问:
vector space generated by $\{I,A,A^2,\dots,A^{2n}\}$
$A$ be $n\times n$ matrix then the  dimension of vector space generated by $\{I,A,A^2,\dots,A^{2n}\}$ is atmost $n$ right? as $c_0I+c_1A+\dots+c_nA^n=0$ with some nonzero co efficient(cayley hamilton). so $\{I,\dots A^n\}$ will be linearly depenedent so others element $\{A^{n+1},\dots,A^{2n}\}$ may form atmost $n$ dimensional vector space. am I right?  </p>

答: Once you can express some power $A^k$ as linear combination of lower powers $A^i$ with $0\leq i<k$, there is no hope finding any higher powers either that are linearly independent of those lower powers. That's proved an easy induction: $A^{m+1}=AA^m$ and if $A^m$ is a linear combination of $A^0,\ldots,A^{k-1}$ then $A^{m+1}$ is a linear combination of $A^1,\ldots,A^k$, but by assumption we can express that final term $A^k$ as  a linear combination of $A^0,\ldots,A^{k-1}$, and this does the job. (This is the linear algebra version of the periodicity phenomenon, the fact that if some power of a group element $g$ gives the identity, then every further power of $g$ will be equal to a lower power of $g$ already seen.) Moreover, the first $k$ for which the above happens satisfies $k\leq n$ where $n$ is the dimension of the vector space acted upon; this follows from the Cayley-Hamilton theorem, but can also be shown by a fairly direct induction on the dimension without using that theorem. So the subspace of the vector space of all $n\times n$ matrices spanned by all the powers of $A$ (no need to stop at $A^{2n}$) has dimension $k\leq n$ and has $A^0=I,A^1,\ldots,A^{k-1}$ as basis.


问:
Proving that $\operatorname{rank}(\mathbf{A}) = 1$ if and only if $\mathbf{A}$ is the product of a non-zero column vector and a non-zero row vector
$\newcommand{\mn}[1]{\mathbf{#1}}$
$\newcommand{\vn}[1]{\mathbf{#1}}$
$\newcommand{\rk}[1]{\operatorname{rank}(#1)}$
$\newcommand{\t}{\mathrm{T}}$

As a beginner in linear algebra, I encountered an interesting problem
in my textbook.  I would like to have my solution verified.

Problem

For a matrix $\mathbf{A}$, prove that
$$\operatorname{rank}(\mathbf{A}) = 1$$
if and only if
$$\mathbf{A} = \mathbf{a}\mathbf{b}^{\mathrm{T}}$$
for some non-zero column vector
$\mathbf{a}$ and non-zero row vector $\mathbf{b}^{\mathrm{T}}$.

My proof

Let:

$m \times n$ be the dimensions of $\mn A$;

$\mn F$ be the $m \times n$ canonical matrix with rank $1$;

$\vn i_k$ be the $k$-dimensional unit vector
$(1, 0, \dotsc, 0)^\t$.

Hence, $\mn F = \vn i_m \vn i_n^\t$.

The only if direction

It is known that $\rk{\mn A} = 1$.
Therefore, there exists invertible matrices $\mn P_{(m \times m)}$ and $\mn Q_{(n \times n)}$, such that
$$ \mn P \mn A \mn Q = \mn F. $$
Thus,
$$ \mn A = \mn P^{-1} \mn F \mn Q^{-1} = \mn P^{-1} \vn i_m \vn i_n^\t \mn Q^{-1}.$$
Let $\vn a = \mn P^{-1} \vn i_m$ and $\vn b^\t = \vn i_n^\t \mn Q^{-1}$, so that $\mn A = \vn a \vn b^\t$.

Because $\det \mn P^{-1} \neq 0$, no column of $\mn P^{-1}$ consists exclusively of zeros.
$\vn a$ is just the first column of $\mn P^{-1}$, so $\vn a \neq \vn 0$.
Similarly, $\vn b^\t \neq \vn 0$.
Therefore, $\mn A = \vn a \vn b^\t$ for non-zero vectors $\vn a$ and $\vn b^\t$.
$\blacksquare$

The if direction

It is known that $\mn A = \mn a \mn b^\t$ for non-zero vectors $\mn a$ and $\mn b$.
Let $a_i$ be one non-zero element of $\vn a$.
By swapping $a_1$ and $a_i$ (possibly $i = 1$), we transform $\vn a$ into $\vn a'$.
Then, build the block matrix
$$
\mn C = \begin{bmatrix}
\vn a' & \begin{matrix} \vn 0 \\ \mn I_{m-1} \end{matrix}
\end{bmatrix}
$$
and swap its first row with its $i$th row to obtain $\mn D$,
so that the first column of $\mn D$ is equal to $\vn a$.
Hence,
$$ \det \mn D = \pm \det \mn C = \pm a_i \neq 0. $$
Let $\mn P = \mn D^{-1}$,
so
$$ \vn a = \mn D \vn i_m = \mn P^{-1} \vn i_m. $$
By the same process,
we find an invertible matrix $\mn Q$ such that
$$ \vn b^\t = \vn i_n^\t \mn Q^{-1}. $$
Therefore,
$$ \mn A = \vn a \vn b^\t
= \mn P^{-1} \vn i_m \vn i_n^\t \mn Q^{-1}
= \mn P^{-1} \mn F \mn Q^{-1}, $$
so
$$ \mn P \mn A \mn Q = \mn F, $$
which shows that $\rk{\mn A} = 1$.
$\blacksquare$

Concerns

Is my proof correct?  Did I misuse any theorem?

Is my proof clear and concise?  Does any step warrant further
explanation?  Can the notation be improved?

Are there simpler beginner-accessible solutions?

What else can be improved?

答:
Alternative proof :

Denote by $L_1, ..., L_n$ the rows of $A$.

Then $\mathrm{rank}(A)=1$ means that all the rows are proportional to a a nonzero one, i.e. that there exists $i \in \lbrace 1, ..., n \rbrace$ such that for every $k \in \lbrace 1, ..., n \rbrace$, there exists $\alpha_k$ such that $L_k=\alpha_k L_i$.

Now let $a=L_i$ and $b=(\alpha_1, ..., \alpha_n)$ and you are done.

ContentLicense: CC BY-SA 4.0


问:
Why should this function has a fixed point?
My problem:

Suppose $D \subset \mathbb{R}^{m+1}$ is the unitary open ball. Let $f : \overline{D} \to \mathbb{R}^{m+1}$ be a continuous function. Suppose also that $\langle x,f(x) \rangle <1$ for all $x \in \mathbb{S}^{m}$. I would like to prove that $f$ has a fixed point.

My attempt:

I tried defining $f_n=\frac{f}{n}$ where $n \geq N=\sup\limits_{x \in \overline{D}} \|f(x)\|$ and taking $x_n$ the fixed point of $f_n$. Taking a subsequence we may assume $x_n \to x$.

Thus from $n \langle x_n,x_n \rangle <1$ we get $x_n \to 0$. But I cannot conclude.

答:
Hint Consider $g\colon\overline{D}\to\overline{D}$ defined by
$$
g(x):=\begin{cases}
f(x) & f(x)\in\overline{D}\\
\frac{f(x)}{\lvert f(x)\rvert} & f(x)\notin\overline{D}
\end{cases}
$$


问:
Extrema of $f:[-1,1]\to\mathbb{R}$ defined as $f(x)=1+|x|$
Let $f:[-1,1]\to\mathbb{R}$ defined by $f(x)=1+|x|$, find the extrema of $f$ in its domain.

I did it in this way: the extrema are reached in internal stationary points, in the points such that $f'$ doesn't exist and in the points on the boundary of $[-1,1]$.

It is $\text{Int}([-1,1])=(-1,1)$, and $f'$ exists for all $x\ne0$ since $|x|$ is not differentiable in $x=0$; so for $x\in(-1,1) \setminus \{0\}$ it is $f'(x)=\frac{|x|}{x}=0 \iff x=0$, but since $0 \notin(-1,1) \setminus \{0\}$ there aren't internal stationary points.

The only internal point where $f$ is not differentiable is $x=0$, and it is $f(0)=1$.

It is $\partial([-1,1])=\{-1,1\}$, and it is $f(-1)=f(1)=2$; since $f(0) \leq f(-1)=f(1)$ it follows that $f$ has maximum $2$ for $x=1$ or $x=-1$ and $f$ has minimum $1$ for $x=0$.

Is this correct? I have a doubt when I evaluate $f'$: one of the hypothesis of Fermat's theorem on stationary points is that $f$ is defined in an open interval $(a,b)$, but in this case I have $(-1,1) \setminus \{0\}$ which is not an interval (if I'm not wrong, a union of interval in general is not an interval). So I'm not sure about my work in that point. My work is wrong and I can save it by considering the two cases $f_1(x)=1-x$ for $x \in (-1,0)$ and $f_2(x)=1+x$ for $x\in(0,1)$, in this way  $f_1$ and $f_2$ are defined on  intervals so all is fine and I just have to take the union of the results I obtain in the single intervals $(-1,0)$ and $(0,1)$? Or my work is correct as well? Thanks.

答:
Why so complicated ?

We have $f(x) \ge 1$ for all $x$ and $f(0)=1.$

Furthermore: $f(x) \le 2 $ for all $x \in [-1,1]$ and $f( \pm 1)=2.$


问:
Compute the limit of the sequence $x_n=(2-\sqrt{2})(2-\sqrt[3]{2})...(2-\sqrt[n]{2})$
At the end of the book where I got the problem it says that the limit is $0$; therefore, I have tried to majorize with a sequence having also limit $0$. Then I have looked at the series obtained by taking $\log(x_n)$ and trying to prove that it goes to $-\infty$.

答:
Taking logs, any term of $$\sum _{k=2}^{\infty } \log \left(2-2^{1/k}\right)$$ is negative and decreasing.

In other words, as
$$\log (2-2^x)=-x \log (2)+O\left(x^2\right);\;\text{ as } x\to 0$$
we have
$$\log \left(2-2^{1/k}\right)\sim -\frac{\log (2)}{k};\;\text{ as }k\to\infty$$
The sum is equivalent to $$(-\log 2)\sum_{k=2}^{\infty}\frac1k$$
diverges to $-\infty$ thus the product converges to $0$.


问:
Dominant subset in bipartite graph NP-complete proof
Let the dominant subset problem (which is known to be NP-complete) be:

Given a graph $G=(V_G,E_G)$ and an integer $k$, is there a subset $W \subset V_G$ with at most $k$ vertex so that any other vertex $u \in V_G \setminus W $ is connected to some $w \in W$?

I am asked to prove that if $G$ is bipartite, it's still NP-complete, and I'm having trouble doing the polynomial reductions: taking an instance of a general graph and an integer $(G,k)$, to a bipartite graph and another integer $(B, k')$ (in polynomial time) so that $G$ has a dominant subset of size $k$ iff $B$ has a dominant subset of size $k'$.

I thought about replacing each edge $e=(u,v)$ in $G$ for a vertex $e'$ connected to $u$ and $v$ in order to create $B$. However, I'm stuck at figuring out the $k'$ that would make the iff true. Maybe I should do that replacement in some edges and not all?

答:
Although the construction you mention is often useful in reducing graph problems to bipartite graphs, in this case it doesn't seem to work. Another common way of constructing a bipartite graph from $G$ is to take copies $A,B$ of $V_G$, so that for each vertex $v \in V_G$ there is a pair of vertices $v_A,v_B$; and draw an edge between $u_A$ and $v_B$ if either $u = v$ or $uv$ is an edge of $G$.

Let $H = (A,B,E)$ be the bipartite graph obtained in this way and let $H'$ be the (still bipartite) graph we get by adding an extra vertex $x$ to $A$, along with edges $xv$ for all $v \in B$. Now you should show that

there is a smallest dominating set of $H'$ of the form $\{x\} \cup B'$ where $B' \subseteq B$, and

dominating sets of this form correspond bijectively to dominating sets of $G$. That is, if $\{x\} \cup B'$ dominates $H'$, then the vertex set $V' \subseteq V_G$ corresponding to vertices in $B'$ dominates $G$, and vice versa.


问:
Given a spanning set, what is the span of the 'transpose' of the set?
Given $$sp\left \{
\begin{pmatrix}
a_1\\
a_2\\
a_3
\end{pmatrix}
,\begin{pmatrix}
b_1\\
b_2\\
b_3
\end{pmatrix}
,\begin{pmatrix}
c_1\\
c_2\\
c_3
\end{pmatrix}
\right \} = \mathbb{R}^3$$

What is
$$sp\left \{
\begin{pmatrix}
a_1\\
b_1\\
c_1
\end{pmatrix}
,\begin{pmatrix}
a_2\\
b_2\\
c_2
\end{pmatrix}
,\begin{pmatrix}
a_3\\
b_3\\
c_3
\end{pmatrix}
\right \}\subseteq \mathbb{R}^3$$

I'm thinking of row spaces and column spaces but with not much luck,
a leading hint would be appreciated more than the final answer.

答:
Hint: The row rank and the column rank of a matrix is the same.

Another solution

Hint: The set of $n$ vectors spans $\mathbb{R}^n$ if and only if the determinant is non-zero.

Hint: The determinant of matrix $A$ is equal to the determinant of matrix $A^t$.

ContentLicense: CC BY-SA 3.0


答:
Note that writing

$f'(x) = \dfrac{|x|}{x} = 0 \iff x = 0$

is not correct. ($\impliedby$ does not make sense.)

What you have done otherwise is fine.

However, this can be done in a much easier manner. First, noting that $|x| \ge 0$ gives $f(x) \ge 1$. On the other hand, $1$ is indeed attained since $f(0) = 1$. This finishes the proof.

Title: Extrema of $f:[-1,1]\to\mathbb{R}$ defined as $f(x)=1+|x|$

Body:
Let $f:[-1,1]\to\mathbb{R}$ defined by $f(x)=1+|x|$, find the extrema of $f$ in its domain.

I did it in this way: the extrema are reached in internal stationary points, in the points such that $f'$ doesn't exist and in the points on the boundary of $[-1,1]$.

It is $\text{Int}([-1,1])=(-1,1)$, and $f'$ exists for all $x\ne0$ since $|x|$ is not differentiable in $x=0$; so for $x\in(-1,1) \setminus \{0\}$ it is $f'(x)=\frac{|x|}{x}=0 \iff x=0$, but since $0 \notin(-1,1) \setminus \{0\}$ there aren't internal stationary points.

The only internal point where $f$ is not differentiable is $x=0$, and it is $f(0)=1$.

It is $\partial([-1,1])=\{-1,1\}$, and it is $f(-1)=f(1)=2$; since $f(0) \leq f(-1)=f(1)$ it follows that $f$ has maximum $2$ for $x=1$ or $x=-1$ and $f$ has minimum $1$ for $x=0$.

Is this correct? I have a doubt when I evaluate $f'$: one of the hypothesis of Fermat's theorem on stationary points is that $f$ is defined in an open interval $(a,b)$, but in this case I have $(-1,1) \setminus \{0\}$ which is not an interval (if I'm not wrong, a union of interval in general is not an interval). So I'm not sure about my work in that point. My work is wrong and I can save it by considering the two cases $f_1(x)=1-x$ for $x \in (-1,0)$ and $f_2(x)=1+x$ for $x\in(0,1)$, in this way  $f_1$ and $f_2$ are defined on  intervals so all is fine and I just have to take the union of the results I obtain in the single intervals $(-1,0)$ and $(0,1)$? Or my work is correct as well? Thanks.


答:
We have
\begin{align}
\ln(x_n) &=\sum_{i=1}^n\ln(2-2^{\frac{1}{n}}) \\
\iff\frac{1}{n}\ln(x_n) &=\frac{1}{n}\sum_{i=1}^n\ln(2-2^{\frac{1}{n}})  \tag{1}\\
\end{align}
And we have
\begin{align}
\frac{1}{n}\sum_{i=1}^n\ln(2-2^{\frac{1}{n}}) \longrightarrow \int_0^1\ln(2-2^x)dx \tag{2}\\
\end{align}
Using Mathematica, I found that
$$\int_0^1\ln(2-2^x)dx=\frac{\ln(2)}{2} - \frac{\pi^2}{\ln(4096)} \approx -0.839996 \tag{3}$$
From (1), (2) and (3), we have
$$\ln(x_n) = n \left( \frac{\ln(2)}{2} - \frac{\pi^2}{\ln(4096)} \right) \longrightarrow -\infty $$
or
$$x_n \longrightarrow 0$$

Title: Compute the limit of the sequence $x_n=(2-\sqrt{2})(2-\sqrt[3]{2})...(2-\sqrt[n]{2})$

Body: At the end of the book where I got the problem it says that the limit is $0$; therefore, I have tried to majorize with a sequence having also limit $0$. Then I have looked at the series obtained by taking $\log(x_n)$ and trying to prove that it goes to $-\infty$.


问:
Convergence of $S_p = \sum_{k=1}^{p}\left( \sum_{j=1}^{k}\frac{r^{k-j}}{j}\right)^2$
I'm looking at the following sum

$$ S_p = \sum_{k=1}^{p}\left( \sum_{j=1}^{k}\frac{r^{k-j}}{j}\right)^2$$
where $r \in (0,1), p > 1$. The goal is to inspect whether the sum converges to some constant, as $p \rightarrow \infty$.

Judging solely by the numerical evaluations I'm guessing that the sums should converge to some value, but not sure how to get there analytically.

One idea was to try finding some smaller upper bound for the inner summation to extract the $\frac{1}{j}$ out to $\frac{1}{k^2}$ or something close (if that makes sense), but haven't found the way so far.

Are there any known results that might work here?

答:
The convergence is not hard to see, but we can evaluate the limit using multiple sums. And, even not assuming their convergence yet, all the interchanges of summation below are still valid, since all the terms are nonnegative. The convergence then follows from the final result.

\begin{align*}
\sum_{n=1}^\infty\left(\sum_{k=1}^n\frac{r^{n-k}}{k}\right)^2&=\sum_{n=1}^\infty\sum_{j,k=1}^n\frac{r^{2n-j-k}}{jk}
\\&=\sum_{j,k=1}^\infty\frac{1}{jk}\sum_{n=\max\{j,k\}}^\infty r^{2n-j-k}
\\&=\frac{1}{1-r^2}\sum_{j,k=1}^\infty\frac{r^{2\max\{j,k\}-j-k}}{jk}
\\&=\frac{1}{1-r^2}\left(\sum_{k=1}^\infty\frac{1}{k^2}+2\sum_{k=2}^\infty\sum_{j=1}^{k-1}\frac{r^{k-j}}{jk}\right)
\\&=\frac{1}{1-r^2}\left(\frac{\pi^2}{6}+2f(r)\right)
\end{align*}
where, after replacing $j$ with $k-j$ in the inner sum,
\begin{align*}
f(r)&=\sum_{k=2}^\infty\sum_{j=1}^{k-1}\frac{r^j}{k(k-j)}=\sum_{j=1}^\infty\sum_{k=j+1}^\infty\frac{r^j}{k(k-j)}
\\&=\sum_{j=1}^\infty\frac{r^j}{j}\sum_{k=j+1}^\infty\left(\frac1{k-j}-\frac1k\right)=\color{blue}{\sum_{j=1}^\infty\frac{r^j}{j}\sum_{k=1}^j\frac1k}
\\&=\sum_{k=1}^\infty\frac1k\sum_{j=k}^\infty\int_0^r x^{j-1}\,dx=-\int_0^r\frac{\log(1-x)}{x(1-x)}\,dx
\end{align*}
(the "blue" sum is clearly convergent). Using the dilogarithm, we then have $$\sum_{n=1}^\infty\left(\sum_{k=1}^n\frac{r^{n-k}}{k}\right)^2=\frac{1}{1-r^2}\left(\frac{\pi^2}{6}+\log^2(1-r)+2\operatorname{Li}_2(r)\right).$$


答:
I would suggest that you focus more on the definitions than trying to compute and work with factorizations.

For example, in the if direction, suppose that $A=ab^T$ and let $v$ be any vector.  Then $Av=a(b^Tv)$.  Since $b^Tv$ is just a scalar, we find that $Av$ is a scalar multiple of $a$.  Hence, $Av\in\operatorname{Span}(a)$.  From the definition, this implies that the rank of $A$ is at most one.

Now, if you let $v=b$, then $b^Tb=\|b\|$, which is nonzero if and only if $b$ is nonzero.  Therefore, the image of $A$ is not only the zero vector, i.e., has dimension at least one.

Title: Proving that $\operatorname{rank}(\mathbf{A}) = 1$ if and only if $\mathbf{A}$ is the product of a non-zero column vector and a non-zero row vector

Body:
$\newcommand{\mn}[1]{\mathbf{#1}}$
$\newcommand{\vn}[1]{\mathbf{#1}}$
$\newcommand{\rk}[1]{\operatorname{rank}(#1)}$
$\newcommand{\t}{\mathrm{T}}$

As a beginner in linear algebra, I encountered an interesting problem
in my textbook.  I would like to have my solution verified.

Problem

For a matrix $\mathbf{A}$, prove that
$$\operatorname{rank}(\mathbf{A}) = 1$$
if and only if
$$\mathbf{A} = \mathbf{a}\mathbf{b}^{\mathrm{T}}$$
for some non-zero column vector
$\mathbf{a}$ and non-zero row vector $\mathbf{b}^{\mathrm{T}}$.

My proof

Let:

$m \times n$ be the dimensions of $\mn A$;

$\mn F$ be the $m \times n$ canonical matrix with rank $1$;

$\vn i_k$ be the $k$-dimensional unit vector
$(1, 0, \dotsc, 0)^\t$.

Hence, $\mn F = \vn i_m \vn i_n^\t$.

The only if direction

It is known that $\rk{\mn A} = 1$.
Therefore, there exists invertible matrices $\mn P_{(m \times m)}$ and $\mn Q_{(n \times n)}$, such that
$$ \mn P \mn A \mn Q = \mn F. $$
Thus,
$$ \mn A = \mn P^{-1} \mn F \mn Q^{-1} = \mn P^{-1} \vn i_m \vn i_n^\t \mn Q^{-1}.$$
Let $\vn a = \mn P^{-1} \vn i_m$ and $\vn b^\t = \vn i_n^\t \mn Q^{-1}$, so that $\mn A = \vn a \vn b^\t$.

Because $\det \mn P^{-1} \neq 0$, no column of $\mn P^{-1}$ consists exclusively of zeros.
$\vn a$ is just the first column of $\mn P^{-1}$, so $\vn a \neq \vn 0$.
Similarly, $\vn b^\t \neq \vn 0$.
Therefore, $\mn A = \vn a \vn b^\t$ for non-zero vectors $\vn a$ and $\vn b^\t$.
$\blacksquare$

The if direction

It is known that $\mn A = \mn a \mn b^\t$ for non-zero vectors $\mn a$ and $\mn b$.
Let $a_i$ be one non-zero element of $\vn a$.
By swapping $a_1$ and $a_i$ (possibly $i = 1$), we transform $\vn a$ into $\vn a'$.
Then, build the block matrix
$$
\mn C = \begin{bmatrix}
\vn a' & \begin{matrix} \vn 0 \\ \mn I_{m-1} \end{matrix}
\end{bmatrix}
$$
and swap its first row with its $i$th row to obtain $\mn D$,
so that the first column of $\mn D$ is equal to $\vn a$.
Hence,
$$ \det \mn D = \pm \det \mn C = \pm a_i \neq 0. $$
Let $\mn P = \mn D^{-1}$,
so
$$ \vn a = \mn D \vn i_m = \mn P^{-1} \vn i_m. $$
By the same process,
we find an invertible matrix $\mn Q$ such that
$$ \vn b^\t = \vn i_n^\t \mn Q^{-1}. $$
Therefore,
$$ \mn A = \vn a \vn b^\t
= \mn P^{-1} \vn i_m \vn i_n^\t \mn Q^{-1}
= \mn P^{-1} \mn F \mn Q^{-1}, $$
so
$$ \mn P \mn A \mn Q = \mn F, $$
which shows that $\rk{\mn A} = 1$.
$\blacksquare$

Concerns

Is my proof correct?  Did I misuse any theorem?

Is my proof clear and concise?  Does any step warrant further
explanation?  Can the notation be improved?

Are there simpler beginner-accessible solutions?

What else can be improved?


问:
Existence of solution "because it's real"
Differential equations coming from physics have a very peculiar aspect: given initial conditions that physically make sense, there must exist a (unique) solution.

This is just because we know that in nature the phenomenon actually happens, and in deterministic theories there is only one way in which the phenomenon can evolve from initial conditions.

Are there examples of (partial) differential equations from physics for which is hard to show existence and uniqueness?
As an example, take n planets that interacts following the gravitational law. The associated differential equation will contain distances between points in the denominator, so it is not clear a priori that solutions will not blow up. However, discarding cases in which planets collide, the solution actually exist, but it is not manifest from regularity of the equations. Another example could be taken from the heat flow equations.

In case such an example is found, I would like to know which actual methods are used to prove uniqueness and existence, and if the physical intuition actually plays a role.

答:
Your question, as I understand it, is:

In deterministic models, does there exist a solution (to the equations of motion or the differential equations governing the system e.g. the heat flow equations), and is it unique?

And I would say that this is the very definition of deterministic.

Wikipedia says:

In mathematics, computer science and physics, a deterministic system
is a system in which no randomness is involved in the development of
future states of the system. A deterministic model will thus always
produce the same output from a given starting condition or initial
state.

$$$$
If your question instead is, "under what conditions are the solutions of a mathematical model of a physical system unique?", then this is more interesting mathematically, but probably physcically irrelevant. For example, take the heat flow equations. As long as the initial starting conditions of temperature - as a function of space and time - $T(x,y,z,t),$ are smooth in a differentiable sense, then the time evolution of the system will surely be unique.

However, if the initial starting conditions of temperature, $T(x,y,z,t),$ are discontinuous, then strange stuff might happen. You could even take it to the extreme and make it everywhere discontinuous. Then the time evolution of the system is probably not deterministic because the rules of the game, i.e. what should happen are probably not well-defined (although you could come up with your own rules for what should happen). This corresponds to the situation where every atom has a temperature which is not close to any of it's surrounding atoms - and for this it is probably very difficult, if not computationally impossible to even approximate the time evolution of the system due to the chaotic nature of "everywhere discontinuous".

So basically the differentiable equations in physics require a nice smooth (or at least, continuous or differentiable?) function in order to determine the time evolution of the system. Once this continuity/differentiability breaks down, you can no longer use the differential equations to predict the time evolution: you need to look to other laws (like laws of particle physics) to see what's going on.

Conclusion: Some systems will not be deterministic when the starting conditions are discontinuous, but will be deterministic when the starting system is continuous; other systems will not be deterministic when the starting conditions are continuous, but will be deterministic when the starting system is differentiable; other systems will not be deterministic when the starting conditions are differentiable, but will be deterministic when the starting system is twice differentiable, and so on. Which one of these is the case for a given system depends on what kind of equations govern the physical system and you would have to look at each system on a case-by-case basis...


问:
If $\tau$ is a vector topology on a normed space s.t. a closed ball is $\tau$-compact, is a continuous function on a closed ball even uniformly cont.?
Let $\mathbb K=\mathbb R$ or $\mathbb K=\mathbb C$, $r>0$, $X$ be a normed $\mathbb K$-vector space, $\tau$ be a vector topology on $X$ s.t. $\overline B_r^X(0):=\{x\in X:\left\|x\right\|_X\le r\}$ is $\tau$-compact and $h:E\to\mathbb C$ be $\tau$-continuous on $\overline B_r^X(0)$, i.e. the restriction $\left.h\right|_{\overline B_r^X(0)}$ of $h$ to $\overline B_r^X(0)$ is continuous with respect to $\left.\tau\right|_{\overline B_r^X(0)}=\{\Omega\cap\overline B_r^X(0):\Omega\in\tau\}$.

Are we able to show that $h$ is uniformly $\tau$-continuous on $\overline B_r^X(0)$?

Let $\varepsilon>0$. By assumption, for all $x\in\overline B_r^X(0)$, there is an open $\left.\tau\right|_{\overline B_r^X(0)}$-neighborhood $N_x$ of $x$ with $$\forall y\in N_x:|h(x)-h(y)|<\varepsilon\tag1.$$ Since $\overline B_r^X(0)$ is $\tau$-compact, there is a finite $F\subseteq\overline B_r^X(0)$ with $$\overline B_r^X(0)\subseteq\bigcup_{x\in F}N_x.\tag2$$ But at this point I'm stuck ...

答: It’s a general fact in any uniform space that a continuous function on a compact subset is uniformly continuous on that subset. Nothing specific to topological vector spaces.</p>


问:
Random Poisson Sum of Random Variables with known distribution
I am trying to get a closed form expression for the expected value of the following summation of RVs: $\sum_{i=1}^{Y} X_{i}$, where $Y$ is Poisson distributed with parameter $\lambda$ and $ X_{i} $ follows some known distribution $f_X(x)$. Are there any means to drop the RV $Y$ using $\lambda$?

Thank you for your time and patience.

答:
Yes if your $X_i$'s are iid, then you can use iterated conditioning to prove the Wald's identity.

\begin{align}
\mathbb{E}_X[\sum_{i=1}^Y X_i] &= \mathbb{E}_Y[\mathbb{E}_X[\sum_{i=1}^y X_i | Y=y]]\\
&=\mathbb{E}_Y[ y \mathbb{E}_X[X_1]]\\
&= \sum_{y=0}^{\infty} y \mathbb{E}_X[X_1] e^{-\lambda} \frac{\lambda^y}{y !}\\
&= \lambda \mathbb{E}[X_1] e^{-\lambda} \sum_{y=1}^{\infty}  \frac{\lambda^{y-1}}{(y-1) !}\\
&= \lambda \mathbb{E}X_1
\end{align}

since the sum is an series expansion for $e^{\lambda}$. We're using the iid property to go from the first to the second line. Then it's just the expectation of the Poisson distribution.


问:
Singularities classification
2 unsolved questions:

Prove that all the singular points of rational function $f(z)=\frac{P_1(z)}{P_2(z)}$, where $P_{1,2}$ are polynomials, are either removable singularities or poles.

Consider function $f(z)$, and it's derivative $f'(z)$. Given $z_0$ is simple pole of $f'(z)$, characterize the point  $z_0$ with respect to $f(z)$.

Thanks in advance!

答:
We can just prove that the limit when z tends to the singularity of $f$ exists and is in the compactification of $\mathbb{C}$ (i.e., infinity is an acceptable value for our limit).
This is because singularities of complex functions are characterized by how the limit looks in the singularity. If the limit exists and is finite, then the singularity is removable. If the limit is infinite, then the singularity is a pole. If the limit does not exist, we got an essential singularity. (All converse propositions hold, so this is a great characterization). So, the proof just comes from the fact that polynomials are holomorphic over $\mathbb{C}$, so they admit a Taylor expansion around our singularity $z_0$. We get then that $\lim_{z \to z_0}{f(z)}=\infty$

2.
Since every complex derivable function is differentiable infinite times, $f(z_0)$ can't be continous at $z_0$

Title: Singularities classification


问:
Isomorphism between homotopy groups of CW-complexes
Let $(Y, y_0)$ and $(Y’, y_0)$ pointed CW-complexes, with $Y’$ obtained from $Y$ by attaching $n+1$-cells. Why is it true that $i_{*}: \pi_{q}(Y, y_0) \to \pi_{q}(Y’, y_0)$ is an isomorphism for $q < n$ and an epimorphism for $q=n$?
I am aware of the fact that this is true if $Y$ is the $n$-skeleton of $Y’$, but here I only know that the $n$-skeleton of $Y’$ is contained in $Y$. By the way, this has already been asked here, but I couldn’t quite understand the answer.

As a related question: if $e_{\alpha}^{n+1}$ is a $n+1$-cell in $Y’$, why is it true that $i_{*}(\alpha) =0$?

Note I am using notation as in Switzer, Lemma 9.8 in the chapter dedicated to Brown theorem.

答:
By cellular approximation, up to homotopy, one may assume that $Y'$ is obtained by adding $n+1$-cells to the $n$-skeleton of $Y$, so that $Y^{(n)}= Y'^{(n)}$

Let $f:S^q\to Y'$ be a pointed map, $q\leq n$. Then by cellular approximation, up to pointed homotopy, $f$ lands in the $n$-skeleton of $Y'$. But the $n$-skeleton of $Y'$ is the same as that of $Y$, so $f$ lands in $Y$ up to homotopy. This shows surjectivity for $q\leq n$.

Similarly, if $q<n$, because then $q+1\leq n$,  again by cellular approximation, any homotopy between two such maps can be homotoped to live in $Y$, and so this shows injectivity for $q<n$.

For the second question, if you've added a cell along $\alpha: S^n\to Y$, then that means, by definition, that there is a commutative diagram :

$\require{AMScd}\begin{CD} S^n @>>> Y\\
@VVV @VVV \\
D^{n+1} @>>> Y'\end{CD}$

it follows that $S^n\to Y\to Y'$ extends to $D^{n+1}$, i.e. it is $0$ in homotopy.


问:
On the Moving Plane Method - Conditions for the reflected cap to stay in the domain
In the application of the Moving Plane Method to Poisson's Equation, James Serrin (A Symmetry Problem in Potential Theory, 1971) provides the following preliminary construction for the proof:

Let $T_0$ be a hyperplane not intersecting the (bounded) domain $\Omega$. We suppose this plane to be continuously moved normal to itself to new positions, until ultimatelyit begins to intersect $\Omega$. From that moment onward, at each stage of the motion the resulting plane $T$ will cut off from $\Omega$ a cap $\Sigma(T)$: that is, $\Sigma(T)$ will be that portion of $\Omega$ which lies on the same side of $T$ as the original plane $T_0$. For any cap $\Sigma(T)$ thus formed, we let $\Sigma'(T)$ be its reflection in $T$. Evidently $\Sigma'(T)$ will be contained in $\Omega$ at the beginning of the process; and indeed as $T$ advances into $\Omega$ the resulting cap $\Sigma'(T)$ will stay within $\Omega$ at least until one of the following two events occur: ($i$) $\Sigma'(T)$ becomes internally tangent to the boundary of $\Omega$ at some point $P$ not on $T$, or ($ii$) $T$ reaches a position where it is orthogonal to the boundary of $\Omega$ at some point $Q$.

Now, I am having a bit of trouble interpreting these conditions for $\Sigma'(T)$ to leave $\Omega$. I understand condition ($i$). However, can it happen that $\Sigma'(T)$ leaves $\Omega$ in such a way that only ($ii$) happens?

I have thought a lot about this. My current understanding is that ($ii$) is not a limit situation for $\Sigma'(T)$ to stay inside $\Omega$, but rather that if ($ii$) occurs, then either $\Sigma'(T)$ is still inside $\Omega$ or ($i$) is occuring as well. I came to think this because this orthogonality is used when it first appears when the plane is moving.

Is my interpretation correct, or is there a positive answer to the question above (boldfaced)?

I tried to be the most clear, but please let me know if it is not sufficient. I will try harder.

Thanks in advance and kindest regards.

答:
Your interpretation of the statement is correct and the answer to your bold question is no. This is not the goal of the statement in the cited article, anyway.

Let me go into some more detail. The construction of Serrin has the goal to prove $\Omega=\Sigma(T')\cup\Sigma'(T')$, where $T'$ denotes the first position of the hyperplane in which either (i) or (ii) occurs. In this case, (ii) always implies (i), due to easy symmetry considerations. However, in the general setting, (ii) can of course occur even if (i) doesn't (so if $\Sigma'(T)\subset \Omega$ even if $T$ moves past $T'$), which is what I understand you were asking.


问:
Finding Variance using MGF with sums of R.V's
Consider a r.v. X with MGF $m(t) = \frac{e^{k(e^t-1)}}{(1-bt)^a} $

$(a,b,k) ∈ \Bbb{R}$

Differentiating twice we retrieve
 $m''(t) = (a+1)ab^2e^{k(e^t-1)}(1-bt)^{-(a+2)}+2abke^{k(e^t-1)+t}(1-bt)^{-(a+1)}+ke^{k(e^t-1)+t}(ke^t+1)(1-bt)^{-a}$

Evaluating at t=0, we retrieve $VarX$

$VarX =m''(0)=a^2b^2+ab^2+2abk+k^2+k$

Now we observe that $m(t) =e^{k(e^t-1)}(1-bt)^{-a} = m_y(t)m_z(t) $

and our original r.v. $X = Y+ Z$

For two r.v.'s $Y$ and $Z$, where  $Y$~$Pois(k)$ and $Z$~$\Gamma(a,b)$

(From the reproductive property of MGF's and comparing our equation with  commonly known MGFs)

Now since $X=Y+Z$, we can again find the variance:

$Var(X)=Var(Y+Z)=Var(Y)+Var(Z)=ab^2+k$

As $Var(Y) = k$ and $Var(Z) = ab^2$

Which contradicts the first expression we found for the variance.

Can anyone please explain to me find the error I have made?

The differentiation seems like the obvious candidate for my error, but I have checked a few times and also done it using Wolfram Alpha.

      https://www.wolframalpha.com/input/?i=second+derivative+of+(e%5E(11(e%5Et-1))*(1-bt)%5E(-a)

Note that I used 11 instead of k and then just put k back in where multiples of 11 appeared, because Wolfram doesn't seem to allow for 3 constants when it differentiates.

I also have assumed that Y and Z are independent implicitly, but I am unsure of how to test this.

答:
Just for completeness

You actually found $E[X^2] =m''(0)=a^2b^2+ab^2+2abk+k^2+k$, as Priyanka said

You could also have found $E[X] =m'(0)=ab+k$

$\mathrm{Var}(X)=E[X^2] -(E[X])^2= a^2b^2+ab^2+2abk+k^2+k - a^2b^2-2abk-k^2 $ $= ab^2+k$, as in your second expression


问:
Whether the subspaces are homeomorphic or not
From topology without tears,

Let $X$ be a unit circle in $\mathbb{R}^2$,that is , $X=\{(x,y):x^2+y^2=1\}$ and has subspace topology.  $Y$ be a subspace in  $\mathbb{R}^2$ given by  $Y=\{(x,y):x^2+y^2=1\} \cup \{(x,y):(x-2)^2+y^2=1\} $
and $Z$ be a subspace in  $\mathbb{R}^2$ given by  $Z=\{(x,y):x^2+y^2=1\} \cup \left\{(x,y):\left(x-\frac{3}{2}\right)^2+y^2=1\right\} $

I have two questions:

1)How to show whether $Y$ is homeomorphic to a interval?

I have theorem:Let $f:(X, \tau_1) \rightarrow (Y, \tau_2)$ be a homeomorphism. Let $a \in X$, so that $X \setminus \lbrace a \rbrace$ is a subspace of $X$ and has induced topology $\tau_3$. Also, $Y\setminus\lbrace f(a)\rbrace$ is subspace of $Y$ and has induced topology $\tau_4$. Then $(X \setminus \lbrace a \rbrace, \tau_3)$ is homeomorphic to $(Y\setminus\lbrace f(a)\rbrace, \tau_4)$.

Using the above theorem  I can show $X$ is not homeomorphic to any interval say $(a,b)$.Because if it is,then $X\setminus\{a\}$ must be homeomorphic to $(a,b)\setminus\{f(a)\}$ for all $a \in X$ by above theorem. But $X\setminus\{a\}$ is always connected but $(a,b)\setminus\{f(a)\}$ is not connected. Therefore $X$ cannot be homeomorphic  to any interval.

But how should I show whether $Y$ is homeomorphic to any interval or not.Because here $Y\setminus\{(1,0)\}$ is not connected.So I cannot use the same argument as above and show that $Y$ is not homeomorphic to any interval.Is it homeomorphic?How should I show that?

2)To show that $Z$ is not homeomorphic to $X$ and $Y$

Using the above theorem I can show that $Z$ is not homeomorphic to $Y$,Since $Z-\{a\}$ is always connected but $Y\setminus\{(1,0)\}$ is not connected so I can define $f(a)=(1,0)$ and show $Z\setminus\{a\}$ cannot be homeomorphic to  $Y\setminus\{f(a)\}$ for  some $a$. And hence $Z$ cannot be homeomorphic to $X$

But how should I show $Z$ is not homeomorphic to $X$,because $Z\setminus\{a\}$ and $X\setminus\{b\}$ is always connected for all $a\in X$ and $b\in Z$. Connectedness cannot be used here.So how should I prove this?

Any hint will be of big help.

Thanks in advance!

答:
$X$ has no cutpoints at all

$Y$ has a unique cutpoint.

$Z$ has the property that no removal of two points simultaneously makes $Z$ disconnected. (And it has no cutpoints).

This topologically distinguishes these spaces, as cutsets and cutpoints are preserved by homeomorphisms.

Note that intervals have at most two non-cutpoints.


问:
If $\varphi_{i}^{j}:V\to \mathbb{R}$ smooth, then $(\varphi_{i}^{j}(p))_{1\leq i,j\leq k}$ is smooth.
Could you help me with the following question, please?

Let $M$ a smooth manifold and $V\subseteq M$ a open set. The maps $\varphi_{i}^{j}:V\to \mathbb{R}$ are smooth for all $i, j\in\{1,...,k\}$. Is the map $F:V\to M(k\times k,\mathbb{R})$ defined by $p\mapsto (\varphi_{i}^{j}(p))_{1\leq i,j\leq k}$ smooth?

Here
\begin{equation}(\varphi_{i}^{j}(p))_{1\leq i,j\leq k}=\begin{pmatrix}\varphi_{1}^{1}(p) & \varphi_{2}^{1}(p) & \cdots & \varphi_{k}^{1}(p)\\ \varphi_{1}^{2}(p) & \varphi_{2}^{2}(p) & \cdots & \varphi_{k}^{2}(p)\\ \vdots & \vdots & \ddots & \vdots\\ \varphi_{1}^{k}(p) & \varphi_{2}^{k}(p) & \cdots & \varphi_{k}^{k}(p) \end{pmatrix}.\end{equation}

I know that $p \mapsto(\varphi_{1}^{i}(p), \varphi_{2}^{i}(p),  \cdots, \varphi_{k}^{i}(p))$ is a smooth map, and I tried to relate this idea to the matrix but to no avail, and I don't know if this idea is correct.

答:
Recall that $M(k\times k, \mathbb{R})$ is a smooth manifold diffeomorphic to $\mathbb{R}^{k^2}$. More precisely, there is a smooth atlas consisting of a single chart $(M(k\times k, \mathbb{R}), \Psi)$ where $\Psi : M(k\times k, \mathbb{R}) \to \mathbb{R}^{k^2}$ maps a matrix $A = [a_{ij}]$ to the vector $(a_{11}, \dots, a_{1k}, a_{21}, \dots, a_{2k}, \dots, a_{k1}, \dots, a_{kk})$.

The map $F : V \to M(k\times k, \mathbb{R})$ is smooth if and only if $\Psi\circ F : V \to \mathbb{R}^{k^2}$ is smooth, and a map to Euclidean space is smooth if and only if each of its component functions are smooth. The result now follows as the component functions of $\Psi\circ F$ are $\varphi_i^j$.


问:
Finding the determinant of $2A+A^{-1}-I$ given the eigenvalues of $A$

Let $A$ be a $2\times 2$ matrix whose eigenvalues are $1$ and $-1$. Find the determinant of $S=2A+A^{-1}-I$.

Here I don't know how to find $A$ if eigenvectors are not given. If eigenvectors are given, then I can find using $A=PDP^{-1}$. Please solve for $A$; the rest I can do.

答: If $u,v$ are eigenvectors of $A$ for the eigenvalues $1,-1$ respectively then $Su=(2A+A^{-1}-I)u=2Au+A^{-1}u-u=2u+u-u=2u$. Similarly $Sv=-4v$. Therefore the eigenvalues of $S$ are $2,-4\ldots$</p>


答:
Here is an alternative answer, which doesn't give an explicit expression for the limit, but I think requires less technology and is more flexible.

Since all the summands are nonnegative, is is enough to find a summable upper bound. Let:

$$u_k := \sum_{j=1}^k \frac{r^{k-j}}{j}.$$

Note that $r^{k-j}$ is small when $j$ is close to $1$, so when $1/j$ is large. Conversely, $r^{k-j}$ is large when $j$ is close to $k$, so when $1/j$ is small. This is something we shall have to use: a naive approach to split the sum à la Cauchy-Schwarz won't work (it will only give that $u_k$ is bounded by a constant). Indeed, if we replace $u_k$ with

$$v_k := \sum_{j=1}^k \frac{r^j}{j},$$

it is immediate that $(v_k^2)$ converges to a positive constant as $k$ goes to infinity, so that $\sum_{k=1}^{+\infty} v_k^2 = +\infty$.

A standard idea, in this situation, is to split the domain into two sub-domains, one where $r^{k-j}$ is small, and one where $1/j$ is small. For all $m \leq k$:

$$u_k = \sum_{j=1}^m \frac{r^{k-j}}{j} + \sum_{j=m+1}^k \frac{r^{k-j}}{j}
\leq \sum_{j=1}^m r^{k-j} + \frac{1}{m+1} \sum_{j=m+1}^k r^{k-j} \leq \frac{r^{k-m}}{1-r} + \frac{1}{(m+1)(1-r)}.$$

Now, we need to choose a good value for $m$. Here, $m = \lfloor k/2 \rfloor$ is enough:

$$u_k \leq \frac{1}{1-r} \left( r^{k/2} + \frac{2}{k}\right).$$

From there, proving that $\sum_{k=1}^{+\infty} u_k^2 < +\infty$ is straightforward.

Title: Convergence of $S_p = \sum_{k=1}^{p}\left( \sum_{j=1}^{k}\frac{r^{k-j}}{j}\right)^2$

Body:
I'm looking at the following sum

$$ S_p = \sum_{k=1}^{p}\left( \sum_{j=1}^{k}\frac{r^{k-j}}{j}\right)^2$$
where $r \in (0,1), p > 1$. The goal is to inspect whether the sum converges to some constant, as $p \rightarrow \infty$.

Judging solely by the numerical evaluations I'm guessing that the sums should converge to some value, but not sure how to get there analytically.

One idea was to try finding some smaller upper bound for the inner summation to extract the $\frac{1}{j}$ out to $\frac{1}{k^2}$ or something close (if that makes sense), but haven't found the way so far.

Are there any known results that might work here?


问:
Simple exercise in fundamental groups and identification degree
I know this is simple, but I'm not being able to do this.

a) Let $\Phi : \mathbb{S}^1 \times \mathbb{S}^1 \to \mathbb{S}^1$ be the multiplication of complex numbers, $\Phi (z_1, z_2) = z_1z_2$ and let $\omega_j: \mathbb{I} \to \mathbb{S}^1$ be any two loops based at 1. Define $(ω_1 • ω_2)(t) = \Phi ◦ (ω_1(t), ω_2(t))$

i) Show that $[ω_1 • ω_2] = [ω_1 ∗ ω_2]$.

ii) Under the identification deg : $π_1(\mathbb{S}^1) \cong \mathbb{Z}$, show that $Φ_\# : π_1(\mathbb{S}^1 × \mathbb{S}^1) → π_1(\mathbb{S}^1)$ is given by $Φ_\#(a, b) = a + b$.

答:
Let $(\Bbb S^1,\times)$ be the topological group with identity element $1$, where $\times$ denotes the usual multiplication of complex numbers. For any two continuous functions $\alpha,\beta:[0,1]\to \Bbb S^1$, define $\alpha\bullet\beta:[0,1]\to \Bbb S^1$ as $\alpha\bullet\beta(t)=\alpha(t)\times\beta(t)$ for all $t\in [0,1]$. Let $C_1$ be the constant loop in $\Bbb S^1$ based at $1$. Also, $*$ stands for concatenation of two paths.

For two loops $\gamma,\delta$ in $\Bbb S^1$ based at $1$ we have $\gamma*C_1\simeq_{\text{rel }1}\gamma\simeq_{\text{rel }1} C_1*\gamma$ as $C_1$ represents the identity element of $\pi_1(\Bbb S^1,1)$, so that  $$\gamma*\delta=(\gamma*C_1)\bullet (C_1*\delta)\simeq_{\text{rel }1}\gamma\bullet \delta\simeq_{\text{rel }1}(C_1*\gamma)\bullet  (\delta*C_1)=\delta*\gamma.$$

In the above, to check the "$=$" just plugin an arbitrary $t\in [0,1]$ both left and right-hand sides. So, we completed part $(i)$.

Note that $\deg:\pi_1(\Bbb S^1,1)\to \Bbb Z$ is a group isomorphism.

Also, $\pi_1\big(\Bbb S^1\times \Bbb S^1,(1,1)\big)\cong\pi_1(\Bbb S^1,1)\times \pi_1(\Bbb S^1,1)$ via projections induced maps. So, write an element of $\pi_1\big(\Bbb S^1\times \Bbb S^1,(1,1)\big)$ as $\big([\omega_1],[\omega_2]\big)$ for some $[\omega_1],[\omega_2]\in \pi_1(\Bbb S^1,1)$. Then, $$\Phi_\#\big([\omega_1],[\omega_2]\big)=[\omega_1\bullet\omega_2]=[\omega_1*\omega_2].$$ So, using degree isomorphism $$\Phi_\#\big(\deg[\omega_1],\deg[\omega_2]\big)=\deg[\omega_1*\omega_2]=\deg[\omega_1]+\deg[\omega_2].$$


问:
What's the minimum number of balls to pick to ensure at least two are of the same color? Given a bag of balls of two colors
If you have a bag of jelly beans with a large number of two different colours, what's the minimum you need to pick out to guarantee that you have at least $2$ of the same colour?

I found an answer online - "$3$, and for the general case picking $m$ with the same colour from $n$ classes you'd need: $m \cdot n + 1$".

But I don't understand the reasoning behind it. Please advise.

Thanks in advance.

答: The reasoning is the following: if you have a large number of jelly beans of 2 different colours in the bag, let's say blue and red, and you draw one at random from your bag, it would be blue or red. Let's suppose to do that and to draw at first a red jelly. Then, if we draw one other jelly, two things could happen: or we are lucky and we draw an other red jelly (in this lucky case we have two jelly of the same colour, as we wanted) and we finish the "game", or we are unlucky and we draw a blue jelly. In this unlucky case, we have two jellies: one blue and one red so we must draw one other jelly, the third: if it will blue, we will have in total 2 blue jellies and one red jelly, instead if it will be red, we will have in total 2 red jellies and one blue jelly. As you see, in both of the case, the third jelly that we draw will make us have at least two jellies of the same colour, doesn't matter how many jellies are in the bag (they could be 10, 100 or infinite, you can repeat this reasoning) provided that they are divided in only two colours. We saw also that when we have to solve this type of problems we have to examine the number of trying that we have to do in the most unlucky case possible to "win".</p>


问:
Minima of $\sin (1+\sin x)$
I want to find the minima of $f(x) = \sin (1+ \sin x)$ for $0<x<6$

So a minimum value of $\sin \theta$ occurs at $\frac{3 \pi }{2}$, so I would have thought to set $1+\sin x = \frac{3 \pi}{2}$, but this is not valid.

The answers are two minima at $\frac{\pi}{2}$ and $\frac{3 \pi}{2}$ but I can’t seem to justify them. I can work out the maxima.

I don’t want to use calculus.

Any help would be appreciated.

答: Notice that $0\leq 1 +\sin x\leq 2$ ($<\pi$). If $0<w<2$, then $\sin w >0$ and if $w=0$, then $\sin w = 0$. Thus the (global) minimum will occur when $1+\sin x =0$, i.e., $\sin x =-1$. Since $0<x<6$, this means $x=\frac{3\pi}{2}$.

Title: Minima of $\sin (1+\sin x)$


答:
$$-1\le\sin x\le 1,\forall x\in\mathbb{R}$$
Thus we know that $0\le\sin x +1\le 2$ therefore
$$\sin (\sin x+1)\ge 0$$
More precisely $\sin (\sin x+1)=0$ at $x=\frac32\pi$ where $\sin x=-1$ and we have the global minimum.

To get the local minimum at $x=\frac{\pi}{2}$ consider that

$\sin x$ increases for $0<x<\frac{\pi}{2}$ and decreases for $\pi>x>\frac{\pi}{2}$

$\sin  (\sin x+1)$ is decreasing when  $\frac{\pi}{2}<x<2$, reaches a minimum $\sin 2$ when $\sin x = 1$ at $x=\frac{\pi}{2}$ then increases again up to $1$ when $x=\pi$

Hope it is clear. It's quite complicated to explain  without derivatives  :)

a graph can help

$$...$$

Title: Minima of $\sin (1+\sin x)$

Body:
I want to find the minima of $f(x) = \sin (1+ \sin x)$ for $0<x<6$

So a minimum value of $\sin \theta$ occurs at $\frac{3 \pi }{2}$, so I would have thought to set $1+\sin x = \frac{3 \pi}{2}$, but this is not valid.

The answers are two minima at $\frac{\pi}{2}$ and $\frac{3 \pi}{2}$ but I can’t seem to justify them. I can work out the maxima.

I don’t want to use calculus.

Any help would be appreciated.


问:
Prove that MGF defined on an open set is infinitely differentiable
I am doing an exercise in the book "Applied Stochastic Analysis" by E-Li-Vanden-Eijnden, and I meet this problem (on Page 26):

(Exercise 1.19) Prove that if the moment generating function $M_X(t)$ can be defined
on an open set $U$, then $M_X(t)\in C^{\infty}(U)$.

Here is my approach:

Use Taylor's expansion at $t=t_0$,
$$\frac{1}{t}\mathbb{E}[e^{(t_0+t)X}-e^{t_0X}]=\mathbb{E}[e^{t_0X}X]+o(t)\mathbb{E}[e^{t_0X}]$$
Let $t\to 0$,
$$\frac{d}{dt}M_X(0)=\mathbb{E}[e^{t_0X}X]$$
Repeat differentiation $n$ times,
$$\frac{d^n}{dt^n}M_X(0)=\mathbb{E}[e^{t_0X}X^n]$$
Since $n$ is arbitrary, as $n\to\infty$, we prove the claim.

Can anyone help me check whether my approach is valid? I'm not quite confident, since the exercise allows only on open sets. Alternative approaches are also fine. Thank you.

Edit:
OKay, I think I figure this out. Choose $U=(-\epsilon,\epsilon)$. For $|t|<\epsilon$,
$$\frac{|t|^n|X|^n}{n!}\leq e^{t|X|}\leq e^{tX}+e^{-tX}<\infty$$
and since
$$\left|\sum_{k=0}^{n}\frac{(tX)^k}{k!}\right|\leq\sum_{k=0}^{n}\frac{|tX|^k}{k!}\leq e^{|tX|}\leq e^{tX}+e^{-tX}$$
By dominated convergence theorem, $$M_X(t)=\mathbb{E}\left[\sum_{k=0}^{\infty}\frac{(tX)^k}{k!}\right]=\sum_{k=0}^{\infty}\frac{t^k}{k!}\mathbb{E}[X^k]$$
$M_X(t)$ coincides with the power series on $U=(-\epsilon,\epsilon)$, and hence it is infinitely differentiable on $U$, and $M_X^{(k)}(0)=\mathbb{E}[X^k]$.

答: Answer for the case $U=(-r,r)$:  First note that $Ee^{t|X|} \leq E(e^{tX}+e^{-tX}) <\infty$ for $0< t <r$.  Let us show that $M_X$ is differentiable. Using similar argumnets and induction we can show that it is infinitely differentiable. Let $|t|<r$ and choose $u$ with $|t|<u<r$. Consider $M_X'(t)=\lim_{h \to 0}  E{e^{tX}}\frac {e^{hX-1}} h$. By MVT we get $ |{e^{tX}}\frac {e^{hX-1}} h| \leq |X| e^{u|X|}$ provided $|h| <u-|t|$. If we show that $E|X|e^{u|X|} <\infty $ it follows by DCT that $M_X$ is differentiable are $t$ and $M_X'(t)=EXe^{tX}$. Now pick $v$ such that $u <v<r$ and note that there exists a finite  constant $C$ such that $|x|e^{u|x|} \leq Ce^{v|x|}$ for all real numbers $x$. It follows that $E|X|e^{u|X|} <\infty $ so we are done.


答:
You cannot say much about the relation between the two spans, except that they have the same dimension (for instance if one is the whole space, the other must be so as well). One way to see that you cannot expect a strict relation is to change the vectors in a way that certainly leaves their span unchanged, for instance permuting them or adding a multiple of one to another. This modifies the "transposed" vectors in a way that is not guaranteed to leave their span unchanged. To simplify your example, start with two vectors equal to zero.

Now that the question says that the original vectors $\vec a,\vec b,\vec c$ do indeed span the space, let me hint at an argument that the transposed ones do so as well. The vectors $\vec a,\vec b,\vec c$ form a basis of $\Bbb R^3$. Thus the map $\phi:\Bbb R^3\to\Bbb R^3$ that sends $(x,y,z)$ to the linear combination $x\vec a+y\vec b+z\vec c$ is an isomorphism (invertible linear map). If you interpret the transposed vectors as $1\times 3$ matrices $(a_i~b_i~c_i)$  (that is, you undo the transposition) then they describe the linear maps $\Bbb R^3\to\Bbb R$ defined by applying $\phi$ and then taking coordinate$~i$ (for $i=1,2,3$, repectively). Given this, can you argue that any $1\times 3$ matrix must be a linear combination of these three matrices, or equivalently that any linear map $\Bbb R^3\to\Bbb R$must be a linear combination of these three linear maps? (This is certainly true for the three coordinate maps themselves.)

Title: Given a spanning set, what is the span of the 'transpose' of the set?

Body:
Given $$sp\left \{
\begin{pmatrix}
a_1\\
a_2\\
a_3
\end{pmatrix}
,\begin{pmatrix}
b_1\\
b_2\\
b_3
\end{pmatrix}
,\begin{pmatrix}
c_1\\
c_2\\
c_3
\end{pmatrix}
\right \} = \mathbb{R}^3$$

What is
$$sp\left \{
\begin{pmatrix}
a_1\\
b_1\\
c_1
\end{pmatrix}
,\begin{pmatrix}
a_2\\
b_2\\
c_2
\end{pmatrix}
,\begin{pmatrix}
a_3\\
b_3\\
c_3
\end{pmatrix}
\right \}\subseteq \mathbb{R}^3$$

I'm thinking of row spaces and column spaces but with not much luck,
a leading hint would be appreciated more than the final answer.


问:
Eliminating time from parametric questions
Is there a general technique to eliminate a parameter from two parametric equations? E.g. given the following two parametric equations dictating the motion of a point how can I eliminate parameter $t$
(standing for time)?

\begin{align}
x_1(t)&=\sin(2\pi t)+2\cos(2\pi t)\\
x_2(t)&=\cos(2\pi t)+2\sin(2\pi t)
\end{align}

Intuitively, I think that these are the parametric equations of a rotated ellipsis. Using Matlab I get the following figure

答: There isn't really a general method, I think. You have to judge each case separately. Here it looks like identities such as $\sin^2(a) + \cos^2(a)= 1$ are going to be helpful in eliminating $t$ terms.


问:
What is the correct phase portrait for this linear system? Does the order of eigenvalues affect the solution curves?
I was attempting this problem in Lawrence Perko,

Since the order of eigenvalues do not matter, I found eigenvectors such that the system is diagonalized with 4 then 2.

It is quite easy and my answer matches the one given in the solution booklet.

The problem lies in how the phase portrait is drawn.

See, in both cases, we have different transformation matrices P since the order of the eigenvectors is different. So I agree the official solution is correct and the rotation of the phase portrait makes sense given that particular P.

But my phase portrait looks different now... since its not a simple rotation anymore. I transforms to P when moving from y to x. So the cross and star I marked should move to the circle and starInCircle. I am not clear what went wrong here...

Is my phase portrait wrong? Or does changing the order of eigenvectors affect the axis somehow?

答:
Excellent work on your part and it is great to see you questioning results.

If you do a parametric plot of $(c_1e^{2t},c_2 e^{4t})$, versus $(c_1e^{4t},c_2 e^{2t})$, they will be different.

Hint: are you sure you are drawing the correct phase portrait in the $y-$space?

Hint: The books result is correct.


问:
Is it always possible to construct this integer sequence?
Let $A_n$ be a sequence of monotonically increasing, positive integers. How can one construct another sequence $a_n$, again positive integers, such that $$\lim_{n \to \infty} \frac{A_n}{\displaystyle \prod_{k=1}^n a_k} = 1$$

Is it even always possible to construct such a sequence?

The obvious thing is to choose $a_{n+1}$ to be $\frac{A_{n+1}}{A_n}$, but its not always guaranteed that this will be an integer. I tried to use $a_{n+1} = \lfloor \frac{A_{n+1}}{A_n} \rfloor$, but I cant seem to work it out. For sequences like $A_n = k^n$ it is very easy to construct $a_n$ if $k$ is an integer itself. What about $A_n =\lfloor \alpha^n \rfloor $ if $\alpha$ is any number greater than 1?

答:
It is not always possible.

Let $A_n=n$ and $p(n)=a_1.a_2....a_n$. Then suppose that, for $n\ge N$, $$0.9<\frac{n}{p(n)}< 1.1$$

Then $\frac{2n}{p(2n)}= \frac{2}{A}\frac{n}{p(n)}$ for some integer $A$. For this to be in the assumed range we require $A=2$ but then $$\frac{n}{p(n)},\frac{2n}{p(2n)},\frac{4n}{p(4n)},...$$  are all constant and so do not tend to $1$.

The only remaining possibility is that, for all sufficiently large $n$, $\frac{n}{p(n)}=1$. But then $p(n)$ does not divide $p(n+1)$.


问:
Archimedean property and Well-Ordering Principle
Bit from book

Definitely me missing something easy here but say I take $a=4$ and $b=9$ then wouldn’t the least element be $b-ma=1$ where $m=2$ and how could  $b-(m+1)a$ also lie in $S$ because wouldn’t this be $9-(3)4=-3$? I’m just not seeing why the notice of $b-(m+1)a$ comes about.
Thanks

答:
What you have noticed is the contradiction which is used in the proof.

The assumption was that all $b-ma$ are positive. This is false as proved in the book and as is illustrated by your example.


答:
$$\begin{cases}
x_1=\sin (2 \pi  t)+2 \cos (2 \pi  t)\\
x_2=2 \sin (2 \pi  t)+\cos (2 \pi  t)\\
\end{cases}
$$
set $$\sin (2 \pi  t)=s;\;\cos(2 \pi  t)=c$$
we know that
$$\cos^2(2 \pi  t)+\sin^2(2 \pi  t)=1\to c^2+s^2=1$$

the system becomes
$$\begin{cases}
x_1=s+2 c\\
x_2=2s+c\\
\end{cases}
$$
whence
$$\begin{cases}
c=\frac{1}{3} (2 x_1-x_2)\\
s= \frac{1}{3} (2 x_2-x_1)\\
\end{cases}
$$
now remember that $$c^2+s^2=1$$
$$\left[\frac{1}{3} (2 x_1-x_2)\right]^2+\left[ \frac{1}{3} (2 x_2-x_1)\right]^2$$
After some simplification we get
$$5 x_1^2-8 x_1x_2+5 x_2^2=9$$

Title: Eliminating time from parametric questions

Body:
Is there a general technique to eliminate a parameter from two parametric equations? E.g. given the following two parametric equations dictating the motion of a point how can I eliminate parameter $t$
(standing for time)?

\begin{align}
x_1(t)&=\sin(2\pi t)+2\cos(2\pi t)\\
x_2(t)&=\cos(2\pi t)+2\sin(2\pi t)
\end{align}

Intuitively, I think that these are the parametric equations of a rotated ellipsis. Using Matlab I get the following figure


问:
Cocompact group action of a deck-transformation
Let $\Sigma$ be an orientable surface and $x\in \Sigma$. Let $\alpha\in \pi_1(\Sigma,x)$ be a non-trivial element and $f:(\Bbb S^1,1)\to (\Sigma,x)$ represents $\alpha$. Consider the lift  $\require{AMScd}$
\begin{CD}
 (\Bbb R,0) @>\displaystyle\ell>> (\widetilde \Sigma,\widetilde x)\\
@V  V V @VV  V\\
(\Bbb S^1,1) @>>\displaystyle f> (\Sigma,x)\\
\end{CD}

All unlabeled maps are the universal covers. Let $\varphi$ be a deck-transformation for the universal cover $\widetilde \Sigma\to \Sigma$ of infinite order i.e. for each $n\in \Bbb N$ the $n$-times composition of $\varphi$ with itself, denoted by $\varphi^{(n)}$ is not identity.

Is it true that there is a
$K\subseteq_\text{compact}\text{image}(\varphi\circ \ell)$ such that
$\text{image}(\varphi\circ \ell)=\bigcup_{m\in\Bbb Z}\varphi^{(m)}(K)$ i.e. $\langle \varphi\rangle$ acts cocompactly on $\text{image}(\varphi\circ \ell)$?

$\bullet$ Actually any non-trivial deck-transformation of an orientable surface$(\text{open/closed})$ has infinite order as the fundamental group of such surface does not have torsion element.

$\bullet$ $\ell$ is a proper map, so that image of $\ell$ is closed.

$\bullet$ If $\Sigma$ has a Riemannian metric and $f$ has the shortest length among all elements in its free homotopy class$($of loops$)$, then $\ell$ is an embedding of $\Bbb R$. Then the above problem can be thought of as the standard $\Bbb Z$-action on $\Bbb R$ by the translation which is cocompact as $\Bbb R/\Bbb Z\cong \Bbb S^1$.

I am not too much confident about the last bullet.

答:
This is not true.

Consider $\Sigma = T^2 = S^1\times S^1$ which has universal cover $\tilde{\Sigma}$. The universal covering map $\pi : \mathbb{R}^2 \to S^1\times S^1$ is given by $\pi(x, y) = (e^{2\pi i x}, e^{2\pi i y})$; fix $\tilde{x} = (0, 0)$ and $x = (1, 1)$. The deck transformations of this covering are translations by elements of the integer lattice $\mathbb{Z}^2$, so the group is generated by $\varphi_1$ and $\varphi_2$ where $\varphi_1(x, y) = (x + 1, y)$ and $\varphi_2(x, y) = (x, y + 1)$.

Now note that the fundamental group of $T^2 = S^1\times S^1$ is generated by $\alpha$ and $\beta$ where $f(z) = (z, 1)$ and $g(1, z)$ are representatives of $\alpha$ and $\beta$ respectively. In this case the map $\ell$ is given by $\ell(x) = (x, 0)$; i.e. the image of $\ell$ is the $x$-axis in $\mathbb{R}^2$. Now $(\varphi_2\circ\ell)(x) = (x, 1)$ which has image the line $y = 1$, while for any integer $m$ the image of $\varphi_2^{(m)}\circ \ell$ is the line $y = m$. Let $K \subseteq \operatorname{image}(\varphi_2\circ\ell) = \mathbb{R}\times\{1\}$ be any subset and write $K = K'\times\{1\}$ where $K' \subseteq \mathbb{R}$. Then $\bigcup_{m\in\mathbb{Z}}\varphi_2^{(m)}(K) = K'\times\mathbb{Z} \neq \mathbb{R}\times\{1\} = \operatorname{image}(\varphi_2\circ\ell)$. Note that $K'\times\mathbb{Z}$ is just the union of all copies of $K$ shifted by an integer amount in the vertical direction.


问:
Finding the value of $ax^4+by^4$

If $\quad a+b=23 , \quad ax+by=79,\quad ax^2+by^2=217,\quad
 ax^3+by^3=691\quad$ find the value of $ax^4+by^4$.

Here is my attempt:

$$(a+b)(x+y)=ax+by+ay+bx\rightarrow 23(x+y)=79+(ay+bx)$$
$$(ax+by)(x+y)=ax^2+by^2+axy+bxy\rightarrow79(x+y)=217+23 xy$$
In each equation I have two unknowns it seems that doesn't work.

Title: Finding the value of $ax^4+by^4$

Body:
Your method is not in vain. We have:

$$(ax+by)(x+y) = ax^2 + by^2 +axy + bxy \to 79(x+y) = 217 + 23xy$$

$$(ax^2 + by^2)(x+y) = ax^3 + by^3 + ax^2y + bxy^2 \to 217(x+y) = 691 + 79xy$$

hence $x + y = 1, xy = -6$. Now:

$$(ax^3 + by^3)(x+y) = ax^4 + by^4 + ax^3y + bxy^3 = ax^4 + by^4 + xy(ax^2 + by^2)$$

$$691\times 1 = ax^4 + by^4 -6 \times 217$$

thus $ax^4 + by^4 = 1993$.


问:
What will the result be if I multiply a non-diagonal metric by a non-diagonal inverse metric?
I know that multiplying a diagonal metric and its inverse is kronocker delta. But I wonder whether this holds up if tensors are not diagonal or not?</p>

答: If $g_{ij}$ are the components of a metric and $g^{ij}$ denote the inverse components, then $g^{ij}g_{jk} = \delta^i_k$ regardless of whether or not $[g_{ij}]$ is diagonal. This is just the statement that for an invertible matrix $A$ we have $A^{-1}A = I$ regardless of whether or not $A$ is diagonal.


答:
I understand that you do not want to use calculus but it seems to be the only way how to identify all minima:

A first derivative of $f(x)$ is $f'(x) = \cos(x)\cos(1+\sin(x))$. From $f'(x)=0$ we have two equations:

$$
\cos(x) = 0
$$

$$
\cos(1+\sin(x))=0
$$

On given interval the first equation has solution $x \in \{\pi/2; 3\pi/2\}$.

The second one can be turned to $1+\sin(x) = \frac{\pi}{2}$ since cosinus is zero in $\pi/2$, hence we have either
$$
x = \arcsin(\pi/2-1)
$$
or
$$
x = \pi - \arcsin(\pi/2-1)
$$

You can show that $\pi/2$ adn $3\pi/2\$ are minima and others are maxima.

Title: Minima of $\sin (1+\sin x)$


答:
That's the point: if the Archimedean property fails for two positive integers $a$ and $b$, then the members of $S = \{ b - na \mid \mbox{$n$ is a positive integer}\}$ would all be positive, but that can't happen.  In particular, it doesn't happen for $a = 4$ and $b = 9$, as your calculations show.

If you know about division, then you know that for any positive integers $a$ and $b$, there exist non-negative integers $q$ and $r$ such that $0 \le r < a$ and
$$
b = qa + r
$$
and then you can take $n = q + 1$ to give an $n$ such that $na > b$. The proof given in the book avoids knowledge of division (and could be used to develop the theory of division).

Title: Archimedean property and Well-Ordering Principle

Body:
Bit from book

Definitely me missing something easy here but say I take $a=4$ and $b=9$ then wouldn’t the least element be $b-ma=1$ where $m=2$ and how could  $b-(m+1)a$ also lie in $S$ because wouldn’t this be $9-(3)4=-3$? I’m just not seeing why the notice of $b-(m+1)a$ comes about.
Thanks


问:
Non closed connected subgroup of $SU(3)$
I’m working on a problem which asks me to show there exist non closed connected subgroup of $SU(3)$. I believe there should be some well-known standard example for this and I would like to remember such an example. If not, then I really have no clues on how to prove this by contradictions, since the only lemma I know about this is a sufficient condition which said maximal abelian Lie sub algebra corresponds to a unique closed connected subgroup. Please give some advices, thanks!

答: Take a look the section headed "Non-example" from https://en.wikipedia.org/wiki/Lie_group</p>


问:
How many three digit strings can be formed from numbers in a set such that the number represented by the string is divisible by three?
Consider the set of numbers $\{0,1,2,3,...,9 \}$ , now we want the number of ways to find three digit strings such that the string is divisible by three (with repetition).

For example,

012 is a good because twelve is divisible by three

009 is good because it is also divisible by three

031 is not because it is not divisible by three

My thought was to use divisibility rules, we can find that the problem can be found equivalent to finding the number of solutions to the equation:

$$ a+b+c = 3n$$
Where $n$ is some integer and $(a,b,c)$ represents the number we picked from the set. For example, $(0,1,2)$ would mean the number 012 and that's divisible by three because

$$ 0 + 1 + 2 = 3n$$

Which is true when $n=1$

Now, a way to solve this would be to use brute force counting but I'm hoping for a better way to count using combinatronics

答:
HINT

Split the digits up into $\{0,3,6,9\},\{1,4,7\},\{2,5,8\}$.

You must use three from the same set or one from each set.


问:
Hemicompact spaces
I want to know if hemicompactness is preserved during homeomorphisms. In other words, can we say that hemicompactness is a topological property?</p>

答:
The rule of thumb for me is "every property that can be defined purely in terms of set theory notions and open sets is a topological property". So yes.

Formally we can check this by proving:

Let $f: X \to Y$ be a homeomorphism. If $X$ is hemicompact then so is $Y$.

So let $K_n$, $ n \in \Bbb N$ be a cofinal compact sequence in $X$. Then define $K'_n=f[K_n] \subseteq Y$, which is compact as $f$ is continuous.

If then $C \subseteq Y$ is compact, then $f^{-1}[C]$ is compact in $X$ (as $f^{-1}$ is continuous) and so $f^{-1}[C] \subseteq K_m$ for some $m$ and then $C=f[f^{-1}[C] \subseteq f[K_m]=K'_m$ and so $\{K'_n: n \in \Bbb N\}$ is a cofinal compact sequence in $Y$ and so $Y$ is hemicompact.

Title: Hemicompact spaces


问:
Hasse Principle Violation example on Birch Swinnerton-Dyer-Dyer paper
I have come across the following surface as an example of a violation to the Hasse Principle.

(1) $uv = x^2 - 13y^2$

(2) $(u+v)(u+2v)= x^2 - 13z^2$

I am trying to show it has solutions in all fields of p-adic numbers and I can't see what is wrong with the following method:

$x^2 - 13z^2 = (u+v)(u+2v) = u^2 + 3uv +2v^2$.

So we have that

(3) $x^2 - 13z^2 = u^2 +3uv +2v^2$

By (1)

$3uv = 3x^2 - 39y^2$

Hence plugging this in to (3)

$x^2 - 13z^2 = u^2 + 3x^2 - 39y^2 + 2v^2$

So we have

(4) $39y^2 - 13z^2 = u^2 +2x^2 +2v^2$

So the surface defined by (1),(2) has solutions if and only if (4) does.

However (4) has solution $u=10, v=1, x=1, y=2, z=2$. Which isn't a solution to (1),(2).

答:
You write that
"So the surface defined by (1),(2) has solutions if and only if (4) does."
but you've only used equation (1) to simplify (2) to (4), so it looks like you've only shown
"So the surface defined by (1),(2) has solutions if and only if (1),(4) does.".
You can't forget about equation (1) just because you plugged it in, it still has to be satisfied by any solution.

As you can see your solution does not satisfy (1).


Title: Finding the value of $ax^4+by^4$

Body:
If $\quad a+b=23 , \quad ax+by=79,\quad ax^2+by^2=217,\quad
 ax^3+by^3=691\quad$ find the value of $ax^4+by^4$.

Here is my attempt:

$$(a+b)(x+y)=ax+by+ay+bx\rightarrow 23(x+y)=79+(ay+bx)$$
$$(ax+by)(x+y)=ax^2+by^2+axy+bxy\rightarrow79(x+y)=217+23 xy$$
In each equation I have two unknowns it seems that doesn't work.

Body:
Hint:

$$(a+b)(ax^2+by^2)-(ax+by)^2=\cdots=ab(x-y)^2\ \ \ \ (1)$$

$$(a+b)(ax^3+by^3)-(ax+by)(ax^2+by^2)=\cdots=ab(x-y)^2(x+y)\ \ \ \ (2)$$

On division, we get $x+y$

$$(a+b)(ax^3+by^3)-(ax^2+by^2)^2=\cdots=abxy(x-y)^2\ \ \ \ (3)$$

$(1)/(3)$ will give $xy$

$$(a+b)(ax^4+by^4)-(ax+by)(ax^3+by^3)=\cdots=ab(x-y)^2((x+y)^2-xy)\ \ \ \ (4)$$

$(4)/(3)$ will give $$\dfrac{23(ax^4+by^4)-79\cdot691}{23(691)-217^2}=\dfrac{xy}{(x+y)^2-xy}$$

Now replace the values of $x+y$ and $xy$


问:
Doubts about preparing for the IMO.
Would any of the following books be any good for preparing for IMO?

Schilling - Measures, Integrals and Martingales

Rudin- Real and Complex Analysis

E.J. Barbeau - Polynomials

F.Klein - Idk any good ones of F.Klein?

And any other books you may suggest. Also, I've already read EGMO and

I suck at geo

I only learnt how to bash from it.

答:
I wouldn't really recommend those, except for maybe some books on elementary Analysis as there are occasionally some sequences problems that come up for which a knowledge of analysis is useful.

Some books I would absolutely recommend are:

Combinatorial problems in mathematical competitions (Zhang)

Solving problems in geometry (Han Wang)

Graph Theory (Zhongyi) - good for some combinatorics problems

104 Number Theory Problems (USA IMO prep book)

Good luck!!

Title: Doubts about preparing for the IMO.


问:
Reflection on Hyperplane
Let $(V, \langle,\rangle) $ be an euclidean
Vector space. For $w \in V, w\ne 0$, the map $s_w: V \rightarrow V$, $s_w (v): = v -2 \tfrac{\langle v,w \rangle}{\langle w,w \rangle}$
is defined (the mapping $s_w$ is the reflection at the hyperplane ${v ∈ V |
\langle v,w \rangle = 0}$).

I want to proof that every orthogonal Endomorphism is a composition of such reflections.

First I formalized the statement:
For every orthogonal Endomorphism $f: V\rightarrow V$, there are $w_1,\dots, w_r \in
V, r ≤ n$ with $f = s_{w_1} ◦. . . ◦ s_{w_r}$.

I am not sure but in order to proof this statement, I had in mind that I could find a $w_1 \in V$ such that $\widetilde{f} = s_{w1} ◦ f$ is a straight line
through the origin and then consider the restriction of
$\widetilde{f}$ on the orthogonal complement of the straight line.

Maybe someone can share some thoughts on this or suggest an easier way. Thanks for any kind of advice!

答:
Proof: For $f = id_V$ the assertion is clear ("composition of 0 reflections"). Let
now $v \in V$ with $f(v) \neq v$. Set $w_1 := f(v) - v$. Set $\widetilde{f}:= s_{w_1} ◦ f$. Then $\widetilde{f}(v) = v$ is true.

Inductively, we can conclude that the assertion holds for the orthogonal complement of $v$. (Due to the orthogonality of $f$, $f$ induces an endomorphism on the orthogonal complement of $v$.)

We can thus write $\widetilde{f}|_{L(v)^⊥} = s_{w_2} ◦ \dots ◦ s_{w_r}$.
The reflections $s_{w_i}$ on $L(v)^⊥$ can be extended to reflections on $V$ by writing $s_{w_i}(v) = v$ for all $ i \in \{2, . . . , r\}$. With $s_{w_1} = s_{w_1}^{−1}$
we obtain:
$$f = s_{w_1} ◦ s_{w_2} ◦ \dots ◦ s_{w_r}$$

Title: Reflection on Hyperplane


问:
Is the punctured plane a normal space? Is a normal space without a point normal?
At first I thought is was obvious that $\mathbb R^2 \setminus \{ P \}$ is normal as a topological subspace of $\mathbb R^2$. But what about if we have two closed sets in $\mathbb R^2$, $F$ and $E$, such that $F \cap E = \{ P \}$. How can we show in general that we can separate $F \setminus \{ P \}$ from $E \setminus \{ P \}$ with open sets in $\mathbb R^2 \setminus \{ P \}$? Is this even true?

To generalise, is it true in general that the subspace obtained from a normal space by removing a point stays normal?

答: Hint: all metric spaces are normal spaces.


问:
Prime divisibility in a prime square bandtwidth
I am seeking your support for proving (or fail) formally the following homework:

Let $p_j\in\Bbb P$ a prime, then any $q\in\Bbb N$ within the interval $p_j<q<p_j^2$ is prime, if and only if all:

$$p_{i}\nmid q$$ with $1\le i<j$

There should be a simple sieving argument behind of this that I can not fish.

I hope I got it now correct.

Your help is welcome

答:
The condition of $q \nmid p_{j-1}$ is absurd, since $q > p_{j-1}$.

So let's assume that we flip the order.

Even then, the statement is still not true. For example, consider $p_4 = 7$, then 8 is a valid counter example.

I believe that the statement you want it:

$q$ is a prime if and only if $ p_i \nmid q$ for all values of $i<j$.

This is a true statement that is easy to show.

Hint: If $q$ is not a prime number, then it must have a prime factor that is less than $\sqrt{q} < p_j$. If $q$ is a prime, then clearly no smaller number divides it.


问:
It is about finding the coefficients of $x^n$
$$p(x) = \left(\frac{1-x^{10}}{1-x}\right)^3$$
I am to find the coefficient of $x^{21}$ in this. I am dumfounded, can anyone help me? I know that it uses combinatorics but can't exactly find the equation.

Body:
Hint :

$$p(x)=(1+x+...+x^9)^3 = (1+x+...+x^9) \times (1+x+...+x^9) \times (1+x+...+x^9)$$


答:
Answer to OP's comment:

For solving the following system:

$$79(x+y)=23xy+217$$$$ 217(x+y)=79xy+691$$

We first eliminate $(x+y) $ from both equations:

$$ (x+y) = \frac{1}{79} (23 xy + 217)\tag{1}$$

And,

$$ x+y = \frac{1}{217} (79 xy + 691) \tag{2}$$

We get:

$$  \frac{1}{217} (79 xy + 691)= \frac{1}{79} (23 xy + 217)$$

We get the result
$$ y = \frac{-6}{x} \tag{3}$$

Now we can substitute (3) into (1), and then if we multiply the whole equation by $x$ , we can solve for the quadratic in x

Title: Finding the value of $ax^4+by^4$


答:
From my own experience the by far most valuable preparation for competitions such as the IMO is to look at past problems. IMO problems are generally designed to have a relatively low threshold regarding to required knowledge: It is all about knowing the ideas how to use the few theorems and techniques. (An exception can be sometimes made in the field of Geometry and inequalities, where knowing a lot of exotic theorems can sometimes help: But usually this then is not the expected solution.)

Because of the specific way the problem selection process for the IMO works there are shortlists of around 8 problems ordered by difficulty of each of the four core disciplines for each year. They are all featured (including solutions) in the

IMO Compendium, a huge and yellow book.

In my opinion, one of the most effective ways of preparing is to simply pick out problems from the short list (the younger, the better: The IMO was way different 50 years ago.) Usually, you will not be able to solve them. But don't be afraid: Then look at the solutions (or at least a part of it) and try to understand how they work, and especially, how one would come up with it. If they utilize theoretical background you don't understand, try to understand in the context of the solution of the exercise.

And, I believe the most important thing during this is: Have fun! Having fun while preparing for a competition makes it so much easier. :)

Title: Doubts about preparing for the IMO.

Body:
Would any of the following books be any good for preparing for IMO?

Schilling - Measures, Integrals and Martingales

Rudin- Real and Complex Analysis

E.J. Barbeau - Polynomials

F.Klein - Idk any good ones of F.Klein?

And any other books you may suggest. Also, I've already read EGMO and

I suck at geo

I only learnt how to bash from it.


问:
Metric Dimension of Lattice Graph
I am sorry, I am beginner in this course about metric dimension of graph. I have a problem to determine metric dimension of lattice on graph theory. When I choose resolving set with $2,3,4,5$ vertices, then there are other vertices have same coordinates. It means metric dimension is greater than 5. Unfortunately, I can't check anymore for $6,7,...,36$ vertices. Any ideas or any lemmas that can help me to prove or determine the metric dimension of picture below? Thank you very much.

答:
Adding an answer so for completeness.

The metric dimension is $3$. Any set of three corner points will be a resolving set, which can be seen by writing out the triple of distances for each point on the grid.

Furthermore, no pair of points is a resolving set. For any pair of points $\{x,y\}$ with distance $d$, you show there will exist two points whose distances to $x$ and $y$ are the same:

If $x$ and $y$ are on the same diagonal (meaning the slope of the line connecting there centers is $\pm 1$), then there exist two points whose distances to $x$ and $y$ are both $d$.

If $x$ and $y$ are adjacent, there will exist two points whose distances to $x$ and $y$ are both $1$.

In all other cases, there will exist two points whose distances to $x$ is  $1$ and whose distance to $y$ is $d-1$.

Title: Metric Dimension of Lattice Graph


答:
Hint:

You obtain the desired coefficient by the long division

$$\frac{x^{30}-3x^{20}+3x^{10}-1}{(x-1)^3}.$$ Only the term $x^{30}$ plays a role and you can perform three successive divisions by $x-1$. Each time, the coefficients are the prefix sum of the coefficients of the dividend.

$$\frac{x^{30}}{x-1}=x^{29}+x^{28}+x^{27}+x^{26}+\cdots$$

$$\frac{x^{30}}{(x-1)^2}=x^{28}+2x^{27}+3x^{26}+4x^{25}+\cdots$$

$$\frac{x^{30}}{(x-1)^3}=x^{27}+3x^{26}+6x^{25}+10x^{24}+\cdots$$

The coefficients are triangular numbers, and that of $x^{21}$ is $T_7$.

Title: It is about finding the coefficients of $x^n$

Body:
$$p(x) = \left(\frac{1-x^{10}}{1-x}\right)^3$$
I am to find the coefficient of $x^{21}$ in this. I am dumfounded, can anyone help me? I know that it uses combinatorics but can't exactly find the equation.


问:
Simple Expect Value Exercise
Question: We have $9$ coins, $1$ of them is false (lighter). We divide them up in pairs (with one left) and weigh them (that is taking two in a balance and seeing if one of them is lighter). What is the expected value of number of weighings to find the false one?

I suspect this is a rather elementary problem, yet, whenever I solve these (or rather - when I think i solved them), I am never sure at all with my result being correct.

My solution The probability of the false coin being in any pair is $2/9$, therefore the probability $\mathbb{P}(X=t)$ that the false coin will be found is $2/9$ for $t=1,2,3$ and $3/9$ for $t=4$ since if it didn't show up in the last weighing, it must be the one left. Therefore:
$$\mathbb{E}[X]=\frac{2}{9}(1+2+3)+4\frac{3}{9}=\frac{24}{9}$$

Is that correct and if not, why?

Thanks for any help!

答:
Your solution is correct for the technique you're using to find the coin.

The nine coins problem is a classic puzzle, it's usually phrased as a challenge to find the light coin with as few weighings as possible.

Your solution to that problem is not optimal.
Instead for your first weighing put three coins on each side of the scale.
So you have sets groups of three coins, and you may deduce which set the coin is in.  Then you can find the coin with a second weighing.

Title: Simple Expect Value Exercise


问:
Show every compact subspace of $m$-manifold has topological dimension at most $m$.
Notation $\bar{U}$ is the closure of $U$.

Definition $U$ is a neighborhood of $x$ means that $U$ is an open set containing $x$.

Definition An $m$-manifold is a Hausdorff space $X$ with a countable basis such that each point $x$ of $X$ has a neighborhood that is homeomorphic with an open subset of $\mathbb{R}^m$.

答:
Theorem 1 Every compact subspace of $\mathbb{R}^N$ has topological dimension at most $N$.

Theorem 2 Let $X$ be a space having finite dimension. If $Y$ is a closed subspace of $X$, then $Y$ has finite dimension and dim $Y\leq$ dim $X$.

Theorem 3 Let $X=Y\cup Z$, where $Y$ and $Z$ are closed subspaces of $X$ having finite topological dimension. Then dim $X$ = max{dim $Y$, dim $Z$}.

Lemma 1 Let $X$ be a space; suppose that $h:X\rightarrow Z$ is an imbedding of $X$ in the compact Hausdorff space $Z$. Then there exists a corresponding compactification $Y$ of $X$; it has the property that there is an imbedding $H:Y\rightarrow Z$ that equals $h$ on $X$. The compactification $Y$ is uniquely determined up to equivalence.

My efforts:

Let $C$ be a compact subspace of $X$. For each point $x\in C$, we have a neighborhood $U_x$ of $x$ homeomorphic with an open subset of $\mathbb{R}^m$. The collection of $U_x\cap C$ is an open covering of $C$. Since $C$ is compact, we have a finite open covering whose members are denoted as $U_i\cap C$.

Let $f_i$ be the homeomorphism mapping $U_i$ to $f_i(U_i)$ of $\mathbb{R}^m$. Then by Theorem 1, $\bar{f}_i(U_i)$ has topological dimension at most $m$.

We view $U_i$ as a space and view $\bar{f}_i(U_i)$ as compact Hausdorff space. The $f_i$ is an imbedding of $U_i$ in $\bar{f}_i(U_i)$. By lemma 1, there exists a corresponding compactification $V_i$ of $U_i$; it has the property that there is an imbedding $F_i:V_i\rightarrow \bar{f}_i(U_i)$ that equals $f_i$ on $U_i$. The compactification $V_i$ is uniquely determined up to equivalence.

Then $V_i$ has topological dimension at most $m$.

Since $C$ is closed, we can write it as $C=\bigcup(\bar{U}_i\cap C)$. Due to equivalence of $V_i$ and $\bar{U}_i$, $\bar{U}_i$ has topological dimension at most $m$. By theorem 2, $\bar{U}_i\cap C$ has topological dimension at most $m$. By theorem 3, $C$ has topological dimension at most $m$.


问:
Can the following expression be related to the multinomial formula?
The following formula
$$(n,p,q,r\ \text{odd})\quad \sum_{\genfrac{}{}{0pt}{1}{p\leq q \leq r}{p+q+r =n}} \frac{n!}{p!\, q!\, r!} \times \begin{cases} 1 & \text{if}\ p< q <r\\
 \frac{1}{2} & \text{if exactly two indices are equal}\\
 \frac{1}{3!}& \text{if}\ p=q=r= \frac{n}{3} \end{cases}$$
reminds me very much of the "multinomial formula"
$$ \left( x_1 + x_2 + \cdots + x_d \right)^n = \sum_{\left|\boldsymbol{\alpha}\right|=n} \genfrac{(}{)}{0pt}{0}{\left|\boldsymbol{\alpha}\right|}{\boldsymbol{\alpha}}\, \mathbf{x}^{\boldsymbol{\alpha}}
$$
where $\boldsymbol{\alpha}\in \mathbb{N}^d,\ \left|\boldsymbol{\alpha}\right| = \alpha_1 + \alpha_2 + \cdots + \alpha_d $  and
$$ \begin{split}
 \genfrac{(}{)}{0pt}{0}{\left|\boldsymbol{\alpha}\right|}{\boldsymbol{\alpha}} & = \genfrac{(}{)}{0pt}{0}{n}{\alpha_1} \genfrac{(}{)}{0pt}{0}{n-\alpha_1}{\alpha_2}\, \cdots \, \genfrac{(}{)}{0pt}{0}{n-\alpha_1 -\alpha_2 - \cdots - \alpha_{d-1}}{\alpha_d}= \frac{n!}{\alpha_1 !\, (n-\alpha_1)!} \frac{(n-\alpha_1)!}{\alpha_2 !\, (n-\alpha_1-\alpha_2)!}\, \cdots\, \frac{(n-\alpha_1 -\alpha_2 - \cdots - \alpha_{d-1})!}{\alpha_d !\, 0 !} \\
 & = \frac{n!}{\alpha_1 !\, \alpha_2 ! \, \cdots \, \alpha_d!}
 \end{split}
$$
is the number of times the facteur $x^{\boldsymbol{\alpha}}:= x_1^{\alpha_1} x_2^{\alpha_2} \cdots x_d^{\alpha_d}$ appears in the expansion of $\left( x_1 + x_2 + \cdots + x_d \right)^n$.

In our case $d=3,\ \boldsymbol{\alpha}= (p,q,r)$.

There could be other possibilities but I'm thinking of taking $(x_1,x_2,x_3)= \left(1, \frac{1}{2}, \frac{1}{3} \right)$ then I need to find a "simple" function of $\boldsymbol{\alpha}$, $f:\mathbb{N}^3 \to \mathbb{N}^3 $  which would take value $(1,0,0)$ if $p<q<r$ (i.e. if $\boldsymbol{\alpha}$ view as a function $1\mapsto \alpha_1=p,\ 2\mapsto \alpha_2=q,\ 3\mapsto \alpha_3=r$ is injective), take value $(0,1,0)$ if exactly two indices are equal (or if $\boldsymbol{\alpha}$ view as a the previous function takes only two values) and $(0,1,1)$ if $p=q=r$.

If we find such a function then the first sum can be rewritten
$$\sum_{\genfrac{}{}{0pt}{1}{p\leq q \leq r}{p+q+r =n}} \frac{n!}{p!\, q!\, r!} \mathbf{x}^{f(\boldsymbol{\alpha})}$$

The final step, if possible would be to relate $\mathbf{x}^{f(\boldsymbol{\alpha})}$ to $\mathbf{x}^{\boldsymbol{\alpha}}$ by integration or derivation or maybe something different.

答:
I leave here my first answer but there is in fact a formula for the sum, cf. the end...

by Fimpellizieri's answer, one can get rid of the $p\leq q\leq r$ condition in the summation.

In addition, the condition $p+q+r=n$ (each quantity being odd and $n\geq 3$) is equivalent to $p'+q'+r'=n'$ with $p=2\, p' +1,\ q=2\, q' +1, r=2\, r' +1$ and $n'= \frac{n-3}{2}$.

Formally (or algorithmically), one should transform each summand of the following sum as:
$$(1+1+1)^{n'}= \sum_{p'+q'+r'=n'} \frac{n'!}{p'!\, q'!\, r'!} \quad \longrightarrow \sum_{p'+q'+r'=n'}\frac{(2\,n'+3)!}{(2\,p'+1)!\, (2\,q'+1)!\, (2\,r'+1)!}= ??$$
This looks like applying a linear form (on suitable spaces of functions of $(x,y,z)$ or polynomials in $(X,Y,Z)$). Let us start with $\mathcal{A}:=\mathbb{R}[X]$ the algebra of polynomials in $X$ (one variable). The transformations I mentionned in the question e.g. evaluation at $X:=1$ or multiple derivations and then evaluation, or integration from $a$ to $b\in \mathbb{R}$ are all linear maps from $\mathbb{R}[X]$ to $\mathbb{R}$. We want to define a map (let us denote it $L_X$) which to a basis element $X^{p'}\mapsto \frac{p'!}{(2\,p'+1)} $. And in fact repeat it for $\mathcal{A}:=\mathbb{R}[X,Y,Z]$. We are thus looking for
$$ \frac{(2\, n' + 3)!}{n'!}\ L_X \circ L_Y \circ L_Z \left( (X+Y+Z)^{n'}\right)$$

Comments/Parenthesis: One may wonder how a general linear map looks like. For $\mathcal{A}:=\mathcal{C}(X),\ X$ compact, the positive linear forms are given by so-called positive radon measure. For any other $\mathcal{A}$ containing $\mathbb{R}[X]$, the question of whether a linear form is completely defined by its value on $X^p$ is known as the moment problem. For $\mathcal{A}:=\mathcal{C}^{\infty}_c(\Omega)$ with suitable topology, the continuous linear maps are given by distributions

Edit: in fact the idea of considering linear forms on spaces of functions (specifically polynomials) is known as umbral calculus!!!

Let us give an (almost non-trivial) example of a linear map from $\mathbb{R}[X]$ to itself (one can always evaluate at some point to have a map to $\mathbb{R}$): translation
$$P(X) \longmapsto P(X+a) = \left. e^{a \frac{d}{dt}} P(X+t) \right)|_{t:=0}$$
can also be defined on each basis elements by
$$ X^p \longmapsto (X+a)^p = X^p + p\, X^{p-1} a + {p \choose 2}\, X^{p-2}\, a^2 + \cdots$$
An explicit answer to the initial question would be provided by an expression of $L_X$ analogous to translation, or even in the form $e^{a \frac{d}{dt}}$ instead of an algorithm for each $X^p$.

Using another hammer tool to solve this little problem... Our map which could be written
$$ L_X: P(X) \longmapsto \sum_{k=0}^{\infty} \frac{1}{(2k+1)!} \left.\frac{d^k P(X)}{dX^k} \right|_{X:=0}$$
commutes with translation and can thus be written as a convolution product, i.e. there exists some function $F$ (or maybe some distribution?) such that
$$ L_X\Big( P(X) \Big) = \left. \int_{\mathbb{R}} F(x-u)\, P(u)\, du \right|_{x=??}$$

Let us call $S_{\text{odd}}(n,3)$ the original sum. By an analogue the recurrence relation for Stirling numbers of the second kind, one finds for $n\geq 3$ odd
$$ S_{\text{odd}}(n+2,3) = 3\, S_{\text{odd}}(n,3) + 2\big( S(n,3)+ S(n,2)  -S_{\text{odd}}(n,3) \big) + 1 =  S_{\text{odd}}(n,3) +\, 3^{n-1} $$

$$\Longrightarrow\quad S_{\text{odd}}(2 p +1,3)= \frac{3^{2p}-1}{8}$$


问:
Can we generalise the statement, "any well-defined function with domain the integers is continuous" beyond just the integers?
Let us use the following definition for continuity:

For a subset $A \subset \mathbb{R}$, $f : A \to \mathbb{R}$ is continuous if $\forall_{\epsilon > 0} \exists_{\delta >0} \forall_{x,y \in \mathbb{Z}}$,  $|x-y|< \delta \implies  |f(x) - f(y)| < \epsilon$.

The statement holds for $\mathbb{Z}$, the proof goes as follows:

we consider the function $g: \mathbb{Z} \to \mathbb{R}$: given any $\epsilon$, we can just let $\delta = \frac{1}{2}$ (any value in $(0,1)$ works). Then $|x-y|< \frac{1}{2} \implies x = y \implies |g(x) - g(y)| = |g(x) - g(x)| = 0 < \epsilon$, as required.

It is easy to see that this argument holds for any codomain set $B$ such that $h: \mathbb{Z} \to B$ is well-defined, since $h(x) - h(x) = 0$ for all $h(x) \in B$.

Hence $h$ is continuous.

I believe a similar argument also works for, say, the set $C(a) = \{a^n : n \in \mathbb{N}\}$ $\forall_{a \in \mathbb{R}\setminus\{0\}}$ (in this case, however, $\delta$ is not constant; it depends on $\epsilon$).

The common property of $\mathbb{Z}$ and $C(a)$ is that they are countable, so one may think that the argument holds for all countable sets, however this isn't the case. We cannot apply this argument to the rationals, $\mathbb{Q}$, since we cannot find a $\delta$ such that $|x-y|<\delta \implies x = y$ $\forall_{x,y \in \mathbb{Q}}$, as we could in the previous examples since given any $\delta$, there exists some $n \in \mathbb{N}$ such that $\frac{1}{10^n} < \delta$. Clearly, $\frac{1}{10^n} \in \mathbb{Q}$, so we can define $y = x + \frac{1}{10^n}$ and then $|x-y| = |x - (x + \frac{1}{10^n})| = \frac{1}{10^n} < \delta$ but $x \neq x + \frac{1}{10^n} = y$, at which points the previous argument falls apart.

So then what would be the generalisation of the statement? My intuition would be to think about functions containing only isolated points, but I am not sure. Thanks!

答: The correct generalization is to domains which are discrete, i.e. topological spaces $(X,T)$ with $T$ being the power set of $X$. This means they do indeed consist of isolated points.


问:
What is the single most influential book every mathematician should read?
If you could go back in time and tell yourself to read a specific book at the beginning of your career as a mathematician, which book would it be?</p>

答:
This question does not have a unique answer. I will concur with Jonathan in that Jayne's "Probability Theory: the logic of science" is a great book.

This book changed my life as a scientist, converting me into a fervent Bayesian. For me it was a truly irreversible experience when I, for the first time, understood and comprehended that probability (as applied to understanding the real physical world) essentially stems from our lack of knowledge, our incomplete information, of reality. Fantastic book, although I admit that Jayne's style might not suit everyone's taste.

ContentLicense: CC BY-SA 2.5


答:
$p(x)= (1-x^{10})^3 (1-x)^{-3}$.

Now use the general binomial formula $$(1-x)^{-3}= \sum_{k=1}^\infty \binom{3+k-1}{k}x^k$$

(see here e.g.) for the second

The first part is just the standard binomial $(1-\binom{3}{1}x^{10} + \binom{3}{2}x^{20} - x^{30})$.

So to get the coefficient of $x^{21}$ in the product, we get $$1 \cdot \binom{23}{21} + (-3)\cdot \binom{13}{11} + 3\binom{3}{1}$$ in total.


问:
Strong deformation retraction between CW-complexes, Brown theorem proof
I am currently reading Lemma 9.11 on Switzer, in the chapter dedicated to Brown representability theorem. However I am stuck on a few points of the proof.

Let $F$ a contravariant functor from pointed CW complexes up to homotopy to pointed sets. Let $Y$ be a pointed CW complex, $u \in F(Y)$ and $(X, A, x_0)$ a CW-pair. Let $g : (A, x_0) \to (Y, y_0)$ a cellular map. Let $T =( I^{+} \wedge A) \vee X \vee Y$, by identifying $[0,a] \in I^{+} \wedge A$ with $a \in X$ and $[1,a] \in I^{+} \wedge A$ with $g(a) \in Y$. Let $A_1 =( [0, 1/2]^{+} \wedge A) \cup X$ and $A_2 =( [1/2, 1]^{+} \wedge A) \cup Y$. Now $A_1 \cup A_2 = T, A_1 \cap A_2$ is isomorphic to $A$. Until this point it is all clear. What I don’t understand is the following:

There is a strong deformation retraction $f: A_1 \to X$ and $Y$ is a strong deformation retract of $A_2$. Thus there are $\bar{v} \in F(A_1)$ with $\bar{v}|_{X} = v$ and a $\bar{u} \in F(A_2)$ with $\bar{u}|_{Y}= u$. Why is that?

答:
Your space $T$ is nothing else than the reduced double mapping cylinder of the maps $\iota : A \hookrightarrow X$ and $g : A \to Y$. To see that, note that $I^+ \wedge A = (I \times
A)/(I \times \{a_0\})$.

The subspaces $A_1$ and $A_2$ of $T$ are copies of the reduced mapping cylinders of $\iota$ and of $g$. Thus you get the usual strong deformation retractions $f : A_1 \to X$ and $r : A_2 \to Y$. Hence the inclusions $i_1 : X \to A_1$ and $i_2 : Y \to A_2$ are homotopy equivalences and induce bijections $i_1^* : F(A_1)  \to F(X)$ and $i_2^* : F(A_2)  \to F(Y)$.

Now consider $v \in F(X)$ and $u \in F(Y)$.

Let $\bar v = (i_1^*)^{-1}(v) \in F(A_1)$. Then by definition $\bar v  \mid_X = i^*_1(\bar v) = i^*_1( (i_1^*)^{-1}(v)) = v$.

Let $\bar u = (i_2^*)^{-1}(u) \in F(A_2)$. Then $\bar u  \mid_Y = u$.


问:
Athletes compete in javelin throwing, one after the other, throwing it at independent distances that are all exponentially
Athletes compete in javelin throwing, one after the other, throwing it at independent distances that are all exponentially distributed with parameter $\lambda$. The first competitor will lead the competition until the $N$ opponent behind him throws the handlebars to a greater distance and takes the lead. Calculate $E(N)$.

What I tried: Let $X_i$ be the result of competition $i$. We know that $X_i\sim\exp(\lambda)$. I tried to use $E\left(N\right)=E\left(E\left(N|X_{1}=x\right)\right)$. But how do I represent $N$ to be related to $X_1$? As I understand I want the first player so $X_N>X_1$.

Really struggling with this question. First time I'm seeing a random variable in an index. How do I solve it?

答:
Think in the following way:

You are first and  you get a distance $x$

Conditioned to your distance, $N$ follows a geometric distribution with parameter $P(Y>x)=e^{-\lambda x}$

That is

$$(N|X=x)\sim Geo(e^{-\lambda x})$$

with mean $E(N|X=x)=e^{\lambda x}$

Using Expectation's properties you get

$$E(N)=E[E(N|X=x)]=E(e^{\lambda x})=\int_0^{\infty}e^{\lambda x} \lambda e^{-\lambda x}dx=\infty$$


问:
Maximal Ideals in the Ring of Complex Entire Functions
Let $X = \mathcal{C}([0,1],\mathbb{R})$ be the ring of all continuous real-valued functions $f:[0,1] \to \mathbb{R}$. For $x \in [0,1]$, let $M_{x} = \\{ f \in M \ | \ f(x)=0\\}$. One can show by using compactness of $[0,1]$ that every maximal ideal is of this form.

Extending the Question to Entire functions: Let $\mathsf{C}(z)$ be the ring of complex entire functions. For $ \lambda \in \mathsf{C}$ let $M_{\lambda}$ denote the set of all entire functions which have a zero at $\lambda$. Then is $M_{\lambda}$ a maximal ideal in $\mathsf{C}(z)$, and does every ideal happen to be of this form? I don't know how to prove this!

答:
No.

Let $f(z)=1/\Gamma(z)$ where $\Gamma$ is the gamma function.
Then $f$ vanishes at $0$, $-1,-2,\ldots$ and nowhere else. For integers
$m$ let $f_m(z)=f(z+m)$. Then $f_m$ vanishes at $-m,-m-1,\ldots$.
Let $I$ be the ideal of the ring of the entire functions generated by the $f_m$.
Then $I$ is proper, since otherwise finitely many of the $f_m$
would generate an ideal containing the constant function $1$. But
any finite set of $f_m$ have a common zero, so this doesn't hold.
So $I\subseteq J$ for some maximal ideal $J$ (using the standard Zorn's
lemma argument). But $J\ne M_\lambda$ for any $\lambda$, as for each $\lambda$
there is some $m$ with $f_m(\lambda)\ne0$, and $f_m\in J$.

OwnerDisplayName: anonymous


问:
Transitive groups
Someone told me the only transitive subgroup of $A_6$ that contains a 3-cycle and a 5-cycle is $A_6$ itself.

(1) What does it mean to be a "transitive subgroup?"  I know that a transitive group action is one where if you have a group $G$ and a set $X$, you can get from any element in $X$ to any other element in $X$ by multiplying it by an element in $G$.  Is a transitive subgroup just any group that acts transitively on a set?  And if so, does its transitiveness depend on the set it's acting on?

(2) Why is $A_6$ the only transitive subgroup of $A_6$ that contains a 3-cycle and a 5-cycle?

Thank you for your help :)

答:
A subgroup of the symmetric group $S_n$ is said to be "transitive" if it acts transitively on the set $\{1,2,\ldots,n\}$ via the natural action (induced by $S_n$).

So here, you would be looking at a subgroup of $A_6$ that acts transitively on $\{1,2,3,4,5,6\}$, in the natural way (i.e., via the permutation action).

Yes, the transitivity of an action depends on the set being acted on. $S_n$ acts transitively on $\{1,2,\ldots,n\}$ via its natural action; but $S_n$ also acts on $\{1,2,\ldots,n\}\times\{n+1,n+2,\ldots,2n\}$ (by acting on $\{1,2,\ldots,n\}$ via the natural action, and acting on $\{n+1,\ldots,2n\}$ by letting $\sigma$ map $n+i$ to $n+\sigma(i)$). This action is not transitive, since $1$ is never mapped to $n+1$. But in the case of groups that have "natural actions", one usually speaks of "transitive", the action being understood.

Suppose $H\lt A_6$ is transitive, and without loss of generality that it contains $(1,2,3,4,5)$ (it does, up to an automorphism of $A_6$). If the $3$-cycle in $H$ fixes $6$, then the $3$-cycle and the $5$-cycle generate the copy of $A_5$ inside of $A_6$. Conjugating by appropriate elements of $H$ you get copies of $A_5$ fixing each of $1$, $2,\ldots,6$ sitting inside of $H$, so $H$ contains all $3$-cycles, hence $H=A_6$.

If the $3$-cycle does not fix $6$, say the $3$-cycle is $(i,j,6)$; there is an element $h\in H$ that maps $6$ to $1$. If $h$ does not map $i$ nor $j$ to $6$, then conjugating $(i,j,6)$ by $h$ drops us to the previous case. If $h$ does map $i$ or $j$ to $6$, then conjugating the $3$-cycle by an appropriate power of $(1,2,3,4,5)$ gives us a $3$-cycle $(i',j',6)$ such that $h(i',j',6)h^{-1}$ fixes $6$, and we are back in the previous case again.

Title: Transitive groups


问:
Schur decomposition of an $n-$by$-n$ matrix
$(\lambda, x)$ is a simple (with multiplicity 1) eigenpair of $A\in \mathbb C_n$ with $x^Hx=1$, $H$ denotes Hermitian.

Use Schur decomposition to show that there exists a nonsingular matrix $(x\ \ X)$ such that
$$
(x\ X)^{-1}A(x\ X)=\left(\matrix{\lambda \ \ \ \ 0\\ \ 0\ \ \  M}\right)
$$
where $M\in\mathbb C_{n-1}$

答:
Hint: Suppose for some invertible $P$ we have
$$
P^{-1}AP = \pmatrix{\lambda&b^H\\ 0&M}
$$
for some $b\in\mathbb{C}^{n-1}$ and $M\in M_{n-1}(\mathbb{C})$. For any vector $u\in\mathbb{C}^{n-1}$, we have
$$
\pmatrix{1&-u^H\\ 0&I}\pmatrix{\lambda&b^H\\ 0&M}\pmatrix{1&u^H\\ 0&I}
=\pmatrix{\lambda&b^H - u^H(M-\lambda I)\\ 0&M}.
$$
Since $\lambda$ is not an eigenvalue of $M$ (why?), we can choose $u$ such that $b^H - u^H(M-\lambda I)=0$ (why?).


问:
Prove that the following set of formulas is contradictory
Prove that the following set of formulas is contradictory:
$$ \left \{ \forall x(Px \implies Qx), \forall x(Qx\implies \exists yRx,y), \exists xPx, \forall x \forall y \neg Rx,y \right \} $$

答: There exists some $x_0$ such that $Px_0$, which implies $Qx_0$, which implies $(Rx_0,y_0)$ for some $y_0$. But because of $\forall x \forall y \neg (Rx,y)$, we simultaneously have $\neg(Rx_0,y_0)$, so you have a contradiction.


问:
Find the volume of a sphere with triple integral
Consider this figure:

As shown in the figure. The center of the sphere is at $(0,0,-2)$ and the radius of the sphere is $4$. My question is how to calculate the volume of the sphere when $z\geq 0$ (upper half space). How can i apply triple integral on this problem?

My idea is shifting the sphere to above up to $2$. First i need to find the angle between the cone and $xy$ plane. That is:
$$\arcsin\left(\frac 12\right) = \frac{\pi}{6}$$
It means the complement of the angle is $\frac{\pi}{3}$ and calculating this integral:

$$\int_0^{\frac{\pi}{3}} \int_0^{2\pi} \int_0^{\frac{\pi}{6}} \rho^2 \sin(\phi) \,\Bbb d\rho\,\Bbb d\theta\, \Bbb d\phi$$

Finally, the volume of the sphere when $z\geq 0$ is that integral minus the volume of the cone. Anyway, do you have the shortest way to solve this problem using triple integral that requires one step only? (Mine is 2 steps, calculate the integral and the cone)

答: You can do it with cylindrical coordinates. Note that\begin{align}x^2+y^2+(z+2)^2\leqslant16&\iff\rho^2+z^2+4z\leqslant12\\&\iff\rho^2\leqslant12-4z-z^2.\end{align}So, compute$$\int_0^{2\pi}\int_0^2\int_0^\sqrt{12-4z-z^2}\rho\,\mathrm d\rho\,\mathrm dz\,\mathrm d\theta.$$You will get $\frac{40\pi}3$.


问:
$L^\infty$ and Sobolev spaces
I am studying Sobolev spaces, and my professors give us the following exercise, but I do not understand how is possible conclude it.

Let $1\leq p< \infty$ and $v\in C_c^\infty(\mathbb{R})$. If
$$ |v(x)|^p\leq p || v ||_p^{p-1} || v' ||_p ,$$
then there existe $C>0$, such that
$$ || v ||_\infty \leq C ||v||_{W^{1,p}}.$$

How is possible proof it ?

答:
Recall Young's inequality: for $\alpha, \beta \in [0,1]$ with $\alpha + \beta = 1$, one has $a^{\alpha}b^{\beta} \leq \alpha a + \beta b$ for any non-negative real numbers $a$ and $b$. Take in this case $\alpha = (p-1)/p$, $\beta = 1/p$, $a = ||v||^{p}_{p}$, $b = ||v'||^{p}_{p}$.

From the  condition and the above inequality, you have that:
\begin{equation}
|v(x)|^p \leq p||v||^{p \times \frac{p-1}{p}}_{p} ||v'||^{p \times \frac{1}{p}}_{p} \leq p \left( \frac{p-1}{p} ||v||_{p}^{p} + \frac{1}{p} ||v'||_{p}^{p} \right) \leq C(p)||v||_{W^{1,p}}^{p},
\end{equation}
where $C(p) = p \max\{ (p-1)/p, 1/p\}$.
By continuity and compactness, take the $p$-th root, then the supremum over all $x$ to conclude.

Title: $L^\infty$ and Sobolev spaces


问:
How many good choices are there in this card problem?

You have a deck of
n
cards with face values
1
,
…
,
n
, out with you pick
m
cards u.a.r. without replacement. You want to know what is the probability that there are at least
m −
1
cards whose face values are consecutive among the
m
that you picked. We have $4\leq n \leq 20$ and $4\leq m \leq n$

I solved this problem a while ago, but I can't remember why my solution works. So I thought we count the good choices and divide by the number of total possible choices. The total possible choices are $n \choose m$, this makes sense. However for the good choices back then I found that there are $(n-m+1)^2$ many. Why? I can't remember why this makes sense.

答:
There are $n-m+2$ ways to pick a run of length $m-1$. For each of these we can pick the remaining card in $n-m+1$ ways. This gives a total of $$(n-m+2)(n-m+1).$$

However each of the $n-m+1$ cases where we have a run of $m$ cards has been counted twice in this process so we must subtract $n-m+1$ giving
$$(n-m+1)^2.$$