问:
Compact metric space group $\operatorname{Iso}(X,d)$ is also compact
Could you tell me how to prove that if metric space $(X,d)$ is compact, then the group $\operatorname{Iso}(X,d)$ is also compact?

The group $\operatorname{Iso}(X,d)$ is considered with topology determined by a metric $\rho$ on $\operatorname{Iso}(X,d)$ such that
$\lim _{n \rightarrow \infty} \rho(h_n, h) =0 \iff \forall x \in X: \lim _{n \rightarrow \infty} d(h_n(x), h(x))=0$

答:
Without the precise definition of the metric $\rho$, I'm not sure how to answer this definitively, but let me outline an approach that you might be able to take. For this approach, I take as a hidden assumption that the Principle of Dependent Choices holds, so to show that $Y:=\text{Iso}(X,d)$ is compact (considered as a metric space with metric $\rho$), it suffices to show that it is complete and totally bounded.

To show that $\langle Y,\rho\rangle$ is a complete metric space, suppose that we have a $\rho$-Cauchy sequence of functions $h_n\in Y.$ Show that for each $x\in X,$ we have that $h_n(x)$ is a $d$-Cauchy sequence of points of $X$. Since $X$ is a compact space under the $d$-metric topology, then it is complete, so $h_n(x)$ converges in $X$. Define $h:X\to X$ by $h(x)=\lim_{n\to\infty}h_n(x).$ Show that $h\in Y,$ and that $\lim_{n\to\infty}\rho(h_n,h)=0,$ so $\langle Y,\rho\rangle$ is a complete metric space.

To show that $\langle Y,\rho\rangle$ is a totally bounded metric space, we must show that for all $\epsilon>0$ there exist $f_1,...,f_n\in Y$ such that $Y$ is covered by the open $\rho$-balls about $f_1,...,f_n$ of radius $\epsilon$. Without a definition for $\rho,$ I'm afraid I have no suggestions for how to proceed with this.

Edit: If julien is correct in his comment above, that you're simply considering $Y$ in the topology of pointwise convergence (which, in retrospect, it seems that you are), then I'm not certain that it's necessarily induced by a metric. Fortunately, we can proceed without worrying about such a metric.

Let's proceed as follows, instead. Denote the set of all functions $X\to X$ by $X^X$. Given a point $x\in X$ and an open set $U$ in $X,$ we define $$P(x,U):=\{f\in X^X:f(x)\in U\},$$ and let $\mathcal B$ be the set of all sets of the form $$P(x_1,B_1)\cap\cdots\cap P(x_k,B_k)$$ for some positive integer $k$, with the $x_j\in X$ and the $B_j$ open balls in $X$. You can show that $\mathcal B$ is a basis for a topology on $X^X$, say $\mathcal T$, which is the set of unions of $\mathcal B$-sets. We call $\mathcal T$ the topology of pointwise convergence, because of the following

Lemma: If $h_n$ is a sequence of functions $X\to X$ and $h:X\to X$, then $h_n\to h$ with respect to the topology $\mathcal T$ if and only if $h_n(x)\to h(x)$ for each $x\in X$.

Proof: Suppose $h_n(x)\to h(x)$ for all $x\in X$, and take any $\mathcal B$-basic neighborhood of $h$, which will be of the form $$P(x_1,B_1)\cap\cdots\cap P(x_k,B_k)$$ for some $x_j\in X$ and some open balls $B_j$ with $h(x_j)\in B_j$. Since $h_n(x_j)\to h(x_j)$, then there exists $n_j$ such that $h_n(x_j)\in B_j$ for $n\ge n_j$. Putting $N=\max\{n_1,...,n_k\}$, we therefore have for $n\ge N$ that $$h_n\in P(x_1,B_1)\cap\cdots\cap P(x_k,B_k).$$ Thus, $h_n\to h$ with respect to $\mathcal T$.

On the other hand, suppose that $h_n\to h$ with respect to $\mathcal T$. Take any $x\in X$ and any open ball $B$ in $X$ such that $h(x)\in B$. Since $P(x,B)$ is a $\mathcal T$-neighborhood of $h$ and $h_n\to h$, then there is some $N$ such that $h_n\in P(x,B)$ for $n\ge N$, meaning $h_n(x)\in B$ for all $n\ge N$. Thus, $h_n(x)\to h(x)$ for all $x\in X$. $\Box$

Now, julien refers to Tychonoff Theorem (a very useful theorem that requires the Axiom of Choice to prove) in his comment above. We really only need the following special case:

Theorem: $X^X$ is compact in the topology of pointwise convergence if (and only if) $X$ is compact.

The "only if" part is fairly easy to prove, and if you're interested, I can prove the other direction (though it can be a bit of a bear to prove without just using another form of the Tychonoff Theorem).

Hence, all that is left for you to do is to show that $Y$ is a closed subspace of $X^X$, from which the result follows.


问:
The Fourier Transform of a Bessel function: $J_a(be^{-x}), a,b>0$
Currently we require the Fourier transform  $g(p)$ of  $f(x)=J_a(be^{-x}), a,b>0$
$$g(p)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}e^{ipx} f(x) dx$$
We could not find it in Tables and Handbooks. Mathematica gives it in terms of Gamma functions and Gauss Hyper geometric function. A closer look suggested that Hypergeometric function was actually a hoax. As we were finally interested in $|g(p)|^2$, it was a pleasant surprise to get
$$|g(p)|^2=\frac{1}{2\pi(p^2+a^2)},~~~~~(1)$$
which is independent of $b$ !, even the verification of (1) by

Mathematica in analytic mode  for various sets of $a,b,p$ has been time consuming.

The question is how to prove (1) and get $g(p)$ by hand?

EDIT:
By Integrate of  Mathematica, I finally get
$$g(p)=\frac{1}{2\sqrt{2\pi}} \left(\frac{2}{b}\right)^{ip} \frac{\Gamma[(a+ip)/2]}{\Gamma[1+(a-ip)/2]}$$
EDIT:
Here is the Mathematica out put, see the pre=factor of hypergeometric function is 0.

g(p)=1/2 Gamma[
1/2 (a + I p)] ((2^(I p) (1/b^2)^((I p)/2))/
Gamma[1/2 (2 + a - I p)] +
2^-a (-(1/b^2)^(-a/2) + b^
a) HypergeometricPFQRegularized[{1/2 (a + I p)}, {1 + a,
1/2 (2 + a + I p)}, -(b^2/4)])

答:
A Bessel function (of any kind) $f(z)$ satisfies the ODE
$$
w^2 f''(w) + w f'(w) + (w^2 - a^2) f(z) = 0.
$$
Making the variable transformation $w = be^{-x}$ leads to the following ODE satisfied by $f(x) = J_a \left(be^{-x}\right)$:
$$
f''(x) + \left(b^2 e^{-2x} - a^2 \right) f(x) = 0
$$
Now we can take the Fourier transform of this ODE. Assuming that the Fourier transform $g(p) = \mathcal{F}\left[ f(x) \right]$ can be analytically continued into the complex plane as
$$
g(z) = \int_{-\infty}^{\infty} e^{izx} f(x) \, dx,
$$
the Fourier transform of $e^{-2x} f(x)$ is
$$
\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{ipx} e^{-2x} f(x) \, dx
= \int_{-\infty}^{\infty} e^{i(p+2i)x} f(x) \, dx = g(p+2i).
$$
With this ODE is transformed into
$$
-p^2 g(p) + b^2 g(p+2i) - a^2 g(p) = 0
$$
or more concisely
$$
g(p+2i) = \frac{p^2 + a^2}{b^2} g(p).
$$
This recurrence can be solved with a little guidance from Mathematica. Letting $g(p) = h\left(\frac{p}{2i}\right)$ and $a = 2c$, the recurrence for $g$ implies the following for $h$:
$$
h(z+1) = 4 \frac{c^2-z^2}{b^2} h(z)
$$
Assuming separability of $h(z) = h_1(z) h_2(z) h_3(z)$ where each satisfies a recurrence $h_1(z+1) = \frac{4}{b^2} h_1(z)$, $h_2(z+1) = (c+z) h_2(z)$, and $h_3(z+1) = (c-z) h_3(z)$,
each of these has the solution, for non-negative $z$ and $c$ and sufficiently large $z$ (i.e. $z\geq c+1$),
$$
h_1(z+1) \propto \left( \frac{4}{b^2} \right)^z \\
h_2(z+1) \propto (z+c-1)! \\
h_3(z+1) \propto (-1)^z (z-c-1)!
$$
or for non-positive integral $z$ and $c$ the alternate possibilities
$$
h_2(z+1) \propto (-1)^z \frac{1}{(-c-z)!}, \quad
h_3(z+1) \propto \frac{1}{(c-z)!}
$$
This implies that there is no unique solution to this recurrence relation. Though technically this gives two possible solutions for integral ranges of $z$ and $c$, one using the first pair of $h_{2,3}$ solutions and the second using the second pair, the possible forms for $h_2$ and $h_3$ can be mixed and matched to still yield an expression that formally satisfies the recurrence. Any resulting such expression can be analytically continued to the complex plane. Edit: the possibilities for $h_{2,3}$, including a term $(-1)^z$ that alternates in sign across the integers, are somewhat anomalous and have less natural or unique analytic continuations to the plane.

As an example, take the very non-arbitrary choice of the first $h_2$ solution and second $h_3$ solution:
$$
h(z) \propto \left( \frac{4}{b^2} \right)^z \frac{\Gamma(c+z)}{\Gamma(c-z+1)}
$$
and hence
$$
g(p) \propto \left(\frac{b}{2} \right)^{ip} \frac{\Gamma\left(\frac{a-ip}{2}\right)}{\Gamma\left(1+\frac{a+ip}{2}\right)}
$$
The proportionality factor can be fixed using the fact that $g(0) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} J_a(be^{-x})\, dx = \int_0^{\infty} \frac{J_a(u)}{u} \, du$. This is independent of $b$ and equal to $\frac{1}{a\sqrt{2\pi}}$, yielding the final result
$$
g(p) = \frac{1}{2\sqrt{2\pi}} \left(\frac{b}{2} \right)^{ip} \frac{\Gamma\left(\frac{a-ip}{2}\right)}{\Gamma\left(1+\frac{a+ip}{2}\right)}
$$

Note that this is identical to the OP's proposed solution up to the substitution $p \mapsto -p$; see below comment on whether there is a possible sign disagreement. This result also satisfies the cross check that the $L^2$ norm of $g(p)$ seems to equal to that of $f(x)$, based on some numerical integrals for a few different choices of $a$ and $b$.

However, a few major limitations are that

The solutions to the recurrence are not unique, and mixing and matching the solutions for $h_2$ and $h_3$ could lead to well-defined alternatives. I happened (after trial and error and multiple edits to the answer to streamline it) to pick the one of the four possibilities that corresponded to the answer provided by the OP (up to a sign change for $p$). I also can't guarantee that there aren't other functional forms that formally solve the recurrence.

The original ODE has two free parameters that are fixed to the specify the Bessel $J$ functions, which must translate to two degrees of freedom in $g(p)$ -- but I can only identify one of them (the proportionality factor), and I'm curious where the other one went.

The OP has obtained an actual result for the Fourier transform which disagrees with my results, but does, in fact, satisfy the recurrence
$$
g(p-2i) = \frac{p^2+a^2}{b^2} g(p).
$$
This leads me wonder if I have a sign error in the above work, but I can't find any. However, the rescaling argument made by @skbmoore does imply that the entire dependence of $g(p)$ on $b$ should be through a term like $b^{ip}$ (and not $b^{-ip}$), by making the change of variables $x = x' + \log(b)$ that maps $J_a(be^{-x})$ to $J_a(e^{-x'})$, and making this substitution in the Fourier transform creates a constant term $e^{ip\log(b)}$.

ContentLicense: CC BY-SA 4.0


问:
Maximum value of modulus in exponential form
I am trying to find the maximum value of
$\left|e^{i\theta}-2\right|+\left|e^{i\theta}+2\right|\mbox{ for }0\le\theta\le2\pi$.

I can replace $e^{i\theta}$ with $a+ib$ and then proceed to get a function of $a$;
Namely;

$$\sqrt{5+4a}+\sqrt{5-4a}$$
I can then use calculus to find the maximum value of this function of $a$.

My question is: Are there any more efficient ways to solve this problem?

答:
$f(x)=\sqrt{x}, f''(x)=-\frac{1}{4x^{3/2}}<0$ so $f$ is strictly concave. By Jensen's inequality,

$$f(5+4a)+f(5-4a)\le 2f\left(\frac{(5+4a)+(5-4a)}{2}\right)=2f(5)=2\sqrt{5}$$

We have equality at $a=0, \theta  = \frac{\pi}{2} \text{ or } \frac{3\pi}{2}$.


问:
Proof of fixed point theorem, or "covering interval theorem"
I took a discrete dynamical systems class last semester, and there is still something bothering me. We had a homework question asking us to prove the following (the "covering interval theorem"):

\begin{gather*}
F: \mathbb{R} \rightarrow \mathbb{R} \; , \; F \in C^{\geq 1} \; , \; I \subset \mathbb{R} \\
F(I) \supset I \implies \exists x \in I \; | \; F(x) = x
\end{gather*}
or in other words, for a continuous real-valued function $F$, show that if there is some interval $I$ where the image of $I$ under $F$ covers $I$, that is $I$ is a subset of $F(I)$, then there exists a fixed point.

I found this problem very unsettling because it seems so obvious, but my professor insisted that there were multiple cases and how it was not trivial, but he never revealed the solution.

My idea was that since $F$ is continuous and $F(I) \supset I$, then

\begin{gather}
\exists J \subset I \; | \; F(J) = I
\end{gather}

and low and behold you can use the intermediate value theorem on that fact:

\begin{gather*}
J = [a,b] \; , \; I = [c,d]\\
J \subset I \implies c \leq a < b \leq d \\
F(J) = I \implies c \leq F(a) \leq d \; \land \; c \leq F(b) \leq d\\
G(x)\equiv F(x) - x \in C^{\geq 1} \implies  G(a) \geq 0 \; \land \; G(b) \leq 0
\end{gather*}
so that completes the proof since if $G = 0$ at $a$ or $b$, then those are the fixed points, and if the strict inequalities hold, by the IVT we have $G = 0$ somewhere on the interval.

So my question is, why is this not a complete proof, or is it? Also does anyone know how to prove the statement about the existence of $J$?

Sorry if this is similar to other questions, one came up that was quite similar, but I'm looking for more explanation here.

Thanks!

答:
You are right, but in fact even your argument can be made a bit shorter:

Write $I=[c,d]$. Since $F(I) \supset I$, there exist $a,b \in I$ such that $F(a) = c \le a$ and $F(b)=d\ge b$. Since $F(a) − a \le 0$ and $F(b) − b \ge 0$, the continuous function $F(x) − x$ has a zero in $I$.


问:
If $f$ is Schwartz, does $f=\mathcal{O}\left(e^{-a|x|^{\epsilon}}\right)$ for some $a,\epsilon>0$?
If $f$ is Schwartz, i.e. $f\in\mathcal{S}\left(\mathbb{R}\right)$, is it true that$f=\mathcal{O}\left(e^{-a|x|^{\epsilon}}\right)$ for some $a,\epsilon>0$ as $|x|\to\infty$?

Functions in the Schwartz class, as well as as their derivatives of all orders, decay faster than any negative power of a monomial. I would expect that the answer to the question is yes, but I am not sure how to start thinking about the positivity or negativity of the statement.

The standard examples of Schwartz functions, such as $e^{-x^2}$, $x^2e^{-x^2}$, and $e^{-x^4}$ all satisfy this bound.

We do know that since $f$ is Schwartz, $f=\mathcal{O}\left(\frac{1}{|x|^m}\right)$, for all $m\in\mathbb{N}$. Additionally, $e^{-a|x|^{\epsilon}}=\mathcal{O}\left(\frac{1}{|x|^m}\right)$ for all natural $m$ and $a, \epsilon > 0$. This means that both $\mathcal{S}$ and $e^{-a|x|^{\epsilon}}$ are contained in $\displaystyle \bigcap_{m=1}^\infty \mathcal{O}\left(\frac{1}{|x|^m}\right)$ (again, for all $a, \epsilon > 0$). This inclusion at least lets us understand where these functions live a little better, but does not answer the initial question. What is an approach that could shed light on whether the statement in the question holds true?

(As a reminder, we say that $f(x)\in \mathcal{O}\left(g(x)\right)$, or equivalently $f(x)= \mathcal{O}\left(g(x)\right)$, if, beyond a sufficiently large value of $x$, $|f|$ is at most a constant times $g$.)

答:
No.  For example, if $h(x)$ is a smooth "bump function" with support in $(0,1)$ you could
take $$f(x) = \sum_{n=2}^\infty n^{-\log n} h(x-n)$$


问:
Variational calculus for functional equation results in under-determined system.
I was attempting to use the calculus of variations to find the unique function $f$ such that

f is smooth

$f(1) = 1, f(2) = 3$

$f(x+1) - f(x) = (x+1)$

$f$ minimizes its arclength from $1$ to $2$.

Now I know that in general $$f(x) = \frac{1}{2}x(x+1) + \sum_{n=0}^{\infty} \left[ a_n \sin(2\pi nx) \right] $$ (for $a_n \in \mathbb{R} $) can be made to satisfy conditions 1 through 3. But I imagine that condition (4) would cut down our solution space strictly to $\frac{1}{2}x(x+1)$.

So I begin setting up our problem. We have a constraint $g(x,f)=0$ given by:

$$ g(x,f) = f(x+1) - f(x) - (x+1)$$

So we define our lagrangian

$$ \mathcal{L}(x,f) = \sqrt{1+(f')^2} - \lambda(x)(f(x+1) - f(x) - (x+1)) $$
So we setup a calculus of variations problems with lagrange multiplier $\lambda(x)$ that is we want to find a smooth choice of $f(x), \lambda(x)$ which minimizes

$$ \int_{1}^{2} \sqrt{1 + (f'(x))^2}  - \lambda(x)(f(x+1) - f(x) - (x+1)) dx$$

The stationary solutions to this will be those satisfying

$$ \frac{\delta \mathcal{L}}{\delta f} = 0 \\ \frac{\delta \mathcal{L}}{\delta \lambda} = 0 $$

Recall that:

$$\frac{\delta \mathcal{L}}{\delta f} = \frac{\partial \mathcal{L} }{\partial f} - \frac{d}{dx} \left[ \frac{\partial \mathcal{L} }{\partial f'}\right] + \frac{d^2}{dx^2} \left[ \frac{\partial \mathcal{L} }{\partial f''}\right] - .... $$

And Recall that:
$$f(x+1)-f(x)= f'(x) + \frac{1}{2!}f''(x) + \frac{1}{3!}f'''(x)  .... $$

So putting this together and substituting into our integral we have

$$\frac{\delta \mathcal{L}}{\delta f} = \frac{d}{dx} \left[ \frac{f'}{\sqrt{1+(f')^2}} \right] +\lambda'(x) - \frac{1}{2!} \lambda''(x) + \frac{1}{3!} \lambda'''(x) ...  = 0$$

$$ \frac{\delta \mathcal{L}}{\delta \lambda} = f(x+1)-f(x)-(x+1)=0$$

And now here is the problem. For ANY choice of $f$, it will always be possible to find a choice of $\lambda$ so that the first equation is true. And the second equation, after adding the boundary conditions on integer $f$ still has the same large solution set we mentioned at the start of this problem.

Where did I go wrong here? I know that these solutions CANNOT ALL be locally optimal since I can perturb any one solution of our original functional equation by a small SPECIFIC test function: $\epsilon \sin(2\pi nx) $ and get another solution, and these solutions will have obviously different arclengths.

答:
You are looking for a thing which is simply not there.

Obviously, the non-smooth solution is the piecewise linear function ($2x-1$ for $1<x\leqslant2$, then $3x-3$ for $2<x\leqslant3$, and so on). Also obviously, we can approximate it with smooth functions on $[1,2]$ as good as we want, both in terms of $f$ and the arc length. Together, these facts mean that the desired minimum is never reached in the class of smooth functions. This is not uncommon for a variational problem.

So it goes.


问:
Check the pointwise and the uniform convergence of $(f_n)_n:D\rightarrow \mathbb{R}$
Check the pointwise and the uniform convergence of $(f_n)_n:D\rightarrow \mathbb{R}$ with

$\displaystyle{f_n(x)=\frac{nx}{1+n^2x^2} \ \text{ for } x\in D=[-1,1]}$

$\displaystyle{f_n(x)=\sqrt{\frac{1}{n^2}+x^2} \ \text{ for } x\in D=\mathbb{R}}$

$\displaystyle{f_n(x)=\sqrt[n]{x} \ \text{ for } x\in D=(0,1)}$

$$$$

I have done the following:

If $x=0$ then we have that \begin{equation*}f(0)=\lim_{n\rightarrow +\infty}f_n(0)=\lim_{n\rightarrow +\infty}\frac{n\cdot 0}{1+n^2\cdot 0^2}=\lim_{n\rightarrow +\infty}\frac{0}{1}=0\end{equation*}
If $x\neq 0$ then we have that
\begin{align*}f(x)&=\lim_{n\rightarrow +\infty}f_n(x)=\lim_{n\rightarrow +\infty}\frac{nx}{1+n^2x^2}=\lim_{n\rightarrow +\infty}\frac{n^2\cdot \frac{x}{n}}{n^2\left (\frac{1}{n^2}+x^2\right )}\\ & =\lim_{n\rightarrow +\infty}\frac{ \frac{x}{n}}{\frac{1}{n^2}+x^2}=\frac{ 0}{0+x^2}=0\end{align*}
So we have pointwise convergence on $[-1,1]$ to $f(x)=0$.
The derivative is equal to $f_n'(x) =\frac{n-n^3x^2}{(1+n^2x^2)^2}$.
It holds that $f_n$ is decreasing on $\left [-1, -\frac{1}{n} \right )$ and $\left (\frac{1}{n}, 1\right ]$ and increasing on $\left [-\frac{1}{n}, \frac{1}{n} \right ]$.
So we have that $\max f_n=f_n\left(\frac1n\right)=\frac{1}{1+1}=\frac{1}{2}$.
It holds that $\displaystyle{\lim_{n\to\infty}\max f_n=\frac{1}{2}\neq 0}$.
So we don't have uniform convergence on $[-1,1]$ to $f(x)=0$.

We have that \begin{equation*}f(x)=\lim_{n\rightarrow +\infty}f_n(x)=\lim_{n\rightarrow +\infty}\sqrt{\frac{1}{n^2}+x^2}=\sqrt{x^2}=|x|\end{equation*}
So we have pointwise convergence on $\mathbb{R}$ to $f(x)=|x|$.

We have that \begin{equation*}f(x)=\lim_{n\rightarrow +\infty}f_n(x)=\lim_{n\rightarrow +\infty}\sqrt[n]{x}=\lim_{n\rightarrow +\infty}x^{\frac{1}{n}}=1\end{equation*}
So we have pointwise convergence on $(0,1)$ to $f(x)=1$.

Is everything correct so far? Could you give me a hint what we could do for the uniform convergence of the sequences of functions at 2 and 3 ?

Do we consider a function $g(x)=f_n(x)-f(x)$ and check if its maximum converges to $0$ ?

答:
What you did is fine, except that, in the case of the third sequence, you forgot that $0$ dos not belong to $(0,1)$. On this set $(f_n)_{n\in\Bbb N}$ converges pointwise to $1$.

In the case of the second sequence, the convergence is uniform. Differentiating, you can check that $\sqrt{\frac1{n^2}+x^2}-x$ is decreasing on $[0,\infty)$. So, its maximum is attained at $0$, where it takes the value $\frac1n$. By a similar argument, the maximum of $\sqrt{\frac1{n^2}+x^2}+x$ on $(-\infty,0]$ is also $\frac1n$. So,$$\sup\left|\sqrt{\frac1{n^2}+x^2}-|x|\right|=\frac1n$$and therefore the convergence is uniform

The third sequence doesn't converge uniformly to $1$, because$$\lim_{n\to\infty}f_n\left(\frac1{n^n}\right)=\lim_{n\to\infty}\frac1n=0.$$


问:
How do I calculate the number of elements that fit into a container on average?
This is a geometry question.

Given that I pour a bucket of small items with a given shape into a smaller bucket, how do I know how many will fit in there on average (obviously depends on how they fall into the smaller bucket) beforehand?

The items all have the same shape.

Help would be greatly appreciated I'm sort of lost here...

答: This is a hard problem in general and can typically not be solved exactly. For instance, if the objects are shaped perfectly spherically one is naturally led to the sphere packing problem, which is notoriously difficult. So your best bet is to just try it in real life, or somehow simulate it.</p>


问:
How the information about acute angles be used to find the length in a segment of a triangle?
The problem is as follows:

In an acute angled triangle $ABC$ $(AB<BC)$, the perpendicular
bisector mediatrix of
$AC$ intersects $BC$ at point $N$, and the height $BH$ intersects $AN$
at point $E$, If $\textrm{AE = 3 m}$ and $\textrm{BC = 13 m}$. Find
$BN$.

The alternatives in my book are:

$\begin{array}{ll}
1.&\textrm{5 m}\\
2.&\textrm{4 m}\\
3.&\textrm{4.5 m}\\
4.&\textrm{3.5 m}\\
\end{array}$

This problem did not come with a diagram or a drawing so the best that I could do is to make an interpretation of what it is being mentioned and this is seen below:

Now here it comes the part on How do I use properly the information which has been mentioned?

First off is my problem with this word: mediatrix, it seems that the aluded word seems to indicate that it is a median and an angle bisector at the same time. But is this okay?. Or is it just a perpendicular bisector and only that?.

Other than that,

I have currently ran out of ideas.

The only thing which had come to my mind is that when a small angle opposes to one particular side that side must be also small when compared to a greater angle. But how does this can be used to solve this problem?.

Since I'm not good on this subject. It would help me a lot if someone could help me here?.

The intended approach I am looking is to solve this relying only in euclidean geometry.

Is my drawing a correct interpretation?. If not please feel free to correct it in such a way that it is right.

Please if you can try to include a drawing in your answer because that could help me better to spot if a construction is needed.

答:
It is trivial to show $AN=NC$

Slightly more complicated, but still easy $\angle NBE=\angle NEB$, therefore $NE=NB$.

In your figure, move the "3m" label onto $AE$, not $EN$
Then $$BN+NC=13\\AN-NE=AE=3\\NC-BN=3$$ Therefore $BN=5$

EDIT
Here are some additional information that was pointed out in the comments:
In step 2, I used $\angle NCA=\angle CAN$ from step 1, then $\angle AEH=90^\circ-\angle CAN$. Opposite angles $\angle AEH$ and $\angle BEN$ are equal. In the right angle triangle $BCH$, the angle $\angle CBH=90^\circ-\angle BCH$. Therefore $\angle CBH=\angle BEN$, so $\triangle BEN$ is isosceles. You can get to the same conclusion if you draw a parallel to $AC$ through $N$, that intersects $BH$ at $M$. You know $MN\perp BH$, and then you look to prove $\angle MNE=\angle MNB$, to show that the two triangles are congruent and $EN=NB$.

The other question that showed up was why is it necessary for the triangle to be acute? It does not need to be (with caveats). Angle $\angle ABC$ can be obtuse. To prove this, start with a very obtuse isosceles triangle $ANC$, where $AN=NC=8$. Now choose point $E$ on $AN$ such that $AE=3$ and $EN=5$. Draw the perpendicular from $E$ to $AC$. Point $B$ is at the intersection of $NC$ and $EH$. You can arbitrarily increase angle $\angle ANC$ towards $180^\circ$, and then $\angle ABC$ will tend to the same value. However, if angle $BAC$is obtuse, the geometry will change. Angle $BCA$ can't be obtuse and have $AB<BC$


问:
deciding if a chain is a composition series (sanity check)
A small sanity check related to Question 2 from here: proof of the Krull-Akizuki theorem (Matsumura)

Let $C$ be an $A$-module, with $A$ commutative ring and suppose that there exists a chain of submodules $C=C_0 \supset C_1 \supset \cdots \supset C_m=0$ such that $C_i/C_{i+1} \cong A/m_i$ where $m_i$ is a maximal ideal of $A$. It seems to me that this implies that this chain is actually a compoosition series and so $C$ has finite length, since every quotient is a field and hence a simple $A$-module. Is this assertion correct?

答:
The only thing that's wrong is the "$C_i/C_{i+1}$ is a field" part. That quotient is a simple $A$ module, yes, but it is not necessarily a field. It is module isomorphic to $A/M$, but not ring isomorphic. (Added: of course, the quotient can be given a field structure by idenfitication with $A/M$, but it does not match the multiplication of the rng quotient $M/N$. Since the identification does not match, I discard it as a natural thing.)

For example, you can take a commutative Artinian ring whose ideals are linearly ordered to see why the quotients aren't fields. If $M/N$ was a field for proper ideals $M$ and $N$, that would mean that the identity of $M/N$ lifts to be an identity of $M$, but unfortunately $M$ does not contain any idempotent elements besides 0.

ContentLicense: CC BY-SA 3.0


问:
For $z \in \mathbb{C}\setminus \{0\}$, is $|1/z| = 1/|z|$?
I have been trying to prove that for $z \in \mathbb{C}$ and $z \neq 0$ that $|1/z| = 1/|z|$ but with no success.  Not all the properties of real numbers can be applied to complex numbers, of course.

答:
Suffices to show $|z|\cdot |1/z|=1.$ To do this, use that law that for $w,z\in \mathbb{C},$
$$
|z|\cdot |w|=|z\cdot w|.
$$


答:
Yes.

Note that $1/|z|$ is, by definition, the real multiplicative inverse of $|z|$, so by uniqueness of such an inverse, since

$$|1/z|\times |z|=|z/z|=1,$$

we have $|1/z|=1/|z|$.

Title: For $z \in \mathbb{C}\setminus \{0\}$, is $|1/z| = 1/|z|$?

Body: I have been trying to prove that for $z \in \mathbb{C}$ and $z \neq 0$ that $|1/z| = 1/|z|$ but with no success.  Not all the properties of real numbers can be applied to complex numbers, of course.


问:
Notion of fractal dimension which distinguishes point and square from square
I am new to fractal dimensions and am looking for a notion of fractal dimension which would help me distinguish between the following two sets on the cartesian plane:

\begin{align}A&=[0,1]\times[0,1] \cup (2,2)\\B&=[0,1]\times[0,1]\end{align}

In other words, I am asking:

Is there some form of a fractal dimension, say $D$, which would give $D(B)=2$, as $B$ is a 2-d square, but $D(A)\not=2$, as $A$ is a 2-d square joined with a 0-d point?

答:
For any point $x$ in any metric space $X$ you can consider the "local dimension at $x$ with radius $r$," namely the Hausdorff dimension of the closed ball

$$B_r(x) = \{ y \in x : d(x, y) \le r \}$$

of radius $r$ around $x$. If this dimension is constant for all sufficiently small $r$ you can consider the "local dimension at $x$" to be this constant value. Probably a similar definition can be written down for local topological dimensions.

A solid square $[0, 1] \times [0, 1]$ has local dimension $2$ at every point, and the isolated point in $A$ has local dimension $0$. So $A$ and $B$ can be distinguished by the fact that the local dimension of every point in $B$ is $2$, whereas this is not true for $A$.

This seems to me to be more or less the obvious thing to do and I'm surprised this isn't standard. Dimensions are inherently local. In other words, the dimension of a geometric object $X$ is inherently not a number but a function on $X$ given by the local dimension at $x \in X$. It happens that we often consider $X$ such that the local dimension at every point is the same number $d$, in which case it would make sense to say that $X$ has dimension $d$. But this is not true in your case, so I'm suggesting that in your case we instead consider the more refined invariant given by local dimension.


答:
Maybe the easiest way to see that is to use exponential form of the complex number $z = \rho e^{i\phi}$. Thus:

$$
\frac{1}{z} = \frac{1}{\rho} e^{-i \phi} \implies \left|\frac{1}{z}\right| = \frac{1}{\rho}
$$
On the other hand:
$$
|z| = \rho \implies \frac{1}{|z|} = \frac{1}{\rho}
$$


问:
How to Explicitly Round to a Given Number of Significant Digits
I have a given number: $N$, and I would like to explicitly round it to the nearest 2 significant digits

I have the following working equation that seems very redundant to me:
$$\begin{align*}
x &= \left\lfloor\log(N)\right\rfloor-1\\
y &= N 10^{-x}\\
\text{Answer} &=\left\lfloor y + (y \bmod 1)\right\rfloor 10^{x}
\end{align*}$$
I chose to use the $Floor$ operation here because I know that I can explicitly break down any floor function into $$\lfloor x\rfloor = x - (x\bmod1)$$

I know that calculating $x$ is probably necessary because I need to find the position of second most significant digit, but I can't help but feel that there is a much simpler way of performing the rounding as dropping down to a tens' digit and raising it back to it's proper position seems rather redundant to me.

Thank you in advance for all of the help

答:
Take a number like $5.678758$. Then multiply it by $100$ to get $567.8758$. You can see that it places the two decimal places in the integer area now, so that only the ones place, which was the original number's hundredths place, will be affected by the round function, which is what happens when you round something to two decimal places.

The round function will make it so that when when the tenths place of our new number is $5$ or above $5$, the ones place of our number increases by $1$, otherwise, it doesn't change. This is exactly what we want, because we want a number like $5.687$ to round up to $5.69$, and a number like $5.683$ to round down to $5.68$. So, then use round() function on this number to get $568$.  Now divide by $100$ to get $5.68$.

So, in short you want to program a function like this for a number $x$: $f(x) = round(100x)/100$.


答:
The main problem that you have here is that you have been trying to re-invent the wheel, rather than actually consulting a Complex Analysis textbook.  The following results are based on Chapter 1 of "An Introduction To Complex Function Theory" [Bruce Palka].

For $z,w \in \mathbb{C}, |z| \times |w| = |z \times w|.$

For $z = (x + iy) \in \mathbb{C}$, let $\overline{z}$ denote $(x - iy).$

Then $\frac{1}{z} = \frac{\overline{z}}{z \times \overline{z}}.$

$z \times \overline{z} = |z|^2.$

$|\overline{z}| = |z|$.

The result that you are trying to prove is a direct consequence of the above intermediate results.


问:
Proof of Egoroff's Theorem
Let $\{f_n \}$ be a sequence of measurable functions, $f_n \to f$ $\mu$-a.e. on a measurable set $E$, $\mu(E) < \infty$. Let $\epsilon>0$ be given. Then $\forall \space n \in \mathbb{N} \space \exists A_n \subset E$ with $\mu(A_n) <\frac{\epsilon}{2^n}$ and $\exists N_n$ such that $\forall \space x \notin A_n$ and $k \ge N_n \space |f_k(x) - f(x)| < \epsilon$. That is: if we define $A = \cup_{n=1}^{\infty} A_n$ with $\mu(A) < \epsilon $ then ${f_n}$ converges uniformly on $E \setminus A$.

$\mathbf{Proof}$: (taken from Royden's Real Analysis)

Let $A = \cup_{n=1}^{\infty} A_n \Rightarrow A \subset E$ and $\mu(A) < \sum_{n=1}^{\infty} \frac{\epsilon}{2^n} = \epsilon. \mathbf{Q1}$.

choose $n_0$ such that $\frac{1}{n_0} < \epsilon$. If $x \notin A$ and $k \ge N_{n_0}$ we have $\space |f_k(x) - f(x)| < \frac{1}{n_0} < \epsilon \space$. $ \square$

$\mathbf{Q1}$: First off, I don't see how $\sum_{n=1}^{\infty} \frac{\epsilon}{2^n} = \epsilon$. It's a geometric series: $ \epsilon \sum_{n=1}^{\infty} \frac{1}{2^n} = \epsilon \frac{1}{1-\frac12} = 2 \epsilon$. Am I wrong?

$\mathbf{Q2}$: The idea behind Egoroff is in order to turn almost sure convergence into uniform convergence on $E$ we only need to take away a really small set, right? Interestingly, as $\epsilon \to 0$ (that is $f_n$ is getting closer to $f$), the measure of the set $A$ is getting proportionally smaller ($\mu(A) \to 0$). So are we ultimately taking away a set of zero measure?

答:
A2: You are correct, that for arbitrarily small $\epsilon$ there is a set $A$, such that $\mu(A)<\epsilon$, where uniform convergence fails. So the measure of $A$ can be arbitrarily small. However, also keep in mind that uniform convergence requires a finite $N$ such that given $\delta>0$, for all $k>N$, $|f_k - f|<\delta$. Imagine that as $\epsilon$ gets smaller and smaller, for a fixed $\delta$ this $N$ may get larger and larger. Then in the limit as $\epsilon\to 0$, $N \to\infty$ and uniform convergence would fail.

My favorite text for Egoroff's theorem and related topics is Lieb and Loss's Analysis book.

Title: Proof of Egoroff's Theorem


问:
For some sequentially compact space $X$, is $X^{\omega_1}$ not sequentially compact?
When we assume the continuum hypothesis, for $X=\{0, 1\}$, the $\omega_1$ product $X^{\omega_1} = X^{\mathfrak{c}}$ is not sequentially compact while $X$ is sequentially compact.
So the proposition in the title is relative consistent from ZFC.
(the assumpution CH can be weaken to $\mathfrak{s} = \aleph_1$, where $\mathfrak{s}$ is the splitting number.)

Then, is it proved in ZFC that for some sequentially compact space $X$, $X^{\omega_1}$ is not sequentially compact?

I suspect that an cardinal $\kappa$ that has $\aleph_1$ cofinality and that is greater than $\mathfrak{c}$ suffices (with the order topology).

答:
According to the Handbook of Set-theoretic Topology (a book that any serious general topologist should have IMO) in the chapters by Vaughn and by van Douwen the fact is mentioned (and partially proved, plus references tot the original papers can be found there as well) that a product of $<\mathfrak{t}$ many sequentially compact (Hausdorff) spaces is again sequentially compact, where $\mathfrak{t}$ is the cardinal invariant known as the tower number (defined on page 115 in the handbook). It is consistent with ZFC that $\mathfrak{c} = \aleph_2 = \mathfrak{t}$ (e.g. by theorem 5.1 of van Douwen's chapter).

In such a model no such $X$ as desired could exist. So the answer is no.


问:
another exponential distribution probability example
I have an exponential variable with $\mu = 169$ hours which represents time to failure for a part in a machine.  I know that the probability that it will last for 100 hours or less is about 0.45.

Given that the part has been used for 100 hours, what is the probability that it will last for 100 more?  I think the memoryless principle applies here, but I can't figure out how.

答:
Use Bayes' rule:

$$P(t>200 | t>100) = \frac{P(t>200) P(t>100 | t>200)}{P(t>100)}$$

where I think the notation is rather straightforward and of course $P(t>100 | t>200) = 1$.

Perhaps this figure will help:


问:
limits and the definition
I am working on proving if a limit exists.

the sequence $s_n=|\frac{n+1}{n}-1|$ has a limit as $n$ approaches infinity, and this limit is s=1.

My work:

A sequence $(s_n)$ of real numbers is said to converge to the real number $s$ provided that
for each $ε > 0$ there exists a number $N$ such that $n>N$ implies $|s_n −s|<ε$.

impose $|s_n-s|<\varepsilon$. which implies $n>1/\varepsilon$ $|\frac{n+1}{n}-1|< \varepsilon$ which means $n>1/\varepsilon$

Now based on our definition, we want $N>1/\varepsilon$. My first misunderstanding is here. I was told that this means for all natural numbers bigger than $n$, the distance between the corresponding element of the sequence and your candidate limit is less than the epsilon given. What does this mean?

Here's the proof:

let $\varepsilon>0$. choose $N>1/\varepsilon$
in this case, if $s=1$,
$|sn-s|=|\frac{n+1}{n}-1|=|1/n|=1/n\tag{2}$
but if $n>N$, then $1/n<1/N$, and also by choice of $N$, $1/n<\varepsilon$, so $|sn-s|<\varepsilon$

My second problem is here. I don’t understand $(2)$. Please help. Thank you.

答:
It would help if your formatting was clearer; it is somewhat confusing to read as it is now.

But we want to show that the sequence $$s_n=\frac{n+1}{n}=1+\frac{1}{n}$$ converges to $1$ as $n\to\infty$.

Let $\varepsilon>0$. By the Archimedean property of the real numbers, there exists an $N\in\mathbb{N}$ so that $0<\frac{1}{N}<\varepsilon$. Now, for all $n\geq N$, we have $$|s_n-1|=\bigg|\bigg(1+\frac{1}{n}\bigg)-1\bigg|=\bigg|\frac{1}{n}\bigg|=\frac{1}{n}\leq \frac{1}{N}<\varepsilon $$ since $n\geq N\Longleftrightarrow \frac{1}{n}\leq\frac{1}{N}$.

No matter what 'epsilon tolerance' one chooses, one can 'look' far out enough into the tail of the sequence and make the distance between $1$ and the elements of $(s_n)$ less than the given epsilon tolerance for all elements of $(s_n)$ after we pass some $s_N$ in the sequence.

Title: limits and the definition


问:
Find eigenvalues without having the matrix.
Let $\{v_1,v_2,v_3,v_4\} $ be a vector basis of $\mathbb{R}^4$ and $A$ a constant matrix of $\mathbb{R}^{4\times 4}$ so that:
$$A v_1=-2v_1,\hspace{0.5cm} Av_2=-v_1,\hspace{0.5cm} Av_3=3v_4,\hspace{0.5cm}Av_4=-3v_3$$
Can I find the eigenvalues of the matrix A?
I know that $\lambda_1=-2$ is a trivial eigenvalue but I don't know how to calculate the others.

答:
Notice that this linear map/matrix is basically two $R^2\to R^2$ maps joined together: One map consists of a linear map from span$\{v_1,v_2\}$ to itself, the other a linear map from span$\{v_3,v_4\}$. Respectively, these have matrix representations
$$
\begin{bmatrix}{-2 \: -1 \\\quad 0 \quad 0}\end{bmatrix}
$$
and
$$
$$
\begin{bmatrix}{\quad 0 \: +3 \\ -3 \quad\:  0}\end{bmatrix}
$$
$$
It is pretty easy to find eigenvalues of these one.


答:
Try to prove that $\overline{\frac 1 z} = \frac 1{\overline z}$.

Then $|\frac 1z| = \sqrt{\frac 1z\cdot \overline{\frac 1z}}=$

$\sqrt{\frac 1 z\cdot \frac 1{\overline z} } =\sqrt{\frac 1{z\overline z}}=$

$\sqrt{\frac 1{|z|^2}} = \sqrt{(\frac 1{|z|})^2}=$

$|\frac 1{|z|}| = \frac 1{|z|}$.

(bear in mind you do know that $|z|$ is a positive real number)

That's probably way more detain than you need.

.....

Alternatively

$|\frac 1{z}| = |\frac 1z\cdot \frac {\overline z}{\overline z}|=$

$|\frac {\overline z}{z\overline z}|= |\frac {\overline z}{|z|^2}|=$

$\frac 1{|z|^2}|\overline z|=\frac 1{|z|^2}|z|=$

$\frac 1{|z|}$.

Bear in mind $|z| = |\overline z|$ and $|z|^2 = z\overline z \in \mathbb R^+$ and $|z|\in \mathbb R^+$.


问:
Find if the function is surjective $f(x) = (x^2+1)/(x^2 + 2)$
I found that the following function from $f:\Bbb R \to\Bbb R$ is not one to one:

$$ f(x) = \frac{x^2+1}{x^2 + 2} $$

I found answer online that it's not surjective as the range is not equal to the codomain, where the range is $\left\{y\mid\frac12 \le y \lt 1 \right \} = \left [\frac12, 1\right)$. Can you please explain how we can get this range?

答:
Just to show that $f$ has a limited range, which is a subset of $\mathbb R$:

Taking $f(0) = \frac12$ temporarily out of consideration, by the mediant inequality if $x\ne 0$,

$$\frac{1}{2} < \frac{1+x^2}{2+x^2} < \frac {x^2}{x^2} = 1$$

Together with $f(0)$,

$$\frac 12 \le f(x) < 1$$

(this does not prove that $f(x)$ takes every value in $\left[\frac12, 1\right)$)


问:
Proof of Alternating Binomial Coefficient Identity relating
I am looking for a proof of the following binomial identity. I encountered it in an article on Euler’s derivation of the gamma function. Euler begins by evaluating the integral:

$$\int_0^1 x^a(1-x)^n\,dx$$

He performs a binomial expansion on the integrand and makes use of the following identity involving alternating binomial coefficients:

$$\sum_{k=0}^{n} (-1)^k {n \choose k} \frac{1}{a+k+1} = n! \prod_{k=0}^{n}\left( \frac{1}{a+k+1} \right).$$

It is not obvious to me on inspection that this formula is true, but I have found it cited elsewhere in the mathematical literature, without proof. I have verified it by hand for the cases of n=2 and n=3, but I am unable to prove the identity in general. I am looking for a good combinatorial or inductive proof of this very interesting fact.

I am aware that this formula relates to the beta function, but I am trying to derive this relationship without reference to this topic.

Thank you for enlightening me.

答:
I think that the "clearest" way to demonstrate the identity is through
the finite differences and
the Falling and Rising Factorial as follows

The finite difference (unitary step) of a function wrt to the variable $z$ is defined as
$$
\Delta _{\,z} \,f(z) = f(z + 1) - f(z)
$$
and its iteration becomes
$$
\Delta _{\,z} ^{\,n} \,f(z) = \Delta _{\,z} \,\left( {\Delta _{\,z} ^{\,n - 1} \,f(z)} \right)
 = \sum\limits_k {\left( { - 1} \right)^{n - k}
 \left( \matrix{  n \cr   k \cr}  \right)\;f(z + k)}
$$

Now, concerning the function $1/z$ we have
$$
\eqalign{
  & \Delta _{\,z} \left( {{1 \over z}} \right) = {1 \over {z + 1}} - {1 \over z}
 = {{ - 1} \over {z\left( {z + 1} \right)}}  \cr
  & \Delta _{\,z} ^{\,2} \left( {{1 \over z}} \right)
 = \Delta _{\,z} \left( {\Delta _{\,z} \left( {{1 \over z}} \right)} \right)
 = {{\left( { - 1} \right)^{\,2} 2} \over {z\left( {z + 1} \right)\left( {z + 2} \right)}}  \cr
  & \;\quad  \vdots   \cr
  & \Delta _{\,z} ^{\,n} \left( {{1 \over z}} \right)
 = \Delta _{\,z} ^{\,n} \left( {\left( {z - 1} \right)^{\,\underline {\, - 1\,} } } \right)
 = \left( { - 1} \right)^{\,\underline {\,n\,} } \left( {z - 1} \right)^{\,\underline {\, - 1 - n\,} }
  = {{\left( { - 1} \right)^{\,n} n!} \over {z^{\,\overline {\,n + 1\,} } }} \cr}
$$
where the last line follows from the fundamental properties of the Falling and Rising Factorial
represented respectively with  $x^{\,\underline {\,k\,} } ,\quad x^{\,\overline {\,k\,} } $.

Putting the two together
$$
\Delta _{\,z} ^{\,n} \left( {{1 \over z}} \right)
 = {{\left( { - 1} \right)^{\,n} n!} \over {z^{\,\overline {\,n + 1\,} } }}
 = \sum\limits_k {\left( { - 1} \right)^{n - k} \left( \matrix{  n \cr  k \cr}  \right)\;{1 \over {z + k}}}
 \quad \left| {\;0 \le n \in Z} \right.
$$


问:
Equivalence class for the following relation
A relation of $\mathbb{R}$ is defined as $a\sim b : a^4-b^2=b^4-a^2$

Show that $\sim$ is equivalence relation (I have done this part)

Determine the equivalence class $[-1]_\sim$

Prove or disprove: Every equivalence class in $\mathbb{R}/\sim$ contains exactly 2 real numbers.

I am facing difficulty in second and third part of the question.

For the second one I think it is $[-1]_\sim=\{-1,1\}$ as $-1\sim x: (-1)^4-x^2 =x^4-(-1)^2 $ so both $-1$ and $1$ are proving the equality.

I think the third statement is true but how can I prove it?
Please help.

答:
Rearrange: $a^4 - b^4 = b^2 - a^2$

But $a^4 - b^4 = (a^2 - b^2)(a^2 + b^2)$ (DOTS)

So $(a^2 - b^2)(a^2 + b^2) = b^2 - a^2$

(1) Assume $a^2 - b^2 \neq 0$

Then divide both sides by $a^2 - b^2$

$a^2 + b^2 = -1$

Since $a^2$ and $b^2$ are both positive in $\Bbb R$, this is a contradiction.

(2) Therefore $a^2 - b^2 = 0$, or $a^2 = b^2$.

Therefore $b = \pm a$ and each equivalence class of a comprises $\{a, -a \}$.

The equivalence class of $-1$ is $\{-1, 1 \}$.


问:
Topological analysis of $A=\{\frac{n+\sqrt 2}{3n+\sqrt 3} | n\in \mathbb{N}\} \cup (-3,0]$ with respect to the standard topology
I am asked to find the interior, the closure, the derived set and the boundary of $A=\{\frac{n+\sqrt 2}{3n+\sqrt 3} | n\in \mathbb{N}\}\cup (-3,0]$ (this is not a homework problem, I am solving past real analysis exams). Then I have to say whether $A$ is connected or compact.

I will write here my solution because I want to see if I approach this problem correctly.  Let $x_n=  \frac{n+\sqrt 2}{3n+\sqrt 3}$ for all $n\in \mathbb{N}$.

We firstly prove that the interior of $A$ is $(-3,0)$.

Since $(-3,0)\subset A$ and $(-3,0)$ is open, we get that it is contained in the interior set. Since the interior is a subset of $A$, we are left to prove that  $x_n$ is not in the interior for any $n$ and that $0$ is also not an interior point.

Suppose there is some $n\in \mathbb{N}$ such that $x_n$ is an interior point. Then there is some $r>0$ such that $(x_n-r, x_n+r)\subset A$. This implies that $(x_n-r, x_n+r)\cap (x_n, x_n+r)\subset A\cap (x_n, x_n+r)$. But $(x_n-r, x_n+r)\cap (x_n, x_n+r)=(x_n, x_n+r)$ is an infinite set, whilst $A\cap (x_n, x_n+r)$ is a finite set and we have reached a contradiction.

Now let's suppose that $0$ is an interior point of $A$. Then there is some $r>0$ such that $(-r,r)\subset A \implies (-r,r)\cap (0, x_1) \subset A \cap (0,x_1)$. The first set contains $r$, whilst the second one is empty (we are using the fact that the sequence $x_n$ is decreasing) and now we are done.

Now we prove that $\overline{A}=\{\frac{n+\sqrt 2}{3n+\sqrt 3} | n\in \mathbb{N}\}\cup [-3,0]\cup \{\frac{1}{3}\}$.

We know that $A\subset \overline{A}$, so we only wish to prove that $-3$ and $\frac{1}{3}$ are in $\overline{A}$.

Since $x_n \to \frac{1}{3}$ we easily get that $\frac{1}{3}\in \overline{A}$.

$-3 \in \overline{A}$ because for any sets $\overline{X\cup Y}=\overline{X}\cup\overline{Y}$ and now we are done.

So, $A\subset B:=\{\frac{n+\sqrt 2}{3n+\sqrt 3} | n\in \mathbb{N}\}\cup [-3,0]\cup \{\frac{1}{3}\}\subset \overline{A}$. Because $\displaystyle \mathcal{C}_B=(-\infty, -3)\cup (0, \frac{1}{3})\cup \bigcup_{n=1}^{\infty}(x_{n+1}, x_n)\cup (x_1, \infty)$, this set is open, so $B$ is closed and we get that $\overline{A}=B$.

Now we show that $A'=[-3,0]\cup \{\frac{1}{3}\}$. We don't have too much to do, because $\overline{A}=A\cup A'$, so we instantly get that $\frac{1}{3},-3\in A'$. Now we just want to show that $x_n$ is not in $A'$ for any $n\in \mathbb{N}$.

Suppose there is some $n\in \mathbb{N}$ such that $x_n \in A'$. Let $r=\min\{x_n-x_{n+1}, x_{n-1}-x_n\}$. Then we have that $(x_n-r, x_n+r)\cap A \setminus \{x_n\}=\emptyset$, so we've reached a contradiction.

Thus, we may say that $A'=[-3,0]\cup\{\frac{1}{3}\}$.

Now, the boundary of $A$ is $\overline{A}\setminus \operatorname{int}A=\{\frac{n+\sqrt 2}{3n+\sqrt 3} | n\in \mathbb{N}\}\cup \{-3,0,\frac{1}{3}\}$.

$A$ is not closed, so it is not compact, and $A$ is not an interval, so it is not connected and now we are finally done.

Are my approaches correct? I am sure that some parts can be shortened, but that's my reasoning and I want to see if it is right.

答:
$A$ is basically $(-3,0]$ union with the set of terms of a convergent sequence outside it with limit $\frac{1}{3}$. So the closure of the union being the union of the closures (for finite unions) tells us the closure is $A \cup \{-3,\frac13\}$ as you rightly concluded. The interior is clearly $(-3,0)$ as the sequence part are just isolated points, so contains no open sets of $\Bbb R$, and the interior of $(-3,0]$ is $(-3,0)$. The boundary is the difference between closure and interior so equals $\{-3,0,\frac{1}{3}\}$ unioned with the terms of the sequence. The derived set is basically the closure minus the isolated points here so $[-3,0] \cup \{\frac13\}$.

$A$ is not compact as it is not closed and not connected as it has isolated points which are relatively clopen (all sequence terms singletons).

I don't think all the details you give are necessary. The eventual conclusions are indeed correct.


答: Let $X$ be an exponential random variable. The memoryless property says that $$ P(X > t+s | X >s) = P(X>t) $$ for any $t\geq 0$ and $s \geq 0$. Recall that the vertical bar "$|$" in the probability statement here can be read as "given" or "conditional on". In your question, we have both $t=100$ and $s=100$. Can you figure it out from here?

Title: another exponential distribution probability example

Body:
I have an exponential variable with $\mu = 169$ hours which represents time to failure for a part in a machine.  I know that the probability that it will last for 100 hours or less is about 0.45.

Given that the part has been used for 100 hours, what is the probability that it will last for 100 more?  I think the memoryless principle applies here, but I can't figure out how.


问:
What is the partition of an empty set?

Definition 11.1.10 (Partitions). Let I be a bounded interval. A partition
  of $I$ is a finite set $P$ of bounded intervals contained in $I$, such that
  every $x$ in $I$ lies in exactly one of the bounded intervals $J$ in $P$.

It is trivial to prove an empty set is the partition of an empty set. However, I found a set whose only element is an empty set could be also a partition of an empty set. Here is my proof, based on the definition of a partition in Analysis I of Tao.

Assume P is not a partition of an empty set, then P could contain some set that is not contained in an empty set. It means an empty set is not a subset of itself, which in turn means that there is some element in an empty set that is not in itself. However an empty set contains nothing, so an empty set must be a subset of itself. Now, the only other case is that there exists some element in the empty set such that it is either not found in any element of P or belongs to more than one elements of P. But again an empty set does not contain anything, so that proves P which contains an empty set is a partition of that empty set.

Could you help to point out if I am correct?

Thanks

答:
One option is to decide that the empty set is not a bounded interval, so this definition does not apply to it!

If we pretend that the empty set is a bounded interval and apply the definition of partition to the empty set, then indeed, both $\emptyset$ and $\{\emptyset\}$ would be partitions, although your proof of the latter case is unnecessarily convoluted: it is vacuously true that, for all $x \in \emptyset$, there is some $J \in P = \{\emptyset\}$ such that $x \in J$, since there are no elements $x$ to check.

While playing pretend is fun, it is worth noting that this would make the quoted definition of partition inconsistent with other uses of that term; typically the parts in a partition are required to be non-empty.

[Some credit is due to Oscar Cunningham; I wrote this answer after a discussion on the subject held on the Category Theory Zulip forum.]

Title: What is the partition of an empty set?


问:
What is the additive identity of a quotient module?
Given a module $M$ and a submodule $S$, the additive identity of $M/S$ is $0+S=S$. Does this mean that there are two additive identities? $0$ and $S$? Why must we write the equation $0+S=S$ instead of just $S$?

答:
The elements of the quotient module $M/S$ are cosets of the submodule $S$, i.e. sets of the form
$$m+S:=\{m+s:s\in S\}$$
where $m\in S$. (Two elements $m,n$ determine the same coset iff $m-n\in S$.)

So, $0$ is the zero element of $M$, and its coset, $0+S=S$ is the zero element of $M/S$.


问:
Does the subspace topology of rationals have the cofinite topology?
Just to be clear, we are talking about the subspace topology of all rational numbers as a subset of $\bf{R}$. The answer is clearly no right? Because every open set in Q is the intersection of open intervals in $\bf{R}$ with Q and their complement are the the remaining rationals which are infinite.

答: The cofinite topology on any infinite set is not Hausdorff (all non-empty open subsets intersect). A metric space like $\Bbb Q$ in the subspace topology is always Hausdorff. So no. Alternatively, the cofinite topology on an infinite set would also be connected and compact and $\Bbb Q$ is neither.


问:
Do $R'$ integral over $R$ imply $\operatorname{Frac}(R')$ algebraic over $\operatorname{Frac}(R)$?
This is Theorem 6.6 from Janusz, Algebraic Number Fields, it says:

Let $R\subseteq R'$ be Dedekind domains with $R'$ integral over $R$ and $\mathfrak{p}$ a nonzero prime ideal of $R$. Suppose
$$\mathfrak{p}R'=\mathfrak{P}_1^{e_1}\cdots\mathfrak{P}_g^{e_g}$$
with the $\mathfrak{P}_i$ distinct prime ideals of $R'$ and $e_i>0$. Let $f_i=f(\mathfrak{P}_i|R)$ be the relative degree of $\mathfrak{P}_i$. Then:

1) $\sum_ie_if_i$ is the dimension of $R'/\mathfrak{p}R'$ over $R/\mathfrak{p}$

2) if $K$ and $L$ are the quotient fields of $R$ and $R'$ respectively and the dimension $[L:K]$ is finite then $\sum_ie_if_i\leq[L:K]$

3) if $S$ is the complement of $\mathfrak{p}$ in $R$ and if $R'_S$ is finitely generated over $R_S$ then $\sum_ie_if_i=[L:K]$

Proof.

My problem is with point 3). It says: suppose $R'_S$ finitely generated module over $R_S$ and take $x_1,\ldots,x_n$ as a minimal generating set, for $R'_S$ over $R_S$. It shows that such $x_i$'s are elements of $L$ linearly indipendent over $K$, and all is ok for me. Then it says:....now that we know that the $x_i$'s are linearly indipendent, we prove that they gives a basis for $L$ over $K$. Suppose not. Then $\sum Kx_i$ is a $K$-vector space properly contained in $L$, thus there exists an element $y\in L$ such that
$$Ky\cap\sum Kx_i=0$$
But $y$ satisfies a monic polynomial over $K$, hence.........and from now on i can understand. So my question is: why do $y$ satisfy a polynomial over $K$? We don't know a priori that $L$ is algebraic, or finite degree, over $K$. So why Janusz state it as an obvious fact? Do this follow from $R'$ integral over $R$? Or from $R'_S$ finitely generated over $R_S$?

Could someone help me please?

答: Your title question has an affirmative answer for any integral extension $R' \supset R$ of domains: see e.g. Proposition 14.10 of these notes.


问:
Striking applications of linearity of expectation
Linearity of expectation is a very simple and "obvious" statement, but has many non-trivial applications, e.g., to analyze randomized algorithms (for instance, the coupon collector's problem), or in some proofs where dealing with non-independent random variables would otherwise make any calculation daunting.

What are the cleanest, most elegant, or striking applications of the linearity of expectation you've encountered?

答:
One I enjoy is:

Enumerate $n$-many balls, place in an urn and draw $k$-many at random without replacement. Let $S = X_1+ \ldots X_k$ be their sum where $X_i$ is the value on the $i^{th}$ ball. Then by linearity of expectation we have

$\displaystyle E[S]=  E \bigg [ \sum_{j = 1}^k X_j \bigg ] = \sum_{j = 1}^k E[X_j] = k \cdot E[X_j] = k \cdot \frac{n+1}{2}$.

And as we draw without replacement we do not have independence.

cf. Expectation of sum of numbered balls


问:
proof about linear transformation
Although I know the properties of linear transformations and the concept of rank, I could not prove this. Can you help me?

I chose an element from the kernel and said this element b. I saw that this element is a rangein element L (b) = sb for some numbers.

答:
Remember that for a finite dimensional vector space $V$ we have $dim(V)=dim(ImF)+dim(kerF)$, where $F$ Is a linear map.

In your case $dim(ImF)=dimV-dim(KerF)=n-(n-1)=1\implies dim(KerF)=n-1$ and this means that the imagine of the map $F$ is generated by a vector $u$ such that $ImF=\langle u \rangle$.

For this reason, $\forall v\in V, F(v)\in \langle u \rangle\implies F(v)=\lambda u$, for $\lambda \in \mathbb K$.

This also shows that the composition $(F\circ F)(v)=F(F(v))=F(\lambda u)=\lambda F(u)$, which is still an element of $\langle u \rangle$.

Since the Kernel of the function is the eigenspace associated to te eigenvalue $t=0$, it could be interesting studying the diagonal form of the representative matrix of the map.

Title: proof about linear transformation


问:
Solution verification about equivalence classes.

I'm absolutely not sure about the notations and the correct terminology and I was wondering if everything checks out..

Question 5

a.

$4^{2\ }-1$ = 15

Therefore we have 15 different partitions.

b.

The number of partition is the same as the number of equivalence relations.
Therefore we have 15 different relations.

c.

By limitation of being an equivalent relation, we have three independent partitions in $\left\{\left\{7,4,2\right\},\left\{1.3\right\},\left\{5\right\}\right\}$

with $\left\{6\right\}$ free to join any partition, or be a partition on it's own.

Therefore we have 4 equivalence relations that satisfy the condition.

The largest one is:

$R=\left\{\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right),\left(5,5\right),\left(6,6\right),\left(7,7\right),\left(7,4\right),\left(4,7\right),\left(7,2\right),\left(2,7\right),\left(7,6\right),\left(6,7\right),\left(4,2\right),\left(2,4\right),\left(2,6\right),\left(6,2\right),\left(3,1\right),\left(1,3\right),\left(6,4\right),\left(4,6\right)\right\}$

答:
My check on the answers with discussion:

a) How many different partitions of the set $\{3,19,27,40\}$ are there?

As JMoravitz comments, the number of partitions is given by the Bell numbers, but for a set of size $4$ it's not difficult to work out de novo, by largest part size $P_M$:

$P_M = 1$: All parts are size $1$, #options $=1$.

$P_M = 2$: Other parts size $1$: #options $=\binom 42 = 6$.

$\phantom{P_M = 2}$: Both parts size $2$ (symmetric): #options $=\frac12\binom 42 = 3$.

$P_M = 3$: Other part must be size $1$: #options $=\binom 43 = 4$.

$P_M = 4$: Whole set is one part, #options $=1$.

Total partitions: $15$

b) How many different equivalence relations can be defined on this set?

As you stated, each equivalence relation can be mapped to & from a partition of the set, so again $15$.

c) How many equivalence relations on $\{1,2,3,4,5,6,7\}$ contain the set $\{(7,4),(6,6),(1,3),(4,2)\}$ but are disjoint from the set  $\{(5,4),(7,3),(3,5)\}$?

The partition implied by the first condition implies that one part has a subset of $\{2,4,7\}$  and one part has a subset of $\{1,3\}$. The exclusions given in the second condition imply that these two parts are distinct and that $5$ is also not a member of either of them, giving us a third part that has $\{5\}$ as a subset. So the only remaining choice is where to put $6$, which is in any of the three parts identified or in a new part of its own, giving $4$ different equivalence relations.

- Out of the equivalence relations for which the conditions hold, [state which] has the largest number of pairs in it.

Adding $6$ as an isolated equivalence class,  $\{\{2,4,7\},\{1, 3\},\{5\},\{6\}\}$ would give the $7$ reflexive pairs with $6+2$ more pairs in the first two classes. Combining $6$ into another class would add pairs in proportion to the pre-existing size of that class, meaning that combining it into the largest class will give most pairs: $\{\{2,4,6,7\},\{1, 3\},\{5\}\}$ which has $7+12+2=21$ pairs.


问:
Calculate a limit of probability
Assume $ \left(X_{k}\right)_{k} $ is a sequence of independent random variables with distribution $ Ber\left(\frac{1}{2}\right) $ (all of them).

Calculate the limit:

$ \lim_{n\to\infty}\mathbb{P}[\prod_{k=1}^{n}\left(1+X_{k}\right)\leq a^{n}] $

for any $ a\geq 1 \in \mathbb {R} $.

Here's what I've tried:

$ \mathbb{P}[\prod_{k=1}^{n}\left(1+X_{k}\right)\leq a^{n}]=\mathbb{P}[\ln\left(\prod_{k=1}^{n}\left(1+X_{k}\right)\leq n\ln a\right) $

$ =\mathbb{P}\left(\sum_{k=1}^{n}\ln\left(1+X_{k}\right)\leq n\ln a\right) $

$ =\mathbb{P}\left(\sum_{k=1}^{n}\frac{\ln\left(1+X_{k}\right)}{\ln2}\leq\frac{n\ln a}{\ln2}\right) $

Now notice that each $ \frac{\ln\left(1+X_{k}\right)}{\ln2} $ distributed $ Ber(1/2) $, and also it is the sum of independent random variables (as measurable functions of independent random variables), so the sum is distributed $ Bin(n,1/2) $. Thus:

Note that if $ a>2 $ then the probability would just be $ 1 $, so im gonna assume $ 1<a<2$.

$ \mathbb{P}\left(\sum_{k=1}^{n}\frac{\ln\left(1+X_{k}\right)}{\ln2}\leq\frac{n\ln a}{\ln2}\right)=\sum_{k=1}^{\left\lfloor n\frac{\ln a}{\ln2}\right\rfloor }\binom{n}{k}\left(\frac{1}{2}\right)^{k}\left(\frac{1}{2}\right)^{n-k}=\frac{1}{2^{n}}\sum_{k=1}^{\left\lfloor n\frac{\ln a}{\ln2}\right\rfloor }\binom{n}{k} $

All I have left to do now is to calcualte

$ \lim_{n\to\infty}\frac{1}{2^{n}}\sum_{k=1}^{\left\lfloor n\frac{\ln a}{\ln2}\right\rfloor }\binom{n}{k}$

Which I dont know how.

Any ideas would be highly appreciated. Thanks in advance.

答: Let $$Y_n = \sum_{k=1}^n X_k \sim \operatorname{Binomial}(n, 1/2).$$  Then it is obvious that $$\prod_{k=1}^n (1 + X_k) = 2^{Y_n}.$$  So $$\Pr[2^{Y_n} \le a^n] = \Pr\left[Y_n \le n \frac{\log a}{\log 2}\right] = \Pr\left[\frac{Y_n - n/2}{\sqrt{n/4}} \le (2c - 1) \sqrt{n}\right],$$ where $c = \frac{\log a}{\log 2}$.  Then as $n \to \infty$, by the central limit theorem, $(Y_n - n/2)/\sqrt{n/4}$ tends to a standard normal distribution.  Consequently the desired probability is $$\lim_{n \to \infty} \Pr[2^{Y_n} \le a^n] = \begin{cases} 0, & a < \sqrt{2} \\ 1/2, & a = \sqrt{2} \\ 1, & a > \sqrt{2}. \end{cases}$$

Title: Calculate a limit of probability


答: If you know SLLN you don't have to compute anything. $P(\frac 1n \sum\limits_{k=1}^{n} \ln (1+X_k) \leq \ln a) \to P(\ln (1+X_1) \leq \ln a)=P(\frac 1 2 ln 2 \leq \ln a)=0$ if $a <\sqrt 2$ and and lmit is $1$ if $a >\sqrt 2$. You can use CLT to show that the limit is $\frac 1 2$ if $a=\sqrt 2$.

Title: Calculate a limit of probability

Body:
Assume $ \left(X_{k}\right)_{k} $ is a sequence of independent random variables with distribution $ Ber\left(\frac{1}{2}\right) $ (all of them).

Calculate the limit:

$ \lim_{n\to\infty}\mathbb{P}[\prod_{k=1}^{n}\left(1+X_{k}\right)\leq a^{n}] $

for any $ a\geq 1 \in \mathbb {R} $.

Here's what I've tried:

$ \mathbb{P}[\prod_{k=1}^{n}\left(1+X_{k}\right)\leq a^{n}]=\mathbb{P}[\ln\left(\prod_{k=1}^{n}\left(1+X_{k}\right)\leq n\ln a\right) $

$ =\mathbb{P}\left(\sum_{k=1}^{n}\ln\left(1+X_{k}\right)\leq n\ln a\right) $

$ =\mathbb{P}\left(\sum_{k=1}^{n}\frac{\ln\left(1+X_{k}\right)}{\ln2}\leq\frac{n\ln a}{\ln2}\right) $

Now notice that each $ \frac{\ln\left(1+X_{k}\right)}{\ln2} $ distributed $ Ber(1/2) $, and also it is the sum of independent random variables (as measurable functions of independent random variables), so the sum is distributed $ Bin(n,1/2) $. Thus:

Note that if $ a>2 $ then the probability would just be $ 1 $, so im gonna assume $ 1<a<2$.

$ \mathbb{P}\left(\sum_{k=1}^{n}\frac{\ln\left(1+X_{k}\right)}{\ln2}\leq\frac{n\ln a}{\ln2}\right)=\sum_{k=1}^{\left\lfloor n\frac{\ln a}{\ln2}\right\rfloor }\binom{n}{k}\left(\frac{1}{2}\right)^{k}\left(\frac{1}{2}\right)^{n-k}=\frac{1}{2^{n}}\sum_{k=1}^{\left\lfloor n\frac{\ln a}{\ln2}\right\rfloor }\binom{n}{k} $

All I have left to do now is to calcualte

$ \lim_{n\to\infty}\frac{1}{2^{n}}\sum_{k=1}^{\left\lfloor n\frac{\ln a}{\ln2}\right\rfloor }\binom{n}{k}$

Which I dont know how.

Any ideas would be highly appreciated. Thanks in advance.


问:
Would $ \lim_{\sigma\rightarrow0^+}\int_{-\sigma}^{\sigma}\frac{1}{\sigma}f(x)dx$ be equal to zero?
Suppose a limit of the form

$$ \lim_{\sigma\rightarrow0^+}\int_{-\sigma}^{\sigma}\frac{1}{\sigma}f(x)dx$$

where $f(x)$ is an integrable function.

I think that this limit would be zero, since we would be taking an integral at a single point, even if the argument was divergent. Is this correct?

答:
I think its also worth noting that the limit you are looking for is given by the Lebesgue differentiation theorem which I will state in the context of the Lebesgue measure on the Euclidean metric space.

Let $f\in L^1$ then for almost every $x \in \mathbb{R}^n$ it follows that $$\lim_{r\downarrow 0}\frac{1}{|B_r(x)|}\int_{B_r(x)}f(y)dy = f(x).$$


问:
Find $E[Y\mid X=x]$
Let $(X,Y)$
$$f(x,y)=\frac{x^2}{4}\hspace{1cm} 0\leq y\leq x\leq 2$$
Find $E[Y\mid X=x]$

I know
$$\dfrac{f(x,y)}{f_X(x)}=\frac{x^2/4}{x^3/4}=\frac{1}{x}$$
My question is about the limits of integration,

I don't know if they are $[0,2]$
$$E[Y\mid X=x]=\int_0^2y\cdot\frac{1}{x}\,dy$$

or  $[0,x]$

$$E[Y\mid X=x]=\int_0^x y\cdot\frac{1}{x}\,dy$$

答:
What you calculated is
$$
\frac{f(x,y)}{f_X(x)}=\frac{1}{x},\quad \color{red}{0\le y\le x\le 2}
$$

Hence for $x>0$,
$$
E(Y|X=x)=\int_0^2 \frac{yf(x,y)}{f_X(x)}dy=\int_0^x\frac{y}{x}dy=\frac{x}{2}.
$$
If $x<0$, then $f(x,y)=0$, and thus
$$
E(Y|X=x)=0.
$$

Title: Find $E[Y\mid X=x]$


问:
Special binary string
Imagine a binary string of increasing length, up to infinity.

What makes it so special? Well, just a simple "rule":

for any given length (odd or even), if one folds the string in half, there is at least one couple of ones that overlaps.

The question is now about the distribution of ones (zeros), as a function of string length. What are its properties? Are there any?

A trivial example that follow the rule is the string $111010101010....$.

答:
We can view the string as a subset $A\subseteq \mathbb N_0$ (positions of $1$s). The bit string has the folding property if and only if each $n\in \mathbb N_\color{red}0$ can be written as the sum of two $\color{red}{\text{(not necessarily distinct)}}$ elements of $A$.

The lexically first string is almost the one you gave: $1\overline{10} $.
If one redundantly replaces the first $0$ with a $1$ and then tries to stay lexically low again, one obtains $11\overline{100}$ and so on.

EDIT: I changed two places to accomodate a different interpretation of "overlap" in case of odd substrings.
Without these red changes, the lexically first string is
$111\overline{010}$.

Title: Special binary string


问:
incomplete vector space of continuous functions.
Let $C([-1,1])$ be the vector space of continuous functions define on $[-1,1]$.

Show in the norm

$$\vert \vert f \vert \vert = \int_{-1}^1 \vert f(x) \vert dx$$

this vector space is not complete.

Does this mean to find a Cauchy sequence in $C([-1,1])$ that does not converge in $C([-1,1])$ under the above norm? Sort of lost then, so would my elements be continuous functions,

$$f: [-1,1] \rightarrow \mathbb{R}$$

any hints? can this be generalized to any closed and bounded $[a,b]$???

can I use $f_n:= x^n$?

答: No, you cannot use $x^{n}$ becasue it converges to the zero function in this norm. Consider a piece-wise linear function which is $1$ for $x <\frac 1 2 $, $0$ for $x>\frac 1  2+\frac1  n$. The limit in norm in this case is a discontinuous function.


问:
Limit of a integral
For $f:[0,1]\rightarrow \mathbb{R}$ continuous, let $a >0.$

\begin{eqnarray*}
L=\lim_{\varepsilon \rightarrow 0}\int_{\varepsilon a}^{\varepsilon b} \frac{f(x)}{x} dx
\end{eqnarray*}

Show that $L=f(0)\ln(\frac{b}{a})$.The first I thought was in mean value theorem for integrals. but  I don't know how can start, if can give me some hint to start. I will be gratefull.

答: By the substitution $y=\frac x {\epsilon}$ the integral becomes $\int_a^{b} \frac {f(\epsilon x)} x dx$. Can you finish?

Title: Limit of a integral


问:
What led Hausdorff to consider the possibility of "unerreichbar" or "weakly inaccessible" cardinals?
I would like to know how it was that Felix Hausdorff came to consider the possibility of, if not indeed necessity of, what he was the first to call "unerreichbare" or "unreachable" cardinals, or what we call, somewhat less clearly, "[weakly] inaccessible cardinals" today. To me such a move clearly constitutes a huge leap of imagination, much like the earlier one Cantor made when he proved that not all infinite sets have the same size. [even if that should have been clear already from the elementary proof that there can be no bijection between ANY set and its powerset] I suspect there was something in Hausdorff's very rich research project, mostly on the ordinals at the time, that "compelled" him to hypothesize such, to his day, very "large" cardinals; indeed cardinals much larger than, and "unreachable" from, the entire series of Beth cardinals, [at least from Beth1 to Bethω as were already known at the time]? ... I also haven't yet found much written about what the reception of Hausdorff's unreachable cardinals was by his contemporaries. I assume many of the latter must have thought the notion completely preposterous, but even that I haven't been able to confirm.  Any direction anyone can provide to me about such things will be greatly appreciated!</p>

答:
As bof said, I think the notion is actually natural enough that it would be weird to not ask about it. But in fact there is a good reason for Hausdorff to explicitly consider it.

The big question in set theory at the time was of course the continuum hypothesis. In $1904$, Konig had proved an interesting technical result about the continuum, namely $$cf(2^\omega)>\omega$$

(actually he proved something more general). Indeed this was the only nontrivial result about the continuum's cardinality known when Hausdorff wrote his book in $1908$, and as Cohen would observe later the only nontrivial result provable in $\mathsf{ZFC}$ at all. Moreover, Hausdorff had a role in this result: it was extracted from a failed argument against $\mathsf{CH}$ put forth by Konig, whose error was detected by Hausdorff (and independently Konig himself). See here for more on this.

So by the time Hausdorff is writing in $1908$, there's already a known connection between the biggest question around and the notion of cofinality, and Hausdorff himself has been involved in forming this connection. Understanding the role of cofinality in cardinal arithmetic is therefore a very natural topic for him to think about, and among the first questions which arise is: "Are the uncountable regular cardinals exactly the uncountable successor cardinals?" So right there the notion of weak inaccessibility emerges naturally.

(That said I don't know of a source which definitely says that this is why Hausdorff introduced them. But it was definitely "in the air" at the time.)

It's also worth noting that a certain amount of the idea "weakly inaccessible cardinals are really big" took a while to become clear. Specifically, slogans like

weakly inaccessible cardinals are so big we can't prove they exist

are rather common, even if they're fairly misleading (e.g. since we can have $2^\omega$ be weakly inaccessible if weakly inaccessibles are consistent in the first place). Now remember that the idea behind such slogan comes from combining two theorems of Godel, and one communal agreement:

$\mathsf{ZFC}$ captures "what we can prove about set theory."

$\mathsf{ZFC}$ cannot prove that $\mathsf{ZFC}$ has a model (let alone a transitive model!) unless $\mathsf{ZFC}$ is inconsistent.

If $\alpha$ is weakly inaccessible, then $L_\alpha\models\mathsf{ZFC}$.

But $\mathsf{ZFC}$ wouldn't even be introduced (let alone essentially universally accepted) in its modern formulation until the late $20$s, while Godel wouldn't prove $(2)$ and $(3)$ until the early $30$s and $50$s respectively.

OK fine, we can easily sidestep $(2)$ above: it would be enough to show instead that $L_\alpha\models$ "There are no weakly inaccessible cardinals" if $\alpha$ is the least weakly inaccessible cardinal, which follows immediately from the appropriate absoluteness of weak inaccessibility. But as far as I'm aware of, even though it's easier that sort of argument post-dated the second incompleteness theorem, so the point about historicity still stands. EDIT: it turns out that my history is a bit off; absoluteness proofs that $\mathsf{ZFC}$ doesn't prove the existence of a strongly inaccessible cardinal were given in the $20$s. Still, this significantly postdates Hausdorff's work.

And finally, observe that point $(3)$ is actually surprisingly subtle. For strong inaccessibility things are very simple: it's almost trivial to show $V_\alpha\models\mathsf{ZFC}$ whenever $\alpha$ is strongly inaccessible. So as soon as we think about absoluteness, or learn the second incompleteness theorem, it's obvious that strongly inaccessible cardinals reach beyond $\mathsf{ZFC}$. However, this argument breaks down for weak inaccessibility: it need not be the case that $V_\alpha\models\mathsf{ZFC}$ if $\alpha$ is weakly inaccessible, and in fact - as noted above - it's possible that $2^\omega$ itself is weakly inaccessible. Instead, some genuine work is needed here: I don't offhand see a way to show that weakly inaccessible cardinals imply the consistency of $\mathsf{ZFC}$ without developing the basic theory of $L$, which is highly nontrivial.

So our modern understanding of the logical strength of weak inaccessibility wasn't available to Hausdorff in $1908$. Of course it's clear from the definition that weakly inaccessible cardinals are pretty big, but there isn't that much evidence at the time that they're particularly special. So I suspect that their reception was much more muted than a reader brought up on the modern story of large cardinals might expect.


问:
Maximum value of modulus in exponential form
I am trying to find the maximum value of
$\left|e^{i\theta}-2\right|+\left|e^{i\theta}+2\right|\mbox{ for }0\le\theta\le2\pi$.

I can replace $e^{i\theta}$ with $a+ib$ and then proceed to get a function of $a$;
Namely;

$$\sqrt{5+4a}+\sqrt{5-4a}$$
I can then use calculus to find the maximum value of this function of $a$.

My question is: Are there any more efficient ways to solve this problem?

答: Let $f(\theta) = \left|e^{i\theta}-2\right|+\left|e^{i\theta}+2\right|$. Note that $f(\theta)\ge0$. We have $$f^2(\theta) = |e^{i\theta}-2|^2 + |e^{i\theta}+2|^2 + 2|e^{2i\theta}-4| = (e^{i\theta}- 2)(e^{-i\theta}- 2) + (e^{i\theta}+2)(e^{-i\theta}+2) + 2|e^{2i\theta}-4| = 1 -4\cos(\theta) +4 +1+4\cos(\theta)+4 +2|e^{2i\theta}-4| = 10+ 2|e^{2i\theta}-4|$$Also we know that $$|e^{2i\theta}-4|\le |e^{2i\theta}| +|4| = 5$$So we have $$f^2(\theta)\le10+2\times 5 = 20$$We can conclude that $$0\le f(\theta)\le \sqrt{20} = 2\sqrt{5}$$If we can find $\theta_0$ such that $f(\theta_0) =  2\sqrt{5}$, we are done. Indeed, we have $f(\frac{\pi}{2}) = 2\sqrt{5}$. Also, this can be seen by $$f^2(\theta) = 20 \implies |e^{2i\theta} - 4| = 5 \implies |e^{2i\theta} - 4|^2 = 25 \implies 1 - 8\cos(2\theta) + 16 = 25 \implies  8\cos(2\theta) = -8 \implies \cos(2\theta) = -1$$Which has solutions $\theta = \frac{\pi}{2},\frac{3\pi}{2}$ in $0\le \theta \le 2\pi$.


问:
Linear Algebra Question
My question is; How can I find an equation relating $a,b$, and $c$ so that the linear system

$$\begin{cases}2x+y-z=a\\ x-2y-3z=b\\ -3x-y+2z=c\end{cases}$$

is consistent for any values of $a,b$, and $c$ that satisfy that equation.

Thanks,

答:
Dear MAxcoder, if you're gonna read and use this text as a part of a homework exercise or an exam, you should indicate that you did so.

You effectively want an equation relating $a,b,c$ that is consistent with the three equations you wrote. In most cases, the 3 equations you wrote are independent, so it must be possible to derive the fourth equation involving $a,b,c$ only out of the three equations only.

How do you do it? Note that we want an equation that only includes $a,b,c$ - it doesn't include $x,y,z$. So we must eliminate all variables $x,y,z$. Clearly, the right combination of the 3 equations of yours is linear.

So take $a_1$ times the first equation, $a_2$ times the second, $a_3$ times the third, add these three terms, and you must get an expression such that all $x,y,z$ terms cancel. For a general collection of $3\times 3$ coefficients in front of $x,y,z$ in the three equations (I will avoid the term matrix because I don't feel that you know how to use it), that wouldn't be possible.

However, for the particular equations you wrote, it's possible. You need to find $\vec v=(a_1,a_2,a_3)$ such that the inner product of $\vec v$ with $(2,1,-3)$ as well as with $(1,-2,-1)$ as well as with $(-1,-3,2)$ vanishes. The three vectors are the columns from the 3 equations, and the three conditions are needed to cancel $x,y,z$.

In general, this would be an overdetermined system but in this case it's not. I can get $\vec v$ e.g. as the cross product of $(1,-2,-1)$ and $(-1,-3,2)$ which is
$$ (1,-2,-1) \times (-1,-3,2) = (-2\times 2-(-1)(-3), (-1)(-1)-2\cdot 1, 1(-3)-(-2)(-1)) $$
That's
$$(-4-3,1-2,-3-2) = (-7,-1,-5) $$
As a check, this vector is also orthogonal to $(2,1,-3)$ - the inner product is $-14-1+15=0$. So all conditions are satisfied. I forgot to say, a cross product is clever because $(\vec u\times \vec v)\cdot \vec u = 0$ automatically.

So now we must take the corresponding combination of the 3 equations you wrote and the result is
$$ 0 = -7a -b-5c $$
in which the left hand sides, i.e. $x,y,z$, have completely cancelled, by design. Normal people would probably write it as $7a+b+5c=0$.

Title: Linear Algebra Question


问:
Geometric intuition behind the Uniform Boundedness Principle
Is there a way to visualize why the Uniform Boundedness Principle should be true? I understand the statement of the theorem but I'm having a hard time seeing a picture of it in my head.</p>

答: You have a collection $(T_i)_{i\in I}$ of bounded operators, and $(e_j)_j$ a collection of directions. Given a direction, the "orbit" $\{T_ie_j,i\in I\}$ is bounded. This means that there is a bounded set $S_j$ such that applications of the bounded operators to $e_j$ cannot make you leave this set. If the space is "without holes", this was because the operators "had a bounded amplitude". </p>


问:
Solve system of linear equations $Mp_1 = q_1 \\ Mp_2 = q_2$
I need to solve a system of linear equations where $p_1,p_2,q_1,q_2$ are two dimensional vectors and $M$ is a $2\times 2$ matrix.

$$Mp_1  = q_1 \\ Mp_2 = q_2$$

If all the variables were scalars I would solve for $M$ in the first equation and put it into the second equation. But how would I solve for $M$ in the first equation in this case?

答:
I am assuming you want to solve for M in terms of the coefficients of $p_1,p_2,q_1$ and $q_2$. Write

$M = \begin{pmatrix} a &b\\ c &d\end{pmatrix}$,

$p_1 = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}$,
$p_2 = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix}$,
$q_1 = \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}$,
and $q_2 = \begin{pmatrix} d_1 \\ d_2 \end{pmatrix}$,

Then your two equations become:

$\begin{pmatrix} aa_1 +ba_2\\ ca_1+da_2\end{pmatrix} = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix}$

and

$\begin{pmatrix} ac_1 +bc_2\\ cc_1+dc_2\end{pmatrix} = \begin{pmatrix} d_1 \\ d_2 \end{pmatrix}$

respectively

Then you have four equations in the four unknowns a,b,c, and d (two vectors, or matrices more generally, are equal iff they have the same respective entries), so you can solve for them in terms of the coefficients of $p_1,p_2,q_1$ and $q_2$.

However, if you want to solve for the coefficients of M in terms of less variables, I can't really guarantee that can happen and/or will have a unique solution with how the problem is set up.


问:
Impose PDE itself as Boundary Condition?
Consider, for example, the elliptic PDE $u_{x}+u_y+u_{xx}+u_{yy}=0$ for $(x,y)\in[0,\infty)^2$.

Solution methods often require me to impose boundary conditions. Often, these arise naturally from applications (physics, biology, economics etc.). But sometimes, one may not know ex ante how the function should behave on a particular boundary.

Suppose my modeling setting imposes a behavior on $u$ as $x\to\infty$, $y=0$ and $y\to\infty$. However, my application in mind does not give me any priors on what $u$ does as $x=0$. Nonetheless, I have to impose something on this boundary. Can I simply impose that, as $x\to0$, the following inhomogeneous ODE holds
$$u_x(0,y)+u_y(0,y)+u_{xx}(0,y)+u_{yy}(0,y)=0.$$
Because we are on the lower boundary of the $x$-domain, we can impose this conditions via finite (forward) differences,
\begin{align*}
\frac{u_{1,j}-u_{0,j}}{\Delta x}+\frac{u_{0,j+1}-u_{0,j-1}}{2\Delta y}+\frac{u_{2,j}-2u_{1,j}+u_{0,j}}{(\Delta x)^2}+\frac{u_{0,j+1}-2u_{0,j}+u_{0,j-1}}{(\Delta y)^2}=0,
\end{align*}
where $u_{i,j}=u(x_i,y_j)$.

So, instead of coming up with some Dirichlet or Neumann conditions, I essentially just impose the PDE itself on one (or several) of the boundaries. Is this something one has heard of? Do you know of problems with similar boundary conditions?

答:
Since you're using a finite difference method, you're turning the the PDE into a set of algebraic equations, I'm sure you know this.

It seems to me that you won't ever get a solution because you are just adding an equation (expanding the dimensions of the matrix $A$ in the equation $Ax = b$). So your issue is that you don't have the first or last entries of the $b$ vector in the above equation. So you're just finding the kernel of the system, but if the system is non-singular (which I think it is) then the kernel is just zero.

So just a few ideas as far as options go:

since you're working in a semi-infinite domain, maybe try other methods of solution like similarity or characteristics.

If you need to use a finite difference approach, then maybe use some scaling arguments. Does it really matter what value $u$ takes on at that boundary, or is it arbitrary and the relevant physical phenomena is really happening far away from that boundary. Maybe try a bunch of values for $u$ or $\frac{\partial u}{\partial n}$ at the boundary. It looks like you're in luck since your governing equation is linear, so there should be some pattern established when you vary the prescribed boundary condition.

Hope this helps!


问:
How can we say topology on set is collection of open sets?
I have recently started studying topology on my own. Topology $\tau$ on set X is defined as some 3 properties

(1)$\phi, X \in \tau$

(2)Arbitrary union is closed

(3)Finite intersection is closed in $\tau$

But then how can we say that
(1) topology is collection of "open" sets?

(2)for every $x \in X, \exists U_x : x \in U_x, U_x \in \tau$

How to prove (1) and (2)?

答:
Given a topology $\tau$ on a set $X$, a set $V\subseteq X$ is open in $(X,\tau)$ if and only if $V\in\tau$. This is the definition of open and there is nothing to prove.

$X\in \tau$ and $\forall x\in X, x\in X$, therefore the property $[\forall x\in X,\exists U_x, (x\in U_x\land U_x\in \tau)]$ is satisfied by $U_x:=X$.


答:
Let $0<a<b$. By direct calculation, for any $t>0$, $\int_{ta}^{tb}\frac{f(0)}{x}dx=f(0)\ln(b/a).$

Let $\varepsilon>0$ be given. We choose $\delta\in(0,1)$ such that $|f(x)-f(0)|<\varepsilon\cdot\frac{1}{\ln(b/a)}$
whenever $x\in[0,\delta]$. (Note that $\ln(b/a)>0$.) Let
$t_{0}=\frac{\delta}{b}.$ Let $t\in(0,t_{0})$ be arbitrary.
Note that $x\in[ta,tb]\Rightarrow x\in[0,\delta]$, so we have
estimation
\begin{eqnarray*}
 &  & \left|\int_{ta}^{tb}\frac{f(x)}{x}dx-f(0)\ln(b/a)\right|\\
 & = & \left|\int_{ta}^{tb}\frac{f(x)}{x}dx-\int_{ta}^{tb}\frac{f(0)}{x}dx\right|\\
 & \leq & \int_{ta}^{tb}\left|f(x)-f(0)\right|\cdot\frac{dx}{x}\\
 & \leq & \varepsilon\cdot\frac{1}{\ln(b/a)}\cdot\int_{ta}^{tb}\frac{dx}{x}\\
 & = & \varepsilon.
\end{eqnarray*}
This shows that $\lim_{t\rightarrow0+}\int_{ta}^{tb}\frac{f(x)}{x}dx=f(0)\ln(b/a).$

Title: Limit of a integral

Body:
For $f:[0,1]\rightarrow \mathbb{R}$ continuous, let $a >0.$

\begin{eqnarray*}
L=\lim_{\varepsilon \rightarrow 0}\int_{\varepsilon a}^{\varepsilon b} \frac{f(x)}{x} dx
\end{eqnarray*}

Show that $L=f(0)\ln(\frac{b}{a})$.The first I thought was in mean value theorem for integrals. but  I don't know how can start, if can give me some hint to start. I will be gratefull.


问:
Number of grid points outside a rook circuit
When I draw a circuit on a 8x8 chess board that passes through all squares, no matter how many turns I draw, I always end up with 18 grid points outside my circuit (and 31 grid point inside). It looks to me that I get also constant values for any $n$ x $m$ board. Has anybody ever seen a formula of how many grid points are inside/outside for an $n$ x $n$ board? It looks obvious that there are (n/2 - 1) * (n - 2) points outside if $n$ is even, but I am searching for a proof.

If you could point me to some literature, that would be great!

答:
For an $m$ by $n$ array with $mn$ even, consider a lattice on the centres of squares. The path passes through $mn$ lattice points and contains no internal points. So, by Pick's theorem it contains an area of $\frac{mn}{2}-1$.

This area consists of unit squares centred on the 'grid points' of the question and so there are $\frac{mn}{2}-1$ enclosed grid points.


问:
If $\alpha$ is algebraic, then so is $\alpha^2 + \alpha$
Let $\alpha\in L$ be algebraic over $K$ in a field extension $L\supseteq K$, i.e. there exists a polynomial $0\neq f\in K[X]$ which suffices $f(\alpha)=0$.

Is there an elementary way to prove that in this case, $\alpha^2+\alpha$ is also algebraic?

答:
Hint Let $\beta=\alpha^2+\alpha$.

$\alpha$ is algebraic implies that
$$[K(\alpha):K]=n< \infty$$
for some $n$. Now show that $1,\beta, \beta^2,..., \beta^n \in K[\alpha]$ cannot be linearly independent over $K$.


问:
Expansion of a modified Struve function in terms of modified Bessel functions
The Wolfram functions site includes (without reference) the identity (source)

$$L_0(z)=\frac{4}{\pi}\sum_{k=0}^\infty \frac{I_{2k+1}(z)}{2k+1}$$
where $L_0(z)$ is the modified Struve function of order zero and $I_n(z)$ is the $n$th modified Bessel function of the second kind. But from my work in another question, I think this is incorrect as written and should include $(-1)^k$, i.e., it should be alternating. Can someone provide a reference to either result or a direct proof of such?

答:
Here is a direct proof. We recall the Jacobi-Anger expansion
$$e^{z \sin \theta}=I_0(z)+2\sum_{k=0}^\infty (-1)^k I_{2k+1}(z)\sin((2k+1)\theta)+2\sum_{k=1}^\infty (-1)^k I_{2k}(z)\cos(2k\theta).$$
Evaluating the first $\theta$-derivative at $\theta=0$, we obtain the identity

$$z=\frac{d}{d\theta}\left(e^{z\sin \theta}\right)_{\theta=0}
=2\sum_{k=0}^\infty (-1)^k (2k+1)I_{2k+1}(z).$$

We now recall that the modified Bessel function $I_n(z)$ satisfies the differential equation

$$x^2 I_n''(x)+x I_n'(x) -(x^2+n^2)I_n(x)=0.$$ As such, we have

\begin{align}
\frac{2}{\pi}z
&=\frac{4}{\pi}\sum_{k=0}^\infty (-1)^k (2k+1)I_{2k+1}(z)\\
&=(x^2 D_x^2 +x D_x -x^2)\sum_{k=0}^\infty \frac{4}{\pi}(-1)^k (2k+1)^{-1}I_{2k+1}(z)
\end{align}
It follows that the sum
$$y(z):=\frac{4}{\pi}\sum_{k=0}^\infty (-1)^k (2k+1)^{-1}I_{2k+1}(z)$$
is a solution of the inhomogeneous Bessel equation $$x^2 y''(x)+xy'(x)+x^2 y(x)=\frac{2}{\pi}x.$$ Since all odd-integer modified Bessel functions vanish at zero, it follows that $y$ also vanishes at zero. We now note that the zeroth modified Struve function $L_0(z)$ is defined as the unique solution of the above inhomogeneous Bessel equation which vanishes at $z=0$. Therefore we conclude that indeed

$$L_0(z)=\frac{4}{\pi}\sum_{k=0}^\infty (-1)^k \frac{I_{2k+1}(z)}{2k+1}.$$


问:
Green's function. Basic
Can anyone give some advice about books where I could find introductory information about Green's function. What are the methods of constructing Green's function.
Actually, Green's function for 3D diffusion equation in unbounded domain is of interest.
I am novice in this topic (Green's function for ODE and PDE) and any help is appreciated.
Thank you in advance! </p>

答: I think that "Boundary Value Problems and Green's Functions" by Stakgold and Holst is a good text. </p>

Title: Green's function. Basic


问:
Assume $f:[a,b] \to \mathbb R$ is continuous and for every $x \in \mathbb R :f(x) \ge 0$,if $M:=\sup\left\{f(x): x \in [a,b]\right\}$
Assume $f:[a,b] \to \mathbb R$  is continuous and for every $x \in \mathbb R :f(x) \ge 0$,if $M:=\sup\left\{f(x): x \in [a,b]\right\}$,show that the sequence $$\left\{\left(\int_{a}^{b}f^{n}\left(x\right)dx\right)^{\frac{1}{n}}\right\}_n$$ Is convergent to $M$.

If $M=\sup\left\{f(x): x \in [a,b]\right\}$ then exists $x \in [a,b]$ such that foe every $\epsilon>0:M - \epsilon < f(x) \le M$,so :

$$\left(M-\epsilon\right)^{n}\left(b-a\right)=\int_{a}^{b}\left(M-\epsilon\right)^{n}dx\le\int_{a}^{b}f^{n}\left(x\right)dx\le\int_{a}^{b}M^{n}dx=M^{n}\left(b-a\right)$$
Taking the $n$th root of the both sides yields:
$$\left(M-\epsilon\right)\left(b-a\right)^{\frac{1}{n}}\le\left(\int_{a}^{b}f^{n}\left(x\right)dx\right)^{\frac{1}{n}}\le M\left(b-a\right)^{\frac{1}{n}}$$

Taking the limit of both sides:

$$\left(M-\epsilon\right) \le \lim_{n \to \infty} \left(\int_{a}^{b}f^{n}\left(x\right)dx\right)^{\frac{1}{n}} \le M$$

I have some questions:

How do we know that $M-\epsilon$ is not negative?

We used that $(M - \epsilon )^n< f^n(x)\le M^n $ and from here $$\int_{a}^{b}\left(M-\epsilon\right)^{n}dx\le\int_{a}^{b}f^{n}\left(x\right)dx\le\int_{a}^{b}M^{n}dx$$

Why that's true??

答:
You are proving that $||f||_p \rightarrow ||f||_\infty$ as $p\rightarrow \infty$.
See my old solution in this post: How to show that $\|f\|_{L^p} \to \|f\|_{L^\infty}$ as $p \to \infty$ if $f \in L^\infty \cap L^{p_0}$ for some $0 < p_0 < \infty$?


问:
Find the area of triangle $\triangle LMN$.

In triangle $\triangle LMN$. $LO$ is median. Also $LO$ is the bisector of angle. If $LO=3\text{ cm}$ and $LM=5\text{ cm}$. Then find the area of triangle $\triangle LMN$.

Now here we will use similarity between $\Delta LMO$ and $\Delta LON$ triangles but I could not proceed further.

答: Its'  isosceles and acute. Because a acute is smaller than a right angle, and there is two of the sides are the same.</p>


问:
distribution of linearly transformed variable
Let $X$ be a random variable and it takes values from $\ell_{2}$, which is the space of square-summable sequences. Next, let us define random variable $U=a(X - b)$, for some $a \in \mathbb{R}_{+}$ and $b \in \ell_{2}$, also both $a$ and $b$ are not all zeros.

Next, assume that for some random variables $V$ and $Y$, taking values from $\ell_{2}$, defined on, possibly, another probability space we know that $V = a(Y - b)$ and $V\stackrel{d}{=}U$.

Does this mean that

$$
Y\stackrel{d}{=}X
$$
?

Does the result depend on the space which the variables take values from?

答:
I do not believe that the result is dependent on which spaces we define things.

I believe we can even prove a more general result than what you are looking for. Let $\varphi$ be a measurable transformation and define $U = \varphi (X)$ and $V = \varphi(Y)$ and assume that $U \overset{d}{=} V$. For a measurable set $B$ we have
\begin{align*}
\mathbb{P}_{U}(B) &= \mathbb{P}_{\varphi (X)}(B) = \mathbb{P}(X \in \varphi^{-1}(B)) = \mathbb{P}_{X}(\varphi^{-1}(B)) \\
\mathbb{P}_{V}(B) &= \mathbb{P}_{\varphi (Y)}(B) = \mathbb{P}(Y \in \varphi^{-1}(B)) = \mathbb{P}_{Y}(\varphi^{-1}(B))
\end{align*}
So we may conclude that
\begin{align*}
\mathbb{P}_{X}(\varphi^{-1}(B)) = \mathbb{P}_{Y}(\varphi^{-1}(B))
\end{align*}
Implying that $\mathbb{P}_{X} = \mathbb{P}_{Y}$, meaning $X\overset{d}{=}Y$.


问:
Meaning of expression similar to cross product
Given are 2 vectors $\vec{x},\vec{y}$ in $\mathbb{R}^4$. I have a function  $f(\vec{x},\vec{y})$ that looks quite symmetrically. Therefore I think that there must be a meaning or any algebraic rule. What is the meaning of $f$?

$$f(\vec{x},\vec{y})=(x_1y_2-x_2y_1)^2
+(x_2y_3-x_3y_2)^2
+(x_3y_4-x_4y_3)^2
+(x_4y_1-x_1y_4)^2
+(x_1y_3-x_3y_1)^2
+(x_2y_4-x_4y_2)^2$$

The first 4 summands look like the squared norm of a cross product. Of course such a cross product has not properties as in $\mathbb{R}^3$. By cross product I just mean the application of the construction rule of a cross product in  $\mathbb{R}^3$ that is analogically extended to  $\mathbb{R}^4$.

Another interpretation would be the squared norm of a (so far unknown) vector product $\mathbb{R}^4\times\mathbb{R}^4\to\mathbb{R}^6$.

Edit:

For simplicity above was given  only a part but the general form looks like
$$f(\vec{x},\vec{y},\vec{z})=\\
(x_1y_2-x_2y_1)^2
+(x_2y_3-x_3y_2)^2
+(x_3y_4-x_4y_3)^2
+(x_4y_1-x_1y_4)^2
+(x_1y_3-x_3y_1)^2
+(x_2y_4-x_4y_2)^2+\\
(x_1z_2-x_2z_1)^2
+(x_2z_3-x_3z_2)^2
+(x_3z_4-x_4z_3)^2
+(x_4z_1-x_1z_4)^2
+(x_1z_3-x_3z_1)^2
+(x_2z_4-x_4z_2)^2+\\
(y_1z_2-y_2z_1)^2
+(y_2z_3-y_3z_2)^2
+(y_3z_4-y_4z_3)^2
+(y_4z_1-y_1z_4)^2
+(y_1z_3-y_3z_1)^2
+(y_2z_4-y_4z_2)^2
$$

答:
As Ted says in the comments, $f(x, y)$ is the squared norm of the wedge product or exterior product

$$\mathbb{R}^4 \times \mathbb{R}^4 \ni v, w \mapsto v \wedge w \in \wedge^2(\mathbb{R}^4) \cong \mathbb{R}^6$$

which is given explicitly in coordinates by exactly the maps $x_i y_j - x_j y_i$ that you are squaring to compute $f$. This follows from writing $v = \sum x_i e_i, w = \sum y_j e_j$ where $e_i$ is the standard basis and computing that

$$v \wedge w = \sum_{i, j} x_i y_j (e_i \wedge e_j) = \sum_{i < j} (x_i y_j - x_j y_i)(e_i \wedge e_j)$$

since $e_i \wedge e_i = 0$ and $e_j \wedge e_i = - e_i \wedge e_j$. Next we invoke the existence of an inner product on $\wedge^2(V)$ for $V$ an inner product space with the property that if $\{ e_i \}$ is an orthonormal basis for $V$ then $\{ e_i \wedge e_j \}, i < j$ is an orthonormal basis for $\wedge^2(V)$; this inner product can be constructed in a few different ways, for example as a determinant

$$\langle v_1 \wedge w_1, v_2 \wedge w_2 \rangle = \det \left[ \begin{array}{cc} \langle v_1, v_2 \rangle & \langle v_1, w_2 \rangle \\ \langle w_1, v_2 \rangle & \langle w_1, w_2 \rangle \end{array} \right].$$

The definition of the wedge product in coordinates, while concrete, obscures many important features of the wedge product, such as its associativity and how it behaves under change of coordinates.

$f(x, y, z)$ is, also as Ted says, $\| x \wedge y \|^2 + \| y \wedge z \|^2 + \| z \wedge x \|^2$. It can be described more conceptually by reinterpreting $x, y, z \in \mathbb{R}^4$ as the columns of a $4 \times 3$ matrix $M$, thought of as a linear transformation $M : \mathbb{R}^3 \to \mathbb{R}^4$ such that $M(a, b, c) = ax + by + cz$. Then $M$ is the squared Frobenius norm of the wedge product

$$\wedge^2(M) : \wedge^2(\mathbb{R}^3) \to \wedge^2(\mathbb{R}^4).$$

$\wedge^2(M)$ is given explicitly in coordinates by a $6 \times 3$ matrix whose entries are the $2 \times 2$ minors of $M$, which again are exactly the maps that you are squaring to compute $f$. And again this definition, while concrete, obscures many important features of the wedge product, such as its multiplicativity: we have $\wedge^2(MN) = \wedge^2(M) \wedge^2(N)$ but working in coordinates makes this very far from obvious.


问:
Simplified expression for Frobenius norm of hessian of matrix function
Let $X \in \mathbb R^{n \times d}$ and $y \in \mathbb R^{n \times 1}$ and $u \in \mathbb R^{k \times 1}$. Define the function $F:\mathbb R^{d \times k} \to \mathbb R$ by $F(W) := y^\top \psi(XW)u$ for some twice-differentiable function $\psi:\mathbb R \to \mathbb R$, and $\psi(A)$ is the matrix with $ij$th entry $\psi(a_{i,j})$.

Question. What is a simplified expression for the Frobenius norm of the hessian $\|\nabla^2 F(W)\|_{Fro}^2$ ?

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}\def\R{{\mathbb R}}\def\v{{\rm vec}}\def\d{{\rm diag}}\def\D{{\rm Diag}}$Let a colon denote the trace/Frobenius product, i.e.
$$\eqalign{
A:B &= {\rm Tr}(A^TB) = B:A \\
A:A &= \big\|A\big\|_F^2\\
}$$
Define the matrices
$$\eqalign{
Z &= XW \qquad&\implies\quad dZ = X\,dW \\
P &= \psi(Z)  \qquad&\implies\quad dP = Q\odot dZ \\
Q &= \psi'(Z) \qquad&\implies\quad dQ = R\odot dZ \\
R &= \psi''(Z) \\
}$$
where $(\odot)$ denotes the elementwise/Hadamard product and $\,(\psi',\psi'')\,$ denote the ordinary first and second derivatives of the $\psi$ function.

Write the cost function in terms of the above, then calculate its gradient.
$$\eqalign{
F &= yu^T:P \\
dF &= yu^T:dP &({\rm differential}) \\
  &= yu^T:Q\odot X\,dW \\
  &= Q\odot yu^T:X\,dW \\
  &= X^T\left(Q\odot yu^T\right):dW \\
\p{F}{W}
  &= X^T\left(yu^T\odot Q\right) \;\doteq\; G\qquad&({\rm gradient}) \\
}$$
Next, calculate the differential of $G$.
$$\eqalign{
dG
 &= X^T\left(yu^T\odot dQ\right) \\
 &= X^T\left(R\odot yu^T\odot X\,dW\right) \\
}$$
The Hessian is going to be a fourth-order tensor, so we're stuck.

One way to proceed is to note that the component-wise self derivative of $W$ is
$$\eqalign{
\p{W}{W_{ij}} &= E_{ij} \;\in\,\R^{d\times k} \\
}$$
where $E_{ij}$ is the matrix with the $(i,j)$ component equal to one and all others equal to zero.

Then the components of the Hessian ${\cal H}$ can be calculated as
$$\eqalign{
{\cal H}_{ijk\ell} &= \p{G_{ij}}{W_{k\ell}} \\
  &= E_{ij}:\Big(X^T\left(R\odot yu^T\odot XE_{k\ell}\right)\Big) \\
  &= x_i^T\big(r_j\odot y\odot x_k\big)\;u_j\delta_{j\ell} \\
}$$
where the vectors $(x_k,r_k)$ are the $k^{th}$ columns of the $(X,R)$ matrices.

Finally, the Frobenius norm can be calculated for this tensor
$$\eqalign{
\big\|{\cal H}\big\|^2_F
 &= \sum_{h=1}^d\sum_{i=1}^k\sum_{j=1}^d\sum_{\ell=1}^k\;{\cal H}_{hij\ell}^2 \\
 &= \sum_{h=1}^d\sum_{i=1}^k\sum_{j=1}^d\;\Big(x_h^T\big(r_i\odot y\odot x_j\big)\;u_i\Big)^2  \\\\
}$$
UPDATE

Since you're only interested in the norm, the $(G,W)$ matrices can be flattened into vectors $(g,w)$. This in turn flattens the Hessian tensor into a matrix (i.e. ${\cal H}\to H$) and allows all calculations to be carried out in matrix notation without explicit summations over the components.

For typing convenience, define the variables
$$\eqalign{
I &\in\R^{k\times k}&&\big({\rm Identity\,matrix}\big) \\
A &= I\otimes X\quad\quad&\implies\quad&C = AA^T = I\otimes XX^T \\
B &= \D(b)\quad&\implies\quad&b = \v(R\odot yu^T) \\
g &= \v(G)\quad&\implies\quad&w = \v(W) \\
}$$
Then
$$\eqalign{
dG &= X^T(R\odot yu^T\odot X\,dW) \\
dg &= (I\otimes X)^T\;\big(b\odot\v(X\,dW)\big) \\
   &= (I\otimes X)^TB\;\v(X\,dW) \\
   &= (I\otimes X)^TB\,(I\otimes X)\;dw \\
\p{g}{w} &= A^TBA \;\doteq\; H \qquad\big({\rm Hessian\,matrix}\big) \\
}$$
Now compute the matrix norm as usual.
$$\eqalign{
\big\|H\big\|_F^2
 &= A^TBA :A^TBA  \\
 &= B:CBC \\
 &= b:\d(CBC) \\
 &= b:(C\odot C)b \\
 &= b^T\left(C\odot C\right)b \\
}$$


问:
REVISTED$^1$: Circumstantial Proof: $P\implies Q \overset{?}{\implies} Q\implies P$
To prove that if a matrix $A\in M_{n\times n} ( F )$ has $n$ distinct eigenvalues, then $A$ is diagonalizable is enough to show that the opposite holds? That is, if $A$ is diagonalizable, then $A$ has distinct eigenvalues?

Please don't answer the actual question. I want to do it myself. I just want to know if it suffice to show the opposite in this particular case.

EDIT$^1$:

If $A$ is similar to a diagonlizable matrix, then $A$ is diagonalizable, right?

答:
To help on this point of logic, let me try something I suggest to my students. You take the sentence

If it's raining, then the ground is wet.

as a reasonable one. And the contrapositive

If the ground is dry, then it is not raining.

seems equivalent and totally reasonable, as well. (So these are logically equivalent statements.)

However, the converse

If the ground is wet, then it is raining.

is certainly not believable. It may have rained hours ago. Or someone spilled some water.

Concrete examples may help you deal with the logical structure in the mathematical theorems.