问:
Prove that the set $A’ = \{x \in \mathbb{R}^k \mid x \in \partial (A/\{x\})\} $ is closed
Let $A$ be a set in $\mathbb{R}^k$.

Prove that the following set:

$$
A’ = \{x \in \mathbb{R}^k \mid x \in \partial (A/\{x\})\}
$$

is a closed set.

What i tried:

Lets try to look at every convergent sequence in $A’$, we will show that it converge to an element $\in A’$, therefore the set is closed. (Its not working...)

The problem is that i dont know how to prove it:

Assume that $\{x_n\}$ is a convergent set in $A’$, now lets assume that it converges to a point $L \notin A’$. Now i dont see any problem that the points will become closer to a point not in $A’$...

Another way is to look at the complement set, if i prove that the complement is open, i proved what is neede, yet here it is very confusing, it satisfies:

$$
\overline{A} = \{ x \in \mathbb{R}^k \mid x \notin \partial (A/\{x\}) \}
$$
This is very confusing and i dont know how to decide if its open or close set, can someone give me a hint?

Thank you.

答:
Let's take the complement of $A'$ as you defined it: $x \notin A'$ iff $x \notin \partial (A\setminus\{x\})$, which means by definition that there is some ball $B(x,r)$ around $x$ that does not intersect $A \setminus \{x\}$ or the complement of $A\setminus\{x\}$. But $x \in B(x,r)$ always and $x$ is in the complement of $A\setminus\{x\}$ so that the second options does not hold.

So $B(x,r) \cap (A\setminus\{x\}) = \emptyset$ holds for some $r>0$.

Now, I claim that in fact $B(x,r)$ is a subset of the complement of $A'$ so that $x$ is an interior point of $\Bbb R^k \setminus A'$ and so $A'$ is closed, as its complement is open.

To see the claim, it suffices to show that any $y \in B(x,r)$ that is not equal to $x$ is not in $A'$. (for $x$ we already know this).
So let $y \neq x$ be given in $B(x,r)$. Then let $r' = \min(d(x,y), r-d(x,y)) > 0$

Then the choice of $r'$ ensures that $x \notin B(y,r')$ and $B(y, r') \subseteq B(x,r)$ and so $B(y,r') \cap (A\setminus \{y\}) = \emptyset$. So This ball witnesses that $y \notin \partial (A\setminus\{y\})$ and hence $y \notin A'$. QED.


问:
Wolframalpha yields different results of matrix multiplication based on parentheses placement
I was trying to multiply some matrices using wolframalpha when I found that the results didn't match up. Specifically, matrix expression like $CX^2C^{-1}$ produced different result than $CXC^{-1}CXC^{-1}$, although these should be the same. I found that this happens because wolframalpha calculates the product differently based on how the parentheses are placed. Here are two examples:

First expression

```<code>({{1,1},{-1,0}}*{{2,1},{1,2}}*{{1,1},{-1,0}}^-1)*{{1,1},{-1,0}}*{{2,1},{1,2}}*{{1,1},{-1,0}}^-1
</code>```

Second expression:

```<code>{{1,1},{-1,0}}*{{2,1},{1,2}}*{{1,1},{-1,0}}^-1*{{1,1},{-1,0}}*{{2,1},{1,2}}*{{1,1},{-1,0}}^-1
</code>```

Is there an adequate reason for this or is it a bug?

答:
As it was mentioned in a comment, you are missing a dot in the first expression. When you don't use the dot (or use *) Wolfram will compute the product as
$$ (A*B)_{ij} = a_{ij} b_{ij}$$
when you really wanted it to compute
$$(A.B)_{ij} = \sum_{k=1}^n a_{ik} b_{kj}$$

In the second expression, you always use "*", so you are not computing what you expect.

Computers don't do what we want, they do what we ask.

Also consider that, even though this does not seem to be a problem for wolframalpha, in Wolfram Mathematica  $\texttt{A^(-1)}$ does not represent the inverse... you must use $\texttt{Inverse[A]}$.

This post was probably a better fit for
http://mathematica.stackexchenge.com

ContentLicense: CC BY-SA 4.0


问:
Show that $n^2\log\left(1+\frac{1}{n}\right)$ does not converge to $1$

How to show that $n^2\log\left(1+\dfrac{1}{n}\right)\to 1$ is false?

I have to show that $\left(1+\dfrac{1}{n}\right)^{n^2}$ doesn't tend to $e.$

ContentLicense: CC BY-SA 3.0

Body: $$\log\left(1+\frac1n\right)=-\log\left(1-\frac1{n+1}\right)\geqslant\frac1{n+1}$$


问:
Direct sum of dual space
Let $E$ be a finite vector space and $L_1$, $L_2$ are subspaces.
I want to show that :

if $E=L_1\oplus L_2$ than $E^*=L_1^*\oplus L_2^*$.

I attempted:

$\forall x\in E, x=x_1+x_2$ where $x_i\in L_i$

now if $\phi\in E^*$, we have $\phi(x)= \phi(x_1)+\phi(x_2) $.

here I want to consider a forme $\phi_1$ that take $\phi (x_1)$ as image, and the same for $\phi_2$ but I had some strugle.

Furthermore, I search here for a response but I found just this post, so I' wondering : Do I need other conditions to show this?

答:
Note that with finite-dimensional spaces, there is an isomorphism $E \cong E^*$. Suppose it is over a field $\mathbb{F}$, so $L_1 \cong \mathbb{F}^m$, $L_2 \cong \mathbb{F^n}$ and $E \cong \mathbb{F}^{n+m}$.

Then, $L_1^* \cong L_1 \cong \mathbb{F}^n$ and similar for $L_2$.

Hence, $E^* \cong \mathbb{F}^{n+m} \cong \mathbb{F}^n \oplus \mathbb{F}^m \cong L_1^* \oplus L_2 ^*$

Title: Direct sum of dual space


问:
A simple $n\times n$ system of linear equations
Can anyone provide the analytical solution of the following $n\times n$ system of linear equations

$\underbrace{\begin{pmatrix}
 a & -b & -b & ... & -b\\
 -b & a & -b & ... & -b\\
 -b & -b & a & ... & -b\\
  : &  : & : & ... & : \\
  : & : & : & ... &: & \\
 -b & -b & -b & ... & a \\
 \end{pmatrix}}_{A}\underbrace{\begin{pmatrix}x_1\\
x_2\\
x_3\\
:\\
:\\
x_n
\end{pmatrix}}_{X}=\underbrace{\begin{pmatrix}y_1\\
y_2\\
y_3\\
:\\
:\\
y_n
\end{pmatrix}}_{Y}
$
where $A_{(n\times n)}$ is a symmetric matrix and $a$, $b$, $y_i$ are positive real numbers.

Note: I would appreciate if someone could provide more details about the theory of eigenvalues-eigenvectors related to the solutuion of a linear system of eaquations.

答:
The matrix is of type: linear combination of the identity and a rank $1$ matrix, the latter being the matrix all of whose entries are $-b$, call it $B$. Then $B$ is easily diagonalisable, with an $n-1$ dimensional kernel (eigenspace for $0$) given by the equation $x_1+\dots+x_n=0$, and a $1$ dimensional other eigenspace, spanned by $(1,1,\ldots,1)$, whose eigenvalue for $B$ equals $-nb$. We have $A=B+(a+b)I_n$; the eigenspaces of $B$ are also eigenspaces for $A$, but with eigenvalues $\lambda=a+b$ respectively $\mu=-nb+a+b=a+(1-n)b$.

The linear problem is easy to solve on each eigenspace: on the $n-1$ dimensional eigenspace $A$ acts by multiplication by $\lambda$ and, provided $\lambda\neq0$, the solution will be $X=\frac1\lambda Y$, while on the $1$-dimensional eigenspace $A$ acts by multiplication by $\lambda$ and, provided $\mu\neq0$, the solution will be $X=\frac1\mu Y$. For the general solution one can decompose $Y$ into its projections $Y_1$ and $Y_2$ on the two eigenspaces, and provided both $\lambda=a+b$ and $\mu=a+(1-n)b$ are nonzero one can find the solution by adding the solutions associated to $Y_1$ and $Y_2$. One has $Y_2=\frac{(y_1+\cdots+y_n)}n(1,1,\ldots,1)$ and $Y_1=Y-Y_2$; the general solution then is $X=\frac1\lambda Y_1+\frac1\mu Y_2$. Working this out explicitly is a bit messy, but straightforward.

Here is a concrete example if you find this too abstract. Take $n=5$, $a=2$ and $b=1$. Then neither $\lambda=a+b=3$ nor $\mu=a+(1-n)b=-2$ is zero, so the problem is solvable for all $Y$. Take for instance $Y=(3,-1,2,5,-4)$, it decomposes as the sum of a multiple $Y_2$ of $(1,1,1,1,1)$ by the scalar $\frac{3-1+2+5-4}5=1$, and a vector $Y_1=Y-Y_2=(2,-2,1,4,-5)$ whose sum of coordinates is $0$. The solution for $X$ then is
$\frac1\lambda Y_1+\frac1\mu Y_2=\frac13(2,-2,1,4,-5)-\frac12(1,1,1,1,1)=\frac16(1,-7,-1,5,-13)$.


问:
How to describe the Caratheodory extension induced by this step function?
Define a function $$F(x) = \begin{cases} 1
\text{ if } x > 0\\ 0 \text{ if } x \leq 0  \end{cases}$$
This left continuous function induce a measure $\mu$ on $\mathbb{R}$, defined by
$$
\mu(A) = \inf \Big\{ \sum_{i=1}^n F(a_i) - F(b_i) \mid A \subset \bigcup_{i=1}^{n}(a_i, b_i] \Big\}
$$

what is the $\mu(A)$ for  $A \subset \mathbb{R}$ when $A$ is measurable?

For an interval of the $(a, b]$, it is obvious that $\mu((a,b]) = 1$ if $b > 0$, and $\mu((a,b]) = 0$ if $b \leq 0$. But what about some more complicated cases?

My naive guess is that $\mu(A) = 0$ if $A$ contains a elements bigger than $0$, but I am not sure if it is correct.

答: $\mu$ is simply the 'delta measure at $0$': $\mu(A)=1$ if $0 \in A$ and $\mu(A)=0$ otherwise.


问:
Find the character table of $\mathbb{Z} / 3 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z} $
I am trying to create a character table for $\mathbb{Z} / 3 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z} $ and I am having some difficulties.

I know that the number of irreducible complex representations of a finite group G is equal to the number of it's distinct conjugacy classes. As $G=\mathbb{Z} / 3 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z} $ is finite, and in particular it has 6 conjugacy classes (which must be (0,0), (1,0), (0,1), (1,1), (2,0) and (2,1), right?) thus there must be 6 irreducible complex representations. Additionally, as G is abelian then all the irreducible complex representations are 1-dimensional. This add up nicely, as the number of such representations is then equal to the order of G.

Now, I am not not sure how to use the fact that they are 1-dimensional to create a character table, as I am not sure what the representations actually are. I have tried following an example, but I am a bit confused on how to do this. In the example the representations has been denoted $T_i$ but it is not quite clear to me one would obtain the $T_i$'s and then fill out the table in this case.

答: Notice that $G=\mathbb{Z}/6\mathbb{Z}$. Take a generator $g\in G$ (for example the class of $1$). Then on any irreducible representation $g$ acts as multiplication by a $6$th root of unity. The $6$ possible roots of unity give you the $6$ possible irreducible representations.


问:
How to get the upper bound of expectation of its characteristic function $E(e^{tX})\le?$
For a random variable $X=0$ or $1$, we know that
$$P(X=1)\ge 1-\gamma$$
(i.e. $P(X=0)\le \gamma$).

How to get the upper bound of expectation of its characteristic function
$$E(e^{tX})\le?$$

答:
I'm not quite certain what you mean but I think you're saying that the random variable $X$ can only take the values $0$ or $1$. If this is the case then if $a:=\mathrm{Pr} (X=0)$ we have $\mathrm{Pr}(X=1) = 1-a$ for $0\leqslant a \leqslant 1$. Then \begin{align*}
\mathbb{E} e^{tX} &= a +(1-a)e^t.
\end{align*} You can see that if $a<1$ then $\mathbb{E} e^{tX}$ is unbounded.


问:
Is it common to denote zero map (e.g. $\forall x, \phi(x)=0$) by $\phi = 0$?
Is it common to denote zero map (e.g. $\forall x, \phi(x)=0$) just by $\phi = 0$? I found similar notation on my textbook but I'm not sure.

答:
As the comments say, $f(x)=0$, $f=0$ and $f\equiv 0$ are all common. To explain the idea behind the notations:

$f\equiv 0$ means "identically zero", meaning "is $0$ at all values of $x$". I mostly think of real/complex functions when I see this.

With $f=0$, we regard $0$ as a function rather than a value; the function $f$ is the function $0$ (or rather the zero function as we usually call it).


问:
Geometric interpretation of $\operatorname{SO}(3)/\operatorname{SO}(2)=S^2$
I understand (in a hand-waving sense) the argument that if an element of $\operatorname{SO}(3)$ is a general rotation in 3D characterized by a direction in 3D and a rotation angle, then "dividing out" the degree of freedom associated with the rotation ($\operatorname{SO}(2)$) leaves a direction in space, and the set of all possible directions can be mapped to the surface of a sphere.

However, can I understand this statement by choosing a particular set of rotations about the $z$-axis (which is $\operatorname{SO}(2)$) as my subgroup of $\operatorname{SO}(3),$ and then forming all cosets of it ? Each coset would be a specific rotation in 3D applied to the set of all rotations about the $z$-axis. I think that I should be able to see a geometrical way that each coset can be mapped to one (and only one) point on the surface of the sphere ($S^2$). However, I can't quite make a geometrical connection.

答:
The group $SO(3)$ naturally acts on $3$ dimensional space, and preserves the unit sphere within this. Furthermore, it acts transitively on this unit sphere, so picking a point $x$ on $S^2$, mapping $g\in SO(3)$ to $g.x$, we have a continuous surjective map $SO(3)\rightarrow S^2$. One may check that the fibre over a point is a copy of the circle group $SO(2)$, a conjugate of the stabiliser of $x$ (or a coset of the stabiliser, if you prefer), so in this sense, each point on $S^2$ gets a copy of $SO(2)$, and together these fill out all of $SO(3)$.

Morally, this is just the orbit stabiliser theorem of group theory, after noticing that the proof holds in the category of topological spaces, if your group action is nice.

We can also notice that our fibres seem to change and depend on our point $x$ chosen, and this reflects that although this map is locally a product, it is not globally so. That is, as topological spaces, we do not have $SO(3)\cong SO(2)\times S^2$, even though we can find isomorphisms when we restrict to the fibres over each hemisphere of $S^2$.


问:
Natural permutation representation definition?
I was wondering about a definition I encountered towards the beginning of Gruson-Serganova:

for G = $S_n$, and $k$ a field, and $s \in S_n$ and $(x_1, ..., x_n) \in k^n$, the natural permutation representation is defined as $\rho_s$ $(x_1, ..., x_n)$ = $(x_{s^{-1} (1)},...,x_{s^{-1}(n)})$.

I understand the permutation representation, but why do the inverses show up here? What makes this 'natural' if the permutation representation isn't? I asked my professor about why the inverses show up, and he said that otherwise it wouldn't be a homomorphism, but I see no reason why it wouldn't end up being a homomorphism.

I feel like I'm missing something pretty basic here.

答:
For $\rho$ to be a representation you want that $\rho_{gh}=\rho_g\rho_h$.

If $e_1,\ldots,e_n$ be the standard basis of $k^n$

The action of $g\in S_n$ is defined by $\rho_g(e_i)=e_{g(i)}$. You can check this gives a representation since $\rho_g\rho_h(e_i) = \rho_g(e_{h(i)}) = e_{gh(i)}=\rho_{gh}(e_i)$. (This does not work if you replace it with the inverse!)

Take $n=3$ and $g=(123)$ for example.
Then $g(e_1)=e_2$ means that $\rho_g(1,0,0)=(0,1,0)$.
In general, the action of $g$ in coordinates is
$\rho_g(x_1,x_2,x_3)= (x_3,x_1,x_2)=(x_{g^{-1}(1)},x_{g^{-1}(2)},x_{g^{-1}(3)})$. This is were the inverses come from

OwnerDisplayName: anon


问:
How many different nonempty baskets can I make consisting of
I have $5$ identical apples, $8$ identical oranges and $13$ identical bananas. How many different nonempty baskets can I make consisting of

a) $5$ fruits?

b) $7$ fruits?

My solution : First of all , I thought that this can be solved by combination with repetition but , i then relaized it cant be ,so i tried to use generating functions , i found that

$a=21$ ,$b=33$ .Is my solution correct ? Do you have any other suggestion to solve this?

Note: For part $a$ , the result of generating fuction and combination with repetition are not same , why did it occur ?

答:
$21$ and $33$ are correct.

One way of seeing $21$ is correct for $5$ fruit is a stars and bars calculation of ${5+3-1 \choose 3-1}$

For $7$ fruit the stars and bars calculation would be ${7+3-1 \choose 3-1}=36$, but this overcounts as there are not enough apples: in particular we cannot have (i) $6$ apples and $1$ orange, or (ii) $6$ apples and $1$ banana, or (iii) $7$ apples.  So the answer is $36-3=33$


问:
Proving $\Bbb{H\otimes_R C }\cong M_2(\Bbb{C})$ versus proving $\Bbb{H\otimes_R C }\cong_{\mathbb R} M_2(\Bbb{C}).$
1- Is there a difference between saying  I want to prove this $\Bbb{H\otimes_R C }\cong M_2(\Bbb{C})$ and I want to prove this $\Bbb{H\otimes_R C }\cong_{\mathbb R} M_2(\Bbb{C})$?

I see the OP in the question titled " Why is $\mathbb{H} \otimes \mathbb{C} \cong \text{End}_{\mathbb{C}} (\mathbb{H})$ " here on this site just asked about  $\mathbb{H} \otimes \mathbb{C} \cong \text{End}_{\mathbb{C}} (\mathbb{H})$ without specifying the isomorphism he required is of $\mathbb R-$algebras or $\mathbb C-$algebras and he got an isomorphism as complex algebras of $\Bbb{H\otimes_R C }\cong M_2(\Bbb{C})$ .

My second question is there a different isomorphism as $\mathbb R-$ algebras of $\Bbb{H\otimes_R C }\cong M_2(\Bbb{C})$ or can I consider the same isomorphism?

答:
Those two statements are technically different, and the second (an iso as $\mathbb{R}$ algebras) implies the first, so we might expect it to be harder to prove.

However in this instance, any isomorphism you think of will be an isomorphism of $\mathbb{R}$ algebras. It can be helpful to instead think of this kind of thing as:

Find an isomorphism between your structures, preserving as little as possible.

Find out what extra properties your isomorphism preserves.

In this case, once you do step 1, you just need to check whether the complex vector space structure is preserved.

One can give isomorphisms of algebras that dont respect all the potential choices of "underlying scalars" however, eg, consider pointwise complex conjugation on $2\times 2$ complex matrices. This is a morphism of $\mathbb{R}$ algebras, but not of $\mathbb{C}$ algebras.


问:
1 and 2 norm inequality
While looking over my notes, my lecturer stated the following inequality; $$\|x\|_2 \leq \|x\|_1 \leq \sqrt{n}\|x\|_2$$ where $x \in \mathbb{R^n}.$ There was no proof given, and I've been trying to prove it for a while now. I know the definitions of the $1$ and $2$ norm, and, numerically the inequality seems obvious, although I don't know where to start rigorously.

Thank you.

答:
$\lVert x \rVert_2 \le \lVert x \rVert_1$ is equivalent to $\lVert x \rVert_2^2 \le \lVert x \rVert_1^2$ (norms are non-negative) which can be shown in an elementary way:

$$\lVert x \rVert_2^2 =  \sum_i \lvert x_i\rvert^2 \le \left( \sum_i \lvert x_i \rvert \right)^2 =  \lVert x \rVert_1^2$$

By expanding the product $\left(\sum_i z_i \right)^2 = \sum_i z_i^2 + \sum_{i \neq j} z_i z_j$ where the second sum of cross-terms is $\ge 0$ since all $z_i$'s are $\ge 0$.

Intuition for inequalities: if $x$ has one component $x_0$ much larger (in magnitude) than the rest, the other components become negligible and $\lVert x \rVert_2 \approx (\sqrt{x_0})^2 = \lvert x_0 \rvert \approx \lVert x \rVert_1   $. On the other hand, if the components of $x$ are about equal (in magnitude), $\lVert x \rVert_2 \approx \sqrt{n x_i^2} = \sqrt n \lvert x_i \rvert$, while $\lVert x \rVert_1 \approx n \lvert x_i \rvert$.

Title: 1 and 2 norm inequality


问:
Questions regarding finite variation and local martingale
Let $A_t:=\int_0^tf(B_s)ds,\quad t\geq0$

with $f$ continuous and $B$ standard Brownian motion.

What is the correct argument that $A$ is of finite variation? Because it can be written as an integral?

And why is $\int_0^t f(B_s)dB_s$ a local martingale? Because $B$ is a local martingale and $f$ is continous?

答:
Since $B_s$ has continuous paths, $f(B_s)$ has continuous paths and hence, $A_t$ has $C^1$ paths (that's the fundamental theorem of calculus). However, any $C^1$ process has locally bounded variation (some authors omit this word) since any $C^1$-function has locally bounded variation.

As for why $\int_0^t f(B_s)\textrm{d}Bs$ is a local martingale, that simply follows from the construction of the Itô integral.


问:
Find highest weight of dual representation of $\mathfrak{sl}(3,\mathbb{C})$
I’m trying to approach the highest weight of the dual representation $(\pi^*,V^*)$ of a given irreducible representation $(\pi,V)$ of $\mathfrak{sl}(3,\mathbb{C})$, whose highest weight is supposed to be $(m_1,m_2)$. The partial order defined on weights is given by $$\mu_1\ge\mu_2  \iff \mu_1-\mu_2=a\alpha_1+b\alpha_2 \\a\ge 0 \\b\ge 0$$ where $\alpha_1=(-1,2),\alpha_2=(2,-1)$ are positive simple roots of $\mathfrak{sl}(3,\mathbb{C})$. The notes I read claims that $(m_2,m_1)$ will be the highest weight of $(\pi^*,V^*)$. The fact I realized is that weights of $(\pi^*,V^*)$ are negatives of $(\pi,V)$, so I tried to prove the claim by direct computation. If we let $(n_1,n_2)$ be a weight of $(\pi,V)$ lower than $(m_1,m_2)$, such that $$(m_1,m_2)-(n_1,n_2)=a_0\alpha_1+b_0\alpha_2$$
then $(-n_1,-n_2)$ will be a weight of $(\pi^*,V^*)$, so if I can prove $(m_2,m_1)\ge(-n_1,-n_2)$, then I’m done. But the computation seems not clear why this is true. $$(m_2,m_1)+(n_1,n_2)=(m_1+m_2,m_1+m_2)-(m_1-n_1,m_2-n_2)\\=(m_1+m_2)(\alpha_1+\alpha_2)-a_0\alpha_1-b_0\alpha_2\\=(m_1+m_2-a_0)\alpha_1+(m_1+m_2-b_0)\alpha_2$$
So to finish the prove, I need to show
$$m_1+m_2\ge a_0\\m_1+m_2\ge b_0$$
But I have no idea why this is true,any helps?

Aside: I know there are some powerful general results to solve this problem, but I’m quite new in the weight theory of representation, so I really want to prove this by just using definitions.

答:
You should use the fact that the set of weights is stable under the action of the Weyl group.

The two simple reflections are $s_1,s_2$. $s_1$ acts on a weight by $s_1(n_1,n_2)=(-n_1,n_1+n_2)$.
You can check that the longest element $w_0=s_1s_2s_1$ acts by $w_0(n_1,n_2)=(-n_2,-n_1)$.

From this you get that $(n_1,n_2)$ is a weight of $V$ if and only if $(n_2,n_1)$ is a weight of $V^*$.


问:
Closed subspace of $\mathbb{R}\times\mathbb{Q}$
In a proof, it says that the subspace $\{(m+\frac{1}{2n},\frac{r_n}{m}) | m,n\in \mathbb{N} \}$ of $\mathbb{R}\times \mathbb{Q}$ is closed, where $(r_n)_{n\in \mathbb{N}}$ is a strictly decreasing sequence in $\mathbb{Q}$ converging to $\sqrt{2}\in\mathbb{R}$. How can I see this?

答:
One possible way to show this is to see $\mathbb{R\times Q}$ as a subspace of $\mathbb{R\times R}$. You can see that the set
$$A:=\left\{\left(m+\frac{1}{2n},\frac{r_n}{m}\right) : m,n\in\mathbb{N}\right\}\cup \{(m,\sqrt{2}/m):m\in\mathbb{N}\},$$
which is a topological closure of the given set over $\mathbb{R\times R}$, is closed.
Moreover, our set is equal to $A\cap \mathbb{R\times Q}$. Could you conclude why the given set is closed in $\mathbb{R\times Q}$?


答:
We need to count the number of non-negative solutions to the equation
$$
n_1+n_2+n_3=n
$$
where $n=5 (a)$ or $7 (b)$ with the restrictions $n_1<6,n_2<8,n_3<13$.

Obviously for the part $(a)$ the restrictions play no role and the answer by stars and bars is simply:
$$
\binom{5+3-1}5=21.
$$

In the case $(b)$ only the restriction on $n_1$ matters, and applying the same method as before plus inclusion-exclusion principle we obtain:
$$
\binom{7+3-1}{3-1}-\binom{7+2-1-6}{2-1}-\binom{7+2-1-7}{2-1}=33,
$$
where we subtracted the cases with 6 and 7 apples.

Body:
I have $5$ identical apples, $8$ identical oranges and $13$ identical bananas. How many different nonempty baskets can I make consisting of

a) $5$ fruits?

b) $7$ fruits?

My solution : First of all , I thought that this can be solved by combination with repetition but , i then relaized it cant be ,so i tried to use generating functions , i found that

$a=21$ ,$b=33$ .Is my solution correct ? Do you have any other suggestion to solve this?

Note: For part $a$ , the result of generating fuction and combination with repetition are not same , why did it occur ?


Title: Show that $n^2\log\left(1+\frac{1}{n}\right)$ does not converge to $1$

Body:
How to show that $n^2\log\left(1+\dfrac{1}{n}\right)\to 1$ is false?

I have to show that $\left(1+\dfrac{1}{n}\right)^{n^2}$ doesn't tend to $e.$

Body:
$$ \log( (1 + 1/n)^n ) \to \log(e) = 1$$ So

$$ n^2 \log( 1 + 1/n) = n \log((1 + 1/n)^n) \to \infty$$


问:
What is the domain of this composite function $\sin^2 x$
I have two functions.

$f(x)=x^2$ with restricted domain $[0, \infty)$

$g(x)=\sin x$ with domain $\mathbb{R}$

I am having trouble determining the domain of $f \circ g .$ I understand that $f(x)$ will not process any negative values, but how do I express the domain $[0, \pi]\cup [2\pi, 3\pi]$ etc.

Thanks !

<img src="https://i.stack.imgur.com/qaBg7.png" alt="Graph of $f(g(x))$" />

答:
The domain $D$ of $f \circ g$ is

$$D=\{x \in \mathbb R: \sin x \ge 0\}.$$

Hence

$$D=\bigcup_{n \in \mathbb Z} [2n\pi, (2n+1)\pi],$$

as Kavi Rama Murthy wrote in his comment.


问:
Find a 5-dimensional vector space over $\Bbb Z_5$ and determine its basis.
Find a 5-dimensional vector space $V$ over $\Bbb Z_5$ and determine its basis.

Attempt:
I pick $V = P_4(\Bbb R)$, the set of all polynomials of degree less or equal than $4$ with the coefficients are in $\Bbb R$.
The basis is $B:=\{1,x,x^2,x^3,x^4\}$. Here's why $B$ be the basis.

Spanning: Since each polynomials $p \in P_4$ can be written as
$p = a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$, which is a linear combination of $1,x,x^2,x^3,x^4$, then
$\text{span}(B) = V$.

Linearly independent: Denote $p_0=1,p_1=x,p_2=x^2,p_3=x^3, p_4 = x^4$. To show that the only coefficient satisfying the vector equation $a_0p_0+a_1p_1+a_2p_2+a_3p_3+a_4p_4 = 0 ... (0)$ are $a_i = 0$ for all $i\in \{0,1,2,3,4\}$. But, the equation (0) is equivalent to the statement that
$a_0+a_1x+a_2x^2+a_3x^3+a_4x^4 = 0 ... (1)$ for all $x \in (-\infty, \infty)$. So, we must show that this is true iff each coefficient in $(1)$ is zero. But, each coefficient in $(1)$ must be zero. Otherwise, the left side of the equation would be a nonzero polynomial with infinitely many roots. On the other hand, a nonzero polynomial of degree $n$ has at most $n$ distinct roots. Thus, the equation $(0)$ has only trivial solution.

Therefore, $B$ is a basis of $V$. In particular, $\dim(V) = 5$.

Am the above true ? What if I pick $V = P_4(\Bbb Z_5)$ ? Firstly, I pick it and know that the standard basis (as above) not form a basis, since $0=5=10$, etc, and hence it's not linearly independent.

答: Any such will be isomorphic to $\Bbb F_{5^5}$.  You could take the quotient $\Bbb F_5[x]/p(x)$ where $p(x)$ is an irreducible fifth degree polynomial. Again a basis would be $\{1,x,x^2,x^3,x^4\}$.


问:
Unipotent upper triangular matrices with integer entries is Zariski dense
Let $N$ be the group of matrices $\begin{bmatrix} 1 & z \\ 0 & 1 \end{bmatrix}$ for $z \in \mathbb{C}$, let $\Gamma$ be the subgroup of $N$ with $z \in \mathbb{Z}$.
I wish to show that $\Gamma$ is Zariski dense in $N$.
Polynomials in $\mathcal{O}[N]$ can be regarded as polynomials on $\mathbb{C}$, as there is a bijection between $N$ and $\mathbb{C}$. Thus if $f \in \mathcal{O}[N]$ vanishes on $\Gamma$, then by the fundamental theorem of algebra $f$ must vanish on $N$, so $\mathcal{O}[N] = \mathcal{O}[\Gamma]$ (the ideal that we quotient $\mathcal{O}[GL(2, \mathbb{C})]$ by is the same). Thus $N = \overline{\Gamma}$. Is this correct?

答:
Yes. But rather that just a bijection, you should check that there exists an isomorphism of algebraic variety between $N$ and $\mathbb{C}$.

Another way of seeing it: the Zariski closure of a set $X$ is the intersection of all $Z(f)$, for $f$ vanishing on $X$.

So $\bar{\Gamma} \supseteq N$.


问:
A group such that all its subgroups are finitely generated
I have been thinking about the following question:

Let $G$ be an infinite group with the property that every proper subgroup of $G$ is finitely generated. Can we say that $G$ is always finitely generated?

Of course, I do not expect a positive answer to this, but I cannot think about any examples regarding this claim. Any counterexample would be appreciated.

答:
Consider the group
$$C(p^\infty)=\cup_{n\in N} C(p^n).$$
Each subgroup of this group is a finite group, but it is not finitely generated.

This group has many different names: the Prüfer p-group or the p-quasicyclic.


答:
Denote $V = \mathbb R \times \mathbb Q$. We have

$$S=\{(m+\frac{1}{2n},\frac{r_n}{m}) | m,n\in \mathbb{N} \} = \bigcup_{m \in \mathbb N} S_m$$

where $$S_m=\{(m+\frac{1}{2n},\frac{r_n}{m}) | n\in \mathbb{N} \}.$$

Each of the $S_m$ is a set of discrete points without any limit point in $\mathbb R \times \mathbb Q$. Hence is closed.

Also for all $m \in \mathbb N$, you have $S_m \subseteq [m, m+1/2] \times \mathbb Q = T_m$. As the $T_m$ are closed and their mutual distances are at least $1/2$, a limit point of $S$ has to be a limit point of one of the $S_m$.

As we proved above that the $S_m$ have no limit points in $V$, we can conclude that $S$ doesn't have any limit points in $V$ and hence is closed.

Title: Closed subspace of $\mathbb{R}\times\mathbb{Q}$

Body: In a proof, it says that the subspace $\{(m+\frac{1}{2n},\frac{r_n}{m}) | m,n\in \mathbb{N} \}$ of $\mathbb{R}\times \mathbb{Q}$ is closed, where $(r_n)_{n\in \mathbb{N}}$ is a strictly decreasing sequence in $\mathbb{Q}$ converging to $\sqrt{2}\in\mathbb{R}$. How can I see this?


问:
Matrix equation involving a Pauli matrix
I should solve the following problem:
find the matrix $A$ that satisfies the following equation:
$$\sin(\pi A)+\cos(\pi A)^2= \left( \begin{array}{ccc}
0 & 1 \\
1 & 0 \end{array} \right)$$
How can I solve the problem?
Thanks in advance.</p>

答:
Note: diagonalization is essential here. One possibility is to change the basis back and forth. That's what I do below. A more elegant way is to simply work with the spectral projections without ever writing the four coefficients of the matrices. That's Muphrid's approach.

First diagonalize the rhs matrix by $P$ orthogonal:

$$
PSP^*=P\pmatrix{0&1\\1&0}P^*=\pmatrix{1&0\\0&-1}=D\qquad \mbox{where}\quad P=\pmatrix{\frac{\sqrt{2}}{2}& \frac{\sqrt{2}}{2}\\-\frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2}}.
$$

Then observe, via the series representations of $\sin$ and $\cos $ and $PB^nP^*=(PBP^*)^n$, that
$$
\sin (\pi A)+\cos(\pi A)^2=S\iff P(\sin (\pi A)+\cos(\pi A)^2)P^*=D
$$
$$
\iff \sin (\pi PAP^*)+\cos(\pi PAP^*)^2=D.
$$
So we need to solve $(E): \;\sin (B)+\cos(B)^2=D$ and make $A=\frac{1}{\pi}P^*BP$ in the end.

If $B$ is a solution of $(E)$, $D$ and $B$ commute. In particular, $B$ must leave the eigenspaces of $D$ invariant. Since they are one-dimensional, this forces $B$ to be diagonal. Whence $(E)$ is equivalent to
$$
B=\pmatrix{\lambda&0\\0&\mu}\qquad \pmatrix{\sin \lambda+\cos^2\lambda&0\\0&\sin \mu+\cos^2\mu}=\pmatrix{1&0\\0&-1}
$$
This yields $\sin \lambda=0$ or $\sin\lambda =1$, and $\sin\mu=-1$. So the solutions of $(E)$ are
$$
B=\pmatrix{k\pi&0\\0&-\frac{\pi}{2}+2l\pi}\qquad B=\pmatrix{\frac{\pi}{2}+2k\pi&0\\0&-\frac{\pi}{2}+2l\pi}.
$$
To obtain the solutions of your initial equation, just use the formula $A=\frac{1}{\pi}P^*BP$:
$$
A=\pmatrix{\frac{-1+2k+4l}{4} & \frac{1+2k-4l}{4}\\ \frac{1+2k-4l}{4}&\frac{-1+2k+4l}{4} }\qquad A=\pmatrix{k+l & \frac{1}{2}+k-l\\\frac{1}{2}+k-l&k+l }
$$
where $k$ and $l$ range over $\mathbb{Z}$.

Conclusion: this can be rewritten
$$
A=\left(\frac{-1+2k+4l}{4}\right) I_2+\left(\frac{1+2k-4l}{4}\right) S\qquad A= (k+l)I_2+\left(\frac{1}{2}+k-l \right)S
$$
where $S$ is your original Pauli matrix on the rhs of your equation.


问:
Let $A$ be a set of $n$ elements and $B$ sets of $n-1$ elements such that $n \in \mathbb{N}$
Let $A$ be a set of $n$ elements and $B$ set of $n-1$ elements such that $n \in \mathbb{N}$. How many surjective functions there are, from $A$ towards $B$?

My attempt :

We can use Pigeonhole Principle to take two elements from $n$ optional X's and map them to one $y$ in the range, then we continue to map one-to-one the other elements in the domain.
So therefore the answer is $\binom{n}{2}
\cdot (n-1)\cdot (n-2)!$. Am I right?

答:
You choose one $y$ out of the $n-1$ elements of $B$, so this makes $n-1$ options. Than you pick two out of the the $n$ elements of $A$, which get mapped to $y$, this gives you the $n$ over 2. So far so good, now you just have to map the remaining $n-2$ elements of $A$ to the $n-2$ elements of $B$. To do so there are exactly $(n-2)!$ many possibilites, so your answer is right.

But why do you assume $A$ and $B$ to be groups? You do not use this in your explanation, you just view them as sets.


问:
It's possible to rearrange each column of a matrix to satisfy a certain condition. (APMC 1984)
Let $A$ be an $m \times n$ array of real numbers, each with absolute value at most $1$. Suppose that the sum of elements of each column of $A$ equals zero. Show that we can rearrange the numbers in each column of $A$ so that the absolute value of the sum of the elements of each row of the rearranged array is smaller than $2$.

Source : Austrian-Polish Mathematical Competition 1984 Problem 7

Added question : Can the bound $2$ be lowered? If it can, what is the smallest possible value $t=t(m,n)$ such that we can arrange the numbers in each column so that the absolute value of the sum of the elements of each row of the rearranged array is smaller than or equal to $t$?

My attempt :

For $n=2$, the claim is almost trivial.  Suppose that $x_1,x_2,...,x_m$ and $y_1,y_2,...,y_m$ be the numbers in the first and the second columns, resp.  We may rearrange the numbers so that $x_1\ge x_2 \ge ... \ge x_m$ and $y_1\le y_2 \le ... \le y_m$.

Suppose $x_k\ge 0>x_{k+1}$ and $y_{l}\le 0<y_{l+1}$.  If $k=l$, then $$|x_i+y_i|\le \max\{|x_i|,|y_i|\}\le 1<2$$
for all $i=1,2,..,m$.  If $k<l$, then
$x_{k+1}+y_{k+1},x_{k+2}+y_{k+2},...,x_l+y_l$ is negative.  Any of the $l-k$ numbers can have absolute value $\ge 2$ only when it equals $-2$.  In this case there exists $j\in\{k+1,k+2,...,l\}$ st $x_j=-1$ and $y_j=-1$.  So $x_j=x_{j+1}=...=x_{m}=-1$ and $y_1=y_2=...=y_j=-1$.

Because $x_1+x_2+...+x_m=0$, we have $x_1+x_2+...+x_{j-1}=m-j+1$.  Similarly, $y_{j+1}+y_{j+2}+...+y_{m}=j$.  So: $$j-1 \ge |x_1+x_2+...+x_{j-1}| =m-j+1$$ and
$$m-j \ge |y_{j+1}+y_{j+2}+...+y_m| =j.$$
That is, $m+2\le 2j \le m$. This is a contradiction.  So, all $x_{k+1}+y_{k+1},x_{k+2}+y_{k+2},...,x_l+y_l$ have absolute values $<2$.  Other sums $x_i+y_i$ for $i=1,2,...,k,l+1,l+2,...,m$ obviously have absolute values $<2$.  We are done.

Similarly if $k>l$, we have the same conclusion.

答:
A "great contender" for consideration is the permutation that minimizes the sum of the absolute value of the sum of each rows.

As it turns out, this works, and you just have to check the algebra.

Claim: For a given permutation of the columns, let the sum of the rows be $s_1, s_2, \ldots s_m$. The permutation $S$ that minimizes $ \sum |s_i|$ will also satisfy $|s_i | < 2$.

Proof by contradiction: Suppose not. WLOG $s_1 \geq 2$.

Since $ \sum s_i = 0$, WLOG $s_2 < 0$.

The idea here is that we want to reduce positive $s_1$ and increase negative $s_2$, and hope that the sum decreases.

$s_1 = \sum b_{1i} \geq 2 > 0$, $s_2 = \sum b_{2i} < 0$.

So there exists a $k$ such that $b_{1k} > b_{2k}$, which we can use to make the switch.

Observe that $ b_{1k} - b_{2k} \leq 1 - (-1)    = 2$.

Consider the permutation $S'$ that in addition switches $b_{1k}, b_{2k}$.

Show that $ |s'_1| + |s'_2| < |s_1 | + |s_2|$. (This should be "obvious", since we're lowering a positive number and increasing a negative number, by a "small enough amount".)

This contradicts the minimality of $ \sum |s_i|$.


问:
What should I learn next for pure math?
I can do basic single variable calculus which is essentially all you do at A Level in the UK. I also just read "What is mathematics?" by Richard Courant which I found very good. I would like to know where to go after I finish my proof writing and basic discrete maths books I am currently reading. Should I study real analysis? What book? Linear algebra?

I would appreciate advice on where to go next and what book would be good. I am currently 14 and my end goal would to become a pure mathematician.

The reason I ask this question is because I was trying to create a structure and I thought about the Gerard t'Hooft physics one or even the Pure mathematician or statiscian plan but they don't seem particularly specific in what you are meant to do where i.e they call a topic calculus and then a later one vector calculus and then analysis but as a pure mathematician I want to go straight into analysis, of reals obviously. In conclusion, I am asking this because I am going to go by steps now and choose the next topic as it comes.

Thank you in advance.

EDIT:

Would this work:

In order...

Elementary Discrete Maths,
Real Analysis,
Linear Algebra,
ODE's,
Probability,
Fourier Analysis,
Complex Analysis,
PDE's,
Graduate stuff which I will get to when needed.

I'm not sure if probability is necessary but I think it'll be interesting.

答: I would agree with many of the answers above that real analysis will strongly propel you forward in the world of pure math.  Where I went to school it was required to study multivariable calculus before moving onto analysis (it probably would have been manageable to study both concurrently).  I did find that multivariate calculus and linear algebra helped me conceptually when thinking about topological spaces which I found beneficial upon entering analysis (the actual course title was "Advanced Calculus").</p>

Title: What should I learn next for pure math?


问:
How to approximate a sharply peaked function with deltas?
I have the function

$$
f(x)=\cases{\frac{1}{\sqrt{\sinh^2(a)-\sinh^2(ax)}} & $-1 \leq x \leq1$ \\ 0 & otherwise}
$$

Where $a \in \mathbb{R}^+$. Here is a plot of $f$:

It would be very convenient if I could justify (in any way: maybe distributionally?) that

$$
f(x)=A\left[ \delta(x-1)+\delta(x+1) \right]+B+\mathcal{O}(?) \quad ; \quad a \to \infty \tag{1}
$$

For some $x$ independent constants $A$ and $B$. I am uncertain of how to do this, or how to estimate the error.

What I've tried

My thought was to look at the Fourier series of $f$, and associate terms with the Fourier series of (2). By expanding (1) around $x= \pm1$, I can find a Laurent series in $(x \pm 1)$, and perform the Fourier integral term by term. The result is messy, and worse: does not allow easy comparison to the Fourier series of (2). This may not be a fruitful approach.

By looking at the wiki page about mollifiers, it seems I need to claim something like: $a f(x) \to \delta (x \pm1), \ a \to \infty$. Superficially, this appears to work, but I'm not sure how to make it concrete, or if there should be any other factors next to $af(x)$ (other than a factor to ensure normalization to unity).

Questions

How can I estimate the error and find $A$ and $B$ in (1)?

If $A$ and $B$ cannot be found, how can I find the ratio $A/B$?

Background

$f(x)$ will eventually appear beneath an integral with a Green's function $G$. I'd like to formulate this for arbitrary $G$. We can take some liberties about the 'niceness' of $G$, as necessary. In particular, it has a well behaved Fourier transform.

答:
One way to interpret the idea behind the question is to ask how much mass concentrates in the boundaries. A clean argument interprets $f$ as density or respectively as a measure and embeds those measures in an appropriate space with a norm to define "closeness" or approximation. Accordingly, I show that the integral of $f$ converges to zero. To me this is strong indication that any definition of the limit of $f$ should not contain $\delta$ distributions. In terms of the original questions this means: $A=0.$

To avoid any worries about the singularity of the integrand we will integrate from $-1+\epsilon$ to $1-\epsilon$ and then argue about the limit $\epsilon\rightarrow 0$ and due to the symmetry of the integrand we only need to deal with the positive part.

$$ \int_{0}^{1-\epsilon}\frac{dx}{\sqrt{\sinh^2 a - \sinh ^2 ax}}=\frac{1}{a}\int_0^{a(1-\epsilon)}\frac{dz}{\sinh a\sqrt{1 - (\frac{\sinh z}{\sinh a})^2}}.$$

Now substitute $y = \frac{\sinh z}{\sinh a}$ with $z=\text{arsinh}(y \sinh a)$ and $dz = \frac{\sinh a dy}{\sqrt{1 + (y \sinh a)^2}}$ and use $\sqrt{1 + (y \sinh a )^2} \geq 1$ to obtain
$$ 0
\leq \frac{1}{a} \int_0^{\frac{\sinh a(1-\epsilon)}{\sinh a}} \frac{dy}{\sqrt{1 + (y \sinh a )^2}\sqrt{1 - y^2}}
\leq \frac{1}{a} \int_0^{\frac{\sinh a(1-\epsilon)}{\sinh a}} \frac{dy}{\sqrt{1 - y^2}} = \frac{1}{a}\arcsin\left(\frac{\sinh a(1-\epsilon)}{\sinh a}\right).$$

Ultimately $$\lim_{\epsilon\rightarrow 0} \lim_{a\rightarrow\infty}\frac{1}{a}\arcsin\left(\frac{\sinh a(1-\epsilon)}{\sinh a}\right) =  \lim_{a\rightarrow\infty}\lim_{\epsilon\rightarrow 0}\frac{1}{a}\arcsin\left(\frac{\sinh a(1-\epsilon)}{\sinh a}\right)=0.$$


问:
Prove that $\measuredangle\gamma= 90^{\circ}$

Given a triangle, if $\sin^{2}\alpha+ \sin^{2}\beta=\!\sin\gamma, \max\!\left ( \measuredangle\alpha- l\measuredangle\beta, \measuredangle\beta- l\measuredangle\alpha \right )\leq 90^{\circ}, \left | l \right |\leq 3$ so
$$\measuredangle\gamma= 90^{\circ}$$
with three angles $\alpha, \beta, \gamma$ and $l= constant.$

I have posted that triangle equality for so long.. also had a case proof$,\quad l= 0$ is the only successful one. Now let me show. By substitution$,\quad\sin\alpha:= \frac{2a\left ( a+ 1 \right )}{2a\left ( a+ 1 \right )+ 1}, \sin\beta:= \frac{2b+ 1}{2b^{2}+ 2b+ 1}$ with positives $a, b$
$$\Rightarrow\sin^{2}\alpha+ \sin^{2}\beta- \left ( \sin\gamma \right )_{= 1}= \frac{4\left ( a- b \right )\left ( a+ b+ 1 \right )\left ( 2ab+ a+ b \right )\left ( 2ab+ a+ b+ 1 \right )}{\left ( 2a^{2}+ 2a+ 1 \right )^{2}\left ( 2b^{2}+ 2b+ 1 \right )^{2}}\Rightarrow$$
$$\Rightarrow a= b\Rightarrow\sin^{2}\alpha+ \sin^{2}\beta= 1\Rightarrow\measuredangle\gamma= 90^{\circ}$$

答:
This answer proves that

if there is a triangle with angles $\alpha,\beta$ and $\gamma$ such that $\sin^2\alpha+\sin^2\beta=\sin\gamma$ and $\max(\alpha−l\beta, \beta−l\alpha)\le\frac{\pi}{2}$ for some $l$ with $|l|\le 3$, then $\gamma=\frac{\pi}{2}$.

Proof :

Since $\gamma=\pi-\alpha-\beta$, we see that $\sin^{2}\alpha+ \sin^{2}\beta=\sin\gamma$ is equivalent to $$\sin^{2}\alpha+ \sin^{2}\beta=\sin(\alpha+\beta)\tag1$$

Here, let $k:=\alpha+\beta$. Since $\beta=k-\alpha$, we see that $(1)$ is equivalent to
$$\begin{align}&\sin^2\alpha+\sin^2(k-\alpha)=\sin k
\\\\&\iff \sin^2\alpha+(\sin k\cos\alpha-\cos k\sin\alpha)^2=\sin k
\\\\&\iff \sin^2\alpha+\sin^2k\cos^2\alpha-2\sin k\cos\alpha\cos k\sin\alpha+\cos^2k\sin^2\alpha=\sin k
\\\\&\iff \sin^2\alpha+\sin^2k(1-\sin^2\alpha)-2\sin k\cos\alpha\cos k\sin\alpha+\cos^2k\sin^2\alpha=\sin k
\\\\&\iff (1-\sin^2k+\cos^2k)\sin^2\alpha-2\sin k\cos\alpha\cos k\sin\alpha=\sin k-\sin^2k
\\\\&\iff 2\cos^2k\sin^2\alpha-2\sin k\cos\alpha\cos k\sin\alpha=\sin k-\sin^2k
\\\\&\iff 2\cos k\sin\alpha\bigg(\cos k\sin\alpha-\sin k\cos\alpha\bigg)=\sin k-\sin^2k
\\\\&\iff 2\cos k\sin\alpha\sin(\alpha-k)=\sin k-\sin^2k
\\\\&\iff 2\cos k\bigg(-\frac 12\bigg)\bigg(\cos(2\alpha-k)-\cos k\bigg)=\sin k-\sin^2k
\\\\&\iff \cos(2\alpha-k)\cos k=1-\sin k
\end{align}$$

Here, let $m:=2\alpha-k$.

Then, since $\alpha,\beta,\gamma$ are written as $$\alpha=\frac{k+m}{2},\qquad\beta=\frac{k-m}{2},\qquad\gamma=\pi-k$$ our conditions

$$\begin{cases}\alpha\gt 0,\quad \beta\gt 0,\quad \gamma\gt 0
\\\alpha+\beta+\gamma=\pi
\\\alpha- l\beta\le \frac{\pi}{2}
\\\beta- l\alpha\leq \frac{\pi}{2}
\\\left | l \right |\leq 3
\\\sin^2\alpha+\sin^2\beta=\sin\gamma\end{cases}$$

can be written as

$$\begin{cases}0\lt k\lt \pi
\\-k\lt m\lt k
\\ -(kl-k+\pi)\le (l+1)m\le kl-k+\pi
\\\left | l \right |\leq 3
\\\cos m\cos k=1-\sin k\end{cases}$$

So, our problem is reduced to proving that

if there is $(k,l,m)$ such that$$\begin{cases}0\lt k\lt \pi
\\-k\lt m\lt k
\\ -(kl-k+\pi)\le (l+1)m\le kl-k+\pi
\\\left | l \right |\leq 3
\\\cos m\cos k=1-\sin k\end{cases}$$ then $k=\frac{\pi}{2}$.

Case 1 : $-3\le l\lt -1$

$$\begin{cases}0\lt k\lt \pi
\\-k\lt m\lt k
\\ \frac{kl-k+\pi}{l+1}\le m\le \frac{-(kl-k+\pi)}{l+1}
\\\cos m\cos k=1-\sin k\end{cases}$$

Suppose that $0\lt k\lt\frac{\pi}{2}$. Then, we have $$\begin{align}\cos k\lt \cos m&\implies\cos k\lt \frac{1-\sin k}{\cos k}
\\\\&\implies \cos^2k\lt 1-\sin k
\\\\&\implies 1-\sin^2k\lt 1-\sin k
\\\\&\implies \sin k(\sin k-1)\gt 0
\\\\&\implies \sin k\lt 0\end{align}$$
which contradicts that $0\lt k\lt\frac{\pi}{2}$.

Suppose that $k=\frac{\pi}{2}$. Then, there is no $m$ such that $\frac{\pi}{2}=\frac{kl-k+\pi}{l+1}\le m\le \frac{-(kl-k+\pi)}{l+1}=-\frac{\pi}{2}$.

Suppose that $\frac{\pi}{2}\lt k\lt \pi$. Then, $0\le\frac{-(kl-k+\pi)}{l+1}\lt\frac{\pi}{2}$ implies $\cos\bigg(\frac{-(kl-k+\pi)}{l+1}\bigg)\le \frac{1-\sin k}{\cos k}$ which is impossible since LHS is positive while RHS is negative.

So, we see that there is no triangle with angles $\alpha,\beta$ and $\gamma$ such that $\sin^2\alpha+\sin^2\beta=\sin\gamma$ and $\max(\alpha−l\beta, \beta−l\alpha)\le\frac{\pi}{2}$ for some $l$ with $-3\le l\lt -1$.

Case 2 : $l=-1$

$$\begin{cases}0\lt k\le \frac{\pi}{2}
\\-k\lt m\lt k
\\\cos m\cos k=1-\sin k\end{cases}$$

$k=\frac{\pi}{2}$ is sufficient.

Suppose that $0\lt k\lt \frac{\pi}{2}$. Then, we have $\cos k\lt \cos m$ which is impossible as seen in Case 1.

Case 3 : $-1\lt l\le 1$

$$\begin{cases}0\lt k\lt \pi
\\-k\lt m\lt k
\\ -\frac{kl-k+\pi}{l+1}\le m\le \frac{kl-k+\pi}{l+1}
\\\cos m\cos k=1-\sin k\end{cases}$$

$k=\frac{\pi}{2}$ is sufficient.

Suppose that $0\lt k\lt \frac{\pi}{2}$. Then, we have $\cos k\lt \cos m$ which is impossible as seen in Case 1.

Suppose that $\frac{\pi}{2}\lt k\lt\pi$. Then $0\le\frac{kl-k+\pi}{l+1}\le \frac{\pi}{2}\implies \cos\bigg( \frac{kl-k+\pi}{l+1}\bigg)\le \frac{1-\sin k}{\cos k}$ which is impossible since RHS is negative while LHS is non-negative.

Case 4 : $1\lt l\le 3$

$$\begin{cases}0\lt k\lt \pi
\\-k\lt m\lt k
\\ -\frac{kl-k+\pi}{l+1}\le m\le \frac{kl-k+\pi}{l+1}
\\\cos m\cos k=1-\sin k\end{cases}$$

$k=\frac{\pi}{2}$ is sufficient.

Suppose that $0\lt k\lt \frac{\pi}{2}$. Then, we have $\cos k\lt \cos m$ which is impossible as seen in Case 1.

Suppose that $\frac{\pi}{2}\lt k\lt\pi$. Then, $k\gt \frac{kl-k+\pi}{l+1}\ge 0$ implies $$\begin{align}&\cos\bigg( \frac{kl-k+\pi}{l+1}\bigg)\le \frac{1-\sin k}{\cos k}
\\\\&\implies \cos\bigg( \frac{3k-k+\pi}{3+1}\bigg)\le\cos\bigg( \frac{kl-k+\pi}{l+1}\bigg)\le \frac{1-\sin k}{\cos k}
\\\\&\implies \cos\bigg(\frac k2+\frac{\pi}{4}\bigg)\le \frac{1-\sin k}{\cos k}
\\\\&\implies \frac{1}{\sqrt 2}\cos\frac k2-\frac{1}{\sqrt 2}\sin\frac k2\le\frac{1-\sin k}{\cos k}
\\\\&\implies (\sqrt{1-s^2}-s)(1-2s^2)\ge \sqrt 2(1-2s\sqrt{1-s^2})
\\\\&\implies (1-2s^2+2\sqrt 2\ s)\sqrt{1-s^2}+2s^3-s-\sqrt 2\ge 0
\end{align}$$ where $s:=\sin\frac k2$ with $\frac{1}{\sqrt 2}\lt s\lt 1$. Now, let $$f(s):=(1-2s^2+2\sqrt 2\ s)\sqrt{1-s^2}+2s^3-s-\sqrt 2$$ In the following, let us prove $f(s)\lt 0$ for $\frac{1}{\sqrt 2}\lt s\lt 1$. Since we have
$$f'(s)=\frac{(6 s^2 - 1)\sqrt{1 - s^2}+6 s^3 - 4\sqrt 2\ s^2  - 5 s + 2\sqrt 2}{\sqrt{1-s^2}}$$There is no $s$ such that $f'(s)=0$ and $\frac{1}{\sqrt 2}\lt s\lt 1$ since $$\begin{align}f'(s)=0&\implies 6 s^3 - 4\sqrt 2 s^2  - 5 s +2\sqrt 2=\sqrt{1 - s^2}\ (1-6 s^2)
\\\\&\implies (6 s^3 - 4\sqrt 2 s^2  - 5 s + 2\sqrt 2)^2=(1-s^2)(1-6s^2)^2
\\\\&\implies 72 s^6 - 48\sqrt 2\ s^5 - 76 s^4 + 64\sqrt 2\ s^3 + 6 s^2 - 20\sqrt 2\ s + 7=0
\\\\&\implies \bigg(s-\frac{1}{\sqrt 2}\bigg)^3 (2 s + \sqrt 2) (6 s + \sqrt 2 - 4) (6 s + \sqrt 2 + 4)=0
\\\\&\implies s=\pm\frac{1}{\sqrt 2},\frac{-\sqrt 2\pm 4}{6}\end{align}$$none of which satisfies $\frac{1}{\sqrt 2}\lt s\lt 1$. Since we have $f'(s)\lt 0$ with $f(1/\sqrt 2)=0$, we finally get $f(s)\lt 0$.

Therefore, having $f(s)\ge 0$ is a contradiction.


问:
How would you go about proving this with a formal proof?

For a set $\mathscr{F}$, define

$$\bigcup!\mathscr{F}=\{x\mid\exists!A\in\mathscr{F}\, x\in A\}$$

(a) If $\mathscr{F}=\big\{\{1,2,3\},\{2,3,4\},\{3,4,5\}\big\}$, what is $\bigcup!\mathscr{F}$?

(b) Suppose $\mathscr{F}$ is a set and $\bigcup!\mathscr{F}\cap\bigcap\mathscr{F}\ne\varnothing$. Show that $\mathscr{F}$ has exactly one element.

I do not know how to write a formal proof to these, please could someone teach me how to write these types of formal proofs? I am pretty sure it has to do with existence and uniqueness.

答:
For (b). Assume $[\cup !\mathscr F ]\cap [\cap \mathscr F]\ne \emptyset.$

If $f,g\in\mathscr F$ with $f\ne g$ then $\cup !\mathscr F\subseteq (f\setminus g)\cup (g\setminus f)$ and $\cap\mathscr F)\subseteq f\cap g.$ But for any $f,g$ we have $[(f\setminus g)\cup (g\setminus f)]\,\cap \, [f\cap g]=\emptyset.$ Therefore $\mathscr F$ has at most 1 member.

If $\mathscr F=\emptyset$ then $[\cup !\mathscr F ]\cap [\cap \mathscr F]\subseteq\cup !\mathscr F=\emptyset.$ Therefore $\mathscr F$ has at least 1 member.


问:
Find all real $x$ and $y$ that satisfy $(x^3+y^3)(x^2+y^2)=2(x+y)=2$
I need help solving.

Find positive real numbers $x$ and $y$ that satisfies the equation:
$$(x^3+y^3)(x^2+y^2)=2(x+y)=2$$

I think that in these type of equations I have try to factor out in one side of the equation and look the prime factorization of the number on the right side; and, after that, divide into cases. But I can't seem to quite do it in this one.

答: For positive $x,y$, $(x^3+y^3) < (x+y)^3$ and $(x^2+y^2) < (x+y)^2$, then use the nice property that $x+y=1$.


问:
A seemingly simple integral
How would one show the following equality,

$$\int_{\mathbb{R}^d}|x|e^{-\frac\beta 2 |x|^2}dx = c_d\beta^{-\frac{d+1}{2}}.$$

Where $c_d$ is a constant only depending on $d$. This is problem is relatively easy to do when $d=1$ by either substitution or integration by parts, when we know the value of the Gaussian integral. My problem is that I'm not aware of any substitution techniques for higher dimensions and integration by parts (i.e. Greens identities) leads me to having to compute a higher dimensional surface integral, which, in arbitrary dimensions I do not know how to compute.

Finally tried brute forcing this and splitting it into $d$ integrals, each integral having the form,
$$ \int_\mathbb R \sqrt{(x^2+a)}e^{-x^2}dx.$$

Which annoyingly wolfram alpha doesn't give me a useful answer for...

答:
As @A rural reader and @projectilemotion mentioned, the appropriate way to evaluate this integral is to switch to spherical coordinates

$$I=\int_{\mathbb{R}^d}|x|e^{-\frac\beta 2 |x|^2}dx = \int..\int_{-\infty}^\infty\sqrt{x_1^2+...+x_d^2}\,e^{-\frac\beta 2 (x_1^2+...+x_d^2)}dx_1...dx_d=$$ $$= \int{d}\Omega\int_{0}^\infty{r}\,e^{-\frac\beta 2 r^2}r^{d-1}dr$$

where $\int{d}\Omega$ is integration over all angles in spherical coordinates, and $\int_0^\infty...{d}r$ is integration over radius.

We can find $\int{d}\Omega$ from

$\int..\int_{-\infty}^{\infty}e^{-x_1^2-x_2^2-...-x_d^2}dx_1...dx_d=$$\int{d}\Omega\int_0^{\infty}e^{-r^2}r^{d-1}dr \Rightarrow \pi^{\frac{d}{2}}=$$\frac{1}{2}$$\int{d}\Omega$$\int_0^{\infty}$$e^{-t}t^{\frac{d}{2}-1}dt$

Integral over all angles $\int{d}\Omega=\frac{2\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})}$

$$I=\frac{2\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})}\int_{0}^\infty{r}\,e^{-\frac\beta 2 r^2}r^{d-1}dr=\frac{2\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})}(\frac{\beta}{2})^{-\frac{d+1}{2}}\int_{0}^\infty{e}^{-t^2}t^ddt=$$ $$=\frac{2\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})}(\frac{\beta}{2})^{-\frac{d+1}{2}}\frac{1}{2}\int_{0}^\infty{e}^{-x}x^{\frac{d-1}{2}}dx$$

$$I=\pi^{\frac{d}{2}}\frac{\Gamma(\frac{d+1}{2})}{\Gamma(\frac{d}{2})}(\frac{\beta}{2})^{-\frac{d+1}{2}}$$

Title: A seemingly simple integral


问:
How do I simplify these logs?
$$a\log\left(3+x\right)\cdot\frac{1}{3+x}=\log\left(3+x\right)+0.5r\log\left(3+x\right)$$
How do I solve for $x$? Can I not just divide through by $\log(3+x)$?

答:
The equation is only defined for $x> -3.$

Case 1: $x=-2.$ Then $\log(3+x)= \log 1=0,$ hence $x=-2$ is a solution.

Case 2: $x \ne -2$. Then $\log(3+x) \ne 0.$ It follows

$$a\frac{1}{3+x}=1+0.5r.$$

Can you proceed ?

Title: How do I simplify these logs?


答:
Yes you can do this but you have to be careful. You have to exclude the case $\log(3+x)=0$, since you are not allowed to divide by zero. That means that
$$a\cdot\frac{1}{3+x}=1+0.5r$$
if $\log(3+x)\neq0$. You should be able to sove for $x$ in the equation above.

For what $x$ is  $\log(3+x)=0$? Since $\log(1)=0$, $x$ must be $-2$. And
$$a\cdot0\cdot\frac{1}{3+x}=0+0.5r\cdot0$$
simplifies to $0=0$, which is obviously correct, so $-2$ is also a solution to your equation.
$$\mathbb{L}=\left\{-2,\frac{a}{1+0.5r}-3\right\}$$

Addendum: It depends if you work with real or complex numbers. If you include complex numbers, $\mathbb{L}$ is correct. If you want $x$ to be real, you have to put further conditions on $r$ and $a$ so that $$\frac{a}{1+0.5r}-3>-3.$$

Title: How do I simplify these logs?

Body:
$$a\log\left(3+x\right)\cdot\frac{1}{3+x}=\log\left(3+x\right)+0.5r\log\left(3+x\right)$$
How do I solve for $x$? Can I not just divide through by $\log(3+x)$?


问:
A rigid exponential question
This question was given by my friend

$$x^{x^{20}}=\sqrt[\sqrt 2]{2}$$

even after applying log this question dosen't simplifies

$x^{20}$ $ln$ $x$ $=$ $\frac{1}{\sqrt{2}}$ $ln$ $2$

if this question is put in wolfram alpha it gives and if I do a approx form it gives 1.067 how does this came

$x=e^{\frac{1}{20} W\left(10 \sqrt{2} \ln (2)\right)}$

How to solve this question

Title: A rigid exponential question

Body:
It happens that this expression
can also be presented in a simple closed form, $x=\sqrt[8]2$:

\begin{align}
(2^{1/8})^{(2^{1/8})^{20}}
&=
(2^{1/8})^{2^{20/8}}
\\&=
(2^{1/8})^{2^{2+1/2}}
\\&=
(2^{1/8})^{4\sqrt2}
\\&=
2^{{4\sqrt2}/8}
\\&=
2^{1/\sqrt2}
\\&=
\sqrt[\sqrt2]2
.
\end{align}


问:
Orthonormal Basis - Angle of Rotation with respect to Standard Orthonormal Basis
I have an orthonormal basis ${\bf{b}}_1$ and ${\bf{b}}_2$ in $\mathbb{R}^2$. I want to find out the angle of rotation. I added a little picture here. I essentially want to find $\theta$

I know that I can compute the angle between two vectors but then there are $4$ combinations here

$\theta_{11} = \arccos({\bf{b}}_1^\top{\bf{e}}_1)$

$\theta_{12} = \arccos({\bf{b}}_1^\top{\bf{e}}_2)$

$\theta_{21} = \arccos({\bf{b}}_2^\top{\bf{e}}_1)$

$\theta_{22} = \arccos({\bf{b}}_2^\top{\bf{e}}_2)$

How would one know which angle is correct? Importantly, here I used the labels $b_1$, $b_2$ in the same order as $e_1$ and $e_2$ but that's not necessarily the same order geometically!

答:
Let $\mathbf{b}_1 = (x_1, y_1)$.  Then, $\theta = \operatorname{atan2}(y_1, x_1)$.

$\operatorname{atan2}(y_1, x_1)$ is the two-argument form of arcus tangent, equivalent to $\arctan(y_1 / x_1)$ except that the two-argument form takes the quadrant (signs of both $x_1$ and $y_1$) into account.

If you do not know which of $\mathbf{b}_1 = (x_1, y_1)$ and $\mathbf{b}_2 = (x_2, y_2)$ to use, then
$$\begin{aligned}
d &= x_1 y_2 - x_2 y_1 \\
\theta &= \begin{cases}
\operatorname{atan2}(y_1, x_1), & d \ge 0 \\
\operatorname{atan2}(y_2, x_2), & d \lt 0 \\
\end{cases} \end{aligned}$$
Here, $d \gt 0$ if $\mathbf{b}_2$ is counterclockwise from $\mathbf{b}_1$, and $d \lt 0$ if clockwise.  This uses the vector which is the clockwise one of the pair.

Note that for arbitrary coordinate system $\mathbf{e}_1$, $\mathbf{e}_2$, you can use
$$\begin{aligned}
x_1 &= \mathbf{b}_1^T \mathbf{e}_1 \\
y_1 &= \mathbf{b}_1^T \mathbf{e}_2 \\
x_2 &= \mathbf{b}_2^T \mathbf{e}_1 \\
y_2 &= \mathbf{b}_2^T \mathbf{e}_2 \\
\end{aligned}$$
above.


问:
What is the fastest way to find the characteristic polynomial of a matrix?
Finding the characteristic polynomial of a matrix of order $n$ is a tedious and boring task for $n > 2$.

I know that:

the coefficient of $\lambda^n$ is $(-1)^n$,

the coefficient of $\lambda^{n-1}$ is $(-1)^{n-1}(a_{11} + a_{22} + \dots + a_{nn})$,

the constant term is $\det{A}$.

When finding the coefficient of the linear term $\lambda$ of the characteristic polynomial of a $3\times 3$ matrix, one has to calculate the determinant of the matrix $A - \lambda I_n$ anyway. (But you don't have to sum all the terms, only the linear terms.)

Does anybody know a faster way?

答:
You can use the Cayley-Hamilton theorem, which says that the matrix $A$ is a root of the minimal polynomial, which divides the characteristic polynomial. In facts, the minimal polynomial has the same roots than the characteristic, except those roots may have lower multiplicity. In particular, if the eigenvalues of $A$ are distinct, then the two polynomials are equals.

You compute the $n=\dim$ powers: $I, A, A^2, A^3,...,A^n$, and then look for coefficients $c_i$ such that $A^n = c_{n-1}A^{n-1}+...+c_1A^1+c_0I$, then the minimal polynomial is $\det(\lambda I - A) = P(X) = \lambda^n-c_{n-1}\lambda^{n-1}-...-c_1\lambda-c_0$.

For example, if $A=\begin{pmatrix} 0&1&0\\1&0&0\\0&0&2\end{pmatrix}$, then $A^2=\begin{pmatrix} 1&0&1\\0&1&0\\0&0&4\end{pmatrix}$, and $A^3=\begin{pmatrix} 0&1&0\\1&0&0\\0&0&8\end{pmatrix}$,

so that $A^2-I=\begin{pmatrix} 0&0&0\\0&0&0\\0&0&3\end{pmatrix}$ and $A^3-A=\begin{pmatrix} 0&0&0\\0&0&0\\0&0&6\end{pmatrix}$, thus $A^3-A=2(A^2-I)$, and

the minimal polynomial is $\lambda^3-\lambda-2(\lambda^2-1)=(\lambda^2-1)(\lambda-2)=(\lambda+1)(\lambda-1)(\lambda-2)$. All the roots being simple, the minimal polynomial is also the characteristic polynomial, and $\chi_A(\lambda)=\det(\lambda I-A)=\lambda^3-2\lambda^2-\lambda+2$.

The search of the coefficients $c_i$ may be done in a systematic way, similar to the Gauss method. However the system is overdetermined, with $n \times n$ equations, and is usually quicker to solve than you think, especially for matrices found in homework.

The eigenvalues will almost always be distinct (so the characteristic and minimal polynomials will be the same), except maybe for matrices found in homework. For these, is is better to look to the power matrix $A^k, k \le n$ as you compute them, and look if they are linear combinations of the previous ones.

For example if $A=\begin{pmatrix} 0&1&1\\1&0&1\\1&1&0\end{pmatrix}$, then $A^2=\begin{pmatrix} 2&1&1\\1&2&1\\1&1&2\end{pmatrix}$, and $A^2=A+2I$. So the minimal polynomial is $\lambda^2-\lambda-2 = (\lambda+1)(\lambda-2)$. The characteristic polynomial being a polynomial of degree 3 with the same roots, it can either be $(\lambda+1)^2(\lambda-2)$ or $(\lambda+1)(\lambda-2)^2$. The multiplicity $\nu_i$ of $(x-\lambda_i)$ in $\chi_A(x) = \prod (x-\lambda_i)^{\nu_i}$, is the dimension of the associated eigenspace $E_{\lambda_i}=\ker (A-\lambda_i I)=\{x\mid Ax=\lambda_i x\}$. In our case, $E_{-1} = \ker(A+I) = \ker \begin{pmatrix} 1&1&1\\1&1&1\\1&1&1\end{pmatrix}=2$, so $\chi_A(\lambda)=(\lambda+1)^2(\lambda-2)$.

I don't know if the method is less tedious, but I find it less boring.


答:
Don't be intimidated or get tripped up by the $\log\left(3+x\right)$. Try to let $z=\log\left(3+x\right)$. Then we have

$$az\cdot\frac{1}{3+x}=z+0.5rz$$

Hopefully this seems like something you've seen before.

Now before dividing by something in an equality, we have to consider that we're dividing by zero.

So we take cases $z=0$ and $z \ne 0$.

Case 1: For $z \ne 0$

$$a\cdot\frac{1}{3+x}=1+0.5r$$

Then

$$a\cdot\frac{1}{1+0.5r}=3+x$$

Case 2: For $z=0$:

$0=\log(3+x) \iff 3+x=1$

So, it seems we have 2 possible values for $x$ as solutions, and they come from 2 possible cases of values for $z$ (zero or non-zero).

Finally, we must have $3+x > 0$ so that the expression '$\log(3+x)$' is defined.

Title: How do I simplify these logs?


答:
$$3y^2\frac{\partial z}{\partial x}-\left(z-x\right)\frac{\partial z}{\partial y}=3y^2$$
You correctly wrote the Charpit-Lagrange characteristic ODEs :
$$\frac{dx}{3y^2}=\frac{dy}{-z+x}=\frac{dz}{3y^2}$$
Don't forget that those ODEs are valid only on the characteristic curves, not elsewhere. Thus solving them leads to chacracteristic curves equations. Other equations of curves are not characteristic curves.

You correctly solved $\frac{dx}{3y^2}=\frac{dz}{3y^2}$  giving the equation of a first characteristic :
$$x-z=C_1$$

In order to find a second characteristic you tried to solve $\frac{dy}{-z+x}=\frac{dz}{3y^2}\quad\implies\quad 3y^2dy=\left(-z+x\right)dz$

If you suppose $z$=constant this ODE is not valid because $z=$constant is outside the characteristic curves. Thus $\int z\,dx\neq zx$ and so :  $\int (-z+x)dx\neq -zx+\frac12 x^2$ as you wrote. That was the mistake.

This would be corrct in integrating on the characteristic $x-z=C_1$ that is $z=x-C_1$ :
$$3y^2dy=\left(-(x-C_1)+x\right)dz$$
$$3y^2dy=C_1dz$$
$$z-\frac{y^3}{C_1}=C_2$$
The general solution of the PDE expressed on the form of implicit equation $C_2=F(C_1)$ is : $z-\frac{y^3}{C_1}=F(C_1)$
$$\boxed{z=\frac{y^3}{x-z}+F(x-z)}$$
$F$ is an arbitrary function.

Condition : $z(t)=0\:;\:x(t)=t^{1:2}\:;\:y(t)=t^{1/3}$
$$0=\frac{t}{t^{1/2}-0}+F(t^{1/2}-0)$$
Let $X=t^{1/2}$
$$F(X)=-X$$
Now the function $F(X)$ is known. We put it into the above general solution where $X=x-z$ :
$$z=\frac{y^3}{x-z}-(x-z)$$
$$\frac{y^3}{x-z}-x=0$$
Solving for $z$ gives the particular solution of the ODE which satisfies the specified condition.
$$z=x-\frac{y^3}{x}$$

Title: Solving $3y^2\frac{\partial z}{\partial x}-\left(z-x\right)\frac{\partial z}{\partial y}=3y^2$, I keep getting the same wrong answer

Body:
For the conditions  $x\left(t\right)=t^{\frac{1}{2}},\:y\left(t\right)=t^{\frac{1}{3}},\:z\left(t\right)=0,\:t>0$

I get this:

$$\frac{dx}{3y^2}=\frac{dy}{-z+x}=\frac{dz}{3y^2}$$

$$(I)\:\:\: \frac{dx}{3y^2}=\frac{dz}{3y^2} $$

$$x=z+C_1$$

$$C_1 = x-z$$

$$(II)\:\:\: \frac{dy}{-z+x}=\frac{dz}{3y^2} $$

$$3y^2dy\:=\:\left(-z+x\right)dz$$

$$y^3=-\frac{z^2}{2}+zx+C_2$$

$$C_2=y^3+\frac{z^2}{2}-zx$$

Putting in conditions

$$C_1=t^{\frac{1}{2}}$$

$$C_2=t$$

$$C_2=C_1^2$$

This means that

$$y^3+\frac{z^2}{2}-zx = \left(x-z\right)^2$$

$$2y^3+z^2-2zx = 2 \left(x-z\right)^2$$

$$2y^3+\left(z^2-2zx+x^2-x^2\right)\:=\:2\:\left(x-z\right)^2$$

$$2y^3-x^2+\left(x-z\right)^2\:=\:2\:\left(x-z\right)^2$$

$$2y^3-x^2=\:\left(x-z\right)^2$$

$$\sqrt{2y^3-x^2}=x-z$$

$$x-\sqrt{2y^3-x^2}=z$$

But it's still wrong

So where's the mistake?


答:
$(a)$ is $\{1,5\}$, the full union is $\{1,2,3,4,5\}$ but $2,3$ and $4$ occur in two sets in $\mathcal{F}$, so are not in $\bigcup !\mathcal{F}$.

$(b)$: Let $x \in \bigcup !\mathcal{F}$ and $x \in \bigcap \mathcal{F}$.
Suppose that $F_1, F_2 \in \mathcal{F}$ exist with $F_1 \neq F_2$.
Then $x \in F_1 \cap F_2$ as $x$ lies in all members of $\mathcal{F}$ by assumption. But then $x$ cannot be in $\bigcup ! \mathcal{F}$ by definition as it is not in a unique member of $\mathcal{F}$, but in at least two. This contradiction shows that $\mathcal{F}$ cannot have two distinct elements and as no elements is ruled out too (an empty union is always empty, also with $!$) $\mathcal{F}$ has a single member, as required.

Title: How would you go about proving this with a formal proof?

Body:
For a set $\mathscr{F}$, define

$$\bigcup!\mathscr{F}=\{x\mid\exists!A\in\mathscr{F}\, x\in A\}$$

(a) If $\mathscr{F}=\big\{\{1,2,3\},\{2,3,4\},\{3,4,5\}\big\}$, what is $\bigcup!\mathscr{F}$?

(b) Suppose $\mathscr{F}$ is a set and $\bigcup!\mathscr{F}\cap\bigcap\mathscr{F}\ne\varnothing$. Show that $\mathscr{F}$ has exactly one element.

I do not know how to write a formal proof to these, please could someone teach me how to write these types of formal proofs? I am pretty sure it has to do with existence and uniqueness.


问:
Meaning of $\vee$ in context of $\sigma-$ Algebras
Consider i.i.d random variables $X(t)$ for $t=0,...,T$ and there corresponding $\sigma-$algebra $\sigma(X(1),...,X(t))$.

Why can I state the following:  $$\sigma(X(1),...,X(t+1))= \sigma(X(1),...,X(t)) \vee \sigma(X(t+1))$$

What does $\vee$ mean in this context?

答:
For 2 $\sigma$-algebras $A$ and $B$, $A \vee B = \sigma(A \cup B)$.

In this case, we have

$$\sigma(X(1),...,X(t)) \vee \sigma(X(t+1)) = \sigma(\sigma(X(1),...,X(t)) \cup \sigma(X(t+1))) = \sigma(X(1),...,X(t+1))$$


问:
Right derivative increasing implies convex
I want to prove the following: If we have a function $f$ (not necessarily differentiable), but with right and left derivative $$f'_{\pm}(x) \equiv\lim_{y \rightarrow x^{\pm}} \frac{f(y)-f(x)}{y-x}$$ existing at all $x$ in $\mathbb{R}$ with $f'_+(x) \leq f'_+(y)$ $\forall x \leq y$, then $f$ is convex.

This is easily proven if $f$ is differentiable using the mean value theorem, but nothing like the same strategy (as far as I can see) works for this example.  Any help would be massively appreciated!

答:
One can show that
$$
\frac{f(y)-f(x)}{y-x} \le f_+'(y) \le \frac{f(z)-f(y)}{z-y}
$$
for $x < y < z$, which implies that $f$ is convex.

For the right inequality, fix $y$ and consider the function
$$
 h(z) = f(z) - f(y) - f_+'(y) (z-y)
$$
for $z \ge y$. Then $h$ has a right derivative with
$$
 h'_+(z) = f'_+(z) - f'_+(y) \ge 0 \, .
$$
It follows that $h$ is non-decreasing, compare $f : (0,1) \rightarrow \mathbb{R}$ countinous with non-negative right-hand derivative is non-decreasing. So $h(z) \ge h(y) = 0$, and that implies the desired estimate.

The proof of the left inequality works similarly.


问:
Deriving Reduction formula - Indefinite integration using integration by parts
I was working on finding the reduction formula for :

$\int \frac{dx}{(x^2+a^2)^n}$
By using integration by parts formula ( $\int f(x) g(x) dx = f(x) \int g(x)dx -\int (f'(x)\int g(x))$ considering 1.dx as second function :

Let $I_n$ =  $\int \frac{dx}{(x^2+a^2)^n} = \frac{dx}{(x^2+a^2)^n}.\int dx -\int \{\frac{dx}{(x^2+a^2)^n}.\int \}dx$

=$\frac{dx}{(x^2+a^2)^n}-\int (x^2+a^2)^{n-1}.2x^2dx$

=$\frac{dx}{(x^2+a^2)^n} +2n\int \frac{x^2}{(x^2+a^2)^{n+1}}dx$

=$\frac{dx}{(x^2+a^2)^n} + \frac{2n}{(x^2+a^2)}\int \frac{dx}{(x^2+a^2)^{n-1}}-\frac{dx}{(x^2+a^2)^n}$

Therefore , $I = \frac{x}{(x^2+a^2)^n}+2nI_n-2n^2I_{n+1}$

Please guide further how to proceed and conclude..Thanks

答: If  $$I_n=\int \frac{dx}{(x^2+a^2)^n}$$ then setting $$u=\frac{1}{(x^2+a^2)^n},~~dv=dx$$ leads us to have $$du=\frac{-2nxdx}{(x^2+a^2)^{n+1}},~~v=x$$ and the method of integrating by parts gives us $$I_n=\frac{x}{(x^2+a^2)^n}+2n\int\frac{x^2dx}{(x^2+a^2)^{n+1}}=...=\frac{x}{(x^2+a^2)^n}+2nI_n-2na^2I_{n+1}$$ Now we see the integrating of $I_{n+1}$ will be dependent to integrating of $I_n$ and....</p>


问:
The set $\mathbb{N}\times\mathbb{N}$ is denumerable.
Let $f:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ be given by $f(m,n)=2^{m-1}(2n-1)$.

1-1: Suppose $f(m,n)=f(a,b)\implies 2^{m-1}(2n-1)=2^{a-1}(2b-1)$.
Since $2n-1$ and $2b-1$ are always odd, they will never by powers of 2. Thus,
$2^{m-1}(2n-1)=2^{a-1}(2b-1)\implies 2^{m-1}=2^{a-1}\implies m-1=a-1\implies m=a$.

Thus, we have $2^{m-1}(2n-1)=2^{m-1}(2b-1)\implies 2n-1=2b-1\implies n=b$.
Thus, $(m,n)=(a,b)$.

Onto: Let $y=2^{m-1}(2n-1)\in\mathbb{N}\implies 2^{m-1}\in\mathbb{N}$(even) and $2n-1\in\mathbb{N}$(odd)
$\implies m\in\mathbb{N}(m\geq1)$ and $n\in\mathbb{N}(n\geq1)$
$\implies (m,n)\in\mathbb{N}\times\mathbb{N}$.
Thus for every $y\in\mathbb{N}$ there exists and $(m,n)\in\mathbb{N}\times\mathbb{N}$ such that
$f(m,n)=y$.

Hence, $f:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ is 1-1 and onto, which implies that
$\mathbb{N}\times\mathbb{N}\approx\mathbb{N}\implies \mathbb{N}$ is denumerable.

I just wanted to know if this proof was sufficient.

答:
Every natural number $n$ has a unique representation as product of a power of $2$ (which can be $1= 2^0$ too) and an odd number. Algorithmically, just keep dividing $n$ by $2$ until we have an odd number left. By the unique factorisation (a weak form) the exponent of $2$ is uniquely determined and hence so is the odd remainder.

The power of two can be uniquely written as $2^{m-1}$ for $m \in \Bbb N$, and the odd number as $2n-1$ with $n \in \Bbb N$ as well. This is all the proof of $f$ being a bijection.


问:
Finding mathematical expression of expected value
I am trying to find the closed form expression of the following integral:
$$\int_{0}^{b}e^{\frac{-1}{a^c}(\frac{1}{x}-\frac{1}{b})^c} \left ( \frac{1}{x}-\frac{1}{b}\right )^{c-1} \frac{1}{x} \, dx,$$
where
$b>0,a>0, 0<x1$

Basically this integral is the expected value of pdf:
$$f(x)=c*\frac{1}{a^c}e^{\frac{-1}{a^c}(\frac{1}{x}-\frac{1}{b})^c} \left ( \frac{1}{x}-\frac{1}{b}\right )^{c-1} \frac{1}{x^2}$$

Any suggestions are appreciated.

答:
Basically this integral is the expected value of pdf:

$$f(x)=\underbrace{\frac{-c}{a^c}}_{<0}\times \underbrace{e^{\frac{-1}{a^c}\left(\frac{b-x}{xb}\right)^c}}_{>0}\times\underbrace{ \left(\frac{b-x}{xb}\right)^{c-1}}_{>0}\times\underbrace{\frac{1}{x^2}}_{>0}\cdot{1}_{(0;b)}(x)$$

As you can see, your pdf cannot be a density, being negative $\forall x$

I expect that after correcting your pdf, with a suited transformation it can be transformed in a Weibull distribution... but I wait your amendments before begin to think about it.


问:
Name of topology related to cofinite topology
Let $\tau_C$ be the cofinite topology on a set $X$. Let $x\in X$. Does the topology
$$\tau=\tau_C\cup\{\{x\}\}$$
have a special name?

答:
I do not think there is a specific name in the literature, but in case you were writing a thesis or some article where this topology appears many times, you may want to define some notation and you may want to call it the "x-cofinite topology".

As Henno Brandsma in noticing in a comment, to be consistent with already existing notation, the "$\{x\}-$extension of the cofinite topology" is also working.


问:
Most random points on a circle
Please assist with this problem.

Suppose 3 (distinct) points are uniformly and independently distributed on a circle of unit length (smaller than a unit circle!). This is really circle and not disc. Call one of these points $B$. Let $M$ be the minimum distance between any 2 of the points.

Find the pdf of $M$. (Well there's no measure theory for this problem, but I assume this pdf exists. Of course we can see for ourselves by computing the cdf $F_M(m)= P(M \le m)$ 1st and then hope the cdf is absolutely continuous.)

My model: The circle is bijective with $[0,1)$, so let's call these 3 points $A,B,C$ s.t. they are iid $\sim \ Unif(0,1)$ (or $[0,1)$ or whatever).

Question: Well, I hope to find the pdf of $M$ via its cdf, which I think I'm able to compute if I know what $M$ is. What is $M$?

I think $M=\min\{X,Y,1-Y,1-X$ $,|X-Y|,1-|X-Y|\}$, where $X$ and $Y$ are the anti-clockwise distances from $B$ to, resp, $A$ and $C$. Somehow $X$ and $Y$ are iid Unif(0,1).

How would I compute $M$'s distribution?

These questions are all related, but I hope I made each self-contained

Random points on a circle

More random points on a circle

Most random points on a circle

Modelling random points on a circle

Remodelling random points on a circle: Arc length between points distributed on circle is uniform?

答:
Three points on a circle divide it in three arcs. One from the arcs is the shortest one. Let its length be $m$. Where can we place the third point? It cannot be closer than $m$ to any of both points. It means it cannot be on the shorter arc connecting the points, and it cannot be on the parts of the larger arc which are closer than $m$ to one of the two first points. Hence in total the arc of the length $3m$ is not accessible for the third point and there remains only the arc of $1-3m$ for it.
Therefore the pdf for the shortest distance is:
$$\rho(m)=
\begin{cases}
a(1-3m),&m<\frac13\\
0,&m>\frac13
\end{cases}
$$
The simplest way to compute the constant $a$ is to use the condition:
$$
\int_0^1\rho(m)\,dm=1
$$
which results in the value $a=6$.

Title: Most random points on a circle


问:
Time, Speed and Distance : Returning back to pick up problem
$A$,$B$ and $C$ started out in a journey to a place which was $120$ km away. $A$ and $B$ went by car at the speed of $50$ kmph while $C$ travelled by tonga at $10$ kmph. After a certain distance $B$ got off and travelled the rest distance by another tonga at $10$ kmph, while $A$ went back for $C$ and reached the destination at the same time as $B$ arrived. The number of hours required for the trip was ____ $?$

I tried to approach the problem by equating the time taken by $A$ and $B$ as both of them has reached the destination at the same time after starting together. That means they must have taken the same amount of time.

Time taken by $B=\frac{d}{50} + \frac{120-d}{10}$ ; where $d$ is the distance travelled by $A$ and $B$ together on a car together.

Time take by $A=\frac{d}{50}$ + Time taken to return back an pick up $C$ + Time taken to reach the destination after picking up $C$

Time taken to return back an pick up C $\Rightarrow$ this I am not able to figure out as when A will return back to pick up C and until them meet, the C will be moving forward for some distance while A would be coming towards him and here the concept of relative distance will come into the picture. Getting confused here!!!

Please help me out.
Answer Provided : 4.8 hours

答:
B and C cover the same distance in the same time using a mix of the two modes of transport. Therefore the amount of time they go by car must be the same, as must the amount of time in the tonga.

Let $t_c$ be the time B (or C) spends in the car, and $t_t$ the time spent in a tonga. We want to know the total time $t_c+t_t$.

Person A drops off person B after $t_c$ hours, and starts travelling back until he meets person C who has now been travelling by tonga for $t_t$ hours. Person A therefore moves backwards for $t_t-t_c$ hours. The net distance he has travelled is $50t_c - 50(t_t-t_c)$, which is the same as the distance C has travelled, $10t_t$. So:

$$50t_c - 50(t_t-t_c) = 10t_t$$

which simplifies to:

$$5t_c = 3t_t$$

We also know that B (or C) traveled 120km in total by car and tonga, so

$$50t_c+10t_t = 120$$

Substituting the first equation into the second, we get

$$30t_t+10t_t = 120\\
t_t = 3$$

So
$$t_c = \frac{3t_t}5 = \frac95$$

And the total time is

$$t_t+t_c = 3 + \frac95 = 4 + \frac45 = 4.8$$


答:
Using mean value inequality

A version of the mean value inequality is:

Let $f : [a, b ] \to \mathbb R$ be a continuous, right (or left) differentiable map with $m \le f_+^\prime(x) \le M$ for all $x \in [a,b]$. Then

$$m(y-x) \le f(y) - f(x) \le M(y -x)$$ for all $x,y \in [a,b]$.

You can see here for proof. See third approach where you'll realize that the proof can be adapted without major change for right differentiable maps only.

Based on this version of the mean value inequality, the result you're looking for follows in a straightforward way. And by the way, the mean value inequality is interesting by itself!

Title: Right derivative increasing implies convex

Body:
I want to prove the following: If we have a function $f$ (not necessarily differentiable), but with right and left derivative $$f'_{\pm}(x) \equiv\lim_{y \rightarrow x^{\pm}} \frac{f(y)-f(x)}{y-x}$$ existing at all $x$ in $\mathbb{R}$ with $f'_+(x) \leq f'_+(y)$ $\forall x \leq y$, then $f$ is convex.

This is easily proven if $f$ is differentiable using the mean value theorem, but nothing like the same strategy (as far as I can see) works for this example.  Any help would be massively appreciated!


问:
Transitive action of $S^1$ on $SO(3)$
I'm working to understand why is $\pi : SO(3) \rightarrow S^2$ wich associates to a matrix $A \in SO(3)$ it's first column,  is an  $S^1$- principal bundle, with the right action of $S^1$ on SO(3) given by

$ A.\lambda = A \begin{pmatrix}
1 & 0 & 0 \\
0 & \cos \alpha &  \sin \alpha \\
0 & -\sin \alpha & \cos \alpha
\end{pmatrix} $ for $\lambda = \cos \alpha + i \sin \alpha \in S^1$.

Please help, I'm stuck  to prove that this action is transitive on the fibers!

答:
Assume $B\in SO(3)$ is in the same fiber as $A$, that is, $\pi(B)=\pi(A)$.

Let $a_1,a_2,a_3$ and $b_1,b_2,b_3$ denote the columns of $A$ and $B$ respectively.

Then $\pi(A)=\pi(B)$ means $a_1=b_1$, and thus ${\rm span}(b_2,b_3)=b_1^\perp=a_1^\perp={\rm span }(a_2,a_3)$ because both $A$ and $B$ are orthogonal matrices.

Consequently, $b_2$ is a rotation of $a_2$ within the plane $a_1^\perp$, and because both matrices are special orthogonal, $b_3$ must be obtained by the same rotation from $a_3$ within $a_1^\perp$.

Title: Transitive action of $S^1$ on $SO(3)$


问:
Procedure or technique or reasoning to get $2+4h$?
I have solved for my students of an high school this simple trigonometric equation:

$$\tan(\pi+6x)=-\tan(2x)\tag 1$$

The $(1)$ is equivalent to (I remember also that $\tan(\alpha)=\tan(\mathbb Z\pi+\alpha)$)

$$\tan(\pi+6x)=\tan(-2x)\tag 2 \iff x=\frac{\pi}{8}k^*, \quad k^*=k-1\in\Bbb Z$$

But the solution of the textbook is $x=k\pi/8$ with $k\neq 4h+2$.

How can I find the value $\color{red}{4h+2}$?

If I calculate the domain I will have

$$\begin{cases}
 x \neq -\dfrac \pi{12}+\Bbb Z\dfrac \pi{6}\\[0.5em] \tag 3
 x\neq \dfrac \pi4+\Bbb Z\dfrac\pi2
\end{cases}$$

I have done the tests and the condition $4h+2$ is equivalent to $(3)$. Just a curiosity looking the 2nd negation of the $(3)$ I have the denominator $4$ and $2$.

答:
Another approach:

For $x=\pi+\frac{\pi}8$:

LHS: $\tan [\pi +6(\pi+\frac{\pi}8)]=\tan(\pi+\frac{3\pi}4)=\tan(-\frac{\pi}4)$

RHS: $-2(\pi+\frac{\pi}8)=- \frac{\pi}4$

So general form of solution can be:

$x=(2m+1)\pi+\frac{\pi}8=[8(2m+1)+1]\frac{\pi}8$

Let $2(2m+1)=h$, we have:

$x=(4h+1)\frac{\pi}8$

So $k=(4h+1)\neq(4h+2)$


问:
Every sequence in $A$ has a Cauchy subsequence $\implies$ $A$ is totally bounded (Carothers, Theorem 7.5)

Theorem 7.5: A set $A$ is totally bounded iff every sequence in $A$ has a Cauchy subsequence.

The $[\Rightarrow]$ direction of the proof is easy, I have trouble understanding the $[\Leftarrow]$ direction.

First, here are the definitions/characterizations of totally bounded I know:

$A\subset (M,d)$ is said to be totally bounded if for all $\epsilon>0$, there exist finitely many points $x_1,\ldots,x_n\in M$ such that $A\subset\bigcup_{i=1}^n B(x_i,\epsilon)$. Equivalently, $A\subset (M,d)$ is said to be totally bounded if for all $\epsilon>0$, there exist finitely many points $x_1,\ldots,x_n\in A$ such that $A\subset\bigcup_{i=1}^n B(x_i,\epsilon)$.

Also, $A$ is totally bounded iff for all $\epsilon>0$,  there are finitely many sets $A_1,A_2,...,A_n\subset A$ with $\text{diam}(A_i) < \epsilon$ for all $1\le i\le n$ such that $A\subset \bigcup_{i=1}^n A_i$.

I will reproduce the main parts of the proof of $[\Leftarrow]$ direction now.

Suppose $A$ is not totally bounded. Then there is some $\epsilon > 0$ such that $A$ cannot be covered by finitely many $\epsilon$-balls. Thus, by induction, we can find a sequence $(x_n)$ in $A$ such that $d(x_n,x_m) \ge \epsilon$ whenever $m\ne n$.

How? Where can I use induction? It is not obvious!

But then, $(x_n)$ has no Cauchy subsequence.

This makes sense! Once I have found an $(x_n)$ with the above properties, I know that it can't have a Cauchy subsequence. To see this, suppose it did - let $(x_{n_k})$ be the Cauchy subsequence of concern. Then we would want, $\forall \epsilon > 0, \exists N\in \Bbb N$ such that $p,q\ge N \implies d(x_{n_p},x_{n_q}) < \epsilon$. Sadly, we have $d(x_{n_p},x_{n_q}) \ge \epsilon$ for all $p\ne q$. Contradiction!

So, only the fact that such a $(x_n)$ exists bothers me - could I get some help in completing that part of the proof by induction or otherwise? Thanks!

答:
It's really by recursion, not induction (which is a proof technique).

So we have $\varepsilon >0$ such that $A$ cannot be covered by finitely many balls of radius $\varepsilon$.

Let $x_1 \in A$ to start the recursion. Then $A$ is not a subset of $B(x_1, \varepsilon)$ by assumption. So pick $x_2 \in A$ such that $x_2 \notin B(x_1, \varepsilon)$. Note that this means that $d(x_1,x_2) \ge \varepsilon$.

Now suppose we have already points $x_1, \ldots x_{m}$ so that $d(x_i, x_j) \ge \varepsilon$ for all $1\le i,j \le m, i \neq j$. By the assumption on $A$ we have

$$A \nsubseteq \bigcup_{i=1}^m B(x_i, \varepsilon)$$

or $A$ would have been coverable by finitely many $\varepsilon$-balls.
So pick $x_{m+1} \in A$ that is not in the union, and note that this means that $d(x_{m+1}, x_i) \ge \varepsilon$ for all $1 \le i \le m$.

So now we have $m+1$ many points $x_1,\ldots, x_m, x_{m+1}$ obeying $d(x_i, x_j) \ge \varepsilon$ for all $i \neq j$ so far. So we can do the recursion step going from $m$ to $m+1$ points, for every $m$.

So we have constructed a sequence $(x_n)_n$ by recursion, all of whose distances are $\ge \varepsilon$ between distinct indices.

The proof that it doesn't have a Cauchy subsequence is fine.


问:
Smoothness of isotropy representation
In Introduction to Riemannian Manifolds John Lee states:

Corollary 3.18. If a Lie group G acts smoothly and transitively on a smooth manifold M with compact isotropy groups, then there exists a G-invariant Riemannian
metric on M.

From what I understand the proof of this corollary follows like this:

There exist a point $p\in M$ such that $G_p$ is compact, therefore because the isotropy representation $I_p:G_p \rightarrow GL(T_pM)$ is smooth then $I_p(G_p)$ is compact and because of theorem 3.17 there exists a G-invariant Riemannian
metric on M.

Is this proof correct? And if so why is the isotropy representation smooth or even the isotropy group a submanifold with differential structure?

答: The proof is correct. By construction (as a stabilizer of a point), the isotropy group $G_p$ is closed in $G$ and hence a Lie subgroup by the closed subgroup theorem. Hence you can restrict the action of $G$ to $G_p$, which defines a smooth map $G_p\times M\to M$. Smoothness of the tangent map of this easily implies that via a "partial derivative" you get a smooth map $G_p\times TM\to TM$. Restricting this to $G_p\times T_pM$, the values lie in $T_pM$ and what you get is exactly $(g,X)\mapsto I_p(g)(X)$, so the latter map is smooth by construction. Applying this to a basis of $T_pM$ and expanding the result in this basis, you see that the matrix describing $I_p(g)$ depends smoothly on $G$ and since these matrices are all invertible, $I_p$ is smooth as a map to $GL(T_pM)$.

Title: Smoothness of isotropy representation


答:
Let $x\in SO(3)$. Write it as $(x_1,x_2,x_3)$ with $x_i\in\mathcal{M}_{3,1}(\mathbb{R})$.

To say that $x\in SO(3)$, means that $(x_i | x_j)=\delta_{i,j}$ where $(.|.)$ is the usual scalar product on $\mathbb{R}^3$.

In particular, $(x_2,x_3)$ is a direct orthonomal basis of the plane $Vect(x_1)^{\perp} \subset\mathbb{R}^3$.

The $S_1$-action on $\pi^{-1}(x_1)$ considered is the rotation in the place $Vect(x_1)^\perp$. This action is the natural action of $S_1 = SO(2)$ on  the set of direct orthonmal basis : it's therefore transitive.

Title: Transitive action of $S^1$ on $SO(3)$

Body:
I'm working to understand why is $\pi : SO(3) \rightarrow S^2$ wich associates to a matrix $A \in SO(3)$ it's first column,  is an  $S^1$- principal bundle, with the right action of $S^1$ on SO(3) given by

$ A.\lambda = A \begin{pmatrix}
1 & 0 & 0 \\
0 & \cos \alpha &  \sin \alpha \\
0 & -\sin \alpha & \cos \alpha
\end{pmatrix} $ for $\lambda = \cos \alpha + i \sin \alpha \in S^1$.

Please help, I'm stuck  to prove that this action is transitive on the fibers!


答: Suppose that $A$ is not totally bounded. Then, fix $\varepsilon>0$ such that, if $a_1,\ldots,a_n\in A$, then $A\not\subset\bigcup_{i=1}^nB(a_i,\varepsilon)$. Take $x_1\in A$. Since $A\not\subset B(x_1,\varepsilon)$, you can take $x_2\in A\setminus B(x_1,\varepsilon)$. Since $A\not\subset B(x_1,\varepsilon)\cup B(x_2,\varepsilon)$, there is some $x_3\in A\setminus B(x_1,\varepsilon)\cup B(x_2,\varepsilon)$. And so on… Then $m\ne m\implies d(x_m,x_n)\geqslant\varepsilon$.

Title: Every sequence in $A$ has a Cauchy subsequence $\implies$ $A$ is totally bounded (Carothers, Theorem 7.5)

Body:
Theorem 7.5: A set $A$ is totally bounded iff every sequence in $A$ has a Cauchy subsequence.

The $[\Rightarrow]$ direction of the proof is easy, I have trouble understanding the $[\Leftarrow]$ direction.

First, here are the definitions/characterizations of totally bounded I know:

$A\subset (M,d)$ is said to be totally bounded if for all $\epsilon>0$, there exist finitely many points $x_1,\ldots,x_n\in M$ such that $A\subset\bigcup_{i=1}^n B(x_i,\epsilon)$. Equivalently, $A\subset (M,d)$ is said to be totally bounded if for all $\epsilon>0$, there exist finitely many points $x_1,\ldots,x_n\in A$ such that $A\subset\bigcup_{i=1}^n B(x_i,\epsilon)$.

Also, $A$ is totally bounded iff for all $\epsilon>0$,  there are finitely many sets $A_1,A_2,...,A_n\subset A$ with $\text{diam}(A_i) < \epsilon$ for all $1\le i\le n$ such that $A\subset \bigcup_{i=1}^n A_i$.

I will reproduce the main parts of the proof of $[\Leftarrow]$ direction now.

Suppose $A$ is not totally bounded. Then there is some $\epsilon > 0$ such that $A$ cannot be covered by finitely many $\epsilon$-balls. Thus, by induction, we can find a sequence $(x_n)$ in $A$ such that $d(x_n,x_m) \ge \epsilon$ whenever $m\ne n$.

How? Where can I use induction? It is not obvious!

But then, $(x_n)$ has no Cauchy subsequence.

This makes sense! Once I have found an $(x_n)$ with the above properties, I know that it can't have a Cauchy subsequence. To see this, suppose it did - let $(x_{n_k})$ be the Cauchy subsequence of concern. Then we would want, $\forall \epsilon > 0, \exists N\in \Bbb N$ such that $p,q\ge N \implies d(x_{n_p},x_{n_q}) < \epsilon$. Sadly, we have $d(x_{n_p},x_{n_q}) \ge \epsilon$ for all $p\ne q$. Contradiction!

So, only the fact that such a $(x_n)$ exists bothers me - could I get some help in completing that part of the proof by induction or otherwise? Thanks!


答:
Supposing, as in your quote, that $A$ is not totally bounded, and that $\varepsilon$ is a witness to this. Then we can construct a sequence as follows:

Start with any point $x_0$ in $A$. We know that $A$ is not in a $\varepsilon$ ball around $x_0$ (else it would be totally bounded), so there is some point $x_1$ in $A \setminus B(x_0,\varepsilon)$.

Repeat this process: after any finite number of steps $k$, we know that $A$ is not contained in $\bigcup_{i=0}^k B(x_i,\varepsilon)$, else it would be totally bounded, so we can choose $x_{k+1}$ to be any element of $A\setminus \bigcup_{i=0}^k B(x_i,\varepsilon)$, and we will have $d(x_{k+1},x_i) > \varepsilon$ for all $i \leq k$ by construction.

Since every point that we add is at least $\varepsilon$ from every previous point, all pairs of distinct points in our sequence are at least $\varepsilon$ apart from one another.


问:
When is the area of a triangle whose side lengths are consecutive integers also an integer?
Consider a triangle with side lengths 3, 4, and 5. By Heron's formula, its area is $\sqrt{6(6 - 5)(6-4)(6 - 3)} = \sqrt{6(1)(2)(3)} = \sqrt{36} = 6$. Are there any other triangles like this?</p>

答:
The general formula for the area of a triangle whose side lengths are consecutive integers can be derived as follows: let x equal the length of the smallest side. The other two sides are then x + 1 and x + 2, and s (half the sum of the sides) is $\frac{3x + 3}{2}$. Subtracting each of the three sides from s and multiplying out gives

$$\sqrt{\frac{(3x + 3)(x + 3)(x - 1)(x + 1)}{16}} = \frac{\sqrt{3(x+3)(x-1)(x+1)^2}}{4}.$$

As we are adding and subtracting only odd numbers in this expression, only odd numbers can possibly give an even number (i.e. divisible by 4) under the radical, and therefore only odd x can yield an integer.

Continuing on this line of reasoning, $(x + 1)^2$ will always be a square no matter what x is. Additionally, since it will always be even (i.e. divisible by 2 at the very least), squaring it will always give a number divisible by 4 (eliminating any risk that we will get a perfect square in the numerator that is not divisible by 4) This means that the other three factors determine the nature of the triangle's area. Obviously, they must be a square for the radical to yield an integer. So we can conclude:

Given a triangle whose side lengths are consecutive integers, in which x is the length of the shortest side, the area will be an integer only if x is odd and $3(x+3)(x-1)$ is a perfect square.

Having tried nearly every odd number under 200, I have found only four such triangles, with side lengths 3 (area 6), 13 (area 84), 51 (area 1170), and 193 (area 16296). It would still be good to know if there is something else that can generate these numbers.


问:
How to prove an that if $\frac{a}{b}=\frac{c}{d}$ then, $a\cdot \:d=b\cdot \:c$
For example if we simplify $\frac{2+x}{2x}=\frac{1}{2}$ then we get $\left(2+x\right)\cdot \:2=2x\cdot \:1$ (this equation has no solution).

So this means that if $\frac{a}{b}=\frac{c}{d}$ then, $a\cdot \:d=b\cdot \:c$. How can I prove that?

答:
''So this means that if a/b=d/c then, a⋅d=b⋅c. How can I prove that?''

You mean $a/b = c/d$.

Just  multiply the equation both with $b$ and $d$. The equality remains valid and gives $ad=bc$.


问:
Linear programming algorithm that minimizes number of non-zero variables?
I have real world problems I'm trying to programmatically solve in the form of

$$Z = c_1 x_1 + c_2 x_2 + \cdots + c_n x_n$$

Subject to

\begin{align}
& a_{11} x_1 + a_{21} x_2 + \cdots + a_{n1} = b_1 \\[6pt]
& 0 \le a_{12} x_1 \le b_2 \\[6pt]
& 0 \le a_{23} x_2 \le b_3 \\
& {}\qquad\vdots \\
& 0 \le a_{nm} x_n \le b_m
\end{align}

Right now I'm using the simplex method. Because of the first constraint there are many solutions and I don't really care to min/max to any particular coefficients. What I really care about is to have as few non-zero variables in the objective function as possible.

A good way to look at it is, I have packages to deliver and $x$s are trucks. I don't care how long they take, I just want to use as few trucks as possible.

One strategy I came up with was to minimize $Z$ and set the objective coefficients to something like:

$$c_n = 10n$$

so that:

$$Z = 10 x_1 + 20 x_2 + \cdots + 10 n x_n$$

This mostly works but is a little hacky and, depending on what $a_{nm}$ is, at times inconsistent.

I'm not married to the simplex method, but it's fast and gives me a solution so I'm using it.

答:
Your variables are x[1..n]. Generate a corresponding set of binary variables b[1..n].

The constraint on each b[y] is as follows:

b[y] * bigM >= x[y]

Big M is a constant that is will always be larger than any value of x.

Your optimisation is now to minimise the sum of the b array.

Note: this is now "integer programming", not "linear programming", and so optimal solutions will take longer to find. You might be able to avoid this if you can reconstruct the problem so that x itself will either be a constant value or zero. Another alternative that might work in some scenarios is to run the linear version of the model multiple times with various combinations of variables missing (or constrained to zero) just to see whether that particular set of variables alone can make a feasible solution.


问:
Show that $n^2\log\left(1+\frac{1}{n}\right)$ does not converge to $1$

How to show that $n^2\log\left(1+\dfrac{1}{n}\right)\to 1$ is false?

I have to show that $\left(1+\dfrac{1}{n}\right)^{n^2}$ doesn't tend to $e.$

答:
By Bernoulli's inequality

$$\left(1+\dfrac{1}{n}\right)^{n^2} > 1+n^2\frac{1}{n}=1+n$$


问:
Is a line, that's contained in a plane, parallel to it?
Have got into a pretty heated debate with a friend, and looking online there's lacking proof
Line contained in a plane

Is a line that is contained within a plane, considered parallel to it? By my understanding it is parallel , if at all points the line has equal distance to the plane, which in this case is always 0, but they are  debating that it intersects the plane at infinitely many points, so it cannot be considered parallel, is there some mathematical proof that would prove or disprove either statement?

I see there is heated debate online about whether a line that matches another line is parallel to it and I'm wondering if there's a clean answer about this.

答:
I assume you are talking about planes and lines in $\mathbb{R^3}$. If so, then all planes in are defined by a normal vector and a point, while all lines are defined by a direction vector and a point. It follows that some line will be parallel to some plane if that line's direction vector is perpendicular to that plane's normal vector, which reduces to checking their dot product.

P.S. This is just a preliminary answer. Do let me know if you require a more in-depth one and I will write it for you.

P.P.S Welcome to Math SE! If this answer has helped clear your doubts, do give it an upvote and the tick!


问:
Do there exist elements in cyclotomic fields with certain absolute value?
Consider the $m$-th cyclotomic field $K$ over $\mathbb Q$. For an integer $n\ge2$, we know it is possible that there does not exist integer $\alpha\in\mathcal O_K$ such that $\alpha\bar\alpha=n$. What if we let $\alpha\in K$? Does it hold for arbitrary $n$ and $m$?

I suppose this is not a hard problem, but I can't find references. Thank you in advance.

答:
We can prove that if $(p)$ is prime in $\mathcal{O}_K$, there isn't an $\alpha \in K$ such that $|\alpha|^2=p$.

Fix $m$, and choose a prime $p$ such that the ideal $(p)$ is inert in $\mathbb{Z}[\zeta_m]. \ \ $ Thus $(p)$ is prime in $\mathbb{Z}[\zeta_m]. \ $ If there were $\alpha \in \mathbb{Q}(\zeta_m)$ such that $ \alpha \overline{\alpha}=p$, then we may choose $k \in \mathbb{N}$ minimal, such that $k\alpha \in \mathbb{Z}[\zeta_m]$. Set $\beta = k \alpha$, then $\beta \overline{\beta}=k^2p \in (p).$ But $\beta,\overline{\beta} \notin (p).$ Thus the ideal $(p)$ is not prime. This is a contradiction.


问:
Is there infinitely many "complex units"
As we know, $i$ = $\sqrt{-1}$, a simple complex unit. In complex space of two dimensions, you graph an axis of $a+bi$ where $i$ is your second dimension axis.

Now, you also know, in three and four dimensional space, you use the quaternions, such that $i^2 = j^2 = k^2 = ijk = -1$. Then, in three dimensional complex space, you can have it in form $a+bi+cj$.

Yet, are there "hexternions" such that $i^2 = j^2 = k^2 = l^2 = m^2 = ijklm=-1$

And also, are there k-ternions such that $t_1^2 = t_2^2 = ... = t_k^2 = t_1 t_2 ... t_k = -1$? And you can have any finite amount of dimensions in complex space?

答: You can build tessarines with any number of dimensions of the form $2^n$, and they still will be commutative and associative.


问:
Necessary and sufficient conditions that a cubic equation has three positive real roots
Given the cubic equation
$$x^3+px^2+qx+r=0$$

What are the necessary and sufficient conditions that this equation has three positive real roots?

My attempt:

From this answer, the necessary and sufficient conditions that a cubic equation has three real roots is
$$-27r^2 + 18 pqr - 4 q^3 - 4 p^3 r + p^2 q^2 \ge 0 \tag{1}$$

In order to make the roots positive, the necessary conditions are
$$p <0\tag{2}$$
$$q >0\tag{3}$$
$$r <0\tag{4}$$
but are these conditions sufficient?

PS: Finally, I found the answer.

答:
if a cubic should have 3 positive real root, lets say the roots are $a$, $b$ and $c$
then the cubic equation can be written as $(x-a)\cdot(x-b)\cdot(x-c)$, now if you expand it
$$x^3-(a+b+c)\cdot x^2+(a.b+a.c+b.c)\cdot x-a.b.c$$
if truly $a ,b, c$ are positive real, then coeffient of $x^2$ is $ < 0$, coefficient of $x $ is $> 0 $ and coefficient of $x^0 $ is $< 0 $ as you have clearly stated, yes it's the sufficient conidition


问:
Number of vectors required to get the whole null space of a matrix
I'm interested in finding the null space of the below matrix,

$A=\left[\begin{array}{rrrr}1 & 2 & 2 & 2 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 8 & 10\end{array}\right]$

Reducing to echelon form we've
$U=\left[\begin{array}{llll}1 & 2 & 2 & 2 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 0\end{array}\right]$

Using two free variables, I get two vectors in the null space as
$\mathbf{x}=\left[\begin{array}{r}-2 \\ 1 \\ 0 \\ 0\end{array}\right]$ and $\mathbf{x}=\left[\begin{array}{r}2 \\ 0 \\ -2 \\ 1\end{array}\right]$.

Their linear combination will also be in the null space but how do I know that their combination will fill the entire null space of matrix $A$ ?

答: By the rank-nullity theorem we have that the rank plus the nullity equals the dimension of the whole space. In this case the rank of the matrix is $2$ (number of pivots in the reduced matrix) and the full space is $4$ so the nullity must be $4-2=2$.


答:
Let $a,b,c$ be three roots of the cubic equations. We will prove that if $(2),(3)$ and $(4)$ hold, $a,b,c$ must be all positive.

Suppose the contradiction, from $(4)$ we can suppose that $a<0,b<0$ and $c>0$.

From $(2)$, we have $c> -a-b$.

From $(3)$, we have
$$ab + c(a+b)>0 \implies ab >c(-a-b)>(-a-b)^2 = a^2 +2ab+b^2$$
or $a^2+ab+b^2 <0$ (contradiction).

So, $a,b,c$ must be all positive.

Then,  $(1),(2),(3)$ and $(4)$ are necessary and sufficient conditions for having all 3 positive roots.

Title: Necessary and sufficient conditions that a cubic equation has three positive real roots

Body:
Given the cubic equation
$$x^3+px^2+qx+r=0$$

What are the necessary and sufficient conditions that this equation has three positive real roots?

My attempt:

From this answer, the necessary and sufficient conditions that a cubic equation has three real roots is
$$-27r^2 + 18 pqr - 4 q^3 - 4 p^3 r + p^2 q^2 \ge 0 \tag{1}$$

In order to make the roots positive, the necessary conditions are
$$p <0\tag{2}$$
$$q >0\tag{3}$$
$$r <0\tag{4}$$
but are these conditions sufficient?

PS: Finally, I found the answer.


问:
Find the integral of $P(X<Y)$
Let X,Y be r.v following a normal distribution ($\mu_X,\mu_Y$ and $\sigma_X^2,\sigma_Y^2 >0$) that are independent. Let $z:=X-Y$. What integral shows $P(X<Y)$. Do not compute it

I can not find the integral but I am fairly sure on how to compute it. What is the integral? Is it just the integral of the normal distribution from $\pm \infty$? I know the linear property applies for $Z$.

答:
Simply calculate $P(Z<0)$

$$Z\sim N(\mu_X-\mu_Y;\sigma_X^2+\sigma_Y^2)$$

To calculate it you need the tables but you can express it in terms of $\Phi$

The text gives you an important hint: find the distribution of $Z=X-Y$

As per mean and variance properties:

$$E(aX+bY)=aE(X)+bE(Y)$$

$$V(aX+bY)=a^2V(X)+b^2V(Y)$$

and for gaussian's properties $Z$ has the distribution I showed you thus

$$\mathbb{P}[Z<0]=\int_{-\infty}^0 f_Z(z)dz=\Phi\left(\frac{\mu_Y-\mu_X}{\sqrt{\sigma_X^2+\sigma_Y^2}}\right)$$

Title: Find the integral of $P(X<Y)$