问:
How to sort a permutation cycle in the mapping form?
Consider a cycle $(6\;  1\;  4\; 3\; 5\; 2  )$ of a symmetric group $S_n$ of degree $n$. This implies $6$ maps to $1$, $1$ maps to $4$, $4$ maps to $3$, $3$ maps to $5$, $5$ maps to $2$ and $2$ maps to $6$. Therefore there is a $2 \times 6 $ associated matrix $$\left(\begin{array}{ll}
1\;2\;3\;4\;5\;6  \\ 4\;6\;5\;3\;2\;1
\end{array}
\right).$$    By using Mathematica, I want to write the same cycle in associated matrix form $$\left(\begin{array}{ll}
1\;2\;3\;4\;5\;6  \\ 4\;6\;5\;3\;2\;1
\end{array}
\right).$$
I have to do the same for each cycle $(a\;\;  b\;\;  c\;\; d\;\; e\;\; f  )$ of length $6$ in $S_6.$
 Please suggest Mathematica commands or provide a Mathematica code.

答:
After a nice discussion with Rohit Kumar, which can be found in the chat linked in the comments, we came up with a solution which I have extended in order to generate all desired permutations in a loop.

We use Permutations to generate all permutations of the set $\{1,\cdots,n\}$ and use ToCycles in order to convert each permutation to cycle notation. If the length is equal to $n$, we have found a match Here is the code:

```<code>f[n_Integer]:= Module[{l={}},
    Do[
      If[Length[j]==n,AppendTo[l,{Range[n], i}]], {i, Permutations[Range[n]]}, {j,ToCycles[i]}];
{l}
]
</code>```

Edit: You might need to use

Needs["Combinatorica`"] // Quiet

at the very beginning.

Example: Let us use $n=4$ for an easier example. We simply call the above function: $f[4]$ which yields to

{{{{1,2,3,4},{2,3,4,1}},{{1,2,3,4},{2,4,1,3}},{{1,2,3,4},{3,1,4,2}},{{1,2,3,4},{3,4,2,1}},{{1,2,3,4},{4,1,2,3}},{{1,2,3,4},{4,3,1,2}}}}

Converting each entry via MatrixForm to a visiualized matrix representation yields to:

$$
\begin{pmatrix}
1 &2 &3 &4  \\
2 & 3 & 4 & 1
\end{pmatrix}
\begin{pmatrix}
1 &2 &3 &4  \\
2 & 4 & 1 & 3
\end{pmatrix}
\begin{pmatrix}
1 &2 &3 &4  \\
3 & 1 & 4 & 2
\end{pmatrix}
$$

$$
\begin{pmatrix}
1 &2 &3 &4  \\
3 & 4 & 2 & 1
\end{pmatrix}
\begin{pmatrix}
1 &2 &3 &4  \\
4 & 1 & 2 & 3
\end{pmatrix}
\begin{pmatrix}
1 &2 &3 &4  \\
4 & 3 & 1 & 2
\end{pmatrix}
$$

With corresponding cyclic notations:

$$(2\ 3\ 4\ 1), \ (2\ 4\ 3\ 1), (3\ 4\ 2\ 1), (3\ 2\ 4\ 1), (4\ 3\ 2\ 1), (4\ 2\ 3\ 1)$$

Notice that there are exactly $(n-1)!$ solutions.

For $n=6$, $f[6]$ yields to the following (truncated) output:

{{{{1,2,3,4,5,6},{2,3,4,5,6,1}},{{1,2,3,4,5,6},{2,3,4,6,1,5}},{{1,2,3,4,5,6},{2,3,5,1,6,4}},{{1,2,3,4,5,6},{2,3,5,6,4,1}},
⋯112⋯
,{{1,2,3,4,5,6},{6,5,2,1,4,3}},{{1,2,3,4,5,6},{6,5,2,3,1,4}},{{1,2,3,4,5,6},{6,5,4,1,3,2}},{{1,2,3,4,5,6},{6,5,4,2,1,3}}}}

Using Length on $f[6]$ yields to $120=(n-1)!=(6-1)!=5!$.

Body:
After a nice discussion with Rohit Kumar, which can be found in the chat linked in the comments, we came up with a solution which I have extended in order to generate all desired permutations in a loop.

We use Permutations to generate all permutations of the set $\{1,\cdots,n\}$ and use ToCycles in order to convert each permutation to cycle notation. If the length is equal to $n$, we have found a match Here is the code:

f[n_Integer]:= Module[{l={}},
    Do[
      If[Length[j]==n,AppendTo[l,{Range[n], i}]], {i, Permutations[Range[n]]}, {j,ToCycles[i]}];
{l}
]

Edit: You might need to use

Needs["Combinatorica`"] // Quiet

at the very beginning.

Example: Let us use $n=4$ for an easier example. We simply call the above function: $f[4]$ which yields to

{{{{1,2,3,4},{2,3,4,1}},{{1,2,3,4},{2,4,1,3}},{{1,2,3,4},{3,1,4,2}},{{1,2,3,4},{3,4,2,1}},{{1,2,3,4},{4,1,2,3}},{{1,2,3,4},{4,3,1,2}}}}

Converting each entry via MatrixForm to a visiualized matrix representation yields to:

$$
\begin{pmatrix}
1 &2 &3 &4  \\
2 & 3 & 4 & 1
\end{pmatrix}
\begin{pmatrix}
1 &2 &3 &4  \\
2 & 4 & 1 & 3
\end{pmatrix}
\begin{pmatrix}
1 &2 &3 &4  \\
3 & 1 & 4 & 2
\end{pmatrix}
$$

$$
\begin{pmatrix}
1 &2 &3 &4  \\
3 & 4 & 2 & 1
\end{pmatrix}
\begin{pmatrix}
1 &2 &3 &4  \\
4 & 1 & 2 & 3
\end{pmatrix}
\begin{pmatrix}
1 &2 &3 &4  \\
4 & 3 & 1 & 2
\end{pmatrix}
$$

With corresponding cyclic notations:

$$(2\ 3\ 4\ 1), \ (2\ 4\ 3\ 1), (3\ 4\ 2\ 1), (3\ 2\ 4\ 1), (4\ 3\ 2\ 1), (4\ 2\ 3\ 1)$$

Notice that there are exactly $(n-1)!$ solutions.

For $n=6$, $f[6]$ yields to the following (truncated) output:

{{{{1,2,3,4,5,6},{2,3,4,5,6,1}},{{1,2,3,4,5,6},{2,3,4,6,1,5}},{{1,2,3,4,5,6},{2,3,5,1,6,4}},{{1,2,3,4,5,6},{2,3,5,6,4,1}},
⋯112⋯
,{{1,2,3,4,5,6},{6,5,2,1,4,3}},{{1,2,3,4,5,6},{6,5,2,3,1,4}},{{1,2,3,4,5,6},{6,5,4,1,3,2}},{{1,2,3,4,5,6},{6,5,4,2,1,3}}}}

Using Length on $f[6]$ yields to $120=(n-1)!=(6-1)!=5!$.


问:
Jacobson radical of unitary subring
This question is taken from "Graduate Course in Algebra for Martin Isaacs" which asking whether a statement is true or not.
The question says
if $S \subset R$ is a unitary subring (so it contains the same unity $1$ as in $R$), then determine whether the following statements is true or not:

(a) $J(S) \subset J(R)$

(b) $J(R) \cap S\subset J(S)$.

Here, $J(S)$ is the intersection of all maximal ideals in $R$.
So I think (a) is wrong because the passing the intersection reverse the inclusion, right?
I don't know if I am right and if so, is there a counterexample support this?

In (b), I believe it is true, as the inclusion reversed.

答:
a) Consider any local domain $S$ with nonzero maximal ideal, and let $R$ its field of fractions.

b) Consider $S=F[x]$ and $R=F[x]_{(x)}$ ($F[x]$ localized at the prime ideal $(x)$).

ContentLicense: CC BY-SA 4.0

Title: Jacobson radical of unitary subring


问:
sum of the eigenvalues = trace($A$)?
Is it true that for a square matrix $A$, all of whose eigenvalues exist in the base field, sum of the eigenvalues = trace($A$)?

The result holds in all the matrices I've studied.

答: Yes. Just look at the characteristic polynomial (say of degree n). Trace=-the coefficient of the term of $x^{(n-1)}$ which is also the sum of the roots of the characteristic polynomial (the coefficient of the term $x^{(n-1)}$ of any monic polynomial of degree $n$ is the sum of its roots with a minus sign.).</p>

ContentLicense: CC BY-SA 3.0

Title: sum of the eigenvalues = trace($A$)?


问:
Hint for solving this problem about constant function on $\mathbb{R^2}$
Can anyone give me some hints to solve this problem?

Assume Lebesgue measure on $\mathbb{R}^2$ and $\mathbb{R}$. Suppose that $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ is a measurable function such that for almost all $x_1 \in \mathbb{R}$ the function $t \rightarrow f(x_1,t)$ is constant and also for almost all $x_2 \in \mathbb{R}$ the function $s \rightarrow f(s,x_2)$ is constant. Show that the function $f$ is constant almost everywhere .

答:
Let $E=\{x\in\mathbb{R}: t\mapsto f(x,t)\text{ is constant}\}$ and $F=\{y\in\mathbb{R}: t\mapsto f(t,y)\text{ is constant}\}$. Then $E,F$ are conull sets, i.e. $m(\mathbb{R}\setminus E)=m(\mathbb{R}\setminus F)=0$. Now take the cartesian product $E\times F$. Note that
$$\mathbb{R}^2\setminus (E\times F)=(\mathbb{R}\setminus E)\times(\mathbb{R}\setminus F)\cup E\times(\mathbb{R}\setminus F)\cup(\mathbb{R}\setminus E)\times F $$
and this is a disjoint union. From this we conclude that $m_2(\mathbb{R}^2\setminus (E\times F))=0$ (because when taking product measures we make the convention $0\cdot\infty=0$) so It suffices to show that $f$ is constant when restricted to $E\times F$.

Let $(x,y)$ and $(x',y')$ be points in $E\times F$. Therefore we have that $f(x,y)=f(x,t)$ for all $t\in\mathbb{R}$, so $f(x,y)=f(x,y')$. But $f(x,y')=f(t,y')$ for all $t\in\mathbb{R}$, so $f(x,y')=f(x',y')$, so we have that $f(x,y)=f(x',y')$.


问:
Help with writing a proof about $\liminf$ and $\limsup$
I need to prove

$\liminf a_n+\liminf b_n≤\liminf(a_n+b_n)$

In trying to prove this thing, I’ve come up with the following proof:

$\limsup a_n+\limsup b_n \ge \limsup(a_n+b_n)$

$A+B \ge L - \epsilon$

Where $A= \limsup( {a_n: n>N})$, $B= \limsup {b_n: n>N})$, $L- \epsilon = \limsup ({a_n +b_n})$, and $S_N= \sup ({a_n +b_n: n>N})$

Now for every $\epsilon>0$ there exists $M>0$
s.t. if $N>M$ then  $|\sup(a_n+b_n: n>N)-L|<\epsilon$ $\implies$ $S_N> L-\epsilon$ for every $N>M$ which means there exists $n>N$ s.t. $a_n+b_n>L-\epsilon$

I know this proof isn't the one I need but it has to be similar. I could use some help with proving $\liminf a_n+\liminf b_n≤\liminf(a_n+b_n)$.

答:
Since for all $m\geq n$,
$$\inf_{k\geq n} a_k+\inf_{k\geq n}b_k\leq a_m+b_m,$$
you get $$\inf_{k\geq n} a_k+\inf_{k\geq n}b_k\leq \inf_{k\geq n}(a_k+b_k).$$
Taking $n\to \infty $ gives the wished result.


问:
Find the supremum and the infimum of $A=\{x \in \Bbb R \mid 2x^2 < x^3 + x\}$.
Find the supremum and the infimum of $A=\{x \in \Bbb R \mid 2x^2 < x^3 + x\}$.

Attempt:
\begin{align*}
A &= \{x \in \Bbb R \mid 2x^2 < x^3+x \}
\\
&= \{x \in \Bbb R \mid x(x^2-2x+1) > 0\}
\\
&= \{x \in \Bbb R \mid x(x-1)^2 > 0 \} \\
&= (0,1) \cup (1, \infty).
\end{align*}

Clearly, $\sup(A)$ doesn't exist since $A$ doesn't have an upper bound.

Now, for all $x \in A, x > 0$ and so, $0$ is a lower bound of $A$. Let $m$ be a lower bound of $A$. Want to show that $m\le 0$. Suppose that $m>0$ with $m \ne 1$. Let $r$ be a rational number with $0 < r < m$. Then, $r \in A$ and $m>r$,  contradiction, since $m$ is a lower bound of $A$. Therefore, $m \le 0$ and so, $\inf(A)=0$.

Am I true?

答:
Shortcut approach.

As your work indicates, the inequality is satisfied if and only if $(x)(x-1)^2 > 0,$ and that this inequality will never be satisfied for either $x=1$ or $x \leq 0.$

All that remains is to show (or simply state) that for any $x> 0$ such that $x \neq 1$, each of the pertinent factors $x$ and $(x-1)^2$ must be strictly positive.


问:
Am I considered to be cancelling a solution in here when I did the another method which is squaring both sides or not?

Solve $2\ln(3x+5)=4$

The second method is:

$\ln(3x+5)^2=4$,

$\implies \ln(3x+5)=2$,

$\implies e^2=3x+5$,

$\implies x= \frac{e^2-5}{3}$

which equals only $ 0.796$, and there is not another answer (assuming if it is not rejected)
This is not an exam's question; I already have the answer and know it, but what I am asking now is that: when I squared both sides, am I considered to be canceling another solution, which is the $-4.13$ (assuming if it was not to be rejected as it is known that no $\ln$ for negative)? And if yes, I am considered to be canceling a solution; why is this possible?
And here is a picture of the whole thing I explained to be clearer.

答:
I don't know any situation in which I would say that squaring an equation "cancels" a solution. It sometimes introduces extraneous "solutions" that you have to reject.

I suppose you could say that squaring sometimes forces you to "cancel a solution" later in the process.

In this case the rule "there is no $\ln$ of a negative number" does not apply,
since the thing you apply $\ln$ to is $3x + 5,$ which is positive as long as
$x > -5/3.$
Instead, you reject the negative square root because it implies that
the output of $\ln$ is negative, and in this case you know that
$\ln(3x+5)>0$ since it is multiplied by a positive number and produces a positive number.

So in the end you do "cancel a solution", not at the time when you square
$\ln(3x+5),$ but rather at the time when you take the square root of $4$
and write only the positive square root.


问:
How to fit a smooth polynomial?
Previously asked on Stats.SE

We have $N$ samples of the unknown function $f(x)$ on the finite interval $[a, b]$. The samples are subject to white noise of known variance. We want to approximate the function $f(x)$ by a polynomial $P_m(x)$ of order $m$. It is, however, known, that the original function $f(x)$  is smooth in some sence. For example, let's say we know that the original $f(x)$ is Lipschitz-continuous on the interval $[a, b]$ with some known constant $K$. Is there a standard procedure to select the best-fitting polynomial $P_m(x)$ while requiring that the polynomial satisfies the smoothness requirement.

I am not bound to Lipschitz continuity. I am happy to switch to another metric of smoothness if it is more convenient for this problem. However, the smoothness metric should be minimax. This means that the bound must apply to the smoothness of the worst point of the function, which is not necessarily the average such as L2.

Note: The question is about how to solve this problem in practice. Namely:

Is there an analytical solution?

Is there a known algotithm used in practice to solve this problem?

Recommend an existing library that can perform this sort of fitting.

答:
This can be done with standard Linear Regression tools. So let the polynomial be given by $p(x) = \sum_{k=0}^m w_k x^k$. We use the fact that for smooth functions, the Lipschitz constant is equal to maximum-norm of the derivative. Hence,  we need to find an upper bound for $\|p'\|_{L^{\infty}}$ on $[a,b]$ w.r.t. the coefficients $w_k$.

Towards this end, it is useful to expand $p'$ in terms of Chebyshev polynomials of the first kind, since these are minimal w.r.t. the $\infty$-norm. We have:

$$\|w_k T_k(x)\|_{L^\infty([-1,+1])} \le \sum_{k}|w_k|\cdot\|T_k(X)\|_{\infty} = \sum_{k}|w_k| = \|w\|_1$$

Thus, you can try to solve your problem with a cleverly constructed LASSO method, using the linear model

$$
\hat y(x, w)
= b +\sum_{k=0}^{m-1} w_k \int T_k(x)dx
= b + \sum_{k=0}^{m-1}w_ k\cdot \underbrace{\tfrac{1}{2} \Big({\tfrac {T_{k+1}(x)}{(k+1)}}-{\tfrac {T_{k-1}(x)}{(k-1)}}\Big)}_{=:\phi_k(x)}
= \phi(x)^T w + b
$$

Map the $x$-data into the interval $[-1, +1]$.

Construct the feature matrix $\Phi(X) = \begin{pmatrix}\phi_k(x_i)\end{pmatrix}_{ik}$

Solve the LASSO problem

$$(1)\qquad  \min_w \tfrac{1}{2N}\|\Phi(x)w + b - y \|_2^2 + \lambda \|w\|_1$$

Then, you are guaranteed that $\operatorname{Lip}(\hat y)=\|\hat y'\|_{L^\infty([-1,+1])} = \|\sum_{k=0}^{m-1}\hat w_k T_k(x)\| \le \|\hat w\|_1$. The degree of Lipschitz continuity is controlled by $\lambda$. By solving the whole Lasso path (i.e. over a range of values of $\lambda$) you can find the best fit that satisfies your condition. This functionality is provided for example in scikit-learn: Lasso and Elastic Net, Lasso Path

Title: How to fit a smooth polynomial?


问:
Group theory book recommendation
I am an undergraduate student and recently took an introductory course containing properties of  groups, homomorphism, normal subgroups, group action, sylows theorem and a bit about simple groups. Does anybody think starting the book "Finite groups: An introduction" by jean pierre serre be a bad idea with this amount of knowledge ?</p>

答:
Why not try one of the following:

John S. Rose, A Course on Group Theory
Derek J.S. Robinson, A Course in The Theory of Groups
Dummit and Foote, Abstract Algebra.
Each of these books has a lot of good examples and exercises!

Title: Group theory book recommendation


问:
Sub-algebra generating a $\sigma$-algebra is dense in this latter
Let $(\Omega,\mathscr{F},\mu)$ be a measured set with $\mu$ a probability measure.
Using $1$ as the indicator function we define a distance
$$ d(A,B) = \int_\Omega{|1_A - 1_B | d\mu} ~~ , ~~ A,B\in\mathscr{F}.$$

I wanted to know if we have a $\sigma$-algebra $\mathscr{A}$ which generates $\mathscr{F}$ then $\mathscr{A}$ is dense in $\mathscr{F}$ according to $d$.
I thought it would be useful to show that the set $\{ F \in \mathscr{F} | \forall \varepsilon > 0 \exists A \in \mathscr{A}, d(A,B) < \varepsilon \}$ is equal to $\mathscr{F}$.
One way is obvious.
The other one I don't see what to do. Any hint ?

Body: $d$ is not a metric. Indeed, take $\Omega =\{1,2,3,4\}$ and $\mu\{1\}=\mu\{2\}=\frac{1}{2}$. Then $d(\{3\},\{4\})=0$ but $\{3\}\neq \{4\}$.


问:
Non-self-mapping automorphism implies abelian
Suppose $\sigma\in\text{Aut}(G)$. If $\sigma^2=1$ and $x^{\sigma}\neq x$ for $1\neq x\in G$, show that if $G$ is finite, it must be abelian.

There's a hint to show that the set $\{x^{-1}x^{\sigma}\mid x\in G\}$ is the whole group $G$. I have already proved this hint ($x^{-1}x^{\sigma} = y^{-1}y^{\sigma}\rightarrow (xy^{-1})^{\sigma} = xy^{-1}\rightarrow x=y$ since $x^{\sigma}=x$ only for $x=1$), but I could not use it to solve the problem.

答:
For each $x\in G$ we have $(x^{-1}x^\sigma)^\sigma=(x^{-1})^\sigma x=(x^{-1}x^\sigma)^{-1}$, but $x^{-1}x^\sigma$ runs over $G$ as $x$ does, so for each $x\in G$ we have $x^\sigma=x^{-1}$. Thus, for any $x,y\in G$ we have

$$(xy)^\sigma=x^\sigma y^\sigma=x^{-1}y^{-1}=(yx)^{-1}=(yx)^\sigma$$

and hence $xy=yx$.


问:
A quite difficult limit: $\lim _{x\to 0}\frac{\exp(x^2)-\cos x-x\sin x}{\sinh^2x}$
I tried this limit with Taylor series. However, proceeding in this manner I obtained a series over another series. Furthermore, I do not know how to continue as long as we have $x$ tends to $0$. Were it to be $x$ tends to infinite, this limit would be simpler.

This is what I got

$$\begin{align}&
\lim _{x\to 0}\frac{\exp(x^2)-\cos x-x\sin x}{\sinh^2x} \\[6pt]
=\;&\frac{\frac12x^2+\frac{5}{8}x^4+\frac{23}{144}x^6+\frac{241}{5760}x^8+\frac{3359}{403200}x^{10}+\cdots}{{x^2+\frac{1}{3}x^4+\frac{2}{45}x^6+\frac{1}{315}x^8+\frac{2}{14175}x^{10}+\cdots}}
\end{align}$$
I can't continue.

答:
It is possible to use the Hopital rule even though, in order to obtain the result, it has to be used twice. For sake of personal preference I will expand the function $\sinh x$ as a sum of exponential functions.
$$\sinh^{2}x=\left(\frac{e^{x}-e^{-x}}{2}\right)^{2}=\frac{e^{2x}+e^{-2x}-2}{4}$$
By substitution, the following limit is obtained where, in the second step, the 4 has been factored out from the limit.
$$\lim_{x \to 0} \frac{e^{x^{2}}-\cos x-x\sin x}{\sinh^{2}x}=4\lim_{x \to 0} \frac{e^{x^{2}}-\cos x-x\sin x}{e^{2x}+e^{-2x}-2}=\left[\frac{0}{0}\right]$$
By obtaining an indeterminate form such as $\left[\frac{0}{0}\right]$ it is possible to use the Hopital rule, so by differentiating the numerator and the denominator we have:
$$4\lim_{x \to 0}\frac {2xe^{x^{2}}+\sin x-\sin x-x\cos x}{2e^{2x}-2e^{-2x}}=\left[\frac{0}{0}\right]$$
Reiterating the application of the same principle stated above we the following result is obtained.

$$4\lim_{x \to 0}\frac{2e^{x^{2}}+4x^{2}e^{x^{2}}+x\sin x-\cos x}{4e^{2x}-4e^{-2x}}=4\frac{2-1}{4+4}=\frac{1}{2}$$


问:
Cube question without coordinates, cross sections and planes
We have a cube with corners $ABCDA_1B_1C_1D_1$. The points $A_1B_1C_1D_1$ lie above the square $ABCD$. We also have a plane that goes through the middle of line $BC$, through the middle of square $ABA_1B_1$, through the middle of $A_1B_1C_1D_1$. At what ratio does the plane cut line $AB$.

What I did, I've put the cube into the coordinate system $B(0,0,1), A_1 (0,1,0) \text{ and } A(0,0,0)$, $D(1,0,0)$ and then it was pretty easy to solve. I just got the equation of the plane and then solved for intersection of the line that goes through AB, to get the point. Then I knew the ratio.

My question is, how to solve this problem without putting the cube into the coordinate system?

SOLUTION: solution should be that the plane cuts AB in ratio 3:1

答:
Fairly quickly:

(1) the plane (call it $\Pi$) also passes through the centre $M_{ABC_2D_2}$ of square $ABC_2D_2$ ($C_2,D_2$ are the reflection of $C,D$ in line $AB$).  This is just extending $M_{A_1B_1C_1D_1}M_{ABB_1A_1}$ to meet the bottom plane $ABCD$.

(2) So the line $M_{BC}M_{ABC_2D_2}$ lies in $\Pi$.  This line intersects AD at $X$ where $AX=\frac32 AD$ and $X$ is opposite to $D$.

Now look at the configuration $A,B,X,M_{BC}$ and you have similar (right-angled) triangles $\triangle AXY$ and $\triangle YBM_{BC}$ where $Y$ is the intersection of diagonals, i.e., intersection of $AB$ with $\Pi$.  So $AY:YB=AX:BM_{BC}=3:1$.


问:
On the substitution of inverse trigonometric functions
$$
\begin{array}{rcl}
V&=&\displaystyle{4\cdot2\int_{-1}^y2\sqrt{1-y^2}\ \mathrm{d}y} \\[0.7em]
 &=&\displaystyle{16\int_\frac{-\pi}{2}^{\arcsin(y)}\cos^2t \ \mathrm{d}t} \\[0.7em]
 &=&\displaystyle{\left[8\left(\arcsin(y)+y\sqrt{1-y^2}\ \right)+4\pi\right]}
\end{array}
$$

Here's an equation on my textbook. I don't know the details of the transformation between them, how the integral interval is transformed (I'm not very familiar with anti trigonometric functions)

答:
While the integral is correct as written, it's a bit confusing, because the $y$ in the upper limit of integration and the $y$ int the integrand are not the same variable.  In the upper limit, $y$ is an independent variable, representing the height of the liquid.  In the integrand, $y$ is a dummy variable, that we can replace by any other.

Doing the substitution is clearer if we first replace the dummy $y$ by some other dummy variable, say $u$.  $$V=16\int_{-1}^y\sqrt{1-u^2}\,\mathrm{d}u$$

Now make the substitution, $u=\sin t$.  Then $\mathrm{d}u=\cos t\mathrm{d}t$, $\sqrt{1-u^2}=\sqrt{1-\sin^2t}=\cos t$.  (Because $t=\arcsin u$, we have $-\frac\pi2\leq t\leq\frac\pi2$, and $\cos t\geq0$, so we don't need to take the absolute value.)

To transform the limits of integration, just note that when $u=-1$, $t=\arcsin(-1)=-\frac\pi2$ and when $u=y$, $t=\arcsin y$. It's this very last point that may be a bit unclear if we don't change the dummy variable.  It would say, "When $y=y$, ...".

The formula for integration by substitution says $$\int_a^bf(\phi(t))\phi'(t)\,\mathrm{d}t=\int_{\phi(a)}^{\phi(b)}f(u)\,\mathrm{d}u,$$ assuming all hypotheses are satisfied.  In this case, we have $$\begin{align}
f(u)&=\sqrt{1-u^2}\\
\phi(t)&=\sin t\\
a&=-\frac\pi2\\
b&=\arcsin y
\end{align}$$


问:
Finding the mgf of $X_1 - X_{(1)}$ from Basu's Theorem
Suppose that $X_1, \ldots, X_n$ where $n \geq 2$ are iid random variables with pdf
$$f_X(x) = e^{-(x-\theta)}I(x \geq \theta).$$

I need to find the moment-generating function of the statistic $U = X_1 - X_{(1)}$ where $X_{(1)}$ is the sample minimum. Prior to this, I have found that the distribution function of $T = X_{(1)}$ is $$
f_T(t) = ne^{-n(t-\theta)}I(t\geq\theta).$$

I have also shown that $X_{(1)}$ is complete sufficient and that $U$ is ancillary, and as such, by Basu's theorem, I know that these two statistics are independent.

Now I am supposed to use this information to find the mgf and cdf of $U$. Since $U$ is bounded I know that I can determine the moment-generating function by taking
$$E[U^k] = E[(X_1 - X_{(1)})^k]$$
rather than directly computing the moment generating function (since I do not know the distribution of $U$, just that $U$ is ancillary w.r.t. $\theta$). However, I do not know how to compute this expectation.

I was hoping there would be an algebraic trick that would allow me to use the independence of $T$ and $U$ in computing this expectation, but I cannot find one (I believe that this is how the exercise is intended to be solved, but I am stumped). I did try writing
$$E[U^k] = E\left[\sum_{j=1}^k \binom{k}{j} X_1^k (-X_{(1)})^{k-j}\right] = \sum_{j=1}^k \binom{k}{j} E\left(X_1^j)\right)E\left(-X_{(1)}^{k-j}\right) $$
but this has not helped me compute the expectation, as I do not think that $E(X_1^j)$ has an elementary integral.

If anyone can help me compute this expectation or figure out what trick to use, it would be appreciated.

答:
Due to independence of $U=X_1-X_{(1)}$ and $X_{(1)}$ by Basu's theorem, whenever the expectations exist,

\begin{align}
E\left[e^{tX_1}\right]&=E\left[e^{t\left(U+X_{(1)}\right)}\right]
\\&=E\left[e^{tU}\right]E\left[e^{tX_{(1)}}\right]
\end{align}

So provided MGF of both $X_1$ and $X_{(1)}$ exist and $E\left[e^{tX_{(1)}}\right]\ne 0$, we must have

$$E\left[e^{tU}\right]=\frac{E\left[e^{tX_1}\right]}{E\left[e^{tX_{(1)}}\right]}$$

As $(X_i-\theta)$'s are i.i.d $\text{Exp}(1)$, distribution of $X_{(1)}-\theta$ is exponential with mean $\frac1n$.

This means $$E\left[e^{t\left(X_1-\theta\right)}\right]=\frac1{1-t}\quad, \, t<1$$

and $$E\left[e^{t\left(X_{(1)}-\theta\right)}\right]=\frac{n}{n-t}\quad, \, t<n$$

Hence MGF of $U$ for $t<1$ is

$$E\left[e^{tU}\right]=\frac{e^{t\theta}}{1-t}\cdot\frac{n-t}{ne^{t\theta}}=\frac{n-t}{n(1-t)}$$


问:
How many right angled triangles with co-prime integer sides and base of length $28cm$ are there?
How many right angled triangles with co-prime integer sides and base of
length $28cm$ are there? Please help me.

My working: I tried assuming that once that base side is $2m = 28$ as $ 2m,m^2-1,m^2+1  $ form Pythagorean triplets. I got one set of possible answer which was $28, 195$ and $197$. Please help me to find the other sets.

答:
We begin with Euclid's formula shown here as
$$ \quad A=m^2-k^2 \qquad B=2mk \qquad C=m^2+k^2\quad$$
We can find all triples with $B=28$ by solving the $B$-function for $k$ and testing a defined range of $m$-values to see which, if any, yield integers.

\begin{equation}
B=2mk\implies k=\frac{B}{2m}\qquad\text{for}\qquad \bigg\lfloor \frac{1+\sqrt{2B+1}}{2}\bigg\rfloor \le m \le \frac{B}{2}
\end{equation}
The lower limit ensures $m>k$ and the upper limit ensures $m\ge 2$
$$B=28\implies\bigg\lfloor \frac{1+\sqrt{56+1}}{2}\bigg\rfloor =4 \le m \le \frac{28}{2}=14\\
\quad \text{and we find} \quad m\in\{7,14\}\implies k\in\{2,1\}$$
$$F(7,2)=(45,28,53)\qquad \qquad F(14,1)=(195,28,197)\qquad $$

If we go with non-primitives for side-$B$, we find
$$7\times F(2,1)=7\times(3,4,5)=(21,28,35)$$
All $B$-values are multiples of $4$ so there are no other triples.

A similar technique for side-$A$ shows just one non-primitive triple
$4\times F(4,3)=4\times(7,24,25)\space = \space F(8,6)=(28,96,100)$


问:
Vector integral of the type $\frac{1}{||x||}$
How can I solve the following integral?

$$\int_0 ^L \frac{1}{||\mathbf{x}-(\mathbf{a}+\mathbf{T} s)||}ds$$

The vector in the denominator represents the distance between a point in space ($\mathbf{x}$) and a straight line, over which we are integrating. The line is parameterized by its initial point ($\mathbf{a}$), its tangent vector $\mathbf{T}$ and the coordinate s, for $s \in (0,L)$

答:
$$\int_0^L \frac{1}{\sqrt{(\mathbf{x}-\mathbf{a} - \mathbf{T}s)^2}}\text{d}s
= \int_0^L \frac{1}{\sqrt{(\mathbf{x}-\mathbf{a})^2 - 2s\mathbf{T}\cdot(\mathbf{x}-\mathbf{a})+ T^2s^2}}\text{d}s.$$

Now we do the substitution $s:= t + \frac{\mathbf{T}\cdot(\mathbf{x}-\mathbf{a})}{T^2}$, d$s$=d$t$ (write $h:=\frac{\mathbf{T}\cdot(\mathbf{x}-\mathbf{a})}{T^2}$) and find

$$=\int_{h}^{L+h}\frac{1}{\sqrt{T^2t^2 - T^2h^2 + (\mathbf{x}-\mathbf{a})^2}}\text{d}t.$$

Due to Cauchy-Schwarz, we know that $T^2h^2\le (\mathbf{x}-\mathbf{a})^2$, hence

$$=\frac{1}{\sqrt{(\mathbf{x}-\mathbf{a})^2 - T^2h^2}}\int_h^{L+h} \frac{1}{\sqrt{1 + t^2\frac{T^2}{(\mathbf{x}-\mathbf{a})^2-T^2h^2}}}\text{d}t.$$

Now, we substitute $u:=t\sqrt{\frac{T^2}{(\mathbf{x}-\mathbf{a})^2-T^2h^2}}$ and obtain

$$=\frac{1}{T}\int_{\sqrt{\frac{T^2h^2}{(x-a)^2-T^2h^2}}}^{\sqrt{\frac{T^2(L+h)^2}{(x-a)^2-T^2h^2}}}\frac{1}{\sqrt{1+u^2}}\text{d}u = \frac{1}{T} (arsinh(..)-arsinh(..)).$$ but I'm sure there is a nice way to write it out in terms of logarithms or similar.


问:
Geometric Meaning of the Partial Derivative
Could anyone explain the geometric meaning behind the partial derivative?

I know that for a "normal" derivative, its geometric meaning is the slope of the tangent of a curve. From what I understand about the partial derivative, it is the slope of the tangent of a cross section of a function with two or more variables. If anyone could expand upon the definition and clarify, I would be grateful. Thank you.

答:
For $f:\mathbb{R}^n\to \mathbb{R}$:
$$
\frac{\partial f}{\partial x_i} = \nabla f\cdot e_i
$$
is the directional derivative along the direction of $e_i$, $e_i$ is $1$ in the $i$-th component and $0$ in others.

If the partial derivative is evaluated at a point $x = (x_1,\ldots,x_n)$, define a mapping:
$$f_i: \mathbb{R} \to \mathbb{R}, \;t\mapsto f(x_1,\ldots,x_i(t),\ldots,x_n),$$
where the $i$-th component is parametrized by $x_i = a + t$, then
$$
\frac{\partial f}{\partial x_i} = \frac{d f_i}{d t}
$$
which is the slope of the tangent to the curve $f_i(t)$ at the corresponding point.


问:
Is the convex hull of product channels strictly contained in the set of all channels
For two finite dimensional Hilbert spaces $\mathcal{H},\mathcal{H'}$ let $T(\mathcal{H},\mathcal{H'})$ be the set of all CPTP maps (quantum channels) $L(\mathcal{H})\to L(\mathcal{H}')$.

Let $\mathcal{H},\mathcal{H_1},\mathcal{H_2}$ be finite dimensional Hilbert spaces, I am wondering whether the convex hull of $T(\mathcal{H_1},\mathcal{H})\otimes T(\mathcal{H_2},\mathcal{H})$ is strictly contained in $T(\mathcal{H_1}\otimes\mathcal{H_2},\mathcal{H}\otimes\mathcal{H})$ (it seems that it does but a proof eludes me).

If so, is there a way to express this set, or an as small as possible set containing this set, as the solution set of a linear matrix inequality?

答:
Yes, this set is strictly contained in the other.

Note that by the main body of your question you are asking whether or not the convex hull of product channels (not separable) is strictly contained in the set of all channels. Separable channels are defined as those for which their Kraus operators have a tensor product form.

To see a counterexample we can look at Example 6.21 from [1] you can find a free version on the author's website. Define a channel $\Phi$ by the action,
$$
\Phi(|i\rangle \langle j| \otimes |k \rangle \langle l|) = \begin{cases}
|i \rangle \langle i | \otimes |i \rangle \langle i | \qquad &\text{if } i=j \text{ and } k=l \\
0 &\text{otherwise}
\end{cases}
$$

Now let us suppose we could write $\Phi = \sum_a p(a) \Phi_a \otimes \Psi_a$ (i.e. a convex combination of product channels). Well then we have for any normalized state $\rho$
$$
\begin{aligned}
\Phi(|0\rangle \langle 0 | \otimes \rho) &= |0 \rangle \langle 0|\otimes |0\rangle \langle 0 | \\
&= \sum_a p(a) \Phi_a(|0\rangle\langle 0|) \otimes \Psi_a(\rho)
\end{aligned}
$$
and then taking the partial trace over the first subsystem we may conclude that
$$
\sum_a p(a) \Psi_a(\rho) = |0\rangle \langle 0|.
$$
However, if we repeat these steps with the input state $|1 \rangle \langle 1| \otimes \rho$ we will instead conclude that $\sum_a p(a) \Psi_a(\rho) = |1\rangle \langle 1 |$. Thus we run into a contradiction and our assumption that our channel can be written as a convex combination of product channels is false.

[1]: Watrous, John, The theory of quantum information, Cambridge: Cambridge University Press (ISBN 978-1-107-18056-7/hbk; 978-1-316-84814-2/ebook). viii, 590 p. (2018). ZBL1393.81004.


问:
Reference request: measure induced by a distribution.
I am looking for a reference that explains the connection between a measure $\mu$ and a distribution as described for instance here:

$$\langle F,g\rangle =\int_{\Omega} g  \;d\mu$$

where $g$ is some test function and $\langle.,.\rangle$ denotes the bilinear pairing between distributions and test functions.
I tried looking inside the humongous 5 volumes book by Gel'fand and Shilov but I wasn't able to find anything (although I am sure it must be there, somewhere).

So if you know where I can find this I would be glad!
Thanks in advance.

答: Uhm... Measures induce distributions, but not the other way around. If you're working on a nice-enough space, measures represent the dual of the space of continuous functions (that are zero at infinity), but distributions represent the dual of the smooth functions (that are zero at infinity)... Or Schwartz functions... Anyway, since you can "do more" with smooth functions (like: differentiate and evaluate), there are "more" distributions than measures.</p>


问:
Maximum value of $(x−1)^2+ (y−1)^2+ (z−1)^2$ with constraint $x^2+y^2+z^2 ≤2 , z≤1$
So the problem is that I have $D(f)=\{(x,y,z), x^2+y^2+z^2 ≤2 , z≤1\}$ and I have to determine the maximum value for the function $(x−1)^2+ (y−1)^2+ (z−1)^2$ in $D$.

I'm just confused as I don't actually know if $z\le1$  counts as a constraint as well, or is it just for me to sketch the area, which is actually a part of the question.
Furthermore, I know that I have to use Lagrange multiplier method, but I honestly don't know how because $\le$ is making the question hard for me. Do I just calculate as usual and count $\le$ the same as $=$?

appreciate all the feedback

Edit: I have calculated the grad f =0 which is = $D(f)=(2(x-1), 2(y-1), 2(z-1))$ where I've got that $x=y=z= 1$ and $f(1, 1, 1)=0$. (I don't know what to do with this though). Then I calculated $L(x, y, x, λ) = (x−1)^2+(y−1)^2+(z−1)^2 +λ(x^2+y^2+z^2-2)$, then the four cases where I got the same value which is $-2λ= 2(x-1)/x = 2(y-1)/y = 2(z-1)/z$. Which means that $x=y=z$, put it in the $D$ function $x^2+ x^2+ x^2$ and ended up with $x=y=z= −+√2/√3$. I took the minus sign for the maximum distance from $(1,1,1)$. which means that the answer is $x=y=z= −√2/√3$. Is it correct?

答:
Using Lagrange multipliers method.

Calling $p = (x,y,z),\ p_0=(1,1,1),\ p_1 = (0,0,1), f = \|p-p_0\|^2$ and using $e_1, e_2$ as slack variables to avoid the inequalities we have

$$
L = f + \lambda(\|p\|^2-2+e_1^2)+\mu(p\cdot p_1-1-e_2^2)
$$

The stationary points are the solutions for

$$
\nabla L = \cases{2(1+\lambda)p-2p_0-\mu p_1=0\\
\|p\|^2-2+e_1^2=0\\
p\cdot p_1-1-e_2^2=0\\
\lambda e_1=0\\
\mu e_2 = 0}
$$

obtaining

$$
\left[
\begin{array}{cccccc}
f & x & y & z & e_1^2 & e_2^2\\
0 & 1 & 1 & 1 & 1 & 0 \\
 3+2 \sqrt{2} & -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 1 & 0 & 0 \\
 3-2 \sqrt{2} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 1 & 0 & 0 \\
\end{array}
\right]
$$

NOTE

Null $e_k$'s represent actuating restrictions.


问:
Prove that $S^1{\setminus\{(1,0)\}}$ is homeomorphic to $]0,1[$
I'm currently working on this book called "Topology Without Tears".
One of the exercises on the chapter of non-homeomorphic spaces. I'm asked to prove that $S^1{\setminus\{(1,0)\}}$ is homeomorphic to $]0,1[$, but I couldn't find any homeomorphism.  Plus I was told that I should use the stereographic projection On $\mathbb{R}$, but I don't know what that is.

答:
What about $f : ]0,1[ \rightarrow \mathcal{S}^1$ defined by
$$f(t)=(\cos(2\pi t),\sin(2\pi t)) \quad ?$$

(where you see $\mathcal{S}^1$ as the subset $\lbrace (x,y), x^2+y^2=1 \rbrace \subset \mathbb{R}^2$)


问:
Why is the kernel of a group homomorphism so called?
I took a course in Linear Algebra last fall. I have come to associate the word kernel with the nullspace of matrices/linear transformations - and now that I am studying Group Theory, similar intuition doesn't seem to carry forward.

Why would anyone name it kernel (in group theory) if it really isn't the same thing as that in linear algebra? Perhaps there is a relation or connect between the two definitions that I am missing. Could someone help me better understand why the kernel of a homomorphism is so defined?

To document the definitions:

(Linear Algebra): Consider a linear transformation $T:V\to W$. $$\ker T = \{x\in V: T(x) = 0\}$$
(Group Theory): Let $\phi:G\to H$ be a group homomorphism. $$\ker\phi = \{x\in G: \phi(x) = 1_H\}$$ where $1_H$ is the identity in $H$.

Thanks!

答:
The kernel of a linear map and of a group homomorphism is just the subset of the domain consisting of all elements which are mapped to the trivial element in the codomain. This encapsulates their general role.

When given a linear map $T\colon V\to W$ or a group homomorphism $\phi\colon G\to H$ we have to make precise what the trivial element in the codomain is. In case of vector spaces it is the zero vector and in case of groups it is the identity element, here $0_W$ and $1_H$, respectively. However, a vector space is in the end just an (abelian) group where the zero vector is the identity element. Hence, the two notions conincide in this case (note that scalar multiplication is not relevant here as $\varphi(av)=a\varphi(v)$ for every $a$ in the base field).

So, viewing $V$ and $W$ as additive groups $(G,+,1_G)=(V,+,0_V)$ and $(H,+,1_H)=(W,+,0_W)$ the group-theoretic kernel of $\phi\colon G\to H\triangleq T\colon V\to W$ becomes the null space:

$$\ker\phi=\{x\in G\,|\,\phi(x)=1_H\}=\{x\in V\,|\,T(x)=0_W\}=\ker T$$

(Here's a lot of notational changes going on; I hope it's not too confusing.)

I will just mention that the concept of a kernel can be generalized beyond this simple example to include the kernels of ring homomorphisms, or module homomorphisms and more related notions. This is best done using some category-theoretic language and universal properties to move to arbitarty categories. This general concept (a special case of a so-called equalizer) will be exactly the same if we then consider vector spaces and linear maps or groups and group homomorphisms. However, this requires some additional work as the definition is not in set-theoretic terms per se.


问:
How to solve this system of equation.
$x^2-yz=a^2$

$y^2-zx=b^2$

$z^2-xy=c^2$

How to solve this equation for $x,y,z$. Use elementary methods to solve (elimination, substitution etc.).

Given answer is:$x=\pm\dfrac{a^4-b^2c^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}\,$, $y=\pm\dfrac{b^4-a^2c^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}\,$ ,$z=\pm\dfrac{c^4-b^2a^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}$

Title: How to solve this system of equation.

Body:
Observe that $$(a^2)^2-b^2\cdot c^2=(x^2-yz)^2-(y^2-zx)(z^2-xy)=x(x^3+y^3+z^3-3xyz)$$

Similarly, $$b^4-c^2a^2=y(x^3+y^3+z^3-3xyz)\text{ and }c^4-a^2b^2=z(x^3+y^3+z^3-3xyz)$$

So, $$\frac x{a^4-b^2c^2}=\frac y{b^4-c^2a^2}=\frac z{c^4-a^2b^2}=\frac1{x^3+y^3+z^3-3xyz}=k\text{(say)}$$

Now, $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=(x+y+z)\frac{\{(x-y)^2+(y-z)^2+(z-x)^2\}}2$

$x+y+z=k(a^4-b^2c^2+b^4-c^2a^2+c^4-a^2b^2)=\frac k2\{(a^2-bc)^2+(b^2-ca)^2+(c^2-ab)^2\}$

and $x-y=k\{a^4-b^2c^2-(b^4-c^2a^2)\}=k(a^2+b^2+c^2)(a^2-b^2)$

$\implies (x-y)^2+(y-z)^2+(z-x)^2=k^2(a^2+b^2+c^2)^2\{(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2\}=k^2(a^2+b^2+c^2)^2\cdot 2(a^4-b^2c^2+b^4-c^2a^2+c^4-a^2b^2)$

$\implies x^3+y^3+z^3-3xyz=k^3\{(a^2+b^2+c^2)(a^4-b^2c^2+b^4-c^2a^2+c^4-a^2b^2)\}^2=k^3(a^6+b^6+c^6-3a^2b^2c^2)^2 $

$$\implies\frac1k=x^3+y^3+z^3-3xyz=k^3(a^6+b^6+c^6-3a^2b^2c^2)^2$$

$$\implies k^2=\frac1{a^6+b^6+c^6-3a^2b^2c^2}$$ as $a^6+b^6+c^6-3a^2b^2c^2=(a^2+b^2+c^2)\frac{\{(bc-a^2)^2+(ca-b^2)^2+(ab-c^2)^2\}}2\ge 0$ for real $a,b,c$


问:
Algebraic solution to $\frac{a}{1+bc}+ \frac{b}{1+ac}+ \frac{c}{1+ab} \le 2$
With $0 \le a \le b \le c \le 1$.

I know that this question has been answered here : If $0\le a,b,c\le 1$, then $\frac a{1+bc}+\frac b{1+ac}+\frac c{1+ab}\le 2$.

But I’m looking for a solution using some simple algebra .

My attempt:

We know that :
$$a^2+1 \geq 2a, a^2\le bc \implies \frac{bc+1}{a}\geq 2$$
$$\iff \frac{a}{bc+1}\le \frac{1}{2}$$

But this trick doest works on:
$$ \frac{b}{1+ac} $$
Because $b^2\geq ac$

答:
Well if you find the answer by sciona difficult  and you want to avoid the use of $\sum_{cyc} a(1-b)(1-c)\ge 0$  here is an alternative approach

After sciona's  work we have to prove $$\frac{a+b+c}{1+abc}\le 2$$ $$ \iff a(1-2bc)+b+c\le 2\tag1$$  Now if

$1-2bc\ge 0$ $$a(1-2bc)+b+c\le 1-2bc+b+c$$  Now if $1-2c\ge 0$ then $1-2bc+b+c=b(1-2c)+1+c\le 1-2c+1+c=2-c\le 2$

If $1-2c\le 0$ then $1-2bc+b+c=b(1-2c)+c+1\le c+1 \le 2$

$1-2bc\le 0$  then $$a(1-2bc)+b+c\le b+c\le 2$$

Done!
Although this seems a lot of work the cases we took are simply intuitive

Note:  The best way to prove is that notice Inequality (1) is a linear function in $a$ and hence its maximum  and minimum  will occur at the boundary and similarly the for $b$ and $c$ so it suffices to check when $a=b=c=0,a=1b=c=0,a=b=1c=0,a=b=c=1$


问:
upper semi-continuous of fuzzy set
Let $u:\mathbb{R^n}\to [0,1]$ be a fuzzy set. (fuzzy set is a set of ordered pairs $(x,u(x)), x\in \mathbb{R^n})$.
Please give an example such that  $u(x)$ be upper semi-continuous. thanks

答:
Let $\mathbf{x} = \{x_i\}^n_{i=1}$. The indicator function of the unit ball defined as
$$u(\mathbf{x}) = \begin{cases} 1 & \text{if $\Vert \mathbf{x} \Vert_2 = \left( \sum^{n}_{i=1} x_i^2 \right)^{1\over2} \leqslant 1$} \\ 0 & \text{otherwise} \end{cases}$$
is upper semi-continuous. Another example would be
$$v(\mathbf{x}) = \begin{cases} 1 & \text{if $0 \leqslant x_i \leqslant 1$ for all $i$}\\ 0 & \text{otherwise} \end{cases}$$
which would be the indicator function of a unit (hyper-)cube, also upper semi-continuous (thanks to the $\leqslant$ in the expression)

Note: this makes $(\mathbf{x},u(\mathbf{x}))$ and $(\mathbf{x},v(\mathbf{x}))$ examples of "crisp" fuzzy sets. They map $\mathbb{R}^n\to \{0,1\}$ which is a special case of $\mathbb{R}^n\to [0,1]$. If you need something with "more fuzziness", I can offer
$$w(\mathbf{x}) = \begin{cases} \frac{1}{1+(\Vert \mathbf{x} \Vert_2)^2}  & \text{if $\Vert \mathbf{x} \Vert_2 \leqslant 1$} \\ \frac{1}{1+4(\Vert \mathbf{x} \Vert_2)^2} & \text{otherwise} \end{cases}$$

Title: upper semi-continuous of fuzzy set


问:
What is the area of the portion of 1/8 of an sphere cut off by two parallel planes?
So the problem that I'm trying to solve is as follows:

Assume 1/8 of a sphere with radius $r$ whose center is at the origin (for example the 1/8 which is in $R^{+}$). Now two parallel planes are intersecting with this portion where their distance is $h$ (Note:$h$ < $r$) and one of the planes passes through the origin.What is the area cut off by these two planes on the 1/8 of the sphere?
You may assume anything that you think is required to calculate this area, as given, my suggestion would be the angles at which the parallel planes cross xyz planes.

Obviously I'm not interested in trivial cases for example when the plane which passes through origin is one of xy, xz or zy planes.

I hope I explained everything clearly. I wish I could draw a picture for this but I don't how. Let me know if you need more clarification. Any hint or help about how to find this area is highly appreciated.

答:
I don't care to make the calculation for you, but I think(?) this is the region you intend:

Addendum.
Here is an idea to avoid computing a complex integral.
Let me assume that the slice "fits" in one octant, as I drew above
(as opposed, e.g., to exiting through the bottom of the octant).
First, imagine that you wanted, not the area in an octant, but the area in a halfspace,
demarcated by one of the coordinate planes.
Then, from a side view, we have this:

Your two planes are $p_1$ and $p_2$, with $p_1$ through the sphere center.
The halfspace boundary is coordinate plane $q$.
Now note that the difference between the half of the $2 \pi h$ area of the slice
and the truncated area is just the area of two spherical triangles (one illustrated,
one in back), whose angles
are all known.

So I believe the area in your octant differs from a quarter of $2 \pi h$ by two
spherical triangles, one at either end of the octant.  The dimensions of both triangles
are known from the orientation of the planes $p_1$ and $p_2$.


问:
Quotient space Hausdorff/$T_2$?
Let $n$ natural number and $X = [0,1] \times \{1,...,n\}$ with the topology $\tau \times \tau_D$, where $\tau$ is the usual topology of $\mathbb{R}$ induced by $[0,1]$ and $\tau_D$ is the discrete topology in $\{1, . . . , n\}$.
We define the equivalence relation of $ (x,n) \sim (x',n')$ iff both points are equal or $x=x'=0.$

My question is, is $X/\sim$ Hausdorff/$T_2$ space? I can't think of how to start, any help is appreciated!

答:
I think you can prove it distinguishing cases:

Let $(x,n), (x',n')\in X$ be to points such that $n\neq n'$ and $x,x'\neq0$, then the sets $(0,1]\times \{n\}$ and $(0,1]\times \{n'\}$ are open neighborhoods of each of them respectively (the interval $(0,1]$ is open in $\tau_U|_{[0,1]}$ and the sets $\{n\}$ and $\{n'\}$ are open in $\tau_D$) and their images under the canonic projection of the equivalence relationship are disjoint (again, quite trivially, as the points of the different sets have different second coordinate, and so are not equal, and have a non-$0$ first coordinate, so they can't be in the same equivalence class).

Let $(x,n), (x',n)\in X$, with $x\neq x'$. We can assume that $x<x'$ without any loss of generality. Since $\tau_U|_{[0,1]}$ is $T_2$, we can take disjoint neighborhoods of each of them contained in $[0,1]\times \{n\}$ and they will again have images with empty intersection under the projection.

The case where $(x,n), (0,n')\in X$, $n\neq n'$ and $x\neq0$ is almost the same as the first one, now taking the neighborhoods $(0,1]\times \{n\}$ and $[0,1]\times \{n'\}$.

Title: Quotient space Hausdorff/$T_2$?


问:
Notation for "function from a subset of $X$ into $Y$"?
The notation $f: X \to Y$ implies that $X$ is the domain of $f$. It is very regularly convenient to say something like:

Let $U \subset X$ and $f: U \to Y$.

but in situations where the actual set $U$ isn't that important, just that it's a subset of $X$. Usually the subset will not be arbitrary, but will be implied by the context. The kind of thing I'm thinking of is:

Let $\alpha : \mathbb{R} \hookrightarrow \mathbb{R}^{2}$ be a plane curve.

Is there a common choice for notation in this case? I imagine there must be (it might even be $\hookrightarrow$ but I'm not certain, I think that might mean injective).

The notation does not have to encode that the required subset is connected or open or whatever the situtation might require. That should be implied by context. I only care that it's a subset.

答:
In algebraic geometry circles a dotted arrow is common notation for partially defined functions, see https://tex.stackexchange.com/q/38670/65997.

I tend to use $X\rightsquigarrow Y$ because I find it easy to miss the dots etc.. Also the other arrows mentioned above tend to have different meanings in analysis, so they end up confusing the writer/reader. $X\nrightarrow Y$ and the like seems to me to signify "not an arrow" or else they are very close to $\mapsto$, which I tend to reserve for the pointwise map.

Probably macroing $X\xrightarrow{\tiny\mbox{ loc.}} Y$ or $X\xrightarrow{\tiny\mbox{ partial}} Y$ would prove more effient for more writers/readers.

Alternatively one could make the categorical switch of considering $X\to Y$ to signify a partial function. I guess in your context this won't do because it matters to distinguish the local and the global.


问:
How does stability analysis of Lotka-Volterra system differ from stability analysis on a normalized version?
How does stability analysis of Lotka-Volterra system differ from stability analysis on a normalized version?

Normalized LV system is such where the coefficients have been "aggregated" so that the system has less variable parameters. I find it confusing to infer how it alters critical points though, since:

Since it's essentially describing the same system, then it should have same critical points.

However the new parametrization may give different critical points than the original parametrization due to them having been shifted by the aggregating of coefficients.

I saw text that did stability analysis on normalized system and made it seem as if it's "as informative" as studying the original system. And I didn't understand why:

https://www.maths.dur.ac.uk/%7Ektch24/term1Notes(10).pdf, p. 2

答:
The rescaled equation does not give new critical points. Notice that
$$\hat x = 1 = \frac d c x \implies x = \frac c d,$$
$$\hat y = 1 = \frac b a y \implies y = \frac a b$$
and $(x,y) = (\frac c d , \frac a b)$ is a steady state of the non-normalize equations.

The normalized system gives the same information because essentially, the parameters $a,b,c,d$ are "summarized" in $\gamma$ (I am using the notations from your reference) and $\hat x, \hat y$ are a percentage, essentially.


问:
Show that such a finite plane graph doesn't exist
Show that there is no finite plane graph whose every face is a hexagon and each two neighboring faces have exactly one edge in common.

So this is one of the questions I am asked to solve for my discrete math 2 class. I believe I have to solve this using the concept of dual planar graph but I might be wrong

答:
For contradiction suppose there exists a graph $G$ which is planar, and whose every face is a hexagon and each two neighboring faces have exactly one edge in common.

The dual of $G$ (call it $G'$) is a planar graph.

Every face of $G$ is adjacent to exactly 6 other faces by construction, so every vertex of $G'$ has degree 6.

There is a theorem that for all planar graphs with 3 or more vertices, $e\leq3v-6$ where $e$ is the number of edges and $v$ is the number of vertices. By handshake lemma, $G'$ has $3v$ edges since all its vertices have degree 6. This is a contradiction so $G$ cannot exist.


问:
Sequences with the following properties...
Suppose that $\lim_{n\to\infty}\frac{a_n}{b_n}$=1. And $\sum^{\infty}_{n=1}a_n$ converges, $\sum^{\infty}_{n=1}b_n$ diverges. Are such sequences $(a_n)$, $(b_n)$ exist? </p>

答:
If we assume that $\{a_k\}$ and $\{b_k\}$ are positive, no such sequences exist:

To see why, suppose that

$$\lim_{n \to \infty} \frac{a_n}{b_n}=1,$$

we may choose $N>0$ sufficiently large so that $b_n < 2a_n$ for all $n \geq N$.  Then the tails of our series satisfy
$$\sum_{k=N}^\infty b_k<\sum_{k=N}^\infty 2a_k,$$
so that $\sum b_k$ converges by the comparison test.

Wikipedia has more information here.


问:
Disjointness in proof of Tarski's Theorem that $|X\times X|=|X|$ for all infinite sets implies Choice
In the Wikipedia proof of Tarski's Theorem about Choice, one assumes that the infinite set $B$ and the ordinal $\beta$ are disjoint.

While it is clear that this assumption can be made, is it actually necessary in the rest of the proof? I cannot locate where this assumption is used.

答:
No, it's not necessary. But it makes it a bit easier when we talk about $\gamma\in\beta$ and $\gamma\notin B$. It is also a bit simpler conceptually to separate the two sets instead of dealing with their intersection.

Sometimes adding a small and unnecessary assumption can clarify the concept behind the proof, and when the unnecessary assumption does not hurt the generality, this is even more welcome.

Let me remark that it is unfortunate that the proof is presented the way that it is. Instead the correct way is to extract the following lemma:

Lemma 1. Suppose that $X$ is a set and $\lambda$ is an ordinal. If $|X\cup\lambda|=|X\times\lambda|$, then $|X|$ and $\lambda$ are comparable.

This, in turn, can be improved and generalised into,

Lemma 2. Suppose that $f\colon C\cup D\to A\times B$ is a surjection, then $C$ maps onto $A$ or $D$ maps onto $B$.

Although it is in fact enough to consider the injective version of this lemma,

Lemma 3. Suppose that $f\colon A\times B\to C\cup D$ is injective, then $A$ injects into $C$ or $B$ injects into $D$.

It is not hard to see why Lemma 2 implies Lemma 3, and Lemma 3 implies Lemma 1 by taking $A=D=\lambda$ and $B=C=X$.


问:
Calculus 2 question
Consider the function $y = x^2$
Estimate the distance along this curve from $x = 0$ to $x = 2$ using
straight line distances on the intervals $[0, 1]$ and $[1, 2]$.
Is this question asking me to calculate approximating area?

答:
Welcome to MSE. You're being asked to approximate arc length, not area. The length of those two segments will be an approximation for the actual arclength of $y = x^2$ from $x = 0$ to $x = 1$.

Visually, you're using the two black segments $AB$ and $BC$ as a stand-in for the green parabola between $A$ and $C$.

Title: Calculus 2 question


问:
Find the relative extrema of $3x^4−36x^3+140x+200$
Find the relative extrema of the function:

$$3x^4−36x^3+140x+200$$

So I found the 1st derivative:

$$12x^3-108x^2+140$$

Then I got lost from here because I don't think it can be factorized, I'm trying to get the critical points then find the critical values. I'm sorry if my equations are not too good, I still don't know the most suitable equation editors for this site

答:
The derivative of $140^x$ is not $140$. Keep in mind that there are four distinct formulae for evaluating the derivative of a number raised to another number:
\begin{align}
\frac{d}{dx}(a^b) &= 0 \\
\frac{d}{dx}(x^a) &= ax^{a-1} \\
\frac{d}{dx}(a^x) &= a^x \log a \\
\frac{d}{dx}(x^x) &= x^x \left(\log x + 1\right)
\end{align}
The first two formulae should be familiar to you: the derivative of a constant is zero, and the derivative of $x^a$, where $a$ is a constant, simply involves the application of the power rule. The last two might not be so familiar. But it is the third, not the second formula that applies here:
$$
\frac{d}{dx}(140^x) = 140^x \log(140) \, .
$$
If you use this formula, then should be able to find the extrema of the function is question by setting the first derivative equal to $0$.


问:
Solving a puzzle: Graph where each node has degree 3
I am trying to solve this puzzle:

Problem:
The  Park is covered by a network of hiking paths.

Goal: Find an algrithm by which you'll find (in finite time) the crossing at which the restaurant is located.

Rules:

The paths form a finite connected graph in which each crossing (that is, each crossing of paths) is of degree 3.

The crossings are indistinguishable one from another: it is not possible to know which crossing one has arrived at, to identify the paths meeting there, or to recognize whether one has been there before.

Strict rule against leaving markings in the park. It is not ok to mark crossings or path in any way.

If you enter a crossing by one path you must leave it by a different path. You have initially arrived at some node by one of its incident paths.

My questions:

For rule 1: I cannot find any graph where every node has a degree of 3. At least one node has degree 2.

2nd question: If I cannot mark the paths where I have been, what can I do to find a sufficient solution? Just go right, right right and one time left?

答:
Define a hypothesis to consist of the following:

a $3$-regular graph representing a hypothetical map of the park;

a vertex $r$ in that graph representing a hypothetical location of the restaurant;

a vertex $s$ in that graph representing a hypothetical starting location;

an edge incident to $s$ representing the hypothetical path by which you arrived at $s$.

There are infinitely many hypotheses, but countably many: for each natural number $n$, there are only finitely many hypotheses in which the park has $n$ vertices. So we may number the hypotheses $H_1, H_2, H_3, \dots$.

One of the hypotheses represents the true map of the park, location of the restaurant, and your starting position: call that hypothesis $H_N$. You don't know which hypothesis this is, but $N$ is some finite number.

Now, here is the strategy that will get you to the restaurant. For each $i=1,2,3,\dots$, you:

Assume hypothesis $H_i$ is true for the time being.

Given the steps you've taken so far, determine the location in $H_i$'s graph where you would now be, if hypothesis $H_i$ were true.

From that location, find directions in $H_i$'s graph that would take you to $H_i$'s restaurant location $r$.

Follow those directions in the actual physical park.

For the first $N-1$ iterations of this procedure (for $i=1, 2, \dots, N-1$), you're going to be following directions based on a false premise, and they're unlikely to get you to the restaurant except by chance.

On the $N^{\text{th}}$ iteration of the restaurant, you'll assume a true hypothesis: the graph you take will be the correct graph, with the restaurant and your starting location and starting direction marked correctly. Then the location you work out in step 2 will happen to be your actual location in the park, and the directions you work out in step 3 will actually take you to the restaurant.


问:
What is the probability that there more rabbits than chickens in each of these three cages?
Suppose we have $4$ chickens and $12$ rabbits, and put them into $3$ cages. What is the probability that there more rabbits than chickens in each of these three cages?

I was trying to solve it by using P(more rabbits than chicken) = 1-P(more chicken than rabbits). Since the number of chickens is far less than the number of rabbits and it would be easier to calculate.

Then, the cardinality of the sample space would be $3^{12+4}$. But I don't know how to continue.

Body:
Let $C_i$ and $R_i$ be the number of chickens and rabbits in cage $i$.

$$P[C_1=i_1,C_2=i_2]={4\choose{i_1}}{{4-i_1}\choose{i_2}}\left(\frac{1}3 \right)^4$$
$$i_1,i_2,i_1+i_2 \in (0,1,2,3,4)$$
and

$$P[R_1=j_1,R_2=j_2]={12\choose{j_1}}{{12-j_1}\choose{j_2}}\left(\frac{1}3 \right)^{12}$$
$$j_1,j_2,j_1+j_2 \in (0,1,2,3,...,12)$$

Then, note that $12-j_1-j_2>4-i_1-i_2 \implies 7+i_1+i_2-j_1\ge j_2$.

So,

$$P[R_1>C_1,R_2>C_2,12-R_1-R_2>4-C_1-C_2]$$
$$=\sum_{i_1=0}^4 \sum_{i_2=0}^{4-i_1} \sum_{j_1=i_1+1}^{12} \sum_{j_2=i_2+1}^{7+i_1+i_2-j_1}P[C_1=i_1,C_2=i_2]P[R_1=j_1,R_2=j_2]$$
$$=\frac{112706}{177147}\approx 0.636$$

p[i1_, i2_, r1_, r2_] =
 Binomial[4, i1] Binomial[4 - i1, i2] Binomial[12, j1] Binomial[12 - j1, j2]/
(3^4 3^12)

Sum[p[i1, i2, j1, j2], {i1, 0, 4}, {i2, 0, 4 - i1},
{j1, i1 + 1, 12}, {j2, i2 + 1, 7 + i1 + i2 - j1}]
112706/177147

N[%]
0.636229


问:
How to proof that adjacency matrices of DAG can be permuted to triangular matrices?
I've come across the statement that the adjacency matrices of directed acyclic digraphs can always be permuted in such a way that we get a triangular matrix. Although I think I kind of get why, I am unsure how to prove the statement formally.

I have seen the following post: Directed acyclic graph and adjacency matrix Unfortunately it's still not clear to me.

Does anyone have a tipp for me?

答:
Because the graph is a DAG, there exists a topological ordering on the vertices. After relabelling the vertices with this ordering, we have labelled vertices $v_1, v_2, \cdots, v_n$ such that if $v_i v_j$ is a directed edge, then $i < j$.

Now, entry $A_{ij}$ of the adjacency matrix is nonzero when there is an edge from $v_i$ to $v_j$. We want to show that $A$ is an (upper) triangular matrix. Formally, this means its entries below the main diagonal are zero. Moreover, entries $A_{ij}$ below the main diagonal are precisely the elements with $i > j$. Can you put it all together now?


问:
Quick Polar Equations Question
So I was given the following prompt:

"A polar curve is given by the equation $r=\frac{100}{\theta^2+1}$ for $\theta \ge0$. What is the instantaneous rate of change of $r$ with respect to $\theta$ when $\theta=2$?"

I guess I'm a bit confused about what $\frac{dy}{dx}$ is going to look like here. I thought I was able to find the derivative of the equation by plugging the function for $r$ into the equations of $y=r\sin\theta$ and $x=r\cos\theta$, but after finding the value of the derivative that I ended at, at $\theta=2$, I'm getting a weird answer. Any help would be appreciated!

答:
Hint.

All you need to find is
$$\frac{dr}{d\theta}\bigg|_{\theta=2}$$

There are no Cartesian coordinates in this problem.

In general, if the quantity $z$ is a function of another quantity $w$ where $z=f(w)$ and $f$ is differentiable, then the "instantaneous rate of change of $z$ with respect to $w$ when $w=a$" is given by $f'(a)$.

Title: Quick Polar Equations Question


问:
closure of domain is compact implies finitely many holes in the domain
In many complex analysis theorems, we make the assumption that the closure of the domain $\bar{\Omega}$ is compact. Does that mean that the domain can only contain finitely many holes? I was thinking this must be true, since the boundary is closed and bounded, so it must also be compact, and if we have infinitely many holes we cannot find a finite subcover.

答: No. Take$$\Omega=D(0,2)\setminus\left(\{0\}\cup\overline{D\left(1,\frac18\right)}\cup\overline{ D\left(\frac12,\frac1{16}\right)}\cup\overline{D\left(\frac14,\frac1{32}\right)}\cup\ldots\right).$$Then $\Omega$ is open and connected, but $\overline\Omega^\complement$ has infinitely many connected components.


问:
Why is $\lim_{t \to 0} \frac{f(x + t(y-x)) - f(x)}{t} = \nabla f(x)^T\cdot (y-x)$?

What I want to understand

I am trying to understand why the following holds:

$$\lim_{t \to 0}  \frac{f(x + t(y-x)) - f(x)}{t} = \nabla f(x)^T\cdot (y-x)$$

with $f: \mathbb{R}^p \mapsto \mathbb{R}$, and $x,y \in \mathbb{R}^p$, $t \in \mathbb{R}^+$

Also I am using the following definition of the gradient:

$\nabla{f(x)} = \begin{bmatrix}
\frac{\partial f}{\partial x_1}(x) &
\frac{\partial f}{\partial x_2}(x) &
\dots &
\frac{\partial f}{\partial x_p}(x)
\end{bmatrix}^\intercal$

with $x_i$ being the $i$-th element of $x$.

Why I want to understand it

It is used when proving that a convex function always lies above its tangent line (see p. 5-6 of this example if you want)

What I know

The solution probably has something to do with the definition of the derivative as the limit of a difference quotient:

$$\lim_{\Delta z \to 0}  \frac{g(z + \Delta z) - g(z)}{\Delta z} = g'(z)$$

With $z, \Delta z \in \mathbb{R}$, and $g: \mathbb{R} \mapsto \mathbb{R}$

However, I do not understand exactly how (or if) I can use this to arrive at this higher dimensional limit at the top of my question from this definition (or what additional information I need)

Thank you very much!

答:
I'm adding this as an answer so as to not have too many comments.

Thanks for editing your questions several times. It now (as far as I can tell) makes perfect mathematical sense, although there is still a missing link: you did not give a definition of the gradient.

I know I'm being very annoying, but there is a reason for that: the expression of the gradient in your question can actually itself be taken as a definition of the gradient. This is called the Gâteaux derivative of a function (see https://en.wikipedia.org/wiki/Gateaux_derivative).

Any answer to you question needs to know what definition you are considering, of course.

Best,

EDIT :

Thanks for adding the gradient definition. Note that your definition can be rephrased as :
$$ (\nabla f)_i(x) = \nabla ^T f(x)\cdot e_i = \frac{\partial f}{\partial x_i}(x) = \frac{d}{dt} f(x+t*e_i) = \lim_{t->0} \frac{f(x+t*e_i) - f(x)}{t} $$
Where $(e_i)$ is the base in which you defined the coordinates.

So that at list when $y-x$ is one of the basis vectors, you have your answer.
Now if you write $ z = y-x = \sum z_i* e_i$, and use the chain rule, you will get your result.


问:
How to calculate $ \int_0^{\pi/2}\log(1+\sin(x))\log(\cos(x)) \,dx $?
How to calculate $$ \int_0^{\pi/2}\log(1+\sin(x))\log(\cos(x)) \,dx \,\,?$$
I tried to use the Fourier series of log sine and log cos and I got that the integral is equal to :
$$ \frac{\pi^2}{24}-\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{n+k}}{k(4k^2-(2n-1)^2)}$$
has anyone a idea to how to find the closed-form of the last series or how to start out differently with the integral?

答:
We have a closed form for the inner sum
$$S_k=\sum_{n=1}^{\infty}\frac{(-1)^{n+k}}{k(4k^2-(2n-1)^2)}$$
$$S_k=(-1)^k \frac{\Phi \left(-1,1,\frac{1}{2}-k\right)-\Phi
   \left(-1,1,\frac{1}{2}+k\right)}{8 k^2}$$ where appears the Hurwitz-Lerch transcendent function. This can rewrite
$$S_k=(-1)^k \frac{-\psi \left(\frac{1}{4}-\frac{k}{2}\right)+\psi
   \left(\frac{3}{4}-\frac{k}{2}\right)+\psi
   \left(\frac{k}{2}+\frac{1}{4}\right)-\psi
   \left(\frac{k}{2}+\frac{3}{4}\right) }{16 k^2 }$$


问:
Find $k$ so that $|a_n\;|≤Cn^{-k}$,$\,$ $|b_n\;|≤Cn^{-k}$ => $f∈C^1(T)$ ($a_n$,$b_n$ are the Fourier coefficients of f)
I'm trying to find k that when $|a_n\;|≤Cn^{-k}$,$|b_n\;|≤Cn^{-k}$, ($|a_0\;|≤C$),  f satisfies $f∈C^1(T)$ which can proof Fourier series is converged. I feel k may be $1$ but I don't know how to proof it.

答:
If $f$ is in $L^{2}$ then $\sum a_k^2 + b_k^2 <\infty$
Since derivative of $f$ has Fourier coefficients $k*a_k$ if the derivative belongs to $L^2$ then $\sum (k*a_k)^2 + (k*b_k)^2 < \infty$ .
Now, using Cauchy-Schwartz one gets this estimate
$$\sum |a_k| + |b_k| = \sum k*|a_k|/k + k*|b_k|/k <= (\sum (k*a_k)^2 + (k*b_k)^2) * (\sum 1/k^2)$$
So, if $\sum (k*a_k)^2 + (k*b_k)^2 < \infty$ then $\sum |a_k| + |b_k| < \infty$ which implies that $f$ has absolutely convergent Fourier series and is continuous.
In short, if $|k*a_k| < 1/k^{1+s}$ and $|k*b_k| < 1/k^{1+s}$ where $s > 1/2$ then $f$ is in $C^{0}(T)$
Using this trick multi times, one gets - if $|k*a_k| < 1/k^{m+s}$ and  $|k*b_k| < 1/k^{m+s}$ then $f$ is in $C^{m-1}(T)$.


问:
"For no element" symbol
The symbol $\forall$ is known to account every element on given set. Is there such symbol as "for no element", something like a crossed $\forall$?

答:
"There is no $x$ such that $P(x)$" is the same with "(There is $x$ such that $P(x)$) does not hold." Hence Your quantifier is simply the negation of the existential quantifier $\exists$. We already have a symbol for this: $\nexists$. However, I rarely see this symbol in practice. (I can say that logicians, at least, use $\lnot\exists$ more than $\nexists$.)

Another way to state $\nexists x P(x)$, or instead, $\lnot\exists x P(x)$ is $\forall x \lnot P(x)$. Informally, it means "Every $x$ satisfies the negation of $P(x)$." It follows from de Morgan's law, and interestingly, the equivalence between $\lnot\exists xP(x)$ and $\forall x\lnot P(x)$ does not require the law of excluded middle. Hence the equivalence is constructively valid, unlike that between $\lnot\forall x P(x)$ and $\exists x \lnot P(x)$.

Title: "For no element" symbol


问:
Solving equations involving modulo operator
In computer programming languages we have an operator called % which expresses the remainder between two numbers. For example $123\%100 = 23$.

I have an equation evolving this operator, namely,

$$\frac{5}{3}(N\%36) - \frac{2}{3}(N\%6) + 2(N\%25) - (N\%5) = (2N)\%100$$

Is there some easy way to solve this equation in $N$ or at least count the solutions within some range of numbers? (I would like some techniques from number theory rather than a brute force solution, which I have already calculated myself.)

答:
Here's a non-brute force proof that: $$N\equiv 0,1,2,3,4\textrm{ or }5\  (\operatorname{mod 30}),$$ i.e. $N=30a+b$ for some integers $a$ and $b$, with $b = 0,1,2,3,4,$ or $5$. Let $N\%d$ denote the minimal nonnegative remainder after division of $N$ by $d$, such that $0\leq N\%d < N$, where $N$ and $d$ are nonnegative integers. Then
$$N\%36 = N\%6 + 6k,$$
where $k = 1, 2, 3, 4,$ or $5$. Similarly:
$$N\%25 = N\%5 + 5l$$
where $l = 1, 2, 3,$ or $4$, and
$$(2N)\%100 = (2N)\%25 + 25m$$
where $m = 1, 2$, or $3$. Note furthermore that we have
$$(2N)\%25 = 2(N\%25) - 25r,$$
where $r=0\textrm{ or }1$.

The above formulae substitute into your equation as follows:
$$\frac{5}{3}(N\%36) - \frac{2}{3}(N\%6) + 2(N\%25) - (N\%5) = (2N)\%100$$
$$\frac{5}{3}\left[ N\%6 + 6k \right]  -  \frac{2}{3}(N\%6)  +  2\left[ N\%5 + 5l \right] - (N\%5) = 2(N\%25) - 25r + 25m$$
$$N\%6 + 10k + N\%5 + 10l = 2\left[ N\%5 + 5l \right] - 25r + 25m$$
$$N\%6 - N\%5 = -10k - 25r + 25m$$
$$N\%6 - N\%5 = 5(5m -2k - 5r). \tag{$\star$}$$

Eq. $(\star)$ implies that $N\%6 - N\%5$ is divisible by five. However, remember that $\%$ was defined such that $0\leq N\%d < N$: thus $N\%6 \leq 5$ and $N\%5 \leq 4$. Thus, the only way for this difference to be divisible by $5$ is if
$$N\%6 - N\%5 = 0\textrm{ or }5.$$
Suppose that $N\%6 - N\%5 = 0$. Then $N\%6 = N\%5$, and since $0 \leq N\%5\leq 4$, we have:
$$N\%6 = N\%5 = 0,1,2,3,\textrm{ or }4. \tag{$*$}$$
You can convince yourself$^1$ that Eq. $(*)$ implies
$$N\%30 = 0,1,2,3,\textrm{ or }4,$$ i.e.
$$N\equiv 0,1,2,3\textrm{ or }4\  (\operatorname{mod 30}).$$
The other case, $N\%6 - N\%5 = 5$ implies $N\%5=0$ and $N\%6=5$, which implies $N\%30=5$, i.e. $N \equiv 5\ (\operatorname{mod 30})$.
Q.E.D.

This type of logic can probably be extended to give the complete solution to your problem, but I don't want to flog it to death. One first next step would be to show that the case $N\%6=5$, $N\%5=0$ can't occur.

This should anyway give you an idea of how basic arithmetic arguments can be brought to bear on this kind of problem. Sometimes things come to this when all else fails in research-level elementary number theory, for example in the proofs that an odd perfect number would have to have more than $k$ distinct prime factors, for various values of k.

$^1$One way to convince yourself is via the Chinese remainder theorem, but it isn't the only way.


问:
Finding a Set of Generators for the Module of Relations to Compute a Presentation Matrix for an $R$-module
Here is what I know.

Let $R = \mathbb Z[\delta],$ where $\delta = \sqrt{-5}.$ If we denote by $I = (2, 1 + \delta),$ define the surjective map $\varphi: R^2 \rightarrow I$ by $(r,s) \mapsto 2r + (1 + \delta)s.$ We have that $$\operatorname{ker}\varphi =
\left(\begin{matrix}-\frac 1 2(1 + \delta)s\\s\end{matrix}\right)=\left(\begin{matrix}-\frac 1 2(1+\delta)\\1\end{matrix}\right)s,$$ hence a generator for $\operatorname{ker}\varphi$ is $\left(\begin{matrix}-\frac 1 2(1+\delta)\\1\end{matrix}\right).$

But below in a solution of exercise 14.5.1 of Algebra (Second Edition) by Artin, the author did not use this generator when the author needed to use a set of generators of the module of relations to be expressed as a linear combination of the set of generators of $I$ to find a presentation matrix as an $R−$module for the ideal $I.$ So, my question is this.
Does it make any difference in the set of generators I should find that we are working in the ring $R = \mathbb Z[\sqrt{-5}]$?

Here is the solution I referenced above.

Exercise 14.5.1. Let $R=\mathbb{Z}[\delta]$ where $\delta=\sqrt{-5}$. Determine a presentation matrix as an $R$-module for the ideal $(2,1+\delta).$

Proof. Let $\varphi:R^2\longrightarrow I$ that sends $(x,y)\rightsquigarrow2x+(1+\delta)y$. Now
$$\ker\varphi=\left\{(x,y)\in R^2\,\middle|\,x=-\frac{1+\delta}{2}y\right\}\leftrightarrow\left\{y\in R^2\,\middle|\,\frac{1+\delta}{2}y\in R\right\}$$
where $\leftrightarrow$ denotes bijection. But $y=2,1-\delta$ are the elements in $R$ with smallest norm satisfying this condition, and moreover the corresponding vectors $(-3,1-\delta),(-1-\delta,2)\in R$ are independent over $R$ since $2r\ne1-\delta$ for any $r\in R$. Thus, $\ker\varphi$ is generated by these two vectors, and by pp. 424-425 we have a presentation matrix
$$A=\left(\begin{array}{cc}
-3\phantom{-} & -1-\delta\\
1-\delta & \phantom{-}2
\end{array}\right).\qquad\square$$

Also, why do we need the $s$ satisfying the condition of  $\operatorname{ker}\varphi$ to be of minimum norm. Could someone clarify this to me please?

EDIT: Also, is the solution in the picture above correct?

Note that this question here helped me a lot in understanding the idea of the presentation matrix.

答:
There is an immediate problem with your generator of $\ker \varphi$: $2$ is not a unit of $R,$ and $2$ does not divide $1 + \sqrt{-5},$ hence the expression $-\frac{1 + \sqrt{-5}} 2$ is not an element of $R.$

Ultimately, the ring $R$ makes all the difference in finding a set of generators for $\ker \varphi$ in the sense that you must pay attention to the structure of $R$ when finding the generators of $\ker \varphi.$ If $R$ were $\mathbb Q(\sqrt{-5}),$ it would make sense to say that $(r, s)$ is in $\ker \varphi$ if and only if $2r + (1 + \sqrt{-5})s = 0$ if and only if $2r = -(1 + \sqrt{-5})s$ if and only if $r = -\frac{1 + \sqrt{-5}} 2$ because the elements $-\frac 1 2$ and $1 + \sqrt{-5}$ belong to the ring $\mathbb Q(\sqrt{-5}).$ Unfortunately, things are not as simple in $\mathbb Z[\sqrt{-5}].$


答:
(Apologies if this has already been said - MSE is telling me an answer already exists but I can't see it...)

No, $f=\sum_1^\infty\frac{\cos(nx)}n$ shows that $k=1$ does not even imply $f$ is continuous. (Because it's clear that $\sigma_n(f,0)\to\infty$.) Similarly, if $f=\sum\frac{\sin(nx)}{n^2}$ then $f$ is certainly continuous, but not continuously differentiable; thus $k=2$ is not enough.

Otoh if $k>2$ then $f\in C^1$, roughly  because $\sum |n|^{-k+1}<\infty$.

Title: Find $k$ so that $|a_n\;|≤Cn^{-k}$,$\,$ $|b_n\;|≤Cn^{-k}$ => $f∈C^1(T)$ ($a_n$,$b_n$ are the Fourier coefficients of f)

Body: I'm trying to find k that when $|a_n\;|≤Cn^{-k}$,$|b_n\;|≤Cn^{-k}$, ($|a_0\;|≤C$),  f satisfies $f∈C^1(T)$ which can proof Fourier series is converged. I feel k may be $1$ but I don't know how to proof it.


问:
Flux of a vector field that is a partial derivative

Let $X = \partial_x$ and $Y = x \partial_y$ be vectorfields on $M =  \mathbb{R}^2$

I want to calculate the flux of $X$ and $Y$. I am confused on how to approach this problem.

The formula to calculate the flux of a vector field is:
$$\Phi = \int_S F \cdot dS$$
However, If I use $F = \partial_x$ we obtain:
$$\int_S \partial_x \cdot dS$$
which doesn't make a lot of sense to me

Edit: $S$ refers to a surface, meaning that I am using a surface integral

答:
This is ALMOST not an answer, but a follow-up comment.

In the language of vector calculus: when $S$ is a region in the plane, the flux $\Phi$ of a vector field $F$ is calculated across the boundary $\partial  S$ of $S$:
$$ \Phi = \int_{\partial S} F\cdot n\, ds,$$
where $n$ is the outward normal, and $ds$ is the length element (i.e., the integral is one dimensional).

But, by the divergence/Gauss theorem, one can also calculate the flux $\Phi$ by the formula
$$ \Phi= \int_S \mathop{\text{div}} F\, dA,$$
where the integral is now an area integral (i.e., a double integral), and $dA$ is an area element, and $\mathop{\text{div}} F$ is the divergence of $F$ (which is often also denoted $\nabla\cdot F$).

Typically one needs to know $S$ (or $\partial S$). However! In your case, the divergence theorem is pretty relevant (i.e., hint: calculate the divergence!) - I am assuming you know it.

Hope this helps.


问:
Solve differential equation $y\cdot (y')^2+(x-y)\cdot y'-x=0$
I'm working on differential equations and I saw this exercise on my book. I have been trying with no success...

The exercise is the following:

Solve
$$y\cdot (y')^2+(x-y)\cdot y'-x=0$$

and say which is the open interval I in which each solution is defined.

It is not linear, nor exact, not homogeneous either...

Thanks for the assistance and best regards.

答:
You can factor this equation as
$$
(y'-1)(yy'-x)=0.
$$
Any solution can be subdivided into segments where one of these factors is zero,
$$
y(x)=x+A\text{ or }y(x)=\pm\sqrt{x^2+B}.
$$
A switch between factors is possible where both factors are zero at the same time, that is, on the line $y=x$. This does not give new solutions.


问:
The value of $\lim_{n \to \infty}\int_{0}^{1}nx^ne^{x^2}dx$
The value of $\lim_{n \to \infty}\int_{0}^{1}nx^ne^{x^2}dx$ is _____

I tried by taking the odd $n$ values as in that case, the integral I suppose was easier to calculate.
So, denote $I_n=n\int_{0}^{1}e^{x^2}x^ndx$, then we have : $I_1=\frac{e-1}{2},\\I_3=\frac{3}{2},\\I_5=5(\frac{e}{2}-1),\\I_7=7(3-e)$

Then I tried calculating (random) values using calculator as integration was getting cumbersome. $I_{31}=2.488$, $I_{51}=2.57$.

I do not see any kind of recurrence so that I can find a general term for odd $n$. I also tried the method given here. The idea was that since integral doesn't depend on $n$ considering $I(n)=\int_{0}^{1}e^{x^2}x^ndx$, then $I'(n)=\int_{0}^{1}x^n\ln(x)e^{x^2}dx$, but this also doesn't lead me to any conclusion as well.

I believe I couldn't get to a proper approach to tackle this question. Can someone please help with the idea that involves in solving these type of questions?

答:
A bit late answer because it was a duplicate somewhere else.

Simple partial integration gives:

\begin{eqnarray*}I_n =\int_0^1nx^ne^{x^2}dx
& = & \frac n{n-1}\int_0^1(n-1)x^ne^{x^2}dx \\
& = & \frac n{n-1}\left(\left.x^ne^{x^2}\right|_0^1-2\int_0^1x^{n+1}e^{x^2}dx\right)\\
& = & \frac n{n-1}\left(e-J_n\right)
\end{eqnarray*}

with

$$0\leq J_n \leq 2e\int_0^1x^{n+1}dx =\frac{2e}{n+1}\stackrel{n\to\infty}{\longrightarrow}0$$

Hence, $\boxed{\lim_{n\to\infty}I_n = e}$.


问:
Boundary points of union of a semi-closed and an open interval
Let R be the metric space with usual metric. Then what are the boundary points of S=(1 2]U (3 4)?

I think they should be {1 2 3 4}.
But if i use the following two points-

Closure of S is union of S and boundary of S

Closure of given set is- [1 2] U [3 4]
(I used accumulation points to find closure)

If i remove S from closure then boundary points seems to be only {1 3 4}.

Please clarify these 2 answers that which one is correct and why?

And also what will be the boundary points if i take the whole space of complex numbers instead of Real numbers?

答:
Recall:
Let $A \subseteq \Bbb R$ . A point $x$ in $\Bbb R$ is called a boundary point of $A$ if every neighborhood of $x$ intersect both $A$ and its complement. With this definition, $\partial S=\{1,2,3,4\}$

Note that $\overline{S}=S \cup \partial(S)$ is correct. But $S$ and $\partial S$ are need not be disjoint.

So remove $S$ from $\overline{S}$ need not equal $\partial S$ !


问:
What are the possible values of chromatic number of G?
The vertex set of a graph G can be partitioned into two disjoint subsets X and Y such that X induces a clique in G (i.e. a complete subgraph of G) and Y is an independent set. Let m = |X|. What are the possible values of the chromatic number of G in terms of m? Explain your answer.

So this is a question I have to solve for my discrete math 2 class.

The chromatic number of a graph G is the smallest number of colors needed to color the vertices of G so that no two adjacent vertices share the same color. I am not sure how to find the solution of this question.

答: Hints: How many colors are needed just for the vertices in $X$?  What would force an additional color for a vertex in $Y$?


问:
Spivak Prologue, Chapter 2, problem 7: prove $\sum_{i=1}^n k^p$ can always be written as sum of powers of n
This is problem 7 of Spivak's Calculus, 4th Edition, Prologue Chapter 2:

Show that $$\sum_{i=1}^n k^p$$ can always be written in the form $$\frac{n^{p+1}}{p+1} + An^{p} + Bn^{p-1} + Cn^{p-2} + ...$$

There is also a hint to use the method from problem 6.

I have a few questions on some specific details about the solution in the solution book, which I reproduce in what follows:

Let A be the set of all natural numbers for which $$\sum_{i=1}^n k^p$$ can be written in the form $$\frac{n^{p+1}}{p+1} + An^{p} + Bn^{p-1} + Cn^{p-2} + ...$$.

Using induction on $p$, for $p=1$ we have

$$\sum_{k=1}^n k = \frac{n(n+1)}{2} = \frac{n^2}{2} + n$$

so $p=1$ is in $A$.

Suppose the statement is true for all natural numbers smaller than $p$ (actually the solution book has smaller than or equals; so here is the first question: is it $<$ or $\leq$ here?).

The Binomial Theorem yields: $$(k+1)^{p+1}-k^{p+1} = (p+1)k^p + terms\ involving \sum_{k=1}^n k^r for\ r < p\tag{1}$$

By our induction assumption, we can write each $\sum_{k=1}^n k^r$ as an expression involving powers $n^s$ with $s\leq p$.

Writing out $(1)$ for $k=1,...,n$ we obtain:
$$2^{p+1}-1^{p+1}=(p+1)1^p +\ ...$$
$$3^{p+1}-2^{p+1}=(p+1)2^p +\ ...$$
$$(...)$$
$$(n+1)^{p+1}-n^{p+1}=(p+1)n^p +\ ...$$

Adding the equations, we obtain:

$$(n+1)^{p+1} - 1 = (p+1)\sum_{k=1}^n k^p +\ terms\ involving\ \sum_{k=1}^n k^r,\ for\ r<p\tag{2}$$

Now at this point, the solution actually says we obtain something else: $$\frac{(n+1)^{p+1}}{p+1}=\sum_{k=1}^n k^p+\ terms\ involving\ \sum_{k=1}^n k^r,\ for\ r<p\tag{3}$$

So the second question is: what assumptions are involved in obtaining $(3)$? Ie, in (2) does the term 1 on the lefthand side cancel with one of the terms on the righthand side? It is also true that each term on the righthand side has a (p+1) term, so is that term the one dividing the lefthand side?

Finally, as we noted above, because of the induction assumption for all natural numbers less than $p$, we can write each $\sum_{k=1}^n k^r$ as an expression involving powers of $n^s$ with $s\leq p$. Therefore, isolating $\sum_{k=1}^n k^p$ we get: $$\sum_{k=1}^n k^p = \frac{(n+1)^{p+1}}{p+1}+\ terms\ involving\ powers\ of\ n\ less\ than\ p+1$$.

Therefore, if $s$ is in $A$ for $s<p$ then $p$ is in $A$; therefore by induction $A=N$.

Final question: is this considered a rigorous proof?

答:
I think Spivak has just been a little lax, probably intentionally since it was not his point of focus. If we go a little more precisely into things, we can be far more direct : we can get $A,B,C$ etc.

Indeed, fix $n$ and $N$ positive integers. We have by the binomial formula:
$$
(N+1)^{n+1} = \binom{n+1}{0}N^0 + \binom{n+1}{1} N^{1} + \ldots + \binom{n+1}{n}_{(=n+1)} N^{n} + \binom{n+1}{n+1}_{(=1)} N^{n+1} \\ \implies (N+1)^{n+1} - N^{n+1} = \binom{n+1}{0}N^0 + \binom{n+1}{1} N^{1} + \ldots +(n+1) N^{n} \\ \implies \frac{(N+1)^{n+1} -N^{n+1}}{n+1} = \frac{\binom{n+1}{0}}{n+1}N^0 + \frac{\binom{n+1}{1}}{n+1} N^1 + ... + N^n \\
\implies N^n = \frac{(N+1)^{n+1} -N^{n+1}}{n+1} -\frac{\binom{n+1}{0}}{n+1}N^0 - \frac{\binom{n+1}{1}}{n+1} N^1 - \cdots - \frac{\binom{n+1}{n-1}}{n+1} N^{n-1}
$$

and now when we sum from $N=0$ to $K$, the first term on the right has telescoping cancellations and the rest sum up . The following emerges :
$$
\sum_{N=0}^K N^n = \frac{(K+1)^{n+1}}{n+1} - \frac 1{n+1} - \frac{\binom{n+1}{0}}{n+1}\sum_{N=0}^K N^0 - \frac{\binom{n+1}{1}}{n+1} \sum_{n=0}^K N^1 - \cdots - \frac{\binom{n+1}{n-1}}{n+1} \sum_{n=0}^K N^{n-1}
$$

which provides the exact relation between the sums in question, the super-rigorous and exact proof. This can be used to calculate the first few sum formulas as well, and also provides intuition for why the Bernoulli numbers come into the general formula, from their recurrence relation. You can see here for more details.

Of course, note that Spivak did not want to mention all this because he felt that writing out the details was not his intention there, he just wanted to use induction. If you want a proof that does just that, without giving you an explicit formula, then your formulation will be slightly different , and you will have the following :

For each $n = 1,2,...$ there exist rational numbers $C_{n,0},C_{n,1},...,C_{n,n}$ such that for every $K=1,2,...$ , we have :
$$
\sum_{N=0}^K N^n = \frac{(N+1)^{n+1}}{n+1} + C_{n,n}N^n + ... + C_{n,0}
$$

and how to prove this? Now go by induction, and $n=1$ is clear, and for induction, assume truth for $1,...,n-1$ and you want to prove the statement for $n$. All you need to do is repeat the above argument , but replace each exact quantity (like the binomial coefficients) except $\binom{n+1}{n}$ and $\binom{n+1}{n+1}$ by saying there exist rational numbers in those places. So for example, you would use binomial theorem to say : there exist rational numbers $r_0,...,r_{n-1}$ such that :
$$
(N+1)^{n+1} = r_0N^0 + r_1 N^{1} + \ldots + (n+1)N^{n} +  N^{n+1}
$$   Then you just have to do the summation, perform the manipulations exactly as done before, and you will land up with the same sum, except with $r_i$ instead of the binomial coefficients, which does it. (super-rigorously, if you like!)


问:
The equation of a plane orthogonal to another plane, when the intersection of the planes is a line from the xOy plane
Determine the equation of a plane orthogonal to the plane:
$ (P) : 3x-y+3z-2=0$
and whose intersection with $(P)$ is a line from the $xOy$ plane.

My idea is the following: since the planes are orthogonal it means that the dot product of the normal vectors is $0$.I think that since the line is from $xOy$, the dot product would be $3a-b=0$, but I'm not sure. Could you please show me what a line from the $xOy$ plane looks like?

答: Let us first see what's the line which belongs both to $P$ and to the plane $xO$. For that we solve the system$$\left\{\begin{array}{l}3x-y+3z=2\\z=0,\end{array}\right.$$which is equivalent to$$\left\{\begin{array}{l}3x-y=2\\z=0\end{array}\right.$$Two solutions of this system are $(1,1,0)$ and $(0,-2,0)$. So, you are after a plane which $ax+by+cz=d$ which contains these two points and which is orthogonal to $P$. So, you solve the system$$\left\{\begin{array}{l}a+b=d\\-2b=d\\3a-b+3c=0.\end{array}\right.$$One solution is $a=\frac32$, $b=-\frac12$, $c=-\frac53$, and $d=1$. So, take the plane$$\frac32x-\frac12y-\frac53z=1\left(\iff9x-3y-10z=6\right).$$


问:
graph colouring
The graph G has been obtained from the complete graph Kn by replacing each edge $e = uv$ by a 2-edge path $ux_ev$, where each $x_e$ is a new vertex. Find chromatic index of the new graph?

can anyone explain what happens to the chromatic index because in some cases it increases by one and some it stays the same

答:
The chromatic index $\chi'(G)$ is always either the maximum degree $\Delta(G)$, or else $\Delta(G)+1$.

Often it is hard to determine which it is, but for bipartite graphs - such as the one in this question - we always have $\chi'(G) = \Delta(G)$.

Title: graph colouring


Title: How to calculate $ \int_0^{\pi/2}\log(1+\sin(x))\log(\cos(x)) \,dx $?

Body:
How to calculate $$ \int_0^{\pi/2}\log(1+\sin(x))\log(\cos(x)) \,dx \,\,?$$
I tried to use the Fourier series of log sine and log cos and I got that the integral is equal to :
$$ \frac{\pi^2}{24}-\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{n+k}}{k(4k^2-(2n-1)^2)}$$
has anyone a idea to how to find the closed-form of the last series or how to start out differently with the integral?

Body:
Substitute $t=\tan\frac x2$

\begin{align} &\int_0^{\pi/2}\ln(1+\sin x)\ln(\cos x) \,dx \\
=&\>2\int_0^{1}\frac{\ln\frac{(1+t)^2}{1+t^2}\ln \frac{1-t^2}{1+t^2} }{1+t^2}\,dt
=\>4I_1 +4 I_2 -2I_3- 6I_4+2I_5
\end{align}
where, per the results

\begin{align}
I_1 &= \int_0^1 \frac{\ln (1+t)\ln(1-t)}{1+t^2} dt
= -G \ln 2-K+\frac{3 \pi ^3}{128}+\frac{3\pi}{32} \ln ^22\\
 I_2 &= \int_0^1 \frac{\ln^2(1+t)}{1+t^2} dt
= -2 G \ln 2-4 K+\frac{7 \pi ^3}{64}+\frac{3\pi}{16} \ln ^22 \\
I_3 &= \int_0^1 \frac{\ln (1+t^2)\ln(1-t)}{1+t^2} dt
=  -\frac{1}{2} G \ln 2+4 K -\frac{5 \pi ^3}{64}+\frac{\pi}{8}  \ln ^22 \\
 I_4 &= \int_0^1 \frac{\ln (1+t^2)\ln(1+t)}{1+t^2} dt
=  -\frac{5}{2} G \ln 2-4 K+\frac{7 \pi ^3}{64}+\frac{3\pi}{8} \ln ^22\\
 I_5 &= \int_0^1 \frac{\ln^2(1+t^2)}{1+t^2} dt
= -2 G \ln 2+4 K-\frac{7 \pi ^3}{96}+\frac{7\pi}{8} \ln ^22
\end{align}

with $K= \Im\text{Li}_3\left(\frac{1+i}{2}\right)$. Together
$$ \int_0^{\pi/2}\ln(1+\sin x)\ln(\cos x) \,dx
=4\Im\text{Li}_3\left(\frac{1+i}{2}\right)-\frac{11\pi^3}{96}+\frac{3\pi}8\ln^22
$$


问:
Connection between semimartingales in Itô calculus and measures in integration theory
I'm currently learning about stochastic calculus for the first time. I have taken a first course in Real Analysis covering Lebesgue integration.

According to Wikipedia, we write out the Itô stochastic integral as

$$Y_t = \int_0^{t}H_s \,dX_s$$

where
"$H$ is a locally square-integrable process adapted to the filtration generated by $X$, which is a Brownian motion or, more generally, a semimartingale." I'm confused because $X_s$ is in the same location in our integral as a measure would be in Lebesgue integration. According to Wikipedia, we write the Lebesgue integral as

$$\int f \,d\mu$$

where $\mu$ is a measure on the domain of $f$ and $f$ is measurable with respect to $\mu$.

In finance, we often use $X_s$ to model the price of a security. $X_s$, or changes in $X_s$, can be negative. This makes it difficult to connect semimartingales and measures.

How is $X_s$ related to measures? Are there ways to interpret $X_s$ as a measure?

答:
As noted in the comments, the concept of a stochastic integral is much more resembling the Lebesgue-Stieljes integral

$$\int _{a}^{b}f(x)\,dg(x)$$
which is defined for  ${\displaystyle f:\left[a,b\right]\rightarrow \mathbb {R} }$  Borel-measurable and bounded and  ${\displaystyle g:\left[a,b\right]\rightarrow \mathbb {R} }$  having bounded variation in $[a, b]$.
I want to emphasize that the condition on $g$ to have bounded variation is crucial here, otherwise this integral is ill-defined. As it turns out, this condition is often not satisfied in the context of stochastic processes.

Rather than considering a general semimartingale $(X_t)_{t \geq 0}$, let's take a much more concrete object to make this point clear.  Let $B:=(B_t)_{t \geq 0}$ be the one-dimensional standard Brownian motion, for example. It is well-known and not so hard to show that $B$ indeed has infinite variation. Hence,  in order to make sense of an expression like
$$
\int_{0}^{t} d B_s\
$$
which intuitively tries to capture the idea of integrating with respect to the Brownian motion $B$, we need to provide some additional justification as this transcends the assumptions we impose on $g$ in the Lebesgue-Stieljes-integral above.

Rather than explaining the construction of a stochastic integral from stratch, I refer to LeGall's succinct and precise exposition in "Brownian Motion, Martingales, and Stochastic Calculus" Ch.4 and Ch.5. In particular, 4.1.1. will help you to gain an understanding of why, as you phrase, the $X_s$ appear at the same place as $\mu$.

Finally, to address your concern that $X$ may be negative, but $\mu$ not, I encourage you to familiarize with the notion of signed measures. This is a generalization of the concept of measure by allowing it to have negative values.


答:
The boundary points of $S=(1,2] \cup (3,4)$ are indeed $\{ 1,2,3,4 \}$ as you say. It is also true that the closure of (any set) $S$ is $S \cup \partial S$. But nobody says this union has to be disjoint. $\partial S$ can overlap with $S$, and it does in this case: $S \cap \partial S = \{ 2 \}$. So that is why $\partial S \neq \overline S \setminus S$. In this case you only have $\overline S \setminus S \subsetneq \partial S $. In general this inclusion does hold (however it might be strict as it is in this case), because we have even $\partial S = \overline S \setminus Int(S)$.

If you see $S$ as a subset of $\mathbb C$, the answer stays the same: $\partial S = \{ 1,2,3,4 \}$. To see this, one way is to use the characterisations of closure and interior with sequences inside $S$, only this time sequences of complex numbers. But since $S$ is on the $O_x$ axis, any the imaginary part of all elements of all such sequences will be $0$ (sequences have to stay inside $S$).

A quicker way to see it if you know a bit of topology and if you defined closure and interior with closed and open sets instead of sequences is: $S$ is included in $\mathbb R$ and the subspace topology on $\mathbb R$ inherited from $\mathbb C = \mathbb R^2$ (the subspace topology on $X \subseteq Y$ is just the topology where open sets are intersections of open sets in $Y$ with $X$) is the same as the usual topology of $\mathbb R$. Also, if you have the inclusion $X \subseteq Y \subseteq Z$, then the topology $X$ inherits from $Y$ (the topology on $Y$ being the one inherited from $Z$) is the same as the topology $X$ iherits directly from $Z$. So $S$ has the same topology inside $\mathbb R$ as it does inside $\mathbb C$, hence it has the same boundary and closure.

Title: Boundary points of union of a semi-closed and an open interval

Body:
Let R be the metric space with usual metric. Then what are the boundary points of S=(1 2]U (3 4)?

I think they should be {1 2 3 4}.
But if i use the following two points-

Closure of S is union of S and boundary of S

Closure of given set is- [1 2] U [3 4]
(I used accumulation points to find closure)

If i remove S from closure then boundary points seems to be only {1 3 4}.

Please clarify these 2 answers that which one is correct and why?

And also what will be the boundary points if i take the whole space of complex numbers instead of Real numbers?


问:
every integers from 1 to 121 can be written as 5 powers of 3
We have a two-pan balance and have 5 integer weights with which it is possible to weight
exactly all the weights integers from 1 to 121 Kg.The weights can be placed all on a plate but you can also put some in a dish and others with the goods to be weighted.

It's asked to find the 5 weights that give us this possibility. It also asks you to prove that the group of five weights is the only one that solves the problem.

Easily i found the 5 weights: 1 , 3 , 9 , 27 , 81

but i can' demonstrate that this group of five weights is the only one that solves the problem.

Can you help me ?

Thanks in advance !

答:
A more general result says, given weights $w_1\le w_2 \le \dots \le w_n$, and if $S_k = \sum_{i=1}^k w_k$; $S_0 = 0$, then everything from $1$ to $S_n$ is weighable iff each of the following inequalities hold:

$$S_{k+1} \le 3S_k + 1 \text{ for } k = 0,\dots,n-1$$

Note that this is equivalent to $w_k \le 2 S_k +1$ for each $k$.

If you wish, a proof is given here.


问:
The expectation of $e^X \left(1-(1-e^{-X}\right)^n)$ when $X$ has Exponential Distribution
To my surprise, I was able to evaluate the following expression in Mathematica:

$$E\left[e^X \left(1-(1-e^{-X}\right)^n) \right] = \frac{y}{y-1} \left(1-\frac{1}{\binom{n+y-1}{y-1}}\right)\quad X\sim\text{Exp}(y)$$
with the right hand side being equal to $\text{HarmonicNumber}(n)$ in the particular case $y=1$, and equal to $n$ when $y=0$.

If we define $\binom{n}{k} = \frac{\Gamma(n+1)}{\Gamma(n-k+1)\Gamma(k+1)}$ the results seems to hold for all $x,y\in\mathbb R$, though I mostly care about $n$ and $y$ as positive integers.

I have no idea how to prove this by hand.
I tried a series expansion of the exponentials without realizing much.
I also tried rewriting in terms of the uniform distribution, since $e^{-X}$ has the same distribution as $U^{1/y}$ for $U\sim\text{Uniform}(0,1)$.

Are there any tricks or properties I'm missing?

Body:
Denote
$$I = E\left[e^X \left(1-(1-e^{-X}\right)^n) \right] =\int_0^{+\infty}e^x(1-(1-e^{-x})^n)ye^{-yx}dx$$
Let's make a change of variables $z = 1-e^{-x}$ then $x = -\ln(1-z)$ and $dx = \frac{1}{1-z}dz$. The integral is from $z = 0$ to $z=1$.
We have:

\begin{align}
I & = \int_0^{+\infty}y(1-(1-e^{-x})^n)e^{-(y-1)x}dz \\
& = \int_0^{1}y(1-z^n)(1-z)^{y-1}\frac{1}{1-z}dz \\
& = \int_0^{1}y(1-z)^{y-1}(\sum_{k=0}^{n-1} z^k)dz \\
& = \sum_{k=0}^{n-1} \left( \int_0^{1}y(1-z)^{y-1}z^kdz  \right)   \\
\end{align}

By using Mathematica, we have
$$\int_0^{1}y(1-z)^{y-1}z^kdz = \frac{\Gamma(1+k)\Gamma(y+1)}{\Gamma(1+k+y)}$$
Hence,
$$I = \sum_{k=0}^{n-1} \left( \frac{\Gamma(1+k)\Gamma(y+1)}{\Gamma(1+k+y)}  \right)$$

I resolve the case you asked where $y\in \mathbb{N}^*$, we have

\begin{align}
I &= \sum_{k=0}^{n-1}  \frac{y!k!}{(k+y)!} \\
 &= y!\sum_{k=0}^{n-1}  \frac{1}{(k+1)...(k+y)} \\
&= \frac{y!}{y-1}\sum_{k=0}^{n-1}  \left( \frac{1}{(k+1)...(k+y-1)} - \frac{1}{(k+2)...(k+y)} \right) \\
&= \frac{y!}{y-1} \left( \frac{1}{(y-1)!} -  \frac{n!}{(n+y-1)!} \right) \\
&=\frac{y}{y-1}\left(1-\frac{1}{\binom{n+y-1}{y-1}}\right)
\end{align}


问:
Smoothly connecting PDEs with finite differences

A PDE with non-smooth inhomogeneity

Let $\mathcal{L}$ be a second-order, linear, elliptic differential operator acting on $\mathcal{C}^2([0,2]^2)$.

I'm numerically solving the inhomogeneous PDE
\begin{align*}
\mathcal{L}u(x,y)+(x-1)^+=0,
\end{align*}
where $(\cdot)^+$ denotes the positive part.

Put differently, I solve two PDEs which need to be connected at $x=1$:
\begin{align*}
\begin{cases}
\mathcal{L}u(x,y)=0 & \text{for }(x,y)\in[0,1]\times[0,2], \\
\mathcal{L}u(x,y)+x-1=0& \text{for }(x,y)\in(1,2]\times[0,2].
\end{cases}
\end{align*}

Approximating  all partial derivatives by central differences, I get the nine-point stencil
\begin{align*}
c_1 u_{i-1,j-1} + c_2 u_{i,j-1} + c_3 u_{i+1,j-1} + c_4 u_{i-1,j} + c_5 u_{i,j} &\\
  + c_6 u_{i+1,j}+ c_7 u_{i-1,j+1} + c_8 u_{i,j+1} + c_9 u_{i+1,j+1} + (x_i-1)^+ &=0.
\end{align*}
Thus, $u$ is the solution to a system of linear equations.

Problem

Plotting the solution $u$ for a fixed $y$, it all looks fine and perfect. However, a plot of $\frac{\partial u}{\partial x}$ as a function of $x$ shows that the derivative is not smooth at $x=1$. The above FD scheme works fine with value-matching (the solution $u$ is perfectly continuous) but struggles with smooth-pasting at $x=1$ (the derivative is not smooth).

Question

How do I ensure smooth-pasting with a finite difference scheme at $x=1$?

Some of my failed attempts include

Impose that forward and backward differences at $x=1$ equal each other (= 2nd order central difference is zero).

Use higher order approximations around $x=1$ such as $u_{xx}\approx \frac{-u(-2h)+16u(-h)-30u(0)+16u(h)-u(2h)}{12h^2}$ and $u_{x}\approx \frac{u(-2h)-8u(-h)+8u(h)-u(2h)}{12h}$ and central differences everywhere else.

Approximating partial derivatives using points only from one side of $x=1$ (i.e. only using either $u(0), u(h), u(2h)$ or instead $u(0),u(-h),u(-2h)$).

Imposing that $u_x\approx\frac{u_{i+1,j}-u_{i-1,j}}{2h}$ equals an average of partial derivatives at $1+h$ and $1-h$.

Note: This problem arises as part of a larger system of free boundary problems. Thus, it's necessary to solve the PDE numerically. This question is also posted here.

答: The solution will be continuous in both $u$ and $u_x$, but the latter won't be smooth at $x=1$. This means the centered difference there won't be second order.


问:
parabolic volume integration mistake
While dealing with volume integral of paraboloid bowl, say the volume within the surface $x^2 + y^2 = r^2 $ and $z = r^2$, where $ 0\le r\le \sqrt{3} $. One classic way seems to firstly have infinitesimal volume via slicing area and infinitesimal height:  $$dV =Adz= \pi r^2dz=\pi z\,dz,$$ and then integrate for total volume $$V= \int {dV}= \int_{z=0}^{3} \pi z \,dz =\frac{1}{2} \pi z^2 |_{0}^{3}=\frac{9}{2} \pi .$$
When I practiced integration via cylinder coordinate $(R,\theta,z)$, I got wrong result but unable to spot where I did wrong. The infinitesimal volume I derived: $$dV=dA*dz=(R\,d\theta\, dR)\,dz,$$
and then I got the total volume$$V= \int {dV}= \int_{z=0}^{3}\int_{\theta=0}^{2\pi}\int_{R=0}^{\sqrt{3}} R\,d\theta \,dR\,dz =\frac{1}{2} \pi z^2 |_{0}^{3}=9 \pi.$$
Sincerely hope my mistake can be pointed out.

答:
The correct integral should be

$\displaystyle \int_{z=0}^{3}\int_{\theta=0}^{2\pi}\int_{r=0}^{\sqrt{z}} r \ dr \ d\theta \ dz$

Please note that the radius is not $\sqrt3$ throughout. It increases with $z$ and for any given $z$, the upper bound of radius is $\sqrt z \ $. If you take upper bound of $r$ independent of $z$, you will get volume of the cylinder of radius $\sqrt3$ and of height $3$.

Title: parabolic volume integration mistake


问:
Expectation of sample variance
Let $s^2$ be sample variance, $\sigma^2$ be population variance

$E[\frac{(n-1)s^2}{\sigma^2}] = E[\chi^2_{n-1}] = (n-1) \implies \frac{(n-1)E[s^2]}{\sigma^2} = n-1 \implies E[s^2]=\sigma^2$

But if i do in followng way, i am getting wrong answer

$E[s^2]=E[\frac{1}{n-1} \sum_{i=1}^n (x_i-\bar x)^2] \\ = \frac{1}{n-1} \sum_i (E[x_i^2] + E(\bar x^2) - 2E[\bar x]E[x_i])$

I know that

$E[\bar x] = \mu$, population mean

$E[x_i^2] = var(x_i)+E[x_i]^2 = \sigma^2+\mu^2 \\ E[\bar x^2] = var(\bar x)+E[\bar x]^2 = \frac{\sigma^2}{n}+\mu^2$

Plugging these in above boldface equation

$E[s^2] = \frac{1}{n-1} (n(\sigma^2+\mu^2)+n(\frac{\sigma^2}{n}+\mu^2)-2\mu^2) \\ \frac{n+1}{n-1} \sigma^2$

I know this is wrong. But i donot know where i made the mistake.

答:
the mistake is that
$$
\mathbb{E}[X_i \bar{X}] \neq\mathbb{E}[X_i] \mathbb{E}[\bar{X}]
$$
since this would imply that $X_i$ and $\bar{X}$ are independent, which they clearly are not. To solve this, expand $X_i \bar{X} = X_i \frac{1}{n} \sum_{j=1}^nX_j$ to show that

\begin{align*}
\mathbb{E}[X_i \bar{X}] &= \frac{1}{n} ((n-1)\mathbb{E}[X_1]\mathbb{E}[X_2] + \mathbb{E}[X_1^2])\\
&= \frac{1}{n} ((n-1)\mu^2 + (\sigma^2 + \mu^2))\\
&= \mu^2 + \frac{\sigma^2}{n}
\end{align*}

Title: Expectation of sample variance