问:
is it necessary for a function to be defined everywhere in it's domain to be onto or one-to-one mapping?
[this is question. I want to know about iv.] I want to know that without being defined everywhere ,can a  mapping be onto and one-to-one?
In iv - D has four elements and B has three elements, while in question only three elements are used. So it cannot be function as per definition. Then how it will onto or one to one without function? Is it possible?
1</p>

答:
This is a matter of terminology and notation. In Set  Theory a function $is$ its graph, and when we say "$f$ is a function from $A$ to $B\,$", or write $f: A\to B$ we mean that $f\subseteq A\times B$ and $\forall a\in A\,\,\exists! \,b\in B\,(\,(a,b)\in f)$... (where "$\exists!$" means "there exists exactly one"). And $A$ is called the domain of $f$.

A source of confusion is that strict adherence to this definition is sometimes tedious or inconvenient. E.g. suppose $f:\Bbb R\setminus \{0\}$ where $f(x)= e^{-1/|x|},\,$ and suppose you  say that we extend $f$ to all of $\Bbb R$ by letting $f(0)=0$. No one will find fault with this. But technically there is no such thing as $f(0)$. A pedantic adherence to the def'n would require you to say "Let $g=f\cup \{(0,0)\}.\,$" Just remember that there are no exceptions to the def'n, but that sometimes it is easier to act as  if there were.


问:
Upper Bound of Sobolev norm by $L_2$ norm
A Paper by Madych and Potter states that if a function $f\in W_2^k(\mathbb{R})$ has evenly spaced zeroes (i.e. if $Z(f):=\{x:f(x)=0\}$, is such that $\underset{y\in\mathbb{R}}\sup dist(y,Z(f))=h<\infty$), then we can estimate $$\|f\|_{L_2}\leq Ch^k|f|_{W_2^k}$$

My question is, is there some upper bound going the other way?  That is, can we have $$|f|_{W_2^k}\leq C\|f\|_{L_2}$$
for some absolute constant, or maybe it has to be in terms of $h$.

答:
Their bound is essentially the Poincaré inequality of order $k$: for a function vanishing on the boundary of a domain (the domain being an interval between two zeroes) we can estimate the $L^2$ norm of the function by its Sobolev norm, taking into account the size of the domain.

There is no hope for reverse inequality, because a function being small does not imply that the derivative is small. You can place zeroes wherever you want and make the graph  wiggly however you want.

One situation in which it's possible to bound a derivative by the size of function itself is when the function solves an elliptic PDE.

ContentLicense: CC BY-SA 3.0


问:
Finding particular solution of an ODE
What is the particular solution of the following ODE:

$$y''(t)+y(t)=-\sin{t}|\sin{t}|$$

My idea is to take a Fourier expansion of $|\sin{t}|$ which makes the right side to be
$$-\sin{t}\left[{2\over\pi}-{4\over\pi}\sum_{k=1}^\infty{\cos(2k t)\over 4k^2-1}\right]$$
If so far it's correct, then what to do next?

答:
Not sure that you need to use the Fourier series for this ODE.

For $t \in [0,\pi]$ the equation writes as

$$y^{\prime \prime}(t)+y(t)=-\sin^2{t} = \frac{\cos 2t -1}{2}$$ and a particular solution is

$$y_0(t) = -\frac{1}{2}\left(1 + \frac{1}{3} \cos 2t\right)$$ that you can find using ODE methods by finding a solution that writes as $a \cos 2t +b$.

And the general solution on $[0,\pi]$ is
$$y(t) = A\cos t + B \sin t + y_0(t).$$

You can follow on doing similar things on $[-\pi,0]$ to get global solutions.

ContentLicense: CC BY-SA 4.0

Title: Finding particular solution of an ODE


问:
Compute the power of this test, $f(x;\theta)=\theta x^{\theta-1}$, $0<x<1$ and $\theta>0$
Let $X$ be a random variable with pdf $f(x;\theta)=\theta x^{\theta-1}$, $0<x<1$ and $\theta>0$. Consider $H_0: \theta=1/2$ versus $H_A: \theta=1/4$

a) Derive the most powerful test using $\alpha=0.05$

b) Compute the power of this test.

I have shown the first part of the exercise, but I don't know how to show the second part.

Answer a):
$$T = \frac{L\left(\theta = \frac{1}{2}\right)}{L\left(\theta = \frac{1}{4}\right)} = 2x^{1/4}$$
$$2x^{1/4}<c$$
$$x < \frac{c^4}{16}$$
$$c_2 = \frac{c^4}{16}$$
$$\alpha = P\left( \text{Type I error} \right) = P\left( \text{reject} \space \space H_0\mid H_0 \space \space \text{true} \right) = 0.05$$
$$\alpha = P\left(x < c \mid \theta = 1/2\right)$$
$$=\int_0^{c_2}{f\left(x;1/2\right)dx}$$
$$\sqrt{c_2} = 0.05$$
$$c_2 = 0.0025$$
$$x < 0.0025$$

I don't know how to compute the power of this test or how to start. The only thing I know, is that I have to find the following:

$$1-\beta = P(\text{Reject} \space H_0 \mid H_A \space \text{true})$$
where $\beta$ is the probability of a Type II error.

Thanks in advance

答:
I don't know how to compute the power of this test or how to start

simply using the definition that is

$$\gamma=\mathbb{P}[X \in C|H_1]$$

Where $C$ is the rejection region. Thus

$$\gamma=\int_0^{0.0025} 0.25 x^{0.25-1}dx=\frac{1}{2\sqrt{5}}\approx22.36\%$$


问:
Eigenvalues of powered matrices $A^3$ and $A^3 - 5I$
Question part A

Question Part B

Im not too sure how to go about this question, I just multiplied the matrix by itself three times and then solved for eigenvalues that way, for part B I'm lost as to how to express it, something along the lines of $A^4 = 2A^3 = 4A^2 $ etc. hoping anyone could shed some light onto how to go about this

答:
The eigenvalues of $A$ are $1,1,1,3$ with multiplicity !

Part A:

If $ \lambda$ is an eigenvalue of $A$, then $ \lambda^3$ is an eigenvalue of $A^3$ and $\lambda^3-5$ is an eigenvalue of $A^3-5I.$

Part B: the characteristic polynomial $p$ of $A$  is

$$ p(x)=(x-1)^3(x-3).$$

Cayley Hamilton gives: $p(A)=0.$

Can you proceed ?


问:
What happens after the cardinality $\mathfrak{c}$?
While having measure theory this year the following came in my mind:

  When we go from finite objects to infinite we "lose" a lot of properties. For example
  the summation isn't well defined anymore, when the sum doesn't converge, and operations
  which are abelian in the finite case are not abelian in the infinite case (think of conditional convergent series).

When we go from countable to uncountable we have things like that too. Measures are most time countable additive but not uncountable additive (at least the lebesgue measure). Sums of uncountable indexing sets only can converge when they are in fact countable (meaning nearly all are 0).

So my question is after Arthur Comments slightly rephrased

What happens when you go from the cardinality $\mathfrak{c}$ to a higher cardinality? Are there more intermediate steps, meaning having a cardinality strictly greater than $\mathfrak{c}$ but not "to big"?

答:
I hope I'm answering your question here.

First, the issue with conditionally convergent series is well known. In a sense, we can consider such series to diverge. The sum is highly sensitive to the order of summation and moreover the series can be re-arranged to give any sum at all. That is not the kind of behaviour you would expect of a convergent object. In the theory of integration there is a similar issue. The integral $\int_{i\infty }^\infty \sin(x)$ does not exist, even though, in some sense, it can be argued to be $0$. Moreover, by chopping the interval $(-\infty ,\infty)$ at suitable places, the integral can be 'computed' by splitting the integral as a sum of integrals to have any value you want. This is why often in integration theory a function whose absolute value is not integrable is not considered integrable.

Just to formalize things a bit, I'll concentrate on the following property, phrased in $\mathbb R$ and then generalized. Given a set of non-negative real numbers $S\subseteq [0,\infty ]$, its sum is defined to be the supremum of the set $\{s_1+\cdots s_k\mid k>0, s_i\in S\}$. This definitions agrees with the common definition of the sum of a series of non-negative elements. (If we allow $S\subseteq [-\infty , \infty ]$, then $S$ will have a sum iff it is countable and the elements give rise to an absolutely convergent series (historical remark: this is essentially how Bolzano defined infinite sums).) Now, it is easy to show that if the cardinality of $S$ is greater than $\aleph_0$, then the sum is $\infty $. This is a property of the non-negative reals $[0,\infty ]$ considered as a lattice with a compatible addition. This is the reason why in measure theory only countable additivity is entertained.

Now, in other lattices this cardinality restriction on sums need not hold. That is, consider a lattice $V$ with a notion of addition that is compatible (in whatever reasonable way) with the lattice structure (a common demand is that addition distributed over arbitrary joins and/or meets). For a subset $S\subseteq V$ one can defined its sum in $V$ as the supremum in $V$ of the same set of sums derived from $S$ as above. In an arbitrary lattice it is not the case the the sum will be $\infty $, i.e., the top element in the lattice, if the cardinality of $S$ is larger than $\aleph_0$. It is easy to construct artificial examples that will show that. I'm sure there are more practical situations where this happens, but I can't think of one right now.

This turned out to be a rather lengthy answer, so now I hope even more that it at least addressed what you were asking.


问:
Prove these numbers are composite: $500^{27}-123^{12}$, $12^{75}+21^{75}$, $58^{32}-49^{32}$

I have to prove the next 3 numbers are composite numbers

$500^{27}-123^{12}$

$12^{75}+21^{75}$

$58^{32}-49^{32}$

I have no idea where to begin, if anyone can give me a hint I will be grateful.

Thank You Very Much!

答:
Hint:

Can you write this in the form $a^3-b^3$? Now recall the formula for the difference of two cubes.

$\gcd(12, 21)= 3$, so you can find a common factor.

Use the difference of two squares.


问:
Why isn't every Hamel basis a Schauder basis?
I seem to have tripped on the common Hamel/Schauder confusion.

If $X$ is any vector space (not necessarily finite dimension) and $B$ is a linearly independent subset that spans $X$, then $B$ is a Hamel basis for $X$.

If there exists a sequence $(e_n)$ such that for every $x \in X$ there exists a unique sequence of scalars $(\alpha_n)$ for which $\lim_{n \to \infty} || x - (\alpha_1e_1 + \cdots + \alpha_ne_n)|| = 0$, then $(e_n)$ is a Schauder basis for $X$.

So I'm tempted to think that every Hamel basis is also a Schauder basis; just extened the finite linear combination into an infinite one by adding zero coeeficients. I know I'm wrong, but what am I missing?

答:
Interestingly, once once knows that not every Schauder basis is a Hamel basis, this implies almost immediately that not every Hamel basis is a Schauder basis:

A Schauder basis is clearly linearly independent, so it can be extended to a Hamel basis. If it wasn't a Hamel basis to begin with, then this new Hamel basis contains an element not contained in the original Schauder basis. But this element can be written as a countable linear combination of the Hamel basis in two different ways: Once just one times itself, and once as countable linear combination of elements from the original Schauder basis. So the new Hamel basis cannot be a Schauder basis.


问:
How can I prove that I can find $N$ real elements that is algebraically independent over $\mathbb{Q}$ for any N$?
I was wondering how can one show that there exist $N$ real elements that is algebraically independent over $\mathbb{Q}$ for any $N$? (I was thinking perhaps Lindemann–Weierstrass theorem can be used. If we can find $N$ real linearly independent  over $\mathbb{Q}$ algebraic numbers then the statement follows by Lindemann–Weierstrass. but I wasn't sure how to find such $N$ real algebraic numbers...)

Any comments and suggestions appreciated!

答: If you alread have $\alpha_1,\ldots,\alpha_{N-1}$ that are algebriacally independent, then $F=\Bbb Q(\alpha_1,\ldots,\alpha_n)$ is still countable, hence $F[X]$ is countable, hence there are only countably many real numbers that are algebraic ocer $F$. Pick one of the uncountably many others as your $\alpha_N$.


问:
How valid is the concern over narrow pipe cryptographic hash function designs?
Narrow pipe hash function designs have recently come under fire, particularly in reference to some SHA-3 candidates. Is this criticism valid? Can it be explained more simply than this paper does?</p>

答:
You're asking whether an ongoing research-level discussion has merits or not.  This is difficult to know since experts are actively working on this stuff and you never know what can happen: criticism can fizzle out (as did Courtois's algebraic attacks on AES, after causing a major scare 10 years ago) or pay off (as did Xiaoyun Wang's MD5 attacks).

You want my opinion?  I'll give it: existential proofs for hash function security (which is what the cited paper in your question offers) do not hold much sway.  Unless you can exhibit an actual attack, I don't think the alarm bells should be ringing.  Here's an appropriate example:

Consider some SHA-3 hash function $H$.  (For the uninitiated, a cryptographic hash function is a map from a string of any length to one of some specified fixed length; it's a public function; there is no key.)  Since $H$ has an infinite domain and a finite range, so by the pigeon-hole principle, there exists a collsion.  Therefore $H$ is not collision-resistant and thus it's insecure.

Bogus right?  Existential proofs don't mean much in a computational setting.  Of course there are collisions, but the hope is that we cannot find them.

That said, there are other papers giving unconvincing attacks against narrow-pipe constructions, so even in the absence of concrete practical attacks, sentiment will grow against the approach even if there will never be any reason to reject it.  As sad as it is, cryptography is still partly religious when it comes to designing primitives (meaning we rely a lot on intuition and instinct rather than on proofs, because proofs of security don't exist for hash functions and blockciphers and integer factorization, etc).

Edited to Add: I prefer to remain anonymous here.  My PhD is in cryptography, I have 5-6 papers related to cryptographic hashing, and I'm a professor at a "pretty good" place in the U.S.  You can read the above opinion with this in mind, or you can discard it as "suspect" and "unsubstantiated."  I'll leave that to you.


问:
Existence of Lipschitz reparametrization
Suppose we are given a continuous path,

$$\gamma:[0,1]\rightarrow (X,d)\text{,}$$

in a metric space $(X,d)$. When we deal with differentiable enough paths in Riemann manifolds we can give a parametrization that is $l$-Lipschitz, where $l$ is the length of the path. If have that $\gamma$ has finite length $l$ in the sense that
$$l=\sup\left\{\sum_{i=0}^{n-1}d(\gamma(t_i),\gamma(t_{i+1}))\,|\, n\in \mathbb{N}; \forall i< n,t_i\in[0,1]\text{ and }t_i<t_{i+1};t_0=0;t_n=1\right\}$$
is finite, can we guarantee the existence of a parametrization that makes the path $l$-Lipschitz?

Note: The necessity for this result comes that I need this fact to prove that certain sequences of path in metric spaces have nice properties in order to obtain a limit path. And in this way proving that compact path metric spaces are geodesic metric spaces.

答:
Yes, every rectifiable path can be parametrized by arclength (and a locally rectifiable path too, if you allow the parameter domain to be an unbounded interval). </p>

<p>Let $L(x)$ be the length of $\gamma_{|[0,x]}$. This is a nondecreasing function of $x$. It is constant precisely on the intervals where $\gamma$ is constant. Thus, $\gamma\circ L^{-1}$ is well defined. This is the desired parametrization. </p>

Body:
Yes, every rectifiable path can be parametrized by arclength (and a locally rectifiable path too, if you allow the parameter domain to be an unbounded interval).

Let $L(x)$ be the length of $\gamma_{|[0,x]}$. This is a nondecreasing function of $x$. It is constant precisely on the intervals where $\gamma$ is constant. Thus, $\gamma\circ L^{-1}$ is well defined. This is the desired parametrization.


问:
Inverse of a function $B_r(0) \to \mathbb{R}^n$

Definition. Let $B_r(0) := \{x \in \mathbb{R}^n \mid \|x\| < r\}$ be the open ball in $\mathbb{R}^n$ with radius $r>0$ and center $0:=(0,\ldots,0) \in \mathbb{R}^n$.

Let $s_r : B_r(0) \to \mathbb{R}^n$ be the map, defined by $$s_r(x) = \frac{r}{\sqrt{r^2-\|x\|^2}} \cdot x $$

Does this function have an inverse? If so, how does $s_r^{-1} : \mathbb{R}^n \to B_r(0)$ look like?

答:
So, for injectivity, let $x,y \in B_r$ with $s_r(x)=s_r(y)$. Then, it follows that
$$
        \begin{alignat*}{3}
            && s_r(x) &= s_r(y) \\
            &\iff& \frac{rx}{\sqrt{r^2-\|x\|^2}} &= \frac{ry}{\sqrt{r^2-\|y\|^2}} \\
            &\iff& \underbrace{\frac{1}{\sqrt{r^2-\|x\|^2}}}_{\text{scalar}} \underbrace{x}_{\text{vector}} &= \underbrace{\frac{1}{\sqrt{r^2-\|y\|^2}}}_{\text{scalar}} \underbrace{y}_{\text{vector}}
        \end{alignat*}
$$
Since vectors in $\mathbb{R}^n$ are equal iff all components are equal, it suffices to prove that the scalar factors are the same:
$$
        \begin{alignat*}{3}
            &\iff& \frac{1}{\sqrt{r^2-\|x\|^2}} &= \frac{1}{\sqrt{r^2-\|y\|^2}} \\
            &\iff& \frac{1}{r^2-\|x\|^2} &= \frac{1}{r^2-\|y\|^2} \\
            &\iff& r^2-\|x\|^2 &= r^2-\|y\|^2 \\
            &\iff& \|x\|^2 &= \|y\|^2 \\
            &\iff& \|x\| &= \|y\|
        \end{alignat*}
$$
This should prove the injectivity of $s_r$.


答:
The right side is odd-symmetric, so that there exists an odd-symmetric solution. To get $y(0)=0$ as required for that symmetry use a term of the general solution so that on $[0,\pi]$
$$\begin{align}
y(t)&=-\frac12\left(1+\frac13\cos(2t)-\frac43\cos(t)\right)
=-\frac12\left(\frac43(1-\cos(t))+\frac13(\cos(2t)-1)\right)
\\
&=-\frac12\left(\frac43\frac{\sin^2(t)}{1+\cos(t)}-\frac23\sin^2(t)\right)
\end{align}$$
Now it is easy to change the sign like the sign of $\sin(t)$
$$
y(t)=-\frac13\sin(t)|\sin(t)|·\frac{1-\cos(t)}{1+\cos(t)}
$$
Note that the roots of the denomitor cancel against the roots of the sine terms, one could also transform that back to
$$
y(t)=-\frac13{\rm sign}(\sin t)·(1-\cos(t))^2
$$

Title: Finding particular solution of an ODE

Body:
What is the particular solution of the following ODE:

$$y''(t)+y(t)=-\sin{t}|\sin{t}|$$

My idea is to take a Fourier expansion of $|\sin{t}|$ which makes the right side to be
$$-\sin{t}\left[{2\over\pi}-{4\over\pi}\sum_{k=1}^\infty{\cos(2k t)\over 4k^2-1}\right]$$
If so far it's correct, then what to do next?


问:
Proof $\vec{r} \cdot (\vec{r})'=0\iff ||\vec{r}||=K,$ with $K \in \mathbb{R}$.
I need to proof the following property:

Let $r: \mathbb{R}\to \mathbb{R}^n$ a smooth vectorial function, then

$\vec{r} \cdot (\vec{r})'=0\iff ||\vec{r}||=K,$ with $K
 \in \mathbb{R}$.

I could prove the $\leftarrow$ direction, but i could not with the right direction.

Any hint is appreciated.

答:
Notice that $r \cdot r: \mathbb{R} \to \mathbb{R}$ is differentiable since $r: \mathbb{R} \to \mathbb{R}^n$ is smooth. Differentiating
$$
(r \cdot r)'=r' \cdot r+r \cdot r'=2(r \cdot r')=2 \cdot 0=0.
$$
Hence $r \cdot r=k$ where $k \geq 0$ is a constant. Then
$$
||r||=\sqrt{r \cdot r}=\sqrt{k}.
$$
Let $K=\sqrt{k}$ and $||r||=K$.


问:
Differentiating $x^{x^{x^{...}}}$

How do I differentiate $$ x^{x^{x^{...}}}$$ with respect to $x$? (Note that $x$ is raised infinitely many times.)

My attempt: Let $y = x^{x^{x^{...}}}$. Taking logarithm of both sides we get $\ln y = y \ln x$ and let $f = y \ln x - \ln y$. Now $$\frac{dy}{dx} = -\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}} = \frac{y^2}{x(1 - y \ln x)}$$
Is this approach correct? If not how do I proceed ?

Help would be appreciated. Thanks.

答:
The way I would do this:
$$y=x^{x^{x^{\cdots}}}=x^y=\exp(y\ln x)$$
where I've used the notation $\exp(...)$ instead of $e^{\cdots}$ to make the working neater. Now, differentiating implicitly with respect to $x$, and using the fact that $\frac{d}{dx}e^{f(x)}=f'(x)e^{f(x)}$gives
$$\begin{align}\frac{dy}{dx}&=\left(\frac{dy}{dx}\ln x+\frac{y}{x}\right)\exp(y\ln x)\\
&=\left(\frac{dy}{dx}\ln x+\frac{y}{x}\right)x^y\end{align}$$
Rearrange to isolate $\frac{dy}{dx}$:
$$\frac{dy}{dx}\left(1-x^y\ln x\right)=\frac{yx^y}{x}\implies\frac{dy}{dx}=\frac{yx^y}{x(1-x^y\ln x)}=\frac{y^2}{x(1-y\ln x)}$$

which is the same result that you got; your way seems fine :)

I hope that was useful. If you have any questions please don't hesitate to ask :)

Title: Differentiating $x^{x^{x^{...}}}$


问:
My son's Sum of Some is beautiful! But what is the proof or explanation?
My youngest son is in $6$th grade. He likes to play with numbers. Today, he showed me his latest finding. I call it his "Sum of Some" because he adds up some selected numbers from a series of numbers, and the sum equals a later number in that same series. I have translated his finding into the following equation:
$$(100\times2^n)+(10\times2^{n+1})+2^{n+3}=2^{n+7}.$$

Why is this so? What is the proof or explanation? Is it true for any $n$?

His own presentation of his finding:

Every one of these numbers is two times the number before it.
  $1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$.

  I pick any one of them, times $100$. Then I add the next one, times $10$. Then I skip the next one. Then I add the one after that.

  If I then skip three ones and read the fourth, that one equals my sum!

Body:
Proof by Induction.

Base case for $n = 1$ is $100 \cdot 2^1 + 10 \cdot 2^2 + 2^4 = 200 + 40 + 16 = 256 = 2^8 = 2^{1+7}$.

Now to prove that it is true for $n+1$, given it is true for $n$.

$f(n+1) = 100 \cdot 2^{n+1} + 10 \cdot 2^{n+2} + 2^{n+4}$ $= 2 \cdot (100 \cdot 2^n + 10 \cdot 2^{n+1} + 2^{n+3})$ $= 2 \cdot 2^{n+7} = 2^{(n+1)+7}$.


问:
Differentiating a column with respect to a matrix
Let $\mathbf{X} = [\mathbf{x}_1 | ... | \mathbf{x}_n]$ be a $m \times n$ matrix.
I would like to differentiate $\mathbf{x}_i = \mathbf{X} \mathbf{e}_i$ (where $\mathbf{e}_i \in \mathbb{R}^{n \times 1}$ is the unit vectors with $1$ on the $i$th place and $0$'s in the rest) with respect to $\mathbf{X}$. Then
$$
d\mathbf{x}_i = d(\mathbf{X}\mathbf{e}_i) = (\mathbf{X} + d\mathbf{X})\mathbf{e}_i - \mathbf{X}\mathbf{e}_i = (d\mathbf{X})\mathbf{e}_i
$$
and therefore
$$
\frac{d\mathbf{x}_i}{d\mathbf{X}} = \mathbf{e}_i \in \mathbb{R}^{n \times 1}
$$
However, I suspect that is not consistent dimension-wise. For example: $f(\mathbf{X}) = \mathbf{a} \mathbf{x}_i$ where $\mathbf{a} \in \mathbb{R}^{1 \times m}$ then simply using the result above
$$
\frac{d f(\mathbf{X})}{d\mathbf{X}} = \frac{d(\mathbf{a}\mathbf{x}_i)}{d\mathbf{X}} = \mathbf{a} \mathbf{e}_i \implies \mbox{Dimensions mismatch!}
$$
since $\mathbf{a} \in \mathbb{R}^{1 \times m}$ and $\mathbf{e}_i \in \mathbb{R}^{n \times 1}$.

How to fix this issue? An idea is to put a pseudo identity matrix
$$
\frac{d\mathbf{x}_i}{d\mathbf{X}} = \mathbf{I}_{m \times n} \mathbf{e}_i \in \mathbb{R}^{n \times 1}
$$
such that $\mathbf{X} = \mathbf{X} \circ \mathbf{I}_{m \times n}$ with Hadamard product. But is this the right way to go?

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}\def\E{{\cal E}}$Use
$(\star)$ to denote the dyadic product
and a colon to denote the double-dot product, i.e.
$$\eqalign{
\Gamma &= A\star B \quad&\implies\quad
\Gamma_{ijk\ell} = A_{ij}B_{k\ell} \\
Y &= \Gamma:X \quad&\implies\quad
Y_{ij}= \sum_{k,\ell}\;\Gamma_{ijk\ell}X_{k\ell} \\
}$$
First, rewrite the linear equation $b=Xa\,$ using index notation
$$\eqalign{
b_i &= X_{ik}\,a_k \\
 &= \delta_{ij} X_{jk}\,a_k \\
 &= \delta_{ij} a_k\,X_{jk} \\
}$$
where $\delta_{ik}$ is a Kronecker delta; these are simply the components of the identity matrix $I$.

Rewrite the linear equation using the dyadic and double-dot products, and then calculate its differential and gradient.
$$\eqalign{
b &= (I\star a):X \\
db &= (I\star a):dX \\
\p{b}{X} &= (I\star a) \\
}$$
Finally, substitute $(a=e_i,\;b=Xa=x_i)\;$ to obtain
$$\eqalign{
\p{x_i}{X} &= I\star e_i \\
}$$


问:
Proving independence when Gaussian noise is added
Consider r.v.'s $X$ and $Y$ which are not independent, i.e. $E[X\mid Y]\neq E[X]$. Can we show that if we add a variable $\epsilon$ following $\mathcal N(0,\sigma_{\epsilon})$, for which it holds that $\epsilon \perp X$ and $\epsilon \perp Y$, the following limit holds:

$$\lim_{\sigma_{\epsilon} \to \infty } E[X\mid Y+\epsilon ]=E[X] $$

Any help would be greatly appreciated!

答:
The question has been edited after this answer was posted.

Supppose $X=Y \sim N(0,1)$ and $\epsilon=\sqrt {\sigma_{\epsilon}} X$. Then your conditions are satisfied but $E(X|Y+\epsilon)=E(X|(1+\sqrt {\sigma_{\epsilon}})X)=X $ which does not tend to $EX=0$.


问:
$x^p-c$ has no root in a field $F$ if and only if $x^p-c$ is irreducible?
Hungerford's book of algebra has exercise $6$ chapter $3$ section $6$ [Probably impossible with the tools at hand.]:

Let $p \in \mathbb{Z}$ be a prime; let $F$ be a field  and let $c \in
 F$. Then $x^p - c$ is irreducible in $F[x]$ if and only if $x^p - c$
  has no root  in $F$. [Hint: consider two cases: char$(F) = p$ and
  char$(F)$ different of $p$.]

I have attempted this a lot. Anyone has an answer?

答: Let $\mathrm{char}(F)=p$ and $P(X)=X^p-c$ be reducible in $F[X]$ and $L=K(a)$, where $a$ is a root of $P(X)$. Note that in $L[X]$ can write $P(X)=(X-a)^p$. Let $f(X)$ be an irreducible factor of degree $1\leq n<p$ in $F[X]$. Now, $\gcd(P(X),f(X))=f(X)$ and gcd is independent of the field extension, so $f(X)=(X-a)^n$. Hence, $a^n, na^{n-1} \in F$. Since $n$ is coprime to $p$, it follows $a^{n-1}\in F$ and $a \in F$.


问:
Given that the chosen ticket is a first class ticket, what is the probability that it is a long distance ticket?
The question is:

Train tickets are distributed into three boxes. Each ticket can be classified as either First Class (FC) or Third Class (TC) and as either Long Distance (LD) or Short Distance (SD). Each box contains the following proportions tickets of each type:

\begin{array}{|c|c|c|c|}
\hline
\text{Box}& \text{FC and SD} & \text{FC and LD} & \text{TC and SD} & \text{TC and LD} \\ \hline
\text{Box 1}& 10\% & 25\% & 15\% & 50\%\\ \hline
\text{Box 2}& 24\% & 22\% & 33\% & 21\%\\ \hline
\text{Box 3}& 40\% & 40\% & 10\% & 10\%\\ \hline
\end{array}

A box is randomly selected and a ticket is randomly chosen from that box. Given that the chosen ticket is a first class ticket, what is the probability that it is a long distance ticket?

So far I have the following:

 Let $C_1$ and $C_2$ denote the events of choosing either a first class ticket or a third class ticket, respectively. Further, let $D_1$ and $D_2$ denote the events of choosing a short distance or long distance ticket, respectively. Also, let $B_1,B_2,B_3$ denote the events of choosing the first, second or third box. We are asked what is the probability that it is a long distance ticket, given that the chosen ticket is a first class ticket. I have:

$$\Pr \left( {{D_2}\left| {{C_1}} \right.} \right) = \frac{{\Pr \left( {{D_2} \cap {C_1}} \right)}}{{\Pr \left( {{C_1}} \right)}} = \frac{{\sum\limits_{n = 1}^3 {\Pr \left( {{D_2} \cap {C_1} \cap {B_n}} \right)} }}{{\sum\limits_{n = 1}^3 {\Pr \left( {{C_1} \cap {B_n}} \right)} }}$$

 Is it correct that I'm applying the law of total probability? If so, the numerator is easy to find from the table, but how would I find ${\sum\limits_{n = 1}^3 {\Pr \left( {{C_1} \cap {B_n}} \right)} }$?

答:
Simply looking at your table, the probability to get a First class ticket is

$$\mathbb{P}[FC]=\frac{1}{3}\cdot0.35+\frac{1}{3}\cdot0.46+\frac{1}{3}\cdot0.80$$

similarly, the probability to get a ticket that is FC AND LD is

$$\mathbb{P}[FC \cap LD]=\frac{1}{3}\cdot 0.25+\frac{1}{3}\cdot0.22+\frac{1}{3}\cdot0.40$$

your answer is the ratio of the two above probabilities.

But before calculating it observe that, being the probability to choose a single box uniformly, this can be written off.

thus

$$\mathbb{P}[LD|FC]=\frac{0.25+0.22+0.4}{0.35+0.46+0.8}=\frac{87}{161}$$


问:
Is $x+2$ invertible in given quotient ring?

Given the quotient ring $R=\mathbb{Q}[x]/(x^2-7)$. Is $x+2$ invertible in $R$?

I guess it is, but how to find it?

答:
Hint:

$x^2-7$ and $x+1$ don't have any common roots, so they are coprime in $\Bbb Q[x]$. This means you can use the extended Euclidean algorithm to find $a(x)$ and $b(x)$ such that $1=a(x)(x^2-7)+b(x)(x+1)$. What does this equation look like in your ring?


问:
Converting to plane polar coordinates
I am trying to find the periodic orbits of the following ODE:

$\ddot{x}-(1-x^{2}-\dot{x}^{2})\dot{x}+x=0$

Now in my book it says that it is easiest to see the orbits of this ODE if it is converted to polar coordinates. But if it is just a single ODE (i.e. not a system) is it still possible to create some sort of decoupled system of ODE's from this using polar coordinates?

答:
Multiply with $2\dot x$ and semi-integrate to find
$$
\frac{d}{dt}(\dot x^2+x^2)=2\dot x^2(1-(\dot x^2+x^2))
$$
so it can seem as a good idea to set $r^2=\dot x^2+x^2$ and use the circle equation to define an angle $\theta$ so that
$$
\dot x=r\cosθ,~~x=r\sinθ
$$
where both $r$ and $θ$ are time-dependent functions.

Title: Converting to plane polar coordinates


问:
Interpreting divergence of velocity field
The wikipedia article on divergence describes one interpretation of divergence: "The velocity of the air at each point defines a vector field. While air is heated in a region, it expands in all directions, and thus the velocity field points outward from that region."

If we have a vector field which represents a force, I interpret the divergence as representing the strength of the field at whatever point it's taken at. However I'm confused on how to interpret the divergence of a velocity field. Clearly if the divergence is positive gas is expanding outward and if it's negative it's contracting, however what quantity is actually represented?

If I have a velocity field with $m/s$ units, then the div presumably has $m/s^2$ units. Is the quantity we get the actual acceleration of gas away from that point?

答:
In solid mechanics, the divergence of the velocity field can be seen as the volumetric strain rate of the solid, which is an invariant of the strain rate tensor.
$$
div(v_i) = \frac{\partial v_i}{\partial x_i} = \dot{\varepsilon}_{ii}
$$
where $\dot{\varepsilon}_{ij}$ is the strain rate tensor. P.S.: summation convention holds.


问:
Let $a, b, c$ be the three roots of $x^3-(k+1)x^2+kx+12=0$, where $k$ is a real number. If $(a-2)^3+(b-2)^3+(c-2)^3=-18$, find the value of $k$

Let $a, b, c$ be the three roots of the equation $x^3-(k+1)x^2+kx+12=0$, where $k$ is a real number. If $(a-2)^3+(b-2)^3+(c-2)^3=-18$, find the value of $k$.

Here is a problem from a Hong Kong competition that I found a way to solve, but I am interested in knowing other methods since my brute-force way doesn't seem to be the solution intended.

My attempt

Let $y=x-2$, the equation becomes
$$(2+y)^3-(k+1)(2+y)^2+k(2+6)+12=0$$
$$y^3+(5-k)y^2+(8-3k)y-2k+16=0$$
Let $y_1, y_2, y_3$ be the new roots of the equation. By Vieta's formula, we have
\begin{cases}
y_1+y_2+y_3=-5+k \\
y_1y_2+y_2y_3+y_1y_3=8-3k\\
y_1y_2y_3=2k-16
\end{cases}
And by $y_1^3+y_2^3+y_3^3 = (y_1+y_2+y_3)^3-3(y_1+y_2+y_3)(y_1y_2+y_2y_3+y_1y_3)+3y_1y_2y_3$, we have
\begin{align*}
y_1^3+y_2^3+y_3^3 &= (y_1+y_2+y_3)^3-3(y_1+y_2+y_3)(y_1y_2+y_2y_3+y_1y_3) +3y_1y_2y_3\\
&= (-5+k)^3-3(-5+k)(8-3k)+3(2k-16)\\
&=k^3-6k^2+12k-53
\end{align*}
By $y_1^3+y_2^3+y_3^3=-18$ given in the problem, we have
$$k^3-6k^2+12k-35=0$$
And $k=5$ is the only real solution.

答:
Here is Stinking Bishop's method:

The condition on $a,b,c$ gives $(a^3 + b^3 + c^3) + (-2 \cdot 3)(a^2 + b^2 + c^2) + (4 \cdot 3)(a + b + c) - 3 \cdot 2^3 = -18$ or that $(a^3 + b^3 + c^3) - 6(a^2 + b^2 + c^2) + 12(a + b + c) = 6$.

Then as you said, $(a+b+c)^3 - 3(a+b+c)(ab+bc+ca) + 3abc - 6((a+b+c)^2 - 2(ab+bc+ca))$ $+\ 12(a+b+c) = 6$, so by Vieta, $(k+1)^3 - 3(k+1)(k) + 3(-12) - 6((k+1)^2 - 2k) + 12(k+1) = 6$, or $k^3 + 3k^2 + 3k + 1 - 3k^2 -3k - 36 - 6k^2 - 12k - 6 + 12k + 12k + 12 = 6$, which implies $f(k) = k^3 - 6k^2 + 12k - 35 = 0$.

Descartes' rule of signs implies there are either three or one positive roots, and since $f(-k) = -k^3 - 6k^2 - 12k - 35 = 0$, there are no negative roots. Together with the rational root theorem, the only rational candidates (it is unlikely $k$ is irrational in a contest problem) are the factors of $35$: $35, 7, 5, 1$, and $5$ is a root. Dividing by $k = 5$ gives $k^2 - k + 7$ where $\Delta > 0$, hence $k = 5$ is the only solution.


问:
Proof that the intersection of all conjugates of a subgroup is normal.
I'm trying to prove that if $H \leq G$, then the intersection of all conjugates of $H$ is a normal subgroup of $G$, ie $\bigcap_{g \in G}gHg^{-1} \trianglelefteq G $.

I know this question has been asked before, and I have read an answer here Intersection of conjugate subgroups is normal of which I used the main idea in one of the replies. Basically I'm wondering if the conclusion to my proof, ie that the kernel of the homomorphism is the intersection of all the conjugates, is actually correct, but for completeness I'll upload the entire proof attempt.

Suppose $H \leq G$. Consider the map $\pi : G \rightarrow Sym( \frac{G}{H} )$, where in this context the set $\frac{G}{H}$ indicates the set of all left cosets of $H$ in $G$. For each $g \in G$, $\pi$ associates $g$ to a mapping $\pi_g : \frac{G}{H} \rightarrow \frac{G}{H}$, $\pi_g(fH) := (gf)H \in \frac{G}{H}$. For it to be true that $\pi_g \in Sym(\frac{G}{H})$ we must show that $\pi_g$ is a one to and onto function. To this end, suppose $\pi_g(xH) = \pi_g(yH)$. This gives us $(gx)H = (gy)H$. Applying $g^{-1}$ to the left of this equation gives $xH = yH$ proving $\pi_g$ is injective. If $yH \in \frac{G}{H}$, let $xH = g^{-1}yH$ which means that $\pi_g(xH) = gxH = gg^{-1}yH = yH$ and $\pi_g$ is onto.  Now, it's also easy to show that $\pi$ is a homomorphism. To do this we need to show that given $g,f \in G$, $\pi_{gf}(xH) = \pi_g \circ \pi_f(xH)$, for all $xH \in \frac{G}{H}$. But $\pi_{gf}(xH) = (gfxH) = g(fx)H = g(\pi_f(xH)) = \pi_g \circ \pi_f(xH)$. \

Now, if $ k \in ker(\pi)$, then $\pi_k(gH) = kgH = gH$ for all $gH \in \frac{G}{H} $. But $kgH = gH \iff g^{-1}kg = h_g \in H$ for some $h_g$ that depends on $g$. This means that if $k \in ker(\pi)$, then $k = gh_gg^{-1} \in gHg^{-1} \implies ker(\pi) = \{ k : k \in gHg^{-1} \text{for all } g \in G \} $.  In other words, $ker(\pi) = \bigcap_{g \in G}gHg^{-1}$. Since the kernel of every homomorphism from $G$ is a normal subgroup, the result follows.

答:
You should rather prove that $\pi$ is a group homomorphism.

Then $\pi_{g^{-1}}=(\pi_g)^{-1}$ will immediately prove that $\pi_g$ is a bijection.

You formally only showed $\ker\pi\subseteq\bigcap_g gHg^{-1}$.

However, the other inclusion follows basically by the same argument.


问:
How to check if number is harmonic divisor or not
I need to write a program which checks if the given number is harmonic divisor or not. So I search on the internet and found a definition of it but couldn't understand really what is harmonic divisor.

This is what I found : https://archive.lib.msu.edu/crcmath/math/math/h/h085.htm

There is some functions in the given link. d() and something like q() (I couldn't type that char). I don't know what this functions are.

Thanks for any help or explanation.

(note: I don't know if sharing link is allowed on this site. If it is not allowed please warn me. I will edit the question.)

答:
The harmonic divisor numbers are OEIS sequence A001599
$$   1, 6, 28, 140, 270, 496, 672, 1638, 2970, 6200, 8128, 8190,\dots $$
The Wikipedia article Harmonic divisor number states:

In mathematics, a harmonic divisor number, [...], is a positive integer whose divisors have a harmonic mean that is an integer.

It also states:

The harmonic mean $\,H(n)\,$ of the divisors of any number $\,n\,$ can be expressed as the formula
$$ H(n) = \frac{n\,\sigma_0(n)}{\sigma_1(n)} $$
where $\,\sigma_i(n)\,$ is the sum of the $i$th powers of the divisors of
$\,n:\sigma_0\,$ is the number of divisors, and $\,\sigma_1\,$ is the sum
of the divisors (Cohen 1997).

and this:

$$ H(4) = \frac{3}{1+\frac12+\frac14} = 12/7, $$
$$ H(5) = \frac{2}{1+\frac15} = 5/3, $$
$$ H(7) = \frac{2}{1+\frac17} = 7/4, $$
and $$ H(140) = H(4\cdot 5\cdot 7) = H(4)\cdot H(5)\cdot H(7)
 = \frac{12}7 \cdot \frac53 \cdot \frac74 = 5. $$


问:
Showing a set function is a premeasure

Definition: Let $S$ be a collection of subsets of a set $X$ and $\mu\colon S\to[0,\infty]$ a set function. Then $\mu$ is called a premeasure provided $\mu$ is both finitely additive and countably monotone and, if $\emptyset$ belongs to $S$, then $\mu(\emptyset) = 0$.

Question: Consider the collection $S = \{\emptyset, [0, 1], [0, 3], [2, 3]\}$ of subsets of $\mathbb{R}$ and define $\mu(\emptyset) = 0$, $\mu([0, 1]) = 1$, $\mu([0, 3]) = 1$, $\mu([2, 3]) = 1$. Show that  $\mu\colon S\to[0,\infty]$ is a premeasure.

Firstly, we'll show finite additivity. Notice, $\{\emptyset,[0, 1]\}$, $\{\emptyset,[0, 3]\}$,
$\{\emptyset,[2, 3]\}$, $\{\emptyset\}$, $\{[0, 1]\}$, $\{[0, 3]\}$, $\{[2, 3]\}$ are all the finite collections of disjoint sets in $S$ whose unions of each induvial collection if back in $S$.

Further, notice for each induvial collection,

$$\mu(\bigcup\{\emptyset,[0,1]\}) = \mu([0,1]) = 1  = 0 + 1 = \mu(\emptyset) + \mu(\{0,1\})$$

$$\mu(\bigcup\{\emptyset,[0,3]\}) = \mu([0,3]) = 1  = 0 + 1 = \mu(\emptyset) + \mu(\{0,3\})$$

$$\mu(\bigcup\{\emptyset,[2,3]\}) = \mu([2,3]) = 1  = 0 + 1 = \mu(\emptyset) + \mu(\{2,3\})$$

$$\mu(\bigcup\{\emptyset\}) = \mu(\emptyset) = 0  = \mu(\emptyset)$$

$$\mu(\bigcup\{[0,1]\}) = \mu([0,1]) = 1  = \mu([0,1])$$

$$\mu(\bigcup\{[0,3]\}) = \mu([0,3]) = 1  = \mu([0,3])$$

$$\mu(\bigcup\{[2,3]\}) = \mu([2,3]) = 1  = \mu([2,3]).$$

Therefore, the set function $\mu$ is finitely additive.

Now, we'll show countably monotone. For $\emptyset$ any cover will have a sum of measure equal to $0$ or greater than $0$. For $[0,1]$ any cover will have a sum of measure equal to $1$ or greater than $1$. For $[0,3]$ any cover will have a sum of measure equal to $1$ or greater than $1$. For $[2,3]$ any cover will have a sum of measure equal to $1$ or greater than $1$. So, indeed whenever a set $E\in S$ is covered by a countable collection
$\{E_k\}_{k=1}^{\infty}$ of sets in $S$, then the measure of $E$ is less than the measure of the cover.

Therefore, the set function $\mu$ is countably monotone.

As $\emptyset$ belongs to $S$, and $\mu(\emptyset)=0$ by construction, we have that $\mu$ is a premeasure.

My question is, is the above correct?

答:
Your argument is correct. Just small improvements might be useful.

Definition: Let $S$ be a collection of subsets of a set $X$ and $\mu\colon S\to[0,\infty]$ a set function. Then $\mu$ is called a premeasure provided $\mu$ is both finitely additive and countably monotone and, if $\emptyset$ belongs to $S$, then $\mu(\emptyset) = 0$.

Question: Consider the collection $S = \{\emptyset, [0, 1], [0, 3], [2, 3]\}$ of subsets of $\mathbb{R}$ and define $\mu(\emptyset) = 0$, $\mu([0, 1]) = 1$, $\mu([0, 3]) = 1$, $\mu([2, 3]) = 1$. Show that  $\mu\colon S\to[0,\infty]$ is a premeasure.

Firstly, we'll show finite additivity. Notice, $\{\emptyset,[0, 1]\}$, $\{\emptyset,[0, 3]\}$ and $\{\emptyset,[2, 3]\}$, $\{\emptyset\}$, $\{[0, 1]\}$, $\{[0, 3]\}$, $\{[2, 3]\}$ are essentially all the finite collections of disjoint sets in $S$ whose unions of each induvial collection if back in $S$. (We could add additional "empty sets" to those collections, but they would not change the finite additivity, because $\mu(\emptyset)=0$ ).

For collections with a single set (that is  $\{\emptyset\}$, $\{[0, 1]\}$, $\{[0, 3]\}$, $\{[2, 3]\}$ ), it is immediate the finite additivity.

Since $\mu(\emptyset)=0$, it is also immediate the finite additivity for $\{\emptyset,[0, 1]\}$, $\{\emptyset,[0, 3]\}$ and $\{\emptyset,[2, 3]\}$.

Now, we'll show  $\mu$ is countably monotone. Let $E \in S$.

If $E=\emptyset$, any countable cover will have a sum of measure equal or greater to $0$.

If $E= [0,1]$,   any countable cover will have a sum of measure equal  or greater to $1$.

For $E=[0,3]$ any countable cover will have a sum of measure equal  or greater to $1$.

For $E=[2,3]$ any countable cover will have a sum of measure equal  or greater to $1$.

So, indeed whenever a set $E\in S$ is covered by a countable collection
$\{E_k\}_{k=1}^{\infty}$ of sets in $S$, then the measure of $E$ is less than the sum of  measures of the elements in the cover.

Therefore, the set function $\mu$ is countably monotone.

Finally,  $\emptyset$ belongs to $S$, and $\mu(\emptyset)=0$ by the definition of $\mu$. So, we have that $\mu$ is a premeasure.

Remark: For finite additivity, if you want to work case by case, it would be

$$\mu(\bigcup\{\emptyset,[0,1]\}) = \mu([0,1]) = 1  = 0 + 1 = \mu(\emptyset) + \mu([0,1])$$

$$\mu(\bigcup\{\emptyset,[0,3]\}) = \mu([0,3]) = 1  = 0 + 1 = \mu(\emptyset) + \mu([0,3])$$

$$\mu(\bigcup\{\emptyset,[2,3]\}) = \mu([2,3]) = 1  = 0 + 1 = \mu(\emptyset) + \mu([2,3])$$

$$\mu(\bigcup\{\emptyset\}) = \mu(\emptyset) = 0  = \mu(\emptyset)$$

$$\mu(\bigcup\{[0,1]\}) = \mu([0,1]) = 1  = \mu([0,1])$$

$$\mu(\bigcup\{[0,3]\}) = \mu([0,3]) = 1  = \mu([0,3])$$

$$\mu(\bigcup\{[2,3]\}) = \mu([2,3]) = 1  = \mu([2,3]).$$

Title: Showing a set function is a premeasure


Title: Let $a, b, c$ be the three roots of $x^3-(k+1)x^2+kx+12=0$, where $k$ is a real number. If $(a-2)^3+(b-2)^3+(c-2)^3=-18$, find the value of $k$

Body:
Let $a, b, c$ be the three roots of the equation $x^3-(k+1)x^2+kx+12=0$, where $k$ is a real number. If $(a-2)^3+(b-2)^3+(c-2)^3=-18$, find the value of $k$.

Here is a problem from a Hong Kong competition that I found a way to solve, but I am interested in knowing other methods since my brute-force way doesn't seem to be the solution intended.

My attempt

Let $y=x-2$, the equation becomes
$$(2+y)^3-(k+1)(2+y)^2+k(2+6)+12=0$$
$$y^3+(5-k)y^2+(8-3k)y-2k+16=0$$
Let $y_1, y_2, y_3$ be the new roots of the equation. By Vieta's formula, we have
\begin{cases}
y_1+y_2+y_3=-5+k \\
y_1y_2+y_2y_3+y_1y_3=8-3k\\
y_1y_2y_3=2k-16
\end{cases}
And by $y_1^3+y_2^3+y_3^3 = (y_1+y_2+y_3)^3-3(y_1+y_2+y_3)(y_1y_2+y_2y_3+y_1y_3)+3y_1y_2y_3$, we have
\begin{align*}
y_1^3+y_2^3+y_3^3 &= (y_1+y_2+y_3)^3-3(y_1+y_2+y_3)(y_1y_2+y_2y_3+y_1y_3) +3y_1y_2y_3\\
&= (-5+k)^3-3(-5+k)(8-3k)+3(2k-16)\\
&=k^3-6k^2+12k-53
\end{align*}
By $y_1^3+y_2^3+y_3^3=-18$ given in the problem, we have
$$k^3-6k^2+12k-35=0$$
And $k=5$ is the only real solution.

Body: using the identity $$a^3+b^3+c^3=3abc+\frac{(a+b+c)}{2} ({(a-b)}^2+{(b-c)}^2+{(c-a)}^2)$$ $$-18=3(a-2)(b-2)(c-2)+\frac{a+b+c-6}{2}({(a-b)}^2+{(b-c)}^2+{(c-a)}^2)$$ Now $$x^3-(k+1)x^2+kx+12=(x-a)(x-b)(x-c)$$ $$\to (a-2)(b-2)(c-2)=2k-16$$ now use $${(a-b)}^2+{(b-c)}^2+{(c-a)}^2={(a+b+c)}^2-3(ab+bc+ca)={(k+1)}^2-3k$$ combining these we get the cubic $$k^3-6k^2+12k-35=0$$


问:
$\frac{1}{\sqrt{1+(\frac{\Omega}{\Omega_c})^{2N}}}$ as $\Omega\to\infty$
I have a function: $$\rvert H_a(\Omega)\lvert = \frac{1}{\sqrt{1+(\frac{\Omega}{\Omega_c})^{2N}}}$$

My professor says in the function as $\Omega \rightarrow \infty$, $1+(\frac{\Omega}{\Omega_c})^{2N} \rightarrow (\frac{\Omega}{\Omega_c})^{2N}$ and hence $\rvert H_a(\Omega)\lvert \rightarrow (\frac{\Omega}{\Omega_c})^{-N}$ as $\Omega \rightarrow \infty$. which does not make any sense.

As per my thinking, as $\Omega$ tends to $\infty$, $1+(\frac{\Omega}{\Omega_c})^{2N}$ tends to $\infty$ and hence $\rvert H_a(\Omega)\lvert \rightarrow 0$.

Which is the correct result?and if my professor is right can anyone please help me understand the rationale behind it?

The function$\rvert H_a(\Omega)\lvert$ is the magnitude response of a butterworth filter.

答:
First, "$|H_a(\Omega)|\to (\Omega/\Omega_c)^{−N}$" is not a limit statement, since $\Omega$ still appears. Rather, this describes the asymptotic behavior of the function.

As for why $|H_a(\Omega)|\to (\Omega/\Omega_c)^{−N}$ holds ... Large $\Omega$ eventually outpaces any constant $\Omega_c$, so that $\Omega/\Omega_c$ is itself large. Raising a large number to the $2N$ power makes a LARGE number. Adding $1$ to such a thing has virtually-no effect, so you can reasonably ignore it. Doing so, the expression simplifies:
$$\frac{1}{\sqrt{1+(\Omega/\Omega_c)^{2N}}}\approx \frac{1}{\sqrt{0+(\Omega/\Omega_c)^{2N}}}=\frac{1}{(\Omega/\Omega_c)^{N}}=(\Omega/\Omega_c)^{−N}$$


问:
Operations with spherically symmetric functions
I know that a function $f: \mathbb{R}^n \to \mathbb{R}$ is spherically symmetric if it is invariant under the action of an orthogonal transformation. That is, $f(Ox)=f(x)$, where $O$ is any orthogonal matrix. Now, my questions are:

Given $f$ spherically symmetric function in $\mathbb{R}^n$, is its Fourier Transform spherically symmetric, as well?

Given $f$ and $g$ spherically symmetric functions in $\mathbb{R}^n$, is the product between $f$ and $g$ spherically symmetric, as well?

答:
The answer to (1) is yes; this is because  if $O$ is orthogonal then so is $O^T$, and $O^T$ is hence measure-preserving::

$$\hat f(O\xi)=\int f(x)e^{ix\cdot(O\xi)}\,dx=\int f(x)e^{i(O^Tx)\cdot\xi}\,dx
=\int f((O^T)^{-1}x)e^{ix\cdot\xi}
=\int f(x)e^{ix\cdot\xi}=\hat f(\xi).$$

The second question is much easier.


问:
Irreducible representations and Abelian subgroups
There is a theorem in representation theory which is surprising to me: the dimension of irreducible (complex) representation of finite group is not greater that $(G:H)$ - an index of Abelian subgroup $H\subset G$. Can anyone provide a reference or a proof of this statement?

答:
Note that if $V$ is any irreducible representation of $G$ and $H$ is a subgroup of $G$, then $V$ appears in the induced representation $\mathrm{Ind}(\mathrm{Res}(V))$, and this in fact implies that every irreducible representation of $G$ appears (as a summand of) $\mathrm{Ind}(W)$ for some irreducible representation of $H$.

If $H$ is abelian, every irreducible representation is one dimensional, and $\mathrm{Ind}(W)$ has dimension $[G:H]$, so the claim follows.


问:
Proving that for all numbers $a, b\in\mathbb{R}$, $\min \{a, b\} \le (a+b)/2$
Can anyone help me solve the two cases that are derived from this problem?</p>

答:
Hint: You may as well assume that $a\le b$. (Why is that okay?)

In fact, you should be able to show that $\min\{a,b\}$ is the midpoint of $a,b$ minus half the distance between $a,b$--that is, $$\min\{a,b\}=\frac{a+b}2-\frac{|a-b|}2.$$


问:
Finding a total ordering of function growth rates
For $f,g:\Bbb R_+\to\Bbb R$, let's define $f<g$ to mean $\exists x_0,\forall x>x_0,f(x)<g(x)$.

Observations about $<$:

It's finer than the ordering provided by O-notation. For example, $x<x+\frac 1x<x+1$.

If $m:\Bbb R\to\Bbb R$ is increasing, then $m\circ f<m\circ g\iff f<g$. (In contrast, O-notation distinguishes $\exp(x)$ from $\exp(2x)$ but not $x$ from $2x$.)

If $m:\Bbb R\to\Bbb R$ is increasing and surjective, then $f\mapsto m\circ f$ is an order isomorphism of $<$.

For any functions $a<b$, the interval $\{f:a<f<b\}$ is order-isomorphic to the whole space of functions (via squeezing $\Bbb R$ into $(a(x),b(x))$ for each $x$). So this order has a lot of symmetry and self-similarity.

$<$ is not a total order. In fact, for any $f$, we can find an interval of mutually-comparable functions that are all incomparable with $f$, such as $\{g_\alpha:\alpha\in[-1,1]\}$ where $g_\alpha(x)=f(x)+\sin(x)+\alpha$.

To understand the space of limiting behaviors captured by $<$, I'd like to describe a maximal chain of it.
Beyond maximality, I feel like I should forbid the chain from having "coarse" segments that can be replaced by a larger set of mutually-comparable functions to obtain a "finer" chain. That is, our chain $C$ should have the property that if $(C\setminus X)\cup Y$ is also a chain, then $|X\cap C|\ge|Y\setminus C|$. This implies maximality, so let's call such a chain "finely maximal". (Does this condition seem appropriate? I'm not confident that cardinality is the right tool for the job.)

Then my questions are:

Is there a finely maximal chain of $<$?

Are all finely maximal chains order-isomorphic to each other?

Is there a canonical choice of finely maximal chain? For example, modulo exchanging functions that are eventually identically equal, is there exactly one finely maximal chain of $<$ containing the $0$ function?

(If yes to any of the above) Is there a simple description of this order structure?

答:
This is only a very partial answer:

It is unclear to me whether there exists a finitely maximal chain but if there is one, I highly doubt that there be a unique one modulo identifying functions that are eventually equal to one another.

Using diagonal definitions of functions, one sees that maximal chains are countably saturated as linear orders. In particular, they are uncountable, and any two maximal chains of cardinality $\aleph_1$ are order isomorphic.

On a related notes, you can look up the notion of Hardy fields (see for instance the first pages of the article On numbers, germs, and transseries), which are chains with a structure of differential field. In particular, you may be interested in maximal Hardy fields.


问:
Transposing ligand binding equation
Dear community members,

I am out of school for more than 10 years and now have to transpose an equation for my work to $x$. I already tried to find a solution, but failed. I would be very grateful for some explanations.

$$y = \frac{(Bx)}{(K+x)} + Nx$$

Thank you in advance,

Anton

答:
$$y=\frac{Bx}{K+x}+Nx$$
$$y(K+x)=Bx+Nx(K+x)$$
$$yK+yx=Bx+NKx+Nx^2$$
$$Nx^2+(NK+B-y)x-Ky=0$$

Now use the quadratic formula:

$$x=\frac{-NK-B+y\pm\sqrt{(NK+B-y)^2+4NKy}}{2N}$$

(there will be up to two solutions for $x$, taking the sign $+$ or $-$ in the numerator).

Title: Transposing ligand binding equation


问:
Cauchy Criterion for Sequences as opposed to Series
So through as I've been venturing through baby Rudin I came upon his definition of a cauchy sequence:

A sequence $\{ p_n \}$ in some metric space $X$ is said to be cauchy if
$$
\forall \; \epsilon > 0 \; \exists \; N \in \mathbb{N} \; s.t. d(p_n, p_m) < \epsilon \; \forall \; n,m \ge N
$$
He then talks about the cauchy criterion for convergence being that a cauchy sequence converges to a point in its contained metric space (similarly this metric space would be called complete). He goes on to say that the cauchy criterion for convergence for series can be restated as the following:

A series $\sum a_n$ converges if and only if
$$
\forall \; \epsilon > 0 \; \exists \; N \in \mathbb{N} \; s.t. \left\lvert \sum_{k=n}^{m} a_k \right\rvert \le \epsilon \; \forall \; n,m \ge N
$$
Note that these series live in $\mathbb{R}^k$
I can see how one would get

A series $\sum a_n$ converges if and only if
$$
\forall \; \epsilon > 0 \; \exists \; N \in \mathbb{N} \; s.t. \left\lvert \sum_{k=n}^{m} a_k \right\rvert < \epsilon \; \forall \; n,m \ge N
$$

but I don't understand why the new definition has a $\le$ rather than a $<$

Could anybody advise me as to why?

Thanks in advanced!

答:
The two formulations are equivalent; proving the first from the second is trivial.

For the other, apply the statement of 1 to $\epsilon/2$; then the $N$ obtained will suffice for $\epsilon$ in formulation 2.

Occasionally, allowing equality will simplify one's proof (because it makes the choice of $N = N(\epsilon)$ easier).

In fact, we can do better than that, and give a useful form for determining convergence that allows you to skip right through tedious computations when the sufficiency of $N$ is apparent -- typically of the form "when $N$ satisfies this complicated inequality involving $N$ and $\epsilon$".

For functions $f,g$, define a variant of Big O notation:

$f \in \mathcal O(g; 0)$ iff there exist $\delta$ and $C$ such that $|x|<\delta$ implies $|f(x)| < C\, |g(x)|$.

Now we have the following result (where $\epsilon$ in $\mathcal O(\epsilon;0)$ signifies the identity function):

Let $(p_k)_k$ be a real sequence (or, for that matter, a sequence in any metric space). Then $(p_k)_k$ is Cauchy iff there exists a function $f(\epsilon) \in \mathcal O(\epsilon; 0)$ such that:

$$\forall \epsilon > 0: \exists N: \forall m,n>N: d(p_m,p_n) < f(\epsilon)$$

The proof is not hard; expanding the given property of $f$ is really all there is to it. Of course, we can apply this to the sequence of partial sums of a series to obtain the Cauchy criterion for convergence of series (in a suitable space, e.g. a complete normed vector space, or the more down to earth $\Bbb R^k$).


答:
The first thing I would do is subtract Nx from both sides: $y- Nx= \frac{Bx}{K+ x}$.

Now get rid of that fraction by multiplying both sides by K+ X:  $(y- Nx)(K+ x)= Ky-KNx+ xy- Nx^2= Bx$.

Combine all terms on the left side:
$Bx- KNx- xy- Nx^2+ Ky= 0$

Just because I like my leading term to be positive, multiply both sides by -1:
$Nx^2+ (KN- B- y)x- Ky- 0$.

Finally, since this involves "$x^2$" use the "Quadratic Formula" to solve for x:
$x= \frac{-KN+ B+ Y\pm\sqrt{(KN-B-y)^2+ 4NKy}}{2N}$

Title: Transposing ligand binding equation

Body:
Dear community members,

I am out of school for more than 10 years and now have to transpose an equation for my work to $x$. I already tried to find a solution, but failed. I would be very grateful for some explanations.

$$y = \frac{(Bx)}{(K+x)} + Nx$$

Thank you in advance,

Anton


问:
Can't solve an Integral, how to ask questions.
\begin{equation*}
\int ^{\frac{\pi }{2}}_{-\frac{\pi }{2}}\frac{\cos x}{e^{x} +1} dx
\end{equation*}

Is there a generally accepted format to ask questions on MSE?
Also can someone tell me how to write integrals, fractions and equations from keyboard....
Thanks!

答:
To write your integral,

```<code>$$ \int_{-\pi/2}^{\pi/2} \left( \frac{\cos x}{1+e^{x}} \right) \, dx. $$
</code>```

produces

$$\int_{-\pi/2}^{\pi/2} \left( \frac{\cos x}{1+e^{x}} \right) \, dx.$$

Also appreciated is when you state what you have tried, for example:

$$\begin{aligned} 1.& \text{ substitution} \\ 2.& \text{ integration-by-parts} \\ 3.& \textit{ etc.}\end{aligned}$$

Body:
To write your integral,

$$ \int_{-\pi/2}^{\pi/2} \left( \frac{\cos x}{1+e^{x}} \right) \, dx. $$

produces

$$\int_{-\pi/2}^{\pi/2} \left( \frac{\cos x}{1+e^{x}} \right) \, dx.$$

Also appreciated is when you state what you have tried, for example:

$$\begin{aligned} 1.& \text{ substitution} \\ 2.& \text{ integration-by-parts} \\ 3.& \textit{ etc.}\end{aligned}$$


问:
What is minimal polynomial over a skew field?
I am reading "Polynomial extensions of skew fields" by J. Treur (see here.) What does he mean by "minimal polynomial $p$ of $\theta$ over skew field $K$"? Can we define the minimal polynomial if $K$ is not commutative? If yes, what are its properties? Is it still true that if $g$ is a polynomial over $K$ such that $g(\theta)=0$, then $p$ divides $g$?

Also, what is a left or right basis $B$ for $V$? My guess is that it is left or right linearly independent set, i.e., $\sum_{j=0}^n a_jb_j=0, b_j\in B$ (and the other form corresponding to "right") if and only if $a_j=0$ for all $j$ and also every element $x\in V$ can be written as $$x=\sum_{j=0}^n a_jb_j$$ (and the other form corresponding to "right").

答:
I think it is necessary (though not sufficient) to be acquainted with linear algebra over skew fields (also known as division algebras) in order to understand what this paper is saying. If $K$ is a skew field, much of basic linear algebra can be applied to define vector spaces over$~K$, provided one pays more attention than usual to order of operands, and distinguishes left $K$-vector spaces and right vector-$K$ spaces (putting $K$ to the right of the word vector is my invention; don't expect to find this in the literature).

In a left $K$-vector space $V$ one has, apart from the usual addition of vectors, a scalar multiplication $K\times V\to V$, which, with scalars written to the left of vectors satisfies an "associative" law $\lambda(\mu v)=(\lambda\mu)v$ where the Greeks are in $K$ and the Romans in $V$. An additive map $f:V\to W$ between left $K$-vector spaces is $K$-linear if $f(\lambda v)=\lambda f(v)$ for all $\lambda\in K,v\in V$, no sweat so far. What is a bit troublesome is that scalar multiplication itself does not give a $K$-linear map $V\to V$ (unless the multiplication is by a scalar from the centre of$~K$). This is even true for $K=K^1$, or more generally $K^n$, viewed as left $K$-vector spaces in the natural way (scalar multiplication being left-multiplication on each entry). By contrast, right-multiplication on all entries does define a $K$-linear map (since multiplication in $K$ is associative even if its is not commutative), and this means that $K$-linear maps $K^n\to K^m$ can only be given by right-multiplication by matrices, if one wishes to preserve the usual rules of matrix multiplication without operand-swapping (and ones sanity). Also one is obliged to represent elements of $K^n$ as row vectors to make this work, as a small example should convince you: for the $K$-linear map $f:K^2\to K^3$ with $f((1,0))=(a,b,c)$ and $f((0,1))=(d,e,f)$ one has from the definition of $K$-linearity that $f((x,y))=(xa+yd,xb+ye,xc+yf)$  which corresponds to the vector-matrix multiplication
$$ \pmatrix{x&y}\pmatrix{a&b&c\cr d&e&f}=\pmatrix{xa+yd&xb+ye&xc+yf}.$$
For right vector-$K$ spaces one has a scalar multiplication $V\times K\to V$ satisfying $(v\lambda)\mu=v(\lambda\mu)$, linear-$K$ maps satisfy $f(v\lambda)=f(v)\lambda)$, the set $K^n$ is viewed as column vectors, with left-multiplication by an $m\times n$-matrix giving a linear-$K$ map $K^n\to K^m$ (order of $m,n$ switched), as usual.

It is important to note that given a left $K$-vector space structure on $V$ one cannot in general derive a right vector-$K$ vector space structure on $V$ from it, and that although $K^n$ has both a natural left $K$-vector space structure and a natural right vector-$K$ vector space structure, the two are different in an essential way.

When, as in the paper, one considers extension $L/K$ of skew fields, then $L$ has both the structure of a left $K$-vector space and of a right vector-$K$ vector space, but again the two differ. Therefore, to employ the language of linear algebra (like bases, linear relations) one has to specify which of the two structures one is referring to; I assume (or hope) that the prefixes left- and right- systematically match the structure being referred to. Then $\theta\in L$ generates $L$ as a left-polynomial extension of $K$ if for some $n$ the first $n$ powers of $\theta$ in $L$ form a bases of $L$ as left $K$-vector space. Then expressing $\theta^n$ on this basis $\theta^n=\lambda_0+\lambda_1\theta+\cdots+\lambda_{n-1}\theta^{n-1}$ one gets the left-minimal polynomial $\lambda_0X^0+\lambda_1 X+\cdots+\lambda_{n-1}X^{n-1}$ of $\theta$ over $K$.

Note that the polynomial ring $K[X]$ can be defined as usual (no operand-swapping), which makes $X$ and its powers central elements, and with coefficients written to the left of powers of $X$, it has a natural left $K$-vector space structure. However, substitution of a non-central element $a\in L$ for $X$ defines a map $K[X]\to L$ that is $K$-linear (by definition it is the $K$-linear map sending the basis elements $X^n$ of $K[X]$ to $a^n\in L$) that I will denote $P\mapsto P[a]$, but this map does not respect multiplication ($(PQ)[a]$ and $P[a]Q[a]$ differ in general), and therefore is not a morphism of rings (at it maps the central element $X$ to the non-central $a$, it cannot be so).

Now to answer your questions. Suppose $P\in K[X]$ is any annihilating polynomial for$~\theta$: one has $P[\theta]=0$ in $L$. Then right-multiplying that relation by $\theta^k$ gives that $(X^kP)[\theta]=0$ for any $k\in\Bbb N$ (remember that $X$ is central in $K[X]$) and taking (left) $K$-linear combinations one finds that $(QP)[\theta]=0$ for all $Q\in K[X]$ (note that this works only for left-multiples of $P$). Also in $K[X]$ one can do Euclidean right-division by (for instance) the minimal polynomial$~\mu$: for any $P\in K[Q]$ there are $Q,R$ in $K[X]$ with $\deg(R)<\deg(\mu)$ such that $P=Q\mu+R$. Then since by the above $(Q\mu)[\theta]=0$ one concludes that $P[\theta]=0$ implies $R[\theta]=0$, which by degree and minimality of $\mu$ implies $R=0$ so $P=Q\mu$. So indeed, any annihilator polynomial of $\theta$ is a left-multiple of its minimal polynomial, or equivalently $\mu$ right-divides $P$. But these statements need not hold with left and right interchanged.

Your final paragraph seems correct.


问:
Banach subspace of Banach space
How to prove that subspace of Banach space is also Banach space?

I believe that's only true if subspace is closed. However, I don't understand why we can't take non-closed subspace. Is there any counterexample of non-closed subspaces that will not be Banach although space was Banach?

答: Consider the subset of differentiable functions of the set of continuous functions defined on a compact interval.</p>

Title: Banach subspace of Banach space


答: Assume $m = \min\{a,b\}> \frac{a+b}{2}$. Then $m+m>\frac{a+b}{2}+\frac{a+b}{2}=a+b\ge m+m$, a contradiction. </p>

Title: Proving that for all numbers $a, b\in\mathbb{R}$, $\min \{a, b\} \le (a+b)/2$

Body: Can anyone help me solve the two cases that are derived from this problem?</p>


问:
Conditioning on probability zero as a limit
I'm studying stochastic processes and am trying to relate different notions of conditional expectation/probability. There is an interesting way in which we can condition on an event of probability zero when we have two jointly continuously distributed random variables $X,Y$:

$$\mathbb{P}(X \in B\mid Y=y):=\lim_{\epsilon \to 0}{\mathbb{P}\left( X \in B\mid y-\epsilon<Y<y+\epsilon\right)}$$

Now the Wikipedia article on this claims that such a limit always exists if the marginal density function of $Y$ is positive, and gives the following proof:

For example, if two continuous random variables $X$ and $Y$ have a joint density $f_{X,Y}(x,y)$, then by L'Hôpital's rule:

$$\begin{align*}
\lim_{\epsilon \to 0}P(Y \in U \mid x_0 - \epsilon \lt X \lt x_0 + \epsilon)&=\lim_{\epsilon \to 0}\frac{\int^{x_0+\epsilon}_{x_0-\epsilon}\int_U f_{X,Y}(x,y)\text{d}y\text{d}x}{\int^{x_0+\epsilon}_{x_0-\epsilon} \int_\mathbb{R} f_{X,Y}(x,y)\text{d}y\text{d}x}\\[5pt]
&=\frac{\int_U f_{X,Y}(x_0,y)\text{d}y}{\int_\mathbb{R}f_{X,Y}(x_0,y)\text{d}y}.\\[5pt]
\end{align*}$$

The resulting limit is the conditional property distribution of $Y$ given $X$ and exists when the denominator, the probability density $f_X(x_0)$, is strictly positive.

Could someone explain to me how this is an application of L'Hôpital's rule? I would also like to know why, given this seemingly perfectly adequate definition, we need to bother with conditioning on $\sigma$-algebras at all?

Thanks in advance!

答:
Explain how this is an application of L'Hôpital's rule

$$\begin{align}
\mathbb{P}(X \in B|y-\epsilon<Y<y+\epsilon)
&= \frac {\mathbb{P}(X \in B,y-\epsilon<Y<y+\epsilon)}{\mathbb{P}(y-\epsilon<Y<y+\epsilon)}\\
&=\frac{\int^{x_0+\epsilon}_{x_0-\epsilon} \int_U f_{X,Y}(x,y)\text{d}x\text{d}y}{\int^{x_0+\epsilon}_{x_0-\epsilon} \int_\mathbb{R} f_{X,Y}(x,y)\text{d}x\text{d}y}
\end{align}
$$
Denote $F(\epsilon) = \int^{x_0+\epsilon}_{x_0-\epsilon} \int_U f_{X,Y}(x,y)\text{d}x\text{d}y$ and
$G(\epsilon)=\int^{x_0+\epsilon}_{x_0-\epsilon} \int_\mathbb{R} f_{X,Y}(x,y)\text{d}x\text{d}y$ , we note that $\lim_{\epsilon\rightarrow 0} F(\epsilon) = 0$
and $\lim_{\epsilon\rightarrow 0} G(\epsilon) = 0$, then we want to know
$$
\lim_{\epsilon\rightarrow 0} \frac {F(\epsilon)}{G(\epsilon)}
$$
This is the form $\frac 00$.

To use L'Hôpital's rule , we need know
$F'(0)$ and $G'(0)$ which as above


答:
It is true that if $X$ is a Banach space and $F\subseteq X$ is a subspace. Then $F$ is a Banach space with the restricted norm of $X$ into $F$ if and only if $F$ is closed in $X$.

A counterexample is the following.

Consider $X=\ell_{\infty}$ with the supremum norm given by
$$||x||_{\infty}=\sup\bigl\{|x(n)|:\,n\in \mathbb{N}\bigr\},$$
for every $x=(x(n))_{n}$ in $\ell_{\infty}$. Now, consider as $F$ the space $c_{00}$ of eventually zero sequences, i.e.
$$c_{00}=\bigl\{x\in \ell_{\infty}:\, \exists\, m \in \mathbb{N}\, \text{such that}\, x(n)=0\, \text{for every}\, n\geq m\bigr\}.$$
Then, $c_{00}$ is not a Banach space, for the sequence $(x_k)_k$ given by
$$x_k=\biggl(1,\, \frac{1}{2},\,...,\,\frac{1}{k},\,0,\,0,\,...\biggr)$$
is a Cauchy sequence since $||x_k-x_m||_{\infty}=1/k$, for every $k>m$ but not convergent in $c_{00}$. Indeed, for the last claim observe that $(x_k)_k$ converges to the element
$$x=\biggl(1,\,\frac{1}{2},\,...,\,\frac{1}{k},\,\frac{1}{k+1},\,...\biggr)$$
in $\ell_{\infty}$ which is not in $c_{00}$.

Edit: Why $(x_k)_k$ is not convergent on $c_{00}$?

Let $x$ be the point in $\ell_{00}$ with $x=(1,1/2,...,1/n,...)$. Then, observe that for every $k$,
$$||x-x_k||_{\infty}=||(0,0,...,\frac{1}{k+1},\frac{1}{k+2},...)||_{\infty}=\frac{1}{k+1}.$$
Hence, $||x-x_k||_{\infty}\to 0$ as $k\to \infty$. If ($x_k)_k$ where convergent on $c_{00}$ to some point $y$ we would have
$$||x-y||_{\infty}\leq ||x-x_k||_{\infty}+||x_k-y||_{\infty}\to 0,$$
as $k\to \infty$. In other words, $x=y$. But $x$ does not belong to $c_{00}$.

Title: Banach subspace of Banach space

Body:
How to prove that subspace of Banach space is also Banach space?

I believe that's only true if subspace is closed. However, I don't understand why we can't take non-closed subspace. Is there any counterexample of non-closed subspaces that will not be Banach although space was Banach?


问:
better understanding of incomplete gamma function $\Gamma(0,x)$
By definition  incomplete Gamma function is:$$\Gamma(0,x)=\int_{x}^{\infty}t^{-1}e^{-t}dt $$
I have an expression which includes

$$\Gamma(0,r(A)e^{i\phi(A)}),$$ where $A>0$ is a parameter, and $r(A)$ is almost linear, $\displaystyle \phi(A)\rightarrow\pm\frac{\pi}{2},\ as\ A\rightarrow\infty  $

I want to have more intuition about the influence of parameter $A$.
I will be very appreciate any comments or a good reference.

答:
When you say $r(A)$ is almost linear, do you mean that for large $A$ that $r(A)\approx\rho A+c$ for some constants $\rho$ and $c$?  Also, what do you mean by $\phi(A)\rightarrow\pm\frac{\pi}{2}$?

Without the $\pm$, and with the above assumption about $r(A)$:

$$\frac{d}{dA}\Gamma(0,r(A)e^{i\phi(A)})=-\left[r(A)e^{i\phi(A)}\right]^{-1}e^{-\left[r(A)e^{i\phi(A)}\right]}\left(r'(A)e^{i\phi(A)}+r(A)i\phi'(A)e^{i\phi(A)}\right)$$

For large $A$, $r(A)\approx\rho A+c$, $r'(A)\approx\rho$, $\phi(A)\approx\frac{\pi}{2}$, and $\phi'(A)\approx0$.  And the above reduces to:

$$\frac{d}{dA}\Gamma(0,r(A)e^{i\phi(A)})\approx-\left[(\rho A+c) i\right]^{-1}e^{-\left[(\rho A+c)i\right]}\left(\rho i +0\right)$$

$$\frac{d}{dA}\Gamma(0,r(A)e^{i\phi(A)})\approx-\frac{1}{A+\frac{c}{\rho}}e^{-(\rho A+c)i}$$

So for large $A$, increasing $A$ by a differential $\Delta A$ will "add" about $\frac{\Delta A}{A+\frac{c}{\rho}}$ to your quantity, but in the direction of $-e^{-(\rho A+c)i}$; that is it will add a complex differential of about $\frac{\Delta A}{A+\frac{c}{\rho}}$ in magnitude at an angle of $-\rho A-c+\pi$.

Imagine an angle that spins around the origin (clockwise if $\rho$ is positive) at a linear rate with respect to $A$.  Now add to your quantity in that rotating direction with a magnitudes that decrease roughly like $\frac{1}{A}$.


问:
Find equation of ellipse
I have an ellipse with focal points at $(-2,0)$ and $(2,0)$ and one of the tops is at $(0,4)$. I need to determine an equation of ellipse in form like $ax^2 + by^2 - c = 0$ based on this information. I can't figure out how to do that. Could I have some advise on the problem? Thanks.

Found the solution thanks to @Tavish which is:

$4x^2 + 5y^2 - 80 = 0$

答:
What you describe is a standard ellipse. The foci are located at $(\pm ae,0)$ where $a$ is half the length of the major axis and $e$ is the eccentricity. So, $$ae =2$$ The distance from the origin to your top is $b$, half the length of the minor axis. $$b=4$$

Squaring the first equation and using the relation $e^2 =1-\frac{b^2}{a^2}$, $$ a^2 -b^2 =4 \implies a^2 =20 $$

The equation of the ellipse is $$\frac{x^2}{a^2} +\frac{y^2}{b^2} =1 \\ \frac{x^2}{20} +\frac{y^2}{16} =1 $$

Title: Find equation of ellipse


问:
Standard normal distribution - Calculating the probability of an electron
Figure $1$  shows $Bohr$'s model for the hydrogen atom. We let $(X, Y)$ denote the position of the electron in relation to the proton-core. In quantum mechanics one can never determine $(X, Y)$ exactly, but we are simplifying the model by using probability distribution. We assume $X∼N(0,1)$ and $Y∼N(0,1)$, i.e. standard normal distribution with expectation $E(X)=0$ and variance $V(X)=1$. $X$ and $Y$ are independent.

Figure 1: Bohrs model with an electron in orbit around the proton-core. The black circle has radius r = 2,15.

(A)

What is the probability that $-2,15 <= x <= 2,15$? What is the probability that the electron is inside the black square? Hint: We assume $X$and $Y$ are independent random variables.

My answer:

To determine $P(a <= x <= b)$ when $X$ is a norml random variable with parameters $\mu$ and $\sigma$, we must evaluate

\begin{align}\\
& \int_{a}^b\frac{1}{\sqrt{2\pi}\sigma}e^{\left( \frac{-(x-\mu)^2}{2\sigma ^2} \right)} dx\\
\end{align}

So we have:

\begin{align}\\
& \Rightarrow P(-2,15\le X\le 2,15)=\int_{-2,15}^{2,15}\frac{1}{\sqrt{2\pi }} * {e^{\frac{-x^2}{2}}} dx \\
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\ = \frac{1}{\sqrt{2\pi }} * \int_{-2,15}^{2,15} e^{\frac{-x^2}{2}} dx \\
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\ = \frac{1}{\sqrt{2\pi }} * \int_{-2,15}^{2,15} e^{\left( \frac{x}{2}\right) ^2} dx \\
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\ = \frac{1}{\sqrt{2\pi }} * \int_{-2,15}^{2,15} e^u du \\
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\ = \frac{2}{\sqrt{2\pi }} * \left[ e^u + C \right]_{-2,15}^{2,15} \\
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\ = \frac{2}{\sqrt{2\pi }} * \left( e^{\left( \frac{2,15}{2}\right) ^2} - e^{\left( \frac{-2,15}{2}\right) ^2} \right) \\
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\ = \frac{2}{\sqrt{2\pi }} * \left( 0 \right) \\
\end{align}

I feel like I have done something wrong with the calculations but I donw know what. Hpw do I calculate the probability for if the electron is within the black square?

答:
You cannot integrate the normal density by substitution over a finite interval. You need tables or a computer.

Your answer should have been $\Phi(2.15)-\Phi(-2.15) \approx 0.968$

Among other errors, you have the false steps $$\int_{-2,15}^{2,15} e^{\frac{-x^2}{2}} dx \to \int_{-2,15}^{2,15} e^{(\frac{x}{2})^2} dx$$ $$\int_{-2,15}^{2,15} e^{(\frac{x}{2})^2} dx \to \int_{-2,15}^{2,15} e^{u} du$$ neither of which can be justified


问:
Why are the transition functions of a tensor bundle invertible?
I am doing the following exercise in Lee's "Introduction to Smooth Manifolds."

Exercise 12.18. Show that $T^kT^*M$, $T^kTM$, and $T^{(k,l)}TM$ have natural structures as smooth vector bundles over $M$, and determine their ranks.

I am trying to use the following two theorems.

Lemma 10.6 (Vector Bundle Chart Lemma). Let $M$ be a smooth manifold with or without boundary, and suppose that for each $p\in M$ we are given a real vector space $E_p$ of some fixed dimension $k$. Let $E=\bigsqcup_{p\in M}E_p$, and let $\pi:E\to M$ be the map that takes each element of $E_p$ to the point $p$. Suppose furthermore that we are given the following data:

(i) an open cover $\{U_\alpha\}_{\alpha\in A}$ of $M$

(ii) for each $\alpha\in A$, a bijective map $\phi_\alpha:\pi^{-1}(U_\alpha)\to U_\alpha\times\mathbb R^k$ whose restriction to each $E_p$ is a vector space isomorphism from $E_p$ to $\{p\}\times\mathbb R^k\cong\mathbb R^k$

(iii) for each $\alpha,\beta\in A$ with $U_\alpha\cap U_\beta\ne\emptyset$, a smooth map $\tau_{\alpha\beta}:U_\alpha\cap U_\beta\to\operatorname{GL}(k,\mathbb R)$ (which will be called a transition function) such that the map $\phi_\alpha\circ\phi_\beta^{-1}$ from $(U_\alpha\cap U_\beta)\times\mathbb R^k$ to itself has the form
\begin{equation}
\phi_\alpha\circ\phi_\beta^{-1}(p,v)=(p,\tau_{\alpha\beta}(p)v)
\end{equation}
Then $E$ has a unique topology and smooth structure making it into a smooth manifold with or without boundary and a smooth rank-$k$ vector bundle over $M$, with $\pi$ as projection and $\{(U_\alpha,\phi_\alpha)\}$ as smooth local trivializations.

A Generalization of Proposition 12.10 (Abstract vs. Concrete Tensor Products). Let $F$ be a field. Let $k,l\in\mathbb N=\{0,1,2,\ldots\}$. If $V_1,\ldots,V_k,V_{k+1},\ldots,V_{k+l}$ are finite-dimensional vector spaces over $F$, then there is a unique linear map
\begin{equation}
f:V_1\otimes\cdots\otimes V_k\otimes V_{k+1}^*\otimes\cdots\otimes V_{k+l}^* \to L(V_1^*,\ldots,V_k^*,V_{k+1},\ldots,V_{k+l};F)
\end{equation}
such that
\begin{equation}
f(v_1\otimes\cdots\otimes v_k\otimes\omega^{k+1}\otimes\cdots\otimes\omega^{k+l}) = v_1^{**}\otimes\cdots\otimes v_k^{**}\otimes\omega^{k+1}\otimes\cdots\otimes\omega^{k+l}\text{.}
\end{equation}
Furthermore, $f$ is a vector space isomorphism.

Here, $v^{**}$ is the image of $v\in V$ with respect to the natural map $V\to V^{**}$.

I want to use the vector bundle chart lemma to give $T^{(k,l)}TM$ a vector bundle structure over $M$. For each smooth chart $(U_\alpha,\phi_\alpha)$ for $M$, define $\phi_\alpha:\pi^{-1}(U_\alpha)\to U_\alpha\times\mathbb R^{n^{k+l}}$ by
\begin{equation}
\phi_\alpha(\alpha^{i_1\cdots i_k}_{i_{k+1}\cdots i_{k+l}} \frac{\partial}{\partial x^{i_1}}\Bigr|_p\otimes\cdots\otimes\frac{\partial}{\partial x^{i_k}}\Bigr|_p \otimes dx^{i_{k+1}}|_p\otimes\cdots\otimes dx^{i_{k+l}}|_p) = (p,(\alpha^{i_1\cdots i_k}_{i_{k+1}\cdots i_{k+l}}))\text{.}
\end{equation}
Let $(U_\alpha,x^1,\ldots,x^n)$ and $(U_\beta,y^1,\ldots,y^n)$ be two smooth charts for $M$. Now I have to show that there is a transition function $\tau$ for $\phi_\alpha\circ\phi_\beta^{-1}$. Let $p\in U_\alpha\cap U_\beta$. Then with some calculations using Proposition 12.10, I found that $\tau(p)$ should be the $n^{k+l}\times n^{k+l}$ matrix
\begin{equation}
\left(\frac{\partial y^{j_1}}{\partial x^{i_1}}(p)\cdots\frac{\partial y^{j_k}}{\partial x^{i_k}}(p) \frac{\partial x^{i_{k+1}}}{\partial y^{j_{k+1}}}(p)\cdots\frac{\partial x^{i_{k+l}}}{\partial y^{j_{k+l}}}(p)\right)
\end{equation}
(The above entry runs over all indices $i_1,\ldots,i_{k+l}$ as it goes down from top to bottom row, and $j_1,\ldots,j_{k+l}$ as it goes from left to right column).
But the hypothesis of the vector bundle chart lemma requires that $\tau(p)$ should be invertible, but I do not know why this is invertible.

答:
I think you are missing the point of Lemma 10.6. Let $p \in U_\alpha \cap U_\beta$. from (ii) it is given that
$$\phi_\alpha\circ  \phi_\beta ^{-1}|_{\{p\} \times \mathbb R^k}: \{p\} \times \mathbb R^k \to \{p\} \times \mathbb R^k$$
is a vector space isomorphism. Thus
$$\phi_\alpha\circ  \phi_\beta  ^{-1}(p, v) = (p, \tau _{\alpha\beta} (p) v)$$
for some $\tau_{\alpha\beta}(p) \in GL(k,\mathbb R)$. The main assumption in (iii) is that the map $\tau_{\alpha \beta }: U_\alpha \cap U_\beta \to GL(k,\mathbb R)$ is smooth. (Say, if $\tau_{\alpha\beta}$ is merely continuous, you get only a continuous vector bundle).

Thus to use Lemma 10.6 it is sufficient to construct $\phi_\alpha$ which satisfies (ii) and checks the smoothness of $\tau_{\alpha\beta}$, which should be obvious in this case.


问:
Example of $a_{n}$ such that $\lim_{n\to\infty }a_{n}=0$, and there is a convg subsequence $\left ( S_{k_{n}} \right )$ of $\left ( S_{n}\right )$
I'm having trouble to find an example of a sequence $\left ( a_{n} \right )$ such that $\lim_{n\to\infty }a_{n}=0$, and there is a convergent subsequence $\left ( S_{k_{n}} \right )$ of $\left (  S_{n}\right )$, but $\sum a_{n}$ diverges.

We denote $S_{k}=a_{1}+a_{2}+\cdot \cdot \cdot +a_{k}$

My first approach was finding $\left (  S_{n}\right )$ that is bounded but divergent (because this will lead us to conclude that there is a convergent subsequence). For example, there is $S_{n}=sin(n)$, but then $\lim_{n\to\infty }a_{n}\neq 0$. So, it fails to be our example.

Is there a better way to find such examples?

答:
Hint:

Try to construct $a_n$ such that $S_n$ oscilates from $1$ to $-1$ and back to $1$ and back to $-1$. Each oscilation will have to take more steps, since $a_n$ will have to be smaller and smaller, but you can afford it, you can have $a_n$ stuck on a particular value for how ever long you want.


答:
Hint: Let $f(x)$ be an even function of $x$. Then:
$$
I=\int_{-a}^{a}\frac{f(x)dx}{e^x+1}\stackrel{x\to-t}=\int_{-a}^{a}\frac{f(t)dt}{e^{-t}+1}.
$$

$$
2I=\int_{-a}^{a}\left[\frac{1}{e^x+1}+\frac{1}{e^{-x}+1}\right]f(x)dx=\int_{-a}^{a}f(x) dx.
$$

Can you take it from here?

Title: Can't solve an Integral, how to ask questions.

Body:
\begin{equation*}
\int ^{\frac{\pi }{2}}_{-\frac{\pi }{2}}\frac{\cos x}{e^{x} +1} dx
\end{equation*}

Is there a generally accepted format to ask questions on MSE?
Also can someone tell me how to write integrals, fractions and equations from keyboard....
Thanks!


问:
Dimension of generic fiber of a morphism between varieties
Let $f: X \to Y$ a dominant morphism of varieties
over a field $k$. My 'varieties' are by definition irreducible
and let denote as $g_X \in X$ and $g_Y \in Y$ their generic points. Dominant
means that $f(g_X)=g_Y$. Let $n= \dim(X), m= \dim(Y)$.
Then it is known that the transcendence degree of function fields
of varieties coinside with their dimensions. Therefore
$\operatorname{trdeg}_k \ K(X) =\dim(X)=n$ and $\operatorname{trdeg}_k \ K(Y) =\dim(Y)$.

I want to check that the dimension of the
generic fiber $X_{g_Y} = f^{-1}(g_Y)$ is
$e= n-m$.

My 'proof' has some gaps I would like to plug. Assume that
both varieties are affine: $X= \operatorname{Spec}(A), Y=\operatorname{Spec}(R)$. The map $f$
is dominant, therefore $R \to A$ is injective.
The affine ring of generic fiber $X_{g_Y}$
is $A \otimes_R K(Y)$.
Since $g_X \in X_{g_Y} $ there exist a map
$A \otimes_R K(Y) \to K(X)= \operatorname{Frac}(A)$ und therefore,
since $K(X)$ is a field, we get the induced inclusion of fields
$K(A \otimes_R K(Y)) \subset K(X)$. $K(A \otimes_R K(Y))$ is the function field of generic fiber $X_{g_Y}$.

Question: what do we know about the field extension
$K(A \otimes_R K(Y)) \subset K(X)$? which transcendence degree
does it have?

My idea is obvious: if I can prove that
$\operatorname{trdeg}_{K(A \otimes_R K(Y)} K(X)= m$ then I can
use additivivity formula for transcendence degrees of field extensions
to show that $\operatorname{trdeg}_k K(A \otimes_R K(Y))=n-m$ because of
$\operatorname{trdeg}_k \ K(X) =n$ and additivity for tower of extensions $k \subset K(A \otimes_R K(Y) \subset K(X)$. But I don't know
how I can figure out that
$\operatorname{trdeg}_{K(A \otimes_R K(Y)} K(X)= m$.

答:
As Youngsu pointed out, the fields $K(A\otimes_RK(Y))$ and $K(X)$ are indeed equal. To see this, note that $A\otimes_RK(Y)$ is the localisation of the domain $A$ at its multiplicative subset $R\setminus\{0\}\subseteq A$, and $K(A\otimes_RK(Y))$ is obtained from $A\otimes_RK(Y)$ by localising at the remaining non-zero elements. By transitivity of localisations, this shows that $K(A\otimes_RK(Y))$ can also be written as the localisation of $A$ at all its non-zero elements, which is $K(X)$.

So what goes wrong here? The problem is that the generic fibre $X_{g_Y}$ is usually not a variety over $k$! Suppose, for example, that $X=Y$ and $f\colon X\rightarrow X$ the identity. Then the generic fibre is $\operatorname{Spec} K(X)$, and $K(X)$ is usually not of finite type as a $k$-algebra, unless it is a finite field extension of $k$ (by Hilbert's Nullstellensatz), i.e., unless $X$ is $0$-dimensional.

However, $X_{g_Y}$ is a variety* over $K(Y)$ (can you show that?), and it is indeed true that $X_{g_Y}$ is $(n-m)$-dimensional as a $K(Y)$-variety! So instead of $\operatorname{trdeg}_kK(A\otimes_RK(Y))=n-m$, you need to show that $$\operatorname{trdeg}_{K(Y)}K(A\otimes_RK(Y))=n-m\,.$$ Now your idea of using transitivity of transcendence degrees works: We've seen above that $K(A\otimes_RK(Y))=K(X)$, and $\operatorname{trdeg}_{K(Y)}K(X)=\operatorname{trdeg}_kK(X)-\operatorname{trdeg}_kK(Y)=n-m$.

* Usually, $K(Y)$ won't be algebraically closed, but I think you're working with the scheme-theoretic definition of varieties anyway.


问:
What is $\arcsin(\cos{x})$ (Spivak Calculus exercise 15-18)?
In the first part of the exercise I proved that $\sin(x+\frac{\pi}2)=\cos{x}$.

The second part is asking what $\arcsin(\cos{x})$ is. Presumably, they want a rule for unrestricted $x$.

We have $\arcsin(\cos{x})=\arcsin(\sin(x+\frac{\pi}2))$. It was said that the domain of $\sin$ usually restricted to $\left[-\frac{\pi}2, \frac{\pi}2\right]$, hence I assume that $x+\frac{\pi}2 \in \left[-\frac{\pi}2, \frac{\pi}2\right]$, and this set is also the co-domain of the $\arcsin$ function.

Since $\sin$ is periodic with period $2\pi$, $\sin(x+\frac{\pi}2)=\sin(2k\pi+x+\frac{\pi}2)$, $2k\pi+x+\frac{\pi}2 \in \left[2k\pi-\frac{\pi}2, 2k\pi+\frac{\pi}2\right], k \in \mathbb{Z}$.

And we have that $\arcsin(\sin(2k\pi+x+\frac{\pi}2))=2k\pi+x+\frac{\pi}2$.

But the answer in the solutions manual is $\begin{cases} x-2k\pi+\frac{\pi}2, x \in \left[2k\pi-\pi, 2k\pi\right] \\ 2k\pi+\frac{\pi}2-x, x \in \left[2k\pi, 2k\pi+\pi\right] \end{cases}$

I can plot the graph and see how $2k\pi+x+\frac{\pi}2$ doesn't work, but where did they get this piecewise function?

答:
By periodicity and by symmetry, we have $$\arcsin(\sin(x))=x+2k\pi$$ or

$$\arcsin(\sin(x))=\pi-x+2k\pi,$$

whichever falls in the range $\left[-\dfrac\pi2,\dfrac\pi2\right]$.

In other words,

$$\begin{cases}x\in\left[-\dfrac\pi2,\dfrac\pi2\right]+2k\pi\to x-2k\pi,\\x\in\left[\dfrac{3\pi}2,\dfrac\pi2\right]+2k\pi\to \pi-x-2k\pi.\end{cases}$$

Now replace $x$ by $x-\dfrac\pi2$.


问:
Is there a four-regular plane graph with 25 vertices that meets corresponding conditions?
The first thing I want to say is whether there is a 25th-order  4-regular planar graph. I found some websites on the Internet, trying to find some clues. Indeed, I found some.

4-regular planar graph is unique?

http://www.mathe2.uni-bayreuth.de/markus/reggraphs.html#PLANAR

I think the second website is great！
In the second URL, surprisingly, there is no any relevant information about the  4-regular planar graph with 25 vertices.

I also try to solve it programmatically, but it seems relatively difficult, the number of vertices is a bit large. I tried to derive the contradiction through Euler’s formula, but failed.
plantri

If it does not exist, then the following question seems to become meaningless.

Actually we want to find a simple  $4$-regular plane graph with $25$ vertices, which satisfies the following conditions:

(1) Each face is either a 3-degree face or a 4-degree face.

(2) There is exactly one vertex associated with all 4-degree faces, and all the other vertices are associated with ** one 3-degree face** and three 4-degree face.

Of course it may not exist.

PS: We can derive some information about the number of faces
Combine the following two formulas

$$3 f_3+4 f_4=2m=25\times 4$$

$$25- \frac{4\times 25}{2}+f_3+f_4=2$$

We get the number of three-degree face is $f_3=8$ and the number of three-degree face is $f_4=19$.

答:
[Edit:] A previous version of this question asked about graphs where the 24 nonspecial vertices were each on 3 $3$-degree faces and a single $4$-degree face.

This is impossible.

Let $\Gamma$ be such a graph, and let $u$ be the unique vertex that is on four $4$-degree faces. A simple counting argument shows that $\Gamma$ has 24 $3$-degree faces and $7$ $4$-degree faces. This means that there is at least one $4$-degree face $F$ that does not contain $u$. But the fact that each vertex on $F$ is only on other $3$-degree faces forces the structure of this graph, there will be only 8 vertices.

(I don't have the graph drawing ready, it will be a four-cycle with triangles on each edge, there will be one way to draw the remaining four edges.)


答:
Let $c=\min\{a,b\}$ and set $A=a-c$, $B=b-c$; observe that $A\ge0$ and $B\ge0$. Then $a=A+c$, $b=B+c$, so
$$
\frac{a+b}{2}=\frac{A+c+B+c}{2}=c+\frac{A+B}{2}\ge c
$$
because $A\ge 0$ and $B\ge 0$.</p>


问:
How to prove the Vandermonde's determinant for a $3\times 3$ matrix when rows and columns have been swapped?
The problem is as follows:

The following determinant is named after french mathematician Alexandre-Théophile Vandermonde who lived in the late 18th century.

Prove this determinant

$\left|\begin{matrix}
1 & 1 & 1\\
a & b & c\\
a^2 & b^2 & c^2
\end{matrix}\right|=(b-a)(c-b)(c-a)$

What I've attempted to do was to use this well known identity on matrices:

$\left|\begin{matrix}
1 & 1 & 1\\
a & b & c\\
a^2 & b^2 & c^2
\end{matrix}\right|=1\left|\begin{matrix}b&c\\b^2&c^2\end{matrix}\right|-1\left|\begin{matrix}a&c\\a^2&c^2\end{matrix}\right|+1\left|\begin{matrix}a&b\\a^2&b^2\end{matrix}\right|$

Then working in the right side of this equation I'm getting:

(Here I'm using the notation of capital $V$ for Vandermonde's determinant)

$V=(bc^2-cb^2)-(ac^2-ca^2)+(ab^2-ba^2)$

Factoring the common terms I'm getting

$V=bc(c-b)-ac(c-a)+ab(b-a)$

But now what? I'm stuck, what else should it be done here? Can someone help me here on how to prove this determinant? Could it be that my method wasn't that right? Please help.

Body:
Note that
$$
V=bc(c-b)-ac(c-a)+ab(b-a) = - (a - b)(a - c)(b - c).
$$


问:
Given $\triangle ABC$ with $AB=BC$ and $\angle ABC=80^\circ$, and $O$ such that $\angle OAC=30^\circ$ and $\angle OCA=10^\circ$, find $\angle BOC$.
I encountered the following problem in a math problem book, in a section, entitled Geometry.

In an $\Delta ABC$ with $AB=BC$ and $\angle ABC = 80^\circ$, the point $O$ is chosen so that $\angle OAC = 30^\circ$ and $\angle OCA = 10^\circ$. Find $\angle BOC$.

I struggled repeatedly trying to solve this using some reasonable selection of methods, and failed utterly. I can construct a solution using analytic geometry, but that is not a fair choice of methods for this section.

From simple geometry, I keep writing down systems of equations using various triangles in the picture, and always keep missing one constraint, as it seems there are infinite amount of solutions to the system, which I know is wrong since the analytic geometry approach yields a unique solution.

I am happy to add my infinite solution approach using the angle constraints, if you find it helpful. Please let me know if that's the case.

I would like to solve this using methods from geometry, without using more advanced techniques. Thank you very much, I really appreciate your help.

答:
Let $P$ be the circumcentre of $\triangle AOC$. Extend $AO$ to meet $BC$ at $D$.

Observe that, $\angle APC=80^{\circ}$ and hence $\triangle BAC\cong \triangle PAC$ and thus $BC=PC$.

Also,  notice that $\triangle POC$ is equilateral and thus $OC=PC$.

Coupled with the result obtained earlier,  we get $BC=PC=OC$ and hence $\triangle BOC$ is isosceles.

Therefore, $\angle BOC=\frac {180-40}{2}=\boxed {70^{\circ}}$


问:
Show that $a$ and $b$ have no greatest common divisor in $\mathbb{Z}[\sqrt-5]$

Show that in the ring $R=\mathbb{Z}[\sqrt{-5}]$, $a=3\cdot 7 \cdot(1+2\sqrt{-5})$ and $b=(1+2\sqrt{-5})\cdot 7\cdot(1+2\sqrt{-5})$ have no greatest common divisor.

Since $N(a)=3^3\cdot 7^3$ and $N(b)=3^2\cdot 7^4$ so it is clear that any common divisor of $a$ and $b$ must have norm dividing $3^2\cdot 7^3$. Now we will get many possibilities. Am I on right track or there is some other smart way to solve this problem?

答:
Note that
$$
\gcd\left(3\cdot7\left(1+2\sqrt{-5}\right),7\left(1+2\sqrt{-5}\right)^2\right)=7\left(1+2\sqrt{-5}\right)\gcd\left(3,\left(1+2\sqrt{-5}\right)\right)
$$
Since $N(3)=9$ and $N\!\left(1+2\sqrt{-5}\right)=21$, we must have
$$
\left.N\!\left(\gcd\left(3,\left(1+2\sqrt{-5}\right)\right)\right)\,\middle|\ 3\right.
$$
But the only element whose norm divides $3$ is $1$.

This would say that the $\gcd$ is $7\!\left(1+2\sqrt{-5}\right)$

Because $3\cdot7=\left(1-2\sqrt{-5}\right)\left(1+2\sqrt{-5}\right)$, the $\gcd$ is also
$$
\gcd\left(\left(1-2\sqrt{-5}\right)\left(1+2\sqrt{-5}\right)^2,7\left(1+2\sqrt{-5}\right)^2\right)=\left(1+2\sqrt{-5}\right)^2\gcd\left(1-2\sqrt{-5},7\right)
$$
Since $N\!\left(1-2\sqrt{-5}\right)=21$ and $N(7)=49$, we must have
$$
\left.N\!\left(\gcd\left(1-2\sqrt{-5},7\right)\right)\,\middle|\ 7\right.
$$
But the only element whose norm divides $7$ is $1$.

This would say that the $\gcd$ is $\left(1+2\sqrt{-5}\right)^2$

$N\!\left(7\!\left(1+2\sqrt{-5}\right)\right)=1029$ and $N\!\left(\left(1+2\sqrt{-5}\right)^2\right)=441$, so these numbers are not associates (the only units here are $\pm1$, so this can be seen simply).

This contradiction says that a $\gcd$ does not exist.


问:
Local Contractibility of Homeomorphism Group of $\mathbb{R}^n$
For this question, let $X = \mathbb{R}^n$, and let $Aut(X)$ be the group of self-homeomorphisms with the compact-open topology.

Then I know that $Aut(X)$ is both locally path-connected, and locally contractible.  I have a decent sense of the implications of local path-connectedness: Two homeomorphisms which are sufficiently close in the compact-open topology can be joined by an isotopy of $X$.

But I don't have a good sense of what local contractibility means for this group.  I've read that spaces very similar to $X$, for example Euclidean Neighborhood Retracts, don't necessarily have locally contractible homeomorphism groups.  Why would that be?

What's the geometric interpretation of local contractibility for these
spaces?  How do you use it?

Thanks!  Sorry it's a bit of a 'soft question.'

答:
According to Černavskiĭ, the local contractibility of $Homeo(M)$ for a noncompact topological manifold $M$ implies that $M$ could be the interior of a compact manifold with boundary. Of course this is true when $M=\mathbb{R}^n$. For a statement see p.8, Th.2, of his paper

A. Černavskiĭ, Local Contractibility of the Group of Homeomorphisms of a Manifold, 1969 Math. USSR Sb. 8 287.

Notes: If $M$ is a metrisable topological manifold, then $Homeo(M)$ in the compact-open topology is a metrisable topological group. If $M$ is compact, then $Homeo(M)$ is locally contractible. If $M$ is noncompact, then $Homeo(M)$ may fail to be locally contractible (cf. Kirby-Edwards). The implication is that $Homeo(M)$ for compact $M$ may be an ANR. This is known to be true in dimensions $\leq 2$ (cf. Yagasaki) however it is an open problem for dimensions $\geq3$.


问:
Sketch the curve $\frac{ax + b}{cx + d}$
I am working through a pure maths book out of interest. I am stuck on this problem:

If $a$, $b$, $c$ and $d$ are positive, sketch the curve $y = \frac{ax + b}{cx + d}$ when

i) $ad - bc \lt 0$

ii)  $ad - bc = 0$

iii)  $ad - bc \gt 0$

I can work out the basics. Thus

$x$-intercept:  $f(x) = 0 \implies \frac{ax + b}{cx + d} = 0 \implies x = \frac{-b}{a}$

so intercept is $[\frac{-b}{a}, 0]$

$y$-intercept: $f(0) = \frac{b}{d}$ at point $[0,\frac{b}{d}]$

horizontal asymptote: $\frac{a}{c}$

vertical asymptote: $cx + d = 0 \implies x = \frac{-d}{c}$

I can also see that $\frac{ax + b}{cx + d} = \frac{a}{c}\left(1 + \frac{\frac{b}{a}-\frac{d}{c}}{x + \frac{d}{c}}\right)$

but I cannot put all this together to see how one could sketch the graphs except in one case: where $ad -bc = 0$ because then $\frac{b}{a}-\frac{d}{c} = 0$ and the graph will be a horizontal line equal to $\frac{a}{c}$

I think not knowing the actual magnitudes makes it difficult for me to visualise.

答:
From your last observation, $$\frac{ax+b}{cx+d} =\frac 1c \left( a +\frac{bc-ad}{cx+d} \right)$$

You can think of this as a series of transformations applied to the graph of $y=\frac 1x$.

Starting from it, first shift to the left by $d$. That gives $\frac {1}{x+d}$. Then squish parallel to the $x$ axis by a factor of $c$. This gives $\frac{1}{cx+d}$.

Now, scale along the $y$ axis by a factor of $bc-ad$ (here, the sign comes into play: if $bc-ad\lt 0$, you have to flip the graph over). Then, move the graph up by $a$ units and finally, squish down by a factor of $c$.


Title: How to prove the Vandermonde's determinant for a $3\times 3$ matrix when rows and columns have been swapped?

Body:
The problem is as follows:

The following determinant is named after french mathematician Alexandre-Théophile Vandermonde who lived in the late 18th century.

Prove this determinant

$\left|\begin{matrix}
1 & 1 & 1\\
a & b & c\\
a^2 & b^2 & c^2
\end{matrix}\right|=(b-a)(c-b)(c-a)$

What I've attempted to do was to use this well known identity on matrices:

$\left|\begin{matrix}
1 & 1 & 1\\
a & b & c\\
a^2 & b^2 & c^2
\end{matrix}\right|=1\left|\begin{matrix}b&c\\b^2&c^2\end{matrix}\right|-1\left|\begin{matrix}a&c\\a^2&c^2\end{matrix}\right|+1\left|\begin{matrix}a&b\\a^2&b^2\end{matrix}\right|$

Then working in the right side of this equation I'm getting:

(Here I'm using the notation of capital $V$ for Vandermonde's determinant)

$V=(bc^2-cb^2)-(ac^2-ca^2)+(ab^2-ba^2)$

Factoring the common terms I'm getting

$V=bc(c-b)-ac(c-a)+ab(b-a)$

But now what? I'm stuck, what else should it be done here? Can someone help me here on how to prove this determinant? Could it be that my method wasn't that right? Please help.

Body:
$$\begin{align}
V &= bc^2 - cb^2 - ac^2 + ca^2 + ab^2 - ba^2\\
&=bc^2 \color{fuchsia}{- bac} - cb^2 + ab^2 - ac^2 + ca^2 \color{fuchsia}{+ abc} - ba^2\\
&=b(c^2-ac-bc+ab)-a(c^2-ac+bc+ab)\\
&=(b-a)(c^2-ac-bc+ab)\\
&=(b-a)(c-b)(c-a)
\end{align}$$


问:
Does my set have empty interior?
I am trying to figure out if my set has empty interior or not. It is defined as follows. Let $(q_n)_{n \in \mathbb{N}} = \mathbb{Q}^2$, we define $U = \bigcup_{n \in \mathbb{N}} B_{\frac{1}{2^n}}(q_n) \subset \mathbb{R}^2$ and $A = U^C$. Now we fix $x \in \mathbb{R}\setminus \mathbb{Q}$ and define $C = A \cap (\{x\} \times \mathbb{R})$. Then we can view $C$ as subset of $\mathbb{R}$ and I wonder if $C$ has any interior points in $\mathbb{R}$. I mean I can always find a point $q_n \in U$ which is arbitrary close to $C$ in $\mathbb{R}^2$. But the radius of its ball $\frac{1}{2^n}$ could be even smaller. I can't say if in every neighbourhood of a $y \in C$ is an intersection with a ball $B_{\frac{1}{2^n}}(q_n)$. Does anyone have ideas how to tackle that?

答:
In general, there may be no interior points. Let $(r_n)=\mathbb Q$ be an enumeration of the rationals.
Let me now construct balls that will contain the point $(x,r_n)$.
For each $n$ there is a rational number $s_n$ such that $|s_n-x| \le \frac1{2^{2n+1}}$.
Now set $q_{2n}:=(s_n,r_n)$, and fill all other rational points of $\mathbb R^2$ into $(q_{2n+1})$.
Then $(x,r_n) \in B_{\frac1{2^{2n}}}(q_{2n})= B_{\frac1{2^{2n}}}(s_n,r_n)
$. The points $(x,r_n)$ are dense in $\{x\}\times \mathbb R$ but not in $C$. So $C$ has no interior points.

[This is a similar construction to proof that subsets of separable sets are separable.]

Title: Does my set have empty interior?