问:
Conditions for a regular semigroup to be a group.
This is Exercise 2.6.11 of Howie's "Fundamentals of Semigroup Theory".

The Details:

Let $S$ be a semigroup.

Definition 1: We say $S$ is regular if for all $a\in S$ there exists an $x$ in $S$ such that $a=axa$.

Definition 2: We say $S$ is cancellative if for all $a, b, c\in S$,
$$ac=bc\implies a=b$$ and $$ca=cb\implies a=b.$$

The Question:

Let $S$ be a regular semigroup. Show that the following are equivalent:

(a) $S$ has exactly one idempotent.

(b) $S$ is cancellative.

(c) $S$ is a group.

My Attempt:

Suppose $S$ is regular.

Showing (a) implies (b):

Suppose (a). Suppose $ac=bc$ for $a, b, c\in S$. We have $c=cxc$ for some $x$, so that $cx=cxcx=e$ and similarly $xc=e$. Then $acx=bcx$ implies $ae=be$.

How do I show that $a=b$? The proof  of the left cancellative property should be dual to that of the right.

Showing (b) implies (c):

Suppose (b). Then what?

Showing (c) implies (a):

Suppose (c). Suppose $a^2=a$. Then $a=(aa)a^{-1}=aa^{-1}=1_S$. Thus the only idempotent in $S$ is the identity $1_S$. Thus (a). $\square$

Please help.

答:
A "direct" proof for (b) $\rightarrow$ (c) (using quasigroup + semigroup $\leftrightarrow$ group):

This is to show that $\forall a, b\in S (\exists! x$ s.t. $ax=b)$ for a given semigroup $S$.

It suffices to show that $\forall a, b\in S (\exists x$ s.t. $ax=b)$ since the uniqueness is already given by $S$ being cancellative.

Let $p$ be an inner inverse of $a$: $apa = a$, since the semigroup $S$ is regular, such $p$ always exists. Also note that $ap$ is idempotent.

Now we have $(ap)b = (apap)b = (ap)(apb) \rightarrow b = apb = a(pb)$ so $x=pb$ is a desired solution.


问:
Is this graph ergodic?
I had a long discussion with a friend of mine about if this simple graph is ergodic. I think it is because every state can be reached in an non-endless number of steps. My friend argues that it is not because there are no state transitions from each node to itself. Is that required? If somebody has a few examples for each case, it would perhaps helps me to understand it better --- other than the pure math...

```<code>A <----> B
</code>```

Here the transition matrix:

```<code>     A    B
 A  0.0  1.0
 B  1.0  0.0
</code>```

答:
Ergodic Markov chains are in places explicitly defined as being irreducible and aperiodic, for example in Randall, Dana. "Rapidly mixing Markov chains with applications in computer science and physics." Computing in Science & Engineering 8.2 (2006): 30-41.

Definition 1: A Markov chain is ergodic if it is

irreducible, that is, for all $x, y \in \Omega$, there is a
$t$ such that $P_t(x, y) > 0$, and

aperiodic, that is, for all $x, y \in \Omega$, $\gcd\{t: P_t(x, y)
> 0\} = 1$, where $\gcd$ is the greatest common
divider.

ContentLicense: CC BY-SA 3.0

Title: Is this graph ergodic?

ContentLicense: CC BY-SA 4.0


问:
Symmetric Matrices $B_{n \times n}$ of the form $A^{T}A$ for some matrix $A_{m \times n}$?
Is every symmetric matrix $B_{n \times n}$ of the form $A^{T}A$ for some matrix $A_{m \times n}$?

What I've done so far is:

For any matrix $A_{m \times n}$, the matrix $A^{T}A$ is symmetric because ($A^{T}A)^{T}=A^{T}(A^{T})^{T}=A^{T}A$, where the superscript T denotes the transpose of A. Not sure where to go from here.

答: Hint: for a real matrix $A$, $A^TA$ is always positive semidefinite.</p>


问:
Proving a logic theorem
I am trying to prove this theorem : $((p ⇒ ¬ r) ∧ (s ∧ r ≡ ¬ q) ≢ ¬ ((r ⇒ ¬ s) ≡ q) ⇒ r ∧ p) ≡ p ∧ (p ⇒ r)$, but I am not sure where to start.

$$((p ⇒ ¬ r) ∧ (s ∧ r ≡ ¬ q) ≢ ¬ ((r ⇒ ¬ s) ≡ q) ⇒ r ∧ p)$$
$$=    ⟨ \text{Definition of ≢ with } p, q ≔ (p ⇒ ¬ r) ∧ (s ∧ r ≡ ¬ q), ¬ (r ⇒ ¬ s ≡ q) ⇒ r ∧ p) ⟩$$
$$¬ ((p ⇒ ¬ r) ∧ (s ∧ r ≡ ¬ q) ≡ ¬ ((r ⇒ ¬ s) ≡ q) ⇒ r ∧ p)$$

Can you give me a hint how to continue? I am interested in a proof using logical laws instead a truth table.

答:
Using material implication we can replace $(p \Rightarrow \lnot r)$ by $(\lnot p \lor \lnot r)$. Using the Morgan's law we can then replace it by $\lnot (p\land r)$.

Using material implication we can replace $(r \Rightarrow \lnot s)$ by $(\lnot r \lor \lnot s)$. Using the Morgan's law we can then replace it by $\lnot (r\land s)$. Using commutativity of conjuction, we can then replace it by $\lnot (s\land r)$

We have an expression $(\lnot (s\land r) \equiv q)$. It can be replaced by $((s\land r) \equiv \lnot q)$.

We have an expression $\lnot((s\land r) \equiv \lnot q) \Rightarrow (r \land p)$. Using material implication  it can be replaced by $\lnot\lnot((s\land r) \equiv \lnot q) \lor (r \land p)$. Using double negation it can be replaced by $((s\land r) \equiv \lnot q) \lor (r \land p)$. Using commutativity of conjuction, we can then replace it by $((s\land r) \equiv \lnot q) \lor (p \land r)$.

We have an expression
$$ (\lnot (p \land r) \land ((s\land r) \equiv \lnot q)) ≢ ((s\land r) \equiv \lnot q) \lor (p \land r) $$
which is of the form $ (\lnot x \land y) ≢ (y \lor x)$ where $x = p\land r$, $y = (s\land r) \equiv \lnot q$. By the definition of $≢$, it can be replaced by
$$ \big((\lnot x \land y) \land \lnot(y \lor x)\big) \lor \big(\lnot(\lnot x \land y) \land (y \lor x)\big)$$
Using de Morgan's laws it can be replaced by
$$ \big((\lnot x \land y) \land (\lnot y \land \lnot x)\big) \lor \big((\lnot\lnot x \lor \lnot y) \land (y \lor x)\big)$$

Using assosciativity of conjugation, $(\lnot x \land y) \land (\lnot y \land \lnot x)$ can be replaced by $\lnot x \land (y \land \lnot y) \land \lnot x$, which is a conjuction that includes a alwayys false expression $y \land \lnot y$. Therefore, this whole expression can be replaced by ${\bf 0}$. Using double negation, $\lnot\lnot x$ can be replaced by $x$ so we get
$$ {\bf 0} \lor \big((x \lor \lnot y) \land (y \lor x)\big)$$
which can be replaced by
$$ (x \lor \lnot y) \land (y \lor x)$$
$$ (x \lor \lnot y) \land (x \lor y)$$
$$ x \lor (\lnot y \land y)$$
$$ x \lor {\bf 0}$$
$$ x$$
In this way, the whole expression is now
$$ x \equiv p \land (p \Rightarrow r)$$
that is
$$ p \land r \equiv p \land (p \Rightarrow r)$$
I'll leave proving that to you, it should be simple enough.

Title: Proving a logic theorem


问:
For every measurable set, there is a set in the $\sigma$-algebra with the same outer measure
I am following a measure theory course, and I am currently looking at approximation theorems for the first time. The lecturer starts by saying that given a set $A$ such that

$$
\pi^{*}(A) < \infty \quad A \in \mathcal{M}
$$

there exists a set $F$ such that

$$
F \in \mathcal{F} \quad A \subseteq F \quad \pi^{*}(A) = \pi^{*}(F)
$$

That is, for any measurable set, there is a set containing $A$ in the $\sigma$-algebra which has the same outer-measure as $A$. I cannot work out why this claim holds and don't really know where I should start with a proof.

I will try to define all the notation below:

$\mathcal{F}$ is the $\sigma$-algebra generated by an algebra $\mathcal{A} \subseteq \mathcal{P}(\Omega)$.

$\pi^*(A)$ is defined as follows:

$$
\pi^{*}(A) = \inf \sum_{i \geq 1} \nu(E_{i})
$$

where the infinum is taken over all subsets $\{E_{i}\} \subseteq \mathcal{A}$ which form a covering of $A$, and $\nu$ is a measure defined on the algebra $\mathcal{A}$.

Lastly $\mathcal{M}$, the set of measurable sets, is defined as follows:

$$
A \in \mathcal{M} \iff \forall E \subseteq \Omega, \quad \pi^{*}(E) = \pi^{*}(E \cap A) + \pi^{*}(E \cap A^{c})
$$

答:
By infimum properties, for any $n\in\mathbb N$, there exists a collection of sets $A^*_n=\{A_{nk}\}_{k\in\mathbb N}$ such that $A\subset \bigcup A^*_n$ for all $n$, and $\pi^*(A)+\frac1n>\sum\limits_{k=1}^\infty \nu(A_{nk})$ for all $n$.

Let $B=\bigcap\limits_{n=1}^\infty\bigcup A_n^*$. Clearly, $A\subset B$, and $B$ is in the $\sigma-$algebra. Moreover, if $\pi^*(B)>\pi^*(A)$, then there exists some $m\in\mathbb N$ such that $\frac1m<\pi^*(B)-\pi^*(A)$. But that implies that $$\pi^*(B)\leq \pi^*\left(\bigcup A_m^*\right)<\pi^*(A)+\frac1m<\pi^*(B)$$which is a contradiction. Thus, $\pi^*(B)\leq \pi^*(A)$, and we are done.


问:
Prove that $\Delta ABC$ is an equilateral triangle if and only if the complex coordinates $a$,$b$, and $c$ satisfy a relation.

Let $\triangle ABC$ be a triangle in the complex plane and let $a$,$b$ and $c$, respectively, be the complex coordinates of its vertices.  Suppose that the tiangle is inscribed in the circle $C(0,1)$.  Prove that $\triangle ABC$ is equilateral if and only if $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=0$$ and $$(a+b)(b+c)(c+a)\neq0\,.$$

I tried to convert $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=0$ into a simpler equation by multipling it by $(a+b)(b+c)(c+a)$

In the end I obtained $$a^2+b^2+c^2+3ab+3bc+3ca=0$$ but I don't know how to continue from here.  Can you help me, please?

答:
Given equilateral triangle, i.e.

$$a= e^{i\alpha}, \>\>\>\>\>b = e^{i(\alpha+\frac{2\pi} 3)}, \>\>\>\>\>c= e^{i(\alpha-\frac{2\pi} 3)}
$$

it is straightforward to verify $a^2+b^2+c^2+3ab+3bc+3ca=0$.

Conversely, given

$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=0$$

rewrite the equation in term of $b =a e^{i x}$,  $c =a e^{ i y} $

$$\frac{1}{1+ e^{i x}}+\frac{1}{1+ e^{i y}}+\frac{1}{e^{i x} + e^{i y}}=0\tag1$$

Suppose $(x,y)$ is a solution, then $(y,x)$ is also a solution due to party, as well as $(-x,-y)$ via conjugate. which implies $x = -y$. Substitute into (1) to obtain $\cos x=-\frac12$, or $x= -y= \pm \frac {2\pi}3$. Thus

$$b = a e^{\pm i\frac{2\pi}3}, \>\>\>\>\>c=a  e^{\mp i \frac{2\pi} 3}
$$

hence, equilateral triangles.


问:
Exact differential equation $\frac{y}{(y+x)^2}dx+ ( \frac{1}{y} - \frac{x}{(x+y)^2})dy=0$
I have to solve the differential equation $\dfrac{y}{(y+x)^2}dx+ ( \dfrac{1}{y} - \dfrac{x}{(x+y)^2})dy=0$

Let $ M= \dfrac{y}{(y+x)^2}$ and  $ N=\dfrac{1}{y} - \dfrac{x}{(x+y)^2}$

Since it is an exact differential equation, the solution is a function U(x,y)

However, I started from $U_x(x,y) = M$

When performing the procedure I obtained that the solution $U(x,y) =  - \dfrac{y}{x+y} + ln|y|+C$

On the other hand, if I start from $U_y(x,y) = N$
When performing the procedure I obtained that the solution $U(x,y) =  \dfrac{x}{x+y} + ln|y|+C$

So are these solutions different?

答:
Both solutions are correct. They differ by a constant, which you can see by pulling out $-1$ from the general constant:

$$\frac x{x+y}+C=\frac x{x+y}-1+C=\frac{x-(x+y)}{x+y}+C=-\frac y{x+y}+C$$


问:
Find the polynomial $f(x)$ which have the following property
Find the polynomial $p(x)=x^2+px+q$ for which $\max\{\:|p(x)|\::\:x\in[-1,1]\:\}$ is minimal.

This is the 2nd exercise from a test I gave, and I didn't know how to resolve it. Any good explanations will be appreciated. Thanks!

答:
Here's an informal argument that doesn't use calculus. Notice that $p(x)$ is congruent to $y = x^2$ (for example, simply complete the square). Now suppose that we chose our values for the coefficients $p,q$ carefully, and it resulted in producing the minimal value of $m$. Hence, we can think of the problem instead like this:

By changing the vertex of $y=x^2$, what is the minimal value of $m$ such that for all $x\in [-1,1], -m \le p(x) \le m$?

By symmetry, there are only two cases to consider (based on the location of the vertex).

Case 1: Suppose the vertex is at $(1,-m)$ and that the parabola extends to the top left and passes through the point $(-1,m)$. Using vertex form, we have $p(x)=(x-1)^2-m$ and plugging in the second point yields $m=(-1-1)^2-m \iff 2m=4 \iff m = 2$.

Case 2: Suppose that the vertex is at $(0, -m)$ and that the parabola extends to the top left and passes through the point $(-1,m)$. Using vertex form, we have $p(x)=x^2-m$ and plugging in the second point yields $m=(-1)^2-m \iff 2m = 1 \iff m = 1/2$.

Since $m=1/2$ is smaller, we conclude that $\boxed{p(x)=x^2-\dfrac{1}{2}}$ is the desired polynomial.


问:
$\mathbb{Z}[\sqrt{2}]/(3-\sqrt{2}) \simeq \mathbb{Z}_n$
I am trying to find integer $n$ such that $\mathbb{Z}[\sqrt{2}]/(3-\sqrt{2}) \simeq \mathbb{Z}_n$.

My first guess is using "the 1st ring isomorphism theorem"

Let $\phi : \mathbb{Z}[\sqrt{2}] \rightarrow \mathbb{Z}_n$. I want to construct homomorphism such that whose kernel is $3-\sqrt{2}$

I tried to think of Norm map, i.e., $N(x) = a^2 - 2b^2$ for $x =a+b\sqrt{2} \in \mathbb{Z}[\sqrt{2}]$ but this was not good.

Any ideas?

答:
The answer is $7$. Maybe the easiest way to think about it is to map
$$
\mathbb{Z}[x] \to \mathbb{Z}_7
$$
by sending $x$ to $3$. Then you can observe that $x^2 - 2$ is in the kernel of this map, so there is a well-defined map
$$
\mathbb{Z}[\sqrt{2}] \to \mathbb{Z}_7
$$
which sends $\sqrt{2}$ to $3$. It's clear that $3-\sqrt{2}$ is in the kernel of that map, which is what you want.


答: We rewrite $\Bbb Z[\sqrt 2]$ as $\Bbb Z[T]/(T^2 - 2)$ via the isomorphism sending $T$ to $\sqrt 2$. Then we have: $$\Bbb Z[\sqrt 2]/(3 - \sqrt 2) \simeq \Bbb Z[T]/(T^2 - 2, 3 - T) \simeq \Bbb Z[T]/(7, 3 - T) \simeq \Bbb Z/7.$$

Body:
I am trying to find integer $n$ such that $\mathbb{Z}[\sqrt{2}]/(3-\sqrt{2}) \simeq \mathbb{Z}_n$.

My first guess is using "the 1st ring isomorphism theorem"

Let $\phi : \mathbb{Z}[\sqrt{2}] \rightarrow \mathbb{Z}_n$. I want to construct homomorphism such that whose kernel is $3-\sqrt{2}$

I tried to think of Norm map, i.e., $N(x) = a^2 - 2b^2$ for $x =a+b\sqrt{2} \in \mathbb{Z}[\sqrt{2}]$ but this was not good.

Any ideas?


问:
Poker calculations on winrate
Just to learn I'm doing some calculations on my poker results.
I'm trying to figure out the probability of given winrate.

The available statistics I got in my database are:

Hands: $30 000$

Winrate: $2$ bets per $100$ hands

Expected winrate (what our winrate would be if we had gained all of our EV) : $8$ bets per $100$ hands

StdDev: $8$ bets per $100$ hands

What's the probability of our winrate per $100$ hands being less than $2$ given that our expected winrate is $8$?

I'm thinking like this:

Central limit theorem can be applied and therefore normalized

Use this formula: $t = \dfrac{2-8}{8/(3k^{0.5})}$

And then look up $t$ in a $z$-table? ($t$ will be very close to $z$ because of the large sample).

Am I doing this right?

答:
Looks fine, except that I don't know what $k$ is.

Standard error $=\frac{0.08}{\sqrt{30000}}$,  $z=\frac{0.08-0.02}{\text{standard error}}$

Title: Poker calculations on winrate


问:
Finding the tangent line of a level curve at $(1,1)$
I'm trying to find the point on a level curve through point $(1,1)$ through which the tangent line is vertical given the function $f(x,y) = 3x^2y^2+2x^2-3x+2y^2$

So I was thinking about it, and I thought that this should happen when the slope is approaching infinity.
I decided to calculate the partial derivative $f_y(x,y)$ which is $f_y(x,y) = 4xy^2+4y$ and set it to $0$, to which the only real solution is $y = 0$.
Yet, I'm not really sure what else I need to do in order to find the $x$ value which is supposed to be $\frac{3}{4} + \frac{\sqrt{41}}{4}$.

Would appreciate some help, thanks in advance!

答:
$f(1,1)=3+2-3+2=4$, so you're looking for where the tangent line to the curve

$$3x^2y^2+2x^2-3x+2y^2=4$$

is vertical. Differentiate both sides implicitly and solve for $\frac{\mathrm dy}{\mathrm dx}$:

$$3\left(2xy^2+2x^2y\frac{\mathrm dy}{\mathrm dx}\right)+4x-3+4y\frac{\mathrm dy}{\mathrm dx}=0\\
\frac{\mathrm dy}{\mathrm dx}=\frac{3-4x-6xy^2}{(6x^2+4)y}$$

The derivative is undefined for $y=0$, at which point

$$2x^2-3x=4$$

which is easily solved.


问:
Find the minimum value of $(\tan C – \sin A)^2 + (\cot C – \cos B)^2$ for the following given data

Let $A, B, C$ be real numbers such that

(i) $(\sin A, \cos B)$ lies on a unit circle centered at origin.

(ii) $\tan C$ and $\cot C$ are defined.

Find the minimum value of  $(\tan C – \sin A)^2 + (\cot C – \cos B)^2$.

My multiple attempts are as follows:-

Attempt $1$:

$$\sin^2A+\cos^2B=1$$
$$\tan^2C+\sin^2A-2\sin A\tan C+\cot^2C+\cos^2 B-2\cot C\cos B$$
$$(\tan^2C+\cot^2C)+1-2\left(\dfrac{\sin A\sin C}{\cos C}+\dfrac{\cos C\cos B}{\sin C}\right)$$

$$(\tan^2C+\cot^2C)+1-2\left(\dfrac{\sin A\sin^2 C+\cos^2 C\cos B}{\sin C\cos C}\right)$$

$$(\tan^2C+\cot^2C)+1-2\left(\dfrac{\sin^2C(\sin A-\cos B)+\cos B}{\sin C\cos C}\right)\tag{1}$$

Now from here how to proceed further.

Attempt $2$:

$$\sin^2A+\cos^2B=1$$
$$\sin^2A=\sin^2B$$
$$A=n\pi\pm B$$

Considering only the principal range, $A=B$, $A=-B$, $A=n\pi-B$, $A=n\pi+B$

Case $1$: $A=B,A=-B$

Put $B=A$ or $B=-A$ in equation $(1)$

$$(\tan^2C+\cot^2C)+1-2\left(\dfrac{\sin A\sin^2 C+\cos^2 C\cos A}{\sin C\cos C}\right)$$

$$(\tan^2C+\cot^2C)+1-2\sqrt{\sin^4C+\cos^4C}\cdot\dfrac{\sin(A+\alpha)}{\sin C\cos C}$$
$$(\tan^2C+\cot^2C)+1-2\sqrt{\tan^2C+\cot^2C}\cdot \sin(A+\alpha)$$

So minimum value will be $3-2\sqrt{2}$

Case $1$: $A=n\pi-B,A=n\pi+B$

Put $B=n\pi-A$ or $B=A-n\pi$

$$(\tan^2C+\cot^2C)+1-2\left(\dfrac{\sin A\sin^2 C-\cos^2 C\cos A}{\sin C\cos C}\right)$$

$$(\tan^2C+\cot^2C)+1-2\sqrt{\sin^4C+\cos^4C}\cdot\dfrac{\sin(A-\alpha)}{\sin C\cos C}$$

$$(\tan^2C+\cot^2C)+1-2\sqrt{\tan^2C+\cot^2C}\cdot\sin(A-\alpha)$$

So minimum value will be $3-2\sqrt{2}$

Any other way to solve this question?

答:
If $x=\tan C$, and we have $\cos^2 A=\cos ^2B$, we need to minimize $$(x-\sin A)^2+\left(\dfrac{1}{x}-\cos A\right)^2=x^2+\dfrac{1}{x^2}+1-2\left(x\sin A+\dfrac{\cos A}{x}\right)$$

So, we need to maximize $\left(x\sin A+\dfrac{\cos A}{x}\right)$.

It is known that maximum of $\alpha \sin \theta+\beta \cos \theta$ is $\sqrt{\alpha^2+\beta^2}$ , so it's maximum is $\sqrt{x^2+\dfrac{1}{x^2}}$, so our expression becomes $$x^2+\dfrac{1}{x^2}-2\sqrt{x^2+\dfrac{1}{x^2}}+1=\left(\sqrt{x^2+\dfrac{1}{x^2}}-1\right)^2$$

And clearly by A.M-G.M inequality, $x^2+\dfrac{1}{x^2}\geq 2$, so the minimum is $(\sqrt{2}-1)^2$.

Equality occurs at $x=1=\tan C$.


问:
Solve the system of equations: $32y+32x^3=6x+17$, $16z+32y^3=6y+9$, $8x+32z^3=6z+5$ where $x,y,z\in \mathbb{R}$
Solve the system of equations: $$\begin{cases} 32y+32x^3=6x+17 \\16z+32y^3=6y+9 \\8x+32z^3=6z+5 \end{cases}$$ where $x,y,z\in \mathbb{R}$ (Bulgaria 1960)

I attempted to solve this question as follows:

$32y+32x^3=6x+17$

$y=-x^3+\frac{6x}{32}+\frac{17}{32}$

$y-\frac{1}{2}=-x^3+\frac{3x}{16}+\frac{17}{32}-\frac{1}{2}$

Here it started getting very complex, and hence I don't think it can be solved this way. After this I tried doing something similar with the other two equations, but once again it was ending up way too complex. It is obvious that the solution is $x=y=z=\frac{1}{2}$, but I can't manage to prove it. Could you please explain to me how to solve this question?

答:
Proof by Contradiction:

Let,  $32x^3-6x>1$  then:

$$\begin{align} 32x^3-6x>1 &\Longrightarrow x>\frac 12\\ 32x^3-6x=17-32y>1 &\Longrightarrow y<\frac 12 \\32y^3-6y=9-16z <1 &\Longrightarrow z>\frac 13\\ 2z^3-6z=5-8x >1 &\Longrightarrow x<\frac 12\end{align}$$ which gives a contradiction.

The same method also works for $32x^3-6x<1$. We obtain again a contradiction.

So, we deduce that $$\begin{align} 17-32y=32x^3-6x=1
&\Longrightarrow y=\frac 12 \\ 9-16z=32y^3-6y
&\Longrightarrow z=\frac 12\end{align}$$

$$\begin{align} x=\frac {6z+5-32z^3}{8}=\frac{4}{8}=\frac 12\end{align}$$

Finally, our real solutions are only $$x=y=z=\dfrac 12.$$

I used :

If $32x^3-6x>1$, then $x>\frac 12$. Because,

$$\begin{align}32x^3-6x-1&=32\left(x+\frac 14\right)^2\left(x-\frac 12\right)>0.&\end{align}$$

If $y<\frac 12$, then $32y^3-6y<1$. Because,

$$\begin{align}32y^3-6y-1&=32\left(y+\frac 14\right)^2\left(y-\frac 12\right)<0.&\end{align}$$

If $z>\frac 12$, then $32z^3-6z>1$. Because,

$$\begin{align}32z^3-6z-1&=32\left(z+\frac 14\right)^2\left(z-\frac 12\right)>0.&\end{align}$$


问:
Is $\{a+b\sqrt 2+c\sqrt 3\mid a,b,c\in \mathbb Q\}$ a field?
Set $$K=\{a+b\sqrt 2+c\sqrt 3\mid a,b,c\in \mathbb Q\}.$$
Is $K$ a field ? I think it is, and I tried to prove that every element are invertible by searching the inverse : Let $a+b\sqrt 2+c\sqrt 3$.
$$(a+b\sqrt 2+c\sqrt 3)(\alpha +\beta \sqrt 2+\gamma \sqrt 3)=1$$
$$\iff \begin{cases}a\alpha +b\beta +c\gamma =1\\a\beta +\alpha b=0\\a\gamma +\alpha c=0\\ b\gamma +c\beta =0\end{cases}$$
and then, we solve this system, and everything should works. Nevertheless, I was wondering if there were an easier method. Any idea ?

答:
Hint

The fact that you have a system with 3 variable and 4 constraint rather lok that there are no solution. Anyways :

If $K$ is a field, then it has degree at most $3$. In fact it's exactly 3 here : why ?

Notice that $\sqrt 2$ and $\sqrt 3$ belong to $K$. Therefore, if it's a field, $\mathbb Q(\sqrt 2,\sqrt 3)\subset K$.

I let you conclude.


问:
I would like to know an intuitive way to understand a Cauchy sequence and the Cauchy criterion.
My understanding from the definition in my book (Rudin) is this.

A seq. $\{p_n\}$ in a metric space $X$ (I only really know $\mathbb R^k$) is said to be a Cauchy sequence if for any given $\epsilon > 0$, $\exists N\in \mathbb N$ such that $\forall n,m\ge N$, $d(p_n,p_m)<\epsilon$.

(1) I see it as, given any tiny value $\epsilon$, we can find a natural number $N$ large enough so that the distance between $p_n$ and $p_m$ is less than $\epsilon$. Am I right ?

The reason I'm asking this is because I was trying to understand the proof of how
$$\sum a_nb_n$$
can converge, and the book said this

$$\left\lvert \sum_{n=p}^{q}a_nb_n\right\rvert \leq \epsilon$$

satisfies the Cauchy criterion and therefore it converges.

I read other questions and answers about the Cauchy sequence, but it didn't really help me…

Can someone explain me what's going on?

Edit:

Suppose

a) the partial sums of $A_n = \Sigma a_n$ form a bounded sequence

b) $b_0 \geq b_1 \geq \dotsb$

c) $\lim_{b \to \infty} b_n = 0$

Using the partial summation formula, algebraically the equation in the bottom is proved

$$\left\lvert \sum_{n=p}^{q}a_nb_n \right\rvert \leq \epsilon$$

Algebraically I had no problem, but I don't know why this proves convergence. I thought to show that a sequence is Cauchy, we need to find the distance between two terms in a sequence. That's where I'm confused.

答:
Since you asked specifically how to understand Cauchy sequences "intuitively" (rather than how to do $\epsilon,\delta$ proofs with them), I would say that the best way to understand them is as Cauchy himself might have understood them.  Namely, for all infinite indices $n$ and $m$, the difference $p_n-p_m$ is infinitesmal.  Such formalisations exist, for instance, in the context of the hyperreal extension of the field of real numbers.

As far as the particular series you asked about, what is going on is that the book is considering the sequence $p_n$ of partial sums of the series, and applying the Cauchy criterion to this sequence.  Then the difference $p_n-p_m$ is the expression $\sum_m^n$ that you wrote down (up to a slight shift in index).

Some thoughts on Cauchy can be found here.


问:
locally Lipschitz in the second variable
Let $E$ be a normed vector space(or Banach space), $\Omega$ an open set of $\mathbb{R}\times E$ and $f$ a function from $\Omega$ to $E$. We say that $f$ is Lipschitz in the second variable on a subset $W$ of $\Omega$ if there is a constant $k$ such that :

$$(t, x_1), (t, x_2) \in W \Rightarrow ||f(t, x_1) - f(t, x_2)||\leq k||x_1 - x_2||$$

We say that $f$ is locally Lipschitz in the second variable if all point of $\Omega$ has a neighborhood on which $f$ is Lipschitz in the second variable.

Now we assume that $f$ is locally Lipschitz in the second variable, $K$ is a compact set include in $\Omega$. Then $f$ is Lipschitz in the second variable on a neighborhood of $K$.

Can someone help？ We may assume that $\Omega$ is connected if it's needed.

答: HINT: Use the fact that $K$ can be cover by a finite number of neighborhood on which $f$ is Lipschitz in the second variable.


问:
Definition of Clifford Algebras
I'm currently studying algebras, and in particular Clifford Algebras. Primarily I've been looking at this paper, and on page 7 and 8 the definition of a Clifford Algebra is given. I have a broad understanding of tensor algebras, but I'm struggling to understand the definition of Clifford algebra by the quotient algebra $T(V)$ with the ideal $I_q$ where $$I_q = < v \otimes v + q(x)1_{T(V)}>$$ In particular, it isn't clear to me why $I_q$ is a proper ideal, or how to interpret its elements, and further the elements of $T(V)/I_q$. I know this might be a lot of questions all in one, but any insights would be greatly appreciated!

答:
To try to correct the notation a little, it is usually defined by $$I_q = \langle \{v \otimes v - q(v)1_{T(V)}\mid v\in V\}\rangle$$

Almost every resource on Clifford algebras demonstrates that if you select a basis $e_1, e_2,\ldots e_n$ for $V$, the collection of products of the form $e_{i_1}\cdot e_{i_2}\cdots e_{i_k}$ where $k\geq 0$ and $i_1<i_2<\ldots < i_k\leq n$ (including the empty one that produces $1$ and the ones that are just single elements of the basis) is a basis for the Clifford algebra.  Within that argument is the evidence that the ideal is proper.

This also gives a concrete representation of the elements, as linear combinations of what they call "blades."  It is reminiscent of the tensor algebra itself, except that now it's finite dimensional, of dimension $2^n$.

About interpretation in practice: for one thing, there are multiple interpretations of elements of the Clifford algebra. Some are used to represent what is acted on, some are used to represent what is acting. Depending on the context, one may be using an ordinary linear model, a homogenous coordinate model, or a "conformal model" or something else.  The second thing is that I don't think all elements have interpretations.

Understanding the various ways to use and interpret the algebra elements is a somewhat extensive topic, but fortunately a lot of people have a lot of good texts that attempt to explain the schemes.   I recommend starting with the wiki and then moving on to the numerous open source resources available online.

Title: Definition of Clifford Algebras


问:
Fourier series for $f(x)=(\pi -x)/2$
I need to find the Fourier series for $$f(x)=\frac{\pi -x}{2}, 0<x<2\pi$$

Since the interval isn't symmetric over $0$, I guess I need to consider $f$'s periodic extension to $\mathbb R$. let's call it $g$. Then $$g(x)=\begin{cases}\frac{\pi -x}{2}, \text{ if } 0<x<\pi\\  \frac{-\pi -x}{2}, \text{ if }{-\pi <x<0}\end{cases}$$

Due to $g$ being odd, the fourier coefficients $a_n$ are all $0$.

And $$b_n=\frac{2}{\pi}\int _0^\pi g(x)\sin(nx)dx=\frac{1}{\pi}\int _0^\pi (x-\pi)\sin (nx)dx$$
According to wolfram alpha $b_n=1/n$ so the Fourier series is
$$\sum _{n=1}^{+\infty}\frac{\sin(kx)}{n}$$

But when I check the result of $x=1/2$ I get this which seems to diverge. It also fails for $x=\pi /2$ and a few other $x$'s.

What am I doing wrong?

I just saw this question. it looks like my answer is correct. am I doing anything wrong in my verification?

EDIT: I realised I forgot to divide by $\pi$. I fixed that but it still doesn't work. I also updated the links.

答:
First of all, you should not extend $f$ if you want the result to have period $2\pi$, which is equal to the domain on which the function definition is given. There is no reason why you would want $f$ to be symmetric over $0$.

What you did to obtain $g$ was not extension. An extension should make the domain bigger while staying unchanged in the given domain.

Edit: I realize there's nothing seriously wrong with the computation. It was the terminology that confused me. The computation just had a minor error, and that was pointed out by Matt in the other answer.

Anyway, integrals in the formulas for coefficients of Fourier series do not need to be symmetric. The domain of integration can start anywhere, as long as it covers one period.

Title: Fourier series for $f(x)=(\pi -x)/2$


问:
Slick argument to find free group in $SO(4)$
It is well known that ${\rm SO}(3)$ contains the free group ${\rm F}_2$, so clearly ${\rm SO}(n)$ for $n>3$ does, too. However, the standard proof with ${\rm SO}(3)$  is somewhat annoying (the one sketched here). I've been able to prove in a nice way that if ${\rm SO}(4)$ contains ${\rm F}_2$, then so does ${\rm SO}(3)$, using ${\rm Spin}(4) \cong {\rm Spin}(3) \times {\rm Spin}(3)$ and a lemma about free groups.

It seems there may be a slicker argument that ${\rm SO}(4)$ contains ${\rm F}_2$ compared to ${\rm SO}(3)$, as we have more room to work, in a sense. Is there a nice way to obtain this?

答:
There is a standard argument which works for all algebraic groups - a relation is an algebraic equation, so a generic pair of elements in any algebraic group will generate a free group.

See

Epstein, D. B. A., Almost all subgroups of a Lie group are free, J. Algebra 19, 261-262 (1971). ZBL0222.22012.


问:
Am I wrong in understanding random variables in this way
I am learning stochastic process and is confused about the definition of random variable.
Here is what I understand .
We know that the three basic elements of a probability space is $(\Omega, \mathscr{F}, P)$, and the probability function is defined in $\mathscr{F}$, but the random variable is defined in $\Omega$.
And I understand in this way, for example.
Suppose that the $\Omega = \{1, 2, 3, 4, 5, 6\}$ and the $\mathscr{F}$ is $\{\{1, 2, 3\}, \{4, 5 ,6\},
\Omega, \varnothing\}$. I can define the probability as $P(\{1, 2, 3\}) = P(\{4, 5 ,6\}) = 0.5$, but I can not define the random variable in $\Omega$, because there is no element in $\Omega$ that satisfies $\{\omega : X(w) \leq x \} \in \mathscr{F}$.

答:
A random variable is a function $X: \Omega \to \mathbb{R}$ which is measurable with respect to the measures $\mathcal{F}$ and $\mathcal{B}_{\mathbb{R}}$. This is equivalent to $X$ verifying that for each $x\in \mathbb{R}$, it holds that $X((-\infty , x]) = \{\omega \in \Omega : X(\omega) \leq x\} \in \mathcal{F}$. Note that this definition does not involve your probability function $P$. If you find an $x \in \mathbb{R}$ such that no $\omega$ in $\Omega$ satisfies $X(\omega) \leq x$, then $\{\omega \in \Omega : X(\omega) \leq x\} = \varnothing$. But $\varnothing \in \mathcal{F}$, so finding such an $x$ does not prove that $X$ fails to be a random variable over $(\Omega, \mathcal{F})$.

For an example of a random variable on the sigma algebra that you defined, consider $X$ such that $X(\omega)= 7$ if $\omega \in \{1,2,3\}$ and $X(\omega)= 15$ if $\omega \in \{4,5,6\}$. Then we have that:

$$\{\omega \in \Omega : X(\omega) \leq x\}=\left\{
\begin{array}{c}
\varnothing \quad \text{ if }(x) < 7 \\
\{1,2,3\} \quad \text{ if } 7 \leq x < 15 \\
\Omega \quad \text{ if }  x \geq 15 \\
\end{array}
\right.$$

So the set $\{\omega \in \Omega : X(\omega) \leq x\}$ is measurable for each $x\in \mathbb{R}$ and thus $X$ is measurable.


问:
Need help in showing that $z\mapsto z^k$ is a surjective homomorphism.

Let $$G=\{z\in \Bbb{C}\mid z^n=1 \quad\text{for some }n \in \Bbb{Z}^{+}\}.$$ Show that for any fixed integer $k>1$, the map $z\mapsto z^k$ from $G$ to itself is a surjective homomorphism but not an isomorphism.

Proving that the map is a homomorphism is easy - for any $z_1,z_2 \in G$ we have $\varphi(z_1z_2)=(z_1z_2)^k=z_1^kz_2^k=\varphi(z_1)\varphi(z_2)$. Proving that it is not an isomorphism is also straightforward - since $\operatorname{ker}\varphi=\{z\in G \mid z^k=1\}$ which is not $\{1\}$, we can conclude that the mapping is not an isomorphism.

I just need some hint to show that the mapping is surjective. Thanks in advance!

答:
Every element $z$ in $G$ can be written in the follow form
$$z={\rm exp}\left(\frac{2\pi i a}{n}\right)$$
for some $a,n \in \mathbb N, n\neq 0$, so a preimage is
$${\rm exp}\left(\frac{2\pi i a}{nk}\right).$$


问:
Definition of Lie Algebra
I have just start studying Lie Algebra. I want to know the motivation for the conditions on the Lie Bracket in the Definition of Lie Algebra. Please explain me or tell me some references. Thanks. </p>

答:
$\mathfrak{gl}(n,\mathbb{R})$ are the most natural examples of Lie algebras. However I never feel they explains why we should study Lie algebras with a $multiplication$ rule as strange as the  Lie bracket (it is not commutative nor associative). I only understand this after reading Barry Simon's Representations of finite and compact groups.

This book has a different flavour from most books on Lie algebras in that it is written from an analyst's perspective. As a result, everything is done on $\mathbb{C}$ or $\mathbb{R}$ or $\mathbb{H}$ (so no fields with positive characteristics), and most proof are computational (not conceptual). Since a lot of people doing Lie theory nowadays are trying to apply the theory to number theory or algebraic geometry, Barry left out some important themes, but his book gives a very nice motivation for the study of Lie theory. After all, it is for the study of differential equations that Lie himself invented this theory.

Now come back to your question. Let $M$ be a smooth manifold and $X$ a vector field on $M$. The existence of differential equations says that

For each $p\in M$ we have $\delta_p>0$, and a smooth function \begin{equation}
\phi_p:(-\delta_p,\delta_p)\to M\end{equation} such that $\phi_p(0)=p$ and $\frac{d}{dt}\phi(t)=X(\phi_p(t))$.

That is, given a smooth vector field on a smooth manifold, then at each point we have a local flow that has this vector field as its velocity.

Let's just assume $M$ is really nice and we can take $\delta_p=\infty$ for all $p$ (each flow can be extended forever). Then for fixed $t$ this flow defines the exponential map $\exp(tX):M\to M$ that maps \begin{equation}
p\mapsto \phi_p(t).
\end{equation}  For each $p\in M$, this function tells you where you are after travelling along $X$ after $t$ seconds.

This map on $M$ induces a map $\exp(tX)_*$ on the tangent bundle. For each tangent vector $v_p$ at $p$ you get an tangent vector at $\exp(tX)(p)$ by pushing forward $v_p$ along $X$. Then a general fact from differential geometry says

\begin{equation}
\frac{d}{dt}|_{t=0}\{\exp(tX)_{*}Y\}=X\circ Y-Y\circ X
\end{equation}  where $Y$ is another vector field and $\circ$ denotes teh composition of two vector fields (vector fields acting on smooth functions by differentiation). (For this fact and the push-forward, see the great answer here).

So you see, you have the Lie bracket here. Since the left-hand side is the derivative of $Y$ along $X$, you see the Lie bracket does arise naturally (at least as natural as derivatives).

Now let $M$ have some more structures, say, it is a Lie group, which is just a group and meanwhile a manifold. Then the exponential map has much better structures. For instance, if you look at a very special type of vector fields, the translation-invariant vector fields (just constant vector fields if you are looking at Euclidean spaces, and this vector space $N$ is the Lie algebra we are studying), then the exponential map is a diffeomorphism around the unity of the group and the origin of $N$.

Well, to conclude, my guess is: in the first place we want to study Lie groups (like $\mathbb{R}^2$ or $\mathbb{S}^1$), then we found a very good linear object, invariant vector fields/ Lie algebras, that pretty much characterizes the structure of Lie groups via the exponential map. And the Lie bracket happens to be the most natural operation on this linear structure (it is just taking derivative of a vector field along another).

Title: Definition of Lie Algebra


问:
Infinite set is dense set in cofinite topology
Define $$\tau = \{U \subset X; U^c \ \text{is finite or} \ U=\emptyset\}.$$
Let $X$ be an infinite set and  $Y \subset X$ be an infinite subset of $X$. Prove that $Y$ is dense in the cofinite topology and that $(X,\tau)$ is separable.

My attempt:

Let $x \in X$ and $O$ be an open set that contains $x$. Using the definition we have that $X\setminus O$ is finite. If $Y \cap X=\emptyset$, then $Y \subset X\setminus O$ which is absurd. So $Y$ intersects $O$ and, as $x$ is a neighbourhood of $O$, $x \in \overline{Y}$ and we have $\overline {Y}=X$.

Is this enought or we have some mistake?

For the second part, i don't know to prove that $(X,\tau)$ is separable, any hint?

答:
It is enough (I would change the phrasing "is absurd" - but I am not a native english speaker)

"For the second part, i don't know to prove that $(X,\tau)$ is separable", you just did prove it - think about the definition of separable space.


问:
rank of $E_i$ in the sum $A=\lambda_1E_1+\dots+\lambda_kE_k$ where $\lambda_i$ are all the eigenvalues of A.
I'm trying to solve the following exersice.

Let $A$ be a $\nu\times\nu$ matrix with elements over a field $F$ and let $\chi_A(x)=(-1)^\nu(\lambda_1-x)^{\sigma_1}\dots(\lambda_k-x)^{\sigma_k}$ be the characteristic polynomial of $A$, with $\lambda_i$ distinct from each other. Now if there exist unique $E_1,E_2,\dots,E_k$ $\nu\times\nu$ matrices such that:

$A=\lambda_1E_1+\dots+\lambda_kE_k$.

$E_i^2=E_i$ for $1\leq i\leq k$.

$E_iE_j=0$ if $1\leq i,j\leq k$, $i\neq j$.

$E_1+E_2+\dots+E_k=I_\nu$.

I want to prove that $rank(E_i)=\sigma_i$ for $1\leq i\leq k$.

I have started the solution like this. We know from a theorem that the matrix $A$ is diagonalizable. We know that the dimension of each of the $k$ eigenspaces of $A$ is $\sigma_i$, i.e. $dimV(\lambda_i)=\sigma_i$. Next, we observe that,

$$A E_i=\lambda_iE_i$$

for every $1\leq i\leq k$ and if we denote by $E_i^{(\mu)}$ the $\nu$ columns of $E_i$ it is easy to show that each, non-zero, column $E_i^{(\mu)}$ is an eigenvector of $A$ with eigenvalue $\lambda_i$. So, we just need to prove that the non-zero columns of $E_i$ are linearly independent, and the proof will be finished.

At the moment I can't show that. Any help?

答:
The equation $AE_i = \lambda_i E_i$ shows that any (non-zero) vector in the column space of $E_i$ is an eigenvector of $A$ associated to the eigenvalue $\lambda_i$. Hence $\operatorname{rank}(E_i) \leq \sigma_i$ for all $1 \leq i \leq k$. Assume that $\operatorname{rank}(E_j) < \sigma_j$ for some $1 \leq j \leq k$. Then

$$ n = \operatorname{rank}(I) = \operatorname{rank}(E_1 + \dots + E_k) \leq \sum_{i=1}^n \operatorname{rank} (E_i) < \sum_{i=1}^n \sigma_i = n $$

and we obtain a contradiction.


答:
Your derivation of $ y = 0 $ from $f_y = 0 $ is correct. Having found $ y $, you need to find the corresponding value of $x$ on the line curve.

The equation for the line curve at $(1,1)$ is $f(x,y) = f(1,1)$, i.e. the set of all points where $f$ has the same value as at $(1,1)$.
By substituting $x=1, y=1$, you have $f(1,1) = 4$, so the line curve has this equation: $$3x^2y^2+2x^2-3x+2y^2=4$$
Simply substitute $y=0$ in this curve equation to find $x$.

Title: Finding the tangent line of a level curve at $(1,1)$

Body:
I'm trying to find the point on a level curve through point $(1,1)$ through which the tangent line is vertical given the function $f(x,y) = 3x^2y^2+2x^2-3x+2y^2$

So I was thinking about it, and I thought that this should happen when the slope is approaching infinity.
I decided to calculate the partial derivative $f_y(x,y)$ which is $f_y(x,y) = 4xy^2+4y$ and set it to $0$, to which the only real solution is $y = 0$.
Yet, I'm not really sure what else I need to do in order to find the $x$ value which is supposed to be $\frac{3}{4} + \frac{\sqrt{41}}{4}$.

Would appreciate some help, thanks in advance!


答:
In this answer, I explain why adjoining an algebraic number to $\mathbb{Q}$ generates a field. In the same way you can conclude that $\mathbb{Q}(\sqrt{2}, \sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$ is a field. Note that here elements are linear combinations of the form $a+b\sqrt2+c\sqrt3+d\sqrt6,$ where $a, b, c, d\in\mathbb{Q}.$

In your case $K$ is just the $\mathbb{Q}$-vector space spanned by $\{1, \sqrt{2}, \sqrt{3}\}$ and not yet closed under multiplication. Once you take the multiplicative closure of $K$ by adjoining $\sqrt{6},$ you recover the above field.

Title: Is $\{a+b\sqrt 2+c\sqrt 3\mid a,b,c\in \mathbb Q\}$ a field?

Body:
Set $$K=\{a+b\sqrt 2+c\sqrt 3\mid a,b,c\in \mathbb Q\}.$$
Is $K$ a field ? I think it is, and I tried to prove that every element are invertible by searching the inverse : Let $a+b\sqrt 2+c\sqrt 3$.
$$(a+b\sqrt 2+c\sqrt 3)(\alpha +\beta \sqrt 2+\gamma \sqrt 3)=1$$
$$\iff \begin{cases}a\alpha +b\beta +c\gamma =1\\a\beta +\alpha b=0\\a\gamma +\alpha c=0\\ b\gamma +c\beta =0\end{cases}$$
and then, we solve this system, and everything should works. Nevertheless, I was wondering if there were an easier method. Any idea ?


问:
Subspace of $\mathbb{R}^2$
Give an example of a nonempty set $U$ of $\mathbb{R}^2$ such that $U$ is closed under addition for every $u$ in $U$, $-u$ is also in $U$ but U is not a subspace of $\mathbb{R}^2$

What I understand from the question is I need to give an example such that the set is closed under addition but not under scalar multiplication.

For this I thought of some sets like $\{(x,y)|x\geq0, y\geq 0\}$ but this does not satisfy the statement " $-u$ is also in $U$ "

Can someone please give an example for this question. Thank you.

答:
Take the subset of all pairs of rational numbers in $\mathbb{R}^2$. Then this subset, $\mathbb{Q}^2 \subset \mathbb{R}^2$ contains the $\textbf{0}$ vector and is closed under vector addition. Also, give any rational number, its negative is also a rational number, so therefore if $\textbf{u} \in \mathbb{Q}^2$ then $-\textbf{u} \in \mathbb{Q}^2$

However, $\mathbb{Q}^2$ is not closed under scalar multiplication over $\mathbb{R}$, and so $\mathbb{Q}^2$ is not a subspace of $\mathbb{R}^2$.

For example, $(2,7) \in \mathbb{Q}^2$ but $\pi(2,7) = (2 \pi, 7 \pi) \notin \mathbb{Q}^2$

Title: Subspace of $\mathbb{R}^2$


答:
Well.  Consider:

$\frac 1{m+a\sqrt p}=\frac {m-a\sqrt{p}}{(m+a\sqrt p)(m-a\sqrt p)} = (m-a\sqrt{p})\frac 1{m^2 - a^2p}$.

So $m+a\sqrt p$ is invertible if $m^2 - a^2p$ is.

So $a + b\sqrt 2 + c \sqrt 3$ is invertible if $(a+b\sqrt 2)^2 - 3c^2$ is.

$(a+b\sqrt 2)^2 - 3c^2 = (a^2 + 2b^2 -3c^2) + 2ab\sqrt 2$ and that is invertible if $(a^2 + 2b^2 - 3c^2)^2 -4a^2b^2$ is.

And as the last monster is real it is.

This assumes that none of these terms are zero though....  we should probably do something about that.

If $m,a \in \mathbb Q$ and $a\ne 0$ then

$(m+a\sqrt 2)^{-1} = (m-a\sqrt{w})\frac 1{m^2 - 2a^2}$ and $m^2 - 2a^2$ can't be $0$ as $2 = (\frac ma)^2$ is impossible.  So $\{a + b\sqrt 2\}$ is a field.

If $m \in \{a + b\sqrt 2\}$ and $a \in \mathbb Q; a \ne 0$ then

$(m + a\sqrt 3)^{-1} = (m-a\sqrt 3)\frac 1{m^2 - 3a^2}$.  $m^2 -3a^2$ supposedly can not be $0$ as that would mean $3=(\frac ma)^2$ and supposedly $\sqrt 3\not \in \{a+b\sqrt 2\}$.

We can verify that last assumption fairly easily.  $\sqrt 3= a+b\sqrt 2; a,b\in \mathbb Q$ would imply $\sqrt 3 - a = b\sqrt 2$ which implies $3+a^2 -2a\sqrt 3=2b^2$.  But $\sqrt 3$ is irrational so that's only possible if $a=0$ but then we have $3 = 2b^2$ so $b^2 = \frac 32$ but that is impossible.

======

I guess I should point out that have only shown that $\{a+b\sqrt 2 + c\sqrt 3\}$ is invertible.   I haven't shown that it is closed under multiplication.

I can't show it it is closed under multiplication as it isn't.

$\sqrt 2\sqrt 3 \not \in \{a + b\sqrt 2 + c \sqrt 3\}$ as

$\sqrt 2 \sqrt 3 = a+ b\sqrt 2 + c\sqrt 3 \implies$

$\sqrt 6 -b\sqrt 2 = c\sqrt 3 \implies$

$6 + 2b^2 - 2b\sqrt 6\sqrt 2 = 3c^2 \implies$

$-4b\sqrt 3 = 3c^2 - 6 -4b^2\implies$

either $b \ne 0$ and $\sqrt 3 \in \mathbb Q$  or $b=0$ and $c^2 =2; c\in \mathbb Q$

So $\{a+b\sqrt 2+c\sqrt 3\}$ is not closed under multiplication and it is not a field.


问:
Prove that $\Delta ABC$ is an equilateral triangle if and only if the complex coordinates $a$,$b$, and $c$ satisfy a relation.

Let $\triangle ABC$ be a triangle in the complex plane and let $a$,$b$ and $c$, respectively, be the complex coordinates of its vertices.  Suppose that the tiangle is inscribed in the circle $C(0,1)$.  Prove that $\triangle ABC$ is equilateral if and only if $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=0$$ and $$(a+b)(b+c)(c+a)\neq0\,.$$

I tried to convert $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=0$ into a simpler equation by multipling it by $(a+b)(b+c)(c+a)$

In the end I obtained $$a^2+b^2+c^2+3ab+3bc+3ca=0$$ but I don't know how to continue from here.  Can you help me, please?

答:
During all the proof I will call $a=z_0$, $b=z_1$ and $c=z_2$.

$\Longrightarrow)$

If $z_0$, $z_1$ and $z_2$form a equilateral triangle inscribed in $C(0,1)$, then we can write $z_j$ as $$z_j=m\cdot e^{i\frac{\pi j}{3}};\quad \text{with } |m|=1.$$ Then, since $z_j$ are rotations of the cubic roots of $1$, then $$\sum\limits_{j=0}^2 m\cdot e^{i\frac{\pi j}{3}}=m\cdot \underbrace{\sum\limits_{j=0}^2 e^{i\frac{\pi j}{3}}}_{=0}=0.$$

The other condition is easy to proof since none of the $z_j$ are diametrically opposite (ie, an equilateral triangle inscribed in a circumference can't have two vertices diametrically opposite).

$\Longleftarrow)$

The proof of this statement can be found (with this same notation) here.


答:
Consider  $s$ ={(x,y): x is rational} in $R^2$
All conditions valid that you ask but not a subspace of $R^2$ as not closed under scalar multiplication.

Title: Subspace of $\mathbb{R}^2$

Body:
Give an example of a nonempty set $U$ of $\mathbb{R}^2$ such that $U$ is closed under addition for every $u$ in $U$, $-u$ is also in $U$ but U is not a subspace of $\mathbb{R}^2$

What I understand from the question is I need to give an example such that the set is closed under addition but not under scalar multiplication.

For this I thought of some sets like $\{(x,y)|x\geq0, y\geq 0\}$ but this does not satisfy the statement " $-u$ is also in $U$ "

Can someone please give an example for this question. Thank you.


问:
Coloring dots in a circle with no two consecutive dots being the same color
I ran into this question, it is not homework. :)

I have a simple circle with $n$ dots, $n\geqslant 3$. the dots are numbered from $1\ldots n$.

Each dot needs to be coloured red, blue or green. No dot can be couloured the same as his neighbors. How many options are there to organize the colours?

Hint: we need to use recurrence relation.

答:
Let $A_n$ be the number of ways you can colour your circle of $n$ dots, and let $B_n$ be the number of ways you can colour a line of $n$ dots. Of course $B_n$ is much easier to figure out: we have a choice of $3$ colours for the dot on the left end, $2$ colours for the one next to it, $2$ for the next, and so on, so $B_n=3\cdot 2^{n-1}$ for $n\ge 1$.

Next, observe that $B_n=A_n+A_{n-1}$ for $n\ge 3$. This is because (as explained in the other answers) the number of ways to colour the line of $n$ dots with different colours at the ends is equal to $A_n$, while the number of ways to do it with the same colour at both ends is equal to $A_{n-1}$.

This gives us the recurrence $A_n=B_n-A_{n-1}=3\cdot 2^{n-1}-A_{n-1}$ which is valid for $n\ge 3$. The initial value, of course, is $A_2=6$. The next few values are:
$$A_3=B_3-A_2=12-6=6;$$
$$A_4=B_4-A_3=B_4-B_3+A_2=24-12+6=18;$$
$$A_5=B_5-A_4=B_5-B_4+B_3-A_2=48-24+12-6=30;$$
and so on.

The pattern is clear: $A_n$ is the sum of a geometric progression of $n-1$ terms with first term $a=B_n=3\cdot 2^{n-1}$ and common ratio $r=-\frac 1 2$. Plugging these values into the formula $\frac{a(1-r^n)}{1-r}$ and simplifying, we get $A_n=2^n+2(-1)^n$.


问:
Prove $\overline{s}_T^2+4\sum_{n=0}^{T-1}\overline{s}_n(s_{n+1}-s_n)\leq 4s_T^2$
I am working on a paper which gives a novel/trajectorial proof of Doob's maximal inequalities you can find it here. I tried to prove the following inequality given on page 2 of the paper. $$\overline{s}_T^2+4\sum_{n=0}^{T-1}\overline{s}_n(s_{n+1}-s_n)\leq 4s_T^2 $$ where $s_0,...,s_T$ are real numbers and $\overline{s}_n=\max\limits_{i\in [n]}s_i$. The author suggests to rearrange the terms and to complete the squares, but for some reason I am stuck and I don't know how to do this. Any help is very much appreciated.

答:
Here's what the authors probably had in mind: Let's first rewrite the sum on the left-hand side of your inequality. Let $0=m_0<m_1<\dotsb <m_r\leq T-1$ be those indices for which $s_{m_i}=\overline{s}_{m_i}$ (i.e., the $s_{m_i}$ are the "peaks" in the sequence $s_0,\dotsc,s_{T-1}$). For $0\leq i< r$, $$\sum_{n=m_i}^{m_{i+1}-1}\overline{s}_n(s_{n+1}-s_n)=\sum_{n=m_i}^{m_{i+1}-1}s_{m_i}(s_{n+1}-s_n)=s_{m_i}s_{m_{i+1}}-s_{m_i}^2,$$ where we use that the sum in the middle is a telescoping sum. Similar for $\sum_{n=m_i}^{T-1}\overline{s}_n(s_{n+1}-s_n)$. This allows us to rewrite the left-hand side of your  inequality as follows: \begin{align*}\overline{s}_T^2+4\sum_{n=0}^{T-1}\overline{s}_n(s_{n+1}-s_n)&=\overline{s}_T^2+4\sum_{i=0}^{r-1}s_{m_i}s_{m_{i+1}}+4s_{m_r}s_T-4\sum_{i=0}^{r}s_{m_i}^2\\ &\leq \overline{s}_T^2+4s_{m_r}s_T-2s_{m_0}^2-2s_{m_r}^2,\end{align*} where the estimate follows from $4s_{m_i}s_{m_{i+1}}-2s_{m_i}^2-2s_{m_{i+1}}^2=-2(s_{m_i}-s_{m_{i+1}})^2\leq 0$ applied for $i=0,\dotsc,r-1$. Now distinguish two cases:

$s_T=\overline{s}_T$. Then $\overline{s}_T^2+4s_{m_r}s_T-2s_{m_0}^2-2s_{m_r}^2=3s_T^2-2(s_T-s_{m_r})^2-2s_{m_0}^2\leq 4s_T^2$.

$s_T< \overline{s}_T$. Then necessarily $\overline{s}_T=s_{m_r}$ and we can estimate $$\overline{s}_T^2+4s_{m_r}s_T-2s_{m_0}^2-2s_{m_r}^2=4s_T^2-(s_{m_r}-2s_T)^2-2s_{m_0}^2\leq 4s_T^2.$$

In either case, the desired inequality follows.


问:
Proof that the set of irrational numbers is dense in reals
I'm being asked to prove that the set of irrational number is dense in the real numbers.  While I do understand the general idea of the proof: </p>

<p>Given an interval $(x,y)$, choose a positive rational number (say) $z=\sqrt{2}$.  By density of rationals, there exists a rational number $p$ in the interval $(x/z, y/z)$, which essentially means that $$\frac{x}{\sqrt{2}} < p < \frac{y}{\sqrt{2}}.$$  I find that $pz$ is irrational, since it is the product of a rational and irrational number. However, my instructions besides writing that proof down is to specifically verify that $y=xz$ is irrational.  What does this have to do with anything and how does it prove denseness of irrationals?  Regardless, assume that $x$ is a nonzero rational number and that $z$ is irrational. For the sake of contradiction, assume that $y=xz$ is rational. This should mean that $y/x$ rational as well, and therefore that $z$ is rational, a contradiction. </p>

答:
I find this way really easy to understand so I thought I would share it. I had come across it in the book  Introduction to Real Analysis by William Trench.

Proof: Given that we know that rational numbers are dense in $\mathbb{R}$ . So $\exists$ $r_1 , r_2 \in \mathbb{Q}$ such that:

$x < r_1 < r_2 <y$

Remark: To find a number between $r_1$ and $r_2$ that is irrational, we need to come up with a small enough number that can we can add to $r_1$ that serves two purposes:

Make it irrational

Make sure the sum of this irrational number and $r_1$ is less than $r_2$

To undertand it better, think of this inequality:  $\frac{r_2-r_1}{2} < \frac{r_2-r_1}{\sqrt2} <  r_2 - r_1$

So,  $\frac{r_2-r_1}{\sqrt2}$ is a number that we can add to $r_1$ to make it less than $r_2$ (it serves the two points outlined above).

Let $t = r_1 + \frac{r_2-r_1}{\sqrt2}$

Therefore $x < r_1 < t <r_2<y$

Note - you can further prove that $t$ is irrational by using the facts:

The sum of an irrational and a rational number is irrational.

The product of an irrational and a rational number is irrational.


问:
Determine if a point is between two lines
First of all, I'm not a math whiz and haven't taken a math class in 20+ years, so be gentle with me. I'm trying to figure out how to make a spreadsheet for airplane weight and balance.

I have two straight lines. The first line has a point at 43.75, 1300 and a second point at 52.5, 1600.

The second line has points at 47.5, 1150 and a second point at 60, 1600.

I am trying to figure out a formula where I can plug in a point and determine mathmatically if the point is between the lines. So for instance I want to know if 55.2, 1547 is between the lines or outside the lines.

These lines are part of a moment envelope. The y for the point we want to examine will never be greater than 1600 nor less than 1150.

So it seems I'd need to figure out if x for any given y is greater than what x is for the "left" line at the same y and then less than x for the "right" line at the same y. So I'd need to find what the coordinates of the line are for a given y and the slope of the line. Am I on the right track?

答:
The equation of a line is of the form $y=mx+n$ with $m=\frac{\Delta y}{\Delta x}$
$$m_1=\frac{1600-1200}{52.5-41.25}\approx 35.56 \qquad n_1=1600-52.5m_1 \approx -267$$
$$m_2=\frac{1600-1150}{60-47.5}=36 \qquad n_2=1600-60m_2=-560$$
You want the point to be below the first and above the second line so its coordinates have so satisfy
$$36x-560<y<35.56x-267$$
Since the plot is pretty bad you could consider $m_1\approx m_2$ which would simplify the formula to
$$267<36x-y<560$$


问:
Is a regular Galois extension of constant Galois extension Galois?
Let $V, W$ be two (smooth, irreducible) varieties defined over $\mathbb{Q}$ and let $\phi: V \rightarrow W$ be a morphism defined over $\mathbb{Q}$. Suppose there is a number field $K$, which is Galois over $\mathbb{Q}$, such that $\phi$ is a regular Galois cover over $K$; i.e. $K(V)/\phi^*(K(W))$ is Galois with $\overline{K} \cap K(V)=K$.

Question: Is the extension $K(V)/\phi^*(\mathbb{Q}(W))$ Galois?

This feels like it should be true since it is built out of a constant Galois extension and a regular Galois extension which don't interact and examples suggest this to be true with the Galois group being a semidirect product. I couldn't find any results in the literature though about this mixed case so would be glad of references as well.

As a simple example, one could take $V=W=\mathbb{P}^1$ with $\phi: x \mapsto x^n$; this induces a regular Galois cover over the cyclotomic field $K=\mathbb{Q}(\zeta_n)$ with the Galois group of $K(V)/\phi^*(\mathbb{Q}(W)$ being isomorphic to $C_n \rtimes \operatorname{Aut}(C_n)$.

答:
$K(V)/\Bbb{Q}(V)$ and $K(V)/\phi^*(K(W)))$ both Galois implies that the finite extension $K(V)/(\Bbb{Q}(V)\cap \phi^*(K(W)))$ is Galois.

If also $\overline{K} \cap \Bbb{Q}(V)=\overline{K} \cap \phi^*(\Bbb{Q}(W))$ then $K(V)/\phi^*(\Bbb{Q}(W))=K(V)/(\Bbb{Q}(V)\cap \phi^*(K(W)))$ is Galois.

But $\overline{K} \cap K(V)=K$ is not enough. Try with $K=\Bbb{Q}(\zeta_3),\phi^*(\Bbb{Q}(W))=\Bbb{Q}(x)$, $\Bbb{Q}(V)=K(V)=K(x)[y]/(y^2-x^3-x-\zeta_3)$.

For $K(V)/\phi^*(\Bbb{Q}(W))$ to be Galois we need to find an automorphism $\sigma$ of $K(V)$ fixing $x$ and such that $\sigma(\zeta_3)=\zeta_3^2$. It means that $\sigma(y)$ is a root of $T^2-x^3-x-\zeta_3^2$ ie. $\sigma(K(V))$ is the function field of an elliptic curve with a different $j$-invariant, so it can't be that $\sigma(K(V))=K(V)$.

$K(V)/\phi^*(K(W))=K(x)[y]/(y^2-x^3-x-\zeta_3)/K(x)$ is Galois because it is a quadratic extension, on the underlying smooth projective curves the covering $[x:y:1]\to [x:1]$ has a few ramified points with $x=0$, and removing them from both curves gives your Galois covering of smooth affine varieties.

The morphism $V\to W$ defined over $\Bbb{Q}$ is obtained from $V:y^2-x^3-x-a=a^2+a+1=yu-1=0\subset \Bbb{A^4}$ and sending $(x,y,a,u)\to x$


问:
Computing the Jacobian matrix in symplectic Euler integration
I am reading a book about geometric numerical integration. In the text, they present Euler symplectic integration

\begin{equation}
p_{n+1}=p_n-hH_q(p_{n+1},q_n)\\
q_{n+1} = q_n+hH_p(p_{n+1},q_n)
\end{equation}

for integrating the Hamiltonian system
\begin{align}
\dot{p}=-H_q(p,q)\\
\dot{q}=H_p(p,q)
\end{align}

In the book it is written that differentiating the first system (symplectic integration scheme) with respect to $(p_n,q_n)$ yields

$$\begin{bmatrix}
I+hH_{qp}^T &0\\
-hH_{pp}& I
\end{bmatrix}
(\frac{\partial(p_{n+1},q_{n+1})}{\partial (p_n,q_n)}) =
\begin{bmatrix}
I &-H_{qq}\\
0& I+hH_{qp}
\end{bmatrix}$$

I think I understand where the matrix on the right side comes from since it is the jacobian matrix of $$f(p_n,q_n) =\begin{bmatrix}p_n-hH_q(p_{n+1},q_n)\\
q_n+hH_p(p_{n+1},q_n)\end{bmatrix}$$ with respect to $(p_n,q_n)$ but I don't really understand how we get the expression on the left, i.e. where does

$$\begin{bmatrix}
I+hH_{qp}^T &0\\
-hH_{pp}& I
\end{bmatrix}
(\frac{\partial(p_{n+1},q_{n+1})}{\partial (p_n,q_n)}) $$ come from?

答:
You have to fully write out the total differentials of the equations, then you get
\begin{align}
dp_{n+1}&=dp_n-hH_{qp}dp_{n+1}-hH_{qq}dq_n\\
dq_{n+1}&=dq_n+hH_{pp}dp_{n+1}+hH_{pq}dq_n
\end{align}
Now sort the components with the same index to each side and write it in matrix form
$$
\pmatrix{I+hH_{qp}&0\\-hH_{pp}&I}\pmatrix{dp_{n+1}\\dq_{n+1}}
=
\pmatrix{I&-hH_{qq}\\0&I+H_{pq}}\pmatrix{dp_n\\dq_n}
$$
Then insert the vector field $(\frac\partial{\partial p_n},\frac\partial{\partial q_n})$ to get the claimed identity.


问:
Confused with solving a three variable equation
Here is the problem:

A dietitian has 3 supplement products that can be used.

Each dose of Supplement I contains 5 units of Vitamin B, 4 units of calcium and 4 units of iron.

Each dose of Supplement II contains 2 units of Vitamin B, 3 units of calcium and 2 units of iron.

Each dose of Supplement III contains 1 unit of vitamin B, 5 units of calcium and 2 units of iron.

If the patient needs 13 units of Vitamin B, 16 units of calcium and 12 units of iron each week, how many doses of each supplement should the dietitian prescribe?

I've got:

$5x+2y+Z=13$

$4x+3y+5z=16$

$4x+2y+2z=12$

Then: I have $x=1+z,\quad y=4-3z,\quad$ and $(1+z,4-3z,z)$

Now I'm confused on how to interpret this into doses, if this is the right approach.
Please help.

答:
Sure, you can write it as a system of equations:
$$\begin{cases}
5x + 2y + z = 13 \\
4x + 3y + 5z = 16 \\
4x + 2y + 2y = 12
\end{cases}$$
where $x, y, z$ is the number of doses of supplement I, II, III, respectively.

Use Gaussian elimination to find a solution (if there is one). That, is, write it on matrix form and perform row operations:
$$
\left(
\begin{array}{ccc|c}
5 & 2 & 1 & 13 \\
4 & 3 & 5 & 16 \\
4 & 2 & 2 & 12
\end{array}
\right)
\sim
\left(
\begin{array}{ccc|c}
5 & 2 & 1 & 13 \\
0 & \frac{7}{5} & \frac{21}{5} & \frac{28}{5} \\
0 & \frac{2}{5} & \frac{6}{5} & \frac{8}{5}
\end{array}
\right)
$$

and so on... You might also want to find the possible nullspace of the matrix to find alternative doses.

EDIT: Missed that you already had the solution!
You can plug in values of $z$, the desired amount of doses of supplement III, in the solution $(1+z, 4-3z, z)$. $z=0$ yields $(1, 4, 0)$ and $z=1$ yields $(2, 1, 1)$ - the first number is the amount of doses of supplement I, the second supplement II and the third supplement III.

If you plug in $z=2$ you get $(3, -2, 2)$, which contain negative doses of supplement II. This will happen for all $z \geq 2$, so these solutions are dismissed.


答:
You forgot to divide by $\pi$. WolramAlpha's result is correct, but the coefficients $b_n$ are given by
$$b_n=\frac{1}{n}$$

Your second formula for the series in WolframAlpha is wrong, you should divide by $k$, not multiply. Then everything should be fine: WolframAlpha

Title: Fourier series for $f(x)=(\pi -x)/2$

Body:
I need to find the Fourier series for $$f(x)=\frac{\pi -x}{2}, 0<x<2\pi$$

Since the interval isn't symmetric over $0$, I guess I need to consider $f$'s periodic extension to $\mathbb R$. let's call it $g$. Then $$g(x)=\begin{cases}\frac{\pi -x}{2}, \text{ if } 0<x<\pi\\  \frac{-\pi -x}{2}, \text{ if }{-\pi <x<0}\end{cases}$$

Due to $g$ being odd, the fourier coefficients $a_n$ are all $0$.

And $$b_n=\frac{2}{\pi}\int _0^\pi g(x)\sin(nx)dx=\frac{1}{\pi}\int _0^\pi (x-\pi)\sin (nx)dx$$
According to wolfram alpha $b_n=1/n$ so the Fourier series is
$$\sum _{n=1}^{+\infty}\frac{\sin(kx)}{n}$$

But when I check the result of $x=1/2$ I get this which seems to diverge. It also fails for $x=\pi /2$ and a few other $x$'s.

What am I doing wrong?

I just saw this question. it looks like my answer is correct. am I doing anything wrong in my verification?

EDIT: I realised I forgot to divide by $\pi$. I fixed that but it still doesn't work. I also updated the links.


问:
combine equal pairs into equivalence classes
Given a set $S$ of values, and a set $P \subset S \times S$ of pairwise equivalences, what is an algorithm for partitioning $S$ into equivalence classes?

$P$ is guaranteed to be an equivalence relation.

Assume that $P$ "covers" every member of $S$.

Efficiency isn't all that crucial:  I'm interested in $\#S < 100000$ and $\#P < 10000$.

This is like building up pairs into clusters, but because this is for exact equality rather than mere similarity, I'm finding that clustering algorithms are red herrings.

答: Equivalence classes on the set $S$ are the same as the connected components of the graph whose edges are given by $P$. Fortunately, there are standard algorithms for computing the connected components: here is one example.


问:
Need help with De'Moivre's Theorem, or maybe that is the wrong approach to this problem.

Let ($a_{n}$) and ($b_{n}$) be the sequences of real numbers such that $(2+i)^n=a_{n}+b_{n}i$ for all integers $n\geq 0$, where $i=\sqrt{-1}$. What is $\sum_{n=0}^{\infty}\frac{a_{n}b_{n}}{7^n}$?

I need some help with the question above. I tried using De'Moivre's Theorem to simplify $(2+i)^n$, but got hung up with simplifying $\sqrt{5}^n\operatorname{cis}(n\theta)$. I was attempting to find $a_{n}$ and $b_{n}$ first, then evaluate the summation.

答:
Let $\theta$ be the acute angle satisfying $\tan \theta = \frac{1}{2}$,

$$a_n = \sqrt{5^n} \cos(n \theta)$$

$$b_n = \sqrt{5^n}\sin (n\theta)$$

$$\frac{a_nb_n}{7^n} = \left( \frac57\right)^n\sin(n\theta)\cos(n\theta)=\frac12 \left( \frac57\right)^n \sin (2n\theta)= \frac12 \Im\left[\left(\frac57e^{2i\theta}\right)^n\right]$$

\begin{align}\sum_{n=0}^\infty \frac{a_nb_n}{7^n} &= \frac12 \Im \left[ \sum_{n=0}^\infty \left( \frac57 e^{2i\theta} \right)^n\right] \\
&= \frac12 \Im \left[ \frac{1}{1-\frac57e^{2i\theta}}\right]\end{align}

Try to simplify from here.

Edit:

\begin{align}
\frac12 \Im \left[ \frac1{1-\frac57 e^{2i\theta}}\right] &= \frac12 \Im \left[ \frac{1-\frac57e^{-2i\theta}}{(1-\frac57 e^{2i\theta})(1-\frac57 e^{-2i\theta})}\right] \\
&=\frac12\left[ \frac{\frac57\sin 2 \theta}{1+\frac{25}{49}-\frac{10}7 \cos2\theta}\right] \\
&=\frac12\left[ \frac{\frac{10}7\sin  \theta \cos \theta}{1+\frac{25}{49}-\frac{10}7 (2\cos^2 \theta -1)}\right] \\
&=\frac12\left[ \frac{\frac{10}7\cdot \frac25}{1+\frac{25}{49}-\frac{10}7 (2\cdot \frac45 -1)}\right] \\
&=\frac12\left[ \frac{\frac{10}7\cdot \frac25}{1+\frac{25}{49}-\frac{10}7 \cdot \frac35}\right] \\
&=\frac12\left[ \frac{\frac47}{1+\frac{25}{49}-\frac{6}7}\right] \\
&=\frac12\left[ \frac{28}{49+25-42}\right] \\
&= \frac{14}{32}\\
&= \frac{7}{16}
\end{align}


问:
Number of positive integer solutions of a equation system with restrictions.
The question is how many possible solutions there are for a equation system subject to the condition that $1 \leq x_1 + y_1$, $1 \leq x_2 + y_2$, $1 \leq x_r + y_r$ where $x_i$ and $y_i$ are non-negative integers.

The system is   $$x_1 + x_2 + \ldots + x_r = n.$$
$$y_1 + y_2 + \ldots + y_r = n.$$

I want to do it by the inclusion-exclusion principle but i can't find a way to pose a initial equation.
After that i would suppose $ 1 - y_1 > x_1 $ and try to introduce it in one of those equations.

Any help is greatly appreciated, thanks.

答:
There may be a simpler solution, but you can certainly run this by inclusion-exclusion on a reducing $r$.

I'm thinking of this as being like deployment of $n$ infantry and $n$ artillery into a position with $r$ locations, where you cannot leave any gap in the line. Without the gap restriction we can easily see this is two cases of stars-and-bars for $\binom{n+r-1}{n}^2$

A gap in both categories ($x_i+y_i=0$) would correspond to a deployment into $r{-}1$ positions. The gap can occur in $\binom r1=r$ locations of course. Then we have to account for overcounting the case of $2$ gaps, etc., giving

$$\sum_{k =0}^{r-1}(-1)^k\binom{r}{k}\binom{n+r-1-k}{n}^2  $$


问:
Can you recommend some books on elliptic function?
I plan to study elliptic function.  Can you recommend  some  books? What is the relationship between elliptic function and elliptic curve？Many thanks in advance!</p>

答:
First of all

ever-modern  Course of modern analysis by Whittaker-Watson.

For a more introductory style, I highly recommend

V. Prasolov, Y. Solovyev Elliptic Functions and Elliptic Integrals.

The relation between elliptic curves and elliptic functions can be sketched as follows. Elliptic curve is topologically a torus which can be realized by cutting a parallelogram in $\mathbb{C}$ and identifying its opposite edges. On the other hand, it can be realized  in $\mathbb{CP}^2$ by an algebraic equation of the form
$$y^2=x^3+ax+b.$$
Elliptic functions provide a map between the two pictures, which is also called uniformization. Essentially, $x,y$ are given by some elementary elliptic functions of $z$ (complex coordinate on the parallelogram).

Compare this with a more familiar example: trigonometric functions $\sin$, $\cos$ provide a uniformization of the circle, which can be defined either via an algebraic equation or in a parametric form:
$$x^2+y^2=1\quad \begin{array}{c}\sin,\cos\\ \Longleftrightarrow \\ \;\end{array}\quad \begin{cases}x=\cos t,\\ y=\sin t,\\ t\in[0,2\pi].\end{cases}$$


问:
Almost sure divergence of a sequence of random variables
Let $Y_n = n X_n$, where $X_n$ is a non-negative martingale with bounded positive mean (no assumptions of independence). By the martingale convergence theorem $X_n \to X$ for some bounded r.v. $X$. Assume $X > 0$ WP $1$ (in order to avoid counter-examples, as I'm only interested in the technique involved here).

It seems obvious that $Y_n \to \infty$ a.s., but how to prove this rigorously? In particular, the problem I am facing is that $\inf X_n$ can be $0$, so I can't think of a way to establish that for an arbitrary $M> 0$, we can find $n$ large enough so that $P\{Y_n > M\} = 1$. I don't think there's a way to use Borel-Cantelli here either, as the sequence is not independent, and there's no way to say that $\sum_n P(A_n) < \infty$ for the sequence of events $\{Y_n \leq M\}$.

答:
This is not true in general.  The constant sequence $X_n = 0$ is a non-negative martingale with bounded mean, and has $nX_n = 0$ as well.  For a non-trivial example, if we say $(Z_n)$ is a sequence of i.i.d. Bernoulli random variables with $P(Z_n=0)=P(Z_n=2)=\frac 12$, then $X_n := \prod_{m=1}^n Z_m$ is a non-negative martingale but converges to $0$ almost surely and so does $nX_n$.

If we want to ensure that $\mathbb{E}[\lim X_n] > 0$, let $X_n$ be as in the second example and define $M_0$ to be an independent Bernoulli random variable with $P(M_0 = 0)=P(M_0=1)=\frac 12$.  The sequence $M_n = M_0 + X_n$ is a martingale, and $(M_n) \rightarrow M$ where $P(M=0)=P(M=1) = \frac 12$, so $\mathbb{E}[M] = \frac 12 > 0$.  The sequence $nM_n$ converges to $0$ on the set $\{M_0 = 0\}$, so does not converge to infinity almost surely.

EDIT: The statement that if $(X_n)$ is a sequence of non-negative random variables with $(X_n) \rightarrow X$ a.s. and $X>0$ a.s., then $(nX_n) \rightarrow \infty$ a.s. is true, and needs no assumptions about $(X_n)$ being a martingale.  For almost every $\omega$, there exists $N = N(\omega)$ such that $X_n(\omega) > \frac 12 X(\omega) > 0$ for $n \ge N$, so $\lim n X_n(\omega) \ge \lim \frac 12 n X(\omega) = \infty$.


问:
Example of Quasi-circular domain
Can someone provide me with an example of a quasi-circular domain?

A domain $D\subset\mathbb{C}^n$ is said to be m-quasi-circular, (where $m=(m_1,m_2,..,m_n); m_i$ being positive integers),
if D is invariant under the map $f_{m,\theta}$ for all $\theta \in \mathbb{R}$.
$$f_{m,\theta} : D\rightarrow \mathbb{C}^n$$
$$(z_1,\dots,z_n)\mapsto (e^{im_1\theta}z_1,\dots,e^{im_n\theta}z_n)$$

Note: If $m_1=m_2=m_3=....=m_n=1$, then the domain is called circular domain.

Can you give an example of a quasi-circular domain that is not circular? Say (2,3)-quasi-circular domain in $\mathbb{C}^2$.

答:
Symmetrized bidisc is a quasi-circular domain that isn't a circular domain.

$$\mathbb{G}=\{(z_1+z_2,z_1z_2)\in \mathbb{C}^2: z_1,z_2\in B(0,1)\}$$
Clearly, this is a (1,2)-quasi-circular domain.

To see that this is not a circular domain, notice that $(2,1)\in \bar{\mathbb{G}}$ but $(-2,-1)\not \in \overline{\mathbb{G}}$.

Title: Example of Quasi-circular domain


问:
95 vectors in a cube (APMC 1995)
Let $C$ denote the cube $[-1,1] \times [-1,1] \times [-1,1]$.  Let $v_1,v_2,...,v_{95}$ be points (treated as vectors from the origin to these points) inside or on the boundary of $C$.  Consider the set $W$ of $2^{95}$ vectors of the form $s_1v_1+s_2v_2+...+s_{95}v_{95}$ where each $s_i=1$ or $-1$.

(a) If $d=48$, then show that there exists a vector $w=(a,b,c)$ in $W$ such that $a^2+b^2+c^2 \le d$.

(b) Does there exist $d<48$ that satisfies (a)?

(c) What is the smallest value of $d$ that satisfies (a)?

Source : Austrian-Polish Mathematical Competition 1995 Problem 8

Generalisation : If we have $n$ vectors $v_1,v_2,...,v_n$ instead and consider $2^n$ vectors of the form $s_1v_1+s_2v_2+...+s_n v_n$ where $s_i=1$ or $-1$, what is the smallest number $d=d_n$ such that there is a vector $w=(a,b,c)$ in $W$ st $a^2+b^2+c^2 \le d$?  At least what is a good upper bound for $d_n$?

Attempt (for the generalisation) :

Since $\sum_{s_i=\pm 1} \left|\sum_{j=1}^n s_jv_j\right|^2=2^n \sum_{j=1}^n|v_j|^2$ and $|v_j|\le \sqrt 3$, it follows that for $d\ge 3n$, there exists $w\in W$ st $|w|^2\le d$.  I don't see how to improve this bound.

答:
Lemma: If $v_1, \dots, v_5$ are five vectors in the cube $C = [-1, 1]^3$, then there exist two indices $1\leq i < j \leq 5$ and $s, t \in \{-1, 1\}$ such that $s v_i + t v_j \in C$.

Proof: We cut the cube $C$ into eight small cubes: $[0, 1] \times [0, 1]\times [0, 1]$, $[-1, 0] \times [0, 1] \times [0, 1]$ ... etc.

Two small cubes are called "opposite to each other" if they are images of each other with respect to the symmetry around the origin. E.g. $[0, 1] \times [0, 1] \times [0, 1]$ is opposite to $[-1, 0] \times [-1, 0] \times [-1, 0]$.

The eight small cubes then form four groups, each group being two small cubes that are opposite to each other.

Since we have five vectors, there are two of them that lie in the same group, say $v_i, v_j$. If they live in the same small cube, then $v_i - v_j$ is in $C$; otherwise, they live in opposite small cubes, and $v_i + v_j$ is in $C$.

Lemma: If $n > 4$ and we have $n$ vectors $v_1, \dots, v_n \in C$, then there exists $n - 1$ vectors $w_1, \dots, w_{n - 1} \in C$ such that the set $\{\sum_i t_i w_i: t_i \in \{-1, 1\}\}$ is a subset of $\{\sum_i s_i v_i: s_i \in \{-1, 1\}\}$.

Proof: We apply the previous lemma to $v_1, \dots, v_5$. Without loss of generality, assume that $sa_1 + ta_2$ is in $C$, with $s, t \in \{-1, 1\}$. Then it suffices to choose $w_1 = sa_1 + ta_2, w_2 = a_3, w_3 = a_4, \dots$.

Lemma: If $n \geq 4$ and we have $n$ vectors $v_1, \dots, v_n \in C$, then there exists $4$ vectors $w_1, \dots, w_4\in C$ such that the set $\{\sum_i t_i w_i: t_i \in \{-1, 1\}\}$ is a subset of $\{\sum_i s_i v_i: s_i \in \{-1, 1\}\}$.

Proof: Use the previous lemma and induction on $n$.

Now it is enough to use your argument (applied to $n = 4$) to conclude that $d = 12$ works for any $n \geq 4$.

Title: 95 vectors in a cube (APMC 1995)


问:
homomorphisms of infinite groups

Prove that each of the following is a homomorphism, and describe its kernel:

the function $f: \mathbb{R}^*\to\mathbb{R}_{>0}$ defined by $f(x)=|x|$

My proof step: The kernel of $f$ is the set $$k=\{x\in\mathbb{R}^*:f(x)=e\}.$$
Let $a,b \in k$, $f(a)f(b)=f(ab)=ee=e$. I think I'm missing a lot of steps.

答:
HINT: To find $\ker f$ you need to answer the following questions:

What is the identity in $\Bbb R_{>0}$? In other words, what is the element $e$ of $\Bbb R_{>0}$?

What elements of $\Bbb R^*$ are mapped to that identity by $f$?

To show that $f$ is a homomorphism, you need to show that it has the homomorphism property: for all $x,y\in\Bbb R^*$, $f(xy)=f(x)f(y)$.

All of this requires using the definition of $f$.

Title: homomorphisms of infinite groups


问:
Multivariable Lipschitz continuous functions that can be written as a linear combination of a function
Let $f_i$ ($i=1, \ldots, n$) be globally Lipschitz continuous functions which can be expressed as

$\left[\begin{array}{c} f_1(x_1, \ldots, x_n) \\ \vdots \\ f_n(x_1, \ldots, x_n)\end{array}\right] = \left[\begin{array}{ccc} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn}\end{array}\right] \left[\begin{array}{c} g(x_1) \\ \vdots \\ g(x_n)\end{array}\right].$

where $a_{ii} \neq 0$ for all $i$.

Can we say that the scalar function $g$ is globally Lipschitz continuous?

答:
Yes.

First, the case where there the line $j=1$ has a sum $\sum_i a_{1i}\neq 0$.
Then we study on the diagonal $x := x_1 = x_2 = \dots = x_n$, and the answer is fast.
$$g(x) = \frac {f_1(x,x,\dots,x)} {\sum_i a_{1i}}$$

Else, we study on the pseudo-diagonal $x := x_2 = \dots = x_n$ with $x_1 := 0$, which gives :
$$g(x) = \frac {f_1(0,x,\dots,x) - a_{11}g(0)} {\sum_{i\ge 2} a_{1i}}$$
Where the denominator is not $0$ because it is equal to $-a_{11}$ because the line 1 has null sum.

In both case, we have easy proofs that $g$ has almost any regularity that the $f_i$ have.


问:
How can I find $\int_0^{2\pi}\sin(x)\sin(x+1)$
I'm trying to find $\int_0^{2\pi}\sin(x)\sin(x+1)$, however, I'm having a lot of trouble.

I've tried using integration by parts on it, but when I ended up with $\int \sin(x)\sin(x+1)$ on both the left and right sides, they just cancelled each other out.

I also tried using the trig identity $\sin A\sin B = \frac12 \cos(A-B) - \frac12 \cos(A+B)$, but when I tried to integrate that to reach an answer, I'm apparently doing it wrong somewhere. For reference, the final indefinite integral I reached was:

$\frac12\sin(-1) - \frac14\sin(2x+1)$.

I was hoping someone would know where I went wrong here.
Thanks very much!

答:
Hint

As you said, you can write

$$\sin x\sin (x+1)=\frac12 \cos (1)-\frac12 \cos(2x+1)$$

See that $\cos (1)$ is a constant number. Now, you have to solve

$$\int_{0}^{2\pi}\cos(1)dx=\cos(1)\int_{0}^{2\pi}1dx$$
$$\int_{0}^{2\pi}\cos(2x+1)dx.$$

Can you finish?


问:
Prove that the function $\cos(x)$ is continuous using the topological definition
We know that $f: (\mathbb{R}, \mathcal{T_{\mathbb{R}}}) \to (\mathbb{R}, \mathcal{T_{\mathbb{R}}})$ is continuous $\iff$ $\forall a< b \in \mathbb{R}, f^{-1}(]a,b[)\in \mathcal{T_{\mathbb{R}}} $

I'm trying to apply this to the function : $f: (\mathbb{R}, \mathcal{T_{\mathbb{R}}}) \to (\mathbb{R},\mathcal{T_{\mathbb{R}}}), x \mapsto \cos(x)$.
Let $a\lt b\in\mathbb{R}$.

if $f^{-1}(]a,b[)=\phi$ then $f^{-1}(]a,b[)\in\mathcal{T_{\mathbb{R}}}$ an open subset.

if $f^{-1}(]a,b[)\not=\phi$ then $x\in f^{-1}(]a,b[)$$\iff$$\cos(x)\in]a,b[ \in \mathcal{T_{\mathbb{R}}}$$\implies$$\exists c\lt d\in \mathbb{R}$ : $\cos(x)\in ]c,d[\subseteq ]a,b[$.

Any help Please.

答:
As a hint, you may want to use the following property of $\cos$ (which is easily proved):

For any $x, y \in \Bbb R$, we have $|\cos(x) - \cos(y)| < |x - y|$.

This is known as ($1$-)Lipschitz.


问:
If $|z| < 1$, show that $|\operatorname{Arg} \frac{1 + z}{1 − z}| < \frac\pi2$
If $|z| < 1$, show that
$$\bigg|\operatorname{Arg} \frac{1 + z}{1 − z}\bigg| < \frac\pi2$$

I know how to solve this using the properties but i wanted to try geometric methods.
But for some reasons, I got the exact opposite answer and still i can't understand where is wrong.

Here is my solution.

Just in case if the image is not clear, it says

Let $z=x+iy\in\mathbb{C}$.

$|z|=\sqrt{x^2+y^2}<1\implies x^2+y^2<1,$ so the point is in the circle where the center is $(0,0)$ and $r=1.$

$$\operatorname{Arg}\left(\frac{1+z}{1-z}\right)=\arg(1+z)-\arg(1-z)+2k\pi, \ \ k\in\mathbb{Z}$$

$\arg(1+z)$ is the angle between real axis and $(-1,0)\sim(x,y)$ (let's say A)

$\arg(1-z)$ is the angle between real axis and $(1,0)\sim (x,y)$ (let's say B)

Then I have to prove $A-B<\pi/2$

And then I draw the complex plane.

Since $(x,y)$ is in the circle, the triangle with $(1,0),(-1,0),(x,y)$ is obtuse triangle.

Thus, $A+(\pi-B)<\pi /2\implies A-B<-\pi/2,$ which seems the opposite.

I'm new to complex field/plane so it's hard to get the fallacy here.
Can you guys help me?
Thank you

答:
Here is maybe a simpler approach. Note

$$w=\frac{1+z}{1-z} = \frac{1-|z|^2 + (z-\bar z)}{|1-z|^2}
$$

Thus, $Re (w)=\frac{1-|z|^2}{|1-z|^2}>0$, which leads to
$$|\text{Arg}(w)| <\frac\pi2$$


问:
Find the domain of $f(x)$ $=$ $\sqrt{1 + ln(1- x)}$

My solution

Since the argument of ln must be greater than zero:

$1 - x > 0$

$ - x > -1$

$ x < 1 $

Then, because $f(x)$ must be a real number, we have:

$1 + ln(1-x) \ge 0$

$ln(1-x) \ge -1$

For the next result I used the follwing property: $e^{ln (x)}$ $=$ $x$

$1-x \ge e^{-1}$

Which simplifies to: $x \le 1 - e^{-1}$

Thus, option B it's the answer. Right? If so, then how can we verify that $- \infty \lt x $?

答:
You need $1+\ln(1-x)\geq 0$ in order to plug it into the square root function. Then,
\begin{equation*} 1+\ln(1-x)\geq 0 \iff \ln(1-x)\geq -1 \iff 1-x\geq e^{-1} \end{equation*}
which implies that the domain of your function is $\{x\in\mathbb{R} \mid x\leq 1-e^{-1}\}=(-\infty,1-e^{-1}]$. This equivalence of sets is the crux of what was pointed out in the first two comments.


问:
Beta function's domain
Why is the Beta function only defined for x,y > 0 ?
$$B(x,y)=\int_0^1t^{x−1}(1−t)^{y−1}dt $$

答:
Just to add to what @tommik and @Suman Chakraborty have written above.

Beta function $B(x,y)$ can be analytically continued to all complex numbers $x,y$, including negative ones, but in this case it is not presented by $\int_0^1t^{x−1}(1−t)^{y−1}dt$.

Analytical continuation of Beta function for all $x,y$ has the form
$$B(x,y)=\frac{1}{(1-e^{2 \pi i{x}} )(1-e^{2 \pi i{y}} )}\int_P t^{x-1} (1-t)^{y-1} {\rm d}x$$

where the integration goes along the Pochhammer contour (for example, look here: Confused about Pochhammer contour?)

It can be shown that analytically continued Beta function is related to the analytically continued Gamma function in its usual way: $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$

For $x,y>0$ both views on Beta-function are identical.

It can be show, for example, through integration along Pochhammer contour that for $x\in(-1,0)$ and $y>0$ $B(x,y)=\lim_{r\to0}(\int_r^1t^{x-1}(1-t)^{y-1}dt+\frac{r^x}{x})$

These representations can be useful for integrals evaluation.

Title: Beta function's domain


问:
Given $M_t$ a martingale show $E\left[\left( \left| \int_0^t M_s dB_s \right|^2 \right) \right]^{1/8} \le E[M_t^2]^{1/8} t^{1/8}$
Given $M_t$ a martingale, $\lim_{t \to \infty}E[M_t]^2=0$, and $t \le 1$, show $$E\left[\left( \left| \int_0^t M_s dB_s \right|^2 \right) \right] \le E[M_t^2] t$$

The first step is that I want to use Itô's isometry to show (I am a little unsure about the absolute value here but I think it works like this)

$$E\left[\left( \left| \int_0^t M_s dB_s \right|^2 \right) \right] = E\left[\left(  \int_0^t \left| M_s \right|^2 ds  \right) \right]$$

Now since we have the traditional integral we can move the expected value inside the integral.

$$=\left(  \int_0^t E\left[\left| M_s \right|^2\right] ds  \right) $$

If this has been correct until here, my biggest confusion is now how to evaluate this integral. Considering the desired answer it seems like we treat the integrand here as a "constant" in the traditional sense so I would just have $\int_0^t E\left[\left| M_s \right|^2\right] ds = tE[\left| M_t \right|^2] - 0E[\left| M_0 \right|^2]$ but I doubt it. Can someone help me here? Thanks.

答:
We have
\begin{align}
E\left[\left(  \int_0^t M_s dB_s  \right)^2 \right] &= E\left[\int_0^t M_s^2 ds\right]   \\
&=\int_0^t E\left[  M_s^2   \right] ds \le \int_0^t E\left[  M_t^2   \right] ds =E\left[  M_t^2   \right]t  \\
\end{align}

Here, we use the fact that $E[M_s^2] \le E[M_t^2]$, effectively, applying the Jensen's inequality
$$M_s^2 =E[M_t|\mathcal{F}_s]^2 \le E[M_t^2|\mathcal{F}_s]$$
$$\implies E[M_s^2] \le E[E[M_t^2|\mathcal{F}_s]] = E[M_t^2]$$


答: Let $f(x)$ be $\sqrt{1+ln(1-x)}$. Let $$u=1+ln(1-x)$$, so we have $$f(x)=\sqrt{u}$$, and we know that the square root function is only valid if $$u \ge 0$$, so we have $$1+ ln(1-x) \ge 0$$, which becomes $$ln(1-x) \ge -1$$, which means $$1-x \ge \frac1e$$,  thus, $$x \le 1 - \frac1e$$, or $$-\infty \lt x \le 1 - \frac1e$$

Title: Find the domain of $f(x)$ $=$ $\sqrt{1 + ln(1- x)}$

Body:
My solution

Since the argument of ln must be greater than zero:

$1 - x > 0$

$ - x > -1$

$ x < 1 $

Then, because $f(x)$ must be a real number, we have:

$1 + ln(1-x) \ge 0$

$ln(1-x) \ge -1$

For the next result I used the follwing property: $e^{ln (x)}$ $=$ $x$

$1-x \ge e^{-1}$

Which simplifies to: $x \le 1 - e^{-1}$

Thus, option B it's the answer. Right? If so, then how can we verify that $- \infty \lt x $?


问:
Vector Analysis cross product and dot product
Three vectors $\vec A,\vec B,\vec C$ Compare the value of these 4 questions.(use the parallelepiped)

a) $(\vec A\times \vec C)\cdot\vec B$

b) $\vec A\cdot(\vec C\times\vec B)$

c) $(\vec A\times\vec B)\cdot\vec C$

d) $\vec B\cdot(\hat A \times \vec A)$

Where $\hat A=\frac{1}{A}\vec A$. I solve this question but something makes me confusing.

First answer $a=b<d<c$

Second answer $a=b≠c,d=0$

But I think second answer is best answer because,
Three vectors about sides of parallelepiped can change each other. So,a),b) is volume of parallelepiped but it makes ±V and  c) is decoding order V
Which is the best answer first or second ?
please solve this question.

答:
To get some intuition it's always good to draw a picture, especially for questions like this.

This drawing show the parallelepiped formed by $\vec A,\vec B,\vec C$. The volume is $(\vec A\times\vec B)\cdot \vec C=V$. The volume is positive in my example because these three vectors follow the right hand rule: if I take the cross product of two (for example $\vec A$ and $\vec B$) then the result will point in the general direction of the third vector (vector $\vec C$ ). From this picture we can conclude that if we switch the vectors in such a way that the right hand rule is still obeyed then the volume stays the same. In other words
$$(\vec A\times\vec B)\cdot \vec C=(\vec B\times\vec C)\cdot \vec A=(\vec C\times\vec A)\cdot \vec B$$

The dot product obeys $\vec u\cdot \vec v=\vec v\cdot \vec u$ so we have
$$(\vec A\times\vec B)\cdot \vec C=\vec C\cdot(\vec A\times\vec B)$$

The cross product obeys $$\vec A\times\vec B=-\vec B\times\vec A$$

Using these relations you should be able to relate $a,b,c$ to eachother and they only differ by a minus sign.

Question d) is different. To see how this works look at the picture and imagine what happens if I move $\vec B$ until it points in the direction of $\vec A$. What happens to the volume?

OwnerDisplayName: eeeventually

OwnerDisplayName: AccidentalTaylorExpansion


问:
Compute the minimum spanning tree in hypercube $Q_{k}$
Suppose that in the hypercube $Q_{k}$, each edge whose endpoints differ in coordinate i has weight $2^{i}$. Compute the minimum weight of a spanning tree.

I know I can use Kruskal's algorithm but not sure how to keep track of not having cycles when I proceed the algorithm.

答:
Here is a general outline/hint.

It is easy to see that the first $2^{k-1}$ stages of Kruskal's will include all of the edges with weight $1$, leaving $2^{k-1}$ pairs of vertices joined by an edge. Considering the new graph $Q'_k$ who vertices are these $2^{k-1}$ pairs, where the weight of the edge joining two pairs is the minimum weight edge joining some vertex in one pair to some vertex in the other, it is easy to show that $Q_k'$ is isomorphic to $Q_{k-1}$, except all edge weights are doubled. Furthermore, completing these pairs to a spanning tree for $Q_k$ is exactly like finding a spanning tree for $Q_{k}'\cong Q_{k-1}$. Therefore, you can proceed by induction. The final spanning tree will have $2^{k-1}$ edges of weight $1$, $2^{k-2}$ edges of weight $2$, $2^{k-3}$ edges of weight $4$, etc.


问:
Subspace of $ P_3(\mathbb{R})$
Let $p_0,p_1\in P_3(\mathbb{R})$ such that $p_0(x)=5, p_1(x)=1+x$. Suppose that $U=\operatorname{span}(p_0,p_1)$. Find a subspace $W$ of $P_3(\mathbb{R})$ such that
$$ P_3(\mathbb{R}) = U \oplus W,$$ here $\oplus$ denotes direct sum of subspaces and $P_3(\mathbb{R})$ denotes Polynomials with degree at most $3.$

What I have done:

By $U=\operatorname{span}(p_0,p_1)$, we get $U=\{5c_1  + c_2 (1+x) | c_1,c_2 \in \mathbb{R}\}.$ Then I think we need such $W$ subspace of $P_3(\mathbb{R})$
$ P_3(\mathbb{R}) = U \oplus W.$
Now to show the above statement I think we need to show two things

$ P_3(\mathbb{R}) = U \oplus W$

$U$ and $W$ are linearly independent.

But what appropriate $W$ can I use here. Please help. Thank you.

答: Suppose $P_3(\Bbb R)$ is the vector space of dimension 4 (not 3), then say $W = \{c_3x^2+c_4x^3\mid c_3,c_4\in\Bbb R\}$ is the subspace you're searching.

Title: Subspace of $ P_3(\mathbb{R})$


问:
Prove that $\Delta ABC$ is an equilateral triangle if and only if the complex coordinates $a$,$b$, and $c$ satisfy a relation.

Let $\triangle ABC$ be a triangle in the complex plane and let $a$,$b$ and $c$, respectively, be the complex coordinates of its vertices.  Suppose that the tiangle is inscribed in the circle $C(0,1)$.  Prove that $\triangle ABC$ is equilateral if and only if $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=0$$ and $$(a+b)(b+c)(c+a)\neq0\,.$$

I tried to convert $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=0$ into a simpler equation by multipling it by $(a+b)(b+c)(c+a)$

In the end I obtained $$a^2+b^2+c^2+3ab+3bc+3ca=0$$ but I don't know how to continue from here.  Can you help me, please?

答:
One direction (the direction assuming that $ABC$ is an equilateral triangle) is obvious.  Here's a geometric proof of the converse.

Let $k$ be the given unit circle.  If $D$, $E$, and $F$ are the midpoints of $BC$, $CA$, and $AB$, then note that the complex coordinates of the images $D'$, $E'$, and $F'$ of $D$, $E$, and $F$ under the inversion about $k$ are $\dfrac{2}{\bar{b}+\bar{c}}$, $\dfrac{2}{\bar{c}+\bar{a}}$, and $\dfrac{2}{\bar{a}+\bar{b}}$, respectively.  Note that the origin $O$ is the incenter of the triangle $D'E'F'$.  Your equation says that the centroid of the triangle $D'E'F'$ coincides with its incenter.  This only happens when $D'E'F'$ is an equilateral triangle, whence $ABC$ is also an equilateral triangle.


问:
Express the sum of a geometric series in the form $p + q\sqrt2$
I am working through a maths text book as a hobby and am stuck on the following problem:

An arithmetic series and a geometric series have r as a common difference and the common ratio respectively. The first term of the arithmetic series is 1 and the first term of the geometric series is 2. If the fourth term of the arithmetic series is equal to the sum of the third and fourth terms of the geometric series, find the three possible values of r.
When $|r| \lt 1$  find, in the form of $p + q\sqrt2$, (i) the sum to infinity of the geometric series (ii) the sum of the first ten terms of the arithmetic series.

This is my approach to the part (i), the geometric series:

The three roots are $1, -1 + \frac{1}{2}\sqrt 2$, $ -1 - \frac{1}{2}\sqrt 2$

Of these only $|-1 + \frac{1}{2}\sqrt 2| \lt 1$

$S_{\infty} = \frac{2}{1 -(-1 + \frac{\sqrt2}{2}}$

which I reduced to $\frac{1}{1 - \sqrt\frac{1}{8}}$

The book says the answer to this is $\frac{8}{7} + \frac{2}{7}\sqrt 2$

I cannot see how this is arrived at. If I apply the binomial expansion I get:

$1 + (-1)(-\frac{1}{\sqrt8}) + \frac{(-1)(-2)(-\frac{1}{\sqrt8})}{2!} +...$ so I am obviously on the wrong track.

Body: $$\frac{1}{1-\sqrt{1/8}} = \frac{\sqrt{8}}{\sqrt{8} - 1} = \frac{\sqrt{8}( \sqrt{8} + 1)}{(\sqrt{8} - 1)(\sqrt{8} + 1)} = \frac{8 + \sqrt{8}}{8 - 1} = \frac{8}{7} + \frac{2\sqrt{2}}{7}.$$


问:
Using the Markov inequality
Let $x_k$ is some stochastic process as $k$ progresses. I have a basic question of using Markov inequality in this scenario. For example if I have some constraint

$$
Pr\{|x_k|\le \alpha\}\ge \beta
$$

How can we use Markov inequality to obtain expression in terms of expectation of $|x_k|$, $E\{|x_k|\}$?
Thank You.

答:
Let us use some intuition first. Your constraint says that $|x|$ is smaller than $\alpha$ with a probability of at least $\beta$. Can it give us any information about the expected value of $|x|$? Unfortunately, no. On the one hand, nothing prevents the case that $\mathsf E|x| = 0$ since the case $x\equiv 0$ always satisfy your constraints. On the other hand, the expectation can be as big as possible. Indeed, let us consider
$$
  y =
\begin{cases}
0,&\text{ with probability } \beta
\\
n,&\text{ with probability }1-\beta.
\end{cases}
$$
Clearly,
$
  \mathsf Ey = n(1-\beta)
$ can be any real number if $\beta<1$. As a result, unless you have a trivial case $\beta = 1$ in which case $\mathsf E|x|\leq \alpha$, you constraint does not provide any guarantees for the expectation.

The trick of Markov-inequality-like-bounds is to write the expectation as an integral, and then restrict your attention to the region that appears in the probability. But since we work with a non-negative random variable, only the bounds of the shape
$$
  \mathsf P\{|x|\geq a\}\geq b \tag{1}
$$
are useful. Or alternatively, bounds of the shape
$$
  \mathsf P\{|x|\leq c\}\leq d
$$
since you can restate them as
$$
  \mathsf P\{|x|\geq c\}\geq 1-d.
$$

Provided the constraint of the shape $(1)$ we denote
$$
  A = \{\omega\in \Omega:|x(\omega)|\geq a\}
$$
so that we know that $\mathsf P(A)\geq b$. Moreover, $|x(\omega)|\geq a$ for any $\omega\in A$. As a result:
$$
  \mathsf E|x| = \int_\Omega |x(\omega)|\mathsf P(\mathrm d\omega)\geq\int_A |x(\omega)|\mathsf P(\mathrm d\omega) \geq \int_A a\mathsf P(\mathrm d\omega) \geq a\cdot b.
$$

Bottom line: the constraints that you have provide no guarantees on the expected value. For such guarantees you need something like $(1)$.

Title: Using the Markov inequality


问:
Solving the following ODE.
I came up with the following ODE:

$\frac{dx}{dt}+\frac{dy}{dt}=k_1(a-x-y)(b-x)+k_2(a-x-y)(c-y)$

where $k_1,k_2,a,b,c$ are constants. I know under changes of variable, the equation can be changed into the following form:

$\frac{dx}{dt}+\frac{dy}{dt}=k_1(a-x-y)x+k_2(a-x-y)y$

which is simpler. But I can't go any further. Hope someone could help. Thanks!

答:
If $k_1=k_2 = k$ we have

$$
(x+y)' = k(x+y)(a-(x+y))
$$

now making $u=x+y$

$$
u'=k u(a-u)\Rightarrow x+y = \frac{a e^{a c_0+a k t}}{e^{a c_0+a k t}-1}
$$

Title: Solving the following ODE.


问:
Boundary Points of a Plane in $\mathbb{R}^3$
Why does a plane in $\mathbb{R}^3$ not possess any boundary points? Isn't the set of points of the plane equal to its boundary set? For instance, for $2x+3y-1=0$, the set of points in $y=(-2x+1)/3$ are itself, boundary points?

答: Viewed as a subspace of $\Bbb R^3$, a plane is closed and has no interior points, hence is its own boundary.


问:
Is Spivak wrong about this counterexample? $f$ integrable on $[-1,1]$, $F=\int_{-1}^xf$, $f$ differentiable at $0$, but $F'$ not continuous at $0$
Let $f$ be Riemann integrable on $[a,b]$, let $c\in(a,b)$, and let $F(x)=\int_a^xf$ for $x\in[a,b]$.

Exercise 3b in Chapter 14 of Spivak's Calculus (3rd edition) asks to give a proof or provide a counterexample of this statement:

If $f$ is differentiable at $c$, then $F'$ is continuous at $c$.

In his solutions manual, Spivak claims to give a counterexample. His function is $f:[-1,1]\to\mathbb{R}$ defined by

$$f(x)=\begin{cases}0&x=0\\1&x\in\left\{-1,1\right\}\\\frac{1}{n^2}&\frac{1}{n}\leq |x|<\frac{1}{n-1}, n\geq2\end{cases}$$

Observe that $f$ has jump discontinuities at points of the form $1/n$ for nonzero integers $n$ and is continuous everywhere else.

Now certainly $f$ is Riemann integrable on $[-1,1]$ (because its sets of discontinuities is countable, thus measure zero). Moreover $f$ is indeed differentiable at $0$.

But I don't think $F'$ is discontinuous at $0$. On the contrary, here is my proof that $F'$ is continuous at $0$.

First, since $f$ is differentiable at $0$, $f$ is continuous at $0$; it follows by the fundamental theorem that $F'$ exists at $0$ and that $F'(0)=f(0)$. We know $f(0)=0$, so $F'(0)=0$.

It remains to show that $\lim_{x\to0}F'(x)=0$. Now this is a little delicate because, as Spivak points out, $F'$ is not even defined at $1/n$ for $n$ a nonzero integer. But that doesn't prevent us from calculating the limit of $F'$. The limit simply must be taken through the domain of $F'$.

Indeed, given $\epsilon>0$, take $N$ such that $\frac{1}{N^2}<\epsilon$, and pick $\delta=\frac{1}{N}$.

Then if $0<|x|<\delta$ and $x$ is in the domain of $F'$, it follows that $|F'(x)|\leq\frac{1}{(N+1)^2}<\frac{1}{N^2}<\epsilon$.

Therefore it is indeed true that

$$\lim_{x\to0}F'(x)=F'(0)$$

so $F'$ is continuous at $0$.

What's going on here? Is this just a matter of one's definition of the limit of a function on a set that isn't an interval? Spivak seems to think "If $\lim_{x\to c}g(x)$ exists then $g$ is defined on a deleted interval around $c$."

答:
The answer to your question in the comments is yes: If $f$ is differentiable at $c,$ then $F'$ is continuous at $c$ within the domain of $F'.$

Proof: Suppose $y_n\to c^+$ (the case $y_n\to c^-$ is similar). Assume  $F'(y_n)$ exists for each $n.$ Because $f'(c)$ exists, we can write

$$\tag 1 f(x) = f(c) +f'(c)(x-c) + \epsilon(x)(x-c),$$

where $\epsilon(x)\to 0$ as $x\to c.$ Actually we don't need the $\epsilon\to 0$ property. If you divide both sides of $(1)$ by $x-c,$ you'll see $\epsilon(x)$ is bounded on $[a,b]\setminus \{c\},$ say by $M$ in absolute value.

Fix $n.$ Then for small $h>0,$

$$\tag 2 \frac{F(y_n+h)-F(y_n)}{h} = \frac{1}{h}\int_{y_n}^{y_n+h} f(x)\,dx.$$

By $(1),$ $(2)$ equals

$$\tag 3 f(c) + \frac{1}{h}\int_{y_n}^{y_n+h}(f'(c)+\epsilon(x))(x-c)\,dx.$$

Now the second expression in $(3)$ is bounded above in absolute value by

$$(|f'(c)|+M)\frac{1}{h}\int_{y_n}^{y_n+h}(x-c)\,dx = (|f'(c)|+M)(y_n-c+h).$$

Thus

$$\left |\frac{F(y_n+h)-F(y_n)}{h} - f(c)\right |\le (|f'(c)|+M)(y_n-c+h).$$

Taking the limit as $h\to 0$ then gives

$$|F'(y_n) - f(c)| \le (|f'(c)|+M)(y_n-c).$$

Now letting $n\to \infty,$ we see

$$F'(y_n) \to f(c) = F'(c),$$

and we're done.


问:
Checking if subspace
I'm struggling a bit with this problem:

Is the following set a subspace of ${R}^3$?

$V_3={(a_1,a_2,a_3): (a_2)^2=a_1-a_3}$

So I see that we have the zero-vector. But I'm having a hard time finding out if the subspace is closed under addition due to the square that confuses me. Because if I had two vectors a and b, I would do the following:

$(a_1+b_1), (a_2+b_2), (a_3+b_3)$

But I cant add $(a_2+b_2)$ because the expression I have for them both are squared. What should I do?

答:
This space is not a subspace.
Consider the two vectors  $a=(2,1,1)$ and $b=( 2,-1,1)$ both are in $v_3$
But their sum is the vector  $ (4,0,2)$ is not in $V_3$ which proves that
$V_3$ is not  a subspace  of $R^3$

Title: Checking if subspace


问:
Difference between upper and lower tails of multiplicative Chernoff bounds
I'm struggling with the intuition behind why Chernoff bounds differ in the upper and lower tails. That is, for the lower tail we have:
$$
    Pr(X \le (1 - \delta)\mu)\ \ \le\ \ e^{-\frac{\mu\delta^2}{2}}
$$
Whereas for the upper tail:
$$
    Pr(X \ge (1 + \delta)\mu)\ \ \le\ \ e^{-\frac{\mu\delta^2}{3}}
$$

Both bounds are found in roughly the same manner, so at a high level, why would they differ in value when we don't know that the distribution is necessarily asymmetric?

答:
One way to see the lack of symmetry is to consider the extreme case $\delta = 1$. Then $P(X< (1-\delta)\mu)=0$ while $P(X> (1+\delta)\mu)$ may be positive. This (weakly) suggests that the first tail probability may be smaller. And the bounds conform to that expectation: $\exp(-\mu \delta^2/2)$ is less than $\exp(-\mu \delta^2/3)$. </p>

<p>Except that you have them switched around (cf. wikipedia, lecture notes). </p>

Body:
One way to see the lack of symmetry is to consider the extreme case $\delta = 1$. Then $P(X< (1-\delta)\mu)=0$ while $P(X> (1+\delta)\mu)$ may be positive. This (weakly) suggests that the first tail probability may be smaller. And the bounds conform to that expectation: $\exp(-\mu \delta^2/2)$ is less than $\exp(-\mu \delta^2/3)$.

Except that you have them switched around (cf. wikipedia, lecture notes).


问:
Divisibility by $12$ of $\overline{abcdef}$
How can one find all six digits numbers formed only with even number that are divisible by $12$?

My attempt was to find all six digits numbers $\overline{abcdef}$ with all digits from $\{0,2,4,6,8\}$ that are divisible with $3$ or $4$. For doing this I used inclusion-exclusion principle. How can I continue?

答:
The tens and unit digits of such numbers will be $x0$, $x4$, $x8$. These are $5\cdot 3=15$ in all and five of these are $0 \pmod 3$, five $1 \pmod 3$  and five $2 \pmod 3$.

The first four digits can be written down in $4 \cdot 5^3=500$ ways, each of which is $0,1,2 \pmod 3$.

We do not need to know how many of them are which modulo $3$. Since each block of four digits can be appended with appropriate block of last two digits ($0$ with $0$, $1$ with $2$ and $2$ with $1$ modulo $3$) in $5$ valid ways, desired number of six-digit numbers is
$$500 \times 5 =2500$$


问:
Slick argument to find free group in $SO(4)$
It is well known that ${\rm SO}(3)$ contains the free group ${\rm F}_2$, so clearly ${\rm SO}(n)$ for $n>3$ does, too. However, the standard proof with ${\rm SO}(3)$  is somewhat annoying (the one sketched here). I've been able to prove in a nice way that if ${\rm SO}(4)$ contains ${\rm F}_2$, then so does ${\rm SO}(3)$, using ${\rm Spin}(4) \cong {\rm Spin}(3) \times {\rm Spin}(3)$ and a lemma about free groups.

It seems there may be a slicker argument that ${\rm SO}(4)$ contains ${\rm F}_2$ compared to ${\rm SO}(3)$, as we have more room to work, in a sense. Is there a nice way to obtain this?

答:
Here is a concrete example of a free group in $SO(4)$.
Let $p=\cos\rho+i\sin\rho$ and $q=\cos\rho+j\sin\rho$ two quaternions, where $\rho$ is a transcendental number.
The quaternions $p$ and $q$ generate a free group in the multiplicative group $H$ of quaternions.
Next we should take advantage of the fact that the quaternions are represented by $4\times4$ real matrices, with $p$ and $q$ represented by orthogonal matrices.
Details here: Dekker T. J., Decompositions of sets and spaces. III., Indag. Math 19 (1957): 104-107.