问:
Upper triangular block matrix - Quotient Space
Let $V$ and $W$ be both K-linear vector spaces. Now $Y$ is a linear subspace of $V$ and $Z$ is a linear subspace of $W$. Let $\alpha: V \rightarrow W$ be a $K$-linear function with $\alpha(Y)\subseteq Z$.
Let $(v_1,...,v_n)$ be a Basis of $V$ that contains a Basis $(v_1,...,v_k)$ of $Y$ and let $(w_1,...,w_m)$ be a Basis of $W$ that contains a Basis $(w_1,...,w_l)$ of $Z$.

Now the task is to show that the Matrix of $\alpha$, regarding the bases $(v_1,...,v_n)$ and $(w_1,...,w_m)$, has the form:
$\pmatrix{A&C\\ 0&B}$. I have no idea how to approach this. I guess the Matrix has to look like this:

\begin{pmatrix}
    a_{11} & \dots &a_{1k} &  c_{11} & \dots &c_{1n} \\
    \vdots &  \ddots  &\vdots & \vdots &  \ddots  &\vdots \\
    a_{m1} & \dots &a_{mk} & c_{11} & \dots &c_{1n} \\
0 & \dots &0&  b_{11} & \dots &b_{1n} \\
    \vdots &  \ddots  &\vdots & \vdots &  \ddots  &\vdots \\
    0& \dots & 0 & b_{11} & \dots &b_{1n}
  \end{pmatrix}

But why and how to proof it? I know that $A$ is the Matrix of $Y$ and $B$ is the Matrix of $V/Y$.

答:
First recall that if $\beta : V \to W$ is a $K$-linear function, the matrix of $\beta$ with respect to the bases $(v_1,\dots,v_n)$ and $(w_1,\dots,w_m)$ is the $m \times n$ matrix $B = (b_{ij})$ such that $$(\forall j \in \{1,\dots,n\}) \quad \beta(v_j) = b_{1j}w_1+b_{2j}w_2+\cdots+b_{mj}w_m.$$

Now, actually the matrix of $\alpha$ with respect to the bases $(v_1,\dots,v_n)$ and $(w_1,\dots,w_m)$ looks like
$$A = \begin{pmatrix}
    a_{11} & \dots &a_{1k} &  a_{1(k+1)} & \dots &a_{1n} \\
    \vdots &  \ddots  &\vdots & \vdots &  \ddots  &\vdots \\
    a_{\color{red}l1} & \dots &a_{\color{red}lk} & a_{l(k+1)} & \dots & a_{ln} \\
0 & \dots &0&  a_{(l+1)(k+1)} & \dots &a_{(l+1)n} \\
    \vdots &  \ddots  &\vdots & \vdots &  \ddots  &\vdots \\
    0& \dots & 0 & a_{m(k+1)} & \dots &a_{mn}
  \end{pmatrix}$$
and this is because we have that $$(\forall j \in \{1,\dots,k\}) \quad \alpha(v_j) = a_{1j}w_1+\cdots+a_{lj}w_l+0w_{l+1}+\cdots+0w_m \in Z$$
(since $(v_1,\dots,v_k)$ is a basis for $Y$, $(w_1,\dots,w_l)$ is a basis for $Z$, and $\alpha(Y) \subseteq Z$).


问:
Side Effects in Lambda Calculus
A simplistic method of creating effects in Lambda calculus is by adding constants: functions that create required effects. For example: we can add a rnd function which represents a random Church numeral, where for every j belonging to every natural number the following reduction rule, where Nj is the church numeral relating to j:

```<code>                   rnd -> Nj.
</code>```

How would the presence of such an effect impact reduction strategies (normal order and lazy evaluation) and with confluence.

答:
Notation. $N_j$ is the Church numeral for the natural number $j$, i.e. $N_j = \lambda f. \lambda x. \underbrace{f (f ( \cdots (f}_{j \text{ times}}x) \cdots ))$.

Yes, adding a side effect such as
$$
\mathsf{rnd} \to N_j \quad \text{for every } j \in \mathbb{N}
$$
to the $\lambda$-calculus impacts on reduction strategies and confluence.

First, the resulting calculus is not confluent because, given $i \neq j \in \mathbb{N}$,
$$
N_i \leftarrow  \mathsf{rnd} \to N_j
$$
where $N_i$ and $N_j$ are distinct normal forms and hence do not have a common term to reduce to.
So, not only the calculus is not confluent, but it is even false that every term has at most one normal form.

Moreover, with such a side effect it is not even true that lazy evaluation produces the same results as normal order (this holds, instead, in the pure case without side effects). Indeed, consider the term

$$\tag{1}
(\lambda x. xx) \, \mathsf{rnd}
$$

According to normal order, the argument $\mathsf{rnd}$ is passed to the function $\lambda x.xx$ without being evaluated, so
\begin{align}
 (\lambda x. xx) \, \mathsf{rnd} &\to_\beta \mathsf{rnd} \, \mathsf{rnd}
\\
 &\to N_1 \, \mathsf{rnd}
\\
 &= (\lambda f. \lambda x. fx) \, \mathsf{rnd}
\\
 &\to_\beta \lambda x. \mathsf{rnd} \, x
\\
 &\to \lambda x. N_0 x
\\
 &\to_\beta \lambda x. \lambda y .y
\end{align}

According to lazy evaluation, the argument $\mathsf{rnd}$ is first evaluated and then passed to the function $\lambda x.xx$, so
\begin{align}
 (\lambda x. xx) \, \mathsf{rnd} &\to_\beta (\lambda x. x \, x) N_0
\\
 &\to (\lambda x. N_0 \,x) N_0
\\
 &= (\lambda x.(\lambda f. \lambda y. y)x)N_0
\\
 &\to_\beta (\lambda x. \lambda y. y) N_0
\\
 &\to_\beta \lambda y .y
\end{align}

Therefore, term $(1)$ may reduce to two different results (i.e. normal forms) depending on whether the evaluation is normal order or lazy.

ContentLicense: CC BY-SA 4.0

Title: Side Effects in Lambda Calculus


问:
Is there a Central limit theorem for max
I have multiple IID (bigger then 0) of unknown distribution and I get the max of them as a result of my calculation.

Can I say that the max also have certain gaussian shape distribution?
If not what is the distribution of the result?

I saw this answer which describe the result of IID product in a log-normal distribution. but as far as I understand this is not the same as max.

I tried to run a simulation in MATLAB with 1000*1000 uniform IID in range 1000 and got that the max of each column is not distibuted randomlly but getting close to 1000.
here is the code:

```<code>r = randi(1000,[1000,1000]);
mxR = max(r);
figure;
hist(mxR);
</code>```

After that I tried some other non - caped distribution (normal dist. with \mu = 5000 and \sigma = 500) and I got something similar tolog-normal dist.
distribution result of max IID (normal)

here is the code:

```<code>r = random('Normal',5000,500,[10000,10000]);
mxR = max(r);
figure;
hist(mxR);
</code>```

Is there a rule or a theorem regarding this?

thanks for your help!!
(If you think this is interesting (or not) please let me know)

答:
Yes, it’s called the Generalized Extreme Value Distribution .

The conditions of convergence are given by the Fisher-Tippet-Gnedenko Theorem.

Basically, it says the distribution of an extreme order statistic (max,min) of an iid sample can converge to one of three distributions.


问:
A simple probability reasing to predict rain fall
A friend told me the following about whether it will rain tomorrow (or not):

The probability that it will rain tomorrow is $1/2$ since it will either happen or not. But -even as a non mathematician- I do not find this argument a convincing one, because the same reasoning can be used to predict that I will be struck by a bolt of lighting in five minutes.

Can someone tell me what is mathematically wrong about this argument?

答:
The two hypothesis -it will rain or it will not rain- aren't equiprobable. Separating the possibilities into two sets doesn't implies that the two set are equal, even if one of them must happen.

When you try to compute the probabilty of having an odd number when you roll a dice is computing by addition of the probabilities of having a 2, a 4 or a 6. You guess they are equals because there is no reason it would be otherwise, but if the dice is loaded, it will not be the case. It's not a theorem, it's an assumption.

Computing the probability it will rain or not depends of many factors, and we use statistics and meteorologics models to have a good estimation knowing a certain amount of informations. I would add that the probability depends of the amount of informations you have. If you know that there is no wind and no cloud 3000miles around, you can tell that the probability is very low that tomorrow will be rainy.

ContentLicense: CC BY-SA 3.0


问:
Integral of $x e^{5x^2 + 2}$
I've tried integrating by parts.  But it doesn't seem to work, my first attempt seems more promising, but I don't know where to go after line 3.  Pls help
My working out is posted in this pitcure </p>

Title: Integral of $x e^{5x^2 + 2}$

Body: $$\int xe^{5x^2+2}dx = \frac1{10} \int e^{5x^2+2}10x\,dx = \frac1{10} \int e^{5x^2+2}d(5x^2+2) = \frac1{10} \int e^u\,du$$


问:
Convergence a.s. from convergence in probability (Steele's book)
In Steele's Stochastic Calculus and Financial Applications, on top of p. 114:

The highlighted portion is an argument for surmising almost sure convergence from convergence in probability by passing to a subsequence. I must be missing something, because I don't see how this argument wouldn't imply more generally, meaning that the two modes of convergence are equivalent (which they, of course, are not).

答:
It was proved that $A_n, B_n$ and $C_n$  converges in probability to $I_1, I_2$ and $0$ correspondly.

Let us take $n_k$ such that $A_{n_k}$ converges to $I_1$ a.s. It's possible - see, e.g., math.stackexchange.com/questions/222264

As $B_{n_k}$ converges to $I_2$ in probability we may take $B_{n_{k_j}}$ from $B_{n_k}$ such that $B_{n_{k_j}}
\to I_2$ a.s.  Finally we will take ${n_{k_{j_l}}} $ such that $C_{n_{k_{j_l}}} \to 0$ a.s.


问:
Use the Residue Theorem to evaluate the following integral:
$$\int_{-∞}^{∞} \frac{x^4}{1+x^8} dx$$

I've found the zeros in the upper half plane to be

$$e^{i \pi/8}, e^{i 3 \pi/8}, e^{i 5 \pi/8}, e^{i 7 \pi/8}$$ (right?)

But then the calculation got really ugly, so I think I'm doing something wrong...please help!!

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
The following integration $\ds{\underline{just}}$ need
$\ds{\underline{two}}$ poles. Namely, $\ds{p_{0} \equiv \expo{\pi\ic/8}\ \mbox{and}\ p_{1} \equiv \expo{3\pi\ic/8}}$.

\begin{align}
&\bbox[5px,#ffd]{%
\int_{-\infty}^{\infty}{x^{4} \over 1 + x^{8}}\,\dd x} =
2\,\Re\int_{0}^{\infty}{x^{4} \over 1 + x^{8}}\,\dd x
\\ &\
\mbox{"Close" the integration along a quarter circle}
\\ &\ \mbox{in the complex plane first quadrant.}
\\[5mm] = &\
2\Re\braces{2\pi\ic\sum_{j = 0}^{1}{p_{j}^4 \over 8p_{j}^{7}} - \int_{\infty}^{0}{\ic^{4}y^{4} \over
1 + \ic^{8}y^{8}}\,\ic\,\dd y}
\\[5mm] = &\
-4\pi\,\Im\sum_{j = 0}^{1}{p_{j}^{5} \over 8 \times \pars{-1}}
\\[5mm] = &\
{\pi \over 2}\,\Im\sum_{j = 0}^{1}p_{j}^{5} =
{\pi \over 2}\,\Im\pars{\expo{5\pi\ic/8} + \expo{15\pi\ic/8}}
\\[5mm] = &\
{\pi \over 2}\bracks{\sin\pars{5\pi \over 8} +
\sin\pars{15\pi \over 8}}
\\[5mm] = &\
\bbx{\root{1 - {\root{2} \over 2}}{\pi \over 2}} \approx 0.8501
\\ &\
\end{align}


答:
You have:
$$I=\int xe^{5x^2+2}dx=e^2\int xe^{5x^2}dx$$
now if we let $u=5x^2\Rightarrow du=10x dx\therefore dx=\frac{du}{10x}$ now putting this back in we get:
$$I=\frac{e^2}{10}\int e^udu=\frac{e^2}{10}e^u+C=\frac{e^{5x^2+2}}{10}+C$$

Title: Integral of $x e^{5x^2 + 2}$

Body:
I've tried integrating by parts.  But it doesn't seem to work, my first attempt seems more promising, but I don't know where to go after line 3.  Pls help
My working out is posted in this pitcure </p>


问:
Find natural numbers $u, v$ for which $\displaystyle 2\sqrt{u} + 4\sqrt{v - u} - 7 \geq v$.
As I said in the title, the problem states:

Solve the following inequation in $\mathbb{N}^2$:
$$ 2\sqrt{u} + 4\sqrt{v - u} - 7 \geq v $$

Source: Mathematical challenges for $8^{th}$ grade.

Approach: By very long calculations, replacing $v - 7 = w$, I obtained an inequation like:
$$400u^2 + [40(w^2 - 16w + 112) - 16w^2]u + (w^2 - 16w + 112)^2 \leq 0$$

So, the function admits a maximum, therefore the $a$ coefficient ($f(x) = ax^2 + bx + x$) is negative. But, $400u^2 \geq 0$, so the inequation is, in fact, an equation!

Back into the first inequation, we can replace $\geq$ by $=$.

Now, we can begin solving like a Diophantine Equation:
$$2\sqrt{u} + 4\sqrt{v - u} - 4 = v$$

Because $v \in \mathbb{N}$, $u$ is a perfect square and $v - u$ is also a perfect square.

Noting $u = \alpha^2$ and $v = \beta^2$. The equation looks like:
$$2\alpha + 4\beta - 7 \geq \alpha^2 + \beta^2$$

Equivalent to:
$$(\alpha - 1)^2 + (\beta - 2)^2 + 2 = 0$$

Impossible in $\mathbb{N}$.

Question: My solution finds that no $(u, v)$ natural verify the inequation. Is there any mistake in my solution? Could you find a better approach?

答:
You have a little problem with swapping signs.

If $a= \sqrt{u} $ and $b=\sqrt{v-u}$ then you have to solve $$2a+4b-7\geq a^2+b^2$$ so $$-2 \geq (a-1)^2+(b-2)^2$$ and thus no solution.


问:
$P \otimes P(\left\{(x,y) \in \mathbb{C}^{2d};x=y \right\})=1$
Let $P$ a probability measure on $(\mathbb{C}^d,B(\mathbb{C}^d)),$ such that for all $E \in B(\mathbb{C}^d),P(E)$ is $0$ or $1.$ Let $X=\left\{(x,y) \in \mathbb{C}^{2d};x=y \right\}.$

Prove that $P \otimes P(X)=1.$

$P \otimes P(X)=\int_{\mathbb{C}^d}P(\left\{x \right\})dP(x),$ $P(\left\{x \right\})$ could be $0$ or $1$, any suggestions how to continue?

答:
I'll show you the case where we have $\mathbb{R}$ instead of $\mathbb{C}^d$ (the idea is the same). We have
$$ 1 =\lim_{R\rightarrow \infty} P([-R,R]).$$
As $P([-R,R])\in \{0,1\}$ we get that there exists some $R>0$. Now we define recursively $[a_0,b_0]:=[-R,R]$ and
$$ [a_{n+1}, b_{n+1}]:= \begin{cases}
[a_n, (a_n+b_n)/2],& \text{if } P([a_n, (a_n+b_n)/2])=1,\\
[(a_n+b_n)/2, b_n],& \text{otherwise.}
\end{cases}$$
Note that we have $P([a_n, b_n])=1$. We have for $a=\lim_{n\rightarrow \infty} a_n$
$$1=\lim_{n\rightarrow \infty} P([a_n, b_n])= P(\{a\}).$$
Thus, $P=\delta_a$.

Then we have
$$ (P\otimes P)(\{(a,a)\})=1$$
and we get the desired claim.


问:
Condition number for a least square problem
For an $m×n$ matrix A with $m \geq n$. We define

$ k(A) = \frac{\max_{\lVert x\rVert=1} \lVert Ax \rVert}{\min_{\lVert x\rVert=1} \lVert Ax \rVert} $

For the euclidean norm, that is, $\lVert A \rVert$ is the maximum eigenvalue of $A^tA$, I try to show that $k(A^tA) = (k(A))^2$.

I appreciate your help.

答:
If $A=UDV^\top$ is the SVD of $A$ with singular values $\sigma_1 \ge \cdots \ge \sigma_n$ then $\kappa(A) = \sigma_1 / \sigma_n$.

Then $A^\top A = VD^\top D V^\top$ so the eigenvalues of $A^\top A$ are $\sigma_1^2 \ge \cdots \ge \sigma_n^2$ and $\kappa(A^\top A) = \sigma_1^2 / \sigma_n^2$.


问:
Help for solving first-order differential equations with two unknown functions
I am familiar with first-order differential equations and how to solve them, but only for the case when there is a single unknown function.

For the following case there are two unknown functions and I'm struggling to determine where to even begin. The two equations:

$$ f'(t) = - Af(t)+ Bg(t) \tag{1}$$
$$ g'(t) = - Bg(t)+ Af(t) \tag{2} $$

Where f and g are functions of the variable t, and A & B are constants. The initial conditions are: $f(0) = 0$ and $g(0) = 1$.

I already have knowledge of the answer for f(t), which is:

$$ f(t) = \frac{B}{A+B}(1-e^{-(A+B)t})  \tag{3} $$

I would really like to understand how to tackle questions of this type: where there are two unknown functions. I would attempt to show my working out but I don't know where to start.

Therefore, I would like to ask if anyone could please provide a hint or suggestion for me, something which I can work off. If necessary, I will update my post with original working. Thank you for your time.

答:
This is an example of a linear system:
$$\begin{pmatrix}f'\\g'\end{pmatrix}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}f\\g\end{pmatrix},$$
or $\mathbf F'=M\mathbf F$. The idea is to look for a solution of the form $\mathbf F=\mathbf ve^{\lambda t}$, where $\mathbf v$ is some (constant) vector. This gives $\lambda\mathbf v=M\mathbf v$, i.e. $\lambda$ is an eigenvalue of $M$, and $\mathbf v$ the corresponding eigenvector.

So we have a bunch of solutions, each corresponding to an eigenvalue/vector of the matrix. As the DE is linear and homogenous, we can just take linear combinations of these to get the general solution. Specifying initial conditions gives you an exact solution.

As a concrete example, for your original DE we have
$$\begin{pmatrix}f'\\g'\end{pmatrix}=\begin{pmatrix}-A&B\\A&-B\end{pmatrix}\begin{pmatrix}f\\g\end{pmatrix}.$$
The eigenvalues of the matrix are $0$ and $-(A+B)$, and the respective eigenvectors are $(B,A)^\intercal$ and $(-1,1)^\intercal$. So the general solution is
$$\begin{pmatrix}f\\ g\end{pmatrix}=C\begin{pmatrix}B\\A\end{pmatrix}e^{0t}+D\begin{pmatrix}-1\\1\end{pmatrix}e^{-(A+B)t}.$$
So using the initial conditions:
\begin{align*}
f(0)=0&\implies0=CB-D \\
g(0)=1&\implies1=CA+D
\end{align*}
So $C=\frac{1}{A+B}$ and $D=\frac{B}{A+B}$. This gives the desired solution.


问:
How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$?
How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? My first idea was to write $$1-\cos x=\frac12\left|e^{{\rm i}x}-1\right|^2\tag1,$$ which is true for all $x\in\mathbb R$, but I don't have a suitable lower bound for the right-hand side at hand.

答:
Rewriting the inequality $$2\sin^2(\frac{x}{2})\ge\frac{x^2}3 \iff (\sqrt2 \sin(\frac{x}{2})  - \frac{x}{\sqrt3})(\sqrt2 \sin(\frac{x}{2})  + \frac{x}{\sqrt3}) \ge 0 \tag{1}$$
Assume $0\le x \le 1$. Obviously we have $$\sqrt2 \sin(\frac{x}{2})  + \frac{x}{\sqrt3}\ge 0$$ So we should analyze $$f(x,a) = \sin(\frac{x}{2}) - \frac{x}{a}$$It's interesting to see this visually:
When $a=2$, the line $y_1 = \frac{x}{2}$ touches $y_2 = \sin(\frac{x}{2})$ only at $x = 0$(remember that $|\sin(\frac{x}{2})|\le |\frac{x}{2}|$). Increasing $a$ further, leads to another intersection point(we can actually see the concavity of $\sin(\frac{x}{2})$ here).

Note that $$f(x,a) = \int_{0}^{x}(\frac{1}{2}\cos (\frac{t}{2}) - \frac{1}{a})dt$$Let $a = \sqrt6$. We can show that the integrand is nonnegative when $0\le t \le 1$. So when $0\le x \le 1$ we have $f(x,\sqrt6)\ge 0 $ which implies $$\sqrt2 \sin(\frac{x}{2})  - \frac{x}{\sqrt3}\ge 0$$In order to show nonnegativity, consider $$0\le t \le 1 \implies 0\le \frac{t}{2} \le \frac{1}{2} \lt \frac{\pi}{2} \implies 1 \ge \cos(\frac{t}{2}) \ge \cos(\frac{1}{2}) \implies \frac{1}{2} \ge \frac{1}{2}\cos(\frac{t}{2}) \ge \frac{1}{2}\cos(\frac{1}{2}) \approx 0.439 \gt \frac{1}{\sqrt6} \approx 0.408$$ Now that we have shown $(1)$ holds for $0 \le x \le 1$, let $x = -t$ and conclude it holds for $-1 \le t \le 0$.


问:
What is the rightmost point $x_F$ for uniform distribution $X\sim U(0, 1)$?
The CDF for $X\sim U(0, 1)$ is defined for $X\in R$. Then for the rightmost point $x_F$, is $x_F=1$ or $x_F=\infty$?

答:
The relevant domain of a density function is generally its support. Given a random variable $X\sim \mathcal{U}(0,1)$, the probability density function is

$$ f_X(x) = \begin{cases}1 & x\in[0,1]\\ 0 & x\notin[0,1] \\ \end{cases} ,$$

and the cumulative distribution function is

$$ F_X(x) = \begin{cases} 0 & x < 0 \\ x & x\in[0,1]\\ 1 & x > 1 \\ \end{cases} . $$

The support of $f_X(x)$ is $[0,1]$, and that's where all the interesting stuff is happening. Outside of that range, the density is $0$ everywhere.

The rightmost point that's relevant in any calculations of probabilities is $x=1$. Could you calculate probabilities involving values of $X>1$? Sure, but it's pretty boring, since the PDF is constant at $f_X(x) = 0$ and the CDF is constant at $F_X(x) = 1$.


问:
A question on Dixon's proof for the global Cauchy Formula
I am following Dixon's proof for the global form of the Cauchy Forula and one of the first things one has to prove is that

is continious, where $f$ is analytic. The book I am following (Lang's Complex Analysis) mentions that the proof for the continuity off the diagonal is obvious, however I can't seem to able to prove it.

My attempt:

Select $(a,b)$ close to $(z,\zeta)$ s.t. $|z-a|< \delta _1$ and $|\zeta-b|< \delta_2$ where $\delta _1 , \delta _2 >0$. Then

$$|g(z,\zeta)-g(a,b)|=|\frac{f(\zeta)-f(z)}{\zeta-z}-\frac{f(b)-f(a)}{b-a}|$$

now if I was to use the triangle inequality I can take care of the bottom terms by selecting $\lambda=min(\delta_1, \delta_2)$ but I then can't use the continiuity of $f$ to get that $g$ is continious off the diagonal. Any thoughts of how to proceed?

答: For $z\ne\zeta$, $g$ is the quotient of two continuous (actually analytic) functions, hence continuous.


答:
If you toss a coin you have two possible outcomes that are as likely to happen so the probability of each is one half.

The difference for getting struck by lightening is that the two outcomes are not as likely to happen (in fact the probability you get struck is really close to $0$) so you can't separate the two outcomes in set of same probability.

Title: A simple probability reasing to predict rain fall

Body:
A friend told me the following about whether it will rain tomorrow (or not):

The probability that it will rain tomorrow is $1/2$ since it will either happen or not. But -even as a non mathematician- I do not find this argument a convincing one, because the same reasoning can be used to predict that I will be struck by a bolt of lighting in five minutes.

Can someone tell me what is mathematically wrong about this argument?


问:
$A \overrightarrow{X}$ and $B \overrightarrow{X}$ are independent random vectors iif $A \Sigma B^{\prime}=\overrightarrow{0}$
I need to prove that if $\overrightarrow{X} \sim N_{p}(\overrightarrow{\mu}, \Sigma)$ then $A \overrightarrow{X}$ and $B \overrightarrow{X}$ are independent random vectors iif $A \Sigma B^{\prime}=\overrightarrow{0}$

It was suggested that I should consider $\overrightarrow{Y}=\left[\begin{array}{ll}A \overrightarrow{X} & B \overrightarrow{X}\end{array}\right]^{\prime},$ and then calculate $V[\overrightarrow{Y}]$ and the characteristic function of $\overrightarrow{Y}$. But I'm having difficulties connecting all of these concepts for the proof.

答:
Let $Y_1=AX$ and $Y_2=BX$, then in general:

(I) $E\{Y_1 Y_2^T\}=A\Sigma B^T+A\mu\mu^TB^T$.

For $Y1$ and $Y2$ to be independent, $E\{Y_1 Y_2^T\}=E\{Y_1\}E\{Y_2^T\}$ must hold, i.e.,

(II) $E\{Y_1 Y_2^T\}=E\{AX\}E\{X^T B^T\}=A\mu\mu^TB^T$.

So, (I)=(II) iff $A\Sigma B^T=0$.


问:
Commutative Diagram and Essential Extension
I have a question about commutative diagram.

is a commutative diagram of right R-modules and right R-module homomorphisms. Suppose that f and g
are isomorphisms, and also rows are exact. I want to show  if $\phi'(M) < M_2$ is an essential extension, then $h$ must be injective.

答:
Hint:

Since $M_2$ is an essential extension of $\phi'(M)$, to show $\ker h=\{0\}$, you only have to show that $\phi'(M)\cap \ker h=\{0\}$, i.e. that for all $m\in M$,
$$h\bigl(\phi'(m)\bigr)=0\implies \phi'(m)=0.$$
Set $n=g(m)$ and do some diagram hunting.

Some more details. Here's how you can start a-hunting:

As the right square is commutative, you can check that $\psi'(n)=0$, hence, as the bottom row is exact, there exists $n_1\in N_1$ such that $n=\psi(n_1)$. Now $f$ is an isomorphism, so there's a unique $m_1\in M_1$ such that $f(m_1)=n_1$. What can you say about $\phi(m_1)$, given that $g$ is an isomorphism?


问:
Indiscernible sequence over a non-forking realization
The following is exercise 7.3.1. of Tent and Ziegler's textbook:

Let $T$ be simple, with a monster model $\mathfrak{U}$, and let $\pi(v,w)$ be a collection of $\mathcal{L}_A$-formulas with $b\in\mathfrak{U}$ such that $\pi(v,b)$ is consistent and does not fork over $A$. If $\mathcal{I}$ is an infinite sequence of $A$-indiscernibles containing $b$, then there is a realization $c$ of $\pi(v,b)$ such that $c\perp_A\mathcal{I}$ and such that $\mathcal{I}$ is indiscernible over $A\cup\{c\}$.

Here is my solution: By compactness, it suffices to assume that $\mathcal{I}$ has order type $\omega$, so let $\mathcal{I}=(b_i)_{i\in\omega}$, with $b_0=b$. Now, let $p(v)\in S_n(A\cup\mathcal{I})$ be a non-forking extension of $\pi(v,b)$, with a realization $c$. By construction, we have $\mathfrak{U}\models\pi(c,b)$ and $c\perp_A\mathcal{I}$. We claim that $\mathcal{I}$ is $A\cup\{c\}$-indiscernible. To see this, we want to show $$\varphi(v,b_{i_1},\dots,b_{i_k})\leftrightarrow\varphi(v,b_{j_1},\dots,b_{j_k})\in\text{tp}(c\big/A\cup\mathcal{I})$$ for every $i_1<\dots<i_k\in\omega$ and $j_1<\dots<j_k\in\omega$ and every $\mathcal{L}_A$-formula $\varphi$. To show this, let $n$ be maximal among the $i_l$ and $j_l$. Since $c\perp_A\mathcal{I}$, in particular we have $c\perp_A\{b_0,\dots,b_n\}$. Then $\mathcal{I}$ may be considered as an $A$-indiscernible sequence with first element $(b_0,\dots,b_n)$ in a natural way, and so by corollary 7.1.5. we may find some $c_n\in\mathfrak{U}$ with $c_n\equiv_{A\cup\{b_0,\dots,b_n\}}c$ and such that $\mathcal{I}$ is $A\cup\{c_n\}$-indiscernible. In particular, the desired formula lies in $$\text{tp}(c_n\big/A\cup\{b_0,\dots,b_n\})=\text{tp}(c\big/A\cup\{b_0,\dots,b_n\})\subseteq\text{tp}(c\big/A\cup\mathcal{I}),$$ and so we are done.

I can't see any error in what I've written, but I looked in the back of Tent and Ziegler's book, and they give a hint for a (seemingly) more elaborate solution, which first proves that $\bigcup_{i\in\omega}\pi(v,b_i)$ does not fork over $A$ and then takes a non-forking extension of that collection. Perhaps the ideas are not so different, since the proof of lemma 7.1.5. uses a similar union of pushforward types ranging over an indiscernible sequence, but I'm not sure whether there's a reason Tent and Ziegler didn't just use lemma 7.1.5. in their solution to this exercise. In particular, is there a mistake anywhere in my proof?

答:
In your proof, you  replace the $A$-indiscernible sequence $\mathcal{I}$ with a "thickening": an $A$-indiscernible sequence $\mathcal{I}^n = (b_{(n+1)k},b_{(n+1)k+1},\dots,b_{(n+1)k+n})_{k\in \omega}$.  Then its true that you can find $c_n \equiv_{Ab_0,\dots,b_n} c$ such that $\mathcal{I}^n$ is $Ac_n$-indiscernible. But you cannot conclude from this that  $\mathcal{I}$ is $Ac_n$-indiscernible.

For  example, $b_0$ and $b_1$ may have different types over $Ac_n$. Your construction only ensures that $b_0$, $b_{n+1}$, $b_{2(n+1)}$, etc. have the same type over $Ac_n$.

In general, it does not follow from $\mathfrak{U}\models\pi(c,b)$ and $c\perp_A\mathcal{I}$ that $\mathcal{I}$ is $A\cup\{c\}$-indiscernible. For example, in the theory of the random graph, suppose $\pi(x,y)$ says $xRy$. We can  take $c$ to be any element  not in $\mathcal{I}$, with $cRb_0$, and we will have $c\perp_A\mathcal{I}$. In particular, $c$ can be connected by an edge or not to the other elements of $\mathcal{I}$. So you have to do something more elaborate to prove this (in particular, you have to use simplicity somewhere, which your solution does not - the statement is not true for arbitrary theories).


问:
Multivariate Diophantine zero-sum equation
I have a problem similar in some ways to this one:

(Diophantine?) Equations With Multiple Variables

Where I need to find all $x$ for

$$a_1x_1+a_2x_2+\cdots+a_kx_k = 0$$

with the following constraints: $a_i \in \mathbb{Z}, x_i \in \mathbb{Z}, 0 \le x_i \lt 10, 0 \lt k \le 10$ . $x$ are chosen uniquely from the range [0, 10) but I might have to brute-force that part.

An example of a brute-force solution to this equation form is

$$
\begin{pmatrix} 10000& 1000& 200& 100& 20& 2& 0& -100& -1000& -10000 \end{pmatrix}
\begin{pmatrix} 2\\ 4\\ 7\\ 8\\ 5\\ 0\\ 9\\ 3\\ 6 \end{pmatrix} = 0
$$

I tried to follow along with the solution in the linked question, but I don't think it applies since "dividing into [the right-hand side]" doesn't make much sense when the right-hand side is zero. Perhaps this isn't actually a Diophantine equation? Another way of putting it is that I need to find roots of a polynomial of order one in $x$ subject to the above constraints; or that I need to solve for the right-hand side of the dot product $0 = A \cdot x$ .

答:
(EDITED)
You can use dynamical programming.

Let $F(k, v)$ be the number of solutions to
$a_1 x_1 + \ldots + a_k x_k = v$ with $x_k \in \{0, \ldots, 9\}$, where $v \in \mathbb Z$.  Of course $F(k,v) = 0$ if $v > 9 \sum_{i=1}^k \max(a_i, 0)$ or $v < 9 \sum_{i=1}^k \min(a_i, 0)$, and $F(1,v) = 1$ if $v \in \{a_1, 2a_1, \ldots, 9 a_1\}$, $0$ otherwise.  By conditioning on the value of $x_k$, $$F(k,v) = \sum_{t=0}^9 F(k-1,v-t a_k)$$
We compute and remember each $F(k,v)$ for each $k$ starting at $1$, and each $v$
in the interval where it can be nonzero.

Then work backwards: assuming $F(k,v) \ne 0$, there are solutions where $x_k$ is each
$t$ from $0$ to $9$ such that $F(k-1,v-a_k t) \ne 0$.


问:
Does the improper integral $\int_0^\infty\sin(x)\sin(x^2)\,\mathrm dx$ converge
Does the following improper integral converge?

$$\lim_{B \to \infty}\int_0^B\sin(x)\sin(x^2)\,\mathrm dx$$

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{B}\sin\pars{x}\sin\pars{x^{2}}\dd x}  =
\Im\int_{0}^{B}\expo{\ic x}\sin\pars{x^{2}}\dd x
\\[5mm] = &\
-\,\Im\int_{B}^{0}\expo{\ic\pars{\ic y}}
\sin\pars{\bracks{\ic y}^{2}}\ic\dd y =
-\int_{0}^{B}\expo{-y}\sin\pars{y^{2}}\,\dd y
\\[5mm] \stackrel{{\rm as}\ B\ \to\ \infty}{\to}\,\,\,\,\,&
\underbrace{-\int_{0}^{\infty}\expo{-y}\sin\pars{y^{2}}\,\dd y}
_{\ds{\mbox{It}\ converges\ !!!.}}
\end{align}


问:
How can simplex coordinates simplify this integration?
My textbook defines simplex coordinates in $\mathbb{R}^n$ by:

$\Phi: \mathbb{R}^n\to \mathbb{R}^n , (x_1,\ldots,x_n)=\Phi(u_1,\ldots,u_n)$ where:

\begin{align*}
x_1 & =u_1(1-u_2), & u_1 & =x_1+x_2+\cdots + x_n \\[6pt]
x_2 & =u_1u_2(1-u_3), & u_1 & =\frac{\phantom{x_1 + {}} x_2+\cdots +x_n}{x_1+x_2+\cdots +x_n} \\
&\,\,\, \vdots & & \,\,\, \vdots \\
x_k & =u_1u_2\cdots u_k(1-u_{k+1}), & u_k & =\frac{\phantom{x_{k-1}+{}}x_k+\cdots +x_n}{x_{k-1} + x_k+\cdots + x_n} \\
& \,\,\, \vdots & & \,\,\, \vdots \\
x_n & =u_1u_2\cdots u_n,  & u_n & =\frac{\phantom{x_{n-1}+{}}x_n}{x_{n-1}+x_n}
\end{align*}

Problem:

A tetrahedron ($3$-simplex) is  given by the vertices $(0,0,0)$,
$(1,0,0)$, $(0,2,0)$, $(0,0,3)$. Find $\int _T xz+y^2 \, \mathrm{d}V$.

I know that I have to use the transformation formula. But how do I describe the tetrahedron with  the new coordinate system?

答:
Hint:

Your tetrahedron is given by
$$T = \{(x,2y,3z) : x,y,z \ge 0, x+y+z \le 1\}.$$
New coordinates $(u,v,w)$ are given by
$$\begin{cases} u = x+y+z, \\ v = \frac{y+z}{x+y+z}, \\ w = \frac{z}{y+z}\end{cases} \qquad \begin{cases} x = u(1-v), \\ y=uv(1-w), \\ z=uvw\end{cases}$$
so check that
$$T=\{(x,2y,3z) : (x,y,z) = \Phi(u,v,w),  0 \le u,v,w \le 1\}) = \Psi([0,1]^3)$$

where $$\Psi(u,v,w) = (u(1-v),2uv(1-w),3uvw).$$


问:
Intuition Behind Generalized Stokes Theorem
Consider the Generalized Stokes Theorem:

\begin{equation}
\int_Md\omega = \int_{\partial{M}} \omega
\end{equation}
Here, $\omega$ is a k-form defined on $R^n$, and $d\omega$ (a k+1 form defined on $R^n$) is the exterior derivative of $\omega$. Let M be a smooth k+1-manifold in $R^n$ and $\partial{M}$ (the boundary of M) be a smooth k manifold.

I know that the above theorem is simply a generalization of well-known vector calculus theorems. However, I am looking for the intuition behind the Generalized Stokes Theorem itself.

I started off by defining the exterior derivative at a point p in $R^n$ as:
\begin{equation}
d\omega_p =\lim_{|vol|\to 0}\frac{\int_{\partial{vol}} \omega}{|vol|}
\end{equation}

In this case, "$vol$" represents a k+1 "parallelpiped" in $R^n$ that contains point p (with $|vol|$ being its "volume"). $\partial{vol}$ represents the boundary of this k+1 "parallelpiped", a k "parallelpiped" itself.

With this definition (assuming it is correct), can we say that $\omega$ represents an infinitesimal "flux" element through $\partial{vol}$ which would imply that $d\omega_p$ is simply the "flux density" at a point p?

If the above is true, can we take the idea that (when applying the Generalized Stokes Theorem) the interior "fluxes" through each $\partial{vol}$ within M cancel out leaving us with the total "flux" out of $\partial{M}$ as the intuition behind the Generalized Stokes Theorem?

Any help is much appreciated.

答:
Yes, this is very good intuition for the theorem. All of this can be made precise.

In fact, it is made precise on pages 188-190 of the second edition of Arnold's Mathematical Methods of Classical Mechanics. There Arnold gives the following theorem, where $\omega$ is a given $k$-form on an $n$-dimensional manifold $M$.

First, the geometric setup. The idea is to construct a $(k+1)$-form by computing its value on a given list of $k+1$ vectors.

So let $\xi_1,\ldots,\xi_{k+1}$ be tangent vectors in $T_xM$, where $x\in M$. We can pull these back to $\mathbb{R}^n$ in the following way. Choose a coordinate system $\phi:U\to\mathbb{R}^n$ for $x\in U$ with $\phi(x)=0$. The preimages of the $\xi_i$ under the differential of $\phi^{-1}$ are tangent vectors $\xi_i^*$ in $T_0\mathbb{R}^n$. But we can naturally identify this tangent space with $\mathbb{R}^n$ itself. Let $\Pi^{*}$ be the parallelepiped in $\mathbb{R}^n$ spanned by these vectors. The map $\phi^{-1}$ carries this linear parallelepiped onto a "curvilinear parallelepiped" $\Pi$ in $M$ (this is like what you call "vol"), the boundary of which is a $k$-chain, $\partial\Pi$ (which is like what you call the boundary of "vol"). Define $F$ to be the integral of the given $k$-form $\omega$ on the boundary of this curvilinear parallelepiped:

$$F(\xi_1,\ldots,\xi_{k+1})=\int_{\partial\Pi}\omega$$

Theorem. There is a unique $(k+1)$-form $\Omega$ on $T_xM$ which is the principal $(k+1)$-linear part at zero of $F$, i.e.

$$F(\epsilon\xi_1,\ldots,\epsilon\xi_{k+1})=\epsilon^{k+1}\Omega(\xi_1,\ldots,\xi_{k+1})+o(e^{k+1})$$

as $\epsilon\to0$. The form $\Omega$ does not depend on the choice of coordinates. And this unique form $\Omega$ is precisely $d\omega$ (in the usual calculation).

In this sense, $d\omega$ is indeed a kind of "flux density" of $\omega$.

The proof of the generalized Stokes theorem then follows, as littleO suggests in a comment, by making precise the idea of chopping up the manifold into little parallelepipeds. One just keeps track of all the orientations as one integrates the flux density over these little regions, $\int_{M}d\omega$. When you do so carefully, the interior fluxes cancel, leaving only the flux along the boundary, $\int_{\partial M}\omega$.


问:
How much warehouse space can we use so that all the crates are accessible?
The warehouse has a square floor with a side of 12 m,
divided into 144 squares with a side of 1 m. In each 1 m square can be placed a crate, filling up the square entirely. All the crates have to placed perfectly in lines. You cannot put the crate in the lower right corner, because there is the only door to the warehouse. Each crate must be accesible from at least one side. We move around the warehouse by walking on empty squares, moving from a square
to the adjacent square.

(a) How many boxes can we store (according to the rules above) in the warehouse? Show a solution.

(b) Find a maximum solution to (a) using n crates and prove we cannot store n + 1 crates.

Here are some examples to help.

Pretty good idea is to use the bottom row for a walkway and then store crates in every second column. Even better idea is to use the leftmost column entirely (even the bottom tile) and then leave one empty column, use two columns, then one gap and so on... This gives 89 used squares if I am right.
All the restrictions I can think of are pretty weak e.g. an empty tile is adjacent to no more than 4 used tiles so at least 1/5 tiles are empty.

答:
Here's a solution with $91$ crates:

```<code>X XX XX XX X
X XX XX XX X
X XX XX XX X
X XX XX XX X
X XX XX XX X
X XX XX XX X
X XX XX XX X
X XX XX XX X
X XX XX XX X
X XX XX XX X
X
X XXXXXXXXX.
</code>```

I can prove that at most $96$ crates are possible.

Suppose that there are $x$ crates and $y$ empty squares we can reach from the start; possibly, there are also some inaccessible empty squares, so $x+y \le 144$. In any arrangement, it is possible to give each square except for the entrance a "parent square" such that:

every parent is an empty space;

by starting at any crate and repeatedly going to the parent square, we eventually get to the door of the warehouse.

Excluding the entrance entrance, there are $x+y-1$ squares that must have a parent. However, each of the $y$ accessible empty squares can only be a parent three times, so $x+y-1 \le 3y$, or $x-2y \le 1$. Therefore $3x = 2(x+y) + (x-2y) \le 2(144)+1 = 289$, which means $x \le \frac{289}{3}$. Because $x$ is an integer, we conclude $x \le 96$.

In the solution above, we get fewer crates due to $7$ empty spaces next to a wall (these are inefficient because they can only be a parent twice) and $9$ crates with two adjacent empty spaces (these are inefficient because they get two parents). With these inefficiencies in play, we actually know $x \le \frac{289 - 7 - 9}{3} = 91$.

It's possible that we can have fewer than $7+9$ inefficiencies. (We can't eliminate inefficiencies entirely; for example, the entrance is an empty space next to a wall, which is already $1$ inefficiency). Proving a lower bound bound on the number of inefficiencies probably requires tedious casework.

Body:
Here's a solution with $91$ crates:

X XX XX XX X
X XX XX XX X
X XX XX XX X
X XX XX XX X
X XX XX XX X
X XX XX XX X
X XX XX XX X
X XX XX XX X
X XX XX XX X
X XX XX XX X
X
X XXXXXXXXX.

I can prove that at most $96$ crates are possible.

Suppose that there are $x$ crates and $y$ empty squares we can reach from the start; possibly, there are also some inaccessible empty squares, so $x+y \le 144$. In any arrangement, it is possible to give each square except for the entrance a "parent square" such that:

every parent is an empty space;

by starting at any crate and repeatedly going to the parent square, we eventually get to the door of the warehouse.

Excluding the entrance entrance, there are $x+y-1$ squares that must have a parent. However, each of the $y$ accessible empty squares can only be a parent three times, so $x+y-1 \le 3y$, or $x-2y \le 1$. Therefore $3x = 2(x+y) + (x-2y) \le 2(144)+1 = 289$, which means $x \le \frac{289}{3}$. Because $x$ is an integer, we conclude $x \le 96$.

In the solution above, we get fewer crates due to $7$ empty spaces next to a wall (these are inefficient because they can only be a parent twice) and $9$ crates with two adjacent empty spaces (these are inefficient because they get two parents). With these inefficiencies in play, we actually know $x \le \frac{289 - 7 - 9}{3} = 91$.

It's possible that we can have fewer than $7+9$ inefficiencies. (We can't eliminate inefficiencies entirely; for example, the entrance is an empty space next to a wall, which is already $1$ inefficiency). Proving a lower bound bound on the number of inefficiencies probably requires tedious casework.


问:
Solving the inequality $0< \lceil \frac{x}{2} \rceil - \lceil \frac{x+1}{3} \rceil$ algebraically
I'm not particularly familiar working with the ceiling function, particularly not algebra with it. I'm trying to solve the above inequality algebraically, specifically when $x\in \mathbb{Z}^{+}$. I've plotted the graph $y= \lceil \frac{x}{2} \rceil - \lceil \frac{x+1}{3} \rceil$ and examined where the curve is greater than $0$. The weird part about this curve is that when I plug in $x=5$, algebraically I obtain a difference of 1, but the plot shows me that the difference is 0, which is why I'm hesitant to trust the graph, I also know that in a proof I can't just put the graph. In terms of what I've tried algebraically, I've tried to consider 6 cases, depending on the equivalence class of $x \pmod{6}$, but again I get stuck because I feel as though I'm missing some properties of how the ceiling function behaves. If anyone could help me solve this I'd really appreciate it!

答:
If $x$ is integer then you write $x=6k+r$ where $r\in\{0,1,2,3,4,5\}$. Then you get $$0<3k+\lceil{r\over 2} \rceil -2k- \lceil{r+1\over 3} \rceil$$

Now check what happens when you put $r=0,1,...$

If $r=0$ then we get $0<k+0 - 1 $...

If $r=1$ then we get $0<k+1 - 1 $...

If $r=2$ then we get $0<k+1 - 1 $...

$\vdots $


问:
Yet another intuition question on surface integrals?
This may seem a basic question about surface integrals, but I'm reading Div, Grad, Curl, and All That, and it prompted me some soul-searching.

In most textbooks, they start discussion of a surface integral by saying take a small patch, $S$, of the surface and imagine a force acting through it.

This force acts at some angle relative to the normal vector, as the book illustrations invariably show.

Then, say the books, the formula for the flux of that small area, $S$, is the dot product of the Force and the Normal vector. And of course as $S$ gets infinitesimally small, integration becomes the solution.

So my intuition question, and apologies for the elementary nature of this. Is the reason you take the dot product of the normal and the force because we are projecting the force on to the normal? And is this because the actual force acting on $S$ is not F itself, since it is an angle (kind of like the incline plane problem in basic physics) to the normal, but the projection of F onto N?

In other words: in a simple incline plane problem, the force of someone pulling up the hill has a vertical and a horizontal component, and to find the amount of force in the horizontal direction you can take the projection of F onto the horizontal plane (right?). So in the case of a surface integral, the force in the normal direction, which I am really interested in, is the dot product of F and the normal vector. And that is why you integrate F dot N ds in surface integrals. And, as the vertical force in an inclined plane problem is a bit of a "waste," so, too, flux will be less the greater the angle between the force and the normal vector.

How wrong is that?

答: I think you're missing the main point in the physical interpretation here. We're finding the flux of the vector field across the surface. We want to find the (rate of) fluid flow of $\vec F$ across the surface (think of $\vec F$ as some variant of a velocity vector field of the fluid). Fluid that flows tangent to the surface contributes no flux. Thus, we want the normal component of $\vec F$, multiplied by area, to figure out how much fluid goes across. (If it helps, see part of my YouTube video. There is a small glitch, unfortunately, in the discussion, which I did correct a few minutes later.)


问:
Recurrence relations of the form $T(n) = n T(n-1) + ...$
In the analysis of algorithms, you often get recurrence relations of the form
$$
T(0) = 1 \\
T(n) = n T(n - 1) + f(n)
$$
where the $f(n)$ varies depending on the details. For certain $f$, the solution is well-known. For example if $f(n)$ is in $\Theta(n)$, the solution is $\Theta(n!)$. Or if $f(n) = n!$, the solution is $\Theta((n+1)!)$.

But I sometimes encounter situations where I don't have an answer. For example, I have no idea what happens if $f$ is a higher-degree polynomial, or if $f = (n+1)!$.
When does the switch from $\Theta(n!)$ to $\Theta((n+1)!)$ occur?
How "bad" may $f$ be for $T(n)$ to still stay within the $\Theta((n+1)!)$ bound?

So my question is whether there is a kind of Master Theorem or general rule for this form of recurrence relation.

答:
We have for any $ k\in\mathbb{N}^{*} $ : \begin{aligned}\frac{T\left(k\right)}{k!}=\frac{T\left(k-1\right)}{\left(k-1\right)!}+\frac{f\left(k\right)}{k!}\end{aligned}

Thus for any $ n\in\mathbb{N}^{*} $ : \begin{aligned}\sum_{k=1}^{n}{\left(\frac{T\left(k\right)}{k!}-\frac{T\left(k-1\right)}{\left(k-1\right)!}\right)}&=\sum_{k=1}^{n}{\frac{f\left(k\right)}{k!}}\\ \iff\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{T\left(n\right)}{n!}-1&=\sum_{k=1}^{n}{\frac{f\left(k\right)}{k!}}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T\left(n\right)&=n!\left(1+\sum_{k=1}^{n}{\frac{f\left(k\right)}{k!}}\right)\end{aligned}


问:
Vector Field Generating Variation Along Curve
I'm learning a proof of the fact that length extremising curves are geodesics of the Levi-Civita connection, and have found something I don't understand. The argument states the following.

Suppose $\gamma$ is an arbitrary curve in a manifold $M$ and $N$ is an arbitrary vector field along $\gamma$ which vanishes at the endpoints. Then there must exist a variation $\tilde{\gamma}$ of $\gamma$ such that $\gamma'$ and $N$ are the pushforwards of the coordinate vector fields on the domain of the variation.

I can't see how this is true, because it would immediately imply that all vector fields commute (I think!). Am I missing something or is it false? If it's wrong, does anyone know how to prove that statement that length extremizing implies geodesic.

The relevant lectures notes from which I've taken this argument are here, page 32. Thanks in advance!

答:
The claim is only that if you're given a vector field $N$ along the curve, then you can find a variation of the curve such that the pushforwards of the coordinate vector fields restrict to $\gamma'$ and $N$ along the curve.  This is not the same as being given a vector field on the manifold.

If the dimension of $M$ is 2, and the variation $\tilde\gamma$ happens to be a smooth embedding, then the pushforwards actually will define vector fields on $M$ in a neighborhood of the curve (minus the endpoints), and these vector fields will commute.  The fact that you can do this for an arbitrary $N$ is no contradiction, because in this case $N$ must be nowhere tangent to the curve, and the values of the bracket along the curve are not determined solely by the values of $N$ and $\gamma'$ there; they depend on how the vector fields are extended, because the coordinate formula for the bracket involves differentiating the coefficients of $\gamma'$ in the direction of $N$.

In the example @Ted Shifrin gave with $N=fT$, the above argument doesn't apply, because if $\gamma'$ and $N$ are linearly dependent, then $\tilde \gamma$  cannot be an embedding.  And in dimensions higher than 2, you won't ever get vector fields defined on an open subset of $M$; they'll only be defined on the (at most) 2-dimensional image of $\tilde \gamma$.


答:
The sequence $\left(\frac{T(n)}{n!}\right)_{n\geqslant 0}$ verifies the recursive relation
$$ \frac{T(n)}{n!}=\frac{T(n-1)}{(n-1)!}+\frac{f(n)}{n!} $$
Summing this identity gives
$$ T(n)=n!\left(1+\sum_{k=1}^n\frac{f(k)}{k!}\right) $$
Therefore if $f(n)=(n+1)!$, then $T(n)\underset{n\rightarrow +\infty}{\sim} \frac{n^2}{2}n!$, or if $f(n)=\sum_{i=0}^p a_in^i$ with $a_p\neq 0$, then
$$ T(n)=n!\left(1+\sum_{i=0}^p a_i\sum_{k=1}^n\frac{k^i}{k!}\right)=\Theta(n! ) $$

Title: Recurrence relations of the form $T(n) = n T(n-1) + ...$

Body:
In the analysis of algorithms, you often get recurrence relations of the form
$$
T(0) = 1 \\
T(n) = n T(n - 1) + f(n)
$$
where the $f(n)$ varies depending on the details. For certain $f$, the solution is well-known. For example if $f(n)$ is in $\Theta(n)$, the solution is $\Theta(n!)$. Or if $f(n) = n!$, the solution is $\Theta((n+1)!)$.

But I sometimes encounter situations where I don't have an answer. For example, I have no idea what happens if $f$ is a higher-degree polynomial, or if $f = (n+1)!$.
When does the switch from $\Theta(n!)$ to $\Theta((n+1)!)$ occur?
How "bad" may $f$ be for $T(n)$ to still stay within the $\Theta((n+1)!)$ bound?

So my question is whether there is a kind of Master Theorem or general rule for this form of recurrence relation.


答:
$f(x)=\frac{1-\cos x}{x^2}$ is an even function, so we only look at $x \in [0,1]$. It's easy to prove $f(x)$ is continuous and differentiable at $x=0$, and

$$f'(x) = \frac{x \sin x + 2 \cos x - 2}{x^3}$$

Now $$x \sin x + 2 \cos x - 2 \le 2 \tan \frac x2 \cdot \sin x - 4 \sin^2 \frac x2\\= 2 \tan \frac x2 \cdot 2 \sin \frac x2 \cos \frac x2 - 4 \sin^2 \frac x2=0, \forall x \in [0,1]$$

Hence $f(x) \ge f(1) = 1 - \cos 1\approx .46 > \frac 13, 1-\cos x \ge (1-\cos 1) x^2 > \frac 13 x^2.\blacksquare$

Body: How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? My first idea was to write $$1-\cos x=\frac12\left|e^{{\rm i}x}-1\right|^2\tag1,$$ which is true for all $x\in\mathbb R$, but I don't have a suitable lower bound for the right-hand side at hand.


问:
How must we define multiplication in a nonabelian group of order 6?
In Non-abelian group of order $6$, a nonabelian group of order 6 was constructed as: $ \{1,a,b,ab,ba,aba\} $ where $1$ is the identity and $a$ and $b$ are their own inverses. Since the group is nonabelian, $ab≠ba$. If we define $aba=bab$, then the group is indeed closed under multiplication:

*
1
a
b
ab
ba
aba

1
1
a
b
ab
ba
aba

a
a
1
ab
b
aba
ba

b
b
ba
1
bab=aba
a
baba=ab

ab
ab
aba
a
abab=ba
1
ababa=b

ba
ba
b
bab=aba
1
baba=ab
a

aba
aba
ab
abab=ba
a
ababa=b
1

My question is: why do we have to have $aba=bab$? In the linked question, the answer says $aba=bab$ "by symmetry", but I don't understand what that means. Are we simply defining $aba$ to equal $bab$, or is there something more interesting going on here?

Thanks!

答: Clearly $bab$ is neither $1$ nor $a$ (because then $ba=ab^{-1}=ab$) nor $b$ nor $ab$ nor $ba$. Therefore, it's the last element; by the same logic, so is $aba$.


答:
Working modulo $6$ there are $6$ residue classes $[0]_6,[1]_6,[2]_6,[3]_6,[4]_6,[5]_6$, thus we can write every element in $\mathbb{Z}$ as $6n+r$ for some $n\in\mathbb{Z}$ and $r=0,1,2,3,4,5$.

The ceiling function has the property that it rounds up to the nearest integer thus $\lceil \text{integer} + d \rceil = \text{integer} +1$ where $0\leq d <1$.

$\lceil\frac{6n}{2}\rceil = 3n$

$\lceil\frac{6n+1}{2}\rceil =\lceil3n+\frac{1}{2}\rceil = 3n+1$

$\lceil\frac{6n+2}{2}\rceil =\lceil3n+1\rceil = 3n+1$

$\lceil\frac{6n+3}{2}\rceil =\lceil3n+1+\frac{1}{2}\rceil = 3n+2$

$\lceil\frac{6n+4}{2}\rceil = 3n+2$

$\lceil\frac{6n+5}{2}\rceil =\lceil3n+2+\frac{1}{2}\rceil = 3n+3$

similarly

$\lceil\frac{6n+1}{3}\rceil = 2n+1$

$\lceil\frac{6n+2}{3}\rceil = 2n+1$

$\lceil\frac{6n+3}{3}\rceil = 2n+1$

$\lceil\frac{6n+4}{3}\rceil = 2n+2$

$\lceil\frac{6n+5}{3}\rceil = 2n+2$

$\lceil\frac{6n+6}{3}\rceil = 2n+2$

for $n \in \mathbb{Z}$

and

$3n-2n-1=n-1$

$3n+1-2n-1=n$

$3n+2-2n-2=n$

$3n+3-2n-2=n+1$

From which it is equal to zero on multiple occasions

$\lceil \frac{6}{2} \rceil - \lceil \frac{7}{3} \rceil =0$ ; $x=6$

$\lceil \frac{1}{2} \rceil - \lceil \frac{2}{3} \rceil =0$ ; $x=1$

$\lceil \frac{2}{2} \rceil - \lceil \frac{3}{3} \rceil =0$ ; $x=2$

$\lceil \frac{3}{2} \rceil - \lceil \frac{4}{3} \rceil =0$ ; $x=3$

$\lceil \frac{4}{2} \rceil - \lceil \frac{5}{3} \rceil =0$ ; $x=4$

Maybe you were looking at $x=6$?

Title: Solving the inequality $0< \lceil \frac{x}{2} \rceil - \lceil \frac{x+1}{3} \rceil$ algebraically

Body: I'm not particularly familiar working with the ceiling function, particularly not algebra with it. I'm trying to solve the above inequality algebraically, specifically when $x\in \mathbb{Z}^{+}$. I've plotted the graph $y= \lceil \frac{x}{2} \rceil - \lceil \frac{x+1}{3} \rceil$ and examined where the curve is greater than $0$. The weird part about this curve is that when I plug in $x=5$, algebraically I obtain a difference of 1, but the plot shows me that the difference is 0, which is why I'm hesitant to trust the graph, I also know that in a proof I can't just put the graph. In terms of what I've tried algebraically, I've tried to consider 6 cases, depending on the equivalence class of $x \pmod{6}$, but again I get stuck because I feel as though I'm missing some properties of how the ceiling function behaves. If anyone could help me solve this I'd really appreciate it!


问:
Derivative of an integral with regard to a different variable
Assume I have the following expression:

$$\int_0^x f(t,\theta) dt$$

where $t$ (or $x$) and $\theta$ are some vector-valued variables, and $f$ is some function which does not permit an easy analytical solution to the integral (But can be approximated numerically). Assume now that I seek the partial derivative $\frac{\partial}{\partial \theta}=[\frac{\partial}{\partial \theta_1},...,\frac{\partial}{\partial \theta_N}]^T$ of this expression, i.e.:

$$\frac{\partial}{\partial \theta}\int_0^x f(t,\theta) dt$$

How can I solve this? Is it possible to approach the solution like this: (?)

$$\frac{\partial}{\partial \theta}\int_0^x f(t,\theta) dt=\int_0^x \frac{\partial f(t,\theta)}{\partial \theta} dt$$

If not, is there a different way to solve this?

答: Since the integration is with respect to $t$ and its limits are independent of $\theta$, your approach is correct in this particular case. Derivation and integration are both linear operators and follow the general rules of the interchangeability of linear operators.


问:
A simple proof that a plane similarity without fixed points is an isometry
Is there a (relatively) simple proof of the fact that a plane similarity without fixed points is an isometry?

In the literature it is usually formulated as "a non-isometric similarity has a fixed point" and the proof is somewhat involved, using the concept of directed distance. I am looking for something simpler, to use for the classification of plane similarities.

A simple proof of the above-mentioned classification will also do.

答:
A similarity can be expressed as:
$$
s(\vec x)=rA\vec x+\vec t,
$$
where $r$ is the scale factor, $A$ is an orthogonal matrix and $\vec t$ is a translation vector. If $s(\vec x)=\vec x$ then:
$$
\vec x=(I-rA)^{-1}\vec t,
$$
where $I$ is the identity matrix. Hence a fixed point doesn't exist only if $I-rA$ is a singular matrix. Plugging in the explicit form of $A$ in ${\mathbb R}^2$ it is easy to see that $\det(I-rA)=0$ only if $r=1$.

EDIT.

In the special case $s=i\circ\sigma$, where $i$ is a translation of vector $\vec t$ and $\sigma$ a stretch of scale factor $r$, it is easy to construct a fixed point $P$ of $s$ if $r\ne1$:
$$
P=O+{1\over1-r}\vec t,
$$
where $O$ is the fixed point of $\sigma$. In fact:
$$
P'=\sigma(P)=O+{r\over1-r}\vec t
\quad\text{and}\quad
s(P)=i(P')=P'+\vec t=O+{1\over1-r}\vec t=P.
$$
If $i$ is a glide reflection, we can set up a coordinate system such that $x$ axis is the reflection line, $\vec t=(t,0)$ is the translation vector, while the fixed point of $\sigma$ is on $y$ axis at $O=(0,d)$.
A simple computation shows then that $P=(x,y)$ is a fixed point if:
$$
x={r\over1-r} t,\quad y={1-r\over1+r}d.
$$


问:
On the pH scale, each unit change in pH represents a tenfold increase in acidity or alkalinity.
Trying to solve similar type equation to this.

On the pH scale, each unit change in pH represents a tenfold increase in acidity or alkalinity. According to the diagram, vinegar is how many times as acidic as pure water?

答: Simply put isn't it 10^(3.8) ?</p>

Body: Simply put isn't it 10^(3.8) ?


问:
Puzzle relating squares to 2020 and 2021 and the question when this happens again?
Take the numbers of the current and next year $(2020,2021)$

I noticed that  $ 2020 = 2* 1010 $ and if we take the square of theses divisors added by $2021$ we get the squares $45$ and $1011$.

Or also $$  2021 = 45^2-2^2 $$
and  $$ 2021 = 1011^2 - 1010^2 $$

What is the next year, if there is one, where it will happen that for both squares of the divisor of the old year plus the new year are square?

答:
Well I wrote a program, and the answer is $2204 = 2\times1102$.

Most of the times (I have tested for $n \leq 50000$ only) the only divisor is $2$, so you end up with $n = 2 \times (n/2)$. There are some quite remarkable numbers though :$144=12\times 12$ $455=13\times35 $ , $4900 = 70 \times 70$, $5719=43\times 133$, $26676=78\times 342$, $41040=180\times 228$.

See you in $2204$.


问:
What is known about Lebesgue's universal covering problem in three dimensions?
Lebesgue's universal covering problem is a relatively well-known open problem in geometry, asking for the convex set of minimal area which contains all planar sets of diameter 1. While the problem is not yet fully resolved, we have pretty good bounds on its size (between $0.832$ and $0.84409$) and, as I understand it, a decent sense of what an optimal solution ought to look like (some minor trimming-off from Pál's 1920 solution).

What progress has been made on the analogous problem in three dimensions? As a very crude starting point, we can take a prism of height $1$ whose base is any solution to the covering problem in two dimensions, but this is almost certainly suboptimal; for instance, given any starting solution with an axis of symmetry (such as the one above), we can find two congruent regions of positive volume whose points are all of distance $1$ from each other (by cutting along planes normal to a sort of space diagonal), and then remove one of those regions.

Are there more sophisticated modifications we can make, or entirely different regions to use? (After all, the best 2D solutions do not look much like the 1D solution crossed with a unit interval.)

答:
After perusing section 11.4 of Research Problems in Discrete Geometry, I came across a result of Makeev, Hausel, Makai, Szucs, and G. Kuperberg that the rhombic dodecahedron with unit inradius, which has volume $\frac{\sqrt{2}}2$, is a universal cover for all sets in $\mathbb{R}^3$ of unit diameter.

One can then observe that opposite degree-4 vertices of this rhombic dodecahedron are at distance $\sqrt{2}$. Consider a slab of width 1 perpendicular to the line connecting these vertices whose center coincides with that of the rhombic dodecahedron. There are two square pyramids of height $\frac{\sqrt{2}-1}2$ lying outside the slab; since any two points in opposite pyramids are at distance at least $1$ from each other, a set of unit diameter has points in at most one of them (WLOG, say the bottom one, since we can flip the set within the vertically-symmetric rhombic dodecahedron.) Then we can remove the top pyramid to obtain a slightly smaller universal cover.

In fact, we can repeatedly apply this procedure to each of the three mutually orthogonal axes between two degree-4 vertices on the rhombic dodecahedron, so we can ultimately remove 3 of these pyramids. This leaves us with a universal cover of volume

$$\frac72-2\sqrt{2}\approx 0.67157$$

(It is also mentioned in this book that the regular octahedron of inradius $1$ serves as a universal cover, but as this has volume $\frac{\sqrt{3}}2\approx 0.866$, it is not a promising candidate.)


问:
How can I attack this definite integral - roots problem?
I have this problem:

Let $f(x)$ be continuous on the interval $[0,1] .$ Prove that if
$$
  \int_{0}^{1} f(x) d x=\int_{0}^{1} x f(x) d x=0
  $$
then $f$ has at least two different roots in (0,1)

Is it better to start with parts integration on second one? or triying
$$
  \int_{0}^{1} f(x) d x=\int_{0}^{1} x f(x) d x=0\Rightarrow \int_{0}^{1} f(x) -x f(x) d x=0 ?
 $$

If I use mean value Theorem I coul get
$$
\int_{0}^{1} (1-x) f(x) d x=(1-\xi)\int_{0}^{1}  f(x) d x=0
$$
With $\xi\in[0,1]$

Therefore
$$
  \int_{0}^{1} f(x) d x=0\qquad  or\qquad \xi =1
 $$

But doesn't seems to help, I already know
$$
  \int_{0}^{1} f(x) d x=0
 $$
and $\xi=1$ has no further use

So I'm kind of stuck right now

I'll appreciate any help or suggestion

Thanks in advance

答:
for a (non-zero) function over this range there must be two areas that "cancel out" i.e. there is a $0<x_0<1$ where $f(x_0)=0$ which suggests:
$$0=\int_0^1f(x)dx=\int_0^{x_0}f(x)dx-\int_{x_0}^1f(x)dx$$
If there is one root in a domain then there must be a positive and negative area, and if there are two roots there must be three regions, 2 positive/negative and 1 the opposite.

On the other hand if we define a function $g(x)=xf(x)$ we know that $g(0)=0$ which means for the regions to cancel out again we must have 3 roots in the function $g$, one being $x=0$ which suggests that $f$ has 2 roots.

I am not sure how much sense this makes so I will try and define it slightly differently, lets say we have a function:
$$\begin{matrix}f_n(x)=\prod_{i=1}^n(x-x_i) & 0\le x_i\le1\end{matrix}$$
this represents a polynomial with $n$ roots in the domain $x\in[0,1]$

If we assume that $f=f_1$ we have the expression:
$$\int f(x)dx=\int_0^1x-x_idx=\left[\frac{x^2}{2}-x_ix\right]_0^1=\frac{1}{2}-x_i=0$$
so there is an $x_i$ which satisfies this solution. If we now try and do the second integral:
$$\int xf(x)dx=\int_0^1x^2-x_ixdx=\left[\frac{x^3}{3}-\frac{x_i}{2}x^2\right]_0^1=\frac{1}{3}-\frac{1}{2}x_i=0$$
and whilst this does have a solution that makes this work it does not match the one from the first integral, suggesting that if this is a polynomial it must have at least two roots.


问:
If $A$ is a ring, then why is $\operatorname{Proj} A[t] \cong \operatorname{Spec} A$?
The following is an example I saw in a book on Algebraic Geometry.

Example: Let $A$ be a ring and consider $A[t]$, with the grading given by $\deg t = 1$ and $\deg a = 0$ for $a \in A$. Then the structure map gives an isomorphism $\operatorname{Proj} A[t] \cong \operatorname{Spec} A$

By structure map, the authors mean a morphism of schemes $\varphi : \operatorname{Proj} A[t] \to \operatorname{Spec} A$. I am not sure exactly what this morphism would be however, I assume that we start off with the standard embedding $A \hookrightarrow A[t]$ and then apply the functor $\operatorname{Spec}$ to get a map $\operatorname{Spec} A[t] \to \operatorname{Spec} A$ and then pre-compose that with the inclusion $i : \operatorname{Proj} A[t] \to \operatorname{Spec} A[t]$ to get the desired morphism $\varphi$.

Now firstly, the $\operatorname{Proj}$ construction only works for graded rings. So $A[t]$ should be of the form $A[t] = \bigoplus_{d \geq 0} A_d$. The authors say above "the grading given by $\deg t = 1$ and $\deg a = 0$ for $a \in A$", and I assume that this means that $A_0$ is the image of the standard inclusion of $A$ in $A[t]$, and that $A_1 = (t)$, the ideal generated by $t$ so that "essentially" $A[t] = A_0 \oplus A_1 = A \oplus (t)$. Is this the correct interpretation?

Now onto the claim that the $\varphi$ yields an isomorphism. I don't even see why $\varphi$ is a bijection, let alone a homeomorphism. If I label the embedding $A \hookrightarrow A[t]$ as $j$, then $\varphi$ is given by $\varphi(\mathfrak{p}) = j^{-1}(\mathfrak{p})$ for any $\mathfrak{p} \in \operatorname{Proj} A[t]$ and to say that $\varphi$ is bijective would mean that every prime ideal $\mathfrak{q}$ of $A$ is of the form $\mathfrak{q} = j^{-1}(\mathfrak{p})$ for some unique $\mathfrak{p} \in \operatorname{Proj} A[t]$ and for starters I do not see why this would be the case.

Furthermore I would need to show that the morphism of sheaves $\varphi^\sharp : \mathcal{O}_{\operatorname{Spec} A} \to \varphi_* \mathcal{O}_{\operatorname{Proj} A[t]}$
is also isomorphic and the way one usually does is by looking at basic open sets on $\operatorname{Spec} A$. In more words -  if I wanted to see what $\varphi^\sharp$ would be in this case, then if I chose an $f \in A$ and considered the open subset $D(f) \subseteq \operatorname{Spec} A$ then $\Gamma(D(f), \mathcal{O}_{\operatorname{Spec} A}) \cong A_f$. But $$\Gamma(D(f), \varphi_*\mathcal{O}_{\operatorname{Proj} A[t]}) = \Gamma(\varphi^{-1}(D(f)), \mathcal{O}_{\operatorname{Proj} A[t]})$$ and this may not have such a nice form as far as I know, since $\varphi^{-1}(D(f))$ might not be a basic open subset of $\operatorname{Proj} A[t]$. So I cannot express $\varphi^\sharp_{D(f)} : \Gamma(D(f), \mathcal{O}_{\operatorname{Spec} A}) \to  \Gamma(\varphi^{-1}(D(f)), \mathcal{O}_{\operatorname{Proj} A[t]})$ in any nice way to allow me to check that $\varphi^\sharp$ would indeed be an isomorphism.

答:
For $R$ a graded ring, $\operatorname{Proj} R$ has as its underlying set the homogeneous prime ideals of $R$ which do not contain the irrelevant ideal. In our case, the irrelevant ideal is $(t)$, and so no homogeneous prime ideal in $\operatorname{Proj} A[t]$ contains $t$. So $D_+(t)\subset\operatorname{Proj} A[t]$ is all of $\operatorname{Proj} A[t]$, and therefore by the identification $D_+(t)\cong \operatorname{Spec} A[t]_{(t)}$ (where this denotes the homogeneous localization, or $(A[t]_t)_0$) and the observation that $A[t]_{(t)}=A$, we see that $\operatorname{Proj} A[t]\cong\operatorname{Spec} A$.

If you want to get more hands-on, you can verify that any homogeneous prime $P\subset A[t]$ not containing the irrelevant ideal is exactly specified by it's intersection with $A$. For every homogeneous element $at^n\in P$, we get that either $a\in P$ or $t\in P$ by primality of $P$. By the assumption that $(t)\not\subset P$, the former must be the case, so $P$ is the ideal generated by $P\cap A$. Thus there's a bijection as sets between $\operatorname{Proj} A[t]$ and $\operatorname{Spec} A$, and one can verify that this induces a bijection between basic open sets ($D(a)\subset \operatorname{Spec} A$ and $D_+(f)\subset \operatorname{Proj} A[t]$ where $f$ is homogeneous of positive degree), so it is even a homeomorphism.


问:
Derivative of a Stochastic Integral with respect to Limit & with respect to Integrator
I have recently come across an attempt to differentiate the following function with respect to $t$ and with respect to $W_t$:

$$F(W_t, t):=\int_{h=0}^{h=t}W_hdW_h$$

Is it possible (i.e. is it "well defined") to differentiate $F$ with respect to $t$ and with respect to $W_t$?

My train of thought would be as follows: by applying Ito's lemma to $W_t^2$, we know that $F_t$ evaluates to $\frac{1}{2}(W_t^2-t)$. Differentiating with respect to $W_t$ would then simply yield $W_t$, whilst differentiating with respect to $t$ would then yield $-\frac{1}{2}$:

Question 1: is the above correct? (I have doubts because $\frac{\partial W_t}{\partial t}$ is undefined. But on the other hand, we are effectively carrying out $\frac{\partial F}{\partial t}$, so following the logic of partial derivatives applied to a function of multiple variables, we should be able to treat all variables "not being differentiated with respect to" as constants).

Question 2: In general, is there a "rule" (that would be an "Ito" equivalent of the Fundamental Theorem of Calculus) for differentiating an Ito integral with respect to its upper limit?

Question 3: What if I now define the function $F(W_1(t),W_2(t),t)$ as follows ($W_1$ and $W_2$ are two independent Brownian motions on the same probability space):

$$F(W_1(t),W_2(t),t):=\int_{h=0}^{h=t}W_1(h)dW_2(h)$$

Would $\frac{\partial F}{\partial t}$, $\frac{\partial F}{\partial W_1}$ and $\frac{\partial F}{\partial W_2}$ be (well) defined?

答:
For question 1, you are correct that $F$ is not differentiable with respect to $t$.  It's not exactly clear what it would mean for $F$ to be differentiable with respect to $W_t$ because in general we don't think of $W_t$ changing without $t$ also changing.  So while $F$ is clearly differentiable with respect to its first argument, I'm not sure that's exactly what you want when you talk about $F(W_t,t)$ being differentiable with respect to $W_t$.

For the second question, which also addresses question 3, we have something very similar to the fundamental theorem of calculus: If $H$ is adapted, right-continuous, and bounded then $$\lim_{t \rightarrow 0} \frac{1}{W_t} \int_0^t H_s dW_s = H_0$$ in probability.

Proof: Fix $K, \varepsilon > 0$.  We compute

\begin{align*}
&\phantom{= }P\left( |\frac 1{W_t}\int_0^t H_s dW_s - H_0| > \varepsilon \right)\\
&= P\left( |\frac 1{W_t}\int_0^t (H_s - H_0)dW_s| > \varepsilon \right) \\
&=  P\left( |\frac 1{W_t}\int_0^t (H_s - H_0)dW_s| > \varepsilon, \frac{|W_t|}{\sqrt{t}} > K \right) + P\left( |\frac 1{W_t}\int_0^t (H_s - H_0)dW_s| > \varepsilon, \frac{|W_t|}{\sqrt{t}} \le K \right) \\
&\le P\left( |\frac 1{W_t}\int_0^t (H_s - H_0)dW_s| > \varepsilon, \frac{|W_t|}{\sqrt{t}} > K \right) + P\left(\frac{|W_t|}{\sqrt{t}} \le K \right).
\end{align*}

Now we focus on the first term and use Markov's inequality:

\begin{align*}
P\left( |\frac 1{W_t}\int_0^t (H_s - H_0)dW_s| > \varepsilon, \frac{|W_t|}{\sqrt{t}} > K \right) &\le P\left( |\frac {1}{K\sqrt{t}}\int_0^t (H_s - H_0)dW_s| > \varepsilon, \frac{|W_t|}{\sqrt{t}} > K \right) \\
&\le  P\left( |\frac {1}{K\sqrt{t}}\int_0^t (H_s - H_0)dW_s| > \varepsilon \right) \\
&\le \frac{1}{\varepsilon^2}\mathbb{E}\left[ \frac{1}{K^2 t} |\int_0^t (H_s - H_0)dW_s|^2\right] \\
&= \frac{1}{\varepsilon^2K^2 t} \mathbb{E}\left[\int_0^t |H_s-H_0|^2 ds \right] \\
&\le \frac{1}{\varepsilon^2K^2 t} \mathbb{E}\left[ t \sup_{0 \le s \le t} |H_s-H_0|^2 \right] \\
&= \frac{1}{\varepsilon^2 K^2}\mathbb{E}\left[\sup_{0 \le s \le t} |H_s-H_0|^2 \right].
\end{align*}

For the second term, by the scaling properties of Brownian motion we have $P\left(\frac{|W_t|}{\sqrt{t}} \le K \right) = P(|X| \le K)$ where $X$ is a standard normal random variable.  Combining everything, we have

\begin{align*}
\lim_{t \rightarrow 0} P\left( |\frac 1{W_t}\int_0^t H_s dW_s - H_0| > \varepsilon \right) &\le \lim_{t \rightarrow 0} \frac{1}{\varepsilon^2 K^2}\mathbb{E}\left[\sup_{0 \le s \le t} |H_s-H_0|^2 \right] + P(|X| \le K).
\end{align*}

Since $H$ is bounded, $\sup_{0 \le s \le t} |H_s-H_0|^2$ is also bounded, and since $H$ is right continuous we know that $\lim_{t \rightarrow 0} \sup_{0 \le s \le t} |H_s-H_0|^2 = 0$ almost surely, so by the dominated convergence theorem we have

\begin{align*}
\lim_{t \rightarrow 0} P\left( |\frac 1{W_t}\int_0^t H_s dW_s - H_0| > \varepsilon \right) &\le \lim_{t \rightarrow 0} \frac{1}{\varepsilon^2 K^2}\mathbb{E}\left[\sup_{0 \le s \le t} |H_s-H_0|^2 \right] + P(|X| \le K) \\
&= P(|X| \le K).
\end{align*}

Finally, sending $K \rightarrow 0$ gives $$\lim_{t \rightarrow 0} P\left( |\frac 1{W_t}\int_0^t H_s dW_s - H_0| > \varepsilon \right) = 0.$$

This result is actually still true if $H$ is continuous but unbounded by localizing $H$ first, so that would be the result you need for question 3.  Let me know if you want me to add on the details for that.


问:
Integration of $ \cos x.\cos 2x...\cos nx$
I wanted to integrate $\int \cos x\cos 2x\cdots \cos nx \, dx$.

What I know is that $ \cos x\cos 2x\cdots \cos nx=\dfrac{1}{2^{n-1}}\sum_\pm \cos((n\pm(n-1)\pm\cdots\pm2\pm1)x)$ where the sum is over all $2^{n-1}$ possible $\pm$.

But quite obviously this is hard to integrate.

From this, I came to know about Werner's formula which I think quite less complicated to solve the above problem. But I don't know how to put this formula for an arbitrary $n$ for the given problem.

Thanks for helping me in advance.

答:
Your question is:
$$I_n=\int\prod_{k=1}^n\cos(kx)dx$$
we could try and use the fact that:
$$\cos(kx)=\frac{e^{ikx}+e^{-ikx}}{2}=\frac{e^{-ikx}}{2}\left(e^{2ikx}+1\right)$$
and then say:
$$\prod_{k=1}^n \cos(kx)=\left(\prod_{k=1}^n\frac{e^{-ikx}}{2}\right)\left(\prod_{k=1}^n(e^{2ix})^k+1\right)$$
this first part is quite easy to do:
$$\prod_{k=1}^n\frac{e^{-ikx}}{2}=2^{-n}\exp\left(-ix\sum_{k=1}^nk\right)=2^{-n}e^{-\frac{in(n+1)}{2}x}$$
now the tough part is calculating:
$$\prod_{k=1}^n(e^{2ix})^k+1$$
and then obviously integrating whatever the result is


问:
countable-to-one set cardinality

Let $B$ and $C$ be nonempty sets. Say a map $ \phi : B\to C$ is countable-to-one if for every $c \in C, \phi^{-1}(\{c\})$ is countable.  Show that if such a map exists then $|B| \leq \aleph_0|C|.$

Observe that $B = \cup_{c\in C} \phi^{-1}(\{c\})$ ($\phi^{-1}(\{c\})\subseteq B$ for each $c\in C$ and for $b \in B, \phi(b) \in C,$ so $b \in \phi^{-1}(\{\phi(b)\})$) and this union is disjoint since $\phi^{-1}(\{c\})\cap \phi^{-1}(\{d\}) = \emptyset$ for $c\neq d.$ Since each $\phi^{-1}(\{c\})$ is countable, it is either finite or equipotent to $\mathbb{N}$; in either case there is an injection $f_c : \phi^{-1}(\{c\})\to \mathbb{N}$ for each $c\in C$. Define $f : B\to \mathbb{N}\times C, f(d) = (f_c(d), c)$ for $d \in \phi^{-1}(\{c\}), c \in C.$ Observe that $f$ is an injection. Indeed, for any $d\neq e\in B,$ we have two cases: $1) d , e \in \phi^{-1}(\{c\}), c \in C$ or $2) d\in \phi^{-1}(\{g\}), e \in \phi^{-1}(\{h\}), g\neq h \in C.$ In case $1), $ since $d\neq e$ and $f_c$ is an injection, $f_c(d) \neq f_c(e),$ so $f(e) = (f_c(e), e)\neq (f_c(d), c) = f(d).$ Now in case $2), f(d) = (f_g(d), g)\neq (f_h(e), h)$ as $g\neq h.$ Thus $f$ is an injection so $|B|\leq |\mathbb{N}||C| = \aleph_0|C|.$

Is this incorrect (i.e. some parts are incorrect)?

答:
I like the idea, but I think your notation in cases 1 and 2 is confusing and sometimes is wrong. For example, you say case 2 is about $d\in\phi^{-1}(\{c\}),e\in\phi^{-1}(\{d\})$, but how can $d$ be an element of $B$ (first part of case 2) and an element of $C$ (second part of case 2) at the same time?

In my opinion you're doing it more complicated than it is: You have an injection $f_c:\phi^{-1}(\{c\})\to\Bbb N$ for each $c\in C$, so you can define $f:B\to\Bbb N\times C$ for each $b\in B$ as $(f_{\phi(b)}(b),\phi(b))$. If you have $b,b'\in B$ such that $(f_{\phi(b)}(b),\phi(b))=(f_{\phi(b')}(b'),\phi(b'))$ then $\phi(b)=\phi(b')$ and therefore $f_{\phi(b)}(b)=f_{\phi(b')}(b')=f_{\phi(b)}(b')$. Since $f_{\phi(b)}$ is an injection, then $b=b'$, so $f$ is an injection.

But, as I said, the idea is good and you solved the problem.

Title: countable-to-one set cardinality


答:
Let $c$ be any odd integer $3$ or greater, let $b=(c^2-5)/2$, and consider the numbers $2b$ and $2b+1$. We have $2^2+(2b+1)=c^2$, and $b^2+(2b+1)=(b+1)^2$.

The original example is $c=45$, $b=1010$ giving $2b=2020$, $2b+1=2021$.

Axel's example is $c=47$, $b=1102$ giving $2b=2204$, $2b+1=2205$.

The last previous year pair would have come from $c=43$, $b=922$, $2b=1844$, $2b+1=1845$.

Clearly, one gets an infinity of examples from these formulas.

All of these examples come from using $2$ as one of the divisors of the first year of the pair of consecutive years. It's a harder problem to find examples that don't use $2$ as a divisor.

Title: Puzzle relating squares to 2020 and 2021 and the question when this happens again?

Body:
Take the numbers of the current and next year $(2020,2021)$

I noticed that  $ 2020 = 2* 1010 $ and if we take the square of theses divisors added by $2021$ we get the squares $45$ and $1011$.

Or also $$  2021 = 45^2-2^2 $$
and  $$ 2021 = 1011^2 - 1010^2 $$

What is the next year, if there is one, where it will happen that for both squares of the divisor of the old year plus the new year are square?


答:
From the inequalities $u\le\lceil u\rceil\lt u+1$, we have

$$\left\lceil x\over2\right\rceil-\left\lceil x+1\over3\right\rceil\gt{x\over2}-\left({x+1\over3}+1 \right)={1\over6}(x-8)$$

The OP's desired (strict) inequality thus holds for certain as soon as $x\ge8$. It remains to check the remaining cases, $x=1,2,\ldots,7$. Checking these one at a time, one finds that strict inequality occurs only for $x=7$ and $x=5$:

$$\left\lceil 7\over2\right\rceil-\left\lceil 8\over3\right\rceil=4-3=1\gt0$$

and

$$\left\lceil 5\over2\right\rceil-\left\lceil 6\over3\right\rceil=3-2=1\gt0$$

The other values all turn out to be equal to $0$.

Added later: I recall being dissatisfied when I first wrote this answer on its reliance on explicit evaluations of $\lceil x/2\rceil$ and $\lceil(x+1)/3\rceil$ for $1\le x\le7$. It finally occurs to me there's a way to simplify things, at least a bit:

The ceiling function $\lceil x/2\rceil$ increases by $1$ to the right of $x=2$, $4$, and $6$, while $-\lceil(x+1)/3\rceil$ dereases by $1$ to the right of $x=2$ and $5$. Thus the combined function $\lceil x/2\rceil-\lceil(x+1)/3\rceil$, which equals $1-1=0$ for small positive (real) values of $x$, doesn't change at all until it gets past $x=4$, at which point it goes up and down by $1$ over the next three intervals, so it's $1$ at $x=5$, back to $0$ at $x=6$, and back to $1$ at $x=7$.


答:
Your function $f$ is fine, but in Case $(2)$ you’re using $d$ for two different things at once. You want $d\in\varphi^{-1}(\{c_1\})$ and $e\in\varphi^{-1}(\{c_2\})$ for distinct $c_1,c_2\in C$. Then you have

$$f(d)=\langle f_{c_1}(d),c_1\rangle\ne\langle f_{c_2}(e),c_2\rangle=f(e)\,,$$

since $c_1\ne c_2$. Once you fix that, $f$ is indeed an injection from $B$ to $\Bbb N\times C$.

Title: countable-to-one set cardinality

Body:
Let $B$ and $C$ be nonempty sets. Say a map $ \phi : B\to C$ is countable-to-one if for every $c \in C, \phi^{-1}(\{c\})$ is countable.  Show that if such a map exists then $|B| \leq \aleph_0|C|.$

Observe that $B = \cup_{c\in C} \phi^{-1}(\{c\})$ ($\phi^{-1}(\{c\})\subseteq B$ for each $c\in C$ and for $b \in B, \phi(b) \in C,$ so $b \in \phi^{-1}(\{\phi(b)\})$) and this union is disjoint since $\phi^{-1}(\{c\})\cap \phi^{-1}(\{d\}) = \emptyset$ for $c\neq d.$ Since each $\phi^{-1}(\{c\})$ is countable, it is either finite or equipotent to $\mathbb{N}$; in either case there is an injection $f_c : \phi^{-1}(\{c\})\to \mathbb{N}$ for each $c\in C$. Define $f : B\to \mathbb{N}\times C, f(d) = (f_c(d), c)$ for $d \in \phi^{-1}(\{c\}), c \in C.$ Observe that $f$ is an injection. Indeed, for any $d\neq e\in B,$ we have two cases: $1) d , e \in \phi^{-1}(\{c\}), c \in C$ or $2) d\in \phi^{-1}(\{g\}), e \in \phi^{-1}(\{h\}), g\neq h \in C.$ In case $1), $ since $d\neq e$ and $f_c$ is an injection, $f_c(d) \neq f_c(e),$ so $f(e) = (f_c(e), e)\neq (f_c(d), c) = f(d).$ Now in case $2), f(d) = (f_g(d), g)\neq (f_h(e), h)$ as $g\neq h.$ Thus $f$ is an injection so $|B|\leq |\mathbb{N}||C| = \aleph_0|C|.$

Is this incorrect (i.e. some parts are incorrect)?


问:
If $\lim\limits_{x \to \pm\infty}f(x)=0$, does it imply that $\lim\limits_{x \to \pm\infty}f'(x)=0$?

Suppose $f:\mathbb{R} \rightarrow \mathbb{R}$ is everywhere differentiable and

$\lim_{x \to \infty}f(x)=\lim_{x \to -\infty}f(x)=0$,

there exists $c \in \mathbb{R}$ such that $f(c) \gt 0$.

Can we say anything about  $\lim_{x \to \infty}f'(x)$ and $\lim_{x \to -\infty}f'(x)$?

I am tempted to say that $\lim_{x \to \infty}f'(x)$ = $\lim_{x \to -\infty}f'(x)=0$.

I started with the following, but I'm not sure this is the correct approach, $$\lim_{x \to \infty}f'(x)= \lim_{x \to \infty}\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}.$$

答:
To correct a incorrect attempt,
let $f(x) = e^{-x^2} \cos(e^{x^4})$,
so

$\begin{align}f'(x) &= e^{-x^2} 4 x^3 e^{x^4}(-\sin(e^{x^3}))
-2x e^{-x^2} \cos(e^{x^3})\\
&= -4 x^3 e^{x^4-x^2} \sin(e^{x^3})
-2x e^{-x^2} \cos(e^{x^3})\\
\end{align}
$

The $e^{x^4-x^2}$ term makes
$f'(x)$ oscillate violently
and unboundedly
as $x \to \pm \infty$.


问:
Reference for a Proposition: "if a subgroup $H$ contains a normal $p$-Sylow $ P$, then $H \subset N_{G}{P}$, a normalizer of a $p$-Sylow in $G$"
Let's first introduce the reference I'm trying to reformulate the proposition from:

How to enumerate subgroups of each order of $S_4$ by hand

For example , for a subgroup of order 6 , the writer explains :

A subgroup of order 6 must have a normal Sylow 3-subgroup, so must live inside the normalizer (inside S4) of a Sylow 3-subgroup. The Sylow 3-subgroups are just the various alternating groups of degree 3, and their normalizers are various symmetric groups of degree 3, so are exactly the 4 subgroups of order 6.

I can't really figure out where is this result coming from , what proposition , theorem it refers to , or just logical argument. If I try to reformulate what I understand in the specific case of $S_4$:

If a subgroup $H$ of $S_4$ contains a normal $p$-Sylow $P$, then $ H\subset N_{S_4}(\rho)$, with $\rho$ a $p$-Sylow of $S_4$ and $N_{S_4}(\rho)$ the normalizer of one of these $p$-Sylow in $S^4$.

I'd like either a correction or a reference for this proposition.

I may be formulating it really bad or maybe I don't get the point in this quote, however, the title of the question will be modified when solved for future readers to be able to refer to it if needed.

答:
I am reading that passage from Jack Schmidt's answer as follows.

It calls upon the following

Fact. If $G$ is a group, $H\le G$ a subgroup, and $K\unlhd H$ yet another subgroup of $G$ that also happens to be a normal subgroup of $H$, then $H\le N_G(K)$. In other words, $H$ is contained in the normalizer of $K$ in $G$.

This hardly needs a proof. The assumption $K\unlhd H$ is equivalent to $hK=Kh$ for all $h\in H$, but the normalizer is
$$
N_G(K)=\{x\in G\mid xK=Kx\}
$$
by definition, so $H\le N_G(K)$.

Anyway, I fill in extra details to the steps of that passage.

A subgroup of order $6$ must have a normal Sylow $3$-subgroup, This can be seen using general Sylow theory, but that feels a bit like swatting flies with cannonballs. Your first course on algebraic structures may have had the result that a group of order six is either cyclic, $C_6=\langle c\rangle$, or the symmetric group $S_3$. Both have a normal Sylow $3$-subgroup. Namely $\langle c^2\rangle$ and $A_3=\langle(123)\rangle$ respectively.

so must live inside the normalizer of a Sylow $3$-subgroup (of $S_4$). Here the Fact is applied. If $H\le S_4$ is an arbitrary subgroup of order six then by the previous bullet it has a normal Sylow $3$-subgroup, call it $P$. In particular, $|P|=3$. Because $3^2\nmid 24=|S_4|$, it follows that $P$ is also a Sylow $3$-subgroup of $G=S_4$. As $P\unlhd H$, the Fact tells us that $H\le N_{S_4}(P)$.

The Sylow 3-subgroups are just the various alternating groups of degree 3, Sylow's theorems tell us that $S_4$ can have either $1$ or $4$ Sylow $3$-subgroups. It is easy to exhibit four: $\langle(123)\rangle$, $\langle(124)\rangle$, $\langle(134)\rangle$ and $\langle(234)\rangle$, each stabilizing one of the four numbers. Therefore the number of Sylow $3$-subgroups of $S_4$ is exactly four.

and their normalizers are various symmetric groups of degree 3, Because the Sylow subgroups are all conjugate (in $S_4$), the orbit-stabilizer theorem tells us that their normalizers all have order $24/4=6$. Obviously each of the listed four Sylow subgroups is normalized by its respective copy of $S_3$. For example $N_{S_4}(\langle(134))\supseteq Sym\{1,3,4\})$. Because the normalizer has order six as does the symmetric group, we must have equality here.

so are exactly the 4 subgroups of order 6. We have seen that any subgroup $H\le S_4$ of order six normalizes a Sylow $3$. We have also seen that the normalizer of any Sylow $3$ has exactly six elements. Therefore we can conclude that $S_4$ has exactly four subgroups of order six, and we have accounted for all of them.


问:
Integrate $\int sin(x)^6 $ using integration by parts?
I know how to solve it using the double angle formula for $cos(2x)$. I also know how to integrate $sin(x)^2$ by parts like this:

$$\begin{align}\int \sin(x)\sin(x)dx &= −\cos(x)\sin(x)-\int(-\cos(x))\cos(x)dx
\\&= -\cos(x)sin(x)+\int(1-\sin(x)^2)dx
\\&= -\cos(x)\sin(x)+x-\int \sin(x)^2dx
\\2\int \sin(x)^2dx &= -\cos(x)sin(x)+x
\\\int \sin(x)^2dx &= -\frac 12 \cos(x)\sin(x)+\frac 12 x \end{align}$$

Is it possible to integrate $sin(x)^6$ by parts?

答:
$$\sin^6x=(\sin^2x)^3$$
and we know that:
$$\cos(2x)=1-2\sin^2x\Rightarrow \sin^2x=\frac{1-\cos(2x)}{2}$$
so we have:
$$\sin^6x=\left(\frac{1-\cos(2x)}{2}\right)^3=\frac{1}{8}\left(1-3\cos2x+3\cos^22x-\cos^32x\right)$$
now we can shuffle this around a bit:
$$3\cos^22x=\frac{3(1+\cos4x)}{2}$$
$$\cos^32x=\cos2x(1-\sin^22x)=\cos2x-\sin^22x\cos2x$$
so if we bring this all together we get:
$$\int\sin^6xdx=\frac18\int1-4\cos(2x)+\frac{3(1+\cos(4x))}{2}+\sin^2(2x)\cos(2x)dx$$
and all of the terms here are easy to integrate


问:
Contour Integration: $\int_0^\infty\frac{1}{x^a(1-x)}\,dx$ for $0<a<1$.
I've been trying to calculate $$\int_0^\infty\frac{1}{x^a(1-x)}\,dx\quad\text{with }0<a<1.$$I haven't had much luck. I tried taking the branch cut with of the positive reals and estimating that way, but I wasn't sure how to handle the poles at $z=0$ and $z=1$ when dealing with branch cuts. I also looked at the Wikipedia page to try to get some insight, but I didn't see any similar examples with the contour closing in on poles that are on a branch cut.</p>

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\large An\ Alternative:}}$
\begin{align}
&\bbox[5px,#ffd]{%
\left.{\rm P.V.}\int_{0}^{\infty}{\dd x \over x^{a}\pars{1 - x}}
\,\right\vert_{\,a\ \in\ \pars{0,1}}} =
-\,\Re\int_{0}^{\infty}{x^{-a} \over x - 1 + \ic 0^{+}}\,\dd x
\\[5mm] = &\
\Re\int_{\infty}^{0}{\ic^{-a}\,\,y^{-a} \over \ic y - 1}
\,\ic\,\dd y =
-\,\Im\bracks{\ic^{-a}\int_{0}^{\infty}
{\,y^{\pars{\color{red}{1 - a}} - 1} \over 1 - \ic y}\,\dd y}
\end{align}
I'll evaluate the last integral with the
Ramanujan's Master Theorem. Note that
$\ds{{1 \over 1 - \ic y} = \sum_{k = 0}^{\infty}\pars{\ic y}^{k} = \sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{1 + k}\expo{-k\pi\ic/2}}\,\,{\pars{-y}^{k} \over k!}}$.

Then,
\begin{align}
&\bbox[5px,#ffd]{%
\left.{\rm P.V.}\int_{0}^{\infty}{\dd x \over x^{a}\pars{1 - x}}
\,\right\vert_{\,a\ \in\ \pars{0,1}}}
\\[5mm] = &\
-\,\Im\pars{\ic^{-a}\,\,\Gamma\pars{1 - a}\braces{\Gamma\pars{1 -\bracks{1 - a}}\expo{\pars{1 - a}\pi\ic/2}\,}}
\\[5mm] = &\
-\,\Re\bracks{\expo{-\pi a\ic}\,\,
\Gamma\pars{1 - a}\Gamma\pars{a}} =
-\cos\pars{\pi a}\,{\pi \over \sin\pars{\pi a}}
\\[5mm] = &\
\bbx{-\pi\cot\pars{\pi a}} \\ &
\end{align}


问:
Geometric interpretation of the euclidean product.
I have always been taught about the euclidean product in $\mathbb{R}^n$ the algebraic way, that is $$xy=\sum_1^n x_iy_i$$

Now I find myself having to come to grips with the geometric definition as $xy=\|x\|\|y\|\cos \theta.$

I am trying to understand how to pass from one to the other.

Now if I fix the standard orthonormal basis of $\mathbb{R}^n,$ I have that $x_i=xe_i$ for every $i.$ Then, If I suppose that the geometric definition holds, I have $x_i=xe_i=\|x\|\|e_i\|\cos \theta_i=\|x\|\cos\theta_i, $ and similarly for $y_i,$ hence $$xy=\sum_i\|x\|\|y\|\cos^2 \theta_i=\|x\|\|y\|\sum_i \cos^2 \theta_i$$
How do I proceed from here to conclude the implication geometric $\implies$ algebraic?

And then I need also to show that the algebraic definition implies the geometric one.

答:
Think about the triangle that connects the vectors $x$, $y$, and the origin together. The law of cosines from geometry states that

$$a^2=b^2+c^2-2bc\cos(\alpha)$$

where $\alpha$ is the angle opposite $a$. We will assign $a$ to be the side of the triangle connecting $x$ and $y$, so $\alpha$ is the angle between $x$ and $y$. The equation reads

$$\|x-y\|^2=\|x\|^2+\|y\|^2-2\|x\|\|y\|\cos\alpha$$

The left side can be expanded, since

$$\|x-y\|^2=\|x\|^2-2(x\cdot y)+\|y\|^2$$

and canceling like terms gives

$$\cos\alpha=x\cdot y$$

Hope this helps!


问:
A slight variation on the Pythagorean theorem
Are there any solution to $$a^2+b^2=c^2+1 $$ where $a \not=0$ and $b\not=0$

This is a follow-on from a previous question For what $n$ and $m$ is this number a perfect square?, which ultimately boils down to the above. Any help would be greatly appreciated.

答:
For example, $5^2 + 5^2 = 7^2 + 1$.  More generally, if $b^2 - 1 = u v$ where $u > v$
are both odd or both even (and are not $b+1$ and $b-1$), you can take $c = (u+v)/2$ and
$a = (u-v)/2$.</p>


答:
I'll make the dot product explicit as $\cdot$.

First, let's correct a mistake of yours. While $x_i=x\cdot e_i=|x|\cos\theta_i$ with $\theta_i$ the angle between $x$ and $e_i$, the angle between $y$ and $e_i$ will in general be different, say $\phi_i$, so the angle $\alpha$ between $x$ and $y$ satisfies$$|x||y|\cos\alpha=x\cdot y=\sum_ix_iy_i=|x||y|\sum_i\cos\theta_i\cos\phi_i,$$i.e. $\cos\alpha=\sum_i\cos\theta_i\cos\phi_i$.

Now we'll move onto why $\sum_ix_iy_i=|x||y|\cos\alpha$. Both sides are invariant under rotations about the origin; for the left-hand side, note a rotational matrix $R$ satisfies$$R^TR=I\implies x^Ty=x^TR^TRy=(Rx)^TRy\implies x\cdot y=Rx\cdot Ry,$$and for the right-hand side note lengths are invariant under rotations, as are angles. So we only need check the special case where $x$ runs along the positive $x$-axis. Then $x$ has a single nonzero component, say $x_1=|x|$, while $y_1=|y|\cos\alpha$. (You can verify this with a diagram; bear in mind $x,\,y$ are coplanar, regardless of $n$.) So$$\sum_ix_iy_i=x_1y_1=|x||y|\cos\alpha.$$

Title: Geometric interpretation of the euclidean product.

Body:
I have always been taught about the euclidean product in $\mathbb{R}^n$ the algebraic way, that is $$xy=\sum_1^n x_iy_i$$

Now I find myself having to come to grips with the geometric definition as $xy=\|x\|\|y\|\cos \theta.$

I am trying to understand how to pass from one to the other.

Now if I fix the standard orthonormal basis of $\mathbb{R}^n,$ I have that $x_i=xe_i$ for every $i.$ Then, If I suppose that the geometric definition holds, I have $x_i=xe_i=\|x\|\|e_i\|\cos \theta_i=\|x\|\cos\theta_i, $ and similarly for $y_i,$ hence $$xy=\sum_i\|x\|\|y\|\cos^2 \theta_i=\|x\|\|y\|\sum_i \cos^2 \theta_i$$
How do I proceed from here to conclude the implication geometric $\implies$ algebraic?

And then I need also to show that the algebraic definition implies the geometric one.


问:
A minimal normal group of a solvable group is abelian
Definition: We say that $H$ is a minimal normal group of $G$ if $H\lhd{G}$ and $\forall K\lneq{H}$ , $K\neq{\{e}\}$ it holds that $ K\ntriangleleft{G}$.

Let $G$ be a finite solvable group, and let $H$ be a minimal normal group of $G$.
Prove that $H$ is abelian.

I wrote a solution but it feels too easy and it proves a stronger claim- that $Z(G)=H$ , so I think it might be wrong:

Since every $g\in{Z(G)}$ commutes with every $h\in{H}$ , it holds that $Z(G)\leq{Z(H)}\leq{H}$. Since $Z(G)\lhd{G}$ , and $H$ is a minimal normal group of $G$, we get that $Z(G)=H$ or $Z(G)={\{e}\}$ . But $G$ is solvable, so we get $Z(G)\neq{\{e}\}$ and therefore $Z(G)=H$ and $H$ is abelian.

I'd like to know if this proof is correct, and if it's wrong, some guidence would be much appreciated.

答:
Your argument commits several errors.

For one thing, solvable groups may have trivial centers (for example, $S_3$).

For another, It is false that $Z(G)\leq Z(H)$ for $H\leq G$. What is true is that $Z(G)\cap H\leq Z(H)$, but again, in general, there is absolutely no reason to assume that a central element of $G$ must be in $H$, and so there is no reason to think it will lie in the center of $H$. When you argue that elements of $Z(G)$ commute with every element of $H$, the conclusion is not that $Z(G)\leq Z(H)$, but rather that $Z(G)\leq C_G(H)$, the centralizer of $H$. The centralizer also contains $Z(H)$, but that doesn’t tell you anything about whether $Z(G)$ and $Z(H)$ have any containment relationship with each other.

As for an argument, here are few things you may want to prove:

Subgroups of solvable groups are solvable.

If $K$ is solvable, then $[K,K]\neq K$.

If $N\triangleleft G$, then $[N,N]\triangleleft G$.

Use those facts to show the desired result.


问:
Diagonalizable and finding Complex Matrix
Consider the matrix

$\beta = \begin{pmatrix}
7 &3  &-4 \\
 -2&-1  &2 \\
 6&2  &-3
\end{pmatrix}$
over the complex numbers.

Explain/show that it is Diagonalizable and find a complex matrix $\gamma$ and a diagonal matrix $\delta$ such that $\gamma ^{-1} \beta \gamma=\delta$

Effort so far

I have shown that it is indeed not possible over the real numbers and I know that
to diagonalize a matrix $A$

Find the eigenvalues of $A$ using the characteristic polynomial.

For each eigenvalue $λ$ of $A$ , compute a basis $B λ$ for the $λ$-eigenspace.

If there are fewer than $n$ total vectors in all of the eigenspace bases $B λ$ , then the matrix is not diagonalizable. I am having trouble with showing this over the complex numbers and to find the matrices $\gamma , \delta$

答:
The characteristic polynomial of $\beta$ is $-\lambda^3+3\lambda^2-\lambda+3=(3-\lambda)(\lambda^2+1)$, whose roots are $3$, $i$, and $-i$. If you search for eigenvectors corresponding to these eigenvalues, you will see that, for instance:

$(1-i,2i,2)$ is an eigenvector corresponding to the eigenvalue $i$;

$(1+i,-2i,2)$ is an eigenvector corresponding to the eigenvalue $-i$;

$(1,0,1)$ is an eigenvalue corresponding to the eigenvalue $3$.

So, if you take$$M=\begin{bmatrix}1-i&1+i&1\\2i&-2i&0\\2&2&1\end{bmatrix}$$(the columns of $M$ are the eigenvectors that I got) then$$M^{-1}\beta M=\begin{bmatrix}i&0&0\\0&-i&0\\0&0&3\end{bmatrix}.$$


问:
Limits and Integration Recursion

I was able to obtain an answer of $k/(k+1)$ for part 1, by interchanging the limit and sum and then evaluating the limit of the integrand. However, for part (2), I'm stuck because using the same idea from part (1), I obtain an infinity - infinity indeterminate form. The substitution x = u^(n^-2) didn't help much either. Any help for parts (1) and (2) would be appreciated.

答:
We can use the series expansion

$$\left(1+\frac{t}{n}\right)^n \sim e^t-\frac{t^2e^t}{2n}+O\left(\frac{1}{n^2}\right)$$

for large $n$ and

$$L=\int_0^1 x^{\frac{1}{k}}dx,$$

to get

$$\lim_{n\to\infty} n\left(\int_0^1\left(\frac{x^{\frac{1}{n}}+k-1}{k}\right)^ndx-L\right) = -\int_0^1\frac{\log^2x}{2k^2}x^{\frac{1}{k}}dx$$

$$=-\frac{d^2}{ds^2}\int_0^1\frac{1}{2}x^{\frac{s}{k}}dx \Bigr|_{s=1} = \frac{-k}{(k+1)^3}$$

by Feynman's trick.

Title: Limits and Integration Recursion


问:
Are some irrational numbers closer to rational numbers than others?
I'm not even sure this question is meaningful because any irrational number can be approximated by a rational as close as you want.  But given that there are infinitely many irrationals between any two rationals, surely some must be closer to a rational number than others?

Is there a metric by which irrationals can be ranked, or can we compare two irrationals, based on how far they are from a rational?

I'm having trouble wrapping my head around this, can anyone assist?

答:
No, not with the usual metric. As the question stands, the answer is surely, “it depends on the metric”.

However, I’ll leave the rest of my answer up-it should be clear that I am working in the usual metric.

Between any two real numbers there are infinitely many rational and irrational numbers.

I should probably elaborate. To me, your question is basically, “which of the following is closer to a rational number: $\sqrt{2}, \pi, e, e^2$ and so on?”

The answer is that they’re all “infinitesimally close” to a rational number, although I’d rather not use this dodgy language.

Let $x$ be any of the irrational numbers above - or any other irrational number - and let $\varepsilon > 0$, no matter how small.

Then $\exists \ q(\varepsilon) \in \mathbb{Q}$ such that $q \in (x-\varepsilon, x+ \varepsilon).$ In fact, there exist infinitely many such rational numbers in this interval.

Furthermore, let $y$ be any real number, rational or irrational.

Then $\exists \ p(\varepsilon) \in \mathbb{Q}$ such that $p \in (y-\varepsilon, y+ \varepsilon)$ (in fact, there are infinitely many rational numbers in this interval) and also $\exists \ j \in \mathbb{R} - \mathbb{Q}$ such that $j \in (y-\varepsilon, y+ \varepsilon)$ (in fact, there are infinitely many irrational numbers are in this interval).

Furthermore, this closeness relationship between $\mathbb{Q}$ and $\mathbb{R} - \mathbb{Q}$ can also be summarised as follows:

$ \mathbb{Q}$ is dense in $\mathbb{R}$, $\mathbb{R} - \mathbb{Q}$ is dense in $\mathbb{R}$, but $\mathbb{Q}$ and $\mathbb{R} - \mathbb{Q}$ are disjoint subsets of $\mathbb{R}$.


问:
Is there a formula for the cosine of 1/5 of an angle?
I'm trying to find a formula for $\cos \frac{x}{5} $ as follow.

By the elementary trigonometric identity, $\cos 5x = 16\cos^5 x - 20\cos^3 x + 5\cos x$

By putting $x=  \frac{x}{5}$ , and using the change of variable $ y = \cos \frac{x}{5} $, we get the quintic equation $ 16y^5 - 20y^3 + 5y - \cos x = 0$.

If we can solve this equation by radicals, then we can get a formula for $\cos \frac{x}{5} $

However, I can not see how we can resolve this equation. Is this a solvable quintic equation? If so, how can we solve it by radicals?

答:
p = 3, trisection

p >= 3

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