问:
Dependence of Axioms of Equivalence Relation?
This question is problem 11(a) in chapter 1 in 'Topics in Algebra' by I.N. Herstein.

These are the properties of equivalence relation given in this book.

Prop 1 $a \sim a$

Prop 2 $a \sim b$ implies $b \sim a$.

Prop 3 $(a \sim b$ and $b \sim c)$ imply $a \sim c$.

Statement Property 2 of an equivalence relation states that if $a \sim b$ then $b \sim a$. By property 3, we have the transitivity i.e. $a \sim b$, and $b \sim c$ then $a \sim c$. What is wrong with following proof that property 2 and 3 imply property 1? Let $a \sim b$; then $b \sim a$, whence by property 3 (using $a = c$), $a \sim a$.

I think I can prove this to be wrong. Without proving equivalence relation first, one can not use $a = c$. Right? After all, equality is equivalent to 'equivalence relation' and 'axiom of substitution' are satisfied. If this is right, then I have trouble with the next part of this problem.

Part 2 Can you suggest an alternative of property 1 which will insure us that prop 2 and prop 3 do imply 1?

Can one give such a formulation without using the idea of '=' or otherwise?

EDIT : Italics are my comments. Rest is as it appeared in the book.

Notion of Equality

I have read in Terry 'Analysis 1' in Appendix A.7 published by Hindustan Book Agency that there are four axioms of 'equality'. First 3 are same as equivalence relation where $\sim $ in replaced by $ = $. The fourth one is known as axiom of substitution. Given any two objects $x$ and $y$ of some type, if $ x = y $, then $f(x) = f(y) $ for all functions or operations $f$.

答: There is no difficulty in supplying a "Prop. $0$" strictly weaker than Prop. $1$ such that Prop. $0$ and Props. $2$ and $3$ imply the current Prop. $1$.  Just say that for any $x$ there is a $y$ such that $x \sim y$. (Of course that $y$ might be $x$ itself.)  Then Prop. $0$, together with $2$ and $3$, imply Prop. $1$.  Thus Prop. $0$, $2$, and $3$ give an alternate axiomatization.  Perhaps this is the intended answer.  Or not. </p>


问:
Computing product of random variables
Consider the random variable $X=I\cdot B$, where $I$ is an bernoulli random with parameter $p$ variable and B an arbitrary r.v.

Why does the following hold $P(X=5)= p \cdot P(B=5)$. What do I use here?

答:
Since $I$ is either $0$ or $1$, therefore the event that $X$ is equal to $5$ is the event that $I = 1$ and $B = 5$. If $I$ and $B$ are independent, then...
Can you take it from there ?

ContentLicense: CC BY-SA 4.0

Title: Computing product of random variables


问:
Some doubts in the evaluation of: limit as $(x,y)\to(0,0)$ of $\frac{\sin xy}{x+y}$
I must evaluate
$$\lim_{(x,y)\to(0,0)}\frac{\sin xy}{x+y}$$
My reasoning is the following, can someone tell me if this is correct?
Since $|\sin t| \leq |t|$ for all $t\in\mathbb{R}$ and it is $|xy|\leq\frac{1}{2}(x^2+y^2)$, we have
$$0 \leq \lim_{(x,y)\to(0,0)}\left|\frac{\sin xy}{x+y}\right|\leq\lim_{(x,y)\to(0,0)}\frac{|xy|}{|x+y|}\leq\lim_{(x,y)\to(0,0)} \frac{x^2+y^2}{2|x+y|}$$
Using polar coordinates it is
$$\lim_{(x,y)\to(0,0)}\frac{|xy|}{|x+y|}=\lim_{\rho \to 0^+} \frac{\rho^2}{2|\rho \cos \theta+\rho\ \sin \theta|}=\lim_{\rho \to 0^+} \frac{\rho}{2|\cos \theta+\sin \theta|}$$
Let $g(\theta):=|\cos \theta+\sin \theta|$, $g$ is a continuous function and the interval $[0,2\pi]$ is compact, so for Weierstrass it has a minimum $m>0$. So
$$\lim_{\rho \to 0^+} \frac{\rho}{2|\cos \theta+\sin \theta|} \leq \lim_{\rho \to 0^+} \frac{\rho}{2m}=0$$
And this is independent from $\theta$. So by the squeeze theorem, the limit is $0$.
Here's my doubts:

I said that $m>0$ and not just $m \geq0$ because $|\cos \theta + \sin \theta|=0\Leftrightarrow \cos \theta =-\sin \theta$ and this is not allowed because $\frac{\sin xy}{x+y}$ is not defined for those values, but I'm not sure if this is correct because I am in a limit context.

I know that polar coordinates, to be bijective, must be such that $\theta\in[0,2\pi)$; here I have heavily used that $\theta\in[0,2\pi]$, is this the same? If yes, why can I extend the interval to $[0,2\pi]$ and not lose informations because now polar coordinates aren't bijective anymore?

To say that $|\sin t| \leq |t|$ I've used the fact that the sin function is Lipschitz-continuous with constant $L=1$, that is $|\sin x - \sin y| \leq |x-y|$ \for all $x,y\in\mathbb{R}$ using this inequality with $0$ in place of $y$ since it is valid for all $y$. Is this correct or I'm not allowed to fix one of the two variables?

Why using polar coordinates independent from the angle $\theta$ shows that the limit surely exists? If I'm not wrong, $\theta$, being an angle, represents only the lines passing through the origin and not all the possible paths; so I'm a bit confused because with polar coordinates I've eliminated only the possibilities of line paths. Can someone clarify this please?

答: If you approach the origin along the line $xy=x+y,\, (x,y)\neq (0,0)$ (which is contained in the domain), your function approaches $1$ $\left(\lim\limits_{t\to 0}\dfrac{\sin t}t=1\right)$, and along $x=0$ the limit is zero. Therefore your limit does not exist.


问:
On the inequality $\sum_{i=1}^n|a_i-b_i|\le\big\lfloor \frac{n^2}{2}\big\rfloor$
Given $i,a_i,b_i\in\{1...n\},\space a_i\neq a_j,b_i\neq b_j,\forall i\neq j$ prove that
$$\sum_{i=1}^n|a_i-b_i|\le\big\lfloor \frac{n^2}{2}\big\rfloor$$
This problem was proposed by the new contributor  @user3458994 and it was closed by five users. I find it somewhat challenging (it does not have an immediate answer), but it is sufficiently well posed and, indeed, it can be solved by answering correctly.

There are many possible sums $\sum_{i=1}^n|a_i-b_i|$; in fact there are $n!$ possibilities (number of permutations of the set $\{1,2,\cdots,n\}$). The minimum for these sums is $0$ corresponding to the identity permutation $a_i\rightarrow b_i=a_i$ in which case the inequality is trivially verified. We expose one of these sums having a maximum value $M$ exactly equal to $\big\lfloor \frac{n^2}{2}\big\rfloor$. I believed no other sum has a value greater than $M$ in which case the problem would be false (am I wrong?).

答:
Here's an almost immediate answer.

By expanding each term of $ |a_i - b_i|$ into the corresponding $\pm (a_i - b_i)$, we ge that
$$ \sum |a_i - b_i | = \sum c_i i,  $$ where $c_i \in \{-2, 0, 2 \}$ and $\sum c_i = 0 $.

Note: This is a necessary, but not sufficient condition. In particular, not all combinations of $c_i$ are possible from the absolute value, so we'd subsequently have to ensure that this can be satisfied. However, we're "lucky" enough that this works out for us.

When $n=2m$ is even, the maximum of $\sum c_i i $ is $ -2\times 1 -2 \times 2 \ldots - 2 \times m + 2 \times (m+1) + 2\times (m+2) + \ldots + 2 \times (2m) = 2m^2$.

This is satisfied with $a_i = i, b_i = n+1-i$, so it's the maximum of $ \sum |a_i - b_i|$.

When $n = 2m+1$ is odd, the maximum of $\sum c_i i $ is $ -2\times 1 -2\times 2 \ldots - 2\times m + 2\times (m+2) + 2\times (m+3) + \ldots + 2 \times (2m+1) = 2m(m+1)$.

This is satisfied with $a_i = i, b_i = n+1-i$, so it's the maximum of $ \sum |a_i - b_i|$.

Note: The necessary and sufficient condition is $ \sum_{i=1}^k c_{n+1-i} \geq 0$ for all $ 1 \leq k \leq n$. Once that is satisfied, there's a pretty natural way to assign the values. (Have a think about it.)


问:
Distribution and pdf of sum of random variables
The independent variables $X$, $Y$ have binomial distributions with parameters $m$, $p$ and $n$, $p$, respectively. Find the pdf of $X + Y$ . What distribution does $X + Y$ have?

Can somebody give me some ideas, please?

答:
Recall that if $W\sim\mathsf{Ber}(p)$ then the probability-generating function of $W$ is
\begin{align}
P_W(s) &= \mathbb E[s^W]\\
&= s^0\cdot\mathbb P(W=0)+s^1\cdot\mathbb P(W=1)\\
&= 1-p+ps.
\end{align}
If $W_1,\ldots,W_n$ are independent random variables with probability-generating functions $P_{W_i}(s)$ then the probability-generating function of $\sum_{i=1}^n W_i$ is $\prod_{i=1}^n P_{W_i}(s)$ (this can readily be shown by induction). Since a $\mathsf{Bin}(n,p)$ random variable has the distribution of the sum of $n$ i.i.d. $\mathsf{Ber}(p)$ random variables, it follows that
\begin{align}
P_X(s) &= (1-p + ps)^m\\
P_Y(s) &= (1-p + ps)^n.
\end{align}
It is clear then that the probability-generating function of $X+Y$ is
$$
P_{X+Y}(s) = P_X(s)P_Y(s) = (1-p + ps)^m(1-p + ps)^n = (1-p + ps)^{m+n},
$$
and hence $X+Y$ has $\mathsf{Bin}(m+n,p)$ distribution.


问:
Is $\phi:\mathbb{Z}_5\to\mathbb{Z}_{30},$ $\phi(x)=6x$ a homomorphism?
Is $\phi:\mathbb{Z}_5\to\mathbb{Z}_{30},$ $\phi(x)=6x$ a homomorphism?

They can help me solve this problem. I know the answer is no. And I think it's a trick that I still don't see

答:
Hints: As, $6$ is a idempotent in $\mathbb{Z}_{30}$

$\phi(xy)=6xy=(6x)(6y)=\phi(x) \phi(y) $


答:
For any permutation, there's some $1 \le k \le n$ values of $i$ where

$$a_i - b_i \lt 0 \tag{1}\label{eq1A}$$

Thus, the remaining $n - k$ values of $i$ will be where

$$a_i - b_i \ge 0 \tag{2}\label{eq2A}$$

For simplicity, if need be, adjust the values of $a_i$ and $b_i$ so the $k$ values where \eqref{eq1A} hold are the ones where $1 \le i \le k$. This then gives

$$\begin{equation}\begin{aligned}
\sum_{i = 1}^{n}|a_i - b_i| & = \sum_{i = 1}^{k}|a_i - b_i| + \sum_{i = k + 1}^{n}|a_i - b_i| \\
& = \sum_{i = 1}^{k}(b_i - a_i) + \sum_{i = k + 1}^{n}(a_i - b_i) \\
& = 2\sum_{i = 1}^{k}(b_i - a_i) - \sum_{i = 1}^{k}(b_i - a_i) + \sum_{i = k + 1}^{n}(a_i - b_i) \\
& = 2\sum_{i = 1}^{k}(b_i - a_i) + \sum_{i = 1}^{k}(a_i - b_i) + \sum_{i = k + 1}^{n}(a_i - b_i) \\
& = 2\sum_{i = 1}^{k}(b_i - a_i) + \sum_{i = 1}^{n}a_i - \sum_{i = 1}^{n}b_i \\
& = 2\sum_{i = 1}^{k}(b_i - a_i)
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$

The last line comes from $\sum_{i = 1}^{n}a_i = \sum_{i = 1}^{n}b_i$ so the last $2$ terms of the line before cancel. In \eqref{eq3A}, the maximum value comes from the $b_i$ being the largest allowed $k$ values, i.e., $n - k + 1 \le b_i \le n$, and $a_i$ being the smallest allowed $k$ values, i.e., $1 \le a_i \le k$. Thus,

$$\begin{equation}\begin{aligned}
2\sum_{i = 1}^{k}(b_i - a_i) & \le 2\left(\sum_{i = n - k + 1}^{n}i - \sum_{i = 1}^{k}i \right) \\
& = 2\left(\frac{k((n - k + 1) + n)}{2} - \frac{k(k + 1)}{2}\right) \\
& = k(n - k + 1 + n - k - 1) \\
& = 2k(n - k)
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$

Note $f(k) = 2k(n - k)$ is a concave down parabola with a maximum at $k = \frac{n}{2}$. For even $n$, this value of $k$ is an integer, with it giving a maximum value of \eqref{eq4A} as

$$\begin{equation}\begin{aligned}
2k(n - k) & \le 2\left(\frac{n}{2}\right)\left(n - \frac{n}{2}\right) \\
& = n\left(\frac{n}{2}\right) \\
& = \left\lfloor \frac{n^2}{2} \right\rfloor
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$

For odd $n$, the same maximum value is achieved with $k = \frac{n - 1}{2}$ and $k = \frac{n + 1}{2}$. Using the first value, we get from \eqref{eq4A} that

$$\begin{equation}\begin{aligned}
2k(n - k) & \le 2\left(\frac{n - 1}{2}\right)\left(n - \frac{n - 1}{2}\right) \\
& = (n - 1)\left(\frac{n + 1}{2}\right) \\
& = \frac{n^2 - 1}{2} \\
& = \left\lfloor \frac{n^2}{2} \right\rfloor
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$

This shows the stated inequality always holds. Note Piquito's answer gives an explicit example where the maximum possible value is reached for even $n$.

Title: On the inequality $\sum_{i=1}^n|a_i-b_i|\le\big\lfloor \frac{n^2}{2}\big\rfloor$

Body:
Given $i,a_i,b_i\in\{1...n\},\space a_i\neq a_j,b_i\neq b_j,\forall i\neq j$ prove that
$$\sum_{i=1}^n|a_i-b_i|\le\big\lfloor \frac{n^2}{2}\big\rfloor$$
This problem was proposed by the new contributor  @user3458994 and it was closed by five users. I find it somewhat challenging (it does not have an immediate answer), but it is sufficiently well posed and, indeed, it can be solved by answering correctly.

There are many possible sums $\sum_{i=1}^n|a_i-b_i|$; in fact there are $n!$ possibilities (number of permutations of the set $\{1,2,\cdots,n\}$). The minimum for these sums is $0$ corresponding to the identity permutation $a_i\rightarrow b_i=a_i$ in which case the inequality is trivially verified. We expose one of these sums having a maximum value $M$ exactly equal to $\big\lfloor \frac{n^2}{2}\big\rfloor$. I believed no other sum has a value greater than $M$ in which case the problem would be false (am I wrong?).


问:
Does the direct sum decomposition of a p-group tell you the number of Prüfer subgroups?
My understanding is that every countable Abelian $p$-group can be written as a direct sum of at most countably many finite cyclic groups and at most countably many copies of the Prüfer $p$-group.

Suppose that when such a direct-sum decomposition is written for a particular countable Abelian $p$-group $G$, there are $N$ copies of the Prüfer $p$-group in the decomposition, where $N$ can be $0$, finite, or infinite.  Then my question is, does that mean that there are exactly $N$ subgroups of $G$ that are isomorphic to the Prüfer $p$-group?  Or can there be more than $N$ such subgroups?

答:
(in additives notations)

With $N=0$ then $$G=\prod_{n\ge 1}' \Bbb{Z}/p^{k_n} \Bbb{Z}$$ where $\prod'$ means we consider the 'sequences' $x_n\in  \Bbb{Z}/p^{k_n} \Bbb{Z}$ with only finitely many terms $\ne 0$.

For $x,y\in G$, $x = p^m y$ implies that $x_n=0$ whenever $m\ge k_n$.

Thus there is no non-zero element $x\in G$ such that $\forall m, \exists y\in G,x=p^m y$ ie. there is no copy of $\Bbb{Z}[p^{-1}]/\Bbb{Z}$ in $G$.

With $N=1$ we are looking at $G\times \Bbb{Z}[p^{-1}]/\Bbb{Z}$ and it works more or less the same way.

With $N=2$ we are looking at $G\times \Bbb{Z}[p^{-1}]/\Bbb{Z}\times \Bbb{Z}[p^{-1}]/\Bbb{Z}$ and something different happens.


问:
Division of power series
Let $f(x) =\sum_{n=0}^{\infty} a_n x^n$ and $g(x) =\sum_{n=0}^{\infty} b_n x^n$ in an open interval $(-r, r)$, and $g \neq 0 $ in $(-r, r)$. How can I prove that $h = f/g$ can be expressed as a power series $\sum_{n=0}^{\infty} c_n x^n$ that converges at least in $(-r,r)$?
I know how to 'calculate' the power series expansion of $h$, for example I can use the determinant method in https://en.wikipedia.org/wiki/Power_series#Multiplication_and_division. The question is, how can I prove that the series expansion 'does exist'(that is, converges)?

答:
This is a long comment and should be treated as such.

Well, it turns out that the radius of convergence of $1/g$ depends not only on that of $g$ but also on the region where $g$ does not vanish.

Consider $$g(x) =2-\cos x=1+\frac{x^2}{2!}-\frac{x^4}{4!}+\dots$$ which converges everywhere so that radius of convergence is $\infty$. But its reciprocal $$\frac{1}{g(x)}=1-\frac{x^2}{2}+\dots$$ does not have radius of convergence $\infty $ but actually it is much smaller.

Why? We know that $g(0)$ is non zero and for real $x$ the value $g(x) >0$, but the reals don't count too much and we exactly need to know a number $R>0$ such that $g(z) \neq 0$ for all complex $z$ with $|z|<R$. Then the radius of convergence of $1/g$ will be at least $R$.

The theorems of Apostol's book also use complex numbers to deal with power series.

I am not very much familiar with complex analysis, but the key is that the idea of a function being expressed via power series in its region of convergence is completely equivalent to a function being analytic in a circular region. I hope there is some textbook available which describes analytic function theory starting from power series approach.

Note: This also explains the term radius of convergence as there is actually a circular region  of convergence in the complex plane involved. When restricting to real variables the only part seen is the interval of convergence and the term radius looks inappropriate for indicating the size of that interval.

Title: Division of power series


问:
Obtaining the distribution of a test statistic under $H_0$
I start with a sampling distribution of an unknown discrete probability distribution with PMF $g(x)$. I want to test if this distribution is in fact a geometric distribution, this is my $H_0$. Furthermore I have good reason te believe that "checking for"
$$\hat{g}(1)(1-\hat{g}(1))-\hat{g}(2)=0$$
is a good criterion to do this in my case.

(Note that you obtain $p(1-p)-p(1-p)=0$ if $g(x)$ corresponds to a geometric distribution)

So I want to build a test statistic, using this criterion. And here starts my question:

I have never encountered a test statistic like this, and I want to learn about the art of creating test statistics. When I browse the internet looking for things like this, I encounter the classical test statistics which I do know, but I want to learn more about the techniques and ideas behind the creation of a test statistic.

In this case, starting from  $\hat{g}(1)(1-\hat{g}(1))-\hat{g}(2)$, how to proceed? How can I even derive the distribution of $\hat{g}(1)$ and $\hat{g}(2)$? In each case the test statistic itself will, under $H_0$, be centered around $0$. My estimators for $\hat{g}(1)$ and $\hat{g}(2)$ are MLE and my sample size is really big, so intuitively I would say that my test statistic will be approximately (?) normally distributed with mean $0$.

Thanks for your input!

答:
I think the following is related to what you are asking:

If you have a i.i.d. sample size $n$ from a distribution where $\mathbb P(X_i=1)=p$ and $\mathbb P(X_i=2)=(1-p)p$, and you observe $X_i=1$ $A$ times and $X_i=2$ $B$ times,  then $\mathbb E\left[\frac An \right] = p$ and $\mathbb E\left[\frac Bn \right] = (1-p)p$,

so $\mathbb E\left[\frac An \right]\left(1-\mathbb E\left[\frac An \right]\right) - \mathbb E\left[\frac Bn \right]=0$,

but $\mathbb E\left[\frac An \left(1- \frac An\right)- \frac Bn\right] = -\frac{(1-p)p}{n} \not = 0$.  It does tend towards $0$ as $n$ increases.

The distribution of $\frac An \left(1- \frac An\right)- \frac Bn$ is not normal or even close to normal for small $n$: it is a discrete asymmetric distribution. As $n$ increases, it converges in distribution to $0$; empirically, suitably rescaled it appears to converge in distribution towards a normal distribution with mean $0$ as $n$ becomes large.


问:
If $f, g \in L_{loc}^1(\Omega)$ and $\langle f, \phi \rangle=\langle g, \phi \rangle$ for all $\phi \in D(\Omega)$, then $f=g$ a.e. in $\Omega$

Let $f, g \in L_{loc}^1(\Omega)$ be locally integrable function. Denote $\langle f, \phi\rangle$ for the action of $f$ on the test function $\phi$ as a distribution. Suppose $\langle f, \phi \rangle=\langle g, \phi \rangle$ for all test functions $\phi \in D(\Omega)$, then $f=g$ a.e. on $\Omega$.

[proof] Let $\Omega_\epsilon = \Omega \setminus \cup_{x \in \partial \Omega} \overline{B_\epsilon(x)}$, where $\partial \Omega$ is the boundary of $\Omega$. Let $\phi_x(y)=j_\epsilon(x-y)$ for $x \in \Omega_\epsilon, y \in \Omega$, where $j$ is the standard mollifier. Then
$$\langle f , \phi_x \rangle= \int_\Omega f(y)\phi_x(y)dy=\int_\Omega f(y)j_\epsilon (x-y)dy=\int_{\mathbb{R}^n}f(y)j_{\epsilon}(x-y) \rightarrow f \text{ in } L^1_{loc} (\text{so a.e. on } \Omega)$$
Similarly, $\langle g, \phi_x \rangle \rightarrow g$ in $L^1_{loc}$ (so a.e. on $\Omega$). Hence $f=g$ a.e. on $\Omega$.

I don't get few things in the poof.

$\int_\Omega f(y)j_\epsilon (x-y)dy=\int_{\mathbb{R}^n}f(y)j_{\epsilon}(x-y)$

Here we are extending $f(y)=0$ if $y \in \mathbb{R}\setminus\Omega$, but why does choosing $x \in \Omega_\epsilon$ guarantee that the support of $j_\epsilon(x-y)$ is contained in $\Omega$?

We have $\langle f, \phi_x \rangle=f\ast j_\epsilon \rightarrow f$ in $L^1_{loc}$ as $\epsilon \rightarrow 0$ from approximation by $C^\infty$-function. (I think theorem gives us the convergence in $L_1$, but anyway this still holds), but how can we get the pointwise a.e. convergence? I think the convergence in $L_p$ in general only guarantee the pointwise a.e. convergence of the subsequence. Our condition is even weaker. (the convergence in $L^1_{loc}$)

I appreciate your help in advance.

答:
$j_\epsilon(x-\cdot)$ is radially symmetric with center $x$. The distance from any $x\in\Omega_\epsilon$ to points in the complement of $\Omega$ is $\ge\epsilon$ therefore $j_\epsilon(x-y)=0$ if $y\notin\Omega$.

Choose the subsequence along which $\langle f,
\phi_x\rangle\to f$ pointwise a.e. From that subsequence, extract a subsequence along which $\langle g,
\phi_x\rangle\to g$. Since the limit of a subsequence is always equal to the limit of the mother sequence, $f=g$ a.e.


问:
Find the derivative of the root of cubic equation
Consider the following cubic equation:

\begin{align}
\alpha ^3-2 \alpha ^2- \alpha(6+4x^2) +8 x^3-4x&=0
\end{align}
Where $x\in \mathbb{R}$.

The solutions $\alpha_1,\alpha_2,\alpha_3$ will depend on $x$. Solving this equation on Mathematica or Wolfram Alpha gives me complicated third roots.

I am not interested in the actual value of the roots, I am only looking for the value of $\frac{\mathrm{d}\alpha}{\mathrm{d}x}$. Does a method exist to obtain a "nice" expression for $\frac{\mathrm{d}\alpha}{\mathrm{d}x}$ without having to actually solve the cubic equation?

Edit:
Applying the advice from the answer, I obtain:
$$\frac{\mathrm{d}\alpha}{\mathrm{d}x}=\frac{(4 + 8 \alpha x - 24 x^2)}{(-6 - 4 \alpha + 3 \alpha^2 - 4 x^2)}$$
Is there a way to eliminate the $\alpha$ terms and have an expression $\frac{\mathrm{d}\alpha}{\mathrm{d}x}$ uniquely depending on $x$?

答:
This answer will overlap alot with @Claude_Leibovici 's answer, but also covers the case when the cubic has 3 real roots.

Starting with Nickalls' parameters related to the geometry of the cubic, applied to your cubic:

$$\begin{align*}
        \alpha_N &= \dfrac{-b}{3a} = \dfrac{2}{3} \quad \text{(abscissa of inflection point)}\\
        \\
        \left[\delta(x)\right]^2 &= \dfrac{b^2-3ac}{9a^2} = \alpha_N^2 - \dfrac{c}{3a} = \dfrac{4}{3}x^2+\dfrac{22}{9}  \\
&\quad (\alpha\text{ distance squared from inflection point to turning point})\\
        \\
        y_N(x) &= f(\alpha_N) = \dfrac{2b^3}{27a^2} - \dfrac{bc}{3a} + d = 8x^3-\dfrac{8}{3}x^2-4x-\dfrac{124}{27}\\
&\quad \text{(ordinate of inflection point)}\\
        \\
        h(x) &= 2a\left[\delta(x)\right]^3 =2\left[\delta(x)\right]^3 = 2\left(\sqrt{\dfrac{4}{3}x^2+\dfrac{22}{9}}\right)^3 \\
&\quad \text{(y distance from inflection point to turning point)} \\
\end{align*}$$

The roots of your cubic, $\alpha_k(x)$, $k \in\{0,1,2\}$, can be expressed as a hyperbolic function of a complex hyperbolic angle, $\psi_k\left(\dfrac{-y_N(x)}{h(x)}\right)$, described in my short write-up here, based off of Nickalls' and Holmes' work:

$$\begin{align*}
        \alpha_k(x) &= \begin{cases}
                \alpha_N + 2\delta(x)\cosh\left[\dfrac{1}{3}\psi_k\left(\dfrac{-y_N(x)}{h(x)}\right)\right] \quad h(x) \ne 0\\
                \\
                \lim_{h(x) \to 0} \alpha_N + 2\delta(x)\cosh\left[\dfrac{1}{3}\psi_k\left(\dfrac{-y_N(x)}{h(x)}\right)\right] \\
\\
\quad \quad \quad=  \alpha_N + \sqrt[3]{\dfrac{-y_N(x)}{a}}\exp\left[\dfrac{1}{3}\left(i2k\pi\right)\right] \quad h(x) = 0 \\
        \end{cases} \\
\end{align*}$$

with

$$\begin{align*}
        \psi_k(u) &= i\theta_k(u)+ \phi(u)= \begin{cases}
                i(2k+1)\pi + \cosh^{-1}(-u) &  u \in \mathbb{R}, u < -1\\
                \\
                i2k\pi + i\cos^{-1}(u) & u \in \mathbb{R}, -1 \le u \le 1\\
                \\
                i2k\pi + \cosh^{-1}(u) &  u \in \mathbb{R}, u > 1\\
                \\
                i\left(2k+\dfrac{1}{2}\right)\pi + \sinh^{-1}(-iu) &  u \in i\mathbb{R}\\
        \end{cases} \\
\end{align*}$$

It's worth noting here that the input argument of $\psi_k(u)$, $u = \dfrac{-y_N(x)}{h(x)}$, is actually a discriminant between the case of 3 real roots and the other cases of 1 real root.

So to find your desired derivative

$$\begin{align*}\dfrac{d}{dx}\alpha_k(x)&= \dfrac{d}{dx}\left(\alpha_N+2\delta(x)\cosh\left[\dfrac{1}{3}\psi_k\left(\dfrac{-y_N(x)}{h(x)}\right)\right]\right)\\
\\
&=2\delta'(x)\cosh\left[\dfrac{1}{3}\psi_k\left(\dfrac{-y_N(x)}{h(x)}\right)\right]\\
\\
&\quad + 2\delta(x)\sinh\left[\dfrac{1}{3}\psi_k\left(\dfrac{-y_N(x)}{h(x)}\right)\right]\dfrac{1}{3}\dfrac{d}{d(-y_N/h)}\psi_k\left(\dfrac{-y_N(x)}{h(x)}\right)\dfrac{d}{dx} \dfrac{-y_N(x)}{h(x)}\\
\\
&=2\delta'(x)\cosh\left[\dfrac{1}{3}\psi_k\left(\dfrac{-y_N(x)}{h(x)}\right)\right]\\
\\
&\quad + 2\delta(x)\sinh\left[\dfrac{1}{3}\psi_k\left(\dfrac{-y_N(x)}{h(x)}\right)\right]\dfrac{1}{3}\psi_k'\left(\dfrac{-y_N(x)}{h(x)}\right)\\
\\
&\quad \quad \cdot \dfrac{y_N(x)h'(x)-h(x)y'_N(x)}{\left[h(x)\right]^2}\\
\\
\end{align*}$$

At this point, the computation of $\delta'(x)$, $y'_N(x)$, and $h'(x)$ should be fairly straightforward.

The derivative of the complex hyperbolic angle $\psi_k(u)$ is

$$\begin{align*}
        \psi_k'(u) &= \begin{cases}
                \dfrac{-1}{\sqrt{u^2-1}} &  u \in \mathbb{R}, u < -1\\
                \\
                \dfrac{-i}{\sqrt{1-u^2}} & u \in \mathbb{R}, -1 \le u \le 1\\
                \\
                \dfrac{1}{\sqrt{u^2-1}} &  u \in \mathbb{R}, u > 1\\
                \\
                \dfrac{-i}{\sqrt{1-u^2}}  &  u \in i\mathbb{R}\\
        \end{cases} \\
\end{align*}$$

Then all that's needed is substitution of all of those functions of $x$ into the expression for $\alpha_k'(x), k \in \{0,1,2\}$.

If you don't want to use the hyperbolic function expression for the cubic roots, then from Nickalls' 1993 paper, you can go with

$$\begin{align*}\alpha_0(x) &= \alpha_N + \sqrt[3]{\dfrac{1}{2}\left(-y_N(x)+\sqrt{\left[y_N(x)\right]^2-\left[h(x)\right]^2}\right)} + \sqrt[3]{\dfrac{1}{2}\left(-y_N(x)-\sqrt{\left[y_N(x)\right]^2-\left[h(x)\right]^2}\right)}\\
\\
\alpha_{1,2}(x) &= \alpha_N - \dfrac{1}{2}\alpha_0(x) \pm i\dfrac{\sqrt{3}}{2}\sqrt{\left[\alpha_0(x)\right]^2-4\left[\delta(x)\right]^2}\\
\end{align*}$$

So computing your desired derivatives:

$$\begin{align*}\alpha'_0(x) &= \dfrac{1}{3}\dfrac{\dfrac{1}{2} \left(-y'_N(x) + \dfrac{y_N(x)y'_N(x)-h(x)h'(x)}{\sqrt{\left[y_N(x)\right]^2-\left[h(x)\right]^2}}\right)}{\left[\sqrt[3]{\dfrac{1}{2}\left(-y_N(x)+\sqrt{\left[y_N(x)\right]^2-\left[h(x)\right]^2}\right)}\right]^2} \\
\\
&\quad + \dfrac{1}{3}\dfrac{\dfrac{1}{2} \left(-y'_N(x) - \dfrac{y_N(x)y'_N(x)-h(x)h'(x)}{\sqrt{\left[y_N(x)\right]^2-\left[h(x)\right]^2}}\right)}{\left[\sqrt[3]{\dfrac{1}{2}\left(-y_N(x)-\sqrt{\left[y_N(x)\right]^2-\left[h(x)\right]^2}\right)}\right]^2} \\
\\\end{align*}$$

$$\begin{align*}\alpha'_{1,2}(x) &= -\dfrac{1}{2} \alpha'_0(x)\pm i\dfrac{\sqrt{3}}{2}\dfrac{\alpha_0(x)\alpha'_0(x) -4\delta(x)\delta'(x)}{\sqrt{\left[\alpha_0(x)\right]^2-4\left[\delta(x)\right]^2}}\\
\end{align*}$$

Again, all that's needed is substitution of all of those functions of $x$ into the expression for $\alpha_k'(x), k \in \{0,1,2\}$.


问:
$TM/ D$ is trivial if and only if there is an $1$-form $\alpha$ such that $D = \text{Ker }\alpha$
Let $ D$ a distribution of dimension $n-1$. I'm trying to prove that $D$ is coorientable (i.e. the bundle $TM/D = \sqcup_{x \in M} T_xM/D_x$ has a global trivialization) if and only if $D$ is the kernel of a non-vanishing $1$-form $\alpha$.

MY ATTEMPT:

$\Leftarrow$ Let us suppose that $D = \text{Ker }\alpha = \bigsqcup_{x \in M}\text{Ker }\alpha(x)$. We will construct a global trivialization $\phi$ from $\alpha$. Since $\alpha(x):T_xM \rightarrow \mathbb R$ is surjective (as $\alpha$ is non-vanishing), $\overline{\alpha(x)}:T_xM/\text{Ker }\alpha(x) \rightarrow \mathbb R$ in a linear iso. Let us define
$$\phi: TM/D \rightarrow M \times \mathbb R:(x, [X_x]) \mapsto (x, \overline{\alpha(x)}([X_x])).$$
We clearly see that $\pi_1 \circ \phi = \pi$ (where $\pi_1: M\times \mathbb R \rightarrow M$ is the projection on the first component and $\pi:TM/D \rightarrow M$ is the projection bundle of $TM/D$). Here it only remain to show that $\phi$ is a diffeo, here is my first question, see below. (1)

$\Rightarrow$ Since there exists a global trivialization
$$\phi: TM/D \rightarrow M \times \mathbb R:(x, [X_x]) \mapsto (x, \phi_x([X_x])),$$
we can define $\alpha(x)(X_x) =  \phi_x([X_x])$. We just have to prove that $\alpha$ is an $1$-form and that $D = \text{Ker }\alpha$. First, since $\phi$ is a trivialization, $\phi_x$ is a linear iso therefore $\alpha(x)$ is a linear functionnal on $T_xM$. Moreover it is clear that $\text{Ker }\alpha(x) = D_x$ because $\phi_x$ is an iso, hence the only vectors that are send to zero are the ones in $D_x$. Now we have to prove that $\alpha$ is a smooth section of $M$. We have that $(\pi' \circ \alpha)(x) = \pi'(\phi_x) = x$. Here is my second question (2).

QUESTIONS:

Here are my questions:

(1): I really do not see how to show that it is a diffeo. I even have some difficulties to show that $\phi$ is a smooth function. If both component of $\phi$ are smooth we are good but how can I show that $\overline{\alpha}$ is smooth in $x$ from the fact that $\alpha$ is ?

(2): Here is a bit the same question as above, how can I show that $\alpha:M \rightarrow T^*M$ is smooth from the fact that $\phi_x$ is ?

Is this a right way to show this equivalence ? It seems to work but I have some difficulties with details..

English is not my mother tongue, so there must be a lot of mistakes, sorry for that.

答:
Let's cheat: fix a Riemannian metric ${\tt g}$ for $M$, so that $TM/\mathcal{D} \cong \mathcal{D}^\perp$.

Assume that $\mathcal{D} = \ker \alpha$ for non-vanishing $\alpha \in \Omega^1(M)$. Use ${\tt g}$ to convert $\alpha$ to a non-vanishing vector field $\alpha^\sharp \in \mathfrak{X}(M)$, normal to $\mathcal{D}$ --- then $\alpha^\sharp \in \Gamma(\mathcal{D}^\perp)$ gives the global trivialization $\Phi\colon M \times \Bbb R \to \mathcal{D}^\perp$ by $\Phi(x,\lambda) = \lambda \alpha^\sharp_x$.

Conversely, assume that $\Phi\colon M\times \Bbb R \to \mathcal{D}^\perp$ is a given global trivialization for $\mathcal{D}^\perp$. Define a vector field $X \in \mathfrak{X}(M)$ by $X_x = \Phi(x,1)$, and note that $X$ never vanishes. In particular, $X \in \Gamma(\mathcal{D}^\perp)$. Then use ${\tt g}$ to convert $X$ to a non-vanishing $1$-form $\alpha = X_\flat \in \Omega^1(M)$. Clearly $\mathcal{D} = \ker \alpha$.


答:
There's some ambiguity here.

First of all, the map is a group homomorphism, which is what is suggested by your tags. If, however, you're asking whether it is a ring homomorphism (per the comments), it depends on what definition of ring and ring homomorphism you're using. Specifically, many texts - most, in my experience - include the existence of a multiplicative unit as part of the definition of ring, and correspondingly require ring homomorphisms to preserve that unit. Other texts don't make these requirements. When there's a danger of ambiguity the terms "unital ring (homomorphism)" and "non-unital ring (homomorphism)" are used.

For example, consider the set $2\mathbb{Z}$ of even integers with the usual notions of addition and multiplication. This is a non-unital ring; whether or not it's considered a ring outright depends on what text you're using. Relatedly, consider the function $$f: \mathbb{Z}\rightarrow\mathbb{Z}: x\mapsto 2x.$$ Both domain and codomain are unital rings, but $f$ itself is only a homomorphism in the non-unital sense. Basically, what we have here is a non-unital ring homomorphism between two non-unital rings which happen to have units. :P More jargonily, the category of unital rings is a non-full subcategory of the category of non-unital rings.

But back to your question. The map $$\phi:\mathbb{Z}_5\rightarrow\mathbb{Z}_{30}: x\mapsto 6x$$ is only a ring homomorphism in the non-unital sense, and so - again, in my experience (I'm not an algebraist) - most texts would not consider it to be a ring homomorphism.

Body:
Is $\phi:\mathbb{Z}_5\to\mathbb{Z}_{30},$ $\phi(x)=6x$ a homomorphism?

They can help me solve this problem. I know the answer is no. And I think it's a trick that I still don't see


问:
Limit in a system
I have the following system of differential equations:

I want to find the $\lim_{t\to\infty}a_t$,$\lim_{t\to\infty}b_t$, $\lim_{t\to\infty}c_t$ and $\lim_{t\to\infty}d_t$.

答:
We see that $(ac)'=ca'+ac' = -abc + ac(b-1) = -ac$; thus $ac = C^2e^{-t}$. I choose the integrating constant anticipating the derivations below.

If we take the derivative of $c$ in terms of $a$, we get $c' = -C^2e^{-t}a^{-2}(a'+a)$.

We aim to find an equation only in terms of $a$. Take the derivative of the first equation to get $a''=-ab'-a'b$. Multiply by $a$ to get $aa''=-a^2b'-a'ab=-a^2b'+(a')^2$. Recognize that $a'+b'+c'=-c$, and thus after substituting expressions for $c$ and $c'$ in terms of $a$, $b'=-c-a'-c'=-\left(C^2a^{-1}e^{-t}+a'-C^2e^{-t}a^{-2}(a'+a)\right)$. Substituting this expression into the $a''$ equation and simplifying, we arrive at the following second order nonlinear equation.

$$aa''=-[C^2e^{-t}-a^2]a'+(a')^2$$

I do not think this equation has a simple closed form solution. However, consider a particular solution of form $a=k e^{-\delta t}$. Then $aa'' = k^2 \delta^2 e^{-2\delta t} = (a')^2.$ Substituting this back into the nonlinear differential, we get $C^2e^{-t} = k^2e^{-2\delta t}$, where equating constants gives us a solution $a = Ce^{-t/2}$, and consequently $a=c$. In this scenario, $b=c'/c+1=1/2$, and $d=\int c dt = \gamma_0-2Ce^{-t/2}$. Then we fit back into the initial conditions $a(0)+b(0)+c(0)+d(0)=2C+1/2+\gamma_0-2C=1$ to get $\gamma_0=1/2$

Title: Limit in a system


问:
Does the following graph have a Hamilton circuit?

A Hamilton circuit (or path) is a path that visits each vertex exactly once (except the start/end point) and ends at the starting point.

I've stared at this for quite a while and cannot find a Hamilton circuit yet my guide says that one exists. Where is the path?

答: There is a discussion on Hamilton cycle problem and associated algorithm here. Also, please take a look at the PairViz package of R.</p>

ContentLicense: CC BY-SA 3.0


问:
Calculation of the Covariance of Gaussian Mixtures
I have a Gaussian mixture model, given by:
$$
X \sim \sum_{i = 1}^M \alpha_i N_p(\mu_i, C_i)
$$
such that $\sum_{i=1}^M\alpha_i =1 $.
Is there a way I can compute the overall covariance matrix if $x$? I would like to say "$X$ has a covariance matrix given by $C$".

答:
Motivated by the answer provided by Michael Hardy, a formal solution to such question might be formulated as follows:

By introducing a new hidden variable $I$ to represent the identity of the local model, the probability of the Gaussian mixtures can be decomposed as:
$$p(x|I=i)=\mathcal{N}(\mu_i,C_i)$$
$$p(I=i)=\alpha_i$$
Therefore,
$$E(x)=E[E(x|I=i)]=\sum_{i=1}^{M} \alpha_i \mu_i$$
\begin{align}
Var(x)&=E[Var(x|I=i)]+Var[E(x|I=i)] \\ &=\sum_{i=1}^{M} \alpha_i C_i + \sum_{i=1}^{M} \alpha_i (\mu_i-\bar{\mu})(\mu_i-\bar{\mu})^T
\end{align}
where $\bar{\mu}=E(x)$. Moreover, the Law of total expectation and Law of total variance have been used in above two equations.


问:
Positive definite(ness) of the inner product?
I was working on an inner product proof. From what I've learned, the positive-definite axiom of inner products involves showing that:

Given a vector space $V$ over $F$, for all $v \in V$, $\langle v, v \rangle \in R$ and $\langle v, v\rangle \geq 0$ with $\langle v, v \rangle = 0$ iff $v = 0$. Or alternatively, $\langle v, v \rangle \geq 0$ with equality iff $v = 0$.

While the other axioms are fairly easy to comprehend, I'm a little confused about this one. In particular, what does it expect one to prove?

My thoughts: First, show that $\langle v, v \rangle \in R$ and $\langle v, v \rangle \geq 0$ for all $v \in V$. Then, assume $v = 0$ and show that $\langle v, v \rangle = 0$. Then, assume $\langle v, v \rangle = 0$ and show that $v = 0$.

Is that what it requires? Or, is one supposed to assume that $v = 0$ and show that $\langle v, v \rangle \geq 0$ AND $\langle v, v \rangle = 0$, and then, assume $\langle v, v \rangle \geq 0$ and $\langle v, v \rangle = 0$ and show that $v = 0$?

After seeing some proofs, I'm still unsure. Any explanation is much appreciated.

答:
The first one is what it's saying. In order to satisfy the axiom as presented, one must show:

$\langle v, v \rangle \in \Bbb{R}$ for all $v \in V$,

$\langle v, v \rangle \ge 0$ for all $v \in V$ (note that $\langle v, v \rangle$ can only be considered $\ge 0$ if it is a real quantity),

$\langle v, v \rangle = 0 \implies v = \vec{0}$,

$v = \vec{0} \implies \langle v, v \rangle = 0$, or in other words, $\langle \vec{0}, \vec{0} \rangle = 0$.

That said, two of these points can be deduced from the other axioms. Inner products must also satisfy $\langle v, w \rangle = \overline{\langle w, v\rangle}$ for all $v, w \in V$, hence
$$\langle v, v \rangle = \overline{\langle v, v\rangle},$$
and the only complex numbers equal to their own conjugate are real numbers. So, we actually don't really need to prove the first point.

We also don't need to prove the last point, due to the linearity condition. We know that, for all $v, w \in V$ and $\lambda \in \Bbb{C}$,
$$\langle \lambda v, w \rangle = \lambda \langle v, w \rangle.$$
In particular, if we take $v = w = \vec{0}$ and $\lambda = 0$, then we get
$$\langle \vec{0}, \vec{0}\rangle = \langle 0 \cdot \vec{0}, \vec{0}\rangle = 0\langle \vec{0}, \vec{0}\rangle = 0.$$
So, you really only need to prove the second and third points, which should hopefully align with the proofs you've seen so far.


问:
Cumulative distribution function of Y
How do you get the cumulative distribution function of $Y$ in terms of $X$?

Let's say that $E[X] = μ, Var[X] = σ^2,$ and $Y =a + bX$.

What would be the process to get the cumulative distribution function?

答:
Let's say that $E[X] = μ, Var[X] = σ^2,$ and $Y =a + bX$.

Wouldn't $E[Y]$ = $a + Bμ$ and $Var[Y] = B^2 σ^2$?

It actually would be that.

What would be the process to get the cumulative distribution function?

You start with the cummulative distribution function for $X$.

If $F_{\small X}(x)$ is the CDF for $X$, and $b>0$, then $F_{\small Y}(y)$, the cdf for $Y$, is:$$F_{\small Y}(y) ~=~ F_{\small X}\left(\dfrac{y-a}b\right)$$

Since $\mathsf P(Y\leqslant y) ~{= \mathsf P(a+bX\leqslant y)\\=\mathsf P\left(X\leqslant (y-a)/b\right)}$

Title: Cumulative distribution function of Y


问:
Does $du=(y-x^2)dx+(x+y^2)dy$ is an exact differential or not?
I need to show if the $du=(y-x^2)dx+(x+y^2)dy$ differential is exact or not. But as far as I understand, it is exact if the function $ u (x, y) $ exists such that the total differential is the one shown. So looking for who the role should be:

Since
$\cfrac{\partial u}{\partial x}=y-x^2$ then
$u(x,y)=\int(y-x^2)dx+\phi(y)$
$u(x,y)=xy-\cfrac{1}{3}x^3+\phi(y)$

And partially deriving with respect to y:
$\cfrac{\partial u}{\partial y}=x+\phi'(y)$

and equaling to $ x + y ^ 2 $

$x+\phi'(y)=x+y^2 \iff \phi'(y)=y^2$ and then $\phi(y)=\int y^2dy=\cfrac{1}{3}y^3+C$

And therefore the function $ u (x, y) $ will be:

$u(x,y)=xy-\cfrac{1}{3}x^3+\cfrac{1}{3}y^3+C$

Is that enough to show that it was accurate? Or how is it shown to be exact for before deducing the function $ u (x, y) $?

答: Your work shows it is exact. But in general $a(x,y)dx+b(x,y)dy$ is exact when $\partial a(x,y)/\partial y=\partial b(x,y) / \partial x$ which can be checked directly without finding what you did.


问:
Continued fractions, why the reciprocal?
Why is  the reciprocal of the fractional part (vs the integer part) taken at each step?  Perhaps this is obvious, if so I don't see it.  Appreciate your guidance.

Example from Wiki showing reciprocal taken at each step:

Body:
It's just the Euclidean algorithm.

$$\cfrac{649}{200}=3+\cfrac{49}{200}=3+\cfrac{1}{\cfrac{200}{49}}=3+\cfrac{1}{4+\cfrac{4}{49}}=3+\cfrac{1}{4+\cfrac{1}{\cfrac{49}{4}}}=3+\cfrac{1}{4+\cfrac{1}{12+\cfrac{1}{4}}}$$


问:
Notation for time interval
I'm trying to write the following using sensible notation:

In the time period between now and M units ago, if n > 0 output 1, if n = 0 output 0. Where n is the number of events that have occurred in the time period.

I think I can write that as:

\begin{equation}
 \label{eqn:algdef}
 \mbox{for }[t-M,t]\mbox{ }  B = \left\{
    \begin{array}{lll}
     1 & for & n > 0\\
     0 & for & n = 0
    \end{array}
   \right.
\end{equation}

However, I'm an engineer, not a mathematician and I want to check that that this notation makes sense.

If it doesn't what's the better way to do this?

答:
If you want understandability and not formality here's what I'd go for:

$B(M) =
\begin{cases}
1:  & E_M \neq \emptyset \\
0: & E_M = \emptyset
\end{cases}$

Where $E_M$ is the set of events that happen between time $(t - M)$ and $t$:

$E_M = \{ e_p: e_p \in E \text{ and } t-m \leq p \leq t\}$

Where $e_p$ is an event that occurs at time $p$, and $E$ is the set of all events.

Title: Notation for time interval


问:
Let $\mathbb{Z} \subset \mathbb{R}$ with the discrete topology prove that $X$ is connected if and only if $f: X \rightarrow \mathbb{Z}$ is constant

Let $\mathbb{Z} \subset \mathbb{R}$ with the discrete  topology prove that $X$ is connected if and only if for every continuous function  $f: X \rightarrow \mathbb{Z}$ is constant function.

Attempt

Suppose that $X$ is connected and consider an arbitrary continuous function $f: X \rightarrow \mathbb{Z}$
since $X$ is connected  then $f(X)\subset \mathbb{Z}$ is connected, and $f(X)$ is an open set since we be considering the discrete topology, that is $f(X) \subset \mathbb{Z}$ is open in the subespace topology
if for any $x\in X$ $f(x)=\mathbb{Z} \cap \mathbb{R}$for all $x\in X$.This is $f$ is a constant function.

Conversely assume that for every continuous function $f:X \rightarrow \mathbb{Z}$ is constant we should prove that $X$ is connected.

Assume that $X$ is not conected, then Exists $U,V\subset X$ such that $U,V \neq \emptyset$ and $U \cup V=X$,but $U \cap V = \emptyset$ then consider the function
$h_1:U \rightarrow \mathbb{Z}$ given by $h_1(x)=1$ which is continuous function.
And consider $h_2:V \rightarrow \mathbb{Z}$ given by $h_2(x)=0$ which is continuous function.
Notice that $U\cap V= \emptyset$ implies that $h_1$ and $h_2$ are equal in their points of intersections.
Consider $h:U \cup V=X \rightarrow \mathbb{Z}$ given by
$$h(x)=\begin{cases}
h_1(x) if\, x\in U&\\
h_2(x)if\, x\in V
\end{cases}$$
Which is non constant continuous function then not every continuous function $f:X \rightarrow \mathbb{Z}$ is constant which is a contradiction, then $X$ is connected.
Is my proof right?

答:
For the first part of the proof the fact that $f[X]$ is open in $\Bbb Z$ is irrelevant: all that matters is that the only non-empty, connected subsets of $\Bbb Z$ are the singletons.

The other direction is fine, though it can be done more simply without the pasting lemma. Just prove the contrapositive. If $X$ is not connected, there is a continuous function $f$ mapping $X$ onto the discrete two-point space $\{0,1\}$. Clearly this $f$ can be viewed as a continuous function from $X$ to $\Bbb Z$, and it is not constant.


问:
Confusion about definition of a regular value
So this is the definition I have:

Given a smooth map of differential manifolds, a point in the image manifold is termed a regular value for the smooth map if every point in its inverse image is a regular point, i.e. if the map from the tangent space at any point in the inverse image, is surjective.

This is how I am understanding things.

So taking it part by part, we have a map from one manifold to another. So far so good.
Because this map need not be bijective, multiple points in the domain may map to the image set.

A regular value is one such that, if you are given a point in the target manifold, and all the points that map to it in the domain manifold, then for every tangent plane to each of the points in the domain, there is a bijection from the tangent plane at the domain point to the tangent plane at the target point.

But. isn't that always true? Aren;t 2 planes always trivially bijective? What am I missing?

答:
No that isn't always the case. For example the inclusion $\mathbb{R} \to \mathbb{R}^2, t \mapsto (t, 0)$ isn't regular on its image, since the tangent spaces of $\mathbb{R}$ are one-dimensional while those of $\mathbb{R}^2$ are two dimensional. In general, for $n < m$ a map from a $n$-manifold to a $m$-manifold isn't regular on its image.

The situation above isn't the only reason why a point can fail to be regular though. The map $\mathbb{R}^2 \to \mathbb{R}^2$ given by $(x, y) \mapsto (x, xy)$ isn't regular at $(0, 0)$.


问:
Use contour integration to compute the Fourier transform,
The problem statement is:

Use contour integration to determine the Fourier transform,
$\large \hat f(ξ)=∫_{-\infty}^{\infty}f(x)e^{−iξx}dx$,

of $\large f(x)=\frac{1}{2−2x−x^2}$.

Some issues that I am running into are:

Attempt #1

I originally chose a lower semi-circular contour, since the negative i on the exponential function will help give the needed estimate of zero along the semi-circular contour, as we let it grow to infinity - on the upper semi-circular contour, the positive imaginary axis poses problems for the factor $e^{-i\xi x}$.

But then the two simple poles of the function are lying on the real axis.

So I had thought of perhaps, when integrating along the real axis, perhaps jump over the two poles, and estimate the integrand on those two tiny circles and show it goes to zero.

But then, my contour would enclose...no poles...and my function would be analytic on and inside the closed contour, and by Cauchy's theorem, my integration yields zero.

Attempt #2

Now, in hopes of enclosing the two (or at least one?) poles, I chose a box contour instead.  A box of height 8, so 4 units above and below the real axis.  This is a square box to start, and ensures that both poles are enclosed.  I then parameterized fixing the width of the box and hoping to let the length of the box grow to infinity.  Then perhaps, integration along the top and bottom edges of this box, which will be integration with respect to the x variable, will give me two desired integrals.  Unfortunately, it looks like these desirable integrals are canceling each other out (of course, by the parametrization of the box).

Now I'm not sure what to do.

Any hints or suggestions are welcome.

Thanks,

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
\hat{\on{f}}\pars{\xi} & \equiv
\bbox[5px,#ffd]{%
\mrm{P.V.}\int_{-\infty}^{\infty}{\expo{−\ic\xi x} \over
2 - 2x - x^{2}}\,\dd x} =
-\expo{\ic\xi}\,\mrm{P.V.}
\int_{-\infty}^{\infty}{\expo{−\ic\xi x} \over x^{2} - 3}\,\dd x
\\[5mm] = &\
-{\root{3} \over 6}\expo{\ic\xi}\,\mrm{P.V.}
\int_{-\infty}^{\infty}\expo{−\ic\xi x}\pars{%
{1 \over x - \root{3}} - {1 \over x + \root{3}}}\,\dd x
\\[5mm] = &\
-{\root{3} \over 6}\expo{\ic\xi}\,\sum_{\sigma = \pm}
\sigma\,\mrm{P.V.}\int_{-\infty}^{\infty}
{\expo{−\ic\xi x} \over x - \sigma\root{3}}\,\dd x
\\[5mm] = &\
-{\root{3} \over 6}\expo{\ic\xi}\,\sum_{\sigma = \pm}
\sigma\left[%
\int_{-\infty}^{\infty}
{\expo{−\ic\xi x} \over x - \sigma\root{3} + \ic 0^{+}}\,\dd x
\right.
\\[2mm] &\ \phantom{-{\root{3} \over 6}\expo{\ic\xi}\,\sum_{\sigma = \pm}
\sigma\left[\right.}
+ \left.\int_{-\infty}^{\infty}
\expo{−\ic\xi x}\,\ic\pi\,\delta\pars{x - \sigma\root{3}}\,\dd x
\right]
\\[5mm] = &\
-{\root{3} \over 6}\expo{\ic\xi}\,\sum_{\sigma = \pm}
\sigma\bracks{%
-\bracks{\xi < 0}2\pi\ic\expo{−\ic\xi\pars{\sigma\root{3}}} +
\ic\pi\expo{−\ic\xi\sigma\root{3}}}
\\[5mm] = &\
-{\root{3} \over 6}\expo{\ic\xi}\,\pi\ic\
\overbrace{\braces{-2\bracks{\xi < 0} + 1}}
^{\ds{\on{sgn}\pars{\xi}}}\
\overbrace{%
\sum_{\sigma = \pm}\sigma\expo{−\ic\xi\pars{\sigma\root{3}}}}
^{\ds{-2\ic\sin\pars{\root{3}\xi}}}
\\[5mm] = &\
\bbx{-{\root{3} \over 3}
\,\pi\expo{\ic\xi}\sin\pars{\root{3}\verts{\xi}}} \\ &
\end{align}


问:
Show that the matrix $A^2 + I$ is invertible for all matrices $A$, where $A$ is an $n \times n$ symmetric matrix.
I'm a little stuck on this problem. I know that since $A$ is symmetric, $A=A^{T}$. I'm also pretty sure that $AA^{T}$ is invertible. Therefore $A^2$ would be invertible. I'm not really sure how to account for the identity matrix (In).

Thanks in advance!

答: Eigenvalues of symmetric matrix are purely real, Enough to show $\det(A^2+I)\ne 0$ i.e enough to show that $-1$ is not an eigen value of $A^2$, $A$ symmetric so if $\lambda_i$'s are eigen values  then $\lambda_i\in\mathbb{R}\forall i$ and the eigen values of $A^2$ will be $\lambda_i^2\ge 0$ so indeed $-1$ can not be an eigen value of $A^2$ </p>


问:
Simplifying $ \frac{2x-x\lvert x-1\rvert+x\lvert x\rvert+5}{\lvert x\rvert+1}$
How would one begin to simplify the following?

$$ \frac{2x-x\lvert x-1\rvert+x\lvert x\rvert+5}{\lvert x\rvert+1}$$

I have found the domain as

$$\begin{cases}
\dfrac{x+5}{1-x},  & \ {(-\infty,0]} \\[4pt]
\dfrac{2x^2+x+5}{x+1}, & \ {[0,1)} \\[4pt]
\dfrac{5x+5}{x+1}, & \ {[1,+\infty)} \\
\end{cases}$$

Thank you for any help!

答:
Hints

When dealing with something like |r|, you want to consider two cases:

$r < 0$ and $r \geq 0.$

Here you have 3 conditions to worry about.

Is $(x - 1) \geq 0$ or is $(x - 1) < 0.$

Is $(x) \geq 0$ or is $(x) < 0.$

Is $|x| + 1 = 0.$  If possible, this would make the denominator $= 0,$ which
is not allowed.  However, this situation is impossible because $|x|$ can not equal
$-1$.

Try to "collapse" the cases that represent points 1 and 2 above into 3 distinct
intervals.  Then, for each interval, create a totally distinct function that
applies only to that interval.

I overlooked that the OP has shown work and that therefore, it is okay to complete the problem.

The original function is

$$ f(x) = \frac{2x-x\lvert x-1\rvert+x\lvert x\rvert+5}{\lvert x\rvert+1}$$

The three intervals will be

Interval 1: $x < 0.~$ The specific function will be $f_1(x).$

$$ f_1(x) = \frac{2x ~- ~[(x)(1 - x)] ~+~ [x(-x)] ~+~ 5}{(-x) ~+ ~1}.$$

Interval 2: $0 \leq x < 1.~$ The specific function will be $f_2(x).$

$$ f_2(x) = \frac{2x ~- ~[(x)(1 - x)] ~+~ [x(x)] ~+~ 5}{(x) ~+ ~1}.$$

Interval 3: $1 \leq x.~$ The specific function will be $f_3(x).$

$$ f_3(x) = \frac{2x ~- ~[(x)(x - 1)] ~+~ [x(x)] ~+~ 5}{(x) ~+ ~1}.$$

At this point, none of $~f_1(x), f_2(x),~$ or $~f_3(x)$ are employing any absolute value signs.  Now, you can consider whether there is any simplification that is common to all three functions $~f_1(x), f_2(x),~$ and $~f_3(x).$

From my perspective, other than multiplying the numerators out, I see no other possible simplification that can be applied to all three functions.


问:
Probability puzzle for a fair die
What is the expected number of times we have to roll a fair six-sided die to obtain that the number of times any two faces appear are equal (after the first rolling)?</p>

答:
Too long for a comment.

Staring from @saulspatz's answer, I wonder what it would become for a fair $n$-sided die and worked in the same spirit
$$\frac1{n^{nk}}\frac{(nk)!}{(k!)^n}\sim \sqrt{n}\, (2 \pi )^{\frac{1-n}{2}} k^{\frac {1-n}2 }$$

$$P_n=\sum_{k=1}^\infty\frac1{n^{nk}}\frac{(nk)!}{(k!)^n}\sim \sqrt{n}\, (2 \pi )^{\frac{1-n}{2}} \, \zeta \left(\frac{n-1}{2}\right)$$ which is almost a straight line in a logarithmic scale.

Title: Probability puzzle for a fair die


问:
Question of calculus concepts
Today I learned some concepts of increasing and decreasing functions, but some is so vague, so I have many questions. Here I can have a list of questions:

Functions f(x) is decreasing on an interval (a,b) then f'(x)<0 on (a,b) yes? But what if f'(x)<0. Is a function f(x)decreasing if $f'(x)<0$ on an interval (a,b)

Functions f(x)>f(b) in the interval (a,b). so is f decreasing on (a,b).

Monotonic functions are flat. If it is a random function f(x), if f(a)=f(b), then is f(x) monotonic on the interval (a,b)?

Thanks

答:
If a function is decreasing it means its slope is negative, i.e., $f^\prime (x) < 0$.

No.  $f(x) > f(b)$ does not mean that $f(x)$ is decreasing.  It could go up and down (so long as it is always greater than $f(b)$.

No.  Monotonic means that the function is either always increasing or always decreasing.

Title: Question of calculus concepts


问:
how many 9 digit numbers have the property that the product of their firsts and last digits is even?
How many 9 digit numbers have the property that the product of their firsts and last digits is even?

I tried listing them out, but there's probably way too many. This is a basic counting problem.

答: The product of the first and last digits is even iff they are not both odd. There are $9×10$ ways to choose the first and last digits; of those $5×5$ result in an odd product due to both digits being odd. The remaining seven digits may be chosen arbitrarily, so the answer is $(9×10-5×5)10^7=65\cdot10^7$.


问:
Do Boolean rings always have a unit element?

Let $(B, +, \cdot)$ be a non-trivial ring with the property that every $x \in B$ satisfies $x \cdot x = x$. How does one prove that such a ring $(B, +, \cdot)$ must have a unit element $1_B$? (Or, in case this is not true in general, what is a counterexample?)

BTW, I'm looking for an elementary proof, not requiring anything more than the definition of a ring, the definition of $(B, +, \cdot)$, and, if necessary, the easily shown facts that $x + x = 0$ and $x\cdot y = y\cdot x,\,\forall\, x,y \in B$.

答:
In the ring $\prod_{i=1}^\infty\Bbb Z_2$, consider the ideal $\oplus_{i=1}^\infty\Bbb Z_2$. It is a subrng without identity.

(Ultimately I think this is isomorphic to Boris' example, but here the operations are clear.)


问:
How to solve $\int_0^{\infty}\frac{x^n}{(x^2+1)^n}\,\mathrm dx$ for $n\ge 2$?
I am trying to find some closed form answer for the integral
$$\int_0^{\infty}\frac{x^n}{(x^2+1)^n}\,\mathrm dx,\quad n\ge 2$$
I am not sure if a closed form exists and I have been trying this integral for hours.

Any tips or hints would be appreciated.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{0}^{\infty}
{x^{n} \over \pars{x^{2} + 1}^{n}}
\,\dd x\,\right\vert_{\,\Re\pars{n}\ >\ 1}}
\,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,
{1 \over 2}\int_{0}^{\infty}
x^{\color{red}{n/2 + 1/2} - 1}\,\,\,\,\pars{1 + x}^{-n}\,\,\dd x
\end{align}
Note that
$\ds{\pars{1 + x}^{-n} =
\sum_{k = 0}^{\infty}{-n \choose k}x^{k} =
\sum_{k = 0}^{\infty}{n + k - 1 \choose k}\pars{-1}^{k}x^{k} =
\sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{n + k} \over \Gamma\pars{n}}{\pars{-x}^{k} \over k!}}$.

Then, with Ramanujan's Master Theorem,
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{0}^{\infty}
{x^{n} \over \pars{x^{2} + 1}^{n}}
\,\dd x\,\right\vert_{\,\Re\pars{n}\ >\ 1}} =
{1 \over 2}\,\Gamma\pars{\color{red}{{n \over 2} + {1 \over 2}}}
{\Gamma\pars{n - \bracks{\color{red}{n/2 + 1/2}}} \over \Gamma\pars{n}}
\\[5mm] = &\
\bbx{\Gamma\pars{n/2 + 1/2}\Gamma\pars{n/2 - 1/2} \over 2\Gamma\pars{n}}\\ &
\end{align}


答:
Decreasing does not imply $f’(x)<0$. For example, consider the function $f(x)=-x^3$ at the point $(0,0)$. $f’(0) = 0, $ however the function is strictly decreasing, even at $x=0$. Also constant functions like $f(x)= 4$ are decreasing...However the converse is true: $f’(x) < 0 \implies$ strictly decreasing.

No. Consider $\sin x$ on $[\frac\pi6,\pi]$

Monotonic functions are not necessarily constant functions. But if $f(a)=f(b)$ and $f$ is monotonic on $[a,b]$ then this forces $f$ to be constant. However, I think you could do with clarification. See: https://math.stackexchange.com/a/3907156/29156

Title: Question of calculus concepts

Body:
Today I learned some concepts of increasing and decreasing functions, but some is so vague, so I have many questions. Here I can have a list of questions:

Functions f(x) is decreasing on an interval (a,b) then f'(x)<0 on (a,b) yes? But what if f'(x)<0. Is a function f(x)decreasing if $f'(x)<0$ on an interval (a,b)

Functions f(x)>f(b) in the interval (a,b). so is f decreasing on (a,b).

Monotonic functions are flat. If it is a random function f(x), if f(a)=f(b), then is f(x) monotonic on the interval (a,b)?

Thanks


问:
$1\cdot3\cdot5\cdots(2n-1) < (\frac{2n}{e})^{n+1}, n \in \mathbb{N}, n \geq 2$ proof doesn't seem to work
I found that excercise in an old book without any hint nor solution. Olthough I know a thing ot two about induction, this one seems to be too tricky for me. I know for sure that it's a proof by induction

It is known that: $3e^3 < 64$ (for the base case)
Prove that:

$$1\cdot3\cdot5\cdots(2n-1) < \left( \frac{2n}{e} \right)^{n+1}, \quad n \in \mathbb{N}, n \ge 2$$

The induction that I am trying to do looks fine in terms of calculations, yet it is not proving anything.

答:
Without induction, the problem would be quite simple since you want to prove that
$$\frac{2^n }{\sqrt{\pi }}\Gamma \left(n+\frac{1}{2}\right)<\left( \frac{2n}{e} \right)^{n+1}$$ Taking logarithms and using Stirling approximation
$$\log(\text{rhs - lhs})=\left(\log (n)-1+\frac{\log (2)}{2}\right)+\frac{1}{24 n}+O\left(\frac{1}{n^3}\right)$$
$$\text{rhs - lhs}=\frac{\sqrt{2} n}{e}+\frac{1}{12 \sqrt{2} e}+\frac{1}{576 \sqrt{2} e
   n}+O\left(\frac{1}{n^2}\right)$$


问:
find the number of seven letter "words" that start with a or do not contain a
Find the number of seven letter "words" that start with a or do not contain a. For this problem, I'm not really sure what method I can use. I tried to subtract the "words" that start with a from 657800 (26 choose 7)</p>

答:
"Words" here should be taken to mean arrangements of letters, not actual dictionary words.

The first case is having an "a" at the start. Which means that the remaining six postions can contain any letter (including "a"). This gives $1\cdot 26^6 = 26^6$ possibilities.

The second case is not containing an "a" at all. Which means all seven positions must be filled with one of the remaining $25$ letters in the alphabet. That gives $25^7$ possibilities.

Note that the two cases are mutually exclusive.

Therefore the total number is $26^6 + 25^7 =6412431401$.

EDIT: Addressing your attempted answer, note that $\binom {26}7$ would only be relevant if you're choosing $7$ (unique, non-repeated) letters out of the alphabet to fill the $7$ positions without regard to order. This is not a relevant consideration here.


答:
So, answer should be as follows:

Here we have $7$ spaces.
The total number of words that start with a are :
$1 \cdot 26^6$ ($1$ is representing a here and $26$ representing rest of letters)
And the words that don't contain a are:
$25^7$ (all words other than a).
See there is no intersection between two cases.
Like you can't have a word that start with a and do not contain a. Right?
So, answer will be $26^6+25^7$.
Thanks, hope it helps.

Title: find the number of seven letter "words" that start with a or do not contain a

Body: Find the number of seven letter "words" that start with a or do not contain a. For this problem, I'm not really sure what method I can use. I tried to subtract the "words" that start with a from 657800 (26 choose 7)</p>


问:
How can you show that $f\delta′=f(0)\delta′−f′(0)\delta$ for a function f that is infinitely differentiable?
Assume that $f$ is infinitely differentiable. Let $\delta$ be the (Dirac) delta functional.

I know that $f\delta = f(0)\delta$, but I'm not sure how to derive the equation  $f\delta′=f(0)\delta′−f′(0)\delta$.

答:
I'm assuming you mean the delta function(al) $\delta(x)$.

If so, you should use formal integration by parts.
$$\int g\times (f\delta') =\int -(fg)'\delta = \int -(f'(0)g(x)+g'(x)f(0))\delta = -f'(0) \int g\delta-f(0)\int g'\delta$$

Integrating the last term by parts and stripping off the $\int g\times$ gives the result.


问:
Completeness of the space of Riemann integrable functions under $\left \| . \right \|_{\infty}$ on $\left [ 0,1 \right ]$
I read a proof about this topic at this site and was convinced with it , then I tried to construct a cauchy sequence of functions to see how things apply . Now I'm confused.

The sequence I constructed is : $f_n(x)=\left\{\begin{matrix}
-lnx & if & x> e^{-n}\\
 n& if & x\leq e^{-n}
\end{matrix}\right.$

I proved that this sequence is cauchy and that it belongs to the mentioned space . For the limit I got that $f_n(x)\rightarrow f(x)=-lnx$

almost everywhere. But $f$ is not bounded on $\left [ 0,1 \right ]$

Did I miss or mess with something ??

答:
First observe that the space $R[0,1]$ of all Riemann integrable functions is a subspace of the complete space $B[0,1]$ formed by all bounded functions.  So $R[0,1]$ is complete iff it is closed in $B[0,1]$.

Theorem. $R[0,1]$ is closed in $B[0,1]$, and hence complete.

Proof. Given a sequence $\{f_n\}_n$ in $R[0,1]$ converging to some $f$ in $B[0,1]$, let $D_n$ be the set of points where $f_n$ is discontinuous.

It is well known that $D_n$ has measure zero, and hence so does
$$
D := \bigcup_n D_n.
$$

Notice that if $x$ is not in $D$ then each $f_n$ is continuous at $x$, and hence so is $f$.  In other words the discontinuities of $f$ lie in $D$ so $f$ is Riemann integrable.  QED


问:
Recursive Sequences with Number Theory
Here is a problem that one of cousins asked me. He is currently studying mathematics at IIT. I am not really sure how to solve it, so please help. I am comfortable with algebraic concepts and number theory at AIME level but don't know too much beyond that. How would I go about solving this problem.

For integers q $\geq4$ find all positive reals $x_{1},x_{2},x_{3}...x_{q}$ such that $x_{n}^2=9x_{n+1}+10x_{n+2}+11x_{n+3}$ holds for all $n=1,2,3....q$ with indices taken modulo q.

答:
For the largest number in the sequence, say $A$, you have:
$$A^2 \leqslant 9A+10A+11A \implies A\leqslant 30$$
Similarly, for the smallest number in the sequence, say $a$, you have:
$$a^2 \geqslant 9a+10a+11a \implies a\geqslant 30$$
Since $A \geqslant x_i \geqslant a$, this forces all the numbers in the sequence to be $30$, which will thus be your only solution.

Title: Recursive Sequences with Number Theory


问:
Simple Diffy-Q problem
So as a fun project, I'm trying to work my way through Kreyzig's "Advanced Engineering Mathematics". But I've gotten to a really simple problem:
$$xy' = 2y$$

where I know the solution is $x^2$ but for the life of me I can't figure out how to integrate this really simple problem properly. I keep ending up with:$$\lg(x) = 1/2\lg(2y), $$

but I don't think that's right. Help?

答:
THis is a problem where we may use separation of variables. That is we begin by moving the $x$ terms to the right side and $y$ terms to the left so that we have $$\frac{y'}{y} = \frac{2}{x}.$$

We then integrate both sides  and have $$\int \frac{1}{y} dy = \int \frac{2}{x} dx$$ which results in $$ln(y) =2\ln(x) + C.$$ Solving for $y$ by exponentiating both sides yields $$y=e^{\ln(x^2) + c} = e^{\ln(x^2)}e^c = Ax^2$$ where $A = e^c$.

Title: Simple Diffy-Q problem


问:
Why is it correct to assume $f = e^{\lambda x}$ in differential equation?
I was solving the following differential equation:

$$f'' + f = 0$$

And, as I learned in class, I assumed that $f = e^{\lambda x}$ and then found the roots of the polynomial and found the solution. But then the following came to my mind:

Why is it correct to assume that $f = e^{\lambda x}$? How do we know that by doing this we are indeed finding all the solutions of the equation and we aren't only finding the solutions that take that form? Because doing this looks like a guessing game, we try to guess the solution and then find it by correcting and fixing the values of $\lambda$.

答:
We can't take $~y=e^{\lambda x}~$ as a trial solution to any second order differential equation, but only for the homogenius linear ones with constant coefficients. In fact you should take $~y=e^{\lambda x}~$ as a trial solution for any homogeneous linear ordinary differential equation with constant coefficients.

One possible reason behind this can be explained as follows: The fundamental theorem of algebra says that any $n^{th}$ degree polynomial has exactly $n$ roots (including multilicity). When you assume $~e^{\lambda x}~$ as a trial solution to one of these equations what happens? The equation is of the form: $$a_0y+a_1y'+a_2y''+\cdots+a_ny^{(n)}=0~.$$ Assuming $~y=e^{\lambda x}~$ as the trial solution we have: $$a_0e^{\lambda x}+\lambda a_1e^{\lambda x}+\lambda^2a_2e^{\lambda x}+\cdots+\lambda^na_ne^{\lambda x}=0~.$$ Canceling $~e^{\lambda x}~$ gives: $$a_0+\lambda a_1+\lambda^2a_2+\cdots+\lambda^n a_n=0~.$$ Which is an $n^{th}$ degree polynomial. Therefore we are guaranteed $n$ solutions to this polynomial, and we know that an $n^{th}$ degree linear differential equation has $n$ linearly independent solutions. So, as long as all the roots are distinct, we have solved the problem since $e^{\lambda x}$ and $e^{mx}$ are linearly independent when $~m≠\lambda~$. Otherwise we need to try other solutions as well. The other solutions that work in this case are of the form $~x^α e^{\lambda x}~.$

Note: For non homogeneous linear differential equation obviously $~e^{\lambda x}~$ will not work. In this case we have to choose $~y=x^m~$ which comes from the motivation that a $n$ times differentiable function can be represented as linear combination of power of $~x^m~.$


问:
Is this a metric between subspaces?
Let $E$, $F$ be vector subspaces of $\mathbb{R}^d$. We define:
\begin{equation*}
 \rho (E,F) = \max \left\{ \max_{v \in S_E} d(v, F) \, , \, \max_{w \in S_F} d(w, E) \right\} \, ,
\end{equation*}
where $d(v, F) = \inf_{f \in F} \| v - f \|$, $d(w, E) = \inf_{e \in E} \| w - e \|$ are the usual point-set distances and $S_E$, $S_F$ are the unit spheres of each subspace.

I'm reading a text that implicitly assumes that $\rho$ is a metric between subspaces. At first glance it really resembles the usual Hausdorff metric, but there's a catch, which is: the point-set distances are maximized over the spheres, but calculated in relation to the whole subspaces.

Thus, whenever I try to prove that this is a metric, I get stuck in the same part of the triangle inequality, which is the following. It is not hard to prove by standard metric space arguments that, given $E,F,G$ subspaces:
\begin{equation*}
 d(v, F) \leq d(w,F) + \| v - w \| \qquad \text{ for every $v \in E$, $w \in G$ }
\end{equation*}
Now, the next standard step would force me to restrict $w$ to $S_G$:
\begin{equation*}
 d(v, F) \leq \max_{w \in S_G} d(w, F) + \| v - w \| \leq \rho(G, F) + \| v - w \|
\end{equation*}
yielding
\begin{equation*}
 d(v, F) - \rho(G, F) \leq \| v - w \| \qquad \text{ for every $v \in E$, $w \in S_G$.}
\end{equation*}
However, now I can no longer relate this with the infimum over $w \in G$ to produce $d(v, G)$, since it is in general smaller than the infimum restricted to the sphere. I even proved that $d(v, G) = \inf_{w \in B_G} \| v - w \|$ when $v \in S_E$, but restrict to the ball is also not enough.

Any insight either into the proof or of a simple counterexample would be appreciated!

答:
If $\Bbb{R}^d$ is equipped with the Euclidean metric, then
$$\rho(E, F) = H(B_E, B_F),$$
which easily shows $\rho$ is a metric, assuming you know $H$ is a metric.

Suppose $E$ is a subspace of $\Bbb{R}^d$, and let $P_E$ be the projection operator onto $E$. Then $P_E$ is linear, with $I - P_E = P_{E^\perp}$. Both of these linear operators have norm $1$ or $0$, achieving the latter if and only if the corresponding subspace is non-trivial.

Recall that the norm of a linear map $T : V \to W$ between normed linear spaces can be expressed in the following equivalent ways:
$$\|T\| = \max_{x \in B_V} \|Tx\| = \max_{x \in S_V} \|Tx\|.$$
This comes from the fact that, if $x \in B_V$ maximises $\|Tx\|$, then $x / \|x\| \in B_V$, and
$$\|Tx\| \ge \left\|T\left(\frac{x}{\|x\|}\right)\right\| = \frac{\|Tx\|}{\|x\|} \implies \|x\| \ge 1 \text{ or } \|Tx\| = 0.$$
But, this either means $\|Ty\| \le \|Tx\| = 0$ for all $y$, i.e. $T = 0$ or $\|x\| = 1$, since $x \in B_V$. So, either $x \in S_V$, or $T$ achieves its maximum everywhere on the ball, including the sphere.

Thus, if we consider the restriction $P_{E^\perp}|_F$ of $P_{E^\perp}$ to a subspace $F$, we have
$$\max_{x \in S_F}d(x, E) = \max_{x \in S_F} \|x - P_E(x)\| = \max_{x \in S_F} \|P_{E^\perp}(x)\| = \|P_{E^\perp}|_F\| = \max_{x \in B_F}\|P_{E^\perp}(x)\| = \max_{x \in B_F}d(x, E).$$
Further, if $\|x\| \le 1$, then $\|P_E(x)\| \le \|P\|\|x\| \le 1$, hence $P_E(x) \in B_E$. Hence, if $\|x\| \le 1$, then
$$d(x, B_E) = \min_{y \in B_E} \|x - y\| \le \|x - P_E(x)\| = \min_{y \in E} \|x - y\| = d(x, E) \le d(x, B_E),$$
i.e. $d(x, B_E) = d(x, E)$. Putting this together, we get
\begin{align*}
 \rho (E,F) &= \max \left\{ \max_{v \in S_E} d(v, F), \max_{w \in S_F} d(w, E) \right\} \\
&= \max \left\{ \max_{v \in B_E} d(v, F), \max_{w \in B_F} d(w, E) \right\} \\
&= \max \left\{ \max_{v \in B_E} d(v, B_F), \max_{w \in B_F} d(w, B_E) \right\} \\
&= H(B_E, B_F).
\end{align*}

Title: Is this a metric between subspaces?


问:
is 0 a perfect square
Based on my research, I found that there are many arguments about this statement, the main factor is the true definition of perfect square. Some said they are the squares of the whole numbers, but some said that any number that can be written as a positive integer to the power of two and some said that an integer that can be expressed as the product of two equal integers.

答: A perfect square is the square of an integer – or nonnegative integer, without loss of generality, since $(-x)^2=x^2$. Since $0=0^2$, $0$ is a perfect square.

Title: is 0 a perfect square


问:
Find the limit in probability of the sequence $(Z_{n})_{n \in N}$
the truth is, the probability subject is very difficult for me, I have this other exercise with which. I don't know where to start, could you help me? I would appreciate it very much.

Let $(X_{n})_{n \in}$ be a family of independent random variables with Bernoulli distribution of parameter $\frac{1}{2}$. For each $n \in N$ we define the random variable:
$$Z_{n}=\sum_{k=1}^{n}\frac{X_{k}}{2^k}$$
Find the limit in probability of the sequence $(Z_{n})_{n \in N}$

答: Let $M_{Z_n}(t)$ be the moment generating function of $Z_n$. Then using independence $$M_{Z_n}(t)=\mathbb{E}\big(e^{tZ_n}\big)=\prod_{k=1}^n\mathbb{E}\Big(e^{tX_k/2^k}\Big)=\frac{1}{2^n}\prod_{k=1}^n\Big(1+e^{t/2^k}\Big)$$ You can see here that $M_{Z_n}(t) \longrightarrow \frac{e^t-1}{t}$. The limit corresponds to the moment generating function for a $\mathcal{U}(0,1)$ random variable.


答:
You're missing an $f$ in the left hand side.

Compute $(f\delta)'$ in two different ways.

On one hand $(f\delta)' = f'\delta + f\delta' = f'(0)\delta + f\delta'$.

On the other hand $(f\delta)' = (f(0)\delta)' = f(0)\delta'$.

Hence $f\delta' = f(0)\delta' - f'(0)\delta$.

Title: How can you show that $f\delta′=f(0)\delta′−f′(0)\delta$ for a function f that is infinitely differentiable?

Body:
Assume that $f$ is infinitely differentiable. Let $\delta$ be the (Dirac) delta functional.

I know that $f\delta = f(0)\delta$, but I'm not sure how to derive the equation  $f\delta′=f(0)\delta′−f′(0)\delta$.


问:
Covariance of 3 variables
$\newcommand{Cov}{\operatorname{Cov}}X,Y,Z$ are three random variables each with mean $0$ and variance $20$. $\Cov(X,Y)=\Cov(X,Z)=10$ and $\Cov(Y,Z)=5$. What is $\Cov(3X+Z,3X+Y)$?

答:
$\newcommand{Cov}{\operatorname{Cov}}$Covariance behaves like multiplication in that it distributes over linear combinations of random variables. Thus
$$\Cov(3X+Z,3X+Y)=9\Cov(X,X)+3\Cov(X,Y)+3\Cov(X,Z)+\Cov(Y,Z)=9\operatorname{Var}(X)+3\cdot10+3\cdot10+5=180+30+30+5=245$$

Title: Covariance of 3 variables


问:
Given the following table of values (list of numbers congruent to 1 or 5 mod 6) is there a formula to represent the value at the $i^{th}$ iteration
As the question states, how can we determine the value at the $i^{th}$ iteration of the following list of numbers congruent to 1 or 5 mod 6:

1

5

7

11

13

17

19

...

答:
If you separate them into odd-indexed terms and even-indexed terms, you can observe two arithmetic progression.

$1,7,13,\ldots$ can be written in the form of $g(n)=a_1+(n-1)d_1=1+6(n-1)$.

$5,11,17, \ldots$ can be written in the form of $h(n)=a_2+(n-1)d_2=5+6(n-1)$.

We want to construct the function \begin{align}f(n) &= \begin{cases} g(k) , & n =2k-1 \\ h(k), & n=2k\end{cases}\\ &= \begin{cases} g\left(\frac{n+1}2\right) , & n =2k-1 \\ h\left(\frac{n}2\right), & n=2k\end{cases}\\
&=\begin{cases} a_1+\left(\frac{n-1}2\right)d_1 , & n =2k-1 \\ a_2+\left(\frac{n-2}2\right)d_2, & n=2k\end{cases}\\
&=\begin{cases} 1+\left(\frac{n-1}2\right)6 , & n =2k-1 \\ 5+\left(\frac{n-2}2\right)6, & n=2k\end{cases}\\
&=\begin{cases} 3n-2 , & n =2k-1 \\ 3n-1, & n=2k\end{cases}\\
&=3n-\frac{3}2+(-1)^n\frac12\end{align}


问:
Finding Cauchy Principal Value for $\int_0^\infty \frac{\ln^2x }{(1-x)^2\sqrt{x}(x-4)} dx$
I am trying to find Cauchy Principal value for
$$\int_0^\infty \frac{\ln^2x }{(1-x)^2\sqrt{x}(x-4)} dx$$

Can you please suggest me where to start? Any help would be appreciated.

Thanks!

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

First, I'll evaluate
$\ds{\left.\oint_{\cal C}{z^{\nu} \over z - a}\,\dd z\,\right\vert_{\substack{-1\ <\ \nu\ <\ 0\\[0.75mm] a\ >\ 0}}\,\,\,}$ where $\ds{\cal C}$ is a key-hole contour which "takes care" of the $\ds{z^{\nu}}$-branch cut
$$
z^{\nu} = \verts{z}^{\nu}\expo{\ic\arg\pars{z}\nu}\,,\quad \arg\pars{z} \in \pars{0,2\pi}\,,\quad z \not= 0
$$

\begin{align}
0 & = \int_{0}^{\infty}{x^{\nu} \over x - a + \ic 0^{+}}\,\dd x +
\int_{\infty}^{0}
{x^{\nu}\expo{2\pi\nu\ic} \over x - a - \ic 0^{+}}\,\dd x
\\[5mm] & =
\mrm{P.V.}\int_{0}^{\infty}{x^{\nu} \over x - a}\,\dd x -
\ic\pi a^{\nu}
\\[2mm] & -
\expo{2\pi\ic\nu}\,\mrm{P.V.}
\int_{0}^{\infty}{x^{\nu} \over x - a}\,\dd x -
\ic\pi a^{\nu}\expo{2\pi\ic\nu}
\\[5mm] & =
\pars{1 - \expo{2\pi\ic\nu}}\,\mrm{P.V.}
\int_{0}^{\infty}{x^{\nu} \over x - a}\,\dd x -
\ic\pi a^{\nu}\pars{1 + \expo{2\pi\ic\nu}}
\end{align}

Then
\begin{align}
\mrm{P.V.}
\int_{0}^{\infty}{x^{\nu} \over x - a}\,\dd x & =
\pi\ic a^{\nu}\,{1 + \expo{2\pi\ic\nu} \over
1 - \expo{2\pi\ic\nu}} =
\pi\ic a^{\nu}\,{\expo{-\pi\ic\nu} + \expo{\pi\ic\nu} \over
\expo{-\pi\ic\nu} - \expo{\pi\ic\nu}}
\\[5mm] & =
-\pi a^{\nu}\cot\pars{\pi\nu}
\\[5mm]
\mrm{P.V.}
\int_{0}^{\infty}{\ln^{2}\pars{x} \over x - a}
\,{\dd x \over \root{x}}& =
\left.\partiald[2]{\bracks{-\pi a^{\nu}\cot\pars{\pi\nu}}} {\nu}
\,\right\vert_{\,\nu\ =\ -1/2} =
2\pi^{2}\,{\ln\pars{a} \over \root{a}}
\end{align}

In addition,
\begin{align}
&\mrm{P.V.}
\int_{0}^{\infty}{\ln^{2}\pars{x} \over \pars{x - a}\pars{x - 4}}
\,{\dd x \over \root{x}}
\\[5mm] = &\
{1 \over a - 4}\bracks{%
\mrm{P.V.}
\int_{0}^{\infty}{\ln^{2}\pars{x} \over x - a}
\,{\dd x \over \root{x}} -
\mrm{P.V.}
\int_{0}^{\infty}{\ln^{2}\pars{x} \over x - 4}
\,{\dd x \over \root{x}}}
\\[5mm] = &\
2\pi^{2}\,{\ln\pars{a}/\root{a} - \ln\pars{2} \over a - 4}
\end{align}

Finally,
\begin{align}
&\mrm{P.V.}
\int_{0}^{\infty}{\ln^{2}\pars{x} \over
\pars{x - 1}^{2}\pars{x - 4}}\,{\dd x \over \root{x}}
\\[5mm] = &\
2\pi^{2}\,\totald{}{a}\bracks{%
{\ln\pars{a}/\root{a} - \ln\pars{2} \over a - 4}}_{\,a\ =\ 1}
\\[5mm] = &\
\bbx{{2 \over 9}\,\pi^{2}\ln\pars{2} - {2 \over 3}\,\pi^{2}}
\approx -5.0595 \\ &
\end{align}


问:
Is there a tighter error bound for this convergent version of Stirling's series?
I have written an implementation of the log factorial function which employs a convergent version of Stirling's series.  This is the formula in the bottom of page 12 of R. Schumacher, "Rapidly Convergent Summation Formulas involving Stirling Series", 2016.

Usually, Stirling's approximation diverges, and this divergence is unsuitable for my purposes of exact sampling algorithms, where series expansions must converge in order for the algorithm to halt almost surely.  This is why I implemented Schumacher's convergent version of Stirling's series. In fact, my implementation calculates the log factorial as an interval that bounds the true log factorial from above and below, and this bound gets tighter as more terms are added.

But unfortunately, since the formula from the paper is monotonically increasing and that paper didn't specify the exact rate of convergence or an upper bound on the error incurred when truncating the series (a bound required for my purposes of exact sampling), I employed an ad hoc error bound in my implementation: The error is—$$(|{s_n}| + |{s_{n+1}|}) \times n,$$where n is the number of terms added, $s_n$ is the last term added, and $s_{n+1}$ is the first neglected term.

This error bound is added to the final interval's upper bound. However, experiments show this new upper bound to be rather loose compared to the lower bound.  And this increasingly shows as more terms of the series are added, which suggests to me that the error bound could be improved substantially.

Thus I ask: Is there a tighter error bound for this formula than the one I just gave?

Body:
<a href="https://tio.run/##dVRLb5tAEL7vr5jaUsojOGDJUoVCe6h6yKFSpeTmIouQtY2AJVqgquP4tyczvHaxcS7B883r@2ZmZf18@JjDzyJ/TkRUFTKJMtjWIq6SQpSMlVUis0TsNnEU7zkEsF57YTiYIXC@g7iFFI4MQPKqlgI8SLYgIAjAhZsbxOhLwS7BKWAcvL/jx32LjiutRUjFQvL9cgFNua9TinAM4Xim1aNr/HWLgA1ji@NdyYxZVKeT6IkxlKrIUagzUXrzpCgd63v8aFijB4LjTCPSF9CUe0u6Nyu2Y0vPdipcdTiJElsuRVFnWaLohsrYku2YlgiKIHDhB3gSfHAlmreFRL4JSrBYLIh06@nofWPX1pAS697hGFPbM9GbixfVIwZSS9sobldVteSFyqr1dObZKiyswUwCteJceqqyE9gJr@ZBCC6hrPM8onOBuuQvDd2s2O0itDJWcZlfSOTqqni6KHYABu2vlWorfCaO7aE8qZXa6ZQ8VhsvSKY5/MIs2ByHYjs0hW3S5T7uI5F@LaGSkSixm5yxwaNbYGq@3DSsAljpM7Z@R9V@gf6GMMFBcjZOd6Wsy@bT9/88WIK6fJUF5fAWLjvjbqgiztJshGicLaxip7ow5EirQXBHnDLXVQmzRy4T9Orp@n@r@RHDTjPG9HxaLQ8Bwf9Xm3ZCbXYd7wpdiUdzFpXXolVst@lDJctawh0YKtZRmKnxoeHAUz8ajc/4FHAJnziSxnmiH8NHF77R3PE9x0RkrvA8JK0nrUCMbzqPEfjH2@Eypk3mjf7jcWnT8VbsfB4HuBjFKxzu3vpXnYSlTAfWTIj49B1FkjrCxuusKn21b3RPLe/eYsyP4mT60PwR@WGztMsNzdPs4xM" rel="nofollow noreferrer">Experimentally, the series appears to have geometric or near geometric convergence. This suggests that Shank's transformation may prove useful for getting a more accurate result. Experimentally this appears to be the case, giving more significant figures of accuracy as $n$ increases. This implies you may use the rough error estimate of

$$\epsilon_n\approx\left|\frac{t_{n+1}^2}{t_{n+2}-t_{n+1}}\right|$$

where $t_n$ is the $n$th term in your series, and that

$$S_n-\frac{t_{n+1}^2}{t_{n+2}-t_{n+1}}$$

is more accurate than $S_n$ than your $S_n=\sum_{k\le n}t_n$.


问:
Can a probability CDF be discontinuous?
I was given the following stochastic process:
$ X(t) = V h(t-Z)$ , $t > 0$

Where $f_V (v) = 1/2, 0\leq v\leq 2$ , $f_Z(z) = 2 e^{-2z}, z\geq 0$ , and:

$
  h(t) = \begin{cases}
          1,  & 0 \leq t\leq 1 \\
          0, & \text{otherwise}
       \end{cases}
    \
$

I was asked to find the CDF of $X(1/2)$. What I did is to write:

$ F_{X(1/2)} (k)= P({X(1/2)}\leq k) = P(0 \leq z \leq 1/2)P(V \leq k) + P(Z > 1/2)P(k \geq 0)$

Solving this I get:

$
  F_{X(1/2)} (k) = \begin{cases}
          1,  & k \geq 2 \\
          (1-e^{-1})\frac{k}{2} + e^{-1}, & 0 \leq k \leq 2 \\
          0 & k < 0
       \end{cases}
    \
$

I'm not sure about this result because this function is discontinuous at $k=0$. If I wanted to get, for example, the PDF by derivation I would get: $f_{X(1/2)}(k)=(1-e^{-1})/2$ which has an area $1-e^{-1} \neq 1$.

Is something wrong with my reasoning ?

答:
Yes, it is possible to have a discontinuous CDF.  In fact, every discrete distribution has a discontinuous CDF; however, your question relates to a mixed distribution with a combination of density and discrete probability mass.  Consider the example

$$W \mid X = \min(X, 2), \\ X \sim \operatorname{Exponential}(\lambda = 1).$$  Then $W$ is what we call a right-censored exponential random variable:  $W = X$ if $X < 2$, otherwise $W = 2$.  Hence $$\Pr[W \le w] = \begin{cases} 0, & w < 0 \\ 1 - e^{-w}, & 0 \le w < 2 \\ 1, & w \ge 2. \end{cases}$$  There is a nonzero density for $W$ in the interval $(0,2)$, and then a discrete probability mass at $W = 2$ with value $\Pr[W = 2] = \Pr[X > 2] = e^{-2}$.

That said, we need to take a closer look at your computations.  I will use the indicator function $\mathbb 1$ in place of your $h(t)$.

I get $$\begin{align}
X(1/2) &= V \mathbb 1 (0 \le 1/2 - Z \le 1) \\
&= \begin{cases}V, & |Z| \le 1/2 \\ 0, & |Z| > 1/2 \end{cases} \\
&= \begin{cases} V, & 0 \le Z \le 1/2 \\ 0, & Z > 1/2. \end{cases}
\end{align}$$
This piecewise definition is the origin of the discontinuous nature of the CDF.  Proceeding with our computations,
$$\begin{align}
F_{X(1/2)}(x) &= \Pr[X(1/2) \le x] \\
&= \Pr[V \le x \mid 0 \le Z \le 1/2] \Pr[0 \le Z \le 1/2] + \Pr[0 \le x \mid Z > 1/2] \Pr[Z > 1/2] \\
&= \left( \frac{x}{2} \mathbb 1 (0 \le x \le 2) + \mathbb 1 (x > 2) \right) (1 - e^{-1}) + \mathbb 1(x \ge 0) e^{-1} \\
&= \begin{cases} 0, & x < 0 \\ \frac{x}{2} (1 - e^{-1}) + e^{-1} , & 0 \le x \le 2 \\ 1, & x > 2.\end{cases}
\end{align}$$
This is equivalent to your solution.  While it is true that the density equals $$f_{X(1/2)}(x) = \frac{1 - e^{-1}}{2}$$ for $0 < x < 2$, there is also a discrete probability mass at $x = 0$:  $$\Pr[X = 0] = e^{-1},$$ which we can see by going back to the original definitions:  $$\Pr[X(1/2) = 0] = \Pr[|Z| > 1/2] = \Pr[Z > 1/2] = e^{-2/2} = e^{-1}.$$

Title: Can a probability CDF be discontinuous?


问:
On the inequality $\sum_{i=1}^n|a_i-b_i|\le\big\lfloor \frac{n^2}{2}\big\rfloor$
Given $i,a_i,b_i\in\{1...n\},\space a_i\neq a_j,b_i\neq b_j,\forall i\neq j$ prove that
$$\sum_{i=1}^n|a_i-b_i|\le\big\lfloor \frac{n^2}{2}\big\rfloor$$
This problem was proposed by the new contributor  @user3458994 and it was closed by five users. I find it somewhat challenging (it does not have an immediate answer), but it is sufficiently well posed and, indeed, it can be solved by answering correctly.

There are many possible sums $\sum_{i=1}^n|a_i-b_i|$; in fact there are $n!$ possibilities (number of permutations of the set $\{1,2,\cdots,n\}$). The minimum for these sums is $0$ corresponding to the identity permutation $a_i\rightarrow b_i=a_i$ in which case the inequality is trivially verified. We expose one of these sums having a maximum value $M$ exactly equal to $\big\lfloor \frac{n^2}{2}\big\rfloor$. I believed no other sum has a value greater than $M$ in which case the problem would be false (am I wrong?).

答:
I have already given an answer in that post. I will post it again here. It's somewhat similar to the rearrangement inequality: When $\{a_i\}$ and $\{b_i=i\}$ have opposite order, the sum of the absolute difference reaches maximum (there could be other cases which also reach this maximum). The rest is just easy calculation.

Lemma: If $x>y,z>w$ then $|x-w|+|y-z|\geqslant |x-z|+|y-w|.$

WLOG we can assume $y\geqslant w$. Then $x>w$.

$$|x-w|+|y-z|\geqslant |x-z|+|y-w| \iff x-w+|y-z| \geqslant |x-z|+y-w \\
\iff |x-y|+|y-z|\geqslant |x-z|$$

which follows from the triangle inequality.

WLOG assume $b_i=i$. Then from the lemma the sum of absolute differences obtains its maximum value when $a_i$ is decreasing, i.e. $$\sum_{i=1}^n|a_i-i| \leqslant \sum_{i=1}^n |n+1-2i|.$$

If $n=2m$, $$\sum_{i=1}^n |n+1-2i|=2(2m-1) + 2(2m-3)+\cdots + 2(1)=2m^2 = \lfloor \frac{n^2}{2} \rfloor.$$

If $n=2m+1$, $$\sum_{i=1}^n |n+1-2i|=2(2m) + 2(2m-2)+\cdots + 2(0)=2m(m+1) = \lfloor \frac{n^2}{2} \rfloor.\blacksquare$$


问:
Normal operator implies $||Tv|| = ||T*v||$
I am trying to understand the proof in Sheldon Axler's textbook that $T \in \mathcal{L}(V)$ is a normal operator if and only if $||Tv|| = ||T^* v||$ for all $v \in V$  I have included his proof below, but I then have questions on it.

(Axler's proof) Let $T \in \mathcal{L}(V)$. We will prove both directions of this result at the same time. Note that
\begin{align*}
\text{$T$ is normal } & \iff T^* T - TT^* = 0 \\
& \iff \langle (T^* T - TT^*)v, v  \rangle = 0 \text{ for all $v \in V$} \\
& \iff \langle T^* T v, v \rangle = \langle TT^* v, v \rangle \text{ for all $v \in V$} \\
& \iff ||Tv||^2 = |T^* v|^2 \text{ for all $v \in V$}
\end{align*}
where we used 7.16 to establish the second equivalence (note that the operator $T^* T - TT^*$ is self-adjoint). The equivalence of the first and last conditions above give the desired result.

The first line follows from the definition of normal: $T^{*} T = TT^{*}$, so rearranging gives $T^{*} T = TT^{\star} = 0$. For the next result, Axler cites a theorem which states:

Suppose $T$ is a self-adjoint operator on $V$ such that $\langle Tv, v \rangle = 0$ for all $v \in V$. Then $T = 0$.

There are two things I do not understand here. First, I do not understand why $T^{*} T - TT^{*}$ is self-adjoint and cannot figure out how to prove it. Second, it seems that Axler is using the result in reverse. 7.16 establishes one direction, that if an operator is self-adjoint and $\langle Tv, v \rangle = 0$, then $T = 0$. But he asserts that the result holds in both directions, beginning with the statement that the operator is $0$. Is the converse also true? (If so, does anyone have any tips on how to prove it?)

The next step follows from linearity in the first entry and rearranging. The final step, I believe, uses the definition of adjoint and the fact that $(T^*)^* = T$.

Help on understanding this proof would be appreciated.

答:
$(AB)^{*}=B^{*}A^{*}$ for any two bounded operators $A$ and $B$. Also $T^{**}=T$ Hence $(T^{*}T-TT^{*})^{*}=T^{*}T^{**}-T^{**}T^{*}=T^{*}T-TT^{*}$ provin that $T^{*}T-TT^{*}$ is self-adjoint.

If it is already  proved that $ \langle Tv, v \rangle=0$ for all $v$ and $T$ self-adjoint implies $T=0$ then  what is your second question? Isn't it obvious $T=0$ implies $ \langle Tv, v \rangle=0$ for all $v$?


答:
The forward direction is immediate, since $\langle 0v,v\rangle=0$ for all $v$.

The sum of self-adjoint operators is self-adjoint, and each of $TT^*$ and $T^*T$ is self-adjoint.

The last step uses the fundamental property $\langle Tv,w\rangle=\langle v,T^*w\rangle$.

Title: Normal operator implies $||Tv|| = ||T*v||$

Body:
I am trying to understand the proof in Sheldon Axler's textbook that $T \in \mathcal{L}(V)$ is a normal operator if and only if $||Tv|| = ||T^* v||$ for all $v \in V$  I have included his proof below, but I then have questions on it.

(Axler's proof) Let $T \in \mathcal{L}(V)$. We will prove both directions of this result at the same time. Note that
\begin{align*}
\text{$T$ is normal } & \iff T^* T - TT^* = 0 \\
& \iff \langle (T^* T - TT^*)v, v  \rangle = 0 \text{ for all $v \in V$} \\
& \iff \langle T^* T v, v \rangle = \langle TT^* v, v \rangle \text{ for all $v \in V$} \\
& \iff ||Tv||^2 = |T^* v|^2 \text{ for all $v \in V$}
\end{align*}
where we used 7.16 to establish the second equivalence (note that the operator $T^* T - TT^*$ is self-adjoint). The equivalence of the first and last conditions above give the desired result.

The first line follows from the definition of normal: $T^{*} T = TT^{*}$, so rearranging gives $T^{*} T = TT^{\star} = 0$. For the next result, Axler cites a theorem which states:

Suppose $T$ is a self-adjoint operator on $V$ such that $\langle Tv, v \rangle = 0$ for all $v \in V$. Then $T = 0$.

There are two things I do not understand here. First, I do not understand why $T^{*} T - TT^{*}$ is self-adjoint and cannot figure out how to prove it. Second, it seems that Axler is using the result in reverse. 7.16 establishes one direction, that if an operator is self-adjoint and $\langle Tv, v \rangle = 0$, then $T = 0$. But he asserts that the result holds in both directions, beginning with the statement that the operator is $0$. Is the converse also true? (If so, does anyone have any tips on how to prove it?)

The next step follows from linearity in the first entry and rearranging. The final step, I believe, uses the definition of adjoint and the fact that $(T^*)^* = T$.

Help on understanding this proof would be appreciated.


答:
(I suppose that with $\delta$ you mean the Dirac function.)
Just evaluate $(f(t)\delta(t))'$ in two ways.

First, use Leibniz' rule:
$$
(f(t)\delta(t))'=f(t)\delta'(t)+f'(t)\delta(t)=f(t)\delta'(t)+f'(0)\delta(t).
$$
We also have
$$
(f(t)\delta(t))'=(f(0)\delta(t))'=f(0)\delta'(t).
$$

Since the two expressions must be equal, we have
$$
f(t)\delta'(t)=f(0)\delta'(t)-f'(0)\delta(t).
$$


问:
Realization of subgroups of $\pi_1(M)$ by submanifolds
Let $M$ be a smooth oriented connected 3-manifold, let $G$ be the fundamental group $\pi_1(M)$. If $M$ is compact, by Kneser-Milnor Decomposition Theorem $M$ can be written as a connected sum of prime closed oriented 3-manifolds $$M=M_1 \sharp \ldots \sharp M_k,$$
then we have that
$$\pi_1(M)=\pi_1(M_1) * \ldots * \pi_1(M_k),$$ and each factor of the free product can be realized by a submanifold, i.e. there exists a compact embedded 3-submanifold with boundary whose fundamental group is $\pi_1(M_i)$ (namely, $M_i$ with an open disc removed). Is there a "similar" construction for a generic subgroups of $\pi_1(M)$?

More specifically, let us suppose that $M$ is non-compact and let $H$ be a finitely generated subgroup of $G$. Under which assumptions (if any) there exists a (compact with boundary) embedded 3-submanifold $N$ of $M$ with $\pi_1(N)=H$ and such that the embedding induces an injection between the fundamental groups $\pi_1(N) \hookrightarrow \pi_1(M)$? Is this problem well studied in the literature?

答:
Sometimes one can. For instance, if $H$ is an abelian subgroup: If the subgroup is cyclic, you can, if the subgroup is $Z^2$ there is a precise condition when you can (it is called the "torus theorem"). One can sometimes find reasonable conditions for this when $H$ is the fundamental group of higher genus surface.
This works better if the manifold $M$ is closed. Then whenever you have, say, a decomposition
$$
\pi_1(M)=A\star_{H} B,
$$
and $H$ is isomorphic to the fundamental group of a closed connected surface $S$, then $A, B, H$ are realized by embedded compact submanifolds.

But the best statement in general is the "virtual one":

Suppose that $H<  \pi_1(M)=G$ is a finitely generated group, $G$ is finitely generated. There is a finite-sheeted covering $p: M'\to M$ such that $H< p_*(\pi_1(M'))\le G$ and $H$ is realized as the fundamental group of a compact embedded submanifold in $M'$.

This is not always possible, but it happens surprisingly often, for instance, when $\pi_1(M)$ contains no $Z^2$-subgroups. This is related to an algebraic property called LERF.

See discussion and references in

"3-manifold groups" by Matthias Aschenbrenner, Stefan Friedl, Henry Wilton.

(a more updated version appeared as a book).


问:
Brick wall with maximum height 3
Given n same-sized rectangular bricks. We want to build a wall with these constraints:

All bricks should be horizontal.

We can put a brick on two other bricks, such that the middle of the top brick is on the border of two other bricks.

The maximum height of the wall should be 3.

Bricks on the first row (the ground) should stick together.

For example, if we have 4 bricks, we can build the wall in 3 ways. If we have 6 bricks, we can make the wall in 9 ways.

Some examples of this recursive function are:

$f(1) = {0 \choose 0} = 1$

$f(2) = {1 \choose 1} = 1$

$f(3) = {2 \choose 0} + {1 \choose 1} = 1 + 1 = 2$

$f(4) = {3 \choose 0} + {2 \choose 1} = 1 + 2 = 3$

$f(5) = {4 \choose 0} + {3 \choose 1} + {2 \choose 2} = 1 + 3 + 1 = 5$

$f(6) = {5 \choose 0} + {4 \choose 1} + {3 \choose 2} + {2 \choose 2}{1 \choose 1} = 1 + 4 + 3 + 1 = 9$

This is the illustration of how we can build the wall using 6 bricks:

I tried to find a recurrence relation for this problem, but I didn't succeed.

答:
This is only a partial answer, deriving a sixth-order linear recurrence.

A wall with $k$ bricks in the bottom row can be decomposed into $k$ diagonals with slope $-1$. The first diagonal contains only the leftmost brick in the bottom row. After that each diagonal can be at most one brick longer than the previous one, up to a maximum length of $3$ bricks. Thus, we are in effect counting the sequences $\langle a_1,\ldots,a_k\rangle$ such that

$a_i\in\{1,2,3\}$ for $i=1,\ldots,k$,

$a_1=1$,

$a_{i+1}-a_i\le 1$ for $i=1,\ldots,k-1$, and

$a_1+\ldots+a_k=n$.

Let $b_n$ be the number of such sequences, and for $i=1,2,3$ let $b_{n,i}$ be the number of such sequences ending in $i$. Then

$$\begin{align*}
b_{n,3}&=b_{n-3,2}+b_{n-3,3}=b_{n-3}-b_{n-3,1}\\
b_{n,2}&=b_{n-2}\\
b_{n,1}&=b_{n-1}\,,
\end{align*}$$

so

$$\begin{align*}
b_{n}&=b_{n,1}+b_{n,2}+b_{n,3}\\
&=b_{n-1}+b_{n-2}+b_{n-3}-b_{n-3,1}\\
&=b_{n-1}+b_{n-2}+b_{n-3}-b_{n-4}\,,
\end{align*}$$

with initial values $b_0=1$, $b_1=1$, $b_2=1$, $b_3=2$, $b_4=3$, and $b_5=5$. The next few values are easily computed:

$$\begin{array}{rcc}
n:&0&1&2&3&4&5&6&7&8&9&10\\
b_n:&1&1&1&2&3&5&9&15&26&45&77
\end{array}$$

By the way, if the height were limited to $2$, we would simply be getting the Fibonacci numbers.

Title: Brick wall with maximum height 3


问:
Evaluating $\int\limits_0^\infty \frac{e^x}{1+e^{2x}}\mathrm dx$, alternate methods
Problem:

Evaluate $$\displaystyle\int\limits_0^\infty \frac{e^x}{1+e^{2x}}\mathrm dx$$

My progress:

I have actually solved the problem, but I fear that I may not have used the "desired" methods.

Using substitution $u = e^x$, we can rewrite the integral and get $$ \arctan(e^x)\bigg|_0^\infty $$

Then we need to evaluate the limit $$\displaystyle\left(\lim\limits_{x\to\infty}\arctan(e^x)\right) - \arctan(e^0)$$

Now, this is where I think my solution differs from the intended method (which I don't know what is).

When I evaluate the above limit, I realize that $\tan(x)$ has a vertical asymptote for $x = \frac\pi2$, which means that $\arctan(x)$ has a horizontal asymptote for $y = \frac\pi2$, and is continuously increasing.

Then, since $x\to\infty \Rightarrow e^x\to\infty$, we can conclude that $$\displaystyle\lim\limits_{x\to\infty}\arctan(e^x) = \frac\pi2$$ which gives the final answer $$\displaystyle\int\limits_0^\infty \frac{e^x}{1+e^{2x}}\mathrm dx = \left(\lim\limits_{x\to\infty}\arctan(e^x)\right) - \arctan(e^0) = \frac\pi2 - \frac\pi4 = \underline{\underline{\frac\pi4}}$$

My question:

Is there a more "calculating" way of evaluating the aforementioned limit? I.e. using some general rules of limits rather than intuition? Bear in mind, I like my own solution. It was satisfying. I just feel like there is something else I should know, which I'm missing out on.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\expo{x} \over
1 + \expo{2x}}\,\dd x} =
\int_{0}^{\infty}{\expo{-x} \over 1 + \expo{-2x}}
\,\dd x =
\int_{0}^{\infty}{\expo{-x} - \expo{-3x}\over 1 - \expo{-4x}}
\,\dd x
\\[5mm] \stackrel{4x\ \mapsto\ x}{=}\,\,\,&
{1 \over 4}\int_{0}^{\infty}{\expo{-x/4} - \expo{-3x/4} \over
1 - \expo{-x}}\,\dd x
\\[5mm] = &\
{1 \over 4}\pars{%
\int_{0}^{\infty}{\expo{-x} - \expo{-3x/4} \over
1 - \expo{-x}}\,\dd x -
\int_{0}^{\infty}{\expo{-x} - \expo{-x/4} \over
1 - \expo{-x}}\,\dd x}
\\[5mm] = &\
{1 \over 4}\bracks{\Psi\pars{3 \over 4} -
\Psi\pars{1 \over 4}}
\end{align}
See A & S Table $\ds{\color{black}{\bf 6.3.22}}$ identity.

With the Euler Reflection Formula:
\begin{align}
& \mbox{} \\
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\expo{x} \over
1 + \expo{2x}}\,\dd x} =
{1 \over 4}\bracks{\pi\cot\pars{\pi \over 4}} =
\bbx{\pi \over 4} \\ &
\end{align}