问:
Continuous Markov Chain, Find Average of Served Clients per hour?
I'm stuck in a problem on markov jump process:

A gas station receives cars at a rate of 20 vehicles per hour, the station has only one gas pump. If the pump is empty, it receives one client, if a new client comes and find that the pump is occupied by a client he waits. Every client takes around 6 minutes to fill his vehicle (1/10 th of an hour). If two clients are at the pump, one filling and the other one waiting, new clients would leave the station directly.

Solution:

Modeling this problem we have the states are the number of clients at the pump at any instant $t$ so we would increment by one with a rate of 20, and decrement by one (leave the station) by a rate of 10 (because it takes in average 1/10th of hour to fill the vehicle).

So the Jump rate matrix for the chain with the following states $\{0,1,2\}$ is:

$$\begin{bmatrix}-20 & 20 & 0 \\10 & -30 & 20 \\ 0 &10 &-10  \end{bmatrix}$$

Solving for the stationary distribution we find:

$$(u_1,u_2,u_3) = (1/7,2/7,4/7) $$

Now I'm asked to find the average number of clients served by the gas station every hour. The answer should be 50/7 but I can't find how to get this.

答:
The limiting mean number of customers served per hour is simply the limiting fraction of time that a customer is being served times the number of customers that would be served if the server was always busy:

$$
\left(1-\frac17\right)\cdot10 = \frac{60}7.
$$


问:
Prove that the algebraic expression is greater than zero
Prove that $x^8 - x^5 + x^2 -x + 1 >0$ for $x \in \mathbb{R}$.

It's from the (junior) high-school competition and the idea is that not everyone there knows calculus, so that I'm looking for more "basic" justification.

My idea was to use AM-GM: $x^8 + x^2 \geq 2\sqrt{x^{10}} = 2x^5$. Thus,
$$x^8 - x^5 + x^2 -x + 1 \geq  x^5-x + 1 $$
So that we can now focus on proving that $x^5 - x + 1 > 0 $ but I don't really see how to do so without calculus...

ContentLicense: CC BY-SA 4.0

Body:
$f(x)=x^8-x^5+x^2-x+1$
If $x>1$, then
$$f(x)=x^5(x^3-1)+x(x-1)+1>0$$
If $x<1$, then
$$f(x)=x^8+x^2(1-x^3)+1-x >0$$
ans for $x=1$,
$$f(x)=1>0$$
Hence for all real values $f(x)>0$.


答:
You can rewrite this $ x^8 - x^5 + x^2 -x + 1 >0 $

as:

$x^8+(x^4-x)^2+(x-1)^2+1.$

You can figure out the rest ;)

Title: Prove that the algebraic expression is greater than zero

Body:
Prove that $x^8 - x^5 + x^2 -x + 1 >0$ for $x \in \mathbb{R}$.

It's from the (junior) high-school competition and the idea is that not everyone there knows calculus, so that I'm looking for more "basic" justification.

My idea was to use AM-GM: $x^8 + x^2 \geq 2\sqrt{x^{10}} = 2x^5$. Thus,
$$x^8 - x^5 + x^2 -x + 1 \geq  x^5-x + 1 $$
So that we can now focus on proving that $x^5 - x + 1 > 0 $ but I don't really see how to do so without calculus...


问:
Sandwich rule for continuity
Let $c ∈ X ⊆ R $ . Suppose that $f : X → \mathbb{R}$, $g : X → \mathbb{R}$, and $h : X → \mathbb{R}$ are functions such that $g(x) ≤ f(x) ≤ h(x)$ for all $x ∈ X$.

Suppose that $g$ and $h$ is continuous at $c$ and $g(c) = h(c).$

Question: Show that f is continuous at c.

Attempt:

I have started by writing down the following.

Let $\epsilon >0$. We know that $g$ is continuous hence we have $\delta_g$ such that

$|x-c|< \delta_g$ $\implies$ $|g(x)-g(c)|< \epsilon$.

We know that $h$ is continuous hence we have $\delta_h$ such that

$|x-c|< \delta_h$ $\implies$ $|h(x)-h(c)|< \epsilon$.

How do I connect $f$ to $h$ and $g$?

答:
Presumably, the sandwich theorem for limits is allowed. So by this theorem,

$$\lim_{x\to c}g(x)\le\lim_{x\to c}f(x)\le\lim_{x\to c}h(x).$$

In addition, by continuity of $g$ and $h$,

$$\lim_{x\to c}g(x)=g(c),\\\lim_{x\to c}h(x)=h(c).$$

Hence

$$\lim_{x\to c}f(x)=f(c)=g(c)=h(c)$$ QED.

Note that continuity of $g$ and $h$ is required, otherwise you could very well have

$$\lim_{x\to c}g(x)\ne g(c)\text{ and }\lim_{x\to c}h(x)\ne h(c).$$

Title: Sandwich rule for continuity


问:
Integrating Function with Modulo
Is there a particular method that is used to evaluate integrals that involve modulo arithmetic? For example;

$$\int_0^t x\sin(x)\, (\mathrm{mod \ 2}) dx$$

The problem that I am trying to solve is
$$\int_0^t -\frac{1}{2}+((s+\pi\omega s)\, (\mathrm{mod \ 1}))\, ds$$
When plotted, I was thinking you could break this up into piecewise functions. But this would be really inefficient. Is there a way that this can be solved? Or is it not possible? Also in the above, $\omega$ is simply a constant.

答:
The function $x \mapsto x (\mod 1)$ (or $x\mapsto x - \lfloor x \rfloor]$) is a sawtooth function. If you want to avoid using piecewise functions you could replace it with its Fourier series:

$$g\left(x\right)\ =\frac{1}{2}-\ \sum_{k=1}^{\infty}\left(-1\right)^{k}\cdot\frac{\sin\left(2\cdot\pi\cdot k\cdot\left(x-\frac{1}{2}\right)\right)}{\pi}$$

Title: Integrating Function with Modulo


问:
Cauchy sequence: distance from point of convergence
If $\{a_n\}$ is a Cauchy sequence and we know that $|a_i - a_{i+1}| < \varepsilon$ for some $i$. Is it possible to say anything about $|a_i - a|$, where $a$ is the point to which the sequence converges?

答:
If the sequence is Cauchy, you know rather more than that: you know that for any $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that $|a_k-a_\ell|<\epsilon$ whenever $k,\ell\ge n_\epsilon$. It follows that $|a_k-a|\le\epsilon$ whenever $k\ge n_\epsilon$. To see this, suppose that there is a $k_0\ge n_\epsilon$ such that $|a_{k_0}-a|>\epsilon$, and let $\delta=|a_{k_0}-a|-\epsilon>0$. The sequence converges to $a$, so there is an $n_\delta\in\Bbb N$ such that $|a_k-a|<\delta$ whenever $k\ge n_\delta$. Now let $\ell=\max\{n_\epsilon,n_\delta\}$; then

$$\begin{align*}
|a_{k_0}-a|&\le|a_{k_0}-a_\ell|+|a_\ell-a|\\
&<\epsilon+\delta\\
&=|a_{k_0}-a|\,,
\end{align*}$$

which is absurd.


答:
First note that setting $x=c$ in the inequality implies that $g(c) \le f(c) \le h(c) = g(c)$ and therefore $g(c) = f(c) = h(c)$.

For $\epsilon > 0$ you have found $\delta_g > 0$ and $\delta_h > 0$ such that
$$
|x-c|< \delta_g \implies |g(x)-g(c)| < \epsilon \implies g(x) > g(c) - \epsilon\\
|x-c|< \delta_h \implies  |h(x)-h(c)| < \epsilon \implies h(x) < h(c) + \epsilon
$$
Now set $\delta = \min(\delta_g, \delta_h)$. Then $|x-c| < \delta$ implies
$$
 f(c) - \epsilon = g(c) - \epsilon <  g(x) \le f(x)
$$
and
$$
 f(x) \le h(x) < h(c) + \epsilon =  f(c) + \epsilon
$$
and therefore
$$
 f(c) - \epsilon < f(x) <  f(c) + \epsilon \iff |f(x) - f(c)| < \epsilon \, .
$$

Title: Sandwich rule for continuity

Body:
Let $c ∈ X ⊆ R $ . Suppose that $f : X → \mathbb{R}$, $g : X → \mathbb{R}$, and $h : X → \mathbb{R}$ are functions such that $g(x) ≤ f(x) ≤ h(x)$ for all $x ∈ X$.

Suppose that $g$ and $h$ is continuous at $c$ and $g(c) = h(c).$

Question: Show that f is continuous at c.

Attempt:

I have started by writing down the following.

Let $\epsilon >0$. We know that $g$ is continuous hence we have $\delta_g$ such that

$|x-c|< \delta_g$ $\implies$ $|g(x)-g(c)|< \epsilon$.

We know that $h$ is continuous hence we have $\delta_h$ such that

$|x-c|< \delta_h$ $\implies$ $|h(x)-h(c)|< \epsilon$.

How do I connect $f$ to $h$ and $g$?


问:
Evaluate the limit $\lim_{n\to\infty}\sqrt{n} \int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx$
Evaluate the limit$$\lim_{n\to\infty}\sqrt{n} \int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx$$

I used L'Hopital's Rule

$$\lim_{n\to\infty}\sqrt{n} \int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx=\lim_{n\to\infty}\frac{\int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx}{\frac{1}{\sqrt{n}}}=\lim_{n\to\infty}\frac{e^{-1}\frac{1/n}{\sin (1/n)}.\frac{-1}{n^2}}{-\frac{1}{2n\sqrt{n}}}=0$$

Is it correct

答:
For $n\geq 1$,
\begin{align*}
0 < \sqrt n \int_0^{1/n} {e^{ - nx} \frac{x}{{\sin x}}dx} & < \frac{1}{{\sin 1}}\sqrt n \int_0^{1/n} {e^{ - nx} dx} \\ & < \frac{1}{{\sin 1}}\sqrt n \int_0^{ + \infty } {e^{ - nx} dx}  = \frac{1}{{\sin 1}}\frac{1}{{\sqrt n }}.
\end{align*}
Thus, by the squeeze theorem, the limit is $0$. A bit more interesting problem is
$$
\mathop {\lim }\limits_{n \to  + \infty } n\int_0^{1/n} {e^{ - nx} \frac{x}{{\sin x}}dx} .
$$


答:
An alternative way which, whilst not as elegant as the others, doesn't really require factorisation, which can sometimes be tricky to spot.

Rearrange to get: $x^8 + x^2 + 1 > x^5 + x$. Now consider $x$ in different cases:

If $x \leq 0$, the the RHS is clearly less than or equal to zero (as $x^5+x = x(x^4+1)$, which is either $0$ or a product of positive and negative number), whilst the LHS is always greater than zero (negative numbers to the power of even degree). So the inequality holds for $x \leq 0$.

If $x\geq1$, it's straightforward to argue the inequality holds again.

Final case is $0<x<1$. Now the LHS will always be greater than $1$ (because of the $+1$), and the RHS will always be less than $1$. So again the inequality holds.

This shows it for all $x\in\mathbb{R}$.


问:
Trying to understand correlation and independence geometrically
I am trying to understand correlation and independence of two random variables geometrically, but found it difficult to grasp the intuition and to explain it rigorously.

First, given two uniformly distributed RVs X,Y, and given the distribution of (X,Y) and (-X,Y) are the same, how can we show and explain $\mathbb{E}XY=\mathbb{E}(-X)Y=-\mathbb{E}XY$. From this equality, is there any way to show whether X,Y are independent or not, uncorrelated or not?

Suppose consider a point with cartesian coordinates (X,Y), on the unit circle with uniformly distributed X and Y. Are X,Y uncorrelated and why? Are X,Y independent and why?

My attempt to reason with the unit circle is that if we pick a point outside the circle then $0\not=\mathbb{E}X\mathbb{E}Y\not=\mathbb{E}XY=0$, am I right?

How about on the unit square?

Sorry for so many questions at once, I am trying to understand the concept as a whole.

Thanks.

答:
Correlation is a measure of linear relationship. In other words, the correlation coefficient measures the extent to which the two random variables $X$ and $Y$ fall on a straight line.

Suppose you were to draw a contour plot of the joint density $f(x,y)$ and you observe that the pattern of lighter and darker regions are identical for every value of $x$ then the variables are independent.

Another way to understand independence is to imagine the joint density function as a mountain. The two variables are independent if the cross-section of the mountain is identical for every value of $x$.

ContentLicense: CC BY-SA 3.0


问:
existence of certain continuous function

Let $(X,d)$ be a metric space and $\emptyset \neq A\subseteq X.$ Define for $x\in X,\mathrm{dist}(x,A) := \inf\{d(x,a) : a\in A\}.$

Show that there is a sequence $U_1, U_2,\cdots$ of open subsets in $(X,d)$ for which $\overline{A} = \cap_{n=1}^{\infty} U_n.$

Let $\emptyset\neq K\subset X$ be a compact set with $K\cap \overline{A} = \emptyset.$ Show that there is a continuous function $g : X\to [0,1]$ so that $g(x) = 0$ for $x\in A$ and $g(y) = 1$ for $y \in K.$

For the first part, define for each $n\in\mathbb{N}, U_n := \{x \in X : \mathrm{dist}(x,A) < \frac{1}n\}.$  We claim that each $U_n$ is open. Fix $n\in\mathbb{N}.$ Let $y \in U_n.$ Then $\mathrm{dist}(y, A) =\inf\{d(x,a) : a \in A\} < \frac{1}n,$ so $\exists a \in A$ so that $d(y,a) < \frac{1}n.$ Consider the open ball $B(y, \epsilon), \epsilon := \frac{1}{n} -d(y,a) > 0.$ Then for any $x \in B(y, \epsilon), d(x,a)\leq d(x,y) + d(y,a) < \epsilon + d(y,a) = \frac{1}n$ and hence $\mathrm{dist}(x,A) \leq d(x,a) < \frac{1}n,$ so that $B(y,\epsilon)\subseteq U_n.$

Now we claim that $\cap_{n\in \mathbb{N}}U_n = \overline{A}.$ Let $x \in \cap_{n\in \mathbb{N}} U_n.$ Then for all $n\in\mathbb{N}, \mathrm{dist}(x,A) < \frac{1}n$ and since $\mathrm{dist}(y,A)\geq 0$ for all $y \in X, $ this implies $ \mathrm{dist}(x,A) = 0.$ One can then show using the definition of infimum that $x\in \overline{A}.$ Thus, $\cap_{n\in\mathbb{N}} U_n \subseteq \overline{A}.$ Now let $b\in \overline{A}.$ Then $\mathrm{dist}(b,A) = 0$ so $b \in \cap_{n\in\mathbb{N}}U_n.$ Thus, $\overline{A} = \cap_{n\in\mathbb{N}} U_n.$

Is this correct?

For the second part, I should make use of the properties of compact sets (e.g. being closed and bounded) and the fact that $K$ and $\overline{A}$ are disjoint. I know the function $f : X\to \mathbb{R}, f(x) = \mathrm{dist}(x,A)$ for all $x$ is continuous. However, I'm not sure how to define the function $g.$

答:
A basic fact that is handy in metric spaces $(X,d)$: for a given non-empty set $A$, the function $f_A: X \to \Bbb R, x \to d(x,A)$ is continuous.

This follows from $|f_A(x)-f_A(y)| \le d(x,y)$, which in turn easily follows from triangle inequality arguments. The OP already knows this fact it seems.

Another standard fact is that $x \in \overline{A}$ iff $d(x,A)= 0$, for all $x \in X$.

It follows that for a closed set $A$, so where $A = f_A^{-1}[\{0\}]$, we can write
$$A = \bigcap_n U_n, \text{ where } U_n = f^{-1}[(-\frac1n, \frac1n)] \text{ is open in } X$$

This is quite close to the idea in the OP's proof. But we don't need a separate proof of the openness of te $U_n$ as they are the inverse images of open sets in $\Bbb R$ under a continous function.

If $K \cap \overline{A} = \emptyset$, we can conclude: $f_A(x) >0$ for all $x \in K$, by the above fact and $f_A$ restricted to $K$ thus assumes some minimum $m_0 >0$ (by compactness the continuous function $f_A$ has a minimum and a maximum on $K$) and then we can define a continuous (piecewise linear even) function $h$ (helper function, hence $h$) on $[0, +\infty) \to [0,1]$ that maps  $[0,\frac12m_0]$ to $0$, $[\frac12m_0,m_0]$ linearly from $0$ to $1$, and is $1$ on $[m_0, +\infty)$. Then $g:=h \circ f_A$ is as required: all points in $A$ are mapped to $0$ by $f_A$ and still to $0$ by $h$, and everything in $K$ is mapped into $[m_0,\rightarrow)$ (as $m_0$ was the minimum of $f_A$ on $K$) and then to $1$ by $h$, etc. The composition of continuous maps is continuous so we're done.


问:
$\overline{xy8}$ is divisible by 7. Find the remainder of $10x + 15y + 21$ divided by $7$.
I have the following problem:
$$\overline{xy8}\text{ is divisible by }7$$

Find the remainder of $10x + 15y + 21$ divided by $7$.

How can I do this? Any hint? Thank you!

I tried to write $\overline{xy8} = 100x + 10y + 8$ then division becomes: $(100x + 10y + 8) - 90x + 5y + 13$. But I can't find the remainder.

答:
You have a good start and you can write $100x+10y+8=7n$ to express divisibility by $7$.

As a hint, what is the simplest expression you can find (on the left-hand side) which would be divisible by $7$? "Simplest" is a bit ambiguous, but look for small positive integers as coefficients/constants.


答:
$100x+10y+8\equiv 2x+3y+1\pmod 7 $

$\Rightarrow  2x+3y+1\equiv 0\pmod 7$

$10x+15y+21\equiv 5(2x+3y)\equiv -5\equiv \boxed{2}\pmod 7$

Title: $\overline{xy8}$ is divisible by 7. Find the remainder of $10x + 15y + 21$ divided by $7$.

Body:
I have the following problem:
$$\overline{xy8}\text{ is divisible by }7$$

Find the remainder of $10x + 15y + 21$ divided by $7$.

How can I do this? Any hint? Thank you!

I tried to write $\overline{xy8} = 100x + 10y + 8$ then division becomes: $(100x + 10y + 8) - 90x + 5y + 13$. But I can't find the remainder.


问:
Logic proof need help
I do not know how to conclude this proof, I cannot put the "code" here so I have to attach an image. Please help

答:
You mostly have it.

Firstly, you may use explosion ($\mathsf X$) to immediately derive $E$ from the assumed $M$.

Secondly the derivation of $E$ from $A$ is not irrelevant, it just should not be placed inside that subproof, but in a second subproof on the base level, so another disjunction elimination is set up to derive $E$ there.

$\def\fitch#1#2{~~~~\begin{array}{|l} #1\\\hline #2\end{array}}
\fitch{~~1.~C\lor E\\~~2.~A\lor M\\~~3.~A\to\lnot C\\~~4.~\lnot S\land\lnot M}{~~5.~\lnot S\hspace{12ex}{\land}\mathsf E~4\\~~6.~\lnot M\hspace{11.5ex}{\land}\mathsf E~4\\[1ex]\fitch{~~7.~M}{~~8.~\bot\hspace{9ex}\lnot\mathsf E~7,6\\~~9.~E\hspace{10ex}\mathsf{X}~8}\\[1ex]\fitch{10.~A}{11.~\lnot C\hspace{9ex}{\to}\mathsf E~10,3\\\fitch{12.~C}{13.~\bot\hspace{8ex}\lnot\mathsf E~12,11\\14.~E\hspace{10ex}\mathsf X~13}\\\fitch{15.~E}{}\\16.~E\hspace{12ex}{\lor}\mathsf E~1,12{-}14,15{-}15}\\[1ex]17.~E\hspace{16ex}{\lor}\mathsf E~2,10{-}16,7{-}9}$

Title: Logic proof need help


问:
Fixed point of curvature operator
Let $y=f(x)$ the graph of a real-valued function. We define its curvature by : $$curv(f) = \frac{|f''|}{(1+(f')^2)^{3/2}}$$

I would like to know if there is any function (apart from the trivial anwser $f(x)=0$) whose curvature is itself. So what is the fixed point of $curv$ ?

答: Multiply both sides by $2f'$ to get $$2ff'=\frac{2f''f'}{(1+(f')^2)^{3/2}}$$ which integrates to $$f^2=a-2(1+(f')^2)^{-1/2}$$ Rearranging and substituting $u=f$,$$ \int\frac{1}{\sqrt{4(a-u^2)^{-2}-1}}\,du=x-x_0$$ Perhaps $y=u^2-a$, $$F(y):=\frac{1}{2}\int\sqrt{\frac{y^2}{(4-y^2)(y+a)}}dy$$ so that $$f(x)=\sqrt{2(F^{-1}(x-x_0)+a)}$$ This explains Leibovici's answer in using inverse functions and elliptic functions.

Title: Fixed point of curvature operator


Body:
$$100x + 10y + 8 \equiv (100-98x) + (10-7)y + 8-7 \equiv 2x +3y + 1 \equiv 0 \pmod 7$$

and:

$$10x+15y+21 \equiv (10-7)x + (15-14)y + 21-21 \equiv 3x + y \equiv -4x-6y \pmod 7$$

so $-4x-6y \equiv -2(2x+3y+1) + 2 \equiv 2 \pmod 7$.


问:
Solve an equation with mod operator
Consider the following equation
$$mod(x+100/x,5)=1.$$
How can we solve it with $x>0$?

First of all, the function $x+100/x$ is a continuous function with range $[20, \infty)$, therefore, it must has solutions. Then, it can be checked that the solutions are discrete, but how can we express them?

答:
In this answer we employ the most general view, interpreting $100/x\bmod 5$ as modular fraction whenever it exists, i.e. when $5$ doesn't divide its  denominator when reduced to least terms.

Let  $100/x = a/b\,$ in least terms. If $\,25\nmid x\,$ then then we can cancel at most one factor of $\,5\,$ so $\,5\mid a,\, 5\nmid b\,$ so $\,100/x = a/b := ab^{-1}\equiv 0,\,$ so $\,1\equiv x+100/x\equiv x,\,$ i.e.  $\,x=1,6,11,\ldots$.

Else $\,25\mid x\,$ so $\,x = 25k\,$ and $\,5\nmid k\,$ (else $5\nmid a,\ 5\mid b\,$ so $\,a/b\,$ does not exist $\!\bmod 5)$. Hence  $\,1\equiv x+100/x\equiv 25k+4/k\equiv 4/k\iff k\equiv 4,\,$ so $\,k = 4+5n\,$ thus $\,x = 25k = 25(4+5n) = 100+125n = 100,225,350,\ldots$

Title: Solve an equation with mod operator


问:
Question on derivative
I want to differentiate $H(p(t),q(t))=1 $ with respect to $t$, where $H:\mathbb{R}\times \mathbb{R} \rightarrow \mathbb{R} $ is a convex function.

I think that it is:

$\displaystyle \frac{dp}{dt} \frac{\partial H(p,q)}{\partial p}+ \frac{dq}{dt} \frac{\partial H(p,q)}{\partial q}=0$

right?

Please, thank you.

答:
If we want something we need a detailed notation flowery but enlightening.

First of all use the implicit differentiation theorem:
\begin{align}
\left.\frac{d\!}{dt}\!F\big( q(t), p(t)\big)\right|_{t=t_0}
=
&
\left.\frac{d\!}{dt}\!F\big( q_0, p(t)\big)\right|_{t=t_0}
+
\left.\frac{d\!}{dt}\!F\big( q(t), p_0\big)\right|_{t=t_0}
\end{align}
for $p_0=p(t_0)$ and $q_0=q(t_0)$.

Second, use the chain rule

$\qquad\left.\dfrac{d\!}{dt}\!F\big( q_0, p(t)\big)\right|_{t=t_0}=\left.\dfrac{d}{dt}p(t)\right|_{t=t_0}\!\cdot\; \left.\dfrac{\partial}{\partial p}F(p,q)\right|_{(p,q)=(p_0,q_0)},$

$\qquad\left.\dfrac{d\!}{dt}\!F\big( q(t), p_0\big)\right|_{t=t_0}=\left.\dfrac{d}{dt}q(t)\right|_{t=t_0}\!\cdot\; \left.\dfrac{\partial}{\partial q}F(p,q)\right|_{(p,q)=(p_0,q_0)}$

to get
\begin{align}
 \left.\frac{d\!}{dt}\!F\big( q(t), p(t)\big)\right|_{t=t_0}
=
&
\left.\dfrac{d}{dt}p(t)\right|_{t=t_0}\!\cdot\; \left.\dfrac{\partial}{\partial p}F(p,q)\right|_{(p,q)=(p_0,q_0)}+
\\
+
&
\left.\dfrac{d}{dt}q(t)\right|_{t=t_0}\!\cdot\; \left.\dfrac{\partial}{\partial q}F(p,q)\right|_{(p,q)=(p_0,q_0)}
\end{align}

Title: Question on derivative


问:
Numbers are formed using the digits $1,1,2,2,2,2,3,4,4$ taken all at a ...

Numbers are formed using digits $1,1,2,2,2,2,3,4,4$ taken all at a time. Find number of such numbers so that odd digits occupy even places.

The odd digits are $1,1,3$ and there are 4 even places

So permutation of arranging 3 items in 4 places where two items are alike is.

Now the solution given to me involves use of combinatorics, and I can’t understand how they came up with it, since we are arranging and not selecting. Can I get an explanation on how combinations can be used to solve this (or how to do it without combinations)

答:
All possible numbers that can be formed
= $9!$/$2!4!2!$

No. of numbers such that odd digits occupy even places = $⁴C_3$×($3!$/$2!$)×($6!$/$4!2!$)


问:
Taking Seats on a Plane
This is a neat little problem that I was discussing today with my lab group out at lunch. Not particularly difficult but interesting implications nonetheless

Imagine there are a 100 people in line to board a plane that seats 100. The first person in line, Alice, realizes she lost her boarding pass, so when she boards she decides to take a random seat instead. Every person that boards the plane after her will either take their "proper" seat, or if that seat is taken, a random seat instead.

Question: What is the probability that the last person that boards will end up in their proper seat?

Moreover, and this is the part I'm still pondering about. Can you think of a physical system that would follow this combinatorial statistics? Maybe a spin wave function in a crystal etc...

答:
Here is a simple proof. For $k\ge 2$, let $p_k$ be the probability that the $k$-th person's seat is taken when he/she tries to sit.

We consider the disjoint events that person $j$ takes up this particular seat, for $1\le j \le n$. In order for that to happen, $j$'s seat has to be taken, and then $j$ is left with $n-j+1$ random choices of which $j$ must take $k$'s seat. We can write:
$$p_{k}={1\over n}+p_2{1\over n-1}+p_3{1\over n-2}\dots+p_{k-1}{1\over n-k}.$$

Thus, $p_2={1\over n}$, $p_3={1\over n}+{1\over n(n-1)}={1\over n-1}$. In general, $p_k={1\over (n-k+2)}$.

Title: Taking Seats on a Plane


问:
Envelope of oscillator signal
How do we find envelope ODE of a modulated oscillator obeying

$$y^{''}(t)+y (t)\left(\dfrac{2 \pi}{t-c}\right)^2 = 0 $$

after obtaining solution say for initial conditions $y(0) = 1, y'(0) = 0?$. Does it oscillate indefinitely?

Could not readily apply $\; c-, p-$  discriminant methods. Please help.

Numerical solution with $ c=t _{max} =6$

Envelope sketched by hand.

答:
If I did not make any mistake, the solution is given by
$$y=\frac{1}{k }\sqrt{\frac{c-t}{c}} \left(\sin \left(\frac{k}{2} \log \left(\frac{c}{c-t}\right)\right)+k \cos
   \left(\frac{k}{2}  \log \left(\frac{c}{c-t}\right)\right)\right)$$ where $k=\sqrt{16\pi^2-1}$.

Now, close to $c^-$, $y$ oscillates indefinitely.

Edit

Defining $\alpha=\sqrt{\frac{c-t}{c}}$, we have
$$y=\alpha  \left(\cos [k \log (\alpha )]-\frac{1}{k}\sin [k \log (\alpha )]\right)$$

The first derivative cancels at
$$t_n=c \left(1-e^{\frac{2 \pi  n}{k}}\right)$$ At these points
$$y(t_n)=(-1)^n e^{\frac{ \pi  n}{k}}\qquad \text{and}\qquad y''(t_n)=(-1)^{n+1}\frac{1+k^2}{4c^2}e^{-\frac{2 \pi  n}{k}}$$

So, if $n$ is odd, these points correspond to maxima and, if $n$ is even, they correspond to minima and then the upper and lower envelopes.

Title: Envelope of oscillator signal


问:
Finding a general term for the sequence
Find a general term in simplest form for the sequence: $$2, 1, -4, 7, -10, 13, -16$$

This is what I tried:

$a_n = a_1 + (n - 1)\cdot d$

$a_n = 2 + (n-1)\cdot 3$

$a_n = (-1)^n (2+(n-1)\cdot 3)$

which doesn't work.

I am pretty confused regarding the first two terms of the sequence and how to deal with the lack of a negative number (all of the terms after that seem to follow a pattern of +,-,+,- so I think multiplying by $(-1)^n$ would help there), as well as the different change in value from the first to the second term in the sequence, where as all the following numbers seem to follow a more recognizable pattern to me.

Any help would be greatly appreciated.

答:
There is a pattern in this sequence:$$\frac{a_1 + a_3}{2}=-a_2,\space \space \space \space \frac{a_2+a_4}{2}=-a_3,\space \space \space \space \frac{a_3+a_5}{2}=-a_4$$ Using mathematical induction, it can be deduced that: $$a_{n}+2a_{n+1}+a_{n+2}=0$$ This is indeed a good deduction but it won't help us find the general term of the sequence alone. Hence, we need to go further:

$a_{n+1}-a_{n}=b_{n}, \space \space \space \space b_{n+1}-b_{n}=c_{n}$

$b_1=-1,\space b_2=-5, ...,\space b_6=-29$

$c_1=-4,\space c_2=16,\space c_3=-28,\space c_4=40,\space c_5=-52$

Once again, if you look at the sequence denoted by $c_n$, you can deduce the general term formula which is $c_n=(-1)^n(12n-8)$. Using this general term formula, we can easily write that $$\begin{align}\ b_{n+1}-b_{n}&= \ (a_{n+2}-a_{n+1})-(a_{n+1}-a_{n})\\
&=\ a_{n+2}-2a_{n+1}+a_{n}\\
&=\ (-1)^n(12n-8)\\
\end{align}$$ This last equation combined with our first deduction gives us $$4a_{n+1}=(-1)^n(8-12n)\Rightarrow a_{n+1}=(-1)^n(2-3n)$$ If we replace $n+1$ with $k$, we eventually get $$a_k=(-1)^{k}(3k-5)$$

Title: Finding a general term for the sequence


问:
Theta-space is a deformation retraction of the doubly-punctured plane, how to find equations.
That theta space is given by $S^1\cup(0\times[-1,1]) \subset\mathbb{R}^2$ it is said that this space is a deformation retract of the doubly punctured plane, here is the explanation I found:

The first one I think is ok if you want to find the equations, if you assume that p and q are at the points 1 and -1. You can define the deformation retraction like this:

$(x_1,x_2)(1-t)+2(x_1,x_2)/\|x\|t, \|x\| \ge 2$

$x, \|x\|\le2$. We can see that it is continuous by the pasting lemma, it is well defined, and we have a homotopy between the identity map and the disc of radius 2, so it is a deformation retraction.

But what about the last one? Is it difficult to find the equations? I have seen many cases where they just use intuition here. But there are many things that has to be shown,

That the deformation leaves $S^1\cup(0\times[-1,1])$ fixed(this is clear from the picture)

That the deformation si continuous(not that clear)

That the deformation is homotpic with the identity map, and that it leaves the theta space fixed during this homomorphism. How is this just "seen" from the picture?

Could these things be "seen" or "felt" intuitively? And do you know how to construct the homotpy explicitely in the last case? I am not sure how to find the equations here. Maybe they are too messy?

答:
I will try to answer your questions one by one with the help of certain diagrams that I have made. Before I begin, I would like to tell you that I am not an expert on this subject. In fact, I have just started studying this. So, any suggestions or methods other than the one I am going to explain here are, indeed, welcome.

Since your question is mostly about the second deformation retraction, we will concentrate on that. First, we see what we want to do. We want to map every point of a disc with two holes to a point in $\mathbb{S}^1 \cup \left( \left\lbrace 0 \right\rbrace \times \left[ -1, 1 \right] \right)$. For the sake of simplicity, we will denote by $B_1 = \left\lbrace \textbf{x} \in \mathbb{R}^2 | \| \textbf{x} \| \leq 1 \right\rbrace \setminus \left\lbrace \left( \frac{1}{2}, 0 \right), \left( - \frac{1}{2}, 0 \right) \right\rbrace$. This is the unit disc with two points removed. Notice that you can do the same analysis that I will be doing by considering a disc of radius $2$ with two points removed (as you have mentioned in the beginning of your question).

As shown in the figure, we want to construct a function which brings the blue shaded region of $\mathbb{R}^2$ to the green shared region.

In the figure, the two red dots denote the removed points. In our case they are $\left( \frac{1}{2}, 0 \right)$ and $\left( -\frac{1}{2}, 0 \right)$. To understand how to construct a function as desired, let us first concentrate on the blue region alone.

From the figures we have been provided as intuitions, it is clear that we want to create a function using straight lines. That is, if there is a point $\left( x, y \right)$ in the right half-plane and inside $B_1$, then we want to look at the line joining $\left( x, y \right)$ and $\left( \frac{1}{2}, 0 \right)$. Then, we would want to map this point $\left( x, y \right)$ to the intersection of the line with the theta space.

This intersection can happen in two cases: One where the line intersects with $\mathbb{S}^1$ and the other, when it intersects with the line segment $\left\lbrace 0 \right\rbrace \times \left[ -1, 1 \right]$. We want the intersection that happens first. That is, we would like to study two classes of lines. I have denoted them in the figure by orange and purple colours. The orange lines are those which intersection the line segment $\left\lbrace 0 \right\rbrace \times \left[ -1, 1 \right]$ and the purple lines are those which intersection with the circle $\mathbb{S}^1$.

Let us look at the orange lines first. Of all the possible orange lines, two of them intersect the theta space at $\left( 0, 1 \right)$ and two intersect the theta space at $\left( 0, -1 \right)$. These four line segments make a quadrilateral. Anything inside this quadrilateral will have an intersection of its line with the line segment $\left\lbrace 0 \right\rbrace \times \left[ -1, 1 \right]$, and anything outside of this quadrilateral will have the intersection with $\mathbb{S}^1$.

Formally, these four lines are given by the equations:

$$y = 1 - 2x, y = 2x - 1, y = 2x + 1, y = -2x - 1.$$

I have written them in the order: first quadrant, fourth quadrant, second quadrant, third quadrant.

So, it means that if for a point $\left( x, y \right) \in B_1$ with $0 \leq y \leq 1 - 2x$ or $2x - 1 \leq y \leq 0$ or $0 \leq y \leq 2x + 1$ or $-2x - 1 \leq y \leq 0$, we would expect the line joining $\left( x, y \right)$ with $\left( \frac{1}{2}, 0 \right)$ or $\left( - \frac{1}{2}, 0 \right)$ (depending on the sign of $x$) to intersect the line segment $\left\lbrace 0 \right\rbrace \times \left[ -1, 1 \right]$.

First, we see for $x \geq 0$. The line joining $\left( x, y \right)$ with $\left( \frac{1}{2}, 0 \right)$ is given by the equation:

$$y' = \left( \dfrac{y}{x - \frac{1}{2}} \right) \left( x' - \dfrac{1}{2} \right).$$

Here, we assume that $x \neq \frac{1}{2}$. We will consider this case later. Now, whenever $0 \leq y \leq 1 - 2x$ or $2x - 1 \leq y \leq 0$, this line intersects with $\left\lbrace 0 \right\rbrace \times \left[ -1, 1 \right]$. The point of intersection will be $\left( 0, \frac{-y}{2x - 1} \right)$. Thus, for the points $\left( x, y \right)$ in the right region of the orange quadrilateral, we get a point $\left( 0, \frac{-y}{2x - 1} \right)$ in the theta space.

Similar computation would tell us that the points $\left( x, y \right)$ in the left half region of the orange quadrilateral should be mapped to $\left( 0, \frac{y}{2x + 1} \right)$.

Now, let us see what happens when $x = \frac{1}{2}$. Since $\left( frac{1}{2}, 0 \right)$ is not included in $B_1$, any point of the form $\left( \frac{1}{2}, y \right)$ will lie outside the orange quadrilateral. Therefore, we now look at the intersection of the corresponding line with $\mathbb{S}^1$. Since on the line $x = \frac{1}{2}$, the intersection will be at $\left( \frac{1}{2}, \pm \frac{\sqrt{3}}{2} \right)$, depending on the sign of $y$.

Similar things can be said about $x = - \frac{1}{2}$. Seethe figure for a better understanding. Now, let us only look at the case when $x \neq \pm \frac{1}{2}$ and $\left( x, y \right)$ lies outside the orange region. Let us first consider the right half plane. Now, we have to look at the intersection of the line $y' = \left( \dfrac{y}{x - \frac{1}{2}} \right) \left( x' - \dfrac{1}{2} \right)$ with $\mathbb{S}^1$. Upon calculation, we get a quadratic equation in $\left( x' - \dfrac{1}{2} \right)$, whose solution(s) are given by

$$x' - \dfrac{1}{2} = \dfrac{-1 \pm \sqrt{1 + 3 \left( 1 + \left( \frac{y}{x - \frac{1}{2}} \right)^2 \right)}}{2 \left( 1 + \left( \frac{y}{x - \frac{1}{2}} \right)^2 \right)}$$

Now, $x' \geq 0$ should hold (otherwise the line would first intersect the segment $\left\lbrace 0 \right\rbrace \times [-1, 1]$, which is not desired for now). Therefore, for the purple points, we get the point

$$\left( \dfrac{-1 \pm \sqrt{1 + 3 \left( 1 + \left( \frac{y}{x - \frac{1}{2}} \right)^2 \right)}}{2 \left( 1 + \left( \frac{y}{x - \frac{1}{2}} \right)^2 \right)} + \dfrac{1}{2}, \left( \dfrac{y}{x - \frac{1}{2}} \right) \left( \dfrac{-1 \pm \sqrt{1 + 3 \left( 1 + \left( \frac{y}{x - \frac{1}{2}} \right)^2 \right)}}{2 \left( 1 + \left( \frac{y}{x - \frac{1}{2}} \right)^2 \right)} \right) \right)$$

Similar computations can be done for purple points in the left half part of $B_1$. So, the function $f: B_1 \rightarrow B_1$ which we wanted to construct is as follows:

$$f \left( x, y \right) = \begin{cases}
\left( 0, \frac{-y}{2x - 1} \right), & x \geq 0 \text{ and } \left( 0 \leq y \leq 1 - 2x \text{ or } 2x - 1 \leq y \leq 0 \right) \\
\left( 0, \frac{y}{2x - 1} \right), & x \leq 1 \text{ and } \left( 0 \leq y \leq 2x + 1 \text{ or } -2x - 1 \leq y \leq 0 \right) \\
\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right), & x = \frac{1}{2} \text{ and } y > 0 \\
\left( \frac{1}{2}, - \frac{\sqrt{3}}{2} \right), & x = \frac{1}{2} \text{ and } y < 0 \\
\left( - \frac{1}{2}, \frac{\sqrt{3}}{2} \right), & x = - \frac{1}{2} \text{ and } y > 0 \\
\left( - \frac{1}{2}, - \frac{\sqrt{3}}{2} \right), & x = - \frac{1}{2} \text{ and } y < 0 \\
\left( \dfrac{-1 \pm \sqrt{1 + 3 \left( 1 + \left( \frac{y}{x - \frac{1}{2}} \right)^2 \right)}}{2 \left( 1 + \left( \frac{y}{x - \frac{1}{2}} \right)^2 \right)} + \dfrac{1}{2}, \left( \dfrac{y}{x - \frac{1}{2}} \right) \left( \dfrac{-1 \pm \sqrt{1 + 3 \left( 1 + \left( \frac{y}{x - \frac{1}{2}} \right)^2 \right)}}{2 \left( 1 + \left( \frac{y}{x - \frac{1}{2}} \right)^2 \right)} \right) \right), & x \geq 0 \text{ and } x \neq \frac{1}{2} \text{ and } y > 1 - 2x \text{ and } y < 2x - 1 \\
\left( \dfrac{-1 \pm \sqrt{1 + 3 \left( 1 + \left( \frac{y}{x + \frac{1}{2}} \right)^2 \right)}}{2 \left( 1 + \left( \frac{y}{x + \frac{1}{2}} \right)^2 \right)} - \dfrac{1}{2}, \left( \dfrac{y}{x + \frac{1}{2}} \right) \left( \dfrac{-1 \pm \sqrt{1 + 3 \left( 1 + \left( \frac{y}{x + \frac{1}{2}} \right)^2 \right)}}{2 \left( 1 + \left( \frac{y}{x + \frac{1}{2}} \right)^2 \right)} \right) \right), & \text{otherwise}
\end{cases}$$

As you can see, writing the function explicitly is way too tedious and lengthy. However, with this analytical expression, one can satisfy themselves about the continuity of $f$.

As for your question about homotopy between the identity function and this particular function we have constructed, we look at the following. While constructing this function, we essentially saw lines and "moved" each point along a certain line until we reached the theta space. So, a natural way to deform the disc (with two points removed) into the theta space is that at the beginning, stay where you are and then start moving along the lines we have considered. We know that then each of these lines (segments) would then start at a point in $B_1$ and end at a point in $\mathbb{S}^1 \cup \left( \left\lbrace 0 \right\rbrace \times \left[ -1, 1 \right] \right)$. Use this intuition to define the homotopy as $H: B_1 \times \left[ 0, 1 \right] \rightarrow B_1$,

$$H \left( \textbf{x}, t \right) = \left( 1 - t \right) \textbf{x} + t f \left( \textbf{x} \right).$$

Now, we can easily verify that this is indeed a homotopy between the identity map and the map $f$, we have constructed.


问:
general formula for partial sum of series
im having trouble figuring out how to find the general formula for partial sums of a series.

Is it a trial and error kind of thing where I just have to figure it out?

or is there a systematic way to figure it out?

in that example from pauls calculus, I dont see how he got that general formula for the partial sums.

Thanks for the help.

答:
It’s the formula for the sum of a finite geometric series:

$$\sum_{k=0}^{n-1}x^k=\frac{1-x^n}{1-x}\;,$$

here with $x=\frac13$. Specifically,

$$\sum_{k=1}^n\frac1{3^{k-1}}=\sum_{k=0}^{n-1}\frac1{3^k}=\sum_{k=0}^{n-1}\left(\frac13\right)^k=\frac{1-(1/3)^n}{1-1/3}=\frac32\left(1-\left(\frac13\right)^n\right)=\frac32\left(1-\frac1{3^n}\right)\;.$$


问:
Base for Fourier series in an interval.
INTRODUCTION:

Given the problem $\begin{cases} u_t-u_{xx}=0 & (x,t)\in(0,\pi)\times(0,+\infty) \\ u(x,0)=g(x)&x\in[0,\pi] \\ u_x(0,t)=u(\pi,t) & t>0 \end{cases}$ I was trying to find the solutions with the method of separetion of variables.

After some calculation I find that the solution, trowing the intial condition on the second line, should be $u_k=c_k\cos(\frac{2k+1}{2}x)e^{-(\frac{2k+2}{2})^2t}$ so to satisfy the initial condition a want to find $c_k$ to have a series that express $g(x)$,

I mean solve $\sum_{_{k>0}} c_k\cos(\frac{2k+1}{2}x)=g(x)$ in the interval $[0,\pi]$.

QUESTIONS:

(i) I know that $\{\cos(kx),\sin(kx)\}_k$ is a orthonormal base (respect to the $L^2[-\pi,\pi]$ scalar product) but what can I say about $\cos(\frac{2k+1}{2}x)$ or $\sin(\frac{2k+1}{2}x)$?

(ii) Is it real that $c_k=\frac{2}{\pi}\int g(x)\cos(\frac{2k+1}{2}x)dx$?

(iii) Is there a general rule to find from the standard base other bases?

Here i was thinking that in the theroy of vector spaces, in finite dimension, i know that all the "good" automorphisms are represented by orthogonal matrices, is there any similar thing in infinite dimension using linear operator?

If I miss some details I will add it, if I can, let me know.

Any suggestion, answer, hint.. would be greatful, thanks.

答:
(I)

You can calculate that $\{\cos(\frac{2k+1}{2}x)\}_{k=0}^{\infty}$ is orthogonal system on $L_2[0, \pi]$. This is orthogonal system of eigenfunctions of the Sturm–Liouville problem:
$$X''(x) + \lambda X(x) = 0, X'(0) = X(\pi) = 0.\quad\quad(1)$$
Any function $X$, which satisfy $X'(0) = X(\pi) = 0$, can be represented as $X(x) = \sum_{k=0}^{\infty} A_k \cos(\frac{2k+1}{2}x)$.

You can calculate that $\{\sin(\frac{2k+1}{2}x)\}_{k=0}^{\infty}$ is orthogonal system on $L_2[0, \pi]$. This is orthogonal system of eigenfunctions of the Sturm–Liouville problem:
$$X''(x) + \lambda X(x) = 0, X(0) = X'(\pi) = 0.\quad\quad(2)$$
Any function $X$, which satisfy $X(0) = X'(\pi) = 0$, can be represented as $X(x) = \sum_{k=0}^{\infty} A_k \sin(\frac{2k+1}{2}x)$.

These systems do not produce basis of all space $L_2[0, \pi]$, but they produce basis of vector space functions, which satisfy corresponding boundary conditions.

(II) Yes. Because you have $X_k(x) = \cos(\frac{2k+1}{2}x)$ and $T_n(t) = c_k e^{-(\frac{2k+2}{2})^2t}$. $u$ can be represented as $\sum_{k=0}^{\infty} X_k(x) T_n (t)$.
$u(x, 0) = \sum_{k=0}^{\infty} c_k X_k(x) = g(x)$ or $u(x, 0) = \sum_{k=0}^{\infty} c_k \cos(\frac{2k+1}{2}x) = g(x)$. So $c_k = \frac{\langle g, X_k \rangle}{\langle X_k, X_k \rangle}$, where $\langle a, b \rangle = \int_{0}^{\pi} a(x) b(x) dx$.

(III) System of functions $X_k$ can be obtained as eigenfunctions of Sturm–Liouville problem. In your case $u(x, t) = X(x) T(t)$. After substituting into equation $X(x) T'(t)  - X''(X) T(t) = 0$ or $\frac{T'(t)}{T(t)} = \frac{X''(x)}{X(x)} = -\lambda$. Now you have equations: $X''(x) + \lambda X(x) = 0$ and $T'(t) + \lambda T(t) = 0$. Using $u_x(0, t) = X'(0) T(t) = 0$ we get $X'(0) = 0$, and using $u(\pi, t) = X(\pi) T(t) = 0$, we get $X(\pi) = 0$. In these problems we do not need basis of all space, we need basis of functions that satisfy boundary conditions, and we find it as eigenfunctions of the Sturm–Liouville problem.

Title: Base for Fourier series in an interval.


问:
Let $N_n = \# \{k \le n: X_k = 6\}$. Establish the almost sure convergence of the sequence $(N_n / n)_{n \geq 1}$
Hi I'm having trouble finding the almost sure convergence of this problem:

Consider a succession of tosses of a balanced die. Let $X_k$ be the result of the $k$-th face and $Y_n = \max(X_k), k ≤ n.$

Let $N_n = \# \{k \le n: X_k = 6\}$. Establish the almost sure convergence of the sequence $(N_n / n)_{n \geq 1}$.

I know that in order to prove almost sure convergence I nedd to show that
$$\mathbb{P}\left(\lim_{n\to\infty} X_n(\omega)=X(\omega)\right)=1$$

But I can't seem to find a way to use it

Can someone help me?

答:
Note that $N_n = \frac{1}{n}\sum^n_{k=1} I_k$, where $I_k \sim \mathrm{Bernoulli}(1/6)$ represents the outcome of the $k$th dice toss (i.e. $I_k = 1$ if $X_k = 6$ with probability $1/6$, and $I_k = 0$ otherwise). Each $I_k$ is independent. Furthermore, $\mathbb{E}[I_k] = 1/6$.

Hence, by the Strong Law of Large Numbers, the sample mean of the i.i.d. random variables $I_k$ will converge almost surely to their common mean. That is, $N_n \to \frac{1}{6}$ a.s.

This should align with what $N_n$ intuitively represents -- the ``average'' number of $6$s you see in $n$ tosses.


问:
Unique continuation at the boundary for harmonic functions in the plane
Consider the set $U = (-1,1) \times \{ 0\} \subset \mathbb R^2$ and a continuous function $f : U \rightarrow \mathbb R$.

Then for any domain $\Omega \subset \mathbb R^2$ such that $U \subset \partial\Omega$, there exist many functions $u$ harmonic in $\Omega$ such that $u|_U = f$ (as the Dirichlet problem is not completely determined).

My question is what happens if we want to prescribe the value of $u$ on $U$ and also the value of $\partial_\nu u$ on $U$? Does there exist a (maximal) domain $\Omega$ and function $u$ harmonic in $\Omega$ with the desired behavior in $U$?

I think, in general, there cannot exist a function $u$ and domain $\Omega$ satisfying this. For example, we could ask for $\partial_\nu u|_U=0$ and $u|_U(x)=\max(0,\vert x \vert -\frac 1 2)$. Thus, by unique continuation at the boundary for harmonic functions, since $u$ and its gradient are both $0$ in an open set of the boundary then $u$ must be identically zero. But the prescribed value at $u$ is not identically zero, so there cannot exist any $u$.

I was wondering whether we can find some compatibility conditions on the prescribed values of $u$ and its normal derivative such that we can ensure the existence of some solution. Also, if there exists a solution, what can we say about the maximal domain where it is defined?
Does anybody know about any reference on this kind of problem?

答:
I hope someone still can shed more light on the problem, but I found out some things:

This is called the Cauchy problem for the Laplace equation and it was first investigated by Hadamard.

The article TOWARDS THE CAUCHY PROBLEM FOR THE LAPLACE EQUATION by DINH NHO HÀO, TRAN DUC VAN gives a small survey on this problem and, in particular, Thm 2.1 discusses necessary and sufficient conditions for the problem I posted. Unfortunately, it doesn't include proof and it doesn't discuss the maximal domain for the solution (which I guess ends up being some sort of free boundary problem).


问:
If $Ax = Bx$ for all $x \in C^{n}$, then $A = B$.
Let $A$ and $B$ are $n\times n$ matrices and $x \in C^{n}$.

If $Ax = Bx$ for all $x$ then $A = B$.
To prove this I have selected  $x$ from Euclidean basis B = {$e_{1},e_{2},...,e_{n}$}.

Then $Ae_{i} = Be_{i}$ implies $i^{th}$ column of A = $i^{th}$ column of B for all $1 \leq i \leq n$.

Hence $A = B.$

Is this proof complete?

答:
Let $\langle x|y\rangle:=\sum_{i=1}^n\overline{x_i}y_i$ be the standard inner product on $\mathbf{C}^n$.

To make the title of the question even more obvious, one can use the fact that $\langle e_i|A\cdot e_j\rangle=A_{ij}$ for all $i,j\in\{1,\ldots,n\}$. Thus, if $A\cdot x=B\cdot x$ for all $x\in\mathbf{R}^n$, $A_{ij}=\langle e_i|A\cdot e_j\rangle=\langle e_i|B\cdot e_j\rangle=B_{ij}$ for all $i,j\in\{1,\ldots,n\}$.


问:
Polytope of cone over a rational normal curve
Consider a rational normal curve $C\subset\mathbb{P}^d$ of degree $d$, and let $W\subset\mathbb{P}^{d+1}$ be a cone over $C$.

Since $C$ is a toric variety $W$ is toric as well. I would like to ask what is the polytope in $\mathbb{R}^2$ associated to the toric surface $W$.

Thank you very much.

答:
This variety can be obtained by taking the Hirzebruch surface $\mathbb{P}_{\mathbb{P}^1}(\mathcal{O} \oplus \mathcal{O}(n))$, and contracting the section with self intersection $-n$. This may be done by manipulating the Delzant polytope accordingly.

I try to express this using a concrete example and then I hope the idea will be clear.

The Delzant polytope for the Hirzebruch surface $\mathbb{P}_{\mathbb{P}^1}(\mathcal{O} \oplus \mathcal{O}(1)) = Bl_{p}(\mathbb{P^2})$ is the quadrilateral with vertices $(0,0),(0,2),(1,2),(3,0)$. Then contracting the $-1$-curve, we "add in the vertex" along the edge corresponding to the minus one curve, giving the triangle with vertices $(3,0),(0,0),(0,3)$ i.e. the Delzant polytope of $\mathbb{CP}^2$.

This "adding the corner" procedure is the general way of contracting an (invariant) curve with negative self intersection in a toric surface. Take the edge corresponding this curve and continue the two edges adjacent to it until they meet; then add this region to the polytope. I leave you to apply this to the polytopes associated to the other Hirzebruch surfaces.

The final result with give (up to scale) the triangle in $\mathbb{R}^2$ with vertices $(0,0),(0,1)$ and $(n,0)$.


问:
Let $X$ be the number of $1$s and let $Y$ be the number of $2$s obtained. What is the joint PMF of $X$ and $Y$.
I was reading Introduction to Probability, 2nd Edition and the following problem appears on page 128:

Consider four independent rolls of a 6-sided die.

Let $X$ be the number of $1$s and let $Y$ be the number of $2$s obtained.

What is the joint PMF of $X$ and $Y$?

My reasoning to solve the problem is as follows:

We need $P_{X,Y}(x, y)$.

Let us compute the probability of having $x + y \ $ $1s$ or $2s$ first, it is:
$${}_4 \mathrm{ C }_{x+y} \cdot \left(\frac{2}{6}\right)^{x+y} \cdot \left(\frac{4}{6}\right)^{4-(x + y)}$$

Now out of the $x + y$ $1$s or $2$s the probability of having $x \space 1s$ is:

$${}_{x + y} \mathrm{ C }_{x} \cdot \left( \frac{1}{2} \right)^{x + y}$$

And the final answer is the product of the two probabilities.

Is my reasoning wrong? and if so then what is my mistake?

答:
The two marginals are binomial but they are not independent thus to show the joint distribution is better to  use a table-representation.

As an example, here below is the joint distribution related to 2 independent rolls of a die... you can easy do the same work for $n=4$. The resulting table is a triangular matrix...

Y=0
Y=1
Y=2
Total

X=0
p(0;0)
p(0;1)
p(0;2)
25/36

X=1
p(1;0)
p(1;1)
0
10/36

X=2
p(2;0)
0
0
1/36

Total
25/36
10/36
1/36
1/36

I think it is easy for you to calculate the probabilties in the body of the table which have to sum both per column and per row to the binomial probabilities


问:
Proof by Induction for $\sum_{i = 1}^{n - 1} \frac{1}{i(i+1)}$
$$\sum_{i = 1}^{n - 1} \frac{1}{i(i+1)} = \frac{n- 1}{n}$$

How would you go about proving this by induction? the base case $n=1$ gives you $1/2 = 0$

The only thing I can think of using is $\sqrt{2}$, but that is not a natural number.

答:
When $n = 1$, the sum on the left  side is empty, because the upper index is $n - 1 = 0$, that is  there are no terms to add.  This gives $0 = 0$ as desired.
Suppose that  $$S_n = \sum_{i=1}^{n-1} \frac{1}{i(i+1)},$$ then we have to prove $$S_{n+1} =   \frac{n}{ (n+1)}.$$  So if the induction hypothesis $$S_n = \frac{n - 1}{n}$$ is true,  we have:
$$S_{n+1} = S_n+  \frac{1}{n (n+1)}.$$ that is $$S_{n+1} =\frac{n-1}{n}+  \frac{1}{n (n+1)}=\frac{n}{n+1}.$$
The proof is compelete.


问:
Trace map argument for number fields
Let $K \subset L$ a finite (and therefore inevitably algebraic)
extension of number fields. Let $O_K$ and $O_L$ be the ring of integers
of $K$ and $L$. The goal is to show that that $O_L$ is finite
$O_K$-module. The later means that there exist
$l_1, l_2,..., l_m \in O_L $ such that $O_L = l_1 \cdot O_K
+ l_2 \cdot O_K + ... + l_m \cdot O_K$.

Now in my lecture notes is noted that this result follows from
an 'easy application of the so called trace map argument'.

I have to confess that I nowhere heard about such
'trace map argument' in that context and I would very thankful
if somebody has an idea what it is and elaborate how this
argument is used here.

My ideas: There exist a $K$-bilinear 'trace map' $Tr: L \times L \to K, (x,y)
\mapsto tr(x \cdot y)$ where $x \cdot y$ is regarded as $K$-linear
endomorphism $L \to L$. Since we are dealing with number fields
it is known that $Tr$ is not degenerated (the extension is separated).
The field extension $K \subset L$ is assumed to be finite, so there exist
a $K$-basis $x_1, ..., x_m \in L$ with $L \cong K^m \cong
x_1K \oplus x_2K \oplus ... \oplus x_mK$.

Now seemingly the trick is to find other basis
$y_1, y_2, ..., y_m \in O_L$ (probably by 'clearing denominators
of the $x_i$) and to show that these $y_i$ are still linear independent $(1)$
over $K$ and for every $l \in L$ holds:

$l \in O_L $ iff $l = \sum_{i=1}^m a_i y_i $ with $a_i \in O_K$ $(2)$

I conjecture that $(2)$ can be derived by the mentioned 'trace map
argument' while $(1)$ can be proved independently. But it is also not
clear how. The $x_i$ form a $K$-basis and not a $L$-basis, therefore
multiplying the basis elements by $x_i$ by elements from $L^*$
could destroy their linear independence, or not?

答:
In order to get the basis $y_1,\dots,y_m$ we need to show that $K\mathcal{O}_L=L$, so we can multiply the $x_i$ with some non-zero elements $r_i\in K$ to get elements $y_i=r_ix_i$ in $\mathcal{O}_L$ which will then of course still form a $K$-basis of $L$. To see $K\mathcal{O}_L=L$, let $x\in L$. As $x$ is algebraic over $K$ there are some $a_i$ with $$x^n+a_1x^{n-1}+\dots+a_n=0$$
Let $d\in \mathcal{O}_K$ be a common denominator of the $a_i$. By multiplying the above equation with $d^nx$ we see that $dx$ satisfies a monic polynomial equation over $\mathcal{O}_K$, hence $dx\in \mathcal{O}_L$ and it follows that $x\in K\mathcal{O}_L$.

It follows as argued at the beginning that we may choose a basis $y_1,\dots,y_m$ of $L/K$ in $\mathcal{O}_L$. Let $B$ be the $\mathcal{O}_K$-submodule of $\mathcal{O}_L$ spanned by the $y_i$. Define $B^\#$ to be the dual module of $B$, i.e. $$B^\#=\{x\in L\mid Tr_{L/K}(xy)\in\mathcal{O}_K \text{ for all } y\in B\}$$ By definition of $B^\#$ we have a well defined map $B^\#\to \mathcal{O}_K^n$ given by $x\mapsto (Tr_{L/K}(xy_1),\dots,Tr_{L/K}(xy_m))$. Since $L/K$ is separable the trace form is non-degenerate, hence $B^\#\to\mathcal{O}_K^n$ is injective. As $\mathcal{O}_K$ is noetherian it follows that $B^\#$ is a finite $\mathcal{O}_K$-module. Since $\mathcal{O}_L\subset B^\#$ using again that $\mathcal{O}_K$ is noetherian we see that $\mathcal{O}_L$ is finite over $\mathcal{O}_K$.

(Actually the fact that $\mathcal{O}_K$ is noetherian is usually proved by the same argument, but over $\Bbb Z$)

Title: Trace map argument for number fields


问:
Number of eigen values of an $N \times N$ matrix
What are the number of eigen values of a non-singular matrix and why ?

What would have happen if matrix is singular, lets say have 2 linear dependent rows.

Preparing for interview, stuck on this question, know the answer is because of characteristic polynomial, but do not have good way to put answer. Please help.

答:
No Conclusion

Possible eigenvalues of a non-singular real skew-symmetric matrix are purely imaginary. Consequently such matrices have no eigenvalue in $\mathbb R$ i.e. the number of eigenvalues is $0.$

The roots of the characteristic polynomial of a $n\times n$ real symmetric matrices are all real whence the number of eigenvalues is $n.$

You may form a $n\times n$ matrix which has exactly $i$ (pre-assigned) eigenvalues where $0\le i\le n$ as:

$$A=\begin{pmatrix}1\\&2\\&&3\\&&&\ddots\\&&&&k\\&&&&&0&-1&-1&...&-1\\&&&&&1&0&-1&...&-1\\&&&&&1&1&0&...&-1\\&&&&&...&...&...&\ddots&...\\&&&&&1&1&1&...&0\end{pmatrix}$$

If $n-i$ is even take $k=i$ in $A;$

If $n$ is even & $i$ is odd, take $k=i-1$ in $A;$

If $n$ is odd & $i$ is even and nonzero, take $k=i-1$ in $A;$

For $i=0$ take any non-singular real skew symmetric matrix.

All can be said about your second question is that $0$ is an eigenvalue here.

Added: The point made by @A Walker is relevant in this regards as polynomials factor linearly over any closed field.


问:
To prove $\inf\{\epsilon>0: A\cap S_\epsilon(B)\neq \emptyset\}=\inf\{d(a,b):a\in A,b\in B\}$.
I want to prove the following:

For two nonempty subsets $A,B$ of a metric space $(X,d)$, the following holds $$\inf\{\epsilon>0: A\cap S_\epsilon(B)\neq \emptyset\}=\inf\{d(a,b):a\in A,b\in B\},$$
where $S_\epsilon(B)=\{y\in X:d(y,B)<\epsilon\}=\bigcup\limits_{x\in B}\{y\in X:d(x,y)<\epsilon\}$.

What I tried is:

I proved $\inf\{d(a,b):a\in A,b\in B\}\leq \inf\{\epsilon>0: A\cap S_\epsilon(B)\neq \emptyset\}$. But I am stuck how to prove $\inf\{\epsilon>0: A\cap S_\epsilon(B)\neq \emptyset\}\leq d(a,b)$ for all $a\in A,b\in B$.

答:
If $\inf \{\epsilon >0: A\cap S_{\epsilon} (B) \neq \emptyset\} =0$ there is nothing to prove. Suppose now that this infimum  is a positive number $r$. For $0<\epsilon <r$ we have
$A\cap S_{\epsilon} (B) = \emptyset$ Hence, for every $a \in A, b \in B$ we have $d(a,b) \geq \epsilon$. Letting $\epsilon $ increase to $r$ we conclude that $d(a,b) \geq r$ for all $b \in B$.


问:
Double limits with Polar coordinates
We have to evaluate:

$$\lim_{(x,y) \to (0,0)} \frac{x^{4}-x^{2}y^{2}+y^{4}}{x^{2}+y^{2}+x^{4}y^{4}} \tag{1}$$

If we use polar coordinates and substitute $x=r\cos(\theta) $ and $y=r\sin(\theta)$ and take $r$ common from the expression in the numerator and that in the denominator and further do a bit of canceling we get that (1) is equivalent to,

$$\lim_{r \to 0}\frac{r^2(\cos^4\theta-\cos^2\theta \sin^2\theta + \sin^4\theta)}{\cos^2\theta+\sin^2\theta+r^6\cos^4\theta \sin^4\theta}$$

My teacher is saying that since the above expression has $r$ in front, therefore, the limit will go to zero. I don't think it is rigorous since we have $r$ in the denominator. Any suggestions. Can't seem to factorise these trigonometric expressions. Also tried $y=mx$ substitution to get different limits. They will all go to 0.

Title: Double limits with Polar coordinates

Body:
Another approach (I really dont like polar method)

$$\left|\frac{x^{4}-x^{2}y^{2}+y^{4}}{x^{2}+y^{2}+x^{4}y^{4}}\right|\le \frac{x^{4}+x^{2}y^{2}+y^{4}}{x^{2}+y^{2}+x^{4}y^{4}}\le $$ $$\le x^2 \underbrace{\frac{x^2+y^2}{{x^{2}+y^{2}+x^{4}y^{4}}}}_{\le 1}+y^2 \underbrace{\frac{y^2}{x^{2}+y^{2}+x^{4}y^{4}}}_{\le 1}\le x^2+y^2$$

So the Squezee Theorems guarantees that your limit is 0.


答:
Yes, this sounds reasonable. Starting with a basis $\alpha_1,\ldots,\alpha_n$ of an extension $L / K$, you can multiply them with elements of $\mathcal{O}_K$ to assume that they in fact lie in $\mathcal{O}_L$. After that, one has the following result, whose proof I'll more or less quote.

Lemma (cf. Neukirch, I.2.9). Write $d$ for the discriminant of $\alpha_1,\ldots,\alpha_n$. Then $d\mathcal{O}_L \subseteq \mathcal{O}_K \alpha_1 + \cdots + \mathcal{O}_K \alpha_n$.

Proof: If $\alpha = c_1 \alpha_1 + \cdots + c_n \alpha_n \in \mathcal{O}_L$, with coefficients $c_i$ in $K$, then the $c_i$ are a solution of the system of linear equations
$$\operatorname{Tr}_{L/K}(\alpha_i\alpha) = \sum_j \operatorname{Tr}_{L/K}(\alpha_i\alpha_j)c_j,$$
and as $\operatorname{Tr}_{L/K}(\alpha_i \alpha) \in \mathcal{O}_K$, they are given as the quotient of an element of $\mathcal{O}_K$ by the determinant $\det \operatorname{Tr}_{L/K}(\alpha_i \alpha_j) = d$. Therefore the $dc_j$ live in $\mathcal{O}_K$, and the desired result follows.

I presume this is enough to infer finite generation of $\mathcal{O}_L$ over $\mathcal{O}_K$.

Title: Trace map argument for number fields

Body:
Let $K \subset L$ a finite (and therefore inevitably algebraic)
extension of number fields. Let $O_K$ and $O_L$ be the ring of integers
of $K$ and $L$. The goal is to show that that $O_L$ is finite
$O_K$-module. The later means that there exist
$l_1, l_2,..., l_m \in O_L $ such that $O_L = l_1 \cdot O_K
+ l_2 \cdot O_K + ... + l_m \cdot O_K$.

Now in my lecture notes is noted that this result follows from
an 'easy application of the so called trace map argument'.

I have to confess that I nowhere heard about such
'trace map argument' in that context and I would very thankful
if somebody has an idea what it is and elaborate how this
argument is used here.

My ideas: There exist a $K$-bilinear 'trace map' $Tr: L \times L \to K, (x,y)
\mapsto tr(x \cdot y)$ where $x \cdot y$ is regarded as $K$-linear
endomorphism $L \to L$. Since we are dealing with number fields
it is known that $Tr$ is not degenerated (the extension is separated).
The field extension $K \subset L$ is assumed to be finite, so there exist
a $K$-basis $x_1, ..., x_m \in L$ with $L \cong K^m \cong
x_1K \oplus x_2K \oplus ... \oplus x_mK$.

Now seemingly the trick is to find other basis
$y_1, y_2, ..., y_m \in O_L$ (probably by 'clearing denominators
of the $x_i$) and to show that these $y_i$ are still linear independent $(1)$
over $K$ and for every $l \in L$ holds:

$l \in O_L $ iff $l = \sum_{i=1}^m a_i y_i $ with $a_i \in O_K$ $(2)$

I conjecture that $(2)$ can be derived by the mentioned 'trace map
argument' while $(1)$ can be proved independently. But it is also not
clear how. The $x_i$ form a $K$-basis and not a $L$-basis, therefore
multiplying the basis elements by $x_i$ by elements from $L^*$
could destroy their linear independence, or not?


问:
SOLVED A-Level logarithm question
So theres a question on one of my logarithm revision sheets that i have no idea how to solve, so i thought id come here to ask for your infinite mathematical wisdom.

Given that $\log_a(y)=\cfrac{3}{2}\ $ and $\log_4(a)=b+2$ Show that $y=2^p$ Where p is an expression in terms of b.

Thank you in advance ~

答:
Write the log expressions in exponential form,

$$a^{3/2}=y \;\;\; and \;\;\; 4^{b+2}=a.$$

Substituting $4^{b+2}$ in the first equation for $a$ and using exponent laws quickly yields, $$p=3b+6.$$

(Recall that $(a^{n})^{m}=a^{nm}$)

Title: SOLVED A-Level logarithm question


问:
Origin of Legendre's constant term.
I'm that student who needs to know where does something comes from. I have been studying Differential Equations and Electrodynamics (I'm a physics student), and I was wondering why we (in physics) use $l(l+1)$ as a constant in differential equations (and math...) which then result to be a Legendre's Differential Equation which can be solved through Legendre polynomials.

The equation is:
$$
\frac{1}{Sin(\theta )}\frac{d}{d\theta } \left( Sin(\theta ) \frac{d\Theta }{d\theta } \right) =-k\Theta
$$
If we set (as the book suggests) the constant $k=l(l+1)$ (which is still a constant), the problem is simply solved through Legendre Polynomials. Why does that specific constant works? (As a general rule for Legendre's DE's/Polynomials).

Solving the problem gives...
$$
\frac{d}{d\theta } \left( Sin(\theta ) \frac{d\Theta }{d\theta } \right) =-l(l+1)Sin(\theta )\Theta
$$
$$
\Theta (\theta )=P_l[Cos(\theta )]
$$
Thanks!

答:
If the question is how do you get from your equation to the Legendre equation, from the context, I would expect you have $\theta \in [0,\pi]$.  Then make the substitution $\cos \theta = x$, and the differential equation you have becomes, for $x \in[-1,1]$,
\begin{align}
(1-x^2) \frac {d^2 \Theta}{dx^2} - 2x \frac{d\Theta}{dx} + k\Theta = 0
\end{align}
This is the Legendre equation.  It has been studied extensively, and is described in the Wikipedia article referenced by Mr. Swanson in the comments.  It has solutions known as the Legendre functions for general $k$, denoted $P_\lambda$ and $Q_\lambda$ where, by convention, $k=\lambda(\lambda+1)$ (I am using $\lambda$ instead of $\ell$ to match the article).  The change of expression for $k$ is simply applied to turn your equation into a form that has recognized solutions.

When $\lambda$ is an integer, $P_\lambda$ a polynomial and the  $Q_\lambda$ are known as a Legendre functions of the second kind.  The boundary conditions for your equation will likely help determine which of these functions are relevant to your case.  For example, applying boundary conditions may constrain $k$ to be an integer of the form $\lambda(\lambda+1)$ and you obtain the Legendre polynomials as an orthogonal family of solutions that forms a functional basis.

Title: Origin of Legendre's constant term.


问:
If $F$ is free, then the functor $M \mapsto \operatorname{Hom}_A(F,M)$ is exact.
This is from Serge Lang's Algebra 3rd Edition (see the picture down below if you feel confused). My problem is, how to prove that if $F$ is free, then $M \mapsto \operatorname{Hom}_A(F,M)$ is exact. Indeed I know I can prove this using the fact that free modules are projective. But this statement is inside the declaration of projective module, so it's not a good idea to use properties of projective modules. Besides, it's chapter 3 into the book and many advanced concepts are not introduced. So I my thought is to prove it using properties of free modules. I was thinking about working on the basis of a module but don't know where to start.

I searched the internet but didn't find a suitable one. By suitable I mean the proof can be inserted before the introduction of projective modules.

You can find some discussions of exact functor and projective modules here (turn to 10.4 and 10.5).

答:
Note that $\hom(F,-)$ is always left exact, so the desired result reduces to showing that $\hom(F,-)$ preserves epimorphisms. Fix $p \colon M \to N$ an epi, and let us prove that

$$
p_\ast \colon \hom(F,M) \to \hom(F,N)
$$

is surjective. Given $g \colon F \to N$, we need $h \colon F \to M$ such that $ph = g$. For a given basis $B$ of $F$; one can define $h$ such that $h(b) \in p^{-1}(g(b))$ for each $b \in B$, guaranteeing the former to hold. This makes sense because each preimage is non-empty, here we use that $p$ is surjective.
The existence of $h$ and its well definedness stem from the fact that $F$ is free.


问:
An $8$ oared boat is to be manned by a crew chosen from $14$ men of which $4$ can only steer but can not row & the rest can row but cannot steer.
Full question: An $8$ oared boat is to be manned by a crew chosen from $14$ men of which $4$ can only steer but can not row & the rest can row but cannot steer. Of those who can row, $2$ can row on the bow side. In how many ways can the crew be arranged?

I think in this question we have to choose 1 person (out of 4) who can steer. No. of ways for that =$⁴C_1$

Now 8 seats are left 4 on the bow side & 4 on the non-bow side. We are given two specific person to always be on bow side.
No. of ways of arranging on non bow side = $⁸C_4$×$4!$

No. of ways ways of arranging on bow side= $⁴C_2$×$4!$

My answer=$⁴C_1$×$⁸C_4$×$⁴C_2$×$(4!)²$

But the answer in book is $⁴C_1$×$⁸C_4$×$⁶C_2$×$(4!)²$

答:
The question should be more clear. I think what the question means is that of the $10$ who can row, $2$ can row only on the bow side. Now it does not mean that those two have to be chosen to row but if they get chosen, they must be chosen to row on the bow side.

So as you correctly worked out, number of ways to choose the person to steer is $4 \choose 1$. We then choose $4$ out of $8$ rowers (taking out two who cannot row on non-bow side) for non-bow side, which is $8 \choose 4$.

Now we are left with $6$ rowers to choose from for bow side or choose which two to leave, which is $6 \choose 4$ or $6 \choose 2$.

Then we can arrange on bow side and non-bow side in $(4!)^2$ ways.

That leads to total arrangements of ${4 \choose 1} {8 \choose 4} {6 \choose 2} (4!)^2$


问:
Vector equations with trigonometry
The question

Part a asks to explain why the vector equation for L1 is the written one. I fail to understand why. Any help? It states that L1 passes through the tip and base perimeter of the cone, but shouldn't the Direction Vector be (0j, 1i, 0k ) in that case?

Thank you.

答:
The question states that the tip is at $(0, 5, 0)$, and that the base perimeter fulfills $x^2 + z^2 = 1$.

(Technically speaking, $x^2 + z^2 = 1$ is really a right circular cylinder of infinite length and radius $1$, axis on the $y$ axis.  Here, they obviously mean that the cone base is a circle of radius $1$ on the $x z$ plane, centered at origin.)

Base perimeter is a circle of radius $1$ centered at origin $(0, 0, 0)$.

If you have a line that passes through $(0, 5, 0)$ in direction $(0, 1, 0)$, it will pass outwards from the tip.  It has already passed through origin, $(0, 0, 0)$.  However, origin is at the center of the base; it is not at the perimeter of the base.  (Perimeter is the outer boundary.)

Any line that passes through the tip at $(0, 5, 0)$ and a point on the perimeter of the base, say $(1, 0, 0)$, $(0, 1, 0)$, $(-1, 0, 0)$, or $(0, -1, 0)$, will fulfill the rule.

We can parametrise the points on the perimeter using angle parameter $\theta$, as $(\cos\theta, 0, \sin\theta)$.  (If $\theta = 0$, the point is $(1, 0, 0)$; if $\theta = 90°$, the point is $(0, 1, 0)$; if $\theta = 180°$, the point is $(-1, 0, 0)$, and so on.)

Any line can be parametrised using a real parameter, say $\lambda$, and two points the line passes through, say $\vec{a}$ and $\vec{b}$:
$$\ell = (1 - \lambda) \vec{a} + \lambda \vec{b} = \vec{a} + \lambda (\vec{b} - \vec{a})$$
(This is also the equation for linear interpolation.)  When $\lambda = 0$, $\ell = \vec{a}$.  When $\lambda = 1$, $\ell = \vec{b}$.

Here, $\vec{a} = (0, 5, 0)$ (the tip), and the point on the perimeter $\vec{b} = (\cos\theta, 0, \sin\theta)$, so the equation of the line is
$$\ell = (1 - \lambda)\left[\begin{matrix}0\\5\\0\end{matrix}\right] + \lambda\left[\begin{matrix}\cos\theta\\0\\\sin\theta\end{matrix}\right] = \left[\begin{matrix}0\\5\\0\end{matrix}\right] + \lambda\left[\begin{matrix}\cos\theta\\-5\\\sin\theta\end{matrix}\right]$$

Title: Vector equations with trigonometry


问:
Why is the Kullback-Leibler divergence not symmetric?
As known the Kullback-Leibler Divergence:

$$\operatorname{KL}=\sum_{i=1}^n \ln(\frac{P(i)}{Q(i)})P(i)$$

is not symmetric.
I would like to know how this can be seen from the formula. I am aware that I could just try it out with exchaning Q and P for some special case, but I would like to know the mathematical reason behind it. Also, is actually here "i" the random variable?

Thanks a lot
Miau

答:
In addition to the algebraic reason that Robert Israel gave, there's a very nice "moral reason" that the Kullback-Leibler divergence is not symmetric. Roughly speaking, it's because you should think of the two arguments of the KL divergence as different kinds of things: the first argument is empirical data, and the second argument is a model you're comparing the data to. Here's how it works.

Take a bunch of independent random variables $X_1, \ldots, X_n$ whose possible values lie in a finite set.* Say these variables are identically distributed, with $\operatorname{Pr}(X_i = x) = p_x$. Let $F_{n,x}$ be the number of variables whose values are equal to $x$. The list $F_n$ is a random variable, often called the "empirical frequency distribution" of the $X_i$. What does $F_n$ look like when $n$ is very large?

More specifically, let's try to estimate the probabilities of the possible values of $F_n$. Since the set of possible values is different for different $n$, take a sequence of frequency distributions $f_1, f_2, f_3, \ldots$ approaching a fixed frequency distribution $f$. It turns out** that
$$\lim_{n \to \infty} \tfrac{1}{n} \ln \operatorname{Pr}(F_n = f_n) = -\operatorname{KL}(f, p).$$
In other words, the Kullback-Leibler divergence of $f$ from $p$ lets you estimate the probability of getting an empirical frequency distribution close to $f$ from a large number of independent random variables with distribution $p$.

You can find everything I just said, and more, in the excellent article "Information Theory, Relative Entropy and Statistics," by François Bavaud.

* You can also do this more generally, but I don't know anything about that.

** Using Stirling's approximation, $\ln k! \in k\ln k - k + O(\ln k)$.


答:
You're just forgetting that

$r^2(\cos^4\theta-\cos^2\theta \sin^2\theta + \sin^4\theta)\le Ar^2$ for some $A$ since $0\le |\cos \theta|,|\sin\theta| \le 1$,

$1+r^6\cos^4\theta\sin^4\theta\ge 1$.
Therefore
$$\frac{r^2(\cos^4\theta-\cos^2\theta \sin^2\theta + \sin^4\theta)}{1+r^6\cos^4\theta \sin^4\theta}\le\frac{Ar^2}1=Ar^2.$$

We simply use here that to get an upper bound of a fraction with positive numerator and denominator, we only need an upper bound for the numerator and a nonzero lower bound for the denominator.

Title: Double limits with Polar coordinates

Body:
We have to evaluate:

$$\lim_{(x,y) \to (0,0)} \frac{x^{4}-x^{2}y^{2}+y^{4}}{x^{2}+y^{2}+x^{4}y^{4}} \tag{1}$$

If we use polar coordinates and substitute $x=r\cos(\theta) $ and $y=r\sin(\theta)$ and take $r$ common from the expression in the numerator and that in the denominator and further do a bit of canceling we get that (1) is equivalent to,

$$\lim_{r \to 0}\frac{r^2(\cos^4\theta-\cos^2\theta \sin^2\theta + \sin^4\theta)}{\cos^2\theta+\sin^2\theta+r^6\cos^4\theta \sin^4\theta}$$

My teacher is saying that since the above expression has $r$ in front, therefore, the limit will go to zero. I don't think it is rigorous since we have $r$ in the denominator. Any suggestions. Can't seem to factorise these trigonometric expressions. Also tried $y=mx$ substitution to get different limits. They will all go to 0.


问:
Is the operator norm of the canonical surjection $X\to X/U$ equal to $1$?
Let $X$ be a normed space, $U$ be a closed subspace of $X$ and $\pi:X\to X/U$ denote the canonical projection. By definition, $\pi$ is surjective and it is easy to show that $\pi\left(B^X_1(0)\right)=B^{X/U}_1(0)$, where $B^X_1(0)$ and $B^{X/U}_1(0)$ denote the open unit balls around $0$ in $X$ and $X/U$, respectively.

It should immediately follow that $\pi$ is an open map and the operator norm of $\pi$ is at most $1$. But can we show that the operator norm is precisely equal to $1$ (unless $X=U$, of course)?

答: Fix $\varepsilon \in (0,1)$. If $S$ is proper (and closed) by Riesz's lemma applied to $1-\varepsilon$ there exists $v \in X$ such that $\|v\| = 1$ and $\|\pi(v)\| = d(v,S) \geq 1-\varepsilon$. Hence $\|\pi\| \geq 1-\varepsilon$; now let $\varepsilon \to 0$.


问:
For a differential equation, prove that there exists an integrating factor dependent only on $x$
Consider the differential equation $M(x, y)dx + N(x,y)dy = 0$. Prove that there exists an integrating factor that is dependent on $x$ if and only if $\frac {M_y - N_x}{N} = f(x)$, a function of $x$ only.

I need to prove that in such a case, the integrating factor is:
$$I(x) = e^ { \int {f(x)dx}}.$$

I tried to approach this problem by rewriting $M(x, y)dx + N(x,y)dy = 0$ as $M(x, y) + N(x,y) \frac {dy}{dx} = 0.$ Assuming the differential equation is exact, there exists a potential function $φ$ such that $ \frac {∂φ}{∂x} + \frac{∂φ}{∂y} \frac{dy}{dx} = 0.$

But this implies that $\frac {∂φ}{∂x} = 0$. So $φ(x, y)$ is a function of $y$ only. So we can deduce that $φ(x, y)$ is a function of $x$ only and $φ(x, y) = c$, where $c$ is a constant.

This is where I'm stuck and I'm not sure how to proceed with this proof. Any guidance is greatly appreciated!

答:
Question: Consider the differential equation $M(x, y)dx + N(x,y)dy = 0$. Prove that there exists an integrating factor that is dependent on $x$ if and only if $\frac {1}{N}(M_y - N_x) = f(x)$, a function of $x$ only.

Proof: Here the given general first order differential equation is $$M(x, y)dx + N(x,y)dy = 0\tag1$$
We shall suppose that there exists an integrating factor that depends only on $~x~$: $~\mu=\mu(x)~$

Now if $~\mu~$ is an integrating factor, then  $~M(x, y)\mu(x)dx + N(x,y)\mu(x)dy~$ must be exact. i.e., $$\dfrac{\partial}{\partial y}\left[M(x, y)\mu(x)\right]=\dfrac{\partial}{\partial x}\left[N(x, y)\mu(x)\right]$$ $$\implies\mu\dfrac{\partial M}{\partial y}=\dfrac{\partial\mu}{\partial x}N+\mu\dfrac{\partial N}{\partial x}$$ $$\implies \dfrac{\partial\mu}{\partial x}=\dfrac 1{N}\left(\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}\right)\mu\tag2$$ Now if $~\mu~$ depends only on $~x~$ and not on $~y~$, then necessarily $~\mu~$ depends only on $~x~.$ Thus the RHS of the above equation must be a function of $~x~$ only. i.e., $$\dfrac 1{N}\left(\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}\right)=f(x)$$ Thus the equation $(2)$ becomes., $$\dfrac{\partial\mu}{\partial x}-f(x)\mu=0$$ which is a first order linear differential equation and can be solved and the solution is $$\mu(x)=A \exp\left[\int f(x) dx\right]=A \exp\left[\int \dfrac 1{N}\left(\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}\right) dx\right]$$which gives us an integrating factor for the equation $(1)$ in the form $~\frac 1{N}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)~$ depends on $~x~$ only.

Conversely, let for the equation $(1)$, $$\dfrac 1{N}\left(\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}\right)=f(x)\implies Nf(x)=\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}\tag3$$
Multiplying both side of $(1)$ by $~e^{\int f(x)dx}~$, we have $~M_1~dx+N_1~dy=0~$ where $~M_1=M~e^{\int f(x)dx}~$ and $~N_1=N~e^{\int f(x)dx}~.$
Now $$\dfrac{\partial M_1}{\partial y}=\dfrac{\partial M}{\partial y}e^{\int f(x)dx}\tag4~~~~~~~~~~\text{and}$$
$$\dfrac{\partial N_1}{\partial x}=\dfrac{\partial N}{\partial x}e^{\int f(x)dx}+N~e^{\int f(x)dx}~f(x)=e^{\int f(x)dx}\left\{\dfrac{\partial N}{\partial x}+Nf(x)\right\}$$ $$\implies \dfrac{\partial N_1}{\partial x}=e^{\int f(x)dx}\left\{\dfrac{\partial N}{\partial x}+\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}\right\}~,~~~\text{from $(3)$}$$
$$\implies \dfrac{\partial N_1}{\partial x}=e^{\int f(x)dx}~\dfrac{\partial M}{\partial y}\tag5$$ From equation $(4)$ and $(5)$, it is clear that $$\dfrac{\partial M_1}{\partial y}=\dfrac{\partial N_1}{\partial x}$$which shows that $~M_1~dx+N_1~dy=0~$ is an exact differential equation and hence $~e^{\int f(x)dx}~$ is it's integrating factor that depends only on $~x~$.


问:
Looking for a quick method to evaluate an expression of polynomial roots
So define a polynomial $P(x) = 4x^3 + 4x - 5 = 0$, whose roots are $a, b $ and $c$. Evaluate the value of $(b+c-3a)(a+b-3c)(c+a-3b)$

Now  tried this in two ways (both failed because it was far too messy)

Expand everything out (knew this was definitely not the required answer but I cant think of the quick method myself)

Using sum and product (which is still quite lengthy but better)

Of course the above methods relied on the use of Vieta's sum/product of roots.

Does anyone have an amazing concise solution for me? I know you guys are full of tricks, and I enjoy reading them.

答: Hint: $b+c-3a = (a+b+c)-4a = -4a$ and everything simplifies easily.


答:
The $2$ answers above are indeed excellent, and should be sufficient for your problem. Below I offer a different way of proving this result:
$$\begin{align}\sum_{i=1}^{n-1}\frac{1}{i(i+1)}&=\sum_{i=1}^{n-1}\frac{1}{i}-\frac{1}{i+1}\\
&=\frac{1}{1}-\frac{1}{2}\\
&+\frac{1}{2}-\frac{1}{3}\\
&+\frac{1}{3}-\frac{1}{4}\\
&~~\vdots\\
&+\frac{1}{n-1}-\frac{1}{n}\end{align}$$
You should be able to see that all terms cancel out other than $1$ and $-\frac{1}{n}$
Hence,
$$\sum_{i=1}^{n-1}\frac{1}{i(i+1)}=1-\frac{1}{n}=\frac{n-1}{n}$$
as required.

Title: Proof by Induction for $\sum_{i = 1}^{n - 1} \frac{1}{i(i+1)}$

Body:
$$\sum_{i = 1}^{n - 1} \frac{1}{i(i+1)} = \frac{n- 1}{n}$$

How would you go about proving this by induction? the base case $n=1$ gives you $1/2 = 0$

The only thing I can think of using is $\sqrt{2}$, but that is not a natural number.


答:
Note that:

$b+c-3a=a+b+c-4a=-4a$;

$a+b-3c=a+b+c-4c=-4c$;

$c+a-3b=a+b+c-4b=-4b$.

So, you're after $(-4a)\times(-4c)\times(-4b)=-64abc$. But you know that $abc=\frac54$.

Title: Looking for a quick method to evaluate an expression of polynomial roots

Body:
So define a polynomial $P(x) = 4x^3 + 4x - 5 = 0$, whose roots are $a, b $ and $c$. Evaluate the value of $(b+c-3a)(a+b-3c)(c+a-3b)$

Now  tried this in two ways (both failed because it was far too messy)

Expand everything out (knew this was definitely not the required answer but I cant think of the quick method myself)

Using sum and product (which is still quite lengthy but better)

Of course the above methods relied on the use of Vieta's sum/product of roots.

Does anyone have an amazing concise solution for me? I know you guys are full of tricks, and I enjoy reading them.


答:
We have $a+b+c=0, ab+bc+ca=1, abc=5/4$
Le $y=a+b-3c=a+b+c-4c=-4c \implies y=-4x$
Let us transform $x^3+4x-5=0$ bt $x=y/-4$ to get a the $y$ equation
$4(y/-4)^3+4(-y/4)-5=0 \implies y^3+16y+80=0$
The y$y$ equation will have the required roots.
So the product of roots is $y_1y_2y_3=-80.$


问:
Group as the Union of Subgroups
We know that a group $G$ cannot be written as the set theoretic union of two of its proper subgroups. Also $G$ can be written as the union of 3 of its proper subgroups if and only if $G$ has a homomorphic image, a non-cyclic group of order 4.

In this paper http://www.jstor.org/stable/2695649 by M.Bhargava, it is shown that a group $G$ is the union of its proper normal subgroups if and only if its has a quotient that is isomorphic to $C_{p} \times C_{p}$ for some prime $p$.

I would like to make the condition more stringent on the subgroups. We know that Characteristic subgroups are normal. So can we have a group $G$ such that , $$G = \bigcup\limits_{i} H_{i}$$ where each $H_{i}$'s are Characteristic subgroups of $G$?

答:
Update: While searching on the internet i found this paper, "A remark on Hyperabelian groups" by G.Baumslag where there is this remark

Moreover so is any torsion abelian group of finite rank, since it is the union of finite characteristic subgroups.  Would like to get a reference for this.

OwnerDisplayName: anonymous

LastEditorDisplayName: anonymous

Title: Group as the Union of Subgroups

ContentLicense: CC BY-SA 2.5


问:
How can I solve this non-linear differential equation?
I'm trying to solve the equation $$y' = 1 - y^2$$
Here is my attempt:
$$y' = 1 - y^2$$
Divide by (1-y^2)
$$\frac{y'}{1-y^2} = 1$$
Integrate both sides:
$$\frac{1}{2}\log|\frac{y+1}{y-1}|=t+c$$
Rearrange
$$y = \frac{ke^{2t}+1}{ke^{2t}-1}$$
I'd have thought that solution was right, but we have to figure out a specific solution with y(0) = 0. But this isn't possible with the above equation.</p>

答: Since you want a solution near $y=0$, you should use $1-y$ in the denominator (as it will be positive) and can remove the absolute value signs. This changes some signs in your answer, giving $$y = \frac{ke^{2t}-1}{ke^{2t}+1}$$ and $k=1$ gives $y(0)=0$</p>


问:
Product of two metrizable space is metrizable given by the $d=d_1+d_2$
I need to show that the product of two metrizable topological space $(X, T_1)$ and $(Y, T_2)$ is metrizable and the metric is given $d=d_1+d_2$ where $d_1$ is the metric that induces the topology $T_1$ and $d_2$ induces $T_2$.

I guess showing that product of $(X, d_1)$ and $(Y, d_2) = ((X,Y), d)$. I'm not sure where to start, but I guess we need to show any open ball in the product metric $B((x,y),r)$ contains the product of open balls from $B(x,r)$ and $B(y,r)$ and vice versa.

答:
First note that the projections $\pi_X: X \times Y \to X$ and $\pi_Y: X \times Y \to Y$ are both continuous seen as maps from $(X \times Y,d)$ because e.g.$$d_1(\pi_X((x,y),\pi_X(x',y')) = d_1(x,x') \le d((x,y),(x',y'))$$ so the projections are contractions (so even uniformly continuous, $\delta=\varepsilon$ works globally).

Because the product topology $\mathcal{T}_p$ on $X \times Y$ is defined as the smallest topology on $X \times Y$ that makes both $\pi_X$ and $\pi_Y$ continuous, and the metric topology from $d$ is one such topology, we have that $\mathcal{T}_p \subseteq \mathcal{T}_d$, showing one inclusion.

To show the reverse, take any open set $O$ in $\mathcal{T}_d$ and $(p,q)$ in $O$. We have to show $(p,q)$ is an interior point for $O$ in the product topology. As the point is arbitrary, it follows that $O$ is then product-open and we're done showing the equality of these topologies.

First note that is an open ball of radius $r>0$ around $(p,q)$ such that $B_d((p,q),r) \subseteq O$ because $O$ is $d$-open. Then note that $U=B(p,\frac{r}{2})$ is open in $X$ and $V=B(q, \frac{r}{2})$ is open in $Y$ (being metric balls in the metric topology) and we claim that $(p,q) \in U \times V \subseteq B_d((p,q), r)$, which is quite obvious from the definitions. But $U \times V$ is product open and so indeed $(p,q)$ is an interior point of $O$ for the product topology.


问:
Given that $F(x)=\int^x_0e^{t^2}dt+1$, which of these is true?
Let $F(x)=\int^x_0e^{t^2}dt+1$. Then:

a) $F(x)$ is convex in $(-\infty , \infty )$.

b) The equation $F(x)=0$ has two solutions.

c) $F(x)$ is invertible and $(F^{-1})'(1)=1$.

d) $F(1)=5$.

e) $F(x)=F(-x)$ for every $x$.

My attempt:

a) $F'(x)=e^{x^2}$, $F''(x)=2xe^{x^2}<0$ for $x<0$ so it's not true.

b) I don't know how to disprove this one.

c) I know that $(F^{-1})'(F(x))=\frac {1}{F'(x)}$, I hesitated a lot here, since if I want to get $(F^{-1})'(1)$, I need to substitute $x=0$, but do I substitute also $x=0$ in $F'(x)$ or $x=1$? (this was  critical to know for this question).

d) I got no idea just pure intuition that the area at $x=1$ of $e^{x^2}$ would be less than $5$.

e) That's not true, it changes the integral, and will somewhere affect the $e$ being with $(-)$ in the solution.

The answer was c). I would really appreciate any help in disproving the wrong answers and any tips regarding my confusion about c).

答:
(b) is wrong because $F'(x) = e^{x^2} > 0$, so that $F$ is strictly increasing.

For (c) one can use that $F'(x) = e^{x^2} \ge 1$ implies that the range of $F$ is $\Bbb R$ and $F$ is invertible. Then
$$
 (F^{-1})'(1) = \frac{1}{F'(F^{-1}(1))}=  \frac{1}{F'(0)} = \frac{1}{e^{0^2}} = 1 \, .
$$


答:
If $U\neq X$, then your projection has norm equal to $1$. Assume it doesn't and let $c=\Vert \pi\Vert<1$, then we have for every $x\in X$
$$ \Vert \pi x\Vert \leq c\Vert x\Vert.$$
This means there exists $u_1\in U$ such that
$$ \Vert x + u_1\Vert \leq (1+c)/2 \Vert x \Vert.$$
Now we apply the same argument to $x+u_1$ and find $u_2\in U$ such that
$$ \Vert x +u_2\Vert \leq [(1+c)/2]^2 \Vert x\Vert.$$
Thus, by induction we find a sequence $(u_n)_{n\geq 1}\subseteq U$ that converges to $x$. As $x$ as arbitrary, we get that $U$ is dense and as $U$ is closed, we get $U=X$.

Title: Is the operator norm of the canonical surjection $X\to X/U$ equal to $1$?

Body:
Let $X$ be a normed space, $U$ be a closed subspace of $X$ and $\pi:X\to X/U$ denote the canonical projection. By definition, $\pi$ is surjective and it is easy to show that $\pi\left(B^X_1(0)\right)=B^{X/U}_1(0)$, where $B^X_1(0)$ and $B^{X/U}_1(0)$ denote the open unit balls around $0$ in $X$ and $X/U$, respectively.

It should immediately follow that $\pi$ is an open map and the operator norm of $\pi$ is at most $1$. But can we show that the operator norm is precisely equal to $1$ (unless $X=U$, of course)?


问:
How prove this $\int_{0}^{\infty}\sin{x}\sin{\sqrt{x}}\,dx=\frac{\sqrt{\pi}}{2}\sin{\left(\frac{3\pi-1}{4}\right)}$
Prove that

$$\int_{0}^{\infty}\sin{x}\sin{\sqrt{x}}\,dx=\frac{\sqrt{\pi}}{2}\sin{\left(\frac{3\pi-1}{4}\right)}$$

I have some question. Using this, find this integral is not converge, I'm wrong?

Thank you everyone

答:
Numerical calculation shows that the graph of

$$ y = \int_{0}^{x} \sin t \sin \sqrt{t} \, dt $$

is given by

Though a graph cannot constitute a proof, it strongly suggests that $y$ cannot converge as $x \to \infty$. Indeed, we can show that

$$ y(x) = -\cos x \sin \sqrt{x} + \frac{\sqrt{\pi}}{2} \sin \left( \frac{3\pi - 1}{4} \right) + o(1) $$

as $x \to \infty$. Thus in ordinary sense, the integral

$$ \int_{0}^{\infty} \sin x \sin \sqrt{x} \, dx $$

diverges. But if we understand the integral in Abel summation senas as follows

$$ \int_{0}^{\infty} \sin x \sin \sqrt{x} \, dx := \lim_{s \to 0^{+}} \int_{(0, \infty)} \sin x \sin \sqrt{x} \; e^{-sx} \, dx, $$

then the oscillating part vanishes and we obtain the identity. In fact, it converges to the proposed limit as weak as in Cesaro summation sense.

Now let us show that the integral converges in Able summation sense. By some calculation, we have

\begin{align*}
\int_{0}^{\infty} \sin x \sin \sqrt{x} \; e^{-sx} \, dx
&= \int_{0}^{\infty} 2x \sin x \sin (x^2) \; e^{-sx^{2}} \, dx \\
&= \int_{0}^{\infty} x \Re \left( e^{-i(x^2-x)} - e^{-i(x^2+x)} \right) e^{-sx^{2}} \, dx \\
&= \Re \int_{-\infty}^{\infty} x e^{-(s+i)x^2 + ix} \, dx.
\end{align*}

By noting that

$$ z e^{-(s+i)z^2 + iz} = z \exp \left\{ -(s+i) \left( z - \tfrac{i}{2(s+i)} \right)^2 + \tfrac{i}{4(1-is)} \right\} $$

is an entire function with a nice vanishing speed as $ \left| \Re z \right| \to \infty$, we find that we can shift the contour of integration so that

\begin{align*}
\int_{0}^{\infty} \sin x \sin \sqrt{x} \; e^{-sx} \, dx
&= \Re \left[ e^{\frac{i}{4(1-is)}} \int_{-\infty}^{\infty} \left(x + \frac{i}{2(s+i)} \right) e^{-(s+i)x^2} \, dx \right] \\
&= \Re \left[ e^{\frac{i}{4(1-is)}} \frac{i}{2(s+i)} \int_{-\infty}^{\infty} e^{-(s+i)x^2} \, dx \right] \\
&= \Re \left[ e^{\frac{i}{4(1-is)}} \frac{i}{2(s+i)} \frac{\sqrt{\pi}}{\sqrt{s + i}} \right].
\end{align*}

Taking $s \to 0^{+}$, we find that in Abel summation sense,

\begin{align*}
\int_{0}^{\infty} \sin x \sin \sqrt{x} \, dx
&= \frac{\sqrt{\pi}}{2} \Re \left( e^{\frac{i-i\pi}{4}} \right)
 = \frac{\sqrt{\pi}}{2} \cos \left( \frac{1-\pi}{4} \right)
 = \frac{\sqrt{\pi}}{2} \sin \left( \frac{3\pi - 1}{4} \right)
\end{align*}

as desired.


问:
Numbers are formed using the digits $1,1,2,2,2,2,3,4,4$ taken all at a ...

Numbers are formed using digits $1,1,2,2,2,2,3,4,4$ taken all at a time. Find number of such numbers so that odd digits occupy even places.

The odd digits are $1,1,3$ and there are 4 even places

So permutation of arranging 3 items in 4 places where two items are alike is.

Now the solution given to me involves use of combinatorics, and I can’t understand how they came up with it, since we are arranging and not selecting. Can I get an explanation on how combinations can be used to solve this (or how to do it without combinations)

答:
Total numbers which can be formed : $ \cfrac {9!}{2!4!2!} $

Total odd digits : $3 , i.e. 1,1,3 $

Total even places : $4$

Ways in which you can arrange the digits : $⁴C_3 * \cfrac{3!}{2!}$  $\longrightarrow (1)$

$3! - $total ways of arranging  $ 1,1,3$

$2! - $two $1's$ are alike

Of course, there is no restriction on the other digits,

Ways of arranging the other digits : $ \cfrac {6!} {4!2!}$  $\longrightarrow (2)$

$6! - $total ways of arranging  $ 2,2,2,2,4,4 $

$2! - $two $4's$ are alike

$4! - $four $2's$ are alike

Now you just multiply $(1)$ and $(2)$ on account of multiplication principle.

The final answer is : $⁴C_3 * \cfrac{3!}{2!}$ $*$$ \cfrac {6!} {4!2!}$

Hope this clears your doubt.


问:
Geometrical Interpretation of the general solution of a first order, quasi linear partial differential equation
In a course on partial differential equations I came through this theorem about the general solution of a first order quasi-linear partial differential equation.

The general solution of a first-order, quasi-linear partial differential equation
$$a(x,y,u)u_x + b(x,y,u)u_y = c(x,y,u)$$
is given by $f(ϕ,ψ)=0$,
where $f$ is an arbitrary function of $ϕ(x,y,u)$ and $ψ(x,y,u).$

$ϕ=C1$ and $ψ=C2$ are solution curves of the characteristic equations
$$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{c}.$$

Is there any geometric interpretation of both these points so that I can have a better intuitive understanding of the graphical representation of $f$,$\phi$ and $\psi$ ?

答:
As per usual set-up, we are looking at $\mathbb{R}^3$ coordinates $(x,y,z)$ for surfaces $z(x,y)=u(x,y)$.  Geometrically the PDE says the vector field $\mathbf{v}(x,y,z)=(a(x,y,u(x,y)),b(x,y,u(x,y)),c(x,y,u(x,y)))$ is orthogonal to $(u_x(x,y,u(x,y)),u_y(x,y,u(x,y)),-1)$.

The level sets of $\phi$ (and similarly $\psi$) are therefore integral surfaces of the vector field $\mathbf{v}$ in $\mathbb{R}^3$ from $\frac{\mathrm{d}x}{a}=\frac{\mathrm{d}y}{b}=\frac{\mathrm{d}u}{c}$.  We want $\phi$ and $\psi$ to have the property that $\phi=C_1$ intersect $\psi=C_2$ transversely for $C_1,C_2\in\mathbb{R}$.

Now $f(v,w)=0$ is (assuming regularity conditions) a smooth curve in $vw$-plane.  So $f(\phi,\psi)=0$ defines a smooth union of characteristic curves, so is also an integral surface of $\mathbf{v}$.


问:
Prove that the algebraic expression is greater than zero
Prove that $x^8 - x^5 + x^2 -x + 1 >0$ for $x \in \mathbb{R}$.

It's from the (junior) high-school competition and the idea is that not everyone there knows calculus, so that I'm looking for more "basic" justification.

My idea was to use AM-GM: $x^8 + x^2 \geq 2\sqrt{x^{10}} = 2x^5$. Thus,
$$x^8 - x^5 + x^2 -x + 1 \geq  x^5-x + 1 $$
So that we can now focus on proving that $x^5 - x + 1 > 0 $ but I don't really see how to do so without calculus...

答:
0)For $x\le 0:$  obvious, since all terms are non-negative.

1)$x\ge 1$: $x^8\ge x^5,$ and $x^2 \ge x;$

2)$0<x<1:$ $x^2 >-x^5;$ and

$x^8-x+1= x^8 +(1-x)>0,$ and we are done(Why?)


问:
Is $\tau(x,y)\leq\sqrt{n}d(x,y)$ where $d$ is the Euclidean metric, and $\tau$ the taxicab metric?
Suppose $d$ is the Euclidean metric on $\mathbb{R}^n$, $d(x,y)=\sqrt{\sum_{i=1}^n(x_i-y_i)^2}$, and $\tau(x,y)=\sum_{i=1}^n|x_i-y_i|$ is the taxicab metric on $\mathbb{R}^n$.

When showing this two metrics are equivalent, it is not hard to bound $\tau$ by
$$
\tau(x,y)\leq n d(x,y).
$$

Is it possible to tighten the bound to $\tau(x,y)\leq\sqrt{n}d(x,y)$?

答:
No, it's not possible:
$\tau( (0.4,0.4), (0,0) ) = 0.8 > 0.46 > \sqrt{2} (0.4^2+0.4^2) = \sqrt{2}d( (0.4,0.4), (0,0) )$

edit: As I somehow missed to take the square root, this counterexample is wrong. See below for a correct answer that it is in fact possible to work with $\sqrt{n}$.


问:
Almost sure convergence of $L^p$ convergent series
I am wondering if my following reasoning is correct.

I have a series $\sum_{i=1}^\infty X_i$ that converges in $L^p(\Omega)$ to some $X$. $L^p$ convergence implies a.e. convergence of a subsequence, hence $\sum_{i=1}^{n_k}X_i$ converges almost everywhere to $X$ for some increasing $(n_k)_k$. But by definition the limit of this subsequence is the same as $\sum_{i=1}^\infty X_i$, so this series also converges a.e. to $X$.

答:
The limit in $L^p$ will coincide with the almost-everywhere-limit of the subsequence.
However, this does not imply that the whole sequence converges almost everywhere.

Classical counterexample:
Write $Y_k:=\sum_{i=1}^k X_i$.
Let $\{Y_k\}$ be the sequence of indicator functions
$$\chi_{[0,1/2]},\chi_{[1/2,1]},\chi_{[0,1/4]},\chi_{[1/4,1/2]},\dots$$
This sequence converges in $L^p([0,1])$ to $0$.
It has an almost everywhere convergent subsequence:
$$\chi_{[0,1/2]},\chi_{[0,1/4]},\chi_{[0,1/8]},\dots$$
Yet, it converges nowhere.


答: It is also immediate from $\pi(B_1^X(0))=B_1^{X/U}(0)$. If the norm of $\pi$ were strictly less than $1$, we would have $\pi(B_1^X(0))\subseteq \|\pi\|B_1^{X/U}(0)\subsetneq B_1^{X/U}(0)$.


问:
How to calculate $(n+1)$'th derivative of $\prod_{i=0}^{n}(x-a_i)$ with constants $a_i \in \mathbb{R}$?
Letting $f(x):=\prod_{i=0}^{n}(x-a_i)$, with a real variable $x \in \mathbb{R}$ and $n\in \mathbb{N}$, I want to calculate the $(n+1)$'th derivative of $f$ with respect to $x$. $f(x)$ is a product of $(n+1)$ linear functions in $x$, each of which are $(n+1)$ times differentiable. In this regard I was thinking of maybe applying the general leibniz rule to calculate the derivative, but I don't fully understand the basis of it yet. Another approach that comes to mind is a calculation involving induction on $n$. What would be a desirable approach?

答:
$\prod_{i=0}^n (x-a_i)$ is a polynomial of degree exactly $n+1$, so the derivative is $(n+1)!$ times the leading coefficient (1, in your case).

So the answer is $(n+1)!$


问:
Divisors on Riemann Surface form a group under pointwise addition(doubt)
In (Rick Miranda, page 129) Divisor on a Riemann Surface is defined as :

A divisor on Riemann surface $X$ is a function $D: X \rightarrow \mathbb{Z}$ whose support is a discrete subset of $X$.

Author states that The divisors on $X$ form a group under pointwise addition.

I'm having difficulty in verifying that divisors form a group under pointwise addition. I tried to check the above fact by taking $X = \mathbb{C}$ as Riemann Surface. Let $f, g$ be two divisors on $X$ with supports $\{\frac{1}{n}\}_{i=1}^\infty$ and $\{0\}$ respectively. f and g are defined as follows
$$f(x) = \begin{cases} n \text{ ,if } x = 1/n\\ 0 \text{ ,otherwise}\\
\end{cases}
$$
$$g(x) = \begin{cases} 1 \text{ , if } x = 0 \\ 0 \text{ ,otherwise}\end{cases}$$

Consider $f+g:X \rightarrow \mathbb{C}$, clearly $(f+g)(x) \ne 0 $ when $x \in \{\frac{1}{n}\}_{i=1}^\infty$ or $x = 0$, Hence support of $f+g$ has a limit. So, by definition $f+g$ is not a divisor.

Is there something that I don't understand?

答: The definition of divisor in the book is wrong; I believe this is  in the errata. For a map $D:X\rightarrow\mathbb{Z}$, the support should be $\overline{\{p\in X:D(p)\neq0\}}$ (the closure of what Miranda calls the support) and a divisor is such a map with discrete support. So, with this definition, your $f$ isn't a divisor and there is no issue.


问:
Calculating the improper integral $\int _{1}^{\infty}\frac{e^{\sin x}\cos x}{x}\,dx$
I want to calculate the value of $$\int_{1}^{\infty}\frac{e^{\sin x}\cos x}{x}\,dx$$

I was able to prove using Dirichlet's test that it does converge, but how can I calculate its value?

答:
It is pretty obvious that there cannot be an elementary result for the given integral, nor one using "mild" special functions like Bessel and error functions.

So we will numerically integrate it. The problem, however, is that the integrand is oscillatory as $x\to\infty$ and only decays like $\frac1x$, which means the following approach is needed:

Break up $[1,\infty)$ at the integrand zeros: $\frac\pi2,\frac{3\pi}2,\frac{5\pi}2\dots$

Compute the integral over each segment, using series acceleration techniques to "predict" future values, then sum everything up

In mpmath this can be accomplished as

quadosc(lambda x: exp(sin(x))*cos(x)/x, [1, inf], period=2*pi)
and this returns the accurate result
$$-0.379597645637239886\dots$$


答:
The key to quickly answer these multiple choice questions is to identify the largest commonality, which in this case is the sign of the first derivative.

The fact that $F'(x)=e^{x^2}$ means $F$ is strictly increasing. This immediately eliminates

b) as $F$ crosses the horizontal axis at most once;

e) as no even function is strictly increasing on the entire domain, and

d) as $\int_0^1e^{t^2}\,dt<(1-0)\times e^{1^2}$.

We can eliminate a) by differentiating again to check the sign, which you have correctly done.

With c), we know that $F$ is invertible as it is continuous and its range is the set of real numbers. The inverse derivative formula which is derived from using the chain rule on $F^{-1}(F(x))=x$, can be used to evaluate $(F^{-1})'(1)=[F'(F^{-1}(1))]^{-1}=F'(0)^{-1}$ as $F$ is bijective.

Thus c) is the only correct answer.

Title: Given that $F(x)=\int^x_0e^{t^2}dt+1$, which of these is true?

Body:
Let $F(x)=\int^x_0e^{t^2}dt+1$. Then:

a) $F(x)$ is convex in $(-\infty , \infty )$.

b) The equation $F(x)=0$ has two solutions.

c) $F(x)$ is invertible and $(F^{-1})'(1)=1$.

d) $F(1)=5$.

e) $F(x)=F(-x)$ for every $x$.

My attempt:

a) $F'(x)=e^{x^2}$, $F''(x)=2xe^{x^2}<0$ for $x<0$ so it's not true.

b) I don't know how to disprove this one.

c) I know that $(F^{-1})'(F(x))=\frac {1}{F'(x)}$, I hesitated a lot here, since if I want to get $(F^{-1})'(1)$, I need to substitute $x=0$, but do I substitute also $x=0$ in $F'(x)$ or $x=1$? (this was  critical to know for this question).

d) I got no idea just pure intuition that the area at $x=1$ of $e^{x^2}$ would be less than $5$.

e) That's not true, it changes the integral, and will somewhere affect the $e$ being with $(-)$ in the solution.

The answer was c). I would really appreciate any help in disproving the wrong answers and any tips regarding my confusion about c).