问:
Prove that $\sum \frac{i+1}{i!+1}<3$

Prove that $$\sum_{i=1}^{n}\frac{i+1}{i!+1}<3$$

My approach: If $k=1$ we can see that $$S_{1}=\sum_{k=1}^{1}\frac{k+1}{k!+1}=\frac{1+1}{1!+1}=\frac{2}{2}=1<3$$ Now, suppose that the statement is true for some $n\in \mathbb{N}$, it's to say suppose that $S_{n}=\sum_{k=1}^{n}\frac{k+1}{k!+1}<3$ is true, so we need to show that is true for $n+1$, it's to say we need to show that $S_{n+1}=\sum_{k=1}^{n+1}\frac{k+1}{k!+1}<3$ is also true.

Now, we can see that
\begin{eqnarray*}
S_{n+1}&=&\sum_{k=1}^{n+1}\frac{k+1}{k!+1}\\
&=&\sum_{k=1}^{n}\frac{k+1}{k!+1}+\sum_{k=n+1}^{n+1}\frac{k+1}{k!+1}
\end{eqnarray*}
but, how can I continue from here?

Title: Prove that $\sum \frac{i+1}{i!+1}<3$

Body:
Alternative proof without induction:

Note that $\frac{i+1}{i!+1}$ looks like $\frac{i}{i!} = \frac{1}{(i-1)!}$ and $\sum_{i\ge1}\frac{1}{(i-1)!}=e < 3$. Unfortunately $\frac{i+1}{i!+1}>\frac{i}{i!} $, but we can make some adjustment:

$$\frac{i+1}{i!+1} < \frac{i+1}{i!}=\frac{i+1}{i} \cdot \frac{1}{(i-1)!}$$

Therefore
$$\sum_{i=1}^{n}\frac{i+1}{i!+1} < \frac{1+1}{1!+1}+\frac{2+1}{2!+1}+\sum_{i=3}^\infty\frac{i+1}{i!+1} < 1+1+\sum_{i=3}^\infty\frac{i+1}{i} \cdot \frac{1}{(i-1)!}\\
< 2+ \frac{4}{3}\sum_{i\ge 3}\frac{1}{(i-1)!}=2+\frac 43\cdot (e-2) \approx 2.958 < 3. \blacksquare
$$


问:
A question about functional in normed space and Hahn Banach theory
Let $X\neq {0}$ a normed space, $x_n\in X$ a sequence in $X,x\in X$.

Assume for every $x^*\in X^*$:
$x^*(x_n)\to x^*x$.

Show that:
$\|x\|\leq \liminf_{n\to \infty} \|x_n\|$.

I tried to use Hahn Banach's corollary that says:
If $X$ is a normed space and $0\neq x\in X$, there is a functional $x^*\in X$ such that $\|x^*\|=1$ and $x^*x=\|x\|$.

We can notice that $x$ that satisfies $x^*(x_n)\to x^*x$ is unique.

Because $X\neq 0$ then the $x$ satisfying this is $\neq 0$.
Then, by the corollary mentioned above and the given information we get:
$$\|x\|=x^*x=\lim_{n\to \infty} x^*x_n.$$
Now, how can I use the fact that $\|x^*\|=1$?

答:
Hints:

$1).\ $ Since $(x_n)\rightharpoonup x,\ J(x_n)\to J(x)\in X^{**}$ where $J(x)f=f(x).$

$2).\ $ The Uniform Boundedness Principle applies to the collection $(J(x_n))$.

$3).\ 2).$ implies $(x_n)$ is bounded.

$4).\ $ Hahn-Banach gives us a $f\in X^*$ such that $f(x)=\|x\|$ and $\|f\|=1$.

$5).\ |f(x_n)|\to |f(x)|$

ContentLicense: CC BY-SA 4.0


问:
What is the radix of a base zero and 1 number system.
There are 2 questions:
1.What is the radix of a base zero number system?
2. What is the radix of a base 1 number system?</p>

答:
A Prisimetric is a vector valued function which represents a
Number in Radix of Zero.

0 denotes no value with identity, it represents other values

00 denotes the value 1.

000 denote the value 2.

0000 denotes the value 3.

A prisimetric represents a point triangulum cum basi wide [or with a wide base]

You can represent unary systems of 1 as a Prisimetric with the same notion e.g.

11 denotes the value 1.

111 denotes the value 2.

1111 denotes the value 3.

Thus they can also represent binary systems and exist within them at the same time.

See also
https://en.wikipedia.org/wiki/Repunit

ContentLicense: CC BY-SA 3.0


问:
Determine the area of the ellipse $A=\{(x,y) \in \mathbf{R}^² \mid 3x^2+4y^2\leqslant12 \}$

Determine the area of the ellipse $A=\{(x,y) \in \mathbf{R}^² \mid 3x^2+4y^2\leqslant12 \}$

I tried to use polar coordinates here, but couldn't get it to work.

If I have $x=r\cos(\theta)$ and $y=r\sin(\theta)$ I get that $$3r^2\cos^2(\theta)+4r^2\sin^2(\theta)\leqslant 12 \Longrightarrow \frac{r^2\cos^2(\theta)}{4}+\frac{r^2\sin^2(\theta)}{3} \leqslant 1$$

but I cannot get anywhere from here. Similarly, if I don't do the polar conversion I have that $$\frac{x^2}{4}+\frac{y^2}{3} \leqslant 1$$

but this also doens't seem to get me any furhter. What should I do here?

答:
Given I have seen surface integral / line integral questions from you, I am assuming you are aware of Green's theorem. Using parametrization of the ellipse centered at the origin, here is just an alternate method using Green's theorem trick -

$r(t) = (a \cos t, b \sin t)$. Here $a = 2, b = \sqrt3$. So, $\, r(t) = (2 \cos t, \sqrt3 \, \sin t)$

$r'(t) = (- 2 \sin t, \sqrt3 \cos t)$

As per Green's theorem $\int_C P \, dx + Q \, dy = \iint_D (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dA \,$ for a simple closed curve.

If we choose a vector field for line integral over our ellipse such that $Q = \frac{x}{2}, P = - \frac{y}{2}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1$

So, $A = \int_C P \, dx + Q \, dy = \displaystyle \frac{1}{2} \int_0^{2\pi} x \, dy - y \, dx = \sqrt3\int_0^{2\pi} (sin^2t + cos^2t) \,dt = 2 \sqrt3 \pi$


问:
Prove that $ S=\{0\}\cup\left(\bigcup_{n=0}^{\infty} \{\frac{1}{n}\}\right)$ is a compact set in $\mathbb{R}$.

Prove that $ S=\{0\}\cup\left(\bigcup_{n=0}^{\infty} \{\frac{1}{n}\}\right)$ is a compact set in $\mathbb{R}$, but $\bigcup_{n=0}^{\infty} \{\frac{1}{n}\}$ is not a compact set.

(Can we use definition of compact set for this problem? )

答: Hint: Suppose you have an open cover $\{U_\alpha\}$ of $S$. Now at least one $U_\beta$ contains $0$. Then because $1/n \to 0$ this means $U_\beta$ contains all but finitely elements of $S$. Thus.....</p>


答:
Let us use the corollary you mentioned.

We have $|x^*(x_n)| \le ||x^*|| \cdot |x_n|$ by definition of norm.  As $x^*(x_n) \to x^*(x)$ we have

\begin{gather}
|x| = |x^*(x)| = \lim_n  | x^*(x_n) | = \liminf_n  | x^*(x_n) | \le \\  \liminf_n||x^*|| \cdot |x_n| = \liminf_n |x_n|.
\end{gather}

Title: A question about functional in normed space and Hahn Banach theory

Body:
Let $X\neq {0}$ a normed space, $x_n\in X$ a sequence in $X,x\in X$.

Assume for every $x^*\in X^*$:
$x^*(x_n)\to x^*x$.

Show that:
$\|x\|\leq \liminf_{n\to \infty} \|x_n\|$.

I tried to use Hahn Banach's corollary that says:
If $X$ is a normed space and $0\neq x\in X$, there is a functional $x^*\in X$ such that $\|x^*\|=1$ and $x^*x=\|x\|$.

We can notice that $x$ that satisfies $x^*(x_n)\to x^*x$ is unique.

Because $X\neq 0$ then the $x$ satisfying this is $\neq 0$.
Then, by the corollary mentioned above and the given information we get:
$$\|x\|=x^*x=\lim_{n\to \infty} x^*x_n.$$
Now, how can I use the fact that $\|x^*\|=1$?


问:
Compare number of partitions.
What is more, partitions of the number $N$ into at most $k$ terms, or partitions of the number $N + k$ into exactly $k$ terms?

We consider partitions that differ only in order to be the same.

Looks like pretty easy combinatorial problem, but I have no idea that should work here.

I tried draw Young tableau, and build some bijections. As I know, there no any formula for finding number of partitions, so I need some more ideas/hints for this problem

答:
A partition of $N$ into at most $k$ terms is the same as a partition of $N$ into exactly $k$ terms where terms are allowed to be $0$.

Start with one of those partitions, and add $1$ to each term. You now have a partition of $N+k$ into exactly $k$ (strictly positive) terms. This process is easily reversible by subtracting $1$ from each term.

So there is a bijection between the partitions you ask about, and there is therefore an equal number of partitions of either type.

Title: Compare number of partitions.


问:
Matrices and power series
For any square matrix $A$, we can define $\sin A$ using the formal power series as follows:
\begin{equation}
\sin A=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}A^{2n+1}
\end{equation}
Prove or disprove: there exists a $2\times2$ real matrix $A$ such that
\begin{equation}
\sin A=\begin{bmatrix}1 & 2020 \\0& 1\end{bmatrix}
\end{equation}

答:
Does this help!?:

$$A=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \implies B=A^n=\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} \implies \sin A=\begin{bmatrix} \sin 1 & \cos 1 \\ 0 & \sin 1 \end{bmatrix}.$$

Use $$\sin A= \sum_{n=0}^{\infty}(-1)^n \frac{A^{2n+1}}{(2n+1)!}$$

Title: Matrices and power series


问:
A doubt regarding nature of $F(a)$, where $F(a)$ is the intersection of all fields containing field $F$ and element $a$.
Proof in Herstein:
Let us consider elements like $f_{0}+f_{1}a+f_{2}a^{2}+\dots f_{s}a^{s}$. Here $f_{0},f_{1}\dots f_{s}\in F$. Now consider the quotient field $U$ generated by elements like $f_{0}+f_{1}a+f_{2}a^{2}+\dots f_{s}a^{s}$ dividing one another. Proving that this is a field is elementary.
Also, $U\subset F(a)$ as every field containing $F$ and $a$ has to contain elements like $f_{0}+f_{1}a+f_{2}a^{2}+\dots f_{s}a^{s}$, and $F(a)\subset U$ as $U$ is a field, and $F(a)$ is the smallest field containing $F$ and $a$. Hence $F(a)=U$.

My question is why does $U$ have to be a quotient group? Why can't it simply be the set of all possible $f_{0}+f_{1}a+f_{2}a^{2}+\dots f_{s}a^{s}$? Even in that case, $F(a)=U$!

答:
OK, well consider the quotient field $F(x)$ of polynomials over $F$.

Certainly $x$ is an element of the field, but just polynomials in $x$ generate $F[x]\subsetneq F(x)$. The same can be said about any $a$ which isn't algebraic over $F$.

On the other hand, if $a$ is algebraic over $F$ with minimal polynomial degree $n$, then $F[a]=F(a)$, and in that case, the polynomials over $a$ are already a field.


问:
converging sequence $x_{n+1} = x_n f(x_n)$
let $f: [0,+\infty[ \rightarrow [0,1[$ a continuous fonction and ($x_n$) a sequence defined by
$$x_{n+1} = x_nf(x_n) \text{ and } x_0 = 1$$

show $x_n$ converges

calculate its limit.

What I have done so far:

$$x_{n+1}-x_n  = x_nf(x_n)-x_{n-1}f(x_{n-1}) = f(x_n)f(x_{x-1})x_{n-1} - f(x_{n-1})f(x_{n-2})x_{n-2}$$
$$=...=f(x_n)f(x_{n-1})...f(x_1)x_1 - f(x_{n-1})...f(x_1)f(x_0)x_0 = [f(x_{n-1})...f(x_1)](f(x_n)x_1-f(x_0)x_0)$$
$$ = [f(x_{n-1})...f(x_1)](f(x_n)f(1)-f(1))= [f(x_{n-1})...f(x_1)f(1)](f(x_n)-1)$$
$$0\leq f(x) < 1 \Longrightarrow x_{n+1}-x_n < 0 \Longrightarrow x_n \text{ is a decreasing sequence}$$
Furthermore $0\leq x_n$, hence $x_n$ converges to its infinimum.

for number 2 I'm not sure if the reasoning is correct:

suppose ($x_n) \longrightarrow l$ then
$$l = \lim_{n \longrightarrow \infty} x_n = \lim_{n \longrightarrow \infty} x_{n+1} = lf(l) $$
$f(l) < 1 \Longrightarrow  l=0$

答:
For part $(1)$, I would just note that $\frac{x_{n+1}}{x_n}=f(x_n)\in[0,1)$.

With $x_0=1$ (or simply $>0$), this sequence is always positive and decreasing (unless $\exists t\in \Bbb N : f(x_t)=0$, in which case the sequence immediately becomes a sequence of $0$s thereafter)

Your reasoning for part $(2)$ is absolutely correct.


问:
$4q=a^2+27b^2$ elementary proof of uniqueness
I’m trying to prove that if $q$ is prime then the representation $$4q=a^2+27b^2$$ is unique (up to sign). Any help would be appreciated.

Edit: By elementary I mean only using divisibility rules, modular arithmetic, etc.

答:
Hint:

$\Bbb Z[\omega]$, with $\omega^2+\omega+1=0$, is a UFD.

Suppose $$\frac{a^2+27b^2}{4}=\frac{c^2+27d^2}{4}\tag{1}$$ $\Bbb Z[\omega]$ is a UFD, so factor as $(1)$ $$\left(\frac{a+3b\sqrt{-3}}{2}\right)\left(\frac{a-3b\sqrt{-3}}{2}\right)=\left(\frac{c+3d\sqrt{-3}}{2}\right)\left(\frac{c-3d\sqrt{-3}}{2}\right)$$ Each factor is a prime in $\Bbb Z[\omega]$ since $N\left(\frac{a+3b\sqrt{-3}}{2}\right)=q$, therefore the factors are associates.

Can you take it from here?


问:
Group topology induced by group action
Let $G$ be a group, $X$ a topological space and $G\times X \longrightarrow X$ a group action so that $\begin{align*}
X\longrightarrow X \\
x \mapsto g\cdot x
\end{align*}$ is a homeomorphism $\forall g \in G$. Is there a coarsest topology on $G$ so that:

$G$ is a topological group

The group action is continuous

?

I have showed that the topology induced by the maps $\begin{align*} G \longrightarrow X \\
g\mapsto g\cdot x\end{align*}$ and
$\begin{align*} G \longrightarrow X \\
g\mapsto g^{-1}\cdot x\end{align*}$ $\forall x\in X$ makes inversion, left action and right action continuous on $G$.

答:
If $X$ is hausdorff and has compact and connected neighbourhoods around each point, the answer is the topology induced by the homomorphism $G\longrightarrow \text{Homeo}(X)$ defined by the group action, where $\text{Homeo}(X)$ is the homeomorphism group equipped with the compact-open-topology. To prove it:

Let $\tau$ be a topology on $G$ so that $G$ is a topological group and the group action $G\times X\longrightarrow X$ is continuous. Since $X$ is hausdorff and locally compact, the exponential object $X^X$ exists in the category $\text{Top}$. By the universal property of exponential objects $\exists!$ continuous $G \overset{\phi }{\longrightarrow} X^X$ so that

commutes. $X^X$ can be constructed as $X^X=C(X,X)$ equipped with the compact-open-topology. $\text{Homeo}(X)\subseteq X^X$. $\text{Im }\phi\subseteq\text{Homeo}(X)$. $G\overset{\phi}{\longrightarrow}\text{Homeo}(X)$ is precisely the homomorphism defined by the group action. There's a theorem stating that if $X$ is hausdorff and has compact and connected neighbourhoods around each point, $\text{Homeo}(X)$ equipped with the compact-open-topology is a topological group https://www.jstor.org/stable/pdf/30037630.pdf?refreqid=excelsior%3A0a25ff84e89d996eeaf57120097c8447. Define $\tau_i$ to be the topology on $G$ induced by $G\overset{\phi}{\longrightarrow}\text{Homeo}(X)$. Then $(G,\tau_i)$ is a topological group from the universl property of induced topology and since $\phi$ is a homomorphism. Since $(G,\tau)\overset{\phi}{\longrightarrow} \text{Homeo}(X)$ is continuous, $\tau_i\subseteq \tau$. That means $\tau_i$ is the coarsest topology so the action is continuous and $G$ is a topological group.

Title: Group topology induced by group action


问:
Is it true that every closed formula is decidable? Why?
I was wondering if every closed formula is decidable (in a complete system). Clearly a non closed formula is not decidable, since it can take some values for which it is true, and other for which it is false.
Let's assume the following definiton of decidability:

In logic, a true/false decision problem is decidable if there exists an effective method for deriving the correct answer.

Question 1

If however we add quantifier to the formula, is true that it can be either true or false, but in my opinion how can it be decidable, if I cannot prove it? (for example a conjecture still unproven).

Question 2

Given a specific and known algorithm, the question for every input does the program halts (which should be a closed question) is it decidable or not? Is the set of input a recursive set or not?

答:
The definition of a complete system is that for every sentence (closed formula), either it or its negation is a theorem. Therefore it is "decidable", in an extended sense (see below) with this procedure: simply start churning out all possible theorems, and wait for the formula or its negation to show up.

The definition of decidable is that there is an algorithm to determine whether any given sentence (closed formula) is a theorem. Hence the caveat: for this procedure to be a true algorithm (effective procedure), you need a way to mechanically generate all the axioms. Otherwise, the procedure is computatable only "relative to the set of axioms".

So for example, so-call True Arithmetic (sometimes called Full Arithmetic), the set of all sentences (closed formulas) that are true in the standard model, isn't truly decidable, because we have no way of algorithmically (i.e., computably) generating all the axioms of this system.

The notion of relative computability is well-established in computability theory (aka recursion theory). So the extended sense is pretty natural. However, it's not the standard meaning of "decidable" for theories in logic.


问:
Continuous variable defined over Rational numbers only?
Let $x(t)$ be a solution of some first order ODE, which is continuous over $t$. In this case, is the continuous $x(t)$ defined only over Rational numbers? what is the reason behind this? Please clarify me if I am wrong.</p>

答: Your question may need further clarification. However, I guess you are wondering about the following fact. If $x:\Bbb R\to\Bbb R$ is continuous and we know values of $x$ at rational points, then we know $x$ everywhere (just by taking limits thanks to the continuity of $x$). </p>


Title: $4q=a^2+27b^2$ elementary proof of uniqueness

Body:
I’m trying to prove that if $q$ is prime then the representation $$4q=a^2+27b^2$$ is unique (up to sign). Any help would be appreciated.

Edit: By elementary I mean only using divisibility rules, modular arithmetic, etc.

Body:
Comment:

$a=b=1 \rightarrow 4q=1+27=28$  ⇒ $q=7$

$b=2$, $a=4$  ⇒ $4q=16 +27\times 4$  ⇒ $q=31$

$b=2$, $a=8$  ⇒ $q=43$

$b=2$, $a=32$  ⇒ $q=283$

$b=2$, $a=256$ ⇒ $q=16411$


问:
Finding a concrete diff.eq solution of non-homogeneous eq. using homogeneous eq.
I have three solutions:
$$y_1(x)=xsin(x)+3cos(x)$$
$$y_2(x)=(x+5)sin(x)$$
$$y_2(x)=(x+4)sin(x)$$

of the diff.eq:
$$y''+p(x)y'+q(x)y=g(x)$$

and the start condition:
$$y(0)=y'(0)=0$$

I need to solve this Initial Value Problem.
Now there are a lot of long ways to solve it, looking for a short one, but without any non-homogeneous formulas, because I've learned only homogeneous stuff.

I know that for homogeneous a general solution is a linear combination of the linear independent solutions, but I don't know if it is true for non-homogeneous (I think it's not, because there is non-zero on the right side of the eq.).

I probably could find $p,q$ and $g$, but that seems to be a long way and maybe unnecessary.
How should I solve this eq, i.e find the needed coefficient without solving non-homogeneous, but transforming it to a homogeneous eq. problem ?

答:
The common term of the three solutions is $y_p(x)=x\sin x$, leaving $\sin x$ and $\cos x$ as homogeneous basis solutions. This means for the differential operator on the left side
$$
D^2+1=D^2+pD+q,
$$
so that $p=0$, $q=1$, $r=2\cos x$.

So the solution satisfying the initial condition is the identified particular solution
$$
y(x)=x\sin x.
$$

This all is looking for the most simple pattern, not the most rigorous treatment of the situation. There may be other solutions with non-constant coefficients, there may be a solution where the right side is zero,...

More rigorous contemplations:

The difference $y_3-y_2=\sin x$ is a homogeneous solution, thus any $x\sin x+c\sin x$ is also a solution.

After identifying $x\sin x$ as one of the possible inhomogenous/particular solutions, from the first equation one concludes that also $\cos x$ is a homogeous/complementary solution. Note that $x\sin x$ satisfies the initial conditions. Now as homogeneous solutions both
$$
0=-\sin x+p\cos x+q\sin x\\
0=-\cos x-p\sin x+q\cos x
$$
which as linear system has the unique solution $p=0$, $q=1$, constant for all $x$. Both homogeneous basis solutions have one non-zero value in value and derivative, so that they can not contribute to the initial conditions.


问:
Question about a proof of the Fundamental Theorem of algebra

Theorem: (Fundamental Theorem of Algebra): Given any positive integer
$n \geq 1$ and any choice of complex numbers $a_0,a_1,..., a_n$, such
that $a_n \neq 0$, the polynomial equation

$(1) \quad a_n z_n +···+ a_1 z + a_0 = 0$

has at least one solution $z \in \mathbb{C}$.

Proof extract: For our argument, we rely on the fact that the function
$|f(z)|=|a_n z^n+...+a_1 z+a_0|$ attains its minimum value. Let $z_0 \in \mathbb{C}$ be a point where the minimum is attained. We will show
that if $f(z_0) \neq 0$, then $z_0$ is not a minimum, thus proving by
contraposition that the minimum value of $|f(z)|$ is zero. Therefore,
$f(z_0) = 0$. If $f(z_0) \neq 0$, then we can define a new function $g : \mathbb{C} \to \mathbb{C}$ by setting

$g(z)=\frac{f(z+z_0)}{f(z_0)}$, for all $z \in \mathbb{C}$.

Note that $g$ is a polynomial of degree $n$, and that the minimum of
$|f(z)|$ is attained at $z_0$ if and only if the minimum of $|g|$ is
attained at $z = 0$. Moreover, it is clear that $g(0) = 1$. More
explicitly, $g$ is given by a polynomial of the form

$(2) \quad g(z)=b_n z^n+...+b_k z^k+1$

with $n \geq 1$ and $b_k \neq 0$ for some $1 \leq k \leq n$. Let
$b_k=|b_k| e^{i \theta}$, and consider $z$ of the form

$(3) \quad z=r |b_k|^{-1/k} e^{i (\pi - \theta)/k}$,

with $r>0$. For $z$ of this form we have

$g(z)=1-r^k+r^{k+1} h(r)$

where $h$ is a polynomial.

The rest of the proof is not relevant for my questions.

i) How do we know that $g(z)$ has the form given in $(2)$? We know that $g(0)=1$, so it is clear that the constant must be $1$. It is also clear that the highest power is $n$ since we basically rescale $f(z)$ by a factor of $f(z_0)$ after translating it by $z_0$. But why must $g$ have this specific form? What would go wrong if we would simply let $k=1$? I think the main point is that when doing this transformation from $f$ to $g$ some powers with non-zero coefficients $a_j$ might drop out, but this cannot happen for $a_n$.

ii) The text mentions $b_k=|b_k|e^{i \theta}$, which I think is simply the representation of the complex number $b_k$ using polar coordinates. Similarly, $(3)$ specifies a complex number $z$ by its polar coordinates. Is this understanding correct?

This is the last bit I need to understand to finally get an understanding why the Fundamental Theorem of Algebra is true.

Thanks a lot for any help!

答:
This is Argand's proof of the Fundamental Theorem of Algebra.

(1) $n=k=1$ is allowed, in which case $g(z)$ takes the form $bz+1$, which has the solution $z=-1/b$. We know $g(z)$ has the given form because $g(0)=1$, so the constant term is 1, and $k$ is simply the exponent of the lowest non-zero term other than the constant term.

(2) Yes, you are correct for point (ii).


答:
The relation $\, \sin(A)^2+\cos(A)^2=I$ also hold for matrices

So we would have to find a matrix for cosine such that $C^2=\begin{pmatrix}0&-4040\\0&0\end{pmatrix}$ which is not possible.

Title: Matrices and power series

Body:
For any square matrix $A$, we can define $\sin A$ using the formal power series as follows:
\begin{equation}
\sin A=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}A^{2n+1}
\end{equation}
Prove or disprove: there exists a $2\times2$ real matrix $A$ such that
\begin{equation}
\sin A=\begin{bmatrix}1 & 2020 \\0& 1\end{bmatrix}
\end{equation}


答:
You are right about ii).

To answer question i), notice the requirement $b_k \ne 0$, which might or might not be true for $k=1$. So, that form which is written is not a specific form of $n$th degree polynomial satisfying $g(0)=1$, it's actually equivalent equivalent to the general form. You would start from the general form $g(z) = b_n z^n + ... + b_1 z + 1$ and then you would set $k = \min\{ i \mid 1 \le i \le n, b_i \ne 0\}$.

Title: Question about a proof of the Fundamental Theorem of algebra

Body:
Theorem: (Fundamental Theorem of Algebra): Given any positive integer
$n \geq 1$ and any choice of complex numbers $a_0,a_1,..., a_n$, such
that $a_n \neq 0$, the polynomial equation

$(1) \quad a_n z_n +···+ a_1 z + a_0 = 0$

has at least one solution $z \in \mathbb{C}$.

Proof extract: For our argument, we rely on the fact that the function
$|f(z)|=|a_n z^n+...+a_1 z+a_0|$ attains its minimum value. Let $z_0 \in \mathbb{C}$ be a point where the minimum is attained. We will show
that if $f(z_0) \neq 0$, then $z_0$ is not a minimum, thus proving by
contraposition that the minimum value of $|f(z)|$ is zero. Therefore,
$f(z_0) = 0$. If $f(z_0) \neq 0$, then we can define a new function $g : \mathbb{C} \to \mathbb{C}$ by setting

$g(z)=\frac{f(z+z_0)}{f(z_0)}$, for all $z \in \mathbb{C}$.

Note that $g$ is a polynomial of degree $n$, and that the minimum of
$|f(z)|$ is attained at $z_0$ if and only if the minimum of $|g|$ is
attained at $z = 0$. Moreover, it is clear that $g(0) = 1$. More
explicitly, $g$ is given by a polynomial of the form

$(2) \quad g(z)=b_n z^n+...+b_k z^k+1$

with $n \geq 1$ and $b_k \neq 0$ for some $1 \leq k \leq n$. Let
$b_k=|b_k| e^{i \theta}$, and consider $z$ of the form

$(3) \quad z=r |b_k|^{-1/k} e^{i (\pi - \theta)/k}$,

with $r>0$. For $z$ of this form we have

$g(z)=1-r^k+r^{k+1} h(r)$

where $h$ is a polynomial.

The rest of the proof is not relevant for my questions.

i) How do we know that $g(z)$ has the form given in $(2)$? We know that $g(0)=1$, so it is clear that the constant must be $1$. It is also clear that the highest power is $n$ since we basically rescale $f(z)$ by a factor of $f(z_0)$ after translating it by $z_0$. But why must $g$ have this specific form? What would go wrong if we would simply let $k=1$? I think the main point is that when doing this transformation from $f$ to $g$ some powers with non-zero coefficients $a_j$ might drop out, but this cannot happen for $a_n$.

ii) The text mentions $b_k=|b_k|e^{i \theta}$, which I think is simply the representation of the complex number $b_k$ using polar coordinates. Similarly, $(3)$ specifies a complex number $z$ by its polar coordinates. Is this understanding correct?

This is the last bit I need to understand to finally get an understanding why the Fundamental Theorem of Algebra is true.

Thanks a lot for any help!


问:
Definition of group action.
I’m taking a topology class and the professor decides to introduce a bit of group theory. We’re getting to group action and I find myself getting a bit confused about the language used in the definition. I will try to rephrase an example I found on wikipedia, and would like to ask for correction if necessary.

The example I’m referring to is the following picture in this link:

Now the introduction of the wikipedia article says that:

In mathematics, a group action on a space is a group homomorphism of a given group into the group of transformations of the space.

I assume in this example, the given group is $C_3$ (with group operation I denote $\cdot$), the space is the set of three vertices of the triangle, and the group of transformations of the space is $S_3$ (with group operation I denote $\star$). Then the group action would be a group homomorphism $\alpha: C_3 \times S_3 \to S_3$ that satisfies the identity and compatibility axioms of group action (here I follow the actual definition of group action given in the article).

There’s an intuitive understanding of what $\alpha(\rho, s)$ would do, for $\rho \in C_3$ and $s \in S_3$. However, I find it hard to verify that $\alpha$ is indeed a group homomorphism, as there’s no “formulae” for the group operations $\cdot$ and $\star$. Specifically, given $\rho$ and $\rho’$ in $C_3$, $s$ and $s’$ in $S_3$, and let $\ast$ be the group operation of $C_3 \times S_3$, I want to check that $\alpha((\rho,s) \ast (\rho’,s’)) = \alpha(\rho,s) \star \alpha(\rho’,s’)$. Here I find myself unable to proceed (except checking by hand), since I don’t know how to express the way any of the operations $\cdot$, $\star$ or $\ast$ work (except for $\ast$, where I think of doing thing component-wise, i.e. $(\rho,s) \ast (\rho’,s’)$ would be equal to $(\rho \cdot \rho’, s \star s’)$).

Also, is saying: “The cyclic group $C_3$ consisting of the rotations by $0^o$, $120^o$ and $240^o$ acts on the set of the three vertices” not technically correct? “The set of the three vertices” is not a group of transformations, and, itself not being a group, it doesn’t have a group operation with which I can check that the action is indeed a homomorphism?

Any idea how I can understand the above?

答:
An action is a way to "using" the elements of a group (the "actors") to "move" each element of a set (the "acted upon") to one element of the same set. Therefore, an action is firstly and foremost a map $\mathcal{A}\colon G\times X \to X$. The usual pair of conditions qualifying such a map as an action are nothing else that the generalization to abstract groups of the basic properties ensured by the natural action of the "prototypical" group $\mathrm{Sym}(X)$, namely:

$Id_X(x)=x, \forall x \in X$

$(\sigma\tau)(x)=\sigma(\tau(x)), \forall \sigma,\tau\in \mathrm{Sym}(X), \forall x\in X$

You can prove that, given an action $\mathcal{A}$ as above, the position $\varphi(g)(x):=\mathcal{A}(g,x)$ defines a group homomorphism $\varphi\colon G\to \mathrm{Sym}(X)$. Conversely, given a group homomorphism $\varphi\colon G\to \mathrm{Sym}(X)$, the position $\mathcal{A}(g,x):=\varphi(g)(x)$ defines a $G$-action on $X$.

Title: Definition of group action.


问:
A point on the unit circle that is not an n-th root of unity.
Consider the set of complex numbers with absolute value 1. Some points on this set are n-th roots of unity. But there are points that aren't. Is there a closed-form expression for even one of those points which are not n-th roots of unity?</p>

答:
Since $n^\textrm{th}$ roots of unity are the values of the form $\zeta_{n,m} = e^{2\pi i m/ n}$ for integers $m,n$ where (if considering primitive roots) $\gcd(m,n) = 1$. So there are only countably many $n^\textrm{th}$ roots of unity, while the unit circle is an uncountable set.

So the question becomes can we represent any of those others in a closed form easily. And we can, we can represent any point on the unit circle as $e^{i \theta}$ so for any $\theta \neq 2\pi \frac{m}{n}$, then $e^{i\theta}$ will be a point on the unit circle that isn't a root of unity. So for example, we have $e^{ni}$ is on the unit circle but isn't a root of unity for any integer $n$.


问:
Linear subspace - closed subspace - new inner product
"Let $(H, \rho)$ be a Hilbert space, and let $\mathscr{S} \subseteq H$ be a linear subspace. Let $\mathscr{P}_{\mathscr{S}} s$ denote the orthogonal projection operator onto the closed linear subspace $\overline{\mathscr{S}}^{\rho}$ in $(H, \rho) .$ We also suppose that $\tau$ is an inner product defined over $\mathscr{S}$ such that $(\mathscr{S}, \tau)$ has structure of Hilbert space."

From this statement I would like to clarify the following:

A linear subspace is not necessarily finite dimensional space, it can be infitite.

A "closed" linear subspace is also not necesarilly finite but is dense in $H$? What is the meaning of  the $s$ in $\mathscr{P}_{\mathscr{S}}s$? What is the difference between closed and finite dimensional?

We assume a new inner product in $\mathscr{S}$, but... I thought the inner product whas inerhited by $H$.

答:
Yes, if $H$ is infinite-dimensional, it will have infinite-dimensional subspaces.

Asserting that a subset $S$ of $H$ is dense means that $\overline S=H$. So, the only subset of $H$ which is both closed and dense if $H$ itself.

Yes, $\mathscr S$ inherits an inner product from $H$, but nothing prevents you from defining a new one.


答:
You can see that the second set isn't compact, because, we can take the series in this set $ a_n=\frac{1}{n}$, which converges to 0, which isn't in this set. Therefore, according to the definition, this set isn't compact.

The first set is bounded between [0,1], and it's closed, because every series in this set which converges must converge to a number within the set. Therefore it's compact, because it's closed and bounded.

Title: Prove that $ S=\{0\}\cup\left(\bigcup_{n=0}^{\infty} \{\frac{1}{n}\}\right)$ is a compact set in $\mathbb{R}$.

Body:
Prove that $ S=\{0\}\cup\left(\bigcup_{n=0}^{\infty} \{\frac{1}{n}\}\right)$ is a compact set in $\mathbb{R}$, but $\bigcup_{n=0}^{\infty} \{\frac{1}{n}\}$ is not a compact set.

(Can we use definition of compact set for this problem? )


问:
Minimizing RSS for model with missing observations. Dummy variable vs Dropping observations
Suppose that the relationship: $Y_i = β_1 + β_2*X_i + u_i$ is being fitted and that the value of X is missing for some observations. One way of handling the missing values problem is to drop those observations. Another is to set X = 0 for the missing observations and include a dummy variable D defined to be equal to 1 if X is missing, 0 otherwise. Demonstrate that the two methods must yield the same estimates of β_1 and β_2. Write down an expression for RSS using the second approach, decompose it into the RSS for observations with X present and RSS for observations with X missing, and determine how the resulting expression is related to RSS when the missing value observations are dropped.

My attempt: Suppose we have n total observations and the first k of them are not missing, and all the rest are missing. So we have model 1 where missing x_i will be 0 and model 2 where d_i is 1 if x_i is missing, and 0 otherwise:

We can see that in RSS_1 and RSS_2 second terms are the same. And the textbook, from which I took this task says that β_1 and β_2 which are minimizing RSS for two models will be the same. Namely, this will be β_1 and β_2 which could be obtained by minimizing the first term of RSS_2, but I have no idea why. Is this because the first term of RSS_1 does not depend on x_i?

答:
In model $2$, let's call $\bar{y}_m$ the mean of the $y_i$ for which $x_i$ is missing, and similarly $RSS_m$ their residual sum of squares, while $RSS_p$ is the residual sum of squares where $x_i$ is present.  Clearly $RSS_2=RSS_p+RSS_m$.

$RSS_p=\sum\limits_\text{present} (y_i - \beta_1-\beta_2 x_i)^2$ and this is minimised when equal to $RSS_1$. We can achieve by setting $\beta_1$ and $\beta_2$ as in model $1$

$RSS_m=\sum\limits_\text{missing} (y_i - \beta_1-\beta_3)^2$ and this is minimised when $\bar{y}_m = \beta_1+\beta_3$. We can achieve this minimum for any $\beta_1$ by setting $\beta_3 = \bar{y}_m-\beta_1$

Since the minimum of a sum is at least the sum of the minima:

$RSS_2=RSS_p+RSS_m$ is minimised in model $2$ by using the $\beta_1$ and $\beta_2$ found in model $1$ together with $\beta_3 = \bar{y}_m-\beta_1$,


问:
Understanding partial expectation.
I am new to probability and statistics, I am reading an article on survival analysis and when defining the expectation of a parametric model, log-normal distribution, the author used the partial expectation. I did a research and found nothing significant, or rather nothing that provide me with a good understanding, so what is that partial expectation, definition, properties. Thankful</p>

答:
The partial expectation is a measure concerning a portion of the standard normal distribution; either on the right-hand side or on the left-hand side of the standard normal.
you can read more in this link.</p>

Title: Understanding partial expectation.


问:
Godel Escher Bach: Why is a Godel proof-pair representable in TNT?
On page 441, fundamental fact 2 asserts that:

The property of forming a proof-pair is testable in BlooP, and consequently, it is represented in TNT by some formula having two free variables.

Why is this the case? I get the fact that a BlooP program can mechanically test if a certain derivation is correct, but how can you represent that it is with a formula? If i for example say that 10 is not a prime, a program could divide all the numbers up to 10 and return that there do exist whole numbers when multiplied, gives the product 10, and therefore the statement is false.
But how can a formal system represent that?

答:
I feel like this has been asked before, but this is one of the key notions behind computability theory — there is a 1:1 primitive recursive mapping (many, in fact) between finite sequences of whole numbers and whole numbers themselves. Perhaps the simplest way — although this version isn't one-to-one, it doesn't have to be — is to repeatedly use what's known as a pairing function $\langle \cdot, \cdot\rangle:\mathbb{N}^2\mapsto\mathbb{N}$ that takes pairs of numbers to a single number, and represent the sequence $\{a_1, a_2, \ldots, a_n\}$ as $\langle n, \langle a_1, \langle a_2, \langle\ldots\langle a_{n-1}, a_n\rangle\ldots\rangle$. Since a program is, ultimately, just a finite sequence of numbers, this gives a unique number for each program. And importantly, given a number we can extract the sequence of numbers corresponding to it by using nothing but bounded loops. This is because it's straightforward to make it such that the pairing function always satisfies the condition that $\langle i,j\rangle$ is greater than both $i$ and $j$ ; this means that to extract the original $i$ and $j$ from $n=\langle i,j\rangle$ we can simply test all $i$ and $j$ between $1$ and $n$ to see what works.

For more information, you might want to learn about the Kleene Recursion Theorem; there's a lot of sometimes dauntingly dense formality, but over time you will build an intuition about these sorts of things.


问:
Defining manifold coordinates without an embedding
This is something that's been bugging me for a while now, I understand that one of the most important facts about differential geometry is that it doesn't require embeddings into higher dimensional spaces. My problem is how we are then able to meaningfully define coordinates without making reference to an external coordinate system in which our manifold is embedded.

For example, when we define a coordinate chart on the circle, we can use the projection onto the axes as: $$(x,y)\mapsto x\quad (x,y)\mapsto y \tag{1}.$$ This basically takes a circle centred at the origin and provides a four separate coordinate charts which cover the circle. This is fine, but it seems to require us to specify the $(x,y)$ coordinates in the first place. I've seen a similar construction where we begin with the polar angle $\theta$ and use this to define the coordinates in a similar way. But this again seems to require us to have the circle originally embedded in $\Bbb R^2$?

How can we provide coordinates for a simple manifold like the circle without referencing an embedding?

答: I think the problem here is that we are so used to visualizing the circle as a geometric object that we consider it (intuitively, and our intuition is correct) embedded in $\mathbb R^2.$ But, if we simply define $M=\{(x,y):x^2+y^2=1\}$ where we give $M$ the subspace topology of $\mathbb R^2$, then, with the charts as in your post, $M$ becomes a bonafide manifold in the abstract sense. The fact that we already know inuitively that $M$ is locally homeomorphic to $\mathbb R$ is something that the abstract definition of $M$ is designed to make rigorous. And the fact that $M$ is a $\textit{subset}$ of $\mathbb R^2$ is incidental. An example of an abstract manifold for which geometric intuition at least for me, does not work, wold be a Lie group, $\mathrm{GL}_n(\mathbb{R})$ for example.


问:
Is the limit definition of the derivative symmetric?
I want to show that $f'(x+h)=f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$. When I consider the secant line formed by the two points in the limit, I see no reason to dismiss the idea that the slope of the secant line also approaches the slope of the tangent line at $(x,f(x+h))$ as well. If we let $c=x+h$, substituting into the conventional definition gives $f'(x)=\lim_{c\rightarrow x}\frac{f(c)-f(x)}{c-x}=\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}=f'(c)=f'(x+h)$. My question may boil down to whether the input to the limit is reversible (is ${c\rightarrow x}$ equivalent to ${x\rightarrow c}?)$. The idea that one point is stationary and the other one is "moving" seems like an unnecessary construct. What really matters is that the distance between the points approaches zero, right?

答:
An interesting question. However, I feel you've misled yourself by using $h$ in the point and the limit. Indeed, replacing one with $k$, we see: $$f'(x+k)=\lim_{h\to 0}\frac{f(x+k+h)-f(x+k)}{h}$$ from which we observe $$\lim_{k\to 0}\bigg[\lim_{h\to 0}\frac{f(x+k+h)-f(x+k)}{h}\bigg]=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=f'(x)$$

In other words, $$\lim_{k\to 0} f'(x+k)=f'(x)$$

Using $h$ twice for two different ideas is what caused the inaccurate statement $f'(x+h)=f(x)$.


答:
Suppose that $4q=a^2+27b^2=c^2+27d^2$ and $a,b,c,d> 0$ (equals 0 is not possible).

We have $a^2d^2-b^2c^2\equiv 0 $ mod $q$.

Then $(ad-bc)(ad+bc)\equiv 0$ mod $q$.

If $ad+bc\equiv 0$ mod $q$, then by $16q^2=(a^2+27b^2)(c^2+27d^2)=(ac-27bd)^2+27(ad+bc)^2$, we must have $ad+bc=0$. This is impossible, so we have $ad-bc \equiv 0 $ mod $q$.

By $16q^2=(a^2+27b^2)(c^2+27d^2)=(ac+27bd)^2+27(ad-bc)^2$,

we have $ad-bc=0$. Thus, $a/c=b/d=t$ gives
$a^2+27b^2=t^2(c^2+27d^2)=c^2+27d^2$. Hence, $t^2=1$ and $t=\pm 1$. Since $a,b,c,d> 0$, we must have $a=c$ and $b=d$.


问:
Uniform distribution over area of a dartboard vs over distance from center
A bad player hits a dartboard represented by the unit circle with uniform probability over its area, and a good player has uniform distance distribution over [0,1].

But what's the difference between the two, if every point in the circle is some measure of distance away based on the radius?

答:
The bad player has probability $\frac34$ of being more than half the radius away from the middle.  The good player has probability $\frac12$ of being that far away.

Similarly at different points.  So the good player is likely to be better than the bad player, at least if they throw independently.


答:
A smooth manifold $M$ of dimension $d$ is locally homeomorphic to and open subset of $\mathbb{R}^d$, if $\varphi$ is a homeomorphism of a connected open set $U\subset M$ onto an open subset of $\mathbb{R}^d$ then $x_i = r_i\circ \varphi$ is called a coordinate function, and together $(U,\varphi)$ a coordinate system (here $r_i(a) = a_i$ for $a\in\mathbb{R}^d$). A smooth manifold inherits a differentiable structure $F = \{ (U_\alpha, \varphi_\alpha): \alpha\in A\}$ which is a collection of coordinate systems satisfying
\begin{align*}
a)& \bigcup_{\alpha\in A}U_\alpha = M\\
b)& \varphi_\alpha \circ \varphi_{\beta}^{-1} \text{ is smooth for all }\alpha, \beta \in A\\
c)& F\text{ is maximal with respect to $b)$; if $(U,\varphi)$ is a coordinate system such that }\varphi \circ \varphi_{\alpha}^{-1}\text{ and }\\& \varphi_{\alpha} \circ \varphi^{-1} \text{ are smooth for all $\alpha\in A$ then }(U,\varphi)\in F.
\end{align*}
This is paraphrased from Frank Warner's Foundations of Differentiable Manifolds and Lie Groups ($F$ typically has some script font, but I couldn't find an appropriate one). Here I use smooth to indicate infinite differentiability. Essentially instead of embedding the manifold in some other space, just use it's local Euclidean properties.

Title: Defining manifold coordinates without an embedding

Body:
This is something that's been bugging me for a while now, I understand that one of the most important facts about differential geometry is that it doesn't require embeddings into higher dimensional spaces. My problem is how we are then able to meaningfully define coordinates without making reference to an external coordinate system in which our manifold is embedded.

For example, when we define a coordinate chart on the circle, we can use the projection onto the axes as: $$(x,y)\mapsto x\quad (x,y)\mapsto y \tag{1}.$$ This basically takes a circle centred at the origin and provides a four separate coordinate charts which cover the circle. This is fine, but it seems to require us to specify the $(x,y)$ coordinates in the first place. I've seen a similar construction where we begin with the polar angle $\theta$ and use this to define the coordinates in a similar way. But this again seems to require us to have the circle originally embedded in $\Bbb R^2$?

How can we provide coordinates for a simple manifold like the circle without referencing an embedding?


答: The probability that the bad player lands in some region of the dart board is equal to the area of the region divided by $\pi$. So the probability that the bad player lands in the circle of radius $r$ (where $r < 1$) is $\frac{\pi r^2}{\pi} = r^2$. The probability that the good player lands in the circle of radius $r$ is $r$, since her distance to the center is uniformly distributed on $[0,1]$.

Title: Uniform distribution over area of a dartboard vs over distance from center

Body:
A bad player hits a dartboard represented by the unit circle with uniform probability over its area, and a good player has uniform distance distribution over [0,1].

But what's the difference between the two, if every point in the circle is some measure of distance away based on the radius?


问:
Self-intersection of Canonical divisors on Ruled Surface
Everything on $\Bbb{C}$.

Let $S$ be ruled surface over a curve $C$ and $K$ a canonical divisor on $S$. What is $K^2$ ?

In particular I would like to understand if $K^2=0$ when $C$ is either rational or elliptic.

Thanks.

答:
Consider the case $C$ rational. For example, we could let $S$ be the blowup of $\mathbb P^2$ at a point, which equals the Hirzebruch surface $\mathbb F_1.$ Then we know that $K_S^2=K_{\mathbb P^2}^2-1=9-1=8$ so in this case $K^2\neq0.$ (Actually, I believe we can show $K_{\mathbb F_n}^2=8$ for every Hirzebruch surface, so in this case we will have $K^2=0$ if and only if $S$ is the blowup of a minimal surface at $8$ (for $\mathbb F_n$) or $9$ (for $\mathbb P^2$) points.)

If $C$ is elliptic and $S=\mathbb P^1\times C$ or if $S$ is geometrically ruled over $C,$ then $K_S^2=8(1-g)=0.$ But again, if we take any such surface and blow up some points, we get a surface with $K_S^2<0.$ So to satisfy the desired condition, we must know that $S$ is minimal.


问:
Describe how to find a basis $\beta$ for $R^n$ and a basis $C$ for $R^m$
Let $T : R^n → R^m$ be a linear transformation. Describe how to find a basis
$\beta$

for $R^n$ and a basis $C$ for $R^m$ such that matrix for $T$ relative to $\beta$ and $C$ is an

diagonal matrix $P$ of size $m × n$.

答:
This is the well known canonical form of a matrix.

Take $\{\beta_{k+1}, \dots, \beta_n\}$ (where $0\le k \le n$) as a basis of the kernel of $T$. Then according to the rank-nullity theorem, the dimension of the image of $T$ is $k$. Take $\{\beta_1, \dots, \beta_{k}\}$ such that $$\{C_1=T(\beta_1), \dots, C_{k} = T(\beta_{k})\}$$ is a basis of the image of $T$. Complete $\{C_1, \dots , C_{k}\}$ into a basis $\{C_1, \dots C_m\}$ of $\mathbb R^m$.

The two basis are the one you're looking for.


问:
Proof Check: For a completed filtration, $\mathcal{F}_{t}^{B}$ is right continuous where $B$ is a standard Brownian motion
Let $B$ be a standard Brownian motion on $(\Omega, \mathcal{F}, \mathbb P)$ and further let $(\mathcal{F}_{t}^{B})_{t \geq 0}$ be the natural filtration associated with $B$ such that $\mathcal{F}_{t}^{B}$ for $t \geq 0$ contains all null sets. Show that the filtration is right-continuous.

My approach:

Trivially, we have $\mathcal{F}_{t}^{B}\subseteq \mathcal{F}_{t+}^{B}$.

Now for the "$\mathcal{F}_{t+}^{B}\subseteq \mathcal{F}_{t}^{B}$", we assume that this does not hold:

we choose $A \in \mathcal{F}_{t+}^{B}\setminus \mathcal{F}_{t}^{B}$ and let $N$ be the null set such that $B$ is continuous on $\overline{\Omega}:=\Omega\setminus N$

Then we can construct a sequence $(\varepsilon_{n})_{n \in \mathbb N}\subseteq(0,\infty)$ with $\varepsilon_{n}\downarrow 0$ as $n \to \infty$ such that $A$ is $B_{t+\varepsilon_{n}}-$ measurable for any $n \in \mathbb N$.

Furthermore $B$ is continuous on $A\setminus N_{A}$ where $N_{A}$ is some null set and thus since $A\setminus N_{A}$ is $B_{t+\varepsilon_{n}}-$ measurable for any $n \in \mathbb N$, we have on $A\setminus N_{A}$ that $B_{t+\varepsilon_{n}}\xrightarrow{n \to \infty} B_{t}$ and thus $A \setminus N_{A}$ must be $B_{t}$ measurable. Hence $A = (A \setminus N_{A} )\cup N_{A}$ is $B_{t}$-measurable which implies $A \in \mathcal{F}_{t}^{B}$ which contradicts the initial assumption.

Is my proof correct? Any improvements?

答:
(I'm going to abbreviate $\mathcal F^B_t$ to $\mathcal F_t$, etc.)

You need to show that
$$
E[G\mid\mathcal F_{t+}] = E[G\mid\mathcal F_{t}]\qquad\qquad(\dagger)
$$
for each bounded $\mathcal F$-measurable $G$. Once this is done, consider $A\in\mathcal F_{t+}$ and take $G=1_A$. Then ($\dagger$) implies that $1_A=E[1_A\mid\mathcal F_{t+}] =E[1_A\mid\mathcal F_t]$ a.s. Because  $\mathcal F_t$ contains all the null sets, this shows that $A$ is $\mathcal F_t$-measurable. Therefore $\mathcal F_{t+}\subset\mathcal F_t$.

The identity ($\dagger$) is a consequence of two things: (i) the (right) continuity of the paths of Brownian motion, and (ii) the stationary independent increments of the Brownian motion.

Fix $t>0$. By the monotone class theorem it's enough to show ($\dagger$) for $G$ of the form $H\cdot K_t$, where $H$ is bounded and $\mathcal F_{t}$-measurable, and
$$
K_u:=\prod_{i=1}^m f_i(B_{u+s_i}-B_u),\qquad u\ge 0,
$$
where $m$ is a positive integer, the $s_i$ are strictly positive numbers and the $f_i$ are bounded and continuous. Notice that $u\mapsto K_u$ is (a.s.) continuous, and $u\mapsto E[K_u]$ is constant. Also, $K_u$ is independent of $\mathcal F_u$ because of the independent increments mentioned before.

Now fix an event $C\in\mathcal F_{t+}$. Let $\{t_n\}$ be a strictly decreasing sequence of reals with limit $t$. Then
$$
\eqalign{
E[1_C\cdot G]
&=E[1_CHK_t]=\lim_{n\to\infty}E[1_CHK_{t_n}]\cr
&=\lim_{n\to\infty}E[1_CH]\cdot E[K_{t_n}]\cr
&=E[1_CH]\cdot E[K_{t}]\cr
&=E[1_CH]\cdot E[K_0]\cr
&=E\left[1_CH\cdot E[K_0]\right].
}
$$
(The third equality follows because $C\in \mathcal F_{t_n}$, and $K_{t_n}$ is independent of $\mathcal F_{t_n}$.)
This calculation shows that $E[G\mid\mathcal F_{t+}]=H\cdot E[K_0]$, which is  $\mathcal F_t$-measurable. Thus ($\dagger$) follows.


问:
Complete the square for $f(x) = 2x^2 + 4x - 6$
I'm studying for a math test. This is the question:

$f(x) = 2x^2 + 4x - 6$. complete the square.

This is how much I get out of the question:
$$2x^2 + 4x - 6$$
$$2(x^2 + 2x - 3)$$
$$2(x^2 + 2x + 1^2 - 1^2 - 3)$$
$$2((x + 1)^2 - 4)$$

But I get stuck here. Can someone complete it and explain?

I'm sorry i forgot a important piece off the question
it says: Complete the square and derived to get the minimum value off $f(x)$

答:
You are more-or-less done. I don't know how you are asked to leave your answer, but there are three possibilities.

You could leave it as it is: $2x^2 + 4x-6 \equiv 2[(x+1)^2-4]$.

You could expand the factor of two: $2x^2 + 4x - 6 \equiv 2(x+1)^2 - 8$.

You could take the two inside the square: $2x^2 + 4x -6 \equiv (\sqrt{2}x+\sqrt{2})^2-8.$

EDIT:

You added the fact that you then must find the minimum value of the function. Since $\operatorname{f}(x) \equiv 2(x+1)^2 - 8$ you find that $\operatorname{f}(x) \ge -8$ for all $x$. This is because $2(x+1)^2 \ge 0$ for all $x$. The function takes its minimum value when $2(x+1)^2 = 0$, i.e. when $x=-1$.


问:
Traverse all permutations of pairs, triples, etc. in a minimal number of batches?
I have a collection of test cases $t_1, \ldots, t_n$ for my software. I suspect my tests themselves have a bug, in which some of them share global state and fail if run in the right order.

I would like to find this by running all tests in some order. If the run fails, I'll employ a minimization technique to find a smallest example. Otherwise, my plan is to run the test suite in a different order until I find a failure or give up.

I would like to choose the order in which I run my tests intelligently. For example, if on the second run I run the test cases in the reverse order of the first, the following holds: for every pair of test cases $t_i$ and $t_j$, I have performed one run in which $t_i$ came before $t_j$ and one run in which $t_j$ came before $t_i$.

I would like to achieve something similar for triples of test cases, in $3! = 6$ runs. However, my own exhaustive search suggests that for $n \geq 5$ this is impossible.

What is the smallest number $k$ such that there exists $k$ permutations of $\{1, \ldots, n\}$ containing all triples in all orders between them? Is there a simple scheme for generating such permutations? Is there a scheme which tries all $m$-tuples, for each $m$?

For $n = 4$ the set of permutations $(0, 1, 2, 3), (0, 3, 2, 1), (1, 3, 0, 2), (2, 1, 0, 3), (2, 3, 0, 1), (3, 1, 2, 0)$ tries all triples. One notes that $(0, 1, 2, 3)$ is here but $(3, 2, 1, 0)$ isn't. Is it ever possible to be optimal with respect to both pairs and triples? (i.e. try all orderings of pairs with the first two permutations and all triple-orderings with the first however-many-it-takes permutations)? Is it possible to be optimal with respect to all tuple sizes simultaneously?

答:
There is an obvious greedy approximation algorithm which might do something useful, run each test case at most once every run and only  be moderately slow for modest input sizes ;-)

TL;DR: to compute the next run, repeatedly commit to running a $k$-tuple you haven't run yet (in ascending order of $k$), provided this commitment is compatible with the commitments you have made so far. Stop once the next run is uniquely determined (and reset your set of commitments).

The pedantic version:

The algorithm stores every ordering of test cases it has run. To compute the next run when wanting to cover $k$-tuples, count how many times each $k$-tuple has been run in the past (in $O(runs * n^k)$ time).

Go through the $n!/k!$ tuples in ascending order of past run-count; if the current tuple is compatible with the tuples selected so far, select it too. Compatibility between (e.g. $k = 3$) tuples $(a, b, c)$ and $(d, e, f)$ means there exists some ordering of all tests such that $a < b < c$ and $d < e < f$ where $<$ means "is run before".

To test for compatibility, draw up a graph with test cases as vertices. When selecting a tuple, e.g. $(a, b, c)$, add edges $(a, b)$ and $(b, c)$ to this graph (or in general $k - 1$ edges between adjacent elements). Compatibility with a new tuple means the graph is (still) acyclic if that tuple is selected.

The next run is a topological ordering of this graph of constraints, once this ordering is unique (i.e. its transitive closure contains $n(n-1)/2$ edges).

It will become unique (if $k > 1$): assume otherwise, e.g. that there is some pair $\{v, w\}$ with no edges between them in the transitive closure of the constraint graph. For every such constraint graph there is some $k$-tuple containing $v$, $w$ and some other test cases in an ordering compatible with the constraint graph. Since the algorithm failed to stop before iterating through all $k$-tuples, it must have iterated through the tuple with $v$ and $w$ that was compatible with the constraint graph. But then an edge between $v$ and $w$ must have been added to (the transitive closure of) the constraint graph.

I'm spitballing that you can break past-run-count ties any way you like. I'm partial to selecting a random tiebreaker ordering up front and reusing it. (This random tiebreaker also guarantees a unique solution for $k = 1$: the random tiebreaker order.)

Note that selecting a random first run and its reverse as the second run exactly follows this greedy algorithm, first for 1-tuples and then for 2-tuples.

Instead of tracking past run-counts and sorting by it, one might track a "constraint wish list" containing all $k$-tuples in ascending order of $k$. Remove each tuple from this list when you issue a run containing that tuple.  Every $n$-tuple (if compatible with the constraints selected so far) enforces a unique constraint graph so you eventually terminate with a unique next run, until you've tried all $n!$ permutations. At that point, stop and declare "no order dependence found".


问:
Why does the empty set not get a relation in a cartesian product?
As far as I understand, when $A=\{1,2\}$ and $B=\{1,2\}$, then $A \times B =\{(1,1),(1,2),(2,1),(2,2)\}$.
But $\emptyset \in A$ and $\emptyset \in B$. Are any of these valid? If not, why not?

a. $(\emptyset, 1) \in A \times B$

b. $(1, \emptyset) \in A \times B$

c. $(\emptyset, \emptyset) \in A \times B$

d. $\emptyset \in A \times B$

答: Actually, $\emptyset$ is not element of $A$ (or of $B$). The set $\emptyset$ is a subset of $A$ (and of any other set), but that's irrelevant for your question.


问:
uniformly Cauchy-sequence is uniformly convergent, proof verification
Let $D$ be any set and $(f_n:D \to \mathbb{R})_{n \in \mathbb{N}}$ such that for all $\epsilon>0$ there exist $N(\epsilon) \in \mathbb{N}$ with $|f_n(x)-f_m(x)|<\epsilon$ for all $x\in D$ and $n,m> N$. Since $f_n(x)$ is a Cauchy-sequence in $\mathbb{R}$ we can define $f(x):=\lim_{n\to \infty} f_n(x)$ for all $x \in D$.

I now want to proof that $f_n$ converges uniformly to $f$: Let $\epsilon>0$ and $N(\epsilon)$ as above. By continuity of $|\cdot|$ we get $|f_m(x)-f(x)|=|f_m(x)-\lim_{n \to \infty}f_n(x)|= \lim_{n \to \infty}|f_m(x)-f_n(x)|< \epsilon$ for all $x \in D$ and $m > N(\epsilon)$.

(I used $x_n<\epsilon$ for all $n>N$ implies $\lim_{n \to \infty} x_n<\epsilon$)

答:
Two quibbles: At the start, say "$(f_n(x))$ is Cauchy for each $x$" ("$f_n(x)$" is a number). In the second paragraph, I think your first $f_n$ should be $f_m$.

Other than that, it is correct and succinct.


问:
Convergence and continuity of a martingale
I need some help with the following:

$1$. I have a $M$ continuous martingale, how can I show that if $t_n \underset{n}{\rightarrow}t$ then:
$$
M_{t_n}\overset{L^1}{\underset{n}{\rightarrow}}M_t
$$
assuming that $\mathbb{E}\left[\underset{t\in[0,T]}{\sup}|M_t|\right]<+\infty$

$2$. Given that $M$ is square integrable how to show that
$t \mapsto \mathbb{E}\left[M_t^2\right]$ is continuous.

thanks

答:
$M_t$ is continuous, $M_{t_n}(\omega) \to M(t)(\omega)$, hence we have almost sure convergence.

Further, $|M_{t_n}| \le \sup_{t \in [0,1]} M_t = \eta$.  As $\mathbb{E} \eta < \infty$ we have $M_{t_n} \to  M_t$ in $L_1$ by dominated convergence theorem.

In general case it's not true - so I think you meant that $M$ is countinious in the second problem also.

As in previous case, it's sufficient to prove that $$\sup_{t \in [0,1]} M_{t}^2 \in L_1. \quad (*)$$

Previous version of proof of (*) (it's not correct, the correct version is above)
It follows from Jensen's inequality for conditional expectations. Indeed, we have
$\mathbb{E} (M_T | \mathcal{F}_t) = M_t$. Hence

$$ \eta =  \mathbb{E} ( M_T^2 | \mathcal{F}_t) \ge \Bigl(\mathbb{E} ( M_T | \mathcal{F}_t) \Bigr)^2 = M_t^2 \Longrightarrow \sup_{t \in [0,T]} M_{t}^2 \le \eta$$
and
$\mathbb{E} \eta = \mathbb{E} M_T^2< \infty$, because our martingale is in $L_2$.

New version of proof of (*). $X_t$ is martingale and hence $X_t^2$ is submartingale and $X_t^2 \ge 0$. Hence by Doob's maximal inequality  (see https://en.wikipedia.org/wiki/Doob%27s_martingale_inequality or
https://fabricebaudoin.wordpress.com/2012/04/10/lecture-11-doobs-martingale-maximal-inequalities/) we get

$$\mathbf{E}\left[\sup_{0 \leq s \leq T} |X_s|^p\right] \leq \left(\frac{p}{p-1}\right)^p\mathbf{E}[|X_T|^p]$$ and if $p=2$ we have
$\mathbf{E}\left[\sup_{0 \leq s \leq T} |X_s|^2\right] < \infty$.


答:
Perhaps an example might help.

Consider the definition of projective space $\mathbb R P^n$, which is the quotient space of $\mathbb R^{n+1} - \{0\}$ under the equivalence relation $p\sim q$ if there exists $r \ne 0$ such that $p=rq$. This space is not given to us by an embedding in any Euclidean space.

Nonetheless, we can work out coordinate charts for this space, and verify that the overlap maps are smooth, and so this is a manifold. These coordinate charts are as follows.

Let $[p] \in \mathbb RP^n$ be the equivalence class of $p = (p_0,...,p_n) \in \mathbb R^{n+1} - \{0\}$.

For $i=0,...,n$ let $U_i = \{[p] \mid p_i \ne 0\} \subset \mathbb R P^n$. Then the function $\phi_i : U_i \to \mathbb R^n$ defined by $\phi_i[p] = \frac{1}{|p_i|}(p_0,...,p_{i-1},p_{i+1},...,p_n)$ is well-defined, and it is a coordinate chart for $\mathbb R P^n$. Taking these all together we get an atlas of coordinate charts $\{\phi_i : U_i \to \mathbb R^n \mid 0 \le i \le n\}$.


问:
How to prove a PDE preserves mass?
My general question is: suppose you are given a PDE (possibly with boundary conditions). What does it mean to say the PDE "conserves mass"?

Specifically, if you are given the PDE
$$- \nabla \cdot (a(x) \nabla u(x)) = f(x) \quad \text{in a domain $\Omega$,}$$
$$u = 0 \quad \text{ on } \partial \Omega,$$
how would you determine whether "mass is conserved"?

My guess would be that the mass is the integral $\int_\Omega u(x) \, dx$. But if this is the case, I don't know how to prove whether it is constant or not.

Edit: In this specific problem, there is no time involved, so you can't actually use a time derivative. However, it is the starting problem of a homogenization procedure, so actually it reads
$$- \nabla \cdot (a_\epsilon(x) \nabla u_\epsilon(x)) = f(x) \quad \text{in a domain $\Omega$,}$$
$$u_\epsilon = 0 \quad \text{ on } \partial \Omega.$$

So the mass conservation might be related to the $\epsilon$, but I don't think so (because from the context I gather that determining whether the problem conserves mass is supposed to be done for a fixed $\epsilon$).

Any help is appreciated!

答:
The concept of conservation of mass (or any other meaningful quantity) is usually applied to time dependent equations, like the heat equation
$$
u_t-\Delta u=0,
$$
or the wave equation
$$
u_{tt}-\Delta u=0.
$$
For the heat equation conservation of mass (or heat, to be more precise) means that $\int_\Omega u(x,t)\,dx$ is constant (i.e.  it does not depend on $t$.)</p>

Title: How to prove a PDE preserves mass?


问:
Binomial series differential equation
Show that

$$(1 + x) g'(x) = \ell g(x)$$

where $$g(x)=\sum _{k=0}^{\infty } x^k \binom{\ell}{k}$$

Title: Binomial series differential equation

Body:
$$g(x)=\sum _{k=0}^{\infty } x^k \binom{\ell}{k}=(1+x)^{\ell}$$
$$g'(x)=\ell(1+x)^{\ell-1}$$
$$(1+x)g'(x)=\ell(1+x)(1+x)^{\ell-1}=\ell(1+x)^{\ell}=\ell g(x)$$


问:
On the positivity of solutions of the differential inequality $y'' - y < 0$
In a recent Q&A the following was demonstrated:

Let $f: [0, 1] \to \Bbb R$ be twice differentiable with $f(0) = f(1) = 0$ and $f''(x) - f(x) < 0$ for $0 < x < 1$. Then $f(x) > 0$ for $0 < x < 1$.

There is a simple proof by noticing that $f$ cannot attain its minimum at a point $c \in (0, 1)$.

I then tried a different approach. Set $h = f - f''$ and let us assume for simplicity that $f''$ is continuous, so that $h$ is continuous as well. Then $f$ is a solution of the differential equation
$$ \tag{*}
 y'' - y = -h \, .
$$
The solutions of the homogenous equation $y'' - y = 0$ are $y_1(x) = \cosh(x)$ and $y_2(x) = \sinh(x)$, and the method of variation of constants (or an educated guess) shows that the general solution of $(*)$ is
$$
 y(x) = C_1 \cosh(x) + C_2 \sinh(x) - \int_0^x \sinh(x-t) h(t) \, dt \, .
$$
Together with the conditions $f(0) = f(1) = 0$ we get that
$$ \tag{**}
 \boxed{f(x) = \frac{\sinh(x)}{\sinh(1)} \int_0^1 \sinh(1-t) h(t)\, dt - \int_0^x \sinh(x-t) h(t) \, dt \, . \, }
$$
So the following is true:

Let $h: [0, 1] \to \Bbb R $ be continuous and positive, and define $f: [0, 1] \to \Bbb R$ via $(**)$. Then $f(x) > 0$ for $x \in (0, 1)$

and I wonder if this can be seen directly, without going back the differential inequality $f'' < f$.

In $(**)$ we have the difference of two positive terms (if $0 < x < 1$), so it is not immediately obvious that $f(x) > 0$. If we split the first integral in two parts then we get
$$
 f(x) = \int_0^x \left( \frac{\sinh(x)}{\sinh(1)} \sinh(1-t) - \sinh(x-t)\right) h(t) \, dt
 + \frac{\sinh(x)}{\sinh(1)} \int_x^1 \sinh(1-t) h(t)\, dt
$$
but now it is not obvious (at least not to me) that the first integral is positive for $0< x < 1$.

So my question is:

Given that $h$ is continuous and positive on $[0, 1]$, can $(**)$ be written in a form which allows to conclude that $f(x) > 0$ for $0 < x < 1$?

This would provide an alternative (albeit more complicated) proof of the initial statement.

答:
There are trig-like identities for hyperbolic functions:

$$\sinh(x-t)=\sinh(x)\cosh(t)-\cosh(x)\sinh(t)$$

$$\sinh(1-t)=\sinh(1)\cosh(t)-\cosh(1)\sinh(t)$$

Hence \begin{align}\frac{\sinh(x)}{\sinh(1)} \sinh(1-t) - \sinh(x-t)&=\sinh(x)\cosh(t)-\coth(1)\sinh(x)\sinh(t)\\
&\quad-\sinh(x)\cosh(t)+\cosh(x)\sinh(t)\\[1mm]
&=\sinh(t)(\cosh(x)-\coth(1)\sinh(x))\ge0\end{align} since $\cosh(x)-\coth(1)\sinh(x)=\cosh(x)(1-\tanh(x)/\tanh(1))\ge0$ for $0\le x\le1$.


答:
The issue here is that a circle is most naturally a subspace of $\mathbb{R}^n$, usually $\mathbb{R}^2$.

But we can choose other ways to think about a circle that don't describe it as an embedded submanifold of $\mathbb{R}^n$. For example, $[0,1]$ with the points $0$ and $1$ identified is a circle. The quotient $\mathbb{R}/\mathbb{Z}$ is also a circle. Each description of a circle has its uses, but we usually choose the embedding into $\mathbb{R}^2$ for visualization purposes.

The moral is that some manifolds in some contexts most naturally present as an embedded submanifold of $\mathbb{R}^n$, but they also exist as abstract manifolds (a topological space plus smoothly compatible charts).


问:
Castelnuovo-Mumford regularity over different rings
Let $S = k[x_1, \ldots, x_n, t]$ be the polynomial ring in $n+1$ variables over a field $k$ and let $R = k[x_1, \ldots, x_n]$.
I have stumbled upon the following definition/result.

Let $\{f_1, \ldots, f_r \} \subset S$ be a set of homogeneous polynomials which have finitely many (projective) solutions over the algebraic closure of $k$, and let $J = \langle f_1, \ldots, f_r \rangle.$ Assume that $t \nmid 0$ mod $J$. The ideal
\begin{equation}
H = J + (t)/(t) \subset S/(t) \cong R
\end{equation}
is called an $\textbf{Artinian reduction}$ of $J$. In this case, we have $\mathrm{reg}_S(J) = \mathrm{reg}_{S/(t)}(H)$.

Now, the only  result I know of that looks similar to this is the Independence Theorem of Local Cohomology (Thm. 13.1.6 in Brodmann & Sharps Local Cohomology)

Theorem: Let $S,R$ be graded rings , $I$ an ideal of $S$, $M$ a graded $R$-module and let $f: S \to R$ be a homogeneous homomorphism. Then $H_I^{i}(M) \cong H_{f(I)R}^{i}(M)$ for all $i \in \mathbb{Z}_{\geq 0}$

I tried to take $f:S \to S/(t)$ as the homomorphism setting $t = 0$ and applying the theorem above. But I
don't see, how the conditions in the definition can be used to show the equality of regularity.
What I was wondering, is, if the conditions in the definition of Artinian reduction are necessary for the result that  $\mathrm{reg}_S(J) = \mathrm{reg}_{S/(t)}(H)$, or if this could be true for all modules $H$ of the form $H = J + (t)/(t) \subset S/(t)$?

Am I looking in the right direction with the independence theorem, or are there other results I'd have to consider?

答:
Consider the exact sequences
$$0\to J\to S\to S/J\to 0,$$
$$0\to\frac{J+(t)}{(t)}\to\frac{S}{(t)}\to \frac{S}{J+(t)}\to0,$$

and
$$0\to \bar S(-1)\xrightarrow{t} \frac SJ\to\frac{S}{J+(t)}\to0,$$
where $\bar S=S/(J:t)$.

Since $t$ is a nonzerodivisor on $S/J$, $\bar S=S/(J:t)=S/J$. Furthermore, $S= k[x_1, \ldots, x_n,t]$, so every local cohomology of $S$ except the $n+1$-th one vanishes. Similarly, every local cohomology of $R=S/(t)$ except the $n$-th one vanishes. So for most values of $i$, we have the isomorphisms $$\newcommand{\m}{\mathfrak m}H^i_{\m}(S/J)\cong H^{i+1}_{\m}(J), \,H^i_{\m}(S/J+(t))\cong H^{i+1}_{\m}(J+(t)/(t)).$$ (I will use the same letter to denote the maximal homogeneous ideals. Thanks to the independence theorem, the distinction doesn't matter.) Moreover, from the third exact sequence, we get the corresponding long exact sequence of local cohomologies
$$H^{i-1}_\m(S/J+(t))\to H^{i}_\m(S/J(-1))\to H^{i}_\m(S/J)\to H^{i}_\m(S/J+(t)).$$
Let $m\geq 0$ be an integer and assume that $J+(t)/(t)$ is $m$-regular. Then $H^{i}_{\m}(J+(t)/(t))_d=0$ for all $d\geq m-i+1$. So $$H^{i-1}_{\m}(S/J+(t))_d=0.$$ Since $d\geq m-(i+1)+1$, we also get $$H^i_{\m}(S/J+(t))_d\cong H^{i+1}_{\m}(J+(t)/(t))_d=0.$$
Then the long exact sequence yields an isomorphism $$H^{i}_\m(S/J(-1))_d=H^{i}_\m(S/J)_{d-1}\cong H^{i}_\m(S/J)_d.$$
But $H^{i}_\m(S/J)$ is Artinian, so $H^{i}_\m(S/J)_e=0$ for $e\gg 0$, so iterating the above isomorphism, we see that $H^{i}_\m(S/J)_{d-1}$ itself is $0$. This in turn implies that $H^{i+1}_{\m}(J)_{d-1}=0$ for all $d-1\geq m-i$.

To conclude that $J$ is $m$-regular, we still have to prove the cases that are left out. I hope the reader can take it from here (see here for some details). Finally, one has to prove the converse as well, i.e., $J$ is $m$-regular implies $J+(t)/(t)$ is $m$-regular. A similar argument should work.


问:
Integrable submartingale
Suppose $X_n$ is a uniformly integrable submartingale and $T$ a stopping time. I wanna show that $X_T$ is integrable. To this end, I write $X_T=\lim_{n\to \infty}\sum_{k=1}^n X_k \mathbb{1}_{\{T=k\}}$. Now I would like to apply the dominated convergence theorem, but I  can't find a clearly integrable function that would upper bound each of the partial sums. Perhaps $\sup_n X_n^+$: but is the uniform integrability of $(X_n)$ enough to guarantee that $\sup_n X_n^+$ is integrable?

答:
Note that by uniform integrability, there exists a random variable $X_\infty$ such that $X_n\to X_\infty$ almost surely and in $L^1$. Moreover, for each non-negative integers $\ell$ and $k$, the submartingale property gives
$$
\mathbb E\left[\lvert X_T\rvert\mathbf{1}_{\{T=k\}}\right]=\mathbb E\left[\lvert X_k\rvert\mathbf{1}_{\{T=k\}}\right]\leqslant \mathbb E\left[\lvert X_{k+\ell}\rvert\mathbf{1}_{\{T=k\}}\right]
$$
and the convergence of $X_{k+\ell}$ to $X_\infty$ in $L^1$ shows that
$$
\mathbb E\left[\lvert X_T\rvert\mathbf{1}_{\{T=k\}}\right]\leqslant \mathbb E\left[\lvert X_\infty\rvert\mathbf{1}_{\{T=k\}}\right].$$
Summing all the inequalities over $k$ and using integrability of $X_\infty$ finishes the proof.

Title: Integrable submartingale


问:
What change of variables is this?
I have come across this integral:
$$\int \frac{f(x+tv)-f(x)}{t}g(x)dx = \int -\frac{g(x)-g(x-tv)}{t}f(x)dx$$

which the author claims is justified by a change of variables, but I cannot see what they did. Would anyone be able to elaborate?

答:
This is only correct in the limit as $t \to 0$ for definite integrals, and that's how it is used in the reference you are giving.

For general indefinite integrals and without the limit it's false. Consider for example the case $f(x) = x, \, g(x) = x^2$. The left hand side does not depend on $t$ in that case, but the right hand side is linear in $t$.

Title: What change of variables is this?


答:
I think he is breaking the difference apart and doing a change of variable in one term only, then recombining.

$$\int_{\mathbb R^n}\frac{f(x+tv) - f(x)}{t}\eta(x)\;dx = \int_{\mathbb R^n}\frac{f(x+tv)}{t}\eta(x)\;dx - \int_{\mathbb R^n}\frac{ f(x)}{t}\eta(x)\;dx$$
$$= \int_{\mathbb R^n}\frac{f(x)}{t}\eta(x-tv)\;dx - \int_{\mathbb R^n}\frac{f(x)}{t}\eta(x)\;dx$$
$$= \int_{\mathbb R^n}f(x)\frac{\eta(x-tv)-\eta(x)}{t}\;dx $$

Title: What change of variables is this?

Body:
I have come across this integral:
$$\int \frac{f(x+tv)-f(x)}{t}g(x)dx = \int -\frac{g(x)-g(x-tv)}{t}f(x)dx$$

which the author claims is justified by a change of variables, but I cannot see what they did. Would anyone be able to elaborate?


问:
Please help me understand the solution of this question!
Question : Six bells commence tooling together and toll at intervals 2,4,6,8,10,12 minutes.In 30 hours,how many times do they toll together?

I found the LCM of 2,4,6,8,10,12 minutes in order to find the minimum minutes after they will tool together.

LCM (2,4,6,8,10,12)=120

It means they will toll together after 2 hours that is 120 mins.So in in 30 hours they should too together 15 times( 30/2).

But the answer provided in the textbook is 16 times.Can you please explain how they got 16?

答: You have committed a fencepost (or Obi-wan) error. The question implies that the bells ring together at the very start of the $30$-hour period; $15$ periods of $120$ minutes pass, and then they ring together again at the very end of the $30$ hours. So they ring together $15+1=16$ times.


问:
How to find all possible pairwise combinations of binary vectors where second vector conditionally depend on the first vector?
I want to find all the possible pairwise combination where the 1 in 2nd vector is conditioned upon where 1 is located in the first vector of the pair. If in the first vector, there is a 1 in the 2nd position, then in the second vector, 1 can be either in 2nd or 3rd position but NOT IN first.

An acceptable output for two binary vectors of length 3 each. All vectors consists of a single 1 and rest are 0:

```<code>        [,1] [,2] [,3]
  [1,]    1    0    0
  [2,]    0    1    0

  [1,]    1    0    0
  [2,]    0    0    1

  [1,]    0    1    0
  [2,]    0    0    1

  [1,]    1    0    0
  [2,]    1    0    0

  [1,]    0    1    0
  [2,]    0    1    0

  [1,]    0    0    1
  [2,]    0    0    1
</code>```

Is there any elegant way to do this?

答:
Any valid list of $n$ vectors, each with length $k$, can be represented by a list of $k$ nonnegative integers whose sum is $n$. Each entry in the list represents the number of ones in its corresponding column. For example,

```<code>        [,1] [,2] [,3]
  [1,]    1    0    0       --->      (1, 1, 0)
  [2,]    0    1    0

  [1,]    0    0    1       --->      (0, 0, 2)
  [2,]    0    0    1
</code>```

So, you just need to list all vectors of length $k$ summing to $n$. You can easily doing this recursively, by making all possible choices for the first entry, then recursively filling the rest. After the following pseudo code is, the collection allVectors will contain all possible vectors.

```<code>set L to a list of length k (indexed from 1 to k)
set allVectors to an empty collection

procedure listAllVectors(sumOfRemainingEntries, currentIndex):
    if currentIndex = k:
        set L[currentIndex] to sumOfRemainingEntries
        add L to allVectors
    else:
        for i in {0, 1, ... , n}:
            set L[currentIndex] to i
            do listAllVectors(sumOfEntries - i, currentIndex + 1)

do listAllVectors(n, 1)
</code>```

Body:
Any valid list of $n$ vectors, each with length $k$, can be represented by a list of $k$ nonnegative integers whose sum is $n$. Each entry in the list represents the number of ones in its corresponding column. For example,

        [,1] [,2] [,3]
  [1,]    1    0    0       --->      (1, 1, 0)
  [2,]    0    1    0

  [1,]    0    0    1       --->      (0, 0, 2)
  [2,]    0    0    1

So, you just need to list all vectors of length $k$ summing to $n$. You can easily doing this recursively, by making all possible choices for the first entry, then recursively filling the rest. After the following pseudo code is, the collection allVectors will contain all possible vectors.

set L to a list of length k (indexed from 1 to k)
set allVectors to an empty collection

procedure listAllVectors(sumOfRemainingEntries, currentIndex):
    if currentIndex = k:
        set L[currentIndex] to sumOfRemainingEntries
        add L to allVectors
    else:
        for i in {0, 1, ... , n}:
            set L[currentIndex] to i
            do listAllVectors(sumOfEntries - i, currentIndex + 1)

do listAllVectors(n, 1)


问:
Prove that $\Re(\rho)=1/2$
Given $$\frac{\mid \int_{x=1}^\infty(\frac{1-x+[x]}{x^2})x^{1-\bar{\rho}} dx\mid}{  \mid \int_{x=1}^\infty(\frac{1-x+[x]}{x^2})x^\rho dx\mid  }=1$$ where [x] denotes the greatest integer function and $\rho$ and $\bar{\rho}$ are complex constants such that $0<\Re(\rho)<1$. Prove that $\Re(\rho)=1/2$.

My try: Define $$\phi(x)= \frac{1-x+[x]}{x^2},\quad f(x)= x^{1-\bar{\rho}}\quad\text{and}\quad g(x)= x^{{\rho}},$$
then $\phi(x)\geq 0$.

And f(x) and g(x) are continuous in $[1,\infty)$. So by mean value theorem for improper integrals there exist some c$\in(1,\infty)$ such that
$$\frac{\mid \int_{x=1}^\infty(\frac{1-x+[x]}{x^2})x^{1-\bar{\rho}} dx\mid}{  \mid \int_{x=1}^\infty(\frac{1-x+[x]}{x^2})x^\rho dx\mid  }=\mid\frac{c^{1-\bar{\rho}}}{c^{\rho}}\mid=1$$
$$\implies c^{1-2\Re(\rho)}=1$$
$$\implies(1-2\Re(\rho))\log c=0$$
$c>1$, gives $$\Re(\rho)=1/2.$$

答:
It is not true. For $\Re(s) > 1$ and by analytic continuation for $\Re(s) >0$
$$F(s)=\int_1^\infty (1-x+\lfloor x\rfloor)x^{-s-1}dx=\frac{\zeta(s)}{s}-\frac1{s-1}+\frac1s$$

Then $|F(1/4)/F(1-1/4)| > 1$ and $|F(1/4+2i)/F(1-1/4-2i)| < 1$ thus any curve $1/4\to 1/4+2i$ passes through a point where $|F(s)/F(1-s)|=1$.

Title: Prove that $\Re(\rho)=1/2$


问:
How to show that $\frac{1}{\sqrt{1-x^4-x^8}}$ is differentiable on its entire domain?
I calculated the derivative of $$f(x)=\frac{1}{\sqrt{1-x^4-x^8}}$$ and was wondering how to proof that $f(x)$ is differentiable on its entire domain: $$\left(-\sqrt[4]{(-1+\sqrt 5)/2},\sqrt[4]{(-1+\sqrt 5)/2}\right)$$

答:
$$f'(x)=\frac{2x^3 \left(1+2 x^4\right)}{\left(1-x^4-x^8\right)^{3/2}}$$
has the same domain as $f(x)$, thus the function is differentiable on all its domain.

Edit

Hot to find the derivative

$$f(x)=\frac{1}{\sqrt{1-x^4-x^8}}$$
chain rule $(f(g(x))'=f'(g(x))\cdot g'(x)$

Derivative of $u^{\alpha}$ is $\alpha u^{\alpha -1}$

Now $f(x)=(1-x^4-x^8)^{-1/2}$

then
$$f'(x)=-\frac{1}{2}\,(1-x^4-x^8)^{-1/2-1}\cdot (-4x^4-8x^7)=$$
$$=\frac{2x^4+4x^7}{(1-x^4-x^8)^{3/2}}$$


答:
You have a typo, the ranges should be $1+\sqrt 5$, not $1-\sqrt 5$.

Well: $$\bigg(\frac 1g\bigg)'=\frac{-g'}{g^2}\implies f'(x)=\frac{-(\sqrt{1-x^4-x^8})'}{1-x^4-x^8}=\frac{4x^3+8x^7}{(1-x^4-x^8)^\frac 32}$$ by the chain rule. It is differentiable wherever this function is defined, which is whenever $1-x^4-x^8\neq 0$

The solutions to this equation are the end points of your interval, but since the interval is open, it doesn't contain them and hence the function is defined across that interval, so it is differentiable.

Title: How to show that $\frac{1}{\sqrt{1-x^4-x^8}}$ is differentiable on its entire domain?

Body: I calculated the derivative of $$f(x)=\frac{1}{\sqrt{1-x^4-x^8}}$$ and was wondering how to proof that $f(x)$ is differentiable on its entire domain: $$\left(-\sqrt[4]{(-1+\sqrt 5)/2},\sqrt[4]{(-1+\sqrt 5)/2}\right)$$


问:
Range of values of parameter $\gamma$ in perturbation matrix
I have an orthogonal matrix $A \in \mathbb{R}^{10^{5}\times 10^{5}}$ and a perturbation matrix $\delta A = \epsilon A, \epsilon = 10^{-16}.$How can I calculate the range of values of parameter $\gamma \in \mathbb{R}^{+}$ such that the relative perturbation error to be less than $10^{-7}$?.Given that $||b||_{2}=1$ and $\delta b = [\epsilon,0,\dots,0,\epsilon]^{T}$ and $$\frac{||x + \delta x ||}{ ||x||} \leq \frac{1+\gamma K(A)}{1- \gamma K(A)} $$

where $K(A) = ||A|| * ||A^{-1}||$.

答:
I feel like I'm missing something here but supposing $\kappa_{2}(A) = \|A \|_{2}\|A^{-1}\|_{2}$  and $A$ is orthogonal then $\kappa_{2}(A) = 1$ and your formula becomes

$$ \frac{1+\gamma}{1-\gamma} $$

Now you should realize the left-hand side is non-negative since $\frac{\| x + \delta x \|}{\|x\|}$ is strictly positive since $\| x\|$ can't be $0$. In that case we know that it's $0$ when $\gamma = -1$ and undefined when $\gamma = 1$. If you solve from when the expression is equal we have

$$ \frac{1+\gamma}{1-\gamma} = 10^{-7} $$

we get

$$ (1+\gamma)(10000000) = 1-\gamma  $$

$$ 10000001\gamma = -9999999  \implies \gamma  = \frac{-9999999}{10000001}$$

So the values that work are where $\gamma \in (-1, \frac{-9999999}{10000001})$.


答: If $f(x)=\frac1{g(x)}$ then $$f'(x)=-\frac{g'(x)}{g^2(x)}$$ so that $f$ is differentiable at all points $x$ such that g(x) is differentiable and $g'(x)\neq0$.  In this case, we have $g(x)=\sqrt{1-x^4-x^8},$ which is defined when $1-x^4-x^8\geq0$ and differentiable when $1-x^4-x^8>0$.  Since the domain of $g$ has been chosen so that $1-x^4-x^8>0,$ the assertion follows.


问:
Find maximum value $|z_1+z_2| + |z_1-z_2|$, with $|z_1|=|z_2|=1$
Let $z_1$ and $z_2$ be two complex numbers such that $|z_1|=|z_2|=1$. Then find the max value of
$$|z_1+z_2| + |z_1-z_2|$$

In my opinion the answer should be 4, because the maximum values of both terms is 2. The given answer is $2\sqrt 2$, though. I just want to verify whether I am making a mistake or the given answer is wrong.

答:
You cannot take the maximum value as $4$ because the terms are not independent(try to find when the equality occurs (it doesnt!)).

As ${(a+b)}^2\le 2(a^2+b^2)\iff {(a-b)}^2\ge 0$ $${(|z_1+z_2|+|z_1-z_2|)}^2\le 2(|z_1+z_2|^2+|z_1-z_2|^2)=8$$ $$\Rightarrow |z_1+z_2|+|z_1-z_2|\le 2\sqrt{2}$$


答:
Note that if $f,g\colon[-1,1]\longrightarrow\Bbb R$ are defined by $f(x)=x$ and $g(x)=-x$, then $\max f=\max g=1$, but $\max(f+g)=1$ too, not $2$.

The answer that you were given is correct. let $\omega=\frac wz$ and take $\alpha\in\Bbb R$ such that $e^{i\alpha}=\frac wz$. Then\begin{align}|z+w|+|z-w|&=\left|1+\frac wz\right|+\left|1-\frac wz\right|\text{ (because $|z|=1$)}\\&=|1+\cos\alpha+i\sin\alpha|+|1-\cos\alpha-i\sin\alpha|\\&=\sqrt{2+2\cos\alpha}+\sqrt{2-2\cos\alpha}\\&=2\left|\cos\frac\alpha2\right|+2\left|\sin\frac\alpha2\right|\\&\leqslant2\sqrt2.\end{align}

Title: Find maximum value $|z_1+z_2| + |z_1-z_2|$, with $|z_1|=|z_2|=1$

Body:
Let $z_1$ and $z_2$ be two complex numbers such that $|z_1|=|z_2|=1$. Then find the max value of
$$|z_1+z_2| + |z_1-z_2|$$

In my opinion the answer should be 4, because the maximum values of both terms is 2. The given answer is $2\sqrt 2$, though. I just want to verify whether I am making a mistake or the given answer is wrong.


问:
Example 7.2.19 from Casella & Berger (EM algorithm)
$X_1, ..., X_n \sim Poi(\tau_i), Y_1, ..., Y_n \sim Poi(\beta\tau_i).$ $X$ and $Y $are mutually independent. However, $X_1$ is missing. I would like to use the EM algorithm to estimate $\beta$ and $\tau_i$. The expected complete-data log likelihood is shown on the top of page 329 in the red box

I do not understand why $\beta^{(r+1)} = \frac{\sum_{i=1}^{n}y_i}{\tau_1^{(r)}+\sum_{i=2}^{n}x_i}$. I differentiated the expected complete-data log likelihood function and got $\beta^{(r+1)} = \frac{\sum_{i=1}^{n}y_i}{\sum_{i=1}^{n}\tau_i}$. I am pretty new to EM algorithm and I don't known where I got wrong. Thanks a lot to anyone who can give me a hint.

答:
I am trying to solve this EM problem, and I got the exactly same answer as yours. So, I believe we are correct and the answer in the book is a typo. My answer is

$\beta^{(r+1)} = \frac{\Sigma_{i=1}^{n}y_i}{\Sigma_{i=1}^{n}\tau_i^{(r)}}$

$\tau_1^{(r+1)} = \frac{y_1+\tau_1^{(r)}}{1+\beta^{(r+1)}}$

$\tau_i^{(r+1)} = \frac{y_i+x_i}{1+\beta^{(r+1)} }$ for $i=2,3,\ldots,n$


问:
Question regarding the definition of Local Connectedness
In Munker's Book "Topology 2nd Edition" a Topological Space $X$ is said to be locally connected at $x\in X$ if "every neighborhood of $x$ contains a connected neighborhood of $x$".
That is given $V$ is a neighborhood of $x$ there is a connected set $U\subseteq X$ (which is not necessarily open in $X$) and a subset $W\subseteq U$ such that $W$ is open in $X$ and $x\in W\subseteq U\subseteq V$.

In wikipedia they give a different definition:
A space $X$ is said to be locally connected at $x\in X$  if for every open set $V$ containing $x$ there is a connected open set $U$ with $x\in U\subseteq V$.

The big difference is that the definition in Wikipedia assures that if $X$ is locally connected at $x$ then every neighborhood of $x\in X$ contains an open connected neighborhood of $x$ while the definition in the book only assures every neighborhood of $x$ contains some but not necessarily open connected neighborhood of $x$.

Obviously the Wikipedia definition implies the book's definition but I don't see whether the book's definition necessarily implies the Wikipedia definition..

答: Munkres uses the term "neighborhood of x" to mean "open subset containing x" (see page 96).</p>


答:
Geometrically, $z_1$ and $z_2$ are on a unit circle. Assume they form a angle $2\theta$, then
$$|z_1+z_2| + |z_1-z_2|=2 \sin\theta +2\cos\theta=2\sqrt2\sin(\theta+\frac\pi4)\le 2\sqrt2
$$
with the max value at the right angle subtended by $z_1$ and $z_2$ to the center.


问:
Absolute value in solution of first order linear differential equation
Let’s solve the first order linear equation $ y’ + {1\over{x}}y = 0$.

$ y’/y = -1/x $

$ ln|y| = -ln|x| + C $

$ |y| = {C\over{|x|}} $

So it seems to me, I have to choose every y that equals to $C/|x|$ if I get the absolute value of y.

Both $C/|x|$ and $C/x$ seems to satisfy this requirement. However, my textbook only has $C/x$ as the solution.

Did I do something wrong?

Or is it a convention to just ignore the absolute value? If so, why?

答:
The equation is defined either on the domain with $x>0$ or the one with $x<0$.

$y=0$ is a solution. You need to discuss this case first as to avoid division by zero.

For $x\ne 0$ the equation is Lipschitz in $y$, thus all solutions are unique.

In total, each solution is constrained to the quadrant that contains the initial point, the sign of $x$ and $y(x)$ is constant. Thus you can push the signs as constant factors into the constants.

Of course you could also notice that $(xy)'=xy'+y=0$ so that $xy(x)=C$, which removes any sign discussions.


问:
Is the power set of a subset of rational numbers denumerable? Could you please prove it?
The power set of given subset A of rational numbers is denumerable? How can I prove it?

A = {x ∈ Q : 2 ≤ x ≤ 5}

答: Any infinite subset of the rationals is countably infinite, so can be put in bijection with the naturals.  The power set of the naturals is not countable.</p>


答:
As $\mathbb Q$ is dense, $A$ is countable infinite. According to Cantor theorem, the cardinality of the power set of a set $S$ is greater than the cardinality of $S$.

The cardinality of the power set of $A$ is therefore not countable.

Title: Is the power set of a subset of rational numbers denumerable? Could you please prove it?

Body:
The power set of given subset A of rational numbers is denumerable? How can I prove it?

A = {x ∈ Q : 2 ≤ x ≤ 5}


问:
Doubt on a Lipschitz Inequality
I am following this paper. Towards the top of page 23 the author has:
$$|D(v,t) - D(v_i,t)| \leq \frac{f(a+tv) - f(a+tv_i)}{t}| + |(v-v_i)\cdot \nabla f(a)| \leq C \cdot |v - v_t|$$

where $D$ denotes the total deriative in the direction of $v$ and $C$ comes from the Lipschitz constant. However, how does the $t$ in the denominator vanish?

答:
Since $f$ is Lipschitz: $$\left|\frac{f(a+tv)-f(a+tv_i)}{t}\right| = \left|\frac{f(a+tv)-f(a+tv_i)}{(a+tv)-(a+tv_i)}(v-v_i)\right|\le L|(v-v_i)|$$
where $L \ge 0$ is the Lipschitz constant.

He then puts $C = L +|\nabla f(a)|$, using Cauchy-Schwarz. Hope it is clear.

Title: Doubt on a Lipschitz Inequality


问:
Finding specific solution for a Diophantine equation
The equation given $18 = 3x + 2y + 1/2z$ where $x, y, z$ are positive integers and $20 = x+y+z$.

Now, not really knowing what to do, I multiplied through by $2$ to make it a linear Diophantine equation with three unknowns, this gives $36 = 6x + 4y + z$. Solving this using a technique I found on this site I was able to obtain a solution to the original equation: $18 = 3(36) + 2(-36) + (-36)1/2$.

As I said, that is a solution, but not within the constraints I want. $x, y, z$ are not all positive integers and they also don't sum up to $20$. (The correct solution is $x = 2$ and $y = 2$ and $z = 16$, found with Wolfram Alpha.)

Now, assuming I know how to solve linear Diophantine equations with three unknowns, how would I take the constraint into account that $x,y,z$ need to sum up to a certain positive integer? What would be the approach here to find the correct solution?

答:
$$
\begin{cases}
6 x+4 y+z=36\\
x+y+z=20\\
\end{cases}
$$
Subtract the two equations
$$5x+3y=16\tag{1}$$
which is satified for positive integers only by $x=2,y=2$.

substitute in the second equation $z=20-x-y$ and get $z=16$, so the unique solution is $x=2,y=2,z=16$.


问:
Why it is important to write a function as sum of even and odd functions?
For the function $f(x)$ we can write it as sum of even and odd functions:

$$f(x)=\underbrace{\frac{f(x)+f(-x)}{2}}_{\text{Even}}+\underbrace{\frac{f(x)-f(-x)}{2}}_{\text{Odd}}$$

My question is why it is important for us to write a function as sum of these two even and odd functions? Is there any application of that?

答:
When I was a high school student I thought that the even/odd decomposition you write about seemed kind of peculiar and not so fundamental. After learning more mathematics I realized the method behind it (extracting "symmetric pieces" by averaging and what you might call anti-averaging) is actually a very simple example of two important processes in mathematics: eigenspace decompositions and averaging over a group to extract symmetric pieces of a function (or vector, etc.).  What I write below is not meant to give you new situations where your even/odd decomposition helps solve a calculus problem, but to show you many further examples of the same idea so you see it is quite broadly occurring in mathematics.

In nearly every situation where there is an operation that iterates twice to be the identity operation you get an analogue of the even/odd decomposition. Here are three examples.

The matrix transpose (where $M^{\top\top} = M$) leads to the expression of a square matrix as a sum of matrices that are symmetric ($M^\top = M$) and skew-symmetric ($M^\top = -M$)
$$
A = \frac{A + A^\top}{2} + \frac{A - A^\top}{2}
$$

Complex conjugation (where $\overline{\overline{z}} = z$) gives an "even/odd" type viewpoint on writing a complex number in standard form is $a+bi$, since this is the sum of a
real number (fitting $\overline{w} = w$) and a purely imaginary number (fitting $\overline{w} = -w$):
$$
z = \frac{z + \overline{z}}{2} + \frac{z - \overline{z}}{2} = a + bi
$$
where $z = a + bi$ and $\overline{z} = a - bi$.

The swap operator on functions ($f(x,y) \mapsto f(y,x)$) or tensors ($v \otimes w \mapsto w \otimes v$) leads to the expression of a function or tensor as a sum of symmetric and antisymmetric functions or tensors:
$$
f(x,y) = \frac{f(x,y) + f(y,x)}{2} + \frac{f(x,y) - f(y,x)}{2}
$$
and
$$
v \otimes w = \frac{v \otimes w + w \otimes v}{2} + \frac{v \otimes w - w \otimes v}{2}.
$$
This has a role in quantum mechanics, where it underlies the distinction between bosons (having symmetric wavefunctions) and fermions (having antisymmetric wavefunctions).

I said that in nearly every situation you get something like an even/odd decomposition because sometimes one of those parts is zero and thus uninteresting.  For instance, a 180-degree rotation $R$ of the plane has $R(v) = -v$ for all $v$ in $\mathbf R^2$, so here the whole space "looks odd" under the effect of $R$.  No vector in $\mathbf R^2$ is fixed by a 180-degree rotation except for the origin.

The use of "order $2$" here keeps the algebra very simple, but we  can also consider higher-order symmetries rather than symmetries of order 2.  Consider for each $n \geq 1$ trying to decompose a function $f:\mathbf C \to \mathbf C$ as a sum of functions $f_k(z)$ that get scaled by the $k$th power of each $n$th root of unity when it undergoes interior scaling by each $n$th root of unity: $f_k(\zeta z) = \zeta^k f_k(z)$ for all $n$th roots of unity $\zeta$ (or equivalently just $\zeta = e^{2\pi i/n}$) and all complex numbers $z$, where $0 \leq k \leq n-1$. The case $n=2$ is even/odd functions on $\mathbf C$ ($f_0(-z) = f_0(z)$ means $f_0$ is an even function and $f_1(-z) = -f_1(z)$ means $f_1$ is an odd function).  Taking $n = 4$, we can try to decompose each function  $f:\mathbf C \to \mathbf C$  as a sum of four functions
$$
f(z) = f_0(z) + f_1(z) + f_2(z) + f_3(z)
$$
where $f_0(iz) = f_0(z)$, $f_1(iz) = if_1(z)$, $f_2(iz) = -f_2(z)$, and $f_3(iz) = -if_3(z)$ for all $z \in \mathbf C$.Here are formulas for each of the functions:
$$
f_0(z) = \frac{f(z) + f(iz) + f(-z) + f(-iz)}{4},
$$
$$
f_1(z) = \frac{f(z) - if(iz) - f(-z) + if(-iz)}{4},
$$
$$
f_2(z) = \frac{f(z) - f(iz) + f(-z) - f(-iz)}{4},
$$
$$
f_3(z) = \frac{f(z) + if(iz) - f(-z) - if(-iz)}{4}.
$$
These averaging formulas are generalizations of the formulas you wrote for determining the even/odd parts of a function $\mathbf R \to \mathbf R$. And this is useful in Fourier analysis, since the Fourier transform on functions has order $4$.

The ideas presented here extend even further to the decomposition of a representation of a finite group as a sum of irreducible representations. For the cyclic group of order $2$ there are two irreducible representations, and that is reflected in the appearance of even functions and odd functions in your formula. So the even/odd decomposition for functions in your question is a special case of a really important idea in math. It is not just some "trick" to solve artificial calculus problems.


答:
The ubiquitous antiderivative of $\dfrac1x$, namely $\log|x|$ is highly questionable, because it does not express that you may not cross $x=0$.

If we use your solution, we have

$$y=\pm\frac c{|x|}.$$

Then, plugging this result in the equation, we get

$$\mp\frac c{x^2}\text{sgn}(x)\pm\frac c{x|x|}=0$$ which is indeed an identity.

Anyway, I claim that the general solution is

$$\begin{cases}x<0\to y=\dfrac{c_-}x,\\x>0\to y=\dfrac{c_+}x.\end{cases}$$

Indeed, these two expressions satisfy the ODE, except at $x=0$.

Title: Absolute value in solution of first order linear differential equation

Body:
Let’s solve the first order linear equation $ y’ + {1\over{x}}y = 0$.

$ y’/y = -1/x $

$ ln|y| = -ln|x| + C $

$ |y| = {C\over{|x|}} $

So it seems to me, I have to choose every y that equals to $C/|x|$ if I get the absolute value of y.

Both $C/|x|$ and $C/x$ seems to satisfy this requirement. However, my textbook only has $C/x$ as the solution.

Did I do something wrong?

Or is it a convention to just ignore the absolute value? If so, why?


问:
Let $f$ be a computable and injective function. Is $f^{-1}$ computable and injective?
So I just started learning about computability, undecidability and Turing machines. And I wonder if:

Given a computable and injective function $f$, is $f^{-1}$ also computable and injective?

I don't really know where to start to prove this, any hint or explanation would be appreciated.

答:
This question seems somewhat under-specified, but I think the answer is no.

Consider $λ(x : ℕ). 2\cdot x$. The pre-image of even naturals are distinct, but the pre-image of all odd naturals is empty, so $f^{-1}$ is not injective.

Consider the embedding from the naturals to the reals $f : ℕ → ℝ$. The pre-image of $r$ is $f^{-1}(r) = \{ n \in ℕ\ |\ f(n) = r \}$, but equality of reals is not computable (not even for computable reals).


问:
Prove that we can divide student in k groups
There are 3 clubs. Each club includes exactly k students and each student in one club has atleast $\dfrac{3k}{4}$ friends and in each of the 2 other clubs (i.e. $\frac{3k}{4}$ in one club and $\frac{3k}{4}$ in other one as well).

Prove that we can divide $3\cdot k$ students in k groups of size $3$ ,such that in each group all of the students be friend with each other.

My attempt:

I tried counting in two ways , and my other attempt was to consider two groups and prove there is a perfect matching between each of them(there is and its proven) and try to build those k groups which leads to nothing:)

答:
Use Hall's theorem to find a perfect matching $M$ between two clubs.

Then, form a bipartite graph where one side is the third club, and the other side is the $k$ edges of $M$, where a student of the third club is adjacent to an edge of $M$ if the student is friends with both endpoints.

Apply Hall's theorem again, this time to this bipartite graph.


答:
You are asked to find the maximum of the sum of the lengths of diagonals $h+d$ of a rhombus of side $a=1$.

Where $\begin{cases}z_1\equiv\vec{AB} && z_1+z_2\equiv\vec{AC}\\z_2\equiv\vec{BC}&& z_1-z_2\equiv\vec{DB}\end{cases}$

Since $(h+d)^2=h^2+d^2+2dh=4a^2+4S=4(S+1)$

Then $h+d=2\sqrt{S+1}$ is maximal when the area is maximal.

This happens when the rhombus is a square, i.e. when $S=\sin(\alpha)=1\iff \alpha=\frac{\pi}2$

In this case $h+d=2\sqrt{1+1}=2\sqrt{2}$


问:
Figuring out shape of four points scattered in a plane(Quadrilateral)

Suppose you're given coordinates of four points $x_1,x_2,x_3,x_4$ , given the fact that the points lie in a plane and polygon is convex. Figure out the shape formed by these four.

So, in a 'brute force' way I have two methods. One is sketching the points in space and the second is to figure out the angles and lengths to classify if it is a parallelogram, square etc.

Are there any other faster algorithms / alternate methods to quickly figure out the shape formed by these points?

Allowed shapes:

Parellogram

Rhombus

Rectangle

Square

Trapezoid

答:
With four given points a parabola passing through them can be uniquely defined numerically  by means of its equation:

$$ y = ax+b +\sqrt{ cd+d} $$

Negative sign produces same parabola beyond "nose" $N$ with infinite slope.


问:
Solve $\sin z = 2$

How to solve $\sin(z) = 2$,where $z$ is a complex number.

Another approach I tried:

$$\sin z=\frac{1}{2i}\left(e^{iz}-e^{-iz}\right)=2$$

$$e^{iz}-e^{-iz}=4i$$

Let $x=e^{iz}$ so that $x^2-4ix-1=0$. Then, I would use the quadratic formula to solve would this be correct?

答:
Note
$$2=\sin z= \cos(2\pi k +\frac\pi2 -z)
= \cosh[i(2\pi k +\frac\pi2 -z)]
$$
which leads to $i(2\pi k +\frac\pi2 -z)=\pm \cosh^{-1} 2$
and the solutions
$$z= 2\pi k +\frac\pi2\pm i \cosh^{-1} 2$$

Title: Solve $\sin z = 2$


问:
Basis and dimension of null space
Let $A=\begin{bmatrix}
 0&1  &2  &1  &1 \\
 1&0  &1  &1  &1 \\
 -1&2  &3  &1  &1
\end{bmatrix}$ with the homogeneous system $A \mathbf{x}=\mathbf0$.

Problem 1

Find the parametric form for the solution set $N_{A}\subseteq \mathbb{R}^5 $ (null)

Solution so far

After row operations to find the augmented matrix in reduced echelon form, I get that the parametric form  with $s=x_3, t=x_4, k=x_5$ is
\begin{bmatrix}
-s-t-k\\
-2s-t-k\\
s\\
t\\
k\\
\end{bmatrix}
or
$\mathbf{v}=\begin{bmatrix}
x_1\\
x_2\\
x_3\\
x_4\\
x_5\\
\end{bmatrix}=s\begin{bmatrix}
-1\\
-2\\
1\\
0\\
0\\
\end{bmatrix}+t\begin{bmatrix}
-1\\
-1\\
0\\
1\\
0\\
\end{bmatrix}+k\begin{bmatrix}
-1\\
-1\\
0\\
0\\
1\\
\end{bmatrix}$

Problem 2 (Main)

Specify a basis for the subspace $N_{A}\subseteq \mathbb{R}^5 $. What is the dimension of $N_{A}\subseteq \mathbb{R}^5 $?

My Solution so far

I am thinking the basis would just be the 3 vectors in my parametric form i.e

$\begin{bmatrix}
-1\\
-2\\
1\\
0\\
0\\
\end{bmatrix},\begin{bmatrix}
-1\\
-1\\
0\\
1\\
0\\
\end{bmatrix},\begin{bmatrix}
-1\\
-1\\
0\\
0\\
1\\
\end{bmatrix}$ and then I am thinking the dimension is just the cardinality of all the vectors in the specified solution space so 3?

I am certain this is wrong, though and I would like the approach for the main problem cleared up! Do I have to extend my solutions with the standard basis to get 5 vectors? Thanks!

edit: I have been taught that if i.e $\mathbb{R}^5$ then I would need to extend this system to 5 vectors. Is this not correct?

答: Yes, those three vectors span $N_A$ and, since they are linearly independent, then are a basis of $N_A$. Therefore $\dim N_A=3$.

Title: Basis and dimension of null space