikipedia article is the ratio of the standard deviation $\sigma$ to the mean $\mu$ of a probability distribution.

For the Poisson Distribution, the mean is equal to the variance $\sigma^2$ so that the CV should be $\sigma/\mu=\sigma/\sigma^2=1/\sigma$

However, in the calcium signaling modeling literature I am seeing some papers that state this differently. One such example is in figure 2B of this paper, although similar language is found in other papers by these researchers and their affiliates in models of the $\text{IP}_3$ receptors. Below is the figure and description, where the bold part is my own emphasis. (ISI stand for "Inter-Spike Interval", which is a stochastic time interval measured from the system in question).

B: Relationship between the standard deviation and the mean of experimental ISIs. Data obtained from 14 ASMC in 5 mouse lung slices. The relationship is approximately linear with a slope of 0.66, which implies that an inhomogeneous Poisson process governs the generation of oscillations. The dashed line indicates where the coefficient of variation (CV) is 1 (as it is for a pure Poisson process).

The paper says that the CV for a Poisson process is $1$. But this doesn't make sense given the above information. If they compared the variance to the mean then this should work out fine since their ratio is $1$, but this is done in many papers by these / affiliated researchers (that are some key people in the field), so I am not sure what is going on here.

Am I missing something with the definition of the CV or Poisson distributions? Also, how does this work with quantities with physical units where the mean and variance do not have the same dimensions? How could the mean be equal to the variance in situations like this?

答:
If the spike events are modeled as a Poisson process with constant rate, then the ISIs, i.e. the time between spike events, are exponentially distributed. An exponential random variable has a mean and standard deviation equal to each other, and thus a CV of 1. Since the figure shows an estimated CV less than 1, they state this implies that their Poisson process would be better modeled with a non-constant rate (inhomogenous).

The units of mean and standard deviation for any random variable will be the same irrespective of the application.


问:
Why does this being a finite set mean that it cannot be the case that all element of the sequence are distinct?
I am currently studying the textbook Principles of Program Analysis by Flemming Nielson, Hanne R. Nielson, and Chris Hankin. Chapter 1.3 Data Flow Analysis says the following:

The least solution. The above system of equations defines the twelve sets
$$\text{RD}_\text{entry}(1), \dots, \text{RD}_{\text{exit}}(6)$$
in terms of each other. Writing $\overrightarrow{RD}$ for this twelve-tuple of sets we can regard the equation system as defining a function $F$ and demanding that:
$$\overrightarrow{RD} = F(\overrightarrow{RD})$$
To be more specific we can write
$$F(\overrightarrow{RD}) (F_\text{entry}(1)(\overrightarrow{RD}), F_\text{exit}(1)(\overrightarrow{RD}), \dots, F_\text{entry}(6)(\overrightarrow{RD}), F_\text{exit}(6)(\overrightarrow{RD}))$$
where e.g.:
$$F_\text{entry}(3)(\dots, \overrightarrow{RD}_\text{exit}(2), \dots, \overrightarrow{RD}_\text{exit}(5), \dots) = \overrightarrow{RD}_\text{exit}(2) \cup \overrightarrow{RD}_\text{exit}(5)$$
It should be clear that $F$ operates over twelve-tuples of sets of pairs of variables and labels; this can be written as
$F : (\mathcal{P}(\mathbf{\text{Var}_\star \times \mathbf{\text{Lab}_\star}))}^{12} \to (\mathcal{P}(\mathbf{\text{Var}_\star \times \mathbf{\text{Lab}_\star}))}^{12}$
where it might be natural to take $\mathbf{\text{Var}_\star} = \mathbf{\text{Var}}$ and $\mathbf{\text{Lab}_\star} = \mathbf{\text{Lab}}$. However, it will simplify the presentation in this chapter to let $\mathbf{\text{Var}_\star}$ be a finite subset of $\mathbf{\text{Var}}$ that contains the variables occurring in the program $\mathbf{S_\star}$ of interest and similarly for $\mathbf{\text{Lab}_\star}$. So for the example program we might have $\mathbf{\text{Var}_\star} = \{ x, y, z \}$ and $\mathbf{\text{Lab}_\star} = \{ 1, \dots, 6, ? \}$.

It is immediate that $(\mathcal{P}(\mathbf{\text{Var}_\star \times \mathbf{\text{Lab}_\star}))}^{12}$ can be partially ordered by setting
$$\overrightarrow{\text{RD}} \sqsubseteq \overrightarrow{\text{RD}}^\prime \ \ \ \text{iff} \ \ \ \forall i : \text{RD}_i \subseteq \text{RD}_i^\prime$$
where $\overrightarrow{\text{RD}} = (\text{RD}_1, \dots, \text{RD}_{12})$ and similarly $\overrightarrow{\text{RD}}^\prime = (\text{RD}_1^\prime, \dots, \text{RD}_{12}^\prime)$. This turns $(\mathcal{P}(\mathbf{\text{Var}_\star \times \mathbf{\text{Lab}_\star}))}^{12}$ into a complete lattice (see Appendix A) with least element
$$\overrightarrow{\emptyset} = (\emptyset, \dots, \emptyset)$$
and binary least upper bounds given by:
$$\overrightarrow{\text{RD}} \sqcup \overrightarrow{\text{RD}}^\prime = (\text{RD}_1 \cup \text{RD}_1^\prime, \dots, \text{RD}_{12} \cup \text{RD}_{12}^\prime)$$

It is easy to show that $F$ is in fact a monotone function (see Appendix A) meaning that:
$$\overrightarrow{\text{RD}} \sqsubseteq \overrightarrow{\text{RD}}^\prime \ \ \ \text{implies} \ \ \ F(\overrightarrow{\text{RD}}) \sqsubseteq F(\overrightarrow{\text{RD}})^\prime$$
This involves calculations like
$$\text{RD}_\text{exit}(2) \subseteq \text{RD}_\text{exit}^\prime(2) \ \ \text{and} \ \ \text{RD}_\text{exit}(5) \subseteq \text{RD}_\text{exit}^\prime(5)$$
imply
$$\text{RD}_\text{exit}(2) \cup \text{RD}_\text{exit}(5) \subseteq \text{RD}^\prime_\text{exit}(2) \cup \text{RD}_\text{exit}^\prime(5)$$
and the details are left to the reader.

Consider the sequence $(F^n(\overrightarrow{\emptyset}))_n$ and note that $\overrightarrow{\emptyset} \sqsubseteq F(\overrightarrow{\emptyset})$. Since $F$ is monotone, a straightforward mathematical induction (see Appendix B) gives that $F^n(\overrightarrow{\emptyset}) \sqsubseteq F^{n + 1}(\overrightarrow{\emptyset})$ for all $n$. All the elements of the sequence will be in $(\mathcal{P}(\mathbf{\text{Var}_\star \times \mathbf{\text{Lab}_\star}))}^{12}$ and since this is a finite set it cannot be the case that all elements of the sequence are distinct so there must be some $n$ such that:
$$F^{n+1}(\overrightarrow{\emptyset}) = F^n(\overrightarrow{\emptyset})$$
But since $F^{n + 1}(\overrightarrow{\emptyset}) = F(F^n(\overrightarrow{\emptyset}))$ this just says that $F^n(\overrightarrow{\emptyset})$ is a fixed point of $F$ and hence that $F^n(\overrightarrow{\emptyset})$ is a solution to the above equation system.

I am confused by this part:

All the elements of the sequence will be in $(\mathcal{P}(\mathbf{\text{Var}_\star \times \mathbf{\text{Lab}_\star}))}^{12}$ and since this is a finite set it cannot be the case that all elements of the sequence are distinct so there must be some $n$ such that:
$$F^{n+1}(\overrightarrow{\emptyset}) = F^n(\overrightarrow{\emptyset})$$

Why does $(\mathcal{P}(\mathbf{\text{Var}_\star \times \mathbf{\text{Lab}_\star}))}^{12}$ being a finite set mean that it cannot be the case that all elements of the sequence are distinct? Furthermore, $n$ wasn't even defined, so I'm not even sure what it's supposed to be.

答:
When they say "consider the sequence $(F^n(\overset{\to}{\emptyset}))_n$", they mean the infinite sequence where $n$ ranges over all of $\mathbb N$. Since this is defined for infinitely many values of $n$, and takes only finitely many values (values in that finite set), there must be two different natural numbers $n,n+k$ with the same value.

Now we need to show that there must be two consecutive numbers with the same value. This is because $F^n(\overset{\to}{\emptyset})\sqsubseteq F^{n+1}(\overset{\to}{\emptyset})$, so if the above holds with $k>1$ we have $F^n(\overset{\to}{\emptyset})\sqsubseteq F^{n+1}(\overset{\to}{\emptyset})\sqsubseteq\cdots\sqsubseteq F^{n+k}(\overset{\to}{\emptyset})$; since the first and last terms are equal, they must all be equal.

Finally, by definition the $(n+1)$th term is the image of the $n$th under $F$, so it must be fixed by $F$ (and all subsequent terms are equal).


问:
Prove that if $z^n$ is a palindrome for some $n>0$, then $z$ is also a palindrome for any alphabet E.
Prove that if $z^n$ is a palindrome for some $n>0$, then $z$ is also a palindrome for any alphabet E.

Here's a proof of the statement above:

Let $w = x^n$.  If $n = 1$, then the result is trivial.

Suppose $n > 1$. Then $$x^n = reverse (x^n) = (reverse  (x))^n.$$
Write this as $$x x \dots x = reverse (x) reverse (x) \dots reverse (x).$$ But $x$ and $reverse (x)$ have the same length, so that $x = reverse (x)$.

Can anyone justify this line in the proof?
$$reverse (x^n) = (reverse  (x))^n$$

Or provide an alternative proof.

答:
A simple argument proves the statement. Essentially, it proves explicitly that $\text{rev}(z^n) = (\text{rev}(z))^n$ (if you assume it, there is nothing to prove).

Before showing the argument, some preliminaries. Any word can be written as $z = z_1 \dots z_k$, where $k \geq 0$ is the length of $z$ and the $z_i$'s are single characters of any alphabet $E$.
A word $z = z_1 \dots z_k$ is palindrome if $z = \text{rev}(z)$ where $\text{rev}(z) = z_k \dots z_1$.

Now, let us see the proof. The hypothesis in the statement says that, for some $n > 0$, we have $z^n = \text{rev}(z^n)$. Written more explicitly, for $z = z_1 \dots z_k$, we have

\begin{align}
z^n &= \overbrace{\overbrace{z_1 \dots z_k}^z \ \cdots \ \overbrace{z_1 \dots z_k}^z}^{n \text{ times}}
\\
&=
\\
\text{rev}(z^n) &= z_k \dots z_1 \ \cdots \ z_k \dots z_1
\end{align}

(Note that the first occurrence of $z_k$ in $\text{rev}(z^n)$ corresponds to the last occurrence of $z_k$ in $z^n$; and the first occurrence of $z_{k-1}$ in $\text{rev}(z^n)$ corresponds to the last occurrence of $z_{k-1}$ in $z^n$; and so on. This is the way to construct $\text{rev}(z^n)$ from $z^n$.)

Since $n > 0$, the identity $z^n = \text{rev}(z^n)$ holds in particular for the first $k$ characters of $z^n$ and $\text{rev}(z^n)$, hence
\begin{align}
z = z_1 \dots z_k = z_k \dots z_1 = \text{rev}(z)
\end{align}
Therefore, $z$ is palindrome.

Note that we have not assumed that $\text{rev}(z^n) = (\text{rev}(z))^n$. Actually, the fact that $\text{rev}(z^n) = (\text{rev}(z))^n$ is an immediate consequence of what we have proved.


问:
The symmetric groups $S_4$ and $S_5$
In which symmetric group $S_4$ or $S_5$ is there an element of order 2 and an element of order 3 that commute?

I have worked out the 9 elements of order 2 and 8 elements of order 3 in $S_4$. Is there a more efficient way of working this out or is it a case of going through each element and seeing if they commute?

答:
To add some more context to Mathsingh's comment:

Recall that in the symmetric group $S_n$, two cycles commute if and only if their supports are disjoint. Since you are interested in a 2-cycle and a 3-cycle, this condition means

$(a,b)$ and $(x,y,z)$ commute if and only if

$a,b \not \in \{ x,y,z \}$

$x,y,z \not \in \{a,b \}$

Now notice that, in $S_4$ there is not enough room! By the pigeonhole principle, we cannot have $5$ different values ($a,b,x,y,z$) all non-equal. So any 2-cycle and 3-cycle you choose will have overlapping supports, and thus cannot commute.

In $S_5$ however, we can choose, say, $(1,2)$ and $(3,4,5)$. These cycles have disjoint supports, and thus commute.

I hope this helps ^_^

Title: The symmetric groups $S_4$ and $S_5$


问:
Solve equation $ \cos x+\sin x=0$
I'm trying to solve an equation here but unfortunately I can't.
The equation:
$$
\cos x + \sin x = 0
$$
I'm trying to solve this by replacing $\cos x$ with $(1-t^2)/(1+t^2)$ and $\sin x$ with $2t/(1+t^2), t=\tan x/2, \ $  but I can't get the right solution.
Also I have tried by squaring both sides but still nothing.

Can anyone help me ?

答:
$$\cos x + \sin x = 0$$

Multiply $\dfrac{\sqrt2}{2}$ to both sides:

$$\dfrac{\sqrt2}{2}\cos x + \dfrac{\sqrt2}{2}\sin x = 0$$

Or:

$$\cos\dfrac{\pi}{4}\cos x + \sin\dfrac{\pi}{4}\sin x = 0$$

Then:

$$\cos\left(x - \dfrac{\pi}{4}\right) = 0$$

So:

$$x - \dfrac{\pi}{4} = \dfrac{\pi}{2} + k\pi$$

Therefore:

$$x = \dfrac{3\pi}{4} + k\pi\qquad (k \in \Bbb Z)$$

Title: Solve equation $ \cos x+\sin x=0$


问:
Counting exercise
Three players a,b,c take turns in a game according to the following rules:

At the start A and B play (so C does not play). The winner of the first trial plays against C and so on until one of the players wins two trials in a row.
  Possible outcomes are
  aa,acc,acbb,acbaa,
  bb,bcc,bcaa,bcabb etc.

We have to prove that probability of A winning equals p(A) = 5/14, p(B) = 5/14, p(C) = 2/7

I have been stuck on this problem for a long time. The only thing I have been able to find out so far is that C can never win on even turns.

Re-edit

Players continue to play until one of them wins two times consecutively and all players are equally good at playing the game.

Apologies for leaving such crucial information.

I have finally solved this problem with a different approach

Sample space

aa ,acc ,acbb ,acbaa ,acbacc ,acbacbb ,acbabcaa ,...

bb ,bcc ,bcaa ,bcabb ,bcabcc ,bcabcaa ,bcabcabb ,...

Each point in the sample space has a probability (1/2^k) associated with it where k is the number of turns.

For example probability of point (aa) = 1/4

Now let us construct a table enumerating probabilities -

Let us consider the event that C wins overall. But C can only win if k = 3,6,9,12,...

P(C) = P(C3) + P(C6) + P(C9) ......

where P(Ck) = Probability of C winning overall after k turns.

P(C) = 1/4 + 1/32  .....

P(C) = a / (1 - r)       ( Sum of an infinite GP )

a = 1/4                  ( First Term )

r = 1/8                  ( Ratio )

P(C) = 2/7

P(A) = 5/14 = P(B)

答:
First, let us assume $A$ wins the first game.  Let $a$ be the probability that $A$ wins overall, $b$ the probability that $B$ wins overall, and $c$ the probability that $C$ wins overall.  Then we can write (in this case) $a=\frac 12 + \frac b2$ because $A$ either wins the second game (and wins overall) or loses the second game and is now in $B$'s position.  Similarly, $c=\frac 12a$ because if $C$ wins his first game he is in $A$'s position, while if he loses his first game $A$ wins overall.  Finally $b=\frac 12c$ because $B$ needs $C$ to win and he is now coming in, which is $C$'s position.  This gives
$$a=\frac 12 + \frac b2\\c=\frac a2\\b=\frac c2\\a=\frac 12+\frac c4==\frac 12+\frac a8=\frac 47\\c=\frac 27\\b=\frac 17$$
This is correct for $C$, but we assumed $A$ won the first one.  Clearly $A$ and $B$ have the same winning probability at the start so we can split their total chances evenly, giving $$P(A)=\frac 5{14}, P(B)=\frac 5{14},P(C)=\frac 27$$</p>

Title: Counting exercise


问:
Closed set on $\mathbb{R}^2$ written as disjoint union of sets
Let $\gamma \subset \mathbb{R}^2$ be a closed set. If $\gamma = \bigcup_{i\in\Lambda} U_i $ with $U_i\cap U_j = \emptyset$ $\forall i\neq j$, there exist $j\in\Lambda$ with $U_j$ being a closed set?

答:
Consider the following example:
$$
{\mathbb R}= \coprod_{n\in {\mathbb Z}} [n, n+1).
$$


问:
Monomorphism of a group automorphisms
Let $\mathbb{A}$ and $\mathbb{B}$ be two algebras on the same language $\mathcal{L}.$ How can I prove that there exists a monomorphism of the group

${\rm Aut}\mathbb{A}\times{\rm Aut}\mathbb{B}$ to the group ${\rm Aut}(\mathbb{A}\times\mathbb{B})?$

答:
Hint:

An element of $\text{Aut} (\mathbb{A}) \times \text{Aut}(\mathbb{B})$ looks like $(f,g)$ where $f \in \text{Aut}(\mathbb{A})$ and $g \in \text{Aut}(\mathbb{B})$.

Then consider the function $(f \times g)(a,b) = (f(a),g(b))$.

Can you show $f \times g \in \text{Aut}(\mathbb{A} \times \mathbb{B})$? If yes, then the map $(f,g) \mapsto f \times g$ is your desired group homomorphism.

I hope this helps ^_^

Title: Monomorphism of a group automorphisms


问:
Why we use subtract to represent distance
I am confused about that:
|x-y|= the distance between x and y

why do we use '-' to represent distance? Where does the principle come from?
I know that if y>0, x-y is exactly the distance; if y<0, x-y means you add the distance from the origin to y and x-y is the distance. But why is that? Somebody set this principle that we should use '+' present positive direction and '-' represent negative direction, and x-y represent reversing the negative to the positive then we can get the distance?
Maybe I am confused about '-' itself. Hoping get your answer! Thank you!

答:
The difference is translation-invariant. For any constant $a$,

$$|x-y|=|(x+a)-(y+a)|.$$

It is also quite intuitive that when you are at point $x$ and travel a distance $d$, you get to $y=x+d$, hence $d=y-x$ !


问:
Can we say," The set of all compact metric spaces"?

Fix a compact topological space $X$. Whether $\{(Y,\phi): Y \text{ is a compact metric space and }\phi:X\to Y \text{ is surjective continuous map}\}$ is a set?

If $I$ is a set and for each $\alpha\in I$, $A_\alpha$ is a set. Is the collection $\{A_\alpha:\alpha\in I\}$ a set?

I am reading a proof of RRT for positive linear functionals articulated by V.S.Sunder and there the first  collection was assumed as a set. I couldn't able to prove it. Nonetheless, I have modified  the required set as  $\mathcal{B}:=\{E\subset X: \exists \text{ cpt m.s. }Y, ~\pi:X\to Y \text{ is surjective, and }E=\pi^{-1}(A),~A\in\mathcal{B}_\pi\}$ and enjoyed the nice proof! of prof.V.S.Sunder.

Ref: V.S. Sunder,   The Riesz representation theorem, Indian J. of Pure and Appl.
Math., 39(6), December 2008, 467-481.

答:
Your question reaches down to the foundation of set theory. A common approach is Zermelo–Fraenkel set theory ZF. It does not define a set as a "collection of objects" or something like that (although that is a nice intuitive understanding), but gives a list of axioms which have to be satisfied by things called sets.

However, it turns out that some "collections" of objects which one can easily name, for example

the collection of all sets,

the collection of all compact metric spaces,

cannot be sets (if they were sets, we would get contradictions as Russell's paradox). Such collections are called proper classes, and in a sense they are too big to satisfy all axioms of ZF. For example, there is no set containing as elements all sets and no set of all compact metric spaces.

Concerning your questions:

The compact metric spaces do not form a set. This is not a drama, because it is possible to find a set of representatives. If you are satisfied with compact metrizable toplogical spaces (i.e. if you are not interested in specific metrics because all metrics generating the same compact topology on a given set are uniformly equivalent), then Daniel Wainfleet's comment shows what to do: The closed subspaces of the Hilbert cube $Q$ form a set $\mathfrak C$. All $X \in  \mathfrak C$ are compact metrizable and each compact metrizable space is homeomorphic to some member of $\mathfrak C$. If you want to consider compact metric spaces, then it is just a little more complicated. The previous argument shows that each compact metric space $(X,d)$ admits a bijection $\phi : X \to X'$ to a subset $X'$ of $Q$. This $\phi$ induces a metric $d_\phi$ on $X'$ making $(X,d)$ and $(X',d_\phi)$ isometrically isomorphic. The metrics on a given set $X'$ form a set, thus 2. shows that all pairs $(X',d')$ where $X'$ is a subset of the Hilbert cube (i.e. a member of the power set of $Q$) and $d'$ is a metric on $X'$ also form a set $\mathfrak M$. Not all such $(X',d')$ are compact, but the subset $\mathfrak M_c$ of compact $(X',d') \in \mathfrak M$ gives you the desired set of representatives for the class of compact metric spaces.

Yes. As Berci comments, this is the replacement axiom of ZF.