问:
Rod cutting: requesting a proof for $2^{n-1}$ possible ways
The rod cutting problem states that: Given a rod of length $n$ inches and a table of prices $P_i$ for $i=1,\dots,n$, determine the maximum revenue $R_n$.

In CLRS chapter 15 Rod cutting page 360:

We can cut up a rod of length $n$ in $2^{n-1}$ different ways

I can intuitively understand why this is so; basically we can construct a binary tree of height $\log N$. But can someone provide me a proof of why this is so?

Body:
For Example:- n = 4
 --- --- --- ---
|   |   |   |   |  --> Rod of length 4 inches
 --- --- --- ---       ^ is showing where we can make cuts
    ^   ^   ^

Now, Above given ^ position either we make a cut or not i.e; 0 for not making a cut and 1 for making a cut. We get $ 2^3 $ binary combinations are as follows:

           ---     --- --- ---
1 0 0 --> |   |   |   |   |   |
           ---     --- --- ---
           --- ---     --- ---
0 1 0 --> |   |   |   |   |   |
           --- ---     --- ---
           --- --- ---     ---
1 0 0 --> |   |   |   |   |   |
           --- --- ---     ---
           ---     ---     --- ---
1 1 0 --> |   |   |   |   |   |   |
           ---     ---     --- ---
           ---     --- ---     ---
1 0 1 --> |   |   |   |   |   |   |
           ---     --- ---     ---
           --- ---     ---     ---
0 1 1 --> |   |   |   |   |   |   |
           --- ---     ---     ---
           ---     ---     ---     ---
1 1 1 --> |   |   |   |   |   |   |   |
           ---     ---     ---     ---
           --- --- --- ---
1 1 1 --> |   |   |   |   |
           --- --- --- ---

Hence, A Rod of length "n" inches can be cut into $ 2^{n-1} $ pieces.


问:
$\operatorname{max}(a(x),b(x))=0 \Rightarrow a(x) \leq 0, b(x) \leq 0, a(x)b(x)=0$?
Could you explain to me why we have this equivalence?

This condition is used to transform the obstacle problem $$\left\{\begin{matrix}
\operatorname{max}(Lu,\varphi-u)=0\\u(0,\cdot)=\varphi(0,\cdot)
\end{matrix}\right.$$ Text says that since this problem doesn't admit solutions in $C^2$ we have to use a variational approach to expand the solution space not only to the functions that satisfy the equation $Lu=0$ but also to the functions $f$ that satisfy the inequality $Lu \leq 0$.

答:
Just note that if $\max\{L_r u, \varphi -u\} =0$, both quantities must be $\leq 0$. Also, if the maximum is zero, for each $x$ at least one of them is zero, so $L_r(u) \cdot (\varphi -u) =0$. So, the original problem can be reformulated as
$$
\begin{cases}
L_r u \leq 0, & \textrm{in } S_T\\
\varphi -u \leq 0, & \textrm{in } S_T\\
L_r u (\varphi -u) = 0,  & \textrm{in } S_T
\end{cases}
$$

or, equivalently,
$$
\begin{cases}
L_r u \leq 0, & \textrm{in } S_T\\
u \ge \varphi, & \textrm{in } S_T\\
(\varphi -u)L_r u  = 0,  & \textrm{in } S_T
\end{cases}
$$

ContentLicense: CC BY-SA 4.0


问:
Find the inverse linear transformation of a matrix
Question: Let $T: R^2$ -> $R^2$  be an invertible linear transformation and suppose that $T[1, 2] = [1, 0] $ and $T[1, -2] = [0, 1]$. Find $T^{-1}[1, 2]$.

Note: [x, y] in the question represents a column matrix with x on the top and y on the bottom. I'm not sure how to format matrices here.

At first, I treated this problem as a matrix inversion problem and got [1, 2]. But I soon realized that this is incorrect. Could someone please explain to me how this works? I couldn't seem to locate any examples online.

Thanks in advance!

答:
Hint: Since $T$ is invertible, we have
$$T^{-1} \left[ \begin{array}{c} 1 \\ 0 \end{array} \right] = \left[ \begin{array}{c} 1 \\ 2 \end{array} \right] \text{  and  } T^{-1}\left[ \begin{array}{c} 0 \\ 1 \end{array} \right]= \left[ \begin{array}{c} 1 \\ -2 \end{array} \right].$$
Can you take it from here?


问:
For which values of $a$ is the system asymptotic stable?

Consider the system of differential equations
$$x'=\bigg[ \begin{matrix} a & 2 \\ a & 1
\end{matrix} \bigg] \cdot \bigg[ \begin{matrix} x_1 \\ x_2 \end{matrix} \bigg]$$
For what values of $a$ is the system asymptotic stable?

Investigate if the system is stable for $a=-1$ and for $a \geq 0$

My idea was to find the characteristic polynomial.
$$P(\lambda)=\det(A-\lambda E) =(a-\lambda)(1-\lambda)-2a=\lambda^2-\lambda(1+a)-a$$

If the system has to be asymptotic stable, the solution must satisfy $x(t) \rightarrow0$ for $t \rightarrow 0$

We try to solve the characteristic equation:
$$ \lambda^2-\lambda(1+a)-a =0 $$
and get
$$\lambda=\frac{1}{2} +\frac{a}{2} + \frac{\sqrt{a^2+6a+1}}{2} \: \: \: \text{or} \: \: \: \lambda=\frac{1}{2} +\frac{a}{2} - \frac{\sqrt{a^2+6a+1}}{2}$$

But I'm not sure where to go from here. Can someone help me out?

答:
The solution is asymptotically stable iff the real part of all the eigenvalues are negative (because that part leads to an exponential expression $e^{\mathrm{Re}(\lambda) t}$).

For two variables, this is equivalent to the trace of the matrix being negative and the determinant positive. You can see this by taking the different cases: $$\mathrm{Tr}(A)=\lambda+\mu\quad(=r+iv+r-iv=2r),\qquad \det(A)=\lambda\mu\quad(=r^2+v^2)$$

Thus, stable solutions occur when $$a+1<0,\qquad a-2a>0$$ $$\therefore\quad a<-1$$

Note: Your solution has an error in the characteristic polynomial (sign of last term).


问:
All subspaces $W$ of $V$ are $T$-invariant, is $T = cI:V\to V$?

$V$ is a vector space over $\mathbb{F}$. Any subspace $W$ of $V$ is $T$-invariant, is $T = cI:V\to V$ for some $c \in \mathbb{F}$?

If $T=cI:V\to V$ for some $c \in \mathbb{F}$, then any subspace $W$ of $V$ is $T$-invariant (obviously). Is the converse true, i.e. if any subspace $W$ of $V$ is $T$-invariant, is $T = cI:V\to V$ for some $c \in \mathbb{F}$?

I'm unable to prove this or find a counterexample. Any hints?

答:
Yes, that is true.

Take a vector $v\ne 0$. Since all 1-dimensional subspaces are invariant, then
$$Tv = c_vv$$
If now $0\ne w\ne v$, for the same reason
$$Tw = c_ww$$

If $w,v$ are dependent, then $w=\lambda v$ and $$Tw = \lambda Tv = \lambda c_vv = c_vw$$
so $c_w=c_v$ necessarily. If $w,v$ are independent,
$$T(w+v) = c_{w+v}(w+v) = c_ww + c_vv \implies (c_w - c_{w+v})w + (c_v - c_{w+v})v=0$$
but the coefficient must be zero, so $c_w = c_{w+v} = c_v$.

This proves that $c_w=c_v$ for any couple of nonzero vectors $v,w$, so there's a unique constant $c$ such that $T = cI$


问:
Standard deviation versus variance
Could someobdy explain me briefly why standard deviation is so widely used in statistics? I just wonder, what is standard deviation needed for when we have variance? Wouldn't it give the same result if we just interpreted the variance instead of standard deviation? So basically, what makes standard deviation better than variance? Thanks!</p>

答:
This is a good question.

For me, the reason to prefer standard deviation is because it shares the same dimensions as the quantity of interest.

For example, if a random variable $X$ is measured in meters (m), then the variance will be measured in $m^2$, while the standard deviation will also be measured in meters.

Since most often I can approximate the distribution of $X$ by a normal distribution, this means I can say things like 'X is likely to be within $\pm2$ standard deviations of the mean.'

For the variance this does not make sense, since I cannot just add together meters and meters squared.

Title: Standard deviation versus variance


问:
$\mathbb P[a<S\leq b]$ for $S=\sum_{i=1}^{N} X_i$ where $N \sim \mathrm{Poi}$ and $X_i$ iid $\sim \mathrm{Exp}$
Let

$S=\sum_{i=1}^{N} X_i$ where

$N \sim \mathrm{Poisson}(\mu)$

$X_i$ iid $\sim \mathrm{Exponential}(\lambda)$ and independent of $N$

I want to determine $\mathbb P[S=0]$ and $\mathbb P[a<S\leq b]$ for $0\leq a<b$.

I don't know how or where I should or could use the following hint, but the author states to use the series representation of the Bessel function $I_\nu$, i.e.

$$I_\nu(z)=(z/2)^\nu\sum_{k=0}^{\infty}\frac{(z/2)^{2k}}{k!\Gamma(v+k+1)}$$

I would appreciate any help to solve this problem.

答:
Since $\textbf{P}[S\in(a,b]]$ uniquely determines the distribution of $S$, you might as well calculate the distribution of $S$.

When dealing with these kinds of sums, some knowledge about moment generating functions (MGFs) can be very usefull:

If it exists (in your case it does), the MGF of $S$ is determined by $M_S(t):=\textbf{E}[e^{tS}]$ where $M_S$ has to be well defined on some interval around zero. If this is the case, the distribution of $S$ is uniquely determined by its MGF. Since the MGFs of your to random variables are known (google), you can use them to easily calculate: \begin{align*}M_S(t)=\textbf{E}[e^{tS}]&=\textbf{E}[e^{t\sum^N_{i=1}X_i}]=\textbf{E}[\textbf{E}[e^{t\sum^N_{i=1}X_i}|N]]\\&=\textbf{E}[\prod^N_{i=1}\textbf{E}[e^{tX_i}]]=\textbf{E}[M_{X_1}(t)^N]=M_N(\ln(M_{X_1}(t))).\end{align*}

You can now either guess the distribution of $S$ and show that its MGF is indeed the right one or determine the distribution of $S$ using its MGF, allthough thats a bit more complicated. You probably need the hint for that.


问:
Connections between number theory and abstract algebra.
I haven't taken abstract algebra yet, but I am curious about connections between number theory and abstract algebra.  Do the proofs of things like Fermat's little theorem, the law of quadratic reciprocity, etc. rely upon techniques found in abstract algebra?

Thanks.

答: See for instance Elementary Number Theory: An Algebraic Approach by Bolker.</p>

ContentLicense: CC BY-SA 3.0


问:
Finding the range of $f(x)=2\csc(2x)+\sec x+\csc x$
Hi this is the question:

Find the range of $$f(x)=2\csc(2x)+\sec x+\csc x$$

What I've tried:

I know that the range of $\csc(x)$ which is $R\setminus (-1,1) $, the range of $\sec(x)$ is $R\setminus (-1,1)$ too. And I've managed to simplify the expression to have in terms of $\sec x$ and $\csc x$ as such:
$$f(x)=\sec(x)\cdot \csc(x)+\sec(x)+\csc(x).$$
But then, what do I do when I have 3 terms and not just $\sec$ or $\csc$?

答:
Starting with $$f(x)=\frac{2}{\sin x+ \cos x-1} \implies y=\frac{1+t^2}{t-t^2} \implies (1+y)t^2-yt-1=0,$$
As $t=\tan(x/2)$ lies in $(-\infty, \infty)$, the range will be all values of $y$ when the above quadratic has real roots: $B^2 \ge 4AC$
$$\implies y^2-4y-4 \ge 0 \implies y\ge 2+\sqrt{2} ~or~ y\le2-\sqrt{2}$$


问:
How many parameters does matrix [0, 1, 2] have?
I know it has rank 1 since there is 1 leading entry, but what about the number of parameter?
I want to say 2, but does the number 0 got me confused</p>

答: That matrix doesn't have any parameters.  In case you're asking for the dimension of its kernel, that's certainly $2$.  Don't confuse it with $$\begin{pmatrix}0\\1\\2\end{pmatrix}.$$  The dimension of the kernel of this matrix is $0$.


问:
Find $\lim\limits_{n \to \infty} \frac{\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} + \dots + \frac{1}{\sqrt{n}}}{\ln (n)}$
$\lim\limits_{n \to \infty} \frac{\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} + \dots + \frac{1}{\sqrt{n}}}{\ln (n)}$

Can we apply  Stolz-Cesaro?

$\lim\limits_{n \to \infty}\frac {\frac{1}{\sqrt{n+1}} - \frac{1}{\sqrt{n}}} {{\ln(n+1)-\ln(n)}}$ =
$\lim\limits_{n \to \infty}\frac{\sqrt n - \sqrt {n+1}}{\sqrt n \sqrt{n+1}\ln(1+\frac{1}{n})}$ = $\lim\limits_{n \to \infty}\frac{1-\sqrt{1+\frac{1}{n}}}{\sqrt{n+1}\ln(1+\frac{1}{n})}$

What can I do from here?

Body:
Option:

$n/√n < 1/√1+1/√2....+1/√n;$

$\dfrac{√n}{2 \log (√n)} < \dfrac{1/√1+1/√2+...1/√n}{\log n};$

$(1/2) \lim_{n \rightarrow \infty} \dfrac{√n}{\log √n} = $?


问:
Divided Differences Proof - Show the following formula is True
I am struggling with this problem on my numerical analysis homework. We're talking about polynomial interpolation and we just learned about divided differences like two days ago.

"Show that the following explicit formula is valid for divided differences:"

$$f\left[x_{0}, x_{1}, \ldots, x_{n}\right]=\sum_{i=0}^{n} f\left(x_{i}\right) \prod_{j \neq i, j=0}^{n}\left(x_{i}-x_{j}\right)^{-1}$$

HINT: If two polynomials are equal, the coefficients of $x^n$ in each are equal.

I get divided differences, and how they are defined recursively. I am also very familiar with both the Lagrange and Newton forms of the unique interpolation polynomial. But I just can't figure out where to begin here. The hint doesn't help me either. Any help is greatly appreciated.

答:
The interpolating polynomial can be given by both these formulas:
$$
p_n(x)= \sum_{i=0}^n f(x_i) \prod_{j\ne i}\frac{x-x_j}{x_i-x_j}
$$

$$
p_n(x)=\sum_{i=0}^n f[x_0,\cdots, x_i] \prod_{j=0}^{i-1}(x-x_i)$$

The coefficients of $x^n$ in the second expression is $f[x_0, \cdots , x_n]$, and the same coefficient in the first expression is $\sum_{i=0}^n f(x_i) \prod_{j\ne i} \frac{1}{x_i-x_j}$. The uniqueness of the interpolating polynomial forces these coefficients to be the same.


问:
How to determine the least distance covered by a light crossing a crystal?
The problem is as follows:

The figure shown in the lower part shows a sucrose crystal that is
about to be studied for its optical properties. The crystal is a
pyramid whose base is an equilateral triangle $\triangle\,BCD$ and its
faces are right isosceles triangles with a common vertex at $A$.
Knowing this, a bright red flashing light is passed from vertex $B$
towards a point $P$ on edge $CD$ and from there it goes to a point $Q$
of the edge $AC$ to return to point $B$. If the length of the route is
needed to be minimum. Determine the angle $\angle\,PQA$.

$\begin{array}{ll}
1.&135^{\circ}\\
2.&105^{\circ}\\
3.&120^{\circ}\\
4.&150^{\circ}\\
\end{array}$

I'm very confused exactly on how to achieve the required condition of the minimum path possible for the light to cross the directions indicated in the crystal.

In other words, how to ensure that the path taken by the light is the least possible?. Can this problem be solved relying only using Euclidean postulates?

So far the only thing which I can spot is that:

$\angle ABD= \angle ADB =\angle ADC = \angle ACD =45^{\circ}$

$\angle CBD = \angle BDC = \angle BCD =60^{\circ}$

To me the biggest source of confusion is how to get that angle which is being asked, $\angle PQA?$, I think that since it will be required that is to be maximum then it will be wide enough to reduce the distance covered by the light pulses.

But I don't know what else can be found from these relations. Please include a drawing or diagram in the answer. As because this figure is in 3-D it is isn't easy to spot how to find the least distance. Can someone help me with this?.

答:
Draw the net of the sucrose crystal.

( When the net is folded into the sucrose crystal, $A = A' = A''$, $B = B'$ and $Q = Q'$. )

The minimum length of the route = $BB'$

$\angle PQA = \angle PQ'A' = \frac{180° - (60° + 45° + 45°)}{2} + 60° + 45° = 120°$


问:
Is this sequence of functions pointwise bounded but not uniformly bounded?
Consider the sequence of functions defined by:

$$f_n(x) := n\chi_{(0,\frac{1}{n}]}.$$

My question is, is this sequence of functions pointwise bounded but not uniformly bounded? For context, I have just come across these definitions in my RA class, however, no example was given to separate the two concepts.

答:
Pointwise bounded means that for each point $x$, $f_n(x)$ is a bounded sequence. Indeed, for any $x$, there is some $n$, starting from which $x$ is not in the given interval, and so from that point onwards we have that $f_n(x)=0$, and so the sequence is bounded (as it only attains a finite number of values).

Uniformly bounded means that there is a uniform bound on all such sequences $f_n(x)$ - there is some $M>0$ such that $|f_n(x)|<M$ for all $x$ and for all $n$. To show that the sequence is not uniformly bounded, show that $f_n(x)$ can get arbitrarily large as $n$ grows, if you pick the right $x$. Can you find a sequence $(x_n)$ such that $f_n(x_n)$ gets arbitrarily large?


问:
An $S^{k-1}$-bundle with structure group $O(k)$ comes from a Riemannian vector bundle
Suppose $\pi:E\to M$ is a smooth $S^{k-1}$-bundle (a fiber bundle with fiber $S^{k-1}$), and suppose that the structure group of $E$ can be reduced to the orthogonal group $O(k)$. Then is it true that there is a vector bundle $E'\to M$ with a Riemannian metric such that the unit sphere bundle of $E'$ is $E$?

Since the transition functions of $E\to M$ have values in $O(k)\subset GL(k,\Bbb R)$, we can form a vector bundle $E'\to M$ with the same transition functions as $E$. But I don't see how to define a Riemannian metric on $E'$ to get $S(E')=E$.

答: Suppose that $E'$ is defined by the trivialization $(U_i)$ (we suppose the fibre isomorphic to the vector space $F$)and the transition functions are defined by $h(x,y)=(x,g_{ij}(x)(y)), g_{ij}(x)\in O(k)$ and $y$ is an element of $E'_x$ the fibre of $x$. Let $b$ be the Euclidean metric. We can define the metric $b_i$ on $U_i\times F$ by $b_i^x(u,v)=b(u,v)$ where $u,v\in E_x$ the fibre of $x$. This metric induces a metric on $E$.


问:
Is there a classification of numbers $k$ such that $n^k+k^n$ is not prime for all $n\geq 2$?
Say a number $k$ is cool when the number $n^k+k^n$ is not a prime for all $n \geq 2$.

Question 1: Is there a classification of cool numbers?

For example $k=2$ or $k=15$ are not cool but $k=4$ is cool. Do you know some other cool numbers?

Note that $n^4+4^n=(n^2-2^{(n+1)/2}n+2^n)(n^2+2^{(n+1)/2}n+2^n)$ for odd $n$ and for even $n$ the number $n^4+4^n$ is even and thus $k=4$ is cool.

Question 2: Call $k$ eventually cool when $n^k+k^n$ is prime only for finitely many $n \geq 2$. Is there a classification of eventually cool numbers?

Do you know eventually cool numbers that are not cool?

答:
Certainly all numbers of the form $4^{2m+1}$ are cool.  When $n = 2h+1$ is odd render

$n^{4^{2m+1}}+(4^{2m+1})^n=a^4+4b^4$

$a=n^{4^{2m}}$

$b=2^{h(2m+1) + m}$

And apply the (Sophie Germain) factorization noted in the question for the specific case $k=4$.


问:
Domain whose boundry has non zero volume.
Can There be a domain in $\mathbb{R^n}$, for any $n$ such that some domain has non zero boundry volume? I.E. volume of boundry is non zero?

Motivation:

In some theorems, it is specified that volume of boundary is non zero. But I cannot think of domains where volume of boundry is non zero.

EDIT

If domain by definition is expected to be open subset of $\mathbb{R^n}$, then I would be looking for such open subsets.

Thank You.

答:
Let $(0,1) \cap \mathbb Q = \{ q_1,q_2, \dots\}$. We define $A = (0,1) \cap\bigcup_k (q_k-4^{-k}, q_k+4^{-k})$. By construction we have $\bar A = [0,1]$ but an easy estimate shows that $\lambda( A) < 1$ where $\lambda$ is the Lebesgue measure.

If you want a connected example, you can take $\left(A \times (0,1)\right) \cup \left((0,1) \times (0, 1/2)\right) \subset \mathbb R^2$. This one is actually contractible.


问:
An interesting problem with "decomposing" natural numbers.
Consider the following one-person game:

A player starts with score $0$ and writes the
number $20$ on an empty whiteboard. At each step, she may erase any one integer (call it $a$)
and writes two positive integers (call them $b$ and $c$) such that $b + c = a$. The player then
adds $b × c$ to her score. She repeats the step several times until she ends up with all $1$s on
the whiteboard. Then the game is over, and the final score is calculated.

Example: At the first step, a player erases $20$ and writes $14$ and $6$, and gets a score of
$14 × 6 = 84$. In the next step, she erases $14$, writes $9$ and $5$, and adds $9 × 5 = 45$ to her
score. Her score is now $84 + 45 = 129$. In the next step, she may erase any of the remaining
numbers on the whiteboard: $5$, $6$ or $9$. She continues until the game is over.

Alya and Bob play the game separately. Alya manages to get the highest possible final score.
Bob, however, manages to get the lowest possible final score. What is the difference between
Alya’s and Bob’s final scores?

I tried to "decompose" into a few numbers and I get the same scores. I am not sure how to prove the conjecture that any numbers will yield the same score no matter which path is taken.

答:
Suppose your hypothesis is that starting with $n$ you end up with a score of $\frac12 n(n-1)$.  It is clearly true starting with $n=1$ as there are no moves and so a score of $0$.

Now suppose you know this is true for $1 \le n \le k$ for some $k$, then start with $k+1$ and split it into $a$ and $k+1-a$ where both are between $1$ and $k$.  You get an immediate score of $a(k+1-a)$ plus (by the hypothesis) later scores of $\frac12 a(a-1)$ and  $\frac12 (k+1-a)(k+1-a-1)$.  Add these up and simplify to $\frac12 (k+1)k$. So it is true for $n=k+1$.

Using strong induction, you can conclude the hypothesis is true for all positive integers $n$.


答:
I turn my comment into an answer. Your conjecture is correct, and can be deduced with a proof by induction.

Claim: For $n>1$ the game always ends with the score $n(n-1)/2$.

Proof: Clear for $n=2$. So assume for numbers less than $n$ and start the game at $n$ with $n=a+b$ and score $ab$. You then continue with separate games on $a$ and $b$, which themselves end with score $a(a-1)/2$ and $b(b-1)/2$. So your final score is $ab+a(a-1)/2+b(b-1)/2=n(n-1)/2$.

Title: An interesting problem with "decomposing" natural numbers.

Body:
Consider the following one-person game:

A player starts with score $0$ and writes the
number $20$ on an empty whiteboard. At each step, she may erase any one integer (call it $a$)
and writes two positive integers (call them $b$ and $c$) such that $b + c = a$. The player then
adds $b × c$ to her score. She repeats the step several times until she ends up with all $1$s on
the whiteboard. Then the game is over, and the final score is calculated.

Example: At the first step, a player erases $20$ and writes $14$ and $6$, and gets a score of
$14 × 6 = 84$. In the next step, she erases $14$, writes $9$ and $5$, and adds $9 × 5 = 45$ to her
score. Her score is now $84 + 45 = 129$. In the next step, she may erase any of the remaining
numbers on the whiteboard: $5$, $6$ or $9$. She continues until the game is over.

Alya and Bob play the game separately. Alya manages to get the highest possible final score.
Bob, however, manages to get the lowest possible final score. What is the difference between
Alya’s and Bob’s final scores?

I tried to "decompose" into a few numbers and I get the same scores. I am not sure how to prove the conjecture that any numbers will yield the same score no matter which path is taken.


问:
Closure in topology proof?
I want to prove $A$ is closed iff $\overline{A}=A$;

I need to use the definition of neighbourhoods instead open sets and not use the complement to prove this.

So wondering how can you prove it just using closure I can't see how it is possible.

Edit it asks you to use the definition of closure directly. Suppose that X is a topological space and A is a subset of X. A point $x \in X$ is a closure points of A if for all neighbourhoods N of x, $N\cap A \not= \emptyset$. The set of closure points of A is the closure of A.

答: Recall that $\bar{A}:=A\cap\{B:B\subseteq A\text{ is closed}\}$. Therefore if $A$ is closed, its closure will be $A\cap A=A$ since $A$ will be in the set $\{B:B\subseteq A\text{ is closed}\}$ and everything else will just be a subset of $A$. Conversely, assume that $\bar{A}=A$. This means that $A\cap\{B:B\subseteq A\text{ is closed}\}=A$. This can only be true when the set has $A$ and a bunch of subsets of $A$. And this only happens, by definition, when $A$ is closed.

Title: Closure in topology proof?


答:
If $GCD(n,k) = g \ne 1$, $n^k + k^n$ is never prime because

$$n^k + k^n = g\bigg({n^k + k^n \over g}\bigg)$$

Title: Is there a classification of numbers $k$ such that $n^k+k^n$ is not prime for all $n\geq 2$?

Body:
Say a number $k$ is cool when the number $n^k+k^n$ is not a prime for all $n \geq 2$.

Question 1: Is there a classification of cool numbers?

For example $k=2$ or $k=15$ are not cool but $k=4$ is cool. Do you know some other cool numbers?

Note that $n^4+4^n=(n^2-2^{(n+1)/2}n+2^n)(n^2+2^{(n+1)/2}n+2^n)$ for odd $n$ and for even $n$ the number $n^4+4^n$ is even and thus $k=4$ is cool.

Question 2: Call $k$ eventually cool when $n^k+k^n$ is prime only for finitely many $n \geq 2$. Is there a classification of eventually cool numbers?

Do you know eventually cool numbers that are not cool?


问:
Probability of Error (Noise in Channel)
In this channel a random input $X$ is chosen from $\{-1,0,1\}$ with equal probabilities and an outputs a value $Y = X + Z$.  Where $Z \sim N(0,4)$ is the noise and is independent from $X$ with $\mu=0$ and $\sigma=2$. With the noise, the channel decides the final output $\bar Y$ with the following conditions:

$\bar Y = 1$ if $Y > \frac{1}2$.

$\bar Y = -1$ if $Y < -\frac{1}2$.

$\bar Y = 0$ if $-\frac1{2}<Y < \frac{1}2$.

Find the probability of making an error.

Attempt:

What I understand, to calculate the error I need to:

$$P_{err}(X=0)= P(\bar Y=1)P(\bar Y = 1| X = 0) + P(\bar Y=-1)P(\bar Y = -1| X = 0)$$
$$P_{err}(X=1)= P(\bar Y=0)P(\bar Y = 0| X = 1) + P(\bar Y=-1)P(\bar Y = -1| X = 1)$$
$$P_{err}(X=-1)= P(\bar Y=0)P(\bar Y = 0| X = -1) + P(\bar Y=1)P(\bar Y = 1| X = -1)$$

Then get: $P_{err} = P_{err}(X=0) + P_{err}(X=-1) + P_{err}(X=1) $

And to calculate the conditional probabilities say $P(\bar Y = 1 | X = 0)$ I did:

$$
\begin{aligned} P(\bar Y = 1| X = 0) &= P(Y > \frac{1}2 | X = 0) \\
& =  P(X+Z>\frac{1}2|X=0) \\
&= P(Z>\frac{1}2 |X=0) \\
&= P(Z>\frac{1}2) P(X=0) \\
&= \frac{1}3P(Z>\frac{1}2) \end{aligned}
$$

Where $P(X=x)=P(\bar Y=y) = \frac{1}3$ since they have equal probabilities. With this I found that $P_{err}(1) = P_{err}(-1) = 0.04459$ and $P_{err}(0) = 0.08918$

The total error I got was:

$P_{err}(total) = P_{err}(-1) + P_{err}(0) + P_{err}(1) = 0.17836$

Is this right?

答:
First I do not understand the probability for $X$ to be respectively $\{-1;0;1\}$.

For the sake of simplicity let's suppose that these probability are equal, say the rv $X$ is Discrete Uniform in the given support.

Thus, the error is the following

$$ERR=\frac{1}{3}\mathbb{P}\Bigg\{Y>-\frac{1}{2}|X=-1\Bigg\}+\frac{1}{3}\mathbb{P}\Bigg\{|Y|\geq\frac{1}{2}|X=0\Bigg\}+\frac{1}{3}\mathbb{P}\Bigg\{Y<\frac{1}{2}|X=1\Bigg\}$$

The probability $\mathbb{P}[Y \in A|X=x]$ is very simple because the realization of $X$ just modify the location parameter of Z.

$$\frac{1}{3}\mathbb{P}[Y>-0.5|Y\sim N(-1;4)]+ \frac{1}{3}\mathbb{P}[|Y|>0.5|Y\sim N(0;4)]+\frac{1}{3}\mathbb{P}[Y<0.5|Y\sim N(1;4)] $$

$$\frac{1}{3}\Phi(-0.25)+\frac{2}{3}\Phi(-0.25)+\frac{1}{3}\Phi(-0.25)=\frac{4}{3}\cdot 0.4013\approx 0.5351$$

If $X$ is not uniform distributed, modify its probabilities accordingly

EDITING AND CORRECTION

If you want to write $\{0;1\}$ you have to write a backslash \ before the parenthesis (I edited your text)

You wrote that the output is

$$ \bbox[5px,border:2px solid red]
{
\overline{Y}=-1 \text{   if   } Y<\frac{1}{2}
\
}
$$

I corrected it in
$$ \bbox[5px,border:2px solid lightgreen]
{
\overline{Y}=-1 \text{   if   } Y<-\frac{1}{2}
\
}
$$

Since you wrote did not make sense in the gobal context.

Title: Probability of Error (Noise in Channel)


问:
existence of specific generators of $F_2$
Let $F_2=\langle a,b\rangle$ be the free group generated by $a$ and $b$, $W^*$ the words of $\{a,\overline{a},b,\overline{b}\}$ and $w \in W^*$ be a reduced word (I am actually looking for an "asymmetric word" so you can assume $w=a^k b$ for some large enough $k$ if this helps). The question is then:

Can we find two generators $s,t \in F_2$ satisfying the following
conditions:

$s$ and $t$ are cyclically reduced

$s$ and $t$ have as common start $w$, i.e. we have as reduced words $s=w s'$ and $t=w t'$ with $s',t' \in W^*$ and $s'$ and $t'$ are non-empty and start
with different letters

$s$ and $t$ end with different letters

This question arose in the construction of a counterexample, namely I need in the Cayley-graph of the free group generators whose translation-axes intersect in a most asymmetric way (or are far away, which is not possible for $F_2$). After several (obviously fruitless) attempts to construct such $s$ and $t$ I forward this question now to you.

答:
If by "two generators" you mean "a generating pair", so $\langle s,t\rangle=F(a, b)$, this is impossible.

The (folded) Stallings' graph corresponding to your a pair $(s, t)$ is a $\theta$-graph, with the central line corresponding to $W$ and the two other edges corresponding to $s′$ and $t′$. The assumptions in the question mean that all three edges have length $\geq1$. However, bases of $F(a, b)$ have Stallings' graph equal to the bouquet with a single vertex and two loop edges, labelled $a$ and $b$ respectively, and so $\langle s,t\rangle\neq F(a, b)$ as claimed.


问:
Combinatorial proof of $\sum^{n}_{i=1}\binom{n}{i}i=n2^{n-1}$.

Prove that
  $$\sum^{n}_{i=1}\binom{n}{i}i=n2^{n-1}$$

I can't find counting interpretations for either of the sides. A hint of "if $S$ is a subset of $\{1, . . . , n\}$ and $S^\prime$ is its complement then $|S| + |S^\prime| = n$" was also given, but I still don't know how to begin.

答:
We can interpret this combinatorially as the number of ways to form a committee (of any size) with one chairman out of a group of $n$ people.

From $n$ people we first pick a committee of size $i$, then choose one the $i$ committee members to be the chairman.  There are ${n \choose i}$ options for the members of the committee, after which there are $i$ options for the chairman.  If we sum over all $i$ from $1$ to $n$, that covers committees of all possible (nonzero) sizes.  So, we have $\sum_{i=1}^n {n \choose i}i$.

On the other hand, we could first pick one person from the $n$ people to be the chairman.  Then for each of the remaining unchosen $n-1$ people, they can be either in or out of the committee.  That's $2^{n-1}$ possibilities.  So, we have $n2^{n-1}$.


问:
How to transform $x(π−x)$ to Fourier series?
$(0 < x < \pi) : f(x) = x(\pi - x)$ and I need to prove that $$f(x) = \frac{\pi^2}{6} - \left(\frac{\cos(2x)}{1^2} + \frac{\cos(4x)}{1^2} + \frac{\cos(6x)}{1^3} ...\right)$$ using Fourier series.

I've been trying this problem for 3 days now, I'm getting a correct series but it's different from the series I am to proof. and what I tried so far is:

let $$g( x) =\begin{cases}
f( x) & 0< x< \pi \\
f( -x) & -\pi < x< 0
\end{cases}$$

now $g$ is an even function so $b_n = 0$ and $a_0 = 1/\pi \int_{-\pi}^{\pi}g(x)\mathrm{d}x$ and $ a_n = 1/\pi\int _{-\pi}^{\pi} g(x)\cos(nx)\mathrm{d}x$

$$a_0 = \frac{1}{\pi} \left(\int_{-\pi}^{0} -x(\pi+x)\mathrm{d}x + \int_{0}^{\pi}x(\pi-x)\mathrm{d}x\right) = \frac{\pi^2}{3}$$

$$a_n = \frac{1}{\pi}\left(\int_{-\pi}^{0}-x(\pi+x)\cos(nx)\mathrm{d}x +\int_{0}^{\pi}x(\pi-x)\cos(nx)\mathrm{d}x\right)$$

$$a_n\pi= -\pi\int_{-\pi}^{0}x\cos(nx)\mathrm{d}x - \int_{-\pi}^{0}x^2\cos(nx)\mathrm{d}x + \pi\int_{0}^{\pi}x\cos(nx)\mathrm{d}x - \int_{0}^{\pi}x^2\cos(nx)\mathrm{d}x$$

$x\cos(nx)$ is odd function and $x^2\cos(nx)$ is even function. so the above equation can be reduced as

$$a_n\pi= 2\pi\int_{0}^{\pi}x\cos(nx)\mathrm{d}x - 2\int_{0}^{\pi}x^2\cos(nx)\mathrm{d}x$$

$$\frac{a_n\pi}{2}= \int_{0}^{\pi}(x\pi - x^2)\cos(nx)\mathrm{d}x$$

$$\frac{a_n\pi}{2}= \bigg[(x\pi - x^2)\sin(nx)/n\bigg]_{0}^{\pi} - 1/n\int_{0}^{\pi}\sin(nx)(\pi - 2x)\mathrm{d}x$$

$$\frac{a_n\pi}{2}=0 - 1/n\left(\bigg[-(\pi-2x)\cos(nx)/n\bigg]_{0}^{\pi} - 2/n\int_{0}^{\pi}\cos(nx)\mathrm{d}x \right)$$
Since $\cos(n \pi) = (-1)^n$,

$$\frac{a_n\pi}{2}= - 1/n\left(\left(\frac{\pi}{n}((-1)^n+1)\right) - 2/n\cdot 0 \right)$$

$$a_n = \frac{-2}{n^2}  ((-1)^n+1)$$

substituting $a_0, a_n$ in the series

$$g(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}a_n\cos(nx)$$

$$g(x) = \frac{\pi^2}{6} + \sum_{n=1}^{\infty} \frac{-2}{n^2} ((-1)^n+1) \cos(nx)$$

$$g(x) = \frac{\pi^2}{6} + \frac{-2}{2^2} \cos(2x) + \frac{-2}{4^2} \cos(4x) + \frac{-2}{6^2} \cos(6x) + ...$$

I can confirm that the above answer will converge to to $f(x)$ here on Desmos

答:
Your second-to-last step is correct;  your error is in the last step.  The quantity $(-1)^n + 1$ is equal to 0 for odd $n$, but is equal to 2 for even $n$ (not 1, as you appear to have assumed).  Thus, the terms in your sum are
\begin{align*}
-\frac{4}{2^2} &\cos (2x)- \frac{4}{4^2}  \cos (4x)- \frac{4}{6^2} \cos (6x)+ \dots \\ &= - \frac{2^2}{(2\cdot 1)^2} \cos (2x)- \frac{2^2}{(2\cdot 2)^2} \cos (4x)- \frac{2^2}{(2\cdot 3)^2}\cos (6x)+ \dots \\
&=-\frac{1}{1^2} \cos (2x)- \frac{1}{2^2}  \cos (4x)- \frac{1}{3^2} \cos (6x)+ \dots
\end{align*}
as desired.


问:
If $f$ is continuous and bounded, and $\text{span}\{x \mapsto f(x+k) \mid k \in \Bbb Z\}$ is finite-dimensional, what can we say about $f$?

If $f : \Bbb R \to \Bbb R$ is continuous and bounded, and $\text{span}\{x \mapsto f(x+k) \mid k \in \Bbb Z\}$ is finite-dimensional, what can we say about $f$?

I found this problem difficult (I found examples of functions for which the hypothesis is true like $n$-periodic functions, where $n \in \Bbb N^*$, but I am struggling for the reverse implication). Can someone help me solve it and does someone have a reference from where it comes?

答:
Answer for the revised question  Now you translate only by integers.  This is called a "homogeneous linear difference equation with constant coefficients".

$$
0 = \sum_{k=K_1}^{K_2} a_k f(x+k), \quad a_{K_1}\ne 0, a_{K_2}\ne 0
$$
The solution (on the integers) is well-known.  These solutions are functions
$$
f(x) = \sum_{n=1}^N C_n(x) \;r_n^x
\tag1$$
where $C_n$ are polynomials, and $r_n$ are complex.  Such a solution is bounded (on $\mathbb Z$, positive and negative) only if the polynomials $C_n$ are constants and $|r_n| = 1$.  If we want real values, we get trigonometric expressions for real and imaginary part of $r_n^x$ when $r_n$ is not real.  Such a function need not be periodic, but (if it is bounded) it is a linear combination of periodic functions.  (Allowing irrational periods!)

For solution on $\mathbb R$, take an arbitrary function on a certain "starting interval" $[a,b] = [0,K_2-K_1]$ and extend
(forward and backward) using the difference equation.  So we get something like $(1)$ except that the constant $C_k$ depends on the value in the starting interval.  We choose our function on the starting interval so that it is continuous, and the values at $a$ and $b$ agree to get a solution that is bounded and continuous.

Previous answer

This is a nice question.  Many generalizations are possible.
For example, what can you say about a (measurable) function $f$ satisfying $$f(x) -2 f(x+1) + f(x+\sqrt{2}\,) = 0\tag1$$  What happens if you add "continuous" or "bounded" or "$L^p$" to the conditions?  What settings other than $\mathbb R$ can be considered?

plug

Edgar, G. A.; Rosenblatt, J. M., Difference equations over locally compact Abelian groups, Trans. Am. Math. Soc. 253, 273-289 (1979). ZBL0417.43006.


问:
How to convert a function of $x$ and $y$ into a function based in polar coordinates?

How did my teacher get from $x^2(x^2+y^2)$ to $r\cdot r^2\cos^2(\theta)(r^2)$?

I'm looking into the equations put I'm still at a loss because those usually require just $x$ and $y$ values like R being equal to:

$r = \sqrt{x^2+y^2}$

Thanks in advance.

答: From the definition of polar coordinates, we have that $x=r\cos(\theta)$ and $y=r\sin(\theta)$, can you take it from here?


问:
Lemma used to prove $\left|HK\right|=\frac{\left|H\right|\left|K\right|}{\left|H \cap K\right|}$
Given a group $G$ and $H,K \le G$,then :

$$\left|HK\right|=\frac{\left|H\right|\left|K\right|}{\left|H \cap K\right|}$$

Where $HK:=\left\{hk:h \in H ,k \in K\right\}$

Lemma:

For $h_1,h_2 \in H$

$$hK=h'K \iff h(H \cap K)=h'(H \cap K)$$

We have:

$$HK=\bigcup_{h \in H}hK$$

Not every such left cosets of $K$ in $H$ are distinct, on the other hand the function $\phi:hK \to K$ with $hk \mapsto k$ is a bijection, so the number of elements in $hK$ is the same as that $K$'s  I showed that the set of left cosets (equivalently right cosets) partitions the group.

By this we see that :$$\left|HK\right|=\left|\color{blue}{\text{the set consiting of all distinct left cosets }}hK\right|\left|K\right|$$

One concludes from the lemma that the number of such distinct left cosets is the same as $\left|H: H \cap K\right|$ but I don't know how such a conclusion is possible, how the lemma helps us?

It looks that $hK \ne h^{'} K$ iff $h(H \cap K) \ne h^{'}(H \cap K)$ and the order of the set of all such distinct $h(H \cap K)$ for $h \in H$  is $\left|H: H \cap K\right|$...

Also, it would be appreciated if someone gives me an example where such left cosets $hk$ are identical.

答:
Consider the map $\varphi: H/H\cap K\longrightarrow HK/K$ by $h(H\cap K)\mapsto hK$.

This is a well defined map by your lemma $\impliedby$.

This map is injective by your lemma $\implies$.

This map is surjective by definition of $HK$.

Therefore this is a natural one-to-one correspondence between these cosets, and the Product Formula follows immediately.

I happen to have written about this yesterday, so here's a link for you https://ml868.user.srcf.net/ExpositoryWritings/Groups3.pdf. There are a few typos I haven't fixed but I hope it is readable and somewhat inspiring.


问:
How do I solve this integral using Feynman diagrams?
I have this integral which I want to solve to forth order in $g$:

$$ Z(g) = \int^\infty_{-\infty} \mathrm{d}x \exp\left(\frac{-1}{2}x^2 + \frac{g}{4!}x^4\right) $$

I can solve this integral quite easily by expanding $e^{\frac{g}{4!}x^4}$ and use this formula:
$$\int^\infty_{-\infty} x^{2k} e^{-x^2/2} \mathrm{d}x = (2 \pi)^{1/2} \frac{(2k)!}{2^k k!}.$$

But I want to solve it "using Feynman diagrams" and I'm not sure how to do that. Usually, it's the other way around, I calculate a diagram using integrals. So I'm confused. How does one solve this?

答:
First of all, the one-sided version of the integral identity you mention below is discussed in this MSE post if you happen to be interested in a proof of that fact.

Unfortunately you're not going to find a solution in terms of so-called elementary functions, but we can state the result in terms of the modified Bessel function of the second kind:

$$\int_{-\infty}^\infty \exp\left(\frac{-x^2}{2}+\frac{g}{4!}x^4\right)\mathrm{d}x=\frac{\sqrt{3} e^{-3g/4}K_{1/4}\left(\frac{-3}{4g}\right)}{\sqrt{-g}}$$
For $g<0$.


问:
How I can get a sequence for the differential sheaf on the blow up of $\mathbb{P}^2$ at a single point?
Let $X:=\mathbb{P}^2$ and let $\pi : Y(:=\mathbb{Bl}_p) \rightarrow X$ be the blow-up of $\mathbb{P}^2$ at a point $p$. Then we can get a exact sequence for the differential sheaf $\Omega_Y^1$ of $Y$ as:
$$0 \rightarrow {\cal{O}}_Y(-2f) \rightarrow \Omega_Y^1 \rightarrow {\cal{O}}_Y(-2h-f) \rightarrow 0$$
In the above sequence $f$ is a fibre of ruling (as seen $Y$ is  a ruled surface), while $h$ is the negative intersection section. My question is that how I can get this sequence in detail. Maybe it is an easy question, but I cannot get any idea for the first step. Thank you for your help.

答:
I think it should be something roughly like this:

Write $Y$ as the total space of the projective bundle $\pi : \mathbb{P}(\mathcal{O} \oplus \mathcal{O}(1)) \rightarrow \mathbb{P}^1$, . There is a bundle morphism $TY \rightarrow \pi^{*}(T\mathbb{P}^1)$ which is given by the map on tangent vectors induced by $\pi$. Note that $\pi^{*}(T\mathbb{P}^1) = \mathcal{O}_{Y}(2f)$ because the tangent bundle of $\mathbb{P}^{1}$ has degree 2. Now dualising gives the desired sequence (note that dualising reverses the direction so we have the required terms in the two positions, this was enough to convince me :)).


问:
Existence of a maximum matching such that v is un-saturated
Ture or False: for every graph G without a perfect matching and every vertex v of G there is a maximum matching M of G such that v is M-unsaturated.</p>

答: Consider $P_3$, the path graph on 3 nodes. It does not have a perfect matching but two maximal matchings (by picking either of the two edges present in $P_3$) neither of which leave the center vertex unsaturated.


问:
Can choosing different free variables alter the solution of a linear system?
Consider the following system:

x + 2y - 3z = 1....(1)

```<code>y - 2z = 2.....(2)

2y - 4z = 4....(3)
</code>```

Here the third equation can be eliminated as it is a multiple of the second equation and the echelon form of the above system will be:

x + 2y - 3z = 1.....(1)

```<code>y  - 2z = 2.....(2)
</code>```

The echelon form has two equations in three unknowns, so there is one free variable. Suppose I choose z as a free variable. I get y = 2 + 2z...(3) and x = -3 - z....(4) Since there is one free variable, the solution space will be a line (Am I right?) From (3) and (4), we get y = - 2x - 4 to be the line of solutions.

Now suppose I choose y as a free variable. Now I will have x = - 2 - y/2 and z = y/2 - 1. Clearly, here the solution line will be in xz plane whereas the solution line above is in XY plane. I do not understand what is going on here. Why do these solution lines depend on the choice of the free variable? Aren't the terms pivot variables and free variables just computational conventions? I do understand why the solution space is a line (and hence infinite solutions). But why two different lines?

I think I am failing to understand these concepts perfectly. Please help me out here. Thanks.

答:
$$a=\{t,-2 t-4,-t-3\};\;b=\left\{-\frac{u}{2}-2,u,\frac{u}{2}-1\right\};\;c=\{-s-3,2 s+2,s\}$$
Are three different parametrization of the same line.

To prove this, note that the direction vectors are parallel
$$\vec{a}=\{1,-2 ,-1\};\;\vec{b}=\left\{-\frac{1}{2},1,\frac{1}{2}\right\};\;\vec{c}=\{-1,2 ,1\}$$
indeed $\vec{c}=-\vec{a}$ and $\vec{b}=-\frac{1}{2}\vec{a}$

Furthermore they pass through the point $(0,-4,-3)$.

Line $a$ for $t=0$, line $b$ for $u=-4$ and line $c$ for $s=-3$.

Whatever variable we choose to let "free" the result is always that line.

Hope this helps


问:
How to show $\operatorname{rad}(J) \subset \operatorname{rad}(I)$ implies $V(I) \subset V(J)$?
I would like to show

$\operatorname{rad}(J) \subset \operatorname{rad}(I)$ if and only if $V(I) \subset V(J)$.

Suppose $V(I) \subset V(J)$, i.e., if $P \supset I$, then $P \supset J$. Then $\bigcap_{P \supset I} P \supset \bigcap_{P \supset J} P$ hence $\operatorname{rad}(J) \subset \operatorname{rad}(I)$.

However, I am having trouble showing the converse:

If $\operatorname{rad}(J) \subset \operatorname{rad}(I)$, then $\bigcap_{P \supset J} P \subset \bigcap_{P \supset I} P$. However, this is just an intersection of prime ideals and it does not immediately imply that the sets on the left hand side are in the intersection of the sets on the right hand side.

But, Hartshorne's book just states this as if it follows immediately. I don't see how.

答: If $I\subset Q$ where $Q$ is a prime ideal,  then $\cap_{I\subset P} P\subset Q$,  so $\sqrt{I} \subset Q$,  hence $J\subset \sqrt{J} \subset \sqrt{I} \subset Q$.


答:
Hint: consider the the set of all subsets of $\{1,2,\dots,n\}$ (of which there are $2^n$) and try to find the total sum of the sizes of the subsets in two different ways. For example, the possible subsets of $\{1,2\}$ are $\{\},\{1\},\{2\},\{1,2\}$. Then adding up the sizes of each subset gives $0+1+1+2 = 4$.

More explicitly, if we add up the sizes of all possible subsets of $[n]=\{1,2,\dots,n\}$, we can either:

$1)$ Note that there are $\binom{n}{i}$ subsets of size $i$ which gives that the total sum of sizes is

$$\sum_{i=1}^{n}\binom{n}{i}i$$

$2)$ Observe that each element is in $2^{n-1}$ subsets, and so contributes $2^{n-1}$ to the total sum. Thus the sum equals

$$n2^{n-1}$$

The value of the sum doesn't change regardless of how we do it, so the expressions must be the same.

Title: Combinatorial proof of $\sum^{n}_{i=1}\binom{n}{i}i=n2^{n-1}$.

Body:
Prove that
  $$\sum^{n}_{i=1}\binom{n}{i}i=n2^{n-1}$$

I can't find counting interpretations for either of the sides. A hint of "if $S$ is a subset of $\{1, . . . , n\}$ and $S^\prime$ is its complement then $|S| + |S^\prime| = n$" was also given, but I still don't know how to begin.


问:
How to prove that r is a double root if and only if it is a root of a polynomial and of its derivative.
I don't know how to start the question. The title is self explanatory. How to approach and make a formal proof?</p>

答:
I was trying to solve this and came across this question. This is my solution. I hope it helps.

Suppose $P(x)$ is a polynomial of degree n and r is a root of $P(x)$ and its derivative.

$$P(x)=(x-r)Q(x)$$
$$P ^\prime(x)=(x-r)S(x)$$
where $Q(x)$ and $S(x)$ are also polynomials.
Based on the first equation, we can also write $$P ^\prime(x)=Q(x)+(x-r)Q^\prime(x)$$
For $x \neq r$
$$S(x)=\frac{P ^\prime(x)}{(x-r)}=\frac{Q(x)+(x-r)Q^\prime(x)}{x-r}$$
$$=\frac{Q(x)}{x-r}+Q^\prime(x)$$
$$=\frac{P(x)}{(x-r)^2}+Q^\prime(x)$$
$$\Rightarrow P(x)=(x-r)^2(S(x)-Q^\prime(x))$$
So, r is a double root of $P(x)$.

If r is a double root of $P(x)$, we have
$$P(x)=(x-r)^2T(x)$$
where $T(x)$ is a polynomial. We have
$$P^\prime(x)=2(x-r)T(x)+(x-r)^2T^\prime(x)$$
So, r is also a root of  the derivative of $P(x)$.


问:
Boolos’s proof of the first incompleteness theorem. Predicate $C(x,y)$ and assumption of completeness.
I am trying really hard to understand how Boolos’s proof works, but I keep having doubts about it. Can you spot any logical flaws in my reasoning? This question arose from a previous question, but it is different from it as a consequence of things that were explained to me in response to such previous question.

The first incompleteness theorem states that any consistent theory is not complete. In other words, stated very naively, the result of the proof is: ¬(“true” ↔ “provable”). To avoid making a circular argument, the assumption about the theory that carries out our proof must be “true” ↔ “provable”. If our theory starts with this assumption, it seems to me that it can be showed $C(x,y) → ¬C(x,y)$ and therefore $¬C(x,y)$ can be proved by contradiction (with $C(x,y)$ being the predicate found in Boolos's proof). The reason for this is it seems to me that assuming $C(x,y)$ true allows you to prove both $∃n(∀(()↔=[]))$ and $¬∃n(∀(()↔=[]))$.

$∃n(∀(()↔=[]))$ has to be proved as part of Boolos’s proof (it is asserted to be true along the proof). From the assumption “true” ↔ “provable”, ¬∃n(∀(()↔=[])) can be deduced in a few mechanical steps. If the implication $(∀(()↔=[]) → ¬∀(()↔=[]))$ is true then we also have $¬∃n(∀(()↔=[]))$. It can be seen easily by rewriting $∀(()↔=[])$ as $J(n)$ and using proof by contradiction $(J(n) → ¬J(n)) → ¬J(n)$ and the generalisation inference rule $J(n) → ∀nJ(n)$. Then just substitute $¬∃n¬$ with $∀n$ (they are syntactically the same) to get $∀n¬J(n) ↔ ¬∃n¬¬J(n) ↔ ¬∃nJ(n) ↔ ¬∃n(∀(()↔=[n]))$.

Now, how can we show the implication $(∀(()↔=[]) → ¬∀(()↔=[]))$ to be true? We see that $∀(()↔=[])$ implies $A(n,m)$. Let’s say $F(x)$ is d symbols long, with d<m by construction ($m = 10k$ and $d = 2k + 24$ in Boolos’s proof).
We also have $A(n,m) → ¬(C(n,y) ∧ (y<m)) → ¬C(n,d) → ¬∀(()↔=[n])$.

An objection could be: $¬C(n,d)$ just implies that there is no proof for $∀(()↔=[n])$, it does not imply $¬∀(()↔=[n])$. But if we have to start with the assumption “true” ↔ “provable” we can deduce these two things are exactly the same.

So to summarise, it seems to me that starting by the assumption that your theory is consistent and complete, $¬C(x,y)$ is proved by showing $C(x,y)$ leads to contradiction. Therefore to me it looks like Boolos’s proof cannot be carried out if you start from these assumptions. But these are the assumptions you have to start from to avoid making a circular argument.

答:
IMO the issue is with your use of formula $F(x)$.

According to Boolos' definition,

"a formula $F(x)$ names a number $n$ if the statement $\forall x (F(x) \leftrightarrow x=[n])$ (where $[n]$ is $ss \ldots s0$, with $n$ times $s$) is in the output of algorithm $M$."

Boolos says (page 386): "$n$ is not named by $F$; in other words, $\forall x (F(x) \leftrightarrow x=[n])$ is not in the output of $M$."

You are switching from "$\forall x (F(x) \leftrightarrow x=[n])$ is not in the output of $M$" to "$\lnot \forall x (F(x) \leftrightarrow x=[n])$ is in the output of $M$."

But this is what we are trying to prove: that the system is (negation) complete:

for each sentence (closed formula) $\phi$ of the language of the system either $\phi$ or $¬ \phi$ is a theorem (an output) of the system.

Thus, if $\phi$ is not an output of the system, then $\lnot \phi$ must be.

But this is exactly what $\mathsf {PA}$ is not: it is (negation) incomplete.

See page 383:

a new proof of the Gödel incompleteness theorem in the form: There is no algorithm whose output contains all true statements of arithmetic and no false ones.

And page 384:

if there were an algorithm that printed out all and only [emphasis added] the true statements of arithmetic...

This is the crux of the matter: from the fact that $\forall x (F(x) \leftrightarrow x=[n])$ is not in the output of $M$ we cannot conclude that $\lnot \forall x (F(x) \leftrightarrow x=[n])$ is the output of $M$.


问:
number of regular cardinals in a weakly inaccessible cardinal
Let $\kappa$ ba weakly inaccessible cardinal. Why are there $\kappa$ regular cardinals $\lambda < \kappa$? I've tried a recursive construction, but I don't know what to do in the limit step. Supremum does not work, since then we loose regularity.</p>

答: Suppose that $\kappa$ is weakly inaccessible. Thus, it is a regular limit cardinal. So $\kappa=\aleph_\beta$ for some ordinal $\beta$. Since $\kappa$ is a limit cardinal, it must be that $\beta$ is a limit ordinal. Since $\kappa$ is regular, it cannot be that $\beta<\kappa$. So $\kappa=\aleph_\kappa$. Thus, there are $\kappa$ many cardinals below $\kappa$. All the successor cardinals $\aleph_{\beta+1}$ for $\beta<\kappa$ are regular, and there are $\kappa$ many of these.</p>

ContentLicense: CC BY-SA 2.5


问:
Abstract Algebra proof for quadratic extensions

Let $n$ be a square free integer and alpha belong to $\mathbb{Z}[\sqrt{n}]$. If $N(\alpha)$ is a prime integer, then $\alpha$ is irreducible in $\mathbb{Z}[\sqrt{n}]$.

Here $N(\alpha)$ denotes the norm of $\alpha$.

How can I show this? I'll get back in the morning because it's super late right now!

答: The norm is multiplicative. So, if $\alpha=\beta\gamma$ then $N(\alpha)=N(\beta) N(\gamma)$. If $N(\alpha)$ is prime, then one of $N(\beta)$ or $N(\gamma)$ is $\pm 1$. You now need that $N(\beta)=\pm 1$ iff $\beta$ is a unit, which comes directly from $N(\beta)=\beta\bar{\beta}$.</p>


问:
Algorithm to solve n variable problem with range
During Software designing our team come across the following problem: we have linear equation with n variables with each variable have positive permissible range or allowed range.

In This Problem the constant a_i are the number of items person wants to be bought of type x_i which has minimum value of purchase l_i and maximum value of purchase r_i and now, buyer know about this ranges that why he smartly ask that is this possible within value b. So now we have to choose one solution if possible and give the invoice accordingly.

So in math form,

```<code> a1x1+a2x2+....+anxn = b
 known parameters:
 a1,a2,...,an and b
 unknown parameter x1,x2,....,xn
 but we know that
 x1 belongs to range [l1,r1]
 x2 belongs to range [l2,r2]
 .
 .
 .
 xn belongs to range [ln,rn]
 0<li<ri
 ai>0
 b>0
</code>```

and we need to find values of all xi in above equation(may be float or integer but must be range). I need to solve this problem as soon as possible.
Example :

```<code>  Let n be 5
  And equation is
  10*x_1+15*x_2+10*x_3+11*x_4+17*x_5=2365
   And allowed ranges are
   For
   X_1 [30,60]
   X_2 [20,40]
   X_3 [20,80]
   X_4 [50,120]
   X_5 [10,140]
  So one of the solution will be
  For
  X_1 =40
  X_2 =36
  X_3 =34
  X_4 =60
  X_5 =25
</code>```

Want to find only one solution.
I need solution in bigO(nlogn)
Or nearly linear as possible as.

答:
Keep in mind that in some cases there may not exist a solution. For instance, if $\sum_{i=1}^n a_i r_i < b$ or $\sum_{i=1}^n a_i l_i > b$, no solution will be available. This corresponds to the situations in which taking the upper bound for each variable is not enough to get $b$, or in which taking the lower bound for each variable already exceeds $b$.

Anyway, if $\sum_{i=1}^n a_i r_i \ge b$ and $\sum_{i=1}^n a_i l_i \leq b$, you can just solve this as linear programming problem... The set of restrictions will be
\begin{align*}
a_1x_1 + \cdots +a_n x_n & = b &\\
x_i & \ge l_i \quad i=1, \cdots , n\\
x_i & \leq r_i \quad i=1, \cdots, n
\end{align*}

The objective function is not very relevant here. You can, for instance, minimize $x_1$, yielding the solution with the smallest value for $x_1$, or any other criteria you see fit. In the example you presented, one gets

$$x_1 = 30, x_2=20, x_3=20, x_4=50, x_5 = 1015/17.$$

Note: Since you just want an admissible solution, you can implement a very simple scheme. Initialise all variables $x_i = l_i$ and compute $a \cdot x$. If $a\cdot x > b$ the problem is impossible; If $a\cdot x = b$, you got the solution; If $a\cdot x < b$ start setting $x_i = r_i$ until $a \cdot x$ exceeds b; when that happens, just tune the last $x_i$ so that the equality is attained.


问:
Combinatorics: Number of integer solutions with lower bounds
How many integer solutions does the equation $x_1 + x_2 + x_3 + x_4 = 15$ have, if
we require that $x_{1}\ge 2, x_{2}\ge 3, x_{3}\ge 10, x_{4}\ge −3?$

I need to understand in general what to do. I don't know the answer so can't check myself.

答:
The problem is the same than to find the number of solution of $$x_1+x_2+x_3+x_4=3,$$
where $x_1,...,x_4\geq 0$. And this is a well know problem which it refer to the stars and bars problem. So, there are $\frac{6!}{3!3!}$ solutions.


问:
Determining the specific pmf when given a Density function which constants

Was wondering if anyone could help me out. So far I've used the fact that the density function is always equal to 1. So I have done the integral of the density function between 0 and 1 and worked out my first equation in the steps of finding the value for the constants. Then I used the fact what we are given the value of the Expected Value which is the same as the integral between 0 and 1 of the density function multiplied by 'x'. Which has given me a second equation to figure out the value of the constants. Then I compared the equation to get the answer that a=(3/2) and b=(1/2). Now that I have these values, I know that I have to do an integral of the new density function, but I don't know what boundaries to set. I was assuming it would be 1/2 and + infinity since the question just asks for x to be greater than a half, but I don't know if this is correct.

答:
You want $$\mathbb P(0.2 \lt X)=\int\limits_{0.2}^{+\infty} f_X^{\,}(x)\, dx =\int\limits_{0.2}^1 \left(\frac32 x^2 +\frac12\right)\, dx$$

as the lower limit (from the question) is $0.2$ and upper limit (from the support) is $1$


问:
changing limit of this double integral $\int \int _R^{ }f\left(x,y\right)dA$ with region R bounded by { $y=3x$, $x=3y$, $x+y=4$}
we are asked to use the substitution, $x=3u+v$ and $y=u+3v$. OK so this is what i did....first i plot the region R

I made a mistake, @Kavi Rama Murthy you are right, the area needs to be divided into 2 region, Region 1 can be $\ \int _0^1\ \int _{\frac{y}{3}}^{3y}\ f\left(x,y\right)dxdy\ $ and Region 2 being $\ \int _1^3\int _{\frac{y}{3}}^{4-y}\ f\left(x,y\right)dxdy$

答: The equations for the lines in terms of $u$ and $v$ are $u=0,v=0$ and $u+v=1$. So the limits are $0 < v < 1-u$ and $0<u<1$.


问:
Change the order of integration of the following double integral
$$\int\limits_{\frac{\pi}{2}}^{\frac{5\pi}{2}}  \int\limits_1^{\sin x} f(x,y) dy dx$$

I sketched the region and noticed that if I somehow shift the region by $2\pi$ to the left then I can use the $\arcsin $ function. However the correct answer was $$\int\limits_{-1}^{1}  \int\limits_{\pi - \arcsin y}^{\arcsin y + 2\pi} f(x,y) dx dy$$  How can this be found?

答:
We have the limits of $x : \frac{\pi}{2}$ to $\frac{5\pi}{2}$

and the limits of $y: 1$ to $\sin x$

We need to convert these such that $y$ lies between $2$ constant limits and $x$ has limits which involve $x$ as a function of $y$.

Now, from $x = \frac{\pi}{2}$ to $\frac{5\pi}{2}$, the highest value $y$ can achieve is $\boxed{1}$ and the lowest value is $\sin\frac{3\pi}{2} = \boxed{-1}$  (you can observe this from the graph)

If $y = \sin x$, then $x = \arcsin(y)$. But, $\arcsin$ has a principal range $[-\frac\pi2,\frac\pi2]$ So we must do transformations accordingly (to bring the term in the required range)

First let us consider the branch for which $\frac\pi2\le x\le\frac{3\pi}2 $  (length $=\pi$)

$\frac\pi2\le x\le\frac{3\pi}2 \Rightarrow -\frac{\pi}{2} \le x - \pi\le \frac\pi2$ or $ -\frac{\pi}{2} \le  \pi-x \le \frac\pi2$

So $\pi-x$ lies in our required range.

Now, $y= \sin(x) = \sin(\pi-x) \Rightarrow\pi-x = \arcsin(y) \Rightarrow \boxed{ x = \pi-\arcsin(y)}$

Again consider $\frac{3\pi}2\le x\le\frac{5\pi}2 $  (length $=\pi$)

$\frac{3\pi}2\le x\le\frac{5\pi}2 \Rightarrow -\frac{\pi}{2} \le x - 2\pi\le \frac\pi2$

So $x-2\pi$ lies in our required range.

Now, $y = \sin(x) = -\sin(2\pi-x) = \sin(x-2\pi) \Rightarrow \boxed{x =2\pi+\arcsin(y)} $

Finally subbing the limits:

The integral becomes $\begin{align}\int_{y=-1}^{1}\int_{x= \pi-arcsin(y)}^{2\pi+\arcsin(y)} f(x,y)dxdy\end{align}$


问:
In what sense are self adjoint elements of C*-algebra "closed"?
Let $\mathcal A$ be a unital C*-algebra.

I read the fact, that the self-adjoint elements $\mathcal S:=\{A\in \mathcal A \vert A^*=A\}$ are closed in $\mathcal A$.

In what sense is "closed" meant here? Topologically closed, i.e. every convergent sequence in $\mathcal S$ has a limit in $\mathcal S$ or algebraically closed? What is to prove for algebraic closure, is it closure under vector addition, under algebra multiplication and under scalar multiplication?

答:
If the set of self-adjoint elements was stable under multiplication, we would have $AB=(AB)^*=B^*A^*=BA$, i.e. any two self-adjoint elements would commute. Writing any element $C=\frac{C+C^*}{2}+i\frac{C-C^*}{2i}$ as a linear combination of two self-adjoint elements, we see that this would imply that $\mathcal A$ be commutative. The converse is clear. Hence the set of self-adjoint elements is stable under multiplication if and only if $\mathcal{A}$ is commutative.

Note that $iA$ is not self-adjoint if $A$ is nonzero self-adjoint, as $(iA)^*=-iA$. So that's not a complex subspace. But for $s,t$ real scalars, and $A,B$ self-adjoint, then
$$
(sA+tB)^*=\overline{s}A^*+\overline{t}B^*=sA+tB
$$
is still self-adjoint. So algebraically, that's a real vector subspace.

Topologically, it is norm closed. Indeed, the function $A\longmapsto A^*$ is an isometry, so it is norm-continuous, as well as $f:A\longmapsto A-A^*$. Hence the set of self-adjoint elements $f^{-1}(0)$ is closed in the norm topology.

If $A$ sits in $B(H)$, meaning it is a norm closed $*$-subalgebra of $B(H)$, then the set of self-adjoint elements is even closed for the wot and the sot (weak and strong operator topology). That's because $A\longmapsto A^*$ is wot-continuous.


问:
How to determine the new region after variable change
Consider the double integral

$$\iint_\limits{E} e^\frac{y}{x+y}\, \, dx\, dy,$$
where $E$ is the region bounded by $x=0,y=0 $ and $x+y=1$.

Now if we change the variable by $x+y=u$ and $y=uv$,then how to determine $E'$, the new region over which the new integral is to be taken? It is clear that $J=u$, the Jacobian and hence the differential becomes $ue^v du dv$ but I cannot guess the general process to determine the ranges of $u$ and $v$. Can someone help me understand the general way?

答:
Let consider the graphs for $u=x+y$ and $y=uv$ inside the given region, we can see that

$0\le x+y\le 1 \implies 0\le u\le 1$

and for any fixed segment aligned with the line $u=x+y$

$0\le y\le u \implies 0\le uv\le u \implies 0\le v\le 1$


问:
Reordering task
I'm trying to solve tiny task within a c++ algorithm but don't want to reinvent the wheel and looking for a math way.

Imagine a row of people when some of them are changing their positions by their own and as a result shifting their neighbors. Example is below(pretty random)

First matrix (Initial state)

```<code>0  | Daniel
1  | Alan
2  | Kate
3  | Tom
4  | Sandra
5  | Andrew
6  | Paul
7  | Laura
8  | Sophia
9  | Emma
10 | Carl
</code>```

Second matrix (Final state)

```<code>7  | Daniel
3  | Alan
2  | Kate
4  | Tom
5  | Sandra
1  | Andrew
6  | Paul
8  | Laura
10 | Sophia
0  | Emma
9  | Carl
</code>```

What I need is a list with minimal moving instructions to achieve the final state:

Emma, please go from 9 to 0 (current positions of all others are shifted)

X, please go from your current(shifted by Emma) position' to position2.

...

I guess that this can be solved as matrix task, but at the same time it looks as a sorting task as well when just factors are changed. Anyway my turn to ask an all-knowing community for help ;)

Update 1 (Levenshtein distance/Wagner-Fischer algo)

Initial string 'order-1': 0123456789A

Final string 'order-2': 73245168A09

Where:

0 for Daniel

1 for Alan

...

To achieve the final state we need to apply these instructions to order-1:

replace '0' with '7';

replace '1' with '3';

delete '3';

insert '1';

delete '7';

replace '9' with 'A';

insert '0';

replace 'A' with '9';

Matrix:

```<code>[00,01,02,03,04,05,06,07,08,09,10,11]
[01,01,02,03,04,05,06,07,08,09,09,10]
[02,02,02,03,04,05,05,06,07,08,09,10]
[03,03,03,02,03,04,05,06,07,08,09,10]
[04,04,03,03,03,04,05,06,07,08,09,10]
[05,05,04,04,03,04,05,06,07,08,09,10]
[06,06,05,05,04,03,04,05,06,07,08,09]
[07,07,06,06,05,04,04,04,05,06,07,08]
[08,07,07,07,06,05,05,05,05,06,07,08]
[09,08,08,08,07,06,06,06,05,06,07,08]
[10,09,09,09,08,07,07,07,06,06,07,07]
[11,10,10,10,09,08,08,08,07,06,07,08]
</code>```

This does not work in my case, because I'm just moving rows(removing&inserting) in a ListView, but not replacing.

What I actually need is a minimal moving list:

Take Emma and Put behind Daniel

Take Andrew and Put behind Daniel

Take Kate and Put behind Daniel

Take Alan and Put behind Daniel

Take Tom and Put behind Daniel

Take Sandra and Put behind Daniel

Take Paul and Put behind Daniel

Take Carl and Put behind Sophia

Where:

Take = Remove

Put = Insert

Take&Put = Move

Update 2

Task is solved with permutation cycles.

答: One of the fundamental properties of permutations is that every permutation can be written as the product of disjoint cycles.  So you can factor the final permutation into disjoint cycles and shift the elements in each cycle.  Shifting the elements of a cycle is relatively simple.</p>

Title: Reordering task


问:
Is it possible to evaluate $\lim_{x\to\infty}\frac{\ln(x+e^x+e^{2x})}{x}$ without applying L'Hopital rule?
$$\lim_{x\to\infty}\frac{\ln(x+e^x+e^{2x})}{x}$$
In the book I am reading, evaluated the limit with L'Hopital rule like this:
$$\lim_{x\to\infty}\frac{\ln(x+e^x+e^{2x})}{x}=\frac{\infty}{\infty}\rightarrow \lim_{x\to\infty}\frac{1+e^x+2e^{2x}}{x+e^x+e^{2x}} $$

Then it used equivalence and write $\lim_{x\to\infty}\cfrac{2e^{2x}}{e^2x}=2$.

My approach to evaluate $\lim_{x\to\infty}\frac{\ln(x+e^x+e^{2x})}{x}$ is different:

in the numerator of the fraction and inside the $\ln()$ the function $e^{2x}$ goes faster to infinity than $x$ , $e^x$. therefore we can ignore these two:
$$\lim_{x\to\infty}\frac{\ln(e^{2x})}{x}=\lim_{x\to\infty}\frac{2x}{x}=2$$
Is my approach right? and was it really necessary to use L'Hopital Rule?

答:
Your approach is correct. For a more rigorous argument do the following:

$\frac {\ln (x+e^{x}+e^{2x})} x=\frac {\ln [e^{2x} (xe^{-2x}+e^{-x}+1)} x=\frac {2x+\ln (xe^{-2x}+e^{-x}+1)} x$. Using the fact that $\ln (1+t) \sim t$ as $t \to 0$ we get $\lim \frac {2x+xe^{-2x}+e^{-x}} x=2$.


问:
Inverse $\mathcal{Z}$ transform of $\frac{1}{z^2(z+1)}$ without delta function
While finding inverse $\mathcal{Z}$ transform of $\frac{1}{z^2(z+1)}$ i split it into three parts using partial fractions:

$$\frac{a}{z} + \frac{b}{z^2}+\frac{c}{z+1}$$

and so answer comes as

$$a \delta(n-1) + b\delta(n-2) + c (-1)^{n-1} H(n+1) $$

where $\delta$ is delta function and $H$ is step function

Now I checked with online calculator tool link

It gives answer as $(-1)^{n+1} H(n-3)$. I cannot understand where the $\delta$ went away? And how can my solution arrive at the one given in online calculator, that is, how can I get rid of delta function in my solution

I have verified that online calculator gives correct result also as its $\mathcal{Z}$ transform comes as stated above.

Body: We have $$X(z) = \frac{1}{z^2(z+1)}= \frac{1}{z^2}+\frac{1}{z+1} - \frac{1}{z}$$Let $$X(z) = \sum_{n = 0}^{\infty}x[n]z^{-n}$$Also $\mathcal{Z}\{\delta[n - 1]\} =  \frac{1}{z}$ and $\mathcal{Z}\{\delta[n - 2]\} =  \frac{1}{z^2}$. Note that $$\mathcal{Z}\{a^{n-1}u[n-1]\} = \sum_{n = 1}^{\infty}a^{n-1}z^{-n} = \sum_{n = 1}^{\infty}\frac{1}{a}(\frac{z}{a})^{-n} = \sum_{n = 0}^{\infty}\frac{1}{a}(\frac{z}{a})^{-n} - \frac{1}{a} = \frac{1}{a}\times\frac{1}{1 - \frac{a}{z}} - \frac{1}{a} = $$ $$\frac{1}{a}\times \frac{a}{z-a} = \frac{1}{z-a}$$Here $a = -1$ and we have $$\mathcal{Z}^{-1}\{X(z)\} = -\delta[n-1] + \delta[n-2] + (-1)^{n-1}u[n-1]$$


问:
Subgroup transitive on the subset with same cardinality
Maybe there is some very obvious insight that i miss here, but i've asked this question also to other people and nothing meaningful came out:

If you have a subgroup G of $S_n$(the symmetric group on n elements), you can consider the natural action of G on the subset of $\{1,...,n\}$; my question is this:what are the G that act transitively on the subsets of the same cardinality; that is, whenever $A,B \subseteq \{1,...,n\}$ with $|A|=|B|$, there is  $g \in G$ so that $gA=B$.

答:
By the Livingstone Wagner Theorem, (Livingstone, D., Wagner, A., Transitivity of finite permutation groups on unordered sets. Math. Z. 90 (1965) 393–403),  if $n \ge 2k$ and $G$ is transitive on $k$-subsets with $k \ge 5$, then $G$ is $k$-transitive. Using the classification of finite simple groups, $A_n$ and $S_n$ are the only finite $k$-transitive groups for $k \ge 6$, and the only 4- and 5-transitive groups are the Mathieu groups, so (at least for $n \ge 10$), the only permutation groups that are $k$-homogeneous for all $k$ are $A_n$ and $S_n$.

See also the similar discussion in Group actions transitive on certain subsets


答: No, it is not really necessary to use L'Hopital's rule. You can do it as follows:\begin{align}\lim_{x\to\infty}\frac{\log(x+e^x+e^{2x})}x&=\lim_{x\to\infty}\frac{\log\left(e^{2x}\left(\frac x{e^{2x}}+\frac1{e^x}+1\right)\right)}x\\&=\lim_{x\to\infty}\frac{2x}x+\frac{\frac x{e^{2x}}+\frac1{e^x}+1}x\\&=2+0=2.\end{align}On the other hand, although your ideae is fine, I think that your justification is far from complete.

Title: Is it possible to evaluate $\lim_{x\to\infty}\frac{\ln(x+e^x+e^{2x})}{x}$ without applying L'Hopital rule?

Body:
$$\lim_{x\to\infty}\frac{\ln(x+e^x+e^{2x})}{x}$$
In the book I am reading, evaluated the limit with L'Hopital rule like this:
$$\lim_{x\to\infty}\frac{\ln(x+e^x+e^{2x})}{x}=\frac{\infty}{\infty}\rightarrow \lim_{x\to\infty}\frac{1+e^x+2e^{2x}}{x+e^x+e^{2x}} $$

Then it used equivalence and write $\lim_{x\to\infty}\cfrac{2e^{2x}}{e^2x}=2$.

My approach to evaluate $\lim_{x\to\infty}\frac{\ln(x+e^x+e^{2x})}{x}$ is different:

in the numerator of the fraction and inside the $\ln()$ the function $e^{2x}$ goes faster to infinity than $x$ , $e^x$. therefore we can ignore these two:
$$\lim_{x\to\infty}\frac{\ln(e^{2x})}{x}=\lim_{x\to\infty}\frac{2x}{x}=2$$
Is my approach right? and was it really necessary to use L'Hopital Rule?


答:
You are right thinking of $e^{2x}$ as the term that's going to diverge faster, and the idea of using it is correct. However, if you want a somewhat more rigorous proof I'd recommend the following steps:

$\displaystyle\lim_{x\to+\infty}\dfrac{\ln(e^{2x}+e^x+x)}{x}=\lim_{x\to+\infty}\dfrac{\ln(e^{2x}(1+\frac{1}{e^x}+\frac{x}{e^{2x}}))}{x}=\lim_{x\to+\infty}\dfrac{\ln(e^{2x})+\ln(1+\frac{1}{e^x}+\frac{x}{e^{2x}})}{x}=\lim_{x\to+\infty}\dfrac{\ln(e^{2x})}{x}=\lim_{x\to+\infty}\dfrac{2x}{x}=2$


问:
Finding the range of $f(x)=2\csc(2x)+\sec x+\csc x$
Hi this is the question:

Find the range of $$f(x)=2\csc(2x)+\sec x+\csc x$$

What I've tried:

I know that the range of $\csc(x)$ which is $R\setminus (-1,1) $, the range of $\sec(x)$ is $R\setminus (-1,1)$ too. And I've managed to simplify the expression to have in terms of $\sec x$ and $\csc x$ as such:
$$f(x)=\sec(x)\cdot \csc(x)+\sec(x)+\csc(x).$$
But then, what do I do when I have 3 terms and not just $\sec$ or $\csc$?

答:
The derivative of the secant function is $\sin x/\cos^2x$; the derivative of the cosecant function is $-\cos x/\sin^2x$, so you get
$$
f'(x)=-4\frac{\cos2x}{\sin^22x}+\frac{\sin x}{\cos^2x}-\frac{\cos x}{\sin^2x}
=\frac{\sin^2x-\cos^2x+\sin^3x-\cos^3x}{\sin^2x\cos^2x}
$$
We can disregard the denominator and factor the numerator as
$$
(\sin x-\cos x)(\sin x+\cos x+\sin^2x+\sin x\cos x+\cos^2x)
=(\sin x-\cos x)(1+\sin x)(1+\cos x)
$$
Quite nicer! This is positive where $\sin x>\cos x$. Limiting ourselves to the interval $(0,2\pi)$, but also noting that the function is undefined at integer multiples of $\pi2$, we see that this happens for $\pi/4<x<5\pi/4$. Thus our function is

decreasing over $(0,\pi/4)$

increasing over $(\pi/4,\pi/2)$

increasing over $(\pi/2,5\pi/4)$

decreasing over $(5\pi/4,3\pi/2)$

decreasing over $(3\pi/2,2\pi)$

Thus, taking into account the asymptotes, we see that the branch in $(0,\pi/2)$ has range $(f(\pi/4),\infty)=(2+2\sqrt{2},\infty)$.

The branch in $(\pi/2,2\pi)$ is a bit more complicated, because the function doesn't have asymptotes at $\pi$ and at $3\pi/2$. Indeed,
$$
\lim_{x\to\pi}f(x)=-1=\lim_{x\to3\pi/2}f(x)
$$
If we extend the function by continuity, the range over $(\pi/2,2\pi)$ would be $(-\infty,f(5\pi/4))=(-\infty,2-2\sqrt{2})$.

As the function is not defined at $\pi$ and $3\pi/2$, the range is
$$
(-\infty,-1)\cup(-1,2-2\sqrt{2})\cup(2+2\sqrt{2},\infty)
$$


问:
Calculate residue of $\text{res}\dfrac{\sin z}{e^{z} -1}$
I need to evaluate $\text{res}\dfrac{\sin z}{e^{z} -1}$ at all singular points. Not sure how to handle a removable singularity.</p>

答:
At $z=0$, as you say, the singularity is "removable," meaning that it is not really a singularity.  There, the residue is zero.

For other poles at $z=i 2 \pi k$, $k \in \mathbb{Z}$, the poles are simple and the residue at these poles is simply

$$\frac{\sin{i 2 \pi k}}{e^{i 2 \pi k}} = i \sinh{2 \pi k}$$


问:
Please explain the difference between two short exact sequence
My question is about exact sequence and tensor product of modules. Consider the following exact sequence of $R$-module for a commutative ring $R$ as in here:
\begin{equation} 0 \to N_1 \to N_2 \to N_3 \to 0 \ \ (*)\end{equation}
Taking tensor product with $R$-module $M$, we get the exact sequence
\begin{equation} M \otimes N_1 \to M \otimes N_2 \to M \otimes N_3 \to 0 \ \ (**)\end{equation}
But if $M$ is a flat $R$-module, we get the exact sequence
\begin{equation} 0 \to M \otimes N_1 \to M \otimes N_2 \to M \otimes N_3 \to 0 \ \ (***)\end{equation}
Why is so ? What is the difference between the exact sequences  $(**)$ and $(***)$  ?

What is special about the map $0 \to M \otimes N_1$ ?

答:
The difference is that in the second exact sequence, $M\otimes N_1\to M\otimes N_2$ is injective, whereas in the first one it need not be so.

The best way to understand this type of phenomenon is with an example:

take $0\to \mathbb{Z}\overset{p}\to \mathbb{Z\to Z}/p \to 0$ and tensor it with $\mathbb Z/p$, this gives $\mathbb Z/p\overset{0}\to \mathbb Z/p\to\mathbb Z/p\to 0$, where of course $0$ is not injective.

If you tensor it with a flat $\mathbb Z$-module, for instance $\mathbb Q$, you will get an injection $\mathbb Q\overset{p}\to \mathbb Q$.


问:
The set of all subsequential limits of a sequence is closed.

Let $X$ be a metric space and $\{x_n\}$ be a sequence in $X$. Let $E$ be the set of all subsequential limits of $\{x_n\}$. Then $E$ is closed.

I have a idea of this proof.
Assume $E$ is a infinite set.
Let $p\in X-E$. Then $\exists r>0$ such that $B(p,r)\cap E=\emptyset$. The reason is if $p\in \bar E$, then there is some subsequential limit $y$ close to $p$, and there exists a subsequence $x_{n_k}$ converges to $y$. So, $d(x_{n_k},p)<d(x_{n_k},y)+d(y,p)<\varepsilon$ which is a contradiction.

So, $B(p,r)\subset X-E$ and $E$ is closed.

答:
I'm sorry your proof is not correct as we could have $X\subset E$: think about $(\cos(n))_{n\in\mathbb{N}}$.(see this link)

The most elementary way I know to prove this is by diagonal subsequence extraction.


答:
Your argument is not precise . You want to make sure that $n_k$ is strictly increasing.

Suppose there is no $r>0$ such that $B(p,r)\cap E =\emptyset$. For each $i$ there exists $y_i \in B(p,\frac 1  i)$ such that $y_i \in E$. Since $y_i$ is a subsequential limit of $(x_n)$ there exist infinitely many $n_i $ such that $d(y_i,x_{n_i}) <\frac 1 i$. Inductively we can choose the $n_i$'s such that $n_i >n_{i-1}$. Now $(x_{n_i})$ is  a subsequence of $(x_n)$ convergin g to $p$, a contradiction.

Title: The set of all subsequential limits of a sequence is closed.

Body:
Let $X$ be a metric space and $\{x_n\}$ be a sequence in $X$. Let $E$ be the set of all subsequential limits of $\{x_n\}$. Then $E$ is closed.

I have a idea of this proof.
Assume $E$ is a infinite set.
Let $p\in X-E$. Then $\exists r>0$ such that $B(p,r)\cap E=\emptyset$. The reason is if $p\in \bar E$, then there is some subsequential limit $y$ close to $p$, and there exists a subsequence $x_{n_k}$ converges to $y$. So, $d(x_{n_k},p)<d(x_{n_k},y)+d(y,p)<\varepsilon$ which is a contradiction.

So, $B(p,r)\subset X-E$ and $E$ is closed.


问:
How can a polynomial ring be finitely generated?
I am struggling to see this:

The following are equivalent:
(1) $x \in B$ is integral over $A$;
(2) $A[x]$ is finite over $A$.

Can a polynomial ring be a module and how can a polynomial rings be finitely generated modules? They have powers so how can it be written as an A-linear combination?

答:
Consider $A=\mathbb Z$ and $B=\mathbb Z[\sqrt[3]{2}]$. Then $B$ a finitely generated $A$-module: $\{1, \sqrt[3]{2}, \sqrt[3]{4}\}$ is a generating set.

Note that $B$ is not a polynomial ring; it is the ring of polynomial expressions in $\sqrt[3]{2}$, which is not the same thing because all powers greater than $2$ can be reduced.


问:
How many numbers between 1 and 10,000,000 don't have the sequence 12? Inclusion-exclusion problem
I got the following question:

How many numbers between 1 and 10,000,000 don't have the sequence 12? This is an inclusion-exclusion problem. Sadly I didn't fully understand its concept, so I tried solving it logically. I would like to know if my solution is correct, and if possible, how to solve it using the inclusion-exclusion principle.

Thanks!

So I said lets first find the numbers that do include the sequence 12, and divided it into cases:

Between 1 and 99: $1\cdot 1=1$

Between 100 and 999: $1\cdot 1 \cdot 10$ + $9\cdot 1 \cdot 1=19$

Between 1000 and 9999: $1 \cdot 1 \cdot 10 \cdot 10 + 2\cdot 9 \cdot 10 \cdot 1 \cdot 1 = 280$

Between 10000 and 99999:  $1 \cdot 1 \cdot 10 \cdot 10 \cdot 10 + 3\cdot 9 \cdot 10 \cdot 10\cdot 1\cdot 1=3700$

Between 100000 and 999999: $1 \cdot 1 \cdot 10 \cdot 10 \cdot 10\cdot 10 + 4\cdot 9 \cdot 10 \cdot 10\cdot 10\cdot 1\cdot 1=46000$

Between 1000000 and 9999999: $1 \cdot 1 \cdot 10 \cdot 10 \cdot 10\cdot 10\cdot 10 + 5\cdot 9 \cdot 10 \cdot 10\cdot 10\cdot 10 \cdot 1\cdot 1=550000$

So the total of numbers that don't include the number 12 is:
10000000-550000-46000-3700-280-19=9400001

I don't havea final answer so I don't know if that correct. Obviousy this solution lacks elegance, so I would like to know how solve this problem using the inclusion-exclusion principle.

Thank in advance!

答:
So you have at most a 7-digit number (since 10,000,000 isn't a possibility): X-X-X-X-X-X-X, and the 12 can be in 6 different positions, so you have 1-2-X-X-X-X-X, X-1-2-X....,X-X-X-X-X-1-2. That is a total of $6*10^5 = 600,000$ different numbers. Now the hard part like I said above is finding the overlaps.

Suppose 12 were to appear twice, that means that we have 3 open spaces otherwise. How many ways can we reorder the two 12's and the extra 3 spots? That would be ${5\choose2} = 10$. Since the open spots can be $10^3$ different numbers, we end up with $10*10^3 =10,000$ overlaps.

Suppose 12 were to appear three times, that means we have 1 open space otherwise. How many ways can we reorder the three 12's and the extra spot? ${4\choose1} = 4$. Since the open spots can be 10 different numbers, we end up with $4*10 = 40$ overlaps.

However, if 12 were to appear three times, that means 12 also appears 2 times, and so instead of have $10,000$ overlaps (from the first overlap case), we actually have $10,000 - 40 = 9960$ total overlaps.

With $10,000,000$ different numbers, we end up with (total numbers - total overlaps) = $10,000,000 - (600,000-9960) = 9,409,960$ which I verified with Java.


问:
Value of the series $\sum _{n=0}^\infty e^{-\lambda n}$
Is it possible to calculate the value of
$$\sum _{n=0} ^ \infty e^{-\lambda n}$$
for $\lambda \in \mathbb{R}$?

I have to premise that I don't know if this problem is solvable. It is easy to prove that the series converges (by integral comparison or asymptotic comparison between $e^{-\lambda n}$ and $n^{-2}$), but the difficult question is that is it possible to calculate the value of this series.

答: Set $x= e^{-\lambda}$, your series is $\sum _{n=0}^\infty x^n= {1\over 1-x}$ =${1\over 1-e^{-\lambda}}$


答:
A more direct but conceptual way is the following. Suppose $(x_n)\in X^{\mathbb{N}}$ is a sequence in the metric space $X$. Denote $L$ the set of all limits of the subsequences of $(x_n)$. Then
\begin{align}
L = \bigcap_{k\geqslant 0}\overline{\{x_n ~|~ n \geqslant k\}}
\end{align}
It is an intersection of closed subset, thus it is a closed subset.

This is because a point $l$ is in $L$ if and only if for every $\varepsilon >0$ and every $n \geqslant 0$, there exists $k \geqslant n$ such that $d(l,x_k) \leqslant \varepsilon$.


问:
Coprime cofactors of n'th powers are n'th powers, up to associates, for Gaussian integers & UFD
I'm trying to do the following exercise from Neukirch's Algebraic Number Theory (exercise 2, $\S 1$, chapter 1):

Show that, in the ring $\mathbb{Z}[i]$, the relation $\alpha\beta=\varepsilon\gamma^n$, for $\alpha,\beta$ relatively prime numbers and $\varepsilon$ a unit, implies $\alpha=\varepsilon^{\prime}\xi^n$ and $\beta=\varepsilon^{\prime\prime}\eta^n$, with $\varepsilon^{\prime},\varepsilon^{\prime\prime}$ units.

My attempt: once $\alpha$ and $\beta$ are relatively prime, $\alpha\nmid\beta$ and $\beta\nmid\alpha$. Therefore $N(\alpha)$ and $N(\beta)$ are relatively prime too (where $N(a+ib)=a^2+b^2$). Therefore,
$$N(\alpha\beta)=N(\alpha)N(\beta)=N(\varepsilon)N(\gamma^n)=N(\gamma)^n=(p_1\cdot\ldots\cdot p_r)^n=p_1^n\cdot\ldots\cdot p_r^n$$

And once $N(\alpha)$ and $N(\beta)$ are relatively prime, we have (if necessary, we reorder the $p_i$'s):

$$N(\alpha)=p_1^n\cdot\ldots\cdot p_k^n=(p_1\cdot\ldots\cdot p_k)^n=N(\xi)^n=N(\xi^n)$$
and
$$N(\beta)=p_{k+1}^n\cdot\ldots\cdot p_r^n=(p_{k+1}\cdot\ldots\cdot p_r)^n=N(\eta)^n=N(\eta^n)$$

And the result follows.

But I'm not sure about one step. It is $\alpha$ and $\beta$ relatively prime $\Rightarrow$ $N(\alpha)$ and $N(\beta)$ relatively prime. I know that if $\alpha|\beta$ then $N(\alpha)|N(\beta)$, but the previous assertion is not necesseraly true, and if it is, and don't know how to prove it. If it is true, the result follows, unless I made something wrong in the rest of the proof.

That's it. If you know another way of doing the exercise, please show me.

答: Assume for now that $\gamma$ is a non zero, non unit Gaussian integer ($\gamma=0$ is trivial and I would be repeating the comment above for when $\gamma$ is a unit). Since $\mathbb{Z}[i]$ is a UFD, $\epsilon\gamma=\epsilon\prod_{i}p_i^{a_i}$. Therefore $\alpha\beta=\epsilon^n\prod_{i}p_i^{na_i}$. Now we reorder the primes (they share no common factors up to units so we are safe to do this since none of the primes in the products clash. I hope this makes sense, but basically we can do this because they are coprime) and use the fact that any unit$^n$ is still a unit in the Gaussian integers to obtain the following unique factorisations: $$\alpha=\epsilon'\prod_j p_j^{na_j}=\epsilon'\left(\prod_j p_j^{a_j}\right)^n=\epsilon'\xi^n$$ and $$\beta=\epsilon''\prod_k p_k^{na_k}=\epsilon''\left(\prod_k p_k^{a_k}\right)^n=\epsilon''\zeta^n$$


答:
Allow me to give you an indication of a relation that holds in very general settings. Consider a first countable topological space $(X, \mathscr{T})$, in other words a space in which every point $x \in X$ admits a countable neighbourhood base.

Given an arbitrary family $x$ of objects (sets, ultimately) indexed by set $I$ and an arbitrary subset $J \subseteq I$, let us write $x_J\colon=\{x_i\}_{i \in J}$ for the set of components of indices $i \in J$ of the family $x$. For arbitrary natural number $m \in \mathbb{N}$, let us denote the unbounded natural interval of all natural numbers $n \geqslant m$ by $[m, \rightarrow)$.

Given an arbitrary sequence of points $x \in X^{\mathbb{N}}$, it is not difficult to show that the set of limits of subsequences of $x$ is equal to the set of points adherent to the sequence $x$, set which is by definition expressed as $\displaystyle\bigcap_{n \in \mathbb{N}}\overline{x_{[n, \rightarrow)}}$. As an intersection of closed subsets, this subset is itself closed.

Metric spaces are in particular first countable (for any point $x \in X$ the collection $\left\{\mathrm{B}_q(x)\right\}_{q \in \mathbb{Q_{+}^{\times}}}$ of open balls centered at $x$ and of rational radius is a neighbourhood base), so the above statement particularly applies.


答:
Your sentence in the comment does not quite make sense: "It is obvious I can write $x$ equals to a polynomial with degree $n$".

But on the other hand, suppose you know that $x$ is a root of a monic polynomial over $A$, which means that $x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 = 0$ where the lower term coefficients $a_0,...,a_{n-1}$ are all in $A$.

Then it is obvious that you can write $x^n$ equal to a polynomial of degree $\le n-1$, namely $$x^n = - a_{n-1} x^{n-1} - ... - a_1 x - a_0$$ And then an easy induction shows for each $k \ge n$ that $x^k$ is also equal to a polynomial of degree $\le n-1$. For instance, in the next step of the induction multiply both sides of that equation by $x$, so the new equation has $x^{n+1}$ on the left hand side, and the leading term on the right hand side will be $-a_{n-1} x^n$, and then use the above equation to substitute for $x^n$ on the right hand side.

It follows that $A[x]$ is finitely generated over $A$, the generating set being $1,x,....,x^{n-1}$.

Title: How can a polynomial ring be finitely generated?

Body:
I am struggling to see this:

The following are equivalent:
(1) $x \in B$ is integral over $A$;
(2) $A[x]$ is finite over $A$.

Can a polynomial ring be a module and how can a polynomial rings be finitely generated modules? They have powers so how can it be written as an A-linear combination?


问:
If we anti-differentiate the Laurent series $ \sum_{n=2}^{\infty} a_n(z - a)^{-n}$, how is the radius of convergence affected?
Let $a, a_n \in \mathbb{C}$ ($n \ge 2$). Suppose we are given the Laurent series

$$ \sum_{n=2}^{\infty} a_n(z - a)^{-n}.$$

I know that we can calculate the set on which this expression converges using Hadamard's formula. That is, we have that the series converges for $|z - a| > \frac{1}{R}$, where $\frac{1}{R} = \limsup \sqrt[n]{|a_n|}$. In case $\limsup \sqrt[n]{|a_n|} = 0$, the series converges for $|z-a|> 0$. And in case $\limsup \sqrt[n]{|a_n|} = \infty$, the series is nowhere convergent.

To continue, suppose that we formally anti-differentiate our series to obtain

$$ \sum_{n=1}^{\infty} \frac{-a_{n+1}}{n}(z - a)^{-n}.$$

My question is, how does the radius of convergence of this new series relate to the old one? Maybe the new radius of convergence is potentially larger? I see that the question boils down to just being smart about Hadamard's formula. We need to compare $\limsup \sqrt[n]{|a_n|}$ with $\limsup \sqrt[n]{\frac{|a_{n+1}|}{n}}$, but this seems like a tricky bit of analysis. It's already clear to me that $\limsup \sqrt[n]{\frac{|a_{n+1}|}{n}} \le \limsup \sqrt[n]{|a_{n+1}|}$, so my hope is that I can somehow compare $\limsup \sqrt[n]{|a_{n+1}|}$ with $\limsup \sqrt[n]{|a_n|}$.

Hints or solutions are greatly appreciated.

答:
Ok, I think with the hint from @mrf I am able to answer my own question. First, I was able to establish the following lemma--and I think it's worth trying to prove this for yourself:

Suppose that $a_n \ge 0, b_n > 0$ with $b_n \to \alpha > 0$. Then

$$ \limsup \frac{a_n}{b_n} = \frac{\limsup a_n}{\alpha}.$$

With this lemma in hand, the following equality is proved, and we proceed to establish the inequality:

$$ \limsup \frac{|a_{n+1}|^{\frac{1}{n}}}{n^{\frac{1}{n}}} = \limsup |a_{n+1}|^{\frac{1}{n}} \le \limsup |a_n|^{\frac{1}{n}}.$$

Hence the radius of convergence of the anti-derivative is at least as large as that of the original Laurent series.

Notice that if $ \limsup |a_n|^{\frac{1}{n}} = \infty $ there is nothing to show. Hence we may assume that $ \limsup |a_n|^{\frac{1}{n}} = M < \infty $. Hence each $ \displaystyle \sup_{m \ge n} \{|a_{m+1}|^{\frac{1}{m+1}}\} = M_n < \infty$. And this in turn says that, for each $n$, we have:

$$\sup_{m \ge n} \{|a_{m+1}|^{\frac{1}{m(m+1)}}\}\le \sup_{m \ge n} \{|a_{m+1}|^{\frac{1}{n(m+1)}}\} = M_n^{\frac{1}{n}} \le  \sup_{m \ge n} \{|a_{m+1}|^{\frac{1}{m+1}}\} = M_n < \infty.$$

Now, using the hint given by @mrf, we get

$$0 \le \sup_{m \ge n}\{|a_{m+1}|^{\frac{1}{m}}\} \le M_n \cdot M_n^{\frac{1}{n}} \to M.$$

I think this does the trick. Let me know if I've messed something up.


问:
My proof about $\text{sup}(AB) = \text{inf}(A) \text{inf}(B)$ for real negative nonempty sets $A,B$
Given two nonempty bounded below sets $A,B\subset \mathbb R$ of negative numbers.

Define $AB=\left\{ab:a \in A,b \in B\right\}$

I concluded that $$\text{sup}(AB) =  \text{inf}(A)  \text{inf}(B)$$

$$ab \le \text{inf}(A)b \le\text{inf}(A)\text{inf}(B)$$

So $AB$ is a nonempty subset of $\mathbb R$ bounded above and so its supremum does exist, by the minimality of supremum :

$$  \text{sup}(AB) \le \text{inf}(A)\text{inf}(B)\tag{I}$$

On the other hand:

$$\text{sup}(AB)\ge ab \iff \frac{1}{a}\text{sup}(AB)\le b$$
Which is true for all $a \in A,b \in B$,from the definition of infimum:
$$\frac{1}{a}\text{sup}(AB)\le \text{inf}(B) \iff \frac{1}{\text{inf}(B)}\text{sup}(AB)\le a$$

Again from the definition of infimum:

$$\frac{1}{\text{inf}(B)}\text{sup}(AB)\le \text{inf}(A) \iff \text{sup}(AB) \ge  \text{inf}(A)  \text{inf}(B)\tag{II}$$

From $\text{(I)}$ and $\text{(II)}$ the result follows.

The result looks like the relation $$\text{sup}(AB) =  \text{sup}(A)  \text{sup}(B)$$ for two positive nonempty real subsets $A,B$.

Can someone check the validity of my proof?

答:
Yeah, it looks alright, in terms of logic, but can be improved in terms of style and exposition. Here are some (opinionated) remarks:

The proof can be much, much shorter. If you have a clear audience in mind, then you'll have a clear(er) idea of what details to keep and not.

Indeed, a short proof leaves less doubt as to its validity!

You don't need completeness axiom here, just show the RHS satisfies the two conditions to be a least upper bound! (I.e., the part before equation (I) doesn't add anything.)

It is much better to tell us where $a$ and $b$ live before, rather than after the inequalities they apply to. In fact, you can even fix them at the beginning!

An alternative proof.

I also think with sup/inf questions, choosing the right tool makes things much more straightforward  (I like sequences myself, but the original definition), if you have the lemmas to hand

You showed $s:= \inf(A)\inf(B)$ is an upper bound for $AB$, that was quite nice (although it would be nice if you made it clear exactly where you used that $b$ negative and $a$ negative!). It too may be shortened, because content wise there isn't much to say!

From there on, you just need to show it is a least upper bound (i.e. supremum) — no need to revoke the completeness axiom.

Way using sequences

I absolutely love this result. If you've seen it before, you can use it, in combination with taking limits in inequalities.

For a non-empty set $C \subset \mathbb R$, there always exists a sequence $(x_n)_{n=1}^∞\subset C$ such that $$(x_n)\to \inf (C).$$
(This applies even if $\inf(C) = -∞$, and for $\sup$ in place of $\inf$)

You may well have seen it in an equivalent form, like, for $C \subset \mathbb R$ if the infimum $\inf(C)$ "exists"/is finite; then for all $\epsilon>0$, there is an element of $C$ inside the interval $$(\inf(C) - \epsilon, \inf(C)].$$

In any case, we can prove the following claim:

Claim: For all $U$ upper bounds on $AB$, $$U \geq s.$$

Proof: Let $U$ be an upper bound on $AB$, and let

$(a_n)_{n=1}^∞$ be a sequence of elements in $A$ which converges to $\inf(A).$

$(b_n)_{n=1}^∞$ be a sequence of elements in $B$ which converges to $\inf(B).$

Then

$(a_nb_n)_{n=1}^∞$ converges to $\inf(A)\inf(B) = s$, by the product rule; and

for all $n \in \mathbb N$, $a_nb_n \in AB$,$\ $ so $$U \geq a_nb_n.$$

Combining these two:

$$a_nb_n \leq U \implies \lim_{n\to ∞}(a_nb_n) \leq U,\qquad\text{i.e. }s \leq U.\text{ QED}$$

Everything follows from this: not only is $s$ is not only an upper bound, but the least upper bound, i.e. supremum.)


问:
An estimator for the Exponential distribution with $\lambda\le 10$
I'm trying to solve these two question below:

For the following experiments, define a statistical model and check
whether the parameter of interest is identified.

a. One observes $n$ i.i.d. Poisson random variables with unknown parameter $\lambda$.

b. One observes $n$ i.i.d. exponential random variables with parameter $\lambda$, which is unknown but a priori known to be no larger than
$10$.

I have the following questions:

What does he mean by "parameter of interest", an estimator with no bias or any estimator converging to the parameter is a valid one? in this case for any question asking an estimator of a parameter, that's enough to use WLLN and mapping theorem.

for the item a, using WLLN:

$$\bar X_n\xrightarrow{P}\lambda$$

and $\bar X_n$ is unbiased because $E[\bar X_n]=\frac{X_1+\ldots+X_n}{n}=\frac{n\lambda}{n}=\lambda$

The last part I found a little tricky.

Using the WLLN, mapping theorem and the fact $E[X_i]=1/\lambda$, I showed $$T_n=\frac{1}{\bar X_n}\xrightarrow{P}\lambda$$

But I'm having problems to find the bias of this estimator (I know this $E[\bar X_n]=\frac{1}{E[\bar X_n]}$is not always true by Jensen's inequality) and I don't know how to use the fact $\lambda\le 10$.

答:
They simply ask you to define the Statistical Model for the two experiments.

The statisical model is definded through the following phases

Definition of the Base Model

Bernullian Sampling (with replacement)

1. Base Model

In your cases, the base model is the following:

$$\Big(X, p(x|\theta), \theta \in \Theta\Big)$$

2. Statistical Model

The given Sampling rule induces the following Stat Model

$$\Big(\mathcal{X}^{(n)}, \prod _{i=1}^{n}p(x_i|\theta), \theta \in \Theta\Big)$$

If the question is what you posted

For the following experiments, define a statistical model and check whether the parameter of interest is identified.

no calculations are needed

For the first experiment you have

$$\Big(\mathcal{X}^{(n)}, \frac{e^{-n\lambda}\lambda^{\Sigma_i X_i}}{\Pi_iX_i!}, \lambda>0 \Big)$$

Where

a. $\mathcal{X}^{(n)}$ is the set of all possible n-tuples where any element can takes the values $\{0,1,2,3,...\}$

b. The function $\frac{e^{-n\lambda}\lambda^{\Sigma_i X_i}}{\Pi_iX_i!}$, given the knowledge of $\lambda$, assigns the probability to any possible n-tuple.

c. $\lambda$ is the parameter of interest

Similar reasoning for the 2nd experiment


问:
Mechanics: Car approaching traffic lights
A driverless car registers that the traffic lights change to amber 40m ahead. The amber light is a 2s warning before turning red. The car is travelling at 17m/s and can accelerate at 4m/s^2 or brake safely at 8m/s^2. What options does the car have?</p>

答:
Assumming acceleration is constant you can work out how far it would travel when accelerating or decelerating.

For acceleration:
Using SUVAT to get: $$v=u+at=25$$ then S=2(17m/s + 25m/s)/2=42 (S=t(u+v)/2)
As displacement>40 then car will be able to travel the 40m before the light changes to red

For deceleration:
Using SUVAT $S=ut+\frac{1}{2}at^2$ => $$0=4t^2 -17t+40$$
As no real solutions cannot decelerate in time.


问:
Use stirlings approximation to find $\binom{m+n}{n}$ when $m$ and $n$ are both large
The question says :

Use Stirling's approximation to estimate  $\binom{m+n}{n}$ when $m$ and $n$ are both large.

I did till following but not sure what's next step to do.

(Stirling's approximation is $n!$ ~ $\sqrt{2\pi n}(\frac{n}{e})^n$) .

So, $\binom{m+n}{n}=\frac{(m+n)!}{(m!)(n!)}$ ~ $\frac{\sqrt{2\pi(m+n)}(\frac{m+n}{e})^{m+n}}{\sqrt{2\pi m}(\frac{m}{e})^{m}\sqrt{2\pi n}(\frac{n}{e})^n}=\sqrt{\frac{1}{2\pi m}+\frac{1}{2\pi n}}(1+\frac{n}{m})^m(1+ \frac{m}{n})^n$.

What should be the next step. Thanks.

答:
$$A=\binom{m+n}{n}=\frac{(m+n)!}{m!\,n!}$$ Take logarithms
$$\log(A)=\log \Big[\binom{m+n}{n}\Big]=\log((m+n)!)-\log(m!)-\log(n!)$$

Now use Stirling approximation
$$\log(p!)=p (\log (p)-1)+\frac{1}{2} \log ( p)+\frac{1}{2} \log (2 \pi  )+\frac{1}{12   p}+O\left(\frac{1}{p^2}\right)$$ Apply it three times ignoring the higher order terms. Now, say that $m > n$ and continue with Taylor expansion for large $m$. This should give
$$\log(A)=n \log \left(\frac{m}{n}\right)+ \left(n-\frac{1}{12 n}-\frac{1}{2} \log (2 \pi  n)\right)+\frac{n (n+1)}{2 m}+O\left(\frac{1}{m^2}\right)$$

Use it for $m=100$ and $n=99$. The above twice truncated expansion would give $\log(A)=129.86$ while the exact value is $\log(A)=135.06$.


问:
How to convert cartesian inequalities to spherical inequalities?

I have no clue how this went from $x^2+y^2 <=3$ to whatever that other thing is. I'm looking at conversion this and this is all I'm finding:

And when I replace x with that squared and y with that other thing squared, I'm not getting the same result.

Any ideas how to get from those cartesian inequalities to those spherical inequalities? Thanks.

答:
You can just substitute for $x$ and $y$ and apply trigonometric identities. For $x^2+ y^2 \leq 3$ for example:

$$x^2+y^2 = (\rho \sin \varphi \cos \theta)^2 + (\rho \sin \varphi \sin \theta)^2 = \rho^2 \sin^2 \varphi (\cos^2 \theta + \sin^2 \theta)=\rho^2 \sin^2 \varphi \leq 3$$


问:
Continuity of a two variable function at$(0,0)$
I have to prove or disprove the continuity of the function at $(0,0)$ which is given by:

$$f(x,y)= \begin{cases}\frac{\sin^{-1}(x+2y)}{ \tan^{-1}(2x+4y)}&\text{ if }
 0<\sqrt{x^2+y^2} \\  \frac 12 &\text{ if } (x,y)=(0,0)\end{cases}.$$

I consider $x+2y=t$ where $t$ approaches to $0$. This reduces the two variable function to that of a single variable in $t$. The resulting function is:

$g(t)= \frac{\sin^{-1}(t)}{\tan^{-1}(2t)}$ for non-zero $t$ and $\frac12$ for $t=0$.

This function is continuous at $t=0$. Does it imply the continuity of the original two-variable function at$(0,0)$. What are the other ways of checking continuity at the said point?

答:
In this case the answer is yes. The continuity of $g$ ensures the continuity of $f$. Basically, we are considering the functions $h(x,y)=x + 2y$ and $g(t) = \frac{\sin^{-1}(t)}{\tan^{-1}(2t)}$ and just saying that
$f = g \circ h$. The results on  limits and continuity of composed functions guarantee your claim.


问:
compute the commutant of a compact self-adjoint operator
Suppose $T$ is a compact self-adjoint operator on $H$, then $T=\sum_{i=1}^{\infty}\lambda_i P_i$, where each $\lambda_i$ is an eigenvalue of $T$, $P_i$ is a projection from $H$ onto $ker(T-\lambda_i)$.

I want to calculate the commutant of $T$. Does the following relation hold?
$T'=\sum_{i=1}^{\infty}(\lambda_i P_i)'$

答:
Besides the non-zero eigenvalues of $T$, let us make sure to include $\lambda_0=0$, and let $P_0$ be the orthonormal projection onto
the null space of $T$.  Then clearly
$$
  T = \sum_{i=1}^{\infty}\lambda_i P_i = \sum_{i=0}^{\infty}\lambda_i P_i,
  $$
and also
$$
  H = \bigoplus_{i=0}^{\infty}P_i(H).
  $$

Notice that an operator $S$ commutes with $T$ iff $S$ commutes with every $P_i$:  the "if" part is obvious because
$T=\sum_{i=1}^{\infty}\lambda_i P_i$, and the  "only if" follows from the fact that each $P_i$ may be writen as a limit
of polynomials applied to $T$ by the functional calculus  (provided the $\lambda _i$'s are distinct).

Thus the commutant of $T$ coincides with the set of all operators leaving each $P_i(H)$ invariant.   Viewing these as
block-diagonal operators we then have
$$
  \{T\} ' = \bigoplus_i P_iB(H)P_i.
  $$