问:
Special groups and special linear groups
Is there a connection between a special group (i.e. a p-group with its derived group, center and frattini subgroup all equal) and the special linear group (i.e. group of matrices with determinant=1)? I've seen the first one in finite group theory and the other one in linear algebra so I'm not sure if there's something more to it or not.

Thanks in advance

答: Not really -- there are too few words available to assign a common term for all situations one would like to distinguish. So the defined term is not "special" but "special linear", respectively "special $p$-group".


问:
Extension of probability space
Here I cite [Remark 4.1 p 169, Karatzas, Ioannis, and Steven E. Shreve. "Brownian motion." Brownian Motion and Stochastic Calculus. Springer, New York, NY, 1998. 47-127.]:

"Our first representation theorem involves the notion of the
extension of a probability space. Let $X$ be an $\mathcal{F}_t$ adapted process on some $(\Omega,\mathcal{F},P)$. We may need a $d$-dimensional Brownian motion independent of $X$, but because $(\Omega,\mathcal{F},P)$ may not be rich enough to support this Brownian motion we need to extend the probability space to construct this."

Then he proceed with the construction of the extension: "Let  $(\hat{\Omega},\hat{\mathcal{F}},\hat{P})$ be another probability space on which we consider a d-dimensional $\hat{\mathcal{F}}_t-$Brownian Motion $\hat{B}$. Set $\tilde{\Omega}=\Omega \times \hat{\Omega}$, $\tilde{\mathcal{F}}=\mathcal{F}\otimes \hat{\mathcal{F}}$, $\tilde{P}=P \times \hat{P}$, define a new filtration by $\tilde{\mathcal{F}}_t=\mathcal{F}_t \otimes \hat{\mathcal{F}}_t$ then augment it and make it right continuous.

We may also extend $X$ and $B$ by defining for $(\omega,\hat{\omega}) \in \tilde{\Omega}$
$$\tilde{X}_t(\omega,\hat{\omega})=X_t(\omega)$$
$$\tilde{B}_t(\omega,\hat{\omega})=B_t(\hat{\omega})"$$

My question is why do we need this extension? What does he mean with "$(\Omega,\mathcal{F},P)$ may not be rich enough to support this Brownian motion" ?

Thanks in advance.

答: Your probability space could be something like $\Omega = \{0\}$, in which case there aren't enough elements to define any non-trivial random variables, let alone something like a Brownian motion.

ContentLicense: CC BY-SA 4.0

Title: Extension of probability space


问:
Maximal ideals and principal ideals domain
Let $R$ be a ring, and an ideal $I = ( p(x) )$ of $R[x]$.

Then consider the statement:
" $I$ is a maximal ideal if and only if $p(x)$ is irreducible "

Does that statement need the hypothesis "$R[x]$ is a principal ideals domain" to be true? Or is that statement true anyway?

答:
Suppose $p(x)$ is not the zero polynomial and is reducible, then there exists some non-constant polynomial such that $q(x)\mid p(x)\implies (p(x))\subsetneq (q(x))\ne R[x]$, so $(p(x))$ is not maximal. This proves the forward direction.

For the other direction, suppose $p(x)\subsetneq J$, some proper ideal of $R[x]$, and that $p(x)$ is irreducible. Then we must have $(p(x),q(x))\subset J$ whence $q(x)$ is coprime to $p(x)$. By Bézout's identity, $\exists a(x),b(x)\in R[x]$ such that $a(x)p(x)+b(x)q(x)=1\in J$, but then $J=R[x]$, a contradiction. So $p(x)$ is reducible in fact, which proves the statement.


问:
Ask for two symbols
I find several symbols in Differentiable Manifold. I list them in the following,
$$
\int_{\Omega}*(V)^b.
$$
I wanna ask about the meaning of $V^b$ so that I can calculate it.

For $*$, I guess it means the dual space of $V^b$.

答:
Well, I found an answer in question 2683129, which says that:

You can avoid this problem using the musical isomorphism $\flat$, which is a map between the tangent bundle of $\mathbb{R}^3$ and its cotangent. If this does not sound familiar, you may think that $\flat$ takes a vector field on $\mathbb{R}^3$ in input, and returns a dual form defined on $\mathbb{R}^3$. $\flat$ is defined through the metric $g$ as follows: $\flat(X) \mapsto X^{\flat}$, where $X^{\flat}$ is a 1-form now, and it acts on vector fields like this: $X^{\flat}(Y) = g(X,Y)$.
Using this definition you can compute $V^{\flat} = xdz + ydx$. In fact $V^{\flat}(V) = x^2+y^2 = g(V,V)$. Further, if $U = \sum_{i=1}^3 f_i(x,y,z)\frac{\partial}{\partial x^i}$, then $V^{\flat}(U) = g(V,U)$ (check!).

This paragraph may help me.

Title: Ask for two symbols


问:
Why is that minimum positive element in the ideals of linear combinations is the GCD of it's factors?

Why is that the smallest positive element in the ideals of the form $a_1\mathbb Z + a_2\mathbb Z+...$ is the greatest common divisor of the coefficients $a_1, a_2...$?

I have seen a proof of that minimum positive element divides all of other elements without remainder but i can't understand how it can be the greatest common divisor. I can understand that any element in the ideal will be a multiple of the common divisors of the coefficients since we can factorize them to be so, but i can't see why it is the GCD. I am looking for intuition.

答:
The GCD of the $a_i$s is a linear combination over $\mathbb{Z}$ of the $a_i$ from euclid GCD algorithm, hence $d\mathbb{Z}\subseteq I$, $I$ being your addition ideal $\sum_i a_i\mathbb{Z}$. Also, as $d\mid a_i\;\forall i$, we get $I\subseteq d\mathbb{Z}$. So, $I=d\mathbb{Z}$.

So, GCD is $d$, and it is the smallest linear combination as $d$ being generator of $I$ has to be the smallest element.


问:
Finding the limit of a particular function
Can not understand the following limit in a past paper.

$$\dfrac{1}{[n(e^{\frac{1}{n}}-1)]}\rightarrow 1$$

答:
$$
\lim_{n \to \infty} \frac{1}{n(e^{\frac{1}{n}}-1)} = \lim_{n \to \infty} \frac{1}{n(1 +\frac1n+\frac{(\frac1n)^2}{2!}+\dots-1)} = \lim_{n \to \infty} \frac{1}{1+\frac{\frac1n}{2!}+\dots} = 1
$$
using Taylor expansion for exponential function
$$
e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots
$$</p>

ContentLicense: CC BY-SA 3.0


问:
Finite group generated by two different order $2$ elements is $\cong$ to $\mathbb{Z}_2^2$ or $D_n$
I'm trying to solve this problem from my group theory course:

Given $G$ finite group generated by two different order $2$ elements.
Prove that $G\cong \mathbb{Z}_2^2$ or $G\cong D_n$ for some $n\geq 3$
(being $D_n$ the dihedral group of degree $n$).

The first part is easy, since if I consider $a,b$ these two different order $2$ elements, and suppose $G$ abelian, then I get
$$G=\{1,a,b,ab\},$$
and obviously this is isomorphic to $\mathbb{Z}_2^2$.

I'm having trouble with the second part. I understand the idea of two different reflections from $D_n$ generating the whole $D_n$, but I don't know how to prove it. I know that
$$D_n\cong\mathbb{Z}_n\rtimes_\phi \mathbb{Z}_2$$
for $\phi$ verifying that $\phi(\bar0)=id$ and $\phi(\bar1)$ being the inversion (correct me if I'm wrong), but I don't know if this is even useful here or how to use it in that case. How can I prove this? Any help will be appreciated, thanks in advance.

答:
There are, of course, infinitely many different possibilities for the value of $n$. So the first thing one has to do is to specify which particular $n$ is the correct one.

For this purpose, it helps to have some knowledge of the finite dihedral group $D_n$, starting with this presentation:
$$D_n = \langle \alpha,\beta \mid \alpha^2 = \beta^2 = (\alpha \beta)^n = \text{Id} \rangle
$$
Furthermore, it helps to know that $n$ is, actually, equal to the order of $\alpha\beta$ in $D_n$.

With that in mind, using that your given group $G$ is finite, it follows that its element $ab$ has finite order; let $n$ be the order of $ab$.

At this point, you can now deduce the existence of a surjective homomorphism $\phi : D_n \to G$, given by the formula $\phi(\alpha)=a$, $\phi(\beta)=b$.

The last thing left to do is to prove that $\phi$ is injective. We know that $\text{kernel}(\phi)$ contains neither $\alpha$ nor $\beta$ (since $a$ and $b$ are nontrivial), nor any element conjugate to $\alpha$ or to $\beta$ (since $\text{kernel}(\phi)$ is normal), nor $(\alpha\beta)^k$ when $1 \le k < n$ (since $ab$ has order $n \ge 2$). That is, in fact, the complete list of all nontrivial elements of the group $D_n$. Thus $\text{kernel}(\phi)$ is trivial, so $\phi$ is injective.


问:
Looking for intuition behind this proof of Cauchy-Schwarz inequality on $\mathbb{C}$

The Cauchy Schwartz inequality states that: $\forall x,y\in V$ ($V$ is an inner product space over $\mathbb{C}$), we have $|\langle x,y\rangle| \leq \|x\| \|y\|$. Moreover, equality occurs iff $x = \lambda y$ for some $\lambda \in \mathbb{C}$ (i.e. they are linearly dependent).

The Proof:

We don't have to prove anything if either $x=0$ or $y=0$ (or both). So let $x,y\in V\backslash\{0\}$. Now consider $z\in V$ such that $$z = x - \frac{\langle x,y \rangle}{\|y\|^2} y$$
where $\langle x,y \rangle$ denotes the inner product of $x$ and $y$. Observe that
$$\|z\|^2 = \|x\|^2 - \frac{|\langle x,y\rangle|^2}{\|y\|^2} \geq 0$$ since $\|z\|^2 \geq 0$ by definition. From this it follows that $|\langle x,y\rangle| \leq \|x\| \|y\|$ as desired.

Lack of Intuition:

Where does the $z$ come from? Why have we defined $z$ in this way? Was it really obvious that this is the way to go? I could really not foresee this proof even after seeing the definition of $z$, it came out of the blue. Could someone please help me with the intuition here?

答:
Let $(V,\langle\cdot,\cdot\rangle)$ be an inner product space, complex if you like, and $\{u_j\}_{j\in J}, J \subseteq\mathbb N \cup \{\infty\}$ be an orthonormal system, (not necessarly a (Schauder) basis), then for any vector $x \in V$ the Bessel' inequality is
$$
\|x\|^2 \ge \sum_{j\in J} |\langle x, u_j\rangle|^2 \tag{*}
$$

If the system $\{u_j\}_{j\in J}$ has only one element $u$, then
$$
\|x\| \ge |\langle x, u\rangle|
$$
now multiply $u$ by $\lambda > 0$ to get another vector $y = \lambda u \ne 0$, and simultaneously the LHS of the latter inequality we get
$$
\lambda\|x\| = \|y\|\|x\| \ge |\langle x, \lambda u\rangle| = |\langle x, y\rangle|
$$
since $\lambda$ is the norm of $\lambda u = y$.

The sum on the RHS of $(*)$ is the norm of $x_\parallel = \displaystyle \sum_{j\in J} \langle x, u_j\rangle u_j$, which is the part of $x$ parallel to the subspace spanned by the system. The rest is $x_\perp = x - x_\parallel$, which similar to $z$ in your question, you can check that it is orthogonal to $x_\parallel$, i.e. $\langle x_\perp, x_\parallel \rangle = 0$.


答: Hint: Use the expansion of $e^x = 1 + x + \frac{x^2}{2!} + ..$ and simplify the denominator.</p>

Title: Finding the limit of a particular function

Body:
Can not understand the following limit in a past paper.

$$\dfrac{1}{[n(e^{\frac{1}{n}}-1)]}\rightarrow 1$$


问:
Expression for f(x) as a function of g(x)

(b) Set g(x) = f(x)  +   (1/|f(x)|) for every 0 < |x−c| < δ. Find an
expression for f(x) as a function of g(x) for every 0 < |x − c| < δ.

We haven't really gone over this topic in class, and I can't seem to figure out what does it mean to find an expression for f(x) s a function of g(x)..

答:
$g(x) = \dfrac{f(x) + 1}{f(x)}$

$g(x) \cdot f(x) = f(x) + 1$

$(g(x) - 1) \cdot f(x) = 1$

$f(x) = \dfrac{1}{g(x)-1}$

I guess that's how it's supposed to be solved


问:
Inverse of matrix - definition
Usually the inverse of a square $n \times  n$ matrix $A$ is defined as a matrix $A'$ such that:

$A \cdot A' = A' \cdot A = E$

where $E$ is the identity matrix.

From this definition they prove uniqueness but using significantly the fact that $A'$ is both right and left inverse.

But what if... we define right and left inverse matrices separately. Can we then prove that:

(1) the right inverse is unique (when it exists)

(2) the left inverse is unique (when it exists)

(3) the right inverse equals the left one

I mean the usual definition seems too strong to me. Why is the inverse introduced this way? Is it because if the inverse is introduced the way I mention, these three statements cannot be proven?

答:
Given $n \times n$ matrices $A$ and $B$ for which holds $AB = I$ where $I$ is the $n \times n$ identity matrix. Now we will try to solve

$BX = I$

for an unknown $n \times n$ matrix $X$. We (left-)multiply both sides by $A$ to get

$ABX = AI$

which gives us

$IX = A$

so $X = A$ (because $IX = X$) and therefore $BA = I$. So we just showed $AB = I$ implies

$AB = BA = I$.

So as you can see it is not necessary to introduce a second definition for the inverse because it will lead to the same. Instead of $B$ we write $A^{-1}$.

Existence of a solution to $BX = I$ is because it can be shown $B$ is a bijective mapping ($x \mapsto Bx$) from $\mathbb{R}^{n}$ to $\mathbb{R}^{n}$.

Title: Inverse of matrix - definition


问:
Prove that $x \gt y \gt 0 \implies x \gt x-y$
Prove that $x \gt y \gt 0 \implies x \gt x-y$

I have approached it as below.

$y \gt 0 \implies y=x+(-x)+y \gt 0 \implies x+((-x)+y)=x+(-(x-y)=x-(x-y)>0 \implies x\gt x-y$

What I observe in this proof is that is does not consider the fact that $x \gt y$ and $x \gt 0$.

Is this proof valid?

答:
If the statement were only true when $x > 0$ and if $x > y$ then, yes, it would be very worrisome that the proof didn't use the condition that that are necessary and the proof probably has a mistake in it.

But the proof wouldn't be invalid because it didn't use all its premises. The proof would be invalid because it has a mistake. And we know the proof has a mistake, because as the proof didn't use the necessary premise, the proof actually proved the statement for cases when it wouldn't actually be true.  And any proof that proves a false statement must have an error.

For example: If we wanted to prove.  If $n$ is an even integer, then $n^2$ is even and someone gave a proof that $n^2$ is even but it never used the fact that $n$ was even, then we know the proof must be wrong.  If it never assumed $n$ was even, then it would work even if $n$ were odd.    But we know if $n$ is odd, then $n^2$ is odd.  So the proof must be wrong because it reached a wrong conclusion.

But if the statement is true whether our premise is true or not then the premise is unnecessary and we don't need to use it (although we could).

Consider the following ridiculous example:  If $n$ is an even integer and Santa Claus is fat, prove that $n^2$ is even.  Pf: If $n$ is even then there is an integer $k$ so that $n = 2k$.  So $n^2 = (2k)^2 = 2^2k^2 = 2(2k^2)$.  $2k^2$ is an integer and so $n^2$ is two times an integer so $n^2$ is even.

Now imaging if a professor said:  "That proof is invalid because it didn't use Santa Claus is fat".

Well, if $n$ is even then $n^2$ will be even whether or not Santa is fat or thin, or purple, or, gasp, doesn't exist.  We don't need that premise.

And your statement is much the same.  If $y > 0$ then $x > x-y$.  Period.  That will be true if $x > y$.  That will be true if $x = y$ and then will be true if $x < $ negative five hundred gajillion.

The premise $x > y$ and $x>0$ are not necessary as the statement is true without  them and that are completely irrelevent.

....

And the proof you gave is fine.  Valid.  It's just great.


问:
The dimension of $SU(n)$
$SU(n)$ denotes the special unitary group. I know its dimension should be $n^2-1$. However, I am trying to prove it and get a wrong result. I have no idea what is wrong with my proof. Therefore, I am wondering if someone could point the mistake out. Thank you! The following is my proof.

I consder the determinant function $\det:U(n)\to\mathbb{C}$, where $U(n)$ denotes the unitary group. $\det$ should be a Lie group homomorphism. Hence, it is a smooth map with constant rank. Then, $SU(n)=\det^{-1}(1)$, which actually shows that $SU(n)$ is an embedded Lie subgroup of $U(n)$.

I have already known that $U(n)$ is $n^2$-dimensional. It suffices to figure out the rank of $\det$, since the codimension of $SU(n)$ is the rank of $\det$.

It suffices to check the rank at $I_n$ which denotes the identity matrix, since $\det$ has a constant rank. Pick $B\in T_{I_n}U(n)=M(n,\mathbb{C})$, where $M(n,\mathbb{C})$ denotes the set of all matrices whose entries are complex.

Therefore, the differential of $\det$ is a map $d(\det)_{I_n}:M(n,\mathbb{c})\to\mathbb{C}$. Consider the curve $$\gamma(t)=I_n+tB$$. Clearly, $\gamma'(0)=B$, $\gamma(0)=I_n$.

Then, I compute
\begin{align*}
d(\det)_{I_n}(B)
    &=(\det\circ\gamma)'(0)\\
    &=\frac{d}{dt}|_{t=0}\det(I_n+tB)\\
    &=tr(B)
\end{align*}

Now, I think I can prove the image of $d(\det)_{I_n}$ is $\mathbb{C}$, since given any $c\in\mathbb{C}$ I take the matrix $B$ as $\frac{c}{n}I_n$.

Therefore, I believe that the rank of $\det$ is 2, which is the dimension of $\mathbb{C}$. So I conclude that the dimension of $SU(n)$ is $n^2-2$.

I know I must make a silly mistake. Could someone point it out? Thank you very much!

答:
The tangent space to $U(n)$ at $1$ is the space of skew Hermitian matrices, not all matrices: this follows by differentiating the defining equation

$$A \overline{A}^t=1.$$ Now $\frac{c}{n} I_n$ is not skew Hermitian unless $c$ is imaginary. In fact, the definition of $U(n)$ implies $|\mathrm{det}(A)|=1$ for all $A \in U(n)$ (use that the determinant is multiplicative). So the determinant is a surjection onto the circle.

Title: The dimension of $SU(n)$


问:
Citation for the Gateaux derivative FTC
The Wikipedia page for the Gateaux Derivative contains the following statement:

A version of the fundamental theorem of calculus holds for the Gateaux derivative of $F$, provided $F$ is assumed to be sufficiently continuously differentiable. Specifically:

Suppose that $F:X\to Y$ is $C^1$ in the sense that the Gateaux derivative is a continuous function $dF:U\times X \to Y$. Then for any $u \in U$ and $h\in X$,

$$
F(u+h)-F(u) = \int_0^1 dF(u+th;h)dt \label{a}\tag{1}
$$

I stumbled on this result and proved it myself (in a less general context) around a year ago, but no one I asked about it recognized it. I am now writing up my work (of which this forms a small but key part) for publication. Now that I have found a statement of the result in Wikipedia, I would like to be able to cite it. But the Wikipedia page doesn't provide a citation or link or name for the theorem. I checked the paper by Gateaux, and didn't see it there.

Does anyone known where I can find it?

Thanks for any help...

答:
This answer arises from my comments above, in order to fulfill a request of the Asker. The result above is a consequence of a more general one, the following theorem (found in reference [1] §2.7, theorem 2.7, pp. 34-35; below I use equation numbering as in this reference) pertaining integration of continuous abstract functions of a real variable i.e. functions $\Phi: \Bbb R\to Y$, where $Y$ is a real Banach space:

Theorem. If the abstract function $\Phi(t)$ has continuous derivative on the open interval $]a,b[$ (i.e. $\Phi\in C^1(]a,b[)$), then for any number $\alpha, \beta\in]a,b[$ the formula
$$
\int\limits_{\alpha}^{\beta}\frac{\mathrm{d}}{\mathrm{d}s}\Phi(s)\mathrm{d}s= \Phi(\beta)-\Phi(\alpha). \label{1}\tag{2.6}
$$
is valid.

Sketch of proof: the first step consist in proving the formula

$$
\frac{\mathrm{d}}{\mathrm{d}t}\int\limits_{a}^{t}\Psi(s)\mathrm{d}s= \Psi(t)\quad\forall t\in]a,b[,\label{2}\tag{2.7}
$$
for any continuous abstract function $\Psi:[a,b]\to Y$, by using an inequality for difference quotients ([1], §2.3, theorem 2.2, pp. 28-29).
Then, by putting
$$
\Psi(t) = \int\limits_{a}^{t}\Phi(s)\mathrm{d}s
$$
we get
$$
\frac{\mathrm{d}}{\mathrm{d}t}\left[\int\limits_{\alpha}^{t}\frac{\mathrm{d}}{\mathrm{d}s}\Phi(s)\mathrm{d}s -\Phi(t)\right] = 0.
$$
and, by applying the mean value theorem for abstract functions, the relation
$$
\int\limits_{\alpha}^{t}\frac{\mathrm{d}}{\mathrm{d}s}\Phi(s)\mathrm{d}s = \Phi(t) +C
$$
holds for a given real constant $C$ which,  evaluated by choosing $t=\alpha$, gives equation \eqref{1}. $\blacksquare$

Now equation \eqref{a} is a direct consequence of \eqref{1} obtained by putting
$$
\Phi(t) = F (u+th)\quad \forall u\in U,\,h\in X.
$$

Notes

Equation \eqref{1} is more general than \eqref{a} since its proof requires only the existence and continuity (respect to only to the real variable $t$) of the functional derivative (or better the functional differential)
$$
\frac{\mathrm{d}}{\mathrm{d}t} F(u+th)= \lim_{t\to 0}\frac{F(u+th)-F(u)}{t},
$$
which could also be nonlinear respect to $h\in X$ (and thus it could not be a Gâteaux nor a Frechet derivative, see this Q&A on such matters).

Vainberg, despite citing several Authors like Norbert Wiener, Lawrence Graves, Michael Kerner,  Israil’ Moiseevich Gel’fand and Mark Konstantinovich Gavurin ([1] §2 p. 24), does not give a precise reference for the result, and this perhaps means that it is due to himself, in the work of synthesis inevitably done during the writing of a monograph.

Bibliography

[1] Vaĭnberg, Mikhail Mordukhovich, Variational methods for the study of nonlinear operators. With a chapter on Newton’s method by L.V. Kantorovich and G.P. Akilov, translated and supplemented by Amiel Feinstein, Holden-Day Series in Mathematical Physics. San Francisco-London- Amsterdam: Holden-Day, Inc. pp. x+323 (1964), MR0176364, ZBL0122.35501.

Title: Citation for the Gateaux derivative FTC


问:
Question on vectors regarding plane moving around earth?
A plane is revolving around the earth with a speed 100 km/hr at a constant height from the surface of earth. The change in the velocity as it travels half circle is

Answer is 200km/hr

First things I noticed :
There is no acceleration of the plane given.There is no constant force as well given.Then why is change in velocity simply not 0.

2 Drew a vector 100 j upwards and 100(-j) downwards. When two vectors are opposite they cut each other.Why don’t they do it here ? =0

Please help in this

答: We want the difference  in the two vectors, $v_0 = 100\hat{j}$ and $v_1 = -100\hat{j}$ so $$v_0 + v_1 = 0$$ but the magnitude of the change in velocity is $$|v_0 - v_1| = |200 \hat{j}| = 200(km/hr)$$An massive object which moves in a circle must be accelerating, in fact the magnitude of it's acceleration must be $a = \frac{v^2}{r}$ where $v$ is the linear velocity and $r$ is the distance from the center of the circle.


问:
How is axiom of choice utilized within the given proof?
I (only a beginner in set theory...) want to see a prove that an infinite set must have an infinite countable subset U:

The following answer sounds quite logically for me:
John Wayland Bales, Prove that every infinite set has a countable subset

However, I guess that this proof depends at least on on some weaker variant of axiom of choice.

John Wayland Bales constructs a series of non empty finite sets with increasing number of elements:

For each $N\in\mathbb{N}$, $S$ contains an element distinct from each
element in $U_N=\{x_1,x_2,\cdots,x_n\}$, so define
$U_{N+1}=U_N\cup\{x_{N+1}\}$ where $x_{N+1}$ is an element of $S$
distinct from each element of $U_N$.

Where is AC hidden in this argumentation?
I feel, that the last step is critical:

Let $$U=\bigcup_{N\in\mathbb{N}}U_N$$
Then $U$ is a countable subset of $S$.

I would like to learn, how the proof is done "correctly" by using AC or some weaker weaker variant (axiom of countable choice...).

My problem: Of course I see that recursive choices of "any" elements to create a new set are crucial, but those choices are done sequentially and not on an infinite set of sets that is already "available" and clearly defined. AC (in it's original textbook variant) is about the latter situation and not about a set, that is still in process of being constructed.

答:
This proof is a great example of how subtle the use of the Axiom of Choice can be.

You are kind of right that the last step is the critical step, but in fact, the whole thing is just a big appeal to the Axiom of Choice.

Definitions by recursion, in general, require you to specify a function which takes in as input your "intermediate step" and produces the next step. But here we don't have a function, we have a relation. We simply say "we can add one more element!", but the choice of this element can be very different and it will result in a different input during the next step of the recursion.

So the claim that the set $\{U_n\mid n\in\Bbb N\}$ even exists is appealing to the Principle of Dependent Choice, which is a sort of countable version of Zorn's lemma, if you will, or a slight strengthening of the Axiom of Countable Choice (although some people will mistakenly assume that that is what Countable Choice mean). We can use the Axiom of Choice, as a whole, by first fixing a choice function and then using it to choose the next element in each step.

Removing that assumption, the proof merely shows that for all $n\in\Bbb N$, an infinite set would have a subset of size $n$. In other words, an infinite cardinal is always larger than all finite cardinals.

Now, you're right of course, that the union of the finite sets need not be itself countable, but here we actually have a way around this: every $U_n$ has exactly $n$ elements, and $U_n\subseteq U_m$ for $n\leq m$. So that means $u_n$ is the unique element in $U_{n+1}\setminus U_n$, and that gives us an enumeration of the union, assuming of course that $\{U_n\mid n\in\Bbb N\}$ exists.


问:
Independent vectors and linear map

Let $L\colon (x_1, x_2, x_3, x_4, x_5, x_6, x_7) \mapsto 2x_3 + 3x_4 + 8x_6 + x_7$: Is $L$ a linear map?
If it is, how to find independent vectors in  $\operatorname{Ker}(L)$?

答:
Recall that the linearity condition for any application $L: V \to W$ between vector spaces $V$ and $W$ is that $L(\lambda x + y) = \lambda L(x) + L(y)$ for every $x,y\in V$. Now in your case we have $V = \mathbb{R}^7$ and $W=\mathbb{R}$ and $L$ is given by:

$$L(x_1, x_2, x_3, x_4, x_5, x_6, x_7) = 2x_3 + 3x_4 + 8x_6 + x_7$$

Now, let's try to apply $L$ to some $\lambda x + y$ and see what happens. First, note that if we pick arbitrary $x,y \in \mathbb{R}^7$ they will be written in the form:

$$x = (x_1, x_2, x_3, x_4, x_5, x_6, x_7)$$

$$y = (y_1, y_2, y_3, y_4, y_5, y_6, y_7)$$

In this case $\lambda x + y = (\lambda x_1 + y_1, \dots, \lambda x_7 + y_7)$ so that it's pretty obvious that we can write the following relation for the components $(\lambda x + y)_i = \lambda x_i + y_i$. Now let's apply $L$: note first what $L$ does, it takes the third component of the vector, multiplies by $2$, sum to the fourth component multiplied by $3$, sum to the sixth component multiplied by $8$ and sum to the seventh component alone.

Now, for the vector $(\lambda x + y)$ the $i$-th component is $(\lambda x + y)_i$ which I've already argumented what it is. So we have:

$$L(\lambda x + y) = 2(\lambda x + y)_3 + 3(\lambda x + y)_4 + 8 (\lambda x + y)_6 +(\lambda x + y)_7$$

Substituting we have:

$$L(\lambda x + y) = 2(\lambda x_3 + y_3) + 3(\lambda x_4 + y_4) + 8(\lambda x_6 + y_6) + (\lambda x_7 + y_7)$$

Now reorganize this putting in evidence $\lambda$:

$$L(\lambda x + y) = \lambda(2x_3 + 3x_4 + 8x_6 + x_7)+(2y_3 + 3y_4 + 8y_6 + y_7)$$

Now look inside the parentheses. There's $L(x)$ and $L(y)$, so that we have shown that we have the linearity condition holding $L(\lambda x + y) = \lambda L(x) + L(y)$.

To find the kernel, you impose $L(x) = 0$ and try to find what conditions are imposed over $x$. The set of all those $x$ is the kernel. So you'll have $2x_3 + 3x_4 + 8x_6 + x_7 = 0$, so to impose a condition, you isolate one of the $x_i$. I'll isolate $x_7$ because it has no coefficients, but you can choose another. We have:

$$x_7 = -2x_3 - 3x_4 - 8x_6$$

Now the solutions are obtained by allowing $x_1$ trough $x_6$ vary as they want, but with $x_7$ restricted to this. So to make this cleaner we set $x_i = \alpha_i$ for $i \neq 7$ and we set $x_7 = -2\alpha_3 - 3\alpha_4 - 8\alpha_6$, so that a general solution $x$ will be:

$$x = (\alpha_1, \alpha_2, \alpha_3, \alpha_4, \alpha_5, \alpha_6, -2\alpha_3 - 3\alpha_4 - 8\alpha_6)$$

So that you have $\ker L = \{x \in \mathbb{R}^7 : x = (\alpha_1, \alpha_2, \alpha_3, \alpha_4, \alpha_5, \alpha_6, -2\alpha_3 - 3\alpha_4 - 8\alpha_6), \alpha_i \in \mathbb{R}\}$

Title: Independent vectors and linear map


问:
What happens if the Euler-Lagrange function is always zero?
As titled, I was considering the mimimization problem where $y(x)$ has two endpoints fixed. That is, minimizing $$\int_a^b L(x,y(x),y'(x)) \, dx$$

where $$\ y(a)=m, y(b)=n $$for all $y(x)$.

if the Euler-Lagrange equation is always zero for all functions $y(x)$, I think the original integral $$\int_a^b L(x,y(x),y'(x)) \, dx$$ is constant for all $y(x)$.

Which means if
$$\frac {d}{dx} \frac {\partial L}{\partial y'} = \frac {\partial L}{\partial y}$$ for all $y(x)$, then
$$\int_a^b L(x,y(x),y'(x)) \, dx = C$$ for some constant $C$.

But I don't know how to prove it. I start by using integration by parts, writing $$\int_a^b L(x,y(x),y'(x)) \, dx$$ as $$\left.\ x*L(x,y(x),y'(x))\right|_a^b - \int_a^b x*\frac {d}{dx}L(x,y(x),y'(x)) \, dx$$
But I don't know how to continue from here, or maybe I shouldn't use IBP here.

Any help will be appreciated.

答:
The simplest reference on this is a discussion in section 2 of Landau & Lifschitz's Classical Mechanics, their first volume in their coyrse of theoretical phtsics. This uses the same setup as you do: one spatial and one temporal dimension. They show that a Lagrangian that differs from another Lagrangian by a total temporal derivative,  that is:

$L'(q,\dot{q},t) = L(q,\dot{q},t) + (d/dt)f(q,t)$

Give actions

$S' := \int_T L'  dt = \int_T L dt + f|\partial T = S + f|\partial T$

Where $S$ is the action of $L$, $S:= \int_T L$ and where by $f|\partial T$, I mean the evaluation of f at the boundary, $\partial T$ of the temporal interval $T$. As the latter term is constant we see that the variational conditions, $\delta S'=0$ and $\delta S=0$ give the same Euler-Lagrange equations of motion.

This has a generalisation to higher spatial dimensions where the extra term is then recognised as a divergence. This relates to the inverse variational problem first posed by Helmholtz over a hundred years ago where he asked which systems of differential equations come from variational principles. The mathematical expression of this is the variational bicomplex; or more naively:

Vector Fields $\rightarrow$ Lagrangians $\rightarrow$ Differential Equations $\rightarrow$ Differential Operators

Here, the first arrow is taking divergences, $Div$, and the second, the Euler-Lagrange operator, $EL$, whilst the third is the Helmholtz operator, $H$. It's also what is called an exact sequence or complex, and so the composition of two arrows vanishes, that is $EL•Div=0$ amd $H•EL=0$. The proper expression of this meams we don't just have a complex, but a bicomplex.

In a sense, whilst the ordinary calculus generalises to the differential geometry of manifolds, aka tangent bundles, and the operators of vector analysis both generalise and assemble themselves into the de Rham complex; the variational calculus generalises to the differential geometry of bundles, aka jet bundles, and the variational operators both generalise and assemble themselves not into a complex, but a bicomplex, reflecting the fact that a bundle, in its most general description, is a surjective morphism between two manifolds - and which in a sense, we can call the de Rham bicomplex of a bundle.


问:
Solve the first order Non-homogenous differential equation, $(x +3y - 4)dx + (x + 4y - 5)dy$
A hint is given: Put $x = X + h$ & $y = Y + k$ then $h = 1, k = 1$.

And the final solution is also given which is given as
$$(x - 1)^4 + 4(x - 1)^3 (y - 1) + 4(y - 1)^2 (x - 1)^2 = K$$

I have tried solving it (attached below) but I am not able to get to the final solution as given above.

Answer Page 1 Link

Answer Page 2 Link

Answer Page 3 Link

答:
$$(x +3y - 4)dx + (x + 4y - 5)dy=0$$
$$\dfrac {dy}{dx}=-\dfrac {(x +3y - 4)}{ (x + 4y - 5)}$$
$$\dfrac {dy}{dx}=-\dfrac {(x \color {red}{-1} +3y \color {red}{ - 3})}{ (x  \color {red}{-1} + 4y  \color {red}{- 4})}$$
$$\dfrac {dY}{dX}=-\dfrac {X +3Y}{ X + 4Y}$$
Substitute:
$$Y=TX \implies Y'=T'X+T$$
The differential equation is homogeneous.
$$T'X+T=-\dfrac {3T+1}{ 4T+1}$$
$$T'X+=-\dfrac {(2T+1)^2}{ 4T+1}$$
$$\int \dfrac { 4T+1}{(2T+1)^2} \, dT=-\int \dfrac {dX}{X}$$
After integration we have:
$$\ln |2Y+X|+\dfrac {X}{4Y+2X}=C$$
Where $y-1=Y$ and $x-1=X$.

The answer provided seems not correct to me.


问:
Is $\left(X_1,... ,X_n,\bar{X}\right)$ jointly normal distributed if $\left(X_1,... ,X_n\right)$ is?
Let $X:=\left(X_1,... ,X_n\right)\sim N_n(\mu,\Sigma)$, $\mu\in\mathbb{R}^n$, $\Sigma\in\mathbb{R}^{n\times n}$ symmetric and positive semi-definite and $\bar{X}:=\frac{1}{n}\sum_{i=1}^n X_i$ as usual.
Is now $Z:=\left(X_1,... ,X_n,\bar{X}\right)$ multivariate normal distributed?</p>

答:
Yes. One possible definition of multivariate normal vector is that every linear combination of its components is (univariate) normally distributed. Since $X$ is multivariate, every linear combination of $(X_1, ..., X_n)$ is normally distributed. But every linear combination of components in $Z$ is also a linear combination of $(X_1, ..., X_n)$ because $\bar{X}$ is linearly dependent on $(X_1, ..., X_n)$. Thus, $Z$ is also multivariate normal.

(There are a few definitions of multivariate normal. They're equivalent:
http://en.wikipedia.org/wiki/Multivariate_normal_distribution )


问:
Justifying a solution for a system of equations in linear algebra
I'm trying to understand how to solve systems of linear equations using matrices, however I'm faced with the following problem. Given the following system:

$$
\begin{cases}
x+2y+3z-3w=a \\
2x-5y-3x+12w=b \\
7x+y+8x+5w=c
\end{cases}
$$

What steps do I need to do to confirm and justify that it only has a single solution, which is:

$$
37a +13b = 9c
$$

Appreciate all hints! Thanks.

答:
You can get your answer using simple elementary row operations:

$$\left(
    \begin{matrix}
    1 & 2 & 3 & -3 & | & a\\
    2 & -5 & -3 & 12 & | & b \\
    7 & 1 & 8 & 5 & | & c \\
     \end{matrix}
\right)  >>{\text{$R2=R2-2R1, R3=R3-7R1$}}>>$$ $$\left(
    \begin{matrix}
    1 & 2 & 3 & -3 & | & a\\
    0 & -9 & -9 & 18 & | & b-2a \\
    0 & -13 & -13 & 26 & | & c-7a \\
     \end{matrix}
\right)>>{\text{$R3=R3+(- \frac{13} {9} R2)$}}>>$$

$$\left(
    \begin{matrix}
    1 & 2 & 3 & -3 & | & a\\
    0 & -9 & -9 & 18 & | & b-2a \\
    0 & 0 & 0 & 0 & | & c-7a + [- \frac{13} {9}(b-2a)]\\
     \end{matrix}
\right)$$

Now, notice that the last row of the matrix is a zero row, thus the system has a solution only if the right side of the equation is equal to zero. thus-

$$c-7a + [- \frac{13} {9}(b-2a)] =0$$
$$-\frac{37}{9}a-\frac{13}{9}b+c=0$$
$$-37a -13b +9c=0$$
$$37a+13b=9c$$


问:
Neither NAND nor NOR generates all n-ary functions on $\{0,1\}$
It is said that NAND alone can generate all $n$-ary functions on $\{0,1\}$, for all natural numbers $n$. However, I think the correct statement is that it can generate all $n$-ary functions, where $n\geq 1$. I don't think it generates the nullary functions $0$ and $1$ themselves. So, then, why do logic books make this error? Shouldn't they say that $\{NAND, 0\}$, rather than NAND alone, is functionally complete?

答:
For all propositions $p$ we may render

$p\text{ nand }p=\text{not }p$

$p\text{ nand }(\text{not }p)=1$

$1\text{ nand }1=0.$


答:
First of all, a set $S$ is infinite iff it can't be put in bijection with $\varnothing$ or $I_n=\{1,2,...,n\} \ \ \forall n \in \mathbb{N}$. This implies that if $S$ is infinite ,it is not empty.

Formally speaking the recursion(where $x_1$ is a fixed element of $S$ that exists since $S\neq \varnothing$):
$$
\left\{
\begin{array}{c}
U_{1}=\{x_1\} \\
U_{n+1}=U_n \cup \{x_{n+1}\} \ \ \ x_{n+1}\in S/U_{n}
\end{array}
\right.
$$

Is ill-posed because it doesn't specify what is $x_{n+1}$.For example if $S=\mathbb{R}$ and $U_1=\{\pi\}$, who is $U_2$? It is not univocally determined , it may be $\{\pi,e\}$, it may be $\{\pi,2\}$,...
So this recursion is ill posed. To make this recursion well-posed we need something that takes "in input" $S/U_{n}$ and gives us an element of $S/U_{n}$ univocally for all $n$. The idea is to construct a choice function:
$$\Gamma: \mathcal{P}(S)/\{\varnothing \} \to S
$$
So that we can define the recursion as:

$$
\left\{
\begin{array}{c}
U_{1}=\{x_1\} \\
U_{n+1}=U_n \cup \{\Gamma (S/U_n)\}
\end{array}
\right.
$$

Notice that this recursion is well posed since $S/U_n$ is non-empty for all $n$, so it can be an argument of choice function. Indeed if it was $S/U_n=\varnothing$ then $S=U_n$, but $S$ is infinite so $S\neq U_n$ (contradiction!).

Title: How is axiom of choice utilized within the given proof?

Body:
I (only a beginner in set theory...) want to see a prove that an infinite set must have an infinite countable subset U:

The following answer sounds quite logically for me:
John Wayland Bales, Prove that every infinite set has a countable subset

However, I guess that this proof depends at least on on some weaker variant of axiom of choice.

John Wayland Bales constructs a series of non empty finite sets with increasing number of elements:

For each $N\in\mathbb{N}$, $S$ contains an element distinct from each
element in $U_N=\{x_1,x_2,\cdots,x_n\}$, so define
$U_{N+1}=U_N\cup\{x_{N+1}\}$ where $x_{N+1}$ is an element of $S$
distinct from each element of $U_N$.

Where is AC hidden in this argumentation?
I feel, that the last step is critical:

Let $$U=\bigcup_{N\in\mathbb{N}}U_N$$
Then $U$ is a countable subset of $S$.

I would like to learn, how the proof is done "correctly" by using AC or some weaker weaker variant (axiom of countable choice...).

My problem: Of course I see that recursive choices of "any" elements to create a new set are crucial, but those choices are done sequentially and not on an infinite set of sets that is already "available" and clearly defined. AC (in it's original textbook variant) is about the latter situation and not about a set, that is still in process of being constructed.


问:
Find the Probability mass function of $X$
We select $4$ random digits where in each turn I have $10$ possible output: ${0,1,2,3,...,9}$

let $X$ be the maximum digit that we got, Find the Probability mass function of $X$.

I know that the possibly of getting specific debit is $1/10$ thus the possibility of specific vector of for is $(1/10)^4$ plus:

$P(X=0)=P(0000)=(1/10)^4$

$P(X=1)=1-P(X not = 1)=1-((X=0)+(X=2)+(X=3)+...+(X=9))$

Any suggestions on how to solve this one?

答:
Let the four numbers that you get be $X_1$, $X_2$, $X_3$ and $X_4$. Now
\begin{align}
\Pr\{\max(X_1,X_2,X_3,X_4)\leq x\}&=\Pr\{X_1\leq x,X_2\leq x,X_3\leq x,X_4\leq x\}\\
&=\Pr\{X_1\leq x\}\Pr\{X_2\leq x\}\Pr\{X_3\leq x\}\Pr\{X_4\leq x\}
\end{align}
where the second equality results from independence. Now you should find $\Pr\{X_i\leq x\}$ and continue; recall the $X_i$'s have the same law.


问:
The diffential of commutator map in a Lie group
Leb $G$ be a Lie group and $f:G\times G\rightarrow G$ be the commutator map $:(x,y)\mapsto xyx^{-1}y^{-1}$.

How to obtain the Lie bracket in the associated Lie algebra of $G$ from the derivatives of $f$?

(We know that the Lie bracket is defined via the adjoint representation.)

(I saw $df_{(e,e)}(X,Y)=[X,Y]$ somewhere. Thanks to @John for remarking this is false. But I think there is indeed a relation between the Lie bracket and derivatives of $f$.  )

答:
Let $G$ be a Lie group with the neutral element $e$. Then, given two elements in its Lie algebra, $\xi,\eta\in\operatorname{Lie} G:=T_eG$, represented by (smooth) paths $x,y\colon (-\varepsilon,\varepsilon)\to G$ with $x(0)=y(0)=e$, one can define the Lie bracket as the second derivative of the commutator map: $$[\xi,\eta]=\frac{d^2}{dt^2}(x(t)y(t)x(t)^{-1}y(t)^{-1})\biggr|_{t=0}.$$
Note that the first derivative of the commutator vanishes, which makes the second derivative well-defined as an element of the tangent space $T_eG$.


问:
How to Find an Equation of a Plane Containing Point Q and Perpendicular to Plane P
Given point $Q(1, 0, -1)$ and plane $P: 2x + 3y - 2z = 6$, how do I find an equation of a plane that contains point $Q$ and is perpendicular to plane $P$?

答:
$Q(1,0,-1);\;P:2x + 3y - 2z = 6$

The equation of a plane is $\pi:ax+by+cz=d$

$\pi\perp P\to (2,3,-2)\cdot (a,b,c)=0\to 2a+3b-2c=0$

Furthermore $Q\in\pi\to a\cdot 1+b\cdot 0+(-1)\cdot c=d\to a-c=d$

Thus we have the system
$$
\begin{cases}
2a+3b-2c=0\\
a-c=d\\
\end{cases}
$$
$$
\begin{cases}
a-c=-\frac{3b}{2}\\
a-c=d\\
\end{cases}
$$
the system has solutions only if $d=-\frac{3b}{2},\,c=a+\frac{3b}{2}$

$2 a x+2 b y+(2a+3 b) z+3 b=0$ for instance
$$2 x+2 y+5 z+3=0$$


问:
Prove that $D'_n=\langle x^2\rangle$
I want to check if my solution to one problem from my group theory course is valid. The problem is:

Given $D_n=\{x^iy^j:0\leq i<n,0\leq j<2\}$, prove that $D'_n=\langle x^2\rangle$.

My attempt is:

Given two different elements $g=x^ay^b$ and $h=x^cy^d$ (it's clear that $g,h\in D_n$). We calculate its commutator:

$$[g,h]=x^ay^bx^cy^dy^{-b}x^{-a}y^{-d}x^{-c}$$
Using that $yx^iy=x^{-i}$ (I'll go almost straight to the result avoiding the operations) we get that

$$[g,h]=x^ax^{-c}y^by^dy^{-b}y^{-d}x^ax^{-c}=x^{2a-2c}=(x^2)^{a-c}\in \langle x^2\rangle.$$
So this proves that, $\forall g,h\in D_n$, $[g,h]\in\langle x^2\rangle$, so from this we conclude that
$$\boxed{D'_n=\langle x^2\rangle}$$
Is my solution valid? If not, why? Any help will be appreciated, thanks in advance.

答:
Well, you showed that $D_n'\subseteq \langle x^2\rangle$. But what about the other direction? It is easy though, because $x^2=[x,y]\in D_n'$, and so $\langle x^2\rangle\subseteq D_n'$.

By the way, another way to see that $D_n'\subseteq\langle x^2\rangle$ is to note that $D_n/\langle x^2\rangle$ has either order $2$ or order $4$, depends if $n$ is even or odd. But in either case, $D_n/\langle x^2\rangle$ is abelian, and so $\langle x^2\rangle$ contains all the commutators of the group.

Title: Prove that $D'_n=\langle x^2\rangle$


问:
Question relating to inverse proportionality.
I just came across a question I can't seem to understand, perhaps it it worded wrong that I do not know. The question is:

$y$ is inversely proportional to $x^3y=44$, when $x=a$. Find $y$ when $x=2a$

答:
$$y\propto \dfrac{1}{x^3}$$

means $$y= \dfrac{\text{constant}}{x^3} \quad \text{or,} \quad yx^3 = \text{constant}$$

So $$y \cdot a^3 = y' \cdot (2a)^3 $$
$$\Rightarrow y' = \dfrac{y}{2^3} = \dfrac{44}{8}$$


问:
Show that $\frac{1}{\sqrt{1+kx}}\geq 1-\frac{kx}{2}$

Let $k>0$. Show that $$\frac{1}{\sqrt{1+kx}}\geq 1-\frac{kx}{2}$$ for all $x>0$.

In order to prove the requested result I'm trying to prove that $$\frac{1}{\sqrt{1+kx}} -1+\frac{kx}{2}\geq0$$ for positive $x$. However, I'm stuck at proving the last inequality. Thank you in advance

答:
Hint: Note that $k$ and $x$ always appear as $kx$, so it suffices to prove the result
$$
g(y)\equiv\frac{1}{\sqrt{1+y}}-1+y/2\geq0
$$
for $y>0$. To this end, check that $g'(y)>0$ for $y>0$ and $g(0)=0$.


答:
Take $1+kx=u^2$ with $u\ge 1$ then we have to prove $$f(u)=3u-u^3\le 2$$ Can you show that $f(u)$ is decreasing for $u\ge 1$ ??

I leave it here

Title: Show that $\frac{1}{\sqrt{1+kx}}\geq 1-\frac{kx}{2}$

Body:
Let $k>0$. Show that $$\frac{1}{\sqrt{1+kx}}\geq 1-\frac{kx}{2}$$ for all $x>0$.

In order to prove the requested result I'm trying to prove that $$\frac{1}{\sqrt{1+kx}} -1+\frac{kx}{2}\geq0$$ for positive $x$. However, I'm stuck at proving the last inequality. Thank you in advance


问:
Why $z^2$ doubles the angle?
$z^2$ is not a conformal map at zero because it doubles the angles. Why is that?

答:
Take two complex numbers $z_1,z_2$ and write them in polar form.

$z_1 = |z_1|(\cos \theta + i\sin\theta)$ and $z_2 = |z_2|(\cos\phi + i\sin \phi)$

Multiply them togehter.

$z_1z_2 = |z_1z_2|(\cos\theta\cos\phi - \sin \theta\sin \phi + i(\cos\theta\sin\phi + \sin\theta\cos\phi))$

Notice that these trig expressions fit the double angle formula.

$z_1z_2 = |z_1z_2|(\cos(\theta+\phi) + i\sin(\theta+\phi))$

Multiplying two complex numbers adds the angles.  And squaring a complex number doubles the angles.

Title: Why $z^2$ doubles the angle?


问:
Calculate the new position of a point after rotating it around another point 2D
I have a point, whose location I know in Cartesian coordinates x and y. Now I need to calculate the coordinates of the point if it is rotated with another point as the centre of rotation, whose position I know too, for a known angle. How do I go about it?

This is in plane coordinate system.(i.e. 2D)

答:
Let $(x,y)$ be the point about which $(x_0,y_0)$ is rotating. Suppose the rotation angle is $\theta$ counterclockwise. Then we find the final position is given by
$$(x_1,y_1) = (x,y) + r(\cos(\theta'), \sin(\theta'))$$
Where $r = |(x,y)-(x_0,y_0)|$ and $\theta' = \theta + \arctan(y_0 - y, x_0-x)$


问:
Projections in Type III von Neumann agebra
If $M$ is a type III von Neumann algebra, I wonder how many projections does $M$ possess? Countably infinite or uncountably?

答:
The only von Neumann algebras that do not have uncountably many projections are the commutative finite-dimensional ones, which have finitely many ones. There is no von Neumann algebra with countably many projections.

Already in $M_2(\mathbb C)$, you have the uncountable family
$$
\begin{bmatrix} t&\sqrt{t-t^2}\\ \sqrt{t-t^2}&1-t\end{bmatrix},\qquad t\in[0,1].
$$
In any factor other than $\mathbb C$, take a nontrivial projection $p$, and consider equivalent projections $q_1\leq p$, $q_2\leq 1-p$. If $v$ is the partial isometry realizing the equivalence, $\operatorname{span}\{q_1,v,v^*,q_2\}$ is isomorphic to $M_2(\mathbb C)$ and so the above example applies.

The above can also be done in an arbitrary non-commutative von Neumann algebra.

And if $M$ is commutative and infinite-dimensional, then there is a copy of $\ell^\infty(\mathbb N)$ inside, so there are also uncountably many projections.


答:
First, you have to translate $(x,y)$ to a new origin, that is
$$
(x,y)\mapsto (x-\alpha,y-\beta)
$$
where $(\alpha,\beta)$ is the point around what you want to rotate.

Then you apply the rotation matrix to get
$$
(x-\alpha,y-\beta) \mapsto (cos(\theta)(x-\alpha) -sin (\theta) (y-\beta), sin(\theta)(x-\alpha) +cos (\theta) (y-\beta))
$$
where $\theta$ is the angle you want to rotate.

And finally, you translate back to the origin obtaining that $(x,y)$ gets transformed to

$$
(cos(\theta)(x-\alpha) -sin (\theta) (y-\beta)+\alpha, sin(\theta)(x-\alpha) +cos (\theta) (y-\beta) +\beta)
$$

Title: Calculate the new position of a point after rotating it around another point 2D

Body:
I have a point, whose location I know in Cartesian coordinates x and y. Now I need to calculate the coordinates of the point if it is rotated with another point as the centre of rotation, whose position I know too, for a known angle. How do I go about it?

This is in plane coordinate system.(i.e. 2D)


问:
Showing that if $3x$ is even then $3x+5$ is odd
I'm learning the absolute basics of how to do proofs, and am really struggling.

If 3x is even then 3x+5 is odd.

This is the solution:

I get that even numbers are 2n and odd numbers are 2n+1. For the life of me, I CANNOT get it into that form shown below. I feel so dumb. I tried looking up other answers before posting, but   nothing I found is this basic.

Work:
-Assumptions-
3x = 2n
3x+5 = 2k+1

-Trying to make sense of 3x-
3x+5 = 2k+1
3x = 2k-4

-Plugging in 2k-4 for 3x-
2k-4 = 2n
2k = 2n+4
k = n+2

-Plugging in n+2 for k-
3x+5 = 2(n+2)+1

...This is where I gave up. I don't know where I'm going with this anymore.

答: Ok so we know that $3x$ is even, that means we can write $3x=2n$ for a suitable $n$, since even means that the number is divisible by two without remainder. But then we have $3x+5=2n+5=2n+(4+1)=2n+2\cdot 2+1=2(n+2)+1$ which clearly is odd.


问:
The right "weigh" to do integrals
Back in the day, before approximation methods like splines became vogue in my line of work, one way of computing the area under an empirically drawn curve was to painstakingly sketch it on a piece of graphing paper (usually with the assistance of a French curve) along with the axes, painstakingly cut along the curve, weigh the cut pieces, cut out a square equivalent to one square unit of the graphing paper, weigh that one as well, and reckon the area from the information now available.

One time, when we were faced with determining the area of a curve that crossed the horizontal axis thrice, I did the careful cutting of the paper, and made sure to separate the pieces above the horizontal axis from the pieces below the horizontal axis. Then, my boss suddenly scooped up all the pieces and weighed them all.

I argued that the grouped pieces should have been weighed separately, and then subtract the weights of the "below" group from the "above" group, while my boss argued that we were calculating the total area, and thus, not separating the pieces was justifiable.

Which one of us was correct?

答:
In my opinion, this is not really a math question.  Which procedure is correct depends what you're going to do with your calculation.

As a matter of definition, the integral indeed measures the signed area (positive area minus negative area), as you suggest.  So your approach is computing an approximation to the definite integral $\int_a^b f$.

But maybe you want the total (unsigned) area.  E.g. if you're going to lay concrete along (some real-world space corresponding to) the region bounded by the curve and the x-axis then surely you want the total area -- there's no such thing as negative concrete.

Without knowing what you're using the calculation for, it's impossible to say.  (Essentially, you're asking us: "Which of these is mathematically correct: $A-B+C$ or $A+B+C$?"  Of course it depends upon what you're trying to do.) I would like to think that your boss knew what the point of it all was, so without further information I guess I would trust him.

In fact, your story arouses my curiosity.  I suppose you're not putting us on, but weighing paper cutouts is just about the last method I would ever think of for computing area (aren't you going to need a very sensitive scale or an awfully big piece of paper to get anywhere with this?).  How long ago are we talking?  What was the job?  You don't have to answer these questions, but it would be interesting to know...

ContentLicense: CC BY-SA 2.5

Title: The right "weigh" to do integrals


问:
How do I compute the Laplacian of a function in terms of a given (general) coordinate transformation?
Consider a coordinate transformation $\boldsymbol{x} = \boldsymbol{x}(\boldsymbol{\xi})$ (with Jacobian $\partial \boldsymbol{x}/\partial \boldsymbol{\xi})$, the scalar function $f(\boldsymbol{x}) = f(\boldsymbol{x}(\boldsymbol{\xi})) = \bar{f}(\boldsymbol{\xi})$ and the definitions

$$\boldsymbol{\nabla}_\xi (\cdot) = \partial (\cdot)/\partial \boldsymbol{\xi},$$
$$\boldsymbol{\nabla}_x (\cdot) = \partial (\cdot)/\partial \boldsymbol{x}.$$

I know how to write the gradient of $f(\boldsymbol{x})$, $\partial f(\boldsymbol{x})/\partial \boldsymbol{x}$ in this case: I start with

$$\boldsymbol{\nabla}_\xi \bar{f}(\boldsymbol{\xi}) = \boldsymbol{\nabla}_x f(\boldsymbol{x})\cdot\partial \boldsymbol{x}/\partial \boldsymbol{\xi} = \boldsymbol{\nabla}_x f(\boldsymbol{x})\cdot\boldsymbol{J} = \boldsymbol{J}^{\mathrm{T}}\cdot\boldsymbol{\nabla}_x f(\boldsymbol{x}),$$

so that
$$\boldsymbol{\nabla}_x f(\boldsymbol{x}) = \boldsymbol{J}^{-\mathrm{T}}\cdot\boldsymbol{\nabla}_x \bar{f}(\boldsymbol{\xi}).$$

But how do I obtain an expression for the Laplacian

$$\boldsymbol{\nabla}_x\cdot\boldsymbol{\nabla}_x f(\boldsymbol{x})?$$

This seems basic, but I can't figure out how to do it. I've searched the internet and math.stackexchange but haven't found any answers for the general case.

答:
You should use the Laplace-Beltrami operator on general Riemannian manifolds, for example:

You can find more information  here.


问:
Number of solutions of system of linear equations over finite field
Suppose $F=GF(2^8)$. Let $u_1,u_2\in K=GF(2^4)$ be linearly independent elements. The functions $x\mapsto tr_n (u_ix)$ are linear functions from $F\to GF(2)$, $i=1,2$. Here $tr_n$ denotes the absolute trace.

Since the functions are linear, I know that there are exactly $2^7$ elements $x$ such that $tr_n(u_ix)=0$, for $i=1,2$.

Computationally, I observed that there are exactly $2^6$ elements such that $ tr_n (u_1x)=0$ and $ tr_n (u_2x)=0$. Since $u_i\in K$, I know that for $x\in K$, both of the traces will be equal to $0$. But how do I determine the remaining $2^6-2^4$ elements. How do I prove that the number of solutions is exactly $2^6$. I have a feeling that it has to do with the linear independence of $u_1$ and $u_2$, but I do not know how to use it.

答:
Proving that there are exactly $2^6$ solutions is easy.
Let
$$V_i=\{x\in F\mid tr(u_ix)=0\},$$
$i=1,2$, be the sets of solutions to one of the equations. As you observed they are kernels of a linear transformation $f_i:F\to GF(2), x\mapsto tr(u_ix)$. Because $u_1\neq0\neq u_2$, both are surjective, and, by rank-nullity, have dimension $7$.

Linear independence is equivalent to the assumption that $u_1$ are $u_2$ are distinct and non-zero. The key is the following

Claim. The subspaces $V_1$ and $V_2$ are not equal.

Proof. Assume contrariwise that $V_1=V_2$. Let $y_1,y_2,\ldots,y_7$ be a basis of $V_1$. We can extend it to a basis of $F$ by joining an eighth element $y_8$. As $y_2\notin V_1=V_2$ we have $tr(u_1y_8)=1=tr(u_2y_8)$. Together with the facts $tr(u_iy_j)=0$, $i=1,2, j=1,2,\ldots,7$. This implies that
$$
tr((u_1+u_2)y_j)=0
$$
for all $j=1,2,\ldots,9$. In other words, $x\mapsto tr((u_1+u_2)x)$ vanishes on all of $F$. By basic properties of the trace maps this implies that $u_1+u_2=0$. In other words $u_1=u_2$. A contradiction.

It follows that $V_1\cap V_2$ is a 6-dimensional subspace. We can re-interpret the intersection as the kernel of the restriction of the transformation $f_2$ to $V_1$:
$$
V_1\cap V_2=\operatorname{Ker}(f_2\vert_{V_1})=\{x\in V_1\mid f_2(x)=0\}.
$$
The image of $f_2\vert_{V_1}$ is a subspace of $GF(2)$, so has dimension either $0$ or $1$. Two cases:

If $f_2\vert_{V_1}$ is the zero map, then it follows that $V_1=V_2$ &endash; a case we excluded.

If $f_2\vert_{V_1}$ is surjective then, by rank-nullity, its kernel is 6-dimensional. Implying that $V_1\cap V_2$ has exactly $2^6$ elements.

For generalizations they key is that the (symmetric) bilinear form $B:F\times F\to GF(2)$
$$
B(x,y)=tr(xy)
$$
is non-degenerate. In other words, given a $y\in F$ such that $B(x,y)=0$ for all $x\in F$, we can conclude that we must have $y=0$. For the trace form this is clear.

Whenever we have a non-degenerate bilinear form $B:V\times V\to GF(2)$, $\dim V=n$, it follows that to each subspace $W\subseteq V$, its orthogonal subspace
$$W^\perp=\{z\in V\mid B(x,z)=0 \ \text{for all $x\in W$}\}$$
has the complementary dimension:
$$
\dim W+\dim W^{\perp}=\dim V.
$$
So if next time you have three linearly independent elements $u_1,u_2,u_3$, it is automatic that the system $tr(u_ix)=0$, $i=1,2,3$, has exactly $2^{8-3}=2^5$ solutions.

Undoubtedly you noticed how I used non-degeneracy of the trace form to conclude that $u_1+u_2=0$.


问:
Converges of two different sequences using Cauchy
I've recently started calc 1.
I've had this problem :

let $a_n$ and $b_n$ be two sequences such that $\forall\varepsilon$ $\exists N$ $\in\mathbb{N}$ such that for $\forall m,n > N$ $|a_n - b_m| < \varepsilon $

prove that both $a_n$ and $b_n$ converge to the same limit.

I've been sitting on this one for 3 days now , I can't figure out how to break them up to two different sequences, I've tried to build a new sequence such that $c_n = a_n - b_m$ but didn't have much luck with continuing from there, I think that you need to prove that both sequences are Cauchy sequences and from there you'll be able to conclude that they have the same limit but aging I'm not sure how to get there.

any help would be much appreciated

答:
There is some $N$ s.t. $\forall n,m >N$ we have $$\vert a_{n}-b_{m}\vert <\frac{\epsilon }{2}$$
It follows that $\forall n,m,l >N$ $$\vert b_{m}-b_{l}\vert \leq \vert b_{m}-a_{n}\vert +\vert b_{l}-a_{n}\vert < \epsilon$$...
can you take it from there?

Edit: Once you have that both sequences are Cauchy, I would switch over to a proof by contradiction.


问:
How can I simplify calculation of this gradient?
Goal: Given $f(\alpha) :=\dfrac{1}{2} \|y-X\alpha\|_2^2$ I want to show that $\nabla f(\alpha)=0\iff X^T X\alpha = X^T y$ where: $X\in\mathbb{R}^{n\times p},\alpha\in\mathbb{R}^p,y\in\mathbb{R}^n$.

If I compute the gradient according to definition I have no problem showing this. But in the lecture the professor is using some gradient/vector propreties as shortcuts and I want to be able to recognise and use them as well. But when using shortcut my solution is wrong by a transposition and I want to know why it fails. Here are the steps:
$$ \nabla f(\alpha) = \frac{1}{2}\frac{d}{d\alpha}\Big(\|y\|_2^2-2y^T X\alpha+\alpha^TX^TX\alpha\Big) $$
I know that $\dfrac{d}{dx} x^TAx = (A+A^T)x$ so $\dfrac{d}{d\alpha} \alpha^TX^TX\alpha = 2X^TX\alpha$ since $X^TX$ is symmetric. Also $\dfrac{d}{d\alpha} \|y\|_2^2= 0$ goes by itself. The error lies in the last gradient calcul I have to do. Here's what I've done:

Since for $x\in\mathbb{R}^p$ $\dfrac{d}{dx} (x) = E\in\mathbb{R}^{p\times p}$ the identity matrix, then:  $\dfrac{d}{d\alpha} 2y^TX\alpha = 2y^TX\cdot E = 2y^TX$. It leads to the result:
$$ \nabla f(\alpha)=0\iff X^T X\alpha = y^T X$$
Which is false by a transposition, the dimensions are not matching so it doesn't really make sense. So question is: Why doesn't my shortcut work in this case, and in what other easy way can I make this work ?

Note: The same problem arise when I consider $f:\mathbb{R}^n\to \mathbb{R}$, $f(x) = (1,\dots,1)\cdot \begin{pmatrix}x_1\\ \vdots\\ x_n\end{pmatrix}= x_1+\cdots +x_n$. Obviously $\nabla f(x) = \begin{pmatrix} 1\\ \vdots \\ 1\end{pmatrix}\in\mathbb{R}^n$. But using the same shortcut as above we have:
$$ \frac{d}{dx} \big(\;(1,\dots,1)\cdot \begin{pmatrix}x_1\\ \vdots\\ x_n\end{pmatrix}\;\big)=(1,\dots,1)\cdot E_{n\times n} = (1,\dots,1)\in\mathbb{R}^{1\times n} $$
Same, nearly the right answer but not quite.

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}$
Explicit use of the inner (dot) product is one way to avoid the transposition error.

Define the vector $w=(X\cdot a-y),\,$ then write the function in terms of this vector and calculate its gradient.
$$\eqalign{
 f &= \tfrac 12w\cdot w \\
df &= w\cdot dw \;=\; w\cdot(X\cdot da) \;=\; (X^T\!\cdot w)\cdot da \\
\p{f}{a} &= X^T\!\cdot w \;=\; X^T(X\cdot a-y) \\
}$$
Taking the transpose of a matrix is never an issue, but anytime you find yourself taking the transpose of a vector, proceed with caution. Rewriting the equation using dot products is one fool-proof method. It forces you to arrange the expression on each side of a dot product into dimensionally compatible forms.

Another fool-proof method is to use dyadic products, while very complicated expressions often require index notation for a solution.


问:
Sum of inverse of imaginary part of non trivial zeroes of Riemann zeta function
I would like to know if $$\sum_{k \geq 1} \frac{1}{\gamma_k}$$ is convergent or not. ($\gamma_k$ is the imaginary part of the k-th non trivial zero of $\zeta$).

In fact, I think this series is actually divergent, because of the "Explicit Formula" : $$\psi(x) = x-\sum_\rho \frac{x^\rho}{\rho} - \ln(2 \pi) - \ln(1-x^{-2})/2$$Where $\psi$ is the Chebyshev function.
So we have $$\sum_\rho \frac{x^\rho}{\rho} = x - \psi(x) - \ln(2\pi) - \frac{1}{2} \ln (1-x^{-2})$$But at $x=1$, the function $\ln(1-x^{-2})$ diverges. So the series $\sum_\rho 1/\rho$ must too. However, this "proof" may not be one because the "$\rho$" are $1/2+i \gamma_k$ (iff the RH is true) and not just $\gamma_k$.

Now, what I would like to know, is the value (if it converges) of $$\sum_{k \geq 1} \frac{1}{\gamma_k^s}$$For any $s \in \mathbb{C}$. I'm aware that this series propably diverges for $\Re(s) \leq 1$ but perhaps there is a way to extend this function analytically as in the case of Riemann's zeta function. Even if I can have an asymptotic approximation.

答: Let $N(T)$ be the number of non-trivial zeros of $\zeta$ such that $0<\gamma\leqslant T$ ($\gamma$ being their imaginary part). It is known that $\displaystyle N(T)\underset{T\rightarrow +\infty}{\sim}\frac{T}{2\pi}\log T$, thus if $\gamma_n$ is the imaginary part of the $n$-th non-trivial zero (I rank them such that $0<\gamma_1\leqslant\ldots\leqslant\gamma_n$ and $\gamma_{-n}=-\gamma_n$ for $n\geqslant 1$). Since $N(\gamma_n)=n$, we have $\displaystyle n\underset{n\rightarrow +\infty}{\sim}\frac{\gamma_n}{2\pi}\log\gamma_n$, taking the logarithm gives $\log n\underset{n\rightarrow +\infty}{\sim}\log\gamma_n$ and thus $\gamma_n\underset{n\rightarrow +\infty}{\sim}\frac{2\pi n}{\log n}$. We can then conclude that $\sum_{n\geqslant 1}\frac{1}{\gamma_n^s}$ converges if and only if ${\rm Re}(s)>1$, in particular the series $\sum_{n\geqslant 1}\frac{1}{\gamma_n}$ diverges. As for the precise value of $\sum_{n\geqslant 1}\frac{1}{\gamma_n^s}$ when ${\rm Re}(s)>1$, I don't have the answer (and I'm not sure if we know anything about it).


问:
Finding range of the expression $\sum_{cyc}\frac{|x+y|}{|x|+|y|}$ where $x,y,z\in\mathbb{R}-\{0\}$
I am supposed to find the range of the following. At present I do not know how to get started. Obviously as is the case with numerous symmetrical expressions, one of the extrema occur when all variables are equal, so $3$ would be a value of extrema, but how to prove such an argument rigorously is non-obvious. Any hints would be appreciated. Thanks.

$$\sum_{cyc}\frac{|x+y|}{|x|+|y|} \text{ for }x,y,z\in\mathbb{R}- \{0\}$$

答:
(Fill in the gaps as needed. If you're stuck, show your work and explain why you're stuck.)

Hint: If $xy \geq 0$, show that $ \frac{ | x+y | } { |x| + |y| } = 1$.

If $ xy < 0$, show that $ 0 \leq \frac{ | x+y | } { |x| + |y| } < 1 $.

Hence, conclude that

$$ 1 \leq \sum \frac{ | x+y | } { |x| + |y| } \leq 3. $$


问:
Solving equation with natural numbers
I have the following question:
In a store I want to pay with exactly 50 coins. There are 2€, 1€, 0,5€ 0,2€ and 0,1€ coins.
So I'd have
2a + 1b + 0,5c + 0,2d + 0,1*e = amount I have to pay
And a+b+c+d+e=50 (number of coins)
Since there are no half Euros, a,b,c,d, and e are natural numbers (0-50)

Is there any way to solve this?
Thanks in advance

答: It depends on the amount you want to pay. For example, if the amount is $0.01$, you do not have a coin with that "resolution". Moreover, if the amount you want to pay is $0.1$, you can only pay with your $0.1$ coin, because you don't have any coins with lower value to combine. Finally, there are amounts you do not have the money for: if something costs $\$1000000$, you don't have a coin with a high enough value such that fifty of it will give you a million dollars. In summary, with the information you have given us, it seems that the problem is underspecified.

Title: Solving equation with natural numbers


问:
Solving $\frac{\cos(\alpha+\beta)}{\cos(\alpha-\beta)}=c$ for $\alpha$
I have been struggling with this equation. Is it unsolvable?

$$\frac{\cos(\alpha+\beta)}{\cos(\alpha-\beta)}=c$$

$\beta$ and $c$ are constant. How can I solve for the angle $\alpha$?

PS: As you might imagine, I'm still new to this, so please be easy on me. Thanks for your help.

答:
I think I've found a solution to this problem, solving for $\alpha$.
Following VIVID's answer i arrived at:
$$\cos(\alpha)\cdot\cos(\beta)(1-c)-\sin(\alpha)\cdot\sin(\beta)(1+c)=0$$
following the angle sum formula:
$$\cos(x+y)=\cos(x)\cdot\cos(y)-\sin(x)\cdot\sin(y)$$
finally i can express $\alpha$ in terms of $\beta$ and c, which gives...

$$\alpha(\beta,c) = \arctan\Biggr(\frac{1-c}{1+c}\cdot\frac{1}{\tan(\beta)}\Biggr)$$

I hope there is no big oversight and this is correct. Also, I've never used LaTeX before, so writing this was painfully slow, so I skipped some steps. I thank everyone for their help.


答:
I think you're getting confused by the $3x$ part. The $3x$ plays no role in the problem.

Suppose you're given any number that is even. I'll call it y. Now we want to show $y+5$ is odd.

Therefore by definition

$y=2k$ for some integer $k$

Now

$y+5 = 2k+5$

Now we just need to show that $2k+5$ is 2 times an integer plus 1

$2k+5 = 2(k+2)+1$

So $2k+5$ is odd because it can be written in the form 2*integer +1 where the integer here is $k+2$. So $y+5$ is odd since $y+5 = 2k+5$

So if any number is even. Then that number plus 5 is odd. It doesn't matter if the original number is 3x or 8z or 3x^2-5x+x^3 etc...

Title: Showing that if $3x$ is even then $3x+5$ is odd

Body:
I'm learning the absolute basics of how to do proofs, and am really struggling.

If 3x is even then 3x+5 is odd.

This is the solution:

I get that even numbers are 2n and odd numbers are 2n+1. For the life of me, I CANNOT get it into that form shown below. I feel so dumb. I tried looking up other answers before posting, but   nothing I found is this basic.

Work:
-Assumptions-
3x = 2n
3x+5 = 2k+1

-Trying to make sense of 3x-
3x+5 = 2k+1
3x = 2k-4

-Plugging in 2k-4 for 3x-
2k-4 = 2n
2k = 2n+4
k = n+2

-Plugging in n+2 for k-
3x+5 = 2(n+2)+1

...This is where I gave up. I don't know where I'm going with this anymore.


答: Remember here that $n$ represents ANY natural number. You got to the answer but you didn’t even realize it. That’s probably because you are thinking syntactically rather than semantically. What I mean is the literal string of symbols $2(n+2)$ didn’t register to you as even because it is not the same as the string $2n$. But $n+2$ is a natural number just like $n$ is. So the the strings $2(n+2)$ and $2n$ both represent even numbers, and so $2(n+2) + 1$ is odd just as you have shown in your last line.


问:
Compute inf and sup of a set A
Compute the infimum and supremum of $A=\{f(x)=\frac{2x+1}{x+2}: x>-2\}$.

I try to do these following passages:
since $f$ is derivable I compute $f'(x)=\frac{3}{x+2}>0$ and so from Monotone Convergence Theorem I can conclude that:
$\lim_{x\to-2^+}f(x)=-\infty=infA$ and $\lim_{x\to+\infty}f(x)=2=supA$. All right?

答:
Your sketch seems more or less correct, depending on which of these theorems you actually have access to in the class and how the final proof is written.

That being said, you can also so this straight from the definitions of inf and sup.

Your supposition is probably correct:

Let $N \in \mathbb{N}$ be any number. Can you prove that there exists an $x \in (-2,\infty)$ such that $f(x) < N$? If so, you have proven that the inf is indeed $-\infty$.

Now, for the sup, we first need to prove that 2 is an upper bound. That shouldn't be too hard. Then, you need to show that for any $\varepsilon>0$, there exists an $x \in (-2,\infty)$ such that $\lvert 2-f(x) \rvert < \varepsilon$.

With these two, you don't need all the machinery that your answer gave.

Title: Compute inf and sup of a set A


问:
asymptotics of $\frac{c^{2(\log\log n)+1}-1}{c-1}$
Let $c>0$ be a constant and $n$ go to infinity.

Asymptotically why is it the case that $\frac{c^{2(\log\log n)+1}-1}{c-1}\sim \log n^{2\log c}$ ?

I don't think it changes anything but the actual relation is $$\frac{2c\log\log n}{n}\frac{c^{2(\log\log n)+1}-1}{c-1}\sim\frac{2c\log\log n}{n}\log n^{2\log c}$$ as $n\rightarrow\infty$.

I don't see how this is true.

I've tried to transform the left value but I don't see how we $\log c$ can appear. And Why does $c<1$ versus $>1$ not matter?

Could the relation be true if we replace $\sim$ by $\lesssim$? (i.e. left =$O$(right))?

答:
If $c>1$,
\begin{align}
\frac{2c\log\log n}{n}\frac{c^{2(\log\log n)+1}-1}{c-1}
&= \frac{2c\log\log n}{n}\frac{c\cdot \left(\log n\right)^{2\log c}}{c-1} - \frac{2c\log\log n}{n}\frac{1}{c-1}\\
&= O\left(\frac{\log\log n\left(\log n\right)^{2\log c}}{n}\right) + O\left(\frac{\log\log n}n\right)\\
&= O\left(\frac{\log\log n\left(\log n\right)^{2\log c}}{n}\right).
\end{align}
If $c<1$, we get $O\left(\frac{\log\log n}n\right)$ at last.


答:
You can use the fact that
$$
\frac{2x+1}{x+2} = 2-\frac{3}{x+2}.
$$
Using this you can prove that the function is decreasing for $x>-2$ instead of using the derivative. Also with this you can calculate the limits.

Title: Compute inf and sup of a set A

Body:
Compute the infimum and supremum of $A=\{f(x)=\frac{2x+1}{x+2}: x>-2\}$.

I try to do these following passages:
since $f$ is derivable I compute $f'(x)=\frac{3}{x+2}>0$ and so from Monotone Convergence Theorem I can conclude that:
$\lim_{x\to-2^+}f(x)=-\infty=infA$ and $\lim_{x\to+\infty}f(x)=2=supA$. All right?


问:
Prove that difference in area of circumcircle and polygon is greater than the difference in area of polygon and incircle.
The problem can be equivalently stated as

PROBLEM : A convex $n$ sided polygon has a circumcircle and an inscribed circle, its area is $B$, and the areas of circumcircle and inscribed circle are $A$ and $C$ respectively. Prove that $2B < A+C$.

I think this problem is very difficult. This is my attempt for special case of polygons i.e Regular polygons.

Naming of parameters :

$R$ be the radius of circumcircle of the polygon.

$r$ be inradius of the polygon.

$n$ be the number of sides of polygon.
$\theta$ = $\frac{2\pi}{n}$ = angle subtended by a side of polygon at centre.

$a$ be the length of side of polygon.

Relations between $R,r,a,\theta$ :

$R^2 = \frac{a^2}{4} + r^2$, $a = 2R*sin(\frac{\theta}{2})$ and $r = R*cos(\frac{\theta}{2})$

We need to prove $2B < A+C$

$\Leftrightarrow \frac{2sin(\theta)}{3+cos(\theta)} < \frac{\pi}{n}$

This can be verified by showing that, inequality is true for $n = 3 $ and LHS decreases faster than RHS.

The method I used for regular polygons isn't applicable to all. There is too much freedom and ambiguity. But I don't have any idea to tackle generalized polygon. Can anybody help me?

答:
It's actually simple. If the perimeter of your convex shape is $P$, then your inequality is equivalent to:
$$\pi(r^2+R^2) > Pr,$$
where $R$ and $r$ are the circumradius and inradius, respectively. A trivial upper bound on the perimeter is:
$$P < 2\pi R$$
where the latter is the length of the circumcircle. But now a simple AM-GM finishes the problem:

$$Pr < 2\pi Rr < \pi(r^2+R^2)$$


问:
prove that $n$ is prime or $n=4$ when $n|4((n−1)!+1)$
Let $n \gt 1$ be an integer such that :

$n|4((n-1)!+1)$

Prove that

$n=4$ or $n$ is prime

答:
Since $4((n-1)!+1)$ has an explicit factor of $4$, plainly $2,4$ divide it, noting that $2$ is a prime. If $n\ne 2,4$ then $n|4((n-1)!+1)$ is true when $n|((n-1)!+1)$. Another way of saying that is $((n-1)!+1) \equiv 0 \bmod n$

If $n$ is prime, Wilson's theorem tells us that $(n-1)! \equiv -1 \bmod n$, which would imply the desired result $((n-1)!+1) \equiv 0 \bmod n$

It remains to be shown that if $n$ is not prime, the proposition fails. If $n>4$ is not prime, it is factorable: $n=ab$ where $2\le a,b \le \frac{n}{2}<n-1$. Since both $a,b$ are factors that appear in the factorial product $(n-1)!$, it is the case that $n\mid(n-1)!$ hence $n\not \mid((n-1)!+1)$ and we are done.


答:
You are dealing with a problem where you are given too much information. Here is another way of doing it (we'll use $m$ for an integer).

If $3x$ is even then $x$ must be even, so we can put $x=2m$. [This is a consequence of the fact that $2$ is a prime, or can be proved in various ways]

Then $3x+5=6m+5=6m+4+1=2(3m+2)+1$.

Now we can put $3m=n$, an integer, to get $2(n+2)+1$.

For an alternative proof, we could put $3m+2=n$ and then we get $2n+1$.

Note that the last two sentences are alternatives to one another. They both use $n$, but $n$ is defined differently in the two cases.

What I'm trying to do here is to unpack how the different expressions for the same thing relate to each other. If you get your head round that you will be flying.


问:
Uniqueness of the sum of roots of unity
Everyone knows that the sum of all roots of unity $r_k$ equals to zero:
$$\sum^{n-1}_{k=0} r_k=\sum^{n-1}_{k=0} e^{\frac{i2k\pi}{n}} =0$$

Does anyone know if it possible to prove that the equation below:

$$\sum^{n-1}_{k=0} a_k r_k=\sum^{n-1}_{k=0} a_k e^{\frac{i2k\pi}{n}} =0$$ where $a_i\in N$ (non-negative integers)

has only one one solution: $a_0=a_1=...=a_{n-1}=constant$

Probably you can suggest me which direction to look in? May be Kummer rings theory can help or the simple prove exist.

答:
For $n = 4$, $(a_0, a_1, a_2, a_3) = (1,1,1,1)$ is a solution, and $(1,0,1,0)$ is another solution from Calvin Lin in the comment above.

So the sum of the two solution vectors $(2,1,2,1)$ is another solution that consists of only positive integers.

Title: Uniqueness of the sum of roots of unity


问:
Is cardinal exponentiation strictly monotone in the exponent?
Let $\kappa$ be a cardinal, does $2^\kappa<2^{\kappa^+}$ always hold? It clearly holds if one assumes generalized continuum hypothesis, but does it also hold if one assumes otherwise?

答: This is independent of $\mathsf{ZFC}$. It consistently holds, because it holds under $\mathsf{GCH}$, and it consistently doesn't hold, because a two step iteration of Cohen forcing over a model of $\mathsf{GCH}$ can be used to force $2^\omega=2^{\omega_1}$.


问:
WolframAlpha functionality: Computing nested integrals
Can wolframpalpha support the numerical calculation of the following type:

$$
\int_{x=0}^\infty f(x) e^{\int_{z=0}^\infty g(x,z)dz} dx
$$

Note that the inside integral is in such form where parameter x cannot be separated from it.

Thanks in advance

PS: The respective command would have been:

int(f(x)*exp[int(g(x,z),{z,0,Infinity})],{x,0,Infinity})

答:
I don't see how to evaluate this type of integral in WolframAlpha.

In Mathematica, we could do this type of integral like so:

```<code>h[x_?NumericQ] := NIntegrate[g[x, z], {z, 0, Infinity}];
NIntegrate[f[x] Exp[h[x]], {x, 0, Infinity}]
</code>```

As a concrete example:

```<code>h[x_?NumericQ] := NIntegrate[Sin[Exp[-x*z^2]], {z, 0, Infinity}];
NIntegrate[Exp[-1/x^2] Exp[-x^2] E^h[x], {x, 0, Infinity}]

(* Out: 0.262062 *)
</code>```

However, WolframAlpha simply doesn't support this type of evaluation sequence and I don't see around the way intermediate definition.

More generally, it's really not the case that WolframAlpha supports general Mathematica  syntax.  If you need to evaluate this type of integral with any regularity, I suggest that you get Mathematica, or some general programmatic tool.

Body:
I don't see how to evaluate this type of integral in WolframAlpha.

In Mathematica, we could do this type of integral like so:

h[x_?NumericQ] := NIntegrate[g[x, z], {z, 0, Infinity}];
NIntegrate[f[x] Exp[h[x]], {x, 0, Infinity}]

As a concrete example:

h[x_?NumericQ] := NIntegrate[Sin[Exp[-x*z^2]], {z, 0, Infinity}];
NIntegrate[Exp[-1/x^2] Exp[-x^2] E^h[x], {x, 0, Infinity}]

(* Out: 0.262062 *)

However, WolframAlpha simply doesn't support this type of evaluation sequence and I don't see around the way intermediate definition.

More generally, it's really not the case that WolframAlpha supports general Mathematica  syntax.  If you need to evaluate this type of integral with any regularity, I suggest that you get Mathematica, or some general programmatic tool.


答:
With asymptotics, constant multipliers are irrelevant. You can actually reduce the problem to $$c^{\log\log n}\sim (\log n)^{\log c}$$
which is just $$c^{\log N}\sim N^{\log c}$$ which I believe is an algebraic identity.

Edit: I left out that $$c^{2\log\log n}\sim (\log n)^{2\log c}$$ and $$c^{\log\log n}\sim (\log n)^{\log c}$$ are equivalent since
$$\frac{c^{2\log\log n}}{(\log n)^{2\log c}}=\left( \frac{c^{\log\log n}}{(\log n)^{\log c}}\right)^{2} $$ So, if
$$\frac{c^{2\log\log n}}{(\log n)^{2\log c}}\rightarrow \text{constant}$$ then
$$\frac{c^{\log\log n}}{(\log n)^{\log c}}\rightarrow \text{constant}$$

Title: asymptotics of $\frac{c^{2(\log\log n)+1}-1}{c-1}$

Body:
Let $c>0$ be a constant and $n$ go to infinity.

Asymptotically why is it the case that $\frac{c^{2(\log\log n)+1}-1}{c-1}\sim \log n^{2\log c}$ ?

I don't think it changes anything but the actual relation is $$\frac{2c\log\log n}{n}\frac{c^{2(\log\log n)+1}-1}{c-1}\sim\frac{2c\log\log n}{n}\log n^{2\log c}$$ as $n\rightarrow\infty$.

I don't see how this is true.

I've tried to transform the left value but I don't see how we $\log c$ can appear. And Why does $c<1$ versus $>1$ not matter?

Could the relation be true if we replace $\sim$ by $\lesssim$? (i.e. left =$O$(right))?


问:
Showing that if $3x$ is even then $3x+5$ is odd
I'm learning the absolute basics of how to do proofs, and am really struggling.

If 3x is even then 3x+5 is odd.

This is the solution:

I get that even numbers are 2n and odd numbers are 2n+1. For the life of me, I CANNOT get it into that form shown below. I feel so dumb. I tried looking up other answers before posting, but   nothing I found is this basic.

Work:
-Assumptions-
3x = 2n
3x+5 = 2k+1

-Trying to make sense of 3x-
3x+5 = 2k+1
3x = 2k-4

-Plugging in 2k-4 for 3x-
2k-4 = 2n
2k = 2n+4
k = n+2

-Plugging in n+2 for k-
3x+5 = 2(n+2)+1

...This is where I gave up. I don't know where I'm going with this anymore.

答:
We know that $3x=2n$. So:
$$\color{blue}{3x}+5=\color{blue}{2n}+5$$
Because the "blue quantities" are equal. Now:
$$3x+5=2n+5=2n+4+1$$
In this step I just wrote $5$ as $4+1$:

$$ 3x+5=2n+5=2n+4+1=2n+2\cdot 2+1$$
In this step I just wrote $4$ as $2 \cdot 2$.

$$ 3x+5=2n+5=2n+4+1=\color{green}{2n+2\cdot 2}+1=\color{green}{2(n+2)}+1$$
The last step is valid because the green quantities are equal. In the end:
$$3x+5=2(n+2)+1$$
This means that $3x+5$ is odd because is of the form $2h+1$ with $h$ integer(in particular $h=(n+2)$)


问:
How to Find an Equation of a Plane Containing Point Q and Perpendicular to Plane P
Given point $Q(1, 0, -1)$ and plane $P: 2x + 3y - 2z = 6$, how do I find an equation of a plane that contains point $Q$ and is perpendicular to plane $P$?

答:
Any plane satisfying the requirement can rotate around the line containing $Q(1,0,-1)$ and perpendicular to the plane $2x + 3y - 2z = 6$, while still satisfying the requirement.

So, let the plane cut the $x$-axis at $A(a,0,0)$ where $a$ serves as a parameter. Then, its normal vector is
$$\vec n=\vec{AQ}\times (2,3,-2)=(1-a,0,-1)\times (2,3,-2)=(3,-2a,3-3a)
$$

and the plane satisfying the requirement is
$\vec n\cdot(x-1, y, z+1)=0$.
Simply to obtain its equation parameterized with $a$
$$x-\frac{2a}3 y+(1-a)z=a$$


问:
Do the algorithms of Prim and Krusksal always produce the same minimum spanning tree, given the same tiebreak criterion?
I teach a course in Discrete Mathematics, and part of the subject matter is a coverage of Prim's algorithm and Kruskal's algorithm for constructing a minimum spanning tree on a weighted graph. Students do not actually implement the algorithms in code; only pseudocode is given; students are asked to hand-trace the algorithm behaviors on a number of exercise and assessments.

In order to make the algorithms deterministic (and so exercises each have a unique solution), the textbooks I've seen generally give a further direction of, "assume that the vertices are ordered alphabetically" (e.g., Rosen Sec. 11.4 exercises 13-15). That is, this presents a deterministic tiebreak criterion, assuming edges are identified with letters themselves in alphabetical order. E.g.: if at some step of either algorithm, $\{a, d\}$ and $\{b, c\}$ are both eligible edges to be added with minimum weight, then the $\{a, d\}$ edge will be chosen. (Here's an example student question arising from this direction.)

I've personally implemented both algorithms in C++ following this criterion, using an adjacency matrix for the underlying edge data. I've found that the exercise direction is clarified for students when we step through a minimal-search loop at least once, with the vertices stored in alphabetical order, and find that the selected edge is the first one found with minimal value.

So my question is: Do the algorithms of Prim and Kruskal always produce the same minimum spanning tree, given the tiebreak-by-alphabetical order criterion? (And ignoring any other possible implementation details.)

My first intuition was that I could sketch an example where the two algorithms produce different trees, but after several attempts I've failed to do so. I also wrote a Monte Carlo program to randomly create a few hundred thousand random weighted graphs, of various vertex, edge, and weight numbers, and send them through my implementation of both Prim's and Kuskal's algorithm; in all of my experimentation, they've produced identical minimum spanning trees.

Some things that are obviously known and not part of this question:

Prim and Kruskal will of course produce trees of the same total weight. This question is about whether the trees themselves are identical, using all the same edges.

Prim and Kruskal will of course produce identical trees if the edge weights are all unique. Clearly any counterexample graph requires at least some edges with equal weight.

Prim and Kruskal may add edges in a different order; it's easy to produce examples of that. This question is whether the final resulting spanning tree is always the same.

Here are some related questions, answers, and comments which do not resolve this question. None of these answers assume the deterministic ordering (alphabetic) criterion here, and therefore conclude that the ordering is ambiguous due to different possible underlying implementations:

Can Prims and Kruskals algorithm yield different min spanning tree?

Prims Algorithm and Kruskal Algorithm

答:
Yes, because a tiebreaker can be modeled as changing the edge weights very slightly to make all edge weights unique.

Suppose that edges $e_1, e_2, \dots, e_m$ have weights $w_1, w_2, \dots, w_m \in \mathbb R$. Let $\epsilon$ be the smallest difference between two distinct edge weights. Suppose also that we've written edges $e_1, e_2, \dots, e_m$ in the tiebreaking order: whenever we are choosing between edges $e_i$ and $e_j$ with the same weight, we pick $e_i$ if $i<j$ and $e_j$ otherwise.

This decision process is equivalent to the decision process where edge $e_i$ has weight $w'_i = w_i + \frac{i}{2m} \cdot \epsilon$. The weights $w_1', w_2', \dots, w_m'$ are all unique, but comparing $w_i'$ and $w_j'$ is equivalent to comparing $w_i$ and $w_j$ when $w_i \ne w_j$. Only when $w_i = w_j$ does the extra term $\frac{i}{2m} \cdot \epsilon$ come into effect, and using it to compare is equivalent to using the tiebreaker.

So Kruskal with tiebreaks is equivalent to Kruskal with weights $w_1', \dots, w_m'$. This is equivalent to Prim with weights $w_1', \dots, w_m'$, because there is only one MST when weights are unique. That, in turn, is equivalent to Prim with the same tiebreaks.


问:
Divergence of Curl for a vector space of dimension $n$
It is a fact that the divergence of curl, once the conditions of continuity $f (x, y,z)$ are satisfied, is equal to zero. However, all the demonstrations I found focus on three-dimensional space and use a lot of algebra. Is there any way to generalize the result to dimension $n$ more succinctly?

答:
It's not a lot of algebra, really. It's just equality of mixed second-order partial derivatives. To generalize to more dimensions, you need to realize there is no curl, although there is divergence. But the machinery you need is differential forms and the exterior derivative $d$. $d\circ d = 0$ is the general version of $\text{div curl}\,\vec F = 0$ and $\text{curl}\,\nabla f = \vec 0$.

Ultimately, in $\Bbb R^3$ you can identify a vector field with either a $1$-form (and then $d$ becomes curl) or a $2$-form (and then $d$ becomes div). In $n$ dimensions, you similarly have the identification of a vector field with either a $1$-form or an $(n-1)$-form. For $1<k<n-1$, there is no interpretation of a $k$-form as representing a vector field.


问:
Apples and their volumes
An apple has a peel that is 1cm thick and a total diameter of 12cm. What percentage of volume of the apple is the peel?

I tried
$$\frac{\text{volume}(\text{radius of 6})-\text{volume}(\text{radius of 5})}{\text{volume}(\text{radius of 6})}$$
and got 42%. Is this correct?

Title: Apples and their volumes

Body:
$$
\frac{\frac{ 4\pi6^3}{3} - \frac{4\pi 5^3}{3}}{\frac{4\pi 6^3}{3}} = \frac{6^3 - 5^3}{6^3} = \frac{216-125}{216} = \frac{91}{216} \approx 0.42
$$
You are right.


问:
Show that $\frac{1}{\sqrt{1+kx}}\geq 1-\frac{kx}{2}$

Let $k>0$. Show that $$\frac{1}{\sqrt{1+kx}}\geq 1-\frac{kx}{2}$$ for all $x>0$.

In order to prove the requested result I'm trying to prove that $$\frac{1}{\sqrt{1+kx}} -1+\frac{kx}{2}\geq0$$ for positive $x$. However, I'm stuck at proving the last inequality. Thank you in advance

答:
I suggest proving that, given $ k>0 $, and $ x>0 $, we have : $$ \fbox{$\begin{array}{rcl}\displaystyle\frac{1}{\sqrt{1+kx}}-1+\frac{kx}{2}=\frac{3k^{2}x^{2}}{4}\int_{0}^{1}{\frac{1-y}{\left(1+kxy\right)^{2}\sqrt{1+kxy}}\,\mathrm{d}y}\end{array}$} $$

The result is then obvious.


问:
Merging 2 ordered sets preserving order of each set
There are 2 separate event chains.

The first event chain is: [A, B, C]

The second event chain is: [1, 2, 3]

You cannot change the order of events in a chain. They both must be ordered as [1, 2, 3] and [A, B, C].

Now we want to combine 2 event chains while preserving the order of each chain.

Like [A, 1, B, 2, C, 3] or [A, 1, 2, 3, B, C].

As you can see, we may have lots of combinations. My question is, how many combinations are there to have?

PS: I'm not a mathematician, this problem came to my mind while studying evolution. The evolution of species must be ordered, for example, we had a single cell organism first, then multi-cell organisms. Considering there may be multiple species having their own course of evolution, their evolution order does not change but when you think about multiple species, although their order does not change, the order which one came first can be changed. TLDR: this is a real problem I was curious about.

答: Let there be $n$ elements in the first list and $m$ in the second.  The combined list will have $m+n$ elements and is completely determined by the positions of the elements of the first list.  So there are $$\binom{n+m}{n}$$ possibilities.


答:
Your problem is that $3x + 5 = 2k +1$ is not an assumption.  It is the conclusion you need to prove.

Your one and only assumption is that $3x = 2n$ for some integer $n$.

so you start with

$3x = 2n$.

.... then you do a bunch of steps ....

.... steps .....

.... and get in the end ........

Conclusion: $3x + 5 = 2(????????) + 1$ where $??????$ is some integer you come up with in you steps.

Let's see what happens when we try.  Let's take it nice and slow:

=======

$3x = 2n$.

$3x +5 = 2n + 5$

....hmmm,  we want $2(??????) + \color{red}1$ in the end so let's pull out the $+\color{red}1$ first.....

$3x + 5 = 2n+5 = 2n + (4 + \color{red}1)=(2n+4) +\color{red}1$

.... hmmm, okay that's the $+1$ now we want $2(\color{red}{??????}) + 1$.  To get the So we need to factor then $2$ out of $2n+4$ and see what we have left.... that will bee the $\color{red}{??????}$

$3x + 5 = (\color{red}{2n+4}) + 1$

$3x + 5= 2(\color{red}{n + 2}) + 1$

.... and that's it......

Conclusion:  $3x+5 = 2(\color{red}{n + 2})+1$.

The $??????$ we wanted turns out to be $\color{red}{n+2}$ an we have

$3x + 5 = (3x+4) + 1 = (2n+4) + 1  = 2(\color{red}{n+2}) + 1$.

And because $\color{red}{n+2}$ is an integer if we let $k = n+2$ be that integer $3x+5 = 2k + 1$ and so... $3x + 5$ is odd.

=======

Although if you want to work backwards

Conclusion:  $3x + 5 = 2k +1$ ..... and we want to solve for $k$ to show it is possible...

$3x + 5 -1 =2k + 1-1$

$3x +4 = 2k$

$k = \frac {3x + 4}2 = \frac {3x}2 + 2$.

.... but is $\frac {3x}2 + 2$ an integer?????

Well, $3x$ is even.  So there is an integer $n$ so that $3x = 2n$ so

$k = \frac {3x}2 +2 = \frac {2n}2 + 2 = n+2$.

So $k=n+2$ is the integer we want to conclude $3x+5 =2k +1$.

If we did it this way our proof would go:

$3x$ is even so there is an integer $n$ so that $3x = 2n$.  Let $k = n+2$; that is an integer.

$2k + 1 = 2(n+2)+1 = 2n + 5 = 3x + 5$.

So $3x+5 = 2k +1$ and that is odd.


问:
How many options are there for 15 student to divide into 3 equal sized groups?
How many options are there for 15 students to divide into 3 equal-sized groups?

Now I know the solution is $\;\dfrac {15!}{5!5!5!3!}\;$ but I can't understand why.

Can anyone please enlighten me?

答:
Consider the following points:

Suppose that we line up the students in a line and choose the first 5 students to belong to group 1, the next 5 to group 2 and the last 5 to group 3.

There are 15! ways to line up students in a line.

However, in any particular line up, the order of the students in each group does not matter. Thus, for each line up, we have 5! 5! 5! ways of arranging the students in each group. Therefore, the number of ways to arrange students in 3 equal groups is:
$$\frac{15!}{5! 5! 5!}$$

But, the labeling of the groups also does not matter. We can label the group 1 to be group 2 and group 2 to be group 3 and so on. We have 3! ways to label the groups. Thus, the final number of ways to arrange students in 3 equal groups is:

$$\frac{15!}{5! 5! 5! 3!}$$


问:
Stone compactification vs initial/final structure as in Bourbaki
Does the Stone-Cech compactification resemble final structure (such as quotient topology) as described in Bourbaki's Theory of Sets Ch.IV? I feel so since continuous $f:X \rightarrow K$ implies the existence of continuous $F:\beta X \rightarrow K$. Not sure I am on the right track though (probably not, but why?) I am not familiar with category theory so my feeling could be obviously wrong.

答:
Not really, it's functor from $\mathtt{Tych}$ to $\mathtt{CompHaus}$, so a map $f: X \to Y$ between Tychonoff spaces induces a unique $\beta f: \beta X \to \beta Y$ in a functiorial way (preserving compositions and identities). It is also has extension property as in AR (absolute retracts) theory.

And with a final or initial structure you start with a set and some maps, while with $\beta X$ you start with $X$ and construct both a new set and a new topology. There is no map.

Both things being "categorical" doesn't mean they're related.


答:
We really just need to choose which 3 out of the 6 positions contains an element of [A, B, C].

e.g. if we choose 1, 2, and 5 the sequence will be
[A, B, 1, 2, C, 3]

so the total number of sequences is just...
$$ N =\binom 63 = 20$$

Title: Merging 2 ordered sets preserving order of each set

Body:
There are 2 separate event chains.

The first event chain is: [A, B, C]

The second event chain is: [1, 2, 3]

You cannot change the order of events in a chain. They both must be ordered as [1, 2, 3] and [A, B, C].

Now we want to combine 2 event chains while preserving the order of each chain.

Like [A, 1, B, 2, C, 3] or [A, 1, 2, 3, B, C].

As you can see, we may have lots of combinations. My question is, how many combinations are there to have?

PS: I'm not a mathematician, this problem came to my mind while studying evolution. The evolution of species must be ordered, for example, we had a single cell organism first, then multi-cell organisms. Considering there may be multiple species having their own course of evolution, their evolution order does not change but when you think about multiple species, although their order does not change, the order which one came first can be changed. TLDR: this is a real problem I was curious about.


问:
Proof of Kőnig's line coloring theorem ($\chi'(G) = \Delta(G)$)
I'm trying to find a proof of Kőnig's line coloring theorem, i.e.:

The chromatic index of any bipartite graph equals its maximum degree

But to my surprise, I've only* been able to find two questions touching the subject:

Edge-coloring of bipartite graphs

Edge coloring of a bipartite graph with a maximum degree of D requires only D colors

As graphs are my Achilles' heel, I'm incapable to use the information contained in the above to prove $\chi'(G) = \Delta(G)$ myself.

* I've found many papers referring to it, but none proving, except for page 4 of CH6.pdf from first question, but I don't think it's sufficient.

答:
I will try and give an overview of the prerequisite knowledge, and include sources at each step so that you can sequentially understand. If you don't understand certain parts (like the construction at the end) I recommend you work some small examples.

Let's first introduce Hall's Theorem:

Theorem: (Hall's Theorem) Let $G$ be a bipartite graph with parts $A$ and $B$. Then $G$ has a matching (independent edge set) saturating $A$ (every vertex of $A$ is the endpoint of some edge in the matching) if and only if for every $X \subseteq A$ we have $|X| \le |N(X)|$.

The two sources I recommend for a good view of Hall's Theorem is Diestel's Graph Theory (which, if I recall, gives four proofs) and West's Introduction to Graph Theory.

The significance of Hall's Theorem here is that for $k$-regular bipartite graphs, we can find a perfect matching. This comes from two things:

A $k$-regular bipartite graph is balanced.

A $k$-regular bipartite graph satisfies Hall's condition.

So now we can prove the following:

Lemma: If $G$ is a $k$-regular bipartite graph, then $\chi'(G) = k$.

We can use induction on $k$. By Hall's Theorem, $G$ has a perfect matching $M$. Consider $G-M$, which is $k-1$-regular (why?). By the induction hypothesis, $\chi'(G) = k-1$, and so we can add $M$ back as a new color, hence extending a proper $k-1$-edge-coloring from $G-M$ to a proper $k$-edge-coloring on $G$.

If you're not familiar with induction, here's a different description: Removing a perfect matching from a $k$-regular bipartite graph gives a $k-1$-regular graph, which must also have a perfect matching... Iterate this process $k$ times.

Now for the finish line. We want to prove the result for any bipartite graph $G$.

Result: If $G$ is a bipartite graph, then $\chi'(G) = \Delta(G)$.

If $G$ is regular, then we are done by the Lemma. Otherwise, there is at least one vertex $v$ in $G$ with $\deg(v) < \Delta(G)$. We can construct a graph $R$ such that

$R$ is bipartite.

$R$ is $\Delta(G)$-regular.

$G \subseteq R$.

One construction is as follows. We have $G$ bipartite with parts $A$ and $B$. Take a copy of $G$, say $G'$ with parts $A'$ and $B'$. Then for each vertex $v$ not of degree $\Delta(G)$ in $G$, we add an edge between $v$ and it's copy $v' \in G'$. This newly obtained graph is bipartite with parts $A \cup B'$ and $B \cup A'$. Repeat this process as necessary. You'll notice that at each iteration the gap between the minimum degree and maximum degree decreases, so we must terminate with a $\Delta(G)$-regular graph $R$ as desired. You'll find this construction is the one given by Jon Noel's comment here.

Using the Lemma, $\chi'(R) = \Delta(G)$, and thus there is a proper $\Delta(G)$-edge-coloring of $R$. Since $G \subseteq R$, this proper coloring works for $G$. I.e. $\chi'(G) = \Delta(G)$.

Some notes.

Note we used the general fact that $\chi'(H) \le \chi'(G)$ for $H \subseteq G$ at the end.

One thing I did glance over is if we are allowing multiple edges, but things still work that way. If we allow multiple edges, can you see why the way we've constructed $R$ takes exactly $1$ iteration? I don't believe there's any real reason to exclude use of multiple edges.

One key take away is to think of color classes in an edge coloring as what they are: matchings.


问:
Clarify about direct image and inverse image (of morphisms of sheaves)
Suppose to have a continuous function $\alpha:X\to Y$ between topological spaces, and suppose to have two sheaves, $\mathcal F$ defined on $X$ and $\mathcal G$ defined on $Y$. It's well known how are defined the direct image $\alpha_*\mathcal F$, also called pushforward, and the inverse image $\alpha^{-1}\mathcal G$, also called pullback. Clearly, any open set in $Y$ containing $\alpha(x)$ must be of the form $\alpha (U)$ for some open $U\subset X$ containing $x$; in fact, $(\alpha^{-1}\mathcal G)_x$ is equal to $\mathcal G_{\alpha (x)}$.

My problem is that I don't understand why, fixed $x\in X$, in general $(\alpha_*\mathcal F)_x$ is not equal to $(\mathcal F)_x$. I understand that the open sets of the form $\alpha ^{-1}(V)$, with $V\subset Y$ open set containing $\alpha (x)$, are not all the open set of $X$ containing $x$, however the equivalence classes in $(\mathcal F)_x$ seem to me to be all covered by $(\alpha_*\mathcal F)_x$. What I mean is that, for any section $f\in \mathcal F(U)$, with $U\subset X$ an open set containing $x$ that is not preimage of an open of $Y$, one could take $V\subset Y$ open, containing $\alpha (x)$, and consider the restriction of $f$ to $W:=\alpha ^{-1} (V)$. So $f_x=(f|_W)_x\in(\alpha_*\mathcal F)_x$. Where is the  contradiction \ misunderstanding in my reasoning? Thanks in advance.

答:
You write:

Clearly, any open set in $Y$ containing $\alpha(x)$ must be of the form $\alpha(U)$ for some open $U\subset X$ containing $x$.

That's not true. There's no reason why sets of the form $\alpha(U)$ should even be open, and $\alpha(x)$ may have many many open neighborhoods in $Y$ which are not images of open neighborhoods of $x$.

You also write:

For any section $f\in \mathcal F(U)$, with $U\subset X$ an open set containing $x$ that is not preimage of an open of $Y$, one could take $V\subset Y$ open, containing $\alpha (x)$, and consider the restriction of $f$ to $W:=\alpha ^{-1} (V)$.

To do this, we would have to have $\alpha^{-1}(V)\subset U$. Why do you think that such a $V$ must exist?


问:
Prove that $a|p+1$ if and only if ${\frac{a}{p} = \frac{1}{n} + \frac{1}{m} }$ such that $n,m \in N$ and $p$ is prime
Let $p$ be a prime and let $n$ be a natural number such that $n \gt a$ .

Prove that $a|p+1$ if and only if exists integers $n,m$

Such  that    ${\frac{a}{p} = \frac{1}{n} + \frac{1}{m}}$

答:
$a\mid p+1 \iff a\cdot n=p+1\\ \iff\frac{a\cdot n}{p}=1+\frac{1}{p} \\ \iff\frac{a}{p}=\frac{1}{n}+\frac{1}{n\cdot p}$

Let $m:=n\cdot p$ and we are done


答:
One direction:

Say $${a\over p}= {1\over m}+{1\over n}\implies amn = p(m+n)$$

Let $g= \gcd(m,n)$ then exists $x,y$ relatively prime such that $m= gx$ and $n= gy$. Then we have $$axyg = p(x+y)\implies x\mid p(x+y) \implies x\mid py \implies x\mid p $$

Case 1: $x=p$ then $ayg = p+y \implies y\mid p\implies y= 1 \implies a\mid p+1$

Case 2: $x=1$ then $ayg = p(1+y) \implies y=1\; \text {or}\; y=p$

If $y=1$ then $m=n$ and $2p= an$ then we can take $n=p$ and $a=2$ and the statement is not true!

If $y=p$ then we get $a\mid p+1$ and we are done.

Other direction: $a\mid p+1$ then $p+1 =an$ so ${a\over p} = {1\over p} +{1\over pn}$ and we are done.

Title: Prove that $a|p+1$ if and only if ${\frac{a}{p} = \frac{1}{n} + \frac{1}{m} }$ such that $n,m \in N$ and $p$ is prime

Body:
Let $p$ be a prime and let $n$ be a natural number such that $n \gt a$ .

Prove that $a|p+1$ if and only if exists integers $n,m$

Such  that    ${\frac{a}{p} = \frac{1}{n} + \frac{1}{m}}$