问:
$L^p(\mu)$ space relations
Let $(X,M, \mu)$ be a measure space. Suppose $1 \leq p < r \leq \infty$.

Suppose $f \in L^p (\mu) \cap L^r(\mu)$. Show that $f\in L^s(\mu)$ for $p \leq s \leq r$.

Attempt:

Suppose $f \in L^p (\mu) \cap L^r(\mu)$. Then $\int |f| ^p d\mu < \infty$ and $\int |f|^r d\mu < \infty$.

Isn't it obvious that $\int |f|^s d\mu < \infty$ already?

I'm bit confused on what to show.

Any help will be appreciated!

Thank you for your help!

答:
This is Proposition 6.10 in Folland: If $0< p <q<r \leq \infty$, then $L^p \cap L^r \subset L^s$ and $\|f\|_s \leq \| f\|_p^{\lambda}\|f \|_r^{1- \lambda}$, where $\lambda \in (0,1)$ is defined by $\lambda = \frac{s^{-1} -r^{-1}}{p^{-1} - r^{-1}}$ when $r \ne \infty$, and $\lambda = p/s$ when $r \ne \infty$.

When $r= \infty$, you use the definition of $\| f\|_{\infty}$ to show the inequality.

When $r \ne \infty$, it's just Holder's inequality.

Title: $L^p(\mu)$ space relations


问:
Continuous functions on locally compact Hausdorff spaces
Show every closed Ideal in $C_0(X)$ for a locally compact Hausdorff space X is of the form $K_F=\{f\in C_0(X)\colon f\lvert_F \equiv 0\}$ for some closed subspace $F\subseteq X$.

Here is what I have so far:

Let $I$ be a closed ideal.

I tried to use the Gelfand Transform.

Consider the evaluation map $ev_x: C_0(X) \rightarrow \mathbb{C}$ defined by $ev_x(f)= f(x)$ for all $x \in F$. Then we have $Ker (ev_x)= K_F$

From here, I'm pretty much lost what to do to get the result desired.

Also, I thought about using Urysohn's Lemma here: https://www.math.ksu.edu/~nagy/real-an/1-05-top-loc-comp.pdf . But still didn't anywhere.

Any help will be appreciated.

答:
Given a closed ideal $I\trianglelefteq  C_0(X)$, let
$$
  F=\{x\in X: f(x)=0,  \text{ for all } f\in I\},
  $$
and let us prove that $I=K_F$.

The inclusion  $I\subseteq K_F$ holds for a pretty elementary reason (which nevertheless sounds a bit like a tongue-twister):
every function in $I$ vanishes on any point where all functions in $I$ vanish.

The hard part is to prove the reverse inclusion $K_F\subseteq I$.   For this  pick  $f\in  K_f$.  Fixing  $\varepsilon >0$, recall that the set
$$
  C=\{x\in  X: |f(x)|\geq  \varepsilon \}
  $$
is compact.

Given  $x$ in $C$, we have that $f(x)$ is nonzero, so clearly $x$ is not in $F$.  Consequently not all functions in $I$
vanish at $x$, and hence we may choose  some element $a_x\in  I$, such that $a_x(x)\neq 0$.  The set
$$
  U_x=\{y\in X: a_x(y)\neq 0\}
  $$
is then an open neighborhood of $x$, and hence $\{U_x\}_{x\in C}$ is an open cover of $C$.  By compactness we may pick a
finite subcover $\{U_{x_1},U_{x_2},\ldots ,U_{x_n}\}$.  Next observe that
$$
  g:= \sum_{i=1}^n\overline {a}_{x_i}a_{x_i}
  $$
lies in $I$, and does not vanish on any point of $C$.

There are multiple endings for this story,  and perpahs the slickest way is as follows:  besides an ideal,  $I$ is a
C*-algebra and $g$ is a non-negative element in $I$, so $g^{1/p}$ lies in $I$  for every positive integer $p$.
Moreover the sequence $\{g^{1/p}\}_{p\in {\mathbb N}}$ converges uniformly to 1 on $C$.  It is also easy to see that
$$
  \|g^{1/p}\| =   \|g\|^{1/p} \leq  2,
  $$
for large enough $p$.
Therefore we may find  $p$ such that
$$
  |f(x)-g(x)^{1/p}f(x)| < \varepsilon , \quad\forall x\in  C,
  $$
and
$$
  |f(x)-g(x)^{1/p}f(x)|  \leq  \big (1+|g(x)^{1/p}|\big ) |f(x)|\leq  (1+2)\varepsilon ,
  $$
for every  $x\in  X\setminus C$.
Consequently $\|f - g^{1/p}f\|\leq  3\varepsilon $, so we deduce that $f$ lies in the closure of $I$, and hence also in $I$
because $I$ is closed.

EDIT. Here is another approach,  assuming we already know the answer in case $X$ is compact.

Assuming that $X$ is not compact, let $\hat X=X\cup \{\infty \}$
be the one-point compactification of $X$, and let
$$
  f\in C_0(X)\mapsto \hat f\in  C(\hat X)
  $$
be the natural identification.  One may then prove that $\hat I$ is a closed ideal in $C(\hat X)$, and hence, by the
compact case,
$$
  \hat I=\{g\in C(\hat X) : g|_G=0\},
  $$
where $G$ is a closed subset of $\hat X$.

Notice that necessarily $\infty \in G$, since otherwise one could use Urysohn to
produce some $g\in C(\hat X)$, vanishing on $G$, with $g(\infty )\neq 0$, but this would imply that $g$ lies in
$\hat I\setminus C_0(X)$, which is impossible.

Setting $F=G\cap X$, we have  that $F$ is a closed subset of $X$.   So, upon identifying $C_0(X)$ as the subalgebra of $C(\hat
X)$ formed by the functions vanishing at $\infty $, we have:
$$
  \hat I=\{g\in C(\hat X) : g|_G=0\} = $$$$ =
  \{g\in C(\hat X) : g|_F=0\} \cap  \{g\in C(\hat X) : g(\infty )=0\} = $$$$ =
  \{g\in C(\hat X) : g|_F=0\} \cap  C_0(X) = $$$$ =
  \{g\in C_0(X) : g|_F=0\} = K_F.
  $$

ContentLicense: CC BY-SA 4.0


问:
topology generated by a metric
What is the procedure to prove that a particular given metric generates a particular given topology?

Actually, I am given with a topology $\rho$ and a distance expression $d$ in my homework. Question is

Show that the topology $\rho$ is metrizable by showing that the distance $d$ defines a metric on $\mathcal T$ that generates the topology $\rho$.

I proved that the given expression $d$ is metric on space. But how to proceed next? How to show that the metric generates the topology?

答:
You have to show two things: every $B_\rho(x,r)$ is open in $\mathcal{T}$ (this shows $\mathcal{T}_\rho \subseteq \mathcal{T}$) and vice versa, for every $O \in \mathcal{T}$ and every $x \in O$ there is some $r>0$ such that $B_\rho(x,r) \subseteq O$, which shows that $\mathcal{T} \subseteq \mathcal{T}_\rho$.

An example where I do this is here e.g.

Title: topology generated by a metric


问:
Row and column operations and matrix similarity
Take for example the following matrix:
$$
A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}
$$

The elementary matrix equivalent to changing the first row with the second is
$$
E = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}
$$
multiplied from the left. The elementary matrix equivalent to changing the first column with the second is the same matrix $E$ multiplied from the right.

After a quick check, I found that $E = E^{-1}$.

Given that, I concluded that:
$$
B = \begin{pmatrix} 5 & 4 & 6 \\ 2 & 1 & 3 \\ 8 & 7 &9\end{pmatrix} = EAE^{-1}
$$

and therefore $A\sim B$.

Is this comprehensive? Does changing rows and column necessarily make the outcome similar to the original?

Thanks

答:
The answer is NO. A counterexample is:

$\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}$.

The answer is positive iff the permutation matrix $E$ satisfy $E=E^{-1}$ or $E^2=I$. So for the matrices of order other than two, the similarity condition ceases to hold.


问:
If $\phi$ is in $\mathcal S$ and $\phi(0)=0$, is $\int_0^1 \nabla \phi(tx)dt$ in $\mathcal S$ too?
Let $\phi\in\mathcal S(\mathbb R^n)$ be a Schwartz function vanishing at $0$. Then the FTC implies
$$ \phi(x) = x\cdot \int_0^1 \nabla \phi (tx)\ dt.$$
Let this integral term be $\psi(x) := \int_0^1 \nabla \phi(tx)dt$. Is it true that  $\psi$ is (component-wise) a Schwartz function?

It is at least $C^\infty$. I tried for a while to come up with counterexamples and failed. I have proved it in dimension $1$, but I need a replacement for division in higher dimensions?

The result is false for arbitrary functions $p$ that solve $\phi(x)= x\cdot p(x)$, even if $\phi$ is Schwartz ($0 = \binom x y \cdot \binom {-y}x$). Also, $\psi$ is not Schwartz if the condition $\phi(0)=0$ is dropped, as then (in 1D) $\psi(x)= \frac{\phi(x)-\phi(0)}x\sim\frac{\phi(0)}x$ for large $x$ which is not enough decay. So a positive proof would need to use something about Schwartz functions and something about vanishing at zero (the two assumptions).

I can provide a detailed attempt if wanted; in fact I have deleted many "answers" after noticing errors. In particular I have tried merely using the Schwartz seminorms to bound the integral, but it seems that either many factors of $t^{-1}$ appear which make the integral diverge, or (trying to treat $t\ll 1$ separately) I am unable to achieve large decay. Of course, as previously mentioned, the issue at $t=0$ should be somehow resolved with the vanishing at $x=0$.

This comes out of trying to understand what the correct fix for a typo in M Zworski's Semiclassical Analysis p.32 (Screenshot,Google books link), in a proof that the Fourier Transform is an automorphism of $\mathcal S$. The online errata (screenshot copy) says to set $\phi(0)=0$, but I'm unsure if the preceeding paragraph is correct with this correction (and in any case, it is stated that $\psi$ may not be in $\mathcal S$ without proof.)

Thoughts appreciated!

 Zworski, Maciej, Semiclassical analysis, Graduate Studies in Mathematics 138. Providence, RI: American Mathematical Society (AMS) (ISBN 978-0-8218-8320-4/hbk). xii, 431 p. (2012). ZBL1252.58001.

答:
Now I have come to the conclusion that it is not true:

Take $\phi(x,y) = -y e^{-(x^2+y^2)/2},$ which is in $\mathcal{S}(\mathbb{R}^2)$ and satisfies $\phi(0,0)=0.$ Then $\partial_x\phi(x,y) = xy e^{-(x^2+y^2)/2}$ so
the $x$-component of $\psi$ is
$$
\psi_x(x,y)
= \int_0^1 t^2xy e^{-t^2(x^2+y^2)/2} \, dt
= \{ s:= t\sqrt{x^2+y^2} \}
\\
= xy\int_0^{\sqrt{x^2+y^2}} \frac{s^2}{x^2+y^2} e^{-s^2/2} \, \frac{ds}{\sqrt{x^2+y^2}}
\\
= \frac{xy}{(x^2+y^2)^{3/2}}\int_0^{\sqrt{x^2+y^2}} s^2 e^{-s^2/2} \, ds
.
$$
Now study $\psi_x(x,y)$ along $(x,y)=(r,r),$ i.e.
$$
\psi_x(r,r)
= \frac{2^{-3/2}}{r} \int_0^{\sqrt{2}r} s^2 e^{-s^2/2} \, ds
.
$$
As $r\to\infty$ this will have the asymptotic behavior
$$
\psi_x(r,r)
\sim \frac{2^{-3/2}}{r} \int_0^{\infty} s^2 e^{-s^2/2} \, ds
= \frac{2^{-3/2}}{r} \sqrt{\frac{\pi}{2}}
$$
which does not tend to $0$ fast enough,
so $\psi_x \not\in \mathcal{S}(\mathbb{R}^2).$


问:
Homeomorphism from the whole plane to the open upper half plane.

Find a homeomorphism from the whole plane to the open upper half plane.

So I need to find a bicontinuous map $f:\mathbb{R^2} \to \mathbb{R} \times[0, \infty)$. If I consider $f(x,y)=(x,e^y)$ then certainly $f$ is continuous since this is a $C^\infty$ function? Both component functions also have a continuous inverse, but neither one is defined at $(x,y)=(0,0)$? Is this a valid candidate or am I just mistaken here?

答:
There is no homeomorphism from $\Bbb R^2$ onto $\Bbb R \times [0,+\infty)$. But there is one from $\Bbb R^2$ onto $\Bbb R \times (0,+\infty)$, the latter is the open half plane (it's an open subset of $\Bbb R^2$, while the former half-closed half-plane is not).

And for that latter $f(x,y)= (x, e^y)$ is a perfectly fine homeomorphism, with inverse $g(x,y)=(x, \ln y)$.


问:
Give an $\delta-\epsilon$ proof that the function $f:\mathbb{R}^3\to\mathbb{R}$ defined by $f(x,y,z)=x^2y+2xz^2$ is continuous at $(1,1,1)$

Give an $\delta-\epsilon$ proof that the function $f:\mathbb{R}^3\to\mathbb{R}$ definded by $$f(x,y,z)=x^2y+2xz^2$$ is continuous at $(1,1,1)$.

Let $\epsilon<0$ then,

$$
\begin{equation}
\begin{split}
|x^2y+2xz^2-3| & = |(x^2y-y)+(y-1)+(2xz^2-2x)+(2x-2)| \\
 & \leq |x^2y-y|+|y-1|+|2xz^2-2x|+|2x-2| \\
& = |y||x+1||x-1|+|y-1|+2|x||z+1||z-1|+2|x-1|
\end{split}
\end{equation}
$$

For single variable I know how to handle $|x+1||x-1|$ $(\text{Using }|x+1||x-1|<\epsilon \implies\delta=\min\{1,\frac\epsilon3\})$, but for multivariable how to avoid any dependencies between variables?

It will be great help If anyone provide a legit way/hint/idea to deal this kind of problem(more general will be appreciated).

答:
Alternative approach

Give an $\delta-\epsilon$ proof that the function
$f:\mathbb{R}^3\rightarrow\mathbb{R}$ definded by
$f(x,y,z)=x^2y+2xz^2$ is continuous at $(1,1,1)$.

Assume that $\delta > 0.$

If $|(x,y,z) - (1,1,1)| < \delta$, then you can assume that

$-\delta < x-1 < \delta$

$-\delta < y-1 < \delta$

$-\delta < z-1 < \delta$.

This means that you can assume that

$1 - \delta < x < 1 + \delta$

$1 - \delta < y < 1 + \delta$

$1 - \delta < z < 1 + \delta$.

Therefore, for $x,y,z$ that satisfy these constraints,

the minimum value of $x^2y$ will be $> (1 - \delta)^3$ and

the maximum value of $x^2y$ will be $< (1 + \delta)^3$

Therefore, the minimum and maximum values for $f(x)$
will be $3(1 - \delta)^3$ and $3(1 + \delta)^3.$

Therefore, you need to set $\delta$ in terms of $\epsilon$

so that $[3(1 - \delta)^3 - f(1,1,1)] > -\epsilon$ and
$[3(1 + \delta)^3 - f(1,1,1)] < \epsilon$.

For convenience, I will arbitrarily insist that
$\delta$ will be $\leq (1/10).$

This allows me to assume that $(1 - \delta)^3 > (1 - 4\delta)$

and that $(1 + \delta)^3 < (1 + 4\delta)$.

So, the problem has been reduced to specifying
$\delta$ in terms of $\epsilon$

so that $[3(1 - 4\delta) - 3] > - \epsilon$
and $[3(1 + 4\delta) - 3] < \epsilon$.

These constraints are satisfied when $12\delta < \epsilon$.

So, take $\delta = \min[(\epsilon/15), (1/10)].$


问:
Function that converges in $L^\infty$ but not in $L^p$
I know that convergence in $L^\infty$ implies convergence in $L^p$ in finite measure spaces since if it is finite then Holder's inequality will hold and this can be proven.

I am trying to think of a function that converges in $L^\infty$ but does not converge in $L^p$ for some $p$ or maybe even for all $p$?

答:
Find a sequence of sets $A_n$ with measure going to infinity fast enough so that the functions
$$
f_n = \frac{1}{n}\;\mathbf1_{A_n}
$$
satisfy $\|f_n\|_\infty \to 0$ but $\|f_n\|_p \to \infty$ for all $p$.

added

Let $A_n$ have measure $2^n$.  Then
$$
\|f\|_\infty = \frac{1}{n} \to 0,
\\
\|f\|_p = \left(\int_{A_n} \frac{1}{n^p}\right)^{1/p} = \frac{2^{n/p}}{n}
\to \infty .
$$


答: Take $f_n=\frac{1}{n}1_{[0,n]}$. It converges in $L^{\infty}$ but not in $L^1$.

Title: Function that converges in $L^\infty$ but not in $L^p$

Body:
I know that convergence in $L^\infty$ implies convergence in $L^p$ in finite measure spaces since if it is finite then Holder's inequality will hold and this can be proven.

I am trying to think of a function that converges in $L^\infty$ but does not converge in $L^p$ for some $p$ or maybe even for all $p$?


问:
Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent.
Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent.

Attempt:
Notice that the $p-$ series $\sum_{n=1}^\infty \frac{1}{n^2}$ with $p=2$ is convergent. Then, we have
\begin{align*}
\lim\limits_{n \to \infty} \frac{\ln\left(\cos \frac{1}{n} \right)}{\frac{1}{n^2}} &= -\frac{1}{2} \cdot \lim\limits_{n \to \infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}} \cdot \frac{1}{\cos \frac{1}{n}} \\
&= -\frac{1}{2}.
\end{align*}
Hence, the limit is equal to $-\frac{1}{2} \ne 0$. Since the $p-$ series is convergent with $p=2$, by the limit comparison test, the series is convergent.

Am I true?

Body:
Yes, your approach is valid.

More Fun with Numeric Approximations

Integrating the negative of the series for $\tan(x)$, derived in $(5)$ of this answer
$$
\tan(x)=\sum_{k=1}^\infty\frac{(-1)^{k-1}B_{2k}\left(4^k-1\right)4^kx^{2k-1}}{(2k)!}\tag1
$$
where $B_{2k}$ are the Bernoulli Numbers, we get the Taylor series
$$
\begin{align}
\log(\cos(x))
&=\sum_{k=1}^\infty\frac{(-1)^kB_{2k}\left(4^k-1\right)4^kx^{2k}}{2k(2k)!}\tag{2a}\\
&=-\frac{x^2}2-\frac{x^4}{12}-\frac{x^6}{45}-\frac{17x^8}{2520}-\frac{31x^{10}}{14175}-\frac{691x^{12}}{935550}+O\!\left(x^{14}\right)\tag{2b}
\end{align}
$$
Applying the Euler-Maclaurin Sum Formula, we get that
$$
\begin{align}
\sum_{k=1}^n\log\left(\cos\left(\frac1k\right)\right)
&=C+\frac1{2n}-\frac1{4n^2}+\frac1{9n^3}-\frac1{24n^4}+\frac7{450n^5}-\frac1{90n^6}\\
&+\frac{89}{8820n^7}-\frac{17}{5040n^8}-\frac{482}{127575n^9}-\frac{31}{28350 n^{10}}\\[6pt]
&+\frac{46577}{3430350 n^{11}}-\frac{691}{1871100 n^{12}}+O\left(\frac1{n^{13}}\right)\tag3
\end{align}
$$
The error term is bounded by $\frac1{20n^{13}}$, so for a given $n$, we should get about $p=13\log_{10}(n)+1$ places of precision:
$$
\begin{array}{r|c|l}
n&p&C\\\hline
10&14&-0.94536905472633\color{#AAA}{705334}\\
100&27&-0.945369054726332935266095215\color{#AAA}{8385884}\\
1000&40&-0.9453690547263329352660952154082701981170\color{#AAA}{39142468}
\end{array}
$$


问:
Find all positive $x,y\in\Bbb{Q}$ such that $x+\frac{1}{y}$ and $y+\frac{1}{x}$ are natural numbers.
What I did is, the product is also a natural number.

$xy+\frac{1}{xy}=m$ is also a natural number. Assuming $xy=z$, we get the equation $z^2-mz+1=0$

Hence, $z=\frac{m\pm \sqrt{m^2-4}}{2}$. Since z is rational $\sqrt{m^2-4}$ has to be rational. It is rational only when $m=2$ (not sure). So $z=xy=1$. So, $y=\frac{1}{x}$

Hence $2x$ and $2y$ are natural numbers.

I can't proceed further. Is my way correct? Any help please.

答:
If $x,y$ are rational $x = \frac pq$ and $y = \frac ab$ (assume in lowest terms) so

$\frac pq + \frac ba = \frac {pa + qb}{aq}\in \mathbb Z$

And $\frac ab +\frac qp = \frac {pa+qb}{pb}\in \mathbb Z$

So $q|pa+qb$ so $q|pa$ but $q$ and $p$ are relatively prime so $q|a$.  Likewise $a|pa + qb$ so $a|qb$ so $a|q$ so $a= q$.  And similarly $p = b$.

So $x = \frac 1y$.

So we must have $x + \frac 1y = x + x = 2x$ and $y + \frac 1x = 2y$ are natural numbers.

Whoo boy.

Okay. Case one $x = 1 \in \mathbb Z$ then $y = \frac 11 = 1$ and $x = y = 1$ are obviously solutions.

Case 2: $x \in \mathbb Z$ but $x > 1$.  Then $y=\frac 1x \not \in \mathbb Z$. But $2y=\frac 2x\in \mathbb N$ so $x|2$ so $x = 2$ and $y= \frac 12$ is a solutions ($2+ \frac 1{\frac 12} = 4$ and $\frac 12 + \frac 12 = 1$).

Case 3: $x\not \in \mathbb Z$ then $2x \in \mathbb Z$ so $x = \frac m2$ for some odd $m$.  But then $y =\frac 2m$ and $2y=\frac 4m$ is an integer so $m|4$.  But $m$ is odd so $m =1$.

So those are the only three possibilities.

$x = y = 1;  x = 2; y =\frac 12$ and $x = \frac 12$ and $y = 2$.


问:
Simplifying $a(a-2) = b(b+2)$
I reduced a number theory problem to finding all ordered pairs $(a,b)$ that satisfy the equation $a(a-2) = b(b+2)$ in a certain range. After thinking about this for a while, I figured that either $a = b + 2$, $a = -b$, $a = b = 0$ or $a = 2 \text{ and } b = -2$. It is easy to prove that these values for $a$ and $b$ will all satisfy the equation, but how would I solve this equation had I not come up with these solutions? And how do I know I did not miss one? In short, can this equation be solved (more) rigorously?

Edit:
Silly of me I did not recognize that my last two solutions were already covered by my first two.

答:
Another approach is to note that $a=b+2$ is an "obvious" solution, then to divide by that solution:
$$
\frac{a(a-2)-b(b+2)}{a-(b+2)}=a+b
$$
giving $a(a-2)-b(b+2)=(a-(b+2))(a+b)$. Therefore, the solutions are
$$
a=b+2\quad\text{and}\quad a=-b
$$</p>

ContentLicense: CC BY-SA 3.0

Title: Simplifying $a(a-2) = b(b+2)$


问:
Unique solution to normal equation proof
So I'm trying to prove the following:

Show that a unique solution for linear regression exists i the features are not linearly dependent. Namely,show that a unique solution $argmin_{\theta} (X^{T}X)^{-1}X^{T}y$
exists iff $X$ has full column rank.

What I did was:

Theorem: For every matrix $A\in R^{nxm}$ we have $Null(A) = Null(A^{T}A)$ and $Rank(A) = Rank(A^{T}A)$

then I said:
if $(X^{T}X)^{-1}X^{T}$ is a solution $\rightarrow$ $(X^{T}X)^{-1}$ is invertible $\rightarrow$
$Null(A^{T}A)=\{0\}$ $\rightarrow$ $Null(A) = \{0\}$ $\rightarrow$ $A$ is full rank

But I'm not sure this is the right way, also the opposite direction seems identical(reversed) to the one I wrote.

Would appreciate some help

Edit

So I derived the following expression for the normal equation through the derivation of the MSE loss:

$(X^{T}X)\theta = X^{T}y$

So $\theta$ is the parameter I used to derive with, and $X,y$ are the "training set"

So I want to prove now that the solution for $\theta$ that minimize the loss is unique given $X$ is full rank.

答:
You have the right basic ideas, but the way you write things doesn't quite make sense. I would argue that the following statements are all equivalent:

There is a unique solution to $\text{argmin}_\theta\|X\theta - y\|$,

There is a unique solution to $X^TX \theta = X^Ty$,

$X^TX$ is invertible,

$X^TX$ has full rank,

$X$ has full column rank.

Indeed, these steps make sense in both the forward and backward directions.


问:
Example for simple shooting method
Intro

Simple shooting methods for boundary value problem of the form:
$$
x'(t)=f(x(t),t), \quad r(x(a),x(b))=0
$$
with $f:\Omega \times I \rightarrow\mathbb{R}^m, \quad r: \mathbb{R}^m \times \mathbb{R}^m \rightarrow \mathbb{R}^m, \quad I=[a,b]$

try to find a starting vector $s$ so that the solution for
$$
x'(t)=f(x(t),t), \quad x(a)=s
$$
provides a solution for $x(t)$ with $r(x(a),x(b))=0 $. The first step for that is that if $x(t;s)$ is the solution for $x(t)$ of  $x'(t)=f(x(t),t), \quad x(a)=s$ with start value $x(a)=s$, we have to find the roots of the function
$$
F(s):= r(s;x(b;s))
$$

Task

I have given the system
$$
x''(t)=\gamma x, \quad x(0)=1, \quad x(1)=2
$$
and need to formulate $F(s)$.

My try

I am absolutely clueless. The only thing I would say ist that $a=0, b=1$ but from there on I don't know what to do. Do I really need to simply put down
$$
F(s):= r(s;x(b;s))= r(s;x(1;s))
$$
I cannot imagine that it would be this simple. I don't know how to go further, only using Wikipedia I would assume that
$$
F(s)=x(1;s)-2
$$
would be the solution but I don't really know if this is right or why. I would appreciate your help!

答:
In the full formalism as given you have $s=(s_1,s_2)=(x(0),x'(0))$ and $F(s)=(x(0;s)-1,\,x(1;s)-2)$.

Of course you can reduce that boundary function to a scalar function by always setting $s_1=1$. But this is a less general formulation.

Title: Example for simple shooting method


问:
unit of closure of involution subalgebra
Let $\mathbb{H}$ be a Hilbert space, $A$ is a involution subalgebra of $B(\mathbb{H})$, let $K=[A(\mathbb{H})]\subset \mathbb{H}$, where $[A(\mathbb{H})]$is the closure of set $\{a\xi: a\in A, \xi \in \mathbb{H}\}$, and if $e$ is the orthogonal projection operator from $\mathbb{H}$ to $K$. A fact is that $\forall a\in A, ae=ea=a $. Now let $\overline{A}$ be closure of $A$ with respect to anyone of six topologies(wo, so, *-so, $\sigma$-wo, $\sigma$-so, $\sigma$-*-so), my question is $\forall a\in \overline{A}$, why $ea=ae=a$?

This problem comes from proof of von Neumann double commutant theorem(general case, i.e. subalgebra is degenerate), my attempt is as follows:

because $\overline{A}^{wo}=\overline{A}^{so}=\overline{A}^{*-so}$ and $\overline{A}^{\sigma-wo}=\overline{A}^{\sigma-so}=\overline{A}^{\sigma-\ *-so}$, so we just check for $so$ and $\sigma-so$.  Firstly for $so$ topology, let $a=\lim_i a_i\in \overline{A}^{so}$, we check $ea=a=\lim_i a_i=\lim_i ea_i$, we only need to check $\lim_ie(a-a_i)=0$.

we know that $a=\lim_i a_i$ mean that  $\forall \{\xi\}_k\subset\mathbb{H}$,
$$\sum_k \|(a_i-a)\xi_k\|^2 \rightarrow 0$$
then
$$\sum_k \|e(a_i- a)\xi_k\|^2 \le \sum_k \|e\|^2\|(a_i- a)\xi_k\|^2\le \sum_k \|(a_i-a)\xi_k\|^2 \rightarrow 0$$
It means that  $\lim_ie(a-a_i)=0$.For $\sigma- so$ topology, the proof is similar.

I want to know my proof is right or not? In the proof of von Neumann double commutant theorem, the result is just ONE sentence, i doubt my idea is too complex！

Thanks in advance.

答:
The argument you wrote is the one for the $\sigma$-sot, where you need $\xi_k$ to be square summable. For the sot, you would do $\|e(a_i-a)\xi\|\to0$. And it is fine.

A slightly different way to do it is to note that if $a_j\to a$ in any of your topologies, then $a_j\to a$ in the wot. So
$$
\langle (ea-a)\xi,\eta\rangle=\lim_j\langle (ea-a_j)\xi,\eta\rangle=\lim_j\langle (ea-ea_j)\xi,\eta\rangle=\lim_j\langle(a-a_j)\xi,e\eta\rangle=0.
$$
As $\xi,\eta$ are arbitrary, $ea-a=0$.


问:
Prove that $\|(\delta,1)\|\geq \|(0,1)\|$ for all $\delta\in\mathbb{R}$ where $\|\cdot\|$ is a norm on $\mathbb{R}^2$
Let $\|\cdot\|$ be a norm on $\mathbb{R}^2$.

Prove that
$$\forall\delta\in\mathbb{R},\; \|(\delta,1)\|\geq \|(0,1)\|.$$
Although, intuitively, it seems true, I can't find a way to prove it (it's an step for a bigger proof and I'm no certain the statement is actually true).

I've tried arriving at it through the triangle inequality, and other inequalities of the sort but ran out of ideas. Pleas help providing a proof or a good hint. Thanks in advance.

答: Here is the unit ball of a norm...

Body: Here is the unit ball of a norm...


问:
Why does the limit defining the transition rate for a Markov chain exist?

Background

Let $S$ be a countable (finite or infinite) state space. The text I am reading defines a Markov process as a stochastic process $\{X(t)\}_{t\in\mathbb R_{\ge0}}$ with the Markov property: For any times $t_1<\cdots<t_n<t_{n+1}$ and any $j_1,\cdots,j_{n+1}\in S$ such that $P(X(t_1)=j_1,\cdots,X(t_n)=j_n)>0$,

$\begin{align}
P(X(t_{n+1})=j_{n+1}|X(t_1)=j_1,\cdots,X(t_n)=j_n)=P(X(t_{n+1})=j_{n+1}|X(t_n)=j_n).
\end{align}$

The process is time homogeneous if $P(X(t+\tau)=k|X(t)=j)$ does not depend on $t$ and is irreducible if every state can be reached from any other state.

Transition Rate

Let $X$ an irreducible, time homogeneous Markov process which remains in each state for a positive length of time and cannot pass through an infinite number of states in finite time. The transition rate from state $j$ to $k$ is defined as

$\begin{align}
q(j,k)=\lim_{\tau\to0}\frac{P(X(t+\tau)=k|X(t)=j)}\tau
\end{align}$

for $j\ne k$ and $q(j,j)=0$.

The things that bug me about this definition are the two-sided limit (instead of $\tau\to0^+$), plus I don't see why the limit needs to exist.

My understanding is that the $q(j,k)$, when suitably normalized, should give the probability of transitioning from $j$ to $k$ the next time the process jumps. If that's right then I'm pretty confident that the limit $\tau\to0^-$, when normalized, should be the probability of having come from $k$ the last time the process jumped, given that it's currently at $j$. Which wouldn't be equal to the right-hand limit, in general. So I'm fairly confident that $\tau\to0^+$ is meant here.

More concerning is the existence of the limit. I assume it does exist as otherwise not mentioning that possibility would be a huge omission, but I don't yet see how to prove this.

Questions

Why must the limit defining $q(j,k)$ exist?

And (less importantly):

Am I right in thinking that the limit should be taken from the right only?

What I Tried

I Googled for information on transition rates, but it looks to me like most places define Markov processes in terms of the transition rates and prove that they satisfy the Markov property, instead of defining them in terms of the Markov property and proving that the transition rates are well-defined. So that wasn't much help.

Also, a search of this site ("transition rate" [markov-chains]) didn't turn up anything that looked relevant in the first couple of pages of results.

答: Today I stumbled across a text by Ward Whitt about continuous time Markov chains. On pages 7 and 8 he addresses this question. He doesn't prove that the limit exists (for him it is an assumption), but he does provide a citation to sections II.1 and II.2 of Kai Lai Chung's 1967 book Markov Chains with Stationary Transition Probabilities. I managed to find a copy of the first edition of Chung's book, and the existence of the limit is Theorem 5 in section II.2 there. I won't reproduce the proof here because it builds on some 13 pages of general theory of transition matrices, but Chung's book is easy enough to find online for those interested.


问:
Prove $\exists x_m + x_n + x_p \ge 100$

Let $x_i \in \mathbb {R^+}$ (with $i = \overline{1,2,3,\ldots,100}$)
  such that: $$x_1^2+x_2^2+x_3^2+\ldots+x_{100}^2 > 10^4 \land x_1+x_2+x_3 + \ldots+x_{100}<300$$

Prove that there exists $x_m+x_n+x_p \ge 100$

(with $m,n,q \in \{1,2,3,\ldots,100\} \land (m-n)(m-p)(n-p) \ne 0\ $)

I see this problem on a math forum. It is nice but I don't know where to start, I can't solve it.

Title: Prove $\exists x_m + x_n + x_p \ge 100$

Body:
Suppose $x_1\ge x_2 \ge x_3\ge \dots \ge x_{100}$, and $x_1+x_2+x_3=s < 100$. Then
$$
x_1^2+x_2^2+x_3^2+\dots +x_{100}^2\le x_1^2+x_2^2+x_3^2+\lfloor\frac{300-s}{x_3}\rfloor x_3^2+(300-\lfloor\frac{300-s}{x_3}\rfloor x_3)^2
$$
since it attains the maximal at $(x_1, x_2, x_3, \dots, x_3,x_k,0,\dots)$, and
$$
x_1^2+x_2^2+x_3^2+\lfloor\frac{300-s}{x_3}\rfloor x_3^2+(300-\lfloor\frac{300-s}{x_3}\rfloor x_3)^2 \\
\le (s-2x_3)^2+2x_3^2+ \lfloor\frac{300-s}{x_3}\rfloor x_3^2+(300-\lfloor\frac{300-s}{x_3}\rfloor x_3)^2,
$$
so the problem becomes

If $a\ge b\ge c \ge 0, n\le 98$, $a+2b<100$ and $a+nb+c<300$, then $a^2+n b^2+c^2<100^2$.


答: You can compose the Cayley transform, $\dfrac {z-i}{z+i}$, with your favorite homeomorphism between the disk and the plane.

Title: Homeomorphism from the whole plane to the open upper half plane.

Body:
Find a homeomorphism from the whole plane to the open upper half plane.

So I need to find a bicontinuous map $f:\mathbb{R^2} \to \mathbb{R} \times[0, \infty)$. If I consider $f(x,y)=(x,e^y)$ then certainly $f$ is continuous since this is a $C^\infty$ function? Both component functions also have a continuous inverse, but neither one is defined at $(x,y)=(0,0)$? Is this a valid candidate or am I just mistaken here?


问:
If $f_n\to f$ pointwise on $\mathbb{R}$ and each $f_n$ is bounded, then $f$ is bounded (counter-example) + question on uniform convergence proof

Let $f_n$ be a sequence of functions defined on $\mathbb{R}$ sucht that $f_n\to f$ pointwise on $\mathbb{R}$ as $n\to\infty$ and each $f_n$ is bounded. Is $f$ necessarily bounded? Prove of give a counter-example. I would like to know if the following example holds as a counter-example. In the end I have also a question on it's non-uniform convergence proof.

Suppose $f_n(x)=x$ if $x<n$ and $f_n(x)=n$ if $x\ge n$. $f_n$ converges pointwise to $f(x)=x$
. Moreover, each $f_n$ is bounded (by $n$) and $f(x)$ is unbounded.

Now I would like to know if my proof on not uniform convergence of $f_n$ is correct. To prove it, we have to satisfy the following: $\exists \epsilon>0 \ \forall N \ \exists n\ge N \ \exists x \in \mathbb{R}:|f_n(x)-f(x)|>\epsilon$

Choosing $x=2n>n$ and $\epsilon=\frac{1}{2}$ we have that $|f_n(x)-f(x)|=|n-2n|=n>\frac{1}{2}$. So, $f_n$ doesn't converge uniformly to $f(x)=x$ regardless how $n$ is big.

答:
Those functions $f_n$ are not bounded. Yes, $n$ is an upper bound, but they have no lower bound. However, you can define$$f_n(x)=\begin{cases}n&\text{ if }x>n\\x&\text{ if }x\in[-n,n]\\-n&\text{ if }x<-n,\end{cases}$$and that will work.

And this sequence $(f_n)_{n\in\Bbb N}$ is not uniformly convergent to $f(x)=x$ because$$(\forall n\in\Bbb N):\bigl|f(n+1)-f_n(n+1)\bigr|=1.$$So, if you take $\varepsilon=1$, and if $N\in\Bbb N$, then $\bigl|f(N+1)-f_N(N+1)\bigr|\geqslant\varepsilon$.


答:
Take $f$ to be an unbounded function. Let $\chi_n$ be the characteristic function of the interval $[-n,n]$, $n$ being a positive integer. Define $f_n=\chi_n\cdot f$. This provides a counterexample to the statement.

Your counterexample and its proof is right.

Title: If $f_n\to f$ pointwise on $\mathbb{R}$ and each $f_n$ is bounded, then $f$ is bounded (counter-example) + question on uniform convergence proof

Body:
Let $f_n$ be a sequence of functions defined on $\mathbb{R}$ sucht that $f_n\to f$ pointwise on $\mathbb{R}$ as $n\to\infty$ and each $f_n$ is bounded. Is $f$ necessarily bounded? Prove of give a counter-example. I would like to know if the following example holds as a counter-example. In the end I have also a question on it's non-uniform convergence proof.

Suppose $f_n(x)=x$ if $x<n$ and $f_n(x)=n$ if $x\ge n$. $f_n$ converges pointwise to $f(x)=x$
. Moreover, each $f_n$ is bounded (by $n$) and $f(x)$ is unbounded.

Now I would like to know if my proof on not uniform convergence of $f_n$ is correct. To prove it, we have to satisfy the following: $\exists \epsilon>0 \ \forall N \ \exists n\ge N \ \exists x \in \mathbb{R}:|f_n(x)-f(x)|>\epsilon$

Choosing $x=2n>n$ and $\epsilon=\frac{1}{2}$ we have that $|f_n(x)-f(x)|=|n-2n|=n>\frac{1}{2}$. So, $f_n$ doesn't converge uniformly to $f(x)=x$ regardless how $n$ is big.


问:
How the degree of standard representation of $S_n$ equals $n-1$?
I found this explanation of standard representation of $S_4$. How can I extend this notion and prove it for a vector space of $n$ dimension.(https://i.stack.imgur.com/vz15S.png)
Can someone also please explain how do I calculate character for this representation.

答:
An answer from a non-mathematician here....

The standard representation of $S_n$ deals with the permutation of $n$ objects.  Any individual permutation can be represented as an $n\times n$ matrix showing the $j$-th final location of the $i$-th element.  You might think that you need all of that information to specify a permutation.

Except  that if you know where the first $n-1$ objects end up, you automatically know where the last one ends up.   So you don't need an $n\times n$ matrix, you only need an $(n-1)\times (n-1)$ matrix.

Concretely, you can think of there being $n$ orthonormal vectors $\hat e_i$ matching the $n$ objects.  One way that I have seen the "standard representation"  determined is calculating the matrix that determines how $(\hat e_{i-1}- \hat e_i)$ changes under the permutation.


问:
Prob. 48, Chap. 1, in Royden's REAL ANALYSIS: At what points is this function continuous?
Here is Prob. 48, Chap. 1, in the book Real Analysis by H.L. Royden and P.M. Fitzpatrick, 4th edition:

Define the real-valued function $f$ on $\mathbb{R}$ by setting
$$
f(x) = \begin{cases} x & \mbox{ if $x$ is irrational} \\ p \sin \frac{1}{q} & \mbox{ if $x = \frac{p}{q}$ is in lowest terms}. \end{cases}
$$
At what points is $f$ continuous?

I think the sequential criterion for continuity is what we will have to apply here.

The function is not continuous at $x = 0 = \frac{0}{1}$, because $f(0) = 0$, but if we consider the sequence $\left( \frac{1}{2^n} \right)_{n \in \mathbb{N}}$ converging to $0$, then the image sequence
$\left( \sin \frac{1}{2^n} \right)_{n \in \mathbb{N} }$ does not converge to $f(0)$.

Am I right?

How to check if $f$ is or is not continuous at a point $a \neq 0$?

答:
The trick is to consider irrationals and rationals separately. If $x$ is irrational and $x_n \to x$ with $x_n$ rational and we let $p_n, q_n$ be such that $x_n = \frac{p_n}{q_n}$, then $q_n \to \infty$ (if not then $|x_n - x| > \delta $ where $\delta = \inf_{p \in \mathbb Z}\{|x - \frac {p}{\prod q_n}|\} > 0$) and so
$$
f(x_n) = p_n \sin \frac 1 {q_n} \sim \frac {p_n}{q_n} \to x = f(x)
$$
(You may want to use some algebra of limits arguments to use the small angle approximation to $\sin \frac 1 {q_n}$).

If we had any sequence $x_n \to x$, now perhaps no longer rational, then we can say that
$$
|f(x_n) - f(x)| = \begin{cases} |x_n - x| &\text{if $x_n$ is irrational} \\
|f(x_n) - x| & \text{if $x_n$ is rational}
\end{cases} \to 0
$$
as $n \to \infty$ by considering the two subsequences separately and work directly with the definition of convergence.

If $x$ is rational then by approximating $x$ with irrationals, $x_n \to x$ we can say that $f$ is continuous at $x$ if and only if $f(x) = x$, as $f(x_n) \to x$. But this is not true for $x \neq 0$ as $\sin(\frac 1 q) \neq \frac 1 q$ for $q \in \mathbb Z$.

For $x = 0$, pick a sequence $x_n \to 0$. If $x_n$ has some rational elements $x_n = \frac {p_n} {q_n}$ Then as $|\sin(\frac 1 {q_n})| < |\frac 1 {q_n}|$ for large $q_n$, we have that
$$
|f(x_n)| = |p_n \sin\frac 1 {q_n}| \leq |x_n| \to 0
$$
for large $n$ and so $f$ is continous as 0.


问:
Prove indefinite integral is continuous
I have the proof from my prof and i understand most of the steps separately but i can't completely comprehend the end. The prof doesn't really explain the steps i would be really happy if someone can summarize it.

$f\epsilon R[a,b] F:= \int_{a}^{x}f(t)dt$ is indefinite integral of f. Proove F is continuous on [a,b]

$F(x)=\int_{a}^{x}f = \int_{a}^{w}f + \int_{w}^{x}f = F(w) + \int_{w}^{x}f \Rightarrow F(x)-F(w)=\int_{w}^{x}f \text{ Now, if } \left | f(x) \right | \leq M \text{ on }n [a,b] -M(x-w)\leq \int_{w}^{x}f\leq M(x-w) \Rightarrow \left | F(x)-F(w) \right |\leq \left | \int_{w}^{x}f \right |\leq M\left | x-w \right | $ as $x\rightarrow w, F(x)\rightarrow F(w)$

答:
I'll try to explain the proof in more detail below:

Consider any number $c\in(a,b)$ (the endpoints are considered separately). The definition of $F$ implies that $F(c)=\int_{a}^{c}f(t)dt$, so using standard properties of integrals, we deduce that for $x\neq c$

\begin{align*}
F(x)-F(c) &= \int_{a}^{x}f(t)dt-\int_{a}^{c}f(t)dt\\
&= \int_{c}^{x}f(t)dt
\end{align*}

$f$ was assumed to be Riemann integrable, so it must be bounded. This means that there is a number $M$ such that $|f(x)|\leq M$ for all $x\in[a,b]$. Using the integral inequality $|\int_{a}^{b}f(x)dx|\leq\int_{a}^{b}|f(x)|dx$, we deduce that

\begin{align*}
|F(x)-F(c)| &= \left|\int_{c}^{x}f(t)dt\right|\\
&\leq \int_{c}^{x}|f(x)|dx\\
&\leq \int_{c}^{x}Mdx\\
&= M|x-c|
\end{align*}

The absolute value brackets around $x-c$ are there to compensate for the possibility that $x<c$.

Thus, we have that $|F(x)-F(c)|\leq M|x-c|$. Letting $x\to c$ and applying the squeeze theorem immediately gives $\lim_{x\to c}F(x)=F(c)$, which is what we wanted.

To prove continuity at the endpoints, the same argument applies, the only difference being that we use left and right hand limits for $a$ and $b$, respectively. Thus, $F$ is continuous on $[a,b]$.

The bound $|F(x)-F(c)|\leq M|x-c|$ actually implies that $F$ is Lipschitz continuous, which is stronger than ordinary continuity.

Title: Prove indefinite integral is continuous


问:
Number of possible solutions to $2 \le |x-1||y+3| \le 5$, where $x$ and $y$ are negative integers

If $$2 \le |x-1||y+3| \le 5$$ and both $x$ and $y$ are negative integers, find the number of possible combinations of $x$ and $y$ .

Below is my solution approach :-

As $x$ is a negative integer, hence $|x-1|$ in the $2 \le |x-1||y+3| \le 5$ will be come $-(x-1)$ or $(1-x)$ and the main equation will transform into $2 \le (1-x)|y+3| \le 5$.

$1st$ case when $y+3 \ge 0 \Rightarrow y \ge -3 \Rightarrow y \in \{{-3,-2,-1}\} $ as $y$ is a negative integer :

For $y=-3$ we can see that there is no valid solution for $x$ as $|y+3|$ part will become $0$, hence this case is invalid.

For $y=-2$ we get the solution set for $x$ to be $x \in \{{-4,-3,-2,-1}\} $  and total number of solutions possible in this case is $4$.

For $y=-1$ we get the solution set for $x$ to be $x \in \{{-1}\} $ and total number of solutions possible in this case is $1$.

So for this $1st$ case when $y+3 \ge 0$ we have in total 5 solutions.

$2nd$ case when $y+3 \lt 0 \Rightarrow y \lt -3$ and in this case $y$ will have infinite values and I am not able to proceed from here.

The answer for the total number of solutions provided is $10$ and you can see that I've been able to find out the $5$ solutions in my $1st$ case. Can someone please guide or help me about how to proceed in the 2nd case?

Thanks in advance !

答:
It is not true that $y$ can take infinitely many numbers.

We have  since $x \le -1$, $|x-1|\ge 2,  $

$$\frac1{|x-1|}\le \frac12$$

we must have

$$|y+3| \le \frac{5}{|x+1|}\le \frac52$$

If $y< -3$, then we have $$-y-3 \le \frac52 $$

$$-\frac{11}2 \le y $$

Hence $-5 \le y < -3$, that is $-5 \le y \le -4$.

For $y=-4=-3-1$, by symmetry, the same $x$ for which $y=-3+1=-2$ satisfies the inequality would work,  there are $4$ of them.

For $y=-5=-3-2$, by symmetry, the same $x$ for which $y=-3+2=-1$ satisfies the inequlaity would work, there is $1$ of them.

Hence in total, there are $10$.


问:
Probability of drawing at least J cards out of I cards of interest from a deck of N cards within D draws with replacement
Given a regular deck of 52 playing cards, if you draw from the deck 10 times with replacement, what is the probability that you have drawn at least 3 different aces (any of the 4 aces) at least once each? And how does this generalize to N playing cards, D draws, I many cards that we are interested in, and J many of the cards we are interested in that we must draw at least once.

I came across a simpler version of this problem while working on a hobby project trying to figure out some statistics related to a Pokémon-themed Discord bot gacha game, but I was intrigued with this problem after I found a few unanswered Math StackExchange posts about this variant. (1, 2) Unfortunately those posts didn't get any attention so I figured I would make a post with the additional information that I found.

Formalization

I was able to formalize the problem as $f(n_1, n_2, k_1, k_2, p_1)$, where:

$n_1$ is the number of cards drawn

$n_2$ is the number of cards that we are interested in

$k_1$ is the number of times at least that we want to draw a card we are interested in

$k_2$ is the number of cards at least that we are interested in that we want to draw at least $k_1$ times

$p_1$  is the probability of drawing a specific card

My attempts so far

Binomial distribution of binomial distributions

I had some initial attempts to solve it by modeling a binomial distribution where the events were obtaining at least 1 of a card of interest in $n_1$ draws (that also being a binomial distribution). But I validated that against some simulation code that I wrote, and the formulation came up with different results, which I think were due to the events being not independent (drawing one particular card in a set of draws effects the probability of drawing at least one of another particular card on the same set of draws).

I think that approach should give an upper bound on the answer for the problem. It models drawing at least $k_1$ copies of a card we are interested in by using a binomial distribution. It then (incorrectly) assumes that we can model that for each card we are interested in drawing as a binomial distribution (not true, since those probabilities are not independent).

But I think the lack of independence is always destructive in this case (drawing a particular card prevents you from have drawing any other card in that same draw action), so the actual probability should be upper bounded by this formula. But I'm not 100% sure on that.

$$
  p_2 = \sum_{k=k_1}^{n_1} \binom{n_1}{k} * p_1^{k} * (1 - p_1)^{n_1 - k}
  $$
$$
  \sum_{k=k_2}^{n_2} \binom{n_2}{k} * p_2^{k} * (1 - p_2)^{n_2 - k}
  $$

Multinomial distribution

After looking around on the internet for similar problems or approaches that might work I learned about multinomial distributions. I think you can try modelling the card drawing with replacement as a multinomial distribution and then sum the results of the probability distribution function with the different input sets that make up all the possible winning sets of draws, where there is probably a trick to group those sets into categories of equal probability.

This is my current best guess as to a framework for a solution that might work, but I haven't worked out the details on it and tested it against the simulation yet.

Simpler variant

Before I came across this variant of the problem, in my hobby project I came across the simpler case where $k_2 = n_2$ and $k_1 = 1$, the "collect all of the Pokémon at least once" problem. Unfortunately I haven't solved this question yet either.

Starting with solving this simpler variant might be useful in finding a solution for the more general case. But if we do find a solution for the general case, it should work for this specific case as well.

Simulation code

I posted the simulation code I wrote on a GitHub gist at the link below. You can run it by providing $n_1$, $n_2$, $k_1$, $k_2$, and the number of total cards in the deck. You can also optionally provide an RNG seed and a number of iterations (number of draws of $n_1$ cards to simulate).

```<code>$ python sim.py simulate 10 4 1 3 52 --seed 42
Namespace(command='simulate', iterations=100000, k_1=1, k_2=3, n_1=10, n_2=4, seed=42, total=52)
100%|████████████████████████████████| 100000/100000 [00:01<00:00, 95673.52it/s]
estimated_probability = 0.01499
</code>```

https://gist.github.com/ExcaliburZero/dfc6ed8b3c7e69f6acbb556844d70add

The script also includes a mode to run the "binomial distribution of binomial distributions" approach that I think upper bounds the result.

```<code>$ python mathoverflow_answer.py calculate 10 4 1 3 52
p_1 = 0.019230769230769232
p_exactly = 0.16147234355824405
p_2 = 0.17649104785295508
result = 0.019079344697167273
</code>```

Internet research

While working on this problem I did some searching online to try to find similar problems or potential approaches for solutions. Here are some links I found the seemed potentially relevant.

Math StackExchange posts with the same question that went unanswered. (1, 2)

Resources related to cumulative distribution functions of multinomial distributions.

A page from the book "Chemical Process Dynamics and Controls", which includes a section 2.2 that gives a cumulative distribution function for multinomial distributions. (3)

A 1981 paper on a representation for a multinomial cumulative distribution function, which I don't really understand yet. (4)

Conclusion

Above are my thoughts on and current attempts at solving this problem. My current best lead is to try something with multinomial distributions, but I haven't been able to solidify and test that approach yet.

Does anyone have thoughts on how to go about solving this problem? Does this approach seem reasonable? Are there better approaches that yield a solution to this problem?

答:
First, let me say that I think simulation is the best way to do this problem.  The reason for this will be apparent as I describe how to do it analytically.

First, let's count the number of ways that we can draw three different Aces.  We have $4$ ways to choose the Aces, then $10\cdot9\cdot8$ ways to distribute them among the $10$ draws.  The other $7$ cards can be any card at all, so we have $52^7$ possibilities for them.  If this were the end of it, we would have a probability of $$\frac{4\cdot10\cdot9\cdot8}{52^3}\approx0.0204824761,$$ but we've double-counted cases where we drew more than $3$ Aces.

All the cases where we drew four Aces, at least $3$ of which are distinct have been counted more that once.  We might have drawn each Ace once, or we might have drawn three different Aces, and drawn one of them twice.  In the first case, we will have counted the event $4$ times, and in the second case, we will have counted each event twice.

Let's do these.  For the cases of all four Aces, we have $$\frac{10\cdot9\cdot8\cdot7}{52^4}\approx0.0006893141$$ for the second case, we have $4$ way to choose the $3$ Aces, then $3$ ways to choose the duplicated Ace, then $\frac{10\cdot9\cdot8\cdot7}2$ ways to distribute the Aces, so a probability
$$\frac{10\cdot9\cdot8\cdot7\cdot6}{52^4}\approx0.00413588$$
We need to subtract $3$ times the first of these, and $1$ times the second, giving approximately $0.014278649$.

Now, we have to consider the cases where we drew five Aces.  If we do this, we will find that we subtracted some events too many times, and we will have to add these back in.  Of course, we have to continue this until we deal with the cases where we drew $10$ Aces.  It gets quite tedious, not only because we have to treat all the ways we could have draw $n$ Aces, at least $3$ of which are distinct, but for each of those cases, we have to count how many times it was added or subtracted at the previous stage.

This is the principle of inclusion and exclusion.  The various steps do give bounds on the probability though. These are the Bonferroni inequalities.  If $p$ is the probability we seek, we can say $$
0.014278649 < p < 0.0204824761$$ which agrees nicely with your simulation.

Of course, we could go through only some of the stages, and prove precise upper and lower bounds on the probabilities, if it were considered worthwhile.  If I had to do that for some reason, I would check my answer by simulation.

It would be an interesting exercise to write a computer program to compute the answer precisely.  If you want to try it, I suggest doing it first for the $3$ Aces problem, before trying to generalize it.

EDIT

I thought of what I think will be a better way of computing precise bounds.  Instead of alternately overstating and understating the probability, compute successively better lower bounds, until the possible understatement is neglibilible.

The probability that of drawing exactly $3$ distinct Aces, and no other Ace is $$\frac{4\cdot10\cdot9\cdot8\cdot48^7}{52^{10}}\approx0.01169627$$
The probability of drawing exactly $4$ Aces, all of which are distinct is $$\frac{10\cdot9\cdot8\cdot7\cdot48^6}{52^{10}}\approx0.0004264268$$  The probability of drawing exactly $4$ Aces, $3$ of which are distinct is $$\frac{10\cdot9\cdot8\cdot7\cdot6\cdot48^6}{52^{10}}\approx0.00255856$$
Adding gives $0.01468126$ as a lower bound on the probability.
How much are we missing we if we quit here?  Certainly, less than the probability of drawing at least $5$ Aces.  There are $\binom{10}5$ ways to choose which draws will be Aces, and $4$ choices of which Ace will be drawn, then $52$ choices for each of the other spots.  This gives $$\frac{4^5\binom{10}5}{52^5}\approx0.0006787$$ so we can say $$0.01468126<p<0.01536$$

This approach would be easier to program than inclusion-exclusion, certainly.


问:
Is the zero set of an analytic function closed nowhere dense?
Given a non-constant analytic function $f(x)$ on a domain $D \subseteq \mathbb{R}^n$. I want to prove that

$\mathcal{Z} = \{x \in \mathcal{D} |f(x) = 0 \}$

is closed nowhere dense.

I originally wanted to prove this by contradiction as follows using the identity theorem.

"Assume that $\mathcal{Z}$ is not a closed nowhere dense set, then $\mathcal{Z}$ must contain a limit point", which shows that $f(x)$ is constant zero by the identity theorem and thus contradicts that $f(x)$ is non-constant.

However, I am not sure if the following statement is true:

"Assume that $\mathcal{Z}$ is not a closed nowhere dense set, then $\mathcal{Z}$ must contain a limit point"

If it is true, could you provide one reference?

If not, is $\mathcal{Z}$ closed nowhere dense, and how can I prove it?

答:
We know $Z$is closed. So we want to show $Z$ is nowhere dense. Assume, to reach a contradiction, that the interior of $Z,$ denoted by $\text {int } Z,$ is nonempty.

Let $x\in \text {int } Z.$ Let $y\in D.$ Because $D$ is open and connected, $D$ is path connected. So there is a continuous path $\gamma:[0,1]\to D$ with $\gamma(0)=x, \gamma(1)=y.$

Let $E$ be the set of $t\in [0,1]$ such that $f=0$ in an open ball $B_{\gamma(t)}$ centered at $\gamma(t).$ Clearly $0\in E.$ By continuity of $\gamma,$ $\gamma^{-1}(B_{\gamma(t)})$ is an open neighborhood of $t$ in $[0,1].$ And if $s$ is in this neighborhood, $\gamma(s)\in B_{\gamma(t)},$ hence there is an open ball centered at $\gamma(s)$ where $f=0.$ It follows that $E$ is a nonempty open subset of $FE.$

Furthermore, since $f$ is real analytic in $D,$ $E$ is precisely the set of $t\in [0,1]$ where all derivatives of $f$ vanish at $\gamma(t).$ Thus $E$ is closed in $[0,1].$ It follows that $E$ is nonempty and clopen in $[0,1].$ Since $[0,1]$ is connected, $E=[0,1].$

This implies $f(y)=0.$ Since $y\in D$ was arbitrary, we have $f=0$ in $D.$ This is a contradiction, since $f$ was assumed to be nonconstant in $D.$ Therefore $\text {int } Z$ is empty, and $Z$ is nowhere dense as desired.


问:
Prove that $\frac{10^n-1}{9n}$ is integer when $n=3^k$
I have no idea how to go about proving this. The furthest I've gotten is to say that the sequence equals 1/1, 11/2, 111/3, etc.

So that means that $\frac{10^a-1}{9}$ must be divisible by 3.

$\frac{10^a-1}{9} = 3b \implies 10^a-1 = 27b \implies 27b+1 = 10^a$.

When is that last statement true?

Body:
The statement is equivalent to proving $$10^{3^k}-1\equiv_{3^{k+2}}0$$
Notice that $\phi(3^{k+2})=3^{k+2}-3^{k+1}=2\cdot 3^{k+1}$, so that for a unit $a$ we have $$a^{2\cdot 3^{k+1}}\equiv_{3^{k+2}}1$$ We also have

Claim: $$10^{3^k}\equiv_{3^{k+2}}10^{3^{k+1}}$$

Proof: $$10^{3^k}\equiv_{3^{k+2}}10^{3^{k+1}-2\cdot 3^{k}}\equiv_{3^{k+2}}10^{3^{k+1}}$$

Let $t=10^{3^k}$, then the above claim may be written as $$t^3\equiv_{3^{k+2}}t\iff t(t+1)(t-1)\equiv_{3^{k+2}}0$$ Both $t,t+1$ are units since $t,t+1\not\equiv_{3}0$, therefore we must have $t-1\equiv_{3^{k+2}}0$.


问:
Definition of a formula in propositional logic
The set of formulae in propositional logic is characterised by ( at least in Derek Goldreis's book)

Every propositional variable is a formula

If $a$ is a formula then so is $\neg a$

If $a,b$ are formulae then so are $(a \star b)$ for any connective $\star$

Nothing else is a formula and every formula is obtained by a finite number of applications of 1)-3).

From this, does it immediately follow that for any formula $a$ exactly one of the following holds

1') $a$ is some propositional variable

2') $a$ is $\neg b$ for some formula $b$

3') $a$ is $(b \star c)$ for some formulae $b,c$?

I know this is true ( infact the stronger Unique readability theorem holds) but it isn't clear to me how this immediately follows from the definition of a formula I gave above.

I suspect it is clause 4) that makes this work. So perhaps my question really is what exactly does clause 4) mean...

Any help would be appreciated! Thanks!

答:
Yes, it does follow immediately, and yes, it is the fourth clause that makes this work.

The fourth clause says that every formula is the result of applying the first three clauses a finite number of times. In particular, any formula $a$ is the result of applying one of the first three clauses to existing formulae. If $a$ results from (1), $a$ must be a propositional variable. If it results from (2), it must be $\neg b$ for some formula $b$ that has already been derived. And if it results from (3), it must be $(b*c)$ for some connective $*$ and formulae $b$ and $c$ that have already been derived.

This is a fairly straightforward recursive definition of the set of formulae. The first clause is the base clause: it defines the propositional variables as formulae and so gives us some input for the construction clauses (2) and (3) that define how more complex formulae are to be built from simpler ones. And the fourth clause, which is often omitted but should always be understood, prevents us from throwing random junk into the collection of formulae: every formula must be a propositional variable or derivable from propositional variables by applications of (2) and (3).


问:
Confusion on how extended binomial theorem works
So I just finished learning the standard binomial theorem and I've just come across the extended (newtons binomial theorem).

As expected I am completely baffled about how it works I do not understand how the following holds :

Suppose $n \in \mathbb{R}$ then for every $x$ satisfying $|x| < 1$, we have $$ (1+x)^{n} = \sum_{r=0}^{\infty} \dbinom{n}r x^r \space \space \space \space ; \space \space r = 0,1,2...$$ and the extended binomial coefficient is given as $$\dbinom{n}r = \dfrac{n!}{(n-r)! \space r!}$$

There are a lot of things that trouble me here first I've read online that factorial are not defined for negative integer values yet somehow we are using them to compute the coefficients in the expansion example:.$$\begin{align*}
(b + a)^n &= b^n\left(1 + \frac{a}{b}\right)^n\\
&= b^n \left(\sum_{k = 0}^{n}{n \choose k}\left(\frac{a}{b}\right)^k\right)\\
&= b^n \left(1 + n \left(\frac{a}{b}\right) + \frac{(n)(n-1)}{2}\left(\frac{a}{b}\right)^2 + \cdots\right)
\end{align*}$$

How is it that say $-4!$ is undefined yet in the formula we are using it  is it then not the same as negative factorial but instead some kind of entirely different method that happens to work?.

I would like some basic intuition on how this binomial theorem actually works if it is possible as the book i am reading just states the formula without any explanation as to  how it works if there is no basic explanation maybe a list of topics I need to learn before to understand it would be helpful.

Thanks In advance.

答:
In the extended binomial theorem, the definition of $^nC_r$ is not as simple as it is for the 'vanilla' binomial theorem. If we define
$$
n!=n \cdot(n-1) \cdot (n-2)\cdot\dots\cdot3\cdot 2\cdot 1
$$
then the formula you have provided is indeed meaningless, as $n!$ only makes sense when $n$ is a natural number. The simplest way to address this is to define $^nC_r$ slightly differently when $n$ is a real number:
$$
\binom{n}{r}=\frac{n(n-1)(n-2)\cdots(n-r+1)}{r!} \, ,
$$
meaning that the problem of defining things such as $(-4)!$ does not arise. Notice that this definition is equivalent to the one you gave when $n$ is a natural number. By the way, the numerator can be written as $n^{\underline{r}}$ using falling factorials.


答: I'd interpret "nothing else is a formula" as meaning what you concluded: that every formula is of one of the three given forms. I don't think the "exactly one" part technically follows, but it's implied. We can think of the set of formulas as an inductive type whose elements are syntax trees.</p>

Title: Definition of a formula in propositional logic

Body:
The set of formulae in propositional logic is characterised by ( at least in Derek Goldreis's book)

Every propositional variable is a formula

If $a$ is a formula then so is $\neg a$

If $a,b$ are formulae then so are $(a \star b)$ for any connective $\star$

Nothing else is a formula and every formula is obtained by a finite number of applications of 1)-3).

From this, does it immediately follow that for any formula $a$ exactly one of the following holds

1') $a$ is some propositional variable

2') $a$ is $\neg b$ for some formula $b$

3') $a$ is $(b \star c)$ for some formulae $b,c$?

I know this is true ( infact the stronger Unique readability theorem holds) but it isn't clear to me how this immediately follows from the definition of a formula I gave above.

I suspect it is clause 4) that makes this work. So perhaps my question really is what exactly does clause 4) mean...

Any help would be appreciated! Thanks!


问:
Maximize the line integral in a region
A vector field $\vec F$ is defined on the disk $D$ of radius $5$

$$\vec F=\big[-y^3+y\sin(xy)\big]\hat{x}+\big[4x(1-y^2)+x\sin(xy)\big]\hat{y}$$

Consider the line integral

$$ \int_C  \vec F\,\mathrm{d}r  $$

where $C$ is some closed curve contained in $D$. For which $C$ is this integral the largest?

We used Green's Theorem and found that $-2≤y≤2$, so $C$ is the boundary of the region bounded by $y=2$, $y=-2$ $\cup$ two circular arcs, and this maximizes the line integral. I don't see it.

答:
The question should have provided with the orientation. In fact, the line integral is largest for a clockwise oriented curve.

If $\vec F = (P, Q)$. For a clockwise oriented curve and given vector field, $P_y - Q_x = y^2-4$. We note that $(y^2-4)$ is negative for $|y| \lt 2$ and positive for $ |y| \gt 2$. I will use it later in the answer.

If we take a closed curve which is boundary of the disk, it is easy to see that,

$\displaystyle \int_0^{2\pi} \int_0^5 r (r^2\sin^2\theta - 4) \ dr \ d\theta = \frac{225\pi}{4} \approx 176.715$

[Now we could find a piece-wise smooth simple closed curve oriented clockwise which avoids most of the region between $-2 \leq y \leq 2$ and that would make the integral even larger.]

If we put a restriction that it is a counterclockwise oriented closed curve as the book says in the hint, the integral becomes

$\displaystyle \int_C \vec F \cdot d\vec r = \iint_R (4-y^2) \ dA$

The below diagram depicts the closed curve that would maximize the line integral.

We know that $(4-y^2)$ is negative for $|y| \gt 2$. The closed curve depicted in the diagram avoids any point where $|y| \gt 2$ but also exploits max range of $-5 \leq x \leq 5$.

You can do the integral in couple different ways. I will do it over the rectangle in the first quadrant and then over region between line segment $AB$ and arc $AB$ above x-axis and then multiply the integral by $4$ due to symmetry.

$\displaystyle 4 \int_0^{\sqrt {21}} \int_0^2 (4 - y^2) \ dy \ dx + 4 \int_{\sqrt{21}}^5 \int_0^{\sqrt{25-x^2}} (4 - y^2) \ dy \ dx \approx 104.9293$

Title: Maximize the line integral in a region


问:
Length of curve using polar coordinates
Let $r = r(\varphi)$ be the equation of a plane curve $\alpha$ in polar coordinates. Prove that the length of $\alpha$ is given by

$$ \int_{\varphi_1}^{\varphi_2} \sqrt{ r^2(\varphi) + \dot{r}(\varphi)^2} \, d \varphi$$

I parametrise $\alpha$ as follows

$$ \alpha(t) \enspace = \enspace |\alpha(t)| \cdot \Big( \cos \varphi(t), \sin \varphi(t) \Big)$$

And call $|\alpha(t)| = r(t)$. I then solved it this way:
\begin{align}
\operatorname{len}(\alpha) \enspace &= \enspace \int_a^b |\dot{\alpha}(t)| \, dt \\
&= \enspace \int_a^b \sqrt{ \dot{\alpha}_1^2(t) + \dot{\alpha}_2^2(t)} \, dt \\
&= \enspace \int_a^b \sqrt{ \dot{\varphi}^2(t) \cdot \Big[ \big( \tfrac{\partial r}{\partial \varphi}\big)^2 + r^2(\varphi) \Big] } \, dt \\
&= \enspace \int_{\varphi_1}^{\varphi_2} \sqrt{\dot{r}^2(\varphi) + r^2(\varphi)} \, d\varphi
\end{align}

where I identified $\varphi(a) = \varphi_1$ and $\varphi(b) = \varphi_2$. However, I am struggling to understand one problem that comes with this:

In my approach I used $r(t) = r\big( \varphi(t) \big)$, so that the derivative is $\tfrac{d}{dt} r \big( \varphi(t) \big) = \dot{\varphi}(t) \cdot \tfrac{\partial r}{\partial \varphi}$. But am I really allowed to do this? If not, then how do I solve this? And if I am allowed to do this, why am I?

答:
I would parametrize directly by angle $\varphi$. By recalling the general formula for a length of a curve given in the parametric form
\begin{align}
L=\int_{t_1}^{t_2}\sqrt{\left(\frac{\text{d}x}{\text{d}t}\right)^2+\left(\frac{\text{d}y}{\text{d}t}\right)^2}\,\text{d}t
\end{align}
then by considering $t\equiv\varphi$ and by substitution $x(\varphi)=r(\varphi)\cos\varphi,~y(\varphi)=r(\varphi)\sin\varphi$ one gets
\begin{align}
L=&\int_{\varphi_1}^{\varphi_2}\sqrt{\left(\frac{\text{d}x}{\text{d}\varphi}\right)^2+\left(\frac{\text{d}y}{\text{d}\varphi}\right)^2}\,\text{d}\varphi=\\
&\int_{\varphi_1}^{\varphi_2}\sqrt{\left( r'(\varphi)\cos\varphi-r(\varphi)\sin\varphi\right)^2+\left( r'(\varphi)\sin\varphi+r(\varphi)\cos\varphi\right)^2}\,\text{d}\varphi=\\
&\int_{\varphi_1}^{\varphi_2}\sqrt{\left(r'(\varphi)\right)^2+r^2(\varphi)}\,\text{d}\varphi
\end{align}
Of course you can ask how to derive the first formula, which is however only direct application of the Pythagoras theorem.

Title: Length of curve using polar coordinates


问:
What is wrong with my approach for orthonormal procedure?
I am given a matrix $$A=\left(\begin{matrix}1&1&0&1\\1&2&1&0\\0&1&3&-2\\1&0&-2&4\end{matrix}\right)$$ which defines a scalar product in $\mathbb{R}^4$ as $\langle x,y\rangle=x^TAy$. I have to find an orthonormal system. I try to orthonorm canonical basis using Gram orthonormal procedure, so:

$$v_1=e_1\\v_2=\frac{e_2-\langle e_2,v_1\rangle v_1}{||e_2-\langle e_2,v_1\rangle v_1||}\\v_3=\frac{e_3-\langle e_3,v_2\rangle v_2-\langle e_3,v_1\rangle v_1}{||e_3-\langle e_3,v_2\rangle v_2-\langle e_3,v_1\rangle v_1||}$$ and so on. But when I do like this I get $\langle v_2,v_3\rangle\neq0$, so this method must be wrong for arbitrary scalar products and I have to do some adjustments, but I can't see them. Hope you can help me, thanks in advance!

答:
You should get
\begin{align*}
v_1 &= e_1,\\
v_2 &= (-1,1,0,0)^T,\\
v_3 &= \frac1{\sqrt{2}}(1,-1,1,0)^T.
\end{align*}
Note that, when you normalise a vector $w$, you should not use the Euclidean 2-norm, but calculate the norm as $\|w\|=\sqrt{\langle w,w\rangle}=\sqrt{w^TAw}$.


问:
Are there any famous/notable mathematicians who have their own YouTube channel?
Are there any famous/notable mathematicians who have their own YouTube channel?

I found this amazing  video regarding the book.
The YouTube channel name is The Math Sorcerer,
but I don't know what the background of The Math Sorcerer is. From the analysis of the video I think he is a famous/notable mathematician.

I want to know more about YouTube channels of notable mathematicians.

答:
Daily Challenge with Po-Shen Loh

Math-life balance by Prof .Mura Yakerson Phd in algebraic geometry and  postdoc at ETH Zurich Switzerland


问:
How is Gauss's theorem used in this quantum mechanics problem?
I'm following my teacher's Quantum Mechanics notes, where he solves the free particle Shrodinger equation by means of a Fourier transform. The free particle equation (no potential energy) is
$$i\partial_t\Psi=\frac{-\hbar}{2m}\nabla^2\Psi$$
He uses Gauss's theorem twice, though he only mentions one of those times. After some work we arrive at the equation
$$i\partial_t\mathcal{F}\Psi=\frac{-\hbar}{2m}\int_{\mathbb{R}^3}\left(\nabla\boldsymbol{\cdot}\left(e^{-i\mathbf{k}\boldsymbol{\cdot}\mathbf{r}}\nabla\Psi\right)+ie^{-i\mathbf{k}\boldsymbol{\cdot}\mathbf{r}}\mathbf{k}\boldsymbol{\cdot}\nabla\Psi\right)\mathrm{d}\mu^3(\mathbf{r})$$
The calligraphic F denotes Fourier transform, and $\mathrm{d}\mu^3(\mathbf{r})$ is the typical three dimensional measure, i.e, $\mathrm{d}x\mathrm{d}y\mathrm{d}z$, just expressed in coordinate-free notation. Now if we had some finite integration region $\Omega\subset\mathbb{R}^3$ we could use Gauss's divergence theorem and write the first term as
$$\int_\Omega \nabla\boldsymbol{\cdot}\left(e^{-i\mathbf{k}\boldsymbol{\cdot}\mathbf{r}}\nabla\Psi\right)\mathrm{d}V=\int_{\partial\Omega}e^{-i\mathbf{k}\boldsymbol{\cdot}\mathbf{r}}\nabla\Psi\boldsymbol{\cdot}\mathbf{n}\mathrm{d}S$$
He claims (without proof) that as we take $\Omega\to\mathbb{R}^3$, that $\nabla\Psi\boldsymbol{\cdot}\mathbf{n}$ vanishes, but offers no proof of this. I kind of understand why this might be true because $|\psi(\mathbf{r},t)|\to 0$ as $\Vert\mathbf{r}\Vert\to\infty$ in order to be square integrable, but who's to say $\Vert\mathbf{n}\Vert$ doesn't shoot off to infinity?

Now onto the second usage. We reach the equation
$$\partial_t\mathcal{F}\Psi=\frac{-\hbar}{2m}\int_{\mathbb{R}^3}\mathbf{k}\boldsymbol{\cdot}\left(\nabla\left(e^{-i\mathbf{k}\boldsymbol{\cdot}\mathbf{r}}\Psi\right)+i\mathbf{k}e^{-i\mathbf{k}\boldsymbol{\cdot}\mathbf{r}}\Psi\right)\mathrm{d}\mu^3(\mathbf{r})$$
He again claims, without proof

The first term,
$$\int_{\Bbb{R}^3}\mathbf{k}\boldsymbol{\cdot}\nabla\left(e^{-i\mathbf{k}\boldsymbol{\cdot}\mathbf{r}}\Psi\right)\mathrm{d}\mu^3(\mathbf{r})$$
Vanishes due to Gauss's theorem.

This one is even more mysterious to me. How do we even apply Gauss's theorem in this case?

Any justification of this, even hand wavy justification, would be much appreciated. Thanks.

答:
$\mathbf{n}$ is a unit normal, i.e. $||\mathbf{n}||=1$, so that's not really an issue. However, usually in order to be able to calculate the Fourier transform of a function, the function has to be a rapidly decreasing function (see e.g. here), and in particular the first derivative must vanish faster than any polynomial grows. It seems this is being assumed in the assertion that $\int_{\partial \Omega}e^{-i\mathbf{k}\cdot\mathbf{r}}\nabla\Psi\cdot\mathbf{n}\,\mathrm{d}S\to 0$ as $\Omega\to\mathbb{R}^3$; the normal derivative vanishes faster than the surface area grows (as the square of the radius).

As to your second question, you can simply pull the $\mathbf{k}$ inside the derivative,

$$
\int_{\Bbb{R}^3}\mathbf{k}\boldsymbol{\cdot}\nabla\left(e^{-i\mathbf{k}\boldsymbol{\cdot}\mathbf{r}}\Psi\right)\mathrm{d}\mu^3(\mathbf{r}) = \int_{\Bbb{R}^3}\nabla\cdot\left(\mathbf{k} e^{-i\mathbf{k}\boldsymbol{\cdot}\mathbf{r}}\Psi\right)\mathrm{d}\mu^3(\mathbf{r})$$

and then apply Gauss' Theorem to the second form. It will vanish for similar reasons to above, i.e. $\Psi$ vanishes faster than any polynomial grows.


问:
How can I prove this for $a+b+c=0$
We wish to prove that

$$\left(\frac{a-b}{c}+\ \frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9,$$ for $a+b+c=0$ and $$abc \neq 0 , a\neq b , b\neq c, c \neq a.$$
I tried working with only $2$ fractions at a time, switching some of the variables (ex.$a=-b-c$) and then adding them up, and multiplying first but could not go further.

Thank you in advance.

答:
You can expand directly, by noticing that the first parantheses is
$$\frac{ab(a-b) + bc(b-c) + ca(c-a)}{abc} = -\frac{(a-b)(b-c)(c-a)}{abc}$$

So the expression becomes

$$\frac{a^3+b^3+c^3 + 3abc - a^2b - a^2c - b^2a -b^2c - c^2a - c^2b}{abc}$$

Collecting $a^3, -a^2b, - a^2c $ and similarly this becomes

$$\frac{a^2(a - b -c) + b^2(b-c-a) + c^2(c - a -b)}{abc} + 3$$

Now $a-b-c = 2a$ by the condition so this becomes just

$$2\frac{a^3+b^3+c^3}{abc} + 3$$

But there is an identity $a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca)$, so we get $a^3+b^3+c^3 = 3abc$.

Thus we get that the initial expression is $9$.

Title: How can I prove this for $a+b+c=0$


问:
The contrapositive of the statement "For all positive real number $x$, there exists an integer n such that $\frac{1}{n} < x.$"
I am trying to express contrapositive of the statement "For all positive real number $x$, there exists an integer n such that $\frac{1}{n} < x.$" in formal language.

The universe for both $x$ an $n$ is the set of all real numbers.

I am confused whether the expression for the contrapositive of the statement in formal language should be

\begin{align}
\forall x,(\forall n,(n\in \mathbb{Z} \rightarrow \frac{1}{n} \ge x) \rightarrow (x \notin \mathbb{R}^+)).
\end{align}

or

\begin{align}
\forall x,(\forall n,(n\in \mathbb{Z} \wedge \frac{1}{n} \ge x) \rightarrow (x \notin \mathbb{R}^+)).
\end{align}

答:
The statement, expressed as an implication with free variable $x$, is
$$
x\in\mathbb{R}^{+}\rightarrow \exists{n}:\left(n\in\mathbb{Z}\;\wedge\;\frac{1}{n}<x \right).
$$
Its contrapositive, then, is
$$
\neg\exists{n}:\left(n\in\mathbb{Z}\;\wedge\;\frac{1}{n}<x \right)\rightarrow x\not\in\mathbb{R}^{+},
$$
or
$$
\forall{n}:\left(n\in\mathbb{Z}\rightarrow\frac{1}{n}\ge x \right)\rightarrow x\not\in\mathbb{R}^{+}.
$$


问:
Probability distributions named after Greek letters
There are some probability distributions named after Greek letters, such as Gamma, Beta, Chi, and Zeta distributions.

I wonder if there are some rules of why and how they are named after Greek letters?

Can we make use of the rules (if exist) to help us memorize these distributions?

Are there other such distributions?

Thanks and regards!

答:
There's a general problem with mathematics that every researcher eventually encounters. There aren't enough letters in the alphabet.

In probability (the only discipline I research) if you discover a new distribution then most of the time one of two things has happened.

1.$\qquad$You've proposed an exotic or special formula at the beginning of your paper so you give it an exotic name, like a Greek letter.

2.$\qquad$You've spent ten pages deriving the formula for your distribution and you've run out of normal letters.

So some of the time the new distributions get given Greek letters. Not always true, the $T$ distribution is important, $W$ is usually used for Brownian motion which is more important than all the other distributions put together.

You might find a pattern that helps to remember them, but it will be a coincidence, not a rule.


问:
a limit containing a digamma function
I wonder if the following limit is correct and how to prove it
$$
\lim _{x \rightarrow+\infty} \frac{4}{p^{2}} x^{4}\left(\frac{1}{x^{2}}+\frac{2}{x}(\log 2+\Psi(x))+\Psi^{\prime}(x)\right)>0
$$
where $p>0$

答:
It is sufficient to use the asymptotic expansions you mentioned in the comments. From the asymptotic expansions it follows that as $x\to\infty$
$$
\frac{4}{p^{2}} x^{4}\left(\frac{1}{x^{2}}+\frac{2}{x}(\log 2+\Psi(x))+\Psi^{\prime}(x)\right)\sim\frac{4}{p^2}x^3(1+\log 4+2\log x)+\mathcal O(x^2)\to\infty.
$$
And so for all $p>0$ the limit will diverge to $\infty$ and thus is greater than zero.

Title: a limit containing a digamma function


问:
Finding the sum of possible values
$14$ students have taken part in a math olympiad. Every problem in that olympuad is solved by exactly $7$ students. Furthermore, the first $13$ students have all solved $10$ problems each. If the $14$th student has solved exactly  $k$ problems, then what is the sum of the allpossible values of $k$?

I have managed the diophantine equation: 130+$k$=7$x$ where $x$ is the total number of questions in the question paper....but i have been stuck determining the sum of the values of $k$.

答: Your equation should have $7x$ on the right.  You are counting the total number of solutions and have been told that each question is solved $7$ times.  Now you can find $k \pmod 7$.  $k$ has to be positive.  If $k$ is too large you can't have $6$ of the first $13$ students solve all the problems the last one solved, so that puts an upper limit on $k$.  For example, if $k=1000$, there must be at least $1000$ problems, so there must be at least $7000$ solutions.  But there are only $130+k=1130$ solutions, so $k$ cannot be that large.

Title: Finding the sum of possible values


问:
Math notation for iterating over pairs of edges
I have a graph and I would like to iterate over each pair of edge. Is there a syntactically nicer way of indicating this other that my current solution:

$$
(u, v), (x, y) \in E, (u, v) \neq (x, y)
$$

答:
You can use $\binom{E}{2} = \{\{e_1,e_2\} \subset E | e_1 \neq e_2\}$. It is not totally standard but its quite common to see it especially in graph theory.

see also Notation for binomial coefficient set

The following notation is used by R.P. Stanley in Topics in Algebraic Combinatorics. See section Basic Notation (p.6)

\begin{array}{ccl}
&2^S\qquad&\qquad \text{the set of all subsets of the set }  S\\
&\binom{S}{k}\qquad&\qquad \text{the set of $k$-element subsets of } S\\
\end{array}


问:
Proving that interval is subset of another interval
let

$I_n=[n, \infty)$

Prove that $I_{n+1} \subseteq I_n$.

So although that is pretty obvious I do not know how to prove it nicely.

for $I_{n}$ the minimum of the set would be n, hence all $k \in \mathbb{R} \geq n \in I_n$

for $I_{n+1}$ the minimum of the set would be n+1, hence all $l \in \mathbb{R} \geq n+1 \in I_{n+1}$.

How can I just argue that $I_{n+1} \subseteq I_{n}$?

答:
Let $ n\ge 0$. To prove that a set $ A$ is a subset of $ B$, you should prove that any element of $ A$ is in $ B$.

Take an arbitrary element $ x$ of $ I_{n+1}$.

$$x\in I_{n+1}\implies x\in[n+1,+\infty)$$
$$\implies x\ge n+1$$
$$\implies x\ge n$$
$$\implies x\in I_n$$

thus $$I_{n+1}\subset I_n$$


问:
Eigenvalues of general $2\times2$ matrix
I am wondering what the expression is for eigenvalues of a $2\times2$ matrix in general form

$$A=\left[
\begin{array}{ccc}
 a_{11} & a_{12}\\
a_{21} & a_{22}
\end{array}
\right]$$

I tried to compute $|A-\lambda I| = 0$ but it got very messy.

答:
The characteristic polynomial is $\lambda^2 - t \lambda + d$ where $t = a_{11} + a_{12}$ is the trace and $d = a_{11} a_{22} - a_{12} a_{21}$ the determinant of $A$.
The eigenvalues are the roots of this quadratic, namely
$$ \frac{t \pm \sqrt{t^2-4d}}{2}$$


问:
$L^p(\mathbb{R})$ contains a subspace which is is isometrically isomorphic to $\ell^p$
For $1 \leq p \leq \infty$, is it true that $L^p(\mathbb{R})$ contains a subspace which is is isometrically isomorphic to $\ell^p$ ?

I don't think it's true. But I can't find any counterexamples.

答: Hint: consider functions that are constant on intervals...</p>


问:
If abelian $P\in{\rm Syl}_p(G)$, and $H\le P$ and $H^g\le P$, show $g\in G$ is the left product of an element of $N_G(P)$ with an element of $C_G(H)$

If abelian $P \in{\rm Syl}_p(G)$, and $H \leq P$ and $H^g \leq P$, show $g \in G$ is the left product of an element of $N_G(P)$ with an element of $C_G(H)$

First, I set $g=nc$, where I hope to find that $n \in N_G(P)$ and $c \in C_G(H)$. I am trying to do this by contradiction, assuming either $n \notin N_G(P)$ (which implies $n \notin N_G(H) )$ or $c \notin C_G(H)$. In the first cast ($n \notin N_G(P)$, $c \in C_G(H)$), we have that $H^g = (H^n)^c \neq H^c = H$. However, $g$ is arbitrary (as far as I know) in $G$, so I can not use that $P$ is abelian to find that $H^g = H$. I know that $P$ must be the unique Sylow $p$-subgroup but I can't see how that will help me.

I am not sure where to go from here.

答:
In his (unfortunately deleted) answer, Arturo Magidin proved that $g = cn$ with $c \in C := C_G(H)$ and $n \in N:= N_G(P)$.

I have found a counterexample that shows that it is not necessarily the case that $g=nc$ for $c \in C$ and $n \in N$.

Let $G=A_7$ and $P = \langle (1,2,3), (4,5,6) \rangle \in {\rm Syl}_3(G)$, and let $H = \langle (1,2,3) \rangle \in P$.

Note that I am taking $H^g$ to mean $g^{-1}Hg$ and I compose permutations from left to right.

Then with $g = (1, 6)(2, 4)(3, 5, 7)$, we have $H^g = \langle (4,5,6) \rangle \le P$.

We have $g = cn$ with $c = (1, 3, 2)(4, 6)(5, 7) \in C$ and $n = (1, 4)(2, 5, 3, 6) \in N$, but it can be checked that $g \not\in NC$. (I checked this on the computer. Since $|NC|=|CN| = 144$, it could be tedious to check it by hand.)


问:
Isomorphism of rings induced from schemes
I'm new to schemes and reading Hartshorne. This is a detail in the proof of Proposition 3.2 chapter 2. I think it should be obvious but I don't see why.

Say we have $X=\operatorname{Spec}(A)$ is an affine scheme. Assume there is an open set $U=\operatorname{Spec}(B)$ of $X$. Then for some $f\in A$, $D(f)\subseteq U$. This is true because $D(f)$ form a base for the topology.

Now let $g$ be the image of $f$ in $B$. Hartshorne claims that $A_f \cong B_g$. I don't understand this detail. Sure there is such a map induced from $A\to B$. But I don't see any details that indicate this is an isomorphism

答: consider $D(f)=Spec\,A_f\to U=Spec\,B\to X=Spec A$ any map between affine schemes come from a map on rings so the natural map $A\to A_f$ factors through $B$, clearly $g$ goes to $f$ in $A_f$ so it is invertible in $A_f$, By universal property you get a map $B_g\to A_f$ which is easy to check that this is the inverse of natural map $A_f\to B_g$


问:
Prove $\exists x (P(x) \lor Q(x)) ≡ \exists x P(x) \lor \exists x Q(x)$.
$$
\exists x (P(x) \lor Q(x)) ≡ \exists x P(x) \lor \exists x Q(x)
$$

I have to prove this. I'm extremely new and yet to grasp the whole tautology logics. I believe, that $$ \exists x P(x) \lor \exists x Q(x) $$ requirement is that all integers are even OR all integers are odd (I'm using integers as a universe of discourse), but for $$ \exists x (P(x) \lor Q(x)) $$ For all $x$ where $x$ is an integers, $x$ is even
OR $x$ is odd. I don't know how to further prove the whole statement, what could be the further proof of equivalency?

答:
Let LHS be True.

Then there is $x_0$ in the universe s.t. $P(x_0) \lor Q(x_0)$ holds true.

WLOG, let $P(x_0)$ holds true.

Then the proposition $\exists x P(x)$ holds true for $x_0$.

And hence the proposition $\exists x P(x) \lor \exists x Q(x)$ holds true.

Similarly you can prove the other direction.


答:
Prove that $∃x(P(x) ∨ Q(x)) \implies ∃xP(x) \lor ∃xQ(x)$

Let $x=x_0$ be a value that satisfies $∃x(P(x) ∨ Q(x)) \implies P(x_0) \lor Q(x_0)$$\implies ∃xP(x) \lor ∃xQ(x)$ (namely, for $x=x_0$)

Prove that $∃xP(x) \lor ∃xQ(x) \implies ∃x(P(x) ∨ Q(x))$

$∃xP(x) \lor ∃xQ(x)$ implies

(a) $∃x=x_1$ s.t. $P(x_1) \implies P(x_1) \lor Q(x_1) \implies ∃x(P(x) \lor Q(x))$ (namely, for $x=x_1$)

OR

(b) $∃x=x_2$ s.t. $Q(x_2) \implies P(x_2) \lor Q(x_2) \implies ∃x(P(x) \lor Q(x))$ (namely, for $x=x_2$).

Title: Prove $\exists x (P(x) \lor Q(x)) ≡ \exists x P(x) \lor \exists x Q(x)$.

Body:
$$
\exists x (P(x) \lor Q(x)) ≡ \exists x P(x) \lor \exists x Q(x)
$$

I have to prove this. I'm extremely new and yet to grasp the whole tautology logics. I believe, that $$ \exists x P(x) \lor \exists x Q(x) $$ requirement is that all integers are even OR all integers are odd (I'm using integers as a universe of discourse), but for $$ \exists x (P(x) \lor Q(x)) $$ For all $x$ where $x$ is an integers, $x$ is even
OR $x$ is odd. I don't know how to further prove the whole statement, what could be the further proof of equivalency?


问:
$\{(x,f(x)): x\in E\}$ is compact in $\mathbb R^2 \implies f:E\to\mathbb R$ is continuous

Given $f: E\to\mathbb R$, define $G: E\to\mathbb R^2$ by $G(x) = (x,f(x))$ for all $x\in E$. $E$ is a compact subset of $\mathbb R$. The following are equivalent:

$f$ is continuous.

$G$ is continuous.

The graph of $f$ is a compact subset of $\mathbb R^2$.

First, let $E = [a,b]$. I have shown $1\implies 2$ and $2\implies 3$. Proving $3\implies 1$ would help me complete the argument full circle. Please let me know if the following makes sense (point out errors if any), and help me complete the proof.

$[1\implies 2]$: If $x_n\to x$ and $f$ is continuous, then $f(x_n)\to f(x)$, and hence $G(x_n)\to G(x)$ meaning $G$ is continuous. Does this need more argument?

$[2\implies 3]:$ The continuous image of a compact set is compact. Hence, $G([a,b])$, i.e. the graph of $f$ is compact in $\mathbb R^2$.

I found the following for $[3\implies 1]$:

Let $G([a,b])$ be compact and $x_n\rightarrow x$ be a convergent sequence in $[a,b]$. We show that $f(x_n)$ converges to $f(x)$. Since the graph is compact, $f(x_n)$ has a convergent subsequence,

Q1. $G([a,b])$ is compact, so $(x_n,f(x_n))$ has a convergent subsequence. How does this tell us that $f(x_n)$ has a convergent subsequence? I feel we need to add some details here.

i.e. $f(x_{n_j})\rightarrow y$. That is $(x_{n_j}, f(x_{n_j}))\rightarrow (x, y)$. The graph is closed. That is, the limit of every convergent sequence in $G([a,b])$ is again in $G([a,b])$. Therefore $(x,y)\in G([a,b])$, i.e. $y=f(x)$. Since this is true for every convergent subsequence, we showed that $f(x)$ is the only limit point of $f(x_n)$, i.e. $f(x_n)$ converges to $f(x)$.

Q2. How is the last sentence concluded? We choose an arbitrary convergent subsequence $f(x_{n_j}) \to y$, and found $y = f(x)$. What now?

My attempt for $[3\implies 1]$:

Suppose $G_f = \{(x,f(x)): x\in [a,b]\}$ is compact in $\mathbb R^2$. Define $\phi_1(x,y) = x, \phi_2(x,y) = y$. Suppose $S\subset\mathbb R$ is an arbitrary closed set in $\mathbb R$. $$f^{-1}(S) = \{x\in [a,b]: f(x) \in S\}$$
We can write $$f^{-1}(S) = \phi_1(G_f \cap \phi_2^{-1}(S))$$
The above expression seems good intuitively but I need help establishing it by showing $\subseteq$ and $\supseteq$ inclusions. Next, $S\subset \mathbb R$ is closed tells us that $\phi_2^{-1}(S) \subset \mathbb R^2$ is closed - since $\phi_1,\phi_2$ are continuous. $G_f \cap \phi_2^{-1}(S)\subset G_f$, and $G_f \cap \phi_2^{-1}(S)$ is closed. So, $G_f \cap \phi_2^{-1}(S)$ is compact. $\phi_1$ is continuous, which means $f^{-1}(S)$ is compact. Hence, $f^{-1}(S)$ is closed. Done!

Please help me fill in the gaps in my attempt, and let me know if you've any other ideas to solve the problem, such as $[3\implies 2]$ and $[2\implies 1]$. I'd also like to understand the proof posted above my attempt. Thank you!

答:
I will help you with your attempt for $[3 \implies 1].$ Your proof is quite interesting due to its versatility. You could use it to prove it for most topological spaces (the Hausdorff ones)!

Lemma: $ f^{-1}(S) = \phi_{1}(G_f \ \cap\ \phi_{2}^{-1}(S))$ for all set S.

Let $S$ be a set. We must prove two inclusions $ f^{-1}(S) \subset \phi_{1}(G_f \ \cap\ \phi_{2}^{-1}(S)) $ and $ \phi_{1}(G_f \ \cap\ \phi_{2}^{-1}(S)) \subset f^{-1}(S). $

$ f^{-1}(S) \subset \phi_{1}(G_f \ \cap\ \phi_{2}^{-1}(S)).$

Let $x \in E$ be such that $f(x) \in S,$ i.e., $x\in f^{-1}(S).$ Note that $(x,f(x))$ belongs to $G_f$ and $\phi_{2}(x,f(x))=f(x) \in S$ which yields $(x,f(x)) \in G_f \ \cap\ \phi_2^{-1}(S)$. Consequently, $x \in \phi_1 ( G_f \ \cap\ \phi_2^{-1}(S)).$

$ \phi_{1}(G_f \ \cap\ \phi_{2}^{-1}(S)) \subset f^{-1}(S) $

Let $x \in \phi_{1}(G_f \ \cap\ \phi_{2}^{-1}(S)),$ there must exist an $y \in G_f \ \cap\ \phi_{2}^{-1}(S)$ such that $\phi_1 (y) = x.$ Since $y \in G_f,$ there must exist $x'\in E$ such that $y = (x', f(x'))$ and, due to $y \in G_f \ \cap\ \phi_{2}^{-1}(S)$, $\phi_2(y)=\phi_{2}(x',f(x')) = f(x') \in S$. Hence, $x' \in f^{-1}(S),$ but, since $y = (x', f(x'))$, we have $ x' = \phi_1(x',f(x')) = \phi_1(y) = x $. Hence,  $x \in f^{-1}(S). $

Proposition: $\{(x,f(x)): x\in E\}$ is compact in $\mathbb R^2 \implies f:E\to\mathbb R$ is continuous

Suppose $G_f = \{(x,f(x)): x\in E \}$ is compact. Define $\phi_1(x,y) = x, \phi_2(x,y) = y$ for all $x \in E$ and $y \in \mathbb{R}$. It's enough to show that inverse image (of $f$) of closed set is closed.

Suppose $S\subset\mathbb R$ is an arbitrary closed set in $\mathbb R$. We have that $$f^{-1}(S) = \{x\in E: f(x) \in S\}$$ and we can write $$f^{-1}(S) = \phi_1(G_f \cap \phi_2^{-1}(S)),  $$ due the above lemma.  Now, the continuity of $\phi_2$ implies that $\phi_2^{-1}(S)$ is closed. Since $\phi_2^{-1}(S)$ is closed and intersection of closed set with a compact is compact, $G_f \cap \phi_2^{-1}(S)$ is compact. Since $\phi_1$ is continuous, it maps compact set to compact set, which means $f^{-1}(S)$ is compact. Hence, the inverse image of closed set is always closed, as desired.


问:
Converse and The Negation of The Converse of The Statement: "For all positive real numbers x, there exists an integer n such that $\frac{1}{n} <x.$"
I am trying to express the converse and the negation of the converse of the following statement both in plain English language and in formal language.

"For all positive real numbers $x$, there exists an integer $n$ such that $\frac{1}{n} <x.$"

Given that the universe for both $x$ and $n$ are the set of all real numbers, I have expressed the statement and its negation  in formal language respectively as follow:

\begin{align*}
\forall x,(x\in \mathbb{R}^+ \rightarrow \exists n,(n\in \mathbb{Z} \wedge \frac{1}{n} < x)). \\
\exists x,(x \in \mathbb{R}^+ \wedge \forall n,(n\in \mathbb{Z} \rightarrow \frac{1}{n}\geq x)).
\end{align*}

However, I am stuck on how to express the converse and the negation of the converse of the original statement.

Here is my attempted expressions for the converse and the negation of the converse of the original statement respectively.

\begin{align*}
\exists n,((n\in \mathbb{Z} \wedge \frac{1}{n} < x) \rightarrow \forall x,(x\in \mathbb{R}^+)).\\
\forall n,((n\in \mathbb{Z} \wedge \frac{1}{n} < x) \wedge \exists x,(x \in \mathbb{R}^+)).
\end{align*}

Each of the above statements is expressed in plain English language as:

\begin{align*}
&\text{"If there exists an integer $n$ such that $\frac{1}{n} < x$, then for every $x$, $x$ is a positive real number."}\\
&\text{"Every integer $n$ has some positive real number $x$ such that $\frac{1}{n}<x$."}
\end{align*}

I am not sure whether the above expressions for the converse and the negation of the converse of the original statement both in formal language and in plain English language are correct or not.

答:
The correct answer to the converse of the statement in formal language is:

\begin{align*}
\forall x,(\exists n,(n\in \mathbb{Z} \wedge \frac{1}{n} < x) \rightarrow (x \in \mathbb{R}^+)).
\end{align*}

which is expressed in plain English language as:

\begin{align*}
\text{For every real number $x$, if there exists an integer $n$ such that $\frac{1}{n} < x$, then $x$ is positive.}
\end{align*}

The correct answer to the negation of the converse of the statement in formal language is:

\begin{align*}
\exists x,(\exists n,(n\in \mathbb{Z} \wedge \frac{1}{n} < x) \wedge (x \notin \mathbb{R}^+)).
\end{align*}

which is expressed in plain English language as:

\begin{align*}
\text{There exists a real number $x$ such that there is an integer $n$ which $\frac{1}{n} < x$ and $x$ is not positive.}
\end{align*}


答:
Thought I would throw in the 3 by 3 case, which is worth remembering:

The eigenvalues are the roots of
$$   \lambda^3 - T \lambda^2 + P \lambda - D $$
where trace
$$T = a_{11} +a_{22} +a_{22} $$
and the sum of the two by two principal minors is
$$  P = \left( a_{11} a_{22} -  a_{12} a_{21} \right) +  \left( a_{22} a_{33} -  a_{23} a_{32} \right) +  \left( a_{11} a_{33} -  a_{13} a_{31} \right) $$

Finally the determinant $D$ is given by the "Rule of Sarrus" https://en.wikipedia.org/wiki/Rule_of_Sarrus which is also worth memorizing

The point is that it is not necessary to write out the determinant of $A-\lambda I$ every time. My suggestion is to memorize the coefficients of $\det \lambda I - A$  as sums of certain little determinants. If nothing else, it reduces errors by hand.

Title: Eigenvalues of general $2\times2$ matrix

Body:
I am wondering what the expression is for eigenvalues of a $2\times2$ matrix in general form

$$A=\left[
\begin{array}{ccc}
 a_{11} & a_{12}\\
a_{21} & a_{22}
\end{array}
\right]$$

I tried to compute $|A-\lambda I| = 0$ but it got very messy.


答:
Comment:As commented user376343 we have:

$$B=-A\times A=-[a+b+c-(\frac ab+\frac bc+\frac ca)]^2=-(\frac ab+\frac bc+\frac ca)^2$$

$$\frac ab+\frac bc+\frac ca=\frac{ca^2+ab^2+bc^2}{abc}\geq3$$

This inequality becomes equality if $a=1, b=2$ and $c=3$. I checked it for $a+b+c=0$ and it is not equal to 3. It is a negative value.I think correct constrain is : a, b and c positive and product must have a negative sign and the sign of inequality must be $\geq$.

Title: How can I prove this for $a+b+c=0$

Body:
We wish to prove that

$$\left(\frac{a-b}{c}+\ \frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9,$$ for $a+b+c=0$ and $$abc \neq 0 , a\neq b , b\neq c, c \neq a.$$
I tried working with only $2$ fractions at a time, switching some of the variables (ex.$a=-b-c$) and then adding them up, and multiplying first but could not go further.

Thank you in advance.


问:
Is the stereographic projection a homomorphism between the sphere and the plane?
I am taking a course on complex analysis, and we define the stereographic projection. Isn't this an onto and $1-1$ continuous mapping from the sphere to the plane ? Meaning that it exist a homomorphism between the sphere and the plane ? But if I remember correctly in differential geometry we said that such homomorphism cant exist. At least I know for a fact that they have different Gauss curvature.

答:
No.  The sphere is compact,  and the plane isn't,  so there can be no such homeomorphism.

Stereographic projection is a homeomorphism between the sphere minus a point and the plane.


答:
Let $(E_n)$ be a partition of $\mathbb R$ such that $λ(E_n)<\infty$, where $λ$ denotes the Lebesgue measure. Consider the functions $ u_n = (\frac{1}{λ(E_n)})^{1/p}  \ \textbf{1} _{E_n}$. Notice that $ ||{u_n}||_p =1 $ and $ {u_n} \mid_{E_j}=0$ for $j \neq n$. Show that for any $a_1, \dots , a_n \in \mathbb R$,
$$ ||\sum_{k=1}^{n} a_k u_k||_p^p= \sum_{k=1}^n |{a_k}|^p .$$
Indeed, for $j>n$,

$$\left ({\sum_{k=1}^n a_k u_k } \right)\mid_{E_j}=0 $$
and for $1 \leq j \leq n$,

$$ ({\sum_{k=1}^n a_k u_k }) \mid_{E_j}=\frac{a_j}{\lambda(E_j)^{1/p}} .$$
Hence,
$$||{\sum_{k=1}^{n} a_k u_k}||^p_{p} = \int |\sum_{k=1}^n a_k u_k|^p = \sum_{j=1}^{\infty}  \int_{E_j} |{\sum_{k=1}^n a_k u_k}|^p = \sum_{j=1}^n \int_{E_j} \frac{|a_j|^p}{\lambda(E_j)} = \sum_{j=1}^n |{a_j}| . $$

Conclude that for any $  x =(x(n))_n \in \ell_p $, we have that
$$ ||\sum_{n=1}^{\infty} {x(n) u_n}||^p_{p} = \sum_{n=1}^{\infty} |{x(n)}|^p $$
and thus the map
$$ \ell_p \ni  x \mapsto \sum_{n=1}^{\infty} x(n) u_n \in L_p $$
is an  isometry.

Body:
For $1 \leq p \leq \infty$, is it true that $L^p(\mathbb{R})$ contains a subspace which is is isometrically isomorphic to $\ell^p$ ?

I don't think it's true. But I can't find any counterexamples.


问:
Application of Cauchy product rule
Let be $\sum\limits_{j=0}^{\infty}a_j$ and $\sum\limits_{k=0}^{\infty}2^{-k-1}$ two (absolute) convergent series. Then, the Cauchy product rule (https://en.wikipedia.org/wiki/Cauchy_product#:~:text=In%20mathematics%2C%20more%20specifically%20in,French%20mathematician%20Augustin%20Louis%20Cauchy.) allows us to rearrange as follows:

$\sum\limits_{k=0}^{\infty}2^{-k-1}\sum\limits_{j=0}^{\infty}a_j=\sum\limits_{k=0}^{\infty}\sum\limits_{j=0}^{k}a_j2^{-(k-j)-1}=\sum\limits_{k=0}^{\infty}\sum\limits_{j=0}^{k}2^{j-(k+1)}a_j$.

Hence, $\sum\limits_{k=0}^{\infty}\sum\limits_{j=0}^{k}2^{j-(k+1)}a_j$ converges absolutely.

My tutor said that I mixed up the indices but I don't see where?

答:
Adding an intermediate step we obtain
\begin{align*}
\color{blue}{\sum_{k=0}^\infty 2^{-k-1}\sum_{j=0}^\infty a_j}
&=\sum_{n=0}^\infty\sum_{\substack{k+j=n\\k,j\geq 0}}2^{-k-1}a_j\\
&=\sum_{n=0}^\infty\sum_{j=0}^n2^{-(n-j)-1}a_j\\
&\,\,\color{blue}{=\sum_{k=0}^\infty \sum_{j=0}^k2^{j-(k+1)}a_j}
\end{align*}
in accordance with OPs result. So, the calculation looks fine (+1).

Title: Application of Cauchy product rule


问:
convergence in $\ell_{1}$ implies convergence in $\ell_{p}$
Assume that we have a sequence of non-negative values $x_{n}$ in $\ell_{1}$. Next, assume that the sequence $x_{n}$ converges to $c$ in $\ell_{1}$.

Is the following correct: the sequence is this converge in $\ell_{p}$, for any $p >1$?

Try: the statement is correct, since $||x||_{p} \leq ||x||_{q}$, if $q \leq p$.

答:
Yes, now it is true: it holds that $l_{1}\subsetneqq l_{p}$ for every $p$. Observe that being convergent in $l_{1}$ means the general term $x_{n}$ is infinitesimal as the series $\sum_{n=0}^{+\infty}x_{n}$ is convergent and we apply Cauchy's criterion. This means that for $n$ large enough, $|x_{n}|^{p}<|x_{n}|$, thus if we sum now:
$$ \sum_{n=0}^{+\infty}|x_{n}|^{p} < \sum_{n=0}^{+\infty}|x_{n}| $$
Since the series on the right is convergent, so it must do the series on the left, thus it coverges in $l_{p}$ for every $p$.


问:
Order of an element given a homomorphism

If  $G_1,G_2$ are groups, $\varphi:G_1 \rightarrow G_2$ is a homomorphism and $a \in(G_1)$, is an element of finite order, what can we say for the order $\varphi(a)$?

My attempt so far:
Let $\operatorname{ord}(a)=r$; Then $(a^r)=e_1 $. But can I use this  to claim that: ($\varphi(a)^r$)$=e_2$?

答: Yes, you can claim that because $\varphi$ is a homomorphism, hence $e_2=\varphi(a^r)=(\varphi(a))^r$ and then you can deduce that the order of $\varphi(a)$ divides $r$.


问:
When is a vector field the gradient of the pressure
In Frank White's book Fluid mechanics problem 4.27 (8th ed.) a 2D-velocity field is given as

$$
\vec{v} = (2xy,-y^2)
$$

Using Euler's equation (assuming stationary, incompressible flow and no gravitational field)

$$
\rho\vec{v}\cdot\nabla\vec{v} = -\nabla p
$$

The pressure gradient is determined to be

$$
\nabla p = (2xy^2, 2y^3)
$$

But if we set $\nabla p = (P,Q)$ then

$
\partial_y P \neq \partial_x Q
$

and $\nabla p$ is not conservative and there cannot exist a pressure field $p$ compatible with the provide velocity field. Hence, the velocity field provide cannot be a solution to Euler's equation with the given assumptions. Is this correct (and hence to problem statement wrong)?

答:
The velocity field with components $u = 2xy$ and $v = -y^2$ does satisfy $\nabla \cdot \mathbf{v} = \frac{\partial u}{\partial x} + \frac{\partial v }{\partial y} =0$ and so could represent a steady, two-dimensional  and incompressible flow. The flow is not irrotational, however,  and not  a potential flow, since the vorticity component in the direction perpendicular to the plane of flow does not vanish:

$$ \omega_z =  \frac{\partial v}{\partial x} - \frac{\partial u }{\partial y} = -2x$$

As you determined, it does not provide a solution to the steady, two-dimensional Euler equations since the pressure field would have to satisfy

$$\frac{\partial p}{\partial x} = -\rho\left(u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y}\right)= -2\rho xy^2, \quad \frac{\partial p}{\partial y} = -\rho\left(u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y}\right)= -2\rho y^3,$$

and there is no scalar field $p$ where $\nabla p$ has those components.

What this means is that there can be no steady, two-dimensional, incompressible, rotational  and inviscid velocity field of this form. Physically no such flow with non-zero vorticity could evolve from rest by the imposition of a pressure field. There can be no production or diffusion of vorticity in two-dimensional, incompressible inviscid flow since all of the terms on the RHS of the vorticity equation are zero. That is

$$ \frac{D \mathbf{\omega}}{Dt} = \frac{\partial \mathbf{\omega}}{\partial t}+ \mathbf{v} \cdot \nabla \mathbf{\omega}= \omega \cdot \nabla \mathbf{v} + \nu \nabla^2\mathbf{v} = 0,$$

since in inviscid flow $\nu = 0$ and in two-dimensional flow we have $\mathbf{\omega}\cdot \nabla \mathbf{V} = \omega_z \frac{\partial \mathbf{v}}{\partial z}= 0$ (no vortex line stretching).

Also note that if  the given velocity field was a steady solution of the Euler equations we would have

$$0 = \mathbf{v} \cdot \nabla \mathbf{\omega} = u\frac{\partial \omega_z}{\partial x} + v \frac{\partial \omega_z}{\partial y} = (2xy)(-2) +(-y^2)(0) = -4xy,$$

which is impossible.


问:
Groups acting on polynomials
I am a little confused about actions right now.

Say I have a group $G$ that acts on the roots of a polynomial, say $(x - x_1)(x - x_2)(x - x_3)$.  We know that this group permutes the roots of a polynomial, that is you can get to any of the roots by having an element of the group act on another root.  Does that mean the group fixes the polynomial?

I feel like it should, but I'm not sure how this actually works.  Supposedly we would have to write $g \cdot [(x - x_1)(x - x_2)(x - x_3)] = (x - x_1)(x - x_2)(x - x_3)$.  But is this actually possible?  Because $G$ acts on the set of the roots, not the polynomial.  I'm not sure if we can say a polynomial is fixed by certain things acting on the roots.  When they say the polynomial is fixed, does that mean the coefficients are fixed?

Also, can you say something like $g \cdot (x_1x_2 + x_3)$ = $(g \cdot x_1)(g \cdot x_2) + g \cdot x_3$?

Thanks!

答:
The coefficients of the polynomial are symmetric functions of the roots. In your particular case the polynomial is
$$ x^3 - (x_1+x_2+x_3)x^2 + (x_1x_2 + x_1x_3 + x_2x_3) x + x_1x_2x_3. $$
Here $x$ is a formal indeterminate - your group doesn't act on it. So all that matters is that the symmetric functions are invariant under permutations of their arguments.</p>

Title: Groups acting on polynomials